# On the Well-Posedness of a Fractional Model of HIV Infection

Authors: Mohamed Jleli; Bessem Samet
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we are concerned with the well-posedness of a fractional model of human immunodeficiency virus infection. Namely, using Grönwall’s lemma and Perov’s fixed point theorem, we obtain sufficient conditions for which the considered model admits a unique solution.


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## Body

## 1. Introduction

The human immunodeficiency virus (HIV) is one of the world’s leading infectious diseases. A big number of people have died around the globe due to this disease. HIV infects vital cells in the human immune system, such as CD4+ T cells. In this way, the body becomes progressively more susceptible to opportunistic infections, leading to the development of AIDS (acquired immunodeficiency syndrome) (see, e.g., [1, 2]).Mathematical models play an important role in understanding the dynamics of HIV infection. The dynamical models for HIV usually consist of systems of ordinary differential equations which range from two-component models (see, e.g., [3, 4]) to three-component (see e.g. [5–7]) and four-component models (see, e.g., [8]). In particular, in [3], the following two-cell model was proposed to describe the HIV infection:
(1)u′t=κ−σut−δutvt,v′t=δutvt−τvt,where u is the density of uninfected CD4+ T cells, v is the density of virus-producing cells, κ is the rate of production of CD4+ T cells, σ is their per capita death rate, δ is the rate of infection of CD4+ T cells, and τ is the rate of disappearance of infected cells.Due to the importance of fractional calculus in modeling real-world phenomena (see, e.g., [9–12] and the references therein), a great attention was paid to the study of fractional models of HIV infection of CD4+ T cells (see, e.g., [13–17] and the references therein).In this paper, a fractional model of (1) is investigated. Namely, we are concerned with the system of fractional differential equations
(2)CD0α,ψut=I0μ,ψκ−σu−δuvt,0≤t≤T,CD0β,ψvt=I0ν,ψδuv−τvt,0≤t≤T,subjet to the initial conditions
(3)u0,v0=u0,v0,where T>0, u0,v0≥0, 0<α, β<1, μ≥1−α, ν≥1−β, ψ∈C10,T, ψ′>0, CD0ℓ,ψ, ℓ∈α,β is the ψ-Caputo fractional derivative of order ℓ, I0ϰ,ψ, ϰ∈μ,ν is the ψ-Riemann-Liouville fractional integral of order ù, and κ,σ,δ,τ>0. Using Grönwall’s lemma and Perov’s fixed point theorem, we derive sufficient conditions for which system (2) subject to the initial conditions (3) admits a unique solution. Moreover, we provide a numerical algorithm that converges uniformly to the unique solution. Notice that when ψt=t, α,μ→1−,0+ and β,ν→1−,0+, (2) reduces to (1).The rest of the paper is organized as follows. In Section2, we recall some notions on fractional calculus and Perov’s fixed point theorem. In Section 3, we state and prove our main results. In Section 4, some special cases are discussed.


