A numerical method based on the reproducing kernel Hilbert space ...

A numerical method based on the reproducing kernel Hilbert space method

arXiv:1305.4445v1 [math.NA] 20 May 2013

for the solution of fifth-order boundary-value problems Mustafa Inc, Ali Akg¨ ul and Mehdi Dehghan Department of Mathematics, Science Faculty, Fırat University, 23119 Elazı˘ g / T¨ urkiye Department of Mathematics, Education Faculty, Dicle University, 21280 Diyarbakır / T¨ urkiye Department of Applied Mathematics, Faculty of Mathematics and Computer Science, Amirkabir University of Technology, No. 424, Hafez Ave., Tehran, Iran [email protected] Abstract: In this paper, we present a fast and accurate numerical scheme for the solution of fifth-order boundary-value problems. We apply the reproducing kernel Hilbert space method (RKHSM) for solving this problem. The analytical results of the equations have been obtained in terms of convergent series with easily computable components. We compare our results with spline methods, decomposition method, variational iteration method, Sinc-Galerkin method and homotopy perturbation methods. The comparison of the results with exact ones is made to confirm the validity and efficiency. Keywords: Reproducing kernel method, Series solutions, fifth-order boundary-value problems, Reproducing kernel space. 1. Introduction In this work we consider the numerical approximation for the fifth-order boundaryvalue problems of the form y (v) = f (x) y + g (x) ,

x ∈ [a, b] ,

(1.1)

with boundary conditions y (a) = A0 ,

y ′ (a) = A1 ,

y ′′ (a) = A2 ,

y (b) = B0 ,

y ′ (b) = B1 ,

(1.2)

where the functions f (x) and g (x) are continuous on [a, b] and A0 , A1 , A2 , B0 , B1 are finite real constants. For more details about computational code of boundary value problems, the reader is referred to [1-3].

1

This type of boundary-value problems arise in the mathematical modelling of viscoelastic flows and other branches of mathematical, physical and engineering sciences [4,5] and references therein. Theorems which list the conditions for the existence and uniqueness of solutions of such problems are thoroughly discussed in a book by Agarwal [6]. Khan [7] investigated the fifth-order boundary-value problems by using finite difference methods. Wazwaz [8] applied Adomian decomposition method for solution of such type of boundaryvalue problems. The use of spline function in the context of fifth-order boundary-value problems was studied by Fyfe [9], who used the quintic polynomial spline functions to develop consistency relations connecting the values of solution with fifth-order derivatives at the respective nodes. Polynomial sextic spline functions were used [10] to develop the smooth approximations to the solution of problems (1.1) and (1.2). Caglar et al. [11] have used sixth-degree B-spline functions to develop first-order accurate method for the solution two-point special fifth-order boundary-value problems. Noor and Mohyud-Din [12,13] applied variational iteration and homotopy perturbation methods for solving the problems (1.1) and (1.2), respectively. Khan [14] have used the non-polynomial sextic spline functions for the solution fifth-order boundary-value problems. El-Gamel [15] employed the sinc-Galerkin method to solve the problems (1.1) and (1.2). Lamnii et al. [16] developed and analyzed two sextic spline collocation methods for the problem. Siddiqi et al. [17,18] used the non-polynomial sextic spline method for special fifth-order problems (1.1) and (1.2). Wang et al. [19] attempted to obtain upper and lower approximate solutions of such problems by applying the sixth-degree B-spline residual correction method. In this paper, the RKHSM [20,21] will be used to investigate the fifth-order boundaryvalue problems. In recent years, a lot of attetion has been devoted to the study of RKHSM to investigate various scientific models. The RKHSM which accurately computes the series solution is of great interest to applied sciences. The method provides the solution in a rapidly convergent series with components that can be elegantly computed. The efficiency of the method was used by many authors to investigate several scientific applications. Geng and Cui [22] applied the RKHSM to handle the second-order boundary value problems. Yao and Cui [23] and Wang et al. [24] investigated a class of singular boundary value problems by this method and the obtained results were good. Zhou et al. [25] used the RKHSM effectively to solve second-order boundary value problems. In 2

[26], the method was used to solve nonlinear infinite-delay-differential equations. Wang and Chao [27], Li and Cui [28], Zhou and Cui [29] independently employed the RKHSM to variable-coefficient partial differential equations. Geng and Cui [30], Du and Cui [31] investigated the approximate solution of the forced Duffing equation with integral boundary conditions by combining the homotopy perturbation method and the RKHSM. Lv and Cui [32] presented a new algorithm to solve linear fifth-order boundary value problems. In [33,34], authors developed a new existence proof of solutions for nonlinear boundary value problems. Cui and Du [35] obtained the representation of the exact solution for the nonlinear Volterra-Fredholm integral equations by using the reproducing kernel Hilbert space method. Wu and Li [36] applied iterative reproducing kernel method to obtain the analytical approximate solution of a nonlinear oscillator with discontinuties. Recently, the method was apllied the fractional partial differential equations and multi-point boundary value problems [34-37]. For more details about RKHSM and the modified forms and its effectiveness, see [20-43] and the references therein. The paper is organized as follows. Section 2 is devoted to several reproducing kernel spaces and a linear operator is introduced. Solution represantation in W26 [a, b] has been presented in Section 3. It provides the main results, the exact and approximate solution of (1.1) and an iterative method are developed for the kind of problems in the reproducing kernel space. We have proved that the approximate solution converges to the exact solution uniformly. Some numerical experiments are illustrated in Section 4. There are some conclusions in the last section. 2. Preliminaries 2.1. Reproducing Kernel Spaces In this section, we define some useful reproducing kernel spaces. Definition 2.1. (Reproducing kernel). Let E be a nonempty abstract set. A function K : E × E −→ C is a reproducing kernel of the Hilbert space H if and only if   ∀t ∈ E, K (., t) ∈ H,  ∀t ∈ E, ∀ϕ ∈ H, hϕ (.) , K (., t)i = ϕ (t) .

(2.1)

The last condition is called ”the reproducing property”: the value of the function ϕ at

the point t is reproduced by the inner product of ϕ with K (., t) 3

Definition 2.2.    u(x) | u(x), u′ (x), u′′ (x), u′′′ (x), u(4) (x), u(5) (x)   W26 [0, 1] = are absolutely continuous in [0, 1],     u(6) (x) ∈ L2 [0, 1], x ∈ [0, 1], u(0) = u(1) = u′ (0) = u′ (1) = 0 = u′′ (0) = 0

        

The sixth derivative of u(x) exists almost everywhere since u(5) (x) is absolutely continuous. The inner product and the norm in W26 [0, 1] are defined respectively by hu(x), g(x)i W 6 = 2

5 X

(i)

(i)

u (0)g (0) +

Z

1

0

i=0

u(6) (x)g (6) (x)dx, u(x), g(x) ∈ W26 [0, 1],

and

kukW 6 = 2

q

hu, ui

W26

, u ∈ W26 [0, 1].

The space W26 [0, 1] is a reproducing kernel space, i.e., for each fixed y ∈ [0, 1] and any u(x) ∈ W26 [0, 1], there exists a function Ry (x) such that

u(y) = hu(x), Ry (x)iW 6 . 2

Definition 2.3.    u(x) | u(x) is absolutely continuous in [0, 1]  W21 [0, 1] = ,   u′ (x) ∈ L2 [0, 1], x ∈ [0, 1],

The inner product and the norm in W21 [0, 1] are defined respectively by

hu(x), g(x)i W 1 = u(0)g(0) + 2

Z

1 0

u′ (x)g′ (x)dx, u(x), g(x) ∈ W21 [0, 1],

and

kukW 1 = 2

q

hu, ui

W1 2

, u ∈ W21 [0, 1].

The space W21 [0, 1] is a reproducing kernel space and its reproducing kernel function Tx (y) is given by

4

,

Tx (y)=

   1 + x,       1 + y,

x ≤ y, (2.2) x > y.

Theorem 2.1. The space W26 [0, 1] is a reproducing kernel Hilbert space whose reproducing kernel function is given by,

Ry (x) =

 P 12 i−1 ,   i=1 ci (y) x  

    P12 d (y) xi−1 , i=1 i

where

x ≤ y,

x > y,

c1 (y) = 0,

c2 (y)= 0,

c3 (y) = 0,

c4 (y) =

2461 2461 335 y5 + y6 + y7 42301989 253811934 507623868

−

7325 11725 4 8321 y 10 − y8 − y 10660101228 1776683538 84603978

+

56255 2003 12221 y3 + y9 + y 11 , 169207956 21320202456 21320202456

c5 (y) = −

158419 158419 99305 y5 − y6 + y7 1353663648 8121981888 14213468304

+

335 728021 157481 y 10 − y8 + y4 341123239296 16243963776 2707327296

−

11725 3 196265 2687 y − y9 − y 11 , 84603978 170561619648 42640404912 5

4056701 2011 158419 y5 − y6 + y7 67683182400 126905967 284269366080

c6 (y) =

c7 (y) =

c8 (y) =

c9 (y) =

+

2461 158419 11843 y 10 + y8 − y4 304574320800 284269366080 1353663648

+

2461 168263 8999 y3 − y9 − y 11 , 42301989 1705616196480 1705616196480

2011 158419 4056701 y5 − y6 + y7 406099094400 7614358020 1705616196480

+

11843 2461 158419 y 10 + y8 − y4 1827445924800 1705616196480 8121981888

+

168263 8999 2461 y3 − y9 − y 11 , 253811934 10233697178880 10233697178880

158419 158419 19861 y5 + y6 − y7 284269366080 1705616196480 596965668768

−

67 104003 157481 y 10 + y8 − y4 71635880252160 682246478592 81219818880

+

39253 2687 335 y3 + y9 + y 11 , 507623868 7163588025216 8954485031520

2461 2461 67 y5 + y6 + y7 284269366080 1705616196480 682246478592

−

1465 335 8321 y 10 − y8 − y4 71635880252160 2387862675072 16243963776

+

12221 11251 2003 y3 + y9 + y 11 , 1137077464320 28654352100864 143271760504320

6

c10 (y) = −

c11 (y) =

168263 168263 39253 y5 − y6 + y7 1705616196480 10233697178880 7163588025216

+

11251 196265 38153 y 10 + y8 − y4 85963056302592 28654352100864 170561619648

+

56255 6313 12751 1 y3 − y9 − y 11 − y2, 21320202456 5372691018912 214907640756480 725760

11843 157481 11843 y5 + y6 − y7 304574320800 1827445924800 71635880252160

−

45611 8321 157481 y 10 − y8 + y4 268634550945600 71635880252160 341123239296

−

38153 49001 1 8321 y3 + y9 + y 11 + y, 10660101228 85963056302592 2149076407564800 3628800

c12 (y) = −

8999 8999 2687 y5 − y6 + y7 1705616196480 10233697178880 8954485031520

+

2003 2687 49001 y 10 + y8 − y4 2149076407564800 143271760504320 42640404912

+

12751 725 1 2003 y3 − y9 − y 11 − , 21320202456 214907640756480 236398404832128 39916800

d1 (y) = −

d2 (y) =

1 y 11 , 39916800

1 y 10 , 3628800

d3 (y) = −

1 y9, 725760

7

12221 12221 2461 y3 + y8 + y5 169207956 1137077464320 42301989

d4 (y) =

+

335 8321 2461 y6 + y7 − y 10 253811934 507623868 10660101228

−

11725 4 56255 2003 y + y9 + y 11 , 84603978 21320202456 21320202456

104003 158419 728021 y4 − y7 − y5 2707327296 81219818880 1353663648

d5 (y) =

−

158419 157481 335 y6 + y 10 − y8 8121981888 341123239296 16243963776

−

196265 2687 11725 3 y − y9 − y 11 , 84603978 170561619648 42640404912

4056701 4056701 158419 y5 + y6 + y7 67683182400 406099094400 284269366080

d6 (y) =

+

2461 158419 11843 y 10 + y8 − y4 304574320800 284269366080 1353663648

+

168263 8999 2461 y3 − y9 − y 11 , 42301989 1705616196480 1705616196480

d7 (y) = −

2011 2011 158419 y5 − y6 + y7 6768318240 7614358020 1705616196480

+

2461 158419 11843 y 10 + y8 − y4 1827445924800 1705616196480 8121981888

+

2461 168263 8999 y3 − y9 − y 11 , 253811934 10233697178880 10233697178880

8

158419 158419 19861 y5 + y6 − y7 284269366080 1705616196480 596965668768

d8 (y) =

d9 (y) =

−

67 99305 157481 y 10 + y8 + y4 71635880252160 682246478592 14213468304

+

335 39253 2687 y3 + y9 + y 11 , 507623868 7163588025216 8954485031520

2461 67 2461 y5 + y6 + y7 284269366080 1705616196480 682246478592

−

8321 1465 335 y 10 − y8 − y4 71635880252160 2387862675072 16243963776

−

11251 2003 7325 y3 + y9 + y 11 , 1776683538 28654352100864 143271760504320

d10 (y) = −

168263 168263 39253 y5 − y6 + y7 1705616196480 10233697178880 7163588025216

d11 (y) =

+

11251 196265 38153 y 10 + y8 − y4 85963056302592 28654352100864 170561619648

+

6313 12751 56255 y3 − y9 − y 11 , 21320202456 5372691018912 214907640756480

11843 11843 157481 y5 + y6 − y7 304574320800 1827445924800 71635880252160

−

8321 157481 45611 y 10 − y8 + y4 268634550945600 71635880252160 341123239296

−

8321 38153 49001 y3 + y9 + y 11 , 10660101228 85963056302592 2149076407564800

9

d12 (y) = −

2687 2003 8999 y4 + y3 − y5 42640404912 21320202456 1705616196480

−

2687 2003 8999 y6 + y7 + y8 10233697178880 8954485031520 143271760504320

−

12751 725 49001 y9 − y 11 + y 10 , 214907640756480 236398404832128 2149076407564800

Proof: hu(x), Ry (x)i

W26

5 X

=

u(i) (0)Ry(i) (0) +

i=0

Z

1

0

u(6) (x)Ry(6) (x)dx,

(2.3)


u(x), Ry (x) ∈ W26 [0, 1] ,

Through several integrations by parts for (2.3) we have

hu(x), Ry (x)i

W6 2

=

5 X i=0

+

i h u(i) (0) Ry(i) (0) − (−1)(5−i) Ry(11−i) (0)

5 X

(5−i) (i)

(−1)

u

(1)Ry(11−i) (1)+

i=0

Z

1 0

(2.4)

u(x)R(12) y (x)dx.

Note that property of the reproducing kernel

hu(x), Ry (x)iW 6 = u(y), 2

If  (5) (6)  Ry (0) − Ry (0) = 0,      (4) (7)  Ry (0) + Ry (0) = 0,      (8)   R′′′ (0) − Ry (0) = 0,   y (6)

Ry (1) = 0,

then (2.4) implies that,

               

(7)

Ry (1) = 0, (8)

Ry (1) = 0, (9)

Ry (1) = 0,

10

(2.5)

Ry(12) (x) = δ(x − y), When x 6= y,

Ry(12) (x) = 0, therefore

Ry (x) =

 P 12 i−1 , x ≤ y,    i=1 ci (y)x 

    P12 d (y)xi−1 , x > y. i=1 i

Since

Ry(12) (x) = δ(x − y), we have

∂ k Ry+ (y) = ∂ k Ry− (y), k = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

(2.6)

∂ 11 Ry+ (y) − ∂ 11 Ry− (y) = 1.

(2.7)

and

Since Ry (x) ∈W26 [0, 1], it follows that Ry (0) = 0, Ry (1) = 0, Ry′ (0) = 0, Ry′ (1) = 0, Ry′′ (0) = 0.

(2.8)

From (2.5)-(2.8), the unknown coefficients ci (y) ve di (y) (i = 1, 2, ..., 12) can be obtained. Thus for x ≤ y, Ry (x) is given by,

11

Ry (x) =

2461 2461 335 8321 x3 y 5 + x3 y 6 + x3 y 7 − x3 y 10 42301989 253811934 507623868 10660101228

−

11725 3 4 12221 56255 7325 x3 y 8 − x y + x3 y 3 + x3 y 9 1776683538 84603978 169207956 21320202456

+

2003 158419 158419 99305 x3 y 11 − x4 y 5 − x4 y 6 + x4 y 7 21320202456 1353663648 8121981888 14213468304

+

335 728021 11725 4 3 157481 x4 y 10 − x4 y 8 + x4 y 4 − x y 341123239296 16243963776 2707327296 84603978

−

2687 4056701 2011 196265 x4 y 9 − x4 y 11 + x5 y 5 − x5 y 6 170561619648 42640404912 67683182400 126905967

+

11843 2461 158419 5 4 158419 x5 y 7 + x5 y 10 + x5 y 8 − x y 284269366080 304574320800 284269366080 135366364

+

2461 168263 8999 4056701 x5 y 3 − x5 y 9 − x5 y 11 + x6 y 5 42301989 1705616196480 1705616196480 40609909440

−

158419 11843 2461 2011 x6 y 6 + x6 y 7 + x6 y 10 + x6 y 8 7614358020 1705616196480 1827445924800 170561619

−

2461 168263 8999 158419 x6 y 4 + x6 y 3 − x6 y 9 − x6 y 11 8121981888 253811934 10233697178880 1023369717888

+

158419 158419 19861 157481 7 10 x7 y 5 + x7 y 6 − x7 y 7 − x y 284269366080 1705616196480 596965668768 716358802

+

104003 335 39253 67 x7 y 8 − x7 y 4 + x7 y 3 + x7 y 9 682246478592 81219818880 507623868 7163588025216

+

2461 2461 67 2687 x7 y 11 + x8 y 5 + x8 y 6 + x8 y 7 8954485031520 284269366080 1705616196480 682246478

12

−

8321 1465 335 12221 x8 y 10 − x8 y 8 − x8 y 4 + x8 y 3 71635880252160 2387862675072 16243963776 1137077464320

+

2003 168263 168263 11251 x8 y 9 + x8 y 11 − x9 y 5 − x9 y 6 28654352100864 143271760504320 1705616196480 1023369717

+

39253 38153 11251 196265 x9 y 7 + x9 y 10 + x9 y 8 − x9 y 4 7163588025216 85963056302592 28654352100864 17056161964

+

6313 12751 1 56255 x9 y 3 − x9 y 9 − x9 y 11 − x9 y 2 21320202456 5372691018912 214907640756480 725760

+

11843 157481 45611 11843 x10 y 5 + x10 y 6 − x10 y 7 − x10 y 10 304574320800 1827445924800 71635880252160 2686345509

−

157481 8321 38153 8321 x10 y 8 + x10 y 4 − x10 y 3 + x10 y 9 71635880252160 341123239296 10660101228 859630563025

+

49001 1 8999 8999 y 11 + y− x11 y 5 − x11 y 6 2149076407564800 3628800 1705616196480 10233697178880

+

49001 2003 2687 x11 y 7 + x11 y 10 + x11 y 8 8954485031520 2149076407564800 143271760504320

−

2003 12751 725 2687 x11 y 4 + x11 y 3 − x11 y 9 − . 42640404912 21320202456 214907640756480 236398404832128

3. Solution represantation in W26 [0, 1] In this section, the solution of equation (1.1) is given in the reproducing kernel space W26 [0, 1]. On defining the linear operator L : W26 [0, 1] → W21 [0, 1] as Lu = u(5) (x) − f (x)u(x). Model problem (1.1) changes the following problem:

13

 

Lu = K(x), x ∈ [0, 1]  u(a) = 0, u′ (a) = 0, u′′ (a) = 0, u(b) = 0, u′ (b) = 0.

(3.1)

3.1. The Linear boundedness of operator L.

Lemma 3.1. If u(x) ∈ W26 [a, b], then u(k) (x) L∞ ≤ Mk ku(x)kW 6 , where Mk (k = 2

0, 1, ..., 5) are positive constants.

