The Analytical Solution of a System of Nonlinear Differential Equations

Transcription

1 Int. Journal of Math. Analyss, Vol. 1, 007, no. 10, The Analytcal Soluton of a System of Nonlnear Dfferental Equatons Yunhu L a, Fazhan Geng b and Mnggen Cu b1 a Dept. of Math., Harbn Unversty Harbn, Helongjang, P. R. Chna b Dept. of Math., Harbn Insttute of Technology Weha, Shandong, P. R. Chna 6409 Abstract In ths paper, the analytcal soluton of a system of nonlnear dfferental equatons s obtaned n the reproducng kernel space W [0, 1]. The exact soluton s represented n the form of seres. The n term approxmaton y j,n (x),j =1,,,p are proved to converge to exact soluton y j (x),j =1,,,p. More mportantly, the method s mplementaton requres no addtonal condtons. Some examples are presented to demonstrate the relablty and effcency of the algorthm developed. Keywords: Analytcal soluton, nonlnear system of dfferental equatons, reproducng kernel 1 Introducton Systems of dfferental equatons whch are often encounters n applcatons. Most realstc systems of ordnary dfferental equatons do not have analytcal solutons so that the numercal technque must be used. They can be readly solved by many methods, such as the smple Taylor seres method and fourthorder Runge-Kutta method [1,],the Tau Method[3,4,5] and the Adoman s decomposton method [7,8]. Over the past few years, many new alternatves to the use of tradtonal methods for the numercal soluton of systems of dfferental equatons have been proposed. In ref.[6], the author present the operatonal approach to the Tau Method for the numercal soluton of mxed-order systems of lnear ordnary dfferental equatons wth polynomal or ratonal 1 Ths paper s the result of research tem on Scence and Technology of educaton offce n Helongjang Provnce. Item Number:

2 45 Yunhu L, Fazhan Geng and Mnggen Cu polynomal coeffcents, together wth ntal or boundary condtons. Dogan Kaya deals wth the mplementaton of the Adoman s decomposton method n chemcal applcatons[9]. In ref. [10], the Adoman s decomposton method s appled to ntal problems for systems of ordnary dfferental equatons n both lnear and nonlnear cases. In ths paper, we wll consder general problems manly focus on systems of nonlnear dfferental equatons. Snce every ordnary dfferental equaton of order n can be wrtten as a systems consstng of n frst-order ordnary dfferental equatons, we restrct our study to a system of frst-order dfferental equatons. We consder the followng systems of nonlnear ordnary dfferental equatons: Ly j = dy j dx = f j(x, y 1 (x),y (x),,y p (x)),y j (0) = 0,j =1,,,p, (1.1) where, y 1 (x),y (x),,y p (x) s the soluton of (1.1), and y j (x) W [0, 1],j = 1,,,p. In ths paper, the exact soluton of (1.1) s obtaned n the reproducng kernel space W [0, 1]. The analytcal soluton s represented n the form of seres. The n term approxmaton y j,n (x),j =1,,,p are proved to converge to the exact soluton y j (x),j =1,,,p. Prelmnary In ths secton, the reproducng kernel space s defned to solve (1.1). Space W [0, 1] s defned as W [0, 1] = {u u s absolutely contnuous functon and u L [0, 1],u(0) = 0}. The nner product (, ) and the the norm W [0,1] are taken to be (u(y),v(y)) def = 1 0 (4uv +5u v + u v )dy, u, v W [0, 1] (.1) def u W = (u, u) respectvely. Space W 1[0, 1] s defned by W 1 [0, 1] = {u u s absolutely contnuous functon and u L [0, 1]} equpped wth the nner product u, v def = 1 0 uvdx u v dx, u, v W 1 [0, 1] and norm u W 1 def = u, u

