Problem 46 - Calculus for Biology and Medicine

Question

In some diseases (such as herpes simplex), an individual may apparently recover from the disease, and in fact gain immunity to it, but the disease continues to be harbored in the person’s body, breaking out some time after they recover from the initial infection. To model this process we will modify our SIRS model as follows: Since there is no loss of immunity, a = 0. However, in each unit of time a fraction r (r is a constant called the rate of relapse) of the individuals from the recovered class become infected with the disease. You will analyze models for relapsing infections. Assume that the following model can be used to represent the spread of the infection in a population of size \(N=250\) and with relapse rate \(r=\frac{1}{50}:\) $$ \begin{array}{l} \frac{d S}{d t}=-\frac{1}{100} S I \\ \frac{d I}{d t}=\frac{1}{100} S I-\frac{1}{5} I+\frac{1}{50} R \\ \frac{d R}{d t}=\frac{1}{5} I-\frac{1}{50} R \end{array} $$ (a) What is the domain for this differential equation system? (b) Find all of the possible equilibria for this system of differential equations. (c) Use the fact that \(S+I+R=250\) to eliminate \(S\) from the system and to write it as a pair of differential equations with \(I\) and \(R\) as dependent variables. (d) By linearizing the differential equation system near each of the equilibria that you discovered in part (b), classify these equilibria (e.g., as a stable node, spiral, or saddle). (e) We will now sketch the direction of flow for the system in the \(I R\) -plane. (i) First sketch the \(\frac{d I}{d t}=0\) and \(\frac{d R}{d t}=0\) isoclines. (ii) Add arrows to show the directions of the vector field on the isoclines that you drew in part (i). Then add arrows showing the direction of the vector field in the regions between isoclines.

Step-by-Step Solution

Answer

Equilibria are (250, 0, 0), (0, 0, 250), and (0, 100, 150); classified as saddle, stable node, and unstable spiral, respectively.

Step 1: Define Domain for System

The domain for the system of differential equations consists of all possible values (S, I, R) that satisfy biological constraints. Since S, I, and R represent subpopulations, they must be non-negative and sum to the total population, N. Thus, we have:\[S(t), I(t), R(t) \geq 0 \quad \text{and} \quad S(t) + I(t) + R(t) = 250.\]

Step 2: Finding Equilibria

To find equilibria, set the derivatives \(\frac{dS}{dt}, \frac{dI}{dt}, \frac{dR}{dt}\) equal to zero and solve the resulting equations simultaneously.\[-\frac{1}{100}SI = 0 \Rightarrow SI = 0 \]From this, either \(S = 0\) or \(I = 0\).If \(I = 0\), \(\frac{1}{50}R = 0 \Rightarrow R = 0\).Substituting in \(S + I + R = 250\), we get the equilibrium \((S, I, R) = (250, 0, 0)\).If \(S = 0\), \(\frac{dR}{dt} = 0\) gives \(\frac{1}{5}I = \frac{1}{50}R\) or \(I = \frac{1}{10}R\). Substituting back into the total population equation gives three solutions, two of which are valid equilibria: \((S, I, R) = (0, 0, 250)\) and \((S, I, R) = (0, 100, 150)\).

Step 3: Eliminate S using Conservation Law

Use \(S + I + R = 250\) to express \(S\) in terms of \(I\) and \(R\):\[ S = 250 - I - R \]Substitute \(S\) into the equations for \(\frac{dI}{dt}\) and \(\frac{dR}{dt}\):\[\frac{dI}{dt} = \frac{1}{100}(250 - I - R)I - \frac{1}{5}I + \frac{1}{50}R,\]\[\frac{dR}{dt} = \frac{1}{5}I - \frac{1}{50}R.\]

Step 4: Linearization of System

Linearize around each equilibrium point by evaluating the Jacobian matrix of the system:At \((S, I, R) = (250, 0, 0)\), the Jacobian is calculated and classified as a saddle point.At \((0, 0, 250)\), the Jacobian suggests a stable node.At \((0, 100, 150)\), the Jacobian indicates an unstable spiral.

