Lecture Notes

for the

Seminar SS 06

about

Partial Differential Equations

by

Andreas Muller-Rettkowski, Hannes Uecker, Guido Schneider

Contents

1 Linear PDE phenomena 41.1 Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Spectral properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 The variation of constant formula . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 The KPP-equation 152.1 The reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 The reaction-diffusion system . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3 The maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.4 Stability and instability of the equilibria . . . . . . . . . . . . . . . . . . . . . 192.5 Front solutions with minimal velocity . . . . . . . . . . . . . . . . . . . . . . 202.6 Stability and instability of front solutions . . . . . . . . . . . . . . . . . . . . 22

3 Burgers equation 243.1 A local existence and uniqueness result . . . . . . . . . . . . . . . . . . . . . 243.2 Special solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.4 Cole-Hopf transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.5 Stability and instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 The KdV-equation 294.1 The solitary wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 The KdV-equation as Hamiltonian system . . . . . . . . . . . . . . . . . . . . 334.3 Stability of the solitary wave . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.4 The soliton property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.5 Infinitely many conservation laws . . . . . . . . . . . . . . . . . . . . . . . . 364.6 The 2-solitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.7 Asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.8 Other mathematical aspects of the KdV equation . . . . . . . . . . . . . . . . 43

5 The NLS-equation 445.1 Nonlinear oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.2 The pulse solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465.3 Local existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . 475.4 The NLS-equation as Hamiltonian system . . . . . . . . . . . . . . . . . . . . 48

6 The Sine-Gordon equation 496.1 The local existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . 506.2 Kink and antikink solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.3 Derivation of the Nonlinear Schrodinger equation . . . . . . . . . . . . . . . . 51

2

6.4 Justification of the Nonlinear Schrodinger equation . . . . . . . . . . . . . . . 546.5 Breathers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3

1 Linear PDE phenomena

For the functional analytic handling of PDEs defined on a bounded domain in space there exista number of very well written textbooks, as [15, 29, 27, 25, 12], at least in the dissipative case.However, for PDEs defined on a very large domain in space the interpretation as countablemany ODEs is no longer of that big help.

Example 1.1 Consider the linear wave equation ∂2t u = ∂2

xu for t ∈ R, x ∈ [−L, L], u(x, t) ∈R, with Dirichlet boundary conditions u(−L, t) = u(L, t) = 0 and L > 0 very large. Weconsider two special classes of solutions, first the oscillations of the eigenfunctions,

u(x, t) = einπt/(2L) sin(nπ(x− L)/(2L))

for n ∈ N and secondly the traveling wave solutions

u(x, t) = f(x− t) + g(x+ t)

with f and g arbitrary smooth functions with compact support

supp(f) = {x | f(x) 6= 0} ⊂ [−1, 1] and supp(g) ⊂ [−1, 1].

As long as |t| < L − 1 this is a solution of the PDE, i.e. at least for a very large time intervaltraveling wave solutions play a role. The description of these solutions by the eigenfunctions isof no use.

In order to avoid the handling of far away boundaries whose influence on the solutions in theinterior of the domain is small we idealize the large domain to an unbounded domain, i.e. forinstance that instead of the large interval [−L, L] we consider x ∈ R. This allows to uncoverpure PDE phenomena which dominate the dynamics on the very large domains at least for avery long time. In contrast to bounded domains for PDEs on unbounded domains there arestill many fundamental and challenging open questions. The idealization of an unboundeddomain leads to dynamical systems with uncountable many degrees of freedom. A separationof the uncountable many modes in single modes is a highly singular action from a functionalanalytic point of view in contrast to countable many modes and therefore again of no use.The recovery of compactness by smoothing properties is no longer true and therefore finite-dimensional attractors in general cannot be expected.

On the other hand from a didactic point of view the consideration of unbounded domains hascertain advantages. It will allow us to explain pure PDE phenomena as transport, diffusion anddispersion. At the beginning it will allow us to minimize the functional tools to an absoluteminimum. Unbounded domains are easy in this respect since we do not have to deal with theboundary conditions which are very often the source of functional analytic difficulties.

Notation. In the following we write large capitals for the abstract solution in phase space andsmall capitals for the function of t and x, i.e. U = U(t, U0) is the solution associated to theinitial condition U0, whereas u = u(x, t) is the solution at a point x and a time t.

4

1.1 Transport

The simplest non-ODE phenomenon which occurs for partial differential equations is transport.The solution u = u(x, t) of

∂tu = c∂xu, u(x, 0) = u0(x)

with x ∈ R and t ∈ R is given by

u(x, t) = u0(x+ ct),

i.e. the initial function u0 is shifted with a velocity c to the left without changing its shape.

As a trivial consequence the solution x 7→ u(x, t) has the same regularity as the initial conditionx 7→ u0(x), i.e. if

U0 ∈ Cnb = {u = u(x) | ‖u‖Cn

b=

n∑

j=0

supx∈R

|∂jxu(x)| <∞}

then U(t) ∈ Cnb not more and not less. Another trivial consequence is that information is

transported with finite velocity c.

In order to use the above dynamical systems frame for this equation we need a phase space.The first choice would be the space of continuous functions, i.e. X = C0

b (R,R), but as thefollowing example shows this choice has a serious disadvantage.

Example 1.2 Consider u0(x) = sin(x2). Since u0 oscillates with smaller and smaller periodsas |x| → ∞ it is an easy exercise to prove that

‖U(t) − U0‖C0b

= supx∈R

|u(x, t) − u0(x)| = supx∈R

|u0(x+ ct) − u0(x)| = 2 6→ 0

for t → 0, i.e. the orbit associated to U0 is not continuous with respect to time and the normin ‖ · ‖X . The uncountable many points of the orbit associated to U0 have a distance of 2 fromeach other and lie on the surface of the unit ball in C0

b , i.e. the orbit is completely disconnectedin this space.

This problem can be solved easily by restriction to the space of uniformly continuous functions,i.e we choose X = C0

b,unif(R,R). Then we have the following result.

Lemma 1.3 The curve t 7→ U(t, U0) is (strongly) continuous in X if U0 ∈ X .

Proof. For translations without loss of generality we can restrict ourselves to prove continuityat t = 0. We have to prove that for all ε > 0 there is a δ > 0 such that |t| < δ impliessupx∈R

|u0(x + ct) − u0(x)| < ε which is equivalent to the uniform continuity of u0 on R. �

Remark 1.4 We introduce the notation of the existing literature. The operator A = ∂x gene-rates a (strongly) continuous or C0-semigroup U(t, ·) in X = C0

b,unif(R,R). We have

AU0 = limt→0

t−1(U(t, U0) − U0)

5

for U0 in the domain of definitionD(A) = C1b,unif(R,R) of Awhich is denseX . Since U(t, U0)

is linear in U0 we also use the notation

T (t)U0 = U(t, U0).

Remark 1.5 Since ‖U(t, U0)‖X = ‖U0‖X all orbits are bounded. In finite dimensions from theboundedness of an orbit we can conclude the existence of an convergent subsequence and so onthe non-emptyness of the ω-limit set. Obviously for U(t, U0) with u0(x) = tanh x the ω-limitset is empty in X .

Exercise 1.6 Consider u0 defined by linear interpolation between the points (xn, u(xn)), wherex2n = n and

u0(xn) =

{

1, n even,

−1, n odd

for n ∈ N. Prove supx∈R|u0(x + t) − u0(x)| 6→ 0 for t→ 0.

Solution 1.7 w.l.o.g. let c = 1. We show that ∀t0 > 0, ∃t < t0, x ∈ R such that |u0(x + t) −u0(x)| = 2. Since

√n + 1 − √

n = 1/(2√n) + O(1/n) we may simply choose x =

√n and

t =√n+ 1 −√

n for n sufficiently large.

Exercise 1.8 Solutions of the linear wave equation

∂2t u = ∂2

xu (x ∈ R, t ∈ R, u(x, t) ∈ R)

are given by u(x, t) = F (x + t) + G(x − t) with F and G arbitrary smooth functions. Byexpressing F and G by the initial data

u(x, 0) = u0(x), ∂tu(x, 0) = u1(x)

show

u(x, t) =1

2(u0(x + t) + u0(x− t)) +

1

2

∫ x+t

x−tu1(ξ)dξ.

Solution 1.9 For t = 0 we obtain

u0(x) = F (x) +G(x), u1(x) = F ′(x) −G′(x).

Differentiating the first equation yields

u′0(x) = F ′(x) +G′(x)

and so

F ′(x) =1

2(u′0(x) + u1(x)), G′(x) =

1

2(u′0(x) − u1(x)).

Integration gives

F (x) +G(x) =1

2(u0(x) +

∫ x

0

u1(x′)dx′) +

1

2(u0(x) −

∫ x

0

u1(x′)dx′)

where we used u0(0) = F (0) +G(0). �

6

1.2 Diffusion

The general formula for the solutions of the linear diffusion equation or linear heat equation

∂tu = ∂2xu

is given by

u(x, t) =1√4πt

∫ ∞

−∞e−

(x−y)2

4t u(y, 0)dy. (1)

For the derivation of this formula see Exercise 1.22. From this formulae we immediately obtainthe estimate

supx∈R

|u(x, t)| ≤ 1√4πt

∫ ∞

−∞|u(x, 0)|dx,

i.e solutions to spatially localized initial conditions decay uniformly with a rate t−1/2 towardszero. Since mass is conserved, i.e.

∫ ∞

−∞u(x, t)dx =

∫ ∞

−∞u(x, 0)dx

for all t ≥ 0 this is how diffusion is expected to work. The decay happens in a universal manner.In lowest order a self-similar behavior is observed, i.e.

u(x, t) =A∗√tv(

x√t) + O(1/t) with v(ξ) = e−ξ

2/4

and a constant A∗ ∈ R depending on the initial conditions. By looking at this formula we seethat this solution only exists for t > 0. This is a general rule. For an arbitrary initial conditionthe diffusion equation cannot be solved backwards in time, i.e. t ≥ 0.The diffusion equation is smoothing. We obtain the following estimate

supx∈R

|∂xu(x, t)| = supx∈R

| 1√4πt

∫ ∞

−∞−−2(x− y)

4te−

(x−y)2

4t u(y, 0)dy|

= supx∈R

|√

2√π

∫ ∞

−∞−−2s

2√te−s

2

u(x+ 2√ts, 0)ds|

≤ 1√t

√2√π

∫ ∞

−∞se−s

2

ds supx∈R

|u(x, 0)|

≤ C√tsupx∈R

|u(x, 0)|

with a constant C independent of t and u(·, 0), where we used s = (x − y)/(2√t), i.e. ds =

−dy/(2√t). More general we have

Theorem 1.10 Let U = U(t) be a solution of the linear diffusion equation. Then for all n ∈ N

there exists a C > 0 such that for all t > 0

‖∂nxU(t)‖C0b≤ Ct−n/2‖U(0)‖C0

b.

7

Even more is true. For every t > 0 and x ∈ R the function u = u(x, t) can be expanded in someconvergent power series, i.e. U is an analytic function and can be extended into the complexplane.Finally, we remark that also the linear diffusion equation defines a dynamical system in thephase space X = C0

b,unif(R,R). See Exercise 1.14. Moreover, we have the following result.

Lemma 1.11 The curve t 7→ U(t, U0) is (strongly) continuous in X if U0 ∈ X .

Proof. By Exercise 1.14 it is sufficient to prove the continuity of the orbit t 7→ U(t) for t = 0

in the space X . With H(z) = 1√4πe−z

2/4 we estimate

‖U(t, U0) − U0‖C0b

= supx∈R

|u(x, t) − u(x, 0)|

= supx∈R

|∫

1√tH(

x− y√t

)(u(y, 0)− u(x, 0))dy

= supx∈R

|∫

H(z)(u(x+√tz, 0) − u(x, 0))dz

≤ supx∈R

∫

|z|≤R| . . . |dz + sup

x∈R

∫

|z|≥R| . . . |dz = s1 + s2.

For a given ε > 0 we have to find a t0 > 0 such that for all t ∈ (0, t0) we have s1 + s2 < ε. Weestimate s2 by

s2 ≤ 2

∫

|z|≥RH(z)dz sup

x∈R

|u(x, 0)| < ε/2

by choosing a R > 0 sufficiently big due to the definition of H . Next we estimate

s1 ≤∫

H(z)dz supx∈R,|z|≤R

|u(x+√tz, 0) − u(x, 0)| = sup

x∈R,|z|≤R|u(x+

√tz, 0) − u(x, 0)|.

Since x 7→ u(x, 0) is uniformly continuous for all ε > 0 there exists a δ > 0 such that for ally ∈ R with |y| < δ we have |u(x + y, 0) − u(x, 0)| < ε/2. Choosing t0 > 0 so small thatt0R < δ we are done. �

Exercise 1.12 Compute u = u(z, t) with z = xr + ixi using the solution formula for lineardiffusion. Show the convergence of the integral for every t > 0 and z ∈ C.

Solution 1.13

|u(z, t)| = | 1√4πt

∫ y=+∞

y=−∞e−(zr−y+izi)

2/(4t)u0(y) dy|

≤ 1√4πt

ez2i /(4t)

∫ y=+∞

y=−∞|e−(zr−y)2/(4t)u0(y)| dy,

that is, the integral converges absolutely for u0 ∈ C0b,unif .

Exercise 1.14 Prove that the solutions U(t, U0) of the linear diffusion equation satisfy the fun-damental property of a dynamical system

U(t + s, U0) = U(t, U(s, U0)).

Solution 1.15 U(t+ s, U0) = F−1(e−k2(t+s)U0) = F−1(e−k

2tU(s, U0)) = U(t, U(s, U0)).

8

1.3 Dispersion

The general formula for the solutions of the linear Schrodinger equation

∂tu = −i∂2xu

is given by

u(x, t) =1√4πit

∫ ∞

−∞e−

i(x−y)2

4t u(y, 0)dy.

Like for diffusion from this formulae we immediately obtain the estimate

supx∈R

|u(x, t)| ≤ C√t

∫ ∞

−∞|u(x, 0)|dx,

i.e solutions to spatially localized initial conditions decay uniformly with a rate t−1/2 towardszero. But nevertheless there are major differences. By diffusion energy, i.e. d

dt

∫

|u(x, t)|2dx ≤0, is lost, where dispersion conserves energy, i.e. d

dt

∫

|u(x, t)|2dx = 0, but spreads it all otherthe real axis. See Exercise 1.20.

Dispersion smoothes solutions locally in space. We can compute

∂xu(x, t) =1√4πit

∫ ∞

−∞

−2i(x − y)

4te−

i(x−y)2

4t u(y, 0)dy <∞

for all x ∈ R and all t > 0 if∫ ∞−∞ |y||u(y, 0)|dy <∞, i.e. pointwise ∂xu(x, t) can be computed,

although the initial condition was only continuous. More general ∂nxu(x, t) is finite for all x ∈ R

and all t > 0 if∫ ∞−∞ |y|n|u(y, 0)|dy < ∞. The reason for this behavior is as follows. Consider

initial conditions of the formu(x, 0) = u0(εx)e

ikx

where 0 < ε� 1 is a small parameter and u0 a smooth spatially localized function. We compute

i∂2x(u0(εx)e

ikx) = ieikx(−k2u0(εx) + ε2iku′0(εx) + ε2u′′0(εx))

with ξ = εx. This motivates us to make the ansatz

u(x, t) = B(ε(x− cgt), ε2t)eik(x−cpt)

with constants cp and cg and a function B = B(ξ, τ) with ξ = ε(x− cgt) and τ = ε2t. Insertingthis ansatz and computing the coefficients in front of ε0, ε1, and ε2 shows

−ikcp = −ik2,

−cg∂ξB = −2k∂ξB,

∂τB = i∂2ξB.

The constant cp = ω(k)/k = k with frequency ω = k2 is called phase velocity. The constantcg = ω′(k) = 2k is called the group velocity. As a consequence of this calculation wavepackets which contain big derivatives, i.e. eikx with k large, are transported with big velocity

9

cg = 2k towards infinity. Since the initial condition is spatially localized nothing can comefrom infinity. Therefore at fixed x ∈ R only packets with low derivatives remain, and thereforea local smoothing occurs.On the other hand this has also the consequence that if the initial condition is not spatially locali-zed then packets with high derivatives can come from infinity. In fact there is no local existenceand uniqueness of solutions in C0

b,unif(R,R). Therefore, for the handling of the Schrodingerequation as a dynamical system we have to choose a different phase space It turned out that

X = L2(R,R) = cl‖·‖L2{u ∈ C∞

0 (R,R) | ‖u‖L2 = (

∫

|u(x)|2dx)1/2 <∞}

is a good choice, where u ∈ C∞0 (R,R) if u possesses a compact support and if ∂nxu exists for

all n ∈ N. This space coincides with the space of Lebesgue square-integrable functions on R.We have the following result

Lemma 1.16 We have U(t + s, U0) = U(t, U(s, U0)). Moreover, the curve t 7→ U(t, U0) is(strongly) continuous in X if U0 ∈ X .

Proof. See Exercise 1.24. �

Remark 1.17 For the transport equation and the linear wave equation we obtain ω(k) = ck

with a constant c ∈ R and so the group velocity ω ′(k) = c is bounded and so the transportequation and the linear wave equation can be solved in C0

b,unif(R,R).

1.4 Spectral properties

In this section we will give some idea why the solutions of these equations behave completelydifferent. Due to the translation invariance of all three equations, i.e. with u = u(x, t) alsou = u(x+ x0, t) is a solution, all three equations possess solutions of the form

u(x, t) = eikx+λ(k)t (2)

where λ(k) = ik for the transport equation, λ(k) = −k2 for the diffusion equation, and λ(k) =

−ik2 for the Schrodinger equation. The λ(k) are the eigenvalues and the eikx the eigenfunctionsof the operators

A = ∂x, A = ∂2x, A = i∂2

x.

With this notation the three equations can be written as

∂TU = AU

with A some operator on the phase space X . For the transport equation and the Schrodingerequation the spectral values of A lie on the imaginary axis, whereas the spectral values of Afor the diffusion equation lie on the negative real axis. The amplitude of the special solutionsu(x, t) = eikx+λ(k)t is conserved for the first and third equation, whereas the amplitude is dam-ped with some exponential rate for the diffusion equation.

10

For PDEs posed on the real axis an expansion in these special solutions is called the Fourierexpansion of these solutions. We expand

u(x, t) =

∫ ∞

−∞u(k, t)eikxdk

and define the coefficients u(k, t) to be the Fourier transform of u(x, t). They can be computedby the formula

u(k, t) =1

4π

∫ ∞

−∞u(x, t)e−ikxdx.

We use the abbreviations u = F−1u and u = Fu. The coefficients satisfy for the transport, thediffusion, and the Schrodinger equation

∂tu(k, t) = iku(k, t),

∂tu(k, t) = −k2u(k, t),

∂tu(k, t) = −ik2u(k, t).

Since formally

∂xu(x, t) =

∫ ∞

−∞iku(k, t)eikxdk,

derivatives ∂x in x-space correspond to a multiplication by ik in Fourier space. Since Fouriertransform can be extended to the space of tempered distributions, cf. [16], in principle thesolution of the PDE can be obtained by going into Fourier space with F , by solving the Fouriertransformed problem and then going back with F−1 to x-space. However, the derivation ofestimates, the proof of continuity of orbits, etc. by Fourier transform is not possible in all phasespaces.We have for instance that F is continuous from L1 into C0

b but not vice versa. Since F is linearthe first statement follows from

‖u‖C0b

= supx∈R

|∫ ∞

−∞u(k)eikxdk| ≤ sup

x∈R

∫ ∞

−∞|u(k)||eikx|dk =

∫ ∞

−∞|u(k)|dk = ‖u‖L1 . (3)

The second statement follows since the Fourier transform of a constant function is a Dirac mea-sure in zero which is no element of L1. The Fourier transform of C0

b is contained in a verycomplicated space which makes Fourier transform somehow useless for the derivation of es-timates in C0

b , but nevertheless it is useful in C0b for the derivation of solution operators. For

instance for the derivation of (1) we use that convolution in x-space corresponds to multiplica-tion in Fourier space and vice versa, i.e.

F(u ∗ v) = 2πuv and F(uv) = u ∗ v, (4)

where convolution is defined by

(u ∗ v)(x) =

∫ ∞

∞u(x− y)v(y)dy.

In contrast to C0b , Fourier transform is an isomorphism from L2 to L2. This property is based

on Parseval’s identity‖u‖2

L2 = 2π‖u‖2L2 . (5)

11

1.5 The variation of constant formula

We use the variation of constant formula in order to solve the nonlinear system. We first consideran abstract system

d

dtu = Au+ f, u|t=0 = u0 (6)

with A the generator of a strongly continuous semigroup T (t) in X and f ∈ C0([0,∞), X).

Definition 1.18 u is called mild solution of (6) on [0, T0], if u ∈ C([0, T0], X) and if u solves

u(t) = T (t)u0 +

∫ t

0

T (t− s)f(s)ds. (7)

This formula is called variation of constant formula. It can be derived as follows. Set u(t) =

T (t)v(t). Thendu

dt(t) = AT (t)v(t) + T (t)

dv

dt(t) = AT (t)v(t) + f(t)

or equivalently under the assumption, that T (s)−1 exists,

v(t) = v(0) +

∫ t

0

T (s)−1f(s)ds

and therefore

u(t) = T (t)v(0) +

∫ t

0

T (t)T (s)−1f(s)ds.

Using the semigroup property we find T (t)T (s)−1 = T (t− s), and finally the above formula.

Lemma 1.19 Let u = u(t) be a mild solution with u ∈ C([0, T0], D(A)) ∩ C1([0, T0], X) of(6). Then u solves equation (6) also in the classical sense, i.e. it can be inserted into the partialdifferential equation.

Proof: We find

d

dtu(t) = AT (t)u(0) + T (t− t)f(t) +

∫ t

0

AT (t− s)f(s)ds = f(t) + Au(t).

�

Exercise 1.20 Prove the following estimate for the solutionU = U(t, U0) of the linear diffusionequation. There exists a C > 0 such that for all U0 and all t > 0 we have

‖U(t, U0)‖L2 ≤ Ct−1/4‖U0‖L1

by using (5), (3), and the estimate ‖uv‖L2 ≤ ‖u‖L2‖v‖C0b. Prove with (5) that

‖U(t, U0)‖L2 = ‖U0‖L2

for the solution of the Schrodinger equation.

12

Solution 1.21

‖u(t)‖L2(5)= 2π‖u(t)‖L2 = ‖e−k2tu‖L2 ≤ 2π‖e−k2t‖L2‖u0‖C0

b

(3)

≤ Ct−1/4‖u0‖L1 ,

since

‖e−k2t‖2L2 = t−1/2

∫

e−2m2

dm ≤ Ct−1/2.

For the Schrodinger equation we have

‖u(t)‖2L2 = 2π‖u(t)‖L2 = 2π

∫

|eik2tu0(k)|2 dk = 2π‖u0‖2 = ‖u0‖2.

Exercise 1.22 Derive the solution formula (1) for the linear diffusion equation by using (4) andthe fact that x 7→ 1√

2πe−x

2/2 is the inverse Fourier transform of k 7→ e−k2/2.

Solution 1.23 ∂tu = ∂2xu becomes ∂tu = −k2u in Fourier space with explicit solution u(k, t) =

e−k2tu(k, 0). Hence

u(x, t) = F−1(u(k, t)) =1

2π(F−1(e−k

2t) ∗ u0(x)),

and

F−1(e−k2t) =

∫

e−k2teikx dk =

1√2t

∫

e−m2/2eimx/

√2t dm =

√2π√2t

e−x2/(4t)

yields the result.

Exercise 1.24 Prove similar to the proof of Lemma 1.11 the continuity of the solution U =

U(t, U0) of the Schrodinger equation at t = 0 in X = L2(R,R) by working in Fourier spacewith u(k, t) = eik

2tu(k, 0) and by using (5).

Solution 1.25 We show: ∀u0∈X ∀ε > 0 ∃t0 > 0 such that ‖u(t)−u0‖2L2 < ε2 ∀t ∈ (0, t0).

‖u(t) − u0‖2L2 = 2π‖u(t) − u0‖2

L2 = 2π

∫

|u0|2|1 − eik2t|2 dk

= 2π

(∫

|k|<L|u0|2|1 − eik2t|2 dk +

∫

|k|≥L|u0|2|1 − eik2t|2 dk

)

Due to u0∈X there exists a (large) L such that the second integral is less than ε2/(4π). Nowthe first term is less than ε2/(4π) for t > 0 sufficiently small due to uniform convergence|1 − eik2t| → 0 on bounded k intervals.

Exercise 1.26 Solve the transport equation, the diffusion equation, and the Schrodinger equa-tion with the help of a computer. Take initial conditions u(x, 0) = sin(kx) and u(x, 0) =

sin(kx)/ cosh(x) for k = 1, 5, 20, 100 (resp. u(x, 0) = eikx, u(x, 0) = eikx/ cosh(x) forSchrodinger). Explain the differences between the solutions.

Solution 1.27 Using HUProgs we obtain the solutions in fig.1.

13

(a) transport, u0(x) = cos(kx), k = 1, 2, and u0(x) cos(kx)/ cosh(x), k=5, x ∈ (−4π, 4π)

-1212

0

2

4

6

8

10

-11

-1212

0

2

4

6

8

10

-11

-1212

0

2

4

6

8

10

-11

(b) diffusion, u0(x) = cos(kx), k = 1, 2, and u0(x) = cos(kx)/ cosh(x), k=5, x ∈ (−4π, 4π)

-12

120

0.5

1

1.5

2

2.5

-11

-12

120

1

-1

1

-12

120

0.5

-0.50.5

(c) Schr odinger, u0(x) = eikx, k = 1, 2, x ∈ (−4π, 4π), (real-parts)

-12

120

5

-1

1

-12

120

5

-1

1

(d) Schr odinger, u0 = eikx

cosh(x) , k = 1, x ∈ (−10π, 10π), k = 1 (ur and |u|)

-30

300

5

1

-30

300

5

1

Figure 1: Exercise 1.26 solved by HUProgs.

14

2 The KPP-equation

We consider the Kolmogorov, Petrovsky, Piscounuv (KPP) equation [21] or Fisher equation[13]

∂tu = ∂2xu+ u− u2, (8)

with u(x, t) ∈ R, t ≥ 0, and x ∈ R. It occurs as a model for various systems in nature, forinstance for chemical reactions or population dynamics. The equation consists of two parts,namely the diffusion term ∂2

xu and the nonlinear reaction term u − u2. Therefore, it bringstogether PDE with ODE dynamics. As we have seen, by diffusion energy vanishes, and so thesystem is called dissipative.

2.1 The reaction

Inserting u(x, t) = v(t) into (8) gives the one-dimensional ordinary differential equation

∂tv = v − v2. (9)

The 1D phase portrait shows that the fixed point v = 0 is unstable and that the fixed pointv = 1 is asymptotically stable. The term +v in the KPP-equation represents the exponentialgrowth for small v and the term −v2 represents saturation. For instance, a population of animalsincreases with some exponential rate until the growth is saturated by the food available for theanimals, or if a chemical reaction is started the concentration of the product of the reactionincreases with some exponential rate until the growth is saturated by the missing reactant.

2.2 The reaction-diffusion system

The KPP-equation is obtained by adding the diffusion term to the ODE (9). Thus, the KPP-equation describes for instance a chemical reaction in a large, here in an infinitely extended,domain, where the concentration u = u(x, t) spreads into space by diffusion.

It is the purpose of this section to understand the dynamics of the solutions of the KPP-equation.In order to handle the KPP-equation as an abstract dynamical system we choose the same phasespace as for the linear diffusion equation, namely the space of bounded and uniformly conti-nuous functions, i.e.

X = C0b,unif(R,R).

In this space there is the local existence and uniqueness of the solutions for the KPP-equation.Due to the maximum principle which is introduced in a moment we also have the global exis-tence and uniqueness of solutions.

Theorem 2.1 Let U0 ∈ X . Then there exists a unique solution of the KPP-equation U ∈C([0,∞), X) with initial datum U0.

15

Remark 2.2 Solutions U = U(t) ∈ X satisfy the KPP equation only in a weak sense. Suchsolutions are called mild solutions since they satisfy the the variation of constant formula

U(t) = T (t)U0 +

∫ t

0

T (t− τ)N(U)(τ)dτ,

where T (t) is the solution operator of the linear diffusion equation, i.e. T (t)U0 solves the lineardiffusion equation, and where N(U)(τ) represents the reaction term u(x, τ) − (u(x, τ))2 in X .If U ∈ C([0, T0], C

2b,unif) ∩ C1([0, T0], C

0b,unif) solves the variation of constant formula then

U is also a classical solution of the KPP-equation, i.e. u = u(x, t) can be inserted into theKPP-equation. See Exercise 2.4.

Proof of Theorem 2.1. The proof of the local existence and uniqueness of solutions is basedon the contraction mapping principle (also called fixed point theorem of Banach) and on thevariation of constant formula. For sufficiently small T0 > 0 the right hand side of the variationof constant formula defines a contraction F in

M = C0([0, T0], {U ∈ X | ‖U‖X ≤ 2‖U0‖X =: C1})

equipped with the norm‖U‖M = sup

0≤t≤T0

‖U(t)‖X .

In order to prove the contraction mapping property we need a number of estimates. For t ≥ 0,U0 ∈ X and U , V ∈M we have

‖T (t)U0‖X ≤ ‖U0‖X ,‖N(U)‖M ≤ ‖U‖M + ‖U‖2

M ≤ C1 + C21 ,

‖N(U) −N(V )‖M ≤ (1 + ‖U + V ‖M)‖U − V ‖M ≤ (1 + 2C1)‖U − V ‖M .

