Page 134 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

The model case: Dirichlet integral 121

(iii) Both above results are however false if a =0, a =1 or p = ∞ (see
Exercise 4.3.3) and if p =1 (see Exercise 4.3.4). This is another reason why,
when dealing with partial differential equations or the calculus of variations,
k
Sobolev spaces and Hölder spaces are more appropriate than C spaces.

Proof. We know from the theory developed in Chapter 3 (in particular,
1,2
Exercise 3.2.1) that (P’) has a unique solution u ∈ W (Ω) which satisfies in
0
addition
Z Z
h∇u (x); ∇v (x)i dx = f (x) v (x) dx, ∀v ∈ W 1,2 (Ω) .
0
Ω Ω
We will only show the interior regularity of u, more precisely we will show that
f ∈ W k,2 (Ω) implies that u ∈ W k+2,2 (Ω). To show the sharper result (4.13) we
loc
refer to the literature (see Theorem 8.13 in Gilbarg-Trudinger [49]).
The claim is then equivalent to proving that ϕu ∈ W k+2,2 (Ω) for every
n
ϕ ∈ C ∞ (Ω).We let u = ϕu and notice that u ∈ W 1,2 (R ) andthatitisaweak
0
solution of
∆u = ∆ (ϕu)= ϕ∆u + u∆ϕ +2 h∇u; ∇ϕi
= −ϕf + u∆ϕ +2 h∇u; ∇ϕi ≡ g.

n
2
Since f ∈ W k,2 (Ω), u ∈ W 1,2 (Ω) and ϕ ∈ C ∞ (Ω) we have that g ∈ L (R ).
0 0
n
We have therefore transformed the problem into showing that any u ∈ W 1,2 (R )
which satisfies
Z Z
n
h∇u (x); ∇v (x)i dx = g (x) v (x) dx, ∀v ∈ W 1,2 (R ) (4.14)
R n R n
n
n
is in fact in W k+2,2 (R ) whenever g ∈ W k,2 (R ).We will prove this claim in
two steps. The first one deals with the case k =0, while the second one will
handle the general case.
n
n
2
Step 1. We here show that g ∈ L (R ) implies u ∈ W 2,2 (R ).To achieve
this goal we use the method of difference quotients. Weintroduce thefollowing
n
notations, for h ∈ R , h 6=0,welet
u (x + h) − u (x)
(D h u)(x)= .
|h|
It easily follows from Theorem 1.36 that
∇ (D h u)= D h (∇u) , kD −h uk 2 n ≤ k∇uk 2 n
L (R ) L (R )
n
kD h uk ≤ γ ⇒ u ∈ W 1,2 (R )
L 2 (R n )