Functions of a Complex Variable (S1) Lecture 11. VII. Integral Transforms. Integral transforms from application of complex calculus

Transcription

1 Functions of a Complex Variable (S1) Lecture 11 VII. Integral Transforms An introduction to Fourier and Laplace transformations Integral transforms from application of complex calculus Properties of Fourier and Laplace transforms Applications to differential equations

2 Fourier transformation f(ω) = e iωt f(t) dt FT[f] f(t) = 1 2π e iωt f(ω) dω FT 1 [ f] We see next how to obtain this transformation starting from a complex integral representation of step functions.

3 The step function Θ(t) Consider 1 e iωt 1 ω iε dω (ε R+ ) Evaluate this integral by complex contour integration: for t > Jordan lemma applies to semicircular arc in UHP = integral = Res iε = 1; for t < Jordan lemma applies to semicircular arc in LHP = integral =. Γ Im i ε ω Im i ε ω Re ω Re ω Γ t > t < Thus 1 e iωt 1 ω iε dω = Θ(t) where Θ(t) is the step function: Θ(t) = 1 for t >, Θ(t) = for t <.

4 step function centred at t = τ: Θ(t τ) = 1 e iω(t τ) 1 ω iε dω 1 finite pulse : combination of Θ functions τ t Θ(t τ 1 ) Θ(t τ 2 ) = 1 e iωt e iωτ 1 e iωτ 2 ω iε dω 1 τ 1 τ 2 t

5 Construction of Fourier transform First construct approximation to f(t) by sum of step functions f(t) n 1 k= f(τ k )[Θ(t τ k ) Θ(t τ k+1 )] = 1 n 1 k= f(τ k ) e iωt e iωτ k e iωτ k+1 ω iε dω f(t) τ τ k τ k+1 τ n t Then let each τ k+1 τ k and expand in τ k = i(τ k+1 τ k ) = idτ: = f(t) = 1 2π e iωt f(τ) e iωτ dτ } {{ } f(ω) dω

6 Thus we obtain representation of f as f(t) = 1 2π e iωt f(ω) dω FT 1 [f] where f(ω) is the Fourier transform defined by f(ω) = f(t) e iωt dt FT [f] We will apply this to functions f in L 2 L 1, i.e., such that f 2 <, f <

7 PROPERTIES OF FOURIER TRANSFORMATION linearity : FT[αf +βg] = α FT[f]+β FT[g] scaling : FT[f(λt)] = 1 λ f( ω λ ) shift : FT[f(t a)] = e iωa f FT of derivative : FT[f (n) ] = (iω) n f derivative of FT : i n f(n) = FT[t n f] convolution theorem : where (f g)(t) df = FT[f g] = f g dy f(y) g(t y) FT of product : FT[fg] = 1 2π f g

8 PARSEVAL IDENTITY Take convolution theorem for the particular case g(t) = f( t). = g(ω) = e iωt f( t) dt = e iωt f(t) dt = f(ω) Then f g = FT 1 [ f g] gives dy f(y) f(y t) = 1 2π dω e iωt f(ω) f(ω) t = dy f(y) 2 = 1 2π dω f(ω) 2

9 Example: Compute the FT of f(t) = 1/(1+t 2 ). f(ω) = e iωt 1 1+t 2 dt This integral can be computed by complex contour integration methods, closing the contour as in the figure below depending on the sign of ω. Im z Im z Re z Re z ω> ω< ω > : f(ω) = Res i = πe ω ω < : f(ω) = Resi = πe ω Thus f(ω) = πe ω for any ω.

