0
$\begingroup$

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 am trying to find the equilibrium point of this equation.

I know that the equilibrium point are values of $N^*$ such that $N_{t+1} = N_t = N^*$. But how do I do that? thanks!

$\endgroup$
1
$\begingroup$

Let $x=N^*$ and solve: $$x=\lambda x(1-x/K)$$ so $x=0$ or $$1=\lambda (1-x/K)$$ $$1/\lambda=1-x/K$$ $$x/K=1-1/\lambda$$ $$x=K(1-1/\lambda)$$

So two solutions, $N^*=0,K(1-1/\lambda)$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ how did you solve for those last 2 equations? $\endgroup$ – user123 Feb 2 '18 at 21:20
  • $\begingroup$ I just added some more details $\endgroup$ – Akababa Feb 2 '18 at 21:22
  • $\begingroup$ okay so the equilibrium points of $x=0$ and $x = K(1-1\ \lambda?)$ $\endgroup$ – user123 Feb 2 '18 at 21:22
  • $\begingroup$ okay so are the equilibrium points $x$ or $N^*$ $\endgroup$ – user123 Feb 2 '18 at 21:23
  • 1
    $\begingroup$ x is just shorthand for $N^*$ $\endgroup$ – Akababa Feb 2 '18 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.