sis_model

SIS model

Going from discrete to continuous models here.

flowchart LR
        S["S"] -->|infection| I["I"]
        I -->|recovery| S

$S$: susceptible. $I$: infectious. Recovery returns individuals to $S$, so the total population stays constant:

$N=S+I\Rightarrow S=N-I.$

Discrete version

In one step, a susceptible individual becomes infected with probability $1-(1-\beta {)}^{I_t}$, assuming independent encounters. A fraction $\alpha$ of infectious individuals recovers.

$\Delta I=\underset{\text{new infections}}{\underbrace{S_t[1-(1-\beta {)}^{I_t}]}}-\underset{\text{recoveries}}{\underbrace{\alpha I_t}}.$

Substitute $S_t=N-I_t$:

$\Delta I=I_{t+1}-I_t=(N-I_t)\left[1-(1-\beta {)}^{I_t}\right]-\alpha I_t.$

Taking the time step to zero

For a step of length $h$, use infection probability $\beta h$ and recovery probability $\alpha h$. Here $\alpha$ and $\beta$ are rates per unit time; choose $h$ small enough that both probabilities lie in $[0,1]$.

$$\Delta I_h=I(t+h)-I(t)=S\left[1-(1-\beta h{)}^I\right]-\alpha Ih,$$

where $S=S(t)$ and $I=I(t)$. The change $\Delta I_h\to 0$ as $h\to 0^{+}$. For the rate of change, divide by $h$:

$$\frac{dI}{dt}=\underset{h\to 0^{+}}{\lim }\frac{\Delta I_h}{h}=\underset{h\to 0^{+}}{\lim }\left[S\frac{1-(1-\beta h{)}^I}{h}-\alpha I\right].$$

Holding $S$ and $I$ fixed in this limit,

$$(1-\beta h{)}^I=1-I\beta h+O(h^2),$$

which gives the continuous mean-field model:

$$\frac{dI}{dt}=\beta SI-\alpha I=\beta (N-I)I-\alpha I.$$

Taylor series

Expand $f$ around $x$ to describe its value a small distance $h$ away:

$$f(x+h)=f(x)+f^{\prime }(x)h+f^{\prime \prime }(x)\frac{h^2}{2!}+f^{\prime \prime \prime }(x)\frac{h^3}{3!}+\dots$$

More compactly,

$$f(x+h)=\sum _{n=0}^{\infty }\frac{f^{(n)}(x)}{n!}{h}^n.$$

The infinite series equals $f(x+h)$ when $f$ is analytic and $h$ is within its radius of convergence. Here $f^{(n)}$ is the $n$th derivative, $f^{(0)}=f$, and $0!=1$.

Keeping only the constant and linear terms gives the first-order approximation:

$f(x+h)\approx f(x)+f^{\prime }(x)h.$

Applying Taylor to the infection probability

Let $f(h)=(1-\beta h{)}^I$, holding $I$ fixed. Expand around $h=0$:

$$\begin{array}{rl}f(0)& =1,\\ f^{\prime }(h)& =-\beta I(1-\beta h{)}^{I-1},\\ f^{\prime }(0)& =-\beta I.\end{array}$$

To first order,

$$(1-\beta h{)}^I=f(0)+f^{\prime }(0)h+O(h^2)=1-\beta Ih+O(h^2).$$

So the probability of infection becomes linear in $h$:

$$\begin{array}{rl}p(h)& =1-(1-\beta h{)}^I\\ & =\beta Ih+O(h^2)\approx \beta Ih.\end{array}$$

Multiplying by $S$ gives the expected new infections over the step:

$Sp(h)\approx \beta SIh.$

This goes to zero with $h$. Divide by $h$ to obtain the infection rate:

$$\underset{h\to 0^{+}}{\lim }S\frac{1-(1-\beta h{)}^I}{h}=\underset{h\to 0^{+}}{\lim }S\left[\beta I+O(h)\right]=\beta SI.$$

Including recovery,

$\frac{dI}{dt}=\beta SI-\alpha I=\beta (N-I)I-\alpha I.$

The approximation is linear in the time step $h$; the resulting SIS equation is still nonlinear in $I$.