# Solutions for a Class of Hadamard Fractional Boundary Value Problems with Sign-Changing Nonlinearity

Authors: Keyu Zhang; Zhengqing Fu
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

Using fixed point methods we establish some existence theorems of positive (nontrivial) solutions for a class of Hadamard fractional boundary value problems with sign-changing nonlinearity.


---

## Body

## 1. Introduction

In this paper, using fixed point methods we study the existence of positive (nontrivial) solutions for the Hadamard fractional boundary value problems with sign-changing nonlinearity:(1)-Dαut=ft,ut,t∈1,e,u1=δu1=δue=0,where α∈(2,3] is a real number, Dα is the left-sided Hadamard fractional derivative of order α, δu(t)=tdu(t)/dt, and f∈C([1,e]×R+,R) is a sign-changing function; i.e., there exists a constant M>0 such that(H0)f(t,u)+M≥0 for all (t,u)∈[1,e]×R+.As is known, fractional differential equations have been paid special attention by many researchers for the reason that they serve as an excellent tool for wide applications in various disciplines of science and engineering such as mechanics, electricity, chemistry, and control theory; for more details, we refer to books [1–3]. In recent years, there have been a large number of papers dealing with the existence of solutions of nonlinear initial (boundary) value problems of fractional differential equations by using some techniques of nonlinear analysis, such as fixed-point results [4–13], iterative methods [14–23], the topological degree [24–29], the Leray-Schauder alternative [30, 31], and stability [32].In [4], the authors studied the following abstract evolution of the system for HIV-1 population dynamics, which takes the form in fractional sense:(2)Dtαut+λft,ut,Dtβut,vt=0,Dtγvt+λgt,ut=0,0<t<1,Dtβu0=Dtβ+1u0=0,Dtβu1=∫01DtβusdAs,v0=v′0=0,v1=∫01vsdBs,where f:(0,1)×[0,+∞)3→(-∞,+∞) and g:(0,1)×[0,+∞)→(-∞,+∞) are two semipositone functions. By using the Guo-Krasnosel’skii fixed point theorem, they not only obtained the existence of positive solutions for (2) but also discussed the effect of parameters λ on the existence of solutions.In [14], the authors adopted generalized α-contractive map to study some fractional integro-differential equations with the Caputo-Fabrizio derivation and obtained the approximate solutions for these problems by using of some appropriate Lipschitz conditions for their nonlinearities.In [15], Cui used the convergence of Cauchy sequences in complete spaces to obtain the unique solution for the fractional boundary value problems:(3)Dtpxt+ptft,xt+qt=0,t∈0,1,x0=x′0=0,x1=0,where f is a Lipschitz continuous function, with the Lipschitz constant associated with the first eigenvalue for the relevant operator. This method can also be applied in papers [16, 17] and references therein.However, as a generalization of fractional calculus by Riemann and Liouville, Hadamard fractional equations have seldom been studied in the literature; we only refer to [8–10, 22, 23, 29, 32] and references therein. In [8], the authors used the Guo-Krasnosel’skii fixed point theorem on cones to establish the existence and nonexistence of positive solutions for (1) with nonnegative nonlinearity λa(t)f(u) and considered solvability for the influence of the parameter intervals.In [32], the authors used Banach and Schauder fixed point theorem to obtain the existence and Hyers-Ulam stability of solutions for Hadamard fractional impulsive Cauchy problems of the form (4)HD1+αut=ft,ut,α∈0,1,t∈1,e\t1,t2,…,tm,Δuti=HJ1+1-αuti+-HJ1+1-αuti-=pi,pi∈R,i=1,2,…,m,HJ1+1-αu1+=u0,u0∈R,where f satisfies a Lipschitz condition.In this paper, motivated by works aforementioned, we used fixed point methods to study the existence of solutions for (1) with sign-changing nonlinearity. We have the main results: (i) when the nonlinear term f grows both superlinearly and sublinearly at ∞, we use the fixed point index theory to obtain two existence theorems of positive solutions for (1); (ii) when f satisfies an appropriate Lipschitz condition, we obtain a unique solution for (1) and establish a sequence of iterations uniformly converges to the unique solution.


## 2. Preliminaries

Definition 1 (see [1–3]).
The left-sided Hadamard fractional derivative of orderα∈n-1,n,n∈Z+, of a function f is given by(5)Dαft=1Γn-αtddtn∫1tln⁡tsn-α+1fsdss,1≤t≤e,where Γ(·) is the Gamma function.
We now offer Green’s function for (1). From Lemma 2.1 of [8], (1) is equivalent to the integral equation(6)ut=∫1eGt,sfs,usdss,t∈1,e,where(7)Gt,s=1Γα1-ln⁡sα-2ln⁡tα-1,1≤t≤s≤e,1-ln⁡sα-2ln⁡tα-1-ln⁡tsα-1,1≤s≤t≤e.Lemma 2 (see [10, Lemma 2.2]).
LetG be defined by (7). Then the following inequalities are satisfied:(8)Gt,s≥0,ln⁡tα-1Ge,s≤Gt,s≤Ge,s,∀t,s∈1,e×1,e,where G(e,s)=(1-ln⁡s)α-2-(1-ln⁡s)α-1 and ∀s∈[1,e].Lemma 3.
Letφ(t)=(1-ln⁡t)α-2-(1-ln⁡t)α-1 and ∀t∈[1,e]. Then the following inequalities are satisfied:(9)∫1eln⁡tα-1φtdtt·φs≤∫1eGt,sφtdtt≤∫1eφtdtt·φs,∀s∈1,e.This is a direct result from Lemma 2, so we omit its proof. Moreover, for convenience, let(10)κ1=∫1eln⁡tα-1φtdtt,κ2=∫1eφtdtt.
LetE≔C[1,e], u≔supt∈1,eut, and P≔{u∈E:u(t)≥0,∀t∈[1,e]}. Then (E,·) becomes a real Banach space and P is a cone on E. Define Bρ≔{u∈E:u<ρ} for ρ>0 in the sequel.
DefineP0={u∈P:u(t)≥(ln⁡t)α-1u and ∀t∈[1,e]}. Then P0 is also a cone on E. In what follows, we verify that when u∈P0, u≥M/Γα∫1e(1-ln⁡s)α-2ds/s; we have u(t)-w(t)≥0, ∀t∈[1,e], where w is a solution for the problem(11)-Dαut=M,t∈1,e,u1=δu1=δue=0,where M is defined by (H0). From (1) and (6), w takes the form as follows:(12)wt=M∫1eGt,sdss,t∈1,e,where G is defined by (7). Indeed, when u∈P0, from (7) we have (13)ut-wt≥ln⁡tα-1u-M∫1eGt,sdss≥ln⁡tα-1u-MΓα∫1e1-ln⁡sα-2ln⁡tα-1dss=ln⁡tα-1u-MΓα∫1e1-ln⁡sα-2dss≥0,∀t∈1,e.
For semipositone condition (H0), we need to construct an appropriate operator with which to study problem (1). Hence, we consider the modified problem(14)-Dαut=ft,max⁡ut-wt,0+M,t∈1,e,u1=δu1=δue=0,where w is a solution for (11). Clearly, we are easy to show that if u solves (14), w solves (11), and u(t)-w(t)≥0(≢0) and ∀t∈[1,e]; then u(t)-w(t) is a positive solution for (1). Consequently, we turn to study the modified problem (14). From (1) and (6), (14) is equivalent to the integral equation(15)ut=∫1eGt,sfs,max⁡us-ws,0+Mdss,t∈1,e,u∈E.Hence, we can define an operator A:P→P as follows:(16)Aut=∫1eGt,sfs,max⁡us-ws,0+Mdss,t∈1,e,u∈E.It is not difficult to prove that if Au0=u0, then u0 is a solution for (14). Moreover, from Lemma 2 we easily have A(P)⊂P0.Now, we offer some basic theorems for fixed point methods used in our problem.Lemma 4 (see [33]).
LetE be a real Banach space and P a cone on E. Suppose that Ω⊂E is a bounded open set and that A:Ω¯∩P→P is a continuous compact operator. If there exists ω0∈P\{0} such that (17)ω-Aω≠λω0,∀λ≥0,ω∈∂Ω∩P,then i(A,Ω∩P,P)=0, where i denotes the fixed point index on P.Lemma 5 (see [33]).
LetE be a real Banach space and P a cone on E. Suppose that Ω⊂E is a bounded open set with 0∈Ω and that A:Ω¯∩P→P is a continuous compact operator. If(18)ω-λAω≠0,∀λ∈0,1,ω∈∂Ω∩P, then i(A,Ω∩P,P)=1.


## 3. Positive Solutions for (1)

Letλ1=κ1-1,λ2=κ2-1, and N=(M/Γα)∫1e1-ln⁡sα-2(ds/s). Then we give some assumptions for nonlinear term f.(H1)liminfu→+∞ft,u/u>λ1 uniformly on t∈[1,e],(H2) there existQ:[1,e]→R+ such that(19)ft,max⁡ut-wt,0+M≤Qt,∀t,u∈1,e×0,MΓα∫1e1-ln⁡sα-2dss, where ∫1eφtQtdt/t<N,(H3)limsupu→+∞ft,u/u<λ2 uniformly on t∈[1,e],(H4) there existl∈(1,e), Q~:[1,e]→R+ such that(20)ft,max⁡ut-wt,0+M≥Q~t,∀t,u∈1,e×0,MΓα∫1e1-ln⁡sα-2dss, where ∫1e(ln⁡l)α-1φtQ~tdt/t>N,(H5)f∈C([1,e]×R,R), and there exists k∈(0,1) such that ft,u-ft,v≤kλ2u-v for t∈[1,e],u,v∈R,(H6)f(t,0)≢0 for t∈[1,e].We now state our main results and offer their proofs.Theorem 6.
Suppose that (H0)-(H2) hold. Then (1) has at least a positive solution.Proof.
We first prove that there existR>N large enough such that(21)u-Au≠λφ0∗,∀λ≥0,u∈∂BR∩P,where φ0∗∈P0 is a given element. If false, there exits u∈∂BR∩P,λ0≥0 such that u-Au=λ0φ0∗. Note that A(P)⊂P0; then Au∈P0 for u∈P, and thus u∈P0. This also implies that u(t)≥ for t∈[1,e].
Note thatu∈∂BR∩P; then u=R>N, and u(t)-w(t)≥0 for u∈P0,t∈[1,e]. From (H1) we have(22)liminfu→+∞ft,max⁡u-w,0+Mu-w=liminfu→+∞ft,u-w+Mu-w>λ1uniformly on t∈[1,e]. As a result, there exist ε1>0 and c1>0 such that(23)ft,ut-wt+M≥λ1+ε1ut-wt-c1,∀t∈1,e. This implies that(24)ut≥Aut≥∫1eGt,sλ1+ε1us-ws-c1dss≥λ1+ε1∫1eGt,sus-wsdss-c1κ2. Note that (12) is multiplied by φ(t) on both sides of the above and integrated over [1,e] and use Lemma 3 to obtain(25)∫1eutφtdtt≥∫1eφtλ1+ε1∫1eGt,sus-wsdss-c1κ2dtt≥λ1+ε1λ1∫1eφtut-wtdtt-c1κ22=1+ε1κ1∫1eutφtdtt-1+ε1κ1∫1eφtM∫1eGt,sdssdtt-c1κ22≥1+ε1κ1∫1eutφtdtt-M1+ε1κ1κ22-c1κ22. Consequently, we have(26)∫1eutφtdtt≤ε1κ1-1M1+ε1κ1+c1κ22. Noting that u∈P0, we have(27)u∫1eln⁡tα-1φtdtt≤ε1κ1-1M1+ε1κ1+c1κ22,andu≤ε1-1κ1-2M1+ε1κ1+c1κ22. Therefore, if we choose R>max⁡N,ε1-1κ1-2M1+ε1κ1+c1κ22, then (21) holds true. From Lemma 4 we have(28)iA,BR∩P,P=0.
On the other hand, we prove that(29)u≠λAu,∀λ∈0,1,u∈∂BN∩P.If false, there exist u∈∂BN∩P,λ1∗∈[0,1] such that u=λ1∗Au; this implies u(t)≤(Au)(t), ∀t∈[1,e], and u≤Au. However, from (H2) we have(30)Aut=∫1eGt,sfs,max⁡us-ws,0+Mdss≤∫1eφsQsdss<N=u,foru∈∂BN∩P. This indicates that Au<u for u∈∂BN∩P. This has a contradiction, and thus (29) holds true. From Lemma 5 we have(31)iA,BN∩P,P=1.
From (28) and (31), we obtain(32)iA,BR\B¯N∩P,P=iA,BR∩P,P-iA,BN∩P,P=-1. Therefore the operator A has at least one fixed point u in (BR\B¯N)∩P with u≥N, and then u(t)-w(t) is a positive solution for (1). This completes the proof.Theorem 7.
Suppose that (H0), (H3), and (H4) hold. Then (1) has at least a positive solution.Proof.
We first prove that there existR>N large enough such that(33)u≠λAu,∀λ∈0,1,u∈∂BR∩P.If false, there exist u∈∂BR∩P,λ2∗∈[0,1] such that u=λ2∗Au, and u∈P0 for the fact that Au∈P0 when u∈P. This also implies u(t)≤(Au)(t),t∈[1,e].
Note thatu∈∂BR∩P; then u=R>N, and u(t)-w(t)≥0 for u∈P0,t∈[1,e]. From (H3) we have(34)limsupu→+∞ft,max⁡u-w,0+Mu-w=limsupu→+∞ft,u-w+Mu-w<λ2uniformly on t∈[1,e]. As a result, there exist ε2∈(0,λ2) and c2>0 such that(35)ft,ut-wt+M≤λ2-ε2ut-wt+c2,∀t∈1,e. This implies that(36)ut≤Aut≤∫1eGt,sλ2-ε2us-ws+c2dss≤λ2-ε2∫1eGt,sus-wsdss+c2κ2. Note that (12) is multiplied by φ(t) on both sides of the above and integrated over [1,e] and use Lemma 3 to obtain(37)∫1eutφtdtt≤∫1eφtλ2-ε2∫1eGt,sus-wsdss+c2κ2dtt≤λ2-ε2λ2∫1eφtut-wtdtt+c2κ22≤1-ε2κ2∫1eutφtdtt+1-ε2κ2∫1eφtM∫1eGt,sdssdtt+c2κ22≤1-ε2κ2∫1eutφtdtt+M1-ε2κ2κ22+c2κ22. Consequently, we have(38)∫1eutφtdtt≤ε2κ2-1M1-ε2κ2+c2κ22. Noting that u∈P0, we obtain(39)u∫1eln⁡tα-1φtdtt≤ε2κ2-1M1-ε2κ2+c2κ22,andu≤ε2κ1κ2-1M1-ε2κ2+c2κ22. Taking R>max⁡N,ε2κ1κ2-1M1-ε2κ2+c2κ22, then (33) is satisfied. From Lemma 5 we have(40)iA,BR∩P,P=1.
On the other hand, we prove that(41)u-Au≠λφ1∗,∀λ≥0,u∈∂BN∩P,where φ1∗∈P is a given element. If false, there exist u∈∂BN∩P,λ3≥0 such that u-Au=λ3φ1∗; this implies u(t)≥(Au)(t),t∈[1,e], and thus u≥Au. However, from (H4) we have(42)Au=maxt∈1,eAut≥∫1eGl,sfs,max⁡us-ws,0+Mdss≥∫1eln⁡lα-1φsQ~sdss>N=ufor u∈∂BN∩P. This has a contradiction, and thus (41) holds true. From Lemma 4 we have(43)iA,BN∩P,P=0.
From (40) and (43), we obtain(44)iA,BR\B¯N∩P,P=iA,BR∩P,P-iA,BN∩P,P=1. Therefore the operator A has at least one fixed point u in (BR\B¯N)∩P with u≥N, and then u(t)-w(t) is a positive solution for (1). This completes the proof.From (6), we define an operator T:E→E as follows:(45)Tut=∫1eGt,sfs,usdss,u∈E. Then T is a completely continuous operator, and u is a solution for (1) if and only if u is a fixed point of T.Theorem 8.
Suppose that (H5)-(H6) hold. Then (1) has only a nontrivial solution, denoted by u∗, and for all u0∈E,u0(t)≢0,t∈[1,e], the sequence un=Tun-1(n=1,2,…) uniformly converges to u∗.Proof.
(H6) ensures that0 is not a solution for (1). Then if (1) has a solution, this solution is nontrivial. For all n∈N, from Lemma 3 we have(46)un+1t-unt=Tunt-Tun-1t=∫1eGt,sfs,uns-fs,un-1sdss≤∫1eGt,sfs,uns-fs,un-1sdss≤kλ2∫1eφsuns-un-1sdss=kλ2∫1eφtTun-1t-Tun-2tdtt≤kλ2∫1eφt∫1eGt,sfs,un-1s-fs,un-2sdssdtt≤k2λ2∫1eφtun-1t-un-2tdtt⋮≤knλ2∫1eφtu1t-u0tdtt.
On the other hand, lettingv=0 and f0=maxt∈[1,e]ft,0 in (H5), we have(47)ft,u≤kλ2u+f0,∀u∈R,t∈1,e. Noting that u1=Tu0, we obtain(48)∫1eφtu1t-u0tdtt≤1+k∫1eφtu0tdtt+κ22f0≔β1. Therefore, for all n∈N, we have(49)un+1t-unt≤knβ1λ2,∀t∈1,e. Consequently, for all n,m∈N, we have(50)un+mt-unt≤un+mt-un+m-1t+un+m-1t-un+m-2t+⋯+un+1t-unt≤β1λ2kn+m-1+kn+m-2+⋯+kn≤knβ1λ21-k→0,whenn→∞. This implies {un} is a Cauchy sequence, and from E’s completeness, there exists u∗∈E such that limn→∞un=u∗. Taking the limits for sequence un=Tun-1 and we have Tu∗=u∗; i.e., u∗ is a nontrivial solution for (1).
Next we prove that (1) has only a solution. If u,v∈E are solutions for (1) and u≠v, then Tnu=u and Tnv=v for all n∈N. By (H5) we obtain(51)ut-vt=Tnut-Tnvt=TTn-1ut-TTn-1vt≤∫1eGt,sfs,Tn-1us-fs,Tn-1vsdss≤kλ2∫1eφsTn-1us-Tn-1vsdss≤kλ2∫1eφt∫1eGt,sfs,Tn-2us-fs,Tn-1vsdssdtt≤k2λ2∫1eφtTn-2ut-Tn-1vtdtt⋮≤kn-1λ2∫1eφtTut-Tvtdtt=kn-1λ2∫1eφtut-vtdtt≤kn-1u-v.This implies that(52)u-v≤kn-1u-v,for alln∈N. Noting that k∈(0,1), then there exists N∈N, when n>N, kn-1<1, and thus a contradiction for the above inequality. This obtains the uniqueness of solutions for (1). This completes the proof.In what follows, we offer some examples for our main results. Letα=2.5, l=e. Then κ1=∫1e(ln⁡t)α-1φtdt/t≈0.1227, κ2=∫1eφtdt/t=4/15≈0.2667, and N=M/Γα∫1e(1-ln⁡s)α-2ds/s≈0.5015M.Example 9.
Letf(t,u)=1/Meln⁡tu2-M for all t∈[1,e], u∈R+, and Q≡1.8M. Then liminfu→+∞ft,u/u=liminfu→+∞1/Meln⁡tu2-M/u=+∞>λ1 uniformly on t∈[1,e], and when (t,u)∈[1,e]×[0,N], f(t,u)+M≤1/M0.2515M2+M≈1.2515M≤Q(t). Moreover, ∫1eφtQtdt/t≈0.48M<N. Therefore, (H1) and (H2) hold.Example 10.
Letf(t,u)=6M/e-0.5015Me-u-M for all t∈[1,e], u∈R+, and Q≡5.4M. Then limsupu→+∞ft,u/u=limsupu→+∞6M/e-0.5015Me-u-M/u=0<λ2 uniformly on t∈[1,e], and when (t,u)∈[1,e]×[0,N], f(t,u)+M≥6M/e-0.5015Me-0.5015M-M+M=6M≥Q(t). Moreover, ∫1e(0.5)1.5φt5.4Mdt/t≈0.5093M>N. Therefore, (H3) and (H4) hold.Example 11.
Letf(t,u)=3.7ku+g(t), where k∈(0,1) and g∈C([1,e],R) with g(t)≢0 for t∈[1,e]. Therefore, (H5) and (H6) hold.



---

*Source: 9046472-2019-02-04.xml*