λ n = L φ n = π L eınπx/l, for n Z

Transcription

1 Chapter 32 The Fourier Transform 32. Derivation from a Fourier Series Consider the eigenvalue problem y + λy =, y( L = y(l, y ( L = y (L. The eigenvalues and eigenfunctions are ( nπ λ n = L 2 for n Z + φ n = π L eınπx/l, for n Z The eigenfunctions form an orthogonal set. A piecewise continuous function defined on [ L...L] can be expanded in a series of the eigenfunctions. π f(x c n L eınπx/l n= 539

2 The Fourier coefficients are c n = π = π L eınπx/l f(x L eınπx/l π L L L eınπx/l e ınπx/l f(x dx. We substitute the expression for c n into the series for f(x. We let ω n = nπ/l and ω = π/l. f(x f(x n= ω n= [ L ] e ınπξ/l f(ξ dξ e ınπx/l. 2L L [ L ] e ıωnξ f(ξ dξ e ıωnx ω. L In the limit as L, (and thus ω, the sum becomes an integral. [ ] f(x e ıωξ f(ξ dξ e ıωx dω. Thus the expansion of f(x for finite L f(x c n = n= L c n π L eınπx/l L e ınπx/l f(x dx 54

3 in the limit as L becomes f(x ˆf(ω = ˆf(ω e ıωx dω f(x e ıωx dx. Of course this derivation is only heuristic. In the next section we will explore these formulas more carefully The Fourier Transform Let f(x be piecewise continuous and let f(x dx exist. We define the function I(x, L. I(x, L = L ( f(ξ e ıωξ dξ e ıωx dω. L Since the integral in parentheses is uniformly convergent, we can interchange the order of integration. = = = = π = π ( L f(ξ e ıω(ξ x dω L ] L [f(ξ eıω(ξ x dξ dξ ı(ξ x L ( f(ξ e ıl(ξ x e ıl(ξ x dξ ı(ξ x sin(l(ξ x f(ξ dξ ξ x f(ξ + x sin(lξ ξ dξ. 54

4 In Example we will show that sin(lξ ξ dξ = π 2. Continuous Functions. Suppose that f(x is continuous. f(x = π I(x, L f(x = π f(x sin(lξ dξ ξ f(x + ξ f(x ξ sin(lξ dξ. If f(x has a left and right derivative at x then f(x+ξ f(x is bounded and ξ Riemann-Lebesgue lemma to show that the integral vanishes as L. f(x + ξ f(x sin(lξ dξ as L. π ξ f(x+ξ f(x ξ dξ <. We use the Now we have an identity for f(x. f(x = ( f(ξ e ıωξ dξ e ıωx dω. Piecewise Continuous Functions. Now consider the case that f(x is only piecewise continuous. f(x + 2 f(x 2 = π = π f(x + sin(lξ dξ ξ f(x sin(lξ dξ ξ 542

5 I(x, L f(x+ + f(x = 2 ( f(x + ξ f(x sin(lξ dξ ξ ( f(x + ξ f(x + ξ sin(lξ dξ If f(x has a left and right derivative at x, then f(x + ξ f(x ξ f(x + ξ f(x + ξ is bounded for ξ, and is bounded for ξ. Again using the Riemann-Lebesgue lemma we see that f(x + + f(x 2 = ( f(ξ e ıωξ dξ e ıωx dω. 543

6 Result Let f(x be piecewise continuous with f(x dx <. The Fourier transform of f(x is defined ˆf(ω = F[f(x] = f(x e ıωx dx. We see that the integral is uniformly convergent. The inverse Fourier transform is defined f(x + + f(x 2 If f(x is continuous then this reduces to = F [ ˆf(ω] = f(x = F [ ˆf(ω] = ˆf(ω e ıωx dω. ˆf(ω e ıωx dω A Word of Caution Other texts may define the Fourier transform differently. The important relation is f(x = Multiplying the right side of this equation by = α α yields f(x = α ( f(ξ e ıωξ dξ e ±ıωx dω. ( α f(ξ e ıωξ dξ e ±ıωx dω. 544

