sech(ax) = 1 ch(x) = 2 e x + e x (10) cos(a) cos(b) = 2 sin( 1 2 (a + b)) sin( 1 2

Transcription

1 Math Fourier transform Here are some formulae that you may want to use: F(f)(ω) def = f(x)e iωx dx, F (f)(x) = f(ω)e iωx dω, (3) F(f g) = F(f)F(g), (4) F(e α x ) = α π ω + α, F( α x + α )(ω) = e α ω, (5) F(f(x β))(ω) = e iβω F(f)(ω), (6) F(e αx ) = e ω 4α α (7) F(H(x)e ax )(ω) =, H is the Heaviside function (8) a iω F(pv( x ))(ω) = i sign(ω) = i (H(ω) H( ω)) (9) F(sech(ax)) = a sech( π a ω), def sech(ax) = ch(x) = e x + e x (0) cos(a) cos(b) = sin( (a + b)) sin( (a b)) () Question 0: (i) Let f be an integrable function on (, + ). Prove that for all a, b R, and for all ξ R, F([e ibx f(ax)])(ξ) = ξ+b af(f)( a ). The definition of the Fourier transform together with the change of variable ax x implies F[e ibx f(ax)])(ξ) = = = = ξ+b af(f)( a ). f(ax)e ibx e iξx dx f(ax)e i(b+ξ)x dx a f(x )e i (ξ+b) a x dx (ii) Let c be a positive real number. Compute the Fourier transform of f(x) = e cx sin(bx). Using the fact that sin(bx) = i (eibx e ibx ), and setting a = c we infer that and using (i) and (7) we deduce In conclusion f(x) = i e (ax) (e ibx e ibx ) ˆf(ξ) = i = i e (ax) e ibx i e (ax) e i( b)x a (e 4 ( ξ+b a ) e 4 ( ξ b a ) ). ˆf(ξ) = i c (e 4c (ξ+b) e 4c (ξ b) ). Question 03: (a) Compute the Fourier transform of the function f(x) defined by { if x f(x) = 0 otherwise

2 Math By definition Hence F(f)(ω) = e iξω = = iω (eiω e iω ) F(f)(ω) = π = sin(ω) ω. sin(ω). ω (b) Find the inverse Fourier transform of g(ω) = sin(ω) ω. Using (a) we deduce that g(ω) = πf(f)(ω), that is to say, F (g)(x) = πf (F(f))(x). Now, using the inverse Fourier transform, we deduce that F (g)(x) = πf(x) at every point x where f(x) is of class C and F (g)(x) = π (f(x ) + f(x + )) at discontinuity points of f. As a result: π if x < F π (g)(x) = at x = 0 otherwise Question 04: Find the inverse Fourier transform of π sin(λω) ω sin(λω ω. Since F(S λ (x)) = π, the inverse Forier transform theorem implies that ( ) F sin(λω) if x < λ (x) = π ω if x = λ 0 otherwise. Question 05: Use the Fourier transform technique to solve the following PDE: t u(x, t) + c x u(x, t) + γu(x, t) = 0, for all x (, + ), t > 0, with u(x, 0) = u 0 (x) for all x (, + ). The solution is By taking the Fourier transform of the PDE, one obtains t F(u) iωcf(y) + γf(y) = 0. F(u)(ω, t) = c(ω)e iωct γt. The initial condition implies that c(ω) = F(u 0 )(ω): The shift lemma in turn implies that F(u)(ω, t) = F(u 0 )(ω)e iωct e γt. F(u)(ω, t) = F(u 0 (x ct))(ω)e γt = F(u 0 (x ct)e γt )(ω). Applying the inverse Fourier transform gives: u(x, t) = u 0 (x ct)e γt. Question 06: Solve by the Fourier transform technique the following equation: xx φ(x) x φ(x) + φ(x) = H(x)e x, x (, + ), where H(x) is the Heaviside function. (Hint: use the factorization iω 3 + ω + iω + = ( + ω )( + iω) and recall that F(f(x))( ω) = F(f( x))(ω)).

