Variations not vanishing at the boundary, Lagrangian.

Question

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.

Answer 1

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:

1. Essential/Dirichlet BC.
2. 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.

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$^1$ 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).

Answer 2

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.

Answer 3

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.