# Existence of Solutions for Fractional Boundary Value Problems with a Quadratic Growth of Fractional Derivative

Authors: Yaning Li; Quanguo Zhang; Baoyan Sun
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this paper, we deal with two fractional boundary value problems which have linear growth and quadratic growth about the fractional derivative in the nonlinearity term. By using variational methods coupled with the iterative methods, we obtain the existence results of solutions. To the best of the authors’ knowledge, there are no results on the solutions to the fractional boundary problem which have quadratic growth about the fractional derivative in the nonlinearity term.


---

## Body

## 1. Introduction

It has been seen that fractional differential equations have better effects in many realistic applications than the classical ones. Qualitative theory and its applications in physics, engineering, economics, biology, and ecology are extensively discussed and demonstrated in [1–4] and the references therein. Some recent contributions to the theory of fractional differential equation can be seen in [5–10].Some classical tools such as fixed point theorems [5], the method of upper and lower solutions, and monotone iterative technique [11, 12] have been widely used to study the fractional differential equation. Recently, the study of fractional differential equations has attracted much attention by using variational methods, for example, [7, 8, 13–19]. We also mention that in the recent works [20, 21], the authors have developed a general approach concerting the existence of solutions.Fractional differential equations containing left and right fractional differential operators have received attention from scientists due to their applications in physical phenomena exhibiting anomalous diffusion. In [7], appropriate fractional derivative spaces were defined and existence and uniqueness results for a fractional boundary value problem were proven using the Lax-Milgram theorem.Jiao and Zhou [8] showed the variational structure of a fractional boundary value problem under an appropriate functional space; they used the least action principle and the Mountain Pass theorem to obtain the existence of at least one solution. Sun and Zhang [22] obtained the existence result for a fractional boundary value problem by using the Mountain Pass method and an iterative technique. In [23], the authors discussed the existence of a fractional boundary value problem with linear growth about the fractional derivative in a nonlinearity term. Compared with some integral-order partial differential equations such as [6, 24–31], the fractional derivatives have hereditary and nonlocal properties so that they are much more suitable for describing long-memory processes than the classical integer-order derivatives.Motivated by the above papers, in this paper, we first investigated the existence of solutions for the following fractional boundary value problems:(1)DtTαD0tαut+D0tαDtTαut=ft,ut,D0tαut,a.e.t∈0,T,u0=uT=0,where D0tα and DtTα are the left and right fractional Riemann-Liouville derivatives of order 1/2<α<1, respectively, f:0,T×ℝ×ℝ⟶ℝ.Note that problem (1) is not variational due to the fractional derivative contained in nonlinearity, so we cannot find a functional such that its critical point is the weak solution corresponding to (1). In order to overcome this difficulty, we consider the following fractional boundary value problem which is independent on the fractional derivative of the solution
(2)DtTαD0tαut+D0tαDtTαut=ft,ut,D0tαwt,a.e.t∈0,T,u0=uT=0,where w is an element of fractional Sobolev space Eα. First, by using variational methods, we obtain the existence of solutions for (2). Then, under the assumption that f is linear growth about the fractional derivative and based on iterative methods, we show there exists a solution for (1). Our conditions are weaker than that in [23].We also discuss the following fractional boundary value problem:(3)DtTαD0tαut+D0tαDtTαut=atgut+λD0tαut2,a.e.t∈0,T,u0=uT=0,where a∈L10,T;ℝ+, λ is a parameter, and g∈Cℝ;ℝ. Compared with (1), the nonlinearity of (3) is quadratic growth about a fractional derivative. By using variational methods and an iterative technique, we obtain that there exists solutions for (3) when λ and a satisfy suitable conditions. To the best of the authors’ knowledge, there are no results on the solutions to the fractional boundary problem which have quadratic growth about the fractional derivative in the nonlinearity term .The paper is organized as follows. In Section2, we will list some important properties of the basic functional space. We show the existence results for (1) and (3) in Section 3 and Section 4, respectively.


## 2. Preliminary

Let us briefly recall the property of a fractional derivative which will be used to construct the variational functional.Lemma 1 [1].
For0<β≤1, if f∈Lpa,bℝ with p1−β>1, J0t1−βft is absolutely continuous, and g:a,b,ℝ is absolutely continuous with gb=0, then
(4)∫abDatβftgtdt=∫abDtbβgtftdt.Now, we recall some properties of the basic function space which have been studied in [32].Throughout this paper, let1/2<α<1.The fractional derivative spaceEα is defined by the completion of C0∞0,T with respect to the norm u=∫0Tut2dt+∫0TD0tαut2dt1/2, where D0tα is the α-order left Riemann-Liouville fractional derivative. Then, Eα is a reflexive and separable Hilbert space. And the Riemann-Liouville fractional derivative exists for the elements in Eα [22].Lemma 2 [32].
For allu∈Eα, we have
(5)uL2≤TαΓα+1D0tαuL2,(6)u∞≤Tα−1/2Γα2α−11/2D0tαuL2.According to (5), one can consider Eα with respect to the equivalent norm
(7)uα=D0tαuL2.Lemma 3 [32].
If the sequenceuk converges weakly to u in Eα, i.e., uk⇀u, then uk⟶u in C0,T,ℝN, i.e., u−uk∞⟶0 as k⟶∞.
By the proof of Proposition 4.1 in [8], we have the following property.Lemma 4.
For anyu∈Eα,
(8)cosπα2uα2≤DtTαuL22≤1cosπα2uα2.


