Connection betwen KdV and the Schrodinger Equation: Difference between revisions

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</math></center>
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left(  \lambda-u\right)
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
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left(  w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
<center><math>
\partial_{t}\lambda=0
\partial_{t}\lambda=0
</math></center>
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
and <math>u\left(  x,t\right)  </math> evolves according to the KdV. Many other
and <math>u\left(  x,t\right)  </math> evolves according to the KdV. Many other
properties can be found
properties can be found.
 
Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).


==Scattering Data==
==Scattering Data==
Line 28: Line 35:
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 58:
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 72:
<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 97:
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 108:
^{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 110: Line 115:
+\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(  x+y\right)  }
+\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(
-\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(
t\right)  }{k_{n}+k_{m}}v_{m}\left(  x, t\right)  e^{-k_{m}x}e^{-k_{n}\left(
y+x\right)  }=0
y+x\right)  }=0
</math></center>
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left(  t\right)  ,</math> and the
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<math>e^{-k_{n}y}</math> to obtain
<center><math>
<center><math>
-v_{n}\left(  x\right)  +c_{n}\left(  t\right)  e^{-k_{n}x}-\sum_{m=1}
-v_{n}\left(  x,t\right)  +c_{n}^2\left(  t\right)  e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left(  t\right) c_{m}\left(  t\right)  }{k_{n}+k_{m}}
^{N}\frac{c_{n}^2\left(  t\right)   }{k_{n}+k_{m}}
v_{m}\left(  x\right)  e^{-\left(  k_{m}+k_{n}\right)  x}=0
v_{m}\left(  x,t\right)  e^{-\left(  k_{m}+k_{n}\right)  x}=0
</math></center>
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
which is an algebraic (finite dimensional system) for the unknowns <math>v_{n}.</math>.
can write this as
 
We can write this as
<center><math>
<center><math>
\left(  \mathbf{I}+\mathbf{C}\right)  \vec{v}=\vec{f}
\left(  \mathbf{I}+\mathbf{C}\right)  \vec{v}=\vec{f}
</math></center>
</math></center>
From the equation we can see that
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>
<center><math>
K\left(  x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left(  t\right)  v_{m}\left(
c_{nm} = \frac{c_{n}^2\left(  t\right)}
x\right)  e^{-k_{m}y}
{k_{n}+k_{m}}e^{-\left(  k_{m}+k_{n}\right)  x}
</math></center>
</math></center>
If we substitute this into the equation
This means
<center><math>
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left(  x\right)  e^{-k_{n}y}
v_n(x, t)=\sum_{m=1}^{N}\left(\left(\mathbf{I}+\mathbf{C}\right)  ^{-1}\right)_{nm}f_m
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right)  e^{-k_{n}\left( x+y\right)  }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left(  t\right)  v_{m}\left(  x\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></center>
</math></center>
which leads to
Since <math>\partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n</math>, this gives us
<center><math>
<center><math>\begin{align}
-\sum_{n=1}^{N}c_{n}\left(  t\right)  v_{n}\left( x\right) e^{-k_{n}y}
K\left(  x,x,t\right)  =-&\sum_{n=1}^{N} v_n(x, t)e^{-k_{n}x} \\
+\sum_{n=1}^{N}c_{n}^{2}\left(  t\right) e^{-k_{n}\left(  x+y\right) }
=-&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_me^{-k_{n}x} \\
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right)  c_{n}^{2}\left(
  =&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right)  ^{-1}\right)_{nm}\partial_x\left(\left(\mathbf{I}+\mathbf{C}\right)_{mn}\right) \\
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right)  e^{-k_{m}x}e^{-k_{n}\left(
  =\,&\operatorname{Tr}\left(\left(\mathbf{I}+\mathbf{C}\right)^{-1}\partial_x\left(\mathbf{I}+\mathbf{C}\right)\right) \\
y+x\right) }=0
=\,&\dfrac{\partial_x\left(\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)}{\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)} \\
=\,&\partial_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
\end{align}
</math></center>
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left(  t\right)  ,</math> and the
And therefore
<math>e^{-k_{n}y}</math> to obtain
<center><math>
<center><math>
-v_{n}\left(  x\right) +c_{n}\left( t\right) e^{-k_{n}x}-\sum_{m=1}
u\left(  x,t\right) =2\partial_{x}K\left(x,x,t\right)=2\partial^2_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
^{N}\frac{c_{n}\left( t\right)  c_{m}\left( t\right)  }{k_{n}+k_{m}}
v_{m}\left(  x\right) e^{-\left(  k_{m}+k_{n}\right) x}=0
</math></center>
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
 
