Exact solutions of (1 + 1)-dimensional integro-differential Ito, KP hierarchy, CBS, MCBS and modified KdV-CBS equations.

1 Introduction

The present study computes the Lie symmetries and exact solutions of some problems modeled by nonlinear partial differential equations. The (1 + 1)-dimensional integro-differential Ito, the first integro-differential KP hierarchy, the Calogero-Bogoyavlenskii-Schiff (CBS), the modified Calogero-Bogoyavlenskii-Schiff (CBS), and the modified KdV-CBS equations are some of the problems for which we want to find new exact solutions. We employ similarity variables to reduce the number of independent variables and inverse similarity transformations to obtain exact solutions to the equations under consideration. The sine-cosine method is then utilized to determine the exact solutions.

Nonlinear partial differential equations (NLPDEs) have been utilized to characterize various nonlinear occurrences in mathematical biology, physics, and several other areas of science and engineering. The most important problem in real-world phenomena is computing exact solutions of nonlinear PDEs. The homogeneous balance method [[1]], the Darboux transform method [[2]], the first integral method [[4]], the tanh function method [[6]], the modified simple equation method [[7]], the method of the auxiliary equation [[9]], the (G′/G)-expansion method [[11]], the F-expansion method [[13]], Jacobi elliptic function method [[15]], and Lie symmetry method [[17]–[19]] are some of the important methods available to compute exact solutions of nonlinear PDEs. Although there is no universal strategy for solving nonlinear PDEs, Lie symmetry analysis is one of the most effective and reliable techniques for discovering new exact solutions to nonlinear PDEs arising in applied mathematics and physics. For the application of some well-known methods to compute numerical and exact solutions of differential equations, the interested reader is referred to see [[20]–[25]].

In this paper, the Lie point symmetry method is applied to nonlinear systems. The symmetry reductions related to the nonlocal symmetry can be analyzed in [[26]] where the truncated Painlevé analysis or the Möbious invariant form yields the nonlocal symmetry of the Gardner equation. Here, the extended system locates the nonlocal symmetry to the local Lie point symmetries. So the nonlocal symmetry is used to find possible reductions in symmetry using the localization technique. In [[27]] the nonlocal symmetries for the (2+1)-dimensional Konopelchenko–Dubrovsky equation are determined with the shortened Painlevé method and the Möbious (conformal) invariant form. Here the nonlocal symmetries are reduced to the Lie point symmetries by inserting auxiliary dependent variables. So, we can generate finite symmetry transformations by solving the initial value problem of the prolonged systems. Similarly using the truncated Painlevé approach and the Möbius (conformal) invariant form, we can drive the nonlocal symmetry for the Drinfel'd-Sokolov-Wilson equation in [[28]]. Meanwhile, for the use of symmetry reductions related to nonlocal symmetry, the nonlocal symmetry will be localized to the Lie point symmetry by introducing three dependent variables.

Recently, Li, Tian, Yang, and Fan have done some interesting work in deriving the solutions of the Wadati-Konno-Ichikawa equation and complex short pulse equation with the help of the Dbar-steepest descent method. They solved the long-time asymptotic behavior of the solutions of these equations and proved the soliton resolution conjecture and the asymptotic stability of solutions of these equations. (See: [[29]–[31]]).

The (1 + 1)- dimensional integro-differential Ito equation is a well known NLPDE, It can be governed by

Graph

$$\begin{array}{c}\hfill u_{tt}+u_{xxxt}+3(2u_x{u}_t+u{u}_{xt})+3u_{xx}{\partial }_x^{-1}\left(u_t\right)=0.\end{array}$$

The mathematical representation of first integro-differential KP hierarchy equation is

Graph

$$\begin{array}{c}\hfill u_t-{\displaystyle \frac{1}{2}}{u}_{xxy}-{\displaystyle \frac{1}{2}}{\partial }_x^{-2}\left(u_{yyy}\right)-2u_x{\partial }_x^{-1}\left(u_y\right)-4u{u}_y=0.\end{array}$$

The Calogero—Bogoyavlenskii—schiff (CBS) equation is represented by

Graph

$$\begin{array}{c}\hfill v_t+v_{xxy}+4v{v}_y+2v_x{\partial }_x^{-1}\left(v_y\right)=0.\end{array}$$

The modified Calogero-Bogoyavlenskii-schiff (CBS) equation can be expressed symbolically as

Graph

$$\begin{array}{c}\hfill v_t+v_{xxy}+4v^2{v}_y+4v_x{\partial }_x^{-1}\left(v{v}_y\right)=0.\end{array}$$

The standard form of modified KdV-CBS equation is

Graph

$$\begin{array}{c}\hfill u_t-4u^2{u}_y-2u_x{\partial }^{-1}{\left(u^2\right)}_y+u_{xxy}-6u^2{u}_x+u_{xxx}=0.\end{array}$$

Developing techniques for the exact solutions of these models, which entail systems of nonlinear PDEs, has been important in the study of nonlinear PDEs.

In this article, we compute the exact solutions of five well-known NLPDEs, namely, the (1 + 1)-dimensional integro-differential Ito equation, the first integro-differential KP hierarchy equation, the Calogero-Bogoyavlenskii-Schiff (CBS) equation, the modified Calogero-Bogoyavlenskii-Schiff (CBS) equation, and the modified KdV-CBS equation. The classical Lie point symmetries are utilized to reduce the number of independent variables via similarity transformations, which ultimately give rise to the exact solutions of the prescribed equations. The sine-cosine method is also employed to derive general exact solutions. The exact solutions presented in this study are concise and straightforward and can be used to establish new solutions for other kinds of NLPDEs arising in different areas of mathematical physics.

The paper is organized in the following pattern: In Section 2, basic definitions, important relations, and the fundamental theorem of the sine-cosine method are presented. In Section 3, Lie symmetries and exact solutions of (1 + 1)-dimensional integro-differential Ito equations are constructed. In Section 4, Lie symmetries and exact solutions of the first integro-differential KP hierarchy equation are determined. In Section 5, Lie symmetries and exact solutions of the Calogero-Bogoyavlenskii-Schiff (CBS) equations are analyzed. In section 6, Lie symmetries and exact solutions of the modified Calogero-Bogoyavlenskii-Schiff (CBS) and in section 7, Lie symmetries and exact solutions of the modified Kdv-CBS equations are evaluated. In the last section, we summarize the concluding remarks.

2 Fundamental operators

Consider the following pth order system of differential equations

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$$\begin{array}{c}\hfill E_{\gamma }(x,u,u^{\left(1\right)},u^{\left(2\right)},...,u^{\left(p\right)})=0,\gamma =1,2,3,...,l,\end{array}$$ (2.1)

with m independent variables x = (x1, x2, x3, ..., xm) and n dependent variables u = (u1, u2, u3..., un).

The differential function Eγ(x, u, u(1), ..., u(p)), in (2.1), is a pth order differential invariant of a group G if

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$$\begin{array}{c}\hfill X^{\left[p\right]}\left(E\right)=0,\end{array}$$ (2.2)

where X[p] is the pth prolongation of the Lie-Bäclund or generalized operator X defined by

Graph

$$\begin{array}{c}\hfill X=\left(\frac{\partial }{\partial x^i}\right){\xi }^i(x,u)+\left(\frac{\partial }{\partial u^{\beta }}\right){\eta }^{\beta }(x,u)+\sum _{q\ge 1}{\theta }_{i_1,i_2,i_3,...,i_q}^{\beta }\frac{\partial }{\partial u_{i_1...i_q}^{\beta }},\end{array}$$ (2.3)

where

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${\theta }_{i_1,i_2,i_3,...,i_q}$ can be determined from

Graph

$$\begin{array}{c}\hfill \begin{array}{cc}& {\theta }_i^{\beta }=D_i\left({\eta }^{\beta }\right)-u_j^{\beta }{D}_i\left({\xi }^j\right),\\ & {\theta }_{i_1,i_2}^{\beta }=D_{i_2}\left({\theta }_{i_1}^{\beta }\right)-u_{j{i}_1}^{\beta }{D}_{i_2}\left({\xi }^j\right),\\ & ⋮\\ & {\theta }_{i_1,i_2,i_3,...,i_q}^{\beta }=D_{i_q}\left({\theta }_{i_1,i_2,i_3,...,i_{q-1}}^{\beta }\right)-u_{j{i}_1,i_2,i_3,...i_{q-1}}^{\beta }{D}_{i_q}\left({\xi }^j\right),q>1.\end{array}\end{array}$$

The total derivative operator with respect to xi takes the form,

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$$\begin{array}{c}\hfill D_i=\frac{\partial }{\partial x^i}+u_i^{\beta }\frac{\partial }{\partial u^{\beta }}+u_{ij}^{\beta }\frac{\partial }{\partial u_j^{\beta }}+···.\end{array}$$ (2.4)

The invariants of the differential function Eγ in (2.1) can be obtained by solving characteristic equations derived from Eq (2.2).

