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

Authors: HuiChol Choi; SungHyok Kwon; Kinam Sin; Sunae Pak; Sungryol So
Journal: Abstract and Applied Analysis
 (2019)

---

## Abstract

In this paper, we consider the following two-point boundary value problems of fuzzy linear fractional differential equations:(
D
c
1,1
α
y
)
(
t
)
⊕
b
(
t
)
⊗
(
D
c
1,1
β
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.


---

## Body

## 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:(1)
D
g
H
C
∗
q
u
t
=
f
t
,
u
t
,
0
<
t
<
T
,
u
0
=
λ
u
T
,
u
g
H
′
0
=
u
0
∈
E
C
1
,where D
g
H
C
∗
q is the fuzzy g
H
-fractional Caputo derivative, u
g
H
′ is generalized Hukuhara derivative of u, E
C
1 is the space of fuzzy number, f
:
0
,
T
×
E
c
1
→
E
c
1
∖
R is a continuous fuzzy valued function, q
∈
(
1,2
] is a real number, and λ
∈
(
0,1
)
∪
(
1
,
+
∞
).Gasilov [26] presented a new approach to a nonhomogeneous fuzzy boundary value problem.But researchers who studied fuzzy differential equations by usingr-cut did not consider if the solutions of r-cut equations constitute intervals. So they had to recheck whether the solutions of r-cut equations constitute intervals or not after solving a problem. For instance, in [27] they only consider the existence of solutions of r-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 r-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 correspondingr-cut equation. So, it is necessary to obtain a new r-cut problem that guarantees the existence of fuzzy solutions.Motivated by the results mentioned above, we consider the following fuzzy boundary value problem:(2)
D
c
1,1
α
y
t
⊕
b
t
⊗
D
c
1,1
β
y
t
⊕
c
t
⊗
y
t
=
f
t
,
t
∈
0,1
,
(3)
y
0
=
y
0
,
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, 1
<
β
<
α
≤
2, D
c
1,1
α is fuzzy fractional Caputo derivative, and R
F 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 newr-cut problem which involves inequalities to obtain the conditions of existence of fuzzy solutions and prove that these inequalities guarantee that the solutions of r-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 onR by R
F. A fuzzy number is a mapping μ
:
R
→
[
0,1
] with the following properties:(i)
μ is normal, i.e., ∃
x
0
∈
R; μ
(
x
0
)
=
1
(ii)
μ is a convex fuzzy subset, i.e.,(4)
μ
λ
x
+
1
-
λ
y
≥
m
i
n
μ
x
,
μ
y
,
∀
x
,
y
∈
R
,
∀
λ
∈
0,1
(iii)
μ is upper semicontinuous on R
(iv)
The sets
u
p
p
(
μ
)
¯ is compact in R (where s
u
p
p
(
μ
)
≔
{
x
∈
R
∣
μ
(
x
)
>
0
})
ThenR
F is called the space of fuzzy numbers.Definition 2 (see [27]).
Letu
,
v
∈
R
F, [
u
]
r
≔
[
u
-
(
r
)
,
u
+
(
r
)
], [
v
]
r
≔
[
v
-
(
r
)
,
v
+
(
r
)
]. The distance structure D
:
R
F
×
R
F
→
R
+ is defined as follows: (5)
D
u
,
v
≔
s
u
p
r
∈
0,1
m
a
x
u
-
r
-
v
-
r
,
u
+
r
-
v
+
r
.
Letx
,
y
∈
R
F. If there exists z
∈
R
F such that x
=
y
⊕
z, then z is called the H-difference (Hukuhara difference) of x
,
y and it is denoted as x
⊝
H
y.Definition 3 (see [29]).
Letf
:
I
→
R
F and fix x
0
∈
(
a
,
b
).
(i) We say thatf is (1)-differentiable at x
0, if there exists an element f
G
′
(
x
0
)
∈
R
F such that, for all h
>
0 sufficiently near to 0, f
(
x
0
+
h
)
⊝
H
f
(
x
0
), f
(
x
0
)
⊝
H
f
(
x
0
-
h
) and the limits (6)
l
i
m
h
→
0
+
f
x
0
+
h
⊝
H
f
x
0
h
=
l
i
m
h
→
0
+
f
x
0
⊝
H
f
x
0
-
h
h
=
f
G
′
x
0exist.
(ii) We say thatf is (2)-differentiable at x
0, if there exists an element f
G
′
(
x
0
)
∈
R
F such that, for all h
>
0 sufficiently near to 0, f
(
x
0
)
⊝
H
f
(
x
0
+
h
), f
(
x
0
-
h
)
⊝
H
f
(
x
0
) and the limits (7)
l
i
m
h
→
0
+
f
x
0
⊝
H
f
x
0
+
h
-
h
=
l
i
m
h
→
0
+
f
x
0
-
h
⊝
H
f
x
0
-
h
=
f
G
′
x
0exist.
Iff is (
n
)
-differentiable at x
0, we denote its first derivatives by D
n
(
1
)
f
(
x
0
) for n
=
1,2.Lemma 4 (see [21]).
Letf
:
(
a
,
b
)
→
R
F be fuzzy valued function, where [
f
(
t
)
]
r
≔
[
f
1
(
t
,
r
)
,
f
2
(
t
,
r
)
] for each r
∈
[
0,1
]. (i)
Iff is (1)-differentiable, then f
1
(
t
,
r
) and f
2
(
t
,
r
) are differentiable functions and [
D
1
1
f
(
t
)
]
r
=
[
f
1
′
(
t
,
r
)
,
f
2
′
(
t
,
r
)
].
(ii)
Iff is (2)-differentiable, then f
1
(
t
,
r
) and f
2
(
t
,
r
) are differentiable functions and [
D
2
1
f
(
t
)
]
r
=
[
f
2
′
(
t
,
r
)
,
f
1
′
(
t
,
r
)
].Definition 5 (see [21]).
Letf
:
(
a
,
b
)
→
R
F and n
,
m
=
1,2. We say that f is (
n
,
m
)
-differentiable at x
0
∈
(
a
,
b
), if D
n
(
1
)
f
(
x
0
) exists, and is (
m
)
-differentiable at x
0. The second derivative of f is denoted by D
n
,
m
(
2
)
f
(
x
0
) for n
,
m
=
1,2.Lemma 6 (see [30]).
Letf
:
(
a
,
b
)
→
R
F be fuzzy valued function and denote its r
-level sets by [
f
(
t
)
]
r
≔
[
f
1
(
t
,
r
)
,
f
2
(
t
,
r
)
] for each r
∈
[
0,1
].(i)
IfD
1
(
1
)
f
(
t
) is (1)-differentiable, then f
1
′
(
t
,
r
)
,
f
2
′
(
t
,
r
) are differentiable and(8)
D
1,1
2
f
t
r
=
f
1
′
′
t
,
r
,
f
2
′
′
t
,
r
.
(ii)
IfD
1
(
1
)
f
(
t
) is (2)-differentiable, then f
1
′
(
t
,
r
)
,
f
2
′
(
t
,
r
) are differentiable and (9)
D
1,2
2
f
t
r
=
f
2
′
′
t
,
r
,
f
1
′
′
t
,
r
.Definition 7.
The Riemann-Liouville fractional integral operator of orderα
>
0 of a real function φ
:
I
→
R is defined as(10)
I
a
+
α
φ
t
=
1
Γ
α
∫
a
t
t
-
s
α
-
1
φ
s
d
s
,where Γ
(
·
) is the Euler gamma function.Definition 8.
Letφ
:
I
→
R, the Caputo fractional derivative of order α
>
0, m
-
1
<
α
<
m, m
∈
N
=
{
1,2
,
⋯
}, be defined as(11)
D
c
a
+
α
φ
t
=
1
Γ
m
-
α
∫
a
t
t
-
s
m
-
α
-
1
φ
m
s
d
s
.Definition 9.
The fuzzy fractional Caputo differentiability of fuzzy valued function is defined as follows:(12)
D
c
1,1
α
f
t
=
1
Γ
2
-
α
∫
a
t
t
-
s
1
-
α
D
1,1
2
f
s
d
s
,where 1
<
α
≤
2, t
>
0, f
:
I
→
R
F. Then we say that f is [
(
1,1
)
-
α
]
c-differentiable.Lemma 10 (see [23, 24, 31]).
Letf
:
I
→
R
F, α
∈
(
1,2
), [
f
(
t
)
]
r
≔
[
f
1
(
t
,
r
)
,
f
2
(
t
,
r
)
]. If f is [
(
1,1
)
-
α
]
c
-differentiable, it holds that(13)
D
c
1,1
α
f
t
r
=
D
c
0
+
α
f
1
t
,
r
,
D
c
0
+
α
f
2
t
,
r
.Lemma 11 (see [27]).
Let{
U
r
∣
r
∈
(
0,1
]
} be a family of real intervals such that the following three conditions are satisfied:(i)
U
r is a nonempty compact interval for all r
∈
[
0,1
]
(ii)
if0
<
r
1
<
r
2
≤
1 then U
r
2
⊆
U
r
1
(iii)
given any nondecreasing sequencer
n
∈
(
0,1
] with l
i
m
n
→
∞
r
n
=
r it is U
r
=
⋂
n
=
1
∞
U
r
n
Then there exists a unique fuzzy numberu
∈
R
F such that [
u
]
r
=
U
r for all r
∈
I and [
u
]
0
=
c
l
(
⋃
r
∈
(
0,1
]
U
r
).
Let us consider the following fractional integral equation:(14)
U
t
+
b
t
I
0
+
λ
1
U
t
+
c
t
I
0
+
λ
2
U
t
=
V
t
,
t
∈
I
,where V
,
b
,
c
∈
C
(
I
), λ
1
,
λ
2
>
0.Lemma 12.
There exists a unique solution of (14) in C
(
I
) and a positive real number k
∗ for the solution to satisfy the following inequality:(15)
U
C
I
≤
e
k
∗
1
-
q
V
C
I
,where k
∗ is any positive number that satisfies q
≔
b
C
(
I
)
/
k
∗
λ
1
+
c
C
(
I
)
/
k
∗
λ
2
<
1.Proof.
LetF
(
U
)
(
t
)
≔
V
(
t
)
-
b
(
t
)
I
0
+
λ
1
U
(
t
)
-
c
(
t
)
I
0
+
λ
2
U
(
t
).
Then equation (14) is as follows:(16)
U
t
=
F
U
t
.Now, we use the following k
-norm equivalent to the norm x
C
(
I
)
≔
m
a
x
{
|
x
(
t
)
|
,
t
∈
I
}.
That is(17)
x
k
≔
m
a
x
e
-
k
t
x
t
,
x
∈
C
I
.For all X
,
Y
∈
C
(
I
), we have the following inequalities:(18)
F
X
t
-
F
Y
t
≤
b
C
I
I
0
+
λ
1
X
t
-
I
0
+
λ
1
Y
t
+
c
C
I
I
0
+
λ
2
X
t
-
I
0
+
λ
2
Y
t
≤
b
C
I
k
λ
1
+
c
C
I
k
λ
2
e
k
t
X
-
Y
k
.Therefore, we get (19)
e
-
k
t
F
X
t
-
F
Y
t
≤
b
C
I
k
λ
1
+
c
C
I
k
λ
2
X
-
Y
k
.Thus, we obtain(20)
F
X
t
-
F
Y
t
k
≤
b
C
I
k
λ
1
+
c
C
I
k
λ
2
X
-
Y
k
.Since the number k
∗ satisfies(21)
q
≔
b
C
I
k
λ
1
+
c
C
I
k
λ
2
<
1
,we obtain the following inequality:(22)
F
X
-
F
Y
k
∗
≤
q
X
-
Y
k
∗
.
SinceC
(
I
) is complete, we have(23)
∃
Y
-
∈
C
I
;
Y
-
=
F
Y
-
.
That is, (14) has unique solution U
(
t
)
∈
C
(
I
).
Then the following equation holds:(24)
U
t
+
b
t
I
0
+
λ
1
U
t
+
c
t
I
0
+
λ
2
U
t
=
V
t
,
t
∈
I
,and we obtain the following inequality: (25)
U
t
≤
b
I
0
+
λ
1
U
t
+
c
I
0
+
λ
2
U
t
+
V
t
≤
b
e
k
t
k
λ
1
U
k
+
c
e
k
t
k
λ
2
U
k
+
V
t
,so we have(26)
e
-
k
t
U
t
≤
b
1
k
λ
1
U
k
+
c
1
k
λ
2
U
k
+
e
-
k
t
V
t
.
Therefore, we get(27)
U
k
≤
b
1
k
λ
1
U
k
+
c
1
k
λ
2
U
k
+
V
k
=
b
1
k
λ
1
+
c
1
k
λ
2
U
k
+
V
k
.
So if there is a positive numberk
∗ satisfying (28)
q
≔
b
C
I
k
∗
λ
1
+
c
C
I
k
∗
λ
2
<
1
,then we get (29)
U
k
∗
≤
1
1
-
q
V
k
∗
,and so we obtain (30)
e
-
k
∗
U
C
I
≤
U
k
∗
≤
1
1
-
q
V
k
∗
≤
1
1
-
q
V
C
I
.Thus we have (31)
U
C
I
≤
e
k
∗
1
-
q
V
C
I
.Consider the integral equation(32)
U
t
+
b
I
0
+
α
-
β
U
t
+
c
I
0
+
α
U
t
=
h
t
,
t
∈
0
,
T
,where b
>
0, c
≥
0, 1
<
β
<
α
≤
2, 0
<
α
-
β
<
1.We define an operatorA by A
(
U
)
(
t
)
≔
b
I
0
+
α
-
β
U
(
t
)
+
c
I
0
+
α
U
(
t
).Lemma 13.
If(
I
-
A
)
h
(
t
)
≥
0, then the integral equation (32) has one nonnegative solution, where I is the identity operator.Proof.
By Lemma12, (32) has a unique solution. If there is a positive number k
∗ satisfying(33)
q
≔
b
k
∗
α
-
β
+
c
k
∗
α
<
1
,then A
k
∗
<
1.
Therefore we can know that(34)
I
+
A
-
1
=
I
-
A
+
A
2
-
⋯
+
-
1
n
A
n
+
⋯
,so we have (35)
U
t
=
I
+
A
-
1
h
t
.
Since(
I
-
A
)
h
(
t
)
≥
0, we obtain (36)
U
t
=
I
-
A
h
t
+
A
2
I
-
A
h
t
+
A
4
I
-
A
h
t
⋯
≥
0
.Lemma 14.
(
I
-
A
)
h
(
t
)
≤
0, and then the integral equation (37)
U
t
+
b
I
0
+
α
-
β
U
t
+
c
I
0
+
α
U
t
=
h
t
,
t
∈
Ihas 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:(38)
D
c
1,1
α
y
t
⊕
b
t
⊗
D
c
1,1
β
y
t
⊕
c
t
⊗
y
t
=
f
t
,
t
∈
0,1
,
(39)
y
0
=
y
0
,
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, 1
<
β
<
α
≤
2.Definition 15.
A fuzzy valued functiony
(
t
) is called a (1,1)-solution of the problem (38), (39) if it satisfies (38), (39) and D
c
1,1
α
y
∈
C
(
I
,
R
F
).
By using r-cuts, we can obtain(40)
D
c
1,1
α
y
t
⊕
b
t
⊗
D
c
1,1
β
y
t
⊕
c
t
⊗
y
t
r
=
f
t
r
,
r
∈
0,1and(41)
D
c
1,1
α
y
t
r
+
b
t
D
c
1,1
β
y
t
r
+
c
t
y
t
r
=
f
t
r
.
D
c
0
+
α
y
1
t
,
r
,
D
c
0
+
α
y
2
t
,
r
+
b
t
D
c
0
+
β
y
1
t
,
r
,
D
c
0
+
β
y
2
t
,
r
+
c
t
y
1
t
,
r
,
y
2
t
,
r
=
f
1
t
,
r
,
f
2
t
,
r
.
By the operation of intervals, we can have(42)
D
c
0
+
α
y
1
t
,
r
+
b
t
D
c
0
+
β
y
1
t
,
r
+
c
y
1
t
,
r
=
f
1
t
,
r
,
D
c
0
+
α
y
2
t
,
r
+
b
t
D
c
0
+
β
y
2
t
,
r
+
c
y
2
t
,
r
=
f
2
t
,
r
.
In particular, the inequalities(43)
y
1
t
,
r
≤
y
2
t
,
r
,
D
c
0
+
α
y
1
t
,
r
≤
D
c
0
+
α
y
2
t
,
r
,
D
c
0
+
β
y
1
t
,
r
≤
D
c
0
+
β
y
2
t
,
rimply that y
1
(
t
,
r
) and y
2
(
t
,
r
) constitute interval and it is the same for D
c
0
+
β
y
1
(
t
,
r
)
,
D
c
0
+
β
y
2
(
t
,
r
) and D
c
0
+
α
y
1
(
t
,
r
)
,
D
c
0
+
α
y
2
(
t
,
r
)
.
Let us denote ther-cut representation of y
0
,
y
1 by [
y
0
]
r
≔
[
y
0,1
(
r
)
,
y
0,2
(
r
)
], [
y
1
]
r
≔
[
y
1,1
(
r
)
,
y
1,2
(
r
)
]. Then the cut forms of boundary conditions (39) are (44)
y
1
0
,
r
=
y
0,1
r
,
y
1
1
,
r
=
y
1,1
r
,
y
2
0
,
r
=
y
0,2
r
,
y
2
1
,
r
=
y
1,2
r
.
Now the expression (42)-(44) is called cut problem of (38)-(39).Definition 16.
(
y
1
(
t
,
r
)
,
y
2
(
t
,
r
)
) is called a solution of (42)-(44) if it satisfies (42)-(44) and D
c
0
+
α
y
1
(
·
,
r
)
,
D
c
0
+
α
y
2
(
·
,
r
)
∈
C
(
I
).Theorem 17.
(i) If(
y
1
(
t
,
r
)
,
y
2
(
t
,
r
)
) is a solution of the problem (42)-(44), then (
u
1
(
t
,
r
)
,
u
2
(
t
,
r
)
) satisfies(45)
u
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
t
,
r
+
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
-
c
t
∫
0
1
G
t
,
s
u
1
s
,
r
d
s
=
f
1
t
,
r
,
u
2
t
,
r
+
b
t
I
0
+
α
-
β
u
2
t
,
r
+
c
t
1
-
t
y
0,2
r
+
t
y
1,2
r
-
c
t
∫
0
1
G
t
,
s
u
2
s
,
r
d
s
=
f
2
t
,
r
,
u
1
t
,
r
≤
u
2
t
,
r
,
1
-
t
y
0,2
r
-
y
0,1
r
+
t
y
1,2
r
-
y
1,1
r
+
I
0
+
α
u
2
t
,
r
-
u
1
t
,
r
-
I
0
+
α
u
2
t
,
r
-
u
1
t
,
r
t
=
1
·
t
≥
0
,where D
c
0
+
α
y
1
(
t
,
r
)
=
u
1
(
t
,
r
), D
c
0
+
α
y
2
(
t
,
r
)
=
u
2
(
t
,
r
).
(ii) If(
u
1
(
t
,
r
)
,
u
2
(
t
,
r
)
) satisfies (45), then (
y
1
(
t
,
r
)
,
y
2
(
t
,
r
)
) is a solution of the problem (42)-(44), where (46)
y
1
t
,
r
=
1
-
t
y
0,1
r
+
t
y
1,1
r
-
∫
0
1
G
t
,
s
u
1
s
,
r
d
s
,
y
2
t
,
r
=
1
-
t
y
0,2
r
+
t
y
1,2
r
-
∫
0
1
G
t
,
s
u
2
s
,
r
d
s
,
(47)
G
t
,
s
≔
1
Γ
α
t
1
-
s
1
-
α
,
s
≥
t
,
t
1
-
s
1
-
α
-
1
t
-
s
1
-
α
,
s
<
t
.Proof.
Suppose(
y
1
(
t
,
r
)
,
y
2
(
t
,
r
)
) is a solution of the problem (42)-(44).
