A New Framework for Numerical Techniques for Fuzzy Nonlinear Equations

Abstract

This paper describes a computational method for solving the nonlinear equations with fuzzy input parameters that we encounter in engineering system analysis. In addition to discussing the existence of solutions, the definition and formalization of numerical solutions is based on a new fuzzy computation operation as a transmission average. Error analysis in numerical solutions is described. Finally, some examples are presented to implement the proposed method and its effectiveness compared to other previous methods.

1. Introduction

When the equations are nonlinear, numerical estimation strategies are often required to deal with the system of equations. Solving a nonlinear system of equations is a problem to be avoided as much as possible, and nonlinear systems are usually approximated by a linear system of equations. However, when it is usually not appropriate, the problem must be directly elucidated and addressed. In most applications, nonlinear system parameters and measurements are not represented continuously with crisp numbers, but with fuzzy numbers. Therefore, it is necessary to study numerical methods for solving fuzzy nonlinear equations. The concept of fuzzy numbers stems from the fact that many measurable phenomena cannot be specified in a completely precise numerical form. Therefore, the study of fuzzy equations as a complete mathematical model for real-world problems with uncertainties has attracted the attention of many researchers. The rapid spread of new ideas has led to a proliferation of numerical methods for solving fuzzy nonlinear equations, such as: Newton’s approach to solving the equivalent fuzzy nonlinear equations [1], modified Newton’s methods [2,3,4], Jacobian update formulation for the Shamanskii method with singular Jacobians [5], the Levenberg–Marquardt approach [6], and Broden’s method [7].

Many of these methods are based on well-known explicit nonlinear equation methods, such as fixed point iteration [8], harmonic mean method [9], and Regula Falsi method [10]. However, the performance of these methods is affected by the calculation of the Jacobian or Jacobian approximation in all iterations. In addition, performance may suffer due to the Jacobian singularity [11,12] at the solution point. Several methods have been discussed in Refs. [13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36] to solve fractional differential equations, initial value problem, nonlinear equations, and linear equations. Numerical methods for fuzzy nonlinear equations are based on the following three principles:

• Using fuzzy operations based on the extension principle (in the context of membership functions) or interval calculations (in the context of $\alpha$-cuts);
• Transforms fuzzy nonlinear equations with $\alpha$- into crisp $2\times 2$ nonlinear systems;
• An implementation of one of the known numerical techniques in principle 2.

Using fuzzy operations or interval calculations based on the extension principle, we face the problem of subtraction and division operations. While the revised definition of subtraction and division allows the use of interval calculus for fuzzy operators, it is not efficient due to the dependent effects of the result support and the complexity of determining nonlinear membership functions, although it always exists. Therefore, the transmission average (TA)-based manipulation by Abbasi et al. [37] was applied in the proposed framework because of its unique properties that ultimately lead to more realistic support for the solution of the equations.

By including the kernel of the solution and the kernel of the constant fuzzy numbers in the fuzzy nonlinear equations when applying the new fuzzy arithmetic operation, we obtain a crisp $2\times 3$ nonlinear system of equations (i.e., the system of unknowns for the solution kernel, and its upper and lower membership functions). We will then obtain approximations to the solution kernel by considering $\alpha$ = 1 and applying known crisp numerical techniques, and also by fuzzing the nonlinear equations for the approximation to the solution kernel in the system of nonlinear equations. We obtain non-convergence in Approximation of the upper and lower membership functions of the approximate solution kernel. Finally, we construct upper and lower membership functions that approximate the solution through a generalization process. The strengths and superiority of the proposed method, as well as the different implementation of principles 1 and 2, will ultimately yield more acceptable responses than other previous methods, as shown in several examples. Finally, the error analysis of the proposed method is described and reviewed. In the newly proposed method, errors are introduced into the numerical solution of fuzzy nonlinear equations. It is logical to show the ingenuity of the relationship with respect to the error of the numerical technique used and the newly proposed error.

The rest of the text is organized as follows. Section 2 contains preliminaries and Section 3 covers new approaches to solving fuzzy nonlinear equations and error analysis. The last two sections, Section 4 and Section 5, include some illustrative examples for comparing the proposed method with other approaches, and concluding comments and recommendations in Section 5, respectively, with hints for future work.

2. Preliminaries

In this section, we cite some useful definitions that will be used in article.

Definition 1 

([38]). A fuzzy set S can be represented as,

$$\begin{array}{c}S={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .S_{\alpha },S_{\alpha }=[{\underline{S}}_{\alpha },{\overline{S}}_{\alpha }],\end{array}$$

(1)

where ${\underline{S}}_{\alpha }$ and ${\overline{S}}_{\alpha }$ are non-decreasing and non-increasing functions, respectively, of

$$\begin{array}{c}{S}_{\alpha }=\left\{x\right|{\mu }_S\left(x\right)\ge \alpha \}.\end{array}$$

(2)

$$\begin{array}{c}\alpha .S_{\alpha }=\left\{(x,\alpha )\right|x\in S_{\alpha }\}.\end{array}$$

(3)

It is easy to see that,

$$\begin{array}{c}{\mu }_S\left(x\right)={\displaystyle \underset{{}_{x\in S_{\alpha }}}{sup}}\alpha .\end{array}$$

(4)

Definition 2 

([39]). A fuzzy number $\tilde{S}$ is called a pseudo-trapezoidal if its membership function ${\mu }_{\tilde{S}}\left(x\right)$ can be represented by

$$\begin{array}{c}{\mu }_{\tilde{S}}\left(x\right)=\left\{\begin{array}{c}{l}_{\tilde{S}}\left(x\right),\underline{s}\le x\le s_1,\hfill \\ 1,s_1\le x\le s_2,\hfill \\ r_{\tilde{S}}\left(x\right),s_2\le x\le \overline{s},\hfill \\ 0,otherwise,\hfill \end{array}\right.\end{array}$$

(5)

where $l_{\tilde{S}}\left(x\right)$ and $r_{\tilde{S}}\left(x\right)$ are nondecreasing and non increasing functions, respectively.

A pseudo-triangular fuzzy number is a particular pseudo-trapezoidal fuzzy number when the $s_1=s_2.$

Definition 3 

([37]). Suppose C and D are two the pseudo-trapezoidal fuzzy numbers with the following α-cut forms:

$$C=\bigcup _{\alpha \in (0,1]}\alpha .C_{\alpha },C_{\alpha }=[{\underline{C}}_{\alpha },{\overline{C}}_{\alpha }],\varphi =\frac{c_1+c_2}{2}(\mathrm{i}.\mathrm{e}.,\mathrm{core}\mathrm{center}(\mathrm{c}.\mathrm{c}.\left)\mathrm{of}\mathrm{C}\right),$$

$$D=\bigcup _{\alpha \in (0,1]}\alpha .D_{\alpha },D_{\alpha }=[{\underline{D}}_{\alpha },{\overline{D}}_{\alpha }],\psi =\frac{d_1+d_2}{2}(\mathrm{i}.\mathrm{e}.,\mathrm{core}\mathrm{center}(\mathrm{c}.\mathrm{c}.\left)\mathrm{of}\mathrm{D}\right).$$

Then $TA$-based fuzzy arithmetic operations are the following:

Addition:

$$\begin{array}{c}C+D={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{(C+D)}}_{\alpha },{\overline{(C+D)}}_{\alpha }],\end{array}$$

(6)

where,

$$\begin{array}{c}{\underline{(C+D)}}_{\alpha }=\frac{\varphi +\psi }{2}+\left(\frac{{\underline{C}}_{\alpha }+{\underline{D}}_{\alpha }}{2}\right),\\ {\overline{(C+D)}}_{\alpha }=\frac{\varphi +\psi }{2}+\left(\frac{{\overline{C}}_{\alpha }+{\overline{D}}_{\alpha }}{2}\right).\end{array}$$

(7)

Subtraction:

$$\begin{array}{c}-D={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{(-D)}}_{\alpha },{\overline{(-D)}}_{\alpha }],\end{array}$$

