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Chapter 8 Fourier Transform We introduce the Fourier transform, a special linear integral transformation for differential equations which are defined on unbounded domains. The method is associated with the French physicist and mathematician Joseph Fourier whose work opened up a new way to solve linear PDEs. As we will see later, the Fourier transform reduces a partial differential equation in (t; x)-domain to an ordinary differential equation in (t;!)-domain. 8.1 Definition The Fourier transform is a linear integral transformation of a function f(x) in ^ x-domain to another function f(!) in !-domain. As we will see in sequel, ! can ^ be considered as the frequency variable and thus f(!) represents the frequency distribution embedded in function f(x). 1. (Admissible functions) We first introduce the admissible space of func- tions. Definition 8.1. A function f(x); ¡1 < x < 1 is called integrable if the following relation holds Z 1jf(x)jdx<1: (8.1) ¡1 Afunction f(x);¡1>p f(x)=< x 1 x>0; > > p :¡ ¡x x<0 is not admissible even it satisfies the following property lim Z Rf(x)dx=0;: R!1 ¡R 1 2 Fourier Transform Note that Z R Z R 1 p lim jf(x)jdx=2 lim p dx= lim 4 R=1: R!1 ¡R R!1 0 x R!1 TheFouriertransformisusuallydefinedforadmissiblefunctions. However, as we will see below, the definition can be extended for a wider class of functions than the admissible ones. 2. (Definition of Fourier transform) Suppose f is a function (not neces- sarily admissible) defined on R:(¡1;1). Its Fourier transform Fffg:= ^ f(!) is defined by the following integral ^ Z 1 ¡i!x Fffg:=f(!)= ¡1f(x)e dx; (8.2) as long as the integral exists. The improper integral is understood in the following sense Z 1f(x)e¡i!xdx= lim Z Rf(x)e¡i!xdx: (8.3) ¡1 R!1 ¡R Example 8.1. Consider the following function 81 0 0; ¡1 x<0 is not admissible, and the definition of the Fourier transform fails as ^ Z 0 ¡i!x Z 1 ¡i!x Z 1 f(!)= ¡e dx+ e dx=¡2i sin(!x)dx: ¡1 0 0 4 Fourier Transform Now, let us consider the following function f (x)= e¡"x x>0 : " "x ¡e x<0 The function is admissible and its transform is ^ Z 1 ¡i!x Z 1 ¡"x ! f (!)= f (x)e dx=¡2i e sin(!x)dx=¡2i : " " "2+!2 ¡1 0 ^ Since f (x)!f(x) for "!0, we define f(!) through the following limit " ^ ^ ¡2i f(!)=limf (!)= : " ! "!0 ^ 4. (Inverse transform) Suppose f(!) is a function defined on !2(¡1;1). ^ The inverse Fourier transform of f is defined by the following integral ¡1 ^ 1 Z 1 ^ i!x F (f)=2 ¡1f(!)e d!; (8.4) as long as the integral exists. The integral is understood in the following sense Z 1^ i!x Z R ^ i!x f(!)e d!= lim f(!)e d!: (8.5) ¡1 R!1 ¡R ¡1 The following theorem establishes the relationship between F;F . See the appendix of this chapter for a proof. Theorem 8.1. Assume that f is an admissible function, then + ¡ ¡1 ^ f(x )+f(x ) F (f)(x)= 2 ; (8.6) + ¡ where f(x ) and f(x ) are the right and left limits of f at x respectively. ^ 2(1¡cosw) Example8.3. Letusverifythetheoremforfunctions f= i! . Wehave ¡1 ^ 1 Z R1¡cos! i!x 2 Z R1¡cos! F (f)(x)=i lim ! e d!= lim ! sin(!x)d!; R!1 ¡R R!1 0 where we used the symmetry property for the integration. The following figures shows the value of the integral for R = 20; 100 respectively. It is observed that for R!1 the integral approaches f(x).
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