Principal Algebra, Invariant Solutions and Representations for Optimal Systems of the Burgers–Huxley Equation

Abstract

This manuscript is committed to the investigation of search reduction methodologies, identification and analysis of invariant solutions, and formulation of algebraic representations about the optimal system. These endeavors are conducted within the contextual framework of Lie symmetries, specifically applied to elucidate the complexities inherent in the Burgers–Huxley Equation. A comprehensive analysis is carried out to achieve a thorough classification of the symmetry group for a generalized form of the Burgers–Huxley Equation, leading to the derivation of its principal algebra and the optimal system, as well as the exploration of its reductions, possible invariant solutions, and algebraic representations. These findings offer valuable insights into the underlying mathematical structures and properties of the Burgers–Huxley Equation. The classification and identification of the principal algebra contribute to advancing our understanding of the applications of this differential equation and establishing a solid foundation for future research endeavors.

Introduction

In [1, 2] the following expression is presented

$$\begin{aligned} u_{t}+\alpha _1 u^nu_x-u_{xx}-\beta _1 u(1-u^n)(u^n-\gamma _1)=0,\,\,\,\,\, 0\leqslant x\leqslant 1,\,\,\,t\geqslant 0 \end{aligned}$$

(1)

where $\alpha _1,\beta _1 \geqslant 0$ are real constants and n is a positive integer and $\gamma _1\, \epsilon \, [0,1]$. The expression (1) is known as the generalized Burgers–Huxley equation, which models the interaction between reaction mechanisms, convection effects and diffusion transports [3,4,5,6,7] with a high correlation to the heat conductivity equations [8,9,10]. The diversity of methods that can be found in the literature to present approximate solutions to Eq. (1) is quite rich [11,12,13,14,15,16,17,18,19]. When $n = 1$, $\alpha _1= 1$, $\beta _1\gamma _1=-\gamma$, $(\beta _1+\beta _1\gamma _1)=\delta$ and $\beta _1=-\epsilon$, Eq. (1) is turned into the Burgers–Huxley equation, which has the form:

$$\begin{aligned} u_{t}+uu_{x}-u_{xx}-\gamma u-\delta u^2-\epsilon u^3=0. \end{aligned}$$

(2)

Also, this equation is used to model nerve pulse propagation in nerve fibers as well as wall motion in liquid crystals [2]. It’s important to highlight works [20,21,22] on Eq. (2), in particular the work presented in [21] where a complete classification of the group of symmetries and their respective reductions are presented. We also highlight recent studies involving Lie symmetries and their application in deriving invariant solutions [23,24,25,26].

On the other hand, algebra representations are crucial for understanding their internal structures, such as determining whether a Lie algebra is indecomposable or not. This, in turn, ensures a more comprehensive study of the algebra, as it suffices to examine the associated subalgebras in their decomposition.

The aforementioned motivates us to find the Lie algebras corresponding to the symmetry group associated with the Eq. (2) and investigate whether the optimal system yields a decomposable algebra.

To address the above, in this work we are going use the complete classiffication of the following generalization of (1), which has the form:

$$\begin{aligned} u_{t}+f(u)u_{x}-u_{x x}+g(u)=0, \end{aligned}$$

(3)

where $f,\,\,g\ne 0$ are arbitrary functions. In [27, 28] present a classification of the symmetry properties of the ($1 + 2$)-dimensional reaction−convection-diffusion equation depending on the values of nonlinearities in this equation. In [7, 29,30,31], they study a nonlinear diffusion-convection-reaction equation with a variable coefficient similar to (3), as well as other nonlinear convection-diffusion equations.

Specifically, the proposal of our work is: i) to present a complete classification of the Lie group of symmetries for the Eq. (3), ii) to present the optimal system for Eq. (2), iii) to use some elements of the optimal system for to obtain reductions and some invariant solutions for (2), and finally iv) to show the representations for the Lie algebra associated to (2), corresponding for its symmetry groups and optimal systems.

Complete Classification

In this section we study the Lie symmetry group for (3). The main result of this section can be presented as follows:

Proposition 1

The Lie point symmetry group of the Eq. (3) with f, g arbitrary functions is generated by

$$\begin{aligned} \Pi _1=\partial _t\,\,\,\text {and}\,\,\,\Pi _2=\partial _x\,\,\,\text {Principal algebra,}\end{aligned}$$

(4)

and the other complementary cases are presented in Table 1.

Table 1 Infinitesimal Generators of Eq. (3).

Where $A=-k_1s_1,\,\,B=-s_2k_1+2k_3,\,\,M(t)=4s_4(k_5-k_1)-4s_4s_3k_1t$ with $s_4\ne 0,$ and

Table 2 Generators for the Table 1

with $4s_3c_4+4s_2c_{4,x}-4c_{4,xx}+c_{4,t}=0$, for $\Pi _{23}$ and with $4s_3c_4+4s_2c_{4,x}-4c_{4,xx}+c_{4,t}=M(t)$, for $\Pi _{24}$ (Table 2).

Proof

The general form for the generator operators of a Lie group of a parameter admitted by (3) is: $t \rightarrow t+\xi ^{1}(t, x, u) +O\left( \epsilon ^{2}\right) ,$ $x \rightarrow x+ \xi ^{2}(t, x, u)+O\left( \epsilon ^{2}\right)$ and $u \rightarrow u+\epsilon \eta (t, x, u)+O\left( \epsilon ^{2}\right) ,$ where $\epsilon$ is the group parameter. The vector field associated with this group of transformations is $\Gamma =\xi ^{1}(t, x, u) \frac{\partial }{\partial t}+\xi ^{2}(t, x, u) \frac{\partial }{\partial x}+\eta (t,x,u)\frac{\partial }{\partial u},$ with $\xi ^{1},\xi ^{2},\eta$ differentiable functions in $\mathbb {R}^3$. To find the infinitesimals $\xi ^{1}(t,x,u),\,\,\xi ^{2}(t,x,u)$ and $\eta (t,x,u)$, we applied the third extension operator,

$$\begin{aligned} \Gamma ^{(2)}= & {} \Gamma +\eta _{[t]}\frac{\partial }{\partial {u_t}}+\eta _{[x]}\frac{\partial }{\partial {u_x}}+\eta _{[xx]}\frac{\partial }{\partial {u_{xx}}}, \end{aligned}$$

(5)

to the Eq. (3), obtaining the following symmetry condition

$$\begin{aligned} \eta \left( u_xf_u+g_u\right) +\eta _{[t]}+\eta _{[x]}(f)+\eta _{[xx]}=0, \end{aligned}$$

(6)

where $\eta _{[t]},\eta _{[x]}$ and $\eta _{[xx]}$ are the coefficients in $\Gamma ^{(2)}$ given by:

$$\begin{aligned} \eta _{[t]}= & {} D_{t}[\eta ]-\left( D_{t}\left[ \xi ^{2}\right] \right) u_{x}-\left( D_{t}[\xi ^{1}]\right) u_{t} \nonumber \\ {}= & {} \eta _{t}+u_{t}\left[ \eta _{u}-\xi ^{1}_{t}\right] -u_{t}^{2}\xi ^{1}_{u}-u_{x}\xi ^{2}_{t}-u_{x} u_{t}\xi ^{2}_{u},\nonumber \\ \eta _{[x]}= & {} D_{x}[\eta ]-\left( D_{x}[\xi ^{2}]\right) u_{x}-\left( D_{x}[\xi ^{1}]\right) u_{t} \nonumber \\= & {} \eta _{x}+u_{x}\left[ \eta _{u}-\xi ^{2}_{x}\right] -u_{t}\xi ^{1}_{x}-u_{t} u_{x}\xi ^{1}_{u}-u_{x}^{2}\xi ^{2}_{u}, \nonumber \\ \eta _{[xx]}= & {} D_{x}\left[ \eta _{[x]}\right] -\left( D_{x}\left[ \xi ^{2}\right] \right) u_{x x}-\left( D_{x}\left[ \xi ^{1}\right] \right) u_{x t}\nonumber \\= & {} \eta _{x x}+\left( 2 \eta _{x u}-\xi ^{2}_{x x}\right) u_{x}-\xi ^{1}_{x x} u_{t}-2 \xi ^{1}_{x u} u_{t} u_{x}-2 \xi ^{1}_{x} u_{x t}\nonumber \\{} & {} \quad +\left( \eta _{u u}-2 \xi _{x u}^{2}\right) \left( u_{x}\right) ^{2}+\left( \eta _{u}-2 \xi ^{2}_{x}\right) u_{x x}-\xi ^{2}_{u u}\left( u_{x}\right) ^{3}\nonumber \\{} & {} \quad -3 \xi ^{2}_{u} u_{x} u_{x x}-\xi ^{1}_{u u}\left( u_{x}\right) ^{2} u_{t}-\xi ^{1}_{u} u_{t} u_{x x}-2\xi ^{1}_{u} u_{x} u_{x t}, \end{aligned}$$

(7)

where $D_{x}$ and $D_{t}$ are total derivative operators: $D_{t}=\partial _t+u_{t}\partial _u+u_{t t}\partial _{u_{t}}+u_{t x} \frac{\partial }{\partial u_{x}}+\cdots$ and $D_{x}=\partial _x+u_{x}\partial _u+u_{xx}\partial _{u_{x}}+u_{t x} \frac{\partial }{\partial u_{t}}+\cdots .$ After applying (7) in (6) we have

$$\begin{aligned}{} & {} \eta \left( u_xf_u+g_u\right) +\eta _{t}+u_{t}\left[ \eta _{u}-\xi ^{1}_{t}\right] -u_{x} u_{t}\xi ^{2}_{u}\nonumber \\{} & {} \quad -u_{t}^{2}\xi ^{1}_{u}-u_{x}\xi ^{2}_{t}+(f)\left( \eta _{x}+u_{x}\left[ \eta _{u}-\xi ^{2}_{x}\right] -u_{t}\xi ^{1}_{x}-u_{t} u_{x}\xi ^{1}_{u}-u_{x}^{2}\xi ^{2}_{u}\right) \nonumber \\{} & {} \quad +\eta _{x x}+\left( 2 \eta _{x u}-\xi ^{2}_{x x}\right) u_{x}-\xi ^{1}_{x x} u_{t}-2 \xi ^{1}_{x u} u_{t} u_{x}-2 \xi ^{1}_{x} u_{x t}\nonumber \\{} & {} \quad +\left( \eta _{u u}-2 \xi _{x u}^{2}\right) \left( u_{x}\right) ^{2}+\left( \eta _{u}-2 \xi ^{2}_{x}\right) u_{x x}-\xi ^{2}_{u u}\left( u_{x}\right) ^{3}\nonumber \\{} & {} \quad -3 \xi ^{2}_{u} u_{x} u_{x x}-\xi ^{1}_{u u}\left( u_{x}\right) ^{2} u_{t}-\xi ^{1}_{u} u_{t} u_{x x}-2\xi ^{1}_{u} u_{x} u_{x t}=0, \end{aligned}$$

