Page 206 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 2: Classical methods 193

7.2.4 Hamilton-Jacobi equation
Exercise 2.5.1. We state without proofs the results (they are similar to the
case N =1 and we refer to Gelfand-Fomin [46] page 88, if necessary). Let H ∈
¡ ¢
N
C 1 [a, b] × R × R N , H = H (x, u, v) and u =(u 1 , ..., u N ). The Hamilton-
Jacobi equation is
N
N
S x + H (x, u, S u )= 0, ∀ (x, u, α) ∈ [a, b] × R × R ,
). Jacobi Theorem reads then as
, ..., S u N
where S = S (x, u, α) and S u =(S u 1 ¢
¡
N
follows. Let S ∈ C 2 [a, b] × R × R N be a solution of the Hamilton-Jacobi
equation and
N
N
det (S uα (x, u, α)) 6=0, ∀ (x, u, α) ∈ [a, b] × R × R ,
¡ ¢ ¡ ¢
2 1 N
where S uα = ∂ S/∂α i ∂u j .If u ∈ C [a, b]; R satisfies
1≤i,j≤N
d N
(x, u (x) ,α)] = 0, ∀ (x, α) ∈ [a, b] × R ,i =1, ..., N
[S α i
dx
and if v (x)= S u (x, u (x) ,α) then
⎧
u (x)= H v (x, u (x) ,v (x))
0
⎨
v (x)= −H u (x, u (x) ,v (x)) .
⎩
0
Exercise 2.5.2. The procedure is formal because the hypotheses of Theorem
2.19 are not satisfied. WehaveseeninExercise2.4.3 that
⎧ p
2
⎨ − g (u) − v 2 if v ≤ g (u)
H (u, v)=
⎩
+∞ otherwise.
The Hamilton-Jacobi equation (it is called in this context: eikonal equation)is
then
p 2 2
2
S x − g (u) − S =0 ⇔ S + S = g (u) .
x
u
u
∗ 2
2
Its reduced form is then, for α> 0, g (u) − (S ) = α and this leads to
u
Z u
p
2
S (u, α)= g (s) − α ds .
∗
u 0
We therefore get
Z u
p
2
S (x, u, α)= αx + g (s) − α ds .
u 0