Solution of Second- and Higher-Order Nonlinear Two-Point Boundary-Value Problems Using Double Decomposition Method

Abstract

The present paper makes use of the efficient double decomposition method to propose a method for solving two-point boundary-value problems, featuring second- and higher-order nonlinear ordinary differential equations. The efficacy of the proposed method is demonstrated on numerous test problems. In the end, a high level of exactitude between the obtained approximate solution and the available exact solution is achieved.

1. Introduction

Nonlinear ordinary differential equations of second- and higher-orders that are rendered through different forms of initial and boundary-value problems have been broadly examined in the literature. Of course, this is connected to their vast relevance in modeling diverse processes of engineering and science applications. As an instance, we mention the second-order nonlinear differential model, characterizing the distribution of temperature field in a given media with a thermal conductivity relying on the temperature field [1]. We also mention the fourth-order nonlinear [2] and the eighth-order [3] differential equations for modeling the vibration of Euler–Bernoulli beams and the uniform beams (torsional vibration), respectively; see [4,5] and the references therein for other related studies on various boundary-value problems of a higher order.

The theory of boundary value problems for higher-order differential equations is closely related to the theory of boundary value problems for a system of first-order differential equations [6,7,8].

Additionally, another aspect of great importance in relation to the boundary-value models is their solutions. It is no wonder that solutions of differential equations play an immense role in the understanding of the physicality of the governing model. With this, various analytical methods have been devised in the past decade to seek exact solutions [9,10]. Agarwal has reported certain conditions for the existence and uniqueness of solutions of higher-order boundary-value problems in his book [11]. However, little information is so far known with regard to the numerical analysis of higher-order boundary-value problems [5]. Moreover, one would find the recent numerical examination of higher-order boundary-value problems through the application of the finite difference method and the spline method [12,13,14,15]. To proceed further, we recall one of the powerful semi-analytical methods that tackle numerous functional equations successfully: the Adomian decomposition method (ADM) [16]. This method has an amazing efficacy and has been endorsed by various researchers in mathematical physics. ADM is a powerful decomposition methodology for the practical solution of linear or nonlinear and deterministic or stochastic operator equations, including ordinary differential equations, partial differential equations, integral equations, etc. The method provides the solution in a rapidly convergent series with components that can be computed iteratively.

More on the method’s convergence studies and a comparative examination with some competing numerical methods can be found in [17,18,19,20] and the references therein. Obviously, the Adomian method is very potent in tackling nonlinear boundary-value models. For example, Wazwaz [21] presented a competent computational method based on the standard Adomian method for the solution of higher-order two-point boundary-value problems. The same scheme was equally applied to other boundary-value problems featuring the fifth and sixth orders in [22,23]. We also make mention of the famous Adomian algorithm modification by Duan and Rach [24] which has a universal application on multi-order multi-point boundary-value problems for nonlinear ordinary differential equations.

However, the main objective of this paper is to employ the double decomposition method to solve some second- and higher-order boundary-value problems featuring nonlinear ordinary differential equations. Recall that Adomian and Rach [25], in 1993, initiated the double decomposition method to improve the proficiency of the standard Adomian method. Further, Aminataei and Hosseini [26] compared the double decomposition method with the standard Adomian method on certain boundary-value problems of the second order. Their finding was that the double decomposition method has more virtues, including higher accuracy and faster convergence, against the standard Adomian method. To finish, we arrange the present paper as follows: Section 2 gives the outline of the celebrated methodology, while Section 3 considers some boundary-value test problems. Section 4 discusses the obtained numerical results. Lastly, Section 5 gives the concluding points.

2. Analysis of the Method

To present the method of concern in this study, we apply the consideration to a generalized nonlinear differential equation of the following type:

$$Lu\left(x\right)+Ru\left(x\right)+Nu\left(x\right)=g\left(x\right),$$

(1)

where $L=\frac{d^n}{d{x}^n}$ is the highest linear differential operator that is considered to be effortlessly invertible, $R$ is also a linear operator that follows same assumptions with $L$, but with order less than that of $L$, while $N$ denotes the nonlinear operator. What is more, the function $g\left(x\right)$ on the other side of the above equation is a prescribed source term, which is a given continuous function.

Therefore, upon applying the inverse linear differential operator $L^{-1}$ to both sides of Equation (1), one obtains

$$u\left(x\right)=\mathsf{\Phi }\left(x\right)+L^{-1}g\left(x\right)-L^{-1}Ru\left(x\right)-L^{-1}Nu\left(x\right),$$

(2)

where the $\mathsf{\Phi }\left(x\right)$ denotes the terms emanating as a result of the application of $L^{-1},$ that is, integrating. More so, the Adomian approach [16] consists of decomposing the unknown function $u\left(x\right)$ of any equation and the nonlinear term $Nu\left(x\right)$ into a sum of an infinite number of components defined by the decomposition series

$$u={{\displaystyle \sum }}_{n=0}^{\infty }{u}_n, Nu={{\displaystyle \sum }}_{n=0}^{\infty }{A}_n.$$

(3)

where the components $u_n\left(x\right), n\ge 0$, are to be determined in a recursive manner. The ADM concerns itself by finding the components $u_0,u_1, u_2,\dots$ individually. The determination of these components can be achieved in an easy way through a recursive relation that involves simple integrals.

Note that the polynomials $A_n’\mathrm{s}$ in the above equation are called the Adomian polynomials and are recurrently obtained using the following relation

$$A_n=\frac{1}{n!}\frac{d^n}{d{\lambda }^n}{\left[ N\left({{\displaystyle \sum }}_{k=0}^n{\lambda }^{k }{u}_k\right)\right]}_{\lambda =0}, n \ge 0.$$

(4)

Thus, the overall recurrent solution of Equation (1) follows without further delay from the standard Adomian decomposition method [16].

Now, we proceed further by switching to the double decomposition methodology as presented in [25,26] in order to obtain a more accurate recurrent solution for Equation (1). In light of this, we further decompose the term $\mathsf{\Phi }\left(x\right)$ in Equation (2) via a sum of an infinite series as follows:

$$\mathsf{\Phi }={{\displaystyle \sum }}_{n=0}^{\infty }{\mathsf{\Phi }}_n.$$

(5)

Therefore, Equation (2) can be rewritten as follows:

$${{\displaystyle \sum }}_{n=0}^{\infty }{u}_n={{\displaystyle \sum }}_{n=0}^{\infty }{\mathsf{\Phi }}_n+L^{-1}g-L^{-1}R{{\displaystyle \sum }}_{n=0}^{\infty }{u}_n-L^{-1}{{\displaystyle \sum }}_{n=0}^{\infty }{A}_n$$

(6)

Without loss of generality, for simplicity, we shall consider second-order ordinary differential equations, where $L=\frac{d^2}{d{x}^2}$, with the boundary conditions $u\left({\alpha }_1\right)={\beta }_1, \mathrm{and} u\left({\alpha }_2\right)={\beta }_2, \mathrm{for} {\beta }_1 \mathrm{and} {\beta }_2\in ℝ$; the same procedures can be applied to a higher order, and therefore, the inverse differential operator $L^{-1}$ in the double decomposition method is taken as twofold indefinite integration, so

$${\mathsf{\Phi }}_n=c_{0,n}+c_{1,n}x,$$

(7)

where $c_{0,n}$ and $c_{1,n}$ can be obtained from the boundary conditions, and the double decomposition solution can be derived from Equation (6) as follows:

$$u_0=c_{0,0}+c_{1,0}x+L^{-1}g\left(x\right) ,$$

$$u_1=c_{0,1}+c_{1,1}x-L^{-1}R{u}_0-L^{-1}\left(A_0\right),$$

$$u_2=c_{0,2}+c_{1,2}x-L^{-1}R{u}_1-L^{-1}\left(A_1\right),$$

Then, we have the approximate solution

$${\phi }_1=u_0$$

$${\phi }_2={\phi }_1+u_1=u_0+u_1$$

$${\phi }_3={\phi }_2+u_2=u_0+u_1+u_2$$

$$⋮$$

The constants $c_{0,j}$ and $c_{1,j}$ are determined by matching each partial sum ${\phi }_n\left(x\right)$ to the boundary values conditions.

Therefore, upon utilizing the prescribed boundary conditions $u\left({\alpha }_1\right)={\beta }_1$ and $u\left({\alpha }_2\right)={\beta }_2$, the matching solution for these boundary conditions is thus given by

$${\phi }_{m+1}\left({\alpha }_1\right)={\beta }_1, {\phi }_{m+1}\left({\alpha }_2\right)={\beta }_2$$

Hence, ${\phi }_1=u_0$, and we write ${\phi }_1\left({\alpha }_1\right)={\beta }_1, {\phi }_1\left({\alpha }_2\right)={\beta }_2.$ Then, matching ${\phi }_1$ to the boundary conditions $c_{0,0}$ and $c_{1,0}$, we obtain the following system of linear equations:

$$\left(\begin{array}{cc}1& {\alpha }_1\\ 1& {\alpha }_2\end{array}\right)·\left(\begin{array}{c}{c}_{0,0}\\ c_{1,0}\end{array}\right)=\left(\begin{array}{c}{\beta }_1\\ {\beta }_2\end{array}\right).$$

(8)

What is more, if the determinant of the above coefficient matrix is nonzero, then ${\phi }_1=u_0$ is determined completely. Next, we proceed to the subsequent approximation ${\phi }_2$ by first determining

$$u_1={\mathsf{\Phi }}_1-L^{-1}R{u}_0-L^{-1}N{u}_0,$$

to obtain

$${\phi }_2={\phi }_1+u_1=u_0+u_1.$$

Matching ${\phi }_2$ to the boundary conditions to evaluate the constants, the expression of ${\phi }_2$ is thus determined completely. Continuing in the same manner, we eventually determine $u_2,u_3,\dots$ until a satisfactory ${\phi }_m$ is achieved.

3. Numerical Examples

The present section is committed to demonstrating the efficacy of the presented methodology. In light of this, several two-point boundary-value problems featuring second and higher-order nonlinear ordinary differential equations are examined as test problems.

Example 1.

Let us consider the following two-point boundary-value problem featuring a nonlinear homogeneous differential equation [27]:

$$u^{″}=u^3-u{u}^{\prime }, 1\le x\le 2,$$

$$u\left(1\right)=\frac{1}{2}, u\left(2\right)=\frac{1}{3} ,$$

that satisfies the following exact analytical solution $u\left(x\right)=\frac{1}{x+1}.$

Consequently, applying the inverse operator$L^{-1}$to both sides of the above model, one obtains

$$u=c_0+c_{1}x+L^{-1}\left(u^3\right)-L^{-1}\left(u{u}^{\prime }\right).$$

Next, on using the double decomposition method, the generalized recurrent scheme of components is given by

$$u_0={\mathsf{\Phi }}_0,$$

$$u_n={\mathsf{\Phi }}_n+L^{-1}\left(A_{n-1}\right)-L^{-1}\left(B_{n-1}\right), n \ge 1.$$

where${\mathsf{\Phi }}_n=c_{0,n}+c_{1,n}x$, and Adomian polynomials$A_n$of a nonlinear term$u^3$are given for a few components as follows:

$$A_0=u_0^3,$$

$$A_1=3u_0^2{u}_1,$$

$$A_2=3u_0^2{u}_2+3u_0{u}_1^2,$$

$$A_3 =3u_0^2{u}_3+6u_0{u}_1{u}_2+u_1^3,$$

$$⋮$$

and Adomian polynomials$B_n$of a nonlinear term$u{u}^{\prime }$are given for a few components as follows:

$$B_0=u_0{u}_0^{\prime } ,$$

$$B_1=u_0^{\prime }{u}_1+u_0{u}_1^{\prime } ,$$

$$B_2=u_0^{\prime }{u}_2+u_1^{\prime }{u}_1+u_2^{\prime }{u}_0 ,$$

$$B_3=u_0^{\prime }{u}_3+u_1^{\prime }{u}_2+u_2^{\prime }{u}_1+u_3^{\prime }{u}_0 ,$$

$$⋮$$

Then ${\phi }_1=u_0=c_{0,0}+c_{1,0}x,$which, upon utilizing the prescribed boundary conditions, yields the following constants:

$$c_{0,0}=0.6666667, c_{1,0}=-0.1666667,$$

such that ${\phi }_1=0.6666667-0.1666667x,$

Next, we proceed to determine${\phi }_2$as follows

$${\phi }_2=u_0+u_1.$$

$$u_1=c_{0,1}+c_{1,1}x+L^{-1}\left(A_0\right)-L^{-1}\left(B_0\right),$$

More so, on using the boundary conditions in the above relation, we obtain the constants as follows:

$$c_{0,1}=-0.02175925, c_{1,1}=-0.08541666,$$

where the explicit expression of${\phi }_2$follows as

$$\begin{gathered} {\phi }_2=0.8819444-0.5483796x+0.2037037x^2-0.0416666x^3 \\ +0.0046296x^4-0.0002314x^5. \end{gathered}$$

Continuing in the same manner, we obtain

$$\begin{gathered} {\phi }_3=0.9596595-0.7888668x+0.4923996x^2-0.2168852x^3+0.0653967x^4 \\ -0.01339602x^5+\dots , \end{gathered}$$

and so on.

Thus, in what follows, we report in

Table 1. The absolute error between the exact solution and ${\phi }_5$ in Example 1.

$\mathit{x}$	$\left|\mathit{u}\left(\mathit{x}\right)-{\mathit{\phi }}_5\right|$	Error in [27]
1.0	$1\times {10}^{-30}$	1.5 $\times {10}^{-9}$
1.1	$6.39597863\times {10}^{-8}$	$5.0806\times {10}^{-5}$
1.2	$1.79292118\times {10}^{-7}$	$2.38437\times {10}^{-4}$
1.3	$2.97763241\times {10}^{-7}$	$4.09327\times {10}^{-4}$
1.4	$3.40230480\times {10}^{-7}$.	$4.98607\times {10}^{-4}$
1.5	$2.58977348\times {10}^{-7}$	$4.95070\times {10}^{-4}$
1.6	$7.13085007\times {10}^{-8}$	$4.18656\times {10}^{-4}$
1.7	$1.44920473\times {10}^{-7}$	$3.06744\times {10}^{-4}$
1.8	$2.82527087\times {10}^{-7}$	$2.06991\times {10}^{-4}$
1.9	$2.47483053\times {10}^{-7}$	$1.75182\times {10}^{-4}$
2.0	$4.00000000\times {10}^{-30}$	$1\times {10}^{-9}$

the comparison between the approximate and exact results of ${\phi }_5$ through their respective absolute error differences, while Figure 1 demonstrates this development graphically.

Example 2.

Let us consider the following two-point boundary-value problem featuring a nonlinear nonhomogeneous differential equation [28]

$$u^{\left(iv\right)}=\sin \left(x\right)+{\sin }^2\left(x\right)-{\left(u^{″}\right)}^2$$

$$u\left(0\right)=0, u^{\prime \left(0\right)}=1, u\left(1\right)=\sin \left(1\right), u\prime \left(1\right)=\cos \left(1\right),$$

that satisfies the following exact analytical solution $u\left(x\right)=\sin \left(x\right).$

Accordingly, upon applying$L^{-1}$to both sides of the given model, one obtains

$$u={\mathrm{c}}_0+c_{1}x+\frac{c_2}{2!}{x}^2+\frac{c_3}{3!}{x}^3+L^{-1}\left(\sin \left(x\right)+{\sin }^2\left(x\right)\right)-L^{-1}\left({\left(u^{″}\right)}^2\right)$$

Then, on using the double decomposition method, and the following recurrent components as follows,

$$u_0={\mathsf{\Phi }}_0+L^{-1}\left(\sin \left(x\right)+{\sin }^2\left(x\right)\right),$$

$$u_n={\mathsf{\Phi }}_n-L^{-1}\left(A_n\right), n \ge 1.$$

where${\mathsf{\Phi }}_n=c_{0,n}+c_{1,n}x+\frac{c_{2,n}}{2!}{x}^2+\frac{c_{3,n}}{3!}{x}^3$and Adomian polynomials$A_n$of a nonlinear term${\left(u^{″}\right)}^2$are given for a few components as follows:

$$A_0={(u_0^{″})}^2$$

$$A_1=2u_0^{″}{u}_1^{″}$$

$$A_2=2u_0^{″}{u}_2^{″}+{(u_1^{″})}^2$$

$$A_3 =2u_0^{″}{u}_3^{″}+2u_1^{″}{u}_2^{″}$$

$$⋮$$

Then

$${\phi }_1=u_0$$

Moreover, the unknown constants are determined from the prescribed conditions as follows

$$c_{0,0}=0, c_{1,0}=0, c_{2,0}=0.0148013, c_{3,0}=-0.0599314.$$

So, ${\phi }_1$is obtained as follows

$${\phi }_1=-0.0550993x^2-0.00998857x^3+\frac{1}{48}{x}^4+\sin \left(x\right)+\frac{1}{16}{\sin }^2\left(x\right),$$

Next, we find ${\phi }_2=u_0+u_1,$by firstly obtaining the following solution constants

$$c_{0,1}=0.4801841, c_{1,1}=6.3407677, c_{2,1}=-0.2607643, c_{3,1}=-1.2424602.$$

That is,

$$\begin{gathered} {\phi }_2=0.4801841+6.3426406x-0.1012272x^2-0.217689575x^3 \\ -0.0008315104x^4-0.00011007x^5+\dots , \end{gathered}$$

Continuing using the above process, we obtain

$$\begin{gathered} {\phi }_3=-33.6046752-260.8103058x+0.6125090x^2+3.2966875x^3 \\ +.14144683x^4-0.001472694x^5+\dots , \end{gathered}$$

and so on.

Equally, we report in

Table 2. The absolute error between the exact solution and ${\phi }_5$ in Example 2.

$\mathit{x}$	$\left|\mathit{u}\left(\mathit{x}\right)-{\mathit{\phi }}_5\right|$	Error in [28]
0.0	1.64899441 $\times {10}^{-25}$	9.5923 $\times {10}^{-14}$
0.1	$1.20296527\times {10}^{-10}$	7.7856 $\times {10}^{-8}$
0.2	$4.00798650\times {10}^{-10}$	2.7231 $\times {10}^{-7}$
0.3	$7.30346840\times {10}^{-10}$	$5.2489\times {10}^{-7}$
0.4	$1.00068075\times {10}^{-9}$	7.7730 $\times {10}^{-7}$
0.5	$1.11730964\times {10}^{-9}$	9.7145 $\times {10}^{-7}$
0.6	$1.03044986\times {10}^{-9}$	1.0502 $\times {10}^{-6}$
0.7	$7.64267406\times {10}^{-10}$	$9.6286\times {10}^{-7}$
0.8	$4.16627690\times {10}^{-10}$	6.8407 $\times {10}^{-7}$
0.9	$1.21057128\times {10}^{-10}$	2.7069 $\times {10}^{-7}$
1.0	$1.46487450\times {10}^{-23}$	1.5676 $\times {10}^{-13}$

the comparison between the approximate and exact results of ${\phi }_5$ through their respective absolute error differences, while Figure 2 compares the exact and the obtained results graphically.

Example 3.

Consider the following two-point boundary-value problem featuring a nonlinear homogeneous differential equation [29]

$$u^{\left(5\right)}=e^{-x}{u}^2$$

$$u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(0\right)=1, u\left(1\right)=u^{\prime }\left(1\right)=e,$$

that satisfies the following exact analytical solution $u\left(x\right)=e^x.$

Therefore, we apply$L^{-1}$to both sides of the governing model to obtain

$$u={\mathrm{c}}_0+c_{1}x+\frac{c_2}{2!}{x}^2+\frac{c_3}{3!}{x}^3+\frac{c_4}{4!}{x}^4+L^{-1}\left(e^{-x}{u}^2\right).$$

Then, on using the double decomposition method as described, and the general scheme one obtains

$$u_0={\mathsf{\Phi }}_0,$$

$$u_n={\mathsf{\Phi }}_n+L^{-1}\left(e^{-x}{A}_{n-1}\right), n \ge 1.$$

where${ \mathsf{\Phi }}_n=c_{0,n}+c_{1,n}x+\frac{c_{2,n}}{2!}{x}^2+\frac{c_{3,n}}{3!}{x}^3+\frac{c_{4,n}}{4!}{x}^4,$and Adomian polynomials$A_n$of a nonlinear term$u^2$are given for a few components as follows:

$$A_0=u_0^2,$$

$$A_1=2 u_0{u}_1,$$

$$A_2=2 u_0{u}_2+u_1^2,$$

$$A_3=2 u_0{u}_3+2 u_1{u}_2,$$

$$⋮$$

Then ${\phi }_1=u_0,$together with the following values

$$c_{0,0}=1, c_{1,0}=1, c_{2,0}=1, c_{3,0}=0.9290729, c_{4,0}=1.5224722.$$

So, explicit expression for${\phi }_1$is thus obtained as follows:

$${\phi }_1=1+x+0.5x^2+0.1548454x^3+0.0634363x^4,$$

Then, we obtain ${\phi }_2=u_0+u_1$by firstly determining the following relevant constants

$$\begin{gathered} c_{0,1}=1.3196661\times {10}^5, c_{1,1}=-\mathrm{46,616.4814656}, \\ {c}_{2,1}=\mathrm{13,694.8920702}, c_{3,1}=-3026.5540664 , c_{4,1}=388.8258482. \end{gathered}$$

That is,

$$\begin{gathered} {\phi }_2=1.3196761\times {10}^5-\mathrm{46,615.4814656}x+6847.9460351x^2 \\ -504.2708322x^3+16.26451335x^4+\dots \end{gathered}$$

In the same fashion, we determine${\phi }_3$as follows

$$\begin{gathered} {\phi }_3=4.4179719\times {10}^6-1.2741127\times {10}^{6}x+1.4821108\times {10}^5{x}^2-8308.7846042x^3 \\ +191.8318906x^4+\dots \end{gathered}$$

In the same way, we report in

Table 3. The absolute error between the exact solution and the approximate of ${\phi }_3$ in Example 3.

$\mathit{x}$	$\left|\mathit{u}\left(\mathit{x}\right)-{\mathit{\phi }}_3\right|$	VIMHP [29]
0.0	3.00000000 $\times {10}^{-23}$	0.0
0.1	1.28946045 $\times {10}^{-12}$	1.0 $\times {10}^{-9}$
0.2	8.71505433 $\times {10}^{-12}$	$2.0 \times {10}^{-9}$
0.3	2.40516757 $\times {10}^{-11}$	1.0 $\times {10}^{-8}$
0.4	4.45866002 $\times {10}^{-11}$	2.0 $\times {10}^{-8}$
0.5	6.400647286 $\times {10}^{-11}$	3.1 $\times {10}^{-8}$
0.6	7.43254934 $\times {10}^{-11}$	$3.7 \times {10}^{-8}$
0.7	6.90319620 $\times {10}^{-11}$	4.1 $\times {10}^{-8}$
0.8	4.71067442 $\times {10}^{-11}$	3.1 $\times {10}^{-8}$
0.9	1.70573303 $\times {10}^{-11}$	1.4 $\times {10}^{-8}$
1.0	2.52865000 $\times {10}^{-24}$	0.0

the comparison between the approximate and exact results of ${\phi }_3$ through their respective absolute error differences, while in Figure 3, we compare the exact and the obtained approximate results graphically.

Example 4.

Consider the following two-point boundary-value problem featuring a seventh-order nonlinear nonhomogeneous differential equation [30]:

$$u^{\left(7\right)}=-u{u^{\prime }}^{ }+e^x\left(-35+\left(-13+e^x\right)x-\left(1+2e^x\right)x^2+e^x{x}^4\right)$$

$$u\left(0\right)=0, u^{\prime }\left(0\right)=1, u^{″}\left(0\right)=0, u^{\left(3\right)}\left(0\right)=-3,$$

$$u\left(1\right)=0, u^{\prime }\left(1\right)=-e, u^{″}\left(1\right)=-4e,$$

that satisfies the following exact analytical solution

$$u\left(x\right)=x\left(1-x\right)e^x.$$

Consequently, applying the inverse operator $L^{-1}$to both sides of the above model, one obtains

$$\begin{gathered} u=c_0+c_{1}x+\frac{c_2}{2!}{x}^2+\frac{c_3}{3!}{x}^3+\frac{c_4}{4!}{x}^4+\frac{c_5}{5!}{x}^5+\frac{c_6}{6!}{x}^6 \\ +L^{-1}\left(e^x\left(-35+\left(-13+e^x\right)x-\left(1+2e^x\right)x^2+e^x{x}^4\right)\right)-L^{-1}\left(uu\prime \right). \end{gathered}$$

Next, the method of interest yields

$$\begin{gathered} u_0=c_{0,0}+c_{1,0}x+\frac{c_{2,0}}{2!}{x}^2+\frac{c_{3,0}}{3!}{x}^3++\frac{c_{4,0}}{4!}{x}^4+\frac{c_{5,0}}{5!}{x}^5+\frac{c_{6,0}}{6!}{x}^6+ \\ L^{-1}(e^x\left(-35+\left(-13+e^x\right)x-\left(1+2e^x\right)x^2+e^x{x}^4\right), \end{gathered}$$

while the general scheme takes the following form

$$u_n={\mathsf{\Phi }}_n-L^{-1}\left(A_{n-1}\right), n \ge 1,$$

where Adomian polynomials$A_n$of a nonlinear term $u{u}^{\prime }$are given for some few components as follows:

$$A_0=u_0{u}_0^{\prime },$$

$$A_1=u_0^{\prime }{u}_1+u_0{u}_1^{\prime } ,$$

$$A_2=u_0^{\prime }{u}_2+u_1^{\prime }{u}_1+u_2^{\prime }{u}_0 ,$$

$$A_3=u_0^{\prime }{u}_3+u_1^{\prime }{u}_2+u_2^{\prime }{u}_1+u_3^{\prime }{u}_0 ,$$

$$⋮$$

Then ${\phi }_1=u_0.$

Using boundary conditions, we have

$$\begin{gathered} c_{0,0}=-2.2148437 c_{1,0}=-2.5781250, c_{2,0}=-2.7343750, \\ {c}_{3,0}=-2.531250, c_{4,0}=-1.8757720 , c_{5,0}=-0.8633528, c_{6,0}=-0.0616062. \end{gathered}$$

Thus, we compute${\phi }_1$as follows:

$$\begin{gathered} {\phi }_1=-2.2148437-2.5781250x-1.3671875x^2-0.4218750x^3 \\ -0.0781571x^4+\dots \end{gathered}$$

Accordingly, we find ${\phi }_2=u_0+u_1$where the following constants are obtained via ${\phi }_2$

$$\begin{gathered} c_{0,1}=-8273.2151911,c_{1,1}=-3246.5648917, c_{2,1}=-1095.3101751, \\ {c}_{3,1}=-298.3535744, c_{4,1}=-58.1387206, c_{5,1}=-6.591191412, c_{6,1}=-2.3911602. \end{gathered}$$

Thus, we obtain

$$\begin{gathered} {\phi }_2=-8275.4300349-3249.1430167x-549.0222750596x^2 \\ -50.1474707x^3-2.5006038x^4+\dots \end{gathered}$$

Moreover, we obtain ${\phi }_3$as in the preceding example as follows:

$$\begin{gathered} {\phi }_3=2.9797946\times {10}^8+8.7550752\times {10}^{7}x+1.1399240\times {10}^7{x}^2 \\ +8.224015011\times {10}^5{x}^3+\mathrm{32,677.12844287}{x}^4+\dots \end{gathered}$$

Similarly, we report in

Table 4. The absolute error between the exact solution and the approximate of ${\phi }_3$ in Example 4.

