Problem 34 - Calculus for Biology and Medicine

Question

The spread of a disease through a population of 100 individuals is represented by the following SIRS model: $$ \begin{array}{l} \frac{d S}{d t}=\frac{1}{10} R-\frac{1}{100} S I \\ \frac{d I}{d t}=\frac{1}{100} S I-\frac{1}{2} I \\ \frac{d R}{d t}=\frac{1}{2} I-\frac{1}{10} R \end{array} $$ In this problem we will sketch the directions of the solution in the SI -plane. (a) Eliminate \(R\) to rewrite the equation system as a system of differential equations in the dependent variables \(S\) and \(I\). (b) Draw the zero isoclines of your system from part (a). (c) 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. (d) Add to your plot from part (b) arrows showing the direction of the vector field on the isoclines, and in the regions between the isoclines.

Step-by-Step Solution

Answer

The system equilibrium at \((S, I) = (50, 13.33)\) is a saddle point, indicating instability.

Step 1: Eliminate R from the Equations

To eliminate \(R\), use the fact that the total population remains constant: \(S + I + R = 100\). Hence, \(R = 100 - S - I\). Substitute this expression into the given differential equations: - Replace \(R\) in \(\frac{dS}{dt} = \frac{1}{10}R - \frac{1}{100}SI\), resulting in \[\frac{dS}{dt} = \frac{1}{10}(100 - S - I) - \frac{1}{100}SI.\] - Simplify to obtain \(\frac{dS}{dt} = 10 - \frac{S}{10} - \frac{I}{10} - \frac{1}{100}SI\). Now express the system in terms of only \(S\) and \(I\).

Step 2: Simplify and Express the Differential Equations

Using the substitution for \(R\), simplify the equations:1. \(\frac{dS}{dt} = 10 - \frac{S}{10} - \frac{I}{10} - \frac{1}{100}SI.\) 2. The original given equation for \(I\) remains: \(\frac{dI}{dt} = \frac{1}{100}SI - \frac{1}{2}I.\) This is the system of differential equations in the variables \(S\) and \(I\).

Step 3: Draw Zero Isoclines

Zero isoclines occur where the rate of change is zero. For the isoclines:- Set \(\frac{dS}{dt} = 0\). Solve for \(I\) in terms of \(S\): \[\frac{dS}{dt} = 10 - \frac{S}{10} - \frac{I}{10} - \frac{1}{100}SI = 0.\] Resulting in the isocline: \[I = \frac{1000 - 10S}{10 + S}.\]- Set \(\frac{dI}{dt} = 0\). Solve for \(S\) in terms of \(I\): \[\frac{dI}{dt} = \frac{1}{100}SI - \frac{1}{2}I = 0.\] Resulting in the isocline: \[S = 50.\]

Step 4: Equilibrium Points

Equilibrium points exist where both \(\frac{dS}{dt} = 0\) and \(\frac{dI}{dt} = 0\). - From \(\frac{dI}{dt} = 0\), we have \(S = 50\).- Substitute \(S = 50\) into the \(\frac{dS}{dt} = 0\) isocline equation: \(I = \frac{1000 - 10(50)}{10 + 50} = 13.33.\) Thus, the equilibrium point is \((S, I) = (50, 13.33)\).

Step 5: Linear Stability Analysis

To classify the equilibrium, compute the Jacobian matrix at the equilibrium point. The Jacobian matrix derived from the system is: \[J = \begin{pmatrix} -\frac{1}{10} - \frac{I}{100} & -\frac{1}{10} - \frac{S}{100} \ \frac{I}{100} & \frac{S}{100} - \frac{1}{2} \end{pmatrix}. \]Substitute \(S = 50\) and \(I = 13.33\) into the Jacobian to evaluate the eigenvalues. If the eigenvalues are real and of opposite signs, the equilibrium is a saddle point.

Step 6: Sketch the Vector Field

Add arrows to the plot of isoclines to indicate the vector field's direction:- In regions where \(\frac{dS}{dt} > 0\) or \(\frac{dI}{dt} > 0\), draw arrows indicating increasing behavior.- Conversely, where \(\frac{dS}{dt} < 0\) or \(\frac{dI}{dt} < 0\), indicate decreasing behavior.- Between isoclines and around equilibrium points, use the linearized system's results to determine stable or unstable directions.

Key Concepts

Differential Equations

In the context of the SIRS model for disease spread, differential equations play a crucial role in understanding how different compartments in a population change over time. Differential equations describe relationships where the rates at which variables change are given, rather than their exact values at a point. This allows us to model dynamic processes, such as infections, recoveries, and susceptibilities in populations.

In our exercise, the system of differential equations is presented as follows:

• For susceptibles (S): \[ \frac{d S}{d t} = \frac{1}{10} R - \frac{1}{100} SI \]
• For infected (I): \[ \frac{d I}{d t} = \frac{1}{100} SI - \frac{1}{2} I \]
• For recovered (R): \[ \frac{d R}{d t} = \frac{1}{2} I - \frac{1}{10} R \]

These equations use derivatives (indicated by \frac{d}{dt}) to express the rates of change of S, I, and R over time. Through substitution and simplification, these equations are rewritten to focus only on S and I, since the total population is constant, simplifying the model to be more manageable. By finding where these rates equal zero, or other specific scenarios, we can predict the behavior of disease spreading in the population.

Equilibrium Points

Within the sphere of differential equations, equilibrium points are a primary focus. These points indicate where there's no change in the state variables over time; essentially, the system is in balance.

In the SIRS model, equilibrium points occur when both \(\frac{dS}{dt} = 0\) and \(\frac{dI}{dt} = 0\). In our exercise, we simplified these conditions by:

• Solving \(\frac{dI}{dt} = 0\) to find that \(S = 50\).
• Substituting \(S = 50\) into \(\frac{dS}{dt} = 0\) to solve for \(I\), yielding \(I = 13.33\).

Therefore, the equilibrium point in this model is at \((S, I) = (50, 13.33)\). This point is crucial as it theoretically represents a moment of pause or balance in the disease spread, where infection and recovery rates level out. These points provide insights into long-term trends and the potential containment of disease outbreaks.

Linear Stability Analysis

Once we determine the equilibrium points, the next step is analyzing their stability, which tells us whether small disturbances will die out or escalate. This is done through linear stability analysis.

The process involves calculating the Jacobian matrix from the system of differential equations at the equilibrium point. For our reduced SIRS model, the Jacobian matrix is:

• \[J = \begin{pmatrix} -\frac{1}{10} - \frac{I}{100} & -\frac{1}{10} - \frac{S}{100} \ \frac{I}{100} & \frac{S}{100} - \frac{1}{2} \end{pmatrix} \]

Explanation continues by substituting the equilibrium values \((S, I) = (50, 13.33)\) into this matrix to analyze the eigenvalues:

• If eigenvalues are real and of opposite signs, the point is a saddle point, indicating instability.
• Eigenvalues with specific conditions determine if the equilibrium is a stable node or another classification.

This classification helps in understanding the potential outcomes of disease dynamics in a population and if an initial outbreak could settle into a steady state or explode.