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[SOLVED] Discrete model analysis

dwsmith

Well-known member
Feb 1, 2012
1,673
I am dealing with (below) and r > 0

$$
N_{*} = \frac{rN_*}{(1 + aN_*)^b}
$$
So the steady states are $N_* = 0$ and $N_* = \frac{\sqrt{r} - 1}{a}$.
Let $f(x) = \frac{rx}{(1 + ax)^b}$.
Then $f'(x) = (ax + 1)^{-b}\left(\frac{br}{ax + 1} + (1 - b)r\right)$.
Evaluating the derivative at $N_*$, we obtain
$$
f'(0) = r \quad \text{and} \quad
f'\left(\frac{\sqrt{r} - 1}{a}\right) = \frac{r}{a}\left(\frac{b}{\sqrt{a}} - b + 1\right).
$$
For $N_* = 0$, $0 < f'(0) < 1$ for $0 < r < 1$.
For this range of $r$, $N_* = 0$ is stable monotonically.
For $r > 1$, $f'(0) > 1$ which is unstable monotonically, and at $r = 1$, we have a tangent bifurcation.

Now when I examine $N_* = \frac{\sqrt{r} - 1}{a}$, I need to find when $f'(N_*) = 1$ for the tangent bifurcation and when $f'(N_*) = -1$ for the pitchfork.

However, I have this if the f'(N_*) = 1

$$
r = \frac{a}{\frac{b}{\sqrt{a}}-b+1}
$$

So would this be the correct way to do this or does it need to be done be considering a and b?

For $N_* = \frac{\sqrt{r} - 1}{a}$, $0 < f'\left(\frac{\sqrt{r} - 1}{a}\right) < 1$ for $0 < r < \frac{a}{\frac{b}{\sqrt{a}} - b + 1}$ which is stable monotonically.
If $r > \frac{a}{\frac{b}{\sqrt{a}} - b + 1}$, then $f'\left(\frac{\sqrt{r} - 1}{a}\right) > 1$ which is unstable monotonically, and at $r = \frac{a}{\frac{b}{\sqrt{a}} - b + 1}$, we have a tangent bifurcation.
For $\frac{-a}{\frac{b}{\sqrt{a}} - b + 1} < r < 0$, $-1 < f'\left(\frac{\sqrt{r} - 1}{a}\right) < 0$ so we have a stable oscillations.
When $r < \frac{-a}{\frac{b}{\sqrt{a}} - b + 1}$, there are unstable oscillations, and at $r = \frac{-a}{\frac{b}{\sqrt{a}} - b + 1}$, we have a pitchfork bifurcation.

I could I say this to contend for a and b:
This stability analysis is valid for all $a,b$ such that $r > 0$.
 
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