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For a bounded source region and fixed positive wavenumber, expand $|R\mathbf{\theta }-{\mathbf{r}}^{\prime }|=R-\mathbf{\theta }\cdot {\mathbf{r}}^{\prime }+O(R^{-1})$. In the outgoing Green function for the three-dimensional Helmholtz equation, this gives $G=e^{ikR}{e}^{-ik\mathbf{\theta }\cdot {\mathbf{r}}^{\prime }}/(4\pi R)+O(R^{-2})$. Volume integration gives the outgoing field's far-field pattern with an $O(R^{-2})$ remainder. The estimate is an asymptotic statement for fixed source size and frequency.

With the normal directed out of an obstacle $D$ into the fluid, a field solving $(\Delta +k^2)\psi =-Q$ outside $D$ has the representationHere $G_k$ is the outgoing Green function for the three-dimensional Helmholtz equation and the observation point lies outside the obstacle. The formula follows from Green second identity; the normal to the exterior fluid on its inner boundary is the negative of the stated obstacle normal. The surface term represents the scattered field.

Use time dependence $e^{-i\omega t}$, positive wavenumber $k$ and aperture normal $\mathbf{n}={\mathbf{e}}_x$. Write $F(y^{\prime },z^{\prime })=E_{\mathrm{inc},y}(0,y^{\prime },z^{\prime })$ on the aperture and extend $F$ by zero outside it. DefineThe outgoing branch has $\kappa \ge 0$ for propagating components and $\mathrm{Im}\kappa >0$ for evanescent components.

Fix the Green function normalization by $G_0(\mathbf{r})=e^{ik|\mathbf{r}|}/(4\pi |\mathbf{r}|)$, so $(\Delta +k^2)G_0=-\delta$. Its transverse Fourier transform satisfies $({\partial }_x^2+{\kappa }^2){\hat{G}}_0=-\delta (x)$. An outgoing solution on both sides is proportional to $e^{i\kappa |x|}$, and its derivative jump fixes ${\hat{G}}_0=i{e}^{i\kappa |x|}/(2\kappa )$. Thus the Weyl plane-wave representation for observation at $x>0$ isSince $\mathbf{n}\times {\mathbf{E}}_{\mathrm{inc}}=F{\mathbf{e}}_z$, the given aperture integral becomes a superposition of curls of ${\mathbf{e}}_z{e}^{i\mathbf{q}\cdot \mathbf{r}}$. Each curl supplies $i\mathbf{q}\times {\mathbf{e}}_z=i(\nu ,-\kappa ,0)$. Therefore, with the unit prefactor in that integral,Every vector amplitude is transverse to $\mathbf{q}$, as Maxwell's equations require. The polarizer specifies the incident aperture field; diffraction can regenerate an $x$ component outside the aperture. In contrast, its $z$ component remains zero in this representation. Evanescent plane waves are needed for the complete near field, not just the propagating disk ${\nu }^2+{\eta }^2\le k^2$.

The factor $1/2$ follows explicitly from the standard outgoing Green function for the three-dimensional Helmholtz equation and the printed unit curl prefactor. If the intended kernel is the doubled half-space kernel $2G_0$, or the aperture formula carries the usual compensating factor $2$, delete that $1/2$. The latter normalization makes the tangential output trace equal to $F$. Keeping the convention explicit avoids silently changing the supplied representation; the angular dependence and transversality of an electromagnetic plane wave are the same in either normalization. Reversing the specified normal also reverses the overall sign.

Take the same $e^{-i\omega t}$ convention and an outgoing field. Define the scattering potential with the positive-contrast sign,The outgoing Green function for the three-dimensional Helmholtz equation isThus the Lippmann-Schwinger equation has a plus sign:Replacing the total field inside the integral by the incident field gives the Born approximation for scalar wave scatteringMore generally, iterating the integral equation gives the Born series ${\psi }_s={\sum }_{j\ge 1}{T}^j{\psi }_i$; its first term is the boxed expression. A sufficient convergence condition is $\Vert T\Vert <1$ on the chosen space of fields in $V$, and accurate single-scattering truncation needs the neglected terms to be small. Weak refractive contrast is useful, but for a sufficiently large or resonant scatterer it is not by itself a guarantee of negligible multiple scattering.

