Is the fraction of susceptibles $S(t)$ increasing or decreasing? $S(t)$ is always

.
Under what conditions is the fraction of susceptibles not changing? $S(t)$ is unchanging if $S'(t)=$

. At any instant during which this condition is satisfied, the fraction is susceptibles holding still.

Let's derive an expression that will tell us for what values of $S(t)$ and $I(t)$ the fraction of susceptible stops changing. For simplicity, we'll drop the $t$ from the state variables, and just write them as $S$ and $I$. Given the above equation for $S'(t)$, the condition in terms of $S$ and $I$ is

$=0$. You can get rid of the minus sign to simplify the condition to

$=0$.

Your answer for the fraction of susceptibles being constant (at least for an instant) should be a product of two factors, one involving $S$ and the other $I$. This product can be zero only if one of the factors is zero. Therefore the fraction of susceptibles is holding still if either

$=0$ or if

$=0$ (enter answer involving $S$ first).

We can interpret this condition as meaning that the number of susceptibles will stay the same when either

or

.

Under what conditions is the rate of change in the fraction of infectives, $I'(t)$, equal to zero? When $I'(t)=$

. Plugging in the above expression for $I'(t)$, this condition, in terms of $S$ and $I$, becomes

$= 0$. Factor this expression to rewrite it as

$=0$.
Your answer for the fraction of infectives being constant (at least for an instant) should be a product of two factors, one involving $S$ and the other $I$. This product can be zero only if one of the factors is zero. Therefore there is no change in the fraction of infectives when either $S=$

or if $I=$

.

Under what conditions does the fraction of infectives increase? It increases if $I >0$ and

. (Answer is an inequality involving $S$.)

Under what conditions does the fraction of infectives decrease? It decreases if $I >0$ and

. (Answer is an inequality involving $S$.)

Since we have two differential equations, we cannot use a single phase line to describe the system. We need two dimensions to specify the values of the two state variables $S$ and $I$. To represent our two-dimensional state space, we'll use a phase plane, as shown to the right. The two axes of the phase plane are $S$ and $I$. A point $(S,I)$ on the phase plane corresponds to a particular state of the system.

**Feedback from applet**

direction of vectors:

location of vectors:

nullclines:

Use the phase plane to summarize your above results. First, draw lines corresponding to the values of $(S$ and $I$ where $S$ is not changing, i.e., where $S'(t)=0$. (Online, use the thick sold blue lines.) You derived two conditions for $S'(t)=0$, each of which is the equation for a line. (On the phase plane, $S$ acts like $x$ and $I$ acts like $y$.) Taken together, these lines are called the $S$ *nullcline*.

Next, draw lines corresponding to the values of $(S,I)$ where $I$ is not changing, i.e., where $I'(t)=0$. (Online, use the thin dashed green lines.) The two equations that you derived for the condition $I'(t)=0$ are each an equation for a line. Taken together, these lines are called the $I$ *nullcline*.

The last step is to draw vector (arrows) to represent whether $S$ is increasing or decreasing and whether $I$ is increasing or decreasing. A rightward pointing vector indicates $S$ is increasing and a leftward pointing vector indicates $S$ is decreasing (since $S$ is acting like $x$). A upward pointing vector indicates $I$ is increasing and a downward pointing vector indicates $I$ is decreasing (since $I$ is acting like $y$.) Therefore, a vector pointing downward and to the right indicates that, in that region of the phase plane, $S$ is decreasing while $I$ is increasing.

Since we only care about physical values of $S$ and $I$ (in particular, we can ignore negative values), the nullclines divide the phase plane into two regions. Draw a vector in each of those to regions indicating whether $S$ and $I$ are increasing or decreasing.