Fourier Transforms

More documents

Recommendations

Info

3.3 Applications of Fourier TransformsThe principal uses of Fourier Transforms are the solution of partial differential equations,possessing a particular structure, and the processing and properties of time series. It is alsopossible to invoke transform methods to solve certain types of integral equations and toenable the summation of certain types of slowly-converging series. Some of theseproperties are shown in the following examples.3.3.1 Solution of Ordinary Differential EquationsAlthough not especially useful for the solution of ordinary differential equations (ODEs), itis instructive to show how transform methods work on ODEs before considering thebroader structure of partial differential equations (PDEs). This is best illustrated by anexample.Example 1Obtain a solution of the Ordinary Differential Equation (ODE)d 2 ydx 2 − 2 y gx ,which → 0asx →, assuming that gx is absolutely integrable.The general solution to a linear ODE is often written in the formyx y 1 x y 2 x ,where y 1 x and y 2 x are the Complimentary Function (CF) and Particular Integral (PI)respectively. The complementary function isy 1 x Ae x Be −x ,for arbitrary constants A and B; these solutions do not → 0asx →. Thus only the PI ispotentially capable of meeting this requirement. For convenience we write yx instead ofthe correct y 2 x.Note that it is assumed that ≠ 0. If 0 can occur, then it is necessary to consider thiscase separately and the CF solution is y 1 x Ax B and which is also incapable ofmeeting the large x requirement unless y 1 x ≡ 0.Taking the FT of the equation gives12
−−d 2 ydx 2d 2 y− 2 y e ikx dx gx e ikx dxdx 2 −eikx dx − 2 ye ikx dx − gx e ikx dx−and this step can be taken because the integration operator is linear. Thus it is necessary todetermine the FT of each term of the equation. The FT of the first term is given by (3.2.4)to be−d 2 ydx 2 e ikx dx − k 2 ŷ ,and the FT of each of remaining terms is simply the transform itself, so that the transformedequation is− k 2 2 ŷk ĝkThis must be inverted to obtain the solution yx 12 −21 ŷk − −−ŷk e −ikx dkĝkk 2 2 .ĝkk 2 2 e −ikx dk .The usual approach would then require some knowledge of gx and its transform ĝk forthe integral to be evaluated. However, in this case, another approach is possible since ŷkis the product of ĝk and 1/k 2 2 ; these two quantities are the FTs of gx andexp−|x|/2 respectively, with the latter given as an earlier example. Thus theconvolution (3.2.8) applies and the inversion can be written asyx − 12 −gu e −|x−u| du ,which satisfies all requirements. The modulus sign in the exponential is important andensures that the solution has the correct behaviour as x →. Note that it is often onlypossible to write the solution of a differential equation as an integral representation.3.3.2 Partial Differential EquationsOne of the principal uses of FTs is to obtain the solution of a certain class of PartialDifferential Equations (PDEs). The PDE should be linear and the boundary conditionsshould be of a certain form; this is a large subject and the details cannot be considered here,so we just present a solution of one particular equation.13
Page 2 and 3: −xfx dx f0 (3.1.1)or equivalentl
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: −|e −a|x| | 2 dx 2 0e −2ax d
Page 15 and 16: uk,y Ak e|k|y Bk e −|k|y .Now ap
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

Create successful ePaper yourself

Delete template?

Save as template?