Problem 19
Question
Consider the following discrete logistic model for the change in the size of a population over time: $$N_{t+1}=R_{0} N_{t}-\frac{1}{100} N_{t}^{2}$$ for \(t=0,1,2, \ldots\) (a) Find all equilibria when \(R_{0}=3.5\) and calculate which (if any) are stable. (b) Calculate the first ten terms of the sequence when \(N_{0}=10\) and describe what you see.
Step-by-Step Solution
Verified Answer
The equilibria are \( N^* = 0 \) (unstable) and \( N^* = 250 \) (stable). The sequence with \( N_0 = 10 \) grows towards 250.
1Step 1: Introduction to the Discrete Logistic Model
The given equation for the population model is a discrete logistic model, which is a type of recurrence relation used to describe how the population changes over time. The equation is: \( N_{t+1} = R_0 N_t - \frac{1}{100} N_t^2 \). We are tasked with finding equilibria and checking their stability, as well as computing the first ten terms of the sequence for a specific initial value.
2Step 1: Find Equilibria
An equilibrium occurs when the population size does not change over time, i.e., \( N_{t+1} = N_t = N^* \). Substituting this into the equation gives: \( N^* = R_0 N^* - \frac{1}{100} (N^*)^2 \). Simplifying, we get: \( 0 = (R_0 - 1) N^* - \frac{1}{100} (N^*)^2 \), leading to: \( N^*(R_0 - 1 - \frac{1}{100}N^*) = 0 \). This yields equilibria at \( N^* = 0 \) and \( N^* = 100(R_0 - 1) \).
3Step 2: Calculate Equilibria for \(R_0 = 3.5\)
Substitute \( R_0 = 3.5 \) into the equilibria equations from Step 1. The equilibria become \( N^* = 0 \) and \( N^* = 100(3.5 - 1) = 250 \). Thus, the equilibria when \( R_0 = 3.5 \) are \( N^* = 0 \) and \( N^* = 250 \).
4Step 3: Analyze Stability of Equilibria
To determine stability, consider the derivative of the function \( f(N) = R_0 N - \frac{1}{100} N^2 \) at the equilibria. The derivative is: \( f'(N) = R_0 - \frac{2}{100} N \). At \( N^* = 0 \), \( f'(0) = 3.5 \), indicating instability since \( |f'(0)| > 1 \). At \( N^* = 250 \), \( f'(250) = 3.5 - \frac{500}{100} = -1.5 \), indicating stability since \( |f'(250)| < 1 \).
5Step 4: Calculate First Ten Terms when \(N_0 = 10\)
Set \( N_0 = 10 \) and \( R_0 = 3.5 \), then calculate: \( N_1 = 3.5 \times 10 - \frac{1}{100} \times 10^2 = 33 \). Continue this process iteratively for each subsequent term to obtain the sequence. Here are the first ten terms (rounded to nearest whole number): 10, 33, 109, 261, 491, 828, 1299, 1964, 2773, 3624. Observe how the terms initially grow rapidly past the unstable equilibrium of 0 towards stabilization around the stable equilibrium of 250.
Key Concepts
Equilibria in Discrete Logistic ModelsStability Analysis of EquilibriaUnderstanding Population DynamicsExploring Recurrence Relations
Equilibria in Discrete Logistic Models
An equilibrium in a discrete logistic model represents a population size that remains constant over time. In mathematical terms, this is where the change in population from one period to the next is zero. For the given model, this means finding values of population size, denoted by \(N^*\), such that \(N_{t+1} = N_t = N^*\).
Insert the equilibrium condition into the equation: \\[ N^* = R_0 N^* - \frac{1}{100} (N^*)^2 \] Simplifying leads to: \\[ 0 = (R_0 - 1) N^* - \frac{1}{100} (N^*)^2 \] Factor this equation to find: \\[ N^*(R_0 - 1 - \frac{1}{100}N^*) = 0 \] This indicates two potential equilibria: one where \(N^* = 0\), meaning the population becomes extinct, and another found at \(N^* = 100(R_0 - 1)\), based on the growth rate \(R_0\).
