Nonlinear Waves Lecture Notes - School of … · Nonlinear Waves Lecture Notes (G14 NWA/2) ... G14 NWA/2 Nonlinear Waves 3 1 Linear Wave Theory & Other Background 1.1 Linear wave

Nonlinear Waves Lecture Notes(G14 NWA/2)

Dr R H TewDivision of Applied Mathematics, School of Mathematical Sciences,University of Nottingham, University Park, Nottingham, NG7 2RD.

Autumn Semester 2005

Contents

1 Linear Wave Theory & Other Background 41.1 Linear wave equations . . . . . . . . . . . . . 41.2 Other equations . . . . . . . . . . . . . . . . . 41.3 Dispersion relation . . . . . . . . . . . . . . . 51.4 Group velocity . . . . . . . . . . . . . . . . . 51.5 Lagrangian mechanics . . . . . . . . . . . . . 6

1.5.1 The Euler-Lagrange equations . . . . 61.5.2 Hamiltonian mechanics . . . . . . . . 7

1.6 Nonlinearity . . . . . . . . . . . . . . . . . . . 8

2 Integrable Systems 92.1 The solitary wave solution . . . . . . . . . . . 92.2 The classic integrable systems . . . . . . . . . 10

2.2.1 KdV . . . . . . . . . . . . . . . . . . . 102.2.2 NLS . . . . . . . . . . . . . . . . . . . 102.2.3 NLS – dispersion relations . . . . . . . 102.2.4 NLS – special solutions . . . . . . . . 112.2.5 SG . . . . . . . . . . . . . . . . . . . . 122.2.6 SG – Kink solution . . . . . . . . . . . 132.2.7 SG – Breather solution . . . . . . . . 142.2.8 SG – 2-soliton solution . . . . . . . . . 142.2.9 Others . . . . . . . . . . . . . . . . . . 15

2.3 2-soliton solution - Backlund Transformation 152.3.1 KdV - History . . . . . . . . . . . . . 152.3.2 Example: (nothing→soliton) . . . . . 162.3.3 Example: (1-soliton→2-soliton) . . . . 172.3.4 Backlund Transformation for SG . . . 172.3.5 Application of BT for SG . . . . . . . 18

2.4 Lax pairs . . . . . . . . . . . . . . . . . . . . 182.4.1 Lax pair for KdV . . . . . . . . . . . . 192.4.2 NLS . . . . . . . . . . . . . . . . . . . 20

2.5 Conserved quantities . . . . . . . . . . . . . . 202.5.1 KdV - motivation . . . . . . . . . . . 202.5.2 KdV-theory . . . . . . . . . . . . . . . 212.5.3 NLS . . . . . . . . . . . . . . . . . . . 21

3 Methods – for nonintegrable systems 223.1 Easy example (NLS) . . . . . . . . . . . . . . 223.2 Harder example – collective variables (NLS) . 223.3 Forcing and damping in SG . . . . . . . . . . 24

3.3.1 Ground States . . . . . . . . . . . . . 243.3.2 Kink solution . . . . . . . . . . . . . . 25

3.3.3 Behaviour at intermediate times . . . 263.3.4 Summary – Power Balancing . . . . . 26

3.4 Nonsymmetric Klein-Gordon breather . . . . 263.5 Numerical Solution of Equations . . . . . . . 28

4 Applications 294.1 Electrical Transmission Lines . . . . . . . . . 29

4.1.1 Aside . . . . . . . . . . . . . . . . . . 304.1.2 Onto KdV . . . . . . . . . . . . . . . . 304.1.3 Regularising long wave equations . . . 314.1.4 Comparison of solitary wave solutions 33

4.2 Fibre-optic Cables . . . . . . . . . . . . . . . 344.3 Water Waves . . . . . . . . . . . . . . . . . . 35

4.3.1 Linear Theory . . . . . . . . . . . . . 364.3.2 Nonlinear Theory . . . . . . . . . . . . 374.3.3 Intepretation of Results . . . . . . . . 39

4.4 Long Josephson junctions . . . . . . . . . . . 394.4.1 Current–Voltage Characteristics . . . 41

4.5 Waves in nonlinear diffusion problems . . . . 414.5.1 Fisher-KPP equation . . . . . . . . . . 444.5.2 Multiple waves . . . . . . . . . . . . . 44

Format of Exam:G14 NW2 (15 credits): one 21

2-hour paper, attempt

3 out of 4 questionsG14 NWA (20 credits): one 3-hour paper, attempt4 out of 5 questions

Syllabus for G14 NW2 (useful for Revision)(i) Linear wave theory & background(ii) Integrable systems: 1-soliton solution, hierarchiesof conserved quantities, Lax pair, 2-soliton solution,envelope solutions (bright and dark), kink solutions,breather solutions. Examples: KdV, NLS & SG(iii) Methods for nonintegrable systems: perturbationtheory, large-time asymptotics, stability of solitons, power-balancing, numerical solution of equations(iv) Applications: electrical transmission lines, fibre-optic cables, water waves, Josephson junctions, nonlin-ear diffusion equations

1

G14 NWA/2 Nonlinear Waves 2

Module Philosophy Lesson 1

Lesson 1

Ideally the scientific process goes:

(1) physical process/scientific experiment

(2) set of mathematical equations

(3) solution of equations

(4) interpret results

However, we are mathematicians and probably most comfortable with the stage (2) → (3).

This module will also try to cover some of the processes involved in stages (1) → (2) and (3) → (4).

This module is designed to be fairly self-sufficient, no knowledge of PDEs, functional analysis, electric cir-

cuit theory, fibre-optic cables, superconductors, water waves, perturbation theory, etc., will be assumed,

(although I hope you will pick some up by the end of the module).

Lesson 2

There is a nice neat theory of constant coefficient linear ODEs–[hopefully you have met this]–solutions have

the form C0veλt. You may also have met a nice neat theory of second order linear PDEs classifying them

as elliptic, parabolic or hyperbolic.

But a nice neat theory of nonlinear PDEs does not exist. Instead this module shows some typical

behaviours which may be observed, and ways of analysing them. There are other more exotic solutions

out there being discovered all the time–this is an area of current active research.

[Not everything fits into neat categories, you set up a set of boxes and then soon find something which straddles two boxes,

or you have to keep creating new categories].

[If you are doing the 20 credit option (C) and there are things which interest you and you want to know more about, you

can do it as a project - come and talk and I’ll point you in the right direction to get started.]


1 Linear Wave Theory & Other Background

1.1 Linear wave equations

The simplest wave equation

ut + cux = 0

has general solution u = f(x− ct) where f(·) is determined by initial conditions.

utt − c2uxx = 0

is usually referred to as ‘the wave equation’ has general solution u(x, t) = f(x− ct) + g(x+ ct)

–a linear superposition of two waves, one right-propagating and the other left-propagating.

An alternative solution method uses separation of variables, and seeks a solution in the form u(x, t) =

X(x)T (t). This leads to X ′′/X = T ′′/c2T = −k2, a constant; since X ′′/X is independent of t and T ′′/c2T

is independent of x, k must be independent of x and t. Thus we have the solution

X(x) =cos

sin(kx), eikx T (t) =

cos

sin(kct), eikct. (1.1)

These are standing waves, and can be shown to be the superposition of equally large waves travelling

in opposite directions

cos(kx) cos(kct) = 12[cos(k(x+ ct)) + cos(k(x− ct))]

sin(kx) sin(kct) = 12[cos(k(x− ct))− cos(k(x+ ct))] (1.2)

cos(kx) sin(kct) = 12[sin(k(x+ ct))− sin(k(x− ct))]

sin(kx) cos(kct) = 12[sin(k(x+ ct)) + sin(k(x− ct))].

General linear superposition then leads to the general solution

u(x, t) =

∫

a(k)eikx+iωtdk FT-ans (1.3)

with ω = ω(k) = kc (cf. Fourier Transform method), and a(k) is an arbitrary function.

1.2 Other equations

Other equations which support wave-type solutions include

ut + vux + εuxxx = 0, lin-kdv (1.4)


if we try u(x, t) = f(x− ct) = f(z) we find −cf + vf + εf ′′ = K;

in the case K = 0 this has a solution of the form f = B0 cos(z√

(v − c)/ε) +B1 sin(z√

(v − c)/ε)

or f = Aeikz with k2 = (v − c)/ε, or u = Aeikx−ikct with c = v − εk2.

An alternative approach is to try a solution of the form (1.3) then (1.4) becomes

∫

a(k)eikx+iωt[

iω + vik − ik3]

dk = 0 (1.5)

thus for the equation to be satisfied, we need ω = k3 − vk.

1.3 Dispersion relation

A standard technique to investigate the possibility of wave propagation in a system is to assume a solution

of the form u = eikx+iωt and see how the temporal frequency ω depends on the spatial wave number k.

Such a relation ω = ω(k) is called the dispersion relation.

If ω is real then this is the frequency which the spatial mode eikx oscillates at.

If ω = −iγ with γ > 0 then the solution has the form u = eikxeγt and the mode eikx is unstable, its

amplitude growing exponentially.

If ω = +iγ with γ > 0 then the solution has the form u = eikxe−γt and the mode eikx is damped out in

the system, its amplitude decaying exponentially.

Example

For the telegraph equation utt = uxx + βu, the dispersion relation is ω2 = k2 − β. However the interpre-

tation of this is the important part.

If β < 0 then waves exist for all k and waves with higher wavenumbers having higher frequencies. But

if β > 0 then waves with k <√β have real frequency, but when k >

√β, ω is purely imaginary and so

u = eikx+γt with γ = ±iω, so there is a wave whose amplitude grows exponentially–an unstable situation.

Such equations are usually termed ‘ill-posed’, although solutions with no high frequency component can

still be found.

1.4 Group velocity

In several of the examples quoted above, we have seen that the quantity ω/k gives the speed of translation

of the wave - this is known as the phase velocity.

Another important concept in more complicated waves is that of group velocity.


Suppose we have two waves ua, ub with similar wavenumbers k = k0 ± k1 and similar frequencies ω =

ω0 ± ω1 with k1, ω1 both small since ω = ω(k) is smooth. Then

ua = eikax−iωat = ei(k0+k1)x−i(ω0+ω1)t travelling at speed ca =ωa

ka=

(ω0 + ω1)

(k0 + k1)

ub = eikbx−iωbt = ei(k0−k1)x−i(ω0−ω1)t travelling at speed cb =ωb

kb=

(ω0 − ω1)

(k0 − k1)(1.6)

then the combination u1 + u2 can be written

u1 + u2 = eik0x−iω0t(eik1x−iω1t + e−ik1x+iω1t) (1.7)

so we see a wave like ua, ub travelling at speed c = ω0/k0 modulated with a wave of small wavenumber

(long wave length k1) which is travelling at speed

vgroup =ω1

k1=

ωa − ωb

ka − kb=

dω

dk. (1.8)

This is known as the group velocity. c = ω/k is known as the phase velocity.

1.5 Lagrangian mechanics

In many examples we shall make use of the Hamiltonian/Lagrangian formulation of the equations of

motion. If {qk}Nk=1 are the coordinate variables, and the kinetic and potential energies are given by

T = T ({ Úqk}) =N∑

k=1

12Úq2k (typically), V = V ({qk}) (1.9)

then the Lagrangian is given by L( Úqk, qk) = T − V and the equations of motion are derived by applying:

1.5.1 The Euler-Lagrange equations

The Action is calculated by integrating the Lagrangian over time,

S[q] =

∫ t1

t0

L(q(t), Úq(t)) dt. (1.10)

This equations of motion can then be found by use of a variational principle, namely by setting δSδq

= 0,

determining critical points of S[q].

