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For real $c$ and $L=\Delta +c$ on a bounded regular domain, let $N$ be the zero-boundary kernel. The problem $Lu=f$, $u=\psi$ on the boundary is solvable exactly when ${\int }_{\Omega }fz+{\int }_{\partial \Omega }\psi {\partial }_{\nu }z=0$ for every $z\in N$. The sign follows from Green second identity. After subtracting a boundary extension of $\psi$, this is the usual orthogonality of the forcing to the null space of the self-adjoint Dirichlet realization. All solutions then differ by an element of $N$.

For the positive Laplacian convention $\Delta G=\delta$, the method of images in the first quadrant reflects an interior pole $(a,b)$ across each axis with negative sign and across both with positive sign:The Dirichlet Green function vanishes on both axes. Its outward normal derivative on the horizontal axis isand the vertical-axis kernel follows by exchange of coordinates. Green second identity converts these derivatives into the quadrant Poisson integral for boundary data.

A global relation is an identity connecting transforms or integrals of the boundary traces of a solution. For a formally self-adjoint operator $L=\Delta -k^2$, Green second identity givesfor every homogeneous adjoint solution $Lv=0$. Exponential adjoint families turn this into a spectral identity between the Dirichlet boundary condition and unknown normal derivative.

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.

Retain $a=p_0^{1/\gamma }/{\rho }_0$ from part (b), and use time translation invariance to write $G(\mathbf{x},t;\mathbf{y},\tau )=g(\mathbf{x},\mathbf{y};t-\tau )$. Define the temporal Fourier transform byMultiplying the transformed Green function equation by $-a(\mathbf{x})$ givesThis is a symmetric divergence-form spatial operator, although the original unweighted operator need not be symmetric in the ordinary volume measure.

Let $G_j(\mathbf{x})=\tilde{G}(\mathbf{x},{\mathbf{y}}_j;\omega )$. The product rule gives the bilinear Green second identityThe frequency terms cancel. Integrating over the domain, the right side becomesThe boundary integral is zero for common homogeneous Dirichlet boundary condition, Neumann boundary condition or reciprocal Robin boundary condition conditions. In an unbounded domain, use the same outgoing limiting-absorption prescription for both Green functions; it gives the corresponding vanishing boundary pairing. This identity has no complex conjugation: it proves transpose wave reciprocity, not a Hermitian or time-reversal identity. We obtain the weighted acoustic Green-function reciprocityThe factor is frequency independent, so inverse Fourier transform gives the same relation at equal time lag. Both time arguments below have lag $t-{\tau }_2$, henceReciprocity exchanges source and receiver while preserving elapsed time; it does not turn a causal response into an advanced one.

Fix the convention $\Delta q-m^{2}q=0$, $m>0$, and let $c_m=m^2/4$. With $z=x+iy$, two families of modified Helmholtz adjoint plane waves areBoth satisfy $\Delta E_{\pm }=m^2{E}_{\pm }$, since the products of their $z$ and $\bar{z}$ exponents are $m^2/4$. They obey $E_{-}(z,k)=\overline{E_{+}(z,\bar{k})}$. Green second identity gives the two conjugate global relations for the modified Helmholtz equationFor real boundary traces, ${\mathcal{G}}_{-}(k)=\overline{{\mathcal{G}}_{+}(\bar{k})}$. Thus the second identity is the conjugate spectral companion, rather than an unrelated extra boundary condition. In complex differential form the first relation is equivalentlyIts integrand is a closed one-form: differentiating its coefficients gives $2E_{+}(q_{z\bar{z}}-c_{m}q)=0$. This supplies a direct derivation of the global relation as well as its outward-normal sign convention.

Write the known Dirichlet boundary condition on side $j$ as $f_j(s)$ and the unknown outward normal derivative as $h_j(s)$, with $-1<s<1$. For any linear combination $v$ of the adjoint waves, the global relation isExpanding the $h_j$ in a finite basis and enforcing these identities at selected collocation points for a global relation gives a linear system with a known right side. The square permits especially simple paired tests.

