Self-similar axisymmetric flows with swirl
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Authors Katsaounis, Theodoros;Mousikou, Ioanna;Tzavaras, Athanasios Eprint version Pre-print
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Download date 2024-01-25 17:33:28
Link to Item http://hdl.handle.net/10754/687338
arXiv:2301.11090v1 [math.AP] 26 Jan 2023
Self-similar axisymmetric flows with swirl
Theodoros Katsaounis, Ioanna Mousikou and Athanasios E. Tzavaras
AbstractWe consider an infinite vortex line in a fluid which interacts with a boundary surface as a simplified model for tornadoes. We study self-similar solutions for stationary axisymmetric Navier-Stokes equations and investigate the types of motion which are compatible with this structure when viscosity is non-negative. For viscosity equal to zero, we construct a class of explicit stationary solutions. We then consider solutions with slip discontinuity and show that they do not exist in this framework.
1 Introduction
Tornadoes are considered among the most extreme and violent weather phenomena on Earth. They can occur under appropriate circumstances in all continents expect Antarctic and can be hazardous causing loss of human lives and extensive properties damages.
Meteorologists define as a tornado a rapidly rotating mass of air that extends downward from a cumuliform cloud, i.e. a cloud formed due to vertical motion of air parcels to the ground. There exists several types of tornadoes, such as landspouts and waterspouts. The majority of the most destructive tornadoes are known as supercell since they are generated within supercell thunderstorms [PM10], [MR14].
Theodoros Katsaounis
University of Crete, Heraklion 71409, Greece and Inst. of App. and Comp. Math. (IACM), FORTH, Heraklion 71110, Greece, e-mail:[email protected]
Ioanna Mousikou
King Abdullah University of Science and Technology, Thuwal 23955-6900, Saudi Arabia, e-mail:
[email protected] Athanasios E. Tzavaras
King Abdullah University of Science and Technology, Thuwal 23955-6900, Saudi Arabia, e-mail:
1
Due to the complexity of tornadoes, the current knowledge about them comes mainly from laboratory experiments and numerical models of idealized supercell thunderstorms, as Rotunno (2013) stated in [Rot13]. In 1972, Ward [War72] con- ducted a pioneering laboratory experiment reproducing a tornado-like flow using a simplified model for a steady flow and a fluid with constant density. Based on this work, several experimental and numerical simulations have taken place and provided important information in the field of fluid dynamics of tornadoes, [Rot13]. Further- more, various attempts have been made to analytically model a tornado-like flow.
Assuming that a vortex line resembles the tornado core, these models are derived using the basic motion of equations of fluid dynamics for an axisymmetric flow, i.e.
the axisymmetric Euler and Navier - Stokes equations, for incompressible homoge- neous fluids. A detailed presentation can be found in [KM17] and [GSHB18] and in references therein.
Motivated by the aforementioned vortex models, a different, theoretical approach was introduced by Long (1958, 1961) [Lon58], [Lon61]. Considering the existence of an infinite vortex line in a fluid interacting with a plane boundary surface, he presented the reduction of incompressible axisymmetric Navier-Stokes equations to a system of differential equations. Independently, Goldshtik (1960) showed that a similar reduction of incompressible axisymmetric Navier-Stokes equations to a sys- tem of differential equations leads to a class of exact self-similar solutions, [Gol60].
Serrin (1972) broadened this class of solutions and described the existence of three different solution profiles depending on an arbitrary parameter and the kinematic viscosity, [Ser72]. There are several studies of mathematical aspects of the afore- mentioned system of differential equations under other types of boundary conditions, [GS89], [GS90], [Gol90], and also studies of the related subject of conical flows, [SH99], [FFA00], [Sht12].
Here, we first develop a class of exact stationary solutions for Euler and equations.
Afterwards, we consider the problem of whether such solutions can be connected with slip-type discontinuities. If this was the case, it would provide a relation with
"two-cell" solutions of Serrin, [Ser72]. We show that they do not exist for the given set of boundary conditions. The same holds true for conical flows. This manuscript is an extract of the work presented in [KMT23] where the connection of such Euler and Navier-Stokes solutions is examined using boundary layer analysis.
