# Difference between revisions of "Introduction to the Inverse Scattering Transform"

Nonlinear PDE's Course
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## Introduction

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

\displaystyle{ \begin{align} \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ u(x,0) & =f\left( x\right) \end{align} }

with $\displaystyle{ \left\vert u\right\vert \rightarrow0 }$ as $\displaystyle{ x\rightarrow\pm\infty. }$

The Miura transformation is given by

$\displaystyle{ u=v^{2}+v_{x} \, }$

and if $\displaystyle{ v }$ satisfies the mKdV

$\displaystyle{ \partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0 }$

then $\displaystyle{ u }$ satisfies the KdV (but not vice versa). We can think about the Miura transformation as being a nonlinear ODE solving for $\displaystyle{ v }$ given $\displaystyle{ u. }$ This nonlinear ODE is also known as the Riccati equation and there is a well know transformation which linearises this equation. It we write

$\displaystyle{ v=\frac{\left( \partial_{x}w\right) }{w} }$

then we obtain the equation

$\displaystyle{ \partial_{x}^{2}w+uw=0 }$

The KdV is invariant under the transformation $\displaystyle{ x\rightarrow x+6\lambda t, }$ $\displaystyle{ u\rightarrow u+\lambda. }$ Therefore we consider the associated eigenvalue problem

$\displaystyle{ \partial_{x}^{2}w+uw=-\lambda w }$

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

$\displaystyle{ \partial_{x}^{2}w+uw=-\lambda w }$

has two kinds of solutions for $\displaystyle{ u\rightarrow0 }$ as $\displaystyle{ x\rightarrow\pm\infty. }$ 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 $\displaystyle{ u\rightarrow0 }$ as $\displaystyle{ x\pm\infty }$ 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

$\displaystyle{ u\left( x\right) =\left\{ \begin{matrix} 0, & x\notin\left[ -1,1\right] \\ b, & x\in\left[ -1,1\right] \end{matrix} \right. }$

where $\displaystyle{ b\gt 0. }$

## Case when $\displaystyle{ \lambda\lt 0 }$

If we solve this equation for the case when $\displaystyle{ \lambda\lt 0, }$ $\displaystyle{ \lambda=-k^{2} }$ we get

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} a_{1}\mathrm{e}^{kx}, & x\lt -1\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1\lt x\lt 1\\ a_{2}\mathrm{e}^{-kx} & x\gt 1 \end{matrix} \right. }$

where $\displaystyle{ \kappa=\sqrt{b-k^{2}} }$ where we have assumed that $\displaystyle{ b\gt k^{2} }$ (there is no solution for $\displaystyle{ b\lt k^{2}). }$ We then match $\displaystyle{ w }$ and its derivative at $\displaystyle{ x=\pm1 }$ to solve for $\displaystyle{ a }$ and $\displaystyle{ b }$. This leads to two system of equations, one for the even ($\displaystyle{ a_{1}=a_{2} }$ and $\displaystyle{ b_{2}=0 }$ ) and one for the odd solutions ($\displaystyle{ a_{1}=-a_{2} }$ and $\displaystyle{ b_{1}=0) }$. The solution for the even solutions is

$\displaystyle{ \left( \begin{matrix} \mathrm{e}^{-kx} & -\cos\kappa\\ k\mathrm{e}^{-kx} & \sin\kappa \end{matrix} \right) \left( \begin{matrix} a_{1}\\ b_{1} \end{matrix} \right) =\left( \begin{matrix} 0\\ 0 \end{matrix} \right) }$

This has non trivial solutions when

$\displaystyle{ \det\left( \begin{matrix} \mathrm{e}^{-kx} & -\cos\kappa\\ k\mathrm{e}^{-kx} & - \kappa \sin\kappa \end{matrix} \right) =0 }$

which gives us the equation

$\displaystyle{ - \kappa \sin\kappa \mathrm{e}^{-kx}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0 }$

or

$\displaystyle{ \kappa \tan\kappa=k=\sqrt{b-\kappa^{2}} }$

We know that $\displaystyle{ 0\lt \kappa\lt \sqrt{b} }$ and if we plot this we see that we obtain a finite number of solutions.

Similarly we repeat the above process for the odd solutions.

