# 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 $I(a)$:

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

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

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

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:

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

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:

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

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:

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

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

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

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

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

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:

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

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

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

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

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

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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