# Tony Tsai

## May the force of science be with you

Apr 22, 2017 - 3 minute read - Comments - Influenza

# Derivation of Number of Infected Individuals I(a)

(Hens et al. 2012, 63:41) gives the formula for the total number of infective individuals $I(a)$:

$$$I(a) = \frac{\lambda}{\lambda - \nu}N(0)l(a)\left[e^{-\nu a} - e^{-\lambda a}\right], \tag{1}$$$

which is obtained by integrating following differential equation with respect to age $a$:

$$$\frac{dI(a)}{da}=\lambda S(a) - (\nu + \mu)I(a). \tag{2}$$$

This is also equation (4.11) in (Anderson and May 1992, 67). Here I will present the details of integration.

Equation (2) can be written as follows:

$$$\frac{dI(a)}{da} + (\nu + \mu)I(a)=\lambda S(a). \tag{3}$$$

This equation can be solved using the technique of “integrating factors”1 by following the steps below:

Step 1. Multiply both sides of the equation by $e^{(\nu + \mu)a}$, to obtain the following:

$$$\frac{dI(a)}{da}e^{(\nu + \mu)a} + I(a)(\nu + \mu)e^{(\nu + \mu)a}=\lambda S(a)e^{(\nu + \mu)a}. \tag{4}$$$

Note that, according to the rules of differentiation, the left-hand side of this equation is equivalent to the derivative of $I(a)e^{(\nu + \mu)a}$ with respect to $a$, and so the equation can be rewritten as follows:

$$$\frac{d}{da}(I(a)e^{(\nu + \mu)a})=\lambda S(a)e^{(\nu + \mu)a}. \tag{5}$$$

Step 2. We now integrate both sides of equation (5) between 0 and $a$ to obtain the following:

$$$\int_{0}^{a}\frac{d}{da}(I(a)e^{(\nu + \mu)a})da = \int_{0}^{a}\lambda S(a)e^{(\nu + \mu)a}da. \tag{6}$$$

Since integration is the converse of differentiation, the left-hand side of equation (6) simplifies to:

$$$\left[I(a)e^{(\nu + \mu)a} \right]_{0}^{a} = I(a)e^{(\nu + \mu)a} - I(0) \tag{7}$$$

However, $I(0)=0$ (since no individuals are infected at birth) and therefore the left-hand side of equation (6) simplifies to $I(a)e^{(\nu + \mu)a}$.

Since here we are discussiing the static model (3.8) (Hens et al. 2012, 63:38) with demographic processes ($\mu$ is the constant mortality rate), the survival function $l(a) = e^{-\mu a}$ and $S(a) = N(0)l(a)e^{-\lambda a} = N(0)e^{-(\mu + \lambda)a}$. The right hand side of equation (6) can be writtern as following:

$$$\int_{0}^{a}\lambda S(a)e^{(\nu + \mu)a}da = \lambda N(0)\int_{0}^{a}e^{(\nu - \lambda)a}da. \tag{8}$$$

By the rules of integration, equation (8) simplifies to the following:

$$$\begin{split} \lambda N(0)\int_{0}^{a}e^{(\nu - \lambda)a}da &= \frac{\lambda N(0)}{\nu - \lambda}\left[e^{(\nu - \lambda)a} \right]_{0}^{a} \\ &= \frac{\lambda}{\lambda - \nu}N(0)\left[1 - e^{(\nu - \lambda)a} \right]. \end{split} \tag{9}$$$

Step 3. Equating the expressions obtained from integrating the left-hand and right-hand sides of equation (6) leads to the following:

$$$I(a)e^{(\nu + \mu)a} = \frac{\lambda}{\lambda - \nu}N(0)\left[1 - e^{(\nu - \lambda)a} \right]. \tag{10}$$$

Dividing both sides of this equation by $e^{(\nu + \mu)a}$ leads to the intended equation (1):

$\begin{equation*} \begin{split} I(a) &= \frac{\lambda}{\lambda - \nu}N(0)\left[e^{(-\nu -\mu)a} - e^{(-\lambda -\mu)a} \right] \\ &= \frac{\lambda}{\lambda - \nu}N(0)e^{-\mu a}\left[e^{-\nu a} - e^{-\lambda a} \right] \\ &= \frac{\lambda}{\lambda - \nu}N(0)l(a)\left[e^{-\nu a} - e^{-\lambda a} \right]. \end{split} \end{equation*}$

### PS

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## References

Anderson, Roy M., and Robert M. May. 1992. Infectious Diseases of Humans: Dynamics and Control. Oxford University Press.

Hens, Niel, Ziv Shkedy, Marc Aerts, Christel Faes, Pierre Van Damme, and Philippe Beutels. 2012. Modeling Infectious Disease Parameters Based on Serological and Social Contact Data: A Modern Statistical Perspective. Vol. 63. Springer Science & Business Media.

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