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Put $\tau =T_r\wedge T_R$. First establish almost sure finiteness. The Itô formula gives the martingale $|B_t{|}^2-2t$ for planar Brownian motion. Stop it at $t\wedge T_R$. The stopped path has norm at most $R$, soBy monotone convergence, $\mathbb{E}{T}_R<\infty$, and hence $\tau <\infty$ almost surely.

The function $h(y)=\log |y|$ is a harmonic function away from $0$: in polar coordinates its Laplacian is $h^{\prime \prime }(\rho )+h^{\prime }(\rho )/\rho =-{\rho }^{-2}+{\rho }^{-2}=0$. The Itô formula shows $h(B_{t\wedge \tau })$ is a local martingale. It is bounded between $\log r$ and $\log R$, so it is a true martingale and passage to $\tau$ is justified by dominated convergence. Writing $p={\mathbb{P}}_x(T_r<T_R)$ and using continuity at exit givesSolving yields the planar Brownian annulus hitting probabilityThe stopping argument stays away from the singularity of the logarithm until the calculation is complete.

Let $X={\partial }^{-2}\zeta$, with an inverse spatial Laplacian defined by a boundary or zero-mode prescription. It depends on $\zeta$, not on its time derivative. Differentiation gives the canonical momentumCount $\zeta ,\pi ,X$ as first order. Inverting perturbatively givesFor the Legendre transform in mechanics, write $v=\dot{\zeta }=\pi +\Delta$ with $\Delta$ second order. The quadratic kinetic contribution is $\pi v-v^2/2={\pi }^2/2-{\Delta }^2/2$, whose correction is fourth order; in the cubic terms one may consequently replace $v$ by $\pi$. The Hamiltonian density is thereforeThe cubic interaction Hamiltonian density isIn the free interaction picture, ${\pi }_I={\dot{\zeta }}_I$. Thus here ${\mathcal{H}}_{\mathrm{int}}=-{\mathcal{L}}_3$ with the time derivatives interpreted as free fields. This equality follows from the cubic Legendre transform for scalar derivative interactions; at higher orders additional terms can appear.

Put $V=|v_0|>0$, take $q>0$ without loss of generality, and write $s=R^2+z^2/q^2$. The Newtonian gravitational potential is singular at the origin; all pointwise claims below concern $s>0$. The zero of the Newtonian gravitational potential is the one in the paper, with an implicit fixed length unit inside the logarithm.

The Poisson equation for Newtonian gravity in cylindrical coordinates givesThe two contributions to the Laplacian areConsequently the mass density of the axisymmetric logarithmic gravitational potential isThe coefficient of $R^2$ is positive. The coefficient of $z^2$ determines the density positivity for an axisymmetric logarithmic potential: strict positivity at every noncentral point requires $q^2>1/2$ and $v_0\ne 0$. Nonnegativity permits $q^2=1/2$, where the mass density vanishes on the noncentral symmetry axis. For $q^2<1/2$ it is negative near that axis. Either sign of a nonzero real $q$ gives the same model, and $q=0$ is undefined. Setting $v_0=0$ yields vacuum away from the singular origin, not strictly positive mass density. The $r^{-2}$ central cusp is locally integrable and contains no hidden point mass: the Newtonian gravitational field flux through a shrinking sphere is of order its radius.

An equatorial circular orbit requires inward gravitational acceleration equal to $v_c^2/R$. Thus the circular speed isThis is a flat galaxy rotation curve. The singular center does not define an additional circular orbit at $R=0$.

The specific orbital energy $E=\frac{1}{2}(v_R^2+v_{\varphi }^2+v_z^2)+\Phi$ is conserved because the Newtonian gravitational potential is time independent. The axial specific angular momentum $L_z=R{v}_{\varphi }$ is conserved because the Newtonian gravitational potential is axisymmetric. Hence the Jeans theorem, or directly $dF(E,L_z^2)/dt=0$ in the Collisionless Boltzmann equation, permits a stationary two-integral galactic distribution function. Using $L_z^2$ makes this galactic distribution function even under reversal of the azimuthal velocity, so it describes a nonstreaming population. A general axisymmetric population can have an odd part in $L_z$, or dependence on a third integral of motion; neither is required for this construction.

At each position define $\rho ⟨g⟩=\int gF{d}^{3}v$. The two-integral galactic distribution function is invariant under swapping $v_R$ and $v_z$, since only their sum of squares occurs in $E$. It is also even in each velocity component. Swapping integration variables proves meridional velocity isotropy of a two-integral distribution, while reflecting one variable makes each mixed integrand odd. ThereforeAll mean velocities vanish, so these are also the corresponding statements for the velocity ellipsoid and its centered covariance tensor. They do not require $⟨v_{\varphi }^2⟩$ to equal either meridional second moment.

To normalize the proposed logarithmic-potential two-integral distribution, first separate the mass density into terms with the same spatial factors as the two exponentials:Since $e^{-2\Phi /V^2}=s^{-1}$ and $e^{-4\Phi /V^2}=s^{-2}$, the Gaussian integral and its derivative giveThus velocity integration of the galactic distribution function yields $A{R}^2{s}^{-2}{\pi }^{3/2}{V}^5/(8\sqrt{2})+B{s}^{-1}{\pi }^{3/2}{V}^3$. Matching the independent spatial factors determinesThis verifies both the stationary Collisionless Boltzmann equation and the required Poisson equation for Newtonian gravity. A shift $\Phi \mapsto \Phi +C$ changes the normalizations to $A{e}^{4C/V^2}$ and $B{e}^{2C/V^2}$; the physical mass density is unchanged.

There is an important distinction between positive mass density and a nonnegative galactic distribution function. For $1/2\le q^2\le 1$, both coefficients above are nonnegative. For a prolate model $q^2>1$, $A<0$ and one must test the sum rather than reject the model just because one term is negative. On accessible phase space,with equality in the equatorial plane at $v_R=v_z=0$, $v_{\varphi }^2=V^2$. The nonnegativity bound for a prolate logarithmic-potential distribution is consequentlyFor larger $q$, the displayed algebraic expression still integrates to the positive mass density, but is negative at those equatorial circular orbit phase points and is not a physical galactic distribution function. This extra restriction is distinct from the earlier density-only answer.

Use the unique curvature $-1$ hyperbolic metric on a closed Riemann surface of genus $g\ge 2$, supplied by the uniformization theorem. This is the implicit setting for the partition and spectral theorems; a genus-zero or genus-one surface does not have this hyperbolic metric.

A partition is a pants decomposition by disjoint essential simple closed geodesics, whose complement is a union of pairs of pants. A pair of pants is a sphere with three disks removed, furnished here with geodesic boundary. Its Euler characteristic is $-1$, so there must be $2g-2$ pants. Every pant has three boundaries and every cutting curve occurs twice, giving $3g-3$ cutting curves. The Bers pants decomposition theorem asserts that a constant $B_g$ depending only on $g$ bounds the lengths of all cuffs in some such pants decomposition of every closed genus-$g$ hyperbolic surface.

To see what geometry the cuffs determine, cut each pair of pants along its three perpendicular seams into two congruent right-angled hyperbolic hexagons. The three alternate sides are ${\ell }_1/2,{\ell }_2/2,{\ell }_3/2$, where the ${\ell }_i>0$ are its boundary lengths. The right-angled hyperbolic hexagon identity determines the seam $s_i$ opposite the half-cuff ${\ell }_i/2$ byWe use the standard existence and uniqueness theorem for a right-angled hyperbolic hexagon with prescribed positive alternating side lengths. Thus the three boundary lengths determine the pair of pants up to isometry.

Fix a topological pants decomposition, label its cuffs, and choose reference seam endpoints and orientations. Gluing two boundaries of the same length requires a translation along that boundary; its signed distance is a twist ${\tau }_i$. The gluing graph, the $3g-3$ positive cuff lengths, and the $3g-3$ twists therefore determine the unmarked hyperbolic surface up to isometry. A full boundary translation has period ${\ell }_i$, so for an unmarked gluing one may take $0\le {\tau }_i<{\ell }_i$, with endpoints identified. For a marked surface the whole real twist records the number of Dehn twists, and the Fenchel–Nielsen coordinates areThe topological gluing data are necessary when the decomposition is not fixed. Different markings or decomposition graphs can describe the same unmarked surface.

Teichmüller space ${\mathcal{T}}_g$ consists of pairs $(X,f)$, where $X$ is a closed genus-$g$ Riemann surface and $f:{\Sigma }_g\to X$ is an orientation-preserving marking, modulo the equivalence $(X,f)\sim (Y,h)$ when a conformal orientation-preserving map $a:X\to Y$ has $a\circ f$ homotopic to $h$. The Fenchel–Nielsen coordinates identify it with ${\mathbb{R}}_{>0}^{3g-3}\times {\mathbb{R}}^{3g-3}$, of real dimension $6g-6$. The moduli space of Riemann surfaces forgets the marking by quotienting by the mapping class group.

The Wolpert generic spectral rigidity theorem says that the locus of closed genus-$g$ hyperbolic surfaces possessing an isospectral but nonisometric partner lies in a locally real-analytic exceptional locus of lower dimension in Teichmüller space. Thus a generic surface is determined, up to isometry, by its unmarked length spectrum, equivalently by its Laplacian spectrum. Here an isometry may reverse orientation; the length spectrum cannot distinguish the two orientations of the same metric. The equivalence of the two spectra is the compact hyperbolic Selberg trace formula.

The simplifying result is the Buser finite length spectrum theorem: for every $g\ge 2$ and $\epsilon >0$ there is $L=L(g,\epsilon )$ such that two closed genus-$g$ hyperbolic surfaces with hyperbolic systoles at least $\epsilon$ have the same entire unmarked length spectrum if their length multisets up to $L$ agree. Multiplicities are included. We will count primitive unoriented closed geodesics; counting all iterates is equivalent by successively removing shorter iterates. The theorem is uniform on the thick part of moduli space, not merely a cutoff chosen separately for a particular pair.

Here is a proof using compactness and polynomial trace equations. We state the subsidiary facts and explain their role. First, Mumford's compactness theorem makes the genus-$g$ thick moduli space compact. It can also be seen from the Bers pants decomposition theorem: cuffs in a Bers decomposition lie in $[\epsilon ,B_g]$, there are finitely many pants decomposition graphs, and unmarked twists may be reduced modulo their cuff lengths. These data lie in finitely many compact boxes of Fenchel–Nielsen coordinates and cover the thick moduli space.

Second, these finite compact boxes give a compact family $K$ of marked hyperbolic structures on a fixed smooth surface ${\Sigma }_g$ representing every member of the thick moduli space. We use the standard smooth dependence of the glued metrics and their hyperbolic holonomy representations on Fenchel–Nielsen coordinates. Choose smooth representatives over finitely many parameter neighborhoods and conjugate the hyperbolic holonomy representation using a fixed lifted Riemannian orthonormal frame. Passing to finite compact subboxes gives a compact set of representations, together with a compact family of representative smooth metrics. Changing the markings between boxes causes no problem: all are markings from the same fixed ${\Sigma }_g$, and the union is finite.

Third, uniformization theorem identifies each marked hyperbolic surface with ${\mathbb{H}}^2/\rho ({\pi }_1{\Sigma }_g)$ for a faithful discrete hyperbolic holonomy representation into ${\mathrm{PSL}}_2(\mathbb{R})$. We use the standard lifting fact that a closed orientable hyperbolic holonomy representation admits a lift to ${\mathrm{SL}}_2(\mathbb{R})$, and lifts may be chosen continuously on small parameter neighborhoods. Choose lifts on finitely many such neighborhoods, allowing the finitely many possible sign choices. A representation is described by the matrices of $2g$ standard fundamental group generators. The polynomial encoding of hyperbolic geodesic lengths uses, for each fixed group word $w$, its squared matrix traceis a polynomial in their entries: products are polynomial, and the inverse of a determinant-one matrix is its polynomial adjugate. For a nontrivial hyperbolic word, the hyperbolic translation length obeysThus equality of the positive lengths is exactly equality of these trace polynomials, independent of lift signs.

Fourth, the length spectrum of a closed hyperbolic surface is locally finite with finite multiplicities. We need the uniform finiteness of short geodesic classes across the compact marked family. More uniformly, for each finite $b$, only finitely many unoriented primitive conjugacy classes can have length at most $b$ in any metric of $K$. To justify uniformity, compactness of the smooth metrics gives a common bilipschitz equivalence comparison with one fixed metric $h_0$ on ${\Sigma }_g$. If ${\ell }_h(\alpha )\le b$, the minimizing geodesic has $h_0$ length at most $Cb$, so ${\ell }_{h_0}(\alpha )\le Cb$. Local finiteness for $h_0$ then gives a finite list. Local finiteness itself follows from proper discontinuity of the cocompact group of the hyperbolic holonomy representation: conjugate a geodesic axis to meet a fixed compact fundamental set, and a bounded-length axis gives a group element moving that compact set a bounded distance, of which there are only finitely many.

Finally, the Hilbert basis theorem says that a polynomial ring in finitely many real variables is Noetherian. Consequently a decreasing sequence of sets cut out by polynomial equations eventually stabilizes. Indeed, their vanishing ideals form an increasing chain and stabilize. Finite unions of such sets are still algebraic: one can take all products of one defining polynomial from each component. This finiteness fact is the engine that turns arbitrarily many spectral comparisons into finitely many.

Enumerate the primitive unoriented conjugacy classes as ${\alpha }_1,{\alpha }_2,\dots$. For each $n$ letand let $F_n$ be the finite set of primitive classes that can have length at most $b_n$ anywhere in $K$. Define $Z_n$ in the finite-dimensional space of pairs of generator matrices as follows: it consists of pairs $(\rho ,\sigma )$ for which there exists an injectionand also an injection satisfying the analogous equations with $\rho$ and $\sigma$ interchanged. There are only finitely many injections. Each choice imposes finitely many polynomial equations; taking their finite unions and then intersecting the two directions proves that $Z_n$ is a real affine algebraic set. On determinant-one representations these equations are exactly the desired matching of lengths. Defining the polynomials on the ambient matrix-entry space is harmless; the argument will only apply them to representations in $K$.

Put $W_n=Z_1\cap \cdots \cap Z_n$. The sets $W_n$ form a decreasing chain of affine algebraic sets, hence for some $N$,For representations in $K$, membership in this intersection is equivalent to equality of the full length spectra. One direction follows by matching equal lengths, including their multiplicities. For the other, fix a length $\ell$. All primitive classes of length $\ell$ on the first surface occur in some finite initial segment. Its injective matching shows that the second surface has at least that multiplicity at $\ell$; the reverse injection gives the reverse inequality. Local finiteness makes both multiplicities finite. Doing this for every $\ell$ proves multiset equality.

TakeIf two surfaces in $K$ have equal length multisets up to $L$, each initial segment through $n\le N$ can be matched injectively on the other surface. Its matching classes lie in $F_n$, since the matched lengths are at most $b_n$. The reverse matching is available as well. Thus the pair belongs to $W_N$, hence to every $W_n$, and the preceding paragraph yieldsAll choices of the compact family were made using only $g$ and $\epsilon$, so the resulting cutoff has the asserted dependence. If the thick family is empty the assertion is vacuous. This proves Buser's finite length spectrum theorem without claiming that equal full length spectra always force isometry; exceptional isospectral pairs are compatible with the theorem and are precisely why Wolpert's theorem is a generic statement.

Let ${\mathcal{A}}^{p,q}$ denote the sheaf of differential forms of type (p, q) printed in the PDF. Take the pointwise Hermitian inner product to be linear in its first argument, and use $\mathrm{vol}={\omega }^n/n!$. The bidegrees in the question specify the conjugate-linear Hodge star: it is uniquely characterized byThus $\ast (c\beta )=\bar{c}\ast \beta$, and $\ast$ maps bidegree $(p,q)$ to $(n-p,n-q)$. If $S$ is the complex-linear extension of the real Hodge star operator, then $\ast =S\circ \text{complex conjugation}$. The operator $S$ is real and $S^2=(-1{)}^{k(2n-k)}$ on degree $k=p+q$, soUsing the complex-linear star instead would give a different bidegree, $(n-q,n-p)$; keeping the convention explicit prevents that ambiguity.

The Hermitian Lefschetz operator is exterior multiplication by $\omega$. Define $\Lambda$ pointwise byIt lowers bidegree by $(1,1)$. In a unitary real coframe with $\omega ={\sum }_j{e}^j\wedge f^j$, it is $\Lambda ={\sum }_j{\iota }_{f_j}{\iota }_{e_j}$, the corresponding sum of interior product of a differential form operators. This gives its existence and identifies it as a smooth operator. On a compact manifold, integrate the pointwise equality against $\mathrm{vol}$ to obtainNo integration by parts is needed for this order-zero operator. Thus $\Lambda$ is both the pointwise and the global formal adjoint.

Write $[A,B]=AB-BA$, and let $k$ denote the degree of the input form. The given Lefschetz commutator is $[L,\Lambda ]=(k-n)I$. For $r=1$ the required formula is exactly this identity. If it holds for $r$, the commutator derivation identity givesThe second commutator acts on degree $k+2r$, so the coefficient on the right isThis proves the commutator formula for powers of the Lefschetz operator

For injectivity of powers of the Lefschetz operator, $r=0$ is immediate. Suppose $r\ge 1$ and $r+k\le n$. If $k=0$ or $1$, then $\Lambda \psi =0$, and $L^r\psi =0$ impliesThe scalar is nonzero, since $k+r\le n$ makes $k-n+r-1\le -1$. Repeating with the smaller power shows $\psi =0$. For $k\ge 2$, induct on $r+k$, treating all $r=0$ cases as already established. The same commutator calculation yields, with $c=r(k-n+r-1)\ne 0$,The induction hypothesis applies to $L^{r-1}$ on degree $k$, since $(r-1)+k<r+k$ and $(r-1)+k\le n$. Hence $\psi =L\alpha$ with $\alpha =c^{-1}\Lambda \psi$ of degree $k-2$. Moreover $L^{r+1}\alpha =L^r\psi =0$. Since $(r+1)+(k-2)=r+k-1$, induction also makes $L^{r+1}$ injective on that degree, so $\alpha =0$ and $\psi =0$. All operators preserve restriction to open sets; the argument applies on every open set. Therefore $L^r$ is an injective morphism of sheaves whenever $r+k\le n$.

On a Kähler manifold, the Kähler identities, with the positive form convention used above, areTaking formal adjoints conjugates the scalar and reverses the order in the commutator. Since ${\Lambda }^{\ast }=L$, this givesTo derive the Kähler Laplacian identity, put $P=\partial$ and $B=\bar{\partial }$, so $P^2=B^2=0$ and $PB+BP=0$. Write $\{A,B\}=AB+BA$ for an anticommutator. The Kähler identities say $B^{\ast }=-i[\Lambda ,P]$ and $P^{\ast }=i[\Lambda ,B]$. Thereforebecause expansion leaves only terms containing $P^2$ or $B^2$. Furthermore,Expanding these expressions and using $PB=-BP$ makes them equal. Since $d=P+B$ and $d^{\ast }=P^{\ast }+B^{\ast }$, the mixed anticommutators already vanish, and consequentlyClosedness of the Kähler form implies $[L,P]=[L,B]=0$. Using the adjoint identity above,The equality of the three Laplacians therefore proves each Laplacian commutes with $L$.

For a compact Kähler manifold, the Dolbeault Hodge decomposition is the orthogonal decompositionHere harmonicity for the Dolbeault Laplacian and for the Hodge Laplacian agrees by the preceding identity. If $\bar{\partial }\alpha =0$ and $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$, then $\bar{\partial }{\bar{\partial }}^{\ast }v=0$. Taking its inner product with $v$ gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Thus $\alpha$ is cohomologous to $h$. Conversely, a harmonic form is $\bar{\partial }$-closed, and a harmonic exact form $h=\bar{\partial }u$ satisfies $\Vert h{\Vert }^2=({\bar{\partial }}^{\ast }h,u)=0$. This proves existence and uniqueness of the harmonic representative and henceThese spaces are finite dimensional by the ellipticity of the Dolbeault Laplacian on the compact manifold.

Complex conjugation commutes with the real operator ${\Delta }_d$ and interchanges the bidegrees $(p,q)$ and $(q,p)$. It is therefore a conjugate-linear bijection of the harmonic spaces, proving Hodge symmetry. The real Hodge star operator commutes with ${\Delta }_d$, as does complex conjugation; hence their composition, our conjugate-linear Hodge star, sends harmonic $(p,q)$-forms bijectively to harmonic $(n-p,n-q)$-forms. Its square is the nonzero scalar established above. This proves Hodge duality, and gives

For the final hard Lefschetz isomorphism on Dolbeault cohomology, take $k=p+q\le n$ and $r=n-k$. This restriction is necessary to make the displayed power nonnegative. The Hermitian Lefschetz operator raises bidegree by $(1,1)$, and the Lefschetz operator preserves harmonic forms; thereforeis well defined. It is injective by injectivity of powers of the Lefschetz operator, since $r+k=n$. Hodge symmetry and Hodge duality give $h^{n-q,n-p}=h^{q,p}=h^{p,q}$, so it is a bijection between finite-dimensional spaces of equal dimension. Because $[L,\bar{\partial }]=0$, the map on harmonic representatives agrees with exterior multiplication by $[\omega {]}^{n-k}$ on Dolbeault cohomology. We concludeFor $p+q>n$, the corresponding valid statement is the inverse of the positive-power isomorphism from bidegree $(n-q,n-p)$ to $(p,q)$; a negative exterior-multiplication power is not defined.

For $\mathbf{F}=\nabla \varphi$, the definition of the Laplacian gives $\nabla \cdot \mathbf{F}={\nabla }^2\varphi =f$. Integrating over $V$ and applying the divergence theorem yieldsThis is the flux form of Poisson equation, for an outward-oriented piecewise smooth boundary and the usual smoothness of $\varphi$ on a neighborhood of the volume. In particular, a singularity cannot be ignored when applying this identity.

With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

An acoustic analogy is an exact rearrangement of the fluid equations into a chosen linear propagation operator acting on an acoustic variable, with everything left over placed on the right as effective forcing. The rearrangement becomes a sound-prediction method only after a reference medium, boundary conditions and approximations to the forcing are specified. In particular, a right-hand-side term need not represent independently generated sound.

Use Einstein summation convention and write ${\partial }_i=\partial /\partial x_i$. Differentiating the continuity equation in time and taking the divergence of the momentum equation eliminates ${\partial }_t(\rho u_i)$:Set $q=p-{\hat{p}}_0$ and $W_{ij}=\rho u_i{u}_j-{\sigma }_{ij}$. Since $p+\chi =q+P$, the constant $P$ has zero Laplacian. The prescribed reference mass density and reference speed of sound are independent of time, soCombining the two identities givesOnly conservation of mass and conservation of momentum have been used; no equation of state or energy equation was needed. Spatial variation of ${\hat{c}}_0$ creates no omitted derivative in this identity, because it multiplies a time derivative. The double divergence of the momentum flux tensor has the structure of an acoustic quadrupole.

For a localized flow, choose the reference fields to match the stationary surrounding medium: ${\hat{\rho }}_0={\rho }_0$ and ${\hat{c}}_0^2=c_0^2$ outside the flow, with $P$ chosen so ${\hat{p}}_0=p_0$ there. A uniform surrounding fluid permits ambient constant mass density and adiabatic sound speed, giving the familiar homogeneous wave equation. A nonuniform surrounding fluid calls for its actual stationary reference profiles, extended sensibly through the flow region. This makes the acoustic variable vanish in the unperturbed exterior and minimizes artificial contrast terms. In a uniform isentropic exterior the leading acoustic relation $p^{\prime }=c_0^2{\rho }^{\prime }$ also makes $Q$ vanish to first order. In a stratified exterior, propagation and entropy-advection effects can remain in $Q$, as the next part demonstrates.

For a radial function, the Laplacian is $\Delta \varphi ={\varphi }_{rr}+2{\varphi }_r/r$. Introduce the retarded and advanced null coordinates and the radial reduction of the three-dimensional wave equation:The chain rule gives ${\partial }_t={\partial }_u+{\partial }_v$ and ${\partial }_r=-{\partial }_u+{\partial }_v$. Consequently the inhomogeneous wave equation isMultiplying the radial wave equation by $r$ removes its first-order radial derivative:Thus the particularly useful retarded and advanced null coordinates form isThese formulas hold for $r>0$. At the axis, smooth spherical symmetry imposes ${\varphi }_r(t,0)=0$ and $w(t,0)=0$; equivalently $w$ has a smooth odd extension across $r=0$.

The spatial factor $r^l{Y}_l^m$ is a regular solid harmonic. Besides the Laplace equation ${\nabla }^{2}F=0$, homogeneity gives $\mathbf{r}\cdot \nabla F=lF$. The divergence of the fluid displacement isDefine the scalar dilation amplitude $D_r=r{U}^{\prime }+(l+3)U+(l/r)V^{\prime }$; this is the quantity denoted $\Delta$ in the question, not the notation ${\Delta }_L$ for a Lagrangian pressure perturbation. Then mass conservation gives $\hat{\rho }=-\rho D_r$.

The equilibrium pressure gradient and homogeneity identity giveConsequently the adiabatic equation of state yieldsFor the force equations, the product rule givesEquating the coefficients of $F{\mathbf{e}}_r$ and $\nabla F$ in the self-gravitating adiabatic displacement equations givesFinally, applying the Laplacian to the gravitational perturbation givesso the Poisson equation becomesThese are the required interior equations for uniform-density stellar oscillation.

There is a radial degeneracy at $l=0$: $F$ is spatially constant and $\nabla F=0$, so the displacement is independent of $V$. The second force equation then cannot be inferred by equating independent vectors. For nonzero $\omega$ it may be imposed as an auxiliary definition of $V$, but it is not an additional physical radial equation. At zero frequency the radial equations should be used directly. This distinction matters for interpreting the zero factor in (c).

Let $\mathbf{n}=\mathbf{X}/R$, $\zeta =6\pi \mu a$, and letUse the method of reflections for Stokes flow. At the fixed sphere, the first sphere produces the incident Stokeslet velocity ${\mathbf{u}}_{\infty }(0)=\mathsf{A}\mathbf{V}$ to leading order. In Faxén translation law, set the second sphere's translational velocity to zero. The applied holding force is thereforeThe first sphere's finite-radius potential dipole adds $O(V{a}^3/R^3)$ to the incident velocity, as does the Laplacian term in Faxén translation law at the fixed sphere. Multiplying by $\zeta$ gives the stated next correction. There is no intermediate $O(\mu V{a}^3/R^2)$ term. This is the holding force and torque for a sphere in a distant Stokeslet.

