# Computer-Based Fuzzy Numerical Method for Solving Engineering and Real-World Applications

Authors: Naila Rafiq; Naveed Yaqoob; Nasreen Kausar; Mudassir Shams; Nazir Ahmad Mir; Yaé Ulrich Gaba; Naveed Khan
Journal: Mathematical Problems in Engineering
 (2021)

---

## Abstract

The nonlinear equation is a fundamentally important area of study in mathematics, and the numerical solutions of the nonlinear equations are also an important part of it. Fuzzy sets introduced by Zedeh are an extension of classical sets, which have several applications in engineering, medicine, economics, finance, artificial intelligence, decision-making, and so on. The most special types of fuzzy sets are fuzzy numbers. The important fuzzy numbers are trapezoidal fuzzy and triangular fuzzy numbers, which have several applications. In this research article, we propose an efficient numerical iterative method for estimating roots of fuzzy nonlinear equations, which are based on the special type of fuzzy number called triangular fuzzy number. Convergence analysis proves that the order of convergence of the numerical method is three. Some real-life applications are considered as numerical test problems from engineering, which contain fuzzy quantities in the parametric form. Engineering models include fractional conversion of nitrogen-hydrogen feed into ammonia and Van der Waal’s equation for calculating the volume and pressure of a gas and motion of the object under constant force of gravity. Numerical illustrations are given to show the dominance efficiency of the newly constructed iterative schemes as compared to existing methods in the literature.


---

## Body

## 1. Introduction

One of the ancient problems of science and engineering in general and in mathematics in particular is to approximate roots of nonlinear equations. The nonlinear equations play a major role in the field of engineering, mathematics, physics, chemistry, economics, medicines, and finance. Many times the particular realization of such type of nonlinear problems involves imprecise and nonprobabilistic uncertainties in the parameter, where the approximations are known due to expert knowledge or due to some experimental data. Due to these reasons, several real world applications contain vagueness and uncertainties. Therefore, in most of real world problems, the parameters involved in the system or variables of the nonlinear equations are presented by a fuzzy number. The concepts of fuzzy numbers and arithmetic operation with fuzzy numbers were first introduced and investigated in [1–10]. Hence, it is necessary to approximate the root of fuzzy nonlinear equation:(1)Fx=c.The standard analytical technique like Buckley and Qu method [11–14] cannot be suitable for solving the nonlinear equations such asax6+bx4−cx3+dx−e=f, x+cosx=g, xlnx+ex−1/1+x2+tanx=h, where a,b,c,d,e,f,g, and h are fuzzy numbers and x is a fuzzy variable.We therefore look towards numerical iterative schemes, which approximate the roots of fuzzy nonlinear equations. To approximate the roots of fuzzy nonlinear equations, Abbsbandy and Asady [15] used Newton Raphson method, Sulaiman et al. [16] give Levenberg-Marquest method, and Mosleh [17] used Adomian decomposition method; see also [16, 18–20]. Iterative methods presented by them have low convergence order to approximate the roots of fuzzy nonlinear equations. Iterative methods of lower convergence order have high computational time and cost according to Kung–Traub conjecture [21].Engineers and mathematicians therefore look towards those numerical methods which are more efficient and with high convergence order and low computational cost, or time, to approximate the roots of highly nonlinear fuzzy equations. The main aim of this research article is to propose efficient higher order iterative method as compared to well-known classical methods [15]. Numerical test results, CPU time, and log of residual show the dominance efficiency of our newly constructed method over the existing methods in the literature.This paper is organized as follows: after introduction in Section2, we recall some fundamental results of fuzzy numbers. In Section 3, we propose numerical iterative schemes for approximating roots of fuzzy nonlinear equations and their convergence analysis. In Section 4, we illustrate some real world applications as numerical test examples to show the performance and efficiency of the constructed method and conclusions in the last section.


## 2. Preliminaries

Definition 1.
A fuzzy number is a fuzzy set likex:ℝ⟶I=0,1, which satisfies the following [22, 23]:(1)
x is upper semicontinuous(2)
xa=0 outside some interval a1,a2(3)
The real numbers areb1,b2 such that a1≤b1≤b2≤a2 andxa is monotonic increasing on a1,b1xa is monotonic decreasing on b2,a2xa=1, for b1≤a≤b2We denote byE the set of all fuzzy numbers. An equivalent parametric form is also given in [24] as follows.Definition 2 (see [25]).
A fuzzy numberx in parametric form is a pair xL,xU of function xLτ, xUτ, 0≤τ≤1, which satisfies the following requirements:(1)
xLτ is a bounded monotonic increasing left continuous function(2)
xUτ is a bounded monotonic decreasing left continuous function(3)
xLτ≤xUτ, 0≤τ≤1A popular fuzzy number is the triangular fuzzy number, which is formed by simply takingb1=b2 in Definition 1. Triangular fuzzy number is simply written as x=a1,a2,a3 and defined in the form of membership function as(2)xa=x−a1a2−a1,ifa1<x<a2,a3−xa3−a2,ifa2<x<a3,0,otherwise.The parametric form is given as(3)xLτ=a1+τa2−a1,xUτ=a3+τa2−a3.LetTFℝ be the set of all triangular fuzzy numbers. The addition and scalar multiplication of fuzzy numbers are defined by the extension principle and represented as follows.For arbitraryx=xL,xU, y=yL,yU, and k>0, we define addition x+y and multiplication by scaler k as(4)x+yLτ=xLτ+yLτ,x+yUτ=xUτ+yUτ,kxLτ=kxLτ,kxUτ=kxUτ.


