Take Home Exam I Key

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1 Take Home Exam I Key MA (5 points) Read sections 2.1 to 2.6 in the text (which we ve already talked about in class), and the handout on Solving the Nonhomogeneous Heat Equation. Solution: You did this. 2. (25 points) Find an infinite Fourier type series solution to the heat equation u t 4u xx = for < x < π with zero Neumann boundary conditions u x (, t) =, u x (π, t) = and initial condition u(x, ) = x 2 (π x) 2. Write out the first three terms explicitly. Plot the solution on < x < π for times t =, 1, 5. Comment does it make sense!? Solution: Actually, there are several variants on this one, as I kept botching up the problem statement. Separating variables shows we should take with a k = 2 π a 2 + a k e k2t cos(kx) π If f(x) = x 2 (π 2 x 2 ) the first few terms are f(x) cos(kx) dx = 24(( 1)k 1) k e 4t cos(x) e 16t cos(2x) If f(x) = x 2 (π x) 2 we find e 16t cos(2x).1875e 64t cos(4x). If you use the original f(x) = x 2 (1 x) 2 you get e 4t cos(x) e 16t cos(2x). A plot of times t =, 1, 3 is shown in Figure 1 for f(x) = x 2 (π x) 2. The solution decays to a constant quite rapidly. 3. (3 points) Find a three-term approximation to the solution to u t u xx = tx on < x < 1 with u(x, ) = and u(, t) = u(1, t) =. Plot the solution at times t =, 1, 3. Comment does it make sense? Solution: You should try something of the form c k (t) sin(kπx). 1

2 Figure 1: Solution at times t =, 1, 5. Jam this into the heat equation above and find that we need (c k(t) + k 2 π 2 c k (t)) sin(kπx) = tx. But if we expand, for each fixed t >, the function tx into a Fourier sine series in x, we obtain coefficients a k = 2t( 1)k+1. kπ So really, we need the c k (t) to be chosen so that (c k(t) + k 2 π 2 c k (t)) sin(kπx) = a k sin(kπx). 2

3 This means that c k (t) satisfies the first order ODE c k(t) + k 2 π 2 c k (t) = 2t( 1)k+1. kπ The initial condition u(x, ) = forces (plug t = into the Fourier sum for u) c k () =. Solve the DE above with this initial condition and get ( ) ( 1) 1+k k 2 π 2 t + e k 2 π 2t 1 c k (t) = 2 k 5 π 5 The solution is The first terms look like ( 2 π 2 t + 2 e π2t 2 2 ) sin (π x) π 5 + A plot of times t =, 1, 3 is shown in Figure (4 points) Our goal is to solve the heat equation ( ) ( 1) 1+k k 2 π 2 t + e k 2 π 2t 1 k 5 π 5 sin(kπx). ( ) ( 4 π 2 t e 4 π2t + 1 sin (2 π x) 16π 5 + ) 18 π 2 t + 2 e 9 π2t 2 sin (3 π x) 243 π 5 u t 2u xx = (1) on the interval < x < 1 (I m taking thermal diffusivity α = 2) with the boundary conditions on the left end (Dirichlet) and u(, t) = (2) u x (1, t) = 3u(1, t) (3) on the right, a Robin condition. We ll use some initial data u(x, ) = f(x), to be specified later. (a) (5 points) What s a physical interpretation of the boundary condition (3)? Solution: This is a Newton cooling condition the bar is losing heat to the ambient environment (at degrees). (b) (5 points) The usual separation of variables procedure yields separable solutions to the heat equation (1) of the form ϕ(x, t) = c 1 e 2λ2t cos(λx) + c 2 e 2λ2t sin(λx), for some constants c 1, c 2, which we ve done a million times now. (I m using ϕ temporarily, instead of u). Show that the boundary condition (2) forces c 1 =, so that ϕ(x, t) = c 2 e λ2t sin(λx). Solution: Well, this is pretty obvious. 3

4 Figure 2: Solution at times t =, 1, 3. (c) (5 points) Show that the boundary condition (3) then forces tan(λ) = λ/3. Convince yourself that there are an infinite number of positive solutions to this equation. For example, in Maple try plot([ lambda/3, tan(lambda)], lambda =..2); Solution: Without the cosine term we are down to ϕ(x, t) = e λ2t sin(λx) (ignore c 2 ). need ϕ x (1, t) = 3ϕ(1, t), which forces λ cos(λ) = 3 sin(λ) after dividing out the exponential piece and plugging in x = 1. The above is equivalent to tan(λ) = λ/3. A plot is pretty convincing this equation has infinitely many positive solutions λ k that can be ordered as < λ 1 < λ 2 <. We 4

5 (d) (1 points) Let λ n denote the nth positive solution to tan(λ) = λ/3 (which we will find below). We now know that for each n, the function u n (x, t) = e 2λ2 n t sin(λ n x) satisfies the heat equation (1) and both boundary conditions (2) and (3). By superposition the function c n e 2λ2 n t sin(λ n x) (4) n=1 should also satisfy the heat equation and both boundary conditions, for any choice of the constants c n. All we need to do is determine the c n to obtain u(x, ) = f(x). But first we need the λ n. Compute λ 1 by finding (numerically) the smallest positive root of λ/3 = tan(λ). Use Maple s fsolve command if you like. Then find λ 2 and λ 3. Solution: You find λ , λ , λ (e) (5 points) According to the theory of Section 2.6 (and stuff done in class), the functions sin(λ m x) and sin(λ n x) should be orthogonal, that is, 1 sin(λ m x) sin(λ n x) dx = if m n. Check this is true for all combinations 1 m, n 3 with m n using the λ j you computed above (at least to numerical round-off error). Solution: All of these integrals are zero, to round-off error. (f) (1 points) Let s take f(x) = 5x 4x 2. We can expand f(x) into a Fourier series as by taking f(x) = c n = 1 c n sin(λ n x) dx n=1 1 f(x) sin(λ n x) dx sin 2 (λ n x) dx (this is equation (14) in section 2.6). Compute c 1, c 2, c 3 for this f, then use these values in (4) with n = 3 to approximate the solution to the BVP. Plot the solution at times t =, t =.1, t = 1 (show the scale on the vertical axis). What does the solution do as t? Does it make sense? Solution: You find that c , c 2.52, c The solution at various times looks like that shown in Figure 3. It decays quite rapidly to zero. 5

6 Figure 3: Solution at times t =,.1, 1. 6