Parametric families of area-preserving curves?

Question

I have an interesting PDE problem that I would like to solve. Consider the diagram below.

I would be very interested to hear about whether there are any "nice" families of parametric curves ${\bf r}={\bf r}(s) $ with $s\in[0,s_{\max}]$ (where $s_{\max}$ can itself depend on a parameter) that satisfy the following conditions:

1. For some parameter $p$, the curve is deformed in some way, but where the area of $\Omega$ remains constant.
2. $y(0) = y(s_{\max})$.
3. The curve cannot be self-intersecting, or go outside the box.

The only examples I can think of at the moment are the cases where the curve is a rectangular "notch" in the middle, parametrised by an aspect ratio $r$ keeping the area fixed, some sort of triangular notch where the bottom vertex of the triangle is free to shift horizontally, or a trapezoidal curve:

In all these cases the arc-length parametrisations would be step functions of $s$. I can't help but think that, even those these do satisfy the conditions I have provided, they are not particularly "novel" or interesting. Can anyone shed some light on whether there are any "non-obvious" parametric families of curves that do what I might like?

Answer 1

Here is an example.

We'll use the family $y=P(x)=x(x-1)(x-a)(x-b)$, with $a \ge 1$ and $b \ge 1$.
If needed, the curves can be expressed as $(x(s), y(s))$ by using $x(s)=s$.
As $P(0)=P(1)=0$ and $P(x) < 0$ for $x \in ]0,1[$, we'll take the surface below $0$ between $0$ and $1$.

This surface is:
$S = -\int_0^1 P(x)dx$
$= -\int_0^1 (x^4-(1+a+b)x^3+(a+b+ab)x^2-abx)dx$
$= -\frac 1 5 + \frac {1+a+b} 4 - \frac {a+b+ab} 3 + \frac {ab} 2$
$= \frac 1 {60} (3-5a-5b+10ab)$

To have $S$ constant we need $2ab-a-b$ constant, which we'll call $c$.
So $2ab-a-b=c, \quad b=\frac {c+a} {2a-1}$.
When $a=1, \; b=c+1, \;$ and when $a=c+1, \; b=1$.

In conclusion the family of curves $y=x(x-1)(x-a)(x-\frac {c+a} {2a-1})$, where $c$ is a constant $>0$ and $a$ is a parameter in $[1,1+c]$, has a constant area below $0$ on $[0,1]$; this area is $\frac {5c+3} {60}$.

Answer 2

I am not sure if this is near to what you are looking for, but mentioning what I believe is related. The area and boundary length have to be given/fixed in the solution. Dido's problem in the calculus of variations between two given points belongs to iso-perimetric class of problems.

It may take some convincing to do.. concepts of maximum, minimum and constant of object function (area) have the same connotation in Calculus of Variations...and have the same method of solution ( e.g., EL equn ).

The solution for maximum area and given boundary length constraint $s_{max}$ like what is shown in sketch can be found from formulation of smooth curve

$$ \int y~ dx - \lambda \sqrt{1+y^{'2}}dx $$ resulting in an arc of a circle which can be also parametrized by

$$ x= h+ \lambda \cos (s/\lambda),~ y= k+ \lambda \sin (s/\lambda)~$$ where $(h,k )$ are arbitrary constants.