Then, we can get(48)
I
0
+
α
D
c
0
+
α
y
1
t
,
r
=
I
0
+
α
u
1
t
,
r
,
I
0
+
α
D
c
0
+
α
y
2
t
,
r
=
I
0
+
α
u
2
t
,
r
.
From the boundary condition (44), we have (49)
y
1
t
,
r
=
y
0,1
r
+
c
1
·
t
+
I
0
+
α
u
1
t
,
r
,
y
2
t
,
r
=
y
0,2
r
+
d
1
·
t
+
I
0
+
α
u
2
t
,
r
,
y
1
1
,
r
=
y
0,1
r
+
c
1
+
I
0
+
α
u
1
t
,
r
t
=
1
,
y
2
1
,
r
=
y
0,2
r
+
d
1
+
I
0
+
α
u
2
t
,
r
t
=
1
,
y
1,1
r
=
y
0,1
r
+
c
1
+
I
0
+
α
u
1
t
,
r
t
=
1
,
y
1,2
r
=
y
0,2
r
+
d
1
+
I
0
+
α
u
2
t
,
r
t
=
1
,
y
1,1
r
-
y
0,1
r
-
I
0
+
α
u
1
t
,
r
t
=
1
=
c
1
,
y
1,2
r
-
y
0,2
r
-
I
0
+
α
u
2
t
,
r
t
=
1
=
d
1
.
Thus, we obtain(50)
y
1
t
,
r
=
y
0,1
r
+
y
1,1
r
-
y
0,1
r
-
I
0
+
α
u
1
t
,
r
t
=
1
·
t
+
I
0
+
α
u
1
t
,
r
,
y
2
t
,
r
=
y
0,2
r
+
y
1,2
r
-
y
0,2
r
-
I
0
+
α
u
2
t
,
r
t
=
1
·
t
+
I
0
+
α
u
2
t
,
r
,and so we have (51)
D
c
0
+
β
y
1
t
,
r
=
I
0
+
α
-
β
u
1
t
,
r
,
D
c
0
+
β
y
2
t
,
r
=
I
0
+
α
-
β
u
2
t
,
r
.
Using Green’s function, (50) can be expressed as follows:(52)
y
1
t
,
r
=
1
-
t
y
0,1
r
+
t
y
1,1
r
-
∫
0
1
G
t
,
s
u
1
s
,
r
d
s
,
y
2
t
,
r
=
1
-
t
y
0,2
r
+
t
y
1,2
r
-
∫
0
1
G
t
,
s
u
2
s
,
r
d
s
.
From (43), (51), and (52), we can obtain (45).
On the other hand, sinceD
c
0
+
α
y
1
(
t
,
r
)
≤
D
c
0
+
α
y
2
(
t
,
r
), we have (53)
u
1
t
,
r
≤
u
2
t
,
r
.
At last, from the inequalityy
1
(
t
,
r
)
≤
y
2
(
t
,
r
), we can have (54)
1
-
t
y
0,2
r
-
y
0,1
r
+
t
y
1,2
r
-
y
1,1
r
+
I
0
+
α
u
2
t
,
r
-
u
1
t
,
r
-
I
0
+
α
u
2
t
,
r
-
u
1
t
,
r
t
=
1
·
t
≥
0
.
Conversely, let us suppose that the pair(
u
1
(
t
,
r
)
,
u
1
(
t
,
r
)
) satisfies (45). From (46), we can get (55)
D
c
0
+
α
y
1
t
,
r
=
u
1
t
,
r
,
D
c
0
+
α
y
2
t
,
r
=
u
2
t
,
r
.
Therefore, we haveD
c
0
+
α
y
1
(
t
,
r
), D
c
0
+
α
y
2
(
t
,
r
)
∈
C
(
I
) and so we can get (56)
D
c
0
+
β
y
1
t
,
r
=
I
0
+
α
-
β
u
1
t
,
r
,
D
c
0
+
β
y
2
t
,
r
=
I
0
+
α
-
β
u
2
t
,
r
,and (57)
D
c
0
+
α
y
1
t
,
r
+
b
t
D
c
0
+
β
y
1
t
,
r
+
c
t
y
1
t
,
r
=
f
1
t
,
r
,
D
c
0
+
α
y
2
t
,
r
+
b
t
D
c
0
+
β
y
2
t
,
r
+
c
t
y
2
t
,
r
=
f
2
t
,
r
.
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(58)
u
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
t
,
r
+
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
+
c
t
I
0
+
α
u
1
t
,
r
-
t
c
t
I
0
+
α
u
1
t
,
r
t
=
1
=
f
1
t
,
r
,
u
2
t
,
r
+
b
t
I
0
+
α
-
β
u
2
t
,
r
+
c
t
1
-
t
y
0,2
r
+
t
y
1,2
r
+
c
t
I
0
+
α
u
2
t
,
r
-
t
c
t
I
0
+
α
u
2
t
,
r
t
=
1
=
f
2
t
,
r
.Assumption 1. q
≔
b
/
Γ
(
α
-
β
+
1
)
+
2
c
/
Γ
(
α
+
1
)
<
1.Denote the following:(59)
g
1
t
,
r
≔
f
1
t
,
r
-
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
,
g
2
t
,
r
≔
f
2
t
,
r
-
c
t
1
-
t
y
0,2
r
+
t
y
1,2
r
,
u
1
0
t
,
r
=
0
,
u
2
0
t
,
r
=
0
.We are going to consider a the scheme of successive approximation(60)
u
1
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
n
+
1
t
,
r
+
c
t
I
0
+
α
u
1
n
+
1
t
,
r
=
g
1
t
,
r
+
t
c
t
I
0
+
α
u
1
n
t
,
r
t
=
1
,
u
2
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
u
2
n
+
1
t
,
r
+
c
t
I
0
+
α
u
2
n
+
1
t
,
r
=
g
2
t
,
r
+
t
c
t
I
0
+
α
u
2
n
t
,
r
t
=
1
.Lemma 18.
The sequence(
u
1
n
(
t
,
r
)
,
u
2
n
(
t
,
r
)
) that satisfies (60) is a Cauchy’s sequence in C
(
I
)
×
C
(
I
).Proof.
We can have the following equations fromn
t
h
,
(
n
+
1
)
t
h terms of the scheme of successive approximation(61)
u
1
n
+
1
t
,
r
-
u
1
n
t
,
r
+
b
t
I
0
+
α
-
β
u
1
n
+
1
t
,
r
-
u
1
n
t
,
r
+
c
t
I
0
+
α
u
1
n
+
1
t
,
r
-
u
1
n
t
,
r
=
t
c
t
I
0
+
α
u
1
n
t
,
r
-
u
1
n
-
1
t
,
r
t
=
1
,
u
2
n
+
1
t
,
r
-
u
2
n
t
,
r
+
b
t
I
0
+
α
-
β
u
2
n
+
1
t
,
r
-
u
2
n
t
,
r
+
c
t
I
0
+
α
u
2
n
+
1
t
,
r
-
u
2
n
t
,
r
=
t
c
t
I
0
+
α
u
2
n
t
,
r
-
u
2
n
-
1
t
,
r
t
=
1
.
For the first equation, we can get(62)
u
1
n
+
1
t
,
r
-
u
1
n
t
,
r
≤
b
I
0
+
α
-
β
u
1
n
+
1
t
,
r
-
u
1
n
t
,
r
+
c
I
0
+
α
u
1
n
+
1
t
,
r
-
u
1
n
t
,
r
+
t
c
I
0
+
α
u
1
n
t
,
r
-
u
1
n
-
1
t
,
r
t
=
1
≤
b
Γ
α
-
β
+
1
+
c
Γ
α
+
1
·
u
1
n
+
1
·
,
r
-
u
1
n
·
,
r
+
c
Γ
α
+
1
·
u
1
n
·
,
r
-
u
1
n
-
1
·
,
rand so we obtain (63)
u
1
n
+
1
·
,
r
-
u
1
n
·
,
r
≤
b
Γ
α
-
β
+
1
+
c
Γ
α
+
1
·
u
1
n
+
1
·
,
r
-
u
1
n
·
,
r
+
c
Γ
α
+
1
·
u
1
n
·
,
r
-
u
1
n
-
1
·
,
r
.
If(64)
γ
≔
1
1
-
q
+
c
/
Γ
α
+
1
c
Γ
α
+
1
<
1
,we can get (65)
u
1
n
+
1
·
,
r
-
u
1
n
·
,
r
≤
1
1
-
b
/
Γ
α
-
β
+
1
+
c
/
Γ
α
+
1
c
Γ
α
+
1
·
u
1
n
·
,
r
-
u
1
n
-
1
·
,
r
=
1
1
-
q
+
c
/
Γ
α
+
1
c
Γ
α
+
1
·
u
1
n
·
,
r
-
u
1
n
-
1
·
,
r
≤
γ
u
1
n
·
,
r
-
u
1
n
-
1
·
,
r
.
As the same way, we can prove foru
2
(
t
,
r
).Since the spaceC
(
I
) is complete, we can say that(66)
∃
u
-
1
·
,
r
∈
C
I
;
u
1
n
·
,
r
→
u
-
1
·
,
r
,
∃
u
-
2
·
,
r
∈
C
I
;
u
2
n
·
,
r
→
u
-
2
·
,
r
.Thus, we have(67)
u
-
1
t
,
r
+
b
t
I
0
+
α
-
β
u
-
1
t
,
r
+
c
t
I
0
+
α
u
-
1
t
,
r
=
g
1
t
,
r
+
t
c
t
I
0
+
α
u
-
1
t
,
r
t
=
1
,
u
-
2
t
,
r
+
b
t
I
0
+
α
-
β
u
-
2
t
,
r
+
c
t
I
0
+
α
u
-
2
t
,
r
=
g
2
t
,
r
+
t
c
t
I
0
+
α
u
-
2
t
,
r
t
=
1
.Let us denote the following:(68)
l
e
n
g
t
,
r
≔
g
2
t
,
r
-
g
1
t
,
r
,
l
e
n
u
n
t
,
r
≔
u
2
n
t
,
r
-
u
1
n
t
,
r
,
g
1
t
,
r
≔
f
1
t
,
r
-
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
,
g
2
t
,
r
≔
f
2
t
,
r
-
c
t
1
-
t
y
0,2
r
+
t
y
1,2
r
,
l
e
n
g
t
,
r
=
l
e
n
f
t
,
r
-
c
t
1
-
t
l
e
n
y
0
+
t
l
e
n
y
1
.Assumption 2. (
I
-
A
)
(
l
e
n
(
g
)
)
≥
0.Assumption 3. (
I
-
A
)
(
t
c
(
t
)
)
≥
0.Theorem 19.
