A bacterial population N(t) grows while producing a poison…
A bacterial population $N(t)$ grows while producing a poison $Q(t)$ that kills it back, according to: $$ \dot N = N(\alpha-\beta Q),\qquad \dot Q = \gamma N $$
Explanation of Method
Introducing the specific growth rate $p=\dot N/N$ turns the coupled system into a single separable relation between $p$ and $N$; integrating it and imposing that $N$ has a maximum $M$ (where $p=0$) determines the constant of integration and leads directly to a $\operatorname{sech}^2$ profile.
- Define the specific growth rate $p=\dot N/N$ and find $\dot p$.
- Eliminate time to get a separable relation between $p$ and $N$.
- Integrate and use the maximum condition $p=0$ at $N=M$ to fix the constant.
- Solve the resulting first-order equation for $N(t)$.
- Verify the solution and describe the growth-then-decay behaviour.
Step 1 · Specific growth rate
With $p=\dot N/N=\alpha-\beta Q$, differentiating gives $\dot p=-\beta\dot Q=-\beta\gamma N$.
Step 2 · Eliminate time
Since $\dot N=pN$, $\dfrac{dp}{dN}=\dfrac{\dot p}{\dot N}=-\dfrac{\beta\gamma}{p}$, so $p\,dp=-\beta\gamma\,dN$. Integrating,
$$ \frac{p^2}{2}=-\beta\gamma N+C. $$
Step 3 · Use the maximum condition
At the maximum $N=M$, $\dot N=0\Rightarrow p=0$, giving $C=\beta\gamma M$. Writing $k^2=2\beta\gamma M$ (a natural combination of the rate constants and the maximum population),
$$ p^2=k^2\left(1-\frac{N}{M}\right). $$
Step 4 · Solve for $N(t)$
Since $\dot N=pN$, this gives $\dot N^2=k^2N^2(1-N/M)$, which is solved (with $t$ measured from the maximum) by
$$ N(t)=\frac{M}{\cosh^2(kt/2)}. $$
Step 5 · Verification
Differentiating gives $\dot N=-kN\tanh(kt/2)$, so $p=-k\tanh(kt/2)$ and $p^2=k^2\tanh^2(kt/2)=k^2(1-N/M)$, exactly matching Step 3. ✓
Summary Table
| Quantity | Value |
|---|---|
| Specific growth rate | $p=\dot N/N=\alpha-\beta Q$ |
| Combined constant | $k^2=2\beta\gamma M$ |
| Behaviour | grows for $t<0$, maximum $M$ at $t=0$, decays to $0$ as $t\to\infty$ |
Additional Notes
- The decay constant $k$ is not purely intrinsic to the bacteria/poison rates — it also scales with $\sqrt{M}$, the peak population itself, since $k^2=2\beta\gamma M$.
- The solution $N=M\operatorname{sech}^2(kt/2)$ is even in $t$: it is symmetric in time around the maximum, mirroring growth and decay.
- Verified by direct differentiation and by checking $\dot N(0)=0$ and $\ddot N(0)<0$, confirming $t=0$ is indeed the maximum.
Result
With $t$ measured from the moment of maximum population, the result is: