A new Approach for Solving Linear Ordinary Differential Equations

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, ISSN 974-57X (Onlne), ISSN 974-5718 (Prnt), Vol. ; Issue No. 1; Year 14, Copyrght 13-14 by CESER PUBLICATIONS A new Approach for Solvng Lnear Ordnary Dfferental Equatons Fawz Abdelwahd Department of Mathematcs, Faculty of Scence, Unversty of Benghaz, Benghaz, Lbya; Emal: fawzabd@hotmal.com ABSTRACT Recently a new numercal scheme, based on seres soluton method, has been ntroduced for solvng Volterra ntegral equatons. In ths wor, we mae use of the natural relaton between ordnary dfferental equatons and Volterra ntegral equatons and apply ths approach to ntal value problems of lnear ordnary dfferental equatons. Ths study led to appromate solutons, whch can be used as algorthms to compute numercal solutons for the ntal value problem. Ths study also ntroduced specal case related to the order of the lnear ordnary dfferental equaton and the ernel of the converted Volterra ntegral equaton. At the end, we nvestgate the sgnfcant features of ths approach, by solvng selected numercal eamples and comparng our results wth the results obtaned by other methods. Under certan condtons, as n the case f ntegral equatons, we found, that ths approach leads to the Taylor epanson of the eact soluton. Keywords: Ordnary dfferental equatons, Integral equatons, Numercal methods. Mathematcs Subject Classfcaton: 65N1, 3C3 1. Introducton Numercal ordnary dfferental equatons studes the numercal soluton of ordnary dfferental equatons [1-]. A varety of methods, analytcal, sem-analytcal, appromate and purely numercal are avalable for the soluton of ordnary dfferental equatons such as the dfferental transformaton method [3], Adoman decomposton method [4] and Runge Kutta methods [5]. Recently a recursve method based on seres soluton method has been ntroduced for solvng Volterra ntegral equatons [6,7]. The man am of ths wor s to apply ths approach to fnd appromate solutons for the ntal value problems of lnear ordnary dfferental equatons. To do that, we use the natural relaton between ordnary dfferental equatons and ntegral equatons [8]. The general form of ntal value problems of lnear ordnary dfferental equatons can be wrtten n the form www.ceser.n/jmc.html www.ceserp.com/cp-jour www.ceserpublcatons.com

n y ( n) a ( n ) ( ) ( ) y g, 1 () y () c,,,3,..., n 1 (1.1) where the coeffcents a ( ) are contnuous functons of and c are constants. To convert the ntal value problem (1.1) to Volterra ntegral equatons, we set ( n) y ( ) u ( ), (1.) Net, f we ntegrate the both sdes of (1.) j tmes from to, we can fnd the followng formula j 1 n 1 n j ( n j) ( t) y ( ) u( t) dt c j n j n j, j 1,,..., n (1.3) ( 1)! ( )! Now, by usng (1.) and (1.3), we can convert the ntal value problem (1.1) to the a lnear Volterra ntegral equaton of second nd of the form u ( ) f( ) tutdt (, ) ( ), u() f() (1.4) wth n ( t) t (, ) aj( ) ( j 1)! j 1 j 1 n n 1 j n f( ) g( ) aj( ) c ( j n )! (1.5) j 1 n j. A Numercal Algorthm In ths secton, we follow ref. [6] and apply the recursve method to the ntal value problem (1.1). But for smplcty, let us frst assume that t (, ) (, t),,1,...,. (.1) 1

