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Symmetry and positivity of the rectangular Dirichlet heat operator

Statement

On the two-dimensional rectangular zero-Dirichlet grid of Fully discrete θ heat finite-difference scheme, set $A_{h} : = A_{h}^{0}$ , the centered positive heat matrix on interior values defined there, with $κ_{x}, κ_{y} > 0$ . In the weighted grid inner product of Rectangular grid and Dirichlet flux stencil, $A_{h}$ is self-adjoint and positive semidefinite; explicitly,

⟨ A_{h} v, w ⟩_{h} = h_{x} h_{y} (κ_{x} \sum_{j = 1}^{N_{y} - 1} \sum_{i = 0}^{N_{x} - 1} D_{x} v_{i + 1 / 2, j} D_{x} w_{i + 1 / 2, j} + κ_{y} \sum_{i = 1}^{N_{x} - 1} \sum_{j = 0}^{N_{y} - 1} D_{y} v_{i, j + 1 / 2} D_{y} w_{i, j + 1 / 2}),

so $⟨ A_{h} v, v ⟩_{h} \geq 0$ for every interior grid vector $v$ .

Facts & Assumptions

Given: A finite two-dimensional rectangular grid with positive spacings and directional diffusivities, zero Dirichlet boundary values, and the centered operator and weighted grid inner product defined in Fully discrete θ heat finite-difference scheme and Rectangular grid and Dirichlet flux stencil.

[F1]

Rectangular grid and Dirichlet flux stencil defines the weighted interior inner product, the edge differences, and the horizontal and vertical sums in the Dirichlet energy.

[F2]

Discrete summation by parts, Poincaré inequality and inverse bound: for zero-boundary test functions, the rectangular Dirichlet operator satisfies $⟨ A_{h}^{a} v, w ⟩_{h} = E_{h}^{a} (v, w)$ ; the proof obtains this identity by separate horizontal and vertical telescoping sums.

[F3]

Fully discrete θ heat finite-difference scheme defines the zero-boundary matrix $A_{h}^{0}$ by its centered anisotropic stencil. In the proof set $A_{h} : = A_{h}^{0}$ and write $A_{h, x}$ and $A_{h, y}$ for its horizontal and vertical stencil parts, respectively, so $A_{h} = A_{h, x} + A_{h, y}$ .

Proof

1.1F1F2F3

Let $E_{x} (v, w)$ and $E_{y} (v, w)$ denote the horizontal and vertical summands of $E_{h}^{1} (v, w)$ from [F1]. The horizontal telescoping calculation and its vertical counterpart in [F2] give the corresponding unweighted directional identities. Multiplying them by $κ_{x}$ and $κ_{y}$ yields $⟨ A_{h, x} v, w ⟩_{h} = κ_{x} E_{x} (v, w)$ and $⟨ A_{h, y} v, w ⟩_{h} = κ_{y} E_{y} (v, w)$ , since all boundary values vanish.

2.1F1F3step 1.1algebra

By [F3] and step 1.1, $⟨ A_{h} v, w ⟩_{h} = κ_{x} E_{x} (v, w) + κ_{y} E_{y} (v, w)$ . Each edge sum is symmetric in $v, w$ , so $⟨ A_{h} v, w ⟩_{h} = ⟨ v, A_{h} w ⟩_{h}$ .

3.1F1step 2.1givenalgebra∎

Taking $w = v$ in step 2.1 gives $⟨ A_{h} v, v ⟩_{h} = κ_{x} E_{x} (v, v) + κ_{y} E_{y} (v, v)$ . Every summand is a positive mesh weight times a square, and $κ_{x}, κ_{y} > 0$ , hence this quantity is nonnegative.

Remarks

No choice axiom is used; the identity is a finite summation-by-parts calculation.

Depends on

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Sources

Endre Süli, B6.1 Numerical Solution of Partial Differential Equations: A brief introduction to the theory of finite difference approximation of PDEs (standard reference, not scraped)
Benjamin Seibold, MIT 18.336, Lecture 14: Von Neumann Stability Analysis (Spring 2009) (standard reference, not scraped)
Martin Krosche, Hermann G. Matthies, and Oliver Kayser-Herold, Lecture Notes: Introduction to PDEs and Numerical Methods, Technical University Braunschweig, Revision 1.1 (2008), hosted by University of Jyväskylä (standard reference, not scraped)