# Optimization of One-Step Block Method for Solving Second-Order Fuzzy Initial Value Problems

Authors: Safa Al-Refai; Muhammed I. Syam; Mohammed Al-Refai
Journal: Complexity
 (2021)

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## Abstract

In this article, we present a one-step hybrid block method for approximating the solutions of second-order fuzzy initial value problems. We prove the stability and convergence results of the method and present several examples to illustrate the efficiency and accuracy of the proposed method. The numerical results are compared with the existing ones in the literature.


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## Body

## 1. Introduction

The second-order initial value problems (2IVPs) of the form(1)y″=ft,y,y′,ya=y0,y′a=y1,t∈a,b,where f is a continuous function on a,b, are interesting problems and they have many applications, particularly in engineering, physics, biology, and chemistry fields. In general, it is very difficult to find the exact solution to such problems especially when f is a nonlinear function of t,y, and y′. Therefore, numerical methods can be used to find their approximate solutions. There are several numerical methods to solve these problems such as the one-step block method [1, 2], Taylor series method [3], Adomian decomposition method [4], fourth-order and Butcher’s fifth-order Runge–Kutta methods [5], and the two-step hybrid block method [6]. In this paper, we are interested to study the 2IVPs of a fuzzy type. These problems consist of fuzzy differential equations (FDEs) with fuzzy initial conditions. The fuzzy initial value problems (FIVPs) are often incomplete or ambiguous. For instance, initial conditions or the values of fuzzy differential equations may not be known accurately. In this situation, FDEs appear as a natural way to model dynamical systems under possibilities of uncertainty. To solve these equations, we define the derivative by one of three different approaches (see [7]). The first approach is based on the Hukuhara derivative instituted by Puri-Ralescu in 1983. The second approach is known as Zadeh’s extension principle, and the last approach is strongly generalized differentiability which is presented by Bede and Gal in 2005. In this study, we focus on the Hukuhara derivative in order to define our differential equations. The FIVPs have several applications that have been highlighted in many research areas, such as civil engineering, physics, control theory, economics, population models, and modeling hydraulic [8–10]. Most problems in physics and engineering are modeled by initial value problems (IVPs). They have many applications such as Bagley–Torvik problem [11], Lane–Emden second-order equations [12], and delay IVP [13]. Since the exact solution for such problems is difficult to compute, several researchers use numerical methods to deal with this task. For example, Hossain et al. [5] used the Runge–Kutta method of order four, Ramos et al. [6] used the hybrid block method (HBM), and Jameel et al [14] used the homotopy analysis method (HAM) and the optimum homotopy analysis method (OHAM).Second-order IVPs play an important role in several applications. Rufai et al. [12] solved the Lane–Emden second-order singular differential equations using three off-step points. They did not implement the second-order IVP. The block system they got is different from our block method. They combined the HBM with an appropriate algorithm which is applied to the first subinterval to circumvent the singular behavior at the left endpoint of the integration interval which made his accuracy depend on both methods. Also, they integrated part of the equation to deal with the singularity in his problem.We consider the fuzzy second-order initial value problems of the form(2)y″=ft,y,y′,ya=a^,y′a=b^,t∈a,b,where a^ and b^ are fuzzy numbers and f is the fuzzy function. Since the function and initial conditions are fuzzy, we apply the β-level set operations to obtain the components of the problem which are as follows:(3)ytβ=y1,βt,y2,βt,a^β=a1,a2,b^β=b1,b2,y′tβ=y1,β′t,y2,β′t,y′′tβ=y1,β′′t,y2,β′′t,ft,y,y′β=f1t,y1,βt,y2,βt,y1,β′t,y2,β′t,f2t,y1,βt,y2,βt,y1,β′t,y2,β′t,f1=minft,w,w′:w∈y1,βt,y2,βt,w′∈y1,β′t,y2,β′t,f2=maxft,w,w′:w∈y1,βt,y2,βt,w′∈y1,β′t,y2,β′t.We then solve the above min-max problems directly. These kinds of problems might be difficult to solve directly and obtaining exact solutions is not always possible. Therefore, researchers were interested in obtaining numerical solutions by using different methods, such as the decomposition method [15], the homotopy analysis method [14, 16], the Runge–Kutta method [17, 18], the least-square method [19], the interactive and standard arithmetic [20–22], the Fréchet derivative method [23], and solving delay fuzzy problems [24]. For more references, see [25, 26, 33–36].The decomposition method is investigated in many papers (see [15]). However, it does not work for all types of problems (see [15]). We use the direct method (min-max problem) to find the solution for the problems in [15], and it works for some of them. However, the direct method is sometimes very complicated and it may not be possible to apply it. Thus, we proposed a new method to find numerical solutions for these problems. Our method depends on the one-step hybrid block method. In this method, we try to optimize the local truncation errors in order to find the best choice of the step point. The main advantage of the proposed method is that it is self-starter where we do not need to use other methods to generate more initial starting conditions.The HBM is easy to use and it gives accurate results. We implement it for the second-order fuzzy initial value problems and it shows that the method works accurately. We use only one off-step method and it gives better results than other methods such as HAM and OHAM. This will open a new door for the researchers to use this approach to solve such problems. Its computational cost is small compared to other methods in the literature review. The order of the proposed method is 3 and the method is stable and convergent.The current paper is organized as follows. In Section2, preliminaries of fuzzy concepts and theorems will be presented. In Sections 3 and 4, we present the optimized one-step hybrid block method and some theoretical results. In Section 5, we apply the proposed method to the fuzzy initial value problems of second order, and we present some numerical results in Section 6 to show the efficiency of the proposed method. Finally, in Section 7, the results will be discussed and some conclusions will be presented.


## 2. Preliminaries

In this section, some preliminaries will be presented to be used in this paper.Definition 1.
(see [27]). A fuzzy number is a function ψ:ℝ⟶0,1 that satisfies the following:(1)
There existsc∈ℝ such that ψc=1(2)
ψsa+1−sb≥minψa,ψb for all a,b∈ℝ and s∈0,1(3)
The setx∈ℝ:ψx≥β is a closed subset of ℝ for all β∈0,1(4)
The closure ofx∈ℝ:ψx>0 is a compact set
The set of all fuzzy numbers is denoted byFℝ. If ψ∈Fℝ and β∈0,1, the β-level set is given by(4)ψβ=x∈ℝ:ψx≥β,and the 0-level set is given by(5)ψ0=closurex∈ℝ:ψx>0.
It is easy to see that we can writeψβ=ψ¯β,ψ¯β, where it holds that(1)
ψβ⊆ψγ if 0<γ≤β≤1(2)
If the sequenceβn is an increasing sequence in 0,1 converging to β, then Limn⟶∞ψβn=ψβ(3)
For anyβ∈0,1, −∞<ψ¯β≤ψ¯β<∞Example 1.
Letψ=a1,a2,a3 be a triangular fuzzy number. Then,(6)ψs=s−a1a2−a1,a1≤s≤a2,s−a3a2−a3,a2≤s≤a3,0,otherwise,and ψβ=1−βa1+βa2,1−βa3+βa2. The graph of the symmetric fuzzy triangle ψ=0,1,2 is given in Figure 1.Figure 1
Fuzzy triangle numberψ=0,1,2.Definition 2.
(see [27]). Let a^,b^∈Fℝ, β∈0,1, and c∈ℝ where a^β=a¯β,a¯β and b^β=b¯β,b¯β. Then,(1)
a^⊕b^β=a¯β+b¯β,a¯β+b¯β,(2)
a^⊖b^β=a¯β−b¯β,a¯β−b¯β,(3)
c⊙a^β=ca¯β,ca¯β,c≥0,ca¯β,ca¯β,c<0,(4)
If there existsc^∈Fℝ such that a^=b^+c^, then c^=a^⊖Hb^ is the H-difference of a^ and b^(5)
The Hausdorff matrixΓH:Fℝ×Fℝ⟶0,∞ is defined by(7)ΓHa^,b^=sup0<b≤1maxa¯β−b¯β,a¯β−b¯β,and Fℝ,ΓH is a matrix space.Definition 3.
(see [7]). The fuzzy-valued function is a function h^:V⟶Fℝ, where V be a real vector space and Fℝ be the set of fuzzy numbers. The function h^x can be written as fβx,gβx for all x∈V and any β∈0,1. The functions fβx and gβx are called β-cut functions of the fuzzy-valued function h^.Definition 4.
(see [7]). Let f:a,b⊂ℝ⟶Fℝ be a fuzzy function and x∈a,b. If there exists f′x∈Fℝ such that(8)f′x=limΔ⟶0+fx+Δ⊖HfxΔ=limΔ⟶0+fx⊖Hfx−ΔΔ,then f is called Hukuhara differentiable at x and the Hukuhara derivative of f at x is f′x.Theorem 1.
(see [28]). Let σ∈Fℝ and g:a,b⊂ℝ+⟶0,∞ be differentiable at x0∈a,b. Let f:a,b⟶Fℝ be a fuzzy function defined by fx=σ⊙gx. If g′x0>0, then f is Hukuhara differentiable at x0 and the Hukuhara derivative of f at x0 is f′x0=σ⊙g′x0.Example 2.
Letσ∈Fℝ and f:ℝ⟶Fℝ be a fuzzy function defined by fx=σ⊙x+1. Then, gx=x+1. If x>−1, then(9)gx=x+1>0,g′x=1>0.
Using Theorem1, we have f is Hukuhara differentiable on 0,∞ and the Hukuhara derivative of f is f′x=σ⊙1. For x<−1,Δ>0, but close to 0, and β∈0,1, we have(10)f¯βx+Δ−f¯βx=Δσ¯β,f¯βx+Δ−f¯βx=Δσ¯β,where f=f¯,f¯. Since(11)f¯βx+Δ−f¯βx≰f¯βx+Δ−f¯βx,then fx+Δ⊖Hfx does not exist. Thus, f is not Hukuhara differentiable on −∞,0. For x=−1,Δ>0, but close to 0, and β∈0,1, we have(12)fβ=σ¯βx+1,σ¯βx+1,x>−1,σ¯βx+1,σ¯βx+1,x≤−1,which implies that f is not Hukuhara differentiable at x=0.


## 3. One-Step Hybrid Block Method with One Off-Step Point

In this section, we drive a numerical method based on the one-step hybrid block method with one off-step point (HBM1),tn+k with 0<k<1, to solve the following differential equation of the form(13)y″t=ft,yt,y′t,t≥0,(14)yt0=y0,(15)y′t0=y1.To derive HBM1, we assume thattn=nh where h is the step size. Since we are planning to use one-step hybrid method with one off-step point, we need to interpolate the solution and its first derivative at tn and to collocate the IVP at tn and tn+k. Then, we solve these equations for the coefficients of the approximate solution. Thus, we have five equations. To able to get a unique solution, we assume that the solution has five coefficients. For this reason, we approximate the solution by polynomial of degree four. If we increase the order of the polynomial, we should take more off-step points. We approximate the solution of problem (13)–(15) by a polynomial of degree 4 as follows:(16)yt≃∑j=04ajtj,and its first derivative by(17)y′t≃∑j=14jajtj−1,and its second derivative by(18)y″t≃∑j=24jj−1ajtj−2.Interpolate equations (16) and (17) at the point tn and collocate equation (18) at the points tn,tn+k=tn+kh, and tn+1=tn+h to get the following system:(19)1tntn2tn3tn4012tn3tn24tn30026tn12tn20026tn+k12tn+k20026tn+112tn+12a0a1a2a3a4=ynyn′fnfn+kfn+1,where yn+j≈ytn+j, yn+j′≈y′tn+j, and fn+j≈y″tn+j,j=0,k,1. Solving system (19) after substituting t=tn+wh to get(20)yw≈yn+hαyn′+h2α0fn+αkfn+k+α1fn+1,y′w≈βyn′+hβ0fn+βkfn+k+β1fn+1,where α,α0,αk,α1,β,β0,βk, and β1 are functions of w. We then evaluate the approximation of yw and y′wat w=k and 1, to get(21)yn+j=yn+hαjyn′+h2α0,jfn+αk,jfn+k+α1,jfn+1,(22)yn+j′=βjyn′+hβ0,jfn+βk,jfn+k+β1,jfn+1,where j=k,1, and(23)α1=1,α0,1=13−112k,αk,1=112k−12k2,α1,1=1122+1k−1,αk=k,α0,k=−112k−4k2,αk,k=k−2k212k−1,α1,k=k412k−1,β1=1,β0,1=12−16k,βk,1=16k−6k2,β1,1=163+1k−1,βk=1,β0,k=−16k−3k,βk,k=2k−3k6k−1,β1,k=k36k−1.In the literature review, researchers used a uniform partition to the interval [0, 1] which makes the method of order 2. However, in this paper, we will choose the partition which makes the method have the largest possible order. To do that, we leavek as a parameter and we choose it so that the proposed method has the largest possible order. To maximize the order of the implicit block method (21) when j=1, we minimize the local truncation errors in the formula for yn+1,(24)ℒytn+1;h=5k−2360h5y5tn+5k2+5k−31440h6y6tn+Oh7.To maximize the order, we solve the following equation fork where 0<k<1(25)5k−2360=0.Hence,(26)k=25,and the local truncation errors for yn+1,yn+k,yn+1′, and yn+k′ are(27)ℒytn+1;h=−h6y6tn7200+Oh7=−1.38889∗10−4h6y6tn+Oh7,ℒytn+k;h=14h5y5tn46875+Oh6=2.98667∗10−4h5h5y5tn+Oh6,ℒy′tn+1;h=−h5y5tn360+Oh5=−2.7777∗10−3h5y5tn+Oh6,ℒy′tn+k;h=8h5y5tn5625+Oh6=1.4222210−3h5y5tn+Oh6.Then the order of HBM1 is3,3,3,3 and the error constant is 0,14/46875,−1/360,8/5625. Therefore, the HBM1 is given as follows:(28)yn+1=yn+hyn′+h2729fn+25fn+k+2fn+1,yn+k=yn+2h5yn′+2h2112527fn+20fn+k−2fn+1,yn+1′=yn′+h363fn+25fn+k+8fn+1,yn+k′=yn′+h22539fn+55fn+k−4fn+1.


## 4. Analysis of the Proposed Method

In this section, we study the main properties of the proposed method such as consistency, stability, and convergence. Let us write system (28) in the form(29)yn+2/5=yn+2h5yn′+2h2112527fn+20fn+2/5−2fn+1,yn+1=yn+hyn′+h2729fn+25fn+2/5+2fn+1,hyn+2/5′=hyn′+h222539fn+55fn+2/5−4fn+1,hyn+1′=hyn′+h2363fn+25fn+2/5+8fn+1.Then, we can rewrite system (29) in the matrix form as(30)A0Ym=A1ym+A2Fm,where(31)Ym=yn+2/5yn+1hyn+2/5′hyn+1′,ym=yn−2/5ynhyn−2/5′hyn′,Fm=h2fnh2fn+2/5h2fn+1,A0=1000010000100001,A1=01025010100010001,A2=541125401125−4112597225722723922555225−42253362536836.Following Fatunla’s approach [29], the characteristic equation of HBM1 is(32)detμA0−A1=detμ−10−250μ−10100μ−1000μ−1=μ2μ−12=0,which implies that μ1=μ2=0 and μ3=μ4=1. Then, the multiplicity of the nonzero roots of the characteristic equation is 2 which does not exceed the order of the differential equation. Hence, it is zero stable.FromSection 3, we see that the local truncation error of system (28) is(33)ℒytn;h=ℒytn+2/5;h,ℒytn+1;h,ℒy′tn+2/5;h,ℒy′tn+1;h=∑i=0∞γihiyitn=γ5h5y5tn+∑i=6∞γihiyitn,where γ0=γ1=γ2=γ3=γ4=0 and γ5=0,14/46875,8/5625,1/360T.Thus, system (28) has order 3,3,3,3T. For simplicity, we write the order as 3. Since the order of system (28) is 3>1, then it is consistent. The consistency and the zero stability of system (28) imply that it is convergent [11, 13].To find the region of absolute stability, we consider the following test problemy′=λy where λ<0, then(34)y″=ft,y,y′=λ2y.Substitutef in the following matrix form:(35)B0Yn+1=B1Yn+hB2Yn+1′+h2C0Fn+C1Fn+1,where(36)Yn+1=yn+2/5yn+1,Yn=yn−2/5yn,Yn′=yn−2/5′yn′,Fn=fn−2/5fn,Fn+1=fn+2/5fn+1,B0=1001,B1=0101,B2=02501,C0=05411250972,C1=401125−4112502572272,to get(37)B0Yn+1=B1Yn+hB2λYn+h2C0λ2Yn+C1λ2Yn+1.LetS=λh, then(38)Yn+1=MSYn,where MS=B0−S2C1−1B1+SB2+S2C0. The eigenvalue of MS is(39)0,900+900S+393S2+93S3+11S4900−57S2+2S4.Letf:ℂ⟶ℂ by fS=900+900S+393S2+93S3+11S4/900−57S2+2S4 where S=λh. The region of absolute stability will be all S∈ℂ such that fS<1. This region is given in Figure 2 and the interval of stability is −4.08611,0.Figure 2
Region of absolute stability.


## 5. Fuzzy Initial Value Problem

In this section, we will apply HBM1 on the fuzzy initial value problem. We will present two new theorems. Consider the following fuzzy initial value problem(40)y″t=ft,y,y′,t≥0,(41)y0=a^,(42)y′0=b^.Let theβ-level of the solution yt,a^,b^ and the function ft,y,y′ be given by(43)yt,β=y1t,β,y2t,β,y′t,β=y1′t,β,y2′t,β,y0,β=a1,a2,y′0,β=b1,b2,ft,y,y′,β=f1t,yt,β,y′t,β,f2t,yt,β,y′t,β.Following the technique described in previous sections, the fuzzy HBM1 is given by(44)y1,n+1,β=y1,n,β+hy1,n,β′+h2729f1,n,β+25f1,n+2/5,β+2f1,n+1,β,y1,n+2/5,β=y1,n,β+2h5y1,n,β′+2h2112527f1,n,β+20f1,n+2/5,β−2f1,n+1,β,y1,n+1,β′=y1,n,β′+h363f1,n,β+25f1,n+2/5,β+8f1,n+1,β,y1,n+2/5,β′=y1,n,β′+h22539f1,n,β+55f1,n+2/5,β−4f1,n+1,β,y2,n+1,β=y2,n,β+hy2,n,β′+h2729f2,n,β+25f2,n+2/5,β+2f2,n+1,β,y2,n+2/5,β=y2,n,β+2h5y2,n,β′+2h2112527f2,n,β+20f2,n+2/5,β−2f2,n+1,β,y2,n+1,β′=y2,n,β′+h363f2,n,β+25f2,n+2/5,β+8f2,n+1,β,y2,n+2/5,β′=y2,n,β′+h22539f2,n,β+55f2,n+2/5,β−4f2,n+1,β,where(45)f1,n,β=minftn,w,w′:w∈y1,n,β,y2,n,β,w′∈y1,n,β′,y2,n,β′,f2,n,β=maxftn,w,w′:w∈y1,n,β,y2,n,β,w′∈y1,n,β′,y2,n,β′,f1,n+2/5,β=minftn+2/5,w,w′:w∈y1,n+2/5,β,y2,n+2/5,β,w′∈y1,n+2/5,β′,y2,n+2/5,β′,f2,n+2/5,β=maxftn+2/5,w,w′:w∈y1,n+2/5,β,y2,n+2/5,β,w′∈y1,n+2/5,β′,y2,n+2/5,β′,f1,n+1,β=minftn+1,w,w′:w∈y1,n+1,β,y2,n+1,β,w′∈y1,n+1,β′,y2,n+1,β′,f2,n+1,β=maxftn+1,w,w′:w∈y1,n+1,β,y2,n+1,β,w′∈y1,n+1,β′,y2,n+1,β′.Theorem 2.
Letft,y,y′ be increasing function in y and y′. Then the following are true.(i)
Ify>0 and y′>0, then(46)f1,n+j,β=ftn+j,y1,n+j,β,y1,n+j,β′,f2,n+j,β=ftn+j,y2,n+j,β,y2,n+j,β′,forj=0,25,1.(ii)
Ify>0 and y′<0, then(47)f1,n+j,β=ftn+j,y1,n+j,β,y2,n+j,β′,f2,n+j,β=ftn+j,y2,n+j,β,y1,n+j,β′,forj=0,25,1.(iii)
Ify<0 and y′<0, then(48)f1,n+j,β=ftn+j,y2,n+j,β,y2,n+j,β′,f2,n+j,β=ftn+j,y1,n+j,β,y1,n+j,β′,forj=0,25,1.(iv)
Ify<0 and y′>0, then(49)f1,n+j,β=ftn+j,y2,n+j,β,y1,n+j,β′,f2,n+j,β=ftn+j,y1,n+j,β,y2,n+j,β′,forj=0,25,1.The proof of the theorem follows straight forward. We can generate the functionsfi,n+j,β for i=1,2 and j=0,2/5,1 for the decreasing case in similar way as in Theorem 2.Theorem 3.
Letc^=c1,c2 be a fuzzy number, a,b be a real number and ft,y=ay′+by+c^. Then the fuzzy system of HBM1 becomes(1)
Ifa≥0 and b≥0, then(50)Yn+1=A1−1B1yj,n,β′+C1A1−1yj,n,β+D1A1−1cj,forj=1 and 2, where(51)A1=1−8bh22254bh21125−8ah22254ah21125−25bh2721−2bh272−25ah272−2ah272−55bh2254bh2251−55ah2254ah225−25bh36−8bh36−25ah361−8ah36,B1=6ah2125+2h59ah272+h39ah225+13ah36+1,C1=6bh2125+19bh272+139bh2253bh36,D1=2h225h222h5h,Yn+1=yj,n+2/5,βyj,n+1,βyj,n+2/5,β′yj,n+1,β′.(2)
Ifa≥0 and b<0, then(52)Yn+1=A2−1B2y1,n,β′y2,n,β′+A2−1C2y1,n,βy2,n,β+A2−1D2c1c2,where(53)A2=10−8ah22254ah21125−8bh22254bh211250001−25ah272−2ah272−25bh272−2bh27200001−55ah2254ah225−55bh2254bh2250000−25ah361−8ah36−25bh36−8bh3600−8bh22254bh211250010−8ah22254ah21125−25bh272−2bh2720001−25ah272−2ah272−55bh2254bh22500001−55ah2254ah225−25bh36−8bh360000−25ah361−8ah36,B2=6ah2125+2h509ah272+h039ah225+103ah36+1006ah2125+2h509ah272+h039ah225+103ah36+1,C2=16bh212519bh272039bh22503bh366bh212519bh272139bh22503bh360,D2=2h2250h2202h50h002h2250h2202h50h,Yn+1=y1,n+2/5,βy1,n+1,βy1,n+2/5,β′y1,n+1,β′y2,n+2/5,βy2,n+1,βy2,n+2/5,β′y2,n+1,β′.(3)
Ifa≤0 and b>0, then(54)Yn+1=A3−1B3y1,n,β′y2,n,β′+A3−1C3y1,n,βy2,n,β+A3−1D3c1c2,where(55)A3=1−8bh22254bh211250000−8ah22254ah21125−25bh2721−2bh2720000−25ah272−2ah272−55bh2254bh2251000−55ah2254ah225−25bh36−8bh360100−25ah36−8ah3600−8ah22254ah211251−8bh22254bh211250000−25ah272−2ah272−25bh2721−2bh2720000−55ah2254ah225−55bh2254bh2251000−25ah36−8ah36−25bh36−8bh3601,B3=2h56ah2125h9ah272139ah22513ah366ah21252h59ah272h39ah22513ah361,C3=6bh2125+109bh272+1039bh22503bh36006bh2125+109bh272039bh22503bh36,D3=2h2250h2202h50h002h2250h2202h50h,Yn+1=y1,n+2/5,βy1,n+1,βy1,n+2/5,β′y1,n+1,β′y2,n+2/5,βy2,n+1,βy2,n+2/5,β′y2,n+1,β′.(4)
Ifa≤0 and b<0, then(56)Yn+1=A4−1B4y1,n,β′y2,n,β′+A4−1C4y1,n,βy2,n,β+A4−1D4c1c2,where(57)A4=1000−8bh22254bh21125−8ah22254ah211250100−25bh272−2bh272−25ah272−2ah2720010−55bh2254bh225−55ah2254ah2250001−25bh36−8bh36−25ah36−8ah36−8bh22254bh21125−8ah22254ah211251000−25bh272−2bh272−25ah272−2ah2720100−55bh2254bh225−55ah2254ah2250010−25bh36−8bh36−25ah36−8ah360001,B4=2h56ah2125h9ah272139ah22513ah366ah21252h59ah272h39ah22513ah361,C4=16bh212519bh272039bh22503bh366bh212519bh272139bh22503bh360,D4=2h2250h2202h50h002h2250h2202h50h,Yn+1=y1,n+2/5,βy1,n+1,βy1,n+2/5,β′y1,n+1,β′y2,n+2/5,βy2,n+1,βy2,n+2/5,β′y2,n+1,β′.Proof.
(1) Leta≥0 and b≥0. Then,(58)ftn+j,yi,n+j,β,yi,n+j,β′=ci+byi,n+j,β+ayi,n+j,β′,where i=1,2 and j=0,2/5,1. Thus,(59)y1,n+1,β=y1,n,β+hy1,n,β′+h2729c1+by1,n,β+ay1,n,β′+25c1+by1,n+2/5,β+ay1,n+2/5,β′+2c1+by1,n+1,β+ay1,n+1,β′,y1,n+2/5,β=y1,n,β+2h5y1,n,β′+2h2112527c1+by1,n,β+ay1,n,β′+20c1+by1,n+2/5,β+ay1,n+2/5,β′−2c1+by1,n+1,β+ay1,n+1,β′,y1,n+1,β′=y1,n,β′+h363c1+by1,n,β+ay1,n,β′+25c1+by1,n+2/5,β+ay1,n+2/5,β′+8c1+by1,n+1,β+ay1,n+1,β′,y1,n+2/5,β′=y1,n,β′+h22539c1+by1,n,β+ay1,n,β′+55c1+by1,n+2/5,β+ay1,n+2/5,β′−4c1+by1,n+1,β+ay1,n+1,β′,y2,n+1,β=y2,n,β+hy2,n,β′+h2729c2+by2,n,β+ay2,n,β′+25c2+by2,n+2/5,β+ay2,n+2/5,β′+2c2+by2,n+1,β+ay2,n+1,β′,y2,n+2/5,β=y2,n,β+2h5y2,n,β′+2h2112527c2+by2,n,β+ay2,n,β′+20c2+by2,n+2/5,β+ay2,n+2/5,β′−2c2+by2,n+1,β+ay2,n+1,β′,y2,n+1,β′=y2,n,β′+h363c2+by2,n,β+ay2,n,β′+25c2+by2,n+2/5,β+ay2,n+2/5,β′+8c2+by2,n+1,β+ay2,n+1,β′,y2,n+2/5,β′=y2,n,β′+h22539c2+by2,n,β+ay2,n,β′+55c2+by2,n+2/5,β+ay2,n+2/5,β′−4c2+by2,n+1,β+ay2,n+1,β′.
We can rewrite the last system in the matrix form as(60)A1Y1=B1y1,n,β′+C1y1,n,β+D1c1,A1Y2=B1y2,n,β′+C1y2,n,β+D1c2,where(61)A1=1−8bh22254bh21125−8ah22254ah21125−25bh2721−2bh272−25ah272−2ah272−55bh2254bh2251−55ah2254ah225−25bh36−8bh36−25ah361−8ah36,Y1=y1,n+2/5,βy1,n+1,βy1,n+2/5,β′y1,n+1,β′,Y2=y2,n+2/5,βy2,n+1,βy2,n+2/5,β′y2,n+1,β′,B1=6ah2125+2h59ah272+h39ah225+13ah36+1,C1=6bh2125+19bh272+139bh2253bh36,D1=2h225h222h5h.
Since(62)detA1=43740000a2h2+15309000abh3−306180000ah+1458000b2h4−41553000bh2+656100000656100000≠0,then A1−1 exist. Thus,(63)y1,n+2/5,βy1,n+1,βy1,n+2/5,β′y1,n+1,β′=E1y1,n,β′+E2y1,n,β+E3c1,where(64)E1=A1−1B1=2hha3abh3+16a2+bh2−1200−165bh+450025h60ha2+21bh2−20a+bh2bh2−57+900hh186bh+a−3abh3−5a2+bh2+60+18002h60ha2+21bh2−20a+bh2bh2−57+180022500−h4h8bh2+225a2+3b2bh2+135h2+500a+3bh24bh2−12525h60ha2+21bh2−20a+bh2bh2−57+900h5hbh2+36a2+3bh2+16bh2+20a+3bh9bh2+262+18002h60ha2+21bh2−20a+bh2bh2−57+1800,E2=A1−1C1=6ab2h5+8b4a2−5bh4−75abh3+3754a2+bh2−10500ah+2250025h60ha2+21bh2−20a+bh2bh2−57+900h−5hbh2−24a2−3bbh2+26h2+280a+2bh11bh2+393+18002h60ha2+21bh2−20a+bh2bh2−57+1800−2bhbh2+753bh2+16ah−6025h60ha2+21bh2−20a+bh2bh2−57+900bhbh2+123bh2+5ah+1502h60ha2+21bh2−20a+bh2bh2−57+1800,E3=A1−1D1=2h2ah−153bh2+16ah−6025h60ha2+21bh2−20a+bh2bh2−57+900−h2ah−63bh2+5ah+1502h60ha2+21bh2−20a+bh2bh2−57+1800−2hbh2+753bh2+16ah−6025h60ha2+21bh2−20a+bh2bh2−57+900hbh2+123bh2+5ah+1502h60ha2+21bh2−20a+bh2bh2−57+1800.
Using similar argument, we can prove (2), (13), and (14).
We endSection 5 by the following algorithm to solve problem (40)–(42).(1)
Lety1,0,β=a1,y2,0,β=a2,y1,0,β′=b1,y2,0,β′=b2, and ε>0(2)
Forn=1,2,…, do step 3 and 4(3)
Solve the fuzzy HBM1 to findy1,n,β,y2,n,β,y1,n,β′,y2,n,β′(4)
Ify1,n,β,y2,n,β−y1,n−1,β,y2,n−1,β<ε or n>Nmax, go to step 5, else n=n+1, go to step 3(5)
Stop(6)
Ifn≤Nmax, the approximate solution is y1,n,β,y2,n,β, else the method is divergent


