Let $\Omega\_1$ and $\Omega\_2$ be (simply connected) domains on $\mathbb{R}^2$, with coordinates $(x,y)$ and $(X,Y)$ respectively. Given a (smooth) function $Z(X,Y)$ such that $Z\left(\partial \Omega\_2 \right)=0$, consider the system of PDEs


\begin{align}
\left(\frac{\partial X}{\partial x} \right)^2+\left(\frac{\partial Y}{\partial x} \right)^2+\left(\frac{\partial Z\left(X,Y \right)}{\partial x} \right)^2=1
\end{align}
\begin{align}
\left(\frac{\partial X}{\partial y} \right)^2+\left(\frac{\partial Y}{\partial y} \right)^2+\left(\frac{\partial Z\left(X,Y \right)}{\partial y} \right)^2=&1,
\end{align}


subjected to


\begin{align}
\left(X,Y\right)\left(\partial \Omega\_1 \right)=\partial \Omega\_2.
\end{align}


Does a solution exist ?


This problem comes as a particular case of Chebyshev mappings, where a surface is covered by a net of lines whose tangents are unit vectors at all its points. In this case the boundary of the surface is constrained to lie on plane. Generically, the (local) existence of solutions is shown by obtaining an equivalent system that is explicitly hyperbolic (by cross-differentiation), so the problem is hyperbolic in nature. In this case the condition is not hyperbolic but elliptic, so the existence of solutions is not secured



Hyperbolic system of PDEs with elliptic-like boundary contions
Here is an example for which there is no solution: Let $\Omega\_1$ be defined by $x^2+y^2\le 1$ and $\Omega\_1$ be defined by $X^2+Y^2\le R^2$, where $R>0$ is large. Take $Z(X,Y) = 0$. Then one is asking for a map $f:\Omega\_1\to\Omega\_2$ that preserves lengths of $x$-lines and $y$-lines and carries the circle $x^2+y^2=1$ onto the circle $X^2+Y^2=R^2$. However, because any two points in $\Omega\_1$ can be joined by a piecewise curve composed of $x$-segments and $y$-segments of total length at most $3$ (say), it follows that the image of $f$ must be contained in a disk of radius at most $3$. When $R$ is sufficiently large, there will be points on the boundary of $\Omega\_2$ that are farther apart than that.


(Note that, if one allows $Z(X,Y)$ to be nonzero, this only makes the problem worse.)


*Remark:* To get a sense of what these restrictions look like. Consider the case where $Z(X,Y)=0$ and $\Omega\_1$ is defined by $x^2+y^2\le 1$. Then each solution to the above equations for $f:\Omega\_1\to\mathbb{R}^2$ is of the form
$$
f(x,y) = \bigl(X(x,y),Y(x,y)\bigr) = \bigl(p\_1(x)+q\_1(y),p\_2(x)+q\_2(y)\bigr),
$$
where $p=(p\_1,p\_2):[-1,1]\to\mathbb{R}^2$ and $q=(q\_1,q\_2):[-1,1]\to\mathbb{R}^2$ are unit speed curves. To describe the domains $\Omega\_2$ such that $f(\partial\Omega\_1) = \partial\Omega\_2$, you need to be able to determine which closed curves $\partial\Omega\_2$ are of the form $\bigl\{f(\cos t,\sin t)\ |\ 0\le t\le 2\pi\ \bigr\}$.


