# Solutions for Integral Boundary Value Problems of Nonlinear Hadamard Fractional Differential Equations

Authors: Keyu Zhang; Jianguo Wang; Wenjie Ma
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper using fixed point methods we establish some existence theorems of positive (nontrivial) solutions for integral boundary value problems of nonlinear Hadamard fractional differential equations.


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## Body

## 1. Introduction

In this work we study the following integral boundary value problems of nonlinear Hadamard fractional differential equations(1)DβφpDαut=ft,ut,1<t<e,u1=u′1=u′e=0,Dαu1=0,φpDαue=μ∫1eφpDαutdtt,where α, β, and μ are three positive real numbers with α∈(2,3], β∈(1,2], and μ∈[0,β), φp(s)=sp-2s is the p-Laplacian for p>1,s∈R, and f is a continuous function on [1,e]×R. Moreover, let φp-1=φq with 1/p+1/q=1. In what follows, we offer some related definitions and lemmas for Hadamard fractional calculus.Definition 1 (see [1, Page 111]).
Theαth Hadamard fractional order derivative of a function y:[1,+∞)→R is defined by(2)Dαyt=1Γn-αtddtn∫1tlog⁡tsn-α-1ysdss,where α>0, n=[α]+1, and [α] denotes the largest integer which is less than or equal to α. Moreover, we here also offer the αth Hadamard fractional order integral of y:[1,+∞)→R which is defined by(3)Iαyt=1Γα∫1tlog⁡tsα-1ysdss,where Γ is the gamma function.Lemma 2 (see [1, Theorem 2.3]).
Letα>0, n=[α]+1. Then(4)IαDαyt=yt+c1log⁡tα-1+c2log⁡tα-2+⋯+cnlog⁡tα-n,where ci∈R,i=1,2,…,n.In recent years, there have been some significant developments in the study of boundary value problems for nonlinear fractional differential equations; we refer to [2–11] and the references therein. For more related works, see also [12–49]. For example, by using monotone iterative methods, Wang et al. [3] investigated a class of boundary value problems of Hadamard fractional differential equations involving nonlocal multipoint discrete and Hadamard integral boundary conditions and established monotone iterative sequences, which can converge to the unique positive solution of their problems. Similar methods are also applied in [4, 5, 12–15].For differential equations with thep-Laplacian, see, for example, [6, 7, 15–20] and the references therein. In [6], Wang considered the nonlinear Hadamard fractional differential equation with integral boundary condition and p-Laplacian operator(5)DβφpDαut=ft,ut,t∈1,T,uT=λIσuη,Dαu1=0,u1=0,where f grows (p-1)–sublinearly at +∞, and by using the Schauder fixed point theorem, a solution existence result is obtained. In [7], Li and Lin used the Guo-Krasnosel’skii fixed point theorem to obtain the existence and uniqueness of positive solutions for (1) with μ=0.However, we note that these are seldom considered Hadamard fractional differential equations with thep-Laplacian in the literature; in this paper we are devoted to this direction. We first utilize the Guo-Krasnosel’skii fixed point theorem to obtain two positive solutions existence theorems when f grows (p-1)–superlinearly and (p-1)–sublinearly with the p-Laplacian, and secondly by using the fixed point index, we obtain a nontrivial solution existence theorem without the p-Laplacian, but the nonlinearity can allow being sign-changing and unbounded from below. This improves and generalizes some semipositone problems [21–31].


## 2. Preliminaries

In this section, we first calculate Green’s functions associated with (1) and then transform the boundary value problem into its integral form. For this, we give the following lemma.Lemma 3.
Letα, β, μ, φp, and Dα, Dβ be as in (1). Then (1) can take the integral form(6)ut=∫1eGt,sφq∫1eHs,τfτ,uτdττdss,fort∈1,e,where(7)Gt,s=1Γαlog⁡tα-11-log⁡sα-2-log⁡t-log⁡sα-1,1≤s≤t≤e,log⁡tα-11-log⁡sα-2,1≤t≤s≤e,and(8)Ht,τ=H1t,τ+μβ-μΓβlog⁡tβ-1log⁡τ1-log⁡τβ-1,fort,τ∈1,e,H1t,τ=1Γβlog⁡tβ-11-log⁡τβ-1-log⁡t-log⁡τβ-1,1≤τ≤t≤e,log⁡tβ-11-log⁡τβ-1,1≤t≤τ≤e.Proof.
Usey(t) to replace f(t,u) in (1). Let Dβ(φp(Dαu(t)))=y(t). Then from Lemma 2 we have (9)φpDαut=Iβyt+c1log⁡tβ-1+c2log⁡tβ-2,forci∈R,i=1,2.Note that Dαu(1)=0 implies φpDαu1=0, and then c2=0. Therefore, we obtain(10)φpDαut=Iβyt+c1log⁡tβ-1.Next, we calculate φp(Dαu(e)) and μ∫1eφp(Dαu(t))(dt/t):(11)φpDαue=Iβye+c1=c1+1Γβ∫1e1-log⁡τβ-1yτdττ,and(12)μ∫1eφpDαutdtt=μ∫1eIβytdtt+μc1∫1elog⁡tβ-1dtt=μc1β+μΓβ∫1e∫1tlog⁡t-log⁡τβ-1yτdττdtt.The condition φp(Dαu(e))=μ∫1eφp(Dαu(t))(dt/t) enables us to obtain(13)c1=μββ-μΓβ∫1e∫1tlog⁡t-log⁡τβ-1yτdττdtt-ββ-μΓβ∫1e1-log⁡τβ-1yτdττ.Substituting c1 into (10) gives(14)
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.Note that -φp(Dαu(t))=φp(-Dαu(t)), and hence we obtain(15)-Dαut=φq∫1eHt,τyτdττ,forα∈2,3,t∈1,e.Then, if we let x(t)=φq∫1eH(t,τ)y(τ)(dτ/τ),t∈[1,e], from Lemma 2 we obtain(16)ut=-Iαxt+c1log⁡tα-1+c2log⁡tα-2+c3log⁡tα-3,forci∈R,i=1,2,3.The condition u(1)=u′(1)=0 implies that c2=c3=0. Then we substitute e into the first derivative of u, and we calculate c1 as follows:(17)c1=1Γα∫1e1-log⁡sα-2xsdss.As a result, from (16) we have(18)ut=-1Γα∫1tlog⁡t-log⁡sα-1xsdss+1Γα∫1elog⁡tα-11-log⁡sα-2xsdss=∫1eGt,sφq∫1eHs,τyτdττdss,fort∈1,e.This completes the proof.Lemma 4.
Green’s functionsG,H defined by (7) and (8) have the following properties:
(i)G,H are continuous, nonnegative functions on [1,e]×[1,e],
(ii)log⁡tα-11-log⁡sα-2-1-log⁡sα-1≤Γ(α)G(t,s)≤1-log⁡sα-2-1-log⁡sα-1, for t,s∈[1,e].
From [7, Lemma 7] and [8, Lemma 2.2] we easily obtain this lemma, so we omit its proof.
Let(19)G1t,s=∫1eGt,τHτ,sdττ,ϕs=1Γα∫1e1-log⁡tα-2-1-log⁡tα-1Ht,sdtt,fort,s∈1,e.Then we obtain the following lemma.Lemma 5.
There existκ1=∫1elog⁡tα-1ϕ(t)(dt/t), κ2=∫1eϕ(t)(dt/t) such that(20)κ1ϕs≤∫1eG1t,sϕtdtt≤κ2ϕs,fors∈1,e.Proof.
We only prove the left inequality above. From Lemma4(ii) we have(21)∫1eG1t,sϕtdtt=∫1e∫1eGt,τHτ,sdττϕtdtt≥1Γα∫1e∫1elog⁡tα-11-log⁡τα-2-1-log⁡τα-1Hτ,sdττϕtdtt=κ1ϕs.This completes the proof.LetE=C[1,e] be the Banach space equipped with the norm u=maxt∈[1,e]ut. Then we define two sets on E as follows:(22)P=u∈E:ut≥0,∀t∈1,e,P0=u∈E:ut≥log⁡tα-1u,∀t∈1,e.Consequently, P,P0 are cones on E. From Lemma 3 we can define an operator A on E as follows:(23)Aut=∫1eGt,sφq∫1eHs,τfτ,uτdττdss,foru∈E,t∈1,e.The continuity of G,H,f implies that A:E→E is a completely continuous operator and the existence of solutions for (1) if and only if the existence of fixed points for A.Lemma 6 (see [50]).
LetE be a Banach space and Ω a bounded open set in E. Suppose that A:Ω→E is a continuous compact operator. If there exists u0∈E∖0 such that(24)u-Au≠μu0,∀u∈∂Ω,μ≥0,then the topological degree deg⁡(I-A,Ω,0)=0.Lemma 7 (see [50]).
