EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 3, 2021, 842-862
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Laplace-Somé Blaise Abbo Method for Solving Nonlinear
Coupled Burger’s Equations

Joseph Bonazebi Yindoula
Laboratory of Numerical Analysis, Computer Science and Applications, Faculty of Science
and Technology, Brazzaville Congo

Abstract. Burger’s equations, an extension of fluid dynamics equations, are typically solved by
several numerical methods. In this article, the laplace-Somé Blaise Abbo method is used to solve
nonlinear Burger equations. This method is based on the combination of the laplace transform
and the SBA method. After reminders of the laplace transform, the basic principles of the SBA
method are described. The process of calculating the Laplace-SBA algorithm for determining the
exact solution of a linear or nonlinear partial derivative equation is shown. Thus, three examples
of PDE are solved by this method, which all lead to exact solutions. Our results suggest that this
method can be extended to other more complex PDEs.
2020 Mathematics Subject Classifications: 47H14, 34G20, 4J25, 65J15
Key Words and Phrases: Laplace-SBA Method, Coupled Burgers equation.

1. Introduction

Partial differential equations (PDEs) describe the reality of numerous physical and
natural phenomena. Burger’s equations are such partial differential equations issues from
fluid mechanics. It finds its application in various field of applied mathematics, such as
the modeling of gas dynamics, accoutis or road traffic. However, to translate the realty of
physical and natural phenomena encountered, this PDE takes the from of a coupled system
(PDES). Several methods have been used to investigate these PDES like the iterative
variational method (VIM) [7], the homotopy perturbation method (HPM) [4] , [8], [10],
double Laplace transform method [6] and Adomian Pade Technique [3], [5], [9]. However,
most solutions of urger’s partial differential equations (coupled or not ) by these methods
are rarely exact. Or the Some Blaise Abbo (SBA) method is a powerful to solve non linear
PDES [1], [2], [12], [13], [15], [16], [17], [18]. By determining the exact solutions. Even if
the calculation of the integrals tuns out to be difficult by this method, the combination of
this-ci with the method of transformation of Laplace makes it possible to overcome this
difficulty. In this paper we use the Laplace-SBA method to construct the exact solution
of coupled Burger’s equations.

DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.3932

Email address: bonayindoula@yahoo.fr (J. B.Yindoula )

http://www.ejpam.com 842 © 2021 EJPAM All rights reserved.



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 843

2. The Laplace-SBA method

Suppose that we need to solve the following equation:

Au = f (1)

In a real Hilbert space H, where A : H −→ H, a linear is or a nonlinear operator, f ∈ H
and u is the unknown function. Let’s suppose that we can decompose operator A in the
following form:

A = L+R+N (2)

where L + R is linear, N nonlinear, L inversible in the "sense", of Adomian with L−1 as
inverse. Using that decoposition equation (2) can be rewritten as:

Lu+Ru+Nu = f (3)

let’s note Lt(u) the Laplace transform of the function u with respect to the variable t.
Applying the Laplace transform to equation (3), we obtain:

Lt (Lu) + Lt (Ru) + Lt (Nu) = Lt (f) (4)

We suppose that: L(.) =
∂m

∂tm
(.). Using the properties of the Laplace transform of a

derivative, we get:

Lt
[
∂mu

∂tm

]
= smLt(u)−

m−1∑
k=0

sku(m−k−1)(x, 0) (5)

Equation (4) gives

smLt(u)−
m−1∑
k=0

sku(m−k−1)(x, 0) + Lt (Ru) + Lt (Nu) = Lt (f) (6)

(6)is equivalent to

smLt(u) = −Lt(Ru)− Lt(Nu) +
m−1∑
k=0

sku(m−k−1)(x, 0) + Lt(f) (7)

Using the successive approximations, we get:

smLt
(
uk
)
= −Lt

(
Ruk

)
− Lt

(
Nuk−1

)
+

m−1∑
j=0

sju(m−j−1)(x, 0) + Lt (f) (8)

According to the SBA method, we suppose that the solution of (8) has the following form:

u = lim
k−→+∞

uk (9)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 844

Where

uk =

+∞∑
k=0

ukn ; k ≥ 1 (10)

let’s denote gk−1 = Nuk−1. Substituting uk into (8), we have:

sm
∑
n≥0
Lt
(
ukn

)
= −

∑
n≥0
Lt
(
Rukn

)
− Lt(gk−1) +

m−1∑
j=0

sju(m−j−1)(x, 0) + Lt (f) (11)

and for every k ≥ 1, we get ukn for n ≥ 0, through the following Laplace- SBA algorithm:
smLt

(
uk0
)
= −Lt (gk−1) +

m−1∑
j=0

sju(m−j−1)(x, 0) + Lt (f) ; k ≥ 1

smLt
(
ukn+1

)
= −Lt

(
Ruk−1n

)
; n ≥ 0

(12)

(12) is equivalent to


Lt
(
uk0
)
= − 1

sm
Lt(gk−1) +

1

sm

m−1∑
j=0

sku(m−j−1)(x, 0) + Lt
(

f

sm

)
; k ≥ 1

Lt
(
ukn+1

)
= − 1

sm
Lt
(
Ruk−1n

)
; n ≥ 0

(13)

Applying the inverse Laplace transform L−1t to (13), we obtain:


(uk0) = −L

−1
t

[
1

sm
Lt(gk−1)

]
+ L−1t

 1

sm

m−1∑
j=0

sku(m−j−1)(x, 0)

+ L−1t

(
f

sm

)
; k ≥ 1

(ukn+1) = −L
−1
t

[
1

sm
Lt
(
Ruk−1n

)]
; n ≥ 0

(14)
The SBA principle needs that, for k = 1, we must choose u0 like Nu0 = 0 and for

k > 1, we must verify that Nuk−1 = 0.
For k = 1, we get:


u10 = −L

−1
t

[
1

sm
Lt(g0)

]
+ L−1t

 1

sm

m−1∑
j=0

sku(m−j−1)(x, 0)

+ L−1t

(
f

sm

)
;

(u1n+1) = −L
−1
t

[
1

sm
Lt
(
Ru1n

)]
; n ≥ 0

(15)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 845

If the series

∑
n≥0

(u1n)

 converges, then u1 =
∑
n≥0

(u1n)

For k = 1, we get:


u20 = −L

−1
t

[
1

sm
Lt(g1)

]
+ L−1t

 1

sm

m−1∑
j=0

sku(m−j−1)(x, 0)

+ L−1t

(
1

sm
f

)
;

u2n+1 = −L
−1
t

[
1

sm
Lt
(
Ru2n

)]
; n ≥ 0

(16)

If the series

∑
n≥0

(u2n)

 converges, then u2 =
∑
n≥0

(u2n). The converge of the series has been

proved in [6].

Repeating this same process for k ≥ 3. If the series

∑
n≥0

(ukn)

 converges, then uk =∑
n≥0

(ukn).

Therefore, u = lim
k−→+∞

uk is the solution of the equation (1).

3. Applications

To illustrate the powerfull, the simplicity and efficiency of Laplace-SBA method in
solving non linear coupled Burger’s equations [6]. Here, we consider three systems of these
equations.

3.1. Example 1

Let’s consider the following system of Burger’s equations [2] [3]
∂u(x, t)

∂t
=

∂2u(x, t)

∂x2
+ 2u(x, t)

∂u(x, t)

∂x
− ∂

∂x
(u(x, t)v(x, t))

∂v(x, t)

∂t
=

∂2v(x, t)

∂x2
+ 2v(x, t)

∂v(x, t)

∂x
− ∂

∂x
(u(x, t)v(x, t))

(17)

with initial conditions {
u(x, 0) = sinx

v(x, 0) = sinx
(18)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 846

3.1.1. Résolution by Laplace-SBA method

According to the Laplace-SBA Method, we suppose that:
N1 (u(x, t)v(x, t)) = 2u(x, t)

∂u(x, t)

∂x
− ∂

∂x
(u(x, t)v(x, t))

N2 (u(x, t)v(x, t)) = 2v(x, t)
∂v(x, t)

∂x
− ∂

∂x
(u(x, t)v(x, t))

(19)

Then, the system (17) gives
∂u(x, t)

∂t
=

∂2u(x, t)

∂x2
+N1 (u(x, t)v(x, t))

∂v(x, t)

∂t
=

∂2v(x, t)

∂x2
+N2 (u(x, t)v(x, t))

(20)

Applying the Laplace transform to (20). we obtain:
Lt (u(x, t)) = 1

su(x, 0) +
1
sLt

(
∂2u(x, t)

∂x2

)
+ 1

sLt (N1 (u(x, t)v(x, t)))

Lt (v(x, t)) = 1
sv(x, 0) +

1
sLt

(
∂2v(x, t)

∂x2

)
+ 1

sLt (N2 (u(x, t)v(x, t)))

(21)

From (21), we have:


u(x, t) = u(x, 0)L−1t

(
1
s

)
+ L−1t

(
1
sLt

(
∂2u(x, t)

∂x2

))
+ L−1t

(
1
sLt (N1 (u(x, t)v(x, t)))

)

v(x, t) = v(x, 0)L−1t

(
1
s

)
+ L−1t

(
1
sLt

(
∂2v(x, t)

∂x2

))
+ L−1t

(
1
sLt (N2 (u(x, t)v(x, t)))

)
(22)

(22) equivalent to:


u(x, t) = sinx+ L−1t

(
1
sLt

(
∂2u(x, t)

∂x2

))
+ L−1t

(
1
sLt (N1 (u(x, t)v(x, t)))

)

v(x, t) = sinx+ L−1t

(
1
sLt

(
∂2v(x, t)

∂x2

))
+ L−1t

(
1
sLt (N2 (u(x, t)v(x, t)))

) (23)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 847

Using the successive approximations, we get:
uk(x, t) = sinx+ L−1t

(
1
sLt

(
∂2uk(x, t)

∂x2

))
+ L−1t

(
1
sLt

(
N1

(
uk−1(x, t)vk−1(x, t)

)))

vk(x, t) = sinx+ L−1t

(
1
sLt

(
∂2vk(x, t)

∂x2

))
+ L−1t

(
1
sLt

(
N2

(
uk−1(x, t)vk−1(x, t)

))) k ≥ 1

(24)
According to the SBA method, we suppose that the solution of
(17) has the following form:

u(x, t) = lim
k−→+∞

uk(x, t) (25)

where

uk(x, t) =

+∞∑
n=0

ukn(x, t); k ≥ 1 (26)

and, for every k ≥ 1, we get ukn(x, t) for n ≥ 0, through the following Laplace-SBA
algorithm:


uk0(x, t) = sinx+ L−1t

(
1
sLt

(
N1

(
uk−1(x, t)vk−1(x, t)

)))
k ≥ 1

ukn+1(x, t) = L
−1
t

(
1
sLt

(
∂2ukn(x, t)

∂x2

))
; n ≥ 0


vk0 (x, t) = sinx+ L−1t

(
1
sLt

(
N2

(
uk−1(x, t)vk−1(x, t)

)))
k ≥ 1

vkn+1(x, t) = L
−1
t

(
1
sLt

(
∂2vkn(x, t)

∂x2

))
; n ≥ 0

(27)

For k = 1, we have the following Laplace-SBA algorithm:


u10(x, t) = sinx+ L−1t

(
1
sLt

(
N1

(
u0(x, t)v0(x, t)

)))
u1n+1(x, t) = L

−1
t

(
1
sLt

(
∂2u1n(x, t)

∂x2

))
; n ≥ 0


v10(x, t) = sinx+ L−1t

(
1
sLt

(
N2

(
u0(x, t)v0(x, t)

)))
v1n+1(x, t) = L

−1
t

(
1
sLt

(
∂2v1n(x, t)

∂x2

))
; n ≥ 0

(28)

Let’s suppose that one can find u0 and v0 asN1

(
u0(x, t)v0(x, t)

)
= 0 andN2

(
u0(x, t)v0(x, t)

)
=

0, we remark that, taking u0(x, t) = v0(x, t) = 0 we obtainN1

(
u0(x, t)v0(x, t)

)
= N2

(
u0(x, t)v0(x, t)

)
=



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 848

0

For k = 1, we have the following Laplace-SBA algorithm:




u10(x, t) = sinx

u1n+1(x, t) = L
−1
t

(
1
sLt

(
∂2u1n(x, t)

∂x2

))
; n ≥ 0

(a)


v10(x, t) = sinx

v1n+1(x, t) = L
−1
t

(
1
sLt

(
∂2v1n(x, t)

∂x2

))
; n ≥ 0

(b)

(29)

From (29)(a), we get:



u10(x, t) = sinx

u11(x, t) = −t sinx

u12(x, t) = (−t)2
2! sinx

...
u1n(x, t) = (−t)n

n! sinx

(30)

From (29)(a), we get:



v10(x, t) = sinx

v11(x, t) = −t sinx

u12(x, t) = (−t)2
2! sinx

...
v1n(x, t) = (−t)n

n! sinx

(31)

From (30) and (31), we obtain:
u1(x, t) =

+∞∑
n=0

u1n(x, t) =

(
+∞∑
n=0

(−t)n

n!

)
sinx = e−t sinx

v1(x, t) =
+∞∑
n=0

v1n(x, t) =

(
+∞∑
n=0

(−t)n

n!

)
sinx = e−t sinx

(32)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 849

For k = 2, we have the following Laplace-SBA algorithm:


u20(x, t) = sinx+ L−1t

(
1
sLt

(
N1

(
u1(x, t)v1(x, t)

)))
u2n+1(x, t) = L

−1
t

(
1
sLt

(
∂2u2n(x, t)

∂x2

))
; n ≥ 0


v20(x, t) = sinx+ L−1t

(
1
sLt

(
N2

(
u1(x, t)v1(x, t)

)))
v2n+1(x, t) = L

−1
t

(
1
sLt

(
∂2v2n(x, t)

∂x2

))
; n ≥ 0

(33)

We remark that:



N1

(
u1(x, t)v1(x, t)

)
= 2u1(x, t)

∂u1(x, t)

∂x
− ∂

∂x

(
u1(x, t)v1(x, t)

