Page 197 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

184 Solutions to the Exercises

7.2 Chapter 2: Classical methods

7.2.1 Euler-Lagrange equation

Exercise 2.2.1. The proof is almost identical to that of the theorem. The Euler-
Lagrange equation becomes then a system of ordinary differential equations,
namely, if u =(u 1 , ..., u N ) and ξ =(ξ , ..., ξ ),wehave
1 N
d £ ¤
0 0
(x, u, u ) = f u i (x, u, u ) ,i =1, ..., N.
dx
f ξ i
Exercise 2.2.2. We proceed as in the theorem. We let
n o
n
X = u ∈ C ([a, b]) : u (j) (a)= α j ,u (j) (b)= β , 0 ≤ j ≤ n − 1
j
If u ∈ X ∩ C 2n ([a, b]) is a minimizer of (P) we have I (u + v) ≥ I (u) , ∀ ∈ R
and ∀v ∈ C ∞ (a, b).Letting f = f (x, u, ξ , ..., ξ ) and using the fact that
0 1 n
d
I (u + v)| =0 =0
d
we find
( )
Z n
b ³ ´ X ³ ´
f u x, u, ..., u (n) v + f ξ i x, u, ..., u (n) v (i) dx =0, ∀v ∈ C 0 ∞ (a, b) .
a
i=1
Integrating by parts and appealing to the fundamental lemma of the calculus of
variations (Theorem 1.24) we find
n
X i+1 d i h ³ (n) ´i ³ (n) ´
(−1) f ξ i x, u, ..., u = f u x, u, ..., u .
dx i
i=1
Exercise 2.2.3. (i) Let
© 1 ª
X 0 = v ∈ C ([a, b]) : v (a)= 0 .
2
Let u ∈ X ∩C ([a, b]) be a minimizer for (P), since I (u + v) ≥ I (u) , ∀v ∈ X 0 ,
∀ ∈ R we deduce as above that
Z b
0
0
{f u (x, u, u ) v + f ξ (x, u, u ) v } dx =0, ∀v ∈ X 0 .
0
a
Integrating by parts (bearing in mind that v (a)= 0)we find
b
Z ½∙ ¸ ¾
d
0
f u − f ξ v dx + f ξ (b, u (b) , u (b)) v (b)= 0, ∀v ∈ X 0 .
a dx