LetE be a Banach space and Ω a bounded open set in E with 0∈Ω. Suppose that A:Ω→E is a continuous compact operator. If(25)Au≠μu,∀u∈∂Ω,μ≥1,then the topological degree deg⁡(I-A,Ω,0)=1.Lemma 8 (see [50]).
LetE be a Banach space and P⊂E a cone in E. Assume that Ω1,Ω2 are open subsets of E with 0∈Ω1⊂Ω¯1⊂Ω2, and let A:P∩(Ω¯2∖Ω1)→P be a completely continuous operator such that either
(G1)Au≤u, u∈∂Ω1∩P, and Au≥u, u∈∂Ω2∩P,
or
(G2)Au≥u, u∈∂Ω1∩P, and Au≤u, u∈∂Ω2∩P.
ThenA has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Positive Solutions for (1)

LetBϱ≔{u∈E:u<ϱ} for ϱ>0. Now, we first list our assumptions on f:(H1)f∈C([0,1]×R+,R+),(H2) there existδ1∈(1,e), t0∈(1,e) such that lim⁡infu→+∞(f(t,u)/φp(u))≥φp(N1), lim⁡infu→0+(f(t,u)/φp(u))≥φp(N2), uniformly on t∈[δ1,e], where 2N1-1,N2-1∈0,log⁡δ1α-1∫1eG(t0,s)φq∫δ1eHs,τdτ/τ(ds/s),(H3) there existsρ1>0 such that f(t,u)≤φp(N3ρ1),∀u∈[0,ρ1], t∈[1,e], where N3-1>∫1eG(e,s)φq∫1eHs,τdτ/τ(ds/s),(H4)lim⁡supu→+∞(ft,u/φpu)≤φp(M1), lim⁡supu→0+(f(t,u)/φp(u))≤φp(M2), uniformly on t∈[1,e], where 2M1-1,M2-1>∫1eG(e,s)φq∫1eHs,τdτ/τ(ds/s),(H5) there existρ2>0, δ1∈(1,e), t0∈(1,e) such that f(t,u)≥φp(M3ρ2),∀u∈[log⁡δ1α-1ρ2,ρ2],t∈[δ1,e], where(26)M3-1∈0,∫1eGt0,sφq∫δ1eHs,τdττdss.Lemma 9.
Suppose that (H1) holds. ThenA(P)⊂P0.Proof.
Ifu∈P, from Lemma 4 we have(27)Aut≤1Γα∫1e1-log⁡sα-2-1-log⁡sα-1φq∫1eHs,τfτ,uτdττdss,∀t∈1,e.On the other hand,(28)Aut≥log⁡tα-1·1Γα∫1e1-log⁡sα-2-1-log⁡sα-1φq∫1eHs,τfτ,uτdττdss≥log⁡tα-1Au,∀t∈1,e.This completes the proof.Remark 10.
Our aim is to find operator equationu=Au has fixed points in P, and from Lemma 9, these fixed points must belong to the cone P0. Therefore, our work space can be chosen P0 rather than P.In what follows, we discuss the existence of positive solutions for (1) in P0.Theorem 11.
Suppose that (H1)-(H3) hold. Then (1) has at least two positive solutions.Proof.
From (H3), whenu∈∂Bρ1∩P0, we have(29)Aut≤maxt∈1,e⁡∫1eGt,sφq∫1eHs,τfτ,uτdττdss≤∫1eGe,sφq∫1eHs,τφpN3ρ1dττdss=N3ρ1∫1eGe,sφq∫1eHs,τdττdss<ρ1,∀t∈1,e.Hence, we obtain(30)Au<u,foru∈∂Bρ1∩P0.On the other hand, by the second limit inequality in (H2), there exists r1∈(0,ρ1) such that(31)ft,u≥φpN2u,∀u∈0,r1,t∈δ1,e.Note that if u∈∂Br1∩P0,t∈[δ1,e], from the definition of P0 we have(32)ut≥log⁡δ1α-1u.This, together with (31), implies that(33)Au=maxt∈1,e⁡Aut≥Aut0=∫1eGt0,sφq∫1eHs,τfτ,uτdττdss≥∫1eGt0,sφq∫δ1eHs,τφpN2log⁡δ1α-1udττdss=N2log⁡δ1α-1u∫1eGt0,sφq∫δ1eHs,τdττdss>u,foru∈∂Br1∩P0.By the first limit inequality in (H2), there exist R1>ρ1 and C1>0 such that(34)ft,u≥φpN1u-C1,∀u∈R+,t∈δ1,e.Note that R1 can be chosen large enough, and if u∈∂BR1∩P0, together with (32), there exists C2>0 such that(35)ft,u≥φpN1log⁡δ1α-1R1-C2,∀t∈δ1,e.Combining this and (33), we find(36)Au≥∫1eGt0,sφq∫δ1eHs,τφpN1log⁡δ1α-1R1-C2dττdss=N1log⁡δ1α-1R1-C2∫1eGt0,sφq∫δ1eHs,τdττdss≥2R1-C3,where C3=C2∫1eG(t0,s)φq∫δ1eHs,τdτ/τ(ds/s). Consequently, we have(37)Au>u,for∂BR1∩P0,ifu→∞.In summary, from (30), (33), and (37) with R1>ρ1>r1, Lemma 8 enables us to obtain that (1) has at least two positive solutions in (B¯R1∖Bρ1)∩P0 and (B¯ρ1∖Br1)∩P0. This completes the proof.Theorem 12.
Suppose that (H1), (H4)-(H5) hold. Then (1) has at least two positive solutions.Proof.
Ifu∈∂Bρ2∩P0, we have u=ρ2, and u∈[log⁡δ1α-1ρ2,ρ2],foru∈P0,t∈[δ1,e]. Hence, from (H5) we obtain(38)Au≥∫1eGt0,sφq∫1eHs,τfτ,uτdττdss≥∫1eGt0,sφq∫δ1eHs,τφpM3ρ2dττdss≥M3ρ2∫1eGt0,sφq∫δ1eHs,τdττdss>ρ2.This indicates that(39)Au>u,foru∈∂Bρ2∩P0.On the other hand, by the second limit inequality in (H4), there exists r2∈(0,ρ2) such that(40)ft,u≤φpM2u,∀u∈0,r2,t∈1,e.This, if u∈∂Br2∩P0, implies that(41)Au≤∫1eGe,sφq∫1eHs,τφpM2uτdττdss≤∫1eGe,sφq∫1eHs,τφpM2r2dττdss=M2r2∫1eGe,sφq∫1eHs,τdττdss<r2.This gives(42)Au<u,foru∈∂Br2∩P0.By the first limit inequality in (H4), there exist R2>ρ2 and C4>0 such that(43)ft,u≤φpM1u+C4,∀u∈R+,t∈1,e.Consequently, if u∈∂BR2∩P0 with R2 large enough, we obtain(44)Au≤∫1eGe,sφq∫1eHs,τφpM1R2+C4dττdss=M1R2+C4∫1eGe,sφq∫1eHs,τdττdss≤12R2+C5,where C5=C4∫1eG(e,s)φq∫1eHs,τdτ/τ(ds/s). Hence, we have(45)Au<u,foru∈∂BR2∩P0,ifu→∞.In a word, from (39), (42), and (45) with R2>ρ2>r2, Lemma 8 enables us to obtain that (1) has at least two positive solutions in (B¯R2∖Bρ2)∩P0 and (B¯ρ2∖Br2)∩P0. This completes the proof.Example 13.
Let(46)ft,u=ρ1p-1-γ1N3p-1uγ1,u∈ρ1,+∞,t∈1,e,ρ1p-1-γ2N3p-1uγ2,u∈0,ρ1,t∈1,e,where γ1∈(p-1,+∞), γ2∈(0,p-1), and N3,ρ1 are defined by (H3). Then(47)lim⁡infu→+∞ft,uφpu=lim⁡infu→+∞ρ1p-1-γ1N3p-1uγ1up-1=+∞≥φpN1,lim⁡infu→0+ft,uφpu=lim⁡infu→0+ρ1p-1-γ2N3p-1uγ2up-1=+∞≥φpN2.Moreover, for u∈[0,ρ1], t∈[1,e] we have(48)ft,u≤ρ1p-1-γ2N3p-1ρ1γ2=N3ρ1p-1.Therefore, (H1)-(H3) hold.Example 14.
