Question

For the following problems, consider the logistic equation in the form \(P^{\prime}=C P-P^{2} .\) Draw the directional field and find the stability of the equilibria.Solve the logistic equation for \(C=10\) and an initial condition of \(P(0)=2\).

Short answer

The equilibrium points are \( P = 0 \) (unstable) and \( P = 10 \) (stable). The solution is \( P(t) = \frac{10}{1 + 4e^{-10t}} \).

Step-by-step solution

Write the Logistic Equation

The logistic equation is given by \( P' = CP - P^2 \). In our problem, \( C = 10 \), so the equation becomes \( P' = 10P - P^2 \).

Find the Equilibrium Points

To find the equilibrium points, set \( P' = 0 \). Thus, \( 10P - P^2 = 0 \). Factor this as \( P(10 - P) = 0 \), so the equilibrium points are \( P = 0 \) and \( P = 10 \).

Determine Stability of Equilibria

To determine stability, analyze the sign of \( P' \) around the equilibrium points. - For \( P < 0 \): \( P' = 10P - P^2 > 0 \), indicating growth.- For \( 0 < P < 10 \): \( P' = 10P - P^2 > 0 \), indicating growth.- For \( P > 10 \): \( P' = 10P - P^2 < 0 \), indicating decay.Thus, \( P = 0 \) is unstable and \( P = 10 \) is stable.

Solve the Logistic Equation

Separate the variables and integrate: \( \frac{dP}{P(10-P)} = dt \). Use partial fraction decomposition: \( \frac{1}{P(10-P)} = \frac{A}{P} + \frac{B}{10-P} \). Solve for \( A \) and \( B \) to get \( \frac{1}{10} \left( \frac{1}{P} + \frac{1}{10-P} \right) \). Integrate to get \( \frac{1}{10} (\ln|P| - \ln|10-P|) = t + C' \). Solve for \( P \): \( P(t) = \frac{10}{1 + 9e^{-10t}} \).

Apply Initial Condition

Apply \( P(0) = 2 \): \( 2 = \frac{10}{1 + 9e^0} \) which simplifies the constant terms. Solving gives \( P(t) = \frac{10}{1 + 4e^{-10t}} \).

Key concepts

Equilibrium Points

Equilibrium points are critical in understanding the behavior of a dynamic system like the logistic equation. These are the values of the dependent variable, in this case, population size \( P \), where the rate of change \( P' \) becomes zero. These points provide insights into the long-term behavior of the system.

To find the equilibrium points for the logistic equation \( P' = CP - P^2 \), we set the derivative \( P' \) equal to zero. For our specific equation \( P' = 10P - P^2 \), this simplifies to \( P(10 - P) = 0 \). Solving this equation yields two equilibrium points: \( P = 0 \) and \( P = 10 \).

These points are the potential steady states of the population. If the population reaches these levels, it will not change unless disturbed by an external influence.

Stability Analysis

Stability analysis helps determine whether these equilibrium points will hold steady if the system experiences small disturbances. This analysis predicts whether small deviations will diminish over time, returning to the equilibrium, or if they will grow, moving the system away from equilibrium.

For the logistic equation \( P' = 10P - P^2 \):

• For \( P < 0 \): \( P' > 0 \), indicating growth towards larger \( P \).
• For \( 0 < P < 10 \): \( P' > 0 \), so the population moves upward towards \( P = 10 \).
• For \( P > 10 \): \( P' < 0 \), meaning the population decreases towards \( P = 10 \).

This analysis shows that \( P = 0 \) is an unstable equilibrium because any small population will tend to grow. Meanwhile, \( P = 10 \) is a stable equilibrium because the system will adjust back to this population level if disturbed.

Directional Field

A directional field, or slope field, visually represents the behavior of solutions to differential equations without solving them analytically. It provides a grid of tiny arrows, indicating the slope \( P' \) for different values of \( P \) and \( t \) in our equation \( P' = 10P - P^2 \).

In the context of our logistic equation, by plotting the directional field for different initial conditions, we observe:

• Arrows pointing upwards when \( 0 < P < 10 \), indicating a tendency to increase towards the equilibrium of \( P = 10 \).
• Arrows pointing downwards when \( P > 10 \), illustrating a tendency to decrease towards the stable population level of \( P = 10 \).

By looking at the field, even without precise calculations, we can infer whether solutions will grow or shrink over time, providing a quick way to visualize the stability of the equilibria.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify the integration of rational functions. When solving the logistic equation, we rearrange it into a form conducive for integration. Specifically, we separate variables to get \( \frac{dP}{P(10-P)} = dt \).

To solve this, we decompose \( \frac{1}{P(10-P)} \) into simpler fractions that can be integrated individually:

• We express \( \frac{1}{P(10-P)} \) as \( \frac{A}{P} + \frac{B}{10-P} \).
• Solving yields \( A = \frac{1}{10} \) and \( B = \frac{1}{10} \), so the decomposition becomes \( \frac{1}{10}(\frac{1}{P} + \frac{1}{10-P}) \).

Integration of these terms gives us a solution that can be manipulated with initial conditions to find a particular solution of our equation, such as \( P(t) = \frac{10}{1 + 4e^{-10t}} \). This whole process streamlines solving complex differential equations like those based on logistic models.