Gen. Math. Notes, Vol. 4, No. 2, June 2011, pp. 37-48
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
Numerical Solution of Volterra-Fredholm
Integro-Differential Equation by Block Pulse
Functions and Operational Matrices∗
Leyla Rahmani1 , Bijan Rahimi1 and Mohammad Mordad2
1
Department of Mathematics, Faculty of science, Islamic Azad University,
Takestan branch, Takestan, Iran
E-mail: l.rahmani@tiau.ac.ir
E-mail: bigrahimi@gmail.com
2
Department of Mathematics, Faculty of science, Islamic Azad University,
Islamshahr branch, Islamshahr, Tehran, Iran
E-mail: mmordad25@yahoo.com
(Received:11-5-11/Accepted:20-6-11)
Abstract
In this paper, Block Pulse Functions and their operational matrices are
used to solve Volterra-Fredholm integro-differential equation (VFIDE). First
the equation is integrated over interval [0, x] and then Block Pulse functions
are used to obtain numerical solution. Some theorems will prove convergence
of the method. Some numerical examples are included to illustrate accuracy of
the method.
Keywords: Volterra-Fredholm integro-differential equation, Block Pulse
Functions, Operational matrix.
1
Introduction
Integro-differential equations have been discussed in many applied fields, such
as biological, physical and engineering problems. Therefore, their numerical
treatment is desired. Many numerical methods used to solve such equations
[1, 3, 4, 6, 7, 8, 10, 12, 14]. In this paper, we propose a new numerical method
38
Bijan Rahimi et al.
to solve the linear VFIDE:

Z x
Z 1
 0
f (x) = g(x) +
k(x, t)f (t) dt +
l(x, t)f (t) dt 0 ≤ x, t < 1,
0
0

f (0) = a0 ,
(1)
where f (x) and g(x) are in L2 ([0, 1)) and k(x, t) and l(x, t) belong to L2 ([0, 1)×
[0, 1)). Moreover k(x, t), l(x, t) and g(x) are known and f (x) is unknown. We
assume Eq. (1.1) has a unique solution.
The paper is organized as follows: In section 2, we describe Block Pulse
functions, their properties and their operational matrices [5, 15]. In section 3,
we propose a new numerical method for solving linear VFIDE. In section 4 we
analysis the error. Finally in section 5 we apply the proposed method on some
examples to show the accuracy and efficiency of the method.
2
Review of Block Pulse Functions
A set of Block Pulse functions (BPFs) are usually defined in [0,1) as:

,
 1, mi ≤ t < (i+1)
m
φi (t) =

0,
otherwise,
(2)
where i = 0, 1, 2, . . . , m − 1 and m is an arbitrary positive integer. According
to (2.1), the unit interval [0,1) is divided into m equidistant subintervals and
the ith Block Pulse function φi has only one rectangular pulse of unit hight in
).
the subinterval [ mi , i+1
m
One of the important properties of BPFs is the disjointness of them, which
can directly be obtained from the definition of BPFs. Indeed:

 φi (t), i = j,
φi (t)φj (t) =
(3)

0,
i 6= j,
where i, j = 0, 1, . . . , m − 1.
The orthogonality of BPFs is derived immediately from:
Z 1
φi (t)φj (t) dt = h δij ,
0
where δij is the Kronecker delta and h =
1
.
m
(4)
39
Numerical Solution of Volterra-Fredholm...
The set of BPFs is complete, i.e. for every f ∈ L2 ([0, 1)), Parseval’s identity
holds:
Z 1
∞
X
2
f (t) dt =
fi2 kφi (t)k2 ,
(5)
0
where
1
fi =
h
Z
i=0
1
f (t)φi (t) dt,
i = 0, 1, . . . , m − 1.
(6)
0
An arbitrary function can be expanded in vector form as:
f (t) ' F T Φ(t) = ΦT (t)F.
(7)
where F = [f0 , f1 , ..., fm−1 ]T and Φ(t) = [φ0 (t), φ1 (t), ..., φm−1 (t)]T .
Now let k(x, t) be arbitrary L2 function of two variables on [0, 1) × [0, 1).
It can be expanded by BPFs as:
k(x, t) ' ΦT (x)KΨ(t),
(8)
where Φ(x) and Ψ(t) are m1 and m2 dimensional BPF vectors, respectively,
and K is the (m1 × m2 ) Block Pulse coefficient matrix with entries kij , i =
0, 1, ..., m1 − 1, j = 0, 1, ..., m2 − 1, as follows:
1
Z
1
Z
kij = m1 m2
k(x, t)φi (x)φj (t) dx dt.
0
(9)
0
In this paper for convenience, we put m1 = m2 = m.
Moreover, from the disjointness property

