# For the difference equation $x_{n+1}=ax_n\exp(-x_n)$, find values of $a$ that lead to a period-doubling bifurcation and values that lead to extinction

For the difference equation, $$x_{n+1}=ax_n\exp(-x_n)$$, find values of $$a\in(-1,1)\cup(1,e^2)$$ that lead to a period-doubling bifurcation, and values that lead to extinction

So for $$x_{n+1}=ax_n\exp(-x_n)$$ I was able to obtain the following fixed points, $$x^*=0$$ and $$x^*=\ln(a)$$. Taking derivative, we get $$f'(x)=a\exp(-x)(1-x)$$.

First case, for stability I was able to show $$|f'(0)|=|a|>1$$ is unstable and $$|f'(0)|=|a|<1$$ is stable. For second case, I was able to show $$1 is stable and $$a<1$$ or $$a>e^2$$ is unstable with $$|f'(\ln(a))|$$.

Now, here's my case, since at $$a=1$$ is the only common value where both fixed points change stability, and are opposite ($$a<1$$ stable for $$x=0$$ and $$a<1$$ unstable for $$x=\ln(a)$$), so this is my period-doubling bifurcation parameter.

For extinction, I said $$\forall\,-1 since it contains all points where both fixed points remain stable thus leading to extinction for any initial condition. Let me know if I'm totally off the mark on both answers please!

• I had to edit a few times, but here I am considering using $\forall a\in(-1,1)\cup (1,e^2)$ Commented May 8, 2022 at 2:20
• What does "extinction" mean? That the iteration produces a zero sequence for initial points $x_0>0$ or $x_0\in[0,1]$ or ...? Commented May 8, 2022 at 9:43

At the crossing of the curves of the two fixed points at $$a=1$$, an exchange of stability happens, but not a period doubling.

You get period doubling when $$f_a'(x_a)$$ passes through $$-1$$. So locally the dynamic is oscillating around the fixed-point, the sequence alternating sides of the fixed-point. As long as it is stable, the alternating sequence still contracts to the fixed-point. When it becomes unstable, then in a medium distance there is still alternating-contracting behavior, while close to the fixed-point it is alternating-expanding, giving convergence to a 2-cycle.

To make it more quantitative, a nice normal form for a bifurcation is $$y_{n+1}=\frac{ry_n}{1+y_n^2}.$$ Then the iteration is contracting towards zero as long as $$1+y_n^2>|r|$$, which remains true for $$|r|>1$$ for $$y_n$$ large enough.

• For $$r\approx 1$$ the sign of $$y_n$$ remains constant, the bifurcation is a fork, evolving from one stable fixed point to two stable fixed points $$\pm\sqrt{r-1}$$ with the unstable fixed point $$0$$ in the middle.
• For $$r\approx-1$$, the $$y_n$$ alternate in sign. Passing to $$r<-1$$ this results in a stable 2-cycle $$y_n=(-1)^2\sqrt{-r-1}$$.

If $$x_{n+1}=x_r+f(x_n-x_r)$$ with $$f(u)=ru(1+f_1u+f_2u^2+...)$$, then the normal form can be achieved in a first approximation via $$y_n=\phi(x_n-x_r),~~~ \phi(u)=\frac{bu}{1+cu},$$ where the coefficients are related to the coefficients of $$f$$ via $$c=-\frac{f_1}{1-r}$$ and $$b^2=f_1^2-f_3$$. This is only useful for the behavior $$r\approx-1$$, for $$r\approx 1$$ one would need some other $$\phi$$. Whenever the value of $$b$$ is not real, this line of arguments is not valid for period doubling.

• For $$x_r=0$$, $$r=a$$, $$f(u)=\exp(-u)=1-u+\frac12u^2+...$$ this gives $$c=\frac1{1-r}$$ and $$b^2=\frac12>0$$.
• For $$x_r=\ln(a)$$ around $$a=e^2$$ we get $$u_{n+1}=(\ln(a)+u_n)\exp(-u_n)-\ln(a) =u_n(1-u_n+\tfrac12u_n^2+...)-\ln(a)u_n(1-\tfrac12u_n+\tfrac16u_n^2+..),$$ so that $$r=1-\ln(a)$$, $$f_1=\frac{1-\frac12\ln(a)}{\ln(a)-1}$$, $$f_2=-\frac12\frac{1-\frac13\ln(a)}{\ln(a)-1}$$, so that for $$a\approx e^2$$ the parameters of $$\phi$$ are $$c\approx 0$$ and $$b^2\approx \frac13>0$$.