Page 123 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

110 Direct methods

Appealing to Exercise 3.5.2 we therefore find that

ZZ µZZ ¶ 2
2 1
(det ∇u (x)) dx ≥ det ∇u 0 (x) dx .
meas Ω
Ω Ω
The right hand side being strictly positive, by hypothesis, we have indeed found
that m> 0, which leads to the claimed counterexample.

3.6.1 Exercises

Exercise 3.6.1 Let n =1 and
½ Z 1 ¾
(P) inf I (u)= f (u (x)) dx = m
0
u∈X
0
© ª
where X = u ∈ W 1,∞ (0, 1) : u (0) = α, u (1) = β .
(i)Assumethat thereexist λ ∈ [0, 1], a, b ∈ R such that
⎧
β − α = λa +(1 − λ) b
⎨
⎩ ∗∗
f (β − α)= λf (a)+ (1 − λ) f (b) .
Show then that (P) has a solution, independently of wether f is convex or not
(compare the above relations with Theorem 1.55). Of course if f is convex, the
above hypothesis is always true, it suffices to choose λ =1/2 and a = b = β − α.
(ii) Can we apply the above considerations to f (ξ)= e −ξ 2 (cf. Section 2.2)
and α = β =0?
¡ 2 ¢ 2
(iii) What happens when f (ξ)= ξ − 1 ?
2
Exercise 3.6.2 Apply the theorem to Ω =(0, 1) ,
³ ´ 2
2 4
f (x, u, ξ)= f (ξ)= (ξ ) − 1 +(ξ )
1
2
2
where ξ =(ξ ,ξ ) ∈ R and
1
2
½ ZZ ¾
(P) inf I (u)= f (∇u (x, y)) dxdy : u ∈ W 1,4 (Ω) = m.
0
Ω
2 2×2
Exercise 3.6.3 Let f (ξ)= (det ξ) ,where ξ ∈ R . Show that f ∗∗ (ξ) ≡ 0.