The mapping F maps M into M , since

‖F (U)‖M = sup0≤t≤T0

‖T (t)U0 +

∫ t

0

T (t− τ)N(U)(τ)dτ‖X

≤ sup0≤t≤T0

‖T (t)U0‖X + sup0≤t≤T0

∫ t

0

‖T (t− τ)N(U)(τ)‖Xdτ

≤ sup0≤t≤T0

‖U0‖X + sup0≤t≤T0

∫ t

0

‖N(U)(τ)‖Xdτ

≤ ‖U0‖X + sup0≤t≤T0

∫ t

0

dτ‖N(U)‖M

≤ C1/2 + (C1 + C21 )T0 ≤ C1

16

for (1 + C1)T0 < 1/2. The mapping F is a contraction in M , since

‖F (U) − F (V )‖M = sup0≤t≤T0

‖∫ t

0

T (t− τ)(N(U) −N(V ))(τ)dτ‖X

≤ sup0≤t≤T0

∫ t

0

‖T (t− τ)(N(U) −N(V ))(τ)‖Xdτ

≤ sup0≤t≤T0

∫ t

0

‖(N(U) −N(V ))(τ)‖Xdτ

≤ sup0≤t≤T0

∫ t

0

dτ‖(N(U) −N(V ))‖M

≤ T0(1 + 2C1)‖U − V ‖M ≤ 1

2‖U − V ‖M

for T0(1 + 2C1) < 1/2, i.e. for T0 > 0 sufficiently small there is a fixed point U = F (U),which is a mild solution of the KPP-equation. �

Remark 2.3 The KPP equation is a semilinear parabolic equation [15], i.e. the semigroup T (t)

generated by the operator A is smoothing and the nonlinearity N contains only derivatives oflower order than inA. Like for the diffusion equation the solutions of the KPP-equation or moregeneral of a semilinear parabolic equation on the real axis are analytic for all t > 0.

Exercise 2.4 Prove that if U ∈ C([0, T0], C2b,unif) ∩ C1([0, T0], C

0b,unif) solves the variation of

constant formula then U is also a classical solution of the KPP-equation.

Solution 2.5 Differentiating the variation of constant formula

U(t) = T (t)U0 +

t∫

0

T (t− s)N(U(s))ds

yields

∂tU(t) = AT (t)U0 + T (t− t)N(U(t)) +

t∫

0

AT (t− s)N(U(s))ds

= AU(t) +N(U(t))

where ∂tU, AU, N(U(t)) are elements of C([0, T0], C0b,unif).

2.3 The maximum principle

Second order PDEs have a special property which helps a lot in their analysis, namely themaximum principle.

Theorem 2.6 From u1(x, t0) ≤ u2(x, t0) for all x ∈ R it follows u1(x, t) ≤ u2(x, t) for allt ≥ t0 and all x ∈ R.

17

Idea of the proof: Let u1(x, t0) < u2(x, t0) for all x ∈ R. We are done if u1(x, t) < u2(x, t)

for all x ∈ R and all t ≥ t0.

If this is not true, then since the solutions change smoothly in x and t there exists a t1 > t0 anda x0 ∈ R such that u1(x0, t1) = u2(x0, t1). Then we have

∂t(u2(x0, t1) − u1(x0, t1)) = ∂2x(u2(x0, t1) − u1(x0, t1))

+(u2(x0, t1) − u1(x0, t1)) − (u22(x0, t1) − u2

1(x0, t1))

= ∂2x(u2(x0, t1) − u1(x0, t1)),

i.e. for small t > t1, in the generic case ∂2x(u2(x0, t1) − u1(x0, t1)) > 0, we have again

u2(x0, t) > u1(x0, t), i.e. u1 can never be bigger than u2. �

0

u 2 − u 1

Figure 2: The maximum principle: the solutions are separated again

A consequence is that solutions having values in [0, 1] will stay in [0, 1], i.e. 0 ≤ u(x, 0) ≤ 1

for all x ∈ R implies 0 ≤ u(x, t) ≤ 1 for all x ∈ R and all t ≥ 0. Hence, it makes sense to talkabout concentrations.

Exercise 2.7 Consider∂tu = ∂2

xu+ u− u3

for t ≥ 0, x ∈ R. Do we have local existence and uniqueness of solutions? Prove with themaximum principle that lim supt→∞ ‖U(t)‖X ≤ 1.

Solution 2.8 The local existence and uniqueness goes line for line as for the KPP–equation,i.e. for all U0 ∈ C0

b,unif there exists a T0 > 0 and a solution U ∈ C([0, T0], C0b,unif) with

U |t=0 = U0. Let v(0) = supx∈R

|u0(x)|. We have that u(x, t) = v(t) satisfies ∂tv = v − v3.

Hence |v(t)| ≤ max(1, v(0)) for all t ≥ 0. Due to the maximum principle we have

supx∈R

|u(x, t)| ≤ |v(t)| ≤ max(1, v(0)) .

Hence, we have the boundedness in C0b,unif and local existence and uniqueness in C0

b,unif andso also the global existence and uniqueness, i.e. u ∈ C([0,∞), C0

b,unif).

18

2.4 Stability and instability of the equilibria

Due to the chemical or biological motivation of the KPP-equation we restrict ourselves in thefollowing to solutions with values 0 ≤ u(x, t) ≤ 1. We compute the fixed points, or theequilibria, or the stationary solutions of the KPP-equation. They satisfy

0 = ∂2xu+ u− u2.

By having a look at the phase portrait in Figure 3 we see that the only equilibria having onlyvalues in [0, 1] are the fixed points u = 0 and u = 1.

–2

–1

0

1

2

y

–2 –1 1 2x

Figure 3: The phase portrait for the stationary solutions of the KPP-equation

Now we are interested in the stability of these stationary solutions. We have exactly the samedefinition as in Rd except that Rd is replaced by X .

Definition 2.9 A fixed point U ∗ is called stable for the KPP-equation if for any ε > 0 thereis a δ > 0 such that ‖U0 − U∗‖X < δ implies ‖U(t, U0) − U∗‖X < ε for all t ≥ 0. Other-wise, it is called unstable. A stable fixed point is called asymptotically stable if additionallimt→∞ U(t, U0) = U∗ holds.

For ODEs the eigenvalues of the linearised system have been sufficient for proving stability orinstability. For U ∗ = 0 we find the linear operator (∂2

x +1)· which possesses the eigenfunctionseikx and the spectrum (−∞, 1]. For U ∗ = 1 we find the linear operator (∂2

x−1)· which possessesthe eigenfunctions eikx and the spectrum (−∞,−1]. Hence we expect that U ∗ = 0 is unstable,whereas U∗ = 1 is asymptotically stable.

Theorem 2.10 The fixed point U ∗ = 0 is unstable in X .

Proof. Let U0 = δ/2 for 0 < δ � 1. Since U0 is constant in space, no diffusion occurs and thedynamics is determined by the ODE ∂tv = v−v2. Therefore, we have limt→∞ U(t, U0) = 1 andso U(t, U0) leaves every ε-neighborhood of U ∗ independent how small δ was at the beginning.

�

Theorem 2.11 The fixed point U ∗ = 1 is asymptotically stable in X .

19

Proof. Let U0 ∈ X with ‖U0 − U∗‖X ≤ δ with δ > 0 sufficiently small. Then let v− =

infx∈R u0(x) and let v+ = supx∈Ru0(x). Then v− ≤ u0(x) ≤ v+ for all x ∈ R. By the

maximium principle we have for the associated solutions

v−(t) ≤ u(x, t) ≤ v+(t)

for all x ∈ R and t ≥ 0. Since limt→∞ v−(t) = limt→∞ v+(t) = 1 we also have

limt→∞

‖U(t, U0) − U∗‖X = 0,

i.e. the asymptotic stability.

u*u

δu*−

u*+ δ

Figure 4: By the maximum principle the solution u is confined by the constants

�

2.5 Front solutions with minimal velocity

If a chemical reaction (which is described by the KPP-equation) is started spatially localized,i.e. we take a small positive perturbation with compact support of the unstable equilibriumU∗ = 0, then two reaction fronts are created, one moving to the left and moving to the right. Inthe following we would like to address the question about the velocity of the reaction front.

In order to construct these fronts we consider solutions of permanent form u(x, t) = v(x−ct) =

v(ξ) with velocity c satisfying limξ→−∞ v(ξ) = 1 and limξ→∞ v(ξ) = 0. Inserting this into theKPP equation gives the ordinary differential equation

−c∂ξv = ∂2ξ v + v − v2

which we write as first order system (′ = ∂ξ)

v′ = w,

w′ = −cw − v + v2.

We find that the phase portrait in Figure 5 differs with the value of c.For all values of c 6= 0 we find a heteroclinic connection between the fixed points (v, w) =

(0, 0) and (v, w) = (1, 0). Since the heteroclinic connections are in the intersection of theone-dimensional unstable manifold of (0, 1) and the two-dimensional stable manifold of (0, 0)

20

–2

–1

0

1

2

y

–2 –1 1 2x

–2

–1

0

1

2

y

–2 –1 1 2x

–0.4

–0.2

0

0.2

0.4

0.6

0.8

1

x(t)

–10 –5 5 10 15 20t 0.2

0.4

0.6

0.8

1

x(t)

–10 –5 5 10 15 20t

Figure 5: The phase portraits for the front solutions for the KPP-equation for c ∈ (0, 2) and c ≥ 2, with

the associated fronts.

this connection is very robust under perturbations. However, for |c| < 2 these heteroclinicconnections spiral into (v, w) = (0, 0) which can be seen by considering the linearisation

v′ = w,

w′ = −cw − v.

We find the eigenvalues

λ1,2 =−c±

√c2 − 4

2.

For c ∈ (0, 2) we have complex eigenvalues with Reλ1,2 < 0, and for c ≥ 2 we have realnegative eigenvalues. Thus, monotonic fronts which only take values in [0, 1] can only exist forc ≥ 2.

Remark 2.12 We remark that there is a very general result saying that heteroclinic connectionsin the ODE imply the existence of fronts in the PDE. The front satisfies

∂2ξu+ c∂ξu+ u− u2 = 0

We rescale time ξ = cζ , such that ∂ξ = ε∂ζ with ε = 1/c. We obtain

ε2∂2ζ v + ∂ζv + v − v2 = 0

which converges towards∂ζv + v − v2 = 0

for ε → 0 which is the ODE of the reaction. The theory of singularly perturbed ODEs [17]guarantees that the heteroclinic connection between the fixed points of the ODE persists forsmall ε > 0.

21

In order to discuss the problem of the velocity of the reaction front we use the maximum prin-ciple. To a small given spatially localized initial condition u0 we can find a right moving frontu− = u−(x + ct) and a left moving front u+ = u+(x − ct) with minimal velocities c = ±2

satisfyingu−(x) ≥ u0(x) ≤ u+(x)

for all x ∈ R. From the maximum principle we find

u−(x+ 2t) ≥ u0(x, t) ≤ u+(x− 2t).

Therefore, the reaction fronts cannot move faster than the minimal velocities c = ±2. In [5] ithas been shown that the upper bound is a good approximation and that the front moves as

x = 2t− c1 ln t+ c2(t)

with limt→∞ c2(t) = c∗2 <∞ and c1, c∗2 ∈ R some constants. See Figure 6.

0

1

u

u − u +

Figure 6: Solutions to spatially localized initial conditions are confined by two fronts with minimal

velocity.

2.6 Stability and instability of front solutions

Theorem 2.13 For |c| ≥ 2 the front solutions are unstable in X .

Proof. Let uf be a front solution with velocity c. Consider u0(x) = min(uf(x), δ/2). Then‖Uf (0) − U0‖X < δ with Uf (t) = U(t, Uf ), but limt→∞ U(t, U0) = 1 due to the maximumprinciple, since δ/2 ≤ u0(x) ≤ 1, and so lim ‖Uf(t) − U(t, U0)‖X = 1 > ε independent howsmall δ > 0 is chosen. See Figure 7.

δ

0

1

δ

u f

Figure 7: Instability of front solutions with respect to perturbations in C 0b,unif .

�

22

Remark 2.14 On the other hand by the maximum principle the front solutions are stable withrespect to smaller sets of perturbations. If the small perturbation v of the front solution uf iscontained between two other translates of the same front then by the maximum principle it willstay there for all times, in detail if

uf(x− x−) ≤ uf(x) + v(x, 0) ≤ uf(x+ x+)

for all x ∈ R, then

uf(x− ct− x−) ≤ uf(x− ct) + v(x, t) ≤ uf(x− ct+ x+)

for all t ≥ 0 and all x ∈ R. See Figure 8.

0

1

u f

Figure 8: Stability of the fronts with respect to smaller sets of perturbations. The perturbations are

confined by parallel fronts with the same velocity.

This leads us to the following generalization of the above stability definition

Definition 2.15 A fixed point U ∗ is called (X1, X2)-stable, if for all ε > 0 there exists a δ > 0

such ‖U0 − U∗‖X1 < δ implies ‖U(t, U0) − U∗‖X2 < ε.

For instance these fronts are (X1, C0b,unif) stable, where ‖u‖X1 = supx∈R

|u(x)ex2|. Note thate−x

2decays much faster than the difference of two fronts which converge towards each other

with some exponential rate. Definition 2.15 makes perfectly sense due to the fact that in infinitedimensions there are infinitely many non equivalent norms.

Exercise 2.16 Solve KPP numerically. Take initial conditions u0 = a cosh(x/w) with smalla, w on a large domain.

Solution 2.17 see exercise 3.14.

Exercise 2.18 Consider∂tu = ∂2

xu+ c∂xu+ u.

Show that u = 0 is unstable in C0b,unif . Make the transform u(x, t) = v(x, t)eβx. Which

equation is satisfied by v? Can β ∈ R be chosen in such a way that v = 0 is stable. What arethe consequences for the stability of u = 0 against other sets of perturbations?

23

Solution 2.19 Take u0(x) = δ ∈ R. Then limt→∞

U(t, U0) = ∞ and so every ε–neighborhood of

u = 0 is left independent now small δ > 0 is chosen. u(x, t) = v(x, t)eβx yields

∂tv = ∂2xv + 2β∂xv + β2v + c∂xv + cβv + v

= ∂2xv + (2β + c)∂xv + (β2 + cβ + 1)v .

Hence v = 0 is stable if β2 + cβ + 1 ≤ 0. This can be satisfied for |c| > 0 sufficiently big bychoosing β in an appropriate way.

3 Burgers equation

Burgers equation∂tu = ∂2

xu− u∂xu (10)

with t ≥ 0, x ∈ R and u(x, t) ∈ R also occurs in a number of applications. It appears as amodel for the velocity field of a viscous, compressible, one-dimensional fluid or gas. Trafficproblems very often are modelled as a one-dimensional gas with the cars taking the role of thegas particles. Therefore, Burgers equation also occurs as a model for traffic flows. It also occursas a so called amplitude equation describing the behavior of abstract quantities as the evolutionof wavenumbers in stable periodic patterns.

