Determine a conserved quantity in a dynamical system LotkaVolterra
I have edited the original post to include the correct derivation. My mathematical weakness showed when I confused integration with differentiation!
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Gus
Updated on December 09, 2021Comments

Gus 12 months
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 Pb\cdot PZ=P\left(abZ\right), \\ \frac{dZ}{dt}=c\cdot PZd\cdot Z=Z\left(cPd\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}=cd\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(cd\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}bZcP \end{equation*}
Edit and this quantity is conserved! I had a bad calculus mistake.

Weltschmerz about 12 yearsA nice context introduction would help.

Gus about 12 yearsThanks for the suggestion.

Aryabhata about 12 yearsYou can go ahead and delete the question, I believe.

Isaac Kay 12 monthsThank you for keeping the question up and updating your solution! However, I couldn't help but wonder if the equation before your final solution should actually read as follows: $$πΈ(π,π)=β«(π\frac{1}{π}βπ)βπ+β«(π\frac{1}{π}βπ)βπ$$ $$E(P,Z)=π\ln{π}ππ+πΆ_π(π)+π \ln{π}βππ+πΆ_π(π)$$ $$\therefore πΈ(π,π)=π\ln{π}+ππ+π\ln{π}βππ+\tilde{C}$$
