The outgoing Green function for the three-dimensional Helmholtz equation satisfies
Its normalization follows by integrating over a small ball about : , so its outward derivative has flux . Apply Green's second identity in the variables:
The terms cancel. Thus the surface integral equals for , and zero for outside the closed volume. Equivalently, remove a small source ball and evaluate its flux before taking its radius to zero. A source on the boundary would require a different limiting formula and is not part of this calculation.
For the upper half-space, take a half-ball of radius . Interpret the first far-field condition as , and the radiation limit in its usual direction-uniform sense, namely the Sommerfeld radiation condition
This is the finite-bound interpretation of the PDF's first condition. For fixed and real ,
Indeed , and its radial derivative is . On the large hemisphere the integrand can therefore be rewritten as
Its area is , so its integral tends to zero. The planar integral, taken over expanding disks if needed, consequently equals for and zero for .
The outward unit normal vector on the bottom plane is . Denote this planar integral by
For reflect the source point to . On the plane,
Adding and cancels the boundary values of and gives . Thus the Neumann representation for a radiating Helmholtz half-space is
The negative sign is consistent with using the upward derivative rather than the outward normal derivative. If that upward derivative is zero everywhere on the plane, the integral is zero. Hence throughout the upper half-space in the specified radiating class; taking a difference also gives uniqueness for prescribed Neumann boundary data.

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