Ifu
-
1
(
t
,
r
), u
-
2
(
t
,
r
) satisfy (67), then they satisfy (69)
u
-
1
t
,
r
≤
u
-
2
t
,
r
,
t
∈
I
.Proof.
Let us consider a the scheme of successive approximation(70)
l
e
n
u
0
t
,
r
=
0
,
l
e
n
u
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
l
e
n
u
n
+
1
t
,
r
+
c
t
I
0
+
α
l
e
n
u
n
+
1
t
,
r
=
l
e
n
g
t
,
r
+
t
c
t
I
0
+
α
l
e
n
u
n
t
,
r
t
=
1
.
Whenn
=
0, we can have the inequality l
e
n
(
u
1
)
(
t
,
r
)
≥
0 from l
e
n
(
u
0
)
(
t
,
r
)
=
0, Assumption 2, and Lemma 13.
For anyn
≥
1, let us suppose that l
e
n
(
u
n
)
(
t
,
r
)
≥
0.
Then, we can obtain(71)
I
-
A
l
e
n
g
t
,
r
+
t
c
t
I
0
+
α
l
e
n
u
n
t
,
r
t
=
1
=
I
-
A
l
e
n
g
t
,
r
+
I
0
+
α
l
e
n
u
n
t
,
r
t
=
1
I
-
A
t
c
t
≥
0and so we have (72)
l
e
n
u
n
+
1
t
,
r
≥
0
.
By the limit of inequality (72), the proof is completed.Let us use the following notations:(73)
l
e
n
y
0
t
,
r
≔
y
0,2
r
-
y
0,1
r
,
l
e
n
y
1
t
,
r
≔
y
1,2
r
-
y
1,1
r
,
l
e
n
u
-
t
,
r
≔
u
-
2
t
,
r
-
u
-
1
t
,
r
.Assumption 4. l
e
n
(
f
)
(
t
,
r
)
-
Γ
(
α
+
2
)
(
b
(
t
)
+
Γ
(
α
-
β
+
1
)
)
/
Γ
(
α
+
2
)
Γ
(
α
-
β
+
1
)
-
c
Γ
(
α
-
β
+
1
)
-
b
Γ
(
α
+
2
)
l
e
n
(
g
)
≥
0.Theorem 20.
(
1
-
t
)
·
l
e
n
(
y
0
)
(
t
,
r
)
+
t
·
l
e
n
(
y
1
)
(
t
,
r
)
+
I
0
+
α
l
e
n
(
u
-
)
(
t
,
r
)
-
I
0
+
α
l
e
n
(
u
-
)
(
t
,
r
)
|
t
=
1
·
t
≥
0.Proof.
From equations (67), we can get(74)
l
e
n
u
-
+
b
t
I
0
+
α
-
β
l
e
n
u
-
+
c
t
I
0
+
α
l
e
n
u
-
=
l
e
n
g
+
t
c
t
I
0
+
α
l
e
n
u
-
t
=
1
,
(75)
l
e
n
g
t
,
r
=
l
e
n
f
t
,
r
-
c
t
1
-
t
l
e
n
y
0
+
t
l
e
n
y
1
.
Sincel
e
n
(
u
-
) is the solution of the equation (76)
l
e
n
u
-
=
l
e
n
g
+
t
c
t
I
0
+
α
l
e
n
u
-
t
=
1
-
b
t
I
0
+
α
-
β
l
e
n
u
-
-
c
t
I
0
+
α
l
e
n
u
-and the inequality (77)
G
t
,
s
≤
s
1
-
s
α
-
1
Γ
α
,we have (78)
l
e
n
u
-
≤
l
e
n
g
+
c
·
t
I
0
+
α
l
e
n
u
-
t
=
1
-
I
0
+
α
l
e
n
u
-
+
b
·
I
0
+
α
-
β
l
e
n
u
-
≤
l
e
n
g
+
c
·
∫
0
1
s
1
-
s
α
-
1
Γ
α
d
s
·
l
e
n
u
-
+
b
Γ
α
-
β
+
1
·
l
e
n
u
-
.
We can get(79)
∫
0
1
s
1
-
s
α
-
1
d
s
=
1
α
+
α
2and so we obtain (80)
l
e
n
u
-
≤
l
e
n
g
+
c
Γ
α
+
2
·
l
e
n
u
-
+
b
Γ
α
-
β
+
1
·
l
e
n
u
-
.
By Assumption 1, we can have the inequalities(81)
Q
≔
c
Γ
α
+
2
+
b
Γ
α
-
β
+
1
≤
q
<
1
,
l
e
n
u
-
≤
1
1
-
Q
l
e
n
g
.
We can get(82)
c
t
I
0
+
α
l
e
n
u
-
-
t
c
t
I
0
+
α
l
e
n
u
-
t
=
1
=
l
e
n
g
-
l
e
n
u
-
-
b
t
I
0
+
α
-
β
l
e
n
u
-and since c
(
t
)
≥
0, we have (83)
c
t
1
-
t
·
l
e
n
y
0
t
,
r
+
t
·
l
e
n
y
1
t
,
r
+
I
0
+
α
l
e
n
u
-
t
,
r
-
I
0
+
α
l
e
n
u
-
t
,
r
t
=
1
·
t
≥
0and (84)
c
t
1
-
t
-
l
e
n
y
0
t
,
r
+
t
·
l
e
n
y
1
t
,
r
+
l
e
n
g
-
l
e
n
u
-
-
b
t
I
0
+
α
-
β
l
e
n
u
-
≥
0
.
Sincec
(
t
)
(
(
1
-
t
)
·
l
e
n
(
y
0
)
(
t
,
r
)
+
t
·
l
e
n
(
y
1
)
(
t
,
r
)
)
+
l
e
n
(
g
)
=
l
e
n
(
f
), we have (85)
l
e
n
f
≥
l
e
n
u
-
+
b
t
I
0
+
α
-
β
l
e
n
u
-
.
By equation (81), we have a condition (86)
l
e
n
f
≥
l
e
n
g
1
-
Q
1
+
b
t
Γ
α
-
β
+
1
,that is, (87)
l
e
n
f
t
,
r
-
Γ
α
+
2
b
t
+
Γ
α
-
β
+
1
Γ
α
+
2
Γ
α
-
β
+
1
-
c
Γ
α
-
β
+
1
-
b
Γ
α
+
2
l
e
n
g
≥
0
.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 Theorem17 and equation (43), we can have the following inequalities:(88)
D
c
0
+
α
y
1
t
,
r
≤
D
c
0
+
α
y
2
t
,
r
,
t
∈
I
,
r
∈
0,1
,
y
1
t
,
r
≤
y
2
t
,
r
,
t
∈
I
,
r
∈
0,1
,
D
c
0
+
β
y
1
t
,
r
≤
D
c
0
+
β
y
2
t
,
r
,
t
∈
I
,
r
∈
0,1
.Hence, we can consider the sets of intervals(89)
U
α
t
,
r
≔
D
c
0
+
α
y
1
t
,
r
,
D
c
0
+
α
y
2
t
,
r
,
r
∈
0,1
,
(90)
U
β
t
,
r
≔
D
c
0
+
β
y
1
t
,
r
,
D
c
0
+
β
y
2
t
,
r
,
r
∈
0,1
,
(91)
U
0
t
,
r
≔
y
1
t
,
r
,
y
2
t
,
r
,
r
∈
0,1
.Theorem 21.
The sets of intervals (89)-(91) generate continuous fuzzy valued functions.Proof.
Consider again the notionD
c
0
+
α
y
1
(
t
,
r
)
=
u
1
(
t
,
r
), D
c
0
+
α
y
2
(
t
,
r
)
=
u
2
(
t
,
r
).
Sinceu
1
(
t
,
r
)
,
u
2
(
t
,
r
) satisfy (45), then we can have u
2
(
t
,
r
)
≥
u
1
(
t
,
r
).
Let us proveu
1
(
t
,
r
1
)
,
u
2
(
t
,
r
1
)
⊃
u
1
(
t
,
r
2
)
,
u
2
(
t
,
r
2
), where 0
<
r
1
<
r
2
≤
1.
The proof is equivalent to(92)
u
1
t
,
r
1
≤
u
1
t
,
r
2
,
u
2
t
,
r
2
≤
u
2
t
,
r
1
.
First, let us proveu
1
(
t
,
r
1
)
≤
u
1
(
t
,
r
2
). From the first equation of (45), we can have (93)
u
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
t
,
r
+
c
t
I
0
+
α
u
1
t
,
r
=
g
1
t
,
r
+
t
c
t
I
0
+
α
u
1
t
,
r
t
=
1
,
u
2
t
,
r
+
b
t
I
0
+
α
-
β
u
2
t
,
r
+
c
t
I
0
+
α
u
2
t
,
r
=
g
2
t
,
r
+
t
c
t
I
0
+
α
u
2
t
,
r
t
=
1
.
We need the following notions:(94)
g
1
t
,
r
≔
f
1
t
,
r
-
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
,
g
2
t
,
r
≔
f
2
t
,
r
-
c
t
1
-
t
y
0,2
r
+
t
y
1,2
r
,
Δ
u
1
≔
u
1
t
,
r
2
-
u
1
t
,
r
1
,
Δ
u
2
≔
u
2
t
,
r
2
-
u
2
t
,
r
1
,
Δ
g
1
≔
g
1
t
,
r
2
-
g
1
t
,
r
1
,
Δ
g
2
≔
g
2
t
,
r
2
-
g
2
t
,
r
1
.
From (93), we can have (95)
Δ
u
1
+
b
t
I
0
+
α
-
β
Δ
u
1
+
c
t
I
0
+
α
Δ
u
1
=
Δ
g
1
+
t
c
t
I
0
+
α
Δ
u
1
t
=
1
.
