View source for Connection betwen KdV and the Schrodinger Equation

{{nonlinear waves course
 | chapter title = Connection betwen KdV and the Schrodinger Equation
 | next chapter = [[Example Calculations for the KdV and IST]]
 | previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

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. Many other
properties can be found

==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}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\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>
w\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)  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}+\frac{1}{2\pi}\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==

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)  }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left(  x,y,t\right)  =-\sum_{m=1}^{N}v_{m}\left(  x,t\right)  e^{-k_{m}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left(  x,t\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}v_{m}\left(  x,t\right)  e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left(  t\right)  e^{-k_{n}\left(  y+z\right)  }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}  v_{n}\left(  x,t\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_{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> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left(  x,t\right)  +c_{n}^2\left(  t\right)  e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left(  t\right)   }{k_{n}+k_{m}}
v_{m}\left(  x,t\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}^2\left(  t\right)  e^{-k_{m}x}</math> and
the elements of <math>mathbf{C}</math> are given by
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}^2\left(  t\right)}
{k_{n}+k_{m}}e^{-\left(  k_{m}+k_{n}\right)  x}
</math></center>
This gives us
<center><math>
K\left(  x,y,t\right)  =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right)  ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left(  x,x,t\right)   &  =-\frac{c_{1}^2\left(  t\right)  
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left(  t\right)  }
{2k_{1}}e^{-2 k_{1} x}}\\
&  =\frac{-1}{e^{-2k_{1}x + 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left(  0\right)  .</math> Therefore
<center><math>\begin{align}
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( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right)  ^{2}}\\
&  =\frac{8k_{1}^{2}}{\left(  \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right)  ^{2}}\\
&  =2k_{1}^{2}\mbox{sech}^{2}\left\{  k_{1}\left(  x-x_{0}\right)  -4k_{1}
^{3}t\right\}
\end{align}</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.