# Higher-Order Derivatives in the Lagrangian

I am trying to derive the equations of motion for a Lagrangian which depends on $(q, \dot{q}, \ddot{q}).$ I proceed by the typical route via Hamilton's Principle, $\delta S = 0$ by effecting a variation $\epsilon \eta$ on the path with $\eta$ smooth and vanishing on the endpoints. After some integrating by parts and vanishing of surface terms, I arrive at (to first order in $\epsilon$): $$\delta S = \int\left[\eta\frac{\partial L}{\partial q} - \eta\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial L}{\partial \dot{q}}\right) + \eta\frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(\frac{\partial L}{\partial \ddot{q}}\right) + \frac{\mathrm{d}^2}{\mathrm{d}t^2}\left(\frac{\partial L}{\partial \ddot{q}} \eta \right)\right]\mathrm{d} t.$$

It is clear to me that either the last term in the integral above should vanish, or else I made an error and it ought not to appear at all. If it is the former case, by what argument does this term vanish?

• It's a total derivative, so when you integrate it you get $\frac{d}{dt}\left(\eta\frac{\partial L}{\partial \ddot{q}}\right)$, which vanishes if $\eta=\dot{\eta}=0$ at the endpoints. – coconut Dec 7 '16 at 19:51
• @AccidentalFourierTransform, I understand that "standard argument" here to be that of eq. (2.8) in Landau; there he mandates that it be a total derivative of a function $f = f(q,t),$ which is not the case here. – Diffycue Dec 7 '16 at 19:53
• @Diffycue But it is! It's the total derivative of $f(q,t)=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \ddot q}\eta$. Here $\eta$ is a function of $t$, and the derivative of the Lagrangian is a function of $q$ and it's derivatives. What part don't you like? – Jahan Claes Dec 7 '16 at 19:57
• Thank you all for responding. @JahanClaes The part I do not like is that $q,$ $\dot{q}$, $\ddot{q}$ are to be treated as independent coordinates in the variational formalism; so then I do not believe it is the case that $\frac{\mathrm{d}}{\mathrm{d} t} \left[\frac{\partial L}{\partial \ddot{q}} \eta\right]$ is only a function of $q$ and $t,$ since it is not the case that we can write $\dot{q}$ or $\ddot{q}$ as functions of $q$ and $t$ before effecting the variation. Is the root of my confusion clearer? – Diffycue Dec 7 '16 at 20:08
• If you are dealing with a lagrangian that depends on $q$, $\dot{q}$ and $\ddot{q}$ then I would say that the usual vanishing of total derivatives isn't true unless you impose the extra condition that the derivative of $q$ vanishes on the endpoints. – coconut Dec 7 '16 at 20:16

You have to impose that $\eta(t_0)=\eta(t_1)=\dot{\eta}(t_0)=\dot{\eta}(t_1)=0$ where $t_0$ and $t_1$ are the endpoints of the time interval over which you are integrating. Then, the last term is: \begin{equation} \int_{t_0}^{t_1}\frac{d^2}{dt^2} \left(\frac{\partial L}{\partial\ddot{q}}\eta\right)dt = \left[\frac{d}{dt}\left(\frac{\partial L}{\partial\ddot{q}}\eta\right)\right]_{t_0}^{t_1} = \left[\eta\frac{d}{dt}\left(\frac{\partial L}{\partial\ddot{q}}\right)\right]_{t_0}^{t_1}+ \left[\dot{\eta}\frac{\partial L}{\partial\ddot{q}}\right]_{t_0}^{t_1} = 0 \end{equation} The Euler-Lagrange equation is then: \begin{equation} \frac{\partial L}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) + \frac{d^2}{dt^2}\left(\frac{\partial L}{\partial \ddot{q}}\right) = 0 \end{equation} As a justification for the conditions over $\eta$ and its derivative at the endpoints observe that, in general, $\partial L/\partial\ddot{q}$ may depend on $\ddot{q}$, so the equation of motion will be of fourth order. To obtain a solution, four conditions will be needed. In the case of $L$ depending only on $q$ and $\dot{q}$, for a second order equation we needed two conditions: fixing $q(t_0)$ and $q(t_1)$. In the fourth order case, it is reasonable to fix $q(t_0)$, $q(t_1)$, $\dot{q}(t_0)$ and $\dot{q}(t_1)$.
Therefore, as $\delta q=\epsilon\eta$ and $\delta \dot{q}=\epsilon\dot{\eta}$ we have that $\eta(t_0)=\eta(t_1)=\dot{\eta}(t_0)=\dot{\eta}(t_1)=0$.