6
$\begingroup$

Can we find $\vec{f} : \mathbb{R}\rightarrow \mathbb{R}^3 $ such that $\vec{f}(t) \cdot \frac{d^2 \vec{f}(t)}{dt^2} =0$ and $\vec{f}(0) = \vec{f}(T)$ for some $T >0$ ? Exclude the trivial cases. I want $\frac{d \vec{f}}{dt} \neq 0$ for some part of the trajectory.

$\endgroup$
2
$\begingroup$

Define $$x(t) = C_1e^t\cos t + C_2e^t\sin t + C_3e^{-t}\cos t + C_4e^{-t}\sin t$$ and $$y(t) = C_2e^t\cos t - C_1e^t\sin t - C_4e^{-t}\cos t + C_3e^{-t}\sin t$$ for some constants $C_1,C_2,C_3,C_4 \in \mathbb{R}$.

Differentiating twice yields $$x''(t) = -2C_1e^t\sin t + 2C_2e^t\cos t + 2C_3e^{-t}\sin t - 2C_4e^{-t}\cos t = 2y(t)$$ and $$y''(t) = -2C_2e^t\sin t - 2C_1e^t\cos t - 2C_4e^{-t}\sin t - 2C_3e^{-t}\cos t = -2x(t)$$

So if we set $f(t) = \begin{bmatrix}x(t)\\y(t)\\0\end{bmatrix}$, then $f''(t) = \begin{bmatrix}2y(t)\\-2x(t)\\0\end{bmatrix}$, and thus, $f(t) \cdot f''(t) = 0$, as desired.

If we pick $C_1 = 1$, $C_2 = 1$, $C_3 = -e^{\pi}$, and $C_4 = e^{\pi}$, we have $x(0) = x(\pi)$ and $y(0) = y(\pi)$. Then, $f(0) = f(\pi)$, as needed.

Finally, it is easy to see that $f'(t) \neq 0$ for most values of $t$.

$\endgroup$
2
  • $\begingroup$ Thanks, but now I'm wondering if there is a smooth periodic motion that satisfies all the criteria as well. That is, $f^{(n)}(0) = f^{(n)}(T)$ for all positive integer $n$. $\endgroup$
    – WhatIAm
    Jul 10 '15 at 1:01
  • 1
    $\begingroup$ ^No there is not. Note that $\dfrac{d^2}{dt^2}[\|f\|^2] = \dfrac{d^2}{dt^2}[f \cdot f] = 2f \cdot f'' + 2 f' \cdot f' = 2\|f'\|^2 \ge 0$. Hence, $\|f(t)\|^2$ is concave up with respect to $t$, and thus, $f(t)$ can only attain any particular value for two values of $t$. (Unless of course $f' \equiv 0$, which is a trivial case). $\endgroup$
    – JimmyK4542
    Jul 10 '15 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.