Introduction to the Inverse Scattering Transform: Difference between revisions

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{{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]]
}}
}}


= Miura transform =


== 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 12:
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 28:
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 46:
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==
= Lax Pair =
 
A powerful way to study the KdV equation is via the idea of a '''Lax pair'''.
 
 
A Lax pair consists of two linear operators <math> L </math> and <math> P </math> such that the KdV equation is equivalent to the so-called '''Lax equation''':
 
:<math> \frac{dL}{dt} = [P, L] = PL - LP </math>
 
For the KdV equation, a classical Lax pair is:
 
:<math> L = -\partial_x^2 + u(x,t) </math>
 
:<math> P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u </math>
 
These are differential operators acting on a function <math> \psi(x,t) </math>. The idea is that the evolution of <math> u(x,t) </math> is such that the operator <math> L </math> undergoes an isospectral deformation: its spectrum does not change in time.
 
If <math> L\psi = \lambda \psi </math> and <math> \psi_t = P\psi </math>, then consistency requires:
 
:<math> \frac{d}{dt}(L\psi) = L_t \psi + L\psi_t = PL\psi </math>
 
so
 
:<math> L_t = [P, L] </math>
 
This is the Lax equation.
 
=== Derivation of the KdV Equation from the Lax Pair ===
 
Let us now compute <math> [P, L] = PL - LP </math> explicitly, using:
 
:<math> L = -\partial_x^2 + u </math>
 
:<math> P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u </math>
 
We compute the commutator acting on a test function <math> \psi </math>:
 
'''Step 1:''' Compute <math> PL\psi </math>.
 
<math>
\begin{align}
PL\psi &= P(-\psi_{xx} + u\psi) \\
&= -4\partial_x^3(-\psi_{xx} + u\psi) + 3u\partial_x(-\psi_{xx} + u\psi) + 3\partial_x u (-\psi_{xx} + u\psi)
\end{align}
</math>


The equation
We compute each term:
<center><math>
 
\partial_{x}^{2}w+uw=-\lambda w
* <math> -4\partial_x^3(-\psi_{xx}) = -4\partial_x^3(-\psi_{xx}) = 4\psi_{xxxxx} </math>
</math></center>
* <math> -4\partial_x^3(u\psi) = -4[(u\psi)_{xxx}] </math>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
* <math> 3u\partial_x(-\psi_{xx}) = -3u\psi_{xxx} </math>
first are waves and the second are bound solutions. It is well known that
* <math> 3u\partial_x(u\psi) = 3u(u\psi)_x = 3u^2\psi_x + 3uu_x\psi </math>
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
* <math> 3\partial_x u (-\psi_{xx}) = -3u_x\psi_{xx} </math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
* <math> 3\partial_x u (u\psi) = 3u_x u \psi + 3u u_x \psi = 6u u_x \psi </math>
incident waves.
 
Add all the terms together:
 
<math>
\begin{align}
PL\psi &= 4\psi_{xxxxx} -4(u\psi)_{xxx} -3u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi -3u_x\psi_{xx} + 6uu_x\psi
\end{align}
</math>
 
'''Step 2:''' Compute <math> LP\psi </math>.
 
<math>
\begin{align}
LP\psi &= (-\partial_x^2 + u)P\psi = -\partial_x^2(P\psi) + u P\psi
\end{align}
</math>
 
We focus on the leading terms in <math> P\psi </math>:
 
<math>
P\psi = -4\psi_{xxx} + 3u\psi_x + 3u_x\psi
</math>
 
Then:
 
<math>
\begin{align}
LP\psi &= -\partial_x^2(-4\psi_{xxx} + 3u\psi_x + 3u_x\psi) + u(-4\psi_{xxx} + 3u\psi_x + 3u_x\psi) \\
&= 4\psi_{xxxxx} -3\partial_x^2(u\psi_x) -3\partial_x^2(u_x\psi) -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
 
Compute:
 
* <math> \partial_x^2(u\psi_x) = u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx} </math>
* <math> \partial_x^2(u_x\psi) = u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx} </math>
 
So:
 
<math>
\begin{align}
LP\psi &= 4\psi_{xxxxx} -3(u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx}) -3(u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx}) \\
&\quad -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
 
Simplify:
 
