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SECTION C: SOLITONS

Course text: Solitons: an introduction

P.G. Drazin & R.S. Johnson

Cambridge University Press, 1989

You will find the lecture notes at

http://www.shef.ac.uk/~robertus/kul/

Outline of the course

1. Introduction

2. Waves of permanent form

3. Scattering & inverse scattering

4. The inverse scattering transform for the KdV equation

5. Conservation Laws

6. The Lax method

1. Introduction

Linear PDEs

Consider the PDE

Lu = 0

where L = L(∂/∂x, ∂/∂t) and u = u(x, t).

If L is linear, i.e. if

L(au+ bw) = aLu+ bLw (1)

∀ constants a and b, and ∀ ‘well behaved’ functions u and w, then

Lu = 0 & Lw = 0 =⇒ L(au+ bw) = 0 (3)

i.e. if u and w are solutions of

Lv = 0 (5)

then so is any linear combination of u and w.

This is called the superposition principle

Example: the wave equation

L =∂2

∂t2− c2 ∂

∂x2(6)

We have that if

u(x, t) = f(x− ct) + g(x+ ct) (8)

for f, g ∈ C2(−∞,∞), then

Lu = 0 ∀x, t (9)

If u(x, 0) = F (x) and ut(x, 0) = G(x) then

u(x, t) =1

2[F (x− ct) + F (x+ ct)] +

∫ x+ct

x−ct

G(ξ)dξ (11)

is the unique solution to Lu = 0 with the given initial conditions.

This is called D’Alambert solution.

-5 -2.5 2.5 5 7.5 10

Figure 1: The D’Alambert solution f(x − ct) of the wave equation

at two different times

Consider f(x− ct) and g(x+ ct) separately. They obey the equation

ut + cux and ut − cux.

respectively.

This follows from the identity

L =∂2

∂t2− c2 ∂

∂x2≡(

∂t± c ∂

∂t∓ c ∂

Normal modes

Suppose that D(i∂/∂t,−i∂/∂x) is a linear operator and

Du = 0. (14)

Then try (i.e. guess that there exists) a solution of the form

u(x, t) = aei(kx−ωt)

for constant wavenumber k and frequency ω.

u(x, t) represents a sinusoidal wave of length 2π/k, period 2π/ω

and phase velocity c = ω/k.

A solution of a linear equation depending exponentially on time

is called normal mode

Because D(i∂/∂t,−i∂/∂x) is linear, if u(x, t) is a normal mode

D(i∂/∂t,−i∂/∂x)u = D(i(iω),−i(ik))u = 0, (15)

which implies

D(ω, k) = 0.

This is called dispersion relation.

Example: the wave equation

Consider again the wave equation

Du ≡ ∂2u

∂t2− c2 ∂

∂x2= 0. (16)

Try u ∝ ei(kx−ωt)

(−iω)2u− c2(ik)2u = 0 (17)

that is

D(ω, k) = c2k2 − ω2 = 0 (18)

ω = ±ck. (19)

±c is called phase velocity of mode.

If the phase velocity is not constant different modes travel at

different velocities and eventually the wave (packet) disperses.

The simplest dispersive equation is

ut + ux + uxxx = 0.

Let us try u ∝ ei(kx−ωt), then

D(ω, k) = −iω + ik + (ik)3 = 0 (20)

ω = k − k3 (21)

c = 1− k2 ≤ 1. (22)

Note that long waves (small k) travel faster.

If we add even derivatives to ut + ux, ω = f(k) is a complex

function of k, and the wave dissipates.

Consider

ut + ux − uxx = 0,

then by trying u ∝ ei(kx−ωt)

D(ω, k) = −iω + ik − (ik)2 = 0 (23)

ω = k − ik2 (24)

u(x, t) = e−k2t+ik(x−t) (25)

The above wave decays exponentially.

An initial value problem

Suppose that

D(i∂/∂t,−i∂/∂x)u = 0 (26)

is a linear PDE of first order in ∂/∂t and u(x, 0) is given.

Let us take the Fourier transform of u(x, 0):

a(k) =1

∫ ∞

−∞u(x, 0)e−ikxdx (27)

u(x, 0) =1

∫ ∞

−∞a(k)eikxdk (28)

Suppose that

D(ω, k) = 0 =⇒ ω = f(k). (29)

Then it may be verified that

u(x, t) =

∫ ∞

−∞a(k)ei[kx−f(k)t]dk. (31)

Wave packet and group velocity

A localized solution which is the superposition of waves of ap-

proximately the same length is called a wave packet.

The components have in general slightly different phase velocity

c = ω/k, and therefore spread, i.e. disperse.

It can be shown that asymptotically the wave packet moves with

the group velocity, which is defined by

cg =dω

dk. (33)

It can be shown that any localized disturbance after a long time

propagates at the group velocities, rather than phase velocities of

its components. So physical properties, like energy, have velocity

cg not c. It turns out that in general cg ≤ c.

Example

Consider the following expression:

a cos(k1x− ω1t) + a cos(k2x− ω2t)

= 2a cos

2(k2 − k1)x−

2(ω2 − ω1t)

× cos

2(k2 + k1)x−

2(ω2 + ω1) t

∼ 2a cos

2(k2 − k1) (x− cgt)

cos [k1 (x− ct)]

= A(ǫx, ǫt) cos(k1x− ω1t) as k2 → k1.

Here A(ǫx, ǫt) = 2a cos [ǫ (x− cgt)], ǫ = (k2 − k1)/2 is small and

cg = limk2→k1

ω2 − ω1

k2 − k1and c =

k1=ω2

k2. (35)

10 20 30 40

Figure 2: The ‘wave packet’ (34).

The method of characteristics and nonlinear waves

Consider again the equation ux+cut = 0 . Then

u(x, t) = constant ∀t (36)

on curves with equation dx/dt = c in the (x, t)-plane, because

dt= ut +

dtux = ut + cux = 0. (37)

Using this property, solution to ux+cut = 0 can be easily

constructed.

u(x, 0) = f(x) (38)

for differentiable f , then

u(x, t) = f(x− ct).

Now consider the nonlinear equation

ux + c(u)ut = 0.

Generalizing the previous ideas, u(x, t) is constant on curves given

by the equation

dt= c(u), x(t) = c(u)t+ constant. (39)

If furthermore u(x, 0) = f(x), the solution is given implicitly by

u(x, t) = f (x− c(u)t).

If c(u) increases with u, then the greater u is the faster the

velocity dx/dt so that an initial solution will steep, and may go

on to break.

The solution may be continued beyond breaking by invoking

some additional hypothesis.

Example

A simple example of nonlinear equation which can have

multivalued solution is

ux + (1 + u)ut = 0.

Using the method of characteristics, we have that the general

solution is

u(x, t) = f (x− (1 + u) t),

where f is an arbitrary function.

Example

Now, consider

ux + uut = 0.

If u(x, 0) = cosπx, then

u(x, t) = cos [π (x− ut)] (40)

gives u(x, t) implicitly.

Where do shocks occur?

By differentiating u(x, t) = cos [π (x− ut)] w.r.t. x we have

ux = −π (1− uxt) sin [π (x− ut)] (41)

ux 1− πt sin [π (x− ut)] = −π sin [π (x− ut)] . (42)

Shocks occur when ux =∞, which implies

tπ sin [π (x− ut)] = 1.

The first shock occurs when

t = 1/π and x− ut = 1/2 + 2n for n = 0,±1,±2, . . .

i.e. where u = 0, x = 1/2 + 2n.

We may have nonlinearity and dissipation, e.g.

ut + (1 + u)ux − uxx = 0

or nonlinearity and dispersion,

ut + (1 + u)ux + uxxx = 0

Let us consider the transformations

1 + u→ u, αu, t→ βt, x→ γx (43)

for the equation in the blue box

we obtain

ut +αβ

γuux +

γ3uxxx = 0. (44)

Choosing (for example) α = −6, β = γ = 1 yields

ut−6uux+uxxx = 0.

This is known as Korteweg-de Vries (KdV) equation.

The discovery of Solitary waves.

Figure 3: The diagram of Russel’s experiment to generate solitary

Figure 4: Parameters used in the description of solitary waves

• J. Scott Russel (1834):

c2 = g(h+ a) (45)

• Boussinesq (1871), Rayleigh (1876):

ζ(x, t) = a sech2 [β (x− ct− x0)] . (46)

• Korteweg & de Vries (1895):

Derived the equation for long weekly nonlinear water waves

3ǫ∂ζ

∂χ+ ζ

∂χ+

3σ∂3ζ

∂χ3

The function ζ(χ) = a sech2 (βχ) is a solution to the previous

equation provided

a = 4σβ2 and ǫ = −2σβ2. (48)

The coordinate χ is defined as

χ = x− (gh)1/2(

1− ǫ

t. (49)

Finally, the solitary wave solution becomes

ζ(x, t) = a sech2

x− (gh)1/2(

. (51)

c ∼ (gh)1/2(

Def.: A solitary wave solution of a PDE

∂t,∂

u = 0 (55)

is a travelling wave solution of the form

u(x, t) = f(x− ct) = f(ξ) (56)

where c is constant and f(ξ)→ 0 as ξ → ±∞.

Discovery of Solitons

• Fermi, Pasta and Ulam (1955). Working on a numerical model

of phonons in an anharmonic lattice. No equipartition of

energy among the modes.

• Kruskal and Zabusky (1965).

They considered the following equation

ut + uux + δ2uxxx = 0

They considered periodic boundary conditions

u(x, 0) = cosπx for 0 < x ≤ 2 (57)

u(x+ 2, t) = u(x, t) (58)

ux(x+ 2, t) = ux(x, t) (59)

uxx(x+ 2, t) = uxx(x, t) (60)

δ = 0 · 022. (61)

Figure 5: The solution of the periodic boundary-value problem for

the KdV equation

Defining properties of solitons

• A nonlinear wave of permanent form.

