f(x)e ikx dx. (1) f(k)e ikx dk. (2) e ikx dx = 2sinak. k f g(x) = g f(x) = f(s)e iks ds

Transcription

1 Mathematical Methods: Page 15 2 Fourier transforms 2.1 Integral transforms The Fourier transform is studied in this chapter and the Laplace transform in the next. They are both integral transforms that may used to find solutions to differential, integral and difference equations and may be used to evaluate definite integral and to sum series. 2.2 Definition of Fourier transform The Fourier transform of f(x) is given by The inverse transform is given by f() = 2π Example: f(x) = Ae a x for a,a > f = Ae ax ix dx+ f(x)e ix dx. (1) f()e ix d. (2) Ae ax ix dx = 2Aa a Example: when x < a and f(x) = otherwise f = a 2.3 Properties of Fourier transforms a e ix dx = 2sina. 1. Scaling: if g(x) = f(αx) (α > ), then g() = 1 α f ( ). α 2. Translation 1: if g(x) = e iµx f(x) (µ ) then g() = f( +µ). 3. Translation 2: if g(x) = f(x µ) (µ ) then g() = e iµ f(). 4. Fourier transform of x n f(x) is (i) ndn f d n. 5. Derivatives: Fourier transform of dn f dx n is (i)n f(). 6. Convolution: A convolution of two functions is given by f g(x) = g f(x) = The Fourier transform of the convolution is given by f g = f(x ξ)g(ξ)e ix dξdx = f(x ξ)g(ξ) dξ. f(s)e is ds g(ξ)e iξ dξ = f() g(). c University of Bristol 217. This material is the copyright of the University unless explicitly stated otherwise. It is provided

2 Mathematical Methods: Page 16 Choosing g( ξ) = f(ξ) (here denotes complex conjugation), so that g() = f(), we deduce Parseval s formula f(ξ) 2 dξ = 1 f() 2 d. 2π 2.4 Fourier Cosine and Sine transforms In many problem we are only concerned with f(x) in x > and we are free to extend the definition of f(x) into x < as we please. Fourier Cosine and Sine transforms correspond to constructing the function of interest to be even or odd, respectively. Cosine transform: F C = Sine transform: F S = Example: f(x) = e bx (b > ). f(x)cosx dx and f(x) = 2 π f(x)sinx dx and f(x) = 2 π Using integration by parts we find F C () = 1 b b F S() Then F C = Derivatives b 2 +b 2 and F S = 2 +b 2. F S () = b F C() F C ()cosx d. (3) F S ()sinx d. (4) The transforms of derivatives are straightforwardly evaluated using integration by parts: ( ) df F S = F C (f), dx ( ) df F C = f()+f S (f). dx These relationships may be used recursively for higher derivatives. We will use Fourier Cosine and Sine transforms for solving pde problem in semi-infinite domains. 2.5 Inverting Fourier transforms using contour integration Example: f = ( 2 +1) 1 The inverse transform formula gives 2π eix d, but this can not be evaluated using elementary techniques. Instead we treat integral in complex -plane, where the integrand has simple poles at = ±i. c University of Bristol 217. This material is the copyright of the University unless explicitly stated otherwise. It is provided

3 Mathematical Methods: Page 17 C 1 =i =i = i = i C 2 Figure 1: Integration paths in the complex -plane with curves (i) C 1 and (ii) C 2 to form a closed contour. ( i)e ix The residue at = i is lim i 2 +1 = e x ( +i)e ix and the residue at = i is lim i 2 +1 = ex 2i. 2i We construct a closed contour in the -plane by integrating along the real axis and then closing by a semi-circular arc in either > (curve C 1 ) or < (curve C 2 ). These two curves are parameterised by writing = e iθ where < θ < π for C 1 and π < θ < for C 2. We note that on either of these curves e ix = e xsinθ. If x > then e xsinθ as provided sinθ >. This corresponds to curve C 1. Conversely If x < then e xsinθ as provided sinθ <. This corresponds to curve C 2. First we tacle the case x >. Closing the contour using C 1, we have ( ) lim e ix π 2 +1 d + e ixeiθ 2 e 2iθ +1 ieiθ dθ = 2πi e x 2i, where the right hand side is 2πi multiplied by the sum of residues enclosed by the contour; in this case the only residue comes from the pole at = i. Then evaluating the inverse transform, we find that 2 e x when x >. Next we tacle x <. Closing the contour using C 2 we have ( e ix ) π lim 2 +1 d + e ixeiθ 2 e 2iθ +1 ieiθ dθ = 2πi ex 2i, where the right hand side is 2πi multiplied by the sum of residues enclosed by the contour (poles encircled clocwise); in this case the only residue comes from the pole at = i. Then evaluating the inverse transform, we find that 2 ex when x < Example: f() = e a 2 (a > ) Applying the inverse Fourier transform formula, we have 2π e a2 +ix d. c University of Bristol 217. This material is the copyright of the University unless explicitly stated otherwise. It is provided

