Page 203 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

190 Solutions to the Exercises

Setting

A (β) − A (α) −1
v (x)= (x − a)+ A (α) and u (x)= A (v (x))
b − a
we have from the preceding inequality

¯ ¯ p Z b
¯ A (β) − A (α) ¯ p
0
I (u) ≥ (b − a) ¯ ¯ = |v (x)| dx = I (u)
¯ b − a ¯ a
as claimed.

7.2.2 Second form of the Euler-Lagrange equation
¡ ¢
and start by the simple observation
Exercise 2.3.1. Write f ξ = f ξ 1 , ..., f ξ N
¡
¢
that for any u ∈ C 2 [a, b]; R N
d
0
[f (x, u, u ) − hu ; f ξ (x, u, u )i]
0
0
dx
N ∙ ¸
X d £ ¤
0
= f x (x, u, u )+ u 0 (x, u, u ) − (x, u, u ) .
0
0
i f u i dx f ξ i
i=1
Since the Euler-Lagrange system (see Exercise 2.2.1) is given by
d £ ¤
0 0
(x, u, u ) = f u i (x, u, u ) ,i =1, ..., N
f ξ i
dx
we obtain
d
0
0
0
0
[f (x, u, u ) − hu ; f ξ (x, u, u )i]= f x (x, u, u ) .
dx
Exercise 2.3.2. The second form of the Euler-Lagrange equation is
∙ ¸
d d 1 2
0= [f (u (x) ,u (x)) − u (x) f ξ (u (x) ,u (x))] = −u (x) − (u (x))
0
0
0
0
dx dx 2
= −u (x) − u (x) u (x)= −u (x)[u (x)+ 1] ,
0
0
00
00
0
and it is satisfied by u ≡ 1.However u ≡ 1 does not verify the Euler-Lagrange
equation, which is in the present case
00
u (x)= −1 .