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Consider mappings $f$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ with differential

\begin{align} \mathsf{d} f= \begin{pmatrix} \cos\psi(x) &\cos\phi(y) \\ \sin \psi(x)& \sin\phi(y) \end{pmatrix}, \end{align}

being $\psi(x)$ and $\phi(y)$ arbitrary functions satisfying $0<\psi(x)-\phi(y)<\pi$. (Here $x$ and $y$ are cartesian coordinates.)

Is there an $f$ (with non constant $\psi$ and $\phi$) mapping homeomorphically a disk to a disk (with the same or smaller radius) ?

(After trying with a bunch of possible functions I guess the answer is no, but a general proof should be possible.)

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  • $\begingroup$ Search for results about Chebyshev net --- it might help. $\endgroup$ Commented Oct 18, 2022 at 21:26
  • $\begingroup$ @AntonPetrunin The question indeed comes from Chebyshev nets. They in general map a planar domain into a surface, but the subset of $\mathbb{R}^2\rightarrow \mathbb{R}^2$ maps necessarily has Jacobian of the form I wrote. I have no notice of a comprehensive treatment of such mappings. $\endgroup$ Commented Oct 19, 2022 at 7:28
  • $\begingroup$ It seems that no such maps exist if $\phi$ and $\psi$ are even functions. $\endgroup$ Commented Oct 19, 2022 at 11:11

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Here are a few comments that you might find useful, though they don't completely solve the problem. First, using symmetries of the problem, you can easily reduce to the case that $f$ is mapping the interior of the unit circle $x^2+y^2<1$ diffeomorphically onto the interior of a circle $u^2+v^2 < r^2$ for some $r$. Second, because the Jacobian of the mapping is $\bigl|\sin\bigl(\phi(y)-\psi(x)\bigr)\bigr|\le 1$, it follows that $r\le 1$ with equality if and only if $\phi(y)-\psi(x)\equiv\pm\pi/2$, in which case, $\phi$ and $\psi$ would have to be constant, which is a trivial solution that has been ruled out. Hence, we can assume that $r<1$.

Let us assume that $f(x,y) = \bigl(u(x,y),v(x,y)\bigr)$ extends $C^2$ to the boundary circle $x^2+y^2=1$ (i.e., $\psi$ and $\phi$ extend differentiably to $[-1,1]$) and consider the curve $\gamma(t) = f(\cos t ,\sin t)$, which maps the unit circle to the circle of radius $r<1$, which has curvature $1/r>1$. By the usual formula for the curvature of $\gamma$, the functions $\phi$ and $\psi$ must satisfy the first order relation \begin{align} (1/r)\bigl(1{-}2\,c\,s\,\cos(\psi(c){-}\phi(s))\bigr)^{3/2} &= \sin(\phi(s){-}\psi(c))\\ &\qquad +(c\cos(\psi(c){-}\phi(s))-s)(1{-}c^2)\,\psi'(c)\\ &\qquad - (s\cos(\psi(c){-}\phi(s))-c)(1{-}s^2)\,\phi'(s)\\ \end{align} where, to save writing, I am using $c = \cos t$ and $s=\sin t$.

However, this is a problem because, setting $t=\pi/2$, we get $$ 1/r = \sin(\phi(1)-\psi(0))-\psi'(0), $$ while setting $t=-\pi/2$, we get $$ 1/r = \sin(\phi(-1)-\psi(0))+\psi'(0). $$ Adding these two equations, we get $$ 2/r = \sin(\phi(1)-\psi(0)) + \sin(\phi(-1)-\psi(0)). $$ But, since $r<1$, the left hand side of this equation is greater than $2$, while each of the terms on the right hand side have absolute value less than or equal to $1$. Impossible.

Thus, such an $f$, if it exists, cannot extend twice differentiablly to the boundary circle $x^2+y^2=1$.

Addendum: In the comments, the OP asked whether, if one dropped the assumption that the image of $f$ is a disk, but kept the assumption that $f$ extends $C^1$ to the boundary circle $x^2+y^2=1$, one could still show that the length of the boundary would be at most $2\pi$. I don't know how to do that in full generality, but I can show that, sufficiently near the 'trivial' case, where $\phi(y) \equiv \pi/2$ and $\psi(x)\equiv0$, one has this inequality.

