1 Introduction

Nonlinear evolution equations (NLEEs) have been applied in fluid mechanics, plasma physics and nonlinear optics [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37]. Researchers have proposed certain methods for solving the NLEEs, such as the Darboux transformation, Bäcklund transformation, inverse scattering transformation, Lie group, consistent Riccati expansion, power-series expansion, polynomial expansion, \(\left( \frac{G^{'}}{G}\right) \) expansion and Hirota bilinear methods [38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53]. The Lie group method has been used to look for the group invariance and certain reductions to the NLEEs [38, 54,55,56,57]. Based on the reduced equations, certain solutions for the NLEEs have been constructed [17, 18, 58, 59].

A (2+1)-dimensional generalized Bogoyavlensky–Konopelchenko (gBK) equation in fluid mechanics and plasma physics has been constructed as [60]

$$\begin{aligned}&u_{xt}+k_{1}u_{xxxx}+k_{2}u_{xxxy}+\frac{2k_1k_3}{k_2}u_{x}u_{xx}+k_{3}u_{x}u_{xy}+k_{3}u_{xx}u_{y}\nonumber \\&\quad +\,\gamma _{1}u_{xx} +\gamma _{2}u_{xy}+\gamma _{3}u_{yy}=0, \end{aligned}$$
(1)

where \(u=u(x,y,t)\) is a real function, x and y are the scaled space variables, t is the scaled time variable, \(k_1\), \(k_2\), \(k_3\), \(\gamma _1\), \(\gamma _2\) and \(\gamma _3\) are the real constants and the subscripts denote the partial derivatives. Lump-type and lump solutions for Eq. (1) have been derived via the Hirota bilinear method [60].

Special cases for Eq. (1) in fluid mechanics and plasma physics have been seen:

  • When \(k_3=3\), \(k_2=1\) and \(k_1=\gamma _1=\gamma _2=\gamma _3=0\), by virtue of a dimensional reduction \(u_{y}\)=\(u_{x}\) and a potential function transformation h(x, t) = \(u_{x}(x,t)\), Eq. (1) has been reduced to the Korteweg-de Vries equation,

    $$\begin{aligned} h_{t}+h_{xxx}+6hh_{x}=0, \end{aligned}$$
    (2)

    for certain shallow water waves, stratified internal waves in a fluid or ion-acoustic waves in a plasma [61,62,63,64,65,66].

  • When \(k_2=1\), \(k_3=4\) and \(k_1=\gamma _1=\gamma _2=\gamma _3=0\), Eq. (1) has been reduced to the Bogoyavlenskii’s breaking soliton equation,

    $$\begin{aligned} u_{xt}+u_{xxxy}+4u_{xx}u_{y}+4u_{x}u_{xy}=0, \end{aligned}$$
    (3)

    for the interaction of a Riemann wave propagating along the y axis and a long wave propagating along the x axis in fluid mechanics [67].

  • When \(k_2=1\), \(k_3=3\) and \(k_1=\gamma _1=0\), Eq. (1) has been reduced to the generalized Calogero–Bogoyavlenskii–Schiff equation [68],

    $$\begin{aligned} u_{xt}+u_{xxxy}+3u_{xx}u_{y}+3u_{x}u_{xy}+\gamma _2u_{xy}+\gamma _3u_{yy}=0. \end{aligned}$$
    (4)

However, to our knowledge, the Lie point symmetry generators, Lie symmetry group and symmetry reductions for Eq. (1) have not been discussed. Hyperbolic-function, power-series, trigonometric-function, soliton and rational solutions for Eq. (1) have not yet been investigated via the Lie group method. In Sect. 2, the Lie point symmetry generators and Lie symmetry group for Eq. (1) will be derived. In Sect. 3, symmetry reductions as well as hyperbolic-function, power-series, trigonometric-function, soliton and rational solutions for Eq. (1) will be obtained through the Lie point symmetry generators. In Sect. 4, the conclusions will be given.

2 Lie group analysis for Eq. (1)

2.1 Lie point symmetry generators for Eq. (1)

According to the Lie group method [54], we consider a one-parameter Lie group of the infinitesimal transformations acting on the independent and dependent variables as

$$\begin{aligned}&{\tilde{x}}=x+\epsilon \xi (x,y,t,u)+O(\epsilon ^2),\nonumber \\&{\tilde{y}}=y+\epsilon \eta (x,y,t,u)+O(\epsilon ^2),\nonumber \\&{\tilde{t}}=t+\epsilon \tau (x,y,t,u)+O(\epsilon ^2),\nonumber \\&{\tilde{u}}=u+\epsilon \phi (x,y,t,u)+O(\epsilon ^2), \end{aligned}$$
(5)

where \({\tilde{x}}\), \({\tilde{y}}\), \({\tilde{t}}\), \({\tilde{u}}\), \(\xi \), \(\eta \), \(\tau \) and \(\phi \) are the real functions of x, y, t and u, \(\epsilon \) is a parameter of the infinitesimal transformation and \(O(\epsilon ^2)\) is the infinitesimal of the same order of \(\epsilon ^2\).

Lie point symmetry generators for Eq. (1) are

$$\begin{aligned} V=\xi (x,y,t,u)\frac{\partial }{\partial x}+\eta (x,y,t,u)\frac{\partial }{\partial y}+\tau (x,y,t,u)\frac{\partial }{\partial t}+\phi (x,y,t,u)\frac{\partial }{\partial u} \end{aligned}$$
(6)

with \(\xi \), \(\eta \), \(\tau \) and \(\phi \) satisfying the condition

$$\begin{aligned} Pr^{(4)}V(E)|_{E=0}=0, \end{aligned}$$
(7)

where

$$\begin{aligned} E= & {} u_{xt}+k_{1}u_{xxxx}+k_{2}u_{xxxy}+\frac{2k_1k_3}{k_2}u_{x}u_{xx}+k_{3}u_{x}u_{xy}\nonumber \\&+\,k_{3}u_{xx}u_{y}+\gamma _{1}u_{xx} +\gamma _{2}u_{xy}+\gamma _{3}u_{yy}. \end{aligned}$$
(8)

\(Pr^{(4)}V(\cdot )\) represents the fourth prolongation of V, defined as [69],

$$\begin{aligned} \,\,\,\,Pr^{(4)}V(\cdot )= & {} \xi \frac{\partial }{\partial x}(\cdot )+\eta \frac{\partial }{\partial y}(\cdot )+\tau \frac{\partial }{\partial t}(\cdot )+\phi \frac{\partial }{\partial u}(\cdot )\nonumber \\&+\,\phi ^{x}\frac{\partial }{\partial u_{x}}(\cdot )+\phi ^{y}\frac{\partial }{\partial u_{y}}(\cdot )+\phi ^{xx}\frac{\partial }{\partial u_{xx}}(\cdot )+\phi ^{xy}\frac{\partial }{\partial u_{xy}}(\cdot )\nonumber \\&+\,\phi ^{yy}\frac{\partial }{\partial u_{yy}}(\cdot )+\phi ^{xt}\frac{\partial }{\partial u_{xt}}(\cdot )+\phi ^{xxxx}\frac{\partial }{\partial u_{xxxx}}(\cdot )+\phi ^{xxxy}\frac{\partial }{\partial u_{xxxy}}(\cdot ),\nonumber \\&\!\!\phi ^{x}=D_{x}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{xx}+\eta u_{xy}+\tau u_{xt},\nonumber \\&\!\!\phi ^{y}=D_{y}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{yx}+\eta u_{yy}+\tau u_{yt},\nonumber \\&\!\!\phi ^{t}=D_{t}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{tx}+\eta u_{ty}+\tau u_{tt},\nonumber \\&\!\!\phi ^{xx}=D_{x}^{2}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{xxx}+\eta u_{xxy}+\tau u_{xxt},\nonumber \\&\!\!\phi ^{xt}=D_{x}D_{t}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{xxt}+\eta u_{xyt}+\tau u_{xtt},\nonumber \\&\!\!\phi ^{xy}=D_{x}D_{y}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{xxy}+\eta u_{xyy}+\tau u_{xyt},\nonumber \\&\!\!\phi ^{yy}=D_{y}^{2}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{xyy}+\eta u_{yyy}+\tau u_{yyt},\nonumber \\&\!\!\phi ^{xxxx}=D_{x}^{4}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\xi u_{xxxxx}+\eta u_{xxxxy}+\tau u_{xxxxt},\nonumber \\&\!\!\phi ^{xxxy}=D_{x}^{3}D_{y}(\phi -\xi u_{x}-\eta u_{y}-\tau u_{t})+\,\xi u_{xxxxy}+\eta u_{xxxyy}+\tau u_{xxxyt}, \end{aligned}$$
(9)

where \(D_{x}\), \(D_{y}\) and \(D_{t}\) are the total derivative operators.

Expanding Expression (7) and splitting on the derivatives of u lead to the following expressions:

$$\begin{aligned}&\tau _{x}=\tau _{y}=\tau _{u}=\xi _{u}=\eta _{x}=\eta _{u}=0,\nonumber \\&\phi _{uu}=\phi _{xu}=\phi _{yu}=\xi _{xx}=\xi _{xy}=0,\nonumber \\&\frac{2k_1k_3}{k_2}\phi _{x}-\xi _{t}+k_{3}\phi _{y}+\gamma _{1}\tau _{t}-\gamma _{1}\xi _{x}-\gamma _2\xi _{y}=0,\nonumber \\&\phi _{u}-\xi _{x}+\tau _{t}-\eta _{y}=0,~k_1\tau _{t}-k_2\xi _{y}-3k_1\xi _{x}=0,\nonumber \\&k_3\phi _{x}-\eta _{t}+\gamma _{2}\tau _{t}-\gamma _2\eta _{y}=0,~\phi _{xy}+\phi _{tu}-\xi _{xt}=0,\nonumber \\&\phi _{xt}+\gamma _2\phi _{xy}=0,~\tau _{t}+\xi _{x}-\phi _{u}=0,~\tau _{t}-2\xi _{x}-\eta _{y}=0. \end{aligned}$$
(10)