## 2. Some Preliminaries

Leta,b∈ℝ2 be such that a<b.Givenf∈Ca,b and θ>0, the (left-sided) Riemann-Liouville fractional integral of order θ of f is defined by (see [10])
(4)Iaθft=1Γθ∫att−sθ−1fsdsifa<t≤b,0ift=a,where Γ is the Gamma function. It can be easily seen that
(5)f∈Ca,b⟹Iaθf∈Ca,b.Lemma 1 (see [10], Property 2.6).
Letf∈Ca,bandθ,η>0. Then,(6)IaθIaηft=Iaθ+ηft,a≤t≤b.
Givenf∈C1a,b and 0<θ<1, the (left-sided) Caputo fractional derivative of order θ of f is defined by (see [10])
(7)CDaθft=Ia1−θf′t,a≤t≤b.Lemma 2 (see [10], Lemma 2.22).
Letf∈C1a,band0<θ<1. Then,(8)IaθCDaθft=ft−fa,a≤t≤b.Lemma 3 (see [10], Lemma 2.21).
Letf∈Ca,band0<θ<1. Then,(9)CDaθIaθft=ft,a≤t≤b.
LetT>0 be fixed. We introduce the set of functions
(10)Ψ=ψ∈C10,T:ψ′t>0forall0≤t≤T.
Givenf∈C0,T, θ>0, and ψ∈Ψ, the (left-sided) ψ-Riemann-Liouville fractional integral of order θ of f is defined by (see [10, 18])
(11)I0θ,ψft=1Γθ∫0tψt−ψsθ−1ψ′sfsdsif0<t≤T,0ift=0.
It can be easily seen that(12)f∈C0,T⇒I0θ,ψf∈C0,T.Remark 4.
In the special caseψt=t, one observes thatI0θ,ψ=I0θ.Remark 5.
Using the change of variablez=ψs, one deduces from ((11)) that(13)I0θ,ψft=Iψ0θf∘ψ−1ψt,0≤t≤T.Using Lemma1 and Remark 5, one deduces the following result.Lemma 6.
Letf∈C0,T, θ,η>0andψ∈Ψ. Then(14)I0θ,ψI0η,ψft=I0θ+η,ψft,0≤t≤T.
Givenf∈C10,T, 0<θ<1, and ψ∈Ψ, the (left-sided) ψ-Caputo fractional derivative of order θ of f is defined by (see [18])
(15)CD0θ,ψft=I01−θ,ψf′ψ′t,0≤t≤T.Remark 7.
In the special caseψt=t, one observes thatCD0θ,ψ=CD0θ.Remark 8.
Using the change of variablez=ψs, one deduces from ((15)) that(16)CD0θ,ψft=CDψ0θf∘ψ−1ψt,0≤t≤T.Using Lemma2, Remark 5, and Remark 8, one deduces the following result.Lemma 9.
Letf∈C10,T, 0<θ<1andψ∈Ψ. Then(17)I0θ,ψCD0θ,ψft=ft−f0,0≤t≤T.Similarly, using Lemma3, Remark 5, and Remark 8, one deduces the following result.Lemma 10.
Letf∈C0,T, 0<θ<1andψ∈Ψ. Then,(18)CD0θ,ψI0θ,ψft=ft,0≤t≤T.Now, we recall some concepts on fixed point theory that will be used later. Letn be a positive natural number and define the partial order °n in ℝn by
(19)y=y1y2⋮yn≺¯nz=z1z2⋮zn⇔yi≤zi,i=1,2,⋯,n,for all y,z∈ℝn. We denote by 0ℝn the zero vector in ℝn, i.e.,
(20)0ℝn=00⋮0.LetX be a nonempty set and d:X×X→ℝn be a given mapping. We say that d is a vector-valued metric on X (see, e.g., [19]), if for all x,y,z∈X,
(21)0ℝn≺¯ndx,y,dx,y=0ℝn⇔x=y,dx,y=dy,x,dx,z≺¯ndx,y+dy,z.In this case, the pairX,d is called a generalized metric space. In such spaces, the notions of convergent sequence, Cauchy sequence, and completeness are similar to those for usual metric spaces.Let us denote byMnℝ+ the set of square matrices of size n with nonnegative coefficients. Given M∈Mnℝ+, we denote by ρM its spectral radius.Lemma 11 (Perov’s fixed point theorem).
LetX,dbe a complete generalized metric space andF:X→Xbe a given mapping. Suppose that there existsM∈Mnℝ+withρM<1such that(22)dFx,Fy≺¯nMdx,y,for all x,y∈X. Then,
(i)
the mappingF admits a unique fixed point in X, say x∗(ii)
for allx0∈X, the sequence xm⊂X defined by xm+1=Fxm converges to x∗
We end this section by recalling the Grönwall’s lemma.Lemma 12.
(see [20]) LetK≥0andf, gbe nonnegative functions on0,Tsuch thatf∈L∞0,Tandg∈L10,T. If(23)ft≤K+∫0tgsfsds,0≤t≤T,then,
(24)ft≤Kexp∫0tgsds,0≤t≤T.


## 3. Main Results

Problem (2) and (3) is investigated under the following assumptions:
(i)
T>0 and ψ∈Ψ, where Ψ is the functional space defined by (10).(ii)
u0,v0≥0 and κ,σ,δ,τ>0(iii)
0<α<1 and μ≥1−α(iv)
0<β<1 and ν≥1−β


### 3.1. Integral Formulation of Problem (2) and (3)

Let(25)V=u,v∈C0,T×C0,T:u,v≥0W=u,v∈C10,T×C10,T:u,v≥0.Suppose thatu,v∈W is a solution to problem (2) and (3). Using the first equation in (2), one obtains
(26)I0α,ψCD0α,ψut=I0α,ψI0μ,ψκ−σu−δuvt,0≤t≤T.Hence, by Lemma6 and Lemma 9, it holds that
(27)ut−u0=I0α+μ,ψκ−σu−δuvt,0≤t≤T.Using the initial conditions (3), one obtains
(28)ut=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σus−δusvsds,0≤t≤T.Similarly, using the second equation in (2), one obtains
(29)I0β,ψCD0β,ψvt=I0β,ψI0ν,ψδuv−τvt,0≤t≤T,which yields
(30)vt−v0=I0β+ν,ψδuv−τvt,0≤t≤T.By the initial conditions (3), it holds that
(31)vt=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδusvs−τvsds,0≤t≤T.Therefore, one deduces that, ifu,v∈W is a solution to problem (2) and (3), then u,v∈V is a solution to the system of integral equations
(32)ut=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σus−δusvsds,vt=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδusvs−τvsds,for all 0≤t≤T.Conversely, suppose thatu,v∈V is a solution to (32). By assumptions (iii) and (iv), one deduces that u,v∈W. Moreover, by (32), one has u0=u0 and v0=v0. On the other hand, using the first equation in (32), Lemma 6, and Lemma 10, one obtains
(33)CD0α,ψut=CD0α,ψI0α+μ,ψκ−σu−δuvt=CD0α,ψI0α,ψI0μ,ψκ−σu−δuvt=I0μ,ψκ−σu−δuvt.Similarly, using the second equation in (32), one obtains
(34)CD0β,ψvt=CD0β,ψI0β+ν,ψδuv−τvt=CD0β,ψI0β,ψI0ν,ψδuv−τvt=I0ν,ψδuv−τvt.Therefore, one deduces that, ifu,v∈V is a solution to the system of integral equation (32), then u,v∈W is a solution to problem (2) and (3).From the above study, the following result holds.Lemma 13.
The following statements are equivalent:(I)
u,v∈W is a solution to problem (2) and (3).(II)
u,v∈V is a solution to the system of integral equations (32).By the above lemma, the study of problem (2) and (3) in W reduces to the study of the system of integral equation (32) in V.


### 3.2. Uniqueness

In this part, using Grönwall’s lemma, we shall prove that the system of integral equations (32) admits at most one solution u,v∈V.Proposition 14.
Suppose that the assumptions (i)–(iv) are satisfied. Then the system of integral equation ((32)) admits at most one solutionu,v∈V.Proof.
Suppose thatu1,v1,u2,v2∈V are two solutions to (32). Then, for all 0≤t≤T, one has
(35)u2t−u1t=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σu2s−δu2sv2sds−u0−1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σu1s−δu1sv1sds=σΓα+μ∫0tψt−ψsα+μ−1ψ′su1s−u2sds+δΓα+μ∫0tψt−ψsα+μ−1ψ′su1sv1s−u2sv2sds,which yields
(36)∣u2t−u1t∣≤σC1∫0tψ′s∣u1s−u2s∣ds+δC1∫0tψ′s∣u1sv1s−u2sv2s∣ds=σC1∫0tψ′s∣u1s−u2s∣ds+δC1∫0tψ′su1sv1s−v2s+v2su1s−u1sds≤σC1∫0tψ′s∣u1s−u2s∣ds+δC1C2∫0tψ′s∣v1s−v2s∣ds+δC1C3∫0tψ′s∣u1s−u2s∣ds=C1σ+δC3∫0tψ′s∣u1s−u2s∣ds+δC1C2∫0tψ′s∣v1s−v2s∣ds,where
(37)C1=ψT−ψ0α+μ−1Γα+μ,C2=max0≤x≤Tu1x,C3=max0≤x≤Tv2x.
Hence, one deduces that(38)∣u2t−u1t∣≤C4∫0tψ′s∣u1s−u2s∣+∣v1s−v2s∣ds,where
(39)C4=C1maxσ+δC3,δC2.
Similarly, for all0≤t≤T, one has
(40)v2t−v1t=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδu2sv2s−τv2sds−v0−1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδu1sv1s−τv1sds=δΓβ+ν∫0tψt−ψsβ+ν−1ψ′su2sv2s−u1sv1sds+τΓβ+ν∫0tψt−ψsβ+ν−1ψ′sv1s−v2sds,which yields
(41)∣v2t−v1t∣+τC5∫0tψ′s∣v1s−v2s∣ds≤δC5C6∫0tψ′s∣v1s−v2s∣ds+δC5C7∫0tψ′s∣u2s−u1s∣ds+τC5∫0tψ′s∣v1s−v2s∣ds=C5δC6+τ∫0tψ′s∣v1s−v2s∣ds+δC5C7∫0tψ′s∣u2s−u1s∣ds,where
(42)C5=ψT−ψ0β+ν−1Γβ+ν,C6=max0≤x≤Tu2x,C7=max0≤x≤Tv1x.
Therefore, one obtains(43)∣v2t−v1t∣≤C8∫0tψ′s∣u1s−u2s∣+∣v1s−v2s∣ds,where
(44)C8=C5maxτ+δC6,δC7.
Next, combining (38) with (43), it holds that
(45)∣u2t−u1t∣+∣v2t−v1t∣≤C4+C8∫0tψ′s∣u2s−u1s∣+∣v2s−v2s∣ds.
Finally, using Grönwall ‘s lemma (see Lemma12), it holds that u1,v1=u2,v2.


### 3.3. Well-Posedness

We first fix some notations. Let(46)aT=ψT−ψ0α+μΓα+μ+1,bT=ψT−ψ0β+νΓβ+ν+1.We introduce the functionf:0,∞→0,∞ defined by
(47)fr=aTσ+bTτ+aT+bTδr+aTσ−bTτ+aT−bTδr2+4aTbTδ2r21/2,r≥0.We denote by∥·∥∞ the norm in C0,T defined by
(48)∥ξ∥∞=max0≤x≤Tξx,ξ∈C0,T.In addition to the assumptions (i)–(iv), suppose that(v)
aT≤τ/δκ and bT≤1/√σ2+4δκ−σ(49)4τ≤σ2+4δκ−σ(vi)
There exists(50)2τδ≤r≤σ2+4δκ−σ2δsuch that
(51)τδ≤u0≤r2,0≤v0≤r2,fr<2.Theorem 15.
Suppose that the assumptions (i)–(vii) are satisfied. Then, the system of integral equation ((32)) admits one and only one solutionu∗,v∗∈V. Moreover, for allf0,g0∈Vsatisfyingu0≤f0, ∥f0∥∞≤rand∥g0∥∞≤r, the sequencefn,gn⊂Vdefined by(52)fn+1t=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σfns−δfnsgnsds,gn+1t=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδfnsgns−τgnsds,for all 0≤t≤T, converges uniformly to u∗,v∗.Proof.
We first introduce the functional space(53)Vr=u,v∈V:u0≤u,∥u∥∞≤r,∥v∥∞≤r.
LetF:Vr→C0,T×C0,T be the mapping defined by
(54)Fu,vt=F1u,vt,F2u,vt,0≤t≤T,where
(55)F1u,vt=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σus−δusvsds,F2u,vt=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδusvs−τvsds.
We shall prove that(56)FVr⊂Vr.
Letu,v∈Vr. Then, for all s∈0,T, one has
(57)κ−σus−δusvs≥κ−σr−δr2≔Pr.
On the other hand, an elementary calculation shows that the polynomial functionPr admits two roots
(58)r1=−σ−σ2+4δκ2δ<0<r2=σ2+4δκ−σ2δ.
Since−δ<0 and (by (vii))
(59)0<r≤σ2+4δκ−σ2δ=r2,one obtains
(60)Pr≥0,which yields
(61)κ−σus−δusvs≥0,0≤s≤T.
Therefore, sinceψ∈Ψ and u0≥0 (see (vii)), one deduces that
(62)F1u,v≥u0≥0.
Moreover, for alls∈0,T, one has
(63)δusvs−τvs=vsδus−τ≥vsδu0−τ.
Since by (vii),u0≥τ/δ, one obtains
(64)δusvs−τvs≥0,0≤s≤T.
Using the above inequality and the fact thatv0≥0 (by (vii)), one deduces that
(65)F2u,v≥0.
On the other hand, by (62), for all 0≤t≤T, one has
(66)∣F1u,vt∣=F1u,vt≤u0+κΓα+μ∫0tψt−ψsα+μ−1ψ′sds=u0+κψt−ψ0α+μΓα+μ+1≤u0+κaT.
SinceaT≤τ/δκ (by (v)) and τ/δ≤u0≤r/2 (by (vii)), one has
(67)u0+κaT≤r2+τδ≤r2+r2=r.
Hence, it follows from (66) and (67) that
(68)∥F1u,v∥∞≤r.
Similarly, by (65), for all 0≤t≤T, one has
(69)∣F2u,vt∣=F2u,vt≤v0+δr2Γβ+ν∫0tψt−ψsβ+ν−1ψ′sds=v0+δr2ψt−ψ0β+νΓβ+ν+1≤v0+δr2bT.
Sincev0≤r/2 (by (vii)), bT≤1/√σ2+4δκ−σ (by (v)) and r≤√σ2+4δκ−σ/2δ (by (vii)), one has
(70)v0+δr2bT≤r2+δr212δr=r2+r2=r.
Therefore, it follows from (69) and (70) that
(71)∥F2u,v∥∞≤r.
Hence, by (62), (65), (68), and (71), one deduces that Fu,v∈Vr. This proves (56).
Consider now the self-mapping,(72)F:Vr→Vr.
By the definition ofF, one observes that, if u,v∈Vr is a fixed point of F; then, u,v∈V is a solution to the system of integral equation (32). In order to prove that F admits a fixed point in Vr, we shall use Perov’s fixed point theorem (see Lemma 11). Namely, we define the vector-valued metric d:Vr×Vr→ℝ2 by
(73)du1,v1,u2,v2=∥u1−u2∥∞∥v1−v2∥∞,u1,v1,u2,v2∈Vr.
Notice thatVr,d is a complete generalized metric space. On the other hand, for all u1,v1,u2,v2∈Vr and 0≤t≤T, one has
(74)∣F1u2,v2t−F1u1,v1t∣≤σΓα+μ∫0tψt−ψsα+μ−1ψ′s∣u1s−u2s∣ds+δΓα+μ∫0tψt−ψsα+μ−1ψ′s∣u1sv1s−u2sv2s∣ds≤σaT∥u1−u2∥∞+δΓα+μ∫0tψt−ψsα+μ−1ψ′su1sv1s−v2s+v2su2s−u1sds≤σaT∥u1−u2∥∞+δaTr∥v2−v1∥∞+∥u2−u1∥∞,which yields
(75)∥F1u2,v2−F1u1,v1∥∞≤aTδr+σ∥u1−u2∥∞+δaTr∥v1−v2∥∞.
Similarly, one has(76)F2u2,v2t−F2u1,v1t≤δΓβ+ν∫0tψt−ψsβ+ν−1ψ′su2sv2s−v1s+v1su2s−u1sds+τΓβ+ν∫0tψt−ψsβ+ν−1ψ′sv1s−v2sds≤δbTr∥v2−v1∥∞+∥u2−u1∥∞+τbT∥v1−v2∥∞,which yields
(77)∥F2u2,v2−F2u1,v1∥∞≤δbTr∥u1−u2∥∞+bTδr+τ∥v1−v2∥∞.
Hence, it follows from (75) and (77) that for all u1,v1,u2,v2∈Vr,
(78)dFu1,v1,Fu2,v2°2Mdu1,v1,u2,v2,where
(79)M=aTδr+σδaTrδbTrbTδr+τ.
On the other hand, the characteristic polynomial ofM is given by
(80)Qλ=aTδr+σ−λbTδr+τ−λ−δ2aTbTr2.
An elementary calculation shows that the polynomial functionQ admits two roots (which are the eigenvalues of M) given by
(81)λ1r=aTσ+δr+bTτ+δr2−12aTσ−bTτ+aT−bTδr2+4aTbTδ2r212,λ2r=fr2.
It can be easily seen that(82)0≤λ1r≤λ2r,which yields
(83)ρM=fr2.
Sincefr<2 (by (vii)), it holds that ρM<1. Therefore, by Lemma 11, the mapping F admits a fixed point u∗,v∗∈Vr⊂V, which is a solution to the system of integral equation (32). Moreover, for all f0,g0∈Vr, the Picard sequence fn,gn defined by fn+1,gn+1=Ffn,gn=F1fn,gn,F2fn,gn converges uniformly to u∗,v∗. On the other hand, by Proposition 14, we know that (32) admits at least one solution in V. Hence, one deduces that u∗,v∗ is the only solution to (32) in V.


## 3.1. Integral Formulation of Problem (2) and (3)

Let(25)V=u,v∈C0,T×C0,T:u,v≥0W=u,v∈C10,T×C10,T:u,v≥0.Suppose thatu,v∈W is a solution to problem (2) and (3). Using the first equation in (2), one obtains
(26)I0α,ψCD0α,ψut=I0α,ψI0μ,ψκ−σu−δuvt,0≤t≤T.Hence, by Lemma6 and Lemma 9, it holds that
(27)ut−u0=I0α+μ,ψκ−σu−δuvt,0≤t≤T.Using the initial conditions (3), one obtains
(28)ut=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σus−δusvsds,0≤t≤T.Similarly, using the second equation in (2), one obtains
(29)I0β,ψCD0β,ψvt=I0β,ψI0ν,ψδuv−τvt,0≤t≤T,which yields
(30)vt−v0=I0β+ν,ψδuv−τvt,0≤t≤T.By the initial conditions (3), it holds that
(31)vt=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδusvs−τvsds,0≤t≤T.Therefore, one deduces that, ifu,v∈W is a solution to problem (2) and (3), then u,v∈V is a solution to the system of integral equations
(32)ut=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σus−δusvsds,vt=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδusvs−τvsds,for all 0≤t≤T.Conversely, suppose thatu,v∈V is a solution to (32). By assumptions (iii) and (iv), one deduces that u,v∈W. Moreover, by (32), one has u0=u0 and v0=v0. On the other hand, using the first equation in (32), Lemma 6, and Lemma 10, one obtains
(33)CD0α,ψut=CD0α,ψI0α+μ,ψκ−σu−δuvt=CD0α,ψI0α,ψI0μ,ψκ−σu−δuvt=I0μ,ψκ−σu−δuvt.Similarly, using the second equation in (32), one obtains
(34)CD0β,ψvt=CD0β,ψI0β+ν,ψδuv−τvt=CD0β,ψI0β,ψI0ν,ψδuv−τvt=I0ν,ψδuv−τvt.Therefore, one deduces that, ifu,v∈V is a solution to the system of integral equation (32), then u,v∈W is a solution to problem (2) and (3).From the above study, the following result holds.Lemma 13.
The following statements are equivalent:(I)
u,v∈W is a solution to problem (2) and (3).(II)
u,v∈V is a solution to the system of integral equations (32).By the above lemma, the study of problem (2) and (3) in W reduces to the study of the system of integral equation (32) in V.


