$2cos(x/2)cos(y/2)=1$" />This curve of $2\cos(x/2)\cos(y/2)=1$ looks like a circle squished in from the sides and top and bottom. I know how to parameterize the curve by dividing it into four 90 degree segments, but second derivatives of the parameterized curves are discontinuous at the transition points. It would be nice to find something like $x=r(\theta)\cos(\theta)$ and $y=r(\theta)\sin(\theta)$, similar to what one might find for a circle. I would like at least $C^2$ continuity. Any ideas?
I will try the following and report back later if it works. Center a unit circle on the origin and parameterize it as $\cos\theta$ and $\sin\theta$. Then project a line out from the center to the corresponding point on the curve, $x=r(\theta)\cos\theta$ and $y=r(\theta)\sin\theta$, where $r(\theta)=\sqrt{x(\theta)^2+y(\theta)^2}$.
There may not be a practical way to express $x(t)$ and $y(t)$ as $C^2$ functions over the entire curve, but only as $G^2$ functions. A suitable workaround that effectively eliminates the $G^2$ continuity problem is presented, so the original question will not be marked as solved.
Let two lines through the origin with $±1$ slopes divide the closed curve into four segments, top and bottom, and left and right. In the top and bottom segments $x(t)$ is a linear function of $t$ and $y(t)$ is a nonlinear function of $t$, and in the left and right segments $y(t)$ is a linear function of $t$ and $x(t)$ is a nonlinear function of $t$.
Parameter $t$ is essentially used as a flywheel to carry the $x(t)$ and $y(t)$ curves correctly between the various curve segments. Parameter $t$ is always independent, variable $x$ is always independent in the top and bottom curve segments, and variable $y$ is always independent in the left and right curve segments. A single equation of motion may be used to solve for independent acceleration $\ddot t$ everywhere, a single equation of motion may be used to solve for independent acceleration $\ddot x$ in the top and bottom curve segments, and a single equation of motion may be used to solve for the independent $\ddot y$ in the left and right curve segments.
In each case the computed independent acceleration may be integrated once and twice to obtain corresponding velocity and displacement. A problem associated with using $\ddot t$ is an undesirable jump in dependent variable $\ddot x(t)$ or $\ddot y(t)$ at the curve segment transition points. And a problem associated with solving only for $\ddot x$ or $\ddot y$ is not easily knowing the next curve segment to transition into. The current value assigned to $t$ always indicates the active curve segment associated with the current independent $\ddot x$ or $\ddot y$, and $\dot t$ indicates which curve segment is about to be transitioned into. The currently independent $x(t)=x$ or $y(t)=y$ is a linear function of $t$, so it also holds that $t$ and $\dot t$ are linear functions of the corresponding independent $x$ or $y$ and $\dot x$ or $\dot y$.
In this approach the parameters $t$ and $\dot t$ are incrementally updated and used only for bookkeeping purposes to insure correct transitions between the respective curve segments.
The two curves below show the $x=\theta_e$ and $y=\theta_o$ values plotted against an angle $\theta$ over a $4\pi$ cycle. A single $\sin\theta$ and one inverse cosine was required to obtain both $C^1$ curves. The equation $(1+\cos\theta_e)(1+\cos\theta_o)=1$ was parameterized with $\theta$ to obtain these curves. A slight modification based on the answer to this question will give similar $C^2$ curves.
Best Answer
I'm not sure if @user170231 realized he practically solved the question with his comment. Using his idea one can construct the parametrization
\begin{cases} \left(\arccos(\cos^2(u))+\arccos(\sin^2(u)),\arccos(\cos^2(u))-\arccos(\sin^2(u))\right) & 0\le u\le \frac{\pi}{2} \\ \left(\arccos(\cos^2(u))-\arccos(\sin^2(u)),\arccos(\cos^2(u))+\arccos(\sin^2(u))\right) & \frac{\pi}{2}\le u\le \pi \\ \left(-\arccos(\cos^2(u))-\arccos(\sin^2(u)),\arccos(\sin^2(u))-\arccos(\cos^2(u))\right) & \pi \le u\le \frac{3 \pi}{2} \\ \left(\arccos(\sin^2(u))-\arccos(\cos^2(u)),-\arccos(\cos^2(u))-\arccos(\sin^2(u))\right) & \frac{3 \pi}{2}\le u\le 2 \pi \end{cases}
which satisfies the requirement of been $C^2$.