Functional derivative of integral with boundary contribution

Question

What is the functional derivative of a functional $F$ that is expressed as a volume integral over a region $\Omega\subset\mathbb R^3$ plus a surface integral over the boundary $\partial\Omega$? An example for such a functional is $$ F[c] = \int_\Omega f(c, \nabla c) \, \mathrm{d}V + \oint_{\partial\Omega} g(c) \, \mathrm{d} S \;. $$ I think that inside the domain the functional derivative reads $$ \frac{\delta F}{\delta c} = \frac{\partial f}{\partial c} - \nabla \frac{\partial f}{\partial (\nabla c)} \;, $$ but I do not know how to deal with the boundary. I'm not even sure whether the problem is well-posed (even assuming reasonably nice properties of $\Omega$, $f$, and $g$).

My more general question therefore is how one deals with functionals of the aforementioned structure.

Answer 1

Generally speaking, the boundary terms in a functional only contributes to the boundary conditions, and would not lead changes to the variational derivative.

For your example, repeat the derivation of the classical Euler-Lagrange equation, and \begin{align} \delta F&=\int_{\Omega}\delta f(c,\nabla c)\,{\rm d}V+\int_{\partial\Omega}\delta g(c)\,{\rm d}S\\ &=\int_{\Omega}\left(\frac{\partial f}{\partial c}\delta c+\frac{\partial f}{\partial\left(\nabla c\right)}\cdot\nabla\delta c\right){\rm d}V+\int_{\partial\Omega}g'(c)\delta c\,{\rm d}S\\ &=\int_{\Omega}\left(\frac{\partial f}{\partial c}\delta c+\nabla\cdot\left(\frac{\partial f}{\partial\left(\nabla c\right)}\,\delta c\right)-\nabla\cdot\left(\frac{\partial f}{\partial\left(\nabla c\right)}\right)\delta c\right){\rm d}V+\int_{\partial\Omega}g'(c)\delta c\,{\rm d}S\\ &=\int_{\Omega}\left(\frac{\partial f}{\partial c}-\nabla\cdot\left(\frac{\partial f}{\partial\left(\nabla c\right)}\right)\right)\delta c\,{\rm d}V+\int_{\partial\Omega}\left(\frac{\partial f}{\partial\left(\nabla c\right)}\cdot n+g'(c)\right)\delta c\,{\rm d}S, \end{align} where $n$ denotes the outward unit normal vector on $\partial\Omega$.

Therefore, the $c$ that minimizes $F$ must satisfy \begin{align} \frac{\partial f}{\partial c}-\nabla\cdot\left(\frac{\partial f}{\partial\left(\nabla c\right)}\right)&=0\quad\text{in }\Omega,\\ \frac{\partial f}{\partial\left(\nabla c\right)}\cdot n+g'(c)&=0\quad\text{on }\partial\Omega \end{align} due to the arbitrariness of $\delta c$.

Answer 2

The problem can be solved by calculating the first variation of the functional from scratch. Here, the first variation reads $\delta F = \lim_{\epsilon \rightarrow 0}(F[c + \delta c] - F[c])$ with $\delta c =\epsilon \phi$ and $\phi$ is a arbitrary function. We thus have \begin{align} F[c + \epsilon \phi] &= \int_\Omega f(c + \epsilon\phi, \nabla c + \epsilon \nabla \phi) \mathrm{d}^3 r + \oint_{\partial\Omega} g(c + \epsilon\phi) \mathrm{d}^2 r \end{align} which can be expanded to first order in $\epsilon$, \begin{align} F[c + \epsilon \phi] &= \int_\Omega\left[ f(c, \nabla c) + \frac{\partial f}{\partial c}\epsilon\phi + \frac{\partial f}{\partial (\nabla c)}\epsilon \nabla \phi \right] \mathrm{d}^3 r + \oint_{\partial\Omega} \left[ g(c) + g'(c)\epsilon\phi) \right] \mathrm{d}^2 r \;. \end{align} We thus obtain for the first variation \begin{align} \delta F &= \int_\Omega\left[ \frac{\partial f}{\partial c}\delta c + \frac{\partial f}{\partial (\nabla c)} \nabla \delta c \right] \mathrm{d}^3 r + \oint_{\partial\Omega} \!\!\! g'(c)\delta c \, \mathrm{d}^2 r \end{align} Using integration by parts \begin{align} \delta F &= \int_\Omega\left[ \frac{\partial f}{\partial c}\delta c - \delta c\nabla\frac{\partial f}{\partial(\nabla c)} \right] \mathrm{d}^3 r + \oint_{\partial\Omega}\left[ \frac{\partial f}{\partial (\partial_\alpha c)} n_\alpha \delta c + g'(c)\delta c \right] \mathrm{d}^2 r \;, \end{align} where $n_\alpha$ is the normal vector of the boundary. In particular, the associated Euler-Lagrange equations read \begin{align} 0 &= \frac{\partial f}{\partial c} - \nabla\frac{\partial f}{\partial (\nabla c)} \\ 0 &= \frac{\partial f}{\partial(\partial_\alpha c)} n_\alpha + g'(c) \label{eqn:stationary_point_boundary} \end{align}