# Is the principle of least action fully equivalent to the Euler-Lagrange equations?

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.

• 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 7 '15 at 0:19
• The question (v3) is essentially a duplicate of physics.stackexchange.com/q/38348/2451 – Qmechanic Jan 7 '15 at 0:46
• @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 7 '15 at 0:52
• You may find this interesting: feynmanlectures.caltech.edu/II_19.html – alarge Jan 7 '15 at 10:28
• – Qmechanic Jan 8 '15 at 18:31
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