Darboux Transformations in Integrable Systems: Theory and their Applications to Geometry (Mathematical Physics Studies, 26)

DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS

MATHEMATICAL PHYSICS STUDIES

Editorial Board:

Maxim Kontsevich, IHES, Bures-sur-Yvette, France

Massimo Porrati, New York University, New York, U.S.A.

Vladimir Matveev, Université Bourgogne, Dijon, France

Daniel Sternheimer, Université Bourgogne, Dijon, France

VOLUME 26

Darboux Transformations

in Integrable Systems

Theory and their Applications to Geometry

by

Chaohao Gu

Fudan University, Shanghai, China

Hesheng Hu


and

Zixiang Zhou


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Contents

Preface ix

1. 1+1 DIMENSIONAL INTEGRABLE SYSTEMS 11.1 KdV equation, MKdV equation and their Darboux

transformations 11.1.1 Original Darboux transformation 11.1.2 Darboux transformation for KdV equation 21.1.3 Darboux transformation for MKdV equation 31.1.4 Examples: single and double soliton solutions 61.1.5 Relation between Darboux transformations for

KdV equation and MKdV equation 101.2 AKNS system 11

1.2.1 2 × 2 AKNS system 111.2.2 N × N AKNS system 16

1.3 Darboux transformation 181.3.1 Darboux transformation for AKNS system 181.3.2 Invariance of equations under Darboux

transformations 231.3.3 Darboux transformations of higher degree and

the theorem of permutability 251.3.4 More results on the Darboux matrices of degree

one 301.4 KdV hierarchy, MKdV-SG hierarchy, NLS hierarchy and

AKNS system with u(N) reduction 341.4.1 KdV hierarchy 351.4.2 MKdV-SG hierarchy 401.4.3 NLS hierarchy 461.4.4 AKNS system with u(N) reduction 48

vi DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS

1.5 Darboux transformation and scattering, inversescattering theory 511.5.1 Outline of the scattering and inverse scattering

theory for the 2 × 2 AKNS system 511.5.2 Change of scattering data under Darboux

transformations for su(2) AKNS system 58

2. 2+1 DIMENSIONAL INTEGRABLE SYSTEMS 65

2.1 KP equation and its Darboux transformation 65

2.2 2+1 dimensional AKNS system and DS equation 68

2.3 Darboux transformation 702.3.1 General Lax pair 702.3.2 Darboux transformation of degree one 712.3.3 Darboux transformation of higher degree and

the theorem of permutability 75

2.4 Darboux transformation and binary Darbouxtransformation for DS equation 782.4.1 Darboux transformation for DSII equation 782.4.2 Darboux transformation and binary Darboux

transformation for DSI equation 82

2.5 Application to 1+1 dimensional Gelfand-Dickey system 84

2.6 Nonlinear constraints and Darboux transformation in2+1 dimensions 87

3. N + 1 DIMENSIONAL INTEGRABLE SYSTEMS 103

3.1 n + 1 dimensional AKNS system 1033.1.1 n + 1 dimensional AKNS system 1033.1.2 Examples 106

3.2 Darboux transformation and soliton solutions 1083.2.1 Darboux transformation 1083.2.2 u(N) case 1103.2.3 Soliton solutions 111

3.3 A reduced system on Rn 116

4. SURFACES OF CONSTANT CURVATURE,BACKLUND CONGRUENCES 121

4.1 Theory of surfaces in the Euclidean space R3 122

4.2 Surfaces of constant negative Gauss curvature,sine-Gordon equation and Backlund transformations 126

Contents vii

4.2.1 Relation between sine-Gordon equation andsurface of constant negative Gauss curvature in R3 126

4.2.2 Pseudo-spherical congruence 1294.2.3 Backlund transformation¨ 1324.2.4 Darboux transformation 1354.2.5 Example 139

4.3 Surface of constant Gauss curvature in the Minkowskispace R2,1 and pseudo-spherical congruence 1414.3.1 Theory of surfaces in the Minkowski space R2,1 1414.3.2 Chebyshev coordinates for surfaces of constant

Gauss curvature 1444.3.3 Pseudo-spherical congruence in R2,1 1494.3.4 Backlund transformation and Darboux¨

transformation for surfaces of constant Gausscurvature in R2,1 154

4.4 Orthogonal frame and Lax pair 1744.5 Surface of constant mean curvature 179

4.5.1 Parallel surface in Euclidean space 1794.5.2 Construction of surfaces 1814.5.3 The case in Minkowski space 184

5. DARBOUX TRANSFORMATION AND HARMONIC MAP 1895.1 Definition of harmonic map and basic equations 1895.2 Harmonic maps from R2 or R1,1 to S2, H2 or S1,1 1925.3 Harmonic maps from R1,1 to U(N) 199

5.3.1 Riemannian metric on U(N) 1995.3.2 Harmonic maps from R1,1 to U(N) 2015.3.3 Single soliton solutions 2075.3.4 Multi-soliton solutions 210

5.4 Harmonic maps from R2 to U(N) 2135.4.1 Harmonic maps from R2 to U(N) and their

Darboux transformations 2135.4.2 Soliton solutions 2195.4.3 Uniton 2205.4.4 Darboux transformation and singular Darboux

transformation for unitons 225

6. GENERALIZED SELF-DUAL YANG-MILLS ANDYANG-MILLS-HIGGS EQUATIONS 2376.1 Generalized self-dual Yang-Mills flow 237

viii DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS

6.1.1 Generalized self-dual Yang-Mills flow 2376.1.2 Darboux transformation 2426.1.3 Example 2456.1.4 Relation with AKNS system 247

6.2 Yang-Mills-Higgs field in 2+1 dimensional Minkowskispace-time 2486.2.1 Yang-Mills-Higgs field 2486.2.2 Darboux Transformations 2496.2.3 Soliton solutions 251

6.3 Yang-Mills-Higgs field in 2+1 dimensional anti-de Sitterspace-time 2566.3.1 Equations and their Lax pair 2566.3.2 Darboux transformations 2576.3.3 Soliton solutions in SU(2) case 2606.3.4 Comparison with the solutions in Minkowski

space-time 263

7. TWO DIMENSIONAL TODA EQUATIONS ANDLAPLACE SEQUENCES OF SURFACES 2677.1 Signed Toda equations 2677.2 Laplace sequences of surfaces in projective space Pn−1 2717.3 Darboux transformation 2777.4 Su chain (Finikoff configuration) 2817.5 Elliptic version of Laplace sequence of surfaces in CPn 291

7.5.1 Laplace sequence in CPn 2917.5.2 Equations of harmonic maps from R2 to CPn in

homogeneous coordinates 2927.5.3 Cases of indefinite metric 2957.5.4 Harmonic maps from R1,1 2967.5.5 Examples of harmonic sequences from R2 to CPn

or R1,1 to CPn 296

References 299

Preface

GU Chaohao

The soliton theory is an important branch of nonlinear science. Onone hand, it describes various kinds of stable motions appearing in na-ture, such as solitary water wave, solitary signals in optical fibre etc.,and has many applications in science and technology (like optical signalcommunication). On the other hand, it gives many effective methodsof getting explicit solutions of nonlinear partial differential equations.Therefore, it has attracted much attention from physicists as well asmathematicians.

Nonlinear partial differential equations appear in many scientific prob-lems. Getting explicit solutions is usually a difficult task. Only in cer-tain special cases can the solutions be written down explicitly. However,for many soliton equations, people have found quite a few methods toget explicit solutions. The most famous ones are the inverse scatteringmethod, Backlund transformation etc.. The inverse scattering methodïs based on the spectral theory of ordinary differential equations. TheCauchy problem of many soliton equations can be transformed to solvinga system of linear integral equations. Explicit solutions can be derivedwhen the kernel of the integral equation is degenerate. The Backlund¨transformation gives a new solution from a known solution by solvinga system of completely integrable partial differential equations. Somecomplicated “nonlinear superposition formula” arise to substitute thesuperposition principle in linear science.

However, if the kernel of the integral equation is not degenerate, itis very difficult to get the explicit expressions of the solutions via theinverse scattering method. For the Backlund transformation, the non-¨linear superposition formula is not easy to be obtained in general. In

x DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS

late 1970s, it was discovered by V. B. Matveev that a method given byG. Darboux a century ago for the spectral problem of second order or-dinary differential equations can be extended to some important solitonequations. This method was called Darboux transformation. After that,it was found that this method is very effective for many partial differen-tial equations. It is now playing an important role in mechanics, physicsand differential geometry. V. B. Matveev and M. A. Salle published animportant monograph [80] on this topic in 1991. Besides, an interestingmonograph of C. Rogers and W. K. Schief [90] with many recent resultswas published in 2002.

The present monograph contains the Darboux transformations in ma-trix form and provides purely algebraic algorithms for constructing ex-plicit solutions. Consequently, a basis of using symbolic calculationsto obtain explicit exact solutions for many integrable systems is estab-lished. Moreover, the behavior of simple and multi-solutions, even inmulti-dimensional cases, can be elucidated clearly. The method cov-ers a series of important topics such as varies kinds of AKNS systems inRn+1, the construction of Backlund congruences and surfaces of constantGauss curvature in R3 and R2,1, harmonic maps from two dimensionalmanifolds to the Lie group U(n), self-dual Yang-Mills fields and the gen-eralizations to higher dimensional case, Yang-Mills-Higgs fields in 2 + 1dimensional Minkowski and anti-de Sitter space, Laplace sequences ofsurfaces in projective spaces and two dimensional Toda equations. Allthese cases are stated in details. In geometric problems, the Lax pair isnot only a tool, but also a geometric object to be studied. Many resultsin this monograph are obtained by the authors in recent years.

This monograph is partially supported by the Chinese Major StateBasic Research Program “Frontier problems in nonlinear sciences”, theDoctoral Program Foundation of the Ministry of Education of China,National Natural Science Foundation of China and Science Foundationof Shanghai Science Committee. Most work in this monograph was donein the Institute of Mathematics of Fudan University.

Chapter 1

1+1 DIMENSIONAL INTEGRABLESYSTEMS

Starting from the original Darboux transformation, we first discussthe classical form of the Darboux transformations for the KdV andthe MKdV equation, then discuss the Darboux transformations for theAKNS system and more general systems. The coefficients in the evolu-tion equations discussed here may depend on t. The Darboux matricesare constructed algebraically and the algorithm is purely algebraic anduniversal to whole hierarchies. The Darboux transformations for reducedsystems are also concerned. We also present the relations between Dar-boux transformation and the inverse scattering theory, and show thatthe number of solitons (the number of eigenvalues) increases or decreasesafter the action of a Darboux transformation.

1.1 KdV equation, MKdV equation and theirDarboux transformations

1.1.1 Original Darboux transformationIn 1882, G. Darboux [18] studied the eigenvalue problem of a lin-

ear partial differential equation of second order (now called the one-dimensional Schrodinger equation)

−φxx − u(x)φ = λφ. (1.1)

Here u(x) is a given function, called potential function; λ is a constant,called spectral parameter. He found out the following fact. If u(x)and φ(x, λ) are two functions satisfying (1.1) and f(x) = φ(x, λ0) is asolution of the equation (1.1) for λ = λ0 where λ0 is a fixed constant,

2 DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS

then the functions u′ and φ′ defined by

u′ = u + 2(ln f)xx, φ′(x, λ) = φx(x, λ) − fxff

fφ(x, λ) (1.2)

satisfy−φ′

xx − u′φ′ = λφ′, (1.3)

which is of the same form as (1.1). Therefore, the transformation (1.2)transforms the functions (u, φ) to (u′, φ′) which satisfy the same equa-tions. This transformation

(u, φ) −→ (u′, φ′), (1.4)

is the original Darboux transformation, which is valid for f = 0.

1.1.2 Darboux transformation for KdV equationIn 1885, the Netherlandish applied mathematicians Korteweg and de

Vries introduced a nonlinear partial differential equation describing themotion of water wave, which is now called the Korteweg-de Vries equa-tion (KdV equation)

ut + 6uux + uxxx = 0. (1.5)

In the middle of 1960’s, this equation was found out to be closely relatedto the Schrodinger equation mentioned above [87]. KdV equation (1.5)ïs the integrability condition of the linear system

− φxx − uφ = λφ,

φt = −4φxxx − 6uφx − 3uxφ(1.6)

which is called the Lax pair of the KdV equation. Here u and φ arefunctions of x and t. (1.6) is the integrability condition of (1.5). Inother words, (1.5) is the necessary and sufficient condition for (φxx)t =(φt)xx being an identity for all λ, where (φxx)t is computed from φxx =(−λ− u)φ (the first equation of (1.6)) and (φt)xx is given by the secondequation of (1.6).

Since the first equation of the Lax pair of the KdV equation is justthe Schrodinger equation, the Darboux transformation (1.2) can also beäpplied to the KdV equation, where the functions depend on t. Obvi-ously the transformation keeps the first equation of (1.6) invariant, i.e.,(u′, φ′) satisfies

−φ′xx − u′φ′ = λφ′. (1.7)

Moreover, it is easily seen that (u′, φ′) satisfies the second equation of(1.6) as well. Therefore, u′ satisfies the KdV equation, which is the

1+1 dimensional integrable systems 3

integrability condition of (1.6). In summary, suppose one knows a solu-tion u of the KdV equation, solving the linear equations (1.6) one getsφ(x, t, λ). Take λ to be a special value λ0 and let f(x, t) = φ(x, t, λ0),then u′ = u + 2(ln f)xx gives a new solution of the KdV equation, andφ′ given by (1.2) is a solution of the Lax pair corresponding to u′. Thisgives a way to obtain new solutions of the KdV equation.

This process can be done successively as follows. For a known solutionu of (1.5), first solve a system of linear differential equations (1.6) andget φ. Then explicit calculation from (1.2) gives new special solutionsof the KdV equation. Since φ′ is known, it is not necessary to solve anylinear differential equations again to obtain (u′′, φ′′). That is, we onlyneed algebraic calculation to get (u′′, φ′′) etc.:

(u, φ) −→ (u′, φ′) −→ (u′′, φ′′) −→ · · · . (1.8)

Therefore, we have extended the Darboux transformation for the Schro-dinger equation to the KdV equation. The basic idea here is to get thenew solutions of the nonlinear equation and the corresponding solutionsof the Lax pair simultaneously from a known solution of the nonlinearequation and a solution of its Lax pair by using algebraic and differentialcomputation. Note that the formula is valid only for f = 0. If f = 0,the Darboux transformation will have singularities.

Remark 1 Let ψ1 = φ, ψ2 = φx, Ψ = (ψ1, ψ2)T , then the Lax pair (1.6)can be written in matrix form as

Ψx =

⎛⎝⎛⎛ 0 1

−λ − u 0

⎞⎠⎞⎞Ψ,

Ψt =

⎛⎝⎛⎛ ux 4λ − 2u

−4λ2 − 2λu + uxx + 2u2 −ux

⎞⎠⎞⎞Ψ.

(1.9)

The transformation φ → φ′ in (1.2) can also be rewritten as a trans-formation of Ψ, which can be realized via algebraic algorithm only. Weshall discuss this Darboux transformation in matrix form later.

1.1.3 Darboux transformation for MKdVequation

The method of Darboux transformation can be applied to many otherequations such as the MKdV equation, the sine-Gordon equation etc.[105]. We first take the MKdV equation as an example. General caseswill be considered in the latter sections.

MKdV equationut + 6u2ux + uxxx = 0 (1.10)


is the integrability condition of the over-determined linear system [2, 119]

Φx = UΦ =

⎛⎝⎛⎛ λ u

−u −λ

⎞⎠⎞⎞Φ,

Φt = V Φ

=

⎛⎝⎛⎛ −4λ3 − 2u2λ −4uλ2 − 2uxλ − 2u3 − uxx

4uλ2 − 2uxλ + 2u3 + uxx 4λ3 + 2u2λ

⎞⎠⎞⎞Φ,

(1.11)that is, (1.10) is the necessary and sufficient condition for Φxt = Φtx

being an identity. The system (1.11) is called a Lax pair of (1.10) andλ a spectral parameter. Here Φ may be regarded as a column solutionor a 2 × 2 matrix solution of (1.11).

There are several ways to derive the Darboux transformation for theMKdV equation. Here we use the Darboux matrix method.

For a given solution u of the MKdV equation, suppose we know afundamental solution of (1.11)

Φ(x, t, λ) =

⎛⎝⎛⎛ Φ11(x, t, λ) Φ12(x, t, λ)

Φ21(x, t, λ) Φ22(x, t, λ)

⎞⎠⎞⎞ (1.12)

which composes two linearly independent column solutions of (1.11).Let λ1, µ1 be arbitrary real numbers and

σ =Φ22(x, t, λ1) + µ1Φ21(x, t, λ1)Φ12(x, t, λ1) + µ1Φ11(x, t, λ1)

(1.13)

be the ratio of the two entries of a column solution of the Lax pair (1.11).Construct the matrix

D(x, t, λ) = λI − λ1

1 + σ2

⎛⎝⎛⎛ 1 − σ2 2σ

2σ σ2 − 1

⎞⎠⎞⎞ (1.14)

and let Φ′(x, t, λ) = D(x, t, λ)Φ(x, t, λ). Then it is easily verified thatΦ′(x, t, λ) satisfies

Φ′x = U ′Φ′, Φ′

t = V ′Φ′, (1.15)


where

U ′ =

⎛⎝⎛⎛ λ u′

−u′ −λ

⎞⎠⎞⎞ ,

V ′ =

⎛⎝⎛⎛ −4λ3 − 2u′2λ −4u′λ2 − 2u′xλ − 2u′3 − u′

xx

4u′λ2 − 2u′xλ + 2u′3 + u′

xx 4λ3 + 2u′2λ

⎞⎠⎞⎞(1.16)

with

u′ = u +4λ1σ

1 + σ2. (1.17)

(1.15) and (1.16) are similar to (1.11). The only difference is that theu in (1.11) is replaced by u′ defined by (1.17). For any solution Φ of(1.11), DΦ is a solution of (1.15), hence (1.15) is solvable for any giveninitial data (the value of Φ′ at some point (x0, t0)). In other words,(1.15) is integrable. The integrability condition of (1.15) implies that u′is also a solution of the MKdV equation. Using this method, we obtaina new solution of the MKdV equation together with the correspondingfundamental solution of its Lax pair from a known one.

The above conclusions can be summarized as follows. Let u be a solu-tion of the MKdV equation and Φ be a fundamental solution of its Laxpair. Take λ1, µ1 to be two arbitrary real constants, and let σ be definedby (1.13), then (1.17) gives a new solution u′ of the MKdV equation,and the corresponding solution to the Lax pair can be taken as DΦ. Thetransformation (u, Φ) → (u′, Φ′) is the Darboux transformation for theMKdV equation. This Darboux transformation in matrix form can bedone successively and purely algebraically as

(u, Φ) −→ (u′, Φ′) −→ (u′′, Φ′′) −→ · · · . (1.18)

Remark 2 Both (1.9) and (1.11) are of the form

Φx = UΦ, Φt = V Φ (1.19)

where U and V are N × N matrices and independent of Φ. The inte-grability condition of (1.19) is

UtUU − VxVV + [U, V ] = 0 ([U, V ] = UV − V U). (1.20)

According to the elementary theory of linear partial differential equations,the solution of (1.19) exists uniquely for given initial data Φ(x0, t0) = Φ0

if and only if (1.20) holds identically. Here Φ is an N ×N matrix, or a


vector with N entries. In this case Φ(x, t) is determined by the ordinarydifferential equation

dΦ = (U dx + V dy)Φ (1.21)

along an arbitrary path connecting (x0, t0) and (x, t). (1.20) is also calleda zero-curvature condition.

1.1.4 Examples: single and double solitonsolutions

Starting with the trivial solution u = 0 of the MKdV equation, onecan use the Darboux transformation to obtain the soliton solutions. Foru = 0, the fundamental solution of the Lax pair can be obtained as

Φ(x, t, λ) =

⎛⎝⎛⎛ exp(λx − 4λ3t) 0

0 exp(−λx + 4λ3t)

⎞⎠⎞⎞ (1.22)

by integrating (1.11). Take λ1 = 0 and µ1 = exp(2α1) > 0, then (1.13)gives

σ = σ1 = exp(−2λ1x + 8λ31t − 2α1), (1.23)

hence

D = λI − λ1

cosh v1

⎛⎝⎛⎛ sinh v1 1

1 − sinh v1

⎞⎠⎞⎞ , (1.24)

wherev1 = 2λ1x − 8λ3

1t + 2α1. (1.25)

(1.17) gives the single soliton solution

u′ = 2λ1 sech(2λ1x − 8λ31t + 2α1), (1.26)

of the MKdV equation.(1.26) is called the single soliton solution because it has the following

properties: (i) It is a travelling wave solution, i.e., it is in the formu′ = f(x − ct); (ii) For any t, limx→±∞ u′ = 0. Speaking intuitively, u′is near 0 outside a small region, i.e., |u| < 2|λ1| sech K when |2λ1x −8λ3

1t + 2α1| > K.The solution of the corresponding Lax pair is

Φ′(x, t, λ) = (Φ′ij(x, t, λ))

= D(x, t, λ)

⎛⎝⎛⎛ exp(λx − 4λ3t) 0

0 exp(−λx + 4λ3t)

⎞⎠⎞⎞ (1.27)


where D is given by (1.24).If we take u′ as a seed solution, a new Darboux matrix can be con-

structed from Φ′ and a series of new solutions of the MKdV equationcan be obtained.

We write down the second Darboux transformation explicitly. Sup-pose u is a solution of the MKdV equation (1.10), Φ is a fundamentalsolution of the corresponding Lax pair (1.11). Construct the Darbouxmatrix D = (Dij) according to (1.13) and (1.14) and let σ = σ1. More-over, take constants λ2 = 0 ( λ2 = λ1) and µ2 = exp(2α2). According to(1.13),

σ′2 =

Φ′22(x, t, λ2) + µ2Φ′

21(x, t, λ2)Φ′

12(x, t, λ2) + µ2Φ′11(x, t, λ2)

. (1.28)

Substituting Φ′ = DΦ into it, we have

σ′2 =

D21(Φ12 + µ2Φ11) + D22(Φ22 + µ2Φ21)D11(Φ12 + µ2Φ11) + D12(Φ22 + µ2Φ21)

∣∣∣∣∣∣∣∣∣∣λ=λ2

=D21(λ2) + D22(λ2)σ2

D11(λ2) + D12(λ2)σ2,

(1.29)

where

σ2 =Φ22(x, t, λ2) + µ2Φ21(x, t, λ2)Φ12(x, t, λ2) + µ2Φ11(x, t, λ2)

. (1.30)

Starting from u = 0, (1.26) and (1.27) are the single soliton solutionand the corresponding fundamental solution of the Lax pair. Substitut-ing (1.24), the expression of D, into (1.27), we have

Φ′(x, t, λ) =

⎛⎝⎛⎛ (λ − λ1 tanh v1)eλx−4λ3t − λ1 sech v1e−λx+4λ3t

−λ1 sech v1eλx−4λ3t(λ + λ1 tanh v1)e−λx+4λ3t

⎞⎠⎞⎞ ,

(1.31)hence

σ′2 =

−λ1 sech v1 + (λ2 + λ1 tanh v1) exp(−v2)(λ2 − λ1 tanh v1) − λ1 sech v1 exp(−v2)

, (1.32)

wherev2 = 2λ2x − 8λ3

2t + 2α2, (i = 1, 2). (1.33)

According to (1.17),

u′′ = 4λ1σ1

1 + σ21

+ 4λ2σ′2

1 + σ′22

= 2(λ22 − λ2

1)(λ2 cosh v1 − λ1 cosh v2)(λ2

1 + λ22) cosh v1 cosh v2 − 2λ1λ2(1 + sinh v1 sinh v2)

(1.34)


is a new solution of the MKdV equation. This is called the doublesoliton solution of the MKdV equation. This name follows from thefollowing asymptotic property of the solutions. We shall show that adouble soliton solution is asymptotic to two single soliton solutions ast → ∞.

Suppose λ2 > λ1 > 0, M is a fixed positive number. Let v1 bebounded by |v1| ≤ M , then x ∼ ∞ as t ∼ ∞. Since

v2 =λ2

λ1v1 − 8λ2(λ2

2 − λ21)t + 2α2 − 2λ2α1

λ1, (1.35)

v2 ∼ +∞ as t ∼ −∞, and

u′′ ∼ −2λ1 sech(v1 − v0) (1.36)

as t → −∞ where

v0 = tanh−1 2λ1λ2

λ21 + λ2

2

. (1.37)

If t ∼ +∞, then v2 ∼ −∞, and

u′′ ∼ −2λ1 sech(v1 + v0). (1.38)

Hence, for fixed v1 (i.e., the observer moves in the velocity 4λ21), the

solution is asymptotic to one single soliton solution (corresponding tothe parameter λ1) as t ∼ −∞ or t ∼ +∞. However, there is a phaseshift between the asymptotic solitons as t ∼ −∞ and t ∼ +∞. That is,the center of the soliton (the peak) moves from v1 = v0 to v1 = −v0.

Similarly, if |v2| ≤ M , then

v1 =λ1

λ2v2 + 8λ1(λ2

2 − λ21)t + 2α1 − 2λ1α2

λ2(1.39)

implies that v1 ∼ ±∞ as t ∼ ±∞, and

u′′ ∼ 2λ2 sech(v2 + v0), t ∼ −∞,

u′′ ∼ 2λ2 sech(v2 − v0), t ∼ +∞.(1.40)

Finally, if t ∼ ±∞ and both v1, v2 tend to ±∞ (i.e., the observer movesin the velocity = 4 λ2

1, 4λ22), then u′′ ∼ 0. Therefore, whenever t ∼ +∞ or

t ∼ −∞, u′′ is asymptotic to two single soliton solutions (see Figures 1.1– 1.3.

This fact means that: (i) a double soliton solution is asymptotic totwo single soliton solutions as t → ±∞; (ii) if two single solitons (theasymptotic behavior as t → −∞) interact, they will almost recover later


Figure 1.1. Double soliton solutions of the MKdV equation, t = −1

Figure 1.2. Double soliton solutions of the MKdV equation, t = 0.1

Figure 1.3. Double soliton solutions of the MKdV equation, t = 1

(t → +∞). Both the shape and the velocity do not change. The onlychange is the phase shift. Physically speaking, there is elastic scatteringbetween solitons. This is the most important character of solitons. Thediscovery of this property (first to the KdV equation) greatly promotesthe progress of the soliton theory.

Remark 3 Starting from the trivial solution u = 0, we can also obtainthe single and double soliton solutions of the KdV equation by usingthe original Darboux transformation mentioned at the beginning of thissection. The computation is simpler and is left for the reader.


The Darboux transformation for the MKdV equation can be used toget not only the single and double soliton solutions, but also the multi-soliton solutions. Moreover, this method can be applied to many othernonlinear equations. We shall discuss the general problem in the nextsection.

1.1.5 Relation between Darboux transformationsfor KdV equation and MKdV equation

The Darboux transformation for the MKdV equation can also be de-rived from the “complexification” of the Schrodinger equation (1.1) di-rectly. That is why the transformation given by the matrix D is alsocalled a Darboux transformation.

Take a solution Φ =

⎛⎝⎛⎛ φ1

φ2

⎞⎠⎞⎞ of (1.11), then the first equation of (1.11)

isφ1,x = λφ1 + uφ2,

φ2,x = −uφ1 − λφ2.(1.41)

Let ψ = φ1 + iφ2 and suppose λ is a real parameter, u is a real function,then ψ satisfies

ψxx = λ2ψ − (iux + u2)ψ. (1.42)This is a complex Schrodinger equation with potential (iux +u2). It canbe checked directly that if u is a solution of the MKdV equation, thenw = iux + u2 is a complex solution of the KdV equation wt + 6wwx +wxxx = 0. The transformation from the solution u of the MKdV equationto the solution w of the KdV equation is called a Miura transformation.

Remark 4 Let v = iu, then (1.10) is

vt − 6v2vx + vxxx = 0, (1.10)′

and the Miura transformation becomes w = vx−v2. If v is a real solutionof (1.10)′, then w is a real solution of the KdV equation.

Take a real number λ0 and a solution f = f1 + if2ff of the equation

(1.42) for λ = λ0, then

⎛⎝⎛⎛ f1

f2ff

⎞⎠⎞⎞ is a solution of (1.41) for λ = λ0. Using

the conclusion to the KdV equation, we know that

ψ′ = ψx − (fxff /f)ψ, w′ = w + 2(ln f)xx (1.43)

satisfy (1.42) and w′ is a solution of the KdV equation. Moreover, thereis a corresponding u′ satisfying the MKdV equation. Now we write downthe explicit expression of u′.


Considering (1.41), the first equation of (1.43) can be rewritten interms of the components as

φ′1 + iφ′

2 = λφ1 − iλφ2 − λ0f

f(φ1 + iφ2). (1.44)

If λ and λ0 are real numbers, then φ1 and φ2 can be chosen as realfunctions. (1.44) becomes

⎛⎝⎛⎛ φ′1

φ′2

⎞⎠⎞⎞ =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜λ − λ0

f21 − f2

2ff

f21 + f2

2ff−λ0

2f1f2ff

f21 + f2

2ff

λ02f1f2ff

f21 + f2

2ff−λ − λ0

f21 − f2

2ff

f21 + f2

2ff

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎝⎛⎛ φ1

φ2

⎞⎠⎞⎞ . (1.45)

It should be noted that the matrix in the right hand side of (1.45) is thecounterpart of the Darboux matrix defined by (1.24).

It can be checked that

⎛⎝⎛⎛ φ′1

φ′2

⎞⎠⎞⎞ satisfies

φ′1,x = λφ′

1 + u′φ′2,

φ′2,x = −u′φ′

1 − λφ′2

(1.46)

whereu′ = −u − 4λ0f1f2ff

f21 + f2

2ff. (1.47)

The integrability condition of (1.46) implies that u′ is a solution of theMKdV equation.

Remark 5 The matrix given by (1.14) and that given by (1.45) are dif-

ferent by a left-multiplied factor

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞. Notice that if (u, φ1, φ2)

is a solution of (1.41), then (−u, φ1,−φ2) is also a solution of (1.41).Therefore, the solution (1.17) given by (1.14) is the minus of the solu-tion given by (1.47). Both u and −u satisfy the MKdV equation. Wecan take any one transformation as the Darboux transformation.

The matrix D is very important hereafter. It is called a Darbouxmatrix.

1.2 AKNS system1.2.1 2 × 2 AKNS system

In order to generalize the Lax pair of the MKdV equation, V. E. Zak-harov, A. B. Shabat and M. J. Ablowitz, D. J. Kaup, A. C. Newell,


H. Segur introduced independently a more general system [2, 119] whichis now called the AKNS system. For simplicity, we first discuss the 2×2AKNS system (i.e., the AKNS system of 2 × 2 matrices), and then themore general N × N AKNS system.

2 × 2 AKNS system is the linear system of differential equations

Φx = UΦ = λJΦ + PΦ,

Φt = V Φ =n∑

j=0

VjVV λn−jΦ,(1.48)

where

J =

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ , P =

⎛⎝⎛⎛ 0 p

q 0

⎞⎠⎞⎞ , V =

⎛⎝⎛⎛ A B

C −A

⎞⎠⎞⎞ (1.49)

A =n∑

j=0

aj(x, t)λn−j ,

B =n∑

j=0

bj(x, t)λn−j ,

C =n∑

j=0

cjc (x, t)λn−j ,

(1.50)

p, q, aj , bj , cjc are complex or real functions of x and t, λ is a real orcomplex parameter, called the spectral parameter. As mentioned inRemark 2, the integrability condition of (1.48)

UtUU − VxVV + [U, V ] = 0 (1.51)

should hold for all λ. In terms of the components, (1.51) becomes

Ax = pC − qB,

Bx = pt + 2λB − 2pA,

CxCC = qt − 2λC + 2qA.

(1.52)

Both sides of the above equations are polynomials of λ. Expanding themin terms of the powers of λ, we have

b0 = c0 = 0,

aj,x = pcjc − qbj (0 ≤ j ≤ n),

bj+1 =12bj,x + paj (0 ≤ j ≤ n − 1),

cjc +1 = −12cj,xc + qaj (0 ≤ j ≤ n − 1),

(1.53)


and the evolution equations

pt = bn,x + 2pan,

qt = cn,x − 2qan.(1.54)

(1.53) can be regarded as the equations to determine A, B, C, and (1.54)is a system of evolution equations of p and q. In (1.53), aj , bj , cjc can bederived through algebraic calculation, differentiation and integration.We can see later that they are actually polynomials of p, q and theirderivatives with respect to x (without any integral expressions of p andq), the coefficients of which are arbitrary functions of t. After solvingaj , bj , cjc from (1.53), we get the system of nonlinear evolution equationsof p and q from (1.54).

For j = 0, 1, 2, 3,

a0 = α0(t), b0 = c0 = 0,

a1 = α1(t), b1 = α0(t)p, c1 = α0(t)q,

a2 = −12α0(t)pq + α2(t),

b2 =12α0(t)px + α1(t)p,

c2 = −12α0(t)qx + α1(t)q,

a3 =14α0(t)(pqx − qpx) − 1

2α1(t)pq + α3(t),

b3 =14α0(t)(pxx − 2p2q) +

12α1(t)px + α2(t)p,

c3 =14α0(t)(qxx − 2pq2) − 1

2α1(t)qx + α2(t)q.

(1.55)

Here α0(t), α1(t), α2(t), α3(t) are arbitrary functions of t, which are theintegral constants in integrating a0, a1, a2, a3 from the second equationof (1.53).

Here are some simplest and most important examples.

Example 1.1 n = 3, p = u, q = −1, α0 = −4, α1 = α2 = α3 = 0. Inthis case, a3 = −ux, b3 = −uxx − 2u2, c3 = 2u. (1.54) becomes the KdVequation

ut + uxxx + 6uux = 0. (1.56)

Example 1.2 n = 3, p = u, q = −u, α0 = −4, α1 = α2 = α3 = 0, thena3 = 0, b3 = −uxx − 2u3, c3 = uxx + 2u3. The equation becomes theMKdV equation

ut + uxxx + 6u2ux = 0. (1.57)


Example 1.3 n = 2, p = u, q = −u, α0 = −2i, α1 = α2 = 0, a2 =− i|u|2, b2 = − iux, c2 = − iux. (1.54) is the nonlinear Schr¨dingerëquation

iut = uxx + 2|u|2u. (1.58)

We have seen that for j = 0, 1, 2, 3, aj , bj , cjc are differential poly-nomials of p and q, i.e., polynomials of p, q and their derivatives withrespect to x, whose coefficients are constants or arbitrary functions of t.

Lemma 1.4 aj, bj, cjc given by (1.53) are differential polynomials of pand q.

Proof. Use induction. The conclusion is obviously true for j = 0.Suppose aj , bj , cjc are differential polynomials of p and q for j < l, we

will prove that al, bl, cl are also differential polynomials of p and q.(1.53) implies that bl, cl are differential polynomials of p and q. Hence

it is only necessary to prove that al is a differential polynomial of p andq. For 1 ≤ j ≤ l − 1,

bjcl+1−j − cjc bl+1−j

= bj(qal−j − 12cl−j,x) − cjc (pal−j +

12bl−j,x)

= (qbj − pcjc )al−j − 12(bjcl−j,x + cjc bl−j,x)

= −(ajal−j +12bjcl−j +

12cjc bl−j)x + ajal−j,x

+12(bj,xcl−j + cj,xc bl−j)


12cjc bl−j)x + (paj +

12bj,x)cl−j

−(qaj − 12cj,xc )bl−j


12cjc bl−j)x + bj+1cl−j − cjc +1bl−j .

(1.59)

Summarize for j from 1 to l − 1, we have

b1cl − c1bl = −l−1∑j=1

(ajal−j +12bjcl−j +

12cjc bl−j)x − (b1cl − c1bl), (1.60)

i.e.,

pcl − qbl = −l−1∑j=1

14a0

(2ajal−j + bjcl−j + cjc bl−j)x. (1.61)

Hence

al = −l−1∑j=1

14a0

(2ajal−j + bjcl−j + cjc bl−j) + αl(t) (1.62)


is a differential polynomial of p and q. The lemma is proved.Since aj , bj , cjc are differential polynomials of p and q, we can de-

fine a0j [p, q[[ ], b0

j [p, q[[ ], c0jc [p, q[[ ] recursively so that they satisfy the

recursion relations (1.53) and the conditions a00[0, 0] = 1, a0

j [0, 0] = 0(1 ≤ j ≤ n). Clearly, these a0

j , b0j , c

0jc are uniquely determined as cer-

tain polynomials of p, q and their derivatives with respect to x. From(1.53), we have

Lemma 1.5

b0j [0, q] = 0, c0

jc [p,[[ 0] = 0,

a0j [p,[[ 0] = a0

j [0, q] = 0 (1 ≤ j ≤ n)(1.63)

for any p and q. Moreover, for any aj , bj , cjc satisfying (1.53), thereexist αj(t) (0 ≤ j ≤ n) such that

ak[p, q[[ ] =k∑

j=0

αk−j(t)a0j [p, q[[ ],

bk[p, q[[ ] =k∑

j=0

αk−j(t)b0j [p, q[[ ],

ck[p, q[[ ] =k∑

j=0

αk−j(t)c0jc [p, q[[ ].

(1.64)

Remark 6 For any positive integer n, the first equation of (1.48) (x(( -equation) is fixed, but the second one depends on the choice of α0(t),· · ·, αn(t). Therefore, the evolution equations (1.54) also depend on thechoice of α0(t), · · ·, αn(t). This means that (1.54) is a series of equa-tions, which is called the AKNS hierarchy. If α0(t), · · ·, αn(t) are allconstants, then the evolution equations have the coefficients independentof t and form a series of infinite dimensional dynamical systems. Es-pecially, if α0 = · · · = αn−1 = 0, αn = 1, then we obtain the normalizedAKNS hierarchy, written as⎛⎝⎛⎛ p

q

⎞⎠⎞⎞t

= KnKK

⎡⎣⎡⎡ p

q

⎤⎦⎤⎤ , (1.65)

where KnKK is a nonlinear differential operator defined by

KnKK

⎡⎣⎡ p

q

⎤⎦⎤ =

⎛⎝⎛⎛ b0n,x + 2pa0

n

c0n,x − 2qa0

n

⎞⎠⎞⎞ . (1.66)


From this definition and (1.53), we know that KnKK is derived from KnKK −1

by recursive algorithm.

1.2.2 N × N AKNS systemIn the last subsection, we introduced the 2 × 2 AKNS system. In

order to obtain more nonlinear partial differential equations, the 2 × 2Lax pair should be generalized naturally to the problems of N × Nmatrices. Therefore, we discuss the Lax pair

Φx = UΦ = λJΦ + P (x, t)Φ,

Φt = V Φ =n∑

j=0

VjVV (x, t)λn−jΦ(1.67)

where J is an N × N constant diagonal matrix, P (x, t), VjVV (x, t) areN × N matrices and P (x, t) is off-diagonal (i.e., its diagonal entries areall zero), λ is a spectral parameter. We assume that all the entries of Jare distinct, though the assumption can be released with restrictions onP and V .

The integrability condition of (1.67) is still

UtUU − VxVV + [U, V ] = 0. (1.68)

Using the expressions of U and V in (1.67), we have

PtPP −n∑

j=0

Vj,xVV λn−j +n−1∑

j=−1

[J, VjVV +1]λn−j +n∑

j=0

[P, VjVV ]λn−j = 0. (1.69)

The coefficients of each power of λ on the left hand side should be zero.This leads to

[J, V0VV ] = 0,

[J, VjVV +1] − Vj,xVV + [P, VjVV ] = 0 (0 ≤ j ≤ n − 1),

PtPP − Vn,xVV + [P, VnVV ] = 0.

(1.70)

For any N × N matrix M , we divide it as M = Mdiag + Moff, whereMdiag is the diagonal part of M and Moff = M − Mdiag (hence Moff isoff-diagonal). Since J is diagonal with distinct diagonal entries and P


is off-diagonal, (1.70) is divided into

V off0VV = 0,

V diagj,xVV = [P, V off

jVV ]diag (0 ≤ j ≤ n),

[J, V offjVV +1] = V off

j,xVV − [P, VjVV ]off (0 ≤ j ≤ n − 1),

(1.71)

andPtPP = V off

n,xVV − [P, VnVV ]off. (1.72)

We can solve VjVV (j = 0, · · · , n) from (1.71) by differentiation andintegration. In fact, similar to the 2 × 2 case, VjVV can be obtained bydifferentiation and integration for n = 0, 1, 2, 3. They are differentialpolynomials of the entries of P . For general n, it can be proved byinduction that each entry of VjVV is a differential polynomial of the entriesof P whose coefficients may depend on t. (The proof is omitted here. See[111]). Therefore, (1.72) gives a system of partial differential equations ofthe entries of P . We shall write VjVV [P ] for the VjVV to specify the dependenceon P .

Example 1.6 Let n = 1, J = A = diag(a1, · · · , aN ), V0VV = B =diag(b1, · · · , bN ) with ai = aj and bi = bj (i = j). Take V1VV = Q(x, t)whose diagonal entries are all 0, then, from (1.71),

Qij =bi − bj

ai − ajPijPP (i = j), (1.73)

and the equation (1.72) becomes

PtPP = Qx − [P, Q]off. (1.74)

Written in terms of the components, it becomes

Pij,tPP = cijPij,xPP +∑

k = i,j

(cik − ckj)PikPP PkjPP (1.75)

wherecij =

bi − bj

ai − aj. (1.76)

This is a system of nonlinear partial differential equations of PijPP (i = j),called the N wave equation.

Similar to the discussion in Lemma 1.5, let V 0jVV [P ] be the solution of

(1.71) satisfying V0VV [0] = I, VlVV [0] = 0 (1 ≤ l ≤ n), then the followinglemma holds.


Lemma 1.5′. The general solution of (1.71) can be expressed as

VkVV [P ] =k∑

j=0

αjV0kVV −j [P ]. (1.77)

where α0, · · · , αn are the corresponding integral constants of VjVV [P ],which are diagonal matrices independent of x but may depend on t.

1.3 Darboux transformation1.3.1 Darboux transformation for AKNS system

LetF (u, ux, ut, uxx, · · ·) = 0, (1.78)

be a system of partial differential equations where u is a function or avector valued function. Consider the AKNS system

Φx = UΦ = (λJ + P )Φ,

Φt = V Φ =n∑

j=0

VjVV λn−jΦ,(1.79)

where J , P , VjVV satisfy the condition in the last section and P is a differ-ential polynomial of u. Suppose (1.78) is equivalent to (1.68), the inte-grability condition of (1.79), then (1.79) is called the Lax pair of (1.78).In this case, (1.78) is the evolution equation (1.72). The non-degenerateN × N matrix solution of (1.79) is called a fundamental solution of theLax pair.

In this section, we suppose that the (off-diagonal) entries of P areindependent and (1.78) is the system of differential equations (1.72) inwhich all off-diagonal entries of P are unknown functions. This systemis called unreduced.

We first discuss the Darboux transformation for the unreduced AKNSsystem.

Definition 1.7 Suppose D(x, t, λ) is an N ×N matrix. If for given Pand any solution Φ of (1.79), Φ′ = DΦ satisfies a linear system

Φ′x = U ′Φ′ = (λJ + P ′)Φ′,

Φ′t = V ′Φ′ =

n∑j=0

V ′jVV λn−jΦ′,

(1.80)

where P ′ is an off-diagonal N×N matrix function, then the transforma-tion (P, Φ) → (P ′, Φ′) is called a Darboux transformation for the unre-duced AKNS system, D(x, t, λ) is called a Darboux matrix. A Darbouxmatrix is of degree k if it is a polynomial of λ of degree k.


According to this definition, P ′ satisfies the equation

P ′offtPP − V ′off

n,xVV + [P ′, V ′nVV ]off = 0, (1.72)′

where the entries of V ′nVV are differential polynomials of P ′. Later, we will

see that V ′nVV = VnVV [P ′] when the Darboux matrix is a polynomial of λ. In

this case (1.72)′ and (1.72) are the same partial differential equations.Substituting Φ′ = DΦ into (1.80), we get

U ′ = DUD−1 + DxD−1,

V ′ = DV D−1 + DtD−1.

(1.81)

Proposition 1 If D is a Darboux matrix for (1.79) and D′ is a Darbouxmatrix for (1.80), then D′D is a Darboux matrix for (1.79).

Proof. Since D′ is a Darboux matrix for (1.80), there exists U ′′ =λJ + P ′′ (P ′′ is off-diagonal) and V ′′ =

∑nj=0 V ′′

jVV λn−j such that Φ′′ =D′Φ′ = D′DΦ satisfies

Φ′′x = U ′′Φ′′, Φ′′

t = V ′′Φ′′. (1.82)

Hence, by definition, D′D is a Darboux matrix for (1.79).

Remark 7 Any constant diagonal matrix K independent of λ is a Dar-boux matrix of degree 0, since under its action according to (1.81),

λJ + P → λJ + KPK−1,n∑

j=0

VjVV λn−j →n∑

j=0

KVjVV K−1λn−j .(1.83)

If we do not consider the relations among the entries of P , this kind ofDarboux matrices are trivial.

Now we first consider the Darboux matrix of degree one, which islinear in λ. Suppose it has the form λI − S where S an N × N ma-trix function, I is the identity matrix. According to Proposition 1 andRemark 7, the discussion on the Darboux matrix K(λI − S) (K is anon-degenerate constant matrix which must be diagonal in order to getthe first equation of (1.80)) can be reduced to the discussion on the Dar-boux matrix λI − S. Therefore, to construct the Darboux matrix, it isonly necessary to construct S.


The differential equations of S are derived as follows. From the firstequation of (1.80),

(λJ +P ′)(λI−S)Φ = ((λI−S)Φ)x = (λI−S)(λJ +P )Φ−SxS Φ. (1.84)

It must hold for any solution of (1.79). Comparing the coefficients ofthe powers of λ, we have

P ′ = P + [J, S], (1.85)

This is the expression of P ′.The term independent of λ in (1.84) gives

SxS = P ′S − SP = PS − SP + JS2 − SJS, (1.86)

i.e.,SxS + [S, JS + P ] = 0. (1.87)

This is the first equation which S satisfies.The second equation of (1.80) leads to

n∑j=0

V ′jVV λn−j(λI − S)Φ = ((λI − S)Φ)t

= (λI − S)n∑

j=0

VjVV λn−jΦ − StSS Φ. (1.88)

Comparing the coefficients of λn+1, λn, · · ·, λ, we can determine V ′jVV

recursively by

V ′0VV = V0VV , V ′

jVV +1 = VjVV +1 + V ′jVV S − SVjVV , (1.89)

and get the second equation of S

StSS = V ′nVV S − SVnVV . (1.90)

From (1.89) VjVV ’s can be expressed as

V ′0VV = V0VV ,

V ′jVV = VjVV +

j∑k=1

[VjVV −k, S]Sk−1 (1 ≤ j ≤ n),(1.91)

and (1.90) becomes

StSS + [S,n∑

j=0

VjVV Sn−j ] = 0. (1.92)


Theorem 1.8 λI − S is a Darboux matrix for (1.79) if and only if Ssatisfies

SxS + [S, JS + P ] = 0,

StSS + [S,n∑

j=0

VjVV Sn−j ] = 0.(1.93)

Moreover, under the action of the Darboux matrix λI−S, P ′ = P+[J, S].

Proof. Suppose λI − S is a Darboux matrix, then (1.93) is just (1.87)and (1.92) derived above. Conversely, if (1.87) and (1.92) hold, then forany solution Φ of (1.79), there are the relations (1.84) and (1.88). Hence(1.80) holds for the P ′ determined by (1.85) and the V ′

jVV determinedby (1.89), which means that λI − S is a Darboux matrix.

This theorem implies that we need to solve S from the system ofnonlinear partial differential equations (1.93) to get the Darboux matrix.Fortunately, most of the solutions of (1.93) can be constructed explicitly.The following theorem gives the explicit construction of the Darbouxmatrix of degree one.

Suppose P is a solution of (1.72). Take complex numbers λ1, · · ·, λN

such that they are not all the same. Let Λ = diag(λ1, · · · , λN ). Let hi

be a column solution of (1.79) for λ = λi. H = (h1, · · · , hN ). Whendet H = 0, let

S = HΛH−1, (1.94)

then we have the following theorem.

Theorem 1.9 The matrix λI−S defined by (1.94) is a Darboux matrixfor (1.79).

Proof. hi is a solution of (1.79) for λ = λi, that is, it satisfies

hi,x = λiJhi + Phi,

hi,t =n∑

j=0

VjVV λn−jhi.(1.95)

By taking the derivatives of H = (h1, h2, · · · , hN ) with respect to x andt, (1.95) is equivalent to

HxHH = JHΛ + PH,

HtHH =n∑

j=0

VjVV HΛn−j .(1.96)


Hence

SxS = HxHH ΛH−1 − HΛH−1HxHH H−1 = [HxHH H−1, S]

= [JS + P, S]

StSS = HtHH ΛH−1 − HΛH−1HtHH H−1 = [HtHH H−1, S]

= [n∑

j=0

VjVV Sn−j , S].

(1.97)

Therefore, the matrix S defined by (1.94) is a solution of (1.93). Theo-rem 1.8 implies that λI−S is a Darboux matrix for (1.79). The theoremis proved.

Theorem 1.10 For any given (x0, t0) and the matrix S0SS , (1.93) has asolution satisfying S(x0, t0) = S0SS . That is, the system is integrable.

Proof. First suppose the Jordan form of S0SS is a diagonal matrix. Sup-pose its eigenvalues are λ1, · · ·, λN and the corresponding eigenvectorsare h0i. Let hi be a solution of (1.95) satisfying hi(x0, t0) = h0i. Thenthese hi are linearly independent in a neighborhood of (x0, t0). Theo-rem 1.9 implies that the Darboux matrix exists, i.e., (1.93) has a solution.If the Jordan form of S0SS is not diagonal, then there is a series of matricesS(k)

0SS such that the Jordan form of S(k)0SS is diagonal and S

(k)0SS → S0SS as

k → ∞. Construct S according to (1.94) with initial value S(k)0SS , then

S(k) solves (1.93). The smooth dependency of the solution of (1.93) tothe initial value implies that S(k) → S and S

(k)xS , S

(k)tSS converge in a neigh-

borhood of (x0, t0). Therefore S is a solution of (1.93) with initial valueS0S . Thus (1.93) is solvable for any given initial value, which means thatit is integrable. The theorem is proved.

We can also prove this theorem by direct but tedious calculation.This theorem implies that a Darboux matrix of degree one can be

obtained either by (1.94) or the limit of such Darboux matrices.

Remark 8 hi can be expressed as hi = Φ(λi)li (i = 1, 2, · · · , N), where l1,l2, · · ·, lN are N linearly independent constant column matrices. HenceH in (1.94) can be written as

H = (Φ(λ1)l1, Φ(λ2)l2, · · · , Φ(λN )lN ) . (1.98)

This construction of Darboux matrix was given by [33, 94]. Theo-rem 1.9 and 1.10 implies that (1.94) contains all the matrices S whichare similar to diagonal matrices and λI − S are Darboux matrices of


degree one. A Darboux matrix expressed by (1.94) is called a diagonal-izable Darboux matrix or Darboux matrix with explicit expressions. Itis useful in constructing the solutions because it is expressed explicitly.Hereafter, we mostly use the diagonalizable Darboux matrices and theword “diagonalizable” is omitted.

The “single soliton solution” can be obtained by the Darboux trans-formation from the seed solution P = 0.

For P = 0, the fundamental solution of (1.67) is Φ = eλJx+Ω(λ,t) whereΩ(t) =

∫ ∑nj=0 VjVV [0](t)λn−j dt is a diagonal matrix. For any constants

λ1, · · ·, λN and column matrices l1, · · ·, lN , let

H =(eλ1Jx+Ω(λ1,t)l1, · · · , eλNJx+Ω(λN ,t)lN

), (1.99)

thenP ′ = [J, HΛH−1] (1.100)

is a solution of (1.72) and the fundamental solution of the Lax pair (1.80)is Φ′ = (λI − HΛH−1)Φ.

Remark 9 If VjVV [0] depends on t, then Ω(λi, t) is not a linear functionof t, hence the velocities of the solitons are not constants [45].

The double soliton solution can be obtained from P ′ by applyingfurther Darboux transformation. Since Φ′ is known in this process, P ′′and Φ′′ can be obtained by a purely algebraic algorithm. The multi-soliton solutions are obtained similarly.

For general AKNS system, det H = 0 may not hold for all ( x, t).Therefore, the solutions given by Darboux transformations may not beregular for all (x, t).

1.3.2 Invariance of equations under Darbouxtransformations

We have known that (P ′, V ′jVV ) and (P, VjVV ) satisfy the same recursion

relations (1.71) and (1.72) holds true for the two sets of functions. V ′jVV is

a differential polynomial of P ′ which is expressed by a similar equalityas (1.77), i.e.,

V ′kVV [P ′] =

k∑j=0

α′j(t)V

0kVV −j [P

′]. (1.101)

Here we prove that actually the coefficients α′0(t), · · · , α′

n(t) are the sameas α0(t), · · · , αn(t) respectively. Therefore P ′ and P satisfy the sameevolution equation (1.72).


Theorem 1.11 Suppose VjVV ’s are differential polynomials of P satisfying(1.71). S is a matrix satisfying (1.87), ViVV ’s are defined by (1.89) andP ′ = P + [J, S]. Then ViVV ’s are differential polynomials of P ′ and

V ′jVV [P ′] = VjVV [P ′] (j = 1, 2, · · · , n). (1.102)

Proof. At first we see that the equation P ′ = P + [J, S] is equivalentto P = P ′ − [J, S] and the equation (1.87) is equivalent to

SxS = [P ′ + SJ, S]. (1.103)

Moreover, for arbitrary x-function P ′. This equation admits solutions ina neighborhood around any given point x = x0. Thus we may considerP ′ as an arbitrary off-diagonal matrix-valued function of x.

From (1.89) we have

[J, V ′jVV +1] − V ′

j,xVV + [P ′, V ′jVV ]

= [J, VjVV +1] − Vj,xVV + [P, VjVV ] + ([J, V ′jVV ] − V ′

jVV −1,x + [P ′, V ′jVV −1])S

−S([J, VjVV ] − VjVV −1,x + [P, VjVV −1]) (j = 0, · · · , n − 1).

(1.104)

Using induction, we know that

[J, V ′jVV +1] − V ′

j,xVV + [P ′, V ′jVV ] = 0 (j = 0, · · · , n − 1) (1.105)

from the equations (1.71) for VjVV [P ]. Moreover, we can prove

V ′diagn,xVV − [P ′, V ′

nVV ]diag = 0. (1.106)

This means that V ′jVV and P ′ also satisfy (1.71). Therefore, as mentioned

above, V ′jVV can be expressed as a differential polynomial of P ′: V ′

jVV =V ′

jVV [P ′].Let

∆j [P ′] = V ′jVV [P ′] − VjVV [P ′], (1.107)

(1.89) implies ∆0 = 0. Suppose ∆k = 0, then (1.71) leads to

[J,∆k+1] = ∆offk,x − [P ′, ∆k]off = 0, (1.108)

hence ∆offk+1 = 0. From (1.71),

∆diagk+1,x = [P ′, ∆k+1]off = 0, (1.109)

which means that ∆diagk+1[P

′] is independent of x. We should prove that

∆diagk+1[P

′] is independent of P ′. Denote P ′ijPP be the entries of P ′, P

′(α)ijPP be


the αth derivative of P ′ijPP with respect to x. Suppose the order of the

highest derivatives of P ′ in ∆diagk+1 is r, then

0 =∂∆diag

k+1

∂x=∑i,j

r∑α=0

∂∆diagk+1

∂P′(α)ijPP

P′(α+1)ijPP . (1.110)

In this equation, the coefficient of P′(r+1)ijPP should be 0. Hence ∆diag

k+1 doesnot contain the rth derivative of P ′, which means that it is independentof P ′. Especially, let S = 0, P = P ′, then (1.89) implies ∆diag

k+1 = 0.Thus (1.102) is proved

Theorem 1.11 implies that for the evolution equations (1.72) in theAKNS system, the Darboux transformation transforms a solution of anequation to a new solution of the same equation.

Note that the Darboux transformation

(P, Φ) −→ (P ′, Φ′) (1.111)

defined byP ′ = P + [J, S], Φ′ = (λI − S)Φ (1.112)

can be taken successively in a purely algebraic algorithm and leads toan infinite series of solutions of the AKNS system:

(P, Φ) −→ (P ′, Φ′) −→ (P ′′, Φ′′) −→ · · · (1.113)

1.3.3 Darboux transformations of higher degreeand the theorem of permutability

The Darboux matrices discussed above are all of degree one. In thissubsection, we construct Darboux matrices with explicit expressionswhich are the polynomials of λ of degree > 1. Then we derive thetheorem of permutability from the Darboux matrices of degree two.

Clearly, the composition of r Darboux transformations of degree onegives a Darboux transformation of degree r. On the other hand, we canalso construct the Darboux transformations of degree r directly.

As known above, a Darboux matrix of degree one is D(x, t, λ) = λI−Swhere S is given by (1.94) if it can be diagonalized (Theorem 1.9). Then,SH = HΛ is equivalent to D(x, t, λi)hi = 0 where hi is a column solutionof the Lax pair for λ = λi such that detH = det(h1, · · · , hN ) = 0. Thisfact can be generalized to the Darboux matrix of degree r, that is, wecan consider an N × N Darboux matrix in the form

D(x, t, λ) =r∑

j=0

Dr−j(x, t)λj , D0 = I. (1.114)


Take Nr complex numbers λ1, λ2, · · ·, λNr and the column solutionhi of the Lax pair for λ = λi (i = 1, · · · , Nr). Let

FrFF =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜h1 h2 · · · hNr

λ1h1 λ2h2 · · · λNrhNr

......

. . ....

λr−11 h1 λr−1

2 h2 · · · λr−1Nr hNr

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ (1.115)

which is an Nr ×Nr matrix. The system D(x, t, λi)hi = 0 is equivalentto

r−1∑j=0

Dr−j(x, t)λjihi = −λr

i hi (i = 1, · · · , Nr) (1.116)

and can be written as

(Dr, Dr−1, · · · , D1)FrFF = −(λr1h1, · · · , λr

NrhNr). (1.117)

This is a system of linear algebraic equations for (Dr, Dr−1, · · · , D1).When detFrFF = 0, it has a unique solution ( Dr, Dr−1, · · · , D1). Therefore,when det FrFF = 0, there exists a unique N×N matrix D(x, t, λ) satisfyingD(x, t, λi)hi = 0 (i = 1, · · · , Nr). We write it as D(h1, · · · , hNr, λ) toindicate that D is constructed from h1, · · · , hNr.

The next theorem shows that it is a Darboux matrix and decompos-able as a product of two Darboux matrices of lower degree [52, 74].

Theorem 1.12 Given Nr complex numbers λ1, · · · , λNr. Let hi be acolumn solution of the Lax pair (1.79) for λ = λi (i = 1, · · · , Nr), FrFF bedefined by (1.115). Suppose det FrFF = 0 , then the following conclusionshold.

(1) There exists a unique matrix D(h1, · · · , hNr, λ) in the form (1.114)such that

D(h1, · · · , hNr, λi)hi = 0 (i = 1, 2, · · · , Nr). (1.118)

In this case, D(h1, · · · , hNr, λ) is a Darboux matrix of degree r for (1.79).(2) If det FrFF −1 = 0 , then the above Darboux matrix of degree r can be

decomposed as

D(h1, · · · , hNr, λ)

= D(D(h1, · · · , hN(r−1), λN(r−1)+1)hN(r−1)+1, · · · ,

D(h1, · · · , hN(r−1), λNr)hNr, λ)·

·D(h1, · · · , hN(r−1), λ).

(1.119)


On the right hand side of this equality, the first term is a Darboux matrixof degree one and the second term is a Darboux matrix of degree (r− 1).

(3) The Darboux matrix D(h1, · · · , hNr, λ) of degree r can be decom-posed into the product of r Darboux matrices of degree one.

(4) P = P − [J,D1] is a solution of (1.72).

Proof. We first prove (2). Let

Λk = diag(λN(k−1)+1, · · · , λNk), (1.120)

HkH = (hN(k−1)+1, · · · , hNk), (1.121)

then

FrFF =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜H1 H2HH · · · HrHH

H1Λ1 H2HH Λ2 · · · HrHH Λr

......

. . ....

H1Λr−11 H2HH Λr−1

2 · · · HrHH Λr−1r

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ . (1.122)

Since

FrFF −1 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜H1 H2HH · · · HrHH −1

H1Λ1 H2HH Λ2 · · · HrHH −1Λr−1

......

. . ....

H1Λr−21 H2HH Λr−2

2 · · · HrHH −1Λr−2r−1

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ (1.123)

is non-degenerate, there is a matrix D(h1, · · · , hN(r−1), λ) of degree (r−1)with respect to λ such that

D(h1, · · · , hN(r−1), λi)hi = 0 (i = 1, 2, · · · , N(r − 1)). (1.124)

Let

h′i = D(h1, · · · , hN(r−1), λi)hi (i = N(r − 1) + 1, · · · , Nr). (1.125)

Construct a Darboux matrix D(h′N(r−1)+1, · · · , h′

Nr, λ) from h′i and let

D′(λ) = D(h′N(r−1)+1, · · · , h′

Nr, λ)D(h1, · · · , hN(r−1), λ), (1.126)

then D′(λi)hi = 0 (i = 1, 2, · · · , N(r− 1)). Moreover, for i = N(r− 1) +1, · · · , Nr,

D′(λi)hi = D(h′N(r−1)+1, · · · , h′

Nr, λi)h′i = 0. (1.127)

Hence D′(λ) = D(λ).


Thus, we have decomposed the matrix of degree r determined by(1.116) to the product of a matrix of degree (r − 1) and a matrix ofdegree one, expressed by (1.119). This proves (2).

For D(h1, · · · , hN(r−1), λ), if all the determinants of FrFF −2, FrFF −3, · · · arenon-zero, then D(h1, · · · , hN , λ) can be decomposed to r matrices of de-gree one by repeating the above procedure. For r = 1, D(h1, · · · , hN , λ)is a Darboux matrix. Hence D(h1, · · · , hNr, λ) is also a Darboux ma-trix and it can be decomposed to the product of r Darboux matrices ofdegree one:

D = (λI − SrSS ) · · · (λI − S1). (1.128)

Since detFrFF = 0, we can always permute the subscripts of Λ i and HiHH sothat all the determinants of FrFF −2, FrFF −3, · · · are non-zero. (3) is proved.

Since D is the product of r Darboux matrices of degree one, D itselfis a Darboux matrix. Hence (1) holds.

Note thatD1 = −(S1 + · · · + SrSS ). (1.129)

After the transformation of λI − S1, P → P ′ = P + [J, S1]. Then afterthe transformation of λI−S2, P ′ → P ′′ = P ′+[J, S2], and so on. Hence,after the transformation of D,

P → P + [J, S1 + · · · + SrSS ] = P − [J, D1]. (1.130)

Therefore, P − [J, D1] is a solution of (1.72). (4) is proved. This provesthe lemma.

Darboux transformation has an important property — the theorem ofpermutability. This theorem originated from the Backlund transforma-¨tion of the sine-Gordon equation and there are a lot of generalizationsand various proofs. The proof here is given by [52] (2× 2 case) and [33](N × N case). This proof does not depend on any boundary conditionsand the permutation of the parameters is expressed definitely.

From the solution (P, Φ(λ)), we can construct the Darboux transfor-mation with parameters λ

(1)1 , · · ·, λ

(1)N and the solutions h

(1)i = Φ(λi)l

(1)i

of the Lax pair. Then the solution (P (1), Φ(1)(λ)) is obtained. Here l(1)i ’s

are N constant vectors. Next, construct a Darboux matrix for (P (1),

Φ(1)(λ)) with parameters λ(2)1 , · · ·, λ

(2)N and l

(2)i to get (P (1,2), Φ(1,2)(λ)).

On the other hand, construct the Darboux transformation for (P, Φ(λ))with parameters λ

(2)i and l

(2)i to get (P (2), Φ(2)(λ)). Then construct the

Darboux transformation with parameters λ(1)i and l

(1)i to get (P (2,1),

Φ(2,1)(λ)). The following theorem holds.


Theorem 1.13 (Theorem of permutability) Suppose

det

⎛⎝⎛⎛ H1 H2HH

H1Λ1 H2HH Λ2

⎞⎠⎞⎞ = 0 , (1.131)

then(P (1,2), Φ(1,2)(λ)) = (P (2,1), Φ(2,1)(λ)). (1.132)

Proof. Theorem 1.12 implies that Φ(1,2)(λ) and Φ(2,1)(λ) are both ob-tained from Φ(λ) by the action of the Darboux transformation of degreetwo, and are expressed by

Φ(1,2)(λ) = D(h(1)1 , · · · , h(1)

N , h(2)1 , · · · , h(2)

N , λ),

φ(2,1)(λ) = D(h(2)1 , · · · , h(2)

N , h(1)1 , · · · , h(1)

N , λ).(1.133)

From (1) of Theorem 1.12, we know that the right hand side of the aboveequations are equal. Hence the theorem of permutability holds.

The theorem of permutability can be expressed by the following Bian-chi diagram:

(P, Φ)

Λ(1), L(1)

Λ(2), L(2)

(P (1), Φ(1))

(P (2), Φ(2))

))

))

(P (1,2), Φ(1,2)) = (P (2,1), Φ(2,1))

Λ(2), L(2)

Λ(1)

, L(1)

(1.134)

Here L(1) and L(2) denote the sets l(1)i and l(2)

i respectively.

Remark 10 The Darboux transformation of higher degree is much morecomplicated than the Darboux transformation of degree one. The the-orem of decomposition implies that Darboux transformations of degreeone can generate Darboux transformations of higher degree. Therefore,we can use Darboux transformations of degree one successively instead ofa Darboux transformation of higher degree so as to avoid the calculationof the determinant of a matrix of very high order (of order Nr). Sincethe algorithm for the Darboux transformation of degree one is purely al-gebraic and independent of the seed solution P , it is quite convenient tocalculate the solutions using symbolic calculation with computer. How-ever, some solutions, e.g., multi-solitons can be expressed by an explicitformulae by using Darboux transformations of higher degree [80].


Remark 11 The proof for Theorem 1.13 is for the Darboux transfor-mations with explicit expressions. Since any Darboux transformationwithout explicit expression is a limit of Darboux transformations withexplicit expressions, the theorem of permutability also holds for the Dar-boux transformations without explicit expressions.

Now we compute the more explicit expression of the Darboux matrixof degree two. Suppose it is constructed from (Λ1, H1) and (Λ2, H2HH )which satisfy (1.131). Let SjS = HjH ΛjH

−1jH and denote

Λα = diag(λ(α)1 , · · · , λ(α)

N ), HαHH = (h(α)1 , · · · , h(α)

N ).

After the action of λI−S1, h(2)j is transformed to (λ(2)

j I−S1)h(2)j . Hence

H2HH is transformed to H2HH = H2HH Λ2 − S1H2HH = (S2 − S1)H2HH . The secondDarboux matrix of degree one is λI − S2 where

S2 = H2HH Λ2H−12HH = (S2 − S1)S2(S2 − S1)−1. (1.135)

According to (1.131), S2 − S1 is non-degenerate. The Darboux matrixof degree two is

D(λ) = (λI − S2)(λI − S1)

= λ2I − λ(S22 − S2

1)(S2 − S1)−1 + (S2 − S1)S2(S2 − S1)−1S1.

(1.136)It is easy to check that D(λ) is symmetric to S1 and S2. Therefore, wecan also obtain the theorem of permutability by this symmetry.

1.3.4 More results on the Darboux matrices ofdegree one

In this subsection, we show that the Darboux matrix method in The-orem 1.9 can be applied not only to the AKNS system, but also to manyother evolution equations, especially to the Lax pairs whose U and V arepolynomials of λ. On the other hand, we also show that those Darbouxmatrices include all the diagonalizable Darboux matrices of the formλI − S, and any non-diagonalizable Darboux matrix can be obtainedfrom the limit of diagonalizable Darboux matrices.