Assumption 5. ((
I
-
A
)
Δ
g
1
≥
0
,
(
I
-
A
)
Δ
g
2
≤
0).
Let us consider a the scheme of successive approximation(96)
Δ
u
1
0
=
0
,
Δ
u
1
n
+
1
+
b
t
I
0
+
α
-
β
Δ
u
1
n
+
1
+
c
t
I
0
+
α
Δ
u
1
n
+
1
=
Δ
g
1
+
t
c
t
I
0
+
α
Δ
u
1
n
t
=
1
.
Ifn
=
0, then we can have (97)
Δ
u
1
1
+
b
t
I
0
+
α
-
β
Δ
u
1
1
+
c
t
I
0
+
α
Δ
u
1
1
=
Δ
g
1
,and so from Assumption 5 and Lemma 13, we can obtain Δ
u
1
1
≥
0.
SupposeΔ
u
1
n
≥
0 for any n
≥
1.
Let us consider the equation(98)
Δ
u
1
n
+
1
+
b
t
I
0
+
α
-
β
Δ
u
1
n
+
1
+
c
t
I
0
+
α
Δ
u
1
n
+
1
=
Δ
g
1
+
t
c
t
I
0
+
α
Δ
u
1
n
t
=
1
.By Assumptions 3, 5 and the induction, we can obtain (
I
-
A
)
(
Δ
g
1
+
t
c
(
t
)
I
0
+
α
Δ
u
1
(
n
)
|
t
=
1
)
≥
0 and so we have Δ
u
1
(
n
+
1
)
≥
0 by Lemma 10.
Finally, we haveΔ
u
1
≥
0. We can also prove Δ
u
2
≤
0 as the same way by Lemma 14.
Then, let us consideru
(
t
,
r
)
=
⋂
n
=
1
∞
[
u
1
(
t
,
r
k
)
,
u
2
(
t
,
r
k
)
] for any nondecreasing sequence r
k
∈
(
0,1
] which satisfies l
i
m
k
→
∞
r
k
=
r, i.e., l
i
m
n
→
∞
u
1
(
t
,
r
k
)
=
u
1
(
t
,
r
)
,
l
i
m
n
→
∞
u
2
(
t
,
r
k
)
=
u
2
(
t
,
r
).
From the first equation of (45), we have (99)
u
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
t
,
r
+
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
+
c
t
I
0
+
α
u
1
t
,
r
-
t
c
t
I
0
+
α
u
1
t
,
r
t
=
1
=
f
1
t
,
r
,
u
1
t
,
r
k
+
b
t
I
0
+
α
-
β
u
1
t
,
r
k
+
c
t
1
-
t
y
0,1
r
k
+
t
y
1,1
r
+
c
t
I
0
+
α
u
1
t
,
r
k
-
t
c
t
I
0
+
α
u
1
t
,
r
k
t
=
1
=
f
1
t
,
r
k
.and so we can get (100)
u
1
t
,
r
-
u
1
t
,
r
k
=
h
t
,
r
k
,
r
-
b
t
I
0
+
α
-
β
u
1
t
,
r
-
u
1
t
,
r
k
-
c
t
I
0
+
α
u
1
t
,
r
-
u
1
t
,
r
k
+
t
c
t
I
0
+
α
u
1
t
,
r
-
u
1
t
,
r
k
t
=
1
.
Let us denote the following:(101)
Δ
k
u
1
≔
u
1
t
,
r
-
u
1
t
,
r
k
,
h
t
,
r
k
,
r
≔
f
1
t
,
r
-
f
1
t
,
r
k
-
c
t
1
-
t
y
0,1
r
-
y
0,1
r
k
+
t
y
1,1
r
-
y
1,1
r
k
.
Using the above notions, the equation can be written as(102)
Δ
k
u
1
+
b
t
I
0
+
α
-
β
Δ
k
u
1
+
c
t
I
0
+
α
Δ
k
u
1
=
h
t
,
r
k
,
r
+
t
c
t
I
0
+
α
Δ
k
u
1
t
=
1
.
Now, we use the following scheme of successive approximation(103)
Δ
k
u
1
0
=
0
,
Δ
k
u
1
n
+
1
+
b
t
I
0
+
α
-
β
Δ
k
u
1
n
+
1
+
c
t
I
0
+
α
Δ
k
u
1
n
+
1
=
h
t
,
r
k
,
r
+
t
c
t
I
0
+
α
Δ
k
u
1
n
t
=
1
.
Whenn
=
0, we have (104)
Δ
k
u
1
1
+
b
t
I
0
+
α
-
β
Δ
k
u
1
1
+
c
t
I
0
+
α
Δ
k
u
1
1
=
h
t
,
r
k
,
r
,and then from Lemma 12, we can obtain inequality Δ
k
u
1
1
≤
e
k
∗
/
1
-
q
h
(
·
,
r
k
,
r
).
Sincel
i
m
r
k
→
r
h
(
·
,
r
k
,
r
)
=
0, we can get (105)
l
i
m
r
k
→
r
Δ
k
u
1
1
=
0
.
For anyn
∈
N, suppose l
i
m
r
k
→
r
Δ
k
u
1
n
=
0 and let us consider when n
+
1.
We can have(106)
Δ
k
u
1
n
+
1
+
b
t
I
0
+
α
-
β
Δ
k
u
1
n
+
1
+
c
t
I
0
+
α
Δ
k
u
1
n
+
1
=
h
t
,
r
k
,
r
+
t
c
t
I
0
+
α
Δ
k
u
1
n
t
=
1
,and so we can obtain (107)
l
i
m
r
k
→
r
h
t
,
r
k
,
r
+
t
c
t
I
0
+
α
Δ
k
u
1
n
t
=
1
=
l
i
m
r
k
→
r
h
t
,
r
k
,
r
+
l
i
m
r
k
→
r
t
c
t
I
0
+
α
Δ
k
u
1
n
t
=
1
=
l
i
m
r
k
→
r
h
t
,
r
k
,
r
+
t
c
t
l
i
m
r
k
→
r
1
Γ
α
∫
0
1
Δ
k
u
1
n
s
,
r
k
1
-
s
1
-
α
d
s
=
l
i
m
r
k
→
r
h
t
,
r
k
,
r
+
t
c
t
1
Γ
α
∫
0
1
l
i
m
r
k
→
r
Δ
k
u
1
n
s
,
r
k
1
-
s
1
-
α
d
s
=
0
.
We can get(108)
Δ
k
u
1
n
+
1
≤
e
k
∗
1
-
q
m
a
x
0
≤
t
≤
1
h
t
,
r
k
,
r
+
t
c
t
I
0
+
α
Δ
k
u
1
n
t
=
1
,and so we can have l
i
m
r
k
→
r
Δ
k
u
1
n
+
1
=
0.
Hence, we havel
i
m
r
k
→
r
Δ
k
u
1
=
0 and as the same way, we obtain l
i
m
r
k
→
r
Δ
k
u
2
=
0, where Δ
k
u
2
≔
u
2
(
t
,
r
)
-
u
2
(
t
,
r
k
).
Finally, we proved that the set of intervals{
[
u
1
(
t
,
r
)
,
u
2
(
t
,
r
)
]
}
r
∈
(
0,1
] generates a fuzzy valued function.
Denote the fuzzy valued function generated by the set of intervals{
[
u
1
(
t
,
r
)
,
u
2
(
t
,
r
)
]
}
r
∈
(
0,1
] by a fuzzy valued function y
~
α
(
t
).
For anyt
∗
∈
I, we can get (109)
D
y
~
α
t
,
y
~
α
t
∗
=
s
u
p
r
∈
0,1
m
a
x
D
c
0
+
α
y
1
t
,
r
-
D
c
0
+
α
y
1
t
∗
,
r
,
D
c
0
+
α
y
2
t
,
r
-
D
c
0
+
α
y
2
t
∗
,
r
.
Real-valued functionsD
c
0
+
α
y
1
(
t
,
r
)
,
D
c
0
+
α
y
2
(
t
,
r
) are continuous since y
1
(
t
,
r
)
,
y
2
(
t
,
r
) are solutions of (42)-(44). So we have D
(
y
~
α
(
t
)
,
y
~
α
(
t
∗
)
)
→
0
(
t
→
t
∗
).
Therefore, the fuzzy valued functiony
~
α
(
t
) is continuous.
As the same way, we can easily prove that the set of intervals{
U
β
(
t
,
r
)
}
r generates a fuzzy valued function.
Now let us consider that the set of intervals{
U
0
(
t
,
r
)
}
r generates a fuzzy valued function. The equation (46) can be expressed as follows:(110)
y
1
t
,
r
=
1
-
t
y
0,1
r
+
t
y
1,1
r
+
∫
0
1
-
G
t
,
s
u
1
s
,
r
d
s
,
y
2
t
,
r
=
1
-
t
y
0,2
r
+
t
y
1,2
r
+
∫
0
1
-
G
t
,
s
u
2
s
,
r
d
s
.
Because Green’s function is continuous and the set of intervals{
[
u
1
(
t
,
r
)
,
u
2
(
t
,
r
)
]
}
r
∈
(
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{
[
y
1
(
t
,
r
)
,
y
2
(
t
,
r
)
]
}
r generates a fuzzy valued function.
Let us denote the fuzzy valued functions generated by (89), (90) by y
~
∗
(
t
)
,
y
~
β
(
t
).
Then, we can easily prove that fuzzy valued functionsy
~
∗
(
t
)
,
y
~
β
(
t
) are continuous.From equation (50), we can have (111)
y
1
t
,
r
+
y
0,1
r
·
t
+
I
0
+
α
u
1
t
,
r
t
=
1
·
t
=
y
0,1
r
+
y
1,1
r
·
t
+
I
0
+
α
u
1
t
,
r
,
y
2
t
,
r
+
y
0,2
r
·
t
+
I
0
+
α
u
2
t
,
r
t
=
1
·
t
=
y
0,2
r
+
y
1,2
r
·
t
+
I
0
+
α
u
2
t
,
r
,and we can get (112)
y
~
∗
⊕
y
~
0
⊗
t
⊕
I
0
+
α
y
~
α
t
=
1
⊗
t
=
y
~
0
⊕
y
~
1
⊗
t
⊕
I
0
+
α
y
~
α
.Theorem 22.
D
c
1
,
1
α
y
~
∗
=
y
~
α
,
D
c
1
,
1
β
y
~
∗
=
y
~
β.Proof.
Let us consider again that the set of intervals,(113)
D
c
0
+
α
I
0
+
α
y
1
t
,
r
,
D
c
0
+
α
I
0
+
α
y
2
t
,
r
r
=
y
1
t
,
r
,
y
2
t
,
r
r
=
U
0
t
,
r
r
,generates a fuzzy valued function and the generated fuzzy valued function is y
~
α
. So I
0
+
α
y
~
α is [
(
1,1
)
-
α
]
c-differentiable and D
c
1,1
α
I
0
+
α
y
~
α
=
y
~
α.
This shows that the left side of equation (112) is [
(
1,1
)
-
α
]
c-differentiable.
Then, we can have(114)
D
c
1,1
α
y
~
∗
⊕
D
c
1,1
α
y
~
0
⊗
t
⊕
D
c
1,1
α
I
0
+
α
y
~
α
t
=
1
⊗
t
=
D
c
1,1
α
y
~
0
⊕
D
c
1,1
α
y
~
1
⊗
t
⊕
D
c
1,1
α
I
0
+
α
y
~
α
.
Furthermore, we can haveD
c
1,1
α
(
y
~
0
⊗
t
)
=
0, D
c
1,1
α
(
I
0
+
α
y
~
α
t
=
1
⊗
t
)
=
0, so we obtain (115)
D
c
1,1
α
y
~
∗
=
D
c
1,1
α
I
0
+
α
y
~
α
=
y
~
α
.
As the same way, we can prove thatD
c
1,1
β
y
~
∗
=
y
~
β.Through the above investigations, we can see thaty
~
∗ is the solution of (38) and (39) by the sense of Definition 15.In fact, from equation (112), we can see that (116)
y
~
∗
=
y
~
0
⊕
y
~
1
⊗
t
⊕
I
0
+
α
y
~
α
⊝
H
y
~
0
⊗
t
⊕
I
0
+
α
y
~
α
t
=
1
⊗
t
,and we can see that the boundary condition is concluded for the fuzzy valued function y
~
∗.At last, we have the fuzzy equation(117)
y
~
α
t
⊕
b
t
⊗
y
~
β
t
⊕
c
t
⊗
y
~
∗
t
=
f
t
.Thus, the fuzzy valued functiony
~
∗ 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[
0,1
) by (118)
h
0
t
=
1
m
,
h
i
t
=
1
m
2
j
/
2
,
k
-
1
2
j
<
t
≤
k
-
1
/
2
2
j
,
2
j
/
2
,
k
-
1
/
2
2
j
<
t
≤
k
2
j
,
0
,
e
l
s
e
,where i
=
0
,
1
,
⋯
m
-
1
,
m
=
2
M, M is a positive integer, and j
,
k represent the integer decomposition of the index i.
We need the following notations:(119)
H
t
≔
h
0
t
,
h
1
t
,
⋯
,
h
m
-
1
t
T
,
C
T
≔
c
0
,
c
1
,
⋯
,
c
m
-
1
,
Δ
t
≔
1
m
=
1
2
r
,
t
k
≔
k
-
0.5
Δ
t
,
k
=
1
,
m
¯
.
The following matrix is called Haar wavelet matrix on the set of collocation points{
t
k
}:(120)
H
m
a
t
≔
h
0
t
1
h
0
t
2
⋯
h
0
t
m
h
1
t
1
h
1
t
2
⋯
h
1
t
m
⋮
⋮
⋱
⋮
h
m
-
1
t
1
h
m
-
1
t
2
⋯
h
m
-
1
t
m
.
This is an orthogonal matrix; namely,H
m
a
t
-
1
=
H
m
a
t
T.
Put(
I
0
+
α
H
)
(
t
)
≔
I
0
+
α
h
0
(
t
)
,
I
0
+
α
h
1
(
t
)
,
⋯
,
I
0
+
α
h
m
-
1
(
t
)
T.Definition 24.
m
×
m matrix F
H
α determined by (
I
0
+
α
H
)
(
t
)
≈
F
H
α
·
H
(
t
) is called the operational matrix of the fractional integral of order α.
If we use the matrixF
B
α determined by(121)
F
B
α
≔
1
m
α
1
Γ
α
+
2
1
ξ
1
ξ
2
⋯
ξ
m
-
1
0
1
ξ
1
⋯
ξ
m
-
2
0
0
1
⋯
ξ
m
-
3
⋮
⋮
⋮
⋱
⋮
0
0
⋯
0
1
,the following equation holds:(122)
F
H
α
=
H
m
a
t
F
B
α
H
m
a
t
T
,where ξ
k
=
(
k
+
1
)
α
+
1
-
2
k
α
+
1
+
(
k
-
1
)
α
+
1.
To solve the problem (38), (39), first, we need to solve the fractional integral equation sequence (123)
g
1
t
,
r
≔
f
1
t
,
r
-
c
t
1
-
t
y
0,1
r
+
t
y
1,1
r
,
g
2
t
,
r
≔
f
2
t
,
r
-
c
t
1
-
t
y
0,2
r
+
t
y
1,2
r
,
u
1
0
t
,
r
=
0
,
u
1
0
t
,
r
=
0
,
(124)
u
1
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
n
+
1
t
,
r
+
c
t
I
0
+
α
u
1
n
+
1
t
,
r
=
g
1
t
,
r
+
t
c
t
I
0
+
α
u
1
n
t
,
r
t
=
1
,
u
2
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
u
2
n
+
1
t
,
r
+
c
t
I
0
+
α
u
2
n
+
1
t
,
r
=
g
2
t
,
r
+
t
c
t
I
0
+
α
u
2
n
t
,
r
t
=
1by using operational matrix and get u
1
(
t
,
r
)
,
u
2
(
t
,
r
) approximately and then have y
1
(
t
,
r
)
,
y
2
(
t
,
r
) by the equations(125)
y
1
t
,
r
=
y
0,1
r
+
y
1,1
r
-
y
0,1
r
-
I
0
+
α
u
1
t
,
r
t
=
1
·
t
+
I
0
+
α
u
1
t
,
r
,
y
2
t
,
r
=
y
0,2
r
+
y
1,2
r
-
y
0,2
r
-
I
0
+
α
u
2
t
,
r
t
=
1
·
t
+
I
0
+
α
u
2
t
,
r
.
We first calculate (124)(126)
C
g
1
T
=
g
1
t
1
,
r
,
g
1
t
2
,
r
,
⋯
,
g
1
t
m
,
r
·
H
m
a
t
T
,
C
g
2
T
=
g
2
t
1
,
r
,
g
2
t
2
,
r
,
⋯
,
g
2
t
m
,
r
·
H
m
a
t
T
,to solve (124) by using operational matrix. Using the notion (127)
u
1
n
t
,
r
:
≈
C
u
1
n
H
t
,
u
2
n
t
,
r
:
≈
C
u
2
n
H
t
,
g
1
t
,
r
:
≈
C
g
1
T
H
t
,
g
2
t
,
r
:
≈
C
g
2
T
H
t
,we have the following repeat equations from (124).(128)
u
1
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
u
1
n
+
1
t
,
r
+
c
t
I
0
+
α
u
1
n
+
1
t
,
r
=
g
1
t
,
r
+
t
c
t
I
0
+
α
u
1
n
t
,
r
t
=
1
,
u
2
n
+
1
t
,
r
+
b
t
I
0
+
α
-
β
u
2
n
+
1
t
,
r
+
c
t
I
0
+
α
u
2
n
+
1
t
,
r
=
g
2
t
,
r
+
t
c
t
I
0
+
α
u
2
n
t
,
r
t
=
1
.
C
u
1
n
+
1
H
t
+
b
t
I
0
+
α
-
β
C
u
1
n
+
1
H
t
+
c
t
I
0
+
α
C
u
1
n
+
1
H
t
=
C
g
1
T
H
t
+
t
c
t
I
0
+
α
C
u
1
n
H
t
t
=
1
,
C
u
2
n
+
1
H
t
+
b
t
I
0
+
α
-
β
C
u
2
n
+
1
H
t
+
c
t
I
0
+
α
C
u
2
n
+
1
H
t
=
C
g
2
T
H
t
+
t
c
t
I
0
+
α
C
u
2
n
H
t
t
=
1
,
C
u
1
n
+
1
H
t
+
b
t
C
u
1
n
+
1
I
0
+
α
-
β
H
t
+
c
t
C
u
1
n
+
1
I
0
+
α
H
t
=
C
g
1
T
H
t
+
t
c
t
C
u
1
n
I
0
+
α
H
t
t
=
1
,
C
u
2
n
+
1
H
t
+
b
t
C
u
2
n
+
1
I
0
+
α
-
β
H
t
+
c
t
C
u
2
n
+
1
I
0
+
α
H
t
=
C
g
2
T
H
t
+
t
c
t
C
u
2
n
I
0
+
α
H
t
t
=
1
,
C
u
1
n
+
1
H
t
+
b
t
C
u
1
n
+
1
F
H
α
-
β
H
t
+
c
t
C
u
1
n
+
1
F
H
α
H
t
=
C
g
1
T
H
t
+
t
c
t
C
u
1
n
F
H
α
H
t
t
=
1
,
C
u
2
n
+
1
H
t
+
b
t
C
u
2
n
+
1
F
H
α
-
β
H
t
+
c
t
C
u
2
n
+
1
F
H
α
H
t
=
C
g
2
T
H
t
+
t
c
t
C
u
2
n
F
H
α
H
t
t
=
1
.
This is a point equation.(129)
C
u
1
n
+
1
H
t
k
+
b
t
k
C
u
1
n
+
1
F
H
α
-
β
H
t
k
+
c
t
k
C
u
1
n
+
1
F
H
α
H
t
k
=
C
g
1
T
H
t
k
+
t
k
c
t
k
C
u
1
n
F
H
α
H
t
M
,
k
=
1
,
M
¯
.
