Wiener–Hopf method: Difference between revisions

Revision as of 20:55, 23 January 2016

Revision as of 16:02, 16 March 2019

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

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

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

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

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

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

: <math> \hatwidehat{f}_{+}(k,y)=\int_{0}int_0^{\infty} f(x,y)e^{-ikx}\,\textrm{d}x, </math>

: <math> \hatwidehat{f}_{-}(k,y)=\int_{-\infty}^{0} f(x,y)e^{-ikx}\,\textrm{d}x. </math>

The boundary conditions then give

: <math> \hatwidehat{g}(k)+\hatwidehat{f}_{+}(k,0) = \hatwidehat{f}_{-}(k,0)+\hatwidehat{f}_{+}(k,0) = \hatwidehat{f}(k,0) = C(k)F(k,0) </math>

and, on taking derivatives with respect to <math>y</math>,

: <math> \hat{f}'_{-}(k,0) = \hatwidehat{f}'_{-}(k,0)+\hatwidehat{f}'_{+}(k,0) = \hatwidehat{f}'(k,0) = C(k)F'(k,0). </math>

Eliminating <math>C(k)</math> yields

: <math> \hatwidehat{g}(k)+\hatwidehat{f}_{+}(k,0) = \hatwidehat{f}'_{-}(k,0)/K(k), </math>

where

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

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: <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>

(Note that this sometimes involves scaling <math>K</math> so that it tends to <math>1</math> as <math>k\rightarrow\infty</math>.) We also decompose <math>K^{+}\hatwidehat{g}</math> into the sum of two functions <math>G^{+}</math> and <math>G^{-}</math> which are analytic in the lower and upper half-planes respectively, i.e.,

:: <math> K^{+}(k)\hatwidehat{g}(k)=G^{+}(k)+G^{-}(k). </math>

This can be done in the same way that we factorised <math> K(k). </math>

Consequently,

: <math> G^{+}(k) + K_{+}(k)\hatwidehat{f}_{+}(k,0) = \hatwidehat{f}'_{-}(k,0)/K_{-}(k) - G^{-}(k). </math>

Now, as the left-hand side of the above equation is analytic in the lower half-plane, whilst the right-hand side is analytic in the upper half-plane, analytic continuation guarantees existence of an entire function which coincides with the left- or right-hand sides in their respective half-planes. Furthermore, since it can be shown that the functions on either side of the above equation decay at large {{mvar|k}}, an application of [[Liouville's theorem (complex analysis)|Liouville's theorem]] shows that this entire function is identically zero, therefore

:<math> \hatwidehat{f}_{+}(k,0) = -\frac{G^{+}(k)}{K^{+}(k)}, </math>

and so

: <math> C(k) = \frac{K^{+}(k)\hatwidehat{g}(k)-G^{+}(k)}{K^{+}(k)F(k,0)}. </math>

== See also ==