φ0 (t)
0
 0
φ1 (t)

Φ(t)ΦT (t) =  ..
..
 .
.
0
0
of BPFs, follows:

···
0

···
0

,
..
..

.
.
· · · φm−1 (t)
(10)
1
Z
Φ(t)ΦT (t) dt = h I,
(11)
ΦT (t)Φ(t) = 1,
(12)
Φ(t)ΦT (t)V = Ṽ Φ(t),
(13)
0
40
Bijan Rahimi et al.
where V is an m dimensional vector and Ṽ = diag(V ). Moreover:
b
ΦT (t)BΦ(t) = ΦT (t)B,
(14)
where B is an (m × m) matrix, B̂ is an m dimensional vector with elements
equal to the diagonal entries of matrix B. As represented in [2, 9]:
Z t
Φ(τ ) dτ ' P Φ(t),
(15)
0
where P is the following (m × m) matrix:

1 2 2
 0 1 2
h

P =  0 0 1
2  .. .. ..
 . . .
0 0 0
···
···
···
..
.
2
2
2
..
.
··· 1




,


(16)
called operational matrix of integration. So, the integral of every function f
can be approximated as follows:
Z t
f (τ ) dτ ' F T P Φ(t).
(17)
0
3
A Method to Solve VFIDE
Consider the following VFIDE:


0
Z
f (x) = g(x) +
x
l(x, t)f (t) dt 0 ≤ x, t < 1,
k(x, t)f (t) dt +
0

1
Z
0
(18)
f (0) = a0 ,
First, we integrate from (3.1) in the interval [0, x], we will have:
Z x
Z xZ u
f (x) = a0 +
g(u)du +
k(u, t)f (t)dtdu+
0
0
0
Z
x
Z
1
l(u, t)f (t) dtdu.
0
(19)
0
Now, we approximate functions f , g, k and l with respect to BPFs as follows:

f (x) ' F T Φ(x) = ΦT (x)F,



g(x) ' ΦT (x)G,
(20)
k(x, t) ' ΦT (x)KΦ(t),



l(x, t) ' ΦT (x)LΦ(t),
41
Numerical Solution of Volterra-Fredholm...
where vectors F and G and matrices K and L are BPFs coefficients of f , g, k
and l, respectively.
Substituting (3.3) into (3.2) we will have:
T
T
x
Z
T
T
Φ (x)F ' a0 + Φ (x)P G +
Φ (u)K
Z
0
Z
x
u
T
Φ(t)Φ (t)F dt du+
0
Z
Φ (u)L
1
T
0
Φ(t)ΦT (t)dt F du.
(21)
0
Substituting (2.10), (2.11), (2.12) and (2.14) into (3.4), results:
ΦT (x) F ' a0 ΦT (x) Φ(x) + ΦT (x) P T G+
Z
x
ΦT (u)K F̃ P Φ(u)du + hΦT (x)P T LF.
(22)
0
Put B = K F̃ P and apply (2.13), we obtain:
Z
x
Z
T
Φ (u)BΦ(u)du =
x
b ' ΦT (x)P T B,
b
ΦT (u)Bdu
0
0
b = PTB
b and using (2.13), above relation converts to:
putting A
b = ΦT (x)A
b = ΦT (x)AΦ(x),
ΦT (x)P T B
(23)
b Substituting (3.6) into (3.5), results:
where A = diag(A).
ΦT (x)F ' ΦT (x)(a0 I + A)Φ(x) + ΦT (x)(P T G + hP T LF ).
Now, with C = a0 I + A and using(2.13), we have:
b + ΦT (x)(P T G + hP T LF ),
ΦT (x)F ' ΦT (x)C
or
b + P T G + hP T LF,
F 'C
(24)
b can be written as follows:
where C
b = a0 1 + h2 K 0 F
C
where 1 = (1, 1, · · · , 1)T and
(25)
42
Bijan Rahimi et al.