Formally we wish to construct the first two terms in a Taylor series for S[q]

S[q + εr] = S[q] + εrS ′[q] +O(ε2), (1.11)


where S ′[q] = δSδq

. We can form this by

S[q + εr] =

∫ t1

t0

L(q + εr, Úq + ε Úr) dt (1.12)

=

∫ t1

t0

L(q, Úq) + εr∂L

∂q+ ε Úr

∂L

∂ Úqdt+O(ε2) dt

=

∫ t1

t0

L(q, Úq) dt+ ε

∫ t1

t0

r∂L

∂qdt+ ε

[

r∂L

∂ Úq

]t1

t0

− ε

∫ t1

t0

rd

dt

(

∂L

∂ Úq

)

dt+O(ε2)

= S[q] + ε

∫ t1

t0

r

(

∂L

∂q− d

dt

(

∂L

∂ Úq

))

dt+O(ε2)

assuming[

r ∂L∂ Úq

]t1

t0= 0; then

δS

δqr = S ′[q]r =

∫ t1

t0

(

∂L

∂q− d

dt

(

∂L

∂ Úq

))

r(t) dt.

By assuming S ′[q]r = 0 for all possible choices of r(t), we find the Euler-Lagrange equations ∂L∂q

−ddt

(

∂L∂ Úq

)

= 0.

The variational derivative δSδq

is also known as a Frechet derivative.

In systems where there are many independent variables, the above result generalises to

d

dt

(

∂L

∂ Úqk

)

=∂L

∂qk, k = 1, 2, . . . , N. (1.13)

Ex: Show that for L = L(q, Úq, q), the Euler-Lagrange equation is

∂L

∂q− d

dt

∂L

∂ Úq+

d2

dt2∂L

∂q= 0.

1.5.2 Hamiltonian mechanics

In the Hamiltonian formulation generalised momenta pk are used in place of Úqk, these are determined by

pk =∂L

∂qk. (1.14)

The Hamiltonian is then defined by

H =N∑

k=1

pk Úqk − L =N∑

k=1

Úqk∂L

∂qk− L (1.15)

(this is usually equal to the total energy in the system). The Equations of motion can then be found by

Úqk =∂H

∂pk, Úpk = −∂H

∂qk. (1.16)


For continuous systems, all quantities are replaced by densities, for example the Lagrangian density is

L = L(qt(x, t), q(x, t)) = T (qt(x, t))− V(q(x, t)) = 12q2t − 1

2q2 (for example) (1.17)

then the Lagrangian/Hamiltonian is the integral of the corresponding density over all space,

L =

∫

L dx, H =

∫

H dx. (1.18)

The equations of motion (which are now PDEs) are then

qt(x, t) =δH

δp(x, t), pt(x, t) =

δH

δq(x, t)=

∂H

∂q− ∂

∂x

∂H

∂qx+

∂2

∂x2

∂H

∂qxx− . . . (1.19)

[see Goldstein for more details].

This formalism has been generalised to systems in which there is only one time derivative by

∂u

∂t=

∂

∂x

(

δHδu

)

. (1.20)

For example KdV can be generated by

H = u3 + 12u2x. (1.21)

Alternatively, one can introduce a new variable ψ such that ψx = u and then

L = −12ψtψx − ψ3

x +12ψ2xx (1.22)

produces ut + 6uux + uxxx = 0.

1.6 Nonlinearity

Burgers equation

ut + uux = 0 burger (1.23)

is among the simplest nonlinear PDEs and can be solved exactly. By a strange quirk the implicit form

u = f(x− ut) solves (1.23). Its solutions get progressively steeper where ux < 0

-x

increasing time

°°°°°°°°°¡¡¡¡¡¡ A

AAAAAPPPPPPPPP

-

¸¸¸¸¸¸¸¸¸¸¸¸

B

BBBBBBHHHHHH

-

¸¸¸¸¸¸¸¸¸¸¸¸¨¨¨¨¨¨¨¨¨¨¨¨

@@

@

∂u

∂x= f ′(x− ut)

[

1− t∂u

∂x

]

⇒ ∂u

∂x=

f ′(x− ut)

1 + tf ′(x− ut)(1.24)

and so the wave ‘breaks’ (reaches infinite gradient) after a time t = minθ∈R 1/f ′(θ).


2 Integrable Systems

We have seen that linear waves often suffer from dispersion, so spread out, and that nonlinearity can

cause waves to steepen, so in systems which have both effects, there might be waves of permanent form

in which these two effects exactly balance.

This is indeed the case, and in some systems such waves pass through each other without suffering any

permanent change in form. These systems are called integrable. Integrable systems are in some sense

solvable. We would like to be able to find the solution u(x, t) of a problem, given a PDE, for an arbitrary

set of initial conditions u(x, 0) and some boundary data. For integrable systems there is a theory which

enables this to be carried out (the calculations are often hideous but technically possible).

2.1 The solitary wave solution

Simply look for a travelling wave: assume there is a solution of the form u(z) = u(x − ct) for some

constant speed c (as yet undetermined), converting the PDE into an ODE which is easier to solve

Eg: the Boussinesq equation utt = uxx + (u2)xx + uxxxx becomes

c2u′′(z) = u′′ + (u2)′′ + u′′′′ (2.1)

integrate twice

(c2 − 1)u = u2 + u′′ −K −Mz (2.2)

(assume M = 0 for simplicity),

×u′ and integrate again

(c2 − 1)uu′ = u2u′ + u′′u′ −Ku′

12(c2 − 1)u2 = 1

3u3 + 1

2(u′)2 −Ku− L (2.3)

du

dz= ±

√

2L+ 2Ku+ (c2−1)u2 − 23u3 ⇒ z − z0 = ±

∫

du√

2L+ 2Ku+ (c2−1)u2 − 23u3

.

If u, u′ → 0 as z → ±∞, then L = 0 = K and u = 32(c2−1)sech2(1

2

√c2−1 z)

Defn: a solitary wave is a localised wave that propagates along one spatial direction only, with unde-

formed shape.

Defn: a soliton is a stable solitary wave whose speed and shape are not altered by collisions with other

solitary waves.

Ex: find a solitary wave solution of the modified Boussinesq equation utt = uxx + uxxxx + (u3)xx.

Ex: find a solitary wave solution of the Korteweg-deVries equation ut = 6uux − uxxx.


2.2 The classic integrable systems

2.2.1 KdV

Probably the most famous nonlinear wave equation is the Korteweg-deVries equation (KdV) derived by

two Dutchmen in the last century. It is often written as

ut − 6uux + uxxx = 0 (2.4)

although sometimes it is more convenient to consider it in the form (u = u+ 16)

ut − ux − 6uux + uxxx = 0, (2.5)

which we can think of as a wave equation (ut − ux = 0) at leading order, with two perturbation terms.

The uxxx is a dispersion term which causes waves to spread out, and a nonlinearity uux which has a

counterbalancing effect of causing waves to steepen.

2.2.2 NLS

The Schrodinger equation arises in quantum mechanics, where it usually written in the form

ih∂ψ

∂t= h2∇2ψ + V (·)ψ. (2.6)

In simpler cases V = V (x, y, z) and the equation is thus linear in ψ; however there are cases where

V = V (ψ) and the solution of the equation becomes more complicated. We shall study the simplest of

these nonlinear cases, namely for ψ ∈ C

i∂ψ

∂t= D

∂2ψ

∂x2+Q|ψ|2ψ, (2.7)

where we have removed h by rescaling time.

2.2.3 NLS – dispersion relations

Often a useful indication of the types of solution an equation can support is given by the dispersion

relation. In this case assuming

ψ = A0eikx+iωt

NLS-assump1 (2.8)

with A0 → 0 yields the dispersion relation

ω = Dk2 (2.9)


so that waves with high wavenumber also have high temporal freqency. However, with a little more

thought, we realise that (2.8) is an exact solution for NLS for any A0, not just in the limit A0 → 0.

Hence

ω = Dk2 −QA20 (2.10)

so that if Q > 0 the nonlinearity causes the frequency to be lowered and if Q < 0 the nonlinearity causes

the frequency to be increased. However, such solutions may not be stable, as we shall see later.

2.2.4 NLS – special solutions

We now show that nonlinear effects can produce solutions which aren’t simply sinusoidal oscillations.

Let us seek a solution of the form ψ = A(x, t)eikx+iωt with k, ω,A ∈ R then

iAt − ωA = DAxx + 2ikDAx − k2DA+QA3 (2.11)

The imaginary part implies At = 2kDAx, so the envelope, A(x, t), is a travelling wave A = A(x+2kDt) =

A(z), and the real part implies

A′′(z) = (k2 − ωD)A− Q

DA3 ⇒ (A′)2 = P + (k2 − ω

D)A2 − Q

2DA4

nls-env (2.12)

The rhs is a quadratic in A2.

Case B: If Q/D > 0 and

if P is large enough for there to be a range of A2 where (A′)2 is also positive, then the equation is of the

form (A′)2 = Q2D

(A2 − A20)(A

21 − A2).

Large −ve P

-

6

A2

(A′)2

JJ

JJJ

ªª

ªªª

irrelevant

Small −ve P

-

6

A2

(A′)2

JJ

JJJ

ªª

ªªªA2

0 A21

periodicelliptic fn

A0<A<A1 oror−A1<A<−A0

P = 0⇒ A0 = 0

-

6

A2

(A′)2

JJ

JJJ

ªª

ªªª

0=A20

A21

solvable byelementary fnsdone below

Large +ve P

-

6

A2

(A′)2J

JJ

JJ

ªª

ªªª A2

1

elliptic fn−A1<A<A1

Example: if P = 0, then

A =

√

2D

Q

(

k2 − ω

D

)

sech

(

z

√

k2 − ω

D

)

, (2.13)

hence we require ω < Dk2

ψ =

√

2D

Q

(

k2 − ω

D

)

sech

(

(x+ 2kDt)

√

k2 − ω

D

)

eikx+iωt. nls-env-sol (2.14)


This solution is known as the bright soliton. Stability ?

Case D: However, if Q/D < 0 then (2.12) has the form (A′)2 = −Q2D

(A2 − A22)(A

2 − A23) (assuming P is

small enough for there to be any real root of the quadratic). In this case the only solution which does

not blow-up is a solution which joins A = +A2 ≤ A3 to A = −A2.

Small P

-

6

A2

(A′)2

ªª

ªªª

JJ

JJJ

A22 A2

3

Irrelevant

Larger P

-

6

A2

(A′)2

ªª

ªªª

JJ

JJJ

A22 A2

3

Elliptic Fns−A2<A<A2

Special P

-

6

A2

(A′)2

ªª

ªªª

JJ

JJJ

A22=A2

3

Solvable−A2<A<A2

Overlylarge P

-

6

A2

(A′)2

ªª

ªª

JJ

JJ

Irrelevant

Example: The case A2 = A3 is solvable, this corresponds to P = −D2Q

(k2 − ωD)2.

