In this section, we proceed to considering the symmetry variable for reduction of equation (1.1) to the systems of nonlinear ordinary differential equation (for details, see [37,38,39]) through the characteristic equation
$$ \frac{dx}{X ( x, t, u, w )} = \frac{dt}{T ( x, t, u, w )} = \frac{du}{U ( x, t, u, w )} = \frac{dw}{W ( x, t, u, w )} $$
(3.1)
for the different operators in the optimal system, and then we continue contracting soliton solutions of the main system (1.1) with the help of modified \(( G'/G )\)-expansion method.
Subalgebra \(\varGamma _{1} + \lambda _{1} \varGamma _{2} + \lambda _{2} \varGamma _{3}\)
The solutions of system (1.1) which are invariant under this subalgebra with the coefficient functions
$$\begin{aligned}& \tau _{2} ( t ) = \frac{K_{1}}{1 - \lambda _{ 1}} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr) ^{ \frac{2 \lambda _{ 1} - \lambda _{ 2} + 1}{1 - \lambda _{ 1}}}, \qquad \tau _{3} ( t ) = \frac{K_{2}}{1 - \lambda _{ 1}} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr) ^{\frac{\lambda _{ 1} + 2}{1 - \lambda _{ 1}}}, \\& \kappa _{1} ( t ) = \frac{K_{3}}{1 - \lambda _{ 1}} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr) ^{ \frac{\lambda _{ 1} - \lambda _{ 2}}{1 - \lambda _{ 1}}}, \qquad \kappa _{2} ( t ) = \frac{K_{4}}{1 - \lambda _{ 1}} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr) ^{ \frac{\lambda _{ 1} + 1}{ 1 - \lambda _{ 1}}}, \\& \kappa _{3} ( t ) = \frac{K_{5}}{1 - \lambda _{ 1}} \tau _{1} ( t ) \biggl( \int \kappa _{1} ( t ) \,dt \biggr) ^{ \frac{\lambda _{ 2} + 2}{1 - \lambda _{ 1}}} \end{aligned}$$
(3.2)
are given by the following relations:
$$ \begin{aligned} &u ( x, t ) = \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ \frac{\lambda _{ 1}}{1 - \lambda _{ 1}}} f ( \xi ), \\ &w ( x, t ) = \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ \frac{\lambda _{ 2}}{1 - \lambda _{ 1}}} g ( \xi ), \end{aligned} $$
(3.3)
where \(\xi = x ( \int \tau _{1} ( t ) )^{ \frac{ - 1}{1 - \lambda _{ 1}}}\) and \(f ( \xi )\), \(g ( \xi )\) are the solution of the system of nonlinear ordinary differential equations
$$ \begin{aligned} &\lambda _{ 1} f - f' \xi + (1 - \lambda _{ 1}) f f' + K_{1} g'' + K_{2} f''' = 0, \\ &\lambda _{ 2} g - g' \xi + K_{3} g g' + K_{4} g'' + K_{5} f''' = 0. \end{aligned} $$
(3.4)
Solving ODEs (3.4) for \(f ( \xi )\) and \(g ( \xi )\), substituting in equation (3.3) the main system (1.1) possesses the following solutions:
$$ \begin{aligned} &\begin{aligned} u ( x, t ) &= x \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1} + 12 K_{2} C_{2}^{2} - 4 C_{2}^{2} \sqrt{6} K_{2} \tanh \biggl( C_{1} + C_{2} x \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1} \biggr) \\ &\quad {}- 12 K_{2} C_{2}^{2} \tanh \biggl( C_{1} + C_{2} \biggl( x \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1} \biggr) \biggr) ^{2}, \end{aligned} \\ &\begin{aligned} w ( x, t ) &= \frac{x ( \int \tau _{1} ( t ) \,dt )^{ - 1}}{K_{3}} + \frac{20 \sqrt{6} K_{2}^{2} C _{3}^{2} \tanh ( C_{1} + C_{2} x ( \int \tau _{1} ( t ) \,dt )^{ - 1} )^{2}}{K_{1}} \\ &\quad {}- \frac{80 \sqrt{6} K_{2}^{2} C_{2}^{3} K_{3}}{6 K_{1} K_{3}}, \end{aligned} \end{aligned} $$
(3.5)
where \(K_{4} = - \frac{10 C_{2}^{2} K_{2}^{2} K_{3}}{3 K_{1}}\), \(K_{5} = - \frac{50 C_{2}^{2} K_{2}^{2} K_{3}}{3 K_{1}^{2}}\), \(\lambda _{1} = \lambda _{2} = 0\), and \(K_{1}\), \(K_{2}\), \(K_{3}\) are arbitrary constants.
Subalgebra \(\varGamma _{2} + \lambda _{3} \varGamma _{3}\)
The solutions of system (1.1) which are invariant under this subalgebra with the coefficient functions
$$\begin{aligned}& \tau _{2} ( t ) = K_{1} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ \lambda _{3} - 1}, \qquad \tau _{3} ( t ) = K_{2} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1}, \\& \kappa _{1} ( t ) = K_{3} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ \lambda _{ 3} - 1}, \qquad \kappa _{2} ( t ) = K_{4} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1}, \\& \kappa _{3} ( t ) = K_{5} \tau _{1} ( t ) \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - \lambda _{3}} \end{aligned}$$
(3.6)
are given by the following relations:
$$ \begin{aligned} &u ( x, t ) = \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1} f ( \xi ), \\ &w ( x, t ) = \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - \lambda _{ 3}} g ( \xi ), \end{aligned} $$
(3.7)
where \(\xi = x\) and \(f ( \xi )\), \(g ( \xi )\) are the solution of the system of nonlinear ordinary differential equations
$$ \begin{aligned} &{-} f + f f' + K_{1} g'' + K_{2} f''' = 0, \\ &{-} \lambda _{ 3} g + K_{3} g g' + K_{4} g'' + K_{5} f''' = 0. \end{aligned} $$
(3.8)
Now, under this subalgebra for arbitrary value of \(\lambda _{3}\), we are able to find out trivial solutions of nonlinear system (1.1). Therefore, to establish the nontrivial solution of system (1.1) by using equation (3.7), we have to solve equation (3.8) by the modified \(( G'/G )\)-expansion method [14, 15] with the restriction \(\lambda _{3} = 0\). Let us suppose that system (3.8) admits a solution in the modified \(( G'/G )\)-expansion method as follows:
$$ \begin{aligned} &f(\xi ) = a_{0} + \sum _{i = 1}^{P} \biggl\{ a_{i} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{i} + b_{i} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{ - i} \biggr\} , \\ &g(\xi ) = c_{0} + \sum_{j = 1}^{q} \biggl\{ c_{j} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{j} + d_{j} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{ - j} \biggr\} , \end{aligned} $$
(3.9)
where \(a_{0}\), \(a_{i}\), \(b_{i}\), \(c_{0}\), \(c_{j}\), \(d_{j}\), and μ are constants. The summation indexes p, q in equation (3.9) are positive integers, and function \(G(\xi )\) is the solution of the linear ODE
$$ G''(\xi ) + \mu G(\xi ) = 0. $$
(3.10)
The positive integers p, q may be decided in view of the homogeneous balance among the highest order derivative terms and nonlinear terms from equation (3.8). Therefore, equalizing the highest-order derivatives and nonlinear terms arriving in equation (3.8), we find \(p = 2\), \(q = 2\), and we set down
$$ \begin{aligned} &f(\xi ) = a_{0} + a_{1} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr) + a _{2} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{2} + b_{1} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{ - 1} + b_{2} \biggl( \frac{G'(\xi )}{G( \xi )} \biggr)^{ - 2}, \\ &g(\xi ) = c_{0} + c_{1} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr) + c _{2} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{2} + d_{1} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{ - 1} + d_{2} \biggl( \frac{G'(\xi )}{G( \xi )} \biggr)^{ - 2}. \end{aligned} $$
(3.11)
Further, substituting (3.11) into the system of ODEs (3.8), using linear ODE (3.10), and equating the coefficients of the same powers of \(( G'/G )\) to zero, we arrive at a set of simultaneous algebraic equations among \(a_{0}\), \(a_{1}\), \(a_{2}\), \(b_{1}\), \(b_{2}\), \(c_{0}\), \(c _{1}\), \(c_{2}\), \(d_{1}\), \(d_{2}\), and μ as follows:
$$\begin{aligned} (\mathrm{i})&\quad 24 b_{2} K_{2} \mu ^{3} + 2 b_{2}^{2} \mu = 0, \\ (\mathrm{ii})&\quad 6 d_{2} K_{1} \mu ^{2} + 6 b_{1} \mu ^{3} K_{2} + 3 \mu b_{1} b_{2} = 0, \\ (\mathrm{iii})&\quad 2 d_{1} K_{1} \mu ^{2} + 40 b_{2} \mu ^{2} K_{2} + 2 a _{0} \mu b_{2} + b_{1}^{2} \mu + 2 b_{2}^{2} = 0, \\ (\mathrm{iv})&\quad 8 d_{2} K_{1} \mu + 8 b_{1} \mu ^{2} K_{2} + a_{0} \mu b_{1} + b_{2} a_{1} \mu + 3 b_{1} b_{2} - b_{2} = 0, \\ (\mathrm{v})&\quad 2 d_{1} K_{1} \mu + 16 b_{2} \mu K_{2} + 2 a_{0} b_{2} - 2 b_{2} a_{2} \mu + b_{1}^{2} + 2 a_{2} b_{2} \mu - b_{1} = 0, \\ (\mathrm{vi})&\quad 2 d_{2}K_{2} - 2 a_{1}\mu ^{2}K_{2} + 2 b_{1} \mu K_{2} - a_{0} a_{1} \mu + a_{0} b_{1}+ a_{1}b_{2} - b_{1} a_{2} \mu \\ &\qquad {} + 2 c _{2} K_{1} \mu ^{2} - a_{0} = 0, \\ (\mathrm{vii})&\quad 2 c_{1} K_{1} \mu - 16 a_{2} \mu ^{2}K_{2} - 2 a_{0} a _{2} \mu - a_{1}^{2}\mu - a_{1} = 0, \\ (\mathrm{viii})&\quad 8 c_{2} K_{1} \mu - 8 a_{1} \mu K_{2} - 3 a_{1} a_{2} \mu - a_{0} a_{1} - a_{2} b_{1} - a_{2} = 0, \\ (\mathrm{ix})&\quad 2 c_{1} K_{1} - 40 a_{2} \mu K_{2} - 2 a_{0} a_{2} - a _{1}^{2} - 2 a_{2}^{2} \mu = 0, \\ (\mathrm{x})&\quad {-} 6 a_{1} K_{2} - 3 a_{1} a_{2} + 6 K_{1} c_{2} = 0, \\ (\mathrm{xi})&\quad {-} 24 a_{2} K_{2} - 2 a_{2}^{2} = 0. \end{aligned}$$
(3.12)
Also, when we put (3.11) into the second equation of (3.8), it brings the system of algebraic equations as follows:
$$\begin{aligned} (\mathrm{i})&\quad 24 b_{2} K_{5} \mu ^{3} + 2 K_{3} d_{2}^{2} \mu = 0, \\ (\mathrm{ii})&\quad 6 d_{2} K_{4} \mu ^{2} + 6 b_{1} \mu ^{3} K_{5} + 3 K_{3} \mu d_{1} d_{2} = 0, \\ (\mathrm{iii})&\quad 2 d_{1} K_{4} \mu ^{2} + 40 b_{2} \mu ^{2} K_{5} + 2 c _{0} \mu d_{2} K_{3} + K_{3} d_{1}^{2} \mu + 2 K_{3} d_{2}^{2} = 0, \\ (\mathrm{iv})&\quad 8 d_{2} K_{4} \mu + 8 b_{1} \mu ^{2} K_{5} + K_{3} c_{0} \mu d_{1} + K_{3} d_{2} c_{1} \mu + 3 K_{3} d_{1} d_{2} = 0, \\ (\mathrm{v})&\quad 2 d_{1} K_{4} \mu + 16 b_{2} \mu K_{5} + 2 K_{3} c_{0} d _{2} - 2 K_{3} d_{2} c_{2} \mu + K_{3} d_{1}^{2} + 2 K_{3} c_{2} d _{2} \mu = 0, \\ (\mathrm{vi})&\quad 2 d_{2}K_{4} - 2 a_{1}\mu ^{2}K_{5} + 2b_{1} \mu K_{5} - K_{3}c_{0} c _{1} \mu + K_{3}c_{0}d_{1} + K_{3}c_{1}d_{2} - K_{3}d_{1}c_{2}\mu \\ &\qquad + 2c_{2}K_{4}\mu ^{2} = 0, \\ (\mathrm{vii})&\quad 2 c_{1} K_{4} \mu - 16 a_{2} \mu ^{2}K_{5} - 2 K_{3} c _{0} c_{2} \mu - K_{3} c_{1}^{2}\mu = 0, \\ (\mathrm{viii})&\quad 8 c_{2} K_{4} \mu - 8 a_{1} \mu K_{5} - 3 K_{3} c_{1} c_{2} \mu - K_{3} c_{0} c_{1} - K_{3} c_{2} d_{1} = 0, \\ (\mathrm{ix})&\quad 2 c_{1} K_{4} - 40 a_{2} \mu K_{5} - 2 K_{3} c_{0} c _{2} - K_{3} c_{1}^{2} - 2 K_{3} c_{2}^{2} \mu = 0, \\ (\mathrm{x})&\quad {-} 6 a_{1} K_{5} - 3 K_{3} c_{1} c_{2} + 6 K_{4} c_{2} = 0, \\ (\mathrm{xi})&\quad {-} 24 a_{2} K_{5} - 2 K_{3} c_{2}^{2} = 0. \end{aligned}$$
(3.13)
On solving algebraic equations (3.12) and (3.13) for arbitrary values of \(K_{1}\) and \(K_{3}\), we get only the trivial solutions of the main system (1.1), which is not a physically interesting case. Therefore, we reach the following results:
Case 1:
$$ \begin{aligned} &a_{0} = a_{0}, \qquad b_{1} = 1, c_{0} = c_{0}, \qquad d_{1} = d_{1}\quad \mbox{and} \\ &a_{1} = a_{2} = b_{2} = c_{1} = c_{2} = d_{2} = K_{3} = \mu = 0; \end{aligned} $$
(3.14)
Case 2:
$$ \begin{aligned} &a_{0} = a_{0},\qquad b_{1} = 1,\qquad c_{1} = \frac{2 K_{4}}{K_{3}}\quad \mbox{and} \\ &a_{1} = a_{2} = b_{2} = c_{0} = c_{2} = d_{2} = d_{2} = K _{1} = \mu = 0. \end{aligned} $$
(3.15)
Now, when we substitute equations (3.14), (3.15) and the solution of linear equation (3.10) for \(\mu = 0\) into equation (3.11), we get the solution of nonlinear equation (3.8). Further, using the value of solution \(f ( \xi )\) and \(g ( \xi )\) into equation (3.7), we have rational solutions of equation (1.1) as follows.
Case 1 gives the rational solution of equation (1.1) by the following relation:
$$ \begin{aligned} &u ( x, t ) = \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1} \biggl[ a_{0} + \biggl( \frac{B}{A + B x} \biggr)^{ - 1} \biggr], \\ &w(x, t) = c_{0} + d_{1} \biggl( \frac{B}{A + B x} \biggr)^{ - 1}, \end{aligned} $$
(3.16)
where A and B are arbitrary constants (see Fig. 1).
Similarly, Case 2 gives the rational solution by the following relation:
$$ \begin{aligned} &u ( x, t ) = \biggl( \int \tau _{1} ( t ) \,dt \biggr)^{ - 1} \biggl[ a_{0} + \biggl( \frac{B}{A + B x} \biggr)^{ - 1} \biggr], \\ &w(x, t) = \frac{2 K_{4}}{K_{3}} \biggl( \frac{B}{A + B x} \biggr) ^{2}, \end{aligned} $$
(3.17)
where A, B, \(K_{3}\), and \(K_{4}\) are arbitrary constants.
