Page 83 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

70 Classical methods

Proof. Step 1. We differentiate (2.16) to get

0
v (x)= S xu (x, u (x)) + u (x) S uu (x, u (x)) , ∀x ∈ [a, b] .
0
Differentiating (2.14) with respect to u we find, for every (x, u) ∈ [a, b] × R,
S xu (x, u)+ H u (x, u, S u (x, u)) + H v (x, u, S u (x, u)) S uu (x, u)= 0 .
Combining the two identities (the second one evaluated at u = u (x)) and (2.15)
with the definition of v,wehave

0
v (x)= −H u (x, u (x) ,S u (x, u (x))) = −H u (x, u (x) ,v (x))
as wished.
Step 2. Since S is a solution of the Hamilton-Jacobi equation, we have, for
every (x, u, α) ∈ [a, b] × R × R,

d
[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)= 0 .
Since this identity is valid for every u, it is also valid for u = u (x) satisfying
(2.15) and thus

0
S xα (x, u (x) ,α)+ u (x) S uα (x, u (x) ,α)= 0 .
This last identity can be rewritten as

d
[S α (x, u (x) ,α)] = 0
dx
which is the claim.
The above theorem admits a converse.
1
Theorem 2.19 (Jacobi Theorem).Let H ∈ C ([a, b] × R × R), S = S (x, u, α)
2
be C ([a, b] × R × R) and solving (2.14), for every (x, u, α) ∈ [a, b]×R×R,with
S uα (x, u, α) 6=0, ∀ (x, u, α) ∈ [a, b] × R × R .
If u = u (x) satisfies

d
[S α (x, u (x) ,α)] = 0, ∀ (x, α) ∈ [a, b] × R (2.19)
dx
then u necessarily verifies

u (x)= H v (x, u (x) ,S u (x, u (x) ,α)) , ∀ (x, α) ∈ [a, b] × R .
0
1
1
Thus if v (x)= S u (x, u (x) ,α),then (u, v) ∈ C ([a, b]) × C ([a, b]) is a solution
of (2.17).