# Are there closed curve solutions for these ODE constraints?

1. Mar 25, 2015

### MisterX

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. Mar 26, 2015

### Svein

I do not understand your question. All velocities of type v(t) = a⋅t + v0 are trivial solutions.

3. Mar 26, 2015

### HallsofIvy

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.

4. Mar 26, 2015

### MisterX

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.

5. Mar 26, 2015

### Svein

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 ).

6. Mar 26, 2015

### the_wolfman

Yes non-trivial solutions exist.

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

Here $\vec V \cdot \frac{d^2}{dt^2}\vec V =0$.

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