EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 3, 2019, 1096-1105
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Application of the SBA method to solve the nonlinear
biological population models

Stevy Mikamona Mayembo1, Joseph Bonazebi-Yindoula1, Youssouf Paré2,∗,
Gabriel Bissanga1

1 Département de Mathématiques et Informatiques, Facultés des Sciences et Techniques,
Université Marien Ngouabi, Brazzaville, Congo
2 Département de Mathématiques, UFR/Sciences Exactes et Appliquées, Université Ouaga
I Pr Joseph Ki-Zerbo, Ouagadougou, Burkina-Faso

Abstract. This paper presents numerical solution of some nonlinear degenerate parabolic equa-
tions arising in the spatial diffusion of biological populations. The SBA method based on combina-
tion of Adomian decomposition method, principle of Picard and successive approximations is used
for solving these equations. The analytical obtained solutions show that the SBA method leads to
more accurate results.
2010 Mathematics Subject Classifications: 47H14, 34G20, 47J25, 65J15
Key Words and Phrases: SBA method, ADM method, biological population model,spatial
diffusion of species, Picard principle, successive approximations

1. Introduction

The spatial diffusion of some biological species is described by nonlinear partial differ-
ential equations. In the last years, various numerical powerful methods have been applied
to get the solutions of general degenerate parabolic equations, such as collocation methods
with mesh-free technique [2], variational iteration method [10], Adomian decomposition
method (ADM) [1, 13], homotopy perturbation method [8, 9], homotopy analysis Sumudu
transform method [12], etc.

The objective of this paper is to apply Some Blaise Abbo(SBA) method, which is an
elegant combination of ADM [3], and Picard principle and successive approximations [5]
to find the analytical solution of some degenerate parabolic equations arising in the spatial
diffusion of time fractional biological populations. A particular form of these equations is
given by :

∂u(x, y, t)

∂t
= ∆u2(x, y, t) + g(u), (x, y) ∈ R2, t > 0 (1)

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3409

Email addresses: @yahoo.fr (S.M. Mayembo), bonayindoula@yahoo.fr (J.B-Yindoula) ,
pareyoussouf@yahoo.fr (Y. Paré), bissanga1@yahoo.fr (G. Bissanga)

http://www.ejpam.com 1096 c© 2019 EJPAM All rights reserved.



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1097

with given initial conditions u(x, y, 0) = f(x, y)

The function u(x, y, t) denotes the population density (number of minimal species per-
unit volume at position (x, y) and time t ), g(u) the population supply due to birth and
death of species. In this study, a form of g(u) is :

g(u) = huα(x, y, t)
(

1− ruβ(x, y, t)
)

(2)

where h, α, β, r are real numbers.
The SBA method overcomes the difficulty arising in calculating the Adomian’s poly-

nomials which is an important advantage over the ADM and other numerical methods
[6, 11]

2. The numerical SBA method

Let us consider the following functional equation:

Au = f (3)

Where A : H → H, is an operator not necessarily linear and H is a Hilbert space
adequately chosen given the operator A, f is given function and u the unknown function.

Let :
A = L+R+N (4)

Where L is an invertible operator in the Adomian ”sense”, R the linear remainder and
N a nonlinear operator. The equation (3) therefore becomes :

Lu+Ru+Nu = f

⇔
u = θ + L−1 (f)− L−1 (Ru)− L−1 (Nu) (5)

Where θ is such that L (θ) = 0. The equation (5) is the Adomian canonical form , using
the successive approximations [10]we get :

uk = θ + L−1 (f)− L−1
(
Ruk

)
− L−1

(
Nuk−1

)
; k ≥ 1 (6)

This yields the following Adomian algorithm [4, 7, 14, 15]{
uk0 = θ + L−1 (f)− L−1

(
Nuk−1

)
; k ≥ 1

ukn+1 = −L−1
(
Rukn

)
;n ≥ 0

(7)

The Picard principle is then applied to the equation (7) let u0 be such that N
(
u0
)

= 0,
for k = 1, we get : {

u10 = θ + L−1 (f) + L−1
(
Nu0

)
u1n+1 = −L−1

(
Ru1n

)
;n ≥ 0

(8)



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1098

If the series

(
+∞∑
n=0

u1n

)
converges, then u1 =

+∞∑
n=0

u1n. For k = 2, we get:

{
u20 = θ + L−1 (f) + L−1

(
Nu1

)
u2n+1 = L−1

(
Ru2n

)
;n ≥ 0

(9)

If the series

(
+∞∑
n=0

u2n

)
converges, then u2 =

+∞∑
n=0

u2n. This process is repeated to k.

If the series

(
+∞∑
n=0

ukn

)
converges, then uk =

+∞∑
n=0

ukn, therefore uk = lim
k→+∞

uk is the

solution of the problem.
With the following hypothesize : At the step k, N(uk) = 0, ∀k ≥ 1.

