Pattern formation and self-organization

Start with a ripple on a ring

Imagine two signed numbers attached to each position around a ring. Call them u and v. Initially, u is 0.01 at one reference position, zero a quarter of the way around, and −0.01 halfway around. The second field has the opposite sign and one third of that amplitude. We supply this ripple and ask whether it grows or fades.

The numbers are dimensionless disturbances about zero. They are not measured concentrations, and a negative value is allowed. Position x is measured in metres around the ring; time t is measured in seconds. The two fields may affect each other locally and also spread through diffusion.

Algebra, functions, graphs, and logarithms prepares the exponential and graph ideas; Geometry, trigonometry, and coordinate systems prepares the cosine shape. A first reading can follow the signs and snapshots. Matrices, linear algebra, and eigenvalue problems and Partial differential equations and Green functions are published return paths for the matrix and space-derivative steps, which are introduced locally below.

Separate local coupling from spreading

Supply a ring of length L = 2π√(10/3) m and the exact linear model w_t = J*w + D*w_xx, where w = (u,v)ᵀ. Here J = [[1,2],[−2,−3]] s⁻¹ describes local coupling, and D = diag(1,10) m²/s describes diffusion of the two fields. The subscript t means a time derivative; xx means the second derivative with respect to physical position x.

Written component by component, the local contributions are u + 2v and −2u − 3v per second. The diffusion contributions are (1 m²/s)uₓₓ and (10 m²/s)vₓₓ. A second space derivative has units m⁻² because the fields are dimensionless, so each complete contribution has units s⁻¹.

The ring is periodic: w(0,t) = w(L,t) and wₓ(0,t) = wₓ(L,t). The two drawn endpoints are the same ring position. They are not impermeable end walls. All coefficients are constant in position and time, so no spatial pattern is imposed through a varying coefficient.

This linear equation is the whole mathematical model supplied here, for t ≥ 0 s. In a physical reaction system, a similar equation may describe only small disturbances near a uniform state. Our chosen numbers do not supply the missing nonlinear reactions or validate a particular material.

Test a uniform disturbance first

A uniform disturbance has the same values everywhere, so its space derivatives vanish. Only J acts. Its two eigenvalues are both −1 s⁻¹. A growth rate with a negative real part gives eventual decay; this repeated root needs some care because J does not have two independent eigenvectors.

Let τ = t/(1 s), a dimensionless numerical time. Write J = (−I + N)/(1 s), where I is the identity matrix and N = [[2,2],[−2,−2]]. Direct multiplication gives N² = 0. The exact uniform solution is w(t) = exp(-tau)*(I + tau*N)*w(0).

The exponential eventually beats the factor linear in τ, so every uniform disturbance tends to zero. It need not become smaller at every instant: the coupling can temporarily enlarge a component or the length of the two-component column. Eventual stability and monotone decay are different statements.

Give each spatial shape its own matrix

A periodic cosine with n complete cycles around the ring has wave number kₙ = 2πn/L. Here n is a nonnegative integer; n = 0 is uniform. Differentiating the cosine twice multiplies it by −kₙ². Thus a field of the form cₙ(t)cos(kₙx) keeps that spatial shape, while its two-component amplitude obeys ċₙ = Aₙcₙ with Aₙ = J − Dkₙ².

For this length, kₙ² = (3/10)n² m⁻². Define the dimensionless number z = kₙ²/(1 m⁻²). Then Aₙ = [[1−z,2],[−2,−3−10z]] s⁻¹. Its trace is (−2−11z) s⁻¹ and its determinant is (1−7z+10z²) s⁻².

For a real two-by-two matrix, negative trace and positive determinant put both eigenvalue real parts below zero. A negative determinant instead forces one positive and one negative real eigenvalue. The table applies this test and gives the largest rate exactly; approximate values only help compare their sizes.

The table is not the all-mode proof. Factor the determinant coefficient as (1−5z)(1−2z): it is negative only for 1/5 < z < 1/2. The allowed values are z = 0, 3/10, 6/5, 27/10, and larger. Only n = 1 lies inside that band; every n ≥ 2 has negative trace and positive determinant.

Read two spatial snapshots

Supply the initial condition w(x,0) = (1/100)(1,−1/3)ᵀcos(2πx/L). This is the n = 1 mode. Its rational amplitude column is not exactly an eigenvector, so the solution retains both the growing and the decaying temporal branches.

The two panels show u and v against position at 0 s and 20 s. Both vertical axes use the same dimensionless field scale. Hollow blue circles mark the initial snapshot and filled orange diamonds mark the later one. The lines describe spatial profiles at fixed times; no line connects one time to the other.

Write u(x,t) = U(t)cos(2πx/L) and v(x,t) = V(t)cos(2πx/L). At 20 s, U is approximately +0.0211414 and V is approximately −0.00700340. The larger amplitudes come from the selected seeded solution, not from measured samples or a claim that every possible disturbance grows.

The table uses a₂₀ = U(20 s) and b₂₀ = V(20 s). The cosine is exactly zero at L/4 and 3L/4, so those field values are exactly zero. The values at 0 and L agree because the ring closes.

Verify both temporal branches

For an optional exact calculation, set d = √2889 and r₊ = (−53+d)/20, r₋ = (−53−d)/20. These r values are dimensionless; the physical rates are r₊ s⁻¹ and r₋ s⁻¹. With τ = t/(1 s), exponentials exp(r₊τ) and exp(r₋τ) have dimensionless arguments.