## 3. Construction of Iterative Schemes

Now, our aim is to obtain a solution for fuzzy nonlinear equationFx=c. The parametric form is as follows:(5)FLxL,xU,τ=cLτ,FUxL,xU,τ=cUτ,∀τ∈0,1.Suppose thatx=ξL,ξU is the solution to the system; that is,(6)FLξL,ξU,τ=cLτ,FUξL,ξU,τ=cUτ,∀τ∈0,1.Therefore, ifx0=x0L,x0U is an approximation solution for this system, then ∀τ∈0,1 there exist hLτ,hUτ such that(7)ξL=x0L+hLτ,ξU=x0U+hUτ.Using Taylor series ofFL, FU about x0L,x0U, ∀τ∈0,1, we have(8)FLξL,ξU,τ=FLx0L,x0U,τ+hLFxLLx0L,x0U,τ+hUFxULx0L,x0U,τ+OhL2+hLhU+hU2=cL,FUξL,ξU,τ=FUx0L,x0U,τ+hLFxLUx0L,x0U,τ+hUFxUUx0L,x0U,τ+OhL2+hLhU+hU2=cU.Ifx0L and x0U are near to ξL and ξU, respectively, then hL and hU are small. We assume, of course, that all needed partial derivatives exist and are bounded. Therefore, for enough small hL and hU, we have, ∀τ∈0,1,(9)FLx0L,x0U,τ+hLFxLLx0L,x0U,τ+hUFxULx0L,x0U,τ=cLτ,FUx0L,x0U,τ+hLFxLUx0L,x0U,τ+hUFxUUx0L,x0U,τ=cUτ,and hence, hLτ and hUτ are unknown quantities that can be obtained by solving the following equations, ∀τ∈0,1,(10)Jx0L,x0U,τh0Lτh0Uτ=cLτ−FLx0L,x0U,τcUτ−FUx0L,x0U,τ,where(11)Jx0L,x0U,τ=FxLLx0L,x0U,τFxULx0L,x0U,τFxLUx0L,x0U,τFxUUx0L,x0U,τ.Thus, our method in component form becomes(12)x1Lτx1Uτ=x0Lτx0Uτ+2−βJx0L,x0U,τ+βKy0L,y0U,τ−1∗3−βJx0L,x0U,τ+β−1Ky0L,y0U,τcLτ−FxLLx0L,x0U,τcUτ−FxLUx0L,x0U,τ,y0Lτy0Uτ=x0Lτx0Lτ+h0Lτh0Uτ,Ky0L,y0U,τ=FyLLy0L,y0U,τFyULy0L,y0U,τFyLUy0L,y0U,τFyUUy0L,y0U,τ.For approximate solutions ofxL and xU, we use the following recursive relation:(13)xn+1Lτxn+1Uτ=xnLτxnUτ+2−βJxnL,xnU,τ+βKynL,ynU,τ−1∗3−βJxnL,xnU,τ+β−1KynL,ynU,τcLτ−FxLLxnL,xnU,τcUτ−FxLUxnL,xnU,τ,where(14)ynLτynUτ=xnLτxnLτ+hnLτhnUτ,JxnL,xnU,τhnLτhnUτ=FxLLxnL,xnU,τFxULxnL,xnU,τFxLUxnL,xnU,τFxUUxnL,xnU,τ,KynL,ynU,τ=FyLLynL,ynU,τFyULynL,ynU,τFyLUynL,ynU,τFyUUynL,ynU,τ,∀τ∈0,1.For initial guess, one can use the fuzzy number(15)x0=x0L1,x0U1,x1L1−x0L0,x0U0−x0U1,and in the parametric form(16)x0τ=x0L1+x0L1−x0L0τ−1,x0Uτ=x0U1+x0U0−x0U11−τ.Remark 1.
SequencexnL,xnUn=0∞ converges to ξL,ξU iff ∀τ∈0,1, limn⟶∞xnLτ=ξLτ and limn⟶∞xnUτ=ξUτ.Lemma 1.
LetFξL,ξU=cL,cU and if the sequence of xnL,xnUn=0∞ converges to ξL,ξU to NM method, then(17)limn⟶∞Pn=0,where(18)Pn=sup0≤τ≤1maxhnLτ,hnUτ.Proof.
It is obviously because∀τ∈0,1 in convergent case(19)limn⟶∞hnLτ=limn⟶∞hnUτ=0.Under certain condition, finally it is shown that NM method is cubic convergence for fuzzy nonlinear equationFx=0. Thus, in compact form, we write(20)yn=xn−F′xn−1Fxn,xn+1=xn−Z∗F′xn−1Fxn,∀τ∈0,1,where Z=2−βF′xnL,xnU,τ+βF′ynL,ynU,τ−13−βF′xnL,xnU,τ+β−1F′ynL,ynU,τ and β∈ℝ (set of real numbers).Theorem 1.
Let,∀τ∈0,1, the functions FL and FU be continuously differentiable with respect to xnLτ and xnLτ. Assume that there exist ξLτ,ξUτ∈ℝ2 and α1,α2>0 such that J−1ξL,ξU,τ≤α1,K−1ξL,ξU,τ−1≤α2 and J,K will be Lipschitz continuous with respect to xnL and xnL; then the NM method converges to ξL,ξU and satisfies the following error equation:(21)en+1=2A22+12A3−βA22en3+Oen4,where Aj=1/j!∗Fjxn,τ/F′xn,τ,j=2,3,….Proof.
Leten=xn−ξ and en+1=xn+1−ξ be the errors in xn and xn+1; then, by Taylor series of Fxn,τ in the neighborhood of ξ, if J−1xn,τ exist, then(22)Fx,τ=Fxn,τ+F′xn,τx−xn+12!F″xn,τx−xn2+⋯,and Fξ,τ=0,(23)Fxn,τ=F′x,ξen+A2en2+A3en3+Oen4.
This gives(24)F′xn,τ−1Fxn,τ=en+A2en2+2A2+2A3en3+⋯,yn−ξ=A2en2+−2A2+2A3en3+⋯.
ExpandingF′yn,τ about ξ, we have(25)F′yn,τ=1+2A22en2+2−2A22+2A3en3+⋯,ZF′xn,τ−1Fxn,τ=en2+−2A22−12A3+βA22en3+⋯,xn+1−ξ=xn−ξ−en2+−2A22−12A3+βA22en3+⋯,en+1=2A22+12A3−βA22en3+Oen4.
Hence, the theorem is proved.There are some well-known existing methods in the literature for solving triangular fuzzy nonlinear equations.Fuzzy version of well-known Newton method [15] (abbreviated as NR) for finding roots of triangular fuzzy equation is as follows:(26)xn+1Lτxn+1Uτ=xnLτxnLτ+hnLτhnUτ,where(27)hnLτhnUτ=J−1xnL,xnU,τcLτ−FLxnL,xnU,τcUτ−FUxnL,xnU,τ.Midpoint iterative schemes [26] (abbreviated as NP) for triangular fuzzy equation are as follows:(28)xn+1Lτxn+1Uτ=ynLτynUτ+J−1xnL+ynL2,ynU+xnU2,τcLτ−FLxnL,xnU,τcUτ−FUxnL,xnU,τ,where(29)ynLτynUτ=xnLτxnLτ+hnLτhnUτ.


## 4. Numerical Applications

Here, we present examples to illustrating Newton’s method for fuzzy nonlinear equations. Examples1 and 2 are considered from Buckley and Qu [11, 27, 28]. All the computations are performed using CAS Maple 18 with 64-digit floating point arithmetic with stopping criteria as follows:(30)ien=Fx,τ<∈,iien=xn+1τ−xnτ<∈,where en represents the absolute error. We take ∈=10−15. In all numerical calculations, we used β=0.000001.Figure1 shows computational time in seconds of iterative schemes NM, NR, and NP for nonlinear fuzzy equations in Examples 1–3 (Cases 1 and 2) and 4, respectively (Algorithm 1).Figure 1
Computational time in seconds.Algorithm 1: NM method.
Step 1: transformFx,τ=c into FLxL,xU,τ=cLτ,FUxL,xU,τ=cUτ,∀τ∈0,1.Step 2: solveFLxL,xU,τ=cLτ,FUxL,xU,τ=cUτ. for τ=0 and τ=1 to obtain xL0 and xU0.Step 3: evaluateFx,τ=c at xL0 and xU0 and compute Jacobian matrix JxL,xU,τ.Step 4: use NM to compute next iterations.Step 5: for given∈>0, ien=Fx,τ<∈iien=xn+1τ−xnτ<∈, then stop.Step 6: setk=k+1 and go to Step 1.


