Solution of linear and nonlinear singular value problems using operational matrix of integration of Taylor wavelets

ABSTRACT

In this paper, we present an effective method under Taylor wavelets and collocation technique to find an approximate solution of linear and non-linear second-order singular value differential equations. Application of this method on various problems confirms the efficiency, easy applicability, and rapid computation. The obtained solution is compared with some other existing numerical solutions. For many problems, this method gives a solution same as the exact solution of the problem which confirms the effectiveness and accuracy of this method. Additionally, we include graphs and figures to demonstrate that the Taylor wavelet method offers better accuracy for a number of problems. MATLAB software is used to process relative data and execute calculations.

1. Introduction

An oscillation that resembles a wave and has an amplitude that starts at zero, rises or falls, and then returns to zero one or more times is called a wavelet. The primary characteristics of the wavelets are multiresolution analysis, density, orthogonality, and compact support. Wavelet-based numerical approaches have gained popularity in recent years, particularly in computer and applied mathematics due to their simplicity and ease of use to find the solution of several classes of differential equations. Wavelet analysis has been used in a variety of disciplines including signal analysis, harmonic analysis, temporal frequency analysis, and others. The study of wavelets is a relatively new and developing field in mathematics.

Differential equations are the formulation of scientific theory for many real-world physical problems. In recent decades, nonlinear equations have drawn a lot of attention since they are utilized in several technical and scientific applications, like gas dynamics, atomic structure, chemical processes, and nuclear physics, for instance [1]. Sometimes, it is impossible to obtain the analytical solution to a given singular differential equation using existing (analytical) methods then we use numerical approaches like the Tau method, operational matrix of integration method, etc. Researchers in several branches of science and engineering have given a lot of attention to wavelets, recently. One of the most frequently used numerical method for solving differential equations is the operational matrix of integration with the collocation method. This approach is based on the operational matrix of integration and collocation methodology, which converts the given problem into a set of algebraic equations. The Newton Rapshon technique can be used to find the unknown coefficients in this system of algebraic equations.

A mathematical model of the thermal behavior of a spherical cloud of gas and nuclear physics similarly faces these problems. It is always crucial to obtain the solution of differential equations because several human disease models are represented in terms of differential equations. The second-order ordinary differential equations (ODEs) have been studied by several researchers, like Tunc and Tunc [2–4], and Hasan and Zhu [5]. Additionally, numerous mathematicians have solved singular value differential equations using different wavelet methods like the Legendre wavelet method [6,7], Haar wavelet collocation method [8], Laguerre wavelets method [9], Chebyshev wavelet operational matrix of integration [10], and Hermite wavelet method [11]. Recently, vivek utilized the Fibonacci wavelet method to solve the system of nonlinear differential equations representing the HIV-infection model [12] and Tunc et al. [13] obtained the solution of Intego-differential equations. Some of the existing methods are described in [14–17] for solving number of singular differential equations. We conclude from an in-depth review of the literature and the conclusions of several research publications that researchers have not used Taylor wavelets to solve the following class of differential equations. This motivated us to implement Taylor wavelets to solve singular differential equations.

Consider a class of the general second-order singular differential equation [18,19]: (1) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{p\mathrm{\prime }(\rho )}{p(\rho )}{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=R(\rho ,\mathrm{\Lambda }),\rho \in (0,1].$(1) These types of equations are our focus throughout this research, under the following four conditions: (2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) (3) $\begin{array}{rl}& \mathrm{Type}\mathrm{II}:{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\alpha }_2,\mathrm{\Lambda }(1)={\beta }_2,\end{array}$(3) (4) $\begin{array}{rl}& \mathrm{Type}\mathrm{III}:\mathrm{\Lambda }(0)={\alpha }_3,{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\beta }_3,\end{array}$(4) in addition to the most general combined boundary conditions as: (5) $\begin{array}{rl}\mathrm{Type}\mathrm{IV}:a_1\mathrm{\Lambda }(0)+a_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)& ={\alpha }_4,\\ b_1\mathrm{\Lambda }(1)+b_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& ={\beta }_4,\end{array}$(5) where $a_i,b_i$, i=1,2, and ${\alpha }_i,{\beta }_i$, i = 1, 2, 3, 4 be given finite constants with $p(0)=0$. $\nu (\rho )=\frac{p\mathrm{\prime }(\rho )}{p(\rho )}$ may be discontinuous at $\rho =0$, whereas $p(\rho )$ and R satisfies the following conditions [20]:

• (D${}_1$)$p(\rho )\in C[0,1]\cap C^1(0,1]$ and $p(\rho )>0,\in (0,1]$.

• (D${}_2$)$\frac{1}{p(\rho )}\in L^1(0,1]$ and ${\int }_0^1\frac{1}{p(\rho )}{\int }_{\rho }^{1}p(x)\mathrm{d}x\mathrm{d}\rho <\mathrm{\infty }$. (for boundary condition (Type-I)).

• (D${}_3$)$p(\rho )\in L^1(0,1]$ and ${\int }_0^1\frac{1}{p(\rho )}{\int }_0^{\rho }p(x)\mathrm{d}x\mathrm{d}\rho <\mathrm{\infty }$. (for boundary conditions (Type-II)).

• (D${}_4$)$R(\rho ,\mathrm{\Lambda }),R_{\mathrm{\Lambda }}(\rho ,\mathrm{\Lambda })\in C(\mathrm{\Omega })$ and $R_{\mathrm{\Lambda }}(\rho ,\mathrm{\Lambda })\ge 0$ on Ω, where $\mathrm{\Omega }:=\{(0,1]\times \mathbb{R}\}$.

Existence and uniqueness of the singular boundary value problem defined in (2(2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) ) along with boundary conditions (3(3) $\begin{array}{rl}& \mathrm{Type}\mathrm{II}:{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\alpha }_2,\mathrm{\Lambda }(1)={\beta }_2,\end{array}$(3) ) and (4(4) $\begin{array}{rl}& \mathrm{Type}\mathrm{III}:\mathrm{\Lambda }(0)={\alpha }_3,{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\beta }_3,\end{array}$(4) ) is discussed in [18,21–23].

Now consider the singular value problem (1(1) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{p\mathrm{\prime }(\rho )}{p(\rho )}{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=R(\rho ,\mathrm{\Lambda }),\rho \in (0,1].$(1) ) along with the condition (4(4) $\begin{array}{rl}& \mathrm{Type}\mathrm{III}:\mathrm{\Lambda }(0)={\alpha }_3,{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\beta }_3,\end{array}$(4) ). If $\nu (\rho )$ and R satisfies the following conditions [24]:

• (D${}_1$)ν is measurable on $[0,1]$,

• (D${}_2$$\nu ⩾0$ on $(0,1]$,

• (D${}_3$)${\int }_0^1\mathrm{x\nu }(x)\mathrm{d}x<\mathrm{\infty }$,

• (D${}_4$)R satisfies the Carathéodory conditions and is bounded. Specifically, there exist a, b with $a<{ϵ}_0<b$ and K>0 such that,

  • (d${}_1$)For each $\rho \in [0,1],R(\rho ,\cdot )$ is continuous on $[a,b]$;

  • (d${}_2$)For each $\mathrm{\Lambda }\in [a,b],R(\cdot ,\mathrm{\Lambda })$ is measurable on $[0,1]$; and

  • (d${}_3$)$\underset{(\rho ,\mathrm{\Lambda })\in [0,1]\times [a,b]}{sup}|R(\rho ,\mathrm{\Lambda })|⩽K$.

Then the solution of the initial value problem (2(2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) ) along the conditions (5(5) $\begin{array}{rl}\mathrm{Type}\mathrm{IV}:a_1\mathrm{\Lambda }(0)+a_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)& ={\alpha }_4,\\ b_1\mathrm{\Lambda }(1)+b_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& ={\beta }_4,\end{array}$(5) ) exists (see Theorem (1) of [25]). In addition to the above conditions, (2(2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) ) has a unique solution, if the function R satisfies the Lipschitz condtion in Λ on [0,1] (see Theorem (2) of [25]) [26].

The condition which guaranteed the uniqueness and existence of Strum-Liouville singular BVP [27] defined by problem (2(2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) ) along with the condition (5(5) $\begin{array}{rl}\mathrm{Type}\mathrm{IV}:a_1\mathrm{\Lambda }(0)+a_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)& ={\alpha }_4,\\ b_1\mathrm{\Lambda }(1)+b_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& ={\beta }_4,\end{array}$(5) ) is also discussed by the author. They used the perturbation technique, maximal principle, Schauder's fixed-point theorem, and mixed monotone iterative techniques to prove it. For this purpose, they present two sufficient conditions. Liu and Yu [28] additionally filled the gap which arises to investigate the uniqueness of a positive solution of those BVP in which the Lipschitz condition is not satisfied in the presence of singularity. Chawla and Shivakumar [23] and R. K. Pandey [29–31] published several results on the existence and uniqueness of the singular and initial value problem having the form (1(1) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{p\mathrm{\prime }(\rho )}{p(\rho )}{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=R(\rho ,\mathrm{\Lambda }),\rho \in (0,1].$(1) ) with the conditions (3(3) $\begin{array}{rl}& \mathrm{Type}\mathrm{II}:{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\alpha }_2,\mathrm{\Lambda }(1)={\beta }_2,\end{array}$(3) ).

Following is the overview of the article's structure: We discuss the Taylor wavelet characteristics and function approximation in Section 3. In Section 4, we present the convergence theorem and error bound of the Taylor wavelet expansion of the functions, while, Section 5 introduces the operational integration matrix evaluation process. The application of the operational matrix of integration is described in Section 6. Section 7 presents evaluation-based numerical problems to test the effectiveness of the suggested method, and Section 8 draws an overall conclusion.

2. Wavelets

Wavelets are generated from a single function ζ known as ‘mother wavelet’ by dilating and translating it. When both the translation and dilation variables are continuous then the family of continuous wavelets is defined as follows: (6) ${\zeta }_{\nu ,\eta }(\rho )=|\nu {|}^{-\frac{1}{2}}\zeta \left(\frac{\rho -\eta }{\nu }\right),\nu ,\eta \in \mathbb{R},\nu \ne 0.$(6) In particular, if the variables ν and η are restricted to discrete quantities i.e. $\nu ={\nu }_0^{-\theta },\eta =\omega {\eta }_0{\nu }_0^{-\theta },{\zeta }_0>1$, and ${\eta }_0>1$, where ω and θ are natural numbers. Then, (7) ${\zeta }_{\theta ,\omega }(\rho )={\left|{\nu }_0\right|}^{\frac{\theta }{2}}\zeta ({\nu }_0^{\theta }\rho -\omega {\eta }_0),$(7) defines the discrete family of wavelets ${\zeta }_{\theta ,\omega }$ and form a basis(wavelet basis) of $L^2(\mathbb{R})$. In particular, when ${\nu }_0=2$ and ${\eta }_0=1$, then ${\zeta }_{\theta ,\omega }(\rho )$ forms an orthonormal basis.

2.1. Taylor wavelets

Four arguments exist for Taylor wavelets ${\zeta }_{\omega ,r}(\rho )=\zeta (\theta ,\hat{\omega },r,\rho )$: $\hat{\omega }=\omega -1,\omega =1,2,\dots ,2^{\theta -1}$, where $r\in \mathbb{N}\cup 0$ is the order of the Taylor polynomial [32]. Consider the family of the functions defined below on the interval $[0,1]$: (8) ${\zeta }_{\omega ,r}(\rho )=\{\begin{array}{ll}{2}^{\frac{\theta -1}{2}}{\tilde{L}}_r(2^{\theta -1}\rho -\hat{\omega }),& {\displaystyle \frac{\hat{\omega }}{2^{\theta -1}}}\le \rho <{\displaystyle \frac{\hat{\omega }+1}{2^{\theta -1}}}\\ 0,& \mathrm{otherwise}\end{array}$(8) with ${\tilde{L}}_r(\rho )=\sqrt{2r+1}{L}_r(\rho ).$Then ${\zeta }_{\omega ,r}(\rho )$ defines the family of Taylor wavelets for the Taylor polynomial $L_r(\rho )={\rho }^r$ of order r and $r=0,1,2,\dots ,\mu -1$. The coefficient $\sqrt{2r+1}$ is for normality, the dilation parameter is $\eta =2^{-(\theta -1)}$ and the translation parameter is $\nu =\hat{\omega }{2}^{-(\theta -1)}$.

2.2. Function approximation

We can express any arbitrary function $\mathrm{\Lambda }\in L^2[0,1]$ using the Taylor wavelet basis as follows [8]: (9) $\mathrm{\Lambda }(\rho )=\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho ),$(9) where ${\zeta }_{\omega ,r}$ are the Taylor wavelets and $a_{\omega ,r}=⟨\mathrm{\Lambda }(\rho ),{\zeta }_{\omega ,r}⟩$ are the Taylor wavelet coefficients. Consider the truncated series of approximation for $\mathrm{\Lambda }(\rho )$, (10) $\mathrm{\Lambda }(\rho )\simeq \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )=E^T\zeta (\rho )={\mathrm{\Lambda }}_n(\rho ),$(10) where T denotes transposition and $E,\zeta (\rho )$ are two matrices of order $n\times 1(n=2^{\theta -1}\mu )$ and defined as (11) $\begin{array}{rl}E& =[a_{1,0},a_{1,1},\dots ,a_{1,\mu -1},a_{2,0},\dots ,a_{2,\mu -1},\dots ,\\ & {a_{2^{\theta -1},0},\dots ,a_{2^{\theta -1},\mu -1}]}^T,\\ \zeta (\rho )& =[{\zeta }_{1,0}(\rho ),{\zeta }_{1,1}(\rho ),\dots ,{\zeta }_{1,\mu -1}(\rho ),{\zeta }_{2,0}(\rho ),\dots ,\\ & {{\zeta }_{2,\mu -1}(\rho ),\dots ,{\zeta }_{2^{\theta -1},0}(\rho ),\dots ,{\zeta }_{2^{\theta -1},\mu -1}(\rho )]}^T.\end{array}$(11)

3. Convergence analysis

In this section, we present two theorems to discuss the convergence and error analysis of the proposed method.

Theorem 3.1

[33]

Let $\mathrm{\Lambda }(\rho )\in L^2(\mathbb{R})$ be a continuous function on the interval $[0,1)$ such that it is bounded by δ i.e. $|\mathrm{\Lambda }(\rho )|<\delta$, for every $\rho \in [0,1)$. Then, the Taylor wavelet coefficients of $\mathrm{\Lambda }(\rho )$ in Equation (10(10) $\mathrm{\Lambda }(\rho )\simeq \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )=E^T\zeta (\rho )={\mathrm{\Lambda }}_n(\rho ),$(10) ) are bounded as: (12) $\left|a_{\omega ,r}\right|<\frac{\lambda }{2^{\frac{\theta -1}{2}}}\delta \frac{2}{2r+1},$(12) where δ is a constant and λ is given by $\lambda =\sqrt{2r+1}.$

Proof.

Consider the following series expansion of $\mathrm{\Lambda }(\rho )$ using Taylor wavelets as defined in (10(10) $\mathrm{\Lambda }(\rho )\simeq \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )=E^T\zeta (\rho )={\mathrm{\Lambda }}_n(\rho ),$(10) ), (13) $\mathrm{\Lambda }(\rho )\simeq \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )=E^T\zeta (\rho )={\mathrm{\Lambda }}_n(\rho ),$(13) where E and $\zeta (\rho )$ are given in Equation (11(11) $\begin{array}{rl}E& =[a_{1,0},a_{1,1},\dots ,a_{1,\mu -1},a_{2,0},\dots ,a_{2,\mu -1},\dots ,\\ & {a_{2^{\theta -1},0},\dots ,a_{2^{\theta -1},\mu -1}]}^T,\\ \zeta (\rho )& =[{\zeta }_{1,0}(\rho ),{\zeta }_{1,1}(\rho ),\dots ,{\zeta }_{1,\mu -1}(\rho ),{\zeta }_{2,0}(\rho ),\dots ,\\ & {{\zeta }_{2,\mu -1}(\rho ),\dots ,{\zeta }_{2^{\theta -1},0}(\rho ),\dots ,{\zeta }_{2^{\theta -1},\mu -1}(\rho )]}^T.\end{array}$(11) ) and the coefficients $a_{\omega ,r}$ can be determined as, (14) $\begin{array}{rl}{a}_{\omega ,r}& =⟨\mathrm{\Lambda },{\zeta }_{\omega ,r}⟩={\int }_0^1\mathrm{\Lambda }(\rho ){\zeta }_{\omega ,r}(\rho )\mathrm{d}\rho \\ & =2^{\frac{\theta -1}{2}}\sqrt{2r+1}{\int }_{\frac{\omega -1}{2^{\theta -1}}}^{\frac{\omega }{2^{\theta -1}}}\mathrm{\Lambda }(\rho )L_r(2^{\theta -1}\rho -\omega +1)\mathrm{d}\rho .\end{array}$(14) Using the definition of Taylor wavelets ${\zeta }_{\omega ,r}(\rho )$, we have (15) $\begin{array}{rl}{\zeta }_{\omega ,r}(\rho )& =2^{\frac{\theta -1}{2}}\sqrt{2r+1}{L}_r(2^{\theta -1}\rho -\omega +1),\\ \frac{\omega -1}{2^{\theta -1}}& \le \rho <\frac{\omega }{2^{\theta -1}}.\end{array}$(15) Let $\lambda =\sqrt{2r+1}$. Let $2^{\theta -1}\rho -\omega +1=x$, then Equation (14(14) $\begin{array}{rl}{a}_{\omega ,r}& =⟨\mathrm{\Lambda },{\zeta }_{\omega ,r}⟩={\int }_0^1\mathrm{\Lambda }(\rho ){\zeta }_{\omega ,r}(\rho )\mathrm{d}\rho \\ & =2^{\frac{\theta -1}{2}}\sqrt{2r+1}{\int }_{\frac{\omega -1}{2^{\theta -1}}}^{\frac{\omega }{2^{\theta -1}}}\mathrm{\Lambda }(\rho )L_r(2^{\theta -1}\rho -\omega +1)\mathrm{d}\rho .\end{array}$(14) ) becomes $a_{\omega ,r}=\frac{\lambda }{2^{\frac{\theta -1}{2}}}{\int }_0^1\mathrm{\Lambda }\left(\frac{x+\omega -1}{2^{\theta -1}}\right)L_r(x)\mathrm{d}x.$Therefore, (16) $\left|a_{\omega ,r}\right|\le \frac{\lambda }{2^{\frac{\theta -1}{2}}}{\int }_0^1\left|\mathrm{\Lambda }\left(\frac{x+\omega -1}{2^{\theta -1}}\right)\right||L_r(x)|\mathrm{d}x.$(16) Seeing the properties of Taylor polynomials, we can say that (17) ${\int }_0^1|L_r(v)|\mathrm{d}v<\frac{2}{2r+1},r>0.$(17) Using the assumption $|\mathrm{\Lambda }(\rho )|<\delta$ in Equations (17(17) ${\int }_0^1|L_r(v)|\mathrm{d}v<\frac{2}{2r+1},r>0.$(17) ) and (16(16) $\left|a_{\omega ,r}\right|\le \frac{\lambda }{2^{\frac{\theta -1}{2}}}{\int }_0^1\left|\mathrm{\Lambda }\left(\frac{x+\omega -1}{2^{\theta -1}}\right)\right||L_r(x)|\mathrm{d}x.$(16) ), we have (18) $\left|a_{\omega ,r}\right|<\frac{1}{2^{\frac{\theta -1}{2}}}\delta \frac{2}{\sqrt{2r+1}}.$(18) Thus the proof of the Theorem 3.1 has been completed. Also, boundedness of the function implies absolutely convergent of the series $\mathrm{\Lambda }(\rho )=\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}{a}_{\omega ,r}$. Hence the Taylor wavelet approximation of the function $\mathrm{\Lambda }(\rho )$ is absolutely convergent.

