一类带时滞的分数阶非线性阻尼问题解的存在性与唯一性

刘静 ,  李涛 ,  高欣月 ,  杨占英

中南民族大学学报(自然科学版) ›› 2026, Vol. 45 ›› Issue (6) : 834 -845.

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中南民族大学学报(自然科学版) ›› 2026, Vol. 45 ›› Issue (6) : 834 -845. DOI: 10.20056/j.cnki.ZNMDZK.20260612
数学与数量经济科学

一类带时滞的分数阶非线性阻尼问题解的存在性与唯一性

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Existence and uniqueness of solutions for a class of fractional-order nonlinear damped problems with time delay

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摘要

研究了一类带时滞的分数阶非线性阻尼问题解的存在性和唯一性.利用Banach压缩映射原理、Krasnoselskii不动点定理以及一个广义Henry型Gronwall不等式,给出了几个保证解存在唯一性的充分条件.所得结果推广了已有文献中的相关结论.

Abstract

The existence and uniqueness of solutions to a class of fractional-order nonlinear damped problems with time delay is studied. Several sufficient conditions are established to ensure the existence and uniqueness of solutions. The proofs depend mainly on Banach’s contraction mapping principle, Krasnoselskii’s fixed-point theorem, and a generalized Henry-type Gronwall inequality. The obtained results extend those conclusions in the existing literature.

关键词

分数阶时滞方程 / 非线性阻尼方程 / 存在性 / 唯一性 / 不动点定理

Key words

fractional-order delayed equation / nonlinear damped equation / existence / uniqueness / fixed point theorem

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刘静,李涛,高欣月,杨占英. 一类带时滞的分数阶非线性阻尼问题解的存在性与唯一性[J]. 中南民族大学学报(自然科学版), 2026, 45(6): 834-845 DOI:10.20056/j.cnki.ZNMDZK.20260612

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Fractional-order calculus has garnered considerable attention due to its remarkable applicability in a wide range of fields[1-4]. In contrast to integer-order calculus, fractional-order calculus inherently exhibits infinite memory and enhanced degrees of freedom. Therefore, it can more accurately characterize phenomena in many fields such as physics, economics, medicine, and biology.
Damping is a ubiquitous physical phenomenon in oscillatory or vibratory systems. Its effects cannot be ignored in the analysis of some structures that contain rubbers, polymers, or viscoelastic materials. Over the past several decades, various material damping models have been formulated using fractional-order differential or integral equations. For certain damping materials, fractional calculus offers a particularly efficient modeling tool, yielding simpler representations that involve only a small number of parameters. In recent years, many scholars have established different types of fractional-order damped equations and have reported a wealth of interesting results regarding the properties of their solutions, including existence[5-7], uniqueness[5-6],controllability[8-12], and stability[13-14].
It is well known that the existence and uniqueness of solutions are the prerequisite for analyzing the dynamics of systems. In [6], Boulares et al. studied the existence and uniqueness of positive solutions for the following fractional differential equation:
0C𝒟tγ2ht=ft,ht+𝒟0C𝒟tγ2-1gt,ht,t∈0,T,h0=ϕ0>0,   h'0=ϕ'0>0,      
where 1<γ2≤2. The proof depends mainly on the method of upper and lower solutions, the Schauder fixed point theorem, and the Banach fixed point theorem. In [5], Sheng and Jiang considered the existence and uniqueness of the solution to the following damped system:
0C𝒟tγ2ht-Q𝒟0C𝒟tγ1ht=ft,ht,t∈0,T,h0=ϕ0,   h'0=ϕ'0,
where 0<γ1≤1<γ2≤2. The proofs rely mainly on the Banach fixed point theorem, the Krasnoselskii fixed-point theorem, and the Schauder fixed point theorem. In [15], Wang et al. considered the following nonlinear fractional-order system with delay:
0C𝒟tγ1ht=Q0ht+Q1ht-η+ft,ht,ht-η,      t∈0,T,ht=ϕt,                                                                                           t∈-η,0,
where 0<γ1≤1. The existence and uniqueness of the solution were proved by using the supremum technique and a generalized Gronwall inequality.
Motivated by the above-mentioned works, we are concerned with the following fractional nonlinear damped problem with time delay:
0C𝒟tγ2ht-Q𝒟0C𝒟tγ1ht=ft,ht,ht-η,t∈0,T,ht=ϕt,    h't=ϕ't,                            t∈-η,0,
where 0<γ1≤1<γ2≤2; Q is a constant matrix in Rn×n, η>0 is the time delay; f:0,T×Rn×Rn→Rn is continuous, and ϕ∈𝒞1-η,0,Rn is the given initial function. We establish several sufficient conditions to ensure the existence and uniqueness of the solution to Problem (4). The proofs depend mainly on Banach’s contraction mapping principle, Krasnoselskii’s fixed-point theorem, and a generalized Henry-type Gronwall inequality. It is worth pointing out that the assumption on f covers those in [5,6,15]. Our proof methods differ from those in [6,15]. In particular, Theorems 1-2 in this paper extend those in [5]. In addition, we also establish an existence and uniqueness theorem (Theorem 3) using a generalized Henry-type Gronwall inequality with delay.
The rest of this paper is organized as follows. Section 1 recalls some necessary preliminaries. Section 2 establishes several theorems on the existence and uniqueness of solutions, along with a corollary. In Section 3, we provide an example to show the correctness of our results.

