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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/ 1
Transcript of SECTION C: SOLITONS - University of Sheffield › kul › sectionB.pdf · SECTION C: SOLITONS...
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/
1
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
2
1. Introduction
Linear PDEs
Consider the PDE
Lu = 0
where L = L(∂/∂x, ∂/∂t) and u = u(x, t).
3
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)
4
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
5
Example: the wave equation
Take
L =∂2
∂t2− c2 ∂
2
∂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)
6
If u(x, 0) = F (x) and ut(x, 0) = G(x) then
u(x, t) =1
2[F (x− ct) + F (x+ ct)] +
1
2c
∫ 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.
7
-5 -2.5 2.5 5 7.5 10
0.2
0.4
0.6
0.8
1
(a)
-5 -2.5 2.5 5 7.5 10
0.2
0.4
0.6
0.8
1
(b)
Figure 1: The D’Alambert solution f(x − ct) of the wave equation
at two different times
8
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 ∂
2
∂x2≡(
∂
∂t± c ∂
∂x
)(
∂
∂t∓ c ∂
∂x
)
(13)
9
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.
10
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.
11
Example: the wave equation
Consider again the wave equation
Du ≡ ∂2u
∂t2− c2 ∂
2u
∂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.
12
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.
13
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.
14
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
2π
∫ ∞
−∞u(x, 0)e−ikxdx (27)
and
u(x, 0) =1
2π
∫ ∞
−∞a(k)eikxdk (28)
15
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)
16
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.
17
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.
18
Example
Consider the following expression:
a cos(k1x− ω1t) + a cos(k2x− ω2t)
= 2a cos
[
1
2(k2 − k1)x−
1
2(ω2 − ω1t)
]
× cos
[
1
2(k2 + k1)x−
1
2(ω2 + ω1) t
]
∼ 2a cos
[
1
2(k2 − k1) (x− cgt)
]
cos [k1 (x− ct)]
= A(ǫx, ǫt) cos(k1x− ω1t) as k2 → k1.
(34)
Here A(ǫx, ǫt) = 2a cos [ǫ (x− cgt)], ǫ = (k2 − k1)/2 is small and
cg = limk2→k1
ω2 − ω1
k2 − k1and c =
ω1
k1=ω2
k2. (35)
19
10 20 30 40
-1.5
-1
-0.5
0.5
1
1.5
2
Figure 2: The ‘wave packet’ (34).
20
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
du
dt= ut +
dx
dtux = ut + cux = 0. (37)
21
Using this property, solution to ux+cut = 0 can be easily
constructed.
If
u(x, 0) = f(x) (38)
for differentiable f , then
u(x, t) = f(x− ct).
22
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
dx
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).
23
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.
24
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.
25
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?
26
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.
27
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
28
we obtain
ut +αβ
γuux +
β
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.
29
The discovery of Solitary waves.
Figure 3: The diagram of Russel’s experiment to generate solitary
waves
Figure 4: Parameters used in the description of solitary waves
30
• 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
∂ζ
∂t=
3
2
( g
h
)1/2(
2
3ǫ∂ζ
∂χ+ ζ
∂ζ
∂χ+
1
3σ∂3ζ
∂χ3
)
(47)
31
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
h
)
t. (49)
Finally, the solitary wave solution becomes
ζ(x, t) = a sech2
[
1
2
(a
σ
)1/2
x− (gh)1/2(
1 +a
2h
)
t
]
. (51)
c ∼ (gh)1/2(
1 +a
2h
)
(52)
32
Def.: A solitary wave solution of a PDE
N
(
∂
∂t,∂
∂x
)
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 → ±∞.
33
Discovery of Solitons
34
• 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
35
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)
36
Figure 5: The solution of the periodic boundary-value problem for
the KdV equation
37
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.
38
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
39
Wave propagation in a classical continuum ⇒ the time evolution is
given by the material derivative:
D
Dt=
∂
∂t+ u
(
∂
∂x
)
. (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.
40
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.
41
2. Waves of permanent form
Travelling waves
Consider the nonlinear PDE
N
(
∂
∂t,∂
∂x
)
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.