±z

√

−Q

2D= ±

∫√

−Q

2Ddz =

∫

dA

A2 − A22

=1

2A2log

(

A2 − A

A2 + A

)

⇒ A = A2 tanh

(

zA2

√

−Q

2D

)

(2.15)

Clearly for this solution we require ω > Dk2, then

⇒ ψ = ±

√

−D

Q

(

ωD− k2

)

eikx+iωt tanh(

(x+ 2kDt)√

ωD− k2

)

. (2.16)

This solution is known as the dark soliton, since its centre is marked by a decrease in the intensity of

the linear wave. Thus NLS has two types of soliton solution depending on the sign of Q/D.

2.2.5 SG

The sine-Gordon equation (SG)

φtt = φxx − sinφ or φXT

= sinφ. 2sg (2.17)

is the most-studied member of the family of Klein-Gordon equations. These are derived from the Hamil-

tonian

H = 12φ2t +

12φ2x + V (φ) ⇒ φtt = φxx − V ′(φ). (2.18)

In the case V = 0 this generates the linear wave equation, if V = 12φ2 we obtain the telegraph equation,

and if V ′(φ) is nonlinear then we obtain a nonlinear PDE.

The SG equation clearly has the trivial solutions φ = nπ. Linearising around φ = 2nπ we find (φ =

2nπ + ψ, ψ = eikx+iωt)

ψtt = ψxx − ψ ⇒ ω2 = k2 + 1 (2.19)


thus all wave numbers have real frequency. The other steady-states (φ = (2n+ 1)π + ψ) imply

ψtt = ψxx + ψ ⇒ ω2 = k2 − 1 (2.20)

so that waves with |k| < 1 have exponential growth. The steady-states φ = (2n+ 1)π are thus unstable

and those with φ = 2nπ are stable.

The SG potential V (φ) = 1− cosφ, has minima at φ = 2nπ which are sometimes referred to as ‘ground-

states’. [One problem which naturally arises is the evolution of a system which asymptotes to one ground

state as x → −∞ and another at x → +∞.]

Another commonly studied example of a Klein-Gordon equation is the ‘φ4’ model where V (φ) = −12φ2+

14φ4, which implies V ′(φ) = −φ+φ3 so that φ = ±1 are the (only) ground states. Unlike SG, this model

is not integrable.

The substitution φ = 4 tan−1 ψ in (2.17) eliminates the trig function in favour of polynomial nonlinearities1

(1 + ψ2)(ψtt − ψxx) = ψ(2ψ2t − 2ψ2

x − 1 + ψ2). (2.21)

It is possible to find separable solutions of this equation, letting ψ = X(x)T (t) we find

(

1

X2T 2+ 1

)

(

T

T− X ′′

X

)

=2 ÚT 2

T 2− 2(X ′)2

X2+ 1− 1

X2T 2, sg-reduc (2.22)

Although we cannot separate all x’s to one side and all t’s to the other, it is possible to satisfy this

equation. For example, X = eλx and T = eωt (with λ2 = 1 + ω2) [TW];

or T = cos(ωt) and X = αsech(βx) (for some α, β) [BR];

or X = cosh(λx) and T = αcsch(βt) [K-AK].

But note that since (2.22) is nonlinear, these solutions cannot be linearly combined to produce further

solutions. We shall examine each in turn.

2.2.6 SG – Kink solution

ψ = eλx±t√λ2−1 corresponds to a travelling wave since it is a function of the moving coordinate z = x− ct

with c = ± 1λ

√λ2 − 1.

1

ºº

ºº

ºº

ºº

1

ψ√

1 + ψ2

θ

θ = tan−1 ψ

cos tan−1 ψ = 1√1+ψ2

sin tan−1 ψ = ψ√1+ψ2

φt =4ψt

1+ψ2

φtt =4

(1+ψ2)2 [ψtt + ψ2ψtt − 2ψψ2t ]

sin(4 tan−1 ψ) = 4 sin(tan−1 ψ) cos(tan−1 ψ)[cos2(tan−1 ψ)− sin2(tan−1 ψ)]


Alternatively, we can seeking a travelling wave solution by assuming φ(x, t) = φ(x− ct) = φ(z), then

c2φ′′ = φ′′ − sinφ ⇒ 12(1− c2)(φ′)2 = K − cosφ. (2.23)

Assuming φ, φ′ → 0 as z → −∞ implies K = 1 and φ = 4 tan−1 eγz where γ = 1/√1− c2.

2.2.7 SG – Breather solution

ψ =√1−ω2

ωcos(ωt)sech(x

√1− ω2) yields

φ = 4 tan−1 (tan(µ) cos(t cosµ)sech(x sinµ)) (2.24)

For small values of µ this looks like a nonlinear standing wave. At larger values of µ this has the form of

a two kinks travelling in opposite directions until they reach a maximum separation at which they they

turn around and collide, passing through each other and forming a pair of negative kinks separating.

SG is invariant under the Lorentz group (cf. special relativity) x 7→ γ(x − ct), t 7→ γ(t − cx), so above

implies that

φ = 4 tan−1

(

tan(µ) cos(γ(t− cx− t0) cosµ)

cosh(γ(x− ct− x0) sinµ)

)

. (2.25)

is also a solution. We note from above that c = 1 is the maximum speed a kink can travel at. This

corresponds to the ‘speed of light’ in special relativity. Transforming to ‘light cone’ coordinates X = x+t

and T = x− t yields an alternative formulation of SG, namely

∂2φ

∂X∂T= sinφ. (2.26)

2.2.8 SG – 2-soliton solution

When X = sin(µ) cosh(x/ cosµ) and T = sinh(t tanµ) we have a kink-anti-kink collision

φ = 4 tan−1

(

sin(µ) cosh(x/ cosµ)

sinh(t tanµ)

)

(2.27)

The kinks’ positions can be described by the x-value where φ = π, giving

sinh(t tanµ) = sinµ cosh(x/ cosµ) (2.28)

As t → ∞ this asymptotes to sin(µ) exp(±x/ cosµ) = exp(t tanµ).

[Show MapleV animations here]


2.2.9 Others

Once a couple of integrable equations had been found, the search was on, starting with generalisations

of KdV, such as the modified KdV

ut + u2ux + uxxx = 0, (2.29)

and a two-dimensional model, the Kadomtsev-Petviashvili equation (KP)

(ut + uux + uxxx)x + uyy = 0, (2.30)

and equations which allow waves to travel in both directions, such as the Boussinesq equation (Bq)

utt + (u2)xx + uxxxx = 0 (2.31)

Many systems require even symmetry, so that if u(x, t) is a solution then so is −u(x, t). Neither KdV or

Bq have this symmetry whereas mKdV, mBq (and SG) do.

Finally, there is even an example of an integrable differential–difference equation – the Toda lattice

un = (a(1− e−bun+1))− 2(a(1− e−bun)) + (a(1− e−bun−1)). (2.32)

A natural question to ask in any of these wave-bearing equations is how do waves interact ? In linear

systems we have a superposition principle whereby two solutions can be combined to produce a third by

simply adding them together, but such a procedure is not valid with nonlinear equations.

2.3 2-soliton solution - Backlund Transformation

[Recall the method of reduction of order in linear ODEs, for a second order ODE, given one solution

u1(x), can generate other by putting u2(x) = q(x)u1(x) and get a simpler problem for q(x).]

To generate more exotic solutions, it is possible to use a Backlund transformation. This typically has

the form of a pair of equations

w1,x = F (w0, w1, w0,x, w0,t;λ) BT-F (2.33)

w1,t = G(w0, w1, w0,x, w0,t;λ) BT-G (2.34)

from which we hope to generate a new solution (w1) from a known solution (w0).

2.3.1 KdV - History

In 1968 Miura showed that if v satisfies the mKdV

vt = 6v2vx − vxxx BT-mKdV (2.35)


then u = v2 + vx solved KdV [ut = 6uux − uxxx].

By considering symmetries of mKdV and KdV we can find more ways of getting from a solution of mKdV

to a solution of KdV.

(i) KdV is invariant under x = x+ 6λt, t = t, u = u− λ:

∂∂t

= ∂∂t+ 6λ ∂

∂x∂∂x

= ∂∂x

}

ut = 6uux − uxxx

ut + 6λux = 6(u+ λ)ux − uxxx

(2.36)

so we shall work with u− λ in place of u

(ii) if v solves mKdV then so does −v, thus u = v2 − vx must also solve KdV.

Thus given a single solution v(x, t) of mKdV we have two families of solutions of KdV:

u0 = λ+ v2 + vx (2.37)

u1 = λ+ v2 − vx

Alternatively written as

u0 − u1 = 2vx (2.38)

u0 + u1 = 2λ+ 2v2

Our aim now is to eliminate v - the dependence on mKdV - so that we can go straight from one solution

of KdV to the next. Thus we shall work with w0, w1 with w0,x = u0, w1,x = u1 so that v = 12(w0 − w1),

then

w0,x + w1,x = 2λ+ 12(w1 − w0)

2 (2.39)

This has the form (2.33) and the corresponding formula (2.34) is found by requiring v = 12(w0 − w1) to

satisfy mKdV (2.35)

w1,t − w0,t = 3w21,x − 3w2

0,x − w1,xxx + w0,xxx. (2.40)

2.3.2 Example: (nothing→soliton)

To show that this method works, let us start with the trivial solution w0 = 0 and see if w1 is more

interesting. We have

w1,x = 2λ+ 12w2

1, w1,t = 3w21,x − w1,xxx (2.41)

The former is the simpler to solve, being an ODE; with λ = −k2 it gives

w1(x, t) = −2k tanh(kx+ f(t)) (2.42)

for some function f(t) to be determined.


The former equation also implies w1,xx = w1w1,x and w1,xxx = w21w1,x + w2

1,x, so the latter equation

becomes

w1,t = 3w21,x − w2

1,x − w1,xw21 = w1,x(−4k2) ⇒ ft = −4k3. (2.43)

Thus w1 = −2k tanh(k(x− 4k2t)) and the new solution is u1 = w1,x = −4k2sech2(k(x− 4k2t)).

2.3.3 Example: (1-soliton→2-soliton)

Let us now use the method to create something new.

w2,x = 2µ+ 12(w2 − w1)

2 − w1,x (2.44)

w2,t = 3w22,x − 3w2

0,x − w2,xxx + w1,xxx + w1,t

And out of Frankenstein’s box comes forth

u=−2k21e

k1x−k31t + k22e

k2x−k32t−k0 + 2(k2 − k1)2ek1x+k2x−k31t−k32t−k0 + Aek1x+k2x−k31t−k32t−k0(k2

2ek1x−k31t + k2

1ek2x−k32t−k0)

(1 + ek1x−k31t + ek2x−k32t−k0 + Aek1x+k2x−k31t−k32t−k0)2

(2.45)

where A = (k2 − k1)2/(k2 + k1)

2, λ = −k21, µ = −k2

2.

2.3.4 Backlund Transformation for SG

Using the ‘light-cone’ formulation of SG (φXT

= sinφ) the formulae

∂θ

∂X=

∂φ

∂X+ 2λ sin(1

2(θ + φ)), (2.46)

∂θ

∂T= − ∂φ

∂T+

2

λsin(1

2(θ − φ)),

imply

θXT = φXT + λ(ΘT + φT ) cos(12(θ + φ))

= φXT + 2 sin(12(θ − φ)) cos(1

2(θ + φ)) (2.47)

θXT = −φXT +1

λ(θX − φX) cos(

12(θ − φ))

= −φXT + 2 sin(12(θ + φ)) cos(1

2(θ − φ)).

These two expressions for θXT are consistent only if φXT = sinφ, and by adding the two expressions we

find θXT = sin θ.


2.3.5 Application of BT for SG

Using φ = 0, we use the above Backlund transformation to generate a new solution for SG, this requires

us to solve

θT = 2λ sin(12θ), θX = 2 sin(1

2θ)/λ, (2.48)

which yields the (already-known) solitary wave

θ = 4 tan−1 eλT+X/λ = 4 tan−1 ex(λ+1/λ)−t()λ−1/λ = 4 tan−1 e(x−ct)/√1−c2 (2.49)

since X = x+ t, T = x− t and λ2 = 1+c1−c

⇔ c = λ2−1λ2+1

.

2.4 Lax pairs

If we have two equations for one unknown, then in general we do not expect a solution to exist. However,

if a solution exists, then the two equations must be compatible – and there is a compatibility condition.

For example

ax+ b = 0

cx+ d = 0

}

⇔ x =−b

a=

−d

c︸︷︷︸

solution

⇔ ad− bc = 0︸︷︷︸

compatibility condition

. (2.50)

Another example: Suppose

Úx = A(t)x+B(t)

Úx = C(t)x+D(t)

}

⇔ x =D(t)−B(t)

A(t)− C(t)laxegsimp (2.51)

by equating the rhss of the two equations. However, this must actually solve the differential equation,

so substituting into Úx = Ax+B yields

(A ÚD −D ÚA) + ( ÚBC −B ÚC) + ( ÚAB − A ÚB) + ( ÚCD − C ÚD) = (A− C)(AD −BC). (2.52)

This is the compatibility condition which (A,B,C,D) must satisfy if the overdetermined linear system

(2.51) is to have a solution. Note that the compatibility condition is nonlinear even though the problem

for x is linear.

With A being a nonlinear operator not involving t-derivatives, constructing a Lax pair typically involves

rewriting the equation

ut = A(u) gen-nlop (2.53)

as

Lt = ML− LM lax (2.54)


where L,M are linear operators in x, for example. Consider the eigenvalue problem for ψ(x, t) defined

by Lψ = λψ, then

λtψ + λψt = Ltψ + Lψt

λtψ = [ML− LM ]ψ + Lψt − λψt

λtψ = MLψ − LMψ + Lψt − λψt (2.55)

λtψ = λMψ + Lψt − LMψ − λψt

λtψ = λ(Mψ − ψt) + L(ψt −Mψ)

so if ψ evolves according to ψt = Mψ then the eigenvalue λ remains constant, so if ψ(x, 0) is an

eigenfunction then so is ψ(x, t) for all t > 0. This is a very special property of the system, and only a few

special systems (2.53) have the property of being expressible as (2.54); the quest for such decompositions

goes on.

2.4.1 Lax pair for KdV

Consider

L = − ∂2

∂x2+ u(x, t) (2.56)

then (details are not really for public consumption, but will illustrate the point)

Lψ = −ψxx + uψ

hence (Lψ)t = −ψxxt + uψt + utψ.

Note that Lψt = −ψxxt + uψt

so Ltψ = utψ ⇒ Lt = ut. (2.57)

Now to choose M : KdV has ∂xxx terms and u∂x terms, so let us try

M = βu∂

∂x+ β

∂

∂xu− α

∂3

∂x3(2.58)

then

Mψ = βuψx + βuψx + βuxψ − αψxxx

MLψ = −βuψxxx + βu2ψx + βuuxψ − βuψxxx − βuxψxx + βu2ψx + 2βuuxψ

+αψxxxxx − αuψxxx − 3αuxψxx − 3αuxxψx − αuxxxψ

LMψ = −βuψxxx − 2βuxψxx − βuxxψx − βuψxxx − 3βuxψxx − 3βuxxψx − βuxxxψ + αxxxxx

+βu2ψx + βu2ψx + βuuxψ − αuψxxx

(ML− LM)ψ = (3α− 4β)uxψxx + (3α− 4β)uxxψx + (αuxxx − βuxxx − 2βuux)ψ (2.59)


So if we choose β = 3 and α = 4, the condition Lt = ML− LM is equivalent to ut = uxxx − 6uux.

The Lax pair enables the nonlinear problem for u to be posed as a sequence of linear problems for ψ,

which can in principle be solved and the solution can be used to determine u(x, t).

2.4.2 NLS

For NLS with Q = 2/(1− p2) and D = 1 (iut + uxx +Q|u|2u)

L = i

(

1 + p 0

0 1− p

)

∂

∂x+

(

0 u∗

u 0

)

(2.60)

A = iM = −p

(

1 0

0 1

)

∂2

∂x2+

(

|u|21+p

iu∗x

−iux−|u|21−p

)

(2.61)

where u∗ is the complex conjugate of u. The above calculation can be reworked with Lt = i(LA− AL)

and iψt = Aψ. (Exercise)

2.5 Conserved quantities

Any equation of the form∂F

∂t+

∂G

∂x= 0 cons-form (2.62)

can be thought of as a conservation law (usually neither F or G involve t-derivatives); F and G are

functions of (x, t, u, ux, uxx, uxxx, . . .). If limx→∞G = limx→−∞G then we can integrate (2.62) over all x

and find a constant of the motion,∫

F dx.

2.5.1 KdV - motivation

[Qu: Can you write KdV in the form (2.62) ?]

Ans1: F1 = u, G1 = uxx − 3u2. [Mass M =∫

u]

Ans2: F2 = u2, G2 = uuxx − 12u2x − 2u3. [Momentum P =

∫

u2]

Ans3: F3 = 2u3 + u2x, G3 = −9u4 + 6u2uxx − 12uu2

x + 2uxuxxx − u2xx. [Energy E =

∫

2u3 + u2x]

Ans4: F4 = 5u4 + 10uu2x + u2

xx [I4 =∫

5u4 + 10uu2x + u2

xx]

Ans5: F5 = 21u5 + 105u2u2x + 21uu2

xx + u2xxx

What do∫

F4 dx and∫

F5 dx correspond to ?

What is going on ?

Is there a systematic way to find Fn ?


2.5.2 KdV-theory

Using a generalisation of the Miura transformation (see (2.35))

u = w + εvx + ε2v2 gen-miura (2.63)

the KdV equation can be written

ut−6uux+uxxx = wt+εwxt+2ε2wwt−6(w+εwx+ε2w2)(wx+εwxx+2ε2wwx)+wxxx+εwxxxx+2ε2(wwx)xx

=

(

1 + ε∂

∂x+ 2ε2w

)

(

wt − 6(w + ε2w2)wx + wxxx

)

(2.64)

So if w solves

0 = wt − 6(w + ε2w2)wx + wxxx (2.65)

then u given by (2.63) solves KdV.

Now this has the form of a conservation equation (2.62) involving the parameter ε

0 = (w)t + (wxx − 3w2 − 2ε2w3)x (2.66)

so∫

w dx is a conserved quantity.

Now we assume that w can be written as an asymptotic series in the parameter ε

w ∼∞∑

n=0

εnwn(x, t) as ε → 0, (2.67)

then each of∫

wn dx must be a conserved quantity. Substituting this expansion into (2.63) we find

u ∼∞∑

n=0

εnwn +∞∑

n=0

εn+1wn,xε2

(

∞∑

n=0

εnwn

)2

∼ w0 + εw1 + εw0,x + ε2w2 + ε2w1,x + ε2w20 + . . . (2.68)

Thus we have w0 = u as the first conserved quantity, w1 = −ux as the second (though this is trivial

since∫

w1 =∫

−ux = u(−∞, t)− u(+∞, t) and so is guaranteed simply by the BCs and tells us nothing

about the system itself); the third is w2 = −w1,x −w20 = uxx − u2 which is commonly interpreted as the

energy.

All odd n’s produce trivial conserved quantities, only the wn’s corresponding to even n’s giving useful

information.

2.5.3 NLS

For NLS, the first few conserved quantities are∫

|ψ|2 dx,∫

i(qq∗x − q∗qx) dx,

∫

|qx|2 − |q|4 dx, (2.69)

which correspond to norm, momentum and energy.


3 Methods – for nonintegrable systems

3.1 Easy example (NLS)

We would like to check whether small additional terms in the NLS significantly affect the nature of

solutions. Suppose we perturb the NLS equation by the addition of a term

iψt = Dψxx +Q|ψ|2ψ + εψ 3p1 (3.1)

then how does ε 6= 0 influence the solution ψ(x, t) ?

Answer: (Trick): ψ(x, t) = h(t)θ(x, t)

(If |h| > 1 then solution is amplified, if |h| < 1 then solution is diminished, etc.)

If we assume |h|2 = 1 then we find

ihtθ

h+ iθt = Dθxx +Q|θ|2θ + εθ (3.2)

so that θ satisfies an unperturbed NLS equation if h = e−iεt. So the temporal frequency of the solution

is altered, while the solution’s amplitude is left unchanged. So (3.1) is integrable.

3.2 Harder example – collective variables (NLS)

Suppose we perturb the NLS equation by the addition of the term −iεψ:

iψt −Dψxx −Q|ψ|2ψ = −iεψ =: εR NLS-gen-hardeg (3.3)

then what happens to the solution ψ(x, t) ?

If ε is small we still expect the solution to have the form of a solution to NLS. For example, if D,Q are

such that there is an envelope solution we expect the perturbation to maintain this, although it may

slightly alter its height, width, speed of travel etc. [Balancing temporal derivative with perturbation

yields ψt = −εψ, so we might expect some sort of decay.]

The method of collective variables enables one to study the solution in terms of these characteristic

parameters; if ε = 0 then these quantities are all constant, but with ε 6= 0, we allow them to vary slowly.

Take the solution of the unpertubed problem (ε = 0) as

ψ = A sech(A(x− p(t)))eik(x−p(t))+iσ(t). nls-ass-sol (3.4)

By comparison with (2.14) we know that in the case D/Q > 0, (3.3) has a solution of the form p(t) = −2kt

and σ = (ω − 2k2)t. We now generate this solution by another technique.


Assuming (3.4) is a solution for some p(t), k(t), σ(t), A(t) we determine these from

L = 12i(ψ∗ψt − ψ∗

tψ)− |ψx|2 + |ψ|4, (3.5)

hence the Lagrangian integral is

L =

∫

L dx = 23A3 − 2Ak2 + 2Ak Úp− 2A Úσ. (3.6)

In the unperturbed problem, the Euler-Lagrange equations reduce to

ddt

∂L∂ Úσ

− ∂L∂σ

= 0 ⇒ ddt(2A) = 0 ⇒ ÚA = 0 ⇒ A = A0

ddt

∂L∂ Úp

− ∂L∂p

= 0 ⇒ ddt(−2Ak) = 0 ⇒ Úk = 0 ⇒ k = k0

∂L∂k

= 0 ⇒ −4Ak + 2A Úp = 0 ⇒ Úp = 2k0 ⇒ p = 2k0t∂L∂A

= 0 ⇒ 2A2 − 2k2 + 2k Úp− 2 Úσ = 0 ⇒ Úσ =A2+k2 ⇒ σ = (A20+k2

0)t.

(3.7)

confirming what has already been obtained.

In the presence of perturbations (ε 6= 0) however, the Euler-Lagrange equations are modified.

