- #1
Kreizhn
- 743
- 1
Homework Statement
Consider a smooth manifold M, and smooth functions [itex] H_i: T^*M \to \mathbb R, i=0,1[/itex] on the cotangent bundle. Further, let [itex]z:[t_1,t_2] \subset \mathbb R \to M [/itex] define a trajectory on [itex] T^*M [/itex] by
[tex] \frac{dz}{dt} = \vec H_0(z(t)) + u(t) \vec H_1(z(t)) [/tex]
where [itex] \vec H_i [/itex] is the Hamiltonian vector field corresponding to [itex] H_1 [/itex] That is, [itex] \vec H_i [/itex] is the unique vector on [itex] T(T^*M) [/itex] satisfying
[tex] \iota_{\vec H_i}\omega = dH_i [/itex]
where [itex] \iota [/itex] is the interior product and [itex] \omega [/itex] is the canonical symplectic form on [itex] T^*M [/itex].
If we know that [itex] \left. H_1 (z(t)) = 0 [/itex] show that
[tex] dH_1(z(t)) \vec H_0(z(t)) + u(t) dH_1(z(t)) \vec H_1(z(t)) = 0 [/tex]
The Attempt at a Solution
Firstly, I am confused with the notation. In particular the statement
[tex] dH_1(z(t)) \vec H_0(z(t)) [/tex]
I think that it should be interpreted as the covector field at a fixed instance z(t) applied to the vector field at z(t); that is, in perhaps less ambiguous notation
[tex] (dH_1)_{z(t)}(\vec H_0(z(t))) [/tex]
The next thing that confuses me is how to apply the chain rule to this vector field. I've tried expressing [tex] \vec H_1 [/itex] in a canonical Darboux coordinate system and then differentiating, but it doesn't seem to lead anywhere. Any input would be appreciated.