42
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 → ±∞
43
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)
−1
2cf2 − f3 +
1
2(f ′)
2= Af +B (71)
44
The condition
f, f ′, f ′′ → 0, as ξ → ±∞
implies
A = B = 0. (72)
Therefore, we have
(f ′)2
= f2(2f + c)
45
ξ =
∫
dξ
dfdf =
∫
df
f ′= ±
∫
df
f (2f + c)1/2(73)
Now we make the substitution
f = −1
2c sech2 Θ, (74)
then we obtain
df = −2f tanhΘdΘ. (75)
46
Finally, the integral becomes
ξ = ±∫
2f tanhΘ
f(
−c sech2 Θ + c)1/2
dΘ = ± 2
c1/2Θ+const. if c > 0. (76)
f(ξ) = −1
2c sech2
[
1
2c1/2 (x− ct− x0)
]
(78)
for aribatrary x0 and c ≥ 0.
47
General waves of permanent form
We have found that
1
2(f ′)
2= f3 +
1
2cf2 + Af +B = F (f) (80)
Or equivalently
f ′ = ±√
2F (f)
48
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.
49
-3 -2 -1 1 2 3f
-15
-10
-5
5
10
15
F
Figure 6: F (f) with three simple zeros (cnoidal or periodic wave).
50
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)
51
(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.
52
(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 ′ = ±√
2(
f +c
6
)3/2
(89)
ξ = ±∫
df√
2(
f + c6
)3/2. (90)
Finally, we obtain
f(ξ) = − c6
+2
(ξ − β), (92)
where β is a constant of integration. This solution al ways diverges
at ξ = β.
53
54
• 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.
55
• 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
f2
dξ
dfdf = 2
∫ f3
f2
df√
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).
56
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
1
2(f ′)
2= E − V (f) = F (f) (97)
For some constant of integration (energy) E.
57
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).
58
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.
59
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
2
)
= const. (101)
(c2 − 1)1
2(f ′)2 + 2 sin2 f
2= A (102)
For some arbitrary constant A.
60
In our previous notation we have
1
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
61
-15 -10 -5 5 10 15f
1
2
3
4
F
Figure 7: F (f) against f for the sine-Gordon equation with A = 0
and c2 = 1/2.
62
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.
63
Now, let us make the substitution
v = tan
(
1
4f
)
. (113)
Differentiating w.r.t. ξ yields
v′ =1
4f ′ sec2
(
1
4f
)
= ±1
2
sin(
12f)
√1− c2
sec2
(
1
4f
)
= ±sin(
f4
)
cos(
f4
)
√1− c2
sec2
(
1
4f
)
= ± v√1− c2
.
(114)
64
Finally, we have
f(ξ) = 4 arctan
± exp
[
± (ξ − ξ0)√1− c2
]
(117)
or
f(ξ) = 4 arctan
∓ exp
[
± (ξ − ξ0)√1− c2
]
(118)
If both signs are the same, we have a positive kink, if they differ
we have a negative kink (antikink)
65
-3 -2 -1 1 2 3X
1
2
3
4
5
6
f
(a) A kink
-3 -2 -1 1 2 3X
1
2
3
4
5
6
f
(b) An antikink
Figure 8: Kink and antikink of the sine-Gordon equation with A = 0
and c2 = 1/2
66
Consider the following solution of the sine-Gordon equation:
tan
(
1
4u
)
=c sinh
[
x√1−c2
]
cosh[
ct√1−c2
] , 0 < c2 < 1 (120)
Let us set a = 1/√
1− c2 and v = tan (f/4).
67
We have
v ∼ ceax − e−ax
e−act∼ c
(
ea(x+ct) − e−a(x−ct))
, t→ −∞. (121)
This implies
v ∼
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)
68
Similarly, we have
v ∼ ceax − e−ax
eact∼ c
(
ea(x−ct) − e−a(x+ct))
, t→∞. (123)
This implies
v ∼
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)
69
-15 -10 -5 5 10 15
-1.5
-1
-0.5
0.5
1
1.5
(a) The solution (120) as t →
−∞
-15 -10 -5 5 10 15
-1.5
-1
-0.5
0.5
1
1.5
(b) The solution (120) as t →
∞
Figure 9: The solution (120) of the sine-Gordon equation as t→ −∞and as t→∞.
70
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.