2 Cylindrical Axisymmetric Navier-Stokes Equations
2.1 IntroductionWe consider the system of Navier-Stokes equations for an incompressible homoge- neous fluid formulated as follows:
Self-similar axisymmetric flows with swirl 3 uπ‘+ (uΒ· β)u=ββπ+πΞu, (1a)
β Β·u=0, (1b)
whereu: R3ΓR+ β R3 is the velocity vector of the fluid, π : R3ΓR+ β Ris pressure andπβ₯0 is the coefficient of kinematic viscosity. Motivated by the shape of a tornado, we introduce cylindrical coordinates(π, π, π§)
π₯1 =π cosπ, π₯2=π sinπ, π₯3=π§,
and focus on axisymmetric flows, i.e. a flow where the velocity vectoru=(π’, π£, π€) does not depend on azimuth angleπ. The axisymmetric Navier-Stokes equations take the form
ππ’
ππ‘ +π’ππ’
ππ +π€ππ’
π π§βπ£2 π =πh1
π
π
ππ
πππ’
ππ
+ π2π’
π π§2 β π’ π2 i
βπ π
ππ (2a)
π π£
ππ‘ +π’π π£
ππ +π€π π£
π π§+π’π£ π =πh1
π
π
ππ
ππ π£
ππ
+ π2π£
π π§2 β π£ π2 i
(2b)
π π€
ππ‘ +π’π π€
ππ +π€π π€
π π§ =πh1 π
π
ππ
ππ π€
ππ
+ π2π€
π π§2 i
βπ π
π π§ (2c) 1
π
π
ππ(ππ’) + π π€
π π§ =0 (2d)
2.2 Self-Similar Formulation
The Navier-Stokes equations remain invariant under scaling
uπ(π‘, π, π§)=πu(π2π‘, ππ, ππ§) and ππ(π‘, π, π§)=π2π(π2π‘, ππ, ππ§).
Looking for self-similar solutions and focusing only on stationary flows, we establish the ansatz
π’(π, π§)= 1
ππ(π), π£(π, π§)= 1
ππ(π), π€(π, π§)= 1
ππ(π) and π(π, π§)= 1 π2π(π). Such an ansatz induces a singularity atπ=0 which in the applied math literature is considered as the line vortex resembling the tornado core. For convenience, we also introduce a new variableπ(π), namely we setπ(π)=πβππ, which coincides with the self-similar form of the stream function. After a lengthy calculation, we obtain a system of ordinary differential equations
π2
2 + (1+π2)π β²
=π
ππβ (1+π2)πβ² β²
βππ2 (3a)
πβ²π=πh
3ππβ²+ (1+π2)πβ²β²i
(3b)
π2βππ2 2
β² +π
β²
=πh
ππβπ2πβ²βπ(1+π2)πβ²β²iβ²
(3c)
πβ²=βπ (3d)
This is viewed as a coupled system of π(π),π(π) and π(π) whereπ(π) = βπβ² and π(π) = πβππβ². After imposing boundary conditions, the problem can be reformulated as
π2 2 βπ
(1+π2)πβ²+ππ
=πΊ(π) + πΈ0 π q
1+π2βπ2
(4a) ππβ²β²+ 3ππβπ
1+π2 πβ²=0 (4b)
where
πΊ(π)=π q
1+π2
β« β
π
1 π2(1+π2)32
β« π
0
π π2(π )ππ
ππ .
Here we consider no-slip conditions on r-axis, i.e.u=0 atπ=0, and no-penetration condition on z-axis, i.e.uΒ·n=0 asπβ β. A restriction on swirlπ is also added to close the system. Namely, we takeπ βπβ, asπ β β. System (4) can now be solved numerically. After multiple numerical experiments, we observe that under certain combinations of parametersπ, πβ, πΈ0 there exist three different profiles of solution. In the first case, the flow is directed outward near the plane π§ = 0 and downward near the vortex line. In the second case it is inward near the planeπ§=0 and upward near the vortex line. For the last case, the flow is directed inward near the planeπ§=0 and downward near the vortex line. These are in agreement with results presented in [Ser72]. Under a suitable change of variables, i.e. settingπ₯ = βπ
1+π2
and Β―Ξ(π₯) =βp
1+π2π(π), Β―π(π₯) =π(π), system (4) takes a similar form as Serrin presented in [Ser72] and thus, his results also hold for (4).
3 Stationary Euler Equations
3.1 Continuous SolutionLet us consider the case of inviscid Navier-Stokes system, i.e. the case where kine- matic viscosity is equal to zero. Therefore, settingπ=0 into (3), the system becomes
Self-similar axisymmetric flows with swirl 5 π2
2 + (1+π2)π β²
=βππ2 (5a)
πβ²π=0 (5b)
π2βππ2 2
β² +π
β²
=0 (5c)
Equation (5b) implies that eitherπ(π)is equal to zero orπ(π)is a constant function.
