In this manuscript, we introduce the notion of ℜα-θ-contractions and prove some fixed-point theorems in the sense of ℜ-complete metric spaces. These results generalize existing ones in the literature. Also, we provide some illustrative non-trivial examples and applications to a non-linear fractional differential equation.
Citation: Khalil Javed, Muhammad Arshad, Amani S. Baazeem, Nabil Mlaiki. Solving a fractional differential equation via θ-contractions in ℜ-complete metric spaces[J]. AIMS Mathematics, 2022, 7(9): 16869-16888. doi: 10.3934/math.2022926
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In this manuscript, we introduce the notion of ℜα-θ-contractions and prove some fixed-point theorems in the sense of ℜ-complete metric spaces. These results generalize existing ones in the literature. Also, we provide some illustrative non-trivial examples and applications to a non-linear fractional differential equation.
The spaciousness of fixed-point theory can be glanced in different fields by looking at its applications. Fixed-point theorems say that functions must have at least one fixed point, under some circumstances. We can see that these results are usually beneficial in the region of mathematics and play a prissy character in detecting the existence and uniqueness of solutions of different mathematical models. Some scientists gave circumstances to find fixed points, in this manner, Banach and Caccioppoli gave Banach–Caccioppoli fixed-point theorem, which was started by Banach [6] in 1922 and was proved by Caccioppoli [7] in 1931. Banach–Caccioppoli fixed-point theorem guaranteed that if it seized, the function must have a fixed-point, under some circumstances. After this meritorious result of Banach and Caccioppoli, the fixed-point theory has taken on new elevations.
Branciari [2] proved Banach–Caccioppoli fixed-point theorem on a class of generalized metric spaces. In 2014, Jleli and Samet [1] coined a new concept of Θ-contraction mappings and established numerous fixed-point theorems for such mappings in complete metric spaces (CMS). Samet et al. [5] proved fixed-point theorems for α-ψ-contractive mappings. Ahmad et al. [4] proved fixed-point results for generalized Θ-contractions. Arshad et al. [8] proved some fixed-point results by using generalized contractions via triangular α-orbital admissibility in the sense of Branciari metric spaces.
Baghani et al. [3] (2017) presented a new generalization of the Banach fixed point theorem (BFPT) by defining the notion of ℜ-sets. The ℜ-set is a non-empty set equipped with a binary relation (called ℜ-relation) having a special structure (see [3]). The metric defined on the ℜ-set is called an ℜ-metric space. The ℜ-metric space contains partially ordered metric spaces and graphical metric spaces. Khalehoghli et al. [19] extended the work in [3] to ℜ-metric spaces, Ali et al. [20] extended the work [19] to partial b-metric space and Khalil et al. [15] extended the work in [3] to ordered theoretic fuzzy metric spaces. Further fixed-point results on ℜ-(generalized) metric spaces have been provided by Javed et al. [15] who initiated the notion of an ℜ-structure and established the Banach contraction principle.
We introduce the concept of ℜ-α-Θ-contractions (αR-ΘR-contractions), establish some fixed-point theorems for these contractions in the sense of ℜ-complete metric spaces and some constructive examples and an application are also imparted. After proving that these contractions have fixed points, we give some examples to validate our results. For some necessary definitions and results, please see [9,10,11,12,13,14,15,16,17,18].
This manuscript is organized as follows. In section 2, some rudimentary concepts as ℜ-sequence, Cauchy ℜ-sequence, ℜ-preserving, ℜ-complete, ℜ-continuous, ℜ-convergent, Θ-contraction, α-Θ-contraction, and α-admissible are given. In section 3, the concept of αR-ΘR-contractions is introduced and some fixed point results are proved in the sense of ℜ-CMSs and some constructive examples are also provided. In section 4, an application to non-linear fractional differential equations is provided.
In this section, we recall some definitions that are necessary for the main work.
Definition 2.1. [3] Let (B,R) be an ℜ-set. A sequence {βω} is said to be an ℜ-sequence if
(∀ω,k∈N,βωRβω+k)or(∀ω,k∈N,βω+kRβω). |
Also, {βω} is called a Cauchy ℜ-sequence if for every ε>0 there exists an integer N such that O(βω,βk)<ε if ω≥N and k≥N. It is clear that βωRβkorβkRβω.
Definition 2.2. [3] Let (B,R) be an ℜ-set. A mapping ξR:B→B is called ℜ-preserving if ξRβRξRδ, whenever βRδ.
Definition 2.3. [3] Let (B,R,O) be an ℜ-MS and ℜ be a binary relation over B. Then B is said to be ℜ-regular if for each sequence {βω} such that βωℜβω+1, for all ω∈N, and βω→e, for some e∈B, then βωℜe, for all ω∈N (briefly, (B,R,O) is called ℜ-regular metric space).
Definition 2.4 [3] Let (B,R,O) be an ℜ-MS. Then ξR:B→B is called ℜ-continuous at β∈B if for each ℜ-sequence {βω} in B with βω→β, we have ξRβω→ξRβ. Also, ξR is said to be ℜ-continuous on B if ξR is ℜ-continuous at each β∈B.
Definition 2.5. [3] Let (B,R,O) be an ℜ-MS. Then B is said to be an ℜ-CMS if every Cauchy ℜ-sequence is convergent in B.
Definition 2.6. [4] Let Θ:(0,∞)→(1,∞) be a function satisfying the below circumstances:
(Θ1)Θ is non-decreasing.
(Θ2) For a sequence {βω}⊆R+.
limω→∞Θ(βω)=1⟺limω→∞βω=0. |
(Θ3) There exist k∈(0,1) and l∈(0,∞] such that
limt→∞Θ(t)−1(t)k=l. |
Let (B,O) be a MS. A mapping ξ:B→B is said to be a nΘ-contraction [4] if there exist k∈(0,1) and a function Θ fulfiling (Θ1)–(Θ3) such that
O(ξβ,ξδ)≠0⇒Θ(O(ξβ,ξδ))≤[Θ(O(β,δ)]k∀β,δ∈B. |
Let Ω denote the set of all functions satisfying (Θ1)−(Θ3).
Definition 2.7. [16] Let (B,O) be a MS and ξ:B→B be a self-mapping. We say that ξ is an α-Θ-contraction if there exist k∈(0,1) and two functions α:B×B→[0,∞) and Θ∈Ω such that
O(ξβ,ξδ)≠0⇒α(β,δ)Θ(O(ξβ,ξδ))≤[Θ(d(β,δ)]k∀β,δ∈B. |
In this section, we introduce the concept of αR-ΘR-contractions and some fixed-point results are also imparted in the sense of R-CMSs.
Definition 3.1. Let (B,O) be an R-CMS and ξR:B→B be a mapping. We say that ξR is an αR-ΘR-contraction if there exist k∈(0,1) and two functions αR:B×B→[0,∞) and Θ∈Ω such that
O(ξβ,ξδ)≠0⇒αR(β,δ)Θ(O(ξRβ,ξRδ)≤[Θ(O(β,δ)]k,∀β,δ∈BwithβRδ. |
Definition 3.2. Let ξR:B→B and αR:B×B→[0,∞). We say that ξR is αR-admissible if for all β,δ∈BwithβRδ,
αR(β,δ)≥1⇒αR(ξRβ,ξRδ)≥1. |
Example 3.3. Let B=(0,1]=A∪B=(0,1]∖{14,13,12}∪{14,13,12}. Define ξR:B→B and αR:B×B→[0,∞) by
ξR(β)=53β |
and
αR(β,δ)=1max{β,δ},∀β∈A,δ∈B. |
Define the R-relation :βRδ⟺β≤δ. Here, ξR is αR-admissible. It is not α-admissible by taking β=1 and δ=12.
Example 3.4. Let B=(−2,2]. Define the relation: βRδ⇔β+δ≥0.
Define the function αR:B×B→[0,∞) by
αR(β,δ)={min{β,δ}1+max{β,δ},ifβ,δ∈(0,2]e−max{β,δ},ifβ,δ∈[0,−2)0,otherwise. |
Define the mapping ξR:B→B by
ξR(β)={1ifβ∈[−12,12]min{1,β}1+max{1,β}otherwise. |
Clearly, ξR is αR-admissible. It is not α-admissible. Indeed, for β=0andδ=−1, one has
α(0,−1)=e0=1. |
But,
α(ξ(0),ξ(−1))=α(1,−12)=0. |
Remark 3.5. The above example shows that an αR-admissible mapping need not to be an α-admissible mapping. But the converse holds.
Theorem 3.6. Let (B,R,O) be an R-CMS and ξR be a self-mapping, R-preserving, R-continuous and αR:B×B→[0,∞) be a function. Suppose that the below circumstances fulfill:
ⅰ) Suppose there exist k∈(0,1) and a function Θ∈Ω such that for all β,δ∈BwithβRδ,
O(ξRβ,ξRδ)≠0⇒αR(β,δ)Θ(O(ξRβ,ξRδ))≤[Θ(O(β,δ))]k. | (3.1) |
ⅱ) ξR is αR-admissible.
ⅲ) ξR is R-continuous.
ⅳ) There exists β0∈B such that β0RξRβ0 and αR(β0,ξRβ0)≥1.
Then ξR has a fixed point e∈B. Moreover, if for every two fixed points e,f of ξR we have αR(e,f)≥1, then the fixed point is unique.
