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The chemostat model proposed by monod was given by, $$ \begin{align} \frac{dx}{dt}&=[K(c)-D]x\\ \frac{dc}{dt}&=D[c_0-c]-\frac1yK(c)x \end{align} $$ where $x(t)$ is the population of micro-organisms and $c(t)$ is the concentration of the nutrient, And $K(c)=\frac{Kc}{K+c},y$ and $D$ are growth rate, constant proportionality for consumption of nutrient by micro-organisms and the dilution rate. I got the steady states,

$$ \begin{align} \bar{x}=0,\bar{c}=c_0\text{ (washout state)}\\ K(\bar c)=D,\bar x=y(c_0-\bar c)\text{ (normal state)} \end{align}\tag1 $$

I check the stability of washout state by assuming, $u(t)=A\exp(\lambda t)$ and $v(t)=B\exp(\lambda t)$ then using small perturbation we get,

$$ \begin{align} x(t)&=0+u(t)\\ c(t)&=c_0+v(t) \end{align} $$

gives

$$ \begin{cases} [\lambda+D-K(c_0)]A&=0\\ (\lambda +D)B+\frac1y K(c_0)A&=0 \end{cases}\tag2 $$

eliminating $A$ and $B$ we get the $\lambda=-D$ or $\lambda=K(c_0)-D$. From where, we can say the washout state is stable if $D>K(c_0)$ (because then both $\lambda$ s are negative). But for the normal state, $$ \begin{align} x(t)&=\bar x+u(t)\\ c(t)&=\bar c+v(t) \end{align} $$ gives, $$ \begin{cases} \frac{du}{dt}&=[K(c)-D](\bar x + u(t))\\ c(t)&=D[c_0-\bar c - v(t)]-\frac1y K(c)(\bar x + u(t)) \end{cases}\tag3 $$

And again we can let $u(t)=A\exp(\lambda t)$ and $v(t)=B\exp(\lambda t)$. But this time I couldn't see how to get the values for lambdas. Any help will be appreciated. Thanks in advance.

I was able to get the desire conditions for $(0,c_0)$ point (getting the eigenvalues). My concern was for the second equilibrium point which I called normal state here,$(\bar x+u(t),\bar c+v(t))$ where $u(t),v(t)$ are the perturbation (I'm actually unable to get the linearized system to work with further). In $(1)$ we have found two states, first one $(0,c_0)$ is easy to handle but I was stuck on second one. because we didn't get the explicit point for $x(t)$ and $c(t)$ rather implicit relation $K(\bar c)=D,\bar x=y(c_0-\bar c)$ for $\bar x,\bar c$.

N00BMaster
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  • I have no idea what you are doing. Where did $u,v$ appear from? And what doe they have to do with determining stability? – copper.hat Nov 11 '23 at 19:15
  • I add more context, @copper.hat, please let me know if more context is needed. I was trying to determine condition for which the equilibria are stable (getting condition from $\lambda$ s like for which values $\lambda<0$) to this system using the ansatz $A\exp(\lambda t)$ and $B\exp(\lambda t)$ for $x(t)$ and $c(t)$. – N00BMaster Nov 11 '23 at 19:47
  • As far as I can determine (I make mistakes) the eigenvalues of the linearisation about the equilibrium are $K(c_0)-D,-D$. If neither of these are zero you can determine stability from these. – copper.hat Nov 11 '23 at 19:53
  • I was wondering how to write the linearize version of $(2)$, @copper.hat Is it should be $$u'(t)=[K(c)-D]\bar x-Du(t),v'(t)=D[c_0-\bar cv(t)]-\frac1yK(c)\bar x?$$Now inserting ansatz I couldn't get $\lambda$s. – N00BMaster Nov 12 '23 at 04:11
  • I don't know how you got those linearised dynamics. I get $u'=(K(c_0)-D)u, v'=-{1 \over y} K(c_0)u - Dv$. I am assuming that you are linearising about $(0,c_0)$. I don't know what you are trying to do with the $\lambda$. Only the eigenvalues matter (assuming the real parts are non zero). – copper.hat Nov 12 '23 at 05:58
  • I was able to get the desire conditions for $(0,c_0)$ point. My concern was for the second equilibrium point which I called normal state here,$(\bar x+u(t),\bar c+v(t))$ where $u(t),v(t)$ are the perturbation @copper.hat In $(1)$ we have found two states, first one $(0,c_0)$ is easy to handle but I was stuck on second one. because we didn't get the explicit point for $x(t)$ and $c(t)$ rather implicit relation $K(\bar c)=D,\bar x=y(c_0-\bar c)$ for $\bar x,\bar c$. – N00BMaster Nov 12 '23 at 13:24
  • I guess I messed up with the context, @copper.hat It will be a great help if you help me with this one which doesn't need the background of the problem too much. And thank you very much for your responses. – N00BMaster Nov 12 '23 at 13:33

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