Green identity and Kirchhoff flux cancellation on a metric graph

If $E=\mathrm{\varnothing }$, both sides are empty sums and the identity is $0=0$. Otherwise fix an edge $e$ and first suppose $u_e$ is real-valued. By [F1] and [F4], $u_e^{\mathrm{\prime }}$ and $v_e$ have absolutely continuous representatives on $[0,{\mathrm{\ell }}_e]$. They are bounded there. The product $u_e^{\mathrm{\prime }}{v}_e$ is absolutely continuous: for any disjoint intervals $[a_j,b_j]$, ${\sum }_j\mathrm{\mid }{u}_e^{\mathrm{\prime }}(b_j)v_e(b_j)-u_e^{\mathrm{\prime }}(a_j)v_e(a_j)\mathrm{\mid }\le \mathrm{\parallel }{u}_e^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}{\sum }_j\mathrm{\mid }{v}_e(b_j)-v_e(a_j)\mathrm{\mid }+\mathrm{\parallel }{v}_e{\mathrm{\parallel }}_{\mathrm{\infty }}{\sum }_j\mathrm{\mid }{u}_e^{\mathrm{\prime }}(b_j)-u_e^{\mathrm{\prime }}(a_j)\mathrm{\mid },$ so the defining absolute-continuity condition follows from that of the two factors. At almost every point both derivatives exist, and [F3] gives $(u_e^{\mathrm{\prime }}{v}_e{)}^{\mathrm{\prime }}=u_e^{\mathrm{\prime }\mathrm{\prime }}{v}_e+u_e^{\mathrm{\prime }}{v}_e^{\mathrm{\prime }}$. Both terms are integrable because $u_e^{\mathrm{\prime }\mathrm{\prime }},v_e^{\mathrm{\prime }}$ are integrable derivatives of absolutely continuous functions and $u_e^{\mathrm{\prime }},v_e$ are bounded. Applying [F2] to the product therefore yields ${\int }_0^{{\mathrm{\ell }}_e}{u}_e^{\mathrm{\prime }}{v}_e^{\mathrm{\prime }}d{x}_e=-{\int }_0^{{\mathrm{\ell }}_e}{u}_e^{\mathrm{\prime }\mathrm{\prime }}{v}_{e}d{x}_e+u_e^{\mathrm{\prime }}({\mathrm{\ell }}_e)v_e({\mathrm{\ell }}_e)-u_e^{\mathrm{\prime }}(0)v_e(0)\mathrm{.}$ If $u_e$ is complex-valued, write the absolutely continuous representative $u_e^{\mathrm{\prime }}=p+iq$. By [F1], $p,q$ are real absolutely continuous functions with derivatives $\mathrm{Re}{u}_e^{\mathrm{\prime }\mathrm{\prime }},\mathrm{Im}{u}_e^{\mathrm{\prime }\mathrm{\prime }}$ almost everywhere; these derivatives lie in $L^2$ since their absolute values are bounded by $\mathrm{\mid }{u}_e^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }$. The test function $v_e$ is real by definition of $V_D$. Apply the preceding product-rule and fundamental-theorem calculation separately to $p{v}_e$ and $q{v}_e$ and combine the two identities, using componentwise complex integration. This proves the same displayed identity for complex $u_e$. The endpoint terms are the head and tail outward derivatives paired with their traces.

Multiply the identity of step 1.1 by ${\kappa }_e$ and sum over the finite edge set. At each vertex, all endpoint-incidence values of $v$ equal the common trace $v(z)$ by definition of $V_D$; this includes both incidences of a loop. Regrouping the finite endpoint sum by vertices gives the displayed identity. If $z\in D$, then $v(z)=0$; if $z\mathrm{\notin }D$, the stated Kirchhoff sum is zero. Hence the boundary contribution vanishes under those conditions.