Examples of complicated parametric Jordan curves

Question

For test purposes I need parametric Jordan curves that are complicated in the sense of having many inflection points and ideally no symmetries. When doing online search I always land at complex analysis and curves related to conformal maps.

Explicit formulas or recipes for generating such curves would be great; even better if sample images could be provided.

I need those kind of test curves for checking an algorithm for generating characteristic functions from closed curves and for checking the approximability of those functions.

Answer 1

Some examples and references are mentioned here Examples of plane algebraic curves. You can find many Jordan curves in the family $e^{it}+re^{int}, 0\leq t\leq 2\pi,$ by choosing parameters properly.

To generalize this, take any polynomial $P$ which is univalent (=injective) in the closed unit disk. The image of the unit circle is a Jordan curve. By Riemann's theorem these curves are dense in the set of all Jordan curves, so it can be arranged that they have plenty of inflection points. A huge supply of univalent polynomials is given by the formula $P(z)=z+\epsilon Q(z)$, where $Q$ is any polynomial and $\epsilon<1/\max_{|z|=1}|Q'(z)|$. These curves depend on $\epsilon$ as parameter, and tend to the unit circle when $\epsilon\to 0$.

Another large class of examples are certain lemniscates, they are not given as parametrized curves but their parametrization sometimes can be obtained. Example: $\{ z:|z^n+1|=k\}$, where $k>1$.

In general, a lemniscate is a level set of a complex polynomial $P(z)$. It is Jordan, when the level $k$ is larger than all critical values of the polynomial. (For other $k$ not equal to moduli of critical values, they are disconnected unions of Jordan curves). They may have plenty of inflection points. To obtain a parametrization, you have to be able to invert the polynomial, if $f$ is the inverse, then the parametrization is $f(ke^{it}), 0\leq t\leq 2\pi d$, $d=\deg P$. Like the previous examples they come in continuous families; when $k$ is large, they resemble circles. A theorem of Hilbert says that every Jordan curve can be approximated by lemniscates so the variety of their shapes is unlimited, but of course most of them cannot be explicitly parametrized. Still those which can be give plenty of examples.

Remark. It is not a surprise that closed curves in the plane are mostly mentioned in connection with functions complex variables. Complex variables give the most convenient way to describe and study them.

Answer 2

My answer is similar to the one given by Prof. Eremenko, but with a less classical flavor. The "supercurves", a family of parametric plane curves described by Superformulas and introduced by Johan Gielis around the year 2000, describe many complex plane shapes, including "natural" ones like leaves, flakes and the like: the three dimensional generalization of these object describe "natural" 3D shapes as well.

[1] J. Gielis, D. Caratelli, Y. Fougerolle, P. E. Ricci, I. Tavkelidze, T. Gerats, "Universal Natural Shapes: From Unifying Shape Description to Simple Methods for Shape Analysis and Boundary Value Problems", PLoS ONE, Public Library of Science, (2012), 7, e29324.

Answer 3

An airplane fellow-passenger in the early 1980s(?) (his last name was probably Brown -- sorry, I can't be $100\%$ sure) shared with me his own idea; I believe that he was an engineer or designer, certainly not a professional mathematician.

For every $(a\ b)\in\mathbb Z^2\ $ draw two quarter-circles of radius $\ \frac 12,\ $ inside the square $\ [a;a+1]\times[b;b+1]\ $ such the centers are either

$$ (a\ b)\qquad\qquad\qquad(a\!+\!1\ \ b\!+\!1) $$ or $$ (a\!+\!1\ \ b)\qquad\qquad\qquad(a\!+\!1\ \ b) $$

Each (countable) choice will provide a 1-dimensional submanifold of $\ \mathbb R^2.\ $ For most of the random choices there will be complicated curves.

Remark 1: It's easy to replace the said curves by similar but smooth (infinitely differentiable) curves.

Remark 2: Following my fellow passenger idea, I defined a similar hexagonal version, where instead of two quarter-circles (selected in the two ways inside the integer squares), one would draw three $\ \frac{2\cdot\pi}3-$arcs or two such arc and an interval which connects the middle-points of the opposite edges of the hexagons of the hexagonal lattice.