Problem 19 - 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): $$ \begin{array}{l} \frac{d N}{d t}=3 N\left(1-\frac{N}{10}\right)-2 P N \\ \frac{d P}{d t}=P N-4 P \end{array} $$ (a) Explain why the prey evolves according to $$ \frac{d N}{d t}=3 N\left(1-\frac{N}{10}\right) $$ in the absence of the predator. Investigate the long-term behavior of solutions to (11.87). (b) Find all equilibria of (11.86), and use the eigenvalue approach to determine their stability. (c) Use a graphing calculator to sketch the solution curve of \((11.86)\) in the \(N-P\) plane when \(N(0)=2\) and \(P(0)=2 .\) Also, sketch \(N(t)\) and \(P(t)\) as functions of time, starting with \(N(0)=2\) and \(P(0)=2\).

Step-by-Step Solution

Answer

Prey shows logistic growth with a carrying capacity of 10. Equilibria are \((0,0), (10,0), (8,2)\) with different stabilities based on eigenvalues. Graphs show behavior starting from \(N(0)=2, P(0)=2\).

Step 1: Modeling Prey Growth in Absence of Predators

The differential equation for prey population, \( \frac{dN}{dt} = 3N\left(1-\frac{N}{10}\right) \), describes logistic growth. In the absence of predators (when \( P = 0 \)), prey grows logistically, starting at a rate proportional to its current population \( N \), and slows as it approaches the carrying capacity, which is 10 in this case.

Step 2: Determine Long-term Behavior

For the equation \( \frac{dN}{dt} = 3N\left(1-\frac{N}{10}\right) \), set \( \frac{dN}{dt} = 0 \) to find equilibrium points. This gives \( N=0 \) or \( N=10 \). Over time with no predators, the population \( N \) will approach 10, the carrying capacity, since any initial population greater than 0 and less than 10 will grow towards 10.

Step 3: Finding Equilibrium Points in the Full System

To find equilibria for both predator and prey, set \( \frac{dN}{dt} = 0 \) and \( \frac{dP}{dt} = 0 \). From \( \frac{dN}{dt} = 3N\left(1-\frac{N}{10}\right) - 2PN = 0 \), solve for \( N \) or \( P \). Likewise, from \( \frac{dP}{dt} = PN - 4P = 0 \), solve for \( N \) or \( P \). Solutions are \( (N, P) = (0, 0), (10, 0), (8, 2) \).

Step 4: Stability Analysis Using Eigenvalues

Calculate the Jacobian matrix for the system and evaluate it at each equilibrium point. For example, at \( (N, P) = (0, 0) \), compute eigenvalues of the Jacobian. Repeat for \( (10, 0) \) and \( (8, 2) \). If the real parts of the eigenvalues are negative, the equilibrium is stable; positive real parts indicate instability.

Step 5: Sketching Solution Curves

Using a graphing calculator or software, graph the trajectories in the \( N-P \) plane starting from \( N(0) = 2 \), \( P(0) = 2 \). Plot individual graphs for \( N(t) \) and \( P(t) \) over time. This visualizes how the populations evolve and interact over time given initial conditions.

Key Concepts

Logistic Growth

Logistic growth is a model that describes how a population expands in an environment with limited resources. It reflects a more realistic scenario compared to exponential growth, itself represented by the idea of unlimited resources. In logistic growth, the population starts growing rapidly when it is well below the carrying capacity. The rate of growth eventually decreases as the population approaches the environment's capacity to sustain the individuals naturally.

In mathematical terms, logistic growth can be illustrated by the differential equation \( \frac{dN}{dt} = rN(1 - \frac{N}{K}) \). Here, \(r\) represents the intrinsic growth rate, \(N\) is the population size, and \(K\) is the carrying capacity, which represents the maximum population size that the environment can handle.

Therefore, for the prey in the Lotka-Volterra model, which has its population described by \( \frac{dN}{dt} = 3N(1 - \frac{N}{10}) \), 10 is identified as the carrying capacity. This setup suggests that without predators, the prey population will grow until it reaches the carrying capacity, stabilizing the population.

Equilibrium Points

Equilibrium points are critical for understanding the behavior of a system. They are conditions where the system does not change over time, meaning both derivatives \( \frac{dN}{dt} \) and \( \frac{dP}{dt} \) are zero. Solving these equations leads us to the equilibrium points for the system.

In our specific model claims, we find the equilibria by

• Setting \( \frac{dN}{dt} = 3N(1 - \frac{N}{10}) - 2PN = 0 \)
• Setting \( \frac{dP}{dt} = PN - 4P = 0 \)

Solving these leads to

• (N, P) = (0, 0): both prey and predator populations are extinct.
• (10, 0): prey population at carrying capacity, no predators present.
• (8, 2): an interesting point balancing both predator and prey interactions.

These points add significant insights into the system's life behavior and indicate potential steady states of population levels.

Stability Analysis

Stability analysis helps determine whether an equilibrium point is stable or unstable. It involves examining how slight deviations from these points affect the system. Specifically, we use eigenvalues from the Jacobian matrix, which is formed by the partial derivatives of the system's coupled differential equations.

The analysis of stability involves:

• Calculating the Jacobian matrix for the model at equilibrium points.
• Evaluating eigenvalues: Negative real parts of eigenvalues indicate stability, while positive real parts suggest instability.

At \((N, P) = (0, 0)\), both species being extinct is usually less interesting for long-term implications. Similarly, at \((10, 0)\), with only prey present at carrying capacity, the absence of natural predators poses questions about real-world applicability. The mix at \((8, 2)\), where both species coexist, is dynamic and offers a real insight into interaction stability between species. Understanding this interaction through stability guides us to predict species behavior over time.

Predator-Prey Dynamics

Predator-prey dynamics delve into how populations of two interacting species—one as predator and the other as prey—affect each other's growth through natural interactions. This dynamic is effectively captured in the Lotka-Volterra and similar models.

Unchecked, prey populations grow according to logistic models, until limited by resources. Meanwhile, predator populations depend on prey numbers for food, affecting their own growth rates. The interplay between these factors dictates how both populations change over time, creating cycles of growth and decline for both species.

• High prey availability can lead to predator population growth, as there is abundance to feed on.
• In contrast, a smaller prey population can starve the predators, causing their numbers to fall.

Such dynamics show the importance of balance in ecosystems, demonstrating the natural checks and balances in predator-prey interactions.

Differential Equations

Differential equations form the backbone of modeling change in systems over time. In the context of predator-prey models, these equations describe how the populations evolve interacting with one another.

The systems take the form of coupled differential equations, representing each species' growth:

• \(\frac{dN}{dt}\) for prey changes based on the intrinsic growth rate minus the effects of predation.
• \(\frac{dP}{dt}\) for predators, indicating growth through prey consumption and their own natural decline.

By setting these differences for time derivatives, we can analyze or simulate the future state of populations. Being differential equations means the focus remains on rates of change and not values alone, providing dynamic insight and predictive power into how populations will play out over long intervals under given conditions.