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∗ 2 τ v µ v + ( v) (µ v) I, (4) ≈ v ∇ ∇ − 3 v∇ ·

 ∗  2 τ t µ v + ( v) µ v+ρk I, (5) ≈ t ∇ ∇ − 3 t∇ · ∗ where I is the identity tensor, ( ) denotes a conjugate tensor, µv denotes the dynamic , µt, and k denote the turbulent viscosity and kinetic energy of turbulence, respectively (see, e.g., [1], [20], [105] and references therein). For the sake of simplicity, it is believed that the Stokes hypothesis is valid, i.e. the coefficient of bulk viscosity is negligible. We will also use the viscosity µ = µv + µt. Obviously, if the flow will be laminar, then µt = 0, k = 0 and, hence, µ = µv will be the dynamic viscosity. Recall that the dynamic viscosity, µv, of a gas increases with absolute tem- perature, T , and, in fact, it is independent of pressure and density at a given temperature (see, e.g., [1, p. 186], [14, p. 153], [46, p. 46], [50, p. 635], [104, pp. 26-28]). Let us cite a couple of widely used approximations for the viscosity of dilute gases (see, e.g., [27], [50], [104]). The power law:

n µv ∝ T , (6) where typically n = 0.76. It is, also, assumed that n = 1 for the case of low temperatures. If the temperature is relatively high, then n ' 0.5, as it is evident from the Sutherland’s law: T 1.5 µ ∝ . (7) v T + const

The turbulent viscosity, µt, and the turbulence kinetic energy, k, can be estimated accurately by a k-ε model (see, e.g., [2], [32], [64], [98]) and, hence,

2Let us note that the energy equation in the paper [11] contains a misprint.

2 it can be used as the “principal tool” for the evaluation. In such a case one needs to solve two additional partial differential equations (PDEs). For mildly complex flows, the k-ε model can be reduced to the one-equation model for the kinetic energy of turbulence using Prandtl and Kolmogorov suggestion (e.g., [1, p. 230], [105, p. 74]). It is assumed that the turbulent viscosity is proportional 0.5 to k . In such a case, the turbulent viscosity, µt, can be evaluated as the following.  0.5 µt = Ckρ k L, Ck = const, (8) where L denotes the turbulence length scale. The balance equation for the kinetic energy of turbulence, k, can be written in the following form (e.g., [1], [105]):

1.5 ∂ρk Cµ k + ρkv = µ k + τ t : ( v) ρ , Cµ = const. (9) ∂t ∇ ∇ · ∇ ∇ − L  Thus, in the case of one-equation model, (8) and (9), the length scale of turbulence is a free parameter in contrast to the two-equation models (e.g., k-ε models), which are complete and, hence, can be used for computation of the kinetic energy, k, as well as the turbulence length scale, L, or equivalent [105]. Because of it, most of these models are in widespread use (e.g., [2], [32], [64], [98]). It is interesting to note that, the so called, algebraic (zero-equation [105]) models enjoy widespread use because of their simplicity. For instance, the widely known α-Disc model (e.g., [3], [63], [73], [84]) is algebraic, namely, the turbulent viscosity is evaluated by (8), where, in fact, the square root of kinetic energy is assumed to be equal to the sound velocity and the turbulence length scale is equal to the disk semi-thickness [73]. Interestingly, the α-Disc model is unstable (see, e.g., [3], [63], [67]). By and large a perfect3 gas flow will be discussed in this paper and, hence,

P = ρRT, R = const. (10)

We will in general consider axisymmetric gas flows in cylindrical coordinates, (r, ϕ, z). The gravitational potential is assumed to be as follows

M Φ= G , G,M = const. (11) − √r2 + z2 It will be mainly assumed that the absolute temperature T is a “free” param- eter, i.e. T = T (r,z,t) is a free (or pre-assigned) function of the coordinates in the region. Such an approach permits us to avoid using the energy equa- tion. Nevertheless, the energy equation (3) will sometimes be replaced by the polytropic relation: P = Kργ, K,γ = const. (12)

3The terms perfect gas and ideal gas are used interchangeably in this paper.

3 If the process is polytropic, then one obtains the following simple formula [81, Sec. 5.4.8] for the case of perfect gas: γ cp dq(e) = c − cv dT, (13) v γ 1 − (e) where dq denotes the external heat flow, cv and cp denote the specific heat capacities at constant volume and pressure, respectively. Let, for definiteness, (e) dT > 0. In such a case, if 1 0 (or, respectively, dq(e) < 0), i.e. heat is supplied (or, respectively, heat is released) (e) with an increase in temperature. If γ = cpcv, then dq = 0 and, hence, such a polytropic process will be adiabatic. Thin disk accretion must be highly nonadiabatic, as emphasized in [85]. We will assume in such a case that γ = 6 cpcv. Let us introduce the following dimensional characteristic quantities: t∗, l∗, ρ∗, v∗, p∗, T∗, µ∗, and k∗ for, respectively, time, length, density, velocity, pres- sure, temperature, viscosity, and kinetic energy of turbulence. The characteristic quantity for sound speed cs∗ p∗ρ∗. The following notation will also be used: ≡ p 2 l∗ p∗ cs∗ Sh = , Eu = 2 = 2 , v∗t∗ ρ∗v∗ v∗ 2 v∗l∗ ρ∗v∗l∗ 2k∗ Fr = , Re = , ϑke = 2 , (14) GM µ∗ 3v∗ where Sh, Eu, Fr, and Re denote, respectively, Strouhal, Euler, Froude, and Reynolds numbers. Throughout of this paper, we assume that cs∗ vk∗ ≪ ≡ √GMl∗ (see, e.g., [18], [101]). Hence, if v∗ = vk∗, then Eu 1 and Fr = 1. ≪ Note that since cs∗ vk∗, then EuFr 1 even if v∗ = vk∗. In what follows we assume that ≪ ≪ 6 1 Eu 1, Eu . (15) ≪ ≪ Fr

It is significant that the Reynolds number, Re, is very high (see, e.g., [14, p. 170], [18, p. 70], [21, p. 165]), i.e.

14 Re 10 . (16) ∼ Let ς denote a variable and let ς∗ denote the characteristic quantity for ς, then the transformation from dimensional to dimensionless variables can be written in the following form: ς ς∗ς. (17) → Notice that we use, as a rule, the characteristic quantity µ∗ µ ∗, where µ ∗ ≡ v v denotes the characteristic quantity for the dynamic viscosity, µv. In such a case, in view of (17), µ µ ∗µ µ ∗ (µ + µ ). However, it can, often, be → v ≡ v v t convenient to use the characteristic quantity µt∗ in addition to µv∗. Then, the transformation (17), for the case of viscosity, can be written as follows:

µt∗ µ µv∗µ µv∗ µv + µt . (18) → ≡ µ ∗  v 

4 For the axisymmetrical flow, we have, in view of (14) and (17), the following non-dimensional system of PDEs. The continuity equation:

∂ρ 1 ∂ (rρv ) ∂ (ρv ) S + r + z =0, (19) h ∂t r ∂r ∂z Conservation of momentum: The r-component:

2 ∂ρvr 1 ∂ 2 ∂ ρvϕ ∂ ρ ∂Φ Sh + r ρvr + (ρvrvz) = (EuP ) + ∂t r ∂r ∂z − r −∂r − Fr ∂r  1 ∂ ∂v 2 1 ∂ (rv ) ∂v 2µ r µ r + z + R ∂r ∂r − 3 r ∂r ∂z e     ∂ ∂v ∂v 2µ ∂v v ∂ µ r + z + r r ϑ ρk, (20) ∂z ∂z ∂r r ∂r − r − ke ∂r     The ϕ-component: ∂ρv 1 ∂ ∂ ρv v S ϕ + r (ρv v )+ (ρv v )+ ϕ r = h ∂t r ∂r ϕ r ∂z ϕ z r 1 ∂ ∂ v ∂ ∂v ∂ v µr ϕ + µ ϕ +2µ ϕ , (21) Re ∂r ∂r r ∂z ∂z ∂r r         The z-component:

∂ρvz 1 ∂ ∂ 2 ∂ ρ ∂Φ Sh + r (ρvzvr)+ ρvz = (EuP ) + ∂t r ∂r ∂z −∂z − Fr ∂z  1 ∂ ∂v ∂v ∂ ∂v 2 1 ∂ (rv ) ∂v µ r + z + 2µ z µ r + z + R ∂r ∂z ∂r ∂z ∂z − 3 r ∂r ∂z e        µ ∂v ∂v ∂ r + z ϑ ρk, (22) r ∂z ∂r − ke ∂z   where 1 Φ= . (23) −√r2 + z2

For definiteness sake, it will be assumed that vϕ 0. The following Taylor series of Φ about z = 0 will also be used. ≥

1 2 1 1 2 Φ =Φ0 +Φ2z + ... = + z + ..., z < r. (24) ≡−√r2 + z2 − r 2r3 | |

We will take p∗ = ρ∗RT∗ and, hence, we obtain from (10) that

P = ρT. (25)

5 If the process is polytropic, then, instead of (12), we have the following non-dimensional relation.

γ Kρ∗ P = κργ, κ = = const (26) p∗ We will, in general, consider the outer regions (see, e.g. , [14], [21], [26], [73]) of accretion disks, i.e. r r (say), where r rs + bs, rs denotes the ≥ 0 0 ≫ non-dimensional radius of star, bs denotes the non-dimensional radial extent of boundary layer (e.g., [18, Sec. 6.2]). For the sake of simplicity, we take the char- acteristic quantity, l∗, for length such that r = 1 and, hence, rs +bs 1. Thus, 0 ≪ the gas flows will be considered in the region Ra= (r, z): r [1, ), z [0, ) , due to the axisymmetry of the problem. { ∈ ∞ ∈ ∞ } Let ς denote a dependent variable. It will be used throughout this paper that ς = ς , ˆς ς . (27) |r→∞ |z→∞ ≡ |r→1 Following the widespread view on thin dynamics (see, e.g., [14], [18], [21], [26], [85], [101]) we will, mainly, consider the case when the gas possesses a small inward velocity, i.e. the radial drift velocity is highly subsonic:

vr cs, (28) | | ≪ where cs denotes the sound speed. We will, as usually, assume that a physically correct model must accurately simulate Keplerian disks since the majority of observed disks are in Keplerian (or sub-Keplerian) rotation around their central accreting objects (see, e.g., [25], [29], [58], [92], [95], [106] and references therein). Since we consider symmetric disks, it is, in general, assumed that 0 z H, where H denotes the disk semi-thickness (e.g. [16], [85]). The temperature≤ ≤ and density at the disk surface, i.e.