The Euler operator is defined as

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$$\begin{array}{c}\hfill \frac{\delta }{\delta u^{\beta }}=\frac{\partial }{\partial u^{\beta }}-D_i\frac{\partial }{\partial u_i^{\beta }}+D_i{D}_j\frac{\partial }{\partial u_{ij}^{\beta }}-···,\end{array}$$ (2.5)

where Di is the total derivative operator.

2.1 Sine-Cosine method

There is no particular method that works for all types of nonlinear evolution equations. The sine-cosine method ([[32]–[34]]) is one such technique and it can be used to solve a wide variety of nonlinear evolution equations. The sine-cosine method works for those PDEs that admit translational symmetries. If a PDE possesses translational symmetries, we may convert it to an ODE by introducing a wave variable and assuming that the solution would take the form of a sine function or cosine function. The sine-cosine algorithm is described as follows:

Sine-Cosine algorithm:

• Integrate the ODE Q(u, us, uss, usss, ...) = 0 as many times as possible and set the constants of integration to zero.
• Suppose the solution of the form

•

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$$\begin{array}{c}\hfill u\left(s\right)=\alpha \text{sin}{\left(\beta s\right)}^k,\end{array}$$ (2.6)

• or

•

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$$\begin{array}{c}\hfill u\left(s\right)=\alpha \text{cos}{\left(\beta s\right)}^k,\end{array}$$ (2.7)

• where we need to find the parameters α, β and k.
• Substitute (2.6) or (2.7) in Q(u, us, uss, usss, ...) = 0 and balance the terms of sine functions when (2.6) is used or balance the terms of cosine functions when (2.7) is utilized.
• After defining the value of k, we separate the terms concerning powers of cosine or sine functions to obtain algebraic system of equations in terms of α and β.

After computing the values of α and β and inserting them into the main equation, we get at the solution.

3 Lie symmetries and exact solutions of (1 + 1)- dimensional Integro-differential Ito equatio...

The (1 + 1)- dimensional integro-differential Ito equation is governed by

Graph

$$\begin{array}{c}\hfill u_{tt}+u_{xxxt}+3(2u_x{u}_t+u{u}_{xt})+3u_{xx}{\partial }_x^{-1}\left(u_t\right)=0.\end{array}$$ (3.1)

Eq (3.1) can be expressed as

Graph

$$\begin{array}{c}\hfill u_{tt}+u_{xxxt}+3(2u_x{u}_t+u{u}_{xt})+3u_{xx}v=0,\\ \\ \hfill u_t=v_x,\end{array}$$ (3.2)

where

Graph

$v={\partial }_x^{-1}\left(u_t\right)and{\partial }^{-1}$ denotes the integral with respect to the subscripts.

Using Eq (2.2), the following overdetermined linear system of PDEs is obtained

Graph

$$\begin{array}{c}\hfill {\xi }_t^1=3{\xi }_x^2,{\xi }_u^1=0,{\xi }_v^1=0,{\xi }_x^1=0,{\xi }_t^2=0,{\xi }_u^2=0,\\ \hfill {\xi }_v^2=0,{\xi }_{xx}^2=0,{\eta }^1=-2{\xi }_x^{2}u,{\eta }^2=-4{\xi }_x^{2}v.\end{array}$$

The following Lie point symmetries can be generated by solving determining equations with one component equal to one and the remaining equal to zero

Graph

$$\begin{array}{c}\hfill X_1={\displaystyle \frac{\partial }{\partial x}},X_2={\displaystyle \frac{\partial }{\partial t}},X_3=3t{\displaystyle \frac{\partial }{\partial t}}+x{\displaystyle \frac{\partial }{\partial x}}-2u{\displaystyle \frac{\partial }{\partial u}}-4v{\displaystyle \frac{\partial }{\partial v}}.\end{array}$$

The translational symmetries corresponding to system (3.2) are

Graph

$$\begin{array}{c}\hfill X_1={\displaystyle \frac{\partial }{\partial x}},X_2={\displaystyle \frac{\partial }{\partial t}}.\end{array}$$ (3.3)

The following similarity transformations are obtained using combination of X1 and X2 i.e X = X1 + αX2

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$$r=-\alpha t+x,s=t,u(t,x)=v\left(r\right),u_t=-\alpha u_r,$$

which further implies

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$$\begin{array}{c}\hfill u_{tt}={\alpha }^2{u}_{rr},u_{tx}=-\alpha u_{rr},u_x=u_r,u_{xx}=u_{rr},\\ \hfill u_{xxx}=u_{rrr},v_t=-\alpha v_r,u_{xxxt}=-\alpha u_{rrrr}.\end{array}$$ (3.4)

Substituting (3.4) in system (3.2), we obtain

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$$\begin{array}{c}\hfill {\alpha }^2{u}_{rr}-\alpha u_{rrrr}+6u_r(-\alpha u_r)+3u(-\alpha u_{rr})+3u_{rr}v=0,\\ \\ \hfill v=c_1-\alpha u.\end{array}$$ (3.5)

The system (3.5) gives rise to

Graph

$$\begin{array}{c}\hfill {\alpha }^2{u}_{rr}-\alpha u_{rrrr}-6\alpha u_r^2+3c_1{u}_{rr}-6\alpha u{u}_{rr}=0.\end{array}$$

Integrating twice yields

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$$\begin{array}{c}\hfill ({\alpha }^2+3c_1)u-3\alpha u^2-\alpha u_{rr}=0.\end{array}$$ (3.6)

The particular solution of (3.6) is

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$$\begin{array}{c}\hfill u\left(r\right)=-{\displaystyle \frac{1}{2}}[\alpha tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-r)\right)}^2+{\displaystyle \frac{3}{\alpha }}tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-r)\right)}^2{c}_1-{\displaystyle \frac{{\alpha }^2+3c_1}{\alpha }}],\\ \hfill v\left(r\right)=c_1+{\displaystyle \frac{\alpha }{2}}[\alpha tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-r)\right)}^2+{\displaystyle \frac{3}{\alpha }}tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-r)\right)}^2{c}_1-{\displaystyle \frac{{\alpha }^2+3c_1}{\alpha }}].\end{array}$$ (3.7)

The set of solutions (3.7) expressed in terms of original variables are

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$$\begin{array}{c}\hfill u(t,x)=-{\displaystyle \frac{1}{2}}[\alpha tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-(x-\alpha t))\right)}^2+{\displaystyle \frac{3}{\alpha }}tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-(x-\alpha t))\right)}^2{c}_1-{\displaystyle \frac{{\alpha }^2+3c_1}{\alpha }}],\end{array}$$

and

Graph

$$\begin{array}{c}\hfill v(t,x)=c_1+{\displaystyle \frac{\alpha }{2}}[\alpha tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-(x-\alpha t))\right)}^2+{\displaystyle \frac{3}{\alpha }}tanh{\left({\displaystyle \frac{1}{2\alpha }}\delta (c_1-(x-\alpha t))\right)}^2{c}_1-{\displaystyle \frac{{\alpha }^2+3c_1}{\alpha }}],\end{array}$$

where

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$\delta =\sqrt{{\alpha }^3+3\alpha c_1}.$

Now using the sine-cosine approach [[32]–[34]], we obtain exact solutions of Eq (3.6). Suppose (3.6) has a solution such as

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$$\begin{array}{c}\hfill u\left(r\right)=\lambda co{s}^k\left(\omega r\right).\end{array}$$ (3.8)

Substituting the values of u from (3.8) in (3.6) yields

Graph

$$\begin{array}{c}\hfill {\alpha }^2\lambda co{s}^k\left(\omega r\right)-\alpha {\displaystyle \frac{{\partial }^2}{\partial r^2}}\left(\lambda co{s}^k\left(\omega r\right)\right)-3\alpha {\lambda }^2{\left(co{s}^k\left(\omega r\right)\right)}^2+3\lambda co{s}^k\left(\omega r\right)c_1=0.\end{array}$$ (3.9)

Eq (3.9) is satisfied if

Graph

$$2k=k-2.$$

Substituting k = −2 in (3.9), we obtain

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$$\begin{array}{c}\hfill {\displaystyle \frac{4\lambda \alpha {\omega }^2}{co{s}^2\left(\omega r\right)}}-{\displaystyle \frac{3{\lambda }^2\alpha }{co{s}^4\left(\omega r\right)}}-{\displaystyle \frac{6\lambda \alpha {\omega }^2}{co{s}^4\left(\omega r\right)}}+{\displaystyle \frac{\lambda {\alpha }^2}{co{s}^2\left(\omega r\right)}}+{\displaystyle \frac{3\lambda c_1}{co{s}^2\left(\omega r\right)}}=0.\end{array}$$

Comparing the coefficients of powers of

Graph

${\displaystyle \frac{1}{cos\left(\omega r\right)}}$ , we have

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$$\begin{array}{c}\hfill 4\lambda \alpha {\omega }^2+{\alpha }^2\lambda +3\lambda c_1=0,\\ \\ \hfill -3\alpha {\lambda }^2-6\alpha \lambda {\omega }^2=0.\end{array}$$ (3.10)

Solution of (3.10) gives

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$$\begin{array}{c}\hfill \lambda ={\displaystyle \frac{{\alpha }^2+3c_1}{2\alpha }},\omega =\sqrt{-{\displaystyle \frac{{\alpha }^2+3c_1}{4\alpha }}}.\end{array}$$

Eq (3.8) using the value of λ, ω and k results in

Graph

$$\begin{array}{c}\hfill u\left(r\right)={\displaystyle \frac{{\alpha }^2+3c_1}{2\alpha }}se{c}^2\left(\sqrt{-{\displaystyle \frac{{\alpha }^2+3c_1}{4\alpha }}}r\right),\end{array}$$

where r = x − αt. The solution of (3.6) in terms of original variables can be expressed as

Graph

$$\begin{array}{c}\hfill u(t,x)={\displaystyle \frac{{\alpha }^2+3c_1}{2\alpha }}se{c}^2\left(\sqrt{-{\displaystyle \frac{{\alpha }^2+3c_1}{4\alpha }}}(x-\alpha t)\right),\end{array}$$

Graph

$$\begin{array}{c}\hfill v(t,x)=c_1-{\displaystyle \frac{{\alpha }^2+3c_1}{2}}se{c}^2\left(\sqrt{-{\displaystyle \frac{{\alpha }^2+3c_1}{4\alpha }}}(x-\alpha t)\right),\end{array}$$

which constitute the exact solutions of (1 + 1)- dimensional integro-differential Ito equation (3.2).

4 Lie symmetries and exact solutions of the first integro-differential KP hierarchy equation

The first integro-differential KP hierarchy equation is governed by

Graph

$$\begin{array}{c}\hfill u_t-{\displaystyle \frac{1}{2}}{u}_{xxy}-{\displaystyle \frac{1}{2}}{\partial }_x^{-2}\left(u_{yyy}\right)-2u_x{\partial }_x^{-1}\left(u_y\right)-4u{u}_y=0,\end{array}$$ (4.1)

which can be rewritten as

Graph

$$\begin{array}{c}\hfill u_t-{\displaystyle \frac{1}{2}}{u}_{xxy}-{\displaystyle \frac{1}{2}}{v}_{yy}-2u_x{v}_x-4u{u}_y=0,\\ \\ \hfill u_y=v_{xx},\end{array}$$ (4.2)

where

Graph

$v_x={\partial }_x^{-1}\left(u_y\right)and{\partial }^{-1}$ denotes the integral with respect to the subscripts.

Using Eq (4.2), the following overdetermined linear set of PDEs are obtained

Graph

$$\begin{array}{c}\hfill {\eta }_u^2=0,{\eta }_v^2=-{\displaystyle \frac{1}{2}}{\xi }_t^1,{\eta }_x^2=-{\displaystyle \frac{1}{2}}{\xi }_t^2,{\eta }_{yy}^2=-{\displaystyle \frac{1}{2}}{\xi }_{tt}^3,{\xi }_u^1=0,{\xi }_v^1=0,\end{array}$$

Graph

$$\begin{array}{c}\hfill {\xi }_x^1=0,{\xi }_y^1=0,{\xi }_u^2=0,{\xi }_v^2=0,{\xi }_x^2={\displaystyle \frac{1}{4}}{\xi }_t^1,{\xi }_y^2=0,\end{array}$$

Graph

$$\begin{array}{c}\hfill {\xi }_u^3=0,{\xi }_v^3=0,{\xi }_x^3=0,{\xi }_y^3={\displaystyle \frac{1}{2}}{\xi }_t^1,{\eta }^1=-{\displaystyle \frac{1}{2}}u{\xi }_t^1-{\displaystyle \frac{1}{4}}{\xi }_t^3.\end{array}$$

The following Lie point symmetries can be generated by solving determining equations with one component equal to one and the remaining equal to zero

Graph

$$\begin{array}{c}\hfill X_1={\displaystyle \frac{\partial }{\partial t}},X_2={\displaystyle \frac{\partial }{\partial x}},X_3={\displaystyle \frac{\partial }{\partial y}},X_4=y{\displaystyle \frac{\partial }{\partial v}},\end{array}$$

Graph

$$\begin{array}{c}\hfill X_5={\displaystyle \frac{\partial }{\partial v}},X_6=t{\displaystyle \frac{\partial }{\partial v}},X_7=t{\displaystyle \frac{\partial }{\partial y}}-{\displaystyle \frac{1}{4}}{\displaystyle \frac{\partial }{\partial u}},\end{array}$$

Graph

$$\begin{array}{c}\hfill X_8=ty{\displaystyle \frac{\partial }{\partial v}},X_9=t{\displaystyle \frac{\partial }{\partial x}}-{\displaystyle \frac{1}{2}}x{\displaystyle \frac{\partial }{\partial v}},X_{10}=t\frac{\partial }{\partial t}+{\displaystyle \frac{1}{4}}x{\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{1}{2}}y{\displaystyle \frac{\partial }{\partial y}}-{\displaystyle \frac{1}{2}}u{\displaystyle \frac{\partial }{\partial u}}-{\displaystyle \frac{1}{2}}v{\displaystyle \frac{\partial }{\partial v}}.\end{array}$$

Using the combination of translational symmetries

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$$\begin{array}{c}\hfill X=X_1+X_2+\alpha X_3\end{array}$$

or

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$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}}+\alpha {\displaystyle \frac{\partial }{\partial y}}.\end{array}$$

We obtain the similarity transformations

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$$\begin{array}{c}\hfill r=y-\alpha t,s=x-t,q=t,u_y=u_r,u_t=-\alpha u_r-u_s,u_x=u_s.\end{array}$$

