Variations not vanishing at the boundary, Lagrangian. This might seem an obvious/stupid question, but I can't seem to find the way through it. Sorry in advance if anything I do is not very rigorous, I come from physics !
Say we have an action $S  = \int_a^b L(t,v,\dot{v})dt$. We are familiar with the process of minimazing this action using the calculus of variations, namely varying $v' = v+\epsilon\delta v$ for arbitrarily small $\epsilon$ and $\delta v$ a sufficiently smooth function.
If we compute now $\delta S$ we get, by integration by parts $\delta S = \int_a^b (\partial_vL- \frac{d}{dt}\partial_{\dot{v}}L)\epsilon\delta v dt +\epsilon\partial_{\dot{v}}L\delta v(b)-\epsilon\partial_{\dot{v}}L\delta v(a)$
Now, usually, we set $\delta v(a) = \delta v(b) = 0$, and we get the Euler-Lagrange equation by requiring 0 variation for S. My question is, do we need to enforce vanishing of the variation of the boundary ? Is it something that is embedded somehow in the process of the calculus of variation, or is it just a reasonable choice ?
Could we, for example, set the derivative of the variation on the boundary instead ? If so, is it possible to still derive some kind of differential equation like the Euler-Lagrange one ? 
My guess would be that $\delta v$ on the boundary vanishes is a requirement for this kind of development, but I am unable to justify it to myself.
 A: Look, as you have chosen $v$ as the path with stationary action and $v'$ as any neighbouring path, at the end points $A$ and $B$, both paths must coincide; else they cannot be deemed as possible paths of a particle going from $A$ to $B$. Just because they do not start and end at those points.
Hope this helps you.
A: If the Lagrangian $L(q,\dot{q},t)$ depends on 
$$\dot{q}~\equiv~ \frac{dq}{dt},\tag{A}$$ 
it is necessary to impose boundary conditions (BCs) to have a well-defined functional/variational derivative of the action functional$^1$
$$I[q]~:=~\int_{t_i}^{t_f} \! dt ~L(q,\dot{q},t).\tag{B}$$ 
At each of the two boundary points $t_i$ and $t_f$, for each dynamical variable, there are 2 possibilities:


*

*Essential/Dirichlet BC.

*Natural BC.


This is also e.g. explained in my Phys.SE answer here.
The nature of the variational problem may naturally select one of the two BCs for other reasons.
--
$^1$ In fact the variational problem (B) itself without BCs is often ill-posed. Example: The action for the harmonic oscillator without BCs is unbounded from below (and hence doesn't have a minimum).
A: The boundary conditions may be necessary to an extreme value. However, in the principle of least action,  motivated by optics (and other subjects), the boundary condition at $x_b$ is not restricted by the calculus of variation.
