Necessary and sufficient curvature condition for a regular planar curve to be simple and closed Given a smooth, $2\pi$-periodic function $\kappa(s)$, the associated planar curve $\gamma(s)$ for which $\kappa(s)$ is the (signed) curvature, is uniquely determined up to Euclidean invariance: a canonical parametrization is for example given by
$$
 \gamma(s) = \left( \int_0^{s} \cos \phi(\sigma)\,\text{d}\sigma,\,\int_0^{s} \sin \phi(\sigma)\,\text{d}\sigma \right),\;\phi(\sigma) = \int_0^\sigma \kappa(\tau)\,\text{d}\tau. \tag{1}
$$
The goal is to determine whether $\gamma$ is closed and simple.
In principle, parametrization $(1)$ suffices to check whether $\gamma$ is closed and simple; however, the nested integrals makes this cumbersome in practice. In my case, I have a family of functions $\kappa(s)$ as periodic orbits of a given dynamical system, and I would like to select those $\kappa$-orbits that give rise to a simple closed curve $\gamma$.
A priori, one could consider the total curvature $K = \int_0^{2\pi} \kappa(s)\,\text{d}s$. For a closed curve $\gamma$, the condition $K = 2\pi$ is necessary to avoid self-intersections. However, this condition is unfortunately not sufficient. It is straightforward to construct an example where a homotopy within a curve family for which $K=2\pi$ induces self-intersection:

Are there results from differential geometry that I can use here, is my only option to check the injectivity and periodicity of the explicit parametrization $(1)$?
[Related question for algebraic, non-closed curves: https://mathoverflow.net/questions/170320/conditions-for-a-parametric-curve-to-avoid-self-intersection]
 A: I'll think more about this, but if the curvature is positive the necessary and sufficient conditions for the curve to be closed and without self intersections is that the total curvature equal $2 \pi$ and that if you set $\tilde{\kappa}(\theta)$ to be equal to the value of $\kappa(s)$ whenever $\gamma'(s)$ is perpendicular to $(\cos(\theta),\sin(\theta))$ and the determinant of the pair of vectors is positive (we want the oriented normal),
then
$$
\int_0^{2\pi} e^{-i \theta} \tilde{\kappa}(\theta)\, d\theta = 0. 
$$
This is because in that case you can solve the differential equation
$h''(\theta) + h(\theta) = \tilde{\kappa}(\theta)$ and $h$ will be the support function of an oval. It's not terribly explicit ...
A: The Four-vertex theorem dictates that the curvature function needs to have at least two local minima and maxima. The converse of the four-vertex theorem tells us that there's at least one parametrization, such that the resulting curve is simple and closed. It basically turns the curve into something closely resembling an ellipse.
For arbitrary curvature functions satisfying the Four-vertex condition and a fixed parametrization, I don't believe it is possible to answer the question.
