MHB Fish Pond Challenge: Show Equilibrium Variation with $R_f$

[lfdahl]

Suppose a pond contains $x(t)$ fish at time $t$, and $x(t)$ changes according to the DE:
\[\frac{\mathrm{d} x}{\mathrm{d} t} = x\left ( 1-\frac{x}{x_0} \right )-R_f\]
where $x_0$ is the equilibrium amount with no fishing and $R_f > 0$ is the constant rate of removal due to fishing. Assume $x(0) = \frac{x_0}{2}$.

(a). If $R_f < \frac{x_0}{4}$, solve for $x(t)$ and show that it tends to an equilibrium amount between $\frac{x_0}{2}$ and $x_0$.
(b). What happens if $R_f \geq \frac{x_0}{4}$?

 

[S.G. Janssens]

Spoiler

There was a https://mathhelpboards.com/calculus-10/help-derivative-22779.html on logistic growth with constant harvesting a few months ago.

In particular, if we start with the system given in post #5 there, and we non-dimensionalize by introducing a new time $\tau := (r K) t$ and a new unknown $x(\tau) := P(t(\tau))$, then in these new terms we have
\[
\frac{dx}{d\tau} = \frac{1}{rK}\frac{dP}{dt} = P\left(1 - \frac{P}{K}\right) - \frac{H}{rK} = x\left(1 - \frac{x}{K}\right) - \frac{H}{r K}.
\]
This is the system given in the challenge if we set $K := x_0$, $R_f := \frac{H}{r K}$ and we abuse notation by denoting non-dimensional time again with $t$ instead of $\tau$.

It is nice to see a problem on population dynamics, by the way.

 

[lfdahl]

Spoiler

There was a https://mathhelpboards.com/calculus-10/help-derivative-22779.html on logistic growth with constant harvesting a few months ago.

In particular, if we start with the system given in post #5 there, and we non-dimensionalize by introducing a new time $\tau := (r K) t$ and a new unknown $x(\tau) := P(t(\tau))$, then in these new terms we have
\[
\frac{dx}{d\tau} = \frac{1}{rK}\frac{dP}{dt} = P\left(1 - \frac{P}{K}\right) - \frac{H}{rK} = x\left(1 - \frac{x}{K}\right) - \frac{H}{r K}.
\]
This is the system given in the challenge if we set $K := x_0$, $R_f := \frac{H}{r K}$ and we abuse notation by denoting non-dimensional time again with $t$ instead of $\tau$.

It is nice to see a problem on population dynamics, by the way.

Thankyou very much, Krylov! (Handshake)
May I ask for the answers of (a). and (b). in the challenge, based on the logistic growth model, you refer to?

 

[lfdahl]

Suggested solution:

Spoiler

\[\frac{\mathrm{d} x}{\mathrm{d} t}=x-\frac{x^2}{x_0} - R_f=-\frac{1}{x_0}\left ( x-\frac{x_0}{2} \right )^2+\left ( \frac{x_0}{4}-R_f \right ).\]

Let $\frac{1}{\sqrt{x_0}}\left ( x-\frac{x_0}{2} \right )=y(t)$, so that $\sqrt{x_0}dy = dx$ and $y(0) = 0$.
Also, let $a^2 = \left | \frac{x_0}{4}-R_f \right |$. In these terms the D.E. is

\[-dt = \frac{\sqrt{x_0}dy}{y^2\mp a^2}\]
where $a^2 = 0$ if $R_f = \frac{x_0}{4}$, negative if $R_f < \frac{x_0}{4}$, and positive if $R_f > \frac{x_0}{4}$.
(a). When $R_f < \frac{x_0}{4}$, $-t = \frac{\sqrt{x_0}}{2a}\ln \left ( \frac{a-y}{a+y} \right )+c$. Since $y(0) = 0$, we have $c = 0$, and so
\[e^{\frac{-2at}{\sqrt{x_0}}}= \frac{a-y}{a+y} = \frac{2a}{a+y}-1\]
i.e. \[y = \frac{2a}{1+e^{\frac{-2at}{\sqrt{x_0}}}}-a.\]

As $t \rightarrow \infty$, clearly $y \rightarrow a$, i.e.
\[\frac{1}{\sqrt{x_0}}\left ( x-\frac{x_0}{2} \right )\rightarrow \sqrt{\frac{x_0}{4}-R_f}\]

and so \[x \rightarrow \frac{x_0}{2}+\sqrt{\frac{x_0^2}{4}-R_fx_0}.\]

(b). When $R_f = \frac{x_0}{4}$, the original equation is

\[\frac{\mathrm{d} x}{\mathrm{d} t}=x-\frac{x^2}{x_0} - R_f=-\frac{1}{x_0}\left ( x-\frac{x_0}{2} \right )^2,\]
which has the obvious constant solution $x(t) = \frac{x_0}{2} = x(0)$.
When $R_f > \frac{x_0}{4}$, \[-t = \frac{\sqrt{x_0}}{a}\arctan \left ( \frac{y}{a} \right ) + c.\]

Again $c = 0$. Now, $-\tan \left ( \frac{at}{\sqrt{x_0}}\right )=\frac{y}{a}$, or $-\sqrt{R_f-\frac{x_0}{4}}\tan \left ( \frac{\sqrt{R_f-\frac{x_0}{4}}}{\sqrt{x_0}}t\right ) = \frac{1}{\sqrt{x_0}}\left ( x-\frac{x_0}{2} \right )$, and so
\[x = \frac{x_0}{2}-\sqrt{R_fx_0-\frac{x_0^2}{4}}\: \tan \left ( \sqrt{\frac{R_f}{x_0}-\frac{1}{4}}\: \cdot t\right ).\]
This is a decreasing function of $t$, which becomes $0$, when $\tan \left ( \sqrt{\frac{R_f}{x_0}-\frac{1}{4}}\: \cdot t\right ) = \frac{x_0}{2\sqrt{R_fx_0-\frac{x_0^2}{4}}}$.
So, the fish population becomes $0$ in a finite time.