Citation: Yang Cao, Quan Shi, Sen-Lai Zhu. A relaxed generalized Newton iteration method for generalized absolute value equations[J]. AIMS Mathematics, 2021, 6(2): 1258-1275. doi: 10.3934/math.2021078
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As an extension of the current development and generalizations in the field of fractional calculus [1,2,3], the investigation of solution behaviors and qualitative properties of the solution in classical or fractional differential equations has become a matter of intense interest for researchers. This reflects the extent of its uses in several applied and engineering aspects. It draws amazing applications in nonlinear oscillations of seismic tremors, the detection of energy transport rate, and energy generation rate. The importance of fractional equations has been recognized in many physical phenomena, in addition to its importance in the mathematical modeling of diseases and viruses to limit and reduce their spread.
The relaxation differential equation gives as u′(ϰ)+u(ϰ)=f(t);u(0+)=u0, whose solution is
u(ϰ)=u0exp(−ϰ)+∫ϰ0f(t−τ)exp(−τ)dτ. |
Some recent contributions to the theory of FDEs can be seen in [4]. In [5], the authors studied the following problem
Dκ0+u(ϰ)=f(ϰ,u(ϰ)),ϰ∈(0,1),u(0)=0, |
where 0<κ<1,f:[0,1]×R+→R+ is continuous and f(ϰ,⋅) is non-decreasing for ϰ∈[0,1], by lower and upper (LU) solution method. The existence and uniqueness of solutions of the FDE
Dκ0+u(ϰ)=f(ϰ,u(ϰ)),(0<κ<1;ϰ>0), | (1.1) |
Dκ−10+u(0+)=u0, | (1.2) |
were obtained in [1,2,6], by using the fixed point theorem (FPT) of Banach.
In [7], the authors discussed the existence and uniqueness of solutions of the following FDE
Dκ0+u(ϰ)=f(ϰ,u(ϰ)),ϰ∈(0,ϰ],ϰ1−κu(ϰ)|ϰ=0=u0, | (1.3) |
by using the LU solution method and its associated monotone iterative (MI) method. The problem (1.3) with non-monotone term has been studied by Bai et al. [8].
In [9], a new appraoch of the maximum principle was presented by using the completely monotonicity of the Mittag-Leffler (ML) function.
On the other hand, there are several definitions and generalizations of the fractional operators (FOs) that contributed a lot to the development of this field. The generalization of RL's FOs based on a local kernel containing a differentiable function was first introduced by Osler [3]. Next, Kilbas et al. [1] dealt with some of the properties of this operator. Then, the interesting properties for this operator have been discussed by Agarwal [10]. Recently, Jarad and Abdeljawad [11] achieved some properties in accordance with the generalized Laplace transform with respect to another function.
In this regard, most of the results similar to our current work are covered under the generalized FOs of Caputo [12] and Hilfer [13], for instance, see [14,15,16,17,18,19], whereas, very few considered results related to the dependence on generalized RL's definition. The authors in [20,21], investigated the existence and uniqueness of positive solutions of the fractional Cauchy problem in the frame of generalized RL and Caputo, respectively.
For this end, as an additional contribution and enrichment to this active field, we consider the following linear and nonlinear relaxation equations with non-monotone term under ψ-RL fractional derivatives (ψ -RLFD):
Dκ;ψ0+u(ϰ)+λDδ;ψ0+u(ϰ)=f(ϰ),ϰ∈(0,h], | (1.4) |
(ψ(ϰ)−ψ(0))1−κu(ϰ)|ϰ=0=u0≠0, | (1.5) |
where 0<h<+∞, 0<κ,δ<1,λ≥0,f∈C([0,h],R) and
Dκ;ψ0+u(ϰ)+λu(ϰ)=f(ϰ,u(ϰ)),ϰ∈(0,h], | (1.6) |
(ψ(ϰ)−ψ(0))1−κu(ϰ)|ϰ=0=u0≠0, | (1.7) |
where f∈C([0,h]×R, R), Dκ;ψ0+and Dδ;ψ0+ are RL fractional derivatives of order κ and δ, respectively, with respect to another function ψ∈C1([0,h],R), which is increasing, and ψ′(ϰ)≠0 for all [0,h]. The main contributions of this work stand out as follows:
i) With a new version of Laplace transform, we obtain Hyers-Ulam (HU) and generalized Hyers-Ulam (GHU) stabilities on the finite time interval to check whether the approximate solution is near the exact solution for a ψ-RL linear FDEs (1.4) and (1.5).
ii) We establish a condition to derive the existence and uniqueness of solutions for ψ-RL nonlinear FDEs (1.6) and (1.7), by using LU solution method along with the Banach contraction map (this generalizes the results in [7]).
iii) We formulate a new condition on the nonlinear term to ensure the equivalence between the solution of the proposed problem and the corresponding fixed point. Then in light of that, we discuss the maximal and minimal solutions for (1.6) and (1.7).
Remark 1.1.
(1) Our results remain valid if λ=0 on problems (1.4)–(1.7), which reduce to
Dκ;ψ0+u(ϰ)=f(ϰ),ϰ∈(0,h], | (1.8) |
(ψ(ϰ)−ψ(0))1−κu(ϰ)|ϰ=0=u0≠0. | (1.9) |
and
Dκ;ψ0+u(ϰ)=f(ϰ,u(ϰ)),ϰ∈(0,h], | (1.10) |
(ψ(ϰ)−ψ(0))1−κu(ϰ)|ϰ=0=u0≠0, | (1.11) |
(2) If ψ(ϰ)=ϰ, the problems (1.10) and (1.11) reduces to problem (1.3) considered in [7].
(3) The linear versions (1.4) and (1.5) generalizes that given in Theorem 5.1 by Jarad et al. [11].
Observe that in the preceding works, the nonlinear term needs to fulfill the monotone or other control conditions. Indeed, the nonlinear FDE with a non-monotone term can respond better to generic regulation, so it is vital to debilitate the control states of the nonlinear term.
This work is coordinated as follows. Section 2 provides some consepts of ψ -fractional calculus. Section 3 studies the stability results for the ψ -RL linear FDEs (1.4) and (1.5). In Section 4, we investigate of the the existence and uniqueness results for the ψ-RL nonlinear FDEs (1.6) and (1.7). Moreover, the existence of maximal and minimal solutions is also obtaind. At the end, we provide some examples in the last section.
Given 0≤a<b<+∞ and s>0, and let ψs(ϰ,a):=(ψ(ϰ)−ψ(a))s. Define a set
Cs;ψ[a,b]={u:u∈C(a,b], ψs(ϰ,a)u(ϰ)∈C[a,b]}. |
Clearly, Cs;ψ[a,b] is a Banach space with the norm
‖u‖Cs;ψ=‖ ψs(ϰ,a)u(ϰ)‖C=maxϰ∈[a,b] ψs(ϰ,a)|u(ϰ)|. |
Definition 2.1. [1] Let θ>0, and f:[a,b]→R be an integrable function. Then the generalized RL fractional integral and derivative with respect to ψ is given by
Iκ;ψa+f(ϰ)=1Γ(κ)∫ϰaψ′(ζ)ψκ−1(ϰ,s)f(ζ)dζ, |
and
Dκ;ψa+f(ϰ)=[1ψ′(ϰ)ddϰ]nIn−κ;ψa+f(ϰ), |
respectively, where n=[κ]+1, and ψ:[a,b]→R is an increasing with ψ′(ϰ)≠0, for all t∈[a,b].
Lemma 2.1. ([11], Theorem 5.1) Let 0<κ<1,λ∈R is a constant, and ϕ∈L(0,h). Then the linear version
{Dκ;ψa+u(ϰ)−λu(ϰ)=ϕ(ϰ), ϰ>a,I1−κ;ψa+ u(ϰ)|ϰ=a=c∈R, | (2.1) |
has the following solution
u(ϰ)=c ψκ−1(ϰ,a)Eκ,κ(λψκ(ϰ,a))+∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(λψκ(ϰ,ζ))ϕ(ζ)dζ. |
Lemma 2.2. [1] For 0<κ≤1, the ML function Eκ,κ(−λ(ψ(ϰ)−ψ(0))κ) satisfies
0≤Eκ,κ(−λ(ψ(ϰ)−ψ(0))κ)≤1Γ(κ),ϰ∈[0,∞),λ≥0. |
Lemma 2.3. (see [11], Lemma 4.2) For Re(s)>|λ|1κ−δ, we have
∫∞0e−s[ψ(ϰ)−ψ(0)][ψ(ϰ)−ψ(0)]β−1Eα,β(−λ[ψ(ϰ)−ψ(0)]α)dϰ=sα−βsα+λ. |
Lemma 2.4. [22] Assume that U is an ordered Banach space, u0,v0∈U, u0≤v0, D=[u0,v0], Q:D→U is an increasing completely continuous map and u0≤Qu0,v0≥Qv0. Then, Q has u∗ and v∗ are minimal and maximal fixed point, respectively. If we set
un=Qun−1,vn=Qvn−1,n=1,2,…, |
then
u0≤u1≤u2≤⋯≤un≤⋯≤vn≤⋯≤v2≤v1≤v0,un→u∗,vn→v∗. |
Definition 2.2. We say that v(ϰ)∈C1−κ;ψ[0,h] is a lower solution of (1.6) and (1.7), if it satisfies
Dκ;ψ0+v(ϰ)+λv(ϰ)≤f(ϰ,v(ϰ)),ϰ∈(0,h), | (2.2) |
ψ1−κ(ϰ,0) v(ϰ)|ϰ=0≤u0. | (2.3) |
Definition 2.3. We say that w(ϰ)∈C1−κ;ψ[0,h] is an upper solution of (1.6) and (1.7), if it satisfies
Dκ;ψ0+w(ϰ)+λw(ϰ)≥f(ϰ,w(ϰ)),ϰ∈(0,h), | (2.4) |
ψ1−κ(ϰ,0) w(ϰ)|ϰ=0≥u0. | (2.5) |
Theorem 2.4. [11] Let 0<κ<1. Then, the generalized Laplace transform of ψ-RL fractional derivative is given by
Lψ[Dκ;ψ0+u(ϰ)]=sκLψ[u(ϰ)]−I1−κ;ψ0+u(ϰ)|ϰ=0, |
where
Lψ{f(t)}=∫∞ae−s[ψ(t)−ψ(a)]ψ′(t)f(t)dt. |
Here, we discuss the HU and GHU stability of ψ-RL linear problems (1.4) and (1.5), by using the ψ-Laplace transform. Before proceeding to prove the results, we will provide the following auxiliary lemmas:
Lemma 3.1. Let 0<κ<1, and u∈C1−κ;ψ[0,h]. If
limϰ→0+(ψ(ϰ)−ψ(0))1−κu(ϰ)=u0,u0∈R, |
then
I1−κ;ψ0+u(0+):=limϰ→0+I1−κ;ψ0+u(ϰ)=u0Γ(κ). |
Proof. The proof is obtained by the same technique presented in Lemma 3.2, see [1], taking into account the properties of the ψ function.
Lemma 3.2. Let 0<κ,δ<1,λ≥0 is a constant, and f:[0,h]→R is a continuous function. Then the linear problems (1.4) and (1.5) has the following solution
u(ϰ)=[Γ(κ)+λΓ(δ)]u0ψκ−1(ϰ,0)Eκ−δ,κ(−λψκ−δ(ϰ,0))+∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ−δ,κ(−λψκ−δ(ϰ,ζ))f(ζ)dζ. | (3.1) |
Proof. Taking the generalized Laplace transform of (1.4) as
Lψ[Dκ;ψ0+u(ϰ)]+λLψ[Dδ;ψ0+u(ϰ)]=Lψ[f(ϰ)]. |
Via Theorem 2.4, we have
sκLψ[u(ϰ)]−I1−κ;ψ0+u(ϰ)|ϰ=0+λ[sδLψ[u(ϰ)]−I1−δ;ψ0+u(ϰ)|ϰ=0]=Lψ[f(ϰ)]. |
From (1.5) and Lemma 3.1, we have I1−κ;ψ0+u(ϰ)|ϰ=0=Γ(κ)u0. It follows that
sκLψ[u(ϰ)]−Γ(κ)u0+λ[sδLψ[u(ϰ)]−Γ(δ)u0]=Lψ[f(ϰ)]. |
One has,
Lψ[u(ϰ)]=s−δsκ−δ+λ[[Γ(κ)+λΓ(δ)]u0+Lψ[f(ϰ)]]. | (3.2) |
Taking L−1ψ to both sides of (3.2), it follow from Lemma 2.3 that
u(ϰ)=[Γ(κ)+λΓ(δ)]u0ψκ−1(ϰ,0)Eκ−δ,κ(−λψκ−δ(ϰ,0))+∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ−δ,κ(−λψκ−δ(ϰ,ζ))f(ζ)dζ, |
which is (3.1).
Theorem 3.1. Let 0<κ,δ<1, λ≥0, and f:[0,h]→R is a continuous function. If u∈C1−κ;ψ[0,h] satisfies the inequality
|Dκ;ψ0+u(ϰ)+λDδ;ψ0+u(ϰ)−f(ϰ)|≤ϵ, | (3.3) |
for each ϰ∈(0,h] and ϵ>0, then there exists a solution ua∈C1−κ;ψ[0,h] of (1.4) such that
|u(ϰ)−ua(ϰ)|≤ψκ(h,0)Γ(κ+1)ϵ. |
Proof. Let
Υ(ϰ):=Dκ;ψ0+u(ϰ)+λDδ;ψ0+u(ϰ)−f(ϰ),ϰ∈(0,h]. | (3.4) |
As per (3.3), |Υ(ϰ)|≤ϵ. Taking the ψ-Laplace transform of (3.4) via Theorem 2.4, we have
Lψ[Υ(ϰ)]=Lψ[Dκ;ψ0+u(ϰ)]+λLψ[Dδ;ψ0+u(ϰ)]−Lψ[f(ϰ)]=sκLψ[u(ϰ)]−I1−κ;ψ0+u(ϰ)|ϰ=0+λ[sδLψ[u(ϰ)]−I1−δ;ψ0+u(ϰ)|ϰ=0]−Lψ[f(ϰ)]. |
From (1.5) and Lemma 3.1, I1−κ;ψ0+u(ϰ)|ϰ=0=Γ(κ)u0. It follows that
Lψ[Υ(ϰ)]=sκLψ[u(ϰ)]−Γ(κ)u0+λsδLψ[u(ϰ)]−λΓ(δ)u0−Lψ[f(ϰ)]. |
One has,
Lψ[u(ϰ)]=s−δsκ−δ+λ[[Γ(κ)+λΓ(δ)]u0+Lψ[Υ(ϰ)]+Lψ[f(ϰ)]]. | (3.5) |
Set
ua(ϰ)=[Γ(κ)+λΓ(δ)]u0ψκ−1(ϰ,0)Eκ−δ,κ(−λψκ−δ(ϰ,0))+∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ−δ,κ(−λψκ−δ(ϰ,ζ))f(ζ)dζ. | (3.6) |
Taking the Laplace transform of (3.6), It follow from Lemma 2.3 that
Lψ[ua(ϰ)]=s−δsκ−δ+λ[[Γ(κ)+λΓ(δ)]u0+Lψ[f(ϰ)]]. | (3.7) |
Note that
Lψ[Dκ;ψ0+ua(ϰ)]+λLψ[Dδ;ψ0+ua(ϰ)]=Lψ[Υ(ϰ)]+Lψ[f(ϰ)]=sκLψ[ua(ϰ)]−Γ(κ)u0+λsδLψ[ua(ϰ)]−λΓ(δ)u0=(sκ+λsδ)Lψ[ua(ϰ)]−(Γ(κ)+λΓ(δ))u0. | (3.8) |
Substituting (3.7) into (3.8), we get
Lψ[Dκ;ψ0+ua(ϰ)]+λLψ[Dδ;ψ0+ua(ϰ)]=Lψ[f(ϰ)], |
which implies that ua(ϰ) is a solution of (1.4) and (1.5) due to Lψ is one-to-one. It follow from (3.5) and (3.7) that
Lψ[u(ϰ)−ua(ϰ)](s)=s−δsκ−δ+λLψ[Υ(ϰ)], |
which implies
u(ϰ)−ua(ϰ)=Lψ{ψκ−1(ϰ,0)Eκ−δ,κ(−λψκ−δ(ϰ,0))}Lψ[Υ(ϰ)]={ψκ−1(ϰ,0)Eκ−δ,κ(−λψκ−δ(ϰ,0))}∗ψΥ(ϰ). |
Thus, from (Definition 2.6, [21]) and Lemma 2.2, we obtain
|u(ϰ)−ua(ϰ)|=|{ψκ−1(ϰ,0)Eκ−δ,κ(−λψκ−δ(ϰ,0))}∗ψΥ(ϰ)|≤∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)|Eκ−δ,κ(−λψκ−δ(ϰ,ζ))||Υ(ζ)|dζ≤ϵ∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)|Eκ−δ,κ(−λψκ−δ(ϰ,ζ))|dζ≤ϵΓ(κ)∫ϰ0ψ′(τ)ψκ−1(ϰ,ζ)dζ≤ψκ(h,0)Γ(κ+1)ϵ. |
Remark 3.1. If h<∞, then (1.4) is HU stable with the constant K:=ψκ(h,0)Γ(κ+1).
Corollary 3.1. On Theorem 3.1, let φ:[0,∞)→[0,∞) is continuous function. If we set φ(ϵ)=ψκ(h,0)Γ(κ+1)ϵ, which satisfies φ(0)=0, then (1.4) is GHU stable.
In this section, we prove the existence and uniqueness results for ψ -RL nonlinear FDEs (1.6) and (1.7), by using the LU solution method and the Banach contraction mapping. Moreover, we discuss the maximal and minimal solutions for the problem at hand. The following hypotheses will be used in our forthcoming analysis:
(A1) There exist constants A,B≥0 and 0<s1≤1<s2<1/(1−κ) such that for ϰ∈[0,h],
|f(ϰ,u)−f(ϰ,v)|≤A|u−v|s1+B|u−v|s2,u,v∈R. | (4.1) |
(A2) f:[0,h]×R→R satisfies
f(ϰ,u)−f(ϰ,v)+λ(u−v)≥0,for ˆu≤v≤u≤˜u, |
where λ≥0 is a constant and ˆu,˜u are lower and upper solutions of problems (1.6) and (1.7) respectively.
(A3) There exist constant ℵ>0 such that
|f(ϰ,u)−f(ϰ,v)|≤ℵ|u−v|,ϰ∈[0,h],u,v∈R. |
Remark 4.1. Suppose that f(ϰ,u)=a(ϰ)g(u) with g is a Hölder continuous and a(ϰ) is bounded, then (4.1) holds.
Theorem 4.1. Suppose (A1) holds. The function u solves problems (1.6) and (1.7) iff it is a fixed-point of the operator Q:C1−κ;ψ[0,h]→C1−κ;ψ[0,h] defined by
(Qu)(ϰ)=Γ(κ)u0ψκ−1(ϰ,0)Eκ,κ(−λψκ(ϰ,0))+∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))f(ζ,u(ζ))dζ. | (4.2) |
Proof. At first, we show that the operator Q is well defined. Indeed, for every u∈C1−κ;ψ[0,h] and ϰ>0, the integral
∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))f(ζ,u(ζ))dζ, |
belongs to C1−κ;ψ[0,h], due to
ψ1−κ(ϰ,0)f(ζ,u(ζ))∈C[0,h], and ψ1−κ(ϰ,0)u(ζ)∈C[0,h], |
bearing in mind that
Φ(ϰ):=ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))=∞∑m=0ψ′(ζ)(−λ)m ψ(2κ−1)(m+1)(ϰ,ζ)Γ(κ(m+1)) |
is continuous on [0,h].
By the condition (4.1), we have
|f(ϰ,u)|≤A|u|s1+B|u|s2+C, | (4.3) |
where C=maxϰ∈[0,h]f(ϰ,0).