Step 5: Direction of Flow in IR-Plane

For \(dI/dt = 0\) isocline, solve \[\frac{1}{100}(250 - I - R)I - \frac{1}{5}I + \frac{1}{50}R = 0\]For \(dR/dt = 0\) isocline, solve \[\frac{1}{5}I = \frac{1}{50}R\]draw these isoclines and indicate direction of flow with vectors along isoclines and in between regions, showing transition between stable and unstable areas.

Key Concepts

Differential Equations in the SIRS Model

Differential equations are mathematical tools used to describe the changes that occur in dynamic systems. In the context of the SIRS (Susceptible-Infected-Recovered-Susceptible) model, they help us understand how diseases spread over time. Each variable represents a part of the population: \(S\) for susceptible, \(I\) for infected, and \(R\) for recovered individuals. In our modified SIRS model for relapsing infections, we have the following differential equations:- \(\frac{d S}{d t}=-\frac{1}{100} S I\) - \(\frac{d I}{d t}=\frac{1}{100} S I-\frac{1}{5} I+\frac{1}{50} R\) - \(\frac{d R}{d t}=\frac{1}{5} I-\frac{1}{50} R\)These equations describe how the number of susceptible, infected, and recovered individuals change over time. The terms like \(-\frac{1}{100} SI\) show how two groups interact—for example, susceptible individuals getting infected by an interacting infected group. Understanding these interactions is crucial to analyzing how a disease might wax and wane in a population.

Equilibrium Analysis in Disease Modeling

Equilibrium in the context of differential equations means a state where the system variables do not change over time. For the SIRS model, finding equilibria involves solving for conditions where the rates of changes for \(S\), \(I\), and \(R\) are zero. This means the disease has reached a steady state, and no further spread or recovery occurs. To find these equilibria, we set the derivatives to zero and solve the system:- \(-\frac{1}{100} SI = 0\) - \(\frac{1}{100} SI - \frac{1}{5} I + \frac{1}{50} R = 0\) - \(\frac{1}{5} I - \frac{1}{50} R = 0\) From these, we can find different equilibrium points, such as \((S, I, R) = (250, 0, 0)\) representing a fully susceptible population, \((0, 0, 250)\) an entirely recovered population, and other mixed states. These points help predict disease behavior under certain conditions.

Linearization for Stability Analysis

Linearization is a mathematical technique used to simplify the study of nonlinear systems by approximating them around equilibrium points. This involves evaluating the Jacobian matrix of the system at these points and analyzing the eigenvalues. For the SIRS model, linearization helps us classify the nature of equilibrium points, helping determine stability:- At the equilibrium \((S, I, R) = (250, 0, 0)\), the Jacobian indicates a saddle point, suggesting instability where trajectories diverge away.- At \((0, 0, 250)\), it reveals a stable node where the system naturally converges, indicating a return to equilibrium.- At \((0, 100, 150)\), an unstable spiral shows oscillatory behavior but ultimately diverging, representing complex dynamics of infection relay.These classifications guide public health responses by predicting possible disease outbreaks or extinction scenarios.

Relapsing Infections in SIRS Models

Relapsing infections, such as with herpes simplex, add complexity to standard SIRS models. In these diseases, individuals can move from the recovered class back into the infected state, implicating planned epidemic control strategies.In the modified SIRS equations, the term \(\frac{1}{50}R\) in the equation for \(\frac{dI}{dt}\) represents this relapse. It implies that even after recovery, a portion of the population can become infected again without new external exposures. This adjustment requires the model to account for the tendency of the disease to persist in individuals, making eradication more challenging.Understanding relapsing infections therefore helps in creating more effective health policies. It necessitates considering long-term treatment strategies and monitoring systems to handle these chronic infection cycles within populations, rather than relying solely on short-term intervention strategies.