Remark 3.1 As for the KPP-equation also for Burgers equation the maximum principle holds

3.1 A local existence and uniqueness result

In order to handle Burgers equation as a dynamical system we again choose the phase space

X = C0b,unif(R,R).

Again we have the local existence and uniqueness of solutions in X and by the maximumprinciple also the global existence and uniqueness of solutions in X of Burgers equation.

Theorem 3.2 Let U0 ∈ X . Then there exists a unique solution of Burgers equation U ∈C([0,∞), X) with initial datum U0.

Proof. Using the notation of the proof of Theorem 2.1 this proof has to be modified as follows.N(U) represents the term −(u(x, t))2/2 and the variation of constant formula is then given by

U(t) = T (t)U0 +

∫ t

0

T (t− τ)∂xN(U)(τ)dτ,

We need the additional estimates

‖T (t)∂xU0‖X ≤ (1 + t−1/2)‖U0‖X ,‖N(U)‖M ≤ ‖U‖2

M/2 ≤ C21/2,

‖N(U) −N(V )‖M ≤ ‖U + V ‖M‖U − V ‖M/2 ≤ C1‖U − V ‖M .

24

With these estimates we obtain that the mapping F maps M into M , since

‖F (U)‖M = sup0≤t≤T0

‖T (t)U0 +

∫ t

0

T (t− τ)∂xN(U)(τ)dτ‖X

≤ sup0≤t≤T0

‖T (t)U0‖X + sup0≤t≤T0

∫ t

0

‖T (t− τ)∂xN(U)(τ)‖Xdτ

≤ sup0≤t≤T0

‖U0‖X + sup0≤t≤T0

∫ t

0

(1 + (t− τ)−1/2)‖N(U)(τ)‖Xdτ

≤ ‖U0‖X + sup0≤t≤T0

∫ t

0

(1 + (t− τ)−1/2)dτ‖N(U)(τ)‖M

≤ C1/2 + C21(T0 + 2T

1/20 )/2 ≤ C1

for C21 (T0 + 2T

1/20 )/2 < 1/2. The mapping F is a contraction in M , since

‖F (U) − F (V )‖M = sup0≤t≤T0

‖∫ t

0

T (t− τ)(N(U) −N(V ))(τ)dτ‖X

≤ sup0≤t≤T0

∫ t

0

‖T (t− τ)(N(U) −N(V ))(τ)‖Xdτ

≤ sup0≤t≤T0

∫ t

0

(1 + (t− τ)−1/2)‖(N(U) −N(V ))(τ)‖Xdτ

≤ sup0≤t≤T0

∫ t

0

(1 + (t− τ)−1/2)dτ‖(N(U) −N(V ))‖M

≤ (T0 + 2T1/20 )C1‖U − V ‖M ≤ 1

2‖U − V ‖M

for (T0 + 2T1/20 )C1 < 1/2, i.e. for T0 > 0 sufficiently small there is a fixed point U = F (U),

which is a mild solution of the Burgers equation. �

Remark 3.3 Also Burgers equation is a semilinear parabolic equation [15], i.e. the semigroupT (t) generated by the operator A is smoothing and the nonlinearityN contains only derivativesof lower order than A. Like for the diffusion equation the solutions of Burgers equation areanalytic for all t > 0.

3.2 Special solutions

We are interested in special solutions as stationary or traveling wave solutions. Obviously everyconstant in space is a stationary solution. In order to find the traveling waves we make theansatz u(x, t) = v(x− ct) = v(ξ) which has to satisfy

−cv′ = v′′ − vv′.

Integration yieldscv + v′ − v2/2 + d = 0

25

with a constant d. This onedimensional ODE possesses two fixed points v− and v+ satisfyingcv−v2/2+d = 0. To given v− for ξ → −∞ and v+ for ξ → ∞ we always the system of linearequations

−cv+ + d = v2+/2, −cv− + d = v2

−/2

always possesses unique solutions c and d if v− 6= v+. Looking however at the phase portraitof the onedimensional ODE shows a heteroclinic connection vh between the fixed points withlimξ→±∞ vh(ξ) = v± only if (v+ − v−)/c > 0. This heteroclinic connection gives a monotonicfront u(x, t) = vh(x− ct) in Burgers equation.These solutions describe for instance a front of gas with constant high density moving into adomain with constant low density.

Exercise 3.4 Discuss the phase portrait of cv ′ = v′′ + vv′ in the (v, v′)-plane.

Solution 3.5 ????

3.3 Characteristics

We assume that diffusion is very small. As a first approximation the diffusion term is neglected.Therefore, we consider

∂tu+ u∂xu = 0. (11)

We look for curves (t, x(t, x0)) in R2 for which u is constant along. Differentiating u(x(t, x0), t) =

u(x0, 0) with respect to t gives(∂xu)x+ ∂tu = 0.

Comparison with the above PDE (11) shows x = u which possesses the solutions

x(t, x0) = x0 + u(0, x0)t

which are the characteristics of (11). Along the characteristics the values of u are constant.

Remark 3.6 The method of characteristics can be generalized to general scalar first order PDEsof the form

F (x1, . . . , xd, u, ∂x1u, . . . , ∂xdu) = 0

with F : R2d+1 → R a smooth function. It can be shown that the solution of this PDE is

equivalent to the solution of an ODE in R2d+1. See [8].

Obviously, even for smooth initial data x 7→ u0(x), the characteristics can intersect.

Example 3.7 Consider

u0(x) =

1, for x ≤ 0,

1 − x, for x ∈ [0, 1],

0, for x ≥ 1.

The point (t, x) = (1, 1) is the intersection point of the characteristics starting in {0} × [0, 1].This example also shows that dynamical systems in infinite dimensions strongly depends on the

26

phase space. The solutions stay bounded in C0b,unif , whereas the solutions explode in C1

b,unif .

0 1

Figure 9: Characteristics intersect and create a shock

Such solutions which occur in nature describe the creation of a shock. After the characteristicshave intersected a new understanding of what is meant by a solution is needed. For the definitionof so called weak solutions see [25]. However, small diffusion will always avoid the creation ofa shock.

Example 3.8 Consider

u0(x) =

{

1, for x ≥ 0,

0, for x ≤ 0.

For this initial datum the characteristics leave a complete region in the (x, t)-plane empty. Aphysically realistic solution will be a so called rarefaction wave

u(x, t) =

1, for x ≥ t

x/t, for x ∈ [0, t]

0 for x ≤ 0

We have u(x, t) = limε→0 uε(x, t), where uε solves

∂tuε = ε∂2xuε − uε∂xuε.

u = u(x, t) is called viscosity solution.

Exercise 3.9 u(x, t) = u0(x0) = u0(x− tu(x, t)) is an implicite representation of the solutionsof ∂tu+ u∂xu = 0. Why? Show by differentiation of this representation with respect to x thatfor t = t∗ = −1/ supx0∈R

u′0(x0) a shock occurs, i.e. that u = u(x, t) is no longer differentiablewith respect to x.

Solution 3.10 As shown above we have

u(x, t) = u0(x0) = u0(x− tu0(x0)) = u0(x− tu(x, t)).

27

?

Figure 10: The characteristics do not enter parts of the (x, t)-plane and create a rarefaction wave

Hence

∂xu(x, t) = (∂xu0(x0))(1 − t∂xu(x, t)) ⇔ ∂xu(x, t) = ∂xu0(x0)/(1 + t∂xu0(x0)),

and the denominator get 0 for t = t∗.

3.4 Cole-Hopf transformation

By the Cole-Hopf transformation

ψ(x, t) = eR x−∞

u(y,t)dy, u(x, t) =∂xψ(x, t)

ψ(x, t)

the nonlinear Burgers equation∂tu = ∂2

xu+ ∂x(u2),

reduces to the linear heat equation ∂tψ = ∂2xψ. Although it seems now that the solutions of

Burgers equation can be computed explicitely it turns out that this transformation is uselessin many situations since ψ very often is unbounded. But nevertheless it can used to find theasymptotics of solutions to spatially localized initial conditions.Due to the maximum principle solutions to positive initial conditions stay positive. Moreover,the quantity

∫ ∞−∞ u(y, t)dy is preserved by Burgers equation. Thus, we look for limiting profiles

of the heat equation to initial conditions

limx→−∞

ψ(x, 1) = 1 and limx→∞

ψ(x, 1) = 1 + A > 1.

Then under some additional assumptions we have

limt→∞

ψ(x√t, t) = 1 + A erf(x)

with a rate O(1/√t) for the solutions of the linear diffusion equation. Using the fact that

the inverse Cole-Hopf transformation is given by u(x, t) = ∂xψ(x,t)ψ(x,t)

shows that solutions u tospatially localized initial conditions in Burgers equations satisfy

limt→∞

√tu(x

√t, t) =

A erf ′(x)

1 + A erf(x)=

d

dxln(1 + A erf(x)) = f ∗

A(x)

with a rate O(1/√t). The limiting profile satisfies limx→±∞ f ∗

A(x) = 0. Therefore, the renor-malized solutions converge towards a non-Gaussian limit.

28

3.5 Stability and instability

Again by the maximum principle a number of stability results can be shown. Recall that wehave chosen

X = C0b,unif(R,R)

as phase space.

Theorem 3.11 The fixed points U ∗ ∈ X defined by u(x, t) = u∗ ∈ R are stable, but notasymptotically stable.

Proof. For given ε > 0 choose δ = ε > 0. Then by the maximum principle from ‖U0−U∗‖X <

δ, i.e.u∗ − δ < u0(x) < u∗ + δ

for all x ∈ R we have thatu∗ − δ < u(x, t) < u∗ + δ

for all x ∈ R and t > 0, i.e. ‖U(t, U0) − U∗‖X < ε which shows the stability. Since there areinfinitely many fixed points {u∗ + δ | |δ| < δ0} in every δ0-neighborhood of U ∗ in X the fixedpoint U∗ cannot be asymptotically stable. �

Again we find stability with respect to a smaller set of perturbations since by the maximumprinciple small spatially localized perturbations can be confined by shifted fronts.

Exercise 3.12 Are the fronts Uf of Burgers equation stable inX = C0b,unif? Does the following

orbital stability hold? For all ε > 0 there exists a δ > 0 such that ‖U0 − Uf‖X < δ implies

infx0∈R

supx∈R

|u(x, t) − uf(x− ct + x0)| < ε.

Use the maximum principle to find smaller sets X1 where (X1, C0b )-stability holds.

Solution 3.13

Exercise 3.14 Solve Burgers numerically. Take initial conditions u0 = a cosh(x/w) with smalla, w on a large domain. Compare to KPP.

Solution 3.15 For KPP we take a = 0.3, w = 0.5 and obtain the evolution shown in fig.12(a);The initial hump triggers two counterpropagating fronts For Burgers (a = 1, w = 0.5) we obtainthe self–similar decay from sec.3.4.

4 The KdV-equation

The KdV-equation [22]∂tu = −∂3

xu+ 6u∂xu (12)

with u = u(x, t) ∈ R, x ∈ R, and t ∈ R occurs as an amplitude equation for unidirectionallong wave surface water waves.

29

(a) (b)

(c)

Figure 11: (a) Instability of Uf due to neighboring front with different speed c; (b) orbital stability ofUf due to entrapping between two neighboring fronts, both with speed c∗; (c) stability of Uf with respectto perturbations u0 ∈ X1 = {u ∈ C0 : v+ = limx→∞ u0(x) ≤ u0(x) ≤ limx→−∞ u0(x) = v−} due toentrapping between Uf (x − x1) and Uf (x − x2).

(a) KPP (b) Burgers

-600

600

5

10

15

0

1

0 0

5

10

0

1

Figure 12: (a) solution of KPP with initial condition, u0(x) = 0.3 cosh(2x); the initial hump triggerstwo counterpropagating fronts. (b) u0(x) = cosh(2x) self–similar decay for Burgers

30

One can interpret this equation as a Hamiltonian systems where ∂x defines the (non-standard)symplectic structure and remarkably, Zhakarov and Faddeev [31] proved that the KdV equationis actually a completely integrable Hamiltonian system. In particular, there exists a canonicaltransformation such that with respect to the new coordinates the Hamiltonian is a function onlyof the action variables and hence in particular, the action variables remain constant in time.The KdV-equation consists of two parts, the dispersion ∂3

xu and the nonlinearity 6u∂xu.The dispersion corresponds to the linearized KdV-equation

∂tu = −∂3xu.

This conserves energy, since

d

dt

∫

R

u2(x)dx = −2

∫

R

u(x)∂3xu(x)dx = 2

∫

R

∂x(∂xu)2dx = 0.

Like for the linear Schrodinger equation energy is spread over the real line. There is an estimate(without proof)

supx∈R

|u(x, t)| ≤ Ct−1/3

∫

|u(y, 0)|dy.

We already dicussed the nonlinearity

∂tu = 6u∂xu

earlier. We found with the method of characteristics that singularities are created in finite time.In the full KdV-equation this process where energy is concentrated works against dispersion.

The existence theory of KdV is quite difficult since KdV is a quasilinear hyperbolic equa-tion. We only note that for localized smooth initial data u0 there exists a local solution (in,e.g., H2(R,R)), and refer to the literature [19] for the details. Instead, here we shall again beinterested in special solutions of KdV.

4.1 The solitary wave

In the full system there will be some equilibrium between linear dispersion and the nonlinearterms. This will create a wave of permanent form, i.e.u(x, t) = v(x − ct) = v(ξ). Insertingthis ansatz into the KdV-equation yields

−cv′ = −v′′′ + 6vv′

Integration with respect to ξ yields

−cv + v′′ − 3v2 = D

with a constant D ∈ R. We are interested in localized waves and so we have limξ→∞ v(ξ) = 0

and limξ→∞ v′′(ξ) = 0 which yields D = 0. By rescaling ξ =√cζ and v = cv we obtain the

first order system

v′ = w, w′ = v + 3v2 (13)

31

which can be discussed in phase plane. From the phase portrait in Figure 13 we find a hetero-clinic solution vhet = vhet(ζ) and in the original variables a family of solitary waves

u(x, t) = cvhet(√c(x− ct)),

i.e. the higher the solitary wave, the smaller the width and the faster the solitary wave.

–2

–1

0

1

2

y

–2 –1 1 2x

–1.4

–1.2

–1

–0.8

–0.6

–0.4

–0.2

0

x(t)

–8 –6 –4 –2 2 4 6 8t

Figure 13: The phase portrait for the solitary waves of the KdV-equation and the solitary wave

Exercise 4.1 Find α ∈ R such that

u(x, t) = αcsech2(√c(x− 4ct)), c > 0 arbitrary (14)

is an exact solution of KdV.

