3Inverse scattering transform
II Integrable Systems
3.6 Infinitely many first integrals
As we’ve previously mentioned, we are expecting our integrable PDE’s to have
infinitely many first integrals. Recall we can construct ϕ = ϕ(x, k, t) such that
Lϕ = k
2
ϕ,
and we had
ϕ(x, k, t) =
(
e
−ikx
x → −∞
a(k, t)e
−ikx
+ b(k, t)e
ikx
x → ∞
.
But when we looked at the evolution of the scattering data, we can actually
write down what
a
and
b
are. In particular,
a
(
k, t
) =
a
(
k
) is independent of
t
.
So we might be able to extract some first integrals from it. We have
e
ikx
ϕ(x, k, t) = a(k) + b(k, t)e
2ikx
as x → ∞.
We now take the average over [
R,
2
R
] for
R → ∞
. We do the terms one by one.
We have the boring integral
1
R
Z
2R
R
a(k) dx = a(k).
For the b(k, t) term, we have
1
R
Z
2R
R
b(k, t)e
2ikx
dx = O
1
R
,
So we have
a(k) = lim
R→∞
1
R
Z
2R
R
e
ikx
ϕ(x, k, t) dx
= lim
R→∞
Z
2
1
e
ikRx
ϕ(Rx, k, t) dx.
So can we figure out what this thing is? Since
ϕ
=
e
−ikx
as
x → −∞
, it is
“reasonable” to write
ϕ(x, k, t) = exp
−ikx +
Z
x
−∞
S(y, k, t) dy
for some function S. Then after some dubious manipulations, we would get
a(k) = lim
R→∞
Z
2
1
exp
Z
Rx
−∞
S(y, k, t) dy
!
dx
= exp
Z
∞
−∞
S(y, k, t) dy
. (†)
Now this is interesting, since the left hand side
a
(
k
) has no
t
-dependence, but
the right-hand formula does. So this is where we get our first integrals from.
Now we need to figure out what S is. To find S, recall that ϕ satisfies
Lϕ = k
2
ϕ.
So we just try to shove our formula of ϕ into this equation. Notice that
ϕ
x
= (S − ik)ϕ, ϕ
xx
= S
x
ϕ + (S − ik)
2
ϕ.
We then put these into the Schr¨odinger equation to find
S
x
− (2ik)S + S
2
= −u.
We have got no
ϕ
’s left. This is a famous type of equation — a Ricatti-type
equation. We can make a guess
S(x, k, t) =
∞
X
n=1
S
n
(x, t)
(2ik)
n
.
This seems like a strange thing to guess, but there are indeed some good reasons
for this we will not get into. Putting this into the equation and comparing
coefficients of k
−n
, we obtain a recurrence relation
S
1
= −u
S
n+1
=
dS
n
dx
+
n−1
X
m=1
S
m
S
n−m
.
This is a straightforward recurrence relation to compute. We can make a
computer do this, and get
S
2
= −u
x
, S
3
= −u
xx
+ u
2
, S
4
= ···
Using the expression for S in (†), we find that
log a(k) =
Z
∞
−∞
S(x, k, t) dx
=
∞
X
n=1
1
(2ik)
n
Z
∞
−∞
S
n
(x, t) dx.
Since the LHS is time-independent, so is the RHS. Moreover, this is true for all
k. So we know that
Z
∞
−∞
S
n
(x, t) dt
must be constant with time!
We can explicitly compute the first few terms:
(i) For n = 1, we find a first integral
Z
∞
−∞
u(x, t) dx
We can view this as a conservation of mass.
(ii) For n = 2, we obtain a first integral
Z
∞
−∞
u
x
(x, t) dx.
This is actually boring, since we assumed that
u
vanishes at infinity. So
we knew this is always zero anyway.
(iii) For n = 3, we have
Z
∞
−∞
(−u
xx
(x, t) + u(x, t)
2
) dx =
Z
∞
−∞
u(x, t)
2
dx.
This is in some sense a conservation of momentum.
It is an exercise to show that
S
n
is a total derivative for all even
n
, so we
don’t get any interesting conserved quantity. But still, half of infinity is infinity,
and we do get infinitely many first integrals!