Fourier Transforms

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many solutions to expect for such equations.Fredholm equations with a kernel of convolution form.Consider the particular formAfx ft gx − t dt hx ,−where A is a real constant and the functions gx and hx are known; the kernel g appearsas a convolution and the range of integration is infinite. Note that by a change of variablein the integral, the equation can also be written asAfx fx − t gt dt hx ,−and so the two equations can be considered as equivalent. This equation is of type 2 ifA ≠ 0 and of type 1 if A 0.Take the complex Fourier Transform of the equation, assuming that f,g and h all possessthe required properties for the transform to exist,and this can be rearranged to giveAf̂k f̂k ĝk ĥk A ĝk f̂k ĥkf̂k ĥkA ĝk .Thus the solution is given by the inverse Fourier Transformfx 12 f̂k e −ikx dk 1−2−ĥkA ĝk e−ikx dkand it is assumed that this can be integrated, if only numerically. If A ≠ 0 and both ĥkand ĝk are absolutely integrable, then the integral exists.The nonlinear equation: special case fx gxIf fx gx, then the equation becomesAfx ft fx − t dt hx .−This is no longer of Fredholm type, since the Fredholm equation is linear whereas this oneis not.Fourier techniques do not work generally on nonlinear equations but will work in this case16
ecause of the convolution structure. The transformed equation becomesf̂k2 Af̂k − ĥk 0and this is a quadratic equation that can be solved to givef̂k − A2 A22 ĥk .Both roots are important. It is sometimes possible to deduce what the solution should be bylooking at the structure. Write the two solutions asf̂1k − A 2 A22 ĥk , f̂2k − A 2 − A22 ĥk .If ĥk is absolutely integrable, then ĥk → 0ask → and sof̂1k → 0,f̂2k → −A.Thus if we require f̂k to be absolutely integrable, then only f̂1k is acceptable. If A 0,then both solutions can be important and an example has been previously given.ExampleUse the FT to solve the integral equation−fx − t ft dt exp − x22and check your answer by direct substitution.Outline of solution onlyStep 1: From (3.2.7), the integral is the convolution f ∗ fx. Take the FT of the equationand use (3.2.8) to obtainf̂k f̂k F exp − x 22 f̂k F exp − x22 gk sayStep 2: Invert transform using (3.2.9)fx 12−gk e −ikx dkStep 3: Determine fx, using contour integration if necessary. The solution should be17
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Fourier Transforms

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