EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 2, 2019, 519-532
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

General solution of linear partial differential equations
modeling homogeneous diffusion-convection-reaction

problems with Cauchy initial condition

Minoungou Youssouf1, Bagayogo Moussa1, Youssouf Paré1,∗

1 Département de Mathématiques, UFR/Sciences Exactes et Appliquées, Université Ouaga
I Pr Joseph Ki-Zerbo, Ouagadougou, Burkina-Faso

Abstract. In this paper, we propose the general solution of diffusion-convection-reaction homo-
geneous problems with condition initial of Cauchy, using the SBA numerical method. This method
is based on the combination of the Adomian Decompositional Method(ADM), the successive ap-
proximations method and the Picard principle.

2010 Mathematics Subject Classifications: 47H14, 34G20, 47J25, 65J15

Key Words and Phrases: SBA method, Adomian Decompositional Method(ADM), homoge-
neous Diffusion-convection-reaction problem.

1. Introduction

Most physical, medical, biological,...,phenomena are modeled by integral equations,
integro-differential equations, ordinary differential equations or by partial differential equa-
tions. Generally it is very difficult, or impossible to determine their analytical solutions.
In this paper, we propose a general solution of linear homogeneous diffusion, convection
and reaction equations with Cauchy initial condition, using the SBA numerical method.

2. The numerical SBA method

Let us consider the following fonctional equation:

Au = f (1)

Where A : H → H, is an operator not necessarly linear and H is a Hilbert space
adequatly chosen given the operator A, f is given function and u the unnown function.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3411

Email addresses: m.youl@yahoo.fr (Y. Minoungou), moussabagayogo94@gmail.com (M. Bagayogo) ,
pareyoussouf@yahoo.fr (Y. Paré)

http://www.ejpam.com 519 c© 2019 EJPAM All rights reserved.



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 520

Let :
A = L+R+N (2)

Where L is an invertible operator in the Adomian ”sense”, R the linear remainder and
N a nonlinear operator. The equation (2) therefore becomes :

Lu+Ru+Nu = f

⇔
u = θ + L−1 (f)− L−1 (Ru)− L−1 (Nu) (3)

Where θ is such that L (θ) = 0. The equation (3) is the Adomian canonical forme , using
the successsive approximations [3] we get :

uk = θ + L−1 (f)− L−1
(
Ruk

)
− L−1

(
Nuk−1

)
; k ≥ 1 (4)

This yields the following Adomian algorithm [5]{
uk0 = θ + L−1 (f)− L−1

(
Nuk−1

)
; k ≥ 1

ukn+1 = −L−1
(
Rukn

)
;n ≥ 0

(5)

The Picard principle is then applied to the equation (5) 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

(6)

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

(7)

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 u2 =

+∞∑
n=0

ukn, therefore u = lim
k→+∞

uk is the

solution of the equation (2)
At each stape k ≥ 1 , we make sure that : N

(
uk
)

= 0.

2.1. A diffusion model

Let us consider the following diffusion model Cauchy initial condition [1, 6] ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂x2
, 0 < ε� 1

u (0, x) = sinωx, ω > 0
(8)



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 521

where (t, x) ∈ Ω = [0,+∞[× R and u ∈ C2 (Ω) .
Appliying the SBA method to (8) at the step k ≥ 0, we obtain the following algorithm

[2, 4]

(PSBA) :


uk0 (t, x) = sinωx

ukn+1 (t, x) = ε

∫ t

0

∂2ukn (s, x)

∂x2
ds, n ≥ 0

(9)

Let us calculate the following terms: uk1 (t, x) , uk2 (t, x) , uk3 (t, x) , ...

uk0 (t, x) = sinωx
uk1 (t, x) = −εω2t sinωx

uk2 (t, x) =

(
εω2t

)2
2!

sinωx

uk3 (t, x) =

(
−εω2t

)3
3!

sinωx

... ...

ukn (t, x) =

(
−εω2t

)n
n!

sinωx

Then we obtain :

uk (t, x) = sinωx
+∞∑
n=0

(
−εω2t

)n
n!

