The question is as follows:
Use the contraction mapping theorem to show that the following system of equations:
$x_1$ + $\dfrac{1}{10}\cos(2x_1 + x_2)$ = $1$
$x_2$ + $\dfrac{1}{10}\sin(x_1 + x_2)$ = $2$.
has a unique solution $(x_1, x_2)$ on $\mathbb{R^2}$.
I'm a little familiar with the contraction mapping theorem in terms of one variable, but this is a whole different beast....any push in the right direction is appreciated!
On
Get $x_1$ and $x_2$ into their own equations and then find the Jacobian matrix. Then apply the $\infty$-norm to find $q$ which will be between zero and one, not inclusive.
Edit (by popular demand):
I wrote this on my phone whilst picking up dinner, so please do excuse my terse response.
Here is what I mean by getting $x_1$ and $x_2$ by themselves. You want \begin{align*}x_1&=\frac{1}{10}[1-x_2-\sin(x_1+x_2)]\\x_2&=\frac{1}{10}[2+x_1+\cos(x_1-x_2)]\end{align*} and then you will find the Jacobian matrix by taking partial derivatives with respect to $x_1$ and $x_2$. Then use the $\infty$-norm of the Jacobian to find what $q$ is.
The way to apply the contraction theorem is to write it in the form $f(\overrightarrow{X})=\overrightarrow{X}$ for $\overrightarrow{X}=(x_1, x_2).$ The form of the equations sort of gives away what $f$ should be; you then just need to show that it is a contraction. That is, show that $|f(\overrightarrow{X})-f(\overrightarrow{Y})|\leq c|\overrightarrow{X}-\overrightarrow{Y}|$ for all pairs of vectors, and for $c<1$. From there you get that $f$ has a unique fixed point, which is what you want.