Proof that any curve in two dimensions has a modelling differential equation.

Question

I was wondering if anyone knew of a proof stating that any smooth, continuous line drawn in two dimensions must have a differential equation that models it.

Best,

James.

Answer 1

I post this as an answer because it is too long: I will work on $\Pi=[0,2\pi]$ out of habit but you can reparametrize if you prefer other intervals: For given $k\in C^0(\Pi)$ there exists a curve $c\in C^2(\Pi,\mathbb R^2)$ parametrized by arc length having $k$ as its signed curvature, i.e. $c''(t)\cdot Jc'(t)=k$, where $J=\left(\begin{smallmatrix}0& -1\\ 1 & 0\end{smallmatrix}\right)$. You can retrieve $c$ from $k$ by computing $$ c(t):=\int_0^t\begin{pmatrix}\cos \varphi (s)\\ \sin\varphi(s)\end{pmatrix}\mathrm ds $$ and $\varphi'(s)=k(s)$, $\varphi(0)=0$. Choosing different integrating constants will change the position of the curve and rotate it by a certain angle with respect to the curve I have written down. Up to this Euclidean motions, $k$ determines $c$ uniquely. Note that the speed $|c'|$ is not enough to determine $c$ since many different curves can be for example arc length parametrized and have hence the same speed. Curvature is in contrast really a geometric quantity as it is parameter-invariant.

You should also note that this holds only for plane curves. In higher dimensions the determination of a curve by its curvature fails rather dramatically and you need an additional datum (torsion) to get essential uniqueness.