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I am citing from Landau and Lifschitz, this statement that will seem to you well-known, trivial, etc:

"Between these positions, (i.e. $q_1$ and $q_2$) the system moves then in such a way that the integral $S = \int_{t_1}^{t_2} L(q, \dot q, t) dt \ $ have the smallest possible value."

My question: Are the Euler-Lagrange equations for this action functional, i.e. the differential laws of Lagrangian mechanics, fully equivalent to the integral statement of the principle of least action?

That the principle of least action implies that classical solutions fulfill the Euler-Lagrange equation is easy to see. But what about the converse implication - is every solution to the E-L equations a minimum of the action?

Can somebody show whether there is equivalence between the minimization of the integral and the differential equation, i.e. that the implication is in both directions?

I saw previous related questions and answers, but I am asking here a question of equivalence.

ACuriousMind
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Sofia
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    Could you make clearer what the question you'd like to have answered is? I see several questions in here, and I think they are quite distinct. – ACuriousMind Jan 07 '15 at 00:19
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    The question (v3) is essentially a duplicate of http://physics.stackexchange.com/q/38348/2451 – Qmechanic Jan 07 '15 at 00:46
  • @CuriousOne : you go too far with the philosophy. I am judging simply, we try to understand the nature. The dependence of the future sounds as if we are planned robots, without the freedom of what we will do next second. See the free will theorem in QM, Kochen and Specker. – Sofia Jan 07 '15 at 00:52
  • @Sofia: Who are Kochen and Specker and why should I care? Or, better, why should the definition of science care about Penn and Teller 2.0? – CuriousOne Jan 07 '15 at 01:19
  • @Sofia Regarding question 2) (v3): Are you asking whether the principle of least action is really one of least action or could it also be of largest action? if yes, the answer is just that it depends on the second derivative of S and whether it changes sign at some point along the trajectory, then the minimum turns into a maximum, such points can potentially be found because Euler-Lagrange's equation corresponds only to a differential $S$ that depends linearly on $\delta r$, i.e. higher order variations (functional derivatives) are omitted ($\mathcal{O}(\delta r^2)$).) – Ellie Jan 07 '15 at 01:24
  • @CuriousOne : To your knowledge: Kochen and Specker are the physicists that formulated the free-will theorem in QM. The concept of contextuality, the free will issues, that revolutioned QM, are based on their work. But, as you say, you shouldn't care, you only should give me marks on my elementary understanding, based on the fact that you don't know. – Sofia Jan 07 '15 at 01:26
  • @Sofia: In other words: they are quantum mystics that didn't get the message from nature that there is no wizard and no curtain. :-) – CuriousOne Jan 07 '15 at 01:28
  • @CuriousOne : did you read the free-will theorem? I stop the conversation here, I don't discuss with people that speak in base of lack of knowledge, and even more, give grades in base of lack of knowledge. – Sofia Jan 07 '15 at 01:32
  • @Sofia: I went trough my quantum mystic phase three decades ago, before becoming a real physicist who actually cares about reality. If you keep working at your definitions, you will get there, too. – CuriousOne Jan 07 '15 at 01:35
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    You may find this interesting: http://www.feynmanlectures.caltech.edu/II_19.html – alarge Jan 07 '15 at 10:28
  • @alarge : yes, the Feynman lectures are interesting. But why shouldn't you post an answer? Different people jumped on my question with minuses, and essentially scolding me why do I ask. To ask is the soul of the science. – Sofia Jan 08 '15 at 16:11
  • @CuriousOne Just fyi, the free-will theorem defines free will as "a lack of pre-determined outcome" – Jim Jan 08 '15 at 17:50
  • Comment to the question (v7): 1. Are you just asking about the difference between minimum action and stationary action? (That would basically be a duplicate of e.g. this Phys.SE post.) 2. Or are you asking about an equivalence between stationary action principle and EL eqs? In the latter case please edit the question accordingly. – Qmechanic Jan 08 '15 at 17:59
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    Let's continue in linked chat room. – Qmechanic Jan 08 '15 at 18:31
  • @Qmechanic , no, I asked whether we can do the inverse proof than it's done in Landau & Lifschitz. They introduce a small change in the variables of the Lagrangian, i.e. $q + \delta q$ and $\dot q + \delta \dot q$. So, they calculate $\delta S$, and after some mathematical manipulations, they obtain the Euler-Lagrange (E.L.) equations. What I ask is if the proof can be done in inverse way, i.e. starting from the E.L. eqs., obtain the integral form, then recover the condition of minimum. It's question of the equivalence between the integral and the diff. formulation of the classical mechanics. – Sofia Jan 08 '15 at 19:20
  • A guess is that if you find a lagrangian that leads to an ODE with non-unique solution, the equivalence might not hold. Maybe in that case the solution is not an extremum of S. – jinawee Jan 08 '15 at 19:29
  • Please move your comments to the chat room, thanks. – Qmechanic Jan 08 '15 at 19:33

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