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| | e^{-2k_{1}x}}{1+\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}}\\ | | {2k_{1}}e^{-2 k_{1} x}}\\ |
| − | & =\frac{-1}{1+e^{-2k_{1}x + 8k_{1}^{3}t-\alpha}} | + | & =\frac{-1}{e^{-2k_{1}x + 8k_{1}^{3}t-\alpha} + 1/2k_1} |
| | \end{matrix}</math></center> | | \end{matrix}</math></center> |
| − | where <math>e^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore | + | where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore |
| | <center><math>\begin{align} | | <center><math>\begin{align} |
| | u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\ | | u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\ |
Revision as of 02:58, 24 September 2010
If we substitute the relationship
[math]\displaystyle{
\partial_{x}^{2}w+uw=-\lambda w
}[/math]
into the KdV after some manipulation we obtain
[math]\displaystyle{
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
}[/math]
where [math]\displaystyle{ 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
[math]\displaystyle{
\partial_{t}\lambda=0
}[/math]
provided that the eigenfunction [math]\displaystyle{ w }[/math] is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and [math]\displaystyle{ 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
[math]\displaystyle{
w_{n}\left( x\right) =c_{n}\left( t\right) e^{-k_{n}x}
}[/math]
as [math]\displaystyle{ x\rightarrow\infty }[/math] with
[math]\displaystyle{
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
}[/math]
The continuous spectrum looks like
[math]\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
}[/math]
[math]\displaystyle{
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
}[/math]
where [math]\displaystyle{ r }[/math] is the reflection coefficient and [math]\displaystyle{ a }[/math] is the transmission
coefficient. This gives us the scattering data at [math]\displaystyle{ t=0 }[/math]
[math]\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)
}[/math]
The scattering data evolves as
[math]\displaystyle{
k_{n}=k_{n}
}[/math]
[math]\displaystyle{
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
}[/math]
[math]\displaystyle{
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
}[/math]
[math]\displaystyle{
a\left( k,t\right) =a\left( k,0\right)
}[/math]
We can recover [math]\displaystyle{ u }[/math] from scattering data. We write
[math]\displaystyle{
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}dk
}[/math]
Then solve
[math]\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) dz=0
}[/math]
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find [math]\displaystyle{ u }[/math] from
[math]\displaystyle{
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
}[/math]
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]\displaystyle{ r\left( k,0\right) =0 }[/math] for all values of [math]\displaystyle{ k }[/math] for some [math]\displaystyle{ u. }[/math] In this case
[math]\displaystyle{
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
}[/math]
then
[math]\displaystyle{
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]
From the equation we can see that
[math]\displaystyle{
K\left( x,y,t\right) =-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}y}
}[/math]
If we substitute this into the equation,
[math]\displaystyle{
-\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) }dz=0
}[/math]
which leads to
[math]\displaystyle{
-\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]
and we can eliminate the sum over [math]\displaystyle{ n }[/math] and the
[math]\displaystyle{ e^{-k_{n}y} }[/math] to obtain
[math]\displaystyle{
-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]
which is an algebraic (finite dimensional system) for the unknows [math]\displaystyle{ v_{n}. }[/math].
We can write this as
[math]\displaystyle{
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
}[/math]
where [math]\displaystyle{ f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x} }[/math] and
the elements of [math]\displaystyle{ mathbf{C} }[/math] are given by
[math]\displaystyle{
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]
This gives us
[math]\displaystyle{
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
}[/math]
We then find [math]\displaystyle{ u(x,t) }[/math] from [math]\displaystyle{ K }[/math].
Single Soliton Example
If [math]\displaystyle{ n=1 }[/math] (a single soliton
solution) we get
[math]\displaystyle{ \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]
where [math]\displaystyle{ e^{-\alpha}=1/c_{0}^{2}\left( 0\right) . }[/math] Therefore
[math]\displaystyle{ \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( 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_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align} }[/math]
where [math]\displaystyle{ \theta=k_{1}x-4k^{3}t-\alpha/2 }[/math] and [math]\displaystyle{ \sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
} }[/math]. This is of course the single soliton solution.
