Page 78 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Hamiltonian formulation 65

Step 4. We therefore have that the functions

(x, u, v) → ξ (x, u, v) ,f x (x, u, ξ (x, u, v)) ,f u (x, u, ξ (x, u, v))
1
2
are C . We then immediately obtain (2.7), (2.8), and thus H ∈ C . Indeed we
have, differentiating (2.11),
⎧
⎪ H x = vξ − f x − f ξ ξ =(v − f ξ ) ξ − f x = −f x
x
x
x
⎪
⎪
⎪
⎨
H u = vξ − f u − f ξ ξ =(v − f ξ ) ξ − f u = −f u
u
u
u
⎪
⎪
⎪
⎪
⎩
H v = ξ + vξ − f ξ ξ =(v − f ξ ) ξ + ξ = ξ
v v v
and in particular
ξ = H v (x, u, v) .
This achieves the proof of the lemma.
The main theorem of the present section is:
2
Theorem 2.10 Let f and H be as in the above lemma. Let (u, v) ∈ C ([a, b])×
2
C ([a, b]) satisfy for every x ∈ [a, b]
⎧
u (x)= H v (x, u (x) ,v (x))
0
⎨
(H)
⎩ 0
v (x)= −H u (x, u (x) ,v (x)) .
Then u verifies
d
0
0
(E) [f ξ (x, u (x) ,u (x))] = f u (x, u (x) ,u (x)) , ∀x ∈ [a, b] .
dx
2
Conversely if u ∈ C ([a, b]) satisfies (E) then (u, v) are solutions of (H) where
0
v (x)= f ξ (x, u (x) ,u (x)) , ∀x ∈ [a, b] .
Remark 2.11 The same remarks as in the lemma apply also to the theorem.
Proof. Part 1.Let (u, v) satisfy (H). Using (2.10) and (2.8) we get
0
u = H v (x, u, v) ⇔ v = f ξ (x, u, u )
0
v = −H u (x, u, v)= f u (x, u, u )
0
0
and thus u satisfies (E).
Part 2. Conversely by (2.10) and since v = f ξ (x, u, u ) we get the first
0
equation
u = H v (x, u, v) .
0