Infinite Spatial Domains and the Fourier Transform

Infinite Spatial Domains and the Fourier Transform 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 [Dec 5, 2005] Consider the heat equation on an “infinite rod” ut = κuxx , u (x, 0) = f (x) , −∞ < x < ∞, t>0 −∞ < x < ∞. Since there is no boundary, we don’t have so-called boundary conditions. However, we assume |u (x, t)| and |f (x)| remain bounded as |x| → ∞ (we will make the re­ quirements more precise soon). We proceed by separation of variables, u (x, t) = X (x) T (t) On substitution into the PDE, we obtain, as before, X ′′ T′ = = −λ κT X and λ is constant since the l.h.s. depends only on t and the middle only on x. The individual problems for X and T are X ′′ + λX = 0 (no BCs) T ′ = −κλT Solving for T gives T (t) = ce−κλt We have not said yet whether λ is negative, positive, or zero. Since we the energy in the rod will stay the same, and will dissipate, we expect |u (x, t)| < ∞ for all times t > 0. If λ < 0, T (t) → ∞ as t → ∞ and X (x) blows up too, and hence λ must be non-negative. If λ = 0, then T (t) = c, X = Ax + B, and A = 0 to avoid a blowup. 1 In fact, since we assumed u → 0 as |x| → ∞, we must have A = 0, so that X = 0. Thus λ = 0 leads to a trivial solution. It turns out that we can relax the conditions on f (x) to allow λ = 0. For λ > 0, we have √  √  λx + B sin λx X = A cos It is customary to let λ = ω 2 (where ω > 0), so that for each λ = ω 2 > 0, we have a solution 2 u (x, t; ω) = (A (ω) cos (ωx) + B (ω) sin (ωx)) e−ω κt Since ω ∈ R, we must integrate, instead of summing, over all possible ω, to obtain the full solution: Z ∞ 2 u (x, t) = (A (ω) cos (ωx) + B (ω) sin (ωx)) e−ω κt dω (1) 0 Imposing the initial condition u (x, 0) = f (x) gives Z ∞ f (x) = (A (ω) cos (ωx) + B (ω) sin (ωx)) dω (2) 0 The concept of orthogonality for a discrete set of eigen-functions can be generalized provided f (x) satisfies the following properties, f (x) is piecewise smooth on every interval [a, b] of the real line Z ∞ |f (x)| dx < ∞. −∞ The second condition ensures f (x) decreases fast enough for large |x|. If these prop­ erties of f (x) hold, then A (ω) and B (ω) are given by Z Z 1 ∞ 1 ∞ A (ω) = f (s) cos (ωs) ds, B (ω) = f (s) sin (ωs) ds. π −∞ π −∞ Thus, 1 u (x, t) = π Z 0 ∞ −ω 2 κt e  cos (ωx) + sin (ωx) Z Z ∞ ∞ −∞ f (s) cos (ωs) ds  (3) f (s) sin (ωs) ds dω −∞ Z Z 1 ∞ ∞ −ω2 κt = e f (s) (cos (ωx) cos (ωs) + sin (ωx) sin (ωs)) dsdω π 0 −∞ Z Z 1 ∞ ∞ −ω2 κt = e f (s) cos (ω (x − s)) dsdω π 0 −∞  Z ∞ Z ∞ 1 −ω 2 κt = e cos (ω (x − s)) dω f (s) ds π 0 −∞ Z ∞ = K (s, x, t) f (s) ds (4) −∞ 2 where Z 1 ∞ −ω2 κt K (s, x, t) = e cos (ω (x − s)) dω π 0 is called the Heat Kernel. √ The integral for K can be calculated by defining z = ω κt, so that Z ∞ 1 x−s 2 K (s, x, t) = √ e−z cos (bz) dz, b= √ . π κt 0 κt Let I (b) = Z ∞ (5) 2 e−z cos (bz) dz (6) 0 You can find I (b) from Tables, e.g. Table of Integrals, Series & Products by Grad­ shteyn & Ryzhik. But here’s how to calculate it: first, differentiate in b, Z ∞  2 ′ I (b) = −e−z z sin (bz) dz 0 2 Integrating by parts with U = sin bz, dV = −ze−z , gives " 2 #∞ Z −z e b ∞ −z2 ′ I (b) = sin (bz) − e cos (bz) dz b 2 0 0 b = 0 − I (b) 2 Thus dI b = − db I 2 Integrating yields I (b) = I (0) e−b and from (6), I (0) = Z ∞ −z 2 e 2 /4 dz = √ 0 Therefore, π . 2 √ π −b2 /4 e 2 Substituting I (b) into (5) gives the Heat Kernel I (b) = (x − s)2 1 1 exp − K (s, x, t) = √ I (b) = √ 4κt π κt 4πκt and temperature, from (4), Z ∞ Z u (x, t) = K (s, x, t) f (s) ds = −∞ ∞ −∞ ! f (s) (x − s)2 √ exp − 4κt 4πκt (7) ! ds. (8) Since K (s, x, t) decays exponentially as |x| , |s| → ∞ , then a sufficient condition for a solution is that |f (s)| is bounded above by some constant as |s| → ∞. 3 1 Hot spot on an infinite rod As an example, consider the initial hot spot ( u (x, 0) = f (x) = u0 /ε, |x| ≤ ε/2 0, |x| > ε/2 where ε > 0. The temperature distribution is ! (x − s)2 f (s) √ exp − ds u (x, t) = 4κt 4πκt −∞ ! Z ε/2 u0 (x − s)2 = √ exp − ds 4κt ε 4πκt −ε/2 Z ∞ To evaluate this, we note that the error function is defined as Z x 2 2 erf (x) = √ e−x dx π 0 √ Let z = (x − s) / 4κt. Then u0 u (x, t) = − √ ε π Z √ (x−ε/2)/ 4κt √ (x+ε/2)/ 4κt
exp −z 2 dz # "Z Z (x−ε/2)/√4κt 0

u0 exp −z 2 dz exp −z 2 dz + = − √ ε π (x+ε/2)/√4κt 0 "Z # √ √ Z (x+ε/2)/ 4κt (x−ε/2)/ 4κt

u0 = √ exp −z 2 dz − exp −z 2 dz ε π 0 0      x − ε/2 u0 x + ε/2 − erf √ = erf √ 2ε 4κt 4κt Plots are given below in Figures 1 to 4. The solution u (x, t) is just the sum of shifted erfx functions, and hence is straightforward to plot once you know what erf(x) √ 2 looks like (see Figures 1, 2). The error function is 2/ π times the area under e−s √ from s = 0 to s = x (Figure 1). Thus as s → ∞, the area is π/2 since erf (∞) = 1. To obtain Figure 3, note that ! !# "   ε (x + ε/2)2 ε (x − ε/2)2 u0 exp − − x− exp − x+ ut = − √ 2 4κt 2 4κt 4ε πκt3/2 and ut < 0, |x| < ε/2 4 1 1 0.8 0.5 erf(x) e −x 2 0.6 0.4 −0.5 0.2 0 −5 0 0 x −1 −5 5 0 x 5 0 x 5 2 Figure 1: Plots of e−x and erf(x). 1 u(x,t)/u0( ε=1, κ t=1) u(x,0)/u0 (ε=1) 1 0.8 0.6 0.4 0.2 0 −5 0 x 0.8 0.6 0.4 0.2 0 −5 5 Figure 2: Plot of u(x, t) for t = 0 and κt = 1, with ε = 1. For |x| > ε/2, u increases and then decreases. Thus ut = 0 when ln x + ε/2 εx = x − ε/2 2κt and hence when t = t∗ = 2 εx x+ε/2 2κ ln x−ε/2 . Fourier Transform Ref: Guenther & Lee, §5.4 and for background, see §3.4 (in particular p. 78 and 79) Ref: Myint-U & Debnath §11.1 – 11.4 [Dec 7, 2005] 5 u(x0,t)/u0( ε=1) 1 0.5 0 0 0.5 1 1.5 2 2.5 κt 3 3.5 4 4.5 5 Figure 3: Plot of u(x, t) for x = 0 (solid), x = ±1/4 (dash) and x = ±3/4 (dot-dash), for ε = 1. 5 4.5 0.05 4 3.5 κt 3 2.5 0.1 2 1.5 0.25 1 0.5 0 −5 0.2 0.1 −4 −3 −2 0.5 −1 0 x 0.05 1 2 3 4 5 Figure 4: Plot of the level curves of u(x, t), i.e. the curves on which u(x, t) = const. Numbers indicate the value of u(x, t)/u0 on the level curve. Here ε = 1. 