3 The analytcal soluton of a system 453 respectvely. In reference [11] the authors proved that W [0, 1],W1 [0, 1] are reproducng kernel spaces wth the kernels R x (y) = { R1 (x, y),y x; R (x, y),y >x; (.) where and R 1 (x, y) = e x y (9e 4 +7e 6 7e 4x 8e 3+x +8e 3+3x 9e +4x ) 1( 7 9e +9e 4 +7e 6 ) + e x+y ( 9e 4 7e 6 +7e 4x +8e 3+x 8e 3+3x +9e +4x ) 1( 7 9e +9e 4 +7e 6 ) + e x+y (4e 3 7e 3x 9e +x +7e 6+x +9e 4+3x 4e 3+4x ) 6( 7 9e +9e 4 +7e 6 ) + e x y ( 4e 3 +7e 3x +9e +x 7e 6+x 9e 4+3x +4e 3+4x ) 6( 7 9e +9e 4 +7e 6 ) R (x, y) = e x+y ( 1+e x )(7+9e +7e x 8e 3+x +9e +x ) 1( 7 9e +9e 4 +7e 6 ) + e3 x y ( 1+e x )(9e+7e 3 8e x +9e 1+x +7e 3+x ) 1( 7 9e +9e 4 +7e 6 ) + e x y ( 4e+9e x 9e 3x 7e 4+x +7e 4+3x +4e 1+4x ) 6( 7 9e +9e 4 +7e 6 ) + e x+y (4e 3 +7e x 7e 3x 9e 4+x +9e 4+3x 4e 3+4x ) 6( 7 9e +9e 4 +7e 6 ) 1 R x (y) = [cosh(x + y 1) + cosh( x y 1)]. snh(1) Then, for any u W [0, 1],v W 1 [0, 1] and a fxed x, t follows that u(x) =(u(ξ),r x (ξ)),v(x) = v(ξ), R x (ξ). 3 The method In ths secton, the analytcal soluton of (1.1) s gven n the reproducng kernel space W [0, 1]. It s clear that L : W [0, 1] W 1 [0, 1] s a bounded lnear operator. Let ϕ (x) =R x (x), R x (y) s the reproducng kernel of W 1 [0, 1], and ψ (x) =L ϕ (x), L s the conjugate operator of L. Theorem 3.1. If {x } =1 s dense on [0, 1], then {ψ (x)} =1 system of W [0, 1]. s the complete Proof. For u(x) W [0, 1], let (ψ (x),u(x)) = 0, It follows ϕ (x),lu(x) = Lu(x )=0,,=1,, Note that {x } =1 s dense on [0,1], hence, Lu(x) = 0. It follows that u(x) =0 from the exstence of L.

4 454 Yunhu L, Fazhan Geng and Mnggen Cu Practse Gram-Schmdt orthonomalzaton for {ψ (x)} =1, ψ (x) = β k ψ k (x). (3.1) k=1 Then { ψ (x)} =1 s the complete orthonormal bass of W [0, 1]. Theorem 3.. Let {x } =1 be dense on [0, 1], f the soluton of (1.1) s unque, then the soluton satsfes the form y j (x) = β k f j (x k,y 1 (x k ),y (x k ),y p (x k )) ψ (x),j =1,,,p. =1 k=1 (3.) Proof. Note that u(x),ϕ (x) = u(x ) and { ψ (x)} =1 of W [0, 1], hence s an orthonormal bass y j (x) = =1 (y j(x), ψ (x)) ψ (x) = =1 (y j(x), k=1 β kψ k (x)) ψ (x) = =1 k=1 β k(y j (x),ψ k (x)) ψ (x) = =1 k=1 β k Ly j (x),ϕ k (x) ψ (x) = =1 k=1 β kf j (x k,y 1 (x k ),y (x k ),,y p (x k )) ψ (x), 1 j p. (3.3) 4 Implementaton In ths secton, a new method of solvng (1.1) s presented. (3.) can be denoted by y j (x) = =1 A (j) ψ (x),j =1,,,p, (4.1) where A (j) = β k f j (x k,y 1 (x k ),y (x k ),,y p (x k )). k=1 Let x 1 = 0, t follows that y 1 (x 1 ),y (x 1 ),,y p (x 1 ) are known from the ntal condtons. So f j (x 1,y 1 (x 1 ),y (x 1 ),,y p (x 1 ))(j =1,, p) are known. Consderng the numercal computaton, we put y 1,0 (x 1 )=y 1 (x 1 ),y,0 (x 1 )=

5 The analytcal soluton of a system 455 y (x 1 ),,y p,0 (x 1 )=y p (x 1 ) and the n term approxmaton to y j (x), j= 1,,,p by y j,n (x) = n =1 ψ (x),j =1,,p (4.) where 1 = β 11 f j (x 1,y 1,0 (x 1 ),y,0 (x 1 ),,y p,0 (x 1 )), y j,1 (x) = 1 ψ 1 (x), = k=1 β kf j (x k,y 1,k 1 (x k ),y,k 1 (x k ),,y p,k 1 (x k )), y j, (x) = =1 B(j) ψ (x), B n (j) = n k=1 β nkf j (x k,y 1,k 1 (x k ),y,k 1 (x k ),,y p,k 1 (x k )), (4.3) j =1,,,p. Lemma 4.1. If y j (x) W [0, 1],j =1,,,p, then there exsts a constant C > 0, such that y j (x) C y j (x) W, y j(x) C y j (x) W,j = 1,,,p. Proof. For any y j (x) W [0, 1], y j (x) = (y j (ξ),r x (ξ)) y j (x) W R x (ξ) W then there exsts a C j > 0, such that Note that y j (x) C j y j (x) W,j =1,,,p, y j(x) =(y j (ξ), d dx R x(ξ)) y j (x) W d dx R x(ξ) W,j =1,,,p, Hence there exsts a C j > 0, such that y j(x) C j y j W. Put C = max{c j,c j,j =1,,,p}, then the proof s complete. Lemma 4.. If y j,n (x) W y j (x)(n ), and y j,n (x),j =1,,,p are bounded, then f j (x n,y 1,n 1 (x n ),y,n 1 (x n ),,y j,n 1 (x n )) converges to f j (x, y 1 (x), y (x),, y p (x))(n ),j =1,,,p.