Solving Expressions (10), we have the following results:

$$\begin{aligned} \tau= & {} 2r_2t+r_3,\nonumber \\ \eta= & {} r_2(-2\gamma _{2}t+4y)+k_{3}r_{1}t+r_5,\nonumber \\ \phi= & {} r_{1}x+r_{2}u+F_{1}^{'}(t)y+F_{2}(t),\nonumber \\ \xi= & {} -r_2 x+\frac{5k_1r_2}{k_2}y+k_3F_{1}(t)+\frac{2k_3r_1k_1-5\gamma _2r_2k_1+3\gamma _1r_2k_2}{k_2}t+r_4, \end{aligned}$$
(11)

where \(r_1\), \(r_2\), \(r_3\), \(r_4\) and \(r_5\) are the real constants, \(F_1(t)\) and \(F_2(t)\) are the real function of t, and \(F_{1}^{'}(t)\) denotes derivative of \(F_{1}(t)\) with respect to t. We derive the Lie point symmetry generators for Eq. (1) as follows:

$$\begin{aligned}&V_1=\frac{\partial }{\partial x}, ~V_2=\frac{\partial }{\partial y}, ~V_3=\frac{\partial }{\partial t},~V_4=x\frac{\partial }{\partial u}+k_3t\frac{\partial }{\partial y}+\frac{2k_1k_3t}{k_2}\frac{\partial }{\partial x},\nonumber \\&V_5=2t\frac{\partial }{\partial t}-x\frac{\partial }{\partial x}+\frac{5k_1y}{k_2}\frac{\partial }{\partial x}+\frac{(-5k_1\gamma _2+3\gamma _1k_2)t}{k_2}\frac{\partial }{\partial x}+(-2\gamma _2t+4y)\frac{\partial }{\partial y}+u\frac{\partial }{\partial u},\nonumber \\&V_{[F_{1}(t)]}^{[1]}=k_3F_{1}(t)\frac{\partial }{\partial x}+F_{1}^{'}(t)y\frac{\partial }{\partial u},~V_{[F_{2}(t)]}^{[2]}=F_{2}(t)\frac{\partial }{\partial u}. \end{aligned}$$
(12)

Motivated by Ref. [55], commutation relations among Generators (12) are shown in Table 1, where the entries in row i and column j are represented by the commutators \([V_{i}, V_{j}]\), which are given by

$$\begin{aligned}{}[V_{i}, V_{j}]=V_{i}V_{j}-V_{j}V_{i}, (i,j=1,2,3,4,5,F_{1},F_{2}). \end{aligned}$$
(13)
Table 1 Commutator table of the Lie algebra for Eq. (1)
Full size table

Motivated by Refs. [58, 59], we take \(F_1(t) = r_6t\), \(F_2(t) = r_7t+r_8\) with the real constants \(r_6\), \(r_7\) and \(r_8\).

Thus, we derive the Lie point symmetry generators for Eq. (1) as follows:

$$\begin{aligned} \Gamma _1= & {} \frac{\partial }{\partial x}, \Gamma _2=\frac{\partial }{\partial y}, \Gamma _3=\frac{\partial }{\partial t}, \Gamma _4=t\frac{\partial }{\partial u}, \Gamma _5=\frac{\partial }{\partial u},\nonumber \\ \Gamma _6= & {} k_3t\frac{\partial }{\partial x}+y\frac{\partial }{\partial u}, \Gamma _7=x\frac{\partial }{\partial u}+k_3t\frac{\partial }{\partial y}+\frac{2k_1k_3t}{k_2}\frac{\partial }{\partial x},\nonumber \\ \Gamma _8= & {} 2t\frac{\partial }{\partial t}-x\frac{\partial }{\partial x}+\frac{5k_1y}{k_2}\frac{\partial }{\partial x}+\frac{\left( -5k_1\gamma _2+3\gamma _1k_2\right) t}{k_2}\frac{\partial }{\partial x}+(-2\gamma _2t+4y)\frac{\partial }{\partial y}+u\frac{\partial }{\partial u}.\nonumber \\ \end{aligned}$$
(14)

2.2 Lie symmetry group for Eq. (1)

In order to obtain the group transformation, which is generated by the infinitesimal generators \(\Gamma _{i}\), we need to solve the following initial problems:

$$\begin{aligned}&\frac{\text {d}{\bar{x}}(\epsilon )}{\text {d}\epsilon }=\xi [{\bar{x}}(\epsilon ) ,{\bar{y}}(\epsilon ),{\bar{t}}(\epsilon ),{\bar{u}}(\epsilon )],~{\bar{x}}|_{\epsilon =0}=x,\nonumber \\&\frac{\text {d}{\bar{y}}(\epsilon )}{\text {d}\epsilon }=\eta [{\bar{x}}(\epsilon ) ,{\bar{y}}(\epsilon ),{\bar{t}}(\epsilon ),{\bar{u}}(\epsilon )],~{\bar{y}}|_{\epsilon =0}=y,\nonumber \\&\frac{\text {d}{\bar{t}}(\epsilon )}{\text {d}\epsilon }=\tau [{\bar{x}}(\epsilon ) ,{\bar{y}}(\epsilon ),{\bar{t}}(\epsilon ),{\bar{u}}(\epsilon )],~{\bar{t}}|_{\epsilon =0}=t,\nonumber \\&\frac{\text {d}{\bar{u}}(\epsilon )}{\text {d}\epsilon }=\phi [{\bar{x}}(\epsilon ) ,{\bar{y}}(\epsilon ),{\bar{t}}(\epsilon ),{\bar{u}}(\epsilon )],~{\bar{u}}|_{\epsilon =0}=u. \end{aligned}$$
(15)

Then, the Lie symmetry group \(g_{i}^{,}s\) generated by \(V_{i}\) can be derived as

$$\begin{aligned} g_1:(x,y,t,u)\rightarrow & {} (x+\epsilon ,y,t,u),~g_2:(x,y,t,u)\rightarrow (x,y+\epsilon ,t,u),\nonumber \\ g_3:(x,y,t,u)\rightarrow & {} (x,y,t+\epsilon ,u),~g_4:(x,y,t,u)\rightarrow (x,y,t,u+t\epsilon ),\nonumber \\ g_5:(x,y,t,u)\rightarrow & {} (x,y,t,u+\epsilon ),~g_6(x,y,t,u)\rightarrow (x+k_3t\epsilon ,y,t,u+y\epsilon ),\nonumber \\ g_7:(x,y,t,u)\rightarrow & {} \left( x+\frac{2k_1k_3}{k_2}t\epsilon ,y+k_3t\epsilon ,t,u+x\epsilon \right) ,\nonumber \\ g_8:(x,y,t,u)\rightarrow & {} \left[ \frac{(y-\gamma _2t)(e^{4\epsilon }-e^{-\epsilon }) +\gamma _{1}k_2t\left( e^{2\epsilon }-e^{-\epsilon }\right) }{k_2}\nonumber \right. \\&\left. \quad +\,xe^{-\epsilon },\gamma _2te^{2\epsilon }+(y-\gamma _2t)e^{4\epsilon }, te^{2\epsilon },ue^{\epsilon }\right] . \end{aligned}$$
(16)

On account of Lie Symmetry Group (16), if \({\bar{f}}(x, y, t)\) is a known solutions for Eq. (1), the corresponding solutions can be obtained as

$$\begin{aligned}&u^{(1)}={\bar{f}}(x-\epsilon ,y,t),\nonumber \\&u^{(2)}={\bar{f}}(x,y-\epsilon ,t),\nonumber \\&u^{(3)}={\bar{f}}(x,y,t-\epsilon ),\nonumber \\&u^{(4)}=t\epsilon +{\bar{f}}(x,y,t),\nonumber \\&u^{(5)}=\epsilon +{\bar{f}}(x,y,t),\nonumber \\&u^{(6)}=y\epsilon +{\bar{f}}(x-k_3t\epsilon ,y,t),\nonumber \\&u^{(7)}=x\epsilon +{\bar{f}}\left( x-\frac{2k_1k_3}{k_2}t\epsilon ,y-k_3t\epsilon ,t\right) ,\nonumber \\&u^{(8)}=e^{\epsilon }{\bar{f}}\left[ xe^{-\epsilon }+\frac{(y-\gamma _2t)\left( e^{-4\epsilon }-e^{\epsilon }\right) +\gamma _{1}k_2t(e^{-2\epsilon }-e^{\epsilon })}{k_2},\nonumber \right. \\&\left. (y-\gamma _{2})e^{-4\epsilon }+\gamma _{2}te^{-2\epsilon },te^{-2\epsilon }\right] . \end{aligned}$$
(17)

3 Symmetry reductions and analytic solutions for Eq. (1)

In this section, we use the combination of Generators (14) to derive the reduction equations and construct some analytic solutions for Eq. (1).

Case 1: For the Lie point symmetry \(V_2= \partial _ y\), we have the following group-invariant solutions:

$$\begin{aligned} u=H(x_1,t_1), \end{aligned}$$
(18)

where \(x_1 = x\), \(t_1 = t\) and H is a function of \(x_1\) and \(t_1\). Substituting Expression (18) into Eq. (1) gives rise to the following reduced equation:

$$\begin{aligned} H_{x_1t_1}+k_1H_{x_1x_1x_1x_1}+\frac{2k_1k_3}{k_2}H_{x_1}H_{x_1x_1}+\gamma _1H_{x_1x_1}=0. \end{aligned}$$
(19)

We suppose that the solutions for Eq. (1) be as follows:

$$\begin{aligned} H(x_1,t_1)=\frac{ae^{cx_1+dt_1+k}}{b+e^{cx_1+dt_1+k}}, \end{aligned}$$
(20)

where a, b, c, d and k are the real constants. Substituting Expression (20) into Eq. (19), we can obtain

$$\begin{aligned} a=\frac{6ck_2}{k_3},~b_1=-k_1c^{3}-\gamma _1c. \end{aligned}$$
(21)

Therefore, soliton solutions for Eq. (1) can be derived as

$$\begin{aligned} u=\frac{6ck_2e^{cx+\left( -k_1c^{3}-\gamma _1c\right) t+k}}{k_3\left[ b+e^{cx+\left( -k_1c_{1} -\gamma _1c\right) t+k}\right] }. \end{aligned}$$
(22)

      Case 2: For the Lie point symmetry \(V_3= \partial _ t\), we have the following group-invariant solutions:

$$\begin{aligned} u=P(x_2,y_2), \end{aligned}$$
(23)

where \(x_2 = x\) and \(y_2 = y\), while P is a function of \(x_2\) and \(y_2\). Substituting Expression (23) into Eq. (1) gives rise to the following reduced equation:

$$\begin{aligned}&k_{1}P_{x_2x_2x_2x_2}+k_{2}P_{x_2x_2x_2y_2}+\frac{2k_1k_3}{k_2}P_{x_2}P_{x_2x_2}+k_{3}P_{x_2}P_{x_2y_2}\nonumber \\&\quad +k_{3}P_{x_2x_2}P_{y_2}+\gamma _{1}P_{x_2x_2}+\gamma _{2}P_{x_2y_2}+\gamma _{3}P_{y_2y_2}=0. \end{aligned}$$
(24)

Applying the Lie group method on Eq. (24), we obtain

$$\begin{aligned} \xi _1=s_1x_2+\frac{5k_1s_1}{k_2}y_2+s_2,~\eta _1=4s_1y_2+s_3,~\phi _1=s_1P-\frac{\gamma _1s_1}{k_1}x_1+F_3(y_2), \end{aligned}$$
(25)

where \(s_1\), \(s_2\) and \(s_3\) are the real constants, while \(F_3(y_2)\) is a real function of \(y_2\). Motivated by Refs. [58, 59], we take \(F_3(y_2) = s_4\) with a real constant \(s_4\). Thus, we derive the Lie point symmetry generators for Eq. (24) as follows:

$$\begin{aligned} \Gamma _1= & {} \frac{\partial }{\partial x_2},~\Gamma _2=\frac{\partial }{\partial y_2},~\Gamma _3=\frac{\partial }{\partial P},~\nonumber \\ \Gamma _4= & {} \left( \frac{5k_1y_2}{k_2}-x_2\right) \frac{\partial }{\partial x_2}+4y_2\frac{\partial }{\partial y_2}+\left( P-\frac{\gamma _1x_2}{k_1}\right) \frac{\partial }{\partial P}. \end{aligned}$$
(26)

For the Lie point symmetry \(\Gamma _1 +\Gamma _2\), the symmetry produces the following invariants:

$$\begin{aligned} f=x_2-y_2,~P=\Phi (f), \end{aligned}$$
(27)

where \(\Phi \) is a real function of f. Substituting Expressions (27) into Eq. (24) gives rise to the following reduced equation:

$$\begin{aligned} (k_1-k_2)\Phi _{ffff}+\left( \frac{2k_1k_3}{k_2}-2k_3\right) \Phi _{f}\Phi _{ff}+(\gamma _1-\gamma _2+\gamma _3)\Phi _{ff}=0. \end{aligned}$$
(28)

We suppose that some solutions for Eq. (28) have the following form:

$$\begin{aligned} \Phi =\sum _{i=0}^{m}a_i\left( \frac{G^{'}}{G}\right) ^{i}, \end{aligned}$$
(29)

where m is a positive integer, while \(a_{i}\)’s are the real constants. Here, G satisfies the second-order linear ordinary differential equation, i.e.,

$$\begin{aligned} G^{''}+BG^{'}+AG=0, \end{aligned}$$
(30)

where \(G^{'}=\frac{\mathrm{d}G}{\mathrm{d}f}\) and \(G^{''}=\frac{\mathrm{d}^2G}{\mathrm{d}f^2}\), while A and B are the real constants. m can be determined via homogeneous balance method between the highest order derivative term and the nonlinear term appearing in Eq. (28). We get \(m=1\). Substituting Eq. (29) into Eq. (28) with Constraint (30) and setting the coefficients of \((\frac{G^{'}}{G})\) equal to zero, we obtain

$$\begin{aligned} a_1=\frac{6k_2}{k_3},~A=\frac{B^2k_1-B^2k_2+\gamma _1-\gamma _2+\gamma _3}{4(k_1-k_2)}. \end{aligned}$$
(31)

When \(\sqrt{B^2-4A}>0\), we derive some solutions for Eq. (1) as

$$\begin{aligned} u(x,y,t)= & {} \frac{6k_2}{k_3}\left\{ \sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_2-k_1)}} \frac{C_1\cosh \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_2-k_1)}}\right] +C_2\sinh \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_2-k_1)}}\right] }{C_2\cosh \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_2-k_1)}}\right] +C_1\sinh \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_2-k_1)}}\right] }\right\} \nonumber \\&+\,a_0-\frac{3Bk_2}{k_3}, \end{aligned}$$
(32)

where \(C_1\) and \(C_2\) are the real constants.

When \(\sqrt{B^2-4A}=0\), we derive some solutions for Eq. (1) as

$$\begin{aligned} u(x,y,t)=a_0+\frac{6k_2}{k_3}\left[ -\frac{B}{2}+\frac{C_4}{C_3+C_4(x-y)}\right] , \end{aligned}$$
(33)

where \(C_3\) and \(C_4\) are the real constants.

