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Use complex amplitudes with time dependence $e^{i\omega t}$, so the outgoing two-dimensional Helmholtz equation kernel has phase $e^{-i{k}_{0}r}$. Dividing the harmonic form of Lighthill acoustic analogy by $c_0^2$ and integrating the two source derivatives by parts gives an amplitude proportional toOnly the magnitude is needed here; the specified kernel omits its constant phase and normalization. In the acoustic compact-source approximation, $\int {\tilde{T}}_{ij}{d}^{2}y=O({\rho }_0{U}^2{\ell }^2)$. With $\omega =O(U/\ell )$, hence $k_0\ell =O(m)$, the two-dimensional compact quadrupole scaling isThe half-power difference from three dimensions comes from cylindrical spreading, including its $k_0^{-1/2}$ factor. As in part (i), the Mach number power is stated with the geometrical range factor separated; $k_{0}r\gg 1$ is still required.

Use the Fourier transform convention $\hat{p}(k,y)={\int }_{\mathbb{R}}{e}^{ikx}p(x,y)dx$. Let $\gamma (k{)}^2=k^2-k_0^2$, choosing $\mathrm{Re}\gamma >0$ on the real contour with $\mathrm{Im}\omega <0$. For time dependence $e^{i\omega t}$, this is the decaying continuation of the outgoing Sommerfeld radiation condition.

The transformed Helmholtz equation has upper and lower solutions $P(k)e^{-\gamma y}$ and $Q(k)e^{\gamma y}$. The two kinematic traces give $-\gamma P={\rho }_0{\omega }^2\hat{\eta }=\gamma Q$, henceThe elastic membrane equation gives $[\hat{p}]=(m{\omega }^2-T{k}^2)\hat{\eta }$. Therefore the acoustic wave on a tensioned massive membrane has dispersion relationEquivalently, $T{k}^2={\omega }^2(m+2{\rho }_0/\gamma )$. The fluid on both sides supplies a positive added mass of an evanescent fluid layer, which is a useful independent check on the sign. In particular, the incident pressure isThe symbol $m$ here denotes elastic membrane mass per area, rather than the fluctuating Mach number used in Question 1.

Linearity permits subtraction of the incident acoustic membrane wave. The scattered pressure solves the homogeneous Helmholtz equation, and the kinematic relation defines its line displacement ${\eta }^{\prime }$ on both halves. On $x<0$, subtracting the incident elastic membrane equation gives the scattered dynamic equation. On $x>0$, there is no elastic membrane, so the total pressure jump must vanish, giving $[p^{\prime }]=-[p_I]$. The pinned endpoint has $\eta (0)=0$, hence ${\eta }^{\prime }(0)=-1$, rather than zero.

Use full and Half-range Fourier transforms with the same $e^{ikx}$ convention as part (a), and putWrite $P^{-}$ for the left transform of the upper scattered pressure, $P^{+}$ for its right transform, and ${\eta }^{\pm }$ for the corresponding transforms of ${\eta }^{\prime }$. Outgoing antisymmetry in $y$ impliesThe second relation comes from the prescribed pressure cancellation on $x>0$.

The endpoint terms in the left Fourier transform of ${\eta }_{xx}^{\prime }$ are crucial:Thus $2P^{-}=a{\eta }^{-}+TA+iT(k+k_I)$. Eliminating ${\eta }^{-}$ gives the Wiener-Hopf equationThere is a defect in the printed right side: with the stated incident exponential, pressure jump, and pinned total displacement, it omits the incident endpoint term and has the opposite sign on the incident-pressure term. The displayed corrected equation follows directly from both boundary traces. The given right side cannot be derived with these definitions. Using $a(k_I)=-2\beta /{\gamma }_I$, an especially useful equivalent form isThe correction is necessary for the reconstructed pressure to satisfy the original physical boundary conditions.

For Wiener-Hopf factorization, take $K=K^{+}{K}^{-}$ with the plus factor analytic and nonzero in the upper half-plane and the minus factor analytic and nonzero in the lower half-plane. Singularities listed next refer to continuation out of each factor's own analytic half-plane. Let $b=\omega \sqrt{m/T}$, so $a=-T(k-b)(k+b)$ and $b$ lies below the real axis. Generically $K^{+}$ inherits the pole at $b$ and the square-root branch point at $k_0$, while $K^{-}$ inherits the pole at $-b$ and the branch point at $-k_0$. The kernel is finite but nonanalytic at the acoustic branch points: its local behavior is a constant plus a square-root term, not an inverse-square-root divergence. Coincident branch points and poles require a limiting treatment.

