Mean-field joint removal of two chemical species (${\dot{\mu }}_i=\lambda -\beta {\mu }_i-C{\mu }_1{\mu }_2$)

Equal constant production, individual linear removal, and joint bimolecular removal give $d({\mu }_1-{\mu }_2)/dt=-\beta ({\mu }_1-{\mu }_2)$. For $\beta >0$ the unique stable equilibrium is symmetric and solves $\lambda =\beta \mu +C{\mu }^2$. For $\beta =0<C$, every point on ${\mu }_1{\mu }_2=\lambda /C$ is an equilibrium: the difference is conserved and trajectories approach the equilibrium on their own constant-difference line.

// Target: mathematical-biology.bigb

If individual removal is absent, simultaneous removal of one molecule from each species leaves $X_1-X_2$ unchanged. Independent constant-rate births change it by plus or minus one, each at rate $\lambda$. The difference is therefore a symmetric continuous-time random walk on the integers with variance growing as displayed. It has no normalizable stationary law. This explains why a deterministic attracting line of equilibria does not provide a stationary stochastic distribution.

// Target: probability-and-statistics.bigb

For $E=C\mu /(\beta +C\mu )<1$ and mean lifetime $\tau =(\beta +C\mu {)}^{-1}$, the normalized restoring matrix is $M={\tau }^{-1}\left(\begin{array}{cc}1& E\\ E& 1\end{array}\right)$ and the diffusion matrix is $D=(\tau \mu {)}^{-1}\left(\begin{array}{cc}2& E\\ E& 2\end{array}\right)$. Solving the normalized stationary fluctuation-dissipation relation for a reaction network gives the displayed covariance. The sum mode remains damped, but the difference mode's restoring rate is $\beta =(1-E)/\tau$. Its noise grows without bound relative to that restoring rate as $E\to 1$, giving strong negative correlations and invalidating small-fluctuation closure near exclusive joint removal.

// Target: mathematical-biology.bigb