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 $(r,t,z)$ due to the presence of the term $\frac{1}{r}$, 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.

1 Introduction

Suppose that the Euler equations have the form [1–4]

$$ \begin{gathered} \frac{\partial w}{\partial t} +w \frac{\partial w}{\partial r} +u \frac{\partial w}{\partial z} - \frac{v^{2}}{r} + \frac{\partial p}{\partial r} =0, \\ \frac{\partial v}{\partial t} +w \frac{\partial v}{\partial r} +u \frac{\partial v}{\partial z} - \frac{vw}{r} =0, \\ \frac{\partial u}{\partial t} +w \frac{\partial u}{\partial r} +u \frac{\partial u}{\partial z} + \frac{\partial p}{\partial z} =0, \\ \frac{\partial w}{\partial r} + \frac{w}{r} + \frac{\partial u}{\partial z} =0. \end{gathered} $$

(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 coordinates $(r,\phi ,\text{ and }z)$, and $p ( r,t,z )$ is the pressure. The flow is called axisymmetric flow if 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]. Manipulation of the results in most applications needs explicit solutions. The Lie symmetry analysis 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 applications [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) to the system of 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:

$$ \textstyle\begin{cases} X_{1} = \frac{\partial }{\partial t} + f_{1} ( t ) \frac{\partial }{\partial z} + f_{1} ' ( t ) \frac{\partial }{\partial u} + ( - f_{1}^{\prime \prime } ( t ) z+ f_{2} ( t ) ) \frac{\partial }{\partial p},\\ X_{2} = f_{3} ( t ) \frac{\partial }{\partial z} + f_{3} ' ( t ) \frac{\partial }{\partial u} + \frac{1}{r^{2} v} \frac{\partial }{\partial v} + ( \frac{-1}{r^{2}} - f_{3}^{\prime \prime } ( t ) z+ f_{4} ( t ) ) \frac{\partial }{\partial p},\\ X_{3} =t \frac{\partial }{\partial t} + f_{5} ( t ) \frac{\partial }{\partial z} + (f_{5} ' ( t ) -u) \frac{\partial }{\partial u} -w \frac{\partial }{\partial w} -v \frac{\partial }{\partial v} +(-2p- f_{5}^{\prime \prime } ( t ) z+ f_{6} ( t ) ) \frac{\partial }{\partial p},\\ X_{4} =r \frac{\partial }{\partial r} + ( z+ f_{7} ( t ) ) \frac{\partial }{\partial z} + ( u+ f_{7} ' ( t ) ) \frac{\partial }{\partial u} +w \frac{\partial }{\partial w} +v \frac{\partial }{\partial v} ( 2p- f_{7}^{\prime \prime } ( t ) z+ f_{8} ( t ) ) \frac{\partial }{\partial p}. \end{cases} $$

(2)

There are an infinite number of possibilities for these vectors as the presence of arbitrary functions $f_{i} ( t )$, $i=1\dots 8$. Using the commutative product between these infinitesimals listed in Table 1 authorizes 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

$$ \begin{gathered} a_{1} =-z f_{3}^{\prime \prime \prime } + f_{4} ' - f_{1} f_{3}^{\prime \prime } + f_{3} f_{1}^{\prime \prime }, \\ a_{2} = f_{5} ' -t f_{1} ', \\ a_{3} = f_{5}^{\prime \prime } - f_{1} ' -t f_{1}^{\prime \prime }, \\ a_{4} =-z f_{5}^{\prime \prime \prime } + f_{6} ' - f_{1} f_{5}^{\prime \prime } + f_{5} f_{1}^{\prime \prime } +2z f_{1}^{\prime \prime } - 2 f_{2} + tz f_{1}^{\prime \prime \prime } -t f_{2} ', \\ a_{5} = f_{7} ' + f_{1}, \\ a_{6} = f_{7}^{\prime \prime } + f_{1} ', \\ a_{7} =-z f_{7}^{\prime \prime \prime } + f_{8} ' - f_{1} f_{7}^{\prime \prime } -z f_{1}^{\prime \prime } +2 f_{2} + f_{7} f_{1}^{\prime \prime }, \\ a_{8} = t f_{3}^{\prime \prime } - f_{3} ', \\ a_{9} = -f_{3} f_{5}^{\prime \prime } +2z f_{3}^{\prime \prime } -2 f_{4} + \frac{2}{r^{2}} +tz f_{3}^{\prime \prime \prime } -t f_{4} ', \\ a_{10} = \frac{-4}{r^{2}} + f_{7} f_{3}^{\prime \prime } - f_{3} f_{7}^{\prime \prime } -z f_{3}^{\prime \prime } +2 f_{4}, \\ a_{11} =t f_{7} ' + f_{5}, \\ a_{12} =t f_{7}^{\prime \prime } + f_{7} ' + f_{5} ', \\ a_{13} =-t zf_{7}^{\prime \prime \prime } + tf_{8} ' + f_{7} f_{5}^{\prime \prime } - f_{5} f_{7}^{\prime \prime } -z f_{5}^{\prime \prime } +2 f_{6} +2 f_{8} -2z f_{7}^{\prime \prime }. \end{gathered} $$

