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Chapter 2: Problem 4

If the given differential equation is autonomous, identify the equilibrium solution(s). Use a numerical solver to sketch the direction field and superimpose the plot of the equilibrium solution(s) on the direction field. Classify each equilibrium point as either unstable or asymptotically stable. $$ P^{\prime}=0.13 P(1-P / 200) $$

Short Answer

Expert verified
Equilibrium solutions are \( P = 0 \) (unstable) and \( P = 200 \) (asymptotically stable).

Step by step solution

01

Identify the Differential Equation

The given differential equation is \( P' = 0.13 P(1 - P/200) \). This is an autonomous equation because it does not explicitly depend on the independent variable (usually time).
02

Find Equilibrium Solutions

Equilibrium solutions occur where \( P' = 0 \). Setting the equation \( 0.13 P(1 - P/200) = 0 \), we find the equilibrium points by solving \( P(1 - P/200) = 0 \). This gives us \( P = 0 \) and \( P = 200 \) as the equilibrium solutions.
03

Classify Equilibrium Points

Calculate the derivative of the right-hand side: \( f(P) = 0.13 P(1 - P/200) \). The derivative of \( f(P) \) is given by \( f'(P) = 0.13(1 - P/200) - 0.13P(-1/200) \). Evaluating \( f'(P) \) at the equilibrium points, \( f'(0) = 0.13 \) (positive, so unstable) and \( f'(200) = -0.13 \) (negative, so asymptotically stable).
04

Use a Numerical Solver

A numerical solver (such as in Python or MATLAB) can plot the direction field. This involves calculating slopes at various points based on the differential equation and plotting them as short line segments in the plane.
05

Superimpose Equilibrium Solutions

Plot \( P = 0 \) and \( P = 200 \) as horizontal lines over the direction field since they are constant solutions. This will show where solutions of the differential equation are equilibrating.

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Key Concepts

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

Equilibrium Solutions
In autonomous differential equations, equilibrium solutions are vital for understanding system behavior. These solutions occur when the rate of change is zero. For the given exercise, the equation is \( P' = 0.13 P(1 - P/200) \). To find equilibrium solutions, set the equation to zero:
\[0.13 P(1 - P/200) = 0\] This leads to solving \( P = 0 \) and \( 1 - P/200 = 0 \), giving us the equilibrium solutions \( P = 0 \) and \( P = 200 \).
- **When \( P = 0 \)**: The growth rate stops, resulting in a constant population.- **When \( P = 200 \)**: The population reaches a stable size and remains constant.
These solutions help us understand where the population stabilizes or remains constant over time.
Direction Field
A direction field provides a visual representation of the behavior of solutions to a differential equation. It is constructed by plotting slope lines at several points in the plane. For the equation \( P' = 0.13 P(1 - P/200) \), it gives a visual idea of how different initial conditions evolve over time.
To create a direction field, use a numerical solver like Python or MATLAB:
- Calculate the slope \( P' \) for various values of \( P \).- Draw short line segments at these points, representing the direction of change.
By superimposing the horizontal lines \( P = 0 \) and \( P = 200 \) over this field, you can easily visualize where the populations trend toward these equilibrium solutions.
Typically, the slope lines will converge towards the stable equilibrium or diverge from the unstable one.
Stability of Equilibrium Points
The stability of equilibrium points is crucial to predict the system's behavior under small disturbances. Stability indicates whether a system will return to equilibrium after a disturbance. Using the derivative of the function \( f(P) = 0.13 P(1-P/200) \), called \( f'(P) \), helps classify these points.
- **At \( P = 0 \)**: Calculate \( f'(0) \). Here it equals 0.13, which is positive. A positive derivative means the point is unstable; small changes will cause deviations away from this equilibrium.
- **At \( P = 200 \)**: Calculate \( f'(200) \). Here it equals -0.13, which is negative. A negative derivative implies asymptotic stability; any small deviation will eventually return to this equilibrium.
Understanding the stability helps in forecasting long-term behavior, making this analysis highly valuable in biological, ecological, or any systems where such differential equations apply. Equilibria that are asymmetric stable attract neighboring lines, becoming centers where solutions stabilize.

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