### 3.1 Exact solutions to the double sine-Gordon equation

Consider the double sine-Gordon (SG) equation [

27,

28]

${u}_{xt}=sinu+sin2u.$

(7)

In order to apply the first integral method described in Section 2, we first introduce the transformations

$sinu=\frac{v-{v}^{-1}}{2i},\phantom{\rule{2em}{0ex}}sin2u=\frac{{v}^{2}-{v}^{-2}}{2i},\phantom{\rule{1em}{0ex}}v={e}^{iu}.$

(8)

Using (2) and (3), Eq. (

7) becomes

$-c{u}^{\u2033}=\frac{v-{v}^{-1}}{2i}+\frac{{v}^{2}-{v}^{-2}}{2i}.$

(9)

We next use the transformation

$\{\begin{array}{c}v={e}^{iu},\hfill \\ {u}^{\prime}=\frac{{v}^{\prime}}{iv},\hfill \\ {u}^{\u2033}=\frac{1}{i}(\frac{{v}^{\u2033}}{v}-\frac{{({v}^{\prime})}^{2}}{{v}^{2}}).\hfill \end{array}$

(10)

We obtain

$-2cv{v}^{\u2033}+2c{\left({v}^{\prime}\right)}^{2}-{v}^{4}-{v}^{3}+v+1=0.$

(11)

Next, we introduce new independent variables

$X=v$,

$Y=\frac{dv}{d\xi}$ which change (11) to the system of ODEs

$\{\begin{array}{c}\frac{dX}{d\xi}=Y,\hfill \\ \frac{dY}{d\xi}=\frac{2c{Y}^{2}-{X}^{4}-{X}^{3}+X+1}{2cX}.\hfill \end{array}$

(12)

Assume that

$d\xi =X\phantom{\rule{0.2em}{0ex}}d\tau $, then (12) becomes

$\{\begin{array}{c}\frac{dX}{d\tau}=XY,\hfill \\ \frac{dY}{d\tau}={Y}^{2}-a{X}^{4}-a{X}^{3}+aX+a,\hfill \end{array}$

(13)

where $a=\frac{1}{2c}$.

According to the first integral method, we suppose that

$X=X(\tau )$ and

$Y=Y(\tau )$ are nontrivial solutions of Eq. (

13) and

$P(X,Y)=\sum _{i=0}^{m}{a}_{i}(X){Y}^{i}$

is an irreducible polynomial in the complex domain

$\mathbf{C}[X,Y]$ such that

$P(X(\tau ),Y(\tau ))=\sum _{i=0}^{m}{a}_{i}(X(\tau ))Y{(\tau )}^{i}=0,$

(14)

where

${a}_{i}(X)$ (

$i=0,1,\dots ,m$) are polynomials in

*X* and

${a}_{m}(X)\ne 0$. Equation (

14) is called the first integral to Eq. (

13). Applying the division theorem, one sees that there exists a polynomial

$H(X,Y)=h(X)+g(X)Y$ in the complex domain

$\mathbf{C}[X,Y]$ such that

$\frac{dP}{d\tau}=\frac{\partial P}{\partial X}\frac{\partial X}{\partial \tau}+\frac{\partial P}{\partial Y}\frac{\partial Y}{\partial \tau}=(h(X)+g(X)Y)(\sum _{i=0}^{m}{a}_{i}(X){Y}^{i}).$

(15)

Suppose that

$m=1$ in (14), and then, by comparing with the coefficients of

${Y}^{i}$ (

$i=2,1,0$) on both sides of (15), we have

Since

${a}_{1}(X)$ is a polynomial in

*X*, from (16) we conclude that

${a}_{1}(X)$ is a constant and

$g(X)=1$. For simplicity, we take

${a}_{1}(X)=1$. Then Eq. (

17) indicates that

$degh(X)\le deg{a}_{0}(X)$. Thus, from Eq. (

18) we conclude that

$degh(X)=deg{a}_{0}(X)=2$. Now suppose that

$h(X)={A}_{2}{X}^{2}+{A}_{1}X+{A}_{0},\phantom{\rule{2em}{0ex}}{a}_{0}(X)={B}_{2}{X}^{2}+{B}_{1}X+{B}_{0}\phantom{\rule{1em}{0ex}}({A}_{2}\ne 0,{B}_{2}\ne 0),$

(19)

where

${A}_{2}$,

${A}_{1}$,

${A}_{0}$,

${B}_{2}$,

${B}_{1}$,

${B}_{0}$ are all constants to be determined. Substituting Eq. (

19) into Eq. (

17), we obtain

${A}_{2}={B}_{2},\phantom{\rule{2em}{0ex}}{A}_{1}=0,\phantom{\rule{2em}{0ex}}{A}_{0}=-{B}_{0}.$

Then

$h(X)={B}_{2}{X}^{2}-{B}_{0}.$

Substituting

${a}_{0}(X)$,

${a}_{1}(X)$ and

$h(X)$ in (18) and setting all the coefficients of powers

*X* to be zero, we obtain a system of nonlinear algebraic equations, and by solving it, we obtain

${B}_{2}={B}_{1}={B}_{0}=\pm \sqrt{-a}.$

(20)

Substituting (20) in (14), we obtain

$Y\pm \sqrt{-a}({X}^{2}+X+1)=0.$

(21)

Combining Eq. (

21) with (13), second-order differential Eq. (

11) can be reduced to

$\frac{dv}{d\xi}=\mp \sqrt{-a}({v}^{2}+v+1).$

(22)

Solving Eq. (

22) directly and changing to the original variables, we obtain the exact solutions to Eq. (

7):

where ${c}_{1}$ is an arbitrary constant and $a=\frac{1}{2c}$.

Therefore, the exact solutions to the double sine-Gordon (SG) equation can be written as

where ${c}_{1}$ is an arbitrary constant.

### 3.2 Exact solutions to the Burgers equation

The Burgers equation [

29]

${u}_{t}+u{u}_{x}-a{u}_{xx}=0$

(27)

is one of the most famous nonlinear diffusion equations. The positive parameter *a* refers to a dissipative effect.

Using (2) and (3), Eq. (

27) becomes

$-c{u}^{\prime}+u{u}^{\prime}-a{u}^{\u2033}=0.$

(28)

We rewrite (28) as follows:

${u}^{\u2033}=\frac{1}{a}u{u}^{\prime}-\frac{c}{a}{u}^{\prime}.$

(29)

By introducing new variables

$X=u$ and

$Y={u}_{\xi}$, Eq. (

29) changes into a system of ODEs

$\{\begin{array}{c}{X}^{\prime}(\xi )=Y,\hfill \\ {Y}^{\prime}(\xi )=\frac{1}{a}XY-\frac{c}{a}Y.\hfill \end{array}$

(30)

Now, the division theorem is employed to seek the first integral to (30). Suppose that

$X=X(\xi )$ and

$Y=Y(\xi )$ are nontrivial solutions to (30), and

$P(X,Y)={\sum}_{i=0}^{m}{a}_{i}(X){Y}^{i}$ is an irreducible polynomial in

$\mathbf{C}[X,Y]$ such that

$P(X(\xi ),Y(\xi ))=\sum _{i=0}^{m}{a}_{i}(X){Y}^{i}=0,$

(31)

where

${a}_{i}(X)$ (

$i=0,1,\dots ,m$) are polynomials in

*X* and

${a}_{m}(X)\ne 0$. Equation (

31) is called the first integral to Eq. (

30). Due to the division theorem, there exists a polynomial

$H(X,Y)=h(X)+g(X)Y$ in

$\mathbf{C}[X,Y]$ such that

$\frac{dP}{d\xi}=\frac{\partial P}{\partial X}\frac{\partial X}{\partial \xi}+\frac{\partial P}{\partial Y}\frac{\partial Y}{\partial \xi}=(h(X)+g(X)Y)(\sum _{i=0}^{m}{a}_{i}(X){Y}^{i}).$

(32)

Suppose that

$m=1$ in (31). By comparing the coefficients of

${Y}^{i}$ (

$i=2,1,0$) on both sides of (32), we have

Since

${a}_{1}(X)$ is a polynomial in

*X*, from (33) we conclude that

${a}_{1}(X)$ is a constant and

$g(X)=0$. For simplicity, we take

${a}_{1}(X)=1$, and balancing the degrees of

$h(X)$ and

${a}_{0}(X)$, we conclude that

$degh(X)=0$ or 1. If

$degh(X)=0$, suppose that

$h(X)=A$, then from (34), we find

${a}_{0}(X)=-\frac{1}{2a}{X}^{2}+(A+\frac{c}{a})X+B,$

where *B* is an arbitrary integration constant.

Substituting

${a}_{0}(X)$ and

$h(X)$ in (35) and setting all the coefficients of powers

*X* to be zero, we obtain a system of nonlinear algebraic equations, and by solving it, we obtain

$A=0,\phantom{\rule{2em}{0ex}}B\ne 0.$

(36)

Using (36) in (31), we obtain

$Y(\xi )=\frac{1}{2a}{X}^{2}-\frac{c}{a}X-B.$

(37)

Combining Eq. (

37) with the first part of (30), we obtain the exact solutions of Eq. (

29) as follows:

${X}_{1}(\xi )=c+\sqrt{2aB+{c}^{2}}[1-tanh(\frac{\sqrt{2aB+{c}^{2}}}{2a}(\xi +{\xi}_{0}))],$

(38)

where ${\xi}_{0}$ is an arbitrary constant.

Therefore, the exact solutions to the Burgers equation can be written as

${u}_{1}(x,t)=c+\sqrt{2aB+{c}^{2}}[1-tanh(\frac{\sqrt{2aB+{c}^{2}}}{2a}(x-ct+{\xi}_{0}))],$

(39)

where ${\xi}_{0}$ is an arbitrary constant and $a>0$.

Now suppose that

$m=2$. By an application of the division theorem, we can conclude that there exists a polynomial

$H(X,Y)=h(X)+g(X)Y$ in

$\mathbf{C}[X,Y]$ such that

$\frac{dP}{d\xi}=\frac{\partial P}{\partial X}\frac{\partial X}{\partial \xi}+\frac{\partial P}{\partial Y}\frac{\partial Y}{\partial \xi}=(h(X)+g(X)Y)(\sum _{i=0}^{2}{a}_{i}(X){Y}^{i}).$

(40)

Comparing the coefficients of

${Y}^{i}$ (

$i=3,2,1,0$) of both sides of (40) yields

Since

${a}_{2}(X)$ is a polynomial of

*X*, from (41) we conclude that

${a}_{2}(X)$ is a constant and

$g(X)=0$. For simplicity, we take

${a}_{2}(X)=1$, and balancing the degrees of

$h(X)$,

${a}_{0}(X)$ and

${a}_{1}(X)$, we conclude that

$degh(X)=0$ or 1. If

$degh(X)=0$, suppose that

$h(X)=A$. Then from (42), we find

${a}_{1}(X)=-\frac{1}{a}{X}^{2}+(A+\frac{2c}{a})X+B,$

where

*B* is an arbitrary constant of integration. From (43) we have

$\begin{array}{rcl}{a}_{0}(X)& =& \frac{1}{4{a}^{2}}{X}^{4}-(\frac{2}{3a}A+\frac{c}{{a}^{2}}){X}^{3}+(\frac{{A}^{2}}{2}+\frac{3c}{2a}A+\frac{{c}^{2}}{{a}^{2}}-\frac{B}{2a}){X}^{2}\\ +(AB+\frac{cB}{a})X+D,\end{array}$

where

*D* is an arbitrary constant of integration. Substituting

${a}_{0}(X)$ and

$h(X)$ in (44) and setting all the coefficients of powers of

*X* to zero, we obtain a system of nonlinear algebraic equations. Solving these equations, we obtain

$A=0,\phantom{\rule{2em}{0ex}}B\ne 0,\phantom{\rule{2em}{0ex}}D\ne 0.$

(45)

Now using (45) in (31), we get

$Y(\xi )=\frac{1}{2a}{X}^{2}-\frac{c}{a}X-\frac{B}{2}\pm \frac{\sqrt{{B}^{2}-4D}}{2}.$

(46)

Combining Eq. (

46) with the first part of (30), we obtain the exact solutions to Eq. (

29) in the form

$\begin{array}{rcl}{X}_{2}(\xi )& =& c+\sqrt{aB+{c}^{2}\mp a\sqrt{{B}^{2}-4D}}\\ \times [1-tanh(\frac{\sqrt{aB+{c}^{2}\mp a\sqrt{{B}^{2}-4D}}}{2a}(\xi +{\xi}_{0}))],\end{array}$

(47)

where

${\xi}_{0}$ is an arbitrary constant of integration. Therefore, the exact solutions to the Burgers equation can be written as

$\begin{array}{rcl}{u}_{2}(x,t)& =& c+\sqrt{aB+{c}^{2}\mp a\sqrt{{B}^{2}-4D}}\\ \times [1-tanh(\frac{\sqrt{aB+{c}^{2}\mp a\sqrt{{B}^{2}-4D}}}{2a}(x-ct+{\xi}_{0}))],\end{array}$

(48)

where ${\xi}_{0}$ is an arbitrary constant and $a>0$.