Subalgebra \(\varGamma _{3} + \lambda _{4} \varGamma _{4}\)
The solutions of system (1.1), which are invariant under this subalgebra with the coefficient functions
$$\begin{aligned}& \tau _{2} ( t ) = \frac{K_{1}}{\lambda _{4}} \tau _{1} ( t ) \exp \biggl( - \frac{1}{\lambda _{4}} \int \tau _{1} ( t ) \,dt \biggr), \qquad \tau _{3} ( t ) = \frac{K_{2}}{\lambda _{4}} \tau _{1} ( t ), \\& \kappa _{1} ( t ) = \frac{K_{3}}{\lambda _{4}} \tau _{1} ( t ) \exp \biggl( - \frac{1}{\lambda _{4}} \int \tau _{1} ( t ) \,dt \biggr), \qquad \kappa _{2} ( t ) = \frac{K_{4}}{\lambda _{4}} \tau _{1} ( t ), \\& \kappa _{3} ( t ) = \frac{K_{5}}{\lambda _{4}} \tau _{1} ( t ) \exp \biggl( \frac{1}{\lambda _{4}} \int \tau _{1} ( t ) \,dt \biggr), \end{aligned}$$
(3.18)
are given by the following relations:
$$ \begin{aligned} &u ( x, t ) = f ( \xi ), \\ &w ( x, t ) = \exp \biggl( \frac{1}{\lambda _{ 4}} \int \tau _{1} ( t ) \,dt \biggr) g ( \xi ), \end{aligned} $$
(3.19)
where \(\xi = x\) and \(f ( \xi )\), \(g ( \xi )\) are the solution of the system of nonlinear ordinary differential equations
$$ \begin{aligned} &\lambda _{4} f f' + K_{1} g'' + K_{2} f''' = 0, \\ &g + K_{3} g g' + K_{4} g'' + K_{5} f''' = 0. \end{aligned} $$
(3.20)
In this subalgebra, we also seek solutions of equation (3.20) by the modified \(( G'/G )\)-expansion method. Therefore, following the procedure adopted in subalgebra 3.2, we get \(p = 2\), \(q = 2\). So we can suppose that the solution of ODEs (3.20) is of the form
$$ \begin{aligned} &f(\xi ) = a_{0} + a_{1} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr) + a _{2} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{2} + b_{1} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{ - 1} + b_{2} \biggl( \frac{G'(\xi )}{G( \xi )} \biggr)^{ - 2}, \\ &g(\xi ) = c_{0} + c_{1} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr) + c _{2} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{2} + d_{1} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{ - 1} + d_{2} \biggl( \frac{G'(\xi )}{G( \xi )} \biggr)^{ - 2}. \end{aligned} $$
(3.21)
Substituting (3.21) into the system of ODEs (3.20), using linear ODE (3.10), and equating the coefficients of same powers of \(( G'/G )\) to zero, we obtain a set of simultaneous algebraic equations among \(a_{0}\), \(a_{1}\), \(a_{2}\), \(b_{1}\), \(b_{2}\), \(c_{0}\), \(c_{1}\), \(c_{2}\), \(d_{1}\), \(d _{2}\), and μ as follows:
$$\begin{aligned} (\mathrm{i}) &\quad 24 b_{2} K_{2} \mu ^{3} + 2 \lambda _{4} b_{2}^{2} \mu = 0, \\ (\mathrm{ii}) &\quad 6 d_{2} K_{1} \mu ^{2} + 6 b_{1} \mu ^{3} K_{2} + 3 \lambda _{4} \mu b_{1} b_{2} = 0, \\ (\mathrm{iii})&\quad 2 d_{1} K_{1} \mu ^{2} + 40 b_{2} \mu ^{2} K_{2} + 2 \lambda _{4} a_{0} \mu b_{2} + \lambda _{4} b_{1}^{2} \mu + 2 \lambda _{4} b_{2}^{2} = 0, \\ (\mathrm{iv})&\quad 8 d_{2} K_{1} \mu + 8 b_{1} \mu ^{2} K_{2} + a_{0} \lambda _{4} \mu b_{1} + b_{2} \lambda _{4} a_{1} \mu + 3 \lambda _{4} b _{1} b_{2} = 0, \\ (\mathrm{v})&\quad 2 d_{1} K_{1} \mu + 16 b_{2} \mu K_{2} + 2 \lambda _{4} a _{0} b_{2} - 2 \lambda _{4} b_{2} a_{2} \mu + \lambda _{4} b_{1}^{2} + 2 \lambda _{4} a_{2} b_{2} \mu = 0, \\ (\mathrm{vi})&\quad 2 d_{2}K_{2} - 2 a_{1}\mu ^{2}K_{2} + 2b_{1} \mu K_{2} - \lambda _{4}a_{0} a_{1}\mu + \lambda _{4}a_{0}b_{1}+ \lambda _{4} a_{1}b _{2} - \lambda _{4}b_{1}a_{2} \mu \\ &\qquad {} + 2 c_{2} K_{1} \mu ^{2} = 0, \\ (\mathrm{vii})&\quad 2 c_{1} K_{1} \mu - 16 a_{2} \mu ^{2}K_{2} - 2 \lambda _{4} a_{0} a_{2} \mu - a_{1}^{2} \lambda _{4} \mu = 0, \\ (\mathrm{viii})&\quad 8 c_{2} K_{1} \mu - 8 a_{1} \mu K_{2} - 3 \lambda _{4} a_{1} a_{2} \mu - \lambda _{4} a_{0} a_{1} - \lambda _{4} a_{2} b_{1} = 0, \\ (\mathrm{ix})&\quad 2 c_{1} K_{1} - 40 a_{2} \mu K_{2} - 2 \lambda _{4} a _{0} a_{2} - \lambda _{4} a_{1}^{2} - 2 \lambda _{4} a_{2}^{2} \mu = 0, \\ (\mathrm{x})&\quad {-} 6 a_{1} K_{2} - 3 \lambda _{4} a_{1} a_{2} + 6 K_{1} c _{2} = 0, \\ (\mathrm{xi})&\quad {-} 24 a_{2} K_{2} - 2 \lambda _{4} a_{2}^{2} = 0. \end{aligned}$$
(3.22)
Also, when we put (3.21) into the second equation of (3.20), it brings the system of algebraic equations as follows:
$$\begin{aligned} (\mathrm{i})&\quad 24 b_{2} K_{5} \mu ^{3} + 2 K_{3} d_{2}^{2} \mu = 0, \\ (\mathrm{ii})&\quad 6 d_{2} K_{4} \mu ^{2} + 6 b_{1} \mu ^{3} K_{5} + 3 K_{3} \mu d_{1} d_{2} = 0, \\ (\mathrm{iii})&\quad 2 d_{1} K_{4} \mu ^{2} + 40 b_{2} \mu ^{2} K_{5} + 2 c _{0} \mu d_{2} K_{3} + K_{3} d_{1}^{2} \mu + 2 K_{3} d_{2}^{2} = 0, \\ (\mathrm{iv})&\quad 8 d_{2} K_{4} \mu + 8 b_{1} \mu ^{2} K_{5} + K_{3} c_{0} \mu d_{1} + K_{3} d_{2} c_{1} \mu + 3 K_{3} d_{1} d_{2} + d_{2} = 0, \\ (\mathrm{v})&\quad 2 d_{1} K_{4} \mu + 16 b_{2} \mu K_{5} + 2 K_{3} c_{0} d _{2} - 2 K_{3} d_{2} c_{2} \mu + K_{3} d_{1}^{2} + 2 K_{3} c_{2} d _{2} \mu + d_{1} = 0, \\ (\mathrm{vi})&\quad 2 d_{2}K_{4} - 2 a_{1}\mu ^{2}K_{5} + 2b_{1} \mu K_{5} - K _{3}c_{0}c_{1}\mu + K_{3}c_{0}d_{1} + K_{3}c_{1}d_{2} - K_{3}d_{1}c _{2}\mu \\ &\qquad{} + 2c_{2}K_{4}\mu ^{2} + c_{0} = 0, \\ (\mathrm{vii})&\quad 2 c_{1} K_{4} \mu - 16 a_{2} \mu ^{2}K_{5} - 2 K_{3} c _{0} c_{2} \mu - K_{3} c_{1}^{2}\mu + c_{1} = 0, \\ (\mathrm{viii})&\quad 8 c_{2} K_{4} \mu - 8 a_{1} \mu K_{5} - 3 K_{3} c_{1} c_{2} \mu - K_{3} c_{0} c_{1} - K_{3} c_{2} d_{1} + c_{2} = 0, \\ (\mathrm{ix})&\quad 2 c_{1} K_{4} - 40 a_{2} \mu K_{5} - 2 K_{3} c_{0} c _{2} - K_{3} c_{1}^{2} - 2 K_{3} c_{2}^{2} \mu = 0, \\ (\mathrm{x})&\quad {-} 6 a_{1} K_{5} - 3 K_{3} c_{1} c_{2} + 6 K_{4} c_{2} = 0, \\ (\mathrm{xi})&\quad {-} 24 a_{2} K_{5} - 2 K_{3} c_{2}^{2} = 0. \end{aligned}$$
(3.23)
On solving algebraic equations (3.22) and (3.23) for arbitrary value of \(K_{5}\), we get only the trivial solutions of the main system (1.1), which is not a physically interesting case. Therefore, we reach the following results.
Case 1:
$$ \begin{aligned} &a_{2} = a_{2},\qquad c_{0} = c_{0},\qquad d_{1} = - \frac{1}{K_{3}}, \qquad K_{2} = - \frac{1}{12} \lambda _{4} a_{2}, \\ &a_{0} = a_{1} = b_{1} = b_{2} = c_{1} = c_{2} = d_{2} = K_{5} = \mu = 0; \end{aligned} $$
(3.24)
Case 2:
$$ \begin{aligned} &a_{2} = - \frac{12 K_{2}}{\lambda _{4}} , \qquad c_{0} = c_{0} , \qquad d_{1} = - \frac{1}{K _{3}} , \quad \mbox{and} \\ &a_{0} = a_{1} = b_{1} = b_{2} = c_{1} = c_{2} = d_{2} = K _{5} = \mu = 0 . \end{aligned} $$
(3.25)
Now when we substitute equations (3.24), (3.25) and the solution of linear equation (3.10) for \(\mu = 0\) into equation (3.21), we get the solution of nonlinear equation (3.20). Further, using the value of solution \(f ( \xi )\) and \(g ( \xi )\) into equation (3.19), we have rational solutions of equation (1.1) as follows.
Case 1 gives the rational solution of equation (1.1) by the following relation:
$$ \begin{aligned} &u(x, t) = a_{2} \biggl( \frac{B}{A + B x} \biggr)^{2}, \\ &w ( x, t ) = \exp \biggl( \frac{1}{\lambda _{ 4}} \int \tau _{1} ( t ) \,dt \biggr) \biggl[ c_{0} - \frac{1}{K_{3}} \biggl( \frac{B}{A + B x} \biggr)^{ - 1} \biggr], \end{aligned} $$
(3.26)
where A, B, \(K_{3}\), and \(\lambda _{4}\) are arbitrary constants (see Fig. 2).
Similarly, Case 2 gives the rational solution of equation (1.1) by the following relation:
$$ \begin{aligned} &u(x, t) = - \frac{12 K_{2}}{\lambda _{4}} \biggl( \frac{A_{2}}{A_{1} + A_{2} x} \biggr)^{2}, \\ &w ( x, t ) = \exp \biggl( \frac{1}{\lambda _{ 4}} \int \tau _{1} ( t ) \,dt \biggr) \biggl[ c_{0} - \frac{1}{K_{3}} \biggl( \frac{A_{2}}{A_{1} + A_{2} x} \biggr)^{ - 1} \biggr], \end{aligned} $$
(3.27)
where A, B, \(K_{2}\), \(K_{3}\), and \(\lambda _{4}\) are arbitrary constants.
Subalgebra \(\varGamma _{4} + \varGamma _{5}\)
The solutions of system (1.1) which are invariant under this subalgebra with the coefficient functions
$$ \begin{aligned} &\tau _{2} ( t ) = K_{1} \tau _{1} ( t ), \qquad \tau _{3} ( t ) = K_{2} \tau _{1} ( t ), \qquad \kappa _{1} ( t ) = K_{3} \tau _{1} ( t ), \\ &\kappa _{2} ( t ) = K_{4} \tau _{1} ( t ), \qquad \kappa _{3} ( t ) = K_{5} \tau _{1} ( t ) \end{aligned} $$
(3.28)
are given by the following relations:
$$ \begin{aligned} &u ( x, t ) = f ( \xi ), \\ &w ( x, t ) = g ( \xi ), \end{aligned} $$
(3.29)
where \(\xi = x - \int \tau _{1}(t) \,dt\) and \(f ( \xi )\), \(g ( \xi )\) are the solution of the system of nonlinear ordinary differential equations
$$ \begin{aligned} &{-} f' + f f' + K_{1} g'' + K_{2} f''' = 0, \\ &{-} g' + K_{3} g g' + K_{4} g'' + K_{5} f''' = 0. \end{aligned} $$
(3.30)
In this subalgebra, follow the procedure in a manner similar to the preceding subalgebra to solve equations (3.30) by the modified \(( G'/G )\)-expansion method, we get the solution of ODEs (3.30) as follows:
$$ \begin{aligned} &f(\xi ) = a_{0} + a_{1} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr) + a _{2} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{2} + b_{1} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{ - 1} + b_{2} \biggl( \frac{G'(\xi )}{G( \xi )} \biggr)^{ - 2}, \\ &g(\xi ) = c_{0} + c_{1} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr) + c _{2} \biggl( \frac{G'(\xi )}{G(\xi )} \biggr)^{2} + d_{1} \biggl( \frac{G'( \xi )}{G(\xi )} \biggr)^{ - 1} + d_{2} \biggl( \frac{G'(\xi )}{G( \xi )} \biggr)^{ - 2}. \end{aligned} $$
(3.31)
By substituting (3.31) into the system of ODEs (3.30), using linear ODE (3.10), and equating the coefficients of same powers of \(( G'/G )\) to zero we obtain a set of simultaneous algebraic equations among \(a_{0}\), \(a_{1}\), \(a_{2}\), \(b_{1}\), \(b_{2}\), \(c_{0}\), \(c_{1}\), \(c_{2}\), \(d_{1}\), \(d _{2}\), and μ as follows:
$$\begin{aligned} (\mathrm{i})&\quad 24 b_{2} K_{2} \mu ^{3} + 2 b_{2}^{2} \mu = 0, \\ (\mathrm{ii})&\quad 6 d_{2} K_{1} \mu ^{2} + 6 b_{1} \mu ^{3} K_{2} + 3 \mu b_{1} b_{2} = 0, \\ (\mathrm{iii})&\quad 2 d_{1} K_{1} \mu ^{2} + 40 b_{2} \mu ^{2} K_{2} + 2 a _{0} \mu b_{2} + b_{1}^{2} \mu + 2 b_{2}^{2} - 2 b_{2} \mu = 0, \\ (\mathrm{iv})&\quad 8 d_{2} K_{1} \mu + 8 b_{1} \mu ^{2} K_{2} + a_{0} \mu b_{1} + b_{2} a_{1} \mu + 3 b_{1} b_{2} - b_{1} \mu = 0, \\ (\mathrm{v})&\quad 2 d_{1} K_{1} \mu + 16 b_{2} \mu K_{2} + 2 a_{0} b_{2} - 2 b_{2} a_{2} \mu + b_{1}^{2} + 2 a_{2} b_{2} \mu - 2 b_{2} = 0, \\ (\mathrm{vi})&\quad 2 d_{2}K_{2} - 2a_{1} \mu ^{2}K_{2} + 2b_{1}\mu K_{2} - a _{0}a_{1}\mu + a_{0}b_{1} + a_{1}b_{2} - b_{1}a_{2}\mu \\ &\qquad {}+ 2 c_{2}K_{1} \mu ^{2} + a_{1} \mu - b_{1} = 0, \\ (\mathrm{vii})&\quad 2 c_{1} K_{1} \mu - 16 a_{2} \mu ^{2}K_{2} - 2 a_{0} a _{2} \mu - a_{1}^{2}\mu + 2 a_{2} \mu = 0, \\ (\mathrm{viii})&\quad 8 c_{2} K_{1} \mu - 8 a_{1} \mu K_{2} - 3 a_{1} a_{2} \mu - a_{0} a_{1} - a_{2} b_{1} + a_{1} = 0, \\ (\mathrm{ix})&\quad 2 c_{1} K_{1} - 40 a_{2} \mu K_{2} - 2 a_{0} a_{2} - a _{1}^{2} - 2 a_{2}^{2} \mu + 2 a_{2} = 0, \\ (\mathrm{x})&\quad {-} 6 a_{1} K_{2} - 3 a_{1} a_{2} + 6 K_{1} c_{2} = 0, \\ (\mathrm{xi})&\quad {-} 24 a_{2} K_{2} - 2 a_{2}^{2} = 0. \end{aligned}$$
(3.32)
Also, when we put (3.31) into the second equation of (3.30), it brings the system of algebraic equations as follows:
$$\begin{aligned} (\mathrm{i})&\quad 24 b_{2} K_{5} \mu ^{3} + 2 K_{3} d_{2}^{2} \mu = 0, \\ (\mathrm{ii})&\quad 6 d_{2} K_{4} \mu ^{2} + 6 b_{1} \mu ^{3} K_{5} + 3 K_{3} \mu d_{1} d_{2} = 0, \\ (\mathrm{iii})&\quad 2 d_{1} K_{4} \mu ^{2} + 40 b_{2} \mu ^{2} K_{5} + 2 c _{0} \mu d_{2} K_{3} + K_{3} d_{1}^{2} \mu + 2 K_{3} d_{2}^{2} - 2 d _{2} \mu = 0, \\ (\mathrm{iv})&\quad 8 d_{2} K_{4} \mu + 8 b_{1} \mu ^{2} K_{5} + K_{3} c_{0} \mu d_{1} + K_{3} d_{2} c_{1} \mu + 3 K_{3} d_{1} d_{2} - d_{1} \mu = 0, \\ (\mathrm{v})&\quad 2 d_{1} K_{4} \mu + 16 b_{2} \mu K_{5} + 2 K_{3} c_{0} d _{2} - 2 K_{3} d_{2} c_{2} \mu + K_{3} d_{1}^{2} + 2 K_{3} c_{2} d _{2} \mu - 2 d_{2} = 0, \\ (\mathrm{vi})&\quad 2 d_{2}K_{4} - 2 a_{1}\mu ^{2}K_{5} + 2 b_{1} \mu K_{5} - K_{3}c_{0}c_{1}\mu + K_{3}c_{0}d_{1} + K_{3}c_{1}d_{2} - K_{3}d_{1}c _{2}\mu \\ &\qquad {}+ 2 c_{2}K_{4}\mu ^{2} + c_{1}\mu - d_{1} = 0, \\ (\mathrm{vii})&\quad 2 c_{1} K_{4} \mu - 16 a_{2} \mu ^{2}K_{5} - 2 K_{3} c _{0} c_{2} \mu - K_{3} c_{1}^{2}\mu + 2 c_{2} \mu = 0, \\ (\mathrm{viii})&\quad 8 c_{2} K_{4} \mu - 8 a_{1} \mu K_{5} - 3 K_{3} c_{1} c_{2} \mu - K_{3} c_{0} c_{1} - K_{3} c_{2} d_{1} + c_{1} = 0, \\ (\mathrm{ix})&\quad 2 c_{1} K_{4} - 40 a_{2} \mu K_{5} - 2 K_{3} c_{0} c _{2} - K_{3} c_{1}^{2} - 2 K_{3} c_{2}^{2} \mu + 2 c_{2} = 0, \\ (\mathrm{x})&\quad {-} 6 a_{1} K_{5} - 3 K_{3} c_{1} c_{2} + 6 K_{4} c_{2} = 0, \\ (\mathrm{xi})&\quad {-} 24 a_{2} K_{5} - 2 K_{3} c_{2}^{2} = 0. \end{aligned}$$
(3.33)
Similarly solving these algebraic equations as under the above subalgebras yields the following.
Case 1:
$$ \begin{aligned} &a_{0} = - \frac{1}{8} \frac{a_{2} c_{1}^{2} - 8 c_{2}^{2}}{c_{2}^{2}}, \qquad a_{1} = \frac{c_{1} a_{2}}{c_{2}}, \qquad a_{2} = a_{2}, \\ &b_{1} = \frac{1}{16} \frac{a_{2} c_{1}^{3}}{c_{2}^{3}}, \qquad b_{2} = \frac{1}{256} \frac{a_{2} c_{1}^{4}}{c_{2}^{4}}, \qquad c_{0} = - \frac{1}{8} \frac{K_{3} c_{1}^{2} - 8 c_{2}}{K_{3} c_{2}}, \\ &c_{1} = c_{1}, \qquad c_{2} = c_{2}, \qquad d_{1} = \frac{1}{16} \frac{ c_{1}^{3}}{c_{2}^{2}}, \\ &d_{2} = \frac{1}{256} \frac{c_{1}^{4}}{c_{2}^{3}}, \qquad K_{1} = \frac{5}{12}\frac{c_{1} a_{2}^{2}}{c_{2}^{2}}, \qquad K_{2} = - \frac{1}{12}a_{2}, \\ &K_{3} = K_{3}, \qquad K_{4} = \frac{5}{12} c_{1} K_{3}, \qquad K_{5} = - \frac{1}{12} \frac{c_{2}^{2}}{a_{2}} K_{3}, \qquad \mu = - \frac{1}{16} \frac{c_{1}^{2}}{c_{2}^{2}}. \end{aligned} $$
(3.34)
Case 2:
$$ \begin{aligned} &a_{0} = 1, \qquad a_{1} = a_{1}, \qquad a_{2} = 0,\qquad b_{1} = b_{2} = 0, \qquad c_{0} = \frac{c_{1}}{2 K_{4}}, \\ & c_{1} = c_{1}, \qquad c_{2} = 0, \qquad d_{1} = d_{2} = 0, \qquad K_{1} = \frac{a_{1}^{2}}{2 c_{1}}, \\ &K_{2} = 0, \qquad K_{3} = \frac{ 2 K_{4}}{c_{1}}, \qquad K_{5} = 0, \\ &K_{4} = K_{4}, \qquad \mu = \mu . \end{aligned} $$
(3.35)
Now, when we substitute equations (3.34), (3.35) and the solution of linear equation (3.10) into equation (3.31), we get the solution of nonlinear equation (3.30). Further, using the value of solution \(f ( \xi )\) and \(g ( \xi )\) into equation (3.29), we have hyperbolic, trigonometric, and rational function solutions of equation (1.1) as follows.
Case 1:
Now, under Case 1 we have \(\mu = - \frac{1}{16} \frac{c_{1}^{2}}{c _{2}^{2}}\), i.e., (\(\mu < 0\)), then the hyperbolic function solution of the main system (1.1) is given as follows:
$$\begin{aligned}& \begin{aligned} u (x, t) &= - \frac{1}{8} \frac{a_{2} c_{1}^{2} - 8 c_{2}^{2}}{c_{2} ^{2}} \\ &\quad {}+ \frac{c_{1} a_{2}}{c_{2}} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr] \\ &\quad {}+ a_{2} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr]^{2} \\ &\quad {}+ \frac{1}{16} \frac{a_{2} c_{1}^{3}}{c_{2}^{3}} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr]^{ - 1} \\ &\quad {}+ \frac{1}{256}\frac{a_{2} c_{1}^{4}}{c_{2}^{4}} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr]^{ - 2}, \end{aligned} \\& \begin{aligned}[b] w(x, t) &= - \frac{1}{8} \frac{K_{3} c_{1}^{2} - 8 c_{2}}{K_{3} c_{2}} \\ &\quad {}+ c_{1} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr] \\ &\quad {}+ c_{2} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr]^{2} \\ &\quad {}+ \frac{1}{16}\frac{ c_{1}^{3}}{c_{2}^{2}} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr]^{ - 1} \\ &\quad {}+ \frac{1}{256} \frac{c_{1}^{4}}{c_{2}^{3}} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr]^{ - 2}, \end{aligned} \end{aligned}$$
(3.36)
where \(a_{2}\), \(c_{1}\), \(c_{2}\), \(K_{3}\), and A, B are arbitrary constants.
Again under Case 1, if we substitute \(\mu = 0\) in equation (3.34), then we have \(K_{1} = K_{4} = c_{1} = 0\). Therefore, for \(\mu = 0\), we obtained the rational function solutions of equation (1.1) as follows:
$$ \begin{aligned} &u (x, t) = 1 + a_{2} \biggl[ \frac{B}{A + B ( x - \int \tau _{1}(t)\,dt )} \biggr]^{2}, \\ & w(x, t) = \frac{1}{8 K_{3}} + c_{2} \biggl[ \frac{B}{A + B ( x - \int \tau _{1}(t)\,dt )} \biggr]^{2}, \end{aligned} $$
(3.37)
where A, B and \(a_{2}\), \(c_{2}\), \(K_{3}\) are arbitrary constants.
Case 2:
If \(\mu > 0\), the trigonometric function solution of the main equation (1.1) is given by the following relation:
$$ \begin{aligned} &u (x, t) = 1 + a_{1} \biggl[ \sqrt{ \mu } \frac{A \sin ( \sqrt{ \mu } ( x - \int \tau _{1}(t)\,dt ) ) - B \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr], \\ &w(x, t) = \frac{c_{1}}{2 K_{4}} + c_{1} \biggl[ \sqrt{\mu } \frac{A \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t)\,dt ) ) - B \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sin ( \sqrt{\mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr], \end{aligned} $$
(3.38)
where A, B, and \(K_{4}\) are arbitrary constants.
If \(\mu < 0\), then the hyperbolic function solution of the main system (1.1) is given as follows:
$$ \begin{aligned} &u (x, t) = 1 + a_{1} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sin ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sin ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr], \\ &w(x, t) = \frac{c_{1}}{2 K_{4}} + c_{1} \biggl[ \sqrt{ - \mu } \frac{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )}{A \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t) \,dt ) ) + B \sinh ( \sqrt{ - \mu } ( x - \int \tau _{1}(t)\,dt ) )} \biggr], \end{aligned} $$
(3.39)
where A, B, and \(K_{4}\) are arbitrary constants.
When \(\mu = 0\), we obtain the rational solution as follows:
$$ \begin{aligned} &u (x, t) = 1 + a_{1} \biggl[ \frac{B}{A + B ( x - \int \tau _{1}(t)\,dt )} \biggr], \\ &w (x, t) = \frac{c_{1}}{2 K_{4}} + c_{1} \biggl[ \frac{B}{A + B ( x - \int \tau _{1}(t)\,dt )} \biggr], \end{aligned} $$
(3.40)
where A, B, and \(K_{4}\) are arbitrary constants (see Fig. 3).