2. Preliminaries
Following [
8,
9,
10], we denote
I as the unit interval
. Let the binary operation
be a topological commutative monoid with unit 1 such that
whenever
and
. In this case, ∗ is said to be a continuous t-norm. For some examples,
and
are continuous t-norms.
Assume that U is an arbitrary set, ∗ is a continuous t-norm, and M is a fuzzy set on . Then is called a fuzzy metric space (FM space for short), when for every and ,
- (FM1)
for every if and only if ;
- (FM2)
;
- (FM3)
;
- (FM4)
is continuous.
Assume that V is a linear space, ∗ is a continuous t-norm and N is a fuzzy set on . Then, is called a fuzzy normed space (FN space for short), when for every and , we have
- (FN1)
for every if and only if ;
- (FN2)
for any ;
- (FN3)
;
- (FN4)
is continuous.
Let be a FN space. We define Then M is a fuzzy metric on V, which is called the fuzzy metric induced by the fuzzy norm N.
We assume that there exists an such that . Therefore, for , we have . Then the sequence converges to x. If, for every , there is an such that for every , then we say that is a Cauchy sequence. A FM space is complete if every Cauchy sequence is convergent in it. It is well known that for a metric space we say it is separable if there exists a countable dense subset Y of X. Let be a metric space and and let be a given mapping. The sequence with initial point is a Picard sequence (PS) where and .
Assume that for every
, there exists a
(which does not depend on
k) such that the following inequality holds
Lemma 1 ([
8]).
Let be a complete fuzzy metric space such that ∗ satisfies (1). If in U there exits a sequence such that, for all , we have for all , then the sequence is a Cauchy sequence in U. We consider the FM space X. We define a Borel -algebra (B-A) on this space and denote it by . For a given measurable space , we denote by the smallest -algebra on containing all sets (for and ).
Definition 1. We consider two FM spaces X and Y. We also assume that is a measurable space (-MS). The function is a -measurable function (-MF) where and for any . Moreover, the continuous function for is called fuzzy Carathéodory (F-C) function.
For the completeness of the paper, we prove the following theorem obtained in [
11].
Theorem 1. We consider , , , and , which are respectively MS, separable FMS, FMS, and F-CF. With these conditions, G is a -measurable (-M).
Proof. Assume that
D is a countable dense subset of
X and
C is a closed subset of
Y. We consider the subset
, where
is a fuzzy metric on
Y. Then,
if and only if for every
there exists
such that
and
(here
is a fuzzy metric on
X). Hence, we conclude that
which implies that
G is
-measurable. □
Corollary 1. We consider , X, Y, , and which are, respectively MS, separable FMS, FMS, F-CF, and Λ-measurable map. With these conditions, from Γ into Y is a Λ-measurable mapping (-MM).
Proof. Let be given by . Therefore k is -measurable and . Additionally, by Theorem 1 G is -measurable. Therefore, it follows readily that is -measurable. □
Consider a measurable space
, a separable FM space
X and a FM space
Y. If the mapping
from
to
Y for every
-MM
is
-M, then we say
is superpositionally measurable (SUP-M). According to the definition of the SUP-M, we have the following results (see Denkowski-Migórski-Papageorgiou ([
11], Remark 2.5.26)):
Furthermore, if
is
-M, then we say that
is a random operator (RO) for every
. Therefore, every fixed point of the random operator (RO)
F is also a random point (RP) such as the
-MM
(
for every
). Our results can extend some recent ones and improve them to obtain new results (see [
12,
13,
14,
15,
16,
17]).
4. Fixed Point Theorems
We assume that
and
X are two non-empty sets. We consider the mapping
. In the following, we propose and prove some theorems that show the existence of a unique random solution (EURS) for mapping
f in a FMS. We denote all mappings from
to
X by
, so that
X is a FMS and
is a MS. A subset of this set is denoted by
, which contains all
-measurable mappings (
-MMs). We consider
. We say that
g and
h are comparable, if we have
or
where ≾ is a partial relationship in
X. The sequence
in which
, for
and
and with starting point
is called a Picard sequence.
Now we consider the following assumptions:
Hypothesis 0 (H0). For each and considering , , and , which are MS, separable complete ordered FM space (SCO-FMS), and random mapping (RM), respectively, the mapping is a monotone operator. Additionally, ∗ in SCO-FMS applies to (1). Hypothesis 1 (H1). For the function , which is a non-decreasing function, we havefor every and for each . Moreover,for every , , and ; Hypothesis 2 (H2). there exists a mapping with “, for each ” or “, for each ”;
Hypothesis 3 (H3). if is a monotone sequence in X and , then and x are comparable for every .
Remark 1.
(i) (H0) describes the space we use as well as the monotonicity of the mapping .
- (ii)
(H1) states the Matkowski-contraction condition (see [
22]).
