# Liouville's theorem and the Wronskian

Liouville's theorem states that under the action of the equations of motion, the phase volume is conserved. The equations of motion are the flow ODE's generated by a Hamiltonian field $X$ and the solutions to Hamilton's equations are the integral curves of the system. The effect of a symplectomorphism $\phi_t$ is that it would take $(p^i,q_i) \to (p^j,q_j)$ or $\omega \to \omega'$ and that would be the contribution of the Jacobian of the transformation. If the flow $\phi_t$ changes variables from time, say, $t$ to time $0$. The Jacobian would be $$J(t) = \left| \begin{array}{ccc} \frac{\partial p_1(t) }{\partial p_1(0) } & \ldots & \frac{\partial p_1(t) }{\partial q_{n}(0) } \\ \vdots & \ddots & \vdots \\ \frac{\partial p_{n}(t) }{\partial q_1(0) } &\ldots & \frac{\partial q_{n}(t) }{\partial q_{n}(0) } \ \end{array} \right|$$ Now, the derivatives appearing in the Jacobian satisfy a linear system of ordinary differential equations and moreover, viewed as a system of ordinary differential equations, it is easy to see that the Jacobian we need is also the Wronskian $W$ for the system of differential equations. My first question is why is this true? I know that the definition of the Wronskian of two differential functions $f,g$ is $W(f,g)=f'g-g'f$. Then I read that Wroskian's of linear equations satisfy the vector ODE \item Wroskian's of linear equations satisfy the vector ODE \begin{equation*} \frac{d}{dt}{\bf{y}} = {\bf{M}}(t) {\bf{y}} \end{equation*} What exactly is this equation?

Then I read that \begin{equation*} W(t) = W(0) \exp\left( \int_0^t dt \, \text{Tr}{\bf{M}}(0) \right) \end{equation*} and for our case this trace vanishes, thus $J(0)=J(t)=1$ which shows that the contribution of the Jacobian over volume integrals is trivial.

Can you help me to understand the above statements? Unfortunately I have lost the reference out of which I read those.

### Equations of Motion

What you are describing is Hamiltonian's view of the evolution of a dynamical system. $\mathbf q$ is the vector describing system's configuration with generalized coordinates (or degrees of freedom) in some abstract configuration space isomorphic to $\mathbb R^{s}$. For instance $s=3n$ particle coordinates for a system of $n$ free particles in Euclidean space. $\mathbf p$ is the vector of generalized momentum attached to instanteanous time rate of the configuration vector. For example for particles of mass $m$ in non-relavistic mechanics $\mathbf p= m \frac{d}{dt}\mathbf q$. Classical mechanics postulates that the simulteanous knowledge of both $\mathbf q(t)$ and $\mathbf p(t)$ at time $t$ is required to predict the system's temporal evolution for any time $t'>t$ (causality). So the complete dynamical state of the system is in fact described by the phase $\mathbf y =(\mathbf p, \mathbf q)$ which evolves in an abstract space called the phase space isomorphic to $\mathbb R^{2s}$. The fact that the knowledge of this phase is sufficient to predict the evolution demands on a mathematical point of view configuration $\mathbf q$ to evolve according to a system of $s$ ODEs of at most second order in time (known as Lagrange equations) or equivalently that phase evolves according to a system of $2s$ ODEs of the first order in time knwown as Hamilton's equations of motion,

$\frac{d\mathbf p}{dt}=-\frac{\partial H}{\partial \mathbf q}(\mathbf y,t)$

$\frac{d\mathbf q}{dt}=\frac{\partial H}{\partial \mathbf p}(\mathbf y,t)$

given some Hamilton's function $H(\mathbf y,t)$ describing the system. Physically it represents mechanical energy. For non-dissipative systems it does not depend over time explicitely. as a consequence of Liouville's theorem it is a conserved quantity along phase-space curves.

