Application of a Hybrid of the Different Transform Method and Adomian Decomposition Method Algorithms to Solve the Troesch Problem

Abstract

The Troesch problem is a well-known and important problem; the ability to solve it is important due to the practical applications of this problem. In this paper, we propose a method to solve this problem using a combination of two useful algorithms: Different Transform Method (DTM) and Adomian Decomposition Method (ADM). The combination of these two methods resulted in a continuous approximate solution to this problem and eliminated the problems that occur when trying to use each of these methods separately. The great advantages of the DTM method are the continuous form of the solution and the fact that it easy to implement error control. However, in too-complex tasks, this method becomes difficult to use. By using a hybrid of ADM and DTM, we obtain a relatively simple-to-implement method that retains the advantages of the DTM approach.

1. Introduction

Troesch’s problem appears, among others, in the description of phenomena related to the study of confinement of a plasma column by radiation pressure or in the theory related to gas porous electrodes. This problem describes an ordinary second-order differential equation of the form

$$y^{\prime \prime }\left(x\right)=\alpha sinh(\alpha ·y\left(x\right)),$$

(1)

where $\alpha \in {\mathbb{R}}_{+}$ is a parameter for $x\in [0,1]$ with conditions

$$y\left(0\right)=0,y\left(1\right)=1.$$

(2)

The problem in question is difficult to solve, which is why it is so important to find an approximate solution. One of the first researchers of this problem was Troesch (hence the name of the task), who used the so-called shooting method in his research [1]. There are many other methods for finding an approximate solution to this problem. Examples include the homotopy perturbation method [2,3,4], the B-spline collocation method [5], the variational iteration method [6], using Bessel polynomials [7], the Galerkin method [8,9], using Chebychev polynomials [10], using the Laplace transform [11], and many others [12,13,14].

In the following article, we will use a combination of two analytical methods for solving operator equations (also nonlinear): the DTM (see Section 2) and the ADM (see Section 3). This combination aims to eliminate the difficulties associated with using each of these methods separately. In the works [8,11], this method is used individually, and in the work [6], we can read about the problems related to this approach.

2. Different Transform Method

The DTM was developed and its application to certain electronic problems was demonstrated by Zhou [15]. This method quickly became quite popular and was successfully used to solve many different types of problems described by ordinary differential equations and their systems [16,17,18], partial differential equations [19,20], fractional differential equations [21,22], fuzzy differential equations [23], integral equations and their systems [24,25], systems of differential–integral equations, including equations with a shifted argument [26,27,28], and many others [29,30,31,32].

In this work, we will be interested in functions that are originals.

Definition 1.

The function f expanded in a Taylor series about a fixed point $x_0$ is called the original.

The Taylor series mentioned in the above definition is used here in a broader context. Most often, the point $x_0$ is 0, and then we encounter a Maclaurin series.

If we assume that the function f is the original, then for this function, we can write

$$f\left(x\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{f^{\left(n\right)}\left(0\right)}{n!}}{x}^n$$

and based on the uniqueness of this expansion, each original f can be uniquely assigned a function (transformation) $F:\mathbb{N}\cup \left\{0\right\}\to \mathbb{R}$:

$$F\left(n\right)={\displaystyle \frac{f^{\left(n\right)}\left(0\right)}{n!}},\mathbb{Z}\ni n\ge 0.$$

Thanks to this, we can write the original f as

$$f\left(x\right)=\sum _{n=0}^{\infty }F\left(n\right)x^n.$$

The fact that we will be dealing with a special case of Taylor series (Maclaurin series) is not a problem. An easy substitution reduces Taylor series to Maclaurin series, and in addition, we have a theorem connecting Taylor transformations of the function f expanded around the point $x_0\ne 0$ and around the point $x_0=0$.

Theorem 1.

If function $F_{x_0}$ is a Taylor transformation of function f in the neighborhood of point $x_0$, and function F is a Taylor transformation of function f in the neighborhood of point 0, then the following relations hold between these transformations:

$$F\left(n\right)=\sum _{i=n}^{\infty }{F}_{x_0}\left(i\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{n}}\right){(-x_0)}^{i-n},F_{x_0}\left(n\right)=\sum _{i=n}^{\infty }F\left(i\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{n}}\right){\left(x_0\right)}^{i-n}.$$

In the process of solving a given task, we will be able to use known expansions of basic functions f and properties of the transformation, thanks to which the problem under consideration will be reduced to various types of algebraic problems. In the cited literature, one can find many properties of the DTM transformation; we will limit ourselves to giving only those that will help us solve the title problem (in the properties below, we will assume that x is an element of the domain of the considered function f).

Theorem 2.

The Taylor transformation of the function $f\left(x\right)=u\left(x\right)\pm v\left(x\right)$ is

$$F\left(n\right)=U\left(x\right)\pm V\left(x\right),\mathbb{Z}\ni n\ge 0.$$

(3)

Theorem 3.

The Taylor transformation of the function $f\left(x\right)=a·u\left(x\right)$, $a\in \mathbb{R}$ is

$$F\left(n\right)=a·U\left(x\right),\mathbb{Z}\ni n\ge 0.$$

(4)

Theorem 4.

The Taylor transformation of the function ${\displaystyle f\left(x\right)=u^{\left(m\right)}\left(x\right)}$, $m\in \mathbb{N}$, is

$$F\left(n\right)={\displaystyle \frac{(n+m)!}{m!}}U(n+m),\mathbb{Z}\ni n\ge 0.$$

(5)

The situation becomes more complicated if composite functions appear in the problem under consideration. Then, there are no simple descriptions of the form of the transformation $U\left(n\right)$ for the original $u\left(x\right)=f\left(g\right(x\left)\right)$. One such approach is Faà di Bruno’s formula [33,34], but this approach complicates the task from the programming side. An effective countermeasure to this problem will be the combination of the DTM transformation and the Adomian decomposition method (see Section 4).

3. Adomian Decomposition Method

The creator of the ADM method is Adomian, who described this method and applied it in the 1980s [35,36]. Many scientists have noticed the great potential of this method and have applied it to many problems described by nonlinear operator equations. These problems include, among others, the heat conduction equation [37,38,39,40,41], the wave equation [42,43], inverse problems [44,45], and many others [46,47,48,49].

The details of the method can be read in the presented literature, and in our case, the ADM will be applied to equations of the form

$$u\left(x\right)=f\left(x\right)+g\left(u\right(x\left)\right),$$

where u is the searched function, f is known as a function, and g is some nonlinear transformation.

We present the function u as a function series, but we can also expand the nonlinear transformation g into a function series

$$g\left(u\left(x\right)\right)=\sum _{i=0}^{\infty }{A}_i(u_0,u_1,\dots ,u_i),$$

where $A_i$, $i\ge 0$, denotes the polynomials introduced by Adomian (Adomian polynomials).

These polynomials are defined as follows:

$$\begin{array}{cc}\hfill A_0& =g\left(u_0\right),\hfill \\ \hfill A_j(u_0,u_1,\dots ,u_j)& ={\displaystyle \frac{1}{j!}}{\displaystyle \frac{d^j}{d{\lambda }^j}}{\left.\left(g\left(\sum _{i=0}^{\infty }{\lambda }^i{u}_i\right)\right)\right|}_{\lambda =0},j\ge 1,\hfill \end{array}$$

where $\lambda$ is a parameter.

For several initial values of j, these polynomials have the form

$$\begin{array}{cc}\hfill A_0& =g\left(u_0\right),\hfill \\ \hfill A_1& =u_1{g}^{\prime }\left(u_0\right),\hfill \\ \hfill A_2& =u_2{g}^{\prime }\left(u_0\right)+{\displaystyle \frac{1}{2}}{u}_1^2{g}^{\prime \prime }\left(u_0\right),\hfill \\ \hfill A_3& =u_3{g}^{\prime }\left(u_0\right)+u_1{u}_2{g}^{\prime \prime }\left(u_0\right)+{\displaystyle \frac{1}{3!}}{u}_1^3{g}^{\left(3\right)}\left(u_0\right),\hfill \\ \hfill A_4& =u_4{g}^{\prime }\left(u_0\right)+(u_1{u}_3+{\displaystyle \frac{1}{2!}}{u}_2^2)g^{\prime \prime }\left(u_0\right)+{\displaystyle \frac{1}{2!}}{u}_1^2{u}_2{g}^{\left(3\right)}\left(u_0\right)+{\displaystyle \frac{1}{4!}}{u}_1^4{g}^{\left(4\right)}\left(u_0\right),\hfill \\ \hfill A_5& =u_5{g}^{\prime }\left(u_0\right)+(u_2{u}_3+u_1{u}_4)g^{\prime \prime }\left(u_0\right)+{\displaystyle \frac{1}{2!}}(u_1^2{u}_3+u_1{u}_2^2)g^{\left(3\right)}\left(u_0\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{3!}}{u}_1^3{u}_2{g}^{\left(4\right)}\left(u_0\right)+{\displaystyle \frac{1}{5!}}{u}_1^5{g}^{\left(5\right)}\left(u_0\right),\hfill \\ \hfill & ⋮\hfill \end{array}$$

This formula may seem complex, but the generation of successive Adomian polynomials can be described by a fairly simple algorithm.

There is an important connection between Adomian polynomials and Taylor transforms that connects these objects [50,51]: if the function f is a composite function of the form $f\left(x\right)=g\left(u\right(x\left)\right)$, then the Taylor transform F of this function is of the form

$$F\left(k\right)=A_k^{*}(u_0,u_1,\dots ,u_k),k\ge 0,$$

(6)

where $A_k^{*}$ denotes a polynomial created from the Adomian polynomial $A_k$, where in place of $u_k$, we put the k-th value of the Taylor transform of the function u, i.e., $U\left(k\right)$.

Combining the above two properties, we can write

$$\begin{array}{cc}\hfill F\left(0\right)& =A_0^{*}=g\left(Y\left(0\right)\right),\hfill \\ \hfill F\left(1\right)& =A_1^{*}=Y\left(1\right)g^{\prime }\left(Y\left(0\right)\right),\hfill \\ \hfill F\left(2\right)& =A_2^{*}=Y\left(2\right)g^{\prime }\left(Y\left(0\right)\right)+{\displaystyle \frac{1}{2!}}{Y}^2\left(1\right)g^{\prime \prime }\left(Y\left(0\right)\right),\hfill \\ \hfill F\left(3\right)& =A_3^{*}=Y\left(3\right)g^{\prime }\left(Y\left(0\right)\right)+Y\left(1\right)Y\left(2\right)g^{\prime \prime }\left(Y\left(0\right)\right)+{\displaystyle \frac{1}{3!}}{Y}^3\left(1\right)g^{\left(3\right)}\left(Y\left(0\right)\right),\hfill \\ \hfill & ⋮\hfill \end{array}$$

4. Solution Method

Now, let us consider the initial Equation (1). For the purposes of using the Formula (6), we can also write (1) in the form

$$y^{\prime \prime }\left(x\right)=\alpha f\left(x\right),$$

where $f\left(x\right)=g\left(y\left(x\right)\right)=sinh\left(\alpha ·y\left(x\right)\right)$. Now, suppose that the function y is the original and its Taylor transform is $Y\left(k\right)$. Let us also suppose that the Taylor transform of f is $F\left(k\right)$. From the first condition (2), we know that $Y\left(0\right)=0$, and from the Formula (6) and the property (3)–(5), we can, for $k\ge 0$, write Equation (1) in the transformed form:

$$(k+1)(k+2)Y(k+2)=\alpha F\left(k\right).$$

(7)

Putting $k=0$ into the Equation (7), we obtain

$$2Y\left(2\right)=\alpha F\left(0\right)\stackrel{\left(6\right)}{⟹}Y\left(2\right)=0.$$

For $k=1$, we obtain

$$6Y\left(3\right)=\alpha F\left(1\right)\stackrel{\left(6\right)}{=}\alpha Y\left(1\right)g^{\prime }\left(Y\left(0\right)\right)⟹Y\left(3\right)={\displaystyle \frac{1}{6}}{\alpha }^{2}s,$$

where $\mathbb{R}\ni s=Y\left(1\right)$, which we are not able to determine at this stage. Therefore, the subsequent values of $Y\left(k\right)$ for $k\ge 4$ will depend on this unknown parameter s. By inserting subsequent values of $k\ge 2$ into the Equation (7), we will be able to determine subsequent values of the transforms $Y\left(k\right)$. And so, for $k=2$, we have $Y\left(4\right)=0$, and for subsequent values of k,

$$\begin{array}{cccc}\hfill Y\left(5\right)& ={\displaystyle \frac{1}{6}}{\alpha }^{4}s\left(s^2+1\right),\hfill & \hfill Y\left(6\right)& =0,\hfill \\ \hfill Y\left(7\right)& ={\displaystyle \frac{1}{120}}{\alpha }^{6}s\left(s^4+11s^2+1\right),\hfill & \hfill Y\left(8\right)& =0,\hfill \\ \hfill Y\left(9\right)& ={\displaystyle \frac{{\alpha }^{8}s\left(s^6+57s^4+102s^2+1\right)}{5040}},\hfill & \hfill Y\left(10\right)& =0,\hfill \\ \hfill Y\left(11\right)& ={\displaystyle \frac{{\alpha }^{10}s\left(s^8+247s^6+1923s^4+922s^2+1\right)}{362880}},\hfill & \hfill Y\left(12\right)& =0,\dots \hfill \end{array}$$

On this basis, we can construct an approximate solution of $y_n\left(x\right)$ of the form

$$y_n\left(x\right)=\sum _{i=0}^{n}Y\left(i\right)x^i,$$

(8)

which will depend on the unknown parameter s. The value of this parameter can be determined using the second condition (2), i.e., the condition $y\left(1\right)=1$, which in our case will take the form of the equation

$$\sum _{i=0}^{n}Y\left(i\right)=1.$$

(9)

Unfortunately, we will not solve this equation in the general case, because we have two unknowns, $\alpha$ and s, but since $\alpha$ is a parameter of Equation (1); for a specific value of the parameter $\alpha$, Equation (9) can be solved, obtaining the final form of the approximate solution $y_n\left(x\right)$.

5. Numerical Results

Let us take an example $\alpha =0.5$. Then, the approximate solutions (8) for $n=3,5,7,9,11$ will take the form

$$\begin{array}{cc}\hfill y_3\left(x\right)& =sx+{\displaystyle \frac{s{x}^3}{24}},\hfill \\ \hfill y_5\left(x\right)& =sx+{\displaystyle \frac{s{x}^3}{24}}+{\displaystyle \frac{s\left(s^2+1\right)x^5}{1920}}=y_3\left(x\right)+{\displaystyle \frac{s\left(s^2+1\right)x^5}{1920}},\hfill \\ \hfill y_7\left(x\right)& =y_5\left(x\right)+{\displaystyle \frac{s\left(s^4+11s^2+1\right)x^7}{322560}},\hfill \\ \hfill y_9\left(x\right)& =y_7\left(x\right)+{\displaystyle \frac{s\left(s^6+57s^4+102s^2+1\right)x^9}{92897280}},\hfill \\ \hfill y_{11}\left(x\right)& =y_9\left(x\right)+{\displaystyle \frac{s\left(s^8+247s^6+1923s^4+922s^2+1\right)x^{11}}{40874803200}}.\hfill \end{array}$$

For each of these equations, we can solve the Equation (9) in the set of real numbers, obtaining

$$s=0.96,s=0.959079,s=0.959045,s=0.959044,s=0.959044.$$

We now have the final form of the approximate solutions (8) for $n=3,5,7,9,11$. In Figure 1, we present these solutions and the exact solution, which is the solution obtained on the basis of the formula representing the solutions of the task (1) using elliptic functions (see, for example, Formula (3) in the work [7] or Formula (3) in the work [11]). In this figure, we also present the absolute errors $\Delta$ of these solutions determined from the formula

$${\Delta }_n\left(x\right)=\left|y\left(x\right)-y_n\left(x\right)\right|,$$

where y is the exact solution and $y_n$ is the approximate solution (8).

Table 1. Comparison of exact values of y and approximate values of $y_n$, $n=3,7,11$, for $\alpha =0.5$ and for $x\in [0,1]$ (points $x_i=0.1i$, $i=0,1,\dots ,10$).

x	$\mathit{y}\left(\mathit{x}\right)$	${\mathit{y}}_3\left(\mathit{x}\right)$	${\mathit{\Delta }}_3\left(\mathit{x}\right)$	${\mathit{y}}_7\left(\mathit{x}\right)$	${\mathit{\Delta }}_7\left(\mathit{x}\right)$	${\mathit{y}}_{11}\left(\mathit{x}\right)$	${\mathit{\Delta }}_{11}\left(\mathit{x}\right)$
0.0	0	0	0	0	0	0	0
0.1	0.09518	0.09604	0.00086	0.09594	0.09604	0.09594	0.00077
0.2	0.19063	0.19232	0.00169	0.19213	0.0015	0.19213	0.00149
0.3	0.28665	0.28908	0.00243	0.2888	0.00214	0.28879	0.00214
0.4	0.38352	0.38656	0.00304	0.38619	0.00266	0.38619	0.00266
0.5	0.48154	0.485	0.00346	0.48455	0.00301	0.48455	0.00301
0.6	0.581	0.58464	0.00364	0.58413	0.00313	0.58413	0.00313
0.7	0.68224	0.68572	0.00348	0.6852	0.00297	0.6852	0.00297
0.8	0.78557	0.78848	0.00291	0.78802	0.00245	0.78802	0.00244
0.9	0.89137	0.89316	0.00179	0.89286	0.00149	0.89285	0.00149
1.0	1	1	0	1	0	1	0

presents the discrete values of the exact solution y (the results in the table are rounded to the fifth decimal place) and the approximate solutions $y_n$, $n=3,5,7,9,11$, for $\alpha =0.5$ and for $x\in [0,1]$ (points $x_i=0.1i$, $i=0,1,\dots ,10$).

The obtained results show the usefulness of the method and are better than the corresponding results obtained with the Laplace decomposition method (see [11]).

Figure 2 shows approximate solutions of $y_{11}$ for several different values of $\alpha$. In

Table 2. Comparison of exact values y and approximate values $y_{11}$ for $\alpha =1$, $\alpha =3$ and for $x\in [0,1]$ (points $x_i=0.1i$, $i=0,1,\dots ,10$).