## Case when $\displaystyle{ \lambda\gt 0 }$

When $\displaystyle{ \lambda\gt 0 }$ we write $\displaystyle{ \lambda=k^{2} }$ and we obtain solution

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} \mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x\lt -1\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1\lt x\lt 1\\ a\mathrm{e}^{-\mathrm{i}kx} & x\gt 1 \end{matrix} \right. }$

where $\displaystyle{ \kappa=\sqrt{b+k^{2}}. }$ Matching $\displaystyle{ w }$ and its derivaties at $\displaystyle{ x=\pm1 }$ we obtain

$\displaystyle{ \left( \begin{matrix} -\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\ ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\ 0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\ 0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik} \end{matrix} \right) \left( \begin{matrix} r\\ b_{1}\\ b_{2}\\ a \end{matrix} \right) =\left( \begin{matrix} \mathrm{e}^{ik}\\ ik\mathrm{e}^{-ik}\\ 0\\ 0 \end{matrix} \right) }$

## Connection with the KdV

If we substitute the relationship

$\displaystyle{ \partial_{x}^{2}w+uw=-\lambda w }$

into the KdV after some manipulation we obtain

$\displaystyle{ \partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial _{x}wQ\right) =0 }$

where $\displaystyle{ Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right) \partial_{x}w. }$ If we integrate this equation then we obtain the result that

$\displaystyle{ \partial_{t}\lambda=0 }$

provided that the eigenfunction $\displaystyle{ w }$ is bounded (which is true for the bound state eigenfunctions). This shows that the discrete eigenvalues are unchanged and $\displaystyle{ u\left( x,t\right) }$ evolves according to the KdV.

## Scattering Data

For the discrete spectrum the eigenfunctions behave like

$\displaystyle{ w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x} }$

as $\displaystyle{ x\rightarrow\infty }$ with

$\displaystyle{ \int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1 }$

The continuous spectrum looks like

$\displaystyle{ v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx} ,\ \ \ x\rightarrow-\infty }$
$\displaystyle{ v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow \infty }$

where $\displaystyle{ r }$ is the reflection coefficient and $\displaystyle{ a }$ is the transmission coefficient. This gives us the scattering data at $\displaystyle{ t=0 }$

$\displaystyle{ S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right) \right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right) }$

The scattering data evolves as

$\displaystyle{ k_{n}=k_{n} }$
$\displaystyle{ c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t} }$
$\displaystyle{ r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t} }$
$\displaystyle{ a\left( k,t\right) =a\left( k,0\right) }$

We can recover $\displaystyle{ u }$ from scattering data. We write

$\displaystyle{ F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n} x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k }$

Then solve

$\displaystyle{ K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left( x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0 }$

This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko }equation. We then find $\displaystyle{ u }$ from

$\displaystyle{ u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right) }$

## 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 $\displaystyle{ r\left( k,0\right) =0 }$ for all values of $\displaystyle{ k }$ for some $\displaystyle{ u. }$ In this case

$\displaystyle{ F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x} }$

then

$\displaystyle{ K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right) \sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0 }$

From the equation we can see that

$\displaystyle{ K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right) \mathrm{e}^{-k_{m}y} }$

If we substitute this into the equation

$\displaystyle{ -\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y} +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) } +\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right) \mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0 }$

$\displaystyle{ -\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y} +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) } -\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left( t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left( y+x\right) }=0 }$

and we can eliminate the sum over $\displaystyle{ n }$ , the $\displaystyle{ c_{n}\left( t\right) , }$ and the $\displaystyle{ \mathrm{e}^{-k_{n}y} }$ to obtain

$\displaystyle{ -v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1} ^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}} v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0 }$

which is an algebraic (finite dimensional system)\ for the unknows $\displaystyle{ v_{n}. }$ We can write this as

$\displaystyle{ \left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f} }$

where $\displaystyle{ f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x} }$ and

$\displaystyle{ c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x} }$
$\displaystyle{ K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left( \mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y} }$

$\displaystyle{ u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I} +\mathbf{C}\right) \right] }$
Lets consider some simple examples. First of all if $\displaystyle{ n=1 }$ (the single soliton solution) we get
$\displaystyle{ \begin{matrix} K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left( t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left( t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\ & =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}} \end{matrix} }$
where $\displaystyle{ \mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) . }$ Therefore
$\displaystyle{ \begin{matrix} u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\ & =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1} x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\ & =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta} /\sqrt{2k_{1}}\right) ^{2}}\\ & =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\} \end{matrix} }$
where $\displaystyle{ \theta=k_{1}x-4k^{3}t-\alpha/2 }$ and $\displaystyle{ \sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0} } }$. This is of course the single soliton solution.