Page 71 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

58 Classical methods

1
1
holds for every v ∈ C (a, b) (i.e. v ∈ C ([a, b]) and v has compact support in
0
(a, b)). Setting
d
0 0 0
P (x)= f ξξ (x, u, u ) ,Q (x)= f uu (x, u, u ) − [f uξ (x, u, u )]
dx
rewrite the above inequality as
Z
b
£ 02 2 ¤
P (x) v + Q (x) v dx ≥ 0 .
a
Exercise 2.2.6 Show (cf. Case 2.3 and Case 2.4)
© 1 ª
(i) If X = u ∈ C ([0, 1]) : u (0) = u (1) = 0 ,then
½ Z ¾
1 ³ ´ 2
2
(P) inf I (u)= (u (x)) − 1 dx = m =0 .
0
u∈X
0
© 1 ª
(ii) If X = u ∈ C ([0, 1]) : u (0) = 1,u (1) = 0 ,then
1
½ Z ¾
2
(P) inf I (u)= x (u (x)) dx = m =0 .
0
u∈X
0
Exercise 2.2.7 Show (cf. Poincaré-Wirtinger inequality), using Poincaré in-
equality (cf. Theorem 1.47), that for λ ≥ 0 small enough then
½ Z ¾
1 1 h 2 2 2 i
(P λ ) inf I λ (u)= (u (x)) − λ (u (x)) dx = m λ =0 .
0
u∈X 2 0
Deduce that u ≡ 0 is the unique solution of (P λ )for λ ≥ 0 small enough.
2
2
Exercise 2.2.8 Let f (u, ξ)= u (1 − ξ) and
½ Z 1 ¾
(P) inf I (u)= f (u (x) ,u (x)) dx = m
0
u∈X −1
© 1 ª
where X = u ∈ C ([−1, 1]) : u (−1) = 0,u (1) = 1 . Show that (P) has no
solution in X. Prove, however, that
½
0 if x ∈ [−1, 0]
u (x)=
x if x ∈ (0, 1]
1
is a solution of (P) among all piecewise C functions.