Substituting above values in Eq (4.2), we obtain

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$$\begin{array}{c}\hfill -\alpha u_r-u_s-{\displaystyle \frac{1}{2}}{u}_{ssr}-{\displaystyle \frac{1}{2}}{v}_{rr}-2u_s{v}_s-4u{u}_r=0.\\ \\ \hfill u_r=v_{ss}.\end{array}$$ (4.3)

The system (4.3) admits the Lie point symmetries

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$X_1={\displaystyle \frac{\partial }{\partial r}},X_2={\displaystyle \frac{\partial }{\partial s}}.$ Using the combination of symmetries X1 and X2, we have

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$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial r}}+\beta {\displaystyle \frac{\partial }{\partial s}}.\end{array}$$

The similarity transformations are

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$$\begin{array}{c}\hfill g=s-\beta r,h=r,u_r=-\beta u_g,u_s=u_g,u_{rr}={\beta }^2{u}_{gg},u_{ss}=u_{gg}.\end{array}$$ (4.4)

Substituting (4.4) in system (4.3), we obtain

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$$\begin{array}{c}\hfill \alpha \beta u_g-u_g+{\displaystyle \frac{\beta }{2}}{u}_{ggg}-{\displaystyle \frac{{\beta }^2}{2}}{v}_{gg}-2u_g{v}_g+4\beta u{u}_g=0,\\ \\ \hfill v_g=c_1-\beta u.\end{array}$$ (4.5)

The system (4.5) gives rise to

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$$\begin{array}{c}\hfill \alpha \beta u_g-u_g+{\displaystyle \frac{\beta }{2}}{u}_{ggg}+{\displaystyle \frac{{\beta }^3}{2}}{u}_g-2c_1{u}_g+6\beta u{u}_g=0.\end{array}$$

Integration w.r.t g, yields

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$$\begin{array}{c}\hfill {\displaystyle \frac{\beta }{2}}{u}_{gg}+3\beta u^2+(\alpha \beta +{\displaystyle \frac{{\beta }^3}{2}}-2c_1-1)u=0.\end{array}$$ (4.6)

The particular solution of (4.6) is

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$$\begin{array}{c}\hfill u\left(g\right)=-{\displaystyle \frac{1}{4\beta }}[{\beta }^{3}tan{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2+2\alpha \beta tan{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2+{\beta }^3\\ \hfill -4tan{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2{c}_1+2\alpha \beta -2tan{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2-4c_1-2],\\ \hfill v\left(g\right)=-{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\beta }^4}{\delta }}tan\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\beta }^4}{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\left)\right]\\ \hfill -{\displaystyle \frac{\alpha {\beta }^2}{\delta }}tan\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)+{\displaystyle \frac{\alpha {\beta }^2}{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\left)\right]\\ \hfill +{\displaystyle \frac{1}{4}}{\beta }^{3}g+{\displaystyle \frac{2c_1\beta }{\delta }}tan\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)-{\displaystyle \frac{2c_1\beta }{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\left)\right]\\ \hfill +{\displaystyle \frac{1}{2}}\alpha \beta g+{\displaystyle \frac{\beta }{\delta }}tan\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)-{\displaystyle \frac{\beta }{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\left)\right]-{\displaystyle \frac{1}{2}}.\end{array}$$

The above solutions can be finally expressed in terms of original variables as

Graph

$$\begin{array}{c}\hfill u(t,x,y)=-{\displaystyle \frac{1}{4\beta }}[{\beta }^{3}tan{\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\right)}^2+2\alpha \beta tan{\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\right)}^2+{\beta }^3\\ \hfill -4tan{\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1)}{2\beta }}\right)}^2{c}_1+2\alpha \beta -2tan{\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1)}{2\beta }}\right)}^2-4c_1-2],\\ \hfill v(t,x,y)=-{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\beta }^4}{\delta }}tan{\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\right)}^2+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\beta }^4}{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\left)\right]\\ \hfill -{\displaystyle \frac{\alpha {\beta }^2}{\delta }}tan\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\right)+{\displaystyle \frac{\alpha {\beta }^2}{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\left)\right]\\ \hfill +{\displaystyle \frac{2c_1\beta }{\delta }}tan\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\right)-{\displaystyle \frac{2c_1\beta }{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\left)\right]\\ \hfill +{\displaystyle \frac{\beta }{\delta }}tan\left({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\right)-{\displaystyle \frac{\beta }{\delta }}arctan\left[tan\right({\displaystyle \frac{\delta (c_1+\beta y-x-t(\alpha \beta -1))}{2\beta }}\left)\right]\\ \hfill +{\displaystyle \frac{1}{4}}{\beta }^3(-\beta y+x+t(\alpha \beta -1))+{\displaystyle \frac{1}{2}}\alpha \beta (-\beta y+x+t(\alpha \beta -1))+{\displaystyle \frac{1}{2}}\beta y-{\displaystyle \frac{1}{2}}x-{\displaystyle \frac{1}{2}}t(\alpha \beta -1),\end{array}$$

where

Graph

$\delta =\sqrt{{\beta }^4+2\alpha {\beta }^2-4\beta c_1-2\beta }.$

Now, Using the sine-cosine approach [[32]–[34]], we obtain the explicit solutions of Eq (4.6). Suppose (4.6) has a solution such as

Graph

$$\begin{array}{c}\hfill u\left(g\right)=\lambda co{s}^k\left(\omega g\right).\end{array}$$ (4.7)

Substituting the value of u from (4.7) in (4.6) yields

Graph

$$\begin{array}{c}\hfill (\alpha \beta -1+{\displaystyle \frac{1}{2}}{\beta }^3-2c_1)\lambda co{s}^k\left(\omega g\right)+{\displaystyle \frac{1}{2}}\beta \left({\displaystyle \frac{{\partial }^2}{\partial g^2}}\left(\lambda co{s}^k\left(\omega g\right)\right)\right)+3\beta {\lambda }^2{\left(co{s}^k\left(\omega g\right)\right)}^2=0.\end{array}$$ (4.8)

Eq (4.8) is satisfied if 2k = k − 2. Replacing k = −2 in (4.8) to obtain

Graph

$$\begin{array}{c}\hfill -{\displaystyle \frac{2\lambda \beta {\omega }^2}{co{s}^2\left(\omega g\right)}}+{\displaystyle \frac{3\beta {\lambda }^2}{co{s}^4\left(\omega g\right)}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{\lambda {\beta }^3}{co{s}^2\left(\omega g\right)}}+{\displaystyle \frac{3\beta \lambda {\omega }^2}{co{s}^4\left(\omega g\right)}}+{\displaystyle \frac{\lambda \alpha \beta }{co{s}^2\left(\omega g\right)}}-{\displaystyle \frac{2\lambda c_1}{co{s}^2\left(\omega g\right)}}-{\displaystyle \frac{\lambda }{co{s}^2\left(\omega g\right)}}=0.\end{array}$$

Comparing the coefficients of powers of

Graph

${\displaystyle \frac{1}{cos\left(\omega g\right)}}$ , we get

Graph

$$\begin{array}{c}\hfill -2\beta {\omega }^2+{\displaystyle \frac{1}{2}}{\beta }^3+\alpha \beta -2c_1-1=0,\\ \\ \hfill 3\beta \lambda +3\beta {\omega }^2=0.\end{array}$$ (4.9)

Solving (4.9) gives

Graph

$$\begin{array}{c}\hfill \lambda ={\displaystyle \frac{1-\alpha \beta -{\displaystyle \frac{1}{2}}{\beta }^3+2c_1}{2\beta }},\omega =\sqrt{{\displaystyle \frac{\alpha \beta -1+{\displaystyle \frac{1}{2}}{\beta }^3-2c_1}{2\beta }}}.\end{array}$$

Using the values of λ, ω and k in (4.7) result in

Graph

$$\begin{array}{c}\hfill u\left(g\right)={\displaystyle \frac{1-\alpha \beta -{\displaystyle \frac{1}{2}}{\beta }^3+2c_1}{2\beta }}se{c}^2\left(\sqrt{{\displaystyle \frac{\alpha \beta -1+{\displaystyle \frac{1}{2}}{\beta }^3-2c_1}{2\beta }}}g\right).\end{array}$$

The solution of (4.6) in terms of original variables can be expressed as

Graph

$$\begin{array}{c}\hfill u(t,x,y)={\displaystyle \frac{1-\alpha \beta -{\displaystyle \frac{1}{2}}{\beta }^3+2c_1}{2\beta }}se{c}^2\left(\sqrt{{\displaystyle \frac{\alpha \beta -1+{\displaystyle \frac{1}{2}}{\beta }^3-2c_1}{2\beta }}}(x-\beta y+t(\alpha \beta -1))\right)\end{array}$$

and this constitute the exact solutions of the first integro-differential KP hierarchy equation (4.1).