• Localized.

• May interact strongly with other soliton and yet retain its

identity.

Def.: A soliton is a solitary wave which asymptotically pre-

serves its shape and velocity upon nonlinear interaction with

other solitary waves, or more generally with another (arbitrary)

localized disturbance.

Applications

Consider a linear wave motion with dispersion. The dispersion

relation will have the form

ω(k) = kc(k2). (62)

As k → 0ω

k∼ c0 − λk2, λ > 0. (63)

Such dispersion relation is obtained by the equation

ut + c0ux + λuxxx = 0

Wave propagation in a classical continuum ⇒ the time evolution is

given by the material derivative:

∂t+ u

. (64)

The balance of nonlinearity and dispersion gives

ut + c0ux + α(uux + λuxxx) = 0, α small.

Thus we have

uτ + uuξ + λuξξξ = 0; ξ = x− c0t, τ = αt. (65)

This KdV equation is valid if x− c0t = O(1), t = O(α−1) as α→ 0.

Def.: The KdV equation is the characteristic equation govern-

ing weekly nonlinear waves whose phase speed attains a simple

maximum for wave of infinite length.

• Long weekly nonlinear water waves;

• gravity waves in a stratified fluid;

• waves in a rotating atmosphere;

• ion-acustic fluid in a plasma;

• pressure waves in a liquid-gas bubble mixture.

• Examples of other nonlinear equations with a wide application:

non linear Schrodinger equation and sine-Gordon equation.

2. Waves of permanent form

Travelling waves

Consider the nonlinear PDE

∂t,∂

u = 0. (66)

Then guess that

u(x, t) = f(ξ), ξ = x− ct

for some constant c.

NB Solution of this type do not always exist.

Solitary waves

Consider the KdV equation

ut − 6uux + uxxx = 0.

We try a solution of the form

u(x, t) = f(ξ) (67)

and we ask for

f, f ′, f ′′ → 0, as ξ → ±∞

Substituting f(ξ) in the KdV equation yields

−cf ′ − 6ff ′ + f ′′′ = 0. (68)

By integrating we have

−cf − 3f2 + f ′′ = A (69)

−cff ′ − 3f2f ′ + f ′f ′′ = Af ′ (70)

2cf2 − f3 +

2(f ′)

2= Af +B (71)

The condition

f, f ′, f ′′ → 0, as ξ → ±∞

implies

A = B = 0. (72)

Therefore, we have

(f ′)2

= f2(2f + c)

dfdf =

f ′= ±

f (2f + c)1/2(73)

Now we make the substitution

f = −1

2c sech2 Θ, (74)

then we obtain

df = −2f tanhΘdΘ. (75)

Finally, the integral becomes

ξ = ±∫

2f tanhΘ

−c sech2 Θ + c)1/2

dΘ = ± 2

c1/2Θ+const. if c > 0. (76)

f(ξ) = −1

2c sech2

2c1/2 (x− ct− x0)

for aribatrary x0 and c ≥ 0.

General waves of permanent form

We have found that

2(f ′)

2= f3 +

2cf2 + Af +B = F (f) (80)

Or equivalently

f ′ = ±√

2F (f)

Qualitative properties of f

• For real solution f , F (f) ≥ 0;

• f ′ changes its sign only at a zero, f1, of F (f);

• f(ξ) increase or decreases monotonically untill F (f) = 0.

-3 -2 -1 1 2 3f

Figure 6: F (f) with three simple zeros (cnoidal or periodic wave).

Behaviour of f(ξ) in a neighbourhood of the zeros of F (f)

(i) If f1 is a simple zero, then

(f ′)2

= 2(f − f1)F ′(f1) +O[

(f − f1)2]

. (81)

If we differentiate both sides w.r.t. ξ we get

2f ′f ′′ = 2f ′F ′(f1) +O[f ′(f − f1)] (82)

f ′′ = F ′(f1) +O[(f − f1)]. (83)

Therefore f ′′(ξ1) = F ′(f1). Since f ′(ξ1) = 0

f(ξ) = f1 +1

2(ξ − ξ1)2 F ′(f1) +O[(ξ − ξ1)3] (85)

(ii) If f1 is a double zero, then

(f ′)2 = (f − f1)2F ′′(f1) +O[(f − f1)3]. (86)

This equation can be solved only if F (f1)′′ > 0. This time we

obtain

f(ξ)− f1 ∼ α exp[

±ξ (F ′′(f1))1/2]

as ξ → ∓∞ (88)

The solution extends from −∞ to ∞ and can have only one peak.

(iii) If f1 is a triple zero, then f1 = −c/6, A = 3(c/6)2 and

B = (c/6)3. Then we have

f ′ = ±√

ξ = ±∫

f + c6

)3/2. (90)

Finally, we obtain

f(ξ) = − c6

(ξ − β), (92)

where β is a constant of integration. This solution al ways diverges

at ξ = β.

• In the cases (a), (d), (e), (f) the solution is always un-

bounded.

• In case (b), F (f) has a double zero at f1 and a simple

zero at f3. f(ξ) has a simple minimum at f3 and attains

its maximum f1 exponentially as ξ → ±∞. This is the

solitary wave.

• In case (c), F (f) has simple zeros at f3, f2, so f has a

simple maximum at f2 and a simple minimum at f3 with

motion of period

dξ = 2

∫ f3

dfdf = 2

∫ f3

2F (f). (94)

These periodic solution are called cnoidal waves, because

they can be expressed in terms of the Jacobian of the elliptic

function cn (when F is the cubic of the KdV equation).

Consider

f ′′ = −dV/df

for a given ‘potential’ function V (f), where f ′ = df/dξ. By writing

f ′′ =df ′

dff ′ (95)

and integrating w.r.t. f the equation in the red box we obtain

2(f ′)

2= E − V (f) = F (f) (97)

For some constant of integration (energy) E.

General features of f(ξ)

• There exist a periodic solution if F (f) is positive between

two simple zero.

• There exist solitary-wave solutions if F (f) is positive be-

tween a simple zero and a double zero of F (f).

• If F (f) is positive between two double zeros, f1 and f2,

then f(ξ) → f1 as ξ → ±∞ and f(ξ) → f2 as ξ → ∓∞.

These solutions are called kink or topological solitons

(sine-Gordon equation).

Example

Consider the sine-Gordon equation

utt − uxx + sinu = 0

Look for a solution of the form u = f(ξ), where ξ = x− ct for some

given constant c.

Substituting f(ξ) in the sine-Gordon equation yields

c2f ′′ − f ′′ + sin f = 0. (98)

By using the identity f ′′ = f ′df ′/df we obtain

(c2 − 1)d[

12 (f ′)2

df+ sin f = 0. (99)

Integration with respect to f gives

(c2 − 1)1

2(f ′)2 − cos f = const. (100)

(c2 − 1)1

2(f ′)2 −

1− 2 sin2 f

= const. (101)

(c2 − 1)1

2(f ′)2 + 2 sin2 f

2= A (102)

For some arbitrary constant A.

In our previous notation we have

2(f ′)2 = F (f) =

2 sin2 12f −A

1− c2 . (104)

Let us set A = 0

F (f) > 0 only if 0 < c2 < 1

-15 -10 -5 5 10 15f

Figure 7: F (f) against f for the sine-Gordon equation with A = 0

and c2 = 1/2.

Now we have

F ′(f) =2 sin f

2 cos f2

1− c2 (105)

F ′′(f) =cos2 f

2 − sin2 f2

1− c2 =cos f

1− c2 . (106)

F (f) = 0, at f = 2πm, m = ±1,±2, . . . (110)

F ′(2πm) = 0 (111)

F ′′(2πm) =1

1− c2 6= 0 (112)

All the zeros are double. The solutions are kinks.

Now, let us make the substitution

v = tan

. (113)

Differentiating w.r.t. ξ yields

v′ =1

4f ′ sec2

√1− c2

= ±sin(

√1− c2

= ± v√1− c2

Finally, we have

f(ξ) = 4 arctan

± exp

± (ξ − ξ0)√1− c2

f(ξ) = 4 arctan

∓ exp

± (ξ − ξ0)√1− c2

If both signs are the same, we have a positive kink, if they differ

we have a negative kink (antikink)

-3 -2 -1 1 2 3X

(a) A kink

-3 -2 -1 1 2 3X

(b) An antikink

Figure 8: Kink and antikink of the sine-Gordon equation with A = 0

and c2 = 1/2

Consider the following solution of the sine-Gordon equation:

=c sinh

x√1−c2

ct√1−c2

] , 0 < c2 < 1 (120)

Let us set a = 1/√

1− c2 and v = tan (f/4).

We have

v ∼ ceax − e−ax

e−act∼ c

ea(x+ct) − e−a(x−ct))

, t→ −∞. (121)

This implies

cea(x+ct) if x > −ctcea(x+ct) − ce−a(x−ct) if ct < x < −ct−ce−a(x−ct) if x < ct

, t→ −∞. (122)

Similarly, we have

v ∼ ceax − e−ax

eact∼ c

ea(x−ct) − e−a(x+ct))

, t→∞. (123)

This implies

cea(x−ct) if x > ct

cea(x−ct) − ce−a(x+ct) if −ct < x < ct

−ce−a(x+ct) if x < −ct, t→∞. (124)

-15 -10 -5 5 10 15

(a) The solution (120) as t →

−∞

-15 -10 -5 5 10 15

(b) The solution (120) as t →

Figure 9: The solution (120) of the sine-Gordon equation as t→ −∞and as t→∞.

Example

The Gardner equation

ut − 6uux + uxxx = 12δu2ux.

We look for a solution of form u(x, t) = f(ξ), where ξ = x− ct. We

then go through exactly the same procedure as for the KdV

equation.

We have

−cf ′ − 6ff ′ + f ′′′ − 12δf2f ′ = 0 (125)

−cf − 3f2 + f ′′ − 4δf3 = A (126)

−cff ′ − 3f2f ′ + f ′f ′′ − 4δf3f ′ = Af ′ (127)

2cf2 − f3 +

2(f ′)

2+ δf4 = Af +B (128)

2(f ′)

2= δf4+f3+

2cf2+Af+B = F (f) (130)

-3 -2 -1 1 2 3

(a) One kink solution

-2 -1 1 2

(b) One periodic solution

-3 -2 -1 1 2 3

(c) Soliton solutions

-3 -2 -1 1 2 3

(d) Soliton solution

Figure 10: Positive δ

-3 -2 -1 1 2 3

(a) Soliton solution

-2 -1 1 2

(b) Periodic solution

-3 -2 -1 1 2 3

(c) No physical solution

-3 -2 -1 1 2 3

(d) Periodic solution

Figure 11: Negative δ

3.1 The Scattering Problem

Consider the KdV equation

We are looking for a general procedure to integrate this equation.

Let us introduce the Miura transformation

u = v2 + vx

Direct substitution leads to

2vvt + vxt − 6(v2 + vx)(2vvx + vxx)

+ 6vxvxx + 2vvxxx + vxxxx = 0, (1)

which can be rearranged to give

2v +∂

(vt − 6v2vx + vxxx) = 0. (2)

The equation

vt − 6v2vx + vxxx = 0

is called modified KdV equation or mKdV.

The Miura transformation

u = v2 + vx (3)

is also known as Riccati equation for v, and can be linearized by

the substitution v = ψx/ψ:

ψxx − uψ = 0. (5)

We now observe that the KdV equation is Galilean invariant ,

u→ λ+ u(x+ 6λt, t), −∞ < λ <∞. (6)

The equation for ψ now becomes

ψxx + (λ− u)ψ = 0, −∞ < x <∞ (8)

This is the time-independent Scrodinger equation. The

eigenvalue problem defined by the parameter λ is the scattering

problem (or Sturm-Liouville problem).

u(x, 0)scattering−−−−−−→ S(0)

ytime evolution

u(x, t)inverse←−−−−−−

scatteringS(t)

The diagram of the inverse scattering for the KdV equation.

Example

Consider the equation

ut + ux + uxxx = 0.

Suppose we are given the initial-value problem u(x, 0) = f(x).

f(x) =

∫ ∞

−∞A(k)eikxdk and A(k) =

∫ ∞

−∞f(x)e−ikxdx. (9)

Here A(k) plays the role of the

‘scattering data’.

Now, the dispersion relation for the previous equation is

ω(k) = k − k3.

Therefore the solution to our linear PDE is

u(x, t) =

∫ ∞

−∞A(k)ei(kx−ω(k)t)dk. (11)

Use of the Fourier transform to solve linear PDE

1. Initial-value problem u(x, 0) = f(x);

2. apply the Fourier transform to f(x) to determine the ‘scat-

tering data’ A(k);

3. time evolution of the scattering data given by A(k)e−iω(k)t;

4. reconstruct u(x, t) by applying the inverse Fourier trans-

form to A(k)e−iω(k)t.

u(x, 0)F.T.−−−−→ A(k)

ytime evolution

u(x, t)inverse←−−−−F.T.

A(k)e−iω(k)t

Diagram of the use of the Fourier transform to solve linear PDEs

In order that appropriate solutions exist we shall require

∫ ∞

−∞|u(x)| dx <∞, (12)

∫ ∞

−∞(1 + |x|) |u(x)| dx <∞ (13)

and∫ ∞

−∞|ψ(x)|2 dx <∞. (14)

We shall also assume ψ and ψx to be continuous.

λ is called eigenvalue and ψ(x;λ) the relative eigenfunction.

The operator

L =∂2

∂x2− u (15)

is linear. The eigenvalue problem can be written Lψ = −λψ.

The function space H, ψ ∈ H is an infinite dimensional linear space

also called Hilbert space. The scalar product (ψ, φ) of ψ, φ ∈ H is

defined by

(ψ, φ) =

∫ ∞

−∞ψ(x)φ∗(x)dx (16)

Because u is integrable, u→ 0 as x→ ±∞. Therefore

ψxx ∼ −λψ, x→ ±∞ (17)

If λ is negative

ψ(x) ∼

αe(−λ)1/2x as x→ −∞βe−(−λ)1/2x as x→∞.

The λ with this property are discrete. This constitute the dis-

crete spectrum.

If λ is positive the ‘eigenfunctions’ are asymptotically a linear

combination of e±iλ1/2x.

This is the continuum spectrum.

The ‘eigenfunctions’ e±iλ1/2x are not square-integrable!

ψ(x) =1√2π

∫ ∞

−∞a(k)eikxdk, (19)

where λ1/2 = k.

• Discrete spectrum: κn = (−λn)1/2 and κ1 < κ2, . . . < κN .

ψn is characterized by

ψn(x) ∼ cn exp(−κnx), x→∞. (22)

Furthermore∫∞−∞ |ψn|2 dx = 1.

• Continuum spectrum. We shall consider eigenfunctions of

the type

ψ(x; k) ∼

e−ikx + beikx as x→∞ae−ikx as x→ −∞.

-4 -2 2 4x

Incident wave

Reflected wave

Transmitted wave

Figure 1: Schematic representation of incident, reflected and trans-

mitted wave

It can be shown that

• If u(x) ≥ 0, −∞ < x <∞ there is no discrete spectrum;

• if u(x) ≤ 0, −∞ < x < ∞ and u(x) → 0 ‘sufficiently

rapidly’, then there is only a finite number of discrete eigen-

values;

• |a|2 + |b|2 = 1 (in quantum mechanics this is the conserva-

tion of probability).

Consider two different discrete eigenfunctions (for the same u):

ψ′′m − (κ2

m + u)ψm = 0, ψ′′n − (κ2

n + u)ψn = 0. (24)

Therefore, we have

(κ2n − κ2

m)ψnψm = ψmψ′′n − ψnψ

′′m =

dxW (ψm, ψn), (25)

where W (ψm, ψn) is the Wronskian of ψm, ψn. By integrating we

[W (ψm, ψn)]∞−∞ = (κ2n − κ2

∫ ∞

−∞ψmψndx. (26)

This implies∫∞−∞ ψmψndx = 0. That is, ψm and ψn are

orthogonal .

Example

Wave in an inhomogeneous medium

One-dimensional propagation of sound or light in an

inhomogeneous medium is governed by

∇2φ− 1

∂2φ

∂t2= 0. (28)

We then insert in the above equation a normal mode

φ(x, t) = ψ(x)ei(ly+mz−ωt). (29)

This yields to

dx2− (l2 +m2)ψ(x) = −ω

c2ψ(x). (30)

We then define

λ = −(

l2 +m2)

, u(x) = − ω2

c2(x). (31)

Therefore (30) becomes

ψ′′ + (λ− u(x))ψ(x) = 0. (32)

If now

c(x)→ c∞ as x→∞, (33)

we coud redefine

λ =ω2

c2∞−(

l2 +m2)

, u(x) = ω2

c2∞− 1

. (34)

Then we have that

u(x)→ 0 as x→∞. (35)

Example

The δ function potential

Consider

u(x) = −V δ(x),

where δ(x) is Dirac’s delta function.

Integrating once the Sturm-Liouville equation

ψxx + (λ− u)ψ = 0, −∞ < x <∞ (36)

yields

ψ′(ǫ)− ψ′(−ǫ) = −∫ ǫ

(V δ(x) + λ)ψ(x)dx

= −V ψ(0)− λ∫ ǫ

ψ(x)dx.

Finally, as ǫ→ 0 we have

limǫ→0

(ψ′(ǫ)− ψ′(−ǫ)) = −V ψ(0). (39)

ψ′ is discontinuous at the origin.

For x > 0 and x < 0, ψ obeys the free particle equation

ψ′′ + λψ = 0. (40)

Since the eigenfunctions must be square-integrable, we have

ψn(x) =

cn exp (−κnx) if x ≥ 0,

dn exp (κnx) if x < 0.(42)

By continuity at x = 0, cn = dn. Then, the normalization condition

leads to

−∞|cn|2 exp(2κnx)dx+

∫ ∞

|cn|2 exp(−2κnx)dx =

|cn|2κn

Therefore, up to an arbitrary phase, we have

cn =√κn (44)

We can now find the only discrete eigenvalue:

limǫ→0

(ψ′(ǫ)− ψ′(−ǫ)) = −2cnκn limǫ→0

exp(−κnǫ) = −V cn, (45)

cn =√κn.

Finally, we have

κn =V

or equivalently λ1 = −V 2/4.

Continuum spectrum

We have that λ > 0. Consider a wave of unit amplitude incident

from the left:

ψ(x) =

e−ikx + beikx if x > 0

ae−ikx if x < 0as x→∞, (49)

Since ψ is continuous, we have

a = 1 + b (50)

Using again

limǫ→0

(ψ′(ǫ)− ψ′(−ǫ)) (51)

yields

−ik + ibk − (−aik) = −V (1 + b). (52)

Finally, we obtain

b(k) = − V

V + 2ik. (54)

The transmission coefficient a is then equal to

a(k) = 1 + b(k) =2ik

V + 2ik. (55)

Note that

|b(k)|2 + |a(k)|2 =V 2

V 2 + 4k2+

V 2 + 4k2= 1. (57)

Example: u(x) = −2 sech2 x.

