Solving nonlinear partial differential equations using a novel Cham method

Abstract

Nonlinear partial differential equations (NLPDEs) have been of great interest in recent years due to their numerous applications. While there are several methods for finding exact solutions to various NLPDEs, more solutions are still required. This paper first proposes the Cham method, a new method for solving NLPDEs that can generate eight families of solutions. The method is then successfully employed to solve the (2+1)-dimensional Bogoyavlenskii's breaking soliton equations. The dynamic behaviour of these equations and the bifurcation of traveling waves are also discussed. Finally, we graphically depict some solutions corresponding to some discovered solutions with different coefficient values. The Cham method is general, effective, and adaptable to many NLPDEs.

Keywords:

• Cham method
• (2 + 1)-dimensional Bogoyavlenskii's breaking soliton equations
• bifurcation theory
• traveling wave solutions

1. Introduction

Most differential equations in geometry are nonlinear models, hence nonlinear partial differential equations (NLPDEs) are useful for describing complex phenomena in many fields, including electrodynamics, thermodynamics, hydrodynamics, fluid flows, aeronautics, and hydrodynamic applications such as shipbuilding, submarine design, climatology, and meteorology.

NLPDEs have received significant research attention over the last four decades in order to propose effective methods for finding exact solutions. Among these methods are the $(\frac{G^{{}^{\mathrm{\prime }}}(\xi )}{G(\xi )})-$ expansion method [1], the $\exp (-\mathrm{\Phi }(\xi ))$-expansion method [2,3], the double auxiliary equations method [4], the ${\coth }_a(\xi )$ expansion method [5], the F-expansion method [6], the Tan-Cot function method [7], the Sine-Cosine function approach [8–10], and various other methods [11–13]. In some studies, bifurcations and phase portraits are investigated to gain a clear understanding of the behaviour and nature of solutions [14,15].

Despite the fact that there are several approaches to solving NLPDEs, their importance warrants further consideration. As a result, this paper aims to propose the Cham method, a new method that can be used to solve various NLPDEs. In addition to its generality, the method is simple and can find several exact solutions.

Section 2 describes our new Cham method. Section 3 investigates the dynamic behaviour and the bifurcation of traveling waves of a selected NLPDE. The Cham method is applied in Section 4 to obtain exact solutions to the chosen NLPDE. Section 5 illustrates some solutions that correspond to some discovered solutions with varying coefficient values. Finally, Section 6 concludes the paper.

2. The proposed Cham method

The new Cham method for solving NLPDEs is described in this section by considering the following NPDE: (1) $F(z,z_t,z_x,z_{\mathrm{xt}},z_{\mathrm{xx}},z_{\mathrm{tt}},\dots )=0,$(1) in which $z=z(x,t)$ is an unknown function.

Phase 1. Convert Eq.(1(1) $F(z,z_t,z_x,z_{\mathrm{xt}},z_{\mathrm{xx}},z_{\mathrm{tt}},\dots )=0,$(1) ) to an ordinary differential equation by using $z(x,t)=Z(\xi )$ with $\xi =x-\mathrm{vt}$, as follows: (2) $P(Z,Z_{\xi },Z_{\mathrm{\xi \xi }},Z_{\mathrm{\xi \xi \xi }},\dots )=0.$(2) Phase 2. Assume the solution to Equation (1(1) $F(z,z_t,z_x,z_{\mathrm{xt}},z_{\mathrm{xx}},z_{\mathrm{tt}},\dots )=0,$(1) ) is expressed as a polynomial in $(\frac{\tanh (Z(\xi ))}{\tanh (\xi )})$, as shown below: (3) $U\left(\xi \right)=\sum _{i=0}^N{a}_i{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^i,$(3) where $a_i$ are constants for all $(i=0,1,\dots ,N)$ and $a_N\ne 0$. All nonlinear terms in Equation (2(2) $P(Z,Z_{\xi },Z_{\mathrm{\xi \xi }},Z_{\mathrm{\xi \xi \xi }},\dots )=0.$(2) ) are used to balance the highest order derivative term to find the value of $N\in {\mathbb{N}}^{+}$. The expression $(\frac{\tanh (Z(\xi ))}{\tanh (\xi )})$ satisfies the ordinary differential equation given in Equation (4(4) $\begin{array}{rl}{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^{\prime }& =A{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2\\ & +B\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+C.\end{array}$(4) ): (4) $\begin{array}{rl}{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^{\prime }& =A{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2\\ & +B\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+C.\end{array}$(4) Therefore, the following are the solutions to Equation (4(4) $\begin{array}{rl}{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^{\prime }& =A{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2\\ & +B\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+C.\end{array}$(4) ):

