Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011
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Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011

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Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011 Taylor Series Method for Solving Linear Fredholm Integral Equation of Second Kind Using MATLAB Hameed H Hameed H a y d e r M A b b a s Z a h r a a A M o h a m m e d Foundation Of Technical Baghdad University, College Almustansiriya university Education Of Science for women. College Of Science Technical Institute Of Alsuwayra Department Of Mathematics Department Of Astronomy Abstract This paper presents a method to find the approximation solution for linear ferdholm integral b equation : by using Taylor series expansion to approximate the kernel y(x)  f (x)   k(x,t) y(t) a N k(x,t) as a summation of multiplication functions f (x) by g (t) i.e. k(x,t)  f (x)g (t) then use n n  n n n 1 the degenerate kernel idea to solve the fredholm integral equation .In this paper we solve the above integral equation with a  0 and b  1,  is a real number, f(x) and k(x,t) are real continues functions . We have deduced a MATLAB program to solve the above equation, we have used MATLAB (R2008a) to perform this program . The presented method has high accurate when compare its results with the other analytical methods results . 1-Introduction Integral equations, that is, equations involving an unknown function which appear under an integral sign.

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Publié le 04 juin 2014
Nombre de lectures 305
Langue English

Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011

Taylor Series Method for Solving Linear Fredholm
Integral Equation of Second Kind Using MATLAB
Hameed H HameedHayder M bbAsa Zahraa A Mohammed
Foundation Of TechnicalBaghdad University, CollegeAlmustansiriya university
Education OfScience for women.College Of Science
Technical Institute Of AlsuwayraDepartment Of MathematicsDepartment Of Astronomy

Abstract
This paper presents a method to findthe approximation solution for linear ferdholm integral
b
equation :y(x)f(x)k(x,t)y(t)using Taylor series expansion to approximate the kernel by

a
N
nn
k(x,t) as a summation of multiplication functionsf(x)byg(t)i.e.k(x,t)f(x)g(t)then use
n n
n1
the degenerate kernel idea to solve the fredholm integral equation .In this paper we solve the above integral
equation witha0andb1,is a real number,f(x)andk(x,t)are real continues functions .
We have deduced a MATLAB program to solve the above equation, we have used MATLAB
(R2008a) to perform this program .
The presented method has high accurate when compare its results with the other analytical methods
results .

1Introduction
Integral equations, that is, equations involving an unknown function which appear
under an integral sign. Such equations occur widely in divers areas of applied
mathematics ,they offer a powerful technique for using the integral equation rather than
differential equations is that all of the conditions specifying the initial value problems or
boundary value problems for a differential equation can often be condensed into a single
integral equation . So that any boundary value problems can be transformed into
fredholm integral equation involving an unknown function of only once variable.
This reduction of what may represent a complicated mathematical model of
physical situation into a single equation is itself a significant step , but there are other
advantages to be gained by replacing differentiation with integration ,some of these
advantages arise because integration is a smooth process ,a feature which has significant
implication when approximation solution are sough .
2Importance of the work
The main purpose is to produce of this paper a new approximation solution by approximate
the kernelk(x,t) usingTaylor series expansion for the function of two variables and making it
as a degenerate kernel then finding the solution of ferdholm integral equation .
3A Review of previous works
There are many papers deal with numerical and approximate solutions of fredholm integral
equations, Akber and Omid (Zabadi & Fard, 2007) produced an approach via optimization
methods to find approximation solution for non linear fredholm integral equation of first kind,
while Vahidi and Mokhtari produced the system of linear fredholm integral equation of second
kind was handled by applying the decomposition method(Vahidi & Mokathri, 2008).Babolian
and Sadghi proposed the parametric form of fuzzy number to convert a linear fuzzy fredholm