5 Lie symmetries and exact solutions of the Calogero-Bogoyavlenskii-schiff (CBS) equation

The Calogero—Bogoyavlenskii—schiff (CBS) equation is governed by

Graph

$$\begin{array}{c}\hfill v_t+v_{xxy}+4v{v}_y+2v_x{\partial }_x^{-1}\left(v_y\right)=0.\end{array}$$ (5.1)

Eq (5.1) can be expressed as system of following two equations

Graph

$$\begin{array}{c}\hfill v_t+v_{xxy}+4vv{}_y+2v_{x}u=0,\\ \\ \hfill u_x=v_y,\end{array}$$ (5.2)

where

Graph

$u={\partial }_x^{-1}\left(v_y\right)and{\partial }^{-1}$ denotes the integral with respect to the subscripts.

Using Eq (5.2), the following overdetermined linear set of PDEs are obtained

Graph

$$\begin{array}{c}\hfill {\xi }_u^1=0,{\xi }_v^1=0,{\xi }_x^1=0,{\xi }_y^1=0,{\xi }_{ttt}^1=0,{\xi }_u^2=0,\end{array}$$

Graph

$$\begin{array}{c}\hfill {\xi }_v^2=0,{\xi }_y^2=0,{\xi }_{tx}^2={\displaystyle \frac{1}{4}}{\xi }_{tt}^1,{\xi }_{xx}^2=0,{\xi }_u^3=0,{\xi }_v^3=0,\end{array}$$

Graph

$$\begin{array}{c}\hfill {\xi }_x^3=0,{\xi }_y^3=-2{\xi }_x^2+{\xi }_t^1,{\xi }_{tt}^3=0,{\eta }^1={\xi }_x^{2}u-u{\xi }_t^1+{\displaystyle \frac{1}{2}}{\xi }_t^2,{\eta }^2=-2{\xi }_x^{2}v+{\displaystyle \frac{1}{4}}{\xi }_t^3.\end{array}$$

The following Lie point symmetries can be generated by solving determining equations with one component equal to one and the remaining equal to zero

Graph

$$\begin{array}{c}\hfill X_1={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}},X_2={\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{\partial }{\partial y}},X_3={\displaystyle \frac{\partial }{\partial x}}+t{\displaystyle \frac{\partial }{\partial y}}+{\displaystyle \frac{1}{4}}{\displaystyle \frac{\partial }{\partial v}},\end{array}$$

Graph

$$\begin{array}{c}\hfill X_4=t{\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}}+y{\displaystyle \frac{\partial }{\partial y}}-u{\displaystyle \frac{\partial }{\partial u}},X_5=(1+x){\displaystyle \frac{\partial }{\partial x}}-2y{\displaystyle \frac{\partial }{\partial y}}+u{\displaystyle \frac{\partial }{\partial u}}-2v{\displaystyle \frac{\partial }{\partial v}},\end{array}$$

Graph

$$\begin{array}{c}\hfill X_6={\displaystyle \frac{1}{2}}{t}^2{\displaystyle \frac{\partial }{\partial t}}+(1+{\displaystyle \frac{1}{4}}tx){\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{1}{2}}yt{\displaystyle \frac{\partial }{\partial y}}+(-{\displaystyle \frac{3}{4}}tu+{\displaystyle \frac{1}{8}}x){\displaystyle \frac{\partial }{\partial u}}+(-{\displaystyle \frac{1}{2}}vt+{\displaystyle \frac{1}{8}}y){\displaystyle \frac{\partial }{\partial v}}.\end{array}$$

The combination of translational symmetries corresponding to system (5.2) are

Graph

$$\begin{array}{c}\hfill X=X_1+X_2+\alpha X_3.\end{array}$$

or

Graph

$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}}+\alpha {\displaystyle \frac{\partial }{\partial y}}.\end{array}$$

The similarity transformations are obtained using combination of X = X1 + X2 + αX3

Graph

$$\begin{array}{c}\hfill r=y-\alpha t,s=x-t,q=t,u_y=u_r,u_t=-\alpha u_r-u_s,u_x=u_s.\end{array}$$ (5.3)

Substituting values from (5.3) in Eq (5.2), we obtain

Graph

$$\begin{array}{c}\hfill -\alpha v_r-v_s+v_{ssr}+4v{v}_r+2v_{s}u=0,\\ \\ \hfill u_s=v_r.\end{array}$$ (5.4)

The system (5.4) admits the translational symmetries

Graph

$X_1={\displaystyle \frac{\partial }{\partial r}},X_2={\displaystyle \frac{\partial }{\partial s}}.$ We use the combination of X1 and X2, i.e

Graph

$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial r}}+\beta {\displaystyle \frac{\partial }{\partial s}}\end{array}$$

and compute the similarity transformations

Graph

$$\begin{array}{c}\hfill g=s-\beta r,h=r,u_r=-\beta u_g,u_s=u_g,u_{rr}={\beta }^2{u}_{gg},u_{ss}=u_{gg}.\end{array}$$ (5.5)

Substituting (5.5) in Eq (5.4) gives rise to

Graph

$$\begin{array}{c}\hfill (\alpha \beta -1+2u)v_g-\beta v_{ggg}-4\beta v{v}_g=0,\\ \\ \hfill -\beta v_g-u_g=0,\end{array}$$ (5.6)

where

Graph

$$\begin{array}{c}\hfill u=c_1-\beta v.\end{array}$$

From system (5.6), we conclude

Graph

$$\begin{array}{c}\hfill -\beta v_{ggg}-6\beta v{v}_g+(\alpha \beta -1+2c_1)v_g=0.\end{array}$$

Integrating w.r.t g and choosing the constant of integration to zero, we arrive at

Graph

$$\begin{array}{c}\hfill -\beta v_{gg}-3\beta v^2+(\alpha \beta -1+2c_1)v=0,\end{array}$$ (5.7)

which finally yields

Graph

$$\begin{array}{c}\hfill v\left(g\right)=-{\displaystyle \frac{1}{2\beta }}[\alpha \beta tanh{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2+2tanh{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2{c}_1-\alpha \beta -tanh{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2-2c_1+1].\end{array}$$

Using v(g) in system (5.6), we obtain

Graph

$$\begin{array}{c}\hfill u\left(g\right)=c_1+{\displaystyle \frac{1}{2}}[\alpha \beta tanh{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2+2tanh{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2{c}_1-\alpha \beta -tanh{\left({\displaystyle \frac{\delta (c_1-g)}{2\beta }}\right)}^2-2c_1+1].\end{array}$$

We apply the inverse informations (5.3), the solution can be expressed in original variables as

Graph

$$\begin{array}{c}\hfill v(t,x,y)=-{\displaystyle \frac{1}{2\beta }}[\alpha \beta tanh{\left({\displaystyle \frac{\delta (c_1-x+\beta y-t(\alpha \beta -1))}{2\beta }}\right)}^2-\alpha \beta -2c_1+1\\ \hfill +2tanh{\left({\displaystyle \frac{\delta (c_1-x+\beta y-t(\alpha \beta -1))}{2\beta }}\right)}^2{c}_1-tanh{\left({\displaystyle \frac{\delta (c_1-x+\beta y-t(\alpha \beta -1))}{2\beta }}\right)}^2],\\ \hfill u(t,x,y)={\displaystyle \frac{1}{2}}\alpha \beta tanh{\left({\displaystyle \frac{\delta (c_1-x+\beta y-t(\alpha \beta -1))}{2\beta }}\right)}^2-{\displaystyle \frac{1}{2}}\alpha \beta +{\displaystyle \frac{1}{2}}\\ \hfill +tanh{\left({\displaystyle \frac{\delta (c_1-x+\beta y-t(\alpha \beta -1))}{2\beta }}\right)}^2{c}_1-{\displaystyle \frac{1}{2}}tanh{\left({\displaystyle \frac{\delta (c_1-x+\beta y-t(\alpha \beta -1))}{2\beta }}\right)}^2,\end{array}$$

where

Graph

$\delta =\sqrt{\alpha {\beta }^2+2\beta c_1-\beta }$ .

which constitute an exact solution of Eq (5.1).

