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  1. #1
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    epidemic model help

    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
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  2. #2
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    Quote Originally Posted by reivera View Post
    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.
    Last edited by mr fantastic; March 3rd 2008 at 04:56 AM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    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}}
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  4. #4
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    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
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  5. #5
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    Quote Originally Posted by dankelly07 View Post
    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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