Math 602 43
6 Fourier transform
Here are some formulae that you may want to use:
1 Z +∞ Z +∞ F(f)(ω) def= f(x)eiωxdx, F −1(f)(x) = f(ω)e−iωxdω, (3) 2π −∞ −∞ F(f ∗ g) = 2πF(f)F(g), (4) 1 α 2α F(e−α|x|) = , F( )(ω) = e−α|ω|, (5) π ω2 + α2 x2 + α2 F(f(x − β))(ω) = eiβωF(f)(ω), (6)
2 −αx2 1 − ω F(e ) = √ e 4α (7) 4πα 1 1 F(H(x)e−ax)(ω) = ,H is the Heaviside function (8) 2π a − iω 1 i i F(pv( ))(ω) = sign(ω) = (H(ω) − H(−ω)) (9) x 2 2 1 π 1 2 F(sech(ax)) = sech( ω), 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 (−∞, +∞). Prove that for all a, b ∈ R, ibx 1 ξ+b and for all ξ ∈ 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 +∞ F[eibxf(ax)])(ξ) = f(ax)eibxeiξxdx 2π −∞ 1 Z +∞ = f(ax)ei(b+ξ)xdx 2π −∞ Z +∞ 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 √ 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 fˆ(ξ) = √ (e 4 a − e 4 a ). 2i a 4π In conclusion
1 1 − 1 (ξ+b)2 − 1 (ξ−b)2 fˆ(ξ) = √ (e 4c − e 4c ). 2i 4πc
Question 103: (a) Compute the Fourier transform of the function f(x) defined by ( 1 if |x| ≤ 1 f(x) = 0 otherwise Math 602 44
Solution: By definition
1 Z 1 1 1 F(f)(ω) = eiξω = = (eiω − e−iω) 2π −1 2π iω 1 2 sin(ω) = . 2π ω Hence 1 sin(ω) F(f)(ω) = . π ω
sin(ω) (b) Find the inverse Fourier transform of g(ω) = ω . Solution: Using (a) we deduce that g(ω) = πF(f)(ω), 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: π if |x| < 1 −1 π F (g)(x) = 2 at |x| = 1 0 otherwise
1 sin(λω Question 104: Find the inverse Fourier transform of π ω . 1 sin(λω) Solution: Since F(Sλ(x)) = π ω , the inverse Forier transform theorem implies that 1 if |x| < λ 1 sin(λω) F −1 (x) = 1 if |x| = λ π ω 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 ∈ (−∞, +∞), t > 0, with u(x, 0) = u0(x) for all x ∈ (−∞, +∞). Solution: By taking the Fourier transform of the PDE, one obtains
∂tF(u) − iωcF(y) + γF(y) = 0. The solution is F(u)(ω, t) = c(ω)eiωct−γt.
The initial condition implies that c(ω) = F(u0)(ω):
iωct −γt F(u)(ω, t) = F(u0)(ω)e e . The shift lemma in turn implies that
−γt −γt F(u)(ω, t) = F(u0(x − ct))(ω)e = F(u0(x − ct)e )(ω). 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 ∈ (−∞, +∞), where H(x) is the Heaviside function. (Hint: use the factorization iω3 + ω2 + iω + 1 = (1 + ω2)(1 + iω) and recall that F(f(x))(−ω) = F(f(−x))(ω)). Math 602 45
Solution: Applying the Fourier transform with respect to x gives 1 1 (−ω2 + 2iω + 1)F(φ)(ω) = F(H(x)e−x)(ω) = . 2π 1 − iω where we used (8). Then, using the hint gives 1 1 1 1 F(φ)(ω) = = 2π (1 − iω)(−ω2 + 2iω + 1) 2π iω3 + ω2 + iω + 1 1 1 1 = . 2π 1 + ω2 1 + iω We now use again (8) and (5) to obtain 1 1 1 1 F(φ)(ω) = π = πF(e−|x|)(ω)F(H(x)e−x)(−ω). π 1 + ω2 2π 1 − i(−ω)
Now we use F(H(x)e−x)(−ω) = F(H(−x)ex)(ω) and we finally have
F(φ)(ω) = πF(e−|x|)(ω)F(H(−x)ex)(ω).
