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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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) {b1=1}; \node[below=4pt] at (v2) {b2=0}; \node[below=4pt] at (v3) {b3=0}; \node[below=4pt] at (v4) {b4=0}; \end{tikzpicture} \end{center}

Facts & Assumptions

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

[F1]

The stiffness is LC=BCBT; 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 LC=(1−100−1100003−300−33). 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 LC is orthogonal to its kernel: for z=LCx and k∈ker⁡LC, kTz=(LCk)Tx=0. Thus this load is not in the image and the Neumann steady equation has no solution.

Depends on

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