### 4.1. Engineering Applications

Example 1.
Fraction conversion of Nitrogen-Hydrogen feed to ammonia is known as fractional conversion. Here, we consider the value of temperature and pressure as triangular fuzzy number, which results in the following fuzzy nonlinear equation:(31)∇1xτ4−∇2xτ3+∇3xτ2+∇4xτ=∇5,where ∇1=1,2.5,3.7, ∇2=7.7,7.9,8.1, ∇3=14.7,14.8,14.9,∇4=2.5,2.7,2.9, and ∇5=1.6,1.7,2.0 are triangular fuzzy numbers. Without any loss of generality, assume that x is positive; then the parametric form of this equation is as follows:(32)1+1.5τxLτ4−7.7+0.2τxLτ3+14.7+0.1τxLτ2+2.5+0.2τxLτ=1.6+0.1τ,3.7−1.2τxUτ4−8.1−0.2τxUτ3+14.9−0.1τ2xUτ+2.9−0.2τxUτ=2−0.3τ.
Figure2 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 1.
Table1 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 1.
To obtain initial guess, we use the above system forτ=1 and τ=0; therefore,(33)2.5xL41−7.9xL21+14.8xL21+2.7xL1=1.7,2.5xU41−7.9xU21+14.8xU21+2.7xU1=1.7,1xL40−7.7xL30+14.7xL20+2.5xL0=1.6,3.7xU40−8.1xU20+14.9xU0+2.9xU0=2.
ConsequentlyxL0=0.2, xU0=0.2, and xL1=xU1=0.2737. Therefore, initial guess is x0=0.26,0.2737,0.29. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(34)1+1.5τxLτ4−7.7+0.2τxLτ3+14.7+0.1τxLτ2+2.5+0.2τxLτ=1.6+0.1τ,3.7−1.2τxUτ4−8.1−0.2τxUτ3+14.9−0.1τ2xUτ+2.9−0.2τxUτ=2−0.3τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 2
Analytical and numerical approximate solutions of Example1. (a) Analytical solution for Example 1. (b) NM method for Example 1. (c) NR method for Example 1. (d) NP method for Example 1.
(a)(b)(c)(d)Table 1
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.26990.29773.9e − 196.9e − 200.3e − 52.7e − 46.3e − 87.3e − 90.10.27030.29541.3e − 191.3e − 186.2e − 55.2e − 66.2e − 95.9e − 80.20.27070.29312.1e − 177.1e − 185.2e − 57.4e − 51.6e − 96.4e − 70.30.27110.29089.7e − 151.8e − 196.4e − 64.6e − 48.8e − 94.3e − 80.40.27150.28849.5e − 168.5e − 166.1e − 55.3e − 67.8e − 91.4e − 90.50.27180.28601.4e − 151.9e − 155.7e − 78.3e − 57.7e − 90.5e − 80.60.27220.28364.0e − 166.9e − 197.5e − 53.4e − 45.5e − 83.8e − 70.70.27260.28127.6e − 173.9e − 188.1e − 64.8e − 38.7e − 91.2e − 70.80.27290.27870.4e − 188.2e − 195.8e − 70.4e − 68.9e − 82.6e − 90.90.27330.27622.6e − 176.1e − 183.8e − 51.3e − 84.8e − 92.3e − 910.27370.27379.6e − 191.2e − 177.9e − 57.8e − 79.8e − 81.8e − 8Example 2.
A chemical engineering problem is finding the volume of van der Waal’s equation. Van der Waal’s equation interprets real and ideal gas behavior, which results in the following equation:(35)P+A1n2x2x−nA2=nRT.
By using the specific values of parameter, we have the following fuzzy nonlinear equation:(36)Δ1xτ3−Δ2xτ2+Δ3xτ=Δ4,where Δ1=0.3,0.4,0.7, Δ2=1,2.3,3.4, Δ3=6.8,7,8.9,Δ4=4.1,5.2,6.7 are triangular fuzzy numbers and x represents the volume of the gas under observation, P for pressure, R for general gas constant, n for number of mole, and T for temperature, and A1,A2 are used for general parameters. Without any loss of generality, assume that xτ is positive; then the parametric form of this equation is as follows:(37)0.3+0.1τxLτ3−1+1.3τxLτ2+6.8+0.2τxLτ=4.1+1.1τ,0.7−0.3τxUτ3−3.4−1.1τxUτ2+8.9−1.9τxLτ=6.7−1.5τ.
Figure3 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 2.
Figure3 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 2.
Table2 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 2.
To obtain initial guess, we use the above system forτ=1 and τ=0; therefore(38)0.4xL31−2.3xL21+7.0xL1=5.2,0.4xU31−2.3xU21+7.0xU1=5.2,0.3xL30−1.0xL20+6.8xL0=4.1,0.7xU30−3.4xU20+8.9xU0=6.7.
ConsequentlyxL0=0.6, xU0=1.1, and xL1=xU1=1.028015150. Therefore, initial guess is x0=0.6,1.028015150,1.1. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose xτ is negative; we have(39)0.3+0.1τxLτ3−1+1.3τxLτ2+6.8+0.2τxLτ=4.1+1.1τ,0.7−0.3τxUτ3−3.4−1.1τxUτ2+8.9−1.9τxLτ=6.7−1.5τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 3
Analytical and numerical approximate solution of Example2. (a) Analytical solution for Example 2. (b) NM method for Example 2. (c) NR method for Example 2. (d) NP method for Example 2.
(a)(b)(c)(d)Table 2
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.65341.12343.9e − 172.9e − 167.3e − 53.7e − 35.3e − 96.3e − 80.10.67951.11851.3e − 151.3e − 161.2e − 38.2e − 64.2e − 96.9e − 80.20.70711.11343.1e − 164.1e − 161.2e − 57.6e − 65.6e − 98.4e − 90.30.73641.10802.7e − 151.3e − 174.4e − 31.6e − 48.8e − 83.3e − 80.40.76771.10243.5e − 158.3e − 136.0e − 48.3e − 47.2e − 95.4e − 90.50.80131.09651.4e − 141.9e − 156.7e − 60.3e − 57.7e − 78.5e − 90.60.83781.09034.0e − 167.9e − 177.5e − 51.4e − 65.2e − 79.8e − 80.70.87761.08387.6e − 163.9e − 188.6e − 60.8e − 68.7e − 92.2e − 80.80.92181.07708.4e − 188.2e − 166.8e − 64.4e − 63.9e − 70.6e − 90.90.97131.06899.6e − 172.1e − 174.8e − 55.3e − 74.1e − 91.3e − 911.02801.02809.0e − 181.2e − 177.9e − 57.8e − 79.6e − 71.0e − 7Example 3.
In engineering, the problem concerns the motion of object under fuzzy environment, resulting in two cases.Case 1.
Vertical motion of object.
Here, we are concerned with the vertical motion of ball by neglecting air resistance and assume constant acceleration 32 ft/sec2. Positive direction of the object y is upward from earth. Then, the resulting fuzzy nonlinear equation is written as(40)A1∗xτ2+V1∗xτ=Y1,where A1=0.8,1.0,1.2, free parameter, V1=1.3,1.75,2.2, velocity of the moving object under constant force of gravity, and Y1=0.1,0.15,0.20, initial position of the object. By substituting the value of A1,V1, and Y1 in the above equation of motion, we get(41)0.8,1.0,1.2xτ2+1.3,1.75,2.2xτ=0.1,0.15,0.20.
