Networks, criticality, and complex systems

Start with three connected numbers

Imagine three quantities that each tend to fade, while a connection lets one quantity reinforce another. Connect node 1 to node 2 and node 2 to node 3. Start with values 1/10, √2/10 and 1/10. Can the connections maintain this pattern, or even make it grow?

Call the three values u₁, u₂ and u₃. They are supplied dimensionless amplitudes, not measured concentrations, probabilities or masses. Negative values are allowed in the model, although this first pattern is positive. Time t is measured in seconds.

Algebra, functions, graphs, and logarithms prepares the exponential and graph reading used here. A first reading can follow the three component equations and the plotted amplitudes. Matrices, linear algebra, and eigenvalue problems and Ordinary differential equations are published return paths for the matrix and rate notation; the needed steps are introduced locally below.

Turn connections into neighbor input

A network records entities as nodes and relationships as edges. Here the edges are undirected: the same connection enters the rule at both of its ends. The graph is simple, with no self-loop and no parallel edges.

The degree counts edges incident to a node. The degrees are 1, 2 and 1; their sum is 4 because each of the two edges has two ends. In this simple graph, degree also counts distinct neighbors. A degree is not the current amplitude of a node.

List nodes in the order 1, 2, 3. The adjacency matrix is A = [[0,1,0],[1,0,1],[0,1,0]]. Its dimensionless entry Aᵢⱼ is 1 if the nodes are connected and 0 otherwise. The diagonal zeros exclude self-connections.

Multiplying A by the column u = (u₁,u₂,u₃)ᵀ gives Au = (u₂,u₁+u₃,u₂)ᵀ: each entry is the sum of that node’s neighbor amplitudes. The upper diagram is a connection schematic; edge length and circle size do not encode physical distance or activity.

Supply a law for change

Supply a decay rate κ = 1 s⁻¹ and a nonnegative coupling g with units s⁻¹. The component rules are u̇₁ = −κu₁ + gu₂, u̇₂ = −κu₂ + g(u₁+u₃), and u̇₃ = −κu₃ + gu₂. A dot means a derivative with respect to time.

The compact form is du/dt = (-kappa I + g A)u, where I is the three-by-three identity matrix. Every term on the right has units s⁻¹ because u is dimensionless. Multiplying a physical rate by time makes the dimensionless argument needed in an exponential.

This is a chosen continuous-time linear model for all t ≥ 0 s. Neighbor input reinforces the supplied amplitudes. It is not conservative diffusion: there is no matching subtraction of the amount sent to a neighbor, and the sum of the three amplitudes need not stay constant. A physical application would need an appropriate interpretation, coefficients and validation.

Find one collective pattern

Consider v₊ = (1,√2,1)ᵀ. Direct multiplication gives Av₊ = (√2,2,√2)ᵀ = √2v₊. Thus the neighbor input has the same shape as the original column, scaled by √2. This is an eigenvector of A; its eigenvalue √2 is dimensionless.

Write u(t) = a(t)v₊, with a(0) = 1/10. Substitution gives ȧ = ra, where the physical rate is r = −κ + √2g. The exact solution is a(t) = a(0)*exp(r*t). The single amplitude changes while the ratio between the three node values stays 1:√2:1.

For this positive seed, r < 0 gives decay, r = 0 gives a constant pattern, and r > 0 gives growth. The threshold is g_c = κ/√2 = (1/√2) s⁻¹. It is a balance between decay and the neighbor input of this collective pattern, not a statement that the individual degree equals √2.

Compare three couplings

Keep the graph, decay rate and initial column fixed. Compare g = (1/2) s⁻¹, (1/√2) s⁻¹ and 1 s⁻¹. The three selected-pattern rates are (√2/2−1) s⁻¹, 0 s⁻¹ and (√2−1) s⁻¹, respectively.

The lower panel plots the common amplitude a against actual time, rather than three independent node values. Circles, squares and diamonds distinguish the cases. One shared marker records the common initial value; each case has markers at 2 s and 4 s. The connecting curves describe continuous solutions, not measured samples.

Define τ = t/(1 s), a dimensionless numerical time. The table gives exact expressions; the short decimal values help read the graph. To recover the three node values at any displayed time, multiply a by (1,√2,1)ᵀ.

The drawing covers 0 ≤ t ≤ 4 s and 0 ≤ a ≤ 0.6. The mathematical solutions extend to all t ≥ 0 s, but the growing branch eventually leaves this display window. The finite set of markers is not the proof of the long-time behavior.

Check all initial patterns

One selected curve does not describe every possible initial state. Two more eigenvectors are v₀ = (1,0,−1)ᵀ with eigenvalue 0, and v₋ = (1,−√2,1)ᵀ with eigenvalue −√2. Together with v₊, they form an orthogonal basis of the three-dimensional state space.

Each initial column can therefore be written as a sum of these patterns. Their physical rates under the supplied evolution are −κ+√2g, −κ and −κ−√2g. Since g ≥ 0, the first is the largest. Below g_c all three rates are negative, so every state tends to zero. Because the matrix is symmetric, the Euclidean norm also decays at least as fast as the exponential of that largest rate.

