# Solutions for a Singular Hadamard-Type Fractional Differential Equation by the Spectral Construct Analysis

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 consider the existence of positive solutions for a Hadamard-type fractional differential equation with singular nonlinearity. By using the spectral construct analysis for the corresponding linear operator and calculating the fixed point index of the nonlinear operator, the criteria of the existence of positive solutions for equation considered are established. The interesting point is that the nonlinear term possesses singularity at the time and space variables.


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## Body

## 1. Introduction

In this paper, we focus on the existence of positive solutions for the following singular Hadamard-type fractional differential equation:(1)DtαDtβzt=ft,zt,−Dtβzt,1<t<e,z1=σz1=σze=0,Dtβz1=σDtβz1=σDtβze=0,where 2<α,β≤3; σ is a differential operator denoted by td/dt, that is, σzt=td/dtzt; Dtα,Dtβ are the Hadamard fractional derivative of order α,β; and f∈1,e×0,+∞×0,+∞,0,+∞ is a continuous function with singularity at t=0,1 and u=0,v=0.Singularity refers to a point or a domain where the given mathematical object is not defined or not “well-behaved.” Near a singular point or zone, a minor change of the variable will lead to major changes of the property of the target object. Many physical phenomena in natural sciences and engineering often exhibit some singular behaviour. For example, Fisk [1] found that in certain materials the quantum fluctuations at absolute zero may push a system into a different phase or state, as result, the process loses its continuity, and then, the singular behaviour happens near the quantum critical points. In fluid mechanics, when a fluid is subjected to a severe impact to form a fracture, singular points or singular domains also follow the fracture. Normally, at singular points and domains, the extreme behaviour such as blow-up phenomena [2, 3], impulsive influence [4–9], and chaotic system [10–13], often leads to some difficulties for people in understanding and predicting the corresponding natural problems. Hence, the study of singularity for complex systems governed by differential equations [14–27] is important and interesting in deepening the understanding of the internal laws of dynamic system.On the other hand, since the fractional differential operator is nonlocal, some often use it to describe viscoelastic behaviour and memory phenomena in various natural science fields such as the silicone gel with the property of weak frequency dependency [28, 29] and advection dispersion in anomalous diffusion [30–34]. In most cases, some are interested in the qualitative properties of solutions for the corresponding fractional equations; for the detail, see [35–65]. In particular, in order to obtain the qualitative properties of solutions, many nonlinear analysis methods, such as fixed point theorems [66–71], iterative techniques [72–80],variational methods [81–98], and upper and lower solution methods [29, 44], have been developed and employed to study the qualitative properties and numerical results of solutions for various types of differential equations. For example, by using the fixed point index theory, Wang [69] established the existence and multiplicity of positive solutions for the following nonlocal singular fractional differential equation:
(2)D0+αut+ft,ut=0,0<t<1,n−1<α≤n,u0=u′0==un−20=0,D0+βu1=∫0ηatD0+γutdVt,where D0+α denotes the standard Riemann-Liouville fractional derivative, n≥3,0<β<1,0≤γ<α−1,η∈0,1,ft,x may be singular at t=0,1, and x=0,at∈L10,1∩C0,1 is nonnegative. In recent years, to enrich the theory of fractional calculus, the fractional derivative and integral were extended to many different forms such as Hadamard, Erdelyi-Kober, Hilfer derivatives, and integrals. In particular, it is more difficult to obtain the qualitative properties of solutions for Hadamard-type fractional differential equations since Hadamard derivatives possess a singular logarithmic kernel [99–106].In this paper, we focus on the existence of positive solutions for the Hadamard-type fractional differential equation (1) with singularity in space variables. Our work has some new contributions. Firstly, the equation contains a Hadamard-type fractional derivative which has a singular logarithmic kernel. Secondly, the nonlinearity can have strong singularity in time and space variables. Thirdly, a new limit condition of integral type is introduced to overcome the difficulty of singularity. The rest of this paper is organized as follows. In Section 2, we firstly introduce the concept of Hadamard fractional integral and differential operators and then give the logarithmic kernel and Green function of the boundary value problem and their properties. Our main results are summarized in Section 3.


## 2. Preliminaries and Lemmas

Before the main results, we firstly recall the definition of the Hadamard-type fractional integrals and derivatives; for detail, see [107].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
(3)Iaαxt=1Γα∫atlntsα−1xssds,t∈a,b,and the α left Hadamard fractional derivative is defined by
(4)Dtαxt=1Γn−αtddtn∫atlntsn−α−1xssds,t∈a,b.In what follows, we consider the following linear auxiliary problem:(5)−Dtβzt=xt,1<t<e,z1=σz1=σze=0.It follows from [99] that problem (5) has a unique solution
(6)zt=∫1eHt,sxsdss,where
(7)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 equation (5). Now, let xt=−Dtβzt, then the Hadamard-type fractional differential equation (1) reduces to the following convenient form:
(8)−Dtαxt=ft,∫1eHt,sxsdss,xt,t∈1,e,x1=σz1=σxe=0.It follows from (6) that equation (8) is equivalent to the following integral equation:
(9)xt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss,where
(10)Gt,s=1Γαlntα−11−lnsα−2−lnt−lnsα−1,1≤s≤t≤e,lntα−11−lnsα−2,1≤t≤s≤e.As a result, in order to find the positive solutions of equation (1), it is sufficient to search the fixed point of the following operator:
(11)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss.Lemma 1 (see [99]).
Letψit=lnt1−lnti−2,i=α,β,t∈1,e. Then, Green’s functions H,G has the following properties:
(12)H,G∈C1,e×1,e,ℝ+.(i)
For allt,s∈1,e, the following inequalities hold:(13)1Γβlntβ−1ψβs≤Ht,s≤1Γβψβs,1Γαlntα−1ψαs≤Gt,s≤1Γαψαs
LetQ be a cone of Banach space E, for any 0<r<R<∞, define
(14)Q¯r=x∈Q:x<r,∂Qr=x∈Q:x=r,Q¯R\Qr=x∈Q:r≤x≤R.
Now, we state the following lemmas which will be used in the rest of the paper.Lemma 2 (see [108]).
AssumeT:Q¯r→Q is a completely continuous operator.
(i)
If there existsx0∈Q\θ such that(15)x−Tx≠μx0,x∈∂Qr,μ≥0,then, the fixed point index iT,Qr,Q=0(ii)
If(16)Tx≠μx,x∈∂Qr,μ≥1,then, the fixed point index iT,Qr,Q=1Lemma 3 (Krein-Rutmann, see [108]).
LetL:E→E be a continuous linear operator, P be a total cone, and LP⊂P. If there exist ψ∈E\−P and a positive constant c such that cLψ≥ψ, then the spectral radius ρL≠0 and has a positive eigenfunction corresponding to its first eigenvalue λ=ρL−1.Lemma 4 (Gelfand’s formula, see [108]).
For a bounded linear operatorL and the operator norm ·, the spectral radius of Ln satisfies
(17)ρL=limn→+∞Ln1/n.In this paper, we use the following assumption:(B1)f∈C0,1×0,+∞×0,+∞,0,+∞ and for any 0<r<R<+∞,
(18)limn→+∞supu,v∈Q¯RQr∫Ωnψαsfs,us,vsds=0,where Ωn=1,1+1/n∪e−1/n,e.Now, letE=C1,e, then E is a Banach space equipped with the norm x=maxt∈1,e∣xt∣. Let P=x∈E:xt≥0,t∈1,e and
(19)Q=x∈P:xt≥lntα−1x,t∈1,e,then, Q is a cone in the Banach space E and Q¯RQr⊂Q⊂P. Let us define a nonlinear operator T:Q¯R\Qr→P and a linear operator L:E→E:
(20)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss,t∈1,e,Lxt=∫1eGt,sxsdss,t∈1,e.Thus, in order to solve equation (1), we only need to find the fixed point of operator equation x=Tx. To do this, we firstly establish some lemmas.Lemma 5.
