Outset of multiple soliton solutions to the nonlinear Schrödinger equation and the coupled Burgers equation

Let us consider the NLSE of the form (Biswas and milovic 2010)

$$\begin{eqnarray}&&i\ {q}_{t}+a\ {q}_{{xx}}+b\ F\left({\left|q\right|}^{2}\right)q=0,\end{eqnarray} \tag{ 3.1.1 }$$

where F is a real valued algebraic function and it is important to have the smoothness of the complex function $F\left({\left|q\right|}^{2}\right)q:C\,\longmapsto \,C.$ Considering the complex plane C as a two-dimensional linear space R², the function $F\left({\left|q\right|}^{2}\right)q$ is k times continuously differentiable so that

$$\begin{eqnarray}&&F\left({\left|q\right|}^{2}\right)q\in \bigcup _{m,n=1}^{\infty }{C}^{k}\left(\left(-n,n\right)\times \left(-m,m\right);{R}^{2}\right)\ \end{eqnarray} \tag{ 3.1.2 }$$

In equation (3.1.1), q is dependent variable and x, t are the independent variable, and the subscripts represent the partial derivative of q with respect to that variables. The first term in (3.1.1) represents the time evolution term, while the second term is due to the group velocity dispersion and the third term represents nonlinearity, where a and b be the coefficient of second and third term respectively. Equation (3.1.1) is a nonlinear partial differential equation (PDE) of parabolic type which is not integrable, in general. For simplicity here we use the special case, $F\left(s\right)=s$, also known as the Kerr law of nonlinearity. In this article, we consider the subsequent NLSE,

$$\begin{eqnarray}&&i\ {q}_{t}+a\ {q}_{{xx}}+b\ \left({\left|q\right|}^{2}\right)q=0.\end{eqnarray} \tag{ 3.1.3 }$$

The nonlinear Schrödinger equation (NLSE) plays a vital role in various fields of physical, biological and engineering science. It appears in many applied areas, including nonlinear optics (Trikia and Biswasb 2011), plasma physics (Shukla and Eliasson 2010) and protein chemistry (Molkenthin et al 2010). In 1973, Hasegawa and Tappert showed that the NLSE is the appropriate equation to describe nonlinear optical light pulse propagation through optical fibers (Hasegawa and Tappert 1973).

We seek the travelling wave solutions of the NLSE given in equation (3.1.3). In order to search the wave solutions to the NLSE with Kerr law nonlinearity, we introduce the following wave transformation:

$$\begin{eqnarray}&&q\left(x,t\right)=u\left(\xi \right){e}^{i\left(\alpha x+\beta t\right)},\ \xi =k\left(x-2\alpha t\right).\end{eqnarray} \tag{ 3.1.4 }$$

Setting the transformation (3.1.4) into (3.1.3), equation (3.1.3) transforms into the following,

$$\begin{eqnarray}&&i\left(-2\alpha k+2\alpha {ka}\right)u^{\prime} -\left(a{\alpha }^{2}+\beta \right)u+{{ak}}^{2}u^{\prime\prime} +{{bu}}^{3}=0.\end{eqnarray} \tag{ 3.1.5 }$$

Separating real and imaginary parts, equation (3.1.5) yields

$$\begin{eqnarray}&&-2\alpha k+2\alpha {ka}=0\end{eqnarray} \tag{ 3.1.6 }$$

and

$$\begin{eqnarray}&&-\left(a{\alpha }^{2}+\beta \right)u+{{ak}}^{2}u^{\prime\prime} +{{bu}}^{3}=0.\end{eqnarray} \tag{ 3.1.7 }$$

Therefore, from (3.1.6), we obtain

$$\begin{eqnarray}&&a=1.\end{eqnarray} \tag{ 3.1.8 }$$

Therefore, equation (3.1.7) becomes

$$\begin{eqnarray}&&-\left({\alpha }^{2}+\beta \right)u+{k}^{2}u^{\prime\prime} +{{bu}}^{3}=0.\end{eqnarray} \tag{ 3.1.9 }$$

Balancing the highest order derivative term $u^{\prime\prime}$ with the highest power nonlinear term u³, gives n=1.

According to the new auxiliary equation method, from equation (2.4), the solution of equation (3.1.9) is of the form

$$\begin{eqnarray}&&u\left(\xi \right)={c}_{0}+{c}_{1}{a}^{f\left(\xi \right)},\end{eqnarray} \tag{ 3.1.10 }$$

where, $f\left(\xi \right)$ is the solution of the nonlinear equation (2.5).

Now, by means of the solution (3.1.10) and (2.5) from equation (3.1.9), we obtain

$$\begin{eqnarray*}&&\begin{array}{l}\{2{k}^{2}{c}_{1}{r}^{2}+{{bc}}_{1}^{3}\}\{{a}^{f\left(\xi \right)}\}{}^{3}+\left(3{k}^{2}{c}_{1}{qr}+3{{bc}}_{0}{c}_{1}^{2}\right)\{{a}^{f\left(\xi \right)}\}{}^{2}\\ \quad +\,\left(-{c}_{1}{\alpha }^{2}-{c}_{1}\beta +3{{bc}}_{0}^{2}{c}_{1}+2{k}^{2}{c}_{1}{pr}+{k}^{2}{c}_{1}{q}^{2}\right){a}^{f\left(\xi \right)}\\ \quad -\,{c}_{0}{\alpha }^{2}-{c}_{0}\beta +{k}^{2}{c}_{1}{pq}+{{bc}}_{0}^{3}=0.\end{array}\end{eqnarray*}$$

Setting the coefficients $\{{a}^{f\left(\xi \right)}\}{}^{j}\ (j=0,1,2,3)$ to zero leads to a set of algebraic equations

$$\begin{eqnarray*}&&2{k}^{2}{c}_{1}{r}^{2}+{{bc}}_{1}^{3}=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&3{k}^{2}{c}_{1}{qr}+3{{bc}}_{0}{c}_{1}^{2}=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&-{c}_{1}{\alpha }^{2}-{c}_{1}\beta +3{{bc}}_{0}^{2}{c}_{1}+2{k}^{2}{c}_{1}{pr}+{k}^{2}{c}_{1}{q}^{2}=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&-{c}_{0}{\alpha }^{2}-{c}_{0}\beta +{k}^{2}{c}_{1}{pq}+{{bc}}_{0}^{3}=0.\end{eqnarray*}$$

Solving the above algebraic equations with the aid of Maple, yields

$$\begin{eqnarray}&&\alpha =\pm \displaystyle \frac{1}{2}\sqrt{-4\beta +8{k}^{2}{rp}-2{k}^{2}\ {q}^{2}},{c}_{0}=\pm \displaystyle \frac{{qk}}{\sqrt{-2b}},{c}_{1}=\pm 2\displaystyle \frac{{kr}}{\sqrt{-2b}}\end{eqnarray} \tag{ 3.1.11 }$$

For the above values of the constants, we attain the subsequent solutions to the NLSE.

Case 1: When ${q}^{2}-4\ {pr}\lt 0$ and $r\ne 0$,

Making use of (3.1.11) and (2.6), (2.7) into equation (3.1.10), we obtain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{4{pr}-{q}^{2}}{-2b}}\tan \left(\displaystyle \frac{\sqrt{4{pr}-{q}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.12 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{4{pr}-{q}^{2}}{-2b}}\cot \left(\displaystyle \frac{\sqrt{4{pr}-{q}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.13 }$$

Case 2: When ${q}^{2}-4\ {pr}\gt 0$ and $r\ne 0$,

Putting (3.1.11) and (2.8), (2.9) into equation (3.1.10), we achieve

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{{q}^{2}-4{pr}}{-2b}}\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4{pr}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.14 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{{q}^{2}-4{pr}}{-2b}}\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4{pr}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.15 }$$

Case 3: When ${q}^{2}+4\ {p}^{2}\lt 0,r\ne 0$ and $r=-p$,

Applying (3.1.11) and (2.10), (2.11) into equation (3.1.10), we accomplish

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{-{q}^{2}-4{p}^{2}}{-2b}}\tan \left(\displaystyle \frac{\sqrt{-{q}^{2}-4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.16 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{-{q}^{2}-4{p}^{2}}{-2b}}\cot \left(\displaystyle \frac{\sqrt{-{q}^{2}-4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.17 }$$

Case 4: When ${q}^{2}+4\ {p}^{2}\gt 0,r\ne 0$ and $r=-p$,

Exerting (3.1.11) and (2.12), (2.13) into equation (3.1.10), we attain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{{q}^{2}+4{p}^{2}\ }{-2b}}\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}+4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.18 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{{q}^{2}+4{p}^{2}\ }{-2b}}\coth \left(\displaystyle \frac{\sqrt{{q}^{2}+4\ {p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.19 }$$

Case 5: When ${q}^{2}-4\ {p}^{2}\lt 0$ and r = p,

Employing (3.1.11) and (2.14), (2.15) from equation (3.1.10), we ascertain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{-{q}^{2}+4{p}^{2}}{-2b}}\tan \left(\displaystyle \frac{\sqrt{-{q}^{2}+4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.20 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{-{q}^{2}+4{p}^{2}}{-2b}}\cot \left(\displaystyle \frac{\sqrt{-{q}^{2}+4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.21 }$$

Case 6: When ${q}^{2}-4\ {p}^{2}\gt 0$ and r = p,

Making use of (3.1.11) and (2.16), (2.17) from equation (3.1.10), we gain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{{q}^{2}-4{p}^{2}\ }{-2b}}\ \tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.22 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm k\sqrt{\displaystyle \frac{{q}^{2}-4\ {p}^{2}}{-2b}}\coth \left(\displaystyle \frac{\sqrt{{q}^{2}-4{p}^{2}}}{2}\xi \right)\end{eqnarray} \tag{ 3.1.23 }$$

Case 7: When ${q}^{2}=4\ {pr}$,

By means of (3.1.11) and (2.18) from equation (3.1.10), we find out

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{2k}{\xi \sqrt{-2b}}.\end{eqnarray} \tag{ 3.1.24 }$$

Case 8: When rp < 0, q = 0 and $r\ne 0$,

Inserting (3.1.11) and (2.19), (2.20) into equation (3.1.10), we derive

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{2k}{\sqrt{-2b}}\sqrt{-{pr}}\ \tanh \left(\sqrt{-{rp}}\ \xi \right)\end{eqnarray} \tag{ 3.1.25 }$$

or

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{2k}{\sqrt{-2b}}\sqrt{-{pr}}\ \coth \left(\sqrt{-{rp}}\ \xi \right)\end{eqnarray} \tag{ 3.1.26 }$$

Case 9: When q = 0 and $p=-r$,

Setting (3.1.11) and (2.21) into equation (3.1.10), we secure

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{2{kr}}{\sqrt{-2b}}\left\{\displaystyle \frac{1+{e}^{(-2r\xi )}}{-1+{e}^{(-2r\xi )}}\right\}.\end{eqnarray} \tag{ 3.1.27 }$$

When p = r = 0, and p = q = K, r = 0, inserting the values of the constants lead to constant solutions and hence has not been documented here, since constant solutions have no physical significance.

Case 10: When q = r = K and p = 0,

Using (3.1.11) and (2.24) into equation (3.1.10), we obtain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{{Kk}(1+{e}^{K\xi })}{\sqrt{-2b}(1-{e}^{K\xi })}.\end{eqnarray} \tag{ 3.1.28 }$$

Case 11: When q = p + r,

Applying (3.1.11) and (2.25) into equation (3.1.10), we attain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{k\left(p-r\right)\{1+{{re}}^{(p-r)\xi }\}}{\sqrt{-2b}\{1-{{re}}^{(p-r)\xi }\}}.\end{eqnarray} \tag{ 3.1.29 }$$

Case 12: When $q=-(p+r)$,

Inserting (3.1.11) and (2.26) into equation (3.1.10), we secure

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{k\left(p-r\right)\{r+\ {e}^{(p-r)\xi }\}}{\sqrt{-2b}\{r\ -\ {e}^{(p-r)\xi }\}}.\end{eqnarray} \tag{ 3.1.30 }$$

Case 13: When p = 0,

By means of (3.1.11) and (2.27) from equation (3.1.10), we ascertain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{{kq}\ (1+{{re}}^{q\xi })}{\sqrt{-2b}(1-{{re}}^{q\xi })}.\end{eqnarray} \tag{ 3.1.31 }$$

Case 14: When $r=q=p\ne 0$,

Making use of (3.1.11) and (2.28) from equation (3.1.10), we gain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{\sqrt{3}\ {pk}}{\sqrt{-2b}}\ \tan \left(\displaystyle \frac{\sqrt{3}}{2}\ p\ \xi \right)\ .\end{eqnarray} \tag{ 3.1.32 }$$

When r = q = 0, and r = 0 embedding the values of the constants lead to constant solutions and hence has not been written here, since constant solutions have no physical significance.

Case 15: When p = q = 0,

Using (3.1.11) and (2.30) into equation (3.1.10), we carry through

$$\begin{eqnarray}&&u\left(\xi \right)=\pm \displaystyle \frac{2k}{\xi \sqrt{-2b}}.\end{eqnarray} \tag{ 3.1.33 }$$

Case 16: When r = p and q = 0,

Using (3.1.11) and (2.31) from equation (3.1.10), we obtain

$$\begin{eqnarray}&&u\left(\xi \right)=\pm 2\displaystyle \frac{{kp}}{\sqrt{-2b}}\ \tan (p\ \xi ).\end{eqnarray} \tag{ 3.1.34 }$$

It is observe that, by means of the new auxiliary equation method, we obtain abundant closed form traveling wave solutions of the NLSE, which might be useful to analyse various complex phenomena in physical sciences, nonlinear optics, fluid dynamics, mathematical biology, plasma physics and other areas.

In this section, we will examine the homogeneous form of the coupled BE equation. We consider the following system of equations (Jawad et al 2010):

$$\begin{eqnarray}&&{u}_{t}-{u}_{{xx}}+2{{uu}}_{{xx}}+\alpha {\left({uv}\right)}_{{\boldsymbol{x}}}=0,\end{eqnarray} \tag{ 3.2.1 }$$

$$\begin{eqnarray}&&{v}_{t}-{v}_{{xx}}+2{{vv}}_{{xx}}+\beta {\left({uv}\right)}_{{\boldsymbol{x}}}=0,\end{eqnarray} \tag{ 3.2.2 }$$

where α and β are arbitrary constants.

In fluid mechanics, Burgers equation is one of the crucial model equations (Rezazadeh et al 2019). In addition these equations are used to narrate the structure of shock waves travelling in a viscous fluid (Mittal and Rohila, 2018), traffic flow and acoustic transmission.

In order to study equations (3.2.1) and (3.2.2), we introduce the subsequent travelling wave transformations

$u\left(x,t\right)=U(\xi ),v\left(x,t\right)=V(\xi )$ and ξ = x − λ t.

Hence equations (3.2.1) and (3.2.2) take the form

$$\begin{eqnarray}&&-\lambda U^{\prime} -U^{\prime\prime} +2{UU}^{\prime} +\alpha \left({UV}\right)^{\prime} =0,\end{eqnarray} \tag{ 3.2.3 }$$

$$\begin{eqnarray}&&-\lambda V^{\prime} -V^{\prime\prime} +2{VV}^{\prime} +\beta \left({UV}\right)^{\prime} =0.\end{eqnarray} \tag{ 3.2.4 }$$

Integrating equations (3.2.3) and (3.2.4) and considering the integration constants to zero, since we are searching for soliton solutions, the equations (3.2.3) and (3.2.4) become

$$\begin{eqnarray}&&-\lambda U-U^{\prime} +{U}^{2}+\alpha {UV}=0.\end{eqnarray} \tag{ 3.2.5 }$$

$$\begin{eqnarray}&&-\lambda V-V^{\prime} +{V}^{2}+\beta {UV}=0.\end{eqnarray} \tag{ 3.2.6 }$$

Now balancing the highest order derivative term $U^{\prime}$ with the highest order nonlinear term U² and UV gives n = 1, m = 1.

Therefore, according to new auxiliary equation method, the solution of equations (3.2.5) and (3.2.6) take the form

$$\begin{eqnarray}&&U\left(\xi \right)={b}_{0}+{b}_{1}{a}^{f(\xi )},\end{eqnarray} \tag{ 3.2.7 }$$

$$\begin{eqnarray}&&V\left(\xi \right)={c}_{0}+{c}_{1}{a}^{f(\xi )},\end{eqnarray} \tag{ 3.2.8 }$$

where, $f\left(\xi \right)$ is the solution of the nonlinear equation (2.5).

Inserting equations (3.2.7) and (3.2.8) together with equations (3.2.5) and (3.2.6), we achieve

$$\begin{eqnarray*}&&\begin{array}{l}\left(\alpha {b}_{1}{c}_{1}+{b}_{1}^{2}-{b}_{1}r\right)\{{a}^{f\left(\xi \right)}\}{}^{2}+\left(-\lambda {b}_{1}-{b}_{1}q+2{b}_{0}{b}_{1}+\alpha {b}_{0}{c}_{1}+\alpha {b}_{1}{c}_{0}\right){a}^{f(\xi )}\\ \quad -\,\lambda {b}_{0}-{b}_{1}p+{b}_{0}^{2}+{b}_{0}{c}_{0}\alpha =0,\end{array}\end{eqnarray*}$$

and

$$\begin{eqnarray*}&&\begin{array}{l}\left(\beta {b}_{1}{c}_{1}+{c}_{1}^{2}-{c}_{1}r\right)\{{a}^{f\left(\xi \right)}\}{}^{2}+\left(-\lambda {c}_{1}-{c}_{1}q+2{c}_{0}{c}_{1}+\beta {b}_{0}{c}_{1}+\beta {b}_{1}{c}_{0}\right){a}^{f(\xi )}\\ \quad -\,\lambda {c}_{0}-{c}_{1}p+{c}_{0}^{2}+\beta {b}_{0}{c}_{0}=0.\end{array}\end{eqnarray*}$$

Now, setting the coefficients $\{{a}^{f\left(\xi \right)}\}{}^{j}\ (j=0,1,2)$ to zero leads to a set of algebraic equations:

$$\begin{eqnarray*}&&\alpha {b}_{1}{c}_{1}+{b}_{1}^{2}-{b}_{1}r=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&-\lambda {b}_{1}-{b}_{1}q+2{b}_{0}{b}_{1}+\alpha {b}_{0}{c}_{1}+\alpha {b}_{1}{c}_{0}=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&-\lambda {b}_{0}-{b}_{1}p+{b}_{0}^{2}+{b}_{0}{c}_{0}\alpha =0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&\beta {b}_{1}{c}_{1}+{c}_{1}^{2}-{c}_{1}r=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&-\lambda {c}_{1}-{c}_{1}q+2{c}_{0}{c}_{1}+\beta {b}_{0}{c}_{1}+\beta {b}_{1}{c}_{0}=0.\end{eqnarray*}$$

$$\begin{eqnarray*}&&-\lambda {c}_{0}-{c}_{1}p+{c}_{0}^{2}+\beta {b}_{0}{c}_{0}=0.\end{eqnarray*}$$

Solving these algebraic equations with the aid of Maple, we achieve

$$\begin{eqnarray*}&&\lambda =\displaystyle \frac{-4{rp}+{q}^{2}+q\sqrt{{q}^{2}-4{rp}}}{q+\sqrt{{q}^{2}-4{rp}}},{b}_{0}=\displaystyle \frac{(q+\sqrt{{q}^{2}-4{rp}})(\beta -1)}{2(\alpha \beta -1)},{b}_{1}=\displaystyle \frac{r(\beta -1)}{\alpha \beta -1},\end{eqnarray*}$$

$$\begin{eqnarray}&&{c}_{0}=\displaystyle \frac{(q+\sqrt{{q}^{2}-4{rp}})(\alpha -1)}{2(\alpha \beta -1)},\ \ {c}_{1}=\displaystyle \frac{r(\alpha -1)}{\alpha \beta -1}.\end{eqnarray} \tag{ 3.2.9 }$$

Now by using the above values of the constants, we attempt to establish the wave solutions to the coupled Burgers equation.

Case 1: When ${q}^{2}-4\ {pr}\lt 0$ and $r\ne 0$,

Using (3.2.9), (2.6) and (2.7), from equation (3.2.7), we achieve

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)}{2\left(\alpha \beta -1\right)}\left\{\sqrt{{q}^{2}-4{rp}}+\sqrt{4\ {pr}-{q}^{2}\ }\tan \left(\displaystyle \frac{\sqrt{4\ {pr}-{q}^{2}\ }}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.10 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)}{2\left(\alpha \beta -1\right)}\left\{\sqrt{{q}^{2}-4{rp}}-\sqrt{4\ {pr}-{q}^{2}\ }\cot \left(\displaystyle \frac{\sqrt{4\ {pr}-{q}^{2}\ }}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.11 }$$

Using (3.2.9), (2.6) and (2.7) into equation (3.2.8), we accomplish

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}-4{rp}}+\sqrt{4\ {pr}-{q}^{2}\ }\tan \left(\displaystyle \frac{\sqrt{4\ {pr}-{q}^{2}\ }}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.12 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}-4{rp}}-\sqrt{4\ {pr}-{q}^{2}\ }\cot \left(\displaystyle \frac{\sqrt{4\ {pr}-{q}^{2}\ }}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.13 }$$

Case 2: When ${q}^{2}-4\ {pr}\gt 0$ and $r\ne 0$,

Putting (3.2.9), (2.8) and (2.9) into equation (3.2.7), we get

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)\sqrt{{q}^{2}-4{rp}}}{2(\alpha \beta -1)}\left\{1-\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4\ {pr}\ }}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.14 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)\sqrt{{q}^{2}-4{rp}}}{2(\alpha \beta -1)}\left\{1-\coth \left(\displaystyle \frac{\sqrt{{q}^{2}-4\ {pr}\ }}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.15 }$$

Putting (3.2.9), (2.8) and (2.9) into equation (3.2.8), we attain

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)\sqrt{{q}^{2}-4{rp}}}{2(\alpha \beta -1)}\left\{1-\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4\ {pr}\ }}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.16 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)\sqrt{{q}^{2}-4{rp}}}{2(\alpha \beta -1)}\left\{1-\coth \left(\displaystyle \frac{\sqrt{{q}^{2}-4\ {pr}\ }}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.17 }$$

Case 3: When ${q}^{2}+4\ {p}^{2}\lt 0,r\ne 0$ and $r=-p$,

Applying (3.2.9), (2.10) and (2.11), from equation (3.2.7), we ascertain

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}+4{p}^{2}}+\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}\tan \left(\displaystyle \frac{\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.18 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}+4{p}^{2}}-\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}\cot \left(\displaystyle \frac{\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.19 }$$

Applying (3.2.9), (2.10) and (2.11), from equation (3.2.8), we find out

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{\left({q}^{2}+4\ {p}^{2}\right)}+\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}\tan \left(\displaystyle \frac{\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.20 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{\left({q}^{2}+4\ {p}^{2}\right)}-\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}\cot \left(\displaystyle \frac{\sqrt{-\left({q}^{2}+4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.21 }$$

Case 4: When ${q}^{2}+4\ {p}^{2}\gt 0,r\ne 0$ and $r=-p$,

Exerting (3.2.9), (2.12) and (2.13) into equation (3.2.7), we derive

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)\sqrt{{q}^{2}+4\ {p}^{2}\ }}{2(\alpha \beta -1)}\left\{1-\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}+4\ {p}^{2}}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.22 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)\sqrt{{q}^{2}+4\ {p}^{2}\ }}{2(\alpha \beta -1)}\left\{1-\coth \left(\displaystyle \frac{\sqrt{{q}^{2}+4\ {p}^{2}}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.23 }$$

Exerting (3.2.9), (2.12) and (2.13) into equation (3.2.8), we secure

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)\sqrt{{q}^{2}+4\ {p}^{2}\ }}{2(\alpha \beta -1)}\left\{1-\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}+4\ {p}^{2}}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.24 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)\sqrt{{q}^{2}+4\ {p}^{2}\ }}{2(\alpha \beta -1)}\left\{1-\coth \left(\displaystyle \frac{\sqrt{{q}^{2}+4\ {p}^{2}}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.25 }$$

Case 5: When ${q}^{2}-4\ {p}^{2}\lt 0$ and r = p,

Employing (3.2.9), (2.14) and (2.15), from equation (3.2.7), we determine

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}-4{p}^{2}}+\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}\ \tan \left(\displaystyle \frac{\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.26 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}-4{p}^{2}}-\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}\ \cot \left(\displaystyle \frac{\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.27 }$$

Employing (3.2.9), (2.14) and (2.15), from equation (3.2.8), we carry out

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}-4{p}^{2}}+\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}\tan \left(\displaystyle \frac{\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.28 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{{q}^{2}-4{p}^{2}}-\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}\cot \left(\displaystyle \frac{\sqrt{-\left({q}^{2}-4\ {p}^{2}\right)}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.29 }$$

Case 6: When ${q}^{2}-4\ {p}^{2}\gt 0$ and r = p,

Making use of (3.2.9), (2.16) and (2.17), from equation (3.2.7), we carry through

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)\sqrt{{q}^{2}-4{p}^{2}}}{2(\alpha \beta -1)}\left\{1-\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4{p}^{2}}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.30 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{(\beta -1)\sqrt{{q}^{2}-4{p}^{2}}}{2(\alpha \beta -1)}\left\{1-\coth \left(\displaystyle \frac{\sqrt{{q}^{2}-4{p}^{2}}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.31 }$$

Making use of (3.2.9), (2.16) and (2.17), from equation (3.2.8), we carry through

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)\sqrt{{q}^{2}-4{p}^{2}}}{2(\alpha \beta -1)}\left\{1-\tanh \left(\displaystyle \frac{\sqrt{{q}^{2}-4{p}^{2}}}{2}\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.32 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{(\alpha -1)\sqrt{{q}^{2}-4{p}^{2}}}{2(\alpha \beta -1)}\left\{1-\coth \left(\displaystyle \frac{\sqrt{{q}^{2}-4{p}^{2}}}{2}\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.33 }$$

Case 7: When ${q}^{2}=4{pr}$,

By means of (3.2.9) and (2.18), from equation (3.2.7), we obtain

$$\begin{eqnarray}&&U\left(\xi \right)=-\displaystyle \frac{\left(\beta -1\right)}{\xi \left(\alpha \beta -1\right)}.\end{eqnarray} \tag{ 3.2.34 }$$

By means of (3.2.9) and (2.18), from equation (3.2.8), we obtain

$$\begin{eqnarray}&&V\left(\xi \right)=-\displaystyle \frac{\left(\alpha -1\right)}{\xi \left(\alpha \beta -1\right)}.\end{eqnarray} \tag{ 3.2.35 }$$

Case 8: When ${rp}\lt 0,q=0$ and $r\ne 0$,

Inserting (3.2.9), (2.19) and (2.20), from equation (3.2.7), we accomplish

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{\sqrt{-{rp}}\ (\beta -1)}{(\alpha \beta -1)}\left\{1-\tanh \left(\sqrt{-{rp}\ }\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.36 }$$

or

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{\sqrt{-{rp}}\ \ (\beta -1)}{(\alpha \beta -1)}\left\{1-\coth \left(\sqrt{-{rp}\ }\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.37 }$$

Inserting (3.2.9), (2.19) and (2.20), from equation (3.2.8), we accomplish

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{\sqrt{-{rp}}\ \ (\alpha -1)}{(\alpha \beta -1)}\left\{1-\tanh \left(\sqrt{-{rp}\ }\ \xi \right)\right\}\end{eqnarray} \tag{ 3.2.38 }$$

or

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{\sqrt{-{rp}}\ \ (\alpha -1)}{(\alpha \beta -1)}\left\{1-\coth \left(\sqrt{-{rp}\ }\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.39 }$$

Case 9: When q = 0 and $p=-r$,

Setting (3.2.9) and (2.21) into equation (3.2.7), we get

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{2r\ (\beta -1){e}^{(-2r\xi )}}{(\alpha \beta -1)(-1+{e}^{(-2r\xi )})}.\end{eqnarray} \tag{ 3.2.40 }$$

Setting (3.2.9) and (2.21) into equation (3.2.8), we get

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{2r\ (\alpha -1){e}^{(-2r\xi )}}{(\alpha \beta -1)(-1+{e}^{(-2r\xi )})}.\end{eqnarray} \tag{ 3.2.41 }$$

Whenever p = r = 0 and p = q = K, r = 0, embedding the values of the constants lead to steady solutions and thus has not been reported here, since consistent solutions have no physical essentialness.

Case 10: When $q=r=K$ and p = 0,

Exerting (3.2.9) and (2.24) into equation (3.2.7), we attain

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{K\ (\beta -1)}{(\alpha \beta -1)(1-{e}^{K\xi })}.\end{eqnarray} \tag{ 3.2.42 }$$

Exerting (3.2.9) and (2.24) into equation (3.2.8), we ascertain

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{K\ (\alpha -1)}{(\alpha \beta -1)(1-{e}^{k\xi })}.\end{eqnarray} \tag{ 3.2.43 }$$

Case 11: When q = p + r,

Employing (3.2.9) and (2.25), from equation (3.2.7), we gain

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{r(\beta -1)(p-r){e}^{\left(p-r\right)\xi }}{\left(\alpha \beta -1\right)\{1-{{re}}^{\left(p-r\right)\xi }\}}.\ \end{eqnarray} \tag{ 3.2.44 }$$

Employing (3.2.9) and (2.25) into equation (3.2.8), we gain

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{r(\alpha -1)(p-r){e}^{\left(p-r\right)\xi }}{\left(\alpha \beta -1\right)\{1-{{re}}^{\left(p-r\right)\xi }\}}.\end{eqnarray} \tag{ 3.2.45 }$$

Case 12: When $q=-(p+r)$,

Making use of (3.2.9) and (2.26), from equation (3.2.7), we find out

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{r\left(\beta -1\right)(p-r)}{\left(\alpha \beta -1\right)\ (r\ -\ {e}^{(p-r)\xi })}.\end{eqnarray} \tag{ 3.2.46 }$$

Making use of (3.2.9) and (2.26), from equation (3.2.8), we find out

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{r\left(\alpha -1\right)(p-r)}{\left(\alpha \beta -1\right)\ (r-\ {e}^{(p-r)\xi })}.\end{eqnarray} \tag{ 3.2.47 }$$

Case 13: When p = 0,

By means of (3.2.9) and (2.27), from equation (3.2.7), we derive

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{q\left(\beta -1\right)}{\left(\alpha \beta -1\right)\ (1-\ {{re}}^{q\xi })}.\end{eqnarray} \tag{ 3.2.48 }$$

By means of (3.2.9) and (2.27), from equation (3.2.8), we derive

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{q\left(\alpha -1\right)}{\left(\alpha \beta -1\right)\ (1-\ {{re}}^{q\xi })}.\end{eqnarray} \tag{ 3.2.49 }$$

Case 14: When $r=q=p\ne 0$,

Inserting (3.2.9) and (2.28) into equation (3.2.7), we secure

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{p(\beta -1)}{2\left(\alpha \beta -1\right)}\left\{\sqrt{-3}+\sqrt{3}\ \tan \left(\displaystyle \frac{\sqrt{3}}{2}\ p\ \xi \right)\ \right\}.\end{eqnarray} \tag{ 3.2.50 }$$

Inserting (3.2.9) and (2.28) into equation (3.2.8), we secure

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{p(\alpha -1)}{2\left(\alpha \beta -1\right)}\left\{\sqrt{-3}+\sqrt{3}\ \tan \left(\displaystyle \frac{\sqrt{3}}{2}\ p\ \xi \right)\ \right\}.\end{eqnarray} \tag{ 3.2.51 }$$

Whenever r = q = 0 and r = 0 implanting the values of the constants lead to consistent arrangements and subsequently has not been composed here, since steady arrangements have no physical essentialness.

Case 15: When p = q = 0,

Putting (3.2.9) and (2.30) into equation (3.2.7), we determine

$$\begin{eqnarray}&&U\left(\xi \right)=-\displaystyle \frac{\left(\beta -1\right)}{\ \xi \left(\alpha \beta -1\right)}.\ \end{eqnarray} \tag{ 3.2.52 }$$

Putting (3.2.9) and (2.30) into equation (3.2.8), we determine

$$\begin{eqnarray}&&V\left(\xi \right)=-\displaystyle \frac{\left(\alpha -1\right)}{\ \xi \left(\alpha \beta -1\right)}.\end{eqnarray} \tag{ 3.2.53 }$$

Case 16: When r = p and q = 0,

Using (3.2.9) and (2.31) into equation (3.2.7), we carry out

$$\begin{eqnarray}&&U\left(\xi \right)=\displaystyle \frac{p\ (\beta -1)}{2(\alpha \beta -1)}\left\{\sqrt{-4}+2\tan \left(p\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.54 }$$

Using (3.2.9) and (2.31) into equation (3.2.8), we carry out

$$\begin{eqnarray}&&V\left(\xi \right)=\displaystyle \frac{p\ (\alpha -1)}{2(\alpha \beta -1)}\left\{\sqrt{-4}+2\tan \left(p\ \xi \right)\right\}.\end{eqnarray} \tag{ 3.2.55 }$$

It is perceive that by means of the new auxiliary equation method, we have achieved abundant closed form traveling wave solutions to the coupled Burgers equation, which might be helpful to explore simplest nonlinear model of turbulence, the motion of the isolated waves and many fields such as hydrodynamics, plasma physics, nonlinear optic etc.