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Graph diffusion and metric network pde numerics — Examples

1 · Prerequisites

2 · Summary

These stipulated calculations illustrate weighted mass conservation and equilibrium on a three-compartment network, Kirchhoff balance at a metric-graph Y-junction, singularity of a disconnected unanchored Neumann solve, and failure of equal-density routing when edge speeds differ. They are mathematical examples, not empirical claims about physical networks. Each calculation cites the exact A-page interface it uses.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Density continuity can violate conservative advection balance

Statement refuted

A universal junction rule declaring the density equal on every incident edge always conserves mass for advection with unequal edge speeds.

Counterexample model

Use one incoming edge with speed $1$ and two outgoing edges with speeds $1$ and $2$ . Mass flux on an edge is stipulated to be speed times density. Giving all incident edges density $1$ yields incoming flux $1$ and total outgoing flux $3$ , so mass is not conserved. A conservative routing can instead send the full incoming flux to the speed- $1$ outgoing edge and zero to the other.

The arrows indicate the stipulated flow direction and speeds; the exact flux sums in the refutation, not the drawing, establish the failure.

\begin{center} \begin{tikzpicture}[>=stealth, every node/.style={font=\small}] \node[circle,draw] (v) at (0,0) {junction}; \node[left] (in) at (-2.1,0) {in: $v = 1, ρ = 1, q = 1$ }; \node[right] (out1) at (2.2,1.0) {out: $v_{1} = 1, ρ = 1, q_{1} = 1$ }; \node[right] (out2) at (2.2,-1.0) {out: $v_{2} = 2, ρ = 1, q_{2} = 2$ }; \draw[->] (in) -- (v); \draw[->] (v) -- (out1); \draw[->,dashed] (v) -- (out2); \end{tikzpicture} \end{center}

Facts & Assumptions

Given: The stipulated one-in/two-out junction, positive speeds $v_{i n} = 1$ , $v_{1} = 1$ , $v_{2} = 2$ , and mass flux $q = v ρ$ .

[F1]

A conservative routing matrix is nonnegative with unit sums down each incoming column (Conservative nonnegative network advection routing).

[F2]

Such a routing maps any nonnegative incoming flux to a nonnegative outgoing flux with the same total (A conservative routing matrix preserves total flux).

Refutation

Given: The junction velocities and density-to-flux relation in the Counterexample.

1.1givenalgebra

If density is continuous with value $ρ = 1$ on all three incident edges, the incoming flux is $v_{i n} ρ = 1$ , while the outgoing total is $v_{1} ρ + v_{2} ρ = 1 + 2 = 3$ . These are unequal, so this junction rule fails mass conservation in the stipulated model.

2.1F1F2step 1.1algebra∎

The matrix $P = (1, 0)^{T}$ is entrywise nonnegative and its sole incoming column sums to $1$ . With incoming flux $q^{i n} = 1$ , it produces $q^{o u t} = (1, 0)^{T}$ , whose total is $1$ by [F1, F2]. The corresponding outgoing densities are $(1, 0)$ , so this conservative routing does not impose equal densities.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A disconnected unanchored network gives a singular steady solve

Example

Let a closed four-vertex network have edges $(1, 2)$ and $(3, 4)$ with conductances $1$ and $3$ . Its steady stiffness is singular: each disconnected positive-conductance component contributes an independent constant kernel mode. The load $(1, 0, 0, 0)^{T}$ is incompatible with the first component's Neumann balance, so it has no steady solution.

The two disconnected edges and their conductances are shown below; all vertex masses are one, and the right-hand side is $b = (1, 0, 0, 0)^{T}$ .

\begin{center} \begin{tikzpicture}[every node/.style={font=\small}] \node[circle,draw] (v1) at (0,0) { $1$ }; \node[circle,draw] (v2) at (1.7,0) { $2$ }; \node[circle,draw] (v3) at (3.5,0) { $3$ }; \node[circle,draw] (v4) at (5.2,0) { $4$ }; \draw[->] (v1) -- node[above] { $c = 1$ } (v2); \draw[->] (v3) -- node[above] { $c = 3$ } (v4); \node[below=4pt] at (v1) { $b_{1} = 1$ }; \node[below=4pt] at (v2) { $b_{2} = 0$ }; \node[below=4pt] at (v3) { $b_{3} = 0$ }; \node[below=4pt] at (v4) { $b_{4} = 0$ }; \end{tikzpicture} \end{center}

Facts & Assumptions

Given: Vertices ${1, 2, 3, 4}$ , oriented edges $1 \to 2$ and $3 \to 4$ , conductances $1, 3$ , no Dirichlet vertices, and mass matrix $M = I$ .