We generalize the Lax pair (1.79) to

Φx = UΦ,

Φt = V Φ,(1.137)


where U and V are polynomials of the spectral parameter λ:

U(x, t, λ) =m∑

j=0

UjU (x, t)λm−j ,

V (x, t, λ) =n∑

j=0

VjVV (x, t)λn−j ,

(1.138)

UjU ’s and VjVV ’s are N × N matrices.Clearly, the integrability condition of (1.137) is

UtUU − VxVV + [U, V ] = 0. (1.139)

In this subsection, we still discuss the Darboux matrices for the Laxpairs without reductions. That is, we suppose that all the entries of UjUU ’sand VjVV ’s are independent except for the partial differential equations(1.139). This is to say that apart from the integrability condition (1.139),there is no other constraint. Therefore, the nonlinear partial differentialequation to be studied is just (1.139), i.e., the equations given by thecoefficients of each power of λ in (1.139) and the unknowns are the N×Nmatrices UjU and VjVV (j = 0, 1, · · · , n). Compared with Subsection 1.3.1,D = λI − S is a Darboux matrix if and only if there exist

U ′(x, t, λ) =m∑

j=0

U ′jU (x, t)λm−j ,

V ′(x, t, λ) =n∑

j=0

V ′jVV (x, t)λn−j

(1.140)

such that Φ′ = (λI − S)Φ satisfies

Φ′x = U ′Φ′, Φ′

t = V ′Φ′ (1.141)

where Φ is a fundamental solution of (1.137).Clearly, U ′ and V ′ have the expressions

U ′ = DUD−1 + DxD−1,

V ′ = DV D−1 + DtD−1

(1.142)

and they satisfyU ′

tUU − V ′xVV + [U ′, V ′] = 0. (1.143)

The remaining problem is to obtain S so that (1.141) holds. If S isobtained, we have the Darboux transformation

(U, V, Φ) → (U ′, V ′, Φ′). (1.144)


Comparing to Theorem 1.8, 1.9 and 1.10, we have

Theorem 1.8′. λI −S is a Darboux matrix of degree one for (1.137)if and only if S satisfies

SxS + [S, U(S)] = 0, StSS + [S, V (S)] = 0. (1.145)

Here

U(S) =m∑

j=0

UjU Sm−j , V (S) =n∑

j=0

VjVV Sn−j . (1.146)

Now suppose (U, V ) satisfies the integrability condition (1.139). Forgiven constant diagonal matrix Λ = diag(λ1, · · · , λN ), let hi be a columnsolution of (1.137) for λ = λi, H = (h1, · · · , hN ). If det H = 0, letS = HΛH−1, then the following theorems holds.

Theorem 1.9′. The matrix λI − S is a Darboux matrix for (1.137).

Theorem 1.10′. The system (1.145) is integrable.

The proofs are omitted since they are similar to the proofs for thecorresponding theorems above.

Note that for the AKNS system, we can solve ViVV [P ]’s from a system ofdifferential equations by choosing “integral constants” and these ViVV [P ]’sare differential polynomials of P . The remaining equation is only theequation (1.72) for P . In the present case, all the entries of UiUU and ViVVare regarded as independent unknowns satisfying the partial differentialequations (1.139).

The inverse of Theorem 1.9′ also holds.

Theorem 1.14 (1) If λI − S is a Darboux matrix for (1.137) and Sis diagonalized at one point, then there exists a constant diagonal ma-trix Λ = diag(λ1, · · · , λN ) and column solutions hi’s of the Lax pair(1.137) for λ = λi (i = 1, 2, · · · , N) such that H = (h1, · · · , hN ) andS = HΛH−1.

(2) If λI − S is a Darboux matrix for (1.137) but it can not be di-agonalized at any points, then there exist a series of Darboux matricesλI − Sk such that Sk’s and their derivatives with respect to x and tconverge to S and its derivatives respectively.

The proof is similar to that for Theorem 1.10.

Example 1.15 An example of a Darboux matrix which is not diagonal-izable everywhere.


Consider the Lax pair

Φx =

⎛⎝⎛⎛ λ p

q −λ

⎞⎠⎞⎞Φ,

Φt =

⎛⎝⎛⎛ −2iλ2 + ipqi −2iλp − ipi x

−2iλq + iqx 2iλ2 − ipqi

⎞⎠⎞⎞Φ

(1.147)

whose integrability condition leads to the nonlinear evolution equations

ipi t = pxx − 2p2q, − iqt = qxx − 2pq2. (1.148)

This system of equations has a solution

p = α sech(αx)e−iα2t, q = −α sech(αx)eiα2t, (1.149)

which is derived from the trivial solution p = q = 0 by the Darbouxmatrix D = λI − HΛH−1 with

Λ =α

2

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ ,

H =

⎛⎝⎛⎛ e−iα2t/2 0

0 eiα2t/2

⎞⎠⎞⎞⎛⎝⎛⎛ eαx/2 −e−αx/2

e−αx/2 eαx/2

⎞⎠⎞⎞ .

(1.150)

Now we take (1.149) as a seed solution, whose corresponding fundamen-tal solution of the Lax pair (1.147) is

(λI − HΛH−1)

⎛⎝⎛⎛ eλx−2iλ2t 0

0 e−λx+2iλ2t

⎞⎠⎞⎞ . (1.151)

Take

Λ(ε) =

⎛⎝⎛⎛ ε 0

0 0

⎞⎠⎞⎞ , (1.152)

then we can choose

H(ε) =

⎛⎝⎛⎛ h(ε)11 h

(ε)12

h(ε)21 h

(ε)22

⎞⎠⎞⎞ , (1.153)


where

h(ε)11 =

(2ε2

α− ε tanh(αx)

)eθ − α

2sech(αx)e−iα2t−θ,

h(ε)12 = −ε tanh(αx) + sech(αx)e−iα2t,

h(ε)21 = −ε sech(αx)eiα2t+θ +

(ε +

α

2tanh(αx)

)e−θ,

h(ε)22 = −ε sech(αx)eiα2t − tanh(αx),

θ = εx − 2iε2t.

(1.154)

When ε → 0,H(ε)Λ(ε)(H(ε))−1 → S, (1.155)

S =α

∆

⎛⎝⎛⎛ sinh(αx)e−iα2t e−2iα2t

− sinh2(αx) − sinh(αx)e−iα2t

⎞⎠⎞⎞ , (1.156)

p′ =2αe−2iα2t

∆, q′ =

2α sinh2(αx)∆

(1.157)

where∆ = (α + 2 − 2 sech(αx)e−iα2t) cosh2(αx). (1.158)

Note that both eigenvalues of S are zero, but S = 0 . Hence S is notdiagonalizable. However, from the construction of S, we know that λI−S is a Darboux matrix, i.e., it satisfies (1.145).

Finally, the conclusions for the Darboux transformations of higherdegree and the theorem of permutability in Subsection 1.3.3 also holdfor the general Lax pair (1.137). Moreover, when U and V in (1.137)are generalized to rational functions of λ, similar conclusions hold [121].

1.4 KdV hierarchy, MKdV-SG hierarchy, NLShierarchy and AKNS system with u(N)reduction

In the last section, we discussed the Darboux transformations for theAKNS system and more general systems. In those cases, we supposedthat there were no reductions. In particular, there were no restrictionsamong the off-diagonal entries of P . However, in many cases, there areconstraints on P and the Darboux transformation should keep thoseconstraints. This problem is solved in many cases. Nevertheless, itshould be very interesting to establish a systematic method to treatwith reduced problems.


In this section, we first discuss some equations when N = 2 and thereis certain relation between p and q. They are the important special casesof the 2 × 2 AKNS system: (1) KdV hierarchy: p is real and q = −1;(2) MKdV-SG hierarchy: q = −p is real; (3) Nonlinear Schrodinger¨hierarchy: q = −p. These special cases were studied widely (e.g. [82, 88,91, 105, 117, 118]). Here we use a unified method to deal with the wholehierarchy, and the coefficients may depend on t [32, 45]. At the end ofthis section, we discuss the general AKNS system with u(N) reduction.This is a generalization of the nonlinear Schrodinger hierarchy and hasmany applications to other problems.

1.4.1 KdV hierarchyConsider the Lax pair [45]

Φx = UΦ, Φt = V Φ (1.159)

where

U =

⎛⎝⎛⎛ 0 1

ζ − u 0

⎞⎠⎞⎞ , V =

⎛⎝⎛⎛ A B

C −A

⎞⎠⎞⎞ , (1.160)

A, B and C are polynomials of the spectral parameter ζ.Compared with Section 1.2, the integrability condition (1.68) leads to

− Ax + C − B(ζ − u) = 0, Bx + 2A = 0,

−ut − CxCC + 2A(ζ − u) = 0.(1.161)

The first two equations imply

A = −12Bx,

C = ζB − uB − 12Bxx.

(1.162)

Substituting (1.162) into (1.161) we get

ut = −2(ζ − u)Bx + uxB +12Bxxx. (1.163)

Let

B =n∑

j=0

bj(x, t)ζn−j , (1.164)

then (1.163) leads to

b0,x = 0,

bj+1,x = ubj,x + 12uxbj + 1

4bj,xxx (0 ≤ j ≤ n − 1),(1.165)


ut = 2ubn,x + uxbn +12bn,xxx. (1.166)

(1.166) is the equation of u. When n ≥ 2, it is called a KdV equation ofhigher order.

Similar to Lemma 1.5, (1.165) leads to

bk =k∑

j=0

αk−j(t)b0j [u], (1.167)

where b0j [u]’s satisfy the recursion relations (1.165) and b0

0[0] = 1, b0j [0] =

0 (j ≥ 1). Clearly b0j [u]’s are determined by (1.165) uniquely.

The first few b0j ’s are

b00 = 1, b0

1 =12u,

b02 =

18uxx +

38u2, · · · .

(1.168)

The corresponding equations aren = 0: Linear equation

ut = α0(t)ux. (1.169)

n = 1:ut = α0(t)

(14uxxx +

32uux

)+ α1(t)ux. (1.170)

If α0 = constant, α1 = 0, (1.170) is the standard KdV equation.n = 2:

ut = α0(t)( 116

uxxxxx +58uuxxx +

54uxuxx +

158

u2ux

)+α1(t)

(14uxxx +

32uux

)+ α2(t)ux,

(1.171)

which is called the KdV equation of 5th-order.Next we discuss the Darboux transformation for the KdV hierarchy

by using the general results for the AKNS system. It seems that thecalculation is tedious. However, we can see the application of the generalresults more clearly. The method here is valid to the whole hierarchycomparing to the special method in Section 1.1.

The U and V given by (1.160) are different from those of the AKNSsystem. However, the Lax pair can be transformed to a Lax pair in theAKNS system by a similar transformation given by a constant matrixdepending on ζ.


Let

R =

⎛⎝⎛⎛ λ 1

−1 0

⎞⎠⎞⎞ (λ2 = ζ),

Ψ = RΦ,

U = RUR−1 =

⎛⎝⎛⎛ λ u

−1 −λ

⎞⎠⎞⎞ ,

V = RV R−1 =

⎛⎝⎛⎛ λB − A λ2B − 2λA − C

−B A − λB

⎞⎠⎞⎞ ,

(1.172)

then Ψ satisfiesΨx = UΨ, Ψt = V Ψ. (1.173)

This is the Lax pair for the KdV equation in the AKNS form. TheDarboux transformation can be constructed based on the discussion inSection 1.3. Take two constants λ1, λ2 and column solutions h1, h2 ofthe Lax pair when λ = λ1, λ2 respectively. Moreover, we want that thematrices given by the Darboux transformation are still of the form of(1.172). That is,

U ′ =

⎛⎝⎛⎛ λ u′

−1 −λ

⎞⎠⎞⎞ ,

V ′ =

⎛⎝⎛⎛ λB[u′] − A[u′] λ2B[u′] − 2λA[u′] − C[u′]

−B[u′] A[u′] − λB[u′]

⎞⎠⎞⎞ .

(1.174)

This condition (especially that the (2, 1) entry of U ′ is −1) holds onlywhen λ2, λ1, h2 and h1 are specified.

Suppose

⎛⎝⎛⎛ α

β

⎞⎠⎞⎞ is a solution of the Lax pair (1.173) for λ = λ0, then⎛⎝⎛⎛ α + 2λ0β

β

⎞⎠⎞⎞ is a solution of (1.173) for λ = −λ0. Thus we choose

Λ =

⎛⎝⎛⎛ λ0 0

0 −λ0

⎞⎠⎞⎞ , H =

⎛⎝⎛⎛ α α + 2λ0β

β β

⎞⎠⎞⎞ . (1.175)

Let

S = HΛH−1 =

⎛⎜⎛⎛⎝⎜⎜ −λ0 − 1τ

1τ2

+2λ0

τ

−11τ

+ λ0

⎞⎟⎞⎞⎠⎟⎟ (1.176)


where τ = β/α, and

D =

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ (λI − S), (1.177)

then after the action of the Darboux transformation given by the Dar-boux matrix D,

U ′ = DUD−1 + DxD−1 =

⎛⎝⎛⎛ λ u′

−1 −λ

⎞⎠⎞⎞ (1.178)

where

u′ = −u − 2(

1τ2

+2λ0

τ

). (1.179)

According to the general discussion to the AKNS system, V ′ is given bythe second equation of (1.174). Therefore, the Darboux transformationgiven by D in (1.177) is a Darboux transformation from any equation inthe KdV hierarchy to the same equation.

Next we compare the results here with those in Section 1.1. If

⎛⎝⎛⎛ α

β

⎞⎠⎞⎞is a solution of (1.173) for λ = λ0, then the corresponding solution of(1.159) is

R−1(λ0)

⎛⎝⎛⎛ α

β

⎞⎠⎞⎞ =

⎛⎝⎛⎛ −β

α + λ0β

⎞⎠⎞⎞ . (1.180)

Let σ be the ratio of the second and the first components, i.e.,

σ =α + λ0β

−β= −1

τ− λ0, (1.181)

then σ satisfies

σx = λ20 − u − σ2, (1.182)

and

S =

⎛⎝⎛⎛ σ σ2 − λ20

−1 −σ

⎞⎠⎞⎞ . (1.183)


In order to get the Darboux matrix for the Lax pair in the form(1.159), let

D = R−1

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ (λI − S)R =

⎛⎝⎛⎛ −σ 1

λ2 − λ20 + σ2 −σ

⎞⎠⎞⎞

=

⎛⎝⎛⎛ −σ 1

ζ − ζ0ζζ + σ2 −σ

⎞⎠⎞⎞ (ζ0ζζ = λ20).

(1.184)Then

U ′ = DUD−1 + DxD−1

=

⎛⎝⎛⎛ 0 1

ζ − 2ζ0ζζ + u + 2σ2 0

⎞⎠⎞⎞ .(1.185)

Hence the action of D keeps the x part of the Lax pair invariant, andtransforms u to

u′ = 2ζ0ζζ − u − 2σ2 (1.186)

(the same as (1.179)). Theorem 1.11 implies that V ′[u] = V [u′], i.e., theDarboux transformation keeps the t part invariant. Therefore, we have

Theorem 1.16 Suppose u is a solution of (1.166), ζ0ζζ is a non-zero

real constant,

⎛⎝⎛⎛ a

b

⎞⎠⎞⎞ is a solution of the Lax pair (1.159) for ζ = ζ0ζζ ,

σ = b/a, then

D =

⎛⎝⎛⎛ −σ 1

ζ − ζ0ζζ + σ2 −σ

⎞⎠⎞⎞ (1.187)

is a Darboux matrix for (1.159). It transforms a solution u of (1.166)to a new solution

u′ = 2ζ0ζζ − u − 2σ2 (1.188)

of the same equation.

Remark 12 In order to let U ′ and U have the same (2, 1)-entry −1, the

Darboux matrix (1.184) is chosen as D = R−1

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ (λI − S)R,

not R−1(λI − S)R. This guarantees that the transformation transformsa solution of (1.166) to a solution of the same equation (1.166).


Remark 13 Let Φ =

⎛⎝⎛⎛ φ

ψ

⎞⎠⎞⎞, then φ satisfies

φxx = (ζ − u)φ,

φt = Aφ + Bφx.(1.189)

It is similar to the system in Section 1.1. However, here A and B can bepolynomials of ζ of arbitrary degrees, whose coefficients are differentialpolynomials of u. The problem discussed in Section 1.1 was a specialcase.

In Theorem 1.16, b = ax, hence σ = ax/a. The transformation D in(1.184) gives

φ → φ′ = φx − σφ = φx − ax

aφ, (1.190)

and (1.182), (1.186) give the original Darboux transformation

u′ = u + 2(ln a)xx. (1.191)

Remark 14 From (1.165), we can get b0, b1, · · · recursively, whose in-tegral constants can be functions of t. Therefore, the coefficients of thenonlinear equations can be functions of t, as in the examples (1.170)and (1.171). The solutions of the equations whose coefficients dependingon t differ a lot from the solutions of the equations whose coefficientsindependent of t. In the latter case, each soliton moves in a fixed veloc-ity and the soliton with larger amplitude moves faster. However, in theformer case, each soliton can have varying velocity (e.g. oscillates), andthe soliton with larger amplitude may move slower.

1.4.2 MKdV-SG hierarchyConsider the Lax pair [32]

Φx = UΦ =

⎛⎝⎛⎛ λ p

−p −λ

⎞⎠⎞⎞Φ,

Φt = V Φ =

⎛⎝⎛⎛ A B

C −A

⎞⎠⎞⎞Φ,

(1.192)

where A, B and C are polynomials of λ and λ−1 satisfying

A(−λ) = −A(λ), B(−λ) = −C(λ). (1.193)

Moreover, suppose

A =n+m∑j=0

ajλ2n−2j+1 (1.194)


(m ≥ 0, n ≥ 0). Unlike the AKNS system, here A, B and C are notrestricted to the polynomials of λ, but the negative powers of λ areallowed. The term with lowest power of λ in (1.194) is an+mλ−2m+1.

The integrability condition UtUU − VxVV + [U, V ] = 0 leads to

Ax = p(B + C),

pt − Bx − 2pA + 2λB = 0,

pt + CxCC + 2pA + 2λC = 0.

(1.195)

Hence

B + C =Ax

p=

n+m∑j=0

aj,x

pλ2n−2j+1,

B − C =(B + C)x + 4pA4

2λ

=n+m∑j=0

(12

(aj,x

p

)x

+ 2paj

)λ2n−2j ,

(1.196)

(thus B(−λ) = −C(λ) holds automatically) and

pt =12(B − C)x − λ(B + C). (1.197)

Comparing the coefficients of λ in (1.197), we can obtain the recursionrelations among aj ’s. They include two parts. The first part

a0,x = 0,

aj+1,x =14p

((aj,x

p

)x

+ 4ajp

)x

(j = 0, 1, · · · , n − 1)(1.198)

are obtained from the coefficients of positive powers of λ and the secondpart ((

an+m,x

p

)x

+ 4an+mp

)x

= 0,((aj,x

p

)x

+ 4ajp

)x

= 4aj+1,x

p

(j = n + m − 1, · · · , n + 1)

(1.199)

are obtained from the coefficients of negative powers of λ. Moreover, theterm without λ leads to the equation

pt − 14

((an,x

p

)x

+ 4anp

)x

+an+1,x

p= 0. (1.200)


The first few aj ’s (0 ≤ j ≤ n) are

a0 = α0(t),

a1 =12α0(t)p2 + α1(t),

a2 = α0(t)(

14ppxx − 1

8p2

x +38p4)

+12α1(t)p2 + α2(t),

· · · .

(1.201)

If V does not contain negative powers of λ, i.e., m = 0, then from thegeneral conclusion to the AKNS system, all aj ’s are differential polyno-mials of p. The equation (1.200) becomes

pt − 14

((an,x

p

)x

+ 4anp

)x

= 0. (1.202)

This is called the MKdV hierarchy. By using the notion a0j in Section 1.2,

these equations can be written as

pt +n∑

j=0

αj(t)MnMM −j [p[[ ] = 0, (1.203)

where

MlMM [p[[ ] = −14

((a0

l,x

p

)x

+ 4a0l p

)x

(l = 0, 1, · · · , n). (1.204)

Especially, if n = 1 and α0 = −4, α1 = 0, then (1.202) becomes theMKdV equation

pt + pxxx + 6p2px = 0. (1.205)

Next, we consider the negative powers of λ in V . Take p = −ux/2 andsuppose it satisfies the boundary condition: u − kπ and its derivativestend to 0 fast enough as x → −∞ (k is an integer).

The first equation of (1.199) gives((an+m,x

ux

)x

+ an+mux

)x

= 0. (1.206)

Write an+m as a function of u, then the above equation becomes

((an+m,uu + an+m)ux)x = 0. (1.207)

The boundary condition as x → −∞ gives an+m,uu + an+m = 0. Hence

an+m = α cos(u + β) (1.208)


where α and β are constants.Now take a special an+m: a0

n+m = 14 cos u. an+j (j = 1, 2, · · · , m − 1)

can be determined as follows. Let gn+j = a0n+j,x/p, then gn+m = 1

2 sin u,and

a0j−1 =

∫ x

−∞

∫∫pgjg −1 dx + a−j−1 (n + 2 ≤ j ≤ n + m) (1.209)

where a−j−1 is the limit of a0j−1 as x → −∞. From the boundary condition

limx→−∞(gjg −1)x = 0, the recursion relations (1.199) become

14(gjg −1)x + p

(a−j−1 +

∫ x

−∞

∫∫pgjg −1 dx

)=∫ x

−∞

∫∫gjg dx. (1.210)

Moreover, supposelim

x→−∞ gjg −1 = 0, (1.211)

then

gjg −1 + 4∫ x

−∞

∫∫p(ξ)

(a−j−1 +

∫ ξ

−∞

∫∫p(ζ)gjg −1(ζ) dζ

)dξ

= 4∫ x

−∞

∫∫ ∫ ξ

−∞

∫∫gjg (ζ) dζ dξ.

(1.212)

This is an integral equation of Volterra type. It has a unique solution inthe class of functions which tend to zero fast enough together with itsderivatives as x → −∞.

Take a−j−1 = 0 and write the solution of (1.212) as

gjg −1 = Q(gjg ) = Q2(gjg +1) = · · · =12Qn+m−j+1[sinu]. (1.213)

Here Q is the operator to determine gjg −1 from gjg defined by (1.212).gjg −1 is not a differential polynomial of gjg .

If n = 0, α0 = 0, then we obtain the SG hierarchy

pt +12

m−1∑j=0

βjβ (t)Qm−j−1[sinu] = 0 (1.214)

where βjβ (t)’s are arbitrary functions of t.Generally, we have the compound MKdV-SG hierarchy

pt +n∑

j=0

αj(t)MnMM −j [p[[ ] +12

m−1∑j=0

βjβ (t)Qm−j−1[sinu] = 0,

(p = −ux2 ).

(1.215)


Example 1.17 n = 0, m = 2, β0 = 0, β1 = 1, then, g2 = 12 sinu, and

the equation becomes the sine-Gordon equation

uxt = sinu. (1.216)

Example 1.18 n = 1, m = 2, α0 = −4, α1 = 0, β0 = 0, β1 = 1, thenthe equation becomes the equation describing one-dimensional nonlinearlattice of atoms [70]

uxt +32u2

xuxx + uxxxx − sin u = 0. (1.217)

Now we consider the Darboux transformation. If

⎛⎝⎛⎛ α

β

⎞⎠⎞⎞ is a solution

of (1.192) for λ = λ0, then

⎛⎝⎛⎛ −β

α

⎞⎠⎞⎞ is a solution of (1.192) for λ = −λ0.

Therefore, we can choose

Λ =

⎛⎝⎛⎛ λ0 0

0 −λ0

⎞⎠⎞⎞ , H =

⎛⎝⎛⎛ α −β

β α

⎞⎠⎞⎞ , (1.218)

where

⎛⎝⎛⎛ α

β

⎞⎠⎞⎞ is a solution of (1.192) for λ = λ0. Let σ = β/α,

S = HΛH−1 =λ0

1 + σ2

⎛⎝⎛⎛ 1 − σ2 2σ

2σ σ2 − 1

⎞⎠⎞⎞ (1.219)

and denote tanθ

2= σ, then

S = λ0

⎛⎝⎛⎛ cos θ sin θ

sin θ − cos θ

⎞⎠⎞⎞ . (1.220)

From σx = −p(1 + σ2) − 2λ0σ, we have

θx = −2p − 2λ0 sin θ. (1.221)

By direct calculation,

(λI − S)U(λI − S)−1 − SxS (λI − S)−1 =

⎛⎝⎛⎛ λ p′

−p′ −λ

⎞⎠⎞⎞ (1.222)


where

p′ = p + 2λ0 sin θ = −p − θx, (1.223)

or equivalently,

u′ = −u + 2θ (1.224)

for suitable choice of the integral constant.It remains to prove that the Darboux matrix λI − S keeps the re-

duction of MKdV-SG hierarchy. This includes (1) the transformed A′,B′ and C ′ still satisfy A′(−λ) = −A′(λ) and B′(−λ) = −C(λ); (2) thecoefficients αj(t)’s keeps invariant.

Since V T (−λ) = −V (λ), ST = S and (λI +S)T (λI−S) = λ2I−S2 =(λ2−λ2

0)I, it can be verified by direct calculation that V ′T (−λ) = −V ′(λ)holds. This proves (1).

(2) is proved as follows. For aj (j ≤ n), this has been proved forthe AKNS system; for aj (j ≥ n + 1), the conclusion follows from theboundary condition at infinity.

Therefore, the following theorem holds.

Theorem 1.19 Suppose u is a solution of (1.200), λ0 is a non-zero real

number,

⎛⎝⎛⎛ a

b

⎞⎠⎞⎞ is a solution of the Lax pair (1.192) for λ = λ0. Let

θ = 2 tan−1(b/a),

S = λ0

⎛⎝⎛⎛ cos θ sin θ

sin θ − cos θ

⎞⎠⎞⎞ , (1.225)

then λI − S is a Darboux matrix for (1.192). It transforms a solutionp of (1.200) to the solution p′ = p + 2λ0 sin θ of the same equation.Moreover, u′ = −u + 2θ with suitable boundary condition, where p =−ux/2, p′ = −u′

x/2.

Remark 15 For the sine-Gordon equation, the Backlund transformationïs a kind of method to get explicit solutions, which was known in thenineteenth century. In that method, to obtain a new solution from aknown solution, there is an integrable system of differential equations tobe solved (moreover, one can obtain explicit expression by using the the-orem of permutability and the nonlinear superposition formula). UsingDarboux transformation, that explicit expression can be obtained directly.This will be discussed in Chapter 4 together with the related geometricproblems.


1.4.3 NLS hierarchyThe Lax pair for the nonlinear Schrodinger hierarchy (NLS hierarchy)¨

is

Φx = UΦ =

⎛⎝⎛⎛ λ p

−p −λ

⎞⎠⎞⎞Φ,

Φt = V Φ =

⎛⎝⎛⎛ A B

C −A

⎞⎠⎞⎞Φ,

(1.226)

where A, B and C are polynomials of λ (λ, p, A, B and C are complex-valued) satisfying

A(−λ) = −A(λ), B(−λ) = −C(λ) (1.227)

(i.e., V ∗(−λ) = −V (λ) where ∗ refers to the complex conjugate trans-pose of a matrix). This is also a special case of the AKNS system. Weshall construct a Darboux matrix keeping this reduction.

The integrability condition UtUU − VxVV + [U, V ] = 0 is

Ax = pC + pB,¯

Bx = pt + 2λB − 2pA,

CxCC = −pt − 2λC − 2pA.¯

(1.228)

We can use (1.55) to write down the coefficients of the powers of λ inA, B and C. They depend on p, px, · · · and the integral constants αj(t).Moreover, there is a nonlinear evolution equation

pt = bn,x + 2pan. (1.229)

Especially, for n = 2, α0 = −2i, α1 = α2 = 0, the equation is thenonlinear Schrodinger equation

ipi t = pxx + 2|p|2p. (1.230)

The Darboux transformation for the nonlinear Schrodinger hierarchy¨

is also constructed from the choice of Λ and H. Suppose

⎛⎝⎛⎛ α

β

⎞⎠⎞⎞ is a

solution of (1.226) for λ = λ0, then

⎛⎝⎛⎛ −β

α

⎞⎠⎞⎞ is a solution of (1.226) for


λ = −λ0. Hence, we can choose

Λ =

⎛⎝⎛⎛ λ0 0

0 −λ0

⎞⎠⎞⎞ , H =

⎛⎝⎛⎛ α −β

β α

⎞⎠⎞⎞ , (1.231)

S = HΛH−1 =1

1 + |σ|2

⎛⎝⎛⎛ λ0 − λ0|σ|2 (λ0 + λ0)σ

(λ0 + λ0)σ −λ0 + λ0|σ|2

⎞⎠⎞⎞ , (1.232)

where σ = β/α. Since detH = 0, S can be defined globally. It can bechecked that H satisfies

H∗H = |α|2 + |β|2, (1.233)

hence S satisfiesS∗S = |λ0|2,S − S∗ = λ0 − λ0.

(1.234)

Therefore, under the action of the Darboux matrix λI − S, U is trans-formed to

U ′ = (λI − S)U(λI − S)−1 − SxS (λI − S)−1 =

⎛⎝⎛⎛ λ p′

−p′ −λ

⎞⎠⎞⎞ (1.235)

where

p′ = p + 2S12 = p +2(λ0 + λ0)σ

1 + |σ|2 . (1.236)

From the discussion on the AKNS system (Theorem 1.11), we knowthat V ′ = (λI − S)V (λI − S)−1 − StSS (λI − S)−1 is also a polynomialof λ and V ′∗(−λ) = −V ′(λ) holds. Moreover, λI − S gives a Darbouxtransformation from an equation in the nonlinear Schrodinger hierarchy¨to the same equation. This leads to the following theorem.

Theorem 1.20 Suppose p is a solution of (1.229), λ0 is a non-real

complex number,

⎛⎝⎛⎛ a

b

⎞⎠⎞⎞ is a solution of the Lax pair (1.226) for λ = λ0.

Let σ = b/a,

S =1

1 + |σ|2

⎛⎝⎛⎛ λ0 − λ0|σ|2 (λ0 + λ0)σ

(λ0 + λ0)σ −λ0 + λ0|σ|2

⎞⎠⎞⎞ , (1.237)


then λI − S is a Darboux matrix for (1.226). It transforms a solution pof (1.229) to a solution

p′ = p +2(λ0 + λ0)σ

1 + |σ|2 (1.238)

of the same equation.

1.4.4 AKNS system with u(N) reductionFor the nonlinear Schrodinger hierarchy, (1.226) and (1.227) imply¨

U(−λ) = −U(λ)∗, V (−λ) = −V (λ)∗. (1.239)

Here we generalize it to the AKNS system.For the AKNS system (1.226), if U and V satisfy (1.239), then we

say that (1.226) has u(N) reduction, because U(λ) and V (λ) are in theLie algebra u(N) when λ is purely imaginary. This is a very popularreduction.

We want to construct Darboux matrix which keeps u(N) reduction.That is, after the action of the Darboux matrix, the derived potentialsU ′(λ) and V ′(λ) must satisfy

U ′(−λ) = −U ′(λ)∗, V ′(−λ) = −V ′(λ)∗. (1.240)

With this additional condition, Λ and H in (1.94) can not be arbitrary.They should satisfy the following two conditions:

(1) λ1, · · · , λN can only be µ or −µ where µ is a complex number (µis not real).

(2) If λj = λk, thenh∗

jhk = 0 (1.241)

holds at one point (x0, t0).In fact, if (1.241) holds at one point, then it holds everywhere. This

is proved as follows.When λj = λk, λk = −λj , hence

hk,x = U(λk)hk, hk,t = V (λk)hk,

h∗j,x = h∗

jU(λj)∗ = −h∗jU(−λj) = −h∗

jU(λk),

h∗j,t = h∗

jV (λj)∗ = −h∗jV (−λj) = −h∗

jV (λk).

(1.242)

This implies that(h∗

jhk)x = 0, (h∗jhk)t = 0. (1.243)

Therefore, h∗jhk = 0 holds everywhere if it holds at one point.


Theorem 1.21 If λj’s, hj’s satisfy the above conditions (1) and (2),H = (h1, · · · , hN ), then det H = 0 holds everywhere if it holds at onepoint. Moreover, U ′ and V ′ given by (1.81) satisfy

U ′(−λ) = −U ′(λ)∗, V ′(−λ) = −V ′(λ)∗. (1.244)

Proof. Let (x0, t0) be a fixed point. Then by the property of lineardifferential equation, all hα with λα = µ are linearly independent ifthey are linearly independent at (x0, t0). Likewise, all hα with λα = µare also linearly independent if they are linearly independent at (x0, t0).Moreover, (1.241) implies that all h1, · · · , hN are linearly independentif they are linearly independent at (x0, t0). Therefore, det H = 0 andS = HΛH−1 is globally defined.

According to the definition of S,

Shj = λjhj , h∗kS

∗ = h∗kλk. (1.245)

Henceh∗

k(S − S∗)hj = (λj − λk)h∗khj . (1.246)

If λj = µ, λk = −µ, then

h∗k(S − S∗)hj = 0. (1.247)

If λj = λk = µ (or λj = λk = −µ), then

h∗k(S − S∗)hj = (µ − µ)h∗

khj . (1.248)

HenceS − S∗ = (µ − µ)I. (1.249)

On the other hand, from (1.245), we have

h∗kS

∗Shj = λj λkh∗khj . (1.250)

If λj = µ, λk = −µ, then

h∗kS

∗Shj = 0. (1.251)

If λj = λk = µ (or λj = λk = −µ), then

h∗kS

∗Shj = |µ|2h∗khj , (1.252)

Therefore,S∗S = |µ|2I. (1.253)

From (1.249) and (1.253), we obtain

(λI + S)∗(λI − S) = (λ − µ)(λ + µ)I. (1.254)


According to the action of the Darboux transformation on VjVV ,

m∑j=0

V ′jVV λm−j

= (λI − S)m∑

j=0

VjVV λm−j(λI − S)−1 + (λI − S)t(λI − S)−1,

(1.255)

(m∑

j=0

V ′jVV λm−j)∗

= (λI − S)∗−1m∑

j=0

V ∗jVV λm−j(λI − S)∗ + (λI − S)∗−1(λI − S)∗t

= −(λI + S)m∑

j=0

VjVV (−λ)m−j(λI + S)−1 − (λI + S)t(λI + S)−1

= −m∑

j=0

V ′jVV (−λ)m−j .

(1.256)Hence V ′(−λ) = −V ′(λ)∗. Likewise, U ′(−λ) = −U ′(λ)∗. The theoremis proved.

As in Section 1.3, a Darboux transformation of higher degree can bederived by the composition of Darboux transformations of degree one.However, with the u(N) reduction, we have also the following specialand more direct construction [117, 17].

Suppose we take l times of Darboux transformations of degree one.Each Darboux transformation is constructed from Λα, HαHH (α = 1, · · · , l).In each Λα = diag(λ(α)

1 , · · · , λ(αk ), suppose λ

(α)1 = · · · = λ

(α)k = µα,

λ(α)k+1 = · · · = λ

(α)N = −µα. Here k is the same for all α. For each

λ(α)j , solve the Lax pair and get a solution h

(α)j satisfying the orthogonal

relations (1.241).Denote HαHH = (h(α)

1 , · · · , h(α)N ), Hα = (h(α)

1 , · · · , h(α)k ). Let

Γαβ =H∗

α Hβ

µβ + µα, (1.257)

D(λ) =l∏

γ=1

(λ + µγ)

⎛⎝⎛⎛1 −l∑

α,β=1

Hα(Γ−1)αβ H∗β

λ + λβ

⎞⎠⎞⎞ . (1.258)

Now we prove that D(λ) is a Darboux matrix.


Let HαHH = (h(α)k+1, · · · , h(α)

N ). Then HαHH = ( Hα, HαHH ) and H∗αHαHH = 0 for

all α = 1, · · · , l. Hence

D(µα) Hα = 0, D(−µα)HαHH = 0. (1.259)

According to Theorem 1.12, D(λ) is a Darboux matrix.Moreover, the inverse of D(λ) can be written out explicitly as

D(λ)−1 =l∏

γ=1

(λ + µγ)−1

⎛⎝⎛⎛1 +l∑

α,β=1


λ − λα

⎞⎠⎞⎞ . (1.260)

(1.258) gives a compact form of Darboux matrix of higher degree.Although it is special, it is very useful.

1.5 Darboux transformation and scattering,inverse scattering theory

The scattering and inverse scattering theory is an important part ofthe soliton theory. It transforms the problem of solving the Cauchyproblem of a nonlinear partial differential equation to the problem ofdescribing the spectrum and eigenfunctions of the Lax pair. Here weconsider the 2 × 2 AKNS system as an example to show the outline ofthe scattering and inverse scattering theory (see [23] for details). More-over, we discuss the change of the scattering data under Darboux trans-formation for su(2) reduction. For the KdV equation, the problem canbe solved similarly, but the scattering and inverse scattering theory issimpler.

1.5.1 Outline of the scattering and inversescattering theory for the 2× 2 AKNS system

First, we give the definition of the scattering data for the 2×2 complexAKNS system. In order to coincide with the usual scattering theory, letλ = − iζ, then the first equation of the 2×2 AKNS system (1.48) becomes

Φx =

⎛⎝⎛⎛ − iζ p

q iζ

⎞⎠⎞⎞Φ. (1.261)

Suppose p, q and their derivatives with respect to x decay fast enoughat infinity. Let C be the complex plane and R be the real line. BesidesC+ and C− are the upper and lower half plane of C respectively, i.e.,C+ = z ∈ C | Im ζ > 0, C− = z ∈ C | Im ζ < 0.


Property 1. For each one of the following boundary conditions, theequation (1.261) has a unique column solution

(1) ψr(x, ζ) = R(x, ζ)e−iζx, limx→−∞R(x, ζ) =

⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞ ,

( Im ζ ≥ 0),

(1.262)

(2) ψr(x, ζ) = R(x, ζ)eiζx, limx→−∞ R(x, ζ) =

⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞ ,

( Im ζ ≤ 0),

(1.263)

(3) ψl(x, ζ) = L(x, ζ)eiζx, limx→+∞L(x, ζ) =

⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞ ,

( Im ζ ≥ 0),

(1.264)

(4) ψl(x, ζ) = L(x, ζ)e−iζx, limx→+∞ L(x, ζ) =

⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞ ,

( Im ζ ≤ 0).

(1.265)

Moreover, ψr, ψl (resp. ψr, ψl) are continuous for ζ ∈ C+ ∪ R (resp.z ∈ C− ∪ R), and holomorphic with respect to ζ in C+ (resp. C−).These solutions are called Jost solutions.

If ζ ∈ R, then ψl and ψl are linearly independent. Hence, there existfunctions r+(ζ), r−(ζ), r+(ζ) and r−(ζ) such that

ψr = r+ψl + r−ψl,

ψr = r+ψl + r−ψl.(1.266)

Considering the Wronskian determinant between ψr, ψl and the Wron-skian determinant between ψr, ψl, we have

Property 2. For ζ ∈ R,

r−(ζ) = R1(x, ζ)L2(x, ζ) − R2(x, ζ)L1(x, ζ),

r+(ζ) = R2(x, ζ)L1(x, ζ) − R1(x, ζ)L2(x, ζ).(1.267)

r−(ζ) can be holomorphically extended to C+ ∪ R, and r+(ζ) can beholomorphically extended to C− ∪ R. Here R1 and R2 are two compo-nents of the vector R, i.e., R = (R1, R2)T . L1, L2, R1, R2, L1, L2 havethe similar meanings.


The asymptotic properties of the four Jost solutions in Property 1 asx → ±∞ are listed in the next property.

Property 3. (1) The following limits hold uniformly for ζ:

limx→−∞R(x, ζ) =

⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞ , ζ ∈ C+ ∪ R,

limx→−∞ R(x, ζ) =

⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞ , ζ ∈ C− ∪ R,

limx→+∞L(x, ζ) =

⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞ , ζ ∈ C+ ∪ R,

limx→+∞ L(x, ζ) =

⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞ , ζ ∈ C− ∪ R.

(1.268)

(2) The following limits hold uniformly for ζ in a compact subset:

limx→+∞R(x, ζ) =

⎛⎝⎛⎛ r−(ζ)

0

⎞⎠⎞⎞ , ζ ∈ C+,

limx→+∞ R(x, ζ) =

⎛⎝⎛⎛ 0

r+(ζ)

⎞⎠⎞⎞ , ζ ∈ C−,

limx→−∞L(x, ζ) =

⎛⎝⎛⎛ 0

r−(ζ)

⎞⎠⎞⎞ , ζ ∈ C+,

limx→−∞ L(x, ζ) =

⎛⎝⎛⎛ r+(ζ)

0

⎞⎠⎞⎞ , ζ ∈ C−.

(1.269)

(3) The following limits hold uniformly for real ζ ∈ R:

limx→+∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣R(x, ζ) −⎛⎝⎛⎛ r−(ζ)

r+(ζ)e2iζx

⎞⎠⎞⎞∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ = 0, ζ ∈ R,

limx→+∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣R(x, ζ) −⎛⎝⎛⎛ r−(ζ)e−2iζx

r+(ζ)

⎞⎠⎞⎞∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ = 0, ζ ∈ R,

limx→−∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣L(x, ζ) −⎛⎝⎛⎛ −r−(ζ)e−2iζx

r−(ζ)

⎞⎠⎞⎞∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ = 0, ζ ∈ R,


limx→−∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣L(x, ζ) −⎛⎝⎛⎛ r+(ζ)

−r+(ζ)e2iζx

⎞⎠⎞⎞∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ = 0, ζ ∈ R.

Rewrite (1.261) asLΦ = ζΦ, (1.270)

where

L =

⎛⎝⎛⎛ i 0

0 − i

⎞⎠⎞⎞⎛⎝⎛⎛ ddx

−⎛⎝⎛⎛ 0 p

q 0

⎞⎠⎞⎞⎞⎠⎞⎞ , (1.271)

then (1.261) becomes a spectral problem of a linear ordinary differentialoperator. We consider its spectrum in L2(R) × L2(R).

If ζ ∈ C+ and r−(ζ) = 0, then (1.267) implies that ψr and ψl arelinearly dependent. Hence ψr → 0 as x → ±∞. Similarly, if ζ ∈ C−and r+(ζ) = 0, then ψr and ψl are linearly dependent. Hence ψr → 0as x → ±∞. Since r−(ζ) and r+(ζ) are holomorphic in C+ and C−respectively, their zeros are discrete. These zeros are the eigenvalues ofL. The set of all eigenvalues of L is denoted by IPσ(L). If ζ ∈ R, thenit can be proved that (1.270) has a nontrivial bounded solution. σ(L) =R ∪ IPσ(L) is called the spectrum of the operator L. Its complimentC − σ(L) is called the regular set of L.

Property 4. If r−(ζ) = 0 and r+(ζ) = 0 hold for ζ ∈ R, then IPσ(L)is a finite set.

Hereafter, we always suppose r−(ζ) = 0 and r+(ζ) = 0 when ζ ∈ R.First we consider the eigenvalues.

If ζ ∈ IPσ(L), then ψr and ψl are linearly dependent. Suppose

ψr(x, ζ) = α(ζ)ψl(x, ζ) (ζ ∈ C+ ∩ IPσ(L)),

ψr(x, ζ) = α(ζ)ψl(x, ζ) (ζ ∈ C− ∩ IPσ(L)).(1.272)

Denote IPσ(L) ∩ C+ = ζ1, · · · ζdζζ and IPσ(L) ∩ C− = ζ1, · · · ζdζ to

be the set of eigenvalues in C+ and C− respectively. Moreover, supposeζ1, · · ·, ζ

dζ are all simple zeros. Corresponding to each eigenvalue, there

is a constant

CkC = α(ζk)/

dr−(ζk)dζ

(k = 1, · · · , d),

CkC = α(ζk)/

dr+(ζk)dζ

(k = 1, · · · , d).(1.273)


Using these data, we define the functions

Bd(y) = − id∑

k=1

CkC eiζky, Bd(y) = id∑

k=1

CkC e−iζky. (1.274)

Next, we consider the continuous spectrum ζ ∈ R. As is known, theFourier transformation of a Schwarz function φ is

F (φ)(k) =1√2π

∫ +∞

−∞

∫∫φ(s)e−iks ds. (1.275)

It can be extended to L2(R) and becomes a bounded map from L2(R)to L2(R).

Property 5.

L(x, ·) −⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞ ∈ L2(R), L(x, ·) −⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞ ∈ L2(R). (1.276)

Denote

N(x, s) =1√2π

F

⎛⎝⎛⎛L(x, ·) −⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞⎞⎠⎞⎞ (s),

N(x, s) = 1√2π

F

⎛⎝⎛⎛L(x, ·) −⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞⎞⎠⎞⎞ (s), (s ≥ 0),

(1.277)

then

L(x, ζ) =

⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞+∫ +∞

0

∫∫N(x, s)eiζs ds, ∀ζ ∈ C+ ∪ R,

L(x, ζ) =

⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞+∫+∞0

∫∫N(x, s)e−iζs ds, ∀ζ ∈ C− ∪ R,

(1.278)

and the above integrals converge absolutely. Moreover,

p(x) = −2N1NN (x, 0), q(x) = −2N2NN (x, 0), (1.279)

where the subscripts refer to the components.

For ζ ∈ R, denote

b(ζ) =r+(ζ)r−(ζ)

, b(ζ) =r−(ζ)r+(ζ)

. (1.280)


It can be proved that

b, b ∈ L2(R) ∩ L1(R) ∩ C0(R). (1.281)

Hence, we can define

Bc(y) =12π

∫ +∞

−∞

∫∫b(ζ)eiζy dζ,

Bc(y) =12π

∫ +∞

−∞

∫∫b(ζ)e−iζy dζ.

(1.282)

The data

ζk, CkC (k = 1, · · · , d), ζk, CkC (k = 1, · · · , d), b(ζ), b(ζ) (ζ ∈ R) (1.283)

are called the scattering data corresponding to (p, q), denoted by Σ(p, q).We can also call

r−(ζ) (ζ ∈ C+ ∪ R), r+(ζ) (ζ ∈ C− ∪ R) (1.284)

the scattering data, since the data in (1.283) can be obtained from thedata in (1.284).

Define B = Bc+Bd and B = Bc+Bd according to (1.274) and (1.282).

Property 6. N and N satisfy the follow system of linear integralequations (Gelfand-Levitan-Marchenko equations)

N(x, s) + B(2x + s)

⎛⎝⎛⎛ 1

0

⎞⎠⎞⎞ +∫ +∞

0

∫∫N(x, σ)B(2x + s + σ) dσ = 0,

N(x, s) + B(2x + s)

⎛⎝⎛⎛ 0

1

⎞⎠⎞⎞ +∫ +∞

0

∫∫N(x, σ)B(2x + s + σ) dσ = 0.

(1.285)

If the scattering data ζk, CkC , ζk, CkC , b(ζ), b(ζ) are known, N and Nare solved from the above integral equations and (1.279) gives (p, q).

The process to get scattering data from (p, q) is called the scatteringprocess. It needs to solve the spectral problem of ordinary differentialequations. The process to get (p, q) from the scattering data is calledthe inverse scattering process. It needs to solve linear integral equations.

Now we consider the evolution of the scattering data. In the AKNSsystem, p and q are functions of (x, t). Therefore, we should considerthe full AKNS system (with time t)

Φx =

⎛⎝⎛⎛ − iζ p

q iζ

⎞⎠⎞⎞Φ, Φt =

⎛⎝⎛⎛ A B

C −A

⎞⎠⎞⎞Φ, (1.286)


where A, B and C are polynomials of ζ,

A =n∑

j=0

aj(− iζ)n−j ,

B =n∑

j=0

bj(− iζ)n−j ,

C =n∑

j=0

cjc (− iζ)n−j .

(1.287)

We also supposeA|p=q=0 = iω(ζ, t). (1.288)

Lemma 1.5 implies B|p=q=0 = C|p=q=0 = 0.

Property 7. Suppose (p, q) satisfies the equations

pt = bn,x + 2pan, qt = cn,x − 2qan (1.289)

given by the integrability condition, then the evolution of the corre-sponding scattering data is given by

r−(ζ, t) = r−(ζ, 0) ζ ∈ C+ ∪ R,

r+(ζ, t) = r+(ζ, 0) ζ ∈ C− ∪ R,

r+(ζ, t) = r+(ζ, 0) exp(−2i∫ t

0

∫∫ω(ζ, τ) dτ) ζ ∈ R,

r−(ζ, t) = r−(ζ, 0) exp(2i∫ t

0

∫∫ω(ζ, τ) dτ) ζ ∈ R,

(1.290)

andζk(t) = ζk(0),

ζk(t) = ζk(0),

CkC (t) = CkC (0) exp(−2i∫ t

0

∫∫ω(ζ, τ) dτ),

CkC (t) = CkC (0) exp(2i∫ t

0

∫∫ω(ζ, τ) dτ),

b(ζ, t) = b(ζ, 0) exp(−2i∫ t

0

∫∫ω(ζ, τ) dτ),

b(ζ, t) = b(ζ, 0) exp(2i∫ t

0

∫∫ω(ζ, τ) dτ).

(1.291)

(1.290) or (1.291) gives the evolution of the scattering data explicitly.


In summary, the process of solving the initial value problem of non-linear evolution equations (1.289) of (p, q) is as follows. Here the initialcondition is t = 0 : p = p0, q = q0.

For given (p0, q0), first solve the x-part of the Lax pair (1.286) forp = p0, q = q0 and get the scattering data corresponding to p0 and q0.Then, using the evolution of the scattering data (1.291), the scatteringdata corresponding to (p(t), q(t)) are obtained. Finally, solve the integralequations (1.285) to get (p(t), q(t)). Therefore, the inverse scatteringmethod changes the initial value problem of nonlinear partial differentialequations to the problem of solving systems of linear integral equations.This gives an effective way to solve the initial value problem. Especially,when br = br = 0, Bc = Bc = 0, (1.285) has a degenerate kernel. Henceit can be solved algebraically and the soliton solutions can be obtained.Please see [23] for details.

Remark 16 Denote Σ(p, q) to be the scattering data corresponding to(p(x, t), q(x, t)), p0(x) and q0(x) to be the initial values of p and q att = 0, then the procedure of inverse scattering method can be shown inthe following diagram:

t = 0 : (p0, q0)

t = t : (p, q)

scattering

inverse scattering

Σ(p0, q0)

Σ(p, q)

(1.292)

For a linear equation, if “scattering” is changed to “Fourier transfor-mation” and “inverse scattering” is changed to “inverse Fourier trans-formation” in the above diagram, then it becomes the diagram for solvingthe initial value problem by Fourier transformations which has been usedextensively for linear problems. Therefore, the scattering and inversescattering method can be regarded as a kind of Fourier method for non-linear problems.

1.5.2 Change of scattering data under Darbouxtransformations for su(2) AKNS system

For the AKNS system, the scattering data include

ζk, CkC , ζk, CkC , b(ζ), b(ζ). (1.293)

The su(2) AKNS system means that U, V ∈ su(2) for ζ ∈ R, i.e., q = −p,A = −A, C = −B. Therefore, it is just the nonlinear Schrodinger


hierarchy. The Lax pair is

Φx =

⎛⎝⎛⎛ − iζ p

−p iζ

⎞⎠⎞⎞Φ,

Φt =

⎛⎝⎛⎛ A B

C −A

⎞⎠⎞⎞Φ

(1.294)

where A(ζ) = −A(ζ), B(ζ) = −C(ζ). Here we consider the su(2) AKNSsystem instead of general 2 × 2 AKNS system because the Darbouxtransformation may exist globally in this case.

If

⎛⎝⎛⎛ α(ζ)

β(ζ)

⎞⎠⎞⎞ is a solution of (1.294), then

⎛⎝⎛⎛ −β(ζ)

α(ζ)

⎞⎠⎞⎞ is also its so-

lution. This leads to the following property.

Property 8. For the Lax pair (1.294), if ζ ∈ R, then there arefollowing relations among the Jost solutions and the scattering data:

R1 = −R2, R2 = R1,

L1 = L2, L2 = −L1,(1.295)

r+(ζ) = −(ζ), r−(ζ) = +(ζ). (1.296)

By reordering the eigenvalues,

d = d, ζk = ζk, CkC = −CkC , b(ζ) = b(ζ) (ζ ∈ R). (1.297)

Therefore, for the su(2) AKNS system, the scattering data can bereduced to ζk ∈ C+, CkC (k = 1, 2, · · · , d) and b(ζ) (ζ ∈ R).

Now we consider the change of the scattering data under Darbouxtransformations.

From the discussion on the nonlinear Schrodinger hierarchy, we know¨that if p is defined globally on (−∞, +∞), so is the Darboux matrix.In order to use the scattering theory, we want that p and its derivativestend to 0 fast enough at infinity.

Take a constant µ and a column solution of the Lax pair

ψr(ζ0ζζ ) − µψl(ζ0ζζ ) =

⎛⎝⎛⎛ R1(ζ0ζζ )e−iζ0x − µL1(ζ0ζζ )eiζ0x

R2(ζ0ζζ )e−iζ0x − µL2(ζ0ζζ )eiζ0x

⎞⎠⎞⎞

(ζ0ζζ ∈ C+).

(1.298)

Let

σ =R2(ζ0ζζ ) − µL2(ζ0ζζ )e2iζ0x

R1(ζ0ζζ ) − µL1(ζ0ζζ )e2iζ0x(1.299)


be the ratio of the second and the first components. Then the Darbouxmatrix is

− iζI − S

= − iζI − 11 + |σ|2

⎛⎝⎛⎛ − iζ0ζζ − iζ0ζζ |σ|2 (− iζ0ζζ + iζ0ζζ )σ

(− iζ0ζζ + iζ0ζζ )σ − iζ0ζζ − iζ0ζζ |σ|2

⎞⎠⎞⎞ ,

(1.300)

and the solution is transformed by

p′ = p + 2i(ζ0ζζ − ζ0ζζ )σ1 + |σ|2 . (1.301)

The change of the scattering data under Darboux transformation is givenby the following theorem [75].

Theorem 1.22 If the scattering data for (1.294) are r−(ζ) (ζ ∈ C+ ∪R), r+(ζ) (ζ ∈ R) and α(ζk) (k = 1, · · · , d), then, under the action ofthe Darboux matrix (1.300) (µ(( = 0 , ζ0ζζ ∈ C+), the scattering data arechanged as follows:

(1) If ζ0ζζ is not an eigenvalue, then, after the action of the Darbouxtransformation, the number of eigenvalues increase one. All the originaleigenvalues are not changed, and ζ0ζζ is a unique additional eigenvalue.Moreover,

r′−(ζ) =ζ − ζ0ζζ

ζ − ζ0ζζr−(ζ) (ζ ∈ C+ ∪ R),

r′+(ζ) = r+(ζ) (ζ ∈ R),

α′(ζk) = α(ζk) (k = 1, · · · , d),

α′(ζ0ζζ ) = 1/µ,

(1.302)

hence

b′(ζ) =ζ − ζ0ζζ

ζ − ζ0ζζb(ζ) (ζ ∈ R), H

C ′kC =

ζk − ζ0ζζ

ζk − ζ0ζζCkC (k = 1, · · · , d), H

C ′0CC =

ζ0ζζ − ζ0ζζ

µr−(ζ0ζζ ).

(1.303)

(2) If ζ0ζζ is an eigenvalue: ζ0ζζ = ζjζ , and µ = α(ζjζ ), then, after theaction of the Darboux transformation, ζ0ζζ is no longer an eigenvalue.


Moreover,

r′−(ζ) =ζ − ζ0ζζ

ζ − ζ0ζζr−(ζ) (ζ ∈ C+ ∪ R),

r′+(ζ) = r+(ζ) (ζ ∈ R),

α′(ζk) = α(ζk) (k = 1, · · · , d, k = j),

(1.304)

andb′(ζ) =

ζ − ζ0ζζ

ζ − ζ0ζζb(ζ) (ζ ∈ R), H

C ′kC =

ζk − ζ0ζζ

ζk − ζ0ζζCkC (k = 1, · · · , d, k = j).

(1.305)

Proof. (1) ζ0ζζ ∈ IPσ(L).Then, both the numerator and denominator of (1.299) are not 0.

Property 3 implies

limx→−∞σ = ∞, lim

x→+∞σ = 0. (1.306)

Hence

limx→−∞(− iζI − S) =

⎛⎝⎛⎛ − iζ + iζ0ζζ 0

0 − iζ + iζ0ζζ

⎞⎠⎞⎞ ,

limx→+∞(− iζI − S) =

⎛⎝⎛⎛ − iζ + iζ0ζζ 0


⎞⎠⎞⎞ .

(1.307)

Under the action of the Darboux transformation, the Jost solutions arechanged to

ψ′r(x, t, ζ) =

1− iζ + iζ0ζζ

(− iζI − S)ψr(x, t, ζ),

ψ′l(x, t, ζ) =

1− iζ + iζ0ζζ

(− iζI − S)ψl(x, t, ζ).(1.308)

HenceR′ =

1− iζ + iζ0ζζ

(− iζI − S)R. (1.309)

If ζ ∈ C+,

limx→+∞R′ =

⎛⎜⎛⎛⎝⎜⎜ ζ − ζ0ζζ

ζ − ζ0ζζ0

0 1

⎞⎟⎞⎞⎠⎟⎟⎛⎝⎛⎛ r−(ζ)

0

⎞⎠⎞⎞ . (1.310)

Thusr′−(ζ) =

ζ − ζ0ζζ

ζ − ζ0ζζr−(ζ), (1.311)


r′−(ζ) has an additional zero ζ0ζζ than r−(ζ). This means that ζ0ζζ is a neweigenvalue. For ζ ∈ R,

R′ ∼⎛⎜⎛⎛⎝⎜⎜ ζ − ζ0ζζ

ζ − ζ0ζζ0

0 1

⎞⎟⎞⎞⎠⎟⎟⎛⎝⎛⎛ r−(ζ)

r+(ζ)e2iζx

⎞⎠⎞⎞ . (1.312)

Hence r′+(ζ) = r+(ζ), and

b′(ζ) =ζ − ζ0ζζ

ζ − ζ0ζζb(ζ). (1.313)

If ζk is a zero of r−(ζ), then (1.308) implies α′(ζk) = α(ζk), and

C ′kC = α′(ζk)

/dr′−(ζk)

dζ=

ζk − ζ0ζζ

ζk − ζ0ζζCkC . (1.314)

When ζ = ζ0ζζ ,

ψ′r(x, t, ζ0ζζ ) =

11 + |σ|2

⎛⎝⎛⎛ |σ|2 −σ

−σ 1

⎞⎠⎞⎞ψr(x, t, ζ0ζζ ),

ψ′l(x, t, ζ0ζζ )

11 + |σ|2

⎛⎝⎛⎛ |σ|2 −σ

−σ 1

⎞⎠⎞⎞ψl(x, t, ζ0ζζ ),

(1.315)

α′(ζ0ζζ ) =σL1 exp(iζ0ζζ x) − L2 exp(iζ0ζζ x)

σR1 exp(− iζ0ζζ x) − R2 exp(− iζ0ζζ x)=

1µ

, (1.316)

C ′0CC = α′(ζ0ζζ )/

dr′−(ζ0ζζ )dζ

=ζ0ζζ − ζ0ζζ

µr−(ζ0ζζ ). (1.317)

(1) is proved.(2) ζ0ζζ = ζjζ ∈ IPσ(L), µ = α(ζjζ ).Now

σ =R2(ζjζ )R1(ζjζ )

=L2(ζjζ )L1(ζjζ )

, (1.318)

hencelim

x→−∞σ = 0, limx→+∞σ = ∞, (1.319)

limx→−∞(− iζI − S) =

⎛⎝⎛⎛ − iζ + iζ0ζζ 0


⎞⎠⎞⎞ ,

limx→+∞(− iζI − S) =

⎛⎝⎛⎛ − iζ + iζ0ζζ 0


⎞⎠⎞⎞ .

(1.320)


Under the action of the Darboux transformation, the Jost solutions be-come

ψ′r(x, t, ζ) =

1− iζ + iζ0ζζ

(− iζI − S)ψr(x, t, ζ),

ψ′l(x, t, ζ) =

1− iζ + iζ0ζζ

(− iζI − S)ψl(x, t, ζ).(1.321)

For ζ ∈ C+,

limx→+∞R′ =

⎛⎜⎛⎛⎝⎜⎜ ζ − ζ0ζζ

ζ − ζ0ζζ0

0 1

⎞⎟⎞⎞⎠⎟⎟⎛⎝⎛⎛ r−(ζ)

0

⎞⎠⎞⎞ , (1.322)

and for ζ ∈ R,

R′ ∼⎛⎜⎛⎛⎝⎜⎜ ζ − ζ0ζζ

ζ − ζ0ζζ0

0 1

⎞⎟⎞⎞⎠⎟⎟⎛⎝⎛⎛ r−(ζ)

r+(ζ)e2iζx

⎞⎠⎞⎞ . (1.323)

Hence

r′−(ζ) =ζ − ζ0ζζ

ζ − ζ0ζζr−(ζ) (ζ ∈ C+ ∪ R),

r′+(ζ) = r+(ζ) (ζ ∈ R),

(1.324)

andb′(ζ) =

ζ − ζ0ζζ

ζ − ζ0ζζb(ζ). (1.325)

From (1.324) we know that the Darboux transformation removes theeigenvalue ζ0ζζ (= ζjζ ).

If ζ = ζk (k = j), then ψ′r = α(ζk)ψ′

l, hence

α′(ζk) = α(ζk),

C ′kC = α′(ζk)

/dr′−(ζk)dζ

=ζk − ζ0ζζ

ζk − ζ0ζζCkC .

(1.326)

The theorem is proved.We have given the formulae for the change of the scattering data un-

der Darboux transformation in the su(2) case. A Darboux transforma-tion increase or decrease the number of eigenvalues (number of solitons).However, it does not affect the scattering data related to the continu-ous spectrum. Thus we can use the Darboux transformation to changea general inverse scattering problem to an inverse scattering problemwithout eigenvalues.


Remark 17 For the KdV equation, q = 1 (or −1) in the Lax pair (1.286).Since q does not tend to zero at infinity, the above conclusions can notbe applied directly. However, the inverse scattering theory for the KdVequation is actually simpler than the AKNS system (see [23]). The con-clusions similar to Theorem 1.22 for the KdV equation holds true as well[29].

Chapter 2

2+1 DIMENSIONAL INTEGRABLESYSTEMS

This chapter is devoted to the Darboux transformations of 2+1 dimen-sional integrable systems. Starting from the KP equation, we discuss theDarboux transformation for 2+1 dimensional AKNS system and moregeneral systems. Unlike the Darboux matrices in 1+1 dimensions, theDarboux transformations here are given by differential operators (calledDarboux operators). The construction of the Darboux operators is uni-form to all the equations in the system, as in the 1+1 dimensional case.The binary Darboux transformation, which is a kind of Darboux trans-formation in integral form, is introduced briefly. Explicit solutions ofthe DSI equation can be obtained by the combination of Darboux trans-formation and binary Darboux transformation. Moreover, the nonlinearconstraint method is used to separate the differentials in the 2+1 di-mensional AKNS system so that the Darboux transformation in 1+1dimensions can be used to get the localized soliton solutions.

2.1 KP equation and its Darboux transformation

A 2+1 dimensional integrable system has three independent variables(x, y, t) where x and y usually refer to space variables and t refers totime variable. A typical 2+1 dimensional integrable partial differentialequation is the Kadomtsev-Petviashvili equation (KP equation) [68]

uxt = (uxxx + 6uux)x + 3α2uyy, (2.1)

where α = ±1 or ± i. (2.1) is called the KPI equation if α = ±1, and theKPII equation if α = ± i. The KP equation is the natural generalization


of the KdV equation, which describes the motion of two dimensionalwater wave. (2.1) can also be written as

vxt = vxxxx + 6vxvxx + 3α2vyy (2.2)

where v satisfies vx = u. The KP equation has a Lax pair

φy = α−1φxx + α−1uφ, (2.3)

φt = 4φxxx + 6uφx + 3(αvy + ux)φ. (2.4)

We first derive φyt by differentiating (2.3) with respect to t and in-serting the expression of φt. Also, we can derive φty by differentiating(2.4) with respect to y and inserting the expression of φy. The equalityφyt = φty is equivalent to (2.1) when φ = 0. The proof of this fact isdirect, which is left for the reader. Therefore, (2.1) is the integrabilitycondition of the overdetermined system (2.3) and (2.4).

The Darboux transformation for the KP equation is similar with thatfor the KdV equation. It can be constructed as follows. Let h be asolution of the Lax pair (2.3) and (2.4). For any solution φ of (2.3) and(2.4), define

φ′ = φx − (hx/h)φ, (2.5)

then φ′ is a solution of

φ′y = α−1φ′

xx + α−1u′φ′,

φ′t = 4φ′

xxx + 6u′φ′x + 3(αv′y + u′

x)φ′(2.6)

whereu′ = u + 2(hx/h)x, v′ = v + 2hx/h. (2.7)

Comparing (2.6) with (2.3) and (2.4), the only difference is that (u, φ)is changed to (u′, φ′). Hence (2.7) gives a new solution u′ of the KPequation [77].

Similar to 1+1 dimensions, if the seed solution u is simple enough, wecan solve the Lax pair (2.3) and (2.4) to get h, then (2.5) gives a morecomplicated solution of the KP equation. Especially, if u = v = 0, then(2.3) and (2.4) becomes

φy = α−1φxx

φt = 4φxxx.(2.8)

Therefore, for any solution h of (2.8) with h = 0, u′ = 2(hx/h)x gives asolution of the KP equation.


Example 2.1 For α = 1, h can be chosen as

h = eλx+λ2y+4λ3t + 1, (2.9)

where λ is a real constant, then

u′ =λ2

2sech2

(12(λx + λ2y + 4λ3t)

)(2.10)

is a solution of the KPI equation.

Example 2.2 For α = − i, let

h = eλx+iλ2y+4λ3t + e−λx+iλ2y−4λ3t, (2.11)

where λ = a + b i is a complex constant, then we obtain a solution of theKPII equation:

u′ = 2a2 sech2(ax − 2aby + 4(a3 − 3ab2)t). (2.12)

These two solutions are both travelling waves, i.e., they are of formu′ = f(t + a1x + a2y) and u′ is invariant along the line t + a1x + a2y =constant on the (x, y) plane. For fixed t, u′ is a non-zero constant alongcertain lines (for KPI, they are λx+λ2y = constant, while for KPII, theyare ax− 2aby = constant), and u′ tends to zero exponentially at infinityalong other lines. Hence the region where u′ is far from zero forms aband on the (x, y) plane. This kind of solutions are call “line-solitons”.This does not happen in 1+1 dimensions.

Suppose we have known a solution u of the KP equation and a setof solutions φ of the corresponding Lax pair. Let h be a special φ,then u′ = u + 2(lnh)xx is a solution of the KP equation. Moreover,φ′ = φx − (hx/h)φ gives the set of solutions of the Lax pair for u′. Nowwe take a special φ′ as h′, then we can obtain another solution u′′ =u′ + 2(lnh′)xx of the KP equation and the solution φ′′ = φ′

x − (h′x/h′)φ′

of the corresponding Lax pair by constructing Darboux transformationwith h′. Continuing this procedure, we obtain a series of solutions of theKP equation without solving differential equations.

