Characterization of Plane Curves *via* Curvature $\kappa(s)$ or Equal Curvature Curves are Congruent

Question

My engagement with this topic was piqued by this question, in which the OP MathMan was seeking help in applying the principle that two plane curves with identical curvature function (I will make this more precise in what follows) are themselves identical "except for probably their position in $\Bbb R^2$" (sic). As I attempted to answer MathMan's concerns, I became more and more aware that the underlying concept was worthy of address in and of itself. Specifically, I began to wonder just how it might be proved. In particular, I wanted, and still want, an analysis/proof of the assertion that two curves the curvature functions of which are the same are "congruent" in the sense that one may be made pointwise identical to the other by a rigid motion of $\Bbb R^2$. In formulating a precise statement of this result, I searched math.stackexchange.com in the usual manner for related questions, but found nothing which seemed exactly on point, so I am proceeding to ask it here.

Having said these things, I turn to my

Question: Let

$I \subseteq \Bbb R \tag 1$

be an open interval, not necesarily bounded, and let

$\alpha, \beta: I \to \Bbb R^2 \tag 2$

be regular, arc-length parametrized curves with curvatures

$\kappa_\alpha, \kappa_\beta: I \to \Bbb R^+ = \{r \in \Bbb R, \; r > 0 \}, \tag 3$

as defined in the Frenet-Serret equations,

$\dot T_\alpha(s) = \kappa_\alpha(s) N_\alpha(s), \; \dot T_\beta(s) = \kappa_\beta(s) N_\beta(s), \tag 4$

where $N_\alpha(s)$ and $N_\beta(s)$ are the unit normal fields to $\alpha(s)$ and $\beta(s)$, respectively. Then if

$\kappa_\alpha(s) = \kappa_\beta(s), \; \forall s \in I, \tag 5$

it follows that there is an orthogonal transformation $O$ of $\Bbb R^2$ and a vector

$\vec v \in \Bbb R^2 \tag 6$

such that

$\alpha(s) = O\beta(s) + \vec v, \; \forall s \in I. \tag 7$

Answer 1

Throughout, by a curve I will refer to a $C^2$ map $I\to\mathbb{R}^2$ with nonvanishing first and second derivatives, where $I\subseteq\mathbb{R}$ is an open interval. For these curves, the unit tangent and normal $T,N$ are always well defined and continuous. Also, I'll use $$ R_\theta:=\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix} $$ As a shorthand for rotation matrices

Since $T$ and $N$ are nonvanishing, orthogonal, and continuous, it must be the case that $N=R_{\pm\pi/2}T$. We may define the orientation of a curve as counterclockwise if $N=R_{\pi/2}T$ and clockwise if $N=R_{-\pi/2}T$.

Your claim can be proven by showing the resulting ODE has unique solutions, and then using a bit of plane geometry:

Lemma: Fix a strictly positive and continuous function $\kappa:I\to\mathbb{R}^2$, and fix $t_0\in I$, $\gamma_0,\dot{\gamma_0}\in\mathbb{R}^2$, with $\|\dot{\gamma}_0\|=1$. There is a unique counterclockwise (alternately, clockwise) unit speed curve $\gamma:I\to\mathbb{R}^2$ with curvature $\kappa$ which satisfies $\gamma(t_0)=\gamma_0$ and $\dot{\gamma}(t_0)=\dot{\gamma}_0$.

Proof: The condition that $\gamma$ has curvature $\kappa$, along with the fact that $N=R_{\pi/2}\dot{\gamma}$ (since $\gamma$ is unit speed and counterclockwise) allow us to write the conditions as a linear second order initial value problem: $$ \ddot{\gamma}(t)=\kappa(t)R_{\pi/2}\dot{\gamma}(t),\ \ \ \ \ \gamma(t_0)=\gamma_0,\ \ \ \ \ \dot{\gamma}(t_0)=\dot{\gamma}_0 $$ This IVP has a global solution, given by $$ \gamma(t)=\gamma_0+\int_{t_0}^tR_{\theta(\tau)}\dot{\gamma}_0d\tau,\ \ \ \ \ \theta(t)=\int_{t_0}^t\kappa(\tau)d\tau $$ And since the differential equation is locally Lipschitz, this solution is unique. To see this, let $\gamma,\lambda$ be two solutions. The set $S=\{t\in I:(\gamma(t),\dot{\gamma}(t))=(\lambda(t),\dot{\lambda}(t))\}$ is open by the Picard-Lindelöf theorem, but its complement $I\setminus S$ is also open, since $(\gamma,\dot{\gamma})$ and $(\lambda,\dot{\lambda})$ are continuous. Therefore one of these sets must be empty, and since $t_0\in S$ by initial conditions we have $\gamma=\lambda$. The clockwise case is true by exactly the same argument with all of the rotation matrices inverted. $\square$

Completing the proof requires a few more straightforward facts about curves in $\mathbb{R}^2$; namely

• The curvature $\kappa$ is invariant under rigid transformation of the curve.
• for any two initial conditions $(\gamma_0,\dot{\gamma}_0)$ and $(\lambda_0,\dot{\lambda_0})$ with $\|\dot{\gamma}_0\|=\|\dot{\lambda}_0\|=1$, there are exactly two rigid motions which takes $(\gamma_0,\dot{\gamma}_0)$ to $(\lambda_0,\dot{\lambda}_0)$, one orientation-preserving and the other orientation reversing.