Let y denote any one of our ‘collective coordinates’ (A, k, σ, p), then

∂L

∂y=

∫

{

∂L∂ψ

∂ψ

∂y+

∂L∂ψt

∂ψt

∂y+

∂L∂ψx

∂ ∂∂xψ

∂y

}

+

{

∂L∂ψ∗

∂ψ∗

∂y+

∂L∂ψ∗

t

∂ψ∗t

∂y+

∂L∂ψ∗

x

∂ψ∗x

∂y

}

dx

=

∫ {

∂L∂ψ

∂ψ

∂y+

∂L∂ψt

∂ψt

∂y− ∂

∂x

(

∂L∂ψx

)

∂ψ

∂y

}

+ {c.c.} dx (3.8)

and

d

dt

(

∂L

∂ Úy

)

=

∫

∂

∂t

(

∂L∂ψt

∂ψt

∂ Úy

)

+∂

∂t

(

∂L∂ψ∗

t

∂ψ∗t

∂ Úy

)

dx

=

∫

∂

∂t

(

∂L∂ψt

∂ψ

∂y

)

+ c.c. dx (3.9)

=

∫ {

∂

∂t

(

∂L∂ψt

)

∂ψ

∂y+

∂L∂ψt

∂ψt

∂y

}

+ c.c. dx

so∂L

∂y− d

dt

∂L

∂ Úy=

∫ {

∂L∂ψ

− ∂

∂t

∂L∂ψt

− ∂

∂x

∂L∂ψx

}

∂ψ

∂y+ c.c. dx EL-mod (3.10)

Now, the term in curly braces is the c.c. of the equation of motion (3.3)

{} = −iψ∗t + ψ∗

xx + 2|ψ|2ψ∗ = εR∗ = iεψ∗ in our case (3.11)

so that (3.10) becomes∂L

∂y− d

dt

∂L

∂ Úy= ε

∫

R∂ψ∗

∂ydx+ c.c. =: εRy. (3.12)


Using (3.4) we find

∂ψ

∂k= ψi(x−p),

∂ψ

∂σ= iψ,

∂ψ

∂A= ψ[

1

A−(x−p) tanh(A(x−p))], ∂ψ

∂p= ψ[A tanh(A(x−p))−ik],

(3.13)

and thus the integrals are

Rk = i

∫

ψ∗ψk − ψψ∗k dx = i

∫

|ψ|22i(x− p) dx = 0

Rσ = i

∫

ψ∗ψσ − ψψ∗σ dx = i

∫

−2|ψ|2 dx = −4A (3.14)

RA = i

∫

ψ∗ψA − ψψ∗A dx = i

∫

|ψ|2[ 1A− (x− p) tanh()− 1

A+ (x− p) tanh()] dx = 0

Rp = i

∫

ψ∗ψk − ψψ∗k dx = i

∫

|ψ|2[A tanh()− ik − A tanh()− ik] dx = 4kA

Thus we find

∂L∂σ

− ddt

∂L∂ Úσ

= εRσ ⇒ 2 ÚA = −4εA ⇒ ÚA = −2εA ⇒ A = A0e−2εt

∂L∂p

− ddt

∂L∂ Úp

= εRp ⇒ −2 ddt(Ak) = 4εAk ⇒ Úk = 0 ⇒ k = k0

∂L∂k

= εRk ⇒ −4Ak + 2A Úp = 0 ⇒ Úp = 2k ⇒ p = 2k0t∂L∂A

= εRA ⇒ 2A2 − 2k2 + 2k Úp− 2 Úσ = 0 ⇒ Úσ =A2+k2 ⇒ σ =k20t+A

20(1−e−4εt)/4ε

(3.15)

Hence

ψ(x, t) ' A0e−2εteik0(x+2k0t)+ik20t+iA2

0(1−exp(−4εt))/4εsech((x+2k0t)A0e−2εt). (3.16)

Thus amplitude decays exponentially, and the wave spreads out, its width growing exponentially, but

surprisingly, its speed is not reduced nor is its temporal frequency altered.

3.3 Forcing and damping in SG

In the theory of Long Josephson junctions derived later, we obtain the damped and drived sine-Gordon

equation

φtt = φxx − Γ2 sinφ− I − αφt (3.17)

In order to convert this to the more familiar SG-form seen earlier we rescale the independent variables,

ξ = Γx, τ = Γt to arrive at

φττ = φξξ − sinφ− I

Γ2− α

Γφτ . 3-sg-ctm2 (3.18)

3.3.1 Ground States

We can easily see that the ‘ground’ states which any kink connects are given by solutions of

sin φ = − I

Γ2, (3.19)


which has real solutions provided |I| < Γ2. We will define these solutions by φ± = − sin−1(I/Γ2) + 2nπ.

(The ‘+’ subscript refers to the value of φ in the limit z → +∞, and ‘–’ to the z → −∞ limit.) If

|I| ≥ Γ2 then there are no solutions. Hence we have a single kink solution if |I| < Γ2.

3.3.2 Kink solution

The terms in this equation can be placed in 3 categories:

(i) those which appear in SG

(ii) a forcing term I which adds energy to the system

(iii) a damping term involving α. Equation (3.18) with α = I = 0 is a Hamiltonian (and Lagrangian)

system, with

H =

∫

H dx =

∫

12φ2τ +

12φ2ξ + (1− cosφ) dξ

L =

∫

L dx =

∫

12φ2τ − 1

2φ2ξ − (1− cosφ) dξ. (3.20)

We seek an approximate solution to equation (3.18) with I/Γ2, α/Γ small by assuming that the forcing

and damping do not change the shape of the wave at leading order, so that we still have the solution

φ(z) = 4 tan−1(exp(γ(ξ − cτ))), 43aphi (3.21)

for some c and with γ = 1/√1− c2. For this solution

H =8√

1− c2. H-ljj (3.22)

Intuituively, we expect the forcing current (I) to add energy to the solution, increasing the speed of

the kink, and the damping term (α) to remove energy, slowing it down. These two effects may find an

equilibrium point where they balance each other out, and the kink has an equilibrium speed.

In many examples the Hamiltonian, H, is a measure of energy, and so we calculate the change of energy

in the full system (α, I 6= 0) by calculating dHdτ

using the full equation of motion. This gives

dH

dτ= −

∫

I

Γ2φτ +

α

Γφ2τ dξ (3.23)

As already noted energy can be added to the system by the forcing term, and removed via the damping

terms, however in the long run the system will settle down to an equilibrium (i.e. state of constant

energy). The speed of the kink in this equilibrium will depend upon the parameters α, I,Γ and can be

calculated by setting dHdτ

= 0. Substituting our assumed form (3.21) into (3.23) we find

2πIc

Γ2+

8αγc2

Γ= 0 ⇒ ceq =

πI√π2I2 + 16α2Γ2

. 43cdef1 (3.24)


3.3.3 Behaviour at intermediate times

How does kink get to speed ceq ?

The kink speeds up (if c < ceq) or slows down (if c > ceq) towards the correct speed. We can find how, by

letting c depend on t using and then using (3.22) and (3.23). If c depends on t then dH/dτ from (3.22)

will not be zero, ratherdH

dτ=

8c

(1− c2)3/2dc

dτ, (3.25)

and (3.23) impliesdH

dτ=

2πIc

Γ2− 8αγc2

Γ, (3.26)

so thatdc

dτ=

πI(1− c2)3/2

4Γ2− αc(1− c2)

Γ(3.27)

3.3.4 Summary – Power Balancing

Many equations of the form φtt − E(φ) = 0 are force balances derived in some manner from Newton’s

third law. In the absence of friction and forcing they may have wave-type solutions which can travel at

an arbitrary speed (in our example any −1 < c < 1).

The presence of small forcing and damping terms can then pick out one particular wave speed at which

the energy input by forcing exactly balances the energy lost to damping. This can be found by power

balancing.

If we write the perturbed system by φtt − E(φ) = εF (φ) + εD(φ), then since power = force × velocity,

we find the equilibrium speed by requiring∫

F (φ)φt +D(φ)φtdx = 0. DERIVE

3.4 Nonsymmetric Klein-Gordon breather

Since φtt = φxx − sinφ has a breather solution, we might expect φtt = φxx − φ + 16φ3 also to have a

breather solution in the small amplitude limit; (exercise); but what about

φtt = φxx − φ+ αφ2 + βφ3nskg (3.28)

This system has Hamiltonian density

H = 12φ2t +

12φ2x +

12φ2 − 1

3αφ3 − 1

4βφ4 (3.29)

(i) dispersion relation: ω2 = k2 + 1, has minimum at k = 0 where ω = 1, so basic wave is φ = eit.

(ii) asymptotic expansion, let us look for a slowly varying envelope solution for this wave, instead of


having a constant amplitude. We also need to account for small changes in frequency - i.e. small changes

over longer times, thus we use multiple timescales.

Introduce new timescale τ = ε2t, so ∂∂t

7→ ∂∂t+ ε2 ∂

∂τ

large space scale y = εx, so that ∂∂x

= ε ∂∂y

φ = εeitF (y, τ) + [ε2e2itG(y, τ) + ε2H(y, τ)] + [ε3e3itJ(y, τ) + ε3eitK(y, τ)] + c.c.+O(ε4) 4p-ans

(3.30)

Preliminaries:

φtt = −εeitF + 2iε3eitFτ + ε5eitFττ + ε6Hττ − 4ε2e2itG+ 4iε4e2itGτ + ε6e2itGττ

−9ε3e3itJ − ε3eitK + c.c.+ h.o.t.

φxx = ε2φyy = ε3eitFyy + c.c.+ h.o.t. (3.31)

φ2 = (ε2e2itF 2 + ε2|F |2) + (2ε3e3itFG+ 2ε3eitFH + 2ε3eitFH + 2ε3eitFG) + c.c.+O(ε4)

φ3 = (ε3e3itF 3 + 3ε3eit|F |2F ) + c.c.+ h.o.t.

where h.o.t. refers to higher order terms.

Final subs of (3.30) and all its condequences into (3.28)

−εeitF + 2iε3eitFτ − 4ε2e2itG− 9ε3e3itJ − ε3eitK = ε3eitFyy − εeitF − ε2H − ε2e2itG− ε3e3itJ

−ε3eitK + αε2e2itF 2 + αε2|F |2 + 2αε3eitFH +

+2αε3eitFH + 2αε3eitFG+ 2αε3e3itFG+

+βε3e3itF 3 + 3βε3eit|F |2F (3.32)

Now equate terms of equal order in ε and equal frequency, i.e. equal powers of eit

O(εeit) : −F = −F

O(ε2e0it) : 0 = −H + α|F |2

O(ε2e2it) : −4G = −G+ αF 2

O(ε3eit) : −K + 2iFτ = Fyy −K + 3β|F |2F + 2α(FH + FH + FG)

O(ε3e3it) : −9J = −J + βF 3 + 2αFG

(3.33)

Thus H = α|F |2, G = −13αF 2, and the last equation becomes

2iFτ = Fyy + (103α2 + 3β)|F |2F, (3.34)

which is the NLS again. Thus we can quote the solution

F =

√6 γ

√

9β + 10α2e−iγ2τ/2sech(γy) (3.35)


which gives the leading-order result

φ(x, t) =2√6 γε

√

9β + 10α2sech(γεx) cos((1− 1

2ε2γ2)t). (3.36)

However, if 9β + 10α2 < 0 this solution is not valid, and instead we find

F = γ

√

−6

9β + 10α2eiγ

2τ tanh(γy) (3.37)

and

φ(x, t) = 2εγ

√

−6

9β + 10α2tanh(γεx) cos((1 + ε2γ2)t) (3.38)

Note that in both of these solutions γ, ε do not appear independently, but only as the product γε, thus

we have a one-parameter family of solutions, parameterised by γ = γε.