The amplitudes are U(t) = [(1+161/(3*d))*exp(r_plus*tau) + (1-161/(3*d))*exp(r_minus*tau)]/200 and V(t) = -[(1+53/d)*exp(r_plus*tau) + (1-53/d)*exp(r_minus*tau)]/600. These expressions are the sum of two temporal branches with the same cosine shape.

At τ = 0, both exponentials are 1. Adding the coefficients gives U(0) = 1/100 and V(0) = −1/300. Differentiating at t = 0 gives U̇(0) = 1/3000 s⁻¹ and V̇(0) = 0 s⁻¹, matching A₁ times the initial column. Substitution checks the full solution; keeping only the positive-rate branch would alter the supplied initial data.

A sine of the same wave number has the same growth rates, as does a shifted cosine. Our chosen initial condition fixes its phase around the ring. Linear instability does not choose one unique crest position.

Change diffusion or the ring length

Keep J and L but set both diffusion coefficients to 1 m²/s. Each mode matrix is now J − kₙ²(1 m²/s)I, so both eigenvalues become −(1+(3/10)n²) s⁻¹. Every mode eventually decays. Equal diffusion shifts the reaction rates together; it does not create the unequal-diffusion growing branch in this model.

Without diffusion, each mode is acted on by J alone and eventually decays, though its spatial shape is not smoothed by spreading. These two comparisons show why local reaction stability is compatible with instability after unequal diffusion is added. Diffusion does not directly amplify a cosine component; the change in how its two coupled components evolve can create a growing branch.

Domain size matters too. Halving the circumference doubles each nonzero kₙ and quadruples kₙ². The smallest nonzero z is then 6/5, already above the unstable band. With the original unequal diffusion coefficients, no admissible nonuniform mode on that smaller ring grows asymptotically.

Separate onset from a finished pattern

This example demonstrates a diffusion-driven linear instability: the uniform reaction dynamics is asymptotically stable, yet an admissible nonuniform mode has a positive growth rate once unequal diffusion is included. Local coupling and spreading can therefore amplify an existing spatial variation without a position-dependent coefficient.

Exactly zero initial disturbance remains exactly zero in this deterministic linear equation. A seed, forcing or noise must be supplied if a disturbance is to appear. Our seed is declared, and its cosine shape remains fixed while its amplitudes evolve.

The model has no term that stops the unstable branch from growing. A persistent bounded pattern, a selected final wavelength, or an observed self-organizing material needs further nonlinear dynamics, boundary information and physical evidence. Linear growth can diagnose an onset mechanism without establishing those later outcomes.

Worked examples

Try it yourself

1. If u is dimensionless, what units do uₓₓ and (1 m²/s)uₓₓ have? Why is the latter compatible with uₜ?

   Show the answer and reasoning

   uₓₓ has units m⁻²; multiplying by m²/s gives s⁻¹, the same units as uₜ.

   A second derivative divides by position twice. Keep the physical coordinate x rather than silently replacing it by the dimensionless ratio x/L.

2. In the original ring, find z, the trace and the determinant for n = 2. Does this mode grow asymptotically?

   Show the answer and reasoning

   z = 6/5, trace = −76/5 s⁻¹ and determinant = 7 s⁻². Both eigenvalue real parts are negative, so it eventually decays.

   Substitute z = (3/10)×4 into −2−11z and 1−7z+10z²; negative trace and positive determinant give the two-by-two stability test.

3. At x = L/4, what are u and v at 0 s and 20 s for the selected seed? Is a tiny computer cosine residual a nonzero model value?

   Show the answer and reasoning

   All four field values are exactly zero. A tiny floating-point residual is an evaluation artifact, not a nonzero exact value.

   The spatial factor is cos(π/2) = 0 at both times, independently of the two amplitudes.

4. With the original J and unequal diffusion coefficients, halve the circumference. Find k₁² and determine whether any nonuniform integer mode has a positive rate.

   Show the answer and reasoning

   k₁² becomes 6/5 m⁻². No nonuniform integer mode grows asymptotically on the smaller ring.

   Each wave number doubles, so its square quadruples. Every allowed nonzero z is at least 6/5, above the unstable interval 1/5 < z < 1/2.

5. A uniform disturbance has a larger norm at 1 s than at 0 s. Does that observation prove a positive asymptotic eigenvalue?

   Show the answer and reasoning

   No. The example has two eigenvalues −1 s⁻¹ but permits temporary growth before eventual decay.

   The matrix couples components and contains a polynomial transient. Check the long-time rate and the full evolution instead of equating every finite increase with asymptotic instability.

6. Does the positive n = 1 growth rate prove a permanent bounded pattern, or create a pattern from an exactly zero initial field?

   Show the answer and reasoning

   Neither. Exactly zero stays zero, and the supplied linear model has no saturation mechanism.

   The declared seed can grow. A final bounded pattern requires additional nonlinear dynamics and, for a material claim, an appropriate physical model and evidence.

What to carry forward

• A uniform reaction state can be asymptotically stable while unequal diffusion makes an admissible nonuniform mode grow.
• Periodic boundary conditions restrict the allowed wave numbers; checking a few plotted modes is not the full proof.
• Eventual modal decay can coexist with finite transient amplification, and the selected seed may contain both temporal branches.
• A growing linear ripple does not by itself establish a persistent nonlinear pattern or a measured self-organizing system.