Let(49)ft,u=log⁡δ1-α-1γ3ρ2p-1-γ3M3p-1uγ3,u∈log⁡δ1α-1ρ2,+∞,t∈1,e,log⁡δ1-α-1γ4ρ2p-1-γ4M3p-1uγ4,u∈0,log⁡δ1α-1ρ2,t∈1,e,where γ3∈(0,p-1), γ4∈(p-1,+∞), and M3,ρ2 are defined by (H5). Then(50)
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.Moreover, for u∈[log⁡δ1α-1ρ2,ρ2],t∈[δ1,e] we have(51)ft,u≥log⁡δ1-α-1γ3ρ2p-1-γ3M3p-1uγ3=M3ρ2p-1.Therefore, (H1), (H4)-(H5) hold.


## 4. Nontrivial Solutions for (1)

In this section we consider the boundary value problem (1) without the p-Laplacian, i.e., p=2. In this case, (1) can be transformed into its integral form as follows:(52)ut=∫1eGt,s∫1eHs,τfτ,uτdττdss=∫1eG1t,sfs,usdss,fort∈1,e.As said in Section 3, we define an operator, still denoted by A, as follows:(53)Aut=∫1eG1t,sfs,usdss,foru∈E,t∈1,e.In what follows, we aim to find the existence of fixed points of A. For this, we list our assumptions on f:(H6)f∈C([1,e]×R,R),(H7) There exist nonnegative functionsa(t),b(t)∈E with b≢0 and K(u)∈C(R,R+) such that(54)ft,u≥-at-btKu,∀u∈R,t∈1,e.Moreover,(55)limu→∞⁡Kuu=0,(H8)liminfu→∞(f(t,u)/u)>κ1-1, uniformly in t∈[1,e],(H9)liminfu→0(ft,u/u)<κ2-1, uniformly in t∈[1,e].Theorem 15.
Suppose that (H6)-(H9) hold. Then (1) has at least one nontrivial solution.Proof.
From (H9) there existε3∈(0,κ2-1) and r3>0 such that(56)ft,u≤κ2-1-ε3u,∀t∈1,e,u∈0,r3.For this r3, we show that(57)Au≠μu,u∈∂Br3,μ≥1.If otherwise, there exist u1∈∂Br3, μ1≥1 such that(58)Au1=μ1u1,and hence, we obtain(59)u1t=1μ1Au1t≤Au1t≤∫1eG1t,sfs,u1sdss≤κ2-1-ε3∫1eG1t,su1sdss.Multiply both sides of the above inequality by ϕ(t) and integrate from 1 to e and together with Lemma 5 we obtain(60)∫1eu1tϕtdtt≤κ2-1-ε3∫1e∫1eG1t,su1sdssϕtdtt≤κ2-1-ε3κ2∫1eu1tϕtdtt.This implies that ∫1eu1tϕ(t)(dt/t)=0, and u1≡0 for the fact that ϕ(t)≢0, for t∈[1,e], which contradicts u1∈∂Br3. Therefore, (57) is true, and from Lemma 7 we obtain(61)deg⁡I-A,Br3,0=1.On the other hand, by (H8), there exist ε4>0 and X0>0 such that(62)ft,u≥κ1-1+ε4u,∀t∈1,e,u>X0.For every fixed ϵ with bϵ∈(0,ε4), b=maxt∈[1,e]|b(t)|, and from (H7), there exists X1>X0 such that(63)Ku≤ϵu,∀u>X1.Combining the two inequalities above, (H7) enables us to find(64)ft,u≥κ1-1+ε4u-at-btKu≥κ1-1+ε4u-at-ϵbtu≥κ1-1+ε4-bϵu-at,∀u>X1,t∈1,e.If we take C6=(κ1-1+ε4-bϵ)X1+maxt∈1,e,u≤X1ft,u, K∗=maxu≤X1K(u). Then we easily have(65)ft,u≥κ1-1+ε4-bϵu-at-C6,∀u∈R,t∈1,e.Note that ϵ can be chosen arbitrarily small, and we