3 Math Applying the Fourier transform with respect to x gives ( ω + iω + )F(φ)(ω) = F(H(x)e x )(ω) = iω. where we used (8). Then, using the hint gives F(φ)(ω) = ( iω)( ω + iω + ) = = + ω + iω. We now use again (8) and (5) to obtain F(φ)(ω) = π π iω 3 + ω + iω + + ω i( ω) = πf(e x )(ω)f(h(x)e x )( ω). Now we use F(H(x)e x )( ω) = F(H( x)e x )(ω) and we finally have The Convolution Theorem (4) gives F(φ)(ω) = πf(e x )(ω)f(h( x)e x )(ω). F(φ)(ω) = π F(e x (H( x)e x ))(ω). We obtain φ by using the inverse Fourier transform i.e., φ(x) = e x (H( x)e x ), φ(x) = e x y H( y)e y dy and recalling that H is the Heaviside function we finally have φ(x) = { 0 e y x y dy = 4 e x if x 0 ( 4 x)ex if x 0. Question 07: Use the Fourier transform technique to solve the following ODE y (x) y(x) = f(x) for x (, + ), with y(± ) = 0, where f is a function such that f is integrable over R. By taking the Fourier transform of the ODE, one obtains That is ω F(y) F(y) = F(f). F(y) = F(f) + ω. and the convolution Theorem, see (4), together with (5) gives Applying F on both sides we obtain F(y) = πf(f)f(e x ) = F(f e x ). y(x) = f e x = e x z f(z)dz

4 Math That is y(x) = e x z f(z)dz. Question 08: Use the Fourier transform method to compute the solution of u tt a u xx = 0, where x R and t (0, + ), with u(0, x) = f(x) := sin (x) and u t (0, x) = 0 for all x R. Take the Fourier transform in the x direction: F(u) tt + ω a F(u) = 0. This is an ODE. The solution is The initial boundary conditions give and c (ω) = 0. Hence Using the shift lemma, we infer that F(u)(t, ω) = c (ω) cos(ωat) + c (ω) sin(ωat). F(u)(0, ξ) = F(f)(ω) = c (ω) F(t, ω) = F(f)(ω) cos(ωat) = F(f)(ω)(eiaωt + e iaωt ). F(t, ω) = (F(f(x at))(ω) + F(f(x + at))(ω)). Usin the inverse Fourier transform, we finaly conclude that u(t, x) = (f(x at) + f(x + at)) = (sin (x + at) + sin (x at)). Note that this is the D Alembert formula. Question 09: Use the Fourier transform method to compute the solution of u tt a u xx = 0, where x R and t (0, + ), with u(0, x) = f(x) := cos (x) and u t (0, x) = 0 for all x R. Take the Fourier transform in the x direction: F(u) tt + ω a F(u) = 0. This is an ODE. The solution is The initial boundary conditions give and c (ω) = 0. Hence F(u)(t, ω) = c (ω) cos(ωat) + c (ω) sin(ωat). F(u)(0, ξ) = F(f)(ω) = c (ω) F(t, ω) = F(f)(ω) cos(ωat) = F(f)(ω)(eiaωt + e iaωt ). Using the shift lemma (i.e., formula (6)) we obtain u(t, x) = (f(x at) + f(x + at)) = (cos (x + at) + cos (x at)). Note that this is the D Alembert formula. Question 0: Solve the integral equation: f(x) + x (, + ). f(y) (x y) + dy = x +4 + x +, for all

5 Math The equation can be re-written f(x) + f x + = x x +. We take the Fourier transform of the equation and we apply the Convolution Theorem (see (4)) F(f) + F( x + )F(f) = F( x + 4 ) + F( x + ). Using (5), we obtain which gives We then deduce F(f) + e ω F(f) = 4 e ω + e ω, F(f)( + e ω ) = e ω ( e ω + ). F(f) = e ω. Taking the inverse Fourier transform, we finally obtain f(x) = x +. Question : Solve the following integral equation (Hint: solution is short): f(y)f(x y)dy e y f(x y)dy = e x. x R. This equation can be re-written using the convolution operator: f f e x f = e x. We take the Fourier transform and use the convolution theorem (4) together with (7) to obtain F(f) F(f) e ω 4 = F(f) F(f)e ω π + e ω = 0 Taking the inverse Fourier transform, we obtain (F(f) e ω π ) = 0. F(f) = e ω π. f(x) = e x. e ω 4 Question : Solve the integral equation: f(x) + 3 (, + ). The equation can be re-written f(x) + 3 e x f = e x. e x y f(y)dy = e x, for all x We take the Fourier transform of the equation and apply the Convolution Theorem (see (4)) F(f) + 3 F(e x )F(f) = F(e x ). Using (5), we obtain F(f) + 3π π + ω F(f) = π + ω,

6 Math which gives We then deduce F(f) ω ω = π + ω. F(f) = π 4 + ω = F(e x ). Taking the inverse Fourier transform, we finally obtain f(x) = e x. Question 3: Solve the following integral equation e (x y) g(y)dy = e x (, + ), i.e., find the function g that solves the above equation. The left-hand side of the equation is a convolution; hence, for all x By taking the Fourier transform, we obtain e x g(x) = e x. e 4 ω Fg(ω) = e ω. That gives F(g)(ω) = e 4 ω. By taking the inverse Fourier transform, we deduce g(x) = e x = π e x. Question 4: Solve the integral equation: f(y)f(x y)dy = 4 How many solutions did you find? The equation can be re-written f f = 4 x + 4. x +4, for all x (, + ). We take the Fourier transform of the equation and apply the Convolution Theorem (see (4)) Using (5), we obtain which gives F(f) 4 = F( x + 4 ) F(f) = e ω. F(f) = ± e ω. Taking the inverse Fourier transform, we finally obtain f(x) = ± x +. We found two solutions: a positive one and a negative one. Question 5: Use the Fourier transform method to solve the equation t u + t +t x u = 0, u(x, 0) = u 0 (x), in the domain x (, + ) and t > 0. We take the Fourier transform of the equation with respect to x t 0 = t F(u) + F( + t xu) = t F(u) + t + t F( xu) = t F(u) iω t + t F(u).

7 Math This is a first-order linear ODE: The solution is t F(u) F(u) Using the initial condition, we obtain The shift lemma (see formula (6)) implies = iω t + t = iω d dt (log( + t )) F(u)(ω, t) = K(ω)e iω log(+t). F(u 0 )(ω) = F(u)(ω, 0) = K(ω). F(u)(ω, t) = F(u 0 )(ω)e iω log(+t) = F(u 0 (x log( + t ))), Applying the inverse Fourier transform finally gives u(x, t) = u 0 (x log( + t )). Question 6: Solve the integral equation: ( f(y) ) e y + (x y) + y f(x y)dy = + y e dy, x (, + ). (Hint: there is an easy factorization after applying the Fourier transform.) The equation can be re-written f (f e x + x ) = + x e x We take the Fourier transform of the equation and apply the Convolution Theorem (see (4)) ( F(f) F(f) ) F(e x ) F( + x ) = F( + x ) F(e x ) Solution : Using (5), (7) we obtain. which gives F(e x ) = F( + x ) = e ω, e ω 4 = e πω ( F(f) F(f) e πω ) e ω = e ω e πω. This equation can also be re-written as follows and can be factorized as follows: F(f) F(f)e πω F(f) e ω + e ω e πω = 0, (F(f) e πω )(F(f) e ω ) = 0. This means that either F(f) = e πω or F(f) = e ω. Taking the inverse Fourier transform, we finally obtain two solutions f(x) = e x, or f(x) = + x.

