Analytical Approximate Solution of Carleman s Equation by Using Maclaurin Series

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1 Interntionl Mthemticl Forum, 5, 2010, no. 60, Anlyticl Approximte Solution of Crlemn s Eqution by Using Mclurin Series M. Yghobifr 1 Institute for Mthemticl Reserch University Putr Mlysi Serdng 43400, Selngor, Mlysi N. M. A. Nik Long Deprtment of Mthemtics, Fculty of Science University Putr Mlysi Serdng 43400, Selngor, Mlysi Abstrct In this pper we introduce expnsion method for solution of crlemn s eqution, in this method we expnd the known function s Mclurin series nd convert the solution into liner combintion of some elements. we proved this liner combintion is uniformly convergence to the nlytic solution. If the known function be polynomil then we hve exct solution. Mthemtics Subject Clssifiction: 45 Keywords: Wekly singulr integrl eqution, Crlemn s eqution, Mclurin series, Approximtion 1 Introduction In the theory of scttering of coustic, electromgnetic, nd erthquke wves by cylinders, infinite strips, nd slits there rises kind of wekly singulr integrl equtions which is esily inverted s 8 ln x t gtdt = fx x <, Corresponding uthor, E-mil ddress: Yghobifr@pnu.c.ir

2 2986 M. Yghobifr nd N. M. A. Nik Long Where gt is unknown nd fx is known functions. Solution of 1 is9,2 gx = 1 π 2 2 t 2 f tdt 2 x 2 t x + 1 π 2 ln 2 2 x 2 ftdt 2 t 2. 2 Integrl equtions with logrithmic kernels lso rise in the boundry vlue problems for two-dimensionl configurtions 4, 5, 6, 7 nd rised in plne elsticity crck problem, the disloction distribution is tken s unknown function, nd resultnt force s the right hnd term 1. In this pper we consider fx C,, nd fter expnding it in the form of Mclurin series, substitute in eqution 2 nd split tht integrls into some elements. In section 2 we evluted these elements. Anlytic nd pproximte solution in the form of series re obtined in section 3. Convergence of the mentioned series re discussed in section 4. Exct solution when the known function be polynomil found in section 5. 2 Recurrence reltions Let w n = t n dt, then obviously 2 t2 2k 2k π, n =2k, w n = 2 k 0, n =2k Lemm 1 Let I n x = Where m = n 2 I n x = π x n 1 + t n dt 2 t 2 x t then m 2j 2j x n 2j 1 2 j j=1 n 1 1 for n even, nd m = for n odd. nd recurrence reltions 2 I 2k+1 x =xi 2k x w 2k, Proof: I n x cn be written s I n x = I 2k x =xi 2k 1 x. t n x n 2 t 2 x t dt + x n dt 2 t 2 x t. 4

3 Solution of Crlemn s eqution 2987 It is known tht see 3, dt 2 t 2 x t = 1 2 t ln t 2 x 2 x 2 t x c 5 2 t 2 From 5 it follows tht the second integrl is zero for x,. Now using 3 nd fctorizing x t in the first integrl of 4 we get I n x = x n 1 dt 2 t + tdt 2 xn 2 2 t + 2 +x t n 2 dt 2 x 2 + t n 1 dt 2 x 2 = x n 1 w 0 + x n 2 w xw n 2 + w n 1. 6 For finding the recurrence reltions between I 2k+1 x nd I 2k x we consider two cses: First let n =2k, then I 2k x = π x 2k ! 41! 2 x2k k 4 2k 4! 4 k 2 k 2! 2 x3 + 2k 2 2k 2! 4 k 1 k 1! x 2 k 1 2j 2j = π x 2k 1 + x 2k 2j 1 2 j j=1 Secondly, let n =2k +1 I 2k+1 x = π x 2k + 2 2! 41! 2 x2k k 2 2k 2! 4 k 1 k 1! 2 x2 2k 2k! + 4 k k! 2 k 2j 2j = π x 2k + x 2k 2j, k =1, 2,. 2 j j=1 Hence compring these two equlities we obtin the recursive for I n which re I 2k+1 x =xi 2k x w 2k, I 2k x =xi 2k 1 x.

4 2988 M. Yghobifr nd N. M. A. Nik Long Combining of two bove equtions get the desirble result. mnipultions we cn get the results of Gkhov 2 With some x n 1 + m k= k 1 2k x n 2k k 3 Anlyticl pproximte solution Write 2 s gx = 1 π 2 2 x 2 Ax+ 1 ln B 2 7 where Ax = t 2 2 f tdt 2 t 2 x t 8 nd B = ftdt 2 t 2 9 We consider fx C, so it is possible to expnd fx s Mclurin series fx =f0 + f 0x + f 0 x 2 + f 0 x f n 0 x n + 2! 3! 3.1 Evlution of Ax Write t 2 2 f t = f 0 2 f 0 2 t +f 0 f 0 2 t 2 + 2! + f n 1 0 n 2! f n+1 2 t n +

5 Solution of Crlemn s eqution 2989 Substituting in 8 yields Ax = f 0 2 I 0 x f 0 2 I 1 x+f 0 f 0 2 I 2 x+ 2! Where I n x is defined in Evlution of B + f n 1 0 n 2! f n+1 2 I n x+ 10 Substituting Mclurin series of ft in 9 nd using Lemm 1 gives B = f0w 0 + f 0w 1 + f 0 w f 2k 0 w 2k + f 2k+1 0 2! 2k! 2k + 1! w 2k+1 + or B = π f0 + f 0 41! f ! f 2k 0 4 k k! 2 2k Approximtion solution The pproximtion solution is: 1 g n x = π 2 A n x+ 1 2 x 2 ln 2 where B n 12 A n x = f 0 2 I 0 x f 0 2 I 1 x+f 0 f 0 2 I 2 x+ 2! + f n 1 0 n 1! f n+1 2 I n x 13 nd B n = π f0 + f 0 41! f ! f 2k 0 4 k k! 2 2k 14 where n =2k +1orn =2k.

