Page 132 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

The model case: Dirichlet integral 119

and observe that v ∈ C 0 ∞ (B R (x)). An easy computation gives, for r = |x − y|,

n − 1 1−n d £ ¤
∆v = ϕ (r)+ ϕ (r)= r r n−1 0
ϕ (r) .
0
00
r dr
From now on we assume that n ≥ 2,the case n =1 is elementary and is discussed
in Exercise 4.3.1. We next let
d £ ¤
ψ (r)= r n−1 0 (4.9)
ϕ (r) .
dr
Note that ψ ∈ C ∞ ( , R) and
0
Z R
ψ (r) dr =0 . (4.10)

Remark that the converse is also true, namely that given ψ ∈ C 0 ∞ ( , R) satisfying
(4.10) we can find ϕ ∈ C 0 ∞ ( , R) verifying (4.9).
Step 2. Let ψ ∈ C 0 ∞ ( , R), satisfying (4.10), be arbitrary. Define then ϕ and
v as above and use such a v in (4.7). We get, since v ≡ 0 on Ω \ B R (x),
Z Z Z R Z
0= u∆vdy = u∆vdy = ψ (r) r 1−n udσ dr
Ω B R (x) ∂B r (x)
R
Z
= ψ (r) w (r) dr

wherewehaveset
Z
w (r)= r 1−n udσ .
∂B r (x)
We can use Corollary 1.25 to deduce that

w (r)= constant, a.e. r ∈ ( , R) .

We denote this constant by σ n−1 u (x) andweuse thefact that is arbitrary to
write
w (r)= σ n−1 u (x) ,a.e. r ∈ (0,R) . (4.11)
Step 3. We now conclude with the proof of the theorem. We first observe
that (4.11) is nothing else than (4.8) with the fact that u is independent of R.
We next integrate (4.8) and get

Z
1
u (x)= u (z) dz . (4.12)
meas B R (x)
B R (x)