4.1. Fourier integral transform (Continued) Definition. For any f(x) L 1 (IR 1 ), the function. e iµx f(x)dx

Transcription

1 4.. Fourier integral transform (Continued) Definition. For any f(x) L (IR ), the function f (µ) = e iµx f(x)dx is called the inverse Fourier transform of f(x). Note: I do not know how to type set the wedge so that it has the same size as the hat. Right now the wedge looks bigger than the hat. Note: In terms of real functions, we have f (µ) = cos(µx)f(x)dx i sin(µx)f(x)dx. If f(x) is an even function: f( x) =f(x), then the inverse Fourier transform is real and equals to the Fourier transform. The inverse Fourier transform has similar properties as the Fourier transform: [f (x)] (µ) = iµf (µ) ( e βx2) = e µ2 4β ( ) 2β e x2 2 = e µ2 2 [δ(x)] (µ) = (µ) = δ(µ) ( ) = δ(µ). We then see that there holds (δ(x) ) = δ(x). A more appropriate way to write this is δ(x)ˆ = δ(x). In general we have this Fourier inversion theorem. Theorem (Fourier Inversion) For any f(x) L (IR ) such that ˆf(µ) L (IR ), there holds ˆf (x) =f(x). Proof. We shall utilize the transform ˆ(µ) = δ(µ) in the form ˆ(y x) = IR eiµ(y x) dµ = δ(y x).

2 Thus we calculate ( ˆf) (x) = IR [ IR f(y)eiµy dy]e iµx dµ then change order of integration ( ) = IR f(y) IR eiµ(y x) dµ dy use the transform of = IR f(y) δ(y x) dy = f(x) woa! You may note that we do not have to use the factor in front of the Fourier transform. In fact many books use different factors. We choose to use this factor so that the inverse transform has the same factor as the (forward) Fourier transform, the inversion theorem holds, and the following Parseval (or Plancherel ) identity holds without extra factors. Theorem. (Parseval (or Plancherel ) identity) Both Fourier and inverse Fourier transforms are bounded linear transforms that preserve the norm and distance in L 2 (IR ): ˆf L 2 = f L 2 = f L 2. Proof. (Not covered in class) We calculate ˆf 2 L = 2 IR ˆf(µ) ˆf(µ)dµ overline means complex conjugate, this is by definition IR = = IR = IR eiµx f(x) dx IR e iµy f(y) dy dµ ) IR eiµ(x y) dµ f(x)f(y) dx dy IR ( IR IR δ(y x)f(x)f(y) dx dy = IR f(x)f(x) dx = f 2 L. 2 Definition (Convolution) Let f(x) andg(x) beinl (IR ). Then the function (f g)(x) = f(x y)g(y) dy IR is called the convolution of the two functions f and g. We note that we can use a change of variables to find (f g)(x) = f(y)g(x y) dy. IR 2

3 Property b. There holds (f g)ˆ(µ) = fˆ(µ) gˆ(µ) and (f g) (µ) = f (µ) g (µ) Proof. We calculate (f g)ˆ(µ) = IR ( IR f(x y)g(y) dy)eiµx dx change order of integration = [ ] IR IR f(x y)eiµ(x y) dx g(y)e iµy dy change of variable x y = z = [ ] IR IR f(z)eiµz dz g(y)e iµy dy = fˆ(µ) gˆ(µ). We can prove the second formula similarly. Property c. There holds (fg)ˆ = fˆ gˆ and (fg) = f g. Proof. By applying the Fourier transform to the second formula of property b, we have (f g)(x) = (f (µ) g (µ))ˆ. Call F = f,g= g,thenfˆ=f,gˆ=g. Sowehave (Fˆ Gˆ)(x) = (F G)ˆ. Dividing by, and realizing that F and G are arbirary, so we can change them to f and g to find (f g)ˆ = (fˆ gˆ). The proof for the other formula is similar. Examples. Solve the heat conduction problem in an infinite slab: u t 2 u x 2 =0 x IR, t > 0 u(x, 0) = f(x). 3

4 Solution. We use the Fourier transform and notation uˆ(ω,t) = e iωx u(x, t)dx. Since t is independent of x, wehave Also we have t uˆ(ω,t) = ( iω) 2 uˆ(ω,t) = We transform the initial data: e iωx t u(x, t)dx. uˆ(ω,0) = fˆ(ω). e iωx x 2 u(x, t)dx. Now we transform the equation: We multiply the equation with the factor e iωx, integrate the equation in x over IR,wehave t (uˆ) + ω 2 uˆ=0. We can find the solution to the ordinary differential equation (see later): uˆ(ω,t) =fˆ(ω)e tω2. Let g be such a function: gˆ=e tω2. We know that Thus we have g(x) =(e tω2 ) = 2t e x2 4t. uˆ(ω,t) =fˆgˆ. Hence Thus u(x, t) =(uˆ(ω,t)) =(fˆgˆ) = f g = 2t u(x, t) = 4πt 4 f(y)e (x y)2 4t dy. f(y)e (x y)2 4t dy.

5 2. Solve the ordinary differential equation u + a 2 u = f(x) L 2 (IR ). Solution. We have Integrating in x, we find Or u e iµx + a 2 ue iµx = f(x)e iµx. ( iµ) 2 uˆ+a 2 uˆ=fˆ. (µ 2 + a 2 )uˆ=fˆ. Or simply uˆ= µ 2 + a 2 fˆ=gˆfˆ where gˆ= µ 2 + a 2. We will show later that g = e a x 2a. Then we have the solution formula: u(x) = f g = f(x y)e a y dy. 2a End of solution. We find g here. First we can use integration by parts twice to find that We solve for I: Now we verify I = 0 cos(µx)e ax dx = a µ2 a 2 I. I = a a 2 + µ 2. gˆ = e a x IR eiµx 2a dx = e a x IR cos(µx) 2a dx + i 0( odd integrand ) = a 0 cos(µx)e ax dx a I = a 2 +µ = gˆ. 2 5

6 Here we solve the ordinary differential equation t (uˆ) + ω 2 uˆ=0. For convenience, we call y(t) = uˆ. Then the equation becomes Dividing the equation by y, wehave t y + ω 2 y =0. t y y = ω2. Using the chain rule of differentiation [F (g(x))] = F (g(x))g (x), we have t ln y = ω 2. Integrating the equation in t from t =0,wehave ln y(t) ln y(0) = ω 2 t. Taking the exponential on both sides, we have y(t) =y(0)e ω2t. Recall that y(0) = fˆ(ω). So we have finally uˆ(ω,t) =fˆ(ω)e tω2. 6