71
We have
−cf ′ − 6ff ′ + f ′′′ − 12δf2f ′ = 0 (125)
−cf − 3f2 + f ′′ − 4δf3 = A (126)
−cff ′ − 3f2f ′ + f ′f ′′ − 4δf3f ′ = Af ′ (127)
−1
2cf2 − f3 +
1
2(f ′)
2+ δf4 = Af +B (128)
1
2(f ′)
2= δf4+f3+
1
2cf2+Af+B = F (f) (130)
72
-3 -2 -1 1 2 3
5
10
15
20
25
(a) One kink solution
-2 -1 1 2
-2
2
4
6
8
(b) One periodic solution
-3 -2 -1 1 2 3
-5
5
10
15
20
(c) Soliton solutions
-3 -2 -1 1 2 3
-5
5
10
15
20
(d) Soliton solution
Figure 10: Positive δ
73
-3 -2 -1 1 2 3
-6
-4
-2
2
4
(a) Soliton solution
-2 -1 1 2
-8
-6
-4
-2
2
(b) Periodic solution
-3 -2 -1 1 2 3
-25
-20
-15
-10
-5
(c) No physical solution
-3 -2 -1 1 2 3
-20
-15
-10
-5
5
(d) Periodic solution
Figure 11: Negative δ
74
3.1 The Scattering Problem
Consider the KdV equation
ut − 6uux + uxxx = 0.
We are looking for a general procedure to integrate this equation.
Let us introduce the Miura transformation
u = v2 + vx
75
Direct substitution leads to
2vvt + vxt − 6(v2 + vx)(2vvx + vxx)
+ 6vxvxx + 2vvxxx + vxxxx = 0, (1)
which can be rearranged to give
(
2v +∂
∂x
)
(vt − 6v2vx + vxxx) = 0. (2)
The equation
vt − 6v2vx + vxxx = 0
is called modified KdV equation or mKdV.
76
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)
77
We now observe that the KdV equation is Galilean invariant ,
i.e.
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).
78
u(x, 0)scattering−−−−−−→ S(0)
KdV
y
ytime evolution
u(x, t)inverse←−−−−−−
scatteringS(t)
The diagram of the inverse scattering for the KdV equation.
79
Example
Consider the equation
ut + ux + uxxx = 0.
Suppose we are given the initial-value problem u(x, 0) = f(x).
Then
f(x) =
∫ ∞
−∞A(k)eikxdk and A(k) =
1
2π
∫ ∞
−∞f(x)e−ikxdx. (9)
Here A(k) plays the role of the
‘scattering data’.
80
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)
81
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.
82
u(x, 0)F.T.−−−−→ A(k)
PDE
y
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
83
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.
84
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)
85
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→∞.
(18)
The λ with this property are discrete. This constitute the dis-
crete spectrum.
86
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.
87
• 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→ −∞.
(23)
88
-4 -2 2 4x
-1
-0.75
-0.5
-0.25
0.25
0.5
0.75
u
Incident wave
Reflected wave
Transmitted wave
Figure 1: Schematic representation of incident, reflected and trans-
mitted wave
89
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).
90
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 =
d
dxW (ψm, ψn), (25)
where W (ψm, ψn) is the Wronskian of ψm, ψn. By integrating we
have
[W (ψm, ψn)]∞−∞ = (κ2n − κ2
m)
∫ ∞
−∞ψmψndx. (26)
This implies∫∞−∞ ψmψndx = 0. That is, ψm and ψn are
orthogonal .
91
Example
Wave in an inhomogeneous medium
One-dimensional propagation of sound or light in an
inhomogeneous medium is governed by
∇2φ− 1
c2(x)
∂2φ
∂t2= 0. (28)
We then insert in the above equation a normal mode
φ(x, t) = ψ(x)ei(ly+mz−ωt). (29)
92
This yields to
d2ψ
dx2− (l2 +m2)ψ(x) = −ω
2
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)
93
If now
c(x)→ c∞ as x→∞, (33)
we coud redefine
λ =ω2
c2∞−(
l2 +m2)
, u(x) = ω2
(
1
c2∞− 1
c2(x)
)
. (34)
Then we have that
u(x)→ 0 as x→∞. (35)
94
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
95
ψ′(ǫ)− ψ′(−ǫ) = −∫ ǫ
−ǫ
(V δ(x) + λ)ψ(x)dx
= −V ψ(0)− λ∫ ǫ
−ǫ
ψ(x)dx.
(37)
Finally, as ǫ→ 0 we have
limǫ→0
(ψ′(ǫ)− ψ′(−ǫ)) = −V ψ(0). (39)
ψ′ is discontinuous at the origin.