If u(x) = −2 sech2 x then it may be verified that

λ1 = −1, ψ1(x) ∝ sechx (59)

This is the only eigenfunction (N=1).

To normalize, let us set ψ1(x) = a sechx:

∫ ∞

−∞|a|2 sech2 xdx = |a|2 [tanhx]∞−∞ = 2 |a|2 . (60)

Therefore, up to a phase factor, we have a = 1/√

2 It can

also be shown that

a(k) =ik − 1

ik + 1, b(k) = 0 ∀k (62)

u(x) = −2 sech2 x is a reflectionless potential.

Example: u(x) = −6 sech2 x.

If u(x) = −6 sech2 x, it can be shown that N = 2 and

λ1 = −1, ψ1(x) =

2tanhx sechx (66)

λ2 = −4, ψ2(x) =

2sech2 x, (67)

b(k) = 0 ∀k (68)

Example: u(x) = −V sech2 x

The Sturm-Liouville equation is now

ψ′′ + (λ+ V sech2 x)ψ = 0 (70)

In order to solve this equation we make the substitution

y = tanhx, −1 < y < 1 for −∞ < x <∞.

Therefore, we have

dx= sech2 x

dy= (1− y2)

dy(71)

and so

(1− y2)d

1− y2) d

λ+ V(

1− y2)]

ψ = 0. (72)

1− y2) d

(1− y2)

ψ = 0. (74)

This is the associated Legendre equation.

Discrete spectrum

First, suppose that V = N(N + 1) .

If λ = −κ2(< 0), the bound solutions occur when

κn = n, n = 1, 2, . . .N. (75)

The eigenfunctions are proportional to the associated Legendre

functions PnN (y), where

PnN (y) = (−1)n

1− y2)n/2 dn

dynPN (y) (76)

PN (y) =(−1)N

1− y2)N

, (77)

PN (y) being the Legendre polynomial of degree N .

Continuum spectrum

If λ = k2(> 0) we look for solutions which behaves like

ψ(x; k) ∼

e−ikx + b(k)eikx as x→∞a(k)e−ikx as x→ −∞.

For x→ −∞ these are given by

ψ(x; k) = a(k)2ik(sechx)−ikF (a, b; c; (1 + y)/2). (80)

F (α, β, ; γ; z) is the hypergeometric function, which is defined by

the series

F (α, β; γ; z) = 1+

∞∑

α(α+ 1) . . . (α+ n− 1)β(β + 1) . . . (β + n− 1)

γ(γ + 1) . . . (γ + n− 1)n!zn (82)

In our case

2− ik +

2− ik −

c = 1− ik (85)

It is fairly easy to show that

ψ(x; k) ∼ a(k)e−ikx as x→ −∞. (86)

It can also be shown that

ψ(x; k) ∼ aΓ(c)Γ(a+ b− c)Γ(a)Γ(b)

e−ikx

+aΓ(c)Γ(c− a− b)Γ(c− a)Γ(c− b)

eikx as x→∞.

Comparing the previous expressions with (78) yields

a(k) =Γ(a)Γ(b)

Γ(c)Γ(a+ b− c)and b(k) =

a(k)Γ(c)Γ(c− a− b)Γ(c− a)Γ(c− b)

We now want to show that b(k) = 0, i.e. this potential is

reflectionless.

We need the identity

2− z)

cosπz. (88)

Hence observe that

Γ(c− a)Γ(c− b) (89)

= π/ cos

It follows that b(k) = 0 if

= N +1

or equivalently

V = N(N + 1). (94)

The case V 6= N(N + 1)

The two coefficients a(k) and b(k) have poles where Γ(a) and Γ(b)

have poles.

This happens at b = −m, m = 0, 1, . . . or

. (96)

There is a finite number of discrete eigenvalues if

2i.e. V > 0. (98)

Their number is

+ 1 (100)

where [z] denotes the integral part of z ( the greatest integer ≤ z).

The eigenvalues are given by

= µ (102)

The eigenfunctions are the associated Legendre functions

Pµν (y), where ν is a solution of the equation

V = ν(ν + 1). (103)

3.2 The Inverse Scattering Problem

We have seen how from the KdV equation

ut − 6uux + uxxx = 0 (104)

we can get to the Schrodinger equation

ψxx + (λ− u)ψ = 0, −∞ < x <∞ (105)

via the Miura transformation

u = v2 + vx (106)

and the substitution v = ψx/ψ

Suppose that we are given u(x) and

ψ′′ + (λ− u)ψ = 0, −∞ < x <∞ (107)

The direct scattering problem is to deduce the scattering data, i.e.

The eigenfunctions and eigenvalues

λm = −κ2m, ψm(x) m = 1, 2, . . . , N (108)

transmission and reflection coefficients

a(k) and b(k) ∀λ = k2. (109)

The inverse scattering problem is to deduce u(x) from the scat-

tering data .

The Marchenko equation

Gelfand & Levitan (1951) solved the inverse scattering problem

by use of Fourier transform (with deep and difficult arguments).

Their solution was simplified by Marchenko.

Let us consider the wave equation

φxx − φzz = 0. (111)

We then express φ(x, z) in term of its Fourier transform:

φ(x, z) =1

∫ ∞

−∞ψ(x; k)e−ikzdk (112)

ψ(x; k) =

∫ ∞

−∞φ(x, z)eikzdz (113)

Substituting the above integral into the wave equation yields

ψxx + k2ψ = 0. (115)

Further, let us suppose that we are interested in a solution ψ of the

previous equation such that

ψ ∼ eikx, x→∞. (117)

This is obtained by setting

φ(x, z) = δ(x− z) +K(x, z), (118)

where K(x, z) = 0 if z < x, and obeys the classical wave equation.

By taking the Fourier transform of (118) we have

ψ(x; k) = eikx+

∫ ∞

K(x, z)eikzdz (120)

The above wavefunction has the correct asymptotic value.

NB Note that∫ ∞

−∞δ(x− z)eikz = eikx and

∫ ∞

−∞eik(x−z) = δ(x− z) (121)

Which equation will K(x, z) if we slightly modify (115) to

ψ′′ + (k2 − u)ψ = 0? (123)

The boundary conditions are always

ψ(x; k) = eikx +

∫ ∞

K(x, z)eikzdz (124)

K(x, z) = 0 if z < x and

ψ ∼ eikx, x→∞. (125)

It can be shown that

Kxx(x, z)−Kzz(x, z)− u(x)K(x, z) = 0 for z > x (126)

u(x) = −2dK

dx= −2 Kx(x, x) +Kz(x, x) (128)

with the condition

K(x, z), Kz(x, z)→ 0 as z →∞. (129)

Here K(x) = K(x, x).

The equation which one then tries to invert is

ψ = ψ∗ + b(k)ψ. (131)

ψ(x; k) = eikx +

∫ ∞

K(x, z)eikzdz. (132)

ψ has the right asymptotic limit:

ψ(x; k) ∼ e−ikx + b(k)eikx as x→∞. (133)

K(x, z) can be found by solving the integral equation

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

∫ ∞

K(x, y)F (y+z)dy = 0 z > x > −∞,(135)

where K(x, z) = 0 if z < x.

The above equation is the Marchenko equation, is a linear

Fredholm integral equation.

The function F (X) is defined by

F (X) =N∑

c2n exp (−κnX) +1

∫ ∞

−∞b(k)eikXdk (137)

where b(k) is the reflection coefficient, κ2n = −λn and cn are the

normalization coefficients

ψn(x) ∼ cn exp(−κnx), x→∞ (138)

cn = limx→∞

[ψn(x) exp(κnx)] (139)

Inverse scattering problem: summary

Consider the Scrodinger equation

ψ′′ + (λ− u(x))ψ = 0. (140)

Suppose we want to find u(x) and are given the scattering data

ψn(x), λn = −κ2n, n = 1, . . . , N (142)

for the discrete spectrum and the reflection coefficient b(k)

for the continuum spectrum.

u(x) = −2dK

dx= −2 Kx(x, x) +Kz(x, x) (144)

with K(x, z), Kz(x, z)→ 0 as z →∞ and K(x, z) = 0 if z < x.

Moreover K(x, z) satisfies the Marchenko equation

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

∫ ∞

K(x, y)F (y+z)dy = 0 z > x > −∞,(146)

Furthermore,

F (X) =N∑

∫ ∞

−∞b(k)eikXdk (148)

where cn are the coefficients

ψn(x) ∼ cn exp(−κnx), x→∞ (149)

cn = limx→∞

[ψn(x) exp(κnx)] . (150)

Solution of the Marchenko equation

It can be solved by iteration:

K1(x, z) =

−F (x+ z) if z > x

0 if z < x(151)

K2(x, z) = −F (x+ z) +

∫ ∞

−∞K1(x, y)F (y + z)dy (152)

K3(x, z) = −F (x+ z) +

∫ ∞

−∞K2(x, y)F (y + z)dy. (153)

It can be shown that Kn(x, z)→ K(x, z).

This is the Neumann series

Now, suppose that F (x+ z) is a separable function, i.e.

F (x+ z) =N∑

Xn(x)Zn(z). (155)

The Marchenko equation can therefore be written as

K(x, z)+

Xn(x)Zn(z)+

∫ ∞

K(x, y)Xn(y)dy = 0

The solution therefore must take the form

K(x, z) =N∑

Ln(x)Zn(z). (159)

Upon this substitution for K(x, z) we obtain

Ln(x)Zn(z) +

Xn(x)Zn(z)

Zn(z)N∑

∫ ∞

−∞Zm(y)Xn(y)dy = 0. (160)

In order for the equation to be identically zero in the variables x

and z, each term in the external sum must be identically zero.

