Page 62 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Euler-Lagrange equation 49

1
identity holds for every v ∈ C ([a, b]) with v (a)= v (b)= 0
Z b ∙ ¸
d
0
0
− [f ξ (x, u (x) , u (x))] + f u (x, u (x) , u (x)) v (x) dx =0 .
a dx
Applying the fundamental lemma of the calculus of variations (Theorem 1.24)
we have indeed obtained the Euler-Lagrange equation (E).
Part 2.Let u be a solution of (E) with u (a)= α, u (b)= β.Since (u, ξ) →
f (x, u, ξ) is convex for every x ∈ [a, b],we get from Theorem1.50that
f (x, u, u ) ≥ f (x, u, u )+ f u (x, u, u )(u − u)+ f ξ (x, u, u )(u − u )
0
0
0
0
0
0
for every u ∈ X. Integrating the above inequality we get
I (u) ≥ I (u)
Z b
0
+ [f u (x, u, u )(u − u)+ f ξ (x, u, u )(u − u )] dx .
0
0
0
a
Integrating by parts the second term in the integral, bearing in mind that u (a)−
u (a)= u (b) − u (b)= 0,we get
I (u) ≥ I (u)
Z b ∙ d ¸
0
+ f u (x, u, u ) − f ξ (x, u, u ) (u − u) dx .
0
a dx
Using (E) we get indeed that I (u) ≥ I (u), which is the claimed result.
Part 3.Let u, v ∈ X be two solutions of (P) (recall that m denote the value
of the minimum) and let us show that they are necessarily equal. Define
1 1
w = u + v
2 2
and observe that w ∈ X. Appealing to the convexity of (u, ξ) → f (x, u, ξ),we
obtain
µ ¶
1 1 1 1 1 1
0
0
0
0
f (x, u, u )+ f (x, v, v ) ≥ f x, u + v, u + v 0 = f (x, w, w )
2 2 2 2 2 2
and hence
1 1
m = I (u)+ I (v) ≥ I (w) ≥ m.
2 2
We therefore get
b
Z ∙ µ ¶¸
1 1 1 1 1 1
0
0
0
f (x, u, u )+ f (x, v, v ) − f x, u + v, u + v 0 dx =0 .
2 2 2 2 2 2
a