Theorem 3.2

[33]

Let $\mathrm{\Lambda }(\rho )\in L^2(\mathbb{R})$ be a continuous function on the interval $[0,1)$ and $|\mathrm{\Lambda }(\rho )|<\delta$ for every $\rho \in [0,1)$. Let ${\mathrm{\Lambda }}^{\ast }(\rho )=\sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )$ be the Taylor wavelet series expansions where $a_{\omega ,r}$, ${\zeta }_{\omega ,r}(\rho )$ be the Taylor wavelet coefficients and Taylor wavelet basis respectively. Then, the bound of the truncated error $e(\rho )$ is given as: $\begin{array}{rl}\Vert e(\rho ){\Vert }_2& =\Vert \mathrm{\Lambda }(\rho )-{\mathrm{\Lambda }}^{\ast }(\rho )\Vert <{\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}^2\right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}^2\right)}^{\frac{1}{2}},\end{array}$where, $a_{\omega ,r}=\frac{\lambda }{2^{\frac{\theta -1}{2}}}\delta \frac{2}{2r+1},\lambda =\sqrt{2r+1}.$

Proof.

Any function $\mathrm{\Lambda }\in L^2[0,1)$ can be expanded in terms of Taylor wavelets as: $\mathrm{\Lambda }(\rho )=\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=0}^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho ).$If ${\mathrm{\Lambda }}^{\ast }(\rho )$ is the expansion truncated by using Taylor wavelets, then the error obtained by truncating the above function can be computed as: (19) $\begin{array}{rl}e(\rho )& =\mathrm{\Lambda }(\rho )-{\mathrm{\Lambda }}^{\ast }(\rho )=\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )\\ & +\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho ).\end{array}$(19) From Equation (19(19) $\begin{array}{rl}e(\rho )& =\mathrm{\Lambda }(\rho )-{\mathrm{\Lambda }}^{\ast }(\rho )=\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )\\ & +\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho ).\end{array}$(19) ), we can write (20) $\begin{array}{rl}\Vert e(\rho )\Vert & \le \Vert \sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )\Vert \\ & +\Vert \sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )\Vert \\ & ={\left({\int }_0^1{|\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left({\int }_0^1{|\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & \le {\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}.\end{array}$(20) Now using the result of Theorem 3.1, which is $\left|a_{\omega ,r}\right|<\frac{\lambda }{2^{\frac{\theta -1}{2}}}\delta \frac{2}{2r+1},$to reduce Equation (20(20) $\begin{array}{rl}\Vert e(\rho )\Vert & \le \Vert \sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )\Vert \\ & +\Vert \sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )\Vert \\ & ={\left({\int }_0^1{|\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left({\int }_0^1{|\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{a}_{\omega ,r}{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & \le {\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}.\end{array}$(20) ) as, (21) $\begin{array}{rl}\Vert e(\rho ){\Vert }_2& <{\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}.\end{array}$(21) Let us define (22) $d_{\omega ,r}=\frac{\lambda }{2^{\frac{\theta -1}{2}}}\delta \frac{2}{2r+1}.$(22) Then from Equations (21(21) $\begin{array}{rl}\Vert e(\rho ){\Vert }_2& <{\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{\left|a_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}.\end{array}$(21) ) and (22(22) $d_{\omega ,r}=\frac{\lambda }{2^{\frac{\theta -1}{2}}}\delta \frac{2}{2r+1}.$(22) ), we get (23) $\begin{array}{rl}|e(\rho ){\Vert }_2& <{\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{\left|d_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{\left|d_{\omega ,r}\right|}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}.\end{array}$(23) Therefore, (24) $\begin{array}{rl}\Vert e(\rho ){\Vert }_2& <{\left(\sum _{\omega =2^{\theta -1}+1}^{\mathrm{\infty }}\sum _{r=0}^{\mu -1}{d}_{\omega ,r}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}\\ & +{\left(\sum _{\omega =1}^{\mathrm{\infty }}\sum _{r=\mu }^{\mathrm{\infty }}{d}_{\omega ,r}^2{\int }_0^1{|{\zeta }_{\omega ,r}(\rho )|}^2\mathrm{d}\rho \right)}^{\frac{1}{2}}.\end{array}$(24) By the definition of Taylor wavelets, we have (25) $\begin{array}{rl}{\zeta }_{\omega ,r}^2(\rho )& =2^{\theta -1}(2r+1)L_r^2(2^{\theta -1}\rho -\omega +1),\\ \frac{\omega -1}{2^{\theta -1}}& \le \rho <\frac{\omega }{2^{\theta -1}}.\end{array}$(25) Integrating Equation (25(25) $\begin{array}{rl}{\zeta }_{\omega ,r}^2(\rho )& =2^{\theta -1}(2r+1)L_r^2(2^{\theta -1}\rho -\omega +1),\\ \frac{\omega -1}{2^{\theta -1}}& \le \rho <\frac{\omega }{2^{\theta -1}}.\end{array}$(25) ) with respect to ρ, we get (26) $\begin{array}{rl}& {\int }_0^1{\zeta }_{\omega ,r}^2(\rho )\\ & =2^{\theta -1}(2r+1){\int }_{\frac{\omega -1}{2^{\theta -1}}}^{\frac{\omega }{2^{\theta -1}}}{L}_r^2(2^{\theta -1}\rho -\omega +1)\mathrm{d}\rho .\end{array}$(26) Let $2^{\theta -1}\rho -\omega +1=u$, Equation (26(26) $\begin{array}{rl}& {\int }_0^1{\zeta }_{\omega ,r}^2(\rho )\\ & =2^{\theta -1}(2r+1){\int }_{\frac{\omega -1}{2^{\theta -1}}}^{\frac{\omega }{2^{\theta -1}}}{L}_r^2(2^{\theta -1}\rho -\omega +1)\mathrm{d}\rho .\end{array}$(26) ) becomes (27) ${\int }_0^1{\zeta }_{\omega ,r}^2(\rho )\mathrm{d}\rho =(2r+1){\int }_0^1{L}_r^2(u)\mathrm{d}u.$(27) But the standard definition of Taylor polynomial implies that, (28) ${\int }_0^1{L}_r^2(u)\mathrm{d}u={\int }_0^1{u}^{2r}\mathrm{d}u=\frac{1}{2r+1}.$(28) Substituting Equation (28(28) ${\int }_0^1{L}_r^2(u)\mathrm{d}u={\int }_0^1{u}^{2r}\mathrm{d}u=\frac{1}{2r+1}.$(28) ) in Equation (27(27) ${\int }_0^1{\zeta }_{\omega ,r}^2(\rho )\mathrm{d}\rho =(2r+1){\int }_0^1{L}_r^2(u)\mathrm{d}u.$(27) ), we get ${\int }_0^1{\zeta }_{\omega ,r}^2(\rho )\mathrm{d}\rho =1.$Thus the Theorem 3.2 has been proved. This theorem also implies consistency and stability of the approximation.