## 3.2. Uniqueness

In this part, using Grönwall’s lemma, we shall prove that the system of integral equations (32) admits at most one solution u,v∈V.Proposition 14.
Suppose that the assumptions (i)–(iv) are satisfied. Then the system of integral equation ((32)) admits at most one solutionu,v∈V.Proof.
Suppose thatu1,v1,u2,v2∈V are two solutions to (32). Then, for all 0≤t≤T, one has
(35)u2t−u1t=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σu2s−δu2sv2sds−u0−1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σu1s−δu1sv1sds=σΓα+μ∫0tψt−ψsα+μ−1ψ′su1s−u2sds+δΓα+μ∫0tψt−ψsα+μ−1ψ′su1sv1s−u2sv2sds,which yields
(36)∣u2t−u1t∣≤σC1∫0tψ′s∣u1s−u2s∣ds+δC1∫0tψ′s∣u1sv1s−u2sv2s∣ds=σC1∫0tψ′s∣u1s−u2s∣ds+δC1∫0tψ′su1sv1s−v2s+v2su1s−u1sds≤σC1∫0tψ′s∣u1s−u2s∣ds+δC1C2∫0tψ′s∣v1s−v2s∣ds+δC1C3∫0tψ′s∣u1s−u2s∣ds=C1σ+δC3∫0tψ′s∣u1s−u2s∣ds+δC1C2∫0tψ′s∣v1s−v2s∣ds,where
(37)C1=ψT−ψ0α+μ−1Γα+μ,C2=max0≤x≤Tu1x,C3=max0≤x≤Tv2x.
Hence, one deduces that(38)∣u2t−u1t∣≤C4∫0tψ′s∣u1s−u2s∣+∣v1s−v2s∣ds,where
(39)C4=C1maxσ+δC3,δC2.
Similarly, for all0≤t≤T, one has
(40)v2t−v1t=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδu2sv2s−τv2sds−v0−1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδu1sv1s−τv1sds=δΓβ+ν∫0tψt−ψsβ+ν−1ψ′su2sv2s−u1sv1sds+τΓβ+ν∫0tψt−ψsβ+ν−1ψ′sv1s−v2sds,which yields
(41)∣v2t−v1t∣+τC5∫0tψ′s∣v1s−v2s∣ds≤δC5C6∫0tψ′s∣v1s−v2s∣ds+δC5C7∫0tψ′s∣u2s−u1s∣ds+τC5∫0tψ′s∣v1s−v2s∣ds=C5δC6+τ∫0tψ′s∣v1s−v2s∣ds+δC5C7∫0tψ′s∣u2s−u1s∣ds,where
(42)C5=ψT−ψ0β+ν−1Γβ+ν,C6=max0≤x≤Tu2x,C7=max0≤x≤Tv1x.
Therefore, one obtains(43)∣v2t−v1t∣≤C8∫0tψ′s∣u1s−u2s∣+∣v1s−v2s∣ds,where
(44)C8=C5maxτ+δC6,δC7.
Next, combining (38) with (43), it holds that
(45)∣u2t−u1t∣+∣v2t−v1t∣≤C4+C8∫0tψ′s∣u2s−u1s∣+∣v2s−v2s∣ds.
Finally, using Grönwall ‘s lemma (see Lemma12), it holds that u1,v1=u2,v2.


## 3.3. Well-Posedness

We first fix some notations. Let(46)aT=ψT−ψ0α+μΓα+μ+1,bT=ψT−ψ0β+νΓβ+ν+1.We introduce the functionf:0,∞→0,∞ defined by
(47)fr=aTσ+bTτ+aT+bTδr+aTσ−bTτ+aT−bTδr2+4aTbTδ2r21/2,r≥0.We denote by∥·∥∞ the norm in C0,T defined by
(48)∥ξ∥∞=max0≤x≤Tξx,ξ∈C0,T.In addition to the assumptions (i)–(iv), suppose that(v)
aT≤τ/δκ and bT≤1/√σ2+4δκ−σ(49)4τ≤σ2+4δκ−σ(vi)
There exists(50)2τδ≤r≤σ2+4δκ−σ2δsuch that
(51)τδ≤u0≤r2,0≤v0≤r2,fr<2.Theorem 15.
Suppose that the assumptions (i)–(vii) are satisfied. Then, the system of integral equation ((32)) admits one and only one solutionu∗,v∗∈V. Moreover, for allf0,g0∈Vsatisfyingu0≤f0, ∥f0∥∞≤rand∥g0∥∞≤r, the sequencefn,gn⊂Vdefined by(52)fn+1t=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σfns−δfnsgnsds,gn+1t=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδfnsgns−τgnsds,for all 0≤t≤T, converges uniformly to u∗,v∗.Proof.
We first introduce the functional space(53)Vr=u,v∈V:u0≤u,∥u∥∞≤r,∥v∥∞≤r.
LetF:Vr→C0,T×C0,T be the mapping defined by
(54)Fu,vt=F1u,vt,F2u,vt,0≤t≤T,where
(55)F1u,vt=u0+1Γα+μ∫0tψt−ψsα+μ−1ψ′sκ−σus−δusvsds,F2u,vt=v0+1Γβ+ν∫0tψt−ψsβ+ν−1ψ′sδusvs−τvsds.
We shall prove that(56)FVr⊂Vr.
Letu,v∈Vr. Then, for all s∈0,T, one has
(57)κ−σus−δusvs≥κ−σr−δr2≔Pr.
On the other hand, an elementary calculation shows that the polynomial functionPr admits two roots
(58)r1=−σ−σ2+4δκ2δ<0<r2=σ2+4δκ−σ2δ.
Since−δ<0 and (by (vii))
(59)0<r≤σ2+4δκ−σ2δ=r2,one obtains
(60)Pr≥0,which yields
(61)κ−σus−δusvs≥0,0≤s≤T.
Therefore, sinceψ∈Ψ and u0≥0 (see (vii)), one deduces that
(62)F1u,v≥u0≥0.
Moreover, for alls∈0,T, one has
(63)δusvs−τvs=vsδus−τ≥vsδu0−τ.
Since by (vii),u0≥τ/δ, one obtains
(64)δusvs−τvs≥0,0≤s≤T.
Using the above inequality and the fact thatv0≥0 (by (vii)), one deduces that
(65)F2u,v≥0.
On the other hand, by (62), for all 0≤t≤T, one has
(66)∣F1u,vt∣=F1u,vt≤u0+κΓα+μ∫0tψt−ψsα+μ−1ψ′sds=u0+κψt−ψ0α+μΓα+μ+1≤u0+κaT.
SinceaT≤τ/δκ (by (v)) and τ/δ≤u0≤r/2 (by (vii)), one has
(67)u0+κaT≤r2+τδ≤r2+r2=r.
Hence, it follows from (66) and (67) that
(68)∥F1u,v∥∞≤r.
Similarly, by (65), for all 0≤t≤T, one has
(69)∣F2u,vt∣=F2u,vt≤v0+δr2Γβ+ν∫0tψt−ψsβ+ν−1ψ′sds=v0+δr2ψt−ψ0β+νΓβ+ν+1≤v0+δr2bT.
Sincev0≤r/2 (by (vii)), bT≤1/√σ2+4δκ−σ (by (v)) and r≤√σ2+4δκ−σ/2δ (by (vii)), one has
(70)v0+δr2bT≤r2+δr212δr=r2+r2=r.
Therefore, it follows from (69) and (70) that
(71)∥F2u,v∥∞≤r.
Hence, by (62), (65), (68), and (71), one deduces that Fu,v∈Vr. This proves (56).
Consider now the self-mapping,(72)F:Vr→Vr.
By the definition ofF, one observes that, if u,v∈Vr is a fixed point of F; then, u,v∈V is a solution to the system of integral equation (32). In order to prove that F admits a fixed point in Vr, we shall use Perov’s fixed point theorem (see Lemma 11). Namely, we define the vector-valued metric d:Vr×Vr→ℝ2 by
(73)du1,v1,u2,v2=∥u1−u2∥∞∥v1−v2∥∞,u1,v1,u2,v2∈Vr.
Notice thatVr,d is a complete generalized metric space. On the other hand, for all u1,v1,u2,v2∈Vr and 0≤t≤T, one has
(74)∣F1u2,v2t−F1u1,v1t∣≤σΓα+μ∫0tψt−ψsα+μ−1ψ′s∣u1s−u2s∣ds+δΓα+μ∫0tψt−ψsα+μ−1ψ′s∣u1sv1s−u2sv2s∣ds≤σaT∥u1−u2∥∞+δΓα+μ∫0tψt−ψsα+μ−1ψ′su1sv1s−v2s+v2su2s−u1sds≤σaT∥u1−u2∥∞+δaTr∥v2−v1∥∞+∥u2−u1∥∞,which yields
(75)∥F1u2,v2−F1u1,v1∥∞≤aTδr+σ∥u1−u2∥∞+δaTr∥v1−v2∥∞.
Similarly, one has(76)F2u2,v2t−F2u1,v1t≤δΓβ+ν∫0tψt−ψsβ+ν−1ψ′su2sv2s−v1s+v1su2s−u1sds+τΓβ+ν∫0tψt−ψsβ+ν−1ψ′sv1s−v2sds≤δbTr∥v2−v1∥∞+∥u2−u1∥∞+τbT∥v1−v2∥∞,which yields
(77)∥F2u2,v2−F2u1,v1∥∞≤δbTr∥u1−u2∥∞+bTδr+τ∥v1−v2∥∞.
Hence, it follows from (75) and (77) that for all u1,v1,u2,v2∈Vr,
(78)dFu1,v1,Fu2,v2°2Mdu1,v1,u2,v2,where
(79)M=aTδr+σδaTrδbTrbTδr+τ.
On the other hand, the characteristic polynomial ofM is given by
(80)Qλ=aTδr+σ−λbTδr+τ−λ−δ2aTbTr2.
An elementary calculation shows that the polynomial functionQ admits two roots (which are the eigenvalues of M) given by
(81)λ1r=aTσ+δr+bTτ+δr2−12aTσ−bTτ+aT−bTδr2+4aTbTδ2r212,λ2r=fr2.
It can be easily seen that(82)0≤λ1r≤λ2r,which yields
(83)ρM=fr2.
Sincefr<2 (by (vii)), it holds that ρM<1. Therefore, by Lemma 11, the mapping F admits a fixed point u∗,v∗∈Vr⊂V, which is a solution to the system of integral equation (32). Moreover, for all f0,g0∈Vr, the Picard sequence fn,gn defined by fn+1,gn+1=Ffn,gn=F1fn,gn,F2fn,gn converges uniformly to u∗,v∗. On the other hand, by Proposition 14, we know that (32) admits at least one solution in V. Hence, one deduces that u∗,v∗ is the only solution to (32) in V.


