Abstract and Applied Analysis
Volume 2019 (2019), Article ID 5129013, 18 pages
https://doi.org/10.1155/2019/5129013

Research Article

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

HuiChol Choi,¹ SungHyok Kwon,¹ Kinam Sin,¹ Sunae Pak,¹ and Sungryol So²

¹Faculty of Mathematics, Kim Il Sung University, Pyongyang, Democratic People’s Republic of Korea
²Information Technology Institute, University of Sciences, Pyongyang, Democratic People’s Republic of Korea

Correspondence should be addressed to Kinam Sin; 15bf12025@hit.edu.cn

Received 3 February 2019; Revised 10 May 2019; Accepted 22 May 2019; Published 17 June 2019

Academic Editor: Feyzi Başar

Copyright © 2019 HuiChol Choi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

In this paper, we consider the following two-point boundary value problems of fuzzy linear fractional differential equations: , , and , where , , , , and . 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.

1. 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 [3–7].

Currently, the approximative methods for solving fuzzy fractional differential equation include the operational matrix method based on orthogonal functions [8–10], linearization formula [11], Homotopy Analysis Method [12–14], and Variation of constant formula [15].

O’Regan [16] and Lakshmikantham [17] proved that two-point boundary value problems of fuzzy differential equations are equivalent to fuzzy integral equations.

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:

where is the fuzzy fractional Caputo derivative, is generalized Hukuhara derivative of , is the space of fuzzy number, is a continuous fuzzy valued function, is a real number, and .

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.

Motivated by the results mentioned above, we consider the following fuzzy boundary value problem:

where , , , , , , is fuzzy fractional Caputo derivative, and is the space of fuzzy number.

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.

2. Preliminaries and Basic Results

Definition 1 (see [28]). We denote the set of all fuzzy numbers on by . A fuzzy number is a mapping with the following properties:

(i) is normal, i.e., ; (ii) is a convex fuzzy subset, i.e.,

(iii) is upper semicontinuous on (iv)The set is compact in (where )

Then is called the space of fuzzy numbers.

Definition 2 (see [27]). Let , , . The distance structure is defined as follows:

Let . If there exists such that , then is called the H-difference (Hukuhara difference) of and it is denoted as .

Definition 3 (see [29]). Let and fix .
(i) We say that is (1)-differentiable at , if there exists an element such that, for all sufficiently near to , , and the limits

exist.
(ii) We say that is (2)-differentiable at , if there exists an element such that, for all sufficiently near to , , and the limits

exist.
If is differentiable at , we denote its first derivatives by for .

Lemma 4 (see [21]). Let be fuzzy valued function, where for each .

(i)If is (1)-differentiable, then and are differentiable functions and .(ii)If is (2)-differentiable, then and are differentiable functions and .

Definition 5 (see [21]). Let and . We say that is differentiable at , if exists, and is differentiable at . The second derivative of is denoted by for .

Lemma 6 (see [30]). Let be fuzzy valued function and denote its level sets by for each .

(i)If is (1)-differentiable, then are differentiable and

(ii)If is (2)-differentiable, then are differentiable and

Definition 7. The Riemann-Liouville fractional integral operator of order of a real function is defined as

where is the Euler gamma function.

Definition 8. Let , the Caputo fractional derivative of order , , , be defined as

Definition 9. The fuzzy fractional Caputo differentiability of fuzzy valued function is defined as follows:

where , , . Then we say that is -differentiable.

Lemma 10 (see [23, 24, 31]). Let , , . If is differentiable, it holds that

Lemma 11 (see [27]). Let be a family of real intervals such that the following three conditions are satisfied:

(i) is a nonempty compact interval for all (ii)if then (iii)given any nondecreasing sequence with it is

Then there exists a unique fuzzy number such that for all and .
Let us consider the following fractional integral equation:

where , .

Lemma 12. There exists a unique solution of (14) in and a positive real number for the solution to satisfy the following inequality:

where is any positive number that satisfies .

Proof. Let .
Then equation (14) is as follows:

Now, we use the following norm equivalent to the norm .
That is

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 .
Then the following equation holds:

and we obtain the following inequality:

so we have

Therefore, we get

So if there is a positive number satisfying

then we get

and so we obtain

Thus we have

Consider the integral equation

where , , , .

We define an operator by .

Lemma 13. If , then the integral equation (32) has one nonnegative solution, where is the identity operator.

Proof. By Lemma 12, (32) has a unique solution. If there is a positive number satisfying

then .
Therefore we can know that

so we have

Since , we obtain

Lemma 14. , and then the integral equation

has one nonpositive solution.

3. Constructive Existence of (1,1)-Solution of Two-Point Value Problem for the Fuzzy Linear Multiterm Fractional Differential Equation

Let us consider the following fuzzy boundary value problem:

where , , , , , .

Definition 15. A fuzzy valued function is called a (1,1)-solution of the problem (38), (39) if it satisfies (38), (39) and .
By using r-cuts, we can obtain

and

By the operation of intervals, we can have

In particular, the inequalities

imply that and constitute interval and it is the same for and
Let us denote the -cut representation of by , . Then the cut forms of boundary conditions (39) are

Now the expression (42)-(44) is called cut problem of (38)-(39).

Definition 16. is called a solution of (42)-(44) if it satisfies (42)-(44) and .

Theorem 17. (i) If is a solution of the problem (42)-(44), then satisfies

where , .
(ii) If satisfies (45), then is a solution of the problem (42)-(44), where

Proof. Suppose is a solution of the problem (42)-(44).
Then, we can get

From the boundary condition (44), we have

Thus, we obtain

and so we have

Using Green’s function, (50) can be expressed as follows:

From (43), (51), and (52), we can obtain (45).
On the other hand, since , we have

At last, from the inequality , we can have

Conversely, let us suppose that the pair satisfies (45). From (46), we can get

Therefore, we have , and so we can get

and

We can easily obtain the other inequalities and prove the boundary condition (44).

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

Assumption 1. .

Denote the following:

We are going to consider a the scheme of successive approximation

Lemma 18. The sequence that satisfies (60) is a Cauchy’s sequence in .

Proof. We can have the following equations from terms of the scheme of successive approximation

For the first equation, we can get

and so we obtain

If

we can get

As the same way, we can prove for .

Since the space is complete, we can say that

Thus, we have

Let us denote the following:

Assumption 2. .

Assumption 3. .

Theorem 19. If, satisfy (67), then they satisfy

Proof. Let us consider a the scheme of successive approximation

When , we can have the inequality from , Assumption 2, and Lemma 13.
For any , let us suppose that .
Then, we can obtain

and so we have

By the limit of inequality (72), the proof is completed.

Let us use the following notations:

Assumption 4. .

Theorem 20. .

Proof. From equations (67), we can get

Since is the solution of the equation

and the inequality

we have

We can get

and so we obtain

By Assumption 1, we can have the inequalities

We can get

and since , we have

and

Since , we have

By equation (81), we have a condition

that is,

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

Theorem 21. The sets of intervals (89)-(91) generate continuous fuzzy valued functions.

Proof. Consider again the notion , .
Since satisfy (45), then we can have .
Let us prove , where .
The proof is equivalent to

First, let us prove . From the first equation of (45), we can have

We need the following notions:

From (93), we can have

Assumption 5. ().
Let us consider a the scheme of successive approximation

If , then we can have

and so from Assumption 5 and Lemma 13, we can obtain .
Suppose for any .
Let us consider the equation

By Assumptions 3, 5 and the induction, we can obtain and so we have by Lemma 10.
Finally, we have . We can also prove as the same way by Lemma 14.
Then, let us consider for any nondecreasing sequence which satisfies , i.e., .
From the first equation of (45), we have

and so we can get

Let us denote the following:

Using the above notions, the equation can be written as

Now, we use the following scheme of successive approximation

When , we have

and then from Lemma 12, we can obtain inequality .
Since , we can get

For any , suppose and let us consider when .
We can have

and so we can obtain

We can get

and so we can have .
Hence, we have and as the same way, we obtain , where .
Finally, we proved that the set of intervals generates a fuzzy valued function.
Denote the fuzzy valued function generated by the set of intervals by a fuzzy valued function .
For any , we can get

Real-valued functions are continuous since are solutions of (42)-(44). So we have .
Therefore, the fuzzy valued function 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 generates a fuzzy valued function. The equation (46) can be expressed as follows:

Because Green’s function is continuous and the set of intervals 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 generates a fuzzy valued function.
Let us denote the fuzzy valued functions generated by (89), (90) by .
Then, we can easily prove that fuzzy valued functions are continuous.