8 Math Solution : Another solution consists of remarking that the equation with the Fourier transform can be rewritten as follows: F(f) F( e x )F(f) F( + x )F(f) + F( e x )F( + x ) = 0, which can factorized as follows: (F(f) F( e x ))(F(f) F( )) = 0. + x The either F(f) = F( e x )) or (F(f) = F( +x ). The conclusion follows easily. Question 7: Use the Fourier transform technique to solve t u(x, t) + sin(t) x u(x, t) + ( + 3t )u(x, t) = 0, x R, t > 0, with u(x, 0) = u 0 (x). (Use the shift lemma: F(f(x β))(ω) = F(f)(ω)e iωβ and the definition F(f)(ω) := + f(x)eiωx dx) Applying the Fourier transform to the equation gives t F(u)(ω, t) + sin(t)( iω)f(u)(ω, t) + ( + 3t )F(u)(ω, t) = 0 This can also be re-written as follows: t F(u)(ω, t) F(u)(ω, t) = iω sin(t) ( + 3t ). Then applying the fundamental theorem of calculus between 0 and t, we obtain Then the shift lemma gives log(f(u)(ω, t)) log(f(u)(ω, 0)) = iω(cos(t) ) (t + t 3 ). F(u)(ω, t) = F(u 0 )(ω)e iω(cos(t) ) e (t+t3). F(u)(ω, t) = F(u 0 (x + cos(t) )(ω)e (t+t3). This finally gives u(x, t) = u 0 (x + cos(t) )e (t+t3). Question 8: Solve the following integral equation (Hint: x 3xa + a = (x a)(x a)): (f(y) 3 e y )f(x y)dy = e x. x R. This equation can be re-written using the convolution operator: f f 3 e x f = e x. We take the Fourier transform and use (7) to obtain F(f) 3 F(f) e ω 4 = F(f) 3F(f)e ω π + e ω = 0 (F(f) e ω π )(F(f) e ω π ) = 0. e ω 4 either F(f) = e ω π, or F(f) = e ω π.

9 Math 60 5 Taking the inverse Fourier transform, we obtain either f(x) = e x, or f(x) = e x. Question 9: Use the Fourier transform technique to solve t u(x, t) xx u(x, t)+sin(t) x u(x, t)+ ( + 3t )u(x, t) = 0, x R, t > 0, with u(x, 0) = u 0 (x). (Hint: use the definition F(f)(ω) := + f(x)eiωx dx), the result F(e αx )(ω) = α e ω 4α, the convolution theorem and the shift lemma: F(f(x β))(ω) = F(f)(ω)e iωβ. Go slowly and give all the details.) Applying the Fourier transform to the equation gives t F(u)(ω, t) + ω F(u)(ω, t) + sin(t)( iω)f(u)(ω, t) + ( + 3t )F(u)(ω, t) = 0 This can also be re-written as follows: t F(u)(ω, t) F(u)(ω, t) = ω + iω sin(t) ( + 3t ). Then applying the fundamental theorem of calculus between 0 and t, we obtain log(f(u)(ω, t)) log(f(u)(ω, 0)) = ω t iω(cos(t) ) (t + t 3 ). F(u)(ω, t) = F(u 0 )(ω)e ωt e iω(cos(t) ) e (t+t3). Using the result F(e αx )(ω) = α e ω 4α where α = 4t, this implies that π F(u)(ω, t) = t F(u 0)(ω)F(e x 4t )(ω)e iω(cos(t) ) e (t+t3 ) π = t F(u 0 e x 4t )(ω)e iω(cos(t) ) e (t+t3). Then setting g = u 0 e x 4t the convolution theorem followed by the shift lemma gives F(u)(ω, t) = π F(g(x + cos(t) 3) ))(ω)e (t+t. t This finally gives u(x, t) = g(x + cos(t) 3) )e (t+t = e ) (t+t3 t t u 0 (y)e (x+cos(t) y) 4t dy. Question 0: Consider the telegraph equation tt u + α t u + α u c xx u = 0 with α 0, u(x, 0) = 0, t u(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 φ (t) + αφ (t) + (α + λ )φ(t) = 0 is φ(t) = e αt (a cos(λt) + b sin(λt)) Applying the Fourier transform with respect to x to the equation, we infer that 0 = tt F(u)(ω, t) + α t F(u)(ω, t) + α F(u)(ω, t) c ( iω) F(u)(ω, t) = tt F(u)(ω, t) + α t F(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 F(u)(ω, t) = e αt F(g) sin(ωct). ωc

10 Math 60 5 Then using (??), we have F(u)(ω, t) = π c e αt F(g)F(S ct ). The convolution theorem implies that u(x, t) = e αt c g S ct = e αt c g(y)s ct (x y)dy. Finally the definition of S ct implies that S ct (x y) is equal to if ct < x y < ct and is equal zero otherwise, which finally means that u(x, t) = e αt c x+ct x ct g(y)dy. Question : Use the Fourier transform technique to solve t u(x, t) xx u(x, t)+cos(t) x u(x, t)+ ( + t)u(x, t) = 0, x R, t > 0, with u(x, 0) = u 0 (x). (Hint: use the definition F(f)(ω) := + f(x)eiωx dx), the result F(e αx )(ω) = α e ω 4α, the convolution theorem and the shift lemma: F(f(x β))(ω) = F(f)(ω)e iωβ. Go slowly and give all the details.) Applying the Fourier transform to the equation gives t F(u)(ω, t) + ω F(u)(ω, t) + cos(t)( iω)f(u)(ω, t) + ( + t)f(u)(ω, t) = 0 This can also be re-written as follows: t F(u)(ω, t) F(u)(ω, t) = ω + iω cos(t) ( + t). Then applying the fundamental theorem of calculus between 0 and t, we obtain log(f(u)(ω, t)) log(f(u)(ω, 0)) = ω t + iω sin(t) (t + t ). F(u)(ω, t) = F(u 0 )(ω)e ωt e iω sin(t) e (t+t). Using the result F(e αx )(ω) = α e ω 4α where α = 4t, this implies that π F(u)(ω, t) = t F(u 0)(ω)F(e x 4t )(ω)e iω sin(t) e (t+t ) π = t F(u 0 e x 4t )(ω)e iω sin(t) e (t+t). Then setting g = u 0 e x 4t the convolution theorem followed by the shift lemma gives F(u)(ω, t) = π F(g(x ) sin(t)))(ω)e (t+t. t This finally gives u(x, t) = g(x ) sin(t))e (t+t = e ) (t+t t t u 0 (y)e (x sin(t) y) 4t dy. Question : Solve the following integral equation (Hint: (a + b) = a + ab + b ): f(y)f(x y)dy y + f(x y)dy + 4 x = 0 x R. + 4

11 Math This equation can be re-written using the convolution operator: f f ( x + ) f + 4 x + 4 = 0. We take the Fourier transform and use the convolution theorem (??) together with (??) to obtain Taking the inverse Fourier transform, we obtain F(f) F(f)e ω + e ω = 0 F(f) F(f)e ω + e ω = 0 (F(f) e ω ) = 0 F(f) = e ω. f(x) = x +. Question 3: State the shift lemma (Do not prove). Let f be an integrable function over R (in L (R)), and let β R. Using the definitions above, the following holds: F(f(x β))(ω) = F(f)(ω)e iωβ, ω R. Question 4: Use the Fourier transform technique to solve t u(x, t)+t x u(x, t)+u(x, t) = 0, x R, t > 0, with u(x, 0) = u 0 (x). Applying the Fourier transform to the equation gives This can also be re-written as follows: t F(u)(ω, t) + t( iω)f(u)(ω, t) + F(u)(ω, t) = 0 t F(u)(ω, t) F(u)(ω, t) = iωt. Then applying the fundamental theorem of calculus we obtain log(f(u)(ω, t)) log(f(u)(ω, 0)) = iω t t. Then the shift lemma gives F(u)(ω, t) = F(u 0 )(ω)e iω t e t. F(u)(ω, t) = F(u 0 (x t )(ω)e t. This finally gives u(x, t) = u 0 (x t )e t. Question 5: Prove the shift lemma: F(f(x β))(ω) = F(f)(ω)e iωβ, ω R. Let f be an integrable function over R (in L (R)), and let β R. Using the definitions above, the following holds: F(f(x β))(ω) = f(x β)e iωx dx

12 Math Upon making the change of variable z = x β, we obtain F(f(x β))(ω) = which proves the lemma. = e iωβ F(f)(ω). f(z)e iω(z+β) dz = e iωβ f(z)e iωz dz Question 6: Use the Fourier transform technique to solve t u(x, t) + cos(t) x u(x, t) + sin(t)u(x, t) = 0, x R, t > 0, with u(x, 0) = u 0 (x). Applying the Fourier transform to the equation gives t F(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: t F(u)(ω, t) F(u)(ω, t) = iω cos(t) sin(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). F(u)(ω, t) = F(u 0 )(ω)e iω sin(t) e cos(t). Then the shift lemma gives F(u)(ω, t) = F(u 0 (x sin(t))(ω)e cos(t). This finally gives u(x, t) = u 0 (x sin(t))e cos(t).