7 Setting α = and choosing sign in the exponentials gives us the Fourier transform pair Other equally valid pairs are and ˆf(ω = f(x = ˆf(ω = f(x = ˆf(ω = f(x = f(x e ıωx dx ˆf(ω e ıωx dω. f(x e ıωx dx ˆf(ω e ıωx dω, f(x e ıωx dx ˆf(ω e ıωx dω. Be aware of the different definitions when reading other texts or consulting tables of Fourier transforms Evaluating Fourier Integrals Integrals that Converge If the Fourier integral F[f(x] = f(x e ıωx dx, converges for real ω, then finding the transform of a function is just a matter of direct integration. We will consider several examples of such garden variety functions in this subsection. Later on we will consider the more interesting cases when the integral does not converge for real ω. 545

8 Example Consider the Fourier transform of e a x, where a >. Since the integral of e a x is absolutely convergent, we know that the Fourier transform integral converges for real ω. We write out the integral. F [ e a x ] = = = e a x e ıωx dx e ax ıωx dx + e (a ır(ω+i(ωx dx + e ax ıωx dx e ( a ır(ω+i(ωx dx The integral converges for I(ω < a. This domain is shown in Figure 32.. Im(z Re(z Figure 32.: The Domain of Convergence 546

9 Now We do the integration. F [ e a x ] = = = = π e (a ıωx dx + [ ] e (a ıωx + a ıω ( a ıω + a + ıω a π(ω 2 + a 2, e (a+ıωx dx ] [ e (a+ıωx a + ıω for I(ω < a We can extend the domain of the Fourier transform with analytic continuation. F [ e a x ] a = π(ω 2 + a 2, for ω ±ıa Example Consider the Fourier transform of f(x =, α >. x ıα [ ] F = x ıα x ıα e ıωx dx The integral converges for I(ω =. We will evaluate the integral for positive and negative real values of ω. For ω >, we will close the path of integration in the lower half-plane. Let C R be the contour from x = R to x = R following a semicircular path in the lower half-plane. The integral along C R vanishes as R by Jordan s Lemma. C R x ıα e ıωx dx as R. Since the integrand is analytic in the lower half-plane the integral vanishes. [ F x ıα ] = 547

10 For ω <, we will close the path of integration in the upper half-plane. Let C R denote the semicircular contour from x = R to x = R in the upper half-plane. The integral along C R vanishes as R goes to infinity by Jordan s Lemma. We evaluate the Fourier transform integral with the Residue Theorem. [ ] F x ıα = i Res = ı e αω We combine the results for positive and negative values of ω. [ ] F = x ıα ( e ıωx x iα, iα { for ω >, ı e αω for ω < Cauchy Principal Value and Integrals that are Not Absolutely Convergent. That the integral of f(x is absolutely convergent is a sufficient but not a necessary condition that the Fourier transform of f(x exists. The integral f(x e ıωx dx may converge even if f(x dx does not. Furthermore, if the Fourier transform integral diverges, its principal value may exist. We will say that the Fourier transform of f(x exists if the principal value of the integral exists. F[f(x] = f(x e ıωx dx Example Consider the Fourier transform of f(x = /x. ˆf(ω = x e ıωx dx If ω >, we can close the contour in the lower half-plane. The integral along the semi-circle vanishes due to Jordan s Lemma. lim R C R x e ıωx dx = 548

11 We can evaluate the Fourier transform with the Residue Theorem. ˆf(ω = ( ( (i Res 2 x e ıωx, ˆf(ω = ı, for ω >. 2 The factor of /2 in the above derivation arises because the path of integration is in the negative, (clockwise, direction and the path of integration crosses through the first order pole at x =. The path of integration is shown in Figure Im(z Re(z Figure 32.2: The Path of Integration If ω <, we can close the contour in the upper half plane to obtain ˆf(ω = ı, for ω <. 2 For ω = the integral vanishes because x is an odd function. ˆf( = = x dx = 549

12 We collect the results in one formula. ˆf(ω = ı 2 sign(ω We write the integrand for ω > as the sum of an odd and and even function. x e ıωx dx = ı 2 ı cos(ωx dx + sin(ωx dx = ıπ x x The principal value of the integral of any odd function is zero. sin(ωx dx = π x If the principal value of the integral of an even function exists, then the integral converges. sin(ωx dx = π x x sin(ωx dx = π 2 Thus we have evaluated an integral that we used in deriving the Fourier transform Analytic Continuation Consider the Fourier transform of f(x =. The Fourier integral is not convergent, and its principal value does not exist. Thus we will have to be a little creative in order to define the Fourier transform. Define the two functions for x > for x > f + (x = /2 for x =, f (x = /2 for x =. for x < for x < 55

