Page 202 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 2: Classical methods 189

Since any element in X piec can be approximated arbitrarily closed by an
element of X (seetheargumentofExercise2.2.6)wededucethat m =0 in
1
½ Z ¾
(P) inf I (u)= f (u (x) ,u (x)) dx = m.
0
u∈X
−1
1
To conclude it is sufficient to observe that if for a certain u ∈ C ([−1, 1]) we have
I (u)= 0, then either u ≡ 0 or u ≡ 1, and both possibilities are incompatible
0
with the boundary data.
Another possibility of showing that m =0 is to consider the sequence
⎧
0 if x ∈ [−1, 0]
⎪
⎪
⎪
⎪
⎨
¡ 1 ¤
2 3
u n (x)= −n x +2nx 2 if x ∈ 0,
n
⎪
⎪
⎪
⎪ ¡ ¤
x if x ∈ , 1
⎩ 1
n
and observe that u n ∈ X and that
1/n
Z
0
I (u n )= f (u n (x) ,u (x)) dx → 0.
n
0
This proves that m =0, as wished.
Exercise 2.2.9. Note firstthatbyJenseninequality m ≥ 1,where
½ Z 1 ¾
0
(P) inf I (u)= |u (x)| dx = m
u∈X 0
© ª
1
and X = u ∈ C ([0, 1]) : u (0) = 0,u (1) = 1 .Let n ≥ 1 be an integer and
n
observe that u n defined by u n (x)= x belongs to X and satisfies I (u n )= 1.
Therefore u n is a solution of (P) for every n.In fact any u ∈ X with u ≥ 0 in
0
[0, 1] is a minimizer of (P).
Exercise 2.2.10. Set v (x)= A (u (x)). We then have, using Jensen inequality,
Z b
p
I (u)= a (u (x)) |u (x)| dx
0
a
Z Z b
b ¯ ¯ p
1 p
¯
0
= ¯(a (u (x))) u (x)¯ dx = |v (x)| dx
0
¯
p
a a
¯ ¯ p
Z ¯ ¯ p
¯ 1 0 ¯ ¯ v (b) − v (a) ¯
¯ b ¯
≥ (b − a) ¯ v (x) dx¯ =(b − a) ¯ ¯
¯b − a ¯ b − a ¯
a ¯
and hence
¯ ¯ p
¯ A (β) − A (α) ¯
I (u) ≥ (b − a) ¯ ¯ .
b − a
¯ ¯