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The individual rates describe population relaxation: ${\gamma }_{12}$ transfers population from level one to level two, and ${\gamma }_{21}$ transfers it back. With $p_1+p_2=1$ and the conventional qubit coordinate $r_z=p_1-p_2$,Thus $T_1^{-1}={\gamma }_{12}+{\gamma }_{21}$ is the longitudinal relaxation rate. The parameter $\Gamma =T_2^{-1}$ is the transverse relaxation rate: it damps the off-diagonal coherence, or the $x,y$ components of the Bloch vector. In the usual completely positive two-level model,so $\Gamma$ includes both the coherence loss caused by population relaxation and additional pure dephasing.

There is a normalization switch in the displayed qubit Affine Bloch equation. Its constant term is appropriate to the conventional Pauli matrices and unit-radius coordinate $r_k=\mathrm{Tr}(\rho {\sigma }_k^{\mathrm{Pauli}})$. With the orthonormal generators used earlier, $s=r/\sqrt{2}$, the same physical rates would instead give a constant term $({\gamma }_{21}-{\gamma }_{12})/\sqrt{2}$. The drift matrix is unchanged. We interpret the supplied qubit equation in its conventional $r$ coordinates.

With zero controls and $g={\gamma }_{12}+{\gamma }_{21}>0$, $\Gamma >0$, the steady state is unique:The earlier normalized vector is $s_{\ast }=r_{\ast }/\sqrt{2}$. Indeed $r_x,r_y$ decay as $e^{-\Gamma t}$ and $r_z-r_{\ast ,z}$ as $e^{-gt}$.

For vanishing rates, the full equilibrium conditions of the Affine Bloch equation are $\Gamma r_x=\Gamma r_y=0$ and $g{r}_z={\gamma }_{21}-{\gamma }_{12}$, together with $\Vert r\Vert \le 1$. In particular, for nonnegative rates, $g=0$ means both population rates vanish. If then $\Gamma >0$, all diagonal density operators are stationary; if also $\Gamma =0$, every density operator is stationary. The formal case $g>0$, $\Gamma =0$ permits extra transverse equilibrium coordinates, but is excluded by the completely positive rate bound above.

Put $g(z)=s(az)=m{a}^2{z}^2/({\sigma }^2+a^2{z}^2)$. The nullclines areAny steady state has nonnegative coordinates because $g\ge 0$. For $a\ne 0$, $g$ is strictly increasing on the positive half-line. If a steady state had $x>y$, then $x=g(y)<g(x)=y$, a contradiction. The case $y>x$ is identical, so all steady states lie on the diagonal. Write $x=y=z$. Besides $z=0$, the equation $z=g(z)$ givesand henceThus three distinct steady states exist exactly when $|a|m>2\sigma$, with $a\ne 0$ and the positive parameter assumptions stated above. For positive excitatory coupling this is $am>2\sigma$. At equality the two positive states merge; below it only the origin remains. If $a=0$, the origin is again the only steady state.

Both nullclines begin at the origin and approach the limiting activity $m$ along their saturating direction; they are reflections of each other across $x=y$. Flow points to the right where $x<g(y)$ and to the left where $x>g(y)$. It points upwards where $y<g(x)$ and downwards where $y>g(x)$. The plot shows these nullclines and typical trajectories for the parameters of the next part.

The origin and upper diagonal state attract nearby trajectories, while the lower positive state is a saddle. Its stable manifold separates the two attracting regions: trajectories in each basin of attraction approach its stable node, giving bistability. The diagonal is invariant, and on it the scalar activity increases between $z_{-}$ and $z_{+}$ and decreases between zero and $z_{-}$ or above $z_{+}$. Activities remain nonnegative and bounded because $0\le g\le m$. The vector field has constant negative divergence $-2/\tau$, so the Bendixson-Dulac criterion also rules out periodic orbits in the physical quadrant. The network is a switch between persistent activity levels, rather than an oscillator.

The Jacobian matrix at a diagonal steady state isIts eigenvectors $(1,1)$ and $(1,-1)$ represent common and opposing perturbations, with respective eigenvalues $(-1+g^{\prime })/\tau$ and $(-1-g^{\prime })/\tau$.

For a nonzero equilibrium, ${\sigma }^2+a^2{z}^2=m{a}^{2}z$ and ${\sigma }^2/a^2=z(m-z)$. Consequently $g^{\prime }(z)=2(m-z)/m$. The lower root has $g^{\prime }>1$ and is a saddle equilibrium; the upper root has $0<g^{\prime }<1$ and is a stable node. At the origin $g^{\prime }(0)=0$, so both eigenvalues are $-1/\tau$.

The supplied parameters give $\sigma /a=4$, and the nonzero roots solve $z^2-10z+16=0$, giving $2$ and $8$. The numerical stability results areBiologically, the origin is a quiescent state and the upper equilibrium a sustained mutually reinforced activity state. The saddle represents a collective activation threshold: a perturbation crossing its basin boundary can switch the network from low to high activity. This persistent-state mechanism can model a memory or decision variable, but the simplified rate equations do not establish that a particular biological circuit uses it.

A steady state is a time-independent physical state: its derivative vanishes. With $n=N^2-1$, the Affine Bloch equation therefore requiresAs a real linear equation, it has a solution exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. If $s_p$ is one solution, all algebraic solutions are $s_p+\ker A$, an affine set of dimension $n-\mathrm{rank}A$. Physical steady states are its intersection with the set of Bloch vectors corresponding to positive trace-one density operators. An arbitrary affine equation need not have a steady state: $A=0$, $c\ne 0$ is a counterexample.

The equation here comes from a finite-dimensional time-independent Lindblad equation, however, so it always has at least one physical steady state. To prove this, start from any density operator and average its trajectory:These averages remain positive with trace one. That set is a compact set, so some sequence has a limit ${\rho }_{\ast }$. If $\mathcal{L}$ is the time-independent generator, thenbecause density operators are bounded. Continuity gives $\mathcal{L}({\rho }_{\ast })=0$. This proves directly that Finite-dimensional Lindbladians have stationary states.

If $A$ has full rank, its algebraic solution $s_{\ast }=-A^{-1}c$ is unique and existence makes that solution physical. For these Lindblad dynamics the converse also holds, even for a boundary steady state. If $A$ is singular, choose a nonzero traceless Hermitian stationary direction $X$. The two parts of its positive-negative decomposition have the same trace $\tau >0$, so $X=\tau ({\rho }_{+}-{\rho }_{-})$ for two density operators. Time-average their trajectories along a common convergent subsequence. Their limits are stationary by the preceding argument, and their difference remains $X/\tau$, since $X$ itself is stationary. The limits are therefore distinct physical stationary states.

Thus the finite-dimensional physical Bloch equation has a unique steady state exactly when $\mathrm{rank}A=n$. These steady states of an affine Bloch equation must be distinguished from convergence to that steady state, which requires the additional spectral condition in (b). The possibly nonorthonormal coordinate basis does not affect existence or uniqueness, since it still gives an invertible parametrization of the traceless Hermitian space.

Subtract a steady state to obtain $e(t)=s(t)-s_{\ast }$ with $\dot{e}=Ae$, hence $e(t)=e^{At}e(0)$. The steady state attracts every initial condition exactly whenIn other words, $A$ must be a Hurwitz stable matrix. To see sufficiency without assuming diagonalizability, each block in its Jordan normal form contributes terms of the form $t^k{e}^{\mu t}$ with finite $k$. Negative real part makes every such term tend to zero, so the whole matrix exponential tends to zero.

