Solving General Fractional Lane-Emden-Fowler Differential Equations Using Haar Wavelet Collocation Method

Abstract

This paper aims to solve general fractional Lane-Emden-Fowler differential equations using the Haar wavelet collocation method. This method transforms the fractional differential equation into a nonlinear system of equations, which is further solved for Haar coefficients using Newton’s method. We have constructed the higher-order Lane-Emden-Fowler equations. We have also discussed the convergence rate and stability analysis of our technique. We have explained the applications and numerically simulated the examples graphically and in tabular format to elaborate on the accuracy and efficiency of this approach.

1. Introduction

In the past few decades, the study of singular initial value problems has attracted the attention of many physicists and mathematicians. Lane-Emden-Fowler equations are these types of equations. Lane [1] introduced the equation in 1870, and Emden and Fowler [2] generalized the equation further. Lane explained the gravitation potential of a spherically symmetric Newtonian self-gravitating star [3] using these equations. These equations have a vast amount of applications in modeling many problems in physics and dynamics. The general Lane-Emden-Fowler equation is as follows:

$${\xi }^{-\eta }{D}^{\beta }\left({\xi }^{\eta }{D}^{\gamma }\right)q=\omega \left(\xi ,q\left(\xi \right)\right),$$

(1)

where $\eta >0$ is a positive real number called a shape factor, $D=\frac{d}{d\xi }$ is the differential operator, $\omega$ is linear or nonlinear function, and $\beta +\gamma$ gives the order of the equation. These equations are used in magnetic field models [4], classical and quantum mechanics [5], biological systems, geometry [6], and fluid mechanics problems [7,8].

Several forms of fractional initial value problems have been proposed in standard models, and there has been a significant interest in developing numerical schemes for their solutions. Fractional calculus is a generalized form of integer order calculus. There are many applications of fractional calculus. Fractional calculus has been deployed to model the oscillation of earthquakes [9], neural networks [10,11], signal processing [12], economics [13], bioengineering [14], and electromagnetism [15].

Many researchers, such as Riemann, Liouville, Caputo, Hadamard, Grunwald, and others, have published extensively about the applications of fractional calculus. Mathematical models with fractional order derivatives provide more insight because they posses the memory effect. Many techniques have been developed to solve fractional problems, such as decomposition methods [16], collocation methods [17], residual power series methods [18,19], finite differences methods [20], perturbation methods, variational iteration methods [21]. Zhang and Han [22] proposed a quasi wavelet method to solve time-dependent fractional partial differential equations. Jiang et al. [23] presented a predictor–corrector difference scheme for nonlinear fractional differential equations. Yang and Zhang [24] proposed a spectral sinc-collocation method for fourth-order heat models.

Wavelet theory has made distinguished contributions to mathematical studies. It is a powerful tool for engineering. Wavelets are used in signal processing, optimal control, and time-frequency analysis [25]. There are many wavelets, such as Daubechies [26], B-spline [4], Legendre [27], and Haar [28]. The Haar wavelet is an orthonormal wavelet with compact support, introduced by a Hungarian mathematician, Alfred Haar, in 1910. The Haar wavelet gives accurate results for small grid points. It contains members of the Daubechies family, so it is very good for computer implementations and is easily expressed in the programming language. Chen and Hsiao [29] derived a Haar operational matrix of the integrals of Haar functions. They made a great contribution to the use of Haar wavelets in applications of dynamic systems. Lepik [30] solved the differential equations using the Haar wavelet. Islam et al. [31] solved the integro-differential equations using the Haar wavelet. Bujurke et al. [32] compute the eigenvalues and solutions of regular Sturm–Liouville problems using Haar wavelets. Chang et al. [33] describe the designation of Haar wavelet matrices in the numerical solution of ODEs. This article aims to solve general fractional Emden-Fowler-type equations using the Haar wavelet collocation method. We write the highest derivative in linear combinations of Haar functions and calculate other derivatives using the integration of Haar functions. This method transforms the fractional differential equation into a nonlinear system of equations, which is further solved for Haar coefficients using the Newton method. After calculating the Haar coefficients, we can easily determine the solution.

The present study is structured as follows: Section 2 defines the Haar wavelet and recalls the basic definitions of fractional calculus. In Section 3, we discuss the construction of the general equation of the Caputo-type fractional Lane-Emden-Fowler differential equation. In Section 4, we discuss the Haar wavelet method. In Section 5, we discuss the convergence rate, stability and error analysis of the technique. In Section 6, we discuss the examples and in Section 7 we discuss the numerical simulation of all of these examples graphically and in tabular format. In the end, we conclude our results.

2. Preliminaries

This section will recall some necessary definitions of fractional calculus and Haar wavelets. These definitions will assist us in the next sections.

Definition 1.

The Riemann-Liouville fractional integral operator $I^{\sigma }$ of order σ on $L_2[0,1]$ is given by

$$\begin{array}{c}{I}^{\sigma }\left(q\left(\xi \right)\right)=\frac{1}{\mathrm{\Gamma }\left(\sigma \right)}\underset{0}{\overset{\xi }{\int }}\frac{q\left(\kappa \right)}{{\left(\xi -\kappa \right)}^{1-\sigma }}d\kappa ,\hfill \\ I^0\left(q\left(\xi \right)\right)=\left(q\left(\xi \right)\right),\hfill \end{array}$$

where $\mathrm{\Gamma }\left(\lambda \right)=\underset{0}{\overset{\infty }{\int }}{\kappa }^{\lambda -1}{e}^{-\kappa }d\kappa$ is gamma function and

$$I^{\sigma }{\xi }^{\upsilon }=\frac{\mathrm{\Gamma }(\upsilon +1)}{\mathrm{\Gamma }(\sigma +\upsilon +1)}{\xi }^{\sigma +\upsilon }.$$

Definition 2.

The Caputo fractional derivative of order σ is given by

$$D^{\sigma }\left(q\left(\xi \right)\right)=I^{n-\sigma }{D}^n\left(q\left(\xi \right)\right)=\frac{1}{\mathrm{\Gamma }(n-\sigma )}\underset{0}{\overset{\xi }{\int }}\frac{q^{\left(n\right)}\left(\kappa \right)}{{\left(\xi -\kappa \right)}^{-n+1+\sigma }}d\kappa$$

provided the integral exists, where n is the smallest integer such that $n-1<\sigma \le n$. It satisfies the following properties:

$$I^{\sigma }{D}^{\sigma }\left(q\left(\xi \right)\right)=q\left(\xi \right)-\sum _{\epsilon =0}^{n-1}{q}^{\left(\epsilon \right)}\left(0^{+}\right)\frac{{\xi }^{\epsilon }}{\epsilon !}$$

and

$$D^{\sigma }{\xi }^{\upsilon }=\frac{\mathrm{\Gamma }(\upsilon +1)}{\mathrm{\Gamma }(\upsilon -\sigma +1)}{\xi }^{\upsilon -\sigma }.$$

Haar Wavelet and Function Approximations

The family of Haar wavelets consists of piecewise constant functions over the real line. They contain only values $-1,0,1$. They are discontinuous, and therefore not differentiable.

$$h_j\left(\varsigma \right)=\left\{\begin{array}{c}1,{\theta }_1\left(j\right)\le \varsigma <{\theta }_2\left(j\right)\hfill \\ -1,{\theta }_2\left(j\right)\le \varsigma <{\theta }_3\left(j\right)\hfill \\ 0,otherwise.\hfill \end{array}\right.$$

where ${\theta }_1\left(j\right)=\frac{r}{2^{\alpha }}, {\theta }_2\left(j\right)=\frac{r+0.5}{2^{\alpha }}, {\theta }_3\left(j\right)=\frac{r+1}{2^{\alpha }},$ $\alpha =0, 1, 2, \dots J$ and $r=0, 1, 2, \dots 2^{\alpha }-1.$

