1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

--

$^1$ 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).

$\endgroup$
  • $\begingroup$ Thank you for your answer and the link you posted. I read through, but I'm not sure I understand it, so I would need confirmation : - Is it right that the variational method is ill-defined if the B.C. of the problem do not make the surface term vanish ? In other words, do we require the B.C. to make the surface term vanish, to be able to apply the variational method ? In that sense, other B.C. than natural or Dirichlet would be "non-sensical" in the variational problem. If this is correct, then that answers my question, but I might not have understood your answer well ! $\endgroup$ – Frotaur Mar 6 '17 at 8:08
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Mar 6 '17 at 20:31
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I understand what you mean. But in trying to find the stationary path, do I implicitly impose it's starting and finishing points ? I understand that in the case of a particle, it is the sensible thing to do. However, in the integral for the action there is only a statement of the starting and finishing times a and b. I am probably still missing something and I don't know if I'm very clear, but could you elaborate on your answer ? $\endgroup$ – Frotaur Mar 3 '17 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.