Proof. Let β0∈B such that (∀δ∈Bβ0Rδ)or(∀δ∈BδRβ0). By condition (ⅲ), β0RξRβ0 or ξRβ0Rβ0. For ω∈N, consider βω=ξRωβo. Assume that ξRβω=ξRβω+1 for some ω∈N. Then βω is a fixed point of ξR and the proof is completed. Let ξRβω≠ξRβω+1 for all ω∈N. Since ξR is R-preserving, (ξRβωRξRβω+1)or(ξRβω+1RξRβω). Hence, {βω} is an R-sequence. Again, by condition (ⅱ),
αR(βω,ξRβω)=αR(βω,βω+1)≥1∀ω∈N. | (3.2) |
From (3.1) and (3.2), we get
1<Θ(O(βω,βω+1))=Θ(O(ξRβω−1,ξRβω)≤αR(βω−1,βω)Θ(O(ξRβω−1,ξRβω)≤[Θ(O(βω−1,βω))]k. | (3.3) |
By (Θ1), we have
O(βω,βω+1)<O(βω−1,βω). |
Hence, the sequence {O(βω,βω+1)} is decreasing and {O(βω,βω+1)} converges to a non-negative real number r≥0 such that
limω→∞O(βω,βω+1)=randO(βω,βω+1)≥r. | (3.4) |
Then we prove that r=0. Suppose that r>0. Using (Θ1), (3.3) and (3.4), we get
1<Θ(r)=Θ(O(βω,βω+1))≤[Θ(O(βω−1,βω))]k≤…≤[Θ(O(β0,β1))]kω∀ω∈N. | (3.5) |
Letting ω→∞ in (3.5), we get Θ(r)=1 and by using (Θ2), we have r=0. Therefore
limω→∞O(βω,βω+1)=0. | (3.6) |
Assume that there are ω,p∈N such that βω=βω+p. We must prove that p=1. Assume that p>1. Using (3.1) and (3.2), we get
Θ(O(βω,βω+1))=Θ(O(βω+p,βω+p+1))=Θ(O(ξRβω+p−1,ξRβω+p))≤αR(βω+p−1,βω+p)Θ(O(ξRβω+p−1,ξRβω+p))≤[Θ(O(βω+p−1,βω+p))]k. | (3.7) |
Using (Θ1), we get
Θ(O(βω,βω+1))<O(βω+p−1,βω+p) |
and by (3.1), we obtain
Θ(O(βω+p−1,βω+p))=Θ(O(ξRβω+p−2,ξRβω+p−1))≤αR(βω+p−2,βω+p−1)Θ(O(ξRβω+p−2,ξRβω+p−1))≤[Θ(O(βω+p−2,βω+p−1))]k<O(βω+p−1,βω+p). | (3.8) |
By (Θ1), we deduce
O(βω+p−1,βω+p)<O(βω+p−1,βω+p). |
Continuing this process, we obtain
O(βω,βω+1)<O(βω+p−1,βω+p)<O(βω+p−2,βω+p−1)<⋯<O(βω,βω+1), |
which implies that p=1 and that contradict our assumption. Therefore, p=1. Now, we will prove that ξR has a fixed point. We now examine that {βω} is a Cauchy R-sequence and we adopt conflicting that {βω} is not a Cauchy R-sequence. So there exists ε>0 and we take two subsequences of {βω}, which are {βωk} and {βσk} with ωk>σk>k for which,
O(βωk,βσk)≥ε,O(βωk,βσk−1)<εandO(βωk,βσk−1)<ε. | (3.9) |
Using the triangular inequality, we derive
ε≤O(βωk,βσk)≤O(βωk,βσk−1)+O(βσk−1,βσk). | (3.10) |
Letting k→∞ in (3.11), using (3.10) and (3.6), we get
limω→∞O(βωk,βσk)=ε. | (3.11) |
By using (3.1), there exists a positive integer k0 such that
O(βωk,βσk)>0∀ωk>σk>k≥k0,Θ(ε)≤Θ(O(βωk+1,βσk+1))=Θ(O(ξRβωk,ξRβσk))≤αR(βωk,βσk)Θ(O(ξRβωk,ξRβσk))≤[Θ(O(βωk,βσk))]k=[Θ(ε)]k. |
This is a contradiction, since k∈(0,1), {βω} is a Cauchy R-sequence. Thus, there is e∈B such that βω→e as ω→∞, then
e=limω→∞βω+1=limω→∞ξRβω=ξRe. |
So e is a fixed point of ξR.
Now, assume that ξR has two fixed points say e≠f. Hence,
O(e,f)=O(ξRe,ξRf)≤αR(e,f)Θ(O(ξRe,ξRf))≤[Θ(O(e,f))]k<Θ(O(e,f)). |
Which leads us to a contradiction. Thus, the fixed point is unique as required.
Theorem 3.7. Let (B,R,O) be an R-regular R-CMS and ξR be a self-mapping, R-preserving and αR:B×B→[0,∞) be a function. Assume that the below situations hold:
(ⅰ) Assume that there exist Θ∈Ω and k∈(0,1) such that for all β,δ∈B with βRδ,
O(ξRβ,ξRδ)≠0⇒αR(β,δ)Θ(O(ξβ,ξδ)≤[Θ(O(β,δ)]k. | (3.12) |
(ⅱ) ξR is αR-admissible.
(ⅲ) There exists β0∈B such that β0RξRβ0 and αR(β0,ξRβ0)≥1.
(ⅳ) If {βω} is an R-sequence in B such that α(βω,βω+1)≥1 for all ω and βω→β, then there exists an R-subsequence {βωk} of {βω} such that α(βωk,β)≥1 for all k.
Then ξR has a fixed point e∈B. Moreover, if for every two fixed points e,f of ξR we have αR(e,f)≥1, then the fixed point is unique.
Proof. Let β0∈B such that (∀δ∈B,β0Rδ)or(∀δ∈B,δRβ0). By condition (ⅲ), β0RξRβ0 or ξRβ0Rβ0.Forω∈N, consider βω=ξRωβo. Assume ξRβω=ξRβω+1 for some ω∈N. Then βω is a fixed point of ξR and the proof is completed. Let ξRβω≠ξRβω+1 for all ω∈N. Since ξR is R-preserving, (ξRβωRξRβω+1)or(ξRβω+1RξRβω). Hence, {βω} is an R-sequence. By condition (ⅰ),
αR(βω,ξRβω)=αR(βω,βω+1)≥1∀ω∈N. | (3.13) |
From (3.12) and (3.13), we get
1<ΘO((βω,βω+1))=Θ(O(ξRβω−1,ξRβω))≤αR(βω−1,βω)Θ(O(ξRβω−1,ξRβω))≤[Θ(O(βω−1,βω))]k. | (3.14) |
By (Θ1), we have
O(βω,βω+1)<O(βω−1,βω). |
Hence, the sequence {O(βω,βω+1)} is decreasing and {O(βω,βω+1)} converges to a non-negative real number r≥0. We have
limω→∞O(βω,βω+1)=randO(βω,βω+1)≥r. | (3.15) |
Then we prove that r=0. Suppose that r>0. Using (Θ1), (3.14) and (3.15), we get
1<Θ(r)=ΘO(βω,βω+1)≤[Θ(O(βω−1,βω))]k≤…≤[Θ(O(β0,β1))]kω∀ω.∈N. | (3.16) |
Letting ω→∞ in (3.16), we get Θ(r)=1 and by using (Θ2) we have r=0 and therefore,
limω→∞O(βω,βω+1)=0. | (3.17) |
Assume that there are ω,p∈N such that βω,=βω+p. Then we prove that p=1. Assume that p>1. By (3.12) and (3.13), we deduce
Θ(O(βω,βω+1))=Θ(O(βω+p,βω+p+1))=Θ(O(ξRβω+p−1,ξRβω+p))≤αR(βω+p−1,βω+p)Θ(O(ξRβω+p−1,ξRβω+p))≤[Θ(O(βω+p−1,βω+p))]k. | (3.18) |
Using (Θ1), we obtain
Θ(O(βω,βω+1))<O(βω+p−1,βω+p) |
and by using (3.12), we derive
Θ(O(βω+p−1,βω+p))=Θ(O(ξRβω+p−2,ξRβω+p−1))≤αR(βω+p−2,βω+p−1)Θ(O(ξRβω+p−2,ξRβω+p−1))≤[Θ(O(βω+p−2,βω+p−1))]k<(O(βω+p−1,βω+p)). | (3.19) |
By (Θ1),
O(βω+p−1,βω+p)<O(βω+p−1,βω+p). |
Continuing this process, we obtain
O(βω,βω+1)<O(βω+p−1,βω+p)<O(βω+p−2,βω+p−1)<⋯<O(βω,βω+1), | (3.20) |
which implies that p=1 and that contradict our assumption. Therefore, p=1. Now, we will prove that ξR has a fixed point. We now verify that {βω} is a Cauchy R-sequence. We assume conflicting that {βω} is not a Cauchy R-sequence. Then there exists ε>0 and we yield two subsequences of {βω} which are {βωk} and {βσk} with ωk>σk>k for which
O(βωk,βσk)≥εO(βωk,βσk−1)<ε. | (3.21) |
Using the triangular inequality, we obtain
ε≤O(βωk,βσk)≤O(βωk,βσk−1)+O(βσk−1,βσk). | (3.22) |
Letting k→∞ in (3.22) and using (3.21) and (3.17), we obtain
limω→∞O(βωk,βσk)=ε. | (3.23) |
By using (3.12), there exists a positive integer k0 such that
O(βωk,βσk)>0∀ωk>σk>k≥k0. |
So,
Θ(ε)≤Θ(O(βωk+1,βσk+1))=Θ(O(ξRβωk,ξRβσk))≤αR(βωk,βσk)Θ(O(ξRβωk,ξRβσk))≤[Θ(O(βωk,βσk))]k=[Θ(ε)]k, |
which is a contradiction since k∈(0,1). Thus, {βω} is a Cauchy R-sequence. Then there is e∈B such that βω→e as ω→∞ and let U={ω∈N:ξRβω=ξRe}. Then we get the following two cases.
Case 1. Assume that U=∞. Then there is a subsequence {βωk} of {βω} such that βωk+1=ξRβωk=ξRe,∀k∈N. Recall that βω→e, so e=ξRe.
Case 2. Assume U<∞. Then there is ω0∈N such that ξRβω≠ξRe,∀ω≥ω0, in particular, βω≠e and O(βω,e)>0 and also O(ξRβω,ξRe)>0,∀ω≥ω0. Then we know that (βωRe)or(eRβω)∀ω∈N. So, we have
αR(βω,e)≥1∀ω≥ω0 |
and we get
αR(βω,e)Θ(O(ξRβω,ξRe))≤[Θ(O(βω,e))]k∀ω≥ω0. |
Since
limω→∞O(βω,e)=0, |
by (Θ2),
limω→∞Θ(O(ξRβω,ξRe))=1, |
which implies
limω→∞(O(ξRβω,ξRe))=0. |
Thus, ξRe=e. Hence, e is the fixed point of ξR. Similarly, to the proof of Theorem 3.6, we can easily deduce that ξR has a unique fixed point.
Theorem 3.8. Let (B,R,O) be an R-CMS and ξR:B→B be a self-mapping, R-preserving, R-continuous and αR:B×B→[0,∞) be a function. Assume that there exist Θ∈Ω and k∈(0,1) such that
O(ξRβ,ξRδ)≠0⇒Θ(O(ξRβ,ξRδ))≤[Θ(U(β,δ))]k,∀β,δ∈BwithβRδandk∈(0,1). | (3.24) |
U(β,δ)=max{O(β,δ),O(β,ξRβ),O(δ,ξRδ),O(β,ξRβ)O(δ,ξRδ)1+O(β,δ)}. | (3.25) |
(ⅰ) ξR is αR-admissible.
(ⅱ) There exists β0∈B such that β0RξRβ0 and αR(β0,ξRβ0)≥1.
Then ξR has a fixed point e∈B.
Proof. Let β0∈B such that (∀δ∈Bβ0Rδ)or(∀δ∈BδRβ0). By condition (ⅱ), β0RξRβ0 or ξRβ0Rβ0. For ω∈N, consider βω=ξRωβo. Assume that ξRβω=ξRβω+1 for some ω∈N. Then βω is a fixed point of ξR and the proof is completed. Let ξRβω≠ξRβω+1 for all ω∈N. Since ξR is R-preserving, (ξRβωRξRβω+1)or(ξRβω+1RξRβω). Hence, {βω} is an R-sequence. By condition (ⅰ), for all ω∈N,
αR(βω,ξRβω)≥1,αR(ξRβω,ξRβω)≥1. |
So,
Θ(O(ξRβω,ξRβω+1))≤αR(ξRβω−1,ξRβω)Θ(O(ξRβω−1,ξRβω))≤[Θ(U(ξRβω−1,ξRβω))]k. | (3.26) |
From (3.25),
U(ξRβω−1,ξRβω)=max{O(ξRβω−1,ξRβω),O(ξRβω−1,ξRξRβω−1),O(ξRβω,ξRξRβω),O(ξRβω−1,ξRξRβω−1)O(ξRβω,ξRξRβω)1+O(ξRβω−1,ξRβω)}=max{O(ξRβω−1,ξRξRβω−1),O(ξRβω−1,ξRξRβω−1),O(ξRβω,ξRξRβω)}=max{O(ξRβω−1,ξRβω),O(ξRβω,ξRβω+1)}. | (3.27) |
If for some ω∈N,
U(ξRβω−1,ξRβω)=O(ξRβω,ξRβω+1), |
then by (3.26)
Θ(O(ξRβω,ξRβω+1))≤[Θ(O(ξRβω,ξRβω+1))]k, |
which implies that
ln[Θ(O(ξRβω,ξRβω+1))]≤kln[Θ(O(ξRβω,ξRβω+1))]. |
This is a contradiction to k∈(0,1). By (3.27), one writes for all ω∈N,
U(ξRβω−1,ξRβω)=O(ξRβω−1,ξRβω), |
and by (3.26)
Θ(O(ξRβω,ξRβω+1))≤[Θ(O(ξRβω−1,ξRβω))]k≤[Θ(O(ξRβω−2,ξRβω−1))]k2≤⋯≤[Θ(O(β,ξRβ))]kω. |
So, we have
1≤Θ(O(ξRβω,ξRβω+1))≤[Θ(O(β,ξRβ))]kω,∀ω∈N. | (3.28) |
Letting ω→∞ in (3.28), we deduce
Θ(O(ξRβω,ξRβω+1))→1. |
Then from (Θ2),
limω→∞O(ξRβω,ξRβω+1)=0. |
By (Θ3) there exist r∈(0,1) and l∈(0,∞] such that
limω→∞Θ(O(ξRβω,ξRβω+1))−1[O(ξRβω,ξRβω+1)]r=l. |
Assume that l∈(0,∞). In this case, let u=l2. With the help of limit's definition, there exists ω0∈N, such that
|Θ(O(ξRβω,ξRβω+1))−1[O(ξRβω,ξRβω+1)]r−l|≤u,∀ω≥ω0. |
This implies that
Θ(O(ξRβω,ξRβω+1))−1[O(ξRβω,ξRβω+1)]r≥l−u=u,∀ω≥ω0. |
Then,
ω[O(ξRβω,ξRβω+1)]r≤Bω[Θ(O(ξRβω,ξRβω+1))−1],∀ω≥ω0, |
where B=1/u.
Now, suppose that l=∞ and u>0 is a random positive number. With the help of limit's definition, there exists ω0∈N such that
|Θ(O(ξRβω,ξRβω+1))−1[O(ξRβω,ξRβω+1)]r−l|≥u,∀ω≥ω0, |
which implies
ω[O(ξRβω,ξRβω+1)]r≤Bω[Θ(O(ξRβω,ξRβω+1))−1],∀ω≥ω0, |
where B=1/u. In all cases, there exists B>0 such that
ω[O(ξRβω,ξRβω+1)]r≤Bω[Θ(O(ξRβω,ξRβω+1))−1],∀ω≥ω0,limω→∞ω[O(ξRβω,ξRβω+1)]r=0. |
So there exists ω1∈N such that
O(ξRβω,ξRβω+1)≤1ω1/r∀ω≥ω1. | (3.30) |
We take βω≠ξRβσ for every ω,σ∈Nwithω≠σ and
Θ(O(ξRβω,ξRβω+2))≤αR(ξRβω−1,ξRβω+1)Θ(O(ξRβω−1,ξRβω+1))≤[Θ(U(ξRβω−1,ξRβω+1))]k. | (3.31) |
U(ξRβω−1,ξRβω+1)=max{O(ξRβω−1,ξRβω+1),O(ξRβω−1,ξRξRβω−1),O(ξRβω+1,ξRξRβω+1),O(ξRβω−1,ξRξRβω−1)O(ξRβω+1,ξRξRβω+1)1+O(ξRβω−1,ξRβω+1)}. | (3.32) |
We know that Θ is non-decreasing, and so we get from (3.31) and (3.32),
Θ(O(ξRβω,ξRβω+2))≤[max{Θ(O(ξRβω−1,ξRβω+1)),Θ(O(ξRβω−1,ξRβω)),Θ(O(ξRβω+1,ξRβω+2)),O(ξRβω−1,ξRβω)O(ξRβω+1,ξRβω+2)1+O(ξRβω−1,ξRβω+1)}]k. |
That is,
Θ(O(ξRβω,ξRβω+2))≤[max{Θ(O(ξRβω−1,ξRβω+1)),Θ(O(ξRβω−1,ξRβω)),Θ(O(ξβω+1,ξβω+2))}]k. | (3.33) |
Let I be the set of ω∈N such that
Aω=max{Θ(O(ξRβω−1,ξRβω+1)),Θ(O(ξRβω−1,ξRβω)),Θ(O(ξRβω+1,ξRβω+2))}=Θ(O(ξRβω−1,ξRβω+1)). |
If |I|<∞, then there exists ω3∈N such that for every ω≥ω3,
max{Θ(O(ξRβω−1,ξRβω+1)),Θ(O(ξRβω−1,ξRβω)),Θ(O(ξRβω+1,ξRβω+2))}=max{Θ(O(ξRβω−1,ξRβω+1)),Θ(O(ξRβω+1,ξRβω+2))}. |
In this case, we get from (3.33),
Θ(O(ξRβω,ξRβω+2))≤[max{Θ(O(ξRβω−1,ξRβω+1)),Θ(O(ξRβω+1,ξRβω+2))}]k. |
Letting ω→∞ in the above inequality and using (3.29), we deduce
Θ(O(ξRβω,ξRβω+2))→1asω→∞. |
If |I|=∞, then we can find a sequence of {Aω} so that
Aω=Θ(O(ξRβω−1,ξRβω+1))forωlargeenough. |
In this case, we derive from (3.33),
1<Θ(O(ξRβω,ξRβω+2))≤[Θ(O(ξRβω−1,ξRβω+1))]k≤[Θ(O(ξRβω−2,ξRβω))]k2≤⋯≤[Θ(O(β0,ξRβ2))]kω |
for ω large enough.
Letting ω→∞, we get
Θ(O(ξRβω,ξRβω+2))→1asω→∞. | (3.34) |
Using (Θ2), we obtain
limω→∞O(ξRβω,ξRβω+2)=0, |
and by the condition (Θ3), there exists ω2∈N such that
O(ξRβω,ξRβω+2)≤1ω1/r∀ω>ω2. | (3.35) |
Let ω3=max{ω0,ω1}. Then we consider two cases.
Case1. If σ>2 is odd, then σ=2L+1,L≥1 and using (3.30),forallω≥ω3, we get
O(ξRβω,ξRβω+σ)≤O(ξRβω,ξRβω+1)+O(ξRβω,ξRβω+2)+⋯+O(ξRβω+2L,ξRβω+2L+1)≤1ω1/r+1(ω+1)1/r+…+1(ω+2L)1/r. |
Case2. If σ>2 is even, then σ=2L,L≥1 and using (3.30) and (3.35) ∀ω≥ω3, we get
(ξRβω,ξRβω+σ)≤O(ξRβω,ξRβω+2)+O(ξRβω+2,ξRβω+3)+⋯+O(ξRβω+2L−1,ξRβω+2L)≤1ω1/r+1(ω+2)1/r+…+1(ω+2L−1)1/r≤∞∑i=ω1i1/r. |
In both cases, we obtain
O(ξRβω,ξRβω+σ)≤∑∞i=ω1i1/r,∀ω≥ω3andσ≥1. |
From the convergence of the series ∑1i1/r (since 1r>1), we obtain that {ξRβω} is a Cauchy R-sequence. Since (B,R,O) is an R-CMS, there is β∗∈B such that ξRβω→β∗ as ω→∞ and we can suppose that ξRβ∗≠β∗. Assume that O(β∗,ξRβ∗)>0. Using (3.24), we get
Θ(O(ξRβω+1,ξRβ∗))≤[Θ(U(ξRβω,β∗))]K,∀ω∈N, |
where
U(ξRβω,β∗)=max{O(ξRβω,β∗),O(ξRβω,ξRβω+1),O(β∗,ξRβω),O(ξRβω,ξRβω+1)O(β∗,ξRβ∗)1+O(ξRβω,β∗)}. |
Letting ω→∞, we obtain
Θ(O(β∗,ξRβ∗))≤[Θ(O(β∗,β∗))]K<Θ(O(β∗,ξRβ∗)). |
Therefore, β∗=ξRβ∗. It is a contradiction to the hypothesis that ξR does not have a periodic point. Thus, ξR has a periodic point β∗ of period q. Assume that the set of fixed-points of ξR is empty. Then we have q>1 and β∗≠ξRβ∗. Using (1), we deduce
Θ(O(β∗,ξRβ∗))=Θ(O(ξRqβ∗,ξRq+1β∗))≤αR(ξRq−1β∗,ξRqβ∗)Θ(O(ξRqβ∗,ξRq+1β∗))≤[Θ(O(β∗,ξRβ∗))]kq<Θ(O(β∗,ξRβ∗)). |
It is a contradiction. Thus, the set of fixed-point of ξR is non-empty, that is, ξR has at least one fixed-point. Now, presume that u,β∗∈B are two fixed-points of ξR and
(uRβ∗)or(β∗Ru),so(ξRuRξRβ∗)or(ξRβ∗RξRu). |
Then
O(β∗,u)=O(ξRβ∗,ξRu)>0. |
Using (3.24), we obtain
Θ(O(β∗,u))=Θ(O(ξRβ∗,ξRu))≤[Θ(O(β∗,u))]k<Θ(O(β∗,u)), |
which is a contradiction. Then ξR has only one fixed point.
Example 3.9. Consider B=(−2,0] and
O(β,δ)={0,ifβ=δmax{β,δ},otherwise.∀β,δ∈B. |
Take βRδ⇔β+δ≥0. Then (B,R,O) is an R-MS but it is not a metric space. For this, let β=−1andδ=−12, then O(β,δ)=max{−1,−12}=−1, does not belong to [0,+∞).
Define the function αR:B×B→[0,∞) by
αR(β,δ)={1,ifβ,δ∈[0,2]e−min{β,δ}ifβ,δ∈(0,−2)0,otherwise. |
Define the mapping ξR:B→B by
ξR(β)={1ifβ∈[−12,12]∪{1},min{1,β}1+max{1,β}ifβ∈(−2,−12)∪(12,2]∖{1}. |
Then (B,R,O) is an R-CMS, but it is not a CMS. Here, we show that it is not a CMS. For this, assume βω=1ω−2 is a Cauchy sequence, letting limit as ω→+∞ then {βω} converges to −2. Hence, it is not a CMS that is clear from the definition of completeness.
If δRβ⟺δ+β≥0 then it is easy to realize that ξRδRξRβ⟺ξRδ+ξRβ≥0. So, ξR is R-preserving.
Assume {βω} is an R-sequence convergent to β. Then
limω→∞O(βω,β)=limω→∞{0,ifβ=δ=0max{βω,β},otherwise. |
Then clearly, this implies that
limω→∞O(ξRβω,ξRβ)=limω→∞{0,ifξRβ=ξRδ=0max{ξRβω,ξRβ},otherwise. |
It shows that ξR is R-continuous.
Also, ξR is αR-admissible, but not an α-admissible mapping. Here, we show that it is not α-admissible. For this, assume that B is not an R-set and we take β=−1 and δ=−12. Then,
α(−1,−12)=e−min{−1,−12}=e1>1, |
and
α(ξ(−1),ξ(−12))=α(−12,1)=0≱1. |
Given Θ:(0,∞)→(1,∞) as Θ(t)=et.
Note that ξR does not fulfill to be an α-Θ-contraction, but it verifies all the conditions of the αR-ΘR-contraction. Take β=−32 and δ=−1. Then α(−32,−1)=e32. Also,
α(−32,−1)e(O(−34,−12))=(e32)e−12=e≰ekO(−32,−1)=e−k. |
So ξR is not an α−Θ-contraction, but ξR is an αR-ΘR-contraction for each k∈[12,1). Clearly, if there exists β0∈B such that β0RξRβ0, then αR(β0,ξRβ0)≥1. Hence, all conditions of Theorem 3.6 are fulfilled and ξR has a fixed point e=1.
Within this part, we apply Theorem 3.6 to investigate the existence and uniqueness of a solution of a nonlinear fractional differential equation (see [17]) given by
dγπβ(t)=f(t,β(t))(t∈(0,1),γ∈(1,2]), |
with boundary conditions
β(0)=0,β'(0)=II∈(0,1), |
where dγπ means the Caputo fractional derivative of order γ, which is given as
dγπf(t)=1Γ(n−γ)t∫0(t−s)n−γ−1fn(s)ds(n−1<γ<n,n=[γ]+1), |
and f:[0,1]×R→R+ is a continuous function. We consider B=C([0,1],R), from [0,1] into R with supremum |β|=Supt∈[0,1]|β(t)|.
The Riemann-Liouville fractional integral of order γ (see [18]) is given by
Iγf(t)=1Γ(γ)t∫0(t−s)γ−1f(s)ds(γ>0). |
Firstly, we give the reasonable form of a nonlinear fractional differential equation and then inquest the existence of a solution by the fixed-point theorem. Now, we assume the below fractional differential equations
dγπβ(t)=f(t,β(t))(t∈(0,1),γ∈(1,2]), | (4.1) |
with the integral boundary conditions
β(0)=0,β′(0)=I(I∈(0,1)), |
where
ⅰ. f:[0,1]×R→R+ is a continuous function.
