Fish Pond Challenge: Show Equilibrium Variation with $R_f$

In summary, the conversation discusses the dynamics of a pond population over time, represented by the differential equation $\frac{\mathrm{d} x}{\mathrm{d} t} = x\left ( 1-\frac{x}{x_0} \right )-R_f$, where $x(t)$ is the population at time $t$, $x_0$ is the equilibrium amount with no fishing, and $R_f$ is the constant rate of removal due to fishing. If $R_f < \frac{x_0}{4}$, the solution for $x(t)$ shows that the population tends to an equilibrium amount between $\frac{x_0}{2}$ and $x_0$. However, if $R_f
  • #1
lfdahl
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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 $R_f \geq \frac{x_0}{4}$?
 
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  • #2
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.
 
  • #3
Krylov said:
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?
 
  • #4
Suggested solution:

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

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.
 

Related to Fish Pond Challenge: Show Equilibrium Variation with $R_f$

1. What is the Fish Pond Challenge?

The Fish Pond Challenge is a scientific experiment that involves measuring the equilibrium variation of a fish pond when exposed to different levels of flow resistance ($R_f$).

2. How is the equilibrium variation of the fish pond measured?

The equilibrium variation of the fish pond is measured by recording the water level at regular intervals and graphing the data to observe any changes over time.

3. What is the purpose of the Fish Pond Challenge?

The purpose of the Fish Pond Challenge is to demonstrate how changes in flow resistance ($R_f$) can affect the equilibrium of a fish pond, and to illustrate the concept of equilibrium in a real-life scenario.

4. How does $R_f$ affect the equilibrium of the fish pond?

5. What are the potential applications of the Fish Pond Challenge?

The Fish Pond Challenge can be used to study the effects of flow resistance on other bodies of water, such as lakes or rivers. It can also be used to demonstrate the importance of maintaining a balanced ecosystem in fish ponds or other aquatic environments.

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