Proof: For any x ∈ [a, b] it holds that

kRx (y)kW 6 = 2

q

hRx (y), Rx (y)iW 6 = 2

p

Rx (x),

from the continuity of Rx (x), there exists a constant M0 > 0, such that kRx (y)kW 6 ≤ M0 . 2

By (2.1) one gets |u(x)| = hu(y), Rx (y)i

≤ kRx (y)k 6 ku(y)k 6 ≤ M0 ku(y)k 6 . W2 W2 W2 6 W

(3.2)

2

For any x, y ∈ [a, b], there exists Mk (k = 1, 2, ..., 5), such that

(k)

Rx (y)

we have

W26

≤ Mk (k = 1, 2, ..., 5),



D E (k) (k) (k) ≤ Rx (y) 6 ku(y)kW26 ≤ Mk ku(y)kW26 u (x) = u(y), Rx (y) W2 W6

(3.3)

2

(k = 1, 2, ..., 5).

Combining (3.2) and (3.3), it follows that

(k)

u (x)

L∞

≤ Mk ku(y)kW 6 (k = 1, 2, ..., 5). 2

Theorem 3.1. Suppose fi′ ∈ L2 [a, b] (i = 0, 1, 2, 3, 4), Then L : W26 [a, b] → W21 [a, b] is a bounded linear operator. Proof: (i) By the definition of the operator it is clear that L is a linear operator.

14

(ii) Due to the definiton of W21 [a, b] , we have k(Lu)(x)k2W 1 = h(Lu)(x), (Lu)(x)i W 1 2 2 Z b [(Lu)′ (x)]2 dx = [(Lu)(a)]2 + a #2 Z  5 ! ′ 2 " 5 b X X  fi (a)u(i) (a) + fi (x)u(i) (x)  dx. = a

i=0

Z

b

[(Lu)′ (x)]2 dx =

a

Z b" a

i=0

#2 4 X (fi′ (x)u(i) (x) + fi (x)u(i+1) (x)) dx u(6) (x) + i=0

# Z b" Z bh 4 i2 X ′ (i) (i+1) (6) (6) (fi (x)u (x) + fi (x)u (x)) dx u (x) u (x) dx + 2 = a

i=0

(fi′ (x)u(i) (x)

(i+1)

a

+

Z b "X 4 a

+ fi (x)u

#2

(x))

i=0

dx,

where Z bh a

and

Z b" a

≤

2

# 4 X (fi′ (x)u(i) (x) + fi (x)u(i+1) (x)) dx u(6) (x)

Z b h a

i2 u(6) (x) dx ≤ ku(x)k2W 6 ,

i=0

 # 2 2 " 4  i 2  2 Z b X (fi′ (x)u(i) (x) + fi (x)u(i+1) (x)) dx . u(6) (x) dx   a i=0

By lemma (3.1) and fi′ (x) ∈ L2 [a, b], we can obtain a constant N > 0, satisfying Z b "X 4 a

#2

(fi′ (x)u(i) (x) + fi (x)u(i+1) (x))

i=0

dx ≤ N (b − a) ku(x)k2W 6 . 2

Furthermore one gets

Z

b a

p [(Lu)′ (x)]2 dx ≤ ku(x)k2W 6 + 2 N (b − a) ku(x)k2W 6 + N (b − a) ku(x)k2W 6 , 2

2

15

2

  p let G = 1 + 2 N (b − a) + N (b − a) > 0, then b

Z

a

[(Lu)′ (x)]2 dx ≤ G ku(x)k2W 6 . 2

Therefore L is a boundend operator. So we obtain the result as required.



3.2. The normal orthogonal function system of W26 [a, b] ∗ ∗ We choose {xi }∞ i=1 as any dense set in [a, b] and let Ψx (y) = L Tx (y), where L

is conjugate operator of L and Tx (y) is given by (2.2). Furthermore, for simplicity let Ψi (x) = Ψxi (x), namely, def

Ψi (x) = Ψxi (x) = L∗ Txi (x). Now several lemmas are given. 6 Lemma 3.2. {Ψi (x)}∞ i=1 is complete system of W2 [a, b].