- (iii)
Assumption (H3) shows the existence of the partial order relation in the case that F is not F-Carathéodory.
We first assume that F is F-Carathéodory mapping (F-CM), then we prove the theorem with this assumption. Then, we further show our theorem without this assumption.
Theorem 2. We consider the assumptions of(H0)–(H2). We assume that G is a F-CM. Therefore, G has a RFP such as a . Furthermore, if there exists such that for every , u is comparable to x and y, then u is a unique random solution (URS) of G.
Proof. We consider the Picard sequences
and
with starting points
such that
and
are comparable. Then, we prove
Considering that the function
is a monotone operator, for every
and every constant
, we have the comparability of
and
. Now, if
, it is clear that (
5) is established and the proof is complete. Then for any
, we assume
. Then by (H1), we have
for every
,
. Now, when
, according to (
6) and considering the property of
, we have
Clearly, this holds for every
,
. We assume that
is a mapping like the one introduced in hypothesis (H2). If, for each
,
, then
is a RFP of
G. Suppose that, for some
,
. By (H2), we have that
and
are two comparable elements of
. Then, by (
5), if we choose
, we obtain
Now, we show that
is a Cauchy sequence for each
. Let
be fixed. Using (
7) and considering
and
for
, there is a
for
such that
for every
. By Lemma 1, we obtain that
is a Cauchy sequence for every
. Then there exists
such that
Using Corollary 1 and considering that
, we have
for every
. In the following, considering the assumption that
G is a F-CM, we prove that for every
,
. With the assumption considered, we have
From
letting
, we obtain
for every
,
. Thus
for each
, that is,
z is a RFP of
G. Next, we prove the uniqueness of the solution. To begin this proof, first we consider another RFP such as
for
G and assume that these two solutions are comparable; then, using (H1), we obtain
Therefore, assuming that
z and
v are comparable, we have nothing to prove. Now we assume that
z and
v are not comparable and further we assume that
is an element comparable to
z and
v. Considering the Picard sequence
with starting point
and also
and
, we have
Therefore, we conclude that
z is a unique random solution (URS) for
G because using (
8) we have
. □
Now, we change the considered assumptions and with these new assumptions we prove another theorem, which also shows the existence of a RBF for G. We consider condition (H3) instead of condition F-Carathéodory (F-C) and also assume that H is a sup-measurable mapping (SUP-MM).
Theorem 3. We consider assumptions(H0)–(H3)as well as SUP-M mapping H. Then, there is a RFP of H like a mapping .
Proof. We consider and as introduced in Theorem 2. According to the assumption that we considered a SUP-MM like H and according to condition (H3), we conclude that for each and , , and and are comparable.
and
. From (H1) we obtain
Letting , we obtain for every . This means that z is a RFP of H. □
Considering the generalized FM space (GFMS), we consider the previous assumptions to solve the above RFP problem corresponding to this space.
Definition 2. Let ∗ be a continuous t-norm on I. Let and be in . We define a binary operation ⋆ on by Then we call ⋆ a continuous t-norm on .
Definition 3 ([
23]).
Let , where is equipped with the following partial orderAdditionally, denote that and ; and for every .
We define in . Specifically, the zero vector is denoted by .
Definition 4. Assume that , ⋆ is a continuous t-norm on and is a vector-value fuzzy set on . Then, is called a fuzzy vector-valued metric space (FVVM space for short) when for every , ,
- (FVM1)
;
- (FVM2)
for every if and only if ;
- (FVM3)
;
- (FVM4)
;
- (FVM5)
is continuous.
Example 1. Let be a vector-valued metric space. Let and be in . We define,and for every and . Then is a FVVM space.
Definition 5. Assume that X is a vector space, ⋆ is a continuous t-norm on and is a vector-valued fuzzy set on . Then, is called a fuzzy vector-valued norm space (FVVN space for short) when for every , ,
- (FVN1)
;
- (FVN2)
for every if and only if ;
- (FVN3)
for every ;
- (FVN4)
;
- (FVN5)
is continuous.
We assume further that all t–norms satisfy (
1). Let
be the family of all non-decreasing maps
such that
- (i)
for every with ;
- (ii)
and for ;
- (iii)
implies .
Example 2. We consider an -like diagonal matrix. According to A, the function is defined as followsthen . Remark 2. For the completeness, we consider hypotheses (H4) and (H6), which are completely similar to the introduced hypotheses (H0) and (H2), differing only in the space.
Next, we add the following conditions to complete the work.
Hypothesis 4 (H4). Considering , , and , which are MS, separable complete ordered FVVM space (SCO-FVVMS), and RO, respectively, is a monotone operator for every . In the assumed space of , ⋆ in (1) satisfies. Hypothesis 5 (H5). There exists a function such that for each , Hypothesis 6 (H6). There exists a mapping with “, for every ” or “, for every ”.
