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Centered-difference residual bound with exact Dirichlet closure

Statement

Let u:[0,1]→R have four continuous derivatives, with u(0)=u(1)=0, let g=−u′′, and suppose ∣u(4)(x)∣≤M on [0,1] for a constant M≥0. For the grid and matrix of Centered finite differences for the zero-Dirichlet boundary-value problem, put uj=u(xj), gj=g(xj), and τj=(Ahugrid)j−gj for 1≤j≤N. Then

max⁡1≤j≤N∣τj∣≤Mh212.

The endpoint closure residuals are exactly zero: u(0)−0=0 and u(1)−0=0. If τ0=τN+1:=0 records those boundary residuals, then max⁡0≤j≤N+1∣τj∣≤Mh2/12 as well.

Facts & Assumptions

Given: The regularity, boundary values, differential equation and fourth-derivative bound in the statement; N≥1 and h=1/(N+1).

[F1]

Centered finite differences for the zero-Dirichlet boundary-value problem fixes xj=jh, sets U0=UN+1=0, and uses (AhU)j=(−Uj−1+2Uj−Uj+1)/h2 for 1≤j≤N.

[F2]

A uniform derivative bound gives a uniform Taylor remainder bound: on a closed interval, if f has four derivatives and ∣f(4)∣≤M, its degree-three Taylor remainder satisfies ∣R3,af(x)∣≤M∣x−a∣4/24.

[F3]

The triangle inequality: for real x,y, ∣x+y∣≤∣x∣+∣y∣; here apply it to the two Taylor remainders.

Proof

1.1givenF1F2choose

The endpoint values of the exact function agree with the algorithm's prescribed endpoint values, so both boundary closure residuals are 0. Fix an interior index j. Applying [F2] with center xj and targets xj−h and xj+h gives u(xj±h)=uj±hu′(xj)+h22u′′(xj)±h36u′′′(xj)+R±, with ∣R±∣≤Mh4/24. Both closed intervals lie in [0,1], including when j=1 or j=N.

2.1givenF1step 1.1algebra

Substituting these expansions into [F1] cancels the odd derivative terms and gives (Ahugrid)j=−u′′(xj)−(R−+R+)/h2. Since g=−u′′, it follows that τj=−(R−+R+)/h2.

3.1F2F3step 1.1step 2.1algebra∎

By [F3] and step 1.1, ∣τj∣≤(∣R−∣+∣R+∣)/h2≤(Mh4/24+Mh4/24)/h2=Mh2/12. This holds for every interior index, and the endpoint residuals from step 1.1 are zero, proving both bounds.

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