Inverse scattering transform: Difference between revisions

Revision as of 07:31, 25 January 2017

Revision as of 04:35, 1 December 2017

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:<math>L=\frac{d^{2}}{dx^{2}}+u,</math>

where u is the "potential". Comparing the expression <math>\frac{dL}{dt}v+L\frac{dv}{dt}</math> with <math>\frac{\partial}{\partial t}\left(\frac{d^{2}v}{dx^{2}}+uv\right)</math> shows us that <math>\frac{\partial LdL}{\partial tdt}=\frac{du\partial u}{dt\partial t},</math> thus ignoring the first term.

After concocting the appropriate Lax pair it should be the case that Lax's equation recovers the original nonlinear PDE.