For necessity, an eigenmode with positive real part grows, and a mode with zero real part persists or oscillates instead of decaying. For a complex eigenvalue of a real matrix, take the real and imaginary parts of its eigenvector to obtain real nondecaying solutions. Thus any eigenvalue in the closed right half-plane prevents convergence of all deviations. Physical density operators span the trace-one affine space, so convergence from all physical initial states also forces decay on every traceless direction.

This gives global asymptotic stability for the affine linear system and uniqueness of its steady state. A merely negative-semidefinite symmetric part proves only nonincrease of a particular norm, not this necessary-and-sufficient condition.

First fix a source convention needed for physical coordinates. The PDF's displayed definition of ${\sigma }_z$ omits a factor of $i$: half the commutator of the two Hermitian transverse matrices is anti-Hermitian and cannot supply a real Bloch coordinate. The Hermitian choice consistent with the supplied dynamical matrix isWith this convention the three Pauli expectation values use the unit-radius normalization, unlike the orthonormal coordinates of Question 2. For example, writing $K=\left(\begin{array}{cc}0& a\\ b& 0\end{array}\right)$ with $a=1+\lambda$, $b=-\lambda$, the dissipator gives rates $-(a-b{)}^2/2$ for $x$, $-(a+b{)}^2/2$ for $y$, and population equation $\dot{z}=b^2-a^2-(a^2+b^2)z$. The Quantum Hamiltonian adds $2\alpha z$ to $\dot{x}$ and $-2\alpha x$ to $\dot{z}$. This independently reproduces the supplied real Bloch dynamics.

Put $q=2\lambda +1$. Since $y$ decays independently, every steady state has $y_{\ast }=0$. The reduced steady-state equations areTheir coefficient determinant isIt is strictly positive unless $q=0$ and $\alpha =0$. Inverting this two-by-two system givesThus the reduced system has a unique steady state except at $\alpha =0$, $\lambda =-1/2$. At that exceptional point the equations reduce to $\dot{x}=0$, $\dot{z}=-z/2$, so every $(x,0)$ is stationary; physical coordinates restrict $-1\le x\le 1$.

As a consistency check on positivity in the nonsingular case,The calculated state is therefore in the unit Bloch disk. The original PDF correctly labels the second steady-state expression as $z_{\ast }$; the TeX conversion's repeated $x_{\ast }$ is another transcription defect, not a second formula for $x_{\ast }$.

With $q=2\lambda +1$, the reduced coefficient matrix and its symmetric part areIts diagonal entries are both strictly negative exactly when $q\ne 0$, equivalently $\boxed{\lambda \ne -1/2}$. The coherent driving cancels from the symmetric part because it is a rotation generator.

For a steady state $s_{\ast }$ put $e=s-s_{\ast }$. Differentiating the squared Euclidean distance givesFor $q\ne 0$ the derivative is strictly negative whenever $e\ne 0$. Integrating the inequality proves the symmetric-part contraction criterion here explicitly:Thus every nonstationary initial state approaches the unique steady state, with strictly decreasing distance; the steady state itself has constant zero distance. If the decoupled $y$ coordinate is retained, it contributes $-e_y^2$ to the squared-distance derivative and also decays, so the conclusion holds for the full qubit state.

For the target-dependent settings in (e), $q=-\cos {\theta }_d$. Every nonequatorial target is therefore globally attractive. At an equatorial target, $q=\alpha =0$: $x(t)=x(0)$, $z(t)=z(0)e^{-t/2}$, and $y(t)=y(0)e^{-t/2}$. A general state approaches $(x(0),0,0)$ rather than the specified pure target $x=\pm 1$. This proves the equatorial exception, not merely failure of a sufficient estimate.

The parameter dependence matters: $q=0$ alone does not exclude attractivity for every possible driving strength. If $\alpha \ne 0$, the reduced matrix has negative trace and positive determinant $4{\alpha }^2$, so both eigenvalues have negative real parts even though its symmetric part is only semidefinite. The nonattractive case asserted for the equatorial target also uses its prescribed $\alpha =0$.

The species $u$ is the activator, since increasing $u$ increases its own production $u^2/v$; $v$ is the inhibitor, since increasing $v$ reduces that production. The unique positive uniform steady state solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_{\ast }=1/b,v_{\ast }=1/b^2}$.

The reaction Jacobian matrix at this state isIts eigenvalues solve ${\lambda }^2+(1-b)\lambda +b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.

A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\mathrm{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant isA growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1{)}^2>4bd$. Together these conditions reduce to the Turing threshold of the quadratic activator-inhibitor model,The unstable band isAt the Turing instability threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt{2})/d}$. On a bounded spatial interval, an allowed wavenumber must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.

A steady state is a time-independent density operator satisfying $\mathcal{L}({\rho }_{\ast })=0$, equivalently a constant physical Bloch vector $s_{\ast }$ satisfying $\boxed{A{s}_{\ast }=-c}$. Let $K=N^2-1$. By the rank criterion for a linear system, this equation has a solution exactly whenWhen it is consistent, every algebraic solution has the form $s_0+v$ with $v\in \ker A$, so the solution affine space has dimension $K-\mathrm{rank}A$ by the rank-nullity theorem. Physical steady states are its intersection with the positive density operator body. The algebraic equilibrium is unique exactly when $\mathrm{rank}A=K$, in which case $\boxed{s_{\ast }=-A^{-1}c}$.

An arbitrary affine differential equation need not have an equilibrium: $A=0$, $c\ne 0$ is inconsistent. For the finite-dimensional, time-independent Lindblad equation under discussion, however, a physical steady state always exists. Indeed, the averages ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$ remain positive with trace one. Compactness supplies a convergent subsequence as $T\to \infty$, andThe limit is a physical steady state, proving that Finite-dimensional Lindbladians have stationary states and that the rank consistency condition is automatically satisfied for this physical generator.

In this setting, full rank also characterizes uniqueness among physical steady states. To justify the converse when $A$ is singular, a nonzero $v\in \ker A$ gives a fixed traceless Hermitian operator $X={\sum }_k{v}_k{\sigma }_k$. Write its positive-negative decomposition $X=X_{+}-X_{-}$. Time-average the evolutions of the two positive operators along a common convergent subsequence; the limits $Y_{\pm }$ are positive stationary operators and satisfy $X=Y_{+}-Y_{-}$. Their traces are equal because $\mathrm{Tr}X=0$. If there were only one stationary density operator, both $Y_{\pm }$ would be this matrix times the same trace, forcing $X=0$, a contradiction. Hence $\boxed{\text{unique physical steady state}⟺\mathrm{rank}A=N^2-1}$.

A steady state is attractive here if every physical initial density operator approaches it. Subtracting its Affine Bloch equation from the trajectory equation gives $\delta \dot{s}=A\delta s$, so $\delta s(t)=e^{At}\delta s(0)$. The necessary and sufficient condition isSuch an $A$ is a Hurwitz stable matrix. To prove sufficiency without assuming diagonalizability, put it in Jordan normal form. Each Jordan block contributes $e^{\lambda t}{\sum }_{j=0}^{q-1}{t}^j{N}^j/j!$, and the negative real part makes the exponential dominate all polynomial factors. For necessity, an eigenmode with positive real part grows; an eigenmode with zero real part remains constant or oscillates, so it does not tend to zero. A nontrivial Jordan block on the imaginary axis also prevents decay.

The physical Bloch vector body has nonempty interior, so differences of physical initial vectors from $s_{\ast }$ span the full real coordinate space. Thus convergence for every physical initial state also forces $e^{At}\to 0$ on that space. This rules out hiding a nondecaying eigenmode outside the physical set. The strict spectral condition entails uniqueness and global asymptotic stability of the steady state.

The positive steady state satisfies $1+b(N^{\ast }{)}^2=r$, soWith $N_t=N^{\ast }+{\eta }_t$, the linearization isIts characteristic polynomial is ${\lambda }^2-\lambda +d$, where $d=2(r-1)/r$. The roots are inside the unit disk precisely when $0<d<1$: when they are real they are positive with sum one, and when complex conjugate their common modulus is $\sqrt{d}$. Therefore the fixed point is linearly asymptotically stable for $1<r<2$ and unstable for $r>2$. At $r=2$,This proves period six for the neutral linearized oscillations. The stronger printed claim of a nonlinear period-six branch is not valid for the stated recurrence. It is important to distinguish these two assertions.

Here is a local obstruction, which also checks that the qualification is not merely absence of a proof. Normalize by the moving fixed point, let $x=N_{t-1}/N^{\ast }(r)-1$, $y=N_t/N^{\ast }(r)-1$, and definePut $v=(x,y{)}^T$, $\delta =r-2$, and $Q=x^2-xy+y^2>0$ for $v\ne 0$. The quadratic and cubic terms of $F_2$'s second component are $x^2/2-xy+x^{2}y/2$. Composing six times givesIn particular the quadratic terms cancel, while the cubic components are $-(x+2y)Q/2$ and $(2x-3y)Q/2$. If a small nonzero $v$ were a period-six point, invertibility of $B$ and the norm of this equation would imply $|\delta |=O(|v{|}^2)$. Taking its determinant with $Bv$ then givesSince $Q\ge |v{|}^2/2$, this is impossible for sufficiently small nonzero $v$. Thus no nontrivial nonlinear six-cycle bifurcates locally at $r=2$. At the threshold the same expansion gives $Q(F_2^6(v))=Q(v)-2Q(v{)}^2+O(|v{|}^5)$, so the fixed point is actually locally asymptotically stable nonlinearly, with slow decay. The valid intended linear-stability conclusion is the period-six neutral oscillation above; a literal nonlinear assertion requires a changed model or wording. This is an example of the distinction between linear resonance and nonlinear periodic bifurcation.

For a spatially homogeneous positive steady state, $\alpha n(n_0^2-n^2)=0$ and $\beta n-\gamma a=0$, givingWrite $n=n_0+\eta$, $a=a_0+\xi$, and define $A=2\alpha n_0^2>0$. Homogeneous linearization givesThe eigenvalues are $-A,-\gamma$, both negative. Hence this steady state is stable to homogeneous disturbances, including the repeated-eigenvalue case, where a possible $t{e}^{-At}$ factor still decays.

For the Dirichlet Green's function, solve $-D{G}_{xx}=\delta (x-\xi )$ with zero endpoint values. Away from $\xi$, $G$ is linear. Continuity at $\xi$ and the derivative jump $G_x(\xi +,\xi )-G_x(\xi -,\xi )=-1/D$ giveFor nonzero Dirichlet boundary conditions, subtract their linear interpolant. The resulting boundary value problem has solutionIn the final heat equation, a steady state has zero time derivative. Here $l=1$, $f(x)=x$, $\alpha =1/D$, $\beta =2/D$. Integrating $u^{\prime \prime }=-x/D$ twice and imposing the endpoint values givesIts second derivative and both boundary conditions verify the answer directly.

Put $F=\bar{\nu }\Sigma$. In steady state, $r^{1/2}{\partial }_r(r^{1/2}F)$ is constant, soThe inward mass accretion rate and outward viscous torque in an accretion disk are

Positive $F_{\mathrm{acc}}$ denotes inward flow. In steady state, mass conservation gives $d{F}_{\mathrm{acc}}/dr=2\pi rW$. In the remote-feeding approximation, take the incoming flux at large radius inside the feeding region to approach ${\dot{M}}_d$. The cumulative loss exterior to $r\ge r_g$ is ${\dot{M}}_w\sqrt{r_g/r}$, so the steady photoevaporating-disk mass flux isIt is constant inside $r_g$, continuous at $r_g$, and increases outwards towards ${\dot{M}}_d$. Its right derivative is positive and decreases as $r^{-3/2}$. The curve has a change of slope at $r_g$ because the wind turns on there. Physically less mass survives to cross successively smaller radii, while the inner disk has no wind sink. The star accretes at ${\dot{M}}_d-{\dot{M}}_w$.

