Question

Epidemics. The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years many mathematical models have been proposed and studied for many different diseases. Deal with a few of the simpler models and the conclusions that can be drawn from them. Similar models have also been used to describe the spread of rumors and of consumer products. Suppose that a given population can be divided into two parts: those who have a given disease and can infect others, and those who do not have it but are susceptible. Let \(x\) be the proportion of susceptible individuals and \(y\) the proportion of infectious individuals; then \(x+y=1 .\) Assume that the disease spreads by contact between sick and well members of the population, and that the rate of spread \(d y / d t\) is proportional to the number of such contacts. Further, assume that members of both groups move about freely among each other, so the number of contacts is proportional to the product of \(x\) and \(y .\) since \(x=1-y\) we obtain the initial value problem $$ d y / d t=\alpha y(1-y), \quad y(0)=y_{0} $$ where \(\alpha\) is a positive proportionality factor, and \(y_{0}\) is the initial proportion of infectious individuals. (a) Find the equilibrium points for the differential equation (i) and determine whether each is asymptotically stable, semistable, or unstable. (b) Solve the initial value problem (i) and verify that the conclusions you reached in part (a) are correct. Show that \(y(t) \rightarrow 1\) as \(t \rightarrow \infty,\) which means that ultimately the disease spreads through the entire population.

Short answer

In summary, we found two equilibrium points for the given differential equation modeling the spread of a contagious disease: \(y_1 = 0\) and \(y_2 = 1\). The equilibrium point \(y_1 = 0\) is unstable, while \(y_2 = 1\) is asymptotically stable, meaning that the disease will eventually spread through the entire population. We solved the initial-value problem and found a formula for \(y(t)\), which verifies our conclusions from the stability analysis.

Step-by-step solution

Part (a): Finding Equilibrium Points and Stability Analysis

To find the equilibrium points, we must first set the differential equation to zero and solve for y: $$ 0 = \alpha y(1 - y) $$ This gives us two equilibrium points: $$ y_1 = 0 \quad \text{and} \quad y_2 = 1 $$ Now, let's analyze the stability of these equilibrium points. For this, we will compute the derivative of \(dy/dt\) with respect to y: $$ \frac{d^2y}{dt^2} = \frac{d}{dy}\left(\alpha y(1-y)\right) = \alpha - 2 \alpha y $$ Now, evaluate this expression at the equilibrium points \(y_1\) and \(y_2\): $$ \frac{d^2y}{dt^2}(y_1) = \alpha > 0 \quad \Rightarrow \quad y_1 = 0 \quad \text{is unstable} $$ $$ \frac{d^2y}{dt^2}(y_2) = -\alpha < 0 \quad \Rightarrow \quad y_2 = 1 \quad \text{is asymptotically stable} $$ From this analysis, we conclude that \(y_1 = 0\) is an unstable equilibrium point, and \(y_2 = 1\) is an asymptotically stable equilibrium point.

Part (b): Solving the Initial-Value Problem and Verifying Conclusions

To solve the given initial-value problem, we need to find a formula for \(y(t)\). Since the given differential equation is separable: $$ \frac{dy}{y(1-y)} = \alpha dt $$ Now integrate both sides of the equation: $$ \int \frac{dy}{y(1-y)} = \int \alpha dt $$ Let's perform a partial fraction decomposition for \(\frac{1}{y(1-y)}\), and we get: $$ \frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y} $$ Solving for A and B, we find that \(A = -B = 1\). So, we can rewrite the integral above as: $$ \int \left(\frac{1}{y} - \frac{1}{1-y}\right)dy = \alpha \int dt $$ Now, perform the integrations: $$ \ln(\frac{y}{1-y}) = \alpha t + C $$ Let's solve for y(t), taking the exponential of both sides and applying the initial condition \(y(0) = y_0\): $$ y(t) = \frac{y_0 e^{\alpha t}}{1 + y_0 (e^{\alpha t} - 1)} $$ Our conclusions from part (a) were that \(y_1 = 0\) is an unstable equilibrium point and \(y_2 = 1\) is an asymptotically stable equilibrium point. This agrees with the outcome of this solution since: - If \(y_0 > 0\), we have \(y(t) \rightarrow 1\) as \(t \rightarrow \infty\), meaning that the disease spreads to the entire population. - The solution above also confirms the stability analysis done previously. Thus, our conclusions from part (a) were correct, and we've learned that in this simple model, ultimately, the disease will spread through the entire population.

Key concepts

Differential Equations

Mathematical epidemiology often uses differential equations to model the spread of infectious diseases. In our problem, the equation provided is \( \frac{dy}{dt}=\alpha y (1-y) \). This is known as a first-order differential equation. It's called 'first-order' because it involves the first derivative of the variable \( y \), which represents the proportion of infectious individuals in a population.
Differential equations help describe how a quantity changes over time. In epidemics, they illustrate the rate of change and how quickly a disease spreads within a population. Because \( \alpha y (1-y) \) is the product of the susceptible and infectious populations, it naturally ties together the interaction between these groups, showing how the disease moves through them.

These equations are crucial for understanding dynamic processes in which numerous factors change simultaneously. They offer a framework to predict future behavior based on current data. For our exercise, this means estimating how the infectious proportion of the population changes as time progresses.

Stability Analysis

In the study of differential equations, stability analysis is used to determine the behavior of solutions as time approaches infinity. It helps determine if equilibrium points are stable or unstable.
In our problem, the equilibrium points are \( y_1 = 0 \) and \( y_2 = 1 \). By examining the second derivative, \( \frac{d^2y}{dt^2} = \alpha - 2\alpha y \), we analyze their stability. When evaluated at \( y_1 \), we find \( \alpha > 0 \), indicating it is unstable. Conversely, at \( y_2 \), the negative value suggests \( y_2 \) is asymptotically stable.

Stability analysis tells us if small perturbations in a system will die out (leading to stability) or grow (indicating instability). This analysis is critical because it helps predict whether an epidemic will die out (if unstable) or persist (if stable).
For a local analysis, if even one small number (\( y \)) becomes bigger over time, it shows that the infection proportion will increase, confirming \( y_2 \).

Equilibrium Points

Equilibrium points are the values where the differential equation equals zero, meaning no change over time. Setting \( \frac{dy}{dt} = 0 \) allows us to find them. In our exercise, \( y_1 = 0 \) and \( y_2 = 1 \) are equilibrium points.
\( y_1 = 0 \) represents the scenario where no one is infected. However, it is an unstable point because any increase in \( y \) can lead to more infections until \( y_2 = 1 \). On the other hand, \( y_2 = 1 \) represents an entire population infection, showcasing asymptotic stability as the system tends to this point when \( t \to \infty \).

Understanding equilibrium points offers insights into long-term disease behavior. It informs public health strategies, such as vaccination, to keep \( y \) low and close to zero, representing minimal disease presence.

Initial Value Problem

An initial value problem in differential equations gives a complete scenario by including a starting point, typically given as \( y(0)=y_0 \). This kind of problem allows us to find a unique solution for \( y(t) \), as demonstrated by solving \( \frac{dy}{dt}=\alpha y(1-y) \). Using initial conditions, we derive the formula \( y(t) = \frac{y_0 e^{\alpha t}}{1 + y_0 (e^{\alpha t} - 1)} \), showing how \( y \) evolves over time.
The initial value \( y_0 \) is crucial because it sets the system's path. Small differences in \( y_0 \) can lead to vastly different paths \(( y(t) )\) over time.

In epidemiology, initial value problems are essential for planning interventions, understanding disease outbreak progression, and predicting future healthcare needs based on initial infection data.