We consider the flow of an incompressible, viscous fluid with the plane *y* = 0 considering variable viscosity and thermal conductivity. Figure 1 demonstrates a physical illustration of the flow model past a vertical surface that is designed for the present study. In the flow geometry, the vertical plate is fixed along the *x*-direction and the fluid is assumed to flow with the plate; the *y*-axis is taken exactly normal to it. A constant uniform temperature *T*_{w} which is greater than *T*_{∞} is maintained by the plate. The vertical surface moving continually in the positive *x*-direction with a velocity *u* = *U*_{∞}.with the above assumptions the basic equations for steady flow are the corresponding equations, under the above assumptions, describing the two-dimensional viscous fluid motion, are given by the model of Dinesh et al. [30] leading to

$$ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 $$

(1)

$$ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho}\frac{\partial }{\partial y}\left(\mu \frac{\partial u}{\partial y}\right) $$

(2)

$$ u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\frac{\partial }{\partial y}\left(\alpha \frac{\partial T}{\partial y}\right)+{q}^{{\prime\prime\prime} } $$

(3)

The boundary conditions of the problem are

$$ {\displaystyle \begin{array}{l}u=0,v=0,T={T}_w={T}_c+C{x}^a,y=0\\ {}u={U}_{\infty },T={T}_{\infty },y\to \infty \end{array}}\Big\} $$

(4)

where *q*′′′ is the exponential form of internal heat generation which is defined by

$$ q\hbox{'}\hbox{'}\hbox{'}=\frac{U_0\left({T}_w-{T}_c\right)}{2x}{e}^{-\eta } $$

Furthermore, by introducing a dimensional stream function *ψ* defined into Eqs. (1)–(4), we have

$$ u=\frac{\partial \psi }{\partial y},v=-\frac{\partial \psi }{\partial x},\frac{T-{T}_c}{T_w-{T}_c}=\theta, \mu ={\mu}_0{e}^{-{\beta}_1\theta },\alpha ={\alpha}_0\left(1+{\beta}_2\theta \right) $$

then we have

$$ {\displaystyle \begin{array}{c}\frac{\partial \mu }{\partial \theta }=-{\mu}_0{\beta}_1{e}^{-{\beta}_1\theta },T={T}_c+\left({T}_w-{T}_c\right)\theta ={T}_c+c{x}^a\theta, \\ {}\frac{\partial T}{\partial x}= ca{x}^{a-1}\theta +c{x}^a\frac{\partial \theta }{\partial x},\frac{\partial T}{\partial y}=c{x}^a\frac{\partial \theta }{\partial y},\frac{\partial^2T}{\partial {y}^2}=c{x}^a\frac{\partial^2\theta }{\partial {y}^2}\end{array}} $$

From Eq. (2)

$$ {\displaystyle \begin{array}{c}u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho}\frac{\partial \mu }{\partial \theta}\frac{\partial \theta }{\partial y}\frac{\partial u}{\partial y}-\frac{1}{\rho}\mu \frac{\partial^2u}{\partial {y}^2}\\ {}\frac{\partial \psi }{\partial y}\frac{\partial^2\psi }{\partial x\partial y}-\frac{\partial \psi }{\partial x}\frac{\partial^2\psi }{\partial {y}^2}=\frac{1}{\rho }{\mu}_0{\beta}_1{e}^{-{\beta}_1\theta}\frac{\partial \theta }{\partial y}\frac{\partial^2\psi }{\partial {y}^2}-\frac{1}{\rho }{\mu}_0{e}^{-{\beta}_1\theta}\frac{\partial^3\psi }{\partial {y}^3}\end{array}} $$

(5)

From Eq. (3)

$$ {\displaystyle \begin{array}{c}u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\frac{\partial \alpha }{\partial \theta}\frac{\partial \theta }{\partial y}\frac{\partial T}{\partial y}+\alpha \frac{\partial^2T}{\partial {y}^2}+{q}^{{\prime\prime\prime}}\\ {}\frac{\partial \psi }{\partial y}\left( Tw- Tc\right)\frac{\partial \theta }{\partial x}-\frac{\partial \psi }{\partial x}\left( Tw- Tc\right)\frac{\partial \theta }{\partial y}={\alpha}_0{\beta}_2\frac{\partial \theta }{\partial y}\left( Tw- Tc\right)\frac{\partial \theta }{\partial y}+{\alpha}_0\left(1+{\beta}_2\theta \right)\left( Tw- Tc\right)\frac{\partial^2\theta }{\partial {y}^2}+{q}^{{\prime\prime\prime}}\\ {}\frac{\partial \psi }{\partial y}\frac{\partial \theta }{\partial x}-\frac{\partial \psi }{\partial x}\frac{\partial \theta }{\partial y}={\alpha}_0{\beta}_2\frac{\partial \theta }{\partial y}\frac{\partial \theta }{\partial y}+{\alpha}_0\left(1+{\beta}_2\theta \right)\frac{\partial^2\theta }{\partial {y}^2}+{q}^{{\prime\prime\prime}}\end{array}} $$

(6)

The boundary conditions (4) become

$$ {\displaystyle \begin{array}{l}\frac{\partial \psi }{\partial y}=0,\frac{\partial \psi }{\partial x}=0,\theta =1,y=0\\ {}\frac{\partial \psi }{\partial y}\to {U}_{\infty },\theta \to 0,y\to \infty \end{array}}\Big\} $$

(7)

### Lie group transformation

The solution of the system of non-similar partial differential Eqs. (5)–(6) subject to the boundary conditions (7) is analytically not possible. Numerical methods are required. However, even with powerful algorithms, the equations remain challenging and expensive also. Therefore, it is necessary to transform these into self-similar ODEs using Lie group transformations (see [31,32,33,34,35]). This effectively reduces the number of independent variables of the governing partial differential equations. Lie algebra is a powerful analytical approach based on continuous symmetry of mathematical structures and objects which has found many uses in modern theoretical physics and applied mathematics. This theory provides a new methodology for analyzing the continuous symmetries of governing equations of many fluid dynamic systems including non-Newtonian transport phenomena. Defining:

$$ \varGamma :{x}^{\ast }=x{e}^{\varepsilon {\alpha}_1},{y}^{\ast }=y{e}^{\varepsilon {\alpha}_2},{\psi}^{\ast }=\psi {e}^{\varepsilon {\alpha}_3},{\theta}^{\ast }=\theta {e}^{\varepsilon {\alpha}_4} $$

(8)

Substituting (8) into (5) and (6), we get

$$ {\displaystyle \begin{array}{l}{e}^{-\varepsilon \left({\alpha}_1+2{\alpha}_2-2{\alpha}_3\right)}\left(\frac{\partial \psi }{\partial y}\frac{\partial^2\psi }{\partial x\partial y}-\frac{\partial \psi }{\partial x}\frac{\partial^2\psi }{\partial {y}^2}\right)=-\frac{1}{\rho }{\mu}_0{\beta}_1{e}^{\beta_1{\alpha}_4\theta }{e}^{-\varepsilon \left(2{\alpha}_2-{\alpha}_3-{\alpha}_4\right)}\frac{\partial \theta }{\partial y}\frac{\partial^2\psi }{\partial {y}^2}\\ {}-\frac{1}{\rho }{\mu}_0{e}^{\beta_1{\alpha}_4\theta }{e}^{-\varepsilon \left(3{\alpha}_2-{\alpha}_3\right)}\frac{\partial^3\psi }{\partial {y}^3}\end{array}} $$

(9)

and

$$ {e}^{-\varepsilon \left({\alpha}_1+{\alpha}_2-{\alpha}_3-{\alpha}_4\right)}\left(\frac{\partial \psi }{\partial y}\frac{\partial \theta }{\partial x}-\frac{\partial \psi }{\partial x}\frac{\partial \theta }{\partial y}\right)+{\alpha}_0{\beta}_2{e}^{-\varepsilon \left(2{\alpha}_2-2{\alpha}_4\right)}{\left(\frac{\partial \theta }{\partial y}\right)}^2+{\alpha}_0\left(1+{\beta}_2{e}^{-\varepsilon {\alpha}_4}\theta \right){e}^{-\varepsilon \left(2{\alpha}_2-{\alpha}_4\right)} $$

(10)

The transformed Eqs. (9) and (10) are invariant under the Lie group of transformation if the following relations among the transform parameters are satisfied.

$$ {\alpha}_1+2{\alpha}_2-2{\alpha}_3=2{\alpha}_1-{\alpha}_3-2{\alpha}_4=3{\alpha}_2-{\alpha}_3-{\alpha}_4 $$

(11)

and

$$ {\alpha}_1+{\alpha}_2-{\alpha}_3-{\alpha}_4=2{\alpha}_2-2{\alpha}_4=2{\alpha}_2 $$

(12)

and the boundary conditions we obtain at *y* → ∞

$$ {\displaystyle \begin{array}{l}\;\frac{\partial \psi }{\partial y}={U}_{\infty }=>{e}^{\varepsilon \left({\alpha}_2-{\alpha}_3\right)}={U}_{\infty }{e}^0,\\ {}{\alpha}_2-{\alpha}_3=0=>{\alpha}_2={\alpha}_3\end{array}} $$

From (11) and (12)

$$ {\alpha}_4=0,{\alpha}_2={\alpha}_3=\frac{1}{2}{\alpha}_1, $$

(13)

