Lie symmetry analysis and invariant solutions of 3D Euler equations for axisymmetric, incompressible, and inviscid flow in the cylindrical coordinates

Abstract

Through the Lie symmetry analysis method, the axisymmetric, incompressible, and inviscid fluid is studied. The governing equations that describe the flow are the Euler equations. Under intensive observation, these equations do not have a certain solution localized in all directions $(r,t,z)$ ( r , t , z ) due to the presence of the term $\frac{1}{r}$ 1 r , which leads to the singularity cases. The researchers avoid this problem by truncating this term or solving the equations in the Cartesian plane. However, the Euler equations have an infinite number of Lie infinitesimals; we utilize the commutative product between these Lie vectors. The specialization process procures a nonlinear system of ODEs. Manual calculations have been done to solve this system. The investigated Lie vectors have been used to generate new solutions for the Euler equations. Some solutions are selected and plotted as two-dimensional plots.

Full text

Sadatetal.Advances in Difference Equations (2021) 2021:486
https://doi.org/10.1186/s13662-021-03637-w
R E S E A R C H Open Access
Lie symmetry analysis and invariant
solutions of 3D Euler equations for
axisymmetric, incompressible, and inviscid
flow in the cylindrical coordinates
R. Sadat1*, Praveen Agarwal2,R.Saleh
1and Mohamed R. Ali3
*Correspondence:
r.mosa@zu.edu.eg
1Department of Mathematics,
Zagazig Faculty of Engineering,
Zagazig University, Zagazig, Egypt
Full list of author information is
available at the end of the article
Abstract
Through the Lie symmetry analysis method, the axisymmetric, incompressible, and
inviscid fluid is studied. The governing equations that describe the flow are the Euler
equations. Under intensive observation, these equations do not have a certain
solution localized in all directions (r,t,z) due to the presence of the term 1
r, which
leads to the singularity cases. The researchers avoid this problem by truncating this
term or solving the equations in the Cartesian plane. However, the Euler equations
have an infinite number of Lie infinitesimals; we utilize the commutative product
between these Lie vectors. The specialization process procures a nonlinear system of
ODEs. Manual calculations have been done to solve this system. The investigated Lie
vectors have been used to generate new solutions for the Euler equations. Some
solutions are selected and plotted as two-dimensional plots.
Keywords: Euler equations; Axisymmetric flow; Lie point symmetries; Analytical
solutions
1 Introduction
Suppose that the Euler equations have the form [1–4]
∂w
∂t+w∂w
∂r+u∂w
∂z–v2
r+∂p
∂r=0,
∂v
∂t+w∂v
∂r+u∂v
∂z–vw
r=0,
∂u
∂t+w∂u
∂r+u∂u
∂z+∂p
∂z=0,
∂w
∂r+w
r+∂u
∂z=0.
(1)
That describes the dynamics of incompressible, axisymmetric flow with swirl [3], where
w(r,t,z), u(r,t,z), and v(r,t,z) are the components of the velocity in the cylindrical coor-
dinates (r,φ,andz), and p(r,t,z) is the pressure. The flow is called axisymmetric flow if
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 2 of 16
the velocity component and the pressure are independent of φ. Navier–Stokes and Euler
equations in the cylindrical coordinates can describe any pipe fluid flow that has more
applications, especially in the medical field. For example, blood flow in stenoses narrow
artery [5–8]. System (1) had been solved using numerical methods in [1,2,9]. Manipula-
tionofthe resultsinmost applicationsneeds explicitsolutions. TheLiesymmetryanalysis
is one of the most important and powerful methods for obtaining closed-form solutions
[10,11]. The method proves its dependence in the fluid mechanics, turbulence field, and
turbulent plane jet model [12–18]. Other researchers apply the method to other applica-
tions [19–25]. In (2007), Oberlack et al. [3] deduced five Lie point symmetries for Euler
equations. Here, we use the commutative product to explore new Lie infinitesimals for
system (1), then we use the investigated Lie vectors to reduce system (1)tothesystemof
ODEs. By solving these ODEs, we explore new analytical solutions for Euler equations.
2 Investigation of Lie infinitesimals for Euler equations
System (1) possesses Lie infinitesimals as follows:
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
X1=∂
∂t+f1(t)∂
∂z+f
1(t)∂
∂u+(–f
1(t)z+f2(t)) ∂
∂p,
X2=f3(t)∂
∂z+f
3(t)∂
∂u+1
r2v∂
∂v+(–1
r2–f
3(t)z+f4(t)) ∂
∂p,
X3=t∂
∂t+f5(t)∂
∂z+(f
5(t)–u)∂
∂u–w∂
∂w–v∂
∂v+(–2p–f
5(t)z+f6(t)) ∂
∂p,
X4=r∂
∂r+(z+f7(t)) ∂
∂z+(u+f
7(t)) ∂
∂u+w∂
∂w+v∂
∂v(2p–f
7(t)z+f8(t)) ∂
∂p.
(2)
There are an infinite number of possibilities for these vectors as the presence of arbi-
trary functions fi(t), i=1...8. Usingthe commutative productbetween theseinfinitesi-
mals listed in Table 1authorizes us to specialize these vectors through the same procedure
as in [10,26]. Firstly, we generate the commutator table as follows in Table 1,where
a1=–zf 
3+f
4–f1f
3+f3f
1,
a2=f
5–tf 
1,
a3=f
5–f
1–tf 
1,
a4=–zf 
5+f
6–f1f
5+f5f
1+2zf 
1–2f2+tzf 
1–tf
2,
a5=f
7+f1,
a6=f
7+f
1,
a7=–zf 
7+f
8–f1f
7–zf
1+2f2+f7f
1,
a8=tf 
3–f
3,
a9=–f3f
5+2zf 
3–2f4+2
r2+tzf
3–tf 
4,
a10 =–4
r2+f7f
3–f3f
7–zf
3+2f4,
a11 =tf
7+f5,
a12 =tf
7+f
7+f
5,
a13 =–tzf 
7+tf 
8+f7f
5–f5f
7–zf
5+2f6+2f8–2zf 
7.
(3)
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 3 of 16
Table 1 Commutator table
[V1,V2]X1X2X3X4
X10f
3∂
∂z+f
3∂
∂u+a1∂
∂p∂
∂t+a2∂
∂z+a3∂
∂u+a4∂
∂pa5∂
∂z+a6∂
∂u+a7∂
∂p
X2–(f
3∂
∂z+f
3∂
∂u+a1∂
∂p)0 –tf
3∂
∂z+a8∂
∂u–2
r2v
∂
∂v+
a9∂
∂p
f3∂
∂z+f
3∂
∂u+4
r2v
∂
∂v+
a10 ∂
∂p
X3–( ∂
∂t+a2∂
∂z+a3∂
∂u+
a4∂
∂p)
–(–tf
3∂
∂z+a8∂
∂u–
2
r2v
∂
∂v+a9∂
∂p)
0a11 ∂
∂z+a12 ∂
∂u+a13 ∂
∂p
X4–(a5∂
∂z+a6∂
∂u+a7∂
∂p)–(f3∂
∂z+f
3∂
∂u+4
r2v
∂
∂v+
a10 ∂
∂p)
–(a11 ∂
∂z+a12 ∂
∂u+a13 ∂
∂p)0
Table 2 Commutator table after optimization
[V1,V2]X1X2X3X4
X100 X10
X200 –2X24X2
X3–X12X200
X40–4X200
The specialization process generates a nonlinear system of ODEs:
tf 
1+2f
1=f
5,
–zf
5+3zf 
1+tzf
1–tf
2–3f2+f
6–f1f
5+f5f
1=0,
f
7+f
1=0, –zf 
7+f
8–f1f
7–zf
1+2f2+f7f
1=0,
tf 
3–f
3=0,
–f3f
5+tzf
3–tf 
5=0, f7f
3–f3f
7+3zf 
3–2f4=0,
–tzf
7+tf
8+f7f
5–f5f
7–zf
5+2f6+2f8–2zf 
7=0,
tf 
7+f5–f
7=0.
(4)
Through manual calculations this system has been solved, and the results are
f1=1
t,f2=1
t3,f3=f4=0,
f5=1, f6=1
t2,f7=–ln(t), f8=–ln(t)
t2.
(5)
Substituting from (5)into(2), we obtain
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
X1=∂
∂t+1
t∂
∂z–1
t2∂
∂u+(–2
t3z+1
t3)∂
∂p,
X2=1
r2v∂
∂v+(–1
r2)∂
∂p,
X3=t∂
∂t+∂
∂z+–u∂
∂u–w∂
∂w–v∂
∂v+(–2p+1
t2)∂
∂p,
X4=r∂
∂r+(z–ln(t)) ∂
∂z+(u–1
t)∂
∂u+w∂
∂w+v∂
∂v(2p–Z
t2–ln(t)
t2)∂
∂p.
(6)
We use these vectors (6) to reproduce the commutator table (Table 2).
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 4 of 16
3 Reduction of the independent variables in Euler equations
3.1 Using Lie vector X1
To snaffle the similarity variables, we solve the associated Lagrange system
dt
1=dz
1
t=–du
1
t2=dp
(– 2
t3z+1
t3).(7)
The similarity variables of system (1)are
u(r,t,z)=R(y,x)+1
t,w(r,t,z)=F(y,x), v(r,t,z)=G(y,x),
p(r,t,z)=H(y,x)+ z
t2,
where, y=r,x=z–ln(t).
(8)
Substituting from(8)into(1), we get the following system with two independent variables:
y∂F
∂y+y∂R
∂x+F=0,
F∂G
∂yy+R∂G
∂xy+FG =0,
–F∂F
∂yy–R∂F
∂xy+G2–y∂H
∂y=0,
F∂R
∂y+R∂R
∂x+∂H
∂x=0.
(9)
System (9)hasfiveLievectorsasfollows:
V1=∂
∂x,V2=∂
∂H,V3=y∂
∂y+x∂
∂x,
V4=1
y2G
∂
∂G–1
y2
∂
∂H,V5=F∂
∂F+G∂
∂G+2H∂
∂H+R∂
∂R.(10)
3.1.1 Using vector V3
This Lie vector will reduce system (9)to
–ηTdθ
dη+θdθ
dη+dβ
dη=0,
–ηTdE
dη+θdE
dη+ET =0,
ηdT
dη–T–dθ
dη=0,
ηTdT
dη–θdT
dη+E2+ηdβ
dη=0,
(11)
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 5 of 16
where the new dependent variables have been obtained from solving the characteristic
equation that the V3was generated.
E(η)=G(y,x), T(η)=F(y,x), β(η)=H(y,x),
θ(η)=R(y,x), η=x
y.(12)
The solutions for system (11) are as follows:
T(η)=c3η+c41+η2,
θ(η)=–c4sinh–1(η),
E(η)=∓


–c3(c4η3+c3η21+η2+c4η+c4sinh–1(η)1+η2–c21+η2)
1+η2,
β(η)=–1
2c4sinh–1(η)2
–c4c31
2η1+η2–1
2sinh–1(η)–c2sinh–1(η)+1
2c4η+c1.
(13)
Back substitution to the original variables using similarity variables in (8)and(12)leads
to
w(r,t,z)=c3(z–ln(t))
r+c41+(z–ln(t))
r2,
u(r,t,z)=–c4sinh–1(z–ln(t))
r, (14)
v(r,t,z)
=∓


–c3(c4(δ)3+c3(δ)21+(δ)2+c4(δ)+c4sinh–1(δ)1+(δ)2–c21+(δ)2)
1+(δ)2,
p(r,t,z)=–1
2c4sinh–1(δ)2
–c4c31
2(δ)1+(δ)2–1
2sinh–1(δ)–c2sinh–1(δ)+1
2c4(δ)+c1,
where δ=(z–ln(t))
r.