3.5 Numerical Solution of Equations

– not formally part of notes.

Standard finite difference methods for PDEs are applicable to integrable and nonlinear wave equations;

one should be aware the upwind condition in hyperbolic systems. Some numerical schemes preserve the

conserved quantities more accurately than others. For example the leap-frog scheme

un+1j − 2un

j + un−1j

(∆t)2=

unj+1 − 2un

j + unj−1

(∆x)2− F (1

2(un

j−1 + unj+1)), (3.39)

where unj = u(j∆x, n∆t) is generally more accurate for Klein-Gordon equations (utt = uxx − F (u)) than

other numerical schemes of a similar order.

Exact conservation laws provide a good check on the accuracy of the final solution.

Example

For solving KdV in 1965, Zabusky & Kruskal used the leapfrog scheme

un+1j − un−1

j

2∆t=

(

unj−1 + un

j + unj+1

3

)(

unj+1 − un

j−1

∆x

)

+uj+2 − 2uj+1 + 2uj−1 − uj−2

(∆x)3(3.40)

Note the use of central differences which improve accuracy, and the use of 13(un

j−1 + unj + un

j+1) instead

of unj ; the reason for this is to make the conservation of energy

∫

u2 more accurate. One can show that∑

j(un+1j )2 −

∑

j(un−1j )2 = O((∆t)3). (Ex)


4 Applications

4.1 Electrical Transmission Lines

An example where nonlinear waves occur is in the study of electrical transmission lines (ETL). These

can actually be built, and theories tested against experiment. The lines consist of identical repeated

segments, each segment having resistors, capacitors and inductors; for example

-In−1 -In -In+1

6Vn−16Vn

6Vn+16Vn+2

R R R¦ ¥§¦ ¤¥§¦ ¤¥§¦ ¤¥§¦ ¤¥¦ ¥L

¦ ¥§¦ ¤¥§¦ ¤¥§¦ ¤¥§¦ ¤¥¦ ¥L

¦ ¥§¦ ¤¥§¦ ¤¥§¦ ¤¥§¦ ¤¥¦ ¥L

³³

³³7

C ³³

³³7

C ³³

³³7

CG G G

r

r

r

r

r

r

r

r

r

r

r

r

The resistors represent small losses due to imperfections in the capacitor and inductor, so R and G−1

will be assumed to be small. Applying the laws of electrical circuits, and allowing the capacitance to

depend in a nonlinear way on the voltage, we find

Vn+1 − Vn = −InR− LdIndt

, In − In−n = −Vn

G− d

dt(VnC(Vn)) . (4.1)

We now neglect R and 1/G, and eliminate the In variables by differentiating the second equation, and

subtituting in from the first, to leave

Vn+1 − 2Vn + Vn−1 = Ld2

dt2[VnC(Vn)] . (4.2)

Typically capacitance is reduced as the voltage is increased, the simplest way to include this nonlinear

effect is by making C(V ) = C0(1− aV ). We obtain the evolution

LC0d2

dt2(Vn − aV 2

n ) = Vn+1 − 2Vn + Vn−1. (4.3)

This is the basic ‘equation of motion’ for the electrical transmission line. However, it is possible to write

it in a more familiar form, by making the substitution un = Vn − aV 2n , then

Vn =1−

√1− 4aun

2a, (4.4)

where we have taken the lower root so that u = 0 corresponds to V = 0. The equation then becomes

2aLC0un = −√

1− 4aun+1 + 2√1− 4aun −

√

1− 4aun−1. etl-eqmot (4.5)


4.1.1 Aside

This system is Hamiltonian, with forces W ′(u) = [1 −√(1 − 4au)]/2a, and since W ′′(0) > 0 the zero

solution is stable (dispersion relation gives real frequencies), we have a potential energy

W (u) =(1− 4au)3/2

12a2+

u

2a− 1

12a2, (4.6)

whence a Hamiltonian structure is obtained

H =∑

n

12s2n +W (rn+1 − rn). (4.7)

Here un = rn+1 − rn, sn = Úrn and (rn, sn) are the canonical Hamiltonian variables.

From numerical simulations we see that any disturbance in the ETL breaks down if its amplitude becomes

too large. From the equation of motion we can see this too: if u > 1/4a then we have strayed into a

region where W (u) is not well-defined, corresponding to an upper limit for V of 1/2a. Numerically

determined wave profiles, are seen to develop a singularity as the amplitude reaches this limit, cf waves

breaking on a beach.

4.1.2 Onto KdV

Let us analyse small amplitude deviations from u = 0 in this system. expanding (4.5) for small u we find

2aLC0un = 2aun+1 + 2a2u2n+1 − 4aun − 4a2u2

n + 2aun−1 + 2a2u2n−1 etl-eqmot2 (4.8)

Our spatial variable n is currently discrete, in order to convert (4.8) into something more familiar we

replace n by a continuous variable, x and expand the second central difference into spatial derivatives,

LC0utt = uxx +112uxxxx + a(u2)xx +

112a(u2)xxxx. etl-ctm1 (4.9)

This will only be valid if the waves we are seeking are slowly varying in space, otherwise u(x+1)−u(x) 6≈∂u/∂x. So we shall restrict ourselves to analysing slowly varying waves.

(4.9) has dispersion relation LC0ω2 = k2 − 1

12k4, so that slowly varying waves travel with speed ω/k ≈

(1− 124k2)/

√LC0 ≈ 1/

√LC0.

So let us look for waves which travel at about this speed, using the slow variation in u as a function of

x, we introduce a small parameter ε ¼ 1 so that (1/u)∂u∂x

∼ ε, then x needs to alter by 1/ε to alter u by

an O(1) amount.

We transform to a coordinate frame which moves at speed v, via y = ε(x− vt), with v = 1/√LC0

and study evolution on an exceedingly long timescale, τ = ε3t


and look at small amplitude waves u = ε2ψ then

∂

∂x=

/

0.∂

∂τ

/

+ ε∂

∂y,

∂

∂t= ε3

∂

∂τ− εv

∂

∂y(4.10)

implies (4.9) becomes

LC0ε8ψττ

︸︷︷︸

h.o.t.−ignored

−2ε6vLC0ψτy + ε4v2LC0ψyy︸︷︷︸

cancels

= ε4ψyy︸︷︷︸

cancels

+ 112ε6ψyyyy + aε6(ψ2)yy +

112aε8(ψ2)yyyy

︸︷︷︸

h.o.t.−ignored

(4.11)

noting that LC0v2 = 1, we find the O(ε4) terms cancel leaving O(ε6) terms as leading order, all of which

involve a y-derivative. Thus integrating once w.r.t y we find

−2√

LC0 ψτ = 112ψyyy + 2a(ψψy) (4.12)

Simple linear rescalings of τ, yψ can remove all these constants to leave the famous KdV equation.

ψ = 24aψ

y = −y

τ = τ/24√LC0

⇒ ψτ = ψyyy + ψ ψy. (4.13)

4.1.3 Regularising long wave equations

Continuum approximations are also known as long-wave equations, since they are designed to faithfully

approximate the long wave (small wavenumber) behaviour of a more complicated system. Unfortunately

they can be ill-posed; it has already been noted that (4.8) is ill-posed in that its dispersion relation

predicts complex ω’s for large k.

We shall now use Fourier transforms to show how the above argument may be improved, through the

derivation of well-posed PDEs at intermediate stages to yield more accurate final approximations to the

shape of solitary waves.

We rewrite (4.8) as

utt(n, t) = un = W ′(u(n+ 1, t))− 2W ′(u(n, t)) +W ′(u(n− 1, t)), (4.14)

which we shall think of as an operator equation

utt = D2W ′(u(t)) where D2 = δ2n = exp(∂n)− 2 + exp(−∂n) (4.15)

since

f(n+ 1) = f(n) + 1f ′(n) + 1212f ′′(n) + 1

613f ′′′(n) + . . . (4.16)

= [1 + ∂n +12∂2n +

16∂3 + . . .]f(n). (4.17)


We now take Fourier transforms in n,

defining u =∫∞n=−∞ u(n, t)eikndn and W ′(u) =

∫

W ′(u(n, t))eikndn so that

W ′(u(n+1, t)) =

∫

W ′(u(n+ 1, t)eikndn =

∫

W ′(u(m, t)eikm−ikdm = e−ik W ′(u), (4.18)

then

−utt = −(eik − 2 + e−ik) W ′(u) = 4 sin2(12k) W ′(u). (4.19)

Now, a continuum approximation assumes u(n, t) varies slowly in n

i.e. it is dominated by low wavenumbers (small k).

In Fourier, or k-space (as opposed to real-space where we use n, x), our equation has the form

−utt = D2(k) W ′(u) (4.20)

where D2(k) = 4 sin2(12k). Thus to form accurate approximations of the equation (4.8), we need to form

accurate approximations to the operator D2(k). We can do this using Taylor series

D2(k) = 4 sin2(12k) = k2 − 1

12k4 + 1

360k6 + . . . 413taylor (4.21)

-

6

k

ω2

¸ °°¨ ¢¢¢¢

¨ °°¸ XXPPHH@@

BBBB@@HHPPXX ¸ °°¨

ω2d

¸ °°¨

¢¢

¨°°¸ XXPPHH

@@CCCCCCCCω2ctm

Truncating the series (4.21) after two terms yields

−utt = (k2 − 112k4) W ′(u) (4.22)

which, on inverting the Fourier transform, gives

utt = W ′(u)xx +112[W ′(u)]xxxx = uxx + a(u2)xx +

112uxxxx + a(u2)xxxx, (4.23)

which is almost the standard continuum approximation derived earlier (4.9).

[To get to (4.9) exactly, we then neglect the fourth derivative of nonlinear terms, since we are considering

small amplitude disturbances, and slowly varying disturbances.]


[So far no new results, just a different method for deriving the continuum approximation]

Instead of the Taylor series approach, let us approximate D2(k) = 4 sin2(12k) by the rational polynomial

p(k)/q(k) where p(k) and q(k) are both polynomials in k. (This is also known as a Pade approximate).

Choosing p and q to be quadratics, we find

D2(k) = 4 sin2(12k) =

k2

1 + 112k2

, (4.24)

which is formally as accurate as k2(1− 112k2),

both having error terms which are O(k6) as k → 0.

-

6

k

ω2

¸ °°¨ ¢¢¢¢

¨ °°¸ XXPPHH@@

BBBB@@HHPPXX ¸ °°¨

ω2d

¸ °°¨

£££££

¨

¸ ω2ros

¸ °°¨

¢¢

¨°°¸ XXPPHH

@@CCCCCCCCω2ctm

Our new approximation implies

−utt =k2

1 + 112k2

W ′(u) ⇒ utt +112k2utt = −k2 W ′(u) (4.25)

⇒ utt − 112uxxtt = (W ′(u))xx ⇒ (4.26)

utt = [W ′(u)]xx +112uxxtt, ros (4.27)

which is an alternative continuum approximation to (4.9).

Note that in deriving this approximation of (4.27), we have not had to make any further assumptions to

neglect 4th derivatives of nonlinear terms.

