Page 129 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

116 Regularity

1
Theorem 4.5 Let g ∈ C ([a, b] × R) satisfy
(H2) there exist p> q ≥ 1 and α 2 ,α 3 ∈ R such that
q
g (x, u) ≥ α 2 |u| + α 3 , ∀ (x, u) ∈ [a, b] × R .
Let
1 p
f (x, u, ξ)= |ξ| + g (x, u) .
p
1 0 p−2 1
0
Then there exists u ∈ C ([a, b]),with |u | u ∈ C ([a, b]), a minimizer of (P)
and the Euler-Lagrange equation holds everywhere, i.e.
d h p−2 i
0
0
|u (x)| u (x) = g u (x, u (x)) , ∀x ∈ [a, b] .
dx
2
Moreover if 1 <p ≤ 2,then u ∈ C ([a, b]).
If, in addition, u → g (x, u) is convex for every x ∈ [a, b], then the minimizer
is unique.
Remark 4.6 The result cannot be improved in general, cf. Exercise 4.2.2.
Proof. The existence (and uniqueness, if g is convex) of a solution u ∈
W 1,p (a, b) follows from Theorem 3.3. According to Lemma 4.2 we know that
u ∈ W 1,∞ (a, b) and since x → g u (x, u (x)) is continuous, we have that the
Euler-Lagrange equation holds everywhere, i.e.

d h p−2 i
0 0
|u (x)| u (x) = g u (x, u (x)) , x ∈ [a, b] .
dx
0 p−2 0 1 0 p−2 0
We thus have that |u | u ∈ C ([a, b]).Call v ≡ |u | u .We may then infer
that
2−p
0 p−1
u = |v| v.
2−p 1
Since the function t → |t| p−1 t is continuous if p> 2 and C if 1 <p ≤ 2,we
1
obtain, from the fact that v ∈ C ([a, b]), the conclusions of the theorem.
4.2.1 Exercises
Exercise 4.2.1 With the help of Lemma 4.2, prove Theorem 4.3 in the following
manner.
(i) First show that u ∈ W 2,∞ (a, b), by proving (iii) of Theorem 1.36.
(ii) Conclude, using the following form of the Euler-Lagrange equation
d
00
0
0
0
0
0
[f ξ (x, u, u )] = f ξξ (x, u, u ) u + f uξ (x, u, u ) u + f xξ (x, u, u )
dx
= f u (x, u, u ) .
0