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A second-order differential equation on a smooth manifold is a smooth section of the double-tangent bundle that satisfies the canonical flip -- that is, a map $\xi: TM \to TTM$ which satisfies $\xi(q, v) = (q, v, v, w)$.

My question is, is there a second-order ODE on a (hopefully closed, but not necessarily so,) connected Riemannian manifold, say, $S^1$ or $\mathbb{R}$ -- that is to say, a vector field on $\xi: TS^1 \to TTS^1$ (i.e., on the tangent bundle to the circle, which is vector bundle isomorphic to the infinite cylinder) or $\xi: T\mathbb{R} \to TT\mathbb{R}$ (i.e., on the tangent bundle to the real line, which is vector bundle isomorphic to the the plane) -- which obeys the canonical flip and which is simultaneously a gradient vector field and a Hamiltonian vector field?

See this for some background information on second-order ODEs on manifolds. See this for a definition of gradient and Hamiltonian vector fields. I'm not sure how important the Hodge decomposition is, but here is some background on the Hodge decomposition on closed, connected Riemannian manifolds and the Helmholtz decomposition on $\mathbb{R}^3$

Thanks much in advance.

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Unfortunately I only know naive differential geometry, so let me know if I've misunderstood your question.

We consider the vector field $f(x,v) = (v,w)$ on $\mathbb R^2$.

Such a vector field is Hamiltonian if and only if it arises from a Hamiltonian of the form $H(x,v) = U(x) + \frac12v^2$, so $w = -\partial U(x)/\partial x$ is a function of $x$ alone.

As $\mathbb R^2$ is simply connected, the vector field $(v, w)$ is a gradient field if and only if its "curl" $\partial w/\partial x - \partial v/\partial v$ is zero everywhere, so $w = x + c$ for some constant $c$.

Thus the only vector fields on $\mathbb R^2$ that satisfy your conditions are $(x,v)\mapsto(v,x+c)$. Such a vector field

  1. corresponds to the second-order ODE $\ddot x = x+c$,
  2. is the gradient of the scalar field $(x,v)\mapsto (x+c)v$, and
  3. is a Hamiltonian vector field with Hamiltonian $H(x,v) = -\frac12(x+c)^2+\frac12v^2$.
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  • $\begingroup$ I think this is perfect! The function $w=x+c$ is harmonic on ℝ, so that jibes with my other post math.stackexchange.com/questions/2028283/… (to be clarified). Thanks so much for the answer! $\endgroup$ Nov 26 '16 at 5:12
  • $\begingroup$ Glad to help. P.S. You don't have to keep editing your comment. :) $\endgroup$
    – user856
    Nov 26 '16 at 5:25

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