Split the inverse transform into the intervals $|\omega |<\mu$ and $|\omega |>\mu$, and in each case pair the positive-frequency integral with its complex conjugate at negative frequency. The low-frequency pair becomessince $B=0$ and the decaying spatial factor is real. In the high-frequency positive half, replace $A(\omega )$ by $B(-\omega {)}^{\ast }$. The negative half is again its conjugate. Combining the two pieces givesEvery integrand satisfies the massive wave equation because ${\omega }^2=k^2+{\mu }^2$. Linearity, and taking the real part, preserve the equation whenever differentiation under the integrals is justified, or distributionally under the corresponding spectral assumptions. The construction is exactly the real bounded Fourier modes of the Klein-Gordon equation decomposition.

The displayed expression uses ordinary spectral amplitudes. If distributions supported at $\omega =\pm \mu$ are admitted, a further bounded threshold contribution $c_1\cos \mu t+c_2\sin \mu t$, independent of $r$, is possible. It is not a missing propagating or evanescent branch of the open frequency intervals.

Reality requires $\tilde{\sigma }(\omega ,r)=\tilde{\sigma }(-\omega ,r{)}^{\ast }$. The way it constrains the amplitudes depends on whether the spatial wave number is real or imaginary.

For $|\omega |>\mu$, $k$ is real and even. Complex conjugation interchanges the two exponentials, so comparing independent spatial factors givesThe companion equation $B(\omega )=A(-\omega {)}^{\ast }$ is the same constraint with frequency reversed and conjugated. Neither oscillatory spatial branch grows as $r\to \infty$.

For $|\omega |<\mu$, put $k=i\kappa$ with $\kappa >0$. ThenBoundedness excludes the growing exponential, and reality then pairs the remaining amplitudes:These are the two independent low-frequency requirements. At threshold boundedness eliminates the separate linear-in-$r$ solution discussed in (ii). The constraints are understood for a Fourier transform representation in which spatial mode amplitudes are well defined; sufficient amplitude integrability also ensures the spectral integral itself is bounded.

At fixed frequency, the transformed equation has characteristic roots $\pm ik$, whereFor a consistent explicit branch chooseThis branch is even in $\omega$. Away from the two thresholds, the spatial solution is $\tilde{\sigma }=A(\omega )e^{ikr}+B(\omega )e^{-ikr}$, so Fourier inversion givesAt $|\omega |=\mu$, the spatial equation has a double zero characteristic root and its full solution is $C+Dr$. The two exponentials at $k=0$ represent only its constant part. This isolated-frequency issue does not change an integral with ordinary integrable amplitudes; threshold-supported distributional modes require separate treatment.

Transform in time, leaving $r$ as a parameter. Applying the derivative identity twice gives $\widetilde{{\partial }_t^2\sigma }=-{\omega }^2\tilde{\sigma }$. ThereforeThese manipulations can be justified for sufficiently decaying smooth fields, or with distributional Fourier transforms when ordinary integrability is unavailable.

For real $f$, complex conjugation commutes with its absolutely convergent Fourier integral:ThusThis Hermitian symmetry pairs the positive- and negative-frequency components of a real signal; its real part is even and its imaginary part is odd.

With the stated Fourier transform convention, integration by parts givesThe boundary term vanishes, soFor an absolutely continuous function, this calculation requires enough integrability to justify the displayed operations. Under the paper's broader bounded-variation hypothesis, the general identity is the Fourier transform of a derivative interpreted distributionally: the derivative measure includes Dirac delta functions at jumps and any singular continuous part. If one instead differentiates only the smooth pieces and discards jumps, the formula needs the corresponding jump corrections. Absolute integrability and bounded variation ensure the limits of $f$ at both infinities are zero, but do not by themselves make its ordinary derivative capture those singular parts.

First move the order parameter term to the right-hand side:Here $p=x^2$, $r=x$, so the integrating factor obeysChoose the irrelevant positive constant $C=1$. Multiplying the equation by $1/x$ givesThus a Sturm-Liouville theory eigenvalue formulation with the requested eigenvalue isThis is the Sturm-Liouville formulation with Bessel order as eigenvalue. The spectral parameter is the squared order of the Bessel function, with $\beta$ treated as fixed. This differs from the also familiar formulation in which ${\beta }^2$ is the eigenvalue.

The factor $1/x$ is singular at the origin. Consequently the integral defining $w$ cannot literally use lower limit zero; take a positive reference point $x_{\ast }>0$, obtain $w=x_{\ast }/x$, and absorb $x_{\ast }$ into the normalization. The differential identity holds for $x>0$. Boundary conditions and admissibility at zero and at the outer endpoint are additional ingredients needed to make a particular self-adjoint spectral problem.

When $r=p^{\prime }$, write the differential expression in divergence form,Two integrations by parts give the Lagrange identity for a Sturm-Liouville operator:The $q$ terms cancel because $q$ is real. Hence the exact boundary criterion, in the sense of self-adjointness defined in the question, isAt regular endpoints, all separated homogeneous conditions of the following type satisfy it:with real coefficients, not both zero at either endpoint. The allowed value-and-flux pairs at an endpoint are proportional, so their determinant $p(f{g}^{\prime }-f^{\prime }g)$ vanishes. This includes Dirichlet boundary conditions, Neumann boundary conditions, Robin boundary conditions and arbitrary mixtures of these at the two ends.

Coupled conditions can also work: periodic conditions $y(b)=y(a)$ and $p(b)y^{\prime }(b)=p(a)y^{\prime }(a)$ make the two endpoint expressions equal; simultaneous minus signs give the antiperiodic case. Thus requiring each endpoint term separately to vanish is sufficient, but not necessary.

More generally, for regular endpoints set $Y=(y(a),y(b){)}^{\mathsf{T}}$ and $N=(-p(a)y^{\prime }(a),p(b)y^{\prime }(b){)}^{\mathsf{T}}$. All maximal homogeneous self-adjoint boundary conditions can be written $AY+BN=0$ with real two-by-two matrices satisfying $\mathrm{rank}(AB)=2$ and $A{B}^{\mathsf{T}}=B{A}^{\mathsf{T}}$. If an endpoint is singular, the relevant limiting boundary form and admissible domain must instead be specified; the assumption $p>0$ only in the interior does not itself fix those endpoint limits.

The far field is proportional to $r{P}_1(\cos \theta )$, since $P_1(\cos \theta )=\cos \theta$. After matching the growing terms, write the exterior solution asAn added constant has here been excluded by interpreting the far-field condition as $U-v_{0}r\cos \theta \to 0$. Differentiate and impose the radial Neumann boundary condition:The Orthogonality of Legendre polynomials makes each coefficient vanish separately. Thus $B_n=0$ for $n\ne 1$, and $2B_1/r_0^3=v_0$. ConsequentlyIndeed ${\partial }_{r}U=v_0(1-r_0^3/r^3)\cos \theta$, which vanishes at the boundary, while the correction decays at infinity. If the far-field condition is intended to specify only the leading growing term, an arbitrary additive constant is also allowed; its gradient has no effect on the Neumann boundary condition.

Substitute $U=R(r)F(\theta )$ into the axisymmetric Laplace equation and multiply by $r^2/(RF)$. The radial and angular terms must be separate constants:ThusWith $x=\cos \theta$, the angular equation becomesRequiring boundedness on the entire polar interval, and using the supplied characterization of bounded Legendre polynomial solutions, gives $\lambda =n(n+1)$ and $F\propto P_n(\cos \theta )$ for $n=0,1,2,\dots$. Trying $R=r^s$ gives $s(s+1)=n(n+1)$, hence $s=n$ or $s=-n-1$.

The globally angular-regular separated modes and their superposition are thereforeCoefficients and convergence are determined by the radial domain and boundary data. A domain containing the origin excludes negative radial powers for a regular solution. If the angular domain excludes a polar axis of the spherical coordinate system and singular angular behavior is allowed, additional Legendre-function branches are possible; the displayed family is the one regular over the complete sphere, as intended by the hint.

For an integer $n\ge 0$, choose the series whose integer parity agrees with $n$ and set the other initial coefficient to zero. At $m=n$, the numerator in the recurrence vanishes, so $a_{n+2}=0$ and every later coefficient in that integer parity vanishes. With a nonzero coefficient of $x^n$, this gives a polynomial of degree exactly $n$. Normalize it by $P_n(1)=1$.

For $n=0$, the even branch is constant; for $n=1$, the odd branch is proportional to $x$. For $n=2$, the recurrence gives $a_2=-3a_0$, so normalization fixes $a_0=-1/2$. HenceThe opposite-parity branch generally does not terminate; polynomial termination does not mean that both independent solutions become polynomials.

Differentiate the power series term by term inside its radius of convergence, and reindex the second derivative so that every coefficient multiplies $x^m$. The differential equation becomesThe uniqueness of power-series coefficients givesThe even coefficients depend only on $a_0$ and the odd coefficients only on $a_1$. These are the two local solution branches of the Legendre differential equation around its ordinary point $x=0$.

For a finite separation use the Cartesian distance, rather than substituting finite coordinate differences into the local line element. Expanding the two horizontal components givesThe vertical contribution is $(z-z^{\prime }{)}^2$, independently. ThusandEquivalently, the horizontal contribution is $(\rho -{\rho }^{\prime }{)}^2+4\rho {\rho }^{\prime }{\sin }^2[(\varphi -{\varphi }^{\prime })/2]$, which also makes its nonnegativity explicit.

Differentiate the position vector, using the angular factor from the scale factors of orthogonal coordinates:The basis is an orthonormal basis, so all cross terms in its squared length vanish. The Euclidean line element is thereforeIn particular, an angular change $d\varphi$ at fixed radius traverses a physical distance $\rho |d\varphi |$.

Dividing the coordinate tangents by their scale factors of orthogonal coordinates gives an orthonormal basisEvaluate the Cartesian cross products componentwise:ConsequentlyThe ordered cylindrical basis is right-handed: its scalar triple product ${\mathbf{e}}_{\rho }\cdot ({\mathbf{e}}_{\varphi }\times {\mathbf{e}}_z)$ equals $+1$.

Write the position vector as $\mathbf{r}(\rho ,\varphi ,z)=(\rho \cos \varphi ,\rho \sin \varphi ,z)$. Its coordinate tangent vectors areTheir pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, areThus the cylindrical coordinate system is orthogonal wherever it is a nonsingular coordinate chart. At $\rho =0$, the angular tangent vector vanishes and the azimuth is undefined; this is the usual coordinate singularity on the axis, not a failure of the Euclidean geometry.

Let $M_{ijkl}={\int }_V{x}_i{x}_j{x}_k{x}_l{d}^{3}x$. It is an isotropic tensor and is invariant under every permutation of its four indices. Thus the three coefficients in the general rank-four isotropic form must coincide:A double tensor contraction givesHere the three contracted Kronecker delta products contribute $9$, $3$ and $3$. Contracting instead with the four unit vectors yieldsThis agrees with subtracting two instances of the isotropic fourth moment of a ball.

Under inversion $\mathbf{x}\mapsto -\mathbf{x}$, the spherical shell and its volume element are unchanged, while the product of three scalar products changes sign. Its integral is therefore its own negative:Equivalently, its rank-three moment tensor is completely symmetric and cannot have a nonzero multiple of the antisymmetric Levi-Civita symbol.

Introduce the second moment tensor $M_{ij}={\int }_V{x}_i{x}_j{d}^{3}x$. The spherical shell is preserved by every rotation, so this is an isotropic tensor, $M_{ij}=A{\delta }_{ij}$. Its trace determines the constant:Contracting with the components of the two unit vectors givesThis is the second-order case of radial tensor moments in three dimensions.

Use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The unit circle is traversed anticlockwise, and the integral becomesOnly the simple pole $z=i/3$ lies inside the contour. Its residue isHence the residue theorem gives