When \(\sqrt{B^2-4A}<0\), we derive some solutions for Eq. (1) as

$$\begin{aligned} u(x,y,t)= & {} \frac{6k_2}{k_3}\left\{ \sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_1-k_2)}} \frac{C_6\cos \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_1-k_2)}}\right] -C_5\sin \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_1-k_2)}}\right] }{C_5\cos \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_1-k_2)}}\right] +C_6\sin \left[ (x-y)\sqrt{\frac{\gamma _1-\gamma _2+\gamma _3}{4(k_1-k_2)}}\right] }\right\} \nonumber \\&+\,a_0-\frac{3Bk_2}{k_3}, \end{aligned}$$
(34)

where \(C_5\) and \(C_6\) are the real constants.

Case 3: For the Lie point symmetry \(V_{(1)}= V_1+V_2+V_3+V_5\), we have the following group-invariant solutions:

$$\begin{aligned} f_1=x-t,h_1=y-t,u=t+R(f_1,h_1), \end{aligned}$$
(35)

where R is a function of \(f_1\) and \(h_1\). Substituting Expressions (35) into Eq. (1) gives rise to the following reduced equation:

$$\begin{aligned}&-R_{f_1h_1}-R_{f_1f_1}+k_{1}R_{f_1f_1f_1f_1}+k_{2}R_{f_1f_1f_1h_1}+\frac{2k_1k_3}{k_2}R_{f_1}R_{f_1f_1}\nonumber \\&\quad +k_{3}R_{f_1}R_{f_1h_1}+k_{3}R_{f_1f_1}R_{h_1}+\gamma _{1}R_{f_1f_1}+\gamma _{2}R_{f_1h_1}+\gamma _{3}R_{h_1h_1}=0. \end{aligned}$$
(36)

Applying the Lie group method on Eq. (36), we obtain

$$\begin{aligned} \xi _2= & {} -s_5f_1+\frac{5k_1s_5h_1}{k_2}+s_6,\eta _2=4s_5h_1+s_7,\nonumber \\ \phi _2= & {} s_5R+\frac{2{(\gamma _2-1)}s_5f_1}{3k_2} -\frac{s_5\left[ 3(\gamma _1-1)k_2-k_1(\gamma _2-1)\right] h_1}{3k_2^{2}}+s_8, \end{aligned}$$
(37)

where \(s_5\), \(s_6\), \(s_7\) and \(s_8\) are the real constants. Thus, we derive the Lie point symmetry generators for Eq. (36) as follows:

$$\begin{aligned} \Upsilon _1= & {} \frac{\partial }{\partial f_1},~\Upsilon _2=\frac{\partial }{\partial h_1},~\Upsilon _3=\frac{\partial }{\partial R},\nonumber \\ ~\Upsilon _4= & {} \frac{3k_2^2R+2k_2(\gamma _2-1)f_1-h_1[3(\gamma _1-1)k_2-k_1(\gamma _2-1)]}{3k_2^2}\frac{\partial }{\partial R}\nonumber \\&+\frac{5k_1h_1-k_2f_1}{k_2}\frac{\partial }{\partial f_1}+4h_1\frac{\partial }{\partial h_1}. \end{aligned}$$
(38)

For the Lie point symmetry \(n_1\Upsilon _1 +n_2\Upsilon _2\), the symmetry produces the following invariants:

$$\begin{aligned} z=n_2f_1-n_1h_1,~R=Q(z), \end{aligned}$$
(39)

where \(n_1\) and \(n_2\) are the real constants, and Q is a real function of z. Substituting Expressions (39) into Eq. (36) gives rise to the following reduced equation:

$$\begin{aligned}&n_2^2\left( \frac{2k_1k_3n_2-2n_1k_2k_3}{k_2}\right) Q_{zz}Q_{z}+n_2^3(k_1n_2-n_1k_2)Q_{zzzz}\nonumber \\&\quad +\left( -n_2^2\gamma _1+n_1n_2\gamma _2+n_1^2\gamma _3-n_1n_2+n_2^2\right) Q_{zz}=0. \end{aligned}$$
(40)

Seeking the solutions for Eq. (40) in a power series of the form

$$\begin{aligned} Q=\sum _{q=0}^{\infty }c_qz^q, \end{aligned}$$
(41)

and substituting Expression (41) into Eq. (40), we obtain

$$\begin{aligned}&24c_4n_2^3(n_2k_1-n_1k_2)+n_2^3(n_2k_1-n_1k_2)\sum _{q=1}^{\infty }(q+4)(q+3)(q+2)(q+1)c_{q+4}z^q\nonumber \\&\quad +n_2^2\left( \frac{2k_1k_3n_2-2k_3k_2n_1}{k_2}\right) \sum _{q=1}^{\infty }\sum _{k=0}^{q}(k+1)(q-k+2)(q-k+1) c_{k+1}c_{q-k+2}z^q\nonumber \\&\quad +2n_2^2\left( \frac{2k_1k_3n_2-2k_3n_1k_2}{k_2}\right) c_1c_2+2\left( -n_2^2\gamma _1 +n_1n_2\gamma _2+n_1^2\gamma _3-n_1n_2+n_2^2\right) c_2\nonumber \\&\quad +\left( -n_2^2\gamma _1+n_1n_2\gamma _2+n_1^2\gamma _3-n_1n_2+n_2^2\right) \ \sum _{q=1}^{\infty }(q+2)(q+1)c_{q+2}z^q=0, \end{aligned}$$
(42)

where \(c_q\)’s are the real constants. From Expression (42), equating the coefficients of each order of z, we can calculate \(c_{q}\) for the case of \(q= 0\), so that

$$\begin{aligned} c_4=\frac{n_2^2(2k_1k_3n_2-2k_3k_2n_1)c_1c_2+(-n_2^2\gamma _1+n_1n_2\gamma _2+n_1^2\gamma _3-n_1n_2+n_2^2)k_2c_2}{12k_2n_2^3(k_2n_1-k_1n_2)}. \end{aligned}$$
(43)