Then, we find the timexτ of the ball to hit the ground which depends on fuzzy parameter τ∈0,1. Without any loss of generality, assume that xτ is positive; then the parametric form of this equation is as follows:(42)0.8+0.2τxLτ2+1.3+0.45τxLτ=0.1+0.05τ,1.2−0.2τxUτ2+2.2−0.45τxUτ=0.2−0.05τ.
Figure4 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 1).
Figure4 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 1).
Table3 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 3 (Case 1).
To obtain initial guess, we use the above system forτ=0 and τ=1; therefore(43)1xL21+1.75xL1=0.6,1xU21+1.75xU1=0.15,0.8xL20+1.3xL0=0.1,1.2xU20+2.2xU0=0.2.
ConsequentlyxL0=0.0735, xU0=0.0867, and xL1=xU1=0.0818. Therefore, initial guess is x0=0.0735,0.0818,0.0867. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(44)0.8+0.2τxLτ2+1.3+0.45τxLτ=0.1+0.05τ,1.2−0.2τxUτ2+2.2−0.45τxUτ=0.2−0.05τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 4
Analytical and numerical approximate solution of Example3 (Case 1). (a) Analytical solution for Example 3 (Case 1). (b) NM method for Example 3 (Case 1). (c) NR method for Example 3 (Case 1). (d) NP method for Example 3 (Case 1).
(a)(b)(c)(d)Table 3
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.07350.08674.9e − 162.9e − 167.3e − 63.7e − 61.4e − 81.3e − 80.10.07460.08632.3e − 151.3e − 151.2e − 51.2e − 53.2e − 86.6e − 80.20.07560.08593.1e − 150.1e − 161.2e − 51.6e − 55.6e − 85.4e − 80.30.07660.08551.7e − 141.2e − 152.4e − 51.3e − 58.8e − 84.3e − 80.40.07750.08503.5e − 145.3e − 143.0e − 41.0e − 41.2e − 73.4e − 80.50.07830.08466.4e − 151.3e − 153.7e − 50.3e − 51.7e − 72.5e − 80.60.07910.08411.0e − 167.0e − 164.5e − 55.4e − 52.2e − 71.8e − 80.70.07980.08361.6e − 163.9e − 165.6e − 60.2e − 62.7e − 71.2e − 80.80.08050.08302.4e − 131.2e − 164.8e − 71.4e − 73.3e − 77.6e − 90.90.08120.08243.6e − 173.1e − 173.8e − 62.3e − 64.0e − 71.4e − 910.08180.08187.0e − 181.0e − 187.0e − 61.8e − 64.6e − 71.8e − 9Case 2.
Downward motion of object.
Here, we discuss the downward motion of object which results in the following fuzzy nonlinear equation:(45)3,4,5xτ2+1,2,3xτ=1,2,3.
Without any loss of generality, assume thatx is positive; then the parametric form of this equation is as follows:(46)3+τxLτ2+1+τxLτ=1+τ,5−τxUτ2+3−τxUτ=3−τ.
Figure5 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 2).
Figure5 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 2).
Table4 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 3 (Case 2).
To obtain initial guess, we use the above system forτ=0 and τ=1; therefore,(47)4xL21+2xL1=2,4xU21+2xU1=2,3xL20+xL0=1,5xU20+3xU0=3.
ConsequentlyxL0=0.4343, xU0=0.5307, and xL1=xU1=0.5. Therefore, initial guess is x0=0.4343,0.5,0.5307. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(48)3+τxLτ2+1+τxLτ=1+τ,5−τxUτ2+3−τxUτ=3−τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 5
Analytical and numerical approximate solution of Example3 (Case 2). (a) Analytical solution for Example 3 (Case 2). (b) NM method for Example 3 (Case 2). (c) NR method for Example 3 (Case 2). (d) NP method for Example 3 (Case 2).
(a)(b)(c)(d)Table 4
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.43430.53061.4e − 151.1e − 293.0e − 42.0e − 74.4e − 95.3e − 160.10.44410.52586.9e − 242.1e − 292.0e − 61.5e − 71.5e − 132.1e − 160.20.45290.52331.3e − 224.8e − 304.4e − 61.0e − 76.5e − 132.2e − 160.30.46080.52051.1e − 217.2e − 317.2e − 57.3e − 81.9e − 120.1e − 200.40.46800.51766.1e − 210.1e − 321.0e − 54.5e − 84.5e − 126.0e − 170.50.47450.51452.6e − 208.1e − 351.4e − 52.5e − 89.0e − 123.1e − 170.60.48050.51448.3e − 208.7e − 331.9e − 51.1e − 81.5e − 111.0e − 160.70.48590.51122.1e − 199.4e − 332.3e − 54.4e − 92.5e − 116.6e − 170.80.49100.50774.9e − 193.0e − 332.8e − 51.0e − 93.8e − 111.1e − 150.90.49560.50409.9e − 190.1e − 363.3e − 57.5e − 115.3e − 112.1e − 1610.50.51.8e − 182.0e − 363.9e − 50.1e − 157.2e − 110.1e − 18Example 4.
Consider the fuzzy nonlinear equation:(49)3,4,5xτ2+1,2,3sinxτ=1,2,3.
Without any loss of generality, assume thatx is positive; then the parametric form of this equation is as follows:(50)3+τxL2+1+τsinxLτ=1+τ,5−τxU2+3−τsinxUτ=3−τ.
Figure6 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 4.
Figure6 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 4.
Table5 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 4.
To obtain initial guess, we use the above system forτ=0 and τ=1; therefore,(51)4xL21+2sinxL1=2,4xU21+2sinxU1=2,3xL20+sinxL0=1,5xU20+3sinxU0=3.
ConsequentlyxL0=0.4380, xU0=0.5399, and xL1=xU1=0.5071. Therefore, initial guess is x0=0.4380,0.5071,0.5399. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(52)3+τxLτ2+1+τxLτ=1+τ,5−τxUτ2+3−τxUτ=3−τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 6
Analytical and numerical approximate solution of Example4. (a) Analytical solution for Example 4. (b) NM method for Example 4. (c) NR method for Example 4. (d) NP method for Example 4.
(a)(b)(c)(d)Table 5
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.43800.53992.7e − 254.7e − 283.6e − 62.4e − 71.1e − 144.2e − 150.10.44830.53771.2e − 231.7e − 282.4e − 61.4e − 72.8e − 133.3e − 150.20.45750.53741.8e − 225.3e − 299.5e − 66.4e − 75.2e − 132.5e − 160.30.46580.53481.4e − 211.3e − 292.6e − 63.6e − 78..2e − 121.8e − 160.40.47330.53206.8e − 213.0e − 305.8e − 51.8e − 71.1e − 121.3e − 160.50.48010.52902.4e − 204.6e − 311.1e − 50.1e − 81.7e − 114.2e − 170.60.48640.52596.8e − 203.5e − 324.5e − 52.1e − 82.0e − 175.1e − 180.70.49220.52261.6e − 209.4e − 332.8e − 51.2e − 82.5e − 112.7e − 180.80.49750.51913.3e − 193.0e − 344.1e − 53.4e − 82.4e − 111.1e − 190.90.50250.51136.2e − 192.1e − 355.7e − 57.3e − 93.4e − 113.6e − 1910.50710.50711.0e − 193.0e − 367.6e − 52.0e − 103.9e − 115.8e − 19


