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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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  • $\begingroup$ A nice context introduction would help. $\endgroup$ Oct 1, 2010 at 22:42
  • $\begingroup$ Thanks for the suggestion. $\endgroup$
    – Gus
    Oct 1, 2010 at 23:01
  • $\begingroup$ You can go ahead and delete the question, I believe. $\endgroup$
    – Aryabhata
    Oct 2, 2010 at 0:12
  • $\begingroup$ 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: $$𝐸(𝑃,𝑍)=∫(π‘Ž\frac{1}{𝑍}βˆ’π‘)βˆ‚π‘+∫(𝑑\frac{1}{𝑃}βˆ’π‘)βˆ‚π‘ƒ$$ $$E(P,Z)=π‘Ž\ln{𝑍}-𝑏𝑍+𝐢_𝑍(𝑃)+𝑑 \ln{𝑃}βˆ’π‘π‘ƒ+𝐢_𝑃(𝑍)$$ $$\therefore 𝐸(𝑃,𝑍)=π‘Ž\ln{𝑍}+𝑏𝑍+𝑑\ln{𝑃}βˆ’π‘π‘ƒ+\tilde{C}$$ $\endgroup$
    – Isaac Kay
    Dec 9, 2021 at 1:47

1 Answer 1

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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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