Page 220 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 4: Regularity 207

1
Since (H1’) holds and u ∈ C ([a, b]), we deduce that there exists γ > 0 such
5
that
f ξξ (x, u (x) , u (x)) ≥ γ > 0, ∀x ∈ [a, b] .
0
5
The Euler-Lagrange equation can then be rewritten as
f u (x, u, u ) − f xξ (x, u, u ) − f uξ (x, u, u ) u 0
0
0
0
u =
00
f ξξ (x, u, u )
0
2
and hence u ∈ C ([a, b]). Returning to the equation we find that the right
2
3
hand side is then C , and hence u ∈ C . Iterating the process we conclude that
u ∈ C ∞ ([a, b]), as claimed.
Exercise 4.2.2. (i) We have
µ ¶
p −2 p p −2 q 2q−p
0 p−q 00 p−q p−q
u = |x| x and u = − 1 |x| = |x|
p − q p − q
2
1
which implies, since p> 2q> 2,that u ∈ C ([−1, 1]) but u/∈ C ([−1, 1]).
(ii) We find that
0 p−2 0 q(p−2) 2q−p p(q−1)
|u | u = |x| p−q |x| p−q x = |x| p−q x
µ ¶ q−1
p − q p(q−1)
q−2
|u| u = |x| p−q .
p
p(q−1)
If we choose, for instance, =4 (which is realized, for example, if p =8 and
p−q
0 p−2 q−2 2
0
q =10/3), then |u | u , |u| u ∈ C ∞ ([−1, 1]), although u/∈ C ([−1, 1]).
(iii) Since the function (u, ξ) → f (u, ξ) is strictly convex and satisfies all the
hypotheses of Theorem 3.3 and Theorem 3.11, we have that (P) has a unique
minimizer and that it should be the solution of the Euler-Lagrange equation
³ ´
0 p−2 0 0 q−2
|u | u = λ |u| u.
A direct computation shows that, indeed, u is a solution of this equation and
therefore it is the unique minimizer of (P).
7.4.2 The model case: Dirichlet integral
Exercise 4.3.1. We have to show that if u ∈ L 1 (a, b) and
loc
Z
b
u (x) ϕ (x) dx =0, ∀ϕ ∈ C 0 ∞ (a, b)
00
a
then, up to changing u on a set of measure zero, u (x)= αx+β for some α, β ∈ R.
This follows exactly as in Exercise 1.3.6.