56 2. FOUR IMPORTANT LINEAR PDE be constant on the time interval [0,t0] provided the initial and boundary conditions are constant. However, the solution may change at times t t0, provided the boundary conditions alter after t0. The solution will however not respond to changes in boundary conditions until these changes happen. Take note that whereas all this is obvious on intuitive, physical grounds, such insights do not constitute a proof. The task is to deduce such behavior from the PDE. Proof. 1. Suppose there exists a point (x0,t0) ∈ UT with u(x0,t0) = M := max ¯ U T u. Then for all suﬃciently small r 0, E(x0,t0 r) ⊂ UT and we employ the mean-value property to deduce M = u(x0,t0) = 1 4rn E(x0,t0 r) u(y, s) |x0 − y|2 (t0 − s)2 dyds ≤ M, since 1 = 1 4rn E(x0,t0 r) |x0 − y|2 (t0 − s)2 dyds. Equality holds only if u is identically equal to M within E(x0,t0 r). Con- sequently u(y, s) = M for all (y, s) ∈ E(x0,t0 r). Draw any line segment L in UT connecting (x0,t0) with some other point (y0,s0) ∈ UT , with s0 t0. Consider r0 := min{s ≥ s0 | u(x, t) = M for all points (x, t) ∈ L, s ≤ t ≤ t0}. Since u is continuous, the minimum is attained. Assume r0 s0. Then u(z0,r0) = M for some point (z0,r0) on L∩UT and so u ≡ M on E(z0,r0 r) for all suﬃciently small r 0. Since E(z0,r0 r) contains L ∩ {r0 − σ ≤ t ≤ r0} for some small σ 0, we have a contradiction. Thus r0 = s0, and hence u ≡ M on L. 2. Now fix any point x ∈ U and any time 0 ≤ t t0. There exist points {x0, x1,...,xm = x} such that the line segments in Rn connecting xi−1 to xi lie in U for i = 1,...,m. (This follows since the set of points in U which can be so connected to x0 by a polygonal path is nonempty, open and relatively closed in U.) Select times t0 t1 · · · tm = t. Then the line segments in Rn+1 connecting (xi−1,ti−1) to (xi,ti) (i = 1,...,m) lie in UT . According to step 1, u ≡ M on each such segment and so u(x, t) = M.

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