# Shrinking a disk with fixed differential

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.)

• Search for results about Chebyshev net --- it might help. Commented Oct 18, 2022 at 21:26
• @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. Commented Oct 19, 2022 at 7:28
• It seems that no such maps exist if $\phi$ and $\psi$ are even functions. Commented Oct 19, 2022 at 11:11

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 $$L(\gamma) = 2\pi$$.

• 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 ? Commented Oct 30, 2022 at 5:47
• @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. Commented Oct 30, 2022 at 9:30
• @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. Commented Oct 31, 2022 at 15:02
• I see. What was the main argument/tool in the calculation ? Commented Nov 2, 2022 at 11:32
• @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. Commented Nov 2, 2022 at 13:33