= Solution
{c}
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 formulation>
$$
\int_\Omega\nabla u\cdot\nabla v=\int_\Omega fv,\qquad
u\in G+H_0^1(\Omega).
$$
The 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> $\varphi_j$, interior coefficients $U_j$ satisfy
$$
\sum_{j\in I}A_{ij}U_j=F_i-\sum_{j\in B}A_{ij}g_j,\qquad
A_{ij}=\int_\Omega\nabla\varphi_i\cdot\nabla\varphi_j,\quad
F_i=\int_\Omega f\varphi_i.
$$
Here $I$ 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 matrix>
$$
\boxed{A^{(T)}=\frac12
\begin{pmatrix}
1&-1&0\\
-1&2&-1\\
0&-1&1
\end{pmatrix}.}
$$
Translations 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 stencil
$$
\boxed{4U_{ij}-U_{i-1,j}-U_{i+1,j}-U_{i,j-1}-U_{i,j+1}=F_{ij}.}
$$
There is no $h^{-2}$ factor in the assembled variational matrix; that factor belongs to a differently normalized finite-difference equation.
\Image[/past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2009-triangle-stencil.png]
{title=Triangular mesh edges along diagonals have zero stiffness coupling}
If $h=1/N$ and $m=N-1$, lexicographic ordering makes the Dirichlet-reduced matrix block tridiagonal with diagonal blocks $\operatorname{tridiag}(-1,4,-1)$ 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.
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