Proof: For u(x) ∈ W26 [a, b], let hu(x), Ψi (x)i = 0 (i = 1, 2, ...), that is hu(x), L∗ Txi (x)i = (Lu)(xi ) = 0. Note that {xi }∞ i=1 is the dense set in [a, b], therefore (Lu)(x) = 0. It follows that u(x) = 0 from the existence of L−1 .



Lemma 3.3. The following formula holds

Ψi (x) = (LηRx (η)) (xi ) , where the subscript η of operator Lη indicates that the operator L applies to function of η. Proof: Ψi (x) = hΨi (ξ), Rx (ξ)iW 6 [a,b] 2

∗

= hL Txi (ξ) , Rx (ξ)iW 6 [a,b] 2

= h(Txi ) (ξ) , (Lη Rx (η)) (ξ)iW 1 [a,b] 2

= (Lη Rx (η)) (xi ) . 16

This completes the proof.



∞  Remark 3.1. The orthonormal system Ψi (x) i=1 of W26 [a, b] can be derived from

Gram-Schmidt orthogonaliztion process of {Ψi (x)}∞ i=1 ,

Ψi (x) =

i X

β ik Ψk (x), (β ii > 0, i = 1, 2, ...)

(3.7)

k=1

where β ik are orthogonal cofficients. In the following, we will give the represantation of the exact solution of Eq.(1.1) in the reproducing kernel space W26 [a, b]. 3.3. The structure of the solution and the main results Theorem 3.2. If u(x) is the exact solution of Eq.(1.1), then

u(x) =

i ∞ X X

β ik g(xk )Ψi (x),

i=1 k=1

where {xi }∞ i=1 is a dense set in [a, b].

Proof: From the (3.7) and uniqeness of solution of (1.1) (see [32]), we have

u(x) = =

=

=

=

∞ X

 u(x), Ψi (x) W 6 Ψi (x)

i=1 i ∞ X X

i=1 k=1 ∞ X i X

i=1 k=1 ∞ X i X

i=1 k=1 i ∞ X X

2

β ik hu(x), L∗ Txk (x)iW 6 Ψi (x) 2

β ik hLu(x), Txk (x)iW 1 Ψi (x) 2

β ik hg(x), Txk (x)iW 1 Ψi (x) 2

β ik g(xk )Ψi (x).



i=1 k=1

Now the approximate solution un (x) can be obtained by truncating the n− term of the exact solution u(x),

17

un (x) =

n X i X

β ik g(xk )Ψi (x).

i=1 k=1

Theorem 3.3. Assume u(x) is the solution of Eq.(1.1) and rn (x) is the error between the approximate solution un (x) and the exact solution u(x). Then the error sequence rn (x) is monotone decreasing in the sense of k.kW 6 and krn (x)kW 6 → 0 [23]. 2

2

4. Numerical Results In this section, four numerical examples are provided to show the accuracy of the present method. All computations are performed by Maple 13. The RKHSM does not require discretization of the variables, i.e., time and space, it is not effected by computation round off errors and one is not faced with necessity of large computer memory and time. The accuracy of the RKHSM for the fifth-order boundary value problems is controllable and absolute errors are small with present choice of x (see Table 1-4). The numerical results we obtained justify the advantage of this methodology. Example 4.1. ([8,11,12]). We first consider the linear boundary value problem   y (5) (x) = y − 15ex − 10xex , 0 < x < 1 (4.1)  y(0) = 0, y ′ (0) = 1, y ′′ (0) = 0, y(1) = 0, y ′ (1) = −e

The exact solution of (4.1) is

y (x) = x(1 − x)ex .

18

If we homogenize the boundary conditions of (4.1), then the following (4.2) is obtained    5 −3 4 (5) x 2  3  u (x) − u(x) = 1 − 5e 2 − 3x + 3x (2 − ) + 4x ( + )   e 2 e             5 −3 4  x 2  −10e −3 + 6x(2 − ) + 12x ( + )   e 2 e            −3 4 −3 4 5 (4.2) x x + ) − 5e 24( + ) −10e 6(2 − ) + 24x(   e 2 e 2 e             −15ex − 10xex , 0