On the other hand, we can have(130)
C
u
1
n
+
1
H
t
k
+
b
t
k
C
u
1
n
+
1
F
H
α
-
β
H
t
k
+
c
t
k
C
u
1
n
+
1
F
H
α
H
t
k
=
C
u
1
n
+
1
I
+
b
t
k
F
H
α
-
β
+
c
t
k
F
H
α
H
t
k
.
Denote the following:(131)
A
≔
I
+
b
t
1
F
H
α
-
β
+
c
t
1
F
H
α
H
t
1
,
⋯
,
I
+
b
t
M
F
H
α
-
β
+
c
t
M
F
H
α
H
t
M
,
B
≔
C
g
1
T
H
t
1
+
t
1
c
t
C
u
1
n
F
H
α
H
t
M
C
g
1
T
H
t
2
+
t
2
c
t
C
u
1
n
F
H
α
H
t
M
⋮
C
g
1
T
H
t
M
+
t
M
c
t
C
u
1
n
F
H
α
H
t
M
.
Then, the simultaneous equation to determineC
u
1
n
+
1 is as follows:(132)
A
C
u
1
n
+
1
=
B
.Lemma 25.
Ifb
C
(
I
)
+
c
C
(
I
)
<
1, equation (132) has a unique solution.Proof.
Prove that the coefficient matrixA is nonnullity. We can have (133)
A
=
I
+
b
t
1
F
H
α
-
β
+
c
t
1
F
H
α
H
t
1
,
⋯
,
I
+
b
t
M
F
H
α
-
β
+
c
t
M
F
H
α
H
t
M
=
H
t
1
+
b
t
1
F
H
α
-
β
+
c
t
1
F
H
α
H
t
1
,
⋯
,
H
t
M
+
b
t
M
F
H
α
-
β
+
c
t
M
F
H
α
H
t
M
=
H
t
1
,
⋯
,
H
t
M
+
b
t
1
F
H
α
-
β
+
c
t
1
F
H
α
H
t
1
,
⋯
,
b
t
M
F
H
α
-
β
+
c
t
M
F
H
α
H
t
M
=
H
m
a
t
+
A
~
,
A
~
≔
b
t
1
F
H
α
-
β
+
c
t
1
F
H
α
H
t
1
,
⋯
,
b
t
M
F
H
α
-
β
+
c
t
M
F
H
α
H
t
M
,
A
=
H
m
a
t
+
A
~
=
I
+
A
~
H
m
a
t
T
H
m
a
t
.
Thus, the nonnullity of the matrixA is equivalent to the nonnullity of the matrix (
I
+
A
~
H
m
a
t
T
). On the other hand, the inequality A
~
H
m
a
t
T
<
1 guarantees the nonnullity of the matrix (
I
+
A
~
H
m
a
t
T
).
Since the inequality(134)
A
~
H
m
a
t
T
2
≤
A
~
2
H
m
a
t
T
2
=
A
~
2holds, the inequality A
~
2
<
1 is a sufficient condition to A
~
H
m
a
t
T
<
1. And by the notion (135)
A
~
=
b
t
1
F
H
α
-
β
+
c
t
1
F
H
α
H
t
1
,
⋯
,
b
t
M
F
H
α
-
β
+
c
t
M
F
H
α
H
t
M
,we can obtain (136)
A
~
T
=
H
T
t
1
b
t
1
F
H
α
-
β
T
+
c
t
1
F
H
α
T
H
T
t
2
b
t
2
F
H
α
-
β
T
+
c
t
2
F
H
α
T
⋮
H
T
t
M
b
t
M
F
H
α
-
β
T
+
c
t
M
F
H
α
T
=
b
t
1
H
T
t
1
F
H
α
-
β
T
+
c
t
1
H
T
t
1
F
H
α
T
b
t
2
H
T
t
2
F
H
α
-
β
T
+
c
t
2
H
T
t
2
F
H
α
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
+
c
t
M
H
T
t
M
F
H
α
T
.By using the notion of (137)
D
1
≔
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
,
D
2
≔
c
t
1
H
T
t
1
F
H
α
T
c
t
2
H
T
t
2
F
H
α
T
⋮
c
t
M
H
T
t
M
F
H
α
T
,we have (138)
A
~
T
=
D
1
+
D
2
,
A
~
T
2
=
D
1
+
D
2
2
≤
D
1
2
+
D
2
2
.
Now let us consider that(139)
D
1
2
≔
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
2
.
We already knew that(140)
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
T
=
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
b
t
1
F
H
α
-
β
H
t
1
,
b
t
2
F
H
α
-
β
H
t
2
,
⋯
,
b
t
M
F
H
α
-
β
H
t
Mand because (141)
b
t
1
H
T
t
1
F
H
α
-
β
T
,
b
t
1
F
H
α
-
β
H
t
1
=
b
t
1
2
H
T
t
1
F
H
α
-
β
T
F
H
α
-
β
H
t
1
=
b
t
1
2
,
b
t
1
H
T
t
1
F
H
α
-
β
T
,
b
t
2
F
H
α
-
β
H
t
2
=
b
t
1
b
t
2
H
T
t
1
F
H
α
-
β
T
F
H
α
-
β
H
t
2
=
0
,we obtain (142)
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
b
t
1
H
T
t
1
F
H
α
-
β
T
b
t
2
H
T
t
2
F
H
α
-
β
T
⋮
b
t
M
H
T
t
M
F
H
α
-
β
T
T
=
d
i
a
g
b
t
1
2
,
b
t
2
2
,
⋯
,
b
t
M
2
.
Thus, we have(143)
D
1
2
≤
m
a
x
k
b
t
k
2
≤
b
C
I
.
As the same way, we can obtain(144)
D
2
2
≤
m
a
x
k
c
t
k
2
≤
c
C
I
.
Finally, we can see that(145)
A
~
2
=
A
~
T
2
≤
b
C
I
+
c
C
I
.We compute the solutionsu
1
(
t
,
r
)
,
u
2
(
t
,
r
) of (124) by using the limit C
u
1
T
,
C
u
2
T in the repeat equation (132).Now, we consider the algorithm to compute (125) by using the operational matrix. We can rewrite the equation (146)
y
1
t
,
r
=
y
0,1
r
+
y
1,1
r
-
y
0,1
r
-
I
0
+
α
u
1
t
,
r
t
=
1
·
t
+
I
0
+
α
u
1
t
,
r
y
2
t
,
r
=
y
0,2
r
+
y
1,2
r
-
y
0,2
r
-
I
0
+
α
u
2
t
,
r
t
=
1
·
t
+
I
0
+
α
u
2
t
,
rby using the notion (147)
h
1
t
,
r
=
1
-
t
y
0,1
r
+
t
y
1,1
r
h
2
t
,
r
=
1
-
t
y
0,2
r
+
t
y
1,2
ras follows:(148)
y
1
t
,
r
=
h
1
t
,
r
-
I
0
+
α
u
1
t
,
r
t
=
1
·
t
+
I
0
+
α
u
1
t
,
r
y
2
t
,
r
=
h
2
t
,
r
-
I
0
+
α
u
2
t
,
r
t
=
1
·
t
+
I
0
+
α
u
2
t
,
r
.First, we denote the following as(149)
h
1
t
,
r
:
≈
C
h
1
T
H
t
,
h
2
t
,
r
:
≈
C
h
2
T
H
tand compute (150)
C
h
1
T
=
h
1
t
1
,
r
,
h
1
t
2
,
r
,
⋯
,
h
1
t
m
,
r
·
H
m
a
t
T
,
C
h
2
T
=
h
2
t
1
,
r
,
h
2
t
2
,
r
,
⋯
,
h
2
t
m
,
r
·
H
m
a
t
T
.Then,y
1
(
t
,
r
)
,
y
2
(
t
,
r
) are computed by (151)
y
1
t
,
r
≈
C
h
1
T
H
t
-
t
C
u
1
T
F
H
α
H
t
M
+
C
u
1
T
F
H
α
H
t
y
2
t
,
r
≈
C
h
2
T
H
t
-
t
C
u
2
T
F
H
α
H
t
M
+
C
u
2
T
F
H
α
H
tfrom (132).


## 5. Numerical Examples

Example 1.
Consider a FVBP(152)
D
1,1
2
y
t
⊕
b
⊗
D
c
1,1
1.5
y
t
⊕
c
⊗
y
t
=
1
+
t
⊗
7.9,8
,
8.1
,
t
∈
0,1
,
y
0
=
-
0.1,0
,
0.1
,
y
1
=
-
0.1,0
,
0.1
,where b
,
c
∈
R
+, y
∈
C
(
[
0,1
]
,
R
F
) and the boundary values (
-
0.1,0
,
0.1
), (
-
0.1,0
,
0.1
), and (
7.9,8
,
8.1
) are triangular fuzzy numbers. Now, let us consider the conditions for Assumptions 1–5.
A condition for Assumption 1:(153)
q
≔
b
Γ
α
-
β
+
1
+
2
c
Γ
α
+
1
=
1.12838
b
+
c
<
1
.
A condition for Assumption 2: we can have(154)
l
e
n
f
t
,
r
=
f
2
t
,
r
-
f
1
t
,
r
=
0.2
1
-
r
1
+
t
,
l
e
n
g
t
,
r
=
0.2
1
-
r
1
+
t
-
c
0.2
1
-
r
=
0.2
1
-
r
1
+
t
-
cand so we can obtain an inequality (155)
I
-
A
1
+
t
-
c
≥
0for a condition for Assumption 2.
Consequently, we can have(156)
I
-
A
1
+
t
-
c
=
1
+
t
-
c
-
b
I
0
+
α
-
β
1
+
t
-
c
-
c
I
0
+
α
1
+
t
-
c
=
1
+
t
-
c
-
b
Γ
1.5
1
-
c
t
0.5
+
t
1.5
1.5
-
c
2
1
-
c
t
2
+
t
3
3
≥
1
-
c
-
1.12838
b
1
-
c
+
1
1.5
-
c
2
1
-
c
+
1
3
=
0.5
-
2.54858
+
c
-
0.78475
+
c
+
b
-
1.88063
+
1.12838
cand so the condition for Assumption 2 is (157)
0.5
-
2.54858
+
c
-
0.78475
+
c
+
b
-
1.88063
+
1.12838
c
≥
0
.
