Math 602 43 6 Fourier transform Here are some formulae that you may want to use: 1 Z +1 Z +1 F(f)(!) def= f(x)ei!xdx; F −1(f)(x) = f(!)e−i!xd!; (3) 2π −∞ −∞ F(f ∗ g) = 2πF(f)F(g); (4) 1 α 2α F(e−αjxj) = ; F( )(!) = e−αj!j; (5) π !2 + α2 x2 + α2 F(f(x − β))(!) = eiβ!F(f)(!); (6) 2 −αx2 1 − ! F(e ) = p e 4α (7) 4πα 1 1 F(H(x)e−ax)(!) = ;H is the Heaviside function (8) 2π a − i! 1 i i F(pv( ))(!) = sign(!) = (H(!) − H(−!)) (9) x 2 2 1 π 1 2 F(sech(ax)) = sech( !); sech(ax) def= = (10) 2a 2a ch(x) ex + e−x 1 1 cos(a) − cos(b) = −2 sin( 2 (a + b)) sin( 2 (a − b)) (11) Question 102: (i) Let f be an integrable function on (−∞; +1). Prove that for all a; b 2 R, ibx 1 ξ+b and for all ξ 2 R, F([e f(ax)])(ξ) = a F(f)( a ). Solution: The definition of the Fourier transform together with the change of variable ax 7−! x0 implies 1 Z +1 F[eibxf(ax)])(ξ) = f(ax)eibxeiξxdx 2π −∞ 1 Z +1 = f(ax)ei(b+ξ)xdx 2π −∞ Z +1 1 1 0 i (ξ+b) x0 0 = f(x )e a dx 2π −∞ a 1 ξ+b = a F(f)( a ): 2 (ii) Let c be a positive real number. Compute the Fourier transform of f(x) = e−cx sin(bx). 1 ibx −ibx p Solution: Using the fact that sin(bx) = −i 2 (e − e ), and setting a = c we infer that 1 −(ax)2 ibx −ibx f(x) = 2i e (e − e ) 1 −(ax)2 ibx 1 −(ax)2 i(−b)x = 2i e e − 2i e e and using (i) and (7) we deduce 1 1 1 − 1 ( ξ+b )2 − 1 ( ξ−b )2 f^(ξ) = p (e 4 a − e 4 a ): 2i a 4π In conclusion 1 1 − 1 (ξ+b)2 − 1 (ξ−b)2 f^(ξ) = p (e 4c − e 4c ): 2i 4πc Question 103: (a) Compute the Fourier transform of the function f(x) defined by ( 1 if jxj ≤ 1 f(x) = 0 otherwise Math 602 44 Solution: By definition 1 Z 1 1 1 F(f)(!) = eiξ! = = (ei! − e−i!) 2π −1 2π i! 1 2 sin(!) = : 2π ! Hence 1 sin(!) F(f)(!) = : π ! sin(!) (b) Find the inverse Fourier transform of g(!) = ! . Solution: Using (a) we deduce that g(!) = πF(f)(!), that is to say, F −1(g)(x) = πF −1(F(f))(x). Now, using the inverse Fourier transform, we deduce that F −1(g)(x) = πf(x) at every point x where 1 −1 π − + f(x) is of class C and F (g)(x) = 2 (f(x ) + f(x )) at discontinuity points of f. As a result: 8 >π if jxj < 1 −1 < π F (g)(x) = 2 at jxj = 1 :>0 otherwise 1 sin(λω Question 104: Find the inverse Fourier transform of π ! . 1 sin(λω) Solution: Since F(Sλ(x)) = π ! , the inverse Forier transform theorem implies that 8 1 if jxj < λ 1 sin(λω) <> F −1 (x) = 1 if jxj = λ π ! 2 :>0 otherwise: Question 105: Use the Fourier transform technique to solve the following PDE: @tu(x; t) + c@xu(x; t) + γu(x; t) = 0; for all x 2 (−∞; +1), t > 0, with u(x; 0) = u0(x) for all x 2 (−∞; +1). Solution: By taking the Fourier transform of the PDE, one obtains @tF(u) − i!cF(y) + γF(y) = 0: The solution is F(u)(!; t) = c(!)ei!ct−γt: The initial condition implies that c(!) = F(u0)(!): i!ct −γt F(u)(!; t) = F(u0)(!)e e : The shift lemma in turn implies that −γt −γt F(u)(!; t) = F(u0(x − ct))(!)e = F(u0(x − ct)e )(!): Applying the inverse Fourier transform gives: −γt u(x; t) = u0(x − ct)e : Question 106: Solve by the Fourier transform technique the following equation: @xxφ(x) − −x 2@xφ(x) + φ(x) = H(x)e , x 2 (−∞; +1), where H(x) is the Heaviside function. (Hint: use the factorization i!3 + !2 + i! + 1 = (1 + !2)(1 + i!) and recall that F(f(x))(−!) = F(f(−x))(!)). Math 602 45 Solution: Applying the Fourier transform with respect to x gives 1 1 (−!2 + 2i! + 1)F(φ)(!) = F(H(x)e−x)(!) = : 2π 1 − i! where we used (8). Then, using the hint gives 1 1 1 1 F(φ)(!) = = 2π (1 − i!)