# New Exact Solutions of Burgers’ Equation Using Power Index Method

Authors: Khalil Ahmad; Khudija Bibi
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this article, we focus on the new exact solutions of Burger’s equation by using a new technique which is known as the power index method (PIM). In this method, we choose suitable indexes of independent variables and similarity transformation so that the partial differential equation may be converted into ODE. We have obtained analytic solution of the ODE by using symbolic package Maple. We have got exact solution of Burgers’ equation by using analytic solution of ODE and similarity transformation. The proposed method has been effectively employed to find new exact solutions for the nonlinear Burgers’ equation. Finally, the proposed resulting answers are compared with the homotopy perturbation, decomposition, and variational iteration solutions.


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## Body

## 1. Introduction

Burgers’ equation is used to label several kinds of phenomena such as turbulence and the estimated theory of stream through a shockwave wandering in a viscous fluid. Burgers [1] first developed this equation primarily to throw light on turbulence described by the interaction of two opposite effects of convection and diffusion. Hopf [2] in 1950 and Cole [3] in 1951 independently introduced a transformation ux,t=−2νθx/θ to convert Burgers’ equation into a linear heat equation θt=νθxx and solved exactly for an arbitrary initial condition. This transformation is known as Hopf–Cole transformation. Due to its widespread variety of applicability, many researchers have explored Burgers’ equation analytically and numerically. The exact solutions of Burgers’ equation are derived by using several methods like the simplest equation method with the Riccati equation [4, 5], direct integration method, and simplest equation method with the Bernoulli equation [6].Exact solutions to nonlinear PDEs play a central role in science, especially in complex physical science since they can provide much physical information and more insight into the physical aspects of the problem [7]. Recently, several methods have been used to find exact solutions of the nonlinear PDEs (see [6, 8] for details). Lie symmetries [3] are employed to reduce the NLPDEs to nonlinear ordinary differential equations (NLODEs) that may then be reduced to quadratures [9] if sufficient symmetries exist. A variety of powerful methods, such as tanh-sech method [2], extended tanh method [3], hyperbolic function method [10], and the first integral method [11], have been used to solve different types of nonlinear systems of PDEs. Due to modern computer programs like Maple and Mathematica, these methods have become the most powerful and effective algebraic methods for finding exact solutions of PDEs. In this article, we will apply power index method to find new exact solutions of Burgers’ equation.In this paper, twelve new exact solutions of general Burgers’ equation are derived by using the power index method. The proposed derived solutions are compared with the homotopy perturbation solution [8] and other existing exact solutions [5] (see Remark 1 for details).


## 2. General Burgers’ Equation

The standard form of general Burgers’ equation is(1)∂∂tux,t+αux,t∂∂xux,t=ν∂2∂x2ux,t,where x, t are the spatial coordinate and time, ux,t is the unknown real function, α is nonlinear term coefficient, and ν is the coefficient kinematic viscosity of fluids and it is equal to reciprocal of Reynolds number. When the Reynolds number is large, equation (1) describes shock wave behavior [12]. If the viscosity ν=0 or negligible, then (1) becomes inviscid Burgers’ equation, and if ν=1, then (1) is called viscous Burgers’ equation. By applying transformation u=∂φ/∂x to (1), we get(2)φxtx,t+αφxx,tφxxx,t=νφxxxx,t.


## 3. Power Index Method

Consider nonlinear partial differential equation (2), and we want to find its exact solutions; we introduce the variable η in general form(3)η=qx,t≔a1xmtn+a2tρ,and the function transform(4)φt,x=b1xrtsfη+b2xα+b3tβ.Figure1 shows the procedure of transformation of PDE to ODE.Figure 1
Flowchart of solution (7) of the PDE.Now we differentiate (3) and (4) according to (2) and we can find the relation of indexes of x and t in each term by the functions(5)px=c1m+c2r+c3α+c4,pt=d1n+sd2+d3ρ+d4β.We observe coefficients ofpx and pt so that the partial differential equations may be transformed to ordinary differential equations. The best optimal indexes of independent variables x and t are chosen in such a way that only two indexes vary at a time and others are fixed constants. We continue this process with different indexes of x and t so that we find all well-defined transformations. The following ODE is our objective in this method.(6)Qη,f,ff′,f′′,…=0.Solve ODE (6) by using computerized symbolic package like Maple. The exact solution of (1) can be obtained from (4) after replacing unknown function fη.(7)ut,x=∂φ∂x=b1rxr−1tsfη+a1b1mxr+m−1ts+nf′η+b2αxα−1.


