• Ohio State University

• MATH 4557

 No books, notes or electronic devices. Show your work. 1. (10 points) Given the formula for the solution of the diffusion equation on the whole line with initial condition u(x, 0) = u0(x), u(x, t) = 1 √ 4πkt Z ∞ −∞ e −(x−y) 2/(4kt)u0(y) dy , find the formula for ut(x, t). Solution ut(x, t) = 1 √ 4πk  −1 2t 3/2  Z ∞ −∞ e −(x−y) 2/(4kt)u0(y) dy + 1 √ 4πkt Z ∞ −∞  (x − y) 2 4kt2  e −(x−y) 2/(4kt)u0(y) dy . Comment: The intent of this question was to test concepts like “how do you differentiate a function of several variables that is, in addition, given as an integral with respect to a third, dummy variable”; it also tests the notion of differentiating under the integral sign. Many (possibly most) people had difficulty doing the actual differentiation correctly, but lost only a point or two for this, as this was not intended to be a calculus question. 2. (10 points) Given that the energy for the heat equation ut = kuxx is E(t) = Z ∞ −∞ u 2 (x, t) dx , and that u(x, t) → 0, ux(x, t) → 0 and ut(x, t) → 0 as |x| → ∞, show that the energy of a solution u on the entire real line is non-increasing as a function of t. Solution Calculate, assuming differentiation under the integral sign is permitted, dE(t) dt = Z ∞ −∞ 2u(x, t)ut(x, t) dx = Z ∞ −∞ 2ku(x, t)uxx(x, t) dx , and integrate by parts: dE(t) dt = 2ku(x, t)ux(x, t) ∞ −∞ − 2k Z ∞ −∞ u 2 x (x, t) dx . The boundary terms are zero, by assumption, and the remaining integral is non-positive (it may be zero), so dE/dt ≤ 0 and hence E(t) does not increase in time.