Difference between revisions of "Connection betwen KdV and the Schrodinger Equation"

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For the discrete spectrum the eigenfunctions behave like
 
For the discrete spectrum the eigenfunctions behave like
 
<center><math>
 
<center><math>
w_{n}\left(  x\right)  =c_{n}\left(  t\right)  e^{-k_{n}x}
+
w_{n}\left(  x,t\right)  =c_{n}\left(  t\right)  e^{-k_{n}x}
 
</math></center>
 
</math></center>
 
as <math>x\rightarrow\infty</math> with
 
as <math>x\rightarrow\infty</math> with
 
<center><math>
 
<center><math>
\int_{-\infty}^{\infty}\left(  w_{n}\left(  x\right)  \right)  ^{2}dx=1
+
\int_{-\infty}^{\infty}\left(  w_{n}\left(  x,t\right)  \right)  ^{2}\mathrm{d}x=1
 
</math></center>
 
</math></center>
 
The continuous spectrum looks like
 
The continuous spectrum looks like
 
<center><math>
 
<center><math>
v\left(  x,t\right)  \approx \mathrm{e}^{-\mathrm{i}kx}+r\left(  k,t\right)  \mathrm{e}^{\mathrm{i}kx}
+
w\left(  x,t\right)  \approx \mathrm{e}^{-\mathrm{i}kx}+r\left(  k,t\right)  \mathrm{e}^{\mathrm{i}kx}
 
,\ \ \ x\rightarrow-\infty
 
,\ \ \ x\rightarrow-\infty
 
</math></center>
 
</math></center>
 
<center><math>
 
<center><math>
v\left(  x,t\right)  \approx a\left(  k,t\right)  \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
+
w\left(  x,t\right)  \approx a\left(  k,t\right)  \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
 
\infty
 
\infty
 
</math></center>
 
</math></center>
Line 51: Line 51:
 
The scattering data evolves as
 
The scattering data evolves as
 
<center><math>
 
<center><math>
k_{n}=k_{n}
+
k_{n}(t)=k_{n}(0) = k_{n}
 
</math></center>
 
</math></center>
 
<center><math>
 
<center><math>
Line 65: Line 65:
 
<center><math>
 
<center><math>
 
F\left(  x,t\right)  =\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}
 
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
+
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left(  k,t\right)  \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
 
</math></center>
 
</math></center>
 
Then solve
 
Then solve
 
<center><math>
 
<center><math>
 
K\left(  x,y;t\right)  +F\left(  x+y;t\right)  +\int_{x}^{\infty}K\left(
 
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
+
x,z;t\right)  F\left(  z+y;t\right)  \mathrm{d}z=0
 
</math></center>
 
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
+
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
}equation. We then find <math>u</math> from
 
 
<center><math>
 
<center><math>
 
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)
 
</math></center>
 
</math></center>
 
  
 
==Reflectionless Potential==
 
==Reflectionless Potential==
Line 92: Line 90:
 
K\left(  x,y,t\right)  +\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)
 
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)
 
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
+
\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(  y+z\right)  }\mathrm{d}z=0
 
</math></center>
 
</math></center>
 
From the equation we can see that
 
From the equation we can see that
 
<center><math>
 
<center><math>
K\left(  x,y,t\right)  =-\sum_{m=1}^{N}v_{m}\left(  x,t\right)  e^{-k_{m}y}
+
K\left(  x,y,t\right)  =-\sum_{n=1}^{N}v_{n}\left(  x,t\right)  e^{-k_{n}y}
 
</math></center>
 
</math></center>
 
If we substitute this into the equation,
 
If we substitute this into the equation,
Line 103: Line 101:
 
^{2}\left(  t\right)  e^{-k_{n}\left(  x+y\right)  }+\int_{x}^{\infty}
 
^{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}
 
-\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
+
^{2}\left(  t\right)  e^{-k_{n}\left(  y+z\right)  }\mathrm{d}z=0
 
</math></center>
 
</math></center>
 
which leads to
 
which leads to
Line 127: Line 125:
 
</math></center>
 
</math></center>
 
where <math>f_{m}=c_{m}^2\left(  t\right)  e^{-k_{m}x}</math> and
 
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
+
the elements of <math>\mathbf{C}</math> are given by
 
<center><math>
 
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}^2\left(  t\right)}
+
c_{mn} = \frac{c_{n}^2\left(  t\right)}
 
{k_{n}+k_{m}}e^{-\left(  k_{m}+k_{n}\right)  x}
 
{k_{n}+k_{m}}e^{-\left(  k_{m}+k_{n}\right)  x}
 
</math></center>
 
</math></center>
Line 147: Line 145:
 
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)  \\
&  =\frac{4k_{1}e^{-2k_{1}x+8k_{1}^{3}t-\alpha}}{\left( 1+e^{-2k_{1}
+
&  =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x+8k_{1}^{3}t-\alpha}\right)  ^{2}}\\
+
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right)  ^{2}}\\
 
&  =\frac{8k_{1}^{2}}{\left(  \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
 
&  =\frac{8k_{1}^{2}}{\left(  \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
 
{2k_{1}}\right)  ^{2}}\\
 
{2k_{1}}\right)  ^{2}}\\
Line 161: Line 159:
 
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
 
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.
 
}</math>. This is of course the single soliton solution.
 +
 +
 +
== Lecture Videos ==
 +
 +
=== Part 1 ===
 +
 +
{{#ev:youtube|KHGoPCoyP28}}
 +
 +
=== Part 2 ===
 +
 +
{{#ev:youtube|8mVq6MQWO3I}}
 +
 +
=== Part 3 ===
 +
 +
{{#ev:youtube|meTgaaGKsfQ}}
 +
 +
=== Part 4 ===
 +
 +
{{#ev:youtube|AfAQyjzvJUU}}

Latest revision as of 05:30, 14 September 2023

Nonlinear PDE's Course
Current Topic Connection betwen KdV and the Schrodinger Equation
Next Topic Example Calculations for the KdV and IST
Previous Topic Properties of the Linear Schrodinger Equation


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,t\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,t\right) \right) ^{2}\mathrm{d}x=1 }[/math]

The continuous spectrum looks like

[math]\displaystyle{ 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]
[math]\displaystyle{ w\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}(t)=k_{n}(0) = 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}\mathrm{d}k }[/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) \mathrm{d}z=0 }[/math]

This is a linear integral equation called the 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) }\mathrm{d}z=0 }[/math]

From the equation we can see that

[math]\displaystyle{ K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}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) }\mathrm{d}z=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} = \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( 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]

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.


Lecture Videos

Part 1

Part 2

Part 3

Part 4