## 4. New Exact Solutions of Burgers’ Equation

### 4.1. Test Case 1

We introduce the wave variableη=x/tm and transformation(8)φt,x=1αfη.Figure2 shows the solution of (1) and compares ux,t and parameters α and ν.Figure 2
3D plot of solutionU1 given in (12).In this case, we have considered single indexm of variable t; using (8), we get the following ODE:(9)f′′η1t2m−1f′η−mxtm−mf′η−vt2m−1f′′′η=0.To find the optimal index value of the independent variablet so that PDE is converted to ODE in new single independent variable η, we have selected two values like m=0 or m=1/2s in which PDE can be transformed into ODE. For m=0, equation (9) can be reduced to the form(10)νf′′′η−f′ηf′′η=0.The analytic solution of (10) is given below:(11)fη=νlntan2C12νη+C2+1+C3.Equations (8) and (11) are used to find the general solution of Burgers’ equation (1):(12)U1≔ux,t=2c1αtanC12νx+C2.


### 4.2. Test Case 2

Now we choose ofm=1/2, and using (8), equation (9) can be reduced to the form(13)f′′η2f′η−η−2νf′′′η−f′η=0.Figure3 displays graphical illustration of the analytical solution of (16).Figure 3
Graph of solutionU2 given in (16) for α=10 and ν=15.The solution of (13) is given by(14)fη=−2νlnη+η22.By using (8) and (14) and the variable η=x/t, we get the solution of (2):(15)φx,t=−2ναlnxt+x22αt.The solution of Burgers’ equation is(16)U2≔ux,t=−2ναx+xαt.


### 4.3. Test Case 3

Now we choose the new variableη=x2/tm and transformation φt,x=1/αfη; (2) can be transformed into(17)f′′η−4x2t2m−1f′η+mx2tm+6νtm−1+mf′η+4νx2t2m−1f′′′η−2tm−1(f′η)2=0.Since the index power ofx is 2, index powers of variable t must have the following relations 2m−1=m or m=0.Whenm=0, the variable is η=x2. Using (8), equation (17) can be reduced to the form(18)νηf′′′η+32f′′η−f′η12f′η+ηf′′η=0.Figure4 compares ux,t and parameter α for fixed ν=15 in (20).Figure 4
Plot of solutionU3 given in (20) for α=1.5,2,3,4 and ν=15.The analytic solution of (18) is given below:(19)fη=νlntanh2η1/2−C22C1ν−1+νlntanh2η1/2−C22C1ν+1+C3.Using (19), the general solution of Burgers’ equation (1) is(20)U3:=ux=−1C1αxtanh2x−C22C1ν.


### 4.4. Test Case 4

Form=1, we take new variable η=x2/t and transformation φt,x=1/αfη which give us the following differential equation:(21)f′′η−4ηf′η+η+6ν+4νηf′′′η−2(f′η)2+f′η=0.The solution of differential equation (21) is(22)fη=−νlnη2+η2,which is the same solution given in (16).Further, we considerη=x4/tm and the transformation φt,x=1/αfη, and equation (2) has the form(23)42νx8t2mf′′′η+f′′η−42x8t2mf′η+mx6tm+1+62νx4tm+f′η6ν+mx2t−12x4tmf′η=0.Figure5 compares ux,t and parameter ν for fixed α=10 in (20).Figure 5
Plot of solutionU3 given in (20) for α=10 and ν=1,2,2.5,3.In (23), to choose optimal values of index power of variable t, we can choose appropriate values m which are 0 and 2. When m=0, the variable is η=x4. Using (8), equation (23) can be reduced to the form(24)8η2f′′η−νf′′′η−18νηf′′η+3f′η2ηf′η−ν=0.The analytic solution of (24) is given below:(25)fη=νlntanh4η1/4−C22C1ν−1+νlntanh4η1/4−C22C1ν+1+C3.The general solution of Burgers’ equation (1) is(26)U4:=ux,t=−x−3/4c1αtanh4x−C22C1ν.