13 Note that = f (x + f + (x. The Fourier transform of f + (x converges for I(ω <. F[f + (x] = e ıωx dx = e ( ır(ω+i(ωx dx. = [ ] e ıωx ıω = ı for I(ω < ω Using analytic continuation, we can define the Fourier transform of f + (x for all ω except the point ω =. F[f + (x] = ı ω We follow the same procedure for f (x. The integral converges for I(ω >. F[f (x] = = e ıωx dx e ( ır(ω+i(ωx dx = [ ] e ıωx ıω = ı ω. Using analytic continuation we can define the transform for all nonzero ω. F[f (x] = ı ω 55

14 Now we are prepared to define the Fourier transform of f(x =. F[] = F[f (x] + F[f + (x] = ı ω + ı ω =, for ω When ω = the integral diverges. When we consider the closure relation for the Fourier transform we will see that F[] = δ(ω Properties of the Fourier Transform In this section we will explore various properties of the Fourier Transform. I would like to avoid stating assumptions on various functions at the beginning of each subsection. Unless otherwise indicated, assume that the integrals converge Closure Relation. Recall the closure relation for an orthonormal set of functions {φ, φ 2,...}, φ n (xφ n (ξ δ(x ξ. n= There is a similar closure relation for Fourier integrals. We compute the Fourier transform of δ(x ξ. F[δ(x ξ] = δ(x ξ e ıωx dx = e ıωξ 552

15 Next we take the inverse Fourier transform. δ(x ξ δ(x ξ e ıωξ e ıωx dω e ıω(x ξ dω. Note that the integral is divergent, but it would be impossible to represent δ(x ξ with a convergent integral Fourier Transform of a Derivative. Consider the Fourier transform of y (x. Next consider y (x. In general, F[y (x] = y (x e ıωx dx [ ] = y(x e ıωx = ıω y(x e ıωx dx = ıωf[y(x] F[y (x] = F [ ] d dx (y (x = ıωf[y (x] = (ıω 2 F[y(x] = ω 2 F[y(x] F [ y (n (x ] = (ıω n F[y(x]. ( ıωy(x e ıωx dx 553

16 Example The Dirac delta function can be expressed as the derivative of the Heaviside function. { for x < c, H(x c = for x > c Thus we can express the Fourier transform of H(x c in terms of the Fourier transform of the delta function Fourier Convolution Theorem. F[δ(x c] = ıωf[h(x c] δ(x c e ıωx dx = ıωf[h(x c] e ıcω = ıωf[h(x c] F[H(x c] = ıω e ıcω Consider the Fourier transform of a product of two functions. F[f(xg(x] = = = = = f(xg(x e ıωx dx ( ˆf(η e ıηx dη g(x e ıωx dx ( ˆf(ηg(x e ı(η ωx dx dη ( ˆf(η g(x e ı(ω ηx dx dη ˆf(ηG(ω η dη 554

17 The convolution of two functions is defined f g(x = f(ξg(x ξ dξ. Thus F[f(xg(x] = ˆf ĝ(ω = ˆf(ηĝ(ω η dη. Now consider the inverse Fourier Transform of a product of two functions. Thus F [ ˆf(ωĝ(ω] = = = = = ˆf(ωĝ(ω e ıωx dω ( f(ξ e ıωξ dξ ĝ(ω e ıωx dω ( f(ξĝ(ω e ıω(x ξ dω dξ ( f(ξ ĝ(ω e ıω(x ξ dω dξ f(ξg(x ξ dξ F [ ˆf(ωĝ(ω] = f g(x = f(ξg(x ξ dξ, F[f g(x] = ˆf(ωĝ(ω. These relations are known as the Fourier convolution theorem. 555

