I'm studying from a book titled "Mathematical Models in Population Biology and Epidemiology" and we're dealing with SIS models. In a chapter called "Infective Periods of Fixed Length", we get to this differential-difference equation $$I'(t) = \beta I(t) [K - I(t)] - \beta I(t - \tau) [K - I(t - \tau)]$$
We find equilibria by incorporating initial data for $-\tau \leq t \leq 0 $ into the model by writing it in the integrated form, $$I(t) = \int_{t - \tau}^{t} \beta I(x) [K - I(x)] dx$$
The authors get that the equilibrium condition for $I(t)$ is $$I = 0 \text{ or } 1 = \beta \tau (K - I)$$
I met with my professor before he left, and he could not figure out how they came up with this. Perhaps, one you could help me understand how they found this equilibrium condition.
This is not an ODE there is lag in time. .But taking derivative of the integral we can see that indeed this integral satisfies the original DE. Indeed $$ I(t)^\prime_t=\big (\int_{t-\tau}^t \beta I(x)(K-I(x)dx\big)_t^\prime=\beta I(t)(K-I(t)-\beta I(t-\tau)(K-I(t-\tau) $$ Now it is not a solution yet. But we want an equlibrium, namely constant soluiton. So plugging constant $I$ in the integral formula we obtain: $$ I=\beta \tau I(K-I) $$ The above equation has 2 constant solution $I=0$,and $I=\frac{\tau\beta K+1}{\beta \tau}$. The last one satisfies condition $1=\beta \tau(K-I)$