For \(n\ge 1\), we obtain

$$\begin{aligned} c_{q+4}= & {} \frac{1}{(q+4)(q+3)(q+2)(q+4)(k_2n_1-k_1n_2)}\nonumber \\&\left[ \left( -n_2^2\gamma _1+n_1n_2\gamma _2+n_1^2\gamma _3-n_1n_2+n_2^2\right) (q+2)(q+1)c_{q+2}\nonumber \right. \\&\left. +n_2^2\left( \frac{2k_1k_3n_2-2k_3k_2n_1}{k_2}\right) \sum _{k=0}^{q}(k+1)(q-k+2)(q-k+1) c_{k+1}c_{q-k+2}\right] .\nonumber \\ \end{aligned}$$
(44)

Then, we derive the power series solutions for Eq. (1) as

$$\begin{aligned} u(x,y,t)= & {} c_0+c_1[n_2(x-t)-n_1(y-t)]+c_2[n_2(x-t)-n_1(y-t)]^2\nonumber \\&+c_3[n_2(x-t)-n_1(y-t)]^3+\frac{[n_2(x-t)-n_1(y-t)]^4}{12k_2n_2^3(k_2n_1-k_1n_2)}\nonumber \\&\left[ n_2^2(2k_1k_3n_2-2k_3k_2n_1)c_1c_2+\left( -n_2^2\gamma _1+n_1n_2\gamma _2+n_1^2\gamma _3 -n_1n_2+n_2^2\right) k_2c_2\right] \nonumber \\&+\sum _{q=1}^{\infty }\frac{[n_2(x-t)-n_1(y-t)]^{q+4}}{(q+4)(q+3)(q+2)(q+4)(k_2n_1-k_1n_2)}\nonumber \\&\left[ n_2^2\left( \frac{2k_1k_3n_2-2k_3k_2n_1}{k_2}\right) \sum _{k=0}^{q}(k+1)(q-k+2)(q-k+1) c_{k+1}c_{q-k+2}\nonumber \right. \\&\left. +\left( -n_2^2\gamma _1+n_1n_2\gamma _2+n_1^2\gamma _3-n_1n_2+n_2^2\right) (q+2)(q+1)c_{q+2}\right] +t. \end{aligned}$$
(45)

      Case 4: For the Lie point symmetry \(V_{(2)}= V_1+V_2+V_3\), we have the following group-invariant solutions:

$$\begin{aligned} f_2=x-t,h_2=y-t,u=S(f_2,h_2). \end{aligned}$$
(46)

where S is a function of \(f_2\) and \(h_2\). Substituting Expressions (46) into Eq. (1) gives rise to the following reduced equation:

$$\begin{aligned} \begin{aligned}&-S_{f_2h_2}-S_{f_2f_2}+k_{1}S_{f_2f_2f_2f_2}+k_{2}S_{f_2f_2f_2h_2}+\frac{2k_1k_3}{k_2}S_{f_2}R_{f_2f_2}\\&\quad +k_{3}S_{f_2}S_{f_2h_2}+k_{3}S_{f_2f_2}S_{h_2}+\gamma _{1}S_{f_2f_2}+\gamma _{2}S_{f_2h_2}+\gamma _{3}S_{h_2h_2}=0. \end{aligned} \end{aligned}$$
(47)

Applying the Lie group method on Eq. (47), we obtain

$$\begin{aligned} \begin{aligned}&\xi _2=-s_9f_2+\frac{5k_1s_9h_2}{k_2}+s_{10},\eta _2=4s_9h_2+s_{11},\\&\phi _2=s_9S+\frac{2{(\gamma _2-1)}s_9f_2}{3k_2} -\frac{s_9[3(\gamma _1-1)k_2-k_1(\gamma _2-1)]h_2}{3k_2^{2}}+s_{12}, \end{aligned} \end{aligned}$$
(48)

where \(s_9\), \(s_{10}\), \(s_{11}\) and \(s_{12}\) are the real constants. Thus, we derive the Lie point symmetry generators for Eq. (46) as follows:

$$\begin{aligned} \Theta _1= & {} \frac{\partial }{\partial f_2},~\Theta _2=\frac{\partial }{\partial h_2},~\Theta _3=\frac{\partial }{\partial S},\nonumber \\ ~\Theta _4= & {} \frac{3k_2^2S+2k_2(\gamma _2-1)f_2-h_2[3(\gamma _1-1)k_2-k_1(\gamma _2-1)]}{3k_2^2}\frac{\partial }{\partial S}\nonumber \\&+\frac{5k_1h_2-k_2f_2}{k_2}\frac{\partial }{\partial f_2}+4h_2\frac{\partial }{\partial h_2}. \end{aligned}$$
(49)

