# Numerical Solution of Higher Order Boundary Value Problems

Authors: Shahid S. Siddiqi; Muzammal Iftikhar
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

The aim of this paper is to use the homotopy analysis method (HAM),
an approximating technique for solving linear and nonlinear higher order boundary value problems. Using HAM, approximate solutions of seventh-, eighth-, and tenth-order boundary value problems are developed. This approach provides the solution in terms of a convergent series. Approximate results are given for several examples to illustrate the implementation and accuracy of the method. The results obtained from this method are compared with the exact solutions and other methods (Akram and Rehman (2013), Farajeyan and Maleki (2012), Geng and Li (2009), Golbabai and Javidi (2007), He (2007), Inc and Evans (2004), Lamnii et al. (2008), Siddiqi and Akram (2007), Siddiqi et al. (2012), Siddiqi et al. (2009), Siddiqi and Iftikhar (2013), Siddiqi and Twizell (1996), Siddiqi and Twizell (1998), Torvattanabun and Koonprasert (2010), and Kasi Viswanadham and Raju (2012)) revealing  that the present method is more accurate.


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## Body

## 1. Introduction

Higher order boundary value problems occur in the study of fluid dynamics, astrophysics, hydrodynamic, hydromagnetic stability, astronomy, beam and long wave theory, induction motors, engineering, and applied physics. The boundary value problems of higher order have been examined due to their mathematical importance and applications in diversified applied sciences.The seventh-order boundary value problems generally arise in modeling induction motors with two rotor circuits. The induction motor behavior is represented by a fifth-order differential equation model. This model contains two stator state variables, two rotor state variables, and one shaft speed. Normally, two more variables must be added to account for the effects of a second rotor circuit representing deep bars, a starting cage, or rotor distributed parameters. To avoid the computational burden of additional state variables when additional rotor circuits are required, model is often limited to the fifth-order and rotor impedance is algebraically altered as function of rotor speed under the assumption that the frequency of rotor currents depends on rotor speed. This approach is efficient for the steady state response with sinusoidal voltage, but it does not hold up during the transient conditions, when rotor frequency is not a single value. So, the behavior of such models shows up in the seventh order [1].Chandrasekhar [2] investigated that when an infinite horizontal layer of fluid is heated from below and is subject to rotation, the instability sets in. When this instability sets in as overstability, it is represented by an eighth-order ordinary differential equation. If an infinite horizontal layer of fluid is heated from below, with the assumption that a uniform magnetic field is applied as well across the fluid in the same direction as gravity and the fluid is subject to the action of rotation, the instability sets in. When this instability sets in as ordinary convection, it is modeled by tenth-order boundary value problem.Siddiqi and Iftikhar used the variation of parameter method for solving the seventh-order boundary value problems in [3]. Liu and Wu [4] give the general differential quadrature rule (GDQR) for the solution of eighth-order differential equation. Explicit weighting coefficients are formulated to implement the GDQR for eighth-order differential equations. Siddiqi and Akram [5] used nonic spline and nonpolynomial spline technique for the numerical solution of eighth-order linear special case boundary value problems. These have also been proven to be second order convergent. Siddiqi and Twizell [6] presented the solution of eighth-order boundary value problem using octic spline. Inc and Evans [7] presented the solutions of eighth-order boundary value problems using Adomian decomposition method. Golbabai and Javidi [8] used homotopy perturbation method (HPM) to solve eighth-order boundary value problems. Recently, Akram and Rehman presented the numerical solution of eighth-order boundary value problems using the reproducing Kernel space method [9]. Geng and Li [10] construct a reproducing Kernel space and solve a class of linear tenth-order boundary value problems using reproducing Kernel method. Siddiqi et al. [11] used the variational iteration technique for the solution of tenth-order boundary value problem. Siddiqi and Akram [12] presented the numerical solutions of the tenth-order linear special case boundary value problems using eleventh degree spline. Siddiqi and Twizell [13] presented the solutions of tenth-order boundary value problems using tenth degree spline, where some unexpected results, for the solution and higher order derivatives, were obtained near the boundaries of the interval. Lamnii et al. [14] developed a spline collocation method using spline interpolants and analyzed theapproximating solutions of some general linear boundary value problems. Domairry and Nadim in [15] compared the HAM and HPM in solving nonlinear heat transfer equation. HAM is employed to compute approximate solution of the system of differential equations governing the problem [16] and also used to detect the fin excellency of convective straight fins with temperature-dependent thermal conductivity in [17]. Moghimi et al. applied HAM to solve MHD Jeffery-Hamel flows in nonparallel walls [18]. Farajeyan and Maleki [19] used nonpolynomial spline in off-step points to solve special tenth order linear boundary value problems. Khan and Hussain in 2011 applied Laplace decomposition method (LDM) to nonlinear Blasius flow equation to obtain series solutions [20]. Khan and Gondal [21] constructed a new method for the solution of Abel’s type singular integral equations. The two-step Laplace decomposition algorithm (TSLDA) makes the calculation much simpler.Khan et al. [22] proposed a method which efficiently finds exact solution and is used to solve nonlinear Volterra integral equations. Khan et al. [23] proposed the coupling of homotopy perturbation and Laplace transformation for solving system of partial differential equations. Nadeem et al. [24] described the stagnation point flow of a viscous fluid towards a stretching sheet and obtained an analytical solution of the boundary layer equation by HAM.Recently, Shaban et al. [25] presented modification of the HAM for solving nonlinear boundary value problems. Arqub and El-Ajou [26] investigated the accuracy of the HAM for solving the fractional order problem of the spread of a disease in a population. In [27], Russo and Van Gorder discussed the application of HAM to general nonlinear Klein-Gordon type equations.In the present paper, the seventh-, eighth-, and tenth-order boundary value problems are solved using the homotopy analysis method (HAM). The following seventh-, eighth-, and tenth-order boundary value problems are considered:(1)u(m)(x)=f(x,u(x)),a≤x≤b,u(i)(a)=Ai,u(j)(b)=Bj,
where for m=7,i=0,1,2,…,m-4 and j=0,1,…,m-5; for m=8,i=j=0,1,2,…,m-5, and for m=10,i=j=0,1,2,…,m-6. Ai’s and Bj’s are finite real constants. Also, f(x,u(x)) is a continuous function on [a,b].


## 2. Homotopy Analysis Method

Liao was the first to apply homotopy analysis method (HAM) [28–31]. This is a general analytic approach to get series solutions of nonlinear equations, including algebraic equations, ordinary differential equations, partial differential equations, differential-integral equations, differential-difference equation, and coupled equations of them. For a given nonlinear differential equation
(2)N[u(x)]=0,x∈Θ,
where N is a nonlinear operator and u(x) is an unknown function, Liao constructed a one parameter family of equations in the embedding parameter q∈[0,1], called the zeroth-order deformation equation
(3)(1-q)L[U(x,q)-u0(x)]-qhH(x)N[U(x,q)]=0,x∈Θ,q∈[0,1],
where h is a nonzero auxiliary parameter, H(x) is an auxiliary function, L is an auxiliary linear operator, u0(x) is an initial guess, and U(x,q) is an unknown function. The homotopy provides us larger freedom to choose both the auxiliary linear operator L and the initial guess than the traditional nonperturbation methods, as pointed out by Liao [29, 31]. At q=0 and q=1, we have U(x,0)=u0(x) and U(x,1)=u(x), respectively. Thus, as q increases from 0 to 1, the solution U(x,q) varies from the initial guess u0(x) to the solution u(x). Expanding U(x,q) by Taylor series with respect to q, (2) becomes
(4)U(x,q)=u0(x)+∑m=1∞‍um(x)qm,
where
(5)um(x)=1m!∂mU(x,q)∂qm|q=0.
If the auxiliary linear operator, the initial guess, the auxiliary parameter h, and the auxiliary function are properly chosen, series (4) converges at q=1, and then the homotopy series solution
(6)u(x)=u0+∑m=1∞‍um(x)mustbe one of solutions of original equations N[u(x)]=0 [29]. Here, um(x) is governed by a linear differential equation related to the auxiliary linear operator L. According to definition (6), the governing equation can be deduced from the zeroth-order deformation (4). Define the vector
(7)u→N(x)={u0(x),u1(x),…,uN(x)}.
Differentiating equation (3) m times with respect to the embedded homotopy parameter q, then setting q=0, and then finally dividing them by m!, the mth-order deformation equation is obtained as
(8)L[um(x)-χmum-1(x)]-hH(x)Rm(u→m-1)=0,x∈Θ,q∈[0,1],
where
(9)Rm(u→m-1)=1(m-1)!∂m-1N(U(x,q))∂qm-1|q=0,χm={0,m≤1,1,m≥1.
Finally, an N-th order approximate solution is given by
(10)u(x)=u0+∑m=1N‍um(x).
To implement the HAM, several numerical examples are considered in the following section.


## 3. Numerical Examples

Example 1.
Consider the following seventh-order boundary value problem:(11)u(7)(x)=-u(x)-ex(35+12x+2x2),0≤x≤1,u(0)=0,u(1)=0,u(1)(0)=1,u(1)(1)=-e,u(2)(0)=0,u(2)(1)=-4e,u(3)(0)=-3.
The exact solution of Example 1 is u(x)=x(1-x)ex [3].
Using the HAM (3), the zeroth-order deformation is given by
(12)(1-q)L[U(x,q)-u0(x)]=qhH(x)(∂7U(x,q)∂q7+U(x,q)+ex(35+12x+2x2)).
Now, the initial approximation, u0(x), is the solution of (∂7/∂x7)u=0 subject to boundary conditions in (11); that is,
(13)u0(x)=x-x32+12(-17+6e)x4+12(27-10e)x5+12(-11+4e)x6.
The linear operator L normally consists of the homogeneous part of nonlinear operator N, whereas parameter h and function H(x) are introduced in order to optimize the initial guess. Try to choose h in such a way that they get a convergent series. Under the rule of solution expression (4), the auxiliary function H(x) can be chosen as H(x)=1. In this way, good approximations of such problems can be obtained without having to go up to high order of approximation and without requiring a small parameter.
Hence, themth-order deformation can be given by
(14)L[um(x)-χmum-1(x)]=hH(x)Rm(u→m-1),
where
(15)R1(u→0)=∂7um-1(x,q)∂q7+um-1(x,q)+ex(35+12x+2x2),(16)Rm(u→m-1)=∂7um-1(x,q)∂q7+um-1(x,q),m≥2.
Now, the solution of the mth-order deformation equations (15) and (16) for m≥1 becomes
(17)um(x)=χmum-1(x)+hL-1[Rm(u→m-1)].
Consequently, the first few terms of the homotopy series solution are as follows:
(18)u0(x)=x-x32+12(-17+6e)x4+12(27-10e)x5+12(-11+4e)x6,(19)u1(x)=h518918400×(-32691859200+32691859200ex-24389164800x-8302694400exx-9081072000x2+1037836800exx2-2335132800x3+795242426295x4-292929440550ex4-1230096003906x5+452496853920ex5+502831877247x6-184994415376ex6+518918400x7+12870x8-429x10-2652x11+936ex11+1755x12-650ex12-330x13+120ex13).
The Nth-order approximation of u(x) can be expressed by
(20)uN(x)=u0(x)+∑m=1N‍um(x).
Equation (20) is a family of the approximate solutions to problem (11) in terms of the convergence control parameter h.
We choose the value of auxiliary parameter ash=-1 to ensure that the solution series converges.The errors in absolute values obtained using the present method are compared with those obtained using the variation of parameter method [3] for Example 1 given in Table 1, which shows that the present method is quite accurate. Figures 1 and 2 show the comparison of exact with approximate solution and absolute errors for Example 1 solution respectively.
Convergence Theorem.
In this subsection, one proves that if the solution series (6) given by HAM is convergent, it must be an exact solution of the considered problem.
If the series
u
0
(
x
)
+
∑
m
=
1
∞
‍
u
m
(
x
)
converges, where
u
m
(
x
)
is governed by (17) under the
definitions
(15) and (16), it must be an exact solution of problem (11).Table 1
Comparison of numerical results for Example1.
x
Exact solution
Approximate series solution
Absolute error present method
Absolute error Siddiqi and Iftikhar [3]
0
.
0
0.0000
0.0000
0.0000
0.0000
0.1
0.9946
0.9946
5.39291
E
-
14
8.55607
E
-
13
0.2
0.1954
0.1954
4.85167
E
-
14
9.94041
E
-
12
0.3
0.2835
0.2835
3.92464
E
-
14
3.52244
E
-
11
0.4
0.3580
0.3580
2.21489
E
-
14
7.3224
E
-
10
0.5
0.4122
0.4122
3.84137
E
-
14
1.08769
E
-
10
0.6
0.4373
0.4373
2.10831
E
-
13
1.29035
E
-
10
0.7
0.4229
0.4229
1.99785
E
-
13
1.51466
E
-
10
0.8
0.3561
0.3561
3.29736
E
-
13
2.717974
E
-
10
0.9
0.2214
0.2214
1.77622
E
-
12
7.48179
E
-
10
1.0
0.0000
-
1.65159
E
-
12
1.65159
E
-
12
2.1729
E
-
09Figure 1
Comparison of the approximate solution with the exact solution for problem (11). Dotted line: approximate solution; solid line: the exact solution.Figure 2
Absolute errors for problem (11).Proof.
Let the series(21)∑m=0∞‍um(x)
be convergent. Then,
(22)u(x)=∑m=0∞‍um(x),(23)limn→∞un(x)=0.
We have
(24)∑m=1n‍[um(x)-χmum-1(x)]=u1+(u2-u1)+(u3-u2)+⋯+(un-un-1).
Using (22),
(25)∑m=1∞‍[um(x)-χmum-1(x)]=limn→∞un(x)=0,
and applying the operator L, we can write
(26)L∑m=1∞‍[um(x)-χmum-1(x)]=∑m=1∞‍L[um(x)-χmum-1(x)]=0;
using the definition (14),
(27)∑m=1∞‍L[um(x)-χmum-1(x)]=hH(x)∑m=1∞‍Rm(u→m-1),
and since h≠0, H(x)≠0, we have
(28)∑m=1∞‍Rm(u→m-1)=0.
From (15) and (16), the following holds:
(29)∑m=1∞‍Rm(u→m-1)=∑m=1∞‍(∂7um-1(x,q)∂q7+um-1(x,q)+(1-χm)ex(35+12x+2x2)∂7um-1(x,q)∂q7)=0.
This completes the proof.Example 2.
The following seventh-order nonlinear boundary value problem is considered:(30)u(7)(x)=u(x)u′(x)+e-2x(2+ex(x-8)-3x+x2),0≤x≤1,u(0)=1,u(1)=0,u(1)(0)=-2,u(1)(1)=-1e,u(2)(0)=3,u(2)(1)=2e,u(3)(0)=-4.
The exact solution of Example 2 is u(x)=(1-x)e-x.
Using the HAM (3), the zeroth-order deformation is given by
(31)(1-q)L[U(x,q)-u0(x)]=qhH(x)(∂7U(x,q)∂q7+U(x,q)∂U(x,q)∂q+e-2x(2+ex(-8+x)-3x+x2)∂7U(x,q)∂q7).
Now, the initial approximation, u0(x), is the solution of (∂7/∂x7)u=0 subject to boundary conditions in (30); that is,
(32)u0(x)=1-2x+3x22-2x33+(36-12e)x46e+(-66+24e)x56e+(30-11e)x66e.
The linear operator L normally consists of the homogeneous part of nonlinear operator N, whereas parameter h and function H(x) are introduced in order to optimize the initial guess. Try to choose h in such a way that they get a convergent series. Under the rule of solution expression (4), the auxiliary function H(x) can be chosen as H(x)=1. In this way, good approximations of such problems can be obtained without having to go up to high order of approximation and without requiring a small parameter.
Hence, themth-order deformation can be given by
(33)L[um(x)-χmum-1(x)]=hH(x)Rm(u→m-1),
where
(34)R1(u→0)=∂7um-1(x,q)∂q7+um-1(x,q)∂um-1(x,q)∂q+e-2x(2+ex(-8+x)-3x+x2),Rm(u→m-1)=∂7um-1(x,q)∂q7+um-1(x,q)∂um-1(x,q)∂q,m≥2.
Now, the solution of the mth-order deformation equations (34) for m≥1 becomes
(35)um(x)=χmum-1(x)+hL-1[Rm(u→m-1)].
Consequently, the first few terms of the homotopy series solution are as follows:
(36)u0(x)=1-2x+3x22-2x33+(36-12e)x46e+(-66+24e)x56e+(30-11e)x66e,(37)u1(x)=he-2-2x52929676800×(2274322050e2-52929676800e2+x+50655354750e2+2x+1654052400e2x+52929676800e2+xx-102964761900e2+2xx+413513100e2x2+77740462800e2+2xx2-34735100400e2+2xx3-100276694820e2xx4+317578525200e1+2xx4-92426387280e2+2xx4+168299259192e2xx5-582227931918e1+2xx5+188997561564e2+2xx5-72364427100e2xx6+264649706520e1+2xx6-87281523925e2+2xx6+21003840e2+2xx7-9189180e2+2xx8+3208920e2+2xx9-2100384e1+2xx10+160446e2+2xx10+3659760e1+2xx11-1113840e2+2xx11-2864160e1+2xx12+1003340e2+2xx12+1306620e1+2xx13-471240e2+2xx13-440640e2xx14+69360e1+2xx14+83640e2+2xx14+969408e2xx15-626688e1+2xx15+99552e2+2xx15-830790e2xx16+596700e1+2xx16-107100e2+2xx16+326700e2xx17-238590e1+2xx17+43560e2+2xx17-49500e2xx18+36300e1+2xx18-6655e2+2xx18).
The Nth-order approximation can be expressed by
(38)u(x)=u0(x)+∑m=1N‍uM(x).
Equation (38) is a family of the approximate solutions to problem (12) in terms of the convergence control parameter h.
We choose the value of auxiliary parameter ash=-1 to ensure that the solution series converges.In Table 2, the exact solution and the series solution of Example 2 are compared, which shows that the method is quite accurate. Figure 3 shows the comparison of exact solution with approximate solution, and Figure 4 shows the absolute errors for Example 2.Table 2
Comparison of numerical results for Example2.
x
Exact solution
Approximate series solution
Absolute error
0
.
0
1.0000
1.0000
0.0000
0.1
0.814354
0.814354
4.15223
E
-
14
0.2
0.654985
0.654985
4.18332
E
-
13
0.3
0.518573
0.518573
1.21736
E
-
12
0.4
0.402192
0.402192
1.95471
E
-
12
0.5
0.303265
0.303265
2.03731
E
-
12
0.6
0.219525
0.219525
1.37063
E
-
12
0.7
0.148976
0.148976
4.66988
E
-
13
0.8
0.0898658
0.0898658
4.8378
E
-
14
0.9
0.040657
0.040657
6.00561
E
-
14
1.0
0.0000
-
1.29172
E
-
15
1.29172
E
-
15Figure 3
Comparison of the approximate solution with the exact solution for problem (12). Dotted line: approximate solution; solid line: the exact solution.Figure 4
Absolute errors for problem (12).Example 3.
The following seventh-order linear boundary value problem is considered:(39)u(7)(x)=xu(x)+ex(x2-2x-6),0≤x≤1,u(0)=1,u(1)=0,u(1)(0)=0,u(1)(1)=-e,u(2)(0)=-1,u(2)(1)=-2e,u(3)(0)=-2.
The exact solution of Example 3 is u(x)=(1-x)ex [32].
Following the procedure of the previous example, this problem is solved using the convergence control parameterh=-1.
The comparison of the absolute errors obtained by the present method and the absolute errors obtained by the method in [32] is given in Table 3, which shows that the present method is quite accurate. In Figure 5, the comparison of exact solution with approximate solution is shown, and Figure 6 shows absolute errors for Example 3.Table 3
Comparison of numerical results for Example3.
x
Exact solution
Approximateseries solution
Absolute errorpresent method
Absolute errorSiddiqi et al.  [32]
0
.
0
1.0000
1.0000
0.0000
0.0000
0.1
0.0994
0.0994
3.41727
E
-
13
4.6585
E
-
13
0.2
0.9771
0.9771
6.25056
E
-
14
5.7126
E
-
12
0.3
0.9449
0.9449
1.42442
E
-
13
2.1299
E
-
11
0.4
0.8950
0.8950
8.83738
E
-
14
4.6995
E
-
11
0.5
0.8243
0.8243
6.43929
E
-
14
7.4307
E
-
11
0.6
0.7288
0.7288
1.51812
E
-
12
8.9219
E
-
11
0.7
0.6041
0.6041
1.47904
E
-
12
7.9767
E
-
11
0.8
0.4451
0.4451
4.94338
E
-
12
4.6686
E
-
11
0.9
0.2459
0.2459
5.3817
E
-
12
1.0960
E
-
11
1.0
0.0000
1.20811
E
-
11
1.20811
E
-
11
6.9252
E
-
16Figure 5
Comparison of the approximate solution with the exact solution for problem (13). Dotted line: approximate solution; solid line: the exact solution.Figure 6
Absolute errors for problem (13).Example 4.
Consider the following eighth-order boundary value problem:(40)u(8)(x)=-xu(x)-ex(48+15x+2x3),0≤x≤1,u(0)=0,u(1)=0,u(1)(0)=1,u(1)(1)=-e,u(2)(0)=0,u(2)(1)=-4e,u(3)(0)=-3,u(3)(1)=-9e.
The exact solution of Example 4 is u(x)=x(1-x)ex [5, 7, 9].
Using the HAM (3), the zeroth-order deformation is given by
(41)(1-q)L[U(x,q)-u0(x)]=qhH(x)(∂8U(x,q)∂q8+xU(x,q)+ex(48+15x+2x3)∂8U(x,q)∂q8).
Now, the initial approximation, u0(x), is the solution of (∂8/∂x8)u=0 subject to boundary conditions in (40); that is,
(42)u0(x)=x-x32+12(-36+13e)x4+12(84-31e)x5+12(-68+25e)x6+12(19-7e)x7.
The linear operator L normally consists of the homogeneous part of nonlinear operator N, whereas parameter h and function H(x) are introduced in order to optimize the initial guess. Try to choose h in such a way that they get a convergent series. Under the rule of solution expression (4), the auxiliary function H(x) can be chosen as H(x)=1. In this way, good approximations of such problems can be obtained without having to go up to high order of approximation and without requiring a small parameter.
Hence, themth-order deformation can be given by
(43)L[um(x)-χmum-1(x)]=hH(x)Rm(u→m-1),
where
(44)R1(u→0)=∂8um-1(x,q)∂q8+xum-1(x,q)+ex(48+15x+2x3),Rm(u→m-1)=∂8um-1(x,q)∂q8+um-1(x,q),m≥2.
Now, the solution of the mth-order deformation equations (44) for m≥1 becomes
(45)um(x)=χmum-1(x)+hL-1[Rm(u→m-1)].
Consequently, the approximations u0, u1,… of the homotopy series solution are obtained.
TheN-th order approximation can be expressed by
(46)u(x)=u0(x)+∑m=1N‍um(x).
Equation (46) is a family of the approximate solutions to problem (14) in terms of the convergence control parameter h.
We choose the value of auxiliary parameter ash=-1 to ensure that the solution series converges.For problem (14), comparison of the results of the present method with the results of Akram and Rehman [9], Inc and Evans [7], and Siddiqi and Akram [5] is shown in Table 4. It is observed that the errors in absolute values of the present method are better. The comparison of exact solution with approximate solution is shown in Figure 7, and absolute errors are shown in Figure 8, respectively.Table 4
Comparison of absolute errors for Example4.
x
Exact solution
Approximateseries solution
Absolute errorpresent method
Akram and Rehman  [9]
Siddiqi and Akram [5]
Inc and Evans [7]
0
.
1
0.0994654
0.0994654
3.89966
E
-
15
1.63
E
-
10
5.62
E
-
10
3.73
E
-
09
0.2
0.195424
0.195424
9.45355
E
-
14
1.63
E
-
09
4.88
E
-
09
6.61
E
-
09
0.3
0.28347
0.28347
7.04437
E
-
14
4.90
E
-
09
1.37
E
-
08
2.33
E
-
08
0.4
0.358038
0.358038
4.36873
E
-
13
8.46
E
-
09
2.29
E
-
08
5.17
E
-
08
0.5
0.41218
0.41218
1.28897
E
-
13
1.01
E
-
08
2.71
E
-
08
9.76
E
-
08
0.6
0.437309
0.437309
4.01956
E
-
13
8.68
E
-
09
2.38
E
-
08
1.78
E
-
06
0.7
0.422888
0.422888
2.06929
E
-
12
5.15
E
-
09
1.49
E
-
08
4.12
E
-
06
0.8
0.356087
0.356087
2.65915
E
-
12
1.76
E
-
09
5.54
E
-
09
1.83
E
-
04Figure 7
Comparison of the approximate solution with the exact solution for problem (14). Dotted line: approximate solution; solid line: the exact solution.Figure 8
Absolute errors.Example 5.
Consider the following boundary value problem:(47)u(8)(x)=u(x)-8(2xcos(x)+7sin(x)),-1≤x≤1,u(-1)=u(1)=0,u(1)(-1)=u(1)(1)=2sin(1),u(2)(-1)=-u(2)(1)=-4cos(1)-2sin(1),u(3)(-1)=u(3)(1)=6cos(1)-6sin(1).
The exact solution of problem (15) is u(x)=(x2-1)sin(x) [5, 6, 9, 14].
Following the procedure of the previous example, this problem is solved using the convergence control parameter ash=-1.
It is observed that the maximum absolute error values are better than those of Akram and Rehman [9], Lamnii et al. [14], Siddiqi and Akram [5], and Siddiqi and Twizell [6] as shown in Table 5. The comparison of exact solution with approximate solution is shown in Figure 9, and absolute errors are shown in Figure 10, respectively.Table 5
Comparison of maximum absolute errors for Problem (15).
Present method
Akram and Rehman  [9]
Lamnii et al. [14]
Siddiqi and Twizell [6] x∈[x4,xk-4]
Siddiqi and Akram [5]
1
.
89
×
1
0
-
13
4.90
×
1
0
-
9
5.01
×
1
0
-
9
1.20
×
1
0
-
5
1.02
×
1
0
-
8Figure 9
Comparison of the approximate solution with the exact solution for problem (15). Dotted line: approximate solution, solid line: the exact solution.Figure 10
Absolute errors for problem (15).Example 6.
Consider the following boundary value problem:(48)u(8)(x)=-8ex+u(x),0<x<1,u(0)=1,u(4)(0)=-3,u(1)(0)=0,u(5)(0)=-4,u(2)(0)=-1,u(1)(1)=-e,u(3)(0)=-2,u(2)(1)=-2e.
The exact solution of problem (16) is u(x)=(1-x)ex [8, 9, 33, 34].
Following the procedure of the previous example, this problem is solved using the convergence control parameter ash=-1.
It is observed that the errors in absolute values are better than those of Akram and Rehman [9], Golbabai and Javidi [8], He [33], and Torvattanabun and Koonprasert [34] as shown in Table 6. Figures 11 and 12 show the comparison of exact solution with approximate solution and absolute errors, respectively.Table 6
Comparison of absolute errors for Problem (16).
x
Present method|u-u1|
Akram and Rehman [9], |u-u7|
Golbabai and Javidi [8] (N=7)
He [33]
Torvattanabun and Koonprasert [34]
0
.
25
2.55351
E
-
15
3.0291
E
-
10
2.1630
E
-
09
4.578
E
-
09
3.8922
E
-
10
0.50
9.65894
E
-
14
7.7317
E
-
09
1.1571
E
-
07
9.840
E
-
09
1.1571
E
-
07
0.75
5.63438
E
-
13
3.1222
E
-
08
1.0479
E
-
06
1.096
E
-
05
1.0479
E
-
06
1.0
9.23328
E
-
13
4.3979
E
-
08
4.2188
E
-
06
1.861
E
-
04
4.2188
E
-
06Figure 11
Comparison of the approximate solution with the exact solution for problem (16). Dotted line: approximate solution; solid line: the exact solution.Figure 12
Absolute errors for problem (16).Example 7.
Consider the following tenth-order boundary value problem:(49)u(10)(x)=-ex(89+21x+x2-x3)+xu(x),-1<x<1,u(-1)=0,u(1)=0,u(1)(-1)=2e,u(1)(1)=-2e,u(2)(-1)=2e,u(2)(1)=-6e,u(3)(-1)=0,u(3)(1)=-12e,u(4)(-1)=-4e,u(3)(1)=-20e.
The exact solution of problem (17) is u(x)=(1-x2)ex [10–14, 19].
Following the procedure of the previous example, this problem is solved using the convergence control parameter ash=-1.
It is observed that the errors in absolute values of the present method are better than those of Farajeyan and Maleki [19], Geng and Li [10], Lamnii et al. [14], Siddiqi and Akram [12], Siddiqi et al. [11], and Siddiqi and Twizell [13] as shown in Table 7. Figures 13 and 14 show the comparison of exact solution with approximate solution and absolute errors for Example 7, respectively.Table 7
Comparison of maximum absolute errors for Problem (17).
Present method
Geng and Li [10]
Siddiqi et al. [11]
Siddiqi and Akram [12]
Siddiqi and Twizell [13] x∈[x5,xk-5]
Lamnii et al. [14]
Farajeyan and Maleki [19]
6
.
42
×
1
0
-
13
9.08
×
1
0
-
12
1.97
×
1
0
-
6
3.28
×
1
0
-
6
2.07
×
1
0
-
3
1.86
×
1
0
-
8
1.75
×
1
0
-
12Figure 13
Comparison of the approximate solution with the exact solution for problem (17). Dotted line: approximate solution; solid line: the exact solution.Figure 14
Absolute errors for problem (17).Example 8.
Consider the following tenth-order nonlinear boundary value problem:(50)u(10)(x)=141754(x+u(x)+1)11,0<x<1,u(0)=0,u(1)=0,u(1)(0)=-12,u(1)(1)=1,u(2)(0)=12,u(2)(1)=4,u(3)(0)=34,u(3)(1)=12,u(4)(0)=32,u(3)(1)=48.
The exact solution of problem (18) is u(x)=(2/(2-x))-x-1 [35].
Following the procedure of the previous example, this problem is solved using the convergence control parameter ash=-1.
It is observed that the errors in absolute values are better than those of Viswanadham and Raju [35] as shown in Table 8. In Figure 15, the comparison of exact solution with approximate solution is shown, and in Figure 16, absolute errors for Example 8 are shown, respectively.Table 8
Comparison of numerical results for Example8.
x
Exact solution
Absolute errorpresent method
Absolute errorKasi Viswanadham and Raju [35]
0
.
1
-
0.0473684
3.95413
E
-
11
1.322478
E
-
06
0.2
-
0.0888889
7.33317
E
-
10
4.231930
E
-
06
0.3
-
0.123529
7.33317
E
-
09
1.676381
E
-
05
0.4
-
0.15
6.06524
E
-
09
4.245341
E
-
05
0.5
-
0.166667
7.74775
E
-
09
6.663799
E
-
05
0.6
-
0.171429
6.56402
E
-
09
6.940961
E
-
05
0.7
-
0.161538
3.48667
E
-
09
4.750490
E
-
05
0.8
-
0.133333
9.23198
E
-
10
1.643598
E
-
05
0.9
-
0.08181824
5.33521
E
-
11
2.607703
E
-
07Figure 15
Comparison of the approximate solution with the exact solution for problem (18). Dotted line: approximate solution; solid line: the exact solution.Figure 16
Absolute errors for problem (18).


## 4. Conclusion

In this paper, the homotopy analysis method (HAM) has been applied to obtain the numerical solutions of seventh- eighth-, and tenth-order boundary value problems. All computational work was carried out using Mathematica software. The numerical results show that only a few number of approximations can be used for numerical purpose with a high degree of accuracy. It is observed that the absolute errors are better than the methods in [3, 5–14, 19, 32–35]. It is also observed that our proposed method is well suited for the solution of higher order boundary value problems and reduces the computational work. HAM converges to exact solutions more rapidly as compared to the other method. Therefore, the present method is an accurate and reliable analytical technique for boundary value problems.



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