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Green's identities relate volume integrals involving gradients and Laplacians to boundary values and normal derivatives. They follow from the divergence theorem and include Green's first identity, Green second identity and Green's third identity.

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.

A particularly simple constructive choice uses paired collocation points for a global relation. For $n=1,\dots ,N$, putFour suitable sets, one for each side, areThe braces in each set range over all $n$. For large $p_n/k$, compute $r_n^{-1}=k/({\omega }_n+p_n)$ to avoid cancellation. The corresponding $(A,B)$ pairs are, respectively,Each pair has the same exponentially growing normal direction and opposite tangential frequencies. Take the phase-weighted difference of the two collocated global relations for a linear boundary value problem. For example, the bottom pair gives the adjoint testThe other three phase-weighted differences giveThus the four spectral sets supply four real sine-tested equations per mode. We retain these $4N$ equations; requiring every uncombined complex equation as well would instead form an overdetermined finite approximation. For real traces each retained equation is the appropriate real combination of its conjugate pair.

The key advantage is that $v_{Bn},v_{Tn}$ vanish on the vertical sides, while $v_{Ln},v_{Rn}$ vanish on the horizontal sides. Orthogonality then eliminates all off-mode unknowns. Define the known right-hand sidesThe retained global relations for a linear boundary value problem areFor example, with $f_{jn}={\int }_{-1}^1{f}_j^{(M)}(s){\varphi }_n(s)ds$, the bottom and top right-hand sides are explicitlyFor completeness the left and right right-hand sides areAll these integrals use known polynomial approximations or the exact Dirichlet boundary condition.

Scale each row by $e^{-{\omega }_n}$. Every block becomesThere is just one off-diagonal entry in each row. This proves diagonal dominance of paired square global-relation collocation. Therefore the assembled $4N\times 4N$ system is strictly diagonally dominant, with row margin $1-{\delta }_n>1-e^{-\pi }$. This proof applies to the explicitly paired and scaled collocation system, rather than assuming arbitrary raw spectral rows are diagonally dominant. Each block is a positive-definite matrix and have spectral condition number of a positive-definite matrixFor instance, setting ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$ gives the concise answerwith the identical formula for the opposite vertical sides. If the known-data integrals use exact $f_j$, these are exactly the first $N$ sine coefficients of the true normal traces: no discarded unknown mode contributes to a retained row. With a finite Legendre polynomial approximation, the error is solely that of the known-data approximation and the final normal-trace truncation. These square modified Helmholtz Dirichlet-to-Neumann coefficients determine the unknown boundary normal derivative; Green's third identity or the separated-variable construction can then recover the interior solution.

Apply Green second identity to $u$ and the free-space Green function $G_{fs}$ on $V$ with a small ball about ${\mathbf{r}}_0$ removed. Both functions are harmonic there, so only boundary terms remain. The contribution from the small sphere tends to $u({\mathbf{r}}_0)$ becausehas unit delta source, while the remaining small-sphere term vanishes. Taking the radius to zero gives Green's third identity