# Positive Solutions for a Weakly Singular Hadamard-Type Fractional Differential Equation with Changing-Sign Nonlinearity

Authors: Xinguang Zhang; Lixin Yu; Jiqiang Jiang; Yonghong Wu; Yujun Cui
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we focus on the existence of positive solutions for a class of weakly singular Hadamard-type fractional mixed periodic boundary value problems with a changing-sign singular perturbation. By using nonlinear analysis methods combining with some numerical techniques, we further discuss the effect of the perturbed term for the existence of solutions of the problem under the positive, negative, and changing-sign cases. The interesting points are that the nonlinearity can be singular at the second and third variables and be changing-sign.


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## Body

## 1. Introduction

In this paper, we study the existence of positive solutions for the following Hadamard-type fractional differential equation with mixed periodic boundary conditions:(1)−Dtβzt−btDtβzt=ft,zt,−Dtβzt+χt,1<t<e,z1=z′1=z′e=0,Dtβz1=Dtβze,Dtβ+1z1=Dtβ+1ze,where 2<β≤3, Dtβ is the Hadamard fractional derivative of order α,β, b∈Lp1,e and f:1,e×ℝ\0×ℝ0→0,∞ is a continuous function with singularity at the second and third space variables; χ∈Lp1,e,ℝ,p≥1 is a variable sign function.Nonlocal characteristics are the most important property of the fractional differential operator; because of this, the fractional differential equations can describe many viscoelasticities and memory phenomena of natural science. For example, some viscoelastic materials, such as the silicone gel with the property of weak frequency dependency, often involve a complicated strain-stress relationship; if letδt and ρt be the stress and the strain, respectively, then the stress decays after a shear jump is governed by the following fractional order viscoelasticity Kelvin-Voigt equation [1, 2]:
(2)δt=ωταDtαρt+ωτβDtβρt,whereα>β>0, ω,τ are constants and Dtα,Dtβ are Riemann-Liouville fractional derivatives. In practice, the study of the qualitative properties of solutions for the corresponding fractional models such as existence, uniqueness, multiplicity, and stability is necessary to analyze and control the model under consideration [3–21]. In [3], Zhang et al. considered a singular fractional differential equation with signed measure
(3)−Dtαxt=ft,xt,Dtβxt,t∈0,1,Dtβx0=0,Dtβx1=∫01DtβxsdAs,where Dtα,Dtβ are the standard Riemann-Liouville derivatives, ∫01xsdAs is denoted by a Riemann-Stieltjes integral and 0<β≤1<α≤2,α−β>1,A is a function of bounded variation and dA can be a signed measure; the nonlinearity ft,x,y may be singular at both t=0,1 and x=y=0. By using the spectral analysis of the relevant linear operator and Gelfand’s formula combining the calculation of a fixed point index of the nonlinear operator, some sufficient conditions for the existence of positive solutions were established. Recently, by using the fixed point index theory, Wang [10] established the existence and multiplicity of positive solutions for a class of singular fractional differential equations with nonlocal boundary value conditions, where the nonlinearity may be singular at some time and space variables.In the recent years, to improve and develop the fractional calculus, there are several kinds of fractional derivatives and integral operators with different kernels such as Hadamard, Erdelyi-Kober, Caputo-Fabrizio derivatives, Hilfer derivatives, and integrals to be given to enrich the application of the fractional calculus such as the Rubella disease and human liver model [22–25]. In particular, the Hadamard derivative is a nonlocal fractional derivative with singular logarithmic kernel. So the study of Hadamard fractional differential equations is relatively difficult; see [26–32].In this paper, we are interested in the existence of solutions for the Hadamard-type fractional differential equation (1) which involves a singular perturbed term χ; we will further discuss the effect of the perturbed term for the existence of solutions when χ is positive, negative, and changing-sign. Our main tools rely on nonlinear analysis methods as well as some numerical techniques. Thus, in order to make our work be more self-contained, a brief overview for these methods and techniques should be necessary. In recent work [33–41], some fixed point theorems were employed to study the qualitative properties of solutions for various types of differential equations. For obtaining numerical and analytical results, many authors [42–59] also developed iterative techniques to solve some nonlinear problems with practical applications. In addition, variational methods [60–78] and upper and lower solution methods [2, 25, 79, 80] also offered wonderful tools for dealing with various nonlinear ordinary and partial differential equations arising from natural science fields. These analysis and techniques not only improved and perfected the relative theoretical framework of differential equations but also gave some new understand for the corresponding natural phenomena.Our work has some new features. Firstly, the equation contains a Hadamard-type fractional derivative which has a singular logarithmic kernel. Secondly, the nonlinearity involves a perturbed term which can be positive, negative, or changing-sign. Thirdly, the nonlinearity is allowed to have weakly singular at space variables, which is a class of interesting natural phenomena. In the end, the effect of the perturbed term for the existence of solutions of the equation is discussed, and the criteria on the existence of positive solutions are established for all cases of the perturbed term including positive, negative, and changing-sign cases. The rest of this paper is organized as follows. In Section2, we firstly introduce the concept of the Hadamard fractional integral and differential operators and then give the logarithmic kernel and Green function of periodic boundary value problem and their properties. Our main results are summarized in Section 3 which includes three theorems for three different situations. An example is given in Section 4.


## 2. Basic Definitions and Preliminaries

Before starting our main results, we firstly recall the definition of the Hadamard-type fractional integrals and derivatives; for detail, see [81].Letα∈ℂ,Reα>0, n=Reα, and a,b be a finite or infinite interval of ℝ+. The α-order left Hadamard fractional integral is defined by
(4)Iaαxt=1Γα∫atlntsα−1xssds,t∈a,b,and the α left Hadamard fractional derivative is defined by
(5)Dtαxt=1Γn−αtddtn∫atlntsn−α−1xssds,t∈a,b.In what follows, we firstly consider the linear auxiliary problem(6)−Dtβzt=xt,1<t<e,z1=z′1=z′e=0.It follows from [26] that the problem (6) has a unique solution:
(7)zt=∫1eHt,sxsdss,where
(8)Ht,s=1Γβlntβ−11−lnsβ−2−lnt−lnsβ−1,1≤s≤t≤e,lntβ−11−lnsβ−2,1≤t≤s≤e,is Green’s function of (6). Now, let xt=−Dtβzt, then the Hadamard-type fractional differential equation (1) reduces to the following second order integrodifferential equation:
(9)x′′t+btxt=ft,∫1eHt,sxsdss,xt+χt,t∈1,e,x1=xe,x′1=x′e.In order to obtain the solution of the second order integrodifferential equation (9), let us consider the linear periodic boundary value problem
(10)x′′+btx=0,x1=xe,x′1=x′e.Clearly, by Fredholm’s alternative, the nonhomogeneous equationx′′+btu=at has a unique e−1-periodic solution:
(11)xt=∫1eGt,sasds,where Gt,s is the Green function of line equation (10) subject to periodic boundary conditions x1=xe,x′1=x′e.Thus, it follows from (10) that the equation (9) is equivalent to the following integral equation:
(12)xt=∫1eGt,sfs,∫1eHs,τxτdττ,xs+χsds.As a result, in order to obtain the solution of equation (1), it is sufficient to find the fixed point of the following operator:
(13)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xs+χsds.Lemma 1 (see [26]).
Green’s functionH has the following properties:
(i)
H∈C1,e×1,e,ℝ+(ii)
For allt,s∈1,e, the following inequality holds:(14)1Γβlntβ−1lns1−lnsβ−2≤Ht,s≤1Γβlns1−lnsβ−2In order to obtain the properties of the Green functionG, define the best Sobolev constants as
(15)Yζ=2πζe−11+1/ζ22+ζ1−2/ζΓ1/ζΓ1/2+1/ζ2,if1≤ζ<∞,4e−1,ifζ=∞,where Γ is the gamma function. Let b∈Lp1,e and denote b≻0 means that bt≥0 for all t∈1,e and bt>0 for t in a subset of positive measure. Define
(16)p∗=pp−1,if1≤p<∞,p∗=1,ifp=∞.Then, the following lemma is a direct consequence of Theorem 2.1 and Corollary 2.3 in [82]:Lemma 2.
Let1≤p≤∞, assume b∈Lp0,T and bt≻0. If bp≤Y2p∗, then Gt,s>0 for all t,s∈1,e×1,e.


