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Let the (non-canonical) Hamiltonian system be given in the form $$\dot{x}=J(x)DH(x)$$ where $H(x)$ is the Hamiltonian function, $J(x)$ is the skew-symmetric symplectic matrix, and $DH$ denotes the vector of partial derivatives of $H$ (the gradient of $H$). Let $J(x)$ have a nonempty null space, and let a function $C(x)$ satisfy $J(x)DC(x) = 0$ for all $x$. Show that value of $C(x)$ is constant on any trajectory of the system regardless of the choice of $H$.

Here is my proposed solution. I use the fact that $J(x)$ is the skew-symmetric in the fourth equality $\big(J(x)^{-1} = J(x)^T = -J(x)\big)$. The fact that $J(x)DC(x) = 0$ is used in the fifth equality.

\begin{equation} \begin{split} C\dot{(x)} & = \nabla{C}\boldsymbol{\cdot}{x} \\ & = \nabla{C}\boldsymbol{\cdot}J(x)DH(x) \\ & = \Big((J(x))^T(\nabla{C})^T\Big)^T\boldsymbol{\cdot}DH(x) \\ & = \Big(-J(x)DC(x)\Big)^T\boldsymbol{\cdot}DH(x) \\ & = 0\boldsymbol{\cdot}DH(x) \\ & = 0 \end{split} \end{equation}

Therefore, as $C\dot{(x)}=0$, this implies that $C(x)$ is constant along any trajectory.

I'm having trouble justifying the first equality and fourth equality. Is this the right approach to this problem? Is there an alternative approach that would lead to the right answer?

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I think you are thinking on the right line. I have the following solution.

Write, along the solution trajectories of the system

$\begin{align} \frac{d}{dt} H(x(t)) &= \frac{\partial H}{\partial x} \dot{x} \nonumber \\ &= (DH(x))^T J(x) DH(x) \qquad (1)\\ &= DH(x)^T (-J(x))^T DH(x) \quad (2)\\ \text{Adding equations (1) and (2), we get,}\\ 2 \frac{d}{dt} H(x(t)) &= 0\\ \Rightarrow \frac{d}{dt} H(x(t)) &= 0 \end{align}$

Hence, you can write $H(x(t)) = c$, for a constant $c$. So, $H(x)$ is constant along the trajectory of the system.

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  • $\begingroup$ Which two equations did you add to get $2 \frac{d}{dt} H(x(t)) = 0$? Can you please show this inside your answer? $\endgroup$ – Axion004 May 19 at 17:07
  • $\begingroup$ @Axion004 Edited my answer. $\endgroup$ – wootman May 21 at 7:28
  • $\begingroup$ To confirm, you added the two equations because J is skew-symmetric which implies that $J^T=-J$ or $(-J)^T=J$. That is why when you add the two equations together, you get $2\frac{d}{dt}H(x(t))$? $\endgroup$ – Axion004 May 21 at 23:00
  • $\begingroup$ Yes, that's correct. $\endgroup$ – wootman May 22 at 1:30
  • $\begingroup$ I see, then why is $2\frac{d}{dt}H(x(t))=0$? If the Hamiltonian system was canonical, then one could write $\dfrac{dH}{dt}=\dfrac{\partial H}{\partial q}\dfrac{dq}{dt}+ \dfrac{\partial H}{\partial p}\dfrac{dp}{dt} =0$ since $H(q(t),p(t))$. Since the problem assumes a non-canonical system, why does adding $(1)+(2)=0$ on the RHS? $\endgroup$ – Axion004 May 22 at 3:42

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