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Write $Y=B_{\varphi }^2$, $x=R^2/L^2$, and retain the frozen axial field $B_z=(B_0/{\rho }_0)\rho$. The radial magnetostatic equilibrium equation now includes both the gradient of the toroidal magnetic pressure and the inward magnetic tension:For $\rho ={\rho }_1{e}^{-x}$, cylindrical gravity givesMultiplication by the integrating factor $R^2$ therefore producesThe integration constant must vanish for regularity on the axis. Direct integration givesIts Taylor expansion at the axis isso the regular field behaves as $B_{\varphi }=O(R)$. At large radius,Nonnegativity at infinity requiresThis condition is also sufficient: at the limiting value, the braces divided by $K_{\mathrm{eff}}{L}^2$ reduce to $2e^{-x}[1-(1+x)e^{-x}]\ge 0$, and decreasing $L^2$ only increases them. Thus the toroidal magnetic field is real and regular at every radius exactly in the stated range.

LetFlux conservation gives $u=U/H$. Substitution into the extensional-force equation yields the third-order equationSinceone integration, using $H\to 1$ and its derivatives tending to zero on the flat-film side, givesPut $q(H)=H_{\xi }$. After division by $q$, this becomesThe integrating factor $H^{-1/2}$ givesA second integration and $q(1)=0$ therefore give

On the bubble side, matching to its cylindrical shape gives $H\sim {\xi }^2/2$, so $H^{-1/2}{H}_{\xi }\to \sqrt{2}$. Taking $H\to \infty$ in the integrated equation yields $8V/3=\sqrt{2}$. Hence

With the convention in the question, the constant velocity-anisotropy parameter is $\beta =1-{\sigma }_t^2/{\sigma }_r^2$. Since $V_c^2=rd\Phi /dr$, the Spherical Jeans equation becomesFor $\nu ={\nu }_0{r}^{-\alpha }$ and $V_c^2=V_0^2{r}^{2\gamma }$, its general integrating factor solution isThe homogeneous term represents a boundary pressure. For an extended scale-free system the physical boundary condition normally removes it, leavingPositivity requires $\alpha -2\beta -2\gamma >0$. A complete physical model must also have a nonnegative galactic distribution function and sensible inner and outer boundary behaviour; for example, strong radial anisotropy is restricted by density-slope--anisotropy inequalities.

Observationally, the tracer density can be estimated from star counts only after correcting distances, extinction, survey selection, and incompleteness. Spectroscopy supplies mainly line-of-sight velocities; proper motions add transverse information but become less precise for distant halo stars. The equation shows the mass--anisotropy--density degeneracy directly: the same measured ${\sigma }_r$ can result from a larger $V_c$, a steeper tracer slope $\alpha$, or a different $\beta$. Even globally constant power laws therefore do not determine the galactic mass profile unless some of these quantities are independently constrained.

If $\alpha$, $\beta$, or $\gamma$ changes near a break radius, the solution at one radius also depends on the outer boundary integral. A break in observed dispersion may be attributed to a mass-profile feature, a tracer-density break, or a change in orbital anisotropy. Separate tracer populations, full three-dimensional velocities, higher velocity moments, and measurements over a wide radial range help break this degeneracy.

More flexible alternatives model a nonnegative solution of the Collisionless Boltzmann equation itself. An action-based galactic distribution function gives an analytic or parametrized $f(\mathbf{J})$; a Schwarzschild orbit-superposition model assigns nonnegative weights to an orbit library; and a made-to-measure stellar-dynamical model adjusts particle weights to reproduce observations. These methods retain more phase-space information than Jeans moments, although their flexibility introduces model choices and regularization.