For constant kinematic viscosity $\nu$, set $Y=\nu \Sigma \sqrt{r}$. The supplied viscous evolution of an accretion disk relation becomes $Y^{\prime }=F_{\mathrm{acc}}/(6\pi \sqrt{r})$. The viscous torque in an accretion disk is proportional to $Y$, so the zero-stress condition at the idealized origin sets $Y(0)=0$. ThusFor $r<r_g$ the integral is $2({\dot{M}}_d-{\dot{M}}_w)\sqrt{r}$. For $r\ge r_g$, separating the two intervals givesConsequently the surface density of a steady photoevaporating disk isBoth the density and its first derivative match at $r_g$. Let $x=r/r_g$ and take the special case ${\dot{M}}_d={\dot{M}}_w$. Then $\Sigma =0$ throughout the inner disk, while outsideThe profile leaves zero with zero slope at $x=1$, rising as $(x-1{)}^2/8$ locally, and approaches ${\dot{M}}_w/(3\pi \nu )$ from below at large radius. The requested sketches are:

The analytic profiles describe the region interior to a distant feeding boundary. A finite outer radius and the disk outside the feeding point need additional boundary conditions. For example, if the wind is truncated at $R\gg r_g$ and $F_{\mathrm{acc}}(R)={\dot{M}}_d$, the exact interior formulas above replace ${\dot{M}}_d$ by ${\dot{M}}_d+{\dot{M}}_w\sqrt{r_g/R}$; the total wind inside $R$ is ${\dot{M}}_w(1-\sqrt{r_g/R})$. The displayed profiles are their remote-boundary limit. Extending the prescribed wind to infinity is useful for estimating mass loss but is not a complete global angular-momentum boundary condition for a finite feeding radius.

As the viscous supply decreases towards the wind rate, the inner accretion flux approaches zero. Wind removal near $r_g$ can cut off replenishment of the inner disk, which then drains on its own viscous timescale, leaving a gap or inner hole. This is photoevaporative gap opening. If supply falls below wind loss, the formal steady profile would require negative inner density and is unphysical; time-dependent depletion must replace it. The exposed outer edge may then be removed rapidly. Equality of the two rates indicates the onset of rapid clearing, rather than a negative steady surface density.

Take the accretion rate $\dot{M}>0$ to mean inward flow. In a steady state, conservation of mass makes the inward mass flux independent of radius, so $\mathcal{F}=\dot{M}$. Integrating its radial expression givesFor Keplerian rotation, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$ and the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma \sqrt{G{M}_{\ast }r}$. The zero-torque inner boundary condition therefore sets $\bar{\nu }\Sigma =0$ at $r_{\ast }$. It fixes $C=-\dot{M}\sqrt{r_{\ast }}/(3\pi )$, giving the steady density profileThis is the Keplerian accretion disk solution on $r_{\ast }<r<r_{\mathrm{out}}$. It determines the kinematic viscosity–surface density product; a separate viscosity closure is needed to turn it into an explicit power law. If the kinematic viscosity is finite and nonzero at the inner edge, the surface density tends to zero there.

