2
$\begingroup$

I've encountered this Theorem in my mathematical biology notes:

$$\bar{𝑥}\text{ is a stable steady state of }f(x_n) = x_{n+1}\text{ iff }|f^{'}(\bar{𝑥})|<1.$$

The definition of a steady state $\bar{x}$ for first order difference equations is given as:

$$x_{n+1} = x_n = \bar{x}$$

My problem lies in applying this theorem, particularly to solve questions like this in the notes:

Consider the following nonlinear difference equation for population growth: $$x_{n+1} = \frac{kx_n}{b+x_n}; \text{ }b,k > 0$$ Establish whether the equation has a nontrivial steady state and determine its stability.

I'm confused, mostly because we're taking derivatives of discrete functions and it is my assumption that the first derivative of steady states should be 0 (i.e. why is the constraint in the theorem < 1?)

$\endgroup$
0
$\begingroup$

I gave this some thought.

Firstly, I fell into the trap of thinking that $$f'(x_n) = f'(x_{n+1}) = 0 \iff x_n\text{ is a steady state}$$

The issue with this is that between two discrete samples, an arbitrary, differentiable $f$ can do whatever it likes and still manage to satisfy $f'(x_n) = f'(x_{n+1}) = 0$

To solve the example, I did the following: $$\bigg(\exists n:x_{n+1} = x_n = \bar{x}\text{ }\bigg) \implies \bigg(x_n = \frac{kx_n}{b+x_n}\bigg) \implies \bigg(x_n = 0 \text{ } \oplus x_n = k - b\bigg)$$ $$\implies f(x_n) \text{ has a nontrivial steady state assuming } k \neq b$$

Now to determine stability, we can apply the Theorem: $$\bigg(f'(x) = \frac{kb}{(b+x)^2}\bigg) \implies \bigg(f'(k-b) = \frac{b}{k}\bigg)$$

Hence if $b \geq k$ the steady state is not stable

I would appreciate if someone could let me know if this is correct.

$\endgroup$

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.