Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

$\frac{dP}{dt}=a\cdot P-b\cdot PZ=P\left(a-bZ\right)$

$\frac{dZ}{dt}=c\cdot PZ-d\cdot Z=Z\left(cP-d\right)$

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

$\frac{dP}{dt}\frac{1}{PZ}=a\frac{1}{Z}-b$

$\frac{dZ}{dt}\frac{1}{PZ}=c-d\frac{1}{P}$

Multiplying with the original system equation gives:

$\frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dZ}{dt}\left(a\frac{1}{Z}-b\right)$

$\frac{1}{PZ}\frac{dP}{dt}\frac{dZ}{dt}=\frac{dP}{dt}\left(c-d\frac{1}{P}\right)$

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: $\frac{d}{dt}E\left(P,Z\right)=\frac{dZ}{dt}\frac{\partial}{\partial Z}E\left(P,Z\right)+\frac{dP}{dt}\frac{\partial}{\partial P}E\left(P,Z\right)$

so then the following should hold:

$\frac{\partial}{\partial Z}E\left(P,Z\right)=\left(a\frac{1}{Z}-b\right)$ and $\frac{\partial}{\partial P}E\left(P,Z\right)=\left(d\frac{1}{P}-c\right)$

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: $E\left(P,Z\right)=a\ln{Z}+dln{P}-bZ-cP$

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

share|improve this question
    
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
add comment

1 Answer 1

up vote 1 down vote accepted

I have edited the original post to include the correct derivation. My mathematical weakness showed when I confused integration with differentiation!

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.