## 3. Main Results

In this section, we firstly give our work spaceE=C1,e which equips the maximum norm x=max1≤t≤ext. Let
(17)P=x∈E:xt≥0,t∈1,e,then E,· is a real Banach space and P is a normal cone of E with normal constant 1. For convenience, we denote a set of functions as
(18)Λ=b∈Lp1,e:b≻0,bp≤Y2p∗,for1≤p≤∞.Now, let us list the hypotheses to be used in the rest the paper.(F0)b∈Λ(F1) There exist a constantλ∈0,1 and two functions h1,h2∈P with h1,h2≡0 on any subinterval of 1,e, such that
(19)h1tu+v−λ≤ft,u,v≤h2tu+v−λ,u,v∈0,∞×0,∞,t∈1,e.Supposeμt is the unique solution of the following equation:
(20)x′′+btx=χt,t∈1,e,x1=xe,x′1=x′e,then from (10), μt can be written as
(21)μt=∫1eGt,sχsds.Similar to (21), we denote κ1t,κ2t as
(22)κ1t=1+1β−1Γβ−λ∫1eGt,sh1sds,κ2t=∫1eGt,sh2sds.Let(23)μ∗=inf1≤t≤eμt,μ∗=sup1≤t≤eμt,κ1∗=min1≤t≤eκ1t,κ1∗=max1≤t≤eκ1t,κ2∗=min1≤t≤eκ2t,κ2∗=max1≤t≤eκ2t.By Lemma2 and (F0)–(F1), we know that κi∗,κi∗>0,i=1,2.Now we shall divideχ into three cases to discuss its influence for the solution of equation (1).


### 3.1. Positive Caseχt≥0

In this case, we have the following result:Theorem 3.
Assume that (F0)–(F1) hold. Ifμ∗≥0, then the Hadamard-type fractional differential equation (1) has at least one positive solution.Proof.
Letl and L be fixed positive constants; we introduce a closed convex set of cone P:
(24)B=x∈P:l≤xt≤L,t∈1,e.For anyx∈B, it follows from Lemma 1 that
(25)llnsβ−1β−1Γβ≤∫1eHs,τxτdττ≤Lβ−1Γβ.With the help of (25), for any s∈1,e, from (19), one gets
(26)h1sLλ1+1/β−1Γβλ≤fs,∫1eHs,τxτdττ,xs≤h2slλ1+lnsβ−1/β−1Γβλ≤h2slλ.It follows from (26) and (F1) that
(27)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≥1Lλ1+1/β−1Γβλ∫1eGt,sh1sds≥κ1∗Lλ,(28)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≤1lλ∫1eGt,sh2sds+μ∗≤κ2∗lλ+μ∗.Thus, (27), (28), Lemma 1, (F1), and the Arzela-Ascoli theorem guarantee that the operator T:B→P is completely continuous.In the following, we shall choose a suitableL>0 such that T maps the closed convex set B into itself. To do this, we only need to choose 0<l<L such that
(29)κ1∗Lλ≥l,κ2∗lλ+μ∗≤L.In fact, takel=1/L, it follows from κ1∗>0 and 0<λ<1 that there exists a sufficiently large L>1 such that
(30)κ1∗L1−λ≥1,Lλκ2∗+μ∗≤L,which imply that (29) holds and l<L.Thus, from Schauder’s fixed point theorem,T has a fixed point x∗∈B, and hence, the Hadamard-type fractional differential equation (1) has at least one positive solution:
(31)z∗t=∫1eHt,sx∗sdss.


