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For a Lagrangian \(L(q, \dot{q}, \ddot{q}, t)\), the equation motion is of the form :

  • A.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) - \(\frac{\partial L}{\partial q}\) = 0
  • B.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) - \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\) + \(\frac{\partial L}{\partial q}\) = 0
  • C.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) + \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\) + \(\frac{\partial L}{\partial q}\) = 0
  • D.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) - \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\) - \(\frac{\partial L}{\partial q}\) = 0

Answer: B

The equation of motion is the \(\ddot{q}\) term, minus the \(\dot{q}\) term, plus \(\partial L/\partial q\), equal to zero: the Euler-Lagrange (Ostrogradsky) equation for a Lagrangian that depends on \(\ddot{q}\).

Varying the action with \(L(q, \dot{q}, \ddot{q}, t)\) and integrating by parts once for the \(\dot{q}\) term and twice for the \(\ddot{q}\) term gives the Ostrogradsky equation:

\[ \begin{aligned} &\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \\ &\quad + \frac{d^{2}}{dt^{2}}\frac{\partial L}{\partial \ddot{q}} = 0 \end{aligned} \]

The signs alternate because each integration by parts brings a minus sign. Rearranged with the second-derivative term first, this is option B. Option A drops the \(\dot{q}\) term, and options C and D get the alternating signs wrong.

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