Difference between revisions of "Example Calculations for the KdV and IST"
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We have already calculated the scattering data for the delta function | We have already calculated the scattering data for the delta function | ||
− | potential. The | + | potential. 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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,\frac{u_{0}}{2ik-u_{0}}e^{8ik^{3}t},\frac{2ik}{2ik-u_{0}}\right) | ,\frac{u_{0}}{2ik-u_{0}}e^{8ik^{3}t},\frac{2ik}{2ik-u_{0}}\right) | ||
</math></center> | </math></center> | ||
− | |||
==Example 2 A Hat Function== | ==Example 2 A Hat Function== |
Revision as of 10:27, 23 September 2010
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Example1 Delta function potential.
We have already calculated the scattering data for the delta function potential. The scattering data is
The spectral data evolves as
so that
Example 2 A Hat Function
We solve for the case when
For this case we need to solve
which has solution and
which has solution.
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
where [math]\displaystyle{ \nu }[/math] is an appropriate constant to make the function increase in value sufficiently rapidly but not too rapidly.