Now we define the complete FVVM space (CFVVMS). For this purpose, we consider a FVVM space such as and a sequence such as . If for every and every with , there exists an such that for every we have , then we say that the sequence converges to x and therefore it is a Cauchy sequence. A space X is a CFVVMS if every Cauchy sequence converges in it.
In the following, we state a result similar to Theorem 2, which is the third theorem to prove the existence of a RFP for F.
Theorem 4. Assume that the conditions(H4)–(H6) are fulfilled and G as a F-C mapping, then there is a RFP such as for G.
Proof. To prove this theorem, we consider the cases introduced in Theorem 2. For example, we consider the sequences
and
as well as
such that these two comparable members are the starting point of the sequences
and
. Now, we prove that
Therefore, considering that for each
,
and
are comparable and
is a monotone, then using (H5)
for each
,
,
.
Let
and property (i) hold for members of
, then (
9) is true. We consider
as the mapping introduced in (H6). Therefore, the following condition holds for every
and this means that
is a RFT for
G and the proof is finished. Now, we assume that
for some
and we also consider the Picard sequence
with the starting point
. Considering (H6) and choosing two comparable members
and
and using (
9) to choose
, we have
In the following, for the constant
, we show that
is a Cauchy sequence.
and this means that
is a Cauchy sequence. Then, we conclude that there exists
such that
From Corollary 1, for every
, we have
and
. Given that
G is a FC-M, we obtain that
From
letting
, we obtain
for every
,
. Thus
for each
,
, that is,
z is a RFP of
G. □
Remark 3. By adding the assumptions in Theorem 2, we can also prove Theorem 4. For this purpose, we prove the uniqueness of RBF by eliminating duplicate problems.
5. Investigating the Solution for a BVP
In this section, we investigate EURS for (
2). We state our investigations in the form of a theorem.
We consider the following
The measurable space of ;
We consider functions which are of F-C type. Then, for each and where , are measurable and are continuous;
For each and each , we consider a family of F-C functions such as , where is defined as . We denote this set of functions with the letter .
Now we define the integral operator
as follows:
with
where
is a continuous function such that
for every
and
.
Remark 4. For , and () are functions which are of F-C type. For constant , the F-C function, is defined as follows Functions and are measurable functions, because the functions in integrals (11) and (12) are measurable. In fact, these integrals are limits of a finite sum of measurable functions. Consequently, Q is a random operator (RO). At this step, we consider the following assumptions
Hypothesis 7 (H7). There exists a non-decreasing function such thatfor each and for every and all with and ; Hypothesis 8 (H8). For each , there exists a function such thatfor each , and ; Hypothesis 9 (H9). For each fixed, and , , (for every ), are all non-decreasing or all non-increasing operators;
Hypothesis 10 (H10). One of the following conditions holds:or We consider the FVVM
on
as follows
for every
,
.
Theorem 5. We consider assumptions (H7)–(H10). We prove that the random integral operator Q has a unique RFP.
Proof. We first show that
is a continuous operator for constant
. For this purpose, we consider the sequence
on
such that
when
. For
, we have
where
,
,
and
.
For any given
and
, then we have
which implies that
by (H8). By analogous reasoning, one has
Then
as
, implies that
is a continuous operator, for each fixed
and for every
. In addition, for each
and
is a monotone operator. Indeed, consider
such that
; that is,
,
, for every
. For every
, if
and
,
, are non-decreasing operators, then
which implies that
Therefore
. In a similar way, we can conclude that if the functions
and
are non-decreasing functions, then for every
and
, we have
. Next, we have to show that
Q is a contraction operator. At this step, we consider condition (H5) and assume that
and
such that
. We have to show
Again, consider
fixed. Let
be such that
, then
where
,
,
and
.
For any given
and
, then we have
which implies that
by (H8). By analogous reasoning, one has
then
Next, we have to show that (H6) holds. We prove using (H10)
or
that is,
for every
or
for every
. Then, for null random variables
, which are defined as follows
one of the following conditions is true
or
Considering that the condition of uniqueness also exists, then all the conditions of Theorem 4 are satisfied and we conclude that there is a unique FP for Q, which results directly from Theorem 4.
□
Here we have a theorem that gives us a unique random solution (URS) of (
2). This theorem is proposed for a specific selection of FC functions
. Therefore, we consider
, where , for ;
the random integral operator
with
where
is given by (
4).
Theorem 6. We assume that conditions (H7)–(H10)
are fulfilled. Then (2) has a URS. Proof. Here we point out that every RBF of
is a solution of (
2) and also every solution of (
2) is a RBF of
. That is, considering the random variables
, we have
is equivalent to
Such that
,
for
and
is a solution of (
2). Then, by Theorem 5, there exists a unique random solution of the Problem (
2). □