AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 745287 10.1155/2014/745287 745287 Research Article Numerical Solution of Seventh-Order Boundary Value Problems by a Novel Method Inc Mustafa 1 Akgül Ali 2,3 Jafari Hossein 1 Department of Mathematics Science Faculty Fırat University 23119 Elazığ Turkey firat.edu.tr 2 Department of Mathematics Education Faculty Dicle University 21280 Diyarbakır Turkey dicle.edu.tr 3 Department of Mathematics Texas A&M University Kingsville Kingsville, TX 78363 USA tamuk.edu 2014 2332014 2014 25 11 2013 03 02 2014 05 02 2014 23 3 2014 2014 Copyright © 2014 Mustafa Inc and Ali Akgül. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We demonstrate the efficiency of reproducing kernel Hilbert space method on the seventh-order boundary value problems satisfying boundary conditions. These results have been compared with the results that are obtained by variational iteration method (VIM), homotopy perturbation method (HPM), Adomian decomposition method (ADM), variation of parameters method (VPM), and homotopy analysis method (HAM). Obtained results show that our method is very effective.

1. Introduction

Consider the seventh-order boundary value problem : (1)u(7)(x)=N(x,u(x)),0x1, with boundary conditions (2)u(i)(0)=Ai,i=0,1,2,3,u(j)(1)=Bj,j=0,1,2. The analytical solution of seventh-order differential equations are rarely exists in literature. However, there are various numerical methods for the solution of (1)-(2). The aim of this work is to apply reproducing kernel Hilbert space method (RKHSM)  to solve the seventh-order boundary value problems. Numerical results of the seventh-order boundary value problems have been obtained by this method in our work. This study shows that the proposed method can be considered as an alternative technique for solving linear and nonlinear problems in science and engineering .

The paper is organized as follows. Section 2 introduces several reproducing kernel spaces. We provide the main results and the exact and approximate solutions of (1)-(2) in Section 3. 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. Reproducing Kernel Spaces

In this section, we define some useful reproducing kernel spaces.

Definition 1.

We define the space H21[0,1] by (3)H21[0,1]={f(x)L2[0,1],x[0,1].}ffis  absolutely  continuous  in  [0,1],hf(x)L2[0,1],x[0,1]}. The inner product and the norm in H21[0,1] are defined, respectively, by (4)f,gH21=f(0)g(0)+01f(x)g(x)dx,hhhhhhhhhhu,fH21[0,1],fH21=f,fH21,fH21[0,1]. The space H21[0,1] is a reproducing kernel space and its reproducing kernel function rx is given by (5)rx(y)={1+x,xy,1+y,x>y.

Definition 2.

We define the space T28[0,1] by the following: (6)T28[0,1]={ff,f,f′′,f(3),f(4),f(5),f(6),f(7)hhare  absolutely  continuous  in[0,1],hhf(8)L2[0,1],x[0,1],hhf(0)=f(0)=f′′(0)=f(3)(0)=f(1)hh=f(1)=f′′(1)=0}. The inner product and the norm in T28[0,1] are defined, respectively, by (7)f,gT28=i=07f(i)(0)g(i)(0)+01f(8)(x)g(8)(x)dx,f,gT28[0,1],fT28=f,fT28,fT28[0,1]. The space T28[0,1] is a reproducing kernel space; that is, for each fixed y[0,1] and any fT28[0,1], there exists a function Ry such that (8)f=f,RyT28.

Theorem 3.

The space T28[0,1] is a reproducing kernel Hilbert space whose reproducing kernel function Ry is given by (9)Ry(x)={i=116ci(y)xi-1,xy,i=116di(y)xi-1,x>y, where ci(y) and di(y) can be obtained by Maple 16 and proof of Theorem 3 is given in Appendix.

3. Exact and Approximate Solutions of (<xref ref-type="disp-formula" rid="EEq1.1">1</xref>)-(<xref ref-type="disp-formula" rid="EEq1.2">2</xref>) in <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M24"><mml:msubsup><mml:mrow><mml:mi>T</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow><mml:mrow><mml:mn>8</mml:mn></mml:mrow></mml:msubsup><mml:mo mathvariant="bold">[</mml:mo><mml:mn>0,1</mml:mn><mml:mo mathvariant="bold">]</mml:mo></mml:math></inline-formula>

The solution of (1)-(2) is given in the reproducing kernel space T28[0,1]. The linear operator (10)L:T28[0,1]H21[0,1] is bounded. After homogenizing the boundary conditions, we obtain (11)Lv=M(x,v(x)),0x1,v(i)(0)=0,i=0,1,2,3,v(j)(1)=0,j=0,1,2. We choose a countable dense subset P={xi}i=1 in [0,1] and let (12)Ψx(y)=L*rx(y), where L* is conjugate operator of L and rx is given by (5). Furthermore, for simplicity let Ψi(x)=Ψxi(x); namely, (13)Ψi(x)=defΨxi(x)=L*rxi(x). Now one can deduce the following lemmas.