= exp
(
−εω2t

)
sinωx

And the exact solution of (8) is ;

u (t, x) = exp
(
−εω2t

)
sinωx

Proposition 1. The exact solution of the following diffusion problem with Cauchy initial
condition :  ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂x2
, ε > 0

u (0, x) = ϕ (αx) , α 6= 0
(10)

is
u (t, x) = exp

(
−εα2t

)
ϕ (αx) (11)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω), ϕ ∈ C2 (R) and ϕ verifie the relation :

ϕ (x) = A cosx+B sinx, A, B ∈ R. (12)

Proof. Let us consider u (t, x) = exp
(
−εα2t

)
ϕ (αx) .

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω), ϕ ∈ C2 (R)
We obtain :

∂

∂t
exp

(
−εα2t

)
ϕ (αx)− ε ∂

2

∂x2
exp

(
−εα2t

)
ϕ (αx) =

(
−εα2

) (
ϕ (αx) + ϕ′′ (αx)

)
exp

(
−εα2t

)
= 0



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 522

⇔ ϕ (αx) + ϕ′′ (αx) = 0

⇔ ϕ (αx) = A cos (αx) +B sin (αx)

Hence u (t, x) = exp
(
−εα2t

)
(A cos (αx) +B sin (αx)) and u (0, x) = ϕ (αx)

Then u (t, x) = exp
(
−εα2t

)
(A cos (αx) +B sin (αx)) is the general solution of (10)

with ϕ (αx) = A cos (αx) +B sin (αx) , A ∈ R, B ∈ R.

2.2. A convection model

Let us consider the following convection model with Cauchy initial [6–8] ∂u (t, x)

∂t
= λ

∂u (t, x)

∂x
;λ > 0

u (0, x) = cosαx;α 6= 0
(13)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C1 (Ω) et ϕ ∈ C1 (R) .
Appliying the SBA method to (13) at the step k ≥ 0.
We obtain the following algorithm :

(PSBA) :


uk0 (t, x) = cosαx

ukn+1 (t, x) = λ

∫ t

0

∂ukn (s, x)

∂x
ds, n ≥ 0

(14)

Let us calculate the following terms: uk1 (t, x) , uk2 (t, x) , uk3 (t, x) , ...

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

uk0 (t, x) = cosαx
uk1 (t, x) = −tαλ sinαx

uk2 (t, x) = −(tαλ)2

2!
cosαx

uk3 (t, x) =
(αλt)3

3!
sinαx

uk4 (t, x) =
(αλt)4

4!
cosαx

uk5 (t, x) = −(αλt)5

5!
sinαx

uk6 (t, x) = −(αλt)6

6!
cosαx

...

uk2n (t, x) = (−1)n
(αλt)2n

(2n)!
cosαx ; n ≥ 0

uk2n+1 (t, x) = − (−1) n (αλt)2n+1

(2n+ 1)!
sinαx ; n ≥ 0

Then



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 523

uk (t, x) = lim
n→+∞

cosαx

m∑
n=0

(−1)n
(αλt)2n

(2n)!
− lim

n→+∞
sinαx

m∑
n=0

(−1)n
(αλt)2n+1

(2n+ 1)!

We obtain the exact solution of the problem (13)

u (t, x) = cos (α (x+ λt)) . (15)

Proposition 2. The exact solution of the following convection model with Cauchy initial
condition  ∂u (t, x)

∂t
= λ

∂u (t, x)

∂x
, ε > 0, λ > 0

u (0, x) = ϕ (αx)
(16)

is
u (t, x) = ϕ (α (x+ λt)) (17)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C1 (Ω), ϕ ∈ C1 (R) .

Proof. Let us consider u (t, x) = ϕ (α (x+ λt)).
We obtain :

∂

∂t
ϕ (α (x+ λt))− λ ∂

∂x
ϕ (α (x+ λt)) = αλϕ′ (α (x+ λt))− αλϕ′ (α (x+ λt))

= 0

∀ϕ ∈ C2 (R) and u (0, x) = ϕ (αx)
In this case, it is necessary and sufficient that the function ϕ ∈ C1 (R).
Hence u (t, x) = ϕ (α (x+ λt)) is the general solution of (16).

