/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Consider the competition between... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the competition between bluegill and redear mentioned in Problem 6. Suppose that \(\epsilon_{2} / \alpha_{2}>\epsilon_{1} / \sigma_{1}\) and \(\epsilon_{1} / \alpha_{1}>\epsilon_{2} / \sigma_{2}\) so, as shown in the text, there is a stable equilibrium point at which both species can coexist. It is convenient to rewrite the equations of Problem 6 in terms of the carrying capacity of the pond for bluegill \(\left(B=\epsilon_{1} / \sigma_{1}\right)\) in the absence of redear and its carrying capacity for redear \(\left(R=\epsilon_{2} / \sigma_{2}\right)\) in the absence of bluegill. a. Show that the equations of Problem 6 take the form $$\frac{d x}{d t}=\epsilon_{1} x\left(1-\frac{1}{B} x-\frac{\gamma_{1}}{B} y\right), \frac{d y}{d t}=\epsilon_{2} y\left(1-\frac{1}{R} y-\frac{\gamma_{2}}{R} x\right)$$ where \(\gamma_{1}=\alpha_{1} / \sigma_{1}\) and \(\gamma_{2}=\alpha_{2} / \sigma_{2} .\) Determine the coexistence equilibrium point \((X, Y)\) in terms of \(B, R, \gamma_{1},\) and \(\gamma_{2}\) b. Now suppose that an angler fishes only for bluegill with the effect that \(B\) is reduced. What effect does this have on the equilibrium populations? Is it possible, by fishing, to reduce the population of bluegill to such a level that they will die out?

Short Answer

Expert verified
Answer: Reducing the carrying capacity of bluegill fish (B) decreases the equilibrium population of bluegill (X) while increasing the equilibrium population of redear (Y).

Step by step solution

01

Rewrite the given equations in terms of carrying capacity

To begin, we need to rewrite the equations of Problem 6 in terms of the carrying capacity of the pond for bluegill (\(B\)) and redear (\(R\)). The equations of Problem 6 are given as: $$\frac{d x}{d t}=\epsilon_{1} x\left(1-\frac{x}{\sigma_{1}}-\frac{\alpha_{1}}{\sigma_{1}} y\right), \frac{d y}{d t}=\epsilon_{2} y\left(1-\frac{y}{\sigma_{2}}-\frac{\alpha_{2}}{\sigma_{2}} x\right)$$ Where \(\epsilon_{1}\), \(\epsilon_{2}\) are the conversion factors of resources into offspring for bluegill and redear, respectively; \(\sigma_{1}\) and \(\sigma_{2}\) are their respective carrying capacities; and \(\alpha_{1}\), \(\alpha_{2}\) are the interspecific competition coefficients. We are given that \(B = \epsilon_{1} / \sigma_{1}\) and \(R = \epsilon_{2} / \sigma_{2}\). We can rewrite these ratios as \(\frac{1}{B} = \frac{\sigma_{1}}{\epsilon_{1}}\) and \(\frac{1}{R} = \frac{\sigma_{2}}{\epsilon_{2}}\). We are also given that \(\gamma_{1}=\alpha_{1} / \sigma_{1}\) and \(\gamma_{2}=\alpha_{2} / \sigma_{2}\). Now, we can rewrite the given equations as: $$\frac{d x}{d t}=\epsilon_{1} x\left(1-\frac{1}{B} x-\frac{\gamma_{1}}{B}y\right), \frac{d y}{d t}=\epsilon_{2} y\left(1-\frac{1}{R}y-\frac{\gamma_{2}}{R} x\right)$$
02

Determine the coexistence equilibrium point

To determine the coexistence equilibrium point \((X, Y)\), we need to set both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) equal to zero and solve for \(x\) and \(y\). $$\frac{d x}{d t}=0=\epsilon_{1} x\left(1-\frac{1}{B} x-\frac{\gamma_{1}}{B}y\right) \Rightarrow 1-\frac{1}{B} x-\frac{\gamma_{1}}{B}y=0$$ $$\frac{d y}{d t}=0=\epsilon_{2} y\left(1-\frac{1}{R}y-\frac{\gamma_{2}}{R} x\right) \Rightarrow 1-\frac{1}{R}y-\frac{\gamma_{2}}{R} x=0$$ Now, solving for \(x\) and \(y\), we get: $$x(1-\frac{1}{B} x-\frac{\gamma_{1}}{B}y)=0 \Rightarrow x = B - B\gamma_{1}y$$ $$y(1-\frac{1}{R}y-\frac{\gamma_{2}}{R} x)=0 \Rightarrow y = R - R\gamma_{2}x$$ Substitute the value of \(x\) from the first equation into the second equation and solve for \(y\): $$y = R - R\gamma_{2}(B - B\gamma_{1}y)$$ $$\Rightarrow y(1 + R\gamma_{2}B\gamma_{1}) = R - R\gamma_{2}B$$ $$\Rightarrow y = \frac{R - R\gamma_{2}B}{1 + R\gamma_{2}B\gamma_{1}}$$ Now, substitute the value of \(y\) back into the equation for \(x\): $$x = B - B\gamma_{1}\left(\frac{R - R\gamma_{2}B}{1 + R\gamma_{2}B\gamma_{1}}\right)$$ So, the coexistence equilibrium point \((X, Y)\) is given by: $$X = \frac{B(1-R\gamma_{2}\gamma_{1})}{1 + R\gamma_{2}B\gamma_{1}},\quad Y = \frac{R - R\gamma_{2}B}{1 + R\gamma_{2}B\gamma_{1}}$$
03

