Page 111 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

98 Direct methods

Exercise 3.4.2 Let
1 p
f (x, u, ξ)= |ξ| + g (x, u) .
p
Find growth conditions (depending on p and n)on g that improve (H3) and still
allow to derive, as in the preceding exercise, (E w ).

2
Exercise 3.4.3 Let n =2, Ω =(0,π) , u = u (x, t), u t = ∂u/∂t, u x = ∂u/∂x
and
½ ZZ ¾
1 ¡ 2 2 ¢ 1,2
(P) inf I (u)= u − u x dx dt : u ∈ W 0 (Ω) = m.
t
2 Ω
(i) Show that m = −∞.
(ii) Prove, formally, that the Euler-Lagrange equation associated to (P) is
thewaveequation u tt − u xx =0.

3.5 The vectorial case

The problem under consideration is
½ Z ¾
(P) inf I (u)= f (x, u (x) , ∇u (x)) dx : u ∈ u 0 + W 1,p ¡ Ω; R N ¢ = m
0
Ω
where n, N > 1 and

n
- Ω ⊂ R is a bounded open set;
N
- f : Ω × R × R N×n −→ R, f = f (x, u, ξ);
³ ´ 1≤j≤N ³ ´ 1≤j≤N
¡ 1 N ¢ N j N×n ∂u j
- u = u , ..., u ∈ R , ξ = ξ i ∈ R and ∇u = ;
1≤i≤n ∂x i 1≤i≤n
¡
¢
j
j
- u ∈ u 0 + W 1,p Ω; R N means that u ,u ∈ W 1,p (Ω), j =1, ..., N,and
0 0
1,p ¡ N ¢
u − u 0 ∈ W Ω; R (which roughly means that u = u 0 on ∂Ω).
0
All the results of the preceding sections apply to the present context when
n, N > 1.However, while for N =1 (or analogously when n =1) Theorem 3.3
was almost optimal, it is now far from being so. The vectorial case is intrinsically
more difficult. For example the Euler-Lagrange equations associated to (P) are
then a system of partial differential equations, whose treatment is considerably
harder than that of a single partial differential equation.
We will present one extension of Theorem 3.3; it will not be the best possible
result, but it has the advantage of giving some flavours of what can be done. For
the sake of clarity we will essentially consider only the case n = N =2; but we
will, in a remark, briefly mention what can be done in the higher dimensional
case.