Linear Hamiltonian System

Question

Suppose the linear system: $\dot{z} = J \frac{\partial{H}}{\partial{z}} = J S(t) z = A(t) z$, with Hamiltonian $H=H(t,z)=\frac{1}{2} z^T S(t)z$.

How can I prove that:

$$\frac{d}{dt}H(t,\xi(t)) = \frac{\partial}{\partial{t}}H(t, \xi(t))$$

being $\xi(t)$ a solution of the linear system?

Notes: \begin{equation*} J=\left[ \begin{matrix} 0 \ I \\ -I \ 0 \end{matrix} \right] \end{equation*} $A(t) = JS(t)$ are matrices.

Thank you

Answer 1

You need to additionally assume that $S(t)$ is a symmetric matrix for all $t$, and that it has even dimension $2k \times 2k$. Here are some hints (without giving away the whole solution). Let’s first simplify notation (to avoid using the $t$ variable in two different ways, which is confusing to me).

Define:

$$H(u, z) = (1/2)z^T S(u) z$$

Thus: \begin{align} \frac{\partial}{\partial u} H(u,z) &= (1/2)z^TS’(u)z \\ \frac{\partial}{\partial z} H(u,z) &= (1/2)(S(u) + S(u)^T)z = S(u)z \end{align} where the last equality holds by the symmetry assumption $S(u)=S(u)^T$.

You are asked to show that for any solution $\xi(t)$ to your differential equation, we have: $$ \frac{d}{dt} \left[H(t, \xi(t))\right] = \frac{\partial}{\partial u} H(t, \xi(t)) $$ Since we already know $\frac{\partial}{\partial u} H(u,z)$, proving the above equation is equivalent to proving that: $$ \frac{d}{dt} \left[H(t, \xi(t))\right] = (1/2)\xi(t)^TS’(t)\xi(t) $$

To do this, you will need to use the chain rule, and then eventually use properties of the $J$ matrix.

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Recall: The chain rule is as follows: Let $G(u,z)$ be a real-valued function of vectors $u$ and $z$, and let $r(t)$ and $s(t)$ be vector-valued functions of time variable $t$. Then by the chain rule: $$ \frac{d}{dt}\left[G(r(t),s(t))\right] = r’(t)^T\left[\frac{\partial}{\partial u} G(r(t),s(t))\right] + s’(t)^T \left[ \frac{\partial}{\partial z} G(r(t),s(t))\right] $$

Answer 2

I found a property that I think solves the problem without having to use chain rule and is easier:

$\frac{df}{dt}=\frac{\partial f}{\partial t} + \{f, H\}$

where $\{, \}$ is the Poisson Bracket.

$\{f,H\} = \nabla f^T J \nabla H$

As property, $\{F, H\} = 0 $ if and only if F is an integral of $\dot{z}=J\nabla H(t,z)$, and we know that H is an integral of this equation.

This way we have: $\{H,H\} = 0$

$\frac{dH}{dt}=\frac{\partial H}{\partial t} + \{H, H\} = \frac{\partial H}{\partial t}$