Past exam of the approximation theory and finite element methods course of the University of Oxford / 2009 / B / 4

Work in the usual bounded Lipschitz, polygonally triangulated setting in which Sobolev traces and conforming meshes are defined. Choose an $H^1$ lift $G$ of the smooth boundary data. Multiplying the Poisson equation by $v\in H_0^1(\Omega )$ and integrating by parts gives the weak formulationThe test function has zero trace, so the boundary flux term vanishes. This formulation only needs first weak derivatives.

Let $V_h$ be the conforming finite element space of continuous functions affine on each triangle, and $V_h^0$ its zero-boundary subspace. Choose a nodal interpolation $G_h$ of the boundary lift and seek $u_h\in G_h+V_h^0$ satisfying the same identity for every $v_h\in V_h^0$. General smooth boundary data need not be exactly representable by boundary piecewise linear functions; this is the usual discrete boundary approximation.

With nodal shape functions ${\phi }_j$, interior coefficients $U_j$ satisfyHere $I$ and $B$ index interior and boundary vertices. Finite element assembly sums the triangle contributions.

Take a translated lower triangle with local vertices $(0,0),(h,0),(h,h)$. Its local functions are $N_1=1-x/h$, $N_2=(x-y)/h$, $N_3=y/h$. Their gradients are constant and the area is $h^2/2$, giving the stiffness matrixTranslations preserve this matrix, while rotations or a different local vertex order permute its entries. The coupling across the hypotenuse is zero.

Each interior grid vertex belongs to six triangles. Summation gives diagonal value 4 and coupling $-1$ to each of its four horizontal and vertical neighbors. Both positive-diagonal edge couplings vanish. Thus, before prescribed boundary values are removed, an interior equation has the five-point stencilThere is no $h^{-2}$ factor in the assembled variational matrix; that factor belongs to a differently normalized finite-difference equation.

If $h=1/N$ and $m=N-1$, lexicographic ordering makes the Dirichlet-reduced matrix block tridiagonal with diagonal blocks $\mathrm{tridiag}(-1,4,-1)$ and neighboring blocks $-I$. A row with all four neighboring nodes unknown has five nonzeros: 4 once and $-1$ four times. Rows adjacent to the boundary have fewer unknown-neighbor entries; their exact count is $1+1_{i>1}+1_{i<m}+1_{j>1}+1_{j<m}$. Removed boundary contributions appear on the right, rather than as new nonzeros.