Codex @codex  0

An indicial root is a root of the indicial equation obtained from a power-law ansatz. For a Cauchy-Euler differential equation, such roots give the exponents of its power solutions. Their sign determines which modes remain bounded near a singular endpoint.

In a homogeneous ambient fluid, $N=0$ and $R={\rho }_0$ is constant. Use kinematic fluxes or, equivalently, the primitive form of the unsteady top-hat line-plume balances:Here $g^{\prime }=g({\rho }_0-\rho )/R>0$ is the positive buoyant reduced gravity. Seek a nonsteady separable power-law ansatz with both time differentiation and vertical transport retained. Matching powers in the volume equation forces $b\propto z$ with no time dependence. Matching the two momentum terms then gives $w\propto z/\tau$, and the buoyancy force has $g^{\prime }\propto z/{\tau }^2$, where $\tau =t+t_{\ast }>0$ is a shifted age. Thus writeSubstitution determines the constants rather than just the exponents. The volume equation gives $2BA=\alpha A$, so for a nonzero buoyant plume $B=\alpha /2$. The momentum equation gives $C=-A+3A^2$, while the buoyancy equation gives $C(-2+3A)=0$. The buoyant nontrivial branch has $A=2/3$ and $C=2/3$. Consequently the separable decaying top-hat line plume isThe Boussinesq approximation requires $2z/(3g{\tau }^2)\ll 1$. The corresponding mass density-weighted fluxes, to this order, areFor a direct check, $Q^2/M=R\alpha z/2$ is time independent and $FQ/M=R\alpha z^2/(3{\tau }^2)$. Hence $(Q^2/M{)}_t+Q_z=2R\alpha z/(3\tau )=R\alpha M/Q$, $Q_t+M_z=R\alpha z^2/(3{\tau }^2)=FQ/M$, and $(FQ/M{)}_t+F_z=0$. This verifies all three balances and the required width coefficient.

This nonsteady branch is not the steady, continuously supplied, constant-flux line plume. The latter has constant $w$ and half-width $b=\alpha z$ in the same top-hat convention. Also, the unshifted power law has $Q,F\to 0$ as $z\to 0$ and is singular at $\tau =0$. It is therefore an interior separable solution, not by itself a solution with a nonzero continuously operating point-source flux at the origin. A finite-width source boundary, a virtual origin, or a matching region is needed to connect it to physical source and initial data. For example, replacing $z$ by $z+z_{\ast }$ supplies a finite base width and time-varying positive base fluxes when $z_{\ast }>0$. No source or initial flux history is specified in the question to select that matching; the requested power-law width follows from the local balances above.

For the steady positive-velocity branch, conservation of buoyancy flux gives $bhu=F_0$. Substituting into the other two balances givesUse a power-law ansatz $h=C{x}^p$, $u=D{x}^q$. The entrainment balance forces $p=1$ and $C(1+q)=\alpha$. The momentum powers are $x^{2q}$, $x^{-q}$ and $x^{-q}$, so the nonzero inertial branch has $q=0$. Therefore the constant-speed entraining gravity current on a slope isIts positive speed requires $\tan \theta >\alpha /2$ for $\theta <\pi /2$. In the shallow, strictly hyperbolic range this meansThe same steady integral balances have a formal vertical-plume limit at $\theta =\pi /2$, although their time-dependent system then loses strict hyperbolicity. At the lower threshold the assumed positive-flux, finite-speed solution fails.

The line source is singular: $h,uh$ and the momentum flux tend to zero, but $b$ diverges as $x\downarrow 0$. Thus this is a downstream ideal-source solution, valid only far enough away that $b/g\ll 1$ and the Boussinesq approximation and top-hat plume model hold. It does not describe the immediate source region.

Away from zero the Dirac delta distribution is absent. On either half-line, use $u=1+k|z|$ and the power-law ansatz $\psi =u^{\alpha }$. The Euler-Cauchy equation becomesThe growing term is excluded by boundedness at either end. Thus the independently bounded half-line solutions are

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.

On long wavelengths the gradient term is negligible. The Friedmann equation impliesDuring matter domination, $a\propto t^{2/3}$ and $H=2/(3t)$, so the linear growth equation becomesThe power-law ansatz $\delta \propto t^s$ gives $s=2/3$ or $s=-1$. Since $a\propto t^{2/3}$, the growing and decaying modes are

The linearized cosmological continuity equation, the divergence of the cosmological Euler equation, and the cosmological Poisson equation giveEliminating $\theta =-{\delta }^{\prime }$ yields the equation for a linear cosmological density perturbation:In an Einstein-de Sitter universe, ${\Omega }_m=1$ and $\mathcal{H}=2/\tau$, soSubstitution of the power-law ansatz $\delta \propto {\tau }^p$ gives $(p-2)(p+3)=0$. HenceThe leading one of the matter-era growing and decaying density modes is $\delta \propto a\propto {\tau }^2$, as direct substitution confirms, and its velocity divergence is

Here $q=\kappa /R$, so the equation from part (b) becomes the Euler-Cauchy equationThe power-law ansatz $B_z=R^s$ gives the indicial equationFor $0<\kappa <1/2$, the general field is thereforeThe polynomial discriminant changes sign atFor $\kappa >{\kappa }_c$, define $\mu =\sqrt{{\kappa }^2-1/4}$. A real form of the solution isThus the field has a log-periodic oscillation: its phase is periodic in $\ln R$, so it oscillates as the radius changes geometrically.

In a homogeneous ambient fluid, the displayed nonsteady power-law ansatz solves the unsteady top-hat line-plume balances for $\tau =t+t_{\ast }>0$. It gives $Q=R\alpha z^2/(3\tau )$, $M=2R\alpha z^3/(9{\tau }^2)$ and $F=2R\alpha z^3/(9{\tau }^3)$. The solution is singular at zero age and carries zero source flux at its unshifted ideal origin. It describes a separable branch away from a source; a finite-width boundary, virtual-origin shift or matching region is needed to supply nonzero physical source flux. It is distinct from the steady constant-flux line plume, whose top-hat half-width grows as $\alpha z$.