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.
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.
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)
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.