Weird system of PDEs defined on a sphere

Question

Let $x$ and $y$ be functions defined on a simply connected (open or closed) portion of the surface of a (unit) sphere, and consider the following system of PDEs:

\begin{align} \|\nabla x\|^2 = \|\nabla y\|^2 = \| \nabla x \times \nabla y \|^2 \end{align}

(Some suitable conditions may be provided.) Are there non-constant solutions for such a system ? Any ideas about how to reduce it to single equations for $x$ and $y$ ?

Answer 1

Consider the vectors $\nabla x=\vec{u}$ and $\nabla y=\vec{v}$ at some point. The functions $x$ and $y$ are defined on the surface of a sphere, so their gradients must be tangent to the surface of the sphere. So $\vec u\times \vec v$ must point radially out of the sphere. Now if $\theta$ is the angle between $\vec u$ and $\vec v$, then your equations are basically $|u|^2=|v|^2=|u||v|\sin\theta$. These two equations imply that $\theta=\pm 90^o$, or simply, $\nabla x\perp\nabla y$. I guess you can have $x$ be any function you want, and then you can construct $y$ by maybe adding a constant everywhere first and then rotating it appropriately so that its gradient field lies perpendicular to $x$'s gradient everywhere.

EDIT: As pointed out in the comments, there is a silly mistake in this answer that renders it useless. Sorry.

Answer 2

There are no (nonconstant classical $C^1$) solutions.

Since $x$ is continuous on a compact space it has a maximum, and there $\|\nabla x\|=0$. At any point where $\nabla x \ne0$ the same is true of $y$ and from $$ \|\nabla y\|=\|\nabla x \times \nabla y\| \le \|\nabla x\|\,\|\nabla y\| $$ you get that $1 \le \|\nabla x\|$. This contradicts the continuity of $\|\nabla x\|$ unless $x$ is constant.