### 4.5. Test Case 5

Now we considerη=xm/tm/2 in general form and the transformation φt,x=1/αfη; then, equation (2) can be expressed in the form(27)mm−1η23νf′′η−f′η2+m2η3νf′′′η−f′ηf′′η+mη2+2/m2f′′η+12f′η2νm−1m−2η+mη1+2/m.This differential equation has the following solution:(28)fη=−2νmlnη+η2/m2,and using (28), we have obtained the following wave transformation:(29)φt,x=−2νmαlnxmtm/2+x22αt.The above transformation gives the same solution of Burgers’ equation (1), which is given in (16).


### 4.6. Test Case 6

Now we consider the wave variableη=xm and transformation φt,x=1/αfη. We obtained the following ordinary differential equation:(30)η3[(νf′′′η−f′ηf′′η)]+m−1m[3f′′η−f′2η]+m−1m−2m2νηf′η]=0.The exact solution of (30) is given below:(31)fη=νlntanhmη1/m−C22C1ν−1+νlntanhmη1/m−C22C1ν+1+C3,where C1,C2, and C3 are integration constants. By using (31), we have obtained exact solution of (1) of the form(32)U5≔ux,t=−x−m+1/mc1αtanhmx−C22C1ν.


### 4.7. Test Case 7

We introduce the wave variableη=x and transformation(33)φt,x=xαfη.We get the following differential equation:(34)−νηf′′′η+f′′ηη2f′η+ηfη−3ν+2f′ηηf′η+fη=0.Using (33), we get the following solutions:(35)φt,x=−νxln12c1ν[c3coshc1/2νx]−c2sinhc1/2νx],U6:=ux,t=2c1ν3/2[4c2cosh2c1/2νx]−c3sinhc1/2νx−2c3]coshc1/2νx[4c2sinhc1/2νx−2c3].


### 4.8. Test Case 8

We choose the variableη=xm and transformation(36)φt,x=xmfη.The following differential equation is obtained:(37)3m−1mη2f′2η−23f′η−13α+νf′′η+m−1m−2νηf′η+1+m2η3νf′′′η−αf′′η1+f′η=0.For particular value ofm=2 in (37), we get the differential equation(38)4α2ηf′′η+f′η+11+f′η−8νηf′′′η+23f′′η=0.The analytic solution of (38) is given by(39)fη=c3−η+ναlogtanh2C1α2η−C22ν−1,U7:=u=∂φ∂x=2C1αtanhC1αC2−2x2ν.


### 4.9. Test Case 9

We introduce the wave variableη=xm, so that(40)φt,x=x2mfη.We have obtained the differential equation(41)m−1η2mαf′′η−3νmf′′η+8m3−43+mη3αmf′′η+6m−4f′η−mνf′′′η2mαη4mf′′η+4m−2−ηm−1m−2νf′η=0.For particular value ofm=1 in (41), we get the differential equation(42)−νηf′′′η+α2η+f′η2+f′′η=0.The analytic solution of (42) is given by(43)fη=−η2+ναlogtanh2C1αC2−η2ν−1.Using (43), we get the solution of Burgers’ equation:(44)U8:=ux,t=∂φ∂x=C1αtanhC1αC2−x2ν.


### 4.10. Test Case 10

We choose the wave variableη=eat, so that(45)φt,x=x2tnfη.Equation (2) is transformed into the differential equation(46)2αan−1f2η+η(lnη)nf′η−n(lnη)n−1fη=0.The analytic solution of (46) is given by(47)fη=(lnη)n2αan−1lnη+c.The solution of (2) is given by(48)φt,x=anx22αant+c.We get the exact solution of Burgers’ equation (1):(49)U9≔ut,x=2anx2αant+c.Forn=0 in transformation (45), we get the ODE(50)aηf′η+2αf2η=0.The corresponding analytic solutions are(51)fη=a2αlnη+ac,φt,x=x22αt+c,U10:=ut,x=2x2αt+c.Remark 1.
Note that if we choosean=1,α=1/2,c=1 and initial condition ux,0=x in solution (49), we obtain the solution of Burgers’ equation:(52)ut,x=2x2t+1,which was constructed in [5] (see Example 1). If we choose a=1,α=1 and initial condition ux,0=2x in solution (49), we obtain same solution of Burgers’ equation which is given in [8] (Example 1).
Figure6 shows the graphical diagram of the analytical solution of (49).Figure 6
Plot of solutionU9 given in (49) for a=−2, α=1, and c=100.


