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Chapter 3: Problem 5

(a) If a constant number \(h\) of fish are harvested from a fishery per unit time, then a model for the population \(P(t)\) of the fishery at time \(t\) is given by $$\frac{d P}{d t}=P(a-b P)-h, \quad P(0)=P_{0},$$ where \(a, b, h,\) and \(P_{0}\) are positive constants. Suppose \(a=5, b=1,\) and \(h=4 .\) since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves corresponding to the cases \(P_{0}>4,1

Short Answer

Expert verified
(a) For \(P_0 > 4\) or \(1 < P_0 < 4\), the population stabilizes at 4; for \(0 < P_0 < 1\), extinction occurs. (b) Integration and graphing confirm phase results. (c) Extinction occurs finite time for \(0 < P_0 < 1\).

Step by step solution

01

Understand the Given Differential Equation

We are given the differential equation \( \frac{dP}{dt} = P(a - bP) - h \) with initial condition \(P(0) = P_0\), where \(a=5\), \(b=1\), and \(h=4\). This equation models the population of fish with a constant harvesting rate.
02

Analyze Equilibrium Points for Phase Portrait

Set \( \frac{dP}{dt} = 0 \) to find equilibrium points: \( P(a - bP) - h = 0 \). Substituting the given values, we get \( P(5 - P) - 4 = 0 \). Solving gives equilibrium points at \( P = 1 \) and \( P = 4 \).
03

Plot the Phase Line

On the phase line, for \(P > 4\), the derivative \( \frac{dP}{dt} \) is negative, indicating that \(P\) will decrease. For \(1 < P < 4\), \( \frac{dP}{dt} \) is positive, indicating that \(P\) will increase towards 4. For \(0 < P < 1\), \( \frac{dP}{dt} \) is negative, indicating \(P\) will decrease towards 0.
04

Determine Long-Term Behavior for Different Initial Conditions

- For \(P_0 > 4\), \(P(t)\) will decrease towards 4 and stabilize there. - For \(1 < P_0 < 4\), \(P(t)\) will increase towards 4 and stabilize there. - For \(0 < P_0 < 1\), \(P(t)\) will decrease towards 0, leading to extinction.
05

Solve the Initial Value Problem

Separate variables and integrate: \( \int \frac{1}{P(5-P)-4} \, dP = \int dt \). Integrate both sides and use \(P(0) = P_0\) to solve for the constant of integration. The integral is somewhat complex and typically requires a partial fraction decomposition.
06

Graph Solutions Using Graphing Utility

Plot \(P(t)\) for initial conditions from the three intervals: 1. \(P_0 > 4\) 2. \(1 < P_0 < 4\) 3. \(0 < P_0 < 1\). The graphs should reflect the behaviors predicted by the phase analysis.
07

Evaluate Extinction in Finite Time

Examine the solution for \(0 < P_0 < 1\). The function approaches zero, indicating extinction. Solve \(P(t) = 0\) to find the time \(t\) when extinction occurs. Depending on the integration, this could be calculated explicitly or estimated graphically.

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Key Concepts

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

Phase Portraits
Phase portraits are a powerful tool used to visually analyze the behavior of differential equations over time. They represent a graphical method to understand how solutions to a differential equation evolve. For our fishery problem, we used phase portraits to sketch how the fish population changes with different initial conditions. This involves plotting the "direction field" or vectors, representing whether the population grows or declines in certain regions.

The phase portrait for this fishery model gives us clear insights:
  • For population levels above 4, the fish population decreases until it stabilizes at 4.
  • Between 1 and 4, the population grows towards 4, an equilibrium point.
  • Below 1, the population decreases towards zero, indicating possible extinction.
By observing these trajectories, we can predict long-term outcomes based on different initial fish populations.
Equilibrium Points
Equilibrium points are essential in understanding the natural tendencies of a system described by differential equations. They are points where the derivative, or rate of change, of the system becomes zero. In the fishery problem, setting \( \frac{dP}{dt} = 0 \) led us to the equations \( P(a-bP) - h = 0 \).

Solving this equation yields equilibrium points, which are solutions where the system does not change over time:
  • **At \(P = 4\)**: The system stabilizes, meaning the fish population neither grows nor declines.
  • **At \(P = 1\)**: This point is unstable since any small perturbation can lead the system into a different state, either growing towards 4 or declining towards extinction.
These points help predict the behavior of the population and form key aspects of the phase portrait analysis.
Long-term Behavior
Long-term behavior analysis in differential equations is crucial to understand how solutions behave as time approaches infinity. For the fishery differential equation, by examining different initial conditions and their trajectories, we identified the eventual fate of the fish population. The phase portrait and equilibrium points guide this analysis.

Here's what we found:
  • **For \(P_0 > 4\)**: The population decreases and settles at 4, as any surplus gets harvested at a rate faster than reproduction can replenish.
  • **For \(1 < P_0 < 4\)**: The population initially rises and then stabilizes at 4, as it finds balance between growth and harvesting.
  • **For \(0 < P_0 < 1\)**: The fishery is not sustainable. The population decreases steadily toward zero, leading to extinction as growth can't compete with harvesting.
Understanding these outcomes is vital in planning sustainable harvesting practices.
Mathematical Modeling
Mathematical modeling is the process of using equations to represent real-life situations. In this case, the differential equation \( \frac{dP}{dt} = P(a - bP) - h \) models a fish population with constant harvesting. This captures the dynamics of natural growth slowed by a carrying capacity and reduced directly by harvesting.

Key aspects of the fishery model include:
  • **Natural Growth**: Represented by \(a-bP\), it indicates how the population would grow without interference, showing logistic growth.
  • **Harvesting Effect**: The term \(-h\)\, fitted into the model to show the constant reduction due to fishing.
  • **Initial Population**: \(P_0\) serves as the starting point to track changes over time.
By carefully designing the mathematical model, we can simulate and predict outcomes, helping guide management decisions for sustainability and conservation.

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Most popular questions from this chapter

The number \(N(t)\) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially, \(N(0)=500,\) and it is observed that \(N(1)=1000 .\) Solve for \(N(t)\) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000 .

Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure \(3.1 .13,\) with an initial velocity \(v_{0}=300 \mathrm{ft} / \mathrm{s}\). The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. (a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by \(d^{2} s / d t^{2}=-g\) (equation (12) of Section 1.3 ). since \(d s / d t=v(t)\) the last differential equation is the same as \(d v / d t=-g,\) where we take \(g=32 \mathrm{ft} / \mathrm{s}^{2} .\) Find the velocity \(v(t)\) of the cannonball at time \(t\) (b) Use the result obtained in part (a) to determine the height \(s(t)\) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3 when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$\frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h},$$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve the DE if the initial height of the water is \(H . \mathrm{By}\) hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h},\) and \(H\) Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is 10 feet high and has radius 2 feet and the circular hole has radius \(\frac{1}{2}\) inch. If the tank is initially full, how long will it take to empty?

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time \(t\). After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria?

Consider the Lotka-Volterra predator-prey model defined by $$\begin{aligned}&\frac{d x}{d t}=-0.1 x+0.02 x y\\\ &\frac{d y}{d t}=0.2 y-0.025 x y,\end{aligned}$$ where the populations \(x(t)\) (predators) and \(y(t)\) (prey) are measured in thousands. Suppose \(x(0)=6\) and \(y(0)=6 .\) Use a numerical solver to graph \(x(t)\) and \(y(t) .\) Use the graphs to approximate the time \(t>0\) when the two populations are first equal. Use the graphs to approximate the period of each population.
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