Math 257/316 Section 201 Midterm 2 Solutions Problem 1. a) [10 points] For the function f (x) = x2 on [0, 1], sketch (roughly) its odd, and even 2-periodic extensions, and find its Fourier sine series, and its Fourier cosine series, using (if you need it) for k = 1, 2, 3, . . ., Z 0 1 2(−1)k x cos(kπx)dx = 2 2 , k π 2 1 Z x2 sin(kπx)dx = 0 2((−1)k − 1) (−1)k − . k3 π3 kπ Using the hints, for k = 1, 2, 3, . . . , the F.S.S. and F.C.S. coefficients are (respectively) Z 2 1 2 2((−1)k − 1) (−1)k bk = x sin(kπx)dx = 2 − , 1 0 k3 π3 kπ Z 2 1 2 4(−1)k ak = x cos(kπx)dx = 2 2 , 1 0 k π while 2 a0 = 1 so F.S.S. of f = 2 1 Z 0 2 x2 dx = , 3 ∞ X 2((−1)k − 1) k3 π3 k=1 F.C.S. of f = (−1)k − sin(kπx), kπ ∞ 1 4 X (−1)k + 2 cos(kπx). 3 π k2 k=1 b) [15 points] Solve the following problem for the heat equation with non-zero derivative (flux) BCs: ut = uxx 0 < x < 1, t > 0 ux (0, t) = 0, ux (1, t) = 2 t>0 u(x, 0) = 0 0≤x≤1 1 There is no steady-state, so we first seek a particular solution of the form v(x, t) = Ax2 + Bx + Ct. To satisfy the heat equation we require 0 = vt − vxx = C − 2A =⇒ C = 2A, and to satisfy the BCs we require 0 = vx (0, t) = (2Ax + B)|x=0 = B, 2 = vx (1, t) = (2Ax)|x=1 = 2A, so A = 1, B = 0, C = 2, and v(x, t) = x2 + 2t. Then we write u(x, t) = v(x, t) + w(x, t) = x2 + 2t + w(x, t) and notice that wt − wxx = ut − vt − (uxx − vxx ) = ut − uxx − 2 + 2 = 0 wx (0, t) = ux (0, t) − vx (0, t) = 0 − 0 = 0, wx (1, t) = ux (1, t) − vx (1, t) = 2 − 2 = 0 2 w(x, 0) = u(x, 0) − v(x, 0) = 0 − x , and we know the solution to this problem is ∞ w(x, t) = a0 X 2 2 + ak cos(kπx)e−k π t , 2 k=1 where the ak are the F.C.S. coefficients of the function −x2 on [0, 1], which we he have from part (a) (up to a minus sign). So: u(x, t) = x2 + 2t − ∞ 1 4 X (−1)k 2 2 − 2 cos(kπx)e−k π t . 3 π k2 k=1 2 Problem 2. a) [19 points] Solve the following equation describing diffusion with growth, subject to non-zero BCs: ut = α2 uxx + u 0 < x < L, t > 0 u(0, t) = 0, u(L, t) = 1 t>0 u(x, 0) = 0 0≤x≤L but leave any Fourier coefficients in terms of integrals (i.e. do not take time to evaluate these integrals). Hint: first find the steady-state, then use separation of variables to find the remainder. A steady-state v(x) must solve α2 v 00 (x) + v(x) = 0, i.e. v 00 = − α12 v, whose general solution is v(x) = A sin(x/α) + B cos(x/α). The BCs require 0 = v(0) = B, 1 , sin(L/α) 1 = v(L) = A sin(L/α) =⇒ A = sin(x/α) (provided sin(L/α) 6= 0 – let’s assume this) so v(x) = sin(L/α) is the steady-state. Now writing u(x, t) = v(x) + w(x, t) we find that w solves: wt = ut = α2 uxx + u = α2 vxx + v + α2 wxx + w = α2 wxx + w w(0, t) = u(0, t) − v(0) = 0 − 0 = 0, w(L, t) = u(L, t) − v(L) = 1 − 1 = 0 w(x, 0) = u(x, 0) − v(x) = −v(x). Doing separation of variables, w(x, t) = X(x)T (t) leads to T0 1 X 00 − = = const. = −λ2 < 0. α2 T α2 X Indeed, we know this X problem well: X 00 (x) = −λ2 X(x), X(0) = 0 = X(L) only has non-zero solutions for negative separation constant (which is why we wrote it as −λ2 < 0), and these are Xk (x) = sin(kπx/L), λk = kπ/L, k = 1, 2, 3, . . . . The the solutions of the corresponding T problem, T 0 = (1−α2 λ2 )T = (1−α2 k 2 π 2 /L2 )T are Tk (t) = e(1−α 2 k 2 π 2 /L2 )t so the general solution is w(x, t) = ∞ X bk sin(kπx/L)e(1−k k=1 3 2 α2 π 2 /L2 )t , (or any multiple), and it remains to satisfy the IC: −v(x) = w(x, 0) = ∞ X sin(kπx/L), k=1 a Fourier sine series. Thus Z Z L 2 2 L v(x) sin(kπx/L)dx = − sin(x/α) sin(kπx/L)dx, bk = − L 0 L sin(L/α) 0 and the full solution is ∞ u(x, t) = sin(x/α) X 2 2 2 2 + bk sin(kπx/L)e(1−k α π /L )t sin(L/α) k=1 with the above formula for bk . b) [6 points] Determine the long-time (t → ∞) behaviour of the solution (considering all possible positive values of the diffusion rate α2 ). There are two cases: 1. α2 > L2 /π 2 : then all the exponentials are decaying to zero, and so we have convergence to the steady state lim u(x, t) = v(x) t→∞ (diffusion beats growth) 2. α2 < L2 /π 2 : then the first exponential is growing with t, and so u(x, t) becomes infinite as t → ∞ (growth beats diffusion) (recall we excluded the possibility α2 = L2 /π 2 when we assumed sin(L/α) 6= 0). 4

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