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Use positive carrier wavenumber $k$ and retain waves traveling mainly in the positive $x$ direction. Factor out their rapid wave phase by writing $\psi =e^{ikx}E(x,z)$. Substitution into the Helmholtz equation is exact and givesFor a transverse Fourier mode $e^{i\nu z}$ at a small angle, $|\nu |/k\ll 1$ and its forward longitudinal wavenumber isThus the envelope varies along $x$ on scale $k/{\nu }^2$, much more slowly than the carrier wavelength, and $|E_{xx}|/|k{E}_x|=O({\nu }^2/k^2)$. Dropping this smaller term derives the parabolic wave equation,With transform $\hat{E}(x,\nu )=\int E(x,z)e^{-i\nu z}dz$, its free propagation isThis is a one-way paraxial approximation, not a replacement for backward waves or large-angle Helmholtz components. Its Fourier multiplier has unit modulus and therefore does not attenuate any admitted component.

Use the $L^2(S^2)$ data norm. The coefficients of the measurement error are $e_n^m=b_n^m-a_n^m$, so Parseval's identity givesAn unregularized continuation of these measurements to $r=R$ replaces $a_n^m$ by $b_n^m$ in the outgoing spherical harmonic expansion. Whenever this continuation has a finite trace, its error isChoose a measurement error supported on a single normalized spherical harmonic, with $e_N^0=\delta$ and all other coefficients zero. The data error has norm exactly $\delta$, butThis perturbation theory is a finite expansion, so it defines an actual outgoing solution of the Helmholtz equation outside any sphere of positive radius. It is not merely an example of nonconvergent formal coefficients. Taking instead $e_N^0=|d_N{|}^{-1}$ gives data errors tending to zero and trace errors of norm one, which directly proves discontinuity of inversion. General square-summable measurement noise can even violate the spherical far-field range condition, leaving no $L^2$ continuation to the chosen sphere.

For obstacle reconstruction one would continue the field toward the unknown boundary and look for a surface on which $u_i+u_s=0$, the Dirichlet boundary condition. The unbounded spherical far-to-near-field continuation shows why small far-field errors can cause arbitrarily large errors in the field used to estimate that surface. Locally, a differentiable normal displacement $\eta$ of a candidate boundary must obey the linearized condition $\Delta u_s+\eta {\partial }_n(u_i+u_s)=0$. Thus a large continued-field error, or a small normal derivative, defeats an uncontrolled boundary estimate.

There is a necessary qualification to the wording about errors in the scatterer itself. The calculation proves an unbounded field-continuation error. An arbitrary perturbed outgoing field need not correspond to any Dirichlet obstacle, and no admissible class or metric on obstacle shapes is specified. It therefore does not by itself prove an arbitrarily large geometric error in every constrained reconstruction method. Such a claim requires a shape model and an analysis of the boundary-extraction step. The instability displayed here is exactly the mechanism that an unregularized reconstruction must control.

Use polar coordinates with $-\pi <\theta <\pi$; the two faces of the screen are $\theta =\pm \pi$. Take $0<{\theta }_0<\pi$, so the incident wave approaches from the upper half-plane, and retain the specified time dependence $e^{i\omega t}$. Outgoing cylindrical waves then have radial factor $e^{-i{k}_{0}r}$. Define the normalized Fresnel integralThe exact Neumann Sommerfeld half-plane diffraction solution for the total spatial field isOne way to verify the Helmholtz equation is to use parabolic coordinates $\xi =\sqrt{2r}\cos (\alpha /2)$, $\eta =\sqrt{2r}\sin (\alpha /2)$. Each term has the form $e^{i{k}_0({\xi }^2-{\eta }^2)/2}\mathcal{F}(\pm \sqrt{k_0}\xi )$. The operator is $({\xi }^2+{\eta }^2{)}^{-1}({\partial }_{\xi }^2+{\partial }_{\eta }^2)$, and ${\mathcal{F}}^{\prime \prime }(s)+2is{\mathcal{F}}^{\prime }(s)=0$ makes its Helmholtz residual zero. At $\theta =\pm \pi$, the two Fresnel arguments coincide, while the derivatives of both the arguments and plane-wave phases cancel in pairs. Hence ${\partial }_{\theta }\Phi =0$, which is the required normal boundary condition.

Away from the Fresnel transition regions, integration by parts gives the Fresnel tail expansionwhere $H$ is the Heaviside step function. The same expansion follows from the complementary-error-function asymptotics in NIST DLMF. Because $s_{\pm }^2=k_{0}r[1+\cos (\theta \pm {\theta }_0)]$, the tails in both terms have outgoing phase $e^{-i{k}_{0}r}$. The scattered far field is consequentlyThe first term cancels the incident geometrical optics field in its shadow, the second is its reflected plane wave and the last is the diffracted edge wave. The expansion is nonuniform at $\theta ={\theta }_0-\pi$ and $\theta =\pi -{\theta }_0$; the exact Fresnel expression resolves those regions. Incidence from below follows by reflecting $\theta$ and ${\theta }_0$.

Use time dependence $e^{i\omega t}$ and consider fields independent of $y$, with $\mathbf{E}=\hat{\mathbf{y}}{E}_y(x,z)$. The source-free Maxwell's equations are $\nabla \times \mathbf{E}=-i\omega \mu \mathbf{H}$ and $\nabla \times \mathbf{H}=i\omega \epsilon \mathbf{E}$. The first givesSubstitute these into the $y$ component of the second equation. The TE wave equation in a stratified magnetodielectric isBecause its coefficients depend only on $z$, ${\partial }_x{E}_y$ obeys the same equation. Since it is proportional to $\mu H_z$, substitution and division by $\mu$ giveFor a conserved transverse wavenumber $k_x$, replace ${\partial }_x^2$ by $-k_x^2$ to obtain the corresponding ordinary Helmholtz equations.

If “propagating along the $z$ axis” is read as strict normal incidence with no $x$ dependence, then $H_z=0$ identically. The nonzero magnetic component is $H_x$, and its equation isThe $H_z$ equation above applies to the oblique transverse electric polarization used in the next part and also includes the zero normal-incidence solution. This distinguishes the named component from the usual transverse magnetic field.

Set $\psi =e^{ikx+\mu {\chi }_1+O({\mu }^2)}$. Substitution into the Helmholtz equation gives the first Rytov approximation equationFor the forward parabolic approximation, omit ${\partial }_x^2{\chi }_1$. With zero perturbation at $x=0$, the Fresnel propagator then givesA jointly Gaussian random field makes ${\chi }_1$ a centered complex Gaussian variable. The complex Gaussian exponential moment therefore gives the first-Rytov coherent meanThe second moment here has no complex conjugation. It cannot be replaced by the positive intensity variance $\mathbb{E}[|{\chi }_1{|}^2]$.

For example, let $\hat{C}(u,v;\mathbf{\kappa })$ be the transverse spectrum of the covariance between depths $u$ and $v$. The Fourier multiplier of the Fresnel propagator is $e^{-i(x-u)|\mathbf{\kappa }{|}^2/(2k)}$, soThis supplies the requested mean in terms of medium statistics. For the longitudinal white-noise model, $\hat{C}=\delta (u-v)\Phi$, which reduces the double longitudinal integral to one.

The exponential is the mean of the first-order logarithmic model. It is not a complete second-order perturbation expansion of the true mean: the omitted logarithmic term ${\mu }^2{\chi }_2$ can itself have a nonzero mean. Strictly at first order, $\mathbb{E}[\psi ]=e^{ikx}+O({\mu }^2)$. As an independent check, the closed white-noise parabolic model gives exactlyfor plane-wave input. A truncated first-Rytov covariance expression with transverse diffraction need not equal this exact mean; omitted logarithmic terms account for the difference.

Write $k=|{\mathbf{k}}_i|$ for the background wavenumber and ${\psi }_i=e^{i{\mathbf{k}}_i\cdot \mathbf{r}}$. The Helmholtz equation givesThe Born approximation for scalar wave scattering replaces $\psi$ in this source term by ${\psi }_i$. For the question's contrast $V=-k(n-1)$, also linearize $n^2-1\simeq 2(n-1)$. Since $(\Delta +k^2)e^{ikR}/R=-4\pi \delta$, the outgoing scattered field isWithout the small-contrast linearization, replace $V$ by $-k(n^2-1)/2$ in this formula.

