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A three-cell periodic grid suppresses the Brusselator's unstable continuous mode

Statement

Use the Brusselator and torus from A Brusselator with a continuous Turing-unstable torus mode, but discretize the torus by the periodic centered grid with $N = 3$ . The continuous $k = 1$ mode is unstable, while every mode of this three-node linearized grid is Hurwitz.

The diagram is schematic and not to scale; it records only the three cell indices and their periodic nearest-neighbor adjacency. The edge joining $2$ to $0$ is the wraparound face, and each edge is labeled by the physical periodic center spacing $h$ . For three cells the two distinct neighbors of each cell are the other two nodes. The eigenvalue calculation below uses the stencil formula, not the drawing.

Facts & Assumptions

Given: The Brusselator dispersion polynomial and continuous unstable band from A Brusselator with a continuous Turing-unstable torus mode, and the periodic grid symbol from Two-species continuous and discrete dispersion criteria.

[F1]

A Brusselator with a continuous Turing-unstable torus mode: $q (λ) = \frac{1}{5} (λ - 2) (λ - \frac{5}{2})$ , the homogeneous Jacobian has negative trace and determinant $1$ , and the continuous torus has an eigenvalue $9 / 4$ in the unstable interval.

[F2]

Two-species continuous and discrete dispersion criteria: for a three-node periodic grid on length $L$ , its Laplacian eigenvalues are $λ_{ℓ} = 4 h^{- 2} \sin^{2} (π ℓ / 3)$ for $ℓ = 0, 1, 2$ , and the mode matrix is Hurwitz when $q (λ_{ℓ}) > 0$ .

Proof

1.1F1F2algebra

Here $h = L / 3 = 4 π / 9$ . The grid symbol gives $λ_{0} = 0$ and $λ_{1} = λ_{2} = \frac{3}{h^{2}} = \frac{243}{16 π^{2}} < 2,$ where the last inequality follows from $π > 3$ , since $243 / (16 π^{2}) < 243 / 144 = 27 / 16 < 2$ .

2.1F1F2step 1.1algebra∎

At $λ_{0} = 0$ , the reaction Jacobian has trace $- 1 / 2$ and determinant $1$ , so it is Hurwitz by the two-dimensional criterion in [F2]. For $ℓ = 1, 2$ , step 1.1 gives $0 < λ_{ℓ} < 2$ , so both factors in $q (λ_{ℓ}) = \frac{1}{5} (λ_{ℓ} - 2) (λ_{ℓ} - \frac{5}{2})$ are negative and their product is positive. The mode trace is $- 1 / 2 - (1 / 10 + 2) λ_{ℓ} < 0$ , so [F2] makes both modes Hurwitz. These are all three grid modes, even though the continuous mode in [F1] is unstable.

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Sources

Luca Doria, Numerical Algorithms, lecture notes, Johannes Gutenberg-Universität Mainz, 2022 (standard reference, not scraped)