Boundary conditions in variational problems

Question

I'm having trouble understanding how appropriate boundary conditions are obtained as part of deriving the Euler-Lagrange equations for a given variational problem. Using the particular example I'm working on at the moment, suppose we wish to minimise the energy functional $$ E[u] := \int_{\Omega}(u -g)^2 \,dx + \int_{\Omega} \Vert \nabla u \Vert^2 v^2 \, dx + \int_{\Omega} \left\lbrace \epsilon \Vert \nabla v \Vert ^2 + \frac{(1 - v)^2}{\epsilon} \right\rbrace \, dx \,, $$ where $g$ and $v$ are given functions and $\epsilon$ a given parameter. Taking the first variation $\delta E[u;\phi]$ and setting this equal to zero, we eventually arrive at the equation $$ \int_{\Omega} (u - g) \phi \, dx + \int_{\Omega} \nabla \cdot (v^2 \nabla u) \phi \, dx + \int_{\partial \Omega} v^2 \frac{\partial u}{\partial n} \phi \, dx = 0 $$ with $\phi$ assumed to be a member of $C^\infty_0(\Omega)$ as usual. The integration of the boundary arises from applying one of Green's identities in the derivation, which is essentially a generalisation of integration by parts to the multivariate setting. At this point, the book I'm reading simply goes on to assert that the above implies $\frac{\partial u}{\partial n} = 0$ on $\partial \Omega$. I can't understand why this should be the case. Even a heuristic argument based on getting rid of a term for simplification doesn't seem to pass muster, as $\phi$ already guarantees the vanishing of the last integral. It would seem the additional condition is imposed simply on order to make the resulting PDE problem well-posed. If this is the case, should one simply impose boundary conditions in an ad hoc fashion until one arrives at the requisite number? Or is there a systematic way in which boundary conditions arise from variational problems?