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Are there closed curve solutions for these ODE constraints?

  1. Mar 25, 2015 #1
    Are there closed curve solutions for ##\mathbf{v}(t) \in \mathbb{R}^3## satisfying this constraint?
    $$\mathbf{v}(t) \cdot \frac{d^2}{dt^2}\mathbf{v} = 0 $$
    Last edited: Mar 25, 2015
  2. jcsd
  3. Mar 26, 2015 #2


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    I do not understand your question. All velocities of type v(t) = a⋅t + v0 are trivial solutions.
  4. Mar 26, 2015 #3


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    Unless v(t) is identically 0, so there is no motion at all, you equation just says that the acceleration is 0 so, as Svein says, the only solutions are straight line trajectories.
  5. Mar 26, 2015 #4
    Thanks for your replies. To be clear I meant

    $$\sum_{i=1}^3 v_i \frac{d^2 v_i}{d t^2} = 0 $$

    In other words acceleration is orthogonal to position. So ##\mathbf{v}(t) \neq 0## does not obviously (to me) imply ##\frac{d^2}{dt^2}\mathbf{v}(t) = 0##. I may still be that the trivial solutions mentioned above are the only ones, but I'm not sure at the moment.
  6. Mar 26, 2015 #5


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    Ah. That is a different proposition. Of course, the trivial solutions are still solutions, but...

    If an object rotates in a circle around a point, the acceleration is orthogonal to velocity, but I do not know offhand any solution to acceleration orthogonal to position. A logarithmic spiral springs to mind, but there acceleration is never quite orthogonal to position (http://en.wikipedia.org/wiki/Logarithmic_spiral ).
  7. Mar 26, 2015 #6
    Yes non-trivial solutions exist.

    Consider the case where [itex]\frac{d^2}{dt^2} v_x = v_z [/itex] ,[itex] \frac{d^2}{dt^2}v_z = - v_x[/itex], and [itex] v_y=0 [/itex].

    Here [itex]\vec V \cdot \frac{d^2}{dt^2}\vec V =0 [/itex].

    You can solve for [itex] v_x [/itex] and [itex] v_z [/itex]. This amounts to solving the 4-th order ODE [itex]\frac{d^4}{dt^4} v_x + v_x =0 [/itex]. Real solutions to this equation exist. The resulting expression for [itex] v_z [/itex] is also real.
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