Choose the time convention $e^{-i\omega t}$ and the outgoing Sommerfeld radiation condition. Define the outgoing Green function for the three-dimensional Helmholtz equation byAway from ${\mathbf{r}}^{\prime }$ it solves the homogeneous Helmholtz equation. Integrating its radial derivative over a small sphere gives $-1$ in the zero-radius limit, verifying the sign of the Dirac delta distribution. At infinity it satisfies $({\partial }_R-ik)G_k=o(R^{-1})$.

For the positive source on the right-hand side, the outgoing field isApplying the Helmholtz equation operator under this source representation gives $Q$. The minus sign is essential with the displayed Green function convention; defining the Green function to satisfy $LG=+\delta$ instead would absorb it. Here $A$ is the solid ball of radius $r_0$, as required by the volume source equation. Its finite volume and $Q\in L^2(A)$ make $Q$ integrable. The Sommerfeld radiation condition specifies the outgoing solution; without it one could add a homogeneous Helmholtz equation solution.

Write $\mathbf{r}=R\mathbf{\theta }$, $|\mathbf{\theta }|=1$. Uniformly for ${\mathbf{r}}^{\prime }$ in the source ball,For fixed $k$ and $r_0$, the outgoing Green function for the three-dimensional Helmholtz equation therefore has the far-field approximation for an outgoing sourceSubstitution into the source integral yieldsThus the source-to-far-field operator at fixed frequency isIt samples the Fourier transform of the source on the sphere of radius $k$. Its square-integrable kernel makes it a Hilbert-Schmidt operator, hence a compact operator. In factUse complex inner products linear in the first argument: $(f,g)=\int f\overline{g}$. Interchanging the volume and surface integrals givesHence the adjoint source-to-far-field operator isThis is a normalized Herglotz wave function; the change in exponential sign comes from the complex conjugation in the adjoint operator.

Use the outgoing Green function for the three-dimensional Helmholtz equationLet ${\mathbf{n}}^{\prime }$ point out of the obstacle $D$ into the exterior. Apply Green second identity to the field and $G_k$ in a large exterior region. Their Sommerfeld radiation conditions make the outer-sphere contribution vanish. The normal on the exterior region's inner boundary is $-{\mathbf{n}}^{\prime }$, so the Kirchhoff–Helmholtz representation isThe first integral is the incident source field ${\psi }_i$. Therefore the scattered field is the surface term. Since ${\psi }_i$ solves the homogeneous Helmholtz equation within the obstacle, its surface term is zero at exterior observation points; the same representation can be written using only scattered-field traces:The unknown boundary condition determines these traces, but the representation itself does not require choosing that condition.

For unit plane wave incidence ${\psi }_i=e^{i{k}_0{\hat{\mathbf{r}}}_0\cdot \mathbf{r}}$, define the scattering wavevector difference $\mathbf{q}=k_0({\hat{\mathbf{r}}}_0-\hat{\mathbf{r}})$ andThe outgoing Green function for the three-dimensional Helmholtz equation givesThe Born far-field pattern is $A$, a Fourier transform sample of the potential. Expanding the far-field Rytov exponential in $1/r$ gives the same leading outgoing term, while the logarithmic representation preserves its accumulated phase and amplitude correction.

Integral identity and representation. Define $L={\nabla }^2+k^2$ and take $n$ to be the outward unit normal. The product rule givesThe $k^2\varphi \psi$ terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:

Choose the Dirichlet Green function with the sign conventionwhere $\delta$ is the three-dimensional Dirac delta. Substitute $\psi =G$, use $L\varphi =f$, and evaluate the Dirac delta integral. The boundary term containing $G$ vanishes and $\varphi =g$ on $S$, leavingHenceThe choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen $k^2$ is not a Dirichlet eigenvalue of $-{\nabla }^2$; at such an eigenvalue a Green inverse for arbitrary data does not exist.