We manipulate the wavelet by translating and dilating it. $\alpha$ here represents the level of wavelet or dilation parameter level and $\varsigma$ represents the translation parameter. J is the maximum level of resolution and the relationship between $2^{\alpha }$ and r is $j=2^{\alpha }+r+1$. For $j=1$

$$h_1\left(\xi \right)=\left\{\begin{array}{c}1,\xi \in \left[0,1\right]\hfill \\ 0,otherwise.\hfill \end{array}\right.$$

$$h_2\left(\xi \right)=\left\{\begin{array}{c}\begin{array}{c}1, \xi \in \left[0,\frac{1}{2}\right)\hfill \\ -1, \xi \in \left[\frac{1}{2},1\right)\hfill \end{array}\hfill \\ 0, otherwise.\hfill \end{array}\right.$$

In particular, the Haar wavelet is an orthogonal square wave family, generally written as

$$h_j\left(\xi \right)=h_2\left(2^{\alpha }\xi -\frac{r}{2^{\alpha }}\right)$$

for $j\ge 3, j=2^{\alpha }+r+1, \alpha \ge 0, 0\le r\le 2^{\alpha }-1$ and

$$\underset{0}{\overset{1}{\int }}{h}_j\left(\xi \right)h_l\left(\xi \right)d\xi =\left\{\begin{array}{c}{2}^{-\alpha };j=l\hfill \\ 0;j\ne l\hfill \end{array}\right.$$

If $g\left(\xi \right)$ is a function defined on interval $\left[0,1\right]$, then the function is approximated using Haar functions, such as

$$g\left(\xi \right)=\sum _{\tau =1}^{\infty }{\lambda }_{\tau }{h}_{\tau },$$

where ${\lambda }_{\tau }$ are the Haar coefficients.

The generalized fractional integration can be calculated analytically as

$${\rho }_{\tau ,\sigma }\left(\xi \right)=\left\{\begin{array}{c}0,\xi \in [0,{\varsigma }_1)\\ \frac{1}{\mathrm{\Gamma }(\sigma +1)}{(\xi -{\varsigma }_1)}^{\sigma },\xi \in [{\varsigma }_1,{\varsigma }_2)\\ \frac{1}{\mathrm{\Gamma }(\sigma +1)}\left[{(\xi -{\varsigma }_1)}^{\sigma }-2{(\xi -{\varsigma }_2)}^{\sigma }\right],\xi \in [{\varsigma }_2,{\varsigma }_3)\\ \frac{1}{\mathrm{\Gamma }(\sigma +1)}\left[{(\xi -{\varsigma }_1)}^{\sigma }-2{(\xi -{\varsigma }_2)}^{\sigma }+{(\xi -{\varsigma }_3)}^{\sigma }\right],\xi \in [{\varsigma }_3,1)\end{array}\right.,$$

where $\sigma$ is a positive real number.

3. Construction of Lane-Emden-Fowler Equation

Consider the general form of the Lane-Emden-Fowler equation

$${\xi }^{-\eta }{D}^{\beta }\left({\xi }^{\eta }{D}^{\gamma }\right)q=\omega \left(\xi ,q\left(\xi \right)\right).$$

(2)

We can obtain higher-order equations of fourth-, fifth- and sixth-order by taking

$$\beta +\gamma =4,$$

$$\beta +\gamma =5,$$

$$\beta +\gamma =6,$$

respectively. There are three possible choices of fourth-order equations, four possibilities for fifth-order equations and five possible choices for sixth-order Lane-Emden-Fowler equations.

3.1. For Fourth-Order Equations

Case 1: When $\beta =3$, $\gamma =1$, the equation will be

$${\xi }^{-\eta }{D}^3\left({\xi }^{\eta }{D}^1\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(3)

After simplification, Equation (3) will be

$$D^{4}q\left(\xi \right)+\frac{3\eta }{\xi }{D}^{3}q\left(\xi \right)+\frac{3\eta (\eta -1)}{{\xi }^2}{D}^{2}q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)}{{\xi }^3}Dq\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(4)

The Equation (4) in fractional order is taken as

$$D^{4\sigma }q\left(\xi \right)+\frac{3\eta }{{\xi }^{\sigma }}{D}^{3\sigma }q\left(\xi \right)+\frac{3\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{2\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)}{{\xi }^{3\sigma }}{D}^{\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0,$$

(5)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, respectively.

Case 2: When $\beta =2$, $\gamma =2$, the equation will be

$${\xi }^{-\eta }{D}^2\left({\xi }^{\eta }{D}^2\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(6)

After simplification, Equation (6) will be

$$D^{4}q\left(\xi \right)+\frac{2\eta }{\xi }{D}^{3}q\left(\xi \right)+\frac{\eta (\eta -1)}{{\xi }^2}{D}^{2}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(7)

The Equation (7) in fractional order is taken as

$$D^{4\sigma }q\left(\xi \right)+\frac{2\eta }{{\xi }^{\sigma }}{D}^{3\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{2\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(8)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, respectively.

Case 3: When $\beta =1$, $\gamma =3$, the equation will be

$${\xi }^{-\eta }{D}^1\left({\xi }^{\eta }{D}^3\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(9)

After simplification, Equation (9) will be

$$D^{4}q\left(\xi \right)+\frac{\eta }{\xi }{D}^{3}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(10)

The Equation (10) in fractional order is taken as

$$D^{4\sigma }q\left(\xi \right)+\frac{\eta }{{\xi }^{\sigma }}{D}^{3\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(11)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, respectively.

3.2. For Fifth-Order Equations

Case 1: When $\beta =4$, $\gamma =1$, the equation will be

$${\xi }^{-\eta }{D}^4\left({\xi }^{\eta }{D}^1\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(12)

After simplification, Equation (12) will be

$$\begin{array}{c}{D}^{5}q\left(\xi \right)+\frac{4\eta }{\xi }{D}^{4}q\left(\xi \right)+\frac{6\eta (\eta -1)}{{\xi }^2}{D}^{3}q\left(\xi \right)+\frac{4\eta (\eta -1)(\eta -2)}{{\xi }^3}{D}^{2}q\left(\xi \right)\hfill \\ +\frac{\eta (\eta -1)(\eta -2)(\eta -3)}{{\xi }^4}Dq\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.\hfill \end{array}$$

(13)

The Equation (13) in fractional order is taken as

$$\begin{array}{c}{D}^{5\sigma }q\left(\xi \right)+\frac{4\eta }{{\xi }^{\sigma }}{D}^{4\sigma }q\left(\xi \right)+\frac{6\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{3\sigma }q\left(\xi \right)+\frac{4\eta (\eta -1)(\eta -2)}{{\xi }^{3\sigma }}{D}^{2\sigma }q\left(\xi \right)\hfill \\ +\frac{\eta (\eta -1)(\eta -2)(\eta -3)}{{\xi }^{4\sigma }}{D}^{\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0,\hfill \end{array}$$

(14)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, respectively.

Case 2: When $\beta =3$, $\gamma =2$, the equation will be

$${\xi }^{-\eta }{D}^3\left({\xi }^{\eta }{D}^2\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(15)

After simplification, Equation (15) will be

$$D^{5}q\left(\xi \right)+\frac{3\eta }{\xi }{D}^{4}q\left(\xi \right)+\frac{3\eta (\eta -1)}{{\xi }^2}{D}^{3}q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)}{{\xi }^3}{D}^{2}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(16)

The Equation (16) in fractional order is taken as

$$D^{5\sigma }q\left(\xi \right)+\frac{3\eta }{{\xi }^{\sigma }}{D}^{4\sigma }q\left(\xi \right)+\frac{3\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{3\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)}{{\xi }^{3\sigma }}{D}^{2\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(17)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, respectively.

Case 3: When $\beta =2$, $\gamma =3$, the equation will be

$${\xi }^{-\eta }{D}^2\left({\xi }^{\eta }{D}^3\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(18)

After simplification, Equation (18) will be

$$D^{5}q\left(\xi \right)+\frac{2\eta }{\xi }{D}^{4}q\left(\xi \right)+\frac{\eta (\eta -1)}{{\xi }^2}{D}^{3}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(19)

The Equation (19) in fractional order is taken as

$$D^{5\sigma }q\left(\xi \right)+\frac{2\eta }{{\xi }^{\sigma }}{D}^{4\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{3\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0,$$

(20)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, respectively.

