Page 79 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

66 Classical methods

0
Moreover since v = f ξ (x, u, u ) and u satisfies (E), we have
d d
0
0
0
v = [v]= [f ξ (x, u, u )] = f u (x, u, u ) .
dx dx
The second equation follows then from the combination of the above identity
and (2.8).
Example 2.12 The present example is motivated by classical mechanics. Let
2
1
m> 0, g ∈ C ([a, b]) and f (x, u, ξ)= (m/2) ξ − g (x) u. The integral under
consideration is
Z b
I (u)= f (x, u (x) ,u (x)) dx
0
a
and the associated Euler-Lagrange equation is
00
mu (x)= −g (x) ,x ∈ (a, b) .
The Hamiltonian is then
v 2
H (x, u, v)= + g (x) u
2m
while the associated Hamiltonian system is

⎧
⎨ u (x)= v (x) /m
0
v (x)= −g (x) .
⎩
0
Example 2.13 We now generalize the preceding example. Let p> 1 and p =
0
p/ (p − 1),
1 p 1 p 0
f (x, u, ξ)= |ξ| − g (x, u) and H (x, u, v)= |v| + g (x, u) .
p p 0

The Euler-Lagrange equation and the associated Hamiltonian system are
d h 0 p−2 i
|u | u 0 = −g u (x, u)
dx
and
⎧ p −2
0
0
⎨ u = |v| v
⎩
v = −g u (x, u) .
0