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Consider the differential equation $$\begin{pmatrix}\dot p \\ \dot q \end{pmatrix} = \frac{1}{p^2+q^2}\begin{pmatrix} p \\ q \end{pmatrix}$$ where $(p,q)^T\in \mathbb R ^2 - \{0\}$. I want to show that this differential equation is Hamiltonian with Hamiltonian function $H(p,q)=-\operatorname{Im} (\log(p+iq))$ on any simply connected set $U\in \mathbb R^2 -\{0\}$.

I know that on any simply connected complex domain we can define a logarithm. However I am stuck on calculating its derivatives with respect to $p$ and $q$. (This is not homework.)

Edit: We have \begin{align*} -\frac{\partial H}{\partial q}(p,q) &=-\frac{\partial}{\partial q}\left\{ -\operatorname{Im} (\log(p+iq))\right\} \\ &=\operatorname{Im}\frac{\partial \log}{\partial q}(p+iq) \\ &=\operatorname{Im} i \log'(p+iq) \\ &=\operatorname{Im} \frac{i}{p+iq} \\ &=\frac{p}{p^2+q^2}. \end{align*}

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By setting $z(t) = p(t)+i q(t)$ DE can be written as: $$ z'(t) = \frac{1}{p(t)-i q(t)} = \frac{1}{\overline{z(t)}} $$ and I think you can get it from here: $\log(z)$ is a primitive of $\frac{1}{z}$.

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  • $\begingroup$ Thank you for your answer. Unfortunately I don't see how to proceed from your hint. I have updated my question with a calculation. $\endgroup$
    – user228680
    Commented Jun 22, 2016 at 12:11

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