2.1. Test examples

In this section, we present some examples with analytical solution to show the efficiency
of method described in previous section for solving equation (1)

2.2. Example 1

Consider equation (1) with α = β = 1, r 6= 0, h 6= 0 and f(x, y) = exp

(
1
2

√
hr
2 (x+ y)

)
Putting these values in this equation, we write :

∂u(x,y,t)
∂t = ∆u2(x, y, t) + hu(x, y, t) (1− ru(x, y, t)) , (x, y) ∈ R2, t > 0

u(x, y, 0) = exp

(
1
2

√
hr
2 (x+ y)

)
(10)

∂u(x, y, t)

∂t
= hu(x, y, t) + ∆u2(x, y, t)− rhu2(x, y, t) (11)

Let 

Lt(u(x, y, t)) = ∂
∂t (.)

L−1t =

∫ t

0
(.) ds

R(u(x, y, t)) = hu(x, y, t)

N(u(x, y, t)) =
∂2u2(x, y, t)

∂x2
+
∂2u2(x, y, t)

∂y2
− rhu2(x, y, t)

(12)

We obtain :

L(u(x, y, t)) = R(u(x, y, t)) +N(u(x, y, t) (13)

Integrating the equation (13) with respect to t gives the following canonical form:



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1099

u(x, y, t) = exp

(
1

2

√
hr

2
(x+ y)

)
+

∫ t

0
R(u(x, y, s))ds+

∫ t

0
N(u(x, y, s)ds (14)

In Applying the method of successive approximations to equation (14) we obtain:

uk(x, y, t) = exp

(
1

2

√
hr

2
(x+ y)

)
+

∫ t

0
R(uk(x, y, s))ds+

∫ t

0
N(uk−1(x, y, s)ds (15)

If now we Apply the Adomian algorithm to equation (15), we obtain:
uk0(x, y, t) = exp

(
1
2

√
hr
2 (x+ y)

)
+

∫ t

0
N(uk−1(x, y, s)ds

ukn(x, y, t) =

∫
0tR(ukn−1(x, y, s))ds, n ≥ 1

(16)

The solution at each stage is :

uk(x, y, t) =
∞∑
n=0

ukn(x, y, t), k = 1, 2, 3, · · · (17)

First step k = 1
In Applying the Picard principle and if we take u0(x, y, s) such as N(u0(x, y, s)) = 0,
we obtain : 

u10(x, y, t) = exp

(
1
2

√
hr
2 (x+ y)

)

u1n(x, y, t) =

∫ t

0
R(u1n−1(x, y, s))ds, n ≥ 1

(18)

It follows that expressions of function u1n(x, y, t) were :

u10(x, y, t) = exp

(
1
2

√
hr
2 (x+ y)

)

u11(x, y, t) = (ht) exp

(
1
2

√
hr
2 (x+ y)

)

u12(x, y, t) = 1
2! (ht)2 exp

(
1
2

√
hr
2 (x+ y)

)
...

u1n(x, y, t) = 1
n! (ht)n exp

(
1
2

√
hr
2 (x+ y)

)
(19)



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1100

In this representation,u1(x, y, t) form the finite sequence :

u1(x, y, t) =
∞∑
n=0

u1n(x, y, t) =

( ∞∑
n=0

1

n!
(ht)n

)
exp

(
1

2

√
hr

2
(x+ y)

)
(20)

and we can write :

u1(x, y, t) = exp (ht)× exp

(
1

2

√
hr

2
(x+ y)

)
= exp

(
ht+

1

2

√
hr

2
(x+ y)

)
(21)

which is the solution in step 1
Second step k = 2

u20(x, y, t) = exp

(
1
2

√
hr
2 (x+ y)

)
+

∫ t

0
N(u1(x, y, s)ds

u2n(x, y, t) =

∫ t

0
R(u2n−1(x, y, s))ds, n ≥ 1

(22)

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

N(u1) =
∂2
(
u1
)2

∂x2
+
∂2
(
u1
)2

∂y2
− rh

(
u1
)2

=

∂2
(

exp

(
ht+ 1

2

√
hr
2 (x+ y)

))2

∂x2
+

∂2
(

exp

(
ht+ 1

2

√
hr
2 (x+ y)

))2

∂y2
−

rh

(
exp

(
ht+ 1

2

√
hr
2 (x+ y)

))2

=

∂2
(

exp

(√
hr
2 (x+ y) + 2ht

))
∂x2

+

∂2
(

exp

(√
hr
2 (x+ y) + 2ht

))
∂y2

−

rh exp

(
2ht+

√
hr
2 (x+ y)

)

= (rh− rh) exp

(
2ht+

√
hr
2 (x+ y)

)
= 0

(23)

From where in step 2, we have the following SBA algorithm:
u20(x, y, t) = exp

(
1
2

√
hr
2 (x+ y)

)
u2n(x, y, t) =

∫ t

0
R(u2n−1(x, y, s))ds, n ≥ 1

(24)



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1101

Again, we write : 

u20(x, y, t) = exp

(
1
2

√
hr
2 (x+ y)

)

u21(x, y, t) = (ht) exp

(
1
2

√
hr
2 (x+ y)

)

u22(x, y, t) = 1
2! (ht)2 exp

(
1
2

√
hr
2 (x+ y)

)
...

u2n(x, y, t) = 1
n! (ht)n exp

(
1
2

√
hr
2 (x+ y)

)
(25)

So the finite sequence form of u2(x, y, t) is:

u2(x, y, t) =
∞∑
n=0

u2n(x, y, t) =

( ∞∑
n=0

1

n!
(ht)n

)
exp

(
1

2

√
hr

2
(x+ y)

)
(26)

Thus,

u2(x, y, t) = exp

(
ht+

1

2

√
hr

2
(x+ y)

)
(27)

is the solution in step 2
In a recurrent way, for the following steps (k ≥ 3), we obtain:

uk(x, y, t) =
∞∑
n=0

ukn(x, y, t) = exp

(
ht+

1

2

√
hr

2
(x+ y)

)
(28)

Therefore, the exact solution of this equation is:

u(x, y, t) = lim
k→+∞

uk(x, y, t) = exp

(
ht+

1

2

√
hr

2
(x+ y)

)
(29)

2.3. Example 2

Consider equation (1) with α = β = 1, r = 0, h 6= 0 and f(x, y) =
√

cosx cosh y
∂u(x, y, t)

∂t
= hu(x, y, t) +N (u(x, y, t)) , (x, y) ∈ R2, t > 0

u(x, y, 0) =
√

cosx cosh y

(30)

where
N (u(x, y, t)) = ∆u2(x, y, t) (31)



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1102

From (30), we obtain the following canonical Adomian form:

u(x, y, t) =
√

cosx cosh y + h

∫ t

0
u(x, y, s)ds+

∫ t

0
N (u(x, y, s)) ds (32)

From (32), the successive approximations give us

uk(x, y, t) =
√

cosx cosh y + Ñk−1 (u(x, y, t)) + h

∫ t

0
uk(x, y, s)ds (33)

Where

Ñ
(
uk−1(x, y, t)

)
=

∫ t

0
N
(
uk−1(x, y, s)

)
ds (34)

From (33), we have the following algorithm of Adomian :
uk0(x, y, t) =

√
cosx cosh y + Ñk−1 (u(x, y, t))

ukn+1(x, y, t) = h

∫ t

0
ukn(x, y, s)ds, n ≥ 0

(35)

Let us apply to (35), the principle of Picard. We remark that u0(x, y, t) = 0 is a root
of the Ñ

(
u0(x, y, t)

)
= 0

For k = 1, we obtain: 

u10(x, y, t) =
√

cosx cosh y

u11(x, y, t) = (th)
√

cosx cosh y

u12(x, y, t) = 1
2! (th)2

√
cosx cosh y

u13(x, y, t) = 1
3! (th)3

√
cosx cosh y

...
u1n(x, y, t) = 1

n! (th)n
√

cosx cosh y

(36)

Let us put

u1(x, y, t) = u10(x, y, t) + u11(x, y, t) + u12(x, y, t) + · · ·

=
(

1 + (th) + 1
2! (th)2 + 1

3! (th)3 + · · ·
)√

cosx cosh y

=
√

cosx cosh yeth

(37)

Second step
For k = 2, we have:



Y. Paré et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1096-1105 1103

Ñ
(
u1(x, y, t)

)
=

∫ t

0
N
(
u1(x, y, s)

)
ds

=

∫ t

0

(
∂2
((√

cosx cosh y
)
esh
)2

∂x2
+
∂2
((√

cosx cosh y
)
esh
)2

∂y2

)
ds

=

∫ t

0
e2sh

(
∂2 (cosx cosh y)

∂x2
+
∂2 (cosx cosh y)

∂y2

)
ds

=

∫ t

0
e2sh (− cosx cosh y + cosx cosh y) ds

= 0

(38)

From (35),we obtain: 

u20(x, y, t) =
√

cosx cosh y

u21(x, y, t) = (th)
√

cosx cosh y

u22(x, y, t) = 1
2! (th)2

√
cosx cosh y

u23(x, y, t) = 1
3! (th)3

√
cosx cosh y

...

(39)

Therefore,

u2(x, y, t) = u20(x, y, t) + u21(x, y, t) + u22(x, y, t) + · · ·

=
(

1 + (th) + 1
2! (th)2 + 1

3! (th)3 + · · ·
)√

cosx cosh y

=
(√

cosx cosh y
)
eth

(40)

Using the procedure for k ≥ 3, the solution to the k step is

uk(x, y, t) = uk0(x, y, t) + uk1(x, y, t) + uk2(x, y, t) + · · ·

=
(

1 + (th) + 1
2! (th)2 + 1

3! (th)3 + · · ·
)√

cosx cosh y

= eth
√

cosx cosh y

(41)

So the exact solution of equation (30) with initial boundaries u(x, y, 0) =
√

cosx cosh y
is

u(x, y, t) = lim
k→+∞

uk(x, y, t) = eth
√

cosx cosh y (42)



REFERENCES 1104

3. Conclusion

The SBA numerical method permitted us to resolve a few nonlinear partial differen-
tial equations modelling diffusion, convection, reaction problems Cauchy type. The SBA
method permitted us to resolve the problems proposed in this paper. It is then a very
powerful numerical tool of analysis for the resolution of these kinds of problems.

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