TH T (r,z,t) , ρ ρ (r,z,t) ,TH 0, ρ 0, (29) ≡ |z=±H H ≡ |z=±H ≥ H ≥ are pre-assigned functions of r and t. In the following we will sometimes assume, for the sake of convenience, that TH 0 and ρH 0. It is pertinent to note that ≈ ≈ ρ = ρ (r,z,t) > 0, z : z < H. (30) ∀ | | It is assumed that the dependent variables are continuous functions of z at z = 0 and, in this case, the values ρ, P , T , vr, vϕ, are even functions of z, whereas vz is an odd one. We will also use the following asymptotic expansions in the limit z 0: → ρ ρ◦ + ρ′′◦z2 + ...,T T ◦ + T ′′◦z2 + ..., ∼ ∼ P P ◦ + P ′′◦z2 + ..., (31) ∼ ◦ ′′◦ 2 ◦ ′′◦ 2 vr v + v z + ..., vϕ v + v z + ..., ∼ r r ∼ ϕ ϕ ′◦ ′′′◦ 3 vz v z + v z + ..., (32) ∼ z z

6 where the symmetry was taken into account. As usually (e.g., [33]), the series in (31)-(32) may converge or diverge, but their partial sums are good approxi- mations to the dependent variables for small enough z. We will, in general, consider geometrically thin accretion disks and, hence, it assumed (see, e.g., [16, p. 157], [18, p. 87, p. 129], [20, p. 65], [85, p. 432], [101, p. 304]) that the semi-thickness, H, everywhere satisfies H ∂H 1, 1. (33) r ≪ ∂r ≪

Sometimes, for the sake of convenience, we will use the set of hyper-real numbers (∗R), which contains the set of real numbers (R), the set of infinitesi- mals (hyper-small numbers), and the set of infinite (hyper-large) numbers (see, e.g., [13], [33], [36], [65]). Let us recall the terminology. If ι (iota) is such that ι 0, then ι is called infinitesimal or hyper-small number. There| | is only one real number that is infinitesimal and that is 0. If ι will be hyper-small but non-zero, then ω = 1ι will be hyper-large, that is, ω will be greater than any real number. The hyper-large numbers must not be| | con- fused with infinity ( ), which is not a number at all. We will use the following ∞ notation: a ∼= b means that a number a is infinitely close to a number b, i.e. their difference a b is infinitesimal, and a ≇ b means that a number a is not − infinitely close to another one b. The notation a b means that a ∼= b but a = b. Let a be a finite hyper-real number. The real number≃ which is infinitely clos6 e to a is called the standard part of a and denoted by st (a), and, hence, st (a) = ∗ ∼ a. Obviously, st (a) = a if and only if (shortened iff) a R. Let f : ∗R R. The hyper-real number L is the limit4 of f(x) as x ∗R approaches∈ a ∗R→(i.e. ∈ ∈′ lim f(x)= L) if whenever x a, f(x) = L. The hyper-real number f (x) will x→a ∼ ≃∗ be the derivative of f : ∗R R at x ∗R iff → ∈ ′ f (x + ∆x) f (x) f (x) = lim − , ∆x 0. (34) ∆x→0 ∆x ≃ The real number F ′ (x) will be the S-derivative of the f(x) at x ∗R iff ∈ ′ f (x + ∆x) f (x) F (x)= st − , ∆x 0. (35) ∆x ∀ ≃   ′ ′ ′ ′ Notice that f (x) F (x) ∼= 0, i.e. F (x)= st [f (x)]. In connection with− the aforesaid, the assumption (30) should be adjusted. To be precise, we should write that ρ ≇ 0, namely st (ρ) > 0, if z < H. It is inconvenient to make constant reference to hyper-real numbers in| this| paper. So, a phrase such as the above one will often be abbreviated to (30) with any definitions understood implicitly. It should induce no difficulty so long as any more precise definitions are mentioned explicitly when that is vital. We will, in general, use the same symbol, ∂, for partial as well as ordinary derivatives of a function even if the function depends on one variable only. The symbol d will be used for the total (full) derivative of a function. 4It can be written in the following form [65]: L = lim f(x) if ε ∗R, ε > 0, δ ∗R, x→a ∀ ∈ ∃ ∈ δ > 0 : x ∗R, 0 < x a < δ f(x) L <ε ∀ ∈ | − | ⇒ | − |

7 2 Laminar Flow

Only laminar flows are considered in this section and, hence, the turbulent viscosity µt = 0, and the kinetic energy of turbulence k = 0.

2.1 Inviscid Flow

We consider the outer regions of accretion disks, where the damping due to molecular viscosity is very small (e.g., [12, p. 70], [18, p. 70], [26, p. 143]) and, hence, we expect that the friction forces are negligible, i.e. the Reynolds 14 numbers are very high (say, Re 10 , e.g., [14, p. 170], [18, p. 70], [21, p. 165]). Therefore, the solutions∼ of the Navier-Stokes equations will be such that the outer flow obeys the laws of inviscid flow [78]. In this connection, only inviscid flows are considered in this sub-section and, hence, the dynamic viscosity µv = 0. The steady-state version of the PDE system (19)-(22) is thus reduced to the following one.

1 ∂ (rρv ) ∂ (ρv ) r + z =0, (36) r ∂r ∂z

1 ∂ 2 ∂ r ρv + (ρvrvz) r ∂r r ∂z ≡  2 ∂vr ∂vr ρvϕ ∂P ρ ∂Φ ρvr + ρvz = Eu , (37) ∂r ∂z r − ∂r − Fr ∂r

1 ∂ ∂ vϕvr r (ρvϕvr)+ (ρvϕvz)+ ρ r ∂r ∂z r ≡ ∂v ∂v v v ρv ϕ + ρv ϕ + ρ ϕ r =0, (38) r ∂r z ∂z r

∂vz ∂vz ∂P ρ ∂Φ ρvr + ρvz = Eu , (39) ∂r ∂z − ∂z − Fr ∂z −0.5 where Φ is defined by Eq. (23), i.e. Φ = r2 + z2 . In what follows the perfect gas equation of state, (25), will, mainly,− be used, i.e. P = ρT . Taking some liberties with the notation (27), we write

(vr, vϕ, vz) = (vr, vϕ, vz) 0, (40) |r→∞ |z→∞ →

(ρ,P,T ) = (ρ,P,T ) (ρ∞, P∞,T∞)= const. (41) |r→∞ |z→∞ → The following notations will also be used.

◦ ◦ ◦ ρ ρ > 0, P P ,T T , (42) ≡ |z=0 ≡ |z=0 ≡ |z=0

◦ ◦ ′◦ ∂vz v vr , v vϕ , v . (43) r ≡ |z=0 ϕ ≡ |z=0 z ≡ ∂z z=0

8 2.1.1 Circular velocity

Since the values ρ, P , vr, and vϕ are even functions of z, whereas vz is an odd one, the PDE system (36)-(38) at the midplane (z = 0) becomes the following ODE (ordinary differential equation) system in the midplane variables.

1 ∂ (rρ◦v◦) r + ρ◦v′◦ =0, (44) r ∂r z

◦ ◦ ◦ 2 ◦ ◦ ◦ ◦ ∂vr ρ vϕ ∂P 1 ρ ρ vr = Eu 2 , (45) ∂r − r  − ∂r − Fr r ∂rv◦ ρ◦v◦ ϕ =0. (46) r r∂r Eq. (39) is satisfied identically at the midplane. Assuming that v◦ = 0 we find r 6 the following exact solution for the midplane value, vϕ , of circular velocity. |z=0 ◦ ◦ Cϕ ◦ vϕ v = , C = const. (47) |z=0 ≡ ϕ r ϕ ◦ For the sake of convenience, let us represent Cϕ as a function of the Froude number, Fr. Let the point r = rm = const (0 rm 1) be the only point where the vortex motion, (47), will also be Keplerian:≤ ≤ 1 vϕk = . (48) √rFr

◦ ◦ Then C rm =1√rmFr C = √rmFr and, hence, ϕ ⇒ ϕ

◦ √rm vϕ = , r rm, 0 rm 1. (49) r√Fr ≥ ≤ ≤ Note that the exact solution, (47), was found under the only assumption ◦ of non-zero inward drift velocity, i.e. vr vr z=0 = 0. Thus, in the case ◦ ≡ |5 6 of inviscid flow with vr = 0 we have the vortex as the only solution for the 6 ◦ ◦ ◦ midplane circular velocity, vϕ. If, however, vr = 0, then vϕ can be sub-Keplerian or even highly non-Keplerian. For the sake of simplicity, it was considered the flow at the midplane. One can readily see that the same results are valid in a more general case. Actually, let z = h (r) denote a streamline, namely a line that is tangent to the meridional velocity vector, (vr, vz). The kinematic condition (see, e.g., [81, p. 165], [104, p. 50]) at the streamline will be the following.

h h ∂h h h v = v , v vz , v vr . (50) z r ∂r z ≡ |z=h(r) r ≡ |z=h(r) 5 −1 An axisymmetric flow with vϕ ∝ r will be called as vortex. In particular, vortex-sinks and vortex-sources are lumped together as vortices.

9 Let ς = ς (r, z) denote a dependent variable. Let the point (r, z) be at the h streamline z = h (r), then the total (full) derivative of ς ς z=h(r) with respect to r is the following. ≡ | dςh ∂ςh ∂ςh ∂h = + . (51) dr ∂r ∂z ∂r In view of (50) and (51), we obtain from (38):

h h drvϕ h v =0, v vϕ . (52) r dr ϕ ≡ |z=h(r) h If v vr = 0, then, in view of (52), we obtain r ≡ |z=h(r) 6 Ch vh vh = ϕ , Ch = const. (53) ϕv ≡ ϕ r ϕ z=h(r) h Thus, in the case of inviscid flow with vr = 0 we have the vortex (53) as the only solution for the circular velocity at the streamline,6 z = h (r). We emphasize that the exact solution, (53), was found without any assumptions about the equation of state, the temperature distribution in the disk, and the gravitational field. It is significant that the circular velocity may differ from the vortex only if the radial velocity is equal to zero. Let us consider such a possibility, namely, we will find some of the circular velocities, provided vr 0 in the region ≡ Ra (= (r, z): r [1, ), z [0, ) ). In such a case we immediately obtain { ∈ ∞ ∈ ∞ } from Eq. (36) that vz 0, since vz z=0 = 0. Hence, instead of (36)-(39), we shall have the following≡ PDE system.|

2 ρvϕ ∂P ρ ∂Φ = Eu + , (54) r ∂r Fr ∂r

− ∂P ρ ∂Φ 2 2 0.5 Eu + =0, Φ= r + z . (55) ∂z Fr ∂z −

Let the circular velocity, vϕ = vϕ (r, z), be a continuous function of z at ◦ the midplane R (r, z): r [1, ), z =0 . In such a case, vϕ 0 in the a ≡ { ∈ ∞ } ≡ region Ra = (r, z): r [1, ), z [0, ) iff the gas pressure distribution is spherically symmetric,{ i.e.∈ ∞ ∈ ∞ }

P = P (R) , R = r2 + z2. (56) Actually, let (56) be valid. Then, in view ofp (56), we obtain from (55) that ∂P ρ = 2 , z> 0. (57) ∂R −EuFrR

By virtue of (57), we find from (54) that vϕ = 0 for all z > 0 and, hence, in view of the continuity assumption, the circular velocity vϕ 0 in the region Ra. ≡ Now, let vϕ 0 in the region Ra. In such a case we find from (54)-(55) that ≡ ∂P ∂P z r =0. (58) ∂r − ∂z

10 Then we obtain from (58) that (56) is valid, i.e. the gas pressure distribution is spherically symmetric in Ra. Let us also note that the proof was done without any assumptions about the equation of state and the temperature distribution in the disk. We emphasize that the statement, i.e. vϕ 0 in Ra P = P (R) in ≡ ⇐⇒ Ra, has been proven for the whole region Ra. Let Rb be a subregion of Ra (i.e. Rb Ra), then one can readily see that the similar statement, i.e. vϕ 0 in ⊂ ≡ Rb P = P (R) in Rb, will also be valid. One can also readily see from (54)-(55)⇐⇒ that ρ = ρ (R) provided (56) and, hence, the temperature distribution of an ideal gas will be spherically symmetric in Ra (or in Rb). Let us now assume that the gas is ideal, (25), and the temperature T = T (r, z) T∞ > 0 is a pre-assigned function of the coordinates in the region Ra = (r, z≥): r [1, ), z [0, ) . In such a case, in view of (25) and (41), we obtain{ the following∈ ∞ exact∈ solution∞ } to Eq. (55):

z ◦ ξ ◦ ◦ P = P exp 1.5 dξ , P = P (r) . (59) 2 2 − EuFrT (r, ξ) r + ξ  Z0   It is a simple matter to evaluate P ◦(r) for each r by virtue of the boundary conditions (41). Actually, in view of (59) and (41), we have

∞ ◦ ξ ∞ ∞ ∞ P = P (r)Ψ (r) , Ψ (r) exp 1.5 dξ . (60) 2 2 ≡ − EuFrT (r, ξ) r + ξ  Z0    It is significant that P∞ = 0 in (60). Otherwise, as we can see from (60), the ◦ ◦ 6 value P (= P (r)) may be assigned arbitrarily provided that Ψ∞ (r)=0. In such a case the problem (54)-(55) is reduced to the only equation (54), where ◦ the two functions, namely vϕ (r, z) and P (r), are the unknown functions, i.e. the problem is not well-posed in the sense of Hadamard (e.g., [1], [97]). With this in mind, under P∞ = 0, we assume that Ψ∞ (r) = 0 for all r [1, ). ◦ 6 ∈ ∞ Then P (r) 0 and, hence, in view of (59), P (r, z) 0 in the region Ra. ≡ ≡ Consequently ρ (r, z) 0 in Ra, as the gas is ideal and, hence, there is no any ≡ flow in the whole region, Ra = (r, z): r [1, ), z [0, ) , for lack of gas. { ∈ ∞ ∈ ∞ } Thus, assuming that P∞ = 0 in (60), we rewrite the exact solution to Eq. (55) in the following form: 6

∞ ξ P = P∞ exp 1.5 dξ , P∞ = const > 0. (61)  E F T (r, ξ) r2 + ξ2  Zz u r  Let us note that (61) leads to (56) if the gas temperature distribution is spher- ically symmetric in Ra. In view of (61), the Eq. (54) gives the exact solution for the circular velocity: ∞ T (r, z) ∂ ξ r2 v2 = r dξ + . (62) ϕ 2 1.5 2 2 1.5 Fr ∂r T (r, ξ) r2 + ξ Fr (r + z ) Zz  11 For the sake of illustration, let us, first, consider an isothermal flow in the region Ra. Since T T∞ = const > 0, we find, by virtue of (61), that ≡ 1 P = P∞ exp , R = r2 + z2. (63) EuFrT∞R   p Thus, it is apparent that vϕ 0 in Ra, since (56) is valid (cf. [39]). The same ≡ result, i.e. vϕ 0 in Ra given that T const, can be obtained directly from (62). ≡ ≡ We make now use of the well-known power-law model for the temperature distribution (see, e.g., [3], [4], [7], [18], [21], [26], [31], [54], [60], [85], [91], [86], [101]). Let us consider the following generalization of the model:

Tˆ T∞ T = − + T∞, T,Tˆ ∞ = const, Tˆ >T∞, α˜ = const 0. (64) rα˜ √r2 + z2 ≥ We obtain from (61)-(62), in view of (64), that

β p α˜ Tˆ T∞ r P = P∞ 1+ − , βp = ; (65) ∞ α˜ √ 2 2 " T r r + z # EuFr Tˆ T∞ −   α˜ ˆ 2 α˜ r T T 1 T T∞ vϕ = ln ,T = − + T∞. (66) Fr Tˆ T∞ T∞ − √r2 + z2 rα˜ √r2 + z2  −  Taking the midplane circular velocity to be no more than the Keplerian velocity, − i.e. v2 v2 (rF ) 1 for all r [1, ), we obtain an upper bound for ϕ z=0 ϕk r the parameter≤ α ˜ in≡ (64)-(66). The midplane∈ ∞ circular velocity can be written in the following form:

1+˜α ˆ 2 1 T∞r T T∞ vϕ = ζ (r) , ζ (r) α˜ 1+ ln 1+ − 1 . (67) z=0 rF ≡ ˆ T∞r1+˜α − r " T T∞  ! # − Considering α ˜ 0, it is easy to check that ζ (r) is decreasing function of r. Hence, we obtain,≥ in view of (67), that

−1 Tˆ Tˆ 0 α˜ α˜m ln 1 . (68) ≤ ≤ ≡ Tˆ T∞ T∞ − ! − Obviously, ifα ˜ =α ˜m, then the midplane circular velocity coincides with the Keplerian velocity at r = 1. Consequently, v2 will be close to v2 in a ϕ z=0 ϕk small vicinity of the point r = 1. If, however, we assume thatα ˜ = 0, then vϕ 0 in Ra. Let us also show that there exists a subregion of Ra, where the midplane≡ circular velocity differs significantly from the Keplerian velocity even ifα ˜ =α ˜m. We will use the following asymptotic expansion in the limit r : → ∞ 2 Tˆ T∞ Tˆ T∞ 1 Tˆ T∞ ln 1+ −1+˜α = −1+˜α −1+˜α + .... (69) T∞r m ! T∞r m − 2 T∞r m !

12 We obtain, by virtue of (69), the asymptotic expansion for v2 in the limit ϕ z=0 r : → ∞

2 ˆ 2 C∞ 1 2 T T∞ vϕ = + o , C∞ =α ˜m − . (70) z=0 F r2+˜αm r2+˜αm 2T∞ r  

Thus, as we can see from (70), the midplane circular velocity differs significantly from the Keplerian one for sufficiently large r. Thus, the simple, well-known models considered above clearly show that the midplane circular velocity can be sub-Keplerian or even highly non-Keplerian. Hence, the “justification” of the widely known assertion that an inviscid (or laminar viscous) accretion disk must be sub-Keplerian (see, e.g., [12], [16], [18], [20], [59], [85], [91], [101]) is questionable. In this connection, mention should be made of another approach, in which one simply assumes that radial pressure forces, i.e. the first term on the RHS of (54), are negligible (see, e.g., [14, p. 165], [21, p. 153], [26, p. 132]). Hence, in view of (54), the disk is almost Keplerian. The arguments for such a decision are based, to a large extent, on the observations of accretion disks and the processes in them, despite the fact that the motion of an inviscid liquid can differ significantly from the turbulent flows that take place in real disks. The surprising thing is that accretion disks are “proven” to be Keplerian in zeroth approximation6. The “proof” was done (see, e.g., [39], [47], [61], [72], [73], [75], [88], [86], [99]) by asymptotic methods7, generally by the regular perturbation technique, applied to singularly perturbed PDEs. A thorough explanation and exemplification of the fundamental differ- ence between regular and singular perturbations can be found, e.g., in [56], [100]. It turns out that the asymptotic solution to the singularly perturbed PDEs, i.e., in general, the equations with small parameters multiplying derivatives, consists of two parts, namely of two power series in small parameters. One power series is the so called regular part (e.g., [76], [100]), where the coefficients of the series are functions of non-dependent variables. Another part, where the coefficients of the power series are functions of stretched variables, is to be considered as a singular part [76] of the asymptotic solution. This part is often called as the boundary layer part (or series) of the asymptotic solution [100]. N. N. Moiseev [56] pointed out that it is not quite felicitous term for the singular part, since the purpose of singular series (together with the regular one) is not only to satisfy the imposed boundary conditions (cf., e.g., [75]). In many works (see, e.g., [39], [47], [72], [74], [88], [86], [99]) the authors use only the regular part of asymptotics, in spite of the fact that the PDEs being investigated are sin- gularly perturbed. Such an approach can lead to erroneous conclusions. To demonstrate it, let us consider the system (54)-(55) in the region Ra = (r, z): { r [1, ), z [0, ) . We assume that 0 < Eu 1, the gas is ideal, (25), ∈ ∞ ∈ ∞ } ≪ 6In zeroth approximation only the first terms of asymptotic expansions are taken into consideration. 7One of the most effective approach for solving the problem (19)-(22), i.e. equations of mathematical physics containing small parameters, is to apply asymptotic methods (e.g., [30], [55], [56], [100], [103]).

13 and the temperature T = T (r, z) is a pre-assigned function of the coordinates in the region Ra. We assume, at first, that the asymptotic solution to the system consists only of the regular series, as it has been done in many astrophysical publications dealing with the asymptotics. Let

2 2 ρ = ρ0 + Euρ1 + Euρ2 + ..., P = P0 + EuP1 + EuP2 + ...,

2 vϕ = vϕ0 + Euvϕ1 + Euvϕ2 + ..., (71) where ρi = ρi (r, z) , Pi = Pi (r, z) , vϕi = vϕi (r, z) , i = 0, 1, 2,... . Then, substituting (71) into (54)-(55) and equating coefficients of the same powers of Eu in both sides of (54) and (55), we obtain the problems for the terms ρi, Pi, vϕi, i = 0, 1, 2,... . Specifically, we obtain the following equations for zeroth approximation: ρ v2 ρ ∂Φ ρ ∂Φ 0 ϕ0 = 0 , 0 =0. (72) r Fr ∂r Fr ∂z By virtue of (23) and (25) we find from (72):

0, z> 0 0, z> 0 ρ = ∗ P = ∗ , 0 ρ , z =0 ⇒ 0 P , z =0  0  0 ∗ ∗ vϕ0, ρ0 =0 vϕ0 z=0 = ∗ , (73) | vϕk 1√rFr, ρ =0  ≡ 0 6 ∗ ∗ ∗ where ρ0, P0 , and vϕ0 are arbitrary functions of r. Thus, in zeroth approxima- tion, the disk is Keplerian and its thickness is equal to zero. The equations for the first approximation are the following:

2 2vϕ0vϕ1ρ + v ρ ∂P ρ ∂Φ ∂P ρ ∂Φ 0 ϕ0 1 = 0 + 1 , 0 + 1 =0. (74) r ∂r Fr ∂r ∂z Fr ∂z Let ι be hyper-small but non-zero, i.e. ι 0. Then, in view of (34) and (73), the second equation in (74) can be written≃ as the following:

ρ1 ∂Φ ∂P0 ρ ∂Φ =0, z> 0 1 Fr ∂z∗ + −P0 , (75) ∂z Fr ∂z ≡ ( ι =0, z =0

∗ whence P0 = 0, and, in view of (73), P0 0. Therefore we have, in zeroth approximation, that the radial pressure forces≡ are equal to zero. It proves the assumption about negligibility of the radial pressure forces (see [14, p. 165], [21, p. 153], [26, p. 132]). Nontheless, we continue our calculations. We find for high-order approximations:

0, z> 0 ρ = ∗ , Pi 0, vϕi 0, i =1, 2, 3,..., (76) i ρ , z =0 ≡ |z=0 ≡  i

14 ∗ ∗ where ρi (= ρi (r)) is an arbitrary function of r. Consequently, in view of (71), we get the following solution to the system (54)-(55): ∗ ∗ 0, z> 0 vϕ, ρ =0 ρ = ∗ , P 0, vϕ z=0 = ∗ , (77) ρ , z =0 ≡ | vϕk 1√rFr, ρ =0   ≡ 6 ∗ ∗ ∗ 2 ∗ ∗ where ρ = ρ0 + Euρ1 + Euρ2 + ..., vϕ is an arbitrary function of r. Then, in ∗ ◦ view of (25) and (77), we obtain that ρ = 0 if T z=0 T (r) = 0, otherwise, if T ◦ (r) = 0 then ρ∗ is an arbitrary function of| r.≡ Thus, in6 the case when T 0 we obtain that the disk is Keplerian, its thickness is equal to zero, the gas≡ density is an arbitrary function of r, and the disk consists of non-interacting particles (since P 0). This result makes some sense if we are dealing with say a pressureless dust≡ . In the case when T > 0, the assumption that the solution to the system (25), (54)-(55) consists only of the regular series leads to the conclusion that ρ 0 in the whole region Ra = (r, z): r [1, ), z [0, ) ≡ { ∈ ∞ ∈ ∞ } and, hence, there is no any flow in Ra for lack of gas. Now, let us assume that the asymptotic solution to the system (25), (54)- (55) consists of two parts [100], namely regular and singular power series in a small parameter. We will consider the case when T = T (r, z) > 0 in Ra. Let us take advantage of the fact that the radial coordinate r enters the equation (55) as a parameter only. Then, in view of (25), the equation (55) can be seen as singularly perturbed ODE and, hence, it can be solved separately from Eq. (54). For the sake of convenience, we will use, in general, the notation introduced in [100]. The stretched variable, s, is defined as follows z s = , ǫ Eu. (78) ǫ ≡ We will seek the asymptotic expansion forp the solution of Eq. (55) in the following form. P =ΛP (z, Eu) + ΠP (s,ǫ) , (79) where ΛP (z, Eu) and ΠP (s,ǫ) are, respectively, the regular and singular parts of the expansion, namely,

2 ΛP =Λ0P (z)+ EuΛ1P (z)+ EuΛ2P (z)+ ..., 2 ΠP = Π0P (s)+ ǫΠ1P (s)+ ǫ Π2P (s)+ .... (80) By virtue of (23) and (25), we rewrite (55) to read: ∂P zP Eu = ̥ (P,T (r, z) ,r,z) , 0 < Eu 1. (81) ∂z ≡− 2 2 3 ≪ FrT (r, z) (r + z ) q One can readily see, by virtue of Vasil’eva theorem [100, p. 26], that the series (80) will be asymptotic in the interval z > 0 if the function T −1 (r, z) will be infinitely differentiable with respect to z in the interval z 0. Substituting the series (80) into Eq. (81), we obtain: ≥ ∂ΛP ∂ΠP E + ǫ = ̥ (ΛP + ΠP,T (r, z) ,r,z) . (82) u ∂z ∂s

15 Following [100], we rewrite the function ̥ of Eq. (82) in the form similar to (79). Taking into account the linear dependence of ̥ on P , we obtain:

̥ (ΛP + ΠP,T (r, z) ,r,z)= ̥ (ΛP (z, Eu) ,T (r, z) ,r,z)+

̥ (ΠP (s,ǫ) ,T (r,sǫ) ,r,sǫ) Λ̥ + Π̥, (83) ≡ where Λ̥ denotes the expansion of ̥ (ΛP,T (r, z) ,r,z) in terms of powers of Eu with the coefficients depending on z, and Π̥ denotes the expansion of the rest part of ̥ (ΛP + ΠP,T (r, z) ,r,z) in powers of ǫ with the coefficients depending on s. Then, by virtue of (83) and (80), we obtain from (82) that

zΛ P ∂P − zΛ P 0 =0, i 1 = i , i =1, 2,.... (84) 2 2 3 ∂z − 2 2 3 FrT (r + z ) FrT (r + z ) q q Since it is assumed that T = T (r, z) > 0 in Ra, we obtain from (84) that ΛiP (z) 0 for i = 0, 1, 2,... . Hence, ΛP 0 in the region Ra. From the above reasoning≡ it is clear that the asymptotic≡ solution to Eq. (55) consists only of the singular part. In view of (80), the equation (82) can now be written as follows: ∂ [Π P (s)+ ǫΠ P (s)+ ...] s [Π P (s)+ ǫΠ P (s)+ ...] 0 1 = 0 1 . (85) ∂s − 2 2 2 3 FrT (r,ǫs) (r + ǫ s ) q Notice that ΠiP (s) (i =0, 1, 2,...) should be boundary functions, namely, they should approach zero as s [100]. For the leading term of the singular part we have: → ∞ ∂Π0P (s) sΠ0P (s) ◦ = 3 ◦ ,T (r) T (r, z) z=0 . (86) ∂s −Frr T (r) ≡ | Let P ∗ Π P (0). Then, in view of (86), we have that: ≡ 0 s2 Π P (s)= P ∗ exp . (87) 0 −2F r3T ◦ (r)  r  Let us restrict our consideration to the zeroth approximation. We assume that P ∗ = P ◦ P (r, z) . Then, by virtue of (78), (79), and (87), we obtain: ≡ |z=0 z2  ◦ P0 Λ0P (z) + Π0P z Eu = P (r)exp 3 ◦ , (88) ≡ −2EuFrr T (r)  p    where P ◦ (r) is an arbitrary function of r. Recall that we consider the gas flow in the region Ra= (r, z): r [1, ), z [0, ) due to the axisymmetry of the problem, in spite{ of the fact∈ that∞ the∈ system∞ (54)-(55)} is valid for all values of z ( , ). We have, because of the axisymmetry, the additional condition ∈ −∞ ∞ for the pressure at z = 0, namely ∂P ∂z z=0 = 0, which is obviously valid for (88). Notice also that we cannot evaluate| P ◦ (r) by virtue of the boundary condition (41), since P (r, z) 0 in (88). Let us use the exact solution 0 |z→∞ →