6 We now introduce the Fourier Transform and show how it is related to the solution of the Heat Problem on an infinite domain. Consider a function g (x) defined on −∞ < x < ∞ that satisfies the properties listed above, namely, g (x) is piecewise smooth on every interval [a, b] of the real line Z ∞ |g (x)| dx < ∞. (9) (10) −∞ The Fourier Transform G (ω) of the function g (x) is defined as Z ∞ 1 g (x) eiωx dx G (ω) = 2π −∞ (11) The Inverse Fourier Transform g (x) of a function G (ω) is given by Z ∞ g (x) = G (ω) e−iωx dω (12) −∞ 2.1 Fourier transform and the solution to the heat equation Ref: Myint-U & Debnath example 11.3.1 To relate the solution of the Heat Problem on an infinite domain −∞ < x < ∞ to the Fourier Transform, we must make some manipulations to our solution. In particular, Eq. (1) becomes Z ∞ 2 (A (ω) cos (ωx) + B (ω) sin (ωx)) e−ω κt dω Z0 ∞ Z ∞ 2 iωx −ω 2 κt F2 (ω) e−iωx e−ω κt dω, = F1 (ω) e e dω + u (x, t) = 0 (13) 0 where 1 1 (A (ω) − iB (ω)) , F2 (ω) = (A (ω) + iB (ω)) . 2 2 We now allow ω to take all real values, i.e., −∞ < ω < ∞, and continue to manipulate the solution u (x, t). Making the change of variable ω → −ω in the first integral of (13) gives Z 0 Z ∞ 2 −iωx −ω 2 κt F2 (ω) e−iωx e−ω κt dω u (x, t) = F1 (−ω) e e dω + 0 Z−∞ ∞ 2 = F (ω) e−iωx e−ω κt dω, (14) F1 (ω) = −∞ where F (ω) = ( F2 (ω) , ω ≥ 0, F1 (−ω) , ω < 0. 7 Furthermore, at t = 0, we have u (x, 0) = f (x) = Z ∞ F (ω) e−iωx dω. (15) −∞ Comparing Eq. (15) to (12) shows that f (x) is the Inverse Fourier Transform of F (ω), or equivalently, F (ω) is the Fourier Transform of the initial temperature distribution f (x). From (11), we have Z ∞ 1 F (ω) = f (x) eiωx dx. (16) 2π −∞ Therefore, given an initial temperature distribution f (x) that satisfies conditions (9) and (10), we find the Fourier Transform F (ω) of f (x), and the temperature distribution at each point in time is then given by (14). The integral in (16) can generally be found in Fourier Transform tables. 2.2 Fourier transform of the heat equation Ref: Myint-U & Debnath §11.2 We now define the Fourier Transform (FT) of a function u (x, t) as an operator: Z ∞ 1 u (x, t) eiωx dx (17) F [u] = U (ω, t) = 2π −∞ Thus, the FT F maps a function of (x, t) to a function of (ω, t). To transform the Heat Equation, we must consider how the FT maps derivatives. Note that   Z ∞ ∂u 1 ∂u F = (x, t) eiωx dx ∂t 2π −∞ ∂t   Z ∞ 1 ∂ ∂ ∂ iωx u (x, t) e dx = F [u] = U (ω, t) = ∂t 2π −∞ ∂t ∂t Also, integration by parts and the fact that u (x, t) → 0 as |x| → ∞ allow us to calculate the FT of ∂u/∂x,   Z ∞ 1 ∂u ∂u = (x, t) eiωx dx F ∂x 2π −∞ ∂x Z ∞ ∞
