Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Another population model is one in which species compete for resources, such as a limited food supply. Such a model is given by $$ \begin{aligned} &x^{\prime}=a x-b x^{2}-c x y \\ &y^{\prime}=d y-e y^{2}-f x y \end{aligned} $$ In this case, assume that all constants are positive. a. Describe the effects/purpose of each terms. b. Find the fixed points of the model. c. Linearize the system about each fixed point and determine the stability. d. From the above, describe the types of solution behavior you might expect, in terms of the model.

Short Answer

Expert verified
(a) The terms denote intrinsic growth rates, intra-species competition and inter-species competition. (b) Fixed points can be found by setting the right sides to zero and solving: Certainly (0,0) is a fixed point and others need to be found by solving nonlinear equations. (c) Linearize the system using the Jacobian and determine its eigenvalues at each fixed point to decide on stability. (d) The solution behavior needs to be analyzed in the context of the model - it could involve species extinction, coexistence, cyclical behavior, abrupt changes called bifurcations as parameters vary, etc.

Step by step solution

01

Interpret the Model

The terms in the equation have the following effects/purposes: \(a x\) and \(d y\) represent the intrinsic growth rates of the two species when there is no competition which result in an exponential growth; \(b x^{2}\) and \(e y^{2}\) denote intra-species competition which slows down the growth rates of each species; And \(c x y\) and \(f x y\) denote inter-species competition between the two species as they compete for the same resources, further slowing down the growth rates.
02

Find the Fixed Points

Setting the right sides of the equations to \(0\) yields \(x^{\prime}=0\) and \(y^{\prime}=0\). The solutions to these equations are the fixed points: solving for \(x\) yields \(x=\frac{a}{b+c y}\) and solving for \(y\) yields \(y=\frac{d}{e+f x}\). The trivial fixed point is \((0,0)\) and the other fixed points can be found by equating the expressions for \(x\) and \(y\) and solving for \(x\) and \(y\). This would require solving a nonlinear equation.
03

Linearize & Determine Stability

To linearize the system about each fixed point, find the Jacobian matrix of the system, which is the matrix of the first order partial derivatives of the functions on the right-hand sides of the equations. The Jacobian at a fixed point provides the best linear approximation to the system near that point. The nature of the eigenvalues of the Jacobian at a fixed point decides the stability: if all eigenvalues have negative real parts, then the fixed point is stable; if at least one eigenvalue has a positive real part, then the fixed point is unstable. Note that the Jacobian and hence its eigenvalues would depend on the fixed point.
04

Describe Solution Behavior

Based on the stability of each fixed point and the directions of the trajectories near each fixed point, which can be inferred from the linear approximation or by directly analyzing the nonlinear system, describe the solution behavior. For instance, if the only fixed point is at (0,0), other solutions would either spiral in or out or approach/depart from it along specific solution curves. Also look out for bifurcations as parameters vary, where the system behavior changes qualitatively. Finally, interpret these behaviors in the context of the model; for example, species extinction when solutions approach (0,0), species coexistence for stable nontrivial fixed points, cyclical behavior for spiraling solutions, etc.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Differential Equations
In mathematical biology, differential equations are crucial for modeling the dynamics of biological systems, like the population changes of competing species. These equations, which contain derivatives, express the rate of change of a quantity (like population size) in relation to other quantities.

Consider an equation such as \( x' = ax - bx^2 - cxy \). Here, the term \( ax \)is indicative of the natural growth of the population without any limiting factors. The term \( -bx^2 \) represents the effect of intraspecific competition, meaning competition within the same species, which naturally limits the growth. Lastly, \( -cxy \) indicates interspecific competition, which occurs when different species compete for the same limited resource. These equations describe complex biological phenomena using a mathematical language that can be analyzed and solved to predict the behavior of the species in question.
Fixed Points Analysis
A fixed point in the context of a dynamical system, such as our population model, is a point where the system is at equilibrium, meaning the population sizes do not change. Finding these points is essential as they represent possible steady states of the ecosystems, like extinction, stable coexistence, or unbounded growth.

To find fixed points, we set the rate of change expressions to zero (\( x' = 0 \) and \( y' = 0 \)), and solve for the population sizes \( x \) and \( y \). In the given example, setting up these equations provides us with certain equations which, when solved, give us the fixed points. These are crucial for understanding the long-term behavior of the species under study, whether they will stabilize at certain population levels, die out, or continue growing indefinitely.
Jacobian Matrix and Stability
Analyzing stability around fixed points is done using the Jacobian matrix. This matrix is a system of first-order partial derivatives and it represents how small changes in population sizes affect the growth rate near a fixed point.

To construct the Jacobian matrix for our system, we would take the derivatives of each function with respect to each species, resulting in a 2x2 matrix (since we have two species). The matrix is evaluated at the fixed points to assess stability. If the eigenvalues of this matrix have negative real parts, the populations return to the fixed point after a small disturbance, indicating a stable equilibrium. Conversely, if any eigenvalue has a positive real part, there exists instability, indicating that a small change in population size could lead to diverging from the equilibrium. Thus, the Jacobian matrix gives us a detailed insight into how robust or sensitive our ecosystem is to changes.
Species Competition Models
The species competition models depict the interaction between species vying for the same resources. The model given by the equations \( x' = ax - bx^2 - cxy \) and \( y' = dy - ey^2 - fxy \) exemplifies such competition, where \(x\) and \(y\) are populations of two different species.

Each term in the model equations conveys essential information: The first term reflects growth unaffected by competition; the second term highlights the impact of competition within a species; and the third captures the effect of competition between species. By analyzing these models using differential equations and stability theories, ecologists can foretell outcomes like whether one species will outcompete the other, if the species will find a way to coexist, or if they will oscillate in a predator-prey like fashion. These models and their analyses are pivotal in conservation biology and resource management, providing insights on how multiple species can coexist in an ecosystem or how changes within that ecosystem might affect its inhabitants.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each problem, determine equilibrium points, bifurcation points, and construct a bifurcation diagram. Discuss the different behaviors in each system. a. \(y^{\prime}=y-\mu y^{2}\) b. \(y^{\prime}=y(\mu-y)(\mu-2 y)\) c. \(x^{\prime}=\mu-x^{3}\) d. \(x^{\prime}=x-\frac{\mu x}{1+x^{2}}\).

8\. Find the fixed points of the following systems. Linearize the system about each fixed point and determine the nature and stability in the neighborhood of each fixed point, when possible. Verify your findings by plotting phase portraits using a computer. a. $$ \begin{aligned} &x^{\prime}=x(100-x-2 y) \\ &y^{\prime}=y(150-x-6 y) \end{aligned} $$ b. $$ \begin{aligned} x^{\prime} &=x+x^{3} \\ y^{\prime} &=y+y^{3} \end{aligned} $$ C. $$ \begin{aligned} &x^{\prime}=x-x^{2}+x y \\ &y^{\prime}=2 y-x y-6 y^{2} \end{aligned} $$ d. $$ \begin{aligned} &x^{\prime}=-2 x y \\ &y^{\prime}=-x+y+x y-y^{3} \end{aligned} $$.

Find the equilibrium solutions and determine their stability for the following systems. For each case, draw representative solutions and phase lines. a. \(y^{\prime}=y^{2}-6 y-16\). b. \(y^{\prime}=\cos y\). c. \(y^{\prime}=y(y-2)(y+3)\). d. \(y^{\prime}=y^{2}(y+1)(y-4)\).

The Michaelis-Menten kinetics reaction is given by $$ E+S \frac{k_{3}}{k_{1}}+E S \underset{k_{2}}{ } E+P $$ The resulting system of equations for the chemical concentrations is $$ \begin{aligned} \frac{d[S]}{d t} &=-k_{1}[E][S]+k_{3}[E S] \\ \frac{d[E]}{d t} &=-k_{1}[E][S]+\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[E S]}{d t} &=k_{1}[E][S]-\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[P]}{d t} &=k_{3}[E S] \end{aligned} $$ In chemical kinetics, one seeks to determine the rate of product formation \(\left(v=d[P] / d t=k_{3}[E S]\right)\). Assuming that \([E S]\) is a constant, find \(v\) as a function of \([S]\) and the total enzyme concentration \(\left[E_{T}\right]=[E]+[E S] .\) As a nonlinear dynamical system, what are the equilibrium points?

.Derive the first integral of the Lotka-Volterra system, \(a \ln y+d \ln x-\) \(c x-b y=C\).

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free