3.1. Application of the generalized Kudryashov method
Balancing the orders of the linear term \(U^{{}^{\prime\prime}}\)and the nonlinear term\(U^{3}\) in Eq. (1.3), we have \(N=M+1.\)So, the solution of Eq. (1.3) can be put in the following form if we choose \(M=1\):
\(U\left(\xi\right)=\frac{a_{0}+a_{1}R\left(\xi\right)+a_{2}R^{2}\left(\xi\right)}{b_{0}+b_{1}R\left(\xi\right)}\). (3.1)
In Eq. (3.1), the constants \(a_{i}\ \left(i=0,\ 1,\ 2\right)\) and\(b_{j}\ \left(j=0,\ 1\right)\) are to be determined later and\(R(\xi)\) satisfies the ODE specified by Eq. (2.2).
Putting Eq. (3.1) with the aid of Eq. (2.2) into Eq. (1.3) and using some mathematical operations, we get a system of algebraic equations. Solving the emanated system of equations, we obtain the following solution sets:
Set-I: \(a_{0}=0\),\(a_{1}=\mp\frac{1}{2}\frac{pb_{1}l_{1}l_{2}\text{\ ln}\left(a\right)}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\),\(a_{2}={\pm b}_{1}\ln\left(a\right)\ \sqrt{\frac{pl_{1}l_{2}}{qh_{1}}}\),\(b_{0}=0\),
\(\omega=-\frac{1}{2}p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)\), and \(b_{1}\) is free.
Set-II: \(a_{0}=0\),\(a_{1}={\pm 2b}_{0}\ln\left(a\right)\ \sqrt{\frac{pl_{1}l_{2}}{qh_{1}}}\),\(a_{2}={\mp 2b}_{0}\ln\left(a\right)\ \sqrt{\frac{pl_{1}l_{2}}{qh_{1}}}\),\(b_{1}=-2b_{0}\),
\(\omega=p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}-h_{1}h_{2}\right)\), and \(b_{0}\) is free.
Set-III: \(a_{0}=0\),\(a_{1}=\pm\frac{b_{1}\ln\left(a\right)}{2}\frac{\sqrt{\text{pq}h_{1}l_{1}l_{2}}\ }{qh_{1}}\),\(a_{2}=\mp\frac{pl_{1}l_{2}\ln\left(a\right)\ b_{1}}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\),\(b_{0}=0,\)
\(\omega=-\frac{1}{2}p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)\), and \(b_{1}\) is free.
Set-IV:\(a_{0}=\pm\frac{b_{0}\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}{qh_{1}}\),\(a_{1}=\mp\frac{{2b}_{0}\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}{qh_{1}}\),\(a_{2}=\pm\frac{2pl_{1}l_{2}b_{0}\ln\left(a\right)}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\),\(b_{1}=-2b_{0}\),\(\omega=-p\left(2l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+h_{1}h_{2}\right)\), and \(b_{0}\) is free.
Set-V:\(a_{0}=\pm\frac{1}{2}\frac{b_{0}\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}{qh_{1}}\),\(a_{1}=0\),\(a_{2}=\mp\frac{2pl_{1}l_{2}b_{0}\ln\left(a\right)}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\),\(b_{1}=2b_{0}\),
\(\omega=-\frac{1}{2}p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)\), and \(b_{0}\) is free.
Set-VI: \(a_{2}=-\left(4a_{0}+2a_{1}\right)\),\(b_{0}=\pm\frac{2qh_{1}a_{0}}{\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}},\ b_{1}=\pm\frac{2\sqrt{\text{pq}h_{1}l_{1}l_{2}}\left(2a_{0}+a_{1}\right)}{pl_{1}l_{2}\ln\left(a\right)}\),
\(\omega=-\frac{1}{2}p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)\), and \(a_{0}\),\(\ a_{1}\) are free.
Consequently, the above sets (I-VI) retrieve the following optical solutions:
\(Q_{1,2}\left(x,y,t\right)\ =\ \frac{\mp\left(\frac{1}{2}\frac{pb_{1}l_{1}l_{2}\text{\ ln}\left(a\right)}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\frac{1}{\left(1+da^{\xi}\right)}-b_{1}\ln\left(a\right)\ \sqrt{\frac{pl_{1}l_{2}}{qh_{1}}}\frac{1}{\left(1+da^{\xi}\right)^{2}}\right)}{\frac{b_{1}}{\left(1+da^{\xi}\right)}}\times e^{i\left(-h_{1}x\ -h_{2}\text{y\ }-\frac{1}{2}p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)t+\ \theta_{0}\right)}\),
\(Q_{3,4}\left(x,y,t\right)\ =\ \frac{{\pm 2b}_{0}\ln\left(a\right)\ \sqrt{\frac{pl_{1}l_{2}}{qh_{1}}}\left(\frac{1}{\left(1+da^{\xi}\right)}-\frac{1}{\left(1+da^{\xi}\right)^{2}}\right)}{b_{0}\left(1-\frac{2}{\left(1+da^{\xi}\right)}\right)}\times e^{i\left(-h_{1}x\ -h_{2}y+p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}-h_{1}h_{2}\right)t\ +\ \theta_{0}\right)}\),
\(Q_{5,6}\left(x,y,t\right)\ =\ \frac{\pm\left(\frac{b_{1}\ln\left(a\right)}{2}\frac{\sqrt{\text{pq}h_{1}l_{1}l_{2}}\ }{qh_{1}}\frac{1}{\left(1+da^{\xi}\right)}-\frac{pl_{1}l_{2}\ln\left(a\right)\ b_{1}}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\frac{1}{\left(1+da^{\xi}\right)^{2}}\right)}{\left(\frac{b_{1}}{\left(1+da^{\xi}\right)}\right)}\times e^{i\left(-h_{1}x\ -h_{2}y-\frac{p}{2}\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)t\ +\ \theta_{0}\right)}\),
\(Q_{7,8}\left(x,y,t\right)\ =\ \frac{\pm\left(\frac{b_{0}\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}{qh_{1}}-\frac{{2b}_{0}\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}{qh_{1}}\frac{1}{\left(1+da^{\xi}\right)}+\frac{2pl_{1}l_{2}b_{0}\ln\left(a\right)}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\frac{1}{\left(1+da^{\xi}\right)^{2}}\right)}{b_{0}\left(1-\frac{2}{\left(1+da^{\xi}\right)}\right)}\times e^{i\left(-h_{1}x\ -h_{2}\text{y\ }-p\left(2l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+h_{1}h_{2}\right)t\ +\ \theta_{0}\right)}\),
\(Q_{9,10}\left(x,y,t\right)\ =\ \frac{\pm\left(\frac{1}{2}\frac{b_{0}\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}{qh_{1}}-\frac{2pl_{1}l_{2}b_{0}\ln\left(a\right)}{\sqrt{\text{pq}h_{1}l_{1}l_{2}}}\frac{1}{\left(1+da^{\xi}\right)^{2}}\right)}{b_{0}\left(1+\frac{2}{\left(1+da^{\xi}\right)}\right)}\times e^{i\left(-h_{1}x\ -h_{2}\text{y\ }-\frac{1}{2}p\left(l_{1}l_{2}\left(\ln\left(a\right)\right)^{2}+2h_{1}h_{2}\right)t\ +\ \theta_{0}\right)}\),
\(Q_{11,12}\left(x,y,t\right)\ =\ \frac{a_{0}+\frac{a_{1}}{\left(1+da^{\xi}\right)}-\frac{\left(4a_{0}+2a_{1}\right)}{\left(1+da^{\xi}\right)^{2}}}{\pm\left(\frac{2qh_{1}a_{0}}{\ln\left(a\right)\sqrt{\text{pq}h_{1}l_{1}l_{2}}}+\frac{2\sqrt{\text{pq}h_{1}l_{1}l_{2}}\left(2a_{0}+a_{1}\right)}{pl_{1}l_{2}\ln\left(a\right)}\frac{1}{\left(1+da^{\xi}\right)}\right)}\times e^{i\left(-h_{1}x\ -h_{2}\text{y\ }-\frac{h_{1}}{b_{1}^{2}}\left(pb_{1}^{2}h_{2}+2qa_{1}^{2}\right)t\ +\ \theta_{0}\right)}\),
where\(\xi=\ l_{1}x\ +\ l_{2}y+p\ (l_{1}h_{2}\ +\ l_{2}h_{1})t\) for all the optical solutions presented in this subsection.