The solutions have been plotted for different values of time as depicted in Figs. 1–4.
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 6 of 16
Figure 1 Velocity component w(r,t,z)atz=2,c3=1,andc4=1
Figure 2 Positive case of velocity component v(r,t,z)atz= 2 and c4=–1
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 7 of 16
Figure 3 Velocity component u(r,t,z)atz=5,c2=1,andc4=–1
Figure 4 The pressure p(r,t,z)atz=5,c1=1,c2=1,c3=1,andc4=1
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 8 of 16
3.1.2 Using V=V1+V4
This vector produces a system of nonlinear ODEs as follows:
ηdT
dη+T=0,
η2Tdθ
dη–1=0,
–ηTdT
dη+E–ηdβ
dη=0,
η2TdE
dη+2ηTE +2θ=0,
(15)
where the new dependent variables are
E(η)=–2x+y2G(y,x)2
y2,T(η)=F(y,x), β(η)=H(y,x)+ x
y2,
θ(η)=R(y,x)whereη=y.
(16)
By solving system (15), new solutions for Euler equations have been produced:
T(η)=c4
η,
θ(η)=ln(η)
c4+c3,
E(η)=–η2ln(η)+0.5η2–c3c4η2+c2c2
4
(c4η)2,
β(η)=–0.5
c2
4
η2+(ln(η))2
c2
4–ln(η)
c2
4+2c3ln(η)
c4+c2
η2–2c1.
(17)
Using the similarity variables in (8)and(16) leads to back substitution to the original
variables:
w(r,t,z)=c4
r,
u(r,t,z)=ln(r)
c4+c3+t–1,
v(r,t,z)=–r2ln(r)+0.5r2–c3c4r2+c2c2
4
(c4r)2+2
(z–ln(t))
r2,
p(r,t,z)=–0.5
c2
4
r2+(ln(r))2
c2
4–ln(r)
c2
4+2c3ln(r)
c4+c2
r2–2c1
–(z–ln(t))
r2+zt–2.
(18)
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 9 of 16
3.1.3 Using Lie vector V=V1+V5
Through the same previous procedure system (9)hasbeenreducedto
Tdθ
dη+θ2+2β=0,
ηdT
dη+T+ηθ =0,
ηTdE
dη+ηθE+ET =0,
–ηTdT
dη+E2–ηTθ–ηdβ
dη=0,
(19)
where the similarity variables are
E(η)=G(y,x), e–x,T(η)=F(y,x), e–x,
β(η)=H(y,x), e–2x,θ(η)=R(y,x), e–x,η=y.(20)
System (19) has closed form solutions as follows:
T(η)=–c3e–0.5Iη2
c1+c3e0.5Iη2
c1+c4e–0.5Iη2
c1
η,
θ(η)=–I(c3e–0.5Iη2
c1+c3e0.5Iη2
c1–c4e–0.5Iη2
c1)
c1,
E(η)=±2c2
3+2c3c4e–Iη2
c1–2c3c4–c2
3e–Iη2
c1–c2
3eIη2
c1–c2
4e–Iη2
c1
η,
β(η)=2c3(c3–c4)
c2
1.
(21)
Back substitution using the similarity variables in (20)and(8) is as follows:
w(r,t,z)=–c3e–0.5Ir2
c1+c3e0.5Ir2
c1+c4e–0.5Ir2
c1
re(z–ln(t)),
u(r,t,z)=–I(c3e–0.5Iη2
c1+c3e0.5Iη2
c1–c4e–0.5Iη2
c1)
c1e(z–ln(t)) +t–1,
v(r,t,z)=±2c2
3+2c3c4e–Ir2
c1–2c3c4–c2
3e–Ir2
c1–c2
3eIr2
c1–c2
4e–Ir2
c1
re–(z–ln(t)) ,
p(r,t,z)=2c3(c3–c4)
c2
1e(z–ln(t)) +zt–2.
(22)
The solutions have been plotted in Figs. 5–8.
