# Nontrivial Solutions for a Higher Order Nonlinear Fractional Boundary Value Problem Involving Riemann-Liouville Fractional Derivatives

Authors: Keyu Zhang; Donal O’Regan; Jiafa Xu; Zhengqing Fu
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper using topological degree we study the existence of nontrivial solutions for a higher order nonlinear fractional boundary value problem involving Riemann-Liouville fractional derivatives. Here, the nonlinearity can be sign-changing and can also depend on the derivatives of unknown functions.


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## Body

## 1. Introduction

In this paper, we investigate the existence of nontrivial solutions for the higher order nonlinear fractional boundary value problem involving Riemann-Liouville fractional derivatives:(1)D0+αut=-ft,ut,D0+β1ut,D0+β2ut,…,D0+βn-1ut,0<t<1,u0=u′0=⋯=un-20=D0+βu1=0,where n-1<α≤n,i-1<βi≤i,α-βn-1>α-β>1, D0+α,D0+β,D0+βi are the Riemann-Liouville fractional derivatives, i=1,2,…,n-1, and the function f:[0,1]×Rn→R is continuous; here R≔(-∞,+∞).There are a large number of papers in the literature studying fractional differential equations because of applications in various disciplines of science and engineering such as mechanics, electricity, chemistry, and control theory; we refer the reader to [1–35] and the references therein. In [1], using fixed point index theory in cones, the authors studied the existence and multiplicity of positive solutions for the nonlocal fractional differential equation boundary value problem(2)D0+αut+qtft,ut=0,0<t<1,n-1<α≤n,u0=u′0=⋯=un-20=0,u1=∫01usdAs,where f(t,u) is a nonnegative continuous function on [0,1]×R+ and satisfies the conditions:( H 1 ) W a n g sublinear growth case: liminfu→0+f(t,u)/u>λ1,limsupu→+∞f(t,u)/u<λ1, uniformly on t∈[0,1],( H 2 ) W a n g superlinear growth case: liminfu→+∞f(t,u)/u>λ1,limsupu→0+f(t,u)/u<λ1, uniformly on t∈[0,1],whereλ1 is the first eigenvalue for the problem:(3)D0+αut+λqtut=0,0<t<1,u0=u′0=⋯=un-20=0u1=∫01usdAs. Conditions similar to (H1)Wang and (H2)Wang can be found in [2–9]. In [2], the authors studied the existence of positive solutions for the nonlinear fractional differential equation:(4)-D0+αyt=ft,yt,y′t,…,yn-2t,0<t<1,y0=y′0=⋯=yn-20=0,yn-21=λyn-2, where λ[y(n-2)]=∫01y(n-2)(t)dA(t) is a linear functional on C[0,1] given by a Stieltjes integral. The nonlinearity f(t,z1,z2,…,zn-1) satisfies the conditions:(5)liminfz1→0,…,zn-1→0ft,z1,z2,…,zn-1z1+⋯+zn-1>λ1,limsupz1+⋯+zn-1→+∞,zn-1→+∞ft,z1,z2,…,zn-1zn-1<λ1,uniformly  ont∈0,1,and(6)liminfz1→0,…,zn-1→0ft,z1,z2,…,zn-1zn-1<λ1,limsupz1+⋯+zn-1→+∞ft,z1,z2,…,zn-1z1+⋯+zn-1>λ1,uniformly  ont∈0,1.In [10–14], the authors used Lipschitz conditions to study the existence and uniqueness of solutions for various fractional differential equations; for example, in [10], the author used the u0-positive operator method to obtain the unique solution for the fractional boundary value problem:(7)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. In [15], the authors used the mixed monotone method to obtain a unique positive solution for the fractional boundary value problem:(8)-D0+αyt=qtft,yt,y′t,…,yn-2t,0<t<1,y0=y′0=⋯=yn-20=yn-21=0, where the nonlinearity f=g+h and g,h have different monotone properties. In [16, 17], the authors extended the methods in [15] to fractional p-Laplacian equations and established some appropriate iterative sequences which converge to their solutions.There are also some papers in the literature devoted to sign-changing nonlinearities; we refer the reader to [7–14, 18, 19, 22, 23, 26–33, 35]. For example, in [18] using coincidence degree theory, the author investigated the existence of nontrivial solutions for the coupled system of nonlinear fractional differential equations:(9)D0+αut=ft,vt,D0+β-2vt,D0+β-1vt,0<t<1,D0+βvt=gt,ut,D0+α-2ut,D0+α-1ut,0<t<1,u0=0,D0+α-1u0=D0+α-1uη,u1=∑i=1m1αiuηi,v0=0,D0+β-1v0=D0+β-1vξ,v1=∑i=1m2βivξi,where f,g:[0,1]×R3→R satisfy Carathéodory conditions.Motivated by the aforementioned papers and some integer order equations [36–39], in this paper we use topological degree to study the existence of nontrivial solutions for the higher order nonlinear fractional boundary value problem (1). Our nonlinearity f can be sign-changing and can also depend on the derivatives of unknown functions. Moreover, our conditions improve the conditions in (5) and (6); see (H4)-(H5) in Section 3.