### 3.2. Negative Caseχt≤0

For this case, we have the following existence result:then the Hadamard-type fractional differential equation (1) has at least one positive solution.Theorem 4.
Assume that (F0)–(F1) hold. Ifμ∗≤0, and
(32)μ∗≥κ1∗λ2κ2∗λ1/1−λ21−1λ2,Proof.
Firstly, from (26) and (F1), we have
(33)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≥1Lλ1+1/β−1Γβλ∫1eGt,sh1sds+μ∗≥κ1∗Lλ+μ∗,(34)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≤1lλ∫1eGt,sh2sds≤κ2∗lλ.Thus, by (33) and (34), in order to guarantee T:B→B, it is sufficient to choose 0<l<L such that
(35)κ1∗Lλ+μ∗≥l,κ2∗lλ≤L.Let us fixL=κ2∗/lλ; to ensure (35) holds, we only need to choose some l>0 such that
(36)0<l<κ2∗1/1+λ,μ∗≥l−κ1∗κ2∗λlλ2=gl.Obviously, the functiongl in 0,∞ has a minimum at
(37)l~=λ2κ1∗κ2∗λ1/1−λ2.Takingl=l~, since κ1∗≤κ2∗,0<λ2<1, we have
(38)0<l=l~=λ2κ1∗κ2∗λ1/1−λ2<κ2∗κ2∗λ1/1−λ2=κ2∗1/1+λ,and from (32), we also have
(39)μ∗≥κ1∗λ2κ2∗λ1/1−λ21−1λ2=gl~≥gl,which implies that (36) holds if l=l~,L=κ2∗/l~λ. Consequently, we have T:B→B.Thus, from Schauder’s fixed point theorem,T has a fixed point x∗∈B, and hence, the Hadamard-type fractional differential equation (1) has at least one positive solution:
(40)z∗t=∫1eHt,sx∗sdss.Remark 5.
Note that the right side of inequality (32) is negative because of the weak force condition 0<λ<1, so if μ∗=0, the inequality (32) is always valid. Thus, if μ∗=0, we can omit the assumption (32).


