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Apply Green second identity to $u$ and the free-space Green function $G_{fs}$ on $V$ with a small ball about ${\mathbf{r}}_0$ removed. Both functions are harmonic there, so only boundary terms remain. The contribution from the small sphere tends to $u({\mathbf{r}}_0)$ becausehas unit delta source, while the remaining small-sphere term vanishes. Taking the radius to zero gives Green's third identity

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Away from $x=\xi$, the Green function solves $G^{\prime \prime }-G=0$. The solution satisfying the left boundary condition is proportional to $\sinh x$, while the solution decaying at infinity is proportional to $e^{-x}$. Continuity at $\xi$ and the unit derivative jumpgiveor $G=-\sinh (\min \{x,\xi \})e^{-\max \{x,\xi \}}$, the dirichlet half-line Green function for d2 minus 1.

The required solution isEvaluation of the two elementary integrals, split at $\xi =x$, givesIndeed, direct differentiation gives $y^{\prime \prime }-y=e^{-2x}$, and both boundary conditions are immediate.

The causal Green function for the heat operator iswhere $H$ is the Heaviside step function. It vanishes for $t<\tau$, satisfies the homogeneous heat equation away from the source, and approaches $\delta (x-\xi )$ as $t\downarrow \tau$. Superposing the responses to all infinitesimal sources gives Duhamel principle:The lower limit and causality give the required homogeneous initial data.

After separating the conserved horizontal factor as $\psi =e^{ipx}f(z)$, the Helmholtz equation becomeswhere $H$ is the Heaviside step function. The outgoing Green function for $d^2/d{z}^2+q_0^2$ isThe Born approximation at first order replaces $f$ on the right-hand side by the incident profile $f_i(z^{\prime })=e^{-i{q}_0{z}^{\prime }}$. For $z>0$ this givesThe integral is understood with the usual outgoing-wave convergence factor. Sincewe obtainTherefore the Born reflected field is

The quadratic worldsheet action makes each $X^{\mu }$ a free boson. Its Green function in the complex plane inverts the operator in the action and uses $\partial \bar{\partial }\log |z-w{|}^2=2\pi {\delta }^{(2)}(z-w)$ with the corresponding convention. The result isAn additive constant is physically irrelevant because it can be absorbed into the zero mode.

Use the sign convention in the question,Then the total field satisfiesThe outgoing free-space Green function iswith $({\nabla }^2+k_0^2)G_0=-\delta$. Hence the Lippmann-Schwinger equation isReplacing the unknown interior total field by the incident field gives the first Born approximation

For the Rytov approximation, put $\psi ={\psi }_i{e}^{\varphi }$. After division by $\psi$, the wave equation givesNeglecting the quadratic term $(\nabla \varphi {)}^2$ makes ${\psi }_i\varphi$ obey the same inhomogeneous equation as the first Born scattered field. ThusandIts power series beginsso the two approximations agree through first order in the scattering potential.

After integration by parts, the quadratic operator in the action is $K=-{\pi }^{-1}\partial \bar{\partial }$. Its Green function $G(z,w)=⟨X(z,\bar{z})X(w,\bar{w})⟩$ therefore satisfiesUsing the given distributional identity giveswhere the additive constant depends on the infrared convention and cancels from neutral correlators.

For a centered real scattering potential fluctuation $W$ and the imaginary part $b$ of the incident-field-weighted Green function, $\phi (\mathbf{r})=-\int b(\mathbf{r},{\mathbf{r}}^{\prime })W({\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }$. Its mean vanishes, and its covariance function is the double integral of the two deterministic kernels against the covariance function of $W$. The correlation of the refractive index itself is insufficient when the scattering potential depends quadratically on that index.

