Stiffness matrix

Remarks

The function $u_h = \sum_j c_j \varphi_j$ satisfies $a(u_h, v) = \ell(v)$ for all $v \in V_h$ exactly when $K c = F$. If $a$ is symmetric, then $K$ is symmetric. If $a$ is coercive, $a(v, v) \ge \alpha \norm{v}^2$ with $\alpha > 0$, then $c^T K c = a(v, v) > 0$ for $v = \sum_j c_j \varphi_j \ne 0$, so $K$ is positive definite in the symmetric case and invertible in general. If $\varphi_i$ and $\varphi_j$ have disjoint supports and $a$ is an integral of local terms, then $K_{ij} = 0$.

Example: for $-u'' = f$ on $(0, 1)$ with $u(0) = u(1) = 0$, take $a(u, v) = \int_0^1 u' v' \, dx$, nodes $x_i = ih$ with $h = 1/(N + 1)$, and hat functions $\varphi_i$ with $\varphi_i(x_j) = \delta_{ij}$. On each of its two intervals $\varphi_i'$ equals $\pm 1/h$, so $K_{ii} = 2/h$, $K_{i, i \pm 1} = -1/h$, and all other entries are $0$. So $K = h^{-1} \operatorname{tridiag}(-1, 2, -1)$.

Non-example: the matrix $M_{ij} = \int_0^1 \varphi_j \varphi_i \, dx$ is built from the $L^2$ inner product, not from the form $a$ of the problem. It is the mass matrix, which has $M_{ii} = 2h/3$ and $M_{i, i \pm 1} = h/6$ for the hat functions above.