Page 212 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 3: Direct methods 199

q
Case 3: p< n.We now only have u, ϕ ∈ L , ∀q ∈ [1,np/ (n − p)].We
p
0
q
0
therefore should have f u ∈ L , q = q/ (q − 1),and f ξ ∈ L . This happens if
0
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
³ ´
p−1
s 1 s 2 s 3
|f u (x, u, ξ)| ≤ β (1 + |u| + |ξ| ) , |f ξ (x, u, ξ)| ≤ β 1+ |u| + |ξ| .
Exercise 3.4.2. Use the preceding exercise to deduce the following growth
¡
¢
conditions on g ∈ C 1 Ω × R .
Case 1: p>n. No growth condition is imposed on g.
Case 2: p = n.There exist β> 0 and s 1 ≥ 1 such that
s 1
|g u (x, u)| ≤ β (1 + |u| ) , ∀ (x, u) ∈ Ω × R .
Case 3: p<n.There exist β> 0 and 1 ≤ s 1 ≤ (np − n + p) / (n − p),so
that
|g u (x, u)| ≤ β (1 + |u| ) , ∀ (x, u) ∈ Ω × R .
s 1
Exercise 3.4.3. (i) Let N be an integer and
u N (x, t)=sin Nx sin t.

Ω and u N =0 on ∂Ω).
We obviously have u N ∈ W 1,2 (Ω) (in fact u N ∈ C ∞ ¡ ¢
0
An elementary computation shows that lim I (u N )= −∞.
(ii) The second part is elementary.
It is also clear that for the wave equation it is not reasonable to impose an
initial condition (at t =0)and a final condition (at t = π).

7.3.4 The vectorial case

Exercise 3.5.1. Let
µ ¶ µ ¶
2 0 0 0
ξ = ,ξ = .
1 0 0 2 0 2
We therefore have the desired contradiction, namely
µ ¶
1 1 1 2 1 2 1 1
f (ξ )+ f (ξ )= (det ξ ) + (det ξ ) =0 <f ξ + ξ 2 =1 .
2
2
1
1
1
2 2 2 2 2 2
Exercise 3.5.2. We divide the discussion into two steps.
¡ ¢
Step 1. Let u, v ∈ C 2 Ω; R 2 with u = v on ∂Ω.Write u = u (x, y)=
(ϕ (x, y) ,ψ (x, y)) and v = v (x, y)= (α (x, y) ,β (x, y)).Use the fact that
¡ ¢
det ∇u = ϕ ψ − ϕ ψ = ϕψ − (ϕψ )
x y y x y x x y