Neumann representation for a radiating Helmholtz half-space ($u(x)=-\frac{1}{2\pi }{\int }_{{\mathbb{R}}^2}{\partial }_{3}u(y^{\prime },0)\frac{e^{ik|x-(y^{\prime },0)|}}{|x-(y^{\prime },0)|}{d}^2{y}^{\prime }$)

Let $(\Delta +k^2)u=0$ for $x_3>0$, with real $k>0$, $u=O(r^{-1})$ and $r({\partial }_{r}u-iku)\to 0$ uniformly in direction. With sufficient boundary regularity, Green's second identity on an expanding half-ball gives the title formula. The plane integral is interpreted as the limit over expanding disks when it is not absolutely convergent. Its sign uses the upward derivative ${\partial }_{3}u$; the outward normal derivative is $-{\partial }_{3}u$.

For $G=e^{ik|x-y|}/(4\pi |x-y|)$, $G=O(r^{-1})$ and $({\partial }_r-ik)G=O(r^{-2})$ at large $|y|=r$. The hemispherical integrand can be written $u({\partial }_r-ik)G-G({\partial }_r-ik)u$, whose integral tends to zero by the Sommerfeld radiation condition. Add the boundary identities for a source point $x$ and its reflected point $x^{\ast }=(x_1,x_2,-x_3)$. On the plane $G(x,y)=G(x^{\ast },y)$ and their normal derivatives are opposite. The terms involving boundary values of $u$ cancel, leaving twice $G{\partial }_{3}u$ and hence the title formula. In particular, zero Neumann boundary data imply $u=0$ in this radiating class.