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Finding equilibrium solutions. For the following discrete population models find all the equilibrium solutions by setting \(N_{k+1}=N_{k}=N .\) Determine the stability of each of the equilibrium populations. (a) \(N_{k+1}=5 N_{k}\) (b) \(N_{k+1}=0.8 N_{k}-0.1 N_{k}^{2}\)

Short Answer

Expert verified
Model (a) has an unstable equilibrium at \(N=0\). Model (b) has stable equilibria at \(N=0\) and \(N=8\).

Step by step solution

01

Understanding Equilibrium Solutions

An equilibrium solution for a discrete population model is a value of \(N\) where the population does not change over time. Mathematically, this is expressed as \(N_{k+1} = N_k = N\). Our goal is to find these values for both models.
02

Solve for Equilibrium in Model (a)

For equation (a), \(N_{k+1} = 5 N_{k}\), we assume \(N_{k+1} = N_{k} = N\). This gives us the equation \(N = 5N\). Solving this, we rearrange to get \(N - 5N = 0\), which simplifies to \(-4N = 0\). Thus, the equilibrium solution is \(N = 0\).
03

Determine Stability for Model (a)

For stability, consider a small perturbation, \(N_{k} = N + \epsilon\). Substituting into \(N_{k+1} = 5N_k\), we have \(N_{k+1} = 5(N + \epsilon) = 5N + 5\epsilon\). Since \(N = 0\), \(N_{k+1} = 5\epsilon\). If \(|5| > 1\), perturbations grow, meaning \(N = 0\) is an unstable equilibrium.
04

Solve for Equilibrium in Model (b)

For equation (b), \(N_{k+1} = 0.8N_{k} - 0.1N_{k}^{2}\), set \(N_{k+1} = N_{k} = N\). This leads to \(N = 0.8N - 0.1N^{2}\), which simplifies to \(0 = -0.1N^{2} + 0.8N - N\) or \(0 = -0.1N^{2} + 0.8N - N\). Rearranging, we get \(0 = 0.8N - N = 0.1N\) or \(0.1N = 0\) and \(0.1N = 0.8N - N^2\). Solving for \(N\), we find \(N = 0\) or \(N = 8\).
05

Determine Stability for Model (b)

Find the derivative of \(f(N) = 0.8N - 0.1N^2\), i.e., \(f'(N) = 0.8 - 0.2N\). For \(N = 0\): \(f'(0) = 0.8\), since \(|0.8| < 1\), \(N = 0\) is stable. For \(N = 8\): \(f'(8) = 0.8 - 0.2\times8 = -0.8\), and \(|-0.8| < 1\), status is also a stable equilibrium.

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Key Concepts

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

Discrete Population Models
Discrete population models are mathematical tools used to describe the population dynamics of a species over discrete time intervals, such as generations or years. Unlike continuous models, which use differential equations, discrete models rely on difference equations to predict population changes.
Discrete population models are valuable in contexts where populations have specific breeding seasons or other time-dependent stages. They help researchers and ecologists understand how species populations evolve over time, considering factors like birth, death, immigration, and emigration. The kind of predictions they allow can aid in decision-making for wildlife management and conservation efforts.
In this context, the population at any given time step, represented by \(N_{k+1}\), is a function of the population at the current step \(N_k\). The goal is to understand how populations develop under various circumstances by considering the mathematical equation that describes the population growth from one time step to the next.
Stability Analysis
Stability analysis is a vital aspect of studying population models, especially when assessing equilibrium solutions. It determines whether a population will return to a particular equilibrium point after a small disturbance or if it will deviate further away. There are a few key steps involved in stability analysis:
  • First, identify the equilibrium solutions of the model. This means finding values of the population where there is no change over time.
  • Next, analyze small perturbations around these equilibrium points to see if they diminish or grow over time.
  • The magnitude of these perturbations after one iteration can be analyzed using the absolute value of the derivative of the governing equation's function.
For instance, if the absolute value of the derivative function evaluated at an equilibrium is less than 1, the equilibrium is stable—it will resist perturbations. Conversely, if the value is greater than 1, the population will likely diverge from the equilibrium, indicating instability.
Mathematical Modeling
Mathematical modeling in ecology, such as using population models, plays a critical role in helping scientists and researchers simulate and predict real-world interactions within ecosystems. Models provide a framework for understanding the complex interactions and behaviors within ecological systems, allowing us to forecast changes under different scenarios. They rely on empirical data and mathematical tools to construct equations that mimic natural processes.
In population models, parameters like birth rate, death rate, carrying capacity, and competition are used to structure these equations. Such models can range from simple linear equations to more complex nonlinear forms depending on the realism and detail required.
Mathematical models serve not only to predict outcomes but also to develop management strategies for species preservation, control invasive species, or optimize habitats for endangered species. They provide critical insights that guide conservation policies and assist in educational settings to illustrate dynamic ecological processes.
Equilibrium Stability
Equilibrium stability in population models refers to the condition of an equilibrium solution under small disturbances. If a population remains close to its equilibrium following a disturbance, it is said to be stable; otherwise, it is unstable. Analyzing equilibrium requires evaluating the system's response to small changes:
  • In mathematical terms, this is done by considering the derivative of the model's function. This derivative provides insights into how the system's trajectory will behave near the equilibrium point.
  • If the absolute value of this derivative is less than 1, it means the equilibrium point is attractive, or stable. This indicates any small changes will eventually die out, bringing the population back to equilibrium.
  • An absolute derivative value greater than 1 suggests that small perturbations will grow, pushing the population further away, implying instability.
Understanding equilibrium stability is essential for both theoretical research and practical applications, such as in wildlife conservation and management strategies, as it informs how resilient a population is to external changes.