Lemma 4.

{ Ψ i ( x ) } i = 1 is complete system of T28[0,1].

Proof.

For fT28[0,1], let f,Ψi=0  (i=1,2,); that is, (14)f,L*rxi=(Lf)(xi)=0. Note that {xi}i=1 is the dense set in [0,1]; therefore, (Lf)(x)=0. It follows that f(x)=0 from the existence of L-1.

Lemma 5.

The following formula holds: (15)Ψi(x)=(LνRx(ν))(xi), where the subscript ν of operator Lν indicates that the operator L applies to function of ν.

Proof.

Consider the following: (16)Ψi(x)=Ψi(ξ),Rx(ξ)T28=L*rxi(ξ),Rx(ξ)T28=(rxi)(ξ),(LνRx(ν))(ξ)H21=(LνRx(ν))(xi). This completes the proof.

Remark 6.

The orthonormal system {Ψ¯i(x)}i=1 of T28[0,1] can be derived from Gram-Schmidt orthogonalization process of {Ψi(x)}i=1, (17)Ψ¯i(x)=k=1iβikΨk(x),(βii>0,i=1,2,), where βik are orthogonal coefficients.

In the following, we will give the representation of the exact solution of (11) in the reproducing kernel space T28[0,1].

Theorem 7.

If u is the exact solution of (11), then (18)u=i=1k=1iβikM(xk,u(xk))Ψ¯i(x), where {xi}i=1 is a dense set in [0,1].

Proof.

From the (17) and uniqueness of solution of (11), we have (19)u=i=1u,Ψ¯iT28Ψ¯i=i=1k=1iβiku,L*rxkT28Ψ¯i=i=1k=1iβikLu,rxkH21Ψ¯i=i=1k=1iβikM,rxkH21Ψ¯i=i=1k=1iβikM(xk,u(xk))Ψ¯i(x). This completes the proof.

Now the approximate solution un can be obtained by truncating the n-term of the exact solution u as (20)un=i=1nk=1iβikM(xk,u(xk))Ψ¯i(x).

Lemma 8.

Assume u is the solution of (11) and rn is the error between the approximate solution un and the exact solution u. Then the error sequence rn is monotone decreasing in the sense of ·T28 and rn(x)T280.

Proof.

From (18) and (20), we obtain (21)u-unT28=i=n+1k=1iβikM(xk,u(xk))Ψ¯i(x)T28. Thus (22)u-unT280,n. In addition (23)u-unT282=i=n+1k=1iβikM(xk,u(xk))Ψ¯i(x)T282=i=n+1(k=1iβikM(xk,u(xk))Ψ¯i(x)T28)2. Then, u-unT28 is monotonically decreasing in n.

Remark 9.

The seventh-order boundary value problems have come out in construction engineering, beam column theory, and chemical reactions. Therefore solutions of the seventh-order boundary value problems are very important in the literature. The reproducing kernel function for seventh-order boundary value problem has not been calculated till now. All computations are performed by Maple 16. The RKHSM does not require discretization of the variables, that is, time and space, and it is not affected by computational round-off errors and one is not faced with necessity of large computer memory and time. The accuracy of the RKHSM for the seventh-order boundary value problems is controllable and absolute errors are small with present choice of x (see Tables 16 and Figures 16). The obtained numerical results justify the advantage of this methodology. We gave transformations to homogenize the boundary conditions for all examples. Additionally, we improved our programme to find numerical results. As shown in Tables 1, 3, and 5 all the numerical results have been found in very short time.