4. Operational matrix of integration

Let $\zeta (\rho )$ be the vector consisting of the Taylor wavelets [32] described in Equation (11(11) $\begin{array}{rl}E& =[a_{1,0},a_{1,1},\dots ,a_{1,\mu -1},a_{2,0},\dots ,a_{2,\mu -1},\dots ,\\ & {a_{2^{\theta -1},0},\dots ,a_{2^{\theta -1},\mu -1}]}^T,\\ \zeta (\rho )& =[{\zeta }_{1,0}(\rho ),{\zeta }_{1,1}(\rho ),\dots ,{\zeta }_{1,\mu -1}(\rho ),{\zeta }_{2,0}(\rho ),\dots ,\\ & {{\zeta }_{2,\mu -1}(\rho ),\dots ,{\zeta }_{2^{\theta -1},0}(\rho ),\dots ,{\zeta }_{2^{\theta -1},\mu -1}(\rho )]}^T.\end{array}$(11) ), then $\mathrm{I\zeta }(\rho )=\mathrm{S\zeta }(\rho ),$defines the relation between the integration operator I and the operational matrix of integration S of order $n\times n$ where $n=2^{\theta -1}\mu$. Here we will discuss some integration characteristics of Taylor wavelets ${\zeta }_{i,j}(\rho )$ and the Integral operator for $i=1,\cdots ,2^{\theta -1}$ and $j=0,1,\cdots ,\mu -1$. So we have (29) $\begin{array}{rl}& I({\zeta }_{i,j}(\rho ))\\ & =I(2^{\frac{\theta -1}{2}}{\overline{T}}_j(2^{\theta -1}\rho -\hat{i}){\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho ))\\ & =2^{\frac{\theta -1}{2}}I(\sqrt{2j+1}{(2^{\theta -1}\rho -\hat{i})}^j{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )),\end{array}$(29) where ${\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )$ is the standard characteristic function defined as ${\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )=\{\begin{array}{ll}1,& {\displaystyle \frac{\hat{i}}{2^{\theta -1}}}\le \rho \le {\displaystyle \frac{\hat{i}+1}{2^{\theta -1}}},\\ 0,& \mathrm{otherwise},\end{array}$and $\hat{i}=i-1$. Therefore, in Equation (29(29) $\begin{array}{rl}& I({\zeta }_{i,j}(\rho ))\\ & =I(2^{\frac{\theta -1}{2}}{\overline{T}}_j(2^{\theta -1}\rho -\hat{i}){\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho ))\\ & =2^{\frac{\theta -1}{2}}I(\sqrt{2j+1}{(2^{\theta -1}\rho -\hat{i})}^j{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )),\end{array}$(29) ), for i = 1, we have (30) $I({\zeta }_{1,j}(\rho ))=2^{\frac{\theta -1}{2}}\sqrt{(2j+1)}{2}^{(\theta -1)j}I({\rho }^j{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )).$(30) It is known that (31) ${(2^{\theta -1}\rho -\hat{i})}^j=\sum _{l=0}^j\left(\begin{array}{c}j\\ l\end{array}\right)2^{(\theta -1)l}{\rho }^l(-1{)}^{j-l}{\hat{i}}^{j-l}.$(31) Now, substituting Equation (31(31) ${(2^{\theta -1}\rho -\hat{i})}^j=\sum _{l=0}^j\left(\begin{array}{c}j\\ l\end{array}\right)2^{(\theta -1)l}{\rho }^l(-1{)}^{j-l}{\hat{i}}^{j-l}.$(31) ) in Equation (29(29) $\begin{array}{rl}& I({\zeta }_{i,j}(\rho ))\\ & =I(2^{\frac{\theta -1}{2}}{\overline{T}}_j(2^{\theta -1}\rho -\hat{i}){\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho ))\\ & =2^{\frac{\theta -1}{2}}I(\sqrt{2j+1}{(2^{\theta -1}\rho -\hat{i})}^j{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )),\end{array}$(29) ) for $i=2,3,\dots ,2^{\theta -1}$, we have (32) $\begin{array}{rl}I({\zeta }_{i,j}(\rho ))& =2^{\frac{\theta -1}{2}}\sqrt{2j+1}\sum _{l=0}^j\left(\begin{array}{c}j\\ l\end{array}\right)2^{(\theta -1)l}(-1{)}^{j-l}{\hat{i}}^{j-l}I\\ & ({\rho }^l{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )).\end{array}$(32) Now, approximating $I({\rho }^l{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho ))$ by using Taylor wavelets, we have (33) $\begin{array}{rl}I({\rho }^l{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho ))& =h_{i,l}(\rho )\simeq \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}^{\mathrm{il}}{\zeta }_{\omega ,r}(\rho )\\ & =E_{i,l}^T\zeta (\rho ),\end{array}$(33) where $E_{i,l}^T=D^{-1}<h_{i,l}(\rho ),\zeta (\rho )>;D=<\zeta (\rho ),\zeta (\rho )>.$Using Equation (33(33) $\begin{array}{rl}I({\rho }^l{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho ))& =h_{i,l}(\rho )\simeq \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}^{\mathrm{il}}{\zeta }_{\omega ,r}(\rho )\\ & =E_{i,l}^T\zeta (\rho ),\end{array}$(33) ) in Equations (30(30) $I({\zeta }_{1,j}(\rho ))=2^{\frac{\theta -1}{2}}\sqrt{(2j+1)}{2}^{(\theta -1)j}I({\rho }^j{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )).$(30) ) and (32(32) $\begin{array}{rl}I({\zeta }_{i,j}(\rho ))& =2^{\frac{\theta -1}{2}}\sqrt{2j+1}\sum _{l=0}^j\left(\begin{array}{c}j\\ l\end{array}\right)2^{(\theta -1)l}(-1{)}^{j-l}{\hat{i}}^{j-l}I\\ & ({\rho }^l{\chi }_{[\frac{\hat{i}}{2^{\theta -1}},\frac{\hat{i}+1}{2^{\theta -1}}]}(\rho )).\end{array}$(32) ), we get $\begin{array}{rl}I({\zeta }_{1,j}(\rho ))& \simeq 2^{\frac{\theta -1}{2}}(\sqrt{2j+1})2^{(\theta -1)j}\sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}^{1j}{\zeta }_{\omega ,r}(\rho )\\ & =\sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{\mathrm{\Theta }}_{\omega ,r}^{1j}{\zeta }_{\omega ,r}(\rho ),j=0,1,\dots ,\mu -1,\end{array}$and $\begin{array}{rl}I({\zeta }_{\mathrm{ij}}(\rho ))& \simeq 2^{\frac{\theta -1}{2}}\sqrt{2j+1}\sum _{l=0}^j\left(\begin{array}{c}j\\ l\end{array}\right)2^{(\theta -1)l}(-1{)}^{j-l}{\hat{i}}^{j-l}\\ & \times \sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{a}_{\omega ,r}^{\mathrm{il}}{\zeta }_{\omega ,r}(\rho )\\ & =\sum _{\omega =1}^{2^{\theta -1}}\sum _{r=0}^{\mu -1}{\mathrm{\Theta }}_{\mathrm{\omega r}}^{\mathrm{ij}}{\zeta }_{\mathrm{\omega r}}(\rho ),i=2,3,\dots ,2^{\omega -1},\\ & j=0,1,\dots ,\mu -1,\end{array}$where ${\mathrm{\Theta }}_{\omega ,r}^{1j}=2^{\frac{\theta -1}{2}}\sqrt{2j+1}{2}^{(\theta -1)j}{a}_{\omega ,r}^{1j},$and ${\mathrm{\Theta }}_{\omega ,r}^{\mathrm{ij}}=2^{\frac{\theta -1}{2}}\sqrt{2j+1}\sum _{l=0}^j\left(\begin{array}{c}j\\ l\end{array}\right)2^{(\theta -1)l}(-1{)}^{j-l}{\hat{i}}^{j-l}{a}_{\omega ,r}^{\mathrm{il}}.$Therefore, we get $\begin{array}{rl}S& =[\begin{array}{ccccc}{\mathrm{\Theta }}_{1,0}^{10}& \dots & {\mathrm{\Theta }}_{1,\mu -1}^{10}& \dots & {\mathrm{\Theta }}_{2^{\theta -1},0}^{10}\\ {\mathrm{\Theta }}_{1,0}^{11}& \dots & {\mathrm{\Theta }}_{1,\mu -1}^{11}& \dots & {\mathrm{\Theta }}_{2^{\theta -1},0}^{11}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ {\mathrm{\Theta }}_{1,0}^{2^{\theta -1},\mu -1}& \dots & {\mathrm{\Theta }}_{1,\mu -1}^{2^{\theta -1},\mu -1}& \dots & {\mathrm{\Theta }}_{2^{\theta -1},0}^{2^{\theta -1}\mu -1}\end{array}\\ & \begin{array}{cc}\dots & {\mathrm{\Theta }}_{2^{\theta -1},\mu -1}^{10}\\ \dots & {\mathrm{\Theta }}_{2^{\theta -1},\mu -1}^{11}\\ ⋮& ⋮\\ \dots & {\mathrm{\Theta }}_{2^{\theta -1},\mu -1}^{2^{\theta -1}\mu -1}\end{array}].\end{array}$For example, we define some structure of Taylor wavelets and operational matrix of integration for $\theta =1$ and $2^{\theta -1}\mu =n=7$: $\begin{array}{rl}& {\zeta }_{1,0}(\rho )=1,\\ & {\zeta }_{1,1}(\rho )=\sqrt{3}\rho ,\\ & {\zeta }_{1,2}(\rho )=\sqrt{5}{\rho }^2,\\ & {\zeta }_{1,3}(\rho )=\sqrt{7}{\rho }^3,\\ & {\zeta }_{1,4}(\rho )=\sqrt{9}{\rho }^4,\\ & {\zeta }_{1,5}(\rho )=\sqrt{11}{\rho }^5,\\ & {\zeta }_{1,6}(\rho )=\sqrt{13}{\rho }^6.\end{array}$Let $\zeta (\rho )=[{\zeta }_{1,0}(\rho ){\zeta }_{1,1}(\rho ){\zeta }_{1,2}(\rho ){\zeta }_{1,3}(\rho ){\zeta }_{1,4}(\rho ){\zeta }_{1,5}(\rho ){\zeta }_{1,6}(\rho )]$.

Each row of the operational matrix of integration can be obtained as follows;

1. Integrate each basis w.r.t. ρ limit from ‘0’ to ‘ρ’.

2. Express the obtained integration value as a linear combination of the given basis.

$\begin{array}{rl}& {\int }_0^{\rho }{\zeta }_{1,0}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& {\displaystyle \frac{1}{\sqrt{3}}}& 0& 0& 0& 0& 0\end{array}\right]{\zeta }_7(\rho ),\\ & {\int }_0^{\rho }{\zeta }_{1,1}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& 0& {\displaystyle \frac{\sqrt{3}}{2\sqrt{5}}}& 0& 0& 0& 0\end{array}\right]{\zeta }_7(\rho ),\\ & {\int }_0^{\rho }{\zeta }_{1,2}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& 0& 0& {\displaystyle \frac{\sqrt{5}}{3\sqrt{7}}}& 0& 0& 0\end{array}\right]{\zeta }_7(\rho ),\\ & {\int }_0^{\rho }{\zeta }_{1,3}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& 0& 0& 0& {\displaystyle \frac{\sqrt{7}}{4\sqrt{9}}}& 0& 0\end{array}\right]{\zeta }_7(\rho ),\\ & {\int }_0^{\rho }{\zeta }_{1,4}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& 0& 0& 0& 0& {\displaystyle \frac{\sqrt{9}}{5\sqrt{11}}}& 0\end{array}\right]{\zeta }_7(\rho ),\\ & {\int }_0^{\rho }{\zeta }_{1,5}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& 0& 0& 0& 0& 0& {\displaystyle \frac{\sqrt{11}}{6\sqrt{13}}}\end{array}\right]{\zeta }_7(\rho ),\\ & {\int }_0^{\rho }{\zeta }_{1,6}(\rho )\mathrm{d}\rho =\left[\begin{array}{lllllll}0& 0& 0& 0& 0& 0& 0\end{array}\right]{\zeta }_7(\rho )\\ & +{\displaystyle \frac{\sqrt{13}}{8\sqrt{15}}}\overline{{\zeta }_{1,7}}(\rho ).\end{array}$Thus, we have ${\int }_0^{\rho }\zeta (\rho )\mathrm{d}\rho =S_{7\times 7}{\zeta }_7(\rho )+\overline{{\zeta }_7}(\rho )$, where $S_{7\times 7}=\left[\begin{array}{ccccccc}0& {\displaystyle \frac{1}{\sqrt{3}}}& 0& 0& 0& 0& 0\\ 0& 0& {\displaystyle \frac{\sqrt{3}}{2\sqrt{5}}}& 0& 0& 0& 0\\ 0& 0& 0& {\displaystyle \frac{\sqrt{5}}{3\sqrt{7}}}& 0& 0& 0\\ 0& 0& 0& 0& {\displaystyle \frac{\sqrt{7}}{4\sqrt{9}}}& 0& 0\\ 0& 0& 0& 0& 0& {\displaystyle \frac{\sqrt{9}}{5\sqrt{11}}}& 0\\ 0& 0& 0& 0& 0& 0& {\displaystyle \frac{\sqrt{11}}{6\sqrt{13}}}\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]$is defined to be the 'operational matrix of integration of order $7\times 7$ and $\overline{{\zeta }_7}(\rho )=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ {\displaystyle \frac{\sqrt{13}}{8\sqrt{15}}}\overline{{\zeta }_{1,7}}(\rho )\end{array}\right].$Now, by repeating the steps above and integrating the aforementioned basis twice with respect to the ρ limit from 0 to ρ and express them as a linear combination of Taylor wavelets basis and placed in matrix form as ${\int }_0^{\rho }{\int }_0^{\rho }\zeta (\rho )\mathrm{d}\rho \mathrm{d}\rho =S_{7\times 7}^{\prime }{\zeta }_7(\rho )+{\overline{{\zeta }_7}}^{\prime }(\rho )$where, $S_{7\times 7}^{\prime }=\left[\begin{array}{ccccccc}0& 0& {\displaystyle \frac{1}{2\sqrt{5}}}& 0& 0& 0& 0\\ 0& 0& 0& {\displaystyle \frac{\sqrt{3}}{6\sqrt{7}}}& 0& 0& 0\\ 0& 0& 0& 0& {\displaystyle \frac{\sqrt{5}}{12\sqrt{9}}}& 0& 0\\ 0& 0& 0& 0& 0& {\displaystyle \frac{\sqrt{7}}{20\sqrt{11}}}& 0\\ 0& 0& 0& 0& 0& 0& {\displaystyle \frac{\sqrt{9}}{30\sqrt{13}}}\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]$and ${\overline{{\zeta }_7}}^{\prime }(\rho )=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ {\displaystyle \frac{\sqrt{11}}{42\sqrt{15}}}\overline{{\zeta }_{1,7}}(\rho )\\ {\displaystyle \frac{\sqrt{13}}{56\sqrt{17}}}\overline{{\zeta }_{1,8}}(\rho )\end{array}\right].$

5. Application of the operational matrix of integration

Here, we will discuss the application of the operational matrix of integration [32] and the collocation method for the given four types of conditions of the singular value problem defined in Equation (1(1) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{p\mathrm{\prime }(\rho )}{p(\rho )}{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=R(\rho ,\mathrm{\Lambda }),\rho \in (0,1].$(1) ).

Consider the boundary conditions of Type I defined in Equation (2(2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) ). Assume (34) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ).$(34) Integrate Equation (34(34) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ).$(34) ) w.r.t. ρ taking limit from 0 to ρ, we get (35) ${\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}].$(35) Again apply integration from 0 to ρ on both sides of Equation (35(35) ${\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}].$(35) ) with respect to the variable ρ yields, (36) $\mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].$(36) Substitute $\rho =1$ in Equation (36(36) $\mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].$(36) ) gives (37) ${\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}].$(37) Now, apply the conditions given in Equation (2(2) $\begin{array}{rl}& \mathrm{Type}\mathrm{I}:\mathrm{\Lambda }(0)={\alpha }_1,\mathrm{\Lambda }(1)={\beta }_1,\end{array}$(2) ) and substitute the obtained equation in Equation (36(36) $\mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].$(36) ), so that (38) $\begin{array}{rl}\mathrm{\Lambda }(\rho )& ={\alpha }_1+\rho ({\beta }_1-{\alpha }_1+E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])\\ & +E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(38) Further, on substituting the values of the function Λ and all its derivatives obtained by the above procedure, into (1(1) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{p\mathrm{\prime }(\rho )}{p(\rho )}{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=R(\rho ,\mathrm{\Lambda }),\rho \in (0,1].$(1) ) we get an algebraic equation of the following form, (39) $\begin{array}{rl}& E^T\zeta (\rho )+\frac{p^{\mathrm{\prime }}(\rho )}{p(\rho )}[({\beta }_1-{\alpha }_1+E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])\\ & +E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}]]\\ & -R(\rho ,{\alpha }_1+\rho ({\beta }_1-{\alpha }_1+E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])\\ & +E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}])=0.\end{array}$(39) Collocate the above-obtained equation at n number of collocation points defined by ${\rho }_i=\frac{2i-1}{2n}$, where $i=0,1,2,\dots ,n$ to get a system of n number of equations. We can solve these equations for the unknown Taylor wavelets coefficient using Newton's method or fsolve command of MATLAB.

Consider Type II boundary conditions defined in Equation (3(3) $\begin{array}{rl}& \mathrm{Type}\mathrm{II}:{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\alpha }_2,\mathrm{\Lambda }(1)={\beta }_2,\end{array}$(3) ). Assume (40) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ).$(40) Integrate Equation (40(40) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ).$(40) ) w.r.t. ρ taking limit from 0 to ρ, we get (41) $\begin{array}{rl}& {\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}],\end{array}$(41) (42) $\begin{array}{rl}& \mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(42) Substitute $\rho =1$ in Equation (42(42) $\begin{array}{rl}& \mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(42) ), we get (43) $\mathrm{\Lambda }(0)=\mathrm{\Lambda }(1)-{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)-E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}].$(43) Now, apply the conditions given in Equation (3(3) $\begin{array}{rl}& \mathrm{Type}\mathrm{II}:{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\alpha }_2,\mathrm{\Lambda }(1)={\beta }_2,\end{array}$(3) ) and substitute Equation (43(43) $\mathrm{\Lambda }(0)=\mathrm{\Lambda }(1)-{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)-E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}].$(43) ) in Equation (42(42) $\begin{array}{rl}& \mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(42) ), we have (44) $\begin{array}{rl}\mathrm{\Lambda }(\rho )& =({\beta }_2-{\alpha }_2-E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])+\rho {\alpha }_2\\ & +E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(44) Similarly proceed as we have done for case (I) to get our algebraic equation for this case.

Consider Type III boundary conditions defined in Equation (4(4) $\begin{array}{rl}& \mathrm{Type}\mathrm{III}:\mathrm{\Lambda }(0)={\alpha }_3,{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)={\beta }_3,\end{array}$(4) ). Assume (45) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\mathit{\zeta }(\rho ).$(45) Integrate Equation (45(45) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\mathit{\zeta }(\rho ).$(45) ) w.r.t. ρ taking limit from 0 to ρ, we get (46) ${\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}].$(46) Integrate Equation (46(46) ${\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}].$(46) ) w.r.t. ρ taking limit from 0 to ρ, we get (47) $\mathrm{\Lambda }(\rho )={\alpha }_3+\rho {\beta }_3+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].$(47) Consider Type IV boundary conditions defined in Equation (5(5) $\begin{array}{rl}\mathrm{Type}\mathrm{IV}:a_1\mathrm{\Lambda }(0)+a_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)& ={\alpha }_4,\\ b_1\mathrm{\Lambda }(1)+b_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& ={\beta }_4,\end{array}$(5) ). Assume (48) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ).$(48) Integrate Equation (48(48) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ).$(48) ) w.r.t. ρ taking limit from 0 to ρ, we get (49) $\begin{array}{rl}& {\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}],\end{array}$(49) (50) $\begin{array}{rl}& \mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(50) Substitute $\rho =1$ in Equations (49(49) $\begin{array}{rl}& {\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )={\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}],\end{array}$(49) ) and (50(50) $\begin{array}{rl}& \mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(50) ), we have (51) $\begin{array}{rl}{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& =[\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]\\ & +E^T[\mathrm{S\zeta }(1)+\overline{\zeta (1)}]\end{array}$(51) and (52) ${\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}]).$(52) Using these values of ${\mathrm{\Lambda }}^{\mathrm{\prime }}(1)$ and ${\mathrm{\Lambda }}^{\mathrm{\prime }}(0)$ in Equation (5(5) $\begin{array}{rl}\mathrm{Type}\mathrm{IV}:a_1\mathrm{\Lambda }(0)+a_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)& ={\alpha }_4,\\ b_1\mathrm{\Lambda }(1)+b_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& ={\beta }_4,\end{array}$(5) ) which will turn out into the following equations: (53) $\begin{array}{rl}& a_1\mathrm{\Lambda }(0)+a_2(\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}]))\\ & ={\alpha }_4,\end{array}$(53) (54) $\begin{array}{rl}& b_1\mathrm{\Lambda }(1)+a_2([\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]\\ & +E^T[\mathrm{S\zeta }(1)+\overline{\zeta (1)}]+E^T[\mathrm{S\zeta }(1)+\overline{\zeta (1)}])={\beta }_4.\end{array}$(54) Observe that the values of $\mathrm{\Lambda }(0)$ and $\mathrm{\Lambda }(1)$ can be obtained by solving Equations (53(53) $\begin{array}{rl}& a_1\mathrm{\Lambda }(0)+a_2(\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}]))\\ & ={\alpha }_4,\end{array}$(53) ) and (54(54) $\begin{array}{rl}& b_1\mathrm{\Lambda }(1)+a_2([\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]\\ & +E^T[\mathrm{S\zeta }(1)+\overline{\zeta (1)}]+E^T[\mathrm{S\zeta }(1)+\overline{\zeta (1)}])={\beta }_4.\end{array}$(54) ), so that we have a truncated expression of ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho ),{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )$ and $\mathrm{\Lambda }(\rho )$ in terms of Taylor wavelet basis and Taylor wavelet coefficients with the provided boundary conditions (5(5) $\begin{array}{rl}\mathrm{Type}\mathrm{IV}:a_1\mathrm{\Lambda }(0)+a_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)& ={\alpha }_4,\\ b_1\mathrm{\Lambda }(1)+b_2{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)& ={\beta }_4,\end{array}$(5) ) as $\begin{array}{rl}& {\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )=E^T\zeta (\rho ),\\ & {\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=([\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]\\ & +E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}]\end{array}$and $\begin{array}{rl}\mathrm{\Lambda }(\rho )& =\mathrm{\Lambda }(0)+\rho ([\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]\\ & +E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$Substituting the expression of the function Λ and its derivatives obtained by the above procedure in Equation (1(1) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{p\mathrm{\prime }(\rho )}{p(\rho )}{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=R(\rho ,\mathrm{\Lambda }),\rho \in (0,1].$(1) ), we get an associated algebraic equation of the following form: (55) $\begin{array}{rl}& E^T\zeta (\rho )+\frac{p^{\mathrm{\prime }}(\rho )}{p(\rho )}([\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)-(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]\\ & +E^T[\mathrm{S\zeta }(\rho )+\overline{\zeta (\rho )}])\\ & -R(\rho ,\mathrm{\Lambda }(0)+\rho ([\mathrm{\Lambda }(1)-\mathrm{\Lambda }(0)\\ & -(E^T[S^{\prime }\zeta (1)+\overline{{\zeta }^{\prime }(1)}])]+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}])=0.\end{array}$(55) Collocate this equation at n number of grid points ${\rho }_i=\frac{2i-1}{2n}$, where $i=0,1,2,\dots ,n$ to get a set of n algebraic equations. We can solve this system of n equations for the value of vector E (Taylor wavelet coefficients) by a suitable method. The required numerical solution will next be presented by substituting these coefficients in Equation (50(50) $\begin{array}{rl}& \mathrm{\Lambda }(\rho )=\mathrm{\Lambda }(0)+\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)+E^T[S^{\prime }\zeta (\rho )+\overline{{\zeta }^{\prime }(\rho )}].\end{array}$(50) ).

6. Numerical experiments

Problem 1

Take into account the following problem (56) ${({\rho }^8{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho ))}^{\mathrm{\prime }}={\rho }^{8}R(\rho ,\mathrm{\Lambda })$(56) with $\left[\begin{array}{c}\mathrm{\Lambda }(0)=1\\ {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=0\end{array}\right],$where $R(\rho ,\mathrm{\Lambda })=18\mathrm{\Lambda }+4\mathrm{\Lambda log}(\mathrm{\Lambda }).$The exact solution of this problem is $\mathrm{\Lambda }(\rho )=e^{{\rho }^2}$. We obtain a Taylor wavelet solution (TWS) for this problem for different values of n = 6, 7, 9 and 11 with $\theta =1$. Table 1 demonstrates that only a few Taylor wavelet basis functions are required to obtain an approximation that is the same as an exact solution with a full agreement up to 7, 8 Digits, while Figure 1 shows the exact and approximate solution. In Figure 2, error variation can be observed (Table 2 provides the absolute error that occurred in TWS for different values of n).

Figure 1. Graphical comparison between the Taylor wavelet solution (TWS) and the Exact solution for Problem 1.

Figure 2. Graphical representation of the error variation of the TWS and the Exact solution for Problem 1.