More generally, equations of motions appear to be formulated as a system of Euler-Cauchy's ODEs

$\frac{d\mathbf y}{dt} = \mathbf f(\mathbf y,t)$

\with $\mathbf f$ some vector-function over $\mathbb R^{2s}\times\mathbb R$ In accordance with Hamiltonian formalism, $\mathbf f$ appears to be $\mathbf f= (-\frac{\partial H}{\partial \mathbf q}, \frac{\partial H}{\partial \mathbf p})$

### Liouville's theorem

Now consider the phase flow $\mathbf y_t$, that is consider the one-parameter ($t\in \mathbb R$) group of transformations $\mathbf y_0\mapsto \mathbf y_t(\mathbf y_0)$ mapping initial phase at time $t=0$ to current one in phase space. This is another parametrization of the phase using $y_0$ as curvilinear coordinates. Suppose you have now hypothetical system replica with initial phase points distributed to fill some volume $\mathcal V_0$ in phase space. Liouville's theorem says that the cloud of points will evolve such as preserving its density along their curves in phase space, like an incompressible fluid flow, keeping the filled volume unchanged. Since

$\mathcal V(t)=\int_{\mathcal V_0} \det \frac{\partial \mathbf y}{\partial \mathbf y_0}(\mathbf y_0,t)\ \underline{dy_0} = \int_{\mathcal V_0} J(\mathbf y_0,t)\ \underline{dy_0}$

Now compute the volume-change time rate at any instant $t$ introducing Euler's form (1),

$\frac{d\mathcal V}{dt}=\int_{\mathcal V} \frac{\partial \mathbf f}{\partial t}+ \nabla_{\mathbf y_0} .\mathbf f(\mathbf y_0,t)\ \underline{dy_0}$. Setting this time rate to zero gives Liouville's theorem in local form:

$\frac{\partial \mathbf f}{\partial t}+ \nabla_{\mathbf y_0} .\mathbf f(\mathbf y_0,t)=0$.

Applying them to Hamilton's form,

$\frac{\partial }{\partial t}[\frac{\partial H}{\partial \mathbf p}-\frac{\partial H}{\partial \mathbf q}] + \frac{\partial}{\partial \mathbf q}\frac{\partial H}{\partial \mathbf p} - \frac{\partial}{\partial \mathbf p}\frac{\partial H}{\partial \mathbf q}=0$ (1)

For conservative systems in energy, $H(\mathbf y,t)$ does not depend upon time explicitely and the left-most term in (1) vanishes. Liouville's theorem can be generalized to any physical observable depending upon the phase of the system $A(\mathbf y,t)$ which is a conserved along the curves of the phase space,

$\frac{dA}{dt}= \frac{\partial A}{\partial t} + \frac{\partial A}{\partial \mathbf q}\frac{\partial H}{\partial \mathbf p} - \frac{\partial A}{\partial \mathbf p}\frac{\partial H}{\partial \mathbf q}=0$

### The Wronskian

Now consider the situation of Euler's ODE can be linearized around phase $\mathbf y_0$. The vectorial function $\mathbf f$ now expresses as a matrix vector product with the phase. You have now a $2s\times 2s$ square linear system of ODEs.

$\frac{d\mathbf y}{dt} = {\mathbf f}(\mathbf y_0,0)+M_{\mathbf y_0}(t)(\mathbf y-\mathbf y_0)+\ldots$

with $M=\frac{\partial f}{\partial \mathbf y_0}$.

One can solve a new system of the form, $\mathbf y'=M_{\mathbf y_0}(t) \mathbf y$ for any translated phase around $\mathbf y_0$.

Consider $2s$ phase solutions of this system $(\mathbf y^1, \mathbf y^2,\ldots, \mathbf y^{2s})$. Then the wronskian

$W=\det(\mathbf y^1, \mathbf y^2,\ldots, \mathbf y^{2s})$ satisfies the first-order ODE,

$\frac{d}{dt}W= \mathrm{tr}(M_{y_0}) W$

and can be integrated as

$W(t) = W_0\exp \int _0^t\mathrm{tr}(M_{\mathbf y_0}(s)) ds$

Hope this helps.

• where do you show in the part of the Wronskian that the determinant vanishes and thus the volume integrals do not change? – Marion Jan 7 '16 at 15:49
• I do not see the connection between $J$ and $W$ any hint ? – Pete Jan 7 '16 at 15:54