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Chapter 17: Problem 11

Suppose immunity is not permanent in the SIR model, and recovered people become susceptible after six months. Modify the meaning of \(R\) and the SIR equations to account for this possibility.

Short Answer

Expert verified
Modify SIR equations to include a return from R to S.

Step by step solution

01

Understanding the Problem

In the traditional SIR model, individuals move from susceptible (S) to infected (I) to recovered (R). Recovered individuals are assumed to have permanent immunity. In this modified problem, immunity is temporary, and recovered individuals become susceptible again after six months.
02

Modify the Meaning of R

The compartment R, which originally represents permanent recovery, now represents temporary recovery. After six months, individuals in compartment R re-enter the susceptible compartment S.
03

Revise the SIR Equations

To incorporate the temporary immunity, we need to modify the SIR model equations: 1. The equation for S should include a term that accounts for individuals moving from R back to S. Let \( \gamma_2 \) be the rate at which recovered individuals return to being susceptible. 2. The equation for R should be modified to subtract the individuals transitioning back to S.The modified equations are:\[ \frac{dS}{dt} = -\beta SI + \gamma_2 R \]\[ \frac{dI}{dt} = \beta SI - \gamma I \]\[ \frac{dR}{dt} = \gamma I - \gamma_2 R \]
04

Understand the Parameters

In these equations, \( \beta \) is the rate of infection, \( \gamma \) is the recovery rate from infection, and \( \gamma_2 \) is the rate of losing immunity, which corresponds to individuals returning to the susceptible state after six months.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temporary Immunity
In traditional epidemiological models like the SIR (Susceptible, Infected, Recovered) model, after recovering from an infection, individuals are assumed to gain lifelong immunity. However, this is not always the case for all diseases. In some instances, immunity can be temporary. When immunity is temporary, recovered individuals can become susceptible again after a period of time. This is crucial for understanding diseases like the flu where immunity might only last for a few months.
  • In our modified SIR model, the compartment R represents temporary recovery rather than permanent immunity.
  • Once individuals recover, they move to the R compartment. Over time, they gradually lose their immunity.
  • After the immunity wanes, they re-enter the susceptible category (S) and can be reinfected.
This constant cycle emphasizes the importance of vaccination and other control measures to manage diseases with temporary immunity.
Differential Equations
Differential equations are a powerful tool in modeling dynamic systems like epidemiological processes. They describe how variables change over time in relation to one another and are fundamental to predicting disease spread. In the SIR model, differential equations are used to represent the rates of change between the compartments (S, I, and R).
  • The equation \( \frac{dS}{dt} = -\beta SI + \gamma_2 R \) shows that the susceptible population decreases as they are infected (proportional to \( \beta SI \)), but some become susceptible again after immunity is lost (proportional to \( \gamma_2 R \)).
  • The rate of infection is modeled by \( \frac{dI}{dt} = \beta SI - \gamma I \), where new infections increase the infected population and recovery decreases it.
  • Lastly, \( \frac{dR}{dt} = \gamma I - \gamma_2 R \) represents the rate of recovery minus the rate at which immunity is lost, pushing some individuals back to the susceptible state.
Through these equations, the dynamics of diseases with non-permanent immunity can be accurately studied.
Epidemiological Models
Epidemiological models are vital for understanding the spread and dynamics of infectious diseases within populations. They help in predicting outbreaks and assessing the impacts of various interventions. The SIR model is a classic compartmental model that divides the population into three groups: Susceptible (S), Infected (I), and Recovered (R).
  • These models provide insights on how quickly an infection can spread and how long an outbreak might last.
  • Adjustments to simple models, such as incorporating temporary immunity, make predictions more realistic for certain diseases.
  • By understanding the flow between compartments, public health strategies like vaccination campaigns can be effectively planned.
With modern enhancements and real-world data, these models become more precise, allowing for better preparation and response to infectious disease threats.

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Most popular questions from this chapter

Argue that \(y=\tan t \quad\) is the only solution to \(\quad y(0)=0, \quad y^{\prime}(t)=1+y^{2}(t)\) Use the integral formula, \(\int \frac{y^{\prime}(\tau)}{1+y^{2}(\tau)} d \tau=\arctan y(\tau)+C\).