1 Preliminaries

This section presents some necessary notation and recalls several definitions as well as auxiliary lemmas.

Let N0 denote the set of nonnegative integers and R+ the set of nonnegative real numbers. Let T>0 be a constant, and let x=max1≤i≤nxi be the infinite norm in Rn. Denote by 𝒞a,b,Rn the Banach space of all continuous functions mapping a,b into Rn, equipped with the norm hc=maxt∈a,bht. For an arbitrary matrix Q=aij∈Rn×n, let Q=max1≤i≤n∑j=1naij. For 1≤p<∞, Lpa,b,Rn denotes the Lebesgue space of p-integrable functions, endowed with the norm fp=∫abftp dt1/p. 𝒜𝒞a,b,Rn denotes the space of absolutely continuous functions on a,b, and 𝒜𝒞ma,b,Rn denotes the space of functions f∈𝒞m-1a,b,Rn with fm-1∈𝒜𝒞a,b,Rn, where m≥2 is an integer.

Definition 1[16]. Let α∈R+ and f∈L1[0,∞,Rn), the Riemann-Liouville fractional integral of order α is defined by

 Itαft=1Γα∫0tt-τα-1fτ dτ, t>0,

where Γ⋅ denotes the Gamma function.

Definition 2[16]. Let α denote the integer part of α∈R+. Let m=α+1. For f∈𝒜𝒞m[0,∞,Rn), the Riemann-Liouville fractional derivative of order α is defined by

0Ltαft=1Γm-αdmdtm∫0tt-τm-α-1fτ dτ, t>0.

Definition 3[17]. For α∈R+, let m=α+1 when α∉N0. If f∈𝒜𝒞m[0,∞,Rn), then the Caputo fractional derivative of order α is defined by

0Ctαft=1Γm-α∫0tt-τm-α-1fmτ dτ, t>0.

Based on Definitions 2-3, we have

0Ctαft=L𝒟tαft-∑k=0m-1tkk!fk0.

In particular, when α∈N0, we have 0Ctαft=fαt.

Lemma 1 (Krasnoselskii’s Fixed-Point Theorem [18]). Let Ω be a nonempty, bounded, closed, and convex subset of a Banach space X,⋅. Suppose that two operators 𝒫1,𝒫2:Ω→X satisfy:

(i) 𝒫1x+𝒫2y∈Ω for every x,y∈Ω;

(ii) 𝒫1 is continuous and compact;

(iii) 𝒫2 is a contraction with constant κ<1.

Then there exists a point ω∈Ω such that 𝒫1ω+𝒫2ω=ω.

Lemma 2 (Arzelà-Ascoli Theorem [19]). Let K be a compact metric space and let ℱ⊂𝒞K,Rn. If ℱ is uniformly bounded and equicontinuous on K, then ℱ is relatively compact in the Banach space 𝒞K,Rn,⋅c.

Lemma 3 (Banach Fixed-Point Theorem [20]). Let X,d be a complete metric space and let 𝒫:X→X be a contraction, i.e., there exists k∈[0,1) such that

d𝒫x,𝒫y≤k dx,y for all  x,y∈X.

Then 𝒫 admits a unique fixed point in X.

Lemma 4 (Hölder’s Inequality[21]). Letp,q≥1 be conjugate exponents satisfying 1/p+1/q=1. If ϕ∈Lpa,b and φ∈Lqa,b, then the product ϕφ belongs toL1a,b and

ϕφL1a,b≤ϕLpa,bφLqa,b.

Lemma 5[22]. Let α,β>0,ℏ,𝓁,ℑ,μ1,μ2,κ1,κ2∈ 𝒞0,T,R+, and ϖt∈𝒞-τ,0,R+. Let ℏ0=ϖ0. Suppose that ℏ,𝓁,ℑ,μ1,μ2,κ1,κ2 are nondecreasing functions. If w∈𝒞-τ,T,R+ satisfies

wt≤ℏt+lt∫0tt-sα-1Γαμ1sws+κ1sws-τds                                          +ℑt∫0tt-sβ-1Γβμ2sws+κ2sws-τds,                  t∈0,T,wt≤ϖt,                                                                                                           t∈-τ,0,

then

wt≤P1/qtexp{Gt2t/q}, t∈0,T,

where

Zt=3q-1ltΓαtα-1/qμ1tα-1p+11/p+ℑtΓβtβ-1/qμ2tβ-1p+11/pq,Rt=3q-1ltΓαtα-1/qκ1tα-1p+11/p+ℑtΓβtβ-1/qκ2tβ-1p+11/pq,Gt=max{Zt,Rt}, Pt=3q-1ℏqt,
p,q>0,1/p+1/q=1,α>1/q, and β>1/q.

2 Main results

This section focuses on the existence and uniqueness of solutions to Problem (4). Let us first present some assumptions on the nonlinear function f.

(H1) The function f: 0,T×Rn×Rn→Rn is continuous.

(H2) There exists a non-negative function δ∈Lρ1*0,T,R+ such that

ft,α,β≤δt, for all  t∈0,T  and  α,β∈Rn,

where ρ1*>1 is a given constant.

(H3) There exist two functions θ1∈Lρ2*0,T,R+ and θ2∈Lρ3*0,T,R+ such that

ft,α1,β1-ft,α2,β2≤θ1t α1-α2+θ2t β1-β2,

for all t∈0,T and all α1,α2,β1,β2∈Rn, where ρ2*,ρ3*>1 are two given constants.

For convenience in the subsequent analysis, let ρ1,ρ2, and ρ3 be the conjugate exponents corresponding to ρ1*,ρ2*, and ρ3* in (H2) and (H3), respectively. That is,

ρk=ρk*ρk*-1, k=1,2,3.

Define 𝒞ω by

𝒞ω=h∈𝒞-η,T,Rn:ht=ϕt  for  t∈-η,0, hc≤ω,

where ω is a constant such that ω≥ϕ𝒞-η,0,Rn.

By Lemma 3.1 in [7], a function h∈𝒞-η,T,ℝn is a solution of Problem (4) if and only if it satisfies the following Volterra-type integral equation

ht=ϕt,                                                                                                    t∈-η,0,ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1                                                   +QΓγ2-γ1∫0tt-sγ2-γ1-1hsds                                 +1Γγ2∫0tt-sγ2-1fs,hs,hs-ηds,            t∈(0,T].

Lemma 6. The set 𝒞ω is a non-empty, closed, and convex subset of𝒞-η,T,Rn.

Proof. It is clear that 𝒞ω is not empty. Indeed, consider the following function z defined by

zt=ϕt,t∈-η,0,ϕ0,t∈(0,T].

Obviously, it is continuous on -η,T since ϕ is continuous on -η,0. Moreover,

zc=max{ϕ𝒞-η,0, ϕ0}≤ω.

Thus, z∈𝒞ω.

Next, we prove that 𝒞ω is convex. For z1,z2∈𝒞ω and θ∈0,1, put w=θz1+1-θz2. For any t∈-η,0, we have

wt=θϕt+1-θϕt=ϕt.

Moreover, the triangle inequality allows us to obtain

wc≤θz1c+1-θz2c≤θω+1-θω=ω.

Consequently, w∈𝒞ω. This indicates that 𝒞ω is convex.

It remains to show that 𝒞ω is closed. Let {zk}⊂𝒞ω be a sequence such that zk→z in 𝒞-η,T,Rn. It is easy to see that

zt=limk→∞zkt=ϕt,    ∀t∈-η,0.

Additionally, the continuity of the norm implies

zc=limk→∞zkc≤ω.