Apply the Weyl plane-wave representation. For an observation plane wholly above or below $D$, put $\sigma =+1$ or $-1$, respectively, and choose $m=\sqrt{1-p^2-q^2}$ with nonnegative imaginary part. Define $\tilde{V}(\mathbf{\kappa })={\int }_{D}V({\mathbf{r}}^{\prime })e^{-i\mathbf{\kappa }\cdot {\mathbf{r}}^{\prime }}{d}^3{r}^{\prime }$. Substitution givesFor the lower observation plane the vertical phase has the opposite sign. This follows from the absolute separation $|z-z^{\prime }|$ in the Weyl identity; its version with $z-z^{\prime }$ alone applies only above the sources.

Factor the forward carrier as $E=e^{ikx}U$. Substitution into the scalar Helmholtz equation gives the exact envelope equationIf $|U_{xx}|\ll 2k|U_x|$, drop the second longitudinal derivative to obtain the parabolic wave equationTypical sufficient conditions are a transverse width $w$ with $kw\gg 1$, longitudinal envelope and material variation on scales much longer than $k^{-1}$, and weak refractive contrast so that factoring out the free-space carrier leaves a slowly varying envelope. A paraxial ordering has transverse propagation angles $O(\eta )$ and ${ϵ}_{\mathrm{rel}}-1=O({\eta }^2)$, with $U_x=O(k{\eta }^{2}U)$ and $U_{xx}$ higher order. A large refractive contrast may require a different, medium-adapted carrier even when the ray direction is nearly axial.

The paraxial approximation retains transverse diffraction and slowly accumulated refraction, but neglects backward propagation, strong reflection from rapid material changes, wide-angle propagation and, in this scalar model, polarization coupling. The supplied vector Helmholtz equation is itself a model assumption: in a source-free inhomogeneous dielectric, Maxwell equations additionally require $\nabla \cdot (ϵ\mathbf{E})=0$, and generally supply a longitudinal-gradient term when the vector wave equation is expanded. An independent scalar component is justified only with a compatible polarization and the relevant small-gradient approximation.

Use SI units and normalize the condensate wavefunction so $|\psi {|}^2=n$. With mechanical momentum operator $-i\hslash \nabla -qA$, the charge current density isFor uniform $n$ and $\psi =\sqrt{n}{e}^{i\vartheta }$, this becomes $j=(nq/m)(\hslash \nabla \vartheta -qA)$.

Under the gauge transformation $A\mapsto A+\nabla \chi$, the wavefunction transforms as $\psi \mapsto e^{iq\chi /\hslash }\psi$, so $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two added gradients cancel in $\hslash \nabla \vartheta -qA$, while $|\psi {|}^2$ stays fixed. Thus the superconducting current is gauge invariant. The same cancellation in the full displayed current proves it without requiring a constant amplitude.

For uniform carrier density in a region with smooth single-valued phase, taking a curl gives the London equation $\nabla \times j=-n{q}^{2}B/m$. In the time-independent setting Ampère's circuital law gives $\nabla \times B={\mu }_{0}j$, and Gauss's law for magnetism gives $\nabla \cdot B=0$. Thereforeso the Helmholtz equation and London penetration depth areThe equation has the modified, exponentially decaying Helmholtz sign. For a planar surface at $x=0$ with material in $x>0$, the bounded solution is $B(x)=B(0)e^{-x/\ell }$; the exponentially growing solution is excluded in the bulk. Hence a macroscopic region far from the surface has negligible field: magnetic flux is expelled from the bulk, the Meissner effect. The literal assertion of zero flux everywhere is the bulk limit: finite penetration remains within distance $\ell$ of a surface. Vortices or flux trapped through a hole require different phase or topology assumptions and are not ruled out by this smooth-phase derivation.

Write $\mathbf{r}=(y,z)$ and prescribe a deterministic entrance field $E_0(\mathbf{r})$. Substitution of the carrier into the Helmholtz equation givesThe usual weak-index paraxial model drops $E_{xx}$ and linearizes $n^2-1=2\mu W+O({\mu }^2)$, givingAll the mean-field formulas below use this model. If the ${\mu }^2{W}^2$ term in $n^2$ is retained, its extra mean and fluctuations must also be included; Gaussian averaging of the linear random potential alone is then insufficient.

