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A Brusselator with a continuous Turing-unstable torus mode

Statement

Consider the Brusselator field

F(u,v)=(1−52u+u2v, 32u−u2v)

with diagonal diffusivity D=diag⁡(1/10,2). Its homogeneous equilibrium is (u∗,v∗)=(1,3/2). On the one-dimensional torus of length L=4π/3, the continuous reaction-diffusion linearization has an unstable Fourier mode with wave number k=1.

Facts & Assumptions

Given: The reaction field and diffusivities in the Statement, and the dispersion criterion of Two-species continuous and discrete dispersion criteria.

[F1]

Two-species continuous and discrete dispersion criteria: for a mode with Laplacian eigenvalue λ, the linearized matrix is J−λD; with negative trace, it has one positive eigenvalue when its determinant q(λ) is negative.

[A1]

The example uses A=1, B=3/2 in the standard Brusselator F(u,v)=(A−(B+1)u+u2v, Bu−u2v).

Proof

1.1F1givenalgebra

Direct substitution gives F(1,3/2)=(0,0). Differentiating the two components gives J=DF(1,3/2)=(1/21−3/2−1),tr⁡J=−1/2,det⁡J=1.

2.1step 1.1givenalgebra

The coefficient in the dispersion polynomial is βT=d2J11+d1J22=9/10 and d1d2=1/5. Therefore q(λ)=15λ2−910λ+1=15(λ−2)(λ−52).

3.1F1step 2.1algebra∎

For the continuous torus mode k=1, λ=(2π/L)2=9/4. This lies strictly between 2 and 5/2, so step 2.1 gives q(9/4)<0. By [F1], the mode matrix has one positive eigenvalue and one negative eigenvalue. This example proves a linear mode instability only; it does not assert nonlinear pattern existence or stability.

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