Some population scientists have argued that population density can get so low that reproduction will be less than natural attrition and the total population will be lost. Named the Allee effect. after W. C. Allee who wrote extensivelv about it \(^{6}\), this mav be a basis for arguing. for example that marine fishing of a certain species (Atlantic cod on Georges Bank, for example \({ }^{7}\) ee Paul Greenberg, "Four Fish, the Future of the Last Wild Food, The Penguin Press, 2010 . The model is a lot more complicated than we present here.) should be suspended, despite the presence of a small residual population. How should we modify the logistic differential equation, \(u^{\prime}=u(1-u),\) to incorporate such a threshold? Assume a fixed area and uniform density throughout the area and a threshold number, \(\epsilon .\) If the population number is less than \(\epsilon\) the population will decline; if the population number is more than \(\epsilon\) the population will increase. a. Modify the direction field for \(u^{\prime}=u(1-u)\) to account for the Allee effect. That is, a \(u, t\) plane, draw the line \(u=1\) and a threshold line \(u=\epsilon\) where \(\epsilon=0.1,\) say. Arrows below \(u=\epsilon\) should point downward; arrows between \(u=\epsilon\) and \(u=1\) should point upwards. Draw enough direction field arrows to indicate the paths of solutions for a threshold model. b. Draw the the logistic phase plane graph and the phase plane graphs for the following three candidates of a threshold logistic differential equation where \(\epsilon=0.1\). \(u^{\prime}=f(u)=u \times(1-u) \quad\) Logistic \(u^{\prime}=f_{1}(u)=u^{\frac{2}{3}} \times(u-\epsilon)^{\frac{1}{3}} \times(1-u) \quad\) Candidate 1 \(u^{\prime}=f_{2}(u)=u \times \frac{u-\epsilon}{u+\epsilon} \times(1-u)\) Candidate 2 \(u^{\prime}=f_{3}(u)=u \times(u-\epsilon) \times(1-u)\) Candidate 3

Show that the variables are not separable in the equation $$\text { a. } \quad y^{\prime}(t)=\ln (t \times y) \quad \text { b. } \quad y^{\prime}(t)=\ln (t+y)$$

The special case of \(y^{\prime}=f(t, y)\) in which \(f(t, y)=F(t)(f\) is independent of \(y)\) has a familiar solution from the Fundamental Theorem of Calculus I. Check by substitution that $$y(t)=y_{a}+\int_{a}^{t} F(x) d x \quad \text { solves } \quad y(a)=y_{a} \quad \text { and } \quad y^{\prime}(t)=F(t)$$ The differential equation $$y(a)=y_{a} \quad y^{\prime}(t)=F(t)$$ has therefore been completely solved. Henceforth we will consider that \(f\) is dependent on \(y\) and possibly also on \(t\).

Show that each solution satisfies the initial condition and the differential equation. $$\begin{aligned} &\text { Solution } \quad \text { Initial Condition } \quad \text { Differential Equation }\\\ &\text { a. } \quad y(t)=e^{2 t}+e^{t} \quad y(0)=2 \quad y^{\prime}(t)-y(t)=e^{2 t} \end{aligned}$$ $$\text { b. } \quad y(t)=\frac{1}{3} e^{t}+\frac{2}{3} e^{-2 t}, \quad y(0)=1, \quad y^{\prime}(t)+2 y(t)=e^{t}$$ c. \(\quad y(t)=t e^{t} \quad y(0)=0 \quad y^{\prime}(t)-y(t)=e^{t}\) d. \(\quad y(t)=\frac{t^{2}}{3}+\frac{1}{t}, \quad y(1)=\frac{4}{3}, \quad t \times y^{\prime}(t)+y(t)=t^{2}\) e. \(\quad y(t)=\sqrt{t+1} \quad y(0)=1 \quad y(t) \times y^{\prime}(t)=\frac{1}{2}\) f. \(\quad y(t)=\sqrt{1+t^{2}}, \quad y(0)=1, \quad y(t) \times y^{\prime}(t)=t\) \(\begin{array}{lll}\text { g. } y(t)=\sqrt{4+t^{2}} & y(0)=2 & y(t) \times y^{\prime}(t)=t\end{array}\) \(\begin{array}{lll}\text { Solution } & \text { Initial } & \text { Differential Equation }\end{array}\) Condition h. \(y(t)=\frac{1}{t+1}, \quad y(0)=1, \quad y^{\prime}(t)+(y(t))^{2}=0\) i. \(\quad y(t)=0.5+0.5 e^{-0.2 \sin t} \quad y(0)=1, \quad y^{\prime}(t)+0.2(\cos t) y(t)=0.1 \cos t\) j. \(\quad y(t)=\tan t, \quad y(0)=0, \quad y^{\prime}(t)=1+(y(t))^{2}\) k. \(y(t)=3\) \(y(0)=3, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\) l. \(y(t)=5\) \(y(0)=5, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\)
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