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I'm trying to understand the logic underneath the concept of action and lagrangian. I know this kind of questions have been asked many times, but I was unable to find an answer to this one.

I've ever read that the action should be minimized, or at least put in an extremum, in the actual physical trajectory of the system in the configuration space. I've also read that this has to do with a sort of "phase interference" between trajectory having the same non-extremal value of the action.

But what make us sure that an extremal value will exists? And supposing it exists why should be it unique?

I can for example imagine a Lagrangian dependent on two generalized coordinates and not explicit dependent on the parameter $s$, so that $L\equiv L\left(q_1(s),q_2(s)\right)$, but I can choose an initial and a final state $\left(q_1(s_i),q_2(s_i)\right),\left(q_1(s_f),q_2(s_f)\right)$, so that apparently there is not a unique path that minimizes the value of the integral \begin{equation*} \int_{s_i}^{s_f} \text{d}s L\left(q_1(s),q_2(s)\right) \end{equation*} such as in the following picture,in which "everything is symmetric" between the two points $\left(q_1(s_i),q_2(s_i),L\left(q_1(s_i),q_2(s_i)\right)\right),\left(q_1(s_f),q_2(s_f),L\left(q_1(s_f),q_2(s_f)\right)\right)$, so that I can imagine two paths (not represented) with the same minimum value of the action.

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    $\begingroup$ 1. The uniquesness question is a duplicate of physics.stackexchange.com/q/115208/2451 and links therein. 2. The existence question (which should probably be asked separately) seems to belong on Math.SE or MO.SE. $\endgroup$ – Qmechanic Jul 27 '20 at 10:56
  • $\begingroup$ Thank you for the answer! I didn't find this question in my searches. But even reading it, sorry, I continue to don't understand what physically will happen. $\endgroup$ – Rob Tan Jul 27 '20 at 11:13
  • $\begingroup$ I mean, if there is just one path, ok, I know the system will take that one; but with more than one?? What will happen? Hopf bifurcation? Superposition of paths? I really don't understand what will happen. $\endgroup$ – Rob Tan Jul 27 '20 at 11:19
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    $\begingroup$ If there is more than one extremum, the theory is not predictive (see Norton's dome). Of course, in the case where there is gauge symmetry this will always be the case. Take the example of electromagnetism without matter. Then the EOMs are $\partial_\mu F^{\mu\nu}=0$. In terms of the potential this is $\square A_\nu+\partial_\nu(\partial\cdot A)=0$. This equations do not allow to solve for $A$ given some initial condition. One would need to further fix a gauge. $\endgroup$ – Iván Mauricio Burbano Jul 27 '20 at 13:14
  • $\begingroup$ Non predictive means chaotic? $\endgroup$ – Rob Tan Jul 27 '20 at 13:21

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