Problem 18 - Calculus for Biology and Medicine

Question

An unrealistic feature of the Lotka-Volterra model is that the prey exhibits unlimited growth in the absence of the predator. The model described by the following system remedies this shortcoming (in the model, we assume that the prey evolves according to logistic growth in the absence of the predator; the other features of the model are retained): $$ \begin{array}{l} \frac{d N}{d t}=N\left(1-\frac{N}{K}\right)-4 P N \\ \frac{d P}{d t}=P N-5 P \end{array} $$ Here, \(K>0\) denotes the carrying capacity of the prey in the absence of the predator. In what follows, we will investigate how the carrying capacity affects the outcome of this predator-prey interaction. (a) Draw the zero isoclines of \((11.78)\) for (i) \(K=10\) and (ii) \(K=3\). (b) When \(K=10\), the zero isoclines intersect, indicating the existence of a nontrivial equilibrium. Analyze the stability of this nontrivial equilibrium. (c) Is there a minimum carrying capacity required in order to have a nontrivial equilibrium? If yes, find it and explain what happens when the carrying capacity is below this minimum and what happens when the carrying capacity is above this minimum.

Step-by-Step Solution

Answer

Nontrivial equilibrium requires \( K > 5 \). Stable for \( K=10 \). No intersection at \( K=3 \).

Step 1: Determine Zero Isoclines for N

To find the zero isocline for the prey population \( N \), set \( \frac{dN}{dt} = 0 \). The equation becomes: \[ N(1 - \frac{N}{K}) - 4PN = 0. \] Factoring out \( N \), we have two zero isoclines: \( N = 0 \) and \( 1 - \frac{N}{K} = 4P \) or equivalently, \( N = K(1 - 4P) \).

Step 2: Determine Zero Isoclines for P

To find the zero isocline for the predator population \( P \), set \( \frac{dP}{dt} = 0 \). The equation becomes: \[ PN - 5P = 0. \] Factoring out \( P \), we get two zero isoclines: \( P = 0 \) and \( N = 5 \).

Step 3: Draw Zero Isoclines for K=10

With \( K = 10 \), plot the isoclines \( N = 10(1-4P) \) and \( N = 5 \) on a graph with \( P \) on the x-axis and \( N \) on the y-axis. The intersection of these lines will give potential equilibrium points. Additionally, \( N = 0 \) and \( P = 0 \) are axes of the plot.

Step 4: Draw Zero Isoclines for K=3

With \( K = 3 \), plot the isoclines \( N = 3(1-4P) \) and \( N = 5 \) similarly. Ensure to check if and where they intersect, as this indicates potential equilibrium points.

Step 5: Analyze Equilibrium for K=10

With the zero isoclines for \( K = 10 \) intersecting at \( N = 5 \) and \( P = 0.75 \) (derived from setting the equations equal), analyze stability by examining the sign of derivatives around this point or using a Jacobian. This results in a nontrivial stable equilibrium.

Step 6: Determine Minimum K for Equilibrium

For nontrivial equilibrium, the zero isoclines must intersect, meaning \( K(1-4P) = 5 \). Solving for \( K \) when \( 1-4(5/K)=0 \), we find the minimum carrying capacity \( K > 5 \). Below this, the predator population cannot be sustained, and above this, stable equilibrium can exist.

Key Concepts

predator-prey model

The predator-prey model is a mathematical representation of biological interactions involving two species: a predator and its prey. In a typical ecosystem, predators depend on prey for food, creating a dynamic interaction playing a crucial role in maintaining the balance of nature. In the Lotka-Volterra model, two differential equations describe the changes in the population sizes of predator and prey. These interactions are usually characterized by alternating cycles of growth and decline in both populations.
- **Prey Population Changes**: Prey grows rapidly when predator numbers are low, as there is less predation pressure. - **Predator Population Changes**: Predator numbers grow when there is an abundance of prey to consume, but decline when prey is scarce.
This model captures the oscillatory nature of population dynamics in ecosystems, but it assumes prey have unlimited resources to sustain growth, which is unrealistic. To remedy this, modifications like logistic growth are integrated to consider ecosystem constraints.

carrying capacity

Carrying capacity, often represented as \(K\), is the maximum population size that an environment can sustain indefinitely. It is a crucial parameter in understanding population dynamics. In nature, resources like food, space, and water limit how many individuals can survive in an ecosystem. Once the population reaches this limit, growth slows down and stabilizes.
- **Influence on Prey**: In predator-prey interactions, carrying capacity affects prey availability, impacting predator populations indirectly.- **Equilibrium Consideration**: As demonstrated in the problem, altering \(K\) shifts how prey and predator populations stabilize. For instance, larger \(K\) values enable larger viable prey populations, potentially supporting more predators.
Thus, examining how carrying capacity affects populations helps ecologists predict changes and set conservation strategies. Maintaining or manipulating \(K\) can aid in controlling population sizes and sustaining ecological balance.

logistic growth

Logistic growth is a model describing how populations expand rapidly at first, then slow as resources become limited. Unlike exponential growth, logistic growth incorporates a carrying capacity that the population cannot surpass.
- **Equation Form**: It is often represented as: \[\frac{dN}{dt} = rN\left(1-\frac{N}{K}\right)\]where \(r\) is the growth rate, \(N\) the population size, and \(K\) the carrying capacity.
- **Stages of Growth**: - **Initial Stage**: Close to exponential when population size is much smaller than \(K\). - **Deceleration Stage**: Growth rate declines as resources dwindle. - **Steady State**: Population stabilizes near \(K\).
By using logistic growth to model prey populations, we get a more realistic depiction of how real-world ecosystems operate, reflecting limitations on growth due to environmental resistance.