Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / B / 6

For any $v\in H^1(\Omega )$, expand $J(u+ϵv)$ in $ϵ$. Its linear term must vanish at a minimizer, givingThus the required functionals areThe form is continuous, symmetric, and coercive, with $a(v,v)=\Vert v{\Vert }_{H^1}^2$. Conversely, this weak equation implies $J(u+w)=J(u)+\frac{1}{2}a(w,w)$, so its unique solution is indeed the global minimizer.

For smooth $u$, Green's first identity rewrites the weak equation asCompactly supported test functions yield $-\Delta u+u=f$. Since the test space is the full $H^1(\Omega )$ and has unrestricted boundary trace, the remaining identity gives the homogeneous Neumann condition.

For a merely weak solution on a polygon, the normal derivative is understood as its weak flux trace. Here $\nabla u\in L^2$ and $\mathrm{div}\nabla u=u-f\in L^2$, so this trace is defined through Green's formula. It vanishes by the same identity. Thus the natural boundary condition is ${\partial }_{n}u=0$, without assuming global $H^2$ regularity on an arbitrary polygon.

Take $V_h$ to be all continuous functions affine on each triangle of the stated negative-diagonal mesh. No boundary value is fixed, because the Neumann condition is natural. The discrete problem isIts matrix is symmetric positive definite: the reaction contribution $\int u_h{v}_h$ removes the constant null mode that would occur for pure Neumann Poisson equations. Thus the solution exists uniquely.

For every $v_h\in V_h$, expand the quadratic functional about $u_h$ and use the discrete weak equation to remove its linear term:The continuous solution minimizes $J$ over the larger space $H^1$, so