curve with constant curvature

Question

Let $\gamma$: $I\twoheadrightarrow C\subset \mathbb{R}^2$ be a plane curve with constant curvature $\kappa>0$. Show that $C$ is part of a circle with radius $\kappa^{-1}$. How to start this question, I don't have any clue.

Answer 1

Parametrizing the curve in terms of it's arclength, we can assert that $\gamma(s)=(x(s), y(s))$ with $||\gamma'(s)||=1$. Then the normal vector is defined by

$$\mathbf{N}=\frac{\frac{d\mathbf{T}}{ds}}{||\frac{d\mathbf{T}}{ds}||}=\frac{\gamma''}{||\gamma''||}$$ since of course $\mathbf{T}=\gamma'(s)$. Setting $v(s)=||\gamma''(s)||$ we write the only Frenet equation:

$$\frac{d\mathbf{N}}{ds}=\frac{\gamma'''v-\gamma'' v'}{v^2}=-\kappa\gamma'$$

This is a differential equation for $\gamma(s)$. It is pretty difficult to solve in general but in this case if we take a dot product with $\gamma'$ and taking into account that $$\gamma'''\cdot\gamma'=(\gamma''\cdot\gamma')'-\gamma''^2=\frac{1}{2}(\gamma'^2)''-\gamma''^2=-v^2$$ we see that $$-\kappa=-v-\frac{v'}{v^2}\gamma''\gamma'\Rightarrow ||\gamma''(s)||=\kappa$$ So we learned that the norm of the acceleration vector is a constant. Putting everything back in our equation we get the linear 3rd order ODE $$\gamma'''(s)+\kappa^2\gamma'(s)=0$$ which has the general solution $$\gamma(s)=\gamma_0+\frac{\mathbf{A}}{\kappa}\cos(\kappa s)+\frac{\mathbf{B}}{\kappa}\sin(\kappa s)$$ where $\mathbf{A,B}$ are arbitrary vectors. We need to fulfill both conditions on the norm of the acceleration and the velocity however which give $$||\gamma''(s)||=\kappa||\mathbf{A}\cos(\kappa s)+\mathbf{B}\sin(\kappa s)||=\kappa$$ This condition makes it easy to see now that $$||\gamma(s)-\gamma_0||=\frac{||\gamma''(s)||}{\kappa^2}=\frac{1}{\kappa}$$ which is evidently the equation for a circle in $\mathbb{R}^2$. This also offers an explicit parametrization of the circle: The constraint requires for all values of the arclength that $$\frac{\mathbf{A}^2+\mathbf{B}^2}{2}+\frac{\mathbf{A}^2-\mathbf{B}^2}{2}\cos(2\kappa s)+\mathbf{A}\cdot\mathbf{B}\sin(2\kappa s)=1$$ which means that $$||\mathbf{A}||=||\mathbf{B}||=1~,~ \mathbf{A}\cdot\mathbf{B}=0$$ Note that the constraint $||\gamma'(s)||=1$ is automatically satisfied with these constraints on the vectors. This allows us to parametrize the vectors as $$\mathbf{A}=(\cos\theta, \sin\theta)~,~\mathbf{B}=(-\sin\theta, \cos\theta)$$ which in turn give the explicit form $$\gamma(s)=\gamma_0+\frac{1}{\kappa}(\cos(\kappa s+\theta),\sin(\kappa s+\theta))$$ where $\gamma_0, \theta$ are arbitrary.