My question is how to show that the function
$f: \mathbb{R}^2 \to \mathbb{R}^2\;,\;$ $f(x,y) = \dfrac{1}{20 \pi}\bigg(xy + \cos(xy), 1 + y^2 - \sin(xy)\bigg)$ has a fixpoint?
I know that I have to use the Banach fixpoint theorem, which I can apply because $f$ does map $\mathbb{R}^2$ to $\mathbb{R}^2$ and $\mathbb{R}^2$ is complete, so if I can show that $f$ is a contraction, I'm done, but I have troubles to see how to show that.
Any ideas? Thanks in advance.
Yes, your function has a fixed point that is
$(x,y)=\left(\dfrac1{40},20\pi\right)\;.$
I have looked for it on the set
$\;A=\left\{(x,y):\;(x,y)\in\mathbb R^2\;\land\;xy=\dfrac{\pi}2\right\}$
in order to simplify the expression of the function indeed
$\cos(xy)=0\quad$ and $\quad\sin(xy)=1\;.$
$\begin{cases}x=\dfrac1{20\pi}\big( xy+\cos(xy)\big)\\y=\dfrac1{20\pi}\big(1+y^2-\sin(xy)\big)\end{cases}$
$\begin{cases}x=\dfrac1{20\pi}\left( \dfrac{\pi}2\right)\\y=\dfrac1{20\pi}\big(y^2\big)\end{cases}$
$\begin{cases}x=\dfrac1{40}\\y=0\;\;\lor\;\;y=20\pi\end{cases}$
Since $\;\left(\dfrac1{40},0\right)\not\in A\;$ and $\;\left(\dfrac1{40},20\pi\right)\in A\;,\;$ your function has the fixed point $\;\left(\dfrac1{40},20\pi\right)\;.$