By Paul DuChateau, D. Zachmann
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Example text
3-1 CHAP. 2 shows that u is identically equal to M in each BR(Xi ), i = 0, 1, . . , n ; hence, u(x * ) = M. Since x* was arbitra ry, u must be equal to M throughout n and, by continuity of u, throughou t 0 . This shows that if u is not a constant in n, the n u can attain its maximum value only on the boundary of n. The above argument, applied to - U, establishes that if u is nonconstant, it can attain its minimum value only on S. 4 Show that if u has the mean-value property in an open region n, then u is harmonic in n .
See Fig. 4-2. x - at = TJo CHAP. 4] 47 SOLUTIO NS T O EVOLUTION E QUATIONS x Domain of dependence F ig. 4-2 (b) Because F and G each vanish outside Ixl < 1, u(l + a, t) must remain zero so long as the domain of dependence of the point (1 + a, t), [1 + a - at, 1 + a + at], remains disjoint from (-1,1); that is, so long as 1 + a - at 2: 1 or Now, the distance fro m the poin t (1 + a, 0) to the interval (-1,1) is just a units. , the p ropagation speed is a units of distance per unit of time . 12 Consider the following modification of th e n -dimensional wave eq uation : (1) whe re c p .
Max Igl + (eO s eX) max n III n, v::= max Igi and s V2=g Calculating L[ v] = v= + VYY ' we find, since eX 2= (4) on S eO= 1 and 1 2= -max n L[v ] = _e x max n III ~I III, inn (5) n. 12 imply u ~ v in and L [ w] 2= I in n , so that u 2= - v in Consequently, in n. lui ~ v ~ max s Igi + (e Q - 1) fGaX ~ III n. since eX 2= 1 in 'Illis establishes (3), with M = eO - 1. We term (3) an a priori estimate of u. When some knowledge of the solution is incorporated, much sharper estimates are possible. 14 Let n be a bounded regi on.