Take $0<r_1<r_2$ and positive mean kinematic viscosity $\bar{\nu }$. The steady state of the supplied viscous evolution of an accretion disk equation satisfies ${\mathcal{F}}^{\prime }=-2\pi rS(r)$. The outer no-flux condition therefore givesPositive $S$ produces positive inward accretion rate in this sign convention. Set $Y=r^{1/2}\bar{\nu }\Sigma$. In a Keplerian disk, the viscous torque in an accretion disk is proportional to $Y$, so the zero-torque inner boundary condition gives $Y(r_1)=0$. Since $Y^{\prime }=\mathcal{F}/(6\pi r^{1/2})$, a second integration yieldsThis is the steady viscous disk supplied by a radial source. It does not require $\bar{\nu }$ to be independent of radius: the equation directly determines $\bar{\nu }\Sigma$. If viscosity depends on the local thermodynamic state, the displayed formula is an implicit relation completed by a separate closure. For nonnegative integrable $S$ it gives nonnegative surface density of a disk, and the inner accretion rate equals the total supplied mass per unit time, $2\pi {\int }_{r_1}^{r_2}rS(r)dr$. Differentiating both nested integrals recovers the flux and the differential equation.

Write the proposed background as $\mathbf{U}=U(x,z){\mathbf{e}}_y$, $U=-3\Omega x/2+q\Omega z$. It satisfies incompressibility, and $\mathbf{U}\cdot \nabla \mathbf{U}=U{\partial }_y\mathbf{U}=0$. A constant pressure has zero gradient. The Coriolis acceleration is $-2\Omega {\mathbf{e}}_z\times \mathbf{U}=2\Omega U{\mathbf{e}}_x=-(3x-2qz){\Omega }^2{\mathbf{e}}_x$, which exactly cancels the prescribed body force. Thus it is a steady state.

The background is ordinary Keplerian radial shear together with vertical shear: ${\partial }_{x}U=-3\Omega /2$ and ${\partial }_{z}U=q\Omega$. At a given radius, different heights have different azimuthal velocities. The small $q$ in this question measures vertical shear; it is distinct from the orbital shear parameter, whose radial value here remains $3/2$. The vertical shear instability draws on this height-dependent differential rotation.

For uniform specific entropy, the material conservation of cross-helicity density criterion isThus the requested condition isIn the steady state, mass conservation gives $\nabla \cdot \mathbf{u}=-\mathbf{u}\cdot \nabla \log \rho$. An equivalent form, expressed entirely in the flow and thermodynamic variables, isIf “isentropic” means only constant specific entropy along individual trajectories, rather than a homentropic flow, the general criterion instead retains $T\mathbf{B}\cdot \nabla s$ on the right-hand side of the material derivative equation. The distinction matters because a source-free cross-helicity conservation law need not make $h_c$ constant along each trajectory.

In a steady state, mass conservation makes $\mathcal{F}$ independent of radius. Taking the accretion rate $\dot{M}>0$ for inward flow gives $\mathcal{F}=-\dot{M}$. The conservation of angular momentum equation then makes $-\dot{M}h+\mathcal{G}$ constant. The zero-torque inner boundary condition fixes this constant to $-\dot{M}{h}_{\mathrm{in}}$, soThe viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^3\bar{\nu }\Sigma {\Omega }^{\prime }=2\pi qh\bar{\nu }\Sigma$. Combining these expressions gives the steady accretion disk with arbitrary rotation lawFor $x=r/R_g$, the Paczyński-Wiita circular orbit formulas giveConsequentlyIn particular, $f(6)=0$ and $f\to 1$ far from the black hole, recovering the outer Keplerian accretion disk.

Inside the innermost stable circular orbit, the gas enters the plunging region of a black-hole accretion disk. Its rapid inward motion leaves little time for stresses to exchange angular momentum, motivating the zero-torque inner boundary condition. This is a thin-disc approximation; a strong magnetic stress could change it.

Matter supplied from very large radius has negligible specific orbital energy, while matter crossing the inner edge carries ${\epsilon }_{\mathrm{in}}=-\eta c^2$. With zero inner torque, no energy is supplied by a stress at that edge. If the heat released outside it escapes by radiative transfer, rather than being lost through inward advection, the integrated conservation of energy balance gives