Page 127 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

114 Regularity

Proof. We know from Remark 3.12 that the following equation holds

Z b
(E w ) [f u (x, u, u ) v + f ξ (x, u, u ) v ] dx =0, ∀v ∈ C ∞ (a, b) . (4.4)
0
0
0
0
a
We then divide the proof into two steps.
Step 1. Define
ϕ (x)= f ξ (x, u (x) , u (x)) and ψ (x)= f u (x, u (x) , u (x)) .
0
0
0
We easily see that ϕ ∈ W 1,1 (a, b) and that ϕ (x)= ψ (x), for almost every
x ∈ (a, b), which means that
d
0
[f ξ (x, u, u )] = f u (x, u, u ) ,a.e. x ∈ (a, b) . (4.5)
0
dx
Indeed since u ∈ W 1,p (a, b), and hence u ∈ L ∞ (a, b), wededucefrom(H3’) that
1
ψ ∈ L (a, b). We also have from (4.4) that
b b
Z Z
0
ψ (x) v (x) dx = − ϕ (x) v (x) dx, ∀v ∈ C ∞ (a, b) .
0
a a
1
Since ϕ ∈ L (a, b) (from (H3’)), we have by definition of weak derivatives the
claim, namely ϕ ∈ W 1,1 (a, b) and ϕ = ψ a.e..
0
0
Step 2. Since ϕ ∈ W 1,1 (a, b),wehavethat ϕ ∈ C ([a, b]) which means that
there exists a constant α 5 > 0 so that
0
|ϕ (x)| = |f ξ (x, u (x) , u (x))| ≤ α 5 , ∀x ∈ [a, b] . (4.6)
Since u is bounded (and even continuous), let us say |u (x)| ≤ R for every
x ∈ [a, b],wehave from(H1)that

f (x, u, 0) ≥ f (x, u, ξ) − ξf ξ (x, u, ξ) , ∀ (x, u, ξ) ∈ [a, b] × [−R, R] × R .

Combining this inequality with (H2) we find that there exists α 6 ∈ R such that,
for every (x, u, ξ) ∈ [a, b] × [−R, R] × R,

p
ξf ξ (x, u, ξ) ≥ f (x, u, ξ) − f (x, u, 0) ≥ α 1 |ξ| + α 6 .
Using (4.6) and the above inequality we find
0 p
0
0
0
0
0
α 1 |u | + α 6 ≤ u f ξ (x, u, u ) ≤ |u ||f ξ (x, u, u )| ≤ α 5 |u | ,a.e. x ∈ (a, b)
which implies, since p> 1,that |u | is uniformly bounded. Thus the lemma.
0