<math>
\begin{align}
LP\psi &= 4\psi_{xxxxx} -3u_{xx}\psi_x -6u_x\psi_{xx} -3u\psi_{xxx} -3u_{xxx}\psi -6u_{xx}\psi_x -3u_x\psi_{xx} \\
&\quad -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
 
Group like terms:
 
* <math> -3u\psi_{xxx} -4u\psi_{xxx} = -7u\psi_{xxx} </math>
* <math> -6u_x\psi_{xx} -3u_x\psi_{xx} = -9u_x\psi_{xx} </math>
* <math> -3u_{xx}\psi_x -6u_{xx}\psi_x = -9u_{xx}\psi_x </math>
* <math> -3u_{xxx}\psi </math>
 
So:
 
<math>
\begin{align}
LP\psi &= 4\psi_{xxxxx} -7u\psi_{xxx} -9u_x\psi_{xx} -9u_{xx}\psi_x -3u_{xxx}\psi + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
 
'''Step 3:''' Compute <math> [P, L]\psi = PL\psi - LP\psi </math>
 
Subtract term by term (all <math> \psi </math> terms):
 
* The <math> 4\psi_{xxxxx} </math> terms cancel.
* Remaining terms:


===Example: Scattering by a Well===
<math>
\begin{align}
[P, L]\psi &= (-4(u\psi)_{xxx} + 3u\psi_{xxx}) + 3u_x\psi_{xx} + 9u_{xx}\psi_x + 3u_{xxx}\psi + 6uu_x\psi
\end{align}
</math>


The properties of the eigenfunction is prehaps seem most easily through the
Use:
following example
<center><math>
u\left(  x\right)  =\left\{
\begin{matrix}


0, & x\notin\left[  -1,1\right] \\
<math> (u\psi)_{xxx} = u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx} </math>
b, & x\in\left[  -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>


==Case when <math>\lambda<0</math>==
So:


If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
<math>
get
\begin{align}
<center><math>
-4(u\psi)_{xxx} &= -4(u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx}) \\
w\left(  x\right) =\left\{
&= -4u_{xxx}\psi -12u_{xx}\psi_x -12u_x\psi_{xx} -4u\psi_{xxx}
\begin{matrix}
\end{align}
</math>


a_{1}\mathrm{e}^{kx}, & x<-1\\
Now add everything:
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}^{-k} & -\cos\kappa\\
<math>
k\mathrm{e}^{-k} & -\kappa \sin\kappa
\begin{align}
\end{matrix}
[P,L]\psi &= (-4u_{xxx}\psi -12u_{xx}\psi_x -12u_x\psi_{xx} -4u\psi_{xxx}) + 3u\psi_{xxx} + 3u_x\psi_{xx} + 9u_{xx}\psi_x + 3u_{xxx}\psi + 6uu_x\psi \\
\right) \left(
&= (-u_{xxx}\psi -3u_{xx}\psi_x -9u_x\psi_{xx} -u\psi_{xxx}) + 6uu_x\psi
\begin{matrix}
\end{align}
</math>


a_{1}\\
So the final result is:
b_{1}
\end{matrix}
\right)  =\left(
\begin{matrix}


0\\
<math>
0
[P,L]\psi = ( -u_{xxx} + 6uu_x ) \psi
\end{matrix}
</math>
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}


\mathrm{e}^{-k} & -\cos\kappa\\
Thus, we have:
k\mathrm{e}^{-k} & -\kappa \sin\kappa
\end{matrix}
\right)  =0
</math></center>
which gives us the equation
<center><math>
- \kappa \sin\kappa \mathrm{e}^{-k}+\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.


In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
:<math> \frac{dL}{dt} = [P, L] \quad \Rightarrow \quad \frac{du}{dt} = -u_{xxx} + 6uu_x </math>
Similarly we repeat the above process for the odd solutions.


==Case when <math>\lambda>0</math>==
Which is exactly the KdV equation.


When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
=== Summary ===
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}


\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
The KdV equation:
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\\
:<math> u_t + 6uu_x + u_{xxx} = 0 </math>
-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\\
can be written as the Lax equation:
b_{1}\\
b_{2}\\
a
\end{matrix}
\right)  =\left(
\begin{matrix}


\mathrm{e}^{ik}\\
:<math> \frac{dL}{dt} = [P, L] </math>
-ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>


==Connection with the KdV==
where


If we substitute the relationship
:<math> L = -\partial_x^2 + u, \qquad P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u </math>
<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==
This reformulation reveals the integrable structure of the KdV equation and allows powerful solution methods such as the inverse scattering transform.


