Global orthogonal coordinates on the open unit ball Is there any diffeomorphism $x:\mathbb{R}^n\to\text{Im }x=B_1\left(0\right)\subset\mathbb{R}^n$ such that 


*

*$x$ is an orthogonal chart, i.e., the coordinate vector fields $X_i=\partial x/\partial u_i$ satisfy 
$X_i\left(u\right)\cdot X_j\left(u\right)=0$ for all $1\leq i\neq j\leq n$ at every point $u=\left(u_1,\dots,u_n\right)\in\mathbb{R}^n$; 

*the Euclidean norm of the differential $dx$ satisfies $$\left\|dx\left(u\right)\right\|^2\mathrel{\mathop:}=\sum_{i=1}^n\left\|X_i\left(u\right)\right\|^2=\rho^2\left(x\left(u\right)\right)$$ 
at every point $u\in\mathbb{R}^n$, where $\rho:B_1\left(0\right)\to\mathbb{R}$ is given by $\rho\left(x\right)=\frac{1-\left\|x\right\|^2}{2}$. In particular, $dx$ vanishes at infinity.


The above assumptions lead to the following second-order PDE
$$\Delta x-\rho\left(x\right)x=\sum_{i=1}^n\frac{\partial}{\partial u_i}\ln\left(\left\|X_i\right\|^2\right)X_i,$$
which might be useful somehow. 
$\hspace{4pt}$ I believe that such a diffeomorphism does not exist. Actually, I can't even think of such an $x$ satisfying just condition 1. By Liouville's theorem we know that $x$ cannot be conformal. Moreover, radial diffeomorphisms typically used to show that $\mathbb{R}^n$ and $B_1\left(0\right)$ are diffeomorphic do not satisfy condition 1 nor 2.
 A: This problem is just the classical problem of finding global Tchebychev coordinates on hyperbolic $n$-space.  By Hilbert's Theorem, this is impossible when $n=2$.  The problem remains open in higher dimensions, despite years of work.
Here is how one can see the reformulation:  Let $J$ be the Jacobian of $x$ with respect to $u$, i.e., $\mathrm{d}x = J\,\mathrm{d}u$.  Then condition 1 is that $J^TJ$ be diagonal, say, $J^TJ = \mathrm{diag}(\lambda_1^2,\ldots,\lambda_n^2)$, and condition 2 is that 
$$
\lambda_1^2+\cdots+\lambda_n^2 = (1-\|x\|^2)^2/4.
$$
Writing $\lambda_i = \tfrac12(1{-}\|x\|^2) g_i$ yields 
$$
g_1^2+\cdots+g_n^2=1\tag1
$$ 
and the above equations become
$$
(g_1\,\mathrm{d}u_1)^2 + \cdots +(g_n\,\mathrm{d}u_n)^2  
= \frac{4}{(1-\|x\|^2)^2}\bigl(\mathrm{d}x_1^2+\cdots+\mathrm{d}x_n^2\bigr)
\tag2
$$
The right hand side of (2) is the standard form of the conformal hyperbolic metric on the unit ball, so the question is asking for coordinates $u_i$ and functions $g_i>0$ on  hyperbolic $n$-space satisfying equations (1) and (2). This is the classic problem of Tchebychev coordinates.
For example, when $n=2$, one can write $(g_1,g_2)=(\cos f,\sin f)$ for some function $f$ taking values in $(0,\tfrac12\pi)$, and then one is asking when
the metric
$$
ds^2 = \cos^2f\,\mathrm{d}u_1^2 + \sin^2f\,\mathrm{d}u_2^2
$$
has Gauss curvature $-1$.  It is well-known that this holds if and only if $f$ satisfies the Sine-Gordon equation
$$
\frac{\partial^2f}{\partial u_1^2} - \frac{\partial^2f}{\partial u_2^2} 
= \cos f\,\sin f.\tag3
$$
Hilbert proved that there do not exist global coordinates $u_i$ on the hyperbolic plane and a function $f(u_1,u_2)$ taking values in $(0,\tfrac12\pi)$ satisfying (3).
As I wrote, it is still not known whether global solutions (with $g_i>0$) exist in higher dimensions, but, as Cartan showed, there do exist many local solutions.  Essentially, they depend on $n(n{-}1)$ functions of one variable.
Addendum:  Of course, if one only wants global orthogonal coordinates, that is not hard:  If you first apply inversion with respect to a point on the boundary of the ball $B_1(0)$ in $x$-space, you'll get an injective diffeomorphism $v:B_1(0)\to\mathbb{R}^n$ whose image is a half-space. Say $v\bigl(B_1(0)\bigr)$ is defined by $v_n>0$ (after some translation and rotation in $v$-space).  Then define
$$
(u_1,\ldots,u_{n-1},u_n) = \bigl(v_1,\ldots,v_{n-1},\log(v_n)\bigr).
$$
You'll then have $u:B_1(0)\to\mathbb{R}^n$ being a bijective orthogonal diffeomorphism, and its inverse will be the diffeomorphism $x:\mathbb{R}^n\to B_1(0)$ that you desire.
When $n=2$, there are more options:  Choose a conformal diffeomorphism $z:B_1(0)\to R\subset\mathbb{C}$ where $R = I_1\times I_2$ is a rectangle that is a product of two intervals $I_i\subset\mathbb{R}$, not both equal to $\mathbb{R}$,
and let $z = v_1 + i v_2$ and choose any bijective diffeomorphisms $h_i:I_i\to\mathbb{R}$.  Then $u_1 = h_1(v_1)$ and $u_2 = h_2(v_2)$ will provide the desired functions.  Now let $x:\mathbb{R}^2\to B_1(0)$ be the inverse of $u:B_1(0)\to\mathbb{R}^2$.
