Page 110 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Euler-Lagrange equations 97

Note that ξ → f (u, ξ) is convex while (u, ξ) → f (u, ξ) is not. We have seen
that m = −∞ and therefore (P) has no minimizer; however the Euler-Lagrange
equation
2
u + λ u =0 in [0, 1]
00
has u ≡ 0 as a solution. It is therefore not a minimizer.
¡ 2 ¢ 2
Example 3.18 Let n =1, f (x, u, ξ)= f (ξ)= ξ − 1 , which is non convex,
and
½ Z ¾
1
1,4
0
(P) inf I (u)= f (u (x)) dx : u ∈ W (0, 1) = m.
0
0
We have seen that m =0. The Euler-Lagrange equation is
d £ ¡ 02 ¢¤
0
(E) u u − 1 =0
dx
and its weak form is (note that f satisfies (H3))
Z 1
¡ 02 ¢ 1,4
(E w ) u u − 1 ϕ dx =0, ∀ϕ ∈ W (0, 1) .
0
0
0
0
It is clear that u ≡ 0 is a solution of (E) and (E w ), but it is not a minimizer
of (P) since m =0 and I (0) = 1. The present example is also interesting for
another reason. Indeed the function
½
x if x ∈ [0, 1/2]
v (x)=
1 − x if x ∈ (1/2, 1]
1
is clearly a minimizer of (P) which is not C ;itsatisfies (E w )but not (E).

3.4.1 Exercises
Exercise 3.4.1 (i) Show that the theorem remains valid if we weaken the hy-
pothesis (H3), for example, as follows: if 1 ≤ p< n,replace (H3)by:
there exist β> 0, 1 ≤ s 1 ≤ (np − n + p) / (n − p), 1 ≤ s 2 ≤ (np − n + p) /n,
1 ≤ s 3 ≤ n (p − 1) / (n − p) so that the following hold, for every (x, u, ξ) ∈
n
Ω × R × R ,
³ ´
s 1 s 2 s 3 p−1
|f u (x, u, ξ)| ≤ β (1 + |u| + |ξ| ) , |f ξ (x, u, ξ)| ≤ β 1+ |u| + |ξ| .
(ii) Find, with the help of Sobolev imbedding theorem, other ways of weaken-
ing (H3) and keeping the conclusions of the theorem valid.