1  iωx 1 = e u (x, t) −∞ − u (x, t) iωeiωx dx 2π 2π −∞  Z ∞  1 iωx = −iω u (x, t) e dx 2π −∞ = −iωF [u] = −iωU (ω, t) . 8 Thus, the FT of an x-derivative of a function is mapped to −iω times the FT of the function. Hence  2    ∂ u ∂u F = −iωF = (−iω)2 F [u] = −ω 2 U (ω, t) 2 ∂x ∂x Thus the FT of the Heat Equation ut = κuxx is ∂ U (ω, t) = −κω 2 U (ω, t) . ∂t Hence, the Fourier Transform maps the heat equation, a PDE, to a first order ODE! Integrating in time gives 2 U (ω, t) = C (ω) e−κω t (18) where C (ω) is an arbitrary function, due to partial integration with respect to time. Setting t = 0 in (17) and (18) gives Z ∞ Z ∞ 1 1 iωx C (ω) = U (ω, 0) = u (x, 0) e dx = f (x) eiωx dx. 2π −∞ 2π −∞ We could substitute C (ω) back into (18) and use the Inverse Fourier Transform to obtain u (x, t). The solution would be the same as that for separation of variables (recall that uniqueness holds). However, u (x, t) can be obtained almost immediately using a result called the Convolution Theorem. One important FT is that of the Gaussian distribution, h i 1 2 −βx2 e−ω /4β F e =√ 4πβ Thus, the FT of a Gaussian is a Gaussian. In particular, the IFT of a Gaussian is h i rπ 2 −1 −αω 2 F e = e−x /4α (19) α 2.3 Convolution Theorem Ref: Myint-U & Debnath §11.3 Convolution Theorem Suppose that F (ω) and G (ω) are the Fourier Trans­ forms of f (x) and g (x), then Z ∞ Z ∞ 1 1 iωx f (x) e dx, G (ω) = g (x) eiωx dx, F (ω) = 2π −∞ 2π −∞ Z ∞ Z ∞ −iωx f (x) = F (ω) e dω, g (x) = G (ω) e−iωx dω. −∞ −∞ 9 Let H (ω) = F (ω) G (ω). The Inverse Fourier Transform (IFT) of H (ω) is Z ∞ Z ∞ 1 1 h (x) = f (s) g (x − s) ds = g (s) f (x − s) ds 2π −∞ 2π −∞ (20) The first integral in (20) is called the convolution of f (x) and g (x). In other words, the IFT of the product of two FTs is the 1/2π times the convolution of the two functions. Proof: [optional] The IFT of H (ω) is Z ∞ Z ∞ −iωx h (x) = H (ω) e dω = F (ω) G (ω) e−iωx dω −∞ −∞ Substituting the IFT of F (ω) gives  Z ∞ Z ∞ 1 iωs h (x) = f (s) e ds G (ω) e−iωx dω 2π −∞ Z ∞ Z ∞−∞ 1 = f (s) G (ω) e−iωx eiωs dsdω 2π −∞ −∞ Assuming we can interchange the order of integration (we can provided R∞ and −∞ |G (ω)| dω are finite and f (s), G (ω) are smooth) we have Z ∞  Z ∞ 1 −iω(x−s) h (x) = f (s) G (ω) e dω ds 2π −∞ −∞ Note that g (x − s) = and hence Z Z ∞ R∞ −∞ |f (s)| ds G (ω) e−iω(x−s) dω −∞ ∞ 1 f (s) g (x − s) ds. 2π −∞ This is the first integral in (20). To obtain the second, we make the transformation w = x − s, so the integral becomes Z ∞ 1 g (w) f (x − w) dw. h (x) = 2π −∞ h (x) = Since w is a dummy variable, we can replace it with s to obtain the second integral in (20). � To apply the convolution theorem to the Heat Equation, we note that the FT of the solution u (x, t) is 2 U (ω, t) = C (ω) e−κω t p 2 2 Since the IFT of C (ω) is f (x) and the IFT of e−κω t is, by (19), π/ (κt)e−x /(4κt) , then, by the Convolution Theorem, r Z ∞ π −(x−s)2 /(4κt) 1 f (s) e ds u (x, t) = 2π −∞ κt which agrees with the solution we found using separation of variables, Eq. (8). 