New Multiple Solution to the Boussinesq Equation and the Burgers-Like Equation

Abstract

By considering an improved tanh function method, we found some exact solutions of Boussinesq and Burgers-like equations. The main idea of this method is to take full advantage of the Riccati equation which has more new solutions. We found some exact solutions of the Boussinesq equation and the Burgers-like equation.

1. Introduction

In recent years, nonlinear phenomena play a crucial role in applied mathematics and physics. Directly searching for exact solutions of nonlinear partial differential equations (PDEs) has become more and more attractive partly due to the availability of computer symbolic systems like Mathematica or Maple that allow us to perform some complicated and tedious algebraic calculation on a computer as well as help us to find exact solutions of PDEs [1–5] now.

Many explicit exact methods have been introduced in the literature [6–14]. Some of them are Painlevé method, homogeneous balance method, similarity reduction method, sine-cosine method, Darboux transformation, Cole-Hopf transformation, generalized Miura transformation, tanh method, Backlund transformation, and others methods [15, 16].

One of the most effectively straightforward methods constructing exact solution of PDEs is the extended tanh function method [17]. Let us simply describe the tanh function. For doing this, one can consider in two variables general form of nonlinear PDE as follows: 𝐻𝑢,𝑢_𝑡𝑡,𝑢_𝑥,𝑢_𝑥𝑥,…=0 (1) and transform (1) with 𝑢(𝑥,𝑡)=𝑢(𝜉),𝜉=𝑘(𝑥−𝜆𝑡), (2) where 𝑘 and 𝜆 are the wave number and wave speed, respectively. After the transformation, we get a nonlinear ODE for 𝑢(𝜉) as follows: 𝐻^′𝑢^′,𝑢^′′,𝑢^′′′,…=0. (3)

The fact that the solutions of many nonlinear equations can be expressed as a finite series of tanh functions that motivates us to seek for the solutions of (3) in the form 𝑢(𝑥,𝑡)=𝑢(𝜉)=^𝑚_𝑖=0𝑎_𝑖tanh^𝑖(𝜉)=^𝑚_𝑖=0𝑎_𝑖𝐹^𝑖, (4) where 𝐹^𝑖=tanh^𝑖(𝜉), an equation for 𝐹(𝜉), is obtained. 𝑚 is a positive integer that can be determined by balancing the linear term of highest order with the nonlinear term in (1), and 𝑘,𝜆,𝑎₁,𝑎₂,…,𝑎_𝑚 are parameters to be determined.

Substituting solution (4) into (3) yields a set of algebraic equations for 𝐹^𝑖, then all coefficients of 𝐹^𝑖 have to vanish. From these relations 𝑘,𝜆,𝑎₁,𝑎₂,…,𝑎_𝑚 can be determined.

In this work, we will consider to solve general Boussinesq equation and the Burgers-like equation by using the improved tanh function method which is introduced by Chen and Zhang [18].

2. Method and Its Applications

The main idea of this method is to take full advantage of the Riccati equation that tanh function satisfies and uses its solutions 𝐹 to replace tanh𝜉. The required Riccati equation is given as 𝐹^′=𝐴+𝐵𝐹+𝐶𝐹², (5) where 𝐹=(𝑑𝐹/𝑑𝜉) and 𝐴,𝐵, and𝐶 are constant. In the following, Chen and Zhang [18] have given several cases to get the solution of (5) in the form of finite series of tanh functions (4).

Case 1. If 𝐶=0,𝐵≠0, then (5) has the solutions exp(𝐵𝜉)−𝐴𝐵. (6)

Case 2. If 𝐴=1/2,𝐵=0, and 𝐶=−1/2, then (5) has the solutions cotℎ𝜉±cscℎ𝜉, tanℎ𝜉±𝑖secℎ𝜉(𝑖²=−1).

Case 3. If 𝐴=𝐶=±1/2,𝐵=0, then (5) has the solutions sec𝜉±tan𝜉, csc𝜉±cot𝜉.