$\mathit{x}$	$\left|\mathit{u}\left(\mathit{x}\right)-{\mathit{\phi }}_3\right|$	VIMHP [30]
0.0	7.24695891 $\times {10}^{-22}$	0.0
0.1	3.05370438 $\times {10}^{-21}$	5.28944 $\times {10}^{-12}$
0.2	3.23715191 $\times {10}^{-20}$	6.44606 $\times {10}^{-11}$
0.3	1.12228840 $\times {10}^{-19}$	2.38427 $\times {10}^{-10}$
0.4	2.14964709 $\times {10}^{-19}$	5.20559 $\times {10}^{-10}$
0.5	2.80162697 $\times {10}^{-19}$	8.11431 $\times {10}^{-10}$
0.6	2.62088240 $\times {10}^{-19}$	9.55209 $\times {10}^{-10}$
0.7	1.73155372 $\times {10}^{-19}$	8.30543 $\times {10}^{-10}$
0.8	7.37260050 $\times {10}^{-20}$	4.67351 $\times {10}^{-10}$
0.9	1.0113907 $\times {10}^{-20}$	1.04882 $\times {10}^{-10}$
1.0	1.0000000 $\times {10}^{-21}$	3.90259 $\times {10}^{-12}$

the comparison between the approximate and exact results of ${\phi }_3$ through their respective absolute error differences, while Figure 4 compares the exact solution with the obtained approximate graphically.

Example 5.

Let us consider the following two-point boundary-value problem featuring a ninth-order nonhomogeneous nonlinear differential equation [31]:

$$u^{\left(9\right)}-u^2{u}^{\prime }={\cos }^3\left(x\right)$$

$$u\left(0\right)=0, u^{\prime }\left(0\right)=1, u^{″}\left(0\right)=0, u^{\left(3\right)}\left(0\right)=-1, u^{\left(4\right)}\left(0\right)=0,$$

$$u\left(1\right)=\sin \left(1\right), u^{\prime }\left(1\right)=\cos \left(1\right), u^{″}\left(1\right)=-\sin \left(1\right), u^{\left(3\right)}\left(1\right)=-\cos \left(1\right),$$

that satisfies the following exact analytical solution $u\left(x\right)=\sin \left(x\right).$

Consequently, applying the inverse operator $L^{-1}$to both sides of the above model, one obtains

$$\begin{gathered} u=c_0+c_{1}x+\frac{c_2}{2!}{x}^2+\frac{c_3}{3!}{x}^3+\frac{c_4}{4!}{x}^4+\frac{c_5}{5!}{x}^5+\frac{c_6}{6!}{x}^6+ \frac{c_7}{7!}{x}^7+\frac{c_8}{8!}{x}^8 \\ +L^{-1}\left({\cos }^3\left(x\right)\right)+L^{-1}\left(u^{2}u\prime \right). \end{gathered}$$

Next, the method of interest yields

$$\begin{gathered} u_0=c_{0,0}+c_{1,0}x+\frac{c_{2,0}}{2!}{x}^2+\frac{c_{3,0}}{3!}{x}^3++\frac{c_{4,0}}{4!}{x}^4+\frac{c_{5,0}}{5!}{x}^5+\frac{c_{6,0}}{6!}{x}^6+ \frac{c_{7,0}}{7!}{x}^7 \\ +\frac{c_{8,0}}{8!}{x}^8+ L^{-1}\left({\cos }^3\left(x\right)\right), \end{gathered}$$

while the general scheme takes the following form

$$u_n={\Phi }_n+L^{-1}\left(A_{n-1}\right), n \ge 1,$$

where Adomian polynomials $A_n$of a nonlinear term $u^2{u}^{\prime }$are given for some few components as follows:

$$A_0=u_0^2{u}_0^{\prime },$$

$$A_1=2u_0{u}_0^{\prime }{u}_1+u_0^2{u}_1^{\prime } ,$$

$$A_2=u_1^2{u}_0^{\prime }+2u_0{u}_1^{\prime }{u}_1+2u_0{u}_0^{\prime }{u}_2+u_2^{\prime }{u}_0^2 ,$$

$$A_3=2u_1{u}_0^{\prime }{u}_2+u_1^2{u}_1^{\prime }+2u_0{u}_2^{\prime }{u}_1+2u_0{u}_1^{\prime }{u}_2+2u_0{u}_0^{\prime }{u}_3+u_3^{\prime }{u}_0^2 ,$$

$$⋮$$

Then ${\phi }_1=u_0.$

Using boundary conditions, we have

$$\begin{gathered} c_{0,0}=0 , c_{1,0}=0.2499618, c_{2,0}=0,c_{3,0}=-0.2496570, c_{4,0}=0 , \\ {c}_{5,0}=0.2468656, c_{6,0}=0.0010475 , c_{7,0}=-0.2315273 , c_{8,0}=0.0337862 \end{gathered}$$

Thus, we compute ${\phi }_1$as follows

$${\phi }_1=0.24996189x-0.0416095107x^3+0.002057213x^5+0.000001454x^6-0.000045937x^7+\dots$$

Accordingly, we find ${\phi }_2=u_0+u_1$where the following constants are obtained via ${\phi }_2$

$$\begin{gathered} c_{0,1}=-1.4138575\times {10}^7,c_{1,1}=5.98728476\times {10}^6, c_{2,1}=1.3117533440\times {10}^6, \\ {c}_{3,1}=-5.1809017\times {10}^5, c_{4,1}=-\mathrm{70,992.506719}, c_{5,1}=\mathrm{22,828.6778}, \\ {c}_{6,1}=1448.4701549, c_{7,1}=-224.6703088, c_{8,1}=-0.033786257. \end{gathered}$$

Thus, we obtain

$$\begin{gathered} {\phi }_2=-1.413857548\times {10}^7+5.987285004\times {10}^6 x+\mathrm{655,876.6711703}{x}^2 \\ -\mathrm{86,348.4005542}{x}^3-2958.02106691x^4+\dots \end{gathered}$$

Similarly, we report in

Table 5. The absolute error between the exact solution and $\left({\phi }_2\right)$ in Example 5.

$\mathit{x}$	$\left|\mathit{u}\left(\mathit{x}\right)-{\mathit{\phi }}_2\right|$	Error in [31]
0.1	$1.10141984\times {10}^{-20}$	1.862645 $\times {10}^{-7}$
0.2	$2.28012627\times {10}^{-19}$	7.301569 $\times {10}^{-7}$
0.3	$1.02772075\times {10}^{-18}$	9.834766 $\times {10}^{-7}$
0.4	$2.34591176\times {10}^{-18}$	1.221895 $\times {10}^{-6}$
0.5	$3.42488794\times {10}^{-18}$	8.344650 $\times {10}^{-7}$
0.6	$3.41624545\times {10}^{-18}$	$3.874302\times {10}^{-6}$
0.7	$2.25161250\times {10}^{-18}$	$5.662441\times {10}^{-6}$
0.8	$8.22300372\times {10}^{-19}$	$4.887581\times {10}^{-6}$
0.9	$8.62879987\times {10}^{-20}$	$2.861023\times {10}^{-6}$

the comparison between the approximate and exact results of ${\phi }_2$ through their respective absolute error differences, while Figure 5 compares the exact solution with the obtained approximate graphically.

4. Discussion of Results

The present study numerically examines a class of boundary-value problems featuring second and higher-order nonlinear ordinary differential equations via the application of the double decomposition method [25,26]. The double decomposition method is an improved method based on the standard Adomian decomposition method [16] that is proven to efficiently solve dissimilar functional equations. Various test examples were successfully examined in the above section, comprising homogenous and nonhomogeneous models, and involving different differential orders. Additionally, the efficacy of the proposed method is reported in Table 1, Table 2, Table 3, Table 4 and Table 5 via their absolute error differences, while Figure 1, Figure 2, Figure 3, Figure 4 and Figure 5 portrays the respective graphical illustrations the consequential exact solutions that are shown to be in perfect agreement with their approximate solutions. Thus, the double decomposition method has better and faster convergence and is in accordance with lesser error. Additionally, as a high-level of exactitude between the obtained approximate solutions and that of the available exact solutions was recorded, this proposed method is highly recommended for various forms of higher-order functional equations.

5. Conclusions

In conclusion, we examined the two-point boundary-value problems featuring second- and higher-order nonlinear ordinary differential equations by devising a proficient method based on the standard Adomian method, that is, the double decomposition method. Hence, the double decomposition method is an efficient numerical method, this could further be verified by examining the reported absolute error tables. Moreover, the magnitude of computation, computer storage, and run time are occupied more and more in accordance with the standard Adomian method. Therefore, with respect to the highly technical computers developed nowadays, the accuracy of calculation is more important.