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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.

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  • $\begingroup$ solve equation for the right hand side $=0$, namely $\beta I(K-I)-\beta I(K-I)=0$. This is a definition of equilibrium (constant solution). Hmm soemthing is wrong here. Check when $t$ is an argument of the function and when it is a mulitplier. $\endgroup$ – Alexander Vigodner Jun 11 '14 at 20:03
  • $\begingroup$ @Alexander Vigodner - Ok, I understand what you're saying. But how do I apply this to the integral I have? $\endgroup$ – SOULed_Outt Jun 11 '14 at 20:09
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    $\begingroup$ In integral assume that $I$ is constant. Then you will have $I=\tau\beta I(K- I)$ .This equation has 2 roots for $I$: $I=0$ and $1=\beta \tau(K−I)$ $\endgroup$ – Alexander Vigodner Jun 11 '14 at 21:19
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    $\begingroup$ This is not an ODE there is lag in time, I was confused in my first comment.But takin derivative we can se that this integral is indeed a solution. We are just looking for a simplest constant solution which is an eqilibrium. $\endgroup$ – Alexander Vigodner Jun 11 '14 at 21:25
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    $\begingroup$ There's something funny about your original delay-differential equation: it's trivially satisfied by any constant function $I(t) = \hat I$. Now, I'm pretty sure most of those solutions aren't stable against small perturbations, but they're still solutions. $\endgroup$ – Ilmari Karonen Jun 11 '14 at 22:36
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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 $1=\beta \tau(K-I)$

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