For a colored medium, taking expectations gives the exact propagation identityStationarity and Gaussianity alone do not close the last term. To make its dependence precise, let $C(s,\mathbf{r})=⟨W(x+s,{\mathbf{r}}_0+\mathbf{r})W(x,{\mathbf{r}}_0)⟩$ and let $U_W(x,s;\mathbf{r},{\mathbf{r}}^{\prime })$ be the random paraxial propagator. Assuming a jointly Gaussian random field, rather than only Gaussian one-point marginals, the functional derivative of the solution isThe Furutsu–Novikov formula therefore yieldsThis is a propagation equation for the mean, with the remaining correlation displayed rather than silently replacing it by a product of means.

The general solution can be written explicitly as a Gaussian-averaged phase-screen product. Put $\Delta =x/N$, $x_j=j\Delta$, ${\mathbf{r}}_N=\mathbf{r}$ and letbe the integral kernel of the Fresnel propagator. Successive free-propagation and random-phase steps, followed by Gaussian averaging, give the colored Gaussian paraxial mean propagatorFor smooth colored media this is the time-slicing representation, with oscillatory integrals understood by the usual Fresnel regularization. At finite $N$ the exponential follows directly from the characteristic function of the jointly Gaussian phase sum. It exhibits why a general transverse covariance cannot be replaced by its value at zero separation along every diffraction path.

For completeness, the unlinearized paraxial model has $E_x=DE+iaWE+i(k{\mu }^2/2)W^{2}E$ and the corresponding mean equation includes $i(k{\mu }^2/2)⟨W^{2}E⟩$. Its colored-medium solution uses the same time-slicing integral, replacing the Gaussian factor byThis follows by completing the square in the finite-dimensional Gaussian integral, with the determinant branch continuous from $\mu =0$. It supplies the generic colored-medium mean when the $W^2$ term is retained. A square of ideal white noise is not defined by the covariance in part (b); that limit uses the linear random-potential model and its stated drift.

If transverse diffraction of the fluctuations is neglected, the simpler straight-ray approximation givesIndeed $E=E_0\exp [ia{\int }_0^{x}W(s,\mathbf{r})ds]$ and the variance of its phase is $2a^2{\int }_0^x(x-s)C(s,0)ds$. If $W$ is constant transversely, this formula is exact with $E_0$ replaced by the free diffracted field $e^{xD}{E}_0$. For example $W=Z$, a single standard normal variable constant throughout space, satisfies stationarity and unit variance but gives $m=e^{-a^2{x}^2/2}{e}^{xD}{E}_0$. Thus a constant-rate exponential attenuation is not a consequence of part (a)'s assumptions alone. It becomes the closed paraxial result under the longitudinal white-noise assumption in the next part.

Take $e^{-i\omega t}$ time dependence and measure angles from the upward normal. Write $p_i=k\sin {\theta }_i$, $q_i=k\cos {\theta }_i>0$ and choose incident amplitude $A_i$. Separate the flat Dirichlet reflection from the rough correction:Taylor expansion of the boundary condition about $z=0$ givesSince ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{A}_i{e}^{i{p}_{i}x}$, the induced trace isUse $\hat{h}(Q)=\int h(x)e^{-iQx}dx$ and $\beta (\xi )=\sqrt{k^2-{\xi }^2}$ with the outgoing branch: $\beta >0$ for $|\xi |<k$ and $\mathrm{Im}\beta >0$ outside. The outgoing angular spectrum solves the Helmholtz equation with that trace, giving the first-order rough-surface scattered fieldIf “scattered field” includes the specular reflection, add $-A_i{e}^{i{p}_{i}x+i{q}_{i}z}$ to this rough correction. Fourier transforms of an infinite random surface are generalized objects; finite-window transforms give an ordinary integral and are also needed for its far-field intensity. The expansion assumes a fixed regular surface shape scaled to small height. Highly oscillatory profiles can require additional control of slopes and spectral moments; small height alone is not a uniform error estimate for every profile.

Suppress the common time factor $e^{i\omega t}$ and use the Fourier transform conventionA right Half-range Fourier transform is analytic above its convergence line and a left transform below it. For the limiting absorption principle with this time convention take $\omega ={\omega }_r-i\epsilon$, ${\omega }_r>0$, $\epsilon >0$, before taking $\epsilon \downarrow 0$. SetThis is the outgoing acoustic square-root branch: at a propagating real wavenumber, $\gamma \to i\sqrt{{\omega }_r^2-k^2}$. Put $J=[\tilde{\varphi }{]}_{y=-0}^{y=+0}$ and $d^{+}={\partial }_y{\Phi }^{+}(k,0)$. Across the fluid part of $y=0$, the field and its normal derivative are continuous; on the sheet both sides have the same normal velocity. The outgoing Helmholtz equation solutions therefore have equal normal derivatives and opposite scattered traces:The jump $J$ is supported on $x<0$, and so is a minus transform.