)
= 2

(
e−t sinx

)( ∂

∂x

(
e−t sinx

))
− ∂

∂x

(
e−2t sin2 x

)
= 2 (cosx sinx) e−2t − 2 (cosx sinx) e−2t = 0

N1

(
u1(x, t)v1(x, t)

)
= 2u1(x, t)

∂u1(x, t)

∂x
− ∂

∂x

(
u1(x, t)v1(x, t)

)
= 2

(
e−t sinx

)( ∂

∂x

(
e−t sinx

))
− ∂

∂x

(
e−2t sin2 x

)
= 2 (cosx sinx) e−2t − 2 (cosx sinx) e−2t = 0

(34)

and (33) becomes:


u20(x, t) = sinx

u2n+1(x, t) = L
−1
t

(
1
sLt

(
∂2u2n(x, t)

∂x2

))
; ∀n ≥ 0


v10(x, t) = sinx

v2n+1(x, t) = L
−1
t

(
1
sLt

(
∂2v2n(x, t)

∂x2

))
; ∀n ≥ 0

(35)

We remark that (35)is the same algorithm that (29). Thus

u2(x, t) = v2(x, t) = e−t sinx (36)

Using the same the procedure for k ≥ 3, we get:

u1(x, t) = v1(x, t) = u2(x, t) = v2(x, t) = u3(x, t) = v3(x, t) = · · · = uk(x, t) = vk(x, t) = e−t sinx
(37)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 850

and the exact solution of (17) is:
u(x, t) = lim

k−→+∞
uk(x, t) = e−t sinx

v(x, t) = lim
k−→+∞

vk(x, t) = e−t sinx

(38)

3.2. Example 2

Now, we consider the following system of Burger’s equations [10]


∂u(x, t)

∂t
− ∂2u(x, t)

∂x2
+ u(x, t)

∂u(x, t)

∂x
+

∂

∂x
(u(x, t)v(x, t)) = 2t2x3 + t2 − 2t+ x2

∂v(x, t)

∂t
− ∂2v(x, t)

∂x2
+ v(x, t)

∂v(x, t)

∂x
+

∂

∂x
(u(x, t)v(x, t)) =

1

x
− 2

t

x3
− t2

x3
− t2

(39)
with initial conditions {

u(x, 0) = 0

v(x, 0) = 0
(40)

3.2.1. Résolution by Laplace-SBA method

Let’s denote: 
N1 (u(x, t)v(x, t)) = u(x, t)

∂u(x, t)

∂x
+

∂

∂x
(u(x, t)v(x, t))

N2 (u(x, t)v(x, t)) = v(x, t)
∂v(x, t)

∂x
+

∂

∂x
(u(x, t)v(x, t))

(41)

the system (41) can be rewritten as follows
∂u(x, t)

∂t
= x2 − 2t+ 2x3t2 + t2 +

∂2u(x, t)

∂x2
−N1 (u(x, t)v(x, t))

∂v(x, t)

∂t
=

1

x
− 2

t

x3
− t2

x3
− t2 +

∂2v(x, t)

∂x2
−N2 (u(x, t)v(x, t))

(42)

Applying the Laplace transform to (42). we obtain:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 851


Lt (u(x, t)) =

(
1

s2
x2 +

4

s4
x3t2 − 2

s3
+ 2

s4

)
+

1

s
Lt
(
∂2u(x, t)

∂x2

)
− 1

s
Lt (N1 (u(x, t)v(x, t)))

Lt (v(x, t)) =
1

s2x
− 2

s3x3
− 2

s4x3
− 2

s4
+

1

s
Lt
(
∂2v(x, t)

∂x2

)
− 1

s
Lt (N2 (u(x, t)v(x, t)))

(43)
From (43), we have:


u(x, t) = tx2 − t2 + L−1t

(
1

s
Lt
(
∂2u(x, t)

∂x2

))
− L−1t

(
1

s
Lt (N1 (u(x, t)v(x, t)))

)
+ 2

3 t
3x3 + 1

3 t
3

v(x, t) =
t

x
− t2

x3
+ L−1t

(
1

s
Lt
(
∂2v(x, t)

∂x2

))
− L−1t

(
1

s
Lt (N2 (u(x, t)v(x, t)))

)
− 1

3

t3

x3
− 1

3 t
3

(44)
Let’s denote

Ñ1 (u(x, t)v(x, t)) = −L−1t

(
1

s
Lt (N1 (u(x, t)v(x, t)))

)
+ 2

3 t
3x3 + 1

3 t
3

Ñ2 (u(x, t)v(x, t)) = −L−1t

(
1

s
Lt (N2 (u(x, t)v(x, t)))

)
− 1

3

t3

x3
− 1

3 t
3

(45)

From (44) and (45), gives:
u(x, t) = tx2 − t2 + L−1t

(
1

s
Lt
(
∂2u(x, t)

