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I have a flow on $n$-dimensional sphere which has a stabilizing action. The tangential velocity will not be a constant, it will indeed decrease to zero as the desired point is reached. First the equations: $x=(x_1,x_2,\ldots,x_{n+1})\in S^n$, the point of stabilization is $x_2 \in S^n$ and the initial condition is $x_1 \in S^n$. I will restrict the initial conditions to the set $x_1^\top x_2 >0$, hence $\parallel \tanh(t)x_2 + (1-\tanh(t))x_1 \parallel_2>0,\forall\ t>0$. The flow is

$$ x(t) = \frac{\tanh(t)x_2 + (1-\tanh(t))x_1}{\parallel \tanh(t)x_2 + (1-\tanh(t))x_1 \parallel_2} \\ x(0)=x_1,\lim_{t\rightarrow \infty} = x_2$$

The velocity can be shown to decrease to zero, since the system is supposed to asymptotically stabilize.

$$ \dot{x}(t) = \frac{\mathrm{sech}^2(t)\left( x_2 - x_1 -2 (x_1^\top x_2) x_1 +2 \tanh(t)(x_1^\top x_2) \left( \tanh(t)(x_2-x_1) +2 x_1 \right) \right)}{2\parallel \tanh(t)x_2 + (1-\tanh(t))x_1 \parallel_2^3} ; \\ \dot{x}(0)= \frac{ x_2 - x_1 -2 (x_1^\top x_2) x_1}{2 },\lim_{t\rightarrow \infty} \dot{x}(t)= 0 $$

The flow is restricted to a 2-dimensional hyperplane in $\mathbb{R}^{n+1}$, because it can be written as $x(t) = \alpha_1(t) x_1 + \alpha_2(t) x_2$. My questions are

  1. By definition of geodesic, the path is forced to have a constant tangential velocity, which for $S^n$ will keep going on a great circle. This flow also lies on the great circle which is a geodesic from $x_1$ to $x_2$. Can this flow be called a geodesic as a curve rather than using the calculus definition?

  2. Or can the flow be regarded to evolve on the geodesic?

  3. I want references for proof that geodesics on $S^n,n>2$ are great circles ($n=2$ is dealt with in many texts). One place I found a similar result as an exercise, Exponential map on the the n-sphere. But I could not find a reference to the exercise.

Vijay
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  • Welcome to MSE! In your differential equation, you have terms such as $|x_{1}|{2}^{2}$, while earlier you'd written $x{1} \in S^{n}$. Is there some reason not to write $|x_{1}|{2} = 1$? (Perhaps I've misunderstood your notation; elsewhere it seems you've used $x{1}$ to denote both a coordinate of $x$ and a point on the sphere. :) – Andrew D. Hwang Jun 24 '14 at 13:55
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    Yeah, i missed it. I have simplified it now. Thanks – Vijay Jun 25 '14 at 09:08

1 Answers1

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Questions 1. and 2. As you note, a geodesic is usually defined in such a way that the speed is constant, e.g., as a path whose velocity field is auto-parallel; or as a critical point of an energy functional (the norm-squared of the velocity) on the space of piecewise-smooth paths from $x_{1}$ to $x_{2}$.

A reparametrization of a geodesic is sometimes called a "pregeodesic". If you're working exclusively on the sphere, I'd suggest using "(an arc of) a great circle" to describe such a path.

Question 3. The fixed-point set $S$ of an isometry $f$ of a Riemannian manifold $(M, g)$ is totally geodesic, i.e., if $p \in S$ and $v \in T_{p}S$, then the geodesic $\gamma$ in $M$ starting at $p$ with initial velocity $v$ stays in $S$ for all time.

Proof sketch: The curve $f\circ \gamma$ is the image of a geodesic under an isometry, hence is also geodesic. But $f\circ \gamma$ starts at $p$ (because $p$ is a fixed point of $f$) and has initial velocity $v$ (since $v$ is tangent to $S$, which consists entirely of fixed points), so $f\circ\gamma = \gamma$ by uniqueness of geodesics.

In the unit sphere $(S^{n}, g)$, the great circle in the $(u_{1}, u_{2})$-plane is invariant under $(n - 1)$ reflections in each of the remaining coordinates, and is precisely their common fixed point set. The argument of the preceding paragraph shows this great circle is totally geodesic. A similar argument holds for arbitrary geodesics in the round sphere by picking a Cartesian coordinate system in which the chosen great circle lies in the plane determined by the first two coordinates.