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Discrete maximum principle and uniform inverse bound for the Dirichlet stencil

Statement

Let N≥1, h=1/(N+1), xj=jh, and Ah=h−2tridiag⁡(−1,2,−1) on the N interior values, with w0=wN+1=0, as in Centered finite differences for the zero-Dirichlet boundary-value problem. For v∈RN, define ∥v∥∞:=max⁡1≤j≤N∣vj∣. The Thomas recurrence in that definition has pivots pj=(j+1)/j>0 and returns the unique solution of AhU=gh for every admissible exact input; neither linear-solve-failure branch is reached. Also, Ah is invertible; if rj≥0 for every j, then (Ah−1r)j≥0 for every j; and for every r∈RN,

∥Ah−1r∥∞≤18∥r∥∞.

In particular the inverse bound is uniform over all N≥1.

Facts & Assumptions

Given: An integer N≥1, its mesh h=1/(N+1), and r∈RN. The endpoints for a vector are fixed at w0=wN+1=0.

[F1]

Centered finite differences for the zero-Dirichlet boundary-value problem defines the nodes, matrix, and exact Thomas recurrence: “Set h=1/(N+1), xj=jh for 0≤j≤N+1”; “The finite-difference equations are −Uj−1+2Uj−Uj+1h2=gj,1≤j≤N”; and the forward pivots satisfy p1=2, pj=2−1/pj−1.

Proof

1.1F1inductionalgebra

Let p1=2 and pj=2−1/pj−1 for 2≤j≤N, as in [F1]. Induction gives pj=(j+1)/j: it holds at j=1, and if pj−1=j/(j−1) then pj=2−(j−1)/j=(j+1)/j>0. Thus every Thomas pivot is nonzero.

1.2F1algebra

Define qj=xj(1−xj)/2 and q0=qN+1=0. Direct expansion of this quadratic gives qj−1−2qj+qj+1=−h2, so Ahq=1. Also 0≤qj≤1/8, since x(1−x)=1/4−(x−1/2)2 on 0≤x≤1.

2.1F1step 1.1algebra

For the equation Ahw=r, put bj=h2rj and s1=b1. The first row is p1w1−w2=s1. If elimination has produced pj−1wj−1−wj=sj−1, then wj−1=(sj−1+wj)/pj−1; substituting this into −wj−1+2wj−wj+1=bj gives pjwj−wj+1=bj+sj−1/pj−1=sj, exactly the recurrence in [F1].

3.1F1step 1.1step 2.1algebra

Since wN+1=0, backward substitution gives wN=sN/pN and then wj=(sj+wj+1)/pj for j=N−1,…,1. Each pivot is positive by step 1.1, so each value is uniquely determined; substitution into the transformed rows proves it solves the original rows. Hence for every r there is exactly one w with Ahw=r, and Ah−1 exists.

4.1givenstep 3.1choosealgebra

Suppose rj≥0 and let w=Ah−1r. If the finite list w0,…,wN+1 had a negative minimum m at an interior index j, then wj−1,wj+1≥m and h2rj=2m−wj−1−wj+1≤0. Since rj≥0 and h>0, equality must hold, forcing both neighbors to equal m. Repeating this equality to the left reaches w0=m<0, contradicting w0=0. Thus wj≥0 at every interior index.

5.1givenstep 4.1step 1.2algebra

Put R=∥r∥∞. The coordinate bounds −R1≤r≤R1 and step 4.1 show that Ah−1(R1−r)=Rq−Ah−1r≥0 and Ah−1(R1+r)=Rq+Ah−1r≥0. Hence −Rq≤Ah−1r≤Rq coordinatewise.

6.1step 1.2step 5.1algebra∎

Step 5.1 and qj≤1/8 from step 1.2 give ∣(Ah−1r)j∣≤R/8 for every j, so ∥Ah−1r∥∞≤∥r∥∞/8 for every mesh.

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