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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.

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

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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.

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Surely not always possible. For example take $f = \chi_A$ where $A = \mathbb{Q} \cap [a, b]$.

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  • $\begingroup$ Can you tell me for which type of functions it is possible? $\endgroup$
    – MAS
    Apr 15, 2018 at 15:25

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