How to show an iteration converges to a unique fixed point in a given interval?

Question

Let $c\in\left(0,1\right)$ and we look on the iteraion: $x_{n+1}=\frac{1}{2}\left(x_{n}^{2}+c\right)$

1. Show that the iteration have two fixed points $0<\xi_{1}<1<\xi_{2}$.

2. Show that for every $x_{0}\in\left[0,\xi_{2}\right)$ the iteration converges to $\xi_{1}.$

3. How does the iteraion act for different $x_{0}$?

Attempt:

1. Let $g\left(x\right)=\frac{1}{2}\left(x^{2}+c\right)$ and we're looking for a solution to $g\left(\xi\right)=\xi$. Using the quadratic formula for the roots of a general quadratic equation we get the solutions $\xi_{1,2}=1\pm\sqrt{1-c}$ which are in the requested intervals.

2. I know that if we have a fixed point $p\in\left[a,b\right]$ and $g\left(x\right)\in\left[a,b\right]$ for all $x\in\left[a,b\right]$ and $\left|g'\left(x\right)\right|\leq k<1$ for every $x\in\left(a,b\right)$ then for every initial point $x_{0}\in\left[a,b\right]$ the iteration defined by $x_{n+1}=g\left(x_{n}\right)$ converges to the unique fixed point $p\in\left[a,b\right]$. In our case $g\left(x\right)=\frac{1}{2}\left(x^{2}+c\right)$ and for every $x\in\left[0,1+\xi_{2}\right)$ we have $$ \begin{aligned} & g\left(x\right)=\frac{1}{2}\left(x^{2}+c\right)<\frac{1}{2}\left(\left(\xi_{2}\right)^{2}+1\right)=\frac{1}{2}\left(\left(1+\sqrt{1-c}\right)^{2}+1\right)=1+\sqrt{1-c}\\ & g\left(x\right)=\frac{1}{2}\left(x^{2}+c\right)>\frac{1}{2}\left(0+0\right)=0 \end{aligned} $$ So $g\left(x\right)\in\left[0,1+\sqrt{1-c}\right)$ but the problem is with $g'\left(x\right)=x$ which means that for every $x\in\left[1,1+\xi_{2}\right)$ we have that $\left|g'\left(x\right)\right|\geq1$.

How can we show convergenes to $\xi_{1}$ for every initial point $x_{0}\in\left[0,1+\xi_{2}\right)$? and also what happens when $x_{0}\notin\left[0,1+\xi_{2}\right)$?

Answer 1

Notice that:

• If $0 \leq x \leq \xi_1$, then $0 \leq x \leq g(x) \leq \xi_1$.

• If $\xi_1 \leq x \leq \xi_2$, then $\xi_1 \leq g(x) \leq x \leq \xi_2$.

• If $x \geq \xi_2$, then $\xi_2 \leq x \leq g(x)$.

From this, we find that

• If $x_0 \in [0, \xi_1]$, then $(x_n)$ is a non-decreasing sequence in $[0, \xi_1]$, hence converges to some $\ell \in [0, \xi_1]$. Since this $\ell$ satisfies $\ell = g(\ell)$ by taking limit to the recurrence relation, we get $\ell = \xi_1$.

• If $x_0 \in [\xi_1, \xi_2)$, then $(x_n)$ is a non-increasing sequence in $[\xi_1, \xi_2)$. Following the same idea, we know that $(x_n)$ converges to some $\ell \in [\xi_1, \xi_2)$ and this $\ell$ is a fixed point of $g$. So $\ell = \xi_1$.

• If $x_0 = \xi_2$, then clearly $x_n = \xi_2$ for all $n$ and hence $x_n \to \xi_2$.

• If $x_0 > \xi_2$, then $(x_n)$ is a non-decreasing sequence in $(\xi_2, \infty)$. This sequence cannot be bounded, for otherwise we may follow a similar argument as before to prove that $(x_n)$ converges to one of the fixed points of $g$, which is impossible. Then the unboundedness and monotonicity of $(x_n)$ altogether implies that $x_n \to +\infty$ as $n\to\infty$.

These behaviors can be easily understood if we draw the cobweb plot for this iteration. The following plot demonstrates the case when $c = 2/3$ and various choices of the initial points $x_0$.

• When $x_0 < 0$, simply replace $x_0$ by $|x_0|$ and notice that all the other terms remains the same for this replaced initial condition. So the former cases covers this case as well.