For the discrete spectrum the eigenfunctions behave like
== Lecture Videos ==  
<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}+ \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==
=== 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}\mathrm{sech}^{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.

Latest revision as of 00:56, 7 August 2025

Nonlinear PDE's Course
Current Topic Introduction to the Inverse Scattering Transform
Next Topic Properties of the Linear Schrodinger Equation
Previous Topic Conservation Laws for the KdV


Miura transform

The inverse scattering transformation gives a way to solve the KdV equation exactly. You can think about is as being an analogous transformation to the Fourier transformation, except it works for a non linear equation. We want to be able to solve

[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.

Lax Pair

A powerful way to study the KdV equation is via the idea of a Lax pair.


A Lax pair consists of two linear operators [math]\displaystyle{ L }[/math] and [math]\displaystyle{ P }[/math] such that the KdV equation is equivalent to the so-called Lax equation:

[math]\displaystyle{ \frac{dL}{dt} = [P, L] = PL - LP }[/math]

For the KdV equation, a classical Lax pair is:

[math]\displaystyle{ L = -\partial_x^2 + u(x,t) }[/math]
[math]\displaystyle{ P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u }[/math]

These are differential operators acting on a function [math]\displaystyle{ \psi(x,t) }[/math]. The idea is that the evolution of [math]\displaystyle{ u(x,t) }[/math] is such that the operator [math]\displaystyle{ L }[/math] undergoes an isospectral deformation: its spectrum does not change in time.

If [math]\displaystyle{ L\psi = \lambda \psi }[/math] and [math]\displaystyle{ \psi_t = P\psi }[/math], then consistency requires:

[math]\displaystyle{ \frac{d}{dt}(L\psi) = L_t \psi + L\psi_t = PL\psi }[/math]

so

[math]\displaystyle{ L_t = [P, L] }[/math]

This is the Lax equation.

Derivation of the KdV Equation from the Lax Pair

Let us now compute [math]\displaystyle{ [P, L] = PL - LP }[/math] explicitly, using:

[math]\displaystyle{ L = -\partial_x^2 + u }[/math]
[math]\displaystyle{ P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u }[/math]

We compute the commutator acting on a test function [math]\displaystyle{ \psi }[/math]:

Step 1: Compute [math]\displaystyle{ PL\psi }[/math].

[math]\displaystyle{ \begin{align} PL\psi &= P(-\psi_{xx} + u\psi) \\ &= -4\partial_x^3(-\psi_{xx} + u\psi) + 3u\partial_x(-\psi_{xx} + u\psi) + 3\partial_x u (-\psi_{xx} + u\psi) \end{align} }[/math]

We compute each term:

  • [math]\displaystyle{ -4\partial_x^3(-\psi_{xx}) = -4\partial_x^3(-\psi_{xx}) = 4\psi_{xxxxx} }[/math]
  • [math]\displaystyle{ -4\partial_x^3(u\psi) = -4[(u\psi)_{xxx}] }[/math]
  • [math]\displaystyle{ 3u\partial_x(-\psi_{xx}) = -3u\psi_{xxx} }[/math]
  • [math]\displaystyle{ 3u\partial_x(u\psi) = 3u(u\psi)_x = 3u^2\psi_x + 3uu_x\psi }[/math]
  • [math]\displaystyle{ 3\partial_x u (-\psi_{xx}) = -3u_x\psi_{xx} }[/math]
  • [math]\displaystyle{ 3\partial_x u (u\psi) = 3u_x u \psi + 3u u_x \psi = 6u u_x \psi }[/math]

Add all the terms together:

[math]\displaystyle{ \begin{align} PL\psi &= 4\psi_{xxxxx} -4(u\psi)_{xxx} -3u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi -3u_x\psi_{xx} + 6uu_x\psi \end{align} }[/math]

Step 2: Compute [math]\displaystyle{ LP\psi }[/math].

