Problem 18 - Calculus for Biology and Medicine

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 \(R_{0}=2.5\) 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

Answer

Equilibria are 0 (unstable) and 150 (stable). The sequence stabilizes at 150 by term 8.

Step 1: Identify Equilibrium Points

An equilibrium point occurs when the population size doesn't change from one time step to the next, meaning \( N_{t+1} = N_t \). Setting \( N_{t+1} = 2.5N_t - \frac{1}{100}N_t^2 \) equal to \( N_t \) gives: \[ N_t = R_0N_t - \frac{1}{100}N_t^2 \]. Replace \( R_0 \) with 2.5 to obtain: \[ N = 2.5N - \frac{1}{100}N^2 \]. Rearrange and solve as a quadratic: \[ 0 = 1.5N - \frac{1}{100}N^2 \]. Factor out \( N \) to find the solutions: \[ N(1.5 - \frac{1}{100}N) = 0 \]. This yields the equilibria \( N = 0 \) and \( N = 150 \).

Step 2: Determine Stability of Equilibria

Evaluate stability using the derivative \( f'(N) \), where \[ f(N) = 2.5N - \frac{1}{100}N^2 \]. The derivative is \[ f'(N) = 2.5 - \frac{1}{50}N \]. For \( N = 0 \), \( f'(0) = 2.5 \), which is greater than 1, implying instability. For \( N = 150 \), \( f'(150) = 2.5 - 3 = -0.5 \), which has an absolute value less than 1, indicating stability.

Step 3: Calculate Sequence for Given Initial Condition

Starting with \( N_0 = 10 \), calculate the first ten terms of the population using the formula \( N_{t+1} = 2.5N_t - \frac{1}{100}N_t^2 \):- \( N_1 = 2.5 \times 10 - \frac{1}{100} \times 10^2 = 25 - 1 = 24 \)- \( N_2 = 2.5 \times 24 - \frac{1}{100} \times 24^2 = 60 - 5.76 = 54.24 \)- \( N_3 = 2.5 \times 54.24 - \frac{1}{100} \times (54.24)^2 = 135.6 - 29.42 = 106.18 \)- Continue this process to find: \( N_4 = 140.50 \), \( N_5 = 148.2 \), \( N_6 = 149.6 \), \( N_7 = 149.9 \), \( N_8 = 150 \), \( N_9 = 150 \), \( N_{10} = 150 \).

Step 4: Analyze the Sequence Behavior

The sequence initially increases and stabilizes as it approaches one of the equilibrium points. With the chosen initial condition, the population size fluctuates then stabilizes around \( N = 150 \), aligning with our earlier analysis that \( N = 150 \) is a stable equilibrium point.

Key Concepts

Equilibrium Points

To find equilibrium points in the context of a discrete logistic model, you need to identify when the population size remains the same from one time point to the next. This means the population won't increase or decrease—it stays constant over time. Mathematically, this is expressed by setting \( N_{t+1} = N_t \). In our problem, this condition translates to the equation \( N = 2.5N - \frac{1}{100}N^2 \), which simplifies to a quadratic equation. Solving this equation, we find two equilibrium points: when \( N = 0 \) and \( N = 150 \).

• \( N = 0 \): This represents a situation where the population is extinct.
• \( N = 150 \): Here, the population stabilizes at this size.

Finding equilibria in logistic models involves solving polynomial equations, typically quadratic, that arise from setting the population change to zero.

Stability Analysis

Stability analysis involves determining whether small perturbations around an equilibrium point will grow or decay over time. In simpler terms, we check if the population will return to equilibrium or move away if slightly disturbed. To do this, we evaluate the derivative of the population function, \( f(N) = 2.5N - \frac{1}{100}N^2 \), which gives \( f'(N) = 2.5 - \frac{1}{50}N \).

• At \( N = 0 \), the derivative \( f'(0) = 2.5 \) is greater than 1, suggesting that even a small increase in population will grow, making this an unstable equilibrium.
• At \( N = 150 \), \( f'(150) = -0.5 \) has an absolute value less than 1, which indicates that any small disturbance will decay, confirming stability.

Stability in discrete models is essentially about checking whether the equilibrium can resist small changes.

Population Dynamics

Population dynamics deals with how populations change over time under various conditions. In the discrete logistic model, the change is controlled by growth rate and self-limitation due to competition or resource constraints. The formula \( N_{t+1} = 2.5N_t - \frac{1}{100}N_t^2 \) accounts for natural growth and limits it through a density-dependent term.
Understanding these dynamics helps in:

• Modeling realistic scenarios where population does not grow indefinitely but is regulated by carrying capacity.
• Predicting future population sizes and possible scenarios if environmental conditions change.

The logistic model is a cornerstone in understanding ecological systems and resource management.

Sequence Calculation

In sequence calculation, we find specific values of the population at discrete time intervals given an initial condition. Starting with \( N_0 = 10 \), we use the recursive formula to find the next population sizes. The calculations proceed step by step as follows:

• \( N_1 = 24 \)
• \( N_2 = 54.24 \)
• \( N_3 = 106.18 \)
• And so on, until \( N_{10} = 150 \)

Each term is derived by applying the formula to the previous term, iteratively.
As observed, the sequence initially increases, reflecting rapid population growth, but stabilizes as it approaches the stable equilibrium point of \( N = 150 \). Calculating sequences allows us to forecast and study how close initial values will align with equilibrium in real-world conditions.