Now, Using the sine-cosine approach [[32]–[34]], we obtain the explicit solutions of Eq (5.7). Suppose (5.7) has a solution such as

Graph

$$\begin{array}{c}\hfill v\left(g\right)=\lambda co{s}^k\left(\omega g\right).\end{array}$$ (5.8)

Substituting v from (5.8) in (5.7), yields

Graph

$$\begin{array}{c}\hfill (\alpha \beta +2c_1-1)\lambda co{s}^k\left(\omega g\right)-\beta {\displaystyle \frac{{\partial }^2}{\partial g^2}}\left(\lambda co{s}^k\left(\omega g\right)\right)-3\beta {\lambda }^2{\left(co{s}^k\left(\omega g\right)\right)}^2=0.\end{array}$$ (5.9)

Eq (5.9) is satisfied if 2k = k − 2. Substituting k = −2 in (5.9) to obtain

Graph

$$\begin{array}{c}\hfill {\displaystyle \frac{4\lambda \beta {\omega }^2}{co{s}^2\left(\omega g\right)}}-{\displaystyle \frac{3\beta {\lambda }^2}{co{s}^4\left(\omega g\right)}}-{\displaystyle \frac{6\beta \lambda {\omega }^2}{co{s}^4\left(\omega g\right)}}+{\displaystyle \frac{\lambda \alpha \beta }{co{s}^2\left(\omega g\right)}}+{\displaystyle \frac{2\lambda c_1}{co{s}^2\left(\omega g\right)}}-{\displaystyle \frac{\lambda }{co{s}^2\left(\omega g\right)}}=0.\end{array}$$

Comparing the coefficients of powers of

Graph

${\displaystyle \frac{1}{cos\left(\omega g\right)}}$ , we obtain the following system

Graph

$$\begin{array}{c}\hfill 4\beta {\omega }^2+\alpha \beta +2c_1-1=0,\\ \\ \hfill \lambda +2{\omega }^2=0.\end{array}$$ (5.10)

Simple manipulations yield

Graph

$$\begin{array}{c}\hfill \lambda ={\displaystyle \frac{\alpha \beta +2c_1-1}{2\beta }},\omega ={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{1-2c_1-\alpha \beta }{\beta }}}.\end{array}$$

Eq (5.8) with the use of λ, ω and k results in

Graph

$$\begin{array}{c}\hfill u\left(g\right)={\displaystyle \frac{\alpha \beta +2c_1-1}{2\beta }}se{c}^2\left(\sqrt{{\displaystyle \frac{1-2c_1-\alpha \beta }{4\beta }}}g\right).\end{array}$$

Using inverse transformations, the solution of (5.7) in terms of original variables can be expressed as

Graph

$$\begin{array}{c}\hfill u(t,x,y)={\displaystyle \frac{\alpha \beta +2c_1-1}{2\beta }}se{c}^2\left(\sqrt{{\displaystyle \frac{1-2c_1-\alpha \beta }{4\beta }}}(x-\beta y+t(\alpha \beta -1))\right),\end{array}$$

Graph

$$\begin{array}{c}\hfill v(t,x,y)={\displaystyle \frac{1}{\beta }}[c_1-{\displaystyle \frac{\alpha \beta +2c_1-1}{2\beta }}se{c}^2(\sqrt{{\displaystyle \frac{1-2c_1-\alpha \beta }{4\beta }}}(x-\beta y+t(\alpha \beta -1))\left)\right],\end{array}$$

where

Graph

$$\begin{array}{c}\hfill u=c_1-\beta v\end{array}$$

which constitute the exact solutions of Eq (5.1).

6 Lie symmetries and exact solutions of the modified Calogero-Bogoyavlenskii-schiff (MCBS) eq...

The modified Calogero-Bogoyavlenskii-schiff (CBS) equation is regulated by

Graph

$$\begin{array}{c}\hfill v_t+v_{xxy}+4v^2{v}_y+4v_x{\partial }_x^{-1}\left(v{v}_y\right)=0,\end{array}$$ (6.1)

which can be expressed into system of two equations

Graph

$$\begin{array}{c}\hfill v_t+v_{xxy}+4v^2{v}_y+4v_{x}u=0,\\ \\ \hfill u_x=v{v}_y,\end{array}$$ (6.2)

where

Graph

$u={\partial }_x^{-1}\left(v{v}_y\right)and{\partial }^{-1}$ denotes the integral with respect to the subscript.

Using Eq (6.2), the following overdetermined linear set of PDEs are obtained

Graph

$$\begin{array}{c}\hfill {\xi }_u^1=0,{\xi }_v^1=0,{\xi }_x^1=0,{\xi }_y^1=0,{\xi }_{tt}^1=0,{\xi }_u^2=0,\end{array}$$

Graph

$$\begin{array}{c}\hfill {\xi }_v^2=0,{\xi }_y^2=0,{\xi }_{tx}^2=0,{\xi }_{xx}^2=0,{\xi }_t^3=0,{\xi }_u^3=0,\end{array}$$

Graph

$$\begin{array}{c}\hfill {\xi }_v^3=0,{\xi }_x^3=0,{\xi }_y^3=-2{\xi }_x^2+{\xi }_t^1,{\eta }^1={\xi }_x^{2}u-u{\xi }_t^1+{\displaystyle \frac{1}{4}}{\xi }_t^2,{\eta }^2=-v{\xi }_x^2.\end{array}$$

The following Lie point symmetries can be generated by solving determining equations with one component equal to one and the remaining equal to zero

Graph

$$\begin{array}{c}\hfill X_1={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}},X_2={\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{\partial }{\partial y}},X_3=t{\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}}+y{\displaystyle \frac{\partial }{\partial y}}-u{\displaystyle \frac{\partial }{\partial u}},X_4=(x+1){\displaystyle \frac{\partial }{\partial x}}-2y{\displaystyle \frac{\partial }{\partial y}}+u{\displaystyle \frac{\partial }{\partial u}}-v{\displaystyle \frac{\partial }{\partial v}}.\end{array}$$

The combination of translational symmetries

Graph

$$\begin{array}{c}\hfill X=X_1+X_2+\alpha X_3\end{array}$$

or

Graph

$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}}+\alpha {\displaystyle \frac{\partial }{\partial y}}.\end{array}$$

The corresponding similarity transformations are

Graph

$$\begin{array}{c}\hfill r=y-\alpha t,s=x-t,q=t,u_y=u_r,u_t=-\alpha u_r-u_s,u_x=u_s.\end{array}$$

Substituting above values in Eq (6.2), we obtain

Graph

$$\begin{array}{c}\hfill -\alpha v_r-v_s+v_{ssr}+4v^2{v}_r+4v_{s}u=0,\\ \\ \hfill u_s-v{v}_r=0.\end{array}$$ (6.3)

System (6.3) admits the translational symmetries

Graph

$X_1={\displaystyle \frac{\partial }{\partial r}},X_2={\displaystyle \frac{\partial }{\partial s}}.$ Using combination of symmetries