Numerical results for Example 10 (time (s): 1.645).

x Exact solution Approximate solution Absolute error Relative error
0.0 1.0 1.0 0.0 0.0
0.1 1.1051709180756476248 1.1051709180727538232 2.893 × 1 0 - 12 2.618 × 1 0 - 12
0.2 1.2214027581601698339 1.2214027581330885422 2.708 × 1 0 - 11 2.217 × 1 0 - 11
0.3 1.3498588075760031040 1.3498588075089984856 6.700 × 1 0 - 11 4.963 × 1 0 - 11
0.4 1.4918246976412703178 1.4918246975765762858 6.469 × 1 0 - 11 4.336 × 1 0 - 11
0.5 1.6487212707001281468 1.6487212707474912552 4.736 × 1 0 - 11 2.872 × 1 0 - 11
0.6 1.8221188003905089749 1.8221188006600701498 2.695 × 1 0 - 10 1.479 × 1 0 - 10
0.7 2.0137527074704765216 2.0137527079512967073 4.808 × 1 0 - 10 2.387 × 1 0 - 10
0.8 2.2255409284924676046 2.2255409289689961077 4.765 × 1 0 - 10 2.141 × 1 0 - 10
0.9 2.4596031111569496638 2.4596031113394646558 1.825 × 1 0 - 10 7.420 × 1 0 - 11
1.0 2.7182818284590452354 2.7182818284590452354 0.0 0.0

Comparison of absolute error of HPM, VIM, and RKHSM for Example 10.

x Absolute error  Absolute error  Absolute error [RKHSM]
0.0 0.0 0.0 0.0
0.1 2.15 × 1 0 - 8 3.8478 × 1 0 - 12 2.8938016 × 1 0 - 12
0.2 2.45 × 1 0 - 7 1.2366 × 1 0 - 10 2.70812917 × 1 0 - 11
0.3 8.42 × 1 0 - 7 2.7788 × 1 0 - 10 6.70046184 × 1 0 - 11
0.4 1.69 × 1 0 - 6 7.5864 × 1 0 - 10 6.46940320 × 1 0 - 11
0.5 2.42 × 1 0 - 6 1.1571 × 1 0 - 9 4.73631084 × 1 0 - 11
0.6 2.62 × 1 0 - 6 1.3132 × 1 0 - 9 2.695611749 × 1 0 - 10
0.7 2.06 × 1 0 - 6 1.2228 × 1 0 - 9 4.808201857 × 1 0 - 10
0.8 1.05 × 1 0 - 6 6.6023 × 1 0 - 10 4.765285031 × 1 0 - 10
0.9 2.14 × 1 0 - 7 1.6534 × 1 0 - 10 1.825149920 × 1 0 - 10
1.0 4.44 × 1 0 - 16 1.3265 × 1 0 - 11 0.0

Numerical results for Example 11 (time (s): 3.123).

x Exact solution Approximate solution Absolute error
0.0 0.0 0.0 0.0
0.1 0.099465382626808286232 0.099465382626808285898 3.34 E - 19
0.2 0.19542444130562717342 0.19542444130562716451 8.91 E - 18
0.3 0.28347034959096065184 0.28347034959096048526 1.6658 E - 16
0.4 0.35803792743390487627 0.35803792743390483786 3.841 E - 17
0.5 0.41218031767503203670 0.41218031767503200826 2.844 E - 17
0.6 0.43730851209372215398 0.43730851209372205668 9.730 E - 17
0.7 0.42288806856880006954 0.42288806856879998460 8.494 E - 17
0.8 0.35608654855879481674 0.35608654855879479280 2.394 E - 17
0.9 0.22136428000412546974 0.22136428000412540564 6.410 E - 17
1.0 0.0 1.24053 E - 20 1.24053 E - 20

Comparison of absolute error of VPM, ADM, HAM, and RKHSM.

x Absolute error  Absolute error  Absolute error  Absolute error [RKHSM]
0.0 0.0 0.0 0.0 0.0
0.1 8.55607 E - 13 1.23082 E - 13 5.39291 E - 14 3.34 E - 19
0.2 9.94041 E - 12 3.7792 E - 13 4.85167 E - 14 8.91 E - 18
0.3 3.52244 E - 11 2.37421 E - 13 3.92464 E - 14 1.6658 E - 16
0.4 7.3224 E - 10 3.62099 E - 13 2.21489 E - 14 3.841 E - 17
0.5 1.08769 E - 10 9.39249 E - 14 3.84137 E - 14 2.844 E - 17
0.6 1.29035 E - 10 4.82947 E - 13 2.10831 E - 13 9.730 E - 17
0.7 1.51466 E - 10 1.09135 E - 13 1.99785 E - 13 8.494 E - 17
0.8 2.717974 E - 10 1.64868 E - 14 3.29736 E - 13 2.394 E - 17
0.9 7.48179 E - 10 7.25975 E - 13 1.77622 E - 12 6.410 E - 17
1.0 2.1729 E - 09 4.54747 E - 13 1.65159 E - 12 1.24053 E - 20

Numerical results for Example 12 (time (s): 5.234).

x Exact solution Approximate solution Absolute error
0.0 1.0 1.0 0.0
0.1 0.81435367623236361584 0.81435367623236697064 3.35480 × 1 0 - 15
0.2 0.65498460246238548694 0.65498460246237032242 1.516452 × 1 0 - 14
0.3 0.51857275447720250625 0.51857275447711724998 8.525627 × 1 0 - 14
0.4 0.40219202762138358044 0.40219202762145521926 7.163882 × 1 0 - 14
0.5 0.30326532985631671180 0.30326532985631655134 1.6046 × 1 0 - 16
0.6 0.21952465443761057305 0.21952465443760682026 3.75279 × 1 0 - 15
0.7 0.14897559113742285441 0.14897559113743255406 9.69965 × 1 0 - 15
0.8 0.089865792823444318286 0.089865792823449635588 5.317302 × 1 0 - 15
0.9 0.040656965974059911188 0.040656965974059021714 8.89474 × 1 0 - 16
1.0 0.0 2.708848 × 1 0 - 22 2.708848 × 1 0 - 22

Comparison of absolute error of ADM, HAM, and RKHSM.

x Absolute error  Absolute error  Absolute error [RKHSM]
0.0 1.67932 E - 12 0.0 0.0
0.1 2.96696 E - 12 4.15223 E - 14 3.3548 E - 15
0.2 1.26055 E - 12 4.18332 E - 13 1.516452 E - 14
0.3 2.10898 E - 12 1.21736 E - 12 8.525627 E - 14
0.4 6.68926 E - 12 1.95471 E - 12 7.163882 E - 14
0.5 7.21923 E - 12 2.03731 E - 12 1.6046 E - 16
0.6 9.75339 E - 12 1.37063 E - 12 3.75279 E - 15
0.7 2.19552 E - 12 4.66988 E - 13 9.69965 E - 15
0.8 4.24917 E - 12 4.8378 E - 14 5.317302 E - 15
0.9 2.27311 E - 13 6.00561 E - 14 8.89474 E - 16
1.0 4.42298 E - 12 1.29172 E - 15 2.708848 E - 22

Comparison of analytical solution and RKHSM solution for Example 10.

Comparison of absolute error of VIM, HPM, and RKHSM for Example 10.

Comparison of analytical solution and RKHSM solution for Example 11.

Comparison of absolute error of ADM, VPM, HAM, and RKHSM for Example 11.

Comparison of analytical solution and RKHSM solution for Example 12.

Comparison of absolute error of ADM, HAM, and RKHSM for Example 12.

4. Numerical Results

In this section, three numerical examples are provided to show the accuracy of the present method.

Example 10.

We first consider the seventh-order nonlinear boundary value problem: (24)u(7)(x)=e-xu2(x),0<x<1,u(0)=u(0)=u′′(0)=u(3)(0)=1,u(1)=u(1)=u′′(1)=e. The exact solution of (24) is given as  (25)u(x)=ex.

After homogenizing the boundary conditions of (24), we obtain (26)v(7)(x)=e-x[v(x)+1+x+x22+x36+x4(212e-572)hh+x5(872-16e)+x6(132e-533)x22+x36+x4(212e-572)]2,hhhhhhhhhhhhhhhhhhhhh0x1,v(0)=v(0)=v′′(0)=v(3)(0)=0,v(1)=v(1)=v′′(1)=0, where we used the following transformation: (27)v(x)=u(x)-1-x-x22-x36-x4(212e-572)-x5(872-16e)-x6(132e-533). Using the RKHSM for this example we obtain Tables 1-2 and Figures 1-2.

Example 11.

We now consider the seventh-order linear BVP (28)u(7)(x)=-u(x)-ex(35+12x+2x2),0x1,u(0)=0,u(0)=1,u′′(0)=0,u(3)(0)=-3,u(1)=0,u(1)=-e,u′′(1)=-4e. The exact solution of (28) is given as  (29)u(x)=x(1-x)ex.

After homogenizing the boundary conditions of (28), we get (30)v(7)(x)+v(x)=-ex(35+12x+2x2)-x+x32-x4(3e-172)-x5(272-5e)-x6(2e-112),0x1,v(0)=v(0)=v′′(0)=v(3)(0)=0,v(1)=v(1)=v′′(1)=0, where, we used the following transformation (31)v(x)=u(x)-x+x3/2-x4(3e-172)-x5(272-5e)-x6(2e-112). Using RKHSM for this example we obtain Tables 3-4 and Figures 3-4.

