In this section, we bring to bear the further improved (G'/G)expansion method to the (2 + 1)dimensional breaking soliton equation which is dreadfully important nonlinear evolution equations in mathematical physics and have been paid attention by a lot of researchers and the extended tanhfunction method to compare the solutions obtained by the two methods.
On solving the (2 + 1)dimensional breaking soliton equation by the projected method
We start with the (2 + 1)dimensional breaking soliton equation (Darvishi & Najafi 2012; Bekir 2010; Ma et al. 2009; Inan 2010) in the form,
{u}_{x\phantom{\rule{0.12em}{0ex}}t}4\phantom{\rule{0.12em}{0ex}}{u}_{x\phantom{\rule{0.12em}{0ex}}y}{u}_{x}2\phantom{\rule{0.12em}{0ex}}{u}_{x\phantom{\rule{0.12em}{0ex}}x}{u}_{y}{u}_{x\phantom{\rule{0.12em}{0ex}}x\phantom{\rule{0.12em}{0ex}}x\phantom{\rule{0.12em}{0ex}}y}=0.
(6)
This equation was first introduced by Calogero and Degasperis in 1977. The breaking soliton equation describe the (2 + 1)dimensional interaction of the Riemann wave propagation along the yaxis with a long wave propagation along xaxis (Ma et al. 2009). In the recent years, a considerable amount of research works on the breaking soliton equation have been accomplished. For example, its solitary wave solutions, periodic and multiple soliton solutions are found in (Inan 2010). Let us now solve the Eq. (6) by the proposed further improved (G'/G)expansion method. To this end, we perceive that the traveling wave variable (2) permits us in converting Eq. (6) into an ODE and upon integration yields:
V\phantom{\rule{0.12em}{0ex}}u\prime +3\phantom{\rule{0.12em}{0ex}}{\left(u,\prime \right)}^{2}+u\u2034=0
(7)
with zero constant of integration. Considering the homogeneous balance between the highest order derivative and the nonlinear term come out in Eq. (7), we deduce that D(u′)^{2} = D(u‴), where D(u′)^{2} stands for degree of (u′)^{2} and so on. This yield n = 1. Therefore, the solution (4) turns out to be
u\left(\xi \right)={\alpha}_{1}\left(\frac{G\prime}{G}\right)+{\alpha}_{0}
(8)
Substituting (8) together with Eq. (5) into (7), we obtain the following polynomial equation in G:
\begin{array}{l}{\alpha}_{1}q\phantom{\rule{0.12em}{0ex}}\left(V+4\phantom{\rule{0.12em}{0ex}}p\right)\phantom{\rule{0.12em}{0ex}}{G}^{2}+\left(3\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}^{2}{q}^{2}+32\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}r\phantom{\rule{0.12em}{0ex}}p+2\phantom{\rule{0.12em}{0ex}}V\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}r+6\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}{q}^{2}\right)\phantom{\rule{0.12em}{0ex}}{G}^{4}\\ +12\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}q\phantom{\rule{0.12em}{0ex}}\left({\alpha}_{1}+4\right)\phantom{\rule{0.12em}{0ex}}{G}^{6}+12\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}\phantom{\rule{0.12em}{0ex}}{r}^{2}\left({\alpha}_{1}+4\right)\phantom{\rule{0.12em}{0ex}}{G}^{8}=0\end{array}
(9)
Setting each coefficient of the polynomial Eq. (9) to zero, we achieve a system of algebraic equations which can be solved by using the symbolic computation software such as Maple and obtain the following two sets of solutions:
The set 1.
{\alpha}_{1}=4,\phantom{\rule{0.5em}{0ex}}{\alpha}_{0}={\alpha}_{0},\phantom{\rule{0.5em}{0ex}}q=2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r},\phantom{\rule{0.5em}{0ex}}V=4\phantom{\rule{0.12em}{0ex}}p,
(10)
where α_{0}, p and r are arbitrary constants.
The set 2.
{\alpha}_{1}=4,\phantom{\rule{0.5em}{0ex}}{\alpha}_{0}={\alpha}_{0},\phantom{\rule{0.5em}{0ex}}q=0,\phantom{\rule{0.5em}{0ex}}V=16p,
(11)
where α_{0}, p and r are arbitrary constants.
Now for the set 1, we have the following solution:
u\left(\xi \right)=4\phantom{\rule{0.12em}{0ex}}\left(\frac{G\prime}{G}\right)+{\alpha}_{0},
(12)
where
\xi =x+y+4\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.24em}{0ex}}t.
According to the step 2 of section 2, we have the subsequent families of exact solutions:
Family 1. If p > 0, the solution of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{p\phantom{\rule{0.12em}{0ex}}qsec\phantom{\rule{1pt}{0ex}}{h}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{{q}^{2}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{\left(1\pm tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\xi \right)\right)}^{2}}\right]}^{\frac{1}{2}}
(13)
or
G\left(\xi \right)={\left[\frac{p\phantom{\rule{0.12em}{0ex}}qcsc{h}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{{q}^{2}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{\left(1\pm coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\xi \right)\right)}^{2}}\right]}^{\frac{1}{2}}
(14)
In these cases we have the ratio,
\frac{G\prime}{G}=\frac{\sqrt{p}\left\{\phantom{\rule{0.12em}{0ex}}{q}^{2}sinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)2p\phantom{\rule{0.12em}{0ex}}rsinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp 2\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{cosh}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm p\phantom{\rule{0.12em}{0ex}}r\right\}}{{q}^{2}{cosh}^{2}\left(\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+2\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{cosh}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm 2p\phantom{\rule{0.12em}{0ex}}rsinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)p\phantom{\rule{0.12em}{0ex}}r}
(15)
and
\frac{G\prime}{G}=\frac{\sqrt{p}\left\{{q}^{2}sinh\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)2p\phantom{\rule{0.12em}{0ex}}rsinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp 2\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{cosh}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm p\phantom{\rule{0.12em}{0ex}}r\right\}}{{q}^{2}{cosh}^{2}\left(\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+2\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{cosh}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm 2p\phantom{\rule{0.12em}{0ex}}rsinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{q}^{2}p\phantom{\rule{0.12em}{0ex}}r}
(16)
respectively.
Since q=2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}, subsequently, we obtain the following traveling wave solutions,
u\left(\xi \right)=\mp 4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\pm \frac{8\sqrt{p}sec{h}^{2}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{sec{h}^{2}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp 2tanh\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+2}+{\alpha}_{0}
(17)
and
u\left(\xi \right)=4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\frac{8\sqrt{p}sec{h}^{2}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{\phantom{\rule{0.12em}{0ex}}3sec{h}^{2}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+2tanh\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp 2}+{\alpha}_{0},
(18)
where
\xi =x+y+4\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.24em}{0ex}}t
Family 2. If p > 0, r > 0, the solution of Eq. (5) has the form,
\begin{array}{c}\hfill G\left(\xi \right)={\left[\frac{psec{h}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\sqrt{p\phantom{\rule{0.12em}{0ex}}r}tanh\left(\sqrt{p}\xi \right)}\right]}^{\frac{1}{2}}\phantom{\rule{0.12em}{0ex}}\mathrm{or}\phantom{\rule{0.5em}{0ex}}G\left(\xi \right)\hfill \\ \hfill ={\left[\frac{pcsc{h}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\sqrt{p\phantom{\rule{0.12em}{0ex}}r}coth\left(\sqrt{p}\xi \right)}\right]}^{\frac{1}{2}}.\hfill \end{array}
(19)
Then we have the ratio,
\frac{G\prime}{G}=\frac{\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{\pm qsinh\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}\phantom{\rule{0.12em}{0ex}}{cosh}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp \sqrt{p\phantom{\rule{0.12em}{0ex}}r}\right\}}{cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\phantom{\rule{0.12em}{0ex}}\left\{\phantom{\rule{0.12em}{0ex}}qcosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}sinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}}
or
\frac{G\prime}{G}=\frac{\sqrt{p}\left\{\pm qsinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}\phantom{\rule{0.12em}{0ex}}{cosh}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp \sqrt{p\phantom{\rule{0.12em}{0ex}}r}\phantom{\rule{0.12em}{0ex}}\right\}}{sinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\phantom{\rule{0.12em}{0ex}}\left\{qsinh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}cosh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}}.
(20)
Since q=2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}, subsequently, we obtain the following traveling wave solutions:
u\phantom{\rule{0.12em}{0ex}}\left(\xi \right)=\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}+2\phantom{\rule{0.12em}{0ex}}\sqrt{p}tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}\phantom{\rule{0.5em}{0ex}}
or
u\left(\xi \right)=\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}+2\phantom{\rule{0.12em}{0ex}}\sqrt{p}coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}
(21)
where
\xi =x+y+4\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.24em}{0ex}}t
Family 3. If p < 0, r > 0, the solution of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{p\phantom{\rule{0.12em}{0ex}}{sec}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\sqrt{p\phantom{\rule{0.12em}{0ex}}r}tan\left(\sqrt{p}\xi \right)}\right]}^{\frac{1}{2}}
or
G\left(\xi \right)={\left[\frac{p\phantom{\rule{0.12em}{0ex}}{csc}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}cot\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}\right]}^{\frac{1}{2}}.
(22)
Then we have the ratio
\frac{G\prime}{G}=\frac{\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{\sqrt{p\phantom{\rule{0.12em}{0ex}}r}2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}{cos}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm qsin\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cos\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}}{cos\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\phantom{\rule{0.12em}{0ex}}\left\{\phantom{\rule{0.12em}{0ex}}2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}sin\phantom{\rule{1pt}{0ex}}(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \pm qcos\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}}
or
\frac{G\prime}{G}=\frac{\sqrt{p}\left\{\sqrt{p\phantom{\rule{0.12em}{0ex}}r}\mp 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}\phantom{\rule{0.12em}{0ex}}{cos}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp qsin\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)cos\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}}{sin\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\phantom{\rule{0.12em}{0ex}}\left\{2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}cos\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm qsin\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}}.
(23)
Since q=2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}, subsequently, we obtain the following traveling wave solutions:
u\left(\xi \right)=\frac{4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{tan\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\mp 1\phantom{\rule{0.12em}{0ex}}\right\}}{1\pm tan\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+sec\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}+{\alpha}_{0}
or
u\left(\xi \right)=\frac{4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{tan\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\pm 1\right\}}{1\mp tan\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)sec\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}+{\alpha}_{0},
(24)
where
\xi =x+y+4\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.24em}{0ex}}t.
Family 4. If p > 0, Δ = 0, the solution of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{p}{q}\left\{1\pm tanh\phantom{\rule{1pt}{0ex}}\left(\frac{1}{2}\sqrt{p}\xi \right)\right\}\right]}^{\frac{1}{2}}
or
G\left(\xi \right)={\left[\frac{p}{q}\phantom{\rule{0.12em}{0ex}}\left\{1\pm coth\phantom{\rule{1pt}{0ex}}\left(\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}\right]}^{\frac{1}{2}}
(25)
Then we have the ratio
\frac{G\prime}{G}=\frac{\sqrt{p}}{4}\left\{\pm 1tanh\phantom{\rule{1pt}{0ex}}\left(\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}
or
\frac{G\prime}{G}=\frac{\sqrt{p}}{4}\phantom{\rule{0.12em}{0ex}}\left\{\pm 1coth\phantom{\rule{1pt}{0ex}}\left(\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}
(26)
Subsequently, we obtain the following traveling wave solutions:
u\left(\xi \right)=\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{\pm 1tanh\phantom{\rule{1pt}{0ex}}\left(\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}+{\alpha}_{0}
or
u\left(\xi \right)=\sqrt{p}\phantom{\rule{0.24em}{0ex}}\left\{\pm 1coth\left(\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}+{\alpha}_{0},
(27)
where
\xi =x+y+4\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.24em}{0ex}}t
Family 5. If p > 0, the solution of Eq. (5) has the form
G\left(\xi \right)={\left\{\frac{p\phantom{\rule{0.12em}{0ex}}{e}^{\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi}}{{\left({e}^{\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\xi}4\phantom{\rule{0.12em}{0ex}}q\right)}^{2}64\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r}\right\}}^{\frac{1}{2}}
(28)
Then we have the ratio
\frac{G\prime}{G}=\frac{\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{\phantom{\rule{0.12em}{0ex}}{e}^{\pm 4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi}16\phantom{\rule{0.12em}{0ex}}{q}^{2}+64\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r\right\}}{\mp {e}^{\pm 4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi}\pm 8\phantom{\rule{0.12em}{0ex}}q\phantom{\rule{0.24em}{0ex}}{e}^{\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi}\mp 16\phantom{\rule{0.12em}{0ex}}{q}^{2}\pm 64\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r}
(29)
Since q=2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}, subsequently, we obtain the following traveling wave solutions:
u\left(\xi \right)=\frac{4\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}{e}^{\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi}}{\pm 16\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}\mp {e}^{\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi}}+{\alpha}_{0}
(30)
where
\xi =x+y+4\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.24em}{0ex}}t
For the set 2, we have the following solution:
u\left(\xi \right)=4\phantom{\rule{0.12em}{0ex}}\left(\frac{G\prime}{G}\right)+{\alpha}_{0}
(31)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t.
According to the step 2 of section 2, we obtain the subsequent families of exact solutions:
Cohort 1. If p > 0, Δ > 0, the solution of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{2\phantom{\rule{0.12em}{0ex}}p}{\pm \sqrt{\mathrm{\Delta}}cosh\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\xi \right)q}\right]}^{\frac{1}{2}}
(32)
Since q = 0, then r < 0. In this case we have the ratio,
\frac{G\prime}{G}=\sqrt{p}tanh\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)
(33)
Therefore, we obtainthe following traveling wave solution,
u\left(\xi \right)=4\phantom{\rule{0.12em}{0ex}}\sqrt{p}tanh\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0},
(34)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t.
Cohort 2. If p > 0, Δ < 0, the solution of Eq. (5) has the form
G\left(\xi \right)={\left[\frac{2\phantom{\rule{0.12em}{0ex}}p}{\pm \sqrt{\mathrm{\Delta}}sinh\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\xi \right)q}\right]}^{\frac{1}{2}}
(35)
Since q = 0, then r > 0. In this case we have the ratio,
\frac{G\prime}{G}=\sqrt{p}coth\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right).
(36)
Therefore, we obtain the following traveling wave solutions:
u\left(\xi \right)=4\phantom{\rule{0.12em}{0ex}}\sqrt{p}coth\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0},
(37)
where
\xi =x+y+16p\phantom{\rule{0.12em}{0ex}}t.
Cohort 3. If p < 0, Δ > 0, the solutions of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{2\phantom{\rule{0.12em}{0ex}}p}{\pm \sqrt{\mathrm{\Delta}}cos\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)q}\right]}^{\frac{1}{2}}
or
G\left(\xi \right)={\left[\frac{2\phantom{\rule{0.12em}{0ex}}p}{\pm \sqrt{\mathrm{\Delta}}sin\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)q}\right]}^{\frac{1}{2}}
(38)
Since q = 0, then r > 0. Thus we have the ratio,
\frac{G\prime}{G}=\sqrt{p}tan\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)
or
\frac{G\prime}{G}=\sqrt{p}cot\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)
(39)
Therefore, we obtain the following traveling wave solutions:
u\left(\xi \right)=4\sqrt{p}tan\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}
or
u\left(\xi \right)=4\phantom{\rule{0.12em}{0ex}}\sqrt{p}cot\phantom{\rule{1pt}{0ex}}\left(2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}
(40)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t.
Cohort 4. If p > 0, r > 0, the solutions of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{psec{h}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\sqrt{p\phantom{\rule{0.12em}{0ex}}r}tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\xi \right)}\right]}^{\frac{1}{2}}
or
G\left(\xi \right)={\left[\frac{pcsc{h}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}\right]}^{\frac{1}{2}}.
(41)
Since q = 0, we have the ratio,
\frac{G\prime}{G}=\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left[tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right]
(42)
Therefore, we obtain the following traveling wave solution:
u\left(\xi \right)=2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)+coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}+{\alpha}_{0}
(43)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t
Cohort 5. If p < 0, r > 0, the solutions of Eq. (5) has the form
G\left(\xi \right)={\left[\frac{p\phantom{\rule{0.12em}{0ex}}{sec}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\sqrt{p\phantom{\rule{0.12em}{0ex}}r}tan\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\xi \right)}\right]}^{\frac{1}{2}}
or
G\left(\xi \right)={\left[\frac{p\phantom{\rule{0.12em}{0ex}}{csc}^{2}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}{q\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}r}cot\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)}\right]}^{\frac{1}{2}}
(44)
Since q = 0, then we have the ratio,
\frac{G\prime}{G}=\frac{1}{2}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left[cot\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)tan\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right].
(45)
Therefore, we obtain the following traveling wave solutions:
u\left(\xi \right)=2\phantom{\rule{0.12em}{0ex}}\sqrt{p}\phantom{\rule{0.12em}{0ex}}\left\{cot\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)tan\phantom{\rule{1pt}{0ex}}\left(\sqrt{p}\phantom{\rule{0.12em}{0ex}}\xi \right)\right\}+{\alpha}_{0},
(46)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t.
Cohort 6. If p > 0, q = 0, the solution of Eq. (5) has the form,
G\left(\xi \right)={\left[\frac{\pm p\phantom{\rule{0.12em}{0ex}}{e}^{\pm 2\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}\xi}}}{164\phantom{\rule{0.12em}{0ex}}p\phantom{\rule{0.12em}{0ex}}r\phantom{\rule{0.12em}{0ex}}{e}^{\pm 4\phantom{\rule{0.12em}{0ex}}\sqrt{p\phantom{\rule{0.12em}{0ex}}\xi}}}\right]}^{\frac{1}{2}}
(47)
Then we have the ratio,
\frac{G\prime}{G}=\pm \frac{1}{8\phantom{\rule{0.12em}{0ex}}\sqrt{r}}coth\phantom{\rule{1pt}{0ex}}\left(\frac{\pm \xi}{4\phantom{\rule{0.12em}{0ex}}\sqrt{r}}\right),
(48)
where 64 pr = 1.
Therefore, we have the solution:
u\left(\xi \right)=\mp \frac{1}{2\phantom{\rule{0.12em}{0ex}}\sqrt{r}}coth\phantom{\rule{1pt}{0ex}}\left(\frac{\pm \xi}{4\phantom{\rule{0.12em}{0ex}}\sqrt{r}}\right)+{\alpha}_{0},
(49)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t.
Cohort 7. If p = 0, r = 0, then the solution of Eq. (5) has the form,
G\left(\xi \right)=\pm \frac{1}{\sqrt{q}\phantom{\rule{0.12em}{0ex}}\xi}.
(50)
Then we have the ratio,
\frac{G\prime}{G}=\frac{1}{\xi}+{\alpha}_{0}.
(51)
Therefore, we have the solution:
u\left(\xi \right)=\frac{4}{\xi}+{\alpha}_{0},
(52)
where
\xi =x+y+16p\phantom{\rule{0.24em}{0ex}}t.
These are the exact solutions of the breaking soliton equation obtained by the further improved (G'/G)expansion method.
On solving the breaking soliton equation by the extended tanhfunction method
With reference to the wellknown extended tanhfunction method (Abdou 2007; ElWakil et al. 2010; Fan 2000; Wazwaz 2008b; Wazzan 2009; Zayed et al. 2004b; Zhang & Xia 2008), the solution of the breaking soliton Eq. (6) can be written in the form,
u\left(\xi \right)={\alpha}_{1}\phantom{\rule{0.12em}{0ex}}\varphi \left(\xi \right)+{\alpha}_{0},
(53)
where ϕ(ξ) satisfy the Riccati equation
\varphi \prime \left(\xi \right)=A+{\varphi}^{2}\left(\xi \right).
(54)
The Riccati Eq. (54) has the following solutions:

(i)
If A < 0, then
\varphi \left(\xi \right)=\sqrt{A}tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)
or
\varphi \left(\xi \right)=\sqrt{A}coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right).
(55)

(ii)
If A > 0, then
\varphi \left(\xi \right)=\sqrt{A}tan\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)\phantom{\rule{0.5em}{0ex}}\mathrm{or}\phantom{\rule{0.5em}{0ex}}\varphi \left(\xi \right)=\sqrt{A}cot\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)
(56)

(iii)
If A = 0, then
\varphi \left(\xi \right)=\frac{1}{\xi}.
(57)
Substituting (53) together with (54) into (7), we obtain the following polynomial equation in φ:
\begin{array}{c}3\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}\phantom{\rule{0.12em}{0ex}}\left({\alpha}_{1}+2\phantom{\rule{0.12em}{0ex}}\right)\phantom{\rule{0.12em}{0ex}}{\phi}^{4}\left(\xi \right)+{\alpha}_{1}\phantom{\rule{0.12em}{0ex}}\left(V+8\phantom{\rule{0.12em}{0ex}}A+6\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}A\right)\phantom{\rule{0.12em}{0ex}}{\phi}^{2}\left(\xi \right)\\ \phantom{\rule{9.4em}{0ex}}+{\alpha}_{1}\phantom{\rule{0.12em}{0ex}}A\phantom{\rule{0.12em}{0ex}}\left(V+2\phantom{\rule{0.12em}{0ex}}A+3\phantom{\rule{0.12em}{0ex}}{\alpha}_{1}\phantom{\rule{0.12em}{0ex}}A\right)=0\end{array}
(58)
Equating the coefficients of this polynomial to zero and solving the set of algebraic equations with the aid of Maple, we obtain the subsequent solution:
{\alpha}_{1}=2,\phantom{\rule{0.5em}{0ex}}{\alpha}_{0}={\alpha}_{0},\phantom{\rule{0.5em}{0ex}}V=4\phantom{\rule{0.12em}{0ex}}A
(59)
where α_{0} and A are arbitrary constants.
Accordingly the exact solutions of Eq. (6) have the following forms:
When A < 0, the solution takes the form,
\varphi \left(\xi \right)=2\sqrt{A}tanh\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}\phantom{\rule{0.5em}{0ex}}
or
\varphi \left(\xi \right)=2\sqrt{A}coth\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0},
(60)
where
\xi =x+y4A\phantom{\rule{0.24em}{0ex}}t.
When A > 0, the solution takes the form,
\varphi \left(\xi \right)=2\phantom{\rule{0.12em}{0ex}}\sqrt{A}tan\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}\phantom{\rule{0.5em}{0ex}}
or
\varphi \left(\xi \right)=2\sqrt{A}cot\phantom{\rule{1pt}{0ex}}\left(\sqrt{A}\phantom{\rule{0.12em}{0ex}}\xi \right)+{\alpha}_{0}
(61)
where
\xi =x+y4A\phantom{\rule{0.24em}{0ex}}t.
When A = 0, the solution takes the form,
u\left(\xi \right)=\frac{2}{\xi}+{\alpha}_{0},
(62)
where
\xi =x+y4A\phantom{\rule{0.24em}{0ex}}t.
From the above results achieved by the two methods, we can draw the following conclusive remarks:
Remark 1. If we put A = − 4p where p > 0, the results arranged in Eq. (59) are identical to the results (34) and (37) respectively.
Remark 2. If we put A = −4p where p < 0 then the results arranged in Eq. (61) are identical to the results given in (40).
Remark 3: Result given in (62) is alike to the result given in (52).