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

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==Example 2 A Hat Function==
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==Example 2 Hat Function Potential==
  
 
We solve for the case when
 
We solve for the case when

Revision as of 03:14, 24 September 2010

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


Example1 Delta function potential.

We have already calculated the scattering data for the delta function potential. 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]

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]