60 

 

 

  

 

 

New Travelling Wave Solution of Burgers Equations 

 

 

 

 

 

 
 

                                                               

                                               

 

 

 

Abstract  

     In this paper, we studied the travelling wave solving for some models of Burger's 

equations. We used sine-cosine method to solution nonlinear equation and we used a direct 
solution after getting travelling wave equation.                                                                   

                                                                   

Keywords: travelling wave solution, cosine-sine Method, nonlinear differential equations, 

Wave equations.  

 

1.Introduction 
     A large variety of physical, chemical, and biological phenomena are waves, such as 

sound waves, string waves, water waves, etc.  Wave phenomena are very important in 

dispersion, diffusion, reaction, and convection, and such phenomena are represented by 

nonlinear partial differential equations. 

 Moving waves are visible in many linear equations and nonlinear wave modeling, such as 

heat waves, string waves, and the cycle life of some organisms one of the travelling waves, 

etc. One of the most important methods for solving wave equations is the Travelling Wave 

solution (TWS), which many researchers have discussed ( see [1,2], the tanh-coth method 

[3], the tanh [4,5] sine-cosine method [6], and see [7-12]). 

 

2.Infinite Sires Method [5,9] 
       To clarify the travelling wave solutions method, we note the following: 

 Let the nonlinear partial differential equation has the form 

𝑔(𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑥𝑥 , … ) = 0,                                                              (1) 

Ibn Al Haitham Journal for Pure and Applied Science 

Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index 

 

Doi: 10.30526/34.3.2678 

 

Article history: Received 19 January 2021, Accepted 11 April 2021, Published in  2021. 

 

Zainab John 

Department of Mathematical, College of 

Science, Al-Mustansiriyah University, 

Baghdad, Iraq. 

zainabjohn22@uomustansiriyah.edu.iq 

 

   Basim Akhudir Abbas 

Department of Mathematical, College of 

Science, Al-Mustansiriyah University, 

Baghdad, Iraq. 

baasim_math@uomustansiriyah.edu.iq 

 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34(3)2021 
 

61 

 

Where 𝑢 = (𝑥, 𝑡) is a solution of equation (1), and g is a polynomial with respect to u and 

with its derivatives. To solve Eq. (1), we follow the following steps:  

Step 1: First, we change the independent variables x, t, by using one independent variable 𝜉 

that combines the two variables as shown 

 Let 𝑢(𝑥, 𝑡) = 𝑓(𝜉)        𝑤ℎ𝑒𝑟𝑒   𝜉 = 𝑥 − 𝑐𝑡  ,                                            (2) 
Where 𝜉 is a wave variable and 𝑐 is constant.  
 

Step2: derivative equation (2) with respect to 𝑥 𝑎𝑛𝑑 𝑡, we have the following ordinary 
differential equations:  

 

𝑢𝑥 = 𝑢
′ =

𝑑𝑢

𝑑 𝜉
 ; 𝑢𝑥𝑥 = 𝑢

′′ =
𝑑2𝑢

𝑑 𝜉2
 ;   𝑢𝑡 = −𝑐𝑢

′ = −𝑐
𝑑𝑢

𝑑 𝜉
 ;    𝑢𝑡𝑡 = 𝑐

2𝑢′′ = 𝑐2
𝑑2𝑢

𝑑 𝜉2
;  

𝑢𝑥𝑡 = −𝑐𝑢
′′ = −𝑐

𝑑2𝑢

𝑑 𝜉2
                                                                        (3) 

When substitute (3) into (1), we find that equation (1) is transform to (linear or nonlinear) 
ordinary differential equation transformed into 

 

𝐺 (𝑓(𝜉),
𝑑𝑓

𝑑𝜉
,

𝑑2𝑓

𝑑𝜉2
, … )                                                                   (4) 

 

      Where G is a polynomial in 𝑓(𝜉) and with its derivatives,  
 

Step 3: integrate (4) with respect to 𝜉 , and let the constant of integration equal to zero. 

Step 4: (a) we can solve equation (4) directly through many methods that solve ordinary 

differential equations. 

or 

(b) use sine-cosine method to solve equation (4). 