## 2. Preliminaries

We present some basic material involving Riemann-Liouville fractional derivatives; for details we refer the reader to the books [40–42].Definition 1 (see [40–42]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function f:(0,+∞)→(-∞,+∞) is given by (10)D0+αft=1Γn-αddtn∫0tt-sn-α-1fsds,where n=[α]+1, [α] denotes the integer part of number α, provided that the right side is pointwise defined on (0,+∞).Definition 2 (see [40–42]).
The Riemann-Liouville fractional integral of orderα>0 of a function f:(0,+∞)→(-∞,+∞) is given by(11)I0+αft=1Γα∫0tt-sα-1fsds, provided that the right side is pointwise defined on (0,+∞).Lemma 3 (see [3, Lemma  2.2]).
Letα>0, then for u,D0+αu∈C(0,1)∩L(0,1), we have(12)I0+αD0+αut=ut+c1tα-1+c2tα-2+…+cNtα-N,for  someci∈R,i=1,2,…,N, where N is the smallest integer greater than or equal to α.In what follows, we present Green’s function associated with problem (1). The following lemma is from [3]; for completeness we provide the proof.Lemma 4 (see [3, Lemma  2.3]).
Letv(t)=D0+βn-1u(t). Then (1) can be transformed into the following nonlinear integrodifferential equation:(13)D0+α-βn-1vt=-ft,I0+βn-1vt,I0+βn-1-β1vt,…,I0+βn-1-βn-2vt,vt,0<t<1,I0+βn-1-n+2v0=D0+β-βn-1v1=0.Consequently, we obtain that (13) is equivalent to the following Hammerstein type integral equation:(14)vt=∫01K1t,sfs,I0+βn-1vs,I0+βn-1-β1vs,…,I0+βn-1-βn-2vs,vsds,where(15)K1t,s=1Γα-βn-1tα-βn-1-11-sα-β-1-t-sα-βn-1-1,0≤s≤t≤1,tα-βn-1-11-sα-β-1,0≤t≤s≤1.Proof.
Letv(t)=D0+βn-1u(t). Then from [4, equations (2.11)], we have(16)D0+αut=D0+α-βn-1vt,D0+β1ut=D0+βn-1-β1vt,D0+β2ut=D0+βn-1-β2vt,⋯,D0+βn-2ut=D0+βn-1-βn-2vt,D0+βn-1ut=vt. Also we obtain u(n-2)(t)=I0+βn-1-n+2v(t), and D0+βu(t)=D0+β-βn-1v(t). Hence, by u(t)=I0+βn-1v(t), we have (13). Consequently, Lemma 3 implies that (13) can be reduced to an equivalent integral equation:(17)vt=-I0+α-βn-1ft,I0+βn-1vt,I0+βn-1-β1vt,…,I0+βn-1-βn-2vt,vt+C1tα-βn-1-1+C2tα-βn-1-2, for some Ci∈R,i=1,2. From the condition I0+βn-1-n+2v(0)=0, we have C2=0. Then we obtain(18)D0+β-βn-1vt=-1Γα-β∫0tt-sα-β-1f~sds+C1Γα-βn-1Γα-βtα-β-1, where we use f~(t) to replace f(t,·,···,·). Then D0+β-βn-1v(1)=0 solves the above equation, and we obtain(19)C1=1Γα-βn-1∫011-sα-β-1f~sds. Therefore, we have(20)vt=-I0+α-βn-1f~t+1Γα-βn-1∫01tα-βn-1-11-sα-β-1f~sds=-1Γα-βn-1∫0tt-sα-βn-1-1f~sds+1Γα-βn-1∫01tα-βn-1-11-sα-β-1f~sds=∫01K1t,sf~sds.This completes the proof.Lemma 5 (see [3, Lemma  2.4]).
The functionK1 has the following properties:
(i)K1:[0,1]×[0,1]→R+ is continuous,
(ii) for all(t,s)∈[0,1]×[0,1], tα-βn-1-1K1(1,s)≤K1(t,s)≤K1(1,s), where K1(1,s)=(1-s)α-β-1-(1-s)α-βn-1-1/Γ(α-βn-1).
LetE≔C[0,1],v≔maxt∈[0,1]|v(t)|,P≔{v∈E:v(t)⩾0,∀t∈[0,1]},P0={v∈P:v(t)≥tα-βn-1-1v,∀t∈[0,1]}. Then (E,·) is a real Banach space, and P,P0 are cones on E. From Lemma 4, we can define an operator A:E→E as follows:(21)Avt=∫01K1t,sfs,I0+βn-1vs,I0+βn-1-β1vs,…,I0+βn-1-βn-2vs,vsds,forv∈E, where K1 is denoted by (15). Note that our functions K1,f are continuous, so the operator A is a completely continuous operator. Moreover if there is a v¯∈E as a fixed point of A, i.e., (13) has s solution v¯, then from v¯(t)=D0+βn-1u¯(t), we have that u¯ is a solution for (1). Therefore we study the existence of fixed points of the operator A. For n nonnegative real numbers μ1,μ2,…,μn which are not all zero, let(22)Lμ1,μ2,…,μnvt=∫01Kμ1,μ2,…,μnt,svsds,forv∈E, where Kμ1,μ2,…,μn(t,s)=μnKn(t,s)+μn-1Kn-1(t,s)+⋯+μ1K1(t,s) for (t,s)∈[0,1]×[0,1], and Ki(i=2,…,n) are(23)K2t,sK3t,s⋮Kn-1t,sKnt,s=1Γβn-1-βn-2∫0τK1t,ττ-sβn-1-βn-2-1dτ1Γβn-1-βn-3∫0τK1t,ττ-sβn-1-βn-3-1dτ⋮1Γβn-1-β1∫0τK1t,ττ-sβn-1-β1-1dτ1Γβn-1∫0τK1t,ττ-sβn-1-1dτ,fort,s∈0,1. Let r(Lμ1,μ2,…,μn) denote the spectral radius of Lμ1,μ2,…,μn, and from Gelfand’s theorem we can prove r(Lμ1,μ2,…,μn)>0 (the proof is standard; see [6, Lemma 5]).
Define an operatorA1:E→E as follows:(24)A1zt=∫01K1t,szsds,forz∈E. Then from Lemma 5(ii), we have(25)A1P⊂P0.Lemma 6 (see [43, Theorem  19.3]).
LetP be a reproducing cone in a real Banach space E and let L:E→E be a compact linear operator with L(P)⊂P. Let r(L) be the spectral radius of L. If r(L)>0, then there exists φ∈P∖{0} such that Lφ=r(L)φ.
Therefore, from Lemma6 there exists φμ1,μ2,…,μn∈P∖{0} such that(26)Lμ1,μ2,…,μnφμ1,μ2,…,μn=rLμ1,μ2,…,μnφμ1,μ2,…,μn.Hence, we obtain(27)φμ1,μ2,…,μnt=1rLμ1,μ2,…,μnLμ1,μ2,…,μnφμ1,μ2,…,μnt=1rLμ1,μ2,…,μn∫01Kμ1,μ2,…,μnt,sφμ1,μ2,…,μnsds=1rLμ1,μ2,…,μn∫01μnKnt,s+μn-1Kn-1t,s+⋯+μ1K1t,sφμ1,μ2,…,μnsds,t∈0,1.Using Lemma 5(ii) and the definitions of Ki(i=1,2,…,n), we have(28)φμ1,μ2,…,μn≤1rLμ1,μ2,…,μn∫01μnKn1,s+μn-1Kn-11,s+⋯+μ1K11,sφμ1,μ2,…,μnsds, and(29)φμ1,μ2,…,μnt≥tα-βn-1-1rLμ1,μ2,…,μn∫01μnKn1,s+μn-1Kn-11,s+⋯+μ1K11,sφμ1,μ2,…,μnsds≥tα-βn-1-1φμ1,μ2,…,μn. This implies(30)φμ1,μ2,…,μn∈P0.Lemma 7 (see [44]).
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(31)u-Au≠μu0,∀u∈∂Ω,μ⩾0, then the topological degree deg(I-A,Ω,0)=0.Lemma 8 (see [44]).
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(32)Au≠μu,∀u∈∂Ω,μ⩾1, then the topological degree deg⁡(I-A,Ω,0)=1.


## 3. Main Results

Letμ1,μ2,…,μn;μ~1,μ~2,…,μ~n be two families of nonnegative real numbers which are not all zero. Now we list our assumptions for f as follows:(H1)f:[0,1]×Rn→R is continuous,(H2) there existb(t),c(t)∈C[0,1] with b(t),c(t)≥0(≢0) and K(xn,xn-1,…,x1)∈C[Rn,R+] such that (33)ft,xn,xn-1,…,x1⩾-bt-ctKxn,xn-1,…,x1,∀xi∈R,t∈0,1,i=1,2,…,n,(H3)limμn|xn|+μn-1|xn-1|+⋯+μ1|x1|→+∞(K(xn,xn-1,…,x1)/( μ n x n + μ n - 1 x n - 1 + ⋯ + μ 1 x 1 ) ) = 0,(H4)liminfμn|xn|+μn-1|xn-1|+⋯+μ1|x1|→+∞(f(t,xn, xn-1, … , x1)/μnxn+μn-1xn-1+⋯+μ1x1)>λμ1,μ2,…,μn, uniformly for t∈[0,1],(H5)limsupμ~nxn+μ~n-1xn-1+⋯+μ~1x1→0(|(f(t,xn, xn-1, … , x1)|/μ~n|xn|+μ~n-1|xn-1|+⋯+μ~1|x1|)<λμ~1,μ~2,…,μ~n, uniformly for t∈[0,1], where(34)λμ1,μ2,…,μn=r-1Lμ1,μ2,…,μn,λμ~1,μ~2,…,μ~n=r-1Lμ~1,μ~2,…,μ~n.Theorem 9.
Assume that (H1)-(H5) are satisfied. Then (1) has at least one nontrivial solution.Proof.
From (H4), there existε0>0 and X0>0 such that(35)ft,xn,xn-1,…,x1⩾λμ1,μ2,…,μn+ε0μnxn+μn-1xn-1+⋯+μ1x1,for all t∈[0,1],μnxn+μn-1xn-1+⋯+μ1x1>X0.
For any givenε with ε0-cε>0, and using (H3) there exists X1>X0 such that(36)Kxn,xn-1,…,x1⩽εμnxn+μn-1xn-1+⋯+μ1x1,∀μnxn+μn-1xn-1+⋯+μ1x1>X1.Hence, (H2), (35), and (36) enable us to obtain (37)ft,xn,xn-1,…,x1⩾λμ1,μ2,…,μn+ε0μnxn+μn-1xn-1+⋯+μ1x1-bt-ctKxn,xn-1,…,x1⩾λμ1,μ2,…,μn+ε0μnxn+μn-1xn-1+⋯+μ1x1-bt-εctμnxn+μn-1xn-1+⋯+μ1x1⩾λμ1,μ2,…,μn+ε0-εctμnxn+μn-1xn-1+⋯+μ1x1-bt⩾λμ1,μ2,…,μn+ε0-εcμnxn+μn-1xn-1+⋯+μ1x1-b, for all t∈[0,1], μn|xn|+μn-1|xn-1|+⋯+μ1|x1|>X1.
Let(38)CX1=λμ1,μ2,…,μn+ε0-εcX1+max0⩽t⩽1,μnxn+μn-1xn-1+⋯+μ1x1⩽X1ft,xn,xn-1,…,x1,K∗=maxμnxn+μn-1xn-1+⋯+μ1x1⩽X1Kxn,xn-1,…,x1. Then we obtain(39)ft,xn,xn-1,…,x1⩾λμ1,μ2,…,μn+ε0-εcμnxn+μn-1xn-1+⋯+μ1x1-bt-CX1,for all (t,xn,xn-1,…,x1)∈[0,1]×Rn. Note that ε can be chosen arbitrarily small, and let (40)R>max⁡b+cK∗+CX1α-βΓα-βn-1-εMμ1,μ2,…,μnc,λμ1,μ2,…,μn+2ε0-2cεb+cK∗+CX1α-βΓα-βn-1ε0-cε-λμ1,μ2,…,μn+2ε0-2cεεcMμ1,μ2,…,μn/Γα-βn-1, where Mμ1,μ2,…,μn=∫01(1-s)α-β-1μn/Γ(βn-1)∫0s(s-τ)βn-1-1dτ+μn-1/Γ(βn-1-β1)∫0s(s-τ)βn-1-β1-1dτ+⋯+μ1ds.
Now we claim(41)v-Av≠μφμ1,μ2,…,μn,∀v∈∂BR,μ⩾0,where φμ1,μ2,…,μn is the positive eigenfunction of Lμ1,μ2,…,μn corresponding to the eigenvalue λμ1,μ2,…,μn, i.e., φμ1,μ2,…,μn=λμ1,μ2,…,μnLμ1,μ2,…,μnφμ1,μ2,…,μn, and from (30), φμ1,μ2,…,μn∈P0.
Suppose (41) is false. Then there exist v0∈∂BR and μ0>0 such that(42)v0-Av0=μ0φμ1,μ2,…,μn.Let(43)v~t=∫01K1t,sbs+csKI0+βn-1v0s,I0+βn-1-β1v0s,…,I0+βn-1-βn-2v0s,v0s+CX1ds,∀t∈0,1.Then we have(44)v~t⩽∫01K1t,sbs+csεμnI0+βn-1v0s+μn-1I0+βn-1-β1v0s+⋯+μ1v0s+K∗+CX1ds⩽tα-βn-1-1Γα-βn-1∫011-sα-β-1bs+csεμnΓβn-1∫0ss-τβn-1-1v0τdτ+μn-1Γβn-1-β1∫0ss-τβn-1-β1-1v0τdτ+⋯+μ1v0s+K∗+CX1ds≤tα-βn-1-1Γα-βn-1∫011-sα-β-1bs+csεμnΓβn-1∫0ss-τβn-1-1dτ+μn-1Γβn-1-β1∫0ss-τβn-1-β1-1dτ+⋯+μ1v0+K∗+CX1ds⩽tα-βn-1-1b+cK∗+CX1α-βΓα-βn-1+tα-βn-1-1Γα-βn-1εc∫011-sα-β-1μnΓβn-1∫0ss-τβn-1-1dτ+μn-1Γβn-1-β1∫0ss-τβn-1-β1-1dτ+⋯+μ1v0ds≤b+cK∗+CX1α-βΓα-βn-1+εcRΓα-βn-1∫011-sα-β-1μnΓβn-1∫0ss-τβn-1-1dτ+μn-1Γβn-1-β1∫0ss-τβn-1-β1-1dτ+⋯+μ1ds.Consequently, we have(45)v~⩽b+cK∗+CX1α-βΓα-βn-1+εcRΓα-βn-1Mμ1,μ2,…,μn<R. Note from (43) and (25), v~∈P0, and then φμ1,μ2,…,μn∈P0 implies that(46)v0t+v~t=Av0t+μ0φμ1,μ2,…,μnt+v~t=∫01K1t,sfs,I0+βn-1v0s,I0+βn-1-β1v0s,…,I0+βn-1-βn-2v0s,v0s+bs+csKI0+βn-1v0s,I0+βn-1-β1v0s,…,I0+βn-1-βn-2v0s,v0s+CX1ds+μ0φμ1,μ2,…,μnt. Then we have(47)v0+v~∈P0.Therefore, (26), (39), and (43) enable us to obtain (48)Av0t+v~t=∫01K1t,sfs,I0+βn-1v0s,I0+βn-1-β1v0s,…,I0+βn-1-βn-2v0s,v0sds+∫01K1t,sbs+csKI0+βn-1v0s,I0+βn-1-β1v0s,…,I0+βn-1-βn-2v0s,v0s+CX1ds⩾λμ1,μ2,…,μn+ε0-εc∫01K1t,sμnI0+βn-1v0s+μn-1I0+βn-1-β1v0s+⋯+μ1v0sds-∫01K1t,sbs+CX1ds+∫01K1t,sbs+csKI0+βn-1v0s,I0+βn-1-β1v0s,…,I0+βn-1-βn-2v0s,v0s+CX1ds⩾λμ1,μ2,…,μn+ε0-εc∫01K1t,sμnΓβn-1∫0ss-τβn-1-1v0τdτ+μn-1Γβn-1-β1∫0ss-τβn-1-β1-1v0τdτ+⋯+μ1v0sds⩾λμ1,μ2,…,μn+ε0-εcμnΓβn-1∫01K1t,s∫0ss-τβn-1-1v0τdτds+μn-1Γβn-1-β1∫01K1t,s∫0ss-τβn-1-β1-1v0τdτds+⋯+μ1∫01K1t,sv0sds⩾λμ1,μ2,…,μn+ε0-εcμn∫01Knt,τv0τdτ+μn-1∫01Kn-1t,τv0τdτ+⋯+μ1∫01K1t,sv0sds=λμ1,μ2,…,μn+ε0-εc∫01Kμ1,μ2,…,μnt,sv0sds⩾λμ1,μ2,…,μn+ε0-εc∫01Kμ1,μ2,…,μnt,sv0sds. Consequently, we obtain (49)λμ1,μ2,…,μn+ε0-cε∫01Kμ1,μ2,…,μnt,sv0sds=λμ1,μ2,…,μn∫01Kμ1,μ2,…,μnt,sv0s+v~sds+ε0-cε∫01Kμ1,μ2,…,μnt,sv0sds-λμ1,μ2,…,μn∫01Kμ1,μ2,…,μnt,sv~sds=λμ1,μ2,…,μnLμ1,μ2,…,μnv0+v~t+ε0-cε∫01Kμ1,μ2,…,μnt,sv0s+v~sds-λμ1,μ2,…,μn+ε0-cε∫01Kμ1,μ2,…,μnt,sv~sds.From (47), v0(t)+v~(t)⩾tα-βn-1-1v0+v~⩾tα-βn-1-1(v0-v~), t∈[0,1], and from (44), we have(50)ε0-cε∫01Kμ1,μ2,…,μnt,sv0s+v~sds-λμ1,μ2,…,μn+ε0-cε∫01Kμ1,μ2,…,μnt,sv~sds⩾ε0-cεR-v~∫01sα-βn-1-1Kμ1,μ2,…,μnt,sds-λμ1,μ2,…,μn+ε0-cε×b+cK∗+CX1α-βΓα-βn-1+εcRΓα-βn-1Mμ1,μ2,…,μn∫01sα-βn-1-1Kμ1,μ2,…,μnt,sds⩾0.Now from (48), (49), and (50), we find(51)Av0t+v~t⩾λμ1,μ2,…,μnLμ1,μ2,…,μnv0+v~t,t∈0,1.Therefore, using (42) and (51), we have (52)v0+v~=Av0+μ0φμ1,μ2,…,μn+v~⩾λμ1,μ2,…,μnLμ1,μ2,…,μnv0+v~+μ0φμ1,μ2,…,μn⩾μ0φμ1,μ2,…,μn. Define μ∗=sup⁡{μ>0:v0+v~⩾μφμ1,μ2,…,μn}. Then we have μ∗⩾μ0 and v0+v~⩾μ∗φμ1,μ2,…,μn. From φμ1,μ2,…,μn=λμ1,μ2,…,μnLμ1,μ2,…,μnφμ1,μ2,…,μn, we obtain (53)λμ1,μ2,…,μnLμ1,μ2,…,μnv0+v~⩾λμ1,μ2,…,μnLμ1,μ2,…,μnμ∗φμ1,μ2,…,μn=μ∗φμ1,μ2,…,μn. Hence (54)v0+v~⩾λμ1,μ2,…,μnLμ1,μ2,…,μnv0+v~+μ0φμ1,μ2,…,μn⩾μ0+μ∗φμ1,μ2,…,μn, which contradicts the definition of μ∗. Therefore, (41) holds, and from Lemma 7, we obtain(55)deg⁡I-A,BR,0=0.
