91Ó°ÊÓ

Consider a population modeled by the initial value problem $$ \frac{d P}{d t}=(1-P) P+M, \quad P(0)=P_{0} $$ where the migration rate \(M\) is constant. [The model (8) is derived from equation (6) by setting the constants \(r\) and \(P_{*}\) to unity. We did this so that we can focus on the effect \(M\) has on the solutions.] For the given values of \(M\) and \(P(0)\), (a) Determine all the equilibrium populations (the nonnegative equilibrium solutions) of differential equation (8). As in Example 1, sketch a diagram showing those regions in the first quadrant of the \(t P\)-plane where the population is increasing \(\left[P^{\prime}(t)>0\right]\) and those regions where the population is decreasing \(\left[P^{\prime}(t)<0\right]\). (b) Describe the qualitative behavior of the solution as time increases. Use the information obtained in (a) as well as the insights provided by the figures in Exercises 11-13 (these figures provide specific but representative examples of the possibilities). $$ M=2, \quad P(0)=4 $$

Step by step solution

01

Find the equilibrium populations

Since we are trying to find the equilibrium populations, we need to set the rate of change of the population \((dP/dt)\) to zero. Thus, the equation becomes: \(0 = (1-P)P + M\) We need to solve this quadratic equation for P and determine nonnegative equilibrium solutions. \(0 = (1-P)P + 2\) Since we have M = 2, the equation can be written as: \(0 = P^{2} - P + 2\)

02

Solve the quadratic equation for P

To solve the quadratic equation \(P^2 - P + 2 = 0\), we use the quadratic formula: \(P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In our case, a = 1, b = -1, and c = 2. Plug in these values into the quadratic formula: \(P = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(2)}}{2(1)}\) This results in: \(P = \frac{1 \pm \sqrt{1 - 8}}{2}\) Since the discriminant (1-8) is negative, there are no real equilibrium solutions for P. This means that we do not have any equilibrium populations in this case.

03

Analyze the regions where population is increasing or decreasing

Now we will analyze if the population is increasing or decreasing for the given values of M and P(0). The first derivative of P(t) with respect to t is: \(\frac{dP}{dt} = (1-P)P + M\) Plugging in the given values of M and P(0), we get: \(\frac{dP}{dt} = (1-4)4 + 2\) This results in: \(P'(t) = -12 + 2 = -10\) Since \(P'(t) < 0\), the population is decreasing.

04

Qualitative behavior of the solution

Based on the information we obtained in the previous steps, we can describe the qualitative behavior of the solution as time increases: - No equilibrium populations exist for the given values of M and P(0). - The population is decreasing as time increases. Thus, as time goes on, this population will continue to decrease under the conditions given by M = 2 and P(0) = 4.

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Key Concepts

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

Equilibrium Solutions

Equilibrium solutions in differential equations represent the states where the system does not change over time. For our population model, an equilibrium is where the population size remains constant. Steady-state solutions occur when the rate of change of the population is zero. This means setting the derivative \( \frac{dP}{dt} = 0 \).
In the given equation \( 0 = (1-P)P + M \), substituting \( M = 2 \) and finding values of \( P \) that satisfy this will give us equilibrium points. However, after solving the quadratic equation \( P^2 - P + 2 = 0 \), it became apparent that the equation has no real solutions due to a negative discriminant. Thus, in this instance, there are no equilibrium solutions. Without equilibrium, the population does not settle at a stable size and continuously changes with time.

Population Modeling

Population modeling is widely used to describe the dynamics of populations over time using mathematical equations. In our case, we employ a differential equation \( \frac{dP}{dt} = (1-P)P + M \) to depict population changes, incorporating the effect of a constant migration rate \( M \).
Such equations typically consider births, deaths, and, as in this case, migration to inform predictions on future population trends. The balance between these factors determines whether a population grows, shrinks, or remains the same. Understanding how populations respond to changes allows us to predict possible scenarios and helps in decision-making for resource allocation and conservation efforts.

Qualitative Analysis

Qualitative analysis in differential equations involves describing how solutions behave without necessarily finding the actual solution explicitly. This often includes determining whether populations are increasing or decreasing over time.
For the given problem, after evaluating \( \frac{dP}{dt} \, \), we found that at \( P = 4 \), the derivative is negative \( P'(t) = -10 \), indicating a decrease in population size. A qualitative sketch of regions in the \( tP \)-plane where solutions trend upwards (\( P'(t) > 0 \)) or downwards (\( P'(t) < 0 \)) can provide insights into population dynamics. Although direct solutions were not found, knowing that the population decreases constantly enables us to predict continual reduction over time.

Initial Value Problems

In mathematical terms, an initial value problem (IVP) primarily involves finding a function that satisfies a given differential equation along with specific initial conditions. For example, with \( \frac{dP}{dt} = (1-P)P + M \) and an initial population \( P(0) = 4 \), our task is to determine the function \( P(t) \) over time.
IVPs are fundamental in understanding dynamic systems since they provide specific starting conditions. In this problem, we started with \( P(0) = 4 \), indicating the initial population size. By understanding the initial conditions along with the given rate of change \((M=2)\), we analyze the population's subsequent trajectory. While this problem did not have stable equilibrium solutions, our knowledge of its initial state and derivative signs guides our understanding of long-term trends.