Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2023 / 1 / a / ii / Solution

The native paper prints $n\ge 1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\ge 0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.

For the sine fixed-point iteration, after one step $x_1\in [-1,1]$. On this interval, the sine has the same sign as its argument, andFor example, $t-\sin t={\int }_0^t(1-\cos s)ds>0$, since the integrand is positive for $0<s\le 1$. By oddness of the sine, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\le a_n$ for $n\ge 1$. This monotone sequence is bounded below by zero, so it has a limit $a\in [0,1]$. The continuity of the sine gives $a=\sin a$, and the strict inequality above forces $a=0$.

Hence $x_n\to 0$, and $0$ is a fixed point of the sine. It is the only real fixed point: every such point has absolute value at most one, where the preceding strict inequality applies. ThusThe Banach fixed-point theorem does not directly apply to the sine on $[-1,1]$: the ratios $|\sin t-\sin 0|/|t-0|$ tend to one as $t\to 0$, so no uniform contraction mapping constant smaller than one exists there.