Family 1: $\mathrm{AC}\ne 0$ and $4\mathrm{AC}-B^2>0$, (5) $\begin{array}{rl}\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)& =-\frac{\begin{array}{c}\mathrm{Btanh}\left(\xi \right)\\ +\tan \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\\ \sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}\end{array}}{2\mathrm{Atanh}\left(\xi \right)},\\ & {\xi }_0\in \mathbb{R}.\end{array}$(5) Family 2: $\mathrm{AC}\ne 0$ and $4\mathrm{AC}-B^2<0$, (6) $\begin{array}{rl}& \left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)=-\frac{\begin{array}{c}\mathrm{Btanh}\left(\xi \right)\\ -\tanh \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\\ \sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}\end{array}}{2\mathrm{Atanh}\left(\xi \right)},\\ & {\xi }_0\in \mathbb{R}.\end{array}$(6) Family 3: $\mathrm{AC}>0$ and $4\mathrm{AC}-B^2=0$, (7) $\begin{array}{rl}\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)& =-\frac{\begin{array}{c}2\mathrm{AC}{\xi }^2-1\\ -\tanh \left(\frac{\ln (\xi )-\ln (\mathrm{AC\xi }-\sqrt{\mathrm{AC}})+{\xi }_0}{2}\right)\end{array}}{2\mathrm{A\xi }(\sqrt{\mathrm{AC}}\xi -1)},\\ & {\xi }_0\in \mathbb{R}.\end{array}$(7) Family 4: Both A and B are equal to zero, (8) $\begin{array}{r}\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)=\frac{\begin{array}{c}C(\ln (\tanh (\xi )+1)\\ -\ln (\tanh (\xi )-1))\end{array}}{2}-{\xi }_0,{\xi }_0\in \mathbb{R}.\end{array}$(8) Family 5: Only C is equal to zero, (9) $\begin{array}{r}\{\begin{array}{l}\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)=-{\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{\begin{array}{c}\tanh (\xi )(A\exp (B\xi )-{\xi }_0)\\ (\exp (2\xi )+1)\end{array}}},\\ \xi \notin \left\{\frac{1}{B}\ln \left(\frac{{\xi }_0}{A}\right),0\right\},\left|{\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{(\mathrm{Aexp}(\mathrm{B\xi })-{\xi }_0)(\exp (2\xi )+1)}}\right|\\ <1.\end{array}\end{array}$(9) Family 6: Both B and C are equal to zero, (10) $\begin{array}{r}\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)=\frac{1}{{\xi }_0-\mathrm{A\xi }},{\xi }_0\in \mathbb{R}.\end{array}$(10) Family 7: Only A is equal to zero, (11) $\begin{array}{r}\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)=-\frac{C+B{\xi }_0\exp \left(\mathrm{B\xi }\right)}{B},{\xi }_0\in \mathbb{R}.\end{array}$(11) In the subsequent sections, we will apply the new Cham method to solve the (2 + 1)-dimensional Bogoyavlenskii's breaking soliton equation (abbreviated Bogoyavlenskii's equation for short), which is defined as follows: (12) $\{\begin{array}{l}{u}_t+\alpha u_{\mathrm{xxy}}+4{\left(\mathrm{uv}\right)}_x=0,\\ u_y=v_x.\end{array}$(12)

3. Bifurcation and phase portraits of Bogoyavlenskii's equations

Based on the bifurcation theory [16], we assume the following: (13) $u(x,y,t)=v(x,y,t)=U\left(\xi \right);\xi =x+y-\mathrm{\lambda t},\lambda \ne 0$(13) where λ denotes the traveling wave's velocity. Using the traveling wave transformation (14(13) $u(x,y,t)=v(x,y,t)=U\left(\xi \right);\xi =x+y-\mathrm{\lambda t},\lambda \ne 0$(13) ), we can transform the Equation (13(12) $\{\begin{array}{l}{u}_t+\alpha u_{\mathrm{xxy}}+4{\left(\mathrm{uv}\right)}_x=0,\\ u_y=v_x.\end{array}$(12) ) to the following: (14) $-\lambda U_{\xi }+\alpha U_{\mathrm{\xi \xi \xi }}+4{\left(U^2\right)}_{\xi }=0.$(14) We first integrate Equation (15(14) $-\lambda U_{\xi }+\alpha U_{\mathrm{\xi \xi \xi }}+4{\left(U^2\right)}_{\xi }=0.$(14) ) with respect to ξ and then set the integrating constant to zero to obtain the following result: (15) $-\mathrm{\lambda U}+\alpha U_{\mathrm{\xi \xi }}+4U^2=0.$(15) As a result, Equation (16(15) $-\mathrm{\lambda U}+\alpha U_{\mathrm{\xi \xi }}+4U^2=0.$(15) ) corresponds to the following Hamiltonian system: (16) $\frac{\mathrm{d}U}{\mathrm{d}\xi }=V,\frac{\mathrm{d}V}{\mathrm{d}\xi }=\frac{1}{\alpha }(\mathrm{\lambda U}-4U^2).$(16) with the following Hamiltonian function: (17) $H(U,V)=3\alpha V^2+8U^3-3\lambda U^2=h,$(17) where h is the integral constant.

The bifurcations of the dynamical system (17(16) $\frac{\mathrm{d}U}{\mathrm{d}\xi }=V,\frac{\mathrm{d}V}{\mathrm{d}\xi }=\frac{1}{\alpha }(\mathrm{\lambda U}-4U^2).$(16) ) result in the following. When $\lambda =0$, $E_0(0,0)$ is an equilibrium point. When $\lambda \ne 0$, $E_0(0,0)$ and $E_1(\frac{1}{4}\lambda ,0)$ are the equilibrium points on the U axis.

As a result, we can obtain the matrix $M(u_e,v_e)$ of the linearized system of Equation (17(16) $\frac{\mathrm{d}U}{\mathrm{d}\xi }=V,\frac{\mathrm{d}V}{\mathrm{d}\xi }=\frac{1}{\alpha }(\mathrm{\lambda U}-4U^2).$(16) ) at an equilibrium point $E_j(j=1,2)$, as follows: (18) $\begin{array}{rl}M(u_e,v_e)& =\left(\begin{array}{cc}0& 1\\ \frac{1}{\alpha }(\lambda -8u_e)& 0\end{array}\right),\\ J& =\frac{1}{\alpha }(8u_e-\lambda );\alpha \ne 0\end{array}$(18) where $J=det(M(u_e,v_e))$.

Case I. If ($\lambda >0$ and $\alpha >0$) or ($\lambda <0$ and $\alpha <0$), we have a saddle point $E_0(0,0)$ since $J(E_0)=-\frac{\lambda }{\alpha }<0$, a center point $E_1(\frac{1}{4}\lambda ,0)$ because $J(E_1)=\frac{\lambda }{\alpha }>0$ and $\mathrm{Trace}(M(E_1))=0$, and two equilibrium points $E_0(0,0)$ and $E_1(\frac{1}{4}\lambda ,0)$. As a result, Equation (16(15) $-\mathrm{\lambda U}+\alpha U_{\mathrm{\xi \xi }}+4U^2=0.$(15) ) has a solitary wave solution and two families of periodic wave solutions. Figure 1 depicts this case using Maple 2022.

Figure 1. Case I when: (a) $\lambda =4$ and $\alpha =2$, and (b) $\lambda =-4$ and $\alpha =-2$.