Zeros of $K$ are the fluid-loaded dispersion relation roots. Lower-half-plane roots, including the incoming guided root $k_I$, belong to the continued $K^{+}$; upper-half-plane roots, including the reflected root $-k_I$ for the even kernel, belong to the continued $K^{-}$. The roots represent membrane-guided modes; the branch cuts represent radiated acoustic waves. Only roots on the selected outgoing sheet are included, not spurious roots created by squaring the dispersion relation. No explicit factorization is required.

Here is an explicit solution in terms of those factors. DefineThe pole of $B$ at $-b$ has residue $C$; the pole at $b$ is canceled by the pole of $K^{+}$. Thus $B^{\pm }$ have the required respective analyticity. Divide the corrected Wiener-Hopf equation by $K^{+}$ and split its right side as $F^{+}+F^{-}$, whereThe difference quotient in $F^{-}$ is removable at $k_I$ and is analytic below. Moving the plus and minus terms to opposite sides yields the common entire function. With the assumed $E(k)=0$, the solution isCombining $P^{-}$ with the known $P^{+}$ gives the full scattered pressure transformThe requested pressure integral, for $y\ne 0$, is consequentlyThe real contour $\mathcal{C}$ uses the causal continuation $\mathrm{Im}\omega <0$; its undamped limit retains the induced pole and branch-cut prescriptions. The residue theorem shows that upper-half-plane zeros of $K^{-}$ yield left-going scattered elastic membrane modes. On the right, the lower incident-pole residue of the scattered integral cancels $p_I$ at the open line, as required.

Finally reconstruct the left scattered displacement:For a generic unspecified $A$, this has a lower-half-plane pole at the bare elastic membrane wavenumber $b$. Such a pole is incompatible with analyticity of an outgoing left-supported scattered displacement: it would represent an additional right-going incoming elastic membrane contribution. It must be removed. Its residue is $-[2P^{-}(b)-TA-iT(b+k_I)]/(2Tb)$, so the incoming-pole cancellation at a pinned membrane edge condition isSince $P^{-}$ is linear in $A$, this fixes the endpoint slope generically. Explicitly it isThe physical lower incident pole of the total displacement is already prescribed by ${\eta }_I$; it must not be confused with this removable spurious pole of the scattered field.

Use the time convention $e^{-i\omega t}$. The free-space Helmholtz equation, with $\psi =e^{ikx}E$, gives the exact reduced equationFor a forward plane wave at angle $\alpha$, $E_x=O(k{\alpha }^{2}E)$ and $E_{zz}=O(k^2{\alpha }^{2}E)$, whereas $E_{xx}=O(k^2{\alpha }^{4}E)$. Thus the paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$. The reduced field satisfies the parabolic wave equationWith the Fourier transform convention used below and its inverse, this becomes an ordinary differential equation for each transverse wavenumber:Consequently the initial-value solution isEquivalently, evaluating the oscillatory Gaussian integral gives the one-dimensional transverse Fresnel propagation formulaThe square-root branch has $\sqrt{1/i}=e^{-i\pi /4}$. The formula is an oscillatory integral for general data, or the Fresnel propagator acting on square-integrable functions. It approaches the initial field as $x\downarrow 0$. A plane wave has transverse wavenumber $\nu =k\sin \alpha$; the reduced axial phase $-{\nu }^{2}x/(2k)$ agrees with $k(\cos \alpha -1)x$ through order ${\alpha }^2$. This also checks the sign. The paraxial approximation applies to $|\nu |\ll k$.

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.

Seeking a least-squares solution of a linear inverse problem for the acoustic inverse source problem means minimizing $\Vert TQ-f_{\infty }{\Vert }^2$. Its normal equation for a linear inverse problem isFor compatible data, the formal minimum-norm solution can be written with the Moore–Penrose inverse of an operator:The inverse in this expression is defined on the range of the restricted operator, not as a bounded everywhere-defined inverse. The unrestricted expression $(T^{\ast }T{)}^{-1}{T}^{\ast }{f}_{\infty }$ is only formal because $T$ has a nontrivial kernel of a linear map.

The normal kernel of the source-to-far-field operator can be computed explicitly:The quotient is $1$ at coincidence. This smooth kernel already exhibits the loss of information.

There are two distinct causes of ill-posedness. First, a fixed-frequency nonradiating source gives exact nonuniqueness. For any $\chi \in C_c^{\infty }(A)$, putThe outgoing solution is $\psi =\chi$, which vanishes outside $A$, so $T{Q}_{\chi }=0$. Equivalently, integration by parts in the source-to-far-field operator at fixed frequency givessince the exponential solves the homogeneous Helmholtz equation and the boundary terms vanish. This provides an infinite-dimensional family of invisible sources. Single-frequency data determine only the spherical restriction of a three-dimensional Fourier transform.