From equation (50), we can have

and we can get

Theorem 22. .

Proof. Let us consider again that the set of intervals,

generates a fuzzy valued function and the generated fuzzy valued function is . So is -differentiable and .
This shows that the left side of equation (112) is -differentiable.
Then, we can have

Furthermore, we can have , , so we obtain

As the same way, we can prove that .

Through the above investigations, we can see that 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 .

At last, we have the fuzzy equation

Thus, the fuzzy valued function is a solution of the problem (38) and (39).

4. 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 by

where , , 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, .
Put .

Definition 24. matrix determined by 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 approximately and then have 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.

On the other hand, we can have

Denote the following:

Then, the simultaneous equation to determine is as follows:

Lemma 25. If , equation (132) has a unique solution.

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 . On the other hand, the inequality guarantees the nonnullity of the matrix .
Since the inequality

holds, the inequality is a sufficient condition to . And by the notion

we can obtain

By using the notion of

we have

Now let us consider that

We already knew that

and because

we obtain

Thus, we have

As the same way, we can obtain

Finally, we can see that

We compute the solutions of (124) by using the limit 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

as follows:

First, we denote the following as

and compute

Then, are computed by

from (132).

5. Numerical Examples

Example 1. Consider a FVBP

where , and the boundary values , , and are triangular fuzzy numbers. Now, let us consider the conditions for Assumptions 1–5.
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

and so the condition for Assumption 2 is

A condition for Assumption 3: we can easily prove that the equation

is nonnegative.
A condition for Assumption 4: we can have

So, we can see that the inequality

is a sufficient condition for Assumption 4. Finally, we have

A condition for Assumption 5: first, we can have

We can easily obtain for .
As the same way, we can have the same inequality for .
The following inequalities are conditions for the FBVP to have a solution.

①②③④⑤

We show the area of that satisfies the above conditions on Figure 1.

Figure 1: Area of and satisfying ①-⑤ for Example 1.

Example 2. Consider the following fuzzy boundary value problem:

where , , and and are triangular fuzzy numbers.
 Assumption 1.
Assumption 2.

We can have as a condition for Assumption 2.
Assumption 3. .
Assumption 4. Denote ; then we get

So we obtain

Assumption 5. We can obtain as Example 1.
Sufficient conditions are given as follows:

①②③④⑤

Let us consider especially for , . We can obtain that

That is, condition 1 holds. Using the inequalities , we can have

i.e., condition 3 holds.
We can obtain

and since , condition 5 holds.
Through the investigations, we can see that the problem of Example 2 has a solution when , 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 and have been satisfied. Also we can see that and have been satisfied for all and , constitute intervals. We compute by Mathematica6.0.

Table 1: In case of for Example 2.

0.015025
-0.104378
-0.104378
8.04222
8.04222

0.140625
-0.867268
-0.867268
9.80614
9.80614

0.265625
-1.47628
-1.47628
12.0982
12.0982

0.390625
-1.89576
-1.89576
14.7463
14.7463

0.515625
-2.08441
-2.08441
17.7042
17.7042

0.640625
-1.99603
-1.99603
20.9508
20.9508

0.765625
-1.5799
-1.5799
24.4836
24.4836

0.890625
-0.780805
-0.780805
28.3149
28.3149

0.984375
0.105687
0.105687
31.3979
31.3979

Table 2: In case of for Example 2.

0.015625
-0.153774
-0.054983
7.99234
8.0921

0.140625
-0.912267
-0.822269
9.74817
9.86412

0.265625
-1.51779
-1.43476
12.0285
12.1679

0.390625
-1.93488
-1.85664
14.6623
14.8303

0.515625
-2.12245
-2.04637
17.6034
17.805

0.640625
-2.03457
-1.95749
20.8309
21.0706

0.765625
-1.62082
-1.53898
24.3424
24.6248

0.890625
-0.826309
-0.7353
28.1502
28.4796

0.984375
0.0550747
0.1563
31.2141
31.5816

6. 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.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The authors would like to thank Professor Manoj Gauthaman, Editorial Office, for his consideration about our paper.

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