Table 1. Numerical comparison of several approximate solutions and Exact solutions for Problem 1.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11	Exact solution
0.1	1.01005043	1.01005016	1.01005016	1.01005016	1.01005016
0.2	1.04080888	1.04081072	1.04081073	1.04081079	1.04081077
0.3	1.09417043	1.09417405	1.09417405	1.09417438	1.09417428
0.4	1.17350821	1.17351029	1.17350966	1.17351077	1.17351087
0.5	1.28402166	1.28402427	1.28401863	1.28401870	1.28402541
0.6	1.43331935	1.43332752	1.43329693	1.43326406	1.43332941
0.7	1.63231224	1.63230614	1.63218890	1.63191630	1.63231621
0.8	1.89649354	1.89639246	1.89606022	1.89462803	1.89648087
0.9	2.24768095	2.24737587	2.24669612	2.24085833	2.24790798

Table 2. Numerical comparison of absolute error for Problem 1.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 10
0.1	2.6517e−07	7.9223e−10	4.7582e−10	9.5855e−10
0.2	1.8863e−06	4.8008e−08	3.9211e−08	2.1055e−08
0.3	3.8486e−06	2.2579e−07	2.2590e−07	1.0342e−07
0.4	2.6593e−06	5.7133e−07	1.2028e−06	9.3601e−08
0.5	3.7511e−06	1.1382e−06	6.7846e−06	6.7087e−06
0.6	1.0064e−05	1.8862e−06	3.2478e−05	6.5352e−05
0.7	3.9761e−06	1.0077e−05	1.2731e−04	3.9991e−04
0.8	1.2670e−05	8.8414e−05	4.2066e−04	1.8528e−03
0.9	2.2703e−04	5.3211e−04	1.2119e−03	7.0497e−03

Problem 2

Take into account the following problem (57) ${({\rho }^2{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho ))}^{\mathrm{\prime }}={\rho }^{2}R(\rho ,\mathrm{\Lambda }),\rho >0,$(57) with $\left[\begin{array}{c}\mathrm{\Lambda }(0)=0\\ {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=0\end{array}\right],$where $R(\rho ,\mathrm{\Lambda })=-4(2e^{\mathrm{\Lambda }}+e^{\mathrm{\Lambda }/2}).$The exact solution of (57(57) ${({\rho }^2{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho ))}^{\mathrm{\prime }}={\rho }^{2}R(\rho ,\mathrm{\Lambda }),\rho >0,$(57) ) is $\mathrm{\Lambda }(\rho )=-2\log (1+{\rho }^2)$. We apply the proposed Taylor wavelet method for the parameters n = 6, 7, 9, 10 with $\theta =1$. Table 3, represents a comparison between the values obtained by TWM, while Table 4, represents error variation for different values of n. In Figure 3, we have shown the TWS and the exact solution, while in Figure 4 error variation can be observed.

Figure 3. Graphical comparison between the Taylor wavelet solution (TWS) and the Exact solution for Problem 2.

Figure 4. Graphical representation of the absolute error variation of the TWS for Problem 2.

Table 3. Numerical comparison of approximate solution and exact solution for Problem 2.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11	Exact solution
0.1	−0.01990268	−0.01990012	−0.01990073	−0.01990064	−0.01990066
0.2	−0.07844121	−0.07844172	−0.07844176	−0.07844134	−0.07844142
0.3	−0.17235136	−0.17235616	−0.17235462	−0.17235429	−0.17235539
0.4	−0.29684031	−0.29684029	−0.29681292	−0.29682189	−0.29684001
0.5	−0.44629246	−0.44628691	−0.44602264	−0.44611869	−0.44628710
0.6	−0.61497001	−0.61496820	−0.61331983	−0.61392605	−0.61496939
0.7	−0.79756200	−0.79754913	−0.78983358	−0.79267646	−0.79755223
0.8	−0.98943534	−0.98938668	−0.96003946	−0.97085596	−0.98939248
0.9	−1.18644160	−1.18664880	−1.09132102	−1.12645557	−1.18665369

Table 4. Error comparison of TWS for different values of n for Problem 2.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11
0.1	2.0231e−06	5.3906e−07	6.9106e−08	2.0744e−08
0.2	2.1589e−07	2.9991e−07	3.4208e−07	8.4876e−08
0.3	4.0254e−06	7.7289e−07	7.6486e−07	1.1022e−06
0.4	3.0761e−07	2.8525e−07	2.7082e−05	1.8118e−05
0.5	5.3583e−06	1.8425e−07	2.6446e−04	1.6840e−04
0.6	6.1612e−07	1.1973e−06	1.6496e−03	1.0433e−03
0.7	9.7692e−06	3.1015e−06	7.7187e−03	4.8758e−03
0.8	4.2857e−05	5.8018e−06	2.9353e−02	1.8537e−02
0.9	2.1209e−04	4.8864e−06	9.5333e−02	6.0198e−02

Problem 3

Take into account the following problem (Ex. 5.6 [34]) (58) ${(\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho ))}^{\mathrm{\prime }}=\mathrm{\rho R}(\rho ,\mathrm{\Lambda }),\rho \in (0,1)$(58) with $\left[\begin{array}{c}\mathrm{\Lambda }(1)=0\\ {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=0\end{array}\right].$Where $R(\rho ,\mathrm{\Lambda })=\frac{64}{(8-{\rho }^2{)}^2}.$The exact solution of this problem is $\mathrm{\Lambda }(\rho )=2\log (\frac{7}{8-{\rho }^2})$. We solve this problem using Taylor wavelet for n = 6, 7, 8, 9 and $\theta =1$. Taylor wavelet solution (TWS) is presented in Table 5 in comparison with the exact solution. In Table 6, TWS is compared with some other existing methods, while error variation for the same values of n and θ is shown in Table 7. Taylor wavelet solution for $n=9,\theta =1$ and exact solution is graphically shown in Figure 5. Figure 6, shows absolute error variation for multiple values of n.

Figure 5. Graphical comparison between the Taylor wavelet solution (TWS) and the Exact solution for Problem 3.

Table 5. Numerical comparison of approximate solution and Exact solution for Problem 3.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 8	TWS at n = 9	Exact solution
0.1	−0.26457232	−0.26456088	−0.26456116	−0.26456121	−0.26456122
0.2	−0.25704908	−0.25703736	−0.25703764	−0.25703769	−0.25703770
0.3	−0.24444699	−0.24443493	−0.24443521	−0.24443525	−0.24443526
0.4	−0.22666921	−0.22665703	−0.22665731	−0.22665736	−0.22665737
0.5	−0.20357697	−0.20356505	−0.20356533	−0.20356538	−0.20356538
0.6	−0.17498610	−0.17497457	−0.17497483	−0.17497490	−0.17497490
0.7	−0.14066177	−0.14065029	−0.14065057	−0.14065062	−0.14065063
0.8	−0.10031088	−0.10029922	−0.10029951	−0.10029955	−0.10029956
0.9	−0.05357169	−0.05356172	−0.05356199	−0.05356203	−0.05356204

Table 6. Comparison of the TWS with some other existing method's solution for Problem 3.

ρ	Ref. [34]	TWS at n = 9	Exact solution
0.1	−0.26456122	−0.26456121	−0.26456122
0.2	−0.25703770	−0.25703769	−0.25703770
0.3	−0.24443526	−0.24443525	−0.24443526
0.4	−0.22665737	−0.22665736	−0.22665737
0.5	−0.20356538	−0.20356538	−0.20356538
0.6	−0.17497490	−0.17497490	−0.17497490
0.7	−0.14065063	−0.14065062	−0.14065063
0.8	−0.10029956	−0.10029955	−0.10029956
0.9	−0.05356204	−0.05356203	−0.05356204

Table 7. Comparison of the absolute error for Problem 3 at multiple values of n.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 8	TWS at n = 9
0.1	0.1110e−04	0.3339e−06	0.5473e−07	0.7498e−08
0.2	0.1138e−04	0.3334e−06	0.5468e−07	0.7504e−08
0.3	0.1172e−04	0.3346e−06	0.5466e−07	0.7505e−08
0.4	0.1184e−04	0.3345e−06	0.5465e−07	0.7507e−08
0.5	0.1158e−04	0.3347e−06	0.5462e−07	0.7507e−08
0.6	0.1120e−04	0.3348e−06	0.5465e−07	0.7509e−08
0.7	0.1113e−04	0.3342e−06	0.5456e−07	0.7509e−08
0.8	0.1131e−04	0.3374e−06	0.5478e−07	0.7513e−08
0.9	0.1164e−04	0.3373e−06	0.5427e−07	0.7550e−08

Problem 4

Take into account the following problem $\begin{array}{rl}& {\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{1}{2\rho }{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=4{\rho }^2{e}^{\mathrm{\Lambda }}(4{\rho }^4{e}^{\mathrm{\Lambda }}-3.5),\rho \in (0,1],\\ & \left[\begin{array}{c}\mathrm{\Lambda }(0)=\log \left({\displaystyle \frac{1}{4}}\right)\\ \mathrm{\Lambda }(1)=\log \left({\displaystyle \frac{1}{5}}\right)\end{array}\right],\end{array}$with the exact solution $\mathrm{\Lambda }(\rho )=-\log (4+{\rho }^4)$. We solve this problem using the Taylor wavelet for n = 6, 7, 9, 11 and $\theta =1$. In Table 8, numerical values are compared to the exact solution (Figure 6). In Table 9 error variation for the same values of n and θ is shown. Taylor wavelet solution for n = 11, $\theta =1$, and the exact solution is graphically shown in Figure 7, while in Figure 8 error variation can be observed graphically.