6 2990 M. Yghobifr nd N. M. A. Nik Long 4 Error nlysis In the following convergence of A n x toax nd B n to B re proved respectively. 4.1 Error of A n x We re going to prove A n x is uniformly convergence to Ax. From 10 nd 13 it follows tht Ax A n x = f n 0 n 1! f n+2 0 f n+1 0 n + 1! 2 I n+1 x+ f n+3 0 n +2! 2 I n+2 x+ f n 0 2 n 1! f n+2 0 f I n+1 x+ 2 n+1 0 f n+3 0 I n+2 x+ n + 1! n + 2! M 2 1 n 1! 1 1 I n+1 x + n + 1! 1 I n+2 x + n + 2! M 2 1 n 1! 1 1 n +1 n + n + 1! 1 n +2 n+1 + n + 2! M 2 1 n 1! n +1n + 1 n +2n+1 + M 2 1 n 2! + 2 n 1 + n 1! n 1! + 2 n+1 + M 2 n n 2! + n+1 n n 1! + +2 n 1! + n+1 + M 3 n 2 n 2! + n 1 n 1 n 1! + +2 n 1! + n + M 3 n 1 n n 2! n 1! + n + Since f n x M, nd In x n n 1 for n N. After using Tylor theorem we hve Ax A n x M 3 n 1 n 2! + n +,ξ eξ, Since n lim n =0 so A n x is uniformly convergence to Ax.

7 Solution of Crlemn s eqution Error of B n We re going to prove B n is convergence to B. From 11 nd 14 it follows tht B B n = π f 2n n+1 n +1! 2n+2 + f 2n n+2 n + 2! 2n+4 + 2n+2 πm 4 n+1 n + 1! + 2n+4 4 n+2 n +2! + 2 n+1 πm n + 1! + 2 n+2 n + 2! + = πm e n 2! since lim n e n =0 2! B n is convergence to B. 5 Exct solutions for polynomils Let fx =β 0 + β 1 x + β 2 x β n x n, then A n x = 2β 2 2 I 1 x+β 1 3β 3 2 I 2 x+ +β n 1 n +1β n+1 2 I n x nd B n = β 0 w 0 + β 2 w β 2k w 2k where n =2k +1orn =2k. Now we cn obtin the solution from Conclusion In this pper we use polynomil s the known function of right hnd side of Crlemn s eqution nd find nlytic solution for relted integrl eqution lso we found for clss of infinite derivble function n pproximte solution. ACKNOWLEDGEMENTS. The uthors would like to thnk the Universiti Putr Mlysi nd MOSTI for the Fundmentl Reserch Grnt scheme, project No: FR.

8 2992 M. Yghobifr nd N. M. A. Nik Long References 1 Y.K. Cheung nd Y.Z. Chen, New integrl eqution for plne elsticity crck problems, Theore. Appl. Frct. Mech., , F.D. Gkhov, Boundry Vlue Problemsin Russin, Nuk, Moscow, M.I. Isrilov, Approximte-nliticl solution of singulr integrl equtions of first kind using qudrture formul. Numericl integrtion nd djcent problems, Collection of rticles of Acdemy Science of Republic of Uzbekistn, Press FAN, Tshkent. in Russin, 1990, D.L. Jin nd R.P. Knwl, Diffrction of elstic wves by two coplnr nd prllel rigid strips, Int. J. Engng. Sci., , D.L. Jin nd R.P. Knwl, Diffrction of elstic wves by two coplnr Griffith crcks in n elstic medium, Int. J. Solid Structures, , D.L. Jin nd R.P. Knwl, Acoustic diffrction of plne wve by two coplnr prllel perfectly soft or rigid strips, Cn.J. Phys., , D.L. Jin nd R.P. Knwl, Scttering of coustic, electromgnetic nd elstic SH wves by two-dimensionl obstcles, Ann.Phys., , R.P.Knwl, Liner Integrl Equtions Theory nd Technique, Second Edition, Birkhuser Boston A.D. Polynin nd A.V. Mnzhirov, HANDBOOK OF INTEGRAL EQUATIONS, CRC PRESS, Appendix List of some I n x re s below: I 0 x =0. I 1 x = π. I 2 x = πx.

9 Solution of Crlemn s eqution 2993 I 3 x = πx 2 + 2! 41! 2 2. I 4 x = πx 3 + 2! 41! 2 2 x. I 5 x = πx 4 + 2! 41! 2 2 x 2 + 4! 4 2 2! 2 4. I 6 x = πx 5 + 2! 2 x 3 + 4! 4 x. 41! ! 2 I 7 x = π x 6 + 2! 2 x 4 + 4! 4 x 2 + 6! 6. 41! ! ! 2 I 8 x = π x 7 + 2! 2 x 5 + 4! 4 x 3 + 6! 6 x. 41! ! ! 2 I 9 x = π x 8 + 2! 2 x 6 + 4! 4 x 4 + 6! 6 x 2 + 8! 8. 41! ! ! ! 2 I 10 x = π x 9 + 2! 2 x 7 + 4! 4 x 5 + 6! 6 x 3 + 8! 8 x 41! ! ! ! 2 Received: My, 2010.