L:Q→Q is a completely continuous operator with the spectral radius ρL≠0. Moreover, L has a positive eigenfunction ω∗ corresponding to the first eigenvalue μ1=ρL−1.Proof.
Firstly, it follows from Lemma2 that, for any x∈Q, one has
(21)Lx=maxt∈1,e∫1eGt,sxsdss≤1Γα∫1eψαsxsdss,Lxt≥1Γαlntα−1∫1eψαsxsdss,which imply that L:Q→Q. Since Ht,s,t∈1,e×1,e is uniform continuity, then the operator L:Q→Q is completely continuous.
On the other hand, by (10), we know that there exists a τ0∈1,e such that Gτ0,τ0>0. Thus, from the continuity of G, there exists a closed interval c,d⊂1,e such that τ0∈c,d and Gt,s>0 for all t,s∈c,d. Now, we take x∈Q such that xτ0>0 and xt=0 for all t∈c,d. Then, for all t∈c,d, one has
(22)Lxt=∫1eGt,sxsdss≥∫cdGt,sxsdss>0.
Thus, there existsμ>0 such that μLxt≥xt for t∈0,1. From Krein-Rutmann’s theorem, we know that the spectral radius ρL≠0 and T have a positive eigenfunction ω∗ that satisfies μ1Lω∗=ω∗, where μ1=ρL−1 is its first eigenvalue. The proof is completed.Lemma 6.
Suppose that (B1) holds, then the operatorT:Q¯R\Qr→Q is completely continuous.Proof.
Firstly, for anyx∈Q¯R\Qr,t∈1,e, it follows from Lemma 2 that
(23)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss≤1Γα∫1eψαsfs,∫1eHs,τxτdττ,xsdss,that is
(24)Tx≤1Γα∫1eψαsfs,∫1eHs,τxτdττ,xsdss.
Similarly, one also has(25)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss≥1Γαlntα−1∫1eψαsfs,∫1eHs,τxτdττ,xsdss≥lntα−1Tx,which implies TQ⊂Q and then TQ¯R\Qr⊂Q.
On the other hand, from (B1), we know that there exists a natural numberl such that
(26)supx∈Q¯R\Qr∫Ωlψαsfs,∫1eHs,τxτdττ,xsdss<12.
Thus, for anyx∈Q¯RQr and 1+1/l≤t≤e−1/l, we have
(27)ln1+1lα−1r≤xt≤x=R,t∈1,e,(28)ln1+1/lα−1ln1+1/lβ−1rβ−1Γβ+1≤lntα−1lntβ−1Γβ∫1eψβτxτdττx≤∫1eHs,τxτdττ≤1Γβ∫1eψβτxτdττ≤Rβ−1Γβ+1,t∈1,e.
Take(29)M1=maxt,x,y∈Ωl∗ft,x,y,where
(30)Ωl∗=1+1l,e−1l×ln1+1/lα−1ln1+1/lβ−1Rβ−1Γβ+1,rβ−1Γβ+1×ln1+1lα−1r,R.
So, it follows from (26), (27) and (28) that
(31)supx∈Q¯R\Qr∫1eψαsfs,∫1eHs,τxτdττ,xsdss≤supx∈Q¯R\Qr∫Ωlψαsfs,∫1eHs,τxτdττ,xsdss+supx∈Q¯R\Qr∫1+1/le−1/lψαsfs,∫1eHs,τxτdττ,xsdss≤12+M1∫1eψαsdss<+∞,which implies T is uniformly bounded for any bounded set.
Secondly, we shall prove thatT:Q¯R\Qr→Q is continuous. To do this, let xn,x0∈Q¯RQr and ∥xn−x0∥→0 (n→∞). For any ε>0, it follows from (B1) that there exists a natural number m>0 such that
(32)supx∈Q¯RQr∫Ωmψαsfs,∫1eHs,τxτdττ,xsdss<ε4.
