Given the discrete difference equation:
$$N_{t+1}=\frac{aN_t}{(1+bN_t)^c}=H(N_t)$$ where $a,b$ and $c$ are positive parameters, how do you identify conditions for linear stability of the fixed points?
I have computed that if $N^{ \star} $ is a fixed point then:
$$N^{ \star}=0 \text{ or } N^{ \star}= \frac{a^{\frac{1}{c}}-1}{b} $$
These fixed points are unique when (only considering positive solutions): $a>1$
$$H'(N_t)= \frac{a}{(1+bN_t)^c}-\frac{abcN_t}{(1+bN_t)^{c+1}}$$
$$H'(0)=a \text{, } H'(\frac{a^{\frac{1}{c}}-1}{b})=1+c(a^{-\frac{1}{c}}-1)$$
Hence the fixed point $N^{ \star}=0$ is stable when $a<1$.
Now, what do you do for the second fixed point? For its linear stability we require:
$$-1<H'(N^{ \star})<1, N^{ \star}= \frac{a^{\frac{1}{c}}-1}{b}$$
$$ -2<c(a^{-\frac{1}{c}}-1)<0 $$
$$c>0 \text{ so: } 1-\frac{2}{c}< a^{-\frac{1}{c}}<1$$
But I have been struggling on what to do here since raising the left hand side to a power can falsify the inequality if it is negative. appreciate the help
If $0<c< 2$ then the LHS is positive, and you can raise both sides to the appropriate power and simplify.
If $c=2$, this reduces to $0 < a^{-1/2} < 1$, or $0 < a^{-1} < 1$, so that $a>1$.
Finally, if $c>2$, then the LHS is negative and, since $a$ is positive, the inequality is always satisfied.