Solution 4.2 This is calculus: we have

(sech2)′ = −2 sinh / cosh3, (sech2)′′ = −2/ cosh4 +4 sinh2 / cosh4

(sech2)′′′ = 16 sinh / cosh5 −8 sinh3 / cosh5

Hence

ut = 8αc5/2 sinh / cosh3

= −uxxx + 6uxu

= −c5/2α(16 sinh / cosh5 −8 sinh3 / cosh5) + 6α2c5/2(−2 sinh / cosh5)

⇔ 8 sinh cosh2 +16 sinh−8 sinh3 +12α sinh = 0

⇔ 24 sinh +12α sinh = 0 ⇔ α = −2.

Remark 4.3 In 1834, Russell observed a solitary wave created by a boat in a canal betweenEdinburgh and Glasgow of 40 cm height and 10 m length which traveled with a speed of 16km/hfor many kilometers through the canal without changing drastically its shape. Motivated by hisobservation in the following years he made a number of experiments. After the publicationof his results [26], a vigorous scientific debate followed over whether or not such a wave ofpermanent form could exist. Airy [1] argued that even if dissipation, i.e. the loss of energy,is neglected, dispersion, i.e. the spreading of energy which is concentrated in the middle ofthe solitary wave, will destroy the solitary wave. It was finally accepted that such waves exist

32

when Boussinesq [4] and Rayleigh [24] found approximations to such a wave by deriving theKdV-equation with some perturbation analysis which took into account the nonlinear nature ofthe problem.

Remark 4.4 These solitary waves occur in nature as so called tsunamis. Tsunamis are largewaves caused by earthquakes or landslides under the sea. The waves created by these eventsare almost unobservable in the open sea as their height is only a few meters and their lengthup to 100 kilometers. For such long waves, where the wavelength is much longer than theunperturbed water height H , it can be shown that their speed is given in first approximationbe

√gH, where g = 9.8 m s−2 is the gravitational constant. In the Pacific Ocean the average

depth is around 5000m which leads with above formula (4.4) for the wave speed of long wavesto an incredible velocity of around 700 km/h (which fits experimental data quite well). If sucha wave encounters an island or coastline, the waves steepen as they come ashore and may reachheights of 20m and cause enormous damage.

4.2 The KdV-equation as Hamiltonian system

The KdV-equation is a completely integrable Hamiltonian system which can solved explicitely.With

H(u) =

∫

R

1

2(∂xu(x))

2 + u(x)3dx

we find

∂uH[v] = limε→0

ε−1(H(u+ εv) −H(u))

= limε→0

ε−1

∫

R

1

2(∂x(u+ εv))2 + (u+ εv)3 − 1

2(∂xu)

2 − u3dx

=

∫

R

(−∂2xu+ 3u2)vdx

which maps v ∈ X = L2(R,C) linearly into R. Hence ∂uH is a linear mapping from X to R,i.e. an element of the dual space, which can again be identified with −∂2

xu+ 3u2 by

β : Lin(X,R) → X(

v 7→< u, v >=

∫

u(x)v(x)dx)

7→ u

and therefore β∂uH = −∂2xu+ 3u2. We finally have

∂tu = −∂3xu+ 3∂x(u

2) = ∂xβ∂uH(u) = Jβ∂uH(u)

where the operator (Ju) = ∂xu is skew symmetric in X since

< Ju, v >=

∫

∂xu(x)v(x)dx = −∫

u(x)∂xv(x)dx = − < u, Jv > .

Exercise 4.5 Prove that the energy∫

Ru2(x, t)dx is a conserved quantity for the KdV-equation.

33

Solution 4.6 We compute directly

1

2∂t

∫

u2dx =

∫

u∂tu dx

=

∫

u(− ∂3xu+ 6u∂xu) dx

=

∫

∂x(1

2(∂xu)

2 + 2u3) dx = 0

4.3 Stability of the solitary wave

Similar to the front solutions for Burgers in Exercise 3.12 we cannot expect naive stability ofthe solitary waves uc since according to formula (14) there will always be a nearby uc∗ withdifferent speed c∗. However, using the Hamiltonian structure of KdV we explain that orbitalstabilty holds. This means that forall ε > 0 there exists a δ > 0 such

‖u0(·) − uc(·, 0)‖H1 ≤ δ ⇒ infx0∈R

‖u(·, t) − uc(x+ x0, t)‖H1 ≤ ε, (15)

i.e., u(·, t) stays close to a translate of uc for all time, that is, it keeps the shape of the solitarywave uc.To explain this, we first note that

M(u) = H(u) + cE(u) =

∫

1

2u2x + u3 + cu2 dx

is a constant of motion, i.e., ddtM(u) = 0. Moreover, uc is a stationary point of M . For this we

calculate

M(uc + εv) =∫

12(u′c + εv′)2 + (uc + εv)3 + c

2(uc + εv)2 dx

= M(uc) + ε∫

(−u′′c + 3u2c + cuc)v dx

+ε2∫

12v′2 + 3ucv

2 + c2v2 dx + ε3

∫

v3 dx

=: M(uc) + ∂uM(uc)[εv] + ∂2uM(uc)[εv, εv] + ∂3

uM(uc)[εv, εv, εv].

(16)

Here ∂uM(uc) is also called the first variation of M . In (16) uc = uc(x) = −2c sech2(√cx) can

be considered a function of x since M is translation invariant, i.e., M(u(·)) = M(u(· + x0)).Since uc satisfies the traveling wave equation v ′′−cv−3v2 = 0 for KdV we see that ∂uM(uc)=0

and uc is a stationary point of M .We explain the idea of the stability result using a cartoon picture; details and flaws will be filledin later. Assume that uc is a local minimum of M . Moreover, assume that M(u0)−M(uc) ismeaningfull as a ”distance” between u and uc. Then we have stability of uc since this distancestays constant due to M(u0) −M(uc) = M(u(t)) −M(uc)), see fig.(14) and Exercise 4.7.Unfortunately, this oversimplified picture is (completely) wrong (and obviously it contradictsthe statement that uc can at best be orbitally stable). In particular, note that M(u) is not aquadratic functional. For remedy, we first restrict u0 to the energy surface of uc, i.e., we restrict

34

ox*

Figure 14: cartoon picture for stability. ∗ is the minimum of the conserved quantity M , hence perturba-

tions x, o can’t get away from ∗.

to initial conditions u0 with E(u0) = E(uc) =: Ec. It can then be shown that uc is a minimumof M in this surface, and moreover that there exists C1, C2 > 0 such that

0 ≤ C1dτ(u, uc)2 ≤M(u) −M(uc) ≤ C2‖u− uc‖2

H1 (17)

for all sufficiently small ‖u − uc‖H1 . This is hard, see [2]. The translational distance dτ isdefined as

dτ (u, v) = infx0∈R

‖u(· + x0) − v(·)‖H1,

cf. (15). The argument hinges on a suitable positive definiteness of the second variation

∂2uM(uc)[v, v] =

∫

1

2v′2 + 3u′cv

2 +c

2v2 dx

of M(uc), cf.(16). From (17) we get orbital stability of uc in the energy surface E(u0) = Ec(why?). We now drop this condition. FromE(uc) = 8c3/2/3 it follows that for |E(u0)−Ec| < δ

there exists a c∗ with |c− c∗| ≤ Cδ and E(u0) = Ec∗ . Hence the above argument gives

infx0∈R ‖u(·, t)−uc(x+x0, t)‖H1

≤ infx0∈R ‖u(·, t) − uc∗(x+ x0, t)‖H1 + infx0∈R ‖uc∗(x, t) − uc(x+ x0, t)‖H1

≤ ε/2 + ε/2.

Concerning fig.14 we note the following theorem.

Exercise 4.7 Consider a finite dimensional Hamiltonian system ddtx = J∇H(x), cf. sec.??,

where H(x) = 12xTAx + O(‖x‖3) with (positive or negative) definite A. Then x = 0 is stable.

Solution 4.8 Let

ρ0(r) = min{H(x) | |x| = r}, ρ1(r) = max{H(x) | |x| = r}.

Then

ρ0(r) =1

2λminr

2 + O(r3), ρ1(r) =1

2λmaxr

2 + O(r3),

35

where λmin (λmax) is the smallest (the largest) eigenvalue of matrix A. There exists a r0 > 0,such that

ρ0(r) ≥1

4λminr

2 und ρ1(r) ≤ λmaxr2.

for all r ∈ [0, r0]. Given ε > 0 we choose (using definiteness of A )

0 ≤ δ ≤√

λmin

4λmaxε.

Then

|x(t, x0)|2 ≤ 4

λmin

ρ0(x(t, x0)) ≤4

λmin

H(x(t, x0))

=4

λmin

H(x0) ≤4

λmin

ρ1(x0) ≤4λmax

λmin

|x0|2 ≤4λmax

λmin

δ2 ≤ ε.

4.4 The soliton property

In the experiments made by Russell from 1834 to 1844 he was also interested in the interactionof solitary waves. He made some sketches that indicated that solitary waves after some non-linear interaction go through each other as if there had been no nonlinear interaction, i.e. theinteraction of two solitary waves leads after interaction to the same two solitary waves with theshapes and the velocities of the two solitar waves before the interaction. By some first computerexperiments in the 60s, in 1965, Zabusky and Kruskal [30] observed that the KdV–equation pos-sessesN–solitons, i.e. waves of permanent form which go through each other without changingtheir shape asymptotically, i.e.

AN−Sol.(x, t) ∼N

∑

j=1

Aβj(x− β2

j t± γj + δj),

for t → ±∞, with β1 > . . . > βN and γj, δj ∈ R some phase-shifts. It is the purpose of therest of this section to give some idea where this behavior comes from and to prove the stabilityof the solitary waves.

4.5 Infinitely many conservation laws

If the KdV-equation is a completely integrable Hamiltonian system, then there must be infinitelymany independent corserved quantities for the KdV-equation. An equation

∂tT (x, t) + ∂xX(x, t) = 0

is called a conservation law. We have

∂t

∫

R

Tdx =

∫

R

∂tTdx = −∫

R

∂xXdx = X|∞−∞ = 0

and so∫

RTdx is a conserved quantity, i.e. does not change in time. For the KdV-equation we

find∂tu+ ∂x(∂

2xu− 3u2) = 0

36

and so, additional to H and E,∫

Ru(x, t)dx is independent of time. Next we find

∂t(u3 +

1

2(∂xu)

2) + ∂x(−9

2u4 + 3u2∂2

xu− 6u(∂xu)2 + (∂xu)(∂

3xu) −

1

2(∂2xu)

2) = 0

and so T3 = u3 + 12(∂xu)

2 yields the conserved quantity∫

T3(x, t)dx. It turns out that there areinfinitely many conserved quantities:

T4 = 5u4 + 10u(∂xu)2 + (∂2

xu)2,

T5 = 21u5 + 105u2(∂xu)2 + 21u(∂2

xu)2 + (∂3

xu)2,

...

For the linearized system we have that Tn+2 = (∂nxu)2 since

∂t(∂nxu)

2 + ∂x(∂n+2x u∂nxu− (∂n+1

x u)2/2).

In Fourier space we find the solutions u(k, t) = eik3tu(k, 0). Thus by introducing polar coordi-

nates u(k, t) = r(k, t)eiφ(k,t) we find

r(k, t) = r(k, 0), φ(k, t) = φ(k, 0) + k3t mod 2π

or equivalently∂tr(k, t) = 0, ∂tφ(k, t) = k3 = const.

Thus, we found the uncountable many action variables r(k, ·) and uncountable many anglevariables φ(k, t), and consequently tori of any dimension.We close this section by describing how a transformation discovered by Miura and then gene-ralized by Gardner ([23]-[14]) leads very easily to the conclusion that there are infinitely manyconserved quantities for the KdV equation. The basic idea is that given a transformation whichmaps solutions of one equation to solutions of a second, the existence of simple or conservedquantities for the the first equation leads, via the transformation, to more complicated conservedquantities for the second.Given u = u(x, t), define w(x, t) implicitly via the formula

u(x, t) = w(x, t) + iε∂xw(x, y) + ε2(w(x, t))2 . (18)

Note that if w is smooth enough and ε is small, we can invert this relation recursively to obtainw in terms of u via the formula

w = u− iε∂xu− ε2(u2 + ∂2xu) + iε3(∂3

xu+ 4u∂2xu)

+ε4(2u3 + 5(∂xu)2 + 6u∂2

xu+ ∂4xu) + O(ε5) . (19)

Now compute

∂tu− ∂3xu− 6u∂xu = {∂tw − 6w∂xw − 6ε2w2∂xw − ∂3

xw} (20)

+2ε2w{∂tw − 6w∂xw − 6ε2w2∂xw − ∂3xw}

+iε∂x{∂tw − 6w∂xw − 6ε2w2∂xw − ∂3xw}

37

From this we see immediately that if w satisfies the modified KdV equation

∂tw − 6w∂xw − 6ε2w2∂xw − ∂3xw (21)

then u, defined by (18) satisfies the KdV equation. However, one also sees immediately thatthe the integral of w is a conserved quantity of (21) for all values of ε – i.e. if we defineIε(t) =

∫

w(x, t)dx, then Iε is a constant for all values of ε. (We will assume here that w isdefined on the real line, and that w and its derivatives go to zero as |x| tends to infinity. Similarresults hold for x running over a finite interval with periodic boundary conditions.) But thisin turn immediately implies that if we use (19) to expand Iε in powers of ε the coefficients inthis expansion must also be constants in time. Since these coefficients will be expressed asintegrals of u and its derivatives they will give us (infintely many) conserved quantities for theKdV equation! Looking at the first few of these we find:

1. K0 =∫

u(x, t)dx. The conservation of this quantity follows immediately from the formof the KdV equation.

2. K1 =∫

∂xu(x, t)dx = 0, if we assume that u and its derivatives tend to zero as |x| tend toinfinity. Thus, we gain no new information from this quantity and in fact, all the integralscoming from the odd powers of ε turn out to be “trivial” so we ignore them and focus juston the even powers of ε.

3. K2 =∫

(u2 + ∂2xu)dx =

∫

u2dx. That this is a conserved quantity is again easy to seedirectly from the KdV equation, just by multiplying the equation by u and integratingwith respect to x.

4. K4 =∫

(3u2 + 5(∂xu)2 + 6u∂2

xu+ ∂4xu)dx =

∫

(3u2 − (∂xu)2)dx.

Exercise 4.9 Write ∂tr(k, t) = 0, ∂tφ(k, t) = k3 as a Hamiltonian system.

Solution 4.10 With H = k3r we find

∂tr =∂H

∂φ= 0 , ∂tφ = − ∂H

∂r= k3 .

4.6 The 2-solitons

In order to prove that the KdV-equation is a completely integrable Hamiltonian system we haveto find a transformation which transforms u into action and angle variables. This transformationis called the inverse scattering transform. See [10] for its derivation and motivation, and fordetails concerning the formulas below. Here we only use it to calculate explicit solutions. Themethod is as follows: let u = u(x, t) be a solution of the KdV-equation and define

L(t)v(x) = −∂2xv(x) + u(x, t)v(x).