The linear fluid equation $i\omega v_y=-{\partial }_y{\rho }^{\prime }$ and the sheet velocity $v_y=i\omega \eta$ imply ${\partial }_y{\rho }^{\prime }={\omega }^2\eta$ on either face. The incident contribution to the left derivative transform is $\omega \sin {\theta }_0/(k+\omega \cos {\theta }_0)$. Writing ${\eta }^{-}={\int }_{-\infty }^0{e}^{ikx}\eta (x)dx$, it follows thatThe upward net pressure is lower minus upper, hence $P=-J$ in transform space. If $s={\eta }^{\prime }(0)$, the pinned endpoint givesCombining these identities gives the Wiener-Hopf equationThis is the plane-wave forcing of a pinned elastic half-sheet.

Let $b=\omega \sqrt{m/T}$ and assume initially that the distinguished points are distinct. Besides the branch points $\pm \omega$, $K$ has simple poles at $\pm b$. Its physical-sheet zeros $\pm k_s$ solve the fluid-loaded dispersion relationFor real positive frequency there is exactly one positive real root $k_s>\max ({\omega }_r,{\omega }_r\sqrt{m/T})$: the left side increases strictly from zero to infinity on that interval. Equivalently, with $q=\gamma (k_s)>0$, $T{q}^3+(T-m){\omega }_r^{2}q-2{\omega }_r^2=0$. Its other roots have negative real part and do not belong to this outgoing sheet. Continue $k_s$ with small absorption; $+\omega ,+b,+k_s$ lie below the real line and their negative partners above it. The Wiener-Hopf factorization assigns the lower branch cut and pole to $K^{+}$, and the upper branch cut and pole to $K^{-}$. The continued $K^{+}$ has a zero at $+k_s$, and $K^{-}$ at $-k_s$; these are poles of their reciprocals, rather than poles of the factors. Branch cuts can be drawn away from the real line into the corresponding half-planes.

For this standard assignment a common strip of analyticity and nonvanishing isThe strip shrinks to the real contour as absorption is removed. The factors continue meromorphically beyond their designated half-planes; the allocation of the poles and branch points is part of the radiation condition. The incident pole $-\omega \cos {\theta }_0$ belongs above the inversion contour. For incidence with $0<{\theta }_0<\pi /2$ the real contour is a direct choice. Other nongrazing angles are obtained by analytic continuation with the incident pole kept on its prescribed side, deforming the contour locally if necessary. Coincident distinguished points and grazing incidence are treated by the corresponding limits, not by dividing by a coincident pole.

Divide the Wiener-Hopf equation by $K^{+}$ and use pole subtraction in a Wiener-Hopf equation. With $a=\omega \cos {\theta }_0$, the minus part of $F/K^{+}$ isIndeed these are exactly the incident-pole and upper bare-sheet-pole residues. At the lower bare-sheet pole $+b$, the forcing pole is canceled by the pole of $K^{+}$ before splitting. Thus $B^{+}=F/K^{+}-B^{-}$ is analytic on the plus side of the contour. The separated equation isWith the stipulated $E=0$,The scattered field is consequentlyThe full time-dependent density perturbation is the incident field plus $e^{i\omega t}\varphi$. The two traces at $y=0$ follow by limiting from above and below. A constant rescaling of the two factors does not change this field.

To determine the endpoint slope, reconstruct the sheet transform using the fluid condition:At $+b$ the continued $K^{+}$ has residue $-{\omega }^2/(2Tb{K}^{-}(b))$. Therefore the derivative transform $d^{+}$ generically has the lower-plane poleThis pole produces a forbidden incoming bare-sheet wave in the reconstructed displacement: $e^{i\omega t-ibx}$ travels towards the endpoint from $x=-\infty$. A minus transform of the specified outgoing solution must be analytic there. The incoming-pole cancellation at a pinned membrane edge is preciselyFor generic parameters this determinesThe lower pole is removed; the upper zero $-k_s$ of $K^{-}$ can still produce the legitimate outgoing evanescent acoustic surface wave, whose phase travels to negative $x$. This distinction separates the spurious bare-sheet wave from the coupled fluid-loaded mode. No explicit factorization has been used.