Ln(x) +Xn(x) +

∫ ∞

−∞Zm(y)Xn(y)dy = 0. (162)

Reflectionless potentials

Suppose that b(k) = 0 and we have only two discrete eigenvalues

(N = 2):

ψ1(x) ∼ c1 exp (−κ1x) , ψ2 ∼ c2 exp (−κ2x) as x→∞,(2)

κ1 6= κ2.

Then, we obtain

F (X) = c21 exp(−κ1X) + c22 exp(−κ2X). (3)

The Marchenko equation then becomes

K(x, z) + c21 exp [−κ1(x+ z)] + c22 exp [−κ2(x+ z)]

∫ ∞

K(x, y)

c21 exp [−κ1(y + z)] + c22 exp [−κ2(y + z)]

dy = 0

F (X) is obviously separable, therefore we set

K(x, z) = L1(x) exp (−κ1z) + L2(x) exp (−κ2z) , (6)

i.e. Xn(x) = c2n exp(−κnx) and Zn(z) = exp(−κnz)

It follows that L1(x) and L2(x) must satisfy

L1 + c21 exp(−κ1x)

∫ ∞

exp (−2κ1y) dy + L2

∫ ∞

exp [−(κ1 + κ2)y] dy

L2 + c22 exp(−κ2x)

∫ ∞

exp [−(κ1 + κ2)y] dy + L2

∫ ∞

exp (−2κ2y) dy

After having evaluated the integrals the system becomes

Ln+c2n exp(−κnx)+c2n

Lm exp [−(κm + κn)x]

κm + κn= 0, n = 1, 2.

The above system can be written as AL+B = 0, where

L = (L1, L2) and

Bn = c2n exp(−κnx) (11)

The matrix A is a square matrix with elements

Amn = δmn + c2mexp [−(κm + κn)x]

κm + κn, (13)

where δmn is the Kronecker delta.

The solution for L is therefore

L = −A−1B. (14)

Moreover, K(x, x) = ETL where En = exp(−κnx).

We now note that

dxAmn = −c2m exp [−(κm + κn)x] = −BmEn. (15)

Therefore, we obtain

K(x, x) =N∑

EmLm = −N∑

A−1)

A−1 dA

Now, for the 2× 2 case if

, (18)

A−1 =1

ad− bc

d −b−c a

After trivial algebra we obtain

A−1 dA

ad− bc

a′d− bc′ · · ·· · · −cb′ + ad′

. (20)

Finally, we have

A−1 dA

ad− bc (a′d+ d′a− b′c− bc′)

d detA

dxlog detA. (22)

We can now evaluate u(x)

u(x) = −2dK

dx= −2

dx2log detA. (24)

In the case with just two discrete eigenvalues, we have

detA =

1 +c212k1

exp(−2κ1x)

1 +c222k2

exp(−2κ2x)

− c21c22

(κ1 + κ2)2 exp [−2(κ1 + κ2)x] . (25)

If we set c2 = 0, we then obtain

u(x) = − 4κ1c21 exp(−2κ1x)

exp(−2κ1x)]2

= −2κ21 sech2 [κ1x+ x0] ,

where exp(x0) = (2κ1)1/2/c1. If κ1 = 1 and c1 =

√2, then we

recover

u(x) = −2 sech2 x. (28)

Reflection coefficient with one pole

Suppose the scattering data are given by

b(k) = − β

β + ikand ψ(x) ∼ β1/2e−βx as x→∞ (30)

where β > 0. b(k) has a simple pole at k = iβ, therefore there is one

discrete eigenvalue which is κ1 = β and c1 = β1/2. We then have

F (X) = βe−βX − β

∫ ∞

−∞

β + ikdk. (31)

The integral can be calculated easily using Cauchy’s residue

theorem:∫ ∞

−∞

β + ikdk = 2πe−βX , X > 0 (32)

and∫ ∞

−∞

β + ikdk = 0 X < 0. (33)

F (X) becomes simply

F (X) = βe−βXH(−X), (35)

where H(−X) is the Heaviside step function.

Figure 1: The contour of the integral (32) in the complex plain..

From the Marchenko equation

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

∫ ∞

K(x, y)F (y+z)dy = 0 z > x > −∞, (36)

it follows that

K(x, z) = 0 for x+ z > 0. (37)

The Marchenko equation now becomes, for x+ z < 0,

K(x, z) + βe−β(x+z) + β

∫ −z

K(x, y)e−β(y+z)dy = 0 (39)

Integrating by parts (on remembering that F (y + z) = 0 for

y + z > 0) yields

K(x, z) + βe−β(x+z) +K(x, x)e−β(x+z) −K(x,−z)

∫ −z

Ky(x, y)e−β(y+z)dy = 0. (40)

The solution is

K(x, z) = −β.

Finally, we have

K(x, z) = −βH(−x− z) and so K(x, x) = −βH(−2x).

The required potential is therefore

u(x) = 2βd

dxH(−2x) = −2βδ(x). (42)

It coincides with the previous example by setting β = V/2.

4. The initial-value problem for the KdV equation

Recapitulation

The potential function u(x) for the Sturm-Liouville

equation

ψ′′ + (λ− u(x))ψ = 0, −∞ < x <∞ (44)

can be reconstructed from the scattering data.

Scattering data:

• Discrete spectrum (λ < 0):

ψn(x), λn = −κ2n, n = 1, . . . , N (48)

ψn(x) ∼ cn exp(−κnx) as x→∞. (49)

• Continuum spectrum (λ > 0):

ψ(x; k) ∼

e−ikx + b(k)eikx as x→∞a(k)e−ikx as x→ −∞.

k = λ1/2, a(k) and b(k) are the transmission and reflection

coefficients respectively: |b(k)|2 + |a(k)|2 = 1.

Then we showed that

u(x) = −2dK(x, x)

dx, (52)

where K(x, z) is the solution to the Marchenko equation

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

∫ ∞

K(x, y)F (y+z)dy = 0 z > x > −∞, (53)

K(x, z) = 0 if z < x. Moreover, F (X) is defined by

F (X) =N∑

∫ ∞

−∞b(k)eikXdk. (54)

Inverse scattering and the KdV equation

Consider the

Introducing the Miura transformation

u = v2 + vx,

the KdV equation becomes

2v +∂

(vt − 6v2vx + vxxx) = 0. (55)

By making the substitution v = ψx/ψ and by applying the

transformation

u→ λ+ u(x+ 6λt, t), −∞ < λ <∞. (56)

the Miura transformation becomes

ψxx + (λ− u(x, t))ψ = 0, −∞ < x <∞. (57)

ψ(x; t), and therefore also λ(t), depend parametrically on time

because v(x, t) is a solution of the mKdV equation (55) and

therefore u(x, t) of the KdV equation.

Inverse scattering transform for the KdV equation

1. Initial-value problem u(x, 0) = f(x);

2. solve the scattering problem with potential f(x), and de-

termine the scattering data S(0) (cn(0), κn(0) and b(k; 0))

at t = 0;

3. determine the time evolution of the scattering data S(t);

4. reconstruct u(x, t) solving the inverse scattering problem

for S(t).

u(x, 0)scattering−−−−−−→ S(0)

ytime evolution

u(x, t)inverse←−−−−−−

scatteringS(t)

The diagram of the inverse scattering for the KdV equation.

Time evolution of the scattering data

We begin once again from the Sturm-Liouville problem for ψ(x; t):

ψxx + (λ− u(x, t))ψ = 0, −∞ < x <∞. (58)

We then differentiate the above equation w.r.t. x,

ψxxx − uxψ + (λ− u)ψx = 0, (59)

and w.r.t. t,

ψxxt + (λt − ut)ψ + (λ− u)ψt = 0. (60)

Here u(x, t) satisfies the KdV equation.

It is now convenient to define

R(x, t) = ψt + uxψ − 2(u+ 2λ)ψx. (62)

We now construct the identity

∂x(ψxR − ψRx) = ψxx (ψt + uxψ − 2uψx − 4λψx)

− ψ(ψxxt + uxxxψ − 3uxψxx − 2uψxxx − 4λψxxx). (63)

The next step consists of eliminating ψxxt and ψxxx using

equations (59) and (60).

This leads to

∂x(ψxR − ψRx) = ψxx (ψt − 2uψx − 4λψx)−ψ (uxxxψ − 4uxψxx)

− ψ (uψt − λψt − λtψ + utψ) + ψ (2u+ 4λ) (uxψ − λψx + uψx) .

We now use the Schrodinger equation

ψxx + (λ− u(x, t))ψ = 0 (65)

to simplify (64).

This yields

∂x(ψxR − ψRx) = ψ2 (λt − ut + 6uux − uxxx) . (66)

Finally, since u(x, t) satisfies the KdV equation, we have

∂x(ψxR − ψRx) = λtψ

2. (68)

This equation can now be used to obtain the time evolution of

the scattering data.

The discrete spectrum

We now choose λ = −κ2n and ψ = ψn, n = 1, . . .N . By integrating

the equation

∂x(ψnxRn − ψnRnx) = −(κn)2tψ

2n (69)

w.r.t. x, we have

[ψnxRn − ψnRnx]∞−∞ = −(κn)2t

∫ ∞

−∞ψ2

n(x)dx = −(κn)2t . (70)

Now the L.H.S. is zero, therefore

(κn)2t = 0 or κn = constant.

We now want to determine the time evaluation of cn.

Integrating (69) w.r.t. x gives

ψnxRn − ψnRnx = gn(t), (71)

where gn(t) are arbitrary functions of t. But

Rn, ψn → 0 as |x| → ∞. (72)

Therefore gn(t) = 0 ∀n, t.

