Global invertibility of a particular differentiable function of $\mathbb R^2$ in itself

Question

This problem comes from SISSA 2007 Entrance Examinations.

Let $\Phi:\mathbb R^2 \to \mathbb R^2$ be the function $$\Phi(x,y)=\begin{pmatrix}5x+\sin(y) \\5y+\arctan(x)\end{pmatrix}$$ Show that $\Phi$ is bijective.

If we observe that $\Phi$'s jacobian is $$\det (J_{\Phi}(x,y))=\begin{vmatrix}5 &\cos(y) \\ \frac{1}{1+x^2}& 5\end{vmatrix}=25-\frac{\cos(y)}{1+x^2}$$ and its determinant is nonzero for every $(x,y)$, we can conclude from this theorem that the function is invertible in about one second.

Since this was a question from an examination I'd expect something more conceptual, but I can't seem to come up with an easy proof of the function's invertibility. How can we solve this problem with less machinery?

Answer 1

Try using the Banach fixed-point theorem to solve $\Phi(x, y)=(c_1, c_2)$ by writing a map $\mathbb{R}^2\rightarrow\mathbb{R}^2$ as

$(x, y)\rightarrow \left(\dfrac{c_1-\sin y}{5},\dfrac{c_2-\arctan x}{5}\right)$.

Prove that this is a contraction, and conclude that $\Phi$ is bijective.

Answer 2

I know that my answer is not a standard one but here we go!

Suppose on contrary that there are two pairs $(x,y)$ and $(\tilde{x}, \tilde{y})$ such that $\Phi(x,y) = \Phi(\tilde{x}, \tilde{y})$. So $x + \sin y = \tilde{x} + \sin \tilde{y}$. Define $f:(-\epsilon, 1+\epsilon) \to \mathbb{R}$ such that $f(t) = 5(x +t(\tilde{x}-x)) + \sin(y +t(\tilde y- y))$. As $f(0) = f(1)$, there is a number $t_0 \in (0,1)$ such that $f'(t_0)=0$ (mean value theorem). So $5(\tilde x -x)+\cos(y+t_0(\tilde y-y)) = f'(t_0)=0 \implies 5(\tilde x -x)=-\cos(y+t_0(\tilde y-y))$. So we have $|\tilde x -x| \leq \frac{|\tilde y-y|} {5} $. Now define $g:(-\epsilon, 1+\epsilon) \to \mathbb{R}$ such that $g(t) = (y -t(\tilde y -y)) +\arctan (x+t(\tilde x - x)$. Again $g(0)=g(1)$, so there is $t_1\in (0,1)$ such that $g'(t_1)=0 \implies 5(\tilde y -y) + \frac{1}{1+(x-t_1(\tilde x -x))^2}(\tilde x-x) =0 \implies5 |\tilde y - y |= \left| \frac{1}{1+(x-t_1(\tilde x -x))^2}(\tilde x-x)\right| $. Therefore, $|\tilde y - y | \leq \frac{| \tilde x-x|}{5}$. By $|\tilde x -x| \leq \frac{|\tilde y-y|} {5}$ and $|\tilde y - y | \leq \frac{| \tilde x-x|}{5}$, we have $|\tilde x - x | \leq \frac{| \tilde x-x|}{25} \implies | \tilde x-x|=0 \implies \tilde x=x $ and $\tilde y =y$.