Case 4: When $\beta =1$, $\gamma =4$, the equation will be

$${\xi }^{-\eta }{D}^1\left({\xi }^{\eta }{D}^4\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(21)

After simplification, Equation (21) will be

$$D^{5}q\left(\xi \right)+\frac{\eta }{\xi }{D}^{4}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(22)

The Equation (22) in fractional order is taken as

$$D^{5\sigma }q\left(\xi \right)+\frac{\eta }{{\xi }^{\sigma }}{D}^{4\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(23)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4$, where $q^{\left(\sigma \right)},q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, respectively.

3.3. For Sixth-Order Equations

Case 1: When $\beta =5$, $\gamma =1$, the equation will be

$${\xi }^{-\eta }{D}^5\left({\xi }^{\eta }{D}^1\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(24)

After simplification, Equation (24) will be

$$\begin{array}{c}{D}^{6}q\left(\xi \right)+\frac{4\eta }{\xi }{D}^{5}q\left(\xi \right)+\frac{10\eta (\eta -1)}{{\xi }^2}{D}^{4}q\left(\xi \right)+\frac{10\eta (\eta -1)(\eta -2)}{{\xi }^3}{D}^{3}q\left(\xi \right)\hfill \\ +\frac{5\eta (\eta -1)(\eta -2)(\eta -3)}{{\xi }^4}{D}^{2}q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)(\eta -3)(\eta -4)}{{\xi }^5}Dq\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.\hfill \end{array}$$

(25)

The Equation (25) in fractional order is taken as

$$\begin{array}{c}{D}^{6\sigma }q\left(\xi \right)+\frac{4\eta }{{\xi }^{\sigma }}{D}^{5\sigma }q\left(\xi \right)+\frac{10\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{4\sigma }q\left(\xi \right)+\frac{10\eta (\eta -1)(\eta -2)}{{\xi }^{3\sigma }}{D}^{3\sigma }q\left(\xi \right)\hfill \\ +\frac{5\eta (\eta -1)(\eta -2)(\eta -3)}{{\xi }^{4\sigma }}{D}^{2\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)(\eta -3)(\eta -4)}{{\xi }^{5\sigma }}{D}^{\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0,\hfill \end{array}$$

(26)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4, q^{\left(5\sigma \right)}\left(0\right)={\varphi }_5$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}, q^{\left(5\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, respectively.

Case 2: When $\beta =4$, $\gamma =2$, the equation will be

$${\xi }^{-\eta }{D}^4\left({\xi }^{\eta }{D}^2\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(27)

After simplification, Equation (27) will be

$$\begin{array}{c}{D}^{6}q\left(\xi \right)+\frac{4\eta }{\xi }{D}^{5}q\left(\xi \right)+\frac{6\eta (\eta -1)}{{\xi }^2}{D}^{4}q\left(\xi \right)+\frac{4\eta (\eta -1)(\eta -2)}{{\xi }^3}{D}^{3}q\left(\xi \right)\hfill \\ +\frac{\eta (\eta -1)(\eta -2)(\eta -3)}{{\xi }^4}{D}^{2}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.\hfill \end{array}$$

(28)

The Equation (28) in fractional order is taken as

$$\begin{array}{c}{D}^{6\sigma }q\left(\xi \right)+\frac{4\eta }{{\xi }^{\sigma }}{D}^{5\sigma }q\left(\xi \right)+\frac{6\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{4\sigma }q\left(\xi \right)+\frac{4\eta (\eta -1)(\eta -2)}{{\xi }^{3\sigma }}{D}^{3\sigma }q\left(\xi \right)\hfill \\ +\frac{\eta (\eta -1)(\eta -2)(\eta -3)}{{\xi }^{4\sigma }}{D}^{2\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.\hfill \end{array}$$

(29)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4, q^{\left(5\sigma \right)}\left(0\right)={\varphi }_5$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}, q^{\left(5\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, respectively.

Case 3: When $\beta =3$, $\gamma =3$, the equation will be

$${\xi }^{-\eta }{D}^3\left({\xi }^{\eta }{D}^3\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(30)

After simplification, Equation (30) will be

$$D^{6}q\left(\xi \right)+\frac{3\eta }{\xi }{D}^{5}q\left(\xi \right)+\frac{3\eta (\eta -1)}{{\xi }^2}{D}^{4}q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)}{{\xi }^3}{D}^{3}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(31)

The Equation (31) in fractional order is taken as

$$D^{6\sigma }q\left(\xi \right)+\frac{3\eta }{{\xi }^{\sigma }}{D}^{5\sigma }q\left(\xi \right)+\frac{3\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{4\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)(\eta -2)}{{\xi }^{3\sigma }}{D}^{3\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(32)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4, q^{\left(5\sigma \right)}\left(0\right)={\varphi }_5$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}, q^{\left(5\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, respectively.

Case 4: When $\beta =2$, $\gamma =4$, the equation will be

$${\xi }^{-\eta }{D}^2\left({\xi }^{\eta }{D}^4\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(33)

After simplification, Equation (33) will be

$$D^{6}q\left(\xi \right)+\frac{2\eta }{\xi }{D}^{5}q\left(\xi \right)+\frac{\eta (\eta -1)}{{\xi }^2}{D}^{4}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(34)

The Equation (34) in fractional order is taken as

$$D^{6\sigma }q\left(\xi \right)+\frac{2\eta }{{\xi }^{\sigma }}{D}^{5\sigma }q\left(\xi \right)+\frac{\eta (\eta -1)}{{\xi }^{2\sigma }}{D}^{4\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(35)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4, q^{\left(5\sigma \right)}\left(0\right)={\varphi }_5$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)},q^{\left(5\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, respectively.

Case 5: When $\beta =1$, $\gamma =5$, the equation will be

$${\xi }^{-\eta }{D}^1\left({\xi }^{\eta }{D}^5\right)q+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(36)

After simplification, Equation (36) will be

$$D^{6}q\left(\xi \right)+\frac{\eta }{\xi }{D}^{5}q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(37)

The Equation (37) in fractional order is taken as

$$D^{6\sigma }q\left(\xi \right)+\frac{\eta }{{\xi }^{\sigma }}{D}^{5\sigma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.$$

(38)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, q^{\left(3\sigma \right)}\left(0\right)={\varphi }_3, q^{\left(4\sigma \right)}\left(0\right)={\varphi }_4, q^{\left(5\sigma \right)}\left(0\right)={\varphi }_5$, where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}, q^{\left(3\sigma \right)}, q^{\left(4\sigma \right)}, q^{\left(5\sigma \right)}$ denotes the Caputo derivative of order $\sigma$, $2\sigma$, $3\sigma$, $4\sigma$, $5\sigma$, respectively.