Except the first step, this algorithm can be realized by algebraic com-putation and differentiations. Therefore, it can be done by symboliccalculation. The solutions are global for all (x, y, t) if h, h′, h′′ · · · do notequal zero. This process can be expressed as

(u, φ) −→ (u′, φ′) −→ (u′′, φ′′) −→ · · · . (2.13)

The differential operator of order three on the right hand side of (2.4)can be changed to differential operators of arbitrary order, then we get


the KP hierarchy

φy = α−1φxx + α−1uφ,

φt =n∑

j=0

vn−j∂jφ,

(2.14)

(∂ = ∂/∂x). Computing the integrability condition of (2.14) and lettingall the coefficients of the derivatives of φ with respect to x be zero, wehave

2vj+1,x = αvj,y − vj,xx +j−1∑k=0

Cn−jnC −kvk∂

j−ku, (2.15)

ut = αvn,y − vn,xx +n−1∑k=0

vk∂n−ku. (2.16)

In (2.15), vj+1 can be solved by integration. Unlike the 1+1 dimen-sional systems such as the KdV hierarchy, here, in general, vj ’s can notbe expressed as differential polynomials of u. Therefore, (u, v1, · · · , vn)are regarded as a set of unknowns of (2.15) – (2.16). The Darbouxtransformation is still valid for this system. In practical problems, someadditional relations among (v1, · · · , vn, u) should be satisfied. This iscalled a reduction of the original one. In that case, we should chooseproper h so that the relations among v1, · · · , vn, u keeps after the Dar-boux transformation. Usually this is a difficult problem and some specialcases can be solved by certain techniques.

2.2 2+1 dimensional AKNS system and DSequation

2+1 dimensional AKNS system is

Φy = JΦx + PΦ, Φt =n∑

j=0

VnVV −j∂jΦ, (2.17)

where J is an N × N constant diagonal matrix, P (x, y, t) is an off-diagonal N ×N matrix, VjVV (x, y, t)’s are also N ×N matrices, ∂ = ∂/∂x.For simplicity, we assume that the diagonal entries of J are distinct.Moreover, we consider the non-degenerate Φ only.

The integrability condition of (2.17) leads to

[J, V offjVV +1] = V off

j,yVV − JV offj,xVV − [P, VjVV ]off +

j−1∑k=0

Cn−jnC −k(VkVV ∂j−kP )off,(2.18)


V diagj,yVV − JV diag

j,xVV = [P, V offjVV ]diag −

j−1∑k=0

Cn−jnC −k(VkVV ∂j−kP )diag, (2.19)

PtPP = V offn,yVV − JV off

n,xVV − [P, VnVV ]off +n−1∑k=0

(VkVV ∂n−kP )off. (2.20)

Here the superscripts “diag” and “off” refer to the diagonal and off-diagonal part of a matrix respectively.

Usually VjVV ’s are not differential polynomials of P . But they can begenerated from P by differentiation and integration with respect to x.(2.19) and (2.20) are regarded as a system of partial differential equa-tions for P and V diag

jVV ’s (j = 0, 1, · · · , n) where V offjVV ’s (j = 1, · · · , n) are

determined by (2.18). (2.17) is the Lax pair of this system of equations.A typical equation in 2+1 dimensional AKNS system is the Davey-

Stewartson equation (DS equation), which is the natural generalizationof the nonlinear Schrodinger equation in 2+1 dimensions.

Take N = 2, n = 2 in (2.17) and let

J = α−1

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ , P =

⎛⎝⎛⎛ 0 u

−εu 0

⎞⎠⎞⎞ ,

α = ±1 or ± i, ε = ±1,

V0VV = 2iαJ, V1VV = 2iαP,

V2VV = iα

⎛⎝⎛⎛ w1 ux + αuy

−εux + αεuy w2

⎞⎠⎞⎞ .

(2.21)

where u, w1, w2 are complex-valued functions, u is the complex conjugateof u. Then, (2.17) becomes

Φy = α−1

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φx +

⎛⎝⎛⎛ 0 u

−εu 0

⎞⎠⎞⎞Φ,

Φt = 2i

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φxx + 2iα

⎛⎝⎛⎛ 0 u

−εu 0

⎞⎠⎞⎞Φx

+iα

⎛⎝⎛⎛ w1 ux + αuy

−εux + αεuy w2

⎞⎠⎞⎞Φ.

(2.22)


(2.19) and (2.20) lead to

iut = −uxx − α2uyy − αu(w1 − w2),

w1,y − α−1w1,x = ε(|u|2)x + αε(|u|2)y,

w2,y + α−1w2,x = ε(|u|2)x − αε(|u|2)y,

(2.23)

and w2 − w1 = α−2(w2 − w1). Denote

v = −ε|u|2 +12α

(w1 − w2), (2.24)

then (2.23) becomes

iut = −uxx − α2uyy − 2εα2|u|2u − 2α2uv,

vxx − α2vyy + 2ε(|u|2)xx = 0.(2.25)

(2.25) is called the DSI equation if ε = 1, α = ±1, and DSII equation ifε = 1, α = ± i. They describe the motion of long wave and short wavein the water of finite depth [20].

2.3 Darboux transformation2.3.1 General Lax pair

Similar to the KP equation, we also want to construct the Darbouxtransformation for the AKNS system. Here we first discuss the Dar-boux transformation for the following more general Lax pair withoutany reductions.

Consider Lax pair

Φy = U(x, y, t, ∂)Φ, Φt = V (x, y, t, ∂)Φ, (2.26)

where

U(x, y, t, ∂) =m∑

j=0

UmUU −j(x, y, t)∂j ,

V (x, y, t, ∂) =n∑

j=0

VnVV −j(x, y, t)∂j(2.27)

are differential operators with respect to x whose coefficients UjUU ’s andVjVV ’s are N × N matrices. For simplicity, we write U(x, y, t, ∂) = U(∂),V (x, y, t, ∂) = V (∂).

Φyt = Φty can be obtained by differentiating the first equation of(2.26) with respect to t or by differentiating the second equation with


respect to x. Let these two equal, we get

UtUU (∂) − VyVV (∂) + [U(∂), V (∂)] = 0. (2.28)

(2.26) is called integrable if (2.28) holds. (2.28) is the generalization ofthe zero-curvature equations in 1+1 dimensions. It gives a system ofpartial differential equations by equating all the coefficients of ∂ to bezero.

Remark 18 The existence and uniqueness of the solutions of a systemof partial differential equations are very difficult problems. The localsolvability of a system of linear partial differential equations have beenstudied by many authors. In the present case, even (2.28) holds, localsolution of (2.26) near t = t0, y = y0 with initial data Φ(t0, x, y0) =Φ0(x) may not exist. However, if each set of equations in (2.26) islocally solvable and the solutions are smooth enough with respect to theparameters y and t, U(∂) and V (∂) satisfy (2.28), then (2.26) is locallysolvable. This follows from the following consideration. Suppose thatthe initial data (x0, y0, t0, Φ0(x)) are given. First solve the first set ofequations of (2.26) at t = t0 with initial value Φ1(x, y0) = Φ0(x) and getthe solution Φ1(x, y). Using Φ1(x, y) as the initial value, solve the secondset of equations of (2.26) for fixed y and get the solution Φ(x, y, t). Using(2.28) and the second equation of (2.26), we have

(Φy − U(∂)Φ)t = V (∂)(Φy − U(∂)Φ). (2.29)

Therefore, Φy = U(∂)Φ holds identically near (x0, y0, t0) by the unique-ness of the solution.

No matter whether the existence and uniqueness hold, (2.28) is calledthe integrability condition of (2.26). It gives a system of nonlinear partialdifferential equations of U(∂) and V (∂). (2.26) is called the Lax pair ofthis system of nonlinear partial differential equations. It is interestingto see that we can apply Darboux transformation as well provided thatthe set of solutions of (2.26) is not empty.

2.3.2 Darboux transformation of degree oneSimilar to 1+1 dimensional case, we can define Darboux operator for

the integrable nonlinear partial differential equations (2.28) and thereLax pair (2.26).

Definition 2.3 A differential operator D(x, y, t, ∂) with respect to x iscalled a Darboux operator for (2.26) if there exist differential operatorsU ′(∂) and V ′(∂) with respect to x such that for any solution Φ of (2.26),Φ′ = D(∂)Φ satisfies

Φ′y = U ′(∂)Φ′, Φ′

t = V ′(∂)Φ′. (2.30)


The transformation (Φ, U(∂), V (∂)) → (Φ′, U ′(∂), V ′(∂)) given by D(∂)is called a Darboux transformation.

Substituting Φ′ = DΦ into (2.30), we have

Dy(∂) = U ′(∂)D(∂) − D(∂)U(∂),

Dt(∂) = V ′(∂)D(∂) − D(∂)V (∂),(2.31)

andU ′

tUU (∂) − V ′yVV (∂) + [U ′(∂), V ′(∂)] = 0. (2.32)

(2.31) is the necessary and sufficient condition for D(∂) being a Darbouxoperator. Hence, if (U(∂), V (∂)) is a solution of (2.28), so is (U ′(∂),V ′(∂)). This means that the Darboux transformation gives a new solu-tion of (2.28). Our main task is to construct the solution D of (2.31).

We first discuss the most fundamental Darboux operator, the Dar-boux operator of degree one. This is the Darboux operator in the formD(x, y, t, ∂) = ∂−S(x, y, t). The Darboux operator of higher degree willbe discussed later. In order to get the general construction of S, we firstderive the equations that S should satisfy.

For a matrix M(x), we define a sequence of matrices M (j) by M (0) = Iand

M (j+1) = M (j)xMM + M (j)M, (2.33)

then, for any solution Φ of the equation Φx = MΦ, ∂jΦ = M (j)Φ holds.For any differential operator

U(∂) =k∑

j=0

UkUU −j∂j , V (∂) =

k∑j=0

VkVV −j∂j (2.34)

and an N × N matrix S, we define

U(S) =k∑

j=0

UkUU −jS(j), V (S) =

k∑j=0

VkVV −jS(j). (2.35)

Suppose that Φ satisfies Φx = SΦ, then U(∂)Φ = U(S)Φ, V (∂)Φ =V (S)Φ. Notice that U(S) and V (S) are not given by replacing ∂ in U(∂)and V (∂) with S. Actually they are obtained by replacing ∂j with S(j).

Theorem 2.4 ∂ − S is a Darboux operator for (2.26) if and only if Ssatisfies

SySS + [S, U(S)] = (U(S))x,

StSS + [S, V (S)] = (V (S))x.(2.36)


Proof. Suppose ∂ − S is a Darboux operator for (2.26), then the firstequation of (2.31) is

SySS − (∂ − S)U(∂) + U ′(∂)(∂ − S) = 0. (2.37)

Let Ψ be the fundamental solution of Ψx = SΨ, then

SySS Ψ = (∂ − S)U(S)Ψ = (U(S))xΨ − [S, U(S)]Ψ. (2.38)

This gives the first equation of (2.36). The second equation is derivedsimilarly. The necessity of (2.36) is proved.

Conversely, suppose S is a solution of (2.36). Define

U ′(∂) =m∑

j=0

U ′mUU −j∂

j , (2.39)

where U ′jU ’s are determined recursively by

U ′0UU = U0UU ,

U ′jU +1 = UjU +1 + Uj,xU − SUjU +

j∑k=0

Cm−jmC −kU

′kU ∂j−kS.

(2.40)

ThenSySS − (∂ − S)U(∂) + U ′(∂)(∂ − S) (2.41)

does not contain any terms with ∂, i.e., it is a matrix-valued functionof x, y and t. On the other hand, for any fundamental solution Φ ofΨx = SΨ, (2.36) leads to

(SySS − (∂ − S)U(∂) + U ′(∂)(∂ − S))Ψ = 0. (2.42)

Hence, as a matrix,

SySS − (∂ − S)U(∂) + U ′(∂)(∂ − S) = 0. (2.43)

This shows that ∂ − S satisfies the first equation of (2.31). The secondone can be proved similarly. Therefore, ∂ −S is a Darboux operator for(2.26). The theorem is proved.

Theorem 2.5 ∂−S is a Darboux operator for (2.26) if and only if thereexists an N × N non-degenerate matrix solution H of (2.26) such thatS = HxHH H−1 [122].


Proof. First we prove the sufficiency, i.e., to show the S satisfies (2.36).From (2.26),

SySS = HxyHH H−1 − SHyHH H−1 = (U(S)H)xH−1 − SU(S)

= [U(S), S] + (U(S))x,(2.44)

which is the first equation of (2.36). The second one is derived in thesame way.

Now suppose S satisfies (2.36). We shall show that the system ofequations

HxHH = SH, HyHH = U(∂)H, HtHH = V (∂)H, (2.45)

has a solution. Clearly (2.45) is equivalent to

HxHH = SH, HyHH = U(S)H, HtHH = V (S)H. (2.46)

Hence, we only need to verify the integrability conditions of (2.46).Let Ψ be a fundamental solution of Ψx = SΨ. (2.37) implies

(Ψy − U(∂)Ψ)x = (SΨ)y − ∂U(∂)Ψ = S(Ψy − U(∂)Ψ). (2.47)

Hence(VyVV (∂) + V (∂)U(∂))Ψ

= (V (∂)Ψ)y − V (∂)(Ψy − U(∂)Ψ)

= (V (S)Ψ)y − V (S)(Ψy − U(S)Ψ)

= V (S)yΨ + V (S)U(S)Ψ.

(2.48)

Similarly,

(UtUU (∂) + U(∂)V (∂))Ψ = U(S)tΨ + U(S)V (S)Ψ. (2.49)

Since det Ψ = 0, the integrability condition (2.28) givesU(S)t − V (S)y + [U(S), V (S)] = 0. (2.50)

Hence the integrability condition HytHH = HtyHH for (2.46) holds.Theorem 2.4 gives the other two integrability conditions HxyHH = HyxHH

and HxtHH = HtxHH . Hence (2.46) is integrable. For given initial valueH = H0HH at (t, x, y) = (t0, x0, y0), (2.46) has a solution H. If H0HH is non-degenerate, H is also non-degenerate in a neighborhood of (t0, x0, y0).That is, (2.46) has a non-degenerate matrix solution H such that S =HxHH H−1. The theorem is proved.


If there is no reduction, this theorem shows that any Darboux operatorin the form ∂−S can be expressed explicitly by the solutions of the Laxpair. Darboux transformation exists as long as the Lax pair has a non-degenerate N ×N matrix solution. Under the Darboux transformation,UjU is transformed to U ′

jU given by (2.40). V ′jVV ’s have similar expressions.

Thus, we have constructed the Darboux transformation

(U, V, Φ) −→ (U ′, V ′, Φ′). (2.51)

This process can be continued by algebraic and differential operationsto get infinite number of solutions provided that the set of solutions ofthe Lax pair for the seed solution is big enough.

For the AKNS system (2.17), the action of the Darboux operator ∂−Sgives

(∂ − S)(J∂ + P ) − SxS = (J∂ + P ′)(∂ − S). (2.52)

The coefficients of ∂2 on both sides are equal. Comparing the coefficientof ∂, we have

P ′ = P + [J, S]. (2.53)

For practical problems, the entries of U and V often have some con-straint relations. In that case, H in the theorem should also satisfy cer-tain conditions so that (U ′, V ′) and (U, V ) satisfy the same constraints.If so, we can obtain a transformation from a solution of a nonlinearpartial differential equation to a solution of the same equation.

Remark 19 For the KP equation, the construction for the Darboux op-erator is completely the same as in Section 2.1. However, for the Davey-Stewartson equation, it is more difficult because we should consider therelations among the entries of P . We shall discuss it in Section 2.4.

Similar with the 1+1 dimensional case, we can also compose severalDarboux transformations of degree one to a Darboux transformation ofhigher degree. However, they can be constructed directly with explicitformulae.

2.3.3 Darboux transformation of higher degreeand the theorem of permutability

Now we discuss a Darboux operator of higher degree. It is a differen-tial operator in the form

D(∂) =r∑

j=0

Dr−j∂j , D0 = I (2.54)


such thatDy(∂) = U ′(∂)D(∂) − D(∂)U(∂),

Dt(∂) = V ′(∂)D(∂) − D(∂)V (∂).(2.55)

Here U ′(∂) and V ′(∂) are differential operators with respect to x.For simplicity, we only discuss the Darboux operator of degree two.

When r > 2, the Darboux operator can also be written down explicitly,but is more complicated.

Theorem 2.6 Let H1 and H2HH be two N × N non-degenerate matrixsolutions of (2.26). Let F be the block matrix⎛⎝⎛⎛ H1 H2HH

∂H1 ∂H2HH

⎞⎠⎞⎞ . (2.56)

Suppose det F = 0 , then the following conclusions hold:(1) There is a unique differential operator of degree two

D(H1, H2HH , ∂) = ∂2 + D1∂ + D2 (2.57)

satisfyingD(H1, H2HH , ∂)HiHH = 0 (i = 1, 2). (2.58)

It is a Darboux operator.(2) The theorem of permutability holds:

D(H1, H2HH , ∂) = D(H2HH , H1, ∂). (2.59)

(3) There is a decomposition

D(H1, H2HH , ∂) = D(D(H1, ∂)H2HH , ∂)D(H1, ∂). (2.60)

Proof. Since det F = 0, the linear algebraic systemD1∂H1 + D2H1 = −∂2H1, D1∂H2HH + D2H2HH = −∂2H2HH (2.61)

for D1, D2 has a unique solution, which determines D(∂) uniquely andD(∂) satisfies (2.58). Since (2.61) is symmetric with respect to H1 andH2HH , (2) holds.

By the definitions of D(D(H1, ∂)H2HH , ∂) and D(H1, ∂),

D(D(H1, ∂)H2HH , ∂)D(H1, ∂)H1 = 0,

D(D(H1, ∂)H2HH , ∂)D(H1, ∂)H2HH = 0.(2.62)


Hence (2.60) holds. From (2.60) it is seen that D(H1, H2HH , ∂) is a Dar-boux operator because it is the composition of two Darboux operatorsof degree one.

Similar to (1.134), the theorem of permutability can be expressed bythe following diagram:

(U, V, Φ)

H1

H2

(U (1), V (1), Φ(1))

(U (2), V (2), Φ(2))

(U (1,2), V (1,2), Φ(1,2))

(U (2,1), V (2,1), Φ(2,1))

H 2

H1

(2.63)

Example 2.7 For the KP equation, N = 1, we can get the expressionof u after the Darboux transformation. Denote HiHH = hi. Suppose theDarboux operator is

r∑j=0

Dr−j∂j , (2.64)

then Theorem 2.6 implies

r−1∑j=0

Dr−j∂jhi = −∂rhi, (2.65)

i.e.,(Dr, · · · , D1)FrFF = −(∂rh1, · · · , ∂rhr). (2.66)

Solving this system, we have

D1 = −det

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜h1 ∂h1 · · · ∂r−2h1 ∂rh1

......

. . ....

...

hr ∂hr · · · ∂r−2hr ∂rhr

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ ·

·

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜det

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜h1 ∂h1 · · · ∂r−2h1 ∂r−1h1

......

. . ....

...

hr ∂hr · · · ∂r−2hr ∂r−1hr

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟

−1

= −(ln detFrFF )x.

(2.67)


Therefore, for the KP equation, the transformation between two solutionsis

u′ = u − 2D1,x = u + 2(ln detFrFF )xx. (2.68)

Many solutions can be obtained in this way [77, 80, 89].

2.4 Darboux transformation and binary Darbouxtransformation for DS equation

2.4.1 Darboux transformation for DSII equationIn Section 2.2 we introduced the DSI and DSII equations (2.25) and

their Lax pairs (2.17) and (2.21). Since the reductions in DSI equa-tion and in DSII equation are different, the method of solving these twoequations are also quite different.

First, consider the DSII equation, i.e., ε = 1, α = − i [120].

In this case, we should have w2 = w1. Hence v = −|u|2 +i2(w1 − w1),

and J , P , VjVV (j = 0, 1, 2) are

J = i

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ , P =

⎛⎝⎛⎛ 0 u

−u 0

⎞⎠⎞⎞ ,

V0VV = 2J, V1VV = 2P, V2VV =

⎛⎝⎛⎛ w1 ux − iuy

−ux − iuy w1

⎞⎠⎞⎞ .

(2.69)

J , P and VjVV have the properties

J = σJσ−1, P = σPσ−1, VjVV = σVjVV σ−1 (j = 0, 1, 2), (2.70)

where

σ =

⎛⎝⎛⎛ 0 1

−1 0

⎞⎠⎞⎞ , (2.71)

P is the matrix each of whose entry is the complex conjugate of thecorresponding entry of P . Now (2.25) becomes

iut = −uxx + uyy + 2|u|2u + 2uv,

vxx + vyy + 2(|u|2)xx = 0.(2.72)


Its Lax pair is

Φy = i

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φx +

⎛⎝⎛⎛ 0 u

−u 0

⎞⎠⎞⎞Φ,

Φt = 2i

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φxx + 2

⎛⎝⎛⎛ 0 u

−u 0

⎞⎠⎞⎞Φx

+

⎛⎝⎛⎛ w1 ux − iuy

−ux − iuy w1

⎞⎠⎞⎞Φ,

(2.73)

with

v = −|u|2 +i

2(w1 − w1). (2.74)

The Darboux operator for (2.73) is constructed as follows.Suppose (ξ, η)T is a solution of (2.73), then (−η, ξ)T is also its solu-

tion. Hence we can choose

H =

⎛⎝⎛⎛ ξ −η

η ξ

⎞⎠⎞⎞ , (2.75)

S = HxHH H−1 =1

|ξ|2 + |η|2

⎛⎝⎛⎛ ξξx + ηηx ηξ¯ x − ξηx

ξηx − ηξx ξξx + ηηx

⎞⎠⎞⎞ . (2.76)

Since H = σHσ−1, we have S = σSσ−1. The equations

U ′(∂)(∂ − S) = (∂ − S)U(∂) − SySS ,

V ′(∂)(∂ − S) = (∂ − S)V (∂) − StSS(2.77)

imply

U ′ = σU ′σ−1,

V ′ = σV ′σ−1.(2.78)

This means that the Darboux transformation keeps the reduction rela-tions (2.70) invariant.

After the action of the Darboux operator ∂ − S,

P ′ = P + [J, S],

V ′2VV = V2VV + V1VV ,x + 2V0VV SxS + [V0VV , S]S + [V1VV , S].

(2.79)


Figure 2.1. Single line-soliton, t = 0

Hence the new solution of the DSII equation is

u′ = u + 2iS12 = u + 2iηξ¯ x − ξηx

|ξ|2 + |η|2 ,

v′ = v − 2(Re S11)x = v − (ln(|ξ|2 + |η|2))xx.

(2.80)

Example 2.8 Take the seed solution u = 0, then we can choose v = 0(w(( 1 = 0), ξ = ξ(x+ iy, t), η = η(x− iy, t) (i.e., ξ is analytic with respectto x + iy and η is analytic with respect to x − iy) satisfying ξt = 2iξxx,ηt = −2iηxx. For these (ξ, η), (u, v) given by (2.80) are all solutions ofDSII equation. Especially, let ξ = eαx+iαy+2iα2t, η = eβx−iβy−2iβ2t, then

u =2i(α − β)e(α+β)x+i(α+β)y+2i(α2+β2)t

e2Reαx−2Imαy−2Im(α2)t + e2Reβx−2Imβy−2Im(β2)t,

v = −4( Re α − Re β)2e2Re(α+β)x−2Im(α+β)y−2Im(α2+β2)t

(e2Reαx−2Imαy−2Im(α2)t + e2Reβx−2Imβy−2Im(β2)t)2.

(2.81)

When t is fixed, the solution u is a constant along the line with slopex : y = Im(β−α) : Re(β−α), and tends to zero in any other directions.This kind of solution also belongs to “line-soliton”.

Multi-line-solitons can be obtained by successive Darboux transforma-tions. They tend to zero at infinity except for finitely many directions.

Figures 2.1 – 2.4 show the single line-soliton and multi-line-solitons,where the parameters are α1 = 3+2i, β1 = 1+ i, α2 = i, β2 = (2+ i)/4.(For the single line-soliton, only (α1, β1) is used.)


Figure 2.2. Double line-soliton, t = 0

Figure 2.3. Double line-soliton, t = 0.5

Figure 2.4. Double line-soliton, t = 1


Remark 20 Comparing to the general method discussed in the last sec-tion, the key point in the construction of Darboux transformation for theDSII equation is the choice of H in (2.75). Although it is successful tothe DSII equation, it can not be applied to the DSI equation.

2.4.2 Darboux transformation and binaryDarboux transformation for DSI equation

When ε = 1 and α = 1, (2.25) becomes

iut + uxx + uyy + 2|u|2u + 2uv = 0,

vxx − vyy + 2(|u|2)xx = 0,(2.82)

and its Lax pair is

Φy =

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φx +

⎛⎝⎛⎛ 0 u

−u 0

⎞⎠⎞⎞Φ,

Φt = 2i

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φxx + 2i

⎛⎝⎛⎛ 0 u

−u 0

⎞⎠⎞⎞Φx

+i

⎛⎝⎛⎛ w1 ux + uy

−ux + uy w2

⎞⎠⎞⎞Φ,

(2.83)

wherev = −|u|2 +

12(w1 − w2), (2.84)

w1 and w2 are real functions.Since we can not find the solution H of (2.83) like that of (2.75)

to construct the Darboux transformation for (2.82), we should use thebinary Darboux transformation. The binary Darboux transformationwas first introduced by V. B. Matveev et al and has many applications[4, 81, 80, 72, 125, 126]. Here we show its application to DSI equationfor constructing new solutions. For the general case, please refer to [80].

For simplicity, rewrite (2.83) as

Φy = JΦx + PΦ,

Φt = 2iJΦxx + 2iPΦx + iV2VV Φ.(2.85)

Apart from this Lax pair, consider its adjoint equations

Ψy = ΨxJ − ΨP,

Ψt = −2iΨxxJ + 2i(ΨP )x − iΨV2VV .(2.86)


With P ∗ = −P and V ∗2VV = V2VV − 2PxPP , we know that if Φ is a solution of

(2.85), then Ψ = Φ∗ is a solution of (2.86), and vice versa. Therefore, assoon as a solution of (2.85) or (2.86) is known, a solution of its adjointequations is also known.

Similar to Section 2.3, we first take Darboux transformation for theadjoint equation (2.86):

Ψ′ = Ψx − ΨS,

P ′ = P + [J, S],

V ′2VV = V2VV − 2PxPP + 2[P, S] − 2S[J, S] + 2(JSxS + SxS J)

(2.87)

whereS = Ψ−1

0 Ψ0,x, (2.88)

Ψ0 is a non-degenerate 2 × 2 matrix solution of (2.86). Notice that P ′does not satisfy P ′∗ = −P ′, and Ψ′∗ is not a solution of (2.85) withP replaced by P ′. In order to preserve the reduction, the binary Dar-boux transformation is a useful tool. To get a new solution of the DSIequation, it needs the following steps:

Step 1: For a solution Φ of (2.85) and a solution Ψ of the adjointequations (2.86), define 1-form

ω(Ψ, Φ) = ΨΦ dx + ΨJΦ dy + 2i(ΨPΦ + ΨJΦx − ΨxJΦ) dt. (2.89)

It can be verified that ω(Ψ, Φ) is a closed 1-form, that is, its exteriordifferential dω(Ψ, Φ) = 0. Hence, in a simply connected region, theintegral of ω along any closed curve is zero. In R2,1, define

Ω(Ψ, Φ)(x, y, t) =∫ (x,y,t)

(

∫∫x

∫∫0,y0,t0)

ω(Ψ, Φ), (2.90)

which is independent of the path of integration, and ω(Ψ, Φ) = dΩ(Ψ,Φ).Step 2: Let Φ′ = Ψ−1

0 Ω(Ψ0, Φ), then we can verify that Φ′ is a solutionof (2.85) with (P, V2VV ) replaced by (P ′, V ′

2VV ).Step 3: Let Φ′

0 = Ψ−10 Ω(Ψ0, Ψ∗

0). Acting the Darboux operator ∂ −Φ′

0Φ′−10 on Φ′, we get the Darboux transformation

Φ′′ = Φ′x − Φ′

0,xΦ′−10 Φ′ = Φ − Ψ∗

0Ω(Ψ0, Ψ∗0)

−1Ω(Ψ0, Φ),

P ′′ = P ′ + [J,Ψ′0,xΨ′−1

0 ] = P + [J,Ψ∗0Ω(Ψ0, Ψ∗

0)−1Ψ0].

(2.91)

P ∗ = −P leads to P ′′∗ = −P ′′. Therefore, we get a new solution of theDSI equation.


The process in Step 1 and Step 2 is called a binary Darboux trans-formation. For the DSI equation, a new solution is obtained by thecomposition of a Darboux transformation and a binary Darboux trans-formation. It needs differentiation and integration in this procedure.

2.5 Application to 1+1 dimensionalGelfand-Dickey system

In this section, we use Theorem 2.5 to discuss the Darboux transfor-mation for the (1+1 dimensional) Gelfand-Dickey system

λΦ = U(x, t, ∂)Φ,

Φt = V (x, t, ∂)Φ(2.92)

where

U(∂) =m∑

j=0

UmUU −j(x, t)∂j ,

V (∂) =n∑

j=0

VnVV −j(x, t)∂j .

(2.93)

From the first equation of (2.92), we can compute Φt and it should bethe same as that given by the second equation of of (2.92). This givesthe integrability condition

UtUU (∂) + [U(∂), V (∂)] = 0 (2.94)

of (2.92).Let D(x, t, ∂) be a differential operator. If for any solution Φ of (2.92),

Φ′ = D(∂)Φ satisfiesλΦ′ = U ′(x, t, ∂)Φ′,

Φ′t = V ′(x, t, ∂)Φ′,

(2.95)

where U ′ and V ′ are differential operators of the form

U ′(∂) =m∑

j=0

U ′mUU −j(x, t)∂j ,

V ′(∂) =n∑

j=0

V ′nVV −j(x, t)∂j ,

(2.96)

then D(x, t, ∂) is called a Darboux operator for (2.92).For a differential operator D(x, t, ∂) = ∂−S(x, t), we have the follow-

ing theorem.


Theorem 2.9 ∂ − S(x, t) is a Darboux operator for (2.92) if and onlyif S = HxHH H−1, where H is an N ×N non-degenerate matrix solution of

HΛ = U(∂)H,

HtHH = V (∂)H,(2.97)

and Λ is a constant upper-triangular matrix.

Proof. Introduce a new variable y and consider the system

Ψy = U(x, t, ∂)Ψ,

Ψt = V (x, t, ∂)Ψ.(2.98)

If ∂ − S(x, t) is a Darboux operator for (2.92), then there exist U ′(∂)and V ′(∂) such that

0 = (∂ − S)U(∂) − U ′(∂)(∂ − S),

StSS = (∂ − S)V (∂) − V ′(∂)(∂ − S).(2.99)

Since S is independent of y and (2.37) holds, ∂−S is a Darboux operatorfor (2.98) which is independent of y. According to Theorem 2.5, thereexists an N × N non-degenerate matrix solution H0HH of (2.98) such thatS = H0HH ,xH−1

0HH . Here H0HH may depend on y.Let L0 = H−1

0HH H0HH ,y, (2.36) leads to

L0 = H−10HH U(S)H0HH ,

L0,x = −H−10HH H0HH ,xH−1

0HH U(S)H0HH + H−10HH (U(S))xH0HH

+H−10HH U(S)H0HH ,x

= H−10HH (U(S))x − [S, U(S)]H0HH = 0,

and (2.50) leads to

L0,y = −H−10HH H0HH ,yH

−10HH U(S)H0HH + H−1

0HH U(S)H0HH ,y = 0,

L0,t = −H−10HH H0HH ,tH

−10HH U(S)H0HH + H−1

0HH (U(S))tH0HH

+H−10HH U(S)H0HH ,t

= H−10HH (U(S))t + [U(S), V (S)]H0HH = 0.

Hence L0 is a constant matrix. Therefore, there exists a constant upper-triangular matrix Λ and a constant matrix T such that L0 = TΛT−1.According to the definition of L0,

H0HH ,y = H0HH TΛT−1.


HenceH0HH (x, y, t) = H(x, t) exp(Λy)T−1

where H satisfies (2.97) and S = HxHH H−1.Conversely, if H is a solution (2.97) and S = HxHH H−1, then S satisfies

(2.99), i.e., ∂ − S is a Darboux operator for (2.92). The theorem isproved.

Remark 21 (1) When N = 1, H satisfies the Lax pair (2.92).(2) If L0 in the above theorem is diagonalizable, then each column in

H satisfies the Lax pair (2.92) for specific λ.

Example 2.10 The original Darboux transformation for the KdV equa-tion can also be deduced from the above theorem.

The KdV equation

ut + 6uux + uxxx = 0 (2.100)

has the Lax pairλφ = −φxx − uφ,

φt = 2(2λ − u)φx + uxφ.(2.101)

From Theorem 2.9, the Darboux transformation is

φ′ = φx − fxff

fφ (2.102)

where f is a solution of the Lax pair for λ = λ0. The new solution givenby this Darboux transformation is

u′ = u + 2(ln f)xx. (2.103)

Example 2.11 The Boussinesq equation

(uxxx + 6uux)x + 3εutt = 0 (ε = ±1)

has the Lax pair

λφ = φxxx +3u

2φx + wφ,

φt = σφxx + σuφ(2.104)

(σ(( 2 = ε, wx = 3(σuxx + ut)/4σ). Theorem 2.9 also gives the Darbouxtransformation [73]

φ′ = φx − fxff

fφ (2.105)


where f is a solution of the Lax pair for λ = λ0. The new solution ofthe Boussinesq equation is

u′ = u + 2(ln f)xx,

w′ = w +32ux + 3((ln f)xxx + (ln f)x(ln f)xx).

(2.106)

2.6 Nonlinear constraints and Darbouxtransformation in 2+1 dimensions

Now we come back to the 2+1 dimensional AKNS system. In thissection we will use the nonlinear constraint method and the Darbouxtransformation method to solve this system.

The basic idea of the nonlinear constraint method is:(1) Find a suitable nonlinear relation between U and Ψ and express

U as a nonlinear matrix function of Ψ: U = f(Ψ).(2) Substitute U = f(Ψ) into the Lax pair so that the original Lax

pair becomes a system of nonlinear partial differential equations of Ψ.In each equation, the derivative with respect to only one of x, y, t isconcerned.

(3) The constraint U = f(Ψ) is suitable so that the new system ofnonlinear equations has a Lax set (generalized Lax pair).

Then by solving the new system of nonlinear equations and its Laxset, we can get solutions of the original problem.

This idea was first applied in 1+1 dimensional integrable systems[11] and was generalized to the (2+1 dimensional) KP equation [14, 71].Here we pay our attention to the 2+1 dimensional AKNS system so thatwe can get localized soliton solutions. With this method, we can alsoget a lot of non-localized solutions [123, 124]. However, since localizedsolutions are more interesting, here we only consider localized solutions[127, 128].

In order to use the nonlinear constraint method, here we add someconditions on the 2+1 dimensional AKNS system. As in Section 2.2, the2+1 dimensional AKNS system is

Ψy = JΨx + U(x, y, t)Ψ,

Ψt =n∑

j=0

VjVV (x, y, t)∂n−jΨ(2.107)

where ∂ = ∂/∂x, J = diag(J1JJ , · · · , JNJ ) is a constant diagonal N × Nmatrix with distinct entries. U(x, y, t) is off-diagonal. Moreover, herewe want that all JjJ ’s are real and U∗ = −U . In this case, we call (2.107)a hyperbolic u(N) AKNS system. The condition U∗ = −U will imply


that the solutions are globally defined, and the condition JjJJ ’s are realwill guarantee that there exist localized solutions.

As in Section 2.2, the integrability conditions of (2.107) are given by(2.20).

Now we introduce a new linear system

Φx =

⎛⎝⎛⎛ iλI iF

iF ∗ 0

⎞⎠⎞⎞Φ, Φy =

⎛⎝⎛⎛ iλJ + U iJF

iF ∗J 0

⎞⎠⎞⎞Φ,

Φt =

⎛⎝⎛⎛ W X

Y Z

⎞⎠⎞⎞Φ =n∑

j=0

⎛⎝⎛⎛ WjWW XjX

−X∗jX ZjZ

⎞⎠⎞⎞λn−jΦ

(2.108)

where F , WjWW , XjX , ZjZ are N × K, N × N , N × K, K × K matricesrespectively (K ≥ 1) and satisfy W ∗

jWW = −WjWW , Z∗jZ = −ZjZ .

The integrability conditions of (2.108) consists of

FyFF = JFxFF + UF,

iFtFF = Xn,xX + iWnWW F − iFZnZZ ,(2.109)

iXjX +1 = Xj,xX + iWjWW F − iFZjZ (j = 0, 1, · · · , n − 1)

Wj,xWW = − iFX∗jX − iXjX F ∗ (j = 0, 1, · · · , n)

Zj,xZ = iF ∗XjX + iX∗jX F (j = 0, 1, · · · , n)

i[J, WjWW +1] = Wj,yWW − [U, WjWW ] + iJFX∗jX + iXjX F ∗J

(j = 0, 1, · · · , n − 1)

Zj,yZ = iF ∗JXjX + iX∗jX JF (j = 0, 1, · · · , n),

(2.110)

UxUU = [J, FF ∗], (2.111)

UtUU = Wn,yWW − [U, WnWW ] + iJFX∗nX + iXnXX F ∗J. (2.112)

For U = 0, F = 0, (2.110) implies that WjWW (λ) = iΩj(t), XjX = 0,ZjZ = iZ0

jZ (t) where Ωj(t)’s are real diagonal matrices and Z0jZ (t)’s are

real matrices.When Z0

jZ (t) = ζjζ (t)IKI (IKI is the K ×K identity matrix) where ζjζ (t)is a real function of t, (2.109) is just the Lax pair (2.107) for n = 1, 2, 3.(2.110) and (2.112) give the recursion relations to determine WjWW , XjX , ZjZ ,together with the evolution equations corresponding to (2.18)–(2.20),which are the integrability conditions of (2.107). (2.111) gives a nonlin-ear constraint between U and F .


This system includes the DSI equation and the 2+1 dimensional N-wave equation as special cases.

In order to consider the asymptotic behavior of the solution U , herewe suppose Ωj is independent of t and ζjζ = 0. Moreover, denote Ω =∑n

j=0 Ωjλn−j and write Ω = diag(ω1, · · · , ωN ).

Remark 22 (2.108) is a special case of the high-dimensional generalizedAKNS system (3.1). Here we only consider this special system to findlocalized solutions. The general theory will be discussed in the next chap-ter.

The soliton solutions are obtained by Darboux transformations fromU = 0, F = 0. In the present case, the Darboux transformation canbe constructed as in Subsection 1.4.4 with u(n) reduction. However, inorder to get localized solutions, there should be more restrictions on theparameters of Darboux transformations.

Let λα (α = 1, 2, · · · , r) be r non-real complex numbers such thatλα = λβ for α = β and λα = λβ for all α, β. Let

Λα = diag(λα, · · · , λα︸︷︷︷︸︸N

, λα, · · · , λα︸︷︷︷︸︸K

). (2.113)

Considering the orthogonal relation (1.241), we can always take

HαHH =

⎛⎝⎛⎛ exp(Qα) − exp(−Q∗α)C∗

αC

CαCC IKI

⎞⎠⎞⎞ , (2.114)

where CαCC ’s are K × N constant matrices,

Qα = diag(q1, · · · , qN ), qjq = iλαx + iλαJjJ y + iωj(λα, t). (2.115)

According to Section 1.4, the derived solutions are always global.However, in order to get localized solutions, we choose special

CαCC = (0, · · · , 0, καlα

, 0, · · · , 0) (2.116)

where κα is a constant K × 1 non-zero vector being the lα’s column ofCαCC .


The Darboux matrices for such Λα, HαHH can be constructed as fol-lows. Let

D(1)(λ) = λ − H1Λ1H−11 , H

(1)αHH = D(1)(λα)HαHH

(α = 2, 3, · · · , r),D(2)(λ) = λ − H

(1)2HH Λ2H

(1)−12HH , H

(2)αHH = D(2)(λα)H(1)

αHH

(α = 3, 4, · · · , r),· · ·D(r)(λ) = λ − H

(r−1)rHH ΛrH

(r−1)−1rHH ,

(2.117)

D(λ) = D(r)(λ)D(r−1)(λ) · · ·D(1)(λ), (2.118)

then D(λ) is a polynomial of λ of degree r. The permutability (Theo-rem 1.12) implies that if (Λα, HαHH ) and (Λβ , HβH ) are interchanged, D(λ)is invariant.

Let

mj = #α | 1 ≤ α ≤ r, lα = j m = (m1, · · · , mN ) (2.119)

then m1 + · · · + mN = r.Suppose

D(λ) = λr − D1λr−1 + · · · + (−1)rDr. (2.120)

The solution given by this Darboux matrix is

U [m] = i[J, (D1)BN]. (2.121)

Here (D1)BNdenotes the first N × N principal submatrix of D1.

In order to consider the localization, the asymptotic behavior as t →∞ and the asymptotic behavior as the phase difference tends to infinityuniformly, we write

qjq = aαjs + bαj (2.122)

where aαj and bαj are independent of s. Here s can be a linear parameterof a straight line in (x, y) plane, or time t, or any other parameters.

Moreover, denote

ρα = Re(aα,lα), φα = Im(aα,lα),

πα = Re(bα,lα), ψα = Im(bα,lα).(2.123)

In order to prove the following theorem, we need some symbols andsimple facts.

If M1MM , M2MM are j × k matrices, we write M1MM.= M2MM if there is a non-

degenerate diagonal k × k matrix A such that M2MM = M1MM A.


If L is a k × k diagonal matrix, M1MM and M2MM are k × k matrices withM1MM

.= M2MM and detM1MM = 0, then M1MM LM−11MM = M2MM LM−1

2MM .Let

M =

⎛⎝⎛⎛ a −v∗/a

v IKI

⎞⎠⎞⎞ (2.124)

where v = 0 is an K × 1 vector, a = 0 is a number. Let

Λ =

⎛⎝⎛⎛ λ0

λ0IKI

⎞⎠⎞⎞ . (2.125)

Then we have

M−1 =1∆

⎛⎝⎛⎛ a v∗

−av ∆IKI − vv∗

⎞⎠⎞⎞ , (2.126)

MΛM−1 =1∆

⎛⎝⎛⎛ λ0∆ + (λ0 − λ0)|a|2 (λ0 − λ0)av∗

(λ0 − λ0)av λ0∆IKI + (λ0 − λ0)vv∗

⎞⎠⎞⎞(2.127)

where ∆ = v∗v + |a|2. Moreover,

lima→∞MΛM−1 =

⎛⎝⎛⎛ λ0

λ0IKI

⎞⎠⎞⎞ , (2.128)

lima→0

MΛM−1 =

⎛⎜⎛⎛⎝⎜⎜ λ0

λ0IKI + (λ0 − λ0)vv∗

v∗v

⎞⎟⎞⎞⎠⎟⎟ . (2.129)

Theorem 2.12 (1) If there is at most one α (1 ≤ α ≤ r) such thatρα = 0, then lims→∞ U [m] = 0.

(2) If ραj = 0 (j = 1, 2, · · · , q) with αj = αk for j = k, ργ = 0 for allγ = αj (j = 1, · · · , q) and lα1 = · · · = lαq , then lims→∞ U [m] = 0.

(3) If ρα = 0, ρβ = 0 (α = β), ργ = 0 for all γ = α, β, and lα = lβ,then

lims→∞U

[m]abU = 0 for (a, b) = ( lα, lβ) (2.130)

and as s → ∞,

U[m]lUUα,lβ

∼ Bαβ exp(i(ψα − ψβ) + i(φβ − φα)s)

Aαβ cosh(πα + πβ − δ(1)αβ ) + cosh(πα − πβ − δ

(2)αβ )

(2.131)


where Aαβ, δ(1)αβ , δ

(2)αβ are real constants, Aαβ > 0, and Bαβ are complex

constants. Moreover, if K = 1, then Bαβ = 0 if and only if κα = 0 andκβ = 0 .

Proof. First suppose ρα = 0. By (2.128) and (2.129),

limραs→±∞HαHH ΛαH−1

αH = S±∞α (2.132)

where

S+∞α =

⎛⎝⎛⎛ λαINI

λαIKI

⎞⎠⎞⎞ ,

S−∞α =

⎛⎜⎛⎛⎝⎜⎜ λαINI + (λα − λα)Elαlα

λαIKI + (λα − λα)κακ∗

α

κ∗ακα

,

⎞⎟⎞⎞⎠⎟⎟ ,

(2.133)EjkE is an N ×N matrix whose (j, k)th entry is 1 and the rest entries arezero.

For β = α,

(λβ − S±∞α )HβH

.=

⎛⎝⎛⎛ exp(Qβ(s)) − exp(−Qβ(s)∗)C±∗βC

C±βC IKI

⎞⎠⎞⎞ (2.134)

where

C±βC = (0, · · · , 0, κ±

βlβ

, 0, · · · , 0),

κ+β =

λβ − λα

λβ − λακβ ,

κ−β =

⎧⎪⎧⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎨⎨⎪⎪⎪⎪⎪⎪⎩⎪⎪λβ − λα

λβ − λακβ − λα − λα

λβ − λα

κ∗ακβ

κ∗ακα

κα if lβ = lα,

κβ − λα − λα

λβ − λα

κ∗ακβ

κ∗ακα

κα if lβ = lα.

(2.135)

Therefore, if ρα = 0, the action of the limit Darboux matrix λ − S±∞α

on HβH (β = α) does not change the form of HβH , but only changes theconstant vector κβ .

If K = 1, then κ±∗β κ±

γ = 0 implies κ±∗β κ±

γ = 0. When K > 1, thisdoes not hold in general.


Now suppose ρα = 0. Without loss of generality, suppose lα = 1.Then

HαHH.=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

exp(πα) − exp(−πα)κ∗α

1 0. . .

...

1 0

κα 0 · · · 0 IKI

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟. (2.136)

By (2.127),

HαHH ΛαH−1αH =

1∆·

·

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜λα∆ + (λα − λα) exp(πα + πα) (λα − λα) exp(πα)κ∗

α

λα 0

. . ....

λα 0

(λα − λα) exp(πα)κα 0 · · · 0 λα∆IK + (λα − λα)κακ∗α

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟(2.137)

where ∆ = exp(πα + πα) + κ∗ακα.

Part (1) of the theorem is derived as follows. Owing to the permutabil-ity of Darboux transformations, we can suppose ρ1 = 0, · · ·, ρr−1 = 0,ρr = 0. Then, as s → ∞, D(α) tends to a diagonal matrix for α ≤ r− 1.Considering (2.137), the limit of (D(r)(λ))BN

is also diagonal, hence

U [m] = i[J, (D1)BN] → 0. (2.138)

Now we turn to prove part (2). We use the construction of Darbouxmatrices in (1.258). However, the λ in (1.258) should be replaced by iλbecause of its appearance in (2.108).

Let

Hα =

⎛⎝⎛⎛ exp(Qα(s))

CαCC

⎞⎠⎞⎞ , Γαβ =H∗

α Hβ

λβ − λα, (2.139)

then the Darboux matrix is

D(λ) =r∏

α=1

(λ − λα)

⎛⎝⎛⎛1 −r∑

α,β=1


λ − λβ

⎞⎠⎞⎞ (2.140)

and the new solution is

U [m] = i

⎡⎣⎡J,r∑

α,β=1

(Hα(Γ−1)αβ H∗

β

)BN

⎤⎦⎤ . (2.141)


First, suppose q = r and αj = j (j = 1, 2, · · · , r).Since

Hj.=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

1. . .

1

exp(πjπ )

1. . .

1

0 · · · 0 κjlj

0 · · · 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

(2.142)

and l1 = · · · = lq, we have

Γ =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜Γ11 · · · Γ1q

.... . .

...

Γq1 · · · Γqq

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ (2.143)

where Γjk’s are N × N diagonal matrices. Therefore,

Γ−1 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜Σ11 · · · Σ1q

.... . .

...

Σq1 · · · Σqq

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ (2.144)

where Σjk’s are also N ×N diagonal matrices. This implies that U [m] =0.

When r > q, we use the permutability of Darboux transformationsand suppose ρ1 = 0, · · ·, ρr−q = 0, ρr−q+1 = · · · = ρr = 0. Then, afterthe action of D(r−q)(λ) · · ·D(1)(λ), the derived H

(r−q)rHH −q+1 and H

(r−q)rHH have

the same asymptotic form as HrHH −q+1 and HrHH respectively, provided thatthe constant vectors κr−q+1 and κr are changed to κ

(r−q)r−q+1 and κ

(r−q)r .

Therefore, as in the case r = q, the limits of the components of U [m] areall zero. This proved part (2).

Now we prove part (3). First consider the case r = 2. Suppose Hj isgiven by (2.142) (j = 1, 2) and l1 = l2. Denote

θ12 =κ∗

1κ2√κ∗

1κ1κ∗2κ2

, (2.145)


g12 = 1 − 4 Im λ1 Im λ2

|λ2 − λ1|2 |θ12|2 > 0, (2.146)

then, by direct calculation, we have

lims→∞U

[m]lUU1,l2

exp(i(φ2 − φ1)s) =2i(JlJJ 2 − JlJJ 1) Im λ1 Im λ2θ12

λ2 − λ1·

· exp(i(ψ1 − ψ2))√g

√√12 cosh(π1 + π2 − δ1) + cosh(π1 − π2 − δ2)

,

δ1 =12

ln g12 +12

ln(κ∗1κ1κ

∗2κ2) + 2 ln

∣∣∣∣∣∣∣∣∣∣λ2 − λ1

λ2 − λ1

∣∣∣∣∣∣∣∣∣∣ ,δ2 =

12

lnκ∗

1κ1

κ∗2κ2

,

(2.147)

and U[m]µνUU → 0 if (µ, ν) = ( l1, l2).

When r > 2, we still use the permutability of Darboux transforma-tions and suppose ρ1 = 0, · · ·, ρr−2 = 0, ρr−1 = ρr = 0. As in the proofof part (2), after the action of D(r−2)(λ) · · ·D(1)(λ), the derived H

(r−2)rHH −1

and H(r−2)rHH have the same asymptotic form as HrHH −1 and HrHH respectively,

provided that the constant vectors κr−1 and κr are changed to κ(r−2)r−1

and κ(r−2)r . Therefore, as in the case r = 2, the limit of UlUU r−1,lr has the

desired form, and the limits of the other components of U [m] are all zero.The theorem is proved.

Now we can discuss the properties of the solution U [m].

(1) Localization of the solutionsFor the Lax pair (2.108),

Qα = iλα(x + Jy) + iω(λα)t. (2.148)

Consider the limit of the solution as (x, y) → ∞ along a straight linex = ξ + vxs, y = η + vys (v2

x + v2y > 0), then

Qα = iλα(ξ + Jη) + iω(λα)t + iλα(vx + Jvy)s. (2.149)

Now

ρα = Re (iλα(vx + JlJJ αvy)) = − Im λα(vx + JlJJ αvy). (2.150)

If there is at most one ρα = 0, then part (1) of Theorem 2.12 impliesthat U [m] → 0 as s → ∞. If ρα = 0, ρβ = 0 (α = β), then lα = lβsince JlJJ α = JlJJ

βif lα = lβ. Hence, part (2) of Theorem 2.12 also implies

U [m] → 0 as s → ∞. Therefore, we have


Theorem 2.13 U [m] → 0 as (x, y) → ∞ in any directions.

(2) Asymptotic behavior of the solutions as t → ∞Now we use a frame (ξ, η) which moves in a fixed velocity (vx, vy),

that is, let x = ξ + vxt, y = η + vyt, then

Qα = iλα(ξ + Jη) + (iλα(vx + Jvy) + iω(λα))t, (2.151)

ρα = − Im λα(vx + JlJJ αvy) − Im(ωlα(λα)). (2.152)

Suppose that for distinct α, β, γ,

det

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜1 JlJJ α σα

1 JlJJβ

σβ

1 JlJJ γ σγ

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ = 0 (2.153)

whereσα = Im(ωlα(λα))/ Im(λα). (2.154)

Then there are at most two ρα = 0 (α = 1, · · · , r). By Theorem 2.12,U

[m]lUUα,lβ

→ 0 only if ρα = 0, ρβ = 0. This leads to

vx =JlJJ ασβ − JlJJ

βσα

JlJJβ− JlJJ α

, vy =σα − σβ

JlJJβ− JlJJ α

. (2.155)

For U[m]jkU → 0, α, β can take mj and mk values respectively, hence

there are at most mjmk velocities (vx, vy) such that U[m]jkU → 0. Therefore,

we have

Theorem 2.14 Suppose (2.153) is satisfied. Then as t → ∞, theasymptotic solution of U

[m]jkU has at most mjmk peaks whose velocities

are given by (2.155) (l(( α = j, lβ = k). If a peak has velocity (vx, vy),then, in the coordinate ξ = x − vxt, η = y − vyt, limt→∞ UabUU = 0 for all(a, b) = ( j, k), and as t → ∞

U[m]jkU ∼ Bαβ exp(i Re(λα − λβ)ξ + i(λαJjJ − λβJkJJ )η + i(φα − φβ)t)

∆,

∆ = Aαβ cosh( Im(λα + λβ)ξ + Im(λαJjJJ + λβJkJJ )η + δ(1)αβ )

+ cosh( Im(λα − λβ)ξ + Im(λαJjJJ − λβJkJJ )η + δ(2)αβ )

(2.156)where Aαβ, δ

(1)αβ , δ

(2)αβ are real constants, Aαβ > 0, and Bαβ are complex

constants,

φγ = Re λγ(vx + JlJJ γvy) + Re(ωlγ (λγ)) (γ = α, β). (2.157)


Remark 23 The condition (2.153) implies that the velocities of the soli-tons are all different. This is true for the DSI equation. However, forthe 2+1 dimensional N-wave equation, all the solitons move in the samevelocity. We shall discuss this problem later.

Example 2.15 DSI equationLet n = 2, N = 2,

J =

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞ , U =

⎛⎝⎛⎛ 0 u

−u 0

⎞⎠⎞⎞ , ω = −2iJλ2, (2.158)

then we have

FyFF = JFxFF + UF,

FtFF = 2iJFxxFF + 2iUFxFF + i

⎛⎝⎛⎛ |u|2 + 2q1 ux + uy

−ux + uy −|u|2 − 2q2

⎞⎠⎞⎞F,(2.159)

− iut = uxx + uyy + 2|u|2u + 2(q1 + q2)u,

q1,x − q1,y = q2,x + q2,y = −(|u|2)x,(2.160)

(FF ∗)D =12

⎛⎝⎛⎛ q1 0

0 q2

⎞⎠⎞⎞ , [J, FF ∗] = UxUU . (2.161)

(2.160) is the DSI equation.If we construct the solution U [m] as above, then Theorem 2.13 implies

that U [m] → 0 as (x, y) → ∞ in any directions. If Re λα = Re λβ

for α = β and lα = lβ, then, Theorem 2.14 implies that as t → ∞,the derived solution u has at most m1m2 peaks (m(( 1 + m2 = r). From(2.154), σα = −4JlJJ α Re λα, hence (2.155) implies that each peak has thevelocity vx = 2 Re(λα − λβ), vy = 2 Re(λα + λβ) (lα = 1, lβ = 2). Thisis the (m1, m2) solitons [30]. When K = 1, these peaks do not vanish ifand only if all κα’s are non-zero.

Figures 2.5 – 2.7 show the solitons u[1,3], u[2,3] and u[3,3] respectively.The parameters are K = 1, t = 2, λ1 = 1− 2i, λ2 = −3− i, λ3 = 2 + i,λ4 = −1 + 3i, λ5 = 2 + 1.5i, λ6 = −0.5 − 1.5i, C1 = (1, 0), C2CC = (0, 1),C3CC = (0, 2), C4CC = (0,−2), C5CC = (2, 0), C6CC = (−2, 0).

(3) Asymptotic solutions as the phases differences tend toinfinity

For the equations whose solitons move in the same speed, like the2+1 dimensional N-wave equation, the peaks do not separate as t → ∞.


Figure 2.5. u[1,3] of the DSI equation


However, we can still see some peaks in the figures. Here we will get thecorresponding asymptotic properties of the solitons.

Theorem 2.16 Let pα (α = 1. · · · , r) be constant real numbers satisfy-ing

det

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜1 JlJJ α pα/ Im λα

1 JlJJβ

pβ/ Im λβ

1 JlJJ γ pγ/ Im λγ

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ = 0 (2.162)

for distinct α, β, γ. Let dα be complex constant K × 1 vectors, κα =dα exp(pατ) and construct the Darboux transformations as above. Let



x = ξ + vxτ , y = η + vyτ , then, for any j, k with 1 ≤ j, k ≤ N , j = k,limτ→∞ U

[m]jkU = 0 only if (vx, vy) takes specific mjmk values.

Proof. Here

Qα = iλα(ξ + Jη) + iω(λα)t + iλα(vx + Jvy)τ. (2.163)

Hence

Hα.=

⎛⎝⎛⎛ exp(Qα(τ))

Dα

⎞⎠⎞⎞ (2.164)

whereDα = (0, · · · , 0, dα

lα, 0, · · · , 0), (2.165)

Qα(τ) = iλα(ξ + Jη) + iω(λα)t + (iλα(vx + Jvy) − pα)τ. (2.166)

The real part of the coefficient of τ in Qα(τ) is

ρα = − Im λα(vx + Jvy) − pα. (2.167)

Condition (2.162) implies that there are at most two ρα’s such thatρα = 0. According to Theorem 2.12, as τ → ∞, U

[m]jkU → 0 only if there

exist ρα = 0, ρβ = 0, α = β, lα = j, lβ = k. Therefore, the theorem isverified.

When the condition (2.153) holds, this theorem is useless, because theevolution will always separate the peaks. However, when (2.153) doesnot hold, especially when it is never satisfied, this theorem reveals a factof the separation of the peaks.


Example 2.17 2+1 dimensional N-wave equationLet n = 1, ω = Lλ where L = diag(L1, · · · , LN ) is a constant real

diagonal matrix such that Lj = Lk for j = k. Then, the integrabilityconditions (2.109) – (2.112) imply

FyFF = JFxFF + UF, FtFF = LFxFF + V F, (2.168)[J, V ] = [L, U ], UtUU − VyVV + [U.V ] + JVxVV − LUxUU = 0, (2.169)UxUU = [J, FF ∗]. (2.170)

(2.169) is just the 2+1 dimensional N-wave equation.Suppose U [m] is constructed as above, then Theorem 2.13 implies that

U [m] → 0 as (x, y) → ∞ in any directions. Theorem 2.14 cannot beapplied here. The reason is: the condition (2.153) holds only if lα = lβfor α = β. Hence for any j, mj = 0 or 1. This implies that (2.153) doesnot hold generally unless mj = 0 or 1 for all 1 ≤ j ≤ N . Therefore,we apply Theorem 2.16 to the previous problem. Theorem 2.16 impliesthat if we choose pα such that (2.162) is satisfied, then, for each (j, k),limτ→∞ U

[m]jkU has at most mjmk peaks. When K = 1, these peaks do not

vanish if and only if all κα’s are non-zero.

Remark 24 Here τ → ∞ means that the phase differences of differentpeaks tend to infinity. Therefore, the peaks are separated by enlargingthe phase differences.

Here are the figures describing the solutions U [0,1,2] and U [1,1,2] of the3-wave equation. The vertical axis is (|u12|2 + |u13|2 + |u23|2)1/4 so thatall the components are shown in one figure. The parameters are

J =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜1

0

−1

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ L =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜2

−1

1

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟K = 1, t = 10, λ1 = 1 − 2i, λ2 = −3 − i, λ3 = 2 + i, λ4 = −1 + 3i,C1 = (0, 1, 0), C2CC = (0, 0, 1), C3CC = (0, 0, 4096), C4CC = (1, 0, 0). Note thatfor U [0,1,2], only U23UU has two peaks, and for U [1,1,2], U12UU , U13UU , U23UU haveone, two, two peaks respectively.


Figure 2.8. U [0,1,2] of the 3-wave equation

Figure 2.9. U [1,1,2] of the 3-wave equation

Chapter 3

N + 1 DIMENSIONAL INTEGRABLESYSTEMS

In this chapter, we discuss the soliton theory in higher dimensions andgeneralize the AKNS system in 1 + 1 dimensions to the AKNS systemin n + 1 dimensions. The Darboux transformation method can still beused and the algorithm is universal and purely algebraic. The collisionof solitons in higher dimensions is also elastic as in 1+1 dimensions. Forthe Cauchy problems, the initial data of this generalized AKNS systemis given on a straight line. This theory covers many problems, especiallysome differential geometric problems in higher dimensions. It can alsobe applied to 2 + 1 dimensional problems (like the KP, DSI and 2+1dimensional N-wave equations) as in Section 2.6. However, for n = 2,the contents do not cover those in Chapter 2.

3.1 n + 1 dimensional AKNS system3.1.1 n + 1 dimensional AKNS system

The physical space-time is 3 + 1 dimensional. On the other hand, thesoliton theory is developed extensively in 1 + 1 and 2 + 1 dimensions.Here we will pay attention to the integrable systems in higher dimensionsand introduce a kind of higher dimensional solitons with the property ofelastic collision [34, 35, 39, 53].

First, we generalize the AKNS system in Section 1.2 to arbitraryn + 1 dimensions. When n = 2, the system here is different from thatin Chapter 2. Here the system still has spectral parameter while inthe generalized AKNS system the spectral parameter is replaced by adifferential operator.

Suppose that (t, x1, x2, · · · , xn) are the coordinates of Rn+1 where(x1, · · · , xn) denotes spatial coordinates and t denotes the time coordi-


nate. Consider the following linear system:

∂i∂∂ Ψ = UiUU Ψ = (λJiJJ + PiPP )Ψ (i = 1, · · · , n),

∂t∂∂ Ψ = V Ψ =m∑

α=0

VαVV λm−αΨ,(3.1)

where ∂i∂∂ are ∂t∂∂ are the partial derivatives with respect to xi and t respec-tively, PiPP ’s and VαVV ’s are N × N matrix functions, PiPP ’s are off-diagonal,JiJJ ’s are constant diagonal matrices. Since (3.1) contains several equa-tions, we call it Lax set.

The integrability conditions of the first part of (3.1) (its spatial part)are

∂j∂ UiUU − ∂i∂∂ UjU + [UiUU , UjUU ] = 0. (3.2)

Considering the coefficients of the powers of λ, we have

[JiJJ , JjJ ] = 0, (3.3)[JiJJ , PjPP ] = [JjJJ , PiPP ], (3.4)∂j∂ PiPP − ∂i∂∂ PjPP + [PiPP , PjPP ] = 0. (3.5)

Since JiJJ is diagonal, the equations in (3.3) are identities.In what follows, we need two more assumptions: (1) JiJJ ’s are linearly

independent. (2) If a matrix A satisfies [A, JiJJ ] = 0 for all i, then Ais a diagonal matrix. The first condition implies that the system isreally an n + 1 dimensional problem and cannot be reduced to a lowerdimensional system in some sense. In fact, if not, there would be a linearcombination of JiJJ ’s and a linear transformation of spatial variables sothat JnJJ = 0, that is, the essential part of the system is (n − 1) + 1dimensional. The second condition is equivalent to that there is a linearcombination J =

∑aiJiJJ such that the diagonal entries of J are distinct.

Therefore, there is a linear transformation in xi’s so that the diagonalentries of J1JJ are distinct.

From (3.4) and (3.5), we have:(i) There is an N × N off-diagonal matrix function P such that

PiPP = [P, JiJJ ] (i = 1, · · · , n). (3.6)

(ii) P satisfies the spatial constraints

∂j∂ PiPP − ∂i∂∂ PjPP + [PiPP , PjPP ] = 0. (3.7)

n + 1 dimensional integrable systems 105

(iii) The integrability conditions of (3.1) give the equations for VαVV ’s:

[JiJJ , V0VV ] = 0, ∂i∂∂ V0VV = 0, (3.8)[JiJJ , V off

αVV +1] = ∂i∂∂ V offαVV − [PiPP , VαVV ]off, (3.9)

∂i∂∂ V diagαVV +1 = [PiPP , V off

αVV +1]diag, (3.10)

(α = 0, 1, · · · , m − 1),

and the evolution equations

∂t∂∂ PiPP − ∂i∂∂ V offmVV + [PiPP , VmVV ]off = 0. (3.11)

There are many redundant equations in (3.7) – (3.11). In fact, withthe assumptions on J , we can always assume that all the diagonal en-tries of J1JJ are distinct, which can be realized by a linear transformationof (x1, · · · , xn). (3.6) implies that the off-diagonal entries of P are de-termined by P1PP . The following lemma shows that (3.7) – (3.10) can bereplaced by part of the equations in it.

Lemma 3.1 If the diagonal entries of J1JJ are distinct, then (3.7) – (3.11)are equivalent to

∂1PiPP − ∂i∂∂ P1PP + [PiPP , P1PP ] = 0, (3.12)[JiJJ , V off

αVV +1] = ∂i∂∂ V offαVV − [PiPP , VαVV ]off, (3.13)

∂i∂∂ V diagαVV = [PiPP , V off

αVV ]diag, (3.14)∂t∂∂ P1PP − ∂1V

offmVV + [P1PP , VmVV ]off = 0. (3.15)

Proof. (3.13) and (3.14) are the same as (3.9) and (3.10). Hence weonly need to prove that (3.7) and (3.11) are consequence of (3.12) and(3.15) with the help of (3.13) and (3.14). Let

∆ij = ∂j∂ PiPP − ∂i∂∂ PjPP + [PiPP , PjPP ]. (3.16)

By Jacobi identity, we have

[J1JJ , ∆ij ] = ∂j∂ [JiJJ , P1PP ] − ∂i∂∂ [JjJJ , P1PP ] + [J1JJ , [PiPP , PjPP ]]

= [JiJJ , ∆1j ] − [JjJJ , ∆1i].(3.17)

PiPP = [P, JiJJ ] and PjPP = [P, JjJJ ] imply that all the diagonal entries of [PiPP , PjPP ]are 0. Hence all the diagonal entries of ∆ij are 0. Using ∆1i = 0 (i ≥ 2),we know that [J1JJ , ∆ij ] = 0 for any i and j. Hence ∆off

ij = 0, i.e., ∆ij = 0.(3.7) is proved.


(3.11) is proved as follows. For any i > 1, using [PiPP , JjJ ] = [PjPP , JiJJ ], wehave

[∂t∂∂ PiPP − ∂i∂∂ VmVV + [PiPP , VmVV ], J1JJ ]

= [∂1VmVV − [P1PP , VmVV ], JiJJ ] − [∂i∂∂ VmVV − [PiPP , VmVV ], J1JJ ]

= ∂1[VmVV , JiJJ ] − ∂i∂∂ [VmVV , J1JJ ] + [[PiPP , VmVV ], J1JJ ] − [[P1PP , VmVV ], JiJJ ].

(3.18)

With (3.7), (3.9), (3.10) and the Jacobi identity, the last equality equals

∂1[PiPP , VmVV −1] − ∂i∂∂ [P1PP , VmVV −1] − [[VmVV , J1JJ ], PiPP ] + [[VmVV , JiJJ ], P1PP ]

= [∂1PiPP − ∂i∂∂ P1PP , VmVV −1] + [PiPP , [P1PP , VmVV −1]] − [[P1PP , [PiPP , VmVV −1]] = 0.

(3.19)Hence (3.11) holds. The lemma is proved.

From this lemma, the spatial constraints are simplified to (3.12) andthe evolution equations are simplified to (3.15). In general, (3.12) and(3.15) are still over-determined partial differential equations.

In Chapter 1, we have known that VαVV ’s do exist when n = 1, and theyare differential polynomials of P . When n > 1, VαVV ’s should satisfy manymore equations (3.13) and (3.14). Hence we need to show that they doexist and are still differential polynomials of P .

Lemma 3.2 (1) There exist VαVV ’s satisfying (3.13) and (3.14). TheseVαVV ’s are polynomials of P and its derivatives with respect to x, and aredenoted as VαVV = VαVV [P ]. (2) For given diagonal matrices V 0

αVV (t)’s whichare independent of xi’s, there exist unique VαVV [P ] such that VαVV [0] =V 0

αVV (t).

Sketch of proof. V0VV and V1VV can be obtained from (3.13) and (3.14) bydirect calculation. For general VαVV , this can be derived inductively. First,from the result in [111], VαVV ’s are polynomials of P , ∂1P , · · ·, ∂α

1 P if x2,· · ·, xn are regarded as parameters. Then from the spatial constraints(3.12) we can prove that these differential polynomials satisfy all theequations in (3.13) and (3.14). The complete proof is formal and tedious.For the details, see [53].

By this lemma, all the equations derived from the Lax set (3.1) arepartial differential equations of x1, · · ·, xn and t.

