 Dirichlet image kernel for the three-dimensional Helmholtz equation

For the upper half-space and the sign convention $(\Delta +k^2)G_D={\delta }_{r^{\prime }}$, reflect the source $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$ and use the displayed difference. This is the negative of the conventional outgoing kernel, whose source sign is $-\delta$. The method of images makes the Dirichlet boundary condition zero on the plane. With $R=|r^{\prime }-r|$ and $\cos \theta =z^{\prime }/R$, its outward normal derivative there isThe outward unit normal is $-e_z$. Differentiating $e^{ikR}/R$ gives $e^{ikR}(ik/R-1/R^2)$; the real and reflected source contributions add in the normal derivative. Green's second identity therefore expresses a solution of the Helmholtz equation as $\varphi (r^{\prime })={\int }_{z=0}\varphi (r){\partial }_n{G}_D(r,r^{\prime })dS$ when the boundary term at infinity vanishes. Suitable decay or compatible outgoing Sommerfeld radiation conditions are needed; vanishing at infinity without further control is not a general uniqueness condition for a real-frequency Helmholtz problem.