4. Method

Consider the general Equation (1)

$${\xi }^{-\eta }{D}^{\beta }\left({\xi }^{\eta }{D}^{\gamma }\right)q=\omega \left(\xi ,q\left(\xi \right)\right).$$

After simplifying and expanding this equation, we have

$$\begin{array}{c}{D}^{\beta +\gamma }q\left(\xi \right){+}^{\beta }{C}_{\beta -1}\left(\frac{\eta }{\xi }\right)D^{\beta +\gamma -1}q\left(\xi \right){+}^{\beta }{C}_{\beta -2}\left(\frac{\eta (\eta -1)}{{\xi }^2}\right)D^{\beta +\gamma -2}q\left(\xi \right)+\dots \hfill \\ {}^{\beta }{C}_0\left(\frac{\eta (\eta -1)(\eta -2)(\eta -3)\dots (\eta -\beta +1)}{{\xi }^{\beta }}\right)D^{\gamma }q\left(\xi \right)+\omega \left(\xi ,q\left(\xi \right)\right)=0.\hfill \end{array}$$

(39)

we generalize this Equation (39) into fractional form as

$$\begin{array}{c}{D}_{\xi }^{(\beta +\gamma )\sigma }q\left(\xi \right)+{}^{\beta }{C}_{\beta -1}\left(\frac{\eta }{{\xi }^{\sigma }}\right)D_{\xi }^{(\beta +\gamma -1)\sigma }q\left(\xi \right)+{}^{\beta }{C}_{\beta -2}\left(\frac{\eta (\eta -1)}{{\xi }^{2\sigma }}\right)D_{\xi }^{(\beta +\gamma -2)\sigma }q\left(\xi \right)+\dots \hfill \\ {}^{\beta }{C}_0\left(\frac{\eta (\eta -1)(\eta -2)(\eta -3)\dots (\eta -\beta +1)}{{\xi }^{\beta \sigma }}\right)D_{\xi }^{\left(\gamma \right)\sigma }q\left(\xi \right)+\omega (\xi ,q\left(\xi \right))=0,\hfill \end{array}$$

(40)

with initial conditions $q\left(0\right)={\varphi }_0, q^{\left(\sigma \right)}\left(0\right)={\varphi }_1, q^{\left(2\sigma \right)}\left(0\right)={\varphi }_2, \dots q^{\left(\right(\beta +\gamma -1\left)\sigma \right)}\left(0\right)={\varphi }_{\beta +\gamma -1},$ where $q^{\left(\sigma \right)}, q^{\left(2\sigma \right)}$ denotes the caputo derivative of order $\sigma$ and $2\sigma$. $D_{\xi }=\frac{d}{d\xi }$ in (40) represents Caputo fractional differential operator.

Clearly, when $\beta =1$, $\gamma =1$ and $\eta =2$, we have the Equation (63) in example 7. In a similar fashion, we can extract all the examples by applying different values of $\beta , \gamma , \eta$.

Now, we discuss the method.

• Step 1: We approximate the highest-order derivative using Haar functions:

  $$D_{\xi }^{(\beta +\gamma )\sigma }q\left(\xi \right)=\sum _{\tau =1}^{2L}{\lambda }_{\tau }{h}_{\tau }.$$

  (41)
• Step 2: We integrate the Equation (41) again and again, and after applying the initial conditions, we get

  $$D_{\xi }^{(\beta +\gamma -1)\sigma }q\left(\xi \right)={\varphi }_{\beta +\gamma -1}+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left(\xi \right),$$

  (42)

  $$D_{\xi }^{(\beta +\gamma -2)\sigma }q\left(\xi \right)={\varphi }_{\beta +\gamma -2}+\frac{{\xi }^{\sigma }}{\mathrm{\Gamma }(\sigma +1)}{\varphi }_{\beta +\gamma -1}+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left(\xi \right).$$

  and so on. The last term is

  $$q\left(\xi \right)=\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,(\beta +\gamma )\sigma }\left(\xi \right)+\frac{{\xi }^{(\beta +\gamma -1)\sigma }}{\mathrm{\Gamma }\left(\right(\beta +\gamma -1)\sigma +1)}{\varphi }_{\beta +\gamma -1}+\dots {\varphi }_0.$$

  (43)
• Step 3: We collocate the points as

  $${\xi }_{\tau }=\frac{\tau -0.5}{2L},\tau =1, 2, \dots 2L.$$
  Now, we substitute all the values of derivatives into Equation (40) and collocate the points, resulting in the system of differential equations as follows

  $$\begin{array}{c}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+{}^{\beta }{C}_{\beta -1}\left(\frac{\eta }{{{\xi }_{\tau }}^{\sigma }}\right)\left({\varphi }_{\beta +\gamma -1}+{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)\right)+\dots \hfill \\ \omega ({\xi }_{\tau },{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,(\beta +\gamma )\sigma }\left({\xi }_{\tau }\right)+\frac{{{\xi }_{\tau }}^{(\beta +\gamma -1)\sigma }}{\mathrm{\Gamma }\left(\right(\beta +\gamma -1)\sigma +1)}{\varphi }_{\beta +\gamma -1}+\dots {\varphi }_0)=0.\hfill \end{array}$$

  (44)
• Step 4: We solve the system of equations (44) using the Newton method and obtain the values of the Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$.
• Step 5: After substituting the values of Haar coefficients into Equation (43), we obtain the numerical solution of Equation (40).

5. Convergence Analysis

Lemma 1

([34]). Assume that $q\left(\xi \right)\in L_2(\Re )$ and has a bounded first derivative, that is, $|q^{\prime }\left(\xi \right)|\le G,\forall \xi \in (0,1),P>0$ and $q\left(\xi \right)={\displaystyle \sum _{\tau =1}^{\infty }}{\lambda }_{\tau }{h}_{\tau }\left(\xi \right)$. Then, $|{\lambda }_j|\le G{2}^{-(3l-2)/2}$.

Theorem 1

([34]). Suppose $q\left(\xi \right)\in L_2(\Re )$ is a continuous function with a bounded first derivative in (0,1). Then, the error norm at the Jth level satisfies

$$\left|\right|E_J\left|\right|\le \sqrt{\frac{G}{12}}D{2}^{-(3/2)L}$$

where $|q^{\prime }\left(\xi \right)|\le G, \forall \xi \in (0,1)$, $G>0$ and $M=2^J$, where J is maximum resolution.

Proof. 

The proof is straightforward. We can refer to [34]. □

5.1. Numerical Error

The maximum absolute error is given by

$$E_C=Max.|q_{\tau }^{exact}-q_{\tau }^{approx.}|,$$

where $q_{\tau }^{exact}$ and $q_{\tau }^{approx.}$ are exact and approximate solutions at the $\tau$th collocation point.

5.2. Rate of Convergence

Rate of convergence is defined by

$$R_C\left(L\right)=\frac{log\left[E_C(L/2)/E_C\left(L\right)\right]}{log2},$$

where $E_C\left(L\right)$ represents the maximum absolute error at L collocation points.

5.3. Stability

The condition number is significant to measure the stability of an algorithm [35]. For stability, the condition number should be bounded. We consider the system of equations formed in our algorithm as

$$\mathrm{H}\mathrm{A}=\mathrm{{\rm Y}},$$

where $\mathrm{H}$ denotes the Haar weights, $\mathrm{A}$ denotes the unknown Haar coefficients and Y is a known vector. The condition number bound for some examples is given in

Table 1. Condition number bound for Examples 1 and 3.

Resolution	Size	Example 1	Example 3
3	$16\times 16$	1.4133 × 10${}^2$	1.8795 × 10${}^2$
4	$32\times 32$	2.8277 × 10${}^2$	3.8086 × 10${}^2$
5	$64\times 64$	5.6547 × 10${}^2$	7.6413 × 10${}^2$

.

Definition 3

([36]). Let us consider the system of equations to be of type $\mathrm{H}\mathrm{A}=\mathrm{{\rm Y}},$ if the inverse of $\mathrm{H}$ exists and is bounded, then the algorithm is stable; that is,

$$\left|\right|{\mathrm{H}}^{-1}\left|\right|\le \mathrm{Z},$$

where $\mathrm{Z}$ is a constant.

The Condition number is bounded [35]; that is,

$$Cond{\left(\mathrm{H}\right)}_2\le ||\mathrm{H}|{|}^2||{\mathrm{H}}^{-1}|{|}^2.$$

6. Applications

Example 1.

Taking $\eta =2$ in Equation (8) gives us this fourth-order fractional Lane-Emden-Fowler equation:

$$D_{\xi }^{4\sigma }q\left(\xi \right)+\frac{4}{{\xi }^{\sigma }}{D}_{\xi }^{3\sigma }q\left(\xi \right)+\frac{2}{{\xi }^{2\sigma }}{D}_{\xi }^{2\sigma }q\left(\xi \right)=3(12-53{\xi }^4+12{\xi }^8){\left(q\left(\xi \right)\right)}^{-15},$$

(45)

with initial conditions $q\left(0\right)=1, q^{\left(\sigma \right)}\left(0\right)=0, q^{\left(2\sigma \right)}\left(0\right)=0, q^{\left(3\sigma \right)}\left(0\right)=0$.

Now, we apply the method, and using initial conditions, we can write

$$q\left(\xi \right)=1+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,4\sigma }\left(\xi \right).$$

(46)

After substituting all the values of $q\left(\xi \right)$ and its derivatives into (45) and using collocation of points, we obtain a system of nonlinear equations as follows:

$$\begin{array}{c}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{4}{{\xi }_{\tau }^{\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)+\frac{2}{{\xi }_{\tau }^{2\sigma }}\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left({\xi }_{\tau }\right)\hfill \\ -3(12-53{\xi }_{\tau }^4+12{\xi }_{\tau }^8){(1+{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,4\sigma }\left({\xi }_{\tau }\right))}^{-15}=0.\hfill \end{array}$$

(47)

We can easily solve this system (47) using the Newton method. After solving this system, we have the values of the Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$ and substituting these values into (46) gives the approximate solution.

The exact solution of (45) when $\sigma =1$ is $q\left(\xi \right)={(1+{\xi }^4)}^{\frac{1}{4}}.$ which is given in [37].

Example 2.

When we substitute $\eta =2$ in Equation (11), we get this fourth-order fractional Lane-Emden-Fowler equation:

$$D_{\xi }^{4\sigma }q\left(\xi \right)+\frac{2}{{\xi }^{\sigma }}{D}_{\xi }^{3\sigma }q\left(\xi \right)+\xi {\left(q\left(\xi \right)\right)}^{-2}=\omega \left(\xi \right),$$

(48)

with initial conditions $q\left(0\right)=0, q^{\left(\sigma \right)}\left(0\right)=0, q^{\left(2\sigma \right)}\left(0\right)=0, q^{\left(3\sigma \right)}\left(0\right)=0$, where

$$\omega \left(\xi \right)=\mathrm{\Gamma }\left(1+4\sigma \right)+8\frac{\mathrm{\Gamma }\left(4\sigma \right)}{\mathrm{\Gamma }\left(\sigma \right)}+\xi {\left(1+{\xi }^{4\sigma }\right)}^{-2}$$

Now, we apply the method, and using the initial conditions, we can write

$$q\left(\xi \right)=1+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,4\sigma }\left(\xi \right).$$

(49)

After substituting all the values of $q\left(\xi \right)$ and its derivatives into (48) and using collocation of points, we get a system of nonlinear equations as follows:

$$\sum _{\tau =1}^{2L}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{2}{{\xi }_{\tau }^{\sigma }}\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)+{\xi }_{\tau }{(1+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,4\sigma }\left({\xi }_{\tau }\right))}^{-2}-\omega \left({\xi }_{\tau }\right)=0.$$

(50)

We can easily solve this system (50) using the Newton method. After solving this system, we have the values of the Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$, and these values in (49) give the approximate solution.

The exact solution of (48) is $q\left(\xi \right)=1+{\xi }^4,$ when $\sigma =1$.

Example 3.

Taking $\eta =4$ in Equation (14) gives the fifth-order fractional Lane-Emden-Fowler equation:

$$\begin{array}{c}{D}_{\xi }^{5\sigma }q\left(\xi \right)+\frac{16}{{\xi }^{\sigma }}{D}_{\xi }^{4\sigma }q\left(\xi \right)+\frac{72}{{\xi }^{2\sigma }}{D}_{\xi }^{3\sigma }q\left(\xi \right)\hfill \\ +\frac{96}{{\xi }^{3\sigma }}{D}_{\xi }^{2\sigma }q\left(\xi \right)+\frac{24}{{\xi }^{4\sigma }}{D}_{\xi }^{\sigma }q\left(\xi \right)+(1-{\xi }^{5\sigma })\left(q\left(\xi \right)\right)+{\xi }^{10\sigma }=\omega \left(\xi \right),\hfill \end{array}$$

(51)

with initial conditions $q\left(0\right)=0, q^{\left(\sigma \right)}\left(0\right)=0, q^{\left(2\sigma \right)}\left(0\right)=0, q^{\left(3\sigma \right)}\left(0\right)=0, q^{\left(4\sigma \right)}\left(0\right)=0$, where

$$\omega \left(\xi \right)=1+\mathrm{\Gamma }\left(1+5\sigma \right)+80\frac{\mathrm{\Gamma }\left(5\sigma \right)}{\mathrm{\Gamma }\left(\sigma \right)}+180\frac{\mathrm{\Gamma }\left(5\sigma \right)}{\mathrm{\Gamma }\left(2\sigma \right)}+160\frac{\mathrm{\Gamma }\left(5\sigma \right)}{\mathrm{\Gamma }\left(3\sigma \right)}+30\frac{\mathrm{\Gamma }\left(5\sigma \right)}{\mathrm{\Gamma }\left(4\sigma \right)}$$

Now, we apply the method, and using initial conditions, we can write

$$q\left(\xi \right)=1+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,5\sigma }\left(\xi \right).$$

(52)

After putting all the values of $q\left(\xi \right)$ and its derivatives into (51) we get a system of nonlinear equations as:

$$\begin{array}{c}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{16}{{\xi }_{\tau }^{\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)+\frac{72}{{\xi }_{\tau }^{2\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left({\xi }_{\tau }\right)+\frac{96}{{\xi }_{\tau }^{3\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,3\sigma }\left({\xi }_{\tau }\right)\hfill \\ +\frac{24}{{\xi }_{\tau }^{4\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,4\sigma }\left({\xi }_{\tau }\right)+(1-{\xi }_{\tau }^{5\sigma })(1+{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,5\sigma }\left({\xi }_{\tau }\right))+{\xi }_{\tau }^{10\sigma }-\omega \left({\xi }_{\tau }\right)=0.\hfill \end{array}$$

(53)

We can easily solve this system (53) using the Newton method. After solving this system, we have the values of Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$ and putting these values in (52) gives the approximate solution.

The exact solution of (51) is $q\left(\xi \right)=1+{\xi }^{5\sigma }.$

Example 4.

When we put $\eta =4$ in Equation (32) we get this sixth-order fractional Lane-Emden-Fowler equation:

$$\begin{array}{c}{D}_{\xi }^{6\sigma }q\left(\xi \right)+\frac{12}{{\xi }^{\sigma }}{D}_{\xi }^{5\sigma }q\left(\xi \right)+\frac{36}{{\xi }^{2\sigma }}{D}_{\xi }^{4\sigma }q\left(\xi \right)+\frac{24}{{\xi }^{3\sigma }}{D}_{\xi }^{3\sigma }q\left(\xi \right)\hfill \\ -45(-280+17056{\xi }^6-79987{\xi }^{12}+63332{\xi }^{18}-7712{\xi }^{24}+32{\xi }^{30}){\left(q\left(\xi \right)\right)}^{13}=0,\hfill \end{array}$$

(54)

with initial conditions $q\left(0\right)=1, q^{\left(\sigma \right)}\left(0\right)=0, q^{\left(2\sigma \right)}\left(0\right)=0, q^{\left(3\sigma \right)}\left(0\right)=0, q^{\left(4\sigma \right)}\left(0\right)=0,$$q^{\left(5\sigma \right)}\left(0\right)=0$.

Now, we apply the method, and using the initial conditions, we can write

$$q\left(\xi \right)=1+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,6\sigma }\left(\xi \right).$$

(55)

After putting all the values of $q\left(\xi \right)$ and its derivatives into (54), we get a system of nonlinear equations as follows:

$$\begin{array}{c}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{12}{{\xi }_{\tau }^{\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)+\frac{36}{{\xi }_{\tau }^{2\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left({\xi }_{\tau }\right)+\frac{24}{{\xi }_{\tau }^{3\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,3\sigma }\left({\xi }_{\tau }\right)\hfill \\ -45(-280+17056{\xi }_{\tau }^6-79987{\xi }_{\tau }^{12}+63332{\xi }_{\tau }^{18}-7712{\xi }_{\tau }^{24}+32{\xi }_{\tau }^{30}){(1+{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,6\sigma }\left({\xi }_{\tau }\right))}^{13}=0.\hfill \end{array}$$

(56)

We can easily solve this system (53) using the Newton method. After solving this system, we have the values of Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$ and substituting these values into (52) gives the approximate solution.

