30 1. Simple Examples of Propagation
intersection of
˜
Γ( s) with x1 = 0 yields
˜
Γ( 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
nj
2
)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}
Γ
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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