The Convolution Theorem (4) gives 1 F(φ)(ω) = π F(e−|x| ∗ (H(−x)ex))(ω). 2π We obtain φ by using the inverse Fourier transform 1 φ(x) = e−|x| ∗ (H(−x)ex), 2 i.e., Z +∞ 1 φ(x) = e−|x−y|H(−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−y|dy = 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 ∈ (−∞, +∞), with y(±∞) = 0, where f is a function such that |f| is integrable over R. Solution: By taking the Fourier transform of the ODE, one obtains
−ω2F(y) − F(y) = F(f).
That is 1 F(y) = −F(f) . 1 + ω2 and the convolution Theorem, see (4), together with (5) gives 1 F(y) = −πF(f)F(e−|x|) = − F(f ∗ e−|x|). 2 Applying F −1 on both sides we obtain
1 1 Z ∞ y(x) = − f ∗ e−|x| = − e−|x−z|f(z)dz 2 2 −∞ Math 602 46
That is 1 Z ∞ y(x) = − e−|x−z|f(z)dz. 2 −∞
2 Question 108: Use the Fourier transform method to compute the solution of utt − a uxx = 0, 2 where x ∈ R and t ∈ (0, +∞), with u(0, x) = f(x) := sin (x) and ut(0, x) = 0 for all x ∈ R. Solution: Take the Fourier transform in the x direction:
2 2 F(u)tt + ω a F(u) = 0.
This is an ODE. The solution is
F(u)(t, ω) = c1(ω) cos(ωat) + c2(ω) sin(ωat).
The initial boundary conditions give
F(u)(0, ξ) = F(f)(ω) = c1(ω) and c2(ω) = 0. Hence 1 F(t, ω) = F(f)(ω) cos(ωat) = F(f)(ω)(eiaωt + e−iaωt). 2 Using the shift lemma, we infer that 1 F(t, ω) = (F(f(x − at))(ω) + F(f(x + at))(ω)). 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 ∈ R and t ∈ (0, +∞), with u(0, x) = f(x) := cos (x) and ut(0, x) = 0 for all x ∈ R. Solution: Take the Fourier transform in the x direction:
2 2 F(u)tt + ω a F(u) = 0.
This is an ODE. The solution is
F(u)(t, ω) = c1(ω) cos(ωat) + c2(ω) sin(ωat).
The initial boundary conditions give
F(u)(0, ξ) = F(f)(ω) = c1(ω) and c2(ω) = 0. Hence 1 F(t, ω) = F(f)(ω) cos(ωat) = F(f)(ω)(eiaωt + e−iaωt). 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 +∞ 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 ∈ (−∞, +∞). 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−|ω|F(f) = e−2|ω| + e−|ω|, 2 4 2 which gives 1 1 1 F(f)(1 + e−|ω|) = e−|ω|( e−|ω| + 1). 2 2 2 We then deduce 1 F(f) = e−|ω|. 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 +∞ √ Z +∞ 2 2 − y − x f(y)f(x − y)dy − 2 2 e 2π f(x − y)dy = −2πe 4π . ∀x ∈ R. −∞ −∞
Solution: This equation can be re-written using the convolution operator:
√ 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
√ −ω2 1 −ω2 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 −ω2 π −ω2π F(f) − 2F(f)e 2 + e = 0 −ω2 π 2 (F(f) − e 2 ) = 0.
This implies −ω2 π F(f) = e 2 . Taking the inverse Fourier transform, we obtain
√ 2 − x f(x) = 2e 2π .
3 R +∞ −|x−y| −|x| Question 112: Solve the integral equation: f(x) + 2 −∞ e f(y)dy = e , for all x ∈ (−∞, +∞). Solution: The equation can be re-written 3 f(x) + e−|x| ∗ f = e−|x|. 2 We take the Fourier transform of the equation and apply the Convolution Theorem (see (4)) 3 F(f) + 2πF(e−|x|)F(f) = F(e−|x|). 2 Using (5), we obtain 1 1 1 1 F(f) + 3π F(f) = , π 1 + ω2 π 1 + ω2 Math 602 48 which gives ω2 + 4 1 1 F(f) = . 1 + ω2 π 1 + ω2 We then deduce 1 1 1 F(f) = = F(e−2|x|). π 4 + ω2 2 1 −2|x| Taking the inverse Fourier transform, we finally obtain f(x) = 2 e . +∞ 2 1 2 R −(x−y) − 2 x Question 113: Solve the following integral equation −∞ e g(y)dy = e for all x ∈ (−∞, +∞), i.e., find the function g that solves the above equation. Solution: The left-hand side of the equation is a convolution; hence,
−x2 − 1 x2 e ∗ g(x) = e 2 . By taking the Fourier transform, we obtain
1 − 1 ω2 1 − 1 ω2 2π √ e 4 Fg(ω) = √ e 2 . 4π 2π That gives 1 − 1 ω2 F(g)(ω) = √ e 4 . 2π By taking the inverse Fourier transform, we deduce √ r 4π 2 2 2 g(x) = √ e−x = e−x . 2π π
R +∞ 4 Question 114: Solve the integral equation: −∞ f(y)f(x−y)dy = x2+4 , for all x ∈ (−∞, +∞). How many solutions did you find? Solution: The equation can be re-written 4 f ∗ f = . x2 + 4 We take the Fourier transform of the equation and apply the Convolution Theorem (see (4)) 4 2πF(f)2 = F( ) x2 + 4 Using (5), we obtain 2πF(f)2 = e−2|ω|. which gives 1 F(f) = ±√ e−|ω|. 2π Taking the inverse Fourier transform, we finally obtain 1 2 f(x) = ±√ . 2π x2 + 1
We found two solutions: a positive one and a negative one. 2t Question 115: Use the Fourier transform method to solve the equation ∂tu + 1+t2 ∂xu = 0, u(x, 0) = u0(x), in the domain x ∈ (−∞, +∞) and t > 0. Solution: We take the Fourier transform of the equation with respect to x 2t 0 = ∂ F(u) + F( ∂ u) t 1 + t2 x 2t = ∂ F(u) + F(∂ u) t 1 + t2 x 2t = ∂ F(u) − iω F(u). t 1 + t2 Math 602 49
This is a first-order linear ODE: ∂ F(u) 2t d t = iω = iω (log(1 + t2)) F(u) 1 + t2 dt The solution is 2 F(u)(ω, t) = K(ω)eiω log(1+t ). Using the initial condition, we obtain
F(u0)(ω) = F(u)(ω, 0) = K(ω).
The shift lemma (see formula (6)) implies
iω log(1+t2) 2 F(u)(ω, t) = F(u0)(ω)e = F(u0(x − log(1 + t ))),
Applying the inverse Fourier transform finally gives
2 u(x, t) = u0(x − log(1 + t )).
Question 116: Solve the integral equation:
+∞ +∞ √ Z √ y2 1 Z 2 (x−y)2 − 2π − 2π f(y) − 2e − 2 f(x−y)dy = − 2 e dy, ∀x ∈ (−∞, +∞). −∞ 1 + y −∞ 1 + y (Hint: there is an easy factorization after applying the Fourier transform.) Solution: The equation can be re-written
√ 2 √ 2 − x 1 1 − x f ∗ (f − 2e 2π − ) = − ∗ 2e 2π . 1 + x2 1 + x2 We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
√ 2 √ 2 − x 1 1 − x 2πF(f) F(f) − 2F(e 2π ) − F( ) = −2πF( ) 2F(e 2π ) 1 + x2 1 + x2
Solution 1: Using (5), (7) we obtain
ω2 √ x2 √ − πω2 − 1 4 1 − 2F(e 2π ) = 2 e 2π = e 2 q 1 4π 2π 1 1 F( ) = e−|ω|, 1 + x2 2 which gives 2 2 − πω 1 −|ω| 1 −|ω| − πω F(f) F(f) − e 2 − e = − e e 2 . 2 2 This equation can also be re-written as follows
2 2 2 − πω 1 −|ω| 1 −|ω| − πω F(f) − F(f)e 2 − F(f) e + e e 2 = 0, 2 2 and can be factorized as follows:
2 − πω 1 −|ω| (F(f) − e 2 )(F(f) − e ) = 0. 2
πω2 − 2 1 −|ω| This means that either F(f) = e or F(f) = 2 e . Taking the inverse Fourier transform, we finally obtain two solutions
√ 2 − x 1 f(x) = 2e 2π , or f(x) = . 1 + x2 Math 602 50
Solution 2: Another solution consists of remarking that the equation with the Fourier transform can be rewritten as follows:
√ 2 √ 2 2 − x 1 − x 1 F(f) − F( 2e 2π )F(f) − F( )F(f) + F( 2e 2π )F( ) = 0, 1 + x2 1 + x2 which can factorized as follows:
√ 2 − x 1 (F(f) − F( 2e 2π ))(F(f) − F( )) = 0. 1 + x2
√ x2 − 2π 1 The either F(f) = F( 2e )) or (F(f) = F( 1+x2 ). The conclusion follows easily.
Question 117: Use the Fourier transform technique to solve ∂tu(x, t) + sin(t)∂xu(x, t) + (2 + 2 3t )u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0(x). (Use the shift lemma: F(f(x − β))(ω) = iωβ 1 R +∞ iωx F(f)(ω)e and the definition F(f)(ω) := 2π −∞ f(x)e dx)
Solution: Applying the Fourier transform to the equation gives
2 ∂tF(u)(ω, t) + sin(t)(−iω)F(u)(ω, t) + (2 + 3t )F(u)(ω, t) = 0
This can also be re-written as follows: ∂ F(u)(ω, t) t = iω sin(t) − (2 + 3t2). F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −iω(cos(t) − 1) − (2t + t3).
This implies −iω(cos(t)−1) −(2t+t3) F(u)(ω, t) = F(u0)(ω)e e . Then the shift lemma gives
−(2t+t3) F(u)(ω, t) = F(u0(x + cos(t) − 1)(ω)e .
This finally gives −(2t+t3) u(x, t) = u0(x + cos(t) − 1)e .
Question 118: Solve the following integral equation (Hint: x2 − 3xa + 2a2 = (x − a)(x − 2a)):
Z +∞ √ 2 2 − y − x (f(y) − 3 2e 2π )f(x − y)dy = −4πe 4π . ∀x ∈ R. −∞
Solution: This equation can be re-written using the convolution operator:
√ 2 2 − x − x f ∗ f − 3 2e 2π ∗ f = −4πe 4π .
We take the Fourier transform and use (7) to obtain
√ −ω2 1 −ω2 1 2 1 4 1 1 4 1 2πF(f) − 2π3 2F(f) e 2π = −4π e 4π q 1 q 1 4π 2π 4π 4π 2 −ω2 π −ω2π F(f) − 3F(f)e 2 + 2e = 0 −ω2 π −ω2 π (F(f) − e 2 )(F(f) − 2e 2 ) = 0.
This implies −ω2 π −ω2 π either F(f) = e 2 , or F(f) = 2e 2 . Math 602 51
Taking the inverse Fourier transform, we obtain
√ 2 √ 2 − x − x either f(x) = 2e 2π , or f(x) = 2 2e 2π .
Question 119: Use the Fourier transform technique to solve ∂tu(x, t)−∂xxu(x, t)+sin(t)∂xu(x, t)+ 2 (2 + 3t )u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0(x). (Hint: use the definition F(f)(ω) := 2 1 R +∞ iωx −αx2 1 − ω f(x)e dx), the result F(e )(ω) = √ e 4α , the convolution theorem and the 2π −∞ 4πα shift lemma: F(f(x − β))(ω) = F(f)(ω)eiωβ. Go slowly and give all the details.)
Solution: Applying the Fourier transform to the equation gives
2 2 ∂tF(u)(ω, t) + ω F(u)(ω, t) + sin(t)(−iω)F(u)(ω, t) + (2 + 3t )F(u)(ω, t) = 0 This can also be re-written as follows: ∂ F(u)(ω, t) t = −ω2 + iω sin(t) − (2 + 3t2). F(u)(ω, t) Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −ω2t − iω(cos(t) − 1) − (2t + t3).