3. Applications of the travelling wave solution 
3.1: Some Models of Burgers Equations [13] 
 

The viscid Burgers equation to be the nonlinear parabolic PDE has the form 
 

𝑣𝑡 = 𝑣𝑥𝑥 +∈ 𝑣𝑣𝑥 .     ,0 < 𝜖 ≤ 1.                                                                      (5) 
 

 3.1.1. Travelling wave solution                                                    

 The travelling wave solution of Eq.(5) is  

 𝜉 = 𝑥 − 𝑐𝑡   ⇒ 𝑣(𝑥, 𝑡) = 𝑣(𝜉) ,  𝑣𝑥 = 𝑣
′ =

𝑑𝑢

𝑑 𝜉
 ; 𝑣𝑥𝑥 = 𝑣

′′ =
𝑑2𝑣

𝑑 𝜉2
 ;   𝑣𝑡 = −𝑐𝑢

′ = −𝑐
𝑑𝑣

𝑑 𝜉
 .                                                                                                               

(6)                                                                   

Substituting Eq. (6) into Eq. (5), we have got 

 

 𝑣′′ + 𝑐𝑣′+∈ 𝑣𝑣′ = 0                                                                                       (7) 

 𝑣′′ + 𝑐𝑣′+∈ (
𝑣2

2
)

′

= 0                                                                                     (8) 

Integral Eq. (8) with respect to 𝜉 we have got 

𝑣′ + 𝑐𝑣 + 𝜖
𝑣2

2
= 𝑐1                                                                                            (9)   



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62 

 

Let 𝑐1 = 0                                                                                                            

𝑣′ = −(𝑐𝑣 + 𝜖
𝑣2

2
)                                                                                       (10) 

𝑑𝑣

−(𝑐𝑣+𝜖
𝑣2

2
)

= 𝑑𝜉  

 

By solving this equation, we have got 

𝑣(𝜉) =
𝑐

(𝑐𝑒
𝑐(𝜉+𝑐2) −𝜖/2)

                                                                                   (11) 

Where = 𝑥 − 𝑐𝑡 , substituted in (11), we have 𝑣(𝑥, 𝑡) =
𝑐

(𝑐𝑒 ((𝑥−𝑐𝑡)+𝑐2)−𝜖/2)
, which is a  

general solution of Burgers equation (5). 

By drawing the solution function (v(x,t)), we find the following figure  

 

  
Singular kink of u 

 
Figure 1. When c=1, 𝛿=1/3, x=[-10,10], t=[-10,10]. 

 

3.2. KdV-Burgers Equation: 

3.2.1: Cosine Function Method: 
        In this part, we want to find (TWS) of KdV-Burgers equation, by using the cosine 

function method see [6,7]. KdV-Burgers equation [13,14] has the form 

𝑢𝑡 + (𝑢
2)𝑥 − 𝛾𝑢𝑥𝑥 + 𝛿𝑢𝑥𝑥𝑥 = 0,                                                              (12) 

  Where 𝛿 and 𝛶>0 are constants and 𝑢 is a function of spatial variable 𝑥 and time 

variable 𝑡  

let 𝑢(𝑥, 𝑡) = 𝑢(𝜉) ,     𝜉 = 𝑥 − 𝑐𝑡                                                              (13) 

 

Application (TWS) on Eq. (12) yields into the ordinary differential equation  

−𝑐𝑢′ + (𝑢2)′ − 𝛾𝑢′′ + 𝛿𝑢′′′ = 0                                                              (14) 



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63 

 

Integrating (14) we get  

−𝑐𝑢 + 𝑢2 − 𝛾𝑢′ + 𝛿𝑢′′ = 0                                                                     (15) 
Then we have the nonlinear travelling wave equation (15). Solution Eq.+(15) by cosine 

function method has the form. 

𝑢(𝜉) = 𝜆𝑐𝑜𝑠𝛽 (𝜇𝜉)                                                                                    (16) 

Where 𝛌, 𝛃 and μ are unknown parameters to find its, we derivatives (16) we get 

 

𝑢′ = −𝜆𝜇𝛽𝑐𝑜𝑠𝛽−1(𝜇𝜉)sin (𝜇𝜉)                                                                (17) 
𝑢′′ = −𝜆𝛽𝜇2𝑐𝑜𝑠𝛽 (𝜇𝜉) + 𝜆𝜇2𝛽(𝛽 − 1)𝑐𝑜𝑠𝛽−2(𝜇𝜉).                                (18) 

Substituting (16-18) into equation (15) gives 

−𝑐𝜆𝑐𝑜𝑠𝛽 (𝜇𝜉) + (𝜆𝑐𝑜𝑠𝛽 (𝜇𝜉))
2

+ 𝛾𝜆𝜇𝛽𝑐𝑜𝑠𝛽−1(𝜇𝜉) sin(𝜇𝜉) + 𝛿 (−𝜆𝛽𝜇2𝑐𝑜𝑠𝛽 (𝜇𝜉) +

𝜆𝜇2𝛽(𝛽 − 1)𝑐𝑜𝑠𝛽−2(𝜇𝜉)) = 0  

The following algebraic equation system is obtained by equating the exponents and the 

coefficient of each pair of the cosine functions. 

