Problem 17 - 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 \(R_{0}=1.5\) 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

Two equilibria: 0 (unstable) and 50 (stable). The sequence shows growth exceeding equilibrium.

Step 1: Define Equilibrium in the Model

An equilibrium occurs when the population size remains constant over time. For the model \(N_{t+1}=R_{0} N_{t}-\frac{1}{100} N_{t}^{2}\), we set \(N_{t+1} = N_t\) to find equilibrium points. This means solving \(R_{0} N_{t} - \frac{1}{100} N_{t}^{2} = N_t\) for \(N_t\).

Step 2: Solve Equation to Find Equilibria

Set up and solve the equation: \(R_{0} N_{t} - \frac{1}{100} N_{t}^{2} = N_t\). Rearrange terms to get \(-\frac{1}{100} N_{t}^{2} + (R_{0} - 1)N_t = 0\). Factor the expression to find equilibria: \(N_t ( R_{0} - 1 - \frac{1}{100} N_{t} ) = 0\). This gives \(N_t = 0\) or \(N_t = 100(R_0 - 1)\). For \(R_0 = 1.5\), equilibria are at \(N_t = 0\) and \(N_t = 50\).

Step 3: Determine Stability of Equilibria

To determine stability, consider the derivative of the updating function \(f(N_t) = 1.5N_t - \frac{1}{100}N_t^2\) around the equilibria. If the absolute value of the derivative is less than one, the equilibrium is stable. Calculate the derivative: \(f'(N_t) = 1.5 - \frac{1}{50}N_t\). At \(N = 0\), \(f'(0) = 1.5 > 1\) (unstable). At \(N = 50\), \(f'(50) = 1.5 - 1 = 0.5 < 1\) (stable).

Step 4: Calculate Population Sequence

Use the given initial population \(N_0 = 10\) to compute subsequent terms using \(N_{t+1}=1.5 N_{t}-\frac{1}{100} N_{t}^{2}\). Calculate \(N_1, N_2, ..., N_{10}\):1. \(N_1 = 13.5\)2. \(N_2 = 18.1975\) 3. \(N_3 = 24.4379\) 4. \(N_4 = 32.0656\) 5. \(N_5 = 40.7534\) 6. \(N_6 = 49.9403\) 7. \(N_7 = 59.9601\) 8. \(N_8 = 70.0158\) 9. \(N_9 = 79.1735\) 10. \(N_{10} = 86.4431\).

Step 5: Describe Sequence Behavior

The initial population of 10 shows a fast growth towards an equilibrium. The terms show continuous growth beyond the stable equilibrium at 50, indicating that for \(N_0 = 10\), the population does not stabilize at 50 due to overshooting caused by the chosen parameter \(R_0\).

Key Concepts

Population Dynamics

Population dynamics refers to the study of how populations change over time. In the discrete logistic model, the formula \(N_{t+1} = R_{0} N_{t} - \frac{1}{100} N_{t}^{2}\) captures these changes. This equation is used to predict future population sizes based on current sizes and growth parameters. Here, \(R_0\) represents the intrinsic growth rate of the population, while the term \(-\frac{1}{100} N_{t}^{2}\) accounts for the limiting effects of resources—often a form of competition or other density-dependent factors.

The discrete logistic model is essential in understanding population dynamics because it includes the concept of carrying capacity. This is the maximum population size that the environment can sustain. As populations grow, resources become scarcer, slowing growth and eventually stopping it when the carrying capacity is reached. Without this slowing factor, populations would grow indefinitely in a linear fashion, which is unrealistic in natural settings.

In our specific example, when \(R_0 = 1.5\), the logistic equation tries to balance growth with carrying capacity, which directly affects how the population changes from one time step \(t\) to the next.

Equilibrium Analysis

Equilibrium in population models occurs when the population size remains constant over time. For the given equation \(N_{t+1} = R_{0} N_{t} - \frac{1}{100} N_{t}^{2}\), finding equilibria involves setting \(N_{t+1} = N_t\). Solving this leads to the equation \(-\frac{1}{100} N_{t}^{2} + (R_{0} - 1)N_t = 0\).

In simple terms, solving this equation helps us pinpoint the population sizes where the system is in balance, i.e., where the population stops changing because birth and death rates are equal. The expression can be factored to yield \(N_t = 0\) or \(N_t = 100(R_0 - 1)\), providing us with potential equilibria. With \(R_0 = 1.5\), these equilibria are \(N_t = 0\) and \(N_t = 50\).

Each equilibrium signifies a point at which the population would, theoretically, remain constant if it ever reached that size. In practice, understanding these equilibria helps in making predictions and managing resources in ecological and conservation contexts.

Stability of Equilibria

The concept of stability tells us whether a population will remain at an equilibrium after small disturbances. For the discrete logistic model, stability analysis involves taking the derivative of the update function. Here, this function is \(f(N_t) = 1.5N_t - \frac{1}{100}N_t^2\). The derivative, \(f'(N_t) = 1.5 - \frac{1}{50}N_t\), helps determine stability.

A stable equilibrium means that if the population is slightly disturbed, it will return to the equilibrium over time. Conversely, if an equilibrium is unstable, a small disturbance leads to a population drifting away from that equilibrium.

Examining the derivative at each equilibrium: for \(N_t = 0\), \(f'(0) = 1.5\), which is greater than 1, indicating instability. For \(N_t = 50\), \(f'(50) = 0.5\), indicating stability since the absolute value is less than 1. This analysis implies that while a population size of 50 is stable, it is unlikely to stay at 0 if any small fluctuations occur. Stability analysis is critical for understanding how populations respond to changes in environmental conditions and management strategies.