### 3.3. Changing-Sign Case forχt

In order to establish the existence result under the case whereχt is changing-sign, we need the following lemma.Lemma 6.
If0<λ<1 and x∈0,∞, the equation
(41)x1−λ2κ2∗+μ∗xλ1+λ=λ2κ2∗κ1∗,has a unique solution l~ which satisfies
(42)0<l~<ε≔λ2κ2∗κ1∗1/2+2λ.Proof.
Let(43)φx=λ2κ2∗κ1∗−x1−λ2κ2∗+μ∗xλ1+λ,then
(44)φ0=limx→0φx=λ2κ2∗κ1∗>0.
On the other hand, since0<λ<1,κ2∗≥κ1∗, one has
(45)0<ε≤κ2∗1/1+λ.
Therefore,(46)φε=ε2+2λ−ε1−λ2κ2∗+μ∗ελ1+λ=ε1−λ2ε1+λ2−κ2∗+μ∗ελ1+λ<ε1−λ2ε1+λ2−κ2∗1+λ<0.
Moreover,(47)φ′x=−1−λ2x−λ2κ2∗+μ∗xλ1+λ−μ∗λ1+λx1−λ2κ2∗+μ∗xλλxλ−1<0,x∈0,∞.
Thus, it follows from (44) to (47) that equation (41) has a unique positive solution l~∈0,ε.Theorem 7.
Assume that (F0)–(F1) hold. Letl~ be the unique solution of equation (41), if μ∗≤0,μ∗≥0 and satisfy
(48)μ∗≥l~−κ1∗l~2λκ2∗+μ∗l~λλ,then the Hadamard-type fractional differential equation (1) has at least one positive solution.Proof.
In this case, following the strategy and notations of (33) and (34), we have
(49)Txt≥1Lλ1+1/β−1Γβλ∫1eGt,sh1sds+μ∗≥κ1∗Lλ+μ∗,Txt≤1lλ∫1eGt,sh2sds+μ∗≤κ2∗lλ+μ∗.
Thus, to ensureT:B→B, it is sufficient to choose 0<l<L such that
(50)κ1∗Lλ+μ∗≥l,κ2∗lλ+μ∗≤L.
To do this, we fixL=κ2∗/lλ+μ∗. Clearly, if l satisfies
(51)0<l<L=κ2∗lλ+μ∗,(52)μ∗≥l−κ1∗l2λκ2∗+μ∗lλλ,then the inequalities of (50) hold.
Let(53)ψx=x−κ1∗x2λκ2∗+μ∗xλλ,notice that
(54)ψ′x=1−κ1∗λ2xλ2−1κ2∗+λ2μ∗xλλ−λ∗xλ2κ2∗+μ∗xλλ−1xλ−1κ2∗+μ∗xλ2λ=1−κ1∗λ2xλ2−1κ2∗+μ∗xλλ1−μ∗xλκ2∗+μ∗xλ=1−κ1∗κ2∗λ2xλ2−1κ2∗+μ∗xλλ+1,and ψ′0=−∞,ψ′∞=1; consequently, there exists l~ such that
(55)ψ′l~=1−κ1∗κ2∗λ2l~λ2−1κ2∗+μ∗l~λλ+1=0,which implies that l~ solves the equation
(56)x1−λ2κ2∗+μ∗xλ1+λ=λ2κ2∗κ1∗,
i.e.,l~ is the unique solution of equation (41).
On the other hand, since(57)ψ′′x=1−λ2λ2κ1∗κ2∗xλ2−2κ2∗+μ∗xλλ+1+1+λλ3μ∗κ1∗κ2∗xλ2+λ−2κ2∗+μ∗xλλ+2>0,the function ψx gets the minimum at l~, i.e.,
(58)ψl~=minx∈0,∞ψx.
Takingl=l~, then the assumption (48) implies that (52) holds.
Thus, we only need to prove that the inequalityL>l is also satisfied. Notice that 0<l=l~<ε≕λ2κ2∗κ1∗1/2+2λ solves (41), we have
(59)l1−λ2κ2∗+μ∗lλ1+λ=λ2κ2∗κ1∗.
It follows fromL=κ2∗/lλ+μ∗ that
(60)l1+λL1+λ=λ2κ2∗κ1∗,and then
(61)L=1lλ2κ2∗κ1∗1/1+λ.
Thus, it follows from(62)l<λ2κ2∗κ1∗1/2+2λ,that
(63)L=1lλ2κ2∗κ1∗1/1+λ>λ2κ2∗κ1∗1/2+2λ>l.
That is, under the assumptions of Theorem7, (51) and (52) all hold. Thus, according to Schauder's fixed point theorem, the Hadamard-type fractional differential equation (1) has at least one positive solution
(64)z∗t=∫1eHt,sx∗sdss.Remark 8.
In this case, the nonlinearityft,u,v of equation (1) may be singular at u=0 and v=0; moreover, χt can be changing-sign Lp1,e function, which is allowed to be singular at some t∈1,e.


## 3.1. Positive Caseχt≥0

In this case, we have the following result:Theorem 3.
Assume that (F0)–(F1) hold. Ifμ∗≥0, then the Hadamard-type fractional differential equation (1) has at least one positive solution.Proof.
Letl and L be fixed positive constants; we introduce a closed convex set of cone P:
(24)B=x∈P:l≤xt≤L,t∈1,e.For anyx∈B, it follows from Lemma 1 that
(25)llnsβ−1β−1Γβ≤∫1eHs,τxτdττ≤Lβ−1Γβ.With the help of (25), for any s∈1,e, from (19), one gets
(26)h1sLλ1+1/β−1Γβλ≤fs,∫1eHs,τxτdττ,xs≤h2slλ1+lnsβ−1/β−1Γβλ≤h2slλ.It follows from (26) and (F1) that
(27)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≥1Lλ1+1/β−1Γβλ∫1eGt,sh1sds≥κ1∗Lλ,(28)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≤1lλ∫1eGt,sh2sds+μ∗≤κ2∗lλ+μ∗.Thus, (27), (28), Lemma 1, (F1), and the Arzela-Ascoli theorem guarantee that the operator T:B→P is completely continuous.In the following, we shall choose a suitableL>0 such that T maps the closed convex set B into itself. To do this, we only need to choose 0<l<L such that
(29)κ1∗Lλ≥l,κ2∗lλ+μ∗≤L.In fact, takel=1/L, it follows from κ1∗>0 and 0<λ<1 that there exists a sufficiently large L>1 such that
(30)κ1∗L1−λ≥1,Lλκ2∗+μ∗≤L,which imply that (29) holds and l<L.Thus, from Schauder’s fixed point theorem,T has a fixed point x∗∈B, and hence, the Hadamard-type fractional differential equation (1) has at least one positive solution:
(31)z∗t=∫1eHt,sx∗sdss.


