Typically in physics (at least the way I learned mechanics), this is derived using the multi-dimensional divergence theorem on the $2N$-dimensional phase space i.e.
$0=\partial_t \rho + \sum\limits_{i=1}^N \left(\frac{\partial(\rho \dot{q_i})}{q_i} + \frac{\partial(\rho \dot{p_i})}{p_i} \right)$,
and the terms in brackets are simplified using Hamiltonian equation of motion to obtain Liouville's Theorem:
$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \frac{\partial \rho}{\partial t} + \sum\limits_i \left(\frac{\partial \rho}{\partial q_i}\dot{q_i} + \frac{\partial \rho}{\partial p_i}\dot{p_i}\right)=0$.
I am wondering if it is possible to derive this without assuming "physics," i.e. Hamiltonian equation of motion, but instead from maximum (differential) entropy principle.
Suppose $\rho=\rho(\vec{p},\vec{q};t)$ is a phase-space distribution with maximum differential entropy $\mathcal{H}[\rho] = \int -\rho\log{\rho}\, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q}$.
Since the entropy is maximized, we can write:
$\frac{\mathrm{d} \mathcal{H}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\int -\rho\log{\rho} \, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q} = \int\mathrm{d}\vec{p}\,\mathrm{d}\vec{q} \, (-\log\rho - 1) \frac{\mathrm{d}\rho}{\mathrm{d}t} = 0$
However, does this necessarily imply $\frac{\mathrm{d}\rho}{\mathrm{d}t} =0$ ? Or, is there an alternative approach to this problem? One way, perhaps, is to show that the differential entropy $\mathcal{H}[\rho]=\int -\rho\log{\rho}\, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q}$ is (up to a constant) equivalent to Boltzmann's definition of entropy $S = k_B \log\left|\Gamma\right|$ (again, without assuming physics).
(I am basically trying to strip as much physics off from statistical mechanics as possible.)
