Difference between revisions of "Example Calculations for the KdV and IST"

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==Example1 Delta function potential.==
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We consider here the two examples we treated in [[Properties of the Linear Schrodinger Equation]].
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==Example1: <math>\delta</math> function potential==
  
 
We have already calculated the scattering data for the delta function
 
We have already calculated the scattering data for the delta function
potential. The scattering data is
+
potential in [[Properties of the Linear Schrodinger Equation]]. The scattering data is
 
<center><math>
 
<center><math>
 
S\left(  \lambda,0\right)  =\left(  k_{1},\sqrt{k_{1}},\frac{u_{0}}{2ik-u_{0}
 
S\left(  \lambda,0\right)  =\left(  k_{1},\sqrt{k_{1}},\frac{u_{0}}{2ik-u_{0}
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</math></center>
 
</math></center>
  
==Example 2 Hat Function Potential==
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==Example 2: Hat Function Potential==
  
 
We solve for the case when
 
We solve for the case when
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\right.
 
\right.
 
</math></center>
 
</math></center>
 +
We have already solved this case in [[Properties of the Linear Schrodinger Equation]].
 
For the even solutions we need to solve
 
For the even solutions we need to solve
 
<center><math>
 
<center><math>
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<center><math>
 
<center><math>
 
u\left(  x\right)  =\frac{20}{2}\left(  \tanh\left(  \nu\left(  x+1\right)
 
u\left(  x\right)  =\frac{20}{2}\left(  \tanh\left(  \nu\left(  x+1\right)
\right)  -\tanh\left(  \nu\left(  x+1\right)  \right)  \right)
+
\right)  -\tanh\left(  \nu\left(  x-1\right)  \right)  \right)
 
</math></center>
 
</math></center>
 
where <math>\nu</math> is an appropriate constant to make the function increase in value
 
where <math>\nu</math> is an appropriate constant to make the function increase in value
 
sufficiently rapidly but not too rapidly.
 
sufficiently rapidly but not too rapidly.
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{| class="wikitable"
 
{| class="wikitable"
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== Lecture Videos ==
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=== Part 1 ===
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{{#ev:youtube|K0zeheguRKo}}

Latest revision as of 03:38, 1 October 2020

Nonlinear PDE's Course
Current Topic Example Calculations for the KdV and IST
Next Topic Reaction-Diffusion Systems
Previous Topic Connection betwen KdV and the Schrodinger Equation


We consider here the two examples we treated in Properties of the Linear Schrodinger Equation.

Example1: [math]\displaystyle{ \delta }[/math] function potential

We have already calculated the scattering data for the delta function potential in Properties of the Linear Schrodinger Equation. The scattering data is

[math]\displaystyle{ S\left( \lambda,0\right) =\left( k_{1},\sqrt{k_{1}},\frac{u_{0}}{2ik-u_{0} },\frac{2ik}{2ik-u_{0}}\right) }[/math]

The spectral data evolves as

[math]\displaystyle{ k_{1}=k_{1} }[/math]
[math]\displaystyle{ c_{1}\left( t\right) =c_{1}\left( 0\right) e^{4k_{1}^{3}t}=\sqrt{k_{1} }e^{4k_{1}^{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]

so that

[math]\displaystyle{ S\left( \lambda,t\right) =\left( k_{1},\sqrt{k_{1}}e^{4k_{1}^{3}t} ,\frac{u_{0}}{2ik-u_{0}}e^{8ik^{3}t},\frac{2ik}{2ik-u_{0}}\right) }[/math]

Example 2: Hat Function Potential

We solve for the case when

[math]\displaystyle{ u\left( x\right) =\left\{ \begin{matrix} 0, & x\notin\left[ -1,1\right] \\ 20, & x\in\left[ -1,1\right] \end{matrix} \right. }[/math]

We have already solved this case in Properties of the Linear Schrodinger Equation. For the even solutions we need to solve

[math]\displaystyle{ \tan\kappa=\frac{k}{\kappa} }[/math]

where [math]\displaystyle{ \kappa=\sqrt{b-k^{2}} }[/math].

For the odd solutions we need to solve and

[math]\displaystyle{ \tan\kappa=-\frac{\kappa}{k} }[/math]

Recall that the solitons have amplitude [math]\displaystyle{ 2k_{n}^{2} }[/math] or [math]\displaystyle{ -2\lambda_{n} }[/math]. This can be seen in the height of the solitary waves.

We cannot work with a hat function numerically, because the jump in [math]\displaystyle{ u }[/math] leads to high frequencies which dominate the response.. We can smooth our function by a number of methods. We use here the function [math]\displaystyle{ \tanh\left( x\right) }[/math] so we write

[math]\displaystyle{ u\left( x\right) =\frac{20}{2}\left( \tanh\left( \nu\left( x+1\right) \right) -\tanh\left( \nu\left( x-1\right) \right) \right) }[/math]

where [math]\displaystyle{ \nu }[/math] is an appropriate constant to make the function increase in value sufficiently rapidly but not too rapidly.


Animation Three-dimensional plot.
Evolution of [math]\displaystyle{ u(x,t) }[/math].
Evolution of [math]\displaystyle{ u(x,t) }[/math]

Lecture Videos

Part 1