Fibonacci wavelets operational matrix approach for solving chemistry problems

Abstract

The key idea and contribution of this study are to present the innovative functional matrix approach for solving the two chemical, mathematical problems such as the absorption of $C{O}_{2}$ into phenyl glycidyl ether (PGE) and the chemical kinetic problem. These chemical problems are characterized as a system of nonlinear ODEs with initial and boundary conditions. Numerical outcomes are obtained to establish the simplicity and efficacy of the developed scheme. Graphs and tables show how consistently and effectively the developed strategy works. Results obtained show that the newly adopted technique is more precise and effective than other methods such as Adam–Bashforth–Moultan method (ABM), Runge–Kutta method (RK4), Adomian Decomposition method (ADM), and residual method (RM). Mathematical software called Mathematica 11.3 has been used to perform all calculations. Theorems explain the convergence of this approach.

1 Introduction

In real life, many variables and parameters are connected under certain circumstances. When we mathematically represent the relationship between these variables and parameters, we usually arrive at a mathematical model of the problem: a differential equation, an integral equation, a system of differential or integral equations, etc. Mathematical modelling is the best way to formulate real-world problems. It is familiar that numerous mathematical characterizations of enormous growth in physical and chemical sciences are represented by SODEs. In the field of chemistry, the above-considered two models that are nonlinear SODEs can describe different kinds of Dirichlet and Neumann-type boundary conditions. Carbon dioxide is vital in plant photosynthesis and the industrial of carbonated soft drinks, removing caffeine from coffee, used in fire extinguishers, and energizing pneumatic models in robotics [1, 2]. $C{O}_{2}$ is a beneficial gas that forms one carbon atom and two oxygen atoms. In recent times the chemical fascination of $C{O}_{2}$ is a significant research focus since the hazard postured by global warming and the conversion of $C{O}_{2}$ into valuable substances is a tremendously elegant result.

Consider the general form of the SODEs, which represents the $C{O}_{2}$ absorbed into the PGE problem [3]:

$$\left.\begin{array}{c}{y}_{1}^{\prime\prime}(x)=f_{1}\left(t,{y}_{1}(x),\ldots {y}_{n}\left( x\right), {y}_{1 }^{{\prime}}(x),\ldots {y}_{n}^{{\prime}}\left( x\right)\right)\\ {{y}_{2}^{\prime\prime}(x)=f}_{2}\left(t,{y}_{1}(x),\ldots {y}_{n}(x), {y}_{1 }^{{\prime}}(x),\ldots {y}_{n}^{{\prime}}\left( x\right)\right)\\ .\\ .\\ .\\ {{y}_{n}^{\prime\prime}(x)=f}_{n}\left(t,{y}_{1} (x),\ldots {y}_{n} (x), {y}_{1}^{{\prime}} (x),\dots .{y}_{n}^{{\prime}} (x)\right)\end{array}\right\}.$$

(1)

With Dirichlet boundary conditions: ${y}_{i}(x)={\alpha }_{i}$, ${y}_{i}(x)={\beta }_{i}, i=1,2,\ldots n.$

Mixed set of Dirichlet and Neumann boundary conditions: ${y}_{i}^{{\prime}}(x)={\gamma }_{i}, {y}_{i}(x)={\mu }_{i}$. Where ${f}_{1}, {f}_{2}, \dots {f}_{n}$ are homogeneous parts of the equation, ${y}_{n}(x)$ are the concentrations of $C{O}_{2}$ and PGE, respectively, ${\alpha }_{i}$’s, ${\beta }_{i}$’s, ${\gamma }_{i}$’s and ${\mu }_{i}$’s are the specified values at the different boundaries.

The mathematical model of the chemical kinetic problem is one of the well-described nonlinear SODEs that Robertson introduced in 1966. There are three spaces in the model of chemical procedure, which are indicated by P, Q, and R. We can define the three reactions as:

$${\text{P}} \to {\text{Q}}$$

(2)

$${\text{Q}} + {\text{R}} \to {\text{P}} + {\text{R}}$$

(3)

$${\text{Q}} + {\text{Q}} \to {\text{R}}{.}$$

(4)

The concentration of P, Q, and R can be denoted by ${u}_{1}, {u}_{2}\ \textit{and}\ {u}_{3}$ respectively. It is worth considering that these are the combinations of three concentrations in one. Let ${b}_{1}$ denote the reaction rate of Eq. (2), and this means that the rate at which ${u}_{2}$ rises and at which ${u}_{1}$ reduces, due of the above reaction, it will be equivalent to ${b}_{1}{u}_{1}$. R acts as a catalyst in the production of P from Q and ${b}_{2}$ represents reaction rate in the reaction Eq. (3), meaning that the decrease of ${u}_{3}$ and the increase of ${u}_{1}$ this reaction will be equivalent to ${b}_{2}{u}_{2}{u}_{3}$. At last, the rate at which this reaction will be equal to ${b}_{3}{u}_{3}^{2}.$ Due to the production of R from Q will have a constant rate equivalent to ${b}_{3}.$

Consider the general form of the nonlinear SODEs, which represents the chemical kinetic problem: [4, 5]

$${u}_{i}^{{\prime}}(x)={g}_{i}\left(x, {u}_{1}(x),{u}_{2}(x),\dots .{u}_{q}(x)\right)$$

(5)

where (${^{\prime}}$) represents the derivative of the dependent variable ${u}_{i}$ concerning the independent variable $x$ and corresponding initial conditions; ${u}_{i}{(x)}_{x=0}={\alpha }_{i}$, where ${\alpha }_{i}$’s are the specified constant vector and ${u}_{i}(x)$ is the solution vector and $i=1,2,3,\ldots q$. Different published works were carried out on the above models, such as the Chemistry problem by Abbasbandy and Shirzadi [6], the chemistry problem by Matinfar et al. [7], An analytical approximation to the solution of the chemical kinetic problem by Hossein A [4], chemistry problems by Jawary and Raham [8], Kumar et al. [9] implemented a comparative study for fractional chemical kinetics and carbon dioxide $C{O}_{2}$ absorbed into phenyl glycidyl ether problems, $C{O}_{2}$ absorbed into the PGE problem by Robertson [10], Ganji et al. [5] proposed He’s Homotopy Perturbation method for the chemistry kinetic problem, Jawary et al. implemented an effective iteration method and VIM (Variational Iteration Method) for the chemical kinetic problem, and the absorption of carbon dioxide into phenyl glycidyl ether problem respectively [11]. The ADM method was used for the simple steady-state condensations of $C{O}_{2}$ and PGE, and it is presented in [3], Duan et al. [12] successfully attempted the simple steady-state condensations of $C{O}_{2}$ and PGE by the ADM Method, Kaya [13] attempted the Adomian decomposition method to the chemical kinetic problem, Khader [14] applied the Laplace–Pade approximation technique to improve the performance of the Chemical Kinetic Problem (CKP) model, MSH Chowdary implemented a novel iterative method for the CK System [15], Boundary domain integral method and HAM for nonlinear BVP’s were proposed by Jawary et al. [16], Singha and Wazwaz [17] proposed an optimal Homotopy analysis method for steady-state concentrations of absorption of carbon dioxide into phenyl glycidyl ether problem.

