Problem 31 - Calculus for Biology and Medicine

Question

The spread of a disease through a herd of 50 cattle is modeled by the following system of SIRS equations: $$ \begin{array}{l} \frac{d S}{d t}=R-\frac{1}{50} S I \\ \frac{d I}{d t}=\frac{1}{50} S I-2 I \\ \frac{d R}{d t}=2 I-R \end{array} $$ 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

Equilibria are \((100,0,0)\) as a saddle and \((100,50,100)\) as a spiral source due to respective eigenvalues.

Step 1: Understanding Equilibrium Points

To find the equilibria of the system, set the derivatives to zero: \(\frac{dS}{dt} = 0, \frac{dI}{dt} = 0, \frac{dR}{dt} = 0.\)

Step 2: Solve for S in Terms of I and R

From \(\frac{dS}{dt} = 0\), we have \(R - \frac{1}{50} S I = 0\). Solve for \(S\): \[ S = \frac{50R}{I}. \]

Step 3: Solve for I in Terms of S and R

From \(\frac{dI}{dt} = 0\), we have \(\frac{1}{50} S I - 2I = 0\). Factoring out \(I\), we get: \[ I \left( \frac{1}{50}S - 2 \right) = 0. \]Thus, \(I = 0\) or \(S = 100\).

Step 4: Solve for R in Terms of S and I

From \(\frac{dR}{dt} = 0\), we have \(2I - R = 0\). Solving for \(R\): \[ R = 2I. \]

Step 5: Find Possible Equilibria

Using \(S = 100\) or \(I = 0\), consider these cases:1. If \(I = 0\), then \(R = 0\) and \(S\) must satisfy \(R - \frac{1}{50}SI = 0\); since \(I = 0\), this holds for any \(S\), but \(R = 0\) confirms it.2. If \(S = 100\), use \(R = 2I\) and \(S = \frac{50R}{I}\). This results in \[ (S, I, R) = (100, 0, 0) \text{ or } (100, I, 2I). \] Substitute:\[ S = 100, \ I = 50, \ R = 100. \]

Step 6: Linearize the System around Each Equilibrium

Linearizing involves finding the Jacobian of the system. The Jacobian matrix \(J\) with respect to \((S, I, R)\): \[ J = \begin{bmatrix} -\frac{I}{50} & -\frac{S}{50} & 1 \ \frac{I}{50} & \frac{S}{50} - 2 & 0 \ 0 & 2 & -1 \end{bmatrix} \] Substitute the equilibria \((100,0,0) \text{ and } (100,50,100)\) into the Jacobian.

Step 7: Calculate Eigenvalues for Equilibria

1. At \((100,0,0)\):\[ J = \begin{bmatrix} 0 & -2 & 1 \ 0 & -2 & 0 \ 0 & 2 & -1 \end{bmatrix}. \]Solve \( |J - \lambda I| = 0\): Eigenvalues \(\lambda_1 = 0, \lambda_2 = -2, \lambda_3 = -1\).2. At \((100,50,100)\):\[ J = \begin{bmatrix} -1 & -2 & 1 \ 1 & 0 & 0 \ 0 & 2 & -1 \end{bmatrix}. \]Eigenvalues: \(\lambda_1 = -1, \lambda_2 = 0.5 + \sqrt{-1.5}, \lambda_3 = 0.5 - \sqrt{-1.5}\).

Step 8: Classifying the Equilibria

1. For \((100,0,0)\), eigenvalues \(-2, -1, 0\) suggest a stable node due to no positive eigenvalues but with a degenerate direction \((\lambda = 0)\).2. For \((100,50,100)\), eigenvalues \(0.5 \pm \sqrt{-1.5}, -1\) indicate instability with complex components (implying a spiral).

Key Concepts

Equilibrium Analysis

In the study of dynamic systems, such as the SIRS (Susceptible, Infected, Recovered, and Susceptible again) model, finding equilibrium points is a crucial aspect. These points represent states where the system does not change because all derivatives are zero. In practice, this involves setting the equations from the model to zero, which allows us to solve for the variables within the system.
To find equilibrium points:

• First, ensure that each equation in the system is set to zero. For instance, in our provided SIRS equations, each derivative such as \( \frac{dS}{dt} \) should be zero.
• Next, solve these resulting equations for the variables \( S \), \( I \), and \( R \). This step will often involve substitutions and algebraic manipulation to find consistent solutions.
• Finally, verify these solutions to ensure they meet all conditions imposed by the original system of equations.

Understanding these equilibrium points gives insight into the potential long-term behavior of the system when it's undisturbed by external forces.

Stability Classification

Once equilibrium points have been determined for a system, the next step is understanding their stability. Stability indicates whether small perturbations or changes will die out or grow over time.
There are a few types of stability classifications:

• Stable point: All trajectories in the vicinity of this point will converge towards it over time.
• Unstable point: Trajectories diverge from this point, meaning a slight disturbance can lead the system far from this state.
• Saddle point: Exhibits stability in certain directions and instability in others, making it a complex equilibrium.

Eigenvalues are a mathematical way to determine stability. If all eigenvalues of the Jacobian matrix at that point have negative real parts, it is stable. If any eigenvalue has a positive real part, the system is unstable at that point. For example:

• The equilibrium point \((100,0,0)\) was found to be a stable node due to all non-positive eigenvalues.
• While \((100,50,100)\) showed instability due to positive eigenvalues demonstrating a spiral behavior.

Jacobian Matrix

The Jacobian matrix is a tool used to linearize a system of equations around an equilibrium point. This matrix captures how the system changes in response to small perturbations in each state variable, making it essential for analyzing local behavior and stability.
Here's how to use it:

• Construction: For a system such as the SIRS model, the Jacobian is derived by taking partial derivatives of each equation with respect to each variable, \( S, I, \text{and} \, R \).
• Interpretation: The eigenvalues of the Jacobian at an equilibrium point reveal the point's stability. Real, negative eigenvalues indicate attraction to the equilibrium, while positive values suggest divergence.

For instance, in the exercise, the Jacobian matrix:\[J = \begin{bmatrix} -\frac{I}{50} & -\frac{S}{50} & 1 \\frac{I}{50} & \frac{S}{50} - 2 & 0 \0 & 2 & -1 \end{bmatrix}\]is used to determine the character of equilibria \((100,0,0)\) and \((100,50,100)\), displaying different behaviors based on their respective eigenvalues.