30 1. Simple Examples of Propagation intersection of ˜ s) with x1 = 0 yields ˜ s)∩{x1=0} ut ∂1u dt dx2 · · · dxd . The Dirichlet condition implies that ut = 0 on this boundary, so the integral vanishes. The contribution of the sides |x x| = t t yield an integral of n0 e + d j=1 nj ut ∂ju , where (n0,n1,n2,...,nd) is the outward unit normal. Since the cone has sides of slope one, n0 = ( d j=1 n2 j )1/2 = 1 2 . The Cauchy–Schwarz inequality yields d j=1 nj ut ∂ju 1 2 |ut||∇xu| 1 2 e . Thus the integrand from the contributions of sides is nonnegative, so the integral over the sides is nonnegative. Combining yields 0 = ˜∩{0≤t≤s} ut u dt dx φ(s) φ(0) , completing the proof. 1.6.1. The method of images. Introduce the notations x = (x1,x ), x := (x2,...,xd), ξ = (ξ1,ξ ), ξ := (ξ2,...,ξd). Definitions 1.6.2. A function f on R1+d is even (resp., odd) in x1 when f(t, x1,x ) = f(t, −x1,x ), respectively, f(t, −x1,x ) = −f(t, x1,x ). Define the reflection operator R by (Rf)(t, x1,x ) := f(t, −x1,x ) . The even (resp., odd) parts of a function f are defined by f + Rf 2 , respectively, f Rf 2 .
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