Question about SIS model

Question

Modify the following SIS model

S´= -bSI + aI
I´ = bSI - aI

to the situation in which there are two competing strains of the same disease, generating two infective classes I1 , I2 under the assumption that coinfections are not possible. Does the model predict coexistence of the two strains or competitive exclusion?

I believe the modification should look like this:

S´= -b1SI1 -b2SI2 + a1I1 + a2I2
I1´ = b1SI1 - a1I1 
I2´= b2SI2 -a2I2

But the book I am reading doesn’t really talk about coexistence of strains or exclusions so I don’t really know where to go from here. The basic reproduction number R0 is pretty obvious but what do I need to check regarding it?

Thanks in advance

Answer 1

Since the total population $N = S + I_1 + I_2$ is constant in this model, we can substitute $S = N - I_1 - I_2$ and get two equations for $I_1$ and $I_2$: $$ \eqalign{\dfrac{dI_1}{dt}&= b1 I_1 (N - I_1 - I_2) - a_1 I_1 = (b_1 N - a_1) I_1 - b_1 I_1^2 - b_1 I_1 I_2\cr \dfrac{dI_2}{dt} &= b_2 I_2 (N - I_1 - I_2) - a_2 I_2 = (b_2 N - a_2) I_2 - b_2 I_2^2 - b_2 I_1 I_2\cr}$$

If $a_1 b_2 \ne a_2 b_1$, there are three equilibrium points:

1. $I_1 = I_2 = 0$, $S = N$
2. $I_1 = 0$, $I_2 = N - a_2/b_2$, $S = a_2/b_2$
3. $I_2 = 0$, $I_1 = N - a_1/b_1$, $S = a_1/b_1$.

Of course if $a_2/b_2 > N$ or $a_1/b_1 > N$, these are not really possible because you can't have a negative population.

The Jacobian matrix at equilibrium point (1.) is $$ \pmatrix{b_1 N - a_1 & 0\cr 0 & b_2 N - a_2\cr}$$ The Jacobian matrix at equilibrium point (2.) is $$ \pmatrix{ (a_2 b_1 - a_1 b_2)/b_2 & 0 \cr a_2 - N b_2 & b_2 N - a_2\cr} $$ and the Jacobian matrix at equilibrium point (3.) is $$ \pmatrix{ (a_1 b_2 - a_2 b_1)/b_1 & 0 \cr a_1 - N b_1 & b_1 N - a_1\cr} $$ Let's say $a_1/b_1 < a_2/b_2$. There are essentially three possibilities.

If $N < a_1/b_1 < a_2/b_2 $, the only "real" equilibrium point is (1.), and it is stable. The infections always die out as $t \to \infty$.

If $a_1/b_1 < N < a_2/b_2$, the "real" equilibrium points are (1.) and (3.) . (1.) is unstable and (3.) is stable. Strain 2 dies out as $t \to \infty$, and we end up with strain 1 endemic (unless we started with $I_1 = 0$).

If $a_1/b_1 < a_2/b_2 < N$, we have all three equilibrium points, but (1.) and (2.) are unstable and (3.) is stable. Again we end up with strain 1 endemic (unless we started with $I_1 = 0$).

The marginal cases where two or more of $a_1/b_1$, $a_2/b_2$ and $N$ are equal are left as an exercise.

Answer 2

I think that the exclusion of the possibility of getting both diseases means that you don't have to have any $I_1$ term in your $I_2'$ equation, and no $I_2'$ term in the $I_1'$ equation. Likewise you don't have to have a separate $I_1I_2$ equation either. Your modification looks good.

$a_1$ is the recovery rate for people sick with strain $1$. People become sick when they catch strain $1$ or strain $2$. It looks like reinfection is possible in this model, as those who recover are returned to the "susceptible" category.

So long as the recovery rates are high enough that there is no possibility of everyone being infected at once, I don't see why either would die out, as there will always be more available susceptible hosts for either strain. I'm not sure how to prove it though.