Page 189 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

176 Solutions to the Exercises

(ii) We find, if x 6=0,that

x i
0
= f (|x|) ⇒ |∇u (x)| = |f (|x|)| .
0
u x i
|x|
Assume, for a moment, that we already proved that u is weakly differentiable in
B R ,then
Z R
p n−1 p
0
k∇uk p = σ n−1 r |f (r)| dr ,
L
0
which is the claim.
, asabove, isindeedthe weak derivative (with
Let us now show that u x i
respect to x i )of u. Wehavetoprove that,for every ϕ ∈ C ∞ (B R ),
0
Z Z
uϕ dx = − ϕu x i dx . (7.6)
x i
B R B R
Let > 0 be sufficiently small and observe that (recall that ϕ =0 on ∂B R )
Z Z Z
uϕ dx = uϕ dx + uϕ dx
x i x i x i
B R B R \B B
Z Z Z
x i
= − ϕu x i dx − uϕ dσ + uϕ dx
|x| x i
B R \B ∂B B
Z Z Z Z
x i
= − ϕu x i dx + ϕu x i dx + uϕ dx − uϕ dσ .
x i
|x|
B R B B ∂B
1
and uϕ are both in L (B R ), we deduce (this follows
Since the elements ϕu x i
x i
from Hölder inequality if p> 1 or from standard properties of integrals if p ≥ 1,
see Exercise 1.3.7) that
Z Z
lim ϕu x i dx =lim uϕ dx =0 .
→0 →0 x i
B B
Moreover, by hypothesis, we have the claim (i.e. (7.6)), since
¯Z ¯
¯ x i ¯ n−1
¯ uϕ dσ ≤ σ n−1 kϕk L ∞ f ( ) → 0, as → 0 .
¯
¯ |x| ¯
∂B
(iii) 1) The first example follows at once and gives
ψ ∈ L p ⇔ sp < n and ψ ∈ W 1,p ⇔ (s +1) p< n .
2) We find, for every 0 <s< 1/2 and p ≥ 1,that
Z 1/2 Z 1/2 2s−1
sp −1 2(s−1) |log 2|
r |log r| dr < ∞, r |log r| dr = < ∞ .
0 0 1 − 2s