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A quadrature method with variable step for solving linear Volterra integral equations of the second kind
Applied Mathematics and Computation 188 (2007) 549–554 www.elsevier.com/locate/amc
A quadrature method with variable step for solving linear Volterra integral equations of the second kind Jafar Saberi-Nadjafi, Mahdi Heidari
*
Department of Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran
Abstract In usual quadrature methods for solving integral equations, divide the integration interval (a, b) into n equal subintervals of the length h = (b a)/n. In this article, we intend to divide the integration interval into n subintervals of different lengths, which solves linear Volterra integral equations more accurately than usual quadrature methods. For further information on quadrature methods with variable step see [L.M. Delves, J.L. Mohamed, Computational Methods for Integral Equations, Cambridge University Press, 1985; L.M. Delves, J. Walsh, Numerical Solution of Integral Equations, Oxford University Press, 1974]. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Linear Volterra integral equation; Quadrature; Variable step
1. Preliminaries and theory The general form of the linear Volterra integral equations is as follows: Z x kðx; tÞuðtÞdt; a 6 x 6 b: uðxÞ ¼ f ðxÞ þ
ð1:1Þ
a
For approximate the integral in (1.1), we intend to divide interval (a, b) into n subintervals of the lengths h1, h2, . . . , hn such that h1 = khn for constant k > 1. In quadrature methods for solving linear Volterra integral equations, the errors in last points are much more than first points, so we consider last subintervals smaller than first subintervals to decrease errors in last points. For further information on quadrature methods, see [1–9]. In order to do so, we act as follows. We consider the last subinterval of the length a and ith subinterval of the length kadi1 for i = 1, 2, . . . , n, which we will calculate d and a. The length of last subinterval is a, and the other hand is equal to kadn1, so pffiffiffi n1 ð1:2Þ d ¼ ð k Þ1 : *
Corresponding author. E-mail addresses: najafi@math.um.ac.ir (J. Saberi-Nadjafi), [email protected] (M. Heidari).
0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.10.086
550
J. Saberi-Nadjafi, M. Heidari / Applied Mathematics and Computation 188 (2007) 549–554
On the other hand, sum of the lengths of all subintervals is equal to b a, so ka þ kda þ kd 2 a þ þ kd n1 a ¼ b a:
ð1:3Þ
Because 0 < d < 1, therefore, we have a¼ (
ðb aÞðd 1Þ ; kðd n 1Þ ð1:4Þ
h1 ¼ ka; hi ¼ kad i1 ¼ dhi1 ;
i ¼ 2; 3; . . . ; n:
Now, as mentioned above, we will apply this idea to repeated trapezoid quadrature, repeated Simpson quadrature and block-by-block methods. Furthermore, we will compare the results. 2. Repeated trapezoid quadrature method with variable step For solving linear Volterra integral equations, the above idea has been applied Z x uðxÞ ¼ f ðxÞ þ kðx; tÞuðtÞdt; a 6 x 6 b:
ð2:1Þ
a
We have ( x0 ¼ a; xi ¼ xi1 þ hi ; and
8 > < uðx0 Þ ¼ f ðx0 Þ; > : uðxi Þ ¼ f ðxi Þ þ
ð2:2Þ
i ¼ 1; 2; . . . ; n
i R P xj j¼1
xj1
kðxi ; tÞuðtÞdt;
i ¼ 1; 2; . . . ; n:
ð2:3Þ
The approximation of every integral in the above-mentioned system by repeated trapezoid rule will yield the following system (In this system, we use the notation ui = u(xi), fi = f(xi) and kij = k(xi, xj) for simplicity.): 8 u0 ¼ f0 ; > > > > h1 h1 > > > > u1 ¼ f1 þ 2 k 10 u0 þ 2 k 11 u1 ; > h þh > > h h > < u2 ¼ f2 þ 21 k 20 u0 þ 1 2 2 k 21 u1 þ 22 k 22 u2 ; 3 ð2:4Þ 2 k 31 u1 þ h2 þh k 32 u2 þ h23 k 33 u3 ; u3 ¼ f3 þ h21 k 30 u0 þ h1 þh > 2 2 > > > > > . > > > .. > > > : 2 un ¼ fn þ h21 k n0 u0 þ h1 þh k n1 u1 þ þ hn12þhn k n;n1 un1 þ h2n k nn un : 2 Example 1. We apply this method for solving linear Volterra integral equation Z 1 x uðxÞ ¼ x þ xtuðtÞdt; 0 6 x 6 2: 5 0
ð2:5Þ
x3
The exact solution to this equation is uðxÞ ¼ xe15 . Table 1 indicates the numerical solution to this equation with repeated trapezoid quadrature method and repeated trapezoid quadrature method with variable step for k = 2 by using MATLAB v6.5. By comparing the results in Table 1, we see that the error in variable step method increase in first points with respect to repeated trapezoid quadrature method, but it decreases in last points. The common points
J. Saberi-Nadjafi, M. Heidari / Applied Mathematics and Computation 188 (2007) 549–554
551
Table 1 The solution to Eq. (2.5) by repeated trapezoid quadrature method (T) and repeated trapezoid quadrature method with variable step (DT) Nodes
T
T error
Nodes
DT
DT error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0 0.100010 0.200120 0.300571 0.401764 0.504270 0.608828 0.716367 0.828016 0.945135 1.069350 1.202599 1.347201 1.505934 1.682153 1.879931 2.104246 2.361237 2.658523 3.005644 3.414636
0 0.000003 0.000013 0.000030 0.000054 0.000086 0.000126 0.000176 0.000238 0.000315 0.000411 0.000531 0.000683 0.000877 0.001125 0.001446 0.001866 0.002417 0.003146 0.004119 0.005426
0 0.138340 0.271724 0.400330 0.524329 0.643885 0.759159 0.870303 0.977465 1.080789 1.180411 1.276464 1.369076 1.458370 1.544466 1.627477 1.707514 1.784684 1.859090 1.930830 2
0 0.138377 0.272134 0.402143 0.529555 0.655691 0.781967 0.909852 1.040835 1.176417 1.318102 1.467401 1.625839 1.794955 1.976317 2.171521 2.382203 2.610040 2.856754 3.124115 3.413939
0 0.000012 0.000046 0.000097 0.000163 0.000244 0.000339 0.000450 0.000578 0.000726 0.000896 0.001093 0.001318 0.001577 0.001873 0.002212 0.002599 0.003038 0.003535 0.004097 0.004730
in these two methods are points 0 and 2. Thus, by comparing the results in these points, one can see that the method with variable step is more accurate than usual quadrature method at point 2. 3. Repeated Simpson quadrature method with variable step We can apply this idea for solving linear Volterra integral equations by repeated Simpson quadrature method. In order to do so, we divide interval (a, b) into n subinterval with different lengths as mentioned above, then divide every subinterval into two equal subintervals. Hence, we have 8 x ¼ a; > < 0 x2i ¼ x2i2 þ hi ; i ¼ 1; 2; . . . ; n; ð3:1Þ > : x2i2 þx2i x2i1 ¼ 2 ; i ¼ 1; 2; . . . ; n: Therefore, by applying repeated Simpson and 4-point interpolation in odd equations (the equations corresponding to ui, when i is odd), the following triangular system will be obtained: 8 u0 > > > > > u1 > > > > > > u2 > > > > > > < u3
¼ f0 ; ¼^ u2 ; ¼ f2 þ h61 k 20 u0 þ 4h61 k 21 u1 þ h61 k 22 u2 ;
¼ f3 þ h30 k 30 u0 þ h31 k 31 u1 þ h32 k 32 u2 þ h33 k 33 u3 ; 2 k 42 u2 þ 4h62 k 43 u3 þ h62 k 44 u4 ; u4 ¼ f4 þ h61 k 40 u0 þ 4h61 k 41 u1 þ h1 þh > 6 > > > > > u5 ¼ f5 þ h61 k 50 u0 þ 4h61 k 51 u1 þ h61 þ h50 k 52 u2 þ h51 k 53 u3 þ h52 k 54 u4 þ h53 k 55 u5 ; > > > > > .. > > > . > > > : 2 u2n ¼ f2n þ h61 k 2n;0 u0 þ 4h61 k 2n;1 u1 þ h1 þh k 2n;2 u2 þ þ 4h6n k 2n;2n1 u2n1 þ h6n k 2n;2n u2n ; 6
ð3:2Þ
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J. Saberi-Nadjafi, M. Heidari / Applied Mathematics and Computation 188 (2007) 549–554
where h2i+1,j s for i = 1, 2, . . . , n 1 in the system (3.2), can be calculated as follows: Z x2iþ1 ðx x2i1 Þðx x2i Þðx x2iþ1 Þ dx h2iþ1;0 ¼ x2i2 ðx2i2 x2i1 Þðx2i2 x2i Þðx2i2 x2iþ1 Þ Z x2iþ1 4 d3 þ 8 hi ; ¼ 2 ðx x2i1 Þðx x2i Þðx x2iþ1 Þdx ¼ 48 hi ð2 þ dÞ x2i2 Z x2iþ1 3 8 ð2 dÞð2 þ dÞ hi ; ðx x2i2 Þðx x2i Þðx x2iþ1 Þdx ¼ h2iþ1;1 ¼ 3 24ð1 þ dÞ hi ð1 þ dÞ x2i2 Z 3 4 x2iþ1 ð2 þ dÞ h2iþ1;2 ¼ 3 hi ; ðx x2i2 Þðx x2i1 Þðx x2iþ1 Þdx ¼ 48 hi d x2i2 Z x2iþ1 8 dð2 þ dÞ hi : h2iþ1;3 ¼ 3 ðx x2i2 Þðx x2i1 Þðx x2i Þdx ¼ 8ð1 þ dÞ hi dð1 þ dÞð2 þ dÞ x2i2
ð3:3Þ
In odd equations, integration applies in three subintervals of the lengths h2i , h2i and hiþ1 . Thus, when i increases, 2 the length of these subintervals are multiplied at d, by it holds the following relation: h2iþ1;j ¼ dh2i1;j ;
i ¼ 1; 2; . . . ; n 1; j ¼ 0; 1; 2; 3:
ð3:4Þ
Therefore, only by calculating h30, h31, h32, h33 we can calculate all other h2i+1,j s by using (3.4). In addition, to that the following relation between all h2i+1,j s holds: 3 X
h2iþ1;j ¼ hi þ
j¼0
hiþ1 ; 2
i ¼ 1; 2; . . . ; n 1:
ð3:5Þ
^2 is obtained by solving the following system: Also u 8 u0 ¼ f ðx0 Þ; > <^ 1 1 1 x0 þx1 ^ u1 ¼ f ðx0 þx Þ þ d41 kðx0 þx ; x0 Þ^ u0 þ d41 kðx0 þx ; 2 Þ^u1 ; 2 2 2 > : x0 þx1 d1 4d 1 ^ u0 þ 6 kðx2 ; 2 Þ^ u1 þ d61 kðx2 ; x2 Þ^u2 : u2 ¼ f ðx1 Þ þ 6 kðx1 ; x0 Þ^
ð3:6Þ
Table 2 The solution to Eq. (2.5) by repeated Simpson quadrature (ST) and repeated Simpson quadrature with variable step (DST) Nodes
ST
ST error
Nodes
DST
DST error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0 0.100007 0.200107 0.300541 0.401710 0.504184 0.608703 0.716191 0.827778 0.944821 1.068940 1.202069 1.346519 1.505060 1.681030 1.878489 2.102385 2.358829 2.655386 3.001543 3.409227
0 4.167104 E11 3.559175 E09 2.709820 E08 2.885796 E08 1.167401 E07 1.010696 E07 3.038436 E07 2.570011 E07 6.555186 E07 5.613385 E07 1.302965 E06 1.138143 E06 2.505270 E06 2.234378 E06 4.779744 E06 4.353402 E06 9.178252 E06 8.545238 E06 1.789066 E05 1.705739 E05
0 0.138020 0.276040 0.403829 0.531618 0.649934 0.768251 0.877797 0.987344 1.088770 1.190196 1.284104 1.378012 1.464959 1.551906 1.632408 1.712911 1.787445 1.861980 1.930990 2
0 0.138044 0.276427 0.405606 0.536970 0.661940 0.791829 0.918286 1.052777 1.186602 1.331784 1.478788 1.640650 1.806565 1.991049 2.181597 2.394678 2.615647 2.863387 3.120660 3.409220
0 3.976481 E10 3.401411 E08 2.028242 E07 2.018390 E07 6.214605 E07 5.381343 E07 1.199756 E06 1.058781 E06 1.962682 E06 1.789517 E06 2.953309 E06 2.768318 E06 4.217703 E06 4.041914 E06 5.802332 E06 5.662164 E06 7.753995 E06 7.683196 E06 1.011957 E05 1.015914 E05
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Example 2. We solve integral equation (2.5) by repeated Simpson quadrature and repeated Simpson quadrature with variable step methods. The results for n = 10, are given in Table 2. As the trapezoid method in Table 2, one can see that the errors in the last points by using variable step method are less than the errors involved in usual quadrature method. 4. Block-by-block method with variable step One of the best methods for solving linear Volterra integral equations is the block-by-block method. This method solves linear Volterra integral equations by solving the following system. For further information on this method, see [3,4,9]: 8 u0 ¼ f0 ; > > > > > k u þ 8h k u 12h k 12 u2 ; u1 ¼ f1 þ 5h > > 12 10 0 12 11 1 > > > > u2 ¼ f2 þ h3 k 20 u0 þ 4h3 k 21 u1 þ h3 k 22 u2 ; > > > > > > k u 12h k 34 u4 ; k u þ 8h > u3 ¼ f3 þ h3 k 30 u0 þ 4h3 k 31 u1 þ h3 þ 5h 12 32 2 12 33 3 > > < u4 ¼ f4 þ h3 k 40 u0 þ 4h3 k 41 u1 þ 2h3 k 42 u2 þ 4h3 k 43 u3 þ h3 k 44 u4 ; ð4:1Þ > > > .. > > . > > > > > > u2n1 ¼ f2n1 þ h3 k 2n1;0 u0 þ 4h3 k 2n1;1 u1 þ þ h3 þ 5h k u > 12 2n;2n2 2n2 > > > > 8h h > þ 12 k 2n;2n1 u2n1 12 k 2n;2n u2n ; > > > : u2n ¼ f2n þ h3 k 2n;0 u0 þ 4h3 k 2n;1 u1 þ þ 4h3 k 2n;2n1 u2n1 þ h3 k 2n;2n u2n ; where h ¼ ba . 2n By applying this idea and using a process similar to the Simpson method, the following system is obtained in place of system (4.1): 8 u0 ¼ f0 ; > > > > h1 1 > k u þ 8h241 k 11 u1 24 k 12 u2 ; u1 ¼ f1 þ 5h > 24 10 0 > > > > h1 4h1 h1 > > > u2 ¼ f2 þ 6 k 20 u0 þ 6 k 21 u1 þ 6 k 22 u2 ; > > h > h1 4h1 5h2 8h2 h2 > 1 > u 3 ¼ f3 þ 6 k 30 u0 þ 6 k 31 u1 þ 6 þ 24 k 32 u2 þ 24 k 33 u3 24 k 34 u4 ; > > > > > > u4 ¼ f4 þ h1 k 40 u0 þ 4h1 k 41 u1 þ h1 þh2 k 42 u2 þ 4h1 k 43 u3 þ h2 k 44 u4 ; < 6 6 6 6 6 ð4:2Þ . > .. > > > > h > > h 4h > þ 5h24n k 2n1;2n2 u2n2 > u2n1 ¼ f2n1 þ 61 k 2n1;0 u0 þ 61 k 2n1;1 u1 þ þ n1 6 > > > > hn n > þ 8h k u 24 k 2n1;2n u2n ; > > 24 2n1;2n1 2n1 > > > h 4h > > u2n ¼ f2n þ 61 k 2n;0 u0 þ 61 k 2n;1 u1 þ þ ðhn16þhn Þk 2n;2n2 u2n2 > > > : þ 4h6n k 2n;2n1 u2n1 þ h6n k 2n;2n u2n : Example 3. In this example, once again we solve Eq. (2.5) by the block-by-block method and the block-byblock method with variable step. The results are given in Table 3. Table 3 shows that by using the variable step method the errors decrease in the odd statements and increase in the even statements. In general, the variable step method normalizes the errors. The maximum error in the block-by-block method is 6.877889 E05 at point 1.9, but the maximum error in the variable step method is 1.424784 E05 at point 1.930990, which is approximately 4.8 times less than the block-by-block method error.
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J. Saberi-Nadjafi, M. Heidari / Applied Mathematics and Computation 188 (2007) 549–554
Table 3 The solution to Eq. (2.5) by block-by-block method (B) and block-by-block method with variable step method (DB) Nodes
B
B error
Nodes
DB
DB error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0 0.100007 0.200107 0.300540 0.401710 0.504184 0.608703 0.716190 0.827778 0.944818 1.068940 1.202063 1.346518 1.505047 1.681029 1.878465 2.102382 2.358784 2.655379 3.001456 3.409210
0 2.890316 E09 3.557610 E09 7.851064 E08 2.851652 E08 3.712513 E07 9.664237 E08 1.066664 E06 2.305046 E07 2.454951 E06 4.527011 E07 5.056587 E06 7.796112 E07 9.862559 E06 1.198285 E06 1.881378 E05 1.596792 E06 3.579639 E05 1.570539 E06 6.877889 E05 9.970176 E08
0 0.138020 0.276040 0.403829 0.531618 0.649934 0.768251 0.877797 0.987344 1.088770 1.190196 1.284104 1.378012 1.464959 1.551906 1.632408 1.712910 1.787445 1.861980 1.930990 2
0 0.138044 0.276427 0.405605 0.536970 0.661938 0.791829 0.918281 1.052777 1.186595 1.331783 1.478779 1.640649 1.806552 1.991047 2.181581 2.394675 2.615627 2.863382 3.120636 3.409213
0 2.759848 E08 3.397482 E08 5.021472 E07 1.965933 E07 1.570334 E06 4.958104 E07 3.012446 E06 8.963513 E07 4.657821 E06 1.356233 E06 6.429119 E06 1.835364 E06 8.297250 E06 2.295589 E06 1.024338 E05 2.699947 E06 1.224043 E05 3.013138 E06 1.424784 E05 3.203088 E06
5. Conclusion As we mentioned above and by looking at the given examples, one can see that the variable step method normalizes the errors. This normalization has a better result in last points, because the errors in these points are much more with respect to first points. In the examples presented, we consider k = 2, because we observe that the results for k = 2 are better than other values. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
K.E. Atkinson, The Numerical Solution of Integral Equations of the Second Kind, Cambridge University Press, 1997. C.T.H. Baker, G.F. Miller, Treatment of Integral Equations by Numerical Methods, Academic Press Inc., London, 1982. L.M. Delves, J.L. Mohamed, Computational Methods for Integral Equations, Cambridge University Press, 1985. L.M. Delves, J. Walsh, Numerical Solution of Integral Equations, Oxford University Press, 1974. A.J. Jerri, Introduction to Integral Equations with Applications, second ed., John Wiley and Sons, 1999. R.P. Kenwal, Linear Integral Equations Theory and Technique, Academic Press, New York and London, 1971. J. Kondo, Integral Equations, Oxford University Press, Kodansha Ltd., 1991. R. Kress, Linear Integral Equations, Springer-Verlag, Berlin Heidelberg, 1989. P.K. Kyte, P. Puri, Computational Methods for Linear Integral Equations, Birkhauser, Boston, 2002.
A quadrature method with variable step for solving linear Volterra integral equations of the second kind Jafar Saberi-Nadjafi, Mahdi Heidari
*
Department of Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran
Abstract In usual quadrature methods for solving integral equations, divide the integration interval (a, b) into n equal subintervals of the length h = (b a)/n. In this article, we intend to divide the integration interval into n subintervals of different lengths, which solves linear Volterra integral equations more accurately than usual quadrature methods. For further information on quadrature methods with variable step see [L.M. Delves, J.L. Mohamed, Computational Methods for Integral Equations, Cambridge University Press, 1985; L.M. Delves, J. Walsh, Numerical Solution of Integral Equations, Oxford University Press, 1974]. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Linear Volterra integral equation; Quadrature; Variable step
1. Preliminaries and theory The general form of the linear Volterra integral equations is as follows: Z x kðx; tÞuðtÞdt; a 6 x 6 b: uðxÞ ¼ f ðxÞ þ
ð1:1Þ
a
For approximate the integral in (1.1), we intend to divide interval (a, b) into n subintervals of the lengths h1, h2, . . . , hn such that h1 = khn for constant k > 1. In quadrature methods for solving linear Volterra integral equations, the errors in last points are much more than first points, so we consider last subintervals smaller than first subintervals to decrease errors in last points. For further information on quadrature methods, see [1–9]. In order to do so, we act as follows. We consider the last subinterval of the length a and ith subinterval of the length kadi1 for i = 1, 2, . . . , n, which we will calculate d and a. The length of last subinterval is a, and the other hand is equal to kadn1, so pffiffiffi n1 ð1:2Þ d ¼ ð k Þ1 : *
Corresponding author. E-mail addresses: najafi@math.um.ac.ir (J. Saberi-Nadjafi), [email protected] (M. Heidari).
0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.10.086
550
J. Saberi-Nadjafi, M. Heidari / Applied Mathematics and Computation 188 (2007) 549–554
On the other hand, sum of the lengths of all subintervals is equal to b a, so ka þ kda þ kd 2 a þ þ kd n1 a ¼ b a:
ð1:3Þ
Because 0 < d < 1, therefore, we have a¼ (
ðb aÞðd 1Þ ; kðd n 1Þ ð1:4Þ
h1 ¼ ka; hi ¼ kad i1 ¼ dhi1 ;
i ¼ 2; 3; . . . ; n:
Now, as mentioned above, we will apply this idea to repeated trapezoid quadrature, repeated Simpson quadrature and block-by-block methods. Furthermore, we will compare the results. 2. Repeated trapezoid quadrature method with variable step For solving linear Volterra integral equations, the above idea has been applied Z x uðxÞ ¼ f ðxÞ þ kðx; tÞuðtÞdt; a 6 x 6 b:
ð2:1Þ
a
We have ( x0 ¼ a; xi ¼ xi1 þ hi ; and
8 > < uðx0 Þ ¼ f ðx0 Þ; > : uðxi Þ ¼ f ðxi Þ þ
ð2:2Þ
i ¼ 1; 2; . . . ; n
i R P xj j¼1
xj1
kðxi ; tÞuðtÞdt;
i ¼ 1; 2; . . . ; n:
ð2:3Þ
The approximation of every integral in the above-mentioned system by repeated trapezoid rule will yield the following system (In this system, we use the notation ui = u(xi), fi = f(xi) and kij = k(xi, xj) for simplicity.): 8 u0 ¼ f0 ; > > > > h1 h1 > > > > u1 ¼ f1 þ 2 k 10 u0 þ 2 k 11 u1 ; > h þh > > h h > < u2 ¼ f2 þ 21 k 20 u0 þ 1 2 2 k 21 u1 þ 22 k 22 u2 ; 3 ð2:4Þ 2 k 31 u1 þ h2 þh k 32 u2 þ h23 k 33 u3 ; u3 ¼ f3 þ h21 k 30 u0 þ h1 þh > 2 2 > > > > > . > > > .. > > > : 2 un ¼ fn þ h21 k n0 u0 þ h1 þh k n1 u1 þ þ hn12þhn k n;n1 un1 þ h2n k nn un : 2 Example 1. We apply this method for solving linear Volterra integral equation Z 1 x uðxÞ ¼ x þ xtuðtÞdt; 0 6 x 6 2: 5 0
ð2:5Þ
x3
