Applied Mathematical Sciences, Vol. 6, 2012, no. 17, 807 - 814
A Computational Method for Solving Linear Volterra Integral Equations Farshid Mirzaee Department of Mathematics, Faculty of Science Malayer University, Malayer, 65719-95863, Iran
[email protected],
[email protected] Abstract The aim of the present paper is to introduce numerical method for solving linear Volterra integral equations of the second kind.The main idea is based on the adaptive Simpson’s quadrature method .The technique is very effective and simple.We show that our estimates have a good degree of accuracy.
Keywords: Volterra integral equations; Quadrature;Simpson’s quadrature method
1.
Introduction
Modified Simpson’s method for solving integral
xi+2
f (x)dx = xi
xi+2 xi
f (x)dx is as followes:
h [fi + 4fi+1 + fi+2 ] 3 h7 (6) h4 [fi − fi+2 f (ζi ); ]− + 180 1260
ζi ∈ (xi , xi+2 ). (1)
In general for integral [a,b] we have:
−1 N 2
b
f (x)dx = a
i=0
x2i+2 x2i
N
−1 2 4h h f (x)dx f (a) + f2i+1 3 3 i=0
N 2
−1 h4 2h h [f (a) − f (b)], + f2i + f (b) + 3 i=1 3 180
where N is even.
(2)
F. Mirzaee
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2.
Development of modified Simpson’s method
Consider linear Volterra integral equations of the second kind: y(t) = x(t) +
t
a ≤ t ≤ b,
k(t, s)y(s)ds;
(3)
a
where k(t,s) and x(t) are known functions, but y(t) is an unknown function [1-4]. Now ,for solving the equations (3) with repeated modified Simpson’s method ,we consider two cases.
Case 1.The partial derivatives k(t,s) does not exist: In this case ,we solve equations (3) with repeated Simpson’s method, so we have: h [k(t, s2i )y2i y(t) = x(t) + 3 i=0 j−1
+4k(t, s2i+1 )y2i+1 + k(t, s2i+2 )y2i+2 ];
(4)
Hence for t = t0 , t1 , · · · , tN ,we get the following system of equations: h = x2j + [k2j,2i y2i + 4k2j,2i+1 y2i+1 3 i=0 j−1
y2j
+k2j,2i+2 y2i+2 ];
j = 1(1)(
N ). 2
(5)
Set y2i+1
y2i + y2i+2 , 2
(6)
then we have
y2j
j−1 x2j + h3 (k2j,0 + 2k2j,1 )y0 + 2h i=1 (k2j,2i−1 + k2j,2i + k2j,2i+1 )y2i 3 ; = h 1 − 3 (2k2j,2j−1 + k2j,2j ) N j = 1(1)( ), (7) 2
where y(a) = y0 = x(a) = x0 .
Solving linear Volterra integral equations
809
Case 2.The partial derivatives k(t,s) exist: In this case ,we solve equations (3) with repeated modified Simpson’s method, so we have: h [k(t, s2i )y2i y(t) = x(t) + 3 i=0 j−1
+4k(t, s2i+1 )y2i+1 + k(t, s2i+2 )y2i+2 ] h4 [J (t, s0 )y0 + k(t, s0 )y0 + 3J (t, s0 )y0 + 180 +3J(t, s0 )y0 − J (t, s2j )y2j − k(t, s2j )y2j −3J (t, s2j )y2j − 3J(t, s2j )y2j ];
j = 1(1)(
N ), 2
(8)
where
∂ 2 k(t, s) ∂ 3 k(t, s) ∂k(t, s) , J (t, s) = , J (t, s) = ∂s ∂s2 ∂s3 must exist. By using equation (6) and for t = t0 , t1 , · · · , tN ,we get the following system of equations: J(t, s) =
h = x2j + [(k2j,2i + 2k2j,2i+1 )y2i + (2k2j,2i+1 + k2j,2i+2 )y2i+2 ] 3 i=0 j−1
y2j
+
h4 [k0,0 y0 + 3J0,0 y0 + 3J0,0 y0 + J0,0 y0 180
−k2j,2j y2j − 3J2j,2j y2j − 3J2j,2j y2j − J2j,2j y2j ] ; j = 1(1)(
N ). (9) 2
By taking three derivative from equation (3) we obtain
y (t) = x (t) +
t
a ≤ t ≤ b,
H(t, s)y(s)ds + k(t, t)y(t);
(10)
a
⇒ y0 = y (a) = x (a) + k(a, a)y(a),
y (t) = x (t) +
t
(11)
H (t, s)y(s)ds + H(t, t)y(t)
a
+T (t, t)y(t) + k(t, t)y (t);
a ≤ t ≤ b,
(12)
F. Mirzaee
810 ⇒ y0 = y (a) = x (a) + H(a, a)y(a) + T (a, a)y(a) +k(a, a)y (a),
y (t) = x (t) +
t
(13)
H (t, s)y(s)ds +
a
H (t, t)y(t) + V (t, t)y(t) +H(t, t)y (t) + T (t, t)y(t) + 2T (t, t)y (t) +k(t, t)y (t);
a ≤ t ≤ b,
(14)
⇒ y0 = y (a) = x (a) + H (a, a)y(a) + V (a, a)y(a) + H(a, a)y (a) +T (a, a)y(a) + 2T (a, a)y (a) +k(a, a)y (a); where H(t, s) =
∂k(t,s) ∂t
, H (t, s) =
∂ 2 k(t,s) , ∂t2 2
a ≤ t ≤ b.
