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Discrete summation by parts, Poincaré inequality and inverse bound

Statement

On the grid of Rectangular grid and Dirichlet flux stencil, put qi+1/2,jx=ai+1/2,jDxvi+1/2,j and qi,j+1/2y=ai,j+1/2Dyvi,j+1/2. For any grid functions v,w,

⟨Ahav,w⟩h=Eha(v,w)−hy∑j=1Ny−1(qNx−1/2,jxwNx,j−q1/2,jxw0,j)−hx∑i=1Nx−1(qi,Ny−1/2ywi,Ny−qi,1/2ywi,0).

In particular, if w vanishes on the boundary, then ⟨Ahav,w⟩h=Eha(v,w). For every v∈Xh,0,

∥v∥h2≤Lx2 hxhy∑j=1Ny−1∑i=0Nx−1∣Dxvi+1/2,j∣2.

If all horizontal and vertical face values satisfy ai+1/2,j,ai,j+1/2≥a0>0, then the homogeneous Dirichlet matrix Aha is invertible and

∥(Aha)−1∥ℓh2→ℓh2≤Lx2a0,

where ℓh2 is the norm ∥⋅∥h on interior vectors. The constants are independent of Nx,Ny.

Facts & Assumptions

Given: Lx,Ly>0, integers Nx,Ny≥2, the grid, its real grid functions, the inner product, the flux operator Aha, and edge energy Eha defined in Rectangular grid and Dirichlet flux stencil.

[F1]

Rectangular grid and Dirichlet flux stencil: Aha is the negative horizontal and vertical flux divergence; Eha is the weighted sum of the horizontal and vertical edge-gradient products; Xh,0 consists of grid functions vanishing on every boundary node.

Proof

1.1givenF1algebra

Fix an interior row j and write qi=qi+1/2,jx. Telescoping the horizontal contribution gives hxhy∑i=1Nx−1−(qi−qi−1)wij/hx=hy∑i=0Nx−1qi(wi+1,j−wij)−hy(qNx−1/2,jxwNx,j−q1/2,jxw0,j). Summing this identity over j gives the horizontal energy and its two boundary terms.

1.2givenF1algebra

Let v∈Xh,0. For each interior row, vij=∑k=0i−1hxDxvk+1/2,j. Cauchy–Schwarz gives ∣vij∣2≤(ihx)∑k=0i−1hx∣Dxvk+1/2,j∣2≤Lx∑k=0Nx−1hx∣Dxvk+1/2,j∣2. Summing with weights hxhy and using ∑i=1Nx−1hx<Lx proves the displayed discrete Poincaré inequality.

2.1F1step 1.1algebra

Applying the same telescoping sum in each column to the vertical flux gives the vertical energy minus the top and bottom boundary terms. Adding it to step 1.1 proves the displayed two-dimensional identity, with all four boundary contributions present.

3.1step 2.1givenalgebra

If w vanishes at every boundary node, both boundary sums in step 2.1 are zero, so ⟨Ahav,w⟩h=Eha(v,w).

4.1F1step 3.1step 1.2algebra

Apply step 3.1 with w=v. The face lower bound gives Eha(v,v)≥a0hxhy∑j,i∣Dxvi+1/2,j∣2≥(a0/Lx2)∥v∥h2 by step 1.2.

5.1step 4.1algebra

If Ahav=0 for v∈Xh,0, step 4.1 forces ∥v∥h=0, hence every interior value is zero; the boundary values already vanish, so v=0. Thus the square finite-dimensional matrix is injective and therefore invertible.

6.1step 3.1step 4.1step 5.1algebra∎

For Ahav=fh, steps 3.1 and 4.1 imply (a0/Lx2)∥v∥h2≤⟨fh,v⟩h≤∥fh∥h∥v∥h. If v=0 the claimed estimate is immediate; otherwise divide by ∥v∥h to obtain ∥v∥h≤(Lx2/a0)∥fh∥h, which is the asserted inverse-operator bound.

Remarks

No choice axiom is used: the proof uses finite sums, Cauchy–Schwarz, and finite-dimensional injectivity implying surjectivity.

Depends on

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Sources