- #1
Kreizhn
- 743
- 1
Hey All,
Here's a stupid and probably ridiculously easy question, but I want to make sure that I have it right.
Let [itex] G [/itex] be a Lie group with Lie algebra [itex] \mathfrak g [/itex]. Assume that [itex] H \in \mathfrak g [/itex] and [itex] \phi \in \mathfrak g^* [/itex] the algebraic dual. Assume that [itex] X(t) [/itex] is an integral curve satisfying
[tex] \dot X(t) = HX(t) [/tex]
and we have a function defined as [itex] \mathcal H(X,\phi) = \phi(HX(t)) [/itex].
(For anyone familiar with geometric control theory, this is essentially Pontryagin's principle only greatly simplified for non-control theorists)
Now I want to calculate [itex] \frac{d\mathcal H}{dX} [/itex] so my question is as follows: Can we pull the [itex] X(t) [/itex] out of the linear functional since it's only a functional on H?
See, because it's a linear function, there's something that is telling me that differentiating it with respect to X should just give [itex] \phi [/itex]. Something that I can partly corroborate by the fact that we should get
[tex] \frac{d}{dX} \langle \phi, X \rangle = \phi [/tex]
However, if I cannot pull the X outside of the functional, then I end up getting something along the lines of
[tex] \begin{align*}
\frac d{dX} \phi(HX(t)) &= \phi'(HX(t)) H
\end{align*}
[/tex]
and between not knowing what [itex] \phi' [/itex] is and that last statement looking pretty useless, I'm not sure if I've done something wrong.
Edit: Messed up last equation. Fixed it.
Here's a stupid and probably ridiculously easy question, but I want to make sure that I have it right.
Let [itex] G [/itex] be a Lie group with Lie algebra [itex] \mathfrak g [/itex]. Assume that [itex] H \in \mathfrak g [/itex] and [itex] \phi \in \mathfrak g^* [/itex] the algebraic dual. Assume that [itex] X(t) [/itex] is an integral curve satisfying
[tex] \dot X(t) = HX(t) [/tex]
and we have a function defined as [itex] \mathcal H(X,\phi) = \phi(HX(t)) [/itex].
(For anyone familiar with geometric control theory, this is essentially Pontryagin's principle only greatly simplified for non-control theorists)
Now I want to calculate [itex] \frac{d\mathcal H}{dX} [/itex] so my question is as follows: Can we pull the [itex] X(t) [/itex] out of the linear functional since it's only a functional on H?
See, because it's a linear function, there's something that is telling me that differentiating it with respect to X should just give [itex] \phi [/itex]. Something that I can partly corroborate by the fact that we should get
[tex] \frac{d}{dX} \langle \phi, X \rangle = \phi [/tex]
However, if I cannot pull the X outside of the functional, then I end up getting something along the lines of
[tex] \begin{align*}
\frac d{dX} \phi(HX(t)) &= \phi'(HX(t)) H
\end{align*}
[/tex]
and between not knowing what [itex] \phi' [/itex] is and that last statement looking pretty useless, I'm not sure if I've done something wrong.
Edit: Messed up last equation. Fixed it.
Last edited: