In this section we study the generalised KdV equation with two power nonlinearities [43] given by

$$\begin{array}{}{\displaystyle {u}_{t}+(a{u}^{n}-b{u}^{2n}){u}_{x}+{u}_{xxx}=0,\phantom{\rule{2em}{0ex}}n>0,}\end{array}$$(32)

where *a*, *b* are constants. This equation describes the propagation of nonlinear long acoustic-type waves. Once the amplitude is not supposed to be small, Eq. (32) serves as an approximate model for the description of weak dispersive effects on the propagation of nonlinear waves along a characteristic direction [43]. It can be noted that for *n* = 1, (32) represents the standard Gardner equation, or the combined KdV-mKdV equation.

The use of wave transformation (3) reduces (32), after integrating once, to an ODE in the form

$$\begin{array}{}{\displaystyle -cU+\frac{a}{n+1}{U}^{n+1}-\frac{b}{2n+1}{U}^{2n+1}+{U}^{\u2033}=0,}\end{array}$$(33)

where integration constant is considered zero. It is more convenient to replace the exponents of nonlinear terms by integers in order to seek for a closed form solution. Accordingly assuming that *U* = *v*^{1/n}, (33) reduces to

$$\begin{array}{}{\displaystyle -A{v}^{2}+B{v}^{3}-C{v}^{4}-D{v}^{\prime 2}+v{v}^{\u2033}=0,}\end{array}$$(34)

where

$$\begin{array}{}{\displaystyle A=cn,\phantom{\rule{1em}{0ex}}B=\frac{an}{n+1},\phantom{\rule{1em}{0ex}}C=\frac{bn}{2n+1},\phantom{\rule{1em}{0ex}}D=\frac{n-1}{n}.}\end{array}$$(35)

Similarly, we suppose that (34) has solutions in the form of (14). Then, by the homogeneous balance principle we have thus assumed that (34) has the following solution

$$\begin{array}{}{\displaystyle v(\xi )={a}_{0}+{a}_{1}f+{b}_{1}g,}\end{array}$$(36)

where *f* = *f*(*ξ*), *g* = *g*(*ξ*) satisfy (6). Substituting (36) with (6) into (34) and setting the coefficients of *f*^{i}g^{j} (*i*, *j* = 1, 2, …) to zero generates a system of algebraic equations. Solving this system of equations, gives rise to

Case 1

$$\begin{array}{}{\displaystyle {a}_{1}=\frac{(n+1)(n+2)r}{a{n}^{2}},\phantom{\rule{1em}{0ex}}c=\frac{1}{{n}^{2}},}\\ {\displaystyle \frac{{a}^{2}}{b}=\frac{(n+1)(n+2{)}^{2}{r}^{2}}{(2n+1)({r}^{2}+\u03f5){n}^{2}},\phantom{\rule{1em}{0ex}}{a}_{0}={b}_{1}=0.}\end{array}$$(37)

Case 2

$$\begin{array}{}{\displaystyle {a}_{0}=\frac{2(n+1)(n+2)}{a{n}^{2}},\phantom{\rule{1em}{0ex}}{b}_{1}=\pm {a}_{0},\phantom{\rule{1em}{0ex}}c=\frac{4}{{n}^{2}},}\\ {\displaystyle \frac{{a}^{2}}{b}=\frac{4(n+1)(n+2{)}^{2}}{(2n+1){n}^{2}},\phantom{\rule{1em}{0ex}}r=0,\phantom{\rule{1em}{0ex}}{a}_{1}=0.}\end{array}$$(38)

Case 3

$$\begin{array}{}{\displaystyle {a}_{0}=\frac{(n+1)(n+2)}{2a{n}^{2}},\phantom{\rule{1em}{0ex}}{b}_{1}=\pm {a}_{0},\phantom{\rule{1em}{0ex}}c=\frac{1}{{n}^{2}},}\\ {\displaystyle \frac{{a}^{2}}{b}=\frac{(n+1)(n+2{)}^{2}}{(2n+1){n}^{2}},\phantom{\rule{1em}{0ex}}r=\sqrt{-\u03f5},\phantom{\rule{1em}{0ex}}{a}_{1}=0.}\end{array}$$(39)

Case 4

$$\begin{array}{}{\displaystyle {a}_{0}=\frac{(n+1)(n+2)}{2a{n}^{2}},\phantom{\rule{1em}{0ex}}{b}_{1}=\pm {a}_{0},\phantom{\rule{1em}{0ex}}{a}_{1}=\pm {a}_{0}\sqrt{\u03f5},}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}c=\frac{1}{{n}^{2}},\phantom{\rule{1em}{0ex}}\frac{{a}^{2}}{b}=\frac{(n+1)(n+2{)}^{2}}{(2n+1){n}^{2}},\phantom{\rule{1em}{0ex}}r=0.}\end{array}$$(40)

Case 5

$$\begin{array}{}{\displaystyle {a}_{0}=\frac{(n+1)(n+2)}{2a{n}^{2}},\phantom{\rule{-0.6cm}{0ex}}\phantom{\rule{1em}{0ex}}{b}_{1}=\pm {a}_{0},\phantom{\rule{1em}{0ex}}{a}_{1}=\pm {a}_{0}\sqrt{{r}^{2}+\u03f5},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle c=\frac{1}{{n}^{2}},\phantom{\rule{1em}{0ex}}\frac{{a}^{2}}{b}=\frac{(n+1)(n+2{)}^{2}}{(2n+1){n}^{2}}.}\end{array}$$(41)

Case 6

$$\begin{array}{}{\displaystyle {a}_{0}=\frac{(n+1)(n+2)}{2a{n}^{2}},\phantom{\rule{1em}{0ex}}{b}_{1}=\pm {a}_{0},}\\ {\displaystyle {a}_{1}=\pm {a}_{0}\frac{n+1}{n}\sqrt{\frac{-n\u03f5}{3(n-2)}},\phantom{\rule{1em}{0ex}}c=\frac{1}{{n}^{2}},}\\ {\displaystyle \frac{{a}^{2}}{b}=\frac{(n+1)(n+2{)}^{2}}{(2n+1){n}^{2}},\phantom{\rule{1em}{0ex}}r=\pm \frac{(2n-1)}{n}\sqrt{\frac{-n\u03f5}{3(n-2)}}.}\end{array}$$(42)

Inserting these results along with (7)-(9) into (36), various solitary wave solutions of the generalised KdV equation can be created as follows.

From Case 1 we obtain the solutions

$$\begin{array}{}{\displaystyle {u}_{1}={\left\{\frac{(n+1)(n+2)r\alpha}{a{n}^{2}[\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )+\alpha r]}\right\}}^{1/n},}\end{array}$$(43)

when *ϵ* = 1: *α*, *β*, *γ* satisfy *γ*^{2} = *α*^{2} + *β*^{2} and when *ϵ* = −1: *α*, *β*, *γ* satisfy *β*^{2} = *α*^{2} + *γ*^{2}.

$$\begin{array}{}{\displaystyle {u}_{2}={\left\{\frac{(n+1)(n+2)r}{a{n}^{2}[\mathrm{sinh}(\xi )+r]}\right\}}^{1/n},\phantom{\rule{2em}{0ex}}\u03f5=1.}\end{array}$$(44)

$$\begin{array}{}{\displaystyle {u}_{3}={\left\{\frac{(n+1)(n+2)r}{a{n}^{2}[\mathrm{cosh}(\xi )+r]}\right\}}^{1/n},\phantom{\rule{2em}{0ex}}\u03f5=-1.}\end{array}$$(45)

From Case 2 we arrive at the solutions

$$\begin{array}{}{\displaystyle {u}_{4}={\left\{\frac{2(n+1)(n+2)}{a{n}^{2}}\left[1\pm \frac{\beta \mathrm{sinh}(\xi )+\gamma \mathrm{cosh}(\xi )}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )}\right]\right\}}^{1/n},}\end{array}$$(46)

when *ϵ* = 1: *α*, *β*, *γ* satisfy *γ*^{2} = *α*^{2} + *β*^{2} and when *ϵ* = −1: *α*, *β*, *γ* satisfy *β*^{2} = *α*^{2} + *γ*^{2}.

$$\begin{array}{}{\displaystyle {u}_{5}={\left\{\frac{2(n+1)(n+2)}{a{n}^{2}}\left[1\pm \mathrm{coth}(\xi )\right]\right\}}^{1/n},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u03f5=1.\end{array}$$(47)

$$\begin{array}{}{\displaystyle {u}_{6}={\left\{\frac{2(n+1)(n+2)}{a{n}^{2}}\left[1\pm \mathrm{tanh}(\xi )\right]\right\}}^{1/n},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u03f5=-1.\end{array}$$(48)

From Case 3 we find the solutions

$$\begin{array}{}{\displaystyle {u}_{7}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\times \left[1\pm \frac{\beta \mathrm{sinh}(\xi )+\gamma \mathrm{cosh}(\xi )}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )\pm \alpha \sqrt{-\u03f5}}\right]\}}^{1/n},}\end{array}$$(49)

when *ϵ* = 1: *α*, *β*, *γ* satisfy *γ*^{2} = *α*^{2} + *β*^{2} and when *ϵ* = −1: *α*, *β*, *γ* satisfy *β*^{2} = *α*^{2} + *γ*^{2}.

$$\begin{array}{}{\displaystyle {u}_{8}={\left\{\frac{(n+1)(n+2)}{2a{n}^{2}}\left[1\pm \frac{\mathrm{cosh}(\xi )}{\mathrm{sinh}(\xi )\pm i}\right]\right\}}^{1/n},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u03f5=1.\end{array}$$(50)

$$\begin{array}{}{\displaystyle {u}_{9}={\left\{\frac{(n+1)(n+2)}{2a{n}^{2}}\left[1\pm \frac{\mathrm{sinh}(\xi )}{\mathrm{cosh}(\xi )\pm 1}\right]\right\}}^{1/n},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u03f5=-1.\end{array}$$(51)

From Case 4 we acquire the solutions

$$\begin{array}{}{\displaystyle {u}_{10}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}[1\pm \frac{\alpha \sqrt{\u03f5}}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\pm \frac{\beta \mathrm{sinh}(\xi )+\gamma \mathrm{cosh}(\xi )}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )}]\}}^{1/n},}\end{array}$$(52)

when *ϵ* = 1: *α*, *β*, *γ* satisfy *γ*^{2} = *α*^{2} + *β*^{2} and when *ϵ* = −1: *α*, *β*, *γ* satisfy *β*^{2} = *α*^{2} + *γ*^{2}.

$$\begin{array}{}{\displaystyle {u}_{11}={\left\{\frac{(n+1)(n+2)}{2a{n}^{2}}\left[1\pm csch(\xi )\pm \mathrm{coth}(\xi )\right]\right\}}^{1/n},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u03f5=1.\end{array}$$(53)

$$\begin{array}{}{\displaystyle {u}_{12}={\left\{\frac{(n+1)(n+2)}{2a{n}^{2}}\left[1\pm isech(\xi )\pm \mathrm{tanh}(\xi )\right]\right\}}^{1/n},}\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u03f5=-1.\end{array}$$(54)

From Case 5 we obtain the solutions

$$\begin{array}{}{\displaystyle {u}_{13}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}[1\pm \frac{\alpha \sqrt{{r}^{2}+\u03f5}}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )+\alpha r}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\pm \frac{\beta \mathrm{sinh}(\xi )+\gamma \mathrm{cosh}(\xi )}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )+\alpha r}]\}}^{1/n},}\end{array}$$(55)

when *ϵ* = 1: *α*, *β*, *γ* satisfy *γ*^{2} = *α*^{2} + *β*^{2} and when *ϵ* = −1: *α*, *β*, *γ* satisfy *β*^{2} = *α*^{2} + *γ*^{2}.

$$\begin{array}{}{\displaystyle {u}_{14}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}[1\pm \frac{\sqrt{{r}^{2}+1}}{\mathrm{sinh}(\xi )+r}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\pm \frac{\mathrm{cosh}(\xi )}{\mathrm{sinh}(\xi )+r}]\}}^{1/n},\phantom{\rule{2em}{0ex}}\u03f5=1.}\end{array}$$(56)

$$\begin{array}{}{\displaystyle {u}_{15}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\left[1\pm \frac{\sqrt{{r}^{2}-1}}{\mathrm{cosh}(\xi )+r}\pm \frac{\mathrm{sinh}(\xi )}{\mathrm{cosh}(\xi )+r}\right]\}}^{1/n}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}\u03f5=-1.}\end{array}$$(57)

From Case 6 we come by the solutions

$$\begin{array}{}{\displaystyle {u}_{16}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}[1\pm \frac{(n+1)}{n}\sqrt{\frac{-n\u03f5}{3(n-2)}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\times \frac{\alpha}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )\pm \alpha \frac{(2n-1)}{n}\sqrt{\frac{-n\u03f5}{3(n-2)}}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\pm \frac{\beta \mathrm{sinh}(\xi )+\gamma \mathrm{cosh}(\xi )}{\beta \mathrm{cosh}(\xi )+\gamma \mathrm{sinh}(\xi )\pm \alpha \frac{(2n-1)}{n}\sqrt{\frac{-n\u03f5}{3(n-2)}}}]\}}^{1/n},}\end{array}$$(58)

when *ϵ* = 1: *α*, *β*, *γ* satisfy *γ*^{2} = *α*^{2} + *β*^{2} and when *ϵ* = −1: *α*, *β*, *γ* satisfy *β*^{2} = *α*^{2} + *γ*^{2}.

$$\begin{array}{}{\displaystyle {u}_{17}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}[1\pm i\frac{(n+1)}{n}\sqrt{\frac{n}{3(n-2)}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\times \frac{1}{\mathrm{sinh}(\xi )\pm i\frac{(2n-1)}{n}\sqrt{\frac{n}{3(n-2)}}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\pm \frac{\mathrm{cosh}(\xi )}{\mathrm{sinh}(\xi )\pm i\frac{(2n-1)}{n}\sqrt{\frac{n}{3(n-2)}}}]\}}^{1/n},\phantom{\rule{2em}{0ex}}\u03f5=1.}\end{array}$$(59)

$$\begin{array}{}{\displaystyle {u}_{18}=\{\frac{(n+1)(n+2)}{2a{n}^{2}}[1\pm \frac{(n+1)}{n}\sqrt{\frac{n}{3(n-2)}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\times \frac{1}{\mathrm{cosh}(\xi )\pm \frac{(2n-1)}{n}\sqrt{\frac{n}{3(n-2)}}}}\\ {\displaystyle \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\pm \frac{\mathrm{sinh}(\xi )}{\mathrm{cosh}(\xi )\pm \frac{(2n-1)}{n}\sqrt{\frac{n}{3(n-2)}}}]\}}^{1/n},}\end{array}$$(60)

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