View source for Introduction to the Inverse Scattering Transform

{{nonlinear waves course
 | chapter title = Introduction to the Inverse Scattering Transform
 | next chapter = [[Reaction-Diffusion Systems]]
 | previous chapter = [[Conservation Laws for the KdV]]
}}

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{matrix}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u  &  =0\\
u(x,0)  &  =f\left(  x\right)
\end{matrix}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=v^{2}+v_{x}
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left(  \partial_{x}w\right)  }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.

==Properties of the eigenfunctions==

The equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves.

===Example: Scattering by a Well===

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left(  x\right)  =\left\{
\begin{matrix}
[c]
0 & x\notin\left[  -1,1\right] \\
b & x\in\left[  -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

\paragraph{Case when <math>\lambda>0</math>}

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}
[c]
a_{1}e^{kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a_{2}e^{-kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
to solve for <math>a</math> and <math>b</math>. This leads to two system of equation, one for the
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
<center><math>
\left(
\begin{matrix}
[c]
e^{-kx} & -\cos\kappa\\
ke^{-kx} & \sin\kappa
\end{matrix}
\right)  \left(
\begin{matrix}
[c]
a_{1}\\
b_{1}
\end{matrix}
\right)  =\left(
\begin{matrix}
[c]
0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}
[c]
e^{-kx} & -\cos\kappa\\
ke^{-kx} & \sin\kappa
\end{matrix}
\right)  =0
</math></center>
which gives us the equation
<center><math>
e^{-kx}\sin\kappa+\left(  \cos\kappa\right)  ke^{-kx}=0
</math></center>
or
<center><math>
\tan\kappa=-k=-\sqrt{b-\kappa^{2}}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions. 

\paragraph{Case when <math>\lambda>0</math>}

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}
[c]
e^{-ikx}+re^{ikx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
ae^{-ikx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}
[c]
-e^{-ik} & \cos\kappa & -\sin\kappa & 0\\
ike^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -e^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ike^{-ik}
\end{matrix}
\right)  \left(
\begin{matrix}
[c]
r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right)  =\left(
\begin{matrix}
[c]
e^{ik}\\
ike^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>


==Connection with the KdV==

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left(  w\partial_{x}Q-\partial
_{x}wQ\right)  =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left(  \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left(  x,t\right)  </math> evolves according to the KdV. 

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left(  x\right)  =c_{n}\left(  t\right)  e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left(  w_{n}\left(  x\right)  \right)  ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left(  x,t\right)  \approx e^{-ikx}+r\left(  k,t\right)  e^{ikx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left(  x,t\right)  \approx a\left(  k,t\right)  e^{-ikx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left(  \lambda,0\right)  =\left(  \left\{  k_{n},c_{n}\left(  0\right)
\right\}  _{n=1}^{N},r\left(  k,0\right)  ,a\left(  k,0\right)  \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left(  t\right)  =c_{n}\left(  0\right)  e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left(  k,t\right)  =r\left(  k,0\right)  e^{8ik^{3}t}
</math></center>
<center><math>
a\left(  k,t\right)  =a\left(  k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left(  x,t\right)  =\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left(  k,t\right)  e^{ikx}dk
</math></center>
Then solve
<center><math>
K\left(  x,y;t\right)  +F\left(  x+y;t\right)  +\int_{x}^{\infty}K\left(
x,z;t\right)  F\left(  z+y;t\right)  dz=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left(  x,t\right)  =2\partial_{x}K\left(  x,x,t\right)
</math></center>


==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left(  k,0\right)  =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left(  x,t\right)  =\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}x}
</math></center>
then
<center><math>
K\left(  x,y,t\right)  +\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)
e^{-k_{n}\left(  x+y\right)  }+\int_{x}^{\infty}K\left(  x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(  y+z\right)  }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left(  x,y,t\right)  =-\sum_{m=1}^{N}c_{m}\left(  t\right)  v_{m}\left(
x\right)  e^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left(  t\right)  v_{n}\left(  x\right)  e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(  x+y\right)  }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left(  t\right)  v_{m}\left(  x\right)
e^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(
y+z\right)  }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left(  t\right)  v_{n}\left(  x\right)  e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(  x+y\right)  }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left(  t\right)  c_{n}^{2}\left(
t\right)  }{k_{n}+k_{m}}v_{m}\left(  x\right)  e^{-k_{m}x}e^{-k_{n}\left(
y+x\right)  }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left(  t\right)  ,</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left(  x\right)  +c_{n}\left(  t\right)  e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left(  t\right)  c_{m}\left(  t\right)  }{k_{n}+k_{m}}
v_{m}\left(  x\right)  e^{-\left(  k_{m}+k_{n}\right)  x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left(  \mathbf{I}+\mathbf{C}\right)  \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left(  t\right)  e^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left(  t\right)  c_{m}\left(  t\right)
}{k_{n}+k_{m}}e^{-\left(  k_{m}+k_{n}\right)  x}
</math></center>
<center><math>
K\left(  x,y,t\right)  =-\sum_{m=1}^{N}c_{m}\left(  t\right)  \left(
\mathbf{I}+\mathbf{C}\right)  ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left(  x,t\right)  =2\partial_{x}^{2}\log\left[  \det\left(  \mathbf{I}
+\mathbf{C}\right)  \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left(  x,x,t\right)   &  =-\frac{c_{1}\left(  t\right)  c_{1}\left(
t\right)  e^{-k_{1}x}e^{-k_{1}x}}{1+\frac{c_{1}\left(  t\right)  c_{1}\left(
t\right)  }{k_{1}+k_{1}}e^{-\left(  k_{1}+k_{1}\right)  x}}\\
&  =\frac{-1}{1+e^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>e^{-\alpha}=2c_{0}^{2}\left(  0\right)  .</math> Therefore
<center><math>\begin{matrix}
u\left(  x,t\right)   &  =2\partial_{x}K\left(  x,x,t\right)  \\
&  =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left(  1+e^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right)  ^{2}}\\
&  =\frac{-8k_{1}^{2}}{\left(  \sqrt{2k_{1}}e^{\theta}+e^{-\theta}
/\sqrt{2k_{1}}\right)  ^{2}}\\
&  =2k^{2}\sec^{2}\left\{  k_{1}\left(  x-x_{0}\right)  -4k_{1}^{3}t\right\}
\end{matrix}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.