1
k
4 11
0
0


2 00
 X

1
k
0
ki1

4 22

 i=1


 X
3 00
3 00
X

1

ki1
ki2
k
4 33
K0 = 
 i=1
i=2



..
..
..

.
.
.





m 00
m 00
m 00
 X
X
X

ki1
ki2
ki3
i=1
i=2
i=3
···
0
0
···
0
0
···
0
0
..
..
.
..
.
.
···
m
X
00
ki,m−1
1
k
4 mm


























i=m−1
X00
and
means that the first and last terms have factor 21 .
Substituting (3.8) into (3.7) and replacing ' with =, we will have:
(I − hP T L − h2 K 0 )F = (P T G + a0 1).
(26)
Solving the system of equations (3.9) we obtain m unknowns f0 , f1 , · · · , fm−1 .
Also, structure of K 0 shows that we do not need to evaluate kij for j > i.
4
Error Analysis
In this section, we analyse the error when a differentiable function f (x) is represented in a series of BPFs over the interval I = [0, 1). We need the following
theorem.
Theorem: Suppose f is continuous in I, is differentiable in (0, 1), and
0
there is a number M such that |f (x)| ≤ M , for every x ∈ I. Then
| f (b) − f (a)| ≤ M | b − a| ,
for all a, b ∈ I.
Proof. See [11].
43
Numerical Solution of Volterra-Fredholm...
Now, we assume that f (x) is a differentiable function on I such that
|f (x)| ≤ M . We define the error between f (x) and its BPFs expansion over
every subinterval Ii as follows:
0
ei (x) = fi − f (x), x ∈ Ii
where Ii = [ mi , i+1
).
m
It can be shown that:
Z i+1
Z
m
2
2
kei k =
ei (x) dx =
i+1
m
1
fi − f (x) dx =
fi − f (η) 2 , η ∈ Ii ,
i
i
m
m
m
(27)
where we used mean value theorem for integral. Using Eq.(2.5) and the mean
value theorem, we have:
i+1
m
Z
f (x) dx = m
fi = m
i
m
2
1
f (ζ) = f (ζ), ζ ∈ Ii .
m
(28)
Substituting (4.2) into (4.1) and using Theorem, we will have:
M2
M2
1
f (ζ) − f (η) 2 ≤
| ζ − η| 2 ≤ 3 .
kei k =
m
m
m
2
(29)
This leads to:
1
Z
2
Z
2
ke(x)k =
=
1
m−1
X
0
m−1
X
0
0
Z
1
e (x) dx =
2
ei (x) dx + 2
i=0
i=0
XZ
i≤j
ei (x) 2 dx
1
ei (x) ej (x) dx.
0
Since for i 6= j, Ii ∩ Ij = ∅, then
2
ke(x)k =
m−1
XZ 1
i=0
m−1
X
ei (x) dx =
kei k 2 .
2
0
(30)
i=0
Substituting (4.3) into(4.4), we have
ke(x)k 2 ≤
hence, ke(x)k =
O( m1 ),
M2
,
m2
where e(x) = fm (x) − f (x) and fm (x) =
m−1
X
i=0
fi φi (x).
44
5
Bijan Rahimi et al.
Numerical Examples
In this section, we present some examples and their numerical results. Examples 2 and 3 were used in [13] and [1], respectively. All computations are
performed by the Maple 9.5 software package.
Example 1: Consider the following VIDE:

