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I was deriving the equation of motion from Lagrangian, by using the principle of least action. Usually, at this point in derivation,

$$\int dt \frac{\partial L}{\partial \dot{q}} \frac{\partial}{\partial t}\delta q=$$

$$=\left[ \frac{\partial L}{\partial \dot{q}}\delta q \right] - \int \delta q \frac{\partial}{\partial t} \frac{\partial L}{\partial \dot{q}} dt$$

we say that at endpoints, there is no variation in the coordinates, so $\delta q_{i} =0= \delta q_{f}$. And you make the boundary term zero. I am confused, since even if at $t_f$ and $t_i$, the $\delta q$ term is zero, how is the $\frac{\partial L}{\partial \dot{q}}$ term zero as well?

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    $\begingroup$ The question is not whether it is zero as well, but rather "Is it finite?". $\endgroup$ Commented May 14 at 16:02

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The Lagrangian $L$ is a smooth (say atleast of class $C^2$) function. So the boundary terms are \begin{align} \left[\frac{\partial L}{\partial\dot{q}}\delta q\right]_{t_1}^{t_2}&= \frac{\partial L}{\partial\dot{q}}(t_2)\delta q(t_2)- \frac{\partial L}{\partial\dot{q}}(t_1)\delta q(t_1)\\ &= \frac{\partial L}{\partial\dot{q}}(t_2)\cdot 0- \frac{\partial L}{\partial\dot{q}}(t_1)\cdot 0\\ &=0-0\\ &=0. \end{align} You’re just missing the basic fact that $0$ times any real number is again $0$. So, we don’t need to impose any conditions on $\frac{\partial L}{\partial \dot{q}}$ at the endpoints (and shouldn’t in this case).

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That's not how it works: We either impose essential/Dirichlet BC or natural BC at each endpoint $t_i$ and $t_f$, but not both. And that's enough to make the boundary terms vanish. Both BC would lead to $2\times 2=4$ BC, which would be too many conditions for a 2nd-order EL equation.

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