96
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)
97
By continuity at x = 0, cn = dn. Then, the normalization condition
leads to
∫ 0
−∞|cn|2 exp(2κnx)dx+
∫ ∞
0
|cn|2 exp(−2κnx)dx =
|cn|2κn
= 1.
Therefore, up to an arbitrary phase, we have
cn =√κn (44)
98
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
2(47)
or equivalently λ1 = −V 2/4.
99
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)
100
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)
101
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+
4k2
V 2 + 4k2= 1. (57)
102
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).
103
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.
104
Example: u(x) = −6 sech2 x.
If u(x) = −6 sech2 x, it can be shown that N = 2 and
λ1 = −1, ψ1(x) =
√
3
2tanhx sechx (66)
λ2 = −4, ψ2(x) =
√3
2sech2 x, (67)
b(k) = 0 ∀k (68)
105
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 <∞.
106
Therefore, we have
d
dx= sech2 x
d
dy= (1− y2)
d
dy(71)
and so
(1− y2)d
dy
[
(
1− y2) d
dy
]
+[
λ+ V(
1− y2)]
ψ = 0. (72)
or
d
dy
[
(
1− y2) d
dy
]
+
[
V +λ
(1− y2)
]
ψ = 0. (74)
This is the associated Legendre equation.
107
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)
and
PN (y) =(−1)N
N !2N
dN
dyN
(
1− y2)N
, (77)
PN (y) being the Legendre polynomial of degree N .
108
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→ −∞.
(78)
For x→ −∞ these are given by
ψ(x; k) = a(k)2ik(sechx)−ikF (a, b; c; (1 + y)/2). (80)
109
F (α, β, ; γ; z) is the hypergeometric function, which is defined by
the series
F (α, β; γ; z) = 1+
∞∑
k=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
a =1
2− ik +
(
V +1
4
)
(83)
b =1
2− ik −
(
V +1
4
)
(84)
c = 1− ik (85)
110
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→∞.
(87)
111
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.
112
We need the identity
Γ
(
1
2− z)
Γ
(
1
2+ z
)
=π
cosπz. (88)
Hence observe that
Γ(c− a)Γ(c− b) (89)
= Γ
[
1
2−(
V +1
4
)1/2]
Γ
[
1
2+
(
V +1
4
)1/2]
(90)
= π/ cos
[
π
(
V +1
4
)1/2]
(91)
113
It follows that b(k) = 0 if
(
V +1
4
)1/2
= N +1
2(92)
or equivalently
V = N(N + 1). (94)
114
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
k = i
[
(
V +1
4
)1/2
−(
m+1
2
)
]
. (96)
115
There is a finite number of discrete eigenvalues if
(
V +1
4
)1/2
>1
2i.e. V > 0. (98)
Their number is
[
(
V +1
4
)1/2
− 1
2
]
+ 1 (100)
where [z] denotes the integral part of z ( the greatest integer ≤ z).
116
The eigenvalues are given by
κm =
(
V +1
4
)1/2
−(
m+1
2
)
= µ (102)
The eigenfunctions are the associated Legendre functions
Pµν (y), where ν is a solution of the equation
V = ν(ν + 1). (103)
117
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
118
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 .
119
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)
120
We then express φ(x, z) in term of its Fourier transform:
φ(x, z) =1
2π
∫ ∞
−∞ψ(x; k)e−ikzdk (112)
and
ψ(x; k) =
∫ ∞
−∞φ(x, z)eikzdz (113)
Substituting the above integral into the wave equation yields
ψxx + k2ψ = 0. (115)
121
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.
122
By taking the Fourier transform of (118) we have
ψ(x; k) = eikx+
∫ ∞
x
K(x, z)eikzdz (120)
The above wavefunction has the correct asymptotic value.
NB Note that∫ ∞
−∞δ(x− z)eikz = eikx and
1
2π
∫ ∞
−∞eik(x−z) = δ(x− z) (121)
123
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 +
∫ ∞
x
K(x, z)eikzdz (124)
K(x, z) = 0 if z < x and
ψ ∼ eikx, x→∞. (125)
124
It can be shown that
Kxx(x, z)−Kzz(x, z)− u(x)K(x, z) = 0 for z > x (126)
and
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).