Z x
0


f (t) dt,
 f (x) =
0


 f (0) = 1
with the exact solution f (x) = cosh(x). Indeed, in this example, we have
g(x) = 0, l(x, t) = 0 and k(x, t) = 1.
The numerical results are shown in table 1.
x
Table 1: Numerical results for Example 1
Exact solution
Approximate solution
Approximate solution
m = 32
m = 64
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.005004
1.020067
1.045339
1.081072
1.127626
1.185465
1.255169
1.337435
1.433086
1.000244
1.006111
1.020829
1.044527
1.077413
1.136067
1.191664
1.257743
1.334886
1.423773
1.000061
1.005193
1.019166
1.046811
1.080467
1.131772
1.186506
1.251676
1.341668
1.431547
Example 2: Consider the following VIDE [13]:

Z x
0


f (t) dt,
 f (x) = 1 −
0


 f (0) = 0
with the exact solution f (x) = sin(x). In this example, we have g(x) = 1,
l(x, t) = 0 and k(x, t) = −1.
45
Numerical Solution of Volterra-Fredholm...
x
Table 2: Numerical results for Example 2
Exact solution
Approximate solution
Approximate solution
m = 64
m = 128
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0.099833
0.198669
0.295520
0.389418
0.479426
0.564642
0.644218
0.717356
0.783327
0.007812
0.101383
0.194063
0.299980
0.387959
0.486243
0.565904
0.640595
0.720580
0.782319
0.003906
0.097500
0.197901
0.296263
0.391571
0.482844
0.562700
0.646312
0.717892
0.784773
The numerical results are shown in table 2.
Example 3: Consider the following linear FIDE [1]:

Z 1

0
−x
−1

 f (x) = −e + e − 1 +
f (t) dt,
0



f (0) = 1
with the exact solution f (x) = e−x . We have g(x) = −e−x + e−1 − 1, l(x, t) = 1
and k(x, t) = 0. This example was used in [1] by using spline functions with
m = 1 and m = 2, where m equals the order of spline functions. The absolute values of error in [1] with stepsize h = 0.1 at the point 0.4 are given as
1.5 × 10−2 and 1.8 × 10−3 for m = 1 and m = 2, respectively.
The numerical results are shown in table 3.
Example 4: Consider the following linear VFIDE:















0
f (x) = −2sin(x) − x2 sin(2x) + 2sin(2x) − 2xcos(2x) − 2ex
+5e
x−1
x
Z
+ 2x +
1
Z
ex−t f (t) dt,
cos(x + t)f (t) dt +
0
0
f (0) = 0
with the exact solution f (x) = x2 . In this example, we have g(x) = −2sin(x)−
x2 sin(2x) + 2sin(2x) − 2xcos(2x) − 2ex + 5ex−1 + 2x, l(x, t) = ex−t and
46
Bijan Rahimi et al.
x
Table 3: Numerical results for Example 3
Exact solution
Approximate solution
Approximate solution
m = 16
m = 64
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.904837
0.818731
0.740818
0.670320
0.606530
0.548812
0.496585
0.449329
0.406570
0.969719
0.910993
0.804005
0.755324
0.666636
0.588375
0.552766
0.487894
0.458378
0.404606
0.992248
0.903455
0.822608
0.737384
0.671399
0.601842
0.547986
0.498951
0.447261
0.407240
k(x, t) = cos(x + t).
The numerical results are shown in table 4.
x
Table 4: Numerical results for Example 4
Exact solution
Approximate solution
Approximate solution
m = 32
m = 64
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0.01
0.04
0.09
0.16
0.25
0.36
0.49
0.64
0.81
6
0.000493
0.012240
0.041571
0.088486
0.152985
0.266336
0.371862
0.494971
0.635662
0.793932
0.000123
0.010384
0.038226
0.092926
0.158856
0.257994
0.362012
0.483610
0.647692
0.807378
Conclusion
The proposed method for solving linear VFIDE was based on BPFs and
their operational matrix. This method converts equation to a linear system of
algebraic equations. It, s accuracy is shown on some examples. As examples
Numerical Solution of Volterra-Fredholm...
47
show error will decrease as m increases. The method is easy to apply and it, s
accuracy is comparable with other methods.
Acknowledgements
The authors are highly grateful to the Islamic Azad University, Takestan
Branch, Iran, for giving all types of support in conducting this research
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