On the other hand, it follows form the fact thatft,u,v is uniformly continuous onΩm∗=1+1/m,e−1/m×ln1+1/mα−1ln1+1/mβ−1R/β−1Γβ+1,r/β−1Γβ+1×ln1+1/mα−1r,R, (21) that
(33)limn→+∞fs,∫1eHs,τxnτdττ,xns−fs,∫1eHs,τx0τdττ,x0s=0,holds uniformly on s∈1+1/m,e−1/m. Thus, the Lebesgue control convergence theorem ensures
(34)∫1+1/me−1/mψαsfs,∫1eHs,τxnτdττ,xns−fs,∫1eHs,τx0τdττ,x0sdss→0,asn→∞.
In other words, for the aboveε>0, there exists a positive integer N such that n>N, we have
(35)∫1+1/me−1/mψαsfs,∫1eHs,τxnτdττ,xns−fs,∫1eHs,τx0τdττ,x0sdss<ε2.
Thus, by (32) and (35), for any n>N, we have
(36)Txn−Tx0≤2supx∈Q¯R\Qr∫Ωmψαsfs,∫1eHs,τx0τdττ,x0sdss+∫1+1/me−1/mψαsfs,∫1eHs,τxnτdττ,xns−fs,∫1eHs,τx0τdττ,x0sdss<2×ε4+ε2=ε.
Therefore,T:Q¯RQr→Q is continuous.
In the end, we shall prove thatT is equicontinuous. Firstly, it follows from (B1) that for any ε>0, there exists a positive integer k such that
(37)supx∈Q¯R\Qr∫Ωkψαsfs,∫1eHs,τxτdττ,xsdss<ε4.
Take(38)M2=maxt,x,y∈Ωk∗ft,x,y,where
(39)Ωk∗=1+1k,e−1k×ln1+1/kα−1ln1+1/kβ−1rβ−1Γβ+1,Rβ−1Γβ+1×ln1+1kα−1r,R.
Notice thatGt,s is uniformly continuous on 1,e×1,e, so for the above ε>0 and fixed s∈1+1/k,e−1/k, there exists δ>0, when ∣t1−t2∣<δ,t1,t2∈1,e, we have
(40)∣Gt1,s−Gt2,s∣≤2M2e−1−1ε.
It follows from the above argument that, for∣t1−t2∣<δ,t1,t2∈1,e, one has
(41)∣Txt−Txt′∣≤2supx∈Q¯R\Qr∫Ωkψαsfs,∫1eHs,τxτdττ,xsdss+supx∈Q¯R\Qr∫1+1/ke−1/k∣Gt,s−Gt′,s∣fs,∫1eHs,τxτdττ,xsdss<2×ε4+ε2=ε,which implies that T is an equicontinuous operator. According to the Arzela-Ascoli Theorem, T:Q¯RQr→Q is completely continuous. The proof is completed.


## 3. Main Results

We state the main results of this paper as follows.Theorem 7.
Letμ1 be the first eigenvalue of L, assume (B1) holds and
(42)limsupu+v→+∞v→+∞ft,u,vv<μ1<liminfu→0+v→0+ft,u,vu+v,uniformly holds on t∈1,e. Then, equation (1) has at least one positive solution.Proof.
Firstly, by Lemma6, we know that T:Q¯R\Qr→Q is completely continuous. Thus, it follows from the extension theorem of a completely continuous operator (see Theorem 8.3 on page 56 of [108]) that, for any R>0, there exists a completely continuous operator T∗:Q¯R→Q. Thus, without loss of the generality, we shall still write it as T.
Next, it follows from (42) that there exists r>0 such that
(43)ft,u,v≥μ1u+v,∣v∣≤rβ−1Γβ+1,x≤r,t∈1,e.
Since, for anyx∈∂Qr, we have
(44)∫1eHs,τxτdττ≤rβ−1Γβ+1,xs≤r,thus, for x∈∂Qr, from (43) and (44), one gets
(45)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss≥μ1∫1eGt,s∫1eHs,τxτdττ+xsdss≥μ1Lxt,t∈1,e.
Letω∗ be the positive eigenfunction corresponding to μ1, i.e., ω∗=μ1Lω∗. In what follows, we shall use the contradiction method to show
(46)x−Tx≠μω∗,x∈∂Qr,μ≥0.
Firstly, we can suppose thatT has no fixed points on x∈∂Qr (otherwise, the proof is finished). In this case, let us suppose there exist x0∈∂Qr and μ0≥0 such that x0−Tx0=μ0ω∗. This implies that μ0>0 and then
(47)x0=Tx0+μ0ω∗≥μ0ω∗.
Letμ¯=supμ∣x0≥μω∗, then μ¯≥μ0,x0≥μ¯ω∗, μ1Lx0≥μ1μ¯Lω∗=μ¯ω∗. Thus, it follows from (45) that
(48)x0=Tx0+μ0ω∗≥μ1Lx0+μ0ω∗≥μ¯ω∗+μ0ω∗=μ¯+μ0ω∗,which contradicts with the definition of μ¯. So (46) holds and from Lemma 2, we have
(49)iT,Qr,Q=0.
On the other hand, it follows from (42) that there exists R1>r and 0<κ<1 such that
(50)ft,u,v≤κμ1v,for∣u+v∣≥R1,∣v∣≥R1.
LetL~x=κμ1Lx. Obviously, L~:E→E is also a bounded linear operator and L~Q⊂Q. Since μ1 is the first eigenvalue of operator L and 0<κ<1, we have
(51)r−1L~=κμ1rL−1=κ−1>1.
From Gelfand’s formula, we have(52)κ=limn→+∞L~n1/n.
Now, chooseε0=1/21−κ, it follows from (52) that there exists an enough large integer N>0 such that when n≥N, one has
(53)∥L~n∥≤κ+ε0n.
LetL~0=I be the identity operator and define
(54)∥x∥∗=∑i=1Nκ+ε0N−i∥L~i−1x∥,x∈E,then ·∗ is still the norm of E. Let
(55)M=supx∈∂QR1∫1eψαsfs,∫1eHs,τxτdττ,xsdss,then from (31), we have M<+∞. Take
(56)R2>maxR1,2ε0M∗,where M∗=M∗. Notice that x∗>κ+ε0N−1x, so we can choose a large enough R>R2 such that x≥R implies x∗>R2.
Next, we shall show(57)Tx≠μx,x∈∂QR,μ≥1.
Otherwise, there existx1∈∂QR and μ∗≥1 such that Tx1=μ∗x1. Let x~t=minx1t,R1 and
(58)Dx1=t∈1,e:x1t>R1.
Sincex1∈C1,e, x1t≤x1=R, there exists 1<t0≤e such that x1t0=R. So for any t∈1,e, we get x~t=minx1t,R1≤minR,R1=R1 and x~t0=minx1t0,R1=minR,R1=R1, which implies x~t=R1, i.e., x~∈∂QR1. Thus by Lemma 2, we have
(59)μ∗x1=Tx1t=∫1eGt,sfs,∫1eHs,τx1τdττ,x1sdss≤∫Dx1Gt,sfs,∫1eHs,τx1τdττ,x1sdss+∫1,e\Dx1ψαsfs,∫1eHs,τx~τdττ,x~sdss≤κμ1∫1eGt,sx1sdss+∫1eψαsfs,∫1eHs,τx~τdττ,x~sdss≤L~x1t+M,t∈1,e.
So it follows fromL~Q⊂Q and (59), we have
(60)0≤L~jTx1t≤L~jL~x1+Mt,j=0,1,2,⋯,N−1.
SinceQ is a normal cone with normality constant 1 and (60), one has
(61)L~jTx1≤L~jL~x1+M,j=0,1,2,⋯,N−1,which leads to
(62)Tx1∗=∑i=1Nκ+ε0N−iL~i−1Tx1≤∑i=1Nκ+ε0N−iL~i−1L~x1+M=L~x1+M∗.
