Age-structured SIR model $z_a + z_t = (1 - z) \lambda(a,t)$ solution I am trying to figure out a result in a paper, so I can adapt it to my own use.
Suppose the proportion of infected individuals of age $a$ at time $t$, $z(a,t)$ is given by the equation:
$$\frac{\partial z}{\partial a} + \frac{\partial z}{\partial t} = (1 - z) \lambda(a,t),$$
where $\lambda(a,t)$ is an age specific force of infection. The solution to this (I think only for for $t>a$) is,
$$z(a,t) = \int_{0}^{a} \lambda(a-\tau,t-\tau) e^{-\int_{0}^{\tau} \lambda(a-\tau',t-\tau') d\tau'} d\tau.$$
This solution makes intuitive sense to me since it is essentially the product of the force of infection in the past times the fraction susceptible at that time.
My question is, how do we get to this expression for $z(a,t)$ from the PDE above?
Attempts: I have been playing with the method of characteristics, where we consider the solutions on lines where $dt/da=1$. To do so, I define the proportion of infected individuals for a cohort of age $a_0$ at $t_0$, $\bar{z}(a_0+h,t_0+h)$. Then I can rewrite the LHS of the PDE as,
$$\frac{d \bar z}{d h} = (1-\bar z) \bar \lambda,$$
which can then be solved using integrating factors:
$$\frac{d}{d h} (\bar z e^{\int_{0}^{h} \bar \lambda(\tau) d\tau})=\bar \lambda e^{\int_{0}^{h} \bar \lambda(\tau) d\tau},$$
at which point, I get stuck. 
PS, if you are interested it's equation (2) in this paper I am trying to recover ((3) I can get, just not (2)!). 
 A: The initial condition seems to be $z(0,t) = 0$. The method of characteristics gives


*

*$a'(s)=1$, letting $a(0)=0$, we know $a=s$.

*$t'(s)=1$, letting $t(0)=t_0$, we know $t=s+t_0$.

*$z'(s)=(1-z(s))\,\lambda(a(s),t(s))$, letting $z(0)=0$, we know $z=1 - e^{-\int_0^s \lambda(a(\sigma), t(\sigma))\, \text d \sigma}$. This expression is obtained by noticing that $w=z-1$ satisfies $w' = -\lambda w$.
Combining everything, we have
\begin{aligned}
z(a,t) &= 1 - e^{-\int_0^a \lambda(\sigma, t-a+\sigma)\, \text d \sigma} \\
&= 1 - e^{-\int_{0}^{a} \lambda(a-\tau', t-\tau')\, \text d \tau'} .
\end{aligned}
Note that the expression in OP can be integrated explicitly with respect to $\tau$ as
\begin{aligned}
z(a,t) &= \int_{0}^{a} \lambda(a-\tau, t-\tau)\, e^{-\int_{0}^{\tau} \lambda(a-\tau', t-\tau')\, \text d \tau'} \, \text d \tau \\
 &= \left[-e^{-\int_{0}^{\tau} \lambda(a-\tau', t-\tau')\, \text d \tau'}\right]_{0}^{a} \\
&= 1 - e^{-\int_{0}^{a} \lambda(a-\tau', t-\tau')\, \text d \tau'}\, ,
\end{aligned}
which is the same expression as the one derived here.