(3)

The specialization process generates a nonlinear system of ODEs:

$$ \begin{gathered} t f_{1}^{\prime \prime } +2 f_{1} ' = f_{5}^{\prime \prime }, \\ -z f_{5}^{\prime \prime \prime \prime } +3z f_{1}^{\prime \prime } +tz f_{1}^{\prime \prime \prime \prime } -t f_{2} ' -3 f_{2} + f_{6} ' - f_{1} f_{5}^{\prime \prime } + f_{5} f_{1}^{\prime \prime } =0, \\ f_{7}^{\prime \prime } + f_{1} ' =0,\qquad -z f_{7}^{\prime \prime \prime } + f_{8} ' - f_{1} f_{7}^{\prime \prime } -z f_{1}^{\prime \prime } +2 f_{2} + f_{7} f_{1}^{\prime \prime } =0, \\ t f_{3}^{\prime \prime } - f_{3} ' =0, \\ -f_{3} f_{5}^{\prime \prime } +tz f_{3}^{\prime \prime \prime } -t f_{5} ' =0,\qquad f_{7} f_{3}^{\prime \prime } - f_{3} f_{7}^{\prime \prime } +3z f_{3}^{\prime \prime } -2 f_{4} =0, \\ -t zf_{7}^{\prime \prime \prime } + tf_{8} ' + f_{7} f_{5}^{\prime \prime } - f_{5} f_{7}^{\prime \prime } -z f_{5}^{\prime \prime } +2 f_{6} +2 f_{8} -2z f_{7}^{\prime \prime } =0, \\ t f_{7}^{\prime \prime } + f_{5} - f_{7} ' =0. \end{gathered} $$

(4)

Through manual calculations this system has been solved, and the results are

$$ \begin{gathered} f_{1} = \frac{1}{t},\qquad f_{2} = \frac{1}{t^{3}},\qquad f_{3} = f_{4} =0, \\ f_{5} =1,\qquad f_{6} = \frac{1}{t^{2}},\qquad f_{7} =- \ln (t),\qquad f_{8} = \frac{- \ln (t)}{t^{2}}. \end{gathered} $$

(5)

Substituting from (5) into (2), we obtain

$$ \textstyle\begin{cases} X_{1} = \frac{\partial }{\partial t} + \frac{1}{t} \frac{\partial }{\partial z} - \frac{1}{t^{2}} \frac{\partial }{\partial u} + ( - \frac{2}{t^{3}} z+ \frac{1}{t^{3}} ) \frac{\partial }{\partial p},\\ X_{2} = \frac{1}{r^{2} v} \frac{\partial }{\partial v} + ( \frac{-1}{r^{2}} ) \frac{\partial }{\partial p},\\ X_{3} =t \frac{\partial }{\partial t} + \frac{\partial }{\partial z} +-u \frac{\partial }{\partial u} -w \frac{\partial }{\partial w} -v \frac{\partial }{\partial v} +(-2p+ \frac{1}{t^{2}} ) \frac{\partial }{\partial p},\\ X_{4} =r \frac{\partial }{\partial r} + ( z- \ln (t) ) \frac{\partial }{\partial z} + ( u- \frac{1}{t} ) \frac{\partial }{\partial u} +w \frac{\partial }{\partial w} +v \frac{\partial }{\partial v} ( 2p- \frac{Z}{t^{2}} - \frac{\ln (t)}{t^{2}} ) \frac{\partial }{\partial p}. \end{cases} $$

(6)

We use these vectors (6) to reproduce the commutator table (Table 2).

Table 1 Commutator table

Table 2 Commutator table after optimization

3 Reduction of the independent variables in Euler equations

3.1 Using Lie vector $\boldsymbol{{X}_{{1}}}$

To snaffle the similarity variables, we solve the associated Lagrange system

$$ \frac{dt}{1} = \frac{dz}{\frac{1}{t}} = - \frac{du}{\frac{1}{t^{2}}} = \frac{dp}{ ( - \frac{2}{t^{3}} z+ \frac{1}{t^{3}} )}. $$

(7)

The similarity variables of system (1) are

$$ \begin{gathered} u ( r,t,z ) =R ( y,x ) + \frac{1}{t},\qquad w ( r,t,z ) =F ( y,x ),\qquad v ( r,t,z ) =G ( y,x ), \\ p ( r,t,z ) =H ( y,x ) + \frac{z}{t^{2}}, \\ \text{where, }y=r, x=z- \ln ( t ) . \end{gathered} $$

(8)

Substituting from (8) into (1), we get the following system with two independent variables:

$$ \begin{gathered} y \frac{\partial F}{\partial y} + y \frac{\partial R}{\partial x} +F=0, \\ F \frac{\partial G}{\partial y} y+ R \frac{\partial G}{\partial x} y+FG=0, \\ -F \frac{\partial F}{\partial y} y- R \frac{\partial F}{\partial x} y+ G^{2} -y \frac{\partial H}{\partial y} =0, \\ F \frac{\partial R}{\partial y} + R \frac{\partial R}{\partial x} + \frac{\partial H}{\partial x} =0. \end{gathered} $$

(9)

System (9) has five Lie vectors as follows:

$$ \begin{gathered} V_{1} = \frac{\partial }{\partial x},\qquad V_{2} = \frac{\partial }{\partial H},\qquad V_{3} =y \frac{\partial }{\partial y} +x \frac{\partial }{\partial x},\\ V_{4} = \frac{1}{y^{2} G} \frac{\partial }{\partial G} - \frac{1}{y^{2}} \frac{\partial }{\partial H}, \qquad V_{5} =F \frac{\partial }{\partial F} +G \frac{\partial }{\partial G} +2H \frac{\partial }{\partial H} +R \frac{\partial }{\partial R}. \end{gathered} $$

(10)

3.1.1 Using vector $\boldsymbol{V}_{\boldsymbol{3}}$

This Lie vector will reduce system (9) to

$$ \begin{gathered} -\eta T \frac{d\theta }{d\eta } +\theta \frac{d\theta }{d\eta } + \frac{d\beta }{d\eta } =0, \\ -\eta T \frac{dE}{d\eta } +\theta \frac{dE}{d\eta } +ET=0, \\ \eta \frac{dT}{d\eta } - T- \frac{d\theta }{d\eta } =0, \\ \eta T \frac{dT}{d\eta } -\theta \frac{dT}{d\eta } + E^{2} +\eta \frac{d\beta }{d\eta } =0, \end{gathered} $$

(11)

where the new dependent variables have been obtained from solving the characteristic equation that the $V_{3}$ was generated.

$$ \begin{gathered} E(\eta )=G(y,x),\qquad T(\eta )=F(y,x),\qquad \beta (\eta )=H(y,x),\\ \theta (\eta )=R(y,x),\qquad \eta = \frac{x}{y}. \end{gathered} $$

(12)