(8)

where,

$$\begin{array}{c}{\underline{(-D)}}_{\alpha }=-2\psi +{\underline{D}}_{\alpha },\\ {\overline{(-D)}}_{\alpha }=-2\psi +{\overline{D}}_{\alpha }.\end{array}$$

(9)

Therefore,

$$C-D=C+(-D),$$

$$\begin{array}{c}C-D={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{(C-D)}}_{\alpha },{\overline{(C-D)}}_{\alpha }],\end{array}$$

(10)

where,

$$\begin{array}{c}{\underline{(C-D)}}_{\alpha }=\frac{\varphi -3\psi }{2}+\left(\frac{{\underline{C}}_{\alpha }+{\underline{D}}_{\alpha }}{2}\right),\\ {\overline{(C-D)}}_{\alpha }=\frac{\varphi -3\psi }{2}+\left(\frac{{\overline{D}}_{\alpha }+{\overline{D}}_{\alpha }}{2}\right).\end{array}$$

(11)

Multiplication:

$$\begin{array}{c}C.D={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{(C.D)}}_{\alpha },{\overline{(C.D)}}_{\alpha }],\end{array}$$

(12)

where,

$$[{\underline{(C.D)}}_{\alpha },{\overline{(C.D)}}_{\alpha }]=\left\{\begin{array}{l}[\left(\frac{\psi }{2}\right){\underline{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\underline{D}}_{\alpha },\left(\frac{\psi }{2}\right){\overline{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\overline{D}}_{\alpha }],\varphi \ge 0,\psi \ge 0,\\ [\left(\frac{\psi }{2}\right){\overline{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\underline{D}}_{\alpha },\left(\frac{\psi }{2}\right){\underline{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\overline{D}}_{\alpha }],\varphi \ge 0,\psi \le 0,\\ [\left(\frac{\psi }{2}\right){\overline{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\overline{D}}_{\alpha },\left(\frac{\psi }{2}\right){\underline{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\underline{D}}_{\alpha }],\varphi \le 0,\psi \le 0,\\ [\left(\frac{\psi }{2}\right){\underset{¯}{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\stackrel{̲}{D}}_{\alpha },\left(\frac{\psi }{2}\right){\stackrel{̲}{C}}_{\alpha }+\left(\frac{\varphi }{2}\right){\underset{¯}{D}}_{\alpha }],\varphi \le 0,\psi \ge 0,\end{array}\right.$$

(13)

Division:

$$\begin{array}{c}{D}^{-1}={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{\left(D^{-1}\right)}}_{\alpha },{\overline{\left(D^{-1}\right)}}_{\alpha }],\end{array}$$

(14)

where,

$$\begin{array}{c}{\underline{\left(D^{-1}\right)}}_{\alpha }=\left(\frac{1}{{\psi }^2}\right){\underline{D}}_{\alpha },\\ {\overline{\left(D^{-1}\right)}}_{\alpha }=\left(\frac{1}{{\psi }^2}\right){\overline{D}}_{\alpha }.\end{array}$$

(15)

Therefore,

$$\begin{array}{c}C.D^{-1}={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{(C.D^{-1})}}_{\alpha },{\overline{(C.D^{-1})}}_{\alpha }],\end{array}$$

(16)

where,

$$[{\underline{(C.D^{-1})}}_{\alpha },{\overline{(C.D^{-1})}}_{\alpha }]=\left\{\begin{array}{l}[\left(\frac{1}{2\psi }\right){\underline{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\underline{D}}_{\alpha },\left(\frac{1}{2\psi }\right){\overline{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\overline{D}}_{\alpha }],\varphi \ge 0,\psi >0,\\ [\left(\frac{1}{2\psi }\right){\overline{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\underline{D}}_{\alpha },\left(\frac{1}{2\psi }\right){\underline{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\overline{D}}_{\alpha }],\varphi \ge 0,\psi <0,\\ [\left(\frac{1}{2\psi }\right){\overline{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\overline{D}}_{\alpha },\left(\frac{1}{2\psi }\right){\underline{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\underline{D}}_{\alpha }],\varphi \le 0,\psi <0,\\ [\left(\frac{1}{2\psi }\right){\underset{¯}{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\stackrel{̲}{D}}_{\alpha },\left(\frac{1}{2\psi }\right){\stackrel{̲}{C}}_{\alpha }+\left(\frac{\varphi }{2{\psi }^2}\right){\underset{¯}{D}}_{\alpha }],\varphi \le 0,\psi >0.\end{array}\right.$$

(17)

Remark 1. 

Division operation on fuzzy number $\tilde{0}=(s_1,-s,s,s_2,\underline{0}\left(x\right),\overline{0}\left(x\right))$ cannot be defined. TA-based fuzzy operations on pseudo-triangular are special operations that, $c_1=c_2=c$ and $d_1=d_2=d$, that is, $\varphi =c,\psi =d$.

3. Solving Fuzzy Nonlinear Equations and Error Analysis

The more fundamental pursuit of the analysis and treatment of complex problems in many research fields (e.g., robotics, radiative transfer, chemistry, economics, etc.) is being shaped by solving nonlinear equations. In some cases, the coefficients of nonlinear equations are given in the form of fuzzy numbers, in which case we will deal with numerical methods for solving fuzzy nonlinear equations. Previous strategies for solving fuzzy nonlinear equations have been very limited because very strong conditions are usually imposed on the equations to solve them. These facts motivate people to solve fuzzy nonlinear equations with a new attitude. Generally speaking,

$$\begin{array}{c}F\left(S\right)=B.\end{array}$$

(18)

Our new method presents the following two situations:

• All coefficients of the nonlinear equation except the crisp coefficients are quasi-triangular fuzzy numbers;
• At least one coefficient of the nonlinear equation is pseudo-trapezoidal.

3.1. All Coefficients of the Nonlinear Equation except the Crisp Coefficients Are Quasi-Triangular Fuzzy Numbers

By considering the operations involved on both sides of the fuzzy nonlinear equation based on transmission averaging, we obtain two crisp nonlinear equations with arbitrary cut $\alpha$ ($0<\alpha \le 1$). According to the application of TA-based fuzzy operations, each of the two equations contains unknowns, including the solution kernel and the points (solutions) of its left and right edges for arbitrary cutting, that is, two equations and three unknowns. Then, by considering $\alpha$ = 1 and applying known clear numerical techniques, we will get an approximation to the solution kernel that gives an approximation of the membership function that does not converge to an approximate solution kernel. Finally, we will obtain membership functions that approximate the solution through a generalization process.

Consider the $\alpha$-cut form of Equation (18) as follows:

$$\begin{array}{c}\left\{\begin{array}{c}\underline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha })={\underline{B}}_{\alpha },\hfill \\ \overline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha })={\overline{B}}_{\alpha },\hfill \end{array}\right.\end{array}$$

(19)

where

$$S=\bigcup _{\alpha \in (0,1]}\alpha .S_{\alpha },S_{\alpha }=[{\underline{S}}_{\alpha },{\overline{S}}_{\alpha }],c.c.\left(S\right)=s,$$

$$F\left(S\right)=\bigcup _{\alpha \in (0,1]}\alpha .F{\left(S\right)}_{\alpha },F{\left(S\right)}_{\alpha }=[\underline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha }),\overline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha })],c.c.\left(F\left(S\right)\right)=f\left(s\right),$$

$$B=\bigcup _{\alpha \in (0,1]}\alpha .B_{\alpha },B_{\alpha }=[{\underline{B}}_{\alpha },{\overline{B}}_{\alpha }],c.c.\left(B\right)=b.$$

By putting $\alpha$ = 1 in Equation (19), we obtain

$$\begin{array}{c}f\left(s\right)=b.\end{array}$$

(20)

We now use a known crisp numerical technique on $f\left(s\right)=b.$ Suppose that $s_1$ is an approximation of the solution using the numerical technique, that is,

$$\begin{array}{c}f\left(s_1\right)\simeq b.\end{array}$$

(21)

If we put the approximate solution $s_1$ in Equation (19), then we will get ${\underline{S_1}}_{\alpha }$ and ${\overline{S_1}}_{\alpha }$ as approximations of ${\underline{S}}_{\alpha }$ and $\overline{S_{\alpha }}$, respectively, that is,

$$\begin{array}{c}\underset{\alpha \to 1}{lim}{\underline{S_1}}_{\alpha }=\underset{\alpha \to 1}{lim}{\overline{S_1}}_{\alpha }=s_2.\end{array}$$

(22)

It worth mentioning that $s_2\ne s_1$. In other words, we obtained an approximation of the membership functions that do not converge to the approximate solution core, i.e., $s_1$. So, we obtain the membership functions of the approximate solution core by generalization process. Since

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 1}{lim}}\frac{s_1}{s_2}{\underline{S_1}}_{\alpha }={\displaystyle \underset{\alpha \to 1}{lim}}\frac{s_1}{s_2}{\overline{S_1}}_{\alpha }=s_1,\end{array}$$

(23)

eventually, we introduce an approximate solution as follows:

$$\begin{array}{c}S\simeq {\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .S_{\alpha }^{*},S_{\alpha }^{*}=[\frac{s_1}{s_2}{\underline{S_1}}_{\alpha },\frac{s_1}{s_2}{\overline{S_1}}_{\alpha }],c.c.\left(S^{*}\right)=s_1.\end{array}$$

(24)

3.2. At Least One Coefficient of the Nonlinear Equation Is a Pseudo-Trapezoidal Fuzzy Number

It is worth mentioning that if at least one constant coefficient in the fuzzy nonlinear equation is quasi-trapezoidal, then we will get a quasi-trapezoidal fuzzy solution. In this case, when dealing with it, the details are slightly changed, which is similar to the previous one. In this case,

$$\begin{array}{c}\left\{\begin{array}{c}\underline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha })={\underline{B}}_{\alpha },\hfill \\ \overline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha })={\overline{B}}_{\alpha },\hfill \end{array}\right.\end{array}$$

(25)

where,

$$S=\bigcup _{\alpha \in (0,1]}\alpha .S_{\alpha },S_{\alpha }=[{\underline{S}}_{\alpha },{\overline{S}}_{\alpha }],S_1=[\underline{s},\overline{s}],s=\frac{\underline{s}+\overline{s}}{2},$$

$$F\left(S\right)=\bigcup _{\alpha \in (0,1]}\alpha .F{\left(S\right)}_{\alpha },F{\left(S\right)}_{\alpha }=[\underline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha }),\overline{F}(s,{\underline{S}}_{\alpha },{\overline{S}}_{\alpha })],$$

$$F{\left(S\right)}_1=[\underline{f}(s,\underline{s},\overline{s}),\overline{f}(s,\underline{s},\overline{s})],f\left(s\right)=\frac{\underline{f}(s,\underline{s},\overline{s})+\overline{f}(s,\underline{s},\overline{s})}{2},$$

$$B=\bigcup _{\alpha \in (0,1]}\alpha .B_{\alpha },B_{\alpha }=[{\underline{B}}_{\alpha },{\overline{B}}_{\alpha }],B_1=[\underline{b},\overline{b}],b=\frac{\underline{b}+\overline{b}}{2},$$

that by putting, $\alpha$ = 1 in Equation (25), we obtain

$$\begin{array}{c}\left\{\begin{array}{c}\underline{f}(s,\underline{s},\overline{s})=\underline{b},\hfill \\ \overline{f}(s,\underline{s},\overline{s})=\overline{b}.\hfill \end{array}\right.\end{array}$$

(26)

By summing the two sides of the above equations:

$$\begin{array}{c}f\left(s\right)=b.\end{array}$$

(27)

Now, we use a known numerical technique in Equation (27). Suppose that $s_1$ is an approximation of the solution using the numerical technique, that is,

$$\begin{array}{c}f\left(s_1\right)\simeq b.\end{array}$$

(28)

If, we put the approximate solution $s_1$ in Equation (25) then, we will get ${\underline{S_1}}_{\alpha }$ and ${\overline{S_1}}_{\alpha }$ as approximations of ${\underline{S}}_{\alpha }$ and ${\overline{S}}_{\alpha }$ respectively, that is,

$$\begin{array}{c}\left\{\begin{array}{c}\underset{\alpha \to 1}{lim}{\underline{S_1}}_{\alpha }=s_2,\hfill \\ \underset{\alpha \to 1}{lim}{\overline{S_2}}_{\alpha }=s_3.\hfill \end{array}\right.\end{array}$$

(29)

It worth mentioning that $s_2<s_3$ and $\frac{s_2+s_3}{2}\ne s_1$, in other words, we obtained an approximation of the membership functions that do not converge to the left and right edges of the approximate solution, i.e., $s_1$. So, we reach the membership functions for approximate solutions by the generalization process. Since,

$$\begin{array}{c}\left\{\begin{array}{c}\underset{\alpha \to 1}{lim}\left(\frac{2s_1+s_2-s_3}{2s_2}\right){\underline{S_1}}_{\alpha }=s_1-\left(\frac{s_3-s_2}{2}\right),\hfill \\ \underset{\alpha \to 1}{lim}\left(\frac{2s_1-s_2+s_3}{2s_3}\right){\overline{S_1}}_{\alpha }=s_1+\left(\frac{s_3-s_2}{2}\right),\hfill \end{array}\right.\end{array}$$

(30)

eventually, we introduce an approximate solution as follows:

$$\begin{array}{c}S\simeq {\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .S_{\alpha }^{*},S_{\alpha }^{*}=[\left(\frac{2s_1+s_2-s_3}{2s_2}\right){\underline{S_1}}_{\alpha },\left(\frac{2s_1-s_2+s_3}{2s_3}\right){\overline{S_1}}_{\alpha }],\\ {}_{}{S}_1^{*}=[s_1-\left(\frac{s_3-s_2}{2}\right),s_1+\left(\frac{s_3-s_2}{2}\right)],c.c.\left(S^{*}\right)=s_1.\end{array}$$

(31)

Remark 2. 

It is worth noting that in Section 3.1 and Section 3.2 if we have an exact value instead of the approximate value of the solution core, then we will be guided to a fuzzy exact solution. Of course, the main purpose of this paper is to obtain an approximate solution for fuzzy nonlinear equations.

3.3. Error Analysis

Evaluating the accuracy of computational results is an important goal of numerical analysis. We distinguish one of several error types that may limit the accuracy:

• Input data error;
• Rounding error;
• Approximate error.

Since we deal with numerical methods, approximation errors will be discussed. In this subsection, an error will be introduced in the numerical solution due to the vague thinking structure. With regard to the errors of the numerical techniques used and the newly proposed errors, a logical relationship is shown.

Lemma 1. 

For each α in the equations of Section 3.1,

$$\begin{array}{c}{\displaystyle \underset{s_1\to s}{lim}}\frac{s_1}{s_2}{\underline{S_1}}_{\alpha }={\underline{S}}_{\alpha },\\ {\displaystyle \underset{s_1\to s}{lim}}\frac{s_1}{s_2}{\overline{S_1}}_{\alpha }={\overline{S}}_{\alpha }.\end{array}$$

(32)

Proof. 