10 2.4 Heat equation on a semi-infinite domain Ref: Guenther & Lee p. 171, Myint-U & Debnath example 11.2.1 Consider the heat equation on a semi-infinite domain 0 ≤ x < ∞, ut = κuxx , u (x, 0) = f (x) , 0≤x<∞ 0≤x<∞ and f (x), u (x, t) approach zero fast enough as |x| → ∞ so that the integral Z ∞ |u (x, t)| dx −∞ is finite. Also, a boundary condition is imposed at x = 0. Either a Type I BC (fixed temperature), u (0, t) = 0, t > 0, (21) or Type II BC (insulated), ux (0, t) = 0, t > 0. (22) To solve this problem, we recall the temperature distribution on the infinite do­ main −∞ < x < ∞ due to an initial temperature fe(x) was given by Z ∞ u (x, t) = K (s, x, t) fe(s) ds −∞ where the Heat Kernel is defined as (x − s)2 1 exp − K (s, x, t) = √ 4κt 4πκt ! . Note that at x = 0, the Heat Kernel is even in s, i.e. K (s, 0, t) = K (−s, 0, t). Also, Z ∞ u (0, t) = K (s, 0, t) fe(s) ds −∞ Thus u (0, t) = 0 if fe(s) is odd. Therefore, the solution to the Heat Problem on the semi-infinite domain 0 ≤ x < ∞ with zero temperature at x = 0 (u (0, t) = 0) is Z ∞ u (x, t) = K (s, x, t) fe(s) ds −∞ where fe(s) is the odd extension of f (s), i.e.   s>0  f (s) fe(s) = 0 s=0   −f (−s) s < 0 11 Similarly, note that the x-derivative of the Heat Kernel is odd at s = 0,   s s2 √ Kx (s, 0, t) = exp − = −Kx (−s, 0, t) , 4κt 2κt 4πκt and hence the solution to the Heat Problem on the semi-infinite domain 0 ≤ x < ∞ with an insulated BC at x = 0 (ux (0, t) = 0) is Z ∞ u (x, t) = K (s, x, t) fe(s) ds −∞ where fe(s) is the even extension of f (s), i.e.    f (s) s > 0 e f (s) = 0 s=0   f (−s) s < 0 3 Fourier Transform solution to Laplace’s Equa­ tion Ref: Guenther & Lee §8.2, problem 5, Myint-U & Debnath example 11.2.1 Suppose the temperature of an infinite wall is kept at f (x), for −∞ < x < ∞. Find the steady-state temperature in the region adjoining the wall, y > 0. The steady-state temperature satisfies Laplace’s equation, ∇2 uE = 0, −∞ < x < ∞, y > 0. uE = f (x) , −∞ < x < ∞, y = 0, The BCs are lim uE (x, y) = 0, lim uE (x, y) = 0. y→∞ |x|→∞ We employ the Fourier transform in x, 1 F [g (x, y)] = 2π Z ∞ g (x, y) eiωx dx −∞ We define UE (ω, y) = F [uE (x, y)]. As before, we have F [uExx ] = −ω 2 F [uE ] = −ω 2 UE (ω, y) , F [uEyy ] = ∂2 ∂2 F [u ] = UE (ω, y) . E ∂y 2 ∂y 2 Hence Laplace’s equation for the steady-state temp uE (x, y) becomes ∂2 UE (ω, y) − ω 2 UE (ω, y) = 0 2 ∂y 12 Solving the ODE and being careful about the fact that ω can be positive or negative, we have UE (ω, y) = c1 (ω) e−|ω|y + c2 (ω) e|ω|y where c1 (ω), c2 (ω) are arbitrary functions. Since the temperature must vanish as y → ∞, we must have c2 (ω) = 0. Thus UE (ω, y) = c1 (ω) e−|ω|y (23) Imposing the BC at y = 0 gives 1 c1 (ω) = UE (ω, 0) = F [uE (x, 0)] = 2π Z ∞ iωx uE (x, 0) e −∞ Note that the IFT of e−|ω|y is Z ∞ Z  −|ω|y  −1 −|ω|y −iωx F e = e e dω =  −∞ ω(y−ix) 0  0 ω(y−ix) e −∞ −ω(y+ix) e e + y − ix −∞ − (y + ix) 1 1 2y = + = 2 y − ix y + ix x + y2 = 1 dx = 2π ∞ dω + Z ∞ Z ∞ f (x) eiωx dx. −∞ e−ω(y+ix) dω 0 0 Therefore, applying the Convolution Theorem to (23) with F −1 [c1 (ω)] = f (x) and   F −1 e−|ω|y = 2y/ (x2 + y 2 ) gives Z ∞ 1 2y ds. uE (x, y) = f (s) 2π −∞ (x − s)2 + y 2 13