2.3. A reaction model

Let us consider the following reaction model Cauchy type [6] ∂u (t, x)

∂t
= γu (t, x) , γ > 0

u (0, x) = sinαx
(18)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C1 (Ω) et ϕ ∈ C1 (R) .
Appliying the SBA algorithm to (18) at the step k ≥ 0, we obtain the following

algorithm :

(PSBA) :

 uk0 (t, x) = sinαx

ukn+1 (t, x) = γ

∫ t

0
ukn (s, x) ds, n ≥ 0

(19)



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 524

Let us calculate the following terms: uk1 (t, x) , uk2 (t, x) , uk3 (t, x) , ...

uk0 (t, x) = sinαx
uk1 (t, x) = γt sinαx

uk2 (t, x) =
(γt)2

2!
sinαx

uk3 (t, x) =
(γt)3

3!
sinαx

uk4 (t, x) =
(γt)4

4!
sinαx

uk5 (t, x) =
(γt)5

5!
sinαx

... ...

ukn (t, x) =
(γt)n

n!
sinαx

uk (t, x) = uk0 (t, x) + uk1 (t, x) + uk2 (t, x) + ...

uk (t, x) = sinαx

+∞∑
n=0

(γt)n

n!

Then

uk (t, x) = exp (γt) sinαx

we obtain the exact solution of the problem (18)

u (t, x) = exp (γt) sinαx (20)

Proposition 3. The exact solution of the following reaction problem Cauchy type: ∂u (t, x)

∂t
= γu (t, x) , γ > 0

u (0, x) = ϕ (αx) , α 6= 0
(21)

is
u (t, x) = exp (γt)ϕ (αx) (22)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C1 (Ω), ϕ ∈ C1 (R) .

Proof. Let us consider u (t, x) = exp (γt)ϕ (αx), we obtain :

∂

∂t
exp (γt)ϕ (αx)− γ exp (γt)ϕ (αx) = exp (γt) (γ − γ)ϕ (αx)

= 0

and u (0, x) = ϕ (αx) ⇐⇒ ∀ϕ ∈ C1 (R) .
In this case,it is necessary and sufficient that the function ϕ ∈ C1 (I) and I ⊂ R or

I = R , hence the general solution of (21) is u (t, x) = exp (γt)ϕ (αx) .



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 525

2.4. A diffusion-convection model

Let us consider the following type of Cauchy linear equation :

(D) :

 ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂x2
+ λ

∂u (t, x)

∂x
; 0 < ε� 1, λ > 0, t > 0, x ∈ R

u (0, x) = sinx

Applying the algorithm SBA to (D) , we have :

(PSBA) :


uk0 (t, x) = sinx

ukn+1 (t, x) =

∫ t

0

(
ε
∂2ukn (s, x)

∂x2
+ λ

∂ukn (s, x)

∂x

)
ds ; n ≥ 0

(23)

Let us determinate the following terms: uk0 (t, x) , uk1 (t, x) , uk2 (t, x) , uk3 (t, x) , ..., ukn (t, x) .



uk0 (t, x) = sinx
uk1 (t, x) = tλ cosx− tε sinx
uk2 (t, x) = 1

2! (sinx) t2λ2 − (cosx) t2λε+ 1
2! (sinx) t2ε2

uk3 (t, x) = −1
6 (cosx) t3λ3 + 1

2 (sinx) t3λ2ε+ 1
2 (cosx) t3λε2 − 1

6 (sinx) t3ε3

uk4 (t, x) = 1
4! (sinx) t4λ4 + 1

3! (cosx) t4λ3ε− 1
4 (sinx) t4λ2ε2−

1
3! (cosx) t4λε3 + 1

4! (sinx) t4ε4

uk5 (t, x) = 1
120 (cosx) t5λ5 − 1

24 (sinx) t5λ4ε− 1
12 (cosx) t5λ3ε2+

1
12 (sinx) t5λ2ε3 + 1

24 (cosx) t5λε4 − 1
120 (sinx) t5ε5

uk6 (t, x) = 1
720 (sinx) t6λ6 − 1

120 (cosx) t6λ5ε+ 1
48 (sinx) t6λ4ε2 + 1

36 (cosx) t6λ3ε3−
1
48 (sinx) t6λ2ε4 − 1

120 (cosx) t6λε5 + 1
720 (sinx) t6ε6

...

Step by step, we then deduct :



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 526

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

uk (t, x) ' sinx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)

−(λt)2

2!
sinx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)

+
(λt)4

4!
sinx

(
1− εt+

(εt)2

2!
− (εt)3

3!
+ ...

)

−(λt)6

6!
sinx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)
... ...

+λt cosx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)

−(λt)3

3!
cosx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)

+
(λt)5

5!
cosx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)
... ...

Then, we obtain


uk (t, x) ' sinx

(
1− εt+

(εt)2

2!
− (εt)3

3!
+ ...

)(
1− (λt)2

2!
+

(λt)4

4!
+ ...

)
+

cosx

(
1− εt+

(εt)2

2!
−−(εt)3

3!
+ ...

)(
λt− (λt)3

3!
+

(λt)5

5!
− ...

)

In a recurcive way, we obtain :

uk (t, x) = lim
n→+∞

sinx

n∑
p=0

(−εt)p

p!

n∑
p=0

(−1)p
(λt)2p

(2p)!
+ cosx

n∑
p=0

(−εt)p

p!

n∑
p=0

(−1)p
(λt)2p+1

(2p+ 1)!

Therefore, we get

uk (t, x) = exp (−εt) (sinx cosλt+ sinλt cosx) .

⇒
uk (t, x) = exp (−εt) sin (x+ λt)

So, the exact solution exact of (D) is

u (t, x) = exp (−εt) sin (x+ λt) .



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 527

Proposition 4. The exact solution of the following reaction problem Cauchy type:

(P4) :

 ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂x2
+ λ

∂u (t, x)

∂x
; 0 < ε� 1, λ > 0, t > 0, x ∈ R

u (0, x) = ϕ (x)

is
u (t, x) = exp (−εt)ϕ (x+ λt) (24)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω), ϕ ∈ C2 (R) .

Proof. Let us consider u (t, x) = exp (−εt)ϕ (x+ λt) We obtain :

∂

∂t
exp (−εt)ϕ (x+ λt)− ε ∂

2

∂x2
exp (−εt)ϕ (x+ λt)− λ ∂

∂x
exp (−εt)ϕ (x+ λt)

= exp (−εt)
(
−εϕ (x+ λt) + λϕ′ (x+ λt)− εϕ′′ (x+ λt)− λϕ′ (x+ λt)

)
= 0

⇒ ϕ′′ (x+ λt) + ϕ (x+ λt) = 0

⇒ ϕ (x+ λt) = A cos (x+ λt) +B sin (x+ λt)

where A, B ∈ R and u (0, x) = ϕ (x) ⇐⇒ ∀ϕ ∈ C1 (R)
In this case, it is necessary and sufficient that the function ϕ ∈ C1 (I) where I ⊂ R or

I = R, hence the general solution of (P4) is u (t, x) = exp (γt)ϕ (αx) .

2.5. A reaction model

Proposition 5. The exact solution of the following reaction problem Cauchy type:

(E)

 ∂u (t, x)

∂t
= γu (t, x) , γ > 0

u (0, x) = ϕ (αx) , α 6= 0
(25)

is
u (t, x) = exp (γt)ϕ (αx) (26)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C1 (Ω), ϕ ∈ C1 (R) .

Proof. Let us consider u (t, x) = exp (γt)ϕ (αx).
We obtain :

∂

∂t
exp (γt)ϕ (αx)− γ exp (γt)ϕ (αx) = exp (γt) (γ − γ)ϕ (αx)

= 0

and u (0, x) = ϕ (αx)
⇐⇒ ∀ϕ ∈ C1 (I) where I ⊂ R or I = R
In this case,it is necessary and sufficient that the function ϕ ∈ C1 (I) where I ⊂ R or

I ⊂ R, hence the general solution of (E) is u (t, x) = exp (γt)ϕ (αx) .



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 528

2.6. A diffusion-reaction model

Let us consider the following diffision-reaction problem Cauchy type:

(F )

 ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂x2
+ γu (t, x) , 0 < ε� 1, γ > 0

u (0, x) = sinαx, α 6= 0
(27)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω) et ϕ ∈ C2 (R) .
Appliying the SBA method at the step k ≥ 0, we obtain the following algorithm :

(PSBA) :

 uk0 (t, x) = sinαx

ukn+1 (t, x) = ε

∫ t

0

∂2u (s, x)

∂x2
ds+ γ

∫ t

0
ukn (s, x) ds; n ≥ 0

(28)

Let us calculate some terms:
We obtain at the same way :

uk (t, x) = exp
((
γ − εα2

)
t
)

(sinαx cosαλt+ cosαx sinαλt)

we obtain the exact solution of the problem (F ) :

u (t, x) = exp
((
γ − εα2

)
t
)

sinα (x+ λt) (29)

Proposition 6. The exact solution of the following diffision-reaction problem Cauchy type

(P6)

 ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂x2
+ γu (t, x) ; ε > 0, λ > 0

u (0, x) = ϕ (αx)
(30)

is
u (t, x) = exp

((
γ − εα2

)
t
)
ϕ (α (x+ λt)) (31)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω), ϕ ∈ C2 (R) and ϕ verifie the relation :

ϕ (x) = A cosx+B sinx; A, B ∈ R (32)

Proof. Let us consider u (t, x) = exp
(
−εα2t

)
ϕ (α (x+ λt)) .

We obtain :

∂

∂t
exp

(
−εα2t

)
ϕ (α (x+ λt))− ε ∂

2

∂x2
exp

(
−εα2t

)
ϕ (α (x+ λt))

−λ ∂

∂x
exp

(
−εα2t

)
ϕ (α (x+ λt))

= exp
(
−εα2t

) (
−εα2ϕ (α (x+ λt)) + αλϕ′ (α (x+ λt))

−εα2ϕ′′ (α (x+ λt))− λαϕ′ (α (x+ λt))
)



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 529

= 0

⇔ ϕ (α (x+ λt)) + ϕ′′ (α (x+ λt)) = 0
⇒ ϕ (α (x+ λt)) = A cos (α (x+ λt)) +B sin (α (x+ λt)) , A, B ∈ R
Hence the general solution of (P6) is of the form :

u (t, x) = exp
(
−εα2t

)
ϕ (α (x+ λt)),

with ϕ (α (x+ λt)) = A cos (α (x+ λt)) +B sin (α (x+ λt)) and u (0, x) = ϕ (αx) .

2.7. A diffusion-convection-reaction problem Cauchy type

Let us consider the following diffision-convection-reaction problem Cauchy type:

(H)

 ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂2x
+ λ

∂u (t, x)

∂x
+ γu (t, x) , ε > 0, λ > 0, γ > 0

u (0, x) = sinαx
(33)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω) et ϕ ∈ C2 (R) .
Appliying the SBA method at the step k ≥ 0, we obtain the following algorithm :

(PSBA) :

 uk0 (t, x) = sinαx

ukn+1 (t, x) = ε

∫ t

0

∂2u (s, x)

∂2x
ds+ λ

∂u (t, x)

∂x
+ γ

∫ t

0
ukn (s, x) ds, n ≥ 0

(34)

Let us calculate the following terms: uk1 (t, x) , uk2 (t, x) , uk3 (t, x) , ...
Let us consider the following Cauchy linear equation :

(P3) :

 ∂u (t, x)

∂t
= ε

∂u (t, x)

∂x
+ µu (t, x) ; 0 < ε� 1, µ > 0, t > 0, x ∈ R

u (0, x) = cosx

Applying the algorithm SBA to (P3) , we have :

PSBA :


uk0 (t, x) = cosx

ukn+1 (t, x) =

∫ t

0

(
ε
∂ukn (s, x)

∂x
+ µukn (s, x)

)
ds ; n ≥ 0

(35)

Let us calculate the following terms :



Y. Minoungou, M. Bagayogo, Y. Paré / Eur. J. Pure Appl. Math, 12 (2) (2019), 519-532 530

uk0 (t, x) , uk1 (t, x) , uk2 (t, x) , uk3 (t, x) , uk4 (t, x) , uk5 (t, x) , ...