Since ${\underline{S_1}}_{\alpha }$ and ${\overline{S_1}}_{\alpha }$ are approximate values of ${\underline{S}}_{\alpha }$ and ${\overline{S}}_{\alpha }$ in Equation (19) for the approximate value of $s_1$ (instead of s), therefore for each $\alpha$:

$$\begin{array}{c}{\underline{S_1}}_{\alpha }\to {\underline{S}}_{\alpha },\\ {\overline{S_1}}_{\alpha }\to {\overline{S}}_{\alpha }.\end{array}$$

(33)

By approaching $\alpha$ to 1 in Equation (33) and according to Equation (22), we obtain

$$\begin{array}{c}{s}_2\to s.\end{array}$$

(34)

Thus,

$$\begin{array}{c}{\displaystyle \underset{s_1\to s}{lim}}\frac{s_1}{s_2}{\underline{S_1}}_{\alpha }={\displaystyle \underset{s_1\to s}{lim}}\frac{s}{s}{\underline{S}}_{\alpha }={\underline{S}}_{\alpha },\\ {\displaystyle \underset{s_1\to s}{lim}}\frac{s_1}{s_2}{\overline{S_1}}_{\alpha }={\displaystyle \underset{s_1\to s}{lim}}\frac{s}{s}{\overline{S}}_{\alpha }={\overline{S}}_{\alpha }.\end{array}$$

(35)

□

Lemma 2. 

For each α in the equations of Section 3.2,

$$\begin{array}{c}{\displaystyle \underset{s_1\to s}{lim}}\left(\frac{2s_1+s_2-s_3}{2s_2}\right){\underline{S_1}}_{\alpha }={\underline{S}}_{\alpha },\\ {\displaystyle \underset{s_2\to s}{lim}}\left(\frac{2s_1-s_2+s_3}{2s_3}\right){\overline{S_1}}_{\alpha }={\overline{S}}_{\alpha }.\end{array}$$

(36)

Proof. 

Since ${\underline{S_1}}_{\alpha }$ and ${\overline{S_1}}_{\alpha }$ are approximate values of ${\underline{S}}_{\alpha }$ and ${\overline{S}}_{\alpha }$ in Equation (25) for the approximate value of $s_1$ (instead of s), therefore for each $\alpha$:

$$\begin{array}{c}{\underline{S_1}}_{\alpha }\to {\underline{S}}_{\alpha },\\ {\overline{S_1}}_{\alpha }\to {\overline{S}}_{\alpha }.\end{array}$$

(37)

By approaching $\alpha$ to 1 in Equation (37) and according to Equation (29), we obtain

$$\begin{array}{c}{s}_2\to \underline{s},\\ s_3\to \overline{s}.\end{array}$$

(38)

Thus,

$$\begin{array}{c}{\displaystyle \underset{s_1\to s}{lim}}\left(\frac{2s_1+s_2-s_3}{2s_2}\right){\underline{S_1}}_{\alpha }={\displaystyle \underset{s_1\to s}{lim}}\left(\frac{2s+\underline{s}-\overline{s}}{2\underline{s}}\right){\underline{S}}_{\alpha }={\underline{S}}_{\alpha },\\ {\displaystyle \underset{s_1\to s}{lim}}\left(\frac{2s_1-s_2+s_3}{2s_3}\right){\overline{S_1}}_{\alpha }={\displaystyle \underset{s_1\to s}{lim}}\left(\frac{2s-\underline{s}+\overline{s}}{\overline{s}}\right){\overline{S}}_{\alpha }={\overline{S}}_{\alpha }.\end{array}$$

(39)

□

Definition 4 

(Error). The proposed approach error of the numerical solution for Section 3.1 and Section 3.2 respectively, are defined as follows:

$$\begin{array}{c}{E}_{3.1}=max\{ma{x}_{{}_{\alpha \in (0,1]}}|\frac{s_1}{s_2}{\underline{S_1}}_{\alpha }-{\underline{S}}_{\alpha }|,ma{x}_{{}_{\alpha \in (0,1]}}|\frac{s_1}{s_2}{\overline{S_1}}_{\alpha }-{\overline{S}}_{\alpha }\left|\right\},\\ and,\\ E_{3.2}=max\{ma{x}_{{}_{\alpha \in (0,1]}}|\left(\frac{2s_1+s_2-s_3}{2s_2}\right){\underline{S_1}}_{\alpha }-{\underline{S}}_{\alpha }|,ma{x}_{{}_{\alpha \in (0,1]}}|\left(\frac{2s_1-s_2+s_3}{2s_3}\right){\overline{S_1}}_{\alpha }-{\overline{S}}_{\alpha }\left|\right\}.\end{array}$$

(40)

It is worth noting that according to Lemmas 1 and 2, there is a logical relationship between the numerical technique’s error used and the new proposed error, that is, the smaller the error of the numerical method, the less the error of the proposed new approach. In other words,

$$\begin{array}{c}\forall ϵ>0,\exists \delta \left(ϵ\right)>0;\forall s_1,|s_1-s|<\delta \left(ϵ\right)\Rightarrow E_{3.1}<ϵ,\\ \mathrm{and},\\ \forall ϵ>0,\exists \delta \left(ϵ\right)>0;\forall s_1,|s_1-s|<\delta \left(ϵ\right)\Rightarrow E_{3.2}<ϵ.\end{array}$$

(41)

4. Illustrative Examples

In this section, several examples to illustrate the application of the proposed method are provided to clarify the reliability and effectiveness of the novel technique and the superiority of our method compared to the proposed methods from Refs. [1,17].

Example 1 

([1]). Consider nonlinear equation $A{X}^2+BX=C$ such that

$$\begin{array}{c}A={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .A_{\alpha },A_{\alpha }=[4\alpha -1,8-5\alpha ],a=c.c\left(A\right)=3,\end{array}$$

(42)

$$\begin{array}{c}B={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .B_{\alpha },B_{\alpha }=[\alpha +1,3-\alpha ],b=c.c\left(B\right)=2,\end{array}$$

(43)

$$\begin{array}{c}C={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .C_{\alpha },C_{\alpha }=[2\alpha -1,4-3\alpha ],c=c.c\left(C\right)=1.\end{array}$$

(44)

Since all the coefficients of the nonlinear equation in the example are triangular fuzzy numbers, we therefore suppose

$$X=\bigcup _{\alpha \in (0,1]}\alpha .X_{\alpha },X_{\alpha }=[{\underline{X}}_{\alpha },{\overline{X}}_{\alpha }],x=c.c.\left(X\right).$$

Without any loss of generality, assume that x is positive then, based on the fuzzy operations of the TA Definition 3 and since $a>0,b>0,$ we obtain

$$\begin{array}{c}{X}^2={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{X^2}}_{\alpha },{\overline{X^2}}_{\alpha }],\\ {\underline{X^2}}_{\alpha }=\frac{1}{2}(x.{\underline{X}}_{\alpha }+x.{\underline{X}}_{\alpha }),\\ {}_{}=x.{\underline{X}}_{\alpha }\\ {\overline{X^2}}_{\alpha }=\frac{1}{2}(x.{\overline{X}}_{\alpha }+x.{\overline{X}}_{\alpha }),\\ {}_{}=x.{\overline{X}}_{\alpha }.\end{array}$$

(45)

$$\begin{array}{c}c.c.\left(X^2\right)=\frac{1}{2}({\underline{X^2}}_1+{\overline{X^2}}_1)\\ {}_{}=\frac{1}{2}\{x.{\underline{X}}_1+x.{\overline{X}}_1\}\\ {}_{}=x.\left\{\frac{1}{2}({\underline{X}}_1+{\overline{X}}_1)\right\}\\ {}_{}=x.x\\ {}_{}=x^2,\end{array}$$

(46)

and,

$$\begin{array}{c}A{X}^2={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{A{X}^2}}_{\alpha },{\overline{A{X}^2}}_{\alpha }],\\ {\underline{A{X}^2}}_{\alpha }=\frac{1}{2}(ax{\underline{X}}_{\alpha }+x^2.{\underline{A}}_{\alpha }),\\ {\overline{A{X}^2}}_{\alpha }=\frac{1}{2}(ax{\overline{X}}_{\alpha }+x^2.{\overline{A}}_{\alpha }).\end{array}$$

(47)

$$\begin{array}{c}c.c.\left(A{X}^2\right)=\frac{1}{2}\{\frac{1}{2}(ax{\underline{X}}_1+x^2.{\underline{A}}_1)+\frac{1}{2}(ax{\overline{X}}_1+x^2.{\overline{A}}_1)\}\\ {}_{}=\frac{1}{2}\{ax(\frac{{\underline{X}}_1+{\overline{X}}_1}{2})+x^2(\frac{{\underline{A}}_1+{\overline{A}}_1}{2})\}\end{array}$$

(48)

$$\begin{array}{c}{}_{}=\frac{1}{2}\{ax.x+x^{2}a\}\\ {}_{}=a{x}^2,\end{array}$$

and,

$$\begin{array}{c}BX={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{BX}}_{\alpha },{\overline{BX}}_{\alpha }],\\ {\underline{BX}}_{\alpha }=\frac{1}{2}(b{\underline{X}}_{\alpha }+x.{\underline{B}}_{\alpha }),\\ {\overline{BX}}_{\alpha }=\frac{1}{2}(b{\overline{X}}_{\alpha }+x.{\overline{B}}_{\alpha }).\end{array}$$

(49)

$$\begin{array}{c}c.c.\left(BX\right)=\frac{1}{2}\{\frac{1}{2}(b{\underline{X}}_1+x.{\underline{B}}_1)+\frac{1}{2}(b{\overline{X}}_1+x.{\overline{B}}_1)\}\\ {}_{}=\frac{1}{2}\{b(\frac{{\underline{X}}_1+{\overline{X}}_1}{2})+x(\frac{{\underline{B}}_1+{\overline{B}}_1}{2})\}\\ {}_{}=\frac{1}{2}\{bx+xb\}\\ {}_{}=bx.\end{array}$$

(50)

Thus,

$$\begin{array}{c}A{X}^2+BX={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{A{X}^2+BX}}_{\alpha },{\overline{A{X}^2+BX}}_{\alpha }],\\ {\underline{A{X}^2+BX}}_{\alpha }=\frac{1}{2}\{bx+\frac{1}{2}(ax{\underline{X}}_{\alpha }+x^2.{\underline{A}}_{\alpha })+a{x}^2+\frac{1}{2}(b{\underline{X}}_{\alpha }+x.{\underline{B}}_{\alpha })\},\\ {\overline{A{X}^2+BX}}_{\alpha }=\frac{1}{2}\{bx+\frac{1}{2}(ax{\overline{X}}_{\alpha }+x^2.{\overline{A}}_{\alpha })+a{x}^2+\frac{1}{2}(b{\overline{X}}_{\alpha }+x.{\overline{B}}_{\alpha })\}.\end{array}$$

(51)

$$\begin{array}{c}c.c.(A{X}^2+BX)=\\ \frac{1}{2}\{\frac{1}{2}\{bx+\frac{1}{2}(ax{\underline{X}}_1+x^2.{\underline{A}}_1)+a{x}^2+\frac{1}{2}(b{\underline{X}}_1+x.{\underline{B}}_1)\}+\frac{1}{2}\{bx+\frac{1}{2}(ax{\overline{X}}_1+x^2.{\overline{A}}_1)+a{x}^2+\frac{1}{2}(b{\overline{X}}_1+x.{\overline{B}}_1)\}\}\\ {}_{}=\frac{1}{2}\{\frac{1}{2}\{2bx+ax(\frac{{\underline{X}}_1+{\overline{X}}_1}{2})+x^2(\frac{{\underline{A}}_1+{\overline{A}}_1}{2})+2a{x}^2+b(\frac{{\underline{X}}_1+{\overline{X}}_1}{2})+x(\frac{{\underline{B}}_1+{\overline{B}}_1}{2})\}\}\\ {}_{}=\frac{1}{4}\{2bx+axx+x^{2}a+2a{x}^2+bx+xb\}\\ {}_{}=\frac{1}{4}\{4bx+4a{x}^2\}\\ {}_{}=a{x}^2+bx.\end{array}$$

(52)

So, the α-cut form of this equation is as follows:

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}\{bx+\frac{1}{2}(ax{\underline{X}}_{\alpha }+x^2.{\underline{A}}_{\alpha })+a{x}^2+\frac{1}{2}(b{\underline{X}}_{\alpha }+x.{\underline{B}}_{\alpha })\}={\underline{C}}_{\alpha },\hfill \\ \frac{1}{2}\{bx+\frac{1}{2}(ax{\overline{X}}_{\alpha }+x^2.{\overline{A}}_{\alpha })+a{x}^2+\frac{1}{2}(b{\overline{X}}_{\alpha }+x.{\overline{B}}_{\alpha })\}={\overline{C}}_{\alpha },\hfill \end{array}\right.\end{array}$$

(53)

By substituting the α-cut forms of A, B, and C from the example, we have

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}\{2x+\frac{1}{2}(3x{\underline{X}}_{\alpha }+x^2.(4\alpha -1))+3x^2+\frac{1}{2}(2{\underline{X}}_{\alpha }+x.(\alpha +1))\}=2\alpha -1,\hfill \\ \frac{1}{2}\{2x+\frac{1}{2}(3x{\overline{X}}_{\alpha }+x^2.(8-5\alpha ))+3x^2+\frac{1}{2}(2{\overline{X}}_{\alpha }+x.(3-\alpha ))\}=4-3\alpha ,\hfill \end{array}\right.\end{array}$$

(54)

which is simplified as follows:

$$\begin{array}{c}\left\{\begin{array}{c}(\frac{3}{2}x+1){\underline{X}}_{\alpha }=(4\alpha -2)-\left(\frac{5+\alpha }{2}\right)x-\left(\frac{5+4\alpha }{2}\right)x^2,\hfill \\ (\frac{3}{2}x+1){\overline{X}}_{\alpha }=(8-6\alpha )-\left(\frac{7-\alpha }{2}\right)x-\left(\frac{14-5\alpha }{2}\right)x^2.\hfill \end{array}\right.\end{array}$$

(55)

By α = 1 in the above system, we obtain

$$3x^2+2x=1.$$

(56)

Now, if we consider the analytical solution in the above equation i.e., $x=\frac{1}{3}$ in Equation (55) then,

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{X}}_{\alpha }=\frac{65\alpha -56}{27},\hfill \\ {\overline{X}}_{\alpha }=\frac{109-100\alpha }{27}.\hfill \end{array}\right.\end{array}$$

(57)

Therefore, a fuzzy exact solution is as follows:

$$X=\bigcup _{\alpha \in (0,1]}\alpha .X_{\alpha },X_{\alpha }=[\frac{65\alpha -56}{27},\frac{109-100\alpha }{27}],x=\frac{1}{3}.$$

(58)

However, if the approximate solution $x=0.31$ is considered in Equation (56), then we get from Equation (55):

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{X_1}}_{\alpha }=\frac{1}{1+\frac{3}{2}\left(0.31\right)}[(4\alpha -2)-\left(\frac{5+\alpha }{2}\right)\left(0.31\right)-\left(\frac{5+4\alpha }{2}\right){\left(0.31\right)}^2],\hfill \\ {\overline{X_1}}_{\alpha }=\frac{1}{1+\frac{3}{2}\left(0.31\right)}[(8-6\alpha )-\left(\frac{7-\alpha }{2}\right)\left(0.31\right)-\left(\frac{14-5\alpha }{2}\right){\left(0.31\right)}^2].\hfill \end{array}\right.\end{array}$$

(59)

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 1}{lim}}{\underline{X_1}}_{\alpha }={\displaystyle \underset{\alpha \to 1}{lim}}{\overline{X_1}}_{\alpha }=0.43.\end{array}$$

(60)

Thus, a fuzzy approximate solution by generalization process is as follows:

$$\begin{array}{c}X\simeq {\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .X_{\alpha }^{*},X_{\alpha }^{*}=[\frac{31}{43}{\underline{X_1}}_{\alpha },\frac{31}{43}{\overline{X_1}}_{\alpha }],x=c.c.\left(X\right)\simeq 0.31.\end{array}$$

(61)

Now suppose that x is negative then, based on the fuzzy operations of the TA Definition 3 and since $a>0,b>0,$ we obtain

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}\{bx+\frac{1}{2}(ax{\overline{X}}_{\alpha }+x^2.{\underline{A}}_{\alpha })+a{x}^2+\frac{1}{2}(b{\underline{X}}_{\alpha }+x.{\overline{B}}_{\alpha })\}={\underline{C}}_{\alpha },\hfill \\ \frac{1}{2}\{bx+\frac{1}{2}(ax{\underline{X}}_{\alpha }+x^2.{\overline{A}}_{\alpha })+a{x}^2+\frac{1}{2}(b{\overline{X}}_{\alpha }+x.{\underline{B}}_{\alpha })\}={\overline{C}}_{\alpha },\hfill \end{array}\right.\end{array}$$

(62)

By substituting the α-cut forms of A, B and C from the example, we have

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}\{2x+\frac{1}{2}(3x{\overline{X}}_{\alpha }+x^2.(4\alpha -1))+3x^2+\frac{1}{2}(2{\underline{X}}_{\alpha }+x.(3-\alpha ))\}=2\alpha -1,\hfill \\ \frac{1}{2}\{2x+\frac{1}{2}(3x{\underline{X}}_{\alpha }+x^2.(8-5\alpha ))+3x^2+\frac{1}{2}(2{\overline{X}}_{\alpha }+x.(\alpha +1))\}=4-3\alpha ,\hfill \end{array}\right.\end{array}$$

(63)

which is simplified as follows:

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{X}}_{\alpha }+\frac{3}{2}x{\overline{X}}_{\alpha }=(4\alpha -2)-\left(\frac{7-\alpha }{2}\right)x-\left(\frac{5+4\alpha }{2}\right)x^2,\hfill \\ \frac{3}{2}x{\underline{X}}_{\alpha }+{\overline{X}}_{\alpha }=(8-6\alpha )-\left(\frac{5+\alpha }{2}\right)x-\left(\frac{14-5\alpha }{2}\right)x^2.\hfill \end{array}\right.\end{array}$$

(64)

By α = 1 in the above system, we obtain

$$3x^2+2x=1.$$

(65)

If we consider the analytical solution in the above equation i.e., $x=-1$ in Equation (64), then

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{X}}_{\alpha }=\frac{24\alpha -34}{10},\hfill \\ {\overline{X}}_{\alpha }=\frac{3\alpha -8}{5}.\hfill \end{array}\right.\end{array}$$

(66)

Since ${\overline{X}}_{\alpha }$ is not a bounded monotonic decreasing the left continuous function, the fuzzy negative root does not exist. The solution is more realistic compared to the method of Abbasbandy et al. [1], which is shown in Figure 1.

Example 2 

([1]). Consider nonlinear equation $A{X}^3+B{X}^2+C=D$ such that

$$\begin{array}{c}A={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .A_{\alpha },A_{\alpha }=[1+\alpha ,3-\alpha ],a=c.c\left(A\right)=2,\end{array}$$

(67)

$$\begin{array}{c}B={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .B_{\alpha },B_{\alpha }=[2+\alpha ,4-\alpha ],b=c.c\left(B\right)=3,\end{array}$$

(68)

$$\begin{array}{c}C={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .C_{\alpha },C_{\alpha }=[3+\alpha ,5-\alpha ],c=c.c\left(C\right)=4,\end{array}$$

(69)

$$\begin{array}{c}D={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .D_{\alpha },D_{\alpha }=[5+3\alpha ,13-5\alpha ],d=c.c\left(D\right)=8.\end{array}$$

(70)

Since all the coefficients of the nonlinear equation in the example are triangular fuzzy numbers, we therefore suppose

$$X=\bigcup _{\alpha \in (0,1]}\alpha .X_{\alpha },X_{\alpha }=[{\underline{X}}_{\alpha },{\overline{X}}_{\alpha }],x=c.c.\left(X\right).$$

Without any loss of generality, assume that x is positive then, based on the fuzzy operations of the TA Definition 3 and since $a>0,b>0,$ we obtain

$$\begin{array}{c}{X}^2={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{X^2}}_{\alpha },{\overline{X^2}}_{\alpha }],\\ {\underline{X^2}}_{\alpha }=x.\underline{X},{\overline{X^2}}_{\alpha }=x.{\overline{X}}_{\alpha }.\end{array}$$

(71)

$$\begin{array}{c}c.c.\left(X^2\right)=x^2,\end{array}$$

(72)

and,

$$\begin{array}{c}B{X}^2={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{B{X}^2}}_{\alpha },{\overline{B{X}^2}}_{\alpha }],\\ {\underline{B{X}^2}}_{\alpha }=\frac{1}{2}(x^2.{\underline{B}}_{\alpha }+b.x{\underline{X}}_{\alpha }),{\overline{B{X}^2}}_{\alpha }=\frac{1}{2}(x^2.{\overline{B}}_{\alpha }+b.x{\overline{X}}_{\alpha }).\end{array}$$

(73)

$$\begin{array}{c}c.c.\left(B{X}^2\right)=b{x}^2,\end{array}$$

(74)

and,

$$\begin{array}{c}{X}^3={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{X^3}}_{\alpha },{\overline{X^3}}_{\alpha }],\\ {\underline{X^3}}_{\alpha }=\frac{1}{2}(x.x{\underline{X}}_{\alpha }+x^2.{\underline{X}}_{\alpha }),{\overline{X^3}}_{\alpha }=\frac{1}{2}(x.x{\overline{X}}_{\alpha }+x^2.{\overline{X}}_{\alpha }).\end{array}$$

(75)

$$\begin{array}{c}c.c.\left(X^3\right)=x^3,\end{array}$$

(76)

and,

$$\begin{array}{c}A{X}^3={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{A{X}^3}}_{\alpha },{\overline{A{X}^3}}_{\alpha }],\\ {\underline{A{X}^3}}_{\alpha }=\frac{1}{2}(x^3.{\underline{A}}_{\alpha }+a.x^2{\underline{X}}_{\alpha }),{\overline{A{X}^3}}_{\alpha }=\frac{1}{2}(x^3.{\overline{A}}_{\alpha }+a.x^2{\overline{X}}_{\alpha }).\end{array}$$

(77)

$$\begin{array}{c}c.c.\left(A{X}^3\right)=a{x}^3,\end{array}$$

(78)

and,

$$\begin{array}{c}A{X}^3+B{X}^2={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{A{X}^3+B{X}^2}}_{\alpha },{\overline{A{X}^3+B{X}^2}}_{\alpha }],\end{array}$$

(79)

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{A{X}^3+B{X}^2}}_{\alpha }=\frac{1}{2}\{b{x}^2+\frac{1}{2}(x^3.{\underline{A}}_{\alpha }+a.x^2{\underline{X}}_{\alpha })+a{x}^3+\frac{1}{2}(x^2.{\underline{B}}_{\alpha }+b.x{\underline{X}}_{\alpha })\},\hfill \\ {\overline{A{X}^3+B{X}^2}}_{\alpha }=\frac{1}{2}\{b{x}^2+\frac{1}{2}(x^3.{\overline{A}}_{\alpha }+a.x^2{\overline{X}}_{\alpha })+a{x}^3+\frac{1}{2}(x^2.{\overline{B}}_{\alpha }+b.x{\overline{X}}_{\alpha })\}.\hfill \end{array}\right.\end{array}$$

(80)

$$\begin{array}{c}c.c.(A{X}^3+B{X}^2)=a{x}^3+b{x}^2.\end{array}$$

(81)

Hence,

$$\begin{array}{c}A{X}^3+B{X}^2+C={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .[{\underline{A{X}^3+B{X}^2+C}}_{\alpha },{\overline{A{X}^3+B{X}^2+C}}_{\alpha }],\end{array}$$

(82)

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{A{X}^3+B{X}^2+C}}_{\alpha }=\frac{1}{2}[\frac{1}{2}\{b{x}^2+\frac{1}{2}(x^3.{\underline{A}}_{\alpha }+a.x^2{\underline{X}}_{\alpha })+a{x}^3+\frac{1}{2}(x^2.{\underline{B}}_{\alpha }+b.x{\underline{X}}_{\alpha })\}+\hfill \\ {}_{}c+{\underline{C}}_{\alpha }+a{x}^3+b{x}^2],\hfill \\ {\overline{A{X}^3+B{X}^2+C}}_{\alpha }=\frac{1}{2}[\frac{1}{2}\{b{x}^2+\frac{1}{2}(x^3.{\overline{A}}_{\alpha }+a.x^2{\overline{X}}_{\alpha })+a{x}^3+\frac{1}{2}(x^2.{\overline{B}}_{\alpha }+b.x{\overline{X}}_{\alpha })\}+\hfill \\ {}_{}c+{\overline{C}}_{\alpha }+a{x}^3+b{x}^2].\hfill \end{array}\right.\end{array}$$

(83)

So, the α-cut form of this equation is as follows:

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}[\frac{1}{2}\{b{x}^2+\frac{1}{2}(x^3.{\underline{A}}_{\alpha }+a.x^2{\underline{X}}_{\alpha })+a{x}^3+\frac{1}{2}(x^2.{\underline{B}}_{\alpha }+b.x{\underline{X}}_{\alpha })\}+\hfill \\ {}_{}c+{\underline{C}}_{\alpha }+a{x}^3+b{x}^2]={\underline{D}}_{\alpha },\hfill \\ \frac{1}{2}[\frac{1}{2}\{b{x}^2+\frac{1}{2}(x^3.{\overline{A}}_{\alpha }+a.x^2{\overline{X}}_{\alpha })+a{x}^3+\frac{1}{2}(x^2.{\overline{B}}_{\alpha }+b.x{\overline{X}}_{\alpha })\}+\hfill \\ {}_{}c+{\overline{C}}_{\alpha }+a{x}^3+b{x}^2]={\overline{D}}_{\alpha },\hfill \end{array}\right.\end{array}$$

(84)

By substituting the α-cut forms of A, B, C, and D from the example, we have

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}[\frac{1}{2}\{3x^2+\frac{1}{2}(x^3.(1+\alpha )+2.x^2{\underline{X}}_{\alpha })+2x^3+\frac{1}{2}(x^2.(2+\alpha )+3.x{\underline{X}}_{\alpha })\}+\hfill \\ {}_{}4+(3+\alpha )+2x^3+3x^2]=5+3\alpha ,\hfill \\ \frac{1}{2}[\frac{1}{2}\{3x^2+\frac{1}{2}(x^3.(3-\alpha )+2.x^2{\overline{X}}_{\alpha })+2x^3+\frac{1}{2}(x^2.(4-\alpha )+3.x{\overline{X}}_{\alpha })\}+\hfill \\ {}_{}4+(5-\alpha )+2x^3+3x^2]=13-5\alpha ,\hfill \end{array}\right.\end{array}$$

(85)

which is simplified as follows:

$$\begin{array}{c}\left\{\begin{array}{c}(\frac{1}{2}{x}^2+\frac{3}{4}x){\underline{X}}_{\alpha }=(3+5\alpha )-\left(\frac{20+\alpha }{4}\right)x^2-\left(\frac{13+\alpha }{4}\right)x^3,\hfill \\ (\frac{1}{2}{x}^2+\frac{3}{4}x){\overline{X}}_{\alpha }=(17-9\alpha )-\left(\frac{22-\alpha }{4}\right)x^2-\left(\frac{15-\alpha }{4}\right)x^3.\hfill \end{array}\right.\end{array}$$

(86)

By α = 1 in the above system, we obtain

$$2x^3+3x^2=4.$$

(87)

However, if the approximate solution $x=0.91$ is considered in Equation (87), then we get from Equation (86):

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{X_1}}_{\alpha }=\frac{1}{\frac{1}{2}{\left(0.91\right)}^2+\frac{3}{4}\left(0.91\right)}[(3+5\alpha )-\left(\frac{20+\alpha }{4}\right){\left(0.91\right)}^2-\left(\frac{13+\alpha }{4}\right){\left(0.91\right)}^3],\hfill \\ {\overline{X_1}}_{\alpha }=\frac{1}{\frac{1}{2}{\left(0.91\right)}^2+\frac{3}{4}\left(0.91\right)}[(17-9\alpha )-\left(\frac{22-\alpha }{4}\right){\left(0.91\right)}^2-\left(\frac{15-\alpha }{4}\right){\left(0.91\right)}^3].\hfill \end{array}\right.\end{array}$$

(88)

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 1}{lim}}{\underline{X_1}}_{\alpha }={\displaystyle \underset{\alpha \to 1}{lim}}{\overline{X_1}}_{\alpha }=0.92.\end{array}$$

(89)

Thus, a fuzzy approximate solution by the generalization process is as follows:

$$\begin{array}{c}X\simeq {\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .X_{\alpha }^{*},X_{\alpha }^{*}=[\frac{91}{92}{\underline{X_1}}_{\alpha },\frac{91}{92}{\overline{X_1}}_{\alpha }],x\simeq c.c.\left(X^{*}\right)=0.91.\end{array}$$

(90)

The solution is more realistic compared to the method of Abbasbandy et al. [1], which is shown in Figure 2.

Example 3 

([17]). Consider nonlinear equation $A{T}^2+V_{0}T+Y_0=Y$ such that

$$\begin{array}{c}U={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .U_{\alpha },U_{\alpha }=[-16,-16],a=c.c\left(U\right)=-16,\end{array}$$

(91)

$$\begin{array}{c}{V}_0={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .V_{0\alpha },V_{0\alpha }=[62+2\alpha ,66-2\alpha ],v_0=c.c.\left(V_0\right)=64,\end{array}$$

(92)

$$\begin{array}{c}{Y}_0={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .Y_{0\alpha },Y_{0\alpha }=[94+2\alpha ,98-2\alpha ],y_0=c.c.\left(Y_0\right)=96,\end{array}$$

(93)

$$\begin{array}{c}Y={\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .Y_{\alpha },Y_{\alpha }=[140+3\alpha ,148-3\alpha ],Y_1=[143,145]y=c.c.\left(Y\right)=144.\end{array}$$

(94)

Since all the coefficients of the nonlinear equation in the example are triangular fuzzy numbers, we therefore suppose

$$T=\bigcup _{\alpha \in (0,1]}\alpha .T_{\alpha },T_{\alpha }=[{\underline{T}}_{\alpha },{\overline{T}}_{\alpha }],T_1=[\underline{t},\overline{t}]t=c.c.\left(T\right)$$

Without any loss of generality, we assume that t is positive, then, based on the fuzzy operations of the TA Definition 3 and since $a<0,v_0>0,$ we obtain

$$A{T}^2+V_{0}T+Y_0=\bigcup _{\alpha \in (0,1]}\alpha .[{\underline{A{T}^2+V_{0}T+Y_0}}_{\alpha },{\overline{A{T}^2+V_{0}T+Y_0}}_{\alpha }],$$

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}[a{t}^2+v_{0}t+{\underline{Y_0}}_{\alpha }+y_0+\frac{1}{2}\{v_{0}t+\frac{1}{2}(at{\overline{T}}_{\alpha }+t^2.{\underline{U}}_{\alpha })+a{t}^2+\hfill \\ {}_{}\frac{1}{2}(v_0{\underline{T}}_{\alpha }+t.{\underline{V_0}}_{\alpha })\left\}\right],\hfill \\ \frac{1}{2}[a{t}^2+v_{0}t+{\overline{Y_0}}_{\alpha }+y_0+\frac{1}{2}\{v_{0}t+\frac{1}{2}(at{\underline{T}}_{\alpha }+t^2.{\overline{U}}_{\alpha })+a{t}^2+\hfill \\ {}_{}\frac{1}{2}(v_0{\overline{T}}_{\alpha }+t.{\overline{V_0}}_{\alpha })\left\}\right].\hfill \end{array}\right.\end{array}$$