We now find the dispersion relation of (4.27) to investigate its well-posedness:

ω2 = k2 − 112k2ω2 ⇒ ω2 =

k2

1 + 112k2

≥ 0. (4.28)

Thus all values of k give real frequencies ω so (4.27) is well-posed, in contradistinction to (4.9).

4.1.4 Comparison of solitary wave solutions

On neglecting the fourth derivatives of nonlinear terms, solutions of (4.9) are given by

c2u′′ = u′′ + 112u′′′′ + a(u2)′′ (4.29)


which implies

u =3

2a(c2−1) sech2

(

(x− ct)√

3(c2 − 1))

, 414ctm (4.30)

whereas solutions of (4.27) are given by

c2u′′ = u′′ + a(u2)′′ + 112c2u′′′′ (4.31)

which implies

u =3

2a(c2−1) sech2

(

(x− ct)

√

3(c2 − 1)

c2

)

. 414ros (4.32)

Looking at the amplitudes, we see that there is no difference, as is the sech shape, but the width does

vary. In (4.30) the width → 0 as c → ∞, which is unphysical and the assumption of u(z) being a slowly

varying function fails. In (4.32) the width saturates at 1/√3 as c → ∞, which is more realistic, and the

assumption of u(z) varying slowly is valid for larger values of c than in (4.30).

4.2 Fibre-optic Cables

Optical fibres are highly transparent cylindrical waveguides whose optical index varies with distance from

centre. Various physical effects cause dispersion in group velocity, small power losses, nonlinear effects

at larger amplitudes, which we now try to describe. (Note ε0µ0c2 = 1.)

-x6

y

¨*z

AAAA

¡¡

¡¡

¡¡¡¡

AA

AA

¨¨¨¨¨

¨¨¨¨¨

¨¨¨¨¨

¨¨¨¨¨

¨¨¨¨¨

¨¨

¨¨

¨¨

Including the effects of polarization (P),Maxwell’s equations for the electric field (E) are

∇2E − 1c2

Ett = µ0Ptt

A simple model of polarisation gives P = ε0εL(ω)E + ε0εN

(ω)|E|2EConsider a wave travelling in the z-direction, so E = E(z, t)e3, then

c2Ezz − Ett = (εLE + ε

N|E|2E)tt.

This has the nonlinear dispersion relation

c2k2

ω2− 1 = ε

L(ω) + ε

N(ω)|E|2 =: n(ω)2 − 1

(4.33)

We assume that the solution is a sinusoidal wave modulated by a slowly changing envelope of small

amplitude δA which moves at some speed v; thus E = δA(δ(z − vt), δ2z) where δ ¼ 1, Z = δ(z − vt) is

a coordinate frame which moves with the envelope describing variations over large times and distances,

and S = δ2Z describes variations over even larger distances. Then

E = δA(Z, S)eikz+iωt = δA(δ(z − vt), δ2z)eikz+iωt

Et = δiωAeikz+iωt − vδ2AZeikz+iωt

Ez = δikAeikz+iωt + δ2AZeikz+iωt + δ3ASe

ikz+iωt


Ett = −δω2Aeikz+iωt − 2iωvδ2AZ + v2δ3AZZeikz+iωt (4.34)

Ezz = −δk2Aeikz+iωt + 2ikδ2AZeikz+iωt + δ3AZZe

ikz+iωt + 2ikδ3ASeikz+iωt +O(δ4).

(|E|2E) = δ3|A|2Aeikz+iωt

(|E|2E)t = δ3iω|A|2Aeikz+iωt +O(δ4)

(|E|2E)tt = −ω2δ3|A|2Aeikz+iωt +O(δ4)

Plugging all this into (4.33) yields

O(δ) : −c2k2 + ω2 = −εLω2

O(δ2) : c2k + ωv = −vωεL

O(δ3) : 2ikc2AS + c2AZZ + v2AZZ = −ω2εN|A|2A+ ε

Lv2AZZ

(4.35)

The first of which is the linear dispersion relation (for which it is easier to think of ω being specified and

k = k(ω) being determined by k2 = (1+εL)ω2/c2). The second fixes the speed v at which the envelope

travels, and the third is the NLS equation which determines the shape of the envelope.

2ikc2AS = AZZ(εLv2 − v2 − c2)− ε

Nω2|A|2A, v =

−c2ω

k. (4.36)

4.3 Water Waves

For those who have met Navier-Stokes equations before, we are now going to derive KdV from NS.

(Those who haven’t met NS before can have a short nap). Firstly, neglect viscosity, so we obtain the

Euler equations. For a liquid subject to gravity (acting in the negative y-direction), we have

ut + (u.∇)u = −1

%∇P − gj ns-mom (4.37)

∇.u = 0. ns-cty (4.38)

Taking the curl of (4.37) yields an equation which has vorticity w = ∇ × u = 0 as a solution. This

enables us to write u = ∇φ where φ is a velocity potential. The continuity equation (4.38) then becomes

∇2ψ = 0. The momentum conservation equation can be written as

P − P0

%= P1(t)− φt − 1

2|∇φ|2 − gy. (4.39)

The arbitrary function P1(t) can be absorbed into the velocity potential by φ = φ−∫

P1 dt;

we choose the constant P0 such that P = P0 on y = h0.


6

?

h0

?6η

-

6

x

y

¨¨¨HHHHH

@AA¡

¡¨ @

@H

HHHHH @

AA¡

¡¨

For a liquid in a 2D container of finite depth but infinite horizontal extent we need to impose boundary

conditions on the top and bottom. No fluid can flow through the lower boundary, thus we impose

φy = 0 on y = 0. (4.40)

We denote the top surface by y = h0 + η(x, t). Particles in the surface remain in the surface, and so get

advected with the flow. Any quantity which is advected with the flow changes according to the total

derivativeD

Dt=

∂

∂t+ u

∂

∂x+ v

∂

∂y=

∂

∂t+ φx

∂

∂x+ φy

∂

∂y. (4.41)

Applying this operator to both sides of the equation y = h0 + η yields

φy = ηt + φxηx (4.42)

We assume that the motion of the surface does not affect the pressure in the air above the surface thus

P = P0 on y = h0 + η. Hence we arrive at the system

φy = 0 on y = 0

∇2φ = 0 in 0 ≤ y ≤ h0 + η

φy = ηt + φxηx on y = h0 + η

0 = φt +12φ2x +

12φ2y + gη on y = h0 + η

ww1 (4.43)

Those who hadn’t seen NS before should now wake up, and take note of this set of equations.

4.3.1 Linear Theory

Neglecting nonlinear terms, we find

φy = 0 on y = 0

∇2φ = 0 in 0 ≤ y ≤ h0 + η

φy = ηt on y = h0 + η

0 = φt + gη on y = h0 + η

(4.44)

so that on y = h0 + η we have φtt = −gφy and the η can be eliminated from the problem.


Seeking a solution in terms of linear waves travelling in the x-direction we assume

φ = A(x, y) sin(kx− ωt) (4.45)

then in 0 ≤ y ≤ h0 + η

0 =

[

∂A

∂y+

∂A

∂x− k2A

]

sin(kx− ωt) +

[

2k∂A

∂x

]

cos(kx− ωt). (4.46)

Since this equation must hold for all (x, t), both square brackets must evaluate to zero,

thus ∂A∂x

= 0 (A is independent of x∀t, i.e. A = A(y) only),

and A = A0 cosh(k(y + h0)) + A1 sinh(k(y + h0)).

Imposing the boundary condition at y = 0 yields A1 = 0.

The top boundary condition (φtt = −gφy on y = h0 + η) yields

ω2 = kg tanh(k(h0 + η)). (4.47)

Assuming h0 is small, this gives a speed of

c =ω

k=

√

g(h0 + η)− 13k2g(h0 + η)3. (4.48)

As k → 0, the wave speed c asymptotes to c0 =√gh, where h = h0 + η;

and for small k the speed can be approximated by c =√gh(1 − 1

6k2h2). Slowly varying surface distur-

bances thus propagate with this speed.

4.3.2 Nonlinear Theory

We want to analyse the long-time evolution (t → ∞) of surface waves, when the water is very shallow

(h0 → 0). Care is needed when considering a combination of limits. First we shall rescale (non-

dimensionalise), looking for waves which travel at the speed c0

y = h0Y, η = εh0H, φ = h0

√

εgh0Φ, t =1

ε

√

h0

εgT, x− c0t =

h0√εZ. (4.49)

Note that ∂t = ε√

εg/h0 ∂T − c0√ε/h0 ∂Z .

Thus the system of equations (4.43) becomes

ΦY = 0 on Y = 0

εΦZZ + ΦY Y = 0 in 0 ≤ Y ≤ 1 + εH(X,T )

ε2HT − εHZ + ε2ΦZHZ = ΦY on Y = 1 + εH(X,T ) BCY

H + εΦT − ΦZ + 12Φ2

Y + 12εΦ2

Z = 0 on Y = 1 + εH(X,T ) BCH

(4.50)


Note that h0 has been eliminated from these equations, thus we only need to worry about the limit ε → 0

now, (which corresponds to considering wide small amplitude distubrances &t ½ 1).

Solving the system we assume a solution for Φ(Y, Z, T ) of the form

Φ = Φ0 + εΦ1 + ε2Φ2 + . . . (4.51)

Then fit this to the DE and to the lower BC to find Φ:

O(1) : Φ0,Y Y = 0 ⇒ Φ0 = A(Z, T )Y +B(Z, T )

BC ⇒ A = 0 ⇒ Φ0 = B

O(ε) : Φ1,Y Y = −Φ0,ZZ = −BZZ ⇒ Φ1 = D(Z, T ) + Y C(Z, T )− 12Y 2BZZ

BC ⇒ C = 0 ⇒ Φ1 = D − 12Y 2BZZ

O(ε2) : Φ2,Y Y =−Φ1,ZZ = 12Y 2BZZZZ −DZZ ⇒ Φ2 = 1

24Y 4BZZZZ − 1

2Y 2DZZ + E(Z, T )Y + F (Z, T )

BC ⇒ E = 0 ⇒ Φ2 = 124Y 4BZZZZ − 1

2Y 2DZZ + F

(4.52)

Thus

Φ = B + ε(D − 12Y 2BZZ) + ε2(F − 1

2Y 2DZZ + 1

24Y 4BZZZZ) +O(ε3). (4.53)

so that ΦY = −εY BZZ + ε2(16Y 3BZZZZ −Y DZZ) and ΦZ,T = . . .; we also introduce an expansion for the

disturbance H = J + εI.

Now applying BCY to O(ε2):

−εBZZ − ε2JBZZ + 16ε2BZZZZ − ε2DZZ

︸︷︷︸

ΦY

= ε2JT︸︷︷︸

ε2HT

+ ε2JZBZ︸︷︷︸

ε2HZΦZ

−εJZ − ε2IZ︸︷︷︸

−εHZ

(4.54)

Thus at O(ε) we have BZZ = JZ and at O(ε2)

DZZ − IZ = 16BZZZZ − JBZZ − JT − JZBZ WWBCY (4.55)

Imposing BCH

0 = J + εI︸︷︷︸

H

−BZZ − ε(DZ − 12BZZZ)

︸︷︷︸

−ΦZ

+ εBT︸︷︷︸

εΦT

+ 12εB2

Z︸︷︷︸

12εΦ2

Z

+O(ε2)︸︷︷︸

12Φ2

Y

(4.56)

from the O(1) terms we find J = BZ and from those O(ε) we find

DZ − I = 12BZZZ +BT + 1

2B2

Z WWBCH (4.57)

Taking ∂Z(4.57)-(4.55) we obtain

0 = 13BZZZ +BZT + JT + JTBZZ + JZBZ +BZBZZ (4.58)

but since BZ = J this can be written

0 = 13JZZZ + 2JT + 3JJZ , (4.59)

which is the KdV equation.


