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bifurcations, steady states, model analysis

dwsmith

Well-known member
Feb 1, 2012
1,673
$N_{t + 1} =\begin{cases}rN_t^{1 - b}, & N_t > K\\
rN_t, & N_t < K
\end{cases}$


The steady states are when $N_{t + 1} = N_t = N_*$.
$$
N_{*} =\begin{cases}rN_*^{1 - b}, & N_* > K\\
rN_*, & N_* < K
\end{cases}
$$
So the steady states are $N_* = \sqrt{r}$ and $N_* = 0$.

I am not sure how to check for stability and bifurcations values for a piece wise defined model.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Look at three cases:

1) Suppose $N_t< 0$. What is $N_{t+1}$? Is it larger than $N_t$ so that the sequence is heading toward 0 or is it smaller so the sequence is heading away from 0?

2) Suppose $0< N_t<\sqrt{r}$. What is $N_{t+1}$? Is it less than $N_t$ so the sequence is heading toward 0 or is it larger so it is heading toward $\sqrt{r}$?

3) Suppose $\sqrt{r}< N_t$. What is $N_{t+1}$? Is it less than $N_t$ so the sequence is heading toward $\sqrt{r}$ or is it larger so it is heading away?

If in 1 and 2 you said that the sequence was heading toward 0, then 0 is stable. If in 2 and 3 you said that the sequence was heading toward $\sqrt{r}$ then that is stable. Notice that is is not possible for both 0 and $\sqrt{r}$ both to be stable- if in 2, the sequence is heading toward 0, it cannot be heading toward $\sqrt{r}$. It is, however, possible for them both to be unstable.