18 Example Using the convolution theorem and the table of Fourier transform pairs in the appendix, we can find the Fourier transform of f(x = x 4 + 5x We factor the fraction. f(x = (x 2 + (x From the table, we know that [ ] 2c F = e c ω for c >. x 2 + c 2 We apply the convolution theorem. First consider the case ω >. [ ] 2 4 F[f(x] = F 8 x 2 + x = ( e η e 2 ω η dη 8 = ( e η e 2 ω η dη + 8 F[f(x] = ( ω e 2ω+3η dη + e 2ω+η dη + 8 = ( 8 3 e 2ω + e ω e 2ω + 3 e ω = 6 e ω 2 e 2ω e η e 2 ω η dη ω e 2ω 3η dη 556

19 Now consider the case ω <. F[f(x] = ( ω e 2ω+3η dη + e 2ω η dη + 8 ω = ( 8 3 eω e 2ω + e ω + 3 e2ω = 6 eω 2 e2ω We collect the result for positive and negative ω. e 2ω 3η dη A better way to find the Fourier transform of is to first expand the function in partial fractions Parseval s Theorem. F[f(x] = 6 e ω 2 e 2 ω f(x = x 4 + 5x f(x = /3 x 2 + /3 x F[f(x] = [ ] 6 F 2 2 [ ] x 2 + F 4 x = 6 e ω 2 e 2 ω Recall Parseval s theorem for Fourier series. If f(x is a complex valued function with the Fourier series n= c n e ınx then π c n 2 = f(x 2 dx. n= 557 π

20 Analogous to this result is Parseval s theorem for Fourier transforms. Let f(x be a complex valued function that is both absolutely integrable and square integrable. f(x dx < The Fourier transform of f( x is ˆf(ω. [ ] F f( x We apply the convolution theorem. We set x =. and = = = = ˆf(ω F [ ˆf(ω ˆf(ω] = This is known as Parseval s theorem. ˆf(ω ˆf(ω e ıωx dω = ˆf(ω ˆf(ω dω = ˆf(ω 2 dω = f(x 2 dx < f( x e ıωx dx f(x e ıωx dx f(x e ıωx dx f(ξf( (x ξ dξ f(ξf(ξ x dξ f(ξf(ξ dξ f(x 2 dx 558

21 Shift Property. The Fourier transform of f(x + c is The inverse Fourier transform of ˆf(ω + c is F[f(x + c] = f(x + c e ıωx dx = f(x e ıω(x c dx F[f(x + c] = e ıωc ˆf(ω F [ ˆf(ω + c] = = ˆf(ω + c e ıωx dω ˆf(ω e ı(ω cx dω Fourier Transform of x f(x. The Fourier transform of xf(x is F [ ˆf(ω + c] = e ıcx f(x F[xf(x] = = = ı ω ( xf(x e ıωx dx ıf(x ω (e ıωx dx f(x e ıωx dx 559

22 Similarly, you can show that F[xf(x] = ı ˆf ω. F[x n f(x] = (i n n ˆf ω n Solving Differential Equations with the Fourier Transform The Fourier transform is useful in solving some differential equations on the domain (... with homogeneous boundary conditions at infinity. We take the Fourier transform of the differential equation L[y] = f and solve for ŷ. We take the inverse transform to determine the solution y. Note that this process is only applicable if the Fourier transform of y exists. Hence the requirement for homogeneous boundary conditions at infinity. We will use the table of Fourier transforms in the appendix in solving the examples in this section. Example Consider the problem We take the Fourier transform of this equation. y y = e α x, y(± =, α >, α. ω 2 ŷ(ω ŷ(ω = We take the inverse Fourier transform to determine the solution. α/π ω 2 + α 2 α/π ŷ(ω = (ω 2 + α 2 (ω 2 + = α ( π α 2 ω 2 + ω 2 + α 2 = ( α/π α 2 ω 2 + α α /π 2 ω

23 Example Consider the Green function problem We take the Fourier transform of this equation. We use the Table of Fourier transforms. We use the convolution theorem to do the inversion. y(x = e α x α e x α 2 G G = δ(x ξ, y(± =. ω 2 Ĝ Ĝ = F[δ(x ξ] Ĝ = F[δ(x ξ] ω 2 + Ĝ = πf [ e x ] F[δ(x ξ] G = π e x η δ(η ξ dη G(x ξ = 2 e x ξ The inhomogeneous differential equation has the solution y y = f(x, y(± =, y = 2 f(ξ e x ξ dξ. When solving the differential equation L[y] = f with the Fourier transform, it is quite common to use the convolution theorem. With this approach we have no need to compute the Fourier transform of the right side. We merely denote it as F[f] until we use f in the convolution integral. 56