16 (61) to Eq. (55) for the evaluation of P ◦ (r) in (88). In such a case we obtain from (88) that

∞ 1 ξ z2 ∞ P0 = P exp 1.5 dξ 3 ◦ . (89) EuFr  T (r, ξ) r2 + ξ2 − 2r T (r)   Z0     Using the exact solution (61) to Eq. (55), we find the relative error δP (P P ) P : ≡ | − 0 | z 1 ξ z2 δP = 1 exp 1.5 dξ ◦ . (90) − E F  2 2 − 2r3T (r)  u r Z T (r, ξ) r + ξ  0     Obviously, the value of δP will be comparatively small in a small vicinity of z = 0 (in the boundary layer [100, p. 13]), since EuFr 1. In this connection, it should be noted that the singular part of the asymptotic≪ expansion is usually necessary for correct solution of singularly perturbed problems like (54)-(55).

2.1.2 Instability of inviscid disk

8 We consider the axisymmetric steady-state flows in the region Ra = (r, z): r [1, ), z [0, ) . If the process in question is polytropic, i.e. (26){ is valid, then∈ this∞ process∈ is∞ described} by the balance equations, (36)-(39), as well as by the equation of state:

P = P (ρ,T ) , P =0. (91) |T =0 For instance, if the gas is perfect, then Eq. (91) is reduced to Eq. (25). If the process is non-polytropic, then the steady-state axisymmetric version of the energy balance equation, (3), should be used instead of Eq. (26). We will mainly consider the ODE system, (44) - (46), in the midplane variables ◦ ς ς z=0, where ς = ς (r, z) denotes a dependent variable. It is assumed that ◦ ≡ | ◦ ◦ q (r) rρ vr is continuous (e.g. [65, p. 106]) at all r [1, ). First≡ of all we prove that any non-vortex flow9 is unstable.∈ ∞ We consider the steady-state accretion disk as input-output [24] (or cause-effect [53]) system. The boundary conditions, namely (40), (41), as well as

q = Q,Q = Q (z) ,Q (z) 0, q rρvr, (92) |r=1 |z→∞ → ≡

vϕ = V, V = V (z) , V (z) 0, (93) |r=1 |z→∞ → are considered as inputs and the vector-functions ρ (r, z) , P (r, z) ,T (r, z) , { vr (r, z) , vϕ (r, z) , vz (r, z) are considered as outputs. Recall that we consider } 8Recall that the flows considered in this section are assumed to be laminar and inviscid. 9 −1 Recall that an axisymmetric flow with vϕ ∝ r , (47), is to be called as vortex. In particular, vortex-sinks and vortex-sources are lumped together as vortices.

17 the gas flow in the region Ra = (r, z): r [1, ), z [0, ) due to the axisymmetry and, hence, we have:{ ∈ ∞ ∈ ∞ }

vz =0. (94) |z=0 The mapping F : input output is defined by the steady-state balance equa- tions, (36)-(39), the steady-state→ axisymmetric version of the energy balance equation, (3), as well as by the equation of state, (91). Following, e.g., [24], [48], and [53] we can say that the input-output system is unstable if an infinitesi- mal increment in the input triggers a finite increment in the output. Such an approach was already used, [11], to investigate instability of magnetic as well as non-magnetic accretion disks. In our case this approach gives the result in a few simple steps. For instance, it is assumed (see [11]) that vr = 0 to have a possibility for vϕ be, e.g., sub-Keplerian. Then, as the input, we take an infinitesimal increment in vr. In such a case we obtain that vϕ will be a vortex, i.e., we obtain a finite increment in the output. In this subsection the instability of gas-dynamic (non-magnetic) accretion disks will be considered in more de- tails. It is already shown above (see Subsection 2.1.1) that the circular velocity may differ from the vortex velocity only if the radial velocity is equal to zero. Accordingly, we assume that q rρvr = 0 (st (ρ) > 0) at all (r, z) Ra and, hence, ≡ ∈ Q =0, z [0, ) . (95) ∀ ∈ ∞ ◦ ◦ Let vϕ = vϕ (r) be a certain midplane circular velocity. For instance, we may assume that the circular velocity will be “Keplerian with a great precision”, as it is argued in, e.g, [14], [18], [20], [21], [26], [91], [101]. An important point is ◦ ◦ that vϕ = vϕ (r) is assumed to be significantly different from the vortex velocity ◦ ◦ R◦ vϕv = Cϕr, i.e. for any interval (a,b) a (r, z): r [1, ), z =0 with a ≇ b we have ⊂ ≡{ ∈ ∞ }

C◦ v◦ v◦ v◦ ϕ ≇ 0, C◦ = const. (96) ϕ − ϕv ∞,(a,b) ≡ ϕ − r ∀ ϕ ∞,(a,b)

Here f ∞ = f ∞,S sup f (r) : r S denotes the Chebyshev norm of a boundedk k functionk k f (r)≡ defined{| on a| set∈S.} Let us change the value of Q, (95), in the boundary condition (92), namely, let Q Q: → ◦ Q = ιaq, ι = const 0, aq = aq (z) ,ea aq =0, (97) ≃ q ≡ |z=0 6 where the bounded function a (z) is such that a 0. All other boundary e e q e e q z→∞e e conditions are unchanged. The change of the boundary| → condition leads to the ◦ ◦ modification of the output, i.e.e ς ς , whereeς denotes a dependent variable. For instance, we obtain from (44) and→ (97) that

r e ◦ ◦ ∂v˜ q◦ 0 q˜◦ = ιa◦ ξρ˜ v˜′◦dξ, ρ ρ , v˜′◦ z . (98) ≡ → q − z ≡ |z=0 z ≡ ∂z Z z=0 1

e e e 18 ◦ ◦ ◦ ◦ Sinceq ˜ =q ˜ (r) is continuous at all r [1, ) andq ˜ r=1 = ιaq = 0, (97), ∈ ∞ ◦ | 6 then there exists a real number r∗ > 1 such thatq ˜ = 0 at all r (1, r∗). We 6 ∈ find from (46) and (93) that e ◦ ◦ Cϕ ◦ v = , r [1, r∗) , C = const = V . (99) ϕ r ∀ ∈ ϕ |z=0 Thus, we obtaine that the infinitesimal increment in the input, i.e. Q Q, see (95) and (97), triggers the finite increment in the output (i.e. v◦ v→◦ ), since ϕ → ϕ v◦ v◦ ≇ 0 in view of (96), which is to say that the flow is unstable.e ϕ − ϕ ∞,(1,r∗) For the sake of illustration, let us consider the axisymmetric steadye -state flow withe the circular velocity represented by the power-law model [101, p. 374]. First, we assume that

q rρvr =0, (r, z) Ra, st (ρ) > 0, (100) ≡ ∀ ∈ ◦ and the midplane circular velocity, vϕ, is represented as the following: ◦ ◦ C ◦ ◦ v = ϕ , r R , C = const, 0.5 κ = const < 1, (101) ϕ rκ ∀ ∈ a ϕ ≤ R◦ where a (r, z): r [1, ), z =0 . We emphasize that st (κ) < 1, i.e. the midplane circular≡ { velocity∈ given∞ by (101)} is assumed to be significantly different ◦ ◦ from the vortex velocity vϕv = Cϕr. The boundary conditions are the same as above, namely: (40), (41), (92), and (93). In view of (100) and (101), we have (95) and, respectively