## 4.1. Engineering Applications

Example 1.
Fraction conversion of Nitrogen-Hydrogen feed to ammonia is known as fractional conversion. Here, we consider the value of temperature and pressure as triangular fuzzy number, which results in the following fuzzy nonlinear equation:(31)∇1xτ4−∇2xτ3+∇3xτ2+∇4xτ=∇5,where ∇1=1,2.5,3.7, ∇2=7.7,7.9,8.1, ∇3=14.7,14.8,14.9,∇4=2.5,2.7,2.9, and ∇5=1.6,1.7,2.0 are triangular fuzzy numbers. Without any loss of generality, assume that x is positive; then the parametric form of this equation is as follows:(32)1+1.5τxLτ4−7.7+0.2τxLτ3+14.7+0.1τxLτ2+2.5+0.2τxLτ=1.6+0.1τ,3.7−1.2τxUτ4−8.1−0.2τxUτ3+14.9−0.1τ2xUτ+2.9−0.2τxUτ=2−0.3τ.
Figure2 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 1.
Table1 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 1.
To obtain initial guess, we use the above system forτ=1 and τ=0; therefore,(33)2.5xL41−7.9xL21+14.8xL21+2.7xL1=1.7,2.5xU41−7.9xU21+14.8xU21+2.7xU1=1.7,1xL40−7.7xL30+14.7xL20+2.5xL0=1.6,3.7xU40−8.1xU20+14.9xU0+2.9xU0=2.
ConsequentlyxL0=0.2, xU0=0.2, and xL1=xU1=0.2737. Therefore, initial guess is x0=0.26,0.2737,0.29. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(34)1+1.5τxLτ4−7.7+0.2τxLτ3+14.7+0.1τxLτ2+2.5+0.2τxLτ=1.6+0.1τ,3.7−1.2τxUτ4−8.1−0.2τxUτ3+14.9−0.1τ2xUτ+2.9−0.2τxUτ=2−0.3τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 2
Analytical and numerical approximate solutions of Example1. (a) Analytical solution for Example 1. (b) NM method for Example 1. (c) NR method for Example 1. (d) NP method for Example 1.
(a)(b)(c)(d)Table 1
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.26990.29773.9e − 196.9e − 200.3e − 52.7e − 46.3e − 87.3e − 90.10.27030.29541.3e − 191.3e − 186.2e − 55.2e − 66.2e − 95.9e − 80.20.27070.29312.1e − 177.1e − 185.2e − 57.4e − 51.6e − 96.4e − 70.30.27110.29089.7e − 151.8e − 196.4e − 64.6e − 48.8e − 94.3e − 80.40.27150.28849.5e − 168.5e − 166.1e − 55.3e − 67.8e − 91.4e − 90.50.27180.28601.4e − 151.9e − 155.7e − 78.3e − 57.7e − 90.5e − 80.60.27220.28364.0e − 166.9e − 197.5e − 53.4e − 45.5e − 83.8e − 70.70.27260.28127.6e − 173.9e − 188.1e − 64.8e − 38.7e − 91.2e − 70.80.27290.27870.4e − 188.2e − 195.8e − 70.4e − 68.9e − 82.6e − 90.90.27330.27622.6e − 176.1e − 183.8e − 51.3e − 84.8e − 92.3e − 910.27370.27379.6e − 191.2e − 177.9e − 57.8e − 79.8e − 81.8e − 8Example 2.
A chemical engineering problem is finding the volume of van der Waal’s equation. Van der Waal’s equation interprets real and ideal gas behavior, which results in the following equation:(35)P+A1n2x2x−nA2=nRT.
By using the specific values of parameter, we have the following fuzzy nonlinear equation:(36)Δ1xτ3−Δ2xτ2+Δ3xτ=Δ4,where Δ1=0.3,0.4,0.7, Δ2=1,2.3,3.4, Δ3=6.8,7,8.9,Δ4=4.1,5.2,6.7 are triangular fuzzy numbers and x represents the volume of the gas under observation, P for pressure, R for general gas constant, n for number of mole, and T for temperature, and A1,A2 are used for general parameters. Without any loss of generality, assume that xτ is positive; then the parametric form of this equation is as follows:(37)0.3+0.1τxLτ3−1+1.3τxLτ2+6.8+0.2τxLτ=4.1+1.1τ,0.7−0.3τxUτ3−3.4−1.1τxUτ2+8.9−1.9τxLτ=6.7−1.5τ.
Figure3 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 2.
Figure3 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 2.
Table2 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 2.
To obtain initial guess, we use the above system forτ=1 and τ=0; therefore(38)0.4xL31−2.3xL21+7.0xL1=5.2,0.4xU31−2.3xU21+7.0xU1=5.2,0.3xL30−1.0xL20+6.8xL0=4.1,0.7xU30−3.4xU20+8.9xU0=6.7.
ConsequentlyxL0=0.6, xU0=1.1, and xL1=xU1=1.028015150. Therefore, initial guess is x0=0.6,1.028015150,1.1. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose xτ is negative; we have(39)0.3+0.1τxLτ3−1+1.3τxLτ2+6.8+0.2τxLτ=4.1+1.1τ,0.7−0.3τxUτ3−3.4−1.1τxUτ2+8.9−1.9τxLτ=6.7−1.5τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 3
Analytical and numerical approximate solution of Example2. (a) Analytical solution for Example 2. (b) NM method for Example 2. (c) NR method for Example 2. (d) NP method for Example 2.
(a)(b)(c)(d)Table 2