Case II. If ($\lambda >0$ and $\alpha <0$) or ($\lambda <0$ and $\alpha >0$), $E_0(0,0)$ and $E_1(\frac{1}{4}\lambda ,0)$ are two equilibrium points. Further, $E_0(0,0)$ is a center point since $\mathrm{Trace}(M(E_0))=0$ and $J(E_0)=-\frac{\lambda }{\alpha }>0$. At $E_1(\frac{1}{4}\lambda ,0)$ we have $J(E_1)=\frac{\lambda }{\alpha }<0$, indicating that $E_1(\frac{1}{4}\lambda ,0)$ is a saddle point. Thus, a single solitary wave solution to Equation (16(15) $-\mathrm{\lambda U}+\alpha U_{\mathrm{\xi \xi }}+4U^2=0.$(15) ) is found, as well as two families of periodic wave solutions, which are depicted in Figure 2 using Maple 2022.

Figure 2. Case II when: (a) $\lambda =4$ and $\alpha =-2$, and (b) $\lambda =-4$ and $\alpha =2$.

4. Solving the Bogoyavlenskii's equations

We apply our new Cham approach to solve Equation (13(12) $\{\begin{array}{l}{u}_t+\alpha u_{\mathrm{xxy}}+4{\left(\mathrm{uv}\right)}_x=0,\\ u_y=v_x.\end{array}$(12) ). Assuming the following: (19) $u(x,y,t)=v(x,y,t)=U\left(\xi \right);\xi =x+y-\mathrm{\lambda t},\lambda \ne 0,$(19) results in the following equation: (20) $-\mathrm{\lambda U}\left(\xi \right)+\alpha U^{{}^{\mathrm{\prime \prime }}}\left(\xi \right)+4U^2\left(\xi \right)=0.$(20) Balancing both $U^2(\xi )$ and $U^{{}^{\mathrm{\prime \prime }}}(\xi )$, we obtain m = 2. As a result, we can assume that the solutions to Equation (21(20) $-\mathrm{\lambda U}\left(\xi \right)+\alpha U^{{}^{\mathrm{\prime \prime }}}\left(\xi \right)+4U^2\left(\xi \right)=0.$(20) ) are the following: (21) $U\left(\xi \right)=a_0+a_1\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+a_2{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2,$(21) where $a_i(i=0,1,2)$ are constants with $a_2\ne 0$. To find the algebraic system, we first substitute Equation (22(21) $U\left(\xi \right)=a_0+a_1\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+a_2{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2,$(21) ) into Equation (21(20) $-\mathrm{\lambda U}\left(\xi \right)+\alpha U^{{}^{\mathrm{\prime \prime }}}\left(\xi \right)+4U^2\left(\xi \right)=0.$(20) ) to represent the left side as polynomials in $(\frac{\tanh (Z(\xi ))}{\tanh (\xi )}{)}^j(j=0,1,2,\dots )$, and then set their coefficients to zero: $\begin{array}{rl}& (\mathrm{BC}{a}_1+2C^2{a}_2)\alpha +4a_0^2-a_0\lambda =0,\\ & (6\mathrm{BC}{a}_2+a_1(2\mathrm{AC}+B^2))\alpha +8a_1(a_0-\frac{\lambda }{8})=0,\\ & ((8\mathrm{AC}+4B^2)a_2+3\mathrm{AB}{a}_1)\alpha +(8a_0-\lambda )a_2+4a_1^2=0,\\ & (2A^2{a}_1+10\mathrm{AB}{a}_2)\alpha +8a_1{a}_2=0,\\ & 6A^2{a}_2\alpha +4a_2^2=0.