Case 4. If 𝐴=1, 𝐵=0, and𝐶=−1, then (5) has the solutions tanℎ𝜉, cotℎ𝜉.

Case 5. If 𝐴=𝐶=1, 𝐵=0, then (5) has the solutions tan𝜉.

Case 6. If 𝐴=𝐶=−1, 𝐵=0, then (5) has the solutions cot𝜉.

Case 7. If 𝐴=𝐵=0, 𝐶≠0, then (5) has the solutions −1/(𝑐𝜉+𝑐₀).

The solutions of (1) can be expressed in the form 𝑢(𝑥,𝑡)=𝑢(𝜉)=^𝑚_𝑖=0𝑎_𝑖𝐹^𝑖, (7) where 𝜉=𝑘𝑥−𝑘𝜔𝑡, 𝑘 and 𝜔 are the wave number and the wave speed, respectively, 𝑛 is a positive integer that can be determined by balancing the linear term of highest order with the nonlinear term in (1), and 𝑎₀,𝑎₁,…,𝑎_𝑛 are parameters to be determined.

Introducing the similarity variable 𝜉=𝑘𝑥−𝑘𝜔𝑡, the travelling wave solutions 𝑢(𝜉) satisfy 𝐻^′𝑢^′,𝑢^′′,𝑢^′′′,…=0. (8)

Balancing the highest order of linear term with nonlinear term in (3), we can determine 𝑛 in (4).

We illustrate the method by considering the Boussinesq equation and Burgers-like equtions.