The exact solution to this equation is uðxÞ ¼ xe15 . Table 1 indicates the numerical solution to this equation with repeated trapezoid quadrature method and repeated trapezoid quadrature method with variable step for k = 2 by using MATLAB v6.5. By comparing the results in Table 1, we see that the error in variable step method increase in first points with respect to repeated trapezoid quadrature method, but it decreases in last points. The common points
J. Saberi-Nadjafi, M. Heidari / Applied Mathematics and Computation 188 (2007) 549–554
551
Table 1 The solution to Eq. (2.5) by repeated trapezoid quadrature method (T) and repeated trapezoid quadrature method with variable step (DT) Nodes
T
T error
Nodes
DT
DT error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0 0.100010 0.200120 0.300571 0.401764 0.504270 0.608828 0.716367 0.828016 0.945135 1.069350 1.202599 1.347201 1.505934 1.682153 1.879931 2.104246 2.361237 2.658523 3.005644 3.414636
0 0.000003 0.000013 0.000030 0.000054 0.000086 0.000126 0.000176 0.000238 0.000315 0.000411 0.000531 0.000683 0.000877 0.001125 0.001446 0.001866 0.002417 0.003146 0.004119 0.005426
0 0.138340 0.271724 0.400330 0.524329 0.643885 0.759159 0.870303 0.977465 1.080789 1.180411 1.276464 1.369076 1.458370 1.544466 1.627477 1.707514 1.784684 1.859090 1.930830 2
0 0.138377 0.272134 0.402143 0.529555 0.655691 0.781967 0.909852 1.040835 1.176417 1.318102 1.467401 1.625839 1.794955 1.976317 2.171521 2.382203 2.610040 2.856754 3.124115 3.413939
0 0.000012 0.000046 0.000097 0.000163 0.000244 0.000339 0.000450 0.000578 0.000726 0.000896 0.001093 0.001318 0.001577 0.001873 0.002212 0.002599 0.003038 0.003535 0.004097 0.004730
in these two methods are points 0 and 2. Thus, by comparing the results in these points, one can see that the method with variable step is more accurate than usual quadrature method at point 2. 3. Repeated Simpson quadrature method with variable step We can apply this idea for solving linear Volterra integral equations by repeated Simpson quadrature method. In order to do so, we divide interval (a, b) into n subinterval with different lengths as mentioned above, then divide every subinterval into two equal subintervals. Hence, we have 8 x ¼ a; > < 0 x2i ¼ x2i2 þ hi ; i ¼ 1; 2; . . . ; n; ð3:1Þ > : x2i2 þx2i x2i1 ¼ 2 ; i ¼ 1; 2; . . . ; n: Therefore, by applying repeated Simpson and 4-point interpolation in odd equations (the equations corresponding to ui, when i is odd), the following triangular system will be obtained: 8 u0 > > > > > u1 > > > > > > u2 > > > > > > < u3
¼ f0 ; ¼^ u2 ; ¼ f2 þ h61 k 20 u0 þ 4h61 k 21 u1 þ h61 k 22 u2 ;
¼ f3 þ h30 k 30 u0 þ h31 k 31 u1 þ h32 k 32 u2 þ h33 k 33 u3 ; 2 k 42 u2 þ 4h62 k 43 u3 þ h62 k 44 u4 ; u4 ¼ f4 þ h61 k 40 u0 þ 4h61 k 41 u1 þ h1 þh > 6 > > > > > u5 ¼ f5 þ h61 k 50 u0 þ 4h61 k 51 u1 þ h61 þ h50 k 52 u2 þ h51 k 53 u3 þ h52 k 54 u4 þ h53 k 55 u5 ; > > > > > .. > > > . > > > : 2 u2n ¼ f2n þ h61 k 2n;0 u0 þ 4h61 k 2n;1 u1 þ h1 þh k 2n;2 u2 þ þ 4h6n k 2n;2n1 u2n1 þ h6n k 2n;2n u2n ; 6
ð3:2Þ
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where h2i+1,j s for i = 1, 2, . . . , n 1 in the system (3.2), can be calculated as follows: Z x2iþ1 ðx x2i1 Þðx x2i Þðx x2iþ1 Þ dx h2iþ1;0 ¼ x2i2 ðx2i2 x2i1 Þðx2i2 x2i Þðx2i2 x2iþ1 Þ Z x2iþ1 4 d3 þ 8 hi ; ¼ 2 ðx x2i1 Þðx x2i Þðx x2iþ1 Þdx ¼ 48 hi ð2 þ dÞ x2i2 Z x2iþ1 3 8 ð2 dÞð2 þ dÞ hi ; ðx x2i2 Þðx x2i Þðx x2iþ1 Þdx ¼ h2iþ1;1 ¼ 3 24ð1 þ dÞ hi ð1 þ dÞ x2i2 Z 3 4 x2iþ1 ð2 þ dÞ h2iþ1;2 ¼ 3 hi ; ðx x2i2 Þðx x2i1 Þðx x2iþ1 Þdx ¼ 48 hi d x2i2 Z x2iþ1 8 dð2 þ dÞ hi : h2iþ1;3 ¼ 3 ðx x2i2 Þðx x2i1 Þðx x2i Þdx ¼ 8ð1 þ dÞ hi dð1 þ dÞð2 þ dÞ x2i2
ð3:3Þ
