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So, I was reading the chapter of small oscillations in Landau and Lifshitz's book of Mechanics. We assume solutions of the equations of motion that are in the form of $X_a=Ae^{iω_at}$ where $A$ is an complex constant. We need real and positive solutions (normal frequency) to the characteristic equation in order to end up with oscillating solutions which are equivalent to the stability around the equilibrium point. In the case the solutions are complex we end up with a term that exponentially decreases and one that exponentially increases. Intuitively, I understand the this leads to instability for random coefficients for each of the terms but what I don't understand is that this violates the conservation of energy. More specifically, why the oscillating terms leads at a Lagrangian that does not explicitly depends on time and the exponential ones leads to one that does , thus the conservation of the Jacobi integral cannot be applied ?

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  • $\begingroup$ Is your question about why exponential solutions are ignored, or why the Lagrangian has an explicit time dependence in the case of oscillatory solutions? Regarding the latter, I'm not really sure if the Lagrangian does have an explicit time dependence. What do you find it to be? $\endgroup$
    – Philip
    Commented Sep 1, 2020 at 17:39
  • $\begingroup$ My question is why exponential solutions are ignored and if the Lagrangian has an explicit time dependence $\endgroup$
    – Konstantina-Dimitra P.
    Commented Sep 1, 2020 at 18:50

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By definition the system is conservative, so that $E = T + U = \textrm{cte}$ . This is compatible with an oscillating trajectory, as long as the increase in the kinetic energy is exactly compensated by a decrease in the potential energy, and vice versa. This is exactly what happens in a harmonic oscillator, and also in any oscillatory solution of a 1D attractive field that depends only on the particle position $V=V(x)$.

If the position - or its absolute value - started to increase in time exponentially, this would lead to a simultaneous increase in $T$ and $U$, because an exponential increase of position in time is also an exponential increase in velocity in time. $T$ and $U$ cannot increase simultaneously without increasing $E$ - so $E$ would not be conserved.

On the other hand, if position started to decrease exponentially, from $|x|>0$ in $t=t_0$ to $x \rightarrow 0$ when $t \rightarrow \infty$, then the velocity would also decrease exponentially, so that both $U$ and $T$ would decrease simultaneously.In this case, $E$ would decrease and it would not be conserved.

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  • $\begingroup$ So for a solution with real and positive roots we end up at a solution X(t)=Acos(ωt+φ) for position and u(t)=-Bsin(ωt+φ) . If we apply these to the Lagrangian it depends explicitly on time ? What I understand is that with a solution like that we do not have simultaneous decrease or increase of T and U , thus the energy can be conserved ? It is not really clear to me why it actually does conserve . $\endgroup$
    – Konstantina-Dimitra P.
    Commented Sep 1, 2020 at 19:04
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    $\begingroup$ Once you find the solution of x and v as function of time and make the substitution, you have the extremal lagrangian only as a function of time, so there is no meaning in defining partial derivative. You have only the total time derivative of the Lagrangian, and that is all you can compute. The theorem of conservation of energy states that the partial derivative of the Lagrangian must be zero and that the total derivative of the energy must be zero. The total time derivative of the Lagrangian may not be zero. $\endgroup$
    – Marco Antonio Ridenti
    Commented Sep 2, 2020 at 20:31

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