The solutions for system (11) are as follows:

$$ \begin{gathered} T ( \eta ) = c_{3} \eta + c_{4} \sqrt{1+ \eta ^{2}}, \\ \theta ( \eta ) =- c_{4} \sinh ^{-1} ( \eta ), \\ E ( \eta ) =\mp \sqrt{\frac{-c_{3} ( c_{4} \eta ^{3} + c_{3} \eta ^{2} \sqrt{1+ \eta ^{2}} + c_{4} \eta + c_{4} \sinh ^{-1} ( \eta ) \sqrt{1+ \eta ^{2}} - c_{2} \sqrt{1+ \eta ^{2}} )}{\sqrt{1+ \eta ^{2}}}}, \\ \beta ( \eta ) = \frac{-1}{2} \bigl( c_{4} \sinh ^{-1} ( \eta ) \bigr)^{2} \\ \hphantom{\beta ( \eta ) =} {}- c_{4} \biggl( c_{3} \biggl( \frac{1}{2} \eta \sqrt{1+ \eta ^{2}} - \frac{1}{2} \sinh ^{-1} ( \eta ) \biggr) - c_{2} \sinh ^{-1} ( \eta ) + \frac{1}{2} c_{4} \eta \biggr) + c_{1}. \end{gathered} $$

(13)

Back substitution to the original variables using similarity variables in (8) and (12) leads to

$$\begin{aligned}& w ( r,t,z ) = c_{3} \frac{(z- \ln ( t ) )}{r} + c_{4} \sqrt{1+ \biggl( \frac{(z- \ln ( t ) )}{r} \biggr)^{2}}, \\& u ( r,t,z ) =- c_{4} \sinh ^{-1} \biggl( \frac{(z- \ln ( t ) )}{r} \biggr), \\& v ( r,t,z ) \\& \quad =\mp \sqrt{\frac{-c_{3} ( c_{4} ( \delta )^{3} + c_{3} ( \delta )^{2} \sqrt{1+ ( \delta )^{2}} + c_{4} ( \delta ) + c_{4} \sinh ^{-1} ( \delta ) \sqrt{1+ ( \delta )^{2}} - c_{2} \sqrt{1+ ( \delta )^{2}} )}{\sqrt{1+ ( \delta )^{2}}}}, \\& p ( r,t,z ) = \frac{-1}{2} \bigl( c_{4} \sinh ^{-1} ( \delta ) \bigr)^{2} \\& \hphantom{p ( r,t,z ) =}{} - c_{4} \biggl( c_{3} \biggl( \frac{1}{2} ( \delta ) \sqrt{1+ ( \delta )^{2}} - \frac{1}{2} \sinh ^{-1} ( \delta ) \biggr) - c_{2} \sinh ^{-1} ( \delta ) + \frac{1}{2} c_{4} ( \delta ) \biggr) + c_{1}, \end{aligned}$$

(14)

where $\boldsymbol{\delta}= \frac{(z- \ln ( t ) )}{r}$.

The solutions have been plotted for different values of time as depicted in Figs. 1–4.

Figure 1

Velocity component $w ( r,t,z )$ at $z=2$, $c_{3} =1$, and $c_{4} =1$

Figure 2

Positive case of velocity component $v ( r,t,z )$ at $z=2$ and $c_{4} =-1$

Figure 3

Velocity component $u ( r,t,z )$ at $z=5$, $c_{2} =1$, and $c_{4} =-1$

Figure 4

The pressure $p ( r,t,z )$ at $z=5$, $c_{1} =1$, $c_{2} =1$, $c_{3} =1$, and $c_{4} =1$

3.1.2 Using $\boldsymbol{V} = \boldsymbol{V}_{\boldsymbol{1}} + \boldsymbol{V}_{\boldsymbol{4}}$

This vector produces a system of nonlinear ODEs as follows:

$$ \begin{gathered} \eta \frac{dT}{d\eta } +T=0, \\ \eta ^{2} T \frac{d\theta }{d\eta } -1=0, \\ -\eta T \frac{dT}{d\eta } +E -\eta \frac{d\beta }{d\eta } =0, \\ \eta ^{2} T \frac{dE}{d\eta } +2\eta TE+2\theta =0, \end{gathered} $$

