88 2. FOUR IMPORTANT LINEAR PDE for a solution of the initial/boundary-value problem ⎧ ⎨ ⎩ ut − uxx = 0 in R+ × (0, ∞) u = 0 on R+ × {t = 0} u = g on {x = 0} × [0, ∞). (Hint: Let v(x, t) := u(x, t) − g(t) and extend v to {x 0} by odd reflection.) 16. Give a direct proof that if U is bounded and u ∈ C1 2(UT ) ∩ C( ¯T U ) solves the heat equation, then max ¯ U T u = max ΓT u. (Hint: Define uε := u − εt for ε 0, and show uε cannot attain its maximum over ¯ U T at a point in UT .) 17. We say v ∈ C1 2(UT ) is a subsolution of the heat equation if vt − Δv ≤ 0 in UT . (a) Prove for a subsolution v that v(x, t) ≤ 1 4rn E(x,t r) v(y, s) |x − y|2 (t − s)2 dyds for all E(x, t r) ⊂ UT . (b) Prove that therefore max ¯ U T v = maxΓT v. (c) Let φ : R → R be smooth and convex. Assume u solves the heat equation and v := φ(u). Prove v is a subsolution. (d) Prove v := |Du|2 + ut 2 is a subsolution, whenever u solves the heat equation. 18. (Stokes’ rule) Assume u solves the initial-value problem utt − Δu = 0 in Rn × (0, ∞) u = 0, ut = h on Rn × {t = 0}. Show that v := ut solves vtt − Δv = 0 in Rn × (0, ∞) v = h, vt = 0 on Rn × {t = 0}. This is Stokes’ rule. 19. (a) Show the general solution of the PDE uxy = 0 is u(x, y) = F (x) + G(y)

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.