Infinite space. Put $r=|x-x^{\prime }|$. Away from the source, spherical symmetry reduces the Helmholtz equation toFor $k\ne 0$ its general radial solution isIntegrate $LG=\delta$ over a ball of radius $\epsilon$ about the source. The divergence theorem yieldsSince $G=A/r+Bk+O(r)$, the volume term is $O({\epsilon }^2)$ and the flux tends to $-4\pi A$. Therefore $A=-1/(4\pi )$, while $B$ remains free:The $B$ term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence $e^{-i\omega t}$, the outgoing Green function for the three-dimensional Helmholtz equation is $-e^{ikr}/(4\pi r)$, corresponding to $B=-i/(4\pi )$.

Zero wavenumber. For fixed finite $B$, $\sin (kr)/r\to 0$ as $k\to 0$, andThis is the decaying Green function of the Laplacian. Directly at $k=0$, the general radial solution is $A/r+C$; source normalization fixes $A=-1/(4\pi )$, and decay at infinity removes $C$. If the free parameter depends on $k$ as $B(k)\sim C/k$, the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

The outgoing Green function for the three-dimensional Helmholtz equation satisfiesIts normalization follows by integrating over a small ball about $x$: $G_F\sim 1/(4\pi |y-x|)$, so its outward derivative has flux $-1$. Apply Green's second identity in the $y$ variables:The ${\omega }^2$ terms cancel. Thus the surface integral equals $-\Psi (x)$ for $x\in V$, and zero for $x$ 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 $\rho$. Interpret the first far-field condition as $\Psi =O({\rho }^{-1})$, and the radiation limit in its usual direction-uniform sense, namely the Sommerfeld radiation conditionThis is the finite-bound interpretation of the PDF's first condition. For fixed $x$ and real $\omega$,Indeed $|y-x|=\rho -\hat{y}\cdot x+O({\rho }^{-1})$, and its radial derivative is $1+O({\rho }^{-2})$. On the large hemisphere the integrand can therefore be rewritten asIts area is $2\pi {\rho }^2$, so its integral tends to zero. The planar integral, taken over expanding disks if needed, consequently equals $-\Psi (x)$ for $x_3>0$ and zero for $x_3<0$.

The outward unit normal vector on the bottom plane is $-e_3$. Denote this planar integral byFor $x_3>0$ reflect the source point to $x^{\ast }=(x_1,x_2,-x_3)$. On the plane,Adding $I(x)=-\Psi (x)$ and $I(x^{\ast })=0$ cancels the boundary values of $\Psi$ and gives $2\int G_F{\partial }_3\Psi =-\Psi (x)$. Thus the Neumann representation for a radiating Helmholtz half-space isThe negative sign is consistent with using the upward derivative ${\partial }_3\Psi$ rather than the outward normal derivative. If that upward derivative is zero everywhere on the plane, the integral is zero. Hence $\Psi =0$ throughout the upper half-space in the specified radiating class; taking a difference also gives uniqueness for prescribed Neumann boundary data.

The $k_0^{2}uv$ terms cancel in $u{L}_{H}v-v{L}_{H}u$. For a ball $B_R$, the product rule and divergence theorem give Green's second identity in the formThus the whole-space integral vanishes when this boundary term tends to zero. For example, uniform bounds $u,v=O(R^{-1-ϵ})$ and $\nabla u,\nabla v=O(R^{-1-ϵ})$, with $ϵ>0$, make it $O(R^{-2ϵ})$.

Pointwise vanishing alone, as stated in the source, is insufficient. Smooth fields agreeing with $u=\sin (k_{0}r)/r$ and $v=\cos (k_{0}r)/r$ outside a ball both tend to zero, but their distant boundary integral is $-4\pi k_0$. One may smooth them inside the ball without changing that flux. The requested zero identity therefore needs sufficient decay or boundary/radiation conditions that actually eliminate the boundary term.