Let $p_n=n\pi /2$, ${\eta }_n=\sqrt{m^2+p_n^2}$, and ${\phi }_n(s)=\sin [p_n(s+1)]$ for $n\ge 1$. Define four real adjoint solutionsThey satisfy the modified Helmholtz equation because ${\eta }_n^2-p_n^2=m^2$. Each is a linear combination of two adjoint exponentials, hence is obtained from the two spectral global relations. More explicitly, put ${\rho }_n^{\pm }=({\eta }_n\pm p_n)/2$, so ${\rho }_n^{+}{\rho }_n^{-}=c_m$. For the $E_{+}$ convention above, the required spectral pairs areAppropriate phase-weighted differences produce $\sin [p_n(s+1)]$. For example the top test is $e^{-{\eta }_n}[e^{i{p}_n}{E}_{+}({\rho }_n^{-})-e^{-i{p}_n}{E}_{+}({\rho }_n^{+})]/(2i)$. The conjugate relation ensures a real system for real data. This is sine collocation of square modified Helmholtz global relations.

The functions ${\phi }_n$ are an orthonormal Fourier sine basis on $[-1,1]$. Set $h_{j,n}={\int }_{-1}^1{h}_j{\phi }_{n}ds$ and compute the known quantity $R_{j,n}={\int }_{\partial \Omega }f{\partial }_n{v}_{j,n}ds$ by numerical integration. Since ${\phi }_n(\pm 1)=0$, each adjoint test vanishes on both adjacent sides. On its own side it equals ${\phi }_n$, and on the opposite side it equals ${\delta }_n{\phi }_n$, where ${\delta }_n=e^{-2{\eta }_n}$. Thus the square modified Helmholtz Dirichlet-to-Neumann coefficients obeyEach block is inverted explicitly: for opposite sides $j,j^{\prime }$,Compute these coefficients for $1\le n\le N$ and reconstruct $h_j^{(N)}={\sum }_{n=1}^N{h}_{j,n}{\phi }_n$. This is a semi-analytical scheme: the spectral basis integrals and two-by-two inverses are explicit, while the known boundary integrals are evaluated numerically. Increase $N$ and the quadrature resolution until the desired convergence is observed; compatible smooth side data have convergent normal-trace expansions, while corner singularities require the usual weaker trace interpretation and more careful quadrature.

The requested weak interaction between sides is particularly clear: the adjacent-side unknown traces contribute exactly zero, and the opposite-side coefficient is $e^{-2\sqrt{m^2+(n\pi /2{)}^2}}\to 0$. The own-side coefficient stays equal to one. Thus diagonal dominance of paired square global-relation collocation is genuine after pairing; individual uncombined exponential samples need not have the same conditioning. The eigenvalues of each block are $1\pm {\delta }_n$, so its condition number is $(1+{\delta }_n)/(1-{\delta }_n)$.

There is also a direct localization check for an individual adjoint wave. On side $j$ with outward unit normal $n_j$ and tangent $t_j$, the normalized test $e^{\eta (n_j\cdot r-1)+ip{t}_j\cdot r}$ has $\eta =\sqrt{m^2+p^2}$. Its opposite-side magnitude is $e^{-2\eta }$, and its magnitude integrates to at most $1/\eta$ along either adjacent side. This side localization of modified Helmholtz plane waves explains why suitable large spectral parameters suppress remote-side effects even before the exact sine cancellation.

Finally the interior solution can be evaluated from the computed traces using the two-dimensional modified Helmholtz fundamental solutionHere $K_0$ is the Modified Bessel function of the second kind, and $(-\Delta +m^2){\Gamma }_m=\delta$. Replace $h$ by its computed finite Fourier sine series and use numerical integration; at interior points the boundary kernels are smooth. The sign follows from Green second identity with this fundamental-solution convention. This completes the numerical integration of the boundary value problem with a Dirichlet boundary condition, rather than stopping at an equation for its unknown boundary derivatives.

The product rule for the divergence givesSubtracting cancels the two gradient products. ThereforeTo obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is $\hat{\mathbf{r}}$; on the inner sphere it is $-\hat{\mathbf{r}}$. Write ${\partial }_r=\hat{\mathbf{r}}\cdot \nabla$ on both spheres. Then the resulting Green second identity isThe minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

With the positive logarithmic fundamental solution used here, a Dirichlet Green function for an interior pole ${\mathbf{r}}_0$ satisfiesEquivalently, $G-G_0$ is harmonic near the pole, $G$ is harmonic away from it, and it has the specified logarithmic singularity. Regularity up to the remaining boundary is understood; on an unbounded domain impose the appropriate condition at infinity as well.