A condition for Assumption 3: we can easily prove that the equation(158)
I
-
A
t
=
t
-
t
α
-
β
α
-
β
α
-
β
+
1
-
t
α
α
α
+
1
=
t
-
4
t
0.5
3
-
t
2
6is nonnegative.
A condition for Assumption 4: we can have(159)
Γ
α
+
2
b
+
Γ
α
-
β
+
1
Γ
α
+
2
Γ
α
-
β
+
1
-
c
Γ
α
-
β
+
1
-
b
Γ
α
+
2
=
6
b
+
0.886227
5.31736
-
6
b
-
0.886227
c
0.2
1
-
r
1
+
t
-
6
b
+
0.886227
5.31736
-
6
b
-
0.886227
c
0.2
1
-
r
2
-
c
=
0.2
1
-
r
1
+
t
-
6
b
+
0.886227
2
-
c
5.31736
-
6
b
-
0.886227
c
≥
0.2
1
-
r
1
-
6
b
+
0.886227
2
-
c
5.31736
-
6
b
-
0.886227
c
.
So, we can see that the inequality(160)
1
-
6
b
+
0.886227
2
-
c
5.31736
-
6
b
-
0.886227
c
≥
0is a sufficient condition for Assumption 4. Finally, we have (161)
5.31736
+
18
b
-
4.431135
c
-
6
b
c
≤
0
.
A condition for Assumption 5: first, we can have(162)
f
1
t
,
r
2
-
f
1
t
,
r
1
=
1
+
t
7.9
+
0.1
r
2
-
1
+
t
7.9
+
0.1
r
1
=
0.1
r
2
-
r
1
1
+
t
,
y
0,1
r
2
-
y
0,1
r
1
=
0.1
r
2
-
0.1
-
0.1
r
1
-
0.1
=
0.1
r
2
-
r
1
,
y
1,1
r
2
-
y
1,1
r
1
=
0.1
r
2
-
r
1
,
Δ
g
1
=
0.1
r
2
-
r
1
1
+
t
1
-
c
.
We can easily obtain1
-
3
b
/
π
-
1.5
c
≥
0 for (
I
-
A
)
Δ
g
1
≥
0.
As the same way, we can have the same inequality for(
I
-
A
)
Δ
g
2
≤
0.
The following inequalities are conditions for the FBVP to have a solution.①
1.1283
8
b
+
c
<
1
②
1
-
c
≥
0
③
0.5
(
-
2.54858
+
c
)
(
-
0.78475
+
c
)
+
b
(
-
1.88063
+
1.1283
8
c
)
≥
0
④
5.31736
+
1
8
b
-
4.43113
5
c
-
6
b
c
≤
0
⑤
1
-
3
b
/
π
-
1.5
c
≥
0We show the area ofb
,
c that satisfies the above conditions on Figure 1.Figure 1
Area ofb and c satisfying ①-⑤ for Example 1.Example 2.
Consider the following fuzzy boundary value problem:(163)
D
1,1
2
y
t
⊕
b
t
⊗
D
c
1,1
1.5
y
t
⊕
c
t
⊗
y
t
=
1
+
t
⊗
7.9,8
,
8.1
,
t
∈
0,1
,
y
0
=
-
0.1,0
,
0.1
,
y
1
=
-
0.1,0
,
0.1
,where b
,
c
∈
C
(
I
), y
∈
C
(
[
0,1
]
,
R
F
), b
(
t
)
,
c
(
t
)
≥
0 and (
-
0.1,0
,
0.1
) and (
7.9,8
,
8.1
) are triangular fuzzy numbers.
Assumption 1. q
≔
b
/
Γ
(
α
-
β
+
1
)
+
2
c
/
Γ
(
α
+
1
)
=
1.12838
b
+
c
<
1
.
Assumption 2.
(164)
l
e
n
f
t
,
r
=
f
2
t
,
r
-
f
1
t
,
r
=
0.2
1
-
r
1
+
t
,
c
t
1
-
t
l
e
n
y
0
+
t
l
e
n
y
1
=
c
t
0.2
1
-
r
,
l
e
n
g
t
,
r
=
0.2
1
-
r
1
+
t
-
c
t
0.2
1
-
r
=
0.2
1
-
r
1
+
t
-
c
t
.We can have (
I
-
A
)
(
1
+
t
-
c
(
t
)
)
≥
0 as a condition for Assumption 2.
Assumption 3. (
I
-
A
)
(
t
c
(
t
)
)
≥
0.
Assumption 4. Denote |
|
|
c
|
|
|
≔
m
i
n
t
∈
[
0,1
]
|
c
(
t
)
|; then we get (165)
0.2
1
-
r
1
+
t
-
6
b
t
+
0.886227
5.31736
-
6
b
-
0.886227
c
0.2
1
-
r
2
-
c
=
0.2
1
-
r
1
+
t
-
6
b
t
+
0.886227
2
-
c
5.31736
-
6
b
-
0.886227
c
≥
0.2
1
-
r
1
-
6
b
t
+
0.886227
2
-
c
5.31736
-
6
b
-
0.886227
c
.
So we obtain(166)
1
≥
6
b
t
+
0.886227
2
-
c
5.31736
-
6
b
-
0.886227
c
.
Assumption 5. We can obtain 1
-
c
≥
0
,
1
-
3
b
/
π
-
1.5
c
≥
0 as Example 1.
Sufficient conditions are given as follows:①
1.12838
b
+
c
<
1
②
(
I
-
A
)
(
1
+
t
-
c
(
t
)
)
≥
0
③
(
I
-
A
)
(
t
c
(
t
)
)
≥
0
④
1
≥
6
(
b
(
t
)
+
0.886227
)
(
2
-
|
|
|
c
|
|
|
)
/
5.31736
-
6
b
-
0.886227
c
⑤
1
-
3
b
/
π
-
1.5
c
≥
0
Let us consider especially forc
(
t
)
=
0.5
t
/
(
1
+
t
2
), b
(
t
)
=
0.2
(
1
-
t
). We can obtain that (167)
c
=
0.25
,
b
=
0.2
,
c
=
0
.
1.12838
b
+
c
=
1.12838
·
0.2
+
0.25
=
0.475676
<
1
,That is, condition 1 holds. Using the inequalities 0.75
+
t
≤
c
(
t
)
≤
1
+
0.7
5
t, we can have (168)
I
-
A
1
+
t
-
c
t
=
1
+
t
-
c
t
-
A
1
+
t
-
c
t
=
1
+
t
-
0.5
t
1
+
t
2
-
0.2
1
-
t
I
0
+
α
-
β
1
+
t
-
0.5
t
1
+
t
2
-
0.5
t
1
+
t
2
I
0
+
α
1
+
t
-
0.5
t
1
+
t
2
≥
t
+
0.75
-
0.2
1
-
t
I
0
+
0.5
1
+
0.75
t
-
0.5
t
1
+
t
2
I
0
+
2
1
+
0.75
t
=
t
+
0.75
-
0.2
1
-
t
2
t
0.5
+
t
1.5
Γ
0.5
-
0.5
t
1
+
t
2
0.5
t
2
+
0.125
t
3
Γ
2
,
≥
t
+
0.75
-
0.1
2
t
0.5
+
t
1.5
-
1
+
0.75
t
0.5
t
2
+
0.125
t
3
≥
t
+
0.55
-
0.1
t
1.5
-
1
+
0.75
t
0.5
t
2
+
0.125
t
3
≥
0
,i.e., condition 3 holds.
We can obtain(169)
I
-
A
t
c
t
=
t
c
t
-
A
t
c
t
=
0.5
t
2
1
+
t
2
-
A
0.5
t
2
1
+
t
2
=
0.5
t
2
1
+
t
2
+
0.01
t
I
0
+
α
-
β
t
2
1
+
t
2
-
0.01
I
0
+
α
-
β
t
2
1
+
t
2
-
0.25
t
1
+
t
2
I
0
+
α
t
2
/
1
+
t
2
≥
0.5
t
2
1
+
t
2
+
0.005
t
I
0
+
α
-
β
t
2
-
0.01
I
0
+
α
-
β
t
2
-
0.25
t
1
+
t
2
I
0
+
α
t
2
=
0.5
t
2
1
+
t
2
+
0.005
·
0.601802
t
3.5
-
0.01
·
0.601802
t
2.5
-
0.25
t
5
12
1
+
t
2
=
-
0.00601802
t
2.5
+
0.00300901
t
3.5
+
0.5
t
2
1
+
t
2
-
0.0208333
t
5
1
+
t
2
≥
0
,
t
∈
0,1and since 1
-
3
b
/
π
-
1.5
c
=
1
-
0.338514
-
0.375
=
0.286486
>
0, condition 5 holds.
Through the investigations, we can see that the problem of Example2 has a solution when c
(
t
)
=
0.5
t
/
(
1
+
t
2
), b
(
t
)
=
0.2
(
1
-
t
) and we can find out the solution approximately by using scheme of successive approximation.
We have presented the numerical result of Example2 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 y
(
t
) for Example 2. For Tables 1 and 2, we can know that y
1
(
t
,
r
)
≤
y
2
(
t
,
r
) and y
1
′
′
(
t
,
r
)
≤
y
2
′
′
(
t
,
r
) have been satisfied. Also we can see that y
1
(
t
,
r
)
≤
y
2
(
t
,
r
) and y
1
′
′
(
t
,
r
)
≤
y
2
′
′
(
t
,
r
) have been satisfied for all r
∈
(
0,1
] and y
1
(
t
,
r
), y
2
(
t
,
r
) constitute intervals. We compute by Mathematica6.0.Table 1
In case ofr
=
1 for Example 2.
t
y
1
(
t
,
1
)
y
2
(
t
,
1
)
y
1
′
′
(
t
,
1
)
y
2
′
′
(
t
,
1
)
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
31.3979Table 2
In case ofr
=
0.5 for Example 2.
t
y
1
(
t
,
1
)
y
2
(
t
,
1
)
y
1
′
′
(
t
,
1
)
y
2
′
′
(
t
,
1
)
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.



---

*Source: 5129013-2019-06-17.xml*