Example 1 (The Boussinesq equation). Let’s consider 𝑢_𝑡𝑡=𝑐₁𝑢+𝑐₂𝑢^𝑛_𝑥𝑥+𝑢_{𝑥𝑥𝑡𝑡}=0. (9)
If we accept that 𝑐₁≥0, 𝑐₂≤0 and 𝑛=2, 𝑐=−1, we conclude (10) by (9) as follows: 𝑢_𝑡𝑡−𝑢_𝑥𝑥+2𝑢⋅𝑢_𝑥𝑥−𝑢_{𝑥𝑥𝑡𝑡}=0 (10) for doing this example. We could use transformation with (1) for the Boussinesq equation. Let us consider the Boussinesq equation as follows: 𝑢(𝑥,𝑡)=𝑢(𝜉),𝜉=𝑘𝑥−𝑘𝑤𝑡. (11)
Substituting (11) into (10), we get 𝑤²𝑢^′′−𝑢^′′+2𝑢^′2+2𝑢⋅𝑢^′′−𝑘²𝑤²𝑢⁴=0, (12) and integrating (12) we deduce the following equation: 𝑤²𝑢^′−𝑢^′+2𝑢⋅𝑢^′−𝑘²𝑤²𝑢^′′′=0. (13)
Balancing (𝑢⋅𝑢^′) with (𝑢^′′′) gives 𝑚=2. Therefore, we may choose 𝑢=𝑎₀+𝑎₁𝐹+𝑎₂𝐹². (14)
Substituting (14) into (13) along with (5) and using Mathematica yield a system of equations 𝑤,𝑡, and 𝐹^𝑀. Setting the coefficients of 𝐹^𝑀 in the obtained system of equations to zero, we can deduce the following set of algebraic polynomials with respect to unknowns 𝑎₀, 𝑎₁, and 𝑎₂𝑢^′=𝑎₁𝐴+𝑎₁𝐵𝐹+𝑎₁𝐶𝐹²+2𝑎₂𝐴𝐹+2𝑎₂𝐵𝐹²+2𝑎₂𝐶𝐹³=0,𝑢⋅𝑢^′=𝑎₀𝑎₁𝐴+𝑎₀𝑎₁𝐵𝐹+𝑎₀𝑎₁𝐶𝐹²+2𝑎₀𝑎₂𝐴𝐹+2𝑎₀𝑎₂𝐵𝐹²+2𝑎₀𝑎₂𝐶𝐹³+𝑎²₁𝐴𝐹+𝑎²₁𝐵𝐹²+𝑎²₁𝐶𝐹³+2𝑎₁𝑎₂𝐴𝐹²+2𝑎₁𝑎₂𝐵𝐹³+2𝑎₁𝑎₂𝐶𝐹⁴+𝑎₁𝑎₂𝐴𝐹²+𝑎₁𝑎₂𝐵𝐹³+𝑎₁𝑎₂𝐶𝐹⁴+2𝑎²₂𝐴𝐹³+2𝑎²₂𝐵𝐹⁴+2𝑎²₂𝐶𝐹⁵,𝑢^′′′=𝑎₁𝐵²𝐴+𝑎₁𝐵³𝐹+7𝑎₁𝐵²𝐶𝐹²+8𝑎₁𝐴𝐵𝐶𝐹+12𝑎₁𝐵𝐶²𝐹³+2𝑎₁𝐴²𝐶+8𝑎₁𝐴𝐶²𝐹²+6𝑎₁𝐶³𝐹⁴+6𝑎₂𝐴²𝐵+14𝑎₂𝐴𝐵²𝐹+52𝑎²𝐴𝐵𝐶𝐹²+16𝑎₂𝐴²𝐶𝐹+40𝑎₂𝐴𝐶²𝐹³+8𝑎₂𝐵³𝐹²+38𝑎²𝐵²𝐶𝐹³+54𝑎₂𝐵𝐶²𝐹⁴+24𝑎₂𝐶³𝐹⁵,𝐹⁰∶𝑤²𝑎₁𝐴−𝑎₁𝐴+2𝑎₀𝑎₁𝐴−𝑘²𝑤²𝑎₁𝐵²𝐴−2𝑘²𝑤²𝑎₁𝐴²𝐶−6𝑘²𝑤²𝑎₂𝐴²𝐹𝐵=0,₁∶𝑤²𝑎₁𝐵+2𝑤²𝑎₂𝐴−𝑎₁𝐵−2𝑎₂𝐵+2𝑎₀𝑎₁𝐵+4𝑎₀𝑎₂𝐴+2𝑎²₁𝐴−𝑘²𝑤²𝑎₁𝐵³−8𝑘²𝑤²𝑎₁𝐴𝐵𝐶−14𝑘²𝑤²𝑎₂𝐴𝐵²−16𝑘²𝑤²𝑎₂𝐴²𝐹𝐶=0,₂∶𝑤²𝑎₁𝐶+2𝑤²𝑎₂𝐵−𝑎₁𝐶−2𝑎₂𝐵+2𝑎₀𝑎₁𝐶+4𝑎₀𝑎₂𝐵+2𝑎²₁𝐵+6𝑎₁𝑎₂𝐴−7𝑘²𝑤²𝑎₁𝐵²𝐶−8𝑘²𝑤²𝑎₁𝐴𝐶²−52𝑘²𝑤²𝐴𝐵𝐶−8𝑘²𝑤²𝑎₂𝐵³𝐹=0,₃∶2𝑤²𝑎₂𝐶−2𝑎₂𝐶+4𝑎₀𝑎₂𝐶+2𝑎²₁𝐶+4𝑎₁𝑎₂𝐵+2𝑎₁𝑎₂𝐵+4𝑎²₂𝐴−12𝑘²𝑤²𝑎₁𝐵𝐶²−40𝑘²𝑤²𝑎₂𝐴𝐶²−38𝑘²𝑤²𝐵²𝐹𝐶=0,₄∶−6𝑎₁𝑎₂𝐶+4𝑎²₂𝐵−6𝑘²𝑤²𝑎₁𝐶³−54𝑘²𝑤²𝑎₂𝐵𝐶²𝐹=0,₅∶4𝑎²₂𝐶−24𝑘²𝑤²𝑎₂𝐶³=0. (15) From the solutions of the system, we can find 𝐵=0,𝑎₀=4𝑘²𝑤²1𝐴𝐶−2𝑤²+12,𝑎₁=0,𝑎₂=6𝑘²𝑤²𝐶²,1𝑤=4𝑘²,𝐴𝐶+1 (16) and with the aid of Mathematica, we find the following.
(i) When we choose 𝐴=1, 𝐵=0, and𝐶=1 in (16), then we can deduce 𝑎₀=4𝑘²𝑤²−12𝑤²+12,𝑎₂=6𝑘²𝑤². (17)
Therefore, the solution can be found as 𝑢(𝑥,𝑡)=4𝑘²𝑤²−12𝑤²+12+6𝑘²𝑤²tan²(𝑘𝑥−𝑘𝑤𝑡). (18)
(ii) In this case, if we take 𝐴=−1, 𝐵=0, and 𝐶=−1 in (16), then we have 𝑎₀=4𝑘²𝑤²−12𝑤²+12,𝑎₂=6𝑘²𝑤²,𝑢(𝑥,𝑡)=4𝑘²𝑤²−12𝑤²+12+6𝑘²𝑤²cot[𝑘𝑥−𝑘𝑤𝑡]. (19)
(iii) Again, when we choose 𝐴=1, 𝐵=0, 𝐶=−1 then from (16) is 𝑎₀=−4𝑘²𝑤²−12𝑤²+12,𝑎₂=6𝑘²𝑤²,𝑢(𝑥,𝑡)=−4𝑘²𝑤²−12𝑤²+12+6𝑘²𝑤²tanℎ²(),𝑢()=𝑘𝑥−𝑘𝑤𝑡𝑥,𝑡−4𝑘²𝑤²−12𝑤²+12+6𝑘²𝑤²cotℎ²(𝑘𝑥−𝑘𝑤𝑡). (20)
(iv) When we choose 𝐴=(1/2), 𝐵=0, and 𝐶=(1/2), then we can find the coefficients of (16) as 𝑎₀=𝑘²𝑤²−12𝑤²+12,𝑎₂=32𝑘²𝑤², (21) and using the coefficients, the solutions can be found as 𝑢(𝑥,𝑡)=𝑘²𝑤²−12𝑤²+12+32𝑘²𝑤²×[sec(𝑘𝑥−𝑘𝑤𝑡)+tan(𝑘𝑥−𝑘𝑤𝑡)]²,𝑢(𝑥,𝑡)=𝑘²𝑤²−12𝑤²+12+32𝑘²𝑤²×[csc(𝑘𝑥−𝑘𝑤𝑡)+cot(𝑘𝑥−𝑘𝑤𝑡)]². (22)
(v) When we choose 𝐴=−(1/2), 𝐵=0, and𝐶=−(1/2), then we can find the coefficients of (16) as follows: 𝑎₀=𝑘²𝑤²−12𝑤²+12,𝑎₂=32𝑘²𝑤²,𝑢(𝑥,𝑡)=𝑘²𝑤²−12𝑤²+12+32𝑘²𝑤²×[sec(𝑘𝑥−𝑘𝑤𝑡)+tan(𝑘𝑥−𝑘𝑤𝑡)]²,𝑢(𝑥,𝑡)=𝑘²𝑤²−12𝑤²+12+32𝑘²𝑤²×[csc(𝑘𝑥−𝑘𝑤𝑡)+cot(𝑘𝑥−𝑘𝑤𝑡)]². (23)
(vi) When we choose 𝐴=(1/2), 𝐵=0, and 𝐶=−(1/2), then we can find the coefficients of (16) as follows: 𝑎₀=−𝑘²𝑤²−12𝑤²+12,𝑎₂=32𝑘²𝑤²,𝑢(𝑥,𝑡)=−𝑘²𝑤²−12𝑤²+12+32𝑘²𝑤²×[coth(𝑘𝑥−𝑘𝑤𝑡)±cscℎ(𝑘𝑥−𝑘𝑤𝑡)]²,𝑢(𝑥,𝑡)=−𝑘²𝑤²−12𝑤²+12+32𝑘²𝑤²×[tanh(𝑘𝑥−𝑘𝑤𝑡)±𝑖secℎ(𝑘𝑥−𝑘𝑤𝑡)]². (24)
(vii) When we choose 𝐴=0, 𝐵=0, and 𝐶≠0, then we can find the coefficients of (16) as follows: 𝑎₀1=−2𝑤²+12,𝑎₂=6𝑘²𝑤²𝐶²,𝑢()=−1𝑥,𝑡2𝑤²+12+32𝑘²𝑤²𝐶²−1(𝑘𝑥−𝑘𝑤𝑡)+𝐶₀². (25)
Figure 1 gives to us 2D and 3D graphics for (25).

(a)

(b)

Figure 1 

The solution (25) is shown at 𝑘=𝑤=1, 𝐶=0.5, and 𝐶0=0.1, and the second graph represents the exact analytical solution of (25) for 𝑡=0.5.

Example 2 (Burger-like equations). Let’s consider 𝑢_𝑡+𝑢_𝑥+𝑢𝑢_𝑥+𝜆𝑢_𝑥𝑥=0, (26) where 𝜆=1, and in order to obtain Burger-like solution of equation, we get 𝑢(𝑥,𝑡)=𝑢(𝜉),𝜉=𝑘(𝑥−𝑤𝑡). (27)
Substituting (27) into (26), we get −𝑤𝑢^′+𝑢^′+𝑢𝑢^′+𝑘𝑢^′′=0. (28)
Balancing (𝑢𝑢^′) with 𝑢^′′ gives 𝑚=1. Therefore, we may choose the following ansatz: 𝑢=𝑎₀+𝑎₁𝐹. (29)
Substituting (29) into (28) along with (5) and using Mathematica yield the following set of algebraic polynomials with respect to unknowns 𝑎₀, 𝑎₁, 𝑎₂, and 𝑎₃𝑢^′=𝑎₁𝐴+𝑎₁𝐵𝐹+𝑎₁𝐶𝐹²+2𝑎₂𝐴𝐹+2𝑎₂𝐵𝐹²+2𝑎₂𝐶𝐹³=0,𝑢𝑢^′=𝑎₀+𝑎₁𝐹𝑎₁𝐴+𝑎₁𝐵𝐹+𝑎₁𝐶𝐹²,𝑢^′′′=𝑎₁𝐵𝐴+𝑎₁𝐵²𝐹+𝑎₁𝐶𝐴𝐹+3𝑎₁𝐵𝐶𝐹²+2𝑎₁𝐶²𝐹³,𝐹⁰∶𝑤²𝑎₁𝐴−𝑎₁𝐴+2𝑎₀𝑎₁𝐴−𝑘²𝑤²𝑎₁𝐵²𝐴−2𝑘²𝑤²𝑎₁𝐴²𝐶−6𝑘²𝑤²𝑎₂𝐴²𝐵=𝐹0,₁∶𝑤²𝑎₁𝐵+2𝑤²𝑎₂𝐴−𝑎₁𝐵−2𝑎₂𝐵+2𝑎₀𝑎₁𝐵+4𝑎₀𝑎₂𝐴+2𝑎²₁𝐴−𝑘²𝑤²𝑎₁𝐵³−8𝑘²𝑤²𝑎₁𝐴𝐵𝐶−14𝑘²𝑤²𝑎₂𝐴𝐵²−16𝑘²𝑤²𝑎₂𝐴²𝐶=𝐹0,₂∶𝑤²𝑎₁𝐶+2𝑤²𝑎₂𝐵−𝑎₁𝐶−2𝑎₂𝐵+2𝑎₀𝑎₁𝐶+4𝑎₀𝑎₂𝐵+2𝑎²₁𝐵+6𝑎₁𝑎₂𝐴−7𝑘²𝑤²𝑎₁𝐵²𝐶−8𝑘²𝑤²𝑎₁𝐴𝐶²−52𝑘²𝑤²𝑎₂𝐵³=𝐹0,₃∶2𝑤²𝑎₂𝐶−2𝑎₂𝐶+4𝑎₀𝑎₂𝐶+2𝑎²₁𝐶+4𝑎₁𝑎₂𝐵+2𝑎₁𝑎₂𝐵+4𝑎²₂𝐴−12𝑘²𝑤²𝑎₁𝐵𝐶²−40𝑘²𝑤²𝑎₂𝐴𝐶²−38𝑘²𝑤²𝐵²𝐶=0. (30)(31) From the solutions of the system, we can find 𝑎₀=𝑤−1,𝑎₁=−2𝑘𝑐, (32) and we obtain the following multiple solution and triangular periodic solutions of (26).(i)When we choose 𝐴=𝐶=1, 𝐵=0 in (31), then 𝑎₁=−2𝑘,𝑎₀=𝑤−1. (33) Therefore, the solution can be found as 𝑢(𝑥,𝑡)=(𝑤−1)−2𝑘tan[𝑘(𝑥−𝑤𝑡)]. (34)(ii)In the case if we take 𝐴=𝐶=−1, 𝐵=0 in (31), then we have 𝑎₁=2𝑘,𝑎₀𝑢()=()[𝑘(.=𝑤−1,𝑥,𝑡𝑤−1+2𝑘cot𝑥−𝑤𝑡)] (35)(iii)Again, when we choose 𝐴=1, 𝐵=0, and 𝐶=−1, 𝑎₀=𝑤−1,𝑎₁𝑢()=()[𝑘(,𝑢()=()[𝑘(.