Insert the equilibrium condition into the equation: \\[ N^* = R_0 N^* - \frac{1}{100} (N^*)^2 \] Simplifying leads to: \\[ 0 = (R_0 - 1) N^* - \frac{1}{100} (N^*)^2 \] Factor this equation to find: \\[ N^*(R_0 - 1 - \frac{1}{100}N^*) = 0 \] This indicates two potential equilibria: one where \(N^* = 0\), meaning the population becomes extinct, and another found at \(N^* = 100(R_0 - 1)\), based on the growth rate \(R_0\).
Stability Analysis of Equilibria
Stability analysis helps us determine whether a population will remain around an equilibrium over time or deviate from it. This is crucial for understanding long-term population behavior.
To assess stability, we examine the derivative \(f'(N)\) of the function \(f(N) = R_0 N - \frac{1}{100} N^2\). Derive: \[ f'(N) = R_0 - \frac{2}{100} N \] Evaluate \(f'(N)\) at the equilibria:
To assess stability, we examine the derivative \(f'(N)\) of the function \(f(N) = R_0 N - \frac{1}{100} N^2\). Derive: \[ f'(N) = R_0 - \frac{2}{100} N \] Evaluate \(f'(N)\) at the equilibria:
- At \(N^* = 0\), the derivative is \(f'(0) = 3.5\). Since \(|f'(0)| > 1\), this equilibrium is unstable.
- At \(N^* = 250\), substitute into the derivative to find \(f'(250) = 3.5 - \frac{500}{100} = -1.5\). As \(|f'(250)| < 1\), this equilibrium is stable.
Understanding Population Dynamics
Population dynamics in a discrete logistic model reveal patterns of growth and behavior over time. For the given exercise, examining how the population evolves allows us to understand its response to initial conditions and its path toward equilibrium.
Starting with an initial population \(N_0\) such as 10, we calculate subsequent population sizes using the recurrence relation:\( N_{t+1} = R_0 N_t - \frac{1}{100} N_t^2 \). The first ten terms give us insights into the dynamics:
Starting with an initial population \(N_0\) such as 10, we calculate subsequent population sizes using the recurrence relation:\( N_{t+1} = R_0 N_t - \frac{1}{100} N_t^2 \). The first ten terms give us insights into the dynamics:
- Rapid Growth: Initially, the population grows quickly beyond the unstable equilibrium of \(N^* = 0\).
- Fluctuation and Stabilization: The population continues to rise and will fluctuate as it nears the stable equilibrium \(N^* = 250\).
- Long-term Stability: Eventually, it stabilizes at or around the stable point, assuming conditions remain consistent.
Exploring Recurrence Relations
Recurrence relations are mathematical expressions that show how a sequence evolves based on its previous terms. In the context of population models, they are crucial for representing consecutive changes in population over discrete time.
For our logistic model, the recurrence relation is: \\[ N_{t+1} = R_0 N_t - \frac{1}{100} N_t^2 \] This equation helps track how each population size (\(N_{t+1}\)) depends directly on the previous size (\(N_t\)) and other parameters like the intrinsic growth rate \(R_0\). Using a recurrence relation:
For our logistic model, the recurrence relation is: \\[ N_{t+1} = R_0 N_t - \frac{1}{100} N_t^2 \] This equation helps track how each population size (\(N_{t+1}\)) depends directly on the previous size (\(N_t\)) and other parameters like the intrinsic growth rate \(R_0\). Using a recurrence relation:
- Enables predictions about future population size, given an initial population \(N_0\).
- Allows for observing how adjustments in \(R_0\) affect overall growth and approach to equilibrium.
- Models the biological constraints perfectly, especially when considering carrying capacity limits.
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