### 4.11. Test Case 11

We choose the variableη=eaxn, so that(53)φt,x=fη.Equation (2) is transformed into the differential equation(54)3nn−1alnηa2−3/n−α3f′2η+νf′′ηη+νηf′η+n2a2−αf′2η+3νf′′ηη2+−αf′ηf′′η+νf′′′ηη3+νηf′ηlnηa3−3/n+νn−1n−2ηlnηa1−3/nf′η=0.Forn=1 in transformation (53), we get the ODE(55)−2αη(f′′η)2−αη2fηf′η+3νηf′′η+νη2f′′′η+νf′η=0.The analytic solution of (55) is given by(56)fη=ναlntanhαc1lnη−c22νc1−1+ναlntanhαc1lnη−c22νc1+1.The exact solution of (2) and Burgers’ equation (1) is given by(57)φt,x=ναlntanhαc1ax−c22νc1−1+ναlntanhαc1ax−c22νc1+1,U11:=ut,x=ac1αtanhαc1−ax+c22νc1.Forn=2 in transformation (53), we get the ODE(58)η2lnη−αf′ηf′′η+νf′′′η−αηlnη+12f′η2+νlnη+32f′η+3νlnη+12f′′η=0.We get the following analytic solution of the ODE:(59)fη=ναlogtanh2i2αlnηc1+c2νν+1+c3.The exact solution of Burgers’ equation (1) is given by(60)U12:=ut,x=ic18aναtani2aαc1x+c2νν.Figure7 shows the graphical illustration of solution (60).Figure 7
Plot of solutionU12 given in (60) for a=−1, α=1, and ν=1.


## 4.1. Test Case 1

We introduce the wave variableη=x/tm and transformation(8)φt,x=1αfη.Figure2 shows the solution of (1) and compares ux,t and parameters α and ν.Figure 2
3D plot of solutionU1 given in (12).In this case, we have considered single indexm of variable t; using (8), we get the following ODE:(9)f′′η1t2m−1f′η−mxtm−mf′η−vt2m−1f′′′η=0.To find the optimal index value of the independent variablet so that PDE is converted to ODE in new single independent variable η, we have selected two values like m=0 or m=1/2s in which PDE can be transformed into ODE. For m=0, equation (9) can be reduced to the form(10)νf′′′η−f′ηf′′η=0.The analytic solution of (10) is given below:(11)fη=νlntan2C12νη+C2+1+C3.Equations (8) and (11) are used to find the general solution of Burgers’ equation (1):(12)U1≔ux,t=2c1αtanC12νx+C2.


## 4.2. Test Case 2

Now we choose ofm=1/2, and using (8), equation (9) can be reduced to the form(13)f′′η2f′η−η−2νf′′′η−f′η=0.Figure3 displays graphical illustration of the analytical solution of (16).Figure 3
Graph of solutionU2 given in (16) for α=10 and ν=15.The solution of (13) is given by(14)fη=−2νlnη+η22.By using (8) and (14) and the variable η=x/t, we get the solution of (2):(15)φx,t=−2ναlnxt+x22αt.The solution of Burgers’ equation is(16)U2≔ux,t=−2ναx+xαt.


## 4.3. Test Case 3

Now we choose the new variableη=x2/tm and transformation φt,x=1/αfη; (2) can be transformed into(17)f′′η−4x2t2m−1f′η+mx2tm+6νtm−1+mf′η+4νx2t2m−1f′′′η−2tm−1(f′η)2=0.Since the index power ofx is 2, index powers of variable t must have the following relations 2m−1=m or m=0.Whenm=0, the variable is η=x2. Using (8), equation (17) can be reduced to the form(18)νηf′′′η+32f′′η−f′η12f′η+ηf′′η=0.Figure4 compares ux,t and parameter α for fixed ν=15 in (20).Figure 4
Plot of solutionU3 given in (20) for α=1.5,2,3,4 and ν=15.The analytic solution of (18) is given below:(19)fη=νlntanh2η1/2−C22C1ν−1+νlntanh2η1/2−C22C1ν+1+C3.Using (19), the general solution of Burgers’ equation (1) is(20)U3:=ux=−1C1αxtanh2x−C22C1ν.


