Page 204 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 2: Classical methods 191

7.2.3 Hamiltonian formulation
Exercise 2.4.1. The proof is a mere repetition of that of Theorem 2.10 and
we skip the details. We just state the result. Let u =(u 1 , ..., u N ) and ξ =
¡ ¢
N
(ξ , ..., ξ ). Weassumethat f ∈ C 2 [a, b] × R × R N , f = f (x, u, ξ),and
1 N
that it verifies
³ ´
2 N N
D f (x, u, ξ)= f ξ i ξ j > 0, for every (x, u, ξ) ∈ [a, b] × R × R
1≤i,j≤N
f (x, u, ξ) N
lim =+∞, for every (x, u) ∈ [a, b] × R .
|ξ|→∞ |ξ|
If we let
H (x, u, v)= sup {hv; ξi − f (x, u, ξ)}
ξ∈R N
N
then H ∈ C 2 ¡ [a, b] × R × R N ¢ and, denoting by
(x, u, v))
H v (x, u, v)= (H v 1 (x, u, v) ,..., H v N
and similarly for H u (x, u, v),we also have
H x (x, u, v)= −f x (x, u, H v (x, u, v))

H u (x, u, v)= −f u (x, u, H v (x, u, v))
H (x, u, v)= hv; H v (x, u, v)i − f (x, u, H v (x, u, v))
v = f ξ (x, u, ξ) ⇔ ξ = H v (x, u, v) .

The Euler-Lagrange system is

d £ ¤
0 0
(x, u, u ) = f u i (x, u, u ) ,i =1, ..., N.
f ξ i
dx
and the associated Hamiltonian system is given, for every i =1, ..., N,by
⎧
0
i
⎨ u = H v i (x, u, v)
⎩
0
i
v = −H u i (x, u, v) .
Exercise 2.4.2. i) The Euler-Lagrange equations are, for i =1..., N,
⎧
00 (t, u)
i
⎪ m i x = −U x i
⎪
⎪
⎪
⎨
00
i
m i y = −U y i (t, u)
⎪
⎪
⎪
⎪
⎩
00
i
m i z = −U z i (t, u) .