Page 99 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

86 Direct methods

true with p =1. This weakening of (H2) leads to the following counterexample.
p 2
2
Let n =1, f (x, u, ξ)= f (u, ξ)= u + ξ and
½ Z ¾
1
(P) inf I (u)= f (u (x) ,u (x)) dx : u ∈ X = m
0
0
© 1,1 ª
where X = u ∈ W (0, 1) : u (0) = 0,u (1) = 1 . Let us prove that (P) has no
solution. We first show that m =1 and start by observing that m ≥ 1 since
Z 1 Z 1
0 0
I (u) ≥ |u (x)| dx ≥ u (x) dx = u (1) − u (0) = 1 .
0 0
To establish that m =1 we construct a minimizing sequence u ν ∈ X (ν being an
integer) as follows

⎧ £ 1 ¤
0 if x ∈ 0, 1 −
⎨ ν
u ν (x)=
¡ ¤
⎩ 1
1+ ν (x − 1) if x ∈ 1 − , 1 .
ν
We therefore have m =1 since
Z 1 q
2 2
1 ≤ I (u ν )= (1 + ν (x − 1)) + ν dx
1
1−
ν
1 p
2
≤ 1+ ν −→ 1,as ν →∞ .
ν
Assume now, for the sake of contradiction, that there exists u ∈ X a minimizer
of (P). We should then have, as above,
Z Z 1
1 p
02
2
0
1= I (u)= u + u dx ≥ |u | dx
0 0
Z 1
0
≥ u dx = u (1) − u (0) = 1 .
0
This implies that u =0 a.e. in (0, 1). Since elements of X are continuous we
have that u ≡ 0 and this is incompatible with the boundary data. Thus (P) has
no solution.
Example 3.8 (Weierstrass example). We have seen this example in Section
2
2.2. Recall that n =1, f (x, u, ξ)= f (x, ξ)= xξ and
½ Z ¾
1
0
(P) inf I (u)= f (x, u (x)) dx : u ∈ X = m X
0