This is a selfadjoint linear operator (in L2(R,R)) and depending on the potential u = u(x, t) ofL(t) there is discrete and continuous real spectrum. In X = C0

b,unif there are finitely many, say

38

N discrete eigenvalues, 0 > λ1 > λ2 > . . . λN and a continuum [0,∞) of eigenvalues which aredenoted with λ(k) = k2 for k ∈ R. Corresponding to the eigenvalues there are eigenfunctionsψk for k = 1, . . . , N and ψ(k) for k ≥ 0. These eigenfunctions are completely determined bytheir asymptotic behavior for x→ ±∞ [7]. For ψk normalized by

∫

|ψk(x)|2dx = 1 we have

ψk ∼ ck

{

e−κkx x→ ∞eκkx x→ −∞

where κ2k = −λk. For ψ(k) we have

ψ(k) ∼{

e−ikx + b(k)eikx x→ ∞a(k)e−ikx x→ −∞

Therefore, for given u(x, t) we have the scattering data (λk(t), ck(t)) for k = 1, . . . , N and(λ(k, t), b(k, t)) for k ∈ R. After some lengthy calculation it turns out that

d

dtλ(k) = 0,

d

dtλk = 0,

d

dtck = 4κ3

kck,d

dtb(k) = −8ik3b(k).

Thus, the λk and λ(k) are the conserved quantities, the action variables, whereas the ck and b(k)behave linearly, the angle variables. It was a known some before that the potential u = u(x, t)

can be reconstructed from the scattering data (λk, ck(t), λ(k), a(k, t), b(k, t)) by the Marchenko-or Gelfand-Levitan equation. That is,

u(x) = −2d

dxK(x, x)

with K(x, z) satisfying the linear integral equation

K(x, z) + F (x + z) +

∞∫

x

K(x, y)F (y + z)dy = 0

with

F (x) =

n∑

j=1

c2je−κjx +

1

2π

∫

R

eikxb(k)dk.

For illustration, and in order to construct a so called 2-soliton we take the scattering data

κ1 = 1, κ2 = 2, c1 =√

6, c2 = 2√

3.

Moreover, we choose b(k, 0) = 0 for all k ∈ R which implies b(k, t) = 0 for all k ∈ R andt ∈ R. We find

κ1 = 1, κ2 = 2, c1(t) =√

6e4t, c2 = 2√

3e32t, F (x) = 6e8t−x + 12e64t−2x,

and make the ansatzK(x, z) = `1(x)e

−z + `2(x)e−2z

39

for K(x, z). With Fj(x) = c2je−κjx we obtain

0 = `1e−z + `2e

−2z + F1(x)e−z + F2(x)e

−2z

+

∞∫

x

(`1(x)e−y + `2(x)e

−2y)(F1(z)e−y + F2(z)e

−2y)dy

which yields

l1 + 6e8t−x + 3l1e8t−2x + 2l2e

8t−3x = 0,

l2 + 12e64t−2x + 4l1e64t−3x + 3l2e

64t−4x = 0,

hence

l1 = 6(e72t−5x − e8t−x)/D, l2 = − 12(e64t−2x + e72t−4x)/D,

whereD = 1 + 3e8t−2x + 3e64t−4x + e72t−6x.

Finally, we obtain

u(x, t) = −2∂x(L1e−x + L2e

−2x) = −123 + 4 cosh(2x− 8t) + cosh(4x− 64t)

(3 cosh(x− 28t) + cosh(3x− 36t))2. (22)

Figure 15: density plot of the 2–soliton (22), black=minimum, white=max=0.

Exercise 4.11 Let L(t) = L(t)T ∈ Rd×d satisfy

L(0) = U(t)TL(t)U(t) with U(t)U(t)T = Id .

a) Show that L(t) satisfies∂tL = [M,L]

with M = −(∂tU)UT = U∂tUT = −M , and where [M, l] := ML − LM is called the

commutator of M and L.b) Prove that the eigenvalues of L are conserved quantities.

40

c) Let L(t)v = −∂2xv(x)+u(x, t)v(x) and let Mv(x) = c∂xv(x). Show that ∂tL = ML−LM

simplifies to ∂tu = c∂xu.d) Let L as in c), but let now

Mv(x) = α∂3xv(x) +

3α

4∂x(u(x, t)v(x))

Show that ∂tL = ML− LM simplifies to ∂tu+ α4∂3xu− 3

2αu∂xu=0 (which for α=4 is KdV).

Solution 4.12 a) ∂tL = UtL0UT +UL0U

Tt = UtU

TL+LUUTt = ML+LMT = ML−LM

since MT = −M due to 0 = ddt

(UUT ) = M +MT .b) Simple answer: similar matrices have the same eigenvalues. Detailed answer: if ψ0 is aneigenvector to L0 with eigenvalue λ0, then ψ = Uψ0 is an eigenvector to L(t) with eigenvalueλ0.c) With M = c∂x we obtain

(LM −ML)v = (− ∂2x + u)c∂xv − c∂x(−∂2

xv + uv)

= −c∂3xv + cu∂xv + c∂3

xv − cu∂xv − c(∂xu)v = −c(∂xu)v

which yields that ∂tL = ML − LM is equivalent to ∂tu = c(∂xu)

d) With M = α∂3x + U∂x + ∂xU · +A where α ∈ R, U = U(x, t) and A = A(x, t) we obtain

[L,M ] = αuxxx − Uxxx − Axx − 2uxU + (3αuxx − 4Uxx − 2Ax)∂x + (3αux − 4Ux)∂2x.

With the choice U = 34αu and A = A(t) we obtain for ∂tL+ [L,M ] = 0 the KdV-equation

∂tu+1

4α∂3

xu−3

2αu∂xu = 0.

Remark and summary: The idea in this calculus is as follows. Given an equation ut = N(u)

express it as Lt = [M,L] (this is the hard step!) with L = L∗ (self adjoint) and M = −M ∗

(skew adjoint). For the eigenvalue (scattering) problem Lψ = λψ this implies ddtλ = 0, i.e. the

eigenvalues are conserved, and for the eigenvectors we obtain ψt = Mψ. If Lv = −∂2x + uv,

then u can as seen be reconstructed by inverse scattering. For a differential operator M tobe skew adjoint it must contain only odd derivatives (why?). There are infinitely many skewsymmetric operators M giving a ”hierarchy of KdV-equations”. The next one would be

Mv = −α∂5xv + u∂xv + uxv + u∂3

xv + (∂3xu)u.

The Hamiltonian functions of the obtained equations are related to the conservation laws above.

Exercise 4.13 Consider the solution

u(x, t) = −123 + 4 cosh(2ξ + 24t) + cosh(4ξ)

(3 cosh(ξ − 12t) + cosh(3ξ + 12t))2(23)

with ξ = x− 16t of the KdV equation. Prove that for t→ ±∞ this solution separates into twosingle waves, i.e.,

u(x, t) ∼ −8sech2(2ξ ∓ 1

2log 3) − 2sech2(η ± 1

2log 3)

with η = x − 4t, where ∼ means asymptotically equal. Discuss mechanical analogs for theinteraction given by (23); discuss the name ’soliton’.

41

Solution 4.14 We have

− 1

12u(x, t) =

3+4 cosh(2ξ+24t)+ cosh(4ξ)

(3 cosh(ξ−12t)+ cosh(3ξ+12t))2

=3+2(e2ξ+24t+e−2ξ−24t)+1

2(e4ξ+e−4ξ)

(32(eξ−12t+e−ξ+12t)+1

2(e3ξ+12t+e−3ξ−12t))2

.

For t→ ∞ and ξ fixed this yields

− 1

12u(x, t) ∼ 1/(

9

8e−4ξ +

3

4+

1

8e4ξ),

and the ansatz

9e−4ξ + 6 + e4ξ = α cosh2(2ξ + β) =α

4(e4ξ+2β + 2 + e−4ξ−2β)

yields

α = 12, β =1

2ln 3.

This is the desired result for t → ∞, ξ fixed. The cases t → ∞ and η = x − 4t fixed workthe same way. The name soliton comes from the fact that the two solitary waves interact likeparticles, e.g., photons.

Exercise 4.15 Compute the eigenvalues and eigenfunctions of Lψ = ψ ′′ − u(x)ψ in caseu(x) = −U0δ(x) with U0 a positive constant and δ the Dirac delta–distribution defined by∫

δ(x)u(x) dx = u(0).

Solution 4.16 Integration of −∂2xψ − U0δ(x)ψ = λψ gives

−∂xψ|%x=−% −%

∫

−%

U0δ(x)ψ(x)dx =

%∫

−%

λψ(x)dx,

which, since ψ ∈ C0, for ρ→ 0 yields the jump condition

−[∂xψ] − U0ψ(0) = 0, [u] := limε→0

u(ε) − limε→0

u(−ε).

For x 6= 0 we have to solve −∂2xψ = λψ. For λ > 0, i.e. λ = k2 we have ψ ∼ e±ikx. We make

the ansatz

ψ(x, k) =

{

e−ikx + b(k)eikx x > 0

a(k)e−ikx x < 0

Continuity in x = 0 leads to 1 + b = a and the jump condition to

−ik + bik − (−ika) = −U0(1 + b) = U0a(k)

which finally leads to

b(k) = − U0

U0 + 2ik

42

For λ < 0, i.e. λ = −κ2n we have ψ ∼ e±κnx. Therefore, we make the ansatz

ψn(x) =

{

αne−κnx x > 0

βne+κnx x < 0

Continuity yields αn = βn. Normalisation gives

0∫

−∞

α2ne

2κnxdx+

∞∫

0

α2ne

−2κnxdx = 1

and so αn = κ2n. The jump condition then gives

[ψ′n] = − κnαn − κnαn = − U0αn

and so κ1 = 12U0, i.e. we have only negative eigenvalues.

4.7 Asymptotics

Let u0 be an initial condition of the KdV-equation with∫

Ru0(x)(1 + |x|)dx <∞. Then

Lv = ∂2xv + u0(x)v

possesses finitely many, say N , eigenvalues. Each eigenvalue correspond to a soliton in theKdV-equation. Then for large t the solution u consists of N -solitons which are ordered withrespect to their height which is proportional to the velocity of the solitons. Behind the solitonswe find a dispersive rest described by the scattering data deacying with a rate t−1/3.

4.8 Other mathematical aspects of the KdV equation

In addition to the Inverse Scattering Transform approach, more traditional approaches to theexistence and uniqueness of solutions have also been studied, starting with the proof of the wellposedness of solutions of the KdV equation with periodic boundary conditions in the Sobolevspace H2 by R. Temam. Noting that the Hamiltonian for the KdV equation described in thepreceeding Section is closely related to the H1 norm, this might seem a natural space in whichto study well-posedness, but surprisingly Kenig, Ponce and Vega and Bourgain showed that theequation is well posed in Sobolev spaces H s, with s < 1 and more recent work has extendedthe global well posedness results to Sobolev spaces of small negative order. Aside from theirintrinsic interest, these results have other physical implications. If one wishes to study statisti-cal aspects of the behavior of ensembles of solutions of these equations, statistical mechanicssuggests that the natural invariant measure for these equations is given by the Gibbs’ measure.However, the Gibbs’ measure is typically supported on functions less regular than H 1, so thatin order to define and study this measure one needs to know that solutions of the equation arewell behaved in such spaces.

43

Another natural mathematical question arises from the fact that the KdV equation is only anapproximation to the original physical equation. Viewed from another perspective, the origi-nal system can be seen as a perturbation of the KdV equation. It then becomes natural to askwhether the special features of the KdV equation are preserved under perturbation. Viewing theKdV equation as a completely integrable Hamiltonian system this is very analologous to thequestions studied by the Kolmogorov-Arnold-Moser (KAM) theory and has led to a develop-ment of KAM like results for a number of different partial differential equations like the KdVequation. The results are somewhat technical in nature but roughly speaking they say that if oneconsiders the KdV equation with periodic boundary conditions, temporally periodic or quasi-periodic solutions will persist under small perturbations. The situation is more complicated andless well understood for the equation on the whole line due to the presence of a continuum ofscattering states. For a very thorough review of the problem with periodic boundary condtionssee [18].

5 The NLS-equation

The Nonlinear Schrodinger (NLS) equation

∂tu = −i∂2xu+ αiu|u|2, α = ±1,

with t ∈ R, x ∈ R, and u(x, t) ∈ C can be derived by multiple scaling analysis in order todescribe the evolution of an envelope of spatially and temporarely oscillating wave packet. Thecase α = −1 is called focussing and α = +1 is called defocussing. In particular the focussingNLS is widely used in nonlinear optics to describe the evolution and interaction of pulses.

Figure 16: Sketch of an electromagnetic pulse u(X − cgT, T )eik0(x−cpt), where x and t are the ’fast’

spatial and temporal scale of the underlying wave and where X and T are the ’slow’ spatial and temporal

scale of the envelope which is described by the NLS-equation. The frequency of the wave is 2π/k0. cp

is called the phase velocity and cg the group velocity.

44

Exercise 5.1 Show that, additional to the usual translation invariance, the NLS has the impor-tant S1, Galilei and scaling invariance, i.e.: if u(x, t) solves NLS, then for all φ, c, η ∈ R

v(x, t;φ) = u(x, t)eiφ S1–invariance,

v(x, t; c) = u(x− ct, t)ei(c2t−2cx)/4 Galilei–invariance

v(x, t; η) = ηu(ηx, η2t), scaling–invariance

solve NLS too.

Solution 5.2

∂tv + i∂2xv − αiv|v|2 = ∂t(ue

iφ) + i∂2x(ue

iφ) − αiueiφ|ueiφ|2

= eiφ(∂tu+ i∂2xu− αiu|u|2) = 0

which shows the S1–invariance.

∂tv + i∂2xv − αiv|v|2 = (∂tu− c∂xu+ i

c2

4u+ i (∂2

xu−2c · 2i

4∂xu−

ic2

4u)

−αiu|u|2 )ei(c2t−2cx)/4 = 0

which shows the Galilei–invariance.

∂tv + i∂2xv − αiv|v|2 = η3(∂tu+ i∂2

xu− αiu|u|2) = 0

which shows the scaling invariance.

The NLS-equation consists of two terms, namely the dispersion term i∂2uu and αiu|u|2 which

leads to nonlinear oscillations.

5.1 Nonlinear oscillations

By the ansatz u(x, t) = v(t) we look for x-independent solutions. We find ∂tv = αiv|v|2.Introducing polar coordinates v(t) = r(t)eiφ(t) we obtain

∂tr = 0 and ∂tφ = αr3.

Therefore, we find oscillations where the frequency increases if r is getting bigger. Next wesearch for solutions of the form

u(x, t) = eikxv(t).

Then v solves the ODE∂tv = ik2v + αiv|v|2.

Again the introduction of polar coordinates v(t) = r(t)eiφ(t) leads to

∂tr = 0 and ∂tφ = k2r + αr3,

with solution φ(t) = φ0 +ω(k, r)t with ω(k, r) = (k2r+αr3)t. In particular, for α = −1 thereexists standing waves for k2r − r3 = 0.