125
The equation which one then tries to invert is
ψ = ψ∗ + b(k)ψ. (131)
with
ψ(x; k) = eikx +
∫ ∞
x
K(x, z)eikzdz. (132)
ψ has the right asymptotic limit:
ψ(x; k) ∼ e−ikx + b(k)eikx as x→∞. (133)
126
K(x, z) can be found by solving the integral equation
K(x, z)+F (x+z)+
∫ ∞
x
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.
127
The function F (X) is defined by
F (X) =N∑
n=1
c2n exp (−κnX) +1
2π
∫ ∞
−∞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)
or
cn = limx→∞
[ψn(x) exp(κnx)] (139)
128
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.
129
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)+
∫ ∞
x
K(x, y)F (y+z)dy = 0 z > x > −∞,(146)
130
Furthermore,
F (X) =N∑
n=1
c2n exp (−κnX) +1
2π
∫ ∞
−∞b(k)eikXdk (148)
where cn are the coefficients
ψn(x) ∼ cn exp(−κnx), x→∞ (149)
or
cn = limx→∞
[ψn(x) exp(κnx)] . (150)
131
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)
and
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
132
Now, suppose that F (x+ z) is a separable function, i.e.
F (x+ z) =N∑
n=1
Xn(x)Zn(z). (155)
The Marchenko equation can therefore be written as
K(x, z)+
N∑
n=1
Xn(x)Zn(z)+
N∑
n=1
Zn(z)
∫ ∞
x
K(x, y)Xn(y)dy = 0
(157)
133
The solution therefore must take the form
K(x, z) =N∑
n=1
Ln(x)Zn(z). (159)
Upon this substitution for K(x, z) we obtain
134
N∑
n=1
Ln(x)Zn(z) +
N∑
n=1
Xn(x)Zn(z)
+N∑
n=1
Zn(z)N∑
m=1
Lm(x)
∫ ∞
−∞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) +
N∑
m=1
Lm(x)
∫ ∞
−∞Zm(y)Xn(y)dy = 0. (162)
135
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)
136
The Marchenko equation then becomes
K(x, z) + c21 exp [−κ1(x+ z)] + c22 exp [−κ2(x+ z)]
+
∫ ∞
x
K(x, y)
c21 exp [−κ1(y + z)] + c22 exp [−κ2(y + z)]
dy = 0
(4)
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)
137
It follows that L1(x) and L2(x) must satisfy
L1 + c21 exp(−κ1x)
+ c21
L1
∫ ∞
x
exp (−2κ1y) dy + L2
∫ ∞
x
exp [−(κ1 + κ2)y] dy
= 0
L2 + c22 exp(−κ2x)
+c22
L1
∫ ∞
x
exp [−(κ1 + κ2)y] dy + L2
∫ ∞
x
exp (−2κ2y) dy
= 0.
(8)
138
After having evaluated the integrals the system becomes
Ln+c2n exp(−κnx)+c2n
2∑
m=1
Lm exp [−(κm + κn)x]
κm + κn= 0, n = 1, 2.
(10)
The above system can be written as AL+B = 0, where
L = (L1, L2) and
Bn = c2n exp(−κnx) (11)
139
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).
140
We now note that
d
dxAmn = −c2m exp [−(κm + κn)x] = −BmEn. (15)
Therefore, we obtain
K(x, x) =N∑
m=1
EmLm = −N∑
m,n=1
Em
(
A−1)
mnBn
= Tr
(
A−1 dA
dx
)
.
(17)
141
Now, for the 2× 2 case if
A =
a b
c d
, (18)
then
A−1 =1
ad− bc
d −b−c a
(19)
142
After trivial algebra we obtain
A−1 dA
dx=
1
ad− bc
a′d− bc′ · · ·· · · −cb′ + ad′
. (20)
Finally, we have
Tr
(
A−1 dA
dx
)
=1
ad− bc (a′d+ d′a− b′c− bc′)
=1
detA
d detA
dx=
d
dxlog detA. (22)
143
We can now evaluate u(x)
u(x) = −2dK
dx= −2
d2
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)
144
If we set c2 = 0, we then obtain
u(x) = − 4κ1c21 exp(−2κ1x)
[
1 +c2
1
2κ1
exp(−2κ1x)]2
= −2κ21 sech2 [κ1x+ x0] ,
(27)
where exp(x0) = (2κ1)1/2/c1. If κ1 = 1 and c1 =
√2, then we
recover
u(x) = −2 sech2 x. (28)
145
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
2π
∫ ∞
−∞
eikX
β + ikdk. (31)
146
The integral can be calculated easily using Cauchy’s residue
theorem:∫ ∞
−∞
eikX
β + ikdk = 2πe−βX , X > 0 (32)
and∫ ∞
−∞
eikX
β + ikdk = 0 X < 0. (33)
F (X) becomes simply
F (X) = βe−βXH(−X), (35)
where H(−X) is the Heaviside step function.