The exact solution of (54) when $\sigma =1$ is $q\left(\xi \right)=\frac{1}{\sqrt{1+{\xi }^6}}.$

Example 5.

Consider the fractional Lane-Emden Fowler equation, [38]:

$$D_{\xi }^{2\sigma }q\left(\xi \right)+\frac{1}{{\xi }^{\sigma }}{D}_{\xi }^{\sigma }q\left(\xi \right)+\left(1+{\xi }^{\sigma }\right){\left(q\left(\xi \right)\right)}^5=\omega \left(\xi \right),$$

(57)

with initial conditions $q\left(0\right)=3,q^{\left(\sigma \right)}\left(0\right)=0$, where

$$\omega \left(\xi \right)=\mathrm{\Gamma }\left(1+2\sigma \right)+\frac{\mathrm{\Gamma }\left(1+2\sigma \right)}{\mathrm{\Gamma }\left(1+\sigma \right)}+\left(1+{\xi }^{\sigma }\right){\left(3+{\xi }^{2\sigma }\right)}^5.$$

The exact solution of (57) is $q\left(\xi \right)=\left(3+{\xi }^{2\sigma }\right).$

Now, we apply the method, and using the initial conditions, we can write

$$q\left(\xi \right)=3+\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left(\xi \right).$$

(58)

After putting all the values of $q\left(\xi \right)$ and its derivatives into (57), we get a system of nonlinear equations as follows:

$$\begin{array}{c}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{1}{{\xi }_{\tau }^{\sigma }}{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)+\left(1+{\xi }_{\tau }^{\sigma }\right){(3+{\displaystyle \sum _{\tau =1}^{2L}}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left({\xi }_{\tau }\right))}^5\hfill \\ -\mathrm{\Gamma }\left(1+2\sigma \right)-\frac{\mathrm{\Gamma }\left(1+2\sigma \right)}{\mathrm{\Gamma }\left(1+\sigma \right)}-\left(1+{\xi }_{\tau }^{\sigma }\right){\left(3+{\xi }_{\tau }^{2\sigma }\right)}^5=0.\hfill \end{array}$$

(59)

We can easily solve this system (59) using the Newton method. After solving this system, we have the values of Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$ and putting these values in (58) gives the approximate solution.

Example 6.

Consider this third-order fractional Lane-Emden-Fowler equation:

$$D_{\xi }^{3\sigma }q\left(\xi \right)+\frac{3}{{\xi }^{\sigma }}{D}_{\xi }^{2\sigma }q\left(\xi \right)-{\left(q\left(\xi \right)\right)}^3=\omega \left(\xi \right),$$

(60)

with initial conditions $q\left(0\right)=0, q^{\left(\sigma \right)}\left(0\right)=0, q^{\left(2\sigma \right)}\left(0\right)=0$, where

$$\begin{array}{c}\omega \left(\xi \right)=-{\xi }^{9\sigma }{e}^{\xi }+{\xi }^{3\sigma }{e}^{\xi }+9{\sigma }^2{\xi }^{3\sigma -1}{e}^{\xi }+\frac{{\left(3\sigma \right)}^2{(3\sigma -1)}^2}{2}{\xi }^{3\sigma -2}{e}^{\xi }\hfill \\ +3{\xi }^{2\sigma }{e}^{\xi }+18{\sigma }^2{\xi }^{2\sigma -1}{e}^{\xi }+9{\sigma }^2(3\sigma -1)(2\sigma -1){\xi }^{2\sigma -2}{e}^{\xi }\hfill \\ +\frac{{\left(3\sigma \right)}^2{(3\sigma -1)}^2{(3\sigma -2)}^2}{6}{e}^{\xi }.\hfill \end{array}$$

Clearly, when $\sigma =1$, Equation (60) becomes

$$q^{‴}\left(\xi \right)+\frac{3}{\xi }{q}^{″}\left(\xi \right)-{\left(q\left(\xi \right)\right)}^3=24e^{\xi }+36\xi e^{\xi }+12{\xi }^2{e}^{\xi }+{\xi }^3{e}^{\xi }-{\xi }^9{e}^{3\xi },$$

which is given in [39].

Now, we apply the method, and using the initial conditions, we can write

$$q\left(\xi \right)=\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,3\sigma }\left(\xi \right).$$

(61)

After putting all the values of $q\left(\xi \right)$ and its derivatives into (60), we get a system of nonlinear equations as follows:

$$\sum _{\tau =1}^{2L}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{3}{{\xi }_{\tau }^{\sigma }}\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)-{(\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,3\sigma }\left({\xi }_{\tau }\right))}^3-\omega \left({\xi }_{\tau }\right)=0.$$

(62)

We can easily solve this system (62) using the Newton method. After solving this system, we have the values of Haar coefficients ${\lambda }_{\tau }, \tau =1, 2, \dots 2L$ and substituting these values into (61) gives the approximate solution.

The exact solution of (60) is $q\left(\xi \right)={\xi }^{3\sigma }{e}^{\xi }.$

Example 7.

Consider the fractional Lane-Emden-Fowler equation, which is given in [40]

$$D_{\xi }^{2\sigma }q\left(\xi \right)+\frac{2}{{\xi }^{\sigma }}{D}_{\xi }^{\sigma }q\left(\xi \right)+8e^{q\left(\xi \right)}+4e^{q\left(\xi \right)/2}=0,$$

(63)

with initial conditions $q\left(0\right)=0, q^{\left(\sigma \right)}\left(0\right)=0$. The exact solution of (63) when $\sigma =1$ is $q\left(\xi \right)=-2log(1+{\xi }^2).$

Now, we apply the method

$$D_{\xi }^{2\sigma }q\left(\xi \right)=\sum _{\tau =1}^{2L}{\lambda }_{\tau }{h}_{\tau }.$$

(64)

Integrating (64) and applying the initial condition gives us

$$D_{\xi }^{\sigma }q\left(\xi \right)=\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left(\xi \right).$$

In the same way, we can write

$$q\left(\xi \right)=\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left(\xi \right).$$

(65)

After substituting all the values of $q\left(\xi \right)$ and its derivatives into (63) and after collocation of points, we get a system of nonlinear equations as follows:

$$\sum _{\tau =1}^{2L}{\lambda }_{\tau }{h}_{\tau }\left({\xi }_{\tau }\right)+\frac{2}{{\xi }^{\sigma }}\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,\sigma }\left({\xi }_{\tau }\right)+8e^{\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left({\xi }_{\tau }\right)}+4e^{\frac{\left(\sum _{\tau =1}^{2L}{\lambda }_{\tau }{\rho }_{\tau ,2\sigma }\left({\xi }_{\tau }\right)\right)}{2}}=0.$$

(66)

Using the Newton method, we can easily solve this system (66). After solving this system, we have the values of the Haar coefficients ${\lambda }_{\tau },\tau =1,2,\dots 2L$ and putting these values in (65) gives the approximate solution.