v◦ = V ◦ V = C◦ . (102) ϕ r=1 ≡ |z=0 ϕ   By virtue of (100) and (94) we obtain from 36 that vz 0 in the region Ra. ≡ Recall that the condition (100), i.e. vr 0 in Ra, was taken to examine a non-vortex flow, which is possible only if the≡ radial velocity is equal to zero. If, however, we assume that vz 0 in Ra, then vr 0 in Ra, provided (95). Let us change the value of≡Q, (95), in the boundary≡ condition (92), namely, let Q Q˜: → ιaq =0, z =0 Q˜ = 6 , ι 0. (103) 0 z > 0 ≃  Givenv ˜z 0 in the region Ra,e we obtain from 36, in view of (103), that ≡ ιaq =0, z =0 q˜(r, z) rρv˜r = 6 , r [1, ) , ι 0. (104) ≡ 0 z > 0 ∈ ∞ ≃  e We find from (46) ande (104) that C˜◦ v◦ = ϕ , r [1, ) , C˜◦ = const. (105) ϕ r ∀ ∈ ∞ ϕ ˜◦ ◦ ◦ ◦ We assume that Cϕ =e Cϕ, see (105) and (102). In such a case vϕ r=1 = vϕ r=1. Thus, we obtain that the infinitesimal increment in the input (Q Q˜), i.e. → e 19 ◦ ◦ st Q˜∗ Q∗ = 0, triggers the finite increment in the output (v v ), since − ϕ → ϕ the midplane circular velocity given by (101) is assumed to be significantly different from the vortex velocity (97). Hence, the flow in question ise unstable with respect to infinitesimal perturbations in the input (92). Let us now examine the stability of the regular part of asymptotic solution, (77), to the singularly perturbed system (36)-(39), provided vr 0 and, hence, ≡ vz 0 in the region Ra. Recall that in the case when T > 0, the assumption that≡ the solution consists only of the regular series, (71), leads to the conclusion that ρ 0 in the whole region Ra and, hence, there is no any flow in the whole ≡ region Ra for lack of gas. Otherwise, i.e. T 0, we obtain that the disk is Keplerian (48), its thickness is equal to zero≡ (i.e. ρ = 0 for all z > 0), the gas density at the midplane is an arbitrary function of r, and the disk consists of non-interacting particles (since P 0). We will assume that st (ρ) > 0 for all z = 0. It is easy to see that this≡ solution satisfies the degenerate system (see, e.g., [55], [100]) corresponding to (36)-(39), i.e. the system obtained from (36)-(39) by putting Eu = 0. The system (36)-(39) at the midplane (z = 0) becomes the following ODE system in the midplane variables. 1 ∂ (rρ◦v◦) r =0, (106) r ∂r ◦ ∂ ◦ ◦ 2 ◦ ◦ 2 ρ rρ (vr ) ρ vϕ = , (107) ∂r − −Frr h i ◦  ◦ ◦ ∂rv rρ v ϕ =0. (108) r ∂r The boundary conditions are the following:

◦ ◦ ◦ ◦ ◦ ◦ q = Q , q rρ v , v 0, (109) |r=1 ≡ r r |r→∞ → ◦ ◦ ◦ ◦ ◦ v = V , v 0, ρ ρ∞ = const > 0. (110) ϕ r=1 ϕ r→∞ → |r→∞ → ◦ ◦ ◦ ◦ Obviously, if Q = 0 in (109), then, in view of (106), q rρ v = 0 at all r r [1, ) and, hence, we find, by virtue of (107), that the≡ disk is Keplerian ∈ ∞ ◦ ◦ (48). It is possible if V = 1√Fr in (110). Let us change the values of Q and V ◦ in the boundary conditions (109) and, respectively, (110), namely, let Q◦ Q˜◦ and V ◦ V˜ ◦: → → ◦ ◦ ◦ Q˜ = ιaq, ι 0, aq = const =0, st V˜ = st (V ) . (111) ≃ 6 ◦ ◦ ◦ In view of (111), we obtain from (106) thatq ˜ rρ˜ v˜ = ιaq at all r [1, ). e e ≡ r ∈ ∞ Since V ◦ V˜ ◦, we find from (108), by virtue of (110), that → ◦ e ◦ V˜ ◦ v = , r [1, ) , V˜ = const. (112) ϕ r ∀ ∈ ∞ ◦ Notice that the vortexe velocity (112) does not satisfy Eq. (107), sinceq ˜ ◦ ◦ ◦ ≡ rρ˜ v˜r = ιaq. Obviously, st (˜q ) = st (ιaq) = st (ι) st (aq) = 0. For the sake of

e 20e e −1 convenience, we assume that st (r) = r, st (Fr) = Fr, and st Fr = 0. It is ◦ ◦ ◦ 6 ◦ ◦ also assumed above that st (˜ρ ) = 0. Consequently, st (rρ˜ v˜r ) = rst (˜ρ ) st (˜vr ) = 0 and, hence, 6  ◦ st (˜vr )=0. (113) By virtue of (106), (107), (109), and (112), we find:

2 ◦ ◦ V˜ ◦ ∂v˜r 1 v˜r = 3 2 ∂r  r  − Frr ⇒

2 ◦ ◦ 2 V˜ (˜vr ) 1 = 2 , r [1, ) . (114) 2 Frr −  2r  ∀ ∈ ∞ It is easily seen that the equality (114) cannot be valid for all r [1, ) due to (113). Actually, we obtain from (114), by virtue of (113), that ∈ ∞

1 (V ◦)2 =0, r [1, ) , (115) Fr − 2r ∀ ∈ ∞

◦ ◦ where V = st V˜ = 1√Fr. Now it is clear that the equality (115) and, hence, the equality  (114) cannot be valid for all r [1, ). Consequently, the vortex velocity (112) does not satisfy Eq. (107).∈ Thus,∞ the infinitesimal increment in the input (Q◦ Q˜◦, V ◦ V˜ ◦) does not trigger a finite increment in the output and, hence, the→ motion in→ question is not unstable. An important remark is in order at this point. The assumption that the asymptotic solution to the singularly perturbed PDEs consists only of the regular series (such as (71), see, e.g., [39], [47], [72], [74], [88], [86], [99]) can lead to a wrong conclusion about the stability of the system. So, we have proven that any axisymmetric steady-state flow (excluding the vortex (47)) is unstable. The proof that the vortex motion is unstable can be found in [11] (see also [96, Sec. 3.4], [101, Sec. 7.2.4]).

2.2 Viscous flow

So far the inviscid model of accretion disk was examined. It has been proven that the vortex, (47), will be the only exact solution for the circular velocity ◦ ◦ field of midplane flow if the radial velocity vr =0. If vr = 0, then there exists a wide range of solutions (including Keplerian6 and sub-Keplerian rotations) for the midplane circular velocity. It has also been proven that any steady-state solution to this model that does not coincide with the vortex will be unstable (see Sec. 2.1.2). The vortex will be linearly unstable [11] in the region under consideration. It is also vital to note that the majority of observed disks are in Keplerian (or sub-Keplerian) rotation around their central accreting objects (see, e.g., [25], [29], [58], [92], [95], [106] and references therein). Since we consider the gas flows, it is easy to show that the laminar viscous disk can be approximated with a great precision by the vortex motion, (47).

21 The steady-state version of the PDE system (20)-(22) is thus reduced to the following one. The r-component:

2 1 ∂ 2 ∂ ρvϕ ∂P ρ ∂Φ r ρvr + (ρvrvz) = Eu + r ∂r ∂z − r − ∂r − Fr ∂r  1 ∂ ∂v 2 1 ∂ (rv ) ∂v 2µ r µ r + z + R ∂r v ∂r − 3 v r ∂r ∂z e     ∂ ∂v ∂v 2µ ∂v v µ r + z + v r r , (116) ∂z v ∂z ∂r r ∂r − r     The ϕ-component (after simple transformations):

1 ∂ ∂ 1 ∂ ∂ω 1 ∂ ∂ω r3ρv ω + (ρv ω)= µ r3 + µ , (117) r3 ∂r r ∂z z r3R ∂r v ∂r R ∂z v ∂z e   e    The z-component: 1 ∂ ∂ r (ρv v )+ ρv2 = r ∂r z r ∂z z

∂P ρ ∂Φ 1 ∂ ∂v r ∂vz Eu + µ + + − ∂z − F ∂z R ∂r v ∂z ∂r r e     ∂ ∂v 2 1 ∂ (rv ) ∂v µ ∂v ∂v 2µ z µ r + z + v r + z , (118) ∂z v ∂z − 3 v r ∂r ∂z r ∂z ∂r      2 2 where Φ = 1√r + z , ω denotes the angular velocity, i.e. vϕ = ωr. Much attention− is given in Sec. 2.1.1 to the investigation of inviscid flow, (36)-(39), on condition that vr 0, namely Eqs. (54)-(55) in the region Ra = ≡ (r, z): r [1, ), z [0, ) . In the case of viscous flow (provided vr 0) {we also have∈ (54)-(55),∞ ∈ instead∞ } of (116), (118), and the following ϕ-component:≡

1 ∂ ∂ω ∂ ∂ω µ r3 + µ =0, v = ωr. (119) r3 ∂r v ∂r ∂z v ∂z ϕ     It is readily seen that we obtain, in general, analogous results to the inviscid problem. To take an example, for the isothermal flow (T T∞ = const > 0) ≡ of ideal gas, the circular velocity vϕ 0 in Ra. As a further example, assume that the asymptotic solution to the system≡ consists only of the regular series, (71). If T > 0, then we obtain (see Sec. 2.1.1) that ρ 0 in the whole region ≡ Ra. Hence, any solution, ω = ω (r, z), to Eq. (119) is admissible for lack of gas. If, however, T 0, then we obtain that the disk is Keplerian, its thickness is equal to zero, the≡ gas density is an arbitrary function of r, and the disk consists of non-interacting particles (see Sec. 2.1.1). Notice that µv T =0 = 0, in view of (6) or (7), and, hence, Eq. (119) will also be satisfied. |

22 2.2.1 Average equations Let us now use the the well known approach for the investigation the accretion disk dynamics (see, e.g., [14, Sec. 12.1]). We will average Eqs. (117) and (36) over the depth of disk. In the following we will sometimes assume that

TH T (r, z) z=H 0 and ρH ρ (r, z) z=H 0, where H = H (r) denotes the≡ disk semi-thickness.| ≈ Let z ≡= H (r)| be the≈ streamline. For the sake of convenience, we will use the following scheme:

ρ (r, z) > 0, z

H 1 f f (r, z) dz, (122) ≡ 2H −ZH where f (r, z) denotes an integrable function, H = H (r) denotes the disk semi- thickness. With the notation (122), we obtain:

∂f 1 ∂ f + f − ∂H = 2Hf z=H z= H . (123) ∂r 2H ∂r − 2H ∂r  It is assumed, (33), that ∂H∂r 1. In such a case the last term in (123) | | ≪ can be seen as negligible provided f z=±H is small enough. We will consider the case when f 0, i.e. | |z=±H ≈

∂f 1 ∂ 2Hf . (124) ∂r ≈ 2H ∂r  We will use the well known (and natural for thin disks) assumption that the variation of vϕ (and, hence, ω) with z is negligible (see, e.g., [18]). If f (z) does not change sign on the interval ( H,H), then this assumption leads to the following equality. −

H H H fωdz = ω fdz ω fdz, (125) |z=ξ ≈ −ZH −ZH −ZH where ξ [ H,H]. In a similar manner we obtain that ∈ − H H ∂ω ∂ω f dz fdz. (126) ∂r ≈ ∂r −ZH −ZH

23 We will also use the assumption that the variation of vr with z is negligible (see, e.g., [14]). Then we obtain the equalities similar to (125), (126).