ⅱ. β(t):[0,1]→R is continuous,
so that
|f(t,β)−f(t,δ)|≤L|β−δ|, |
for all t∈[0,1] and for all β,δ∈B such that β(t)−δ(t)≥0,L is a constant with LЛ<1 where
Л=1Γ(γ+1)+2kγ+1Γ(γ)(2−k2)Γ(γ+1). |
Here, ξR is αR-admissible. Also, there exists β0(t)∈B such that β0(t)RξRβ0(t) and αR(β0(t),ξRβ0(t))≥1. Then the differential equation (4.1) has a unique solution.
Proof. We take the below R relation on B:
β(t)Rδ(t)iffβ(t)+δ(t)≥0forallt∈[0,1]. |
The given function O(β,δ)=Supt∈[0,1]|β(t)−δ(t)|∀β,δ∈B is an R-CM. We define a mapping ξR:B→B by
ξRβ(t)=1Γ(γ)t∫0(t−s)γ−1f(s,β(s))ds+2t(2−k2)Γ(γ)k∫0(s∫0(s−m)γ−1f(m,β(m))dm)ds, | (4.2) |
for all t∈[0,1]. Equation (4.1) has a solution a function β∈B iff β(t)=ξRβ(t) for all t∈[0,1]. For the purpose to check the existence of a fixed point of ξR, we are going to examine that ξR is R-preserving, an R-contraction and R-continuous.
Let for all t∈[0,1] so that β(t)Rδ(t), which means that β(t)+δ(t)≥0, and clearly from Eq (4.2),
ξRβ(t)+ξRδ(t)≥0. |
This implies that
ξRβ(t)RξRδ(t). |
Hence, ξR is R-preserving. For all t∈[0,1] and β(t)Rδ(t), we get
ξRβ(t)−ξRδ(t)=1Γ(γ)t∫0(t−s)γ−1f(s,β(s))ds+2t(2−k2)Γ(γ)k∫0(s∫0(s−m)γ−1f(m,β(m))dm)ds−[1Γ(γ)t∫0(t−s)γ−1f(s,δ(s))ds+2t(2−k2)Γ(γ)k∫0(s∫0(s−m)γ−1f(m,δ(m))dm)ds]. |
Next, we show that ξR is an R-contraction. For t∈[0,1] so that β(t)Rδ(t), we obtain
|ξRβ(t)−ξRδ(t)|=|1Γ(γ)t∫0(t−s)γ−1f(s,β(s))ds+2t(2−k2)Γ(γ)k∫0(s∫0(s−m)γ−1f(m,β(m))dm)ds−1Γ(γ)t∫0(t−s)γ−1f(s,δ(s))ds+2t(2−k2)Γ(γ)k∫0(s∫0(s−m)γ−1f(m,δ(m))dm)ds|≤1Γ(γ)t∫0(t−s)γ−1|f(s,β(s))−f(s,δ(s))|ds+2t(2−k2)Γ(γ)k∫0(s∫0(s−m)γ−1|f(m,β(m))−f(m,δ(m))|dm)ds≤L|β−δ|Γ(γ)t∫0(t−s)γ−1ds+2L|β−δ|Γ(γ)k∫0(s∫0(s−m)γ−1dm)ds≤L|β−δ|Γ(γ+1)+2kγ+1L|β+δ|Γ(γ)(2−k2)Γ(γ+2)≤L|β−δ|(1Γ(γ+1)+2kγ+1Γ(γ)(2−k2)Γ(γ+2))=LЛ|β−δ|. |
From the fact LЛ<1. Let us take Θ(t)=etet,∀t>0. Then
αR(β(t),δ(t))Θ(d(ξRβ,ξRδ)=αR(β(t),δ(t))e(d(ξRβ,ξRδ))ed(ξRβ,ξRδ)≤αR(β(t),δ(t))e(LЛd(β,δ))eLЛd(β,δ)≤αR(β(t),δ(t))e(kd(β,δ))ekd(β,δ)≤[e(d(β,δ))ed(β,δ)]k=[Θ(d(β,δ)]k, |
where k=LЛ and k∈(0,1). This implies that ξR is an R-contraction.
Suppose {βn} is an R-sequence in B such that {βn} converge to β∈B. Because ξR is R-preserving, {βn} is an R-sequence for each n∈N. Because ξR is an R-contraction, we have
αR(β(t),δ(t))Θ(d(ξRβn(t),ξRβ(t)))≤[Θ(d(βn(t),β(t))]k. |
As limn→∞d(βn(t),β(t))=0 for all τ>0, then it is clear that
limn→∞d(ξRβn(t),ξRβ(t))=0. |
Hence, ξR is R-continuous. Thus, all circumstances of Theorem 3.6 are fulfilled. This implies that β(t) is the fixed point of ξR.
In this manuscript, the notion of the concept of αR-ΘR-contractions is introduced and some fixed-point results are proved in the sense of ℜ-CMSs by using an αR-ΘR-contraction. Some constructive examples and applications to the fractional differential equation are also imparted. This work can also be extended in the sense of ℜ-extended metric spaces, ℜ-controlled metric spaces, ℜ-double controlled metric spaces, ℜ-triple controlled metric spaces, and many other structures.
The authors declare that they have no competing interests regarding the publication of this paper.
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