(15)

where the new dependent variables are

$$ \begin{gathered} E ( \eta ) = \frac{-2x+ y^{2} G ( y,x )^{2}}{y^{2}},\qquad T ( \eta ) =F ( y,x ),\qquad \beta ( \eta ) =H ( y,x ) + \frac{x}{y^{2}}, \\ \theta (\eta )=R(y,x)\quad \text{where }\eta =y. \end{gathered} $$

(16)

By solving system (15), new solutions for Euler equations have been produced:

$$ \begin{gathered} T ( \eta ) = \frac{c_{4}}{\eta }, \\ \theta ( \eta ) = \frac{\ln (\eta )}{c_{4}} + c_{3}, \\ E ( \eta ) = \frac{- \eta ^{2} \ln ( \eta ) +0.5 \eta ^{2} - c_{3} c_{4} \eta ^{2} + c_{2} c_{4}^{2}}{( c_{4} \eta )^{2}}, \\ \beta ( \eta ) =-0.5 \biggl( \frac{c_{4}^{2}}{\eta ^{2}} + \frac{ ( \ln ( \eta ) )^{2}}{c_{4}^{2}} - \frac{\ln ( \eta )}{c_{4}^{2}} +2 \frac{c_{3} \ln ( \eta )}{c_{4}} + \frac{c_{2}}{\eta ^{2}} -2 c_{1} \biggr). \end{gathered} $$

(17)

Using the similarity variables in (8) and (16) leads to back substitution to the original variables:

$$ \begin{gathered} w ( r,t,z ) = \frac{c_{4}}{r}, \\ u ( r,t,z ) = \frac{\ln (r)}{c_{4}} + c_{3} + t^{-1} , \\ v ( r,t,z ) = \sqrt{\frac{- r^{2} \ln ( r ) +0.5 r^{2} - c_{3} c_{4} r^{2} + c_{2} c_{4}^{2}}{( c_{4} r)^{2}} +2 \biggl( \frac{ ( z- \ln ( t ) )}{r^{2}} \biggr)}, \\ p ( r,t,z ) =-0.5 \biggl( \frac{c_{4}^{2}}{r^{2}} + \frac{ ( \ln ( r ) )^{2}}{c_{4}^{2}} - \frac{\ln ( r )}{c_{4}^{2}} +2 \frac{c_{3} \ln ( r )}{c_{4}} + \frac{c_{2}}{r^{2}} -2 c_{1} \biggr)\\ \hphantom{p ( r,t,z ) =}{} - \biggl( \frac{ ( z- \ln ( t ) )}{r^{2}} \biggr) +z t^{-2}. \end{gathered} $$

(18)

3.1.3 Using Lie vector $\boldsymbol{V} = \boldsymbol{V}_{\boldsymbol{1}} + \boldsymbol{V}_{\boldsymbol{5}}$

Through the same previous procedure system (9) has been reduced to

$$ \begin{gathered} T \frac{d\theta }{d\eta } + \theta ^{2} +2 \beta =0, \\ \eta \frac{dT}{d\eta } +T+\eta \theta =0, \\ \eta T \frac{dE}{d\eta } + \eta \theta E +ET=0, \\ -\eta T \frac{dT}{d\eta } + E^{2} -\eta T\theta -\eta \frac{d\beta }{d\eta } =0, \end{gathered} $$

(19)

where the similarity variables are

$$ \begin{gathered} E ( \eta ) =G ( y,x ),\qquad e^{-x},\qquad T ( \eta ) =F ( y,x ),\qquad e^{-x}, \\ \beta ( \eta ) =H ( y,x ),\qquad e^{-2x},\qquad\theta (\eta )=R(y,x),\qquad e^{-x},\qquad \eta =y. \end{gathered} $$

(20)

