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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 $y$ ?

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

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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.

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  • $\begingroup$ Thanks. Con one say something similar if the domain in not the whole sphere but just a connected (open or closed) portion of it ? $\endgroup$ Sep 26, 2022 at 17:52
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    $\begingroup$ @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. $\endgroup$ Sep 26, 2022 at 19:14
  • $\begingroup$ So in light of that, can we say a non-constant solution $may$ (or does) exist ? $\endgroup$ Sep 26, 2022 at 19:23
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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.

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    $\begingroup$ 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. $\endgroup$ Sep 25, 2022 at 0:46
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    $\begingroup$ @TedShifrin Woah. Can't believe I messed that up like that! $\endgroup$ Sep 25, 2022 at 1:01

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