The wavelet theory is a relatively renewed and emerging theory in mathematical research. Wavelet analysis is a great tool that significantly impacts study and engineering. The primary criteria that draw researchers to the wavelet are orthogonality and effective localization. With the help of wavelets that are mathematical operations, data may be divided into several frequency components, each of which can then be analyzed at a different resolution. It is possible to employ wavelets as a mathematical tool to extract information from various data formats. Wavelet acquired the attention of many mathematics researchers for solving various types of highly nonlinear typical problems. Different wavelet collocation schemes applied for some of the common mathematical problems are Hermite, Legendre, and Laguerre wavelets [18,19,20,21,22,23,24,25,26], Bernoulli wavelets [28], Fractional delay differential equations [27], Fractional Riccati differential equation [29]. Few wavelet collocation methods generally used for solving various engineering, physical and biological models include Legendre wavelets [30], Haar wavelets [31,32,33], Chebyshev wavelets [34], Bernoulli wavelets [27, 43,44,45], and Ultraspherical wavelets [35, 36].

Fibonacci wavelets produced by Fibonacci polynomials are a new accumulation to the field of wavelet families. It has an added advantage in contrast with the other wavelet methods as follows [39]:

• The number of terms of the Fibonacci polynomials ${P}_{m}(x)$ is less than the number of the terms of the Legendre polynomials ${L}_{m}(x).$ It helps to reduce CPU time.

• Error components in the OMI representing Fibonacci polynomials are less than that of Legendre polynomials.

• Fibonacci polynomials ${P}_{m}(x)$ have less coefficient of individual terms than in Legendre polynomials ${L}_{m}(x).$ Computational errors can be reduced using this property.

• Using the Mathematica command Fibonacci [m, x], the coefficients of the Fibonacci polynomials can be easily obtained in computer programs.

Due to its superior properties and advantages over other wavelets, it grabs the attention of many researchers. As a consequence, researchers start using this package to solve mathematical problems such as the Nonlinear Hunter–Saxton Equation [48], fractional order optimal control problems [37], time-varying delay problems [36], nonlinear Stratonovich Volterra integral equations [38], time-fractional telegraph equations [39], time-fractional bioheat transfer model [40], nonlinear Volterra integral equations [41], Dual-phase lag heat transfer model [42]. Also, some articles referred to the improvisation of concept [49,50,51].

This paper is organized as follows. Section 2, named “Preliminaries,” defines Fibonacci wavelets. OMI of Fibonacci wavelets carried out in Sect. 3. The method of solution and application of the proposed scheme is explained in Sects. 4 and 5, respectively. Finally, Sect. 6 gives the conclusion of the article.

2 Preliminaries of the Fibonacci wavelets

On the interval $[0,1]$, Fibonacci wavelets are defined as [36, 39],

$${\varphi }_{n, m}(x)=\left\{\begin{array}{ll}\frac{{2 }^{\frac{k -1}{2}}}{\sqrt{{S}_{m}}} {P}_{m}\left({2}^{k-1}x-\widehat{n}\right), &\quad\frac{\widehat{n}}{{2}^{ k-1}}\le x<\frac{\widehat{n} +1}{ {2}^{ k-1}}\\ 0,&\quad \text{Otherwise},\end{array}\right.$$

with

$${S}_{m}=\int\limits_{0}^{1}{\left({P}_{m}(x)\right)}^{2}dx,$$

where ${P}_{m}(x)$ is the Fibonacci polynomial of degree $m=0,1,2\dots , M-1$, translation parameter $n=1, 2, \dots , {2}^{k-1}$ and $k$ represents the level of resolution $k=1, 2, \dots$, respectively. The quantity $\frac{1}{\sqrt{{S}_{m}}}$ is a normalization factor. The Fibonacci polynomials are defined as follows in the form of the recurrence relation for every $x\in {R}^{+}$:

$${P}_{m+2}(x)=x{P}_{m+1}(x)+{P}_{m}(x),\quad \forall m\ge 0,$$

with initial conditions ${P}_{0}(x)=1, { P}_{1}(x)=x$.

Fibonacci wavelets are compactly supported wavelets formed by Fibonacci polynomials over the interval [0,1].

Basic Definitions [46, 47]:

Convergence: A real sequence $\{{a}_{n}\}$ converges to a limit $a$ if for given $\epsilon >0$, there exists $N\in {\mathbb{N}}$ such that if $n>N$ then $\left|{a}_{n}-a\right|<\epsilon .$

Uniform Convergence: Let $E$ be a nonempty subset of ${\mathbb{R}}$. A sequence of functions ${f}_{n}:E\to {\mathbb{R}}$ is said to converge uniformly on $E$ to a function $f$ if and only if for every $\epsilon >0$ there is an $N\in {\mathbb{N}}$ such that

$$n\ge {\mathbb{N}}\text{ implies }\left|{f}_{n}(x)-f(x)\right|<\epsilon\quad \forall x\in E.$$

Absolute Continuity: A real-valued function $f$ defined on $[a,b]$ is said to be absolutely continuous on $[a,b]$ if, given $\epsilon >0,$ there is a $\delta >0$ such that

$$\sum_{i=1}^{n}\left|f\left({x}_{i}^{{\prime}}\right)-f({x}_{i})\right|<\epsilon \quad\text{whenever}\quad \sum_{i=1}^{n}\left({x}_{i}^{{\prime}}-{x}_{i}\right)<\delta .$$

Outer Measure: Let $A$ be any set of real numbers, then the outer measure of $A$ written as ${m}^{*}A$ is defined by ${m}^{*}A={\mathrm{inf}}_{A\subseteq \bigcup {I}_{n}}\sum_{n=1}^{\infty }l({I}_{n})$, where infimum is taken over the countable covering of $A$ by open intervals.

Theorem 1

[27] Let ${L}^{2}[0,1]$ be the Hilbert space generated by the Fibonacci wavelet basis. Let $\eta (x)$ be the continuous bounded function in ${L}^{2}[0,1]$. Then the Fibonacci wavelet expansion of $\eta (x)$ converges with it.

Proof

Let $\eta :[0,1]\to R$ be a continuous function and $\left|\eta (x)\right|\le \mu$, where $\mu$ be any real number. Then Fibonacci wavelet dilation of $y(x)$ can be expressed as,

$$\eta (x)=\sum_{n=1}^{{2}^{\frac{k -1}{2}}}\sum_{m=0}^{M-1}{ a}_{n,m}{ \varphi }_{n,m} (x).$$

${a}_{n,m}-=\langle \eta ( x ),{ \varphi }_{n,m} ( x )\rangle$ denotes inner product:

$${a}_{n,m}-={\int }_{0}^{1}\eta \left( x \right){ \varphi }_{n,m}\left( x \right) dx.$$

Since ${\varphi }_{n,m}$ are the orthogonal basis:

$${a}_{n,m}={\int }_{I}\eta (x){F}_{m}\left({2}^{k-1}x-n+ 1\right)dx \quad\text{where }I=\left[\frac{n -1}{{2}^{k-1}}\right., \left.\frac{n}{{2}^{k-1}}\right).$$

Then substitute ${2}^{k-1}x-n+1=y$ then we get,

$${a}_{n,m}={2}^{\frac{k-1}{2}}{\int }_{0}^{1}\eta \left(\frac{y + n-1}{{2}^{k-1}}\right){F}_{m}(y)\frac{dy}{{2}^{k-1}}$$

$${a}_{n,m} ={ 2 }^{\frac{1-k}{2}}{\int }_{.0}^{1}\eta \left(\frac{y + n-1}{{2 }^{k-1}}\right) \sqrt{\int\limits_0^1{S}_{m}^{2}\left(\frac{y+ n.-1}{{2 }^{k-1}}\right)\frac{dy}{{2}^{k-1}}}{\beta }_{m}\left(y\right) dy.$$

By generalized mean value theorem,

$${a}_{n,m} =\frac{{2 }^{\frac{-k+1}{2}}}{\sqrt{\frac{(-1{)}^{ m-1}(m !{)}^{2}{ \alpha }_{ 2m}}{(2m)!}}}\eta \left(\frac{\xi +n-1}{{2 }^{k- 1}}\right){\int }_{0}^{1}{\beta }_{m}\left( y \right)dy \quad\text{for some } \xi \in \left(0,1\right).$$

Since ${\beta }_{m}(y)$ is a bounded continuous function. Put ${\int }_{0}^{1}{\beta }_{m}(y)dy=h$

$$\left|{a}_{n,m}\right| =\left|\frac{{2}^{\frac{-k+1}{2}}}{\sqrt{\frac{(-1{)}^{ m-1}(m !{)}^{2}{\alpha }_{2m}}{(2 m)!}}}\right|\left|\eta \left(\frac{\xi +n- 1}{{2}^{k-1}}\right)\right|h.$$

Since $\eta$ remains bounded.

Hence, $\left|{a}_{n,m}\right| =\left|\frac{{2 }^{\frac{-k+1}{2}}\mu h}{\sqrt{\frac{(-1{) }^{m-1}(m !{)}^{2}{ \alpha }_{2m}}{(2m)!}}}\right|.$

Therefore, ${\sum }_{n,m=0}^{\infty }{a}_{n,m}$ is absolutely convergent. Hence the Fibonacci wavelet series expansion $\eta (x)$ converges uniformly to it.

Theorem 2

[27] Let $I\subset R$ be a finite interval with length $m(I)$. Furthermore, $f(x)$ is an integrable function defined on $I$ and $\sum_{i=0}^{M-1}\sum_{j=1}^{{2}^{k-1}}{a}_{i,j} {\varphi }_{i,j}(x)$ be a good Fibonacci wavelet approximation of $f$ on $I$ with for some $\epsilon >0$, $\left|f(x)-\sum_{i=0}^{M-1}\sum_{j=1}^{{2}^{k-1}}{a}_{i,j} {\varphi }_{i,j}(x)\right|\le \epsilon , \forall x \in I$. Then $-\epsilon m\left(I\right)+{\int }_{I}\sum \sum {a}_{i,j} {\varphi }_{i,j}(x) dx \le$ ${\int }_{I}f(x)dx \le \epsilon m\left(I\right)+{\int }_{I}\sum \sum {a}_{i,j} {\varphi }_{i,j}(x) dx$.

3 Operational matrix of integration (OMI)

At k = 1 and M = 10, the Fibonacci wavelet basis is examined as follows:

$${\varphi }_{1,0}(x)=1,$$

$${\varphi }_{1,1}\left( x \right)=\sqrt{3}x,$$

$${\varphi }_{1,2}\left( x \right)=\frac{1}{2}\sqrt{\frac{15}{7}} (1+{x}^{2}),$$

$${\varphi }_{1,3}(x)=\sqrt{\frac{105}{239}} x(2+{x}^{2}),$$

$${\varphi }_{1,4}(x)=3\sqrt{\frac{35}{1943}} (1+3{x}^{2}+{x}^{4}),$$

$${\varphi }_{1,5}(x)=\frac{3}{4}\sqrt{\frac{385}{2582}} x(3+4{x}^{2}+{x}^{4}),$$

$${\varphi }_{1,6}(x)=3\sqrt{\frac{5005}{1268209}} (1+6{x}^{2}+5{x}^{4}+{x}^{6}),$$

$${\varphi }_{1,7}(x)=3\sqrt{\frac{5005}{2827883}} x(4+10{x}^{2}+6{x}^{4}+{x}^{6}),$$

$${\varphi }_{1,8}(x)=\frac{3}{2}\sqrt{\frac{85085}{28195421}} (1+10{x}^{2}+15{x}^{4}+7{x}^{6}+{x}^{8}),$$

$${\varphi }_{1,9}(x)=3\sqrt{\frac{1616615}{5016284989}} x(5+20{x}^{2}+21{x}^{4}+8{x}^{6}+{x}^{8}),$$

$${\varphi }_{1,10}(x)=3\sqrt{\frac{1616615}{11941544471}} (1+15{x}^{2}+35{x}^{4}+28{x}^{6}+9{x}^{8}+{x}^{10}),$$

$${\varphi }_{1,11}(x)=\frac{3}{8}\sqrt{\frac{37182145}{10276002038}} x\left(6+35{x}^{2}+56{x}^{4}+36{x}^{6}+10{x}^{8}+{x}^{10}\right),$$

where,

$${\varphi }_{10}(x)={[{\varphi }_{1,0}(x),{\varphi }_{1,1} \left(x \right),{\varphi }_{1,2 }\left(x \right),{\varphi }_{1,3}\left(x \right),{\varphi }_{1,4} \left(x \right),{\varphi }_{1,5}\left(x \right),{\varphi }_{1,6}\left(x \right),{\varphi }_{1,7}(x),{\varphi }_{1,8}(x),{\varphi }_{1,9}(x)]}^{T}.$$

Integrating the first ten bases described above for the range of $x$ limits from 0 to $x$ in the form of a linear combination of Fibonacci wavelet basis. We obtain,

$$\underset{0}{\overset{x}{\int }}{\varphi }_{1, 0}(x)dx=\left[\begin{array}{cccccccccc}0& \frac{1}{\sqrt{3}}& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{1,1}(x)dx=\left[\begin{array}{cccccccccc}-\frac{\sqrt{3}}{2}& 0& \sqrt{\frac{7}{5}} & 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,2}}(x)dx=\left[\begin{array}{cccccccccc}0& \frac{\sqrt{5}}{6\sqrt{7}}& 0& \frac{\sqrt{239}}{42}& 0& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,3}}(x)dx=\left[\begin{array}{cccccccccc}-\frac{\sqrt{105}}{2\sqrt{239}}& 0& \frac{7}{2\sqrt{239}}& 0& \frac{\sqrt{1943}}{4\sqrt{717}}& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,4}}(x)dx=\left[\begin{array}{cccccccccc}0& 0& 0& \frac{\sqrt{717}}{5\sqrt{1943}}& 0& \frac{4\sqrt{2582}}{5\sqrt{21373}}& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,5}}(x)dx=\left[\begin{array}{cccccccccc}-\frac{\sqrt{385}}{4\sqrt{2582}}& 0& 0& 0& \frac{\sqrt{21373}}{24\sqrt{2582}}& 0& \frac{\sqrt{1268209}}{24\sqrt{33566}}& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,6}}(x)dx=\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& \frac{4\sqrt{33566}}{7\sqrt{1268209}}& 0& \frac{\sqrt{2827883}}{7\sqrt{1268209}}& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,7}}(x)dx=\left[\begin{array}{cccccccccc}-\frac{3\sqrt{5005}}{4\sqrt{2827883}}& 0& 0& 0& 0& 0& \frac{\sqrt{1268209}}{8\sqrt{2827883}}& 0& \frac{\sqrt{28195421}}{4\sqrt{48074011}}& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,8}}(x)dx=\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{48074011}}{18\sqrt{28195421}}& 0& \frac{\sqrt{5016284989}}{18\sqrt{535712999}}\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}{\varphi }_{{1,9}}(x)dx=\left[\begin{array}{cccccccccc}-3\sqrt{\frac{323323}{25081424945}} & 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{535712999}}{5\sqrt{5016284989}}& 0\end{array}\right]{\varphi }_{10}(x)+\frac{\sqrt{11941544471}}{10\sqrt{5016284989}}{\varphi }_{{1,10}}(x).$$

Hence,

$$\underset{0}{\overset{x}{\int }}\varphi (x)dx={|\!\!\!\text{B}}_{10\times 10}{\varphi }_{10}(x)+\overline{{\varphi }_{10}(x)}$$

(6)

where,

$${|\!\!\!\text{B}}_{10\times 10}=\left[\begin{array}{cccccccccc}0& \frac{1}{\sqrt{3}}& 0& 0& 0& 0& 0& 0& 0& 0\\ -\frac{\sqrt{3}}{2}& 0& \sqrt{\frac{7}{5}} & 0& 0& 0& 0& 0& 0& 0\\ 0& \frac{\sqrt{5}}{6\sqrt{7}}& 0& \frac{\sqrt{239}}{42}& 0& 0& 0& 0& 0& 0\\ -\frac{\sqrt{105}}{2\sqrt{239}}& 0& \frac{7}{2\sqrt{239}}& 0& \frac{\sqrt{1943}}{4\sqrt{717}}& 0& 0& 0& 0& 0\\ 0& 0& 0& \frac{\sqrt{717}}{5\sqrt{1943}}& 0& \frac{4\sqrt{2582}}{5\sqrt{21373}}& 0& 0& 0& 0\\ -\frac{\sqrt{385}}{4\sqrt{2582}}& 0& 0& 0& \frac{\sqrt{21373}}{24\sqrt{2582}}& 0& \frac{\sqrt{1268209}}{24\sqrt{33566}}& 0& 0& 0\\ 0& 0& 0& 0& 0& \frac{4\sqrt{33566}}{7\sqrt{1268209}}& 0& \frac{\sqrt{2827883}}{7\sqrt{1268209}}& 0& 0\\ -\frac{3\sqrt{5005}}{4\sqrt{2827883}}& 0& 0& 0& 0& 0& \frac{\sqrt{1268209}}{8\sqrt{2827883}}& 0& \frac{\sqrt{28195421}}{4\sqrt{48074011}}& 0\\ 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{48074011}}{18\sqrt{28195421}}& 0& \frac{\sqrt{5016284989}}{18\sqrt{535712999}}\\ -3\sqrt{\frac{323323}{25081424945}} & 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{535712999}}{5\sqrt{5016284989}}& 0\end{array}\right],$$

$$\overline{{\varphi }_{10}(x)}=\left[\begin{array}{c}\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \frac{\sqrt{11941544471}}{10\sqrt{5016284989}}{\varphi }_{{1,10}}(x)\end{array}\end{array}\right].$$

Again, integrating the above ten basis, we obtain,

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,0}}(x)dxdx=\left[\begin{array}{cccccccccc}-\frac{1}{2}& 0& \sqrt{\frac{7}{15}} & 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,1}}(x)dxdx=\left[\begin{array}{cccccccccc}0& -\frac{1}{3}& 0& \frac{\sqrt{239}}{6\sqrt{35}}& 0& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,2}}(x)dxdx=\left[\begin{array}{cccccccccc}-\frac{\sqrt{5}}{2\sqrt{21}}& 0& \frac{1}{4}& 0& \frac{\sqrt{1943}}{168\sqrt{3}}& 0& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,3}}(x)dxdx=\left[\begin{array}{cccccccccc}0& -\frac{5\sqrt{35}}{12\sqrt{239}}& 0& \frac{2}{15}& 0& \frac{\sqrt{2582}}{5\sqrt{7887}}& 0& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,4}}(x)dxdx=\left[\begin{array}{cccccccccc}-\frac{\sqrt{35}}{2\sqrt{1943}}& 0& \frac{7\sqrt{3}}{10\sqrt{1943}}& 0& \frac{1}{12}& 0& \frac{\sqrt{1268209}}{30\sqrt{277849}}& 0& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,5}}(x)dxdx=\left[\begin{array}{cccccccccc}0& -\frac{\sqrt{385}}{4\sqrt{7746}}& 0& \frac{\sqrt{2629}}{40\sqrt{7746}}& 0& \frac{2}{35}& 0& \frac{\sqrt{2827883}}{168\sqrt{33566}}& 0& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,6}}(x)dxdx=\left[\begin{array}{cccccccccc}-\frac{\sqrt{5005}}{4\sqrt{1268209}}& 0& 0& 0& \frac{\sqrt{277849}}{42\sqrt{1268209}}& 0& \frac{1}{24}& 0& \frac{\sqrt{28195421}}{28\sqrt{21559553}}& 0\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,7}}(x)dxdx=\left[\begin{array}{cccccccccc}0& -\frac{\sqrt{15015}}{4\sqrt{2827883}}& 0& 0& 0& \frac{\sqrt{16783}}{7\sqrt{5655766}}& 0& \frac{2}{63}& 0& \frac{\sqrt{5016284989}}{72\sqrt{913406209}}\end{array}\right]{\varphi }_{10}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,8}}(x)dxdx=\left[\begin{array}{cccccccccc}-\frac{3\sqrt{17017}}{8\sqrt{140977105}}& 0& 0& 0& 0& 0& \frac{\sqrt{21559553}}{144\sqrt{28195421}}& 0& \frac{1}{40}& 0\end{array}\right]{\varphi }_{10}(x)+\frac{\sqrt{11941544471}}{180\sqrt{535712999}}{\varphi }_{{1,10}}(x),$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{x}{\int }}{\varphi }_{{1,9}}(x)dxdx=\left[\begin{array}{cccccccccc}0& -\sqrt{\frac{969969}{25081424945}} & 0& 0& 0& 0& 0& \frac{\sqrt{913406209}}{90\sqrt{5016284989}}& 0& \frac{2}{99}\end{array}\right]{\varphi }_{10}(x)+\frac{4\sqrt{10276002038}}{55\sqrt{115374554747}}{\varphi }_{{1,11}}(x).$$

Hence,

$$\int_0^{x}\int_0^{x}\varphi (x)dxdx={|\!\!\!\text{B}}^{{\prime}}{\varphi }_{10}(x)+\overline{\varphi }{^{\prime}}_{10}(x)$$

(7)

where

$${|\!\!\!\text{B}}_{10\times 10}^{{\prime}}=\left[\begin{array}{cccccccccc}-\frac{1}{2}& 0& \sqrt{\frac{7}{15}} & 0& 0& 0& 0& 0& 0& 0\\ 0& -\frac{1}{3}& 0& \frac{\sqrt{239}}{6\sqrt{35}}& 0& 0& 0& 0& 0& 0\\ -\frac{\sqrt{5}}{2\sqrt{21}}& 0& \frac{1}{4}& 0& \frac{\sqrt{1943}}{168\sqrt{3}}& 0& 0& 0& 0& 0\\ 0& -\frac{5\sqrt{35}}{12\sqrt{239}}& 0& \frac{2}{15}& 0& \frac{\sqrt{2582}}{5\sqrt{7887}}& 0& 0& 0& 0\\ -\frac{\sqrt{35}}{2\sqrt{1943}}& 0& \frac{7\sqrt{3}}{10\sqrt{1943}}& 0& \frac{1}{12}& 0& \frac{\sqrt{1268209}}{30\sqrt{277849}}& 0& 0& 0\\ 0& -\frac{\sqrt{385}}{4\sqrt{7746}}& 0& \frac{\sqrt{2629}}{40\sqrt{7746}}& 0& \frac{2}{35}& 0& \frac{\sqrt{2827883}}{168\sqrt{33566}}& 0& 0\\ -\frac{\sqrt{5005}}{4\sqrt{1268209}}& 0& 0& 0& \frac{\sqrt{277849}}{42\sqrt{1268209}}& 0& \frac{1}{24}& 0& \frac{\sqrt{28195421}}{28\sqrt{21559553}}& 0\\ 0& -\frac{\sqrt{15015}}{4\sqrt{2827883}}& 0& 0& 0& \frac{\sqrt{16783}}{7\sqrt{5655766}}& 0& \frac{2}{63}& 0& \frac{\sqrt{5016284989}}{72\sqrt{913406209}}\\ -\frac{3\sqrt{17017}}{8\sqrt{140977105}}& 0& 0& 0& 0& 0& \frac{\sqrt{21559553}}{144\sqrt{28195421}}& 0& \frac{1}{40}& 0\\ 0& -\sqrt{\frac{969969}{25081424945}} & 0& 0& 0& 0& 0& \frac{\sqrt{913406209}}{90\sqrt{5016284989}}& 0& \frac{2}{99}\end{array}\right],$$

$$\overline{\varphi }{^{\prime}}_{10}(x)=\left[\begin{array}{c}\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \frac{\sqrt{11941544471}}{180\sqrt{535712999}}{\varphi }_{{1,10}}(x)\\ \frac{4\sqrt{10276002038}}{55\sqrt{115374554747}}{\varphi }_{{1,11}}(x)\end{array}\end{array}\right].$$

In the same way, we can generate matrices of different sizes for our handiness.

4 Fibonacci wavelet method

Consider the two non-linear reactions in normalized form

$$\left.\begin{array}{l}{y}^{\prime\prime}\left(x \right)=\frac{{\eta }_{1}y (x) z (x)}{1+{\xi }_{1} y(x)+{\xi }_{2 }z(x)}\\ {z}^{\prime\prime}\left(x \right)=\frac{{\eta }_{2}y(x)z(x)}{1+{\xi }_{1}y(x)+{\xi }_{2}z(x)}\end{array}\right\}$$

(8)

and boundary conditions will become:

$$y\left(0\right)=\alpha , y\left(1\right)=\beta , {z}^{{\prime}}\left(0\right)=\gamma , z\left(1\right)=\mu$$

(9)

where ${\eta }_{1}, {\eta }_{2}, {\xi }_{1},{\xi }_{2}$ are normalized arguments, $z(x)$ denotes the condensation of PGE and $y(x)$ denotes the condensation of $C{O}_{2}$. Here, we would like to bring the solution of system (8) that is extracted from the $C{O}_{2}$ absorbed into the PGE problem (1) in the Fibonacci wavelet space.

Assume that,

$$\frac{{d}^{2}y}{{d}^{2}x}\approx {A}^{T}\varphi (x)$$

(10)

$$\frac{{d}^{2}z}{{d}^{2}x}{\approx B}^{T}\varphi (x)$$

(11)

where, ${A}^{T}=[ a_{{1,0}},\ldots {a}_{1, M-1}, a_{{2,0}},\ldots{a}_{2,M-1},a_{{2}^{k-1},0},\ldots a_{{2}^{k-1},M-1}],$

$${B}^{T}=[{ b}_{{1,0}},\ldots {b}_{1,M-1}, {b}_{{2,0}},\ldots {b}_{2,M-1},{b}_{{2}^{k-1},0},\ldots {b}_{{2}^{k-1},M-1}],$$

$$\varphi (x)=[\varphi (x)_{{1,0}},\ldots \varphi (x)_{1,M-1},\varphi (x)_{{2,0}},\ldots\varphi (x)_{2,M-1},\varphi (x)_{{2}^{k-1},0},\ldots\varphi (x)_{{2}^{k-1},M-1}].$$

Integrating the Eqs. (10) and (11) with respect to ‘$x$’ from ‘$0$’ to ‘$x$’. We get

$$y{^{\prime}}(x)\approx y{^{\prime}}(0)+\underset{0}{\overset{x}{\int }}{A}^{T}\varphi (x)dx$$

$${z}^{{\prime}}(x)\approx {\mathrm{z}}^{{\prime}}\left(0\right)+\underset{0}{\overset{x}{\int }}{B}^{T}\varphi (x)dx.$$

Approximating ${z}^{{\prime}}\left(0\right)=\gamma$ by the Fibonacci wavelets as ${z}^{{\prime}}\left( 0 \right)={C}^{T}\varphi (x)$, we get

$$\left.\begin{array}{l}{y}^{{\prime}}(x) \approx y{^{\prime}}(0) +{A}^{T}[\,|\!\!\text{B} \varphi (x) + \overline{\varphi }(x)]\\ {z}^{{\prime}}(x) \approx {C}^{T}\varphi (x) +{B}^{T}[\,|\!\!\text{B} \varphi (x{^{\prime}}) + \overline{\varphi }(x)]\end{array}\right\}.$$

(12)

Integrating Eq. (12) concerning ‘$x$’ between ‘$0$’ and ‘$x$’ and approximating $y(0)=\alpha$ by the Fibonacci wavelets as $y\left(0\right)={D}^{T}\varphi (x).$ We obtain

$$\left.\begin{array}{l}y(x)\approx {D}^{T}\varphi (x)+\text{x} \text{y}{^{\prime}}(0) +{A}^{T}[\,|\!\!\text{B} \varphi {^{\prime}}(x) + \overline{\varphi }{^{\prime}}(x)]\\ z(x)\approx z\left(0\right)+\text{x} {C}^{T}\varphi (x) +{B}^{T}[\,|\!\!\text{B} \varphi {^{\prime}}(x) + \overline{\varphi }{^{\prime}}(x)]\end{array}\right\}.$$

(13)

We again express $y\left(1\right)$ and $z\left(1\right)$ in the Fibonacci wavelet as $y\left(1\right)={E}^{T}\varphi (x)$ and $z\left( 1 \right)={F}^{T}\varphi (x):$

$$\left.\begin{array}{l}y{^{\prime}}\left(0\right)\approx {E}^{T}\varphi (x)-{D}^{T}\varphi (1) -{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + \overline{\varphi }{^{\prime}}\left(1\right)\right]\\ z\left(0\right)\approx {F}^{T}\varphi (x)-{C}^{T}[\,|\!\!\text{B} \varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)] -{B}^{T}[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)]\end{array}\right\}.$$