By Lemma 2.2, for u(ϰ)∈C1−κ;ψ[0,h], we have
|ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))f(ζ,u(ζ))dζ|≤ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))|f(ζ,u(ζ))|dζ≤ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))(A|u|s1+B|u|s2+C)dζ≤ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ)){Aψ(κ−1)s1(ζ,0)[ψ1−κ(ζ,0)|u(ζ)|]s1+Bψ(κ−1)s2(ζ,0)[ψ1−κ(ζ,0)|u(ζ)|]s2+C}dζ≤A(‖u‖C1−κ;ψ)s1 ψ1−κ(ϰ,0)Γ(κ)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ψ(κ−1)s1(ζ,0)dζ+B(‖u‖C1−κ;ψ)s2 ψ1−κ(ϰ,0)Γ(κ)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ψ(κ−1)s2(ζ,0)dζ+CΓ(κ+1)ψ1(ϰ,0)≤A(‖u‖C1−κ;ψ)s1 Γ((κ−1)s1+1)Γ((κ−1)s1+κ+1)ψ(κ−1)s1+κ+1−κ(ϰ,0)+B(‖u‖C1−κ;ψ)s2 Γ((κ−1)s2+1)Γ((κ−1)s2+κ+1)ψ(κ−1)s2+κ+1−κ(ϰ,0)+CΓ(κ+1)ψ1(ϰ,0)≤Γ[(κ−1)s1+1]Aψ(κ−1)s1+1(h,0)Γ[(κ−1)s1+κ+1](‖u‖C1−κ;ψ)s1+Γ[(κ−1)s2+1]Bψ(κ−1)s2+1(h,0)Γ[(κ−1)s2+κ+1](‖u‖C1−κ;ψ)s2+CΓ(κ+1)ψ1(h,0). |
Thus, the integral exists and belongs to C1−κ;ψ[0,h].
The previous inequality and the hypothesis 0<s1≤1<s2<1/(1−κ) imply that
limϰ→0+ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))f(ζ,u(ζ))dζ=0. |
Since limϰ→0+Eκ,κ(−λψκ(ϰ,0))=Eκ,κ(0)=1/Γ(κ) it follows that
limϰ→0+ψ1−κ(ϰ,0)(Qu)(ϰ)=u0. |
.
The above arguments concerted along with Lemma 2.1 yields that the fixed-point of Q solves (1.6) and (1.7). And the vice versa. The proof is complete.
Next, we consider the compactness of Cs;ψ[0,h]. Let F⊂Cs;ψ[0,h] and X={g(ϰ)=ψs(ϰ,0)h(ϰ)∣h(ϰ)∈F}, then X⊂C[0,h]. It is obvious that F is a bounded set of Cs;ψ[0,h] iff X is a bounded set of C[0,h].
Thus, to prove that F⊂Cs;ψ[0,h] is a compact set, it is sufficient to show that X⊂C[0,h] is a bounded and equicontinuous set.
Theorem 4.2. Let f:[0,h]×R→R is a continuous and (A1) holds. Then Q is a completely continuous.
Proof. Given un→u∈C1−κ;ψ[0,h], with the definition of Q and condition (A1), we get
‖Qun−Qu‖C1−κ;ψ=‖ψ1−κ(ϰ,0)(Qun−Qu)‖∞=max0≤ϰ≤h|ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))[f(ζ,un)−f(ζ,u)]dζ|≤1Γ(κ)max0≤ϰ≤hψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)[A|un−u|s1+B|un−u|s2]dζ≤1Γ(κ)[Amax0≤ϰ≤hψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ψ−s1(1−κ)(ζ,0) ψs1(1−κ)(ζ,0)|un−u|s1dζ+Bmax0≤ϰ≤hψ1−κ(ϰ,0)∫ϰ0∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ψ−s2(1−κ)(ζ,0) ψs2(1−κ)(ζ,0)|un−u|s2dζ]≤1Γ(κ)[A(‖un−u‖C1−κ;ψ)s1 max0≤ϰ≤hψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ψ−s1(1−κ)(ζ,0)dζ+B(‖un−u‖C1−κ;ψ)s2 max0≤ϰ≤hψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ψs2(1−κ)(ζ,0)dζ]≤A(‖un−u‖C1−κ;ψ)s1Γ[1−s1(1−κ)]Γ[1−s1(1−κ)+κ]ψ1−s1(1−κ)(h,0)+B(‖un−u‖C1−κ;ψ)s1Γ[1−s2(1−κ)]Γ[1−s2(1−κ)+κ] ψ1−s2(1−κ)(h,0)→0,(n→∞). |
Thus, Q is continuous.
Assume that F⊂C1−κ;ψ[0,h] is a bounded set. Theorem 4.1 shows that Q(F)⊂C1−κ;ψ[0,h] is bounded.
Finally, we show the equicontinuity of Q(F). Given ϵ>0, for every u∈F and ϰ1,ϰ2∈[0,h],ϰ1≤ϰ2,
|[ψ1−κ(ϰ,0)(Qu)(ϰ)]ϰ=ϰ2−[ψ1−κ(ϰ,0)(Qu)(ϰ)]ϰ=ϰ1|≤[Γ(κ)u0Eκ,κ(−λψκ(ϰ,0))]ϰ2ϰ1+[ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)×Eκ,κ(−λψκ(ϰ,ζ))f(ζ,u(ζ))dζ]ϰ2ϰ1≤[Γ(κ)u0Eκ,κ(−λψκ(ϰ,0))]ϰ2ϰ1+ψ1−κ(ϰ2,0)Γ(κ)∫ϰ2ϰ1ψ′(ζ)ψκ−1(ϰ2,ζ)|f(ζ,u(ζ))|dζ+ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)∫ϰ10ψ′(ζ)[ψκ−1(ϰ2,ζ)−ψκ−1(ϰ1,ζ)]|f(ζ,u(ζ))|dζ:=[Γ(κ)u0Eκ,κ(−λψκ(ϰ,0))]ϰ2ϰ1+I1+I2. |
As Eκ,κ(−λψκ(ϰ,0)) is uniformly continuous on [0,h]. Thus
[Γ(κ)u0Eκ,κ(−λψκ(ϰ,0))]ϰ2ϰ1→0; as ϰ2→ϰ1, |
I1:=ψ1−κ(ϰ2,0)Γ(κ)∫ϰ2ϰ1ψ′(ζ)ψκ−1(ϰ2,ζ)|f(ζ,u(ζ))|dζ≤ψ1−κ(ϰ2,0)Γ(κ)∫ϰ2ϰ1ψ′(ζ)ψκ−1(ϰ2,ζ)(A|u|s1+B|u|s2+C)dζ≤p1A(‖u‖C1−κ;ψ)s1Γ(p1+κ)ψ1−κ(ϰ2,0)ψp1(ϰ2,ϰ1)+p2B(‖u‖C1−κ;ψ)s2Γ(p2+κ)ψ1−κ(ϰ2,0)ψp2(ϰ2,ϰ1)+CΓ(κ+1)ψ1(ϰ2,ϰ1)ψ1−κ(ϰ2,0)→0, as ϰ2→ϰ1, |
where p1=(κ−1)s1+1 and p2=(κ−1)s2+1,
I2:=ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)∫ϰ10ψ′(ζ)[ψκ−1(ϰ2,ζ)−ψκ−1(ϰ1,ζ)]|f(ζ,u(ζ))|dζ≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)∫ϰ10ψ′(ζ)[ψκ−1(ϰ2,ζ)−ψκ−1(ϰ1,ζ)]×(A|u|s1+B|u|s2+C)dζ:=J1+J2+J3−J4−J5−J6→0, |
where
J1≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)A(‖u‖C1−κ;ψ)s1∫ϰ10ψ′(ζ)ψκ−1(ϰ2,ζ)ψp1(ζ,0)dζ→0 |
J2≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)B(‖u‖C1−κ;ψ)s2∫ϰ10ψ′(ζ)ψκ−1(ϰ2,ζ)ψp2(ζ,0)dζ→0 |
J3≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)C∫ϰ10ψ′(ζ)ψκ−1(ϰ2,ζ)dζ→0 |
J4≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)A(‖u‖C1−κ;ψ)s1∫ϰ10ψ′(ζ)ψκ−1(ϰ1,ζ)ψp1(ζ,0)dζ→0 |
J5≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)B(‖u‖C1−κ;ψ)s2∫ϰ10ψ′(ζ)ψκ−1(ϰ1,ζ)ψp2(ζ,0)dζ→0 |
J6≤ψ1−κ(ϰ2,0)−ψ1−κ(ϰ1,0)Γ(κ)C∫ϰ10ψ′(ζ)ψκ−1(ϰ1,ζ)dζ→0 |
as ϰ2→ϰ1 along with the continuity of ψ. To summarise,
|[ψ1−κ(ϰ,0)(Qu)(ϰ)]ϰ=ϰ2−[ψ1−κ(ϰ,0)(Qu)(ϰ)]ϰ=ϰ1|→0, as ϰ2→ϰ1. |
Thus, Q(F) is equicontinuous. The proof is complete.
Theorem 4.3. Let f:[0,h]×R→R is a continuous, (A1) and (A2) hold, and v,w∈C1−κ;ψ[0,h] are lower and upper solutions of (1.6) and (1.7), respectively, such that
v(ϰ)≤w(ϰ),0≤ϰ≤h. | (4.4) |
Then, the problems (1.6) and (1.7) has x∗ and y∗ as minimal and maximal solution, respectively, such that
x∗=limn→∞Qnv,y∗=limn→∞Qnw. |
Proof. Obviously, if functions v,w are lower and upper solutions of problems (1.6) and (1.7), then there are v≤Qv, and w≥Qw. Indeed, by the definition of the lower solution, there exist q_(ϰ)≥0 and ϵ≥0 such that
Dκ;ψ0+v(ϰ)+λv(ζ)=f(ϰ,v(ϰ))−q_(ϰ),ϰ∈(0,h),ψ1−κ(ϰ,0)v(ϰ)=u0−ϵ. |
Using Theorem 4.1 and Lemma 2.2, we obtain
v(ϰ)=Γ(κ)(u0−ϵ)ψκ−1(ϰ,0)Eκ,κ(−λψκ(ϰ,0))+∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λψκ(ϰ,ζ))[f(ζ,v(ζ))−q_(ζ)]dζ≤(Qv)(ϰ). |
By the definition of the upper solution, there exist ¯q(ϰ)≥0 such that
Dκ;ψ0+w(ϰ)+λw(ζ)=f(ϰ,w(ϰ))+¯q(ϰ),ϰ∈(0,h),ψ1−κ(ϰ,0)w(ϰ)=u0+ϵ. |
Similarly, there is w≥Qw.