At g = g_c, the v₊ contribution persists while the other two contributions decay. Zero is stable but not asymptotically attracting: an arbitrarily small nonzero v₊ seed does not tend to zero. Above g_c, states with a nonzero v₊ contribution grow in magnitude in the long run, but specially chosen states with no such contribution still decay.

For example, v₀ decays as e⁻ᵗᵃᵘv₀ even when g = 1 s⁻¹. An exactly zero state stays zero. The positive growth rate establishes instability of the zero state; it does not force every initial column to grow.

Change the graph or the decay time

Remove only the edge between nodes 2 and 3. The new adjacency is A′ = [[0,1,0],[1,0,0],[0,0,0]]. Its eigenvalues are 1, −1 and 0: nodes 1 and 2 form a pair, and node 3 is isolated. The largest evolution rate is now −κ+g, so its threshold is g = κ = 1 s⁻¹.

The same coupling g = 1 s⁻¹ is above the original graph’s threshold but exactly at the new one. Node 3 has no neighbor input and decays at rate −κ. Reanalyzing the new graph matters; an edge change is not just a cosmetic change to the picture.

Below the original graph’s threshold, define the relaxation time of the v₊ pattern as T_rel = 1/(κ−√2g). The amplitude then loses a factor e over that positive time. For g = (1/2) s⁻¹, T_rel = (2+√2) s. It becomes arbitrarily large as g approaches g_c from below. At or above the threshold, this formula does not give a positive decay time.

Separate a threshold from criticality

This finite model has a collective threshold: a connection-dependent rate passes through zero. For every fixed finite time, its matrix-exponential solution varies smoothly with g. The sharp distinction between eventual decay, persistence and growth concerns long-time behavior, not a finite-time jump in the plotted amplitude.

In equilibrium statistical physics, a thermodynamic phase transition involves a different limiting question. For example, a finite Ising partition function is a finite sum of analytic terms; the thermodynamic limit lets the number of sites tend to infinity and can admit a nonanalytic transition. That equilibrium result does not forbid thresholds in finite dynamical equations.

A long relaxation time in this chosen equation is not evidence by itself for a measured critical exponent, universal scaling, avalanches or a scale-free network. Nor does the linear growing branch approach a saturated final state: this model supplies no nonlinear term that could stop it.

A network describes relationships, and a dynamical rule specifies what those relationships do. Using both can reveal collective behavior. To explain a particular complex system, the states, interaction rules, domain and physical evidence must fit that system; the three-node calculation establishes only its declared model conclusions.

Worked examples

Try it yourself

1. On the original graph, change the state to u = (−1,0,1)ᵀ. What are the three degrees and the neighbor-input column Au?

   Show the answer and reasoning

   The degrees remain (1,2,1); Au = (0,0,0)ᵀ.

   Degrees depend on the edge set. The two opposite endpoint amplitudes cancel in node 2’s neighbor sum, while each endpoint has a zero-valued neighbor.

2. Set g = 0 and u(0) = (1,0,0)ᵀ. Find u at 1 s. Can the zero-valued nodes become nonzero under this uncoupled law?

   Show the answer and reasoning

   u(1 s) = (1/e,0,0)ᵀ. The other two entries remain zero.

   Each component obeys u̇ᵢ = −uᵢ/(1 s) independently, so no neighbor term transfers a contribution to another node.

3. At g = 1 s⁻¹ on the original graph, start with u(0) = (−1,0,1)ᵀ. Does the state grow despite the positive largest rate?

   Show the answer and reasoning

   No. It is u(t) = e⁻ᵗᵃᵘ(−1,0,1)ᵀ and tends to zero.

   This state has Au = 0 and no v₊ contribution. Instability allows growing perturbations but does not make every special initial state grow.

4. At g = (1/√2) s⁻¹ on the original graph, start with u(0) = (1,0,1)ᵀ. Use u(0) = (v₊+v₋)/2 to find the long-time limit.

   Show the answer and reasoning

   The limit is (1/2,√2/2,1/2)ᵀ.

   The v₊ part has zero rate and remains v₊/2. The v₋ part has rate −2 s⁻¹ and decays. A zero largest rate does not imply attraction to zero.

5. With κ = 1 s⁻¹, compare the original graph’s threshold with the graph after edge 2–3 is removed. At g = 1 s⁻¹, which graph has a positive largest rate?

   Show the answer and reasoning

   The thresholds are (1/√2) s⁻¹ and 1 s⁻¹. Only the original graph has a positive largest rate at g = 1 s⁻¹.

   The adjacency’s largest eigenvalue changes from √2 to 1. The new graph is exactly at threshold, with a persistent pair mode rather than a positive-rate one.

6. The original graph’s relaxation time tends to infinity as g approaches its threshold from below. Does this prove universal critical scaling or a bounded final pattern in a material?

   Show the answer and reasoning

   No. It is a conditional property of the supplied finite linear model.

   The calculation neither validates a material model nor supplies a nonlinear saturation rule, a thermodynamic-limit theory or comparisons between different systems.

What to carry forward

• Connections and a dynamical law play different roles: the same graph can support different kinds of evolution.
• A collective growth threshold must be checked across all independent patterns; a chosen seed can behave differently from the largest rate.
• Finite linear thresholds and long relaxation times need further evidence before supporting claims about empirical criticality or nonlinear final states.