Fourier Transforms

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Consider the PDE∂ 2 u∂x 2 ∂2 u∂y 2 0on the domain defined by − x and y ≥ 0. This equation is called thetwo-dimensional Laplace Equation and the domain is often referred to as the upper halfplane. The solution is required to satisfy ux,y gx on y 0,for some given functiongx, andux,y → 0asy → . The latter condition known as a Far-field or Boundednesscondition.The strategy is to take the FT of the PDE and the boundary condition. This gives an ODEthat we attempt to solve subject to the transformed boundary condition. It is hoped that thefinal solution can be obtained by inversion but from previous experience we know that anintegral equation may arise.As the boundary conditions are given in terms of y, wetaketheFTinx. Define uk,y −ux,y e ikx dx gk −gx e ikx dx uk,0 .The FT of the two terms in the PDE are−−∂ 2 u∂x 2 e ikx dx − k 2 uk,y using (3.2.4)∂ 2 u∂y 2 e ikx dx ∂2∂y 2−ux,y e ikx dx ∂2 u∂y 2 .We regard the resulting equation as one in which y is the variable and and k is a constant,so partial differentiation in y can be replaced by full differentiation. Thus we need to solvethe equationd 2 u − k 2 u 0,dy 2subject to uk,0 gk and uk,y → 0asy → .It may seem that the solution of this ODE is straightforward to obtain, since equations ofthis type are associated with expontentially increasing or decreasing solutions. However,the parameter k lies within the range − k and so there are three cases: k 0,k 0and k 0. The first and third can be combined by using k 2 |k| 2 and then the questionarises as to how k 0 should be accommodated.For k ≠ 0, the solution of this ODE is easily shown to be14
uk,y Ak e|k|y Bk e −|k|y .Now apply the boundary and far-field conditions. The far-field condition can only besatisfied if Ak 0 and the condition on y 0givesBk gk, thus uk,y gk e−|k|yandbyinversionux,y 12 gk exp−ikx |k|y dk .−In fact the transform uk,y is of convolution form and results obtained previously can beemployed to determine the functions involved. The details are not presented here and thefinal form isux,y y −gkx − k 2 y 2 dk .and this is known as Poisson’s formula.3.3.3 Integral EquationsIntegral equations occur when the unknown function occurs as part of an integrand ratherthan in a differential equation. If the integral is over a fixed domain - such as the finitedomain a,b or an infinite one −,, then the integral equation is called a FredholmEquation. If the domain varies with the unknown x, then the equation is known as aVolterra Equation. Only equations of Fredholm type are considered here.Fredholm equations are divided into two main categories, known as the First Type and theSecond Type; there is a Third Type but this can be re-written as one of second type and isnot considered here. The general form of these equations types, for unknown fx, isType 1 hx abkx,t ft dtType 2fx hx abkx,t ft dtwhere is a constant. The function kx,t is called the Kernel and hx is called the FreeTerm. Ifhx 0 in type 2, then the equation is sometimes said to be homogeneous.Both of these definitions apply only to linear equations. Thus if f 1 x and f 2 x are bothsolutions of either a type 1 or type 2 equation, then fx Af 1 x Bf 2 x is also a solutionfor constants A and B. This comment is made despite the fact that we do not yet know how15
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Page 4 and 5: fx Hx a 1 − Hx − aSignum (Sig
Page 6 and 7: f̂k a−a1 e ikx dxa 1ike ikx −a
Page 9 and 10: fx fx −2 12−f̂k e −ikx dk
Page 11 and 12: −|e −a|x| | 2 dx 2 0e −2ax d
Page 13: −−d 2 ydx 2d 2 y− 2 y e ikx
Page 17 and 18: ecause of the convolution structure
Page 19 and 20: complex FS of f on −L ≤ ≤
Page 21 and 22: 3.4 Fourier Transforms and Generali
Page 23 and 24: function and kgk 0, then gk is a c

Fourier Transforms

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