(14)

Substituting Eqs. (14) into (13), we obtain

$$\left.\begin{array}{l}y(x)\approx {{D}^{T}\varphi (x)+x(E}^{T}\varphi (x)-{D}^{T}\varphi (1)-{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}} \left(1\right)\right])+{A}^{T}\left[{|\!\!\!\text{B}}^{{\prime}}\varphi (x) + \overline{\varphi }(x)\right]\\ z\left( x \right)\approx {F}^{T}\varphi (x)-{C}^{T}[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)]-{B}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}} \varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)\right]+{B}^{T}\left[\,|\!\!\text{B} \varphi (x)+ \overline{\varphi } (x)\right]\end{array}\right\}.$$

(15)

Now collocate (8) after substituting the above equations by following grid points ${x}_{i}=\frac{2i-1}{{2}^{k}M},i= 1, 2\ldots M$. The system with ${2}^{k} M$ algebraic equations is obtained as shown below,

$$\left.\begin{array}{c}{A}^{T}\varphi ({x}_{i}) =\frac{{\eta }_{1}({{D}^{T}\varphi ({x}_{i})+{x}_{i}(E}^{T}\varphi \left({x}_{i}\right)-{D}^{T}\varphi (1)-{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left(1\right)+ {\overline{\varphi } }^{{\prime}} \left(1\right)\right])+{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left({x}_{i}\right)+ \overline{\varphi }\left({x}_{i}\right)\right]) ({F}^{T}\varphi ({x}_{i})-{C}^{T}[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)]-{B}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}} \varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)\right]+{B}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+ \overline{\varphi } \left({x}_{i}\right)\right]) }{1+{\xi }_{1}({{D}^{T}\varphi ({x}_{i})+{x}_{i}(E}^{T}\varphi \left({x}_{i}\right)-{D}^{T}\varphi (1)-{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left(1\right)+ {\overline{\varphi } }^{{\prime}} \left(1\right)\right])+{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left({x}_{i}\right)+ \overline{\varphi }\left({x}_{i}\right)\right]) +{\xi }_{2 }({F}^{T}\varphi ({x}_{i})-{C}^{T}[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)]-{B}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}} \varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)\right]+{B}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+ \overline{\varphi } \left({x}_{i}\right)\right])}\\ {B}^{T}\varphi ({x}_{i}) =\frac{{\eta }_{2}({{D}^{T}\varphi ({x}_{i})+{x}_{i}(E}^{T}\varphi \left({x}_{i}\right)-{D}^{T}\varphi (1)-{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left(1\right)+ {\overline{\varphi } }^{{\prime}} \left(1\right)\right])+{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left({x}_{i}\right)+ \overline{\varphi }\left({x}_{i}\right)\right]) ({F}^{T}\varphi ({x}_{i})-{C}^{T}[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)]-{B}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}} \varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)\right]+{B}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+ \overline{\varphi } \left({x}_{i}\right)\right]) }{1+{\xi }_{1}({{D}^{T}\varphi ({x}_{i})+{x}_{i}(E}^{T}\varphi \left({x}_{i}\right)-{D}^{T}\varphi (1)-{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left(1\right)+ {\overline{\varphi } }^{{\prime}} \left(1\right)\right])+{A}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}}\varphi \left({x}_{i}\right)+ \overline{\varphi }\left({x}_{i}\right)\right]) +{\xi }_{2 }({F}^{T}\varphi ({x}_{i})-{C}^{T}[\,{|\!\!\!\text{B}}^{{\prime}}\varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)]-{B}^{T}\left[\,{|\!\!\!\text{B}}^{{\prime}} \varphi (1) + {\overline{\varphi } }^{{\prime}}\left(1\right)\right]+{B}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+ \overline{\varphi } \left({x}_{i}\right)\right])}\end{array}\right\}$$

Utilizing the Secant method, solve the aforementioned equations to determine the values of the unknown Fibonacci wavelet coefficients. The numerical solution of the system (8) using the Fibonacci wavelet is obtained by substituting these coefficient values into (13).

Next, we will consider the system of ODEs, which represents the Chemical kinetic problem as follows:

$$\left.\begin{array}{l}{u}_{1}^{{\prime}}(x)=-{b}_{1}{u}_{1}(x)+{b}_{2}{ u}_{2}{(x) u}_{3}(x)\\ {u}_{2}^{{\prime}} (x)={b}_{1}{u}_{1}(x)-{b}_{2} {u}_{2}(x) {u}_{3}(x)-{b}_{3} {u}_{2}\, {(x)}^{2}\\ {u}_{3}^{{\prime}} (x)={b}_{3} {u}_{2}{ (x)}^{2}\end{array}\right\}$$

(16)

$$\text{and the initial conditions are}:{u}_{1}\left(0\right)={\delta }_{1}, {u}_{2}\left(0\right)={\delta }_{2}, {u}_{3}\left(0\right)={\delta }_{3}.$$

(17)

We want to bring the solution of the SODEs in the Fibonacci wavelet space that is extracted from a chemical kinetic problem (5).

Assume that,

$${u}_{1}{^{\prime}}(x)\approx {A}^{T}\varphi (x)$$

(18)

$${u}_{2}{^{\prime}}(x)\approx {B}^{T}\varphi (x)$$

(19)

$${u}_{3}{^{\prime}}(x)\approx {C}^{T}\varphi (x)$$

(20)

where, ${A}^{T}=[ {a}_{{1,0}},\ldots {a}_{1, M-1},{ a}_{{2,0}},\ldots {a}_{2,M-1},{a}_{{2}^{k-1},0},\ldots {a}_{{2}^{k-1},M-1}],$

$${B}^{T}=[{ b}_{{1,0}},\ldots {b}_{1,M-1}, {b}_{{2,0}},\ldots {b}_{2,M-1},{b}_{{2}^{k-1},0},\ldots {b}_{{2}^{k-1},M-1}],$$

$${C}^{T}=[ {c}_{1,0},\ldots {c}_{1,M-1},{c}_{2,0},\ldots {c}_{2,M-1},{c}_{{2}^{k-1},0},\ldots {c}_{{2}^{k-1},M-1}],$$

$$\varphi (x)=[\varphi (x)_{{1,0}},\ldots \varphi (x)_{1,M-1},\varphi (x)_{{2,0}},\ldots\varphi (x)_{2,M-1},\varphi (x)_{{2}^{k-1},0},\ldots\varphi \left(x)_{{2}^{k-1},M-1}\right].$$

We obtain the following set of equations on integrating the Eqs. (18)–(20) with respect ‘$x$’ from ‘$0$’ to ‘$x$’.