## 4. Some Special Cases

Some special cases of Theorem15 are discussed in this section. We first consider the case α+μ=β+ν≔ξ. In this case, one has
(84)aT=bT=ψT−ψ0ξΓξ+1≔cT.Corollary 16.
Suppose that the following conditions are satisfied:(85)A1T>0,ψ∈ΨA20<α,β<1,α+μ=β+ν≔ξ≥1A3κ,σ,δ,τ>0A44τ≤σ2+4δκ−σA5u0=τδ,0≤v0≤u0A6CT<minτδκ,1σ2+4δκ−σ,2σ+5τ+σ−τ2+16τ2Then, the system of integral equation (32) admits one and only one solution u∗,v∗∈V. Moreover, for all f0,g0∈V satisfying u0≤f0, ∥f0∥∞≤2τ/δ and ∥g0∥∞≤2τ/δ, the sequence fn,gn⊂V defined by (52) converges uniformly to u∗,v∗.Proof.
The results follow immediately by takingr=2τ/δ and α+μ=β+ν in Theorem 15.
Consider now the special case of Corollary16 when ψt=t.Corollary 17.
Suppose that the following conditions are satisfied:(86)B10<α,β<1,α+μ=β+ν≔ξ≥1B2κ,σ,δ,τ>0B34τ≤σ2+4δκ−σB4u0=τδ,0≤v0≤u0B40<T<Γξ+1minτδκ,1σ2+4δκ−σ,2σ+5τ+σ−τ2+16τ21/ξ
Then, the system of integral equation (32) admits one and only one solution u∗,v∗∈V. Moreover, for all f0,g0∈V satisfying u0≤f0, ∥f0∥∞≤2τ/δ and ∥g0∥∞≤2τ/δ, the sequence fn,gn⊂V defined by (52) converges uniformly to u∗,v∗.Proof.
In this case, one has(87)CT=TξΓξ+1.
Therefore, the assumptionA6 in Corollary 16 reduces to the assumption B5. Hence, by Corollary 16, the desired results follow.
We take nowψt=t and consider the limit cases α,μ→1−,0+ and β,ν→1−,0+. Notice that in this case, the system of integral equation (32) reduces to
(88)ut=u0+∫0tκ−σus−δusvsds,0≤t≤Tvt=v0+∫0tδusvs−τvsds,0≤t≤T,which corresponds to the integral representation of the system of ordinary differential equations (1) subject to the initial conditions (3).
Takingξ=1 in Corollary 17, one deduces the following results.Corollary 18.
Suppose that the following conditions are satisfied:(89)C1κ,σ,δ,τ>0C24τ≤σ2+4δκ−σC3u0=τδ,0≤v0≤u0C40<T<minτδκ,1σ2+4δκ−σ,2σ+5τ+σ−τ2+16τ2
Then, the system of integral equation (88) admits one and only one solution u∗,v∗∈V. Moreover, for all f0,g0∈V satisfying u0≤f0, ∥f0∥∞≤2τ/δ and ∥g0∥∞≤2τ/δ, the sequence fn,gn⊂V defined by
(90)fn+1t=μ0+∫0tκ−σfns−δfnsgnsds,0≤t≤T,gn+1t=v0+∫0tδfnsgns−τgnsds,0≤t≤T,converges uniformly to u∗,v∗.



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