(8)

after substituting the expression $u_{t}$ by (3) and analyze the coefficients with respect to the independent variables $u_x,u_{x}^2,u_{x}^3,u_xu_{xx},u_x^2u_{xx},u_{xx}^2,\cdots$ and 1 we get the following system of determining equations:

$$\begin{aligned} \xi ^{1}_x=\xi ^{1}_u=\xi ^{2}_u=\eta _{uu}=0, \end{aligned}$$

(9a)

$$\begin{aligned} 2\xi ^{2}_{x}-\xi ^{1}_t=0,&\end{aligned}$$

(9b)

$$\begin{aligned} \eta f_u+f( -\xi ^{2}_x+\xi ^{1}_t)-2\eta _{xu}+\xi ^{2}_{xx}-\xi ^{2}_t=0,&\end{aligned}$$

(9c)

$$\begin{aligned} \eta g_u+g(\xi ^{1}_{t}-\eta _{u})+f\eta _{x}-\eta _{xx}+\eta _t=0,&\end{aligned}$$

(9d)

It’s clear that with f and g arbitrary functions, we have $\xi ^{1}=k_1,\,\,\,\xi ^{2}=k_2$ and $\eta =0.$ therefore, the group of symmetries is $\Pi _1=\partial _t,\,\,\,\Pi _2=\partial _x.$ Solving (9a) we have:

$$\begin{aligned} \xi ^{1}=c_1(t),\,\,\,\xi ^{2}=c_2(t,x)\,\,\, \text {and}\,\,\,\eta =uc_3(t,x)+c_4(t,x), \end{aligned}$$

(10)

where $c_1,c_2,c_3$ and $c_4$ are arbitrary functions. From (9b) we have $2c_{2,x}-c_{1,t}=0$ then differentiating with respect to x we get $c_{2,xx}=0$ thus $c_2(t,x)=xd_1(t)+d_2(t)$ where $d_1,d_2$ are arbitrary functions. Hence $d_1(t)=\frac{c_{1,t}}{2}.$ From (10)

$$\begin{aligned} \xi ^{1}=c_1(t),\,\,\,\xi ^{2}=x(\frac{c_{1,t}}{2})+d_2(t)\,\,\, \text {and}\,\,\,\eta =uc_3(t,x)+c_4(t,x), \end{aligned}$$

(11)

From (11) into (9c)

$$\begin{aligned} c_3(uf_u)+c_4f_u+f(\frac{c_{1,t}}{2})-2c_{3,x}-\frac{xc_{1,tt}}{2}-d_{2,t}=0. \end{aligned}$$

(12)

Differentiating (12) twice with respect to x and once with respect to u we have respectively

$$\begin{aligned} c_{3,x}(uf_u)+c_{4,x}f_u-2c_{3,xx}-\frac{c_{1,tt}}{2}&=0, \end{aligned}$$

(13a)

$$\begin{aligned} c_{3,xx}(uf_u)+c_{4,xx}f_u-2c_{3,xxx}&=0, \end{aligned}$$

(13b)

$$\begin{aligned} c_{3,xx}(uf_u)_u+c_{4,xx}f_{uu}&=0. \end{aligned}$$

(13c)

From (13c), two cases should be considered $f_{uu}=0$ and $f_{uu}\ne 0.$

Case I.) Suppose $f_{uu}=0$ in (13c) then $f(u)=s_1u+s_2$ where $s_1,s_2$ are arbitrary constants and $u(2s_1c_{3,xx})+s_2c_{3,xx}=0.$ Analyzing the coefficients u and 1 in the previous expression we have

$$\begin{aligned} s_1c_{3,xx}= & {} 0, \end{aligned}$$

(14)

$$\begin{aligned} s_2c_{3,xx}= & {} 0. \end{aligned}$$

(15)

From (14), two cases should be considered $s_1\ne 0$ and $s_1= 0.$

Case I.1.) Suppose $s_1\ne 0$ in (14) then $c_{3,xx}=0$ implying $c_3(t,x)=xd_3(t)+d_4(t)$ where $d_3,d_4$ are arbitrary functions. From (13b) we obtain $c_{4,xx}=0$ then $c_4(t,x)=xd_5(t)+d_6(t)$ where $d_5,d_6$ are arbitrary functions. From (13a) we have $d_3=0$ and $d_5=\frac{c_{1,tt}}{2s_1}.$ From (12) we obtain $d_4=-\frac{c_{1,t}}{2}$ and $d_6=-\frac{s_2c_{1,t}}{2s_1}+\frac{d_{2,t}}{s_1}.$ From (11):

$$\begin{aligned} \xi ^{1}=c_1(t),\,\,\,\xi ^{2}=x(\frac{c_{1,t}}{2})+d_2(t)\,\,\,\eta =-u\left( \frac{c_{1,t}}{2}\right) +x(\frac{c_{1,tt}}{2s_1})-\frac{s_2c_{1,t}}{2s_1}+\frac{d_{2,t}}{s_1}. \end{aligned}$$

(16)

Using (16) into (9d) we have

$$\begin{aligned} g_u\left( u(-\frac{c_{1,t}}{2})+x(\frac{c_{1,tt}}{2s_1})-\frac{s_2c_{1,t}}{2s_1}+\frac{d_{2,t}}{s_1}\right) +\frac{3}{2}gc_{1,t}+x(\frac{c_{1,ttt}}{2s_1})+\frac{d_{2,tt}}{s_1}=0. \end{aligned}$$

(17)

Differentiating in (17) with respect to x and u we have respectively

$$\begin{aligned} g_u\left( c_{1,tt}\right) +c_{1,ttt}&=0, \end{aligned}$$

(18a)

$$\begin{aligned} g_{uu}\left( c_{1,tt}\right)&=0. \end{aligned}$$

(18b)

From (18b), two cases should be considered $g_{uu}\ne 0$ and $g_{uu}= 0.$

Case I.1.2) Suppose $g_{uu}\ne 0$ in (18b) then $c_1(t)=tk_1+k_2$ where $k_1,k_2$ are arbitrary constants. From (17)

$$\begin{aligned} g_u\left( u(-\frac{k_1}{2})-\frac{s_2k_1}{2s_1}+\frac{d_{2,t}}{s_1}\right) +\frac{3k_1}{2}g+\frac{d_{2,tt}}{s_1}=0. \end{aligned}$$

(19)

Differentiating in (19) with respect to t and u we have respectively $g_u\left( \frac{d_{2,tt}}{s_1}\right) +\frac{d_{2,ttt}}{s_1}=0$ and $d_{2,tt}=0$ then $d_2(t)=tk_3+k_4$ where $k_3,k_4$ are arbitrary constants. From (18a)

$$\begin{aligned} g_u\left( u(-\frac{k_1}{2})-\frac{s_2k_1}{2s_1}+\frac{k_3}{s_1}\right) +\frac{3k_1}{2}g=0. \end{aligned}$$

(20)

Solving for (20) with respect to g we get $g(u)=s_3\left( Au+B\right) ^3$ with $A=-k_1s_1$ and $B=-s_2k_1+2k_3.$ From (16) we have $\xi ^{1}=tk_1+k_2,\,\,\,\xi ^{2}=x(\frac{k_1}{2})+tk_3+k_4$ and $\eta =-u\left( \frac{k_1}{2}\right) -\frac{s_2k_1}{2s_1}+\frac{k_3}{s_1}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\,\Pi _2=\partial _x,$ $\Pi _3=k_3(t\partial _x+\frac{\partial _u}{s_1})$ and $\Pi _4=t\partial _t+\frac{x}{2}\partial _x-(\frac{u}{2}+\frac{s_2}{2s_1})\partial _u$ with $f(u)=us_1+s_2,$ $g(u)=s_3\left( Au+B\right) ^3.$

Note that if $s_1=1,\,\,s_2=0$ and $k_3=0$ implying $B=0$ and $\epsilon =s_3k_1^3$ we have the symmetries presented by [21].

Case I.1.2) Suppose $g_{uu}=0$ in (18b) then $g(u)=s_3u+s_4$ where $s_3,s_4$ are arbitrary constants. From (18a) we get

$$\begin{aligned} s_3\left( c_{1,tt}\right) +c_{1,ttt}=0, \end{aligned}$$

(21)

From (21), two cases should be considered $s_3\ne 0$ and $s_3= 0.$

Case I.1.2.1) Suppose $s_3\ne 0$ in (21) then we have $c_1(t)=\frac{k_1e^{-ts_3}}{s_3^2}+k_3t+k_2$ where $k_1,k_2,k_3$ are arbitrary constants. From (17) we get $k_1=k_3=0$ and $d_2(t)=\frac{k_4e^{-ts_3}}{s_3}+k_5$ where $k_4,k_5$ are arbitrary constants. From (16) we obtain $\xi ^{1}=k_2,\,\,\,\xi ^{2}=\frac{k_4e^{-ts_3}}{s_3}+k_5$ and $\eta =-\frac{k_4e^{-ts_3}}{s_1}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\,\Pi _2=\partial _x$ and $\Pi _5=e^{-ts_3}\left( \partial _x-\frac{s_3}{s_1}\partial _u\right)$ with $f(u)=us_1+s_2,\,\,\,g(u)=s_3u+s_4.$ Note that if $s_1=1,\,\,s_2,s_4=0$ and $\gamma =-s_3$ we have the symmetries presented by [21]

Case I.1.2.2) Suppose $s_3=0$ in (21) then $c_{1,ttt}=0$ implying $c_1(t)=\frac{k_1}{2}t^2+tk_2+k_3$ where $k_1,k_2,k_3$ are arbitrary constants. From (17) we get $d_2(t)=-\frac{s_4s_1k_1}{4}t^3-\frac{3s_4s_1k_2}{4}t^2+tk_4+k_5$ where $k_4,k_5$ are arbitrary constants. From (16) we obtain $\xi ^{1}=\frac{k_1}{2}t^2+tk_2+k_3,\,\,\,\xi ^{2}=x(\frac{(tk_1+k_2)}{2})-\frac{s_4s_1k_1}{4}t^3-\frac{3s_4s_1k_2}{4}t^2+tk_4+k_5$ and $\eta =-u\left( \frac{(tk_1+k_2)}{2}\right) +x(\frac{k_1}{2s_1})-\frac{s_2(tk_1+k_2)}{2s_1}-\frac{3s_4k_1t^2}{4}-\frac{3s_4k_2t}{2}+\frac{k_4}{s_1}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\,\Pi _2=\partial _x,$ $\Pi _6=t\partial _t+\frac{x}{2}\partial _x-\frac{u}{2}\partial _u,$ $\Pi _7=t^2\partial _t+xt\partial _x+(\frac{x}{s_1}-xt)\partial _u,$ $\Pi _8=-\frac{s_4s_1t^3}{4}\partial _x-\frac{3s_4t^2}{4}\partial _u,$ $\Pi _9=-\frac{3s_4s_1t^2}{4}\partial _x-\frac{3s_4t}{2}\partial _u,$ $\Pi _{10}=-\frac{s_2(tk_1+k_2)}{2s_1}\partial _x$ and $\Pi _{11}=t\partial _x+\frac{1}{s_1}\partial _u$ with $f(u)=us_1+s_2,\,\,\,g(u)=s_4.$ Note that if $s_1=1,\,\,s_2,s_4=0$ and $\nu =1$ we have the Burgers equation and the symmetries presented by [32, 33].

Case I.2.) Suppose $s_1=0$ in (14) then from (15) we have $s_2c_{3,xx}=0.$ We know that $s_2\ne 0$ then $c_{3,xx}=0$ implying $c_3(t)=xd_3(t)+d_4(t)$ where $d_3,d_4$ are arbitrary functions. From (13a) we obtain $c_{1,tt}=0$ implying $c_1(t)=tk_1+k_2$ where $k_1,k_2$ are arbitrary functions. From (12) we get $d_3(t)=\frac{s_2k_1}{4}-\frac{d_{2,t}}{2}$ thus $c_3(t)=x\left( \frac{s_2k_1}{4}-\frac{d_{2,t}}{2}\right) +d_4(t).$ From (11) we have $\xi ^{1}=tk_1+k_2,\,\,\,\xi ^{2}=x(\frac{k_1}{2})+d_2(t)$ and $\eta =u\left( x\left( \frac{s_2k_1}{4}-\frac{d_{2,t}}{2}\right) +d_4(t)\right) +c_4(t,x).$ From (9d):

$$\begin{aligned}{} & {} g_u\left( u\left( x(\frac{s_2k_1}{4}-\frac{d_{2,t}}{2})+d_4\right) +c_4\right) +g\left( k_1-\left( x(\frac{s_2k_1}{4}-\frac{d_{2,t}}{2})+d_4\right) \right) \nonumber \\{} & {} \quad +u\left( \frac{s_2^2k_1}{4}-\frac{s_2d_{2,t}}{2}+x(-\frac{d_{2,tt}}{2})+d_{4,t}\right) +s_2c_{4,x}-c_{4,xx}+c_{4,t}=0 \end{aligned}$$

(22)

Differentiating twice with respect to x and once with respect to u we have respectively

$$\begin{aligned}{} & {} g_u\left( u\left( \frac{s_2k_1}{4}-\frac{d_{2,t}}{2}\right) +c_{4,x}\right) +g\left( \frac{d_{2,t}}{2}-\frac{s_2k_1}{4}\right) -u\left( \frac{d_{2,tt}}{2}\right) \nonumber \\{} & {} \quad +s_2c_{4,xx}-c_{4,xxx}+c_{4,tx}=0 \end{aligned}$$

(23)

$$\begin{aligned} g_{uu}\left( u\left( \frac{s_2k_1}{4}-\frac{d_{2,t}}{2}\right) +c_{4,x}\right) -\frac{d_{2,tt}}{2}&=0 \end{aligned}$$

(24a)

$$\begin{aligned} g_{uu}\left( c_{4,xx}\right)&=0 \end{aligned}$$

(24b)

From (24b), two cases should be considered $g_{uu}\ne 0$ and $g_{uu}= 0.$

Case I.2.1) Suppose $g_{uu}\ne 0$ in (24b) then $c_4(t)=xd_5(t)+d_6(t)$ where $d_5,d_6$ are arbitrary constants. From (24a) we obtain

$$\begin{aligned} ug_{uu}\left( s_2k_1-2d_{2,t}\right) -g_{uu}(4d_5)-2d_{2,tt}=0. \end{aligned}$$

(25)

Differentiating in (25) with respect to u we obtain

$$\begin{aligned} \left( ug_{uu}\right) _u\left( s_2k_1-2d_{2,t}\right) -g_{uuu}(4d_5)=0.\end{aligned}$$

(26)

From (26), two cases should be considered $g_{uuu}\ne 0$ and $g_{uuu}= 0.$

Case I.2.1.1) Suppose $g_{uuu}\ne 0$ in (26) then we get

$$\begin{aligned} \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\left( s_2k_1-2d_{2,t}\right) -(4d_5)=0.\end{aligned}$$

(27)

Differentiating in (27) with respect to u we obtain

$$\begin{aligned} \left( \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\right) _u\left( s_2k_1-2d_{2,t}\right) =0. \end{aligned}$$

(28)

From (28), two cases should be considered $\left( \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\right) _u\ne 0$ and $\left( \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\right) _u= 0.$

Case I.2.1.1.1) Suppose $\left( \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\right) _u\ne 0$ in (28) then $d_2(t)=t\frac{s_2k_1}{2}+k_3$ where $k_3-$arbitrary constant and from (27) we obtain $d_5=0.$ From (22)

$$\begin{aligned} g_u\left( ud_4+d_6\right) +g\left( k_1-d_4\right) +u\left( d_{4,t}\right) +d_{6,t}=0. \end{aligned}$$

(29)

Differentiating twice in (29) with respect to u we obtain respectively

$$\begin{aligned} g_{uu}\left( ud_4+d_6\right) +k_1g_u+d_{4,t}&=0, \end{aligned}$$

(30a)

$$\begin{aligned} g_{uuu}\left( ud_4+d_6\right) +g_{uu}(d_4+k_1)&=0. \end{aligned}$$

(30b)

Dividing by $g_{uuu}$ in (30b) and differentiating twice in (30b) with respect to u we obtain respectively

$$\begin{aligned} d_4+(d_4+k_1)\left( \frac{g_{uu}}{g_{uuu}}\right) _u&=0, \end{aligned}$$

(31a)

$$\begin{aligned} (d_4+k_1)\left( \frac{g_{uu}}{g_{uuu}}\right) _{uu}&=0. \end{aligned}$$

(31b)

From (31b), two cases should be considered $\left( \frac{g_{uu}}{g_{uuu}}\right) _{uu}\ne 0$ and $\left( \frac{g_{uu}}{g_{uuu}}\right) _{uu}= 0.$

Case I.2.1.1.1.1) Suppose $\left( \frac{g_{uu}}{g_{uuu}}\right) _{uu}\ne 0$ in (31b) then $d_4=-k_1$ and from (31a) we have $d_4=0$ inplying $k_1=0.$ From (30a) and (30b) we obtain $d_6=0.$ Thus $\xi ^{1}=k_2,\,\,\,\xi ^{2}=k_3$ and $\eta =0.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ $\left( \frac{g_{uu}}{g_{uuu}}\right) _{uu}\ne 0$ and $\left( \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\right) _u\ne 0.$

Case I.2.1.1.1.2) Suppose $\left( \frac{g_{uu}}{g_{uuu}}\right) _{uu}= 0$ in (31b) then we have

$$\begin{aligned} \left( \frac{g_{uu}}{g_{uuu}}\right) _{u}= m_1, \end{aligned}$$

(32)

where $m_1-$arbitrary constant. From (32), two cases should be considered $m_1\ne 0$ and $m_1= 0.$

Case I.2.1.1.1.2.1) Suppose $m_1\ne 0$ in (32) then we get

$$\begin{aligned} g_{uu}=m_3(um_1+m_2)^{\frac{1}{m_1}}, \end{aligned}$$

(33)

where $m_2,m_3$ are arbitrary constants. From (33) we have to consider two cases $m_1\ne -1$ and $m_1=-1.$

Case A.):

Suppose $m_1\ne -1$ in (33) then we obtain

$$\begin{aligned} g_{u}=\frac{m_3(um_1+m_2)^{\frac{1+m_1}{m_1}}}{m_1+1}+m_4, \end{aligned}$$

(34)

where $m_4$ is arbitrary constants. From (34) we have to consider two cases $m_1\ne -\frac{1}{2}$ and $m_1=-\frac{1}{2}.$

Case A.1) Suppose $m_1\ne -\frac{1}{2}$ in (34) then we obtain $g(u)=\frac{m_3(um_1+m_2)^{\frac{1+2m_1}{m_1}}}{(1+m_1)(1+2m_1)}+um_4+m_5$ where $m_5-$arbitrary constant. From (31a) we get $d_4=-\frac{m_1k_1}{(1+m_1)}.$ From (30b) we have $d_6=-\frac{m_2k_1}{1+m_1}.$ From (30a) we get

$$\begin{aligned} k_1m_4=0 \end{aligned}$$

(35)

From (35) we have to consider two cases $k_1\ne 0$ and $k_1=0.$

Case A.1.1) Suppose $k_1\ne 0$ in (35) then $m_4=0.$ Thus $g(u)=\frac{m_3(um_1+m_2)^{\frac{1+2m_1}{m_1}}}{(1+m_1)(1+2m_1)}+m_5.$ From (29) we have $m_5=0.$ Hence $\xi ^{1}=tk_1+k_2,\,\,\,\xi ^{2}=x(\frac{k_1}{2})+t\frac{s_2k_1}{2}+k_3$ and $\eta =-u\frac{m_1k_1}{(1+m_1)}-\frac{m_2k_1}{1+m_1}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\Pi _2=\partial _x,\,\,\,\Pi _{12}=t\partial _t+\frac{x}{2}\partial _x-\frac{um_1}{1+m_1}\partial _u$ and $\Pi _{13}=\frac{s_2t}{2}\partial _x-\frac{m_2}{1+m_1}\partial _u$ with $f(u)=s_2,$ and $g(u)=\frac{m_3(um_1+m_2)^{\frac{1+2m_1}{m_1}}}{(1+m_1)(1+2m_1)}.$

Case A.1.2) Suppose $k_1=0$ in (35) then $d_4=d_6=0.$ Thus $\xi ^{1}=k_2,$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ and $g(u)=\frac{m_3(um_1+m_2)^{\frac{1+2m_1}{m_1}}}{(1+m_1)(1+2m_1)}+um_4+m_5.$

Case A.2) Suppose $m_1=-\frac{1}{2}$ in (34) then $g_{u}=4m_3(-u+2m_2)^{-1}+m_4$ implying $g(u)=-4m_3\ln {(2m_2-u)}+um_4+m_5$ where $m_5$ is arbitrary constant. From (31a) we have $d_4=k_1.$ From (30b) we get $d_6=-2m_2k_1.$ From (30a) we obtain

$$\begin{aligned} k_1m_4=0. \end{aligned}$$

(36)

From (36) we have to consider two cases $k_1\ne 0$ and $k_1=0.$

Case A.2.1) Suppose $k_1\ne 0$ in (36) then $m_4=0.$ From (29) we have $m_3=0$ then $g(u)=m_5$ and so we have a contradiction with $g_{uu}\ne 0$

Case A.2.2) Suppose $k_1=0$ in (36) then we have $\xi ^{1}=k_2$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ and $g(u)=-4m_3\ln {(2m_2-u)}+um_4+m_5.$

Case B.):

Suppose $m_1=-1$ in (33) then $g_{uu}=m_3(-u+m_2)^{-1}.$ From (31a) we have $k_1=0.$ From (30b) we get $d_4=1$ and $d_6=-m_2.$ From (30a) we have $m_3=0$ then $g_{uu}=0$ and so we have a contradiction with $g_{uu}\ne 0$

Case I.2.1.1.1.2.2) Suppose $m_1= 0$ in (32) then $g(u)=s_3p_1^2e^{\frac{u}{p_1}}+s_4u+s_5$ where $p_1,s_3,s_4,s_5$ are arbitrary constants. From (31a) we obtain $d_4=0.$ From (30b) we get $d_6=-p_1k_1.$ From (30a) we have

$$\begin{aligned} k_1s_4=0. \end{aligned}$$

(37)

From (37) we have to consider two cases $k_1\ne 0$ and $k_1=0.$

Case C.1):

Suppose $k_1\ne 0$ in (37) then $g(u)=s_3p_1^2e^{\frac{u}{p_1}}+s_5.$ From (29) we get $s_5=0.$ Thus $\xi ^{1}=tk_1+k_2,\,\,\,\xi ^{2}=x\frac{k_1}{2}+\frac{ts_2k_1}{2}+k_3$ and $\eta =-p_1k_1.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\Pi _2=\partial _x$ and $\Pi _{14}=t\partial _t+\frac{(x+s_2t)}{2}\partial _x-p_1t\partial _u$ with $f(u)=s_2,$ and $g(u)=s_3p_1^2e^{\frac{u}{p_1}}.$

Case C.2):

Suppose $k_1= 0$ in (37) then $\xi ^{1}=k_2$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ and $g(u)=s_3p_1^2e^{\frac{u}{p_1}}+us_4+s_5.$

Case I.2.1.1.2) Suppose $\left( \frac{\left( ug_{uu}\right) _u}{g_{uuu}}\right) _u= 0$ in (28) then we obtain

$$\begin{aligned} \frac{\left( ug_{uu}\right) _u}{g_{uuu}}= p_1,\,\,p_1\,\,\text{ constant. } \end{aligned}$$

(38)

From (38) we have to consider two cases $p_1= 0$ and $p_1\ne 0.$

Case DI.):

Suppose $p_1= 0$ in (38) then $g(u)=p_2u\ln {(u)}+s_3u+s_4$ where $p_2,s_3,s_4$ are arbitrary constants. From (27) we get $d_5=0.$ From (25) we have $d_{2,tt}+p_2d_{2,t}-\frac{p_2s_2k_1}{2}=0,$ then solving this expression we obtain $d_2(t)=\frac{s_5e^{-p_2t}}{p_2}+\frac{k_1s_2t}{2}+s_6$ where $s_5,s_6$ are arbitrary constants. From (23) we obtain

$$\begin{aligned}&s_5(p_2-1)=0,&\end{aligned}$$

(39)

$$\begin{aligned}&s_5s_4=0.&\end{aligned}$$

(40)

From (40) we have to consider two cases $s_5= 0$ and $s_5\ne 0.$

Case DI.1) Suppose $s_5= 0$ in (40) then $d_2(t)=\frac{k_1s_2t}{2}+s_6.$ From (22) we obtain

$$\begin{aligned} d_6=k_1&=0, \end{aligned}$$

(41a)

$$\begin{aligned} p_2d_4+d_{4,t}&=0, \end{aligned}$$

(41b)

$$\begin{aligned} s_4d_4&=0. \end{aligned}$$

(41c)

From (41c) we have to consider two cases $s_4= 0$ and $s_4\ne 0.$

Case DI.1.1) Suppose $s_4= 0$ in (41c) then $g(u)=p_2u\ln {(u)}+s_3u$, $d_2=s_6$ and $d_4(t)=p_3e^{-p_2t}$ where $p_3-$arbitrary constant. Thus $\xi ^{1}=k_2,\,\,\,\xi ^{2}=s_6$ and $\eta =up_3e^{-p_2t}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\Pi _2=\partial _x$ and $\Pi _{15}=e^{-p_2t}\partial _u$ with $f(u)=s_2,$ and $g(u)=p_2u\ln {(u)}+s_3u.$

Case DI.1.2) Suppose $s_4\ne 0$ in (41c) then $d_6=k_1=d_4=0$ and $d_2=s_6.$ Thus $\xi ^{1}=k_2$ and $\xi ^{2}=s_6.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ and $g(u)=p_2u\ln {(u)}+s_3u+s_4.$

Case DI.2) Suppose $s_5\ne 0.$ in (40) then $s_4=0,p_2=1,$ $d_2(t)=s_5e^{-t}+\frac{k_1s_2t}{2}+s_6$ and $g(u)=u\ln {(u)}+s_3u.$ From (22) we have $k_1=d_6=0$ and $d_4(t)=s_7e^{-t}-\frac{s_2s_5e^{-t}}{2}$ where $s_7-$arbitrary constant. Thus $\xi ^{1}=k_2,\,\,\,\xi ^{2}=s_5e^{-t}+s_6$ and $\eta =u\left( x\left( \frac{s_5e^{-t}}{2}\right) +s_7e^{-t}-\frac{s_2s_5e^{-t}}{2}\right) .$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\Pi _2=\partial _x,$ $\Pi _{16}=e^{-t}\partial _x+ue^{-t}(\frac{x-s_2}{2})\partial _u$ and $\Pi _{17}=ue^{-t}\partial _u$ with $f(u)=s_2,$ and $g(u)=u\ln {(u)}+s_3u.$

Case DII.):

Suppose $p_1\ne 0$ in (38) then $g(u)=p_2u\ln {(u-p_1)}-p_1p_2\ln {(u-p_1)}+s_3u+s_4$ where $p_2,s_3,s_4$ are arbitrary constants. From (27) $d_5(t)=\frac{p_1s_2k_1}{4}-\frac{p_1d_{2,t}}{2}.$ From (25) we obtain $d_{2,tt}+p_2d_{2,t}-\frac{p_2s_2k_1}{2}=0$ and solving this expression we get $d_2(t)=\frac{s_5e^{-p_2t}}{p_2}+\frac{k_1s_2t}{2}+s_6$ where $s_5,s_6$ are arbitrary constants. From (23) we have $s_5=0$ then $d_2(t)=\frac{k_1s_2t}{2}+s_6$ and $d_5(t)=0.$ From (22) we get $d_4=d_6=k_1=0.$ Thus $\xi ^{1}=k_2$ and $\xi ^{2}=s_6.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ and $g(u)=p_2u\ln {(u-p_1)}-p_1p_2\ln {(u-p_1)}+s_3u+s_4$.

Case I.2.1.2) Suppose $g_{uuu}= 0$ in (26) then $d_{2,t}=\frac{s_2k_1}{2}$ implying $d_{2}(t)=\frac{s_2k_1t}{2}+s_3$ and $g(u)=\frac{u^2s_4}{2}+us_5+s_6$ where $s_3,s_4,s_5,s_6$ are arbitrary constants. From (25) we get $d_5=0$ then $c_4=d_6(t).$ From (22) we have $d_4=-k_1,\,\,d_6=-\frac{s_5k_1}{s_4}$ and

$$\begin{aligned} k_1(2s_6s_4-s_5)=0. \end{aligned}$$

(42)

From (42) we have to consider two cases $k_1= 0$ and $k_1\ne 0.$

Case I.2.1.2.1):

Suppose $k_1= 0$ in (42) then $d_4=d_6=0.$ Thus $\xi ^{1}=k_2$ and $\xi ^{2}=s_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2,$ and $g(u)=\frac{u^2s_4}{2}+us_5+s_6$.

Case I.2.1.2.2):

Suppose $k_1\ne 0$ in (42) then $s_6=\frac{s_5}{2s_4}.$ Thus $\xi ^{1}=tk_1+k_2,\,\,\,\xi ^{2}=x(\frac{k_1}{2})$ $+\frac{s_2k_1t}{2}+s_3$ and $\eta =-uk_1-\frac{s_5k_1}{s_4}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\Pi _2=\partial _x,$ $\Pi _{18}=t\partial _t+\frac{x}{2}\partial _x-u\partial _u$ and $\Pi _{19}=\frac{s_2t}{2}\partial _x-\frac{s_5}{s_4}\partial _u$ with $f(u)=s_2,$ and $g(u)=\frac{u^2s_4}{2}+us_5+\frac{s_5}{2s_4}$.

Case I.2.2) Suppose $g_{uu}= 0$ in (24b) then $g(u)=s_3u+s_4$ where $s_3,s_4$ are arbitrary constants. From (24a) we have $d_{2,tt}=0$ implying $d_2(t)=k_3t+k_4$ where $k_3,k_4$ arbitrary constants. From (23) we get

$$\begin{aligned} 4s_3c_4+4s_2c_{4,x}-4c_{4,xx}+c_{4,t}&=M(t), \end{aligned}$$

(43a)

$$\begin{aligned} s_4(2k_3-s_2k_1)&=0. \end{aligned}$$

(43b)

From (43b) we have to consider two cases $s_4= 0$ and $s_4\ne 0.$

Case I.2.2.1):

Suppose $s_4= 0$ in (43b) then from (43a) and (22) we obtain $M(t)=0$ and $d_4=-t\left( s_3k_1+\frac{s2}{4}(k_1s_2-2k_3)\right) +k_5.$ Thus $\xi ^{1}=tk_1+k_2,$ $\xi ^{2}=\frac{xk_1}{2}+k_3t+k_4$ and $\eta =u\left( x\left( \frac{s_2k_1}{4}-\frac{k_3}{2}\right) -t\left( s_3k_1+\frac{s_2}{4}(k_1s_2-2k_3)\right) +k_5\right) +c_4(t,x).$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x,\,\Pi _{20}=\partial _u,$ $\Pi _{21}$$=t\partial _t+\frac{x}{2}\partial _x+u\left( \frac{xs_2}{4}-\frac{t(4s_3+s_2^2)}{4}\right) \partial _u,$ $\Pi _{22}=t\partial _x+u(\frac{ts_2-x}{2})\partial _u$ and $\Pi _{23}=c_4(t,x)\partial _u$ with $4s_3c_4+4s_2c_{4,x}-4c_{4,xx}+c_{4,t}=0,$ $f(u)=s_2,$ and $g(u)=s_3u$.

Case I.2.2.2):

Suppose $s_4\ne 0$ in (43b) then $k_3=\frac{s_2k_1}{2}.$ from (43a) and (22) we obtain $M(t)=4s_4(k_5-k_1)-4s_4s_3k_1t$ and $d_4=-s_3k_1t+k_5.$ Thus $\xi ^{1}=tk_1+k_2,\,\,\,\xi ^{2}=\frac{xk_1}{2}+\frac{ts_2k_1}{2}+k_4$ and $\eta =u\left( -s_3k_1t+k_5\right) +c_4(t,x).$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x,\,\Pi _{20}=\partial _u$ $\Pi _{24}=t\partial _t+\frac{(x+ts_2)}{2}\partial _x-s_3ut\partial _u,$ and $\Pi _{23}=c_4(t,x)\partial _u$ with $4s_3c_4+4s_2c_{4,x}-4c_{4,xx}+c_{4,t}=M(t),$ $f(u)=s_2,$ and $g(u)=s_3u+s_4$.

Case II.)Suppose $f_{uu}\ne 0$ in (13c) then we obtain

$$\begin{aligned} c_{3,xx}\left( \frac{(uf_u)_u}{f_{uu}}\right) +c_{4,xx}=0. \end{aligned}$$

(44)

Differentiating in (44) with respect to u we have

$$\begin{aligned} c_{3,xx}\left( \frac{(uf_u)_u}{f_{uu}}\right) _u=0. \end{aligned}$$

(45)

From (45), two cases should be considered $\left( \frac{(uf_u)_u}{f_{uu}}\right) _u=0$ and $\left( \frac{(uf_u)_u}{f_{uu}}\right) _u\ne 0.$

Case II.1)Suppose $\left( \frac{(uf_u)_u}{f_{uu}}\right) _u=0$ in (45) then we have

$$\begin{aligned} \frac{(uf_u)_u}{f_{uu}}=s_1,\,\,s_1\,\,\text{ arbitrary } \text{ constant. } \end{aligned}$$

(46)

From (46) we can consider two cases $s_1=0$ and $s_1\ne 0.$

Case II.1.1)Suppose $s_1=0$ in (46) then $f(u)=s_2\ln {(u)}+s_3$ where $s_2,s_3$ are arbitrary constants. From (44) we have $c_{4,xx}=0$ inplying $c_4(t,x)=xd_5(t)+d_6(t).$ where $d_5,d_6$ are arbitrary functions. From (13b) we get $c_3(t,x)=\frac{4e^{\frac{xs_2}{2}}}{s_2^2}d_7(t)+xd_8(t)+d_9(t).$ From (13a) we obtain $d_5=0$ and $d_8=\frac{c_{1,tt}}{2s_2}.$ From (12) we have $d_6=0,\,\,d_9=\frac{d_{2,t}}{s_2}$ and $c_1=k_1-$arbitrary constant. From (11):

$$\begin{aligned} \xi ^{1}=k_1,\,\,\,\xi ^{2}=d_2(t)\,\,\, \text {and}\,\,\,\eta =u\left( \frac{4e^{\frac{xs_2}{2}}}{s_2^2}d_7(t)+\frac{d_{2,t}}{s_2}\right) , \end{aligned}$$

(47)

Using (47) into (9d) we have

$$\begin{aligned} ug_u(4d_7)-g(4d_7)+u\ln {(u)}(2s_2^2d_7)+u(2d_7s_2s_3-s_2^2d_7+4d_{7,t})&=0, \end{aligned}$$

(48a)

$$\begin{aligned} ug_u(d_{2,t})-g(d_{2,t})+u(d_{2,tt})&=0. \end{aligned}$$

(48b)

Differentiating twice with respect to u in (48b) we obtain respectively

$$\begin{aligned} (ug_u)_u(d_{2,t})-g_u(d_{2,t})+d_{2,tt}&=0, \end{aligned}$$

(49a)

$$\begin{aligned} (ug_u)_{uu}(d_{2,t})-g_{uu}(d_{2,t})&=0. \end{aligned}$$

(49b)

From (49b) we can consider two cases $g_{uu}=0$ and $g_{uu}\ne 0.$

Case II.1.1.1)Suppose $g_{uu}=0$ in (49b) then $g(u)=us_4+s_5$ where $s_4,s_5$ are arbitrary constants. From (49a) we obtain $d_{2,tt}=0$ implying $d_2(t)=tk_2+k_3$ where $k_2,k_3$ are arbitrary constants. From (48b) we have

$$\begin{aligned} s_5k_2=0. \end{aligned}$$

(50)

From (49b) we can consider two cases $s_5=0$ and $s_5\ne 0.$

Case II.1.1.1.1)Suppose $s_5=0$ in (50) then $g(u)=us_4.$ From (48a) we get $d_7=0.$ From (47) we have $\xi ^{1}=k_1,\,\,\,\xi ^{2}=tk_2+k_3$ and $\eta =\frac{uk_2}{s_2}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x$ and $\Pi _{25}=t\partial _x+\frac{u}{s_2}\partial _u,$ with $f(u)=s_2\ln {(u)}+s_3,$ and $g(u)=s_4u$. Case II.1.1.1.2)Suppose $s_5\ne 0$ in (50) then $k_2=0.$ From (48a) we obtain $d_7=0.$ From (47) we have $\xi ^{1}=k_1$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2\ln {(u)}+s_3,$ and $g(u)=s_4u+s_5$. Case II.1.1.2)Suppose $g_{uu}\ne 0.$ in (49b) then we have

$$\begin{aligned} \left( \frac{(ug_u)_{uu}}{g_{uu}}\right) d_{2,t}-d_{2,t}=0. \end{aligned}$$

(51)

Differentiating in (51) with respect to u we get

$$\begin{aligned} \left( \frac{(ug_u)_{uu}}{g_{uu}}\right) _ud_{2,t}=0. \end{aligned}$$

(52)

From (52) we can consider two cases $\left( \frac{(ug_u)_{uu}}{g_{uu}}\right) _u=0$ and $\left( \frac{(ug_u)_{uu}}{g_{uu}}\right) _u\ne 0.$

Case II.1.1.2.1)Suppose $\left( \frac{(ug_u)_{uu}}{g_{uu}}\right) _u= 0$ in (52) then we obtain

$$\begin{aligned} \frac{(ug_u)_{uu}}{g_{uu}}=p_1, \,\,p_1 \text{ is } \text{ arbitrary } \text{ constant. } \end{aligned}$$

(53)

From (53) we can consider two cases $p_1=0$ and $p_1\ne 0.$

Case II.1.1.2.1.1)Suppose $p_1= 0$ in (53) then $g(u)=up_2+p_3\ln {(u)}+p_4$ where $p_2,p_3,p_4$ are arbitrary constants. From (51) we have $d_{2,t}=0$ implying $d_2=k_3-$arbitrary constant. From (48a) we get $d_7=0.$ From (47) we obtain $\xi ^{1}=k_1$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2\ln {(u)}+s_3,$ and $g(u)=up_2+p_3\ln {(u)}+p_4$.

Case II.1.1.2.1.2)Suppose $p_1\ne 0$ in (53) then from (51)

$$\begin{aligned} d_{2,t}(p_1-1)=0. \end{aligned}$$

(54)

From (54) we can consider two cases $p_1=1$ and $p_1\ne 1.$

Case II.1.1.2.1.2.1)Suppose $p_1=1$ in (54) then from (53) we obtain $g(u)=p_2u\ln {(u)}-p_3+up_4$ where $p_2,p_3,p_4$ are arbitrary constants. From (49a) we have $d_{2,tt}+p_2d_{2,t}=0$ implying $d_2(t)=\frac{k_4e^{-p_2t}}{p_2}+k_5$ where $k_4,k_5$ are arbitrary constants. From (48b) we obtain

$$\begin{aligned} p_3k_4=0. \end{aligned}$$

(55)

From (55) we can consider two cases $p_3=0$ and $p_3\ne 0.$

Case A.):

Suppose $p_3=0$ in (55) then $g(u)=p_2u\ln {(u)}+up_4.$ From (48a) we have $d_7=0.$ From (47) we get $\xi ^{1}=k_1,\,\,\,\xi ^{2}=\frac{k_4e^{-p_2t}}{p_2}+k_5$ and $\eta =u\left( \frac{-k_4e^{-p_2t}}{2}\right) .$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\,\Pi _2=\partial _x$ and $\Pi _{26}=e^{-p_2t}\left( \frac{1}{p_2}\partial _x-\frac{u}{2}\partial _u\right)$ with $f(u)=s_2\ln {(u)}+s_3,$ and $g(u)=p_2u\ln {(u)}+up_4.$

Case B.):

Suppose $p_3\ne 0$ in (55) then $k_4=0$ implying $d_2(t)=k_5.$ From (48a) we obtain $d_7=0.$ From (47) we have $\xi ^{1}=k_1$ and $\xi ^{2}=k_5.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2\ln {(u)}+s_3,$ and $g(u)=p_2u\ln {(u)}+up_4-p_3.$

Case II.1.1.2.1.2.2)Suppose $p_1\ne 1$ in (54) then $d_{2,t}=0$ implying $d_2=k_3.$ From (53) we have $g(u)=\frac{(up_2+p_3)}{1-p_1}+\frac{p_3}{p_1(p_1-1)}+s_5u^{p_1}$ where $p_2,p_3,s_5$ are arbitrary constants. From (48a) we get $d_7=0.$ From (47) we obtain $\xi ^{1}=k_1$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2\ln {(u)}+s_3,$ and $g(u)=\frac{(up_2+p_3)}{1-p_1}+\frac{p_3}{p_1(p_1-1)}+s_5u^{p_1}.$

Case II.1.1.2.2)Suppose $\left( \frac{(ug_u)_{uu}}{g_{uu}}\right) _u\ne 0$ in (52) then $d_{2,t}=0$ implying $d_2=k_3-$arbitrary constant. Differentiating in (48a) with respect to u we obtain

$$\begin{aligned} ug_{uu}(4d_7)+2s_2^2d_7\ln {(u)}+2d_7s_2s_3+4d_{7,t}=0. \end{aligned}$$

(56)

Differentiating in (56) with respect to u we obtain

$$\begin{aligned} d_7\left( 2(ug_{uu})_u+s_2^2u^{-1}\right) =0. \end{aligned}$$

(57)

From (57) we can consider two cases $d_7=0$ and $d_3\ne 0.$

Case II.1.1.2.2.1)Suppose $d_7=0$ in (57) then From (47) we get $\xi ^{1}=k_1$ and $\xi ^{2}=k_3.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_2\ln {(u)}+s_3$ and $\left( \frac{(ug_u)_{uu}}{g_{uu}}\right) _u\ne 0$.

Case II.1.1.2.2.2) Suppose $d_7\ne 0$ in (57) then $2(ug_{uu})_u+s_2^2u^{-1}=0$ and solving this expression for g we have $g(u)=-\frac{s_2^2(\ln {(u)})^2}{4}-m_1\ln {(u)}+m_2$ where $m_1,m_2$ arbitrary constants. From (56) we obtain $2d_{7,t}+(2m_1+s_2s_3)d_7=0$ implying $d_7(t)=m_3e^{-\frac{t(2m_1+s_2s_3)}{2}}.$ From (48a) we have $d_7=0,$ but this is a contradiction with $d_7\ne 0.$ Therefore in this case we do not have symmetries.

Case II.1.2)Suppose $s_1\ne 0$ in (46) then $f(u)=s_4(u-s_1)^{-1}$ where $s_4$ is arbitrary constant. From (44) we have $c_{3,xx}s_1+c_{4,xx}=0.$ From (13b) we get $c_{3,xx}=0$ and $c_{4,xx}=0$ implying $c_3=xd_3(t)+d_4(t)$ and $c_4=xd_5(t)+d_6(t)$ where $d_3,d_4,d_5,d_6$ are arbitrary functions. From (13a) we obtain $d_3=d_5=c_{1,tt}=0$ then $c_1(t)=tk_3+k_4$ where $k_3,k_4$ are arbitrary constants. From (12) we obtain $d_4=d_6=k_3=d_{2,t}=0$ then $d_2=k_5-$arbitrary constant. From (11) we have $\xi ^{1}=k_4$ and $\xi ^{2}=k_5.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $f(u)=s_4(u-s_1)^{-1}$ and $g-$arbitrary function.

Case II.2) Suppose $\left( \frac{(uf_u)_u}{f_{uu}}\right) _u\ne 0$ in (45) then $c_{3,xx}=0$ implying $c_3=xd_5(t)+d_6(t)$ where $d_5,d_6$ are arbitrary functions. From (44) we have $c_{4,xx}=0$ then $c_4=xd_7(t)+d_8(t)$ where $d_7,d_8$ are arbitrary functions. From (13a) we obtain

$$\begin{aligned} f_u(ud_5+d_7)-\frac{c_{1,tt}}{2} =0, \end{aligned}$$

(58)

From (58) we can consider two cases $d_5=0$ and $d_5\ne 0.$

Case II.2.1) Suppose $d_5=0$ in (58) then $f_u(d_7)-\frac{c_{1,tt}}{2} =0$ and differentiating this expression with respect to u we have $f_{uu}d_7=0$ but we know that $f_{uu}\ne 0$, therefore $d_7=0.$ Hence from (58) we obtain $c_{1,tt}=0$ then $c_1(t)=k_3t+k_4$ where $k_3,k_4$ are arbitrary constants. From (12) we get

$$\begin{aligned} f_u(ud_6+d_8)+f(\frac{k_3}{2})-d_{2,t}=0. \end{aligned}$$

(59)

From (59) we can consider two cases $d_6=0$ and $d_6\ne 0.$

Case II.2.1.1) Suppose $d_6=0$ in (59) then $(d_8)f_u+(\frac{k_3}{2})f-d_{2,t}=0$ and differentiating this expression with respect to u we have

$$\begin{aligned} f_{uu}d_8+\frac{k_3}{2}f_u=0. \end{aligned}$$

(60)

From (60) we can consider two cases $k_3=0$ and $k_3\ne 0.$

Case II.2.1.1.1) Suppose $k_3=0$ in (60) then $d_8=0$ implying $d_{2,t}=0$ and so $d_2=k_5-$arbitrary constant. From (11) we get $\xi ^{1}=k_4$ and $\xi ^{2}=k_5.$ Therefore the group of symmetries is $\Pi _1=\partial _t$ and $\Pi _2=\partial _x$ with $\left( \frac{(uf_u)_u}{f_{uu}}\right) _u\ne 0$ and $g-$arbitrary function.

Case II.2.1.1.2) Suppose $k_3\ne 0$ in (60) then $f(u)=\frac{2s_2d_8}{k_3}e^{-\frac{uk_3}{2d_8}}+s_3$ where $s_2,s_3$ are arbitrary constants. Thus $d_{2,t}=\frac{k_3s_3}{2}$ implying $d_2=t\frac{k_3s_3}{2}+k_5$ where $k_5$ is arbitrary constant. From (11) we have $\xi ^{1}=tk_3+k_4,\,\,\,\xi ^{2}=\frac{xk_3}{2}+t\frac{k_3s_3}{2}+k_5$ and $\eta =d_8.$ From (9d) we obtain $d_8g_u+k_3g+d_{8,t}=0$ and differentiating this expression with respect to u we have $g_{uu}d_8+k_3g=0$ then solving we obtain $g(u)=\frac{s_4d_8}{k_3}e^{-\frac{uk_3}{d_8}}+s_5$ where $s_4,s_5$ are arbitrary constants. Thus $d_{8,t}=-k_3s_5$ then $d_{8}=k_6-tk_3s_5.$ Hence $\xi ^{1}=tk_3+k_4,\,\,\,\xi ^{2}=\frac{xk_3}{2}+t\frac{k_3s_3}{2}+k_5$ and $\eta =-tk_3s_5+k_6.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x,\,\Pi _3=\partial _u$ and $\Pi _{27}=t\partial _t+\frac{(x+s_3t)}{2}\partial _x-ts_5\partial _u$ with $f(u)=\frac{2s_2(k_6-tk_3s_5)}{k_3}e^{-\frac{uk_3}{2(k_6-tk_3s_5)}}+s_3$ and $g(u)=\frac{s_4(k_6-tk_3s_5)}{k_3}e^{-\frac{uk_3}{(k_6-tk_3s_5)}}+s_5$.

Case II.2.1.2) Suppose $d_6\ne 0$ in (59) then we have

$$\begin{aligned} f_u(ud_6+d_8)+f(\frac{k_3}{2})-d_{2,t}=0. \end{aligned}$$

(61)

From (61) we can consider two cases $k_3=0$ and $k_3\ne 0.$

Case II.2.1.2.1) Suppose $k_3=0$ in (61) then $f_u(ud_6+d_8)-d_{2,t}=0$ and differentiating this expression with respect to u we get $f_u(d_6)+f_{uu}(ud_6+d_8)=0$ and solving this expression with respect f we obtain $f(u)=\frac{s_1}{d_6}\ln {(ud_6+d_8)}+s_2$ where $s_1,s_2$ are arbitrary constants. From (61) we obtain $d_{2,t}=s_1$ implying $d_2(t)=ts_1+s_3$ where $s_3-$arbitrary constant. From (11) we have $\xi ^{1}=k_4,\,\,\xi ^{2}=ts_1+s_3$ and $\eta =ud_6+d_8.$ From (9d) we get

$$\begin{aligned} (ud_6+d_8) g_u+g(-d_6)+ud_{6,t}+d_{8,t}=0, \end{aligned}$$

(62)

Differentiating twice in (62) with respect to u we have respectively

$$\begin{aligned} (ud_6+d_8) g_{uu}+d_{6,t}&=0, \end{aligned}$$

(63a)

$$\begin{aligned} g_{uu}d_6+ (ud_6+d_8) g_{uuu}&=0. \end{aligned}$$

(63b)

From (63b) we can consider two cases $g_{uuu}=0$ and $g_{uuu}\ne 0.$

Case II.2.1.2.1.1) Suppose $g_{uuu}=0$ in (63b) then $g(u)=\frac{s_5u^2}{2}+us_6+s_7$ where $s_5,s_6,s_7$ are arbitrary constants. From (63b) we get $s_5d_6=0$ implying $s_5=0.$ From (63a) we have $d_{6,t}=0$ then $d_6=k_5.$ From (62) we obtain

$$\begin{aligned} d_{8,t}+s_6d_8(t)-s_7k_5=0. \end{aligned}$$

(64)

From (64) we can consider two cases $s_6=0$ and $s_6\ne 0.$

Case II.2.1.2.1.1.1) Suppose $s_6=0$ in (64) then $d_{8,t}-s_7k_5=0$ implying $d_8(t)=ts_7k_5+s_8.$ Thus $\xi ^{1}=k_4,\,\,\xi ^{2}=ts_1+s_3$ and $\eta =uk_5+ts_7k_5+s_8.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x,\,\Pi _{20}=\partial _u,$ $\Pi _{28}=ts_1\partial _x$ and $\Pi _{29}=(u+ts_7)\partial _u$ with $f(u)=\frac{s_1}{k_5}\ln {(uk_5+ts_7k_5+s_8)}+s_2$ and $g(u)=s_7$.

Case II.2.1.2.1.1.2) Suppose $s_6\ne 0$ in (64) then $d_8(t)=s_8e^{-s_6t}+\frac{s_7k_5}{s_6}$ where $s_8-$arbitrary constant. Thus $\xi ^{1}=k_4,\,\,\xi ^{2}=ts_1+s_3$ and $\eta =uk_5+s_8e^{-s_6t}+\frac{s_7k_5}{s_6}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x,\,\Pi _{30}=t\partial _u,$ $\Pi _{31}=(u+\frac{s_7}{s_6})\partial _u$ and $\Pi _{32}=s_8e^{-s_6t}\partial _u$ with $f(u)=\frac{s_1}{k_5}\ln {(uk_5+s_8e^{-s_6t}+\frac{s_7k_5}{s_6})}+s_2$ and $g(u)=us_6+s_7$.

Case II.2.1.2.1.2) Suppose $g_{uuu}\ne 0$ in (63b) then we get

$$\begin{aligned} \left( \frac{ g_{uu}}{g_{uuu}}\right) d_6+ (ud_6+d_8) =0. \end{aligned}$$

(65)

Differentiating in (65) we have

$$\begin{aligned} d_6\left( \left( \frac{ g_{uu}}{g_{uuu}}\right) _u+1\right) =0, \end{aligned}$$

(66)

From (66) we get $\left( \frac{ g_{uu}}{g_{uuu}}\right) _u+1=0$ and solving for g we have $g(u)=s_4\left( -(p_2-u)\ln (p_2\right.$$\left. -u)\right) -s_4u+s_5u+s_6$ where $p_2,s_4,s_5,s_6$ are arbitrary constants. From (65) we obtain $d_8=-p_2d_6.$ From (63a) we get $s_4d_6+d_{6,t}=0$ implying $d_6(t)=s_7e^{-s_4t}$ and $d_8=-p_2s_7e^{-s_4t}.$ From (62) we have $s_4-s_5 =0$ and $p_2s_4-s_6 =0$ implying $s_4=s_5$ and $s_6=p_2s_5.$ Hence $\xi ^{1}=k_4,\,\,\xi ^{2}=ts_1+s_3$ and $\eta =us_7e^{-s_5t}-p_2s_7e^{-s_5t}.$ Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x,\,\Pi _{33}=t\partial _x,$ and $\Pi _{34}=e^{-s_5t}(u-p_2)\partial _u$ with $f(u)=\frac{s_1e^{s_5t}}{s_7}\ln {(us_7e^{-s_5t}-p_2s_7e^{-s_5t})}+s_2$ and $g(u)=s_5\left( -(p_2-u)\ln {(p_2-u)}\right) +s_6$.

Case II.2.1.2.2) Suppose $k_3\ne 0$ in (61) then we get

$$\begin{aligned} f_u(ud_6+d_8)+f(\frac{k_3}{2})-d_{2,t}=0. \end{aligned}$$

(67)

Differentiating in (67) with respect to u we get

$$\begin{aligned} f_u(d_6+\frac{k_3}{2})+f_{uu}(ud_6+d_8)=0. \end{aligned}$$

(68)

Differentiating twice in (68) with respect to u we have respectively

$$\begin{aligned} \left( \frac{ f_u}{f_{uu}}\right) _u(d_6+\frac{k_3}{2})+ d_6&=0, \end{aligned}$$

(69a)

$$\begin{aligned} \left( \frac{ f_u}{f_{uu}}\right) _{uu}(d_6+\frac{k_3}{2})&=0. \end{aligned}$$

(69b)

From (69b) we get $\left( \frac{ f_u}{f_{uu}}\right) _{uu}=0$ implying $\left( \frac{ f_u}{f_{uu}}\right) _{u}=p_1.$ From (69a) we obtain $d_6=-\frac{p_1k_3}{2(p_1+1)}.$ From (68) we have $d_8=\frac{p_2p_1k_3}{2(p_1+1)}-p_2k_3.$ From (67) we get $s_2=p_2=d_{2,t}=0$ implying $d_2=k_5,\,d_8=0$ and $f(u)=\frac{s_1(up_1)^{\frac{1}{p_1}+1}}{(p_1+1)}.$ From (11) we have $\xi ^{1}=tk_3+k_4,\,\,\xi ^{2}=\frac{xk_3}{2}+k_5$ and $\eta =u\left( -\frac{p_1k_3}{2(p_1+1)}\right) .$ From (9d) we obtain $g(u)=m_1u^{\frac{(3p_1+2)}{p_1}}$ where $m_1$ is arbitrary constant. Therefore the group of symmetries is $\Pi _1=\partial _t,\,\Pi _2=\partial _x$ and $\Pi _{35}=t\partial _t+\frac{x}{2}\partial _x-\frac{p_1u}{2(p_1+1)}\partial _u$ with $f(u)=\frac{s_1(up_1)^{\frac{1}{p_1}+1}}{(p_1+1)}$ and $g(u)=m_1u^{\frac{(3p_1+2)}{p_1}}$.

Case II.2.2) Suppose $d_5\ne 0$ in (58) then

$$\begin{aligned} f_u(ud_5+d_7)-\frac{c_{1,tt}}{2}=0. \end{aligned}$$

(70)

Differentiating with respect u in (70) we have $f_u(d_5)+f_{uu}(ud_5+d_7)=0$ which is equivalent to

$$\begin{aligned} \left( \frac{f_u}{f_{uu}}\right) (d_5)+ud_5+d_7=0. \end{aligned}$$

(71)

Differentiating twice with respect to u in (71) we have respectively

$$\begin{aligned} \left( \frac{f_u}{f_{uu}}\right) _u(d_5)+d_5&=0, \end{aligned}$$

(72a)

$$\begin{aligned} \left( \frac{f_u}{f_{uu}}\right) _{uu}(d_5)&=0. \end{aligned}$$

(72b)

From (72b) we get $\left( \frac{f_u}{f_{uu}}\right) _{uu}=0$ implying $\left( \frac{f_u}{f_{uu}}\right) =um_1+m_2$ where $m_1,m_2$ are arbitrary constants. From (72a) we have $d_5(m_1+1)=0$ implying $m_1=-1.$ From (71) we obtain $d_7=-m_2d_5.$ Thus $f(u)=w_1\ln {(m_2-u)}+w_2$ where $w_1,w_2$ are arbitrary constants. From (70) we get $m_2=0$ and $c_{1,tt}=w_1d_5.$ From (12) we obtain $c_{1,t}=0$ but this is a contradiction with $0=w_1d_5$ because we know that $w_1,d_5\ne 0,$ so in this case we have no symmetries. Therefore we have finished all possible cases, and this ends the proof of the Proposition 1. $\square$

Remark 1

Note that if we consider the Case I.1.2 of Proposition 1 in Table 1, by choosing $s_1=1,\,s_2=0$ and $k_3=0$, we obtain $B=0$ and $\epsilon =s_3k_1^3$. This leads us to derive the symmetry group for equation (2), which coincides with the findings presented in [21]. The resulting symmetry group is as follows:

$$\begin{aligned} X_{1}=\partial _{t},\quad X_{2}=\partial _{x},\quad X_{3}=x \partial _{x}+2t \partial _{t}-u \partial _{u}, \,\,\,\epsilon \ne 0\, \textit{and} \, \gamma , \delta =0. \end{aligned}$$

(73)

Remark 2

Note that if we consider the Case I.1.2.1 of Proposition 1 in Table 1, by selecting $s_1=1$, $s_2,s_4=0$ and $\gamma =-s_3$, we derive the symmetry group for equation (2). Remarkably, this symmetry group aligns precisely with the findings presented in [21]. The resulting symmetry group is as follows:

$$\begin{aligned} X_{1}=\partial _{t},\quad X_{2}=\partial _{x},\quad X_{3}=e^{t\gamma } \partial _{x}+\gamma e^{t\gamma }\partial _{u},\quad \gamma \ne 0 \,\,\textit{and}\,\, \delta , \epsilon =0. \end{aligned}$$

(74)

Remark 3

Note that if we consider the Case I.1.2.2 of Proposition 1 in Table 1, by selecting $s_1=2$, $s_2,s_4=0$ and $\nu =1$, we obtain the symmetry group for the Burgers equation that coincides with the findings presented in [32, 33].

Optimal System

In this section, we present the optimal system associated to the symmetry group of (2) using the symmetry group (73). Taking into account [32, 34,35,36,37], we have a systematic approach to obtaining potential reductions.

To begin with, we should calculate the corresponding commutator table, which can be derived from the operator.

$$\begin{aligned}{}[\Pi _{\omega },\Pi _{\tau }]=\Pi _\omega \Pi _\tau -\Pi _\tau \Pi _\omega =\sum _{i=1}^n\left( \Pi _\omega (\xi _{\tau }^i)-\Pi _\tau (\xi _{\omega }^i) \right) \frac{\partial }{\partial x^i}, \end{aligned}$$

(75)

where $i=1,2,3$, with $\omega , \tau =1,2,3$ and $\xi _{\omega }^i,\xi _{\tau }^i$ are the corresponding coefficients of the infinitesimal operators $\Pi _{\omega },\Pi _{\tau }.$ After applying the operator (75) to the symmetry Group of (73), we obtain the operators that are shown in the following table

Next we calculate the adjoint action representation of the symmetries (73) and to do that, we use Table 3 and the operator

$$\begin{aligned} {\text {Ad}}\left( e^{\lambda \Pi }\right) G=\sum _{n=0}^{\infty } \frac{\lambda ^n}{n!}({\text {Ad}}(\Pi ))^nG\,\,\,\text {for the symmetries} \,\,\Pi \,\, \text {and}\,\,G. \end{aligned}$$

Making use of this operator, we can construct the Table 4, which shows the adjoint representation for each $\Pi _i.$

Table 3 Commutators table associated to the symmetry group of (73)

Table 4 Adjoint representation of the symmetry group (73)

Proposition 2

The optimal system corresponding to the symmetry group (73) is described by the vector fields

$$\begin{aligned} \Pi _3\,\,;\,\,\,b_1\Pi _1+b_2\Pi _2\,\,\,\text {with}\,\,b_2>0\,\,;\,\,\,b_3\Pi _1\,\,\,\text {with}\,\,b_3>0. \end{aligned}$$

Proof

We initiate by considering a generic nonzero vector and the symmetry generators (73). Let

$$\begin{aligned} G=a_{1} \Pi _{1}+a_{2} \Pi _{2}+a_{3} \Pi _{3}. \end{aligned}$$

(76)

The aim is to streamline as many coefficients $a_i$ as feasible by means of adjoint mappings to G and using Table 4.

• 1) Assuming $a_3 = 1$ in (76), we have that $G = a_{1} \Pi _{1}+a_{2} \Pi _{2}+\Pi _{3}$. Applying the adjoint operator to $(\Pi _1,G)$ we get

  $$\begin{aligned} G_{1}={\text {Ad}}\left( \exp \left( \lambda _{1} \Pi _{1}\right) \right) G=\left( a_1-2\lambda _{1}\right) \Pi _1+a_2\Pi _2+\Pi _3. \end{aligned}$$

  Using $\lambda _1=\frac{a_1}{2}$, then $\Pi _1$ is eliminated, therefore $G_1 =a_2\Pi _2+\Pi _3$. Now, Applying the adjoint operator to $(\Pi _2,G_1)$ we get

  $$\begin{aligned} G_{2}={\text {Ad}}\left( \exp \left( \lambda _{2} \Pi _{2}\right) \right) G_1=(a_2-\lambda _2)\Pi _2+\Pi _3. \end{aligned}$$

  Using $\lambda _2=a_2$, then $\Pi _2$ is eliminated, then $G_2=\Pi _3$. Finally applying the adjoint operator to $(\Pi _3,G_2)$ we don’t have any reduction then we have the first element of the optimal algebra:

  $$\begin{aligned} G_2=\Pi _3. \end{aligned}$$

  This is how a reduction of the generic element (76) ends.

• 2) Assuming $a_3 = 0$ and $a_2=1$ in (76), we have that $G = a_{1} \Pi _{1}+ \Pi _{2}$. Applying the adjoint operator to $(\Pi _1,G), (\Pi _2,G)$ we don’t have any reduction, finally applying the adjoint operator to $(\Pi _3,G)$ we get

  $$\begin{aligned} G_{3}={\text {Ad}}\left( \exp \left( \lambda _{3} \Pi _{3}\right) \right) G=a_1e^{2\lambda _3}\Pi _1+e^{\lambda _3}\Pi _2. \end{aligned}$$

  Then, using $b_1=a_1e^{2\lambda _3}$ and $b_2=e^{\lambda _3}$, with $b_2>0$, we have other element of the optimal algebra:

  $$\begin{aligned} G_3=b_1\Pi _1+b_2\Pi _2 \,\,\,;\,\,\text {with}\,\,b_2>0. \end{aligned}$$

  This is how other reduction of the generic element (76) ends.

• 3) Assuming $a_3= a_2 = 0$ and $a_1=1$ in (76), we have that $G = \Pi _{1}$. Applying the adjoint operator to $(\Pi _1,G)$ and $(\Pi _2,G)$ we don’t have any reduction, on the other hand applying the adjoint operator to $(\Pi _3,G)$ we get

  $$\begin{aligned} G_{4}={\text {Ad}}\left( \exp \left( \lambda _{4} \Pi _{3}\right) \right) G= e^{2\lambda _4}\Pi _1. \end{aligned}$$

  (77)

  We don’t have any reduction, then using $\lambda _{4}=\frac{1}{2}ln(b_3)$, with $b_3>0$, we have other element of the optimal algebra

  $$\begin{aligned} G_4=b_3\Pi _1,\,\,\,\,\,\text {with} \,b_3>0. \end{aligned}$$

  (78)

  This is how other reduction of the generic element (76) ends.

$\square$

Reductions by Generators of the Optimal System

In this section, we characterize all possible reductions taking into account some operators that generate the optimal system presented in Proposition 2. For this purpose, we use the method of invariant surface condition in [38] (presented in section 3.1.3), which is given by the following equation

$$\begin{aligned} Tu_t+Xu_x=U , \end{aligned}$$

(79)

using the element $b_1\Pi _1+b_2\Pi _2$ from Proposition 2, using $b_1,b_2=1$ and the symmetries (73), under the condition (79), we obtain that $u_t+ u_x=0$, then solving this PDE, we have $u(t, x)=y(x-t)$, which is a explicit transformation for (2) as $y'-yy'+y''+\epsilon y^3=0$, which is third-order nonlinear ordinary differential equation. Using an analogous procedure with all of the elements of the optimal algebra (Proposition 2), we obtain both implicit and explicit transformations that are shown in the Table 5, with y being a function.

Table 5 Reductions for (2) using invariant surface condition and Proposition 2

Note that the equation in item 2 of the table above is an Lienard equation and can be solved as follows. Take $w(z)=y'$ and $z=\frac{1}{2}y^2.$ Thus, the equation of item 2 becomes $w(z) w'(z)-w(z)+2\epsilon z=0$ which is an Abel equation of the of the second type and this type of equation has a parameterized solution of the form

$$\begin{aligned} z=Ce^{-\int \frac{\tau d\tau }{\tau ^2-\tau +2\epsilon }}\,\,\,\,\text{ and }\,\,\,\,w=C\tau e^{-\int \frac{\tau d\tau }{\tau ^2-\tau +2\epsilon }},\,\,\,\,\text{ where } C \text{ arbitrary } \text{ constant, } \end{aligned}$$

this procedure can be found at Polyanin Handbook (see [39], section $2.2.3-2$, item 2). Carrying out an analogous procedure for the group of symmetries in (74), we obtain the commutators (Table 6) and its adjoint representation (Table 7), and we obtained the optimal system presented in Proposition 3. Finally we got the reductions associated (Table 8).

Table 6 Commutators table associated to the symmetry group of (74)

Table 7 Adjoint representation of the symmetry group (74)

Proposition 3

Let be $\gamma \ne 0$. The optimal system associated to the symmetry group (74) is given by the vector fields

$$\begin{aligned}{} & {} \Pi _1+b_3\gamma \Pi _3\,\,;\,\,\,a_1\Pi _1+a_2\Pi _2\,\,\,\text {with}\,\,a_1\ne 0\,\,; \\{} & {} a_2\Pi _2+b_1\Pi _3\,\,\,\text {with}\,\,b_1>0;\,\,a_1\Pi _1+\Pi _2+b_2\gamma \Pi _3. \end{aligned}$$

Table 8 Reductions for (2) using invariant surface condition and Proposition 3

Note that the equation in item 1 of the table above is an Lienard equation and can be solved as follows. Take $w(z)=y'$ and $z=\frac{1}{2}y^2.$ Thus, the equation of item 2 becomes $w(z) w'(z)-w(z)+\gamma z=0$ which is an Abel equation of the of the second type and this type of equation has an implicit solution of the form

$$\begin{aligned} z=w+\gamma \ln {(w-\gamma )}+c_1\,\,\,\,\text{ where } c_1 \text{ arbitrary } \text{ constant, } \end{aligned}$$

this procedure can be found at Polyanin Handbook (see [39], section $2.2.3-2$, item 2). Also note that the equation in item 3 of the table above is separable and can be solved as follows:

$$\begin{aligned} y=c_1e^{\frac{1-\gamma }{\gamma }\left( \frac{e^{-\gamma r}}{\gamma r}\right) }\,\,\,\,\text{ where } c_1 \text{ arbitrary } \text{ constant. } \end{aligned}$$

By the reductions outlined in Tables 5,8, and the expressions of some of the presented solutions, the following assertions and physical interpretations are delineated concerning the study equation (2):

The term $u_{xx}$ corresponds to small-scale diffusion, whereas the term $uu_{x}$ contributes to non-linear transport. The coefficients $\gamma$, $\delta$, and $\epsilon$ introduce additional dissipative and non-linear effects that impact the temporal evolution of the solution. By assigning different values to these coefficients, the aim is to comprehend the physical relationships between the concepts of diffusion, non-linear transport, and dissipation. It is observed that when one concept maintains a constant behavior, the others may exhibit polynomial, exponential, or logarithmic behaviors. This inverse perspective facilitates the exploration of diverse application scenarios and provides insights into how variations in the coefficients influence the overall dynamics of the system modeled by the Burgers–Huxley equation (2).

Classification of Lie Algebra

The classification of Lie algebras provides a deep and systematic mathematical framework that is essential for understanding and analyzing groups and dynamical systems across various realms of mathematics and physics. In particular, the classification of Lie algebras into semisimple and solvable algebras yields substantial benefits in the following aspects [40, 41]:

• i Group Structure: The classification enables the decomposition of related groups into more manageable components, such as semidirect products of semisimple and solvable groups. This decomposition sheds light on the internal structure of groups, simplifying their study and facilitating the analysis of fundamental properties.

• ii Problem Solvability: The distinction between semisimple and solvable algebras has practical applications in solving specific problems. In certain contexts, the structure of solvable algebras can be exploited to address particular problems more effectively.

• iii Development of Methods and Techniques: The classification of Lie algebras serves as a foundation for the development of advanced algebraic methods and techniques. These methods are employed to study dynamical systems, representation theory, and symmetry problems in various scientific fields.

• iv Applications in Theoretical Physics: In theoretical physics, the classification of Lie algebras is essential for formulating and solving differential equations, understanding symmetries in quantum field theory, and describing elementary particles. It provides a unified algebraic structure for comprehending the underlying symmetries in the fundamental laws of physics.

In this way, the study of Lie algebra enables the extension of differential equation analysis to other fields of research and mathematical interest. This provides a profound and systematic mathematical foundation that is indispensable for comprehending and analyzing groups and dynamic systems across various scientific domains. Moreover, it has practical applications in solving specific problems, advancing algebraic methods, and representing algebras in other abstract fields. This generates interest among other investigators, offering a pathway for future researchers to leverage the symmetries associated with the equations under study.

Option 1

We proceed to classify the optimal system presented in the Proposition 2.

Let be $Z_1=\Pi _3=x\partial _{x}+2t\partial _{t}-u\partial _{u}\,\,;\,Z_2=b_1\Pi _1+b_2\Pi _2=b_1\partial _{t}+b_2\partial _{x}\,\,\text {with}\,\,b_2>0\,\,;\,Z_3=b_3\Pi _1=b_3\partial _{t}\,\,\text {with}\,b_3>0.$

First, we consider the following Table 9:

Table 9 Commutators table associated to the optimal system A

Option 2

We proceed to classify the optimal system presented in the Proposition 3.

Let be $Z_1=\Pi _1+\gamma \Pi _3=\partial _{t}+\gamma e^{\gamma t}\partial _{x}+\gamma ^2 e^{\gamma t}\partial _{u}\,\,;\,Z_2=\Pi _1+\Pi _2=\partial _{t}+\partial _{x} \,;\,Z_3=\Pi _2+\Pi _3=(1+e^{\gamma t})\partial _{x}+\gamma e^{\gamma t}\partial _{u}\,\,;\,\,Z_4=\Pi _1+\Pi _2+\gamma \Pi _3=\partial _{t}+(1+\gamma e^{\gamma t})\partial _{x}+\gamma ^2 e^{\gamma t}\partial _{u}.$

First, we consider the following Table 10:

Table 10 Commutators table associated to the optimal system B

Classification of Lie Algebra

Generically a finite dimensional Lie algebra in a field of characteristic 0 is classified by the Levi’s theorem, which state that any finite dimensional Lie algebra can be write as a semidirect product of a semisimple Lie algebra and a Solvable Lie algebra. Then we have two important classes to be considered, namely the solvable and the semisimple one. [42,43,44].

The next definition and proposition it will be used in the classification [45, 46].

Definition 1

Let $\mathfrak {g}$ be a finite-dimensional Lie algebra over an arbitrary field k. Choose a basis ${e_{j}}$, $1\le i \le n$, in $\mathfrak {g}$ where $n= \text {dim } \mathfrak {g}$ and set $[e_i,e_j]=C_{ij}^{k}e_{k}$. Then the coefficients $C_{ij}^{k}$ are called structure constants.

Proposition 4

Let $\mathfrak {g}_{1}$ and $\mathfrak {g}_{2}$ be two Lie algebras of dimension n. Suppose each has a basis with respect to which the structure constant are the same. Then $\mathfrak {g}_{1}$ and $\mathfrak {g}_{2}$ are isomorphic.

Lie Algebra Option 1

Let us call $\mathfrak {g}$ the Lie algebra generated by the infinitesimals (73). Considering the following non vanish relations. $[\Pi _1,\Pi _2]=2\Pi _1$,

$[\Pi _2,\Pi _3]=\Pi _2$. the derived algebra is $\mathfrak {g}^{(1)}=span \{2\Pi _1,\Pi _2\}$, then $\mathfrak {g}^{(2)}=\{0\}$. Consequently this Lie algebra is solvable. Furthermore, according by the Bianchi classification, this Lie algebra it isomorphic to the solvable Bianchi V Lie algebra $\mathfrak {g}_{3,3}$.

Lie Algebra Option 1, Optimal System

Taking in account the generators of the optimal system and the relations given by the table (9) this is a indecomposable solvable Lie algebra.

In fact we obtain that $\mathfrak {g}^{(1)}=span \{-2Z_2-\frac{b_1}{b_3}Z_3,-2 Z_3\}$, and $\mathfrak {g}^{(2)}=span \{0\}$.

Lie Algebra Option 2

Let us call $\mathfrak {g}$ the Lie algebra generated by the infinitesimals (74). Considering the following non vanish relations. $[\Pi _1,\Pi _3]=\gamma \Pi _3$.

We remark that the only solvable two dimensional Lie algebra is $\mathfrak {aff}(\mathbb {R})$ with the relations $[e_1,e_2]=e_1$ for a chosen basis $\{e_1,e_2\}$.

The derived algebra is $\mathfrak {g}^{(1)}=span \{\gamma \Pi _3\}$, and so $\mathfrak {g}^{(2)}=\{0\}$. Then this Lie algebra is solvable. Furthermore, it is a decomposable solvable Lie algebra. In fact, we can write this Lie algebra as $\mathfrak {g}=\mathbb {R}\oplus \mathfrak {g}_2$, where $\mathfrak {g}_2$ is isomorphic to the two solvable non nilpotent Lie algebra $\mathfrak {aff}(\mathbb {R})$. Now, by doing the next assignment, $e_{1}:=\Pi _{3}$, and $e_{2}=:\frac{1}{\gamma }\Pi _1$, we obtain $[e_{1},e_{2}]=e_{1}$ and by proposition (4) the Lie algebra $\mathfrak {g}_2$ is isomorphic with $\mathfrak {aff}(\mathbb {R})$. Consequently the Lie algebra associated with the symmetries (74) is isomorphic to $\mathbb {R}\oplus \mathfrak {aff}(\mathbb {R})$.

Lie Algebra Option 2, Optimal System

Taking in account the generators of the optimal system and the relations given by the table (10), we obtain that this Lie algebra is a decomposable solvable Lie algebra. Now, By doing the following assignment, $e_{1}:=Z_1$,   $e_{2}:=Z_2-Z_4$,   $e_{3}:=Z_1-Z_4$,   $e_{4}=Z_2+\gamma Z_3+Z_4$, we obtain $[e_1,e_2]=\gamma e_2$, and by doing again a change of basis, $h_{1}:=-\frac{1}{\gamma }e_1$, and $h_2=e_{2}$, it holds that $[h_2,h_1]=h_2$. By using the proposition 4 with the structure constants of the Lie algebra $\mathfrak {aff}(\mathbb {R})$ we conclude that this Lie algebra is isomorphic to the Lie algebra $\mathbb {R}^2\oplus \mathfrak {aff}(\mathbb {R})$.

Conclusions

By applying the Lie symmetry group (see Propositions 1), we successfully computed the principal algebra for a generalization of the Burgers–Huxley equation. Through the analysis of Lie symmetries and the optimal system (for the Burgers–Huxley equation (2)), we obtained various reductions and invariant solutions (see Sect. 4). Remarkably, these reductions and solutions are not documented in the existing literature to date. Also, the symmetry classification (3) was presented for the first time in the literature.

Additionally, we conducted a comprehensive classification of the Lie algebra for the Lie group and the associated optimal system linked to Burgers–Huxley equation (2). Consequently, we have effectively accomplished our original objective.

The representations obtained in Sect. 5, provide a strong motivation for future studies aimed at understanding the structures governed by the Lie algebras associated with these groups. In particular, the description of the optimal algebras for option 1, which is indecomposable, and the decomposable option 2 are of significant interest. The latter option ensures a more comprehensive study as it suffices to investigate the subalgebras associated with its decomposition.

In the same vein, concerning the representation of the Lie algebra of the symmetry group (73) and (74), further exploration can be conducted in future studies on geometry and solution spaces linked to the algebraic structure associated with the transformation group.

Moreover, in future investigations, the theory of equivalence groups could be considered to obtain preliminary classifications associated with a comprehensive classification of Equation (2). Additionally, we aim to explore nonclassical symmetries as part of our research.