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I want to work out the dynamics of this equation (by that I mean find the fixed points of the system and then the stability of the fixed points):

$\frac{dN}{dt}= \sin(\alpha t)\left(rN\left(c-N\right)-\mu N\right)$

I would normally start by making the above equation equal to $f(N)$ to find the fixed points, then differentiating with respect to $N$ to find the stability of the fixed points.

If the $\sin(\alpha t)$ term wasn't in this equation I could work it out easily. It's the time variable that is bugging me. How would I go about working out the fixed points and stability? and how would I go about plotting this?

Thanks, Ben.

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    $\begingroup$ The equation is separable, you are only one partial fraction decomposition away from the exact solution. You can also treat is as Bernoulli equation and substitute $u=N^{-1}$, $$-u'=N^{-2}N'=\sin(αt)(r(cu−1)−μu),$$ which now is linear in $u$ of first order (and again separable without needing partial fractions). $\endgroup$
    – Lutz Lehmann
    Feb 20, 2019 at 18:14
  • $\begingroup$ If I had an equation $ \frac{dN}{dt} = r(c-N)-\mu N = f(x), $ the fixed point of this is at $ N = \frac{rc}{r+\mu} $ and to find the stability we have to differentiate f(x). $ f'(x) = -r-\mu, f'\left(\frac{rc}{r+\mu}\right) < 0. $ As it's always less than 0 at $\frac{rc}{r+\mu}$ the fixed point is always stable no matter the values of r,c or $\mu$. I was just wondering how to do this when we have a time - dependent forcing. $\endgroup$
    – benharmer
    Feb 23, 2019 at 12:04
  • $\begingroup$ The point is that after integration you get $$\ln|r(cu−1)−μu|=\frac{rc-μ}{α}\cos(αt)$$ which results in a periodic behavior of $u$ and thus $N$. There is no convergence to or divergence from the fixed points. $\endgroup$
    – Lutz Lehmann
    Feb 23, 2019 at 12:56
  • $\begingroup$ Ah I get you! Thank you! $\endgroup$
    – benharmer
    Feb 23, 2019 at 16:35

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