can write this as
=== Single Soliton Example ===
<center><math>
 
\left(  \mathbf{I}+\mathbf{C}\right)  \vec{v}=\vec{f}
If <math>n=1</math> (a single soliton
</math></center>
where <math>f_{m}=c_{m}\left(  t\right)  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}}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}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
solution) we get
<center><math>\begin{matrix}
<center><math>\begin{matrix}
K\left(  x,x,t\right)  &  =-\frac{c_{1}\left(  t\right)  c_{1}\left(
K\left(  x,x,t\right)  &  =-\frac{c_{1}^2\left(  t\right)   
t\right)  e^{-k_{1}x}e^{-k_{1}x}}{1+\frac{c_{1}\left(  t\right)  c_{1}\left(
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left(  t\right)  }
t\right)  }{k_{1}+k_{1}}e^{-\left(  k_{1}+k_{1}\right)  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}}\\
&  =2k_{1}^{2}\mbox{sech}^{2}\left\{  k_{1}\left(  x-x_{0}\right)  -4k_{1}
&  =2k_{1}^{2}\operatorname{sech}^{2}\left\{  k_{1}\left(  x-x_{0}\right)  -4k_{1}
^{3}t\right\}
^{3}t\right\}
\end{align}</math></center>
\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}
where <math>\theta=k_{1}x-4k_{1}^{3}t-\alpha/2</math> and <math>\sqrt{2k_{1}}e^{-\alpha/2}=e^{-k_{1}x_{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 08:42, 5 October 2025

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 we obtain

[math]\displaystyle{ \int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial _{x}wQ\right)\Bigg|^\infty_{-\infty}=0 }[/math]

and, since [math]\displaystyle{ w^2\gt 0 }[/math] except for some isolated points,

[math]\displaystyle{ \partial_{t}\lambda=0 }[/math]

provided that the eigenfunction [math]\displaystyle{ w }[/math] vanishes at [math]\displaystyle{ x=\pm\infty }[/math] (which is true for the bound state eigenfunctions, i.e. those from the discrete spectrum). 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.

Some more details on the content of this lecture can be found in Dunajski, Maciej (2009). Solitons, Instantons, and Twistors (Chapters 2.2 and 2.3).

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, t\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 unknowns [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_{nm} = \frac{c_{n}^2\left( t\right)} {k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x} }[/math]

This means

[math]\displaystyle{ v_n(x, t)=\sum_{m=1}^{N}\left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_m }[/math]

Since [math]\displaystyle{ \partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n }[/math], this gives us

[math]\displaystyle{ \begin{align} K\left( x,x,t\right) =-&\sum_{n=1}^{N} v_n(x, t)e^{-k_{n}x} \\ =-&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_me^{-k_{n}x} \\ =&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}\partial_x\left(\left(\mathbf{I}+\mathbf{C}\right)_{mn}\right) \\ =\,&\operatorname{Tr}\left(\left(\mathbf{I}+\mathbf{C}\right)^{-1}\partial_x\left(\mathbf{I}+\mathbf{C}\right)\right) \\ =\,&\dfrac{\partial_x\left(\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)}{\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)} \\ =\,&\partial_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right) \end{align} }[/math]

And therefore

[math]\displaystyle{ u\left( x,t\right) =2\partial_{x}K\left(x,x,t\right)=2\partial^2_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right) }[/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}\operatorname{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_{1}^{3}t-\alpha/2 }[/math] and [math]\displaystyle{ \sqrt{2k_{1}}e^{-\alpha/2}=e^{-k_{1}x_{0} } }[/math]. This is of course the single soliton solution.

Lecture Videos

Part 1

Part 2

Part 3

Part 4