Page 122 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Relaxation theory 109

We conclude this section with two examples.

Example 3.30 Let us return to Bolza example (Example 3.10). We here have
n =1,
¡ 2 ¢ 2 4
f (x, u, ξ)= f (u, ξ)= ξ − 1 + u
1
½ Z ¾
1,4
0
(P) inf I (u)= f (u (x) ,u (x)) dx : u ∈ W 0 (0, 1) = m.
0
We have already shown that m =0 and that (P) has no solution. An elementary
computation (cf. Example 1.53 (ii)) shows that
⎧
⎨ f (u, ξ) if |ξ| ≥ 1
f ∗∗ (u, ξ)=
u if |ξ| < 1 .
⎩ 4
Therefore u ≡ 0 is a solution of

1
½ Z ¾
¡ ¢ 1,4
0
P inf I (u)= f ∗∗ (u (x) ,u (x)) dx : u ∈ W (0, 1) = m =0.
0
0
1,4
The sequence u ν ∈ W 0 (ν ≥ 2 being an integer) constructed in Example 3.10
satisfies the conclusions of the theorem, i.e.
u ν u in W 1,4 and I (u ν ) → I (u)= 0,as ν →∞ .
2
Example 3.31 Let Ω ⊂ R be aboundedopen setwithLipschitz boundary. Let
¡ 2 ¢
1,4
u 0 ∈ W Ω; R be such that
ZZ
det ∇u 0 (x) dx 6=0 .
Ω
2
Let, for ξ ∈ R 2×2 , f (ξ)= (det ξ) ,
½ ZZ ¾
1,4 ¡ 2 ¢
(P) inf I (u)= f (∇u (x)) dx : u ∈ u 0 + W Ω; R = m
0
Ω
½ ZZ ¾
1,4 ¡ 2 ¢
(P) inf I (u)= f ∗∗ (∇u (x)) dx : u ∈ u 0 + W Ω; R = m.
0
Ω
We will show that Theorem 3.28 is false, by proving that m> m. Indeed it is
easy to prove (cf. Exercise 3.6.3) that f ∗∗ (ξ) ≡ 0, which therefore implies that
m =0. Let us show that m> 0. Indeed by Jensen inequality (cf. Theorem 1.51)
1,4 ¡ 2 ¢
we have, for every u ∈ u 0 + W Ω; R ,
0
ZZ µ ZZ ¶ 2
2 1
(det ∇u (x)) dx ≥ meas Ω det ∇u (x) dx .
Ω meas Ω Ω