integral equation of second kind to a linear of integral equation of the second kind in crisp case
(Babolian & Goghory, 2005).
Hana and others considered the problem of numerical inversion of fredholm integral
equation of the first kind via piecewise interpolation(Hannaet al.,Maleknejad and others 2005).
proposed to use the continuous legender wavelets on the interval0,1to solve the linear second
kind integral equation (Maleknejadet al., 2003),the numerical methods to approximate the
solution of system of second kind fredholm integral equation were proposed by Debonis and
Laurita (Debonis & Laurita , 2008).
Chanet al.,on polynomial interpolation to approximate matricespresented a scheme based
Afrom the discretizetion the integral operators(Chanet al.,2002) and cubic spline interpolations
has been proposed to solve integral equations by Kumar and Sangal (Kumar and Sangal, 2004)
3 Separate or degenerate kernel
A kernelk(x,t) is called separable if it can be expressed as the sum of a finite number of
terms ,each of which is the product of a function of only and a function oftonly i.e.
n

k(x,t)g(x)h(t)(Raisinghania, 2007).
i i
i1
4 Solution of ferdholm integral equation of second kind with degenerate kernel
(Raisinghania, 2007).
Consider the non homogenous fredholm integral equation of second kind
b
y(x)f(x)k(x,t)y(t)dt..................................................(1)

a
Since the kernelk(x,t)is degenerate or separate we take
n

k(x,t)f(x)g(t).........................................................(2)
i i
i1
Where the functionsf(x) assumedto be linearly independent, using (2) and (1) reduces
i
b
n
toy(x)f(x)[f(x)g(t)]y(t)dt...............................(3)
 
i i
i1
a
b
n
ory(x)f(x)f(x)g(t)y(t)dt...................(4)
 
i i
i1
a
n

using (4) ,(3) reduces toy(x)f(x)C f(x)........................................(5)
i i
i1
where constantsC(i1,2,3,........,n)are to be determined in order to find the solution of (1) in
i
the form given by (5) . We now proceed to evaluateC'sas follows:
i
n

from (5) we havey(t)f(t)C f(t)........................................(6)
i i
i1
substituting the values ofy(x)andy(t)given in (5) and (6) respectively in (3) , we have
b
n nn
f(x)C f(x)f(x)f(x)g(t){f(t)C f(t)}dt
  
i ii ii i
i1i1i1
a
b b
n nn
orC f(x)f(x){g(t)f(t)dtC g(t)f(t)dt}............................(7)
  
i ii ij ij
i1i1j1
a a

Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011

b
b
Now, letand( )f(t)dt......
g(t)f(t)dt ijgitj.... .......................(8)

i i
a a
Whereandare known constant, then (7) may simplify as
i ij
n nn nn
i ii iij ji iiij j
C f(x)f(x){ C}orf(x){C C}0, but the
i1i1j1i1j1
n

functionsf(x)are linearly independent ,thereforeC C0 i1,2,3,...,nor
i ii ijj
j1
n
iij ji
C Ci1,2,3,...,n .......................(9)
j1
Then we obtain the following system of linear equations to determineC,C,...,C
1 2n
(1)CC...C
11 112 212n1
C(1)C...C
21 122 22n n2
:
:
CC...(1)C
n1 1n2 2nn nn
The determinateD( )of system
1
11 121n
1 
21222
n
D()   ...........................(10)

 
121
n nnn

Which is a polynomial inof degree at most (n) ,D( )is not identically zero ,since when
0,D()1.to discuss the solution of (1) , the following situation arise:
Situation I : when at least on right member of the system(), (),....,()is non zero
1 2n
,the following two cases arise under this situation :
(i) ifD( )0,then a unique non zero solution of system(), (),....,()exist
1 2n
and so (1) has unique non zero solution given by (5) .
(ii) ifD( )0,then the equations(), (),....,()have either no solution or they
1 2n
possess infinite solution and hence (1) has either no solution or infinite solution.
Situation II: whenf(x)0 ,then (8) shows that0forj1,2,...,n.Hence the
j
equations(), (),....,()reduce to a system of homogenous linear equation .Thefollowing
1 2n
two cases arises under this situation
(i) ifD()0,then a unique zero solutionCC...C0of the system
1 2n
(), (),....,()exist and so from (5) we see that (1) has unique zero solution
1 2n
y(x)0.

