= Solution
Apply the <product rule> to the <vector field> $\Phi\nabla\Psi-\Psi\nabla\Phi$:
$$
\nabla\cdot(\Phi\nabla\Psi-\Psi\nabla\Phi)
=\Phi\Delta\Psi-\Psi\Delta\Phi.
$$
The gradient cross terms cancel. Integrating and applying the <divergence theorem> with outward unit normal $\mathbf n$ proves <Green's second identity>:
$$
\boxed{\int_V(\Phi\Delta\Psi-\Psi\Delta\Phi)\,dV
=\oint_S(\Phi\partial_n\Psi-\Psi\partial_n\Phi)\,dS.}
$$
Choose the <Dirichlet Green function> $G_D(\mathbf r,\mathbf r')$ with $\Delta_{\mathbf r}G_D=\delta(\mathbf r-\mathbf r')$ and $G_D=0$ on $S$. Taking $\Psi=G_D$ and using the <Dirac delta distribution>, the left side becomes $\Phi(\mathbf r')$ while the second boundary term vanishes. Hence
$$
\boxed{\Phi(\mathbf r')=\oint_S f(\mathbf r)\,\partial_{n_{\mathbf r}}G_D(\mathbf r,\mathbf r')\,dS.}
$$
The plus sign is tied to the convention $\Delta G_D=+\delta$, whose free-space <fundamental solution> is negative in three dimensions. The singularity may equivalently be handled by excising a small sphere about $\mathbf r'$ before applying the smooth identity.
Back to article page