∂x2

))
+ Ñ1 (u(x, t)v(x, t))

v(x, t) =
t

x
− t2

x3
+ L−1t

(
1

s
Lt
(
∂2v(x, t)

∂x2

))
+ Ñ2 (u(x, t)v(x, t))

(46)

Applying successive approximations to (46), we have:


uk(x, t) = tx2 − t2 + L−1t

(
1

s
Lt
(
∂2uk(x, t)

∂x2

))
+ Ñ1

(
uk−1(x, t)vk−1(x, t)

)

vk(x, t) =
t

x
− t2

x3
+ L−1t

(
1

s
Lt
(
∂2vk(x, t)

∂x2

))
+ Ñ2

(
uk−1(x, t)vk−1(x, t)

) (47)

We look for the solution of (47) in the following form:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 852


uk(x, t) =

+∞∑
n=0

ukn(x, t); k ≥ 1

vk(x, t) =
+∞∑
n=0

vkn(x, t); k ≥ 1

(48)

From (47), we obtain the following Laplace-SBA algorithm:


uk0(x, t) = tx2 − t2 + Ñ1

(
uk−1(x, t)vk−1(x, t)

)
k ≥ 1

ukn+1(x, t) = L
−1
t

(
1

s
Lt
(
∂2ukn(x, t)

∂x2

))
; n ≥ 0


vk0 (x, t) =

t

x
+ Ñ2

(
uk−1(x, t)vk−1(x, t)

)
k ≥ 1

vkn+1(x, t) = −
t2

x3
+ L−1t

(
1

s
Lt
(
∂2vkn(x, t)

∂x2

))
; n ≥ 0

(49)

For k = 1, we have the following Laplace-SBA algorithm:


u10(x, t) = tx2 − t2 + Ñ1

(
u0(x, t)v0(x, t)

)
k ≥ 1

u1n+1(x, t) = L
−1
t

(
1

s
Lt
(
∂2u1n(x, t)

∂x2

))
; n ≥ 0


v10(x, t) =

t

x
+ Ñ2

(
u0(x, t)v0(x, t)

)
k ≥ 1

v1n+1(x, t) = −
t2

x3
+ L−1t

(
1

s
Lt
(
∂2v1n(x, t)

∂x2

))
; n ≥ 0

(50)

Let’s suppose that one can find u0 and v0 as Ñ1

(
u0(x, t)v0(x, t)

)
= 0 and Ñ2

(
u0(x, t)v0(x, t)

)
=

0, we obtain the following Laplace-SBA algorithm:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 853





u10(x, t) = tx2 − t2

u11(x, t) = t2

...

u1n(x, t) = 0 ∀ n ≥ 2



v10(x, t) =
t

x

v11(x, t) = −
t2

x3
+

t2

x3
= 0

...

v1n(x, t) = 0 ∀ n ≥ 2

(51)

Therefore, we get: 
u1(x, t) = u10(x, t) + u11(x, t) + · · · = tx2

v1(x, t) = v10(x, t) + v11(x, t) + · · · =
t

x

(52)

For k = 2, we have the following Laplace-SBA algorithm:


u20(x, t) = tx2 − t2 + Ñ1

(
u1(x, t)v1(x, t)

)
k ≥ 1

u2n+1(x, t) = L
−1
t

(
1

s
Lt
(
∂2u2n(x, t)

∂x2

))
; n ≥ 0


v20(x, t) =

t

x
+ Ñ2

(
u1(x, t)v1(x, t)

)
k ≥ 1

v2n+1(x, t) = −
t2

x3
+ L−1t

(
1

s
Lt
(
∂2v2n(x, t)

∂x2

))
; n ≥ 0

(53)

We remark that:
Ñ1 (u(x, t)v(x, t)) = −1

6 t
3
(
4x3 + 2

)
+ 2

3 t
3x3 + 1

3 t
3 = 0

Ñ2 (u(x, t)v(x, t)) =
1
6
t3

x3

(
2x3 + 2

)
− 1

3

t3

x3
− 1

3 t
3 = 0

(54)

and (53) becomes:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 854





u10(x, t) = tx2 − t2

u11(x, t) = t2

...

u1n(x, t) = 0 ∀ n ≥ 2



v10(x, t) =
t

x

v11(x, t) = −
t2

x3
+

t2

x3
= 0

...

v1n(x, t) = 0 ∀ n ≥ 2

(55)

We remark that (55) is the same algorithm that (51). Thus
u2(x, t) = tx2

v2(x, t) =
t

x

(56)

Using the same the procedure for k ≥ 3, we get:
u2(x, t) = u3(x, t) = · · · = x2t

v2(x, t) = v3(x, t) = · · · = t

x

(57)

Thus, the solution of example 2 is:
u(x, t) = lim

k−→+∞
uk(x, t) = x2t

v(x, t) = lim
k−→+∞

vk(x, t) =
t

x

(58)

