Green's second identity on a glued elementary solid region

Both $u$ and $v$ are $C^2$ on $O$, hence also $C^1$ there by [F3]. So [L1] applies as it stands and gives ${\iiint }_E(⟨\mathrm{\nabla }u,\mathrm{\nabla }v⟩+u\mathrm{\Delta }v)={\iint }_{\mathrm{\partial }E}u⟨\mathrm{\nabla }v,n⟩$; and it applies again with the roles of the two functions exchanged, which is legitimate exactly because both are $C^2$, giving ${\iiint }_E(⟨\mathrm{\nabla }v,\mathrm{\nabla }u⟩+v\mathrm{\Delta }u)={\iint }_{\mathrm{\partial }E}v⟨\mathrm{\nabla }u,n⟩$.

All the integrands appearing in step 1.1 are continuous: $\mathrm{\nabla }u$ and $\mathrm{\nabla }v$ have continuous components by [F1] and [F3], $\mathrm{\Delta }u$ and $\mathrm{\Delta }v$ are continuous by [F1] and [F3], and each boundary integrand is a continuous function on a compact Jordan parameter region by [F4]. So every one of them is integrable over the relevant set by [L3], and differences of them may be taken inside the integrals by [L2].

Subtract the second identity of step 1.1 from the first, using step 2.1 to combine the integrals. By the symmetry of the inner product in [F2] the two terms $⟨\mathrm{\nabla }u,\mathrm{\nabla }v⟩$ and $⟨\mathrm{\nabla }v,\mathrm{\nabla }u⟩$ are equal and cancel, leaving ${\iiint }_E(u\mathrm{\Delta }v-v\mathrm{\Delta }u)$ on the left and ${\iint }_{\mathrm{\partial }E}(u⟨\mathrm{\nabla }v,n⟩-v⟨\mathrm{\nabla }u,n⟩)$ on the right.