According to the selection ofR2, we have M∗<ε0/2R2. Thus, it follows from the fact x≥R implies ∥x∥∗>R2 and (53), (54), and (62) that
(63)μ∗x1∗=Tx1∗≤∥L~x1∥∗+M∗=∑i=1Nκ+ε0N−i∥L~ix1∥+M∗=κ+ε0∑i=1N−1κ+ε0N−i−1∥L~ix1∥+∥L~Nx1∥+M∗≤κ+ε0∑i=1N−1κ+ε0N−i−1∥L~ix1∥+κ+ε0N∥x1∥+M∗=κ+ε0∑i=1Nκ+ε0N−i∥L~i−1x1∥+M∗≤κ+ε0∥x1∥∗+ε02R2≤κ+ε0∥x1∥∗+ε02∥x1∥∗=14κ+34∥x1∥∗.
Notice thatμ∗≥1, we have 1/4κ+3/4≥1, and then κ≥1, which is a contradiction with 0<κ<1. So (57) is indeed valid and it follows from Lemma 2 that
(64)iT,QR,Q=1.
Thus, (49) and (64) lead to
(65)iT,QR\Q¯r,Q=iT,QR,Q−iT,Qr,Q=1,which implies T has at least one fixed point on QR\Q¯r. Consequently, equation (1) has at least one positive solution.Theorem 8.
Letμ1 be the first eigenvalue of L, suppose (B1) holds, and
(66)limsupu→0+v→0+ft,u,vv<μ1,liminfu+v→+∞ft,u,vu+v>μ~1uniformly on t∈1,e, where μ~1 is any eigenvalues of L. Then, equation (1) has at least one positive solution.To prove Theorem8, we need some preliminaries and lemma. For any enough small 0<ε<1, define
(67)Lεxt=∫1+εe−εGt,sxsdss,t∈1,e.By Lemma5, we know Lε:Q→Q is still a completely continuous linear operator with the spectral radius ρLε≠0, and Lε has a positive eigenfunction ωε corresponding to its first eigenvalue με=ρLε−1.Lemma 9.
There exists an eigenvalueμ~1 of L such that
(68)limε→0+με=μ~1.Proof.
Letεm→0m→∞ satisfying ε1≥ε2≥⋯≥εm≥⋯. Thus, for ω∈Q and k>m, one has
(69)Lεmωt≤Lεkωt≤Lωt,t∈1,e,and then
(70)Lεmnωt≤Lεknωt≤Lnωt,t∈1,e,n=2,3,⋯,where Lεmn=LLεmn−1,n=2,3,⋯. It follows from the fact that Q is a normal cone with normality constant 1 that
(71)∥Lεmn∥≤∥Lεkn∥≤∥Ln∥,n=1,2,⋯.
By Gelfand’s formula, we haveμεm≥μεk≥μ1, where μ1 is the first eigenvalue of L, which implies μεm is monotonous with lower boundedness μ1. Let
(72)limm→+∞μεm=μ~1,we assert μ~1 is an eigenvalue of L.
In fact, letωεm be positive eigenfunctions of Lεm corresponding to μεm with ∥ωεm∥=1,m=1,2,⋯. So we have
(73)ωεmt=μεmLεmωεmt=μεm∫1+εme−εmGt,sωεmsdss,t∈1,e.
Since(74)Lεmωεm=max1≤t≤e∫1+εme−εmGt,sωεmsdss≤1Γα∫1eψαsdss<+∞,m=1,2,⋯,
Lεmωεm⊂Q is uniformly bounded.
Moreover, for anym and t1,t2∈1,e, one has
(75)∣Lεmωεmt1−Lεmωεmt2∣≤∫εm1−εm∣Gt1,s−Gt2,s∣ωεmsdss.