## 3.2. Negative Caseχt≤0

For this case, we have the following existence result:then the Hadamard-type fractional differential equation (1) has at least one positive solution.Theorem 4.
Assume that (F0)–(F1) hold. Ifμ∗≤0, and
(32)μ∗≥κ1∗λ2κ2∗λ1/1−λ21−1λ2,Proof.
Firstly, from (26) and (F1), we have
(33)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≥1Lλ1+1/β−1Γβλ∫1eGt,sh1sds+μ∗≥κ1∗Lλ+μ∗,(34)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsds+μt≤1lλ∫1eGt,sh2sds≤κ2∗lλ.Thus, by (33) and (34), in order to guarantee T:B→B, it is sufficient to choose 0<l<L such that
(35)κ1∗Lλ+μ∗≥l,κ2∗lλ≤L.Let us fixL=κ2∗/lλ; to ensure (35) holds, we only need to choose some l>0 such that
(36)0<l<κ2∗1/1+λ,μ∗≥l−κ1∗κ2∗λlλ2=gl.Obviously, the functiongl in 0,∞ has a minimum at
(37)l~=λ2κ1∗κ2∗λ1/1−λ2.Takingl=l~, since κ1∗≤κ2∗,0<λ2<1, we have
(38)0<l=l~=λ2κ1∗κ2∗λ1/1−λ2<κ2∗κ2∗λ1/1−λ2=κ2∗1/1+λ,and from (32), we also have
(39)μ∗≥κ1∗λ2κ2∗λ1/1−λ21−1λ2=gl~≥gl,which implies that (36) holds if l=l~,L=κ2∗/l~λ. Consequently, we have T:B→B.Thus, from Schauder’s fixed point theorem,T has a fixed point x∗∈B, and hence, the Hadamard-type fractional differential equation (1) has at least one positive solution:
(40)z∗t=∫1eHt,sx∗sdss.Remark 5.
Note that the right side of inequality (32) is negative because of the weak force condition 0<λ<1, so if μ∗=0, the inequality (32) is always valid. Thus, if μ∗=0, we can omit the assumption (32).