For a free massive Proca field in the mostly-plus Minkowski metric, the covariant inverse of the quadratic action and the canonical time-ordered product of vector potentials differ locally. With Fourier transform $e^{iq\cdot (x-y)}$, the covariant Proca propagator is $-i({\eta }_{ab}+q_a{q}_b/m^2)/(q^2+m^2-i0)$, while the literal canonical two-point correlation function subtracts $i{\delta }_a^0{\delta }_b^0/m^2$ in momentum space. The time component is nondynamical; its physical-mode numerator is $|\mathbf{q}{|}^2/m^2$, rather than $-1+(q^0{)}^2/m^2$. The difference is an instantaneous Dirac delta function, not another propagating state. Both expressions agree away from coincidence. This distinction matters when identifying a covariant Green function with a canonical correlator in a constrained field theory.

For receiver-dependent Green functions $g_{\mathbf{\zeta }}(\mathbf{z})$ and aperture weight $a(\mathbf{\zeta })$, define ${\mathcal{W}}_A(\mathbf{Z},\mathbf{p})=\int a(\mathbf{\zeta })W[g_{\mathbf{\zeta }},g_{\mathbf{\zeta }}](\mathbf{Z},\mathbf{p})d\mathbf{\zeta }$. Its inverse separation Fourier transform is the kernel of the time reversal operator on the source plane.

Let $P$ be a monic polynomial of positive degree $m\ge 1$ with distinct roots of a polynomial $r_j$. The partial fraction decompositionfollows by matching the simple pole residues; the difference is entire and vanishes at infinity. Comparing its expansion there gives ${\sum }_j{r}_j^{\ell }/P^{\prime }(r_j)=0$ for $0\le \ell <m-1$ and ${\sum }_j{r}_j^{m-1}/P^{\prime }(r_j)=1$. These identities give the continuity and derivative jumps needed for constant-coefficient Green functions.

An Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$ changes $\partial \cdot A$ by $\square \lambda$. Consequently gauge fixing by a quadratic function of $\partial \cdot A$ retains transformations satisfying $\square \lambda =0$, subject to the same boundary conditions. Eliminating those residual transformations is part of specifying an inverse with a particular Green function prescription.

For $\mathrm{Re}p>0$, let $r_{-}$ and $r_1,r_2$ be the characteristic roots of $r^3+r+p=0$, split by the signs of their real parts. Put $d_j=3r_j^2+1$. The real-line Green function of $p+{\partial }_x+{\partial }_x^3$ isThe reciprocal-polynomial root identities imply continuity of $R_p,R_p^{\prime }$ and jump $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. The distributional derivative therefore gives $(p+{\partial }_s+{\partial }_s^3)R_p={\delta }_0$. The transport term changes the roots: the pure Airy resolvent kernel with $r^3+p=0$ cannot be substituted unchanged.

For a positive quadratic bending operator $K$ with a real orthonormal basis of eigenfunctions, the equipartition theorem gives $C(x,y)=k_{B}T{\sum }_n{W}_n(x)W_n(y)/{\mu }_n$, where $K{W}_n={\mu }_n{W}_n$. This is the inverse-operator Green function multiplied by Boltzmann constant and temperature. Unconstrained zero-energy filament modes prevent a normalizable canonical ensemble, and negative modes signal an unstable quadratic model.

For both ends clamped and $K=A{\partial }_x^4$, the diagonal is $C(x,x)=k_{B}T{x}^3(L-x{)}^3/(3A{L}^3)$. This differs from a clamped-free tip variance. The center variance is $k_{B}T{L}^3/(192A)$.

In a reciprocal medium, appropriately normalized source and receiver responses can be exchanged: the scalar Green function satisfies $G(\mathbf{r},{\mathbf{r}}^{\prime })=G({\mathbf{r}}^{\prime },\mathbf{r})$. Reciprocity alone does not imply absence of wave absorption.

For an operator $c^{-2}{\partial }_t^2-a^{-1}\nabla \cdot (a\nabla )$ with positive stationary coefficients and reciprocal boundary conditions, its delta-normalized Green function satisfiesMultiplying its frequency-domain equation by $a$ and using the bilinear Green second identity proves the relation. Both Green functions must use the same outgoing or causal prescription. Exchanging source and receiver preserves time lag, and involves no complex conjugation.