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.65341.12343.9e − 172.9e − 167.3e − 53.7e − 35.3e − 96.3e − 80.10.67951.11851.3e − 151.3e − 161.2e − 38.2e − 64.2e − 96.9e − 80.20.70711.11343.1e − 164.1e − 161.2e − 57.6e − 65.6e − 98.4e − 90.30.73641.10802.7e − 151.3e − 174.4e − 31.6e − 48.8e − 83.3e − 80.40.76771.10243.5e − 158.3e − 136.0e − 48.3e − 47.2e − 95.4e − 90.50.80131.09651.4e − 141.9e − 156.7e − 60.3e − 57.7e − 78.5e − 90.60.83781.09034.0e − 167.9e − 177.5e − 51.4e − 65.2e − 79.8e − 80.70.87761.08387.6e − 163.9e − 188.6e − 60.8e − 68.7e − 92.2e − 80.80.92181.07708.4e − 188.2e − 166.8e − 64.4e − 63.9e − 70.6e − 90.90.97131.06899.6e − 172.1e − 174.8e − 55.3e − 74.1e − 91.3e − 911.02801.02809.0e − 181.2e − 177.9e − 57.8e − 79.6e − 71.0e − 7Example 3.
In engineering, the problem concerns the motion of object under fuzzy environment, resulting in two cases.Case 1.
Vertical motion of object.
Here, we are concerned with the vertical motion of ball by neglecting air resistance and assume constant acceleration 32 ft/sec2. Positive direction of the object y is upward from earth. Then, the resulting fuzzy nonlinear equation is written as(40)A1∗xτ2+V1∗xτ=Y1,where A1=0.8,1.0,1.2, free parameter, V1=1.3,1.75,2.2, velocity of the moving object under constant force of gravity, and Y1=0.1,0.15,0.20, initial position of the object. By substituting the value of A1,V1, and Y1 in the above equation of motion, we get(41)0.8,1.0,1.2xτ2+1.3,1.75,2.2xτ=0.1,0.15,0.20.
Then, we find the timexτ of the ball to hit the ground which depends on fuzzy parameter τ∈0,1. Without any loss of generality, assume that xτ is positive; then the parametric form of this equation is as follows:(42)0.8+0.2τxLτ2+1.3+0.45τxLτ=0.1+0.05τ,1.2−0.2τxUτ2+2.2−0.45τxUτ=0.2−0.05τ.
Figure4 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 1).
Figure4 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 1).
Table3 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 3 (Case 1).
To obtain initial guess, we use the above system forτ=0 and τ=1; therefore(43)1xL21+1.75xL1=0.6,1xU21+1.75xU1=0.15,0.8xL20+1.3xL0=0.1,1.2xU20+2.2xU0=0.2.
ConsequentlyxL0=0.0735, xU0=0.0867, and xL1=xU1=0.0818. Therefore, initial guess is x0=0.0735,0.0818,0.0867. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(44)0.8+0.2τxLτ2+1.3+0.45τxLτ=0.1+0.05τ,1.2−0.2τxUτ2+2.2−0.45τxUτ=0.2−0.05τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 4
Analytical and numerical approximate solution of Example3 (Case 1). (a) Analytical solution for Example 3 (Case 1). (b) NM method for Example 3 (Case 1). (c) NR method for Example 3 (Case 1). (d) NP method for Example 3 (Case 1).
(a)(b)(c)(d)Table 3
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.07350.08674.9e − 162.9e − 167.3e − 63.7e − 61.4e − 81.3e − 80.10.07460.08632.3e − 151.3e − 151.2e − 51.2e − 53.2e − 86.6e − 80.20.07560.08593.1e − 150.1e − 161.2e − 51.6e − 55.6e − 85.4e − 80.30.07660.08551.7e − 141.2e − 152.4e − 51.3e − 58.8e − 84.3e − 80.40.07750.08503.5e − 145.3e − 143.0e − 41.0e − 41.2e − 73.4e − 80.50.07830.08466.4e − 151.3e − 153.7e − 50.3e − 51.7e − 72.5e − 80.60.07910.08411.0e − 167.0e − 164.5e − 55.4e − 52.2e − 71.8e − 80.70.07980.08361.6e − 163.9e − 165.6e − 60.2e − 62.7e − 71.2e − 80.80.08050.08302.4e − 131.2e − 164.8e − 71.4e − 73.3e − 77.6e − 90.90.08120.08243.6e − 173.1e − 173.8e − 62.3e − 64.0e − 71.4e − 910.08180.08187.0e − 181.0e − 187.0e − 61.8e − 64.6e − 71.8e − 9Case 2.
Downward motion of object.
Here, we discuss the downward motion of object which results in the following fuzzy nonlinear equation:(45)3,4,5xτ2+1,2,3xτ=1,2,3.
Without any loss of generality, assume thatx is positive; then the parametric form of this equation is as follows:(46)3+τxLτ2+1+τxLτ=1+τ,5−τxUτ2+3−τxUτ=3−τ.
Figure5 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 2).
Figure5 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 3 (Case 2).
Table4 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 3 (Case 2).
To obtain initial guess, we use the above system forτ=0 and τ=1; therefore,(47)4xL21+2xL1=2,4xU21+2xU1=2,3xL20+xL0=1,5xU20+3xU0=3.