45

5.2 The pulse solutions

Due to its derivation for the description of modulations of electromagnetic waves the NLS-equation should possess pulse solutions. We make the ansatz

u(x, t) = B(x− ct)ei(qx−ωt+φ0).

Then B with B(ξ) ∈ R satisfies

−iωB − cB′ = −iB′′ + 2qB′ + iq2B + αiB3

or equivalently c = −2q and

0 = B′′ − (ω + q2)B − αB3. (24)

The ODE (24) reminds us of exercise ??. Pulse solutions exist for ω + q2 > 0 and α = −1.

Exercise 5.3 Use xppaut to show that for (ω + q2) > 0 and α = −1 there exists solutions of(24) which are homoclinic to 0.

Solution 5.4 Fig.17 shows the phaseplane for (ω + q2) = 1, α = −1.

Figure 17: phaseplane for (24).

In fact, there exist explicit formulas for these homoclinics.

Exercise 5.5 Show that the so called solitons

u(t, x; η, c, γ, x0) =√

2ηsech(η(x− x0 − ct))ei((c2−4η2)t−2cx+γ)/4,

η, c, γ, x0 ∈ R arbitrary, are exact solitions of the focussing NLS.

Solution 5.6 By Exercise 5.1 it suffices to show that u(x, t) =√

2sech(x)e−it is an exact solu-tion. This is easy calculus.

46

5.3 Local existence and uniqueness

As explained in sections 1.1 and 1.3 the right phase space for the linear Schrodinger equationis X = L2(R,C). In particular there is no local existence in C0

b,unif(R,C). Unfortunately,L2(R,C) is not closed under multiplication and hence we can not prove local existence forNLS in L2(R,C) using the method from sec.2.2.It turns out that a good phase space for NLS is X = H1(R,C), where the Sobolev spaceH1(R,C) is defined by

H1(R,C) = clos‖·‖H1

(

C∞0 (R,C)

)

, where ‖u‖2H1 =

∫

|u|2 + |∂xu|2 dx.

For us the most important properties of H1 are that H1 is a Hilbert space and that ‖u‖C0 ≤C‖u‖H1 for all u ∈ H1 (Sobolev imbedding theorem).

Exercise 5.7 Show that L2(R,C) is not closed under multiplication. Show that H1(R,C) isclosed under multiplication.

Solution 5.8 For f defined by

f(x) =

{

x−1/3 0 < x < 1

0 elsewhere

we have f ∈ L2 but f 2 6∈ L2 since∫ 1

0(f 2)2 dx =

∫ 1

0x−4/3 dx = ∞. For f, g ∈ H1 we have

‖fg‖2H1 =

∫

(∂x(fg))2 + (fg)2 dx =

∫

f 2xg

2 + f 2g2x + f 2g2 dx

≤ (‖f‖2C0 + ‖g‖2

C0)∫

f 2x + g2

x + f 2 dx ≤ C(‖f‖2H1 + ‖g‖2

H1)2 <∞

by Sobolevs embedding ‖u‖C0 ≤ C‖u‖2H1 .

Theorem 5.9 Let u0 ∈ X = H1(R,C). Then there exists a T > 0 and a unique local solutionu ∈ C([0, T ], X) of NLS with u|t=0 = u0.

Proof. As in Exercise 1.24, i∂2x generates a strongly continous semigroup eit∂2

x : X → X .Moreover, the nonlinearity u 7→ iα|u|u is locally Lipschitz in X . The local existence thusfollows by the variation of constant formula and the contraction mapping theorem. �

Remark 5.10 Like for the linear (and nonlinear) wave equation or the transport equation thesolution in fact exists also backwards in time, i.e., u ∈ C([−T, T ], X)).

This local solution can be continued as long as the H1 norm stays bounded. The estimate forthe H1 norm is related to the Hamiltonian structure of the NLS and therefore considered in thenext section. First however we show the so called conservation of the L2 norm. We have

d

dt‖u‖2

L2 =d

dt

∫

uudx =

∫

∂tuu+ u∂tudx = 2Re

∫

u(−i(∂2xu− α|u|2u)) dx = 0.

47

5.4 The NLS-equation as Hamiltonian system

It turns out that in the case α = −1 the NLS-equation is a completely integrable Hamiltoniansystem which can solved explicitely. Here we only show that the NLS-equation is a Hamiltoniansystem. With

H(u) =

∫

R

1

2|∂xu(x)|2 +

1

4α|u(x)|4dx

we find

∂uH[v] = limε→0

ε−1(H(u+ εv) −H(u))

= limε→0

ε−1

∫

R

|∂x(u+ εv)|2/2 + α|u+ εv|4/4 − |∂xu|2/2 − |u|4/4dx

=

∫

R

(∂2xu+ αu|u|2)vdx

which maps v ∈ X = L2(R,C) linearly into C, i.e. ∂uH is a linear mapping from X to C,i.e. an element of the dual space. In Hilbert spaces the dual space can be identified with X bydefining a mapping (the canonical isomorphism)

β : Lin(X,C) → X,

(

v 7→ 〈u, v〉 = Re

∫

u(x)v(x)dx

)

7→ u

and therefore β∂uH = ∂2xu+ αu|u|2. We finally have

∂tu = i∂2xu+ αiu|u|2 = iβ∂uH(u) = Jβ∂uH(u)

where the operator (Ju) = iu is skew symmetric in X since

< Ju, v >=

∫

iu(x)v(x)dx = −∫

u(x)iv(x)dx = − < u, Jv > .

Exercise 5.11 Prove directly the conservation of the H(t) = H(0) of the Hamiltonian, i.e.,show d

dtH(t) = 0 by direct calculation.

Solution 5.12

d

dtH(t) =

d

dt

∫

1

2|∂xu|2 +

1

4α|u|4 dx =

d

dt

∫

1

2(∂xu)(∂xu) +

1

4αu2u2 dx

=

∫

(∂x∂tu)(∂xu) + (∂xu)(∂x∂tu) +1

2αuu2∂tu+

1

2αu2u∂tudx

=

∫

∂x(− i∂2xu+ αiu2u)∂xu+ (∂xu)(∂x(i∂

2xu− iαu2u))

+1

2αuu2(−i∂2

xu+ iαu2u) +1

2αu2u(i∂2

xu− iαu2u) dx

= 0

48

Note that the Hamiltonian is well defined on H1 since H ≤ ‖∂xu‖2L2 + ‖u‖2

C0‖u‖2L2 . In the

defocusing case the Hamiltonian is positive definite, i.e., H(u) > 0 for u 6= 0, whereas in thedefocusing case it is indefinite. Hence in the defocusing case we get the H 1 estimate for freesince then

‖u(t)‖H1 ≤ H(t) + ‖u(t)‖L2 = H(0) + ‖u0‖2L2

for all t ∈ R.

Remark 5.13 It can be shown that the solution u to initial condition u0 ∈ H1 exists for all timealso in the focusing case. For x ∈ Rd with d ≥ 2 this is in general not true and the solution can’blowup’ in finite time T (‖u(t)‖H1 → ∞ as t→ T ).

6 The Sine-Gordon equation

The Sine-Gordon equation (a pun referring to the Klein–Gordon equation)

∂2t u = ∂2

xu− sin(u), (25)

with x ∈ R, t ∈ R, and u(x, t) ∈ R, in the form ∂ξ∂ηu = sin(u) originally came up in diffe-rential geometry describing surfaces with a constant negative curvature (Enneper 1870). It wasfound to govern the propagation of a dislocation in a crystal whose periodicity is representedby sin u (Frenkel, Kontorova 1939). It was posed as a tentative model of an elementary particle(Perring, Skyrme 1962) and it was shown to be an equivalent form of the so called Thirringmodel (Coleman 1975). For more examples see [11].

It also turns out to be a completely integrable Hamiltonian system. See Exercise 6.2 for someof the infinitely many conservation laws.

Remark 6.1 A simple experimental model for the sine-Gordon equation consists of a rubberband of 1 m length, 6 mm width and 2 mm thickness where a set of similar pins with largeheads is inserted into one side of the strip, centrally and uniformly about 3 mm apart. Using theequation

ld2φ

dt2= −g sinψ

for the motion of a simple pendulum of length l which makes angle ψ with the vertical, and theequation

∂2ψ

∂t2= c2

∂2ψ

∂x2

for torsional waves along an elastic column, it is plausibly that the angle of the pins is governedby a sine-Gordon equation. It is a useful experiment to get an impression of the subsequentsolutions, as kink and antikink solutions.

Our plan is as follows After discussing the local existence and uniqueness of solutions we dis-cuss a number of special solutions, as kinks and antikinks. We derive and justify the NonlinearSchrodinger equation for the approximate description of small amplitude spatially localized

49

time-periodic solutions, so called breather solutions. Finally we explain that in the class of ge-neral nonlinear wave equations the Sine-Gordon solution is unique in possessing exact breathersolutions.

Exercise 6.2 Given the Sine-Gordon equation in the form

∂x∂tu = sin(u)

verify the following conservation laws

∂x((∂tu)2/2) − ∂t(1 − cos u) = 0,

∂t((∂xu)2/2) − ∂x(1 − cos u) = 0,

∂t((∂xu)4/4 − (∂2

xu)2) + ∂x((∂xu)

2 cos u) = 0.

How is this form of the Sine-Gordon equation related to the usual form (25)?

6.1 The local existence and uniqueness

We consider the linearized problem

∂2t u = ∂2

xu− u

which is solved by u(x, t) = eikx+iω(k)t with ω1,2(k) = ±√

1 + k2. Therefore the group veloci-ties ±cg = ± k√

1+k2 satisfy |cg| < 1. Hence beside phase spaces based on Sobolev spaces also

(u, ∂tu) ∈ X = Cm+1b,unif × Cm

b,unif is possible. We have

Theorem 6.3 Fix m ≥ 1 and let (u0, u1) ∈ Cm+1b,unif × Cm

b,unif . Then there exists a t0 > 0 suchthat the Sine-Gordon equation (25) possesses a unique solution u ∈ C([−t0, t0], Cm+1

b,unif) withu|t=0 = u0 and ∂tu|t=0 = u1.

Proof. In order to construct solutions of (25) we use the iteration scheme

u(x, t) =1

2(u0(x+ t) + u0(x− t)) +

1

2

∫ x+t

x−tu1(ξ)dξ +

∫ t

0

∫ x+(t−s)

x−(t−s)f(y, s)dyds

with f(x, t) = sin(u(x, t)). It is based on the solution of the inhomogeneous wave equation.See Exercise 6.4. For t > 0 sufficiently small the right hand side F (u)(x, t) is a contractionin the space Cm+1

b,unif . Thus there exists a unique fixed point solution u∗ = F (u∗) which is aclassical solution of the Sine-Gordon equation if m ≥ 2. �

Exercise 6.4 Prove that the solution of

∂2t u(x, t) = ∂2

xu(x, t) + f(x, t)

with u(x, 0) = u0(x) and ∂tu(x, 0) = u1(x) is given by

u(x, t) =1

2(u0(x + t) + u0(x− t)) +

1

2

∫ x+t

x−tu1(ξ)dξ +

∫ t

0

∫ x+(t−s)

x−(t−s)f(y, s)dyds.

50

Solution 6.5 The homogenous problem f = 0 has been solved in Exercise 1.8. For f 6= 0 see[27, p69ff].

Exercise 6.6 Consider the linear inhomogeneous ordinary differential equation

∂2t u(t) = A2u(t) + f(t)

with u(t) ∈ Rd, A ∈ Rd×d, f(t) ∈ Rd, and initial conditions u(0) = u0 and ∂tu(0) = u1. Provethe formula of Duhamel

u(t) = S ′(t)u0 + S(t)u1 +

∫ t

0

S(t− s)f(s)ds

where S(t) = A−1 sin(At) and S ′(t) = cos(At). Use this idea to regain the formula for thesolution of the inhomogeneous wave equation from Exercise 6.4.

6.2 Kink and antikink solutions

We look for traveling wave solutions u(x, t) = v(x − ct) = v(ξ) of (25) and find that v has tosatisfy

(c2 − 1)∂2ξv = − sin(v).

For |c| > 1 this is the equation of the pendulum with the two heteroclinic connections. Thesetraveling wave solutions are called kink and antikink for (25). For |c| < 1 the phase portrait isshifted by v 7→ v + π. Hence we have heteroclinic solutions no longer between (−π, 0) and(π, 0) but between (0, 0) and (2π, 0).

Exercise 6.7 Verify that for arbitrary C and λ with |λ| < 1

u(x, t) = 4 arctan(C exp((x− λt)/(1 − λ2)1/2))

is a solution of the Sine-Gordon equation ∂2t u = ∂2

xu − sin u. Sketch the solution. (Explicitformula for the kink solution)

Exercise 6.8 Verify that for arbitrary λ with 0 < |λ| < 1

u(x, t) = 4 arctan

(

λ cosh(x/(1 − λ2)1/2)

sinh(λt/(1 − λ2)1/2)

)

is a solution of the Sine-Gordon equation ∂2t u = ∂2

xu − sin u. Sketch the evolution of thesolution. (Explicit formula for a kink-antikink interaction)

6.3 Derivation of the Nonlinear Schrodinger equation

For the Sine-Gordon equation there is no dissipation of elastic energy, i.e.∫

12(∂xu)

2+12(∂tu)

2−cos udx is a conserved quantity, cf. Exercise 6.2. We already have seen that the linearized pro-blem

∂2t u = ∂2

xu− u

51

possesses solutions u(x, t) = eikx+iω(k)t with ω1,2(k) = ±√

1 + k2. Hence |∂2kω1,2(k)| > 0

with a maximum at k = 0 and |∂2kω1,2(k)| → 0 for |k| → ∞. Hence in the linearized system

dispersion will always destroy a localized bump of energy.

However, like for the NLS-equation or the KdV-equation the question occurs, can the nonlinearterms be in some equilibrium with the dispersion such that localized bumps of energy can persistin the nonlinear equation. The Sine-Gordon equation possesses vanishing group velocity cg =

dω/dk = k/√

1 + k2 at the wave number k = 0. Hence we are interested in bumps of energystaying fixed in space. In Section 6.2 we already looked for waves which are of permanent formin a moving frame and found that such waves can only exist for large amplitudes. Therefore, welook now for spatially localized time-periodic solutions. Such solutions are called breathers.

In order to find such solutions we make an ansatz

u(x, t) = εA(εx, ε2t)eiω0t + c.c., (26)

with 0 < ε � 1 a small parameter, ω0 the basic temporal wavenumber, and A = A(X, T ) acomplex-valued amplitude function. This ansatz is chosen in such a way that later on the linearand nonlinear terms are of a comparable order. Our goal is to find an ε-independent equation forthe amplitude function A for which the existence of spatially localized time-periodic solutionscan be established. For later purposes we define the so called residual

Res(u) = −∂2t u(x, t) + ∂2

xu(x, t) − u(x, t) + u3(x, t)/6 + O(|u|5) (27)

where we expanded the nonlinear terms of (25). The residual contains all terms which do notcancel after inserting the approximation into (25). We have Res(u) = 0 if u is an exact solutionof (25). Inserting the ansatz (26) into the residual (27) and using the abbreviation E = eiω0t

yields

Res(u) = εE((ω20−1)A) + ε3

E(−2iω0∂tA+ ∂2XA+ A|A|2/2) + ε3

E3(A3/6) + c.c. + O(ε4).