147
Figure 1: The contour of the integral (32) in the complex plain..
148
From the Marchenko equation
K(x, z)+F (x+z)+
∫ ∞
x
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
x
K(x, y)e−β(y+z)dy = 0 (39)
149
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
x
Ky(x, y)e−β(y+z)dy = 0. (40)
The solution is
K(x, z) = −β.
150
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.
151
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.
152
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→ −∞.
(50)
k = λ1/2, a(k) and b(k) are the transmission and reflection
coefficients respectively: |b(k)|2 + |a(k)|2 = 1.
153
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)+
∫ ∞
x
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∑
n=1
c2n exp (−κnX) +1
2π
∫ ∞
−∞b(k)eikXdk. (54)
154
Inverse scattering and the KdV equation
Consider the
ut − 6uux + uxxx = 0.
Introducing the Miura transformation
u = v2 + vx,
the KdV equation becomes
(
2v +∂
∂x
)
(vt − 6v2vx + vxxx) = 0. (55)
155
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.
156
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).
157
u(x, 0)scattering−−−−−−→ S(0)
KdV
y
ytime evolution
u(x, t)inverse←−−−−−−
scatteringS(t)
The diagram of the inverse scattering for the KdV equation.
158
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.
159
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).
160
This leads to
∂
∂x(ψxR − ψRx) = ψxx (ψt − 2uψx − 4λψx)−ψ (uxxxψ − 4uxψxx)
− ψ (uψt − λψt − λtψ + utψ) + ψ (2u+ 4λ) (uxψ − λψx + uψx) .
(64)
We now use the Schrodinger equation
ψxx + (λ− u(x, t))ψ = 0 (65)
to simplify (64).
161
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.
162
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.
163
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.
164
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
1
2(ψn)2t +
(
uψ2n − 2ψ2
nx + 4κ2nψ
2n
)
x= hnψ
2n. (76)
165
We now integrate (76) w.r.t. x:
1
2
d
dt
(∫ ∞
−∞ψ2
ndx
)
= hn
∫ ∞
−∞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).
166
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.
167
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.
168
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)
169
Therefore, we have that
R(x, t; k) ∼(
da
dt+ 4ik3a
)
e−ikx, x→ −∞. (86)
As a consequence
ψxR− ψRx → 0, as x→ −∞. (87)
Thus g(t; k) = 0 for all t.
170
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
da
dt+ 4ik3a = ha. (91)
171
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
db
dteikx + 4ik3(e−ikx − eikx) = h(e−ikx − eikx). (94)
172
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
db
dt= 8ik3b (96)
whose solutions are
a(k; t) = a(k; 0) and b(k; t) = b(k; 0) exp(8ik3t).
173
Evolution of the scattering data: summary
• κn = constant, cn(t) = cn(0) exp(4κ3nt);
• b(k; t) = b(k; 0) exp(8ik3t).
174
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).
175
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).
176
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).
177
Step 3: We now want to solve the Marchenko equation
K(x, z; t) +F (x+ z; t) +
∫ ∞
x
K(x, y; t)F (y+ z; t)dy = 0, (98)
with
F (X ; t) =N∑
n=1
cn(0)2 exp(
8κ3nt− κnX
)
+1
2π
∫ ∞
−∞b(k; 0) exp(8ik3t+ ikX)dk (99)
178
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).
179
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.
180
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)
181
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
∫ ∞
x
K(x, y; t)e8t−(y+z)dy = 0. (104)
182
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
∫ ∞
x
e−2ydy = 0. (105)
The above equation can be solved directly to yield
L(x, t) = − −2e8t−x
1 + e8t−2x. (107)
183
The potential is then given by
u(x, t) = 2∂
∂x
(
2e8t−2x
1 + e8t−2x
)
= − 8e2x−8t
(1 + e2x−8t)2
= −2 sech2(x− 4t).
