epidemic model help

**reivera** · March 2nd 2008, 11:27 PM

Hi I'am new here just looking for some help with this model, I dont know how to begin.

ok

So I have -
(x should be lamder and a alpha..)

dn/dt = kn(1-xn^a)

for no. of flies n(t) where k, x and a are const.
(with 0<a<1,0<x<1)

Integrate to find an equation for n(t) given n(0) = ((n0) meaning first term..

I need to show that 1/n(1-xn^a) = 1/n + xn^(a-1)/1-xn^a, but just dont know where to begin..

I

**mr fantastic** · March 3rd 2008, 04:06 AM

Originally Posted by reivera

Hi I'am new here just looking for some help with this model, I dont know how to begin.

ok

So I have -
(x should be lamder and a alpha..)

dn/dt = kn(1-xn^a)

for no. of flies n(t) where k, x and a are const.
(with 0<a<1,0<x<1)

Integrate to find an equation for n(t) given n(0) = ((n0) meaning first term..

I need to show that 1/n(1-xn^a) = 1/n + xn^(a-1)/1-xn^a, but just dont know where to begin..

I

OK, here's a beginning:

Assume that $\frac{1}{n(1 - \lambda n^{\alpha})} = \frac{A}{n} + \frac{f(n)}{1 - \lambda n^{\alpha}} = \frac{A(1 - \lambda n^{\alpha}) + n f(n)}{1 - \lambda n^{\alpha}}$ .

Then $1 \equiv A(1 - \lambda n^{\alpha}) + n f(n) = A - A \lambda n^{\alpha} + n f(n)$ .

So it is required that:

$1 = A$ .... (1)

$0 = -A \lambda n^{\alpha} + n f(n)$ .... (2)

Substitute (1) into (2): $0 = -\lambda n^{\alpha} + n f(n) \Rightarrow f(n) = \frac{\lambda n^{\alpha}}{n} = \lambda n^{\alpha - 1}$ .

So $\frac{1}{n(1 - \lambda n^{\alpha})} = \frac{1}{n} + \frac{\lambda n^{\alpha - 1}}{1 - \lambda n^{\alpha}}$ .

So the differential equation becomes:

$\frac{dt}{dn} = \frac{1}{k} \, \frac{1}{n(1 - \lambda n^{\alpha})} = \frac{1}{k} \left(\frac{1}{n} + \frac{\lambda n^{\alpha - 1}}{1 - \lambda n^{\alpha}}\right)$ .

Therefore $t = \frac{1}{k} \left( \ln n - \frac{1}{\alpha}\ln (\lambda n^{\alpha} - 1) \right) + C$ .

**mr fantastic** · March 3rd 2008, 04:44 AM

Originally Posted by mr fantastic

OK, here's a beginning:

Assume that $\frac{1}{n(1 - \lambda n^{\alpha})} = \frac{A}{n} + \frac{f(n)}{1 - \lambda n^{\alpha}} = \frac{A(1 - \lambda n^{\alpha}) + n f(n)}{1 - \lambda n^{\alpha}}$ .

Then $1 \equiv A(1 - \lambda n^{\alpha}) + n f(n) = A - A \lambda n^{\alpha} + n f(n)$ .

So it is required that:

$1 = A$ .... (1)

$0 = -A \lambda n^{\alpha} + n f(n)$ .... (2)

Substitute (1) into (2): $0 = -\lambda n^{\alpha} + n f(n) \Rightarrow f(n) = \frac{\lambda n^{\alpha}}{n} = \lambda n^{\alpha - 1}$ .

So $\frac{1}{n(1 - \lambda n^{\alpha})} = \frac{1}{n} + \frac{\lambda n^{\alpha - 1}}{1 - \lambda n^{\alpha}}$ .
[snip]

Alternatively, so that Krizalid's ulcer doesn't get any bigger:

$\frac{1}{n(1 - \lambda n^{\alpha})} = \frac{(1 - \lambda n^{\alpha}) + \lambda n^{\alpha}}{n(1 - \lambda n^{\alpha})} = \frac{(1 - \lambda n^{\alpha})}{n(1 - \lambda n^{\alpha})} + \frac{\lambda n^{\alpha}}{n(1 - \lambda n^{\alpha})} = \frac{1}{n} + \frac{\lambda n^{\alpha}}{n(1 - \lambda n^{\alpha})}$

$= \frac{1}{n} + \frac{\lambda n^{\alpha - 1}}{1 - \lambda n^{\alpha}}$

**dankelly07** · March 6th 2008, 04:28 PM

sorry I meant to post it here, how would you determine the form for n(t) as t -> infinity? I've been asked in my uni work to do that a few times, but this one looks difficult

**mr fantastic** · March 6th 2008, 08:12 PM

Originally Posted by dankelly07

sorry I meant to post it here, how would you determine the form for n(t) as t -> infinity? I've been asked in my uni work to do that a few times, but this one looks difficult

As t --> +oo, kt --> +oo so it's required that the right hand side --> +oo.

At a glance, this will happen when the argument of the log is equal to zero, that is, $\lambda n^{\alpha} - 1 = 0$ ....

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