H (t, s) =
(15)
∂ 3 k(t,s) , ∂t3
T (t, t) = dk(t,t) , T (t, t) = d Tdt(t,t) , V (t, t) = dH(t,t) ,x (t), x (t), x (t) must 2 dt dt exist. We solve equations (10) ,(12) and (14) with repeated modified Simpson’s method.By using (6) and for t = t0 , t1 , · · · , tN , we obtain h + [(H2j,2i + 2H2j,2i+1 )y2i + (2H2j,2i+1 + H2j,2i+2 )y2i+2 ] 3 i=0 j−1
y2j
=
x2j +
h4 [H0,0 y0 + L0,0 y0 + 3L0,0 y0 + 3L0,0 y0 180
−L2j,2j y2j − 3L2j,2j y2j − 3L2j,2j y2j − H2j,2j y2j ]
+k2j,2j y2j ;
=
x2j
N ). 2
(16)
h + [(H2j,2i + 2H2j,2i+1 )y2i + (2H2j,2i+1 + H2j,2i+2 )y2i+2 ] 3 i=0 j−1
y2j
j = 1(1)(
Solving linear Volterra integral equations
+
811
h4 y0 + 3M0,0 y0 + H0,0 y0 [M y0 + 3M0,0 180 0,0
−M2j,2j y2j − 3M2j,2j y2j − 3M2j,2j y2j − H2j,2j y2j ] ; j = 1(1)( +H2j,2j y2j + T2j,2j y2j + k2j,2j y2j
N ). 2
(17)
h + [(H2j,2i + 2H2j,2i+1 )y2i + (2H2j,2i+1 + H2j,2i+2 )y2i+2 ] 3 i=0 j−1
y2j
x 2j
=
+
h4 [D y0 + 3D0,0 y0 + 3D0,0 y0 + H0,0 y0 180 0,0
−D2j,2j y2j − 3D2j,2j y2j − 3D2j,2j y2j − H2j,2j y2j ] y2j + V2j,2j y2j + H2j,2j y2j + T2j,2j y2j +H2j,2j +2T2j,2j y2j + k2j,2j y2j ; j = 1(1)(
Where L(t, s) = M(t, s) =
∂ 2 k(t,s) ∂t∂s
, L (t, s) =
∂ 3 k(t,s) , ∂s∂t2 4
∂ 3 k(t,s) , ∂s2 ∂t ∂ 4 k(t,s) , ∂s2 ∂t2
M (t, s) =
N ). 2
L (t, s) =
∂ 4 k(t,s) , ∂s3 ∂t
M (t, s) =
∂ 5 k(t,s) , ∂s3 ∂t2
5
(18)
6
∂ k(t,s) ∂ k(t,s) D(t, s) = ∂∂sk(t,s) 2 ∂t2 ,D (t, s) = ∂s3 ∂t2 ,D (t, s) = ∂s4 ∂t2 must exist. For i = 1(1)( N2 ) from systems (9) ,(16) ,(17) and (18) we obtain a system with 2N equations and 2N unknowns. By solving system, the approximate solution of equation (3) ,is obtained.
3.
Numerical examples
In this section, we intend to compare this method with other methods such as repeated Simpson’s (S), repeated modified trapezoid (MT), and Pouzet (P) methods (Table 1).We solve these example by using MATLAB v7.1. example 1.In this example we solve equation [5]: 1 y(t) = t + 5
t
tsy(s)ds; 0
0 ≤ t ≤ 2,
(19)
F. Mirzaee
812 t3
where exact solution is y(t) = te 15 and numerical results are shown in Table 2.