(109)
This is the solitary wave solution of amplitude −2 at speed of
propagation 4.
184
Two solitons solution
The initial profile is now taken to be
u(x, 0) = −6 sech2 x.
The Sturm-Liouville equation is now
ψxx +(
λ+ 6 sech2 x)
ψ = 0.
185
The discrete spectrum of this equation is given by
κ1 = 1 and κ2 = 2 (112)
ψ1(x) =
√
(
3
2
)
tanhx sechx and ψ2(x) =
√3
2sech2 x
(113)
The asymptotic behaviour of these solutions is given by
ψ(x) ∼√
6e−x, 2√
3e−2x as x→∞. (114)
186
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)
187
The Marchenko equation is therefore
K(x, z; t) + 6e8t−(x+z) + 12e64t−2(x+z)
+
∫ ∞
x
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
188
L1 + 6e8t−x + 6e8t
(
L1
∫ ∞
x
e−2ydy + L2
∫ ∞
x
e−3ydy
)
= 0
L2 + 12e64t−x + 12e64t
(
L1
∫ ∞
x
e−3ydy + L2
∫ ∞
x
e−4ydy
)
= 0
(122)
Evaluating the integrals yields
L1 + 6e8t−x + 3L1e8t−2x + 2L2e
8t−3x = 0 (123)
L2 + 12e64t−2x + 4L1e64t−3x + 3L2e
64t−4x = 0. (124)
189
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∂
∂x
(
L1e−x + L2e
−2x)
= 12∂
∂x
[(
e8t−2x + e72t−6x − 2e64t−4x)]
/D
(127)
190
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→ ±∞.
191
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.
192
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→ ±∞.
193
Figure 2: Time evolution of the two solitons solution
194
Figure 3: The paths of the wave crests of the two solitons
195
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.
196
N solitons solution
The initial profile is given by
u(x, 0) = −N(N + 1) sech2 x.
The Sturm-Liouville equation is now
ψxx +[
λ+N(N + 1) sech2 x]
ψ = 0.
197
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.
198
The time evolution of the normalization coefficients is given by
cn(t) = cn(0) exp(
4n3t)
(6)
The function F (X ; t) is now
F (X ; t) =N∑
n=1
cn(0)2 exp(
8n3t− nX)
. (8)
199
The Marchenko equation now becomes
K(x, z; t) +N∑
n=0
cn(0)2 exp[
8n3 − n(x+ z)]
+
∫ ∞
x
K(x, y; t)
N∑
n=1
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∑
n=1
Ln(x, t)e−nx. (10)
200
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)
(11)
201
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)
202
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[
n(
x− 4n2t)
∓ xn
]
, t→ ±∞. (18)
203
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
[
n(
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→ ±∞.
204
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
205
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∑
n=1
cn(0)2 exp(
8κ3nt− κnX
)
+1
2π
∫ ∞
−∞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.
206
The integral
I(X ; t) =1
2π
∫ ∞
−∞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.
207
The main contribution as t→∞ comes from where φ(k) is
stationary and
I(t) ∼ 2Γ(
1n
)
(n!)1/n
n[
t∣
∣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.
208
Therefore, our integral becomes
I(X ; t) ∼ Γ (1/3) eiπ/6
π
√
2
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.
209
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.
210
Figure 2: Solution to the Kdv equation with positive sech2 x initial
condition.
211
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.
212
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
[
(
V +1
4
)1/2
− 1
2
]
+ 1. (28)
Therefore, we have two discrete eigenvalues.
213
Figure 4: Initial condition given by u(x, 0) = −4 sech4 x.
214
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→∞.
215
• (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
1
2π
∫ ∞
−∞b(k, t)eikXdk (30)
of F (X ; t) in the Marchenko equation.
216
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.
217
5. Conservation laws
Consider the equation of continuity of a compressible fluid,
∂ρ
∂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 .
218
Now, suppose that ρu→ const. as |x| → ∞, and ρ and (ρu)x are
integrable.
The continuity equation implies that
d
dt
∫ ∞
−∞ρdx =
∫ ∞
−∞
∂ρ
∂tdx = −
∫ ∞
−∞
∂ (ρu)
∂xdx = − [ρu]∞−∞ = 0.
(33)
Therefore the integral
∫ ∞
−∞ρdx = constant (35)
represents the conservation of the total mass in the system.