Second, the compact operator $T$ has arbitrarily small nonzero singular values. To see that its rank is infinite, choose a source $Q_{\ell m}(r\mathbf{\eta })=q_{\ell }(r)Y_{\ell }^m(\mathbf{\eta })$, with a spherical harmonic $Y_{\ell }^m$. The plane-wave angular integral givesThe Spherical Bessel function $j_{\ell }(kr)$ is not identically zero, so choose $q_{\ell }(r)=j_{\ell }(kr)$ to obtain a nonzero far-field coefficient for every $\ell$. The mutually orthogonal angular modes prove infinite rank. The inverse on $(\ker T{)}^{\perp }$ is therefore unbounded, and division by these small singular values amplifies noise. Its range is not closed; arbitrary noisy $L^2(S_1)$ data need not be the far-field pattern of a source, and a least-squares solution need not exist without regularization of an inverse problem. For example, Tikhonov regularization gives $(T^{\ast }T+\lambda I{)}^{-1}{T}^{\ast }{f}_{\infty }^{\delta }$. It stabilizes the recovered visible component but cannot identify the missing fixed-frequency nonradiating source component.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

Since $x=r\cos \theta$, the function on the left of the defining expansion is $e^{ix}$. Thereforewhich is the two-dimensional Helmholtz equation.

In polar coordinates,Substitute the defining Fourier cosine series and equate the coefficient of each $\cos (n\theta )$. Its angular second derivative contributes $-n^2$, so every Bessel function $J_n$ satisfiesor equivalently the Bessel differential equation

Expand the exponential through cubic order and useThis givesComparing with $J_0+2{\sum }_{n\ge 1}{i}^n{J}_n\cos (n\theta )$ yields

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives

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.

Let $V=k_0^2(n^2-1)$ be the scattering potential supported in $D$. The background Helmholtz equation and its outgoing kernel give the scalar Lippmann-Schwinger equationReplacing the unknown field in the integral by the incident field gives the Born approximation for scalar wave scattering:For the Rytov approximation, work where ${\psi }_i\ne 0$ and set $\psi ={\psi }_i{e}^{\varphi }$. Dividing the wave equation by the total field and subtracting the incident equation givesNeglect the quadratic gradient term. If ${\varphi }_1={\psi }_{s,B}/{\psi }_i$, direct differentiation shows $\Delta {\varphi }_1+2\nabla \log {\psi }_i\cdot \nabla {\varphi }_1=-V$, soExpanding the exponential in the potential gives ${\psi }_R={\psi }_i+{\psi }_{s,B}+O(V^2)={\psi }_B+O(V^2)$. Thus both approximations have the same first-order field, although exponentiation retains a particular family of higher powers.

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

For a slowly varying envelope $E=\psi e^{-ikx}$, the paraxial approximation to the Helmholtz equation isBecause $n=1+\mu W$ and ${\mu }^2\ll 1$, passage through a sufficiently thin phase screen producesIts modulus is one at the screen exit. Beyond the screen, $n=1$, so the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. A Taylor expansion in propagation distance givesSincewe findIt follows thator, equivalently, $|E{|}^2=1-x{\varphi }^{\prime \prime }/k+O(x^2)$. Thus random phase curvature produces local focusing and defocusing: free-space diffraction converts phase fluctuations into amplitude fluctuations immediately after the screen.

DefineIf $A$ and $B$ commute, the Helmholtz equation operator factorizes asWhen $n$ varies with $x$, the exact product also contains the commutator $[A,B]$; neglecting it assumes longitudinal changes are slow. The forward-propagating factor isbecause it admits $e^{i{k}_{0}x}$ in a uniform medium.

Put $\psi =E{e}^{i{k}_{0}x}$. The one-way equation becomesForthe Taylor expansion $(1+Q{)}^{1/2}=1+Q/2+O(Q^2)$ givesThus the parabolic wave equation is

The expansion is accurate under the paraxial approximation: transverse wavenumbers satisfy $|k_z|/k_0\ll 1$, the envelope varies slowly on the carrier scale, the refractive-index contrast is weak enough for $Q^2$ to be negligible, and $n$ varies slowly in $x$ so $[A,B]$ is small. The one-way factor discards backward propagation and reflection; the square-root expansion additionally discards large-angle and higher-order diffraction, and it does not accurately represent strongly evanescent components or abrupt longitudinal interfaces.