Watching the trajectory of a double pendulum, I caught myself wondering if it would be possible to prove that the path the second pendulum makes contains "cusps" or singular points. Upon investigating a bit on conditions of these singular points, I came upon this Wikipedia article that states that a parametric curve $(x(t),y(t))$ has a singular point $t_s$ when both derivatives vanish at that point:

$$\frac{dx(t)}{dt}\Bigg|_{t=t_s} = \frac{dy(t)}{dt}\Bigg|_{t=t_s} = 0$$

If we describe the motion of a pendulum using the canonical angles of the pendulum with the normal, we have that the curve the second pendulum describes is given by

$$ x(t) = l_1 \sin(\theta_1(t)) + l_2\sin(\theta_2(t))$$ $$ y(t) = -l_1 \cos(\theta_1(t)) - l_2\cos(\theta_2(t))$$

It is straight forward to prove that if at a certain point $t_s$,

$$\frac{d\theta_1(t)}{dt}\Bigg|_{t=t_s} = \frac{d\theta_2(t)}{dt}\Bigg|_{t=t_s} = 0$$

the derivatives of $x$ and $y$ vanish. Therefore, if we show that the derivatives of the angles both vanish at the same time, we have a singular point.

Now, for the double pendulum, the angles satisfy a second order system of nonlinear differential equations, which are very hard to treat analytically. Nonetheless, these equations are basically combinations of derivatives and trigonometric functions of the angles. This leads me to believe that some sort of mean value analysis can be performed.

For reference, let us look at the simple pendulum. We know that its position in time is given by

$$x(t) = l\sin(\theta(t))$$ $$y(t)=-l\cos(\theta(t))$$

Its singular points happen when $\dot{x}$ and $\dot{y}$ vanish, which in this case is equivalent to $\dot{\theta}$ vanishing. We can find analytically that $\theta$ passes through $0$ several times (at multiples of one quarter of the period). The mean value theorem can be used to show that, $\theta$ being continuous, $\dot{\theta}$ is vanishes for some points. We can find analytically that these points are the initial angle $\theta_0$ and its negative, and they occur at multiples of the half period.

This analysis uses the fortuitous coincidence that the condition for $\dot{x}$ and $\dot{y}$ to vanish is equivalent to the one-dimensional quantity $\dot{\theta}$ vanishing. For two pendulums, the condition is two dimensional, and this kind of argument cannot be used, since when we apply the mean value theorem for each coordinate separately, there is no guarantee that their derivatives will vanish at the same time.

However, the nature of the problem, and the fact that it involves highly "repetitive" motion, gives me the feeling that for some point, both derivatives will coincide in vanishing.

It would be very interesting if anyone had a different approach to this problem, specially using the system of DEs that relate both angles.

  • $\begingroup$ Why do you feel that there are singular points along many trajectories instead of a set of measure $0$? $\endgroup$ – Douglas Zare Mar 17 '14 at 20:54
  • $\begingroup$ Well, in the single pendulum we have that singular points occur for a countable (measure zero) amount of times. For the double pendulum, we "see" singular points happen (can't be sure if they are truly singular) as well, albeit less frequently. I believe that the set will be measure zero, countable and with a lower natural density than that of the single pendulum case. $\endgroup$ – cako Mar 17 '14 at 21:17
  • $\begingroup$ Nothing indicates to me that the set of trajectories that have at least $1$ singular point has positive measure. You are conjecturing that every trajectory hits this codimension $2$ set where $2$ derivatives vanish simultaneously, and moreover does so infinitely often? If that's false, you will have a hard time proving it. $\endgroup$ – Douglas Zare Mar 17 '14 at 21:21
  • $\begingroup$ I think you misunderstood. I'm not claiming anything on the set of trajectories; my question is simpler. Given a certain initial condition, can we establish that the trajectory will have a singular point? $\endgroup$ – cako Mar 17 '14 at 21:39
  • $\begingroup$ I think you misunderstand. I know you are trying to prove that, and I tried to explain why it seems implausible. Good luck. $\endgroup$ – Douglas Zare Mar 17 '14 at 23:38

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