3.1.2 ExamplesIn (3.1), m, n, N (N ≥ n) and JiJJ are quite arbitrary. Hence (3.1)

leads to a lot of nonlinear partial differential equations. Here we showtwo simple examples.


Example 3.3 n = N = 2, m = 2,

J1JJ =

⎛⎝⎛⎛ 1 0

0 0

⎞⎠⎞⎞ , J2JJ =

⎛⎝⎛⎛ 0 0

0 − i

⎞⎠⎞⎞ , P =

⎛⎝⎛⎛ 0 p

q 0

⎞⎠⎞⎞ ,

p, q are complex-valued functions. Then,

P1PP = [P, J1JJ ] =

⎛⎝⎛⎛ 0 −p

q 0

⎞⎠⎞⎞ , P2PP = [P, J2JJ ] = i

⎛⎝⎛⎛ 0 −p

q 0

⎞⎠⎞⎞ ,

and the Lax set is

∂1Ψ = (λJ1JJ + P1PP )Ψ, ∂2Ψ = (λJ2JJ + P2PP )Ψ,

∂t∂∂ Ψ = (V0VV λ2 + V1VV λ + V2VV )Ψ.

The spatial constraints∂2P1PP − ∂1P2PP = 0

becomes∂1p + i∂2p = 0, ∂1q + i∂2q = 0,

i.e., p, q are holomorphic functions of the complex variable z = x1 + ix2,and ∂1p = pz, ∂1q = qz. From the recursion relations,

V0VV = V 00VV =

⎛⎝⎛⎛ a1 0

0 a2

⎞⎠⎞⎞ ,

V1VV =

⎛⎝⎛⎛ b1 (a2 − a1)p

(a1 − a2)q b2

⎞⎠⎞⎞ ,

V2VV =

⎛⎝⎛⎛ (a1 − a2)pq + c1 (a2 − a1)px + (b2 − b1)p

(a2 − a1)qx + (b1 − b2)q (a2 − a1)pq + c2

⎞⎠⎞⎞ ,

where a1, b1, c1, a2, b2, c2 are functions of t only. Moreover, we get thenonlinear evolution equations

∂p

∂t+ (a2 − a1)pzz + (b2 − b1)pz

+2(a2 − a1)p2q + (c2 − c1)p = 0,∂q

∂t+ (a1 − a2)qzz + (b2 − b1)qz

+2(a1 − a2)pq2 + (c1 − c2)q = 0.


Example 3.4 n = N = 3, m = 2,

J1JJ =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜i

0

0

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , J2JJ =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜0

i

0

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , J3JJ =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜0

0

i

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ .

Now

V0VV [P ] = V 00VV (t),

V1VV [P ] = [P, V0VV ] + V 01VV (t),

V2VV [P ] = −∑

JjJ

[∂P

∂xj, V0VV

]+ [P, [P, V0VV ]]off + [P, V 0

1VV ]off

−([P, V0VV ]P )diag + V 02VV (t).

We get a system of second order quasi-linear partial differential equa-tions. Its soliton solutions will be discussed later.

3.2 Darboux transformation and soliton solutions3.2.1 Darboux transformation

In this subsection we discuss the Darboux transformation for thehigher dimensional AKNS system.

We have known in Chapter 1 that in all Darboux matrices those ofdegree one are the most fundamental ones. The Darboux matrices ofhigher degree can be produced from the Darboux matrices of degree oneinductively. Hence we only discuss the Darboux matrix in the form

D(x, t, λ) = λI − S(x, t). (3.20)

Let Ψ′ = (λI − S)Ψ and substitute it into

∂i∂∂ Ψ′ = (λJiJJ + P ′iPP )Ψ′, (3.21)

we getP ′

iPP = PiPP + [JiJJ , S], (3.22)

∂i∂∂ S + SPiPP = P ′iPP S. (3.23)

From (3.22) and P ′iPP = [P ′, JiJJ ], we have

P ′ = P − S + arbitrary diagonal matrix.

For simplicity, letP ′ = P − S. (3.24)


Then (3.23) becomes

∂i∂∂ S = [PiPP , S] + [JiJJ , S]S. (3.25)

Substitute Ψ′ = (λI − S)Ψ into

∂t∂∂ Ψ′ =m∑

α=0

V ′αVV λm−αΨ′, (3.26)

and expand both sides in terms of the powers of λ, we get

V ′0VV = V0VV ,

V ′αVV +1 = VαVV +1 − SVαVV + V ′

αVV S (α = 0, 1, · · · , m − 1),(3.27)

and

∂t∂∂ S =

[m∑

α=0

VαVV Sm−α, S

]. (3.28)

(3.23) and (3.28) are partial differential equations for S. It can beverified directly that they are completely integrable. Therefore, for theinitial value of S at t = t0, xi = xi0, the solution S exists uniquely insome region around t = t0, xi = xi0.

What is more, we need to prove

V ′αVV = VαVV [P ′]. (3.29)

so that P ′ and P satisfy the same evolution equation (3.11) (or (3.15)).For α = 0, 1, 2 · · ·, (3.29) can be verified directly. For general α, theproof is quite tedious. It is similar to that for Theorem 1.11 and can befound in [53].

From the above discussion and the general method of constructingDarboux matrix, we have

Theorem 3.5 Suppose P is a solution of (3.15). For given constantdiagonal matrix Λ = diag(λ1, · · · , λN ), let hi be a column solution of(3.1) with λ = λi, H = (h1, · · · , hN ). When det H = 0 , let S = HΛH−1.Then λI − S is a Darboux matrix for (3.1).

Proof. The proof is similar to that for Theorem 1.9.Therefore, with Darboux transformation, we can get a series of solu-

tions(P (0), Ψ(0)) −→ (P (1), Ψ(1)) −→ (P (2), Ψ(2)) −→ · · ·

from a given solution P (0) and the corresponding solution Ψ(0) of theLax set (3.1). We have also seen that the formula

S = HΛH−1 (3.30)

appears in many cases.


3.2.2 u(N) caseIf P satisfies some extra constraints, usually the Darboux transfor-

mation constructed above can not keep these constraints. In this case,corresponding constraints should be added on the Darboux transforma-tion. This would be quite complicated in general. Here we will discussan interesting and simple case in which

JjJJ = iej = i diag(0, · · · , 0, 1j, 0, · · · , 0), (3.31)

and P ∈ u(N) (i.e. P ∗ = −P ). Suppose a solution P ∈ u(N) of (3.5) isknown, let Φ be the fundamental solution of its Lax set. Now

P ∗iPP = [P, JiJJ ]∗ = [J∗

iJJ , P ∗] = [JiJJ , P ] = −PiPP . (3.32)

In solving (3.8) – (3.10), take the “integral constant” (function of t)for VαVV to be a diagonal matrix V 0

αVV (t) in u(N), i.e., V 0αVV (t) are purely

imaginary diagonal matrices, then

VαVV ∈ u(N), α = 0, 1, · · · , m. (3.33)

In fact, V0VV = V 00VV (t) ∈ u(N). Suppose VlVV ∈ u(N), then from (3.9) for

α = l, we have [JjJJ , V offlVV +1] ∈ u(N). This implies V off

lVV +1 ∈ u(N). From (3.10)for α = l, we have ∂i∂∂ V diag

lVV +1 ∈ u(N). Therefore, V diaglVV +1 ∈ u(N) if it holds

at one point. (3.33) is proved by induction.Now we construct the Darboux transformation (P, Φ) → (P ′, Φ′) so

that P ′ ∈ u(N). As in Subsection 1.4.4, let λα (α = 1, · · · , N) take onlytwo mutually conjugate complex numbers:

λα = µ or µ (µ = ¯ µ). (3.34)

Let h0α’s satisfy

h0∗β h0

α = 0 (µβ = µα).

Solve

∂i∂∂ hα = (λαJiJJ + PiPP )hα, ∂t∂∂ hα =m∑

l=0

VlVV λm−lα hα

for initial data hα = h0α at some point to get the column solution hα.

Then when λα = λβ (i.e. λβ = λα), ∂i∂∂ (h∗βhα) = 0 and ∂t∂∂ (h∗

βhα) = 0hold. Hence

h∗βhα = 0 (3.35)

holds everywhere. Moreover, suppose that all the solutions h0α corre-

sponding to λα = µ are linearly independent, and all the solutions h0β


corresponding to λβ = µ are linearly independent. Then all these solu-tions are linearly independent. Hence (3.35) implies that h1, · · · , hNare linearly independent. This implies detH = 0, and S = HΛH−1 isdefined globally on Rn+1.

According to the discussion in Subsection 1.4.4, we have

S + S∗ = (µ + µ)I, (3.36)

andS∗S = |µ|2I. (3.37)

We have also

(λI − S)∗(λI − S) = (λ − µ)(λ − µ)I.

Moreover, from Subsection 1.4.4, V ′αVV ∈ u(N). Therefore, we get the

construction of Darboux transformation for the AKNS system with u(N)reduction. The Darboux transformations for many other systems withu(N) reduction can be constructed similarly.

3.2.3 Soliton solutionsNow we construct single and multi-soliton solutions. For simplicity,

only the case n = 3, N = 3 and m = 2 will be considered. It will alsobe proved that the soliton interaction in multi-soliton solution is elastic.

Take

V 00VV =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜a1 i 0 0

0 a2 i 0

0 0 a3 i

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , V 01VV =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜b1 i 0 0

0 b2 i 0

0 0 b3 i

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ ,

V 02VV =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜c1 i 0 0

0 c2 i 0

0 0 c3 i

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ ,

where ai, bi, ci (i = 1, 2, 3) are real constants. As before, take the seedsolution P = 0, then

Ψ0 =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ei(λx1+φ1(λ)t) 0 0

0 ei(λx2+φ2(λ)t) 0

0 0 ei(λx3+φ3(λ)t)

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , (3.38)

whereφi(λ) = aiλ

2 + biλ + ci (i = 1, 2, 3). (3.39)


Letλ1 = µ, λ2 = λ3 = µ (µ = ¯ µ), (3.40)

ζiζζ = µxi + φi(µ)t. (3.41)

Let

H =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜eiζ1 −aeiζ1 −beiζ1

aeiζ2 eiζ2 0

beiζ3 0 eiζ3

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , (3.42)

then λi’s and H satisfy the conditions in (3.34) and (3.35), and

det H = ei(ζ1+ζ2+ζ3)∆, (3.43)

∆ = 1 + |a|2ei[(ζ2−ζ2)−(ζ1−ζ1)] + |b|2ei[(ζ3−ζ3)−(ζ1−ζ1)] > 0. (3.44)

The derived solution consists of three waves. They are

p = S23 = ab(µ − µ)ei[(ζ2−ζ1)−(ζ3−ζ1)]/∆,

q = S13 = (µ − µ)bei(ζ1−ζ3)/∆,

r = S12 = (µ − µ)aei(ζ1−ζ2)/∆.

(3.45)

According to (3.41) and (3.45), |p|, |q| and |r| are travelling waves withvelocities

vi = −φi(µ) − φi(µ)µ − µ

(i = 1, 2, 3) (3.46)

respectively. In fact,

|p| = |ab| |µ − µ|ei(µ−µ)[(ξ2−ξ1)+(ξ3−ξ1)]/2/∆,

|q| = |b| |µ − µ|ei(µ−µ)(ξ3−ξ1)/2/∆,

|r| = |a| |µ − µ|ei(µ−µ)(ξ2−ξ1)/2/∆,

(3.47)

where∆ = 1 + |a|2ei(µ−µ)(ξ2−ξ1) + |b|2ei(µ−µ)(ξ3−ξ1), (3.48)

ξi = xi − vit (i = 1, 2, 3). (3.49)

For fixed t, |p|, |q| and |r| are all bounded. Moreover, when |x| →∞, they tend to zero rapidly along all directions except the followingdirections:

(i) for |p| : |x2 − x3| → +∞, 2x1 − x2 − x3 → +∞,

(ii) for |q| : |x3 − x1| → +∞, 2x2 − x1 − x3 → +∞,

(iii) for |r| : |x2 − x1| → +∞, 2x3 − x1 − x2 → +∞.

(3.50)


Here we suppose i(µ − µ) > 0. The solution we derived here is called asingle-soliton solution. Each solution consists of three waves p, q and r.If i(µ− µ) < 0, the inequalities 2x1−x2−x3 → +∞ · · · in (3.50) shouldbe replaced by 2x1 − x2 − x3 → −∞ · · · .

These solutions are not localized, since p, q and r are constants alongthe straight line x2 − x1 = constant, x3 − x1 = constant. They are akind of line solitons appeared in Chapter 2.

The other components of S are

S11 = 1∆(µ + µ|a|2ei(µ−µ)(ξ2−ξ1) + µ|b|2ei(µ−µ)(ξ3−ξ1)),

S22 = 1∆(µ + µ|a|2ei(µ−µ)(ξ2−ξ1) + µ|b|2ei(µ−µ)(ξ3−ξ1)),

S33 = 1∆(µ + µ|a|2ei(µ−µ)(ξ2−ξ1) + µ|b|2ei(µ−µ)(ξ3−ξ1)).

(3.51)

These expressions will be useful in considering multi-soliton solutions.

Remark 25 If ai’s, bi’s and ci’s are functions of t rather than constants,then the soliton solution has variable velocities. The term φi(λ)t in(3.41) should be changed to

ωi(λ, t) =∫ t

t

∫∫0

∫∫φi(λ, τ) dτ =

∫ t

t

∫∫0

∫∫[ai(τ)λ2 + bi(τ)λ + ci(τ)] dτ. (3.52)

Especially, |p|, |q| and |r| are constants along the curve

xi +ωi(µ, t¯ ) − ωi(µ, t)

µ − µ= constant.

Different choice of the functions ai’s, bi’s and ci’s leads to different be-havior of the waves. For instance, if the integrals of ai’s, bi’s and ci’s areperiodic functions, we get the waves which oscillate periodically. If ai’s,bi’s and ci’s increase fast at infinity, the waves have large accelerationand will move apart soon.

k-multi-soliton solution can be obtained by k times of Darboux trans-formation of degree one from the seed solution P = 0. Take the param-eters µ0, µ1, · · ·, µk−1 (µa = µb, µa = ¯ µb when a = b), then

Ψl(λ, x, t) = (λI − Sl−1) · · · (λI − S0SS )Ψ0 (k = 1, · · · , l − 1). (3.53)

Here the matrices Sk (k = 0, 1, · · · , l − 1) are constructed from Φk byusing (3.30) with (3.34) and (3.35) successively. Then we obtain thesolution

P l = −(S0SS + S1 + · · · + Sl−1) (3.54)


of the nonlinear equation. Moreover, for considering the behavior of themulti-soliton solutions, we assume

vi(µl) (i = 1, 2, · · · , n; l = 0, 1, · · · , k − 1) (3.55)

are distinct.Multi-soliton solutions have the following property.

Theorem 3.6 When t → ±∞, a k-multi-soliton is asymptotic to ksingle solitons. These single solitons for t → −∞ and t → +∞ have thesame amplitude.

Proof. We first consider the case k = 2. Choose the moving coordinatewith velocity vi(µ1), that is, keep ξ1

i = xi−vi(µ1)t finite and t → ±∞.Then

ξ0i − ξ0

jξ = xi − xj − (vi(µ0) − vj(µ0))t → ±∞ (i = j), (3.56)

S0SS ∼

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜σ±

1

σ±2

σ±3

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ (t → ±∞) (3.57)

where σ±i = µ0 or µ0. Hence,

Ψ1 ∼

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜(λ − σ±

1 )ei(λx1+φ1(λ)t)

(λ − σ±2 )ei(λx2+φ2(λ)t)

(λ − σ±3 )ei(λx3+φ3(λ)t)

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ . (3.58)

Choose H1 so that

H1 ∼

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜eiζ1

1 −a1eiζ11 −b1eiζ1

1

a1eiζ12 eiζ1

2 0

b1eiζ13 0 eiζ1

3

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , (3.59)

whereζ1iζζ = µ1xi + φi(µ1)t − ln(µ1 − σ±

i )i. (3.60)

(3.59) and (3.60) are similar to (3.42) and (3.41). For the asymp-totic behavior of S1, the equalities similar to (3.45), (3.47) and (3.51)hold, provided that a, b, µ and ζiζζ ’s are changed to a1, b1, µ1 and ζ1

iζζ ’srespectively. Hence, when t → ±∞ and ξ1

i ’s keep finite,

P 2 = −Soff0SS − Soff

1


Figure 3.1. p wave of the double soliton solution, t = −1

Figure 3.2. p wave of the double soliton solution, t = −0.5

is asymptotic to a single soliton solution.The asymptotic behavior of p1, q1 and r1 are similar when t → +∞

and t → −∞ respectively. However, the extra term − ln[(µ1 − σ±i )i in

(3.60) are different for t → +∞ and t → −∞. The difference of thesetwo terms represents the phase shift. This means that the “center” ofthe wave ξ1

i = 0, xi − vi(µ1)t = 0 shifts differently for t → +∞ andt → −∞.

When t → ±∞, ξ0i = xi − v(µ0)t keep finite (then ξ1

i → ±∞), thesame result can be obtained by the theorem of permutability. Hence thetheorem is proved for k = 2.

Similar to Chapter 1, multi-soliton solution can be obtained by severalDarboux transformations of degree one. Using induction, we can provethat when t → ±∞, a multi-soliton solution splits up into several singlesoliton solutions. The interactions of these single solitons are elastic.That is, the shape and velocity of the norm of these solitons will notchange from t → −∞ to t → +∞, but there is a phase shift. The phaseshift is given by the difference of σ±

i in (3.60) for t → −∞ and t → +∞.In summary, the interactions of the solitons are elastic, and the phase

shifts of the solitons can be computed explicitly. This implies that thehigher dimensional solitons also have this basic property of 1+1 dimen-sional solitons. However, these solitons are not local. They do notapproach to zero along every direction.

Figures 3.1 – 3.5 describe the p wave of the double soliton solution.


Figure 3.3. p wave of the double soliton solution, t = 0

Figure 3.4. p wave of the double soliton solution, t = 0.5

Figure 3.5. p wave of the double soliton solution, t = 1

3.3 A reduced system on Rn

As an example, we consider a special AKNS system on Rn. Let

e1 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜1

0

0

. . .

0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟, e2 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

1

0

. . .

0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟,


· · · en =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

0

0

. . .

1

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟(3.61)

be n diagonal matrices. Let

EiEE =

⎛⎝⎛⎛ ei 0

0 −ei

⎞⎠⎞⎞ (3.62)

be n 2n × 2n matrices. Consider the system

∂Ψ∂xi

= (λEiEE + PiPP )Ψ, (3.63)

where PiPP ’s are real 2n× 2n matrices. Note that if ai (i = 1, 2, · · · , n) aren real numbers so that |ai| are distinct, then the entries of

∑aiEiEE are

distinct. Hence there exists 2n × 2n matrix P such that

PiPP = [P, EiEE ]. (3.64)

What is more, we want that P satisfies the following extra conditions.This is a reduction of (3.63). Let P be written as

P =

⎛⎝⎛⎛ P11PP P12PP

P21PP P22PP

⎞⎠⎞⎞ (3.65)

where P11PP , P12PP , P21PP and P22PP are n × n matrices. It is required that

P11PP = −P22PP is symmetric,

P12PP = −P21PP is anti-symmetric.(3.66)

This is an AKNS system with strong reduction, which is derived from ageometric problem [7].

The Darboux transformation will be constructed as follows [47]. Bythe general theory (Section 3.2), let Λ = diag(λ1, · · · , λ2n), ψa be acolumn solution of (3.63) for the eigenvalue λ = λa. Let

H = (ψ1, · · · , ψ2n), (3.67)

S = HΛH−1. (3.68)


From the general method for constructing the Darboux matrix, we knowthat λI − S is a Darboux matrix, and the Darboux transformation(P, Ψ) → (P ′, Ψ′) is given by

Ψ′ = (λI − S)Ψ, P ′ = P − S. (3.69)

In order that P ′ still satisfies the conditions (3.65) and (3.66), we needspecial choice of λa’s and ψa’s. First, we prove the following lemma.

Lemma 3.7 Suppose P satisfies (3.66), and

ψ(λ0) =

⎛⎝⎛⎛ ψ1(λ0)

ψ2(λ0)

⎞⎠⎞⎞ (3.70)

is a column solution of (3.63) with λ = λ0 where ψ1, ψ2 are n-vectors,then,

ψ(λ0) =

⎛⎝⎛⎛ ψ2(λ0)

ψ1(λ0)

⎞⎠⎞⎞ (3.71)

is a column solution of (3.63) with λ = −λ0.

Proof. Let P11PP = F , P21PP = G, then

PiPP =

⎛⎝⎛⎛ [F, ei] G, eiG, ei [F, ei]

⎞⎠⎞⎞ =

⎛⎝⎛⎛ FiFF Gi

Gi FiFF

⎞⎠⎞⎞ , (3.72)

whereFiFF = [F, ei], Gi = G, ei = Gei + eiG. (3.73)

(3.63) can be written as

∂ψ1

∂xi= λeiψ

1 + FiFF ψ1 + Giψ2,

∂ψ2ii

∂xi= −λeiψ

2 + Giψ1 + FiFF ψ2

(3.74)

which is invariant under the transformation

ψ =

⎛⎝⎛⎛ ψ1

ψ2

⎞⎠⎞⎞ −→ ψ =

⎛⎝⎛⎛ ψ2

ψ1

⎞⎠⎞⎞ , λ → −λ (3.75)

The lemma is proved.Now take a non-zero real number µ. Let

λi = µ, λn+i = −µ, (i = 1, 2, · · · , n). (3.76)


Let ψi = ψi(µ) (i = 1, 2, · · · , n) be n linearly independent column solu-tions of (3.63) with λ = µ, and

ψn+i = ψi (3.77)

where ψi (i = 1, 2, · · · , 2n) are defined by (3.72). Using (3.74) and P ∗iPP =

−PiPP we know

∂

∂xi(ψ∗

αψβ) = 0 (λα = λβ) (α, β = 1, 2, · · · , 2n). (3.78)

Hence ifψ∗

αψβ = 0 (λα = λβ) (3.79)

hold at one point, they hold everywhere. We shall prove later that thereare linearly independent ψα’s so that (3.77) and (3.79) hold.

From the construction of S, SH − HΛ = 0, i.e.,

Sψβ = λβψβ, ψ∗γS∗ = λγψ∗

γ . (3.80)

Henceψ∗

γS∗Sψβ = λβλγψ∗γψβ . (3.81)

When λβ = λγ , ψ∗γψβ = 0. When λβ = λγ , λβλγ = µ2. Thus,

ψ∗γS∗Sψβ = µ2ψ∗

γψβ (3.82)

for all β and γ. Therefore, S∗S = µ2I. On the other hand,

ψ∗γ(S∗ − S)ψβ = (λγ − λβ)ψ∗

γψβ = 0. (3.83)

So S satisfiesS∗ = S, S∗S = µ2I. (3.84)

Rewrite S as

S =

⎛⎝⎛⎛ S11 S12

S21 S22

⎞⎠⎞⎞ (SabS (a, b = 1, 2) are n × n matrices). (3.85)

Since S∗ = S, S11 and S22 are symmetric matrices, and S12 = S∗21.

Moreover, since Λ =

⎛⎝⎛⎛ µI 0

0 −µI

⎞⎠⎞⎞, by the construction of H we know

thatS11 = −S22, S12 = −S21. (3.86)

This implies that the matrix S satisfies the conditions (3.65) and (3.66)for P . Hence P ′ = P − S also satisfies these conditions. It is remained


to prove that there are linearly independent column vectors ψα(λ) sat-isfying (3.79). In fact, it is sufficient to choose their initial value to belinearly independent and to satisfy (3.79). From S∗S = µ2I,

S211 − S2

12 = µ2I, [S12, S11] = 0. (3.87)

Letσ = µ−1(S11 + S12), (3.88)

thenσ∗ = µ−1(S11 − S12), (3.89)

and σ∗σ = I, i.e., σ is an n × n orthogonal matrix.Take σ0 to be an arbitrary constant orthogonal matrix. Let

S011 =

µ

2(σ0 + σ0∗), S0

12 =µ

2(σ0 − σ0∗), (3.90)

then(S0

11)2 − (S0

12)2 = µ2I, [S0

12, S011] = 0. (3.91)

Let

S0 =

⎛⎝⎛⎛ S011 S0

12

−S012 −S0

11

⎞⎠⎞⎞ . (3.92)

S0 is a symmetric matrix. Since S0∗S0 = (S0)2 = µ2I, the eigenvaluesof S0 are ±µ. If ψ0 is the eigenvector with eigenvalue µ, then ψ0 is theeigenvector with eigenvalue −µ. Therefore, ψ0

i ’s and ψ0i (i = 1, · · · , n)

form a set of 2n linearly independent vectors ψ0α (α = 1, · · · , 2n) with

ψn+i = ψ0i (i = 1, · · ·n). These are the initial values of ψα’s to be

required and the Darboux matrix S which keeps the reduction is con-structed.

Theorem 3.8 The Darboux transformation which keeps the reduction(3.66) exists and is obtained from the above construction of S.

Remark 26 (3.63) can be generalized to

∂Ψ∂xi

= (λEiEE + PiPP +Qi

λ)Ψ. (3.93)

In this case the above construction of Darboux transformation is stillvalid. The transformation of Qi is given by

Q′i = SQiS

−1. (3.94)

Using the Darboux transformation in (3.66) and (3.93), we can solve thegeometric problem in [7] and give the explicit solution of the B¨cklund¨transformation presented there.

Chapter 4

SURFACES OF CONSTANTCURVATURE,BACKLUND CONGRUENCES ANDDARBOUX TRANSFORMATION

There are many important partial differential equations originatingfrom classical differential geometry. The famous sine-Gordon equationis one of them.

The non-Euclidean geometry was initiated in the early nineteenthcentury. Afterwards, it was found that the surfaces of constant negativeGauss curvature realize the non-Euclidean geometry locally. Therefore,these surfaces were studied extensively. The sine-Gordon equation andits Backlund transformation appeared in that period. A surface of con-stant negative Gauss curvature corresponds to a non-zero solution ofthe sine-Gordon equation, and the Backlund transformation provides a¨way to construct a new solution of the sine-Gordon equation from aknown one, and a way to construct a new surface of constant negativeGauss curvature from a known surface of constant negative Gauss curva-ture too. Since the middle of the twentieth century, the transformationsof the solutions of some partial differential equations become effectivemethods in the soliton theory. The role of differential geometry in thesoliton theory becomes more and more important.

Since the Backlund transformation depends on solving a system ofintegrable nonlinear partial differential equations, usually its solutionscan not be expressed explicitly, except for some very special cases. How-ever, as in the previous chapters, the Darboux transformation is a wayto obtain explicit expressions. In this chapter, we combine the Darbouxtransformation and the classical Backlund transformation to realize theconstruction of the Backlund congruences and the surfaces of constantnegative Gauss curvature. Hence a serious of such congruences and sur-faces can be obtained explicitly by purely algebraic algorithm. Besides,we also discuss the surfaces of constant Gauss curvature in the Minkow-


ski space R2,1. The existence and explicit construction of the general-ized Backlund congruence are studied as well. However, the situation¨becomes more complicated since a surface or a line congruence may betime-like or surface-like. Especially, we give the geometric meaning ofthe Backlund transformation between α = sin α and α = sinh α, theexplicit construction of the time-like and space-like surfaces of constantnegative Gauss curvature and the Backlund congruences. The Darbouxtransformations are also used to construct surfaces of constant meancurvature.

In summary, in these geometric problems, the Lax pair is not only atool for solving related partial differential equations, but also the geo-metric objects that we want to study. With the idea of Darboux trans-formation, some of the geometric objects can be constructed by purelyalgebraic algorithm. So it becomes much easier to constructing them bycomputer.

There are also other geometric problems related with integrable sys-tems which we will not discuss in this book. For example, the solitonsurfaces introduced by A. Sym [96, 97, 17] and the topics in [90].

For the convenience of the readers we give a sketch of the basic factsof the theory of surfaces in R3 and R2,1 respectively by using the differ-ential forms.

4.1 Theory of surfaces in the Euclidean space R3

We use the moving frame method to introduce the basic theory ofsurfaces in Euclidean space. In R3, r represents the position vector ofa point. The length of r is r =| r |= √

x2 + y2 + z2. As is known,a surface in R3 is a two dimensional differential manifold embedded(or immersed) in R3. It can be covered by some open subsets (surfacecharts) which are homeomorphic to connected open regions of a plane.In each surface chart, r can be represented by a vector-valued functionof two parameters as r = r(u1, u2) and in the intersection of two surfacecharts, the parametric representations (u1, u2) and (v1, v2) are linked

by differentiable relations va = φa(u1, u2) (a = 1, 2) with∂(φ1, φ2)∂(u1, u2)

=

0. ra =∂r∂ua

(a = 1, 2) form a basis of the tangent plane at r, and

n =r1 × r2

| r1 × r2 | is the normal vector of the surface. r1, r2,n form

a frame at r. As a basis of the global theory of surfaces, the localtheory of surfaces discusses the properties of the surface charts with theabove-mentioned parametrization. We use the term surface to indicate

Surfaces of constant curvature, B¨cklund congruences¨ 123

a surface chart, since we only consider the local theory. The frame(r1, r2,n) is called a natural frame because it is obtained naturally fromthe parametrization of the surface. Differentiate r1, r2, n and write themas linear combinations of ra,n, we obtained the fundamental equationsof a surface

dr = duara,

dra = ωbarb + ω3

an,

dn = ωb3rb, (a, b = 1, 2),

(4.1)

where ωba, ω

3a, ω

b3 are 1-forms of u1, u2. In (4.1) and hereafter, the summa-

tion convention is used, i.e., the symbol Σ is omitted for double indices.The first fundamental form of the surface is

I = ds2 = dr · dr = gab dua dub, (4.2)

wheregab = ra · rb = gba. (4.3)

Sincen · dra + ra · dn = 0,

ωb3 and ωa

3 are related by

ω3a = −gabω

b3. (4.4)

Letω3

a = babdub. (4.5)

From d2r = 0, ω3a ∧ dua = 0. We have bab = bba. The quadratic form

II = −dr · dn = −gabωb3 dua = bab dua dub (4.6)

is the second fundamental form of the surface with coefficients

bab = − ∂r∂ua

· ∂n∂ub

=∂2r

∂ua∂ub· n. (4.7)

Two principal curvatures are eigenvalues of the second fundamental formwith respect to the first fundamental form, i.e., they are two roots of

det(bαβ − λgαβ) = 0.

Since the first fundamental form is positive definite, two principal cur-vatures are both real.

Now ωba can be written as

ωba = Γb

ac duc, (4.8)


where Γbac’s are the Christoffel symbols of the surface. From d2r = 0,

we obtain dua ∧ ωba = 0, hence Γc

ab = Γcba. Moreover, ra · rb = gab and

ra · drb + dra · rb = dgab imply

dgab = gacωcb + gcbω

ca.

Therefore,

Γcab =

12gcd(∂gad

∂ub+

∂gbd

∂ua− ∂gab

∂ud

). (4.9)

Exterior differentiating the fundamental equations (4.1) gives the inte-grability conditions of (4.1). Using d2r = 0 and the exterior differentia-tions of the second and third equations of (4.1), we obtain

dωba + ωb

c ∧ ωca = ω3

a ∧ ωb3, (4.10)

dω3a + ω3

b ∧ ωba = 0. (4.11)

(4.10) and (4.11) are called Gauss equations and Codazzi equations ofthe surface respectively. The left hand side of (4.10) are usually writtenas

dωba + ωb

c ∧ ωca =

12Rb

acdduc ∧ dud, (4.12)

where

Rbacd =

∂Γbad

∂uc− ∂Γb

ac

∂ud+ Γb

ecΓead − Γb

edΓeac (4.13)

is the Riemannian curvature tensor. It is fully determined by the co-efficients of the fundamental form gab and their derivatives of first andsecond order. From (4.10), (4.4) and (4.5),

12Rb

acd duc ∧ dud = −bac duc ∧ gbebed dud. (4.14)

HenceRb

acd = −gbe(bacbed − badbec). (4.15)

DenoteRbacd = gbeR

eacd, (4.16)

thenRbacd = bbcbad − bacbbd,

where the subscripts a, b, c, d take the values 1 or 2. In fact, there isonly one independent equation, i.e.,

R1212 = b11b22 − b212. (4.17)


This is another form of the Gauss equation. It can also been written as

R1212

g11g22 − g212

=b11b22 − b2

12

g11g22 − g212

= K (4.18)

where K is called the total curvature or Gauss curvature of the surface.The left hand side of (4.18) is determined by the first fundamental form.Before Gauss’ work, K is expressed by the first fundamental form to-gether with the second fundamental form (second equality of (4.18)).The first equality in (4.18) is the famous Gauss Theorem. It impliesthat the Gauss curvature is actually determined by the first fundamen-tal form of the surface. The properties which can be determined by thefirst fundamental form are call intrinsic properties.

The above discussion is summarized as follows. For a given surfaceS, its first and second fundamental forms satisfy the Gauss-Codazziequations. Conversely, given two differential forms I = gab dua dub

(a, b = 1, 2, gab is positive definite), II = bab dua dub, and supposethat the Gauss-Codazzi equations hold, then there exists a surface chartwhose first and second fundamental forms are I and II respectively. Thissurface chart is uniquely determined in a simply connected region up torigid motions and reflections. This is the fundamental theorem of sur-faces. In fact, the surface chart r = r(u, v) is determined by solvingthe system of linear equations (4.1). Since the integrability condition of(4.1) is just the Gauss-Codazzi equations, (4.1) is completely integrable.Hence the fundamental theorem of surfaces holds.

The tangent vectors r1 and r2 can be replaced by their linear combina-tions. Suppose e1 and e2 are linear combinations of r1, r2 and e1, e2 areorthogonal with each other, then e1, e2,n form an orthogonal frameof the surface at r. The fundamental equations of the surface can bewritten as

dr = ωaea,

dea = ωbaeb + ω3

an,

dn = ωa3ea.

(4.19)

There are also the following relations among ωa and ωba:

dωa + ωab ∧ ωb = 0 (a, b = 1, 2), (4.20)

ωji + ωi

j = 0 (i, j = 1, 2, 3). (4.21)

ωab (a, b = 1, 2) are uniquely determined by (4.20) and the relation ωa

b +ωb

a = 0. In the orthogonal frame, the Gauss-Codazzi equations aresimplified as

dω12 = ω3

2 ∧ ω13 (4.22)


anddω3

a + ω3b ∧ ωb

a = 0. (4.23)

Especially, the Gauss equation can also be written as

dω12 = R1212ω

1 ∧ ω2 = Kω1 ∧ ω2. (4.24)

Hereafter, we will mainly use the orthogonal frames. Moreover, theorientation of the frame e1, e2,n will not always be fixed, that is, nmay be replaced by −n.

4.2 Surfaces of constant negative Gausscurvature, sine-Gordon equation andBacklund transformations¨

4.2.1 Relation between sine-Gordon equation andsurface of constant negative Gausscurvature in R3

Suppose S is a surface of constant negative Gauss curvature in R3.By using a scaling transformation of S, we can also suppose K = −1.Take the lines of curvature as coordinate curves and let e1 and e2 be theunit tangent vectors of the lines of curvature. Denote

dr =∂r∂u

du +∂r∂v

dv = A due1 + B dve2, (4.25)

ω1 = A du, ω2 = B dv. (4.26)

The first fundamental form of the surface is

ds2 = A2 du2 + B2 dv2, (4.27)

and the second fundamental form is

II = k1A2 du2 + k2B

2 dv2 = k1(ω1)2 + k2(ω2)2, (4.28)

where k1 and k2 are principal curvatures. The Gauss curvature K = k1k2

(= −1). On the other hand, (4.6) implies

II = ω31ω

1 + ω32ω

2. (4.29)

Comparing with (4.28), we get

ω31 = k1ω

1 = k1A du, ω32 = k2ω

2 = k2B dv. (4.30)

From (4.20),dωa + ωa

b ∧ ωb = 0,


henceω2

1 = −ω12 = −Av

Bdu +

Bu

Adv. (4.31)

The Codazzi equationdω3

1 + ω32 ∧ ω2

1 = 0

leads to(k1A)v = k2Av,

i.e.,(k1 − k2)Av + k1vA = 0. (4.32)

Since K = k1k2 = −1, we can set k1 = tanα

2, k2 = − cot

α

2, (0 < α < π),

thenk1 − k2 =

1

sinα

2cos

α

2

.

Substituting it into (4.32), we get

(log A)v −(

log cosα

2

)v

= 0.

HenceA = cos

α

2U(u).

Similarly,B = sin

α

2V (v).

Here U(u) and V (v) are functions depending on u and v respectively. Letdu1 = U(u) du, dv1 = V (v) dv, then u1 and v1 become new parametersand will still be written as u, v. Hence

A = cosα

2, B = sin

α

2,

ω1 = cosα

2du, ω2 = sin

α

2dv,

ω31 = sin

α

2du, ω3

2 = − cosα

2dv,

ω21 =

12(αv du + αu dv) = −ω1

2.

(4.33)

Substituting them into Gauss equation (4.10), we get

dω12 = ω3

2 ∧ ω12 = k1k2ω

1 ∧ ω2 = −ω1 ∧ ω2,

from which it is seen that α satisfies the sine-Gordon equation

αuu − αvv = sin α. (4.34)


The coordinates we are using are called the Chebyshev coordinates, andthe corresponding frame is called the Chebyshev frame. It can be checkeddirectly that the Codazzi equations are consequence of the sine-Gordonequation (4.34). Hence the fundamental theorem of surfaces leads to thefollowing theorem.

Theorem 4.1 For any solution α of the sine-Gordon equation (4.34)(0 < α < π), one can construct a surface of constant negative Gauss cur-vature by solving the fundamental equation of surfaces (4.19), in whichthe coefficients ωa, ωa

b , ω3b are given by (4.33).

Note that under the Chebyshev coordinates, the first fundamentalform of the surface is

ds2 = cos2α

2du2 + sin2 α

2dv2, (4.35)


II = cosα

2sin

α

2(du2 − dv2). (4.36)

Hence two systems of asymptotic curves are real and the directions du :dv = 1 : ±1 are the asymptotic directions of the surface. The cosine ofthe angle between these two directions is

cos θ = cos2α

2− sin2 α

2= cos α. (4.37)

Therefore, the geometric meaning of α is the angle between two asymp-totic curves of the surface.

Remark 27 Theorem 4.1 is a local result. On the other hand, the clas-sical Hilbert Theorem says that there is no complete surface of constantnegative Gauss curvature in R3 [56]. Here a complete surface means atwo dimensional open manifold on which each geodesic can be extendedinfinitely.

Remark 28 On a surface of constant positive Gauss curvature (not asphere), there are also Chebyshev coordinates. In this case,

ω1 = coshα

2du, ω2 = sinh

α

2dv,

ω31 = sinh

α

2du, ω3

2 = coshα

2dv,

ω21 = −ω1

2 =12(−αv du + αu dv).

(4.38)


The construction of the surface depends on the solution of the negativesinh-Laplace equation

α = − sinh α. (4.39)

The proof of these facts is similar with the case of constant negativeGauss curvature.

Remark 29 Suppose a solution of the sine-Gordon equation or the neg-ative sinh-Laplace equation is known. The construction of surface ofconstant negative Gauss curvature or constant positive Gauss curvatureis reduced to solving the fundamental equations of the surface. Theseare completely integrable systems of linear partial differential equations.The solutions of this system can be obtained by solving ordinary differ-ential equations. However, the construction of explicit solutions is stillnot easy. The Darboux transformation will provide an efficient methodfor the explicit construction.

4.2.2 Pseudo-spherical congruenceLine congruences originated from the study of refraction and reflec-

tion of light. A two-parametric family of straight lines is called a linecongruence. For example, all the normal lines of a surface constitutea line congruence which is called the normal congruence of the surface.However, in general, a line congruence may not be a normal congruence,that is, there may not exist a surface which is orthonormal to all thestraight lines in the congruence.

Locally, a line congruence can be expressed as follows. Suppose S isa surface expressed as

X = X(u, v).

Given a unit vector ξ(u, v) at each point of the surface, let

Y(u, v, λ) = X(u, v) + λξ(u, v). (4.40)

When (u, v) are fixed and λ changes, Y forms a straight line. Hence Yrepresents a line congruence. The surface S is called the reference sur-face. Clearly, the reference surface of a line congruence is quite arbitrary.For a curve C on the reference surface S, the straight lines in the congru-ence passing through C form a ruled surface. Suppose the equations ofC are u = u(t), v = v(t). Substituting them into (4.40), we can get theequations of the ruled surface parametrized by t and λ. If there exists

λ = λ(t) such thatdYdt

and ξ are parallel, then the ruled surface becomes

a developable surface, and the curve Y(u, v, λ) = Y(u(t), v(t), λ(t)) isthe line of regression of the developable surface. Now

Y = Y(t) = X(u(t), v(t)) + λ(t)ξ(u(t), v(t)),


anddYdt

=dXdt

+dλ

dtξ + λ(t)

dξ

dt= µξ,

that is, ξ,dXdt

anddξ

dtare linearly dependent. Hence, the ruled surface

is developable if and only if u = u(t), v = v(t) satisfy

det(dX

dt,dξ

dt, ξ)

= 0. (4.41)

The differential form of this condition is

det(Xu du + Xv dv, ξu du + ξv dv, ξ ) = 0.

Write this equation as

A du2 + 2B du dv + C dv2 = 0. (4.42)

This is the quadratic equation of du : dv. When B2 −AC > 0, there aretwo different real roots for du : dv which correspond to two developablesurfaces in the line congruence. Thus for each line in the congruencethere are two developable surfaces passing through it.

As is known, a developable surface is generated by the tangent lines ofa spatial curve named line of regression. When B2 − AC > 0, each linein the congruence are tangent to a line of regression of each developablesurface. The tangent points are called focal points. The set of all focalpoints form the focal surfaces. A focal surface is also obtained from thelines of regression. Therefore a line congruence can be regarded as theset of all common tangent lines of two focal surfaces.

We only consider the case B2 − AC > 0. Suppose S and S∗ aretwo focal surfaces of a line congruence. Suppose PP ∗ is a line in thecongruence, which is the common tangent line of two focal surfaces, andP , P ∗ are the tangent points. The correspondence between P and P ∗leads to the correspondence between S and S∗. In differential geometry,this is called a Laplace transformation between the surfaces S and S∗ (itis different from the terminology Laplace transformation in analysis).

Suppose n and n∗ are unit normal vectors of S at P and of S∗ at P ∗respectively. Let τ be the angle between n and n∗, and l be the distancebetween P and P ∗, i.e.,

n · n∗ = cos τ (sin τ = 0) , (4.43)

dppd ∗ = l (l = 0) . (4.44)

When τ and l are both constants, the congruence is called a pseudo-spherical congruence.


Historically, a surface of constant negative Gauss curvature is calleda pseudo-sphere. This is the origin of the name pseudo-spherical con-gruence. A pseudo-spherical congruence is also called a Backlund con-¨gruence.

Theorem 4.2 (Backlund Theorem) Two focal surfaces¨ S and S∗ of apseudo-spherical congruence are both the surfaces of constant negativeGauss curvature with the same K = − sin2 τ/l2.

Proof. Let P be a general point on S, with position vector r(u, v). Its

corresponding point P ∗ has position vector r ∗(u, v).−→

PP ∗ is a common

tangent line of S and S∗. Let e1 be the unit vector along−→

PP ∗. Lete1, e2,n be the orthogonal frame of S at P and e ∗

1 , e ∗2 ,n ∗ be the

orthogonal frame of S∗ at P ∗ with e ∗1 = e1. From the above assumptions,

r ∗ = r + le1, (4.45)

e ∗2 = cos τe2 + sin τn,

n ∗ = − sin τe2 + cos τn.(4.46)

The fundamental equations of S and S∗ are

dr = ωaea,

dea = ωbaeb + ω3

an,

dn = ωa3ea,

(4.47)

anddr ∗ = ω∗ae ∗

a ,

de ∗a = ω∗b

a e ∗b + ω∗3

a n,

dn ∗ = ω∗a3 ea

(4.48)

respectively. From (4.45) – (4.48), we get

dr ∗ = ωaea + lde1 = (ωa + lωa1)ea + lω3

1n

= ω∗ae ∗a = ω∗1e1 + ω∗2(cos τe2 + sin τn).

Henceω∗1 = ω1, ω∗2 =

l

sin τω3

1. (4.49)

Rewrite (4.45) and (4.46) so that r, e1, e2 are expressed by r∗, e∗1, e∗2,then we have

ω2 =l

sin τω∗3

1 . (4.50)


By (4.46) and (4.48),

de ∗2 = cos τ(ω1

2e1 + ω32n) + sin τ(ω1

3e1 + ω23e2)

= ω∗12 e ∗

1 + ω∗32 n ∗ = ω∗1

2 e1 + ω∗32 (− sin τe2 + cos τn),

which leads toω∗3

2 = ω32.

The Gauss curvature K∗ of S∗ is determined by

dω∗21 = ω∗3

1 ∧ ω∗23 = −K∗ω∗1 ∧ ω∗2. (4.51)

Denote ω∗32 = ω3

2 = bω1 + cω2, then

ω∗31 ∧ ω∗2

3 =sin τ

lω2 ∧ (−bω1) = b

sin τ

lω1 ∧ ω2. (4.52)

On the other hand, ω3a ∧ ωa = 0 implies that ω3

1 has the form

ω31 = aω1 + bω2.

Henceω1 ∧ ω2 =

1bω1 ∧ ω3

1 =1b

sin τ

lω∗1 ∧ ω∗2. (4.53)

Substituting (4.52) and (4.53) into (4.51), we get K∗ = −sin2 τ

l2. Simi-

larly, K = −sin2 τ

l2holds. The theorem is proved.

4.2.3 Backlund transformationTheorem 4.1 implies that a solution α (α = 0) of the sine-Gordon

equation leads to a surface of constant negative Gauss curvature. TheBacklund theorem implies that two focal surfaces of a pseudo-spherical¨congruence have the same constant negative Gauss curvature, and thesetwo focal surfaces correspond to two solutions of the sine-Gordon equa-tion. However, the above proof of the Backlund theorem does not implythe existence of the Backlund congruences. We are going to constructthe Backlund congruence from a given surface S of constant negativeGauss curvature with K = −1.

Suppose (e1, e2,n) is the Chebyshev frame of S at r, and (u, v) arethe Chebyshev coordinates. Let

r ∗ = r + le = r + l(cos θe1 + sin θe2) (4.54)

be the transformation from the surface S to the surface S∗. Let P andP ∗ be the corresponding points on S and S∗ with position vectors r and


Figure 4.1.

r ∗ respectively. It is required that the lines PP ∗ form a pseudo-sphericalcongruence. Here e = cos θe1 + sin θe2, or θ, is to be determined.

Suppose S corresponds to a solution α of the sine-Gordon equation.Differentiating (4.54) and using (4.33), we have

dr ∗ = dr + l(cos θde1 + sin θde2) + l(− sin θe1 + cos θe2)dθ

=[cos

α

2du − l sin θdθ − l sin θ

(αv

2du +

αu

2dv)]

e1

+[sin

α

2dv + l cos θdθ + l cos θ

(αv

2du +

αu

2dv)]

e2

+[l sin

α

2cos θ du − l cos

α

2sin θ dv

]n.

(4.55)

Since the angle τ between the normal vector n ∗ of S∗ at P ∗ and thenormal vector n of S at P is a constant, and PP ∗ is tangent to S∗ atP ∗, we have

n ∗ = sin τ sin θe1 − sin τ cos θe2 + cos τn. (4.56)

n ∗ is the normal vector of S∗, hence it must satisfy dr ∗ · n ∗ = 0. Thus

l sin τdθ − sin τ(

cosα

2sin θ du − sin

α

2cos θ dv

)+l sin τ

(αv

2du +

αu

2dv)

−l cos τ(

sinα

2cos θ du − cos

α

2sin θ dv

)= 0.

(4.57)

If (4.57) can be solved for θ, then PP ∗ generates exactly a pseudo-spherical congruence. From the Backlund theorem, the surfaces¨ S and

S∗ have the same K = K∗ = −sin2 τ

l2. Without loss of generality,

suppose l = sin τ . Let θ = α1/2, then dθ =12

(∂α1

∂udu+

∂α1

∂vdv). (4.57)


can be written as a system of partial differential equations:

12

sin τ(∂α1

∂u+

∂α

∂v

)= sin τ sin

α1

2cos

α

2+ cos τ cos

α1

2sin

α

212

sin τ(∂α1

∂v+

∂α

∂u

)= − sin τ cos

α1

2sin

α

2− cos τ sin

α1

2cos

α

2.

(4.58)

This system is the original form of the Backlund transformation ofsine-Gordon equation. Regarded as a system of α1, its integrabilitycondition is the sine-Gordon equation

αuu − αvv = sinα.

Since it holds already, the solution α1 of (4.58) exists uniquely for agiven value α10 (0 < α10 < π) of α1 at (u0, v0). From αuv = αvu, wesee that α1 also satisfies the sine-Gordon equation. This transformationfrom α to α1 is called a Backlund transformation between two solutionsöf the sine-Gordon equation. Thus we have obtained

Theorem 4.3 Let S be a surface of constant negative Gauss curvatureK = − sin2 τ/l2 in R3, l and sin τ be non-zero constants and e0 be aunit tangent vector of S at P ∈ S. Then there exists a surface S∗ ofK = − sin2 τ/l2 such that the common tangent lines of SS∗ constitute apseudo-spherical congruence with parameters (l, sin τ) and the line fromP in the direction e0 belongs to the congruence.

This theorem indicates the method of the construction of pseudo-spherical congruences.

In differential geometry, the surface S∗ is called a Backlund transfor-¨mation of the surface S. For S∗, (4.55) can be rewritten as

dr ∗ = A∗ due ∗1 + B∗ dve ∗

2 . (4.59)

We shall prove that the Chebyshev coordinates (u, v) of S are also theChebyshev coordinates of S∗. Since e ∗

1 and e ∗2 are unit vectors, (4.48)

and (4.55) lead to

A∗ = cosα1

2, B∗ = sin

α1

2,

e ∗1 = (cos

α

2cos

α1

2− cos τ sin

α

2sin

α1

2)e1

+(cosα

2sin

α1

2+ sin τ sin

α

2cos

α1

2)e2 + sin τ sin

α

2n,

e ∗2 = (sin

α

2cos

α1

2+ cos τ cos

α

2sin

α1

2)e1

+(sinα

2sin

α1

2− cos τ cos

α

2cos

α1

2)e2 − sin τ cos

α

2n.

(4.60)


Henceω∗

1 = cosα1

2du, ω∗

2 = sinα1

2dv. (4.61)

Computing dn ∗ from (4.56) and expanding it in terms of e ∗1 , e ∗

2 , weget

dn ∗ = ω∗13 e ∗

1 + ω∗23 e ∗

2 ,

ω∗13 = − sin

α1

2du, ω∗2

3 = cosα1

2dv.

(4.62)

Comparing with (4.33), we conclude that (u, v) are the Chebyshev co-ordinates of S∗. The difference of sign can be eliminated by using −n∗instead of n∗.

Theorem 4.4 Suppose S∗ is the Backlund transformation of a surface¨S of constant negative Gauss curvature, and (u, v) are the Chebyshevcoordinates of S, then (u, v) are also the Chebyshev coordinates of S∗.

Thus one can apply the Backlund transformation successively to ob-¨tain a sequence of pseudo-spherical congruences and surfaces of constantnegative Gauss curvature.

4.2.4 Darboux transformationIn the last subsection, we have got the following well-known facts.

Suppose α is a solution of the sine-Gordon equation and S is the corre-sponding surface of constant negative Gauss curvature. If α1 is a solutionof (4.58), then (4.54) gives an explicit expression of the Backlund trans-¨formation of S where θ = α1/2. In order to get the explicit expression ofS∗, we need to use Darboux transformation to get the explicit expressionof α1. Let

ξ =u + v

2, η =

u − v

2, (4.63)

then (4.58) becomes

(α1 + α)ξ = 2β sinα1 − α

2,

(α1 − α)η =2β

sinα1 + α

2,

(4.64)

whereβ =

1 − cos τ

sin τ= 0 . (4.65)

The sine-Gordon equation becomes

αξη = sin α. (4.66)


In Chapter 1, the Darboux transformation for the sine-Gordon equa-tion has already been introduced. The construction is as follows. Let αbe a solution of the sine-Gordon equation, then the Lax pair

Φη =

⎛⎝⎛⎛ λ −αη

2αη

2−λ

⎞⎠⎞⎞Φ,

Φξ =14λ

⎛⎝⎛⎛ cos α sinα

sinα − cos α

⎞⎠⎞⎞Φ

(4.67)

is completely integrable. It can be checked that if

⎛⎝⎛⎛ φ1(λ)

φ2(λ)

⎞⎠⎞⎞ is a column

solution of the Lax pair (4.67), then

⎛⎝⎛⎛ φ2(λ)

−φ1(λ)

⎞⎠⎞⎞ is a column solution

of the Lax pair with λ replaced by −λ. Hence

⎛⎝⎛⎛ φ2(−λ)

−φ1(−λ)

⎞⎠⎞⎞ is also a

column solution of the Lax pair (4.67). Φ(λ) can be chosen as

Φ(λ) =

⎛⎝⎛⎛ φ1(λ) φ2(−λ)

φ2(λ) −φ1(−λ)

⎞⎠⎞⎞ . (4.68)

Take λ = λ1 = 0. Let⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ be a column solution of the Lax pair

(4.67) with λ = λ1, and

h1 = φ1(λ1) + b1φ2(−λ1), h2 = φ2(λ1) − b1φ1(−λ1),

where b1 is a constant. Then

⎛⎝⎛⎛ −h2

h1

⎞⎠⎞⎞ is a solution of the Lax pair

(4.67) with λ = −λ1. Let

H =

⎛⎝⎛⎛ h1 −h2

h2 h1

⎞⎠⎞⎞ ,

then detH = h21 + h2

2, which is not zero provided that

⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ is not

a trivial solution. (From the property of linear ordinary differential


equations, h1 = h2 = 0 holds everywhere if it holds at one point.) Let

S = H

⎛⎝⎛⎛ λ1 0

0 −λ1

⎞⎠⎞⎞H−1 =λ1

h21 + h2

2

⎛⎝⎛⎛ h21 − h2

2 2h1h2

2h1h2 h22 − h2

1

⎞⎠⎞⎞

=λ1

1 + σ2

⎛⎝⎛⎛ 1 − σ2 2σ

2σ −1 + σ2

⎞⎠⎞⎞ = λ1

⎛⎜⎛⎛⎝⎜⎜ cosψ

2sin

ψ

2sin

ψ

2− cos

ψ

2

⎞⎟⎞⎞⎠⎟⎟ ,

(4.69)

whereσ =

h2

h1, ψ = 4 tan−1 σ, (4.70)

cosψ

2=

1 − σ2

1 + σ2, sin

ψ

2=

2σ

1 + σ2. (4.71)

The matrixD(λ) = λI − S (4.72)

is the Darboux matrix. Let

Φ1(λ) = D(λ)Φ(λ), (4.73)

then Φ1(λ) satisfies the Lax pair (4.67) when α is replaced by certainα1. This can be verified directly as follows. From

Φ1η = −SηSS Φ + (λI − S)Φη

=

(− SηSS + (λI − S)

⎛⎝⎛⎛ λ −αη/2

αη/2 −λ

⎞⎠⎞⎞)Φ(λ)

=

⎛⎝⎛⎛ λ −α1η/2

α1η/2 −λ

⎞⎠⎞⎞ (λI − S)Φ

and det Φ(λ) = 0, we obtain

−SηSS + (λI − S)

⎛⎝⎛⎛ λ −αη/2

αη/2 −λ

⎞⎠⎞⎞ =

⎛⎝⎛⎛ λ −α1η/2

α1η/2 −λ

⎞⎠⎞⎞ (λI − S).

Now compare the coefficients of the powers of λ in both sides. The termswith λ2 are equal identically. The terms with λ lead to

α1η − αη = −4λ1 sinψ

2. (4.74)

The term with λ0 leads to

α1η = −αη + ψη. (4.75)


Similarly, considering the part of the Lax pair with derivative to ξ, weget

α1 = ψ − α, (4.76)

ψξ =1λ1

sinα − α1

2. (4.77)

(4.76) gives the new solution of the sine-Gordon equation explicitly by

Darboux transformation. Let β = − 12λ1

, then (4.75) – (4.77) lead to

(4.64). Therefore, α1 is a solution of the partial differential equations(4.64) in the Backlund transformation. More precisely, we have

Theorem 4.5 Take λ1 = − 12β

, then α1 derived by the Darboux trans-

formation (Φ, α) → (Φ1, α1) is a solution of (4.64).

The explicit construction of the surface S∗ of constant negative Gausscurvature is as follows.

Theorem 4.6 Suppose α (0 < α < π) is a solution of the sine-Gordonequation, and Φ is the fundamental solution of its Lax pair. Let S bethe surface of constant negative Gauss curvature corresponding to α (ex-pressed in Chebyshev coordinates). Then the Darboux transformationgives a new solution α1 of the sine-Gordon equation, and the transfor-mation (4.54) gives the surface of constant negative Gauss curvaturecorresponding to α1.

According to this theorem, we can get a series of surfaces of constantnegative Gauss curvature by applying Darboux transformation (4.54) toa solution α of the sine-Gordon equation, the fundamental solution Φ ofits Lax pair and the corresponding surface S of constant negative Gausscurvature with Chebyshev coordinates. This can be illustrated by thefollowing diagram.

α

solution Φ of Lax pair −→ (α1, Φ1) −→ (α2, Φ2) −→↓ ↓

surface S −→ S∗ −→ S∗∗ −→· · ·

In the diagram, the arrows from α to Φ and from α to S are realized bysolving a system of linear integrable partial differential equations. Otherarrows can be realized by purely algebraic operations. Here the algebraicoperations include exponential, trigonometric and inverse trigonometricfunctions. In the construction of Backlund transformation, we only needto express cos

α1

2and sin

α1

2in terms of σ, cos

α

2and sin

α

2. These

expressions are purely algebraic.


4.2.5 ExampleStarting from α = 0, a series of non-trivial solutions of the sine-

Gordon equation can be obtained by using Darboux transformation.With α = 0, the Lax pair is

Φη =

⎛⎝⎛⎛ λ 0

0 −λ

⎞⎠⎞⎞Φ, Φξ =14λ

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞Φ,

hence the fundamental solution is

Φ =

⎛⎝⎛⎛ eλη+ ξ4λ 0

0 e−λη− ξ4λ

⎞⎠⎞⎞ .

Take λ = λ1. From (4.70) we have

σ = −be−2λ1η− ξ2λ1 .

For simplicity, take b = −1, then

α1 = 4 tan−1 σ = 4 tan−1(e−2λ1η− ξ2λ1 ).

The Darboux matrix is

D(λ) =

⎛⎝⎛⎛ λ + λ1 tanh γ −λ1 sech γ

−λ1 sech γ λ − λ1 tanh γ

⎞⎠⎞⎞ ,

whereγ = −2λ1η − ξ

2λ1.

Then

Φ1(λ) = D(λ)Φ(λ)

=

⎛⎝⎛⎛ (λ + λ1 tanh γ)eλη+ ξ4λ −λ1 sech γe−λη− ξ

4λ

−λ1 sech γeλη+ ξ4λ (λ − λ1 tanh γ)e−λη− ξ

4λ

⎞⎠⎞⎞ .

When λ = λ2,

φ′1(λ2) = (λ2 + λ1 tanh γ)eλ2η+ ξ

4λ2 ,

φ′2(λ2) = (−λ1 sech γ)eλ2η+ ξ

4λ2 ,

σ′ =(−λ1 sech γ)eλ2η+ ξ

4λ2 + b(λ2 − λ1 tanh γ)e−λ2η− ξ4λ2

(λ2 + λ1 tanh γ)eλ2η+ ξ4λ2 + b(−λ1 sech γ)e−λ2η− ξ

4λ2


and the new solution

α2 = 4 tan−1 σ′ − α1.

On the other hand, by the fundamental equations of the surface forα = 0,

dr = e1 du, de1 = 0, de2 = −ndv, dn = e2 dv,

we haver = (u, 0, 0), e1 = (1, 0, 0),

e2 = (0, cos v, sin v), n = (0, sin v,− cos v).

It is not a surface, but a line together with a two-parametric familyof orthonormal frames along the line. However, since the fundamen-tal equations hold, we can still use it to construct surface of constantnegative Gauss curvature. Now

θ =α1

2= 2 tan−1 σ,

cos θ = − tanh γ, sin θ = sech γ.

According to (4.54), the equation for the surface S∗ is

r ∗ = r + l(

cosα1

2e1 + sin

α1

2e2

),

orx1 = u − l tanh γ,

x2 = l sech γ cos v,

x3 = l sech γ sin v,(l =

−4λ1

1 + λ21

, γ = −2λ1η − ξ

2λ1

).

In order to construct the second surface S∗∗, we first write down theexpressions of e ∗

1 and e ∗2 . From (4.60) and α = 0,

e ∗1 = cos

α1

2e1 + sin

α1

2e2,

e ∗2 = cos τ

(sin

α1

2e1 − cos

α1

2e2

)− sin τn.

Hence the equation for the second surface is

r ∗∗ = ue1 − 2λ1

1 + λ21

(cos

α1

2e1 + sin

α1

2e2

)− 2λ2

1 + λ22

(cos

α2

2e ∗

1 + sinα2

2e ∗

2

).

These surfaces of constant negative Gauss curvature can be plottedby computer. In [85], there are figures for some interesting surfacesincluding several surfaces of constant negative Gauss curvature.


4.3 Surface of constant Gauss curvature in theMinkowski space R2,1 and pseudo-sphericalcongruence

4.3.1 Theory of surfaces in the Minkowski spaceR2,1

As the Euclidean space R3, the Minkowski space R2,1 is a three di-mensional flat space. A vector l in R2,1 has three coordinates l1, l2and l3. In an orthonormal coordinate system, the inner product of twovectors l = (l1, l2, l3) and m = (m1, m2, m3) in R2,1 is given by

l · m = l1m1 + l2m2 − l3m3, (4.78)

and the square of the norm of a vector l is

l2 = l21 + l22 − l23, (4.79)

which is not positive definite. According to the sign of l2, there are threetypes of non-zero vectors:

l2 > 0 space-like,

l2 < 0 time-like,

l2 = 0 light-like.

(4.80)

A light-like vector is also called a null vector. All the null vectors forma light cone l21 + l22 − l23 = 0. A space-like vector points to the exteriorof the light cone, whereas a time-like vector points to the interior of thelight cone. Moreover, according the sign of l3, time-like and light-likevectors are divided into past-oriented and future-oriented vectors. Allthe vectors orthogonal to a time-like vector are space-like. A vectororthogonal to a space-like vector may be space-like, time-like or light-like. The vectors orthogonal to a light-like vector may be light-like orspace-like, and all these vectors form the tangent planes of the light cone.A light-like vector is always orthogonal to itself.

A point (x, y, z) in R2,1 can be expressed by a position vector r. In thespecial theory of relativity, the space and time are expressed uniformly bya four dimensional Minkowski space-time. The study of the geometry ofR2,1 may help understanding the space-time in the relativity. The termsspace-like, time-like, light-like etc. also originate from the relativity.

For a surface in R2,1, normal vector n can also be defined as the unitvector orthogonal to the tangent plane. When n is time-like, all thevectors on the tangent plane are space-like, and the surface is calledspace-like. When n is space-like, the vectors on the tangent plane may


be space-like, time-like or light-like, and the surface is called time-like.There is also light-like surface, whose normal vector is light-like. In thiscase, the normal vector locates in the tangent plane. A mixed surfaceis a connected surface which includes space-like, time-like and light-likeparts. In this book, we only consider the space-like surfaces and time-likesurfaces.

Similar to R3, the fundamental equation of a space-like or time-likesurface can be written as

dr = ωaea,

dea = ωbaeb + ω3

an,

dn = ωa3ea.

(4.81)

The integrability conditions are

dωa + ωab ∧ ωb = 0, ω3

a ∧ ωa = 0,

dωab + ωa

c ∧ ωca = −ωa

3 ∧ ω3b (Gauss equation),

dω3a + ω3

b ∧ ωba = 0 (Codazzi equation).

(4.82)

Space-like and time-like surfaces should be considered separately.

(1) space-like surfaceFor the orthogonal frame e1, e2,n at r,

ea · eb = δab, ea · n = 0, n2 = −1, (4.83)

andω1

1 = ω22 = 0, ω1

2 = −ω12, ω3

a = ωa3 . (4.84)

The first fundamental form of the surface is

I = dr · dr = (ω1)2 + (ω2)2 = gabωaωb, (4.85)

whereg11 = g22 = 1, g12 = g21 = 0.

Letω3

a = bacωc (bac = bca), (4.86)

then the second fundamental form of the surface is

II = −dr · dn = −ω3aω

a = −bacωaωc. (4.87)

The Gauss equation is

dωab = 1

2Rabcdω

c ∧ ωd = −ωa3 ∧ ω3

b

= −12(bacbbd − badbbc)ωc ∧ ωd.


As before, it contains only one independent equality

R1212 = −(b11b22 − b212) (R1212 = R1

212). (4.88)

The intrinsic definition of Gauss curvature is

K =R1212

g11g22 − g212

= R1212. (4.89)

Hence the Gauss equation can be written as

K = −(b11b22 − b212), (4.90)

whiledω1

2 = −ω13 ∧ ω3

2 = R1212ω1 ∧ ω2 = Kω1 ∧ ω2. (4.91)

(4.90) differs from the corresponding formula in Euclidean space bya sign. The principal curvatures of the surface are still the eigenvaluesof the second fundamental form with respect to the first fundamentalform.

(2) Time-like surfaceTake the orthogonal frame e1, e2,n at r, so that e1 is space-like, e2

is time-like, ande 2

1 = 1, e 22 = −1, n 2 = 1. (4.92)

Thenω1

1 = ω22 = 0, ω1

2 = ω21,

ω13 = −ω3

1, ω32 = ω3

2.(4.93)

The first fundamental form

ds2 = dr · dr = (ω1)2 − (ω2)2 = gabωaωb (4.94)

is not positive definite. Here g11 = 1, g22 = −1, g12 = 0. Write

ω3a = bacω

c (bac = bca), (4.95)

then the second fundamental form is

II = −dr · dn = −ω1ω13 + ω2ω2

3

= ω1ω31 + ω2ω3

2 = bacωaωc.

(4.96)

The intrinsic definition of Gauss curvature is

K =R1212

g11g22 − g212

= −R1212, (4.97)


The Gauss equation is

dω12 = R1

212ω1 ∧ ω2 = R1212ω

1 ∧ ω2 = −ω13 ∧ ω3

2

= (b11b22 − b212)ω

1 ∧ ω2 = −Kω1 ∧ ω2.(4.98)

The right hand side of the above formula differs from (4.91) by a sign.Note that if the frame satisfies e2

1 = −1, e22 = 1, then (4.98) will be

changed to

dω12 = R1

212ω1 ∧ ω2 = −R1212ω

1 ∧ ω2 = −ω13 ∧ ω3

2

= −(b11b22 − b212)ω

1 ∧ ω2 = Kω1 ∧ ω2.(4.98′)

Comparing the formulae for the surfaces in Euclidean space with thespace-like or time-like surfaces in Minkowski space, there are differenceson sign (especially for the expressions of K).

4.3.2 Chebyshev coordinates for surfaces ofconstant Gauss curvature

The surfaces of constant Gauss curvature in R2,1 may be space-like,time-like or light-like, also may have positive or negative curvature (K =±1). Moreover, for time-like surface of constant positive Gauss curva-ture, the eigenvalues of the second fundamental form with respect to thefirst fundamental form (i.e., the principal curvatures) may be real, imag-inary, or be repeated eigenvalues with only one dimensional eigenspace.Therefore, we need to consider these cases separately. In each case,appropriate frame and coordinates are chosen and the surfaces are con-structed with the help of the solutions of the corresponding integrableequations. First of all, we write down the fundamental equations andthe Gauss equations under Chebyshev coordinates in each case.