This implies −ω2t −iω(cos(t)−1) −(2t+t3) F(u)(ω, t) = F(u0)(ω)e e e . 2 −αx2 1 − ω 1 Using the result F(e )(ω) = √ e 4α where α = , this implies that 4πα 4t
r 2 π − x −iω(cos(t)−1) −(2t+t3) F(u)(ω, t) = F(u )(ω)F(e 4t )(ω)e e t 0
r 2 π − x −iω(cos(t)−1) −(2t+t3) = F(u ∗ e 4t )(ω)e e . t 0
2 − x Then setting g = u0 ∗ e 4t the convolution theorem followed by the shift lemma gives r 1 π 3 F(u)(ω, t) = F(g(x + cos(t) − 1))(ω)e−(2t+t ). 2π t This finally gives
r r Z +∞ 2 1 −(2t+t3) −(2t+t3) 1 − (x+cos(t)−1−y) u(x, t) = g(x + cos(t) − 1)e = e u0(y)e 4t dy. 4πt 4πt −∞
2 2 Question 120: Consider the telegraph equation ∂ttu + 2α∂tu + α u − c ∂xxu = 0 with α ≥ 0, u(x, 0) = 0, ∂tu(x, 0) = g(x), x ∈ R, t > 0 and boundary condition at infinity u(±∞, t) = 0. Solve the equation by the Fourier transform technique. (Hint: the solution to the ODE φ00(t) + 2αφ0(t) + (α2 + λ2)φ(t) = 0 is φ(t) = e−αt(a cos(λt) + b sin(λt)) Solution: Applying the Fourier transform with respect to x to the equation, we infer that
2 2 2 0 = ∂ttF(u)(ω, t) + 2α∂tF(u)(ω, t) + α F(u)(ω, t) − c (−iω) F(u)(ω, t) 2 2 2 = ∂ttF(u)(ω, t) + 2α∂tF(u)(ω, t) + (α + c ω )F(u)(ω, t) Using the hint, we deduce that
F(u)(ω, t) = e−αt(a(ω) cos(ωct) + b(ω) sin(ωct)).
The initial condition implies that a(ω) = 0 and F(g)(ω) = ωcb(ω); as a result, b(ω) = F(g)(ω)/(ωc) and sin(ωct) F(u)(ω, t) = e−αtF(g) . ωc Math 602 52
Then using (??), we have π F(u)(ω, t) = e−αtF(g)F(S ). c ct The convolution theorem implies that Z ∞ −αt 1 −αt 1 u(x, t) = e g ∗ Sct = e g(y)Sct(x − y)dy. 2c 2c −∞
Finally the definition of Sct implies that Sct(x − y) is equal to 1 if −ct < x − y < ct and is equal zero otherwise, which finally means that
1 Z x+ct u(x, t) = e−αt g(y)dy. 2c x−ct
Question 121: Use the Fourier transform technique to solve ∂tu(x, t)−∂xxu(x, t)+cos(t)∂xu(x, t)+ (1 + 2t)u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0(x). (Hint: use the definition F(f)(ω) := 2 1 R +∞ iωx −αx2 1 − ω f(x)e dx), the result F(e )(ω) = √ e 4α , the convolution theorem and the 2π −∞ 4πα shift lemma: F(f(x − β))(ω) = F(f)(ω)eiωβ. Go slowly and give all the details.)
Solution: Applying the Fourier transform to the equation gives
2 ∂tF(u)(ω, t) + ω F(u)(ω, t) + cos(t)(−iω)F(u)(ω, t) + (1 + 2t)F(u)(ω, t) = 0
This can also be re-written as follows: ∂ F(u)(ω, t) t = −ω2 + iω cos(t) − (1 + 2t). F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −ω2t + iω sin(t) − (t + t2).
This implies −ω2t iω sin(t) −(t+t2) F(u)(ω, t) = F(u0)(ω)e e e .