By Theorem 4.2, Q:C1−κ;ψ[0,h]→C1−κ;ψ[0,h] is increasing and completely continuous. Setting D:=[v,w], by the use of Lemma 2.4, the existence of x∗,y∗ is gotten. The proof is complete.
Theorem 4.4. Let f:[0,h]×R→R is a continuous and (A3) hold. Then problems (1.6) and (1.7) has a unique solution ˜u in the sector [v0,w0] on [0,h], provided
ℵΓ(κ+1)ψκ(h,0)<1, | (4.5) |
where v0,w0 are lower and upper solutions, respectively, of (1.6) and (1.7) , and v0(ϰ)≤w0(ϰ).
Proof. Let ˜u is a solution of (1.6) and (1.7). Then v0≤˜u≤w0. Consider the operator Q:C1−κ;ψ[0,h]→C1−κ;ψ[0,h] defined by (4.2). For any u1,u2∈C1−κ;ψ[0,h], we have
‖Qu1−Qu2‖C1−κ;ψ=‖ψ1−κ(ϰ,0)(Qu1−Qu2)‖∞×max0≤ϰ≤h|ψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)Eκ,κ(−λ(ψκ(ϰ,ζ))×[f(ζ,u1)−f(ζ,u2)]dζ|≤1Γ(κ)max0≤ϰ≤hψ1−κ(ϰ,0)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ)ℵ|u1−u2|dζ≤max0≤ϰ≤hℵΓ(κ)∫ϰ0ψ′(ζ)ψκ−1(ϰ,ζ) ‖u1−u2‖C1−κ;ψdζ≤ ℵΓ(κ+1)ψκ(h,0)‖u1−u2‖C1−κ;ψ. |
From (4.5), we obtain
‖Qu1−Qu2‖C1−κ;ψ<‖u1−u2‖C1−κ;ψ. |
According to Banach's contraction mapping [23], Q has a unique fixed point, which is unique solution.
In this part, we provide two examples to illustrate main results.
Example 5.1. Consider the following problem
Dκ;ψ0+u(ϰ)+110u(ϰ)=1+ϰ−κΓ(1−κ)+sinπϰΓ(1−κ)√π(|u(ϰ)|0.5+|u(ϰ)|1.5),ψ1−κ(ϰ,0)u(ϰ)|ϰ=0=12≠0. |
Here ϰ∈(0,1], λ=110,Dκ;ψ0+ is the ψ-RL fractional derivative of order 0<κ<1. Obviously,
f(ϰ,u)=1+ϰ−κΓ(1−κ)+sinπϰΓ(1−κ)√π(|u|0.5+|u|1.5), |
and for ϰ∈[0,1],u,˜u∈[0,∞), we have
|f(ϰ,u)−f(ϰ,˜u)|≤|sinπϰ|Γ(1−κ)√π[|u|0.5+|u|1.5−|˜u|0.5−|˜u|1.5]≤1Γ(1−κ)√π[|u|0.5−|˜u|0.5+|u|1.5−|˜u|1.5]≤1Γ(1−κ)√π[|u−˜u|0.5+|u−˜u|1.5]=A|u−˜u|0.5+B|u−˜u|1.5, |
for 0<s1=0.5<s2=1.5<1/(1−κ)=2, for κ=12, here A=B=1Γ(1−κ)√π. Moreover, we have
|f(ϰ,u)|≤|sinπϰ|Γ(1−κ)√π[|u|0.5+|u|1.5]≤1Γ(1−κ)√π[1+|u|0.5+|u|1.5]=A|u|0.5+B|u|1.5+C, |
where C=maxϰ∈[0,1]f(ϰ,0)=1Γ(1−κ)√π. Also, for ϰ∈[0,1],u,˜u∈[0,∞], we have
f(ϰ,u)−f(ϰ,˜u)=sinπϰΓ(1−κ)√π(|u|0.5−|˜u|0.5+|u|1.5−|˜u|1.5)≥sinπϰΓ(1−κ)√π(|u|1.5−|˜u|1.5)≥−1Γ(1−κ)√π|u−˜u|, |
where λ=1Γ(1−κ)√π>0. From the foregoing, we conclude that (A1), (A2) and (4.3) are satisfied. Hence, problem (3.4) has a solution on [0,1].
