In class our teacher told us that, if a Lagrangian contain $\ddot{q_i}$ (i.e., $L(q_i, \dot{q_i}, \ddot{q_i}, t)$) the energy will be unbounded from below and it can take any lower values (in other words be unstable). In this type of systems can we show that the energy is conserved ? Or in such system does energy conservation is applicable ?
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$\begingroup$ Does this answer your question? Lagrangian and conservation of energy $\endgroup$– user3517167Commented Mar 10, 2021 at 16:18
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$\begingroup$ @user3517167 My lagrangian contains $\ddot{q_i}$ $\endgroup$– seVenVo1dCommented Mar 10, 2021 at 16:20
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$\begingroup$ There's this paper arxiv.org/abs/astro-ph/0601672 in which at section 2 there's a detailed discussion on the Ostrogradsky instability. Hope it helps. $\endgroup$– NooneCommented Mar 10, 2021 at 16:31
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/610562/2451 , physics.stackexchange.com/q/489969/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Mar 10, 2021 at 16:58
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Not enough reputation to comment, sorry. It should still be true that if there is no explicit $t$-dependence and the potential is a function of $q$, then the Lagrangian conserves energy.
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2$\begingroup$ What does it mean for a Lagrangian to conserve energy? Do you mean that energy is conserved along the solutions of the EOM associated to the Lagrangian? $\endgroup$– NDewolfCommented Mar 10, 2021 at 18:51
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$\begingroup$ Yeah, that's definitely a better way to word it. $\endgroup$ Commented Mar 10, 2021 at 18:56