Exact Solutions to the Boussinesq-Burgers Equations

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1. Introduction

In the present paper, we will investigate the well-known Boussinesq-Burgers equations in the form

${u}_{t}+2u{u}_{x}-\alpha {v}_{x}=0,$ (1)

${v}_{t}+2{\left(uv\right)}_{x}-\alpha {u}_{xxx}=0,$ (2)

where $\alpha $ = constant. Equations ((1) & (2)) emerge in the investigation of fluid flow, and describe the proliferation of shallow water waves. $u=u\left(x,t\right)$ represents horizontal velocity; $v=v\left(x,t\right)$ represents the height of the water surface above horizontal at bottom. Equations ((1) & (2)) have been investigated by many authors with different methods (see [1] - [6] and references therein).

In this paper, we will propose a somewhat different method to find various exact solutions of Equations ((1) & (2)). First of all, a nonlinear transformation from the solution of a linear equation to the solution of Equations ((1) & (2)) is derived by using the simplified homogeneous balance method [7] [8] [9] ; then based on the nonlinear transformation, various exact solutions of Equations ((1) & (2)) can be obtained by the given various solutions of the linear equation.

2. Derivation of Nonlinear Transformation

Considering the homogeneous balance between $u{u}_{x}$ and ${v}_{x}$ in Equation (1), and between ${\left(uv\right)}_{x}$ and ${u}_{xxx}$ in Equation (2):

$2m+1=n+1,m+n+1=m+3\Rightarrow m=1,n=2,$

according to the simplified homogeneous balance method [7] [8] [9] , we can suppose that the solution of Equations ((1) & (2)) is of the form

$u\left(x,t\right)=A{\left(\mathrm{ln}\varphi \right)}_{x},$ (3)

$v\left(x,t\right)=B{\left(\mathrm{ln}\varphi \right)}_{xx},$ (4)

where the constants A and B, as well as the function $\varphi =\varphi \left(x,t\right)$ are to be determined later. Substituting (3), (4) into the left hand sides of Equations ((1) & (2)), yields

${u}_{t}+2u{u}_{x}-\alpha {v}_{x}=A\frac{\partial}{\partial x}\left[\frac{{\varphi}_{t}-\alpha \frac{B}{A}{\varphi}_{xx}}{\varphi}+\left(A+\alpha \frac{B}{A}\right){\left(\frac{{\varphi}_{x}}{\varphi}\right)}^{2}\right]\mathrm{,}$ (5)

${v}_{t}+2{\left(uv\right)}_{x}-\alpha {u}_{xxx}=B\frac{{\partial}^{2}}{{\partial}_{x}^{2}}\left[\frac{{\varphi}_{t}-\alpha \frac{A}{B}{\varphi}_{xx}}{\varphi}+\left(A+\alpha \frac{A}{B}\right){\left(\frac{{\varphi}_{x}}{\varphi}\right)}^{2}\right]\mathrm{.}$ (6)

In expressions (5), (6), setting the coefficients of ${\left(\frac{{\varphi}_{x}}{\varphi}\right)}^{2}$ to zero, yields

$A+\alpha \frac{B}{A}=0,A+\alpha \frac{A}{B}=0\Rightarrow A=\pm \alpha ,B=-\alpha .$ (7)

Using the results (7), the expressions (3), (4) becomes

$u\left(x,t\right)=\pm \alpha {\left(\mathrm{ln}\varphi \right)}_{x},$ (8)

$v\left(x,t\right)=-\alpha {\left(\mathrm{ln}\varphi \right)}_{xx},$ (9)

and the expressions (5), (6) can be simplified as

${u}_{t}+2u{u}_{x}-\alpha {v}_{x}=\pm \alpha \frac{\partial}{\partial x}\left[\frac{{\varphi}_{t}\pm \alpha {\varphi}_{xx}}{\varphi}\right]=0,$ (10)

${v}_{t}+2{\left(uv\right)}_{x}-\alpha {u}_{xxx}=-\alpha \frac{{\partial}^{2}}{{\partial}_{x}^{2}}\left[\frac{{\varphi}_{t}\pm \alpha {\varphi}_{xx}}{\varphi}\right]=0,$ (11)

provided that the function $\varphi =\varphi \left(x,t\right)$ satisfies the linear equation

${\varphi}_{t}\pm \alpha {\varphi}_{xx}=0.$ (12)

From (8)-(12), we come to the conclusion that if $\varphi =\varphi \left(x,t\right)$ is a solution of the Equation (12), substituting it into the expressions (8), (9), we have the solution of Equations ((1) & (2)). Thus the expressions (8), (9) and Equation (12) together comprise a nonlinear transformation that from the solution of Equation (12) to the solution of Equations ((1) & (2)). According to the nonlinear transformation composed of expressions (8), (9) and Equation (12), in order to obtain the solution of Equations ((1) & (2)), we have to give the solution of Equation (12). Many solutions of Equation (12) are easily given by

${\varphi}_{1}=1+{\text{e}}^{kx\mp \alpha {k}^{2}t},$

${\varphi}_{2}=1\mp \alpha xt+\frac{1}{6}{x}^{3},$

${\varphi}_{3}=1+{\text{e}}^{\mp \alpha t}\left(A\mathrm{cosh}x+B\mathrm{sinh}x\right),$

${\varphi}_{4}=1+{\text{e}}^{\pm \alpha t}\left(A\mathrm{cos}x+B\mathrm{sin}x\right),$

A and B are constants, $\cdots $ , and so on.