For the generalized plane wave eigenfunctions, their orthogonality is distributional:Here $\int e^{iqx}dx=2\pi \delta (q)$ is the Fourier representation of the Dirac delta function. These eigenfunctions are not ordinary square-integrable functions or decaying test fields.

Expand $\Phi =\int A_k{\psi }_k{d}^{3}k$ and define $V_k=\int {\psi }_k^{\ast }(r^{\prime })V(r^{\prime })d^3{r}^{\prime }$. Since $L_H{\psi }_k=(k_0^2-k^2){\psi }_k$, projection of the source equation gives $(k_0^2-k^2)A_k=V_k$. Formally dividing and substituting back givesAt the shell $k=k_0$, a pole prescription is required; homogeneous solutions can otherwise be added. The formula is an inverse of the Helmholtz equation with that convention still to be selected.

Let $H$ denote the kernel in this source formula, so $L_{H}H=+{\delta }^{(3)}$. Put $l=|r-r^{\prime }|$ and align the Fourier polar axis with $r-r^{\prime }$. Angular integration givesConsequently the radial Fourier representation of the Helmholtz Green function givesThe last Green function written in the PDF has the opposite denominator. It is obtained by setting $G=-H$, the common convention $L_{H}G=-{\delta }^{(3)}$:Thus the two displayed source and Green formulas are consistent only when this minus sign is retained; $G$ cannot also be the positive-delta inverse kernel.

Finally the integral is not an ordinary improper integral across $k=k_0$. A Cauchy principal value yields $G_{\mathrm{PV}}(l)=\cos (k_{0}l)/(4\pi l)$. The outgoing prescription $k^2-k_0^2-i0$ yields the outgoing Green function for the three-dimensional Helmholtz equation, $G_{+}(l)=e^{i{k}_{0}l}/(4\pi l)$; the opposite sign gives the incoming wave. The Sokhotski–Plemelj formula accounts for their difference. The source does not specify which prescription is intended, so the boxed radial expression must be read with one chosen consistently.

Impose a homogeneous Dirichlet boundary condition on the Green function in its first variable, $G(r,r^{\prime })=0$ for $r\in S$. The second boundary term then vanishes, givingIn a bounded domain, a Dirichlet Green function requires $k^2$ not to be a Dirichlet eigenvalue of $-\Delta$; otherwise a resonance can obstruct its existence. The representation assumes such a Green function exists.

In all space, put $\rho =|r-r^{\prime }|$ and try $G=A{e}^{ik\rho }/\rho$. For $\rho >0$, direct differentiation gives $({\rho }^2{G}^{\prime }{)}^{\prime }+k^2{\rho }^{2}G=0$, so the Helmholtz equation holds away from the source. Integrate its inhomogeneous version over a small source-centred ball. The $k^{2}G$ volume integral tends to zero, while the flux isIt must equal the integrated delta, one. ThereforeThis is the negative of the convention used in the outgoing Green function for the three-dimensional Helmholtz equation, where the source is $-\delta$. The sign here follows the present $+\delta$ convention. The factor $e^{ik\rho }$ chooses the outgoing solution, with time convention $e^{-i\omega t}$ and the Sommerfeld radiation condition; one could add homogeneous solutions without a radiation prescription.

The first scattered correction to the incident wave isInserting it into the amplitude gives a second, positive contribution because the two minus signs multiply:Thus the amplitude through second Born order is $f^{(1)}+f^{(2)}+O(V^3)$. The intervening outgoing Green function for the three-dimensional Helmholtz equation describes propagation between the two interactions; replacing it by a real standing-wave kernel would discard the imaginary part needed for unitarity.

For $(\Delta +k^2)\psi =Q$, the outgoing Green function for the three-dimensional Helmholtz equation yields the far-field patternThe sign corresponds to the Green function satisfying $(\Delta +k^2)G=-\delta$. This Hilbert-Schmidt operator samples the source Fourier transform on the sphere $|\mathbf{\xi }|=k$; for a ball of radius $r_0$, its squared Hilbert-Schmidt norm is $r_0^3/3$.