Page 199 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

186 Solutions to the Exercises

Applying the implicit function theorem we can find 0 > 0 and a function t ∈
1
C ([− 0 , 0 ]) with t (0) = 0 such that
G ( , t ( )) = 0, ∀ ∈ [− 0 , 0 ]

which implies that u + v + t ( ) w ∈ X.Note also that

G ( , t ( )) + G h ( , t ( )) t ( )= 0, ∀ ∈ [− 0 , 0 ]
0
and hence
t (0) = −G (0, 0) .
0
Sinceweknowthat

F (0, 0) ≤ F ( , t ( )) , ∀ ∈ [− 0 , 0 ]

we deduce that
0
F (0, 0) + F h (0, 0) t (0) = 0
and thus letting λ = F h (0, 0) be the Lagrange multiplier we find

F (0, 0) − λG (0, 0) = 0
or in other words

b
Z
0
0
0
0
0
0
{[f ξ (x, u, u ) v + f u (x, u, u ) v] − λ [g ξ (x, u, u ) v + g u (x, u, u ) v]} dx =0 .
a
Appealing once more to the fundamental lemma of the calculus of variations and
to the fact that v ∈ C ∞ (a, b) is arbitrary we get
0
½ ¾
d d
0
[f ξ (x, u, u )] − f u (x, u, u )= λ [g ξ (x, u, u )] − g u (x, u, u ) .
0
0
0
dx dx
1
Exercise 2.2.5. Let v ∈ C (a, b), ∈ R and set ϕ ( )= I (u + v).Since u is a
0
minimizer of (P) we have ϕ ( ) ≥ ϕ (0), ∀ ∈ R, and hence we have that ϕ (0) = 0
0
(which leads to the Euler-Lagrange equation) and ϕ (0) ≥ 0. Computing the
00
last expression we find
b
Z
© 2 02 ª 1
f uu v +2f uξ vv + f ξξ v dx ≥ 0, ∀v ∈ C (a, b) .
0
0
a
¡ ¢ 0
Noting that 2vv = v 2 and recalling that v (a)= v (b)= 0,we find
0
Z ½ µ ¶ ¾
b
02 d 2 1
f ξξ v + f uu − f uξ v dx ≥ 0, ∀v ∈ C (a, b) .
0
a dx