Energy, kernel, and weighted mass balance for a finite network

Reversing an edge orientation replaces its incidence column $b_e$ by $-b_e$, but the contribution $c_e{b}_e{b}_e^{\mathsf{T}}$ to $BC{B}^{\mathsf{T}}$ does not change. For the zero extension $\bar{x}$, [F1] gives $b_e^{\mathsf{T}}\bar{x}={\bar{x}}_{\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{d}(e)}-{\bar{x}}_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}(e)}$. Expanding ${\bar{x}}^{\mathsf{T}}BC{B}^{\mathsf{T}}\bar{x}={\sum }_e{c}_e(b_e^{\mathsf{T}}\bar{x}{)}^2$ proves the displayed energy identity. It is nonnegative term by term, so $L_D$ is symmetric positive semidefinite.

If $L_{D}x=0$, step 1.1 gives a sum of nonnegative terms equal to $x^{\mathsf{T}}{L}_{D}x=0$. For every $e\in E_{+}$ its endpoint difference must therefore be zero. Along every path in $G_{+}$, $\bar{x}$ has equal values at successive vertices, hence is constant on each positive-conductance component; it is zero on any such component meeting $D$ because $\bar{x}=0$ there. Conversely, if these componentwise conditions hold, every edge with $c_e>0$ has zero endpoint difference, while an edge with $c_e=0$ contributes zero regardless of its difference. Thus $BC{B}^{\mathsf{T}}\bar{x}=0$ and $L_{D}x=0$.

Suppose $D=\mathrm{\varnothing }$. From [F1], every column of $B$ sums to zero, so ${\mathbf{1}}^{\mathsf{T}}B=0$ and therefore ${\mathbf{1}}^{\mathsf{T}}{L}_C=0$. Multiplying $M{u}^{\mathrm{\prime }}+L_{C}u=f$ on the left by ${\mathbf{1}}^{\mathsf{T}}$ gives ${\mathbf{1}}^{\mathsf{T}}M{u}^{\mathrm{\prime }}+{\mathbf{1}}^{\mathsf{T}}{L}_{C}u={\mathbf{1}}^{\mathsf{T}}f$. The second term vanishes, and the first is the derivative of ${\mathbf{1}}^{\mathsf{T}}Mu$, proving the balance law and its homogeneous conservation case.