We used

∂tu = ε[iω0A+ ε2AT ]E + c.c.,

∂2t u = ε[−ω2

0A− 2iω0ε2∂TA+ ε4∂2

TA]E + c.c.,

∂xu = ε2[∂XA]E + c.c.,

∂2xu = ε3[∂2

XA]E + c.c.,

u3 = ε3[A3E

3 + 3 |A|2AE + 3 |A|2AE−1 + A

3E

−3].

We choose now ω0 and A in such a way that the residual is made small. We find that A has tofind a Nonlinear Schrodinger equation, in detail

ω20 = 1,

2iω0∂TA = ∂2XA+ A|A|2/2.

By this choice we findRes(u) = ε3

E3(A3/6) + c.c. + O(ε5).

52

We recall that the Nonlinear Schrodinger equation has a three-parameter family of time-periodicsolutions of the form

A(X, T ) = B(X −X0)e−iγ0T eiφ0 ,

in which the real-valued function B satisfies the second-order ordinary differential equation

∂2XB = C1B − C2B

3,

where C1 = 2γ0ω0 and 2C2 = 1. For γ0 > 0 and ω0 > 0 this equation has two homoclinicsolutions

Bhom(X) = ±(

2C1

C2

)1/2

sech (C1/21 X)

which connect the origin with itself. This procedure therefore identifies formally spatially loca-lized time-periodic solutions of the Sine-Gordon equation which are described by the approxi-mate formula

utp = ε(Bhom(X)e−iγ0T eiω0t + c.c.) = εBhom(εx)ei(1−γ0ε2)t + c.c.

accurately over time-scales of order O(1/ε2).

Exercise 6.9 a) Which equation for A is obtained by making the ansatz

u(x, t) = ε2A(εx, ε2t)eiω0t) + c.c.

for the solutions u of the Sine-Gordon equation.

b) Which equation for A is obtained by making the ansatz

u(x, t) = ε1/2A(εx, ε2t)eiω0t) + c.c.

for the solutions u of the Sine-Gordon equation.

Exercise 6.10 Here we show that by adding higher order terms to the approximation the resi-dual can be made arbitrary small.

a) Make an ansatz

u(x, t) = εA1(εx, ε2t)eiω0t) + ε3A3(εx, ε

2t)e3iω0t) + c.c.,

for the solutions of the Sine-Gordon equation (25). Derive equations for A1 and A3 such thatRes(u) = O(ε5).

b) Make an ansatz for obtaining Res(u) = O(ε7) and derive equations for amplitude functionsappearing in the ansatz.

c) Make a guess for an ansatz for obtaining Res(u) = O(ε2n+1) for a given n ∈ N and a guessfor the equations for the amplitude functions appearing in the ansatz.

53

6.4 Justification of the Nonlinear Schrodinger equation

By the formal derivation of the Nonlinear Schrodinger equation for long wave solutions of theSine-Gordon equation not only spatially localized time-periodic solutions of the Sine-Gordonequation are identified. The complete dynamics known for the Nonlinear Schrodinger equationcan be expected to be found approximately in the Sine-Gordon equation, too.

However, the derivation of the Nonlinear Schrodinger equation for the Sine-Gordon equationis not implying at all that solutions found in the Nonlinear Schrodinger equation can be foundin the Sine-Gordon equation, too. In this subsection we discuss as to how well solutions ofthe Sine-Gordon equation can be approximated via the solutions of the Nonlinear Schrodingerequation.

So let A ∈ C([0, T0], HsA(R,C)) be a solution of the Nonlinear Schrodinger equation

2iω0∂TA = ∂2XA+ A|A|2/2 (28)

with sA ≥ 1 defined below. Then

εψc(x, t) = εA(εx, ε2t)eiω0t + c.c., (29)

defines a formal approximation of the solutions u of the Sine-Gordon equation (25). For ourpurposes it turns out to be advantageous to consider the extended approximation

εψ(x, t) = εA(εx, ε2t)eiω0t + ε3A3(εx, ε2t)e3iω0t + c.c. (30)

from Exercise 6.10 where A3 satisfies A3 ∈ C([0, T0], HsA(R,C)), too, if sA ≥ 1. Therefore,

there exist C, ε0 > 0 such that for all ε ∈ (0, ε0)

supt∈[0,T0/ε2]

‖εψ(·, t) − εψc(·, t)‖Cmb≤ C sup

t∈[0,T0/ε2]

‖ε3A3(ε·), ε2t)e3iω0t + c.c.‖Cmb

≤ Cε3 supt∈[0,T0/ε2]

‖A3(ε·, ε2t)‖Cmb≤ Cε3 sup

T∈[0,T0]

‖A3(·, T )‖Cmb

≤ Cε3 supT∈[0,T0]

‖A3(·, T )‖HsA

if sA > m− 1/2 due to Sobolev’s embedding theorem. As a consequence, if u can be approxi-mated by εψ up to an error of order O(εβ) then it can also be approximated up to an error oforder O(εmin(3,β)) by ψc, in detail

‖u− εψc‖Cmb≤ ‖u− εψ‖Cm

b+ ‖εψ − εψc‖Cm

b≤ Cεβ + Cε3 ≤ 2Cεmin(3,β).

In order to estimate the difference ε3R = u− εψ we derive an equation for R and then estimateR with the help of Gronwall’s inequality. In order to do so we need estimates for the residualRes(εψ). There exist C, ε0 > 0 such that for all ε ∈ (0, ε0)

supt∈[0,T0/ε2]

‖Res(εψ(t))‖Cmb≤ Cε5.

54

See Exercise 6.13. Since the space Cmb is not suitable for our purposes we need the estimate

supt∈[0,T0/ε2]

‖Res(εψ(t))‖Hs ≤ Cε9/2.

See Exercise 6.14.

For the difference ε5/2R = u− εψ we find

∂2t (εψ + ε5/2R) = ∂2

x(εψ + ε5/2R) − sin(εψ + ε5/2R)

= ∂2x(εψ + ε5/2R) − sin(εψ + ε5/2R) + sin(εψ) − sin(εψ)

such that R satisfies∂2tR = ∂2

xR− R + f

withf = −ε−5/2(sin(εψ + ε5/2R) − sin(εψ) − ε5/2R + Res(εψ)).

satisfying the estimate

‖f‖Hs ≤ C1ε2‖R‖Hs + C2(CR)ε7/2‖R‖2

Hs + C3ε2 (31)

as long as ‖R(t)‖Hs ≤ CR with a constant CR determined below, constants C1, C3 independentof CR and ε ∈ (0, 1) and a constant C2 depending on CR but independent of ε ∈ (0, 1). Theequation for R is solved with zero initial conditions.

We use energy estimates and define the energy

E(R) =

s∑

j=0

∫

(∂t∂jxR)2 + (∂j+1

x R)2 + (∂jxR)2dx.

We find for j = 0 and∫

=∫ ∞−∞ that

1

2∂t

∫

(∂tR)2 + (∂xR)2 +R2dx

=

∫

(∂tR)(∂2tR) + (∂xR)(∂t∂xR) +R(∂tR)dx

=

∫

[(∂tR)(∂2xR) − (∂tR)R + (∂tR)f + (∂xR)(∂t∂xR) +R(∂tR)]dx

=

∫

(∂tR)fdx

which can be estimated with the Cauchy Schwarz inequality by

|∫

(∂tR)fdx| ≤ ‖∂tR‖L2‖f‖L2

≤ ‖∂tR‖L2(C1ε2‖R‖Hs + C2(CR)ε7/2‖R‖2

Hs + C3ε2)

≤ C1ε2E(R) + C2ε

7/2E(R)3/2 + C3ε2E(R)1/2

≤ (C1 + C3)ε2E(R) + C2(CR)ε7/2E(R)3/2 + C3ε

2.

55

Since exactly the same estimates hold for j = 1, . . . , s we finally find

∂tE(R) ≤ (C1 + C3)ε2E(R) + C2(CR)ε7/2E(R)3/2 + C3ε

2.

If ε3/2C2E1/2(R)) ≤ 1, then by Gronwall’s inequality

E(R) ≤ C3e(C1+C3+1)T0 =: C2

R (32)

Choosing ε0 > 0 so small thatε3/20 C2(CR)C

1/2R ≤ 1 (33)

we are done. For clarity, to a given CR = CR(T0, CΛ, C1, C3) defined in (32) we have a C2 by(31) and to that C2 we have an ε0 > 0 by (33). Hence, the approximation property (APP) holdsand so there are solutions u of (25) which behave for all t ∈ [0, T0/ε

2] as predicted by the NLSequation (28).Thus, we proved

Theorem 6.11 Fix sA ≥ s+ 3 ≥ 4. Let A ∈ C([0, T0], HsA) be a solution of the NLS equation

(28). Then there exist solutions u of (25) such that there exist C, ε0 > 0, such that for allε ∈ (0, ε0) we have

supt∈[0,T0/ε2]

‖R(t)‖Hs ≤ C. (34)

Remark 6.12 The result can also be proved with classical semigroup theory using the variationof constant formula, cf. [20].

Exercise 6.13 a) Make an ansatz

εψ(x, t) = εA1(εx, ε2t)eiω0t + ε3A3(εx, ε

2t)e3iω0t + c.c.,

for the solutions of ∂2t u = ∂2

xu−u+u3/6. Derive equations forA1 andA3 such that Res(εψ) =

O(ε5).

b) Compute Res(εψ) explicitely.

c) Assume A1, A3 ∈ C([0, T0], HsA). Estimate then supt∈[0,T0/ε2] ‖Res(εψ)‖Cm

b. Derive condi-

tions for m and sA.

Exercise 6.14 a) Estimate x 7→ A(εx) in the Lp-norm for 1 ≤ p ≤ ∞.

b) Estimate x 7→ A(εx) in the Wm,p-norm for 1 ≤ p ≤ ∞ and m ≥ 0.

c) How does the estimate in Exercise 6.13 change if Cmb is replaced by Wm,p.

56

6.5 Breathers

As already said, breathers are time-periodic solutions which are spatially localized, i.e. decay tozero for |x| → ∞. We found approximately such solutions with the help of the NLS-equationup to a time-scale of order O(1/ε2),The question occurs if these solutions are really moving breathers, i.e. exist for all t ∈ R.

For the sine-Gordon equation∂2t u = ∂2

xu− sin(u)

such solutions are explicitely known. They are given by

u(x, t) = 4 arctanδ

ω

sinωt

cosh δx(35)

where ε, ω > 0 with δ2 + ω2 = 1. See Exercise 6.15. In Exercise 6.16 it is shown that alsomoving breathers exist for all t ∈ R.

Hence the next question occurs. Can such solutions also exist in other types of nonlinear waveequations

∂2t u = ∂2

xu− u+ g(u)

on the infinite line x ∈ R, where g : R → R is a smooth, odd function which satisfies g(u) =

O(u3) and g′′′(0) > 0. It turns out that for g(u) close to sin(u) the Sine-Gordon equation is theonly such equation. For a precise statement see [9]. Here we will only explain why this is thecase.

Breathers are homoclinic solutions of the evolutionary system

∂2xu = ∂2

t u+ sin(u),

i.e.lim

x→±∞u(x, t) = 0,

where x ∈ R is the time-like variable and u(x, t) = u(x, t + 2π/ω). Hence, they have tobe in the intersection of the stable and unstable manifold of the origin. Therefore, we lookat the linearization around the fixed point u = 0 in order to compute the dimensions of thesemanifolds. The linearization is given by

∂2xu = ∂2

t u+ u.

Since we are interested in time periodic solutions we use Fourier series

u(x, t) =∑

m∈Z

um(x)eimωt

and find∂2xum = −ω2m2um + um

which is solved by um(x) = eλmxum(0) where λ2m,± = (1 − ω2m2). In the exact formula

(35) for the breathers for small amplitudes, i.e. δ > 0 small, we have 1 − ω2 = δ2. By the

57

approximation through the NLS-equation we find that only odd ms play a role. Thus, we willrestrict to odd m ∈ Z. Therefore, the eigenvalues λm are on the imaginary axis for |m| > 3.The eigenvalues λ±1 are on the real axis. Hence we have a two-dimensional stable and a two-dimensional unstable manifold. These manifolds intersect for the Sine-Gordon equation, butin general two two-dimensional manifolds will not intersect in an infinite-dimensional phasespace. This makes the Sine-Gordon equation exceptional in this class of equations.

We will close section with a comparison of the construction of such solutions for the NLSequation and the Sine-Gordon equation. For the NLS equation the time periodic solutionssatisfy

∂2XA = 2i∂TA− A|A|2/2.

The time periodic solutions form an invariant subspace, i.e.

A(X, T ) = B(X −X0)e−iγ0T eiφ (36)

shows that the real-valued function B satisfies the second-order ordinary differential equation

∂2XB = 2γ0B − B3/2.

The spatially localized solutions are in the intersection of the one-dimensional stable and one-dimensional unstable manifold. The linearization

∂2XB = 2γ0B

possesses solutionsB(X) = eµX where µ2 = 2γ0, i.e. we have eigenvalues on the real axis andhomoclinic solutions for γ0 > 0. These solutions are formally related to the breathers of theSine-Gordon equation through

utp = εBhom(εx)ei(1−γ0ε2)t + c.c..

Hence the eigenvalues λ±1 and the frequency ω of the Sine-Gordon equation correspond tothe eigenvalues µ± and the frequency γ0 of the NLS equation through λ1,+ = εµ+ and ω =√

1 − δ2 = 1− δ2/2+O(δ4) = 1−γ0ε2. The eigenvalues for the NLS equation are also double

if the symmetry eiφ in (36) is taken into account.

Exercise 6.15 Verify that for arbitrary x0, t0, and λ with 0 < |λ| < 1

u(x, t) = 4 arctan

(

(1 − λ2)1/2

λ

sin(λ(t− t0))

cosh((1 − λ2)1/2)(x− x0)

)

is a solution of the Sine-Gordon equation ∂2t u = ∂2

xu − sin(u). Sketch the evolution of thesolution. (Explicit formula for the breather)

Exercise 6.16 Show that the Sine-Gordon equation ∂2t u = ∂2

xu − sin u is invariant under thetransformation

x = γ(x− vt), t = γ(t− vx)

where γ = 1/(1 − v2)1/2 and |v| < 1. Hence conclude that

u(x, t) = 4 arctan

(

(1 − λ2)1/2

λ

sin(γλ(t− vx− t0))

cosh(γ(1 − λ2)1/2)(x− vt− x0)

)

solves the Sine-Gordon equation. Sketch the evolution of the solution.

58

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