## 3.3. Changing-Sign Case forχt

In order to establish the existence result under the case whereχt is changing-sign, we need the following lemma.Lemma 6.
If0<λ<1 and x∈0,∞, the equation
(41)x1−λ2κ2∗+μ∗xλ1+λ=λ2κ2∗κ1∗,has a unique solution l~ which satisfies
(42)0<l~<ε≔λ2κ2∗κ1∗1/2+2λ.Proof.
Let(43)φx=λ2κ2∗κ1∗−x1−λ2κ2∗+μ∗xλ1+λ,then
(44)φ0=limx→0φx=λ2κ2∗κ1∗>0.
On the other hand, since0<λ<1,κ2∗≥κ1∗, one has
(45)0<ε≤κ2∗1/1+λ.
Therefore,(46)φε=ε2+2λ−ε1−λ2κ2∗+μ∗ελ1+λ=ε1−λ2ε1+λ2−κ2∗+μ∗ελ1+λ<ε1−λ2ε1+λ2−κ2∗1+λ<0.
Moreover,(47)φ′x=−1−λ2x−λ2κ2∗+μ∗xλ1+λ−μ∗λ1+λx1−λ2κ2∗+μ∗xλλxλ−1<0,x∈0,∞.
Thus, it follows from (44) to (47) that equation (41) has a unique positive solution l~∈0,ε.Theorem 7.
Assume that (F0)–(F1) hold. Letl~ be the unique solution of equation (41), if μ∗≤0,μ∗≥0 and satisfy
(48)μ∗≥l~−κ1∗l~2λκ2∗+μ∗l~λλ,then the Hadamard-type fractional differential equation (1) has at least one positive solution.Proof.
In this case, following the strategy and notations of (33) and (34), we have
(49)Txt≥1Lλ1+1/β−1Γβλ∫1eGt,sh1sds+μ∗≥κ1∗Lλ+μ∗,Txt≤1lλ∫1eGt,sh2sds+μ∗≤κ2∗lλ+μ∗.
Thus, to ensureT:B→B, it is sufficient to choose 0<l<L such that
(50)κ1∗Lλ+μ∗≥l,κ2∗lλ+μ∗≤L.
To do this, we fixL=κ2∗/lλ+μ∗. Clearly, if l satisfies
(51)0<l<L=κ2∗lλ+μ∗,(52)μ∗≥l−κ1∗l2λκ2∗+μ∗lλλ,then the inequalities of (50) hold.
Let(53)ψx=x−κ1∗x2λκ2∗+μ∗xλλ,notice that
(54)ψ′x=1−κ1∗λ2xλ2−1κ2∗+λ2μ∗xλλ−λ∗xλ2κ2∗+μ∗xλλ−1xλ−1κ2∗+μ∗xλ2λ=1−κ1∗λ2xλ2−1κ2∗+μ∗xλλ1−μ∗xλκ2∗+μ∗xλ=1−κ1∗κ2∗λ2xλ2−1κ2∗+μ∗xλλ+1,and ψ′0=−∞,ψ′∞=1; consequently, there exists l~ such that
(55)ψ′l~=1−κ1∗κ2∗λ2l~λ2−1κ2∗+μ∗l~λλ+1=0,which implies that l~ solves the equation
(56)x1−λ2κ2∗+μ∗xλ1+λ=λ2κ2∗κ1∗,
i.e.,l~ is the unique solution of equation (41).
On the other hand, since(57)ψ′′x=1−λ2λ2κ1∗κ2∗xλ2−2κ2∗+μ∗xλλ+1+1+λλ3μ∗κ1∗κ2∗xλ2+λ−2κ2∗+μ∗xλλ+2>0,the function ψx gets the minimum at l~, i.e.,
(58)ψl~=minx∈0,∞ψx.
Takingl=l~, then the assumption (48) implies that (52) holds.
Thus, we only need to prove that the inequalityL>l is also satisfied. Notice that 0<l=l~<ε≕λ2κ2∗κ1∗1/2+2λ solves (41), we have
(59)l1−λ2κ2∗+μ∗lλ1+λ=λ2κ2∗κ1∗.
It follows fromL=κ2∗/lλ+μ∗ that
(60)l1+λL1+λ=λ2κ2∗κ1∗,and then
(61)L=1lλ2κ2∗κ1∗1/1+λ.
Thus, it follows from(62)l<λ2κ2∗κ1∗1/2+2λ,that
(63)L=1lλ2κ2∗κ1∗1/1+λ>λ2κ2∗κ1∗1/2+2λ>l.
That is, under the assumptions of Theorem7, (51) and (52) all hold. Thus, according to Schauder's fixed point theorem, the Hadamard-type fractional differential equation (1) has at least one positive solution
(64)z∗t=∫1eHt,sx∗sdss.Remark 8.
In this case, the nonlinearityft,u,v of equation (1) may be singular at u=0 and v=0; moreover, χt can be changing-sign Lp1,e function, which is allowed to be singular at some t∈1,e.


