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Let's say I have the following nonlinear dynamical system:

\begin{align} \ x' &= -xy \\\\ \ y' &= -xy \\\\ \ z'&=xy \end{align}

For reference, this comes from a chemical system. Namely, it models concentration changes of the three chemical species when reactants X and Y combine to form product Z. Now, I know from conservation of mass that such a dynamical system does have a conserved quantity: X + Y + 2Z. I also know that for an arbitrary dynamical system to be conservative, its divergence has to be zero. In my case, I calculated my system's divergence out to be: \begin{align} \ div(X) &= -x-y \\\\ \end{align} This is clearly not zero except when both X and Y are zero or when X = -Y (but you can't have negative masses so this case is physically impossible). Yet, I know my system is conservative so how can it have this non-zero divergence? Is there something I'm missing?

Антон Тараскин
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1 Answers1

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A conservative vector field has divergence zero.

A conservative dynamical system has equations which admit autonomous single-valued first integrals. (Conserved quantities under some symmetry of the equations.)

Your sentence "I also know that for an arbitrary dynamical system to be conservative, its divergence has to be zero." Conflates these two uses of "conservative".

Eric Towers
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  • In light of this, how should I interpret Marco Disce's answer in the question below? https://math.stackexchange.com/questions/1705447/how-to-determine-whether-dynamical-systems-are-dissipative-conservative-or-expa – Антон Тараскин Aug 25 '20 at 21:00
  • Or section 7.4 of this page on dynamical systems: https://def.fe.up.pt/dynamics/dynamical_systems.html. Both seem to suggest looking for a 0 divergence even for dynamical systems. – Антон Тараскин Aug 25 '20 at 21:10
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    Note that, unlike the question, that answer does not contain the word "conservative". Perhaps @MarcoDisce was choosing his words carefully, avoiding the over-loaded terminology "conservative". – Lee Mosher Aug 25 '20 at 21:11