uk0 (t, x) = cosx
uk1 (t, x) = tµ cosx− tε sinx

uk2 (t, x) =
(tµ)2

2
cosx− µεt2 sinx− (tε)2

2
cosx

uk3 (t, x) =
(tµ)3

6
cosx− 1

2
t3µ2ε sinx− 1

2
t3µε2 cosx+

(tε)3

6
sinx

uk4 (t, x) =
(tµ)4

24
cosx− 1

6
t4µ3ε sinx− 1

4
t4µ2ε2 cosx+

1

6
t4µε3 sinx+

(tε)4

24
cosx

uk5 (t, x) =
(tµ)5

120
cosx− 1

24
t5µ4ε sinx− 1

12
t5µ3ε2 (cosx) +

1

12
t5µ2ε3 sinx+

1

24
t5µε4 cosx− (tε)5

120
sinx

...

Step by step, we then deduct
uk (t, x) '

(
1− (εt)2

2
+

(εt)4

4!
− ...

)(
1 + µt+

(µt)2

2!
+

(µt)3

3!
+ ...

)
cosx−(

εt− (εt)3

3!
+

(εt)5

5!
− ..

)(
1 + µt+

(µt)2

2!
+

(µt)3

3!
+ ...

)
sinx

In a recurcive way, we otain:

uk (t, x) = lim
n→+∞

n∑
p=0

(−1)p
(εt)2p

(2p)!

n∑
p=0

(µt)p

p!
cosx+

n∑
p=0

(−1)p
(εt)2p+1

(2p+ 1)!

n∑
p=0

(µt)p

p!
sinx

then,we get :
uk (t, x) = exp (µt) cos (εt+ x)

And the exact solution of (P3) is

u (t, x) = lim
k→+∞

uk (t, x) = exp (µt) cos (εt+ x)

Proposition 7. The exact solution of the following diffision-convection problem Cauchy
type

(P7)

 ∂u (t, x)

∂t
= ε

∂2u (t, x)

∂2x
+ λ

∂u (t, x)

∂x
+ γu (t, x) , ε > 0, λ > 0, γ > 0

u (0, x) = ϕ (αx)
(36)



REFERENCES 531

is
u (t, x) = exp

((
γ − εα2

)
t
)
ϕ (α (x+ λt)) (37)

where (t, x) ∈ Ω = [0,+∞[× R, u ∈ C2 (Ω), ϕ ∈ C2 (R) and ϕ verifie the relation :

ϕ (x) = A cosx+B sinx, A, B ∈ R (38)

Proof. Let us consider u (t, x) = exp
((
γ − εα2

)
t
)
ϕ (α (x+ λt))

We get :
∂ exp

((
γ − εα2

)
t
)
ϕ (α (x+ λt))

∂t
− ε

∂2 exp
((
γ − εα2

)
t
)
ϕ (α (x+ λt))

∂2x
−

λ
∂ exp

((
γ − εα2

)
t
)
ϕ (α (x+ λt))

∂x
− γ exp

((
γ − εα2

)
t
)
ϕ (α (x+ λt))

= exp
((
γ − εα2

)
t
) [(

γ − εα2
)
ϕ (α (x+ λt)) + αλϕ′ (α (x+ λt))

]
+

exp
((
γ − εα2

)
t
) [
εα2ϕ′′ (α (x+ λt))− λαϕ′ (α (x+ λt))− γϕ (α (x+ λt))

]
= 0
⇒ εα2 (ϕ′′ (α (x+ λt)) + ϕ (α (x+ λt))) ∗ 0
⇒ ϕ′′ (α (x+ λt)) + ϕ (α (x+ λt)) = 0
or ϕ′′ (x) + ϕ (x) = 0
⇒ ϕ (x) = A cosx+B sinx;A,B ∈ R
And u (0, x) = ϕ (αx) .

3. Conclusion

The SBA numerical method permitted us to resolve a few linear partial differential
equations modelling diffusion, convection, reaction problems Cauchy type. The SBA
method pemitted 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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