Graph

$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial r}}+\beta {\displaystyle \frac{\partial }{\partial s}},\end{array}$$

we find similarity transformations

Graph

$$\begin{array}{c}\hfill g=s-\beta r,h=r,u_r=-\beta u_g,u_s=u_g,u_{rr}={\beta }^2{u}_{gg},u_{ss}=u_{gg}.\end{array}$$

Substituting above values in system (6.3), results in

Graph

$$\begin{array}{c}\hfill -\beta v{v}_g-u_g=0,\\ \\ \hfill {\displaystyle \frac{-\beta v^2}{2}}-u+c_1=0.\end{array}$$ (6.4)

From (6.4), we conclude

Graph

$$\begin{array}{c}\hfill -\beta v_{ggg}-6\beta v^2{v}_g+(\alpha \beta -1+4c_1)v_g=0.\end{array}$$

Integrating w.r.t g and choosing constant of integration to zero gives rise to

Graph

$$\begin{array}{c}\hfill -\beta v_{gg}-2\beta v^3+(\alpha \beta -1+4c_1)v=0.\end{array}$$ (6.5)

Now using the sine-cosine approach [[32]–[34]], we obtain the explicit solutions of Eq (6.5). Suppose (6.5) has a solution such as

Graph

$$\begin{array}{c}\hfill v\left(g\right)=\lambda co{s}^k\left(\omega g\right).\end{array}$$ (6.6)

Substituting the values of v from (6.6) in (6.5) yields

Graph

$$\begin{array}{c}\hfill -2\beta {\lambda }^3{\left(co{s}^k\left(\omega g\right)\right)}^3+\lambda co{s}^k\left(\omega g\right)\beta k^2{\omega }^2-{\displaystyle \frac{\lambda co{s}^k\left(\omega g\right)\beta k^2{\omega }^2}{co{s}^2\left(\omega g\right)}}+\lambda co{s}^k\left(\omega g\right)\alpha \beta \\ \hfill +{\displaystyle \frac{\lambda co{s}^k\left(\omega g\right)\beta k{\omega }^2}{co{s}^2\left(\omega g\right)}}+4\lambda co{s}^k\left(\omega g\right)c_1-\lambda co{s}^k\left(\omega g\right)=0.\end{array}$$ (6.7)

Eq (6.7) is satisfied if 3k = k − 2. Replacing k = −1 in (6.7) to obtain

Graph

$$\begin{array}{c}\hfill -{\displaystyle \frac{2\beta {\lambda }^3}{co{s}^3\left(\omega g\right)}}+{\displaystyle \frac{\lambda \beta {\omega }^2}{cos\left(\omega g\right)}}-{\displaystyle \frac{2\lambda \beta {\omega }^2}{co{s}^3\left(\omega g\right)}}+{\displaystyle \frac{\lambda \alpha \beta }{cos\left(\omega g\right)}}+{\displaystyle \frac{4\lambda c_1}{cos\left(\omega g\right)}}-{\displaystyle \frac{\lambda }{cos\left(\omega g\right)}}=0.\end{array}$$

Comparing the coefficients of powers of

Graph

${\displaystyle \frac{1}{cos\left(\omega g\right)}}$ , we have

Graph

$$\begin{array}{c}\hfill -2\beta {\lambda }^3-2\lambda \beta {\omega }^2=0,\\ \\ \hfill \lambda \beta {\omega }^2+\lambda \alpha \beta +4\lambda c_1-\lambda =0.\end{array}$$ (6.8)

Solving (6.8) gives

Graph

$$\begin{array}{c}\hfill \lambda =\sqrt{{\displaystyle \frac{\alpha \beta +4c_1-1}{\beta }}},\omega =\sqrt{{\displaystyle \frac{-\alpha \beta -4c_1+1}{\beta }}}g.\end{array}$$

Using values of λ, ω and k in (6.6) results in

Graph

$$\begin{array}{c}\hfill v\left(g\right)=\sqrt{{\displaystyle \frac{\alpha \beta +4c_1-1}{\beta }}}sec\left(\sqrt{{\displaystyle \frac{-\alpha \beta -4c_1+1}{\beta }}}g\right),\end{array}$$

Graph

$$\begin{array}{c}\hfill u\left(g\right)=c_1-{\displaystyle \frac{\alpha \beta +4c_1-1}{2}}se{c}^2\left(\sqrt{{\displaystyle \frac{-\alpha \beta -4c_1+1}{\beta }}}g\right).\end{array}$$

The solution of (6.5) in terms of original variables can be expressed as

Graph

$$\begin{array}{c}\hfill v(t,x,y)=\sqrt{{\displaystyle \frac{\alpha \beta +4c_1-1}{\beta }}}sec\left(\sqrt{{\displaystyle \frac{-\alpha \beta -4c_1+1}{\beta }}}(x-\beta y+t\right(\alpha \beta -1\left)\right)),\\ \\ \hfill u(t,x,y)=c_1-{\displaystyle \frac{\alpha \beta +4c_1-1}{2}}sec{\left(\sqrt{{\displaystyle \frac{-\alpha \beta -4c_1+1}{\beta }}}(x-\beta y+t(\alpha \beta -1))\right)}^2.\end{array}$$ (6.9)

Eq (6.9) constitute the exact solution of the modified Calogero-Bogoyavlenskii-schiff (CBS) Eq (6.2).

7 Lie symmetries and exact solutions of the modified KdV-CBS equation

The modified KdV-CBS equation is expressed as

Graph

$$\begin{array}{c}\hfill u_t-4u^2{u}_y-2u_x{\partial }^{-1}{\left(u^2\right)}_y+u_{xxy}-6u^2{u}_x+u_{xxx}=0.\end{array}$$ (7.1)

Eq (7.1) can be re-written as

Graph

$$\begin{array}{c}\hfill u_t-4u^2{u}_y-2u_{x}v+u_{xxy}-6u^2{u}_x+u_{xxx}=0,\\ \\ \hfill v_x-2u{u}_y=0,\end{array}$$ (7.2)

where

Graph

$v={\partial }_x^{-1}{\left(u^2\right)}_{y}and{\partial }^{-1}$ denotes the integral with respect to the subscripts.

Using Eq (7.2), we obtain the following set of overdetermined linear PDEs

Graph

$$\begin{array}{c}\hfill {\xi }_u^1=0,{\xi }_v^1=0,{\xi }_x^1=0,{\xi }_y^1=0,{\xi }_{tt}^1=0,{\xi }_u^2=0,\\ \hfill {\xi }_v^2=0,{\xi }_x^2=-{\displaystyle \frac{1}{3}}{\xi }_y^2+{\displaystyle \frac{1}{3}}{\xi }_t^1,{\xi }_{ty}^2=0,{\xi }_{yy}^2=0,{\xi }_t^3=0,\\ \hfill {\xi }_u^3=0,{\xi }_v^3=0,{\xi }_x^3=0,{\xi }_y^3={\displaystyle \frac{1}{3}}{\xi }_t^1+{\displaystyle \frac{2}{3}}{\xi }_y^2,{\eta }^1={\displaystyle \frac{1}{3}}u({\xi }_y^2-{\xi }_t^1),\\ \hfill {\eta }^2=-{\displaystyle \frac{1}{3}}(3u^2+v){\xi }_y^2-{\displaystyle \frac{2}{3}}v{\xi }_t^1-{\displaystyle \frac{1}{2}}{\xi }_t^2.\hfill \end{array}$$ (7.3)

We get the following Lie point symmetries

Graph

$$\begin{array}{c}\hfill X_1={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}},X_2={\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{\partial }{\partial y}},X_3=t{\displaystyle \frac{\partial }{\partial t}}+(1+{\displaystyle \frac{1}{3}}x){\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{1}{3}}y{\displaystyle \frac{\partial }{\partial y}}-{\displaystyle \frac{1}{3}}u{\displaystyle \frac{\partial }{\partial u}}-{\displaystyle \frac{2}{3}}v{\displaystyle \frac{\partial }{\partial v}},\\ \hfill X_4=(1-{\displaystyle \frac{1}{3}}x+y){\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{2}{3}}y{\displaystyle \frac{\partial }{\partial y}}+{\displaystyle \frac{1}{3}}u{\displaystyle \frac{\partial }{\partial u}}-({\displaystyle \frac{1}{3}}v+u^2){\displaystyle \frac{\partial }{\partial v}}.\hfill \end{array}$$