For the Lie point symmetry \(n_3\Theta _1 +\Theta _2\), the symmetry produces the following invariants:

$$\begin{aligned} z_1=f_2-n_3h_2,~L=S(z_1), \end{aligned}$$
(50)

where \(n_3\) is a real constant and L is a real function of \(z_1\). Substituting Expressions (50) into Eq. (47) gives rise to the following reduced equation:

$$\begin{aligned}&\left( \frac{2k_1k_3-2n_3k_2k_3}{k_2}\right) L_{z_1z_1}L_{z_1}+(k_1-n_3k_2)L_{z_1z_1z_1z_1}\nonumber \\&\quad +(\gamma _1-n_3\gamma _2+n_3^2\gamma _3+n_3-1)L_{z_1z_1}=0. \end{aligned}$$
(51)

We suppose that the solutions for Eq. (51) have the following form:

$$\begin{aligned} L=b_0+\sum _{j=1}^{M}b_jW(z_1)^{j}+\sum _{j=1}^{M}d_jW(z_1)^{-j}, \end{aligned}$$
(52)

where M is a positive integer, \(b_0\), \(b_{j}\)’s and \(d_{j}\)’s are the real constants, W satisfies

$$\begin{aligned} \frac{dW}{dz_1}=W^{2}+p_1z_1+p_2, \end{aligned}$$
(53)

while \(p_1\) and \(p_2\) are the real constants. M can be determined via homogeneous balance method between the highest order derivative term and the nonlinear term appearing in Eq. (51). We get \(M=1\). Substituting Expression (52) into Eq. (51) with Constraint (53) and setting the coefficients of \(W(z_1)\) equal to zero, we obtain the following results:

Case 4.1:

$$\begin{aligned} b_1=p_1=0,~d_1=\frac{k_2(\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1)}{2k_3(k_1-n_3k_2)}, ~p_2=\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{4(k_1-n_3k_2)}. \end{aligned}$$

Hereby, the solutions for Eq. (1) are obtained as

$$\begin{aligned} u=b_0+\frac{k_2(\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1)\cot \left\{ [c+x-t-n_3(y-t)]\sqrt{\frac{\gamma _3n_3^2 -\gamma _2n_3+\gamma _1+n_3-1}{4(k_1-n_3k_2)}}\right\} }{2k_3(k_1-n_3k_2) \sqrt{\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{4(k_1-n_3k_2)}}},\nonumber \\ \end{aligned}$$
(54)

where c is a real constant.

Case 4.2:

$$\begin{aligned} d_1=p_1=0,~b_1=-\frac{6k_2}{k_3}, ~p_2=\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{4(k_1-n_3k_2)}. \end{aligned}$$

Hereby, the solutions for Eq. (1) are obtained as

$$\begin{aligned} u=b_0-\frac{6k_2\sqrt{\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{4(k_1-n_3k_2)}}\tan \left\{ [c+x-t-n_3(y-t)]\sqrt{\frac{\gamma _3n_3^2 -\gamma _2n+\gamma _1+n_3-1}{4(k_1-n_3k_2)}}\right\} }{k_3 }.\nonumber \\ \end{aligned}$$
(55)

Case 4.3:

$$\begin{aligned} \begin{aligned}&p_1=0,~p_2=\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{16(k_1-n_3k_2)},\\&b_1=-\frac{6k_2}{k_3},~d_1=\frac{3k_2(\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1)}{8k_3(k_1-n_3k_2)}. \end{aligned} \end{aligned}$$

Hereby, the solutions for Eq. (1) are obtained as

$$\begin{aligned} \begin{aligned} u&=b_0-\frac{6k_2\sqrt{\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{16(k_1-n_3k_2)}}\tan \left\{ [c+x-t-n_3(y-t)]\sqrt{\frac{\gamma _3n_3^2 -\gamma _2n_3+\gamma _1+n_3-1}{16(k_1-n_3k_2)}}\right\} }{k_3 }\\&\quad +\frac{3k_2(\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1)\cot \left\{ [c+x-t-n_3(y-t)]\sqrt{\frac{\gamma _3n_3^2 -\gamma _2n_3+\gamma _1+n_3-1}{16(k_1-n_3k_2)}}\right\} }{8k_3(k_1-n_3k_2) \sqrt{\frac{\gamma _3n_3^2-\gamma _2n_3+\gamma _1+n_3-1}{16(k_1-n_3k_2)}}}. \end{aligned} \end{aligned}$$
(56)

4 Conclusions

In this paper, a (2+1)-dimensional gBK equation in fluid mechanics and plasma physics, i.e., Eq. (1), has been investigated. Lie Point Symmetry Generators (14) and Lie Symmetry Group (16) for Eq. (1) have been derived via the Lie group method. Symmetry Reductions (19), (28), (40) and (51) for Eq. (1) have been obtained from Cases 1-4. Soliton Solutions (22), Hyperbolic-Function Solutions (32), Rational Solutions (33), Power-Series Solutions (45) as well as Trigonometric-Function Solutions (34) and (54)–(56) for Eq. (1) have been derived.