(95)

So, the α-cut form of this equation is as follows:

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}[a{t}^2+v_{0}t+{\underline{Y_0}}_{\alpha }+y_0+\frac{1}{2}\{v_{0}t+\frac{1}{2}(at{\overline{T}}_{\alpha }+t^2.{\underline{U}}_{\alpha })+a{t}^2+\frac{1}{2}(v_0{\underline{T}}_{\alpha }+t.{\underline{V_0}}_{\alpha })\}]={\underline{Y}}_{\alpha },\hfill \\ \frac{1}{2}[a{t}^2+v_{0}t+{\overline{Y_0}}_{\alpha }+y_0+\frac{1}{2}\{v_{0}t+\frac{1}{2}(at{\underline{T}}_{\alpha }+t^2.{\overline{U}}_{\alpha })+a{t}^2+\frac{1}{2}(v_0{\overline{T}}_{\alpha }+t.{\overline{V_0}}_{\alpha })\}]={\overline{Y}}_{\alpha },\hfill \end{array}\right.\end{array}$$

(96)

By substituting the α-cut forms of U, $V_0,Y_0$, and Y from the example, we have

$$\begin{array}{c}\left\{\begin{array}{c}\frac{1}{2}[-16t^2+64t+(94+2\alpha )+96+\frac{1}{2}\{64t+\frac{1}{2}(-16t{\overline{T}}_{\alpha }-16t^2)-16t^2+\frac{1}{2}(64{\underline{T}}_{\alpha }+t.(62+2\alpha ))\}]\hfill \\ {}_{}=140+3\alpha ,\hfill \\ \frac{1}{2}[-16t^2+64t+(98-2\alpha )+96+\frac{1}{2}\{64t+\frac{1}{2}(-16t{\underline{T}}_{\alpha }-16t^2)-16t^2+\frac{1}{2}(64{\overline{T}}_{\alpha }+t.(66-2\alpha ))\}]\hfill \\ {}_{}=148-3\alpha ,\hfill \end{array}\right.\end{array}$$

(97)

which is simplified as follows:

$$\begin{array}{c}\left\{\begin{array}{c}16{\underline{T}}_{\alpha }-4t{\overline{T}}_{\alpha }=(90+4\alpha )-\left(\frac{223+\alpha }{2}\right)t+28t^2,\hfill \\ -4t{\underline{T}}_{\alpha }+16{\overline{T}}_{\alpha }=(102-4\alpha )-\left(\frac{225-\alpha }{2}\right)t+28t^2.\hfill \end{array}\right.\end{array}$$

(98)

By α = 1 and the summing two sides of the above equations, we obtain

$$-16t^2+64t=48.$$

(99)

Now, if we consider the analytical solutions in the above equation, i.e., $t=1$ and $t=3$ in Equation (98), then

for the$t=1$:

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{T}}_{\alpha }=\frac{87+21\alpha }{120},\hfill \\ {\overline{T}}_{\alpha }=\frac{153-21\alpha }{120},\hfill \end{array}\right.\end{array}$$

(100)

for the$t=3$:

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{T}}_{\alpha }=\frac{159+5\alpha }{56},\hfill \\ {\overline{T}}_{\alpha }=\frac{177-5\alpha }{56}.\hfill \end{array}\right.\end{array}$$

(101)

Thus, the fuzzy exact solutions, respectively, are as follows:

$$T=\bigcup _{\alpha \in (0,1]}\alpha .T_{\alpha },T_{\alpha }=[\frac{87+21\alpha }{120},\frac{153-21\alpha }{120}],T_1=[0.9,1.1].$$

(102)

and

$$T=\bigcup _{\alpha \in (0,1]}\alpha .T_{\alpha },T_{\alpha }=[\frac{159+5\alpha }{56},\frac{177-5\alpha }{56}],T_1=[2.93,3.07].$$

(103)

However, if the approximate solutions $t=0.97$ and $t=3.03$ are considered in Equation (99), then we get from Equation (97):

for the$t\simeq 0.97$:

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{T_1}}_{\alpha }=\frac{28.23+34.94\alpha }{\left(120.48\right)},\hfill \\ {\overline{T_1}}_{\alpha }=\frac{137.86-34.94\alpha }{\left(120.48\right)}.\hfill \end{array}\right.\end{array}$$

(104)

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 1}{lim}}{\underline{T_1}}_{\alpha }=0.52,\\ {\displaystyle \underset{\alpha \to 1}{lim}}{\overline{T_1}}_{\alpha }=0.85,\end{array}$$

(105)

for the$t\simeq 3.03$:

$$\begin{array}{c}\left\{\begin{array}{c}{\underline{T_1}}_{\alpha }=\frac{183.99+4.82\alpha }{\left(54.55\right)},\hfill \\ {\overline{T_1}}_{\alpha }=\frac{201.39-4.82\alpha }{\left(54.55\right)}.\hfill \end{array}\right.\end{array}$$

(106)

$$\begin{array}{c}{\displaystyle \underset{\alpha \to 1}{lim}}{\underline{T_1}}_{\alpha }=3.46,\\ {\displaystyle \underset{\alpha \to 1}{lim}}{\overline{T_1}}_{\alpha }=3.60,\end{array}$$

(107)

Thus, the fuzzy approximate solutions by the generalization process, respectively, are as follows:

$$\begin{array}{c}T\simeq {\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .T_{\alpha }^{*},T_{\alpha }^{*}=[\left(\frac{2\left(0.97\right)+0.52-0.85}{2\left(0.52\right)}\right){\underline{T_1}}_{\alpha },\left(\frac{2\left(0.97\right)-0.52+0.85}{2\left(0.85\right)}\right){\overline{T_1}}_{\alpha }],\\ T_1^{*}=[0.81,1.13],\end{array}$$

(108)

and

$$\begin{array}{c}T\simeq {\displaystyle \bigcup _{\alpha \in (0,1]}}\alpha .T_{\alpha }^{*},T_{\alpha }^{*}=[\left(\frac{2\left(3.03\right)+3.46-3.60}{2\left(3.46\right)}\right){\underline{T_1}}_{\alpha },\left(\frac{2\left(33.03\right)-3.46+3.60}{2\left(3.60\right)}\right){\overline{T_1}}_{\alpha }],\\ T_1^{*}=[2.96,3.1].\end{array}$$

(109)

The solutions are better, practical, and more realistic compared to the method of Buckley and Qu [17], as shown in Figure 3 and Figure 4.

5. Conclusions

The extension principle is one of the most basic ideas in fuzzy set theory. It provides a general approach to extend crisp mathematical concepts to solving fuzzy quantities, such as practical algebraic operations on fuzzy numbers. These operations are computationally efficient generalizations of interval analysis. In most methods for solving fuzzy nonlinear equations, the equations often consider extreme conditions, which shows the limitations of these methods. These facts prompt people to solve fuzzy nonlinear equations with a new attitude. In this paper, by changing the fuzzy arithmetic operation in 1, under the TA-based operation framework of Abbasi et al. [37], instead of the extension principle or the interval algorithm, we get a different system of clear nonlinear equations than other previous methods, with unknown three instead of unknown two. Using $\alpha$ = 1 and a known crisp numerical technique, we obtain an approximation to an unknown, called the kernel of the fuzzy solution, and with this approximation we will arrive at an approximate kernel that does not converge to the membership function of the approximation. Then, we will go through the generalization process to reach membership function convergence to approximate the solution kernel. The strengths and superiority of the proposed method mentioned with different implementations of principles 2 and 3 can ultimately lead to a more acceptable solution than other previous methods, which has been demonstrated in some examples. In the newly proposed method, a bug is introduced due to the ambiguous thinking structure. Therefore, a computational method is proposed to solve the fuzzy nonlinear equations performed in TA-based operations to bring it closer to reality. In addition to using the proposed technique in the numerical solution of the fuzzy nonlinear equations used in this paper, it can be considered in the future in systems of fuzzy nonlinear equations and fuzzy differential equations.