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The model was described many times before, so I keep the details concise. We have $$ \frac{\partial}{\partial t} \rho(t,a) + \frac{\partial}{\partial a} \rho(t,a) = -\delta(a) \rho(t,a) $$ where $\rho$ is population (number) density and $\delta$ is instantaneous mortality. Moreover, $$ \Pi(a) = \exp \left( - \int_{0}^{a} \delta (\alpha) d\alpha \right) $$ is the probability of surviving to age $a$. According to Cushing (1998) one can redefine $$ \psi(t,a) = \frac{\rho(t,a)}{\Pi(a)} $$ such that the McKendrick PDE above simplifies to $$ \frac{\partial}{\partial t} \psi(t,a) + \frac{\partial}{\partial a} \psi(t,a) = 0 $$ leading to a traveling-wave solution $\psi(t,a) = \varphi(t-a)$.

What I'm wondering is how this kind of simplification is possible when there is a maximum age $a_{M} < \infty$? Because then $\Pi(a_{M}) = 0$, leaving $\psi(t,a)$ undefined. Cushing doesn't mention anything in that regard, so am I missing something trivial?

Durden
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2 Answers2

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There is absolutely no issue in writing a model in which the instantaneous mortality rate contains a singularity at $a=a_M$. Dividing by $\Pi(a)$ still works for $a<a_M$ and the resulting equation should be used only in the half strip $(0,\infty)\times(0,a_M)$. One can explicitly solve the model, and the general solution is given by

$$\rho(t,a)= \exp\left({-\int_0^a\delta(a')da'}\right)\varphi(t-a)$$

You can check this solves the PDE for any function $\varphi$. Now one can use appropriate BC's - usually by specifying the functions $\rho(0,a), \rho(t,0)$ - to find a solution in the half strip that vanishes precisely at $a=a_M$. This is hardly a surprising feature, it just says that the population density is zero at the maximum possible age. For a specific example, choose say $\delta(a)=1/(a_M-a)$ in which case it is easy to see that the general solution has a zero exactly at the maximum age for sufficiently regular $\varphi$:

$$\rho(t,a)=\left(1-\frac{a}{a_M}\right)\varphi(t-a)$$

Thus it can be seen that simple poles in the instantaneous mortality naturally implement a $C^0$ profile for the population density with a boundary (with the reasonable physical assumption that $\rho=0, a\geq a_M$). Of course, certain singular boundary conditions can in principle produce spurious behavior at the boundary, but these should be ruled out when one considers only the space of physically sensible ones.

K. Grammatikos
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  • Could you please elaborate on your last point? The "interesting" boundary condition of the McKendrick model is the renewal boundary $\rho(t,0) = \int_{0}^{a_{M}} \beta(\alpha) \rho(t,\alpha) d\alpha$ where $\beta(a)$ is a birth kernel. Is this the one potentially causing issues? – Durden Apr 11 '23 at 21:20
  • I had simpler boundary conditions in mind when I wrote that comment, and none of the boundary conditions I could think of satisfied the continuity condition $\rho(0,a_M)=0$. There seems to be no reason that this renewal boundary condition should cause trouble near the boundary, as long as the kernel $\beta$ does not diverge at $a_M$. Note however, that the kernel $\beta$ needs to be picked very carefully, since not all homogeneous Fredholm kernels have an eigenfuction with eigenvalue 1. (This reduces to a hom. Fredholm equation of the 2nd kind for $\varphi$) – K. Grammatikos Apr 11 '23 at 22:23
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$\Pi$ cannot have any zeros if it is defined as

$$\Pi(a)=\exp\left(-\int_0^a \delta(a)da\right)$$

since the exponential function has no zeros. That being said, in real life, there is presumably an age after which the probability of living is zero. If you want to include that possibility in your model, then $\delta(a)$ will not be well-defined, since it is given by

$$\delta(a)=\Pi'(a)/\Pi(a)$$

In this case, the original McKendrick equation cannot be posed.

Joshua Tilley
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  • Well, $\lim_{a \to a_{M}} \Pi(a) = 0$ if $\int_{0}^{a_{M}} \delta(\alpha) d\alpha = +\infty$, which many math biology books pose (including Cushing, see his conditions (2.7) in the link above). It makes sense biologically: survival probability to age $a_{M}$ is zero if instantaneous mortality $\delta(a_{M})$ is infinite. But how could one then divide by $\Pi(a_{M})$? – Durden Apr 10 '23 at 23:35