=−2𝑘,𝑥,𝑡𝑤−1+2𝑘tanh𝑥−𝑤𝑡)]𝑥,𝑡𝑤−1+2𝑘coth𝑥−𝑤𝑡)] (36)(iv)When we choose 𝐴=(1/2), 𝐵=0, and 𝐶=−(1/2), 𝑎₀=𝑤−1,𝑎₁𝑢()=()[[𝑘([𝑘(,𝑢()=()[[𝑘([𝑘(.=−𝑘,𝑥,𝑡𝑤−1−𝑘sec𝑥−𝑤𝑡)]+tan𝑥−𝑤𝑡)]]𝑥,𝑡𝑤−1−𝑘cos𝑒𝑐𝑥−𝑤𝑡)]+cot𝑥−𝑤𝑡)]] (37)(v)When we choose 𝐴=𝐶=−(1/2), 𝐵=0, then we can find the coefficients of (31) as follows: 𝑎₀=𝑘,𝑎₀𝑢()=()[[𝑘([𝑘(,𝑢()=()[[𝑘([𝑘(.=𝑤−1,𝑥,𝑡𝑤−1+𝑘sec𝑥−𝑤𝑡)]−tan𝑥−𝑤𝑡)]]𝑥,𝑡𝑤−1+𝑘cos𝑒𝑐𝑥−𝑤𝑡)]−cot𝑥−𝑤𝑡)]] (38)(vi)When we choose 𝐴=(1/2), 𝐵=0, and 𝐶=−(1/2), then we can find the coefficients of (31) as follows: 𝑢()=()[[𝑘([𝑘(,𝑢()=()[[𝑘([𝑘(.𝑥,𝑡𝑤−1+𝑘coth𝑥−𝑤𝑡)]±cos𝑒𝑐ℎ𝑥−𝑤𝑡)]]𝑥,𝑡𝑤−1+𝑘tanh𝑥−𝑤𝑡)]±𝑖secℎ𝑥−𝑤𝑡)]] (39)Figure 2 gives to us 2D and 3D graphics for (39).

(a)

(b)

Figure 2 

The solution (39) is shown at 𝑘=𝑤=1, and the second graph represents the exact analytical solution of (39) for 𝑡=0.5.

3. Conclusion

We have presented a generalized tanh function method and used it to solve the Boussinesq equation and the Burgers-like equation. In fact this method is readily applicable to a large variety of nonlinear PDEs.

Firstly, all the nonlinear PDEs which can be solved by other tanh function method can be solved easily by this method. Secondly we have used only the special solutions of (4). If we use only the special solutions of (4), we can obtain more solutions. We are also aware of the fact that not all fundamental equations can be treated with the method.

We also obtain some new and more general solutions at the same time. Furthermore, this method is also computerizable, which allows us to perform complicated and tedious algebraic calculation on a computer.