Unstable points of dynamic systems

Question

I am just starting dynamical systems and came across the following problem in my textbook.

Considering the discrete time logistic growth model,

$$N_{t+1} = \lambda N_t\left(1-\frac{N_t}{K}\right)$$

where $\lambda = 1+ b -d >0$ is the net reproductive rate and $K>0$ is a parameter that affects how the population grows when the population is large.

I have determined the two equilibrium points of this system which are,

$$N^*=0$$

$$N^* = K\left(1-\frac{1}{\lambda}\right)$$

Now I am trying to find the conditions for each of the equilibria that I found above to be unstable.

Here is my explanation below,

Computing the derivative $f'(x)$ and evaluating it in the fixed points. If $|f'(x^*)|>1$ then $x^*$ is an unstable fixed point.

$$f'(x) = \lambda -\frac{2x\lambda}{K}$$

Using $x=0$

$$f'(0) = \lambda$$

So it is unstable when $\lambda>1$

But then how do I use $x = K\left(1-\frac{1}{\lambda}\right)$? Am I proceeding correctly?

You do realize that the plot of system is symmetric, right? i.e. a change of variables can swap the two equilibrium points. — polfosol, Feb 03 '18 at 15:43

score 1 · Accepted Answer · answered Feb 03 '18 at 15:43

1

Plug in and obtain

$$f'\left(K\left(1- \frac{1}{\lambda} \right)\right) = \lambda -2\lambda\left(1- \frac{1}{\lambda}\right)=-\lambda+2$$

answered Feb 03 '18 at 15:43

user

154,566

so $|f'(x^*)| >1$ when $\lambda >1$? is this correct? – user123 Feb 03 '18 at 15:48
how would I work that out? – user123 Feb 03 '18 at 15:54
you need to solve $|2-\lambda|>1\implies \lambda^2-4\lambda+3>0$ – user Feb 03 '18 at 16:03
so when $x>3$ and $x<1$? – user123 Feb 03 '18 at 16:08
it is stable for $1<\lambda<3$ and unstable for $\lambda>3$ – user Feb 03 '18 at 16:20
but what about $\lambda < 1$ for example $|2-0.5|>1$ – user123 Feb 03 '18 at 16:31
yes it should be unstable also for $0<\lambda<1$ – user Feb 03 '18 at 16:40
but even for $\lambda<0$ it is true, for example $|2-(-0.5)|<1$ so it should be for $x<1$ – user123 Feb 03 '18 at 16:52
I was assuming $\lambda >0$ and in this case to have a positive value of N for the equilibrium point we need to have $\lambda>1$ or $\lambda <0$. So if you need to have positive value for N you must consider some limitaion for $\lambda$. – user Feb 03 '18 at 17:01
sorry I forgot that condition in the problem, thanks for your help! – user123 Feb 03 '18 at 17:56
@user123 You are welcome! Bye – user Feb 03 '18 at 17:57
so then its unstable for $\lambda >3, 0 < \lambda < 1$ – user123 Feb 03 '18 at 17:57
@user123 if you want $N^* = K\left(1-\frac{1}{\lambda}\right)$ to be positive it is unstable for $\lambda \ge 3$ and $\lambda <0$ since in the interval $0 < \lambda < 1$ we have that $N^* = K\left(1-\frac{1}{\lambda}\right)<0$ – user Feb 03 '18 at 18:02

Unstable points of dynamic systems

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