(ii) ifD( )0,then the system(), (),....,()posses infinite non zero
1 2n
solutions and so (1) has infinite non zero solutions , those value offor which
D() 0are known as the eigenvalues and any nonzero solution of the
b
homogenous fredholm integral equationy(x)k(x,t)y(t)dt is known as a

a
corresponding eigenfunction of integral equation .
Situation III: whenf(x)0but
b bb
g(x)f(x)dx0,g(x)f(x)dx0,...,g(x)f(x)0i.e.f(x)is orthogonal to all the
 
1 2n
a aa
functionsg(t),g(x),...,g(x),then (8) shows that,,...,reduce to a system of
1 2n1 2n
homogenous linear equations. The following two cases arise under this situation .
(i) ifD()0,then a unique zero solutionCC...C0then (1) has
1 2n
only unique solutiony(x)0.
(ii) IfD() 0then the system(), (),....,()possess infinite nonzero
1 2n
solutions and (1) has infinite nonzero solutions .The solution corresponding to the
eigenvalues of.
41 Example (1) : find the analytical solution of the following integral equation

1

y(x)1(13xt)y(t)dt

Solution :sincek(x,t)13xtthat mean
0
k(x,t)separated functionf(x)1,f(x)3x g(t)1,g(t)t,f(x)1,1, from
1 21 2
equation (6) we obtainy(x)1[C3xC],then
1 2
C1C
111 12   11 121 1
1 1
  
      
1C1C
21 212 2 21 212 2
1
1 11
3 1
2
x
11d1, 3xdx ,xdx , 3x dx 1
 
12 2122
2 2
0 00 0
3
1 1
0C1
115
2
dx1 ,xdx ,then  implies thatC,
 1
1 2 
1
1
2
0 02C2 3
2
2
2 5
Candy(x)1[2x].
2
3 3
5 Taylor series of function with two variables (Karris, 2004)
Letf(x,y)is a continuous function of two variablesandy,then the Taylor series
expansion of functionfat the neighborhood of any real numberawith respect to the variable
yis :
n n
(ya)
n
taylor(f,y,a)f(x,ya)
non!y

Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011

m nn
(ya)th
andtaylor(f,y,a,m)f(x,ya)that mean themterms of Taylor
n
non!y
expansion to the function at the neighborhoodawith respect to the variabley
51 Examples
xy
Example (2) :The five terms ofTaylor series expansion of the functionf(x,y)eat
1)a0and 2)a3as the following :
1 11
2 23 34 4
1)taylor(f,y,0,5)1xyy xy xy x
2 6 24
2)
3x3x12 23x13 33x134 4x
taylor(f,y,3,5)e(y3)xe(y3)x e(y3)x e(y3)x e
2 6 24
xy
Example (3):Compare the values of the function f(x,y)e at the point (2,4) with its
Taylor expansion of three terms .
xy8
Solution: f(x,y)e andf(2,4)e2980.9
2
y
2y y2 2y
the three terms ofTaylor expansion istaylor(f,x,2,3)ey(x2)e(x2)e,
2
then the Taylor expansion at (2,4) is 2981.
6Remark:The Taylor series must be calculated at the point or close to the point that we want
the value of the function at that point as shown in example (3).

7Our work :since any continuous functionk(x,t)of two variables can be approximated by the
Taylor expansion therefore , then this function can be separated as a summation of product terms
n
off(x) byg(t) i.e.k(x,t)f(x)g(t)

i ii i
i1
xt
71 Example (4) : iff(x,t)e,then the Taylor expansion with respect the variabletata0
1 11
2 23 34 4
with five terms istaylor(f,t,0,5)1txt xt xt x,that mean
2 6 24
1 11
2 34
f(x)1,f(x)x,f(x)x,f(x)x,f(x)x,and
1 23 4 5
2 6 24
2 3 4
g(t)1,g(t)t,g(t)t,g(t)t,g(t)t
1 2 34 5

711 The Algorithm of separation kernel and solution of fredholm integral equation
a input the kernelk(x,t)
b input the functionf(x)
c input the value of
d input the valuesaandb
e input the number of Taylor series' termsN
f calculate the Taylor expansion ofk(x,t)with respectt,
i i
N
(ta)
taylor(f,t,a,N)f(x,ta)
i
ioi!y
g fromffindf(x)andg(t),i0,1,,N
i i

b b
h calculateg(x)f(x)dx i,j1,2,,Nandg(x)f(x)dx
 
ij ij ii
a a
,j1,2,,N
1 
 11121N
 
1 
21222N
 
i calculate the matrixA
 
  
 
1
N1N2NN
j calculate the determinateD(A)of matrixA
k iff(x)0go to step n
l ifD(A)0the system has infinite number of solutions ,go to step s
m the system has unique solutionCC...C0,go to step s
1 2N
n if0go to step r
i
o ifD(A)0, the system has infinite number of solutions ,go to step s
p the system has unique solutionCC...C0
1 2N
q ifD(A)0,the system has no real solution, go to step s
n
1T

r the solution of system isCAtheny(x)f(x)C f(x)
i iji ii
i1
s end
712 Numerical results
In this section we present numerical results by solve the ferdholm integral equation by our
approximation solution then comparison it with analytical solution
7121 Examples
1
Example (5) :integral equationthe approximation solution ofy(x)1sin(xt)dtas

0
2 34
t tt
following :taylor(sin(xt),t,5)sin(x)tcos(x)sin(x)cos(x)sin(x), that
2 624
implies
11 1
f(x)sin(x),f(x)cos(x),f(x)sin(x),f(x)cos(x),f(x)sin(x)
1 23 45
2 624
and
2 3 4
g(t)1,g(t)t,g(t)t,g(t)t,g(t)t, by using the previous algorithm and
1 2 34 5
the related MATLAB program the solution isy13.9878sin(x)2.3833cos(x),
alfa =
0.4597 0.84150.2298 0.14020.0192
0.3012 0.38180.1506 0.06360.0125
0.2232 0.23910.1116 0.03990.0093
0.1771 0.17170.0885 0.02860.0074
0.1467 0.13310.0733 0.02220.0061
beta =[1.0000 0.5000 0.3333 0.2500 0.2000]
A =
0.5403 0.84150.2298 0.14020.0192

Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011

0.3012 0.6182 0.1506 0.06360.0125
0.2232 0.23911.1116 0.03990.0093
0.1771 0.17170.0885 1.02860.0074
0.1467 0.13310.0733 0.0222 0.9939
C = [4.83872.6109 1.7935 1.36551.1020]
Y(x) =1+3.9878*sin(x)+2.3833*cos(x)
While the analytical solution by using the degenerate kernel was in Raisinghania. (2007)
y14.01sin(x)2.404 cos(x).
The following table shows the analytical and approximate results
Table (1) comparison betweenthe analytical solution and theapproximation solution of
1
y(x)1sin(xt)dt

0

Analytical solutionApproximate solution
X y1=1+4.01*sin(x)+2.404*cos(x)y2=1+3.9878*sin(x)_2.3833*cos(x) Error=abs(y1y2)
6.28318 3.4040052423.383305213 0.020700029
5.65487 5.301897875.272102379 0.029795491
5.02655 5.556612395.529102297 0.027510093
4.39823 4.070856554.056139771 0.014716779
3.76991 1.4121383551.415836197 0.003697843
3.14159 1.4040026211.383302606 0.020700015
2.51327 3.3018966743.272101186 0.029795487
1.88496 3.5566130743.529102973 0.027510101
1.25664 2.0708588542.056142058 0.014716796
0.62832 0.5878586020.584160778 0.003697823
0 3.4043.3833 0.0207
0.628318 5.3018954775.272099993 0.029795484
1.256637 5.5566137595.529103649 0.02751011
1.884955 4.0708611584.056144345 0.014716813
2.513274 1.4121444421.415842246 0.003697803
3.141592 1.4039973791.383297394 0.020699985
3.76991 3.301894283.2720988 0.02979548
4.398229 3.5566144433.529104325 0.027510118
5.026547 2.0708634632.056146632 0.014716831
5.654866 0.5878525140.58415473 0.003697784
6.283184 3.4039947583.383294787 0.020699971

The following figure shows comparison between of the two results


Fig(1) the analytical and approximation solutions results of integral equation
1
y(x)1sin(xt)dt

0
Example (6) :The approximation solution of the integral equation
1 1
2
y(x)x{xt(xt) }dtas the following:

0
1
2
k(x,t)xt(xt)
1 11 11
1 121354
2 22 22
taylor(k,t,1,5)xx(xx)(t1)x(t1)x(t1)x(t1)
2 8 16 128
That implies
1 11 11
11 15
2 22 22
f(x)xx,f(x)xx,f(x)x,f(x)x,f(x)x
1 23 45
2 8 16128
2 3 4
g(t)1,g(t)(t1),g(t)(t1) ,g(t)(t1) ,g(t)(t1) .
1 23 4 5
By using the algorithm and the MATLAB program we obtain the solution is
1
2
y3.6601x2.3743x
alfa =
1.16670.8333 0.08330.0417 0.0260
0.43330.3000 0.03330.0167 0.0104
0.23570.1595 0.01900.0095 0.0060
0.15160.1008 0.01270.0063 0.0040
0.10720.0703 0.00920.0046 0.0029
beta =[ 0.50000.1667 0.08330.0500 0.0333]
A =
0.16670.8333 0.08330.0417 0.0260
0.43331.3000 0.03330.0167 0.0104
0.23570.1595 1.01900.0095 0.0060
0.15160.1008 0.01271.0063 0.0040
0.10720.0703 0.00920.0046 1.0029
C =[3.0452 1.1206 0.60550.3874 0.2729]

Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(19): 2011

Y = 3.6601*x+2.3743*x^(1/2)
1
96 60
2
While the analytical solution was([ 9]yxx
26 26
Table (2) comparison between the analytical solution and the approximation
1 1
2
solution ofy(x)x{xt(xt) }dt

0
Analytical solutionApproximate solution
x Error=abs(y1y2)
y1=(90/26)x+(60/26)x^.5 y2=3.6601x+2.3743x^.5
0 00 0
0.5 3.4779387263.508933631 0.030994905
1 66.0344 0.0344
1.5 8.3647958578.398061748 0.033265891
2 10.6481851410.67796726 0.029782117
2.5 12.8795511512.90434792 0.024796778
3 15.073963415.09270823 0.01874483
3.5 17.2403739117.25225857 0.011884659
4 19.3846153819.389 0.004384615
4.5 21.5107392521.50710089 0.003638363
5 23.6216953323.6095962 0.012099134
5.5 25.7197104925.69877707 0.020933423
6 27.8065147927.7764235 0.030091295
6.5 29.8834840529.84395102 0.039533039
7 31.9517337931.90250734 0.049226457
7.5 34.0121833633.95303834 0.059145014
8 36.0656010635.99633452 0.069266535
8.5 38.112636838.03306454 0.07957226
9 40.1538461540.0638 0.090046154

The following figure shows the comparison between the two results


Fig (2) the analytical and approximation solutions results of integral equation

1 1
2
y(x)x{xt(xt) }dt

0
713 Remark: We find Taylor expansion of the kernel at the pointaat1 instead
adivision by zero.0 toavid the
Conclusion and future work
The method of approximate kernel by Taylor expansion is a new method to solve
the fredholm integral equation of second kind, and it has high accurate results, in this
paper we have approached to solve the fredholm integral equation with integration limits
from 0 to 1 just.
In future work we hope to solve the fredholm integral equation of second kind with
integration limits froma tob whatever the values ofaandb.
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