Page 205 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

192 Solutions to the Exercises

y
z
x
In terms of the Hamiltonian, if we let u i =(x i ,y i ,z i ) ,ξ =(ξ ,ξ ,ξ ) and
i
i
i
i
y
x
z
v i =(v ,v ,v ),for i =1..., N,we find
i i i
( )
N ∙ ¸
X 1 2
H (t, u, v)= sup hv i ; ξ i − m i |ξ | + U (t, u)
i i
ξ∈R 3N 2
i=1
N 2
X |v i |
= + U (t, u) .
2m i
i=1
ii) Note that along the trajectories we have v i = m i u , i.e.
0
i
y
x
z
0
v = m i x ,v = m i y ,v = m i z i 0
0
i
i
i
i
i
and hence
N
1 X
0
H (t, u, v)= m i |u | + U (t, u) .
i
2
i=1
Exercise 2.4.3. Although the hypotheses of Theorem 2.10 are not satisfied in
the present context; the procedure is exactly the same and leads to the following
analysis. The Hamiltonian is
⎧ p 2
⎨ − g (x, u) − v 2 if v ≤ g (x, u)
H (x, u, v)=
⎩
+∞ otherwise.
2
We therefore have, provided v <g (x, u),that
⎧ v
0
⎪ u = H v = p
⎪ 2
⎪ g (x, u) − v
⎨
⎪ 1
⎪ g u
⎩ v = −H u = p .
0
⎪
2 g (x, u) − v 2
0
We hence obtain that 2vv = g u u and thus
0
£ 2 ¤ 0
v (x) − g (x, u (x)) + g x (x, u (x)) = 0.
If g (x, u)= g (u),weget (c being a constant)
2
v (x)= c + g (u (x)) .