(1) space-like surface of constant positive Gauss curvature(K = +1)

e1, e2 and n satisfy (4.83), and e1, e2 are tangent to the lines ofcurvature. Take

ω1 = cosα

2du, ω2 = sin

α

2dv. (4.99)

Similar to the surface of constant negative Gauss curvature in the Eu-clidean space, we still have

ω21 = −ω1

2 =12(αv du + αu dv). (4.100)

Takeω3

1 = sinα

2du, ω3

2 = − cosα

2dv, (4.101)


then it can be verified easily that the Codazzi equation

dω31 + ω3

2 ∧ ω21 = 0, dω3

2 + ω31 ∧ ω1

2 = 0 (4.102)

holds, andb11 = tan

α

2, b22 = − cot

α

2, b12 = 0. (4.103)

HenceK = −b11b22 = 1. (4.104)

The first fundamental form is

ds2 = (ω1)2 + (ω2)2 = cos2α

2du2 + sin2 α

2dv2, (4.105)


II = cosα

2sin

α

2(du2 − dv2). (4.106)

The Gauss equation becomes the negative sine-Gordon equation

αuu − αvv = − sinα. (4.107)

By exchanging the parameters u and v, it is the same as the usual sine-Gordon equation.

(2) space-like surface of constant negative Gauss curvature(K = −1)

Under the assumption that there are no umbilic points, e1, e2 and nsatisfy (4.83), and e1, e2 are tangent to the lines of curvature. Take

ω1 = coshα

2du, ω2 = sinh

α

2dv, (4.108)

thenω1

2 = −ω21 =

12(αv du − αu dv). (4.109)

Takeω3

1 = sinhα

2du, ω3

2 = coshα

2dv, (4.110)

thenb11 = tanh

α

2, b22 = coth

α

2, b12 = 0. (4.111)

HenceK = −b11b22 = −1. (4.112)

The first fundamental form is

ds2 = (ω1)2 + (ω2)2 = cosh2 α

2du2 + sinh2 α

2dv2, (4.113)



II = coshα

2sinh

α

2(du2 + dv2). (4.114)

The Gauss-Codazzi equations become

α = sinhα. (4.115)

(3) Time-like surface of constant positive Gauss curvature(K = +1)

Under the assumption that there are no umbilic points, this case isdivided into the following three subcases.

(3a) Principal curvatures are realLet e1, e2 and n satisfy (4.92). Take

ω1 = coshα

2du, ω2 = sinh

α

2dv, (4.116)

thenω1

2 = ω21 =

12(αv du + αu dv). (4.117)

Letω3

1 = sinhα

2du, ω3

2 = − coshα

2dv, (4.118)

thenb11 = tanh

α

2, b22 = − coth

α

2, (4.119)

K = −b11b22 = +1. (4.120)

The Codazzi equation holds. The first fundamental form is

ds2 = (ω1)2 − (ω2)2 = cosh2 α

2du2 − sinh2 α

2dv2, (4.121)


II = coshα

2sinh

α

2(du2 − dv2). (4.122)

The Gauss equation becomes

αuu − αvv = − sinhα. (4.123)

(3b) Principal curvatures are imaginaryTake the frame so that

e21 = 0, e2

2 = 0, n · ea = 0 (a = 1, 2),


and

e1 · e2 =12eα > 0.

From the structure equations (4.81), we get

ω12 = ω2

1 = 0,

ω31 +

12eαω2

3 = 0, ω32 +

12eαω1

3 = 0.

Choose coordinates (u, v) [51] so that

ω1 = du − e−αdv, ω2 = −e−αdu − dv,

and

ω13 =

12(du − eαdv), ω3

2 =12(eαdu + dv).

Then

ω31 =

12(du − eαdv), ω3

2 =12(eαdu + dv).

d2r = 0 leads toω1

1 = αudu, ω22 = αvdv.

Hence the first fundamental form and the second fundamental form are

I = dr2 = −du2 − 2 sinhα du dv + dv2,

II = −dr · dn = −2 cosh α du dv.

Therefore, K = 1 and two principal curvatures are imaginary. From theGauss equation dωa

b + ωac ∧ ωc

b = −ωa3 ∧ ω3

b , we get

αuv = cosh α.

All the Codazzi equations are also the consequences of the cosh-Gordonequation. Therefore, from a solution of the cosh-Gordon, a time-likesurface with Gauss curvature 1 and two imaginary principal curvaturescan be obtained by solving the fundamental equations of the surface. Theparameters (u, v) are called asymptotic Chebyshev coordinates, since uand v are asymptotic lines.

(3c) Repeated principal curvature with only one principal directionTake e1, e2 satisfying

e21 = 0, e2

2 = 0, e1 · e2 =12eα.


Take the parameters (u, v) so that

ω1 = du − e−αdv, ω2 = −dv,

ω13 = −du − e−αdv, ω2

3 = dv.

The first and second fundamental forms are

I = −eα du dv + dv2,

II = −eα du dv.

The Gauss-Codazzi equations become the Liouville equation

αuv =12eα.

From any solution of this equation, one can get a surface of this kind bysolving the fundamental equations of surface.

(4) Time-like surface of constant negative Gauss curvature(K = −1)

Let e1, e2 and n satisfy (4.92),

ω1 = cosα

2du, ω2 = sin

α

2dv, (4.124)

thenω1

2 = ω21 =

12(−αv du + αu dv). (4.125)

Takeω3

1 = sinα

2du, ω3

2 = cosα

2dv, (4.126)

then the first fundamental form is

ds2 = (ω1)2 − (ω2)2 = cos2α

2du2 − sin2 α

2dv2, (4.127)


II = sinα

2cos

α

2(du2 + dv2), (4.128)

andb11 = tan

α

2, b22 = cot

α

2.

The Gauss curvature K = −1. The Gauss-Codazzi equation

dω12 = −ω1

3 ∧ ω32

becomes α = sin α. (4.129)


In summary, we have the following theorem.

Theorem 4.7 The Gauss-Codazzi equations of various kinds of surfaces(space-like or time-like) of constant Gauss curvature (K(( = ±1) can besine-Gordon equation, sinh-Gordon equation, cosh-Gordon equation orLiouville equation in a suitable parametrization. The construction ofthese surfaces is reduced to solve these equations and to integrate thefundamental equations of surfaces.

This result can be listed as follows:

First fundamental form: cos2 α2du2 + sin2 α

2dv2

K = 1 Second fundamental form: cos α2

sin α2(du2 − dv2)

Space- Gauss equation: αuu − αvv = − sin α

like First fundamental form: cosh2 α2du2 + sinh2 α

2dv2

K = −1 Second fundamental form: cosh α2

sinh α2(du2 + dv2)

Gauss equation: α = sinh α

(a) First fundamental form: cosh2 α2du2 − sinh2 α

2dv2

K = 1 Second fundamental form: cosh α2

sinh α2(du2 − dv2)

Gauss equation: αuu − αvv = − sinh α

(b) First fundamental form: −du2 − 2 sinh αdu dv + dv2

Second fundamental form: −2 cosh α du dv

Time- Gauss equation: αuv = cosh α

like (c) First fundamental form: −eα dudv + dv2

Second fundamental form: −e−α du dv

Gauss equation: αuv = 12eα

First fundamental form: cos2 α2

du2 − sin2 α2dv2

K = −1 Second fundamental form: cos α2

sin α2(du2 + dv2)

Gauss equation: α = sin α

There are still two problems left. (1) How to get explicit solutions ofthese partial differential equations? (2) How to get the explicit expres-sions of these surfaces. In order to answer these problems, we need toconsider the Backlund transformation and pseudo-spherical congruencesin R2,1 and use the Darboux transformation.

4.3.3 Pseudo-spherical congruence in R2,1

In R2,1, a line congruence may be space-like, time-like, light-like ormixed. Here we want to study time-like line congruences (i.e., all thelines in it are time-like) and space-like line congruences (i.e., all the


lines in it are space-like). Under the assumption that they have twofocal surfaces, these line congruences can be classified into followingfour types.

(a) The line congruence is space-like, and two focal surfaces are space-like;

(b) The line congruence is space-like, and two focal surfaces are time-like;

(c) The line congruence is time-like, and two focal surfaces are alsotime-like;

(d) The line congruence is space-like, and one focal surface is space-like, another focal surface is time-like.

Let S and S∗ be two focal surfaces of a line congruence Σ, PP ∗ bethe lines in Σ which are common tangent lines of S and S∗, and P ∈ S,P ∗ ∈ S∗ are the tangent points. Let n and n ∗ be the normal vectors ofS and S∗ at P and P ∗ respectively. If

|PP ∗| = l, n · n ∗ = k (4.130)

are non-zero constants, then Σ is called a pseudo-spherical congruence.In [66, 84], the Backlund transformation for various pseudo-spherical¨congruences has been discussed. However, the discussion there is to becompleted and the method of explicit construction is insufficient.

Here we shall consider all the possible cases and use the Darbouxtransformation to get explicit expressions of the surfaces and congru-ences.

Theorem 4.8 (Generalized Backlund theorem) Two focal surfaces of apseudo-spherical congruence in R2,1 are surfaces of the same constantGauss curvature K. In cases (a), (b) and (c), K is positive. In case (d),K is negative.

Proof. The theorem should be proved for all four cases.Case (a): The congruence is space-like, and two focal surfaces S and

S∗ are space-likeTake the orthogonal frames e1, e2,n and e ∗

1 , e ∗2 ,n ∗ of S at r and

of S∗ at r∗ respectively so that e1 = −e ∗1 is the unit vector parallel to

PP ∗. Here r and r∗ are the position vectors of P and P ∗ respectively.Then

e 21 = e 2

2 = e ∗21 = e ∗2

2 = 1, n 2 = n ∗2 = −1,

ande ∗

2 = (cosh τ)e2 + (sinh τ)n,

n ∗ = (sinh τ)e2 + (cosh τ)n.(4.131)


Here τ = constant = 0. Moreover,r ∗ = r + le1 (l = constant = 0) . (4.132)

Differentiating this equation and comparing it with dr ∗ = ω∗1e ∗1 +ω∗2e ∗

2 ,we get

ω1 = −ω∗1,

ω2 + lω21 = (cosh τ)ω∗2,

lω31 = (sinh τ)ω∗2.

(4.133)

Hence(cosh τ)ω3

1 − (sinh τ)ω21 =

sinh τ

lω2. (4.134)

From

ω∗31 = (−de ∗

1 ) · n ∗ = −ω2

lsinh τ,

ω∗32 = (−de ∗

2 ) · n ∗ = ω32,

(4.135)

we have

−ω∗13 ∧ ω∗3

2 = ω32 ∧ (−ω2

lsinh τ) =

sinh τ

lω3

1 ∧ ω1

=sinh2 τ

l2ω∗1 ∧ ω∗2,

(4.136)

Hence the Gauss curvatures of S and S∗ are the same constant

K∗ = K =sinh2 τ

l2.

Case (b): The congruence is space-like, and two focal surfaces S andS∗ are time-like

Now

e 21 = e ∗2

1 = 1, e 22 = e ∗2

2 = −1, n 2 = n ∗2 = 1, (4.137)

and

e ∗1 = −e1,

e ∗2 = cosh τe2 + sinh τn,

n ∗ = sinh τe2 + cosh τn, (n · n ∗ = cosh τ = constant).

(4.138)

From r ∗ = r + lr1 (l = constant),

ω1 = −ω∗1,

ω2 + lω21 = cosh τω∗2,

lω31 = sinh τω∗2.

(4.139)


(4.138) and (4.139) are the same as (4.131) and (4.133). Moreover,

ω∗31 = (de ∗

1 ) · n ∗ = −ω2

lsinh τ, ω∗3

2 = (de ∗2 ) · n = ω3

2. (4.140)

Hence

−ω∗13 ∧ ω∗3

2 =sinh τ

lω3

2 ∧ ω2 = −sinh τ

lω3

1 ∧ ω1

= −sinh2 τ

l2ω∗1 ∧ ω∗2.

(4.141)

This means that S and S∗ are surfaces of the same constant positiveGauss curvature

K∗ = K =sinh2 τ

l2. (4.142)

Case (c): The congruence is time-like, and two focal surfaces S andS∗ are time-like

Now

e 21 = e ∗2

1 = −1, e 22 = e ∗2

2 = 1, n 2 = n ∗2 = 1, (4.143)

ande ∗

1 = −e1,

e ∗2 = (cos τ)e2 + (sin τ)n,

n ∗ = −(sin τ)e2 + (cos τ)n

(n · n ∗ = cos τ = constant = 0) .

(4.144)

Differentiating r ∗ = r + le1, we get

ω1 = −ω∗1,

ω2 + lω21 = (cos τ)ω∗2,

lω31 = (sin τ)ω∗2.

(4.145)

(4.144) implies

ω∗31 = −ω2

lsin τ,

ω∗32 = ω3

2,(4.146)

which leads to

−ω∗13 ∧ ω∗3

2 = ω32 ∧ (−ω2

lsin τ) =

sin τ

lω3

1 ∧ ω1

=sin2 τ

l2ω∗1 ∧ ω∗2.

(4.147)


(4.98)′ implies that

K∗ =sin2 τ

l2, K =

sin2 τ

l2. (4.148)

Hence S and S∗ are surfaces of the same constant positive Gauss curva-ture.

Case (d): The congruence is space-like, and one focal surface S istime-like, another focal surface S∗ is space-like

Now

e 21 = e ∗2

1 = 1, e 22 = −1, e ∗2

2 = 1, n 2 = 1, n ∗2 = −1, (4.149)

ande ∗

1 = −e1,

e ∗2 = (sinh τ)e2 + (cosh τ)n,

n ∗ = (cosh τ)e2 + (sinh τ)n

(n · n ∗ = sinh τ = constant).

(4.150)

Differentiating r ∗ = r + le1 (l = constant), we get

ω1 = −ω∗1,

ω2 + lω21 = (sinh τ)ω∗2,

lω31 = (cosh τ)ω∗2.

(4.151)

Moreover,

ω∗31 =

cosh τ

lω2, ω∗3

2 = ω32 (4.152)

leads to

−ω∗13 ∧ ω∗3

2 =cosh τ

lω3

2 ∧ ω2 = −cosh τ

lω3

1 ∧ ω1

= −cosh2 τ

l2ω∗1 ∧ ω∗2.

(4.153)

Hence, from (4.91),

K∗ = −cosh2 τ

l2. (4.154)

On the other hand,

−ω13 ∧ ω3

2 = −ω∗32 ∧ cosh τ

lω∗2 =

cosh τ

lω∗3

1 ∧ ω∗1

=cosh2 τ

lω1 ∧ ω2.

(4.155)


Hence, from (4.98),

K = −cosh2 τ

l2. (4.156)

The theorem is proved.This theorem does not confirm that all four cases can occur, because

it does not contain the proof of existence. This will be completed in thenext subsection.

4.3.4 Backlund transformation and Darbouxtransformation for surfaces of constantGauss curvature in R2,1

Now we shall give the method for constructing new surfaces of con-stant Gauss curvature in R2,1 from a known one by using pseudo-sphericalcongruence [61, 63]. This also gives the proof of existence of variouspseudo-spherical congruences. It is the generalization of the classicalBacklund transformation. Using Darboux transformation, this analyt-ïcal process can be realized algebraically. The method is similar withthat in Section 4.2. However, in each case the method has some specialfeature.

(1) S is a space-like surface of constant positive Gauss cur-vature, the congruence is space-like

Let e1, e2,n be the Chebyshev frames at r. If there exists a Backlund¨transformation such that its another focal surface S∗ is space-like, then

r ∗ = r + l(cos θe1 + sin θe2), (l = constant = 0) . (4.157)

The normal vector n ∗ of S∗ is time-like and orthogonal to−→

PP ∗. Hence

n ∗ = sinh τ(− sin θe1 + cos θe2) + cosh τn, (τ = constant = 0) .(4.158)

Differentiating r ∗, using the fundamental equations of surface, and usingthe equations (4.99) – (4.101) on space-like surfaces of constant positiveGauss curvature, we get

dr ∗ =[

cosα

2− l sin θ

(θu +

αv

2

)]e1 + l cos θ

(θu +

αv

2

)e2

+l cos θ sinα

2n

du +− l sin θ

(θv +

αu

2

)e1

+[sin

α

2+ l cos θ

(θv +

αu

2

)]e2 − l sin θ cos

α

2n

dv.

(4.159)


Let θ = α1/2 and suppose K = 1, then l = sinh τ . The conditiondr ∗ · n ∗ = 0 implies

l(α′u + αv) = 2 sin

α1

2cos

α

2+ 2 cosh τ cos

α1

2sin

α

2,

l(α′v + αu) = −2 cos

α1

2sin

α

2− 2 cosh τ sin

α1

2cos

α

2.

(4.160)

(4.160) is a system of partial differential equations for α1. Its integra-bility condition is exactly (4.107)

αuu − αvv = − sinα. (4.161)

Therefore, (4.160) is solvable. This gives the following theorem.

Theorem 4.9 For a given space-like surface S of constant positive Gausscurvature, there exists a space-like pseudo-spherical congruence whose fo-cal surfaces are S and another space-like surface S∗ of the same constantGauss curvature.

Using the method in Section 4.2, we can prove that (u, v) are also theChebyshev coordinates of S∗. Using the Darboux transformation forthe sine-Gordon equation, a series of surfaces of constant positive Gausscurvature can be obtained. If a solution α of the sine-Gordon equationcorresponding to the given surface of constant positive Gauss curvaturetogether with a fundamental solution of its Lax pair are known, then theconstruction is purely algebraic. Note that in this case u and v should be

interchanged (i.e. ξ =u − v

2, η =

u + v

2) so that (4.161) can be changed

to the standard sine-Gordon equation (4.66).

(2) S is a time-like surface of constant positive Gauss curva-ture

There are following three subcases.(2a) Two principal curvatures of S are real and distinctTake the Chebyshev coordinates, then

ω1 = coshα

2du, ω2 = sinh

α

2dv,

ω12 = ω2

1 =12(αv du + αu dv),

ω31 = sinh

α

2du, ω3

2 = − coshα

2dv,

(4.162)

ande2

1 = 1, e22 = −1, n2 = 1.


(2a1) Space-like Backlund transformation¨The equation of S∗ is

r ∗ = r + l(cosh θe1 + sinh θe2) (4.163)

and the normal vector is

n ∗ = sinh τ(sinh θe1 + cosh θe2) + cosh τn. (4.164)

Suppose K = 1, then the proof of Theorem 4.8 leads to l = sinh τ . Wehave

dr ∗ =[

coshα

2+ l sinh θ

(θu +

αv

2

)]e1 + l cosh θ

(θu +

αv

2

)e2

+l cosh θ sinhα

2n

du +l sinh θ

(θv +

αu

2

)e1

+[sinh

α

2+ l cosh θ

(θv +

αu

2

)]e2 − l sinh θ cosh

α

2n

dv.

(4.165)From dr ∗ · n ∗ = 0, we have

l(θu +

αv

2

)= sinh θ cosh

α

2+ cosh τ cosh θ sinh

α

2,

l(θv +

αu

2

)= − cosh θ sinh

α

2− cosh τ sinh θ cosh

α

2.

(4.166)

Regarded as a system of partial differential equations for θ = α1/2, itsintegrability condition is

αvv − αuu = sinh α. (4.167)

This equation holds because it is the Gauss equation of S. It is easy tosee that S∗ is also time-like. This proves the existence of the pseudo-spherical congruence. By Theorem 4.8, the curvature of S∗ is also +1.

Therefore, we have the following theorem.

Theorem 4.10 There exist space-like pseudo-spherical congruences withtwo time-like focal surfaces of constant positive Gauss curvature.

Similarly, we can prove that (u, v) are also the Chebyshev coordinatesof S∗, and can write down the explicit expressions of e ∗

1 , e ∗2 .

In order to construct S∗ explicitly in terms of the Darboux transforma-tion, we want to discuss the Darboux transformation for the sinh-Gordonequation. Let

ξ =u + v

2, η =

v − u

2, (4.168)

then (4.123) becomesαξη = sinh α. (4.169)


It has a Lax pair

Φξ =λ

2

⎛⎝⎛⎛ 0 e−α

eα 0

⎞⎠⎞⎞Φ, Φη =12

⎛⎜⎛⎛⎝⎜⎜ −αη1λ

1λ

αη

⎞⎟⎞⎞⎠⎟⎟Φ, (4.170)

that is, the integrability condition of (4.170) is just (4.169).

Let

⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ be a column solution of the Lax pair (4.170) with λ = λ1,

then

⎛⎝⎛⎛ h1

−h2

⎞⎠⎞⎞ is a column solution of (4.170) with λ = −λ1. Let

H =

⎛⎝⎛⎛ h1 h1

h2 −h2

⎞⎠⎞⎞ and suppose h1, h2 = 0, then the Darboux matrix is

D(λ) = I − λH

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜1λ1

0

0 − 1λ1

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟H−1 = I − λ

λ1

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 0h1

h2h2

h10

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ . (4.171)

It can be checked directly that (α1, Φ1) defined by

eα1 = e−α(h2

h1

)2, Φ1 = D(λ)Φ (4.172)

still satisfy the Lax pair (4.170). Hence α1 is a solution of (4.169), and(α,Φ) −→ (α1, Φ1) is the Darboux transformation for (4.169).

Now we prove that α1 obtained by (4.172) provides a solution θ =α1/2 of (4.166).

Let θ = α1/2. Adding and subtracting (4.168) and (4.166), we get

l(α1ξ

2+

αξ

2

)= (1 − cosh τ) sinh

α1 − α

2,

l(α1η

2− αη

2

)= −(1 + cosh τ) sinh

α1 + α

2.

(4.173)

Since

⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ satisfies the Lax pair (4.170),

(ln |h1|)ξ =λ1

2e−α h2

h1, (ln |h1|)η = −αη

2+

12λ1

h2

h1,

(ln |h2|)ξ =λ1

2eα h1

h2, (ln |h2|)η =

12λ1

h1

h2+

12αη.

(4.174)


Choose

⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ so thath2

h1is positive (if it is negative, the discussion is

similar), then (4.172) leads to

α1ξ + αξ = (2 ln |h2|)ξ − (2 ln |h1|)ξ

= λ1

(eα h1

h2− e−α h2

h1

)= −2λ1 sinh

α1 − α

2.

(4.175)

Likewise,

α1η + αη = 2αη +1λ1

(e−(α1+α)/2 − e(α+α1)/2),

or equivalently

α1η − αη = − 1λ1

sinhα1 + α

2, (4.176)

where λ1 =−1 + cosh τ

l. Since l = sinh τ , λ−1

1 =1 + cosh τ

sinh τ, (4.175)

and (4.176) become (4.173). Thus by using Darboux transformation wecan obtain the solution α1 of (4.176) explicitly. Besides, in realizingthe Backlund transformation (4.163), we only need to use cosh

α1

2and

sinhα1

2, hence the algorithm is purely algebraic.

Now we turn to examples. Starting from α = 0 and solving the Laxpair

Φξ =λ

2

⎛⎝⎛⎛ 0 1

1 0

⎞⎠⎞⎞Φ, Φη =12λ

⎛⎝⎛⎛ 0 1

1 0

⎞⎠⎞⎞Φ, (4.177)

we get

Φ(λ) =

⎛⎝⎛⎛ cosh γ sinh γ

sinh γ cosh γ

⎞⎠⎞⎞ ,

(γ =

λ

2ξ +

12λ

η

). (4.178)

Let

h1 = cosh γ1, h2 = sinh γ1,

(γ1 =

λ1

2ξ +

12λ1

η

). (4.179)

According to (4.171), the Darboux matrix is

D(λ) = I − λ

λ1

⎛⎝⎛⎛ 0 coth γ1

tanh γ1 0

⎞⎠⎞⎞ . (4.180)


Hence

Φ1(λ) = D(λ)Φ(λ)

=

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ cosh γ − λ

λ1coth γ1 sinh γ sinh γ − λ

λ1coth γ1 cosh γ

sinh γ − λ

λ1tanh γ1 cosh γ cosh γ − λ

λ1tanh γ1 cosh γ

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ .

(4.181)

Take λ2 > 0 (λ2 = λ1), γ2 =λ2

2ξ +

12λ2

η, and let

h1 = cosh γ2 − λ2

λ1coth γ1 sinh γ2,

h2 = sinh γ2 − λ2

λ1tanh γ1 cosh γ2,

(4.182)

then

eα2 = e−α1

( h2

h1

)2

gives a new solution α2. This process can be done successively to getα1, α2, α3, · · ·.

The surfaces are constructed as follows. Starting from α = 0 andsolving the fundamental equations of the surface, we obtain a family offrames along a line. In suitable coordinates, they are

r = (u, 0, 0), e1 = (1, 0, 0),

e2 = (0, sinh v, cosh v), n = (0,− cosh v,− sinh v).(4.183)

Then by Theorem 4.8 and formula (4.163), α1 gives a time-like surfaceof Gauss curvature +1. Using α2, α3, · · ·, we can construct a series ofsurfaces in the same way.

(2a2) Time-like Backlund transformation¨From the same S, the time-like Darboux transformation is

r ∗ = r + l(sinhα1

2e1 + cosh

α1

2e2),

n ∗ = sin τ(coshα1

2e1 + sinh

α1

2e2) + cos τn,

(4.184)

where l and τ are constants, sin τ = l = 0.Computing dr∗ and using the condition n∗ · dr∗ = 0, we get the

equations

l(α1u

2+

αv

2

)= − cosh

α1

2cosh

α

2+ cos τ sinh

α1

2sinh

α

2,

l((α1v

2+

αu

2

)= sinh

α1

2sinh

α

2+ cos τ cosh

α1

2cosh

α

2,

(4.185)


which is similar with (4.166). This is the Backlund transformation fromα to α1. Its integrability condition

αuu − αvv = − sinhα (4.186)

already holds. Hence the Backlund transformation from S to S∗ can berealized by solving this integrable system. If the expression of S and thesolution of the Lax pair corresponding to the sinh-Gordon equation areknown, then S∗ can be obtained by Darboux transformation in algebraicalgorithm.

(2b) Two principal curvatures of S are imaginaryWhen two principal curvatures of S are imaginary, the frame e1, e2,n,

parameters (u, v) and ω1, ω2, ωij (i, j = 1, 2, 3) have been chosen in the

case (3b) of Subsection 4.3.2.(2b1) Time-like Backlund transformation¨Suppose that the expression of S∗ is

r∗ = r + l(ae1 + be2),

where l is a non-zero constant and abeα = −1. The last condition meansthat PP ∗ is a time-like straight line and (r∗ − r)2 = −l2. If we wantthat

n∗ = (ae1 − be2) cos τ + n sin τ

is the normal vector of S∗ where sin τ = n · n∗ = constant, then byn∗ · dr∗ = 0, we get

r∗u = (1 − labu

b)e1 + (−e−α + lbu)e2 +

l

2(a − a−1)n,

r∗v = (−e−α + lav)e1 + (−1 − lbav

a)e2 +

l

2(b + b−1)n

(4.187)

where l = sin τ . From n∗ · dr∗ = 0,

2b−1bu = −µ(a − a−1), 2a−1av = − 1µ

(b + b−1), (4.188)

with µ = sec τ − tan τ .Let α1 be a function such that

a = expα1 − α

2, b = − exp

−α1 − α

2,

then

(α1 + α)u = 2µ sinhα1 − α

2, (α1 − α)v =

2µ

coshα1 − α

2. (4.189)


This is the Backlund transformation for the cosh-Gordon equation. Theintegrability condition for α1 is that α satisfies the cosh-Gordon equationwhich is the Gauss equation of S and hence is satisfied by α. Therefore,α1 exists and S∗ is obtained via α1. From the Backlund theorem weknow that S∗ is time-like and has constant positive Gauss curvatureK = 1. Moreover, by tedious calculation we can verify that (u, v) arethe asymptotic Chebyshev coordinates of S∗.

Now we want to get the expression of α1 by using Darboux transfor-mation.

Lemma 4.11 The Lax pair of cosh-Gordon equation is

Φu = UΦ =λ

2

⎛⎝⎛⎛ 0 e−α

eα 0

⎞⎠⎞⎞Φ, Φv = V Φ =12

⎛⎝⎛⎛ −αv λ−1

λ−1 αv

⎞⎠⎞⎞Φ.

(4.190)

Proof. The proof follows from direct calculations, showing that theintegrability condition of (4.190)

UvUU − VuVV + [U, V ] = 0

is equivalent to the cosh-Gordon equation.

Lemma 4.12 (i) If

⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ is a solution of the Lax pair for λ = λ0,

then

⎛⎝⎛⎛ −h1

h2

⎞⎠⎞⎞ is a solution of the Lax pair for λ = −λ0.

(ii) Suppose λ is purely imaginary and h2/h1 is purely imaginary atone point (u0, v0), then h2/h1 is purely imaginary in a neighborhood of(u0, v0).

Proof. (i) is obvious. Now we prove (ii).From the Lax pair (4.190),

(h2

h1

)u

=λ0

2

(eα + e−α

(h2

h1

)2)

,(h2

h1

)v

=1

2λ0+ αv

(h2

h1

)− 1

2λ0

(h2

h1

)2

.


Let A =h2

h1+

h2

h1, then

Au =λ0

2e−α

(h2

h1− h2

h1

)A,

Av = αvA +1

2λ0

(h2

h1− h2

h1

)A.

Hence if A = 0 at (u0, v0), then it holds in a neighborhood of (u0, v0).The lemma is proved.

By the general formulae for constructing Darboux matrix, we have

Φ1 =

⎛⎝⎛⎛I − λ

λ0H

⎛⎝⎛⎛ λ−10 0

0 −λ−10

⎞⎠⎞⎞H−1

⎞⎠⎞⎞Φ

=

⎛⎝⎛⎛I +λ

λ0

⎛⎝⎛⎛ 0 h1/h2

−h2/h1 0

⎞⎠⎞⎞⎞⎠⎞⎞Φ,

where

H =

⎛⎝⎛⎛ h1 −h1

h2 h2

⎞⎠⎞⎞ .

From

Φ1u =λ

2

⎛⎝⎛⎛ 0 −eα1

eα1 0

⎞⎠⎞⎞Φ1,

we get

eα1 = −(

h2

h1

)2

e−α.

This is the explicit expression of α1. The explicit expression of S∗ canbe obtained from r∗ = r + l(ae1 + be2) by using this α1. It can also beproved that α1 is a solution of the Backlund transformation.¨

Therefore, the explicit formula for the Backlund transformation of Sis obtained.

(2b2) Space-like Backlund transformation¨The expression of S∗ is still

r∗ = r + l(ae1 + be2),

but abeα = 1 now. The normal vector of S∗ is

n∗ = (ae1 − be2) sinh τ + n cosh τ.


Then

r∗u = (1 − labu

b)e1 + (−e−α + lbu)e2 +

l

2(a + a−1)n,

r∗v = (−e−α + lav)e1 + (−1 − lbav

a)e2 +

l

2(b − b−1)n.

(4.191)

Let l = sinh τ . n∗ · dr∗ = 0 leads to

2bu

b= µ(a + a−1),

2au

a= − 1

µ(b − b−1),

where µ = cosech τ − coth τ . Let

a = exp−α1 − α

2, b = exp

α1 − α

2,

then we get

(α1 − α)u = 2µ coshα1 + α

2, (α1 + α)v =

2µ

sinhα1 − α

2, (4.192)

which is the Backlund transformation for the cosh-Gordon equation. If¨ µis changed to µ−1, this is just (4.189). Using Darboux transformation, wecan get the explicit expression of α1, which gives the explicit Backlund¨transformation from S to S∗.

(2c) Two principal curvatures of S are equal and there is only oneprincipal direction

Choose the frame (e1, e2,n) and the parameters (u, v) as the case (3c)in Subsection 4.3.2, then

e21 = e2

2 = 0, e1 · e2 =12eα,

ω1 = du − e−αdv, ω2 = −dv,

ω13 = −du − e−αdv, ω2

3 = dv.

HenceI = −eα du dv + dv2, II = −eα du dv.

The Gauss-Codazzi equations become the Liouville equation

αuv =12eα.

Take the time-like Backlund transformation for¨ S as

r∗ = r + l(ae1 + be2), abeα = −1,


thenn∗ = (ae1 − be2) cos τ + n sin τ.

From n∗ · dr∗ = 0, we get

(α1 + α)u = −µ expα − α1

2, (α1 − α)v =

2µ

coshα1 + α

2.

Considering the Lax pair

Φu =λ

2

⎛⎝⎛⎛ 0 0

eα 0

⎞⎠⎞⎞Φ, Φv =12

⎛⎝⎛⎛ −αv λ−1

λ−1 αv

⎞⎠⎞⎞Φ (4.193)

of the Liouville equation, we can use Darboux transformation to get theexplicit expression of S∗.

The space-like Backlund transformation is given by

r∗ = r + l(ae1 + be2), abeα = 1,

and S∗ can be obtained by Darboux transformation in a similar way.In summary, we have the following theorem.

Theorem 4.13 Suppose S is a time-like surface of constant positiveGauss curvature without umbilic points, then there are time-like andspace-like Backlund congruences such that¨ S is one of its focal surface,and another focal surface is also a time-like surface of constant positiveGauss curvature.

(3) Space-like pseudo-spherical congruence with one space-like focal surface and one time-like focal surface

Suppose that S is a time-like surface of constant negative Gauss cur-vature −1. Choose the Chebyshev coordinates and the correspondingframe as (4) in Subsection 4.3.2, then (4.92) and (4.124) – (4.126) hold.Apply the Backlund transformation

r ∗ = r + l(cosh θe1 + sinh θe2), (l = cosh τ),

n ∗ = cosh τ(sinh θe1 + cosh θe2) + sinh τn.(4.194)

By the fundamental equations of S, the integrability condition, togetherwith (4.124) – (4.126), we have

dr ∗ =[

cosα

2+ l sinh θ

(θu − αv

2

)]e1 + l cosh θ

(θu − αv

2

)e2

+l cosh θ sinα

2n

du +l sinh θ(θv +

αu

2

)e1

+[sin

α

2+ l cosh θ

(θv +

αu

2

)]e2 + l sinh θ cos

α

2n

dv.

(4.195)


From n ∗ · dr ∗ = 0, we have

sinh θ(

cosα

2+ l sinh θ

(θu − αv

2

))−l cosh2 θ

(θu − αv

2) + sinh τ cosh θ sin

α

2= 0,

l sinh2 θ(θv +

αu

2

)− cosh θ

(sin

α

2+ l cosh θ

(θv +

αu

2

))+ sinh τ sinh θ cos

α

2= 0,

(4.196)

i.e.,

l(θu − αv

2

)= sinh θ cos

α

2+ sinh τ cosh θ sin

α

2,

l(θv +

αu

2

)= − cosh θ sin

α

2+ sinh τ sinh θ cos

α

2,

(4.197)

where l = cosh τ . The integrability condition for θ is

α = sin α, (4.198)

which is satisfied already. Hence for any given initial condition θ = θ0

at u = u0 and v = v0, the solution θ of (4.197) exists, and it satisfies

α1 = sinhα1,

where α1 = θ/2. This implies that the corresponding pseudo-sphericalcongruence exists, and r ∗ is a space-like surface of constant negativeGauss curvature.

Theorem 4.14 There exists a space-like pseudo-spherical congruencewith one time-like and one space-like focal surfaces of constant negativeGauss curvature.

We shall prove that (u, v) are also the Chebyshev coordinates of thespace-like surface of constant negative Gauss curvature S∗. In fact,according to (4.195),

dr ∗ = ω∗1e ∗1 + ω∗2e ∗

2 , (4.199)

ω∗1 = cosh θ, ω∗2 = sinh θ, (4.200)

where

e ∗1 =

1cosh θ

[cos

α

2+ l sinh θ

(θu − αv

2

)e1

+l sinh θ(θu − αv

2

)e2 + l sinh θ sin

α

2n,

e ∗2 =

1sinh θ

l sinh θ

(θv +

αu

2

)e1

+[sin

α

2+ l cosh θ

(θv +

αu

2

)]e2 + l sinh θ cos

α

2n.

(4.201)


From (4.197),e ∗2

1 = 1, e ∗22 = 1, e ∗

1 · e ∗2 = 0. (4.202)

Hencedn ∗ = ω∗3

1 e ∗1 + ω∗3

2 e ∗2 , (4.203)

andω∗3

1 = sinh θ du, ω∗32 = cosh θ dv. (4.204)

Let θ = α1/2, then (u, v) are the Chebyshev coordinates of S∗. More-over, α1 satisfies

α1 = sinhα1. (4.205)

This can also be derived from the differentiation of (4.197).We can do the same procedure conversely, i.e., starting from a space-

like surface of constant negative Gauss curvature S∗, the reverse proce-dure leads to a time-like surface of constant negative Gauss curvatureS.

The Backlund transformations described above are not transforma-tions between two solutions of the same sine-Gordon equation or thesame sine-Laplace equation. In fact, they transform a solution of thesine-Laplace equation to a solution of the sinh-Laplace equation, andvice versa (the differential form of these transformations has been givenin [91]). Applying the transformations twice, we can get a transforma-tion of solutions of the same equation. Darboux transformation will givean explicit form of these transformations.

Introduce the complex variables

ζ =u + iv

2, ζ =

u − iv2

, (4.206)

then∂

∂ζ=

∂

∂u− i

∂

∂v,

∂

∂ζ=

∂

∂u+ i

∂

∂v,

∂2

∂ζ∂ζ=

∂2

∂u2+

∂2

∂v2= .

(4.207)

The integrability condition of the Lax pair

Φζ =λ

2

⎛⎝⎛⎛ 0 −e−α

eα 0

⎞⎠⎞⎞Φ,

Φζ =12

⎛⎝⎛⎛ −αζ 1/λ

−1/λ αζ

⎞⎠⎞⎞Φ

(4.208)

is α = αζζ = sinh α. (4.209)


If α can be complex, this equation is called the complex sinh-Laplaceequation. If α is real, it is the usual sinh-Laplace equation. If α is purelyimaginary (α = iβ, β is real), it becomes the sine-Laplace equation

β = βζζ = sin β.

For the complex sinh-Laplace equation, the Darboux transformation(α,Φ) −→ (α1, Φ1) can be constructed similarly as the real sinh-Laplaceequation. However, the Lax pair (4.208) is slightly different from (4.170)for simplifying the calculations.

Take λ1 = 0. Let⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ be a column solution of the Lax pair for

λ = λ1, then

⎛⎝⎛⎛ −h1

h2

⎞⎠⎞⎞ is a column solution of the Lax pair for λ = −λ1.

Let

H =

⎛⎝⎛⎛ h1 −h1

h2 h2

⎞⎠⎞⎞ ,

then

S = H

⎛⎝⎛⎛ 1/λ1 0

0 −1/λ1

⎞⎠⎞⎞H−1 =1λ1

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 0h1

h2h2

h10

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ ,

D(λ) = I − λS =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 1 − λ

λ1

h1

h2

− λ

λ1

h2

h11

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ .

Hence the Darboux transformation is

Φ1(λ) = D(λ)Φ(λ).

Here α1 is determined by

eα1 = −e−α(

h2

h1

)2

. (4.210)

Suppose α is real, then α1 is purely imaginary if and only if∣∣∣∣∣∣∣∣∣∣h2

h1

∣∣∣∣∣∣∣∣∣∣2 e−α = 1. (4.211)


It is remained to find h1 and h2 such that (4.211) is satisfied. Takeλ = λ1 with |λ1| = 1. From the Lax pair,

h1ζ = −λ1

2e−αh2, h2ζ =

λ1

2eαh1,

h1ζ = −12

2αζh1 +

12λ1

h2, h2ζ = − 12λ1

h1 +αζ

2h2.

(4.212)

Take the complex conjugation, we get

h1ζ = −12αζ h1 +

12λ1

h2, h2ζ = − 12λ1

h1 +αζ

2h2,

h1ζ = − λ1

2e−αh2, h2ζ =

λ1

2eαh1.

(4.213)

Let

A =∣∣∣∣∣∣∣∣∣∣h2

h1

∣∣∣∣∣∣∣∣∣∣2 e−α − 1 =h2h2

h1h1e−α − 1, (4.214)

then

Aζ = −h2h2

h1h1e−ααζ + e−α

[h2h2ζ + h2ζ h2

h1h1− h2h2(h1h1ζ + h1ζ h1)

(h1h1)2

].

Using (4.212) and (4.213),

Aζ =λ1

2

(h2

h1e−α − h2

h1

)A. (4.215)

Similarly,

Aζ =λ1

2

(h2

h1− h2

h1e−α

)A. (4.216)

If we choose h1 and h2 so that A = 0 holds at one point (initial value),then A = 0 holds identically. Hence the following theorem is true.

Theorem 4.15 Suppose α is real, λ1 is a complex number with |λ1| = 1.Take h1 and h2 satisfying (4.211) at some point. Then α1 derived bythe Darboux transformation is purely imaginary.

This is the Darboux transformation from a solution of the sinh-Lap-lace equation to a solution of the sine-Laplace equation.

Now suppose α is purely imaginary. From (4.211), α1 is real if andonly if (

h2

h1

)2

e−α =(

h2

h1

)2

e−α < 0,


since eα1 > 0. From the above equality,(h2h1

h1h2e−α

)2

= 1,

henceh2h1

h1h2e−α = ±1.

On the other hand, (h2

h1

)2

e−α =h2h2

h1h1

h2h1

h1h2e−α.

Hence we should choose h1 and h2 such that

h2h1

h1h2e−α = −1. (4.217)

Similar to the proof of Theorem 4.15, by taking λ1 such that |λ1| = 1,(4.217) holds everywhere if it holds at one point. This leads to thefollowing theorem.

Theorem 4.16 Suppose α is a purely imaginary solution of (4.209), λ1

is a complex number with |λ1| = 1. Take h1 and h2 satisfying (4.217) atsome point. Then the Darboux transformation provides a real α1 whichis a solution of the sinh-Laplace equation.

Using the above two kinds of Darboux transformations, we get thefollowing series of Darboux transformations

(α,Φ) −→ (α1, Φ1) −→ (α2, Φ2) −→ · · · .If α is purely imaginary and β = − iα is a solution of the sine-Laplaceequation, then α1, α3, · · ·, α2n+1, · · · are solutions of the sinh-Laplaceequation, and β2 = − iα2, β4ββ = − iα4, · · ·, β2n = − iα2n, · · · are solutionsof the sine-Laplace equation. On the other hand, if α is a solution ofthe sinh-Laplace equation, then all α2n (n = 1, 2, 3, · · ·) are solutionsof the sinh-Laplace equation, and all β2n+1 = − iα2n+1 (n = 0, 1, 2, · · ·)are solutions of the sine-Laplace equation. This can be shown by thefollowing figure:

α0 α1 α2 α3 α4 · · ·

α0 α1 α2 α3 α4 · · ·


αi and αi−1 are real for odd i and purely imaginary for even i.Now we prove that this Darboux transformation is consistent with the

Backlund transformation (4.197).¨Rewrite (4.210) as

e(α1+α)/2 =h2

h1. (4.218)

Differentiating (4.218) with respect to ζ and using (4.212), we get

(α1 + α)ζ

2= λ1 sinh(α − α1). (4.219)

Similarly, differentiating (4.218) with respect to ζ and using (4.212), weget

(α1 − α)ζ

2=

1λ1

sinhα + α1

2. (4.220)

Let θ =α1

2, λ1 =

1 + i sinh τ

cosh τ, then from (4.197), we obtain (4.219) and

(4.220). Therefore, the function α1 derived from the Darboux trans-formation is a solution of (4.219) and the construction of the Backlund¨congruence from a time-like surface of constant negative Gauss curvatureis completed. This gives the following theorem.

Theorem 4.17 Starting from one focal surface, the Backlund congru-ence of type (d) together with its another focal surface can be constructedby using Darboux transformation (4.210) and Backlund transformation(4.194),

Example:Take the trivial solution α = 0. The Lax pair is

Φζ =λ

2

⎛⎝⎛⎛ 0 −1

1 0

⎞⎠⎞⎞Φ, Φζ =12

⎛⎝⎛⎛ 0 1/λ

−1/λ 0

⎞⎠⎞⎞Φ. (4.221)

Its fundamental solution is

Φ(λ) =

⎛⎝⎛⎛ eγ e−γ

− ieγ ie−γ

⎞⎠⎞⎞ , γ =(λ

2ζ − 1

2λζ)i. (4.222)

Using the Darboux transformation, we can get α1. Since α = 0 canbe regarded as a real or a purely imaginary solution, we can get purelyimaginary or real α1 from α = 0.


First we want to find real α1. Let λ = λ1 with |λ1| = 1, then

γ1 =(λ1

2ζ − 1

2λ1ζ)i (4.223)

is real. Let

⎛⎝⎛⎛ h1

h2

⎞⎠⎞⎞ be a column solution of the Lax pair for λ = λ1,

i.e.,h1 = eγ1 + be−γ1 , h2 = − ieγ1 + b ie−γ1 . (4.224)

Take b to be real, thenh2h1

h1h2= −1.

Suppose b is positive, then it can be taken as 1 by adding a constant onγ1. From (4.210),

eα1/2 = ih2

h1=

eγ1 − e−γ1

eγ1 + e−γ1= tanh γ1, (γ1 > 0),

ande−α1/2 = coth γ1,

coshα1

2= coth(2γ1), sinh

α1

2= − cosech(2γ1). (4.225)

When γ1 < 0, the right hand sides of (4.225) should change signs.α = 0 does not correspond to a time-like surface, but only a system of

orthogonal frames along a straight line. In fact, from (4.124) – (4.126),

dr = due1, de1 = 0, de2 = dvn, dn = dve2, (4.226)

hence

r = (u, 0, 0), e1 = (1, 0, 0),

e2 = (0, sinh v, cosh v), n = (0, cosh v, sinh v).(4.227)

According to (4.194), the space-like surface of constant negative Gausscurvature is

r ∗ = (u + l coth(2γ1),−l cosech(2γ1) sinh v,

−l cosech(2γ1) cosh v).(4.228)

According to the general theory, α1 = −2 tanh−1( sech(2γ1)) is a solutionof the sinh-Laplace equation. It is defined on the (u, v) plane except for


the straight line γ1 = 0. When γ1 → 0, the point on the surface tendsto infinity.

Now we apply the Darboux transformation again.

Φ1(λ) = D(λ)Φ(λ)

=

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜I − λ

λ1

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 0h1

h2h2

h10

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟⎛⎝⎛⎛ eγ e−γ

− ieγ ie−γ

⎞⎠⎞⎞

=

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜(

1 +λ

λ1coth γ1

)eγ

(1 − λ

λ1coth γ1

)e−γ

− i(

1 +λ

λ1tanh γ1

)eγ i

(1 − λ

λ1tanh γ1

)e−γ

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ .

(4.229)Take λ2 with |λ2| = 1 and let λ2/λ1 = eiµ (µ is real), then

h′1 = (1 + eiµ coth γ1)eγ2 + b(1 − eiµ coth γ1)e−γ2 ,

h′2 = i(−1 − eiµ tanh γ1)eγ2 + ib(1 − eiµ tanh γ1)e−γ2 ,

(4.230)

whereγ2 =

(λ2

2ζ − 1

2λ2ζ)i (4.231)

is real. In order that the derived solution α2 is purely imaginary andβ = − iα2 satisfies the sine-Laplace equation, (4.211) should hold. Thiscan be done if b is chosen to be a real number. In fact, when b is real,

h′1 = ih′

2 coth γ1eiµ, (4.232)∣∣∣∣∣∣∣∣∣∣h′2

h′1

∣∣∣∣∣∣∣∣∣∣2 e−α1 =1

coth2 γ1coth2 γ1 = 1.

In (4.210), if α is replaced by α1,h2

h1is replaced by

h′2

h′1

, then β = − iα2

is a solution of the sine-Laplace equation. From the explicit expressions

of cosβ

2, sin

β

2, the Backlund transformation can be constructed explic-

itly and the time-like surface of constant negative Gauss curvature isobtained.

Now we construct purely imaginary α1 from α = 0. By the funda-mental solution (4.222) of the Lax pair corresponding to α = 0, take the

column solution

⎛⎝⎛⎛ h1

h1

⎞⎠⎞⎞ such that

∣∣∣∣∣∣∣∣∣∣h2

h1

∣∣∣∣∣∣∣∣∣∣2 = 1, (4.233)


i.e.,e2γ1 + |b|2e−2γ1 − (b + b)e2γ1 + |b|2e−2γ1 + (b + b)

= 1. (4.234)

Hence b must be purely imaginary. Since |b| can be transformed to 1 byadding a constant to γ1, we can choose b = ± i. Suppose b = i, then

h1 = eγ1 + ie−γ1 , h2 = eγ1 − ie−γ1 . (4.235)

Henceeα1/2 = i

h2

h1= sech(2γ1) + i tanh(2γ1),

e−α1/2 = − ih1

h2= sech(2γ1) − i tanh(2γ1).

Let α1 = iβ1, then

cosβ1

2= sech(2γ1), sin

β1

2= tanh(2γ1),

β1

2= cos−1( sech(2γ1)),

(4.236)

β1 is a solution of the sine-Laplace equation. The Darboux matrix is

D(λ) = I − λ

λ1

⎛⎝⎛⎛ 0 ie−α1/2

− ieα1/2 0

⎞⎠⎞⎞ (4.237)

and the fundamental solution is

Φ1(λ) = D(λ)Φ(λ)

=

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ eγ(1 − λ

λ1e−α1/2) e−γ(1 +

λ

λ1e−α1/2)

− ieγ(1 − λ

λ1eα1/2) ie−γ(1 +

λ1

λ1eα1/2)

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ .(4.238)

Take λ2 such that |λ2| = 1. Let λ2/λ1 = eiµ (µ is real) and γ2 as before.Let

h′1 = eγ2

(1 − eiµe−α1/2

)+ be−γ2

(1 + eiµe−α1/2

),

h′2 = − ieγ2

(1 − eiµeα1/2

)+ ibe−γ2

(1 + eiµeα1/2

).

(4.239)

Take b to be purely imaginary, then

h′1 = −h′

1e−iµeα1/2, h′

2 = h′2e

−iµe−α1/2, (4.240)

henceh′

2h′1

h′1h

′2

e−α1 = −1.


This implies that

eα2/2 = ih′

2

h′1

e−α1/2, e−α2/2 = ih′

1

h′2

eα1/2 (4.241)

are real positive functions, and α2 satisfies the sinh-Laplace equation.From α1 and α, one can use the similar method as in the last subsectionto construct Backlund congruences as well as the space-like and time-like¨surfaces of constant negative Gauss curvature.

4.4 Orthogonal frame and Lax pairIn this section, we consider the relation between the orthogonal frame

of a surface of constant negative Gauss curvature in R3 and the Lax pairof the sine-Gordon equation. The geometric meaning of the Lax pair iselucidated clearly.

The group SU(2) consists of all 2 × 2 matrices A satisfying A∗A = Iand detA = 1. Its general element is of form

A =

⎛⎝⎛⎛ a0 + a1 i a2 + a3 i

−a2 + a3 i a0 − a1 i

⎞⎠⎞⎞ (4.242)

where a0, a1, a2, a3 are real numbers with a20 + a2

1 + a22 + a2

3 = 1. The Liealgebra su(2) consists of all 2× 2 matrices A satisfying A∗ + A = 0 andtrA = 0. Its general element is of form

A =

⎛⎝⎛⎛ β i γ + δ i

−γ + δ i −β i

⎞⎠⎞⎞ (4.243)

where β, γ, δ are real numbers.The group SO(3) consists of all 3 × 3 real orthogonal matrices with

determinant 1. Its Lie algebra so(3) consists of all 3× 3 anti-symmetricreal matrices. The correspondence

σ :12

⎛⎝⎛⎛ β i γ + δ i

−γ + δ i −β i

⎞⎠⎞⎞←−−→

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜0 β δ

−β 0 γ

−δ −γ 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ (4.244)

is an isomorphism between su(2) and so(3), i.e.,

[σA, σB] = σ[A, B] (4.245)

for all A, B ∈ su(2). This isomorphism of Lie algebras can be lifted toan isomorphism of the group SU(2) to the double covering of SO(3) (seeLemma 4.18 and Lemma 4.19).


We turn to discuss the geometrical meaning of the fundamental so-lution Φ to the Lax pair of the sine-Gordon equation. According to(4.33), under the Chebyshev coordinates, there exist orthogonal framesof a surface S of constant negative Gauss curvature satisfying⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟u

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

αv

2sin

α

2−αv

20 0

− sinα

20 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ ,

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟v

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

αu

20

−αu

20 − cos

α

20 cos

α

20

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ .

(4.246)

These frames are called Chebyshev frames. Using the isomorphism σ−1,we derive a system of equations in 2 × 2 matrices

Ψu =12

⎛⎜⎛⎛⎝⎜⎜ αv

2i i sin

α

2i sin

α

2−αv

2i

⎞⎟⎞⎞⎠⎟⎟Ψ,

Ψv =12

⎛⎜⎛⎛⎝⎜⎜ αu

2i − cos

α

2cos

α

2−αu

2i

⎞⎟⎞⎞⎠⎟⎟Ψ,

(4.247)

where Ψ is a 2 × 2 matrix in SU(2).Let

ξ =u + v

2, η =

u − v

2. (4.248)

We have

Ψξ = Ψu + Ψv =12

⎛⎜⎛⎛⎝⎜⎜ αξ

2i −e−αi/2

eαi/2 −αξ

2i

⎞⎟⎞⎞⎠⎟⎟Ψ,

Ψη = Ψu − Ψv =12

⎛⎜⎛⎛⎝⎜⎜ −αη

2i eαi/2

−e−αi/2 αη

2i

⎞⎟⎞⎞⎠⎟⎟Ψ.

(4.249)

Define

Φ(ξ, η) =

⎛⎝⎛⎛ e−αi/4 0

0 eαi/4

⎞⎠⎞⎞Ψ, (4.250)


(4.249) becomes

Φξ =12

⎛⎝⎛⎛ 0 −e−αi

eαi 0

⎞⎠⎞⎞Φ,

Φη =12

⎛⎝⎛⎛ −αη i 1

−1 αη i

⎞⎠⎞⎞Φ.

(4.251)

Let ξ = µξ1, η =1µ

η1 and denote ξ1 and η1 by ξ and η again, we

can introduce a spectral parameter µ in (4.251) formally. Thus we havean alternative form of the Lax pair of the sine-Gordon equation withspectral parameter µ:

Φξ =µ

2

⎛⎝⎛⎛ 0 −e−αi

eαi 0

⎞⎠⎞⎞Φ,

Φη =12

⎛⎝⎛⎛ −αη i 1/µ

−1/µ αη i

⎞⎠⎞⎞Φ.

(4.252)

Remark 30 By using the isomorphism of su(2)⎛⎝⎛⎛ i 0

0 − i

⎞⎠⎞⎞ −→⎛⎝⎛⎛ 0 1

−1 0

⎞⎠⎞⎞ ,

⎛⎝⎛⎛ 0 1

−1 0

⎞⎠⎞⎞ −→⎛⎝⎛⎛ i 0

0 − i

⎞⎠⎞⎞ ,⎛⎝⎛⎛ 0 i

i 0

⎞⎠⎞⎞ −→⎛⎝⎛⎛ 0 − i

− i 0

⎞⎠⎞⎞ ,

and changing λ toi

2µ, the Lax pair (4.67) becomes (4.252).

In applying Darboux transformation we can use the Lax pair (4.252)as well as (4.67), but one should note that the λ in (4.67) and µ in

(4.252) are related by λ =i

2µ, i.e., if we use real λ1 to construct the

Darboux matrix for (4.67), then we should use purely imaginary µ1 for(4.252).

In order to find out the relation between the Chebyshev frames andthe fundamental solution of the Lax pair, we need the map from SU(2)to SO(3) corresponding to the isomorphism (4.244) between their Liealgebras.


Lemma 4.18 Let A ∈ SU(2) be defined by (4.242). The map

τ(A) =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜a20+a2

2−a23−a2

1 2(a0a1−a2a3) 2(a0a3+a1a2)

−2(a0a1+a2a3) a20+a2

3−a22−a2

1 2(a0a2−a1a3)

2(−a0a3+a1a2) 2(−a0a2−a3a1) a20−a2

3−a22+a2

1

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ (4.253)

is an isomorphism from SU(2) to the double covering of SO(3). If Ψsatisfies (4.249), then τ(Ψ) satisfies (4.246).

Proof. It is necessary to prove:(i) τ(A) ∈ SO(3);(ii) τ is a homomorphism, i.e., τ(AB) = τ(A)τ(B);(iii) τ(A) = τ(B) if and only if B = ±A;(iv) τ : SU(2) → SO(3) induces an isomorphism σ between their Lie

algebras. Moreover, if A is a matrix function of certain parameters, then(dτ(A))(τ(A))−1 = σ((dA)A−1).

These are the basic facts on the relationship between SU(2) andSO(3). They can all be verified by direct calculations.

Lemma 4.19 Suppose τ(A) = (aij) ∈ SO(3), then A is given by

a20 =

14(1 + a11 + a22 + a33), a2

1 =14(1 + a33 − a11 − a22),

a22 =

14(1 + a11 − a22 − a33), a2

3 =14(1 + a22 − a11 − a33),

a0a1 =14(a12 − a21), a0a2 =

14(a23 − a32),

a0a3 =14(a13 − a31).

(4.254)

If the sign of any one of ai (say a0) is fixed, the signs of the other ai’sare also fixed. Hereafter, unless otherwise stated, τ−1(A) is referred toone branch of this double covering.

Now we can discuss the relation between the solution of the Lax pairrelated to the solution α of the sine-Gordon equation and the Chebyshevframe of the surface S of constant negative Gauss curvature correspond-ing to α.

Let S be the surface with K = −1 and related with the solutionα(ξ, η) of the sine-Gordon equation, (e1, e2,n) be the Chebyshev framesand Φ(1, ξ, η) be the fundamental solution of the Lax pair with µ = 1.


From the above discussion, we have

Φ(1, ξ, η) =

⎛⎝⎛⎛ e−αi/4 0

0 eαi/4

⎞⎠⎞⎞ τ−1

(⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟)

. (4.255)

and ⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ = τ

(⎛⎝⎛⎛ e−αi/4 0

0 eαi/4

⎞⎠⎞⎞Φ(1, ξ, η)

). (4.256)

(4.255) and (4.256) give the algebraic relation between Φ(1, ξ, η) andthe Chebyshev frame, so we can say Φ(1, ξ, η) is an SU(2) representationof the Chebyshev frame.

We turn to the geometrical meaning of the fundamental solution

Φ(µ, ξ, η). From (4.252) it is easily seen that Φ = Φ(µ,

ξ

µ, µη)

(µ is

real and non-zero) satisfies

Φξ =12

⎛⎝⎛⎛ 0 −e−αµ i

eαµ i 0

⎞⎠⎞⎞ Φ, Φη =12

⎛⎝⎛⎛ −αµη i 1

−1 αµη i

⎞⎠⎞⎞ Φ. (4.257)

Here αµ = α(µξ,

η

µ

)is another solution of the sine-Gordon equation.

The corresponding surface SµS −1 with K = −1 is called the Lie transfor-mation of S [26]. Comparing (4.257) with (4.251) we obtain

Theorem 4.20 Φ(µ,

ξ

µ, µη)

is an SU(2) representation of the surface

SµS −1.

Remark 31 When α and Φ(1, ξ, η), or equivalently the Chebyshev frame,is known, the surface r = r(ξ, η) can be determined by direct integration.In fact, the right hand side of the equation

dr = cosα

2due1 + sin

α

2dve2, (4.258)

is known already and its exterior derivative is zero. Hence r can bedetermined by the integration of (4.258).

In conclusion, if a solution α of the sine-Gordon equation is known,then a surface S with K = −1 and the fundamental solution of the Lax


pair together with the Lie transformation SµS can be obtained by solvinga system of linear partial differential equations, or simply by integration.Having these data, by using Darboux transformation, a series of surfacesof K = −1 together with their Lie transformations can be obtained bypurely algebraic algorithm.

Similarly, for various surfaces of constant Gauss curvature in R2,1, wecan get the relation between the Chebyshev frames and the solutions ofthe Lax pair, i.e. the relation between the Lax pair and the Chebyshevframes of the family of surfaces obtained by the Lie transformation fromthe seed surface. In this case, we need the isomorphism between the Liealgebras su(1, 1) and so(2, 1). Here su(1, 1) is the set of 2 × 2 matrix Asatisfying

A

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞+

⎛⎝⎛⎛ 1 0

0 −1

⎞⎠⎞⎞A∗ = 0,

and so(2, 1) is the Lie algebra of the Lorentz group on R2,1 with deter-minant 1.

4.5 Surface of constant mean curvatureRecently, the surface of constant mean curvature is studied widely.

For example, the surface of constant mean curvature immersed in Eu-clidean space and homeomorphic to torus is constructed which gives ananswer to the Hopf conjecture [110]. In this section we will use Dar-boux transformation to construct a series of surfaces of constant meancurvature from a known surface of constant mean curvature by purelyalgebraic algorithm.

4.5.1 Parallel surface in Euclidean spaceSuppose S is a surface in R3. By moving each point on S a distance l

along the normal direction, we obtain another surface, which is a parallelsurface of S. Take the orthogonal frame such that e1, e2 are tangent tothe lines of curvature, then the fundamental equations of S are

dr = ωaea (a = 1, 2),

dea = ωbaeb + ω3

an,

dn = ωa3ea.

(4.259)

Since e1 and e2 are tangent to the lines of curvature,

ω31 = b11ω

1, ω32 = b22ω

2, (4.260)

where b11 and b22 are principal curvatures of the lines of curvature.


Suppose S(l) is the parallel surface of S whose distance to S is l, thenits position vector is

r ∗ = r + ln. (4.261)

By differentiation, we get

dr ∗ = ω∗aea, (4.262)

whereω∗a = ωa + lωa

3 = (1 − lbaa)ωa. (4.263)

Hence e1 and e2 are still the tangent vectors of S(l), and n is the normalvector. When 1 − lb11 = 0 and 1 − lb22 = 0 hold, the surface S(l) isregular, and e1, e2,n is an orthonormal frame of S(l) at r∗. Moreover,

ω∗12 = ω1

2, ω∗31 = ω3

1 = b11ω1, ω∗3

2 = ω32 = b22ω

2. (4.264)

On the other hand, ω∗31 and ω∗3

2 can be written as the linear combinationsof ω∗1 and ω∗2:

ω∗31 = b∗11ω∗1 + b∗12ω∗2 = b∗11ω1 − lb∗11ω3

1 + b∗12ω∗2,

ω∗32 = b∗22ω2 − lb∗22ω3

2 + b∗21ω∗1.(4.265)

Hence

b∗11 − lb∗11b11 = b11, b∗22 − lb∗22b22 = b22, b∗12 = b∗21 = 0, (4.266)

which gives

b∗11 =b11

1 − lb11, b∗22 =

b22

1 − lb22. (4.267)

This means that e1 and e2 are still unit vectors tangent to the lines ofcurvature on S(l), and the principal curvatures are b∗11 and b∗22. TheGauss curvature and the mean curvature of the parallel surface S(l) are

K∗ =b11b22

(1 − lb11)(1 − lb22)=

K

1 − 2lH + l2K, (4.268)

H∗ =H − lK

1 − 2lH + l2K, (4.269)

respectively, where H = 12(b11 +b22) is the mean curvature of the surface

S. The most simple surfaces of constant mean curvature are spheres andcylinders. We shall not consider these trivial cases.

(4.269) implies

H(−2lH∗ − 1) = −lK − H∗(1 + l2K). (4.270)


Suppose that K is a constant, H is not a constant. We want to find S(l)such that H∗ is a constant. The right hand side of (4.270) is a constant.Hence

H∗ = − 12l

, 1 − l2K = 0. (4.271)

Therefore, K > 0, l = ±√

1K

. Conversely, if the mean curvature H of

the surface S is a constant, then from (4.268),

K(1 − l2K∗) = K∗(1 − 2lH). (4.272)

If K is not a constant and K∗ is a constant, then the right hand sideof (4.272) is a constant. However, since K is not a constant, we should

have l =1

2H, K∗ =

1l2

. This gives the proof of the well-known facts:

Theorem 4.21 If the surface S of constant positive Gauss curvature

K =1l2

is not a sphere, then on each side of S, there is a parallel

surface of constant mean curvature H∗ = ± 12l

whose distance to S is l.

If the surface S of constant mean curvature H =12l

is neither spherenor cylinder, then there is a parallel surface of constant positive Gauss

curvature K =1l2

whose distance to S is l.

Therefore, in the Euclidean space, the construction of the surface ofconstant mean curvature and the construction of surface of constantpositive Gauss curvature are equivalent in local sense.

4.5.2 Construction of surfacesWe shall consider the construction of surfaces of constant positive

Gauss curvature. Although there is no Backlund congruence, Darbouxtransformation can still be used.

As mentioned before, if we choose the Chebyshev coordinates of asurface of constant positive Gauss curvature, then (4.38) holds, hence

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟u

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

αv

2sinh

α

2−αv

20 0

− sinhα

20 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (4.273)


⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟v

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0 −αu

20

αu

20 cosh

α

20 − cosh

α

20

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ . (4.274)

Its integrability condition is

α = − sinhα.

Using the isomorphism (4.244) between so(3) and su(2), the system(4.273) and (4.274) is equivalent to

Ψu =12

⎛⎜⎛⎛⎝⎜⎜ − iαv

2i sinh

α

2i sinh

α

2iαv

2

⎞⎟⎞⎞⎠⎟⎟Ψ,

Ψv =12

⎛⎜⎛⎛⎝⎜⎜ iαu

2cosh

α

2− cosh

α

2− iαu

2

⎞⎟⎞⎞⎠⎟⎟Ψ.

(4.275)

Let

ζ =u + iv

2, ζ =

u − iv

2, (4.276)

then

Ψζ = Ψu − iΨv =12

⎛⎜⎛⎛⎝⎜⎜ αζ

2− ie−α/2

ieα/2 −αζ

2

⎞⎟⎞⎞⎠⎟⎟Ψ,

Ψζ = Ψu + iΨv =12

⎛⎜⎛⎛⎝⎜⎜ −αζ

2ieα/2

− ie−α/2αζ

2

⎞⎟⎞⎞⎠⎟⎟Ψ.

(4.277)

Write

Φ =

⎛⎝⎛⎛ e−α/4 0

0 eα/4

⎞⎠⎞⎞Ψ, (4.278)

then

Φζ =12

⎛⎝⎛⎛ 0 − ie−α

ieα 0

⎞⎠⎞⎞Φ,

Φζ =12

⎛⎝⎛⎛ −αζ i

−i αζ

⎞⎠⎞⎞Φ.

(4.279)


Take the constant λ such that |λ| = 1, then 1/λ = λ. Rewrite ζ asζ

λand ζ as λζ, then we get the Lax pair with spectral parameter

Φζ =λ

2

⎛⎝⎛⎛ 0 − ie−α

ieα 0

⎞⎠⎞⎞Φ,

Φζ =12

⎛⎝⎛⎛ −αζ i/λ

− i/λ αζ

⎞⎠⎞⎞Φ,

(4.280)

whose integrability condition is still the negative sinh-Laplace equation(4.39). Thus we have the Lax pair for the sinh-Laplace equation.

The surface of constant positive Gauss curvature can be constructedexplicitly as follows.

For any complex α, sinh(π i + α) = − sinhα. By changing λ to − iλ,α to π i + α, the Lax pair (4.208) of the sinh-Laplace equation becomesthe Lax pair of the negative sinh-Laplace equation. Thus the solutionof the sinh-Laplace equation (and the fundamental solution of its Laxpair) and the solution of the negative sinh-Laplace equation (and thefundamental solution of its Lax pair) can be changed with each other.We shall use the Darboux transformation for the sinh-Laplace equationto construct the solution of the negative sinh-Laplace equation.

Suppose α is a solution of the negative sinh-Laplace equation and Φis the fundamental solution of its Lax pair, the algorithm is shown inthe following diagram.