In odd equations, integration applies in three subintervals of the lengths h2i , h2i and hiþ1 . Thus, when i increases, 2 the length of these subintervals are multiplied at d, by it holds the following relation: h2iþ1;j ¼ dh2i1;j ;
i ¼ 1; 2; . . . ; n 1; j ¼ 0; 1; 2; 3:
ð3:4Þ
Therefore, only by calculating h30, h31, h32, h33 we can calculate all other h2i+1,j s by using (3.4). In addition, to that the following relation between all h2i+1,j s holds: 3 X
h2iþ1;j ¼ hi þ
j¼0
hiþ1 ; 2
i ¼ 1; 2; . . . ; n 1:
ð3:5Þ
^2 is obtained by solving the following system: Also u 8 u0 ¼ f ðx0 Þ; > <^ 1 1 1 x0 þx1 ^ u1 ¼ f ðx0 þx Þ þ d41 kðx0 þx ; x0 Þ^ u0 þ d41 kðx0 þx ; 2 Þ^u1 ; 2 2 2 > : x0 þx1 d1 4d 1 ^ u0 þ 6 kðx2 ; 2 Þ^ u1 þ d61 kðx2 ; x2 Þ^u2 : u2 ¼ f ðx1 Þ þ 6 kðx1 ; x0 Þ^
ð3:6Þ
Table 2 The solution to Eq. (2.5) by repeated Simpson quadrature (ST) and repeated Simpson quadrature with variable step (DST) Nodes
ST
ST error
Nodes
DST
DST error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0 0.100007 0.200107 0.300541 0.401710 0.504184 0.608703 0.716191 0.827778 0.944821 1.068940 1.202069 1.346519 1.505060 1.681030 1.878489 2.102385 2.358829 2.655386 3.001543 3.409227
0 4.167104 E11 3.559175 E09 2.709820 E08 2.885796 E08 1.167401 E07 1.010696 E07 3.038436 E07 2.570011 E07 6.555186 E07 5.613385 E07 1.302965 E06 1.138143 E06 2.505270 E06 2.234378 E06 4.779744 E06 4.353402 E06 9.178252 E06 8.545238 E06 1.789066 E05 1.705739 E05
0 0.138020 0.276040 0.403829 0.531618 0.649934 0.768251 0.877797 0.987344 1.088770 1.190196 1.284104 1.378012 1.464959 1.551906 1.632408 1.712911 1.787445 1.861980 1.930990 2
0 0.138044 0.276427 0.405606 0.536970 0.661940 0.791829 0.918286 1.052777 1.186602 1.331784 1.478788 1.640650 1.806565 1.991049 2.181597 2.394678 2.615647 2.863387 3.120660 3.409220
0 3.976481 E10 3.401411 E08 2.028242 E07 2.018390 E07 6.214605 E07 5.381343 E07 1.199756 E06 1.058781 E06 1.962682 E06 1.789517 E06 2.953309 E06 2.768318 E06 4.217703 E06 4.041914 E06 5.802332 E06 5.662164 E06 7.753995 E06 7.683196 E06 1.011957 E05 1.015914 E05
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Example 2. We solve integral equation (2.5) by repeated Simpson quadrature and repeated Simpson quadrature with variable step methods. The results for n = 10, are given in Table 2. As the trapezoid method in Table 2, one can see that the errors in the last points by using variable step method are less than the errors involved in usual quadrature method. 4. Block-by-block method with variable step One of the best methods for solving linear Volterra integral equations is the block-by-block method. This method solves linear Volterra integral equations by solving the following system. For further information on this method, see [3,4,9]: 8 u0 ¼ f0 ; > > > > > k u þ 8h k u 12h k 12 u2 ; u1 ¼ f1 þ 5h > > 12 10 0 12 11 1 > > > > u2 ¼ f2 þ h3 k 20 u0 þ 4h3 k 21 u1 þ h3 k 22 u2 ; > > > > > > k u 12h k 34 u4 ; k u þ 8h > u3 ¼ f3 þ h3 k 30 u0 þ 4h3 k 31 u1 þ h3 þ 5h 12 32 2 12 33 3 > > < u4 ¼ f4 þ h3 k 40 u0 þ 4h3 k 41 u1 þ 2h3 k 42 u2 þ 4h3 k 43 u3 þ h3 k 44 u4 ; ð4:1Þ > > > .. > > . > > > > > > u2n1 ¼ f2n1 þ h3 k 2n1;0 u0 þ 4h3 k 2n1;1 u1 þ þ h3 þ 5h k u > 12 2n;2n2 2n2 > > > > 8h h > þ 12 k 2n;2n1 u2n1 12 k 2n;2n u2n ; > > > : u2n ¼ f2n þ h3 k 2n;0 u0 þ 4h3 k 2n;1 u1 þ þ 4h3 k 2n;2n1 u2n1 þ h3 k 2n;2n u2n ; where h ¼ ba . 