2 −αx2 1 − ω 1 Using the result F(e )(ω) = √ e 4α where α = , this implies that 4πα 4t
r 2 π − x iω sin(t) −(t+t2) F(u)(ω, t) = F(u )(ω)F(e 4t )(ω)e e t 0
r 2 π − x iω sin(t) −(t+t2) = F(u ∗ e 4t )(ω)e e . t 0
2 − x Then setting g = u0 ∗ e 4t the convolution theorem followed by the shift lemma gives r 1 π 2 F(u)(ω, t) = F(g(x − sin(t)))(ω)e−(t+t ). 2π t
This finally gives
r r Z +∞ 2 1 −(t+t2) −(t+t2) 1 − (x−sin(t)−y) u(x, t) = g(x − sin(t))e = e u0(y)e 4t dy. 4πt 4πt −∞
Question 122: Solve the following integral equation (Hint: (a + b)2 = a2 + 2ab + b2):
Z +∞ Z +∞ 2 4 f(y)f(x − y)dy − 2 2 f(x − y)dy + 2π 2 = 0 ∀x ∈ R. −∞ −∞ y + 1 x + 4 Math 602 53
Solution: This equation can be re-written using the convolution operator: 2 4 f ∗ f − 2( ) ∗ f + 2π = 0. x2 + 1 x2 + 4 We take the Fourier transform and use the convolution theorem (??) together with (??) to obtain
2πF(f)2 − 4πF(f)e−|ω| + 2πe−2|ω| = 0 F(f)2 − 2F(f)e−|ω| + e−2|ω| = 0 (F(f) − e−|ω|)2 = 0
This implies F(f) = e−|ω|. Taking the inverse Fourier transform, we obtain 2 f(x) = . x2 + 1
Question 123: State the shift lemma (Do not prove). Solution: Let f be an integrable function over R (in L1(R)), and let β ∈ R. Using the definitions above, the following holds:
iωβ F(f(x − β))(ω) = F(f)(ω)e , ∀ω ∈ R.
Question 124: Use the Fourier transform technique to solve ∂tu(x, t)+t∂xu(x, t)+2u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0(x).
Solution: Applying the Fourier transform to the equation gives
∂tF(u)(ω, t) + t(−iω)F(u)(ω, t) + 2F(u)(ω, t) = 0
This can also be re-written as follows: ∂ F(u)(ω, t) t = iωt − 2. F(u)(ω, t) Then applying the fundamental theorem of calculus we obtain 1 log(F(u)(ω, t)) − log(F(u)(ω, 0)) = iω t2 − 2t. 2 This implies iω 1 t2 −2t F(u)(ω, t) = F(u0)(ω)e 2 e . Then the shift lemma gives 1 F(u)(ω, t) = F(u (x − t2)(ω)e−2t. 0 2 This finally gives 1 u(x, t) = u (x − t2)e−2t. 0 2
Question 125: Prove the shift lemma: F(f(x − β))(ω) = F(f)(ω)eiωβ, ∀ω ∈ R. Solution: Let f be an integrable function over R (in L1(R)), and let β ∈ R. Using the definitions above, the following holds:
1 Z +∞ F(f(x − β))(ω) = f(x − β)eiωxdx 2π −∞ Math 602 54
Upon making the change of variable z = x − β, we obtain
1 Z +∞ 1 Z +∞ F(f(x − β))(ω) = f(z)eiω(z+β)dz = eiωβ f(z)eiωzdz 2π −∞ 2π −∞ = eiωβF(f)(ω). which proves the lemma.
Question 126: Use the Fourier transform technique to solve ∂tu(x, t) + cos(t)∂xu(x, t) + sin(t)u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0(x).
Solution: Applying the Fourier transform to the equation gives
∂tF(u)(ω, t) + cos(t)(−iω)F(u)(ω, t) + sin(t)F(u)(ω, t) = 0
This can also be re-written as follows: ∂ F(u)(ω, t) t = iω cos(t) − sin(t). F(u)(ω, t)
Then applying the fundamental theorem of calculus, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = iω sin(t) + cos(t) − 1.
This implies iω sin(t) cos(t)−1 F(u)(ω, t) = F(u0)(ω)e e . Then the shift lemma gives
cos(t)−1 F(u)(ω, t) = F(u0(x − sin(t))(ω)e .
This finally gives cos(t)−1 u(x, t) = u0(x − sin(t))e .