I am having some difficulties in finding the stability of the following equilibrium solutions:
If:
$\frac{dN}{dt}=rN(1-a(N-b)^{2}$
To find all equilibrium solutions I simply did:
$\frac{dN}{dt}=rN(1-a(N-b)^{2}=0$
I found the following solutions: $0,\frac{-\sqrt(a)+ab}{a},\frac{\sqrt(a)+ab}{a} $
And these are the equilibrium solutions.
Then to check stability, I simply calculated the derivative of $\frac{dN}{dt}=rN(1-a(N-b)^{2}$ and I obtained: $-r(-1+a(N-b)^{2})$.
To check stability I know that I can plug in my equilibrium solutions in here: $-r(-1+a(N-b)^{2})$
For Equilibrium solution $0$ it gives: $-r(-1+a(0-b)^{2})=r-ab^{2}r$
For Equilibrium solution $\frac{-\sqrt(a)+ab}{a}$ it gives: $-r(-1+a((\frac{-\sqrt(a)+ab}{a})-b)^{2})=0$
For Equilibrium solution $\frac{\sqrt(a)+ab}{a}$ it gives: $-r(-1+a((\frac{\sqrt(a)+ab}{a})-b)^{2})=0$
I saw that because the last two give the value of $0$ I need to do further analysis by analyzing values at the right and left side. But as this is not a linear problem( with actual values) how can I check stability in this case?
Can anyone help me on this and let me know if what I did is correct?
Thanks
Let the right-hand side function be $f(N) = rN\left(1 - a\left(N - b\right)^2\right)$. In what follows, suppose $r\neq 0$; otherwise $f(N) = 0$ for $r=0$. We first examine some simple cases.
For the remaining of the analysis, we will suppose $a>0$ so that we have three equilibrium solutions. You are correct that you need to find $f'(N)$ to determine the stability of the equilibrium solutions, but you computed $f'(N)$ wrongly. Observe that $f(N)$ is the product of $rN$ and $\left(1 - a\left(N - b\right)^2\right)$ so we use Product Rule to find $f'(N)$: \begin{align*} f'(N) & = \underbrace{r\left(1 - a\left(N - b\right)^2\right)}_\textrm{This is what you had} + \underbrace{rN\left(-2a(N-b)\right)}_\textrm{This is what you were missing} \\ & = r\left(1 - a\left(N - b\right)^2\right) - 2raN(N-b). \end{align*} For $N_1 = 0$, we have $$ f'(0) = r\left(1 - ab^2\right). $$ For $N = \dfrac{1}{\sqrt{a}} + b = N_2$, we have $N - b = \dfrac{1}{\sqrt{a}}$ and so \begin{align*} f'\left(\frac{1}{\sqrt{a}} + b\right) & = r\left(1 - a\left(\dfrac{1}{a}\right)\right) - 2ra\left(\frac{1}{\sqrt{a}} + b\right)\left(\frac{1}{\sqrt{a}}\right) \\ & = -2ra\left(\frac{1 + \sqrt{a}b}{a}\right) \\ & = -2r\left(1 + \sqrt{a}b\right). \end{align*} For $N = -\dfrac{1}{\sqrt{a}} + b = N_3$, we have $N - b = -\dfrac{1}{\sqrt{a}}$ and so \begin{align*} f'\left(-\frac{1}{\sqrt{a}} + b\right) & = r\left(1 - a\left(\dfrac{1}{a}\right)\right) - 2ra\left(-\frac{1}{\sqrt{a}} + b\right)\left(-\frac{1}{\sqrt{a}}\right) \\ & = 2ra\left(\frac{-1 + \sqrt{a}b}{a}\right) \\ & = 2r\left(-1 + \sqrt{a}b\right) \\ & = -2r\left(1 - \sqrt{a}b\right). \end{align*}
We now perform a case analysis. Observe that $f'(N) = 0$ for all three equilibrium solutions $N_1, N_2, N_3$ if $1-ab^2 = 0$, so suppose not. Also observe that $ab^2>0$ since we assume $a>0$ which means that $1-ab^2$ can either be negative or positive. For notational convenience, let $\lambda = \sqrt{a}b$. Then \begin{align*} f'\left(N_1\right) & = r\left(1 - \lambda^2\right) \\ f'\left(N_2\right) & = -2r\left(1 + \lambda\right) \\ f'\left(N_3\right) & = -2r\left(1 - \lambda\right) \\ \end{align*}
Suppose $1 - \lambda^2 > 0$, then $\lambda^2 - 1 < 0 $ and this implies $-1 < \lambda < 1$, or $$ 1 + \lambda > 0 \ \ \textrm{ and } \ \ 1 - \lambda > 0. $$
Suppose $1 - \lambda^2 < 0$, then $\lambda^2 - 1 > 0$ and this implies $\lambda < -1$ and $\lambda > 1$, or $$ 1 + \lambda < 0 \ \ \textrm{ and } \ \ 1 - \lambda < 0. $$
We can finally determine the stability of each of the equilibrium solutions.
1) For $1 - \lambda^2 > 0$,
If $r > 0$, then $N_1$ is unstable and both $N_2, N_3$ are stable.
If $r < 0$, then $N_1$ is stable and both $N_2, N_3$ are unstable.
2) For $1 - \lambda^2 < 0$,
If $r > 0$, then $N_1$ is stable and both $N_2, N_3$ are unstable.
If $r < 0$, then $N_1$ is unstable and both $N_2, N_3$ are stable.