## 6. Numerical Examples

In this section, we discuss four examples. Two are linear fuzzy initial value problems while the others are nonlinear.Example 3.
Consider the second-order fuzzy linear initial value problem(65)y″=−yx,y0=0,y′0=γ^β,where γ^β=0.9+0.1β,1.1−0.1β and h=0.1. Let yx=y1x,β,y2x,β be a fuzzy solution and y′x=y1′x,β,y2′x,β.
By implementing theβ-level sets, the problem will be(66)y1′′x,β,y2′′x,β=−y2x,β,−y1x,β,y10,β,y20,β=y1′0,β,y2′0,β=β,2−β.
Using HBM1, we get(67)y1,n+1,β=y1,n,β+hy1,n,β′−h2729y2,n,β+25y2,n+2/5,β+2y2,n+1,β,y1,n+2/5,β=y1,n,β+2h5y1,n,β′−2h2112527y2,n,β+20y2,n+2/5,β−2y2,n+1,β,y1,n+1,β′=y1,n,β′−h363y2,n,β+25y2,n+2/5,β+8y2,n+1,β,y1,n+2/5,β′=y1,n,β′−h22539y2,n,β+55y2,n+2/5,β−4y2,n+1,β,y2,n+1,β=y2,n,β+hy2,n,β′−h2729y1,n,β+25y1,n+2/5,β+2y1,n+1,β,y2,n+2/5,β=y2,n,β+2h5y2,n,β′−2h2112527y1,n,β+20y1,n+2/5,β−2y1,n+1,β,y2,n+1,β′=y2,n,β′−h363y1,n,β+25y1,n+2/5,β+8y1,n+1,β,y2,n+2/5,β′=y2,n,β′−h22539y1,n,β+55y1,n+2/5,β−4y1,n+1,β.
We can rewrite the above system in the matrix form as(68)Ym=A−1Bym+A−1Cym′,where(69)A=10002h27225h272000100−4h2112540h211250000108h3625h36000001−4h22555h225002h27225h272001000−4h2112540h211250001008h3625h36000010−4h22555h225000001,B=1−9h2721−54h211250−3h360−39h225−9h2721−54h211251−3h360−39h2250,C=h02h5010100h02h50101,Ym=y1,n+1y1,K+ny1,n+1′y1,k+n′y2,n+1y2,K+ny2,n+1′y2,k+n′,ym=y1,ny2,n,ym′=y1,n′y2,n′.
The error in the approximation ofy1,n and y2,n for β=0,0.25,0.5,0.75,1 is given in Tables 1 and 2 , respectively, where the exact solution is given by(70)0.1β−0.1sinhx+sinx,0.1−0.1βsinhx+sinx.
The approximate solutions ofy1 and y2 for β=0,0.25,0.5,0.75,1 are given in Figures 3 and 4, respectively. Also, in Figure 5, the approximate solutions for y1 and y2 become close to the crisp solution as β approaches to 1.Table 1
The absolute error in approximatingy1,n for h=0.1.
y1,nxβ00.250.50.7510000000.14.50043×10−124.39764×10−124.29483×10−124.19204×10−124.08924×10−120.22.49082×10−82.56047×10−82.63012×10−82.69978×10−82.76943×10−80.37.40714×10−87.6176×10−87.82805×10−88.03851×10−88.24896×10−80.41.46254×10−71.50506×10−71.54759×10−71.59012×10−71.63264×10−70.52.39628×10−72.46812×10−72.53997×10−72.61182×10−72.68366×10−70.63.51805×10−73.62763×10−73.73722×10−73.8468×10−73.95639×10−70.74.79859×10−74.95509×10−75.11159×10−75.26808×10−75.42458×10−70.86.20368×10−76.41718×10−76.63068×10−76.84418×10−77.05768×10−70.97.6945×10−77.9762×10−78.2579×10−78.5396×10−78.82131×10−719.22809×10−79.59051×10−79.95292×10−71.03153×10−61.06777×10−6Table 2
The absolute error in approximatingy2,n for h=0.1.
y2,nxβ00.250.50.7510000000.13.67807×10−123.78086×10−123.88367×10−123.98646×10−124.08926×10−120.23.04804×10−82.97839×10−82.90874×10−82.83908×10−82.76943×10−80.39.09078×10−88.88032×10−88.66987×10−88.45941×10−88.24896×10−80.41.80275×10−71.76023×10−71.7177×10−71.67517×10−71.63264×10−70.52.97104×10−72.8992×10−72.82735×10−72.75551×10−72.68366×10−70.64.39473×10−74.28515×10−74.17556×10−74.06598×10−73.95639×10−70.76.05057×10−75.89408×10−75.73758×10−75.58108×10−75.42458×10−70.87.91169×10−77.69819×10−77.48469×10−77.27119×10−77.05768×10−70.99.94812×10−79.66642×10−79.38471×10−79.10301×10−78.82131×10−711.21274×10−61.1765×10−61.14026×10−61.10402×10−61.06777×10−6Figure 3
The approximate solutiony1 for β=0,0.25,0.5,0.75,1.Figure 4
The approximate solutiony2 for β=0,0.25,0.5,0.75,1.Figure 5
The approximate solutiony1 and y2 and the crisp solution for β=0,0.25,0.5,0.75.
(a)(b)(c)(d)Example 4.
Consider the following nonlinear second-order fuzzy initial value problem(71)y″x=−y′x2,y0=γ^β,y′0=θ^β,0≤x≤0.1,where γ^β=β,2−β,θ^β=1+β,3−β, and h=0.01. Let yx=y1x,β,y2x,β be a fuzzy solution and y′x=y1′x,β,y2′x,β.
Usingβ-level sets, the problem will have the following form:(72)y1′′x,β,y2′′x,β=−y1x,β,y2x,β2,y10,β,y20,β=β,2−β,y1′0,β,y2′0,β=1+β,3−β,where the exact solution is given by(73)lnβeβ+eβx+eβ,ln3e2−β−βe2−βx+e2−β.
In Tables3 and 4 , we present the absolute error of results obtained by the current (HBM1) method and the ones obtained in [14].Table 3
Absolute error fory1,0.1.
y1,0.1βErrorError fory1,0.1Error forh1 HAM in [14]Error forh2 HAM in [14]Error for OHAM in [14]02.42502×10−101.53529×10−73.98956×10−81.59889×10−100.25.79712×10−104.51332×10−73.91122×10−103.91122×10−100.41.20466×10−91.1207×10−63.79328×10−93.79328×10−90.62.25974×10−92.4597×10−64.47279×10−82.61947×10−90.83.92067×10−94.9128×10−66.92457×10−86.69669×10−816.39707×10−99.10987×10−61.53606×10−71.11097×10−8Table 4
Absolute error fory2,0.1.
y2,0.1βErrorError fory2,0.1Error forh1 HAM in [14]Error forh2 HAM in [14]Error for OHAM in [14]04.07084×10−89.6735×10−56.7076×10−61.26440×10−70.22.98335×10−86.4822×10−53.6619×10−69.18889×10−80.42.13238×10−84.2133×10−51.8656×10−67.34552×10−80.61.48046×10−82.6432×10−58.76047×10−73.52946×10−80.89.93257×10−91.5907×10−53.76985×10−71.51728×10−816.39707×10−99.10987×10−61.53606×10−71.11097×10−8Example 5.
Consider the second-order fuzzy linear initial value problem(74)y″=yx,0<x<1,y0=θ^β,y′0=γ^β,where θ^β=γ^β=0.75+0.25β,1.125−0.125β and h=0.1. Let yx=y1x,β,y2x,β be a fuzzy solution.
By implementing theβ-level sets, the problem will be(75)y1′′x,β,y2′′x,β=y1x,β,y2x,β,y10,β,y20,β=y1′0,β,y2′0,β=0.75+0.25β,1.125−0.125β,where the exact solution is given by(76)0.75+0.25βex,1.125−0.125βex.
In Tables5 and 6 , we present the absolute error of results obtained by the current (HBM1) method and the ones obtained in [30, 31] at x=1.Table 5
Absolute error ofy1,β at x=1 for h=0.1.
y1,ββErrorError fory1,βError for method in [30]Error for method in [31]01.20×10−160.0030.0030.11.30×10−120.0030.0030.21.60×10−120.0030.0030.32.10×10−120.010.010.42.20×10−120.010.010.51.90×10−120.0080.0040.61.80×10−120.0040.0040.71.30×10−120.0040.0030.81.60×10−120.010.0020.91.70×10−120.010.0111.50×10−120.0040.004Table 6
Absolute error ofy2,β at x=1 for h=0.1.
y2,ββErrorError fory2,βError for method in [30]Error for method in [31]01.30×10−160.0020.0020.12.30×10−120.0010.0010.22.50×10−120.0010.0010.33.10×10−120.0050.0050.42.90×10−120.010.010.52.80×10−120.0010.0050.62.90×10−120.010.010.72.50×10−120.010.010.82.40×10−120.0050.0050.92.70×10−120.010.0112.30×10−120.0040.004Example 6.
Consider the second-order fuzzy linear initial value problem(77)y″x=x2⊙y′x⊕2x⊙yx⊕θ^⊙x,x≥0,y0=γ^,y′0=0,where θ^β=2+2β,6−2β,γ^β=1+β,3−β, and h=0.1. Let yx=y1x,β,y2x,β be a fuzzy solution and y′x=y1′x,β,y2′x,β.
By implementingβ-level sets, the problem becomes(78)y1′′x,β,y2′′x,β=x2⊙y1′x,β,y2′x,β⊕2x⊙y1x,α,y2x,α⊕x⊙2+2β,6−2β,y10,β,y20,β=1+β,3−β,y1′0,β,y2′0,β=0,where the exact solution is given by(79)2ex3/3−11−β,2ex3/3−13−β.
In Table7, we present the absolute error of the results obtained by the current (HBM1) method and the ones obtained in [32] at x=1.Table 7
Absolute error ofy1,β and y2,β at x=1 for h=0.1.
ErrorβError fory1,βError fory1,β in [32]Error fory2,βError fory2,β in [32]01.20×10−159.90×10−82.30×10−153.0×10−70.12.30×10−121.10×10−72.1×10−122.90×10−70.22.40×10−121.20×10−73.1×10−122.80×10−70.32.60×10−121.30×10−74.0×10−122.70×10−70.42.90×10−121.40×10−74.3×10−122.60×10−70.53.10×10−121.50×10−74.7×10−122.50×10−70.63.20×10−121.60×10−75.0×10−122.40×10−70.73.40×10−121.70×10−75.1×10−122.30×10−70.83.50×10−121.80×10−75.3×10−122.20×10−70.93.70×10−121.90×10−75.5×10−122.10×10−714.0×10−122.0×10−75.7×10−122.0×10−7


## 7. Conclusion

In this paper, we approximated the solution for fuzzy second-order initial value problems using HBM1. This method is characterized by high precision and stability. Besides, this method has an order of 3 which implies that it is consistent and convergent. Also, we drew the region of absolute stability. We solved four examples to show the efficiency and the accuracy of the proposed method. We note from the tables that our results are accurate even when we use only one off-step point. Moreover, our results with other results show that our results are more accurate. The computational cost is very reasonable since we used one off-step point. For researchers who are interested to get more accuracy, they can use two or three off-step points. The absolute error in this case will be almost zero. In addition, the proposed method is self-starter. We generate the approximate solutions using the initial conditions only. The results show that the proposed method is promising with fuzzy initial value problems. From the figures, we notice that the influence of the cut level on the lower solution is that as the cut level is increasing, the profile solution is increasing. Also, we notice that the influence of the cut level on the upper solution is that as the cut level is increasing, the profile solution is decreasing. Finally, we see that the crisp solution is bounded by lower and upper solutions and they become close to the crisp solution as the cut level becomes close to one.



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