Notice thatGt,s is uniformly continuous on 1,e×1,e; then, (75) implies Lεmωεm⊂Q is equicontinuous. Thus, from the Arzela-Ascoli theorem and limm→+∞μεm=μ~1, we know ωεm→ω0 as m→+∞, which yields ω0=1. Take the limit for two sides of (73), we obtain
(76)ω0t=μ~1∫1eGt,sω0sdss,t∈1,e,that is, ω0=μ~1Lω0, and μ~1 is an eigenvalue of L.Proof of Theorem 8.
Firstly, it follows from (66) that for any t∈1,e, there exists r>0 such that
(77)ft,u,v≤μ1v,∣u∣≤rβ−1Γβ+1,∣v∣≤r.
Thus, for anyx∈∂Qr, since
(78)∫1eHs,τxτdττ≤rβ−1Γβ+1,∣xs∣≤x=r,we have
(79)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss≤μ1∫1eGt,sxsdss=μ1Lxt,t∈1,e.
In the following, we prove that(80)Tx≠μx,foranyx∈∂Qr,μ≥1.
Firstly, we may suppose thatT has no fixed point on ∂Qr (otherwise, the proof is completed). If (80) is not true, i.e., there exist x0∈∂Qr and μ0≥1 satisfying Tx0=μ0x0. Thus μ0>1, it follows from (79) that
(81)μ0x0=Tx0≤μ1Lx0.
From induction, we have(82)μ0nx0≤μ1nLnx0,n=1,2,⋯,which implies that
(83)∥Ln∥≥∥Lnx0∥∥x0∥≥μ0n∥x0∥μ1n∥x0∥=μ0nμ1n.
Thus, by Gelfand’s formula, we have(84)rL=limn→∞∥Ln∥n≥μ0μ1>1μ1,which contradicts with rL=μ1−1. Therefore, (80) is valid and from Lemma 2, we get
(85)iT,Qr,Q=1.
On the other hand, for any fixed small enough0<ε<1, take
(86)ρε=Γα−1ln1+εβ−1β−1Γα+β−2+ln1+εα−1.
By (66) and limε→0+με=μ~1, we can find a sufficiently small ε>0 and R>r such that
(87)ft,u,v≥μεu+v,u+v≥ρεR,t∈1,e,where με is the first eigenvalue of Lε.
Letωε be the positive eigenfunction of Lε corresponding to με, which implies ωε=μεLεωε. For any x∈∂QR,s∈1+ε,e−ε, we have
(88)∫1eHs,τxτdττ+xs≥1Γβlnsβ−1∫1eψβτlnτα−1dττx+lnsα−1x≥Γα−1ln1+εβ−1β−1Γα+β−2+ln1+εα−1R=ρεR.
Thus, it follows from (87) and (88) that
(89)Txt=∫1eGt,sfs,∫1eHs,τxτdττ,xsdss≥∫1+εe−εGt,sfs,∫1eHs,τxτdττ,xsdss≥με∫1+εe−εGt,s∫1eHs,τxτdττ+xsdss≥με∫1+εe−εGt,sxsdss=μεLεxt,t∈1,e.
Similar to the proof of Theorem7, we obtain
(90)x−Tx≠μωε,x∈∂QR,μ≥0.
According to Lemma2, we have
(91)iT,QR,Q=0.
Combining (80) and (91), one has
(92)iT,QR\Q¯r,Q=iT,QR,Q−iT,Qr,Q=−1.
Hence,T has at least one fixed point on QR\Q¯r, and then equation (1) has at least a positive solution. The proof is completed.


## 4. Conclusion

Singular behaviour is a class of important natural phenomena in many physical science, mathematics, engineering, and bioscience. So, the study for singularity is an interesting and challenging problem. In this paper, we consider the existence of positive solutions for a Hadamard-type fractional differential equation with singular nonlinearity by introducing a new limit-type growth condition. The main advantage of the assumption is that it provides an effective method for handling the singularity at space variables. This assumption is valid and reasonable and easier to get the solution of the target equation.



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*Source: 8392397-2020-09-30.xml*