In light of the new omicron variant and breakthrough infections in people who have been vaccinated or previously infected, I was asked to discuss what a model would predict. The simplest model that includes reinfection is an SIRS model, where R, which stands for recovered, can become susceptible again. The equations have the form
I have ignored death due to infection for now. So like the standard SIR model, susceptible, S, have a chance of being infected, I, if they contact I. I then recovers to R but then has a chance to become S again. Starting from an initial condition of S = N and I very small, then S will decrease as I grows.
The first thing to note that the number of people N is conserved in this model (as it should be). You can see this by noting that the sum of the right hand sides of all the equations is zero. Thus and thus the integral is a constant and given that we started with N people then there will remain N people. This will change if we include births and deaths. Given this conservation law, then the dynamics have three possibilities. The first is that it goes to a fixed point meaning that in the long run the numbers of S, I and R will stabilize to some fixed number and remain there forever. The second is that it oscillates so S, I, and R will go up and down. The final one is that the orbit is chaotic meaning that S, I and R will change unpredictably. For these equations, the answer is the first option. Everything will settle to a fixed point.
To show this, you first must find an equilibrium or fixed point. You do this by setting all the derivatives to zero and solving the remaining equations. I have always found the fixed point to be the most miraculous state of any dynamical system. In a churning sea where variables move in all directions, there is one place that is perfectly still. The fixed point equations satisfy
There is a trivial fixed point given by S = N and I = R = 0. This is the case of no infection. However, if is large enough then this fixed point is unstable and any amount of I will grow. Assuming I is not zero, we can find another fixed point. Divide I out of the second equation and get
Solving the third equation gives us
which we can substitute into the first equation to get back the second equation. So to find I, we need to use the conservation condition S + I + R = N which after substituting for S and R gives
which we then back substitute to get
The fact that and must be positive implies is necessary.
The next question is whether this fixed point is stable. Just because a fixed point exists doesn’t mean it is stable. The classic example is a pencil balancing on its tip. Any small perturbation will knock it over. There are many mathematical definitions of stability but they essentially boil down to – does the system return to the equilibrium if you move away from it. The most straightforward way to assess stability is to linearize the system around the fixed point and then see if the linearized system grows or decays (or stays still). We linearize because linear systems are the only types of dynamical systems that can always be solved systematically. Generalizable methods to solve nonlinear systems do not exist. That is why people such as myself can devote a career to studying them. Each system is its own thing. There are standard methods you can try to use but there is no recipe that will always work.
To linearize around a fixed point we first transform to a coordinate system around that fixed point by defining , , , to get
So now s = h = r = 0 is the fixed point. I used lower case h because lower case i is usually . The only nonlinear term is , which we ignore when we linearize. Also by the definition of the fixed point the system then simplifies to
which we can write as a matrix equation
, where and . The trace of the matrix is so the sum of the eigenvalues is negative but the determinant is zero (since the rows sum to zero), and thus the product of the eigenvalues is zero. With a little calculation you can show that this system has two eigenvalues with negative real part and one zero eigenvalue. Thus, the fixed point is not linearly stable but could still be nonlinearly stable, which it probably is since the nonlinear terms are attracting.
That was a lot of tedious math to say that with reinfection, the simplest dynamics will lead to a stable equilibrium where a fixed fraction of the population is infected. The fraction increases with increasing or and decreases with . Thus, as long as the reinfection rate is much smaller than the initial infection rate (which it seems to be), we are headed for a situation where Covid-19 is endemic and will just keep circulating around forever. It may have a seasonal variation like the flu, which is still not well understood and is beyond the simple SIRS equation. If we include death in the equations then there is no longer a nonzero fixed point and the dynamics will just leak slowly towards everyone dying. However, if the death rate is slow enough this will be balanced by births and deaths due to other causes.
2 thoughts on “The dynamics of breakthrough infections”
If you make beta time-dependent, e.g. sinuosoidal with a one-year period to reflect seasonality of infectiousness, does the whole system inherit those seasonal dynamics? I imagine that is what we will end up with, like any other virus.
Yes, if beta were periodic (and slow enough) then the fixed point would follow beta. The seasonality of flu is most likely due to some time dependent forcing, we just don’t really know what it is exactly. My guess is that cold and dryness makes the nose more susceptible to infection as well as allowing the virus to be more stable outside of the host.