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Chapter 3: Problem 44

A model that describes the population of a fishery in which harvesting takes place at a constant rate is given by $$\frac{d P}{d t}=k P-h$$ where \(k\) and \(h\) are positive constants. (a) Solve the DE subject to \(P(0)=P_{0}\) (b) Describe the behavior of the population \(P(t)\) for increasing time in the three cases \(P_{0}>h / k, P_{0}=h / k,\) and \(00\) such that \(P(T)=0\). If the population goes extinct, then find \(T\)

Short Answer

Expert verified
(a) \(P(t) = \frac{h}{k} + (P_0 - \frac{h}{k})e^{kt}\); (b) Varies with \(P_0\); (c) Extinction occurs if \(P_0 < \frac{h}{k}\), at \(T = \frac{1}{k} \ln\left(\frac{h/k}{h/k - P_0}\right)\).

Step by step solution

01

Recognize the form of the equation

The given differential equation is \(\frac{d P}{d t} = k P - h\), which is a linear first-order ordinary differential equation. The goal is to solve this equation subject to the initial condition \(P(0) = P_0\).
02

Set up the method for solving

This differential equation is separable. It can be rewritten as \(\frac{dP}{kP - h} = dt\). The left side is a function of \(P\) and the right side is a function of \(t\).
03

Integrate both sides

Integrate both sides of the equation: \[\int \frac{dP}{kP - h} = \int dt\]. The integral of the left side with respect to \(P\) is \(\frac{1}{k} \ln|kP - h|\), and the integral of the right side with respect to \(t\) is \(t + C\).
04

Solve for P(t)

Using the integration results, express \(P(t)\): \[\frac{1}{k} \ln |kP - h| = t + C\]. Solving for \(P\), we exponentiate both sides and rearrange: \[|kP - h| = e^{k(t + C)} = Ae^{kt}\], where \(A = e^{kC}\). By solving, we find \(P(t) = \frac{h}{k} + Ce^{kt}\).
05

Apply initial condition

Apply the initial condition \(P(0) = P_0\): \(P_0 = \frac{h}{k} + C\). This gives \(C = P_0 - \frac{h}{k}\). Substitute back into \(P(t)\): \[P(t) = \frac{h}{k} + (P_0 - \frac{h}{k})e^{kt}\].
06

Analyze P(t) for different initial scenarios

Evaluate the behavior of \(P(t)\) for three cases:- For \(P_0 > \frac{h}{k}\): \(C > 0\). Thus, \(P(t) > \frac{h}{k}\) and it grows exponentially without going extinct.- For \(P_0 = \frac{h}{k}\): \(C = 0\). Thus, \(P(t) = \frac{h}{k}\), a constant equilibrium.- For \(0 < P_0 < \frac{h}{k}\): \(C < 0\). \(P(t)\) decreases exponentially and could reach 0 in finite time.
07

Calculate extinction time T

For \(P_0 < \frac{h}{k}\), set \(P(T) = 0\) and solve:\[0 = \frac{h}{k} + (P_0 - \frac{h}{k})e^{kT}\]\[-\frac{h}{k} = (P_0 - \frac{h}{k})e^{kT}\]Solve for \(T\):\[T = \frac{1}{k} \ln\left(\frac{h/k}{h/k - P_0}\right)\].

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Key Concepts

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

Population Dynamics
Population dynamics is the study of how populations of living organisms change over time and space. In mathematical modeling, such scenarios often involve the use of differential equations to describe the rate of change in a population, considering various influencing factors. In the context of this exercise, we discuss a fishery population subject to harvesting.

Key elements in this model include:
  • The natural growth rate of the population, represented by the term involving constant \(k\).
  • The reduction in population due to harvesting, reflected by the constant \(h\).
This allows us to analyze behaviors like growth, stability, and extinction scenarios for different initial populations. Understanding population dynamics provides insights for ecology, conservation, and resource management.
Separable Differential Equations
Separable differential equations are those that can be rewritten to separate the variables, typically in the form \( f(y)dy = g(x)dx \). In our problem, we have the equation \(\frac{dP}{dt} = kP - h\).

The method involves rearranging to get \( \frac{dP}{kP-h} = dt \). This lets us integrate each side separately to find a solution. When dealing with population dynamics, this method is useful because it allows us to see how the rate of change (derivative) of the population relates to time.
  • Integrate both sides to solve for the variables.
  • Integration leads to finding a function that gives us \( P(t) \).
This approach is straightforward, making it easier to handle models involving growth or decay characteristics.
Exponential Growth and Decay
Exponential growth and decay describe how populations increase or decrease at a constant relative rate. This is a common pattern in natural systems and is crucial in understanding the dynamics of populations like the fishery in our model.

When initial conditions are applied, different outcomes appear based on the relationship between the initial population and \( \frac{h}{k} \):
  • If \( P_0 > \frac{h}{k} \), the population grows without limits. This is a case of exponential growth due to adequate resources and low harvesting.
  • If \( P_0 = \frac{h}{k} \), the system is at equilibrium where growth is balanced by harvesting.
  • If \( 0 < P_0 < \frac{h}{k} \), the population declines, potentially reaching extinction over time, indicating exponential decay.
These patterns help in predicting long-term behavior under different environmental and management scenarios.
Initial Value Problem
An initial value problem involves finding a function that satisfies a differential equation and meets an initial condition at a particular point. In this exercise, the starting population is given by \( P(0) = P_0 \).

Initial conditions are vital because:
  • They allow us to determine the specific form of the solution that fits the particular scenario we are analyzing.
  • They help calculate constants of integration to provide a complete solution.
With our solution \(P(t) = \frac{h}{k} + (P_0 - \frac{h}{k})e^{kt}\), the initial value \( P_0 \) helped determine the constant \( C \) in our equation. Initial value problems are common in modeling exercises where predictions over time are needed, providing a tailored approach to understanding specific circumstances.

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Most popular questions from this chapter

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