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Line 1: {{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
Line 12:
Line 10: 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
Line 28:
Line 26: 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>
Line 46:
Line 44: 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}^{-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)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}
\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & - \kappa \sin\kappa
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
- \kappa \sin\kappa \mathrm{e}^{-kx}+\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.
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}\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}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
}</math>. This is of course the single soliton solution.
[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