3.3. Example 3

Let’s consider the following non homogeneous form of coupled Burger’s equations [6]:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 855


∂u(x, t)

∂t
− 1

x
∂
∂x

(
x
∂u(x, t)

∂x

)
− 2u(x, t)

∂u(x, t)

∂x
+

∂

∂x
(u(x, t)v(x, t)) =

(
−x2 − 4

)
e−t

∂v(x, t)

∂t
− 1

x
∂
∂x

(
x
∂v(x, t)

∂x

)
− 2v(x, t)

∂v(x, t)

∂x
+

∂

∂x
(u(x, t)v(x, t)) =

(
−x2 − 4

)
e−t

(59)
with initial conditions {

u(x, 0) = x2

v(x, 0) = x2
(60)

3.3.1. Résolution by Laplace-SBA method

Let’s denote: 
N1 (u, v) = 2u(x, t)

∂u(x, t)

∂x
− ∂

∂x
(u(x, t)v(x, t))

N2 (u, v) = 2v(x, t)
∂v(x, t)

∂x
− ∂

∂x
(u(x, t)v(x, t))

(61)

the system (59) can be rewritten as follows
∂u(x, t)

∂t
=
(
−x2 − 4

)
e−t + 1

x
∂
∂x

(
x
∂u(x, t)

∂x

)
+N1 (u, v)

∂v(x, t)

∂t
=
(
−x2 − 4

)
e−t + 1

x
∂
∂x

(
x
∂v(x, t)

∂x

)
+N2 (u, v)

(62)

Applying the Laplace transform to the system (62) with respect to variable t, we have:


sLt (u(x, t)) = u(x, 0) + Lt

((
−x2 − 4

)
e−t
)
+ Lt

(
1
x

∂
∂x

(
x
∂u(x, t)

∂x

))
+ Lt (N1 (u, v))

sLt (v(x, t)) = v(x, 0) + Lt
((
−x2 − 4

)
e−t
)
+ Lt

(
1
x

∂
∂x

(
x
∂v(x, t)

∂x

))
+ Lt (N2 (u, v))

(63)
(63) is equivalent to


Lt (u(x, t)) = 1

sx
2 + 1

s

(
−x2+4

s+1

)
+ 1

sLt
(

1
x

∂
∂x

(
x
∂u(x, t)

∂x

))
+ 1

sLt (N1 (u, v))

Lt (v(x, t)) = 1
sx

2 + 1
s

(
−x2+4

s+1

)
+ 1

sLt
(

1
x

∂
∂x

(
x
∂v(x, t)

∂x

))
+ 1

sLt (N2 (u, v))

(64)



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 856

Using the inverse Laplace transform to (64), we get


u(x, t) = L−1t

(
1
sx

2
)
+ L−1t

(
1
s

(
−x2+4

s+1

))
+ L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u(x, t)

∂x

)))
+ L−1t

(
1
sLt (N1 (u, v))

)

v(x, t) = L−1t

(
1
sx

2
)
+ L−1t

(
1
s

(
−x2+4

s+1

))
+ L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v(x, t)

∂x

)))
+ L−1t

(
1
sLt (N2 (u, v))

)
(65)

From (65), we have:


u(x, t) = e−tx2 + 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u(x, t)

∂x

)))
+ L−1t

(
1
sLt (N1 (u, v))

)

v(x, t) = e−tx2 + 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v(x, t)

∂x

)))
+ L−1t

(
1
sLt (N2 (u, v))

)
(66)

Applying the method of successive approximations to the system (66), we obtain:


uk(x, t) = e−tx2 + 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂uk(x, t)

∂x

)))
+ L−1t

(
1
sLt

(
N1

(
uk−1, vk−1

)))

vk(x, t) = e−tx2 + 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂vk(x, t)

∂x

)))
+ L−1t

(
1
sLt

(
N2

(
uk−1, vk−1

)))
(67)

We look for the solution of (67)in the following form:
uk(x, t) =

+∞∑
n=0

ukn(x, t); k ≥ 1

vk(x, t) =

+∞∑
n=0

vkn(x, t); k ≥ 1

(68)

From (67), we obtain the following Laplace-SBA algorithm:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 857





uk0(x, t) = e−tx2 + L−1t

(
1
sLt

(
N1

(
uk−1, vk−1

)))
uk1(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂uk0(x, t)

∂x

)))

ukn+1(x, t) = +L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂ukn(x, t)

∂x

)))
; n ≥ 0



vk0 (x, t) = e−tx2 + L−1t

(
1
sLt

(
N1

(
uk−1, vk−1

)))
vk1 (x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂vk0 (x, t)

∂x

)))

vkn+1(x, t) = +L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂vkn(x, t)

∂x

)))
; n ≥ 0

k ≥ 1 (69)

Foor k = 1, we have the following Laplace-SBA algorithm:



u10(x, t) = e−tx2 + L−1t

(
1
sLt

(
N1

(
u0, v0

)))
u11(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u10(x, t)

∂x

)))
u1n+1(x, t) = +L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u1n(x, t)

∂x

)))
; n ≥ 0



v10(x, t) = e−tx2 + L−1t

(
1
sLt

(
N1

(
u0, v0

)))
v11(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v10(x, t)

∂x

)))
v1n+1(x, t) = +L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v1n(x, t)

∂x

)))
; n ≥ 0

(70)

Let’s suppose that one can find u0 and v0 asN1

(
u0(x, t)v0(x, t)

)
= 0 andN2

(
u0(x, t)v0(x, t)

)
=

0, we obtain the following Laplace-SBA algorithm:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 858





u10(x, t) = e−tx2

u11(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u10(x, t)

∂x

)))
u1n+1(x, t) = L

−1
t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u1n(x, t)

∂x

)))
; ∀ n ≥ 2



v10(x, t) = e−tx2

v11(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v10(x, t)

∂x

)))
v1n+1(x, t) = L

−1
t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v1n(x, t)

∂x

)))
; ∀ n ≥ 2

(71)

From (71), we obtain:


u10(x, t) = e−tx2

u11(x, t) = 4e−t − 4 + 4− 4e−t = 0

u1n+1(x, t) = 0 ; ∀ n ≥ 2


v10(x, t) = e−tx2

v11(x, t) = 4e−t − 4 + 4− 4e−t = 0

v1n+1(x, t) = 0 ; ∀ n ≥ 2

(72)

Therefore, we get: u1(x, t) = u10(x, t) = x2e−t

v1(x, t) = v10(x, t) = x2e−t
(73)

For k = 2, we have the following Laplace-SBA algorithm:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 859





u20(x, t) = e−tx2 + L−1t

(
1
sLt

(
N1

(
u1, v1

)))
u21(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u20(x, t)

∂x

)))
u2n+1(x, t) = +L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u2n(x, t)

∂x

)))
; n ≥ 0



v20(x, t) = e−tx2 + L−1t

(
1
sLt

(
N1

(
u1, v1

)))
v21(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v20(x, t)

∂x

)))
v2n+1(x, t) = +L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v2n(x, t)

∂x

)))
; n ≥ 0

(74)

We remark that:

N1

(
u1, v1

)
= 2u1(x, t)

∂u1(x, t)

∂x
− ∂

∂x

(
u1(x, t)v1(x, t)

)
= 2

(
x2e−t

) ∂

∂x

(
x2e−t

)
− ∂

∂x

(
x4e−2t

)
= 4x3e−2t − 4x3e−2t = 0

N2

(
u1, v1

)
= 2v1(x, t)

∂v1(x, t)

∂x
− ∂

∂x

(
u1(x, t)v1(x, t)

)
= 2

(
x2e−t

) ∂

∂x

(
x2e−t

)
− ∂

∂x

(
x4e−2t

)
= 4x3e−2t − 4x3e−2t = 0

(75)

and (74) becomes:



J. B. Yindoula / Eur. J. Pure Appl. Math, 14 (3) (2021), 842-862 860





u20(x, t) = e−tx2

u21(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u20(x, t)

∂x

)))
u2n+1(x, t) = L

−1
t

(
1
sLt

(
1
x

∂
∂x

(
x
∂u2n(x, t)

∂x

)))
; ∀ n ≥ 2



v20(x, t) = e−tx2

v11(x, t) = 4e−t − 4 + L−1t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v20(x, t)

∂x

)))
v2n+1(x, t) = L

−1
t

(
1
sLt

(
1
x

∂
∂x

(
x
∂v2n(x, t)

∂x

)))
; ∀ n ≥ 2

(76)

We remark that (76) is the same algorithm that (71). Thus obtain:u2(x, t) = x2e−t

v2(x, t) = x2e−t
(77)

Using the same the procedure for k ≥ 3, we get:u1(x, t) = u2(x, t) = · · · = uk(x, t) = x2e−t

v1(x, t) = v2(x, t) = · · · = vk(x, t) = x2e−t
(78)

Thus, the solution of example 3 is:
u(x, t) = lim

k−→+∞
uk(x, t) = x2e−t

v(x, t) = lim
k−→+∞

vk(x, t) = x2e−t
(79)

4. Conclusion

The results of this paper show that the use of this method allowed to obtain the
exact solutions of the coupled Burger’s equations. Consequently, Laplace-SBA method
is a powerful mathematical tool for solving any system of Burger’s equations with initial
conditions. However, further research is needed to verify the relevance and effectiveness of
this method in the resolution of various nonlinear differential equations.