## 4.4. Test Case 4

Form=1, we take new variable η=x2/t and transformation φt,x=1/αfη which give us the following differential equation:(21)f′′η−4ηf′η+η+6ν+4νηf′′′η−2(f′η)2+f′η=0.The solution of differential equation (21) is(22)fη=−νlnη2+η2,which is the same solution given in (16).Further, we considerη=x4/tm and the transformation φt,x=1/αfη, and equation (2) has the form(23)42νx8t2mf′′′η+f′′η−42x8t2mf′η+mx6tm+1+62νx4tm+f′η6ν+mx2t−12x4tmf′η=0.Figure5 compares ux,t and parameter ν for fixed α=10 in (20).Figure 5
Plot of solutionU3 given in (20) for α=10 and ν=1,2,2.5,3.In (23), to choose optimal values of index power of variable t, we can choose appropriate values m which are 0 and 2. When m=0, the variable is η=x4. Using (8), equation (23) can be reduced to the form(24)8η2f′′η−νf′′′η−18νηf′′η+3f′η2ηf′η−ν=0.The analytic solution of (24) is given below:(25)fη=νlntanh4η1/4−C22C1ν−1+νlntanh4η1/4−C22C1ν+1+C3.The general solution of Burgers’ equation (1) is(26)U4:=ux,t=−x−3/4c1αtanh4x−C22C1ν.


## 4.5. Test Case 5

Now we considerη=xm/tm/2 in general form and the transformation φt,x=1/αfη; then, equation (2) can be expressed in the form(27)mm−1η23νf′′η−f′η2+m2η3νf′′′η−f′ηf′′η+mη2+2/m2f′′η+12f′η2νm−1m−2η+mη1+2/m.This differential equation has the following solution:(28)fη=−2νmlnη+η2/m2,and using (28), we have obtained the following wave transformation:(29)φt,x=−2νmαlnxmtm/2+x22αt.The above transformation gives the same solution of Burgers’ equation (1), which is given in (16).


## 4.6. Test Case 6

Now we consider the wave variableη=xm and transformation φt,x=1/αfη. We obtained the following ordinary differential equation:(30)η3[(νf′′′η−f′ηf′′η)]+m−1m[3f′′η−f′2η]+m−1m−2m2νηf′η]=0.The exact solution of (30) is given below:(31)fη=νlntanhmη1/m−C22C1ν−1+νlntanhmη1/m−C22C1ν+1+C3,where C1,C2, and C3 are integration constants. By using (31), we have obtained exact solution of (1) of the form(32)U5≔ux,t=−x−m+1/mc1αtanhmx−C22C1ν.


## 4.7. Test Case 7

We introduce the wave variableη=x and transformation(33)φt,x=xαfη.We get the following differential equation:(34)−νηf′′′η+f′′ηη2f′η+ηfη−3ν+2f′ηηf′η+fη=0.Using (33), we get the following solutions:(35)φt,x=−νxln12c1ν[c3coshc1/2νx]−c2sinhc1/2νx],U6:=ux,t=2c1ν3/2[4c2cosh2c1/2νx]−c3sinhc1/2νx−2c3]coshc1/2νx[4c2sinhc1/2νx−2c3].


## 4.8. Test Case 8

We choose the variableη=xm and transformation(36)φt,x=xmfη.The following differential equation is obtained:(37)3m−1mη2f′2η−23f′η−13α+νf′′η+m−1m−2νηf′η+1+m2η3νf′′′η−αf′′η1+f′η=0.For particular value ofm=2 in (37), we get the differential equation(38)4α2ηf′′η+f′η+11+f′η−8νηf′′′η+23f′′η=0.The analytic solution of (38) is given by(39)fη=c3−η+ναlogtanh2C1α2η−C22ν−1,U7:=u=∂φ∂x=2C1αtanhC1αC2−2x2ν.


