Page 207 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

194 Solutions to the Exercises

The equation S α = β (where β is a constant) reads as

Z u(x) ds
x − α p = β
g (s) − α 2
u 0
which implies
αu (x)
0
1 − p =0 .
g (u (x)) − α 2
Note that, indeed, any such u = u (x) and
p
v = v (x)= S u (x, u (x) ,α)= g (u (x)) − α 2

satisfy
⎧ √
g(u(x))−α 2
⎪ 0
⎪ u (x)= H v (x, u (x) ,v (x)) =
⎨ α
0
0
⎪ g (u(x))u (x)
⎪ 0
⎩ v (x)= −H u (x, u (x) ,v (x)) = √ .
2 g(u(x))−α 2
Exercise 2.5.3. The Hamiltonian is easily seen to be
v 2
H (u, v)= .
2a (u)
The Hamilton-Jacobi equation and its reduced form are given by
2 ∗ 2 2
(S u ) (S ) α
u
S x + =0 and = .
2a (u) 2a (u) 2
p
0
Therefore, defining A by A (u)= a (u),we find
α 2
∗
S (u, α)= αA (u) and S (x, u, α)= − x + αA (u) .
2
p
Hence, according to Theorem 2.19 (note that S uα = a (u) > 0)the solution is
given implicitly by

S α (x, u (x) ,α)= −αx + A (u (x)) ≡ β = constant .

Since A is invertible we find (compare with Exercise 2.2.10)

u (x)= A −1 (αx + β) .