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Given a differentiable function $k(s)$, $s\in I$, show that the parametrized plane curve having $k(s)=k$ as curvature is given by $$ \alpha (s) = \left( \int \cos\theta(s)ds + a, \int \sin\theta(s)ds + b \right) $$ where $$ \theta(s)= \int k(s)ds + \varphi $$ and that the curve is determined up to a translation of the vector $(a,b)$ and a rotation of the angle $\varphi$.

This exercise is from Do Carmo Differential Geometry of Curves and Surfaces, section 1.5. My problem is that I'm not even sure where to start. It is not clear for me what is required for this type of proofs. I mean, the proof is based on some constructive procedure? Or the usual way is, instead, start as "if such $\alpha$ exist, it must verify such and such..."

Карпатський
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2 Answers2

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You have to show two things. First, you have to show that the function $\alpha(s)$ has curvature $k(s)$. So compute the curvature of $\alpha$ and show that it turns out to be $k$.

Then you have to show that any other curve with the same curvature differs from $\alpha$ by a translation and rotation. One way of doing this is by writing down a differential equation that $\alpha$ must satisfy and appealing to uniqueness of solutions of second-order ODEs given initial conditions. The initial conditions will be the curve's starting point and initial velocity, which will give the translation and angle.

Neal
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  • Thanks a lot! This differential equation is related with Frenet-Serret formulas, right? If is not, can you be a little more detailed about that second part? – Карпатський Sep 13 '15 at 15:52
  • @Карпатський Right. The Frenet-Serret formulas are essentially differential equations. – Neal Sep 13 '15 at 16:36
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In this case, you can do it constructively. (In fact, it's a bit of both.)

The (signed) curvature of a curve at a point is the number satisfying (with a dot denoting $d/ds$) $$ \dot{T} = kN, $$ with $T$ the tangent vector $\dot{\alpha}$, and then $N$ is the perpendicular to $T$ taken in the positive direction. To actually do the calculation, we can avoid getting out matrix exponentials and other such perversions by working in complex numbers: if $T=(u,v) = u+iv $, then $N=(-v,u) = -v+iu =iT$, and the equation reads $$\dot{T}=ikT.$$ We can then write down the solution as $$ T = \exp{i\left( \int k(s) \, ds + \varphi \right)}, $$ where $\varphi$ is a constant of integration determined by the initial direction/argument of $T$. Using the definition of $\theta$ in the question, we have, converting back to real vectors by taking real and imaginary parts, $$ T = (\cos{\theta(s)},\sin{\theta(s)}). $$

You can then apply the uniqueness theorem for ordinary differential equations to obtain the result (i.e. use that if $\dot{x}(s)=f(s)$, then $x(s)=\int_{s_0}^s f(t) \, dt +x(s_0)$).

Chappers
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  • I appreciate your answer, but I know almost nothing about functions with complex numbers. – Карпатський Sep 13 '15 at 16:17
  • If you prefer, you can think about the matrix $J=\begin{pmatrix} 0 & -1 \ 1 & 0\end{pmatrix}$, which acts on vectors $(u,v)$ in the same way that multiplication by $i$ acts on $u+iv$. Then the relevant fact is that the differential equation $v' = kJv$ is solved by $v(t) = \exp{\left(J\int k , dt\right)} v_0 $, where $\exp{A}$ is the matrix exponential given by $I + A+A^2/2!+A^3/3! + \dotsb$. It is easy to show that for $J$, $\exp{tJ}$ is the rotation matrix $J=\begin{pmatrix} \cos{t} & -\sin{t} \ \sin{t} & \cos{t} \end{pmatrix}$, and you get the same result as with complex numbers. – Chappers Sep 13 '15 at 16:27