My question states: Prove that the following coordinate transformation is invertible everywhere, at all values of $(x, y)$ .
$$u = \arctan(x – y)$$
$$v = \sinh(3x) + 2\sinh(y)$$
That is x and y can be functions of both $u$ and $v$: (i.e. $x(u,v) = x, y(u,v) = y$ )
So, I say $$g(u,v) = (\arctan(x-y), \sinh(3x) + 2\sinh(y))$$ . The inverse function theorem says that g will be locally invertible near a specific point a iff its Jacobian Matrix, [Dg], is invertible at a.
So naturally, since I want to prove that $g(x,y)$ is invertible everywhere, I thought to show that [Dg] is invertible everywhere.
Since the Jacobian of $g(x,y)$ is a $2\times2$ matrix, it's clear that any matrix A equal to: \begin{bmatrix}a&b\\c&d\end{bmatrix}
is not invertible if $ad = bc$.
So, since my Jacobian is the matrix of partial derivatives, A is equal to:
\begin{bmatrix}(1/(1+(x-y)^2)&-1/(1+(x-y)^2)\\3cosh(3x)&2cosh(y)\end{bmatrix}
I naturally set up $$ad = bc: 2\cosh(y)/(1+(x-y)^2) = -3\cosh(3x)/(1+(x-y)^2)$$which, by simple multiplication, yields:
$$2\cosh(y) = -3\cosh(3x).$$
Now I don't have any experience with hyperbolic cosine functions but graphing
$$2\cosh(y) = -3\cosh(3x)$$
yields no solutions, which would prove that $g(x,y)$ is invertible everywhere. Proving it is difficult, though.
My question is: how do I prove that the system $2\cosh(y) = -3\cosh(3x)$ has no solutions for any $(x,y)$?
My apologies for the lackluster formatting; I'm new to math stackexchange and don't know a ton of coding.
Thank you in advance!
Best Answer
Just use the definition of cosh: $$ cosh(z) := \frac{e^{z}+e^{-z}}{2} > 0 \Rightarrow 2cosh(y)+3cosh(3x) > 0$$ The image of the exponential function is always greater zero. This contradicts: $$\exists x,y \in \mathbb{R}:2cosh(y)=-3cosh(3x) \Leftrightarrow 2cosh(y)+3cosh(3x) =0 $$