# Determine a conserved quantity in a dynamical system Lotka-Volterra

I have a two state dynamical system. The two state variables are $P$ and $Z$ and $a,b,c,d$ are parameters. The system equations are:

\begin{equation*} \frac{dP}{dt}=a\cdot P-b\cdot PZ=P\left(a-bZ\right), \\ \frac{dZ}{dt}=c\cdot PZ-d\cdot Z=Z\left(cP-d\right). \end{equation*}

Multiplying both sides by $\frac{1}{PZ}$ gives:

\begin{equation*} \frac{dP}{dt}\frac{1}{PZ}=a\frac{1}{Z}-b, \\ \frac{dZ}{dt}\frac{1}{PZ}=c-d\frac{1}{P}. \end{equation*}

Multiplying with the original system equation gives:

\begin{equation*} \frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right), \\ \frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dP}{dt}\left(c-d\frac{1}{P}\right). \end{equation*}

Taking the difference of the last two equations gives: $0=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right)+\frac{dP}{dt}\left(d\frac{1}{P}-c\right)$

By the chain rule, the total time derivative of my conserved quantity is:

\begin{equation*} \frac{d}{dt}E\left(P,Z\right)=\frac{dZ}{dt}\frac{\partial}{\partial Z}E\left(P,Z\right)+\frac{dP}{dt}\frac{\partial}{\partial P}E\left(P,Z\right) \end{equation*}

so then the following should hold:

\begin{equation*} \frac{\partial}{\partial Z}E\left(P,Z\right)=\left(a\frac{1}{Z}-b\right)~\text{and}~ \frac{\partial}{\partial P}E\left(P,Z\right)=\left(d\frac{1}{P}-c\right) \end{equation*}

so

$E\left(P,Z\right)=\int\left(a\frac{1}{Z}-b\right)\partial Z=\int\left(d\frac{1}{P}-c\right)\partial P$ $=a\ln{Z}-bZ+C_{Z}\left(P\right)=d\ln{P}-cP+C_{P}\left(Z\right)$

Finally:

\begin{equation*} E\left(P,Z\right)=a\ln{Z}+dln{P}-bZ-cP \end{equation*}

Edit and this quantity is conserved! I had a bad calculus mistake.

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A nice context introduction would help. – Weltschmerz Oct 1 '10 at 22:42
Thanks for the suggestion. – Gus Oct 1 '10 at 23:01
You can go ahead and delete the question, I believe. – Aryabhata Oct 2 '10 at 0:12