This question is a follow-on of this answer of mine. This answer provides a numerical solution ; the initial question was about possible closed form expressions.
Q1 : Is there a reference in the existing literature about these orthogonal curves ?
Q2 : These orthogonal curves look symmetrical with respect to second diagonal line with equation $x+y=1$. But is it the case ?
Q3 : Let us consider one of the orthogonal functions $(x(t),y(t))$. Let $z(t)$ be any or $x(t)$, $y(t)$ or $z(t)=x(t)+i y(t)$. This function verifies (by converting the first order system into a second-order ODE) :
$$z''(t)+\frac{2}{\sin(2t)}z'(t)+z(t)=0 \tag{*}$$
(with appropriate initial conditions) or the equivalent equation written under the Sturm-Liouville form :
$$\frac{d}{dt}(\tan(t) \ z'(t))+\tan(t) \ z(t)=0 \tag{**}$$
but with eigenvalue $\lambda = 0$ (?). Is it possible to take advantage of these forms (*) or (**) in order to have a different approach to this issue ?
Remark : If the initial points $I_k$ (notations of my text) are replaced by their symmetrical points on the "Eastern" side of square $S$
(symmetry with respect to diagonal line with equation $x+y=1$) and if we replace instruction yp=[Y/ta;X*ta] by yp=[-Y/ta;-X*ta], we get (numericaly speaking) the same curves.
Hello Jean Marie!
I think the thing is slightly asymmetric. I took one point, $(\frac{4}{5},\frac{3}{5} ).$ This is along the circle, so the normal to the curve passes through the origin.
The reflected point is $(\frac{2}{5},\frac{1}{5} ).$ I decided to numerically find the exponent $p$ that works. Then to check whether the normal line passes through $(1,1).$
My programmable calculator decided that $p \approx 0.563895524.$ The gradient vector ($x^p + y^p = 1$) divided by $p$ is $(x^{p-1}, y^{p-1})$ and at $(\frac{2}{5},\frac{1}{5} )$ this came out $(1.491225834, 2.017548332).$ The second divided by the first comes out $1.352946204,$ which is not exactly $4/3.$ With my figures, the normal line passes very near to $(1,1)$ but not through it.
Probably some way to repeat this with high precision in, say, Pari-gp, which I have.