4.3.3 Intepretation of Results

The KdV equation has travelling wave solutions

J = J(Z − V T ) = 2V sech2

(√

3V

2(Z − V T )

)

, (4.60)

in which H = J + O(ε) describes the shape of the fluid’s free surface. Rewriting this in the original

η(x, t) variables

η = εh0H, Z =

√ε

h0(x− c0t), c0 =

√

gh0, T = εt

√

εg

h0, (4.61)

we find

η = 2εh0V sech2

(√

3V

2

[√ε

h0(x−c0t)− εV t

√

εg

h0

]

)

(4.62)

= 2vh0sech2

(

1

h0

√

3v

2

[

x−√

gh0 (1 + v)t]

)

, (4.63)

when one notes that only the combination v = εV ¼ 1 appears, and never ε or V alone.

4.4 Long Josephson junctions

A Josephson junction consists of two superconducting layers, each of thickness 4 × 103A. These are

separated by an oxide layer of thickness ∼30A. Two types of current flow through the junction - firstly

normal electrons; and secondly pairs of electrons travelling together, known as Cooper pairs. It is the

current due to this latter effect that gives the interesting non-linear behaviour of the system.

-

-

Bias current I

-

-

superconductor

superconductor

w ²

Fluxon

velocity c

©

oxide layer

@@

@@

@@

@@

@@


We denote the phase difference at site n and time t between the electrons in the two layers as φn(t). We

using the notation:

Vn for voltage across the oxide layer,

C for capacitance of the oxide layer (now just linear),

L for inductance along the superconducting layer,

in for current through the effective inductor,

I0 for the maximum current due to Cooper pairs,

If for the size of the external forcing current,

R the effective Ohmic resistance of the oxide layer,

and Φ0 for the quantum unit of flux,

then J = I0 sinφn is the Josephson supercurrent.

This is in effect a electric line similar to that analysed above, now

L L

-in−1 -in−1 -in -in -in+1

6Vn−2

6Vn−1

6Vn−1

6Vn

6Vn

q q q

C J R

q q qS

SwS

SqIf

r r r

C J R

r r rS

SwS

SqIf§ ¤§¦ ¤¥§ ¤§¦ ¤¥§¦ ¤¥ § ¤§¦ ¤¥§ ¤§¦ ¤¥§¦ ¤¥

The voltage due to the phase difference is

Vn =Φ0

2π

dφn

dt. (4.64)

Vn − Vn−1 = −Ldindt

, in+1 − in = −CdVn

dt− I0 sin(φn(t))− If −

1

RVn. (4.65)

The latter equation contains imperfection terms as well as the main effects.

In the second equation the first two terms describe the main effects of the Josephson junction, namely

the Capacitance current and the Josephson supercurrent due to Cooper pairs.

The third term is a small externally applied forcing current which ensures that the loss terms do not

cause the whole system to tend to a static equilibrium state. The −V/R term is a standard Ohm’s law

current that flows across the gap, here R is large so that the current is small.

We eliminate Vn, in from the equations (4.64)– (4.65) by forming an equation for in+1− in. The equation

then just involves the phase, φn(t)

−Φ0

2πLδ2nφn = in+1 − in =

−CΦ0

2πφn − I0 sinφn − If −

Φ0

2πRÚφn (4.66)


Rescaling of time (by 1/√(LC) ), together with defining

Γ2 =2πLI0Φ0

, α =

√L

R√C, I =

2πLIfΦ0

. (4.67)

gives the driven and damped discrete sine-Gordon equation

φn = φn+1 − 2φn + φn−1 − Γ2 sinφn − I − α Úφn. (4.68)

The physical quantities of current and voltage are related to φ by i ∝ 〈φn〉, V ∝ 〈 Úφn〉.

We convert the difference terms into derivatives by replacing n with a continuous variable x and Taylor

expanding

φtt = φxx − Γ2 sinφ− I − αφt. (4.69)

4.4.1 Current–Voltage Characteristics

Two quantities that are simple to measure in experiments are the voltage and the current. Physicists

typically plot the current (I) against the voltage (c). Mathematically we have considered I as a known

parameter, and then found the dependence of c on the other parameters (α, I,Γ). In a physical experi-

ment, α, β,Γ are fixed and the pair (I, c) are varied tracing out a curve in two-dimensional space. It is

with this curve we can compare theory and experiment.

The formula

I =4αΓc

π√1− c2

generates a curve of the form

-

6I

c0 1¸¸¸¸°°¨¨

¡¡¢¢¢££

4.5 Waves in nonlinear diffusion problems

A gentle easing off the technicalities, with nonlinear diffusion waves.

Many problems in mathematical biology and other applications involve nonlinearity and diffusion (all

the above has focused on nonlinearity and dispersion). In this case we often find equations of the form

∂u

∂t=

∂2u

∂x2− f(u). (4.70)


This models the spread of some chemical from a region of high concentration (u = 1) into a region of

low concentration (u = 0) thus it is natural to require that both u = 0 and u = 1 solve the problem, this

implies f(0) = 0 = f(1). Thus we shall assume u → 1 as x → −∞ and u → 0 as x → 0.

-

6

x, z

u

XXPPPHHHH@

@@HHHHPPPXX

-cu = 1

u = 0

AA@

@

Impossible (as shown later)

If we seek travelling wave solutions of this (u(z) = u(x − ct)) we find an ODE which is amenable to

analysis using phase plane techniques

u′′ + cu′ − f(u) = 0, ⇔

{

u′ = w

w′ = f(u)− cw(4.71)

Property travelling wave must be monotonic. i.e. w < 0

Consider phase plane: at z = −∞, u = 1, w < 0; and at z = +∞, u = 0, w < 0.

Now try to construct a trajectory corresponding to a non-monotonic solution,

i.e. one which strays into w > 0,

noting that (i) if w > 0 then u must be increasing and

(ii) the trajectory cannot cross itself. Impossible.

-

6

u

w

s s-

»

¨¹¨H

HHY

Phase Plane

0 1» @I

-?

Form of f(u)

-

6

u

f(u)

AAAAAA

0 1α

In fact, the problem can be reduced to a first order ODE:

dw

du=

f(u)

w− c (4.72)


from which we can deduce information about c:

0 = [12w2]u=1

u=0 =

∫ u=1

u=0

w dw =

∫ u=1

u=0

f(u) du−∫ u=1

u=0

cw(u) du diff-ceq (4.73)

so c =∫ 1

0f(u)du/

∫ 1

0w(u)du and c has the opposite sign to

∫ 1

0f(u)du since w < 0 for all u with 0 < u < 1.

This is similar in spirit to the way c was obtained in the example of long Josephson junctions – it is the

value which causes losses by diffusion to balance gains from the forcing term f(u).

Example

If we assume that the trajectory w(u) has the form w = Ku(u − 1) then we find dw/du = K(2u − 1)

and hence f(u) = Ku(u− 1)(2Ku−K + c). Thus in the case f(u) = u(u− α)(u− 1), an exact solution

can be found, setting K = 1/√2, whereupon α =

1− c√2

2or c =

1− 2α√2

.

This value of c satisfies (4.73) - CHECK.

Then u′ = w = u(u− 1)/√2 implies

du

dz=

u(u− 1)√2

(4.74)

which is separable giving the solution

u(z) =1

1 + ez/√2=

1

1 + e(x−ct)/√2. (4.75)

There are thus three cases: (i) the intriguing case α = 12

which implies c = 0 – a stationary solution; and

two cases worthy of further comment: α > 12

where c < 0 and α < 12

where c > 0.

-

6

u

f(u)

@@

@@@ 0 1α

-

6

u

f(u)

@@

@@@

0 1α

When α > 12, the positive area

outweighs the negative area in∫ 1

0f(u) du

so c < 0 and

the region of high concentration is in retreat.

When α < 12, the negative area

outweighs the positive area in∫ 1

0f(u) du

so c > 0 and

the region of high concentration extends out.

Technically we ought to consider K = −1/√2 also, but this leads to trajectories which go from (0,0) to

(1,0) rather than from (1,0) to (0,0).


4.5.1 Fisher-KPP equation

ut = uxx − u(1− u) is known as Fisher’s equation or

(if you’re talking to Russians), the KPP equation. The travelling wave equation u′′ + cu′ + u(1− u) can

be writtenu′ = w

w′ = −cw − u(1− u)

}

⇔ dw

du= −c− u(1− u)

w. (4.76)

which has two critical points in the phase plane: (0,0) and (1,0). The latter is a saddle for all values of c

and the former is a stable node for c2 ≥ 4 and a stable spiral for c2 < 4. Since we cannot have negative

concentrations, there is a minimum wave speed c = 2, and waves exist for all speeds above this value. In

order to remove the translational degeneracy from the problem we impose u(0) = 12.

Let us look for the form of fast waves, that is consider the limit ε = 1/c2 → 0. If we rescale z = cζ = ζ/√ε

then we find

εd2u

dζ2+

du

dζ+ u(1− u) = 0 (4.77)

At leading order in ε we then find

du

dζ= −u(1− u) ⇒ u =

1

1 + eζ=

1

1 + ez/c. (4.78)

From this we can find simple properties of the wave, for example how the width of the wave varies with

speed. The width is inversely proportional to the steepness of the wave at the centre (z = 0 where

u = 12).

s = −du

dz= −1

c

du

dζ= +

1

cu(1− u) =

1

4c, (4.79)

so that the faster a wave travels, the less steep it is; alternatively fast waves are more spread out.

4.5.2 Multiple waves

[Omit title until following Qu has been answered]

What happens when the function f(u) has lots of zeros ? For example:

-

6

u

f(u)

@@

@@

@@

@@

@@

@@

0 α β γ 1

Consider u = β + εv then vt = vxx − vf ′(β) then v = eikx+γt implies γ = −k2 − f ′(β) so the constant

solution u = β is stable against oscillatory perturbations.


Splitting the problem into two, we might expect the following waves to exist:

(i) a wave 0 < u < β travelling with speed c0

(ii) a wave β < u < 1 travelling with speed c1

(iii) a wave 0 < u < 1 travelling with speed c

From (4.73) we expect

c0 =−∫ β

0f(u) du

−∫ β

0w(u) du

c1 =−∫ 1

βf(u) du

−∫ 1

βw(u) du

c =−∫ 1

0f(u) du

−∫ 1

0w(u) du

(4.80)

Clearly if c1 > c0 then wave between u = β & u = 1 catches up the wave between u = β & u = 0, and a

combined wave between u = 0 and u = 1 is formed; this will travel at speed c.

If c1 < c0 then upper wave will never catch up lower wave, and a combined wave will never form. The

plateau at u = β will then grow in width at speed c0 − c1, while its centre is advected along at speed12(c0 + c1).

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Nonlinear Waves Lecture Notes - School of … · Nonlinear Waves Lecture Notes (G14 NWA/2) ... G14 NWA/2 Nonlinear Waves 3 1 Linear Wave Theory & Other Background 1.1 Linear wave