\end{array}$The solution to this algebraic system for $a_0$, $a_1$, $a_2$, and λ yields two sets: Set 1: (22) $\begin{array}{rl}& a_0=-{\displaystyle \frac{\left(3\mathrm{AC\alpha }\right)}{2}},a_1=-{\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}},\\ & a_2=-{\displaystyle \frac{\left(3A^2\alpha \right)}{2}},\lambda =-\alpha (4\mathrm{AC}-B^2)\end{array}$(22) We substitute Equation (23(22) $\begin{array}{rl}& a_0=-{\displaystyle \frac{\left(3\mathrm{AC\alpha }\right)}{2}},a_1=-{\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}},\\ & a_2=-{\displaystyle \frac{\left(3A^2\alpha \right)}{2}},\lambda =-\alpha (4\mathrm{AC}-B^2)\end{array}$(22) ) into Equation (22(21) $U\left(\xi \right)=a_0+a_1\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+a_2{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2,$(21) ) to get: (23) $\{\begin{array}{l}U\left(\xi \right)=-{\displaystyle \frac{\left(3\mathrm{AC\alpha }\right)}{2}}-{\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}}\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)-{\displaystyle \frac{\left(3A^2\alpha \right)}{2}}{\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)}^2,\xi =x+y+\alpha (4\mathrm{AC}-B^2)t.\end{array}$(23) Set 2: (24) $\begin{array}{rl}& a_0=-\frac{\alpha (2\mathrm{AC}+B^2)}{4},a_1=-\frac{3\mathrm{AB\alpha }}{2},\\ & a_2=-\frac{3A^2\alpha }{2},\lambda =\alpha (4\mathrm{AC}-B^2)\end{array}$(24) Equation (25(24) $\begin{array}{rl}& a_0=-\frac{\alpha (2\mathrm{AC}+B^2)}{4},a_1=-\frac{3\mathrm{AB\alpha }}{2},\\ & a_2=-\frac{3A^2\alpha }{2},\lambda =\alpha (4\mathrm{AC}-B^2)\end{array}$(24) ) is substituted into Equation (22(21) $U\left(\xi \right)=a_0+a_1\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+a_2{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2,$(21) ) to get: (25) $\{\begin{array}{l}U\left(\xi \right)=-{\displaystyle \frac{\alpha (2\mathrm{AC}+B^2)}{4}}-{\displaystyle \frac{3\mathrm{AB\alpha }}{2}}\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)-{\displaystyle \frac{3A^2\alpha }{2}}{\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)}^2,\xi =x+y-\alpha (4\mathrm{AC}-B^2)t.\end{array}$(25) As a result, using Equations (24(23) $\{\begin{array}{l}U\left(\xi \right)=-{\displaystyle \frac{\left(3\mathrm{AC\alpha }\right)}{2}}-{\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}}\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)-{\displaystyle \frac{\left(3A^2\alpha \right)}{2}}{\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)}^2,\xi =x+y+\alpha (4\mathrm{AC}-B^2)t.\end{array}$(23) ), (26(25) $\{\begin{array}{l}U\left(\xi \right)=-{\displaystyle \frac{\alpha (2\mathrm{AC}+B^2)}{4}}-{\displaystyle \frac{3\mathrm{AB\alpha }}{2}}\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)-{\displaystyle \frac{3A^2\alpha }{2}}{\left({\displaystyle \frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}}\right)}^2,\xi =x+y-\alpha (4\mathrm{AC}-B^2)t.\end{array}$(25) ), and solutions to Equation (4(4) $\begin{array}{rl}{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^{\prime }& =A{\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)}^2\\ & +B\left(\frac{\tanh (Z(\xi ))}{\tanh \left(\xi \right)}\right)+C.\end{array}$(4) ), the exact solutions to Equation (13(12) $\{\begin{array}{l}{u}_t+\alpha u_{\mathrm{xxy}}+4{\left(\mathrm{uv}\right)}_x=0,\\ u_y=v_x.\end{array}$(12) ) are obtained as follows:

