Pulliam Lab - Summer 2014 in the Pulliam Lab @ UF

Dushoff et al. 2004 PNAS

Calculating the endemic equilibrium

The equations from the paper are:

$\frac{dS}{dt} = \frac{N-S-I}{L} - \frac{\beta S I}{N}$ $\frac{dI}{dt} = \frac{\beta S I}{N} - \frac{I}{D}$

By definition, the system is at equilibrium when $\frac{dS}{dt} = 0$ and $\frac{dI}{dt} = 0$, which is to say that

$\frac{N-S^*-I^*}{L} - \frac{\beta S^* I^*}{N} = 0$ $\frac{\beta S^* I^*}{N} - \frac{I^*}{D} = 0$

We now need to solve these two equations for the two unknowns, $S^$ and $I^$. Let’s start by looking at the second equation, which can be rewritten as:

$I^* \left(\frac{\beta S^*}{N} - \frac{1}{D}\right) = 0$

It should be clear that one solution of this equation is $I^*=0$ , but we’re interested in the other solution (since we’re interested in the endemic equilibrium). So, we want to find the value of $S^*$ for which

$\left(\frac{\beta S^*}{N} - \frac{1}{D}\right) = 0$

To find $S^*$, we can add $1/D$ to both sides of the equation, giving: $\frac{\beta S^*}{N} = \frac{1}{D}$

Now, if we multiply both sides by $N/\beta$, we get:

$S^* = \frac{N}{\beta D}$

We can now substitute the expression for $S^*$ into equation (1) to give

$\frac{N-\frac{N}{\beta D}-I^*}{L} - \frac{\beta \frac{N}{\beta D} I^*}{N} = 0$

or,

$\frac{N-\frac{N}{\beta D}-I^*}{L} = \frac{\beta \frac{N}{\beta D} I^*}{N}$

Notice that we can cancel some terms on the right hand side of this equation to get:

$\frac{N-\frac{N}{\beta D}-I^*}{L} = \frac{I^*}{D}$

And we can re-write the left hand side of this equation to get:

$\frac{N}{L}-\frac{N}{\beta D L}-\frac{I^*}{L} = \frac{I^*}{D}$

Adding $\frac{I^*}{L}$ to both sides gives

$\frac{N}{L}-\frac{N}{\beta D L} = \frac{I^*}{L} + \frac{I^*}{D}\\$ $= I^* \left( \frac{1}{L} + \frac{1}{D} \right)$

By dividing both sides by $\left( \frac{1}{L} + \frac{1}{D} \right)$, we have an expression for $I^*$:

$I^* = \frac{\frac{N}{L}-\frac{N}{\beta D L}}{\frac{1}{L} + \frac{1}{D}}$

But it’s not pretty! Let’s clean it up a bit by multiplying the numerator and the denominator both by $L$, then canceling terms where we can:

$I^* = \frac{\frac{LN}{L}-\frac{LN}{\beta D L}}{\frac{L}{L} + \frac{L}{D}}$ $= \frac{N-\frac{N}{\beta D}}{1 + \frac{L}{D}}$

which can also be written as

$I^* = \frac{N-S^*}{1 + \frac{L}{D}}$