let(66)R3≥maxκ1-1+2ε4-bϵ∫1eWsas+bK∗+C6ds/sε4-bϵΓα-bϵκ1-1+2ε4-bϵ∫1eWsds/s,∫1eWsas+bK∗+C6ds/sΓα-bϵ∫1eWsds/s,where W(s)=∫1e1-log⁡τα-2H(τ,s)(dτ/τ), for s∈[1,e]. Now we prove that(67)u-Au≠μϕ,∀u∈∂BR3,μ≥0,where ϕ is defined by (19). Indeed, if (67) is not true, then there exists u2∈∂BR3 and μ0>0 such that(68)u2-Au2=μ0ϕ.Let u~(t)=∫1eG1(t,s)[a(s)+b(s)K(u2(s))+C6](ds/s). Then u~∈P0 and(69)u~t=∫1eG1t,sas+bsKu2s+C6dss≤∫1e∫1e1Γαlog⁡tα-11-log⁡τα-2Hτ,sdττas+bsKu2s+C6dss≤1Γαlog⁡tα-1∫1e∫1e1-log⁡τα-2Hτ,sdττas+bsKu2s+C6dss=1Γαlog⁡tα-1∫1eWsas+bsKu2s+C6dss.Consequently, we have(70)u~≤1Γα∫1eWsas+bsKu2s+C6dss≤1Γα∫1eWsas+C6dss+bΓα∫u2≤X1WsKu2sdss+∫u2>X1WsKu2sdss≤1Γα∫1eWsas+bK∗+C6dss+bϵΓα∫u2>X1Wsu2sdss≤1Γα∫1eWsas+bK∗+bϵR3+C6dss.Plus u~ into (68) gives(71)u2t+u~t=Au2t+u~t+μ0ϕt=∫1eG1t,sfs,u2s+as+bsKu2s+C6dss+μ0ϕt.Note that f(s,u2(s))+a(s)+b(s)K(u2(s))+C6∈P,s∈[1,e] and ϕ∈P0. Lemma 9 enables us to know that u2+u~∈P0. From (65) we have(72)Au2t+u~t=∫1eG1t,sfs,u2s+as+bsKu2s+C6dss≥∫1eG1t,sfs,u2s+as+C6dss≥∫1eG1t,sκ1-1+ε4-bϵu2sdss≥∫1eG1t,sκ1-1+ε4-bϵu2sdss.On the other hand, we have(73)κ1-1∫1eG1t,su2s+u~sdss+ε4-bϵ∫1eG1t,su2sdss-κ1-1∫1eG1t,su~sdss≥κ1-1∫1eG1t,su2s+u~sdss.This inequality holds if(74)ε4-bϵ∫1eG1t,su2sdss-κ1-1∫1eG1t,su~sdss≥0.Indeed, u2+u~∈P0 implies that u2(t)+u~(t)≥log⁡tα-1u2+u~≥log⁡tα-1(u2-u~), for t∈[1,e]. Consequently,(75)ε4-bϵ∫1eG1t,su2s+u~sdss-κ1-1+ε4-bϵ∫1eG1t,su~sdss≥ε4-bϵR3-u~∫1eG1t,slog⁡sα-1dss-κ1-1+ε4-bϵΓα∫1eWsas+bK∗+bϵR3+C6dss∫1eG1t,slog⁡sα-1dss≥0.As a result, we have(76)Au2t+u~t≥κ1-1∫1eG1t,su2s+u~sdss≔κ1-1Tu2+u~t,∀t∈1,e,where (Tu)(t)=∫1eG1(t,s)u(s)(ds/s), for u∈E, t∈[1,e]. Using (68) we obtain(77)u2+u~=Au2+u~+μ0ϕ≥κ1-1Tu2+u~+μ0ϕ≥μ0ϕ.Define(78)μ∗=sup⁡μ>0:u2+u~≥μϕ.Note that μ0∈μ>0:u2+u~≥μϕ, and then μ∗≥μ0, u2+u~≥μ∗ϕ. From Lemma 5 we have(79)κ1-1Tu2+u~≥μ∗κ1-1Tϕ≥μ∗ϕ,and hence(80)u2+u~≥κ1-1Tu2+u~+μ0ϕ≥μ0+μ∗ϕ,which contradicts the definition of μ∗. Therefore, (67) holds, and from Lemma 6 we obtain(81)deg⁡I-A,BR3,0=0.This, together with (61), implies that(82)deg⁡I-A,BR3∖B¯r3,0=deg⁡I-A,BR3,0-deg⁡I-A,Br3,0=-1.Therefore the operator A has at least one fixed point in BR3∖B¯r3, and (1) has at least one nontrivial solution. This completes the proof.Example 16.
Letf(t,u)=au-bk(u), k(u)=ln⁡(u+1), u∈R,t∈[1,e], where a∈(κ1-1,+∞) and b∈(a,a+κ2-1). Then limu→+∞(k(u)/u)=0, and limu→+∞((au-bln⁡(u+1))/u)=a>κ1-1, limu→0(au-bln⁡u+1/u)=a-b<κ2-1. Therefore, (H6)-(H9) hold.



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*Source: 2193234-2018-11-01.xml*