Net, we dfferentate (1.4) tmes wth respect of. Hence for 1,, 3, 4, we have u ( ) f ( ) ( u, ) ( ) ( tutdt, ) ( ) (.) 1 u ( ) f ( ) ( u, ) ( ) ( u, ) ( ) ( u, ) ( ) ( tutdt, ) ( ) (.3) 1 u ( ) f ( ) (, ) u ( ) (, ) u ( ) (, ) u( ) (, ) u( ) 1 (, ) u ( ) (, ) u( ) (, t) u( t) dt 1 3 (.4) u ( ) f ( ) (, ) u ( ) 3 (, ) u ( ) 3 (, ) u ( ) (, ) u( ) (, ) u( ) (, ) u ( ) (, ) u ( ) 1 1 1 ( u, ) ( ) ( u, ) ( ) ( u, ) ( ) ( tutdt, ) ( ) 3 4 (.5) To fnd the 1 st appromate soluton of (1.4), we substtute u 1 ( ) E E 1, wth E f() nto (.) and then settng. Ths leads to a formula for the unnown parameter E 1 of the form (1!) E f () (, ) E (.6) 1 Smlarly, to fnd the nd appromate soluton of (1.4), we substtute u ( ) E E E 1 nto (.4) and then settng. Ths leads to a formula for the unnown parameter E of the form (!) E f () (,) E (,) E (,) E (.7) 1 1 Hence, by repeatng the same procedure above, we can fnd the followng formulas for the unnown parameters E 3 and E4 (3!) E f () (,) E (,) E (,) E 3 1 (,) E (,) E (,) E 1 1 1 (.8) 13

(4!) E f () 6 (,) E 6 (,) E 3 (,) E 4 3 1 (,) E (,) E (,) E (,) E 1 1 1 1 (,) E (,) E (,) E 1 3 (.9) Ths leads to 4 th appromate soluton of the ntegral equaton (1.5) of the form u ( ) E E E E E (.11) 3 4 4 1 3 4 Note that, the appromate soluton (.1) can be used as an algorthm to fnd numercal soluton for the ntegral equaton (1.5) and hence the ntal value problem (1.1). At the end, f we suppose that coeffcents of the ntal value problem (1.1) and g ( ) are analytcal functons, then t s clear that f ( ) and t (, ) wll be analytc functons wth respect to all ther arguments. In ths case, f u ( ) h ( ) s an eact soluton of the ntegral equaton (1.5), whch we assume t s nfntely dfferentable functon n the neghborhood of, then ths procedure obtans the Taylor epanson of the eact soluton u h( ) around, and hence the Taylor epanson of the eact soluton of ntal value problem (1.1). Ths result can be proven by usng the same procedure of refs. [6,7]. 3. Specal Case In ths secton, we nvestgate ths procedure for frst order lnear ordnary dfferental equaton, or ndependently of the ntal value problem (1.1), for the case when the ernel of the Volterra ntegral equaton (1.4) s a functon of only, say t (, ) H ( ). In ths case, we can rewrte the formulas (.6-9) n the forms (1!) E f () H() E (3.1) 1 (!) E f () H () E H() E (3.) 1 (3!) E f () 3 H () E 3 H () E H() E (3.3) 3 1 14

(4!) E f () 4 H () E 6 H () E 8 H () E 6 H() E (3.4) 4 1 3 In ths case we can ntroduce a close form formula for the parameters E ; 1,,3,...,. To do that, we compare the coeffcents of the equatons (3.1-4) wth the bnomal coeffcents. Ths comprsng study ntroduces the followng recursve formula for the parameters E f () 1!, 1,,3,... (3.5) ( ) ( ) E H E 1!! 1 ( )( )! Ths leads to the th appromate soluton u ( ) E E (3.6) 1 whch can be used as an algorthm to compute the numercal solutons for the converted ntegral equaton. Furthermore, n ths case, f the coeffcents of the ntal value problem (1.1) and g ( ) are analytcal functons, then (3.6), wth coeffcents gven by the formula (3.5), leads to the Taylor epanson of the eact soluton of converted ntegral equaton. 4. Numercal Applcatons In ths secton, we demonstrate the sgnfcant features of ths method. To do that, we apply ths technque on selected lnear ordnary dfferental equatons and compare our results wth results obtaned by other methods. Eample (1) consder the ntal value problem y y ; y() 3 (4.1) The eact soluton of the ntal value problem (3.1) s y ( ) 1/ (7/) e, whch has the Taylor seres epanson 7 7 4 7 6 7 8 y ( ) 3... (4.) 4 1 48 15

Now (4.1) can be converted to an ntegral equaton of the form (1.4) wth t (, ) and f ( ) 7. Hence by usng the recursve formula (3.5) wth E and H( ), we were able to fnd E1 7, E, E3 7, E4, E5 7/, E6, E7 7/6, E8 and E9 7/4. Ths leads to the followng 9 th appromate soluton 7 7 7 6 4 3 5 7 9 u 9( ) 7 7, (4.3) Now by usng (1.) and the ntal condtons, we can wrte the 1 th the ntal value problem (4.1) n the form appromate soluton of 7 7 7 7 7 ( ) 3 4 1 48 4 4 6 8 1 y 1 (4.4) At the end, wth the help of the Mathematca Pacages, we can wrte the soluton of the ntal value problem (4.1) n the form th appromate 1 7 1 4 1 6 1 8 1 1 ( 1) y ( ) (1... ) (4.5) 6 4 1! whch can be wrtten n the form 1 7 ( 1) y ( ) (4.6)! Hence, for ths eample, we can deduce the eact soluton as 1 7 ( 1) 1 7 y ( ) Lm y ( ) e! Eample (): Consder the ntal values problem y 6y 3y 63y, y ( ) y () y () 1 (4.7) 16

Now a 1 ( ) 6, a ( ) 3, ( ) 63 3 a, c c1 c 1, g ( ) and n 3. Usng ths data and (1.4) and (1.5), we can convert the ntal value problem (4.7) to an ntegral equaton of the form u( ) 433 47 315 6 3( t) 315( t) u( t) dt (4.8) Hence, f () 433, f () 47, f () 63, () f () ; (,) 6, 1 (,) 33, (,) 63 and (,) ; ( 3,4,...). To fnd th appromate soluton of the ntegral equaton (4.8), we used ths data wth E 433. Now, the formulas (.6-9) leads to the 4 th appromate soluton 7361 36881 3 474353 4 u ( ) 433 11665 (4.9) 4 6 4 In addton, f mae use of (1.) and the ntal condtons, we can wrte the 7 th soluton of the ntal value problem (4.7) n the form appromate 1 433 3 11665 4 7361 5 y ( ) 1 7 6 4 1 36881 6 474353 7 (4.1) 7 54 At the end, we can use (4.1) as an algorthm to fnd a numercal soluton of the ntal value problem (4.7). Furthermore, the graphs (4.1) and (4.), show the plottng of the eact 7 9 1 soluton (red) ( y( ) 1e 7e 16e ), aganst the appromate solutons y 3 ( ) and y ( ) 4 respectvely. 17

Graph (4.1) Plottng the eact soluton aganst y 3 ( ) Graph (4.) Plottng the eact soluton aganst y 4 ( ) Eample (3): Consder the ntal values problem y 3y y 3, y () = 1, y () = (4.11) In ref.[9], ths ntal value problem has been solved by usng the dfferental transformaton method. Ths soluton was gven n term of the power seres 1 1 3 1 4 1 5 y ( ) 1... (4.1)! 3! 4! 5! Now by usng a 1 ( ) 3, a ( ), c 1 c, g ( ) 3 convert the ntal value problem (4.11) to an ntegral equaton of the form and n, we can 18

u ( ) 1 3 ( t) utdt ( ) (4.13) Usng the same prevous proceedng we can fnd f () 1, f (), () f (), (,) 3, (,) 1, and (,) ; (,3,...). Hence, the formulas (.6-9) wth E 1, gves E1 1, E 1/, E3 1/6 and E4 1/4. Ths leads to the 4 th appromate soluton 1 1 3 1 4 u ( ) 1 (4.14) 4 6 4 At the end, f mae use of (1.) and the ntal condtons, we can wrte the 6 th soluton of the ntal value problem (4.11) n the form appromate 1 1 3 1 4 1 5 1 6 y ( ) 1 (4.15) 6 6 4 1 7 Wth the help of the Mathematca Pacages, ths proceedng enables us to wrte the appromate soluton of the ntal value problem (4.11) n the form th y 1 1 3 1 4 1 1 ( ) 1...! 3! 4!!! (4.16) Hence, t s clear that the eact soluton s gven by 1 y( ) Lm y ( ) e (4.17)! Eample (4): Consder the ntal value problem y y, (4.18) y() d =, 1 y () = d (4.19) 19

The equaton (4.18), whch s nown as the Ary equaton, has no soluton n term of elementary functons, as many other mportant ordnary dfferental equatons. The most mportant soluton of ths equaton s the Ary functon of the frst nd A(, ) whch ts propertes have been etensvely studed [1,11]. Adoman [4] appled the decomposton method technque to fnd the seres soluton of the Ary equaton (4.18), however, after hard wor, the seres soluton method gves the seres soluton n the form æ 3 ö ç å = 1 ( 3)(5 6) ((3-1) (3 )) y ( ) = d 1+ + è ø æ 3+ 1 ö 1 + ç å = 1 (3 4)(6 7) ((3 ) (3+ 1)) d è ø (4.) Now a 1 ( ), a ( ), c d c d1, g ( ) and n. Usng ths data, we can convert the ntal value problem (4.18-19) to an ntegral equaton of the form u ( ) d d ( tutdt ) ( ) 1 (4.1) As n the prevous eample, we can calculate f (), f () d, f () d1 () f (), (, ), (, ), 1 (, ) and (,) ; ( 3,4,...). Hence the formulas (.6-9) wth E, gves E 1 d, E d1, E3 and E d /6 4. Ths leads to the 4th appromate soluton d 4 u ( ) d d (4.) 4 1 6 Now, f mae use of (1.) and the ntal condtons, we can wrte the 6 th soluton n the form appromate d 3 d 1 4 d 6 y ( ) d d (4.3) 6 1 6 1 18 Ths can be wrtten n the form

1 3 1 6 1 4 y ( ) d 1 6 d ( 3) ( 3)(5 6) 1 (4.4) (3 4) whch agrees wth the seres soluton (4.). Note that, ths study showed the accuracy, the relablty and the effcency of ths method comparng wth dfferental transformaton method, the Adoman method and the seres soluton method. 5. Concluson In ths wor, we studed the soluton of the ntal value problems of ordnary dfferental equatons by applyng a recursve method technque. Ths study led to appromate solutons, whch can be used as algorthms to compute numercal solutons for the ntal value problem. In addton, ths study enables us to ntroduce a formula for specal case related to the order of the ordnary dfferental equatons and the ernel of the converted ntegral equaton. To demonstrate the sgnfcant features of ths approach, we appled ths technque on selected lnear ordnary dfferental equatons and compare our results wth the eact soluton and the results obtaned by dfferent technques such as the dfferental transformaton method the Adoman method and the seres soluton method. Ths comparson study shows the accuracy, the relablty and the effcency of ths method. Certanly much remans to be done on ntroducng the same procedure for other types of dfferental equatons, whch can be vewed as a fascnatng challenge for further research. 6. References [1] J. C. Butcher, Numercal Methods for Ordnary Dfferental Systems, John Wley & Sons, 3. [] J. C. Butcher, Numercal methods for ordnary dfferental equatons n the th century, Journal of Computatonal and Appled Mathematcs 15 () 1-9. [3] S. Chang and I. Chang, A new algorthm for calculatng one-dmensonal dfferental transform of nonlnear functons, Appled Mathematcs and Computaton 195 (8) 799 88. [4] G. Adoman, A revew of the decomposng method and some recent results for non-lnear equatons, Math. Comput. Model. 13(7) (199) 11-143. [5] D. Zngg, T. Chsholmm, Runge Kutta methods for lnear ordnary dfferental equatons, Appled Numercal Mathematcs 31 (1999) 7 38. 1

[6] A. Tahmasb, O. S. Fard, Numercal Soluton of Nonlnear Volterra Integral Equaton of the Second Knd by Power Seres, Journal of Informaton and Computng Scence, Vol. 3, No. 1 (8) 57-61. [7] A. Tahmasb, O. S. Fard, A Numercal Scheme to solve Nonlnear Volterra Integral Equatons System of the Second Knd, Journal of Informaton and Computng Scence, Vol. 3, No. (8) 97-13. [8] A. M. Wazwaz, A Frst Course n Integral Equatons, WSPC, New Jersey, (1997). [9] K. Batha and B. Batha, A New Algorthm for Solvng Lnear Ordnary Dfferental Equatons, World Appled Scences Journal 15 (1): 1774-1779, 11. [1] M. Abramowtz and I. Stegun, Handboo of mathematcal functons, Dover boos n mathematcs, 1965. [11] Danel Zwllnger, Handboo of dfferental equatons, 3rd edton, Academc Press, 1997.