219
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
d
dt
∫∫∫
Ω
ρ(x, t)dx =
∫∫∫
Ω
∂ρ(x, t)
∂tdx
= −∫∫∫
Ω
∇ · j dx−∫∫
∂Ω
j · n dS = 0.
(37)
This implies that the total charge∫∫∫
Ω
ρ(x, t)dx = constant. (38)
220
A general form of 1-dimensional conservation law is
∂T
∂t+∂X
∂x= 0, (40)
where T (x, t, u, ux, . . .) is the density and X(x, t, u, ux, . . .) is the
flux
N.B. They cannot depend on ut.
T and Xx are integrable and
X → constant as |x| → ∞. (41)
221
The conservation law implies
∫ ∞
−∞Tdx = constant. (43)
The integral of T is conserved, or is a constant of motion.
222
The KdV equation
Consider the KdV equation,
ut − 6uux + uxxx = 0.
It is already in the form of conservation law:
∂u
∂t+
∂
∂x
(
uxx − 3u2)
= 0. (45)
223
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.
224
If we multiply the KdV equation by u we obtain another
conservation law
∂(
12u
2)
∂t+
∂
∂x
(
uuxx −1
2u2
x − 2u3
)
= 0. (50)
Therefore we have
∫ ∞
−∞u2 dx = constant. (52)
The above integral expresses momentum conservation.
225
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
∂
∂t
(
u3 +1
2u2
x
)
+∂
∂x
(
−9
2u4 + 3u2uxx − 6uu2
x + uxuxxx −1
2u2
xx
)
= 0. (55)
226
The previous expression is in a form of conservation law. Therefore,
we have
∫ ∞
−∞
(
u3 +1
2u2
x
)
dx = constant. (57)
This integral expresses the energy conservation for water waves.
227
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)
228
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!
229
The previous equation can be rewritten as
∂w
∂t+
∂
∂x
(
wxx − 3w2 − 2ǫ2w3)
(62)
and therefore
∫ ∞
−∞w dx = constant. (64)
230
We now set
w(x, t; ǫ) ∼∞∑
n=0
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)
231
• We insert the asymptotic expansion for w in the Gardner
transformation;
• We equates coefficients of ǫn for each n = 0, 1, . . ..
∞∑
n=0
ǫnwn ∼ u− ǫ∞∑
n=0
ǫnwnx − ǫ2( ∞∑
n=0
ǫnwn
)2
(70)
232
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
x
. . .
(72)
233
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)
234
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 =
∫ ∞
−∞
(
u3 +1
2u2
x
)
dx = constant. (82)
235
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.
236
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.
237
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 =∞∑
n=1
cnφn, (84)
where cn are constants and φn is an appropriate basis. Hmust also be equipped with a scalar product (φ, ψ).
238
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.
239
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
2π
∞∑
n=−∞cne
inx. (93)
The scalar product is defined by
∫ 2π
0
g∗(x)f(x)dx. (94)
240
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.
241
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)
242
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).
(96)
By taking the scalar product with ψ of the above expression, we
have
(ψ, ψ)λt = (ψ, (L− λ)(ψt −Mψ))
= (((L− λ)ψ, (ψt −Mψ))
= (0, (ψt −Mψ)) = 0.
(97)
(L− λ is self-adjoint.)
243
It follows that λt = 0 and therefore
λ = constant.
From eq. (96) with λt = 0 it also follows that
(L− λ)(ψt −Mψ) = 0.
244
(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.
245
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.
246
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)
247
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)
248
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.
249
Consider the simplest choice
M = c∂
∂x(108)
for c constant, then
[L,M ] = c
(
− ∂2
∂x2+ u(x, t)
)
∂
∂x− c ∂
∂x
(
− ∂2
∂x2+ u(x, t)
)
= −cux
(109)
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)
250
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.
251
We now choose M so that it involves at most a third-order
differential operator,
M = −α ∂3
∂x3+ U
∂
∂x+
∂
∂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)
∂2
∂x2
(113)
252
It follows that [L,M ] is a multiplicative operator if
U =3
4αu and A = A(t) (115)
The Lax equation now becomes
Lt + [L,M ] = ut −3
2αuux +
1
4αuxxx = 0. (117)
253
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)
254
The previous equation can be recast, on using the Schrodinger
equation
ψxx + (λ− u)ψ = 0, (121)
as
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.
255
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+
n∑
m=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).
256
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)
257