REFERENCES 861

References

[1] K. Abbaoui, Y. Cherruault,Convergence of Adomian method applied to differen-
tial equations, Mathematical and computer Modellings 20(5), (1994) 103-109. DOI:
https://doi.org/10.1016/0895-1221(94)00144-8

[2] M.A. Abdoua, A. A. Solimanb, "Variational iteration method for solving Burger’s and
coupled Burger’s equations" Journal of computational and Applied Mathematic 181
(2005) 245-251. DOI: https://doi.org/10.1016/j.cam.2004.11.032

[3] M.Al-Mazmumy, The Modified Adomian Decomposition Method for Solving Nonlinear
Coupled Burger’s Equations, Nonlinear Analysis and Differential Equations, 3(2015)
111-122.DOI: https://dx.doi.org/10.12988/nade.2015.41226

[4] J .Biazar, M. Eslami, H. Ghazvini. Homotopy pertubation method for systems of
partial differential equations.Int.J. Nonlinear Sci. Num. Simul-8(3) (2003) 411-416.
DOI: 10.1515/IJNSNS.2007.8.3.413

[5] M.Dehghan, A Hamidi, M.Shakou Rifar, The solution of coupled Burger’s equa-
tions using Adomian-pade technique . Appl. Math. Comput.189(2007) 1034-1047.
DOI:10.1016/j.amc.2006.11.179

[6] Hassan Eltayeb, Said Mesloub, Adem Kilieman, Anote on a singular coupled Burgers
equation ad double Laplace transform method .Journal of Nonlinear Sciences and
Applications 11(2018) 635-643. DOI: http://dx.doi.org/10.22436/jnsa.011.05.05

[7] J.H.He, Variational iteration method a kind of nolinear analytical technique: some
examples. Int J.Nonlinear Mech .34 (1999) 699-08

[8] A.A Hemeda Homotopy Perturbation Method for Systems of Nonlinear Coupled Equa-
tions. Appplied Mathematical Sciences 6(96) (2012) 4787-4800.

[9] Hongqing Zhu, Huazhong Shu, Meiyu Ding, Numerical solutions of two-
dimensional Burger’s equations by discrete Adomian decomposition method
Computers and Mathematics with Applications 60(2010) 840-848. DOI:
https://doi.org/10.1016/j.camwa.2010.05.031

[10] M.Mirzazadeh, Z. Ayati. New homotopy perturbation method for system of
Burgers equations. Alexandria Engineering Journal (55) (2016) 1619-1624.DOI:
https://doi.org/10.1016/j.aej.2016.02.003

[11] WM. Moslem, R Sabry, Zakharov-Kuz, Netsov-Burger’s equation for dust ion acoustic
waves. Chaos solitons Fractals 36(2008) 628-634. Doi:10.1016/j.chaos.2006.06.097

[12] G. D. Nkaya, F. Bassono, R. Yaro, J. Bonazebi Yindoula, G.Bissanga, "Application
of the SBA Method for Solving the partial Differential Equation ", Journal of Math-
ematics Research.10(6) 2018. DOI:10.5539/jmr.v10n6p98.



REFERENCES 862

[13] Y. Pare, Y. Minoungou, A. Wassiha Nébié, "Resolution of nonlinear convection-
diffusion-reaction equations of Cauchy’s kind by the Laplace SBA method",
European Journal of Pure and Applied Mathematical. 12(3), (2019) 771-789.
DOI:10.29020/nybg.ejpam.v12i3.3430

[14] M.M. Rashidi, E. Erfani, New analytical method for solving Burger’s and nonlinear
heat transfer equations and comparison with HAM. Comput.Phys Commun.180(2009)
1539-1544. DOI:10.1016/j.cpc.2009.04.009

[15] Joseph Bonazebi Yindoula, Youssouf Pare, Gabriel Bissanga, Francis Bassono. Appli-
cation of the Adomian Decomposition Method (ADM) and the SOME BLAISE ABBO
(SBA) method to solving Diffusion-Reaction equation, Advances in Theoretical and
Applied Mathematics; 9(2014), N◦2, 97-104.

[16] B. ABBO, Nouvel algorithme numérique de résolution des Equations Différentielles
Ordinaires (EDO) et des Equations aux dérivées Partielles (EDP) non linéaires. Thèse
de doctorat unique, Université de de Ouagadougou, janvier 2007, UFR/SEA; Départe-
ment de Mathématiques et Informatique (Burkina Faso).

[17] Stevy Mikamona Mayembo, Joseph Bonazebi-Yindoula, Youssouf Pare, Gabriel Bis-
sanga. Application of the SBA method to solve the nonlinear biological population
models. European journal of pure and applied mathematics. Vol. 12, No. 3,2019, 1-10.
ISSN 1307-5543 -www.ejpam.com

[18] Pare Youssouf, Résolution de quelques équations fonctionnelles par la méthode SBA
(Somé Blaise-Abbo), Diss, Université de Ouagadougou, Mai 2010.