= Chemostat steady state
{title2=$\mu(s_*)=D,\quad x_*=Y(s_{\mathrm{in}}-s_*)$}
A <chemostat steady state> has constant nutrient <concentration> and <biomass concentration> even while nutrient and cells flow through the vessel. It is an <equilibrium> of the <chemostat> <ordinary differential equations>, rather than a culture in which all growth has stopped. Assume $D>0$ and the <Monod equation>. Setting both <derivatives> to zero gives the always-existing <chemostat washout> <equilibrium> $(s_*,x_*)=(s_{\mathrm{in}},0)$. A positive <biomass concentration> instead requires growth to balance removal, $\mu(s_*)=D$, whence
$$
s_* = \frac{DK_s}{\mu_{\max}-D},\qquad
x_* = Y(s_{\mathrm{in}}-s_*).
$$
The positive <chemostat steady state> exists exactly when
$$
0<D<\mu(s_{\mathrm{in}})=\frac{\mu_{\max}s_{\mathrm{in}}}{K_s+s_{\mathrm{in}}}.
$$
The weaker condition $D<\mu_{\max}$ only makes the denominator positive; it does not ensure $s_*<s_{\mathrm{in}}$ or $x_*>0$. Because the <Monod equation> is increasing, the feed must permit a <specific growth rate> greater than $D$. At a positive <chemostat steady state>, increasing feed nutrient <concentration> at fixed $D$ increases the <biomass concentration> without changing $s_*$, within the one-nutrient assumptions.
For a concrete <chemostat>, let $\mu_{\max}=1\,\mathrm{h}^{-1}$, $K_s=1\,\mathrm{g}\,\mathrm{L}^{-1}$, $s_{\mathrm{in}}=9\,\mathrm{g}\,\mathrm{L}^{-1}$ and $Y=0.5\,\mathrm{g}\text{ biomass}/(\mathrm{g}\text{ nutrient})$. At $D=0.5\,\mathrm{h}^{-1}$,
$$
s_* = \frac{0.5\times1}{1-0.5}=1\,\mathrm{g}\,\mathrm{L}^{-1},\qquad
x_* = 0.5(9-1)=4\,\mathrm{g}\,\mathrm{L}^{-1}.
$$
The nutrient supply minus outflow is $D(s_{\mathrm{in}}-s_*)=4\,\mathrm{g}\,\mathrm{L}^{-1}\,\mathrm{h}^{-1}$, matching nutrient consumption $\mu(s_*)x_*/Y=4\,\mathrm{g}\,\mathrm{L}^{-1}\,\mathrm{h}^{-1}$. Biomass production and removal both equal $2\,\mathrm{g}\,\mathrm{L}^{-1}\,\mathrm{h}^{-1}$. The <chemostat washout> threshold is $D_c=\mu(9)=0.9\,\mathrm{h}^{-1}$. For example, $D=0.95\,\mathrm{h}^{-1}<\mu_{\max}$ formally gives $s_*=19\,\mathrm{g}\,\mathrm{L}^{-1}$ and $x_*=-5\,\mathrm{g}\,\mathrm{L}^{-1}$, which cannot describe a physical culture.
\Image[/chemostat-washout-equilibria.png]
{title=<Chemostat steady state> nutrient and biomass branches meeting at the <chemostat washout> threshold}
There is a useful exact simplification: the nutrient-equivalent <concentration> $z=s+x/Y$ satisfies $\dot z=D(s_{\mathrm{in}}-z)$, so $z(t)=s_{\mathrm{in}}+[z(0)-s_{\mathrm{in}}]e^{-Dt}$. This combines dissolved nutrient and the nutrient equivalent of biomass; it does not assert that these two variables account for every chemical constituent. In coordinates $(z,x)$, the <Jacobian matrix> at the positive <chemostat steady state> has <eigenvalues> $-D$ and $-\mu'(s_*)x_*/Y$, both negative. Hence the positive <chemostat steady state> is <locally asymptotically stable>. In the worked example the <eigenvalues> are $-0.5\,\mathrm{h}^{-1}$ and $-2\,\mathrm{h}^{-1}$.
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