Table 1 Solution of example 1 with S,MT,P methods (methods of Ref.[5])
Table 3 Nodes t Exact solution t=0 0 t=0.1 0.10001 t=0.2 0.20011 t=0.3 0.30054 t=0.4 0.40171 t=0.5 0.50418 t=0.6 0.60870 t=0.7 0.71619 t=0.8 0.82778 t=0.9 0.94482 t=1 1.06894 t=1.1 1.20207 t=1.2 1.34652 t=1.3 1.50506 t=1.4 1.68103 t=1.5 1.87849 t=1.6 2.10238 t=1.7 2.35882 t=1.8 2.65538 t=1.9 3.00152 t=2 3.40921
S 0 0.10001 0.20011 0.30054 0.40171 0.50418 0.60870 0.71619 0.82778 0.94482 1.06894 1.20207 1.34652 1.50506 1.68103 1.87849 2.10238 2.35883 2.65538 3.00154 3.40922
MT P 0 0 0.10001 0.10001 0.20011 0.20011 0.30054 0.30054 0.40171 0.40171 0.50418 0.50418 0.60870 0.60870 0.71619 0.71619 0.82778 0.82778 0.94482 0.94482 1.06894 1.06894 1.20207 1.20207 1.34652 1.34652 1.50506 1.50506 1.68103 1.68103 1.87849 1.87849 2.10238 2.10238 2.35882 2.35882 2.65537 2.65538 3.00152 3.00152 3.40920 3.40921
Solving linear Volterra integral equations
813
Table 2 Solution of example 1 with modified Simpson’s method
Nodes t t=0 t=0.1 t=0.2 t=0.3 t=0.4 t=0.5 t=0.6 t=0.7 t=0.8 t=0.9 t=1 t=1.1 t=1.2 t=1.3 t=1.4 t=1.5 t=1.6 t=1.7 t=1.8 t=1.9 t=2
4.
Exact solution h=0.1 0 0 0.10000700000 0.10000702101 0.20010700000 0.20010702461 0.30054000000 0.30054009502 0.40171000000 0.40171005021 0.50418400000 0.50418403112 0.60870300000 0.60870301202 0.71619100000 0.71619108761 0.82777800000 0.82777803941 0.94482000000 0.94482003462 1.06893910575 1.06893913112 1.20207000000 1.20207009811 1.34652000000 1.34652008392 1.50506000000 1.50506007302 1.68103000000 1.68103006304 1.87848000000 1.87848002207 2.10238000000 2.10238004226 2.35882000000 2.35882003991 2.65538000000 2.65538005121 3.00153000000 3.00153002628 3.40920973065 3.40920970498
h=0.05 0 0.10000700012 0.20010700011 0.30054000081 0.40171000043 0.50418400022 0.60870300098 0.71619100013 0.82777800062 0.94482000017 1.06893910505 1.20207000042 1.34652000092 1.50506000052 1.68103000092 1.87848000078 2.10238000047 2.35882000063 2.65538000000 3.00153000077 3.40920973185
Conclusions
In this work, we applied an application of modified Simpson’s method for solving the linear Volterra integral equations.According to the numerical results which obtaining from the illustrative examples, we conclude that for sufficiently small h we get a good accuracy, since by reducing step size length the least square error will be reduced.In ith equation of quadrature system in (9) (for using Case 2) the error of approximation of integral given in linear intei gral equation with repeated modified Simpson’s method is 1260 h7 f (6) (ζ), but
F. Mirzaee
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i this, for instance, by using repeated Simpson’s method is 180 h5 f (4) (ζ), and by i h5 f (4) (ζ). This method will be using repeated modified trapeziod method is 720 developed by authors for solving two-dimensional Volterra integral equations and their systems.
References [1] C.T.H.Baker, G.F.Miller, Treatment of Integral Equations by Numerical Methods, Academic Press Inc., London, 1982. [2] L.M.Delves, J.L.Mohamed, Computational Methods for Integral Equations, Cambridge University Press, 1985. [3] A.J.Jerri, Introduction to Integral Equations with Applications, Second ed.,Jhon Wiley and Sons, 1999. [4] R.Kress ,Linear Integral Equations, Springer-Verlag,Berlin Heidelberg,1989. [5] J.Saberi-Nadjafi,M.Heidari, Solving Linear Integral Equations of the Second Kind with repeated modified trapezoid quadrature method, Appl. Math. Comput. 189(2007) 980-985. Received: August 2011