Example 5.2. Consider the following problem
{Dκ;ψ0+u(ϰ)+110u(ϰ)=12ϰ3(ϰ−u(ϰ))3−14ϰ4,ψ1−κ(ϰ,0)u(ϰ)|ϰ=0=1≠0. | (5.1) |
Here ϰ∈(0,1], λ=110,Dκ;ψ0+ is the ψ-RL fractional derivative of order 0<κ<1. Obviously,
f(ϰ,u)=12ϰ3(ϰ−u)3−14ϰ4, |
and for ϰ∈[0,1],u,˜u∈[0,∞),we have
|f(ϰ,u)−f(ϰ,˜u)|≤12ϰ3|(ϰ−u)3−(ϰ−˜u)3|≤12ϰ3|−(u3−˜u3)−3ϰ(˜u2−u2)+3ϰ2(˜u−u)|≤32ϰ5|u−˜u|≤32|u−˜u|=B|u−˜u|1.5, |
for s2=1.5<1/(1−κ)=2, for κ=12, here A=0,B=32. Take v0(ϰ)=0, w0(ϰ)=ψ2(ϰ,0)=[ψ(ϰ)−ψ(0)]2, it is not difficult to verify that v0(ϰ), w0(ϰ) be lower and upper solutions, respectively, of (5.1), and v0(ϰ)≤w0(ϰ). Then for ϰ∈[0,1],
Dκ;ψ0+v0(ϰ)+110v0(ϰ)=0≤12ϰ6−14ϰ4=f(ϰ,v0(ϰ))Dκ;ψ0+w0(ϰ)+110w0(ϰ)=Dκ;ψ0+[ψ(ϰ)−ψ(0)]2+110[ψ(ϰ)−ψ(0)]2=Γ(3)Γ(3−κ)[ψ(ϰ)−ψ(0)]2−κ+110[ψ(ϰ)−ψ(0)]2=83√πϰ32+110ϰ2≥12ϰ6(1−ϰ)3−14ϰ4=f(ϰ,w0(ϰ)), |
where we used ψ(ϰ)=ϰ. In addition, let ϵ>0, q_(ϰ)=ϰ2, and ¯q(ϰ)=ϰ2, and consider
{Dκ;ψ0+v(ϰ)+110v(ϰ)=12ϰ6−14ϰ4−q_(ϰ),ϰ∈(0,h),ψ1−κ(ϰ,0)v(ϰ)=u0−ϵ. | (5.2) |
and
{Dκ;ψ0+w(ϰ)+110w(ϰ)=12ϰ6(1−ϰ)3−14ϰ4+¯q(ϰ),ϰ∈(0,h),ψ1−κ(ϰ,0)w(ϰ)=u0+ϵ. | (5.3) |
By Lemma 4.1, we have
v(ϰ)=Γ(12)(u0−ϵ)1√ϰEκ,κ(−√ϰ10)+∫ϰ01√(ϰ−ζ)E12,12(−√(ϰ−ζ10)(ζ62−ζ44−ζ2)dζ, |
and
w(ϰ)=Γ(12)(u0+ϵ)1√ϰEκ,κ(−√ϰ10)+∫ϰ01√(ϰ−ζ)E12,12(−110√(ϰ−ζ)(ζ6(1−ζ)32−ζ44−ζ210)dζ. |
Thus, all assumptions of Theorem 4.2 are fulfilled. As per Theorem 4.3, problem (5.1) has minimal and maximal solutions u∗∈[v0,w0], ˜u∗∈[v0,w0], which can be obtained by
u∗=limn→∞ vn,˜u∗=limn→∞wn, |
where
vn(ϰ)=Γ(12)u01√ϰE12,12(−√ϰ10)+∫ϰ01√(ϰ−ζ)E12,12(−√(ϰ−ζ10)(12ζ3(ζ−vn−1(ζ))3−14ζ4)dζ, n≥1 |
and
wn(ϰ)=Γ(12)u01√ϰE12,12(−√ϰ10)+∫ϰ01√(ϰ−ζ)E12,12(−√(ϰ−ζ10)(12ζ3(ζ−wn−1(ζ))3−14ζ4)dζ, n≥1. |
In the current work, we have investigated two classes of fractional relaxation equations. Our results were based on generalized Laplace transform, fixed point theorem due to lower and upper solutions method, and functional analysis approaches. The psi- RL fractional operator, which is connected with numerous well-known fractional operators, has been used in our study. Ulam-Hyer's stability of solutions for the linear version has been shown by the generalized Laplace transform approach. Then by establishing the method of lower and upper solutions along with Banach's fixed point technique, we have investigated the existence and uniqueness of iterative solutions for the nonlinear version with the non-monotone term f(ϰ,u(ϰ)), which permits the nonlinearity f to manage the condition (A1) to |f(ϰ,u)|≤A|u|s1+B|u|s2+C. Besides, we have also discussed the maximal and minimal solutions to the nonlinear version. Then, some known results in the literature have been extended. Finally, two examples to illustrate the obtained results have been provided.
This work was supported by the Deanship of Scientific Research, Vice Presidency for Graduate Studies and Scientific Research, King Faisal University, Saudi Arabia [Grant No. 757].
No conflicts of interest are related to this work.
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