Page 77 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

64 Classical methods

2
Step 3. The inverse function theorem, the fact that f ∈ C and the inequality
1
(2.4) imply that ξ ∈ C . But, as an exercise, we will establish this fact again.
First let us prove that ξ is continuous (in fact locally Lipschitz). Let R> 0 be
fixed. From (2.5) we deduce that we can find R 1 > 0 so that

|ξ (x, u, v)| ≤ R 1 ,for every x ∈ [a, b] , |u| , |v| ≤ R.
1
Since f ξ is C ,we can find γ > 0 so that
1
¯ ¡ ¢¯ ¡ ¯ ¯¢
f ξ (x, u, ξ) − f ξ x ,u ,ξ ≤ γ 1 |x − x | + |u − u | + ξ − ξ (2.12)
¯ 0 0 0 ¯ 0 0 ¯ 0 ¯
¯ ¯
0
0
for every x, x ∈ [a, b] , |u| , |u | ≤ R, |ξ| , ξ ≤ R 1 .
¯ 0 ¯
From (2.4), we find that there exists γ > 0 so that
2
f ξξ (x, u, ξ) ≥ γ , for every x ∈ [a, b] , |u| ≤ R, |ξ| ≤ R 1
2
¯ ¯
and we thus have, for every x ∈ [a, b] , |u| ≤ R, |ξ| , ξ ≤ R 1 ,
¯ 0 ¯
¯ ¡ ¢¯ ¯ ¯
f ξ (x, u, ξ) − f ξ x, u, ξ ≥ γ ξ − ξ . (2.13)
¯ 0 ¯ ¯ 0 ¯
2
Let x, x ∈ [a, b] , |u| , |u | ≤ R, |v| , |v | ≤ R.By definition of ξ we have
0
0
0
f ξ (x, u, ξ (x, u, v)) = v
0
0
0
0
0
0
f ξ (x ,u ,ξ (x ,u ,v )) = v ,
which leads to
0
0
f ξ (x, u, ξ (x ,u ,v )) − f ξ (x, u, ξ (x, u, v))
0
0
0
0
0
0
0
0
0
0
= f ξ (x, u, ξ (x ,u ,v )) − f ξ (x ,u ,ξ (x ,u ,v )) + v − v.
Combining this identity with (2.12) and (2.13) we get
0 0 0 0 0 0
γ |ξ (x, u, v) − ξ (x ,u ,v )| ≤ γ (|x − x | + |u − u |)+ |v − v |
2
1
which, indeed, establishes the continuity of ξ.
1
We now show that ξ is in fact C . From the equation v = f ξ (x, u, ξ) we
deduce that ⎧
⎪ f xξ (x, u, ξ)+ f ξξ (x, u, ξ) ξ =0
x
⎪
⎪
⎪
⎨
f uξ (x, u, ξ)+ f ξξ (x, u, ξ) ξ =0
u
⎪
⎪
⎪
⎪
⎩
f ξξ (x, u, ξ) ξ =1 .
v
2
1
Since (2.4) holds and f ∈ C ,wededucethat ξ ∈ C ([a, b] × R × R).