ISSN 1749-3889 (print), 1749-3897 (online)International Journal of Nonlinear Science

Vol.13(2012) No.4,pp.419-421

A Note On The Solutions of Nonlinear Fifth-Order Boundary Value Problems

S. A. Odejide , O. A. AdesinaDepartment of Mathematics, Obafemi Awolowo University, 220005, Ile-Ife, Nigeria.

(Received 21 February 2012, accepted 29 May 2012)

Abstract: In this article, the weighted residual method is used to solve nonlinear fifth-order boundary valueproblems arising in viscoelastic fluid flows. A numerical example is presented to illustrate the accuracy, effi-ciency and reliability of this method.

Keywords: Fifth-order boundary value problems; weighted residual method; viscoelastic; residual function;collocation

1 IntroductionNonlinear fifth-order boundary value problems are one of the most important problems arising in the mathematical mod-eling of viscoelastic fluid flows [5, 6]. In this paper, we consider a fifth-order boundary value problem of the form

yv(x) = f(x; y); 0 < x < b; (1:1)

with boundary conditions

y(0) = A0; y0(0) = A1; y00(0) = A2; y(b) = B0; y0(b) = B1; (1:2)

where a prime denotes differentiation with respect to x, moreover, f(x; y) is a given continuous, nonlinear function ofy = y(x). Ai(i = 0; 1; 2) and Bi(i = 0; 1) are finite constants. Results on the existence and uniqueness of solutions ofthis types of problems are well discussed in [1].

Many works have been done on nonlinear fifth order boundary value problems. For instance, Khan [8] used the finitedifference method in his investigation, while Wazwaz [11] employed the decomposition method. Recently, Erturk [7]investigated the fifth-order boundary value problems (1:1) and (1:2) using differential transformation of the kth derivativeof a function f(x) of the form

F (k) =1

k!

dkf(x)

dxk

x=x0

; (1:3)

and the inverse differential transformation defined by

f(x) =nX

k=0

F (k)(x x0)k: (1:4)

For more expositions on the method, see [7].In this present paper, we shall employ the weighted residual method to solve the fifth-order boundary value problem

which are assumed to have unique solution in the interval of integration. See [1] for further discussions. Both differentialtransformation and weighted residual method are thought to have certain advantages (and disadvantages). Many scientistsare of the opinion that both methods have capacity for further developments. As we shall see later in this work, ourapproach converges faster than that of Erturk [7].

The break down of the remaining part of this paper is as follows. In Section two, we give a brief description of theweighted residual method. Numerical results are presented in Section three while Section four gives the conclusion.

Corresponding author. E-mail address: samdrei@yahoo.com

Copyright cWorld Academic Press, World Academic UnionIJNS.2012.06.30/622

420 International Journal of Nonlinear Science, Vol.13(2012), No.4, pp. 419-421

2 Weighted residual methodThe idea of weighted residual method is to seek an approximate solution to the BVP with a polynomial involving a set ofparameters. A polynomial of the form

y(x) = (x a)nn1Xk=0

Bk(x b)k + (x b)nn1Xk=0

Ak(x a)k; n 2; (2:1)

where Ak and Bk are constants to be determined, which satisfies the given boundary conditions (1:2) is assumed to bethe solution of equation (1:1). The function y(x) which is an approximation to the exact solution of equation (1:1) isthen substituted into equation (1:1) to get the residual function R(x). This function R(x) is made as small as possibleby setting it to zero at some equally spaced collocation points in the given interval. The system of these equations is thensolved to determine the parameters Ak and Bk. Then y(x) is considered as the approximate solution of the boundaryvalue problem. For more details on this method, see [24, 9, 10].

3 Numerical resultsExample 3:1In this section, we consider the nonlinear BVP

yv(x) = exy2(x); 0 < x < 1 (3:1)

with boundary conditionsy(0) = 1; y0(0) = 1; y00(0) = 1; y(1) = e; y0(1) = e; (3:2)

whose theoretical solution is y(x) = ex as contained in [7].With n = 7, a = 0 and b = 1, using equation (2:1), we have

y(x) = x76X

k=0

Bk(x 1)k + (x 1)76X

k=0

Akxk: (3:3)

Substituting equation (3:3) into equation (3:2), we obtain

y(x) = x7(19:0279727992134 16:3096909707543x+B2(x 1)2 +B2(x 1)2+B3(x 1)3 +B4(x 1)4 +B5(x 1)5 +B6(x 1)6)+(x 1)7(1 8x 35:5x2 +A3x3 +A4x4 +A5x5 +A6x6):

(3:4)

We then substitute equation (3:4) into equation (3:1) to obtain the residual function and this function is collocated at nineequally spaced points x =

110 ;

210 ;

310 ;

410 ;

510 ;

610 ;

710 ;

810 ;

910

, and equated to zero. The system of nine equations so

obtained is solved simultaneously to getA3 = 115:66666666667; A4 = 309:208333333330; A5 = 718:966666666677;A6 = 1506:22638888887; B2 = 58:4430593118694; B3 = 161:284721821903;B4 = 377:501388927251; B5 = 787:259722218879; B6 = 1506:22638888915:Hence, the approximate solution for equation (3:4) becomes

y(x) = x7(19:0279727992134 16:3096909707543x+ 58:4430593118694(x 1)2+58:4430593118694(x 1)2 161:284721821903(x 1)3 + 377:501388927251(x 1)4787:259722218879(x 1)5 + 1506:22638888915(x 1)6) + (x 1)7(1 8x35:5x2 115:66666666667x3 309:208333333330x4 718:966666666677x51506:22638888887x6):

(3:5)

From the Table 1, one can see that with n = 7 WRM converges to the exact solution than that obtained by DTM [7]which converges to the exact solution at n = 13.

4 ConclusionsIn this article, the weighted residual method was used to solve the nonliear fifth-order boundary value problems which hasproved to be very simple, accurate, effective and reliable compared with Differential transformation method (DTM) [7].

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S. Odejide, O. Adesina: A Note On The Solutions of Nonlinear Fifth-Order Boundary Value Problems 421

Table 1: Comparison of y(x) solutions by Weighted Residual Method (WRM) with the Differential TransformationMethod DTM [7] and exact for Example 3:1

x WRM(n=7) DTM[7](n=7) Exact0.0 1.0 1.0 1.00.1 1.1051709181 1.1051708175 1.10517091810.2 1.2214027582 1.2214020677 1.22140275820.3 1.3498588076 1.2214020677 1.34985880760.4 1.4918246976 1.4918209843 1.49182469760.5 1.6487212707 1.6487157324 1.64872127070.6 1.8221188004 1.8221120435 1.82211880040.7 2.0137527075 2.0137460312 2.01375270750.8 2.2255409285 2.2255360184 2.22554092850.9 2.4596031112 2.4596011705 2.45960311121.0 2.7182818285 2.7182818285 2.7182818285

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