−𝑐𝜆 − 𝛿𝜆𝛽𝜇2 = 0                                                                                    (19) 

𝜆2 + 𝛿𝜆𝜇2𝛽(𝛽 − 1) = 0                                                                          (20) 

2𝛽 = 𝛽 − 2 ⇒ 𝛽 = −2 , substitute in (19-20) we have  

𝜇 = ∓√
𝑐

2𝛿
     ,   𝜆 = 3𝑐,  substitute in (16) we get the solution  

𝑢(𝜉) = 3𝑐 𝑐𝑜𝑠−2(√
𝑐

2𝛿
  𝜉)  or 𝑢(𝜉) = −3𝑐 𝑐𝑜𝑠−2(√

𝑐

2𝛿
  𝜉) 

𝑢(𝑥, 𝑡) = 3𝑐 𝑠𝑒𝑐2(√
𝑐

2𝛿
(𝑥 − 𝑐𝑡)  )                                                 (21) 

By drawing the solution function u(x,t), we find the following figure  

 

The periodic solution of u 
Figure 2. When c=1, 𝛿=1/3, x=[-10,10], t=[-10,10]. 

 



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64 

 

3.2.2.Sine Function Method: 

 

Solve equation (15) by use (sine-function method), we have the form, 

   𝑢 = 𝜆𝑠𝑖𝑛𝛽 (𝜇𝜉)                                                                                      (22) 

Where 𝛌, 𝛃 and μ are unknown parameters, to find its, we Derivatives (22) we get. 

 

  𝑢′ = 𝜆𝜇𝛽𝑠𝑖𝑛𝛽−1(𝜇𝜉)cos(𝜇𝜉)                                                               (23) 

  𝑢′′ = −𝜆𝛽𝜇2𝑠𝑖𝑛𝛽 (𝜇𝜉) + 𝜆𝜇2𝛽(𝛽 − 1)𝑠𝑖𝑛𝛽−2(𝜇𝜉)                              (24) 

  Substituting (22-24) into equation (15) yield in 

 −𝑐𝜆𝑠𝑖𝑛𝛽 (𝜇𝜉) + (𝜆𝑠𝑖𝑛𝛽 (𝜇𝜉))
2

− 𝛾𝜆𝜇𝛽𝑠𝑖𝑛𝛽−1(𝜇𝜉)cos(𝜇𝜉) + 𝛿(−𝜆𝛽𝜇2𝑠𝑖𝑛𝛽 (𝜇𝜉) +  

𝜆𝜇2𝛽(𝛽 − 1)𝑠𝑖𝑛𝛽−2(𝜇𝜉)) = 0  

  

We find the following system of algebraic and we obtain to parameters  by equating the 

exponents and the coefficient of each pair of the (cosine function). 

 

−𝑐𝜆 − 𝛿𝜆𝛽𝜇2 = 0                                                                                            (25) 

𝜆2 + 𝜆𝜇2𝛽(𝛽 − 1) = 0 ,                                                                                  (26) 
 2𝛽 = 𝛽 − 2 ⇒ 𝛽 = −2                                                                                   (27)                                            
Substitute (27) in (25,26) we have  

𝜇 = ∓√
𝑐

2𝛿
,  𝜆 = 3c 

Then, we have the solution 

𝑢 = −3𝑐 𝑠𝑖𝑛−2(√
𝑐

2𝛿
𝜉)  or 𝑢 = 3𝑐 𝑠𝑖𝑛−2(√

𝑐

2𝛿
𝜉)   

𝑢 = ∓3𝑐 𝑐𝑠𝑐2(√
𝑐

2𝛿
(𝑥 − 𝑐𝑡))  

 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 34(3)2021 
 

65 

 

 
The periodic solutions of u 

Figure 3. When c=1, 𝛿=1/3, x=[-10,10], t=[-10,10]. 
 

 

Conclusion  

     In this work, we conclude that the method of sine and cosine function gives the exact 

solution to the nonlinear partial differential equation, which is difficult to find in other ways. 

 

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