Wiener–Hopf method: Difference between revisions

Revision as of 20:01, 23 January 2016

Revision as of 20:55, 23 January 2016

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derivatives with respect to {{mvar|x}} and {{mvar|y}},

subject to the mixed conditions on {{mvar|y}}=0, for some prescribed

function <{{math>|''g''(''x'')</math>}},

:<math>f=g(x)\text{ for }x\leq 0, \quad f_{y}=0\text{ when }x>0</math>

and decay at infinity i.e. <math>{{mvar|f\rightarrow 0</math>}}→0 as <math>\boldsymbol{x}\rightarrow \infty</math>.

Taking a [[Fourier transform]] with respect to {{mvar|x}} results in the following [[ordinary differential equation]]

: <math>\boldsymbol{L}_{y}\hat{f}(k,y)-P(k,y)\hat{f}(k,y)=0,</math>

where <math>\boldsymbol{L}_{y}</math> is a linear operator containing <math>{{mvar|y</math>}} derivatives only, <{{math>|''P''(''k,y'')</math>}} is a known function of {{mvar|y}} and {{mvar|k}} and

: <math> \hat{f}(k,y)=\int_{-\infty}^{\infty} f(x,y)e^{-ikx}\textrm{d}x. </math>

If a particular solution of this ordinary differential equation which satisfies the necessary decay at infinity is denoted <{{math>| ''F''(''k,y'')</math>}}, a general solution can be written as

: <math> \hat{f}(k,y)=C(k)F(k,y), </math>

where <{{math>|''C''(''k'')</math>}} is an unknown function to be determined by the boundary conditions on <math>{{mvar|y}}=0</math>.

The key idea is to split <math>\hat{f}</math> into two separate functions, <math>\hat{f}_{+}</math> and <math>\hat{f}_{-}</math> which are analytic in the lower- and upper-halves of the complex plane, respectively,

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: <math> K(k)=\frac{F'(k,0)}{F(k,0)}. </math>

Now <math>K(k)</math> can be decomposed into the product of functions <math>K^{-}</math> and <math>K^{+}</math> which are analytic in the upper and lower half-planes respectively.

To be precise, <math> K(k)=K^{+}(k)K^{-}(k), </math> where

: <math> \hbox{log}K^{-} = \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{\hbox{log}(K(z))}{z-k} \textrm{d}z, \quad \hbox{Im}k>0, </math>

: <math> \hbox{log}K^{+} = -\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{\hbox{log}(K(z))}{z-k} \textrm{d}z, \quad \hbox{Im}k<0. </math>