Prove that Ricker map has no 2-cycles for $0<r<2$

Question

The question is related to the Exercise 26, Section 2.12. in Linda Allen's book "An Introduction to Mathematical Biology":

Show that the Ricker model $x_{n+1}= x_n e^{r( 1- \frac{x_n}{K} )} = f(x_n) $, where $r, K>0$, does not have any 2-cycles for $0<r<2$ by showing that $1+ f'(x) \neq 0$ for $x \in (0, \infty)$. That will lead to conclusion that positive equilibrium $K$ is globally asymptotically stable for $0<r<2$.

I first substitute $x_n$ by $\frac{x_n}{K}$ and get the equation $x_{n+1}= x_n e^{r( 1- x_n )} = f(x_n) $. The fixed points are now $0$ and $1$, so we have to show that $1$ is globally asymptotically stable for $0<r<2$. Then I tried to consider the function $g(x)=1+f'(x)= 1+ e^{r(1-x)} - r x e^{r(1-x)} $ . Then $g(0)= 1 + e^r>0$ and $g'(x)= r e^{r(1-x)} ( r x -2)$. Here $g'$ changes sign, so I cannot conclude that $g$ is increasing and therefore positive. How can one prove that $1+ f'(x) \neq 0$ for $x >0$? Thanks in advance.

Answer 1

• $f'(x)=\left(1-\frac{xr}K\right)\cdot \exp\left(r-\frac {rx}K\right)$.
• It's known that $e^t\geq 1+t$ for all $t\in\Bbb R$.
• Therefore $ \exp \left(\frac{rx}K-r\right)\geq \frac{rx}K-r+1>\frac{rx}K-1$, since $r<2$.
• Therefore $f'(x)>-1$.