Chapter 14: Fourier Transforms and Boundary Value Problems in an Unbounded Region

Transcription

1 Chapter 14: Fourier Transforms and Boundary Value Problems in an Unbounded Region 王奕翔 Department of Electrical Engineering National Taiwan University December 25, / 27 王奕翔 DE Lecture 15

2 So far we have seen how to solve boundary-value problems within a bounded region, where the boundary conditions are given at finite boundaries, e.g., (one-dimensional) heat equation, wave equation: x [, L] (two-dimensional) Laplace s equation: (x, y) [, a] [, b] Exception: a BVP on semi-finite plate u = y r 2 u = u = Note: we are able to solve this via Fourier series because the homogeneous boundary conditions are still at finite boundaries x =, a. From the homogeneous boundary conditions, we can conclude that the solution u(x, t) is a Fourier cosine/sine series in x. u = f(x) x 2 / 27 王奕翔 DE Lecture 15

3 What if the homogeneous boundary conditions are given at infinite boundaries, i.e., ±? y u(x, 1) = y u(x, 1) = u = r 2 u = u = F (y) u = r 2 u = u = F (y) x u = x u(x, 1) = Separation of variables: unable to use the homogeneous boundary conditions to find the possible values of the separation constant λ. 3 / 27 王奕翔 DE Lecture 15

4 We introduce Fourier Transforms to deal with this issue. 4 / 27 王奕翔 DE Lecture 15

5 1 Fourier Transforms 2 5 / 27 王奕翔 DE Lecture 15

6 From Fourier Series to Fourier Integral Recall: For a function f(x) defined on ( p, p), its Fourier series is n= ( 1 p ) f(x)e i nπ p x dx e i nπ p x 2p p Let α n := nπ p and α := α n+1 α n = π p. Taking p, α, and the Fourier series becomes 1 2π lim p = 1 2π = 1 2π n= { F p (α n ) {( }}{ p ) f(x)e iαnx dx e iαnx α p } lim Fp(α) p { e iαx dα f(x)e iαx dx α = π p when p } e iαx dα 6 / 27 王奕翔 DE Lecture 15

7 Fourier Integral and Fourier Transform Definition (Fourier Integral and Fourier Transform) The Fourier integral of a function f(x) defined on (, ) is f(x) = 1 F(α)e iαx dα, where F(α) = f(x)e iαx dx. 2π The Fourier transform of f(x) is F {f(x)} = f(x)e iαx dx := F(α). The inverse Fourier transform of a function F(α) is F 1 {F(α)} = 1 F(α)e iαx dα := f(x). 2π 7 / 27 王奕翔 DE Lecture 15

8 Fourier and Inverse Fourier Transforms f(x) F F(α) = f(x)e iαx dx F(α) F 1 f(x) = 1 F(α)e iαx dα 2π Fourier Coefficients and Fourier Series f(x) FS c n = 1 2p p p f(x)e i nπ p x dx c n FS 1 f(x) = n= c n e i nπ p x 8 / 27 王奕翔 DE Lecture 15

9 Examples Example Find the Fourier{ integral representation and the Fourier transform of the 1, < x < 2 function f(x) =, otherwise. The Fourier transform F(α) = F {f(x)} = The Fourier integral f(x) = 1 2π f(x)e iαx dx = 2 e iαx dx = 1 ( e 2iα 1 ) iα F(α)e iαx dα = 1 i ( e 2iα 1 ) e iαx dα 2π α 9 / 27 王奕翔 DE Lecture 15

10 Sufficient Condition of Convergence Theorem (Convergence of Fourier Integral) Let f and f be piecewise continuous on every finite interval and let f be absolutely integrable on (, ) (i.e., f(x) dx converges). Then, its Fourier integral converges to f(x) at a point where f(x) is continuous 1 (f(x+) + f(x )) at a point where f(x) is discontinuous. 2 1 / 27 王奕翔 DE Lecture 15

11 Alternative Form of the Fourier Integral f(x) = 1 2π { = 1 2π = 1 2π = 1 2π = 1 π F(α)e iαx dα = 1 2π F(α) (cos αx + i sin αx) dα + { F(α) (cos αx + i sin αx) dα + F(α) (cos αx + i sin αx) dα { [F(α) + F( α)] cos αx + i [F(α) F( α)] sin αx {[ f(x) cos αx dx ] } F(α) (cos αx + i sin αx) dα } F( α) (cos αx i sin αx) dα } dα [ ] } cos αx + f(x) sin αx dx sin αx dα F(α) + F( α) = F(α) F( α) = f(x)e iαx dx + f(x)e iαx dx f(x)e iαx dx = 2 f(x)e iαx dx = 2i f(x) cos αx dx f(x) sin αx dx 11 / 27 王奕翔 DE Lecture 15

