Page 74 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Hamiltonian formulation 61

Denoting by g ( ) the last integrand, we get
∙ ¸
η
0 y 0
g ( )= η f + f x η − 2 u f ξ η y

y
η y
which leads to
g (0) = λ [−f x ϕ +(u f ξ − f) ϕ ] .
0
0
0

Since by hypothesis u is a minimizer of (P) and u ∈ X we have I (u ) ≥ I (u)
and hence
¯ Z b
d ¯
0
0= I (u ) = λ {−f x (x, u (x) ,u (x)) ϕ (x)
¯
d ¯
=0 a
0
0
0
0
+[u (x) f ξ (x, u (x) ,u (x)) − f (x, u (x) ,u (x))] ϕ (x)} dx
b
Z
0
= λ {−f x (x, u (x) ,u (x))
a
¾
d
+ [−u (x) f ξ (x, u (x) ,u (x)) + f (x, u (x) ,u (x))] ϕ (x) dx .
0
0
0
dx
Appealing, once more, to Theorem 1.24 we have indeed obtained the claim.
2.3.1 Exercises
N
Exercise 2.3.1 Generalize Theorem 2.7 to the case where u :[a, b] → R ,
N ≥ 1.
Exercise 2.3.2 Let
1 2
f (x, u, ξ)= f (u, ξ)= ξ − u.
2
Show that u ≡ 1 is a solution of (2.3), but not of the Euler-Lagrange equation
(E).
2.4 Hamiltonian formulation
Recall that we are considering functions f :[a, b] × R × R → R, f = f (x, u, ξ),
and
Z
b
I (u)= f (x, u (x) ,u (x)) dx .
0
a
The Euler-Lagrange equation is
d
0
0
(E) [f ξ (x, u, u )] = f u (x, u, u ) ,x ∈ [a, b] .
dx