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The five-point Laplacian is exact on a quadratic solution

Example

On any rectangle $Ω = (0, L_{x}) \times (0, L_{y})$ , let $u (x, y) = x^{2} + y^{2}$ , $a \equiv 1$ , $f = - 4$ , and impose the exact sampled Dirichlet boundary values. On every rectangular uniform grid with $N_{x}, N_{y} \geq 2$ , the five-point Dirichlet scheme returns $U_{i j} = u (x_{i}, y_{j})$ at all nodes.

Facts & Assumptions

Given: The five-point stencil and grid inner product of Rectangular grid and Dirichlet flux stencil, the positive face bound $a_{f} = 1$ , and the exact sampled boundary values of $u (x, y) = x^{2} + y^{2}$ .

[F1]

Rectangular grid and Dirichlet flux stencil: when $a \equiv 1$ , $A_{h}^{a}$ is the positive five-point operator $- D_{x}^{+} D_{x}^{-} - D_{y}^{+} D_{y}^{-}$ .

[F2]

Discrete summation by parts, Poincaré inequality and inverse bound: for positive face values, the homogeneous Dirichlet matrix has zero kernel and is invertible.

Verification

1.1givenF1algebra

For every interior node, $D_{x}^{+} D_{x}^{-} x_{i}^{2} = 2$ and $D_{y}^{+} D_{y}^{-} (y_{j}^{2}) = 2$ ; the other coordinate second differences vanish. Thus $(A_{h}^{1} u^{*})_{i j} = - (2 + 2) = - 4 = f (x_{i}, y_{j})$ .

2.1givenstep 1.1algebra

The sampled values $u_{i j}^{*}$ also satisfy the exact Dirichlet boundary rows, so they solve the full discrete system.

3.1F2step 2.1algebra∎

Any two discrete solutions have a difference in $X_{h, 0}$ in the kernel of $A_{h}^{1}$ . By [F2] that kernel is zero, so the solution is unique and equals $u^{*}$ .

Depends on

Used by

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Sources

Endre Süli, Numerical Solution of PDEs, complete lecture notes (standard reference, not scraped)
Benjamin Seibold, MIT 18.336 Lecture 5: Finite Difference Approximation (standard reference, not scraped)