7. Numerical Simulation and Conclusions

This paper uses the Haar wavelet collocation method to find the numerical solution to the general-order fractional Lane-Emden-Fowler equation. This numerical scheme is presented in general order. We have documented many examples of second-, third-, fourth-, fifth-, and sixth-order fractional differential equations. We have numerically simulated those examples graphically and in tabular format. It is clear from the simulations that this method works very nicely. Figure 1 and Figure 2 show the comparison of exact and numerical solutions and the absolute error of Example 1 for $\sigma =1$ respectively. Figure 3 and Figure 4 show the HWCM solution for different values of $\sigma$ and Haar coefficients of Example 1 respectively. Figure 5 and Figure 6 show the comparison of exact and numerical solutions and the absolute error of Example 2 for $\sigma =1$ respectively. Figure 7 and Figure 8 show the HWCM solution for different values of $\sigma$ and Haar coefficients respectively. Figure 9 and Figure 10 show the comparison of exact and numerical solutions and the absolute error of Example 3 for $\sigma =1$ respectively. Figure 11 and Figure 12 show the HWCM solution for different values of $\sigma$ and Haar coefficients respectively. Similarly, Figure 13 and Figure 14 show the comparison of exact and numerical solutions and the absolute error of Example 4 for $\sigma =1$ respectively. Figure 15 and Figure 16 show the HWCM solution for different values of $\sigma$ and Haar coefficients respectively. Figure 17 and Figure 18 show the comparison of exact and numerical solutions and the absolute error of Example 5 for $\sigma =1$ respectively. Figure 19 and Figure 20 show the HWCM solution for different values of $\sigma$ and Haar coefficients respectively. Figure 21 and Figure 22 show the comparison of exact and numerical solutions and the absolute error of Example 6 for $\sigma =1$ respectively. Figure 23 and Figure 24 show the HWCM solution for different values of $\sigma$ and Haar coefficients respectively. Figure 25 and Figure 26 show the comparison of exact and numerical solutions and the absolute error of Example 7 for $\sigma =1$ respectively. Figure 27 and Figure 28 show the HWCM solution for different values of $\sigma$ and Haar coefficients respectively.

Table 2. Comparison of HWCM and exact solution of Example 1 for J = 5.

$\mathit{\xi }$	HWCM	Exact	${\mathit{\psi }}_{10}$ [37]	QBSM [37]	${\mathit{\psi }}_8$ [37]	$\mathit{\sigma }=0.75$	$\mathit{\sigma }=0.85$	$\mathit{\sigma }=0.95$
0.1	1.00002	1.00002	1.00003	1.00003	1.00003	1.00087	1.00021	1.00005
0.4	1.00633	1.00633	1.00634	1.00634	1.00634	1.04867	1.02305	1.00989
0.5	1.01527	1.01527	1.01527	1.01527	1.01527	1.08507	1.04683	1.02262
0.6	1.03093	1.03093	1.03093	1.03093	1.03093	1.12870	1.08112	1.04374
0.9	1.13442	1.13441	1.13440	1.13438	1.13438	1.27334	1.23519	1.16789

,

Table 3. Error Comparison of Example 1.

$\mathit{\xi }$	HWCM Error	${\mathit{E}}_8$ [37]	${\mathit{E}}_{10}$ [37]
0.1	0	6.6342 × 10${}^{-6}$	2.1208 × 10${}^{-6}$
0.4	3.1 × 10${}^{-8}$	4.0829 × 10${}^{-6}$	1.3122 × 10${}^{-6}$
0.5	2.01 × 10${}^{-7}$	6.8710 × 10${}^{-7}$	2.2781 × 10${}^{-7}$
0.6	4.19 × 10${}^{-7}$	4.7993 × 10${}^{-6}$	1.5386 × 10${}^{-6}$
0.9	5.85 × 10${}^{-6}$	3.4277 × 10${}^{-5}$	1.1079 × 10${}^{-5}$

,

Table 4. Comparison of HWCM and exact solution of Example 2 for J = 3.

$\mathit{\xi }/32$	Haar Solution	Exact Solution	$\mathit{\sigma }=0.55$	$\mathit{\sigma }=0.75$	$\mathit{\sigma }=0.95$
1/32	1.00000095	1.00000095	1.00048828	1.00003052	1.00000191
3/32	1.00007725	1.00007725	1.0054744	1.00082397	1.00012402
5/32	1.00059605	1.00059605	1.0168424	1.0038147	1.000864
7/32	1.00228977	1.00228977	1.03530903	1.01046753	1.00310315
9/32	1.00625706	1.00625706	1.0613767	1.02224731	1.00806402
11/32	1.01396275	1.01396275	1.09544077	1.0406189	1.01728711
13/32	1.02723789	1.02723789	1.13783053	1.06704712	1.03261481
15/32	1.04827976	1.04827976	1.18882992	1.10299683	1.05617937
17/32	1.07965183	1.07965183	1.24868963	1.14993286	1.09039325
19/32	1.12428379	1.12428379	1.31763472	1.20932007	1.13794113
21/32	1.18547153	1.18547153	1.39586991	1.28262329	1.20177316
23/32	1.26687717	1.26687717	1.48358329	1.37130737	1.28509911
25/32	1.37252903	1.37252903	1.58094908	1.47683716	1.39138314
27/32	1.50682163	1.50682163	1.68812974	1.60067749	1.52433927
29/32	1.67451572	1.67451572	1.80527765	1.74429321	1.68792719
31/32	1.88073826	1.88073826	1.93253636	1.90914917	1.88634851

,

Table 5. Absolute error of HWCM of Example 2 for $\sigma =1$.

Resolution (J)	3	4	5	6	7	8
HWM error	2.2204 × 10${}^{-16}$	0	0	4.4409 × 10${}^{-16}$	0	0
CPU time (seconds)	1.132563	1.383034	1.861307	4.541592	18.758987	125.318744

,

Table 6. Absolute error of HWCM of Example 2 for different $\sigma$.

Resolution	3	4	5	6	7
$\sigma =0.55$	3.4658 × 10${}^{-11}$	3.7683 × 10${}^{-11}$	3.9526 × 10${}^{-11}$	4.0468 × 10${}^{-11}$	4.0935 × 10${}^{-11}$
$\sigma =0.75$	1.207 × 10${}^{-12}$	1.3634 × 10${}^{-16}$	1.4571 × 10${}^{-16}$	1.5068 × 10${}^{-16}$	1.533 × 10${}^{-16}$
$\sigma =0.95$	2.2204 × 10${}^{-16}$	4.4409 × 10${}^{-16}$	4.4409 × 10${}^{-16}$	2.2204 × 10${}^{-16}$	2.2204 × 10${}^{-16}$

,

Table 7. Comparison of HWCM and exact solution of Example 3 for J = 3.

$\mathit{\xi }/32$	Haar Solution	Exact Solution	$\mathit{\sigma }=0.55$	$\mathit{\sigma }=0.75$	$\mathit{\sigma }=0.95$
1/32	1.00000003	1.00000003	1.00007258	1.00000227	1.00000007
3/32	1.00000724	1.00000724	1.00148909	1.0001396	1.00001309
5/32	1.00009313	1.00009313	1.00606743	1.00094804	1.00014813
7/32	1.00050089	1.00050089	1.01530582	1.00334815	1.00073241
9/32	1.0017598	1.0017598	1.03054949	1.00859205	1.00241652
11/32	1.00479969	1.00479969	1.05304775	1.01823518	1.00626834
13/32	1.01106539	1.01106539	1.08398105	1.03411733	1.01386017
15/32	1.02263114	1.02263114	1.1244767	1.0583485	1.02735087
17/32	1.04231504	1.04231504	1.17561905	1.0932977	1.04956442
19/32	1.0737935	1.0737935	1.23845668	1.14158378	1.08406539
21/32	1.12171569	1.12171569	1.3140076	1.20606767	1.13523194
23/32	1.19181797	1.19181797	1.40326325	1.28984572	1.20832665
25/32	1.2910383	1.2910383	1.50719168	1.39624385	1.30956557
27/32	1.42763075	1.42763075	1.62674006	1.52881239	1.44618555
29/32	1.61127988	1.61127988	1.76283673	1.69132139	1.62651014
31/32	1.85321519	1.85321519	1.916393	1.88775649	1.86001428

,

Table 8. Absolute error of HWCM of Example 3 for $\sigma =1$.

