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While solving the problem in this question, I found cases where the numerical optimization failed, suspecting unboundedness of the function being minimized. The function approximates the action of the system in question.

I decided that this result could be explained by an unbounded from below action. But I'm still in doubt because it may be my implementation problem.

So, the question is: do there really exist such physical systems with finite number of degrees of freedom, where the action is unbounded from below, given fixed values for $q(t_1)$ and $q(t_2)$? If yes, how can one decide whether a given system is of such type?

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Ruslan
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    Of course, there exist. What about mechanics with $S=\int dt,L$ and $L=E_{kin}-E_{pot}$ with some general potential energy unbounded from either side? – Luboš Motl Nov 02 '14 at 11:18
  • My thoughts: Maybe your numerical scheme doesn't implement boundary conditions properly? Ex: If $L=\dot{x}^2-x^2$, then $x(t)=b\cos(t)$ satisfies the Euler Lagrange equations, but $\int L=-b^2 \cos(t)\sin(t)$ can be made arbitrarily negative by increasing $b$. –  Nov 02 '14 at 11:23
  • ($\int L$ being $\int_0^tLdt$ given $x=b\cos(t)$) –  Nov 02 '14 at 11:31
  • @NeuroFuzzy certainly, I suspect something in that area, that's why my question whether it could be something other than a mistake. In your example fixing $x(0)$ and $x(t)$ limits the value of $\int_0^tLdt$. – Ruslan Nov 02 '14 at 11:34
  • @LubošMotl I mean the action with trajectories with fixed endpoints. I don't quite see how your example would lead to unboundedness of the action. – Ruslan Nov 02 '14 at 11:37
  • Dear @Ruslan, it doesn't matter that the endpoints are fixed. If the potential is sufficiently positive/negative, it will dominate the action. Take $V=K/r^2$. Trajectories with fixed endpoints that visit the region $r=+\epsilon$ will clearly be dominated by the $V$ term which may be arbitrarily positive or arbitrarily negative for various $K$, and extreme for small enough $\epsilon$, and the kinetic energy term will be negligible in comparison. – Luboš Motl Nov 03 '14 at 07:50
  • @LubošMotl yeah, thanks for explanation. I've realized this after Qmechanic's answer. – Ruslan Nov 03 '14 at 08:11

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Example: Consider an action functional

$$\tag{1} S[q]~=~\int_{t_i}^{t_f} \! dt ~L, \qquad L~=~\frac{1}{2}m\dot{q}^2-V(q),$$

with Dirichlet boundary conditions (BC)

$$\tag{2} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f,$$

where the potential $V$ has a repulsive pole

$$\tag{3} V(q_0)~=~+\infty$$

at $q=q_0$. Then it is possible to choose a virtual $C^1$-curve $\gamma:[t_i,t_f]\to \mathbb{R}$ that satisfies the BC (2) and sits for a while at the pole $q_0$, so that the action functional $$\tag{4} S[\gamma]~=~-\infty$$ is unbounded from below.

In particular, if the exists a unique stationary path $q_{\rm cl}$ [which satisfies the Euler-Lagrange equation and the BC (2)], it cannot minimize the action functional $S$.

Qmechanic
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  • How can it sit for a while at the pole? Wouldn't the total energy be infinite then? – Ruslan Nov 02 '14 at 13:11
  • Note that $\gamma$ is not supposed to be the actual path. The principle of stationary action does also involve virtual paths. – Qmechanic Nov 02 '14 at 13:14
  • Ah. And it could even just cross that pole, and that might be enough, no need to "sit for a while"? – Ruslan Nov 02 '14 at 13:16
  • Yes, as long as $\int ! dt~ V(\gamma(t))=+\infty$. – Qmechanic Nov 02 '14 at 13:18
  • Thanks, you've opened my eyes :) . It seems we could even take just unbounded analytic potential like $V(q)=q^4$, and it would appear that if we consider paths starting and ending at $q=0$, sinusoidal paths with large enough amplitude will give unbounded from below values to the action, while those satisfying equations of motion will leave action bounded from below. So singularities in the potential appear not even necessary. – Ruslan Nov 02 '14 at 13:50
  • $\uparrow$ Agree. – Qmechanic Nov 02 '14 at 13:55