If we insert (13) into the scaling (8), the set of transformations reduces to a one-parameter group of transformations given by

$$ \varGamma :{x}^{\ast }=x{e}^{\varepsilon {\alpha}_1},{y}^{\ast }=y{e}^{\varepsilon \frac{1}{2}{\alpha}_1},{\psi}^{\ast }=\psi {e}^{\varepsilon \frac{1}{2}{\alpha}_1},\kern0.36em {\theta}^{\ast }=\theta $$

(14)

Expanding Eq. (14) by Tailor’s method and the remaining terms up to *O* (*ε*^{2}) of the one-parameter group, we further get

$$ {x}^{\ast }-x= x\varepsilon {\alpha}_1+o\left({\varepsilon}^2\right),{y}^{\ast }-y=- y\varepsilon \frac{1}{2}{\alpha}_1+o\left({\varepsilon}^2\right),{\psi}^{\ast }-\psi =\psi \varepsilon \frac{1}{2}{\alpha}_1+o\left({\varepsilon}^2\right),{\theta}^{\ast }-\theta =0 $$

(15)

From Eq. (15), one can easily deduce the set of transformation in the form of the following characteristic equations:

$$ \frac{d x}{x{\alpha}_1}=\frac{d y}{\frac{1}{2}y{\alpha}_1}=\frac{d\psi}{\frac{1}{2}\psi {\alpha}_1}=\frac{d\theta}{0} $$

(16)

Integrating the subsidiary equations (16),

$$ \frac{dx}{x{\alpha}_1}=\frac{dy}{\frac{1}{2}y{\alpha}_1}, $$

we get

$$ \frac{1}{2}\ln x-\ln y=\mathrm{constan}t\;\left(\ln \sqrt{\frac{u_{\infty }}{\nu }}-\ln \eta \right)\left(\mathrm{say}\right) $$

or

$$ \eta =y\sqrt{\frac{u_{\infty }}{\nu x}} $$

From the subsidiary equations

$$ \frac{d x}{x{\alpha}_1}=\frac{d\theta}{0}, $$

we get *dθ* = 0,that is *θ*(*η*)= constant=*θ*(say).

Also integrating the equation \( \frac{d x}{x{\alpha}_1}=\frac{d\psi}{\frac{1}{2}\psi {\alpha}_1}, \)

we get \( \frac{\psi }{x^{\frac{1}{2}}}= \) constant\( =\sqrt{u_{\infty}\nu }f\left(\eta \right) \)(say),

i.e., \( \psi =\sqrt{u_{\infty}\nu }{x}^{\frac{1}{2}}f\left(\eta \right) \)

or

$$ \psi =\sqrt{u_{\infty}\nu x}f\left(\eta \right) $$

Thus, the new similarity transformations are obtained as follows: \( \eta =y\sqrt{\frac{u_{\infty }}{\nu x}},\psi =\sqrt{u_{\infty}\nu x}f\left(\eta \right),\theta =\theta \left(\eta \right) \)

We introduce the following transformations

$$ \eta =y\sqrt{\frac{U_{\infty }}{\nu x}},\psi =\sqrt{u_{\infty}\nu x}f\left(\eta \right),\mu ={\mu}_0{e}^{-{\beta}_1\theta },\kern0.48em \alpha ={\alpha}_0\left(1+{\beta}_2\theta \right),\theta =\frac{T-{T}_{\infty }}{T_w-{T}_{\infty }} $$

(17)

$$ u=\frac{\partial \psi }{\partial y}={u}_{\infty }f\hbox{'},v=-\frac{\partial \psi }{\partial x}=\frac{1}{2}\sqrt{\frac{u_{\infty}\nu }{x}}\left(\eta f\hbox{'}-f\right) $$

Using the above Eq. (17), the transformed equations yield the following transformed, dimensionless system of ordinary differential equations:

$$ {f}^{{\prime\prime\prime} }+\frac{e^{\beta_1\theta }}{2}f{f}^{{\prime\prime} }-{\beta}_1\;{f}^{{\prime\prime} }{\theta}^{\prime }=0 $$

(18)

$$ \left(1+{\beta}_2\theta \right){\theta}^{{\prime\prime} }+{\beta}_2{\left({\theta}^{\prime}\right)}^2-\frac{1}{2}\Pr \left(2a{f}^{\prime}\theta -f{\theta}^{\prime}\right)+c\Pr {e}^{-\eta }=0 $$

(19)

where Pr = *υ*/*α* is the Prandtl number, *c* = 0 or 1 corresponding to with and without internal heat generation.

The boundary conditions are converted to:

$$ {\displaystyle \begin{array}{l}f=0,{f}^{\prime }=0,\theta =1\;\mathrm{at}\;\eta =0\\ {}{f}^{\prime }=1,\theta =0\kern0.24em \mathrm{as}\;\eta \to \infty \end{array}}\Big\} $$

(20)