Resolution (J)	3	4	5	6	7	8
HWCM error	2.2204 × 10${}^{-16}$	4.4409 × 10${}^{-16}$	2.2204 × 10${}^{-16}$	4.4409 × 10${}^{-16}$	4.4409 × 10${}^{-16}$	2.2204 × 10${}^{-16}$
CPU time (seconds)	0.947930	1.152302	1.580108	3.506233	14.581024	96.118520

,

Table 9. Absolute error of HWCM of Example 3 for different $\sigma$.

Resolution	3	4	5	6	7
$\sigma =0.55$	9.6373 × 10${}^{-7}$	8.6165 × 10${}^{-7}$	8.8235 × 10${}^{-7}$	8.924 × 10${}^{-7}$	8.9734 × 10${}^{-7}$
$\sigma =0.75$	1.5783 × 10${}^{-7}$	1.6771 × 10${}^{-7}$	1.7277 × 10${}^{-7}$	1.7534 × 10${}^{-7}$	1.7663 × 10${}^{-7}$
$\sigma =0.95$	2.855 × 10${}^{-8}$	3.0805 × 10${}^{-8}$	3.1984 × 10${}^{-8}$	3.2587 × 10${}^{-8}$	3.2891 × 10${}^{-8}$

,

Table 10. Comparison of HWCM and exact solution of Example 5 for J = 3.

$\mathit{\xi }/32$	Haar Solution	Exact Solution	$\mathit{\sigma }=0.55$	$\mathit{\sigma }=0.75$	$\mathit{\sigma }=0.95$
1/32	3.000977	3.000977	3.022097	3.005524	3.001381
3/32	3.008789	3.008789	3.073989	3.028705	3.011136
5/32	3.024414	3.024414	3.129778	3.061763	3.029394
7/32	3.047852	3.047852	3.187907	3.102311	3.055706
9/32	3.079102	3.079102	3.247743	3.149155	3.089800
11/32	3.118164	3.118164	3.308935	3.201541	3.131480
13/32	3.165039	3.165039	3.371255	3.258935	3.180596
15/32	3.219727	3.219727	3.434546	3.320931	3.237022
17/32	3.282227	3.282227	3.498688	3.387212	3.300655
19/32	3.352539	3.352539	3.563591	3.457515	3.371404
21/32	3.430664	3.430664	3.629182	3.531623	3.449192
23/32	3.516602	3.516602	3.695402	3.609350	3.533947
25/32	3.610352	3.610352	3.762200	3.690534	3.625606
27/32	3.711914	3.711914	3.829536	3.775034	3.724113
29/32	3.821289	3.821289	3.897373	3.862724	3.829414
31/32	3.938477	3.938477	3.965679	3.953493	3.941461

,

Table 11. Absolute error of HWCM of Example 5.

Resolution (J)	3	4	5	6	7	8
HWM error	1.3323 × 10${}^{-15}$	1.3323 × 10${}^{-15}$	1.3323 × 10${}^{-15}$	8.8818 × 10${}^{-16}$	8.8818 × 10${}^{-16}$	8.881 × 10${}^{-16}$
CPU time (seconds)	0.984097	1.105506	1.766658	5.013335	23.477339	163.859775

,

Table 12. Comparison of HWCM and exact solution of Example 6 for J = 5.

$\mathit{\xi }$	Haar Solution	Exact Solution	Abs. Error	QBSM [39]	$\mathit{\sigma }=0.75$	$\mathit{\sigma }=0.85$	$\mathit{\sigma }=0.95$
0.1	0.001103955	0.001105170	1.215 × 10${}^{-6}$	0.001086890	0.010800993	0.004553355	0.001757272
0.2	0.009765978	0.009771222	5.244 × 10${}^{-6}$	0.009734551	0.045949298	0.025702835	0.013439484
0.3	0.036433810	0.036446187	1.237 × 10${}^{-5}$	0.036392585	0.115449927	0.075115699	0.046307124
0.4	0.095453799	0.095476780	2.298 × 10${}^{-5}$	0.095408690	0.233442897	0.167418101	0.115047307
0.5	0.206052735	0.206090158	3.742 × 10${}^{-5}$	0.206011349	0.415714680	0.320710379	0.238809970
0.6	0.393521515	0.393577660	5.614 × 10${}^{-5}$	0.393493595	0.683319863	0.557181115	0.442071948
0.7	0.690637639	0.690717178	7.953 × 10${}^{-5}$	0.690635447	1.071858878	0.908280793	0.756777190
0.8	1.139368842	1.139476955	1.081 × 10${}^{-4}$	1.139407772	1.614156801	1.411098928	1.223361386
0.9	1.792908154	1.793050668	1.425 × 10${}^{-4}$	1.793007437	2.350158149	2.109231858	1.891889010

,

Table 13. Absolute error of HWCM of Example 6 for $\sigma$ = 1.

Resolution (J)	3	4	5	6	7	8
HWCM error	2.70 × 10${}^{-3}$	7.0396 × 10${}^{-4}$	1.7755 × 10${}^{-4}$	4.2618 × 10${}^{-5}$	8.5528 × 10${}^{-6}$	1.5229 × 10${}^{-6}$
$R_C\left(L\right)$	—	1.939394	1.987268	2.058690	2.316994	2.489575
CPU time (seconds)	0.985559	1.175652	1.641920	4.269479	19.495244	136.006084

,

Table 14. Comparison of HWCM and exact solution of Example 7 for J = 3.

$\mathit{\xi }$	Haar Solution	Exact Solution	Abs. Error	VIM [41]	$\mathit{\sigma }=0.55$	$\mathit{\sigma }=0.75$	$\mathit{\sigma }=0.95$
0.1	$-0.01990248$	$-0.01990066$	8.95 × 10${}^{-5}$	$-0.01990066$	$-0.24073779$	$-0.08666051$	$-0.02698493$
0.2	$-0.07847532$	$-0.07844142$	3.4 × 10${}^{-5}$	$-0.07844138$	$-0.44642869$	$-0.22954585$	$-0.09871719$
0.3	$-0.17240439$	$-0.17235539$	4.9 × 10${}^{-5}$	$-0.17235319$	$-0.61995038$	$-0.39098635$	$-0.20674133$
0.4	$-0.29695266$	$-0.29684001$	1.12 × 10${}^{-4}$	$-0.29680298$	$-0.76453325$	$-0.55548832$	$-0.34321737$
0.5	$-0.44645232$	$-0.44628710$	1.65 × 10${}^{-4}$	$-0.44596354$	$-0.89619911$	$-0.71772963$	$-0.50067339$
0.6	$-0.61519627$	$-0.61496939$	2.26 × 10${}^{-4}$	$-0.61310592$	$-1.01542342$	$-0.87501502$	$-0.67261708$
0.7	$-0.79786090$	$-0.79755223$	3.08 × 10${}^{-4}$	$-0.78950866$	$-1.11575420$	$-1.02264092$	$-0.85328881$
0.8	$-0.98975390$	$-0.98939248$	3.61 × 10${}^{-4}$	$-0.96127658$	$-1.20678787$	$-1.16109114$	$-1.03825098$
0.9	$-1.18708784$	$-1.18665369$	4.34e-04	$-1.10296039$	$-1.29291002$	$-1.29224026$	$-1.22443125$

and

Table 15. Abs. error of HWCM of Example 7 for $\sigma$ = 1.

Resolution (J)	3	4	5	6	7	8
HWCM error	4.70 × 10${}^{-4}$	1.19 × 10${}^{-4}$	3.0098 × 10${}^{-5}$	7.5548 × 10${}^{-6}$	1.8926 × 10${}^{-6}$	4.7363 × 10${}^{-7}$
$R_C\left(L\right)$	1.963964	1.978709	1.988404	1.992906	1.998703	1.998537
CPU time (seconds)	2.388590	1.868506	2.802558	7.324519	33.503430	196.555354

give us a comparison of the numerical values of Haar and the exact solution and document the numerical values for different values of $\sigma$. From all the tables and graphs, we conclude that the method is quite accurate and gives us good outcomes.