Supposed thatπβ 0 and thusπ(π)is continuous, we have π β‘π0,
whereπ0is a given constant. This yields to a simple system of differential equation which can be solved analytically. In order to define the constants arising after integra- tion, boundary conditions are imposed. Motivated by the structure of the problem, we consider no-penetration boundary conditions on both axes, i.e.uΒ·n=0. In other words, we require that the orthogonal component of the velocity vector is equal to zero on the axes, which implies thatπ =0 atπ = 0 andπ β 0 as π β β. Consequently, an explicit family of solutions that depends on parametersπ0=π(0) andπΈ0=π(0)is derived as follows
π2(π)=2π0π(π) and π(π)=π0 whereπ(π)=πp
1+π2βπ2andπ0 =πΈ0+π
2 0
2 must be a positive constant. Expres- sions forπ, πandπcan easily be calculated using the definition ofπ(π).
It is worth mentioning that ifπis positive, then the flow is directed inward near the planeπ§ = 0 and upward near the vortex line. Conversely, ifπ is negative, the flow has the reverse direction, i.e it is directed outward near the planeπ§ =0 and downward near the vortex line, see Fig.1. Such behaviors also occur when solving Navier - Stokes equations, [Ser72].
0 1 2 3 4
0 1 2 3 4
r
z
0 1 2 3 4
0 1 2 3 4
r
z
Fig. 1 Velocity vector field(π’, π€)in(π , π§)plane forπ0=1 andπΈ0=1. Left,π >0; right,π <0
3.2 Discontinuous Solutions
Although the flow patterns described in the previous section coincide with flows derived using the stationary Navier-Stokes equations, the interesting case whereπ changes sign and thus flow changes direction is not observed. To examine whether this phenomenon is feasible, we assume that a velocity solution of (5a)β(5c) has a discontinuity at some pointπ=π, forπβ (0,β). Hence, we introduce an ansatz
π=
(πβ , πβ (0, π)
π+ , πβ (π,β) and π=
(πβ , πβ (0, π) π+ , πβ (π,β)
and seek for solutions in each domain independently. Under the restriction of continu- ity ofπatπ=π, i.e.π+(π)=πβ(π) =0, and no-penetration boundary conditions on the axes, i.e.πβ(0) = 0, π+(π) β 0 asπ β β, the discontinuous solution becomes
π2 2 =



ο£²


 πβ
"
π(π) βπ(π) βπ(π)
π2 (π2βπ2)
#
, πβ (0, π) π+
π(π) βπ(π)
, πβ (π,β)
(8)
whereπ+, πβare constants.
3.2.1 Weak Formulation
Let (π, π , π, π) be a (generally weak) self-similar solution of Euler equations which satisfies the system of ordinary differential equations (5a) - (5c) in the sense of distributions. Under a suitable choice of test function, the weak form of the system can be defined over a closed interval[π, π] βR+and we obtain
π2(π)
2 + (1+π2)π(π)
π
π
=β
β« π
π
ππ2(π)ππ (9a)
π(π)π(π)
π
π
=β
β« π
π
π(π)π(π)ππ (9b)
π2βπ π2
2 β²
+π(π)
π
π
=0 (9c)
π(π)
π π
=β
β« π
π
π(π)ππ (9d)
From (9d), we infer thatπβ²(π)=πexists and is locally integrable. Recalling (8),π is indeed integrable. This implies thatπ(π)andπ2(π)are absolutely continuous on
Self-similar axisymmetric flows with swirl 7 [π, π]. Using this observation, we can easily conclude that (9) is a good definition of weak solution of (5a) - (5c) in the class of functions of bounded variation. Therefore, there exists a countable setπβ (0,β)consisting of the points of jump discontinuity and the right and left limits of the solution exist at anyπ βπ. In addition, the jump conditions
1 2
π2+βπβ2
+ (1+π2)
π+βπβ
=0 π+π+βπβπβ=0
π+2βππ2+ 2
β²
β
π2ββππβ2 2
β² +
π+βπβ
=0 π+βπβ=0
hold for anyπ βπ. Here, we denote the one-sided limits as(πΒ±, πΒ±, πΒ±, πΒ±, πΒ±). The last equation implies thatπis continuous for anyπβ (0,β). Hence, the jump conditions reduce to
h π
i
=0 (11a)
h
π2βππ πβ² i
=0 (11b)
Ifπ(π)is non-zero, the prior identities are satisfied when all functions(π, π, π , π, π) are continuous for allπ β (0,β). Therefore, it is sufficient to consider that there exists a point of jump discontinuity π β π such that π+(π) = πβ(π) = 0. This impliesπ(π)is continuous for anyπwhileπ(π)andπβ², and thusπ(π)andπ(π), have a jump discontinuity atπ=π.
Proposition 1 (Nonexistence of solutions)Although from the prospective of regu- larity it could be a weak solution, the class of discontinuous solutions(8)does not exist.
Proof Supposeπis given in form (8). From jump condition (11), we have πββπ+
πβ²(π)=2πβπ(π)
π β π+
πβ =1β2 π(π)
π πβ²(π) (12) which provides an additional relation for constantsπ+, πβ, with the right hand-side to be negative. We want to check if this relation is compatible with sign restrictions for constantsπ+, πβ. By constructionπ+is always positive sinceπ(π)is a non-negative function. Therefore, it is sufficient to examine the sign ofπβby finding the sign of π2β. Forπβ (0, π), set
π½(π)=π(π) βπ(π) βπ(π)
π2 (π2βπ2)
We observe thatπ½(0)=π½(π)=0,π½β²(0)=πβ²(0) >0 andπ½β²β²<0. This implies that π½(π)>0βπβ (0, π), and thusπβis also positive. So, we get a contradiction.
4 Conical Flows
Motivated by the study of Euler equations presented in the previous section, we are interested in extending it for the case of axisymmetric conical flows, i.e. for flows in a cone-shaped domain. Suppose there existsπ0 βR, we seek solutions of (5a)β(5c) defined over the interval[π0,β).
4.1 Continuous Solutions
π§
π π=π0>0 π=π
π=π0<0 Fig. 2 Conical shaped domain Let us begin with the case where solutions are con-
tinuous. As before, we assumeπβ 0 andπ0=π(π0). If no-penetration boundary conditions are imposed on both ends of the domain[π0,β), we get the con- ditions:
π(π0) βπ0π(π0)=0 atπ=π0, π(π) β0 asπβ β
Therefore, solutions of (5a)β(5c) for a conical do- mainπβ [π0,β)become
π2 2 =
π02 2 +π΄0
π(π) βπ(π0) 1βπ(π0) π =π0
4.2 Discontinuous Solutions
To investigate now the existence of discontinuous solutions, we consider a solution of (5a)β(5c) with a discontinuity at some pointπ=π, forπ β (π0,β). Under the restriction of continuity ofπatπ=π, i.e.π+(π)=πβ(π) =0, and no-penetration boundary conditions, the discontinuous solution takes the form
π2 2 =



ο£²


 πβ
"
π(π) βπ(π)
βπ(π0) βπ(π) π2
0βπ2 (π2βπ2)
#
, πβ (π0, π)
π+
π(π) βπ(π)
, πβ (π,β)
(14)
whereπ+,πβare constants.
Self-similar axisymmetric flows with swirl 9 Proposition 2 (Nonexistence of solutions)Although from the prospective of regu- larity it could be a weak solution, the class of discontinuous solutions(14)does not exist.
Proof Suppose there existsπexpressed as (14). Because of jump conditions (11), we request
π+
πβ =1β2π(π0) βπ(π)
π02βπ2 π β π0+π 2π
π+ πβ
= π0+π
2π β 1
π(π)
π(π0) βπ(π) π0βπ
(15) As before, it is sufficient to check if this relation is compatible with sign restrictions for constants π+, πβ. Since π is decreasing, it is clear that π+ is positive for all πβ (π,β).
β’ Case 1:π0 >0
If π0 < π, we get that the right hand-side of (15) is negative. Therefore, it is satisfied ifπβis also negative. To find this, we check the sign ofπ2β. Set
π½πππ(π)=π(π) βπ(π) βπ(π0) βπ(π)
π02βπ2 (π2βπ2)=(π2βπ2)
πΉ(π) βπΉ(π0)
whereπΉ(π) = π(π)βπ(π)
π2βπ2 . Using thatπΉ(π) is a decreasing function and π0 <
π, we get that π½(π) is positive. This implies that πβ is positive and leads to contradiction.
β’ Case 2:π0 <0
We consider first the instance where |π0| < π. This is equivalent to case 1 described above. So, let us move to the instance where|π0| > π. From (15), we have that the right hand-side of the above relation is negative. Since π02π+π < 0, (15) is satisfied ifπβis positive. To find this, we check again the sign ofπ2β. It is clear that(π2βπ2)>0 forπβ (β|π0|, π). SinceπΉ(π)is a decreasing function, we conclude thatπ½πππ(π)is negative and as consequenceπβis also negative. This
also leads to contradiction.
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