From (H5), there exist0<ε1<λμ~1,μ~2,…,μ~n and 0<r<R such that (56)ft,xn,xn-1,…,x1⩽λμ~1,μ~2,…,μ~n-ε1μ~nxn+μ~n-1xn-1+⋯+μ~1x1, for all xi∈R,i=1,2,…,n,t∈[0,1] with 0⩽μ~n|xn|+μ~n-1|xn-1|+⋯+μ~1|x1|<r. Consequently, we obtain(57)Avt⩽λμ~1,μ~2,…,μ~n-ε1∫01K1t,sμ~nI0+βn-1vs+μ~n-1I0+βn-1-β1vs+⋯+μ~1vsds≤λμ~1,μ~2,…,μ~n-ε1∫01K1t,sμ~nΓβn-1∫0ss-τβn-1-1vτdτ+μ~n-1Γβn-1-β1∫0ss-τβn-1-β1-1vτdτ+⋯+μ~1vsds=λμ~1,μ~2,…,μ~n-ε1μ~n∫01Knt,τvτdτ+μ~n-1∫01Kn-1t,τvτdτ+⋯+μ~1∫01K1t,svsds=λμ~1,μ~2,…,μ~n-ε1∫01Kμ~1,μ~2,…,μ~nt,svsds=λμ~1,μ~2,…,μ~n-ε1Lμ~1,μ~2,…,μ~nvt,∀t∈0,1,v∈E,v⩽r. Now for this r, we claim(58)Av≠λv,∀v∈∂Br,λ⩾1.Suppose the contrary. Then there exist v0∈∂Br and λ0⩾1 such that Av0=λ0v0. Let ω(t)=v0t. Then ω∈∂Br∩P and (59)ω⩽1λ0λμ~1,μ~2,…,μ~n-ε1Lμ~1,μ~2,…,μ~nω⩽λμ~1,μ~2,…,μ~n-ε1Lμ~1,μ~2,…,μ~nω. By induction, we have ω⩽(λμ~1,μ~2,…,μ~n-ε1)mLμ~1,μ~2,…,μ~nmω, for m=1,2,⋯. As a result, we have (60)ω⩽λμ~1,μ~2,…,μ~n-ε1mLμ~1,μ~2,…,μ~nmω,1⩽λμ~1,μ~2,…,μ~n-ε1mLμ~1,μ~2,…,μ~nm. Therefore, by Gelfand’s theorem, we have (61)λμ~1,μ~2,…,μ~n-ε1rLμ~1,μ~2,…,μ~n=λμ~1,μ~2,…,μ~n-ε1limm→∞Lμ~1,μ~2,…,μ~nmm⩾1. This contradicts (62)λμ~1,μ~2,…,μ~n-ε1rLμ~1,μ~2,…,μ~n=1-ε1rLμ~1,μ~2,…,μ~n<1. Hence (58) holds, and from Lemma 8, we have(63)deg⁡I-A,Br,0=1.Now (55) and (63) together imply that (64)deg⁡I-A,BR∖B¯r,0=deg⁡I-A,BR,0-deg⁡I-A,Br,0=-1. Therefore the operator A has at least one fixed point in BR∖B¯r. Equivalently, (1) has at least one nontrivial solution. This completes the proof.Example 10.
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, where m∈N+, 0<ρ1<λμ1,μ2,…,μn, and ρi≥0,i=2,3,…,m, with ∑i=1m(-1)iρi≥0. Let b(t)=1+t2ln⁡2, c(t)=1+t2, and K(xn,xn-1,…,x1)=∑j=1nμjxj1/3ln∑j=1nμjxj+1. We now show that conditions (H1)-(H5) are satisfied. Indeed, for the function K, we obtain(66)limμnxn+μn-1xn-1+⋯+μ1x1→+∞⁡∑j=1nμjxj1/3ln⁡∑j=1nμjxj+1μnxn+μn-1xn-1+⋯+μ1x1≤limμnxn+μn-1xn-1+⋯+μ1x1→+∞⁡∑j=1nμjxj1/3ln⁡∑j=1nμjxj+1∑j=1nμjxj=0, and hence, (H3) holds. For the function f, it is clear that (H1)-(H2) hold. Also note we have(67)liminfμnxn+μn-1xn-1+⋯+μ1x1→+∞∑i=1mρi∑j=1nμjxji∑j=1nμjxj=+∞, and(68)limsupμnxn+μn-1xn-1+⋯+μ1x1→0∑i=1mρi∑j=1nμjxji∑j=1nμjxj=ρ1<λμ1,μ2,…,μn. Then (H4)-(H5) hold. Therefore, all the conditions in Theorem 9 are satisfied, and thus (1) has at least one nontrivial solution.Now, we consider a special case of this example. Letn=2, μ1=1,μ2=1, and consider the operators(69)Lμ1vt=∫01μ1K1t,svsds,Lμ2vt=∫01μ2K2t,svsds,forv∈E. Then for all m∈N, from Lemma 5(ii), we have(70)Lμ1m=maxt∈0,1∫01​⋯∫01​︸m⁡K1t,s1K1s1,s2⋯K1sm-1,smds1ds2⋯dsm≤∫01​⋯∫01​︸m⁡K11,s1K11,s2⋯K11,smds1ds2⋯dsm=∫01K11,sdsm, and so(71)limm→∞Lμ1mm≤∫01K11,sds. For all m∈N, from Lemma 5(ii), we have(72)Lμ1m=maxt∈0,1⁡∫01​⋯∫01​︸m⁡K1t,s1K1s1,s2⋯K1sm-1,smds1ds2⋯dsm≥maxt∈0,1⁡∫01​⋯∫01​︸m⁡tα-β1-1K11,s1s1α-β1-1K11,s2⋯sm-1α-β1-1K11,smds1ds2⋯dsm=∫01K11,sds∫01K11,ssα-β1-1dsm-1, and so(73)limm→∞Lμ1mm≥∫01K11,ssα-β1-1ds. From Gelfand’s theorem, we obtain that the spectral radius of Lμ1, denoted by r(Lμ1), satisfies(74)∫01K11,ssα-β1-1ds≤rLμ1≤∫01K11,sds.Note the definition ofK2, and from Lemma 5(ii), we have(75)1Γβn-1-βn-2∫0τtα-βn-1-1K11,ττ-sβn-1-βn-2-1dτ≤K2t,s=1Γβn-1-βn-2∫0τK1t,ττ-sβn-1-βn-2-1dτ≤1Γβn-1-βn-2∫0τK11,ττ-sβn-1-βn-2-1dτ. Thus tα-βn-1-1K2(1,s)≤K2(t,s)≤K2(1,s), for t,s∈[0,1]. Using a similar method, we have that the spectral radius of Lμ2, denoted by r(Lμ2), satisfies(76)∫01K21,ssα-β1-1ds≤rLμ2≤∫01K21,sds. Consequently, the spectral radius of the operator Lμ1,μ2 satisfies(77)∫01K11,ssα-β1-1ds+∫01K21,ssα-β1-1ds≤rLμ1,μ2≤∫01K11,sds+∫01K21,sds.Thus(78)1∫01K11,ssα-β1-1ds+∫01K21,ssα-β1-1ds≥λμ1,μ2≥1∫01K11,sds+∫01K21,sds.Now take ρ1∈0,1/∫01K1(1,s)ds+∫01K2(1,s)ds.



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*Source: 2381530-2019-06-19.xml*