$${u}_{1}(x)\approx {u}_{1}\left(0\right)+\int_0^x {A}^{T}\varphi (x)dx$$

$${u}_{2}(x)\approx {u}_{2}\left(0\right)+\int_0^{x}{B}^{T}\varphi (x)dx$$

$${u}_{3}(x)\approx {u}_{3}\left(0\right)+\int_0^{x}{C}^{T}\varphi (x)dx.$$

Using Eq. (6) and physical constraints in Eq. (17) formulated in terms of $\theta (x)$. We obtain,

$$\left.\begin{array}{l}{u}_{1}(x)\approx {D}^{T}\varphi (x) +{A}^{T}[\,|\!\!\text{B} \varphi (x) +\overline{\varphi }(x)]\\ {u}_{2}(x)\approx {E}^{T}\varphi (x) +{B}^{T}[\,|\!\!\text{B} \varphi (x) + \overline{\varphi } (x)]\\ {u}_{3}(x)\approx {F}^{T}\varphi (x) +{C}^{T}[\,|\!\!\text{B} \varphi (x) + \overline{\varphi }(x)]\end{array} \right\}$$

(21)

where F, D, and E are the known vectors. On Substituting Eqs. (18), (19), (20), and (21) in (16), we obtain,

$$\left.\begin{array}{l}{A}^{T}\varphi (x)\approx -{b}_{1}\left( {D}^{T}\varphi (x) +{A}^{T}\left[\,|\!\!\text{B} \varphi (x)+\overline{\varphi } (x)\right]\right)+{b}_{2} \left({E}^{T}\varphi (x) +{B}^{T}\left[\,|\!\!\text{B} \varphi (x) + \overline{\varphi }(x)\right]\right)\\ \left({F}^{T}\varphi (x) +{C}^{T}\left[\,|\!\!\text{B} \varphi (x) + \overline{\varphi }(x)\right]\right)\\ \\ {B}^{T}\varphi (x)\approx {b}_{1}\left( {D}^{T}\varphi (x) +{A}^{T}\left[\,|\!\!\text{B} \varphi (x) +\overline{\varphi }(x)\right]\right)-{b}_{2} \left({E}^{T}\varphi (x)+{B}^{T}\left[\,|\!\!\text{B} \varphi (x) +\overline{\varphi }(x)\right]\right)\\ \left({F}^{T}\varphi (x) +{C}^{T}\left[\,|\!\!\text{B} \varphi (x ) + \overline{\varphi }(x)\right]\right)-{b}_{3} {\left({E}^{T}\varphi (x) +{B}^{T}\left[\,|\!\!\text{B} \varphi (x)+ \overline{\varphi }(x)\right]\right)}^{2}\\ {C}^{T}\varphi (x)\approx {b}_{3} {\left({E}^{T}\varphi (x) +{B}^{T}\left[\,|\!\!\text{B} \varphi (x) + \overline{\varphi }(x)\right]\right)}^{2}\end{array}\right\}.$$

(22)

On collocating all the equations of (22) by using the grid points ${x}_{i}=\frac{2i-1}{{2}^{k}M},i= 1, 2\ldots M$, we arrive the system with ${2}^{k} M$ algebraic equations as below:

$$\left.\begin{array}{l}{A}^{T}\varphi ({x}_{i})\approx -{b}_{1}\left( {D}^{T}\varphi ({x}_{i}) +{A}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+\overline{\varphi }\left({x }_{i}\right)\right]\right)+{b}_{2} \left({E}^{T}\varphi ({x}_{i}) +{B}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+\overline{\varphi }\left({x}_{i}\right)\right]\right)\\ \left({F}^{T}\varphi ({x}_{i}) +{C}^{T}\left[\,|\!\!\text{B} \varphi ({x}_{i}) + \overline{\varphi }\left({x}_{i}\right)\right]\right)\\ \\ {B}^{T}\varphi ({x}_{i})\approx {b}_{1}\left( {D}^{T}\varphi ({x}_{i}) +{A}^{T}\left[\,|\!\!\text{B} \varphi ({x}_{i}) +\stackrel{-}{\varphi}\left({x}_{i}\right)\right]\right)-{b}_{2} \left({E}^{T}\varphi ({x}_{i}) +{B}^{T}\left[\,|\!\!\text{B} \varphi ({x}_{i}) +\overline{\varphi }\left({x}_{i}\right)\right]\right)\\ \left({F}^{T}\varphi ({x}_{i}) +{C}^{T}\left[\,|\!\!\text{B} \varphi ({x}_{i}) + \overline{\varphi }\left({x}_{i}\right)\right]\right)-{b}_{3} {\left({E}^{T}\varphi ({x}_{i}) +{B}^{T}\left[\,|\!\!\text{B} \varphi \left({x}_{i}\right)+ \overline{\varphi }\left({x}_{i}\right)\right]\right)}^{2}\\ {C}^{T}\varphi ({x}_{i})\approx {b}_{3} {\left({E}^{T}\varphi ({x}_{i}) +{B}^{T}\left[\,|\!\!\text{B} \varphi ({x}_{i}) + \overline{\varphi }\left({x}_{i}\right)\right]\right)}^{2}\end{array}\right\}.$$

(23)

Utilizing the Newton–Raphson/Secant method, solve the aforementioned equations to determine the values of the unknown Fibonacci wavelet coefficients. The numerical solution of the system (16) using the Fibonacci wavelet is obtained by substituting these coefficient values into (21).

5 Numerical results and discussion

Example 5.1

The non-linear SODEs of the $C{O}_{2}$ absorbed into phenyl glycidyl ether problem is of the form

$$\left.\begin{array}{l}{y}^{\prime\prime}(x)=\frac{{\eta }_{1}y(x)z(x)}{1+{\xi }_{1}y(x)+{\xi }_{2}z(x)}\\ {z}^{\prime\prime}(x)=\frac{{\eta }_{2}y(x)z(x)}{1+{\xi }_{1}y(x)+{\xi }_{2}z(x)}\end{array}\right\}$$

(24)

$$\text{and the boundary conditions will be}:\ y\left(0\right)=\alpha , y\left(1\right)=\beta , {z}^{{\prime}}\left(0\right)=\gamma , z\left(1\right)=\mu$$

(25)

Case 1: ${\eta }_{1}=1, {\eta }_{2}=2, {\xi }_{1}=1, {\xi }_{2}=3, \alpha =1, \beta =0.5, \gamma =0,$ $\mu =1.$ Tables 1 and 2 and Figs. 1, 2, 3, and 4 depict all the numerical simulations and graphical results of this particular case. We have compared the proposed method with the optimal homotopy analysis method (OHAM) [17], the Adomian decomposition method (ADM) [17], and the ND Solve solution due to the non-availability of the exact solution. It can be observed that the FWCM solution is very much close to the NDSolve solution. The CPU time taken is 1.017 s for the two tables of values together.

Table 1 FWCM solution compared with the various methods for $y(x)$

Table 2 FWCM solution compared with the various methods for $z(x)$

Fig. 1

The plot of FWCM solution at k = 1 for $y(x)$ with different methods of solution

Fig. 2

A plot of FWCM solution at k = 1 for $z(x)$ with different methods of solution

Fig. 3

Absolute error (AE) comparison of $y(x)$ at k = 1 and M = 6

Fig. 4

Absolute error (AE) comparison of $z(x)$ at k = 1 and M = 6

Case 2: ${\eta }_{1}=2, {\eta }_{2}=2, {\xi }_{1}=1, {\xi }_{2}=3, \alpha =1, \beta =0.5, \gamma =0,$ $\mu =1.$ Tables 3 and 4 and Figs. 5, 6, 7 and 8 depict this case’s numerical simulations and graphical results. We have compared the proposed method (FWCM) with the Adomian Decomposition method (ADM), Residual method (RM), and ND solve solution due to the non-availability of the exact solution. It can be observed that the FWCM solution is very close to the NDSolve solution. The CPU time taken is 1.827 s for the two tables of values together.

