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A linear error on the one-dimensional Poisson coarse space is removed exactly

Example

Consider the homogeneous weak Dirichlet form for −u′′ on (0,1), in the sense of Weak Dirichlet solutions for a divergence-form operator and Uniformly elliptic divergence-form operators and their sesquilinear forms, and the uniform mesh with four fine intervals. The continuous piecewise-linear nodal basis has interior hat functions ϕ1,ϕ2,ϕ3. Its stiffness matrix is

Ah=4(2−10−12−10−12).

The coarser mesh has one interior node. Its piecewise-linear prolongation is P=(1/2,1,1/2)T. The vector e=(1,2,1)T=2P is represented exactly on the coarse space, so one exact Galerkin correction removes it.

This calculation uses the homogeneous weak-form branch, whose functional-space interface assumes Countable Choice. It establishes the finite matrix identity without asserting solution existence for a forcing or a finite-element error estimate, and it makes no claim about inhomogeneous trace data.

Facts & Assumptions

Given: The interval (0,1), its zero-boundary weak Poisson form, the four-interval uniform mesh, and its continuous piecewise-linear nodal basis.

[F1]

Weak Dirichlet solutions for a divergence-form operator: in the homogeneous branch, a weak solution is an H01 class satisfying a(u,v)=F(v) for every test function; that branch assumes Countable Choice. Only the definition of the weak form is used here, not an existence theorem.

[F2]

Uniformly elliptic divergence-form operators and their sesquilinear forms: with coefficient a11=1 and b=c=0, the form is a(u,v)=∫01u′(x)v′(x) dx in the real case.

[F3]

Galerkin coarse matrices and the exact coarse-correction error map: AH=PTAhP and C=I−PAH−1PTAh when AH is invertible.

[F4]

Continuous piecewise-affine functions with zero endpoints lie in H1-zero: each continuous piecewise-affine function on [0,1] vanishing at 0 and 1 belongs to H01(0,1).

[A1]

Assume The Axiom of Countable Choice (ACω) for the homogeneous Sobolev interface in [F1]. The finite matrix calculation uses this homogeneous branch.

Verification

1.1givenA1F1F2F4algebra

Each fine nodal hat is continuous piecewise affine and zero at both endpoints, so [F4] places it in H01(0,1), the test space in [F1] under [A1]. On each interval its derivative is constant. For i=j, the integral ∫ϕi′ϕi′ receives 1/h from each of its two adjacent intervals, giving 2/h=8; for adjacent indices, the shared interval contributes −1/h=−4; nonadjacent supports have zero derivative-product integral. Thus the matrix of a(⋅,⋅) is the displayed Ah.

2.1F3step 1.1algebra

With h=1/4, direct multiplication gives AhP=(0,4,0)T and AH=PTAhP=4. Therefore [F3] gives C=I−P(0,1,0), which maps (x1,x2,x3)T to (x1−x2/2,0,x3−x2/2)T.

3.1step 2.1algebra∎

For e=(1,2,1)T, step 2.1 yields Ce=(1−1,0,1−1)T=0. The error is exactly the prolongation of its single coarse value 2, so the exact coarse correction removes it.

Depends on

Used by

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