Analyze the effect of reducing B on the equilibrium populations

We are interested in how reducing the carrying capacity \(B\) (by fishing for bluegill) affects the equilibrium populations. If \(B\) is reduced, the equilibrium population of bluegill \(X\) will be reduced as well. However, the equilibrium population of redear \(Y\) will increase, since the interspecific competition for resources between both species is reduced.
04

Check the possibility of reducing the bluegill population to extinction by fishing

To investigate whether it's possible to reduce the population of bluegill to such a level that they will die out, we should examine the equilibrium point \(X\). As \(B\) approaches 0 (due to intensive fishing), the term \(B(1-R\gamma_{2}\gamma_{1})\) in the numerator of the expression for \(X\) also approaches 0. Meanwhile, the denominator \((1 + R\gamma_{2}B\gamma_{1})\) remains positive. Therefore, as \(B\) approaches 0, we have \(X \approx \frac{0}{(1 + R\gamma_{2}B\gamma_{1})} \approx 0\), meaning that the bluegill population will decrease towards extinction. However, we must also consider factors such as the minimum viable population, reproduction rates, and natural fluctuations in population dynamics before making a definitive conclusion. In practice, it could be possible that fishing alone cannot cause extinction, but it may significantly reduce the bluegill population and alter the ecosystem balance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Point in Population Dynamics
When studying animal populations, such as those of bluegill and redear in a pond, the concept of an equilibrium point becomes a central topic in mathematical ecology. An equilibrium point occurs when population growth rates are balanced, resulting in a state where the population sizes remain constant over time.

To find an equilibrium point in the context of a two-species system, we set the population growth rates to zero. This approach doesn't literally mean that individuals stop reproducing or dying, but rather that the births and deaths are in perfect balance. The significance of such a point is that it indicates a stable configuration of populations where neither species overpowers the other, assuming no external disturbances occur.

Mathematically, as shown in our exercise, once we set the growth rate derivatives equal to zero, we can solve the resulting equations to find the population levels that satisfy this condition. For the bluegill and redear, the equilibrium point is found by intersecting their zero growth isoclines. The resulting values, denoted here as \(X\) and \(Y\), correspond to the stable population sizes at which both species can coexist in the pond ecosystem.
Carrying Capacity and Its Role in Population Equations
The concept of carrying capacity, symbolized as \(K\) in many models, or in our case as \(B\) for bluegill and \(R\) for redear, is an essential factor in understanding population dynamics. It refers to the maximum number of individuals that an environment can support indefinitely without causing the environment to be degraded over the long term.

In differential equations, carrying capacity acts as a limiting factor for population growth, ensuring that the populations do not grow beyond the resources available to them. In the solution to our exercise, carrying capacity plays a critical role by defining the terms that limit the growth of both bluegill and redear populations. When we introduce a factor such as angler fishing, which selectively reduces the carrying capacity \(B\) for bluegill, the equilibrium populations are affected. Reduced \(B\) in turn decreases the bluegill population, denoted as \(X\), which can potentially lead to their extinction if overfished, as well as affects the redear population by potentially increasing it, denoted as \(Y\).
Interspecific Competition and Its Impact
In the world of ecosystem dynamics, interspecific competition is the rivalry between different species for the same resources. This type of competition can significantly impact the growth and survival of the competing species. In our exercise, bluegill and redear are in competition for resources within a pond, which is represented mathematically by the competition coefficients \(\gamma_{1}\) and \(\gamma_{2}\).

The coefficients indicate the intensity of the competition; higher values suggest stronger competition. These parameters are essential as they influence the shape and position of the species' zero growth isoclines in the phase plane. The interaction between these two species determines the collective equilibrium state, which may favor one species over the other, or in some cases, like in this exercise, lead to a stable coexistence.

Fishing for bluegill not only reduces their specific population but can also alter the competitive dynamics between the species, potentially leading to increased growth of the redear population due to reduced competition for shared resources. The complex interplay of these factors underscores the intricacy of managing natural populations and ecosystems.

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Most popular questions from this chapter

A generalization of the undamped pendulum equation is $$ d^{2} u / d t^{2}+g(u)=0 $$ where \(g(0)=0, g(u)>0\) for \(00\) for \(u \neq 0,-k

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-y-0.75 x)}\end{array} $$

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-0.5 y-1.5 x)}\end{array} $$

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r^{2}\left(1-r^{2}\right), \quad d \theta / d t=1 $$

Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (0.5,0.5) , corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of \(\delta a\) and \(\delta b\) for the species \(x\) and \(y,\) respectively. In this case equations ( 3 ) are replaced by $$\frac{d x}{d t}=x(1-x-y)+\delta a, \quad \frac{d y}{d t}=\frac{y}{4}(3-4 y-2 x)+\delta b$$ The question is what effect this has on the location of the stable equilibrium point. a. To find the new critical point, we must solve the equations $$\begin{aligned} x(1-x-y)+\delta a &=0 \\ \frac{y}{4}(3-4 y-2 x)+\delta b &=0 \end{aligned}$$ One way to proceed is to assume that \(x\) and \(y\) are given by power series in the parameter \(\delta ;\) thus $$x=x_{0}+x_{1} \delta+\cdots, \quad y=y_{0}+y_{1} \delta+\cdots$$ Substitute equations (44) into equations (43) and collect terms according to powers of \(\delta\). b. From the constant terms (the terms not involving \(\delta\) ), show that \(x_{0}=0.5\) and \(y_{0}=0.5,\) thus confirming that in the absence of immigration or emigration, the critical point is (0.5,0.5) . c. From the terms that are linear in \(\delta,\) show that \\[ x_{1}=4 a-4 b, \quad y_{1}=-2 a+4 b \\] d. Suppose that \(a>0\) and \(b>0\) so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.

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