	$\mathit{\alpha }=1$			$\mathit{\alpha }=3$
$\mathit{x}$	$\mathit{y}\left(\mathit{x}\right)$	${\mathit{y}}_{\mathbf{11}}\left(\mathit{x}\right)$	${\Delta }_{\mathbf{11}}\left(\mathit{x}\right)$	$\mathit{y}\left(\mathit{x}\right)$	${\mathit{y}}_{\mathbf{11}}\left(\mathit{x}\right)$	${\Delta }_{\mathbf{11}}\left(\mathit{x}\right)$
0.0	0	0	0	0	0	0
0.1	0.081797	0.084661	0.002864	$0.025946$	0.027075	0.001129
0.2	0.164531	0.170171	0.00564	$0.054248$	0.056609	0.002361
0.3	0.249167	0.257394	0.008227	$0.087495$	0.091305	0.003811
0.4	0.336732	0.347223	0.010491	$0.128777$	0.134397	0.00561
0.5	0.428347	0.4406	0.012253	$0.182056$	0.190036	0.00797
0.6	0.525274	0.538534	0.01326	$0.252747$	0.263908	0.011161
0.7	0.628971	0.642129	0.013158	$0.348805$	0.364292	0.015487
0.8	0.741168	0.752608	0.01144	$0.483138$	0.504007	0.020869
0.9	0.86397	0.871363	0.007393	$0.680163$	0.704013	0.02385
1.0	1	1.0	$1.11(-16)$	1	1.0	$2.75(-16)$

and

Table 3. Comparison of exact values y and approximate values $y_{11}$ for $\alpha =4$, $\alpha =10$ and for $x\in [0,1]$ (points $x_i=0.1i$, $i=0,1,\dots ,10$).

	$\mathit{\alpha }=4$			$\mathit{\alpha }=10$
$\mathit{x}$	$\mathit{y}\left(\mathit{x}\right)$	${\mathit{y}}_{\mathbf{11}}\left(\mathit{x}\right)$	${\Delta }_{\mathbf{11}}\left(\mathit{x}\right)$	$\mathit{y}\left(\mathit{x}\right)$	${\mathit{y}}_{\mathbf{11}}\left(\mathit{x}\right)$	${\Delta }_{\mathbf{11}}\left(\mathit{x}\right)$
0.0	0	0	0	0	0	0
0.1	0.081797	0.013276	0.001788	$4.211(-5)$	0.000118	0.000076
0.2	0.164531	0.028707	0.003866	$1.299(-4)$	0.000365	0.000235
0.3	0.249167	0.048805	0.006575	$3.589(-4)$	0.001009	0.00065
0.4	0.336732	0.076875	0.01037	$9.779(-4)$	0.002749	0.001771
0.5	0.428347	0.117628	0.01592	$2.659(-3)$	0.007472	0.004813
0.6	0.525274	0.178181	0.024257	$7.228(-3)$	0.02028	0.013051
0.7	0.628971	0.269837	0.036909	$1.966(-2)$	0.054827	0.035163
0.8	0.741168	0.411406	0.055126	$5.373(-2)$	0.146913	0.093183
0.9	0.86397	0.635472	0.072162	$1.521(-1)$	0.387611	0.235497
1.0	1	1.0	$1.66(-16)$	1	1	$2.22(-16)$

, we present exact results (for $\alpha =10$, these results can be found, e.g., in [6]) obtained by the presented method for $\alpha =1$, $\alpha =3$, $\alpha =4$ and $\alpha =10$.

In this case, the obtained results also show the usefulness of the method and are better than the corresponding results obtained with the above-mentioned method (for $\alpha =1$). For $\alpha =10$, they are comparable to other results or better compared to, e.g., the homotopy perturbation method (see [4]).

6. Conclusions

This paper presents a hybrid method for solving Troesch’s problem. The proposed approach turned out to be effective, also in comparison to other methods, as shown by the presented numerical comparison. An additional advantage of the presented method is, apart from its simplicity, the possibility of its algorithmization. Using a tool that can perform symbolic transformations (the author used the Wolfram Mathematica program in version 14.0; see [52,53]), one can construct a universal procedure dependent on the parameters occurring in the problem and in the method ($\alpha$ and n), thanks to which one can solve Troesch’s problem together with generating results and creating a graphical illustration of the solution. Additionally, as many cited sources in the literature indicate, many methods are sensitive to the value of the parameter $\alpha$ occurring in the problem—for the value $\alpha >1$, many methods generate results that leave much to be desired. For the discussed method, we obtained good results even for $\alpha =10$ (see Table 3), and even larger (see Figure 2).