Page 198 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 2: Classical methods 185

Using the fundamental lemma of the calculus of variations and the fact that v (b)
is arbitrary we find
⎧
d
⎪ 0 0
⎨ [f ξ (x, u, u )] = f u (x, u, u ) , ∀x ∈ [a, b]
dx
⎪
⎩
f ξ (b, u (b) , u (b)) = 0 .
0
We sometimes say that f ξ (b, u (b) , u (b)) = 0 is a natural boundary condition.
0
(ii) The proof is completely analogous to the preceding one and we find, in
addition to the above conditions, that
f ξ (a, u (a) , u (a)) = 0 .
0
2
Exercise 2.2.4. Let u ∈ X ∩ C ([a, b]) be a minimizer of (P). Recall that
( )
Z b
1
X = u ∈ C ([a, b]) : u (a)= α, u (b)= β, g (x, u (x) ,u (x)) dx =0 .
0
a
We assume that there exists w ∈ C ∞ (a, b) such that
0
Z
b
0
0
[g ξ (x, u (x) , u (x)) w (x)+ g u (x, u (x) , u (x)) w (x)] dx 6=0;
0
a
this is always possible if
d
0
0
[g ξ (x, u, u )] 6= g u (x, u, u ) .
dx
By homogeneity we choose one such w so that
Z b
0
[g ξ (x, u (x) , u (x)) w (x)+ g u (x, u (x) , u (x)) w (x)] dx =1 .
0
0
a
Let v ∈ C ∞ (a, b) be arbitrary, w as above and define for , h ∈ R
0
b
Z
0
0
0
F ( , h)= I (u + v + hw)= f (x, u + v + hw, u + v + hw ) dx
a
Z
b
0
0
G ( , h)= g (x, u + v + hw, u + v + hw ) dx .
0
a
Observe that G (0, 0) = 0 and that by hypothesis
Z b
G h (0, 0) = [g ξ (x, u (x) , u (x)) w (x)+ g u (x, u (x) , u (x)) w (x)] dx =1 .
0
0
0
a