We now integrate again

ψnxRn − ψnRnx (73)

and obtain (using integration by parts)

Rn/ψn = hn(t), (74)

where hn(t) (n = 1, 2, . . . , N) are arbitrary functions too.

By multiplying (74) by ψ2n, we obtain

ψn(ψnt + uxψn − 2uψnx + 4κ2nψnx) = hnψ

2n (75)

or, using ψnxx − (κ2n + u(x, t))ψn = 0

2(ψn)2t +

uψ2n − 2ψ2

nx + 4κ2nψ

x= hnψ

2n. (76)

We now integrate (76) w.r.t. x:

(∫ ∞

−∞ψ2

∫ ∞

−∞ψ2

ndx (77)

This implies that hn(t) = 0 ∀n, t.

It follows that

Rn = ψnt + uxψn − 2(

u− 2κ2n

ψnx = 0.

This is the time-evolution equation for ψn(x; t).

The previous equation (red box) can be used to find the time

evolution of the cn(t).

We know that

u→ 0 and ψn(x; t) ∼ cn(t) exp(−κnx) as x→∞. (78)

This asymptotic behaviour inserted in the equation for ψn(x; t)

(red box) gives

dcndt− 4κ3

ncn = 0 or cn(t) = cn(0) exp(4κ3nt), (80)

where cn(0) are the normalization constants determined at t = 0.

The continuous spectrum

We start again from equation

∂x(ψxR − ψRx) = λtψ

2. (82)

By integrating w.r.t. x over R we obtain that

λt = 0 or λ = constant.

By integrating once Eq. (82) gives

ψxR− ψRx = g(t; k),

where R is R evaluated in terms of ψ and g(t; k) is a function of

integration.

R is defined by

R(x, t) = ψt + uxψ − 2(u+ 2λ)ψx. (83)

For the continuous eigenfunctions we have

ψ(x; t, k) ∼ a(k; t)e−ikx as x→ −∞ (84)

Therefore, we have that

R(x, t; k) ∼(

dt+ 4ik3a

e−ikx, x→ −∞. (86)

As a consequence

ψxR− ψRx → 0, as x→ −∞. (87)

Thus g(t; k) = 0 for all t.

By integrating (87) once more we have

R/ψ = h(t; k) or R = hψ. (88)

Now, using again

ψ(x; t, k) ∼ a(k; t)e−ikx as x→ −∞ (89)

we have that

dt+ 4ik3a = ha. (91)

The behaviour of R as x→∞ is given by

R(x, t; k) ∼ db

dteikx + 4ik3(e−ikx − eikx) as x→∞. (92)

where we have used

ψ ∼ e−ikx + b(k; t)eikx. (93)

Substituting (92) into R = hψ yields

dteikx + 4ik3(e−ikx − eikx) = h(e−ikx − eikx). (94)

Because eikx and e−ikx are linearly independent the above equation

impliesdb

dt− 4ik3b = hb and h(t; k) = 4ik3. (95)

Finally, we haveda

dt= 0 and

dt= 8ik3b (96)

whose solutions are

a(k; t) = a(k; 0) and b(k; t) = b(k; 0) exp(8ik3t).

Evolution of the scattering data: summary

• κn = constant, cn(t) = cn(0) exp(4κ3nt);

• b(k; t) = b(k; 0) exp(8ik3t).

Construction of the solution of the KdV equation:

summary

We want to integrate the KdV equation

ut − 6uux + uxxx = 0, t > 0, −∞ < x <∞,

with initial condition u(x, 0) = f(x).

Step 1: We solve the Sturm-Liouville equation

ψxx + (λ− u)ψ = 0, −∞ < x <∞.

with u(x, 0) = f(x).

i.e. We determine the discrete spectrum −κ2n, the normalization

constants cn(0), and the reflection coefficient b(k; 0).

Step 2: The time evolution of the scattering data is given by

• κn = constant;

• cn(t) = cn(0) exp(4κ3nt);

• b(k; t) = b(k; 0) exp(8ik3t).

Step 3: We now want to solve the Marchenko equation

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

∫ ∞

K(x, y; t)F (y+ z; t)dy = 0, (98)

F (X ; t) =N∑

cn(0)2 exp(

8κ3nt− κnX

∫ ∞

−∞b(k; 0) exp(8ik3t+ ikX)dk (99)

Finally, the solution of the KdV equation can be expressed as

u(x, t) = − ∂

∂xK(x, t) and K(x, t) = K(x, x, t). (101)

We have reduced the solution of a nonlinear partial differen-

tial equation to that of solving two linear problems (a second

order ODE and an integral equation).

4.1 Reflectionless potentials

Solitary wave

We obtain the solitary wave by posing a suitable initial-value

problem, without the assumption that the solution takes the form

of a steady progressing wave.

The initial profile is taken to be

u(x, 0) = −2 sech2 x.

The Sturm-Liouville equation at t = 0 is

ψxx +(

λ+ 2 sech2 x)

ψ = 0.

We have already studied this scattering problem. We have

b(k) = 0 and the discrete spectrum has only one eigenvalue

κ1 = 1 and ψ1(x) = 2−1/2 sechx.

Moreover, we have

ψ(x) ∼ 21/2e−x, x→∞, so c1(0) = 21/2 . (102)

Now we have c1(t) = 21/2e4t, and therefore

F (X ; t) = 2e8t−X .

The Marchenko equation now becomes

K(x, z; t)+2e8t−(x+z)+2

∫ ∞

K(x, y; t)e8t−(y+z)dy = 0. (104)

Since F (X ; t) is separable we can set K(x, z; t) = L(x, t)e−z.

Therefore, the equation for L(x, t) becomes

L+ 2e8t−x + 2Le8t

∫ ∞

e−2ydy = 0. (105)

The above equation can be solved directly to yield

L(x, t) = − −2e8t−x

1 + e8t−2x. (107)

The potential is then given by

u(x, t) = 2∂

2e8t−2x

1 + e8t−2x

= − 8e2x−8t

(1 + e2x−8t)2

= −2 sech2(x− 4t).

This is the solitary wave solution of amplitude −2 at speed of

propagation 4.

Two solitons solution

The initial profile is now taken to be

u(x, 0) = −6 sech2 x.

ψxx +(

λ+ 6 sech2 x)

ψ = 0.

The discrete spectrum of this equation is given by

κ1 = 1 and κ2 = 2 (112)

ψ1(x) =

tanhx sechx and ψ2(x) =

2sech2 x

The asymptotic behaviour of these solutions is given by

ψ(x) ∼√

6e−x, 2√

3e−2x as x→∞. (114)

Therefore, we have

c1(0) =√

6 c2(0) = 2√

3 (115)

c1(t) =√

6e4t c2(t) = 2√

3e32t (116)

This potential is reflectionless, therefore b(k; 0) = b(k; t) = 0.

The function F (X ; t) is now

F (X ; t) = 6e8t−X + 12e64t−2X . (118)

The Marchenko equation is therefore

K(x, z; t) + 6e8t−(x+z) + 12e64t−2(x+z)

∫ ∞

K(x, y; t)(

6e8t−(y+z) + 12e64t−2(y+z))

dy = 0 (119)

The solution K(x, z; t) must take the form

K(x, z; t) = L1(x, t)e−z + L2(x, t)e

−2z (120)

Inserting the above expression into (119) and collecting the

coefficients of e−z and e−2z, we obtain the pair of equations

L1 + 6e8t−x + 6e8t

∫ ∞

e−2ydy + L2

∫ ∞

e−3ydy

L2 + 12e64t−x + 12e64t

∫ ∞

e−3ydy + L2

∫ ∞

e−4ydy

Evaluating the integrals yields

L1 + 6e8t−x + 3L1e8t−2x + 2L2e

8t−3x = 0 (123)

L2 + 12e64t−2x + 4L1e64t−3x + 3L2e

64t−4x = 0. (124)

The previous system can be easily solved to give

L1(x, t) = 6(

e72t−5x − e8t−x)

/D (125)

L2(x, t) = −12(

e64t−2x + e72t−4x)

/D, (126)

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

The solution u(x, t) to the KdV equation then becomes

u(x, t) = −2∂

L1e−x + L2e

−2x)

= 12∂

e8t−2x + e72t−6x − 2e64t−4x)]

After a little bit of manipulation it can be simplified to give

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

[3 cosh (x− 28t) + cosh (3x− 36t)]2. (129)

We now want to look at the behaviour of this

solution as t→ ±∞.

We now set ξ = x− 16t, i.e. we follow a wave which moves at speed

16 (if it exists):

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

[3 cosh (ξ − 12t) + cosh (3ξ + 12t)]2 . (130)

Taking the limit as t→ ±∞ yields

u(x, t) ∼ −8 sech2(

2ξ ∓ 12 log 3

as t→ ±∞, ξ = x− 16t.

Similarly, by setting η = x− 4t

u(x, t) ∼ −2 sech2(

2η ± 12 log 3

as t→ ±∞, ξ = x− 4t.

The last two expression can be combined (the error terms are

asymptotically small) to give

u(x, t) ∼ −8 sech2

2ξ ∓ 1

2log 3

− 2 sech2

2η ± 1

2log 3

as t→ ±∞.

Figure 2: Time evolution of the two solitons solution

Figure 3: The paths of the wave crests of the two solitons

Comments

• As t→ −∞ the solution ha the form of two solitons, the taller

travelling faster;

• the taller catches the smaller, they coalesces, they form our

initial profile at t = 0 and then the taller moves away;

• as t→∞ we have two solitons again;

• trace of the nonlinear interaction: after the interaction the two

waves are phase shifted. The taller wave has moved forward,

the shorter backwards.