H H H H ∂vr ∂vr fvrdz vr fdz, f dz fdz. (127) ≈ ∂r ≈ ∂r −ZH −ZH −ZH −ZH

Recall that the dynamic viscosity, µv, of a gas increases with temperature, T , and, in fact, it is independent of pressure and density at a given temperature (see, e.g., [1, p. 186], [14, p. 153], [46, p. 46], [50, p. 635], [104, pp. 26-28]). We can assume, in view of (6), (7), and (121), that µ = 0. Hence, v|z=±H ∂v µ ϕ =0. (128) v ∂z z=±H

After the averaging of Eq. (117) over the depth of disk we obtain:

1 ∂ 3 1 (r vrρω)+ (ρvϕvz) (ρvϕvz) = r2 ∂r 2H |z=H − |z=−H h i 1 ∂ ∂ω 1 ∂v ∂v µ r3 + µ ϕ µ ϕ . (129) r2R ∂r v ∂r 2HR v ∂z − v ∂z e   e "   z=H   z=−H #

Then, by virtue of (120), (122)-(126), (128), and (129), we find that ∂ 1 ∂ ∂ω 2Hr3v ρ ω = 2Hr3µ . (130) ∂r r R ∂r v ∂r e    Analogously, we obtain from Eq. (36) that

1 ∂ (2Hrvrρ) =0, 2Hrvrρ Q = const < 0. (131) r ∂r ⇒ ≡ Let us note that we confine ourselves here to the case when Q = 0. By virtue of (131) and (130) we obtain the following steady-state version6 of the viscous evolution equation (cf. [14, p. 166]):

∂ 2 1 ∂ 3 ∂ω r ωQ = r Mv ,Mv 2Hµ . (132) ∂r R ∂r ∂r ≡ v e    Now we consider the widly used approximation, i.e. that the laminar, yet viscous disk is nearly Keplerian (see, e.g., [14], [18], [26], [101]). Let us use the power-law model [101, p. 374] for the angular velocity ω, i.e. 1 ω ∝ , κ = const, 0.5 κ 1. (133) r1+κ ≤ ≤

Let Mˆ v Mv , then, by virtue of (133) and (132), we find: ≡ |r=1

Mˆ v ReQ 1 Mv = 1 , Q = const < 0. (134) r1−κ − 1+ κ − r1−κ  

24 Remark that Mv = const if κ = 1 (the vortex motion). If Mv = const in (132), then ω ∝ r−2 fulfills Eq. (132). Since we consider the model of laminar viscous disk, it should be mentioned that the dynamic viscosity, µv, is often represented as follows: µv = νρ, where ν denotes the kinematic viscosity (see, e.g., [14], [18], [26], [101]). It is nothing more than the definition of the kinematic viscosity, i.e. ν µ ρ, (see, e.g, [14], ≡ v [23], [50], [81], [104]). Obviously, if the temperature T = const, then µv = const −1 and, hence, ν ∝ ρ (in particular, ν as ρ 0). Using µv = νρ, we obtain after the integrating of Eq. (117) over→ the ∞ depth→ of disk (see, e.g., [14, p. 166]) that Mv = ν , where denotes the surface density, i.e. =2Hρ. Using the mass-weighted averaging, suggested by A. Favre (see the references in [1]), i.e. P P P fρ f , (135) ≡ ρ we obtain νρ µ µ ν = = v ρ ρ ν = v . (136) ρ ρ ⇒ ρ   In view of (136), we find that Mv = ν = (µvρ)2Hρ = 2Hµv. Hence, the product ν in [14, p. 166] does not depend on density (even formally10), since we consider the thin disk accretionP that is highly nonadiabatic [85]. If we assume, forP instance, that ν = const (see, e.g., [14, Sec. 12.2]), then, in light of (136), we find that µv ∝ ρ. Hence, such an assumption can lead to wrong conclusions. Now we can show that a laminar viscous disk tends to be the vortex motion, i.e. approximately ω ∝ r−2. After the integration of Eq. (132) we obtain:

1 ∂ω ωQ cϕQ = 3 , cϕ, Q = const. (137) Re ∂r rMv − r Mv

Since Re 1, we are looking for the asymptotic expansion to the solution of (137) in the≫ following form ω =Λω + Πω, where Λω and Πω are, respectively, the regular and singular parts of the expansion. Let us assume, at first, that the asymptotic solution to Eq. (137) consists only of the regular series. We obtain in the zeroth approximation: c Λ ω = ϕ , c = const. (138) 0 r2 ϕ One can readily see that the regular part, (138), of zeroth approximation to the solution of the viscous problem (132) coincides with the exact solution of the non-viscous system. Let vr vr r=1 and let ω ω r=1. Taking into considera- tion the singular part of the≡ expansion, we find≡ that| the zeroth approximation,

b − b 10If the flow will be adiabatic, then T ∝ ργ 1. In such a case we obtain, formally, that (γ−1)n 0.51 µv ∝ ρ . For instance, in the case of monatomic gas we can write that µv ∝ ρ √ρ, since typically n = 0.76. However, if T is given, then ρ is given too and vice versa. Thus,≈ we cannot vary ρ and, hence, µv without varying T .

25 ω Λ ω + Π ω, can be written as the following: 0 ≡ 0 0

cϕ vrρ ω0 = 2 + ω cϕ exp Re (r 1) , vr < 0. (139) r − ( µv − )   b b b b Thus, in view of (16) and (139), we concludeb that the laminar viscous disk can be approximated with a great precision by the vortex motion (cf., e.g., [14], [18], [26], [101]).

2.2.2 Turbulent flow Let us estimate the value of κ in the power-law model [101, p. 374] for the ◦ midplane circular velocity vϕ: C◦ v◦ = ϕ , r R◦, C◦ , κ = const, 0.5 κ 1, (140) ϕ rκ ∀ ∈ a ϕ ≤ ≤ R◦ where a (r, z): r [1, ), z =0 . The Euler number, Eu, characterizes “losses” [28]≡ {(the pressure∈ loss∞ 11 [37, p.} 84]) in a flow. We assume that the turbulent disk should be such that the losses are minimal, i.e. the Euler number should be as small as possible. Such an approach was already used in [11] to argue that a turbulent disk tends to be Keplerian. In this subsection the arguments will be considered in more details. Using Prandtl and Kolmogorov suggestion [1, p. 230] that the turbulent viscosity, µt, is proportional to the square root of the kinetic energy of turbulence, k, we evaluate, by means of (8), the midplane value of the viscosity µ◦ µ as the following. t ≡ t|z=0 . ◦ ◦ ◦ 0 5 ◦ µ = CκLκρ k , k k , Cκ = const, (141) t ≡ z=0   ◦ where Lκ denotes a turbulence length scale, ρ ρ . ≡ |z=0 The variation of the Euler number, Eu, with the Reynolds number, Re, obtained with some k-ε models, is depicted in [8, Fig. 2]. In the case of laminar 12 flow, as we can see in [8, Fig. 2(b)(Problem 2)] , the smaller Re, the larger Eu and, hence, for sufficiently low values of Re the values of Eu are greater than that for turbulent flow. In such a case we assume that all turbulent terms are eliminated in (20), more over, it is assumed that the midplane velocity ◦ vr vr z=0 = 0 in (20). Let us remind (see Sec. 2.2.1) that a laminar viscous ≡ | ◦ disk tends to be the vortex motion provided that vr = 0. We will also use the following asymptotic expansion 6

∞ ◦ −i T (r) T θir , θi = const, (142) ≡ |z=0 ≈ i=0 X 11 2 The Euler number can be written in the following form: Eu = (p∗ pˇ)  ρ∗v∗, where − pˇ p r→∞ 0. ≡12Problem| ≈ 2 is selected insofar as we consider, in fact, a gas flow around a spherical object with no-slip boundary conditions at its surface.

