It bothers me that many basic books on the classical mechanics don't discuss the following difference between "Newton's laws" and the "Principle of stationary action". Newton's laws can predict the behavior of a system if one sets initial positions and velocities. On the contrary, if we want to construct an action functional we have to set an initial and a final position. From the mathematical point of view, this is the difference between initial and boundary value problem, which have different qualitative behavior (BVPs may have many solutions or no solution).

My question: Which approach is more reasonable - the initial value problem or the boundary value problem?

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    $\begingroup$ More reasonable for what? They are different types of problem and each has its uses. $\endgroup$ – Nathaniel Jan 26 '15 at 12:50
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/38348/2451 and links therein. $\endgroup$ – Qmechanic Jan 26 '15 at 12:53
  • $\begingroup$ Thanks, I don't understand why I missed the other answer! $\endgroup$ – user127911 Jan 26 '15 at 13:02

The Lagrangian boundary value problem is ill defined in general (it is well defined when the initial and final configurations are sufficiently close to each other). There is no existence and uniqueness property of the solutions. Conversely, the local and global problem with initial data, if the kinetic energy of the Lagrangian is positive-defined is always well-posed: The solution exists and is unique, if the Lagrangian is sufficiently regular.

Within this picture the variational principle has to be understood as a procedure to define just the equations of motion rather than the solutions.

  • $\begingroup$ I've always had this intuitive view: Suppose we have a path $q$ of a particle that runs from point $A$ to point $B$. The action principle tells us that the action is stationary along $q$. We therefore conclude that $q$ satisfies the equations of motion. But the converse is not generally true, right? If $q$ satisfies the equations of motion (the EL equations), then there is no need for it to hit $B$ assuming only that it starts at $A$. Is this right? $\endgroup$ – Ryan Unger Jan 26 '15 at 13:44
  • $\begingroup$ If you only know that it starts from A there are infinitely many solutions of such form and, in general, none of them reaches B. $\endgroup$ – Valter Moretti Jan 26 '15 at 19:54
  • $\begingroup$ @ValterMoretti would you mind elaborating a bit how close is sufficiently close? $\endgroup$ – Shing Mar 20 '18 at 10:23
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    $\begingroup$ For instance, if the Lagrangina is made of the only kinetic part only, which determines a Riemannian metric, sufficiently close means that the two points must belong to a geodesically convex neighborhood of that metric. $\endgroup$ – Valter Moretti Mar 20 '18 at 11:08

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