Static condensation is the exact interface problem

Let $f_i=(f_I^{(i)},f_{\mathrm{\Gamma }}^{(i)}{)}^T$ be local load contributions whose assembly is the chosen global Galerkin load vector, as specified above. The global interior rows are local to each subdomain and give $u_I^{(i)}=(K_{II}^{(i)}{)}^{-1}(f_I^{(i)}-K_{I\mathrm{\Gamma }}^{(i)}{u}_{\mathrm{\Gamma }}^{(i)})\mathrm{.}$ The local interface residual contribution is $K_{\mathrm{\Gamma }I}^{(i)}{u}_I^{(i)}+K_{\mathrm{\Gamma }\mathrm{\Gamma }}^{(i)}{u}_{\mathrm{\Gamma }}^{(i)}-f_{\mathrm{\Gamma }}^{(i)}$. Substituting the interior formula makes it $S_i{u}_{\mathrm{\Gamma }}^{(i)}-(f_{\mathrm{\Gamma }}^{(i)}-K_{\mathrm{\Gamma }I}^{(i)}(K_{II}^{(i)}{)}^{-1}{f}_I^{(i)})$, precisely the local Schur contribution in [F3]. The global interface rows are the assembled sum of these contributions, not separate local interface equations.

Sum the reduced equations from step 1.1 over subdomains and identify copies of each shared nodal value as in [F1]. Because both the stiffness and the local load vectors assemble by summing their shared interface components, this gives the assembled interface Schur system with the correspondingly reduced global right-hand side. Any solution of the original global system yields this trace. Conversely, a solution of the assembled interface system and the uniquely defined interior values from step 1.1 satisfy every local interior equation and the assembled sum of interface equations, hence the assembled global system.

The global stiffness is positive definite by [F4], so the original solution is unique. Once the interface trace is solved, each local interior vector is exactly the reconstruction formula in step 1.1, which is unique because $K_{II}^{(i)}$ is invertible. Its trace-dependent homogeneous part is $H_i{u}_{\mathrm{\Gamma }}^{(i)}$ from [F2]. Therefore static condensation is exact in both directions.