Use the scattering potential convention $V=k_0^2(n^2-1)$. With time dependence $e^{-i\omega t}$ the outgoing Green function is $G_0(r,r^{\prime })=e^{i{k}_0|r-r^{\prime }|}/(4\pi |r-r^{\prime }|)$ and satisfies $(\Delta +k_0^2)G_0=-\delta$. The total-field Lippmann-Schwinger equation isReplacing the field in the integral by the incident field gives the first Born approximation for scalar wave scattering:For the first Rytov approximation, write $\psi ={\psi }_0{e}^{\chi }$ on a region where the incident field is nonzero and a continuous complex logarithm can be chosen. The logarithmic wave perturbation equation isThe last dot product is bilinear, not a squared modulus. Dropping it leaves a linear equation for ${\chi }_1$. Multiplication by ${\psi }_0$ shows that ${\psi }_0{\chi }_1$ satisfies the same outgoing inhomogeneous Helmholtz equation as ${\psi }_1$. HenceIntroduce a small multiplier $\lambda$ on $V$. Then ${\psi }_1=O(\lambda )$ andThe two total-field approximations agree to first order. Exponentiating the first logarithmic wave perturbation does not supply the generally missing second-order Born series contribution.

A nonspherical configuration requires the vector form of hydrostatic equilibrium and the Poisson equation, rather than the spherical mass-coordinate equations:For the same index-one polytrope, hydrostatic balance inside the positive-density region gives $\Phi =C-2K\rho$. Thus its interior mass density must obey the Helmholtz equationA positive separated solution with the required boundary values isIts maximum is ${\rho }_c$ at the centre of the cube and it vanishes on every face. Its Laplacian is $-3{\pi }^2\rho /L^2$, so it satisfies the interior equations whenTogether with $P=K{\rho }^2$ and $\Phi =C-2K\rho$, this explicitly constructs the cubic polytropic interior.

The mass integral separates into three elementary sine integrals:ThereforeThis construction satisfies the equations inside the prescribed cube. It does not by itself establish the existence of an isolated self-gravitating star: the interior potential must also match the exterior vacuum field generated by that very mass density. That additional physical requirement is addressed in part (c).

Choose the time dependence $e^{-i\omega t}$ and measure distances in background-wavenumber units. Thus $L_0=\Delta +1$, the incident field satisfies $L_0{\psi }_i=0$, and the scattering potential convention is $V=n^2-1$, giving $L_0\psi =-V\psi$. To match the minus sign in the paper's integral, take the outgoing Green function to satisfy $L_{0}G=\delta$. This is the negative of the frequently used outgoing Helmholtz equation fundamental solution satisfying $L_{0}G=-\delta$. In physical coordinates one restores $k_0^2$ in the scattering potential, or absorbs it in the integral kernel; the sign convention must remain consistent.

On a region where the incident and total fields are nonzero, introduce the logarithmic wave perturbationChoose a continuous logarithm branch connected to the unperturbed field. Substitution into the Helmholtz equation givesThe quadratic term uses the complex bilinear dot product, not the squared modulus of the gradient. Write $V=\epsilon V_1$ and $\chi =\epsilon {\chi }^{[1]}+\cdots$. The first-order Rytov approximation discards that quadratic term. Since $L_0({\psi }_i\chi )={\psi }_i{\mathcal{L}}_i\chi$, the outgoing first correction isand henceThe validity of the first Rytov approximation concerns the omitted logarithmic correction. Its next contribution satisfies ${\mathcal{L}}_i{\chi }_2=-\nabla {\chi }_1\cdot \nabla {\chi }_1$, so a useful explicit criterion is that the outgoing solution ${\chi }_2$ be small in the region of interest. Weak refractive index contrast and small wave phase gradients on a wavelength scale provide the usual perturbative regime, with weak amplitude fluctuations and no strong focusing or zeros that destroy the logarithm. A sufficient local source comparison is $|\nabla {\chi }_1\cdot \nabla {\chi }_1|\ll |V|$ where the scattering potential is nonzero, together with control of propagation of that error. This is not a universal pointwise test at zeros of $V$.