Realizing possible accelerations of paths on a sphere

Question

$\newcommand{\al}{\alpha}$

Let $x,v \in \mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, $w \in \mathbb{R}^{n+1}$ satisfy $\langle x,v \rangle=0, \langle x,w \rangle=-1$.

Does there exist a smooth path $\alpha:(-\epsilon,\epsilon) \to \mathbb{S}^n$ such that $$ \alpha(0)=x, \dot \alpha(0)=v, \ddot \alpha(0)=w? $$

---

The condition $\langle x,w \rangle=-1$ is necessary for the existence of such a path:

Differentiating twice $\langle \alpha,\alpha \rangle=1$ we get

$$ \langle \dot \alpha,\dot \alpha\rangle+\langle \alpha,\ddot\alpha \rangle=0. $$

I tried to play with paths of the form of $$ \al(t)=a(t)x+b(t)v(t), $$ where $v(t) \in T_x\mathbb{S}^n$, but so far my computations did not reach a definite conclusion.

Note that every path $\al$ can be written in the above form. Part of the problem here is that $a(0)=1$, so differentiating $b(t)=\sqrt{1-a^2(t)}$ is not trivial at $t=0$.

Answer 1

$\newcommand{\brak}[1]{\left\langle#1\right\rangle}$tl; dr: Yes.

---

Suppose $x$ and $v$ are orthogonal unit vectors, interpreted as a point of the unit sphere and a tangent vector, let $w^{\perp}$ be an arbitrary vector orthogonal to $x$, and set $w = w^{\perp} - x$, so that $\brak{w, x} = -1$.

Define $$ X(t) = x\cos t + v\sin t + \tfrac{1}{2}t^{2} w^{\perp},\qquad \alpha(t) = \frac{X(t)}{|X(t)|}. $$ Since $x\cos t + v\sin t$ is a unit vector for all $t$, we have \begin{align*} |X(t)|^{2} &= 1 + t^{2}\brak{x\cos t + v\sin t, w^{\perp}} + \tfrac{1}{4}t^{4} |w^{\perp}|^{2} \\ &= 1 + t^{2}\sin t\brak{v, w^{\perp}} + \tfrac{1}{4}t^{4} |w^{\perp}|^{2} \\ &= 1 + O(t^{3}) \end{align*} at $t = 0$. Since $(1 + u)^{-1/2} = 1 - \frac{1}{2}u + O(u^{2})$ at $u = 0$, we have $1/|X(t)| = 1 + O(t^{3})$ at $t = 0$, so $$ \alpha(t) = X(t)[1 + O(t^{3})]. $$ Particularly, $\alpha'(0) = X'(0)$ and $\alpha''(0) = X''(0) = -x + w^{\perp} = w$.

Answer 2

The first condition is not necessary. Let $R(t) \in \mathsf{SO}(3)$ and $\omega(t) = (\omega^1(t), \omega^2(t), \omega^3(t)) \in \mathbb{R}^3.$ Let $$ S(\omega(t)) := \begin{pmatrix} 0 & \omega^3(t) & -\omega^2(t) \\ -\omega^3(t) & 0 & \omega^1(t) \\ \omega^2(t) & -\omega^1(t) & 0 \end{pmatrix} $$ Consider the dynamical system,

$$\begin{aligned} \dot{R}(t) &= S(\omega(t))\, R(t),\\ \dot{\omega}(t) &= u(t) \end{aligned},$$

where $u(t) \in \mathbb{R}^3$ is any arbitrary signal. Since $R(t)\in\mathsf{SO}(3)$ for all $t,$ we have that,

$$ x(t) := R(t)\,e_1 \in \mathsf{S}^{2}, $$ for all $t.$ This is independent of $u(t).$ It is straightforward to see that the velocity of $x$ (so-to-speak) is orthogonal to $x(t)$ with the standard inner product. Observe that,

$$\begin{aligned} \langle x(t), \dot{x}(t) \rangle &= \langle R(t)\,e_1, S(\omega(t))\,R(t)\,e_1 \rangle\\ &= (R(t)\,e_1)^\top S(\omega(t)) (R(t)\,e_1)\\ &= 0, \end{aligned}$$ where the last step follows from the fact that $S(\omega(t))$ is skew-symmetric.

The only constraint on the acceleration profile is that, $$ \langle \ddot{x}(t), {x}(t) \rangle = -\langle \dot{x}(t), \dot{x}(t) \rangle $$ or, equivalently, $$ \frac{\langle \ddot{x}(t), {x}(t) \rangle}{\langle \dot{x}(t), \dot{x}(t) \rangle} = -1. $$ That is, the "projection" of the acceleration on the position is the speed (of traversal) squared.

So, for example, one could pick $u(t) = (100, 100, 100)^\top$ with the initial condition $\omega(0) = 0$ and $x(0) \in \mathsf{S}^2$ and you will find that the acceleration does not satisfy your necessary condition anywhere yet $\langle x(t), \dot{x}(t) \rangle = 0$ and $x(t) \in \mathsf{S}^2$ for all $t.$

Of course, this differential equation can also be used to describe curves with the initial conditions you specified with an appropriate choice for $\omega(0).$