The vorticity of a Stokeslet at displacement $\mathbf{r}$ is $\mathbf{F}\times \mathbf{r}/(4\pi \mu r^3)$. At the fixed sphere, $\mathbf{r}=-\mathbf{X}$, soSet its angular velocity to zero in Faxén rotation law. The applied holding couple isThe displayed sign is the external couple needed to oppose the ambient rotation, rather than the hydrodynamic couple on the sphere.

The leading reflected flow at the first sphere is $\mathsf{G}(\mathbf{X}){\mathbf{F}}_2=-{\mathsf{A}}^2\mathbf{V}$. Apply Faxén translation law there with its prescribed force $\mathbf{F}=\zeta \mathbf{V}$. Since $(I+\mathbf{nn}{)}^2=I+3\mathbf{nn}$, the mobility correction from a fixed distant sphere isThe next correction comes from finite-radius terms in the incident/reflected flow and in Faxén translation law, together with the fixed sphere's induced stresslet and holding-couple rotlet. Each gives $O(V{a}^4/R^4)$ at the first sphere. The symbol $\mathbf{V}$ is its isolated-sphere velocity scale, not its actual velocity in the two-sphere problem.

The first sphere has no applied couple. Its angular velocity follows from Faxén rotation law and the reflected Stokeslet:The component of ${\mathbf{F}}_2$ parallel to $\mathbf{X}$ contributes no vorticity. The next reflected rotlet and stresslet give the stated error order.

Use the Gaussian units convention implicit in the $4\pi$ coefficient. For a uniform dielectric and a counterion-only electrolyte, the Boltzmann distribution and Poisson-Boltzmann equation areThe reference electric potential is $\varphi (a)=0$, so $n_0=n(a)$; it is not a prescribed nonzero bulk concentration at infinity. A salt reservoir with additional ion species would require their charge densities as well. With $u=\log (r/a)$, the cylindrical Laplacian is $r^{-2}{d}^2/d{u}^2$. Since $\beta q\varphi =\Phi +2u$ and $r^2=a^2{e}^{2u}$, the factors of $e^{2u}$ cancel:Here primes mean differentiation with respect to $u$, and the surface condition is $\Phi (0)=0$ in this coordinate.

Multiply by ${\Phi }^{\prime }$ to obtain a first integral:The two specified far-field limits set $K=0$. For a solution tending upwards to infinity, take the positive square root. Writing $b=\sqrt{C/2}$ gives $(e^{\Phi /2}{)}^{\prime }=b$, hence the cylindrical counterion-only Poisson-Boltzmann profileFor $b>0$, $\Phi \to \infty$ and ${\Phi }^{\prime }=2b/(1+bu)\to 0$. The corresponding number density is

Apply Gauss's law to a coaxial surface per unit cylinder length. Define the positive accumulated cylindrical screening charge byThe signed outward electric field is $E_r=2[-\lambda +Q(r)]/(ϵr)$, so its magnitude is $2|\lambda -Q(r)|/(ϵr)$. In the branch here, $Q(r)<\lambda$, the field is inward andIn this display, ${\varphi }^{\prime }$ is a radial derivative. The quantities $Q$ are electric charges per unit axial length, not total charges of an infinite cylinder.

At the surface, $a{\varphi }^{\prime }(a)=2\lambda /ϵ$. Define the charge-$q$ Bjerrum length ${\ell }_B=\beta q^2/ϵ$ and the Manning parameter $\eta =\lambda {\ell }_B/q$. Matching the derivative of the solution givesThus the condensed branch exists only for $\eta >1$, and its normalization isThe sign condition $b>0$ is essential: simply squaring $\eta -1$ would produce a spurious subcritical density. Direct integration using $rn(r)dr=n_0{a}^{2}du/(1+bu{)}^2$ yieldsEquivalently, $E(r)=2{\lambda }_c[1+b/(1+b\log (r/a))]/(ϵr)$: the remaining far-field line-charge magnitude is ${\lambda }_c$. The Manning condensed fraction is $1-1/\eta$.