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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: v1=1, ρ=1, q1=1}; \node[right] (out2) at (2.2,-1.0) {out: v2=2, ρ=1, q2=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 vin=1, v1=1, v2=2, and mass flux q=vρ.

[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 vinρ=1, while the outgoing total is v1ρ+v2ρ=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 qin=1, it produces qout=(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.

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