This is not a textbook problem, at least that I'm aware of. I was doodling around with some curves and came to this reasonable conjecture: Let $A\subseteq\mathbb{R}^n$ and $\gamma, \bar\gamma:[a, b]\rightarrow\mathbb{R}^n$ be two regular simple closed curves (so, both are of class $C^{\infty}$ with non vanishing derivative, injective in $[a, b)$, and $\gamma^{(k)}(a)=\gamma^{(k)}(b)$, $\bar\gamma^{(k)}(a)=\bar\gamma^{(k)}(b)$, $\forall k\geq 0$) such that $\gamma([a, b])=A=\bar\gamma([a, b])$, $||\gamma '(t)||=||\bar\gamma '(t)||$ $\forall t\in [a, b]$, $\gamma (a)=\bar\gamma(a)$ and $\gamma ' (a)=\bar\gamma ' (a)$, then $\gamma=\bar\gamma$. I put the last hypothesis so that $\gamma$ and $\bar\gamma$ travel $A$ in the same direction. I still don't know how to attack this problem or even if these hypothesis are enough, so I would appreciate any help with this. Thank you.
If two closed simple curves have the same trace, same starting point and same starting derivative, then they are equal.
curvesdifferential-geometry
Related Solutions
$\newcommand{\Reals}{\mathbf{R}}$Let $\Gamma_{0}$ denote $\Gamma$ with the point $\gamma(a) = \gamma(b) = \beta(c) = \beta(d)$ removed.
Each of $\gamma:(a, b) \to \Gamma_{0}$ and $\beta:(c, d) \to \Gamma_{0}$ is a diffeomorphism, so $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is a diffeomorphism (as smooth as the less smooth of the two paths),[*] and extends to the endpoints uniquely by continuity. The extended overlap map is a reparametrization.
For the starred point, if you need to go lower-level than this the important point is that $\gamma_{*}:T(a, b) \to T\Gamma$ is an isomorphism of one-dimensional vector spaces at each point because $\gamma$ is regular, and similarly $\beta_{*}:T(c, d) \to T\Gamma$ is an isomorphism of one-dimensional vector spaces at each point. Consequently, $( \beta^{-1}\circ\gamma)_{*} = (\beta^{-1})_{*}\circ\gamma_{*}:T(a, b) \to T(c, d)$ is an isomorphism of one-dimensional spaces at each point.
Added: We know the function $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is bijective and satisfies $\gamma = \beta \circ (\beta^{-1} \circ \gamma)$. As indicated by the comments, we need to show $\beta^{-1} \circ \gamma:[a, b] \to [c, d]$ is smooth and has smooth inverse.
Proposition: If $\gamma(t_{0}) = \beta(\tau_{0})$ for some $t_{0}$ in $(a, b)$ and $\tau_{0} = \beta^{-1} \circ \gamma(t_{0})$ in $(c, d)$, then $\gamma'(t_{0})$ and $\beta'(\tau_{0})$ are proportional, i.e., each is a (non-zero) multiple of the other.
Proof: Pick an orthonormal basis $(e_{j})_{j=1}^{n-1}$ for the hyperplane orthogonal to $\gamma'(t_{0})$ and define the smooth mapping $F:\Reals^{n} \to \Reals^{n}$ by $$ F(u_{1}, \dots, u_{n-1}, t) = u_{1}e_{1} + \cdots + u_{n-1}e_{n-1} + \gamma(t). $$ Since $Df = [e_{1}\ \cdots\ e_{n-1}\ \gamma'(t)]$ is non-singular at $T_{0} := (0, \dots, 0, t_{0})$, the inverse function theorem guarantees there is a neighborhood $U$ of $T_{0}$ in which $F$ is smoothly invertible. Write $V = F(U)$. By construction, the functions $\phi_{j} := u_{j} \circ F^{-1}$ are smooth, vanish simultaneously on the trace of $\gamma$ and nowhere else, and have linearly independent gradients in $V$. That is, the trace of $\gamma$ intersected with $V$ is precisely the zero set of $\Phi := (\phi_{1}, \dots, \phi_{n-1})$, and the kernel of this mapping, $\ker D\Phi(T_{0})$, is one-dimensional. Since $\beta$ has the same trace as $\gamma$, the non-zero vector $\beta'(\tau_{0})$ lies in this same one-dimensional subspace. That is, $\beta'(\tau_{0})$ and $\gamma'(t_{0})$ are proportional. This completes the proof of the proposition.
Write $\gamma = (\gamma_{k})_{k=1}^{n}$ and $\beta = (\beta_{k})_{k=1}^{n}$, and note that $\gamma_{k} = \beta_{k} \circ (\beta^{-1} \circ \gamma)$ for each $k$. Let $t_{0}$ be an arbitrary point of $(a, b)$ and put $\tau_{0} = \beta^{-1} \circ \gamma(t_{0})$. Because $\beta$ is regular, there exists an index $k$ such that $\beta_{k}'(\tau_{0}) \neq 0$. The preceding proposition guarantees $\gamma_{k}'(t_{0}) \neq 0$. By the one-variable inverse function theorem, there exists a $\delta > 0$ such that $\beta_{k}$ is smoothly invertible in $(\tau_{0} - \delta, \tau_{0} + \delta)$. Consequently, $\beta_{k}^{-1} \circ \gamma_{k}$ is smoothly invertible in some neighborhood of $t_{0}$ as a composition of smoothly invertible functions. But $\beta^{-1} \circ \gamma = \beta_{k}^{-1} \circ \gamma_{k}$ (!) in this neighborhood by applying $\beta_{k}^{-1}$ to both sides of $\gamma_{k} = \beta_{k} \circ (\beta^{-1} \circ \gamma)$. Since $t_{0}$ was arbitrary, $\beta^{-1} \circ \gamma$ is locally smoothly invertible at each point.
In particular, the preceding paragraph shows that the bijection $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is continuous. A continuous bijection whose domain is an interval is strictly monotone (by the intermediate value theorem), and a strictly monotone function between bounded intervals extends uniquely by continuity to the endpoints. In summary, we have shown that $\beta^{-1} \circ \gamma:[a, b] \to [c, d]$ is smooth and has smooth inverse.
Best Answer
HINT:
Like Ted Shifrin suggested, the idea would be that the length of the curve from $\gamma(a)$ to $\gamma(x)$ equals $$\int_{a}^x \|\gamma'(t)\| dt$$ (this because the function is injective) so it equals the length from $\gamma(a)= \bar\gamma(a)$ to $\bar\gamma(x)$. Now if they start in the same direction, it follows that $\gamma(x) = \bar\gamma(x)$, for all $x$.