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Most popular questions from this chapter

Modelling insect populations. Many insect populations breed only at specific times of the year and all the adults die after breeding. These may be modelled by a difference equation, such as $$ X_{n+1}=r\left(X_{n}-0.001 X_{n}^{2}\right) $$ Using Maple or MATLAB, investigate what happens as the parameter \(r\) (the growth rate) is varied from \(r=0\) to \(r=3 .\) Sketch all the different types of growth patterns observed, labelled with the corresponding value of \(r\).

Density-dependent births. Many animal populations have decreasing per-capita birth rates when the population density increases, as well as increasing per- capita death rates. Suppose the density-dependent per-capita birth rate \(B(X)\) and density-dependent death rate \(A(X)\) are given by $$ B(X)=\beta-(\beta-\alpha) \delta \frac{X}{K}, \quad A(X)=\alpha+(\beta-\alpha)(1-\delta) \frac{X}{K} $$ where \(K\) is the population carrying capacity, \(\beta\) is the intrinsic per- capita birth rate, \(\alpha\) is the intrinsic per-capita death rate and \(\delta\), where \(0 \leq \delta \leq 1\), is a parameter describing the extent that density dependence is expressed in births or deaths. Show that this still gives rise to the standard logistic differential equation $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right) $$

Chemostat. A chemostat is used by microbiologists and ecologists to model aquatic environments, or waste treatment plants. It consists of a tank filled with a mixture of some medium and nutrients, which microorganisms require to grow and multiply. A fresh nutrient-medium mixture is pumped into the tank at a constant rate \(F\) and the tank mixture is pumped from the tank at the same rate. In this way the volume of liquid in the tank remains constant. Let \(S(t)\) denote the concentration of the nutrient in the tank at time \(t\), and assume the mixture in the tank is well stirred. Let \(x(t)\) denote the concentration of the microorganism in the tank at time \(t .\) (a) Draw a compartmental diagram for the amount of nutrient. (b) In the absence of the organism, suggest a model for the rate of change of \(S(t)\). (c) If the microorganisms' per-capita uptake of the nutrient is dependent on the amount of nutrient present and is given by \(p(S)\), and the per-capita reproduction rate of the microorganism is directly proportional to \(p(S)\), extend the model equation above to include the effect of the organism. (The per-capita uptake function measures the rate at which the organism is able to absorb the nutrient when the nutrient's concentration level is \(S .)\) (d) Now develop an equation describing the rate of change of the concentration of the live organism \(\left(x^{\prime}\right)\) in the tank to derive the second equation for the system. (e) The nutrient uptake function \(p(S)\) can be shown experimentally to be a monotonically increasing function bounded above. Show that a Michaelis-Menten type function $$ p(S)=\frac{m S}{a+S} $$ with \(m\) and a positive, non-zero constants, satisfies these requirements. What is the maximum absorption rate? And why is a called the half-saturation constant? (Hint: The maximum absorption rate is the maximum reached by \(p(S) .\) For the second part, consider \(p(a) .)\)

Predicting population size. In a population, the initial population is \(x_{0}=100\). Suppose a population can be modelled using the differential equation $$ \frac{d X}{d t}=0.2 X-0.001 X^{2} $$ with an initial population size of \(x_{0}=100\) and a time step of 1 month. Find the predicted population after 2 months. (Use either an analytical solution or a numerical solution from Maple or MATLAB.)

Fishing with quotas. In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas that vary depending on estimates of the population at the current time. One harvesting model that takes this into account is $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right)-h X $$ (a) Show that the only non-zero equilibrium population is $$ X_{e}=K\left(1-\frac{h}{r}\right) $$ (b) At what critical harvesting rate can extinction occur? Although extinction can occur with this model, as the harvesting parameter \(h\) increases towards the critical value the equilibrium population tends to zero. This contrasts with the constant harvesting model in Section \(3.3\) and Exercise \(3.7\), where a sudden population crash (from a large population to extinction) can occur as the harvesting rate increases beyond a critical value.

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