System (19) has closed form solutions as follows:

$$\begin{aligned}& \begin{gathered} T ( \eta ) = \frac{- c_{3} e^{- \frac{0.5I\eta ^{2}}{c_{1}}} + c_{3} e^{\frac{0.5I\eta ^{2}}{c_{1}}} + c_{4} e^{- \frac{0.5I\eta ^{2}}{c_{1}}}}{\eta }, \\ \theta ( \eta ) =- \frac{I( c_{3} e^{- \frac{0.5I\eta ^{2}}{c_{1}}} + c_{3} e^{\frac{0.5I\eta ^{2}}{c_{1}}} - c_{4} e^{- \frac{0.5I\eta ^{2}}{c_{1}}} )}{c_{1}}, \\ E ( \eta ) =\pm \frac{\sqrt{2 c_{3}^{2} +2 c_{3} c_{4} e^{- \frac{I\eta ^{2}}{c_{1}}} -2 c_{3} c_{4} - c_{3}^{2} e^{- \frac{I\eta ^{2}}{c_{1}}} - c_{3}^{2} e^{\frac{I\eta ^{2}}{c_{1}}} - c_{4}^{2} e^{- \frac{I\eta ^{2}}{c_{1}}}}}{\eta }, \\ \beta ( \eta ) = \frac{2 c_{3} ( c_{3} - c_{4} )}{c_{1}^{2}}. \end{gathered} \end{aligned}$$

(21)

Back substitution using the similarity variables in (20) and (8) is as follows:

$$ \begin{gathered} w ( r,t,z ) = \frac{- c_{3} e^{- \frac{0.5Ir^{2}}{c_{1}}} + c_{3} e^{\frac{0.5Ir^{2}}{c_{1}}} + c_{4} e^{- \frac{0.5Ir^{2}}{c_{1}}}}{r} e^{ ( z- \ln ( t ) )} , \\ u ( r,t,z ) =- \frac{I( c_{3} e^{- \frac{0.5I\eta ^{2}}{c_{1}}} + c_{3} e^{\frac{0.5I\eta ^{2}}{c_{1}}} - c_{4} e^{- \frac{0.5I\eta ^{2}}{c_{1}}} )}{c_{1}} e^{ ( z- \ln ( t ) )} + t^{-1}, \\ v ( r,t,z ) =\pm \frac{\sqrt{2 c_{3}^{2} +2 c_{3} c_{4} e^{- \frac{Ir^{2}}{c_{1}}} -2 c_{3} c_{4} - c_{3}^{2} e^{- \frac{Ir^{2}}{c_{1}}} - c_{3}^{2} e^{\frac{Ir^{2}}{c_{1}}} - c_{4}^{2} e^{- \frac{Ir^{2}}{c_{1}}}}}{r e^{- ( z- \ln ( t ) )}} , \\ p ( r,t,z ) = \frac{2 c_{3} ( c_{3} - c_{4} )}{c_{1}^{2}} e^{ ( z- \ln ( t ) )} +z t^{-2}. \end{gathered} $$

(22)

The solutions have been plotted in Figs. 5–8.

Figure 5

Velocity component $w ( r,t,z )$ at $z=2$, $c_{1} =1$, $c_{3} =1$, and $c_{4} =2$

Figure 6

Positive case velocity component $v ( r,t,z )$ at $z=2$, $c_{1} =1$, $c_{3} =I$, and $c_{4} =2I$

Figure 7

Velocity component $u ( r,t,z )$ at $z=2$, $c_{1} =1$, $c_{3} =1$, and $c_{4} =2$

Figure 8

The pressure $p ( r,t,z )$ at $c_{1} =1$, $c_{3} =1$, and $c_{4} =2$

3.2 Using Lie vector $\boldsymbol{X} = \boldsymbol{X}_{\boldsymbol{3}} + \boldsymbol{X}_{\boldsymbol{4}}$

By solving the subsidiary equation, we explore the similarity variables

$$ \begin{gathered} u ( r,t,z ) =R ( y,x ) + \frac{1}{t},\qquad w ( r,t,z ) =F ( y,x ),\qquad v ( r,t,z ) =G ( y,x ), \\ p ( r,t,z ) =H ( y,x ) + \frac{z}{t^{2}}, \\ \text{where }y= \frac{t}{r}, x= \frac{z- \ln ( t )}{r}, \end{gathered} $$