(α,Φ(λ)) (α2, Φ2(λ)), α=iπ+α′

λ=iλ′

--- λ′=−iπ

α′=α−iπ

(α′, Φ′(λ′)) twice Darboux transformations−→− (α′2, Φ

′2(λ

′))

In this diagram, (α′, Φ′(λ′)) and (α′2, Φ

′2(λ

′)) are solutions of the sinh-Laplace equation and the fundamental solutions of the correspondingLax pairs, (α,Φ(λ)) and (α2, Φ2(λ)) are solutions of the negative sinh-Laplace equation and the fundamental solutions of the correspondingLax pairs, the “twice Darboux transformations” are

(α′, Φ′(λ′)) −→ (α′1, Φ

′1(λ

′)) −→ (α′2, Φ

′2(λ

′))

where α′1 is purely imaginary while α′ and α′

2 are real. This process hasbeen explained in detail in Section 4.3.


Following the inverse of the procedure from (4.273) and (4.274) to

(4.280), we get the frame

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e′1e′2n′

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ from Φ2. In this case there is no

Backlund congruence, hence we can not get the expression of the surfaceälgebraically. However,

ω′1 = coshα2

2du, ω′2 = sinh

α2

2dv (4.281)

are known, and the integrability condition of

dr′ = ω′1e′1 + ω′2e′2 (4.282)

holds. Hence the surface of constant positive Gauss curvature r′ can beobtained by a direct integration (without solving differential equations).

Therefore, the surfaces of constant mean curvature are parallel withthe surfaces of constant positive Gauss curvature. The new surface ofconstant mean curvature is derived from suitable parallel surface of thesurface of constant positive Gauss curvature (K = 1, l = ±1). Fromb11 = b22, b11b22 = 1, we have 1− b11 = 0, 1 − b22 = 0, hence the parallelsurface exists and is regular.

Theorem 4.22 Suppose the expression of a non-spherical surface ofconstant positive Gauss curvature (resp. surface of constant mean curva-ture) under the Chebyshev coordinates is known and its Lie transforma-tion is also known, then new surface of constant positive Gauss curvature(resp. surface of constant mean curvature) and its Lie transformationcan be obtained by algebraic computation together with an integral.

This process can be continued successively. Whenever we know a fam-ily of non-spherical surfaces of constant positive Gauss curvature (resp.surfaces of constant mean curvature) which are Lie transformation witheach other, a series of such kinds of surfaces can be obtained. Algebraiccomputation together with an integral is needed in this process.

4.5.3 The case in Minkowski spaceIn the Minkowski space, we shall consider both space-like and time-

like surfaces.

(a) Space-like surfaceSuppose S is a space-like surface of constant Gauss curvature. Let e1

and e2 be the unit vectors tangent to the lines of curvature and n bethe unit normal vector. (e1, e2,n) forms an orthogonal frame, and

ω31 = b11ω

2, ω32 = b22ω

2. (4.283)


From e 21 = e 2

2 = 1, n 2 = −1, we have ω31 = ω1

3, ω32 = ω2

3.The position vector of the parallel surface S(l) is

r ∗ = r + ln. (4.284)

By differentiation,

dr ∗ = ωaea + lωa3ea = ω∗aea. (4.285)

Henceω∗1 = ω1 + lω3

1 = (1 + lb11)ω1,

ω∗2 = ω2 + lω32 = (1 + lb22)ω2.

(4.286)

Since ω∗3a = ω3

a,

b∗11 =b11

1 + lb11, b∗22 =

b22

1 + lb22, b∗12 = 0, (4.287)

we have

K∗ = −b∗11b∗22 =

K

1 + 2lH − l2K,

H∗ =12(b∗11 + b∗22) =

H − lK

1 + 2lH − l2K,

(4.288)

andH(2lH∗ − 1) = −lK − (1 − l2K)H∗. (4.289)

Suppose that K, H∗ are constants and H is not a constant, then

H∗ =12l

, K = − 1l2

. (4.290)

Alternatively, if H, K∗ are constants and K is not a constant, thenfrom

K(1 + l2K∗) = K∗(1 + 2lH),

we have

H = − 12l

, K∗ = − 1l2

. (4.291)

This leads to the following theorem.

Theorem 4.23 For each space-like surface of constant Gauss curvature

− 1l2

, two parallel surfaces with distance ±l have constant mean curvature

H = ± 12l

respectively. For each space-like surface of constant mean


curvature − 12l

and non-constant Gauss curvature, the parallel surface

with distance l is of constant negative Gauss curvature − 1l2

.

Therefore, the construction of a space-like surface of constant meancurvature is equivalent to the construction of a space-like surface of con-stant negative Gauss curvature (see Section 4.2). The following theoremholds.

Theorem 4.24 From a known space-like surface of constant mean cur-vature, a series of space-like surfaces of constant mean curvature can beobtained via constructing parallel surfaces, applying Backlund transfor-¨mation and Darboux transformation.

Remark 32 In this case, the Backlund congruence belongs to the case(d) in Subsection 4.3.3. In the construction of B¨cklund transformation,¨time-like surface of constant negative Gauss curvature will appear as anintermediate configuration.

(b) Time-like surfaceNow suppose S is a time-like surface, e 2

1 = 1, e 22 = −1. Similar to

the space-like case, ω31 = −ω1

3, ω32 = ω2

3, and

b∗11 =b11

1 − lb11, b∗22 =

b22

1 + lb22, b12 = 0. (4.292)

K = −b11b22, H =12(b11 − b22) leads to

K∗ =K

1 − 2lH + l2K, H∗ =

H − lK

1 − 2lH + l2K. (4.293)

If K and H∗ are constants, then

H∗ = − 12l

, K =1l2

, (4.294)

while K∗ and H are constants, we have

H =12l

, K∗ =1l2

. (4.295)

This gives the following theorem.

Theorem 4.25 Two parallel surfaces of a time -like surface of constant

positive Gauss curvature K =1l2

with “distance” ±l are surfaces of


constant mean curvature ± 12l

. On one side of a time-like surface of

constant mean curvature H =12l

, there is a parallel surface with distance

l and with constant positive Gauss curvature1l2

.

Therefore, the construction of time-like surface of constant mean cur-vature and the construction of time-like surface of constant positiveGauss curvature are equivalent.

Theorem 4.26 From a known time-like surface of constant mean cur-vature, a series of time-like surfaces of constant mean curvature can beobtained via constructing parallel surfaces, Backlund transformation and¨Darboux transformation.

In this case the related Backlund congruences belong to the case (2)ïn Subsection 4.3.4.

Chapter 5

DARBOUX TRANSFORMATIONAND HARMONIC MAP

Harmonic map is an important subject in differential geometry [25,24, 115] and is closely related with mathematical physics and the solitontheory [24, 57, 46, 48]. At the beginning of this chapter, we introduce thenotion of the harmonic map. Then the harmonic maps from Euclideanplane R2 or Minkowski plane R1,1 to Euclidean sphere S2 in R3, andH2 or S1,1 in Minkowski space R2,1 are elucidated. We shall show thatall these harmonic maps can be obtained from the construction of thesurfaces of constant Gauss curvature in Chapter 4, and show the relationsof these harmonic maps with some special soliton equations. Therefore,we can construct new harmonic maps from known harmonic maps andtheir extended solutions by using purely algebraic algorithm.

Using Darboux transformation, we can also construct the harmonicmaps from R1,1 or R2 to the Lie group U(N) and get explicit expressionsof the solutions. The solitons interact elastically. The unitons are alsoconsidered by using Darboux transformation. Comparing with [102],the construction here is more explicit with purely algebraic algorithmby using Darboux transformation.

5.1 Definition of harmonic map and basicequations

Riemannian manifold and Lorentzian manifold are the generalizationsof the Euclidean space and Minkowski space. A Riemannian manifold ora Lorentzian manifold is an n dimensional differential manifold M witha metric g. In the local coordinates (x1, · · · , xn) of M , the metric g is


expressed as

ds2 =n∑

i,j=1

gij(x)dxi dxj = gij(x)dxi dxj .

If (gij) is positive definite, then M is a Riemannian manifold and ds isthe differential of the arc length of a curve. If the eigenvalues of (gij)have the signs (+, · · · , +,−), then M is a Lorentzian manifold and g iscalled a Lorentzian metric.

Let M and N be Riemannian manifolds or Lorentzian manifolds, φ :M → N be a C2-map. The energy (or the action if M is a Lorentzianmanifold) of the map φ is

E(φ) =∫

M

∫∫e(φ) dVMVV (5.1)

where dVMVV is the volume element of M . e(φ) is called the energy (action)density. In local coordinates,

e(φ) =N

gαβ(φ)∂φα

∂xi

∂φβ

∂xj M

gij(x)

(α, β = 1, · · · , n; i, j = 1, · · · , m).(5.2)

Here n = dim N , m = dim M .M

gij ’s andN

gαβ ’s are contravariant compo-

nents ofM

g and covariant components ofN

g respectively, and (M

gij) is the

inverse of (M

gijM

).

The Euler equation of the functional E(φ) is derived directly as fol-lows. In local coordinates, dVMVV =

√g

√√dmx where g = |det(

M

gijM

)|. Hencethe Euler equation of E(φ) is

∂e(φ)√

g√√

∂φγ− ∂

∂xk

∂e(φ)√

g√√

∂(φγ,k)

= 0. (5.3)

For simplicity, we write gij forM

gijM

, and gαβ forN

gαβ . The partial derivative∂φγ

∂xkis denoted by φγ

,k. Then, the first term of (5.3) is

∂gαβ(φ)∂φγ

φα,iφ

β,jg

ij√g,√√

and the second term is

2∂

∂xk(gαγgkiφα

,i

√g

√√) = 2

(∂gαγ

∂φβgkiφα

,iφβ,k

√g

√√

+ gαγ∂gki

∂xkφα

,i

√g

√√+ gαγgki ∂2φα

∂xi∂xk

√g

√√+ gαγgkiφα

,i

∂√

g√√

∂xk

).

Darboux transformation and harmonic map 191

Using the Christoffel symbolsM

Γkij and

N

Γγαβ of M , N and the well-known

formulae∂gαγ

∂φβ=

N

Γλαβgλγ +

N

Γλγβgαλ, (5.4)

∂gij

∂xk= −glj

M

Γilk − gil

M

Γjlk, (5.5)

∂√

g√√

∂xk=

M

Γiik

√g,

√√(5.6)

we obtain the Euler equation

gik( ∂2φγ

∂xi∂xk−

M

Γjik

∂φγ

∂xj+

N

Γγαβ

∂φα

∂xi

∂φβ

∂xk

)= 0. (5.7)

Definition 5.1 A C2 map φ from M to N is called a harmonic mapif and only if it satisfies (5.7) [24].

(5.7) is the system of partial differential equations for the harmonicmap φ from M to N . If M is a Riemannian manifold, (5.7) is a systemof nonlinear elliptic equations. If M is a Lorentzian manifold, (5.7) isa system of nonlinear hyperbolic equations. The harmonic map from aLorentzian manifold is also called a wave map.

Harmonic map is a very important subject both in mathematics andin physics. In mathematics, there are many known special cases. If Nis the line R, a harmonic map is a harmonic function on M (especially,when M = Rn, it is the ordinary harmonic function); if M is the lineR or the circle S1, the harmonic map is a geodesic or a closed geodesicrespectively. A minimal surface is a conformal harmonic map whichkeeps the angles between two lines invariant.

In physics, there are quite a lot of applications:(1) Nonlinear σ-model (or chiral field) is a harmonic map M → N

where M is a Minkowski space and N is usually a homogeneous space.Especially, if N is a Lie group, it is called a principal chiral field [86].

(2) Ernst equation describes the static axially symmetric solution ofEinstein gravitation in vacuum [27]. It is the Euler equation of theEnergy ∫ 1

φ2

3∑i=1

[( ∂φ

∂xi

)2+( ∂ψ

∂xi

)2]d3x

under the axially symmetric constraint. Therefore, the Ernst equationis an axially symmetric harmonic map from R3 to the hyperbolic plane


H2 which is equipped with the Poincare metric´

ds2 =1φ2

(dφ2 + dψ2).

(3) In particle physics, string is used as a model of hadron. In fourdimensional Lorentzian space-time V , the world surface describing themotion of a classical string is a two dimensional time-like surface. It isdetermined by the equation

φαττ − φα

σσ +V

Γαβγ(φβ

τ φγτ − φβ

σφγσ) = 0,

This is the equation for the harmonic map from two dimensional Min-kowski plane R1,1 to V [36].

(4) Some solutions of the Yang-Mills equation in R4 under R-gaugecan also be obtained from harmonic maps.

(5) The simplest model for liquid crystal is a harmonic map from R2

to S2.In this book, we only discuss harmonic maps which can be constructed

explicitly by Darboux transformation. The starting manifold M is theEuclidean plane R2 = (x, y) or Minkowski plane R1,1 = (t, x) (or apart of them). In these two cases, the equations for the harmonic mapsare

φγxx + φγ

yy +N

Γγαβ(φα

xφβx + φα

y φβy ) = 0 (5.8)

andφγ

tt − φγxx +

N

Γγαβ(φα

t φβt − φα

xφβx) = 0 (5.9)

respectively.

5.2 Harmonic maps from R2 or R1,1 to S2, H2 orS1,1

Let S2 be the sphere of Euclidean space R3, which consists of all thepoints l = (l1, l2, l3) with l 2 = 1.

The coordinates in Euclidean plane R2 are represented by (x, y), andthe differential form of the Euclidean metric is ds2 = dx2 + dy2.

A map from R2 to S2 can be written as l = l(x, y) with l 2 = 1. From(5.7) or (5.8), the equation of l for the harmonic map R2 → S2 is givenby

lxx + lyy + (l 2x + l 2

y )l = 0. (5.10)

It can be derived as follows:

E(l) =∫Ω

∫∫ [( ∂l∂x

)2+( ∂l∂y

)2]dx dy,


where Ω is a region of R2, l satisfies l 2 = 1. Introduce the Lagrangianundetermined multipliers λ, we rewrite E(l) as

E(l) =∫Ω

∫∫ [( ∂l∂x

)2+( ∂l∂y

)2+ λ(l 2 − 1)

]dx dy,

then the Euler equation becomes

λl =∂

∂x

( ∂l∂x

)+

∂

∂y

( ∂l∂y

),

or l = λl.

With l 2 = 1, we have l · lx = 0 and l · ly = 0, hence l · lxx = −l 2x ,

l · lyy = −l 2y . This leads to λ = −(l 2

x + l 2y ), and (5.10) follows.

As is known, the Minkowski plane R1,1 is the simplest two dimensionalLorentzian manifold, whose metric is ds2 = dt2 − dx2. Instead of (5.7),the equation for the harmonic map from R1,1 to S2 is

ltt − lxx + (l 2t − l 2

x )l = 0. (5.11)

The proof is similar to the Euclidean space.In the Minkowski space R2,1, we can define two kinds of “spheres”:

H2 and S1,1. H2 is defined by

l 2 = −1, l3 > 0. (5.12)

It is the upper branch of the biparted rotational hyperboloid in R2,1

and realizes the Lobachevsky geometry globally, while the surface ofconstant negative Gauss curvature in R3 only realizes a certain part ofthe Lobachevsky plane.

S1,1 is given byl 2 = 1. (5.13)

It is a uniparted rotational hyperboloid. It is a time-like surface in theMinkowski space. Its metric is indefinite.

Similar to the above discussions, the equations for the harmonic mapsfrom R2 and H2 to S1,1 are

lxx + lyy − (l 2x + l 2

y )l = 0 (l 2 = −1, l3 > 0) (5.14)

andlxx + lyy + (l 2

x + l 2y )l = 0 (l 2 = 1), (5.15)

respectively. The equations for the harmonic maps from R1,1 to H2 andS1,1 are

ltt − lxx − (l 2t − l 2

x )l = 0 (l 2 = −1, l3 > 0) (5.16)


andltt − lxx + (l 2

t − l 2x )l = 0 (l 2 = 1) (5.17)

respectively. A harmonic map may be defined in a region of R2 or R1,1

instead of the whole plane.The harmonic maps from R2 (or R1,1) are conformal invariant. That

is, suppose h : Ω → N is a harmonic map from a region Ω in R2 (orR1,1) to a Riemannian (or Lorentzian) manifold N , φ is a conformalmap from a region Ω1 in R2 (or R1,1) to Ω, then the map h φ is aharmonic map from Ω1 to N .

This fact can be derived from the general equation or the energyintegral. In fact, when n = 2, the energy integral (5.1) is invariantunder the conformal transformation of R2 (or R1,1). Hence its Eulerequation has the same property. Here we explain it by two cases.

Case 1: Conformal invariance of the harmonic map from R2 to S2

(or H2, S1,1)First we consider R2 as a one dimensional complex space C1 with

complex coordinate z = u + iv and metric ds2 = dz d . Supposez = f(w) = u(u1, v1)+ iv(u1, v1) is a conformal map from a region Ω1 toΩ, l is a harmonic map from Ω to S2, then lf : l = l(x(u1, v1), y(u1, v1))is a harmonic map from Ω1 to S2 (or H2, S1,1). In fact, (5.10) can bewritten as

lzz + (lz · lz)l = 0. (5.18)

Under the transformation z = z(w), (5.18) is transformed to

lww + (lw · lw)l = 0.

If S2 is changed to H2 or S1,1, the proof is similar.Case 2: Conformal invariance of the harmonic map from R1,1 to S2

(or H2, S1,1)Let

ξ =t + x

2, η =

t − x

2(5.19)

be the characteristic coordinates (or light-cone coordinates) of R1,1, thends2 = 4dξ dη. A conformal map φ of R1,1 is of form

ξ = f(ξ1), η = g(η1). (5.20)

(5.11) can be written as

lξη + (lξ · lη)l = 0 (l 2 = 1). (5.21)

l φ is also a harmonic map, since it satisfies

lξ1η1 + (lξ1 · lη1)l = 0.


Similar result holds if S2 is changed to H2 or S1,1.Using the conformal invariance, we can define normalized harmonic

map.First we show that a harmonic map l from R1,1 to S2 satisfies

(l 2ξ )η = 0, (l 2

η )ξ = 0, l 2 = 1. (5.22)

Suppose (5.21) holds, then take the inner product with lξ, we obtain(l 2

ξ )η = 0. Similarly, (l 2η )ξ = 0 holds. This leads to (5.22). Conversely,

suppose (5.22) holds andξ

lξ’s, lη’s are linearly independent, then

lξ · lξη = 0, lη · lξη = 0. (5.23)

Hencelξη = σl.

The inner product with l leads to

σ = l · lξη = (l · lξ)η − lξ · lη = −lξ · lη,Hence (5.21) holds.

From (5.22),l 2ξ = f(ξ), l 2

η = g(η).

When f(ξ) = 0 and g(η) = 0, we can define the transformation ξ = ξ(ξ1)

and η = η(η1) such that(dξ1

dξ

)2= f(ξ),

(dη1

dη

)2= g(η). Rewrite (ξ1, η1)

as (ξ, η), we obtainl 2ξ = 1, l 2

η = 1. (5.24)

The harmonic map satisfying condition (5.24) is called a “normalizedharmonic map” [38]. When l 2

ξ = 0 and l 2η = 0, a harmonic map can

always be transformed to a normalized harmonic map by the transfor-mation ξ = ξ(ξ1), η = η(η1). Thus, some problems on harmonic mapscan be simplified by using conformal transformations.

For a normalized harmonic map,

l 2x + l 2

t = 1, lt · lx = 0 (5.25)

holds [58]. In fact, (5.24) is equivalent to (5.25). Hence a normalizedharmonic map can also be defined by (5.25).

A harmonic map from R2 to S2 (or H2, S1,1) can also be normalized.From (5.18) and l 2 = 1, we have (l 2

ζ )ζ = 0 and (l 2ζ)ζ = 0. Hence l 2

ζ isa holomorphic function of ζ. If

l 2ζ = (l 2

u − l 2v ) − 2ilu · lv = 1, (5.26)


i.e.,l 2u − l 2

v = 1, lu · lv = 0, (5.27)

this harmonic map is called normalized.Suppose l = l(u, v) is a harmonic map defined in a simply connected

region Ω and l 2ζ = 0, then l can be normalized. In fact, take the confor-

mal map ζ = ζ(ζw),

l 2w = l 2

ζ

( dζ

dw

)2. (5.28)

When l 2ζ = 0, the equation ( dw

dζ)2 = l 2

ζ has a single-valued solution

w = f(ζ) in the simply connected region Ω. w maps Ω to Ω1. The mapl f−1 from Ω1 to S2 (or H2, S1,1) has been normalized.

We have seen in Chapter 4 that there are many kinds of surfaces ofconstant Gauss curvature on which Chebyshev coordinates (u, v) and thecorresponding Chebyshev frames (e1, e2,n) exist (under the assumptionthat there are no umbilical points). The map from a region of R2 (orR1,1) to the surface defined by Chebyshev coordinates is called a Cheby-shev map. In this case, the normal vector n defines the Gauss map ofS2 (or H2, S1,1).

Theorem 5.2 The composition of the Chebyshev map and the Gaussmap is a normalized harmonic map, and vise versa [58].

Proof. We prove this theorem in several cases.

(a) R1,1 → S2

Suppose l is a normalized harmonic map. Take e1 and e2 so that e1 isthe unit vector of lt, e2 is the unit vector of lx, then (5.25) implies thate1 is orthogonal to e2. Moreover, there is a function α(x, t) such that

lt = − sinα

2e1, lx = cos

α

2e2. (5.29)

Since l, e1 and e2 form a orthonormal frame,

e1t = sin α2 l + σe2,

e1x = τe2,

e2t = −σe1,

e2x = − cos α2 l − τe1.

(5.30)

Their integrability condition leads to

σ =12αx, τ =

12αt (5.31)


and

αtt − αxx = sinα. (5.32)

Denote l = n, we have

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟t

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

αx

2sin

α

2−αx

20 0

− sinα

20 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ ,

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟x

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0

αt

20

−αt

20 − cos

α

20 cos

α

20

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜

e1

e2

n

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ ,

(5.33)

This is just the equation of Chebyshev frame for a surface of constantnegative Gauss curvature, l = n is its normal vector, l(u, v) is the Gaussmap. Therefore, a normalized harmonic map R1,1 → S2 is the composi-tion of a Chebyshev map and a Gauss map. Converse conclusion followsfrom

n 2ξ = (nt + nx)2 = sin2 α

2+ cos2

α

2= 1,

n 2η = (nt − nx)2 = 1

and the fact that nξ and nη are linearly independent.

(b) R2 → H2

Suppose l(x, y) is a normalized harmonic map, then (5.25) holds. Lete1 and e2 be the unit vectors of lu and lv respectively. We can choosefunction α(x, t) so that

lu = coshα

2e1,

lv = sinhα

2e2.

Similar to the case (a), they satisfy a system of different equations. Theequations are just the equations of the Chebyshev frame for the surface ofconstant negative Gauss curvature in R2,1 (Section 4.3), n is the Gaussmap from this surface to H2.

Other cases can be proved similarly. Here we list the results in thefollowing table.


From To Constant curvature Differential equation

S2 ⊂ R3 + α = − sinh α

R2 H2 ⊂ R2,1 − (space-like) α = sinh α

S1,1 ⊂ R2,1 − (time-like) α = sin α

S2 ⊂ R3 − ∂2α∂u2 − ∂2α

∂v2 = sin α

H2 ⊂ R2,1 + (space-like) ∂2α∂u2 − ∂2α

∂v2 = sin α

R1,1 ∂2α∂u2 − ∂2α

∂v2 = sinh α

S1,1 ⊂ R2,1 + (time-like) ∂2α∂u2 − ∂2α

∂v2 = cosh α

∂2α∂u2 − ∂2α

∂v2 = eα

This theorem implies that the construction of various normalized har-monic maps is closely related to the construction of the Chebyshevframes of various surfaces of constant Gauss curvature. Therefore, thefollowing theorem holds [58–60].

Theorem 5.3 Suppose a normalized harmonic map from R2 or R1,1

to various two dimensional “sphere” is known. Then, using Darbouxtransformation, we can obtain an infinite series of normalized harmonicmaps of the same type. Moreover, normalized harmonic map from R2

to S1,1 (resp. H2) can also be constructed by Darboux transformationfrom a known normalized harmonic map from R2 → H2 (resp. S1,1).

Remark 33 It is know that for a surface in R3 with constant positiveGauss curvature, the Gauss map is a harmonic map [92]. This can bederived from Theorem 5.3 as a corollary. For example, for a surface ofconstant Gauss curvature 1 in R3, the corresponding surface of constantmean curvature (see Section 4.5) is

ω∗1 = ω1 ± ω31, ω∗2 = ω2 ± ω3

2,

i.e.,

ω∗1 = (coshα

2± sinh

α

2)du, ω∗2 = (cosh

α

2± sinh

α

2)dv.

Its first fundamental form is

ds∗2 = (coshα

2± sinh

α

2)2(du2 + dv2)

= e±α(du2 + dv2).


Hence its Chebyshev coordinates are isothermal coordinates. This meansthat the Chebyshev coordinates give a conformal correspondence of sur-faces of constant mean curvature and the Euclidean plane. The aboveconclusion follows from the conformal invariance of the harmonic mapin two dimensional cases.

Remark 34 We can prove that there is a similar conclusion as Remark 33for space-like and time-like surfaces of constant mean curvature in R2,1.In Section 4.5 we have known that a space-like surface of constant meancurvature is a parallel surface of a space-like surface of constant negativeGauss curvature, while a time-like surface of constant mean curvatureis a parallel surface of a time-like surface of constant positive Gausscurvature. Suppose their Gauss curvatures are −1 and +1 respectively,then

ω∗1 = ω1 ± ω31, ω∗2 = ω2 ± ω3

2.

For space-like surface,

ω∗1 = (coshα

2± sinh

α

2)du, ω∗2 = (sinh

α

2± cosh

α

2)dv,

henceds∗2 = (ω∗1)2 + (ω∗2)2 = e±α(du2 + dv2)

where (u, v) are isothermal coordinates. In the time-like case,

ω∗1 = (coshα

2± sinh

α

2)du, ω∗2 = (sinh

α

2± cosh

α

2)dv,

henceds∗2 = (ω∗1)2 − (ω∗2)2 = e±α(du2 − dv2)

where (u, v) are also isothermal coordinates. From the above theorem,we know that a Gauss map of a surface of constant mean curvature is aharmonic map, and a Chebyshev map from (u, v) plane to the surface isa conformal map.

Remark 35 The harmonic map from the Minkowski plane R1,1 was firststudied in [37]. The solution of the Cauchy problem for harmonic mapexists globally if the target manifold is a complete Riemannian manifold.In [38], it was pointed out that when the target manifold is S1,1, globalsolution of the Cauchy problem may not exist.

5.3 Harmonic maps from R1,1 to U(N)

5.3.1 Riemannian metric on U(N)

The group U(N) is composed of all N × N unitary matrices, i.e., itconsists of all N × N matrices g satisfying gg∗ = I. On U(N), there is


a Riemannian metric

ds2 = − tr(dg g−1dg g−1) = − tr(dg g∗dg g∗). (5.34)

Here g ∈ U(N), dg is its differential, the trace (tr) of a matrix is the sumof all its diagonal entries. This metric ds2 is invariant under the left andright translation (i.e., the transformation g → g1 = ag and g → g2 = gafor any fixed element a of U(N)) of the group. In fact,

dg1 g−11 = a dg g−1a−1,

dg2 g−12 = dg g−1

imply

tr(dg g−1dg g−1) = tr(dg1 g−11 dg1 g−1

1 ) = tr(dg2 g−12 dg2 g−1

2 ).

Next we prove that ds2 is actually a Riemannian metric, i.e., (5.34) ispositive definite. Since U(N) is a differential manifold, we take the localcoordinate uα of (N) so that

g = g(uα) (α = 1, 2, · · · , r),where r = N2 is the dimension of U(N). Then

dg =∂g

∂uαduα,

ds2 = − tr( ∂g

∂uαg∗

∂g

∂uβg∗)duα duβ = gαβduα duβ ,

gαβ = − tr( ∂g

∂uαg∗

∂g

∂uβg∗).

(5.35)

Suppose ξα are real and not all zero. Let

Q =∂g

∂uαg−1ξα.

Substituting ξα for duα into ds2, we obtain gαβξαξβ = − tr(Q2). Onthe other hand, by differentiating gg∗ = I, we have

∂g

∂uαg∗ + g

∂g∗

∂uα= 0,

which gives Q + Q∗ = 0. Hence

− tr(Q2) = tr(QQ∗) ≥ 0.

Clearly tr(QQ∗) = 0 if and only if Q = 0, which means ξα = 0. There-fore, gαβξαξβ ≥ 0, and the equality holds if and only if ξα = 0. Thisproves that ds2 is a Riemannian metric.


5.3.2 Harmonic maps from R1,1 to U(N)

Suppose (x, t) are orthonormal coordinates of R1,1. Let ξ = x+t, η =x−t, then (ξ, η) are characteristic coordinates (light-cone coordinates) ofR1,1, and the metric of R1,1 is ds2 = dξ dη. The action of the harmonicmap g(ξ, η) from R1,1 to U(N) is

A[g] =∫

gαβuαξ uβ

ηdξ dη

= −∫

tr( ∂g

∂uαg∗

∂g

∂uβg∗)uα

ξ uβηdξ dη

= −∫

tr(∂g

∂ξg∗

∂g

∂ηg∗)dξ dη.

(5.36)

DenoteA = gηg

∗, B = gξg∗, (5.37)

then A and B are valued in the Lie algebra u(N) of the Lie group U(N).We can write down the Euler equation of A[g] by using the standard

procedure of variation. Denote

AΩ[g] = −∫Ω

∫∫tr(gξ g∗gη g∗)dξ dη,

where Ω is a bounded region. Suppose g depends on another parameter

τ , i.e., g = g(τ, ξ, η) and g(0, ξ, η) = g. Let∂g

∂τ

∣∣∣∣∣∣∣∣∣∣τ=0

= h and h is 0 at

the boundary of Ω. Then

dAΩ[g]dτ

∣∣∣∣∣∣∣∣∣∣τ=0

= −∫Ω

∫∫tr(hξg

∗A + gξh∗A + Bhηg

∗ + Bgηh∗)dξ dη.

Take the partial integration for the first and the third terms, we have

dAΩ[g]dτ

∣∣∣∣∣∣∣∣∣∣τ=0

= −∫Ω

∫∫tr(−hg∗Aξ − hg∗ξA + gξh

∗A

−Bηhg∗ − Bhg∗η + Bgηh∗)dξ dη.

Differentiating gg∗ = I and g∗g = I leads to hg∗ = −gh∗ and h∗g =−g∗h. Hence

gξh∗A = gξh

∗gg∗gηg∗ = −gξg

∗hg∗gηg∗,

Bgηh∗ = gξg

∗gηh∗gg∗ = −gξg

∗gηg∗hg∗,

hg∗ξA = hg∗ξgg∗gηg∗ = −hg∗gξg

∗gηg∗,

Bhg∗η = gξg∗hg∗gg∗η = −gξg

∗hg∗gηg∗.


With the above equation and using the equality

tr(A1A2 · · ·An) = tr(AnA1A2 · · ·An−1),

we obtain

dAΩ[g]dτ

∣∣∣∣∣∣∣∣∣∣τ=0

=∫Ω

∫∫tr(hg∗(Aξ + Bη))dξ dη = 0.

Since hg∗ can be an arbitrary matrix in the Lie algebra u(N) and Aξ+Bη

is also a matrix in the Lie algebra u(N), the Euler equation

Aξ + Bη = 0 (5.38)

is derived. By direct calculation, A and B defined by (5.37) also satisfy

Aξ − Bη + [A, B] = 0. (5.39)

Remark 36 (5.37) is actually gξ = Bg and gη = Ag, and (5.39) is theintegrability condition of (5.37). Hence, in R1

η,1 (or its simply connected

region), A and B which satisfy (5.39) can determine g(ξ, η) uniquelyup to a right-multiplier of a fix element of U(N). Therefore, we canconsider A and B as unknown functions and study the partial differentialequations (5.38) and (5.39) for A and B.

As in many cases considered before, we need to find the Lax pair withspectral parameter for (5.38) and (5.39). Notice that the integrabilitycondition of

Φη = λAΦ, Φξ = λ(2λ − 1)−1BΦ (5.40)

isλAξ − λ(2λ − 1)−1Bη + λ2(2λ − 1)−1[A, B] = 0,

i.e.,λ2(Aξ − Bη + [A, B]) + (λ2 − λ)(Aξ + Bη) = 0.

It should hold true for arbitrary λ, so we get (5.38) and (5.39).Hence, the following theorem holds.

Theorem 5.4 The partial differential equations (5.38) and (5.39) haveLax pair (5.40).

A non-degenerate N×N matrix solution Φ(λ) of the Lax pair is calledits fundamental solution. A fundamental solution is determined by itsvalue at one point, say (0, 0). When λ is real, if Φ(λ)|(0,0) ∈ U(N),


then Φ(λ) ∈ U(N) holds everywhere. Hereafter, we suppose Φ(λ)|(0,0) ∈U(N) for real λ. Then Φ(1)K is a harmonic map for any K ∈ U(N).

Remark 37 We can replace U(N) by any matrix Lie group G. If we use

A = gηg−1, B = gξg

−1

to replace (5.37), then (5.38) and (5.39) are extended as the equationsfor harmonic maps to G, and the Lax pair is still (5.40). In this case Gmay not have an invariant positive definite metric.

Now we use the Darboux transformation to construct explicit solu-tions of (5.38) and (5.39).

First, suppose the group G is a complex (or real) general linear groupGL(N,C) (or GL(N,R)). A(ξ, η) and B(ξ, η) are real (or complex)N × N matrix functions. Suppose Φ(ξ, η, λ) satisfies det Φ = 0, and A,B, Φ are solutions of (5.38)–(5.40).

We want to construct a Darboux matrix D in the form

D(λ) = I − λS (5.41)

so thatΦ1 = D(λ)Φ (5.42)

with suitable A1 and B1 satisfying (5.38)–(5.39). That is, Φ1, A1, B1

satisfyΦ1η = λA1Φ1, Φ1ξ = λ(2λ − 1)−1B1Φ1. (5.40)′

Substituting (5.42) into the above equations and using (5.40), we have

A1 = A − SηSS , B1 = B + SξS (5.43)

andSηSS S = AS − SA, SξS S − 2SξS = SB − BS. (5.44)

Therefore, as soon as we get a nontrivial solution S(ξ, η) of (5.44), then,A1, B1 are obtained from (5.43) and Φ1 is obtained from (5.41), (5.42).Hence, Φ1 satisfies (5.40), and A1, B1 satisfy (5.38), (5.39). This meansthat A1 and B1 give a new solution.

The explicit expression of S is constructed as follows.Take N real (or complex) constants λ1, · · ·, λN , which are not all

the same and λα = 0 , 1/2, 1; α = 1, 2, · · · , N . Take N constant columnvectors l1, · · · , lN so that the N × N matrix

H = (Φ(λ1)l1, · · · , Φ(λN )lN ) (5.45)


is non-degenerate. (Each column Φ(λα)lα in H is a column solution ofthe Lax pair when λ = λα.) Let

Λ = diag(λ1, λ2, · · · , λN ), (5.46)

then we have the following theorem.

Theorem 5.5 The matrix

S = HΛ−1H−1 (5.47)

satisfies (5.44).

Proof. hα = Φ(λα)lα is a column solution of (5.40) for λ = λα, i.e., itsatisfies

hαη = λαAhα, hαξ = λα(2λα − 1)−1Bhα,

(α = 1, 2, · · · , N).(5.48)

HenceHηHH = AHΛ, HξHH = BHΛ(2Λ − 1)−1,

and

SηSS = AHΛΛ−1H−1 − HΛ−1H−1AHΛH−1 = A − SAS−1.

This is the first equation of (5.44). On the other hand,

SξS = BHΛ(2Λ − 1)−1Λ−1H−1

− HΛ−1H−1BHΛ(2Λ − 1)−1H−1

= BH(2Λ − 1)−1H−1 − SBHΛ(2Λ − 1)−1H−1.

It is easy to check that the second equation of (5.44) is satisfied.Therefore, we have the following theorem.

Theorem 5.6 Suppose (A, B,Φ) is a solution of (5.38)–(5.40), S isdefined by (5.47), then (A1, B1, Φ1) defined by (5.43) and (5.42) is alsoa solution of (5.38)–(5.40).

Thus, we have the Darboux transformation

(A, B,Φ) −→ (A1, B1, Φ1).

Moreover, the algorithm is purely algebraic since S, SξS , SηSS can be ob-tained from A, B and Φ with explicit algebraic expressions.


In this construction, we cannot guarantee that the condition detH =0 holds globally. That is, if the seed solution (A, B,Φ) is defined on R1,1

(or its simply connected region Ω), it is not always possible that A1, B1

and Φ1 can be defined on R1,1 (or Ω) globally. What we can be sure isthat detH = 0 holds in a small region since it holds at some point byconstruction. There is another problem. If G is a group, A and B arevalued in the Lie algebra G′ of G, whether will A1 and B1 be still in G′?These are two related difficult problems. We shall give their answers forG = U(N).

As we have known, the group U(N) is composed of all the N × Nmatrices g satisfying g∗ = g−1. A matrix A belongs to the Lie algebrau(N) of U(N) if and only if

A∗ + A = 0. (5.49)

Now suppose (A, B,Φ) is a solution of (5.38)–(5.40) which is definedon whole R1,1 and A, B ∈ u(N). (If they are defined in a simply con-nected region of R1,1, the statements below hold as well.) The Darbouxtransformation is constructed following (5.43) and (5.42). In order tohave A1, B1 ∈ u(N), we need

(S + S∗)η = 0, (S + S∗)ξ = 0. (5.50)

Hence we want to make specific S to satisfy (5.50).Let λ0 be a non-zero complex number and

λα = λ0 or λ0 (α = 1, 2, · · · , N). (5.51)

Choose lα so thath∗

αhβ = 0 (if λα = λβ) (5.52)

holds at one point (say ξ = η = 0) and hα’s are linearly independent. Weshall show that the S constructed from these λα and lα satisfies (5.50).

First, we prove that (5.52) holds everywhere on R1,1 if it holds at onepoint. In fact, (5.48) leads to

(h∗αhβ)η = λαh∗

αA∗hβ + λβh∗αAhβ ,

(h∗αhβ)ξ = λα(2λα − 1)−1h∗

αB∗hβ + λβ(2λβ − 1)−1h∗αBhβ .

Hence(h∗

αhβ)η = (h∗αhβ)ξ = 0

when λα = λβ (i.e. λα = λβ). This means that (5.52) holds everywhereif it holds at one point.


Next, for the same λα = λ0 (or λ0), hα’s are linearly independenteverywhere if they are linearly independent at one point, since (5.48)is linear. Choose the initial value of hα so that they are linearlyindependent and satisfy (5.52), then hα are linearly independent and(5.52) holds everywhere. This means that the solution is globally definedon R1,1.

(5.47) leads to SH − HΛ−1 = 0, i.e.,

Shβ − λ−1β hβ = 0.

Its complex conjugate transpose gives

h∗βS∗ − λ−1

β h∗β = 0.

Henceh∗

β(S∗ + S)hγ = (λ−1β + λ−1

γ )h∗βhγ .

It is zero if λβ = λγ . When λβ = λγ = λ0 (or λ0), it equals( 1λ0

+1λ0

)h∗

βhγ . Hence

h∗β(S∗ + S)hγ = h∗

β

( 1λ0

+1λ0

)Ihγ .

Since hγ (γ = 1, 2, · · · , N) are linearly independent, we have

S∗ + S =( 1λ0

+1λ0

)I. (5.53)

Hence (5.50) holds, and A1, B1 ∈ u(N). Therefore, we have the followingtheorem.

Theorem 5.7 Let G = U(N), A, B ∈ u(N). Suppose (A, B,Φ) is asolution of (5.38)–(5.40) on R1,1, S is defined by (5.47) in which λα

and lα’s are chosen so that (5.51), (5.52) and det H = 0 are satisfied.Then (A1, B1, Φ1) defined by (5.42) is also a global solution of (5.38)–(5.40) on R1,1. Moreover, A1, B1 ∈ u(N).

Now we turn to the Lie group SU(N). A matrix g ∈ SU(N) if andonly if g ∈ U(N) and det g = 1. The Lie algebra su(N) of SU(N)composes of all matrices A such that A ∈ u(N) and trA = 0.

The above Theorem 5.7 also holds for the case of SU(N). In this case,there are additional conditions trA = trB = 0 and tr A1 = trB1 = 0.Thus, we must have tr SηSS = 0 and trSξS = 0. In fact, the first equationof (5.44) can be written as

SηSS = −SAS−1 + A


and trSηSS = 0 follows. From the definition of S, −S + 2I = −H(Λ−1 −2I)H−1. Since λ0 is not real, 2I − S is non-degenerate. The secondequation of (5.44) can be written as

SξS (S − 2I) = (S − 2I)B − B(S − 2I),

which implies trSξS = 0.

Remark 38 Let λ = 1, then Φ(1) satisfies

Φ(1)η = AΦ(1), Φ(1)ξ = BΦ(1).

Comparing with (5.37), we know that g = Φ(1)g0 is a harmonic mapwhere g0 is a constant matrix. This kind of harmonic map has beenstudied in [8]. Here we use Darboux transformation to get more explicitconclusions.

5.3.3 Single soliton solutionsTake the trivial solution as a seed solution, single soliton and multi-

soliton solutions can be obtained by Darboux transformations as men-tioned above. In this process, only algebraic algorithm is necessary. Forsimplicity, we only consider the harmonic map R1,1 → SU(2). However,the following discussions are essentially the same for U(N).

An element of SU(2) can be written as⎛⎝⎛⎛ γ β

−β γ

⎞⎠⎞⎞ ,

where β and γ are complex numbers satisfying γγ + ββ = 1. Take Aand B to be two non-zero constant elements of su(2) with [A, B] = 0.Clearly they satisfy the equation of harmonic map (5.38) and (5.39).Suppose

A =

⎛⎝⎛⎛ ipi 0

0 − ipi

⎞⎠⎞⎞ , B =

⎛⎝⎛⎛ iq 0

0 − iq

⎞⎠⎞⎞ ,

where p and q are non-zero real numbers. The solution of dg0 = (Adη +Bdξ)g0 is

g0 =

⎛⎝⎛⎛ ei(pη+qξ) 0

0 e−i(pη+qξ)

⎞⎠⎞⎞ ,

which is a harmonic map. The corresponding Φ0(λ) is

Φ0(λ) =

⎛⎝⎛⎛ l(λ) 0

0 l−1(λ)

⎞⎠⎞⎞ ,


wherel(λ) = exp

(iλpη +

iλ2λ − 1

qξ).

g0, A, B and Φ0 constitute the seed solution of the Darboux transfor-mation.

Take λ1 = λ0, λ2 = λ0,

H = (h1, h2) =

⎛⎝⎛⎛ l(λ0) bl(λ0)

al−1(λ0) l−1(λ0)

⎞⎠⎞⎞ .

Thenh∗

2h1 = bl(λ0)l(λ0) + al−1(λ0)l−1(λ0).

Sincel(λ0) = exp

(iλ0pη + iλ0

2λ0−1qξ)

= l−1(λ0),

l(λ0) = l−1(λ0),

the equality h∗2h1 = 0 holds if and only if b = −a.

By direct calculations, we have

det H = er + |a|2e−r > 0,

S = HΛ−1H−1

=1

er + |a|2e−r

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜er

λ0+

e−r

λ0|a|2

( 1λ0

− 1λ0

)aeis

( 1λ0

− 1λ0

)ae−is er

λ0+

e−r

λ0|a|2

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ ,

where

r = i(λ0 − λ0)pη + i( λ0

2λ0 − 1− λ0

2λ0 − 1

)qξ,

s = (λ0 + λ0)pη +( λ0

2λ0 − 1+

λ0

2λ0 − 1

)qξ

are real linear functions of ξ and η. The new fundamental solution andthe new harmonic map are

Φ1(λ) = (I − λS)Φ(λ),

g1 = Φ1(1)∣∣∣∣∣∣∣∣∣∣ λ0

λ0 − 1

∣∣∣∣∣∣∣∣∣∣respectively. The above right-multiplier

∣∣∣∣∣∣∣∣∣∣ λ0

λ0 − 1

∣∣∣∣∣∣∣∣∣∣ is used to keep g1 ∈SU(2).


Figure 5.1.

Remark 39 If we use r − ln |a| and s + i ln(a/|a|) instead of r and s,then a in S can be changed to 1.

g1 can be written as

g1 =

⎛⎝⎛⎛ γ1 β1

−β1 γ1

⎞⎠⎞⎞

with

γ1 =

(1 − 1

λ0

)er +

(1 − 1

λ0

)e−r

er + e−rei(pη+qξ)

∣∣∣∣∣∣∣∣∣∣ λ0

λ0 − 1

∣∣∣∣∣∣∣∣∣∣ ,β1 = −

( 1λ0

− 1λ0

) 1er + e−r

ei(s−pη−qξ)

∣∣∣∣∣∣∣∣∣∣ λ0

λ0 − 1

∣∣∣∣∣∣∣∣∣∣ .Write

β1 = ρ1eiθ1 , γ1 = σ1eiτ1 ,

where ρ1 > 0, σ1 > 0, θ1, τ1ττ is real, then

ρ1 =12

∣∣∣∣∣∣∣∣∣∣∣∣∣ λ0 − λ0

λ0(λ0 − 1)

∣∣∣∣∣∣∣∣∣∣∣∣∣ sech r,

σ1 = (1 − ρ21)

1/2.

The figures for ρ1 and σ1 with respect to r are shown in Figure 5.1.Both ρ1 and σ1 are of the shape of solitons. As is known, SU(2) is

a three dimensional manifold, which is actually the three dimensionalsphere

S3 = (x1, x2, x3, x4) |x21 + x2

2 + x23 + x2

4 = 1

in R4. The relation between (x1, x2, x3, x4) and

⎛⎝⎛⎛ γ β

−β γ

⎞⎠⎞⎞ is given by

γ = x1 + ix2, β = x3 + ix4.


Unless r depends on x or t only, the image of the map tends to twocircles x3 = x4 = 0 and x2

1 + x22 = 1 as x → ±∞ for fixed t, or as

t → ±∞ for fixed x. Hence these two circles are limiting circles.

Remark 40 For SU(2), the invariant metric is the standard metric ofthe sphere S3. Therefore, a harmonic map to SU(2) is just a harmonicmap to the sphere S3.

5.3.4 Multi-soliton solutionsMulti-soliton solutions can be obtained by composition of successive

Darboux transformations.Let λ0, λ1, · · ·, λk be k + 1 non-real complex numbers. Suppose that

they are distinct and satisfy

(i) |2λl − 1| (l = 0, 1, · · · , k) are distinct,

(ii) (λl − λl)p ±( λl

2λl − 1− λl

2λl − 1

)q = 0 .

Here p and q are the constants appeared in the seed solution

g0 =

⎛⎝⎛⎛ ei(pη+qξ) 0

0 e−i(pη+qξ)

⎞⎠⎞⎞ .

Let

rl = i(λl − λl)pη + i( λl

2λl − 1− λl

2λl − 1

)qξ,

sl = (λl + λl)pη +( λl

2λl − 1+

λl

2λl − 1

)qξ,

(l = 0, 1, 2, · · · , k).

They are real linear functions of ξ and η. It can be verified that rl/rj ’s(l = j) are not constants if and only if (i) holds. On the other hand, (ii)means that rl really depends on x and t. Here r0 and s0 are just r ands in the construction of the single soliton solutions.

Since we have g0 and Φ0, we can define

Φj(λ) = (I − λSjS −1) · · · (I − λS0S )Φ0(λ),

gjg = (I − SjS −1) · · · (I − S0SS )g0

∣∣∣∣∣∣∣∣∣∣ λ0

λ0 − 1

∣∣∣∣∣∣∣∣∣∣ · · ·∣∣∣∣∣∣∣∣∣∣∣∣∣ λj−1

λj−1 − 1

∣∣∣∣∣∣∣∣∣∣∣∣∣(j = 1, 2, · · · , l),

recursively. Let S0SS be the matrix S in the construction of the singlesoliton solutions, S1, S2, · · · be defined by (5.47) for Φ = Φ1, Φ2, · · ·.


According to the expression of S0SS ,

limr0→±∞S0SS =

⎛⎝⎛⎛ µ0 0

0 µ0

⎞⎠⎞⎞

where

µ0 =

⎧⎪⎧⎧⎨⎪⎪⎪⎨⎨⎩⎪⎪ 1/λ0 as r0 → +∞,

1/λ0 as r0 → −∞.

Lemma 5.8 When rl is bounded and t → +∞,

limt→±∞Sl −→

⎛⎝⎛⎛ µl 0

0 µl

⎞⎠⎞⎞

(µl = 1/λl or 1/λl; l = 0, 1, 2, · · · , k − 1).

Proof. We use induction to prove it. Suppose it holds for l = 0, 1, 2,· · · , i − 1. Now we consider the case l = i.

Since ri is finite, r0, · · ·, ri−1 → ±∞ when t → +∞, hence

Φi ∼⎛⎝⎛⎛ A(λ)l(λ) 0

0 A(λ)l−1(λ)

⎞⎠⎞⎞ ,

where ∼ represents asymptotic value,

A(λ) =i−1∏h=0

(1 − λµh), A(λ) =i−1∏h=0

(1 − λµk).

Since

A(λ) = A(λ),

we have

HiHH ∼⎛⎝⎛⎛ A(λi)l(λi) biA(λi)l(λi)

aiA(λi)l−1(λi) A(λi)l−1(λi)

⎞⎠⎞⎞

where

bi = −ai.


As before, ai and bi can be put into ri and si so that ai = 1, bi = −1.By direct calculation,

SiSS = HiHH Λ−1i H−1

iH ∼ 1|A(λi)|2eri + |A(λi)|2e−ri

·

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜|A(λi)|2

λieri +

|A(λi)|2λi

e−ri

( 1λi

− 1λi

)A(λi)A(λi)eisi

( 1λi

− 1λi

)A(λi)A(λi)e−isi

|A(λi)|2λi

eri +|A(λi)|2

λie−ri

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ .

When t → +∞ and ri → ±∞,

SiSS ∼⎛⎝⎛⎛ µi 0

0 µi

⎞⎠⎞⎞ , µi =1λi

or1λi

.

The lemma is proved.When rk−1 is finite and t → +∞, r0, · · · , rk−2 → ±∞, hence

gk ∼ (I − Sk−1)

⎛⎝⎛⎛ 1 − µk−2

1 − µk−2

⎞⎠⎞⎞ · · ·⎛⎝⎛⎛ 1 − µ0

1 − µ0

⎞⎠⎞⎞

·⎛⎝⎛⎛ ei(pη+qξ)

e−i(pη+qξ)

⎞⎠⎞⎞ ∣∣∣∣∣∣∣∣∣∣ λ0

λ0 − 1· · · λk−1

λk−1 − 1

∣∣∣∣∣∣∣∣∣∣ .According to the asymptotic expression of Sk−1, we know that theasymptotic behavior of gk as rk−1 bounded and t → +∞ is the same asa single soliton solution.

If one of r0, · · · , rk−2 does not tend to ±∞ (say, ri → ±∞ with certaini < k − 1) as t → +∞, then the other rj → ±∞. According to thetheorem of permutability of Darboux transformation, we can change theDarboux matrix with λi to be the last one. The factors should havesome change, but the result is still gk. The asymptotic behavior of gk ast → +∞ along the direction ri → ±∞ is still a single soliton.

If t → −∞, there are similar conclusions, but µi should be changedto its complex conjugate. This means that the soliton has “phase shift”.Therefore, we have the following theorem.

Theorem 5.9 A harmonic map from R1,1 to U(N) (or SU(N)) derivedfrom the trivial solution by k Darboux transformations has the followingproperty. When t → ±∞, the asymptotic solution split up into k singlesolitons. When t → +∞ and t → −∞, these k single solitons arearranged in opposite order and have their own phase shifts.


Thus, the behavior of the multi-soliton solutions of a harmonic mapfrom R1,1 to U(N) (or SU(N)) is quite similar to that of the multi-solitons of KdV equations etc., i.e., all of them have the property ofelastic scattering.

5.4 Harmonic maps from R2 to U(N)

5.4.1 Harmonic maps from R2 to U(N) and theirDarboux transformations

Let (x, y) be the orthonormal coordinates of R2, then the energy ofφ(x, y) from R2 (or a region Ω in R2) to U(N) is

S[φ] = −∫

tr(φxφ−1φxφ−1 + φyφ−1φyφ

−1)dx dy. (5.54)

φ is called a harmonic map from R2 to U(N) if the Euler equation forS[φ] holds. We suppose Ω to be simply connected.

Denoteζ = x + iy, ζ = x − iy (5.55)

and∂

∂ζ=

12

( ∂

∂x+ i

∂

∂y

),

∂

∂ζ=

12

( ∂

∂x− i

∂

∂y

). (5.56)

LetA = φζφ

−1, B = φζφ−1. (5.57)

Similar to Section 5.3, φ is a harmonic map if and only if A and B satisfy

Aζ − Bζ + [A, B] = 0, (5.58)

Aζ + Bζ = 0. (5.59)

Since φ ∈ U(N), i.e., φ∗ = φ−1,

A∗ = φ∗−1φ∗ζ = φ(φ−1)ζ = −φφ−1φζφ

−1 = −B.

Hence the constraint on A and B is

A∗ = −B. (5.60)

Conversely, if A and B satisfy the constraint (5.60) and the equations(5.58), (5.59), then φ can be solved from

φζ = Aφ, φζ = Bφ. (5.61)


Moreover, φ satisfies

(φ∗φ)ζ = φ∗ζφ + φ∗φζ = (φζ)

∗φ + φ∗φζ

= (Aφ)∗φ + φ∗Bφ = φ∗(A∗ + B)φ = 0.(5.62)

Similarly,(φ∗φ)ζ = 0. (5.63)

If φ∗φ = I holds at one point, it holds identically. Hence, (5.58), (5.59)and (5.60) can be regarded as the equations of the harmonic map fromR2 to U(N). These equations uniquely determine the harmonic map upto a right-multiplied constant matrix in U(N).

Now consider the Lax pair

Φζ = λAΦ, Φζ =λ

2λ − 1BΦ (5.64)

(λ = 1 /2) whose integrability conditions are (5.58) and (5.59). Supposethe initial condition Φ|ζ=0 ∈ U(N), then

g = Φ(1) (5.65)

is a harmonic map. g is unique up to a right-multiplied constant matrixin U(N). Here Φ(λ) is called an extended solution of the harmonic map

g [102]. Notice that if |2λ − 1| = 1, then λ =λ

2λ − 1and Φ(λ) ∈ U(N).

The harmonic maps from R2 to U(N) have been discussed in [102].Here we mainly consider their properties related to the Darboux trans-formation.

Remark 41 The symbols used here differ from those in [102]. The maindifferences are:

(i) The A and B are equivalent to 2Aζ and 2Aζ in [102].(ii) In the definition of A, B and the Lax pair (5.64), the order of

multiplication of matrices is different from those used in [102]. There-fore, left multiplications in [102] are changed to right multiplications inmany places.

(iii) The parameters λ andλ

2λ − 1were written as

12(1−λ) and

12(1−

λ−1) in [102]. In our notation we introduce µ = 1−2λ which correspondsto the λ in [102].

As in Section 5.3, we can construct the Darboux matrix D = I − λSand the Darboux transformation

(A, B,Φ) −→ (A1, B1, Φ1) (5.66)


where

Φ1 = (I − λS)Φ, (5.67)

A1 = A − Sζ , B1 = B + Sζ . (5.68)

In order to make (A1, B1, Φ1) satisfy the Lax pair

Φ1ζ = λA1Φ1, Φ1ζ =λ

2λ − 1B1Φ1, (5.69)

S should satisfy

SζS = AS − SA, Sζ(S − 2I) = SB − BS. (5.70)

Moreover, A∗1 = −B1 must hold.

Take N complex numbers λ1, λ2, · · ·, λN ( = 0 , 1/2). Let

hρ = Φ(λρ)lρ (ρ = 1, 2, · · · , N) (5.71)

where lρ’s are constant column vectors so that H = (h1, h2, · · · , hN ) isnon-degenerate, then

S = HΛ−1H−1 (5.72)

with Λ = diag(λ1, λ2, · · · , λN ) satisfies (5.70). The proof for this is thesame as in R1,1 case.

It is necessary to choose λρ and lρ so that A∗1 = −B1 holds. This can

be done as follows. Take ω1 to be a non-real complex number and

ω2 =ω1

2ω1 − 1. (5.73)

We choose

λρ =

⎧⎨⎧⎧⎩⎨⎨ ω1 (ρ = 1, · · · , k)

ω2 (ρ = k + 1, · · · , N)(0 < k < N). (5.74)

Moreover, suppose |2ω1 − 1| = 1 so that ω1 = ω2. Let h1, · · · , hk be thecolumn solutions of the Lax pair for λ = ω1, hk+1, · · · , hN be the columnsolutions of the Lax pair for λ = ω2, i.e.,

ha = Φ(ω1)la, hp = Φ(ω2)lp

(a = 1, 2, · · · , k; p = k + 1, · · · , N)(5.75)

where l1, · · · , lN are linearly independent constant column vectors. Ifh1, · · · , hN are chosen to be orthogonal with each other at one point(say (0, 0)), then

h∗pha = 0 (a = 1, 2, · · · , k; p = k + 1, · · · , N) (5.76)


hold everywhere. This follows from

(h∗p)ζ = (hpζ)∗ =

( ω2

2ω2 − 1Bhp

)∗= −ω1h

∗pA,

(h∗pha)ζ = −ω1h

∗1Aha + h∗

1ω1Aha = 0,(5.77)

and similarly,(h∗

pha)ζ = 0. (5.78)

From the definition of S, we have

Sha =1ω1

ha, Shp =1ω2

hp, (5.79)

h∗aS

∗ =1ω1

h∗a, h∗

pS∗ =

1ω2

h∗p. (5.80)

Henceh∗

a(S∗ − S)hb =

( 1ω1

− 1ω1

)h∗

ahb,

h∗p(S

∗ − S)hq =( 1ω2

− 1ω2

)h∗

phq,

h∗p(S

∗ − S)ha = 0,

h∗a(S

∗ − S)hp = 0

(a, b = 1, · · · , k; p, q = k + 1, · · · , N).

(5.81)

According to (5.73),1ω1

− 1ω1

=1ω2

− 1ω2

. (5.82)

Soh∗

β(S∗ − S)hα = h∗β

( 1ω1

− 1ω1

)Ihα

(α, β = 1, 2, · · · , N)

and1ω1

− 1ω2

is real. Since hα consists of N linearly independent

vectors,

S∗ − S =( 1ω1

− 1ω1

)I. (5.83)

Then

A∗1 + B1 = (A1 − Sζ)

∗ + B + Sζ = A∗ + B − S∗ζ + Sζ = 0,

i.e., A1 and B1 satisfy the constraint (5.60) for U(N). Therefore, wehave the following theorem.


Theorem 5.10 Suppose that λρ’s and hρ’s satisfy (5.74) and (5.76),hρ’s are linearly independent. Then the Darboux transformation (5.66)gives a new harmonic map from R2 to U(N), which is expressed by

g1 = Φ1(1)(Φ1|0)−1

where (Φ1|0) is the value of Φ1 at a fixed point ζ = 0.

This topic has also been discussed by several authors such as [112]and [54].

Now we use the Darboux transformation to derive the projective op-erators π and π⊥ in CN given by [102].

According to (5.83),

S∗ − 1ω1

I = S − 1ω1

I,

so S − 1ω1

I is Hermitian. On the other hand,

S − 1ω1

I = H(Λ−1 − 1

ω1I)H−1,

so S − 1ω1

I has k eigenvalues 0 and N − k real eigenvalues1ω2

− 1ω1

.

Moreover,

π =( 1ω2

− 1ω1

)−1(S − 1

ω1I)

=( 1ω2

− 1ω1

)−1·

·H

⎡⎢⎡⎢⎢⎢⎣⎢⎢⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜

1ω1

IkI 0

01ω2

INI −k

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟−

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜1ω1

IkI 0

01ω1

INI −k

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟⎤⎥⎤⎥⎥⎥⎦⎥⎥H−1

= H

⎛⎝⎛⎛ 0 0

0 INI −k

⎞⎠⎞⎞H−1

(5.84)is an Hermitian matrix with k eigenvalues 0 and N − k eigenvalues 1. πis a projective matrix, and

π⊥ = I − π = H

⎛⎝⎛⎛ IkI 0

0 0

⎞⎠⎞⎞H−1 (5.85)


is the complement of π. The invariant subspace of π⊥ is orthogonal tothat of π. From (5.84) and (5.85),

S =1ω2

π +1ω1

π⊥, (5.86)

and the Darboux matrix

D = I − λS =(1 − λ

ω2

)π +

(1 − λ

ω1

)π⊥ = (π + γπ⊥)

(1 − λ

ω2

)(5.87)

where

γ =ω2(ω1 − λ)ω1(ω2 − λ)

. (5.88)

Remark 42 S can also be written as

S =1ω1

I −( 1ω1

− 1ω2

)π. (5.89)

Substituting it into (5.70), we get

πζ = −ω1πA + ω2Aπ + (ω1 − ω2)πAπ,

πζ = ω1Bπ − ω2πB + (ω2 − ω1)πBπ,(5.90)

These equations are first derived in [102] by the decomposition of loopgroup. Here we use Darboux transformation to give the explicit expres-sions of the solutions of these equations. Moreover, the new harmonicmap is given by

g1 = Φ1(1)/(

1 − 1ω2

)= (π + γ1π

⊥)Φ(1) (5.91)

where

γ1 =1 − 1/ω1

1 − 1/ω2(5.92)

satisfies γ1γ1 = 1. If we are considering the harmonic map to SU(N),then the new harmonic map is

g1 = Φ1(1)(I − Λ−1)−1. (5.93)


5.4.2 Soliton solutionsSimilar to Section 5.3, the soliton solutions of the harmonic map from

R2 → U(N) can be constructed explicitly. For simplicity, we chooseN = 2, so the solution is a special harmonic map from R2 → SU(2).

Take the seed solution

g0 =

⎛⎝⎛⎛ eτ ζ−τζ¯ 0

0 e−(τ ζ−τζ¯ )

⎞⎠⎞⎞

where τ is a non-zero complex number, then

A = g0ζg−10 =

⎛⎝⎛⎛ τ 0

0 −τ

⎞⎠⎞⎞ , B = g0ζg−10 =

⎛⎝⎛⎛ −τ 0

0 τ

⎞⎠⎞⎞ .

The solution of the Lax pair (5.64) is

Φ0 =

⎛⎝⎛⎛ l(λ) 0

0 l−1(λ)

⎞⎠⎞⎞ (5.94)

wherel(λ) = exp

(λτ ζ − λ

2λ − 1τζ¯). (5.95)

Take a complex number ω1 satisfying |2ω1 − 1| = 1. Let

ω2 =ω1

2ω1 − 1,

thenl−1(ω1) = l(ω2), l−1(ω2) = l(ω1). (5.96)

Let

H =

⎛⎝⎛⎛ l(ω1) −al¯ (ω2)

al−1(ω1) l−1(ω2)

⎞⎠⎞⎞ , (5.97)

thendet H = |l(ω1)|2 + |a|2|l(ω2)|−2. (5.98)

By (5.72),

S =1

er + |a|2e−r

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜er

ω1+ |a|2 e−r

ω2

( 1ω1

− 1ω2

)aeis

( 1ω1

− 1ω2

)ae−is er

ω2+ |a|2 e−r

ω1

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (5.99)


wherer = (ω1 − ω2)τ ζ + (ω1 − ω2)τ ζ,¯

s = − i((ω1 + ω2)τ ζ − (ω1 + ω2)τ ζ¯ )(5.100)

are real valued functions. |a|2, a and a can all be eliminated by addingsome constants to r and s. Therefore, we can suppose a = 1. Theharmonic map is

g1 = Φ1(1) = (I − S)

⎛⎜⎛⎛⎝⎜⎜ω1

1 − ω10

0ω2

1 − ω2

⎞⎟⎞⎞⎠⎟⎟ g0(ζ, ζ)

=

⎛⎝⎛⎛ α β

−β α

⎞⎠⎞⎞ ,

(5.101)

where

α =(

er +1γ1

e−r)

eτ ζ−τζ¯ , β =ω2 − ω1

ω1(1 − ω2)eis−(τ ζ−τζ¯ ), (5.102)

γ1 is defined by (5.92). The asymptotic behavior of g1 is similar to thatin R1,1 case. Moreover, we can also construct multi-soliton solutionsexplicitly, and can prove that a k-soliton solution is asymptotic to ksingle solitons as y → ±∞. The proof is similar to the R1,1 case and isomitted here.

5.4.3 UnitonUniton is a special harmonic map from R2 (or a region Ω ⊂ R2) to

U(N) and was introduced by K. Uhlenbeck in [102]. It is an importantnotion because each harmonic map from S2 to U(N) is a uniton. In orderto be compatible with [102], we first change some parameters which areused before.

For Lax pair (5.64), let µ = 1 − 2λ, then

λ =1 − µ

2,

λ

2λ − 1=

1 − µ−1

2. (5.103)

LetΦ(λ) = Φ

(1 − µ

2

)= Ψ(µ), (5.104)

(5.64) becomes

Ψζ =1 − µ

2AΨ, Ψζ =

1 − µ−1

2BΨ. (5.105)


Definition 5.11 Suppose g is a harmonic map from R2 (or Ω) toU(N). If its extended solution Ψ(µ) satisfies the following four con-ditions:

(a) Ψ(µ) =n∑

a=0

TaTT µa (a polynomial of µ), (5.106)

(b) Ψ(1) = I, (5.107)(c) Ψ(−1) = g, (5.108)(d) Ψ(µ)∗ = Ψ(µ−1)−1 (µ = 0) , (5.109)

then g is called a uniton, and Ψ(µ) is called an extended solution ofuniton.

The conditions (b), (c) and (d) above are constraints to the initialvalues. (b) means that Φ(µ)|0 = I for µ = 1. (c) holds if Ψ(−1)|0 ∈U(N). (d) holds everywhere if it holds at z = 0. The last statementcan be proved by differentiating Ψ(µ)∗Ψ(µ−1) and using the Lax pairtogether with the relation A∗ = −B. (a) is an essential condition, basedon the conditions (b), (c) and (d).

For a uniton, the extended solution Ψ(µ) and its degree are not unique.The minimum degree of the extended solutions is called the degree ofthe uniton, or uniton number [102].

Theorem 5.12 Suppose an N × N matrix valued polynomial

Ψ(µ) =n∑

a=0

TaTT µa

satisfies the conditions (b) and (d), then Ψ(µ) can be written as

Ψ(µ) = (π1 + µπ⊥1 ) · · · (πn + µπ⊥

n ), (5.110)

where πa (a = 1, · · · , n) are Hermitian projections, and π⊥a ’s are their

orthogonal complements.

Proof. First, we consider the case n = 1, i.e.,

Ψ(µ) = T0TT + T1TT µ.

By Ψ(1) = I, T0TT + T1TT = I. The condition (d) is

(T0TT + (I − T0TT )µ)∗(T0TT + (I − T0TT )µ−1) = I,


which leads toT ∗

0TT (I − T0TT ) = (I − T0TT )∗T0TT = 0,

T ∗0TT T0TT + (I − T ∗

0TT )(I − T0TT ) = I.

HenceT ∗

0TT T0TT = T ∗0TT = T0TT .

This means that T0TT is the Hermitian projective operator π0 and T1TT =I − T0TT = π⊥.

Now we use mathematical induction to prove the general result. As-sume (5.110) is true for n, and

Ψ(µ) =n+1∑a=0

TaTT µa

satisfies conditions (b) and (d). Without loss of generality, supposeT0TT = 0, TnTT +1 = 0. (Otherwise, the degree of Ψ( µ) is less than n + 1 orΨ(µ) can be written as µΨ1(µ) with deg(Ψ1(µ)) ≤ n. The assumption ofinduction implies that the theorem is true.) det Ψ(µ) is a polynomial ofµ. If it has a non-zero root µ = µ0 = 0, then Ψ( µ0) is degenerate, whichis contradict to condition (d). Hence detΨ(µ) = 0 only when µ = 0.It follows that det T0TT = 0 and T0TT l = 0 has non-zero solutions for l. Allthe solutions l of T0TT l = 0 form a proper subspace P . Let π⊥

n+1 be theHermitian projective operator to P . It is non-trivial and its orthogonalcomplement is denoted by πn+1 (πn+1 = 0, π⊥

n+1 = 0).Let

Ψ(µ) = Ψ(µ)(πn+1 + µ−1π⊥n+1).