24 32.6 The Fourier Cosine and Sine Transform The Fourier Cosine Transform Suppose f(x is an even function. In this case the Fourier transform of f(x coincides with the Fourier cosine transform of f(x. F[f(x] = f(x e ıωx dx = f(x(cos(ωx ı sin(ωx dx = f(x cos(ωx dx The Fourier cosine transform is defined: = π F c [f(x] = ˆf c (ω = π f(x cos(ωx dx f(x cos(ωx dx. Note that ˆf c (ω is an even function. The inverse Fourier cosine transform is F c [ ˆf c (ω] = = = = 2 ˆf c (ω e ıωx dω ˆf c (ω(cos(ωx + ı sin(ωx dω ˆf c (ω cos(ωx dω ˆf c (ω cos(ωx dω. 562

25 Thus we have the Fourier cosine transform pair f(x = F c [ ˆf c (ω] = 2 ˆf c (ω cos(ωx dω, ˆfc (ω = F c [f(x] = π f(x cos(ωx dx The Fourier Sine Transform Suppose f(x is an odd function. In this case the Fourier transform of f(x coincides with the Fourier sine transform of f(x. F[f(x] = f(x e ıωx dx = f(x(cos(ωx ı sin(ωx dx = ı π f(x sin(ωx dx Note that ˆf(ω = F[f(x] is an odd function of ω. The inverse Fourier transform of ˆf(ω is F [ ˆf(ω] = = 2ı ˆf(ω e ıωx dω ˆf(ω sin(ωx dω. Thus we have that f(x = 2ı = 2 ( ı π ( π f(x sin(ωx dx sin(ωx dω f(x sin(ωx dx sin(ωx dω. 563

26 This gives us the Fourier sine transform pair f(x = Fs [ ˆf s (ω] = 2 ˆf s (ω sin(ωx dω, ˆfs (ω = F s [f(x] = π f(x sin(ωx dx. Result The Fourier cosine transform pair is defined: f(x = Fc [ ˆf c (ω] = 2 ˆf c (ω = F c [f(x] = π The Fourier sine transform pair is defined: f(x = Fs [ ˆf s (ω] = 2 ˆf s (ω = F s [f(x] = π ˆf c (ω cos(ωx dω f(x cos(ωx dx ˆf s (ω sin(ωx dω f(x sin(ωx dx 32.7 Properties of the Fourier Cosine and Sine Transform Transforms of Derivatives Cosine Transform. Using integration by parts we can find the Fourier cosine transform of derivatives. Let y be a function for which the Fourier cosine transform of y and its first and second derivatives exists. Further assume that y 564

27 and y vanish at infinity. We calculate the transforms of the first and second derivatives. F c [y ] = y cos(ωx dx π = [ ] y cos(ωx π + ω π = ωŷ c (ω π y( y sin(ωx dx F c [y ] = y cos(ωx dx π = [ y cos(ωx ] π + ω y sin(ωx dx π = π y ( + ω [ ] y sin(ωx π ω2 y cos(ωx dx π = ω 2 ˆfc (ω π y ( Sine Transform. You can show, (see Exercise 32.3, that the Fourier sine transform of the first and second derivatives are F s [y ] = ω ˆf c (ω F s [y ] = ω 2 ŷ c (ω + ω π y(. 565

28 Convolution Theorems Cosine Transform of a Product. Consider the Fourier cosine transform of a product of functions. Let f(x and g(x be two functions defined for x. Let F c [f(x] = ˆf c (ω, and F c [g(x] = ĝ c (ω. F c [f(xg(x] = π = π = 2 π f(xg(x cos(ωx dx ( 2 ˆf c (η cos(ηx dη g(x cos(ωx dx ˆf c (ηg(x cos(ηx cos(ωx dx dη We use the identity cosacosb = (cos(a b + cos(a + b. 2 = π = = ˆf c (η [ π ˆf c (ηg(x ( cos((ω ηx + cos((ω + ηx dx dη g(x cos((ω ηx dx + π ˆf c (η ( ĝ c (ω η + ĝ c (ω + η dη ] g(x cos((ω + ηx dx dη ĝ c (ω is an even function. If we have only defined ĝ c (ω for positive argument, then ĝ c (ω = ĝ c ( ω. = ˆf c (η ( ĝ c ( ω η + ĝ c (ω + η dη 566

