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.

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.
