# Fixed point with time dependence

If I have a growth rate of

$$\frac{dN}{dt} = (1+cos(\alpha t))r(c-N)-\mu N$$

and I used the normal method of finding a fixed point (by making the differential equal to 0) I would get

$$N = \frac{(1+\text{cos}(\alpha t))rc}{\mu+r(1+\text{cos}(\alpha t))}.$$

Which doesn't produce the fixed point. What does this value represent? and how would I find the fixed point(s) of the system?

• benharmer, I think you made an algebra mistake. If $dN/dt=0$ for all $t$, then $r(c-N)-\mu N$=0. Apr 25, 2019 at 15:20
• @irchans Sorry, I have fixed my mistake. I used the wrong formula! Apr 30, 2019 at 10:41
• After you revised your question, I think there are no fixed points if $r\neq0$ and $\mu\neq 0$. Suppose $N(t) = d$ when $a<t<b$. Then $N'(t)=0$ when $t\in(a,b)$. The expression $\cos(\alpha t)$ is changing on the interval, but $N'(t)=0$ on that interval, so that implies $N(t)=c=d$. But then $\mu N(t)$ is non-zero. So basically, there are no fixed points if $r\neq0$ and $\mu\neq 0$. That's not quite a proof, but if you filled in all the details, it would be. May 1, 2019 at 18:16
• @irchans, sorry, I could not understand very well your comment, maybe because is my firts time on Dynamical Systems. If there exists an equilibrium $d$, so it must be s.t. $N(t)\to d$ when $a<t<b$ for an interval $(a,b)\ni d$, I could not understand why did you put $N(t)=d$ at whole interval to construct the contradiction. Thank you so much! Oct 9, 2020 at 0:42
• Hi Na'omi. You are correct. My idea does not prove that $N(t)$ does not converge to $d$ within a neighborhood of $d$. For a formal proof, I think that you could assume that there is a fixed point $d$. Then prove that there exists an $\epsilon>0$ s.t. $| N(t_0) - d| <\epsilon$ implies $| N(t) - d| >\epsilon$ for some $t>t_0$. I'm not sure this works. If I get time, I will look at it later. I did a numerical experiment with r=$\mu$=1/100 and c=1. In that case $N(t)$ oscillated around N=1/2. Oct 9, 2020 at 13:43