EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 3, 2019, 1260-1276
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Comparison of SBA numerical method and method of
separation of variables (Fourier) on wave equations.

Rasmane Yaro1, Youssouf Paré2,∗, Bakari Abbo3

1 Université de Dédougou, Dédougou, Burkina-Faso
2 Département de Mathématiques, UFR/Sciences Exactes et Appliquées,
Université Joseph Ki-Zerbo, Ouagadougou, Burkina-Faso
2 Département de Mathématiques, Facultés des Sciences et Techniques,
Université de Ndjaména, Ndjaména, Tchad

Abstract. In this paper, our aim is to use the SBA numerical method (combination of Adomian
method and Picard successive approximations) and Fourier method or method of separation of
variables to construct the solution of some wave equations. We compare the two methods and
apply them to some wave equations.

2010 Mathematics Subject Classifications: 65L07, 65P40, 40A05, 34F05

Key Words and Phrases: Numerical SBA method, Adomian method, Dynamical Model, Pi-
card’s principle, successive approximations and wave equations

1. Introduction

Many problems are governed by partial differential equations, or by systems of partial
differential equations. It is difficult to find their exact solutions. In this work, the SBA
numerical method, [3, 9] and Fourier method permitted us to find the exact solution of
some wave equations.

2. Description of the methods

2.1. Description of the SBA numerical method

Let’s consider the following functional equation

Au = f (1)

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3471

Email addresses: yarorasmane@yahoo.fr (R. Yaro), pareyoussouf@yahoo.fr (Y. Paré),
bakariabbo@yahoo.fr (B. Abbo)

http://www.ejpam.com 1260 c© 2019 EJPAM All rights reserved.



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1261

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

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.

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 Lθ = 0
Equation (3) is the Adomian canonical form [2, 6–8]
Using the successive approximations [1, 4], we get:

uk = θk + L−1(fk) + L−1(R(uk)) + L−1(N(uk−1)); k ≥ 1 (4)

This yields the following Adomian algorithm [5]{
uk0 = θk + L−1(fk) + L−1(N(uk−1)); k ≥ 1

ukn = L−1(R(ukn−1)); n ≥ 1
(5)

The Picard principle is then applied to equation (5) : let u0 be such that N(u0) = 0
for k = 1, we get :{

u10 = θ1 + L−1(f1) + L−1(N(u0))
u1n = L−1(R(u1n−1)); n ≥ 1

If the series
(∑

n≥0 u
1
n

)
converges, then u1 =

∑
n≥0 u

1
n

For k = 2,we get:{
u20 = θ2 + L−1(f2) + L−1(N(u1))

u2n = L−1(R(u2n−1)); n ≥ 1

If the series
(∑

n≥0 u
2
n

)
converges, then u2 =

∑
n≥0 u

2
n. This process is repeated to k.

If the series
(∑

n≥0 u
k
n

)
converges, then uk =

∑
n≥0 u

k
n.

Therefore u = lim
k→+∞

uk is the solution of the problem.

Thus, given the problem
(p) : Au = f,

we combine ideas from the classical techniques to derive the following appropriate approx-
imate scheme.



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1262

{
uk0 = θk + L−1(fk) + L−1(N(uk−1)); k ≥ 1

ukn = L−1(R(ukn−1)); n ≥ 1
(6)

called SBA algorithm.

2.2. Description of the Fourier method or method of separation of vari-
ables

The Method of separation of variable (also known as Fourier method) is one of several
methods for solving ordinary and partial differential equations.

This method cannot always be used, even when it can be used it will not always be
possible to get the solution of the problem. However, it can be used to easily in the (1−D)
heat equation with no sources, in the (1−D) wave equation and in the (2−D) heat and
wave equations.

Let’s consider the following general functional equation

Au = f (7)

The method of separation of variables relies upon the assumption that a function of
the form:

u(x, t) = X(x)T (t) (8)

if we are in the case of one -dimension x of space(1−D)

and

u(x, y, t) = U(x, y)T (t) = X(x)Y (y)T (t) (9)

if we are in the case of two dimension x and y of space (2 −D) will be a solution to
linear homogeneous partial differential equation in x and t when when we are in (1−D)
and x, y and t when when we are in (2.−D).