The point is to start with the formula for the length of $\gamma$, which is $$ L(\gamma) = \int_0^{2\pi} |\gamma'(t)|\,\mathrm{d}t = \int_0^{2\pi} \bigl(1{-}\sin(2t)\,\cos(\psi(\cos t){-}\phi(\sin t))\bigr)^{1/2}\,\mathrm{d}t $$ Write $\phi(y)=\pi/2+\kappa(y)$ and this becomes $$ L(\gamma) = \int_0^{2\pi} \bigl(1{-}\sin(2t)\,\sin(\psi(\cos t){-}\kappa(\sin t))\bigr)^{1/2}\,\mathrm{d}t. $$ The key is to break this up into four integrals over the four quadrants. For $0\le t\le \pi/2$, set $$ \begin{align} a_1(t) &= \psi(\cos t)-\kappa(\sin t)\\ a_2(t) &= -\psi(\cos(\pi{-}t))+\kappa(\sin(\pi{-}t)) = -\psi(-\cos t) + \kappa(\sin t)\\ a_3(t) &= \psi(\cos(\pi{+}t))-\kappa(\sin(\pi{+}t)) = \psi(-\cos t) - \kappa(-\sin t)\\ a_4(t) &=-\psi(\cos(-t))+\kappa(\sin (-t)) = -\psi(\cos t)+\kappa(-\sin t) \end{align} $$ and note that $a_1+a_2+a_3+a_4=0$. Moreover, we find that $$ L(\gamma) = \sum_{i=1}^4 \int_0^{\pi/2} \bigl(1{-}\sin(2t)\,\sin a_i(t)\bigr)^{1/2}\,\mathrm{d}t $$ For a small parameter $\lambda$ consider the function $$ f(\lambda) = \sum_{i=1}^4 \int_0^{\pi/2} \bigl(1{-}\sin(2t)\,\sin (\lambda a_i(t))\bigr)^{1/2}\,\mathrm{d}t. $$ Because the sum of the $a_i$ vanishes, the Taylor series expansion of $f$ at $\lambda=0$ to second order takes the form $$ f(\lambda)\simeq 2\pi - \lambda^2\int_0^{\pi/2}\frac{\sin^2(2t)}{8}\left(\sum_{i=1}^4 a_i(t)^2\right)\,\mathrm{d}t. $$ Clearly, $f$ has a strict local maximum at $\lambda=0$ unless the $a_i$ vanish, in which case $L(\gamma) = 2\pi$.

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  • $\begingroup$ Thank you. I wonder why one needs only two points in the whole analysis; is it something implicit in this family of mappings ? Also, although a different question, if one releases the constraint on the image to be a circle, is there an evident way to prove that the perimeter of the image domain is smaller than that of the initial one ? $\endgroup$ Commented Oct 30, 2022 at 5:47
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    $\begingroup$ @DanielCastro: Your'e welcome. For the first question, at the points $t=\pm\pi/2$, the coefficient of $\phi'$ goes to zero and the velocity of $\gamma$ goes to $1$, so the differential relation simplifies considerably. A similar thing happens when $t = 0$ or $\pi$, but with the coefficient of $\psi'$ going to zero instead. For the second question, I don't see any obvious reason why the average value of the velocity of $\gamma$ would be less than $1$, but I'll think about it. $\endgroup$ Commented Oct 30, 2022 at 9:30
  • $\begingroup$ @DanielCastro: I thought a little bit about why the length of $\gamma$ would be less than or equal to $2\pi$ even if one didn't require that the image be a circle, and I did a calculation that showed that, at least near the 'trivial' solution that $\phi(y)-\psi(x) \equiv \pi/2$, all the 'nearby' choices of $\phi$ and $\psi$ do yield a lower perimeter. I haven't seen how to prove that it would always be true, through. $\endgroup$ Commented Oct 31, 2022 at 15:02
  • $\begingroup$ I see. What was the main argument/tool in the calculation ? $\endgroup$ Commented Nov 2, 2022 at 11:32
  • $\begingroup$ @DanielCastro: It's a simple idea, but it won't fit into a comment, so I'll append it to the end of my answer above. $\endgroup$ Commented Nov 2, 2022 at 13:33

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