# Curve whose signed curvature is a function

Let $f:[a,b]\rightarrow\mathbb{R}$ be a function. Is it always possible to find a curve whose signed curvature is the function $f$? I know if $f$ is smooth then it can be possible. But I don't know any result for any arbitrary function. Please tell me is there any results. Any type of results of this type will be very helpful.
Thank you.

• It's certainly possible whenever $f$ is continuous, or piecewise-continuous with jump discontinuities (but undefined at the jump points). Apr 15, 2018 at 15:56
• Sir, please tell me how to define a curve whose signed curvature is a given continuous function.
– MAS
Apr 15, 2018 at 16:02

Supposing that $s\in [0,a]$ is the arclength parameter of the curve you seek, you want to solve the Frenet equation $$\frac{d\mathbf T}{ds} = \kappa(s)\mathbf N(s),$$ where $\mathbf T$ is the unit tangent and $\mathbf T(s),\mathbf N(s)$ always form a positively oriented ("right-handed") orthonormal basis for the plane. Once you specify an initial condition (say $\mathbf T(0) = (1,0)$), this system of ordinary differential equations will have a unique solution. You then integrate $$\frac{d\alpha}{ds} = \mathbf T(s)$$ (with an initial condition, say $\alpha(0)=(0,0)$) to get the curve.
Surely not always possible. For example take $f = \chi_A$ where $A = \mathbb{Q} \cap [a, b]$.