# Weird system of PDEs defined on a sphere

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 $$y$$ ?

• Is there a reason you write $(v)^2$ on the left and $\|v\|^2$ on the right? Sep 24, 2022 at 22:52
• @TedShifrin Oh no, my apologies. They are the same. Sep 24, 2022 at 22:53

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.

• Thanks. Con one say something similar if the domain in not the whole sphere but just a connected (open or closed) portion of it ? Sep 26, 2022 at 17:52
• @DanielKatzner A closed subset is compact, but it looks like the argument falls apart because the maximum can be at the boundary with the gradient not zero. Sep 26, 2022 at 19:14
• So in light of that, can we say a non-constant solution $may$ (or does) exist ? Sep 26, 2022 at 19:23

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.

• You forgot the square on the right-hand side of the equations. So we get $\|\nabla x\| = \|\nabla y\| = |\csc\theta|$. I don't think you can resurrect your conclusions. Sep 25, 2022 at 0:46
• @TedShifrin Woah. Can't believe I messed that up like that! Sep 25, 2022 at 1:01