# Model for spread of contagious disease.

1. Dec 3, 2013

### elitewarr

Hello guys, I am quite unsure in how to start the modeling for population. If I define y as the number of infected persons and y' as the rate of spread is it correct? I see the number of contacts between infected and non infected persons to be equal to the number of infected persons. Otherwise I have no idea how to form the variables.

Thank you.

2. Dec 3, 2013

### economicsnerd

If $p\in(0,1)$ is the infected proportion, it may be convenient to model things in terms of $x=\log\dfrac{p}{1-p}$, as the latter (which is in one-to-one correspondence with the former) can take any real value.

It sounds like what is being described is that $\dot p$ is proportional to $e^x$, so say $\dot p = \alpha e^x$ for some $\alpha>0$.

Then we can compute $\dot x e^x = \dfrac{d}{dt} e^x = \dfrac{(1-p)\dot p - p(-\dot p)}{p^2} = \dfrac{\dot p}{p^2}= \dfrac{\alpha e^x}{p^2}$, and so $\dot x = \dfrac{\alpha}{p^2}$. One can further verify that $p = \dfrac{1}{1+e^{-x}}$, and so $\dot x = \alpha(1+e^{-x})^2$.

From the above equation, it's easy to see that:
- $x$ is strictly increasing over time if $-\infty<x<\infty$, and therefore $p$ is strictly increasing if $0<p<1$.
- Therefore, $x$ must converge to something finite or infinite. It's easy to see that it can't converge to anything finite over time. Therefore $x\to\infty$. Thus $p\to 1$ as long as $0<p\leq 1$.

The only remaining case of $p$ starting at $0$ is easy to consider on its own.

3. Dec 3, 2013

### economicsnerd

Oh whoops, it's simpler than that. Disregard.

Number of contacts between infected and uninfected should be proportional to $p(1-p)$ maybe?

4. Dec 3, 2013

### nsaspook

http://arxiv.org/abs/1311.6376

5. Dec 5, 2013

### elitewarr

Sorry for the late reply. Thank you guys. But I think the simpler version should suffice according to the logistic equation.

And.. I don't rawly understand what you wrote economicsnerd haha.. My knowledge isn't that advanced yet..