In the previous section, we have dealt with the similarity reductions and derived the corresponding reduced equations. In this section, we use the tanh method on reduced equations, obtaining some exact solutions of Eq. (1). With the help of exact solutions, we can understand some motion rules of waves of the \((2+1)\)-dimensional integrable coupling of KdV equation.
The main steps of the tanh method [24, 25] are expressed as follows:
1. Consider the following nonlinear differential equations:
$$ \textstyle\begin{cases} F_{1}(u,v,u_{x},v_{x},u_{xx},v_{xx},\ldots ,u_{y},v_{y},\ldots ,u_{t},v_{t}, \ldots )=0, \\ F_{2}(u,v,u_{x},v_{x},u_{xx},v_{xx},\ldots ,u_{y},v_{y},\ldots ,u_{t},v_{t}, \ldots )=0, \end{cases} $$
(20)
where \(F_{1}\), \(F_{2}\) are polynomials of the u, v and their derivatives.
2. By using the wave transformations
$$ \textstyle\begin{cases} u(x,y,t)=\Phi (\xi ), \\ v(x,y,t)=\Psi (\xi ), \end{cases} $$
(21)
where \(\xi =lx+ky+ct\), and l, k, c are unknown constants, and substituting (21) into Eq. (20), we obtain the following nonlinear ordinary differential equations:
$$ \textstyle\begin{cases} F_{1}(\Phi ,\Psi ,l\Phi ',l\Psi ',l^{2}\Phi '',l^{2}\Psi '',\ldots ,k \Phi ',k\Psi ',\ldots ,c\Phi ',c\Psi ',\ldots )=0, \\ F_{2}(\Phi ,\Psi ,l\Phi ',l\Psi ',l^{2}\Phi '',l^{2}\Psi '',\ldots ,k \Phi ',k\Psi ',\ldots ,c\Phi ',c\Psi ',\ldots )=0. \end{cases} $$
(22)
3. Next, we introduce the independent variable
$$ Y=\tanh (\xi ), $$
(23)
which leads to the following changes:
$$\begin{aligned}& \frac{d}{d\xi }=\bigl(1-Y^{2}\bigr)\frac{d}{dY}, \\& \frac{d^{2}}{d\xi ^{2}}=\bigl(1-Y^{2}\bigr)\biggl[-2Y \frac{d}{dY}+\bigl(1-Y^{2}\bigr) \frac{d^{2}}{dY^{2}} \biggr], \\& \frac{d^{3}}{d\xi ^{3}}=\bigl(1-Y^{2}\bigr)\biggl[ \bigl(6Y^{2}-2\bigr)\frac{d}{dY}-6Y\bigl(1-Y^{2} \bigr) \frac{d^{2}}{dY^{2}}+\bigl(1-Y^{2}\bigr)^{2} \frac{d^{3}}{dY^{3}}\biggr]. \end{aligned}$$
4. We assume that the solution of Eq. (22) is written as the following form:
$$ \Phi (Y)=\sum_{i=0}^{n}a_{i}Y^{i},\qquad \Psi (Y)=\sum_{i=0}^{m}b_{i}Y^{i}, $$
(24)
where n, m are positive integers, which are decided by balancing the highest order nonlinear terms with the derivative terms in the resulting equations. After deciding n, m, taking (23) and (24) into (22), we obtain a polynomial concerning \(Y^{i}\) (\(i=0,1,2,\ldots \)). Then we gather all terms of \(Y^{i}\) (\(i=0,1,2,\ldots \)) and make all them equal to zero. Solving these algebraic equations, we get the values of the unknown numbers \(a_{i}\), \(b_{i}\) (\(i=0,1,\ldots \)), l, k and c. Then, putting these values into the equations, we get exact solutions of equations.
Case 4.1. For Eq. (12), substituting Eq. (21) into (12), we get the following equations:
$$ \textstyle\begin{cases} k\Phi '=l^{3}\Phi ^{(3)}+6l\Phi \Phi ', \\ k\Psi '=l^{3}\Psi ^{(3)}-3l^{2}k\Phi ^{(3)}+6l\Phi '\Psi +6l\Phi \Psi '-6k\Phi \Phi '. \end{cases} $$
(25)
Concerning (25), balancing \(\Phi ^{(3)}\) with \(\Phi \Phi '\), we have
$$ 2\times 3+n-3=n+2\times 1+n-1 \quad \Longrightarrow\quad n=2, $$
balancing \(\Phi ^{(3)}\) with \(\Phi '\Psi \), we have
$$ 2\times 3+n-3=2\times 1+n-1+m \quad \Longrightarrow\quad m=2. $$
Hence, according to Eq. (24), the solution of Eq. (12) is assumed to be
$$ \textstyle\begin{cases} \Phi (Y)=a_{0}+a_{1}Y+a_{2}Y^{2}, \\ \Psi (Y)=b_{0}+b_{1}Y+b_{2}Y^{2}. \end{cases} $$
(26)
Then, substituting Eq. (23) and Eq. (26) into Eq. (25), we collect all terms of \(Y^{i}\) and obtain the algebraic equations including unknown numbers \(a_{i}\), \(b_{i}\) (\(i=0,1,2\)), l and k. By solving these equations, we have the following solutions:
$$ \begin{aligned} &l=l,\qquad k=k, \qquad a_{0}=\frac{8l^{3}+k}{6l},\qquad a_{1}=0,\qquad a_{2}=-2l^{2}, \\ &b_{0}= \frac{k(-16l^{3}+k)}{6l^{2}},\qquad b_{1}=0,\qquad b_{2}=4lk. \end{aligned} $$
(27)
Putting (27) into Eq. (12), we obtain the exact solution as follows:
$$ \textstyle\begin{cases} u(x,y,t)=\frac{8l^{3}+k}{6l}-2l^{2}\tanh ^{2}(lx+k(t-\ln y)), \\ v(x,y,t)= \frac{\frac{k(-16l^{3}+k)}{6l^{2}}+4lk\tanh ^{2}(lx+k(t-\ln y))}{y}, \end{cases} $$
(28)
where \(l\neq 0\), k are arbitrary constants.
Figures 1 and 2 depict the exact solution of Eq. (12), which is obtained by taking \(l=1\), \(k=1\) at \(t=1\).
Case 4.2. For Eq. (13), similarly, substituting Eq. (21) into (13), we have the following ordinary differential equations:
$$ \textstyle\begin{cases} l^{3}\Phi ^{(3)}+6l\Phi \Phi '=0, \\ l^{3}\Psi ^{(3)}+3l^{2}k\Phi ^{(3)}+6l\Phi '\Psi +6l\Phi \Psi '+6k \Phi \Phi '=0. \end{cases} $$
(29)
Then, balancing \(\Phi ^{(3)}\) and \(\Phi \Phi '\), \(\Phi ^{(3)}\) and \(\Phi '\Psi '\) for (29), we have \(n=m=2\).
Therefore, on the basis of Eq. (24), the solution of Eq. (13) can be assumed to be
$$ \textstyle\begin{cases} \Phi (Y)=a_{0}+a_{1}Y+a_{2}Y^{2}, \\ \Psi (Y)=b_{0}+b_{1}Y+b_{2}Y^{2}. \end{cases} $$
(30)
Next, substituting Eq. (23) and Eq. (30) into Eq. (29), we make all coefficients of \(Y^{i}\) vanish and obtain the algebraic equations including the unknown numbers \(a_{i}\), \(b_{i}\) (\(i=0,1,2\)), l and k. Solving these equations, we have the following solutions:
$$ \begin{aligned} &l=l,\qquad k=k, \qquad a_{0}=\frac{4l^{2}}{3},\qquad a_{1}=0,\qquad a_{2}=-2l^{2}, \\ &b_{0}= \frac{8lk}{3},\qquad b_{1}=0,\qquad b_{2}=-4lk. \end{aligned} $$
(31)
So, the exact solution of Eq. (13) is
$$ \textstyle\begin{cases} u(x,y,t)=\frac{4l^{2}}{3}-2l^{2}\tanh ^{2}(lx+ky), \\ v(x,y,t)=\frac{8lk}{3}-4lk\tanh ^{2}(lx+ky), \end{cases} $$
(32)
where l, k are arbitrary constants. This solution is a static solution of Eq. (1).
When we take \(l=1\), \(k=1\), the values of u, v are illustrated in Figs. 3 and 4.
Case 4.3. For Eq. (14), equally, substituting Eq. (21) into (14), we get the following ordinary differential equations:
$$ \textstyle\begin{cases} -\Phi '=c^{2}\Phi ^{(3)}+6\Phi \Phi ', \\ -c\Psi '=c^{3}\Psi ^{(3)}+3kc^{2}\Phi ^{(3)}+6c\Phi '\Psi +6c\Phi \Psi '+6k\Phi \Phi '. \end{cases} $$
(33)
Furthermore, balancing \(\Phi ^{(3)}\) and \(\Phi \Phi '\), \(\Phi ^{(3)}\) and \(\Phi '\Psi '\) for (33), we have \(n=m=2\).
Therefore, based on Eq. (24), the solution of Eq. (14) can be assumed to be
$$ \textstyle\begin{cases} \Phi (Y)=a_{0}+a_{1}Y+a_{2}Y^{2}, \\ \Psi (Y)=b_{0}+b_{1}Y+b_{2}Y^{2}. \end{cases} $$
(34)
Next, substituting Eq. (23) and Eq. (34) into Eq. (33), we make all coefficients of \(Y^{i}\) vanish and obtain the algebraic equations including unknown numbers \(a_{i}\), \(b_{i}\) (\(i=0,1,2\)), k and c. Solving these equations, we have the following solutions:
$$ \begin{aligned} &k=k,\qquad c=c,\qquad a_{0}=\frac{8c^{2}-1}{6},\qquad a_{1}=0,\qquad a_{2}=-2c^{2}, \\ &b_{0}= \frac{k(16c^{2}+1)}{6c},\qquad b_{1}=0,\qquad b_{2}=-4kc. \end{aligned} $$
(35)
So, the exact solution of Eq. (14) is
$$ \textstyle\begin{cases} u(x,y,t)=\frac{8c^{2}-1}{6}-2c^{2}\tanh ^{2}(ky+c(x-t)), \\ v(x,y,t)=\frac{k(16c^{2}+1)}{6c}-4kc\tanh ^{2}(ky+c(x-t)), \end{cases} $$
(36)
where \(c\neq 0\) and k are arbitrary constants.
Figures 5 and 6 portray the solution of Eq. (14), which is obtained by taking \(k=-1\), \(c=1\) at \(t=1\).
Case 4.4. For Eq. (15), in the same way, substituting Eq. (21) into (15), we have the following ordinary differential equations:
$$ \textstyle\begin{cases} -k\Phi '=l^{3}\Phi ^{(3)}+6l\Phi \Phi ', \\ -k\Psi '=l^{3}\Psi ^{(3)}+3l^{2}k\Phi ^{(3)}+6l\Phi '\Psi +6l\Phi \Psi '+6k\Phi \Phi '. \end{cases} $$
(37)
Then, balancing \(\Phi ^{(3)}\) and \(\Phi \Phi '\), \(\Phi ^{(3)}\) and \(\Phi '\Psi '\) for (33), we have \(n=m=2\).
Therefore, based on Eq. (24), the solution of Eq. (15) can be assumed to be
$$ \textstyle\begin{cases} \Phi (Y)=a_{0}+a_{1}Y+a_{2}Y^{2}, \\ \Psi (Y)=b_{0}+b_{1}Y+b_{2}Y^{2}. \end{cases} $$
(38)
Next, substituting Eq. (23) and Eq. (38) into Eq. (37), we make all coefficients of \(Y^{i}\) vanish and obtain the algebraic equations including unknown numbers \(a_{i}\), \(b_{i}\) (\(i=0,1,2\)), l and k. Solving these equations, we have the following solutions:
$$ \begin{aligned} &l=l, \qquad k=k,\qquad a_{0}=\frac{8l^{3}-k}{6l},\qquad a_{1}=0,\qquad a_{2}=-2l^{2}, \\ &b_{0}= \frac{k(16l^{3}+k)}{6l^{2}},\qquad b_{1}=0, \qquad b_{2}=-4lk. \end{aligned} $$
(39)
So, the exact solution of Eq. (15) is
$$ \textstyle\begin{cases} u(x,y,t)=\frac{8l^{3}-k}{6l}-2l^{2}\tanh ^{2}(lx+k(y-t)), \\ v(x,y,t)=\frac{k(16l^{3}+k)}{6l^{2}}-4lk\tanh ^{2}(lx+k(y-t)), \end{cases} $$
(40)
where \(l\neq 0\), k are arbitrary constants.
When we take \(l=1\), \(k=-1\) at \(t=0\), the values of u, v are illustrated in Figs. 7 and 8.
Case 4.5. For Eq. (16), likewise, substituting Eq. (21) into (16), we get the following ordinary differential equations:
$$ \textstyle\begin{cases} -c\Phi '=l^{3}\Phi ^{(3)}+6l\Phi \Phi ', \\ -c\Psi '=l^{3}\Psi ^{(3)}-3l^{3}\Phi ^{(3)}+6l\Phi '\Psi +6l\Phi \Psi '-6l\Phi \Phi '. \end{cases} $$
(41)
Then, balancing \(\Phi ^{(3)}\) and \(\Phi \Phi '\), \(\Phi ^{(3)}\) and \(\Phi '\Psi '\) for (41), we have \(n=m=2\).
Therefore, based on Eq. (24), the solution of Eq. (16) can be assumed to be
$$ \textstyle\begin{cases} \Phi (Y)=a_{0}+a_{1}Y+a_{2}Y^{2}, \\ \Psi (Y)=b_{0}+b_{1}Y+b_{2}Y^{2}. \end{cases} $$
(42)
Next, substituting Eq. (23) and Eq. (42) into Eq. (41), we make all coefficients of \(Y^{i}\) vanish and obtain the algebraic equations including unknown numbers \(a_{i}\), \(b_{i}\) (\(i=0,1,2\)), l and c. Solving these equations, we have the following solutions:
$$ \begin{aligned} &l=l,\qquad c=c,\qquad a_{0}=\frac{8l^{3}-c}{6l},\qquad a_{1}=0,\qquad a_{2}=-2l^{2}, \\ & b_{0}=- \frac{16l^{3}+c}{6l},\qquad b_{1}=0,\qquad b_{2}=4l^{2}. \end{aligned} $$
(43)
So, the exact solution of Eq. (16) is
$$ \textstyle\begin{cases} u(x,y,t)=\frac{8l^{3}-c}{6l}-2l^{2}\tanh ^{2}(l(x-y)+c(x-t)), \\ v(x,y,t)=-\frac{16l^{3}+c}{6l}+4l^{2}\tanh ^{2}(l(x-y)+c(x-t)), \end{cases} $$
(44)
where \(l\neq 0\), c are arbitrary constants.
Figures 9 and 10 depict the exact solution of Eq. (16), which is obtained by taking \(l=1\), \(c=1\) at \(t=0\).