For a Keplerian orbit, write $h=K{r}^{1/2}$, so $h^{\prime }=h/(2r)$. Substituting the stated torque laws intogivesWiththe integrating factor is $x^{1/2}{e}^{\lambda \sqrt{x}}$. The boundary condition $y(1)=0$ then givesThusand

Let the source's hydrogen-ionizing photon production rate bewhere $h{\nu }_0$ is the hydrogen ionization energy. The idealized Strömgren sphere has an almost fully ionized interior and a thin ionization front. In photoionization equilibrium, every ionization is balanced by a Case B recombination, so spherical symmetry givesFor pure hydrogen of constant number density $n$, the interior has $n_e=n_p=n$, and the Strömgren radius is therefore

Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust absorption coefficient is $a(r)=f_d{\sigma }_{d}n(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the linear ordinary differential equationIf $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d{\sigma }_d$. Define the dust optical depthMultiplication by the integrating factor $e^{{\tau }_d(r)}$ and integration to the dusty front gives its governing equationThe exponential function weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.

For constant $n$, put $a=f_d{\sigma }_{d}n$ and $\tau =a{R}_d$. The elementary integralreduces the equation toEquivalently, with $y=R_d/R_S$ and ${\tau }_S=a{R}_S$,Its dust-free limit is $y\to 1$, while absorption makes $R_d<R_S$ for nonzero dust abundance.

The odd $X_1$ copy number decreases by two until it reaches one, so $⟨x_1(t)⟩\to 1$. The exact $X_2$ first-moment equation isBecause copy numbers are nonnegative integers, $x_2^2\ge x_2$, and the last mixed moment is nonnegative. HenceThe integrating factor $e^{{\alpha }_{3}t}$, together with $⟨x_1(t)⟩\to 1$, gives the comparison bound

Subtracting the exact $X_4$ first-moment equation from the $X_3$ equation cancels reaction 5:The assumed inequality is preciselyChoose a positive $\epsilon$ smaller than this gap divided by ${\alpha }_6$. The limit superior bound implies that, for all sufficiently large $t$,Thus $⟨x_3⟩-⟨x_4⟩$ grows at least linearly. Since $⟨x_4⟩\ge 0$,so the required species index is $\boxed{i=3}$.

The critical-point equations are $2xy=0$ and $x^2+y^2=1$, so the points are $(\pm 1,0)$ and $(0,\pm 1)$. The Hessian isThus $(0,1)$ is a strict minimum, $(0,-1)$ a strict maximum, and $(\pm 1,0)$ are saddles. The local contour patterns are respectively ellipses, inverted ellipses, and crossing hyperbolas.

The gradient flow has $\dot{x}=-2xy$ and $\dot{y}=1-x^2-y^2$. HencePutting $u=y^2$ gives $x{u}^{\prime }-u=x^2-1$. Multiplication by the integrating factor $1/x$ yieldsand therefore the general trajectory is

Let $z$ increase downward. The slender viscous thread is locally in uniaxial extension. The transverse stress equals the ambient pressure, so the Trouton ratio gives the excess axial stress ${\sigma }_{zz}+p_e=3\mu w_z$. Mass conservation and axial force balance therefore give

In steady flow $Aw=Q$. Dividing the momentum equation by $A=Q/w$ givesWith $w=\hat{w}W$ and $z=\hat{z}Z$, chooseThe dimensionless equation becomesTreating $(W^{\prime }{)}^2$ as a function of $W$ and using an integrating factor yieldsBecause $A=Q/(\hat{w}W)$, the area sketches are the reciprocals of the functions $W$ found below. The dimensional vertical deviatoric stress is

Substitute part (ii) into $\dot{E}+q_{EI}E=q_{SE}{S}_0{e}^{-q_{SE}t}$. Multiplication by the integrating factor $e^{q_{EI}t}$ givesSince $E(0)=0$, integration and rearrangement yield

Complexifying the directional transport equation yields spectral solutions analytic inside and outside the unit circle. An integrating factor built from known attenuation removes its multiplication term. The limiting Green kernels encode weighted line measurements and transverse Hilbert transforms. Their jump gives a Riemann-Hilbert problem; its solution recovers the source from an asymptotic coefficient. A general primary treatment of this spectral construction is www.stat.uchicago.edu/~guillaumebal/PAPERS/AttRadTh.pdf .