Green's identities

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.

For a bounded smooth domain $\Omega \subset {\mathbb{R}}^3$ and $u\in C^2(\overline{\Omega })$, let $\Phi (x,y)=1/(4\pi |x-y|)$, normalized by $-{\Delta }_y\Phi (x,y)={\delta }_x(y)$. For an interior point $x$, Green's third identity isIt represents the function through its volume PDE source term, boundary values and outward normal derivative. For a harmonic function the volume term vanishes, so boundary data alone reconstruct the interior.

To derive it, apply Green's second identity on $\Omega \setminus \overline{B_{\epsilon }(x)}$. The inner sphere has the outward normal of the punctured domain, pointing toward $x$. Its singular fundamental solution term tends to $u(x)$ as $\epsilon \downarrow 0$, giving the stated signs. Replacing $\Phi$ by the Dirichlet Green function makes its boundary value zero and leaves the Poisson kernel multiplying the boundary values of $u$. Thus the identity connects Green functions, boundary integral representations and Dirichlet problems.

For smooth functions $u,v$ on a region $V$,

If $f$ is constant on the boundary and $g$ is a regular harmonic function throughout a volume, the divergence theorem applied to $f\nabla g$ gives ${\int }_V\nabla f\cdot \nabla g={\int }_{\partial V}f{\partial }_{n}g$. The constant boundary value factors out; a second application of the divergence theorem makes the remaining flux ${\int }_V\Delta g=0$. This requires enough regularity to use Green's first identity over the entire volume.

Although $1/r$ is harmonic away from the origin in three dimensions, it is not a regular harmonic function on a ball containing the origin. Its gradient is $-\hat{r}/r^2$, so the displayed improper volume integral follows by canceling the $r^2$ in the spherical volume element. On the annulus $\epsilon <r<a$, Green's first identity gives outer boundary contribution $-4\pi$ and inner contribution $4\pi \epsilon /a$, recovering the same limit. Its global Laplacian contains a Dirac delta distribution, so the regular-harmonic gradient-pairing theorem is inapplicable.