(23)

which reduce system (1) to

$$ \begin{gathered} - \frac{\partial G}{\partial y} + xF \frac{\partial G}{\partial x} + yF \frac{\partial G}{\partial y} -R \frac{\partial G}{\partial x} -FG=0, \\ x \frac{\partial F}{\partial x} +y \frac{\partial F}{\partial y} -F+ \frac{\partial R}{\partial x} =0, \\ - \frac{\partial R}{\partial y} +xF \frac{\partial R}{\partial x} +yF \frac{\partial R}{\partial y} -R \frac{\partial R}{\partial x} - \frac{\partial H}{\partial x} =0, \\ - \frac{\partial F}{\partial y} + xF \frac{\partial F}{\partial x} + yF \frac{\partial F}{\partial y} -R \frac{\partial F}{\partial x} + G^{2} + \frac{\partial H}{\partial x} x+ \frac{\partial H}{\partial y} y=0. \end{gathered} $$

(24)

This system possesses three Lie vectors as follows:

$$ V_{1} = \frac{\partial }{\partial H},\qquad V_{2} =y \frac{\partial }{\partial x} + \frac{\partial }{\partial R},\qquad V_{3} =y \frac{\partial }{\partial y} -F \frac{\partial }{\partial F} -G \frac{\partial }{\partial G} -2H \frac{\partial }{\partial H} -R \frac{\partial }{\partial R}. $$

(25)

• Using $\boldsymbol{V} = \boldsymbol{V}_{\boldsymbol{1}} + \boldsymbol{V}_{\boldsymbol{2}}$

Following the same procedure system (24) will be reduced to

$$ \begin{gathered} - \frac{dE}{d\eta } +\eta T \frac{dE}{d\eta } - ET=0, \\ - \frac{dT}{d\eta } +\eta T \frac{dT}{d\eta } + E^{2} +\eta \frac{d\beta }{d\eta } =0, \\ -\eta \frac{d\theta }{d\eta } -\theta + \eta ^{2} T \frac{d\theta }{d\eta } -1=0, \\ \eta ^{2} \frac{dT}{d\eta } -\eta T-1=0 \end{gathered} $$

(26)

with new variables

$$ \begin{gathered} E ( \eta ) =G ( y,x ),\qquad T ( \eta ) =F ( y,x ),\qquad \beta ( \eta ) =-H ( y,x ) + \frac{x}{y}, \\ \theta ( \eta ) =R ( y,x ) - \frac{x}{y},\qquad \eta =y. \end{gathered} $$

(27)

By solving system (26), we have

$$ \begin{gathered} T ( \eta ) = \frac{-1}{2\eta }, \\ \theta ( \eta ) =-1+ \frac{c_{3}}{\eta ^{2/3}}, \\ E ( \eta ) = c_{2} \eta ^{2/3}, \\ \beta ( \eta ) = \frac{-3 c_{2}^{2}}{2} \eta ^{2/3} - \frac{3}{8\eta ^{2}} + c_{1}. \end{gathered} $$

(28)

Using the similarity variables in (23) and (27) authorizes us to back substitution to the original variables

$$ \begin{gathered} w ( r,t,z ) = \frac{-r}{2t} , \\ u ( r,t,z ) =-1+ \frac{c_{3}}{ ( \frac{t}{r} )^{\frac{2}{3}}} - \frac{z- \ln ( t )}{t} + t^{-1}, \\ v ( r,t,z ) = c_{2} \biggl( \frac{t}{r} \biggr)^{2/3} , \\ p ( r,t,z ) = \frac{-3 c_{2}^{2}}{2} \biggl( \frac{t}{r} \biggr)^{2/3} - \frac{3}{8 ( \frac{t}{r} )^{2}} - \frac{z- \ln ( t )}{t} + c_{1} +z t^{-2}. \end{gathered} $$

(29)

The results have been plotted as shown in Figs. 9–12.

Figure 9

Velocity component $w ( r,t,z )$

Figure 10

Velocity component $v ( r,t,z )$ at $c_{2} =1$

Figure 11

Velocity component $u ( r,t,z )$ at $z=5$ and $c_{3} =-1$

Figure 12

The pressure $p ( r,t,z )$ at $z=1$, $c_{1} =1$, and $c_{4} =1$

4 Conclusions

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 illustrate that the velocity components decrease with increasing the spatial or temporal coordinates. 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].