Since T0TT π⊥n+1 = 0, Ψ(µ) is a polynomial of µ of degree n and condition

(b) holds. The condition (d) holds for Ψ(µ) too, since

Ψ(µ−1)Ψ(µ)∗ = Ψ(µ−1)(πn+1 + µπ⊥n+1)(πn+1 + µ−1π⊥

n+1)Ψ(µ)∗

= Ψ(µ−1)Ψ(µ)∗ = I.

By the assumption of induction, Ψ(µ) =n∏

a=1

(πa + µπ⊥a ), hence

Ψ(µ) =n+1∏a=1

(πa + µπ⊥a ).

The theorem is proved.This theorem shows that the extended solution of uniton can be ex-

pressed as a product (5.110). However, this decomposition is point-wise.


We have not proved the smoothness of πa and π⊥a . This will be obtained

later.In [102], the uniton of degree one is constructed as follows.Suppose Ψ(µ) = π + µπ⊥ is an extended solution of uniton. Let

Ψζ(µ)Ψ−1(µ) = (1 − µ)πζ(π + µ−1π⊥)

= (1 − µ)πζπ +1 − µ

µπζπ

⊥,

Ψζ(µ)Ψ−1(µ) = (1 − µ)πζ(π + µ−1π⊥)

= (1 − µ−1)(−πζπ⊥) + (1 − µ)πζπ.

Comparing with the Lax pair (5.105), we have

A = 2πζπ, B = −2πζπ⊥ (5.111)

and the conditions for π

πζπ⊥ = 0, πζπ = 0. (5.112)

The last two conditions are equivalent, since

πζπ = (ππζ)∗ = (−ππ⊥

ζ )∗ = (πζπ⊥)∗.

From π⊥π = 0, we have 0 = (π⊥π)ζ = π⊥πζ − πζπ. Hence (5.112) isequivalent to

π⊥πζ = 0. (5.112)′

The projective operator π satisfying this condition is constructed asfollows.

If u1, · · · , uk are k linearly independent vector functions satisfying

uaζ =∑

cabub, (5.113)

then for any invertible linear combination

va =∑

qabub (det(qab) = 0) ,

whose coefficients are arbitrary functions, vaζ ’s are also linear combi-nations of v1, · · · , vk. Now suppose v1, · · · , vk are Schmidt orthogonal-ization of u1, · · · , uk, i.e., they are linear combinations of u1, · · ·, uk

satisfyingv∗bva = δba, (5.114)

thenπ =

∑vav

∗a (5.115)


is an Hermitian projective operator of rank k:

π∗ = π, ππ = π, (5.116)

and v1, · · ·, vk (also u1, · · ·, uk) are its invariant vectors. Moreover, from(5.113), vaζ =

∑cabvb where cab’s are suitable functions, and

πζ =∑

cabvbv∗a +

∑vav

∗aζ .

From π⊥va = (I − π)va = va − va = 0 we know that (5.112)′ holds.Conversely, suppose π is an Hermitian projective operator of rank k

satisfying (5.112)′, then for any k linearly independent invariant vectorsua (k = 1, · · · , k),

πua = ua, π⊥ua = 0.

By (5.112)′,

π⊥uaζ = π⊥(πua)ζ = π⊥πζua + π⊥πuaζ = 0.

Hence uaζ is a linear combination of u1, · · ·, uk.Moreover, if ua’s satisfy (5.113), then there exist linearly independent

vectors w1, · · ·, wk which are linear combinations of ua’s and satisfy

waζ = 0.

In fact, this can be done as follows. Let U = (u1, · · · , uk), C = (cab),then U satisfies

UζUU = UC.

Let W = (w1, · · · , wk) = UB where B is a k × k matrix, then

WζWW = UζUU B + UBζ = U(CB + Bζ).

B can be determined by solving the linear equation

Bζ + CB = 0.

Therefore, to construct π, we only need a set of linearly indepen-dent vectors wα whose entries are holomorphic functions of ζ (i.e. anti-holomorphic functions of ζ). This gives the construction of unitons ofdegree one.

Remark 43 Let the set of all k dimensional subspaces of CN be attachedat each point of R2 (or Ω). Then we get a k dimensional plane bundleon R2. A section of this plane bundle is to assign a k dimensionalsubspace at each point of R2 (or Ω). It can also be obtained by assigningk dimensional Hermitian projective operator π = π(ζ, ζ) at each point.


In [102], the section was called holomorphic if π satisfies the conditionπ⊥πζ = 0. Because of the difference of the symbols (See Remark 41), thecondition here should be πζπ

⊥ = 0, or equivalently, π⊥πζ = 0, i.e., thesection is anti-holomorphic. As mentioned above, this kind of sectioncan be constructed by k linearly independent anti-holomorphic vectors.

Example 5.13 Take N = 2. Let f , g be two holomorphic functions ofζ without common zero. Let

u =1

(|f |2 + |g|2)1/2

⎛⎝⎛⎛ f

g

⎞⎠⎞⎞ ,

then

π =1

|f |2 + |g|2

⎛⎝⎛⎛ f

g

⎞⎠⎞⎞ (f , g) =1

|f |2 + |g|2

⎛⎝⎛⎛ |f |2 fg

gf |g|2

⎞⎠⎞⎞ ,

π⊥ = I − π =1

|f |2 + |g|2

⎛⎝⎛⎛ |g|2 −fg

−gf |f |2

⎞⎠⎞⎞ .

The expression of the uniton is

g = π − π⊥ =1

|f |2 + |g|2

⎛⎝⎛⎛ |f |2 − |g|2 2fg

2gf |g|2 − |f |2

⎞⎠⎞⎞ ,

and the extended solution is

Ψ(µ) =1

|f |2 + |g|2

⎛⎝⎛⎛ |f |2 + µ|g|2 (1 − µ)fg

(1 − µ)gf |g|2 + µ|f |2

⎞⎠⎞⎞ .

5.4.4 Darboux transformation and singularDarboux transformation for unitons

We construct the Darboux matrix by using the algorithm in Sec-tion 5.1.

Suppose g is a uniton and Ψ(µ) is its extended solution satisfying thenormalizing condition: Ψ∗(µ−1) = Ψ(µ)−1. Let

ω1 =1 − ε

2, ω2 =

1 − ε−1

2. (5.117)


This is equivalent to putting µ to be ε. Choose the basis of Cn such that

L1 =(

l1 · · · lk

)=

⎛⎝⎛⎛ IkI

0

⎞⎠⎞⎞ ,

L⊥1 = L2 =

(lk+1 · · · lN

)=

⎛⎝⎛⎛ 0

INI −k

⎞⎠⎞⎞ .

Then

HεHH =(

Ψ(ε)L1 Ψ( −1)L2

),

πε = HεHH

⎛⎝⎛⎛ 0 0

0 INI −k

⎞⎠⎞⎞H−1εH , π⊥

ε = HεHH

⎛⎝⎛⎛ IkI 0

0 0

⎞⎠⎞⎞H−1εH .

(5.118)

The Darboux transformation of the extended uniton Ψ(µ) is

Ψ1(µ) = (πε + γπ⊥ε )Ψ(µ)(σ + γ−1σ⊥), (5.119)

where (πε + γπ⊥ε ) is the Darboux matrix, (σ + γ−1σ⊥) is an additional

factor in which σ and σ⊥ are the constant Hermitian projective operatorsto L⊥

1 and L1 respectively. Moreover,

γ = γε(µ) =ω2

(ω1 − 1−µ

2

)ω1

(ω2 − 1−µ

2

) =(µ − ε)(¯− 1)( µ − 1)(1 − ε)

. (5.120)

Clearly,

γε(µ−1) =(µ−1 − ε)(¯− 1)(¯µ−1 − 1)(1 − ε)

=(1 − εµ)(¯− 1)(¯− µ)(1 − ε)

= (γε(µ))−1. (5.121)

Hence

Ψ1(µ−1) = (πε + γ−1π⊥ε )Ψ(µ−1)(σ + γσ⊥),

Ψ∗1(µ

−1)−1 = (πε + γπ⊥ε )Ψ(µ)(σ + γ−1σ⊥) = Ψ1(µ).

(5.122)

This means that Ψ1 also satisfies the normalizing condition.Now consider

Ψ1(µ) = πεΨ(µ)σ + πεΨ(µ)γ−1(µ)σ⊥ + π⊥ε γ(µ)Ψ(µ)σ + π⊥

ε Ψ(µ)σ⊥.(5.123)

On the right hand side, the first and fourth terms are polynomials of µ,the second and third terms are rational functions of µ. µ = ε might be apole of the second term and µ = ε−1 might be a pole of the third term.


However, we can prove that the denominators µ− ε and εµ− 1 in thesetwo terms can be cancelled with the corresponding enumerators. Henceall the terms are polynomials of µ. This fact is proved as follows.

Notice that

H−1εH =

⎛⎝⎛⎛ C1(ε)L∗1Ψ(ε)∗

C2CC (ε)L∗2Ψ( −1)∗

⎞⎠⎞⎞ , (5.124)

where C1(ε) is a k× k matrix and C2CC (ε) is an (N − k)× (N − k) matrix,which are determined by

C1(ε)L∗1Ψ(ε)∗Ψ(ε)L1 = IkI ,

C2CC (ε)L∗2Ψ( −1)∗Ψ( −1)L2 = INI −k

(5.125)

respectively. In (5.125), L∗1Ψ(ε)∗Ψ(ε)L1 is invertible. In fact, suppose

l is a k dimensional vector such that L∗1Ψ(ε)∗Ψ(ε)L1l = 0, then from

l∗L∗1Ψ(ε)∗Ψ(ε)L1l = 0 we obtain Ψ(ε)L1l = 0. It follows that L1l = 0

and hence l = 0. Consequently, C1(ε) is determined by (5.125) uniquely.Similarly, C2CC (ε) also exists uniquely. Therefore,

πε = HεHH

⎛⎝⎛⎛ 0 0

0 INI −k

⎞⎠⎞⎞H−1εH = Ψ( −1)L2C2CC (ε)L∗

2Ψ( −1)∗,

π⊥ε = HεHH

⎛⎝⎛⎛ IkI 0

0 0

⎞⎠⎞⎞H−1εH = Ψ(ε)L1C1(ε)L∗

1Ψ(ε)∗,

(5.126)

and

πεΨ(µ)γ−1(µ)σ⊥ = Ψ( −1)L2C2CC (ε)L∗2Ψ( −1)∗Ψ(µ)γ−1σ⊥. (5.127)

Denote the right hand side of the last equation by γ−1F (µ). Whenµ = ε,

F (ε) = Ψ( −1)L2C2CC (ε)L∗2Ψ( −1)∗Ψ(ε)σ⊥. (5.128)

Using the normalizing condition Ψ( −1)∗Ψ(ε) = I and

σ⊥ =

⎛⎝⎛⎛ IkI 0

0 0

⎞⎠⎞⎞ ,

we get F (ε) = 0. This means that µ = ε is actually not a pole of thesecond term of (5.123). Similar conclusion holds for the third term.

Thus, we have the following theorem.

Theorem 5.14 A Darboux transformation with right-multiplied nor-malizing factor (σ + µ−1σ⊥) transforms an extended solution Ψ(µ) to


an extended solution Ψ1(µ). The degree of Ψ1(µ) with respect to µ can-not exceed the degree of Ψ(µ) with respect to µ.

We have mentioned before that the extended solutions of a uniton arenot unique, and their degrees are not definite either. However, the unitonnumber is defined as the minimum of the degrees of all possible extendedsolutions. The above theorem implies that a Darboux transformationcannot increase uniton number.

In [102], singular Backlund transformation was introduced. For a¨set of usual Backlund transformations with parameter¨ ε, let ε → 0,then the derived transformation is called a singular Backlund trans-¨formation. Since Backlund transformation can be solved explicitly by¨Darboux transformation, singular Darboux transformation (limit of Dar-boux transformations with parameter ε as ε → 0) should give singularBacklund transformation explicitly. Clearly, when¨ ε → 0, γε → µ. Henceif πε → π, π⊥

ε → π⊥, the Darboux transformations will converge to asingular Darboux transformation

Ψ1(µ) = (π + µπ⊥)Ψ(µ).

Here the factor (σ + µ−1σ⊥) is neglected.

Lemma 5.15 Suppose Ψ(µ) is an extended solution of a uniton, thenΨ1(µ) = (π + µπ⊥)Ψ(µ) is an extended solution of a uniton if and onlyif

(2πζ + πA)π⊥ = 0, (5.129)

π⊥Aπ = 0. (5.130)

Proof. By direct calculation,

Ψ1ζΨ−11 =

1 − µ

2(2πζπ + πAπ + π⊥Aπ⊥)

+1 − µ

2µ(2πζπ

⊥ + πAπ⊥) +µ(1 − µ)

2π⊥Aπ

=1 − µ

2A1.

This leads to (5.129) and (5.130).Alternatively,

Ψ1ζΨ−11 =

1 − µ−1

2(−2πζπ

⊥ + πBπ + π⊥Bπ⊥)

+1 − µ

2(2πζπ − π⊥Bπ) +

1 − µ−1

2µπBπ⊥

=1 − µ−1

2B1.


Hence we have2πζπ − π⊥Bπ = 0, (5.129)′

πBπ⊥ = 0. (5.130)′

However, (5.129)′ and (5.130)′ are equivalent to (5.129) and (5.130) re-spectively. In fact, complex conjugate of (5.130) is (5.130)′. Taking thecomplex conjugate of (5.129)′, we have

0 = 2ππζ + πAπ⊥ = −2ππ⊥ζ + πAπ⊥ = 2πζπ

⊥ + πAπ⊥.

This is just (5.129). The lemma is proved.

Remark 44 Let ε → 0 in the Backlund transformation (5.90), then weöbtain (5.129) and (5.130). Hence (5.129) and (5.130) are called theequations of singular Backlund transformation in [102].

Notice that when ε → 0, C1(ε) and C2CC (ε) may not converge, andL∗

1Ψ(ε)∗Ψ(ε)L1 and L∗2Ψ( −1)∗Ψ( −1)L2 may not converge to invertible

matrices. However, by the following method we can eliminate the termwhich tends to infinity as ε → 0 so that πε tends to π. This method canbe regarded as a method of renormalization [49].

The rank of Ψ(ε)L1 is k. From the second equation of (5.126) and thefirst equation of (5.125), we have

π⊥ε Ψ(ε)L1 = Ψ(ε)L1.

Hence each column of Ψ(ε)L1 is an invariant vector of π⊥ε and its entries

are polynomials of ε whose degrees do not exceed n. Consider the termsof the zeroth order of ε in the columns of Ψ(ε)L1 and choose a maximallinearly independent set in it. The corresponding column vectors ofΨ(ε)L1 constitute an ε-dependent matrix

χ0 = χ00 + χ0

1ε + · · · + χ0nεn,

where χ00 is of full rank. The coefficients of ε0 in other columns of

Ψ(ε)L1 are linear combinations of the columns in χ00. Subtract these

other columns by suitable linear combination of the columns in χ0 sothat all the coefficients of ε0 are eliminated. After divided by ε, weobtain a set of columns in the form

b + b1ε + · · · + bn−1εn−1.

Combining these columns to the matrix χ0, we obtain a matrix M1MM (ε).Evidently, the columns of M1MM span the same bundle of subspaces PεPP over


R2 (or Ω) as the columns of Ψ(ε)L1. We can transform the matrix M1MM (ε)to M2MM (ε) in the same way and so on. Finally we obtain a matrix

M(ε) = τ0ττ + τ1ττ ε + · · · + τnττ εn (5.131)

such that τ0ττ is of rank k. In fact, if rankτ0ττ < k, we can make the abovetransformation further, since the columns of M span the bundle PεPP .

LetWεWW = M(ε)(M∗(ε)M(ε))−1M∗(ε). (5.132)

It is easily seen that WεWW is Hermitian and W 2εWW = WεWW . Moreover, WεWW M(ε)

= M(ε), i.e., W (ε) is the Hermitian projective operator to PεPP . Hence

WεWW = π⊥ε . (5.133)

Let ε → 0, we have π⊥ε → π⊥ with

π⊥ = τ0ττ (τ∗0ττ τ0ττ )−1τ∗

0ττ (5.134)

and πε → π = I − π⊥.Thus we have proved the convergence of πε (and τ⊥

εττ ) as ε → 0 andobtained the explicit formulae for π⊥ and π. The convergence of πε,ζ

and πε,ζ as ε → 0 can be obtained from some properties of harmonicmaps (see [50]).

We have the following theorem.

Theorem 5.16 Let π⊥ be defined by (5.134), then

Ψ1 = (π + µπ⊥)Ψ (5.135)

is a new extended solution of uniton. Moreover, π and π⊥ satisfy (5.129)and (5.130).

Thus we have established the singular Darboux transformation whichrealizes the singular Backlund transformation. According to the Theo-¨rem 5.14, the extended solution of uniton

(π + µπ⊥)Ψ(σ + µ−1σ⊥) (5.136)

is of degree ≤ n. Hence the uniton number cannot be increased bysingular Backlund transformation.¨

We shall take n = 1 as the example to see the limiting process of πε

and π⊥ε as ε → 0.

When n = 1,Ψ(µ) = π1 + µπ⊥

1 ,


πε = (π1 + −1π⊥1 )L2C2CC (ε)L∗

2(π1 + ε−1π⊥1 ), (5.137)

π⊥ε = (π1 + επ⊥

1 )L1C1(ε)L∗1(π1 + π⊥

1 ), (5.138)

C1(ε)L∗1(π1 + |ε|2π⊥

1 )L1 = IkI , (5.139)C2CC (ε)L∗

2(π1 + |ε|−2π⊥1 )L2 = INI −k. (5.140)

The existence of the limits of πε and π⊥ε as ε → 0 is discussed as

follows. By using a transformation of the basis, we have

L1 =

⎛⎝⎛⎛ IkI

0

⎞⎠⎞⎞ , L2 =

⎛⎝⎛⎛ 0

INI −k

⎞⎠⎞⎞ . (5.141)

Let

π1 =

⎛⎝⎛⎛ π11 π12

π21 π22

⎞⎠⎞⎞ (5.142)

as a block matrix where π11, π12, π21, π22 are k × k, k × (N − k),(N − k)× k, (N − k)× (N − k) matrices respectively. Since π∗

1 = π1, wehave

π∗11 = π11, π∗

12 = π21, π∗21 = π12, π∗

22 = π22. (5.143)

From π21 = π1,

π211 + π12π21 = π11, π21π11 + π22π21 = π21,

π11π12 + π12π22 = π12, π21π12 + π222 = π22.

(5.144)

Since π11 is an Hermitian matrix, there exists a k × k unitary matrix βsuch that

π11 = β diag(0, · · · , 0, λr+1, · · · , λk)β∗ (λr+1, · · · , λk = 0) . (5.145)

Hence (5.139) becomes

C1(ε)β(

diag(0, · · · , 0, λr+1, · · · , λk)

+|ε|2 diag(1, · · · , 1, 1 − λr+1, · · · , 1 − λk))β∗ = IkI ,

or equivalently,

C1(ε) = β

(diag

(|ε|−2, · · · , |ε|−2,

1λr+1 + |ε|2(1 − λr+1)

, · · · , 1λk + |ε|2(1 − λk)

))β∗,

(5.146)


From (5.126),

π⊥ε =

⎛⎝⎛⎛ π11 + ε(1 − π11)

π21 − επ21

⎞⎠⎞⎞C1(ε)(π11 + (1− π11), π12 − επ12). (5.147)

From (5.145),π11β diag(IrII , 0)β∗ = 0. (5.148)

On the other hand, suppose a k dimensional vector l satisfies π11l = 0,then by (5.144), π12π21l = 0. Hence

(π21l)∗(π21l) = l∗π12π21l = l∗(π11 − π211)l = 0,

which implies π21l = 0. Therefore,

π21β diag(IrII , 0, · · · , 0) = 0. (5.149)

Substituting these relations into (5.147) and letting ε → 0, we get

limε→0

π⊥ε = π⊥

=

⎛⎝⎛⎛ π11

π21

⎞⎠⎞⎞β diag(0, · · · , 0,

1λr+1

, · · · , 1λk

)β∗ ( π11 π12

)

+

⎛⎝⎛⎛ β diag(IrII , 0)β∗ 0

0 0

⎞⎠⎞⎞ ,

(5.150)and

limε→0

πε = π =

⎛⎝⎛⎛ −π12

1 − π22

⎞⎠⎞⎞α

· diag( 1λk+1

, · · · , 1λk+s

, 0, · · · , 0)α∗ ( −π21 1 − π22

)+

⎛⎝⎛⎛ 0 0

0 α diag(0, INI −k−s)α∗

⎞⎠⎞⎞ .

(5.151)

Here α is a suitable (N − k) × (N − k) unitary matrix which can makeπ22 to be diagonal.

Finally, we prove the following theorem.

Theorem 5.17 (Theorem of factorization) Any extended solution ofuniton Ψ(µ) of degree n can be factorized as

Ψ(µ) = (πn + µπ⊥n )(πn−1 + µπ⊥

n−1) · · · (π1 + µπ⊥1 )C0CC , (5.152)


where πi and π⊥i = I − πi (i = 1, 2, · · · , n) are Hermitian projective

operators, being analytic functions of (x, y), C0CC is a constant matrix inU(N) which may be chosen as I.

Before proving this theorem, we first give the following two lemmas.

Lemma 5.18 Suppose Ψ(µ) is an extended solution of uniton of degreen, then there exists an Hermite projective operator π such that

Ψ1(µ) = (π + µπ⊥)Ψ(µ)

is an extended solution of uniton defined on a dense open subset of Ωand the degree of Ψ1(µ) is at most n.

Proof. Suppose

Ψ(µ) = T0TT + T1TT µ + · · · + TnTT µn.

Since the equations for harmonic map is elliptic and U(N) is a real ana-lytic manifold, A and B are both analytic functions of the real variablesx and y, so is T0TT . Suppose the maximum rank of T0TT is k, then the pointsat which the rank of T0TT is k form a dense open subset Ω1 of Ω. More-over, suppose L1 is a constant N × (k + a) matrix (a ≥ 0) such that themaximum rank of T0TT L1 is also k, then the points at with the rank ofT0TT L1 is k also form a dense open subset Ω2 of Ω. We use L = (L1, L2)to construct the Darboux transformation. From (5.125) and (5.126),

π⊥ε = Ψ(ε)L1C1(ε)L∗

1Ψ(ε)∗,

C1(ε)L∗1Ψ1(ε)∗Ψ1(ε)L1 = IkI .

When ε → 0,C1(0) = (L∗

1T∗0TT T0TT L1)−1,

π⊥ = T0TT L1C(0)L−11 T ∗

0TT

holds in Ω2. By Ψ(µ)∗Ψ(µ−1) = I, we have

T ∗0TT TnTT = 0.

Hence π⊥TnTT = 0, and the degree of Ψ1(µ) on Ω2 does not exceed n. Thelemma is proved.

Lemma 5.19 Suppose Ψ(µ) is an extended solution of uniton of degreen, then there exists an Hermitian projective operator τ defined on adense open subset of Ω such that

Ψ(µ) = (τ + µτ⊥)Ψ−1(µ),


where Ψ−1(µ) is an extended solution of uniton whose degree does notexceed n − 1.

Proof. Take τ as π⊥ in Lemma 5.18, then τ⊥ = π. Let

Ψ−1(µ) =(τ⊥ +

1µ

τ)Ψ(µ),

then

Ψ1(µ) = (π + µπ⊥)Ψ(µ) = (π + µπ⊥)(π⊥ + µπ)Ψ−1(µ) = µΨ−1(µ).

Hence the degree of Ψ−1(µ) does not exceed n − 1.On the other hand, it can be verified directly that Ψ−1(µ) is an ex-

tended solution of uniton if and only if

τ⊥Aτ − 2τ⊥τζττ = 0,

τAτ⊥ = 0.

Since τ = π⊥, τ⊥ = π, these are just the equations (5.129) and (5.130)for π and π⊥. The lemma is proved.

Proof of Theorem 5.17. Use mathematical induction. Suppose thetheorem is true for all extended solutions of uniton of degree n − 1.According to Lemma 5.19,

Ψ(µ) = (τnττ + µτ⊥nττ )Ψ−1(µ)

holds in a dense open subset. By the assumption of induction,

Ψ(µ) = (πn + µπ⊥n )(πn−1 + µπ⊥

n−1) · · · (π1 + µπ⊥1 )

holds in a dense open subset where τnττ and τ⊥nττ are denoted by πn and

π⊥n . The left hand side is an analytic function of x and y on Ω, and the

right hand side is an analytic function of x and y in a dense open subsetof Ω. By analytic continuation, we know that the above equality holdsin whole Ω. The theorem is proved.

Remark 45 This theorem was obtained in [102]. Here the constructionof factorization is more general. Especially, the rank of each πi is notfixed. The only restriction is 1 ≤ rankπi ≤ N − ki where ki is given byLemma 5.18. Since singular Darboux transformation can be constructedalgebraically, the factorization here can also be realized algebraically.

Take µ = −1, then Ψ(−1) can be factorized as

Ψ(−1) = (πn − π⊥n )(πn−1 − π⊥

n−1) · · · (π1 − π⊥1 ).


Moreover, the unitons to the Grassmannian have the similar factoriza-tion (see [44]).

Remark 46 In the proof of Lemma 5.19, π and π⊥ (or τ⊥ and τ) areconstructed from Ψ(µ). But we have not constructed them from Ψ−1(µ)algebraically. There was some negligence in [49]. It is still not clearwhether one can obtain all unitons by purely algebraic algorithm fromone unitons. Partial solution has been obtained in [55].

Chapter 6

GENERALIZED SELF-DUALYANG-MILLS EQUATIONSAND YANG-MILLS-HIGGS EQUATIONS

6.1 Generalized self-dual Yang-Mills flowYang-Mills equations are one of the most important equations in the-

oretical physics to describe the fundamental interactions in nature. Self-dual Yang-Mills equations are special case of the Yang-Mills equations.They have also great significance in differential topology. Moreover, alot of soliton equations can be reduced from self-dual Yang-Mills equa-tions. Instead of the general theory of Yang-Mills fields, here we onlyconsider a kind of their generalization in the point view of soliton theory.The Darboux transformation can be applied to this generalized self-dualYang-Mills flow [40, 41]. Furthermore, a reduction of the generalizedself-dual Yang-Mills flow leads to the AKNS system.

6.1.1 Generalized self-dual Yang-Mills flowFirst we introduce briefly the self-dual Yang-Mills fields.Let G be a matrix Lie group, g be its Lie algebra. The Euclidean

space R4 has metric

ds2 = ηijdxi dxj (with summation convention, ηij = δij). (6.1)

Gauge potential is a set of functions Ai’s valued in the Lie algebra g.The strength of the gauge field is the anti-symmetric tensor

FijFF = ∂i∂∂ Aj − ∂j∂ Ai + [Ai, Aj ]. (6.2)

A Yang-Mills field is defined by the gauge potential satisfying the Yang-Mills equations

ηjk(∂FijFF

∂xk+ [Ak, FijFF ]

)= 0. (6.3)


For any G-valued function G(x), the transformation

A′i = G(x)AiG(x)−1 + (∂i∂∂ G(x))G(x)−1

is called a gauge transformation. A gauge transformation does notchange the physical essence of the gauge field. Especially, it does notchange the Yang-Mills equations.

The dual of the gauge intensity F is a tensor ∗F defined by

(∗F )ij =12εijklη

kaηlbFabFF , (6.4)

where

εijkl =

⎧⎪⎧⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎨⎨⎪⎪⎪⎪⎪⎪⎩⎪⎪1,

−1,

0,

(i, j, k, l) is an even permutation of (1, 2, 3, 4)

(i, j, k, l) is an odd permutation of (1, 2, 3, 4)

otherwise,

(6.5)

that is,

(∗F )12 = F34FF , (∗F )23 = F14FF , (∗F )31 = F24FF ,

(∗F )34 = F12FF , (∗F )14 = F23FF , (∗F )24 = F31FF .(6.6)

If ∗F = F , i.e.,

F12FF = F34FF , F23FF = F14FF , F31FF = F24FF , (6.7)

this gauge field is called a self-dual Yang-Mills field. An self-dual Yang-Mills field always satisfies the Yang-Mills equation. Similarly, the gaugefield satisfying ∗F = −F is called anti-self-dual. The self-duality andanti-self-duality can be interchanged by changing the orientation of thespace-time. Therefore, it is enough for us to discuss only the self-dualYang-Mills field.

The self-dual Yang-Mills field on R4 can be complexified to the gaugefield on C4. Besides, R4 can be replaced by R2,2 with the metric

ds2 = dx21 + dx2

2 − dx23 − dx2

4. (6.8)

In [98], the self-duality was extended to R4n. Here we will extend it toR2n.

Suppose (x, p) = (x1, · · · , xn; p1, · · · , pn) are the real coordinates ofR2n (n ≥ 2), Ai(x1, · · · , xn, p1, · · · , pn)’s are a set of N × N (real orcomplex) matrix functions of (x, p). Consider the Lax set

LiΨ ≡(

∂

∂pi− λ

∂

∂xi

)Ψ = −AiΨ, (6.9)

Generalized self-dual Yang-Mills and Yang-Mills-Higgs equations 239

where Ψ is an N ×N matrix function, λ is the spectral parameter which

appears as the coefficient of∂

∂xi.

The integrability conditions of (6.9) are

∂Ai

∂pj− ∂Aj

∂pi− [Ai, Aj ] = 0 (6.10)

and∂Ai

∂xj− ∂Aj

∂xi= 0. (6.11)

When n = 2, this is just the self-dual Yang-Mills equation [116]. In[98], self-dual Yang-Mills field was generalized to the current form with“space” dimension 4n (n ≥ 1). We call Ai satisfying (6.10) and (6.11)a generalized Yang-Mills potential.

Remark 47 We have mentioned that the self-duality depends on the met-ric of the space. For four dimensional space, here we only consider R2,2,

with the metricds2 = dpd 1 dx1 + dpd 2 dx2.

From (6.10), there exists an N × N matrix J such that

Ai = − ∂J

∂piJ−1. (6.12)

Hence, (6.11) becomes

∂

∂xi

(∂J

∂pjJ−1

)− ∂

∂xj

(∂J

∂piJ−1

)= 0. (6.13)

Clearly, (6.13) and (6.12) are equivalent to (6.10) and (6.11).We shall consider a flow of the generalized self-dual Yang-Mills field

by introducing a “time” variable so that Ai’s, J and Ψ depend on t aswell as the “space” variables (x, p). Moreover, suppose that Ψ satisfiesthe evolution equation

∂Ψ∂t

= V Ψ =m+q∑a=0

VaVV λm−aΨ (q ≥ 0), (6.14)

where VaVV ’s are N ×N -matrix valued functions which are independent ofthe spectral parameter λ.


(6.14) and (6.9) form a Lax set in R2n+1. Its integrability conditionsconsist of the recursive relations among VaVV ’s

∂V0VV

∂xi= 0,

∂VaVV

∂xi=

∂VaVV −1

∂pi+ [Ai, VaVV −1], (a = 1, 2, · · · , m),

(6.15)

and∂VmVV +q

∂pi= [VmVV +q, Ai],

∂VaVV −1

∂pi=

∂VaVV

∂xi+ [VaVV −1, Ai], (a = m + q, · · · , m + 2)

(6.16)

together with the evolution equations

∂Ai

∂t+

∂VmVV

∂pi− ∂VmVV +1

∂xi+ [Ai, VmVV ] = 0. (6.17)

A solution of (6.10), (6.11) and (6.17) is called a generalized self-dualYang-Mills flow. The gauge potentials Ai’s satisfy not only the self-dualYang-Mills equation, but also an additional evolution equation (6.17).

Remark 48 There is no direct physical meaning of the generalized self-dual Yang-Mills flow. However, it contains a one parametric family ofself-dual Yang-Mills solutions and it will be seen later that almost allknown soliton equations are the reductions of the generalized self-dualYang-Mills equation.

Theorem 6.1 If Ai’s satisfy the generalized self-dual Yang-Mills equa-tion (6.10) and (6.11), then the equations (6.16) and (6.17) for ViVV ’s arecompletely integrable and all ViVV ’s are expressed by differentiations andintegrations of Ai’s.

Proof. From the first equation of (6.15),

V0VV = V0VV (p, t). (6.18)

The other equations in (6.15) lead to

∂2VaVV

∂xi∂xj=

∂2VaVV −1

∂pi∂xj+

[∂Ai

∂xj, VaVV −1

]+

[Ai,

∂VaVV −1

∂xj

]

=∂2VaVV −2

∂pi∂pj+[∂Aj

∂pi, VaVV −2

]+[Aj ,

∂VaVV −2

∂pi

]+

[∂Ai

∂xj, VaVV −1

]+

[Ai,

∂VaVV −2

∂pj

]+ [Ai, [Aj , VaVV −2]].

(6.19)


Using (6.10), (6.11) and the Jacobi identity, we know that the righthand side of (6.19) is symmetric for i and j. Hence (6.15) is completelyintegrable. Take pi’s as parameters, VnVV ’s can be determined recursivelyas

V0VV = V 00VV (p, t),

VaVV =∑

i

∫ (x,p,t)

(0

∫∫,p,t)

(∂VaVV −1

∂pi+ [Ai, VaVV −1]

)dxi + V 0

aVV (p, t),(6.20)

where V 0aVV (p, t)’s are integral “constants” and the integrals are taken

along the path in the n-dimensional subspace pi = const. In fact, theseintegrals are independent of the path because of the integrability con-ditions. Moreover, we can choose V 0

aVV ’s as arbitrary functions of p andt. If they are independent of t, then the evolution equations (6.17) area system of integro-differential equations which do not depend on t ex-plicitly.

LetWaWW = J−1VaVV J (a = m + q, · · · , m + 1), (6.21)

then (6.16) implies

∂WaWW −1

∂pi=

∂WaWW

∂xi+[J−1 ∂J

∂xi, WaWW

](a = m + q, · · · , m + 2), (6.22)

and∂WmWW +q

∂pi= 0.

HenceWmWW +q = W 0

mWW +q(x, t).Moreover, according to (6.22),

WaWW −1 =∑

i

∫ (x,p,t)

(

∫∫x,

∫∫0,t)

(∂WaWW

∂xi+[J−1 ∂J

∂xi, WaWW

])dpd i + W 0

aWW −1(x, t)

(a = m + q, · · · , m + 2).(6.23)

The integrals are now taken along a path in the n-dimensional subspacexi = constant, and independent of the path. Here xi’s are regarded asparameters.

Therefore, when Ai’s satisfy (6.10) and (6.11), we can write down VaVV(a = 0, · · · , m; a = m+ q,m+ q−1, · · · , m+1) recursively so that (6.15)and (6.16) are satisfied. The theorem is proved.

(6.10) and (6.11) can be regarded as “spatial constraints” among Ai’s,and (6.17) are their evolution equations. The meaning of “spatial con-straints” is that at any moment Ai’s satisfy the generalized self-dualYang-Mills equation.


6.1.2 Darboux transformationLet D be an N × N matrix

D = λI − S, (6.24)

where S is independent of λ. Let

Ψ′ = DΨ (6.25)

and we want

LiΨ′ =(

∂

∂pi− λ

∂

∂xi

)Ψ′ = −A′

iΨ′,

∂Ψ′

∂t=

m+q∑a=0

V ′aVV λm−aΨ′.

(6.26)

As in Section 3.2, direct calculation implies

A′i = Ai − ∂S

∂xi, (6.27)

∂S

∂pi= SAi − A′

iS = [S, Ai] +∂S

∂xiS, (6.28)

V ′0VV = V0VV ,

V ′aVV = VaVV + V ′

aVV −1S − SVaVV −1 (a = 1, 2, · · · , m),(6.29)

V ′mVV +q = SVmVV +qS

−1,

V ′mVV +k = SVmVV +kS

−1 − (VmVV +k+1 − V ′mVV +k+1)S

−1,

(k = q − 1, · · · , 1),

(6.30)

and∂S

∂t+ SVmVV − V ′

mVV S − VmVV +1 + V ′mVV +1 = 0. (6.31)

Here we assume that S is non-degenerate. It can be verified that (6.28)and (6.31) are completely integrable, that is, for any given initial dataS(t0, p0, x) at t = t0, p = p0, there exists S(t, p, x) satisfying (6.28),(6.31) and the initial condition.

Theorem 6.2 Suppose the matrix S is non-degenerate and satisfies theequations (6.28) and (6.31), then D = λI − S is a Darboux matrix for(6.9) and (6.14).

Now we turn to the explicit construction of the matrix S, which isessentially the same as that for the AKNS system. Take λ1, · · ·, λN to


be N complex numbers such that at least two of them are different. Lethα (α = 1, 2, · · · , N) be a column solution of (6.9) and (6.14) for λ = λα,i.e., hα satisfies

∂hα

∂pi= λα

∂hα

∂xi− Aihα,

∂hα

∂t=

m+q∑a=0

VaVV λm−aα hα. (6.32)

LetH = (h1, h2, · · · , hN ), Λ = diag(λ1, · · · , λN ). (6.33)

Suppose H is non-degenerate. Let

S = HΛH−1. (6.34)

Theorem 6.3 Suppose S is defined by (6.34), then D = λI − S is aDarboux matrix.

Proof. From the first equation of (6.32), we have

∂H

∂pi=

∂H

∂xiΛ − AiH. (6.35)

Hence

∂S

∂pi=

∂H

∂xiΛ2H−1 − HΛH−1 ∂H

∂xiΛH−1 − AiS + SAi,

∂S

∂xi=

∂H

∂xiΛH−1 − HΛH−1 ∂H

∂xiH−1.

(6.36)

Therefore,∂S

∂pi= [S, Ai] +

∂S

∂xiS. (6.37)

This is just (6.28). On the other hand, it is easy to see that

∂H

∂t=

m+q∑a=0

VaVV HΛm−a, (6.38)

which implies

∂S

∂t=

m+q∑a=0

VaVV HΛm−a+1H−1 −m+q∑a=0

HΛH−1VaVV HΛm−aH−1

=m+q∑a=0

VaVV Sm−aS − Sm+q∑a=0

VaVV Sm−a.

(6.39)


From (6.29) and (6.30), we know that the right hand side of (6.39) is−SVmVV +V ′

mVV S+VmVV +1−V ′mVV +1. Hence (6.31) holds. The theorem is proved.

If Ψ(λ) is a fundamental solution of the Lax set, then a column so-lution of the Lax set can be written as Ψ(λ)l where l = l(λx + p) is acolumn and can be an arbitrary function of λx1 + p1, · · · , λxn + pn. It isnoted that in the case of AKNS system, l should be a constant column.Hence there is more freedom for constructing the Darboux matrix in thepresent case.

Thus we have constructed explicitly the Darboux transformation

(Ψ, Ai) → (Ψ′, A′i) (6.40)

provided that the gauge group G is GL(N,C). In the construction, thereare only algebraic and differential operations. Moreover, the Darbouxtransformations can be continued successively with the same algorithm.

In many cases, the gauge potentials Ai’s should belong to certainsubalgebra of gl(N), say Ai ∈ u(N), i.e., A∗

i + Ai = 0. We also wantVaVV ∈ u(N).

In the construction of Darboux transformation, the equalities A′∗i +

A′i = 0 and V ′∗

aVV +V ′aVV = 0 should also be satisfied. This can be guaranteed

by the following two constraints:(i) Take

λα = µ or µ (µ is not real), (6.41)

(ii) Take hα’s such thath∗

αhβ = 0 (6.42)

whenever λα = λβ .In fact, when λα = λβ (i.e. λα = λβ), the Lax set leads to(

∂

∂pi− µ

∂

∂xi

)(h∗

αhβ) = 0,

∂

∂t(h∗

αhβ) = 0.(6.43)

Hence, h∗αhβ is a holomorphic function of xi +µpi and independent of t.

If we take the initial values of hα, hβ such that h∗αhβ = 0 at t = 0 and

xi + µpi = 0, then h∗αhβ = 0 holds everywhere.

Similar to the discussion in Section 1.4, we have

S∗ + S = (µ + µ)I, S∗S = |µ|2.(6.27) implies that A′∗

i + A′i = 0 holds. Moreover, we have V ′∗

aVV + V ′aVV = 0

inductively by (6.29) and (6.30). Therefore, the above construction ofthe Darboux matrix keeps the u(N) reduction.


Remark 49 If Ai’s are independent of t, then the above system is re-duced to the generalized self-dual Yang-Mills system, and the Darbouxtransformation is still valid.

6.1.3 ExampleStarting from the trivial solution Ai = 0, we want to construct solu-

tions by Darboux transformation. We have to determine V0VV , V1VV , · · ·, VmVV ,VmVV +1, · · ·, VmVV +q first. When Ai = 0, (6.15) and (6.16) become

V0VV = V0VV (p),∂VaVV

∂xi=

∂VaVV −1

∂pi(a = 1, 2, · · · , m; m + q, · · · , m + 2),

VmVV +q = V 0mVV +q(x).

(6.44)

Hence, the entries of VmVV are polynomials of x of degree ≤ m, the entriesof VmVV +1 are polynomials of p of degree ≤ q−1. Moreover, (6.17) becomes

∂VmVV

∂pi=

∂VmVV +1

∂xi. (6.45)

Thus the entries of VmVV are polynomials of p of degree ≤ q, the entriesof VmVV +1 are polynomials of x of degree ≤ m + 1. Therefore, the entriesof V0VV are polynomials of p of degree ≤ m + q and the entries of VmVV +q

are polynomials of x of degree ≤ m + q. From the expression of V and(6.44), we have

λ∂V

∂xi− ∂V

∂pi= 0. (6.46)

Hence the entries of V are functions of λpi + xi (i = 1, · · · , n). V can beexpressed as

V =1λq

P (λp + x). (6.47)

Here the entries of P (λp+x) are polynomials of λpi +xi (i = 1, · · · , n) ofdegree ≤ m+ q. Therefore, the fundamental solution of (6.9) and (6.14)is

Ψ = exp(

1λq

P (λp + x)t)

. (6.48)

The Darboux transformation

(0, Ψ) → (A′, Ψ′) → (A′′, Ψ′′) → · · · , (6.49)

can be constructed successively so that a series of explicit solutions areobtained. We still call them soliton solutions.


To write down the solutions more explicitly, take N = 2,

V =

⎛⎝⎛⎛ iv(λ) 0

0 − iv(λ)

⎞⎠⎞⎞ , (6.50)

wherev(λ) =

1λq

u(λp + x), (6.51)

u is a real polynomial of degree m + q. Then

Ψ =

⎛⎝⎛⎛ eiv(λ)t 0

0 e−iv(λ)t

⎞⎠⎞⎞ . (6.52)

Let µ = σ + iτ be a complex number where σ, τ are real and τ = 0.Take

H =

⎛⎝⎛⎛ eiv(µ)t −g∗(µ)eiv(µ)t

g(µ)e−iv(µ)t e−iv(µ)t

⎞⎠⎞⎞ , (6.53)

where g(µ) = g(µp + x), g∗(µ) = g(µ). For any holomorphic functiong, the two columns of H are solutions of the Lax set for λ = µ and µrespectively. The condition (6.42) is also satisfied. Moreover,

∆ = det H = ewt + |g(µ)|2e−wt = 0 , (6.54)

where w = i(v(µ) − v(µ)) is a real-valued function.Let

g(µ) = eρ(µ)+iθ(µ),

v(µ) + v(µ) = κ(µ),(6.55)

where ρ(µ), θ(µ) are real-valued functions. Then

S = HΛH−1

=

⎛⎝⎛⎛ σ − iτ tanh(ρ − wt) iτ sech(ρ − wt)ei(−θ+κt)

iτ sech(ρ − wt)ei(θ−κt) σ + iτ tanh(ρ − wt)

⎞⎠⎞⎞ (6.56)

and we get the single soliton solution

Ai = − ∂S

∂xi. (6.57)

This kind of soliton solutions may have very complicated behaviorbecause the choice of v and g has large freedom. The appearance of thefunctions tanh and sech in S implies that if v and g are chosen suitably,


the entries of S tend to constants and the entries of Ai’s tend to 0 whenρ − wt → ∞. However, these solutions are not travelling waves. Multi-soliton solutions can be obtained by successive Darboux transformationswith long expressions.

6.1.4 Relation with AKNS systemLet JiJJ ’s be n N × N constant matrices and

Φ(p, t) = exp(−∑xjJjJJ )Ψ,

PiPP (p, t) = − exp(−∑xjJjJ )Ai exp(∑

xjJjJJ ),

UaUU (p, t) = exp(−∑xjJjJJ )VaVV exp(∑

xjJjJJ ).

(6.58)

Here Ψ, Ai’s and VaVV ’s are the matrix valued functions in the Lax set(6.9) and (6.14). Suppose that Φ, PiPP ’s and UaUU ’s are all independent ofxi’s. Thus, (6.58) gives constraints for the Lax set (6.9) and (6.14).

Under these constraints, (6.9) and (6.14) become

∂Φ∂pi

= (λJiJJ + PiPP )Φ,

∂Φ∂t

=m+q∑a=0

UaUU λm−aΦ.(6.59)

This is the AKNS system on Rn+1 where the space-time variables are(p1, · · · , pn, t). Therefore we have

Theorem 6.4 AKNS system is a kind of constraints of the self-dualYang-Mills flow.

R. S. Ward points out that all known soliton equations in lower di-mensions can be the constraints of the self-dual Yang-Mills equation onR4. This means that the self-dual Yang-Mills equation has profoundcontent. Although the high dimensional AKNS system in Chapter 3cannot become a constraint of the self-dual Yang-Mills equation on R4,it is a constraint of a high dimensional generalized self-dual Yang-Millsflow.


6.2 Yang-Mills-Higgs field in 2+1 dimensionalMinkowski space-time

6.2.1 Yang-Mills-Higgs fieldThe Yang-Mills-Higgs equations in R2,1 are important partial differ-

ential equations in mathematical physics. The system of equations isa reduction of the self-dual Yang-Mills equation on R2,2 and the Higgsfield appears as one of the reduced gauge potential. This means thatthe Higgs field occurs inside the framework of Yang-Mills theory. Thisequation is called Bogomolny equation or monopole equation. Therewere a series of work on the Bogomolny equations [76]. The case of R2,1

has the advantage that there is a real time variable and the metric is ofMinkowski. The system is also integrable. A systematical sketch withbeautiful results on these equations can be found in [108].

Let R2,1 be the Minkowski space-time (x, y, t) with the metric ds2 =dx2 + dy2 − dt2, G be a matrix Lie group, g be its Lie algebra. LetA = (At, Ax, Ay) be a gauge potential valued in g and Φ a g−valuedfunction called Higgs field.

We consider the Yang-Mills-Higgs field satisfying the following Bogo-molny equations

(∗F )i = DiΦ (i = 1, 2, 3, x1 = x, x2 = y, x3 = t). (6.60)

Here F is the field strength, DiΦ is the gauge-derivative of Φ, i.e.,

FijFF = ∂j∂ Ai − ∂i∂∂ Aj + [Ai, Aj ], (6.61)

DiΦ = ∂i∂∂ Φ + [Ai, Φ], (6.62)

and ∗ is the Hodge operator.For the Minkowski space-time R2,1 with oriented coordinate (t, y, x)

and metric ds2 = dx2 + dy2 − dt2, the Bogomolny equation becomes

FxyFF = DtΦ, FtyFF = DxΦ, FxtFF = DyΦ. (6.63)

It admits a Lax pair. In fact, let

L1 = −∂t∂∂ − ∂y∂∂ + λ∂x∂ , L2 = λ∂t∂∂ − λ∂y∂∂ − ∂x∂ , (6.64)

where λ is a parameter. The Lax pair is

L1Ψ = (At + Ay − λAx − λΦ)Ψ,

L2Ψ = (−λAt + λAy + Ax − Φ)Ψ,(6.65)


where Ψ is an N×N matrix-valued function. The integrability conditionof the Lax pair (6.65) is just the Bogomolny equation (6.63) [76]. TheLax pair (6.65) can also be written down in the covariant form

(Dt + Dy − λDx)Ψ = λΦΨ,

(λDt − λDy − Dx)Φ = −ΦΨ(6.66)

where DiΨ = ∂i∂∂ Ψ + AiΨ (i = t, x, y).If Ψ(t, x, y, λ) is an N × N matrix solution of Lax pair (6.65) with

det Ψ = 0, and ( L1Ψ)Ψ−1, (L2Ψ)Ψ−1 are of degree one with respect toλ, then Ψ is called a Lax representation of the Yang-Mills-Higgs field,since At, Ax, Ay and Φ can be determined by Ψ.

6.2.2 Darboux TransformationsAt first we consider the case G = GL(N,C). Let (Ψ, A, Φ) satisfy the

Lax pair. It is required to find a matrix S(t, x, y) such that

Ψ′ = (λI − S)Ψ, (6.67)

together with some A′, Φ′, satisfy the Lax pair (6.65) too. Hence, A′ =(A′

t, A′x, A′

y) and Φ′ give a solution of the Bogomolny equation [42].Substituting Ψ′ in (6.67) into the Lax pair (6.65) with (Ai, Φ) replaced

by (A′i, Φ

′) (i = t, x, y), we obtain

A′t = At − (Φ′S − SΦ), A′

y = Ay − (Φ′S − SΦ),

A′x = Ax − (Φ′ − Φ)

(6.68)

and

−StSS − SySS = AtS − SAt + AyS − SAy − 2(Φ′S − SΦ)S,

StSS − SySS = SAt − AtS + AyS − SAy + 2(Φ′ − Φ),

SxS = SAx − AxS + (Φ′ − Φ)S + (Φ′S − SΦ).

(6.69)

From the last equation of (6.69), we get

Φ′ =12(SxS S−1 − SAxS−1 + Ax + Φ + SΦS−1). (6.70)

Substituting (6.70) into the first two equations of (6.69), we have

(∂t∂∂ + ∂y∂∂ )S = SxS S − (AtS − SAt) − (AyS − SAy)

+AxS2 − SAxS + ΦS2 − SΦS,(6.71)


(∂t∂∂ − ∂y∂∂ )S = SxS S−1 − (AtS − SAt) + (AyS − SAy)

+Ax − SAxS−1 − Φ + SΦS−1.(6.72)

The problem is reduced to find S such that (6.71) and (6.72) are satisfied.This is a complicated system of nonlinear partial differential equations.Fortunately, as before, the solution S can be constructed explicitly by a“universal” formula.

Let λ1, · · · , λN be N complex numbers so that they are not all equal,ha be a column solution of the Lax pair for λ = λa (a = 1, 2, · · · , N).Construct

H = (h1, h2, · · · , hN ) (det H = 0) , (6.73)

andS = HΛH−1, Λ = diag(λ1, · · · , λN ). (6.74)

It is easily seen that

(∂t∂∂ + ∂y∂∂ )H = HxHH Λ − AtH − AyH + AxHΛ + ΦHΛ (6.75)

and

(∂t∂∂ − ∂y∂∂ )H = HxHH Λ−1 − AtH + AyH + AxHΛ−1 − ΦHΛ−1. (6.76)

Consequently, direct calculation shows that (6.71) and (6.72) are satis-fied. Thus, we have

Theorem 6.5 Let S be defined by (6.74). Then λI − S is a Darbouxmatrix for the Lax pair (6.65), and (A′

t, A′x, A′

y, Φ′) given by (6.68) and

(6.70) satisfies the Bogomolny equation (6.63).

The Darboux transformation

(Ψ, A,Φ) → (Ψ′, A′, Φ′)

can be applied successively to obtain a series of solutions by using thesame purely algebraic algorithm.

Remark 50 In general, if G = GL(N,R), the solution (Ψ′, A′, Φ′) isonly a local solution, since we cannot make sure that the conditiondet H = 0 holds globally.

Remark 51 ha (a = 1, · · · , N) may be expressed as

ha = Ψ(λa)la, (6.77)

where la’s are columns satisfying

L1la = 0, L2la = 0 (6.78)


for λ = λa. Moreover, if λa is not real, then each entry of la is aholomorphic function of

ω(λa) = λ2a(t + y) + 2λax + (t − y). (6.79)

When G is not GL(N,C), but its subgroup, some reductions shouldbe added in the construction of Darboux transformation. An interestingcase is G = U(N) (or SU(N)). In this case, (A, Φ) should be valued inthe Lie algebra u(N), i.e.,

Ai + A∗i = 0, Φ + Φ∗ = 0. (6.80)

After Darboux transformation we should have

A′i + A′∗

i = 0, Φ′ + Φ′∗ = 0. (6.81)

This can be realized by choosing λa’s and ha’s such that

λa = µ or µ (µ is not real), (6.82)h∗

ahb = 0 if λa = λb. (6.83)

Theorem 6.6 Suppose Ai’s and Φ are valued in u(N). If we chooseλa’s and ha’s such that (6.82) and (6.83) hold, then after the Darbouxtransformation, A′

i’s and Φ′ are still valued in u(N).

Proof. As in Section 3.2, with the conditions (6.82) and (6.83), we have

S + S∗ = (µ + µ)I, S∗S = µµ∗I. (6.84)

Thus 1µS is valued in U(N), SxS and S − µI are valued in u(N). From

(6.70) it is seen that Φ′ ∈ u(N) and

Φ′S − SΦ =12(SxS − SAx + AxS − ΦS + SΦ) ∈ u(N). (6.85)

Hence A′x, A′

y and A′t are valued in u(N). The proof is completed.

Remark 52 The solution (Ψ′, A′, Φ′) is global if (Ψ, A, Φ) is global.

Remark 53 The theorem also holds for the case of SU(N).

6.2.3 Soliton solutionsStarting with the trivial solution

Ai = Φ = 0,


the first Darboux transformation gives single solitons. Now the Lax pair(6.65) becomes

(−∂t∂∂ − ∂t∂∂ + λ∂x∂ )ψ = 0,

(λ∂t∂∂ − λ∂t∂∂ − ∂x∂ )ψ = 0.(6.86)

The general solution of (6.86) is

ψ = f(ω(λ)) (6.87)

where each entry of f is an arbitrary holomorphic function, ω is de-fined by (6.79). The Darboux transformation is constructed in the lastsubsection and the expressions for Φ′, A′

i are

Φ′ = 12SxS S−1, A′

x = −12SxS S−1,

A′t = −1

2SxS , A′y = −1

2SxS .(6.88)

It is obvious that all solutions are SU(N) solutions.If µ = i, the solution is static. If µ = i, the line ω(µ) = c can be

expressed in the form

x = βt + β0, y = γt + γ0, (6.89)

and the solution (Ψ′, A′, Φ′) is a travelling wave. That is, as functionsof (x, y, t), (Ψ′, A′, Φ′) depends on x − βt and y − γt only.

In the case N = 2, we write down the single solitons more explicitly.Let f1, f2ff be two holomorphic functions without common zero. We have

H =

⎛⎝⎛⎛ f1(ω(µ)) −f2ff (ω(µ))

f2ff (ω(µ)) f1(ω(µ))

⎞⎠⎞⎞ , Λ =

⎛⎝⎛⎛ µ 0

0 µ

⎞⎠⎞⎞ . (6.90)

Then

S = HΛH−1 =1

|F1FF |2 + |F2FF |2

⎛⎝⎛⎛ µ|F1FF |2 + µ|F2FF |2 (µ − µ)F1FF F2FF

(µ − µ)F1FF F2FF µ|F2FF |2 + µ|F1FF |2

⎞⎠⎞⎞ .

(6.91)Here F1FF and F2FF denote f1(ω(µ)) and f2ff (ω(µ)) respectively. Hence

||Φ′||2 = −12

tr Φ′2 = −18

tr(SxS S−1SxS S−1) =4( Im µ)2|F1FF F ′

2FF − F2FF F ′1FF |2

(|F1FF |2 + |F2FF |2)2(6.92)

where F ′jFF denotes f ′

jf (ω(µ)). It is noted that if µ = i, along any straightline on the plane t = constant, ω(µ) → ∞ when x2 + y2 → ∞. It


Figure 6.1. Example (1), ||Φ||2 at t = 10


was shown in [108] that if F1FF = P1PP (ω(µ)), F2FF = P2PP (ω(µ)) where P1PP andP2PP are certain polynomials of ω, then the solution is localized. Moregenerally, if F1FF = P1PP (ω(µ)), F2FF = P2PP (ω(µ)) exp(Q(w(µ))) where Q isanother polynomial of ω with deg Q ≤ maxdeg P1PP , deg P2PP , then alongany straight line on the plane t = t0, we still have SxS → 0 as x2+y2 → ∞.However, the solutions may not be completely localized (see example(3)). This class of single solitons is slightly wider than those constructedin [108], where Q(ω) = 1.

We draw the pictures of ||Φ||2 at t = 10 for the following examples(1) µ = i/2, f1 = 1, f2ff = ω.(2) µ = i/2, f1 = 1, f2ff = (ω − 1)(ω + 6)(ω − 4i).(3) µ = i/2, f1 = 1, f2ff = ωeω/10. The solution Φ does not approachto zero if t = t0 and (x, y) → ∞ along the curves |ω|2eRe ω/10 = constant.



By applying Darboux transformations p times to the trivial solution,we obtain p-soliton solutions. Let µ1, µ2, · · · , µp be p complex numbers(µj = real , µj = i , µj = µk for j, k = 1, 2, · · · , p). The Lax representationof a p-soliton solution is

Ψp = (λI − SpSS )(λI − SpSS −1) · · · (λI − S1) (6.93)

At first we consider the asymptotic behavior of the double soliton

Ψ2 = (λI − S2)(λI − S1) (6.94)

as t → ±∞. Letx = a1t + b1, y = a2t + b2 (6.95)

be a straight line which is different from ω(µ1) = c1. Then along (6.95),ω(µ1) → ∞ as t → ±∞. Hence

λI − S1 ∼ λI − S0S (6.96)

where S0SS is the limit of S as t → ±∞. Hence the asymptotic behaviorof Ψ2 is the Darboux transformation of the constant matrix λI − S0SSby using µ = µ2. Therefore, Ψ2 behaves as the Lax representation ofa single-soliton along the straight line ω(µ2) = constant. Besides, thepotential Ai’s and Higgs field Φ approach to zero along each straight lineif it is neither ω(µ1) = constant nor ω(µ2) = constant..

It is remaining to consider the asymptotic behavior of Ψ2 along thestraight line ω(µ1) = constant.. From the theorem of permutability forDarboux transformation,

Ψ2 = (λI − S2)(λI − S1) = (λI − S′1)(λI − S′

2), (6.97)


Figure 6.4. Splitting of ||Φ||2 for a 2-soliton (t = 0)

where S′1, S′

2 are defined in the same way as S1, S2 but the order of(µ1, l1) and (µ2, l2) is changed. Thus the above argument implies thatΨ2 behaves as the Lax representation of the single-soliton defined by(µ1, l1) along the straight line ω(µ1) = constant. as t → ±∞. Thedouble soliton is splitting up into two single solitons as t → ±∞. Forgeneral p we have following splitting theorem as well.

Theorem 6.7 A p-soliton splits up into p single solitons asymptoticallyas t → ±∞.

We draw a picture for ||Φ||2 showing the splitting of a 2-soliton withµ1 = i/2, µ2 = 2 − i/2 and for both µ1 and µ2, f1 = 1, f2ff = ω.

Remark 54 The expression (6.93) can be written in the form [108]

Ψp = I +p∑

k=1

MkM

λ − µk. (6.98)

Remark 55 Consider the relationship between gauge transformation andDarboux transformation. Let g be a function of (x, y, t) valued in thegroup G and Ψ(x, y, t, λ) be a Lax representation of a solution (A,Φ) ofthe Bogomolny equation (6.63). Then

Ψ → Ψ = gΨ (6.99)

is a gauge transformation. In fact, from the Lax pair it is easily seenthat

Ai = gAig−1 − gxg−1, Φ = gΦg−1. (6.100)


Figure 6.5. Splitting of ||Φ||2 for a 2-soliton (t = 10)

Moreover,

gΨ′ = g(λI − S)Ψ = (λI − S)gΨ = (λI − S)Ψ (6.101)

withS = gSg−1. (6.102)

(6.101) means that the gauge transformations commute with the Darbouxtransformation.

6.3 Yang-Mills-Higgs field in 2+1 dimensionalanti-de Sitter space-time

6.3.1 Equations and their Lax pairThe Bogomolny equation in 2+1 dimensional anti-de Sitter space-

time is also known to be integrable in the sense that it has a Lax pair[108, 107, 109]. Here we use the Darboux transformation method toderive its explicit solutions and discuss the asymptotic behavior of thesolutions as time t → ∞. Moreover, the solutions are compared withthose in the Minkowski space-time when the curvature of the anti-deSitter space-time tends to zero [129, 130].

The 2+1 dimensional anti-de Sitter space-time is the hyperboloid

U2 + V 2 − X2 − Y 2 = ρ2 (6.103)

(ρ > 0) in R2,2 with the metric

ds2 = −dU2 − dV 2 + dX2 + dY 2. (6.104)


It has constant Gauss curvature −1/ρ2. Define the local coordinate

r =ρ

U + X− ρ + 1, x =

Y

U + X, t = − V

U + X, (6.105)

then a part U +X > 0 on the 2+1 dimensional anti-de Sitter space-timeis represented by the Poincare coordinates (r, x, t) with r > −ρ + 1 andthe metric is

ds2 =ρ2

(r + ρ − 1)2(−dt2 + dr2 + dx2) =

ρ2

(r + ρ − 1)2(dr2 + du dv)

(6.106)where u = x + t, v = x − t. Clearly, when ρ → +∞, the metric on theregion r > −ρ + 1 tends to the flat metric of the Minkowski space-time.

Remark 56 Here we take the half space as r > −ρ+1 rather than r > 0so that the solitons can keep in a bounded region as ρ → +∞. This willbe shown in Subsection 6.3.4.

As in the Minkowski space-time, the Bogomolny equation is also

(∗F )i = DiΦ (i = u, v, r). (6.107)

With the metric (6.106) and the orientation (v, u, r), (6.107) becomes

DuΦ =r + ρ − 1

ρFurFF ,

DvΦ = −r + ρ − 1ρ

FvrFF ,

DrΦ = −2(r + ρ − 1)ρ

FuvFF .

(6.108)

It has a Lax pair

((r + ρ − 1)Dr + ρΦ − 2(ρλ − u)Du)ψ = 0,(2Dv +

ρλ − u

r + ρ − 1Dr − ρ(ρλ − u)

(r + ρ − 1)2Φ)

ψ = 0(6.109)

where the action of Dµ on ψ is Dµψ = ∂µ∂ ψ + Aµψ. This Lax pairwas first proposed by [107] to ρ = 1, which can be easily generalized toarbitrary ρ. With the help of this Lax pair, we can construct Darbouxtransformation in next subsection.

6.3.2 Darboux transformationsFirst we consider the case G = GL(N,C). This is a case free of

reduction. Let(λ − u/ρ)I − S(u, v, r) (6.110)


be a Darboux matrix, then there exists (A′u, Av, A

′r, Φ

′) such that forany solution ψ of (6.109), ψ′ = ((λ − u/ρ)I − S)ψ satisfies

((r + ρ − 1)D′r + ρΦ′ − 2(ρλ − u)D′

u)ψ′ = 0,(2D′

v +ρλ − u

r + ρ − 1D′

r −ρ(ρλ − u)

(r + ρ − 1)2Φ′)

ψ′ = 0(6.111)

with D′µ = ∂µ∂ +A′

µ. Here the term u/ρ in (6.110) is used only to simplifythe calculation.

To determine the Darboux transformation, it is sufficient to find thematrix function S.

For given (A, Φ), (A′, Φ′) and an arbitrary matrix function Q, wedefine

∆iQ = ∂i∂∂ Q + A′iQ − QAi (i = u, v, r),

δQ = Φ′Q − QΦ.(6.112)

Substituting ψ = ((λ − u/ρ)I − S)ψ into (6.111) and considering(6.109), we get

∆uI = 0, (r + ρ − 1)∆rI + 2ρ∆uS + ρδI + 2I = 0,

(r + ρ − 1)∆rS + ρδS = 0, ∆rI − ρ

r + ρ − 1δI = 0,

2∆vI − ρ

r + ρ − 1∆rS +

ρ2

(r + ρ − 1)2δS = 0, ∆vS = 0.

(6.113)

This system is equivalent to

∆uI = 0, (6.114)∆vS = 0, (6.115)

∆vI =ρ

r + ρ − 1∆rS, (6.116)

∆rI +ρ

r + ρ − 1∆uS +

1r + ρ − 1

I = 0, (6.117)

∆rI − ρ

r + ρ − 1δI = 0, (6.118)

∆rS +ρ

r + ρ − 1δS = 0. (6.119)

The new solution (A′u, A′

v, Ar, Φ′) is solved from (6.114), (6.115), (6.118)and (6.119) as

A′u = Au, (6.120)


A′v = SAvS

−1 − (∂v∂∂ S)S−1, (6.121)

A′r =

12(SAr − ∂r∂∂ S)S−1 +

12Ar +

ρ

r + ρ − 1(SΦS−1 − Φ), (6.122)

Φ′ =r + ρ − 1

2ρ(SAr − ∂r∂∂ S)S−1 − r + ρ − 1

2ρAr

+12(SΦS−1 + Φ), (6.123)

while (6.116) and (6.117) give the equations that S should satisfy.

Remark 57 The Bogomolny equation is gauge invariant. We choose theform of the Darboux matrix (6.110) so that Au keeps unchanged in Dar-boux transformation.

Using the standard procedure of constructing Darboux transforma-tion, we have

Theorem 6.8 Let Λ = diag(λ1, · · · , λN ) where λ1, · · · , λN are complexnumbers. Let hj be a column solution of the Lax pair (6.109) with λ = λj,H = (h1, · · · , hN ). If det H = 0 , then S = HΛH−1 − (u/ρ)I satisfies(6.116) and (6.117). Hence (λ − u/ρ)I − S is a Darboux matrix for(6.109).

This theorem can also be generalized as follows.

Theorem 6.9 Let Λ = diag(λ1, · · · , λN ) where λj(u, v, r) is a solutionof

∂r∂∂ λ − 2(ρλ − u)r + ρ − 1

∂u∂∂ λ = 0, ∂v∂∂ λ +ρλ − u

2(r + ρ − 1)∂r∂∂ λ = 0. (6.124)

Let hj be a column solution of the Lax pair (6.109) with λ = λj(u, v, r),H = (h1, · · · , hN ). If det H = 0 and define S = HΛH−1 − (u/ρ)I, then(λ − u/ρ)I − S is a Darboux matrix for (6.109).

The general non-constant solution of (6.124) is given by the implicitfunction

v − (r + ρ − 1)2

ρλ − u+

ρ − 1λ

= constant. (6.125)

Remark 58 Clearly Theorem 6.9 is a generalization of Theorem 6.8 be-cause constant λ is a solution of (6.124). We shall show later that con-stant Λ always leads to global solutions on the 2+1 dimensional anti-deSitter space-time. However, usually a non-constant Λ only leads to localsolutions.


According to Theorem 6.8 and Theorem 6.9, we can construct explicitsolutions of (6.108) from a known solution of (6.108) and the correspond-ing solution of the Lax pair (6.109).

When the group G = U(N), the Lie algebra consists of all anti-Hermitian matrices. Hence A∗

µ = −Aµ, Φ∗ = −Φ. To keep this re-duction, there should be more constraints on λj ’s and hj ’s in the con-struction of the Darboux matrix. They are

λj = λ0 or λ0 for certain non-real λ0,

h∗jhk = 0 if λj = λk.