Case 1: When $4\mathrm{AC}-B^2>0$ and $\mathrm{AC}\ne 0$. (26) $\begin{array}{rl}& \{\begin{array}{l}{u}_{1.1}(\xi )=-{\displaystyle \frac{\left(3\mathrm{AC\alpha }\right)}{2}}+{\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}}\\ \left(\frac{\begin{array}{c}B\tanh \left(\xi \right)\\ +\tan \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\\ \sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)\\ -{\displaystyle \frac{\left(3A^2\alpha \right)}{2}}{\left(\frac{\begin{array}{c}B\tanh \left(\xi \right)\\ +\tan \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\\ \sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)}^2,\\ v_{1.1}(\xi )=u_{1.1}(\xi ),\\ \xi =x+y+\alpha (4\mathrm{AC}-B^2)t,{\xi }_0\in \mathbb{R}.\end{array}\end{array}$(26) (27) $\begin{array}{rl}& \{\begin{array}{l}{u}_{2.1}(\xi )=-\frac{\alpha (2\mathrm{AC}+B^2)}{4}+\frac{3\mathrm{AB\alpha }}{2}\\ \left(\frac{\begin{array}{c}B\tanh \left(\xi \right)\\ +\tan \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\\ \sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)\\ -{\displaystyle \frac{3A^2\alpha }{2}}{\left(\frac{\begin{array}{c}B\tanh \left(\xi \right)\\ +\tan \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\\ \sqrt{\tanh {\left(\xi \right)}^2(4\mathrm{AC}-B^2)}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)}^2,\\ v_{2.1}(\xi )=u_{2.1}(\xi ),\\ \xi =x+y-\alpha (4\mathrm{AC}-B^2)t,{\xi }_0\in \mathbb{R}.\end{array}\end{array}$(27) Case 2: When $4\mathrm{AC}-B^2<0$ and $\mathrm{AC}\ne 0$. (28) $\begin{array}{rl}& \{\begin{array}{l}{u}_{1.2}(\xi )=-{\displaystyle \frac{\left(3\mathrm{AC\alpha }\right)}{2}}+{\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}}\left(\frac{\begin{array}{c}B\tanh \left(\xi \right)-\tanh \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)-{\displaystyle \frac{\left(3A^2\alpha \right)}{2}}{\left(\frac{\begin{array}{c}B\tanh \left(\xi \right)-\tanh \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)}^2,\\ v_{1.2}(\xi )=u_{1.2}(\xi ),\\ \xi =x+y+\alpha (4\mathrm{AC}-B^2)t,{\xi }_0\in \mathbb{R}.\end{array}\end{array}$(28) (29) $\begin{array}{rl}& \{\begin{array}{l}{u}_{2.2}(\xi )=-{\displaystyle \frac{\alpha (2\mathrm{AC}+B^2)}{4}}+{\displaystyle \frac{3\mathrm{AB\alpha }}{2}}\left(\frac{\begin{array}{c}B\tanh \left(\xi \right)-\tanh \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)-{\displaystyle \frac{3A^2\alpha }{2}}{\left(\frac{\begin{array}{c}B\tanh \left(\xi \right)-\tanh \left(\frac{\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}({\xi }_0-\xi )}{2\tanh \left(\xi \right)}\right)\sqrt{\tanh {\left(\xi \right)}^2(B^2-4\mathrm{AC})}\end{array}}{2\mathrm{Atanh}\left(\xi \right)}\right)}^2,\\ v_{2.2}(\xi )=u_{2.2}(\xi ),\xi =x+y-\alpha (4\mathrm{AC}-B^2)t,{\xi }_0\in \mathbb{R}.\end{array}\end{array}$(29)