(−!2 + 2i! + 1) 2π i!3 + !2 + i! + 1 1 1 1 = : 2π 1 + !2 1 + i! We now use again (8) and (5) to obtain 1 1 1 1 F(φ)(!) = π = πF(e−|xj)(!)F(H(x)e−x)(−!): π 1 + !2 2π 1 − i(−!) Now we use F(H(x)e−x)(−!) = F(H(−x)ex)(!) and we finally have F(φ)(!) = πF(e−|xj)(!)F(H(−x)ex)(!): The Convolution Theorem (4) gives 1 F(φ)(!) = π F(e−|xj ∗ (H(−x)ex))(!): 2π We obtain φ by using the inverse Fourier transform 1 φ(x) = e−|xj ∗ (H(−x)ex); 2 i.e., Z +1 1 φ(x) = e−|x−yjH(−y)eydy −∞ 2 and recalling that H is the Heaviside function we finally have ( 1 Z 0 1 e−x if x ≥ 0 φ(x) = ey−|x−yjdy = 4 2 1 x −∞ ( 4 − x)e if x ≤ 0: Question 107: Use the Fourier transform technique to solve the following ODE y00(x)−y(x) = f(x) for x 2 (−∞; +1), with y(±∞) = 0, where f is a function such that jfj is integrable over R. Solution: By taking the Fourier transform of the ODE, one obtains −!2F(y) − F(y) = F(f): That is 1 F(y) = −F(f) : 1 + !2 and the convolution Theorem, see (4), together with (5) gives 1 F(y) = −πF(f)F(e−|xj) = − F(f ∗ e−|xj): 2 Applying F −1 on both sides we obtain 1 1 Z 1 y(x) = − f ∗ e−|xj = − e−|x−zjf(z)dz 2 2 −∞ Math 602 46 That is 1 Z 1 y(x) = − e−|x−zjf(z)dz: 2 −∞ 2 Question 108: Use the Fourier transform method to compute the solution of utt − a uxx = 0, 2 where x 2 R and t 2 (0; +1), with u(0; x) = f(x) := sin (x) and ut(0; x) = 0 for all x 2 R. Solution: Take the Fourier transform in the x direction: 2 2 F(u)tt + ! a F(u) = 0: This is an ODE. The solution is F(u)(t; !) = c1(!) cos(!at) + c2(!) sin(!at): The initial boundary conditions give F(u)(0; ξ) = F(f)(!) = c1(!) and c2(!) = 0. Hence 1 F(t; !) = F(f)(!) cos(!at) = F(f)(!)(eia!t + e−ia!t): 2 Using the shift lemma, we infer that 1 F(t; !) = (F(f(x − at))(!) + F(f(x + at))(!)): 2 Usin the inverse Fourier transform, we finaly conclude that 1 1 u(t; x) = (f(x − at) + f(x + at)) = (sin2(x + at) + sin2(x − at)): 2 2 Note that this is the D'Alembert formula. 2 Question 109: Use the Fourier transform method to compute the solution of utt − a uxx = 0, 2 where x 2 R and t 2 (0; +1), with u(0; x) = f(x) := cos (x) and ut(0; x) = 0 for all x 2 R. Solution: Take the Fourier transform in the x direction: 2 2 F(u)tt + ! a F(u) = 0: This is an ODE. The solution is F(u)(t; !) = c1(!) cos(!at) + c2(!) sin(!at): The initial boundary conditions give F(u)(0; ξ) = F(f)(!) = c1(!) and c2(!) = 0. Hence 1 F(t; !) = F(f)(!) cos(!at) = F(f)(!)(eia!t + e−ia!t): 2 Using the shift lemma (i.e., formula (6)) we obtain 1 1 u(t; x) = (f(x − at) + f(x + at)) = (cos2(x + at) + cos2(x − at)): 2 2 Note that this is the D'Alembert formula. 1 R +1 f(y) 1 1 Question 110: Solve the integral equation: f(x) + 2π −∞ (x−y)2+1 dy = x2+4 + x2+1 , for all x 2 (−∞; +1). Math 602 47 Solution: The equation can be re-written 1 1 1 1 f(x) + f ∗ = + : 2π x2 + 1 x2 + 4 x2 + 1 We take the Fourier transform of the equation and we apply the Convolution Theorem (see (4)) 1 1 1 1 F(f) + 2πF( )F(f) = F( ) + F( ): 2π x2 + 1 x2 + 4 x2 + 1 Using (5), we obtain 1 1 1 F(f) + e−|!jF(f) = e−2j!j + e−|!j; 2 4 2 which gives 1 1 1 F(f)(1 + e−|!j) = e−|!j( e−|!j + 1): 2 2 2 We then deduce 1 F(f) = e−|!j: 2 1 Taking the inverse Fourier transform, we finally obtain f(x) = x2+1 . Question 111: Solve the following integral equation (Hint: solution is short): Z +1 p Z +1 2 2 − y − x f(y)f(x − y)dy − 2 2 e 2π f(x − y)dy = −2πe 4π : 8x 2 R: −∞ −∞ Solution: This equation can be re-written using the convolution operator: p 2 2 − x − x f ∗ f − 2 2e 2π ∗ f = −2πe 4π : We take the Fourier transform and use the convolution theorem (4) together with (7) to obtain p −!2 1 −!2 1 2 1 4 1 1 4 1 2πF(f) − 2π2 2F(f) e 2π = −2π e 4π q 1 q 1 4π 2π 4π 4π 2 −!2 π −!2π F(f) − 2F(f)e 2 + e = 0 −!2 π 2 (F(f) − e 2 ) = 0: This implies −!2 π F(f) = e 2 : Taking the inverse Fourier transform, we obtain p 2 − x f(x) = 2e 2π : 3 R +1 −|x−yj −|xj Question 112: Solve the integral equation: f(x) + 2 −∞ e f(y)dy = e , for all x 2 (−∞; +1).