Very simple question:

There is a well-known model in epidemiology called SIR model. It describes the changes in the number of susceptible, infectious and recovered individuals in a population. It is described simply as:

$$\frac{dS}{dt} = -IRS$$ $$\frac{dI}{dt} = IRS-\alpha I$$ $$\frac{dR}{dt} = \alpha I$$

where $S$, $I$, $R$ are the proportions of susceptible, infectious and recovered people, respectively. $\alpha$ is the recovery rate, also described as the inverse of the average infectious period.

My question is the following: Does it make sense to have an equation such as $\displaystyle \frac{dR}{dt} = \alpha I$ where $dR/dt$ is measured in $1/\text{time}$ and consequently, $\alpha I$ is also measured in $1/\text{time}$, but the right-hand side uses days, for example, $\displaystyle \frac{1}{3 \text{ days}}$ and the left-hand side uses other measure of time?. Mathematically, I think, everything looks correct, since the units match both sides. However, the use of this equation looks suspicious to me, because the left-hand side is telling me the variation per unit time and I expect this variation to be measured in a very short time interval. Therefore, when I see that, in a very small time interval, the change of $R$ (of recovered people) is equal to the proportion of infectious people that recovered from disease in a matter of days, it just seems a bit incongruous.

Maybe, the left-hand side also uses days, but then how should I justify that $dR/dt$ would have units of 1/time measured in days?

Thanks in advance

  • $\begingroup$ The important thing as you noted in your nice question is that dimensional speaking we need to balance. The scaling between days hours and seconds is arbitrary, but of course this will be accounted for (un)desirable in the growth rates of the solutions. I would also hasten to have all the differentials on the lhs with same time scales :). $\endgroup$ – Chinny84 Jun 30 '14 at 23:01
  • $\begingroup$ Thank you. That is what I thought. After all, if we integrate a differential equation, it would be really odd to perform operations using days and seconds. I probably have some cognitive dissonance because I always thought about differential in terms of very small intervals, but I guess "time unit" doesn't mean it must be seconds or something else. $\endgroup$ – Robert Smith Jun 30 '14 at 23:13

Mathematics is agnostic to units.

Mathematical models, however, do care very much about units. (The numeral 1 attached to the units 'seconds' and 'hour' give two very different meanings.) So it is extremely important that when you do write down a model and do actual analysis and computations on it, you include conversion factors which makes your unit agree.

To take your model: $R$ is the number of people recovered, and $I$ is the number of people infected; both are either regarded as "unit-less" or "ones". $\alpha$ is the rate of recovery, which has unit "inverse-time". The value of $\alpha$, however, depends on the unit chosen for time: since

$$ 1~\mathrm{s}^{-1} = 3600~\mathrm{hr}^{-1} = 86400~\mathrm{day}^{-1} $$

The fact that the left hand side is written as a differential does not mean that its unit must be "small"! Go back to something simple, like physics. The speed of an object is the "instantaneous displacement" divided by the "instantaneous time change", but on the highways we usually measure speed by kilometers per hour (or miles per hour, if you are in certain English speaking countries).

So a recovery rate of $dR/dt = 1 / \mathrm{day}$ can be analogously interpreted as $dR/dt = \frac1{12} \frac{\text{persons}}{\text{hour}}$: yes, it makes no sense to say that 1/12 of a patient recovered in the last hour, but on average over the course of the day that is the rate at which patients recover.

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  • $\begingroup$ Thank you. The fact that the left hand side is written as a differential does not mean that its unit must be "small" Yes, that part was confusing me. $\endgroup$ – Robert Smith Jul 1 '14 at 15:27

It is possible to solve analytically the system of three ODEs. This could be useful for numerical applications, in order to obtain more accurate numerical results than with the the usual methods of numerical solving of the ODEs.

The solution is presented on a parametric form : For given $R$, compute the corresponding $S , I, t$ with the formulas.

This is not a direct answer of the question raised, but might help anyways.

enter image description here

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  • $\begingroup$ Interesting. Do you know why some sources assert that there is not analytic solution to the SIR model. Maybe, it is quite different to assume R and get S, I, t as you have done, instead of obtaining solution for S, I and R. $\endgroup$ – Robert Smith Jul 1 '14 at 23:15
  • $\begingroup$ The assertion "There is-" or "There is not- analytic solution" might be ambiguous if the set of functions allowed to express the expected solution is not specified. In the present case, it is true to say that there is no analytic solution which can be expressed in terms of a finite number of elementary functions, even in terms of a finite number of referenced special functions. But, as shown, an analytic solution can be expressed in term of an integral, which itself cannot be expressed in terms of a finite number of elementary or referenced special functions. $\endgroup$ – JJacquelin Jul 2 '14 at 5:53
  • $\begingroup$ Moreover, suppose that a new special function be defined such as $R(t)=t^{-1}(R)$ from the above known integral $t(R)$, i.e. the inverse function. Suppose that this new special function be referenced and becommes a standard special function in the lists of special functions and be implemented in the mathematical softwares. Then we could expressed on a direct manner $S(t)$ and $I(t)$. Then we could say that there is an analytical solution which can be expressed in terms of a finite number of standard functions. $\endgroup$ – JJacquelin Jul 2 '14 at 6:10
  • $\begingroup$ This manner to understand the role played by the special functions is discussed in fr.scribd.com/doc/14623310/… $\endgroup$ – JJacquelin Jul 2 '14 at 6:17
  • $\begingroup$ Interesting. I will take a look at your link. Thanks! $\endgroup$ – Robert Smith Jul 2 '14 at 22:14

In the differential equation

  dR/dt =  α·I

the left member has the dimension size of population over time, and the right side has the same dimension if the dimension of α is the inverse of time. This makes the equation unit-independent: one can choose any unit for the common dimension -- the corresponding numeric equations are all equivalent since all the units of a dimension differ by a non-zero numerical factor.

On the other hand, the two members of

    dS/dt = -R·I·S

have different dimensions: the dimension of the right member is (size of population)³ while that of the left member is size of population over time. The meaning of that equation depends on a factor (unit) of dimension size of population squared times time; the equation is not meaningful without the specification of that factor.

Only if both sides of an equation have the same dimension is the equation independent of the choice of units. It is customary in physics to check equations for the same dimension of each member ("dimension calculus") -- other equations cannot represent a law of nature unless the units are spelled out.

Michael Deckers. 2020-03-30

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