Differentiable map sending square to circle

Question

The question is that simple. Is there any known consruction of a differentiable map $\phi \colon U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that sends a closed square $Q\subset U$ to a circle?

Of course, the derivative of $\phi$ at the vertices of $Q$ will not be invertible.

Note that the complex map $z\mapsto z^2$ sends the first quadrant to the upper half-plane. Then it straightens out the right angle.

Answer 1

If we take the square $Q$ with vertices at $(\pm1,\pm1)$ and $(\pm1,\mp1)$, then $\phi\colon Q\to S^1$ defined by

$$\phi(x,y)=\frac{(x,y)}{\sqrt{x^2+y^2}}$$

maps $Q$ to the unit circle $S^1$.

Answer 2

• First translate $Q$ to some square $R$ in the upper half plane, and so $R$ is disjoint from the origin;
• Next compose with the norm map $(x,y) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)$ which takes $R$ onto some arc $A$ of the upper half-circle (you needed $R$ to be disjoint from the origin so that the denominator of this map is never zero on $R$);
• And finally, depending on the total angle $\theta \in (0,\pi)$ of $A$, take a sufficiently high power $z \mapsto z^n$ (chosen so that $n\ge \frac{2\pi}{\theta}$) to stretch $A$ over the whole circle.

If you write down each of these three functions in $(x,y)$ coordinates, and then compose them in the correct order, you'll get your desired function.

Answer 3

Not sure if by square you mean a full-square with interior or just the boundary. (In the second case, others have answered this.)

If you mean a square with interior, then you can consider a map that maps the upper-most edge of the square to a curve going around the circle one and a half revolution in one direction, and then back.

On the lower edge of the square, the map will be constant.

The rest of the square will be a smooth homotopy between those two.

If you mean a map from a full square into a ball (disc) then it's even easier: consider a map, in polar coordinates $r v\mapsto \phi(r) v$ so that $\phi$ is an identity on $[0, \frac{1}{2}]$ and equals to $1$ for $r\geq 1$.

Answer 4

We usually use the polar coordinate formula $r(\theta,n)=(\cos^n\theta+\sin^n\theta)^{-1/n}$ to plot an almost square .