This is called a product solution and provided the boundary conditions are also linear
and homogeneous this will also satisfy the boundary conditions.

However , as noted above this will only rarely satisfy the initial condition, but that is
something for us to worry about in the next section.



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1263

3. Applications

3.1. Problem 1

Let’s consider the following Wave’s model of one-dimension of space:

(P1)



∂2u(x, t)

∂t2
= c2 4 u(x, t), c > 0

u(x, 0) = ϕ(x) = sin
(πx
L

)
ut(x, 0) = φ(x) = π sin

(πx
L

)
u(0, t) = 0

u(L, t) = 0, L > 0

(10)

Solving yhe wave equation involves identifyinf the functions u(x, t) that solve the par-
tial differential equation that represent the amplitude of the wave at any position x at any
time t.
• Solving by SBA method

By Integrating (10) we get the Adomian canonical form [2, 9] of the problem (10) :

u(x, t) = u(0, x) + t
∂u(0, x)

∂t
+ c2L−1tt (4u(x, t)) (11)

Applying successive approximations method to (11), we get :

uk(x, t) = uk(0, x) + t
∂uk(0, x)

∂t
+ c2L−1tt (4uk(x, t)) + Ñ(uk−1(x, t)), k ≥ 1 (12)

where Ñ(uk(x, t)) = 0 ∀ k ∈ N

Applying the SBA algorithm to (12), we get :

(P kSBA)

 uk0(x, t) = uk(x, 0) + t
∂uk(0, x)

∂t
, k ≥ 1

ukn(x, t) = c2L−1tt (4ukn−1(x, t)), n ≥ 1
(13)

For k = 1, We have:

(P 1
SBA)

{
u10(x, t) = u1(x, 0)

u1n(x, t) = c2L−1tt (4ukn−1(x, t)), n ≥ 1
(14)



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1264

We obtain : 

u10(x, t) = ϕ(x) + tφ(x)

u11(x, t) = c2(ϕ2(x)
t2

2!
+ φ2(x)

t3

3!
)

u12(x, t) = c4(ϕ4(x)
t4

4!
+ φ4(x)

t5

5!
)

u13(x, t) = c6(ϕ6(x)
t6

6!
+ φ6(x)

t7

7!
)

...

u1n(x, t) = c2n(ϕ2n(x)
t2n

(2n)!
+ φ2n(x)

t2n+1

(2n+ 1)!
)

We have 
ϕ(x) = sin

(πx
L

)
⇒ ϕ2n(x) = (−1)n

(π
L

)2n
sin
(πx
L

)
φ(x) = π sin

(πx
L

)
⇒ φ2n(x) = π(−1)n

(π
L

)2n
sin
(πx
L

)
Then

u1n(x, t) = sin
(πx
L

)(−1)n

(
cπt

L

)2n

(2n)!
+ (−1)n

L

c

(
cπt

L

)2n+1

(2n+ 1)!


ϕ1
m(x, t) =

m−1∑
n=0

u1n(x, t)

Let’s put
Then the approached solution at the the first step is:

u1(x, t) = lim
m→+∞

ϕ1
m(x, t)

= sin
(πx
L

)
cos

(
cπt

L

)
+
L

c
sin

(
cπt

L

)

such us Ñ(uk(x, t)) = 0, ∀k ≥ 0 at the step k ,we have :

(P kSBA)

 uk0(x, t) = uk(x, 0) + t
∂uk(0, x)

∂t
, k ≥ 1

ukn(x, t) = c2L−1tt (4ukn−1(x, t)), n ≥ 1
(15)



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1265

By unfolding : 

uk0(x, t) = ϕ(x) + tφ(x)

uk1(x, t) = c2(ϕ2(x)
t2

2!
+ φ2(x)

t3

3!
)

uk2(x, t) = c4(ϕ4(x)
t4

4!
+ φ4(x)

t5

5!
)

uk3(x, t) = c6(ϕ6(x)
t6

6!
+ φ6(x)

t7

7!
)

· · ·

ukn(x, t) = c2n(ϕ2n(x)
t2n

(2n)!
+ φ2n(x)

t2n+1

(2n+ 1)!
)

Then

ukn(x, t) = sin(
πx

L
)((−1)n

( cπtL )2n

(2n)!
+ (−1)n

L

c

( cπtL )2n+1

(2n+ 1)!

ϕkm(x, t) =
m−1∑
n=0

ukn(x, t)

Then the approached solution at the the first step is:

uk(x, t) = lim
m→+∞

ϕkm(t)

= sin(
πx

L
)(cos(

cπt

L
) +

L

c
sin(

cπt

L
)

Therefore, we obtain the exact solution of the problem (P1) :

u(x, t) = lim
k→+∞

uk(x, t)

= (cos(
cπt

L
) +

L

c
sin(

cπt

L
)) sin(

πx

L
)

• Solving by Fourier method
We find all solutions of the wave equation (P1) with the general form:

u(x, t) = T (t)X(x) (16)

for some function X(x) that depends on x but not t and some function T (t) that
depends only on t but not x.



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1266

Substitute equation (16) into the one-dimensional equation (10), we get

X(x)T
′′
(t) = c2T (t)X

′′
(x)⇐⇒ T

′′
(t)

c2T (t)
=
X
′′
(x)

X(t)
(17)

Such
T
′′
(t)

c2T (t)
depends on t and

X
′′
(x)

X(t)
depends on x, we can put:

T
′′
(t)

c2T (t)
=
X
′′
(x)

X(t)
= −k2 (18)

Then


T
′′
(t) = −(ck)2T (t)

and

X
′′
(x) = −k2X(x)

⇐⇒


T
′′
(t) + (ck)2T (t) = 0

and

X
′′
(x) + k2X(x) = 0

(19)

Then


T (t) = A cos(ckt) +B sin(ckt)

and
X(x) = C cos(kx) +D sin(kx)

(20)

where A, B, C and D are arbibrairy constantes.
We have

u(x, t) = (A cos(ckt) +B sin(ckt)) (C cos(kx) +D sin(kx)) (21)

Let’s calculate the constantes


u(t, 0) = 0

and
u(t, L) = 0

=⇒
{

C = 0
D sin(kL) = 0

D sin(kL) = 0 =⇒
{

D 6= 0
sin(kL) = 0

sin(kL) = 0⇐⇒ k =
nπ

L
(n ∈ Z)

Then
∀n ≥ 1, we have



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1267

un(x, t) = (An cos(
cnπt

L
) +Bn sin(

cnπt

L
)) sin(

nπx

L
) (22)

Then

u(x, t) =

+∞∑
n=1

(An cos(
cnπt

L
) +Bn sin(

cnπt

L
)) sin(

nπx

L
) (23)

We can noticed that if we choose k2 instead of −k2 the solution of the equation

X
′′
(x) + k2X(x) = 0 (24)

which is

X(x) = Aekx +Be−kx

don’t verify the initial condition:

u(t, 0) = u(t, L) = 0

So choosing k2 is impossible.

We have

∂u(x, t)

∂t
=

+∞∑
n=1

(
−cnπ

L
An cos(

cnπt

L
) +

cnπ

L
Bn sin(

cnπt

L
)

)
sin(

nπx

L
) (25)

For t = 0, we have

+∞∑
n=1

cnπ

L
Bn sin(

nπx

L
) = φ(x) (26)

u(0, x) = ϕ(x)⇐⇒ ϕ(x) =
+∞∑
n=1

An sin
(nπx
L

)
(27)

We have


An =

2

L

∫ L

0
ϕ(z) sin

(nπz
L

)
dz

and

Bn =
2

cnπ

∫ L

0
φ(z) sin

(nπz
L

)
dz

(28)



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1268

3.2. Problem 2

Let’s consider the following wave’s model :

(P2)



∂2u(x, y, t)

∂t2
= c2 4 u(x, y, t), c > 0

u(x, y, 0) = f1(x, y)
ut(x, y, 0) = f2(x, y)
u(0, y, t)) = h1(0, y, t)
u(L, y, t)) = h2(L, y, t)
u(x, 0, t)) = g1(x, 0, t)
u(x, l, t)) = g2(x, l, t)

(29)

Where 

4u(x, y, t) =
∂2u(x, y, t)

∂x2
+
∂2u(x, y, t)

∂y2

f1(x, t) = sin
(πx
L

)
sin
(πy
l

)
f2(x, t) = 0
h1(x, t) = 0
h2(x, t) = 0
g1(x, t) = 0
g2(x, t) = 0

(30)

Solving yhe wave equation involves identifyinf the functions u(x, y, t) that solve the
partial differential equation that represent the amplitude of the wave at any position x
and y at any time t.
• Solving by SBA method

By Integrating (29) we get the Adomian canonical form [2, 9] of the problem (29) :

u(x, y, t) = u(x, y, 0) + t
∂u(x, y, 0)

∂t
+ c2L−1tt (4u(x, y, t)) (31)

Applying successive approximations method to (31), we get :

uk(x, y, t) = uk(x, y, 0)+t
∂uk(x, y, 0)

∂t
+c2L−1tt (4uk(x, y, t))+Ñ(uk−1(x, y, t)), k ≥ 1 (32)

where N∼(uk(x, y, t)) = 0 ∀k ∈ N

Applying the SBA algorithm to (32), we get :



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1269

(P kSBA)

 uk0(x, y, t) = uk(x, y, 0) + t
∂uk(x, y, 0)

∂t
+ Ñ(uk−1(x, y, t)), k ≥ 1, k ≥ 1

ukn(x, t) = c2L−1tt (4ukn−1(x, y, t)), n ≥ 1
(33)

For k = 1, We have:

(P 1
SBA)

 u10(x, y, t) = u1(x, , y, 0) + t
∂u1(x, y, 0)

∂t
u1n(x, t) = c2L−1tt (4ukn−1(x, y, t)), n ≥ 1

(34)

We obtain: 

u10(x, t) = sin(πxL ) sin(πyL )

u11(x, y, t) = −c2( π2

L2 + π2

l2
) sin(πxL ) sin(πyl )

t2

2!

u12(x, y, t) = c4( π
2

L2 + π2

l2
)2 sin(πxL ) sin(πyl )

t4

4!

u13(x, y, t) = −c6( π2

L2 + π2

l2
)3 sin(πxL ) sin(πyl )

t6

6!
...

u1n(x, y, t) = (−1)nc2n6( π
2

L2 + π2

l2
)n sin(πxL ) sin(πyl )

t2n

(2n)!

Let’s put

ϕ1
m(x, y, t) =

m−1∑
n=0

u1n(x, y, t)

ϕ1
m(x, y, t) = sin

(πx
L

)
sin
(πy
l

)m−1∑
n=0

(−1)n

(
cπt

Ll

√
L2 + l2

)2n

(2n)!

Let’s put
Then the approached solution at the the first step is:

u1(x, y, t) = lim
m→+∞

ϕ1
m(x, y, t)

= sin
(πx
L

)
sin
(πy
l

)
cos

(
cπt

Ll

√
L2 + l2

)



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1270

.
such us N∼(uk(x, y, t)) = 0 ∀k ≥ 0 at the step k ,we have :

(P kSBA)

 uk0(x, y, t) = uk(x, , y, 0) + t
∂uk(x, y, 0)

∂t
u1n(x, y, t) = c2L−1tt (4ukn−1(x, y, t)), n ≥ 1

(35)

By unfolding :

uk0(x, t) = sin
(πx
L

)
sin
(πy
L

)
uk1(x, y, t) = −c2

(
π2

L2
+
π2

l2

)
sin
(πx
L

)
sin
(πy
l

) t2
2!

uk2(x, y, t) = c4
(
π2

L2
+
π2

l2

)2

sin
(πx
L

)
sin
(πy
l

) t′4
4!

uk3(x, y, t) = −c6
(
π2

L2
+
π2

l2

)3

sin
(πx
L

)
sin
(πy
l

) t′6
6!

...

ukn(x, y, t) = (−1)nc2n6
(
π2

L2
+
π2

l2

)n
sin
(πx
L

)
sin
(πy
l

) t
′2n

(2n)!

Then
Let’s put

ϕkm(x, y, t) =

m−1∑
n=0

u1n(x, y, t)

ϕkm(x, y, t) = sin(
πx

L
) sin(

πy

l
)
m−1∑
n=0

(−1)n
( cπtLl
√
L2 + l2)2n

(2n)!

Then the approached solution at the the first step is:

uk(x, y, t) = lim
m→+∞

ϕkm(x, y, t)

= sin(
πx

L
) sin(

πy

l
) cos(

cπt

Ll

√
L2 + l2)

Therefore, we obtain the exact solution of the problem (P2) :

u(x, y, t) = lim
k→+∞

uk(x, y, t) (36)



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1271

= sin
(πx
L

)
sin
(πy
l

)
cos

(
cπt

Ll

√
L2 + l2

)

• Solving by Fourier method or separation of variables Method

We find all solutions of the wave equation (P2) with the general form:

u(x, y, t) = T (t)U(x, y) (37)

where U depends on x and y :

U(x, y) = X(x)Y (y) (38)

for some function X(x) that depends on x, some function (y) that depends on y and
some function T (t) that depends only on t but not x and y.

Substitute equation (37) and (38) into the two-dimensional equation (31) we get:



∂2u(x, y, t)

∂t2
= T

′′
(t)U(x, y)

∂2u(x, y, t)

∂x2
= T (t)

∂2U(x, y)

∂x2
∂2u(x, y, t)

∂y2
= T (t)

∂2U(x, y)

∂y2

and

T
′′
(t)U(x, y) = c2T (t)(

∂2U(x, y)

∂x2
+
∂2U(x, y)

∂y2
)

Then

T
′′
(t)

c2T (t)
=

∂2U(x, y)

∂x2
+
∂2U(x, y)

∂y2

U(x, y)
= −λ2 (39)

Then we have

T
′′
(t) + (λc)2T (t) = 0 =⇒ T (t) = α cos(λct) + β sin(λct) (40)

where α and β are arbibrairy constantes.
And we get



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1272

u(x, y, t) = (α cos(λct) + β sin(λct))U(x, y) (41)

Then we have

∂2U(x, y)

∂x2
+
∂2U(x, y)

∂y2

U(x, y)
= −λ2 =⇒ X

′′
Y (y) +X(x)Y

′′
(y)

X(x)Y (y)
= −λ2 (42)

Then

X
′′
Y (y) +X(x)Y

′′
(y) + λ2X(x)Y (y) = 0 (43)

Let’s put

X
′′
(x)

X(x)
=
Y
′′
(y) + λ2Y (y)

Y (y)
= −µ2 (44)

We obtain 
X
′′
(x) + µ2X(x) = 0

and

Y
′′
(y) + ς2X(y) = 0; ς2 = λ2 + µ2

(45)

Then


X(x) = A cos(µx) +B sin(µx)

and
Y (y) = C cos(ςy) +D sin(ςy)

(46)

where A, B, C and D are arbibrairy constantes.
We have

u(x, y, t) = ((α cos(λct) + β sin(λct))X(x)Y (y) (47)

Let’s calculate the constantes

With the initial condition

u(x, y, 0) = sin
(πx
L

)
sin
(πy
l

)
ut(x, y, 0) = 0
u(0, y, t)) = 0
u(L, y, t)) = 0
u(x, 0, t)) = 0
u(x, l, t)) = 0



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1273

we get



αX(x)Y (y) = sin
(πx
L

)
sin
(πy
l

)
X(0) = 0
X(L) = 0
Y (0) = 0
Y (l) = 0
βcλ = 0

=⇒



αX(x)Y (y) = sin
(πx
L

)
sin
(πy
l

)
X(0) = 0
X(L) = 0
Y (0) = 0
Y (l) = 0
βcλ = 0

we obtain by unfoilding
A = 0

sin(µL) = 0
C = 0

sin(ςl) = 0
β = 0

=⇒


A = 0

µ = nπ
L ;∀n ≥ 1
C = 0

ς = nπ
l ;∀n ≥ 1
β

Then we get 
X(x) = sin(nπxL )

and
Y (y) = sin(nπyl )

(48)

Then
∀n ≥ 1, we have

un(x, y; t) = ((αn cos(λnct) + βn sin(λnct)) sin(
nπx

L
) sin(

nπy

l
);∀n ≥ 1 (49)

and

λ2n = (
nπ

Ll
)2(L2 + l2) (50)

Let’s put

Tn(t) = ((αn cos(λnct) + βn sin(λnct));∀n ≥ 1 (51)

We obtain

u(x, y, t) =

+∞∑
n=1

Tn(t) sin(
nπx

L
) sin(

nπy

l
) (52)

Such



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1274


u(x; y, 0) = f1(x, y) = sin

(nπx
L

)
sin
(nπy

l

)
and

ut(x, y, 0) = f2(x, y) = 0

We have 
f1(x, y) =

+∞∑
n=1

αn sin
(nπx
L

)
sin
(nπy

l

)
and

αn =

∫ L

0.

∫ l

0
f1(x, y) sin

(nπx
L

)
sin
(nπy

l

)
dxdy

(53)

Then by unfolding ,we get
α1 =

∫ L

0.
sin2

(πx
L

)
dx

∫ l

0
sin2

(πy
l

)
dy = 1

and

αn =

∫ L

0.

∫ l

0
f1(x, y) sin

(nπx
L

)
sin
(nπy

l

)
dxdy = 0;∀n 6= 0

Therefore, we obtain the exact solution of the problem (P2) :

u(x, y, t) =
+∞∑
n=1

un(x, y, t)

= cos(λ1ct) sin
(πx
L

)
sin
(πy
l

)
where

λ2n = (
nπ

Ll
)2(L2 + l2) (54)

λ1 =
nπ
√
L2 + l2

Ll
(55)

Therefore, we obtain the exact solution of the problem (P2) :

u(x, y, t) = sin
(πx
L

)
sin
(πy
l

)
cos

(
cπt
√
L2 + l2

Ll

)
(56)

We can noticed that if we choose λ2 and µ2 instead of −λ2 and −µ2 the solution of
the equation

don’t verify the initial condition:



R. Yaro, Y. Paré , B. Abbo / Eur. J. Pure Appl. Math, 12 (3) (2019), 1260-1276 1275

u(0, y, t) = u(L, y, 0) = u(x, 0, t) = u(x, l, 0) = 0

So choosing λ2 and µ2 is impossible.

Where 

A1 =
2

L

∫ L

0
sin2

(πz
L

)
dz = 1

An =
2

L

∫ L

0
ϕ(z) sin

(nπz
L

)
dz = 0 if n 6= 1

and

Bn =
2

c

∫ L

0
sin2

(πz
L

)
dz =

L

c

Bn =
2

cnπ

∫ L

0
φ(z) sin

(nπz
L

)
dz = 0 if n 6= 1

We obtain



A1 = 1
An = 0 ∀ n 6= 1

and

Bn =
L

c
Bn = 0 ∀ n 6= 1

Therefore, we obtain the exact solution of the problem (P1) :

u(x, t) =

+∞∑
n=1

(
An cos(

cnπt

L
) +Bn sin(

cnπt

L
)

)
sin
(nπx
L

)
(57)

=

(
cos(

cπt

L
) +

L

c
sin(

cπt

L
)

)
sin
(πx
L

)
3.3. Comparison of the solution

Method SBA method

Problem 1 u(x, t) =

(
cos

(
cπt

L

)
+
L

c
sin

(
cπt

L

))
sin
(πx
L

)
Problem 2 u(x, y, t) = sin

(πx
L

)
sin
(πy
l

)
cos

(
cπt
√
L2 + l2

Ll

)



REFERENCES 1276

Method Fourier method

Problem 1 u(x, t) =

(
cos

(
cπt

L

)
+
L

c
sin

(
cπt

L

))
sin
(πx
L

)
Problem 2 u(x, y, t) = sin

(πx
L

)
sin
(πy
l

)
cos

(
cπt
√
L2 + l2

Ll

)

4. Conclusion

The numerical method SBA and Fourier method or method of separation of variable
permitted us to resolve some partial differential equations in this paper.

In this paper, we showed that using the both methods, we get the same solution.
There are then some very powerful numerical tools of analysis for the resolution of

partial differential equations.

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