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As I understand it, the Euler-Lagrange equation is a necessary but not a sufficient condition to determine if the action integral reaches an extremal through the trajectory described by $q(t)$.

If this is so, what is a sufficient condition?

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  • $\begingroup$ Where have you heard this? I think you are confusing extremum with minimum. The EL equations are necessary and sufficient for extrema but only necessary for minima. $\endgroup$ – Ryan Unger Jan 22 '15 at 20:16
  • $\begingroup$ physics.stackexchange.com/q/160771/66165 second post $\endgroup$ – Stan Shunpike Jan 22 '15 at 20:17
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    $\begingroup$ Well, actually the OP is right. Extremum means either maximum or minimum and EL equations are just necessary for that. They are necessary and sufficient for stationary paths. $\endgroup$ – Valter Moretti Jan 22 '15 at 20:19
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    $\begingroup$ I do not think so, you said that the EL equations are sufficient (and necessary) for extrema. This is false. There are solutions of these equations which are not maxima nor minima. Exactly as it happens for saddle points of functions $\mathbb R^2 \to \mathbb R$. They are stationary points but they are not maxima and they are not minima, in other words they are not extrema. $\endgroup$ – Valter Moretti Jan 22 '15 at 20:30
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    $\begingroup$ It seems I was the one who mixed up definitions. I meant stationary, sorry. I do believe there is a form of the second derivative test that can be applied to functionals, perhaps that will answer OP's question. $\endgroup$ – Ryan Unger Jan 22 '15 at 20:38
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Let me begin with an analogy to high school calculus. We know that a function $f(x)$ has a stationary point $x_0$ if at that point $$\left.\frac{df}{dx}\right|_{x_0}=0$$ There are three possibilities: minimum, maximum and saddle. This is called the first derivative test. To test for these conditions, we use the second derivative test. We check $$\left.\frac{d^2f}{dx^2}\right|_{x_0}$$ If this is positive, $f$ is concave up and thus $x_0$ is a minimum. If it is negative, $x_0$ is concave down and a maximum. If both first and second derivatives are zero at $x_0$, it is a point of inflection, more specifically, a saddle point (a point of inflection which has zero first derivative is a saddle).

Now we come to variational calculus. Path integral quantum mechanics tells us that the action $$S[q]=\int L\,dt$$ is stationary along the classical path, a condition which we write as $$\delta S[q][h]=0$$ where the functional derivative of a functional $G[f]$ is defined as $$\delta G[f][h]=\left.\frac{d}{d\epsilon}G[f+\epsilon h]\right|_{\epsilon=0}$$ The action principle implies the Euler-Lagrange equations $$\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial\dot q}=0$$ But these equations are only necessary for extremal solutions, and saddle point solutions are possible.

So suppose we solve the EL equations and are not sure if it is a minimum, maximum or saddle point. We perform the second derivative test for functionals as follows: The second functional derivative is $$\delta^2 G[f][h]=\left.\frac{d^2}{d\epsilon^2}G[f+\epsilon h]\right|_{\epsilon=0}$$ so we look at the integral $$\delta^2 S[q_\text{c}][h]$$ where $q_\text{c}$ is the curve which solves the EL equations. If it is positive, $q_\text{c}$ is a minimum, etc.

You should take a look at this paper, "When action is not least". It discusses the differences between minima, maxima and saddle points in great detail.

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  • $\begingroup$ The problem with second derivative test is that in more than one dimension, second derivative is not unique but depends on direction chosen. In the case of the functional, the space is infinitely dimensional, which is even more different from the one-dimensional case. The second derivative will depend on the chosen function $h$. But how to try all possible functions and check whether the sign is always the same? $\endgroup$ – Ján Lalinský Jan 22 '15 at 21:26
  • $\begingroup$ Gray and Taylor discuss this in their paper. They show that only for paths small enough can the dependence on the choice of $h$ made negligible. It seems that for arbitrary paths, it is necessary to investigate all functions $h$ and this may be hard. $\endgroup$ – Ján Lalinský Jan 22 '15 at 21:26
  • $\begingroup$ It's not foolproof, granted. However, there are certain situations where it works. I linked the paper for a more thorough discussion. $\endgroup$ – Ryan Unger Jan 22 '15 at 21:28

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