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In my particular problem, the Lagrangian of the system is: $$ L = \frac{m(\dot r^2 + r^2\dot \varphi^2)}{2} + \frac{m\omega^2 (r\sin \varphi)^2}{2} $$

From there, we can derive the equations of motion: $$ \begin{cases} \frac{d}{dt} \frac{\partial L}{\partial \dot r} - \frac{\partial L}{\partial r} = 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot \varphi} - \frac{\partial L}{\partial \varphi} = 0 \end{cases} $$

or

$$ \begin{cases} m\ddot r - mr\dot \varphi^2 - m\omega^2 r\sin^2\varphi = 0\\ mr^2\ddot \varphi - m\omega^2 r^2 \sin \varphi \cos \varphi = 0 \end{cases} $$

The action functional:

$$ S = \int_{t_0}^{t_1} L dt = \int_{t_0}^{t_1} \left( \frac{m(\dot r^2 + r^2\dot \varphi^2)}{2} + \frac{m\omega^2 (r\sin \varphi)^2}{2} \right) dt $$

So, how do you prove that there is only one trajectory that the system can travel from point $q_0$ to point $q_1$ during a fixed time interval $T = t_1 - t_0 > 0$.

I don't understand how to tackle these type of problems. Do I need to solve the equations of motion and show that given the boundary conditions, there is only one solution? Or should I somehow use the action functional?

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  • $\begingroup$ I have appreciated your effort :-) +1 $\endgroup$
    – Sebastiano
    Commented Apr 23, 2021 at 20:58
  • $\begingroup$ see meta.stackoverflow.com/q/284236 $\endgroup$ Commented Apr 23, 2021 at 21:14
  • $\begingroup$ Would "How to use Halmiton's principle to prove the uniqueness of a trayectory?" be better? Or how do you suggest I reformulate my question? $\endgroup$
    – Belen
    Commented Apr 23, 2021 at 21:32
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    $\begingroup$ probably rephrase so your question applies to any Lagrangian, not just the one specific to your question, but use your current Lagrangian as an example. Avoid “any help would be appreciated”-type statements. $\endgroup$ Commented Apr 23, 2021 at 21:40
  • $\begingroup$ Every time I have heard someone mention the principle of least action applied to a problem, they always jump straight into the Euler Lagrange equations. They say "from the principle of least action follows that the Euler Lagrange equations must hold" and go on. I think you could apply this in your case, assuming you have already derived the EL equations from Hamilton's principle. $\endgroup$
    – Johnn.27
    Commented Apr 23, 2021 at 21:50

1 Answer 1

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In general you can't, because it isn't true. Consider the simple pendulum with boundary conditions $\phi(0)=0$ and $\phi(T)=0$ for some $T$. There are clearly an infinity of trajectories which obey the equations of motion and obey these fixed boundary conditions, including the trivial solution $\phi(t)=\phi'(t)=0$, because the pendulum could go around the loop any integer number of times.

If you carefully follow the derivation of the Euler-Lagrange equations, the statement is that if there exists an extremal trajectory $\phi$ with fixed boundary conditions which extremizes the action $S[\phi]$, then

$$\delta S[\phi] = 0 \iff \frac{d}{dt}\left(\frac{\partial L}{\partial \dot \phi}\right) = \frac{\partial L}{\partial \phi}$$

This does not guarantee the existence of such a trajectory, and if it does exist, it is not guaranteed to be unique. The pendulum problem has the latter feature; for an example of the former, consider the Lagrangian $L(x,\dot x) = \frac{1}{2}\dot x^2 + \frac{1}{2}x^2$ and boundary conditions $x(0)=0$, $x(2\pi)=1$.


The system you describe also features non-unique solutions. In Cartesian coordinates, that's just a particle which is free in the $x$-direction and subject to a harmonic potential in the $y$-direction. If our boundary conditions are chosen correctly, we could have unique solutions or non-unique ones. For example, $x(0)=0,y(0)=0$ and $x(2\pi/\omega)=0,y(2\pi/\omega)=0$ has the trivial solution of a particle sitting at the origin as well as the particle oscillating once in the y-direction.

That being said, for almost all choices of boundary condition, the action-extremizing solutions are unique. In the $x$-direction, fixing $x(0)$ and $x(T)$ fixes the velocity uniquely, because it's a free particle. In the $y$-direction, fixing $T$ (along with the initial conditions) tells you what part of the oscillation you end in; from there, you can show with a bit of algebra that only one choice of initial $y$-velocity will work unless $T=2\pi/\omega$ and $y(0)=y(T)=0$.

In mechanics, however, we often don't talk about trajectories with fixed boundary conditions, but rather trajectories with fixed initial conditions. The uniqueness of these trajectories (and possible conditions for non-uniqueness) follows from the theory of second order ODEs.

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  • $\begingroup$ "In mechanics, however, we often don't talk about trajectories with fixed boundary conditions, but rather trajectories with fixed initial conditions." Then what if you are given an initial position at time $t_0$ and a final position at time $t_1$? Isn't that called fixed boundary conditions? $\endgroup$
    – Belen
    Commented Apr 23, 2021 at 23:19
  • $\begingroup$ @Belen Yes, but in my experience that kind of problem arises less often. Hence my use of often, not always. $\endgroup$
    – J. Murray
    Commented Apr 23, 2021 at 23:22
  • $\begingroup$ Okay, I thought I had maybe messed up in the terminology. $\endgroup$
    – Belen
    Commented Apr 23, 2021 at 23:28

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