Apply Green second identity with $f=\psi$ and $g=G$. Since $\psi$ is a harmonic function, the volume integral is $\int \psi {\delta }_{{\mathbf{r}}_0}dA=\psi ({\mathbf{r}}_0)$. The term $G{\partial }_n\psi$ vanishes on the boundary by the Dirichlet boundary condition. ThereforeOne may equivalently excise a small disc around the pole and take its radius to zero. The sign is positive because this question uses ${\nabla }^{2}G=+\delta$, not the alternative negative-Laplacian convention.

The formal adjoint of $\mathcal{L}=\Delta -k^2$ is itself. With the bilinear convention used in Green second identity, take any complex-valued $v$ satisfying $\mathcal{L}v=0$. The product rule for divergence gives the familyEquivalently, the differential one-formis a closed differential form. Arbitrary linear combinations of adjoint solutions give further members of this family. One can also write it as $\nabla \cdot [v^2\nabla (u/v)]=0$ wherever $v\ne 0$, but the first formula has no division and remains valid at zeros of $v$. For a sesquilinear convention replace $v$ by $\overline{v}$; no such conjugation is built into the bilinear identity used below.

Let $L=\Delta +c$ with real $c$, and letThe Fredholm alternative for an elliptic Dirichlet problem says that $N$ is finite-dimensional. If $N=0$, all the prescribed data admit a unique solution. In general the exact boundary compatibility in the self-adjoint Fredholm alternative isThese conditions are necessary and sufficient; when they hold, all solutions form $u_0+N$. Thus in the second branch some data are incompatible, while compatible data have nonunique solutions. The outward normal derivative fixes the displayed sign.

To prove the alternative, put $X=\{w\in C^{2,\alpha }(\overline{\Omega }):w{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The standard zero-boundary Poisson equation theorem makes $\Delta :X\to Y$ an isomorphism. Multiplication by $c$, followed by ${\Delta }^{-1}$, is a compact operator $K:X\to X$: the embedding $C^{2,\alpha }\hookrightarrow C^{0,\alpha }$ is compact, and multiplication is bounded. Writing $u=\psi +w$ givesThe Fredholm alternative for a compact operator implies finite kernel and cokernel of equal dimension, and invertibility exactly when the kernel vanishes.

For the explicit compatibility conditions, use the self-adjoint Dirichlet realization of $L$ on $L^2$. It has compact resolvent: a sufficiently large positive shift of $-L$ is coercive, its inverse gains two derivatives, and the Rellich-Kondrachov compactness theorem makes that inverse compact. The Fredholm solvability condition for a self-adjoint operator is $F\perp N$. Standard boundary elliptic regularity upgrades the resulting weak solution to $C^{2,\alpha }$ for these data and boundary. Thus it gives the same kernel and solvability as the preceding Hölder-space formulation.

Finally Green second identity, with $z=0$ on the boundary and $Lz=0$, givesConsequently $\int zF=\int zf+\int \psi {\partial }_{\nu }z$, proving both necessity and sufficiency with the stated sign. If using complex data and real $c$, replace $z$ by $\overline{z}$ in the pairings.

Remove a small disc about $x$ and apply Green second identity to $u(x_0)$ and $G_0(x;x_0)$. Since ${\nabla }_{x_0}^2{G}_0=\delta (x_0-x)$ and ${\nabla }^{2}u=-\rho$, the shrinking inner boundary contributes $u(x)$ and gives

For the unit disc, the method of images gives the Dirichlet Green functionwhich vanishes on the boundary by the hinted identity. The boundary representation with $u(1,{\theta }_0)=\delta ({\theta }_0-\alpha )$ evaluates the normal derivative of $G_D$ at ${\theta }_0=\alpha$, giving the Poisson kernel

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.

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

For $m>0$, the decaying fundamental solution of $-\Delta +m^2$ in two dimensions is $K_0(mr)/(2\pi )$, with $K_0$ the Modified Bessel function of the second kind. Green second identity reconstructs an interior solution from its Dirichlet boundary condition $f$ and outward normal derivative $h$ as $q(r)={\int }_{\partial \Omega }({\Gamma }_{m}h-f{\partial }_n{\Gamma }_m)ds$.

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.