Table 8. Numerical comparison between the Exact solution and the TWS for Problem 4.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11	Exact solution
0.1	−1.38546454	−1.38634644	−1.38633294	−1.38631790	−1.38631936
0.2	−1.38535906	−1.38673617	−1.38666090	−1.38669150	−1.38669428
0.3	−1.38672156	−1.38837018	−1.38835855	−1.38831346	−1.38831731
0.4	−1.39074297	−1.39273642	−1.39272928	−1.39266935	−1.39267396
0.5	−1.39962224	−1.40186916	−1.40186787	−1.40179356	−1.40179854
0.6	−1.41582428	−1.41825793	−1.41826351	−1.41817570	−1.41818054
0.7	−1.44207284	−1.44466940	−1.44468280	−1.44458278	−1.44458685
0.8	−1.48111174	−1.48387058	−1.48389208	−1.48378159	−1.48378398
0.9	−1.53577013	−1.53826064	−1.53829800	−1.53817890	−1.53817818

Table 9. Absolute error comparison of the TWS for different values of n for Problem 4.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11
0.1	8.5482e−04	2.7088e−05	1.3586e−05	1.4544e−06
0.2	1.3352e−03	4.1890e−05	2.7285e−05	2.7739e−06
0.3	1.6494e−03	5.2873e−05	4.1239e−05	3.8511e−06
0.4	1.9310e−03	6.2453e−05	5.5322e−05	4.6140e−06
0.5	2.1763e−03	7.0613e−05	6.9327e−05	4.9804e−06
0.6	2.3563e−03	7.7385e−05	8.2966e−05	4.8487e−06
0.7	2.5140e−03	8.2556e−05	9.5949e−05	4.0713e−06
0.8	2.6722e−03	8.6602e−05	1.0810e−04	2.3872e−06
0.9	2.4081e−03	8.2457e−05	1.1982e−04	7.1596e−07

Problem 5

Take into account the following problem (59) ${(\rho {\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho ))}^{\mathrm{\prime }}=-\rho e^{\mathrm{\Lambda }}$(59) with $\left[\begin{array}{c}\mathrm{\Lambda }(1)=0\\ {\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=0\end{array}\right]$and the exact solution of this problem is $\mathrm{\Lambda }(\rho )=2\log (\frac{v_0+1}{v_0{\rho }^2+1})$, where $v_0=3-2^{3/2}$. We obtain the Taylor wavelet solution (TWS) for this problem for different values of n = 6, 7, 9 and 10 with $\theta =1$. Table 10 demonstrates that only a few Taylor wavelet basis functions are required to obtain an approximation that is the same as an exact solution with a full agreement up to 5, 6 Digits, while Figure 9 shows the exact and approximate solution. In Figure 10 error variation of TWS can be observed. (Table 11, provides the absolute error for multiple values of n).

Figure 6. Graphical representation of the absolute error variation of the Taylor wavelet solution (TWS) for Problem 3.

Table 10. Numerical comparison of the TWS and Exact solution for Problem 5.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11	Exact solution
0.1	0.31326972	0.31327706	0.31326135	0.31327416	0.31326585
0.2	0.30301916	0.30302650	0.30301081	0.30302361	0.30301542
0.3	0.28605082	0.28605811	0.28604247	0.28605527	0.28604726
0.4	0.26253447	0.26254162	0.26252614	0.26253890	0.26253112
0.5	0.23269989	0.23270676	0.23269167	0.23270431	0.23269678
0.6	0.19682963	0.19683603	0.19682171	0.19683407	0.19682680
0.7	0.15525062	0.15525621	0.15524333	0.15525502	0.15524810
0.8	0.10832490	0.10832916	0.10831879	0.10832900	0.10832276
0.9	0.05644011	0.05644236	0.05643612	0.05644311	0.05643860

Table 11. Absolute error comparison for different values of n for Problem 5.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11
0.1	3.8732e−06	1.1216e−05	4.4913e−06	8.3142e−06
0.2	3.7437e−06	1.1083e−05	4.6114e−06	8.1931e−06
0.3	3.5565e−06	1.0850e−05	4.7878e−06	8.0074e−06
0.4	3.3429e−06	1.0497e−05	4.9777e−06	7.7789e−06
0.5	3.1062e−06	9.9795e−06	5.1112e−06	7.5300e−06
0.6	2.8331e−06	9.2264e−06	5.0871e−06	7.2650e−06
0.7	2.5202e−06	8.1099e−06	4.7673e−06	6.9199e−06
0.8	2.1447e−06	6.3999e−06	3.9730e−06	6.2417e−06
0.9	1.5172e−06	3.7651e−06	2.4770e−06	4.5163e−06

Problem 6

Take into account the following problem, (60) ${\mathrm{\Lambda }}^{\mathrm{\prime \prime }}(\rho )+\frac{(2+\rho )}{\rho }{\mathrm{\Lambda }}^{\mathrm{\prime }}(\rho )=\frac{5{\rho }^3(5{\rho }^5{e}^{\mathrm{\Lambda }}-\rho -6)}{4+{\rho }^5},$(60) with $\left[\begin{array}{c}{\mathrm{\Lambda }}^{\mathrm{\prime }}(0)=0\\ \mathrm{\Lambda }(1)+5{\mathrm{\Lambda }}^{\mathrm{\prime }}(1)=-5-\log (5)\end{array}\right].$Which has the exact solution as $\mathrm{\Lambda }(\rho )=-\log (4+{\rho }^5)$. We solve this problem using Taylor wavelets for n = 6, 7, 9, 11 and $\theta =1$. Table 12 shows that a very less number of basis functions are required to obtain our truncated solution presented by the Taylor wavelets series of the minimum error.  Absolute error variation can be seen for different values inTable 13 (Taylor wavelet approximate solution and absoluter error occured in the proposed method can be graphically observed in Figures 11 and 12 respectively).

Figure 7. Graphical comparison between the Taylor wavelet solution (TWS) and the Exact solution for Problem 4.

Figure 8. Graphical representation of the absolute error variation of the Taylor wavelet solution (TWS) for Problem 4.

Figure 9. Graphical comparison between the Taylor wavelet solution (TWS) and the Exact solution for Problem 5.

Figure 10. Graphical representation of the absolute error variation of the Taylor wavelet solution (TWS) for Problem 5.

Figure 11. Graphical comparison between the Taylor wavelet solution (TWS) and the Exact solution for Problem 6.

Figure 12. Graphical representation of the absolute error variation of the Taylor wavelet solution (TWS) for Problem 6.

Table 12. Numerical comparison of the TWS and the Exact solution for Problem 6.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11	Exact solution
0.1	−1.33039242	−1.41273268	−1.39157664	−1.38423751	−1.38629686
0.2	−1.33047234	−1.41280969	−1.39165476	−1.38431507	−1.38637435
0.3	−1.33100364	−1.41333644	−1.39218293	−1.38484259	−1.38690167
0.4	−1.33295230	−1.41528537	−1.39413346	−1.38679243	−1.38885108
0.5	−1.33817673	−1.42051031	−1.39936013	−1.39201861	−1.39407650
0.6	−1.34965075	−1.43198255	−1.41083293	−1.40349113	−1.40554781
0.7	−1.37153810	−1.45389461	−1.43274048	−1.42539811	−1.42745309
0.8	−1.40904655	−1.49149989	−1.47032412	−1.46297866	−1.46503160
0.9	−1.46798946	−1.55050855	−1.52929171	−1.52193569	−1.52398677

Table 13. Absolute error comparison for different values of n for Problem 6.

ρ	TWS at n = 6	TWS at n = 7	TWS at n = 9	TWS at n = 11
0.1	5.5904e−02	2.6436e−02	5.2798e−03	2.0593e−03
0.2	5.5902e−02	2.6435e−02	5.2804e−03	2.0593e−03
0.3	5.5898e−02	2.6435e−02	5.2813e−03	2.0591e−03
0.4	5.5899e−02	2.6434e−02	5.2824e−03	2.0587e−03
0.5	5.5900e−02	2.6434e−02	5.2836e−03	2.0579e−03
0.6	5.5897e−02	2.6435e−02	5.2851e−03	2.0567e−03
0.7	5.5915e−02	2.6442e−02	5.2874e−03	2.0550e−03
0.8	5.5985e−02	2.6468e−02	5.2925e−03	2.0529e−03
0.9	5.5997e−02	2.6522e−02	5.3049e−03	2.0511e−03

7. Conclusion

Using the Taylor wavelets, we tried to implement an operational integration matrix to find the solution to singular value linear and non-linear differential equations under various given conditions. This approach is crucial for the advancement of fresh studies in the discipline of numerical analysis and is advantageous for beginning researchers. When compared to the several numerical solutions, the proposed method performs very satisfactorily on applying several problems and performs well in comparison with other numerical methods. The results of this analysis are summarized as follows:

1. Comparing the accuracy of the current method to other numerical techniques found in the literature, it is more accurate.

2. The method's steps are rapid and simple to implement in computer programming, and they may be increased to higher orders with very minor methodological modifications.

3. The operational matrix of integration is generalized to make this method simple.

4. The Taylor wavelet technique is a new one that was recently used to construct this operational integration matrix.

5. In future, researchers can also try this approach to solve any class of differential equations of the scope.

CRediT authorship contribution statement

Vivek developed the fundamental concept of the article, wrote the text, and completed all phases of the research's proofs.

Availability of the Data

All the data associated with the manuscript is included in it.

Acknowledgments

All authors are grateful to the reviewers for their outstanding comments, which encouraged us at each step of the well-read proofs of the article to develop a distinct and excellent script.

Disclosure statement

No potential conflict of interest was reported by the author(s).