3. Various Exact Solutions of Equations ((1) & (2))

Substituting the solution ${\varphi}_{1}=1+{\text{e}}^{kx\mp \alpha {k}^{2}t},$ of Equtions (12) into (8), (9), we have the solitary wave solutions of Equations ((1) & (2)):

$u\left(x,t\right)=\pm \frac{1}{2}\alpha k\left\{1+\mathrm{tanh}\left[\frac{1}{2}\left(kx\mp \alpha {k}^{2}t\right)\right]\right\},$ (13)

$v\left(x,t\right)=-\frac{1}{4}\alpha {k}^{2}{\text{sech}}^{2}\left[\frac{1}{2}\left(kx\mp \alpha {k}^{2}t\right)\right].$ (14)

Substituting the solution ${\varphi}_{2}=1\mp \alpha xt+\frac{1}{6}{x}^{3},$ of Equation (12) into (8), (9), we have rational solutions of Equations ((1) & (2)):

$u\left(x,t\right)=\frac{\pm \frac{\alpha}{2}{x}^{2}-\alpha t}{1\mp \alpha xt+\frac{1}{6}{x}^{3}},$ (15)

$v\left(x,t\right)=-\frac{\alpha x}{1\mp \alpha xt+\frac{1}{6}{x}^{3}}+\alpha \frac{{\left(\frac{1}{2}{x}^{2}\mp \alpha t\right)}^{2}}{{\left(1\mp \alpha xt+\frac{1}{6}{x}^{3}\right)}^{2}}.$ (16)

Substituting the solution ${\varphi}_{3}=1+{\text{e}}^{\mp \alpha t}\left(A\mathrm{cosh}x+B\mathrm{sinh}x\right),$ of Equation (12) into (8), (9), we have the exact solutions containing hyperbolic functions of Equations ((1) & (2)):

$u\left(x,t\right)=\pm \alpha \frac{{\text{e}}^{\mp \alpha t}\left(A\mathrm{sinh}x+B\mathrm{cosh}x\right)}{1+{\text{e}}^{\mp \alpha t}\left(A\mathrm{cosh}x+B\mathrm{sinh}x\right)},$ (17)

$v\left(x,t\right)=\frac{-\alpha {\text{e}}^{\mp \alpha t}\left(A\mathrm{cosh}x+B\mathrm{sinh}x\right)}{1+{\text{e}}^{\mp \alpha t}\left(A\mathrm{cosh}x+B\mathrm{sinh}x\right)}+\frac{\alpha {\text{e}}^{\mp 2\alpha t}{\left(A\mathrm{sinh}x+B\mathrm{cosh}x\right)}^{2}}{{\left[1+{\text{e}}^{\mp \alpha t}\left(A\mathrm{cosh}x+B\mathrm{sinh}x\right)\right]}^{2}}.$ (18)

Substituting the solution ${\varphi}_{4}=1+{\text{e}}^{\pm \alpha t}\left(A\mathrm{cos}x+B\mathrm{sin}x\right),$ of Equation (12) into (8), (9), we have the exact solutions containing trigonometric functions of Equations ((1) & (2)):

$u\left(x,t\right)=\pm \alpha {\text{e}}^{\pm \alpha t}\frac{\left(-A\mathrm{sin}x+B\mathrm{cos}x\right)}{1+{\text{e}}^{\pm \alpha t}\left(A\mathrm{cos}x+B\mathrm{sin}x\right)},$ (19)

$v\left(x,t\right)=\frac{\alpha {\text{e}}^{\pm \alpha t}\left(A\mathrm{cos}x+B\mathrm{sin}x\right)}{1+{\text{e}}^{\pm \alpha t}\left(A\mathrm{cos}x+B\mathrm{sin}x\right)}+\frac{\alpha {\text{e}}^{\pm 2\alpha t}{\left(-A\mathrm{sin}x+B\mathrm{cos}x\right)}^{2}}{{\left[1+{\text{e}}^{\pm \alpha t}\left(A\mathrm{cos}x+B\mathrm{sin}x\right)\right]}^{2}}$ (20)

$\cdots $ , and so on.

It is well known that the linear Equation (12) can admit an infinite many solutions, based on the nonlinear transformation composed of (8), (9) and (12), we can obtain more solutions of Equations ((1) & (2)), provided that more solutions of Equation (12) are given.

4. Conclusion

In this paper, the nonlinear transformation composed of (8), (9) and (12) for the Boussinesq-Burgers equations has been derived by using the simplified homogeneous balance method. The important role of the nonlinear transformation is that the problem of solving nonlinear Boussinesq-Burgers equations becomes that of solving a linear equation, and the latter is much easily to solve for the researchers.

Acknowledgements

The authors are very grateful to the referees for their valuable comments. The project supported in part by the National Natural Science Foundation of China (Grant Nos. 11301153, 11601225) and The Doctoral Foundation of Henan University of Science and Technology (Grant No. 09001562) and The Science and Technology Innovation Platform of Henan University of Science and Technology (Grant No. 2015XPT001).

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