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If the per capita birth rate of a population is given by $r[1 − a(N − b)^2]$ where $r$, $a$, and $b$ are positive parameters, write down a population model equation of the form $\frac{d N}{dt} = f(N)$. Nondimensionalise the equation so that the dynamics depend on a single dimensionless parameter $k = b(a/r)^{1/2}$. If $u$ is your nondimensional population, sketch $f(u)$ for $k > 1$ and $k < 1$ and discuss how the qualitative behaviour of the solution changes with $k$ and the initial condition.

I think we have

$$\frac{dN}{dt}=r[1 − a(N − b)^2].$$

I've tried some different ways when nondimensionalising this equation but I cannot find a way to have the model depending on single parameter $k=b(a/r)^{1/2}$. Could you please help me to find it out?

Vân Kami
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    Sorry but $b\sqrt{a/r}$ is not dimensionless. To get a dimensionless evolution, consider $$u=\sqrt a\cdot(N-b)\qquad \tau=(r\sqrt a)\cdot t$$ then $$\frac{du}{d\tau}=1-u^2$$ – Did Sep 18 '18 at 16:59
  • Coul you explain why we know the dimensions of those parameters? Thank you so much! – Vân Kami Sep 19 '18 at 03:43
  • Looking at the ODE, one sees that $b$ must be of the dimension of $N$ (or $N-b$ does not exist), that $a$ must be of the dimension of $1/N^2$ (or $1-a(N-b)^2$ does not exist), and that $r$ must be of the dimension of $N/t$ (or the RHS is not $dN/dt$). – Did Sep 19 '18 at 05:57
  • Oh thank you! I got it. Maybe there is something wrong in this exercise. Thank you a lot! – Vân Kami Sep 19 '18 at 06:11
  • There is a little mistake in your reasoning. You are given the per capita birth rate, not the RHS. The DE must be $\frac{dN}{dt}=r[1-a(N-b)^2]N$. However, this does not solve the main problem. The parameter $k$ is still not dimensionless. – Dmitry Sep 20 '18 at 09:00
  • Ah I see. Thank you so much! That is an exercise in Mathematical Biology book of Murray, my prof asks me to do it but there are a lot of things make me confused here. – Vân Kami Sep 20 '18 at 17:42

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