[F1]

The stiffness is $L_{C} = B C B^{T}$ ; its kernel consists of componentwise constants on positive-conductance components, and its left kernel is the same because it is symmetric (Finite weighted network diffusion data, Energy, kernel, and weighted mass balance for a finite network).

Verification

Given: The disconnected weighted path-pair network and the stated load.

1.1F1givenalgebra

Direct multiplication gives $L_{C} = (\begin{array}{cccc} 1 & - 1 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & 0 & 3 & - 3 \\ 0 & 0 & - 3 & 3 \end{array}) .$ The two independent component indicator vectors $(1, 1, 0, 0)^{T}$ and $(0, 0, 1, 1)^{T}$ lie in its kernel by [F1], so the steady matrix is singular.

2.1F1step 1.1algebra∎

The vector $(1, 0, 0, 0)^{T}$ has dot product $1$ with the first component indicator, while every vector in the image of the symmetric matrix $L_{C}$ is orthogonal to its kernel: for $z = L_{C} x$ and $k \in \ker L_{C}$ , $k^{T} z = (L_{C} k)^{T} x = 0$ . Thus this load is not in the image and the Neumann steady equation has no solution.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A Y-junction satisfies Kirchhoff flux balance

Example

Take three unit edges oriented from a central vertex $v$ to leaves $a, b, c$ , with unit diffusivity and prescribed leaf values $u (a) = 2$ , $u (b) = 0$ , and $u (c) = 1$ . Define on the three edge coordinates $u_{1} (x) = 1 + x$ , $u_{2} (x) = 1 - x$ , and $u_{3} (x) = 1$ for $0 \leq x \leq 1$ . The common central trace is $1$ , and the total outward diffusive flux at $v$ is zero.

The arrows give the coordinate orientations from the central vertex; the drawing is schematic and the proof uses the exact endpoint values below.

\begin{center} \begin{tikzpicture}[>=stealth, every node/.style={font=\small}] \node[circle,draw] (v) at (0,0) { $v : 1$ }; \node[circle,draw] (a) at (-2,1.4) { $a : 2$ }; \node[circle,draw] (b) at (2,1.4) { $b : 0$ }; \node[circle,draw] (c) at (0,-2) { $c : 1$ }; \draw[->] (v) -- node[above left] { $u_{1} (x) = 1 + x$ } (a); \draw[->] (v) -- node[above right] { $u_{2} (x) = 1 - x$ } (b); \draw[->] (v) -- node[right] { $u_{3} (x) = 1$ } (c); \end{tikzpicture} \end{center}

Facts & Assumptions

Given: The three unit intervals with a common central endpoint, their orientation from $v$ to the leaves, unit diffusivity, the displayed edge functions, and zero interior load.

[F1]

A metric graph has edgewise interval coordinates and common endpoint traces at each vertex (Anchored finite metric graph and its H1 energy space).

[F2]

For edgewise H2 functions, the signed endpoint terms in Green's identity group by common vertex trace and outward flux; zero test traces remove Dirichlet endpoints (Green identity and Kirchhoff flux cancellation on a metric graph).

[A1]

The general endpoint-representative interface assumes the Axiom of Choice (The Axiom of Choice, One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives).

Verification

Given: The oriented unit Y-graph and the three edge functions in the Example.

1.1A1F1givenalgebra

At the central endpoint $x = 0$ , all three values equal $1$ . At the leaves $x = 1$ , the values are $2, 0, 1$ , respectively. Hence the functions define a continuous metric-graph function with the specified Dirichlet data.