4 Mathematical Methods: Page 18 C2 =ix/(2a) C1 C C3 Figure 2: Deformation of the integration contour in the complex -plane We complete the square to write a 2 ix = a(( ix/(2a)) 2 +x 2 /(2a) 2 ), so that 2π e ax2 /(4a) e a( ix/(2a))2 d. Since the integrand has no singularities in the complex -plane, we may deform the contour without altering its value. In particular we find that e a( ix/(2a))2 d = e a2 d as, since the contribution from curves C 1 and C 3 vanish (see figure 2). So we find that 2π e x2 /4a e a2 d = 1 4πa e x2 /4a. (5) 2.6 Using Fourier transforms to solve partial differential equations Example: One dimensional conduction along an infinite rod. The temperature θ(x, t) satisfies θ t = κ 2 θ x 2 (t > ), subject to θ(x,) = g(x). To proceed, we tae the Fourier transform with respect to x, so that the governing equation becomes θ t = κ2 θ, subject to θ(,) = g(). This may be integrated straightaway to give θ = ge κ2t. We may invert this by recognising the expression as the product of two transforms (see (5)). Therefore the inverse transform gives a convolution, θ(x,t) = 1 4πκt e (x ξ)2 /(4κt) g(ξ) dξ. c University of Bristol 217. This material is the copyright of the University unless explicitly stated otherwise. It is provided

5 Mathematical Methods: Page Example: One dimensional conduction along a semi-infinite rod. The temperature θ(x, t) satisfies θ t = θ κ 2 (x >,t > ), x 2 subject to θ(x,) = (initial uniform temperature) and θ(,t) = θ (end of rod held at fixed temperature). WeproceedbytaingtheFourierSinetransformwithrespecttox, Θ S (,t) = so that the governing equation becomes Θ S t This may be integrated straightaway to give Inverting the transform gives = κ 2 Θ S +κθ(,t), subject to Θ S (,) =. Θ S (,t) = θ θ(x,t) = 2θ π ( = θ ( ) 1 e κ2 t. (1 e κ2t ) sinx d, ) 1 2 π x/(2 where erfc(z) denotes the complementary error function. κt) e η2 dη, θ(x,t)sinx dx, ( ) x = θ erfc 2, (6) κt t t=.1 t=.1 t= x Figure 3: The temperature θ(x,t) as a function of x at t =.1,.1,1 c University of Bristol 217. This material is the copyright of the University unless explicitly stated otherwise. It is provided

6 Mathematical Methods: Page Example: Two-dimensional fluid flow Potential flow in a region {(x,y) : x >,y > } driven by an inflow along y =. The fluid velocity field u = ( φ/ x, φ/ y) satisfies 2 φ x + 2 φ 2 y =, 2 subject to φ/ x = on x = (impermeable wall) and φ/ y = g(x) on y =. Furthermore we require decay of the velocity field as x,y. We tae the Fourier Cosine transform with respect to x, Φ C (,y) = φcosx dx. Under this transform, Laplace s equation become y Source Fluid flow x 2 Φ C y 2 2 Φ C =, and the solution satisfying the transformed boundary conditions is Φ C = G Ce y, where G C is the Fourier Cosine transform of g(x). Inverting gives φ(x,y) = 2 π G C e y cosx d. 2.7 Multiple Fourier transforms For f(x,y) defined on < x,y <, we may tae Fourier transforms with respect to x and y. Then F(,l) = f(x,y)e i(x+ly) dxdy, (7) while inversion gives f(x,y) = 1 4π 2 F(,l)e i(x+ly) ddl (8) Similar multiple transforms can be performed with Fourier Cosine and Sine transforms and/or combination of them. DefineΦ(,l) = Example: eturn to equation becomes φ(x,y)cosxcosly dxdy,sothatlaplace s ( 2 +l 2 )Φ G C () =. Inverting gives φ(x,y) = 4 G C () cosxcosly ddl. π 2 2 +l2 cosly But 2 +l 2dl = π 2 e y and so we recover the solution of c University of Bristol 217. This material is the copyright of the University unless explicitly stated otherwise. It is provided