6 456 Yunhu L, Fazhan Geng and Mnggen Cu Proof. From the gven condton y j,n (x) W y j (x)(n ),j =1,,,p and Lemma 4.1, t follows that, for x [0, 1] Observng that y j,n 1 (x) y j (x) 0(n ), y j,n 1(ξ) C y j,n 1 (x) W, y j,n 1 (x n ) y j (x) = y j,n 1 (x n ) y j,n 1 (x)+y j,n 1 (x) y j (x), = y j,n 1 (ξ) x n x + y j,n 1 (x) y j (x), by the boundedness of y j,n W, we get y j,n 1 (x n ) y j (x) 0,n, The contnuaton of f j (x, y 1 (x),y (x),,y p (x) mples that f j (x n,y 1,n 1 (x n ),y,n 1 (x n ),,y j,n 1 (x n )) f j (x, y 1 (x), y (x),, y p (x)) (n ), j=1,,,p. Theorem 4.1. Assume y j,n (x) W are bounded n (4.), f {x } =1 s dense on [0, 1], then n term approxmate soluton y j,n (x) converges to the exact soluton y j (x),j =1,,,p of (1.1), and the exact soluton s expressed as y j (x) = =1 ψ (x),j =1,,,p, (4.4) where s gven by (4.3). Proof. Frst we wll prove the convergence of y j,n (x),j =1,,,p. (y j,n (x),y j,n (x)) = ( n =1 ψ (x), n =1 ψ (x)) = n =1 ( ). (4.5) Namely y j,n = n W =1 (B(j) ), t mples that y j,n (x),j =1,,,p are monotoncally ncreasng functons. Due to the condtons that y j,n (x) W s bounded, hence y j,n (x) W s convergent and there exsts a constant c, such that ) = c, j =1,,,p. Hence If m>n, then y j,m (x) y j,n (x) W =1 = ( m =n+1 l,j =1,,,p. ψ (x) W = m ( =n+1 ) 0,m,n,

7 The analytcal soluton of a system 457 Consderng the completeness of W [0, 1], we get y j,n (x) y j (x),n. Second, we wll prove that y j (x),j =1,,,p s the soluton of Eq.(1.1) On takng lmts n (4.) y j (x) = =1 Snce (Ly j )(x n )= =1 B(j) L ψ,ϕ n = 1,,,p, t follows that n β nl (Ly j )(x l )= ( ψ n, β nl ψ l )= l=1 If n = 1, then =1 l=1 ψ (x),j =1,,,p (4.6) =1 B(j) ( ψ,l ϕ n )= =1 =1 B(j) ( ψ,ψ n ),j = ( ψ, ψ n )=,j =1,,,p. (Ly j )(x 1 )=f j (x 1,y 1,0 (x 1 ),y,0 (x 1 ),,y p,0 (x 1 )),j =1,,,p. If n =, then β 1 (Ly j )(x 1 )+β (Ly j )(x )=β 1 f j (x 1,y 1,0 (x 1 ),y,0 (x 1 ),,y p,0 (x 1 )) + β f j (x,y 1,1 (x ),y,1 (x ),,y p,1 (x ))j =1,,,p. It s clear that (Ly j )(x )=f j (x,y 1,1 (x ),y,1 (x ),,y p,1 (x )),j =1,,,p. Moreover, t s easy to see by nducton that n (Ly j )(x l )=f j (x l,y 1,l 1 (x l ),y,l 1 (x l ),,y p,l 1 (x l )),l=1,,j =1,,,p. (4.7) Snce {x } =1 s dense on [0, 1], for z [0, 1], there exsts a subsequence {x nj } such that x nl z, l. Hence, let l n (4.7), by the convergence of y j,n and Lemma 4., we have (Ly j )(z) =f j (z, y 1 (z), y (z),, y p (z)),j =1,,,p. (4.8) From (4.8), t follows that y j,j =1,,,p satsfes (1.1). Snce ψ (x) W [0, 1], clearly, y j,j =1,,,psatsfes the ntal condtons of (1.1). That s, y j,j =1,,,p s the soluton of (1.1). The applcaton of unqueness of soluton of (1.1) then yelds that y j (x) = ψ (x),j =1,,,p. (4.9) =1