(6.126)

The second condition holds identically if it holds on a line u = u0 =constant, v = v0 = constant. In fact, h∗

jhk satisfies

(r + ρ − 1)∂r∂∂ (h∗jhk) − 2(ρλk − u)∂u∂∂ (h∗

jhk) = 0,

2∂v∂∂ (h∗jhk) +

ρλk − u

r + ρ − 1∂r∂∂ (h∗

jhk) = 0.(6.127)

The general solution of this system is h∗jhk = fjkf (ω(λk)) where

ω(λ) = v − (r + ρ − 1)2

ρλ − u+

ρ − 1λ

(6.128)

and fjkf (z) is a holomorphic function of z. Hence fjkf (ω(λk)) ≡ 0 iffjkf (ω(λk)) = 0 on the line u = u0, v = v0. Therefore, h∗

jhk = 0identically.

With the condition (6.126), the Darboux transformation gives Φ′ ∈u(N), A′

µ ∈ u(N), provided that Φ ∈ u(N), Aµ ∈ u(N).

6.3.3 Soliton solutions in SU(2) case

(1) Expressions of the solutionsSingle soliton solutions are given by Darboux transformations from

the trivial seed solution Ai = 0 (i = u, v, r), Φ = 0. Here we onlyconsider the case where all λj ’s are constants. In this case

∂r∂∂ hj − 2(ρζjζ − u)r + ρ − 1

∂u∂∂ hj = 0, ∂v∂∂ hj +ρζjζ − u

2(r + ρ − 1)∂r∂∂ hj = 0. (6.129)

The general solution ishj = fjf (ω(ζjζ )) (6.130)

where fjf is a column matrix whose entries are all holomorphic functionsand ω is defined by (6.128). Then the Darboux matrix is λI − HΛH−1

with H = (h1, · · · , hN ).


Now suppose G = SU(2). Considering the conditions (6.126), we takeΛ = diag(λ0, λ0) where λ0 ∈ C is not real, τ = ω(λ0),

H =

⎛⎝⎛⎛ α(τ) −β(τ)

β(τ) α(τ)

⎞⎠⎞⎞

where α, β are holomorphic functions of τ . Let σ(τ) = β(τ)/α(τ). Then

S =λ0 − λ0

1 + |σ|2

⎛⎝⎛⎛ 1 σ

σ |σ|2

⎞⎠⎞⎞+ λ0 − u

ρ. (6.131)

Φ′ =λ0 − λ0

(1 + |σ|2)2

⎛⎝⎛⎛ (|σ|2)u σ2σu − σu

σ2σu − σu −(|σ|2)u

⎞⎠⎞⎞ , (6.132)

||Φ′||2 = −12

tr Φ′2 =4( Im λ0)2

(1 + |σ|2)2 |∂u∂∂ σ|2. (6.133)

Therefore, a single soliton solution depends on an arbitrary meromor-phic function σ(τ) and ||Φ′|| is given by (6.133).

Multi-soliton solutions can be constructed by successive actions ofDarboux transformations on trivial solution.

(2) Globalness of the solutionsTill now, the solutions are only defined by the local coordinate (u, v, r),

or on the part with U + X > 0 in the 2+1 dimensional anti-de Sitterspace-time (6.103). However, these solutions can be extended to thewhole 2+1 dimensional anti-de Sitter space-time in a natural way.

According to (6.105) and (6.128),

τ =ρλ0(Y + V )(U + X) − 1 − Y 2 + V 2

(ρλ0(U + X) − Y + V )(U + X)+

ρ − 1λ0

=ρλ0(Y + V ) + X − U

ρλ0(U + X) − Y + V+

ρ − 1λ0

.

(6.134)

Denote

ξ = ρλ0(Y + V ) + X − U, η = ρλ0(X + U) − Y + V, (6.135)

then both ξ and η can never be zero on (6.103) when λ0 is not real.Hence τ is a smooth function of U, V, X, Y on (6.103), so are

∂u∂∂ τ = −(r + ρ − 1)2

(ρλ0 − u)2= −ρ2

η2, ∂v∂∂ τ = 1,

∂r∂∂ τ = −2(r + ρ − 1)2

ρλ0 − u= −2ρ

η.

(6.136)


Therefore, all the solutions which are obtained by Darboux transfor-mation from trivial solution can be extended naturally to whole 2+1dimensional anti-de Sitter space-time, provided that λ0 is chosen to bea non-real constant.

(3) Localization of the solutionsThe infinity of the 2+1 dimensional anti-de Sitter space-time is at

r → −ρ + 1. We shall verify that ||Φ′|| → 0 as r → −ρ + 1. In fact,

||Φ′||2 =4( Im λ0)2

(1 + |σ(τ)|2)2 |σ′(τ)|2|∂u∂∂ τ |2

=4( Im λ0)2

(1 + |σ(τ)|2)2 |σ′(τ)|2 (r + ρ − 1)4

|ρλ0 − u|4 .

(6.137)

When r → −ρ + 1, τ → v + (ρ − 1)/λ0. Hence ||Φ′|| → 0. This meansthat the solutions are all localized on the 2+1 dimensional anti-de Sitterspace-time when λ0 is a constant.

(4) Asymptotic behavior of the solutions when t → ∞Without loss of generality, suppose λ0 = i. For general λ0, a trans-

formation of the coordinates will lead to this case.

Example 6.10 σ(τ) = τ , this is just the localized solution (25) of [107],and

||Φ′||2 =4r4

((r2 + x2 − t2)2 + 2x2 + 2t2 + 1)2. (6.138)

Letx = tR cos θ, r = tR sin θ. (6.139)

When t and θ are fixed, ||Φ′||2 is a function of R only. Its maximumappears at R = ±√

t2 + 1/t. Hence as t → ∞, the solution has a ridgewhich is located on the circle x2 + r2 = t2 + 1. This is quite differentfrom the usual solitons which have only some peaks.

Figure 6.6 shows this solution at t = 10. The vertical axis is ||Φ′||1/2.

Example 6.11 σ(τ) is a polynomial of τ of degree k (k ≥ 1) and allthe roots τ1ττ , · · · , τkτ of σ(τ) are simple.

The asymptotic behavior of the solution is roughly as follows [129].(1) If | Im τjτ | << 1, then ||Φ′

j || is not small only near a half circlecentered at r = 0. (This half circle is a geodesic of the Poincar´ plane´for t = constant.) We call such a shape as a ridge.

(2) If Im τjτ >> 1, then ||Φ′j || is not small only near one point. The

shape is a peak.(3) If Im τjτ << −1, then ||Φ′

j || is small everywhere. Nothing can beshown in the figure.


Figure 6.6.

Figure 6.7.

For example, let

σ(τ) = (τ − 2)(τ − 6)(τ + 6)(τ − 2i)(τ − 6i)(τ + 6i), (6.140)

the shape of ||Φ||1/4is shown in Figure 6.7 (t(( = 10). There are threeridges (corresponding to roots 2, 6, 6 of σ(τ)) and two peaks (correspond-ing to roots 2i, 6i of σ(τ)).

Example 6.12 σ(τ) = sin(τ/20). ||Φ||1/4 is shown in Figure 6.8 (t(( =10). Although it looks complicated, it is still localized on the 2+1 dimen-sional anti-de Sitter space-time because it decays as r → 0.

6.3.4 Comparison with the solutions inMinkowski space-time

When ρ → +∞, the space-time r > −ρ + 1 with metric (6.106) tendsto whole R2,1 with the Minkowski metric ds2 = dr2 + du dv. By takingthe limit ρ → +∞ in (6.108) and (6.109), the Bogomolny equation and


Figure 6.8.

its Lax pair tend to

DuΦ = FurFF , DvΦ = −FvrFF , DrΦ = −2FuvFF , (6.141)

and(Dr + Φ − 2λDu)ψ = 0,

(2Dv + λDr − λΦ)ψ = 0,(6.142)

which are the Bogomolny equation (6.63) and the Lax pair (6.66) inthe Minkowski space-time R2,1, with the correspondence (λ, r, u, v) →( 1

λ , x, y + t, y − t). Then the Darboux transformation (λ − u/ρ)I − S =λI − HΛH−1 also tends to the Darboux transformation in R2,1, so arethe solutions (A′

u, A′v, A

′r, Φ

′). The choice of the half space r > −ρ + 1keeps the soliton solutions in a bounded region as ρ → +∞.

However, the asymptotic behavior of the solutions are quite differentfor finite ρ and ρ → +∞. In the Minkowski space-time, the asymptoticbehavior of the solutions is quite simple when σ(τ) is a polynomial withonly simple roots. In this case, each root of σ(τ) corresponds to a peakin its graph. However, for finite ρ, we have shown that different kindof the root of σ(τ) creates different shape in the graph. For σ(τ) =(τ − 2)(τ − 6)(τ + 6)(τ − 2i)(τ − 6i)(τ + 6i), Figure 6.7 has shown thesolution for ρ = 1. Figure 6.9 and Figure 6.10 show the solution for ρ = 2and ρ → +∞ respectively. In Figure 6.9, the ridges are shorter thanthose in Figure 6.7, and in Figure 6.10, each root of σ(τ) corresponds toa peak.

For σ(τ) = sin(τ/20), Figure 6.8 has shown the solution for ρ = 1.Here Figure 6.11 and Figure 6.12 show the solution for ρ = 30 andρ → +∞ respectively.


Figure 6.9.

Figure 6.10.

Figure 6.11.


Figure 6.12.

Chapter 7

TWO DIMENSIONAL TODAEQUATIONS AND LAPLACESEQUENCES OF SURFACESIN PROJECTIVE SPACE

7.1 Signed Toda equationsToda equations are a class of important integrable systems [100]. Be-

sides the original one-dimensional Toda equations, the two dimensionalToda equations also attract many authors in recent years [80, 31, 54, 83,10]. In particular, the elliptic version of two dimensional Toda equationshas many applications in differential geometry, such as minimal surfaces,surfaces of constant mean curvature, harmonic maps, etc.

The hyperbolic two dimensional Toda equations can be traced back toG. Darboux [19]. Consider the hyperbolic partial differential equation

zxt + azx + bzt + cz = 0. (7.1)

Definez1 = zt + bz, (7.2)

then z1 satisfies a hyperbolic equation

z1,xt + a1z1,x + b1z1,t + c1z1 = 0, (7.3)

which is called a Laplace transformation of (7.1) in t-direction. Similarly,the function

z−1 = zx + az, (7.4)

satisfiesz−1,xt + a−1z−1,x + b−1z−1,t + c−1z−1 = 0, (7.5)

and is called a Laplace transformation of (7.1) in x-direction.Applying these Laplace transformations successively, we get a series

of hyperbolic systems

zi,xt + aizi,x + bizi,t + cizi = 0 (i = 0,±1,±2, · · ·), (7.6)


which is called a Laplace sequence of equations starting from (7.1)G. Darboux derived the equations

(lnhi)xt = −hi−1 + 2hi − hi+1 (hi =∂ai

∂t+ aibi − ci). (7.7)

He set hi = eωi and obtained the equations

ωi,xt = −eωi−1 + 2eωi − eωi+1 (i = 1, 2, · · · , n) (7.8)

which are called hyperbolic Toda equations now.The original Toda equations are a system of ordinary differential equa-

tions in the form

d2ωi

dt2= eωi−1 − 2eωi + eωi+1 (i = 1, 2, · · · , n) (7.9)

(in the periodic case, ω0 = ωn, ωn+1 = ω1). It was established byM. Toda [100] in the study of the motion of a chain of particles in whichtwo neighboring particles attract each other in a nonlinear way.

Consider a system of n particles of the same kind with equal massarranged along a line and jointed by identical springs. Let the positionof the n particles be q1, q2, · · ·, qn and q1 < q2 < · · · < qn.

Suppose the ith particle is acted by the i−1th particle and the i+1thparticle only. If the force is dependent of the distance exponentially, thenwe have a system of Toda equations (7.9).

In [64], we noted that hi may be negative. We should set

hi = αieωi . (7.10)

Here αi = ±1 as hi > 0 or hi < 0 respectively. Thus (7.8) is changed to

ωi,xt = −αi−1eωi−1 + 2αie

ωi − αi+1eωi+1 . (7.11)

We call (7.11) two dimensional signed Toda equations.Accordingly, the one dimensional signed Toda equations should be

d2ωi

dt2= αi−1e

ωi−1 − 2αieωi + αi+1e

ωi+1 (i = 1, 2, · · · , n). (7.12)

The physical interpretation is that the system describes the motion ofa chain of two kinds of particles and the interacting force between twoneighboring particles is attractive or repulsive according to that theybelong to the different kinds or the same kind respectively [65].

This is a system consisting of two different kinds of particles, positiveand negative, with equal mass and the interaction between two neighbor-ing particles of the same (resp. different) kind is repulsive (resp. attrac-tive). The magnitudes of the forces are the same exponential function of

Two dimensional Toda equations and Laplace sequences of surfaces 269

the relative displacement qi+1 − qi. Consequently, the acting force fromthe i + 1th particle to the ith particle is

md2qi

dt2= ±eqi+1−qi . (7.13)

Here we have + sign (resp. − sign) if two particles are of different (resp.same) kinds.

Define αi = 1 if the ith particle and the i+1th particle are of differentkinds, αi = −1 if the ith particle and the i+1th particle are of the samekinds, then the force acting on the ith particle should be

md2qi

dt2= αieqi+1−qi − αi−1eqi−qi−1 (i = 2, · · · , n − 1),

md2q1

dt2= α1eq2−q1 , m

d2qn

dt2= −αn−1eqn−qn−1 .

(7.14)

If the particles lie on a closed chain and let q1 = qn+1, q0 = qn, we have

md2qi

dt2= αieqi+1−qi − αi−1eqi−qi−1 (i = 1, · · · , n),

md2qi+1

dt2= αi+1eqi+2−qi+1 − αieqi+1−qi (i = 1, · · · , n).

(7.15)

Then

md2

dt2(qi+1 − qi) = αi+1eqi+2−qi+1 − 2αieqi+1−qi + αi−1eqi−qi−1 . (7.16)

Letqi+1 − qi = ωi, (7.17)

then

md2ωi

dt2= αi+1ewi+1 − 2αiewi + αi−1ewi−1 . (7.18)

Thus we get the signed Toda equations.Consequently, we see that(i) The arrangement of the particles of the two kinds on the line

determine the values of αi (i = 1, 2, · · · , n).(ii) The values αi (= ±1) determine the arrangement of the two kinds

of particles on the line.(iii) If the chain is a closed one (i.e. q1 = qn+1), then the number of

αi which equals 1 should be even.


We return to the two dimensional case and treat the signed Todaequations from the point of view of integrable system. It is known thatthe two dimensional Toda equations can be deduced from their Lax pair

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟t

= λ

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

0 0 · · · 0 p1

p2 0 · · · 0 0

0 p3 · · · 0 0...

.... . .

......

0 0 · · · pn 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (7.19)

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟x

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

σ11λ 0 · · · 0

0 σ21λ · · · 0

0 0 σ3 · · · 0...

......

. . . 1λ

1λ 0 0 · · · σn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (7.20)

i.e., the integrability conditions of (7.19) and (7.20) are

pi,x = pi(σi − σi−1),

σi,t = pi − pi+1.(7.21)

In the case of pi > 0, we can write pi = eωi and (7.21) is equivalent tothe two dimensional Toda equations (7.8).

However, in the general cases, we should put pi = αieωi in (7.21)

(αi = ±1), and the signed Toda equations (7.12) are derived as well. Inparticular, if

p1 = p3 = · · · = p2k−1 = eω,

p2 = p4 = · · · = p2k = e−ω, n = 2k,(7.22)

the equations (7.21) are reduced to sinh-Gordon equation

ωxt = 4 sinh ω. (7.23)

Whenp1 = p3 = · · · = p2k−1 = e−ω,

p2 = p4 = · · · = p2k = −eω,(7.24)

we have the cosh-Gordon equation

ωxt = −4 cosh ω. (7.25)


If x, t are replaced by ζ, ζ, the complex coordinates of the Euclideanplane, then (7.11) becomes the elliptic version of the signed Toda equa-tions

ωi,ζζ = −αi−1eωi−1 + 2αie

ωi − αi+1eωi+1 . (7.26)

Here the sign on the right hand side of the Toda equations (7.8) andthe signed Toda equations (7.11) is different from that in the ordinarydifferential equations (7.9) and (7.12). This is due to our calculation isaccording to Darboux’s original equation (7.8). If we change t to −t,then the sign of (7.8) and (7.11) will be the same as that in (7.9) and(7.12).

7.2 Laplace sequences of surfaces in projectivespace Pn−1

The Laplace sequences of surfaces as an important subject in classicalprojective differential geometry, have been studied extensively [28, 95,63, 62]. In this section, we elucidate the relationship between Laplacesequences of surfaces of period n in Pn−1 and the Toda equations ofperiod n. Different from Darboux, we consider this problem from thepoint of view of integrable system.

We start from the fundamental equations of the periodic Laplace se-quences of surfaces which are written in the form of first-order partialdifferential equations. Multiplying suitable factors to the homogeneouscoordinates of the points of the surfaces and changing an independentvariable of the fundamental equations, we can simplify the fundamen-tal equations of the periodic Laplace sequences of surfaces quite signifi-cantly. We find that there are two types of n periodic Laplace sequencesof surfaces in Pn−1. It is noted that type II occurs only for even nand was not mentioned by Darboux and the researchers in this field.Moreover, both types have the integrability conditions of the same form

∂2ωi

∂x∂t= −αi−1e

ωi−1 + 2αieωi − αi+1e

ωi+1 , (i = 1, 2, · · · , n; ωn+1 = ω1)

(7.27)where αi = ±1.

Let Pn−1 be n − 1 dimensional projective space, (x1, x2, · · · , xn) bethe homogeneous coordinates of a point N ∈ Pn−1 and

N = N(t, x) (i.e. xa = xa(t, x), a = 1, 2, · · · , n) (7.28)

be the equations of a surface of Pn−1 in homogeneous coordinates. Thestraight line determined by N and NtNN is the tangent line of the t curve(i.e. x = constant) and the straight line passing through N and NxNN isthe tangent lines of the x-curve. Suppose that N , NtNN , NxNN are linearly


independent, then two tangent lines at N do not coincide and N , NtNN ,NxNN determine the tangent plane of the surface N . This also means thatthe surface N is regular.

Suppose that there exist parameters (t, x) of the surface N(t, x) suchthat along each x-curves the tangent lines T (t) of the t curves forma developable surface, i.e., there exists a point λN + NtNN on each T (t)generating a curve C(t) parametrized by x such that the tangent line ofC(t) is just T (t). Analytically,

(λN + NtNN )x = NtxNN + λNxNN + λxN (7.29)

is a linear combination of N and NtNN . Hence N(t, x) should satisfy ahyperbolic equation

NtxNN + aNtNN + bNxNN + cN = 0. (7.30)

Conversely, suppose that a surface N(t, x) satisfies hyperbolic equa-tion (7.30). Let

N ′ = NtNN + bN (7.31)

which is a point on the tangent line T (t). From

N ′xNN = (NtNN + bN)x = NtxNN + bNxNN + bxN = −aNtNN + (bv − c)N, (7.32)

it is seen that when x varies N ′ generates a curve whose tangent lineis T (t), i.e., along each x-curve the tangent lines of the t-curves forma developable surface. If a surface N(t, x) has the above property, wesay that the surface N(t, x) admits a conjugate net and (t, x) are calledconjugate parameters.

We have

Theorem 7.1 A surface N(t, x) in Pn−1 admits a conjugate net withconjugate parameters (t, x) if and only if it satisfies a hyperbolic equation(7.30).

Suppose that N(t, x) satisfies (7.30). The surface

N ′ = NtNN + bN (7.33)

is called the Laplace transformation of the surface N in the t-direction.It can be verified by calculation that N ′ satisfies a hyperbolic equationin the form

N ′txNN + a′N ′

tNN + b′N ′xNN + c′N ′ = 0. (7.30)′

Hence N ′ admits a conjugate net with conjugate parameters (t, x) too.The tangent lines T (t) to the surface N are tangent lines of surface N ′


too, i.e., the straight lines NN ′ constitute a line congruence with focalsurfaces N and N ′.

Similarly, we can define the Laplace transformation of N in the x-direction

N ′′ = NxNN + aN (7.34)

and N ′′ satisfies a hyperbolic equation in the form

N ′′txNN + a′′N ′′

tNN + b′′N ′′xNN + c′′N ′′ = 0.

Any regular surface in P3 always admits a conjugate net as seen inthe elementary theory of surfaces in Euclidean space. In Pn, a surfaceN(t, x) satisfying (7.30) or equivalently admitting a conjugate net iscalled a Laplace surface.

Starting from a given Laplace surface S0S , the Laplace transformationsin t and x directions give a sequence of surfaces

· · · , N−NN l, N−NN l+1, · · · , N0NN , N1NN , · · · , NmNN , · · · . (7.35)

It can be easily shown that the Laplace transformation (7.33), (7.34)and the Laplace sequence (7.35) are independent of the change of homo-geneous coordinates N(t, x) → µN(t, x) (µ = 0) geometrically.

The Laplace sequences of period n in the projective space Pn−1 is ofspecial interest. Let a system of surfaces

NiNN = NiNN (t, x) (i = 1, 2, · · · , n) (7.36)

satisfyNi,tNN = µiNiNN + piNiNN −1 (pi = 0) ,

Ni,xNN = σiNiNN + qiNiNN +1 (qi = 0) ,(7.37)

(NnNN +1 = N1NN , N0NN = NnNN ), then the system of surfaces constitute a Laplacesequence of period n, and vice versa. In fact, the line NiNN NiNN +1 is thecommon tangent line of surfaces NiNN and NiNN +1. In other words, (7.37)means that the surfaces NiNN and NiNN +1 are the two focal surfaces of theline congruence NiNN NiNN +1, and hence NiNN +1 (resp. NiNN ) is the Laplacetransformation of NiNN (resp. NiNN +1).

We shall first simplify the fundamental equation (7.37) of Laplace se-quences by multiplying a suitable factor on the homogeneous coordinatesof each surface NiNN = NiNN (t, x).

LetNiNN = ki(x, t)NiNN , (ki(x, t) = 0) (7.38)

we haveNi,tNN = µiNiNN + piNiNN −1,

Ni,xNN = σiNiNN + qiNiNN +1,(7.37)′


whereµi =

ki,t

ki+ µi, pi =

ki

ki−1pi,

σi =ki,x

ki+ σi, qi =

ki

ki+1qi.

(7.39)

From the expression µi, we choose

ki = k0i (x)e−

∫µi dt, (7.40)

then µi = 0 holds true. Here k0i (x) is an arbitrary function of x.

From the integrability condition of (7.37)′, we get qi,t = qi(µi−µi+1) =0. Hence qi’s depend on x only.

Furthermore, let N ′iNN = mi(x)NiNN (mi(x) = 0), we have

N ′i,tNN = p′iN ′

iNN −1,

N ′i,xNN = σ′

iN′iNN + q′iN ′

iNN +1

(7.37)′′

wherep′i =

mi

mi−1pi, σ′

i =mi,x

miσi, q′i =

mi

mi+1qi,

(mn+1 = m1, m0 = mn).(7.41)

From the last equation, we see that

q′1q′2 · · · q′n = q1q2 · · · qn ≡ Q(x) (7.42)

is independent of the choice of mi. When Q > 0 or Q < 0 and n is odd,let

q = Q1/n (7.43)

and take arbitrary m1 = 0,mi+1 = miqiq

−1 (i = 1, 2, · · · , n), (7.44)

then from (7.41) we have

q′i = q (i = 1, 2, · · · , n). (7.45)

We should note that the choice of mn+1 is consistent with mn+1 = m1,since

mn+1 = mnqnq−1 = mn−1qn−1qnq−2 = · · · = m1qnqn−1 · · · q1q−n = m1.

(7.46)When Q < 0 and n is even, let

q = (−Q)1/n (7.47)


and take arbitrary m1 = 0,mi+1 = miqiq

−1 (i = 1, 2, · · · , n − 1), mn+1 = −mnqnq−1. (7.48)

Then we have

q′i(x) = q(x) (i = 1, 2, · · · , n − 1), q′n = −q(x). (7.49)

The choice of mn+1 is still consistent with mn+1 = m1.Thus we have the following theorem.

Theorem 7.2 There are two types of Laplace sequences of surfaces withperiod n in projective space Pn−1. Their fundamental equations can bewritten as

Type I:

⎧⎨⎧⎧⎩⎨⎨ Ni,tNN = piNiNN −1 (pi = 0) ,

Ni,xNN = σiNiNN + qNiNN +1 (q = 0)(7.50)

and

Type II:

⎧⎨⎧⎧⎩⎨⎨ Ni,tNN = piNiNN −1 (pi = 0) ,

Ni,xNN = σiNiNN + ciqNiNN +1 (q = 0)(7.51)

respectively. Here ci = 1 for i = n and cn = −1. However, the Laplacesequence of Type II can occur only for even n.

By a transformation of the variable x, (7.50) and (7.51) can be reducedto

Type I:

⎧⎨⎧⎧⎩⎨⎨ Ni,tNN = piNiNN −1,

Ni,xNN = σiNiNN + NiNN +1

(7.52)

and

Type II:

⎧⎨⎧⎧⎩⎨⎨ Ni,tNN = piNiNN −1,

Ni,xNN = σiNiNN + ciNiNN +1.(7.53)

Remark 59 In the case n being even, sgn(Q) is a projective invariant ofLaplace sequences of surfaces. This is the reason why there are two typesof Laplace sequences of surfaces of period n for even n.

We write

Ψ =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN...

NnNN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ . (7.54)


By the rescaling (t, x, σi, pi) →(λt,

x

λ, λσi, pi

), we can introduce the

spectral parameter in (7.50) and (7.51). Thus we obtain the Lax pair.

Theorem 7.3 The fundamental equations (7.52) of the Laplace se-quence of type I with period n in Pn−1 are the Lax pair of two dimen-sional signed Toda equations.

Note that the matrix form of (7.52) is (7.19) and (7.20).The Laplace sequence of surface of type II should correspond to the

Lax pair

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟t

= λ

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

0 0 · · · 0 p1

p2 0 · · · 0 0

0 p3 · · · 0 0...

.... . .

......

0 0 · · · pn 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (7.55)

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟x

=

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

σ11λ 0 · · · 0

0 σ21λ · · · 0

0 0 σ3 · · · 0...

......

. . . 1λ

− 1λ 0 0 · · · σn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎝⎜⎜ψ1

...

ψn

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (7.56)

(7.56) differs from (7.20) slightly. However, (7.55) and (7.56) can serveas the Lax pair of the signed Toda equations too.

The integrability conditions of (7.55) and (7.56) are slightly differentfrom (7.21) too. In fact, they are

pa,x = pa(σa − σa−1), σa,t = pa − pa+1 (a = 2, · · · , n − 1),

p1,x = p1(σ1 − σn), σ1,t = −p1 − p2,

pn,x = pn(σn − σn−1), σn,t = p1 + pn.

(7.57)

Comparing with (7.21), it is seen that the only change is the expressionsfor σ1,t and σn,t.

If we set

pa = αaeωa (a = 2, · · · , n), −p1 = α1e

ω1 (7.58)

where

αa = sgn(pa) (a = 2, · · · , n), α1 = sgn(−p1), (7.59)


then (7.57) is reduced to the signed Toda equations (7.11).

Theorem 7.2′ The fundamental equations of the Laplace sequences oftype II with period n are the Lax pair of the two dimensional signed Todaequations too.

Remark 60 If the Laplace sequences of surfaces is non-periodic, thereare infinite number of surfaces corresponding to the infinite number ofToda equations and the Lax pair contains matrices of infinite order.

7.3 Darboux transformationIn the previous chapters we have applied Darboux transformation

to construct new solutions of many integrable systems from some seedsolutions. More precisely, the Darboux transformation is an algorithmicmethod to accomplish the transformation

(u, Φ) → (u′, Φ′).

Here u is a known solution and Φ is a fundamental solution of the Laxpair corresponding to u, u′ is the new solution and Φ′ is the fundamen-tal solution of the Lax pair corresponding to u′. In some geometricalproblems, the fundamental solutions are the geometrical objects to beidentified. In this chapter, (u, Φ) are solutions of the two dimensionalsigned Toda equations and the Laplace sequences of surfaces of periodn in Pn−1. It is very interesting to see how to apply the Darboux trans-formation to these objects.

Matveev has provided the Darboux transformation successively al-ready [79]. We can use his method to construct periodic Laplace se-quences of type I. Besides, Hu has modified Matveev’s method to fit thecase of type II. The Darboux matrix of the two cases can be derived byusing the general formula D = I − λS in a unified way too.

At first, let (Ψ, pi, σi) be the functions appeared in the Lax pair (7.19)and (7.20) in which Ψ = (N1NN , · · · , NnNN )T is the Laplace sequence of typeI in Pn−1 and pi = αie

ωi .The Darboux transformation for Laplace sequence of surfaces of type

I is

N ′aN = NaNN − Ψ0

a

Ψ0a−1

NaNN −1 (7.60)

where

Ψ0a = NaiNN li, (7.61)


NaiNN is the ith homogeneous coordinate of NaNN and li is a set of constants.The corresponding

p′a = pa − Ψ0a

Ψ0a−1

, (7.62)

σ′a = σa +

Ψ0a−1

Ψ0a

− Ψ0a

Ψ0a−1

(a = 1, · · · , n). (7.63)

To construct the Darboux transformation for the Laplace sequence oftype II we can still use (7.62) and (7.63) for a = 2, · · · , n − 2. However,σ′

1, σ′n−1 and σ′

n should be changed slightly as

σ′1 = σ1 +

Ψ02

Ψ01

+Ψ0

1

Ψ0n

,

σ′n−1 = σn−1 +

Ψ0n

Ψ0n−1

− Ψ0n−1

Ψ0n−2

,

σ′n = σn − Ψ0

1

Ψ0n

− Ψ0n

Ψ0n−1

.

(7.64)

The validity of all these formulae can be verified by direct calculations.On the other hand, let λ0 be a special value of the spectral parameter,

Ψ0 be a solution of the Lax pair (7.19) and (7.20) (resp. (7.55) and

(7.56)) with λ = λ0. Let ω = e2π in and Ω = diag(1, ω, ω2, · · · , ωn−1).

The Darboux matrix is I − λS with

S = HΛ−1H−1. (7.65)

HereΛ = λ0Ω, H = (Ψ0, ΩΨ0, · · · , Ωn−1Ψ0). (7.66)

The Darboux transformation is

Ψ′ = (λ − λS)Ψ. (7.67)

Direct calculations show that the formulae (7.60) and (7.62)–(7.64) hold.Now we turn to some examples, i.e., use the above Darboux transfor-

mation to construct Laplace sequences.

(1) Laplace sequence of type ITake the trivial solution of two dimensional Toda equation

σa = 0, pa = 1. (7.68)

The fundamental equations (7.52) are reduced to

N1NN ,t = N4NN , N2NN ,t = N1NN , N3NN ,t = N2NN , N4NN ,t = N3NN ,

N1NN ,x = N2NN , N2NN ,x = N3NN , N3NN ,x = N4NN , N4NN ,x = N1NN .(7.69)


Solving the equations, we obtain the Laplace sequence of period 4:

N1NN cosh u sinhu cos v sin v

N2NN sinhu cosh u sin v − cos v

N3NN cosh u sinhu − cos v − sin v

N4NN sinhu cosh u − sin v cos v

(7.70)

Here u = t + x, v = t − x.In (7.70), the rows are homogeneous coordinates of the point NaNN , and

each column is a solution of the Lax pair

∂Ψa

∂t= Ψa−1,

∂Ψa

∂x= Ψa+1 (a = 1, 2, 3, 4; Ψ0 = Ψ4, Ψ5 = Ψ1).

(7.71)It is seen that N1NN and N3NN generate the surface

S1 : x21 − x2

2 = x23 + x2

4 or x2 + y2 + z2 = 1 (7.72)

if we use the inhomogeneous coordinates x = x1/x4, y = x2/x4, z =x3/x4. Similarly, N2NN and N4NN generate the surface

S2 : x22 − x2

1 = x23 + x2

4 or y2 − x2 − z2 = 1. (7.73)

A typical quadrilateral together with S1 and S2 is shown in Figure 7.1.By using Darboux transformation, we obtain a new Laplace sequence

of period 4

N ′1NN = N1NN − Ψ0

1

Ψ04

N4NN , N ′2NN = N2NN − Ψ0

2

Ψ01

N1NN ,

N ′3NN = N3NN − Ψ0

44

3

Ψ02

N2NN , N ′4N = N4NN − Ψ0

11

4

Ψ03

N3NN .

(7.74)

Here Ψ0a is a linear combination of the 4 columns in (7.70), i.e.,

Ψ01 = a cosh u + b sinhu + c cos v + d sin v,

Ψ02 = a sinhu + b cosh u + c sin v − d cos v,

Ψ03 = a cosh u + b sinhu − c cos v − d sin v,

Ψ04 = a sinhu + b cosh u − c sin v + d cos v.

(7.75)

By a long calculation, we obtain

N ′1 b − cz3 + dz2 −a − cz1 − dz4 az3 + bz1 + d −az2 + bz4 − c

N ′2NN −b − cz4 + dz1 a + cz2 + dz3 az4 − bz2 + d −az1 − bz3 − c

N ′3NN b + cz3 − dz2 −a + cz1 + dz4 −az3 − bz1 + d az2 − bz4 − c

N ′4NN −b + cz4 − dz1 a − cz2 − dz3 −az4 + bz2 + d az1 + bz3 − c.

(7.76)


Figure 7.1. Example of Laplace sequences of surfaces of period 4 (Type I): N1 andN3NN are on the surface x2 +y2 + z2 = 1, N2NN and N4NN are on the surface y2 −x2 − z2 = 1

Herez1 = cosh u cos v + sinhu sin v,

z2 = cosh u cos v − sinhu sin v,

z3 = cosh u sin v + sinhu cos v,

z4 = cosh u sin v − sinh u cos v.

(7.77)

It is not difficult to prove that N ′1NN , N ′

2NN , N ′3NN , N ′

4N generate four algebraicvarieties in P3. In fact, among the parameters z1, z2, z3, z4, there arealgebraic relations

z1z2 + z3z4 = 0, z21 − z2

2 − z23 + z2

4 = 0. (7.78)

From these relations and the parametric representation of N ′iNN , we can

eliminate these parameters and find that x = x1/x4, y = x2/x4, z =x3/x4 satisfy an algebraic equation FiFF (x, y, z) = 0. Hence each N ′

iNN lieson an algebraic variety.

On can apply Darboux transformations successively to get an infinitesequence of Laplace sequences of surfaces of period 4 in P3.

(2) Laplace sequences of Type IINow we construct the Laplace sequences of surfaces of type II of period

4 in P3 explicitly. Here we also take λ0 = 1.


For the trivial solution of (7.57), we can take

σ1 = σ2 = σ3 = σ4 = 0, p1 = −1, p2 = p3 = p4 = 1. (7.79)

So the fundamental equations for the Laplace sequences are

N1NN ,t = −N4NN , N2NN ,t = N1NN , N3NN ,t = N2NN , N4NN ,t = N3NN ,

N1NN ,x = N2NN , N2NN ,x = N3NN , N3NN ,x = N4NN , N4NN ,x = −N1NN .(7.80)

Solving these equations we obtain

N1 ev cos u ev sin u e−v cos v e−v sin v

N2NNev(cos u − sin u)√

2

ev(cos u + sin u)√2

− e−v(cos u + sin u)√2

e−v(cos u − sin u)√2

N3NN −ev sin u ev cos u e−v sin u −e−v cos u

N4NN − ev(sin u + cos u)√2

ev(cos u − sin u)√2

e−v(cos u − sin u)√2

e−v(cos u + sin u)√2

(7.81)Here u = (x − t)/

√2, v = (x + t)/

√2, N1NN and N3NN generate the surface

S1 : x1x4 = x2x3 or x = yz, N2NN and N4NN generate the surface S2 : x1x3 =−x2x4 or y = −xz.

A typical quadrilateral N1NN , N2NN , N3NN , N4NN together with the surfaces S1

and S2 is shown in Figure 7.2.By using the formula (7.60), we can get

N ′aN = NaNN − Ψ0

a

Ψ0a−1

NaNN −1 (7.82)

as in the case of Type I. Hence the Darboux transformation gives a newLaplace sequence of period 4. These four surfaces are algebraic surfacestoo.

Remark 61 The surfaces (7.72), (7.73) do not contain straight lines,while the surfaces S1 and S2 appeared in Laplace sequences of surfacesof type II contain two systems of straight lines. This fact reflects that theLaplace sequences of surfaces of type I and those of type II are not equiv-alent in the real projective geometry. The projective invariant sgn(Q) oftype I and type II are +1 and −1 respectively.

7.4 Su chain (Finikoff configuration)There are a class of special Laplace sequences of period 4 in the pro-

jective space P3. They were introduced by P. S. Finikoff and B. Q. Suaround 1930 with a plenty of geometrical properties but without explicitexamples [28, 95]. Su showed that the integrability condition of the fun-damental equations is the sinh-Gordon equation. In this section, we


Figure 7.2. Example of Laplace sequences of surfaces of period 4 (Type II): N1 andN3NN are on the surface z = xy, N2NN and N4NN are on the surface y = −xz

simplify the fundamental equations of the Su chain and introduce thespectral parameter. Then the Darboux transformations are constructedand a series of explicit examples are obtained [62].

Following Su’s notations, the Laplace sequences of the surfaces isN1NN N3NN N2NN N4NN and the fundamental equations are reduced to⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟u

=12

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0 0 −e−φ 0

0 0 0 e−φ

0 eφ 0 0

eφ 0 0 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ , (7.83)

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟v

=12

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜−φv 0 0 1

0 −φv 1 0

−1 0 φv 0

0 1 0 φv

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ . (7.84)


To introduce the spectral parameter, we use the transformation u →u

λ, v → λv and the equations for Su chain lead to the Lax pair

Φu = λUΦ =λ

2

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0 0 −e−φ 0

0 0 0 e−φ

0 eφ 0 0

eφ 0 0 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟Φ, (7.85)

Φv = V Φ =12

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜−φv 0 0 1/λ

0 −φv 1/λ 0

−1/λ 0 φv 0

0 1/λ 0 φv

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟Φ. (7.86)

The integrability condition of the Lax pair is

φuv = sinh φ. (7.87)

We have already known the way of constructing the solutions of (7.87)together with the solution of 2 × 2 matrix Lax pair. Now we shouldconstruct the solution of (7.87) together with 4 × 4 matrix Lax pair(7.85) and (7.86).

Introduce 2 × 2 matrices

a =

⎛⎝⎛⎛ −1 0

0 1

⎞⎠⎞⎞ , b =

⎛⎝⎛⎛ 0 1

1 0

⎞⎠⎞⎞ ,

I =

⎛⎝⎛⎛ 1 0

0 1

⎞⎠⎞⎞ , O =

⎛⎝⎛⎛ 0 0

0 0

⎞⎠⎞⎞ .

(7.88)

Write U and V to be block matrices

U =12

⎛⎝⎛⎛ 0 e−φa

eφb 0

⎞⎠⎞⎞ , V = C +1λ

D, (7.89)

where

C =12

⎛⎝⎛⎛ −φvI 0

0 φvI

⎞⎠⎞⎞ , D =12

⎛⎝⎛⎛ 0 b

a 0

⎞⎠⎞⎞ , (7.90)

then the Lax pair (7.85) and (7.86) take the form

Φu = λUΦ, Φv =(C +

1λ

D)Φ. (7.91)


Consider the 2 × 2 Lax pair

Ψu = λUΨ, Ψv = V Ψ, (7.92)

where

U =12

⎛⎝⎛⎛ 0 e−φ

eφ 0

⎞⎠⎞⎞ , V =12

⎛⎝⎛⎛ −φv 1/λ

1/λ φv

⎞⎠⎞⎞ . (7.93)

This Lax pair has appeared in Section 4.3. Let (h1, h2)T be a columnsolution of (7.92) for λ = λ1, then the Darboux matrix is

D(λ) = I − λ

λ1

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 0h1

h2h2

h10

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 1 − λ

λ1

h1

h2

− λ

λ1

h2

h11

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ , (7.94)

and the new solution φ1 of the sinh-Gordon equation is defined by

eφ1 = e−φ(h2

h1

)2. (7.95)

Now introduce the map E from 2 × 2 matrices to 4 × 4 matrices:

E :

⎛⎝⎛⎛ α β

γ δ

⎞⎠⎞⎞ −→⎛⎝⎛⎛ αI βa

γb δI

⎞⎠⎞⎞ . (7.96)

The Darboux matrix D(λ) is mapped to

D(λ) =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ I −λ1

λ

h1

h2a

−λ1

λ

h2

h1b I

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ . (7.97)

By using the differential equations for h1 and h2, we can see that Φ1(λ) =D(λ)Φ(λ) satisfies

Φ1u(λ) = λU1UU Φ1(λ), Φ1v(λ) = V1VV Φ1(λ), (7.98)

where U1UU and V1VV are derived from replacing φ in U and V by φ1 in(4.175). Φ1 is a Su chain obtained from the known Su chain by Darbouxtransformation. In fact,

Φ1(λ) =

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ I − λ

λ1

h1

h2a

− λ

λ1

h2

h1b I

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟Φ(λ). (7.99)


Differentiating it with respect to u, the first equation of (7.98) is

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ 0 − λ

λ1

(h1

h2

)ua

− λ

λ1

(h2

h1

)ub 0

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟

+λ

2

⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ I − λ

λ1

h1

h2a

− λ

λ1

h2

h1b I

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟⎛⎝⎛⎛ 0 e−φa

eφb 0

⎞⎠⎞⎞

=λ

2

⎛⎝⎛⎛ 0 e−φ1a

eφ1b 0

⎞⎠⎞⎞⎛⎜⎛⎛⎜⎜⎜⎝⎜⎜ I − λ

λ1

h1

h2a

− λ

λ1

h2

h1b I

⎞⎟⎞⎞⎟⎟⎟⎠⎟⎟ .

(7.100)

Equating the terms with λ2, we get the first equation of (4.172)

eφ1 =(h2

h1

)2e−φ.

The terms with λ give

− 1λ1

(h1

h2

)u

+e−φ

2− e−φ1

2= 0,

− 1λ1

(h2

h1

)u

+e−φ

2− e−φ1

2= 0,

(7.101)

which are the equations thath1

h2and

h2

h1in 2×2 Darboux matrix should

satisfy. Hence the first equation of (7.98) holds.The second equation of (7.98) can be verified in a similar way. Hence

the matrix D(λ) derived from the Darboux matrix D(λ) for the 2×2 Laxpair is exactly a Darboux matrix for the Su chain. Thus the followingtheorem holds.

Theorem 7.4 Suppose a solution of the sinh-Gordon equation and afundamental solution of its Lax pair are known. Then a series of Suchains can be constructed by the map E and the Darboux transformationsuccessively. The algorithm is purely algebraic.


Now we give some geometric properties. Let µ =h1

h2, then

D(λ) =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

1 0λ

λ1µ 0

0 1 0 − λ

λ1µ

− λ

λ1

1µ

0 1 0

0 − λ

λ1

1µ

0 1

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟. (7.102)

Su chain obtained from a known one via Darboux transformation is⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N ′

1

N ′2NN

N ′3NN

N ′4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜1 0

1

λ1µ 0

0 1 0 − 1

λ1µ

− 1

λ1

1

µ0 1 0

0 − 1

λ1

1

µ0 1

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

N1

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1 +

µ

λ1N3NN

N2NN − µ

λ1N4NN

N3NN − 11

λ1µN1

N4NN − 1

λ1µN2NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ .

(7.103)When u and v are fixed, N1NN N3NN N2NN N4NN forms a spatial quadrilateral. Afterthe Darboux transformation, N ′

1NN , N ′3NN , N ′

2NN and N ′4N locate on four sides of

the quadrilateral N1NN N3NN N2NN N4NN . We say that the quadrilateral N ′1NN N ′

3NN N ′2NN N ′

4N ’is incident to the quadrilateral N1NN N3NN N2NN N4NN . The following theorem holds.

Theorem 7.5 Under the Darboux transformation, the spatial quadri-lateral N ′

1NN N ′3NN N ′

2NN N ′4N is incident to N1NN N3NN N2NN N4NN .

In the construction of Darboux transformation of the sinh-Gordonequation, we want two parameters. One is assigned value of the spectralparameter λ = µ, another is a constant column vector l such that h =Φ(µ)l. Now let (µA, lA) be n sets of parameters, SC0CC be a given Su chain.By applying the Darboux transformation with parameters (µA, lA) (A =1, 2, · · · , n) successively, we get a series of Su chains

SC0CC → SC1 → · · · → SCnCC −1 → SCnCC .

If the order of the parameters (µA, lA) is changed, we have another seriesof Su chains

SC0CC → SC ′1 → · · · → SC ′

nCC −1 → SC ′nCC .

From the theorem of permutability, SCnCC = SC ′nCC . Since the inverse of

a Darboux transformation is a Darboux transformation too, we have aperiodic chain of Su chains of period 2n

SC0CC

SC1 → SC2CC → · · · → SCnCC −1

SC ′1 ← SC ′

2CC ← · · · ← SC ′nCC −1

SCnCC .


Therefore, we get a series of solutions with period 2n (n = 2, 3, 4, · · ·),i.e., with period 4, 6, 8, · · ·.

Theorem 7.6 There exists a periodic chain of Su chains of period 2nconnected by Darboux transformations and each quadrilateral of these Suchains is incident to the quadrilateral of the neighboring Su chain.

Especially, the series of Su chains of period 4 has four Su chains. Thequadrilaterals of their congruences are incident one another.

Example:The simplest Su chain corresponds to the solution φ = 0. In this case,

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟u

=λ

2

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0 0 −1 0

0 0 0 1

0 1 0 0

1 0 0 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ ,

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟v

=12λ

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜0 0 0 1

0 0 1 0

−1 0 0 0

0 1 0 0

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N1NN

N2NN

N3NN

N4NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ .

(7.104)

The first equation can be written as

N1NN u = −λ

2N3NN , N3NN u =

λ

2N2NN , N2NN u =

λ

2N4NN , N4NN u =

λ

2N1NN ,

hence

N1NN uuuu = −(λ

2

)4N1NN . (7.105)

Solving this ordinary differential equation,

N1NN = A1 cosh β cos β + A2 cosh β sinβ

+A3 sinhβ cos β + A4 sinhβ sin β,(7.106)


where β =√

2λ

4u, and A1, A2, A3, A4 are four dimensional row vectors

which are independent of u. Then,

N3NN = − 2λ

N1NN u =√

22

(− (A2 + A3) cosh β cos β

+(A1 − A4) cosh β sin β − (A1 + A4) sinhβ cos β

+(A3 − A2) sinhβ sin β),

N2NN =2λ

N3NN u = −A4 cosh β cos β

+A3 cosh β sin β − A2 sinh β cos β

+A1 sinh β sinβ,

N4NN =2λ

N2NN u =√

22

((A3 − A2) cosh β cos β

+(A1 + A4) cosh β sin β + (A1 − A4) sinhβ cos β

+(A2 + A3) sinhβ sin β).

(7.107)

From the second equation of (7.104),

N1NN v =12λ

N4NN , N4NN v =12λ

N2NN ,

N2NN v =12λλ

2λN3NN , N3NN v = − 1

2λN1NN ,

and hence

A1v =12λ

√2

2(A3 − A2), A2v =

12λ

√2

2(A4 + A1),

A3v =12λ

√2

2(A1 − A4), A4v =

12λ

√2

2(A2 + A3).

(7.108)

By differentiation, we have Ai,vvvv = −(λ

2

)4Ai (i = 1, 2, 3, 4). Hence

each component of

Ai = (ai, bi, ci, di) (i = 1, 2, 3, 4)

are linear combinations of coshα cos α, cosh α sinα, sinh α cos α and

sinhα sin α with constant coefficients. Here α =√

24

λv.When the initial condition is taken as

N1NN = (1, 0, 0, 0), N2NN = (0, 1, 0, 0),

N3NN = (0, 0, 1, 0), N4NN = (0, 0, 0, 1)(7.109)


at u = v = 0, then

A1 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜cosh α cos α

sinhα sinα√

22 (− sinhα cos α + cosh α sin α)√

22 (cosh α sinα + sinh α cos α)

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

,

A2 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜cosh α sin α

− sinhα cos α√

22 (− cosh α cos α − sinh α sin α)√

22 (− cosh α cos α + sinhα sinα)

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

,

A3 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜sinhα cos α

cosh α sin α√

22 (− cosh α cos α + sinhα sinα)√

22 (cosh α cos α + sinh α sin α)

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

,

A4 =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜sinhα sinα

− cosh α cos α

−√

22 (cosh α sinα + sinhα cos α)

√2

2 (cosh α sinα − sinh α cos α)

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

.

Substituting into (7.107) and letting λ = 1, we can get the first Su

chain. Let α + β = σ, α − β = τ . When λ = 1, σ =√

24

(u + v),

τ =√

24

(u − v).

N1NN =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜cosh σ cos τ

sinhσ sin τ√

22 (cosh σ sin τ − sinhσ cos τ)√

22 (cosh σ sin τ + sinhσ cos τ)

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

,


N2NN =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜− sinhσ sin τ

cosh σ cos τ√


√2

2 (− cosh σ sin τ + sinhσ cos τ)

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

,

N3NN =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

√2

2 (− cosh σ sin τ − sinhσ cos τ)√

22 (− cosh σ sin τ + sinhσ cos τ)

cosh σ cos τ

− sinhσ sin τ

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

,

N4NN =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

√2

2 (− cosh σ sin τ + sinhσ cos τ)√


sinh σ sin τ

cosh σ cos τ

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟T

.

It is easy to see that N1NN and N2NN are the same quadratic surface

x23 − x2

4 = −2x1x2, (7.110)

N3NN and N4NN are the same quadratic surface

x21 − x2

2 = −2x3x4. (7.111)

The second Su chain can be constructed as follows. From (7.94), the2 × 2 Darboux matrix is

I − λ

λ1

⎛⎝⎛⎛ 0 tanh γ1

coth γ1 0

⎞⎠⎞⎞ ,

where γ1 =λ1

2u +

12λ1

v. Hence the Darboux matrix for the Su chain is

I − λ

λ1

⎛⎝⎛⎛ 0 a tanh γ1

b coth γ1 0

⎞⎠⎞⎞ .


Let λ = 1, then

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜N ′

1NN

N ′2NN

N ′3NN

N ′4N

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟ =

⎛⎜⎛⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎜⎜

N1NN +tanh γ1

λ1N3NN

N2NN − tanh1γ1

λ1N4NN

N3NN − coth1γ1

λ1N1NN

N4NN − coth1γ1

λ1N2NN

⎞⎟⎞⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎟⎟. (7.112)

A series of solutions can be obtained successively in this way.

7.5 Elliptic version of Laplace sequence ofsurfaces in CPn

In this section, we define the elliptic Laplace sequences and obtaintheir relations with harmonic sequences. We also get explicit examplesof harmonic sequences [21]. Using Darboux transformation, a series ofexplicit harmonic sequences can be constructed.

7.5.1 Laplace sequence in CPn

CPn is the n dimensional complex projective space. Let Z = (Z1, · · · ,Zn+1) be homogeneous coordinates of a point in CPn where Zi (i =1, 2, · · · , n + 1) are complex numbers which are not all zero. Z = f(ζ, ζ)is a map from R2 (or a region of R2) to CPn. Here ζ = z + iy is thecomplex coordinate of R2. If f satisfies

fζff ζ + afζff + bfζff + cf = 0 (7.113)

and fζff , fζff , f are linearly independent, then f is called an elliptic Laplacesurface in CPn parametrized by (ζ, ζ). We use [f ] to denote the surfaceor the class of functions λf |λ = 0 .

As for the case of real projective space, we can define two kinds ofLaplace transformations LI([f ]) and LII([f ]) of an elliptic Laplace sur-face [f ] by

f1 = fζff + bf, f−1 = fζff + af (7.114)respectively. Geometrically, the transformations LI and LII are inde-pendent of the choice of f in [f ], i.e., if we replace f by λf (λ = 0), thenthe surface LI([f ]) and LII([f ]) are unchanged. A sequence of ellipticLaplace surfaces

· · · , [f−2], [f−1], [f ], [f1], [f2ff ], · · · (7.115)

can be obtained such that

[fiff +1] = LI([fiff ]), [fiff −1] = LII([fiff ]). (7.116)


The sequence (7.115) is called an elliptic Laplace sequence, which isdetermined by [f ] completely, provided that fi,ζff , fi,ff ζ and fiff are linearlyindependent.

An elliptic Laplace sequence of surfaces can be represented in thecanonical form

fi,ζff ζ = σifi,ff ζ + pififf ,

fi,ζff = fiff +1 + σififf ,

fiff +1,ζ = pi+1fiff

(7.117)

andpi+1 = pi − σi,ζ , σi+1 =

pi+1,ζ

pi+1+ σi. (7.118)

In fact, by changing f to f0ff = λf , (7.113) can be written as

f0ff ,ζζ = σ0f0ff ,ζ + p0f0ff (p0 = 0) (7.119)

and (7.114) takes the form

f0ff ,ζ = f1 + σ0f0ff , f0ff ,ζ = p0f−1. (7.120)

All other equations in (7.117) can be obtained by acting LI and LII

successively.

Remark 62 If pi = αieωi , (7.118) is equivalent to the signed Toda equa-

tions of elliptic version

ωi,ζζ = −αi−1eωi−1 + 2αie

ωi − αi+1eωi+1 . (7.121)

7.5.2 Equations of harmonic maps from R2 toCPn in homogeneous coordinates

In Chapter 5 we have seen that the energy of the map from a Rieman-nian manifold (M, g) to another Riemannian manifold (N, a) is definedby

E(φ) =∫

E(φ) dVMVV . (7.122)

Here

E(φ) = gijaαβ∂φα

∂xi

∂φβ

∂xj, dVMVV =

√g

√√dx1 · · ·dxn (7.123)

in local coordinates. The map φ is called a harmonic map if it is a criticalpoint of the energy integral and the Euler-Lagrange equations

∂(E(φ)√

g√√

)∂yα

− ∂

∂xi

(∂(E(φ)

√g

√√)

∂yαi

)= 0 (7.124)


are the equations of harmonic maps.By introducing the Hermitian metric

(Z, W ) =n+1∑α=1

ZαWα, |Z|2 =

n+1∑α=1

ZαZα (7.125)

in Cn+1, CPn becomes a Riemannian manifold with the Fubini-Studymetric [69]

ds2 =|Z|2(dZ, dZ) − (Z,dZ)(dZ, Z)

|Z|4 . (7.126)

Lemma 7.7 The map [f ] : R2 → CPn is harmonic if and only if

fζff ζ = σfζff + pf (7.127)

and

(fζff , f) = 0, (7.128)

σ =(|f |2)ζ

|f |2 (7.129)

hold true for some f ∈ [f ]. Here

p = −(fζff , fζff )|f |2 . (7.130)

The proof can be done by writing out and simplifying the Euler-Lagrange equations based on the Fubini-Study metric.

By using this lemma and the canonical form of elliptic Laplace se-quences, we obtain the following result.

Theorem 7.8 Let [fiff ] be an elliptic Laplace sequence. If one of [fiff ](say [f0ff ]) is harmonic, then all [fiff ] are harmonic maps.

Proof. It is sufficient to prove that f1 and f−1 are harmonic if f0ff = fis harmonic. At first, it is seen that the condition (7.128) means thatf ⊥ f−1 and condition (7.129) means that f ⊥ f1. From f1 = fζff − σf ,it follows that

f1,ζ = fζff ζ − σζf − σfζff = (p − σζ)f. (7.131)

Hence (f1,ζ , f1) = (p − σζ)(f, f1) = 0. Besides,

p1 =(f1,ζζ , f1)

|f1|2 = −(f1,ζ , f1,ζ)|f1|2 = −|p1|2|f |2

|f1|2 . (7.132)


Hence

p1 = −|f1|2|f |2 . (7.133)

From

σ1 =p1,ζ

p1+ σ =

(|f1|2)ζ

|f1|2 − (|f |2)ζ

|f |2 , (7.134)

it follows that condition (7.129) is satisfied. Hence [f1] is harmonic.Now turn to f−1. From fζff = pf−1, it follows (f−1, f) = 0. Hence

σ−1 =(f−1,ζ , f−1)(f−1, f−1)

. (7.135)

Since f is harmonic, we have

p|f |2 = −(fζff , fζff )|f |2 = −|p|2|f−1|2, (7.136)

and hence

p = − |f |2|f−1|2 . (7.137)

Fromσ =

pζ

p+ σ−1, (7.138)

we have

σ−1 =(f−1, f−1)ζ

(f−1, f−1). (7.139)

Hence(f−1,ζ , f−1) = (f−1, f−1)ζ (7.140)

and it follows(f−1, f−1,ζ) = 0. (7.141)

Hence f−1 is harmonic. The theorem is proved.Thus, the Laplace sequences of surfaces becomes the harmonic se-

quences in [16].

Theorem 7.9 Let [fiff ] be an elliptic Laplace sequence. If f−1 ⊥ f0ffand f0ff ⊥ f1, the [fiff ] is a harmonic sequence.

From the properties of harmonic sequences or by direct calculation,there are following facts.

(1) Let

eωi =|fiff +1|2|fiff |2 . (7.142)


The elliptic Toda equations

ωi,ζζ = eωi−1 − 2eωi + eωi+1 (7.143)

are satisfied. Hence from a harmonic sequence, a solution of two-dimen-sional elliptic Toda equations can be constructed.

(2) If f−1 ⊥ f0ff , f0ff ⊥ f1 and f−1 ⊥ f1, then fiff −1 ⊥ fiff , fiff ⊥ fiff +1 andfiff −1 ⊥ fiff +1 holds true for all i, and these fiff ’s are minimal surfaces inCPn.

For the proof of fiff −1 ⊥ fiff +1, it is sufficient to verify f ⊥ f2ff andf ⊥ f−2. From

0 = (f, f−1)ζ = (fζff , f−1)+(f, f−1,ζ) = (f1, f−1)−(σf, f−1)+¯−1(f, f−2),(7.144)

it is seen that (f, f−2) = 0. From

0 = (f1, f)ζ = (f1,ζ , f) + (f1, fζff ) = (f2ff , f) + (σ1f1, f−1) + (f1, f−1),(7.145)

we can see (f2ff , f) = 0. Moreover, the metric on the surface [fiff ] inducedfrom the Fubini-Study metric is proportional to the metric dζ dζ, i.e.,the map [fiff ] is conformal. It is well-known that a harmonic map isminimal if it is conformal [114].

7.5.3 Cases of indefinite metricIf we use the indefinite metric

(Z, W )J =J∑

a=1

ZaWa −

n+1∑b=J+1

ZbWb,

|Z|2J =J∑

a=1

ZaZa −

n+1∑b=J+1

ZbZb

(7.146)

in Cn+1 to replace the positive definite metric (Z, W ), we obtain thesubmanifolds

CPnJ+JJ = [Z] | |Z|2J > 0,

CPnJ− = [Z] | |Z|2J < 0,

CPnJ0JJ = [Z] | |Z|2J = 0,

(7.147)

and the Fubini-Study metric can be extended to CPnJ+JJ and CPn

J− .The above results can be extended to CPn

J+JJ and CPnJ− with some

modifications, provided that |fiff |2J = 0. In these cases the two-dimen-sional signed Toda equations of elliptic version

ωp,ζζ = αp−1eωp−1 − 2αpe

ωp + αp+1eωp+1 (7.148)

can be obtained.


7.5.4 Harmonic maps from R1,1

Instead of R2 (or S2), we take the Minkowski plane R1,1 = (ξ, η)with ds2 = dξ dη, then the Laplace sequence can be written in the form

fi,ξηff = σifi,ηff + pififf ,

fi,ξff = fnff +1 + σififf ,

fiff +1,η = pi+1fiff

(7.149)

and

pi+1 = pi − σi,η,

σi+1 =pi+1,η

pi+1+ σi.

(7.150)

Moreover, it is seen that if fiff −1 ⊥ fiff and fiff ⊥ fiff +1, then fiff is a harmonicmap (wave map) from R1,1 to CPn (or CPn

J+JJ , CPnJ−). However, in

general fiff −1 ⊥ fiff and fiff ⊥ fiff +1 do not imply fiff +1 ⊥ fiff +2, i.e., startingwith a harmonic map, the Laplace sequence may not be a harmonic map.

7.5.5 Examples of harmonic sequences from R2 toCPn or R1,1 to CPn

Example 1. Let

f = (f1, f2, · · · , fn+1)

=(α1 exp

(λ1ξ − ξ

λ1

), · · · , αn+1 exp

(λn+1ξ − ξ

λn+1

)).

(7.151)

We take λk (k = 1, 2, · · · , n + 1) such that |λk|2 = 1 and

|α1|2λ1 + · · · + |αn+1|2λn+1 = 0. (7.152)

Then

· · · , fzff z, fzff , f, fzff , fzzff , · · · (7.153)

is a harmonic sequence in CPn since fζff ζ + f = 0.In particular, if we take

|αk|2 = 1, λk = e2kπn+1

i, (7.154)

then the harmonic sequence is of period n.


Example 2. For a harmonic sequence from R1,1 to CPn−1 (n = 2k)

f = (f1, f2, · · · , fn),

f1 = α1 cos(λ1ξ − η

λ1

), f2 = α1 sin

(λ1ξ − η

λ1

),

f3 = α2 cos(λ2ξ − η

λ2

), f4 = α2 sin

(λ2ξ − η

λ2

),

· · ·f2k−1 = αk cos

(λkξ − η

λk

), f2k = αk sin

(λkξ − η

λk

),

(7.155)

where λ1, · · ·, λk are real numbers, we have

(f, f) = |α1|2 + · · · + |αn|2 = constant,

(f, fξff ) = 0, (f, fηff ) = 0,

fξηff + f = 0.

(7.156)

The Laplace sequence is

· · · , fηηff , fηff , f, fξff , fξξff , · · · . (7.157)

It is easily verified that

(fkff , fkff +1) = 0, (fkff −1, fkff ) = 0. (7.158)

Hence fkff is a harmonic map.

Index

2+1 dimensional AKNS system, 65, 68,69, 87

2+1 dimensional N-wave equation, 89, 97,100

AKNS hierarchy, 15AKNS system, 1, 12, 16, 18, 23, 25, 30,

32, 34–36, 38, 41, 42, 45–48,51, 56, 58, 59, 64, 70, 75, 103,108, 111, 117, 237, 242, 244,247

Backlund congruence, 121, 122, 131, 132,¨164, 170, 174, 181, 184, 186,187

Backlund transformation, 28, 45, 120–¨122, 134, 135, 138, 149, 150,154, 156, 158–164, 166, 170,172, 186, 187, 228–230

binary Darboux transformation, 65, 82–84

Bogomolny equation, 248–250, 255, 257Boussinesq equation, 86

Chebyshev coordinates, 128, 132, 134,135, 138, 144, 147, 155, 156,161, 164–166, 175, 181, 184

Chebyshev frame, 128, 132, 154, 175–179,196–198

continuous spectrum, 55, 63cosh-Gordon equation, 147, 149, 161

Darboux matrix, 1, 4, 7, 11, 18, 19, 21–23,25–28, 30–34, 38, 39, 45–48,50, 51, 59, 60, 65, 90, 92, 93,108, 109, 118, 120, 137, 139,157, 158, 162, 173, 176, 203,212, 214, 218, 225, 226, 242–244, 250, 258–260, 277, 278,284, 285, 290

Darboux operator, 65, 71–73, 75–77, 79,83–86

Darboux transformation, 1–7, 9–11, 18,23, 25, 28–31, 34, 36–40, 44–47, 50, 51, 59–61, 63, 65–68,70–72, 75, 77, 79, 80, 82–84,86, 87, 89, 93–95, 98, 103,108–111, 113, 115, 117, 118,120–122, 129, 135, 136, 138,139, 149, 150, 154–161, 163,164, 166–170, 172, 176, 179,181, 183, 186, 187, 189, 192,198, 203–205, 207, 208, 210,212, 214, 217, 218, 226–228,233, 237, 244, 245, 247, 250–252, 254–260, 262, 264, 277–281, 284–287

Davey-Stewartson equation, 69, 75differential polynomial, 14, 15, 17–19, 23,

24, 32, 40, 42, 43, 68, 69, 106DS equation, 69DSII equation, 70, 78, 80, 82DSI equation, 65, 70, 78, 82–84, 89, 97

elliptic Laplace sequence, 292–294elliptic Laplace surface, 291elliptic Toda equations, 295

focal surface, 130–132, 150–155, 164, 165,170

gauge transformation, 238, 255, 256

harmonic map, 189, 191–196, 198, 199,201, 203, 207, 208, 210, 212–214, 217–221, 230, 233, 292,293, 295–297

harmonic sequence, 294–297Higgs field, 248, 254


integrability condition, 2–5, 11, 12, 16,18, 31–33, 35, 41, 46, 57, 66,68, 71, 74, 84, 88, 100, 104,105, 124, 125, 134, 142, 155–157, 160, 161, 164–166, 182–184, 196, 202, 214, 239–241,249, 270, 274, 276, 283

inverse scattering, 1, 51, 56, 58, 63, 64

Jost solution, 52, 53, 59, 61, 63

KdV equation, 2, 3, 9, 10, 36, 37, 51, 64,66, 86, 213

KdV hierarchy, 35, 36, 38, 68KP equation, 65–67, 70, 75, 77, 78, 87KP hierarchy, 68KPII equation, 67KPI equation, 65, 67

Laplace sequence, 268, 273, 275–282, 294,296, 297

Laplace transformation, 130, 267, 272,273

Lax pair, 2–7, 11, 16, 18, 23, 25, 26, 28,30–37, 39, 40, 45–47, 50, 51,58, 59, 64, 66, 67, 69–71, 75,78, 79, 82, 86–88, 95, 122, 136–139, 155, 157, 158, 160, 161,164, 166–168, 170–172, 174–177, 179, 183, 202–204, 214,215, 219–221, 223, 248–250,252, 255–257, 259, 260, 264,270, 276–279, 283–285

Lax set, 87, 104, 106, 107, 109, 110, 238,240, 244, 246, 247

line congruence, 122, 129, 130, 149, 150

Miura transformation, 10MKdV equation, 3–8, 10, 11, 13, 42MKdV hierarchy, 42MKdV-SG hierarchy, 35, 43, 45

negative sine-Gordon equation, 145negative sinh-Laplace equation, 129, 183nonlinear constraint, 65, 87, 88nonlinear Schrodinger equation, 14, 46, 69¨nonlinear Schrodinger hierarchy, 35, 46–¨

48, 59normalized harmonic map, 195–198

pseudo-spherical congruence, 130–135,149, 150, 154–156, 164, 165

scattering data, 51, 56–60, 63self-dual Yang-Mills equation, 237, 239–

241, 247self-dual Yang-Mills field, 237–239self-dual Yang-Mills flow, 237, 240, 247signed Toda equations, 268–271, 276, 277signed Toda equations of elliptic version,

292, 295sine-Gordon equation, 3, 28, 44, 45, 121,

127–129, 132–136, 138, 139,145, 149, 155, 166, 174–178

sine-Laplace equation, 166–169, 172, 173singular Darboux transformation, 228,

230, 234sinh-Gordon equation, 149, 156, 160, 284–

286sinh-Laplace equation, 166, 167, 169, 171,

174, 183spectral parameter, 1, 4, 12, 16, 31, 35,

103, 176, 183, 202, 239Su chain, 283–287, 289, 290surface of constant Gauss curvature, 121,

144, 154, 179, 184, 185, 189,196, 198

surface of constant mean curvature, 122,179–181, 184, 186, 187, 198,199

surface of constant negative Gauss curva-ture, 121, 126, 128, 129, 131,132, 134, 135, 138, 140, 144,145, 148, 164–166, 170–172,174, 186, 193

surface of constant positive Gauss curva-ture, 128, 144, 146, 154–156,164, 181, 183, 184, 186, 187

theorem of permutability, 25, 28–30, 34,45, 76, 77, 115, 212, 254, 286

Toda equation, 267, 268, 270, 277

uniton, 189, 220–226, 228, 230, 232–235

Yang-Mills equation, 237, 238Yang-Mills field, 237, 238Yang-Mills-Higgs equation, 248Yang-Mills-Higgs field, 248, 249

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