# Why does not the Hamiltonian depend on the derivative of the state?

I am reading "Optimal Control Theory" from Kirk. When solving the optimal control problem, there is defined The Hamiltonian $H(x(t),u(t),p(t), t)$ as $g(x(t), u(t), t) + p [a(x(t), u(t), t]$ where $x(t)$ is the state, $p(t)$ is the costate, $u(t)$ is the control input, $a(x(t), u(t), t)$ describes the system dynamics $\dot x = a(x(t),u(t),t)$, and $g(x(t), u(t), t)$ is the function whose integral is the performance measure. But what if the performance measure contains the first derivative? For example, when I am paying speed penalty?

I am rather confused, because in the book so far, $\dot x$ has always been a part of $g(x,\dot x, u, t)$.

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Your question would probably benefit from more background. I don't think MO has a lot of OC folks yet. – Steve Huntsman Jun 20 '10 at 4:00

In control theory - maybe I should say control engineering - it is reasonable to assess a penalty to the magnitude of the control input $u$ as well as to the state's deviation from the nominal value 0. Also, this formulation has the merit that it leads to solvable problems - at least in the case of linear systems with quadratic costs. If there is a penalty for the speed, than the problem can be reformulated in a way that includes $\dot{x}$ in an expanded state vector. If $\dot{x}$ can be chosen directly then set $u=\dot{x}$ and recover the formulation you are used to.
As the derivative of the state is a function of the state and the controls $\dot x = a(x, u, t)$ you can rewrite any performance Lagrangian $g(x, \dot x, u, t)$ in the standard form $\tilde g(x(t), u(t), t) = g(x, a(x, u, t), u, t)$.