ConsequentlyxL0=0.4343, xU0=0.5307, and xL1=xU1=0.5. Therefore, initial guess is x0=0.4343,0.5,0.5307. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(48)3+τxLτ2+1+τxLτ=1+τ,5−τxUτ2+3−τxUτ=3−τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 5
Analytical and numerical approximate solution of Example3 (Case 2). (a) Analytical solution for Example 3 (Case 2). (b) NM method for Example 3 (Case 2). (c) NR method for Example 3 (Case 2). (d) NP method for Example 3 (Case 2).
(a)(b)(c)(d)Table 4
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.43430.53061.4e − 151.1e − 293.0e − 42.0e − 74.4e − 95.3e − 160.10.44410.52586.9e − 242.1e − 292.0e − 61.5e − 71.5e − 132.1e − 160.20.45290.52331.3e − 224.8e − 304.4e − 61.0e − 76.5e − 132.2e − 160.30.46080.52051.1e − 217.2e − 317.2e − 57.3e − 81.9e − 120.1e − 200.40.46800.51766.1e − 210.1e − 321.0e − 54.5e − 84.5e − 126.0e − 170.50.47450.51452.6e − 208.1e − 351.4e − 52.5e − 89.0e − 123.1e − 170.60.48050.51448.3e − 208.7e − 331.9e − 51.1e − 81.5e − 111.0e − 160.70.48590.51122.1e − 199.4e − 332.3e − 54.4e − 92.5e − 116.6e − 170.80.49100.50774.9e − 193.0e − 332.8e − 51.0e − 93.8e − 111.1e − 150.90.49560.50409.9e − 190.1e − 363.3e − 57.5e − 115.3e − 112.1e − 1610.50.51.8e − 182.0e − 363.9e − 50.1e − 157.2e − 110.1e − 18Example 4.
Consider the fuzzy nonlinear equation:(49)3,4,5xτ2+1,2,3sinxτ=1,2,3.
Without any loss of generality, assume thatx is positive; then the parametric form of this equation is as follows:(50)3+τxL2+1+τsinxLτ=1+τ,5−τxU2+3−τsinxUτ=3−τ.
Figure6 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 4.
Figure6 shows analytical and numerical approximate solution of fuzzy nonlinear equation in Example 4.
Table5 clearly shows the dominance behavior of NM over NR and NP in terms of absolute error on the same number of iterations n = 3 for Example 4.
To obtain initial guess, we use the above system forτ=0 and τ=1; therefore,(51)4xL21+2sinxL1=2,4xU21+2sinxU1=2,3xL20+sinxL0=1,5xU20+3sinxU0=3.
ConsequentlyxL0=0.4380, xU0=0.5399, and xL1=xU1=0.5071. Therefore, initial guess is x0=0.4380,0.5071,0.5399. After 3 iterations, we obtain the solution which the maximum error would be less than 10−3. Now suppose x is negative; we have(52)3+τxLτ2+1+τxLτ=1+τ,5−τxUτ2+3−τxUτ=3−τ.
Forτ=0, we have xU0>xL0, and therefore, negative root does not exist.Figure 6
Analytical and numerical approximate solution of Example4. (a) Analytical solution for Example 4. (b) NM method for Example 4. (c) NR method for Example 4. (d) NP method for Example 4.
(a)(b)(c)(d)Table 5
Comparison of iterative schemes NM, NR, and NP on the same number of iterationsn=3.
τxLxUNMNRNPxn+1−xnfxnxn+1−xnfxnxn+1−xnfxn0.00.43800.53992.7e − 254.7e − 283.6e − 62.4e − 71.1e − 144.2e − 150.10.44830.53771.2e − 231.7e − 282.4e − 61.4e − 72.8e − 133.3e − 150.20.45750.53741.8e − 225.3e − 299.5e − 66.4e − 75.2e − 132.5e − 160.30.46580.53481.4e − 211.3e − 292.6e − 63.6e − 78..2e − 121.8e − 160.40.47330.53206.8e − 213.0e − 305.8e − 51.8e − 71.1e − 121.3e − 160.50.48010.52902.4e − 204.6e − 311.1e − 50.1e − 81.7e − 114.2e − 170.60.48640.52596.8e − 203.5e − 324.5e − 52.1e − 82.0e − 175.1e − 180.70.49220.52261.6e − 209.4e − 332.8e − 51.2e − 82.5e − 112.7e − 180.80.49750.51913.3e − 193.0e − 344.1e − 53.4e − 82.4e − 111.1e − 190.90.50250.51136.2e − 192.1e − 355.7e − 57.3e − 93.4e − 113.6e − 1910.50710.50711.0e − 193.0e − 367.6e − 52.0e − 103.9e − 115.8e − 19


## 5. Conclusion

In this research paper, we construct highly efficient family of two-step numerical iterative method to approximate roots of triangular fuzzy nonlinear equations. A set of real life applications are considered as numerical test problems show the practical performance and dominance efficiency of NM over NP and NR. From Tables1–5 and Figures 1–6, we observe that numerical results of test examples, CPU time, and residual errors corroborate theoretical analysis and illustrate the effectiveness and rapid convergence of our proposed family of iterative method NM as compared to the methods NP and NR.



---

*Source: 6916282-2021-11-01.xml*