N solitons solution

The initial profile is given by

u(x, 0) = −N(N + 1) sech2 x.

ψxx +[

λ+N(N + 1) sech2 x]

ψ = 0.

The discrete spectrum of this equation is given by

κn = n, n = 1, . . .N (3)

ψn(x) ∝ PnN (tanhx) (4)

The discrete eigenfunction take the asymptotic form

ψn ∼ cne−nx as x→∞, (5)

where cn can be found using the normalization condition.

The time evolution of the normalization coefficients is given by

cn(t) = cn(0) exp(

The function F (X ; t) is now

F (X ; t) =N∑

cn(0)2 exp(

8n3t− nX)

The Marchenko equation now becomes

K(x, z; t) +N∑

cn(0)2 exp[

8n3 − n(x+ z)]

∫ ∞

K(x, y; t)

c2n(0) exp[

8n3t− n(y + z)]

dy = 0. (9)

The solution for K(x, z; t) must now take the form

K(x, z; t) =N∑

Ln(x, t)e−nx. (10)

Inserting K(x, z; t) into (9) and collecting the coefficients of enz the

integral equation is replaced by an algebraic system:

AL+B = 0.

L and B are the column vectors with elements Ln and

Bn = cn(0)2 exp(

8n3t− nx)

The N ×N matrix A has elements

Amn = δmn +c2m(0)

m+ nexp

8m3t− (m+ n)x]

. (13)

We now from the inverse scattering theory of reflectionless

potentials that

u(x, t) = −2∂2

∂x2log detA. (15)

The asymptotic form of the solution can be determined by setting

ξn = x− 4κ2nt = x− 4n2t (16)

and then taking the limit t→ ±∞. For fixed ξn we have

u(x, t) ∼ −2n2 sech2[

x− 4n2t)

∓ xn

, t→ ±∞. (18)

The phase xn is given by

exp(2xn) =∏

m=1m6=n

n−mm+m

sgn(n−m)

n = 1, 2, . . . , N (19)

We can combine the previous asymptotic solutions to obtain

u(x, t) ∼ −2∑N

n=1 n2 sech2

x− 4n2t)

∓ xn

, t→ ±∞.

We can combine these solutions as if the equation were linear

because the error that we make in doing so is exponentially small

as t→ ±∞.

Figure 1: The three solitons solution with u(x, 0) = −12 sech2 x. (a)

t = 0; (b) t = 0.05; (c) t = 0.2. −u is plotted against x

4.2 General description of the solution

When b(k) 6= 0 the Marchenko equation cannot be solved for

K(x, z; t) in closed form.

In general the function F (X ; t) is

F (X ; t) =N∑

cn(0)2 exp(

8κ3nt− κnX

∫ ∞

−∞b(k; 0) exp(8ik3t+ ikX)dk. (20)

We now consider the contribution to the solution K(x, z; t) of the

integral in the above expression.

The integral

I(X ; t) =1

∫ ∞

−∞b(k; 0) exp(8ik3t+ ikX)dk. (21)

is of the type

I(t) =

∫ ∞

−∞f(k)eiφ(k)tdk, φ(k) ∈ R (23)

The leading order behaviour of this integral as t→∞ can be

determined using the stationary phase approximation.

The main contribution as t→∞ comes from where φ(k) is

stationary and

I(t) ∼ 2Γ(

(n!)1/n

∣φ(n)(c)∣

]1/nf(c)eitφ(c)±iπ/(2n), t→∞. (25)

Here φ(n)(c) is the first non zero derivative at the stationary point

c, and we use the factor eiπ/(2n) if φ(n)(c) > 0 and the factor

e−iπ/(2n) if φ(n)(c) < 0.

Therefore, our integral becomes

I(X ; t) ∼ Γ (1/3) eiπ/6

3b(0; 0)t−1/3, t→∞ (26)

Let consider for a moment an initial condition u(x, 0) with no

discrete spectrum (e.g. positive δ-function or positive sech2 profile.)

F (x, z; t) and therefore K(x, z; t) are of order O(t−1/3).

As a consequenceu(x, t) is of order O(t−1/3) too.

Therefore, as t→∞, the term −6uux in

ut − 6uux + uxxx = 0 (27)

is negligible and therefore, locally , we have

ut ∼ −uxxx, t→∞.

This is a linear equation whose dispersion relation is ω = −k3.

In the absence of the discrete spectrum u(x, t) behaves like a

wave packet propagating (almost) linearly to the left with group

velocity dω/dt = −3k2 and whose amplitude decays as t−1/3.

Figure 2: Solution to the Kdv equation with positive sech2 x initial

condition.

Now, let us consider a system whose initial condition has one

discrete eigenvalue and b(k) 6= 0 (for example a negative

δ-function).

Figure 3: Initial condition with a δ profile.

Example: negative sech2

Let us choose as initial condition

u(x, 0) = −4 sech4 x.

4 is not of the form N(N + 1), and therefore the reflection

coefficient b(k) is not zero.

The number of eigenvalues is given by

+ 1. (28)

Therefore, we have two discrete eigenvalues.

Figure 4: Initial condition given by u(x, 0) = −4 sech4 x.

Recapitulation

• The solution u(x, t) evolves for t > 0 so that κn = const.,

cn(t) = cn(0) exp(4κ3nt) and b(k; t) = b(k; 0) exp(8ik3t).

• The solution separates into two parts as t→∞:

• (i) There is a precession of N solitary waves of depression.

Each wave has positive velocity which is proportional to its

amplitude. They are essentially nonlinear waves which in-

teract like solitons, changing only their phases. They are

associated with the discrete spectrum and with the sum∑N

n=1 cn(t) exp(−κnX) of F (X ; t) in the Marchenko equa-

tion. In the exceptional case of b(k, 0) = 0 ∀k, the preces-

sion of solitary waves is the complete solution as t→∞.

• (ii) If b(k, 0) 6= 0 for some k, then there exists also an

oscillatory wave train as t → ∞. This train has u(x, t)

of both signs for each t. It is essentially linear,

i.e. ut ∼ −uxxx locally as t→∞. It propagates to the left

with group velocity cg = −3k2 < 0. The train is associated

with the continuous spectrum and the integral

∫ ∞

−∞b(k, t)eikXdk (30)

of F (X ; t) in the Marchenko equation.

New definition of soliton

Def.: A soliton is that component of the solution of a nonlinear

evolution equation which depends only upon one constant

discrete eigenvalue of the underlying scattering problem as t→±∞

This definition clarifies what is meant by the ‘identity’ of the

soliton: it is that property which maintains the constancy of the

discrete eigenvalues.

5. Conservation laws

Consider the equation of continuity of a compressible fluid,

∂t+∂(ρu)

∂x= 0, (32)

where ρ(x, t) is the density and u(x, t) is the x-velocity of the fluid.

The above equation expresses the conservation of mass of

fluid .

Now, suppose that ρu→ const. as |x| → ∞, and ρ and (ρu)x are

integrable.

The continuity equation implies that

∫ ∞

−∞ρdx =

∫ ∞

−∞

∂tdx = −

∫ ∞

−∞

∂ (ρu)

∂xdx = − [ρu]∞−∞ = 0.

Therefore the integral

∫ ∞

−∞ρdx = constant (35)

represents the conservation of the total mass in the system.

Example

The conservation of electric charge is expressed by

∂ρ(x, t)

∂t+∇ · j = 0, (36)

where ρ(x, t) is the charge density and j is the density current.

It follows that

∫∫∫

ρ(x, t)dx =

∫∫∫

∂ρ(x, t)

∂tdx

= −∫∫∫

∇ · j dx−∫∫

j · n dS = 0.

This implies that the total charge∫∫∫

ρ(x, t)dx = constant. (38)

A general form of 1-dimensional conservation law is

∂t+∂X

∂x= 0, (40)

where T (x, t, u, ux, . . .) is the density and X(x, t, u, ux, . . .) is the

N.B. They cannot depend on ut.

T and Xx are integrable and

X → constant as |x| → ∞. (41)

The conservation law implies

∫ ∞

−∞Tdx = constant. (43)

The integral of T is conserved, or is a constant of motion.

The KdV equation

Consider the KdV equation,

It is already in the form of conservation law:

uxx − 3u2)

= 0. (45)

The density and the flux are given respectively by

T (x, t) = u(x, t) (46a)

X(x, t) = uxx − 3u2. (46b)

Therefore, we have

∫ ∞

−∞u dx = constant. (48)

The above integral expresses the mass conservation.

If we multiply the KdV equation by u we obtain another

conservation law

uuxx −1

x − 2u3

= 0. (50)

Therefore we have

∫ ∞

−∞u2 dx = constant. (52)

The above integral expresses momentum conservation.

We now construct the density for energy conservation.

Consider

3u2 × (KdV) + ux ×∂

∂x(KdV). (53)

This gives

3u2 (ut − 6uux + uxxx)+ux

uxt − 6u2x − 6uuxx + uxxxx

= 0, (54)

which can be written as

2u4 + 3u2uxx − 6uu2

x + uxuxxx −1

= 0. (55)

The previous expression is in a form of conservation law. Therefore,

we have

∫ ∞

−∞

dx = constant. (57)

This integral expresses the energy conservation for water waves.

In the 1960s 8 more integral of motion for he KdV were found

by trial, ingenuity and error. What is their physical meaning?

Gardner found an infinity of conservation laws.

We introduce the Gardner transformation.

u = w + ǫ2w + ǫwx (59)

The KdV equation now becomes

ut − 6uux + uxxx = wt + ǫwxt + 2ǫ2wwt − 6(

w + ǫwx + ǫ2w2)

wx + ǫwxx + 2ǫ2wwx

+ wxxx + ǫwxxxx + 2ǫ2(wwx)xx

1 + ǫ∂

∂x+ 2ǫ2w

wt − 6(

w + ǫ2w2)

wx + wxxx

. (60)

u is a solution of the KdV equation if w is a solution of

wt − 6(w + ǫ2w2)wx + wxxx = 0

NB The opposite is not necessarily true!

The previous equation can be rewritten as

wxx − 3w2 − 2ǫ2w3)

and therefore

∫ ∞

−∞w dx = constant. (64)

We now set

w(x, t; ǫ) ∼∞∑

ǫnwn(x, t), ǫ→ 0. (66)

This series is asymptotic and does not need to be convergent. Then

∫ ∞

−∞wn dx = constant, n = 0, 1, . . . (68)

• We insert the asymptotic expansion for w in the Gardner

transformation;

• We equates coefficients of ǫn for each n = 0, 1, . . ..

∞∑

ǫnwn ∼ u− ǫ∞∑

ǫnwnx − ǫ2( ∞∑

w0 = u; w1 = −w0x = −ux; w2 = −w1x − w20 = uxx − u2

w3 = −w2x − 2w0w1 = −(uxx − u2)x + 2uux;

w4 = −w3x − 2w0w2 − w21

2uux − (uxx − u2)x

x− 2u(uxx − u2)− u2

The integrals of exact differentials (in x) are zero.

We have

w2m+1 =∂

∂x(something) . (73)

Therefore trivially∫ ∞

−∞w2m+1 dx = 0 (74)

All the non trivial constants of motion are given by

∫ ∞

−∞w2m dx = constant, m = 0, 1, . . . (76)

w0, w2 and w4 are the constants of motions previously found:

∫ ∞

−∞w0 dx =

∫ ∞

−∞u dx = constant, (80)

∫ ∞

−∞w2 dx =

∫ ∞

−∞u2 dx = constant, (81)

∫ ∞

−∞w4 dx =

∫ ∞

−∞

dx = constant. (82)

6. The Lax pair

Question: Is the KdV equation the only evolution equation

with the special properties that we have studied? (i.e. solution

by inverse scattering theory, constancy of the eigenvalues as-

sociated to the direct scattering problem and therefore simple

time-evolution of the scattering data.)

Lax (1968) found a reason underlying the fact that the eigenval-

ues of the scattering problem are constant while the solution of

the KdV equation evolves in a complicated way. The KdV equa-

tion does not stand alone in the class of evolution equations.

Suppose that we want to solve the initial value problem given by

ut = N(u)

with u(x, 0) = f(x) , with u ∈ Y (Y being some

appropriate function space) and N : Y → Y is a

operator independent of t but in general depending on u, x, and on

the derivatives of u w.r.t. x.

N need not be a partial differential operator.

An Hilbert space H is a vector linear space (possibly infinite

dimensional), complete, i.e. each element ψ ∈ H can be ex-

pressed as

ψ =∞∑

cnφn, (84)

where cn are constants and φn is an appropriate basis. Hmust also be equipped with a scalar product (φ, ψ).

A scalar product is a bilinear function H × H → C with the

following properties

(ψ, φ) = (φ, ψ)∗ (89)

(ψ, φ+ ξ) = (ψ, φ) + (ψ, ξ) (90)

(0, ψ) = 0 (91)

(ψ, ψ) ≥ 0, (ψ, ψ) = 0 iff ψ = 0. (92)

Def.: A linear operator L in H is self-adjoint if

(φ, Lψ) = (Lφ, ψ). All the eigenvalues of L are real.

Examples

(i) Any finite-dimensional vector space with the usual scalar

product is an Hilbert space. Linear operators are given by matrices.

(ii)The space of square-integrable functions in the interval [0, 2π),

L2([0, 2π)), is an Hilbert space. Each function f(x) can be

expanded in a Fourier series

f(x) =1

∞∑

n=−∞cne

inx. (93)

The scalar product is defined by

∫ 2π

g∗(x)f(x)dx. (94)

In our case N(u) is the KdV equation

N(u) = 6uux − uxxx

We now suppose that the evolution equation can be expressed as

Lt = ML− LM ,

i.e. ut −N(u) = Lt + LM −ML = Lt + [L,M ] = 0.

M and L are linear operator acting on a Hilbert space H and

which may depend on u(x, t). We also assume that L is self-

adjoint.

We now introduce the spectral equation

Lψ = λψ for t ≥ 0 and −∞ < x <∞.

Because L depends upon t, in general the eigenvalues λ(t) depend

upon t too.

By differentiating w.r.t. t the spectral equation, we have

Ltψ + Lψt = λtψ + λψt. (95)

By using Lt = ML− LM the previous equation becomes

λtψ = (L− λ)ψt + (ML− LM)ψ

= (L− λ)ψt +Mλψ − LMψ

= (L− λ)(ψt −Mψ).

By taking the scalar product with ψ of the above expression, we

(ψ, ψ)λt = (ψ, (L− λ)(ψt −Mψ))

= (((L− λ)ψ, (ψt −Mψ))

= (0, (ψt −Mψ)) = 0.

(L− λ is self-adjoint.)

It follows that λt = 0 and therefore

λ = constant.

From eq. (96) with λt = 0 it also follows that

(L− λ)(ψt −Mψ) = 0.

(ψt −Mψ) is therefore an eigenvector of L with eigenvalue λ.

Therefore, we have

ψt −Mψ = α(t)ψ

where α(t) is a scalar function of t.

We now set

M ′ = M + α(t)I,

where I is the identity operator.

Since L commutes with α(t)I, from the equation Lt = ML−LM it

follows that

Lt = M ′L− LM ′.

We can therefore redefine M by using M ′: this will not alter

Lt = ML− LM (which is a representation of ut −N(u)).

We therefore have the time-evolution equation for ψ

ψt = Mψ.

These results can be summarized in the following

Theorem: If the evolution equation

ut −N(u) = 0 (101)

can be expressed as the Lax equation

Lt + LM −ML = Lt + [L,M ] = 0 (102)

and if Lψ = λψ, then λt = 0 and ψ evolves according to

ψt = Mψ. (103)

6.1 The Lax KdV hierarchy

We now want to apply the previous idea to the KdV equation,

therefore

N(u) = 6uux − uxxx

The problem is How to choose L and M ?

In our case the obvious choice for L is the Schrodinger operator

L = − ∂2

∂x2+ u(x, t) (105)

It turns out that M must be a skew-symmetric operator

(φ,Mψ) = −(Mφ,ψ)

A natural choice is therefore to construct M from a suitable linear

combination of odd derivatives.

Consider the inner product (φ, ψ) =∫∞−∞ φψdx, then

(Mφ,ψ) =

∫ ∞

−∞

∂nφ

∂xnψdx = −

∫ ∞

−∞φ∂nψ

∂xndx = −(φ,Mψ). (106)

Moreover, Lt + [L,M ] must be a multiplicative operator.

Consider the simplest choice

M = c∂

∂x(108)

for c constant, then

[L,M ] = c

− ∂2

∂x2+ u(x, t)

∂x− c ∂

− ∂2

∂x2+ u(x, t)

= −cux

Note that if ∂/∂x and a(x) are two operators, their composition

the ∂/∂x[a(x)] applied to b(x) gives

∂x(a(x)b(x)) = axb+ abx (110)

Finally, we have

Lt + [L,M ] = ut − cux.

The one-dimensional wave equation ut − cux has an

associated spectral problem with eigenvalues which are constant

of motion.

We now choose M so that it involves at most a third-order

differential operator,

M = −α ∂3

∂x3+ U

∂xU +A. (112)

Here α is a constant, U = U(x, t) and A = A(x, t).

After simple algebra, we have

[L,M ] = αuxxx − Uxxx −Axx − 2uxU

+ (3αuxx − 4Uxx − 2Ax)∂

∂x+ (3αux − 4Ux)

It follows that [L,M ] is a multiplicative operator if

4αu and A = A(t) (115)

The Lax equation now becomes

Lt + [L,M ] = ut −3

2αuux +

4αuxxx = 0. (117)

For α = 4 we recover the KdV equation. The operator M now

becomes

M = −4∂3

∂x3+ 3u

∂x+ 3

∂xu+ A(t) (119)

and so the time-evolution equation for ψ is

ψt = −4ψxxx + 3uψx + 3 (uψ)x + Aψ. (120)

The previous equation can be recast, on using the Schrodinger

equation

ψxx + (λ− u)ψ = 0, (121)

ψt = 4 (λψ − uψ)x + 3uψx + 3 (uψ)x +Aψ

= 2 (u+ 2λ)ψx − uxψ +Aψ(122)

For A = 0, this is the time-evolution equation that we found

for the discrete eigenfunctions; with A = 4ik3 we have the cor-

responding equation for the equation for the continuous eigen-

functions.

The KdV equation is the second example in the Lax-formulation

framework with L being the Scrodinger operator. The proce-

dure adopted can be extended to higher-order nonlinear evolu-

tion equations.

Trial and error leads to

M = −α ∂2n+1

∂x2n+1+

Um∂2m−1

∂x2m−1+

∂2m−1

∂x2m−1Um

+A (124)

where α is a constant, Um = Um(x, t) and A = A(t).

The restriction that [L,M ] must be a multiplicative operator

imposes n conditions on the n unknown function Um.

n = 1 gives the KdV equation. It can be shown that for n = 2 the

evolution equation is

ut +30u2ux− 20uxuxx− 10uuxxx +uxxxxx = 0. (126)

section c: solitons - university of sheffield › kul › sectionb.pdf · section c: solitons...

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