Example 12.

Consider the following seventh-order nonlinear BVP (32)u(7)(x)=u(x)u(x)+e-2x(2+ex(x-8)-3x+x2),hhhhhhhhhhhhhhhhhhhhhhhhh0x1,u(0)=1,u(0)=-2,u′′(0)=3,u(3)(0)=-4,u(1)=0,u(1)=-1e,u′′(1)=2e. The exact solution of (32) is given as  (33)u(x)=(1-x)e-x.

After homogenizing the boundary conditions of (32), we have(34)v(7)(x)-(-2+3x-2x2+4x3(6e-2)+5x4(4-11e)+6x5(5e-116))v(x)-(1-2x+32x2-23x3+x4(6e-2)+x5(4-11e)+x6(5e-116))v(x)=[(1-2x+32x2-23x3+x4(6e-2)+x5(4-11e)+x6(5e-116))(-2+3x-2x2+4x3(6e-2)+5x4(4-11e)+6x5(5e-116))]+v(x)v(x)+e-2x(2+ex(x-8)-3x+x2),0x1,v(0)=v(0)=v′′(0)=v(3)(0)=0,v(1)=v(1)=v′′(1)=0,

where we used the following transformation: (35)u(x)=v(x)+1-2x+32x2-23x3+x4(6e-2)+x5(4-11e)+x6(5e-116).

Using RKHSM for this example we obtain Tables 5-6 and Figures 16.

Remark 13.

Using our method we chose 36 points on [0,1]. In Tables 16, we computed the absolute errors |u(x,t)-un(x,t)| at the points {(xi):xi=i,i=0.0,0.1,,1.0}. The RKHSM tested on three problems, one linear and two nonlinear. A comparison with VIM , HPM , ADM , VPM , and HAM  was made and it was seen that the present method yields good results (see Tables 16 and Figures 16).

5. Conclusion

In this paper, we introduced an algorithm for solving the seventh-order problem with boundary conditions. For illustration purposes, we chose three examples which were selected to show the computational accuracy. It may be concluded that the RKHSM is very powerful and efficient in finding exact solution for a wide class of boundary value problems. The approximate solution obtained by the present method is uniformly convergent. Clearly, the series solution methodology can be applied to much more complicated nonlinear differential equations and boundary value problems. However, if the problem becomes nonlinear, then the RKHSM does not require discretization or perturbation and it does not make closure approximation. Results of numerical examples show that the present method is an accurate and reliable analytical method for the seventh-order boundary value problem.

Appendix Proof of Theorem <xref ref-type="statement" rid="thm2.1">3</xref>.

Let fT28[0,1]. By Definition 2 we have (A.1)f,RyT28=i=07f(i)(0)Ry(i)(0)+01f(8)(x)Ry(8)(x)dx. Through several integrations by parts for (A.1) we have (A.2)f,RyT28=i=07u(i)(0)[Ry(i)(0)-(-1)(7-i)Ry(15-i)(0)]+i=07(-1)(7-i)u(i)(1)Ry(15-i)(1)+01u(x)Ry(16)(x)dx. Note that property of the reproducing kernel (A.3)u,RyT28=u(y). Now, if (A.4)Ry(4)(0)+Ry(11)(0)=0,Ry(5)(0)-Ry(10)(0)=0,Ry(6)(0)+Ry(9)(0)=0,Ry(7)(0)-Ry(8)(0)=0,Ry(8)(1)=0,Ry(9)(1)=0,Ry(10)(1)=0,Ry(11)(1)=0,Ry(12)(1)=0, then (A.2) implies that (A.5)Ry(16)(x)=δ(x-y), when xy, then (A.6)Ry(16)(x)=0, and therefore (A.7)Ry(x)={i=116ci(y)xi-1,xy,i=116di(y)xi-1,x>y. Since (A.8)Ry(16)(x)=δ(x-y), we have (A.9)Ry+(k)(y)=Ry-(k)(y),k=0,,14,Ry+(15)(y)-Ry-(15)(y)=1. Since RyT28[0,1], it follows that (A.10)Ry(0)=Ry(0)=Ry′′(0)=Ry(3)(0)=Ry(1)=Ry(1)=Ry′′(1)=0. From (A.4)–(A.10), the unknown coefficients ci(y) and di(y)  (i=1,2,,16) can be obtained. This completes the proof.

Disclosure

This paper is a part of the Ph.D. thesis of Ali Akgül.

Conflict of Interests

The authors declare that they do not have any competing or conflict of interests.

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