Inverse scattering transform: Difference between revisions

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'''Step 1.''' Determine the nonlinear partial differential equation. This is usually accomplished by analyzing the [[physics]] of the situation being studied.

'''Step 2.''' Employ ''forward scattering''. This consists in finding the [[Lax pair]]. The Lax pair consists of two linear [[Operator (mathematics)|operator]]s, <math>L</math> and <math>M</math>, such that <math>Lv=\lambda v</math> and <math>\frac{dv}{dt}v_t=Mv</math>. It is extremely important that the [[eigenvalue]] <math>\lambda</math> be independent of time; i.e. <math>\frac{d\lambda}{dt}lambda_t=0.</math> Necessary and sufficient conditions for this to occur are determined as follows: take the time [[derivative]] of <math>Lv=\lambda v</math> to obtain

:<math>\frac{dL}{dt}L_t v + L\frac{dv}{dt} v_t = \frac{d\lambda}{dt}lambda_t v + \lambda \frac{dv}{dt}v_t .</math>

Plugging in <math>Mv</math> for <math>\frac{dv}{dt}v_t</math> yields

:<math>\frac{dL}{dt}L_t v + LMv = \frac{d\lambda}{dt}lambda_t v + \lambda Mv.</math>

Rearranging on the far right term gives us

:<math>\frac{dL}{dt}L_t v + LMv = \frac{d\lambda}{dt}lambda_t v + MLv.</math>

Thus,

:<math>\frac{dL}{dt}L_t v + LMv - MLv = \frac{d\lambda}{dt}lambda_t v.</math>

Since <math>v\not=0</math>, this implies that <math>\frac{d\lambda}{dt}lambda_t = 0</math> [[iff|if and only if]]

:<math>\frac{dL}{dt}L_t + LM - ML = 0. \, </math>

This is [[Lax's equation]]. One important thing to note about Lax's equation is that <math>\frac{dL}{dt}L_t</math> is the time derivative of <math>L</math> precisely where it explicitly depends on <math>t</math>. The reason for defining the differentiation this way is motivated by the simplest instance of <math>L</math>, which is the Schrödinger operator (see [[Schrödinger equation]]):

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

where u is the "potential". Comparing the expression <math>\frac{dL}{dt}L_t v + L\frac{dv}{dt} v_t</math> with <math>\frac{\partial}{\partialpartial_t t}\left(\frac{d^v_{2}v}{dx^{2}xx} +uv\right)</math> shows us that <math>\frac{dL}{dt}L_t =\frac{\partial u}{\partial t}u_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.