Problem 30 - Calculus for Biology and Medicine

Question

The spread of a disease through a population of 100 elephants is modeled by the following system of SIRS equations: $$ \begin{array}{l} \frac{d S}{d t}=\frac{R}{10}-\frac{1}{200} S I \\ \frac{d I}{d t}=\frac{1}{200} S I-\frac{I}{10} I \\ \frac{d R}{d t}=\frac{1}{10} I-\frac{R}{10} \end{array} $$ Find all of the equilibria for this model and classify them (e.g." as stable nodes, unstable nodes, or saddles) by analyzing the linearized system.

Step-by-Step Solution

Answer

Equilibrium at (20,0,80) is a saddle point.

Step 1: Understanding the Variables and Parameters

In this SIRS model, the variables are as follows: \( S(t) \) is the susceptible population, \( I(t) \) is the infected population, and \( R(t) \) is the recovered population at time \( t \). The total population \( N = 100 \) elephants is the sum of \( S(t), I(t), \) and \( R(t) \) at any time. The parameters include the rates of infection, recovery, and susceptible replenishment.

Step 2: Set the Time Derivatives to Zero for Equilibrium

Equilibrium points occur when the time derivatives of all variables are zero, meaning no change over time. Set \( \frac{dS}{dt} = 0 \), \( \frac{dI}{dt} = 0 \), and \( \frac{dR}{dt} = 0 \). Solve the equations: \[ \frac{R}{10} - \frac{1}{200} SI = 0 \] \[ \frac{1}{200} SI - \frac{I}{10} = 0 \] \[ \frac{I}{10} - \frac{R}{10} = 0 \].

Step 3: Solve for Equilibrium Points

Start by solving \( \frac{I}{10} = \frac{R}{10} \), which implies \( I = R \). Substitute in other equations.

Step 4: Substitution and Simplification

Substitute \( I = R \) in \( \frac{1}{200} SI = \frac{I}{10} \) to get \( S = 20 \). This equation simplifies because both sides divide by \( I \), assuming \( I eq 0 \). Therefore, \( S = 20 \), \( I = R = 0 \) or solve \( R = \frac{R}{10} \) leading to \( R = 0 \).

Step 5: Analyze the Equilibrium Point

Check \( R + S = N \) at equilibrium gives \( R = 80 \), \( I = 0 \). At \( S = 20 \), \( I = 0 \), \( R = 80 \) is equilibrium.

Step 6: Linearization and Classification

Linearize by setting \( J = \text{Jacobian matrix} \) and testing eigenvalues to classify. Compute around \( (S,I,R) = (20,0,80) \).

Step 7: Compute Jacobian Matrix

The Jacobian involves partial derivatives with respect to \( S, I, R \). Determine \[ J = \begin{bmatrix} -\frac{I}{200} & -\frac{S}{200} & \frac{1}{10} \ \frac{I}{200} & \frac{S}{200} - \frac{1}{10} & 0 \ 0 & \frac{1}{10} & -\frac{1}{10} \end{bmatrix} \].

Step 8: Evaluate and Solve for Eigenvalues

Solve \( \det(J - \lambda I) = 0 \). The eigenvalues dictate stability: Real parts negative imply stability.

Key Concepts

Equilibrium in the SIRS Model

When studying disease dynamics with the SIRS model, one important goal is to find the equilibrium points. An equilibrium is a state where the numbers of susceptible (\(S\)), infected (\(I\)), and recovered (\(R\)) individuals remain constant over time, meaning the system is in balance.
In our exercise, finding the equilibrium involves setting the time derivatives for these populations to zero:

• \(\frac{dS}{dt} = 0\)
• \(\frac{dI}{dt} = 0\)
• \(\frac{dR}{dt} = 0\)

Solving these equations, we are essentially looking for a point where the inflow and outflow for each group equate, ideally finding numerical values for \(S\), \(I\), and \(R\) that satisfy the equations. For instance, by equating terms in these derivative functions and substituting accordingly, it's given that one significant equilibrium occurs at \(S = 20\), \(I = 0\), and \(R = 80\). These values show how the system can stabilize without any infections spreading further, which is critical in understanding how we might manage disease control.

Linearization of the SIRS Model

Linearization is necessary to analyze complex nonlinear systems, like the SIRS model, around equilibrium points. In this method, we approximate the nonlinear system with a linear one by using a Taylor series expansion around a point of interest — specifically around the equilibrium point.
This is done through creating the Jacobian matrix, which consists of partial derivatives of the system with respect to its variables. The Jacobian essentially represents the sensitivity of each equation to changes in the variables and allows us to predict local behavior near the equilibrium.
In our exercise, after we determine that an equilibrium point exists at \(S = 20\), \(I = 0\), and \(R = 80\), we proceed by finding the Jacobian matrix. For instance, each element of the matrix is calculated as a partial derivative of the original system's equations with respect to \(S\), \(I\), and \(R\). Through this, we can capture how small changes in these populations around equilibrium might propagate, helping us linearly approximate the system's behavior.

Stability Analysis Using Eigenvalues

After linearizing the SIRS model via the Jacobian matrix, the next step is to analyze stability using eigenvalues. Stability analysis seeks to determine whether small disturbances at equilibrium will subside or amplify over time.
This is done by solving for the eigenvalues of the Jacobian matrix. Recall that eigenvalues provide clues about the behavior of the system near equilibrium. The nature of the real parts of these eigenvalues is critical in stability:

• If all eigenvalues have negative real parts, the system is asymptotically stable at that equilibrium—perturbations will fade away.
• If any eigenvalue has a positive real part, the equilibrium is unstable, as disturbances will grow.
• If eigenvalues have zero real parts, the analysis is inconclusive and more investigation is needed.

In our scenario, once the Jacobian is evaluated at \(S = 20\), \(I = 0\), and \(R = 80\), we determine the eigenvalues by solving the characteristic equation, \(\det(J - \lambda I) = 0\). These eigenvalues guide us in classifying the equilibrium as a stable node, unstable node, or saddle point, thereby defining the dynamic behavior of the disease spread under the given conditions.