Problem 14 - Calculus for Biology and Medicine

Question

Assume that $$ \begin{array}{l} \frac{d N}{d t}=5 N-P N \\ \frac{d P}{d t}=P N-P \end{array} $$ (a) Show that this system has two equilibria: the trivial equilibrium \((0,0)\), and a nontrivial one in which both species have positive densities. (b) Use the eigenvalue approach to show that the trivial equilibrium is unstable. (c) Determine the eigenvalues corresponding to the nontrivial equilibrium. Does your analysis allow you to infer anything about the stability of this equilibrium? (d) Use a graphing calculator to sketch curves in the \(N-P\) plane. Also, sketch solution curves of the prey and the predator densities as functions of time.

Step-by-Step Solution

Answer

The equilibria are \((0,0)\) (unstable) and \((1,5)\) (neutral stability).

Step 1: Find the Equilibria

To find the equilibria, set both derivatives to zero: \( \frac{dN}{dt} = 0 \) and \( \frac{dP}{dt} = 0 \). This implies \( 5N - PN = 0 \) and \( PN - P = 0 \). From the first equation, \( N(5 - P) = 0 \), leading to \( N = 0 \) or \( P = 5 \). From the second equation, \( P(N-1) = 0 \), leading to \( P = 0 \) or \( N = 1 \). Thus, the equilibria are \((0,0)\) and \((1,5)\).

Step 2: Stability of Trivial Equilibrium

Evaluate the stability at \((0,0)\) by considering the Jacobian matrix. The partial derivatives give the Jacobian matrix: \[ J = \begin{pmatrix} 5 - P & -N \ P & N - 1 \end{pmatrix} \]. At \((0,0)\), this becomes \( \begin{pmatrix} 5 & 0 \ 0 & -1 \end{pmatrix} \). The eigenvalues are \( \lambda_1 = 5 \) and \( \lambda_2 = -1 \). Since one eigenvalue is positive, \((0,0)\) is unstable.

Step 3: Eigenvalues for Nontrivial Equilibrium

For \((1,5)\), substitute into the Jacobian to get \( J = \begin{pmatrix} 0 & -1 \ 5 & 0 \end{pmatrix} \). The characteristic equation is \( \lambda^2 + 5 = 0 \), giving eigenvalues \( \lambda = \pm i \sqrt{5} \). These are purely imaginary, suggesting a center (neutral stability).

Step 4: Graphical Analysis

Using a graphing calculator, plot the equilibrium points \((0,0)\) and \((1,5)\). Sketch the trajectories, noting the spiral or circular patterns around \((1,5)\), indicating oscillatory behavior. Numerical solutions for \(N(t)\) and \(P(t)\) should show predator-prey cycles.

Key Concepts

Equilibrium Analysis

In dynamical systems, equilibrium analysis is crucial for understanding the long-term behavior of the system. An equilibrium point is where the system remains unchanged, meaning the rate of change is zero. In our exercise, the given predator-prey model equations

• \( \frac{d N}{d t}=5 N-P N \)
• \( \frac{d P}{d t}=P N-P \)

have two main equilibria:

• The trivial equilibrium \((0,0)\) where both predator and prey are absent.
• A nontrivial equilibrium at \((1,5)\) where both species coexist in positive densities.

To find these, you set the derivatives to zero and solve:

• \( 5N - PN = 0 \) implies \( N(5-P) = 0 \) resulting in \( N = 0 \) or \( P = 5 \).
• \( PN - P = 0 \) implies \( P(N-1) = 0 \) resulting in \( P = 0 \) or \( N = 1 \).

Thus, these conditions lead to the two equilibrium points. Understanding these helps us predict how populations can stabilize.

Jacobian Matrix

The Jacobian matrix is a tool used in equilibrium analysis to determine the local behavior of a dynamical system around equilibrium points. For a system with two variables \(N\) and \(P\), the Jacobian matrix \(J\) is derived from the partial derivatives of the system's equations:\[ J = \begin{pmatrix} \frac{\partial f}{\partial N} & \frac{\partial f}{\partial P} \ \frac{\partial g}{\partial N} & \frac{\partial g}{\partial P} \end{pmatrix} \]In our example,

• \( \frac{\partial f}{\partial N} = 5 - P \)
• \( \frac{\partial f}{\partial P} = -N \)
• \( \frac{\partial g}{\partial N} = P \)
• \( \frac{\partial g}{\partial P} = N - 1 \)

Thus, the Jacobian matrix is:\[ J = \begin{pmatrix} 5 - P & -N \ P & N - 1 \end{pmatrix} \]Evaluating this matrix at our equilibrium points provides further insight into system behavior. It is especially helpful in assessing stability based on the eigenvalues derived from \(J\).

Eigenvalues and Stability Analysis

To determine the stability of equilibrium points, we analyze the eigenvalues of the Jacobian matrix. Stability is assessed based on whether these eigenvalues are positive, negative, or imaginary.

• If at least one eigenvalue is positive, the equilibrium is unstable.
• If all eigenvalues are negative, the equilibrium is stable.
• Imaginary eigenvalues suggest oscillatory behavior or neutral stability.

At the trivial equilibrium \((0,0)\), the Jacobian is: \[ J = \begin{pmatrix} 5 & 0 \ 0 & -1 \end{pmatrix} \]The eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = -1\). A positive eigenvalue indicates instability, making \((0,0)\) an unstable point.

For the nontrivial point \((1,5)\), the matrix is: \[ J = \begin{pmatrix} 0 & -1 \ 5 & 0 \end{pmatrix} \]Its characteristic equation is \( \lambda^2 + 5 = 0 \) leading to eigenvalues \( \lambda = \pm i\sqrt{5} \). These imaginary roots indicate that the system demonstrates neutral stability with cyclical, oscillatory behavior at this equilibrium, typical in predator-prey dynamics.

Predator-Prey Models

Predator-prey models provide a mathematical framework to study ecological interactions between species. They help us understand how populations of predators and prey fluctuate over time. The given model

• \( \frac{d N}{d t}=5 N-P N \)
• \( \frac{d P}{d t}=P N-P \)

describes such interactions, showing how prey growth, constrained by predators, influences predator survival.

The terms:

• \(5N\) represents exponential growth of prey with no predators.
• \(-PN\) captures the decrease by predation.
• \(PN\) shows increase in predator numbers proportional to prey availability.
• \(-P\) suggests predator decline in the absence of prey.

Analyzing these equations through stability and graphical methods illustrates typical cycles. Preys' growth unchecked by predators leads to more predation, increasing predator numbers. However, as prey diminishes, predator numbers dwindle, allowing prey recovery. Predominantly, this model encapsulates the balance and flux vital in ecosystems.