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Two floating Poisson subdomains and primal anchoring

Example

Take two adjacent unit-square subdomains, each split by the diagonal from its southwest to its northeast corner into two affine triangles, and put coefficient $ρ = 1$ . Before outer Dirichlet values are imposed, each local floating stiffness matrix in the cyclic vertex order $(a, b, c, d)$ is

$K_{i} = \frac{1}{2} (\begin{array}{cccc} 2 & - 1 & 0 & - 1 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ - 1 & 0 & - 1 & 2 \end{array}) .$

The two interface endpoints are primal corners. Impose homogeneous Dirichlet data on the full top exterior side of the two-square domain, a nonempty segment that is a union of mesh sides. The local constant modes are thereby anchored, and the shared primal corners enforce their common interface values; after exact interior elimination, the assembled Schur solve is the original finite-element problem.

Facts & Assumptions

Given: Two matching square subdomains with the two-triangle $P_{1}$ mesh just specified; the cyclic node order $(a, b, c, d)$ ; coefficient one; primal interface endpoints; and homogeneous Dirichlet values on the full top exterior side of the domain.

[F1]

Scalar finite-element substructuring model: each local Poisson energy is the sum of triangle integrals of $∣ \nabla v_{h} ∣^{2}$ .

[F2]

Floating local kernels and positive interior blocks: the floating local kernel on a connected subdomain is the constants.

[F3]

Primal constraints remove local floating kernels: all subdomain vertices primal and a nonempty global Dirichlet boundary suffice for positive definiteness of the partially assembled energy; this is a sufficient general criterion.

[F4]

Static condensation is the exact interface problem: eliminating interiors gives the assembled interface Schur system; the interior values reconstructed from the local loads and interface trace recover the global solve, with the discrete harmonic extension as the trace-dependent homogeneous part.

Verification

1.1givenF1constructalgebra

On triangle $(a, b, c)$ the affine basis gradients are $(- 1, 0), (1, - 1), (0, 1)$ and its area is $1 / 2$ ; on triangle $(a, c, d)$ they are $(0, - 1), (1, 0), (- 1, 1)$ and its area is $1 / 2$ . Adding the two local Gram matrices of these gradients gives the displayed $K_{i}$ .

2.1givenstep 1.1F2algebra

The displayed matrix has zero row sums, and its quadratic form is $\frac{1}{2} [(u_{a} - u_{b})^{2} + (u_{b} - u_{c})^{2} + (u_{c} - u_{d})^{2} + (u_{d} - u_{a})^{2}]$ . It is therefore nonnegative with kernel exactly the constant vector, agreeing with [F2].

3.1givenF3F4step 2.1algebra∎

For a partially assembled trace of zero energy, both nonnegative local energies vanish. By step 2.1 each local vector is constant; its imposed zero top-side values make that constant zero. Hence the partially assembled quadratic form has trivial kernel and is positive definite, directly, without needing the all-vertices-primal criterion in [F3]. The only shared nodes are the primal endpoints, so their equalities enforce full interface continuity. Theorem [F4] then identifies the condensed interface solution and the load-dependent interior reconstruction, whose homogeneous trace-dependent part is harmonic, with the original global finite-element solution.

Depends on

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Sources

Olof B. Widlund, Introduction to the Design and Theory of Domain Decomposition Algorithms, DD20 Tutorial, pp. 35–37 and 83–84 (standard reference, not scraped)