[math]\displaystyle{ \begin{align} LP\psi &= (-\partial_x^2 + u)P\psi = -\partial_x^2(P\psi) + u P\psi \end{align} }[/math]

We focus on the leading terms in [math]\displaystyle{ P\psi }[/math]:

[math]\displaystyle{ P\psi = -4\psi_{xxx} + 3u\psi_x + 3u_x\psi }[/math]

Then:

[math]\displaystyle{ \begin{align} LP\psi &= -\partial_x^2(-4\psi_{xxx} + 3u\psi_x + 3u_x\psi) + u(-4\psi_{xxx} + 3u\psi_x + 3u_x\psi) \\ &= 4\psi_{xxxxx} -3\partial_x^2(u\psi_x) -3\partial_x^2(u_x\psi) -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi \end{align} }[/math]

Compute:

  • [math]\displaystyle{ \partial_x^2(u\psi_x) = u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx} }[/math]
  • [math]\displaystyle{ \partial_x^2(u_x\psi) = u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx} }[/math]

So:

[math]\displaystyle{ \begin{align} LP\psi &= 4\psi_{xxxxx} -3(u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx}) -3(u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx}) \\ &\quad -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi \end{align} }[/math]

Simplify:

[math]\displaystyle{ \begin{align} LP\psi &= 4\psi_{xxxxx} -3u_{xx}\psi_x -6u_x\psi_{xx} -3u\psi_{xxx} -3u_{xxx}\psi -6u_{xx}\psi_x -3u_x\psi_{xx} \\ &\quad -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi \end{align} }[/math]

Group like terms:

  • [math]\displaystyle{ -3u\psi_{xxx} -4u\psi_{xxx} = -7u\psi_{xxx} }[/math]
  • [math]\displaystyle{ -6u_x\psi_{xx} -3u_x\psi_{xx} = -9u_x\psi_{xx} }[/math]
  • [math]\displaystyle{ -3u_{xx}\psi_x -6u_{xx}\psi_x = -9u_{xx}\psi_x }[/math]
  • [math]\displaystyle{ -3u_{xxx}\psi }[/math]

So:

[math]\displaystyle{ \begin{align} LP\psi &= 4\psi_{xxxxx} -7u\psi_{xxx} -9u_x\psi_{xx} -9u_{xx}\psi_x -3u_{xxx}\psi + 3u^2\psi_x + 3uu_x\psi \end{align} }[/math]

Step 3: Compute [math]\displaystyle{ [P, L]\psi = PL\psi - LP\psi }[/math]

Subtract term by term (all [math]\displaystyle{ \psi }[/math] terms):

  • The [math]\displaystyle{ 4\psi_{xxxxx} }[/math] terms cancel.
  • Remaining terms:

[math]\displaystyle{ \begin{align} [P, L]\psi &= (-4(u\psi)_{xxx} + 3u\psi_{xxx}) + 3u_x\psi_{xx} + 9u_{xx}\psi_x + 3u_{xxx}\psi + 6uu_x\psi \end{align} }[/math]

Use:

[math]\displaystyle{ (u\psi)_{xxx} = u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx} }[/math]

So:

[math]\displaystyle{ \begin{align} -4(u\psi)_{xxx} &= -4(u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx}) \\ &= -4u_{xxx}\psi -12u_{xx}\psi_x -12u_x\psi_{xx} -4u\psi_{xxx} \end{align} }[/math]

Now add everything:

[math]\displaystyle{ \begin{align} [P,L]\psi &= (-4u_{xxx}\psi -12u_{xx}\psi_x -12u_x\psi_{xx} -4u\psi_{xxx}) + 3u\psi_{xxx} + 3u_x\psi_{xx} + 9u_{xx}\psi_x + 3u_{xxx}\psi + 6uu_x\psi \\ &= (-u_{xxx}\psi -3u_{xx}\psi_x -9u_x\psi_{xx} -u\psi_{xxx}) + 6uu_x\psi \end{align} }[/math]

So the final result is:

[math]\displaystyle{ [P,L]\psi = ( -u_{xxx} + 6uu_x ) \psi }[/math]

Thus, we have:

[math]\displaystyle{ \frac{dL}{dt} = [P, L] \quad \Rightarrow \quad \frac{du}{dt} = -u_{xxx} + 6uu_x }[/math]

Which is exactly the KdV equation.

Summary

The KdV equation:

[math]\displaystyle{ u_t + 6uu_x + u_{xxx} = 0 }[/math]

can be written as the Lax equation:

[math]\displaystyle{ \frac{dL}{dt} = [P, L] }[/math]

where

[math]\displaystyle{ L = -\partial_x^2 + u, \qquad P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u }[/math]

This reformulation reveals the integrable structure of the KdV equation and allows powerful solution methods such as the inverse scattering transform.

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