Page 219 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

206 Solutions to the Exercises

We have to prove that we can find a constant α> 0 so that

0
|u (x + h) − u (x)| ≤ α |h| , a.e. x ∈ ω
0
c
for every open set ω ⊂ ω ⊂ (a, b) and every h ∈ R satisfying |h| < dist (ω, (a, b) ).
Using (7.17) we have
0 0
γ |u (x + h) − u (x)|
1
¯ ¯
Z 0
¯ u (x+h) ¯
¯ ¯
≤ ¯ f ξξ (x, u (x) ,z) dz¯
u (x)
¯ 0 ¯
0 0
≤ |f ξ (x, u (x) , u (x + h)) − f ξ (x, u (x) , u (x))|
0 0
≤ |f ξ (x + h, u (x + h) , u (x + h)) − f ξ (x, u (x) , u (x))|
0
0
+ |f ξ (x, u (x) , u (x + h)) − f ξ (x + h, u (x + h) , u (x + h))| .
Now let us evaluate both terms in the right hand side of the inequality. Since
we know from Lemma 4.2 that x → f u (x, u (x) , u (x)) is in L ∞ (a, b) and the
0
Euler-Lagrange equation holds we deduce that x → ϕ (x)= f ξ (x, u (x) , u (x))
0
is in W 1,∞ (a, b). Therefore applying Theorem 1.36 to ϕ,wecan find a constant
γ > 0, such that
2
0 0
|ϕ (x + h) − ϕ (x)| = |f ξ (x + h, u (x + h) , u (x + h)) − f ξ (x, u (x) , u (x))|
≤ γ |h| .
2
Similarly since u ∈ W 1,∞ and f ∈ C ,we can find constant γ ,γ > 0,such
∞
4
3
that
0 0
|f ξ (x, u (x) , u (x + h)) − f ξ (x + h, u (x + h) , u (x + h))|
≤ γ (|h| + |u (x + h) − u (x)|) ≤ γ |h| .
3 4
Combining these two inequalities we find
γ + γ 4
2
0 0
|u (x + h) − u (x)| ≤ |h|
γ 1
as wished; thus u ∈ W 2,∞ (a, b).
(ii) Since u ∈ W 2,∞ (a, b), and the Euler-Lagrange equation holds, we get
that, for almost every x ∈ (a, b),
d
0
00
0
0
[f ξ (x, u, u )] = f ξξ (x, u, u ) u + f uξ (x, u, u ) u + f xξ (x, u, u )
0
0
dx
0
= f u (x, u, u ) .