Case 3: Only C is equal to zero, (30) $\begin{array}{rl}& \{\begin{array}{l}{u}_{1.3}(\xi )={\displaystyle \frac{\left(3\mathrm{AB\alpha }\right)}{2}}\left({\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{\tanh (\xi )(\mathrm{Aexp}(\mathrm{B\xi })-{\xi }_0)(\exp (2\xi )+1)}}\right)-{\displaystyle \frac{\left(3A^2\alpha \right)}{2}}{\left({\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{\begin{array}{c}\tanh (\xi )(A\exp (B\xi )-{\xi }_0)\\ (\exp (2\xi )+1)\end{array}}}\right)}^2,v_{1.3}(\xi )=u_{1.3}(\xi ),\xi =x+y-\alpha B^{2}t,{\xi }_0\in \mathbb{R}.\\ \xi \notin \left\{\frac{1}{B}\ln \left({\displaystyle \frac{{\xi }_0}{A}}\right),0\right\},\\ \left|{\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{(\mathrm{Aexp}(\mathrm{B\xi })-{\xi }_0)(\exp (2\xi )+1)}}\right|<1.\end{array}\end{array}$(30) (31) $\begin{array}{rl}& \{\begin{array}{l}{u}_{2.3}(\xi )=-{\displaystyle \frac{\alpha (2\mathrm{AC}+B^2)}{4}}+{\displaystyle \frac{3\mathrm{AB\alpha }}{2}}\left({\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{\tanh (\xi )(\mathrm{Aexp}(\mathrm{B\xi })-{\xi }_0)(\exp (2\xi )+1)}}\right)-{\displaystyle \frac{3A^2\alpha }{2}}{\left({\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{\tanh (\xi )(\mathrm{Aexp}(\mathrm{B\xi })-{\xi }_0)(\exp (2\xi )+1)}}\right)}^2,v_{2.3}(\xi )=u_{2.3}(\xi ),\xi =x+y+\alpha B^{2}t,{\xi }_0\in \mathbb{R}.\\ \xi \notin \left\{\frac{1}{B}\ln \left({\displaystyle \frac{{\xi }_0}{A}}\right),0\right\},\\ \left|{\displaystyle \frac{B(\exp (2\xi )-1)\exp (\mathrm{B\xi })}{(\mathrm{Aexp}(\mathrm{B\xi })-{\xi }_0)(\exp (2\xi )+1)}}\right|<1.\end{array}\end{array}$(31)

5. Graphical illustration

We graphically depict some solutions in Maple 2022. When $\frac{\lambda }{\alpha }>0$ and $\alpha \ne 0$, the Equation (21(20) $-\mathrm{\lambda U}\left(\xi \right)+\alpha U^{{}^{\mathrm{\prime \prime }}}\left(\xi \right)+4U^2\left(\xi \right)=0.$(20) ) has a solitary wave solution $u_{1,2}(\xi )$, and two periodic wave solutions $u_{1,3}(\xi )$ and $u_{2,1}(\xi )$. When $\frac{\lambda }{\alpha }<0$ and $\alpha \ne 0$, the Equation (21(20) $-\mathrm{\lambda U}\left(\xi \right)+\alpha U^{{}^{\mathrm{\prime \prime }}}\left(\xi \right)+4U^2\left(\xi \right)=0.$(20) ) has a solitary wave solution $u_{2,2}(\xi )$, and two periodic wave solutions $u_{1,1}(\xi )$ and $u_{2,3}(\xi )$ (Figure 3–5).

Figure 3. (a) $u_{1.1}(\xi )$ when $A=B=2,C=1,{\xi }_0=0,\alpha =-1$, and y = 0, (b) $u_{2.1}(\xi )$ when A = B = 2, $\alpha =1$, y = 0, ${\xi }_0=0$, and C = 1.

Figure 4. (a) $u_{1.2}(\xi )$ for y = 0, $\alpha =2$, ${\xi }_0=0$, A = C = 1, and $B=2\sqrt{2}$, (b) $u_{2.2}(\xi )$ for A = 1, $B=2\sqrt{2},$C=1$,{\xi }_0=0,\alpha =\frac{1}{2}$, and $y=0$.

Figure 5. (a) $u_{1.3}(\xi )$ when $\alpha =2$, A = B = 1, and $C={\xi }_0=y=0$, (b) $u_{2.3}(\xi )$ for y = 0, $\alpha =-4$, ${\xi }_0=0$, A = B = 1, and C = 0

6. Conclusion

We presented the Cham method as a new and efficient method for obtaining eight solution families to NLPDEs. This method has been used successfully for solving Bogoyavlenskii's equations. We also discussed the dynamic behaviour of these equations and the bifurcation of traveling waves, and illustrated some solutions graphically. We hope that the generality of the proposed method will inspire researchers to use it to solve other complex differential equations.

Disclosure statement

No potential conflict of interest was reported by the author(s).