Table 3 FWCM solution compared with the various methods for $y(x)$

Table 4 FWCM solution compared with the various methods for $z(x)$

Fig. 5

The plot of FWCM solution at k = 1 for $y(x)$ with different methods of solution

Fig. 6

The plot of FWCM solution at k = 1 for $z(x)$ with different methods of solution

Fig. 7

Absolute error (AE) comparison of $y(x)$ at k = 1 and M = 6 with different methods in the literature

Fig. 8

Absolute error (AE) comparison of $z(x)$ at k = 1 and M = 6 with different methods in the literature

Example 5.2

Next, Consider the system of ODEs of Chemical Kinetic Problem as follows:

$${u}_{1}^{{\prime}}(x)=-{b}_{1}{u}_{1}(x)+{b}_{2}{ u}_{2}{(x)u}_{3}(x)$$

$${u}_{2}^{{\prime}}(x)={b}_{1}{u}_{1}(x)-{b}_{2} {u}_{2}(x){u}_{3}(x)-{b}_{3} {u}_{2}{(x)}^{2}$$

$${u}_{3}^{{\prime}}(x)={b}_{3} {u}_{2}{(x)}^{2}.$$

And the physical constraints are: ${u}_{1}\left(0\right)=1, {u}_{2}\left(0\right)=0, {u}_{3}\left(0\right)=0.$

Tables 5, 6 and 7 and Figs. 9, 10, 11, 12, 13 and 14 depict this problem’s numerical simulations and graphical results. We have compared the proposed method (FWCM) solution with the Adam–Bashforth–Moultan method (ABM), Runge–Kutta method (RK4), and ND Solve solution due to the non-availability of the exact solution. It can be observed that the FWCM solution is very much close to the ND Solve solution. It is clear from the Tables and Graphs that the approximate solution obtained from the FWCM is close to the ND Solve solution, which shows the efficiency of the proposed method. Also, we got a better solution by increasing the value of M and k, which can be seen in Tables 8, 9 and 10. Comparison of AE with distinct values of M and k are presented in graphs 15–17. The CPU time taken is 0.891 s for all three tables of values together.

Table 5 FWCM solution compared with the various approaches for ${u}_{1}(x)$

Table 6 FWCM solution compared with the various approaches for ${u}_{2}(x)$

Table 7 FWCM solution compared with the various approaches for ${u}_{3}(x)$

Fig. 9

The plot of the FWCM solution at k = 1 for ${u}_{1}(x)$ with different methods of solution

Fig. 10

The plot of the FWCM solution at k = 1 for ${u}_{2}(x)$ with different methods of solution

Fig. 11

The plot of the FWCM solution at k = 1 for ${u}_{3}(x)$ with different methods of solution

Fig. 12

Absolute error (AE) comparison of ${u}_{1}(x)$ at k = 1 and M = 6 with different methods in the literature

Fig. 13

Absolute error (AE) comparison of ${u}_{2}(x)$ at k = 1 and M = 6 with distinct approaches in the literature

Fig. 14

Absolute error (AE) comparison of ${u}_{3}(x)$ at k = 1 and M = 6 with distinct approaches in the literature

Table 8 Comparison of Absolute error (AE) with different values of k and M for ${u}_{1}(x)$

Table 9 Comparison of Absolute error (AE) with different values of k and M for ${u}_{2}(x)$

Table 10 Comparison of Absolute error (AE) with different values of k and M for ${u}_{3}(x)$

Example 5.3

Perhaps we will consider one more system of nonlinear ODEs in which the exact solution is available to examine the performance of the novel technique FWCM [4]:

$${w}_{1}^{\prime\prime}(x)=x{w}_{1}^{{\prime}}(x)-{w}_{1}+{x}^{3}-2{x}^{2}+6x$$

$${w}_{2}^{\prime\prime}(x)=-x{w}_{1}^{{\prime}}-{w}_{1}{w}_{2}+{x}^{5}-{x}^{4}+2{x}^{3}+{x}^{2}-x+2$$

With boundary conditions: ${w}_{1}\left(0\right)=0, {w}_{1}\left(1\right)=0, {w}_{2}\left(0\right)=0, {w}_{2}\left(1\right)=0.$

The Exact solutions of the above system are: ${w}_{1}(x)={x}^{3}-x, {w}_{2}(x)={x}^{2}-x$. We compared the solutions obtained from the FWCM and ND Solve methods with the Exact solutions in Tables 11 and 12. Also, we brought the Absolute Errors of FWCM and NDSolve method with the exact solutions in the same tables. Graphical comparison of solutions and absolute errors are exposed in Graphs 18–21. Tables and Graphs reveal the precision and efficacy of the developed method. The CPU time taken is 0.594 s for the tables of values together.

Table 11 FWCM solution compared with the various approaches for ${w}_{1}(x)$

Table 12 FWCM solution compared with the various approaches for ${w}_{2}(x)$

6 Conclusion

In this article, we have implemented a new collocation approach called the Fibonacci wavelet collocation method (FWCM). The functional matrix of integration is obtained by using the Fibonacci wavelets. We solved two nonlinear SODEs representing the two familiar chemistry models using this matrix. The accuracy and proficiency of this novel technique were demonstrated by comparing the obtained solutions with other standard methods such as RK4 and ND Solve. Tables and Graphs reveal that the FWCM converges rapidly compared to other existing techniques (ADM, RM, HPM, and VIM) in the literature. Also, numerical illustrations support the claim that only a few Fibonacci wavelets are sufficient to obtain satisfactory results. Hence the proposed method is a very attractive and efficient numerical technique to solve different chemical, biological, and physical problems. From Tables 8, 9, and 10 and Figs. 15, 16, and 17, we can see that the values of k and M incresases the accuracy of the solution. The future recommendation of the proposed scheme is that it is suitable for solutions with sharp edge/jump discontinuities. By slightly modifying the method, the Fibonacci wavelet method can solve the higher-order system of ordinary differential equations and extend to PDEs and other mathematical models with different physical conditions. It is used to obtain the solution of the differential equation in the universal domain by taking the suitable transformation (Figs. 18, 19, 20, 21).

Fig. 15

AE comparison for ${u}_{1}(x)$ at various values of M and k

Fig. 16

AE comparison for ${u}_{2}(x)$ at various values of M and k

Fig. 17

AE comparison for ${u}_{3}(x)$ at various values of M and k

Fig. 18

The plot of the FWCM solution at k = 1 for ${w}_{1}(x)$ with exact and NDsolve solutions

Fig. 19

The plot of the FWCM solution at k = 1 for ${w}_{2}(x)$ with exact and NDsolve solutions

Fig. 20

Absolute error (AE) comparison of ${w}_{1}(x)$ at k = 1 and M = 6 with ND solve

Fig. 21

Absolute error (AE) comparison of ${w}_{2}(x)$ at k = 1 and M = 6 with ND solve