2n By applying this idea and using a process similar to the Simpson method, the following system is obtained in place of system (4.1): 8 u0 ¼ f0 ; > > > > h1 1 > k u þ 8h241 k 11 u1 24 k 12 u2 ; u1 ¼ f1 þ 5h > 24 10 0 > > > > h1 4h1 h1 > > > u2 ¼ f2 þ 6 k 20 u0 þ 6 k 21 u1 þ 6 k 22 u2 ; > > h > h1 4h1 5h2 8h2 h2 > 1 > u 3 ¼ f3 þ 6 k 30 u0 þ 6 k 31 u1 þ 6 þ 24 k 32 u2 þ 24 k 33 u3 24 k 34 u4 ; > > > > > > u4 ¼ f4 þ h1 k 40 u0 þ 4h1 k 41 u1 þ h1 þh2 k 42 u2 þ 4h1 k 43 u3 þ h2 k 44 u4 ; < 6 6 6 6 6 ð4:2Þ . > .. > > > > h > > h 4h > þ 5h24n k 2n1;2n2 u2n2 > u2n1 ¼ f2n1 þ 61 k 2n1;0 u0 þ 61 k 2n1;1 u1 þ þ n1 6 > > > > hn n > þ 8h k u 24 k 2n1;2n u2n ; > > 24 2n1;2n1 2n1 > > > h 4h > > u2n ¼ f2n þ 61 k 2n;0 u0 þ 61 k 2n;1 u1 þ þ ðhn16þhn Þk 2n;2n2 u2n2 > > > : þ 4h6n k 2n;2n1 u2n1 þ h6n k 2n;2n u2n : Example 3. In this example, once again we solve Eq. (2.5) by the block-by-block method and the block-byblock method with variable step. The results are given in Table 3. Table 3 shows that by using the variable step method the errors decrease in the odd statements and increase in the even statements. In general, the variable step method normalizes the errors. The maximum error in the block-by-block method is 6.877889 E05 at point 1.9, but the maximum error in the variable step method is 1.424784 E05 at point 1.930990, which is approximately 4.8 times less than the block-by-block method error.
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Table 3 The solution to Eq. (2.5) by block-by-block method (B) and block-by-block method with variable step method (DB) Nodes
B
B error
Nodes
DB
DB error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0 0.100007 0.200107 0.300540 0.401710 0.504184 0.608703 0.716190 0.827778 0.944818 1.068940 1.202063 1.346518 1.505047 1.681029 1.878465 2.102382 2.358784 2.655379 3.001456 3.409210
0 2.890316 E09 3.557610 E09 7.851064 E08 2.851652 E08 3.712513 E07 9.664237 E08 1.066664 E06 2.305046 E07 2.454951 E06 4.527011 E07 5.056587 E06 7.796112 E07 9.862559 E06 1.198285 E06 1.881378 E05 1.596792 E06 3.579639 E05 1.570539 E06 6.877889 E05 9.970176 E08
0 0.138020 0.276040 0.403829 0.531618 0.649934 0.768251 0.877797 0.987344 1.088770 1.190196 1.284104 1.378012 1.464959 1.551906 1.632408 1.712910 1.787445 1.861980 1.930990 2
0 0.138044 0.276427 0.405605 0.536970 0.661938 0.791829 0.918281 1.052777 1.186595 1.331783 1.478779 1.640649 1.806552 1.991047 2.181581 2.394675 2.615627 2.863382 3.120636 3.409213
0 2.759848 E08 3.397482 E08 5.021472 E07 1.965933 E07 1.570334 E06 4.958104 E07 3.012446 E06 8.963513 E07 4.657821 E06 1.356233 E06 6.429119 E06 1.835364 E06 8.297250 E06 2.295589 E06 1.024338 E05 2.699947 E06 1.224043 E05 3.013138 E06 1.424784 E05 3.203088 E06
5. Conclusion As we mentioned above and by looking at the given examples, one can see that the variable step method normalizes the errors. This normalization has a better result in last points, because the errors in these points are much more with respect to first points. In the examples presented, we consider k = 2, because we observe that the results for k = 2 are better than other values. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
K.E. Atkinson, The Numerical Solution of Integral Equations of the Second Kind, Cambridge University Press, 1997. C.T.H. Baker, G.F. Miller, Treatment of Integral Equations by Numerical Methods, Academic Press Inc., London, 1982. L.M. Delves, J.L. Mohamed, Computational Methods for Integral Equations, Cambridge University Press, 1985. L.M. Delves, J. Walsh, Numerical Solution of Integral Equations, Oxford University Press, 1974. A.J. Jerri, Introduction to Integral Equations with Applications, second ed., John Wiley and Sons, 1999. R.P. Kenwal, Linear Integral Equations Theory and Technique, Academic Press, New York and London, 1971. J. Kondo, Integral Equations, Oxford University Press, Kodansha Ltd., 1991. R. Kress, Linear Integral Equations, Springer-Verlag, Berlin Heidelberg, 1989. P.K. Kyte, P. Puri, Computational Methods for Linear Integral Equations, Birkhauser, Boston, 2002.