1.2F2givenalgebra

The outward derivative at the tail $x = 0$ is $- u_{i}^{'} (0)$ . The three values are $- 1, + 1, 0$ , whose sum is $0$ . Each edge has $u_{i}^{''} = 0$ . Therefore the edge equation has zero load and the free central vertex satisfies Kirchhoff balance.

2.1F2step 1.2given∎

For any test function with zero trace at the three leaves, [F2] shows that the edge bilinear form has only the central boundary contribution, which is its test trace times the sum $- 1 + 1 + 0 = 0$ . Thus the displayed function is also a weak harmonic field for these Dirichlet data.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Three-compartment migration conserves weighted mass

Example

Consider a closed three-vertex path with orientation $1 \to 2 \to 3$ , masses $(1, 2, 3)$ , conductances $(1, 2)$ , homogeneous diffusion, and initial state $u (0) = (2, 1, 4)^{T}$ . Its conserved weighted mass is $16$ . The constant state $(8 / 3, 8 / 3, 8 / 3)^{T}$ is an equilibrium with that same weighted mass, and the exact trajectory from the stated nonnegative initial state stays entrywise nonnegative. No convergence-to-equilibrium claim is made.

The diagram records the orientation, conductances, and vertex masses; edge lengths are schematic and are not used in the matrix calculation.

\begin{center} \begin{tikzpicture}[>=stealth, every node/.style={font=\small}] \node[circle,draw] (v1) at (0,0) { $1$ }; \node[circle,draw] (v2) at (2.1,0) { $2$ }; \node[circle,draw] (v3) at (4.2,0) { $3$ }; \draw[->] (v1) -- node[above] { $c_{1} = 1$ } (v2); \draw[->] (v2) -- node[above] { $c_{2} = 2$ } (v3); \node[below=4pt] at (v1) { $m_{1} = 1$ }; \node[below=4pt] at (v2) { $m_{2} = 2$ }; \node[below=4pt] at (v3) { $m_{3} = 3$ }; \end{tikzpicture} \end{center}

Facts & Assumptions

Given: The oriented path, $M = diag (1, 2, 3)$ , $C = diag (1, 2)$ , $D = \emptyset$ , and the stated initial vector.

[F1]

The closed network equation is $M u^{'} + L_{C} u = 0$ , with $L_{C} = B C B^{T}$ and weighted mass $1^{T} M u$ (Finite weighted network diffusion data).

[F2]

For a closed network, $1^{T} M u$ is constant under homogeneous diffusion; constants on a positive-conductance connected component lie in $\ker L_{C}$ (Energy, kernel, and weighted mass balance for a finite network).

[F3]

The homogeneous exact semigroup is defined by the finite matrix exponential and preserves nonnegative data (Positive graph-diffusion semigroup and Euler maps).

Verification

Given: The three-compartment matrices, initial state, and homogeneous network equation in the Example.

1.1F1givenalgebra

With the stated orientation, $B = (\begin{array}{cc} - 1 & 0 \\ 1 & - 1 \\ 0 & 1 \end{array}), L_{C} = B C B^{T} = (\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 2 & 2 \end{array}) .$ The graph is connected and both conductances are positive.

1.2F1F2givenalgebra

The initial weighted mass is $1^{T} M u (0) = 2 + 2 + 12 = 16$ . By [F2] it is constant for all times for which the homogeneous solution is evaluated.

2.1F1F2step 1.1algebra

The vector $1 = (1, 1, 1)^{T}$ has $L_{C} 1 = 0$ by the displayed matrix in step 1.1, and its weighted mass at value $8 / 3$ is $(8 / 3) (1 + 2 + 3) = 16$ . Thus the stated constant state is an equilibrium with the same weighted mass.

3.1F3givenstep 1.1∎

Since $u (0) \geq 0$ , [F3] implies the exact semigroup trajectory remains entrywise nonnegative for every $t \geq 0$ .

Graph diffusion and metric network pde numerics — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Density continuity can violate conservative advection balance

Statement refuted

Counterexample model

Facts & Assumptions

Refutation

A disconnected unanchored network gives a singular steady solve

Example

Facts & Assumptions

Verification

A Y-junction satisfies Kirchhoff flux balance

Example

Facts & Assumptions

Verification

Three-compartment migration conserves weighted mass

Example

Facts & Assumptions

Verification

Sources