Constructive Existence of (1,1)-Solutions to Two-Point Value Problems for Fuzzy Linear Multiterm Fractional Differential Equations

. In this paper, we consider the following two-point boundary value problems of fuzzy linear fractional differential equations: ( 𝑐 𝐷 𝛼1 ,1 𝑦)(𝑡) ⊕ 𝑏(𝑡) ⊗ ( 𝑐 𝐷 𝛽1,1 𝑦)(𝑡) ⊕ 𝑐(𝑡) ⊗ 𝑦(𝑡) = 𝑓(𝑡) , 𝑡 ∈ (0,1) , 𝑦(0) = 𝑦 0 and 𝑦(1) = 𝑦 1 , where 𝑏, 𝑐 ∈ 𝐶(𝐼) , 𝑏(𝑡), 𝑐(𝑡) ≥ 0 , 𝑦, 𝑓 ∈ 𝐶(𝐼, R F ) , 𝐼 = [0,1] , 𝑦 0 ,𝑦 1 ∈ R F and 1 < 𝛽 < 𝛼 ≤ 2 . Our existence result is based on Banach fixed point theorem and the approximate solution of our problem is obtained by applying the Haar wavelet operational matrix.


Introduction
A lot of researchers have studied fuzzy differential equations especially fuzzy boundary value problems (FBVPs) because they are effective tools for modeling processes.FBVPs arise in many applications such as modeling of fuzzy optimal control problem [1] and HIV infection [2].Many theoretical researches have been carried out on fractional differential equations over the last years [3][4][5][6][7].
Lakshmikantham et al. [18] considered Riemann-Liouville differentiability concept based on the Hukuhara differentiability to solve fuzzy fractional differential equations.Prakash [19] considered initial value problems for differential equations of fractional order with uncertainty.Mazandarania [20] investigated the solution to fuzzy fractional initial value problem (FFIVP) under Caputo-type fuzzy fractional derivatives by a modified fractional Euler method.As we can see, fuzzy initial value problems were studied by many researchers, but few fuzzy boundary value problems were considered in special cases.Nieto [21] considered second order fuzzy differential by the sense of (1,1), (1,2), (2, 1), (2, 2)-derivatives.Also Nieto [22] investigated the existence and uniqueness of solutions for a first-order linear fuzzy differential equation with impulsive boundary value condition.Ngo et al. [23] proved the existence and uniqueness results of the solution to initial value problem of Caputo-Katugampola (CK) fractional differential equations in fuzzy setting and [24] present that a fractional fuzzy differential equation and a fractional fuzzy integral equation are not equivalent in general.Wang [25] considered the existence and uniqueness of solution for a class of FFDEs: fuzzy valued function,  ∈ (1,2] is a real number, and  ∈ (0, 1) ∪ (1, +∞).
Gasilov [26] presented a new approach to a nonhomogeneous fuzzy boundary value problem.
But researchers who studied fuzzy differential equations by using -cut did not consider if the solutions of -cut equations constitute intervals.So they had to recheck whether the solutions of -cut equations constitute intervals or not after solving a problem.For instance, in [27] they only consider the existence of solutions of -cut equation.And the existence of fuzzy solutions was considered in specific example.In that specific example, they noted that the fuzzy solutions do not exist even if the solutions of -cut equation exist.
These facts lead to the following: the existence of solutions of fuzzy problem is not equivalent to the existence of solutions of corresponding -cut equation.So, it is necessary to obtain a new -cut problem that guarantees the existence of fuzzy solutions.
We obtain existence result by using Banach fixed point theorem and obtain its approximate solution by applying the Haar wavelet operational matrix.Also we present a new -cut problem which involves inequalities to obtain the conditions of existence of fuzzy solutions and prove that these inequalities guarantee that the solutions of -cut equations constitute fuzzy solutions.Our paper is organized as follows: In Section 2, we recall some definitions and basic results and prove some lemmas that will be useful to our main results.Section 3 investigated the constructive existence of solutions to our problem.In Section 4, a method to find out the solutions is given.Section 5 presented two examples to illustrate our results.In Section 6, we summarize our main results.

Preliminaries and Basic Results
Definition 1 (see [28]).We denote the set of all fuzzy numbers on R by R F .A fuzzy number is a mapping  : R → [0, 1] with the following properties: (i)  is normal, i.e., ∃ 0 ∈ R; ( 0 ) = 1 (ii)  is a convex fuzzy subset, i.e., (iii)  is upper semicontinuous on R (iv) The set supp() is compact in R (where supp() fl { ∈ R | () > 0}) Then R F is called the space of fuzzy numbers.
Proof.Let ()() fl () − () 0+ ().Then equation ( 14) is as follows: Now, we use the following  − norm equivalent to the norm For all ,  ∈ (), we have the following inequalities: Therefore, we get Thus, we obtain Since the number  * satisfies we obtain the following inequality: Since () is complete, we have That is, (14) has unique solution () ∈ ().
By this theorem, the existence of solutions of (42)-( 44) is equivalent to the existence of solutions of (45).

Now, let us consider the integral equation
Denote the following: We are going to consider a the scheme of successive approximation Proof.We can have the following equations from  ℎ , (+1) ℎ terms of the scheme of successive approximation and so we obtain If we can get As the same way, we can prove for  2 (, ).
For any  ≥ 1, let us suppose that (  )(, ) ≥ 0.Then, we can obtain and so we have By the limit of inequality (72), the proof is completed.
Let us use the following notations: ( Since () is the solution of the equation and the inequality We can get and so we obtain By Assumption 1, we can have the inequalities ()     .
Now let us consider if the solutions of ( 42)-(44) generate a fuzzy valued function and if the generated fuzzy valued function satisfies (38), (39).
From Theorem 17 and equation (43), we can have the following inequalities: Hence, we can consider the sets of intervals We need the following notions: From (93), we can have Let us consider a the scheme of successive approximation If  = 0, then we can have and so from Assumption 5 and Lemma 13, we can obtain Δ 1 1 ≥ 0.
Finally, we have Δ 1 ≥ 0. We can also prove Δ 2 ≤ 0 as the same way by Lemma 14.
From the first equation of (45), we have and so we can get (100) Let us denote the following: Using the above notions, the equation can be written as Now, we use the following scheme of successive approximation Abstract and Applied Analysis When  = 0, we have and then from Lemma 12, we can obtain inequality ‖Δ   1 1 ‖ ≤ (  * /(1 − ))‖ℎ(⋅,   , )‖.
For Therefore, the fuzzy valued function ỹ () is continuous.
As the same way, we can easily prove that the set of intervals {  (, )}  generates a fuzzy valued function.Now let us consider that the set of intervals { 0 (, )}  generates a fuzzy valued function.The equation ( 46) can be expressed as follows: (110) Because Green's function is continuous and the set of intervals {[ 1 (, ),  2 (, )]} ∈(0,1] generates a continuous fuzzy valued function, the set of intervals that is composed of right sides of these equations generates a fuzzy valued function. Hence, the set of intervals {[ 1 (, ),  2 (, )]}  generates a fuzzy valued function.
Then, we can have Abstract and Applied Analysis 11 Furthermore, we can have As the same way, we can prove that    1,1 ỹ * = ỹ .Through the above investigations, we can see that ỹ * is the solution of ( 38) and (39) by the sense of Definition 15.
In fact, from equation (112), we can see that and we can see that the boundary condition is concluded for the fuzzy valued function ỹ * .
At last, we have the fuzzy equation Thus, the fuzzy valued function ỹ * is a solution of the problem (38) and (39).

The Numerical Solution of the Fuzzy Fractional Differential Equations by Using Haar Wavelet Operational Matrix
Definition 23.The orthogonal set of Haar functions are defined in the interval [0, 1) by where  = 0, 1, ⋅ ⋅ ⋅  − 1,  = 2  ,  is a positive integer, and ,  represent the integer decomposition of the index .
We need the following notations: The following matrix is called Haar wavelet matrix on the set of collocation points {  }: This is an orthogonal matrix; namely, Definition 24. ×  matrix    determined by (  0+ )() ≈    ⋅ () is called the operational matrix of the fractional integral of order .
If we use the matrix    determined by the following equation holds: where To solve the problem (38), (39), first, we need to solve the fractional integral equation sequence by using operational matrix and get  1 (, ),  2 (, ) approximately and then have  1 (, ),  2 (, ) by the equations We first calculate (124) to solve (124) by using operational matrix.Using the notion we have the following repeat equations from (124).
This is a point equation.
Proof.Prove that the coefficient matrix  is nonnullity.We can have Thus, the nonnullity of the matrix  is equivalent to the nonnullity of the matrix ( + Ã   ).On the other hand, the inequality ‖ Ã We compute the solutions  1 (, ),  2 (, ) of (124) by using the limit    1 ,    2 in the repeat equation ( 132).Now, we consider the algorithm to compute (125) by using the operational matrix.We can rewrite the equation by using the notion ℎ 1 (, ) = (1 − )  0,1 () + First, we denote the following as and compute Then,  1 (, ),  2 (, ) are computed by from (132).
A condition for Assumption 1: A condition for Assumption 2: we can have and so we can obtain an inequality for a condition for Assumption 2. Consequently, we can have A condition for Assumption 5: first, we can have We can easily obtain 1−3/√−1.5≥ 0 for (−)Δ 1 ≥ 0.
As the same way, we can have the same inequality for ( − )Δ 2 ≤ 0.
The following inequalities are conditions for the FBVP to have a solution.We show the area of ,  that satisfies the above conditions on Figure 1.
where ,  ∈ (),  ∈ ([0, 1], R F ), (), () ≥ 0 and (−0.1, 0, 0.1) and (7.9, 8, 8.1) are triangular fuzzy numbers.0.2(1 − ) and we can find out the solution approximately by using scheme of successive approximation.We have presented the numerical result of Example 2 by using proposed method in Tables 1 and 2. Table 1 is the numerical results for Example 2 when the level value is equal to 1 and Table 2 is the numerical results when the level value is equal to 0.5.The first rows of Tables 1 and 2 are the numeral values of the dependent variables and the second rows are the values of the lower functions and the third rows are the values of upper functions and the fourth and fifth rows are the values of lower and upper functions for (1, 1)derivatives of () for Example 2. For Tables 1 and 2, we can know that  1 (, ) ≤  2 (, ) and   1 (, ) ≤   2 (, ) have been satisfied.Also we can see that  1 (, ) ≤  2 (, ) and   1 (, ) ≤   2 (, ) have been satisfied for all  ∈ (0, 1] and  1 (, ),  2 (, ) constitute intervals.We compute by Mathematica6.0.

Conclusion
We have studied one condition for a two-point boundary value problem to have a solution and investigated one method to calculate the solution.In conclusion, we have to consider the cut-equations and problem (43) with the inequality condition to get the fuzzy solution.The conditions were given by the coefficients, boundary conditions, and nonhomogeneous terms of fuzzy differential equations.The next step in the research is to extend the results of this paper to generalized cases.

A 1 .Figure 1 :
Figure 1: Area of  and  satisfying A-E for Example 1.

Table 1 :
In case of  = 1 for Example 2.

Table 2 :
In case of  = 0.5 for Example 2.