## 4. Examples

In this section, we give some examples with positive, negative, and changing-sign perturbations to demonstrate the application of our main results.We consider the following Hadamard-type fractional differential equation with different perturbations:(65)Dt9/2zt+14Dt5/2zt=392zt+∣Dt5/2zt∣1/2+χt,1<t<e,z1=z′1=z′e=0,Dt5/2z1=Dt5/2ze,Dt9/2z1=Dt9/2ze.Example 9.
Consider the case of equation (65) with positive force term χt=1.Conclusion. The Hadamard-type fractional differential equation (65) has at least one positive solution.Proof.
Takeλ=1/2, h1t=19,h2t=20 and
(66)ft,u,v=392u+v1/2,u,v∈0,∞×0,∞,t∈1,e,then
(67)19u+v1/2≤ft,u,v≤20u+v1/2,u,v∈0,∞×0,∞,t∈1,e.
It is easy to check that the Green function of the equationx′′t+1/4xt=at with periodic boundary conditions x1=xe,x′1=x′e is
(68)Gt,s=sine−1−t+s2sint−s2,1≤s≤t≤e,sine−1−s+t2sins−t2,1≤t≤s≤e.
Thus,Gt,s>0,t,s∈1,e×1,e, and (F1) holds.
Now, let us computeμ∗,μ∗ and κ1∗,κ2∗. Note that
(69)μt=∫1eGt,sχsds=sin12e−12+12cos12e−12−12cos12e−12e=0.1935.
By simple computations, one has(70)μ∗=μ∗=0.1935,κ1∗=κ1∗=3.0004,κ2∗=κ2∗=3.8700.
Now, takeL=15.3615 and let l=1/L=0.0651. Clearly, L satisfies the inequalities
(71)3.0004L1/2≥1,3.87L1/2+0.1935≤L,thus, according to Theorem 3, the Hadamard-type fractional differential equation (65) has at least one positive solution.Example 10.
We consider the case of Hadamard-type fractional differential equation (65) with negative force term χt=−3.Conclusion. The Hadamard-type fractional differential equation (65) has at least one positive solution.Proof.
We still takeλ=1/2, h1t=19,h2t=20, then as Example 9, we have Gt,s>0,t,s∈1,e×1,e, and (F1) holds.
Now, let us computeμ∗,μ∗ and κ1∗,κ2∗. Note that
(72)μt=∫1eGt,sχsds=−3sin12e−12+12cos12e−12−12cos12e−12e=−0.5805.
By simple computations, one has(73)μ∗=μ∗=−0.5805,κ1∗=κ1∗=3.0004,κ2∗=κ2∗=3.8700.
Soμ∗=−0.5805≤0, and
(74)μ∗=−0.5805≥κ1∗λ2κ2∗λ1/1−λ21−1λ2=−0.8295,thus, according to Theorem 4, the Hadamard-type fractional differential equation (65) has at least one positive solution.Example 11.
Consider the case of Hadamard-type fractional differential equation (65) with changing-sign force term
(75)χt=−1,t∈1,2,10000729,t∈2,e.Conclusion. The Hadamard-type fractional differential equation (65) has at least one positive solution.Proof.
It follows from Example9 that Gt,s>0,t,s∈1,e×1,e, and (F1) holds.
Now, let us computeμ∗,μ∗ and κ1∗,κ2∗. Note that
(76)μt=∫1eGt,sχsds=−sine−12−12cose−12−sine−22cos32−t,1≤t<2,10000729−12ecose−12+cose−12+sine−12−sin12cose+22−t,2≤t≤e,κ1t=1+1β−1Γβ−λ∫1eGt,sh1sds=15.5059∫1eGt,sds,κ2t=∫1eGt,sh2sds=20∫1eGt,sds.
By simple computations, we have(77)μ∗=−0.0798,μ∗=1,κ1∗=κ1∗=3.0004,κ2∗=κ2∗=3.8700.
Consequently, equation (41) reduces to
(78)x3/43.8700+x1/23/2=2.9029,which has a unique solution l~=0.2200. Thus, one has
(79)μ∗=−0.0798≥l~−κ1∗l~2λκ2∗+μ∗l~λλ=−0.0969,which implies that (48) holds.
According to Theorem7, the Hadamard-type fractional differential equation (65) has at least one positive solution.


## 5. Conclusion

The force effect from outside in many complex processes often leads to the system governed equations possessing perturbations. In this work, we establish several criteria on the existence of positive solutions for a Hadamard-type fractional differential equation with singularity in all of the cases where the perturbed term is positive, negative, and changing-sign. The main advantage of our assumption is that it provides an effective bound for the perturbed termχ. This classification is valid and reasonable, and it is also easier to get the solution of the target equation by simple calculation.



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