Possible boundary conditions in derivation of Euler-Lagrange equations

Question

Given a Lagrange density

$$\mathcal{L} = g^{ij} \phi_{,i} \phi_{,j} - V(\phi)\tag{1}$$

I have read (e.g. here) that the boundary term that occurs through variation of the action

$$ \delta I = \int_V dV \, \text{EL} \, \delta \phi + \oint_{\partial V} dS \, n^i \, \partial_i \phi \, \delta \phi\tag{2}$$

vanishes for Dirichlet conditions ($\delta \phi = 0$), for homogeneous Neumann conditions ($\mathbf{n}(\phi) = 0$) AND for their combinations, i.e. Robin conditions ($ a \phi + b \mathbf{n}(\phi) = c$).

It is easy to see how Dirichlet conditions and homogeneous Neumann conditions make the boundary integral vanish. However, I do not understand how Robin conditions achieve the same thing.

Answer 1

1. As mentioned in the linked Phys.SE post there are 2 possible boundary conditions (BCs):

   • Essential/Dirichlet BC,

   • Natural BC,

   which can be chosen independently for each connected components of the boundary $\partial V$.

   The BC cannot change discontinuously within a connected boundary component since the field variables are usually assumed to be continuous/sufficiently differentiable due to physical reasons.

   However, if there is more than 1 connected boundary components, then combinations are possible: Essential/Dirichlet BC on some connected boundary component and natural BC on the rest.

   NB: The word combination should not be understood as a linear combination, cf. OP's question.

2. Example: In OP's action (1), the natural BC is a Neumann BC. If we add a boundary term proportional to $$\oint_{\partial V} \!dS ~ \phi^2$$ to OP's action (1), then the natural BC becomes a Robin BC.