Therefore, z∈𝒞ω. We conclude that 𝒞ω is a closed subset of 𝒞-η,T,Rn.

Theorem 1. Suppose that (H1)-(H3) hold. If

QTγ2-γ1Γγ2-γ1+1+Tγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3<1,(7)

then Problem (4) has a unique solution on -η,T.

Proof. Notice that the condition (7) evidently implies QTγ2-γ1Γγ2-γ1+1<1. This allows us to choose a constant ϱ such that

ϱ≥Γγ2-γ1+1Γγ2-γ1+1-QTγ2-γ1ϕ𝒞-η,0,Rn+Tϕ'0+QTγ2-γ1Γγ2-γ1+1ϕ0+Tγ2-1/ρ1*δLρ1*Γγ2(γ2-1ρ1+1)1/ρ1.

Let

𝒞ϱ={h∈𝒞-η,T,Rn:ht=ϕtfort∈-η,0, hc≤ϱ}.

By Lemma 6, 𝒞ϱ equipped with the supremum norm ⋅c is a complete metric space.

The operator F:𝒞ϱ→𝒞-η,T,Rn is defined as

Fht=ϕt,t∈-η,0,ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1    +QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds    +1Γγ2∫0tt-sγ2-1fs,hs,hs-η ds,t∈(0,T].

First, let us prove F𝒞ϱ⊆𝒞ϱ. Let h∈𝒞ϱ. Evidently, Fh∈𝒞-η,T,Rn. For t∈-η,0, it is obvious that Fht=ϕt≤ϕ𝒞-η,0,Rn≤ϱ. For t∈(0,T], by (H2), Hölder’s inequality and hc≤ϱ, we derive

Fht=|ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds+1Γγ2∫0tt-sγ2-1fs,hs,hs-η ds|≤ϕ0+Tϕ'0+QTγ2-γ1Γγ2-γ1+1ϕ0+QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds+1Γγ2∫0tt-sγ2-1δs ds≤ϕ0+Tϕ'0+QTγ2-γ1Γγ2-γ1+1ϕ0+QTγ2-γ1Γγ2-γ1+1ϱ+1Γγ2∫0tt-sγ2-1ρ1 ds1/ρ1δLρ1*≤ϕ0+Tϕ'0+QTγ2-γ1Γγ2-γ1+1ϕ0+ϱ+Tγ2-1/ρ1*δLρ1*Γγ2(γ2-1ρ1+1)1/ρ1≤ϱ.

This indicates Fhc≤ϱ. Hence, F maps 𝒞ϱ into 𝒞ϱ.

Now, let us show that F is a contraction mapping. Suppose that h1,h2∈𝒞ϱ. When t∈-η,0, it is evident that Fh1t-Fh2t=0. For t∈(0,T], we apply (H3) to get

Fh1t-Fh2t≤QΓγ2-γ1∫0tt-sγ2-γ1-1h1s-h2s ds+1Γγ2∫0tt-sγ2-1fs,h1s,h1s-η-fs,h2s,h2s-η ds≤QΓγ2-γ1∫0tt-sγ2-γ1-1h1s-h2s ds+1Γγ2∫0tt-sγ2-1θ1sh1s-h2s ds+1Γγ2∫0tt-sγ2-1θ2sh1s-η-h2s-η ds.

Applying Hölder’s inequality to the integrals involving θ1 and θ2, we obtain

Fh1t-Fh2t≤Qh1-h2cΓγ2-γ1∫0tt-sγ2-γ1-1 ds+h1-h2cΓγ2∫0tt-sγ2-1θ1s ds+∫0tt-sγ2-1θ2s ds≤h1-h2cQtγ2-γ1Γγ2-γ1+1+1Γγ2∫0tt-sγ2-1ρ2 ds1/ρ2θ1Lρ2*+∫0tt-sγ2-1ρ3 ds1/ρ3θ2Lρ3*≤h1-h2cQTγ2-γ1Γγ2-γ1+1+Tγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3.

Therefore,

Fh1-Fh2c≤h1-h2cQTγ2-γ1Γγ2-γ1+1+Tγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3.

Combining this with the condition (7), we conclude that F is a contraction mapping on the complete metric space 𝒞ϱ. According to Banach’s contraction mapping principle, F has a unique fixed point h*∈𝒞ϱ, which is the unique solution to Problem (4) on -η,T. This completes the proof.

Remark 1. When the time delay η is absent in Problem (4), Theorem 1 reduces to Theorem 4.1 in [5].

Theorem 2. Suppose that (H1)-(H3) hold. If

QTγ2-γ1Γγ2-γ1+1<1,

and

Tγ2-1/ρ2*Γγ2θ1Lρ2*(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*Γγ2θ2Lρ3*(γ2-1ρ3+1)1/ρ3<1,

then Problem (4) admits at least one solution on -η,T.

Proof. By (8), we can choose a constant ϱ such that

ϱ≥Γγ2-γ1+1Γγ2-γ1+1-QTγ2-γ1ϕ𝒞-η,0,Rn+Tϕ'0+QTγ2-γ1Γγ2-γ1+1ϕ0+Tγ2-1/ρ1*δLρ1*Γγ2γ2-1ρ1+11/ρ1.

Let

𝒞ϱ=h∈𝒞-η,T,Rn:ht=ϕtfort∈-η,0, hc≤ϱ.

By Lemma 6, 𝒞ϱ is a nonempty, bounded, closed, and convex subset of 𝒞-η,T,Rn. Define the operators F1,F2:𝒞ϱ→𝒞-η,T,Rn by

F1ht=ϕt,t∈-η,0,ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds,t∈(0,T],

and

F2ht=0,t∈-η,0,1Γγ2∫0tt-sγ2-1fs,hs,hs-η ds,t∈(0,T].

We first verify F1x+F2h∈𝒞ϱ for all x,h∈𝒞ϱ. Clearly, F1x+F2h is continuous. For any t∈-η,0, F1x+F2ht=ϕt. Moreover, we get that F1x+F2ht=ϕt≤ϕ𝒞-η,0,Rn≤ϱ. For t∈(0,T], the use of (H2) and Hölder’s inequality yield

F1xt+F2ht=|ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1xs ds+1Γγ2∫0tt-sγ2-1fs,hs,hs-η ds|≤ϕ0+Tϕ'0+QTγ2-γ1Γγ2-γ1+1ϕ0+QTγ2-γ1Γγ2-γ1+1ϱ+1Γγ2∫0tt-sγ2-1δs ds≤ϕ0+Tϕ'0+Tγ2-γ1Qϕ0+ϱΓγ2-γ1+1+1Γγ2∫0tt-sγ2-1ρ1 ds1/ρ1∫0tδsρ1* ds1/ρ1*≤ϕ0+Tϕ'0+Tγ2-γ1Qϕ0+ϱΓγ2-γ1+1+Tγ2-1/ρ1*δLρ1*Γγ2(γ2-1ρ1+1)1/ρ1≤ϱ.

This implies F1x+F2hc≤ϱ. Hence, F1x+F2h∈𝒞ϱ for all x,h∈𝒞ϱ.

Next, we demonstrate that F2 is a contraction mapping. Let h1,h2∈𝒞ϱ. For t∈-η,0, F2h1t-F2h2t=0. For t∈(0,T], using (H3), we obtain

F2h1t-F2h2t=1Γγ2∫0tt-sγ2-1fs,h1s,h1s-η ds-1Γγ2∫0tt-sγ2-1fs,h2s,h2s-η ds≤1Γγ2∫0tt-sγ2-1θ1sh1s-h2s+θ2sh1s-η-h2s-η ds

Applying Hölder’s inequality, we get

F2h1t-F2h2t≤h1-h2cΓγ2∫0tt-sγ2-1θ1s ds+∫0tt-sγ2-1θ2s ds≤h1-h2cΓγ2∫0tt-sγ2-1ρ2 ds1/ρ2∫0tθ1sρ2* ds1/ρ2*+∫0tt-sγ2-1ρ3 ds1/ρ3∫0tθ2sρ3* ds1/ρ3*≤h1-h2cTγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3.

Therefore,

F2h1-F2h2c≤Tγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3h1-h2c.

With the condition (9), F2 is a contraction mapping.

In what follows, we will prove that the operator F1 is continuous and compact.

Let us first discuss the continuity of F1. Suppose that {hn}n=1∞⊂𝒞ϱ is a sequence such that hn→h uniformly on -η,T. For any t∈-η,0, F1hn-F1ht=0. For any t∈0,T, we obtain

F1hnt-F1ht=ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1hns ds-ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds=QΓγ2-γ1∫0tt-sγ2-γ1-1hns-hs ds≤QΓγ2-γ1∫0tt-sγ2-γ1-1hns-hs ds≤QTγ2-γ1Γγ2-γ1+1hn-hc.

Hence,

F1hn-F1hc≤QTγ2-γ1Γγ2-γ1+1hn-hc.

Therefore, we obtain that F1hn-F1hc converges to 0 in 𝒞-η,T,Rn. This implies the continuity of F1.

In order to prove the compactness of F1𝒞ϱ, we first consider the uniform boundedness of F1𝒞ϱ. Let h∈𝒞ϱ. For t∈-η,0, F1ht≤ϕ𝒞-η,0≤ϱ. For t∈(0,T], we obtain

F1ht=ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds≤ϕ0+Tϕ'0+Tγ2-γ1Γγ2-γ1+1Qϕ0+QΓγ2-γ1∫0tt-sγ2-γ1-1hs ds≤ϕ0+Tϕ'0+Tγ2-γ1Γγ2-γ1+1Qϕ0+QTγ2-γ1Γγ2-γ1+1ϱ.

Thus,

F1hc≤ϕ0+Tϕ'0+Tγ2-γ1Γγ2-γ1+1Qϕ0+QTγ2-γ1Γγ2-γ1+1ϱ.

This gives that, F1𝒞ϱ is uniformly bounded.

Now we prove that F1𝒞ϱ is equicontinuous on -η,T. On one hand, for 0≤t1<t2≤T, we have

F1ht2-F1ht1≤ϕ'0t2-t1+ϕ0QΓγ2-γ1+1t2γ2-γ1-t1γ2-γ1+QΓγ2-γ1∫0t2t2-sγ2-γ1-1hs ds-∫0t1t1-sγ2-γ1-1hs ds≤ϕ'0t2-t1+ϕ0QΓγ2-γ1+1t2γ2-γ1-t1γ2-γ1+ϱQΓγ2-γ1∫t1t2t2-sγ2-γ1-1 ds+∫0t1t2-sγ2-γ1-1-t1-sγ2-γ1-1 ds.

Consequently,

F1ht2-F1ht1≤ϕ'0t2-t1+ϕ0QΓγ2-γ1+1t2γ2-γ1-t1γ2-γ1+ϱQΓγ2-γ1+12t2-t1γ2-γ1+t2γ2-γ1-t1γ2-γ1.

Obviously, the term on the right hand side of this inequality is irrelevant to h. As t2 tends to t1, we have F1ht2-F1ht1→0. As a result, F1𝒞ϱ is equicontinuous on 0,T.

On the other hand, for t∈-η,0, F1ht=ϕt. It is evident that F1𝒞ϱ is equicontinuous on -η,0. Moreover, F1𝒞ϱ is equicontinuous on -η,T. Combining this with the uniform boundedness of F1𝒞ϱ, we use the Arzelà-Ascoli theorem to show that F1 is a compact operator.

With the help of Krasnoselskii’s fixed point theorem, we conclude that there exists h*∈𝒞ϱ such that F1h*+F2h*=h*. Consequently, h* is a solution to Problem (4) on -η,T. The proof is complete.

Remark 2. When the time delay η is absent in Problem (4), Theorem 2 reduces to Theorem 4.2 in [5].

To further investigate the uniqueness of the solution to Problem (4), we introduce the following assumption, which is stronger than (H3).

(H4) There exist two nondecreasing functions θ1,θ2∈𝒞0,T,R+ such that

ft,α1,β1-ft,α2,β2≤θ1t α1-α2+θ2t β1-β2,

for all t∈0,T and all α1,α2,β1,β2∈Rn.

Theorem 3. Suppose that (H1), (H2) and (H4) hold. If

QTγ2-γ1Γγ2-γ1+1<1,

and

Tγ2-1/ρ2*Γγ2θ1Lρ2*(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*Γγ2θ2Lρ3*(γ2-1ρ3+1)1/ρ3<1,

then Problem (4) has a unique solution on -η,T.

Proof. According to Theorem 2, there exists at least one solution to Problem (4). Let h1 and h2 be two solutions to Problem (4), and let ξt=h1t-h2t.

For t∈-η,0, h1=h2=ϕt. For t∈0,T, it is obvious that h1 and h2 satisfy the equivalent integral equation (6). Namely,

hit=ϕ0+tϕ'0-tγ2-γ1Qϕ0Γγ2-γ1+1+QΓγ2-γ1∫0tt-sγ2-γ1-1his ds+1Γγ2∫0tt-sγ2-1fs,his,his-η ds,

where i=1,2.

Clearly, ξt=0, t∈-η,0. For t∈0,T,

ξt=QΓγ2-γ1∫0tt-sγ2-γ1-1h1s-h2s ds+1Γγ2∫0tt-sγ2-1fs,h1s,h1s-η-fs,h2s,h2s-η ds≤QΓγ2-γ1∫0tt-sγ2-γ1-1h1s-h2s ds+1Γγ2∫0tt-sγ2-1fs,h1s,h1s-η-fs,h2s,h2s-η ds≤QΓγ2-γ1∫0tt-sγ2-γ1-1ξs ds+1Γγ2∫0tt-sγ2-1θ1sξs+θ2sξs-η ds.

The condition 0<γ1≤1<γ2≤2 implies γ2-γ1>0 and γ2>0. Applying Lemma 5, we get

ξt≤0, t∈0,T.

Hence, ξt=0 for any t∈-η,T. Moreover, h1t≡h2t for t∈-η,T. This completes the proof of Theorem 3. □

In the absence of time delay, Problem (4) reduces to the following form

0C𝒟tγ2ht-Q𝒟0C𝒟tγ1ht=ft,ht,t∈0,T,h0=ϕ0,h'0=ϕ'0,

where 0<γ1≤1<γ2≤2.

In this case, (H1), (H2) and (H4) reduce to the following assumptions.

(H1)' The function f:0,T×Rn→Rn is continuous.

(H2)' There exists a non-negative function δ∈Lρ1*0,T,R+ such that

ft,α≤δt, for all  t∈0,T   and  α∈Rn,

where ρ1*>1 is a given constant.

(H4)' There exists θ∈𝒞0,T,R+ such that

ft,α1-ft,α2≤θt α1-α2

for all t∈0,T and all α1,α2∈Rn.

Corollary 1. Suppose that (H1)', (H2)' and (H4)' hold. If

QTγ2-γ1Γγ2-γ1+1<1   and   θ∞Tγ2Γγ2+1<1,

then Problem (10) has a unique solution on 0,T.

3 Application

Example 1. For Problem (4), the parameters are given as follows: γ1=1/2, γ2=3/2, Q=0.2, T=1, η=0.2, f(t,x,y)=110+150sin x+150sin y, ϕ(t)=1+t2, ϕ'(t)=12. The corresponding form is written as:

0C𝒟t3/2ht-0.2 𝒟0C𝒟t1/2ht=110+150sin ht+150sin ht-0.2,     0<t≤1,ht=1+t2,    h't=12-0.2≤t≤0. 

Obviously, f is continuous on [0,1]×R×R and satisfies the following conditions:

|f(t,x,y)|≤750,

and

|f(t,x1,y1)-f(t,x2,y2)|≤150|x1-x2|+150|y1-y2|.

Let δ(t)=7/50and θ1(t)=θ2(t)=1/50, and let ρ1*=ρ2*=ρ3*=ρ2=ρ3=2. Obviously, δ ∈L20,T,R+, θ1∈L20,T,R+, and θ2∈L20,T,R+.

(1) With the above parameters, we derive

|Q|Tγ2-γ1Γ(γ2-γ1+1)=0.2<1,

and

Tγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3=2⋅1/50Γ(3/2)2≈0.0319<1.

According to Theorem 2, Problem (12) has at least one solution on [-0.2,1].

(2) On the basis of the above parameters, we obtain

QTγ2-γ1Γγ2-γ1+1+Tγ2-1/ρ2*θ1Lρ2*Γγ2(γ2-1ρ2+1)1/ρ2+Tγ2-1/ρ3*θ2Lρ3*Γγ2(γ2-1ρ3+1)1/ρ3≈0.2319<1.

By Theorem 1, Problem (12) has a unique solution on [-0.2,1].

(3) It is easy to see that θ1(t) and θ2(t) are continuous and nondecreasing on [0,1]. Based on the above data, the conditions of Theorem 3 are satisfied. Therefore, by Theorem 3, there exists a unique solution to Problem (12) on [-0.2,1].

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基金资助

国家自然科学基金资助项目(62476292)

湖北省自然科学基金资助项目(2020CFB637)

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