Problem 41 - Calculus for Biology and Medicine

Question

To model the spread of a disease in a population of size \(N\) we derived a differential equation model: $$ \frac{d I}{d t}=(k b-c) I-\frac{k b}{N} I^{2} $$ where \(I(t)\) is the number of infected individuals at time \(t\), and \(k, b\), and \(c\) are all positive coefficients. Assuming that I ( \(t\) ) is modeled by Equation \((8.70)\), you should locate the equilibria of the model, and find which of these equilibria are stable. Draw a vector field plot for each problem. \(k=1, b=1, c=0.5, N=50 .\)

Step-by-Step Solution

Answer

The equilibria are at \( I=0 \) (unstable) and \( I=25 \) (stable).

Step 1: Find the Equilibrium Points

To find equilibria, set \( \frac{dI}{dt} = 0 \). The differential equation is \( \frac{dI}{dt} = (kb - c)I - \frac{kb}{N}I^2 \). Substitute the values \( k = 1, b = 1, c = 0.5, \) and \( N = 50 \):\[ \frac{dI}{dt} = (1 \times 1 - 0.5)I - \frac{1 \times 1}{50}I^2 = 0 \]This simplifies to:\[ 0.5I - \frac{1}{50}I^2 = 0 \]Factor out \( I \):\[ I(0.5 - \frac{1}{50}I) = 0 \]So there are two solutions: \( I = 0 \) and \( I = 25 \).

Step 2: Determine Stability of Equilibria

To determine stability, examine the sign of \( \frac{dI}{dt} \) around \( I = 0 \) and \( I = 25 \). Expand the expression:\[ \frac{dI}{dt} = 0.5I - \frac{1}{50}I^2 \]For values of \( I \) slightly greater than 0, \( \frac{dI}{dt} > 0 \), suggesting instability at \( I = 0 \).For values near \( I = 25 \): Evaluate \( \frac{dI}{dt} = 0.5(25 + \epsilon) - \frac{1}{50}(25 + \epsilon)^2 \) for small \( \epsilon \). Simplified kinetic analysis shows\[ \frac{dI}{dt} = -\frac{1}{50} \epsilon^2 - \frac{1}{2} \epsilon \]such that \( \frac{dI}{dt} \) transitions from negative to positive, marking \( I = 25 \) as stable.

Step 3: Draw a Vector Field Plot

In a vector field plot, show \( \frac{dI}{dt} \) on the vertical axis (as arrows) and \( I \) on the horizontal axis:- For \( I < 25 \), arrows point to the right.- At \( I = 25 \), arrows are stationary \( (\frac{dI}{dt} = 0) \).- For \( I > 25 \), arrows point to the left (back to 25).Thus, arrows move towards \( I = 25 \), showing a stable point.

Key Concepts

Equilibrium Analysis

In mathematical modeling, especially in the study of differential equations, equilibrium analysis is fundamental. Equilibrium points refer to values of the variable in a model where the system remains constant over time. In the context of a disease spread model, these points indicate situations where the number of infected individuals, represented by the variable \( I \), does not change.

To find equilibrium points, we set the first derivative of \( I \) with respect to time, \( \frac{dI}{dt} \), equal to zero. It’s akin to finding the roots of an equation. For the given differential equation \( \frac{dI}{dt} = (kb-c)I - \frac{kb}{N}I^2 \), substituting specific values for the parameters leads to \( 0.5I - \frac{1}{50}I^2 = 0 \). Solving this gives two critical points: \( I = 0 \) and \( I = 25 \).

These equilibrium points, \( I = 0 \) and \( I = 25 \), represent states where the population of infected individuals remains constant, implying that the spread of the disease is neither increasing nor decreasing.

Stability of Equilibria

Determining the stability of an equilibrium point is crucial to understand the long-term behavior of a system. A stable equilibrium means that if the population size is slightly disturbed, it will naturally return to this equilibrium. An unstable equilibrium suggests the opposite.

For stability analysis, we evaluate the sign of \( \frac{dI}{dt} \) near each equilibrium point. For the point \( I = 0 \), if \( I \) is slightly more than zero, \( \frac{dI}{dt} > 0 \), indicating an increase in \( I \). It implies that the equilibrium point at \( I = 0 \) is unstable.

On the other hand, for \( I = 25 \), small deviations (like adding \( \epsilon \)) result in \( \frac{dI}{dt} \) transitioning to remain negative or return to zero, thus indicating that \( I \) reverts back to 25. This point, therefore, is stable. The stability analysis helps in predicting how a population might respond to small perturbations.

Vector Field Plot

A vector field plot provides a visual representation of the behavior of a system governed by differential equations. It helps to illustrate how \( I \), the number of infected individuals in this model, changes with respect to time at different population sizes.

To create the vector field, plot \( \frac{dI}{dt} \) on the vertical axis and \( I \) on the horizontal axis. From the previously determined points:

• When \( I < 25 \), the directional arrows point to the right, indicating an increase in \( I \).
• At \( I = 25 \), arrows are stationary because \( \frac{dI}{dt} = 0 \).
• When \( I > 25 \), arrows point left, showing a decrease in \( I \) towards the equilibrium.

This visualization reflects that \( I = 25 \) is a stable equilibrium where the system tends to settle over time.

Disease Spread Modeling

Disease spread modeling through differential equations helps in understanding how infectious diseases propagate within populations. This model uses parameters to encapsulate real-world conditions such as transmission rates (\( k \)), recovery rates (\( c \)), and population interaction scales (\( b \)).

The core equation \( \frac{dI}{dt} = (kb-c)I - \frac{kb}{N}I^2 \) is a simplified representation that considers initial infection boosts and internal disease dynamics such as herd immunity or saturation effects at larger values of \( I \).

Through an equilibrium analysis revealed by setting \( \frac{dI}{dt} = 0 \), we identify points of stabilization, which tell us under what conditions the disease persists or dies out. Such models are invaluable for public health planning, allowing predictions on epidemic behaviors and guidance on effective intervention strategies.