The combination of translational symmetries

Graph

$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{\partial }{\partial x}}+\alpha {\displaystyle \frac{\partial }{\partial y}}\end{array}$$

rovide the similarity transformations

Graph

$$\begin{array}{c}\hfill r=y-\alpha t,s=x-t,q=t,u_y=u_r,u_t=-\alpha u_r-u_s,u_x=u_s.\end{array}$$ (7.4)

Using change of variables (7.4), system (7.2) transforms to

Graph

$$\begin{array}{c}\hfill -\alpha u_r-u_s+u_{ssr}-4u^2{u}_r-2u_{s}v-6u^2{u}_s+u_{sss}=0,\\ \\ \hfill v_s-2u{u}_r=0.\end{array}$$ (7.5)

Eq (7.5) admits the Lie point symmetries

Graph

$X_1={\displaystyle \frac{\partial }{\partial r}},X_2={\displaystyle \frac{\partial }{\partial s}}$ . We use the combination of symmetries

Graph

$$\begin{array}{c}\hfill X={\displaystyle \frac{\partial }{\partial r}}+\beta {\displaystyle \frac{\partial }{\partial s}},\end{array}$$

to find the canonical variables

Graph

$$\begin{array}{c}\hfill g=s-\beta r,h=r,u_r=-\beta u_g,u_s=u_g,u_{rr}={\beta }^2{u}_{gg},u_{ss}=u_{gg}.\end{array}$$ (7.6)

(7.5) with the help of similarity transformations and then integration w.r.t g, we find

Graph

$$\begin{array}{c}\hfill v=c_1-\beta u^2,\\ \\ \hfill (\alpha \beta -1)u_g+4\beta u^2{u}_g-2u_{g}v-\beta u_{ggg}-6u^2{u}_g+u_{ggg}=0.\end{array}$$ (7.7)

The solution of above system gives the following equation:

Graph

$$\begin{array}{c}\hfill (1-\beta )u_{ggg}+(\alpha \beta -1-2c_1)u_g+6\beta u^2{u}_g-6u^2{u}_g=0.\end{array}$$

Integration w.r.t g gives rise to

Graph

$$\begin{array}{c}\hfill (1-\beta )u_{gg}+(\alpha \beta -1-2c_1)u+2(\beta -1)u^3=0.\end{array}$$ (7.8)

Now, using the sine-cosine approach [[32]–[34]], we obtain the explicit solutions of Eq (7.8). Suppose (7.8) has a solution such as

Graph

$$\begin{array}{c}\hfill u\left(g\right)=\lambda co{s}^k\left(\omega g\right).\end{array}$$ (7.9)

Substituting u from (7.9) in (7.8) yields

Graph

$$\begin{array}{c}\hfill 2\beta {\lambda }^3{\left(co{s}^k\left(\omega g\right)\right)}^3-2{\lambda }^3{\left(co{s}^k\left(\omega g\right)\right)}^3-\lambda co{s}^k\left(\omega g\right)k^2{\omega }^2+\lambda co{s}^k\left(\omega g\right)\beta k^2{\omega }^2-{\displaystyle \frac{\lambda co{s}^k\left(\omega g\right)\beta k^2{\omega }^2}{co{s}^2\left(\omega g\right)}}\\ \hfill +{\displaystyle \frac{\lambda co{s}^k\left(\omega g\right)k^2{\omega }^2}{co{s}^2\left(\omega g\right)}}-{\displaystyle \frac{\lambda co{s}^k\left(\omega g\right)k{\omega }^2}{co{s}^2\left(\omega g\right)}}+\lambda co{s}^k\left(\omega g\right)\alpha \beta -\lambda co{s}^k\left(\omega g\right)(2c_1+1)+{\displaystyle \frac{\lambda co{s}^k\left(\omega g\right)\beta k{\omega }^2}{co{s}^2\left(\omega g\right)}}=0.\end{array}$$ (7.10)

Eq (7.10) is satisfied if 2k = −2. Substituting k = −1 in (7.10), we obtain

Graph

$$\begin{array}{c}\hfill {\displaystyle \frac{2\beta {\lambda }^3}{co{s}^3\left(\omega g\right)}}+{\displaystyle \frac{\lambda \beta {\omega }^2}{cos\left(\omega g\right)}}-{\displaystyle \frac{2\lambda \beta {\omega }^2}{co{s}^3\left(\omega g\right)}}-{\displaystyle \frac{2\lambda c_1}{cos\left(\omega g\right)}}+{\displaystyle \frac{\lambda \alpha \beta }{cos\left(\omega g\right)}}-{\displaystyle \frac{\lambda }{cos\left(\omega g\right)}}-{\displaystyle \frac{\lambda {\omega }^2}{cos\left(\omega g\right)}}+{\displaystyle \frac{2\lambda {\omega }^2}{co{s}^3\left(\omega g\right)}}-{\displaystyle \frac{2{\lambda }^3}{co{s}^3\left(\omega g\right)}}=0.\end{array}$$

Comparing the coefficients of powers of

Graph

${\displaystyle \frac{1}{cos\left(\omega g\right)}}$ , we arrive at

Graph

$$\begin{array}{c}\hfill \beta {\lambda }^2-{\lambda }^2-\beta {\omega }^2+{\omega }^2=0,\\ \\ \hfill \beta {\omega }^2+\alpha \beta -{\omega }^2-2c_1-1=0.\end{array}$$ (7.11)

Solving (7.11) gives

Graph

$$\begin{array}{c}\hfill \lambda =\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}},\omega =\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}g.\end{array}$$

Using the values of λ, ω and k in (7.9) results in

Graph

$$\begin{array}{c}\hfill u\left(g\right)=\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}sec\left(\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}g\right),\end{array}$$

Graph

$$\begin{array}{c}\hfill v\left(g\right)=c_1-\beta \left({\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}\right)se{c}^2\left(\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}g\right).\end{array}$$

Thus,

Graph

$$\begin{array}{c}\hfill u(t,x,y)=\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}sec\left(\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}(x-\beta y+t(\alpha \beta -1))\right),\end{array}$$

Graph

$$\begin{array}{c}\hfill v(t,x,y)=c_1-\beta \left({\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}\right)se{c}^2\left(\sqrt{{\displaystyle \frac{\alpha \beta -2c_1-1}{1-\beta }}}(x-\beta y+t(\alpha \beta -1))\right),\end{array}$$

which constitute the exact solutions of the modified KdV-CBS equation (7.2).

8 Conclusion

The exact solutions of the (1 + 1)- dimensional integro-differential Ito, the first integro-differential KP hierarchy, the Calogero-Bogoyavlenskii-Schiff (CBS), the modified Calogero-Bogoyavlenskii-Schiff (CBS) and the modified KdV-CBS equations have been successfully established by utilizing the similarity transformations and the sine-cosine method. In the evaluation of exact solutions, the reduction in the number of independent variables via similarity variables and the use of inverse similarity transformations have been made. By substituting back, it has been checked that the acquired solutions satisfy the prescribed equations. Furthermore, we have done the Lie symmetry analysis to investigate these solutions. Thus, the proposed approach is more efficient, reliable, and concise by means of computational complexity. It can provide more exact solutions as compared to the other methods that exist in the literature. The obtained solutions are new and innovative to existing ones and therefore, more appropriate to understand. In fact, the proposed method is readily applicable to a large variety of nonlinear evolution equations which frequently appear in mathematical physics and nonlinear sciences.

The author expresses gratitude to the referees for their constructive criticism, which helped to strengthen the paper's substance.

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By Amina Amin; Imran Naeem and Adnan Khan

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