10 Application of FT to boundary value problems Example : 2 u x u y 2 =, y > u(x, ) = G(x) (Dirichlet boundary condition) Taking FT in x at fixed y gives ODE for ũ(ω,y): ω 2 ũ+ 2 ũ y 2 =, ũ(ω,) = G(ω) = ũ(ω,y) = G(ω)e ω y Now use convolution theorem : u(x,y) = G FT 1 [e ω y ] }{{} [πy(1+x 2 /y 2 )] 1 Evaluating the convolution product gives u(x,y) = 1 π dζ y (ζ x) 2 G(ζ) Poisson integral formula +y2

11 Laplace transformation f(z) = e zt f(t) dt LT[f] f(t) = 1 a+i a i e zt f(z) dz LT 1 [ f] Im z contour in complex z plane for inverse Laplace transform a Re z

12 f exponentially bounded, i.e. f(t) < Me µt, M,µ real positive constants f defined for t (set f = for t < ) Consider g(t) = f(t)e µt. Define LT [f] from FT [g] : FT [g] = = g(t) e iωt dt = f(t) e (µ+iω)t dt f(t) e zt dt LT [f] = f(z) (z = µ+iω) Construct inverse Laplace transform from FT 1 [ g] : i.e., f(t) = 1 2π g(t) = 1 2π dω e iωt FT [g] dω e (µ+iω)t f(z) = 1 µ+i µ i dz e zt f(z)

13 PROPERTIES OF LAPLACE TRANSFORMATION linearity : LT[αf +βg] = α LT[f]+β LT[g] scaling : LT[f(λt)] = 1 λ f( z λ ) derivative of LT : ( 1) n f(n) = LT[t n f] LT of derivative : LT[f ] = z f f() LT[f (n) ] = z n f z n 1 f()... f (n 1) () convolution theorem : where (f g)(t) df = t LT[f g] = f g dy f(y) g(t y)

14 EXAMPLE: Laplace transforms of powers LT[f] = f(z) = For integerpowers f(t) = t n, n N dt e zt t n = n! z n+1 f(1) = n! For non integerpowers f(t) = t α, α C = generalization of the factorial to complexvariable : Euler gamma function defined for Re α > as Γ(α) = dt e t t α 1 For α = n, Γ(n) = (n 1)! can be continued to any α in C;poles of order 1 at α =, 1, 2,... Γ(α+1) = αγ(α) (integrate by parts) Γ(1/2) = π (change variable t ρ = t) Thus LT[t α ](z) = Γ(α+1) z α+1

15 Example: Compute the LT 1 of F(z) = 1/ z. f(t) = 1 a+i a i e zt 1 z dz (a > ) This integral can be computed by complex integration methods, closing the contour as in the figure. Γ Im z Γ Then write branch cut along R e zt 1 z dz = by Cauchy theorem Γ = a+i a i + C R + L 1 + Let R, r. Apply Jordan lemma C r + L 2 C R C R, L 1 L 2 Cr a C r. Re z

16 Thus f(t) = 1 = 1 a+i a i dx e zt 1 z dz [ e xt x e iπ/2 e xt ] x e iπ/2 }{{} discontinuity across the branch cut Change integration variable x x. Then f(t) = 1 [ ] e xt dx e xt x ( i) x i = 1 π dx e xt x = 1 π dρ 2 t e ρ2 = 1 πt

17 Solve theequation for y(t) y(t) = h(t) t Linear integral equations K(t τ) y(τ) dτ where h and K are assigned functions Apply LT to the equation and use convolution theorem : ỹ(z) = h(z) K(z)ỹ(z) ỹ(z) = h(z)/(1+ K(z)) Then the solution is given by Laplace inverse transform : y(t) = LT 1 [ỹ] = LT 1 [ h(z)/(1+ K(z))] EXAMPLE Determine the function y(t) obeying the equation y(t) = t+ t sin(t τ) y(τ) dτ h(t) = t h(z) = 1/z 2 ; K(t) = sint K(z) = 1/(1+z 2 ) Then ỹ(z) = h(z)/(1+ K(z)) = (1+z 2 )/z 4 y(t) = LT 1 [ỹ] = LT 1 [1/z 4 +1/z 2 ] = t 3 /3!+t = t 3 /6+t