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


1 Answer 1


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 $\bigl\{f(\cos t,\sin t)\ |\ 0\le t\le 2\pi\ \bigr\}$.

  • $\begingroup$ Thank you. OK maybe with completely arbitrary domains $\Omega_1$ and $\Omega_2$ the posing is too general. If we restrict, for instance $\mathsf{Area}(\Omega_2)\leq\mathsf{Area}(\Omega_1)$ is there some hope of having solutions ? $\endgroup$ Jan 10 at 9:21
  • 2
    $\begingroup$ @DanielCastro: Well, an area bound won' t work because you could just take $\Omega_1$ to be the unit disk and $\Omega_2$ to be an ellipse with a large major axis and sufficiently small minor axis, and you'd have the area bound you propose, but the same argument as above for circles would show that the desired $f$ won't exist. Maybe if you assume that $\Omega_1$ and $\Omega_2$ are convex and the '$XY$-diameter' of $\Omega_2$ is sufficiently small relative to the '$xy$-diameter' of $\Omega_1$, you'd have a chance, but I suspect that it's still hopeless with only these hypotheses. $\endgroup$ Jan 10 at 9:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .