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Assume for this spectral construction that the source and known attenuation are sufficiently smooth and decaying, for instance compactly supported, and that $\mu \ge 0$ for the physical interpretation. The following steps identify both the forward attenuated Radon transform and the route to its inversion.

First write the complex transport operator asFor $k=e^{i\theta }$ this is the real directional derivative ${\mathbf{d}}_{\theta }\cdot \nabla$, with ${\mathbf{d}}_{\theta }=(\cos \theta ,\sin \theta )$. Put ${\mathbf{n}}_{\theta }=(-\sin \theta ,\cos \theta )$ and $x=p{\mathbf{n}}_{\theta }+s{\mathbf{d}}_{\theta }$. The spectral equation on the unit circle becomesIts minus sign fixes the appropriate endpoint condition: use $F(p,+\infty )=0$. The integrating factor givesThus the measured quantity at the opposite end of the line isThe exponent is the attenuation accumulated between the source point and the detector at the negative end. The common convention with detector at the positive end is the same transform after reversing the direction, ${\mathcal{R}}_{\mu }^{+}f(p,\theta )={\mathcal{R}}_{\mu }^{-}f(-p,\theta +\pi )$. Using an incoming zero condition at the negative end while retaining the printed minus sign would instead give a growing integrating factor, not physical attenuation.

Next, for $|k|\ne 1$, the operator $L_k$ is a complex elliptic first-order operator. Its decaying whole-plane Green function isThis is obtained by a real-linear change of variables in the Cauchy-Green operator; its change of orientation explains the sign. Write $G_{k}a$ for convolution with this kernel. Solve $L_{k}h=\mu$, and set $F=e^h\Psi$. This removes the attenuation:The normalized spectral solution is analytic separately inside and outside the unit circle. As $k$ approaches that circle, the Green function's characteristic singularity produces two limiting values. Their relation is computed from the weighted line integrals above together with transverse Hilbert transforms; the known attenuation determines the integrating factor weights. The two spectral limits are not individually just the incoming and outgoing real characteristic solutions: the singular-kernel prescription matters.

Finally formulate the resulting additive Riemann-Hilbert problem on the unit circle. Its jump is determined by the measured attenuated Radon transform and known $\mu$. With the unit circle oriented counterclockwise and jump $J=F_{\mathrm{inside}}-F_{\mathrm{outside}}$, a Cauchy integral formula reconstructs the normalized spectral solution:The normalization at zero supplies the compatibility condition $\int J/\zeta d\zeta =0$. Recover $f$ from $L_{k}F-\mu F$, or from its small-$k$ coefficient: if $F=k{F}_1+O(k^2)$, then $f={\partial }_{\bar{z}}{F}_1$. Equivalently the large-$k$ coefficient gives $f={\partial }_z{F}_{-1}$ when $F=k^{-1}{F}_{-1}+O(k^{-2})$. This spectral reconstruction of an attenuated Radon transform is the analogue of recovering $q$ from a coefficient in (ii). Known attenuation and full directed line data are inputs; one does not determine an arbitrary unknown attenuation and source simultaneously from this argument.

The integrating factor is $e^{-2x}$, so $(e^{-2x}y{)}^{\prime }=e^{(\lambda -2)x}$. Integrating gives the general solutionThe particular solution is only defined up to a homogeneous solution. Subtract $e^{2x}/(\lambda -2)$ from it, absorbing that multiple into the arbitrary constant. The same family can then be writtenHolding $\tilde{C}$ fixed, the quotient tends to $x{e}^{2x}$, the derivative of $e^{\lambda x}$ with respect to $\lambda$ at $2$. Thus the resonant family is $y=e^{2x}(\tilde{C}+x)$. Direct differentiation gives $y^{\prime }-2y=e^{2x}$, so it is the full general solution at resonance. This resonant exponential forcing in a first-order equation limit concerns a reparametrized family, not a fixed value of the original divergent constant $C$. The initial condition gives $\tilde{C}=2e^{-2}-1$, hence

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

The unknown Neumann boundary condition can be removed by the Fokas method. Evaluate the global relation for the half-line free Schrodinger equation at $-k$:Consequently $k{G}_0-i{G}_1=2k{G}_0-{\hat{q}}_0(-k)+e^{i{k}^{2}t}\hat{q}(-k,t)$. In the representation from part (a), the last term has integral ${\int }_{\partial D_{+}}{e}^{ikx}\hat{q}(-k,t)dk=0$, by the Cauchy integral theorem and Jordan lemma. It is analytic in the upper half-plane, and the exponential decays on the closing first-quadrant arc. Hence an expression involving only the given data isOne may replace $t$ by $T$ in the upper limit defining $G_0$: the contribution of boundary times $s>t$ closes to zero in $D_{+}$, since $e^{-i{k}^2(t-s)}$ then decays there. Using $t$ makes causality transparent.

For an explicit proof of uniform convergence, it is useful to apply boundary lifting before inversion. Set $h(x)=e^{-x}$, $c=g_0(0)$, $w_0(x)=q_0(x)-ch(x)$, andThe compatibility condition gives $w_0(0)=0$. The Fourier sine transform of $w=q-g_{0}h$ satisfies $S_t+i{k}^{2}S=Hf$, with initial value $S_0$. The integrating factor therefore gives the equivalent representationThis last integral is absolutely and uniformly convergent for $x\ge 0$, $0\le t\le T$ under concrete sufficient hypotheses $w_0,w_0^{\prime },w_0^{\prime \prime }\in L^1$, decay of the boundary terms, and $g_0\in C^2[0,T]$. Indeed, two integrations by parts give $|S_0(k)|\le \Vert w_0^{\prime \prime }{\Vert }_1/k^2$ for $k\ge 1$. Another integration by parts, this time in $s$, givesThus the initial term is uniformly $O(k^{-2})$ and the forcing term uniformly $O(k^{-3})$. For $0<k<1$, use $|S_0(k)|\le \Vert w_0{\Vert }_1$ and the bounded time integral. An integrable majorant proves the claimed uniform convergence and permits evaluation at both boundaries.

At $t=0$, Fourier sine inversion gives $q(x,0)=c{e}^{-x}+w_0(x)=q_0(x)$. At $x=0$, the integral vanishes, giving $q(0,t)=g_0(t)$, including the compatible corner. To verify the equation, note that $h^{\prime \prime }=h$ and the transformed equation impliesAdding the lifted part gives $i{q}_t+q_{xx}=0$. Under the stated smoothness, this holds classically in the interior; differentiated spectral integrals can first be Gaussian-regularized, or read in the sine-transform $L^2$ sense and then identified with the smooth solution. Uniform convergence of $q$ itself does not require claiming uniform convergence of every differentiated integral at the corner.

The dimensions are $[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[u_0]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Under the stated model, $m,\alpha ,u_0,g$ are the only parameters determining the ascent time. If $m^p{\alpha }^q{u}_0^r{g}^s$ is dimensionless, then $p+q=0$, $r+s=0$, and $-q-r-2s=0$, leaving one independent dimensionless variable. One convenient choice isThus dimensional analysis fixes the form, but Newton's second law must determine the function $f$.

Take upward velocity as positive. Throughout ascent, Newton's second law with linear drag gives $m\dot{v}=-mg-\alpha v$. An integrating factor or direct solution of this linear ordinary differential equation givesThe first zero of this velocity is the highest point, soThis vertical ascent under linear drag has $T<u_0/g$ for $u_0>0$, since $\log (1+\lambda )<\lambda$. The limit $\alpha \to 0$ recovers the no-drag ascent time $u_0/g$; $u_0=0$ gives $T=0$.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

Use $A=R_e^2-R_i^2$, $S=R_e+R_i$, $D=R_e-R_i$, and $Q=Aw=A(0)w_0$. Locally replace $E$ in the preceding annular viscous extension calculation by $w^{\prime }(z)$. Its leading pressure isThe axial liquid stress relative to ambient pressure, together with the two circumferential surface tension forces, transmits a cut forceIt would be incorrect simply to differentiate this and retain an extra gas-pressure term. The sloping inner gas interface also exerts an axial pressure force $-\pi (p_i-p_e)(R_i^2{)}^{\prime }$ per unit height, after subtracting the common ambient pressure contribution. For constant gas pressures these terms cancel. The resulting capillary tensile force of a hollow slender thread therefore givesThus the printed axial equation holds even for unequal constant gas pressures; their influence remains in the radial evolution.

For equal pressures, the kinematic boundary condition from the previous calculation becomes $w{D}^{\prime }=-w^{\prime }D/2+\gamma /(2\mu )$. Multiplying by the appropriate integrating factor givesAt closure $R_i=0$, so $D=\sqrt{A}$ and $D\sqrt{w}=\sqrt{Q}$. HenceThe right side is positive for a genuine annulus. On the gravity-driven quadratic family from the previous part, $w^{1/2}$ grows linearly with height and this integral diverges logarithmically. It must then reach the finite positive threshold at a finite height. This explains the expected capillary closure of a falling hollow thread on that freely falling branch without requiring its full coupled profile.

There is a real boundary-data limitation to that expectation. Neither the printed nozzle speed nor the local equations alone guarantee finite closure for every imposed nozzle tension. To see this within the same leading model, define $T=3\mu Q{w}^{\prime }/w+\gamma S$, so $T^{\prime }=-\rho gQ/w$. For equal pressures and $w^{\prime }>0$, both radii decrease and $S\le S_0$. Ifthen a bootstrap gives $T>T_0/2$ and $w\ge w_0{e}^{cz}$: the maximum accumulated weight is smaller than the assumed tension margin. Consequently the closure integral, including its prefactor, is at most $\gamma /(\mu c{w}_0)$. If this is smaller than $\sqrt{A_0}-D_0$, closure never occurs at finite height. For example, in consistent units choose $\mu =\rho g=w_0=1$, $R_i(0)=\lambda$, $R_e(0)=2\lambda$, $\gamma =\lambda$, $T_0=100{\lambda }^2$. Then $c=47/9$, the weight bound is $27{\lambda }^2/47<50{\lambda }^2$, and $9\lambda /47<(\sqrt{3}-1)\lambda$. Small $\lambda$ makes the initial geometry slender. Thus finite closure is the expected freely falling branch behaviour, not a theorem following from nozzle speed alone. If $\gamma =0$ there is no capillary closure by this mechanism.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

An integrating factor explains the second hint and the first-order inner calculation. Formultiplication by $e^{z^2}$ gives $(e^{z^2}{Y}^{\prime }{)}^{\prime }=e^{z^2}(2a_{1}z+a_2+a_3\mathrm{erf}z)$. The term $a_{1}z$ already produces $2a_{1}z$, and integrating the remaining terms twice gives $a_2{E}_0+a_3{E}_1$, in terms of the Gaussian drift primitives. The homogeneous solutions are a constant and the error function. HenceFor the piecewise forcing in the inner problem, $F$ replaces $a_{1}z$ and provides the continuously matched particular solution for $2zH(-z)$. The coefficients $2c$ and $2d$ then account for the $2Y_0$ forcing.

Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere ${\sigma }_t^2$ is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is $1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)$. These definitions agree; they must not be mixed by inserting an extra factor of two.

The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is $2\nu ({\sigma }_r^2-{\sigma }_t^2)/r$. Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.

By the shell theorem, the gravitational potential of a spherical mass distribution satisfies $d\Phi /dr=GM(<r)/r^2$. Expanding the derivative above, dividing by $\nu$, and using logarithmic derivatives givesThe tracer density $\nu$ need not be the density of the gravitating matter. The circular speed is $V_c^2=rd\Phi /dr=GM(<r)/r$, whether or not the tracers themselves follow circular orbits.

Now let $\beta$ be constant, $\nu ={\nu }_0(r/r_0{)}^{-\gamma }$, and $V_c^2=V_0^2(r/r_0{)}^{2\alpha }$. With $y={\sigma }_r^2$, the Spherical Jeans equation readsFor a power-law ansatz $y=K{V}_c^2$, $y^{\prime }=2\alpha y/r$, so $(\gamma -2\beta -2\alpha )y=V_c^2$. Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.

There is a boundary condition implicit in selecting this solution. The integrating factor $r^{2\beta -\gamma }$ gives, when $D=\gamma -2\beta -2\alpha \ne 0$,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring $r^{2\beta }\nu {\sigma }_r^2\to 0$ at infinity. Under that condition,and convergence requires $D>0$, giving the boxed result. If $D=0$, the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when $D\ne 0$.

To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector ${\mathbf{R}}_{\odot }$ of length $R_{\odot }$, and a tracer at $\mathbf{r}=r\hat{\mathbf{r}}$. Its heliocentric line of sight isIf $\chi$ is the angle between $\hat{\mathbf{\ell }}$ and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues ${\sigma }_r^2,{\sigma }_t^2,{\sigma }_t^2$ in the local spherical basis. Projection onto the line of sight therefore givesAt fixed $r$, ${\sigma }_r$ and $\beta$ are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes $u$ uniform on $[-1,1]$, giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set $a=r^2+R_{\odot }^2$ and $b=2r{R}_{\odot }$, and divide the integrand:The odd term integrates to zero, while ${\int }_{-1}^{1}du/(a-bu)=b^{-1}\ln [(a+b)/(a-b)]$. For $r>0$, $R_{\odot }>0$, and $r\ne R_{\odot }$, this givesThe removable limit at $r=R_{\odot }$ is $A=1/2$. As $r/R_{\odot }\to \infty$, $A\simeq 2R_{\odot }^2/(3r^2)$, while as $r/R_{\odot }\to 0$, $A\to 2/3$. If $R_{\odot }=0$, $A=0$ and the observed velocity dispersion is exactly radial. For isotropic velocities, $\beta =0$, the correction vanishes at every radius.