12 Alternative Form of the Fourier Integral f(x) = 1 π where A(α) = f(x) cos αx dx, B(α) Fourier Integral of an Even Function f(x) = 1 π {A(α) cos αx + B(α) sin αx} dα, = f(x) sin αx dx. { A(α) cos αx + B(α) sin αx } dα, where A(α) = 2 f(x) cos αx dx, B(α) = f(x) sin αx dx =. Fourier Integral of an Odd Function f(x) = 1 π where A(α) = f(x) cos αx dx =, B(α) { A(α) cos αx + B(α) sin αx} dα, = 2 f(x) sin αx dx. 12 / 27 王奕翔 DE Lecture 15

13 Fourier Cosine Integral and Fourier Cosine Transform Definition (Fourier Cosine Integral and Fourier Cosine Transform) The Fourier cosine integral of a function f(x) defined on (, ) is f(x) = 2 π F(α) cos αx dα, where F(α) = The Fourier cosine transform of f(x) is F c {f(x)} = f(x) cos αx dx = F(α). The inverse Fourier cosine transform of a function F(α) is Fc 1 {F(α)} = 2 π F(α) cos αx dα = f(x). f(x) cos αx dx. 13 / 27 王奕翔 DE Lecture 15

14 Fourier Sine Integral and Fourier Sine Transform Definition (Fourier Sine Integral and Fourier Sine Transform) The Fourier sine integral of a function f(x) defined on (, ) is f(x) = 2 π F(α) sin αx dα, where F(α) = The Fourier sine transform of f(x) is F s {f(x)} = f(x) sin αx dx = F(α). The inverse Fourier sine transform of a function F(α) is Fs 1 {F(α)} = 2 π F(α) sin αx dα = f(x). f(x) sin αx dx. 14 / 27 王奕翔 DE Lecture 15

15 Examples Example Find the Fourier integral representation of the function { 1, a < x < a f(x) =., otherwise f(x) is an even function. Hence, its Fourier integral is a cosine integral f(x) = 2 { } f(x) cos αx dx cos αx dα π = 2 { a } cos αx dx cos αx dα π = 2 π sin αa cos αx α dα 15 / 27 王奕翔 DE Lecture 15

16 Examples Example On (, ), represent f(x) = e x (a) by a Fourier cosine integral, and (b) by a Fourier sine integral. (a) f(x) = 2 π (b) f(x) = 2 π { { } e x cos αx dx cos αx dα = 2 1 cos αx dα π 1 + α2 } e x sin αx dx sin αx dα = 2 α sin αx dα π 1 + α2 16 / 27 王奕翔 DE Lecture 15

17 FIGURE Function Fourier defined Transforms on (, ) in Example 3 A y x (a) cosine integral Therefore the cosine integra y f (b) Similarly, we have x B (a) cosine integral y The sine integral of f is then (b) sine integral FIGURE f (a) is the even extension of f; (b) is the odd extension of f Figure shows the gra the two integrals in (13) and 17 / 27 王奕翔 DE Lecture 15

18 Fourier Transforms of Derivatives The Fourier transform has many operational properties, and many of them resemble those of the Laplace transform. In this lecture we only focus on the Fourier transform of derivatives, as it is useful in solving BVP s of PDE s. Fact If f(x), f (x) as x ±, then F {f (x)} = iαf {f(x)}, F s {f (x)} = αf c {f(x)}, F {f (x)} = α 2 F {f(x)} F s {f (x)} = α 2 F s {f(x)} + αf() F c {f (x)} = αf s {f(x)} f(), F c {f (x)} = α 2 F c {f(x)} f () 18 / 27 王奕翔 DE Lecture 15

19 F { f (x) } = = F s { f (x) } = F s { f (x) } = f (x)e iαx dx = [f(x)e iαx] + iα e iαx d (f(x)) f(x)e iαx dx = iαf {f(x)} f(x) as x ± f (x) sin αx dx = sin αx d (f(x)) = [f(x) sin αx] α f(x) cos αx dx = αf c {f(x)} f(x) as x f (x) cos αx dx = cos αx d (f(x)) = [f(x) cos αx] + α f(x) sin αx dx = αf s {f(x)} f() f(x) as x 19 / 27 王奕翔 DE Lecture 15

