Page 84 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Hamilton-Jacobi equation 71

Proof. Since (2.19) holds we have, for every (x, α) ∈ [a, b] × R,

d
0
0= [S α (x, u (x) ,α)] = S xα (x, u (x) ,α)+ S uα (x, u (x) ,α) u (x) .
dx
From (2.14) we obtain, for every (x, u, α) ∈ [a, b] × R × R,
d
0= [S x (x, u, α)+ H (x, u, S u (x, u, α))]
dα
= S xα (x, u, α)+ H v (x, u, S u (x, u, α)) S uα (x, u, α) .
Combining the two identities (the second one evaluated at u = u (x)), with the
hypothesis S uα (x, u, α) 6=0,weget
u (x)= H v (x, u (x) ,S u (x, u (x) ,α)) , ∀ (x, α) ∈ [a, b] × R
0
as wished. We still need to prove that v = −H u .Differentiating v we have, for
0
every (x, α) ∈ [a, b] × R,
0
v (x)= S xu (x, u (x) ,α)+ u (x) S uu (x, u (x) ,α)
0
= S xu (x, u (x) ,α)+ H v (x, u (x) ,S u (x, u (x) ,α)) S uu (x, u (x) ,α) .
Appealing, once more, to (2.14) we obtain, for every (x, u, α) ∈ [a, b] × R × R,

d
0= [S x (x, u, α)+ H (x, u, S u (x, u, α))]
du
= S xu (x, u, α)+ H u (x, u, S u (x, u, α)) + H v (x, u, S u (x, u, α)) S uu (x, u, α) .
Combining the two identities (the second one evaluated at u = u (x))we infer
the result, namely

0
v (x)= −H u (x, u (x) ,S u (x, u (x) ,α)) = −H u (x, u (x) ,v (x)) .
This achieves the proof of the theorem.
1
Example 2.20 Let g ∈ C (R) with g (u) ≥ g 0 > 0.Let
1 2
H (u, v)= v − g (u)
2
be the Hamiltonian associated to
1 2
f (u, ξ)= ξ + g (u) .
2
The Hamilton-Jacobi equation and its reduced form are given by
1 2 1 ∗ 2
S x + (S u ) − g (u)= 0 and (S ) = g (u) .
u
2 2