29 Inverse Cosine Transform of a Product. Now consider the inverse Fourier cosine transform of a product of functions. Let F c [f(x] = ˆf c (ω, and F c [g(x] = ĝ c (ω. F c [ ˆf c (ωĝ c (ω] = 2 = 2 = 2 π = π = = ˆf c (ωĝ c (ω cos(ωx dω ( f(ξ cos(ωξ dξ ĝ c (ω cos(ωx dω π f(ξ f(ξĝ c (ω cos(ωξ cos(ωx dω dξ f(ξĝ c (ω ( cos(ω(x ξ + cos(ω(x + ξ dω dξ ( 2 ĝ c (ω cos(ω(x ξ dω + 2 f(ξ ( g( x ξ + g(x + ξ dξ ĝ c (ω cos(ω(x + ξ dω dξ Sine Transform of a Product. You can show, (see Exercise 32.5, that the Fourier sine transform of a product of functions is F s [f(xg(x] = ˆf s (η ( ĝ c ( ω η ĝ c (ω + η dη. Inverse Sine Transform of a Product. You can also show, (see Exercise 32.6, that the inverse Fourier sine transform of a product of functions is F s [ ˆf s (ωĝ c (ω] = f(ξ ( g( x ξ g(x + ξ dξ. 567

30 Result The Fourier cosine and sine transform convolution theorems are F c [f(xg(x] = Fc [ ˆf c (ωĝ c (ω] = F s [f(xg(x] = F s [ ˆf s (ωĝ c (ω] = ˆf c (η [ ĝ c ( ω η + ĝ c (ω + η ] dη f(ξ ( g( x ξ + g(x + ξ dξ ˆf s (η ( ĝ c ( ω η ĝ c (ω + η dη f(ξ ( g( x ξ g(x + ξ dξ Cosine and Sine Transform in Terms of the Fourier Transform We can express the Fourier cosine and sine transform in terms of the Fourier transform. First consider the Fourier cosine transform. Let f(x be an even function. F c [f(x] = π We extend the domain integration because the integrand is even. = f(x cos(ωx dx f(x cos(ωx dx Note that f(x sin(ωx dx = because the integrand is odd. = f(x e ıωx dx = F[f(x] 568

31 F c [f(x] = F[f(x], For general f(x, use the even extension, f( x to write the result. F c [f(x] = F[f( x ] for even f(x. There is an analogous result for the inverse Fourier cosine transform. [ ] [ ] ˆf(ω = F ˆf( ω F c For the sine series, we have F s [f(x] = ıf [sign(xf( x ] [ ] [ Fs ˆf(ω = ıf sign(ω ˆf( ω ] Result The results: F c [f(x] = F[f( x ] F s [f(x] = ıf[sign(xf( x ] F c F s [ ˆf(ω ] = F [ ˆf( ω ] [ ] [ ˆf(ω = ıf sign(ω ˆf( ω ] allow us to evaluate Fourier cosine and sine transforms in terms of the Fourier transform. This enables us to use contour integration methods to do the integrals Solving Differential Equations with the Fourier Cosine and Sine Transforms Example Consider the problem y y =, y( =, y( =. 569

32 Since the initial condition is y( = and the sine transform of y is ω 2 ŷ c (ω + ω y( we take the Fourier sine π transform of both sides of the differential equation. ω 2 ŷ c (ω + ω π y( ŷ c(ω = We use the table of Fourier Sine transforms. (ω 2 + ŷ c (ω = ω π ω ŷ c (ω = π(ω 2 + y = e x Example Consider the problem y y = e 2x, y ( =, y( =. Since the initial condition is y ( =, we take the Fourier cosine transform of the differential equation. From the table of cosine transforms, F c [e 2x ] = 2/(π(ω ω 2 ŷ c (ω π y ( ŷ c (ω = 2 π(ω ŷ c (ω = π(ω 2 + 4(ω 2 + = 2 ( /3 π ω 2 + /3 ω = 2/π 3 ω /π 3 ω 2 + y = 3 e 2x 2 3 e x 57