8 458 Yunhu L, Fazhan Geng and Mnggen Cu 5 Example In ths secton, some numercal examples wll be tested by usng the method dscussed above. For comparson reasons, the problems have known solutons. All experments were performed n MATHEMATICA 5.0. Example 1 Consder the system where dy 1 dx = y 1 + g 1 (x) dy dx = y 1 y + g (x) dy 3 dx = y + g 3 (x) g 1 (x) = 1+x + x, g (x) = e x ( e x( 1+x)x +( 1+x) x e x (1 3x + x ), g 3 (x) = e x ( 1+x) x +( 1+x)cosx + snx, y (x) W [0, 1], 0 x 1, =1,, 3 subject to boundary condtons y (0) = 0 whch has the exact soluton gven by y 1 (x) = x(x 1),y (x) = x(x 1)e x,y 3 (x) =(x 1)snx. The exact and approxmate solutons and the absolute error are dsplayed n Table 1 Table and Table 3 wth N = 100. Table 1: Node True soluton y 1 (x) Approxmate soluton y 1,100 (x) Absolute error E E E E E E E E E E E E-6 Example Consder the system dy 1 dx = y 1 + y y 3 + g 1 (x) dy dx = y 1 y y 3 y + g (x) dy 3 dx = y + g 3(x)

9 The analytcal soluton of a system 459 Table : Node True soluton y (x) Approxmate soluton y,100 (x) Absolute error E E E E E E E E E E E E-6 Table 3: Node True soluton y 3 (x) Approxmate soluton y 3,100 (x) Absolute error E E E E E E E E E E E E-6

10 460 Yunhu L, Fazhan Geng and Mnggen Cu where g 1 (x) = 1+x + x e x ( 1+x) xsnx, g (x) = e x ( e x( 1+x)x +( 1+x) x e x (1 3x + x ) +e x ( 1+x) xsnx, g 3 (x) = e x ( 1+x) x +( 1+x)cosx + snx, y (x) W [0, 1], 0 x 1, =1,, 3 subject to boundary condtons y (0) = 0 whch has the exact soluton gven by y 1 (x) = x(x 1),y (x) = x(x 1)e x,y 3 (x) =(x 1)snx. The exact and approxmate solutons and the absolute error are dsplayed n Table 4 Table 5 and Table 6 wth N = 100. Table 4: Node True soluton y 1 (x) Approxmate soluton y 1,100 (x) Absolute error E E E E E E E E E E E E-6 Table 5: Node True soluton y (x) Approxmate soluton y,100 (x) Absolute error E E E E E E E E E E E E-5

11 The analytcal soluton of a system 461 Table 6: Node True soluton y 3 (x) Approxmate soluton y 3,100 (x) Absolute error E E E E E E E E E E E E-6 References [1] W.Cheney,D.Kncad,Numercal Mathematcs and Computng,Books/Cole Publshng Company,Calforna,1985. [] C.F.Gerald,P.O.Wheatley,Appled Numercal Analyss, Addson-Wesley, Calforna, [3] M.R.Crsc and E.Russo, An extenson of Ortz recursve formulaton of the Tau Method to certan lnear systems of ordnary dfferental equatons, Maths, Comput.41(1983),7-4. [4] A.E.M.EMsery and E.L. Ortz,Tau-Lnes:Anew hybrd approach to the numercal treatment of crack problems based on the Tau Method,Comp.Meth.n Appl.Mech.and Engng.56(1986),65-8. [5] E.L.Ortz,The Tau Method,SIAM J.Numer Analyss, 6(1969), [6] K.M.Lu C.K.Pan,The Automatc Soluton to Systems of Ordnary Dfferental Equatons by the Tau Method,Computers and Mathematcs wth Applcatons, 38(1999), [7] G.Adoman, Solvng Fronter Problems of Physcs: The Decomposton Method, Kluwer Academc Publshers, Boston, MA, [8] G.Adoman, Arevew of the decompostonmethod appled mathematcs, J.Math.Anal.Appl.135(1988), [9] Dogan Kaya, A relable method for the numercal soluton of the knetcs problems, Appled Mathematcs and Computaton, 156(004),61-70.

12 46 Yunhu L, Fazhan Geng and Mnggen Cu [10] Nuran Guzel, Mustafa Bayram,On the numercal soluton of stff systems,appled Mathematcs and Computaton, 170 (005),30-36 [11] Chun-L L,Mng-Gen Cu,The exact soluton for solvng a class nonlnear operator equaton n reproducng kenel space.appled Mathematcs And Computaton, 143(-3)(003), Receved: December 1, 006