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

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{{nonlinear waves course
 
{{nonlinear waves course
 
  | chapter title = Introduction to the Inverse Scattering Transform
 
  | chapter title = Introduction to the Inverse Scattering Transform
  | next chapter = [[Reaction-Diffusion Systems]]
+
  | next chapter = [[Properties of the Linear Schrodinger Equation]]
 
  | previous chapter = [[Conservation Laws for the KdV]]
 
  | previous chapter = [[Conservation Laws for the KdV]]
 
}}
 
}}
  
 
== Introduction ==
 
  
 
The inverse scattering transformation gives a way to solve the KdV equation
 
The inverse scattering transformation gives a way to solve the KdV equation
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Fourier transformation, except it works for a non linear equation. We want to
 
Fourier transformation, except it works for a non linear equation. We want to
 
be able to solve
 
be able to solve
<center><math>\begin{align}
+
<center><math>\begin{matrix}
 
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u  &  =0\\
 
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u  &  =0\\
 
u(x,0)  &  =f\left(  x\right)
 
u(x,0)  &  =f\left(  x\right)
\end{align}</math></center>
+
\end{matrix}</math></center>
 
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>
 
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>
  
 
The Miura transformation is given by
 
The Miura transformation is given by
 
<center><math>
 
<center><math>
u=v^{2}+v_{x} \,
+
u=-v^{2}-\partial_x v\,
 
</math></center>
 
</math></center>
 
and if <math>v</math> satisfies the mKdV
 
and if <math>v</math> satisfies the mKdV
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then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
 
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
 
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
+
nonlinear ODE is also known as the Riccati equation and there is a well known
 
transformation which linearises this equation. It we write
 
transformation which linearises this equation. It we write
 
<center><math>
 
<center><math>
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in the inverse scattering transformation. Note that this is Schrodinger's equation.
 
in the inverse scattering transformation. Note that this is Schrodinger's equation.
  
==Properties of the eigenfunctions==
+
== Lecture Videos ==  
 
 
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}
 
 
 
0, & x\notin\left[  -1,1\right] \\
 
b, & x\in\left[  -1,1\right]
 
\end{matrix}
 
\right.
 
</math></center>
 
where <math>b>0.</math>
 
 
 
==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}
 
 
 
a_{1}\mathrm{e}^{kx}, & x<-1\\
 
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
 
a_{2}\mathrm{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 equations, 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}
 
 
 
\mathrm{e}^{-k} & -\cos\kappa\\
 
k\mathrm{e}^{-k} & -\kappa \sin\kappa
 
\end{matrix}
 
\right)  \left(
 
\begin{matrix}
 
 
 
a_{1}\\
 
b_{1}
 
\end{matrix}
 
\right)  =\left(
 
\begin{matrix}
 
 
 
0\\
 
0
 
\end{matrix}
 
\right)
 
</math></center>
 
This has non trivial solutions when
 
<center><math>
 
\det\left(
 
\begin{matrix}
 
 
 
\mathrm{e}^{-k} & -\cos\kappa\\
 
k\mathrm{e}^{-k} & -\kappa \sin\kappa
 
\end{matrix}
 
\right)  =0
 
</math></center>
 
which gives us the equation
 
<center><math>
 
- \kappa \sin\kappa \mathrm{e}^{-k}+\left(  \cos\kappa\right)  k\mathrm{e}^{-kx}=0
 
</math></center>
 
or
 
<center><math>
 
\kappa \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.
 
 
 
In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
 
Similarly we repeat the above process for the odd solutions.
 
 
 
==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}
 
 
 
\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
 
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
 
a\mathrm{e}^{-\mathrm{i}kx} & 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}
 
 
 
-\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)
 
</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)  \mathrm{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 \mathrm{e}^{-\mathrm{i}kx}+r\left(  k,t\right)  \mathrm{e}^{\mathrm{i}kx}
 
,\ \ \ x\rightarrow-\infty
 
</math></center>
 
<center><math>
 
v\left(  x,t\right)  \approx a\left(  k,t\right)  \mathrm{e}^{-\mathrm{i}kx},\ \ \ 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)  \mathrm{e}^{4k_{n}^{3}t}
 
</math></center>
 
<center><math>
 
r\left(  k,t\right)  =r\left(  k,0\right)  \mathrm{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)  \mathrm{e}^{-k_{n}
 
x}+\int_{-\infty}^{\infty}r\left(  k,t\right)  \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
 
</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)  \mathrm{d}z=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==
+
=== Part 1 ===
  
In general the IST is difficult to solve. However, there is a simplification
+
{{#ev:youtube|P3uMk9OS8p4}}
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)  \mathrm{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)
 
\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
 
</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)  \mathrm{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)  \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
 
</math></center>
 
which leads to
 
<center><math>
 
-\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
 
</math></center>
 
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left(  t\right)  ,</math> and the
 
<math>\mathrm{e}^{-k_{n}y}</math> to obtain
 
<center><math>
 
-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
 
</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)  \mathrm{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}}\mathrm{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}\mathrm{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)  \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}</math></center>
 
where <math>\mathrm{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}\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}\mathrm{sech}^{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}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
 
}</math>. This is of course the single soliton solution.
 

Latest revision as of 03:26, 15 September 2020

Nonlinear PDE's Course
Current Topic Introduction to the Inverse Scattering Transform
Next Topic Properties of the Linear Schrodinger Equation
Previous Topic 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

[math]\displaystyle{ \begin{matrix} \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ u(x,0) & =f\left( x\right) \end{matrix} }[/math]

with [math]\displaystyle{ \left\vert u\right\vert \rightarrow0 }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty. }[/math]

The Miura transformation is given by

[math]\displaystyle{ u=-v^{2}-\partial_x v\, }[/math]

and if [math]\displaystyle{ v }[/math] satisfies the mKdV

[math]\displaystyle{ \partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0 }[/math]

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

[math]\displaystyle{ v=\frac{\left( \partial_{x}w\right) }{w} }[/math]

then we obtain the equation

[math]\displaystyle{ \partial_{x}^{2}w+uw=0 }[/math]

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

[math]\displaystyle{ \partial_{x}^{2}w+uw=-\lambda w }[/math]

The eigenfunctions and eigenvalues of this scattering problem play a key role in the inverse scattering transformation. Note that this is Schrodinger's equation.

Lecture Videos

Part 1