## 4.9. Test Case 9

We introduce the wave variableη=xm, so that(40)φt,x=x2mfη.We have obtained the differential equation(41)m−1η2mαf′′η−3νmf′′η+8m3−43+mη3αmf′′η+6m−4f′η−mνf′′′η2mαη4mf′′η+4m−2−ηm−1m−2νf′η=0.For particular value ofm=1 in (41), we get the differential equation(42)−νηf′′′η+α2η+f′η2+f′′η=0.The analytic solution of (42) is given by(43)fη=−η2+ναlogtanh2C1αC2−η2ν−1.Using (43), we get the solution of Burgers’ equation:(44)U8:=ux,t=∂φ∂x=C1αtanhC1αC2−x2ν.


## 4.10. Test Case 10

We choose the wave variableη=eat, so that(45)φt,x=x2tnfη.Equation (2) is transformed into the differential equation(46)2αan−1f2η+η(lnη)nf′η−n(lnη)n−1fη=0.The analytic solution of (46) is given by(47)fη=(lnη)n2αan−1lnη+c.The solution of (2) is given by(48)φt,x=anx22αant+c.We get the exact solution of Burgers’ equation (1):(49)U9≔ut,x=2anx2αant+c.Forn=0 in transformation (45), we get the ODE(50)aηf′η+2αf2η=0.The corresponding analytic solutions are(51)fη=a2αlnη+ac,φt,x=x22αt+c,U10:=ut,x=2x2αt+c.Remark 1.
Note that if we choosean=1,α=1/2,c=1 and initial condition ux,0=x in solution (49), we obtain the solution of Burgers’ equation:(52)ut,x=2x2t+1,which was constructed in [5] (see Example 1). If we choose a=1,α=1 and initial condition ux,0=2x in solution (49), we obtain same solution of Burgers’ equation which is given in [8] (Example 1).
Figure6 shows the graphical diagram of the analytical solution of (49).Figure 6
Plot of solutionU9 given in (49) for a=−2, α=1, and c=100.


## 4.11. Test Case 11

We choose the variableη=eaxn, so that(53)φt,x=fη.Equation (2) is transformed into the differential equation(54)3nn−1alnηa2−3/n−α3f′2η+νf′′ηη+νηf′η+n2a2−αf′2η+3νf′′ηη2+−αf′ηf′′η+νf′′′ηη3+νηf′ηlnηa3−3/n+νn−1n−2ηlnηa1−3/nf′η=0.Forn=1 in transformation (53), we get the ODE(55)−2αη(f′′η)2−αη2fηf′η+3νηf′′η+νη2f′′′η+νf′η=0.The analytic solution of (55) is given by(56)fη=ναlntanhαc1lnη−c22νc1−1+ναlntanhαc1lnη−c22νc1+1.The exact solution of (2) and Burgers’ equation (1) is given by(57)φt,x=ναlntanhαc1ax−c22νc1−1+ναlntanhαc1ax−c22νc1+1,U11:=ut,x=ac1αtanhαc1−ax+c22νc1.Forn=2 in transformation (53), we get the ODE(58)η2lnη−αf′ηf′′η+νf′′′η−αηlnη+12f′η2+νlnη+32f′η+3νlnη+12f′′η=0.We get the following analytic solution of the ODE:(59)fη=ναlogtanh2i2αlnηc1+c2νν+1+c3.The exact solution of Burgers’ equation (1) is given by(60)U12:=ut,x=ic18aναtani2aαc1x+c2νν.Figure7 shows the graphical illustration of solution (60).Figure 7
Plot of solutionU12 given in (60) for a=−1, α=1, and ν=1.


## 5. Conclusion

In this study, we have applied PIM for the development of theoretical solutions of Burgers’ equation. The test cases and results show that PIM can solve linear and nonlinear PDEs with parameters. We have discussed the choices of indexes of independent variables in new variable and transform and obtained exact solutions of Burgers’ equation for suitable values of indexes. We have plotted some solutions of Burgers’ equation and also showed graphical effect of parameters involved in Burgers’ equation in its exact solution.



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