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 10 of 16
Figure 5 Velocity component w(r,t,z)atz=2,c1=1,c3=1,andc4=2
Figure 6 Positive case velocity component v(r,t,z)atz=2,c1=1,c3=I,andc4=2I
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Figure 7 Velocity component u(r,t,z)atz=2,c1=1,c3=1,andc4=2
Figure 8 The pressure p(r,t,z)atc1=1,c3=1,andc4=2
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 12 of 16
3.2 Using Lie vector X=X3+X4
By solving the subsidiary equation, we explore the similarity variables
u(r,t,z)=R(y,x)+1
t,w(r,t,z)=F(y,x), v(r,t,z)=G(y,x),
p(r,t,z)=H(y,x)+ z
t2,
where y=t
r,x=z–ln(t)
r,
(23)
which reduce system (1)to
–∂G
∂y+xF ∂G
∂x+yF ∂G
∂y–R∂G
∂x–FG =0,
x∂F
∂x+y∂F
∂y–F+∂R
∂x=0,
–∂R
∂y+xF ∂R
∂x+yF ∂R
∂y–R∂R
∂x–∂H
∂x=0,
–∂F
∂y+xF ∂F
∂x+yF ∂F
∂y–R∂F
∂x+G2+∂H
∂xx+∂H
∂yy=0.
(24)
This system possesses three Lie vectors as follows:
V1=∂
∂H,V2=y∂
∂x+∂
∂R,V3=y∂
∂y–F∂
∂F–G∂
∂G–2H∂
∂H–R∂
∂R. (25)
•Using V=V1+V2
Following the same procedure system (24) will be reduced to
–dE
dη+ηTdE
dη–ET =0,
–dT
dη+ηTdT
dη+E2+ηdβ
dη=0,
–ηdθ
dη–θ+η2Tdθ
dη–1=0,
η2dT
dη–ηT–1=0
(26)
with new variables
E(η)=G(y,x), T(η)=F(y,x), β(η)=–H(y,x)+x
y,
θ(η)=R(y,x)–x
y,η=y.(27)
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 13 of 16
By solving system (26), we have
T(η)=–1
2η,
θ(η)=–1+ c3
η2/3 ,
E(η)=c2η2/3,
β(η)=–3c2
2
2η2/3 –3
8η2+c1.
(28)
Using the similarity variables in (23)and(27) authorizes us to back substitution to the
original variables
w(r,t,z)=–r
2t,
u(r,t,z)=–1+ c3
(t
r)2
3–z–ln(t)
t+t–1,
v(r,t,z)=c2t
r2/3,
p(r,t,z)=–3c2
2
2t
r2/3 –3
8(t
r)2–z–ln(t)
t+c1+zt–2.
(29)
The results have been plotted as shown in Figs. 9–12.
Figure 9 Velocity component w(r,t,z)
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 14 of 16
Figure 10 Velocity component v(r,t,z)atc2=1
Figure 11 Velocity component u(r,t,z)atz= 5 and c3=–1
4Conclusions
We deduce an infinite number of Lie infinitesimals, and through commutative product
properties, we minimize these vectors to four Lie vectors. Through some combinations
between these vectors, we explore exact solutions for Euler equations. The results illus-
tratethat thevelocity componentsdecreasewithincreasingthe spatialor temporalcoordi-
nates. The pressure may be appearing as a negative value, and this is reasonable according
to the human pressure in the case of the tapered artery [6].
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Sadatetal.Advances in Difference Equations (2021) 2021:486 Page 15 of 16
Figure 12 The pressure p(r,t,z)atz=1,c1=1,andc4=1
Acknowledgements
The authors thank the reviewers.
Funding
Not applicable.
Availability of data and materials
Not applicable.
Declarations
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Zagazig Faculty of Engineering, Zagazig University, Zagazig, Egypt. 2Department of
Mathematics, Anand International College of Engineering, Jaipur, 302012, India. 3Department of Basic Science, Faculty of
Engineering at Benha, Benha University, Benha, 13512, Egypt.
Publisher’s Note
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Received: 8 August 2021 Accepted: 19 October 2021
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