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.A population of deer inside a park has a carrying capacity of 200 and a growth rate of \(2 \%\). If the initial population is 50 deer, what is the population of deer at any given time?

Short answer

The population of deer at time \(t\) is \(P(t) = \frac{200}{1+3e^{-0.02t}}\). The equilibria are at 0 (unstable) and 200 (stable).

Step-by-step solution

Identify Equation Parameters

The logistic equation is given by \(P' = CP - P^2\). We need to identify the constants. The problem states a carrying capacity of 200 and a growth rate of \(2\%\). This means the growth rate \(C\) is \(0.02\), and the carrying capacity, which is effectively a cap for the population due to \(-P^2\), suggests the maximum \(b\), derived from \(P^2\), is ~200.

Set Up The Equation

We set up the logistic differential equation using the identified parameters: \[ P' = 0.02P - \frac{1}{200}P^2 \]. The equation reflects the growth and the limiting factor on the population.

Find The Equilibrium Points

Equilibrium occurs when \(P' = 0\). This gives us the equation \(0.02P - \frac{1}{200}P^2 = 0\). Factor out \(P\): \[ P(0.02 - \frac{1}{200}P) = 0 \]. The solutions are \(P = 0\) and \(P = 200\), indicating equilibria.

Evaluate Stability of Equilibria

To determine stability, evaluate the derivative \(P'(P)\): \(P' = 0.02P - \frac{1}{200} P^2\) and \(dP'/dP = 0.02 - \frac{1}{100}P\). Evaluate at equilibrium points: \(dP'/dP|_{P=0} = 0.02 > 0\), so \(P = 0\) is unstable. \(dP'/dP|_{P=200} = -0.02 < 0\), so \(P = 200\) is stable.

Describe The Directional Field

The directional field shows the rate \(P'\) based on the population size \(P\). Above \(P = 200\), \(P'\) becomes negative, indicating a decrease toward \(200\). Below \(200\), \(P'\) is positive, pushing population up towards \(200\), except \(P=0\), which diverges.

Solve the Differential Equation

Using the initial condition \(P(0) = 50\), solve the logistic equation \( \frac{dP}{P(0.02 - \frac{1}{200}P)} = dt\). Separate variables and integrate: \(\int \frac{1}{P(0.02 - \frac{1}{200}P)} dP = \int dt\). Integrate to find \(P(t)\): \(P(t) = \frac{200}{1+3e^{-0.02t}}\), considering constants adjusted for \(P(0) = 50\).

Determine the Population Function

The population at any time \(t\) is given by \(P(t) = \frac{200}{1+3e^{-0.02t}}\). This represents the logistic growth converging towards the carrying capacity with the factors derived before.

Key concepts

Population Dynamics

Population dynamics is the study of how populations change over time. In this context, we're looking at the population of deer within a park.
Several factors influence these changes:

• Birth and death rates: These define how quickly a population can grow or shrink naturally.
• Carrying capacity: This is the maximum number of individuals that an environment can sustainably support. In our exercise, this capacity is 200 deer.
• Environmental factors: These include food supply, habitat space, and predation, which impact the carrying capacity.

The logistic equation used in this exercise models these dynamics by accounting for the intrinsic growth rate and the limiting effects of the environment.
Understanding population dynamics is significant as it can help in planning for wildlife conservation and managing population size effectively.

Differential Equations

Differential equations are a crucial mathematical tool used to describe rates of change. In our exercise on deer populations, we employ the logistic differential equation.
This specific equation is written as: \[ P' = CP - P^2 \]where:

• P: The population size at any given time.
• C: The growth coefficient, reflecting the deer growth rate, here set at 2% or 0.02.

Differential equations help us express complex relationships simply. In this case, it reflects how population size increases rapidly first and then slows as it approaches the park's carrying capacity due to "crowding effects."
By solving differential equations, we can predict future population sizes given current conditions.

Stability Analysis

Stability analysis examines whether a system's equilibrium points are stable or unstable, allowing us to predict long-term behavior. In our logistic model:

• Stable Equilibrium: If small deviations bring the system back to equilibrium. Here, the deer population stabilizes at 200.
• Unstable Equilibrium: If small deviations push the system away from equilibrium. The point of zero population is unstable as any new deer introduce growth.

To assess stability, we differentiate the logistic equation with respect to the population size, \( P \): \[\frac{dP'}{dP} = 0.02 - \frac{1}{100} P\] Analyzing at equilibrium points allows us to see:
- At \( P = 0 \), growth is positive, indicating instability.
- At \( P = 200 \), growth is negative, demonstrating stability.
This analysis helps in determining whether populations will converge towards a steady level or diverge, crucial for long-term planning and management.

Equilibrium Points

Equilibrium points in a system are the values where change stops. For the deer population, these points occur when the growth rate is zero, or \( P' = 0 \).
In our example, solving the equation: \[ 0.02P - \frac{1}{200}P^2 = 0 \] allows us to determine the possible equilibria.

• Factoring: \( P(0.02 - \frac{1}{200}P) = 0 \) provides the solutions \( P = 0 \) and \( P = 200 \).

The point \( P = 0 \) represents no deer, which is purely theoretical.
The point \( P = 200 \) indicates a stable population where the birth rate balances the natural limiting factors.
These equilibrium points are vital to understanding and planning population dynamics over time.