Definition 37.2.1An open subset, U, of ℝ^{n}has a Lipschitz boundary if it satisfies thefollowing conditions. For each p ∈ ∂U ≡U∖U, there exists an open set, Q, containing p,an open interval
(a,b)
, a bounded open box B ⊆ ℝ^{n−1}, and an orthogonal transformationR such that
RQ = B × (a,b), (37.2.26)
(37.2.26)
n
R (Q∩ U ) = {y ∈ ℝ : ^y ∈ B,a < yn < g(^y)} (37.2.27)
(37.2.27)
where g is Lipschitz continuous on B,a < min
{ -}
g(x) : x ∈ B
, and
^y ≡ (y1,⋅⋅⋅,yn−1).
Letting W = Q ∩ U the following picture describes the situation.
PICT
The following lemma is important.
Lemma 37.2.2If U is an open subset of ℝ^{n}which has a Lipschitz boundary, thenit satisfies the segment condition and so X^{m,p}
(U)
= W^{m,p}
(U )
.
Proof:For x ∈ ∂U, simply look at a single open set, Q_{x} described in the above
which contains x. Then consider an open set whose intersection with U is of the form
R^{T}
({y :^y ∈ B,g(^y)− ε < yn < g (^y )})
and a vector of the form εR^{T}
(− en)
where ε is
chosen smaller than min
{ --}
g (x) : x ∈ B
− a. There is nothing to prove for points of
U.
One way to extend many of the above theorems to more general open sets than ℝ^{n} is
through the use of an appropriate extension theorem. In this section, a fairly general one
will be presented.
V− ≡ {(^y,yn) : yn < g(^y)} ,V+ ≡ {(^y,yn) : yn > g (^y)} ,
for g a Lipschitz function of the sort described in this definition. Suppose u^{+}and u^{−}are Lipschitz functions defined on V^{+}and V^{−}respectively and suppose thatu^{+}
(^y,g(^y))
= u^{−}
(^y,g (y^))
for all
^y
∈ B. Let
{ u+ (^y,y ) if (y^, y) ∈ V+
u(^y,yn) ≡ u − (^y,yn) if (^y,ny ) ∈ V−
n n
and supposespt
(u)
⊆ B ×
(a,b)
. Then extending u to be 0 off of B ×
(a,b)
, u iscontinuous and the weak partial derivatives, u_{,i}, are all in L^{∞}
(ℝn)
∩ L^{p}
(ℝn )
for allp > 1 and u_{,i} =
(u+)
_{,i}on V^{+}and u_{,i} =
(u− )
_{,i}on V^{−}.
Proof:Consider the following picture which is descriptive of the situation.
PICT
Note first that u is Lipschitz continuous. To see this, consider
|u(y1)− u (y2)|
where
( )
^yi,yin
= y_{i}. There are various cases to consider depending on whether
y_{n}^{i} is above g
The other cases are similar. Thus u is a Lipschitz continuous function which has compact
support. By Corollary 34.5.4 on Page 4316 it follows that u_{,i}∈ L^{∞}
n
(ℝ )
∩L^{p}
n
(ℝ )
for all
p > 1. It remains to verify u_{,i} =
+
(u )
_{,i} on V^{+} and u_{,i} =
−
(u )
_{,i} on V^{−}. The last claim is
obvious from the definition of weak derivatives.
Lemma 37.2.4In the situation of Lemma 37.2.3letu ∈ C^{1}
Therefore, since the integrand in 37.2.28 has compact support and because of
the Lipschitz continuity of all the functions, the dominated convergence theorem may be
applied to obtain
∫
w(^y,yn)ϕ,i(y)dy =
U
∫
ϕ (y)[− D u(^y,2g(^y)− y )(en−1)+ 2D u (^y,2g(^y)− y )(Dg (^y )en−1)]dy
U 1 n i 2 n i
∫ [ ]
= ϕ(y) − ∂u-(^y,2g(^y)− y )+ 2-∂u-(^y,2g(^y)− y ) ∂g-(y^) dy
U ∂yi n ∂yn n ∂yi
and so
w (y) = ∂u-(^y,2g(^y)− y )− 2-∂u-(^y,2g(^y)− y ) ∂g-(^y-) (37.2.29)
,i ∂yi n ∂yn n ∂yi
(37.2.29)
whenever i < n which is what you would expect from a formal application of the chain
rule. Next suppose i = n.
∫
w (^y,yn)ϕ,n(y)dy
U
∫
u(^y,2g(^y)−-(yn +-h))−-u(^y,2g(^y)−-yn)
= hli→m0 − U h ϕ (y )dy
∫ D2u-(^y,2g(^y)−-yn)h+-o-(h)-
= lihm→0 U h ϕ (y)dy
∫ ∂u
= ----(y^,2g (y^)− yn)ϕ(y)dy
U ∂yn
showing that
w,n(y) = − ∂u-(^y,2g(^y)− yn) (37.2.30)
∂yn
(37.2.30)
which is also expected.
From the definnition, for y ∈ ℝ^{n}∖U ≡
{(^y,y ) : y ≤ g(^y)}
n n
it follows w_{,i} = u_{,i} and
on U,w_{,i} is given by 37.2.29 and 37.2.30. Consider