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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.

The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius $r>R_{\odot }$, let $\vartheta$ be the angle between the line of sight and ${\mathbf{R}}_{\odot }$. Geometry gives ${\sin }^2\chi =(R_{\odot }/r{)}^2{\sin }^2\vartheta$, so that particular weighting gives $A=2R_{\odot }^2/(3r^2)$ exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in $A=⟨{\sin }^2\chi ⟩$; the boxed dispersion relation remains valid.

Invert the equation and regard $x$ as a function of $y$:This is a first-order linear differential equation. Multiplication by the integrating factor $e^{-y}$ givesso $x=e^{2y}+C{e}^y$. The initial condition $x=1$ at $y=0$ gives $C=0$. Hence $x=e^{2y}$ and

The first and third equations form a constant-coefficient linear system of differential equationsThe eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $2$ and $-4$. Applying the initial data givesThe remaining equation becomesAn integrating factor, or the usual exponential and trigonometric particular solutions, givesThese values satisfy all three initial conditions.

Multiplication by the integrating factor $e^{2x}$ givesHence the Sturm-Liouville problem is

The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives

At the moving boundary,Substitution into the result of part (b) gives the Rayleigh collapse equationTreat $v=\dot{a}$ as a function of $a$, so that $\ddot{a}=vdv/da$. With $w=v^2$, the equation becomesMultiplication by the integrating factor $a^3$ after division by $a$ givesUsing $w(a_0)=0$,The collapsing branch has negative radial velocity, so

The closed-box model of galactic chemical evolution assumes a fixed total baryonic mass, no gas inflow or outflow, instantaneous and homogeneous mixing, a constant stellar yield, a fixed initial mass function, and usually the instantaneous recycling approximation. It predicts $Z=y\log (1/\mu )$ for gas fraction $\mu$, but real galaxies violate these assumptions.

Evidence for galactic outflows includes blueshifted absorption, broad or split emission lines, extraplanar ionized and molecular gas, X-ray bubbles, and metal-enriched circumgalactic material. The low baryon fractions and low effective yields of dwarf galaxies, together with the galaxy mass--metallicity relation, provide indirect evidence for preferential gas and metal loss. Sustained star formation for longer than a closed reservoir's depletion time, low-metallicity high-velocity clouds, metallicity dilution during starbursts, the G-dwarf problem, and circumgalactic or intergalactic accretion signatures imply continuing galactic gas inflow.

Let $f(t)=A{e}^{-t/\tau }$ be the inflow rate, let the outflow rate be $\lambda \psi$, and let $R$ be the promptly returned fraction. The total baryonic, gas, and metal masses obeyThe first metal term is newly synthesized material; the second locks existing metals into long-lived stars and removes them in an ambient-composition wind. Expanding the final derivative and substituting ${\dot{M}}_g$ cancels those common terms, leaving

The observed Kennicutt–Schmidt law relates star-formation and gas surface densities approximately by ${\Sigma }_{\mathrm{SFR}}\propto {\Sigma }_g^{1.4}$, with a nearly linear relation to molecular gas over many resolved regimes. For the simple integrated model requested here, take a constant depletion coefficient $S$ so thatThen, with $\alpha =S(1-R+\lambda )$,The integrating factor $e^{\alpha t}$ gives, for $\alpha \ne 1/\tau$,At resonance, $\alpha =1/\tau$, the continuous limit isThe first term is depletion of the initial reservoir; the second is gas supplied by the exponentially declining inflow and subsequently consumed or expelled.

Along the unperturbed photon path, combine the temperature and gravitational-redshift terms and use ${\mathcal{T}}^{\prime }=-\Gamma$:The integrating factor is $e^{-\mathcal{T}}$, and $g=\Gamma e^{-\mathcal{T}}$. Neglecting the exponentially hidden initial boundary term gives the Cosmic microwave background line-of-sight solutionFor instantaneous recombination, $g(\tau )=\delta (\tau -{\tau }_{\mathrm{dec}})$, and the observer potential contributes only an unobservable monopole. ThereforeThe first two terms form the ordinary Sachs-Wolfe effect, the velocity term is the Doppler CMB anisotropy at last scattering, and the integral is the Integrated Sachs-Wolfe effect produced by evolving potentials.

Assume the condition in part (ii), so $m_0<0$. Substituting the wavevector solution into the vertical ray equation gives the linear ordinary differential equationwhereThe integrating factor $e^{\gamma t}$ givesThe bracket is strictly positive, so $z(t)>0$ at every finite time. Moreover,so the improper integral converges to a finite positive value. It follows thatfor some $C>0$. The packet therefore approaches $z=0$ asymptotically and never reaches it at finite time. This is the vertical trapping of an internal gravity wave by planar strain.

LetThe integrating factor identity $(Iy{)}^{\prime }=If$ givesIf $y_1,y_2$ are distinct solutions, their difference is a nonzero homogeneous solution, so every solution is

The coefficient matrix has eigenvector $u_{+}=(1,1{)}^T$ with eigenvalue $3$ and eigenvector $u_{-}=(1,-1{)}^T$ with eigenvalue $-1$. WriteSinceand $y(0)=(1,-2{)}^T=-\frac{1}{2}{u}_{+}+\frac{3}{2}{u}_{-}$, the system diagonalises toUsing an integrating factor givesThe unique fastest term is thereforeConsequently $e^{-nt}y(t)$ diverges for $n<3$, tends to zero for $n>3$, and for $n=3$ tends to the finite nonzero vectorThus the only requested integer isThis is an instance of dominant eigenmode in a forced linear system.

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.