Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / B / 6

Besides $A,B,C$, set $D=(1,2)$, $E=(2,1)$ and $F=(2,2)$. The four triangles are $ABD$, $BCE$, $BDE$ and $DEF$, each of area $1/2$. The nodal function is one at $B$, zero at every other node, and affine on each triangle. The vertex conditions giveThe formulas agree on every shared edge, proving continuity.

Let $V=\{v\in H^1(\Omega ):v=0\text{ on }{\Gamma }_D\}$, with boundary values interpreted as traces. The outward unit normal on ${\Gamma }_N$ is $(-1,-1)/\sqrt{2}$, so the stated derivative condition is precisely ${\partial }_{n}u=0$. integration by parts givesThe Dirichlet boundary has positive length, so the Poincaré inequality makes the form coercive. Let $V_h\subset V$ be the continuous piecewise affine space on the four triangles. The finite element approximation satisfies the same identity for every $v_h\in V_h$.

Nodes $A,C,D,E,F$ have zero values, leaving $V_h=\mathrm{span}\{{\varphi }_B\}$. Although isolated endpoints have zero boundary measure, continuity along the Dirichlet edges forces the nodal values at $A$ and $C$ to vanish.

Write $u_h=U_B{\varphi }_B$. On $ABD,BCE,BDE$, the gradients are $(0,-1)$, $(-1,0)$ and $(-1,-1)$, respectively. ThusThe integral of an affine nodal function over a triangle equals its area times the mean of its three vertex values. Each supported triangle therefore contributes $1/6$ to $\int {\varphi }_B$, giving total $1/2$. The reduced system is $[2][U_B]=[1/2]$, and the complete nodal answer isOnly one algebraic unknown remains despite the two-dimensional domain.