How did the boundary term vanish in deriving equation of motion from Lagrangian? [closed]

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?

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

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).
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.