20 1 Fourier Transforms 2 2 / 27 王奕翔 DE Lecture 15

21 Heat Equation in an Infinite Rod Solve u(x, t) : ku xx = u t, < x <, t > subject to : u(±, t) =, u x (±, t) =, t > u(x, ) = f(x), < x < Step 1: Take the Fourier transform w.r.t. x: Let u(x, t) F U(α, t). The original problem becomes kα 2 U(α, t) = du dt subject to: U(α, ) = F(α) Note: The condition u(±, t) =, u x (±, t) = is used to conclude that u xx F α 2 U(α, t) 21 / 27 王奕翔 DE Lecture 15

22 Heat Equation in an Infinite Rod Solve u(x, t) : ku xx = u t, < x <, t > subject to : u(±, t) =, u x (±, t) =, t > u(x, ) = f(x), < x < Step 2: Solve U(α, t): kα 2 U(α, t) = du dt = U(α, t) = C(α)e kα2t. Plug in U(α, ) = F(α), we get C(α) = F(α). Hence, U(α, t) = F(α)e kα2t. 22 / 27 王奕翔 DE Lecture 15

23 Heat Equation in an Infinite Rod Solve u(x, t) : ku xx = u t, < x <, t > subject to : u(±, t) =, u x (±, t) =, t > u(x, ) = f(x), < x < Step 3: Take inverse Fourier transform to find u(x, t): { } u(x, t) = F 1 {U(α, t)} = F 1 F(α)e kα2 t = 1 ( ) f(x)e iαx dx e iαx e kα2t dα 2π 23 / 27 王奕翔 DE Lecture 15

24 Laplace s Equation in a Semi-Infinite Plate Solve u(x, y) : u xx + u yy =, < x < a, < y < subject to : u x (, y) = f(y), u x (a, y) = g(y), < y < u y (x, ) =, u(x, ) = u y (x, ) =, < x < a Step 1: Take the Fourier cosine transform w.r.t. y: Let u(x, y) F c U(x, α). The original problem becomes d 2 U dx 2 α2 U(x, α) u y (x, ) = s.t. U(, α) = F(α), U(a, α) = G(α) Note: The condition u(x, ) = u y (x, ) = is used to conclude that F c α 2 U(x, α) u y (x, ) u yy 24 / 27 王奕翔 DE Lecture 15

25 Laplace s Equation in a Semi-Infinite Plate Solve u(x, y) : u xx + u yy =, < x < a, < y < subject to : u x (, y) = f(y), u x (a, y) = g(y), < y < Step 2: Solve U(x, α): d 2 U dx 2 u y (x, ) =, u(x, ) = u y (x, ) =, < x < a α2 U(x, α) = = U(x, α) = C 1 (α) cosh αx + C 2 (α) sinh αx. Plug in U(, α) = F(α), U(a, α) = G(α), we get G(α) F(α) cosh αa C 1 (α) = F(α), C 2 (α) =. ( sinh αa G(α) = U(x, α) = F(α) cosh αx + sinh αa F(α) ) sinh αx. tanh αa 25 / 27 王奕翔 DE Lecture 15

26 Laplace s Equation in a Semi-Infinite Plate Solve u(x, y) : u xx + u yy =, < x < a, < y < subject to : u x (, y) = f(y), u x (a, y) = g(y), < y < u y (x, ) =, u(x, ) = u y (x, ) =, < x < a Step 3: Take inverse Fourier cosine transform to find u(x, y): u(x, y) = Fc 1 {U(x, α)} { ( G(α) = Fc 1 F(α) cosh αx + sinh αa { F(α) cosh αx + = 1 2π ( G(α) sinh αa F(α) ) } sinh αx tanh αa F(α) tanh αa where F(α) = F c {f(y)}, and G(α) = F c {g(y)}. ) } sinh αx cos αy dα 26 / 27 王奕翔 DE Lecture 15

27 Remarks 1 Boundary conditions at ±, such as u(±, t) = u x (±, t) = and u(x, ) = u y (x, ) =, are used to guarantee that the Fourier transforms of the second-order partial derivates exist. 2 Which transform to use? Suppose the unbounded range is on x. If the range is (, ), use Fourier transform. If the range is (, ) and at the given condition is on u, use Fourier sine transform. (Because F s {f (x)} = α 2 F s {f(x)} + αf()!) If the range is (, ) and at the given condition is on u x, use Fourier sine transform. (Because F c {f (x)} = α 2 F c {f(x)} f ()!) 27 / 27 王奕翔 DE Lecture 15