Page 200 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 2: Classical methods 187

Exercise 2.2.6. i) Setting
½
x if x ∈ [0, 1/2]
u 1 (x)=
1 − x if x ∈ (1/2, 1]
we find that I (u 1 )= 0. Observe however that u 1 /∈ X where
© 1 ª
X = u ∈ C ([0, 1]) : u (0) = u (1) = 0 .
Let > 0.Since u 1 ∈ W 1,∞ (0, 1),we can find v ∈ C ∞ (0, 1) (hence in particular
0 0
v ∈ X)suchthat
ku 1 − vk W 1,4 ≤ .
¡ 2 ¢ 2
Note also that since f (ξ)= ξ − 1 we can find K> 0 such that
³ ´
3 3
|f (ξ) − f (n)| ≤ K 1+ |ξ| + |η| |ξ − η| .
Combining the above inequalities with Hölder inequality we get (K denoting a
e
constant not depending on )
1
Z
0
0
0 ≤ m ≤ I (v)= I (v) − I (u 1 ) ≤ |f (v ) − f (u )| dx
1
0
Z
1 n³ ´ o
0 3
0 3
≤ K 1+ |v | + |u | |v − u |
0
0
1 1
0
µZ ¶3/4 µZ 1 ¶1/4
1 ³ ´ 4/3
0 3
0 4
0 3
0
≤ K 1+ |v | + |u | |v − u |
1
1
0 0
e
e
≤ K ku 1 − vk W 1,4 ≤ K .
Since is arbitrary we deduce the result, i.e. m =0.
ii) The argument is analogous to the preceding one and we skip the details.
We first let
© 1 ª
X = v ∈ C ([0, 1]) : v (0) = 1,v (1) = 0
© 1 ª
= v ∈ C ([0, 1]) : v (0) = 1,v (1) = 0
X piec
piec
1
where C 1 stands for the set of piecewise C functions. We have already proved
piec
that
1
½ Z ¾
2
0
(P piec ) inf I (u)= x (u (x)) dx =0
u∈X pie c
0
We need to establish that m =0 where
1
½ Z ¾
2
(P) inf I (u)= x (u (x)) dx = m
0
u∈X
0