26 in the limit r . As usually [33], the series in (142) may converge or diverge, but its partial→ ∞ sums are good approximations to T ◦ (r) for large enough ◦ r. Assuming that T (r) 0 as r , we find that θ0 = 0. Thus, we can use the following approximation→ for the→ disk ∞ temperature profile: θ T ◦ (r)= 1 + O r−2 , θ = const. (143) r 1  It is also assumed that the characteristic quantity, T∗, for temperature is taken such that θ1 = 1. Using (31), (32), and the above assumptions, we obtain the following equation, which is the steady-state version of Eq. (20) in the midplane variables:

◦ ◦ 2 ◦ ◦ ρ vϕ ∂P 1 ρ = Eu + 2 , r> 1. (144) r  ∂r Fr r In view of (143), we have, in the first approximation, that the midplane tem- perature T ◦ (r) = r−1. In view of (25), we write P ◦ = ρ◦T ◦. Then, by virtue of (140), we obtain:

◦ ◦ 2 ◦ ρˆ 1 Cϕ 1 ρ = α−1 exp ζ 2κ−1 1 , ζ = , α = , (145) r r − Eu (2κ 1) EuFr   − ◦ ◦ whereρ ˆ ρ r=1 and 0.5 < κ 1. We obtain≡ | from (141), by virtue≤ of (145), that

0.5 ◦ Cρ 1 ◦ ◦ µ = exp ζ 1 , Cρ ρˆ CκLκ k . (146) t rα−1 r2κ−1 − ≡     If vr = 0, then the steady-state version of Eq. (21) can be written in the following form (see, e.g., [73]), provided that µ µ . v ≪ t ◦ ∂ ∂ v µ◦r3 ϕ =0. (147) ∂r t ∂r r    ◦ ◦ Let us find µt = µt (r) such that Eq. (147) will be fulfilled by the power-law model (140). By virtue of (140), we find from (147) that

◦ A µ = µ , A = const. (148) t r1−κ µ

Equating (146) and (148) at r = 1, we find that Cρ r=1 = Aµ. Let us assume, ◦ ◦ ◦ | for the sake of simplicity13, that k k k in the ε-vicinity (ε 1) of ≈ ≡ r=1 ≪ r = 1, i.e. at r (1, 1+ ε). Let us now assume that µ◦ of (146) and µ◦ of (148) b t t ∈ 13We may assume (e.g., [84]) that k c2. Since c2 = ∂P/∂ρ, we find, in view of (25) ◦ ∝ s s and (143), that k 1/r. Hence, we will get a little more cumbersome formulas, but not fundamental difficulties.∝

27 (having, in general, different values) coincide each other in the vicinity of r =1 with accuracy O ε2 . In such a case we obtain that  1 Eu . (149) ∝ 2 κ − As we can see from (149) and (140), Eu min if κ 0.5. Thus, in the frame of our assumptions, we find that the turbulent→ flow tends→ to be Keplerian. Based on this conclusion, we can estimate the turbulent viscosity, µt, provided that the dynamic viscosity µv µt. From the above discussion≪ it follows that the magnitude of the centrifugal force is approximately equal to the gravitational force in the case of turbulent disk. Hence, we assume for the case of turbulent flow:

ρv2 ρ ∂Φ v2 1 1 3 ϕ = ϕ = z2 + ... . (150) r F ∂r ⇒ r F r2 − 2r4 r r   ◦ ′′◦ 2 Taking into account (32), i.e. vϕ = vϕ + vϕ z + ..., we obtain from (150) that ◦ ◦ 1 ′′◦ 3vϕ vϕ = , vϕ = 2 . (151) √rFr − 4r After simple transformations, the steady-state version of Eq. (21), where µ = µv + µt, can be written in the following form. ρv ∂ ∂v 1 ∂ ∂ω 1 ∂ ∂v r (rv )+ ρv ϕ = µ r3 + µ ϕ , (152) r ∂r ϕ z ∂z r2R ∂r t ∂r R ∂z t ∂z e   e   where ω denotes the angular velocity. By virtue of (31), (32), and (151), we can write the momentum equation (152) in the midplane variables, i.e. at z = 0, as the following: ∂ ρ◦v◦ r2ω◦ = r ∂r 1 ∂ ∂ω◦ 3ω◦  µ◦r3 µ◦, ω◦ ω , µ◦ µ . (153) rR ∂r t ∂r − 2R t ≡ |z=0 t ≡ t|z=0 e   e Since the disk tends to be Keplerian, i.e. ω◦ ∝ r−1.5, we obtain from (153) the ◦ following equation in the unknown µt . ∂µ◦ 3µ◦ R ρ◦v◦ t + t + e r =0. (154) ∂r 2r 3 Let z = h (r) denote a streamline, namely a line that is tangent to the meridional velocity vector, (vr, vz), and let hˆ h . We assume that hˆ 1 ≡ |r=1 | |≪ and hˆ (r) 0 if hˆ = 0, i.e. the streamline z = h (r) is neighboring to the ≡ streamline z = 0. It is also assumed that hˆ ≇ 0. Consequently, the kinematic condition (see, e.g., [81, p. 165], [104, p. 50]) at the streamline will be the following. d ∂h (z h)=0 vz = vr . (155) dt − ⇒ |z=h ∂r |z=h

28 Using (32) and the following transformation on h:

h hh,ˆ (156) → we rewrite (155) to read:

h∂hˆ v′◦hhˆ + v′′′◦hˆ3h3 + ... = v◦ + v′′◦hˆ2h2 + ... . (157) z z ∂r r r   From this equation we obtain with the accuracy O(hˆ2), that

∂ ln h v′◦ = v◦ . (158) z r ∂r We remark that (158) is invariant for the inverse transformation on h, i.e. h ← hhˆ . Let us consider the case when h is a linear function of r, i.e. v◦ h = hrˆ v′◦ = r . (159) ⇒ z r In view of (159), we rewrite (44) in the following form:

∂ (ρ◦v◦) 2 r + (ρ◦v◦)=0. (160) ∂r r r ◦ ◦ ◦ Let us note that the equations (154) and (160) are linear in µt and (ρ vr ). We assume, that ◦ ◦ ◦ µt = a (ρ vr )+ b, a = a (r) , b = b (r) . (161) In such a case (154) can be written as follows:

◦ ◦ ∂ (ρ v ) ∂a 3 R ◦ ◦ ∂b 3b r + + + e (ρ v )+ + =0. (162) ∂r ∂r 2r 3a r a∂r 2ar   From (160) and (162) we obtain the following equations for a (r) and b (r).

∂a 3 R 2 + + e = , (163) ∂r 2r 3a r ∂b 3b + =0. (164) a∂r 2ar Then, we find from (163), (164) that

2R a = e r + C◦√r , b = C◦r−1.5, C◦, C◦ = const. (165) − 3 a b a b  Thus, in view of (161) and (165), we find:

◦ 2 ◦ ◦ ◦ ◦ −1.5 ◦ µ = Re ρ v (r + C √r) + C r , v 0. (166) t −3 r a b r ≤  

29 Following Prandtl and Kolmogorov (e.g., [1, p. 230], [105, p. 74]) we expect ◦ that the turbulent viscocity, µt , can be modeled as (141). In view (141), the ◦ ◦ turbulent viscocity is proportional to ρ and, hence, we may assume that Cb = 0. ◦ Assuming that the disk semi-thickness H ∝ r + Ca √r, we obtain from (166):

0.5 ◦ ◦ ◦ ◦ ◦ ◦ ◦ 2(1+ Ca ) ˆ r + Ca √r µt = Rebµρ k H, bµ = , H = H ◦ , (167) 3Hˆ 1+ Ca   ˆ where H = H r=1, and | ◦ ◦ 2 k (v ) , Re 1. (168) ∼ r ≫ We can conclude from (167) and (18) that µ ∗µ ∗ Re (see also, e.g., [6, p. t v ∼ ◦ 213]). It is necessary to stress that the expression (168) for k is nothing more than the estimation of the turbulent kinetic energy provided that all of the above assumptions are valid. The more correct approach is◦ to use the one-equation model (9) to estimate the turbulent kinetic energy, k , in (167). Let us note, if we assume that the turbulent viscosity can be represented in ◦ the form (167) then, using the following transformation on µt

◦ ◦ µ Reµ (169) t −→ t we find that the PDEs are not singularly perturbed, namely, the small param- eter, 1/Re, does not premultiply the highest-order turbulent-viscous terms.

3 Conclusion

By and large axisymmetric steady-state perfect gas flows are investigated. First of all, it is considered the case of inviscid flow (Sec. 2.1.1) with a non-zero radial velocity, vr = 0. It is shown that the only exact solution for the circular 6 velocity, vϕ, can be a vortex. It is important that vϕ may differ from the vortex only if vr = 0. In the case vr = 0, it is used the popular power-law model for the temperature distribution to demonstrate that the midplane circular velocity can differ significantly from the Keplerian one. Special attention is devoted to the “proof” that accretion disks are Keplerian in zeroth approximation. The “proof” was done in many works by the regular perturbation technique, applied to singularly perturbed PDEs. It is demonstrated that such an approach can lead to erroneous conclusion and, hence, the singular part of the asymptotic expansion is usually necessary for correct solution of the singularly perturbed problem. Thus, the well-known arguments that an inviscid accretion disk must be sub-Keplerian are disproved. The instability of inviscid non-vortex flow investigated in Sec. 2.1.2. It is considered the steady-state accretion disk as input-output system. The bound- ary conditions are assumed as inputs. The mapping F : input output is defined by the steady-state balance equations as well as by the→ equation of state. It is shown that an infinitesimal increment in the input triggers the finite increment in the output, which is to say that the flow is unstable. After that,

30 the stability of the regular part of asymptotic solution to the singularly per- turbed system is examined. It is found that the infinitesimal increment in the input does not trigger a finite increment in the output and, hence, the motion in question is stable. Thus, the assumption that the asymptotic solution to the singularly perturbed PDEs consists only of the regular series can lead to a wrong conclusion about the stability of the system. A laminar viscous disk dynamics is investigated in Sec. 2.2.1. The accretion disk is assumed geometrically thin. It is used the well known approach for the investigation of the accretion disk dynamics, namely, the equations of mass and momentum conservation are averaged over the depth of disk. It is assumed that the asymptotic solution consists of the regular as well as singular series The found zeroth approximation shows that the laminar viscous disk tends to be the vortex motion. Thus, the well-known arguments that a laminar viscous accretion disk must be sub-Keplerian are questionable. Turbulent flow is studied in Sec. 2.2.2. It is found that the turbulent flow tends to be Keplerian. This result was obtained on the basis that a turbulent gas tends to flow with minimal losses, i.e., the Euler number should be as small as possible. Based on the fact that the midplane circular velocity is Keplerian, the turbulent viscosity is evaluated. It is demonstrated that turbulent viscosity, µt, is proportional to the Reynolds number, Re. Hence, using the obvious transformation on the turbulent viscosity, the non-singularly perturbed PDEs can be obtained, namely, the small parameter, 1/Re, will not premultiply the highest-order turbulent-viscous terms.

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