Given these facts, and two curves $\gamma,\lambda:I\to\mathbb{R}^2$ with equal curvatures, we may choose a base point $t_0\in I$ and choose a rigid motion $g:\mathbb{R}^2\to\mathbb{R}^2$ which takes $(\gamma(t_0),\dot{\gamma}(t_0))$ to $(\lambda(t_0),\dot{\lambda}(t_0))$, chosen to be orientation preserving if $\lambda$ and $\gamma$ have the same orientation, and reversing otherwise. By the previous lemma, $g\circ\gamma=\lambda$.

Answer 2

Here is an other approach. I want to share it with you for two reasons.

• This approach emphasizes the geometric picture: if the Frenet frames of two curves are the same (modulo orientation), then the curves are congruent.

• The approach easily generalises to the congruence theorem for curves in $\mathbb{R}^3$.

Proof: Take $s_0 \in I$. We assume that $\kappa_\alpha = \epsilon\kappa_\beta$, with $\epsilon = \pm 1$. Consider the unique isometry $F$ such that $F(\alpha(s_0))= \beta(s_0)$, $F_*(T_\alpha(s_0)) = T_\beta(s_0)$ and $F_*(N_\alpha(s_0)) = \epsilon N_\beta(s_0)$. If $\epsilon = 1$, $F$ preserves orientation; if $\epsilon = -1$, $F$ reversed the orientation.

Consider the "image curve" $\gamma = F\circ \alpha$. Note that $\gamma$ also has unit speed and that $\kappa_\gamma = \kappa_\alpha$. The latter fact follows from $F_*(\alpha') = \gamma'$, $F_*(\alpha'') = \gamma''$ and the definition of curvature.

Now consider the function $$ f: I \to \mathbb{R}: s\mapsto T_\beta(s) \cdot T_\gamma (s) + \epsilon N_\beta(s) \cdot N_\gamma (s). $$ By the Cauchy-Schwarz inequality, $f(s)\leq 2$ and equality at a point holds if and only if $T_\beta= T_\gamma $ and $N_\beta= \epsilon N_\gamma $ at that point. By the Frenet formulas and $\kappa_\beta= \epsilon \kappa_\gamma $, we get $$ \begin{align*} f'(s) &= \kappa_\beta N_\beta\cdot T_\gamma + T_\beta\cdot \kappa_\gamma N_\gamma - \epsilon\kappa_\beta T_\beta\cdot N_\gamma - \epsilon N_\beta\cdot \kappa_\gamma T_\gamma \\ &= \kappa_\beta\left( N_\beta\cdot T_\gamma + \epsilon T_\beta\cdot N_\gamma - \epsilon T_\beta\cdot N_\gamma - N_\beta\cdot T_\gamma \right) \\ &= 0, \end{align*} $$ where we omitted the argument $s$ for brevity. So $f$ is constant. Since $f(s_0)=2$, we know that $T_\beta(s)=T_\gamma (s)$ for all $s\in I$, and hence $\beta= \gamma + c$, where $c$ is a constant vector. But $c = \beta(s_0)-\gamma(s_0) = 0$, so the curves $\beta$ and $\gamma = F\circ \alpha$ are equal.

A last remark. This argument works for the congruence theorem for two curves in $\mathbb{R}^3$. Essentially, one now has to use the function to $f(s) = T_\beta \cdot T_\gamma + N_\beta\cdot N_\gamma + B_\beta\cdot B_\gamma$ and distinguish between the cases $\tau_\alpha = \pm \tau_\beta$.

Answer 3

With respect to some generality of geometrical curves and surfaces .. curvature as a function of arc length provides for a natural or intrinsic relation/equation of a curve. Curvature and arc length are entirely dependent on first fundamental form coefficients/ derivatives which are bending invariant, a feature of next for isometry of surfaces and the Flatlanders.

Integrated shapes result in intrinsic congruency but upto Euclidean motions i.e., for any translations and rotations of the curve in the plane.

In 3-space differential geometry the fundamental theorem of space curves states that every regular curve with non-zero curvature has its shape/size completely and congruently determined by curvature and torsion of a curve on a surface. Given boundary conditions during integration determine the rigid space curve's displacement and rotation but cannot alter the inherent/intrinsic curvature.

If bending of surface is to be considered in Riemannian / pseudo-Riemannian geometries in addition, the second fundamental form is brought into play. The Gauss–Codazzi–Mainardi equations are fundamental formulas which link together the induced metric and second fundamental form of a submanifold, its immersion into the Riemannian or pseudo-Riemannian manifold.