Q:

An application of the Banach Fixed Point Theorem to a system of equations

I am solving a question whose first item is to demonstrate the Banach Fixed Point Theorem, and the second item is as follows:
Show that for any parameter $t \in \mathbb{R}$ the system
$$
\begin{cases}
x = \frac{1}{2}\sin(x+y) + t - 1 \\
y = \frac{1}{2}\cos(x-y) - t + \frac{1}{2}
\end{cases}
$$
has a unique solution that depends continuously on the parameter $t$.
Now, this seems clear to me that the question asks one to apply the just proved Banach Fixed Point Theorem to find the existence of fixed points for the map
$$
F(x, y) = (\frac{1}{2}\sin(x+y) + t - 1, \frac{1}{2}\cos(x-y) - t + \frac{1}{2})
$$
However, I am not able to show that such a map is a contraction. Any hints wil be the most appreciated.
Thanks in advance.

A:

By the mean value theorem, for any $a,b \in \mathbb{R}$, $|\sin(a)-\sin(b)| \le |a-b|$. Therefore, $$|(\frac{1}{2}\sin(x+y)+t-1)-(\frac{1}{2}\sin(x'+y')+t-1)| = \frac{1}{2}|\sin(x+y)-\sin(x'+y')|$$ $$\le \frac{1}{2}|(x+y)-(x'+y')| = \frac{1}{2}|(x-x')+(y-y')|.$$ Similarly, $$|(\frac{1}{2}\cos(x-y)-t+\frac{1}{2})-(\frac{1}{2}\cos(x'-y')-t+\frac{1}{2})| = \frac{1}{2}|\cos(x-y)-\cos(x'-y')|$$ $$\le \frac{1}{2}|(x-y)-(x'-y')| = \frac{1}{2}|(x-x')-(y-y')|.$$ Therefore, $$|F(x,y)-F(x',y')| \le \sqrt{\frac{1}{4}|(x-x')+(y-y')|^2+\frac{1}{4}|(x-x')-(y-y')|^2}$$ $$= \frac{1}{2}\sqrt{2|x-x'|^2+2|y-y'|^2} = \frac{\sqrt{2}}{2}|(x,y)-(x',y')|.$$ This shows that $F$ is a contraction with Lipschitz constant at most $\frac{\sqrt{2}}{2}$.

