Determine a conserved quantity in a dynamical system Lotka-Volterra


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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Updated on December 09, 2021


  • Gus
    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 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*}


    $ 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 $ $=a\ln{Z}-bZ+C_{Z}\left(P\right)=d\ln{P}-cP+C_{P}\left(Z\right)$


    \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.

    • Weltschmerz
      Weltschmerz about 12 years
      A nice context introduction would help.
    • Gus
      Gus about 12 years
      Thanks for the suggestion.
    • Aryabhata
      Aryabhata about 12 years
      You can go ahead and delete the question, I believe.
    • Isaac Kay
      Isaac Kay 12 months
      Thank 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: $$𝐸(𝑃,𝑍)=∫(π‘Ž\frac{1}{𝑍}βˆ’π‘)βˆ‚π‘+∫(𝑑\frac{1}{𝑃}βˆ’π‘)βˆ‚π‘ƒ$$ $$E(P,Z)=π‘Ž\ln{𝑍}-𝑏𝑍+𝐢_𝑍(𝑃)+𝑑 \ln{𝑃}βˆ’π‘π‘ƒ+𝐢_𝑃(𝑍)$$ $$\therefore 𝐸(𝑃,𝑍)=π‘Ž\ln{𝑍}+𝑏𝑍+𝑑\ln{𝑃}βˆ’π‘π‘ƒ+\tilde{C}$$