## 3. Existence Result for (1)

We assume thatft,x,y satisfies the following conditions:H0 f:0,T×ℝ×ℝ⟶ℝ is measurable in t∈0,T for every x,y∈ℝ×ℝ and continuous in x,y∈ℝ×ℝ for a.e.t∈0,T, and there exist a∈Cℝ+,ℝ+,b∈L10,T;ℝ+, such that
(9)ft,x,y≤axbt+y2,for all t∈0,T,x,y∈ℝ.H1 There are constants μ>2 and R>0 such that, for x≥R,y∈ℝ,
(10)0<μFt,x,y≤ft,x,yx,where Ft,x,y=∫0xft,s,yds.H2 limx→0ft,x,y/x=0 uniformly for t∈0,T and y∈ℝ.In order to derive a weak solution of (1), we suppose that u is a solution of (1), and multiplying (1) by an arbitrary v∈C0∞0,T and by Lemma 1, we have
(11)∫0TD0tαutD0tαvt+DtTαutDtTαvtdt−∫0Tft,ut,D0tαutvtdt=0.Since (11) is well defined for u,v∈Ea, the weak solution of (1) may be defined as follows.Definition 5.
A weak solution of (1) is a function u∈Eα such that
(12)∫0TD0tαutD0tαvt+DtTαutDtTαvtdt−∫0Tft,ut,D0tαutvtdt=0,for every v∈Eα.Definition 6.
A functionu is called a solution of (1), if D0tαut and DtTαut exist, J−tT1−αD0tαut+J0t1−αDtTαut is derivable for almost every t∈0,T, and u satisfies (1).For a givenw∈Eα, we consider the functional φw:Eα⟶ℝ, defined by
(13)φwu=12∫0TD0tαut2+DtTαut2dt−∫0TFt,ut,D0tαwtdt.In view of assumptionH0, we know that φ is continuously differentiable and
(14)φw′uv=∫0TD0tαutD0tαvt+DtTαutDtTαvtdt−∫0Tft,ut,D0tαwtvtdt,for u, v∈Eα. Hence, a critical point of φw gives us a weak solution of (2).Lemma 7.
Ifu is a weak solution of (1), then u is also a solution of (1).Proof.
Letu be a weak solution of (1), then u∈Eα, so 0Dtαut and tDTαut exist and u0=uT=0. For every v∈C0∞, we have
(15)∫0TD0tαutD0tαvt+DtTαutDtTαvtdt−∫0Tft,ut,D0tαutvtdt=0.
Sincev∈C0∞, we have D0tαvt=J0tαv′t and DtTαvt=J−tTαv′t, so
(16)∫0TJ−tT1−αD0tαut+J0t1−αDttαut−∫0tfs,us,D0sαusdsv′tdt=0.
Then, there exists a constantC such that
(17)J−tT1−αD0tαut+J0t1−αDtTαut−∫0tfs,us,D0sαusds=C,thus,
(18)DtTαD0tαut+D0tαDtTαut=ft,ut,D0tαut,t∈0,T.Lemma 8.
SupposeH0 and H1 hold, then φw satisfies the (PS) condition.Proof.
Letun⊂Eα, φwun is bounded, and φw′un⟶0; we first show that un is bounded.It follows fromH1 that
(19)μφwun−φw′unun=μ2−1∫0TD0tαunt2+DtTαunt2dt+∫t∈0,Tunt≥Rft,unt,D0tαwtunt−μFt,unt,D0tαwtdt+∫t∈0,Tunt<Rft,unt,D0tαwtunt−μFt,unt,D0tαwtdt≥μ2−1∫0TD0tαunt2+DtTαut2dt+∫t∈0,Tunt<Rft,unt,D0tαwtunt−μFt,unt,D0tαwt,which implies that un is bounded.From the reflexivity ofEα, we may extract a weakly convergent subsequence that, for simplicity, we call un, un⇀u then un−u∞⟶0. Next, we will prove that un strongly converges to u. By H0, we know that
(20)∫0Tft,unt,D0tαwt−ft,ut,D0tαwtunt−utdt⟶0asn⟶∞.Note that(21)φw′un−φw′uun−u=∫0TD0tαunt−ut2+DtTaunt−ut2dt−∫0Tft,unt,D0tawt−ft,ut,D0tawtunt−utdt≥1+cosπα2un−uα2−∫0Tft,unt,D0tαwt−ft,ut,D0tαwtunt−utdt.Byφ′un⟶0 and un⇀u, we obtain that
(22)φ′un−φ′uun−u⟶0asn⟶∞.Thus,un−uα⟶0asn⟶∞. Therefore, φw satisfies the (PS) condition.Lemma 9.
Letw∈Eα and suppose H0, H1, and H2 hold, then (2) has at least one nontrivial solution.Proof.
The proof relies on the Mountain Pass theorem [33, 34]. It is clear that φw∈C1Eα,R, φw0=0, and φw satisfies the (PS) condition from Lemma 8. By H2, for all ε>0, there is a δ>0 such that
(23)Ft,x,y≤εx2,x<δ,t∈0,T,y∈ℝ.Letρ=δΓα2α−11/2/2Tα−1/2 and choose uα=ρ; by the above inequality and (6), we have
(24)φwu=12∫0TD0tαut2+DtTαut2dt−∫0TFt,ut,D0tαwtdt≥1+cosπα22uα2−εTu∞2≥1+cosπα22−εT2αΓα22α−1uα2.It suffices to chooseε=1+cosπα2Γα22α−1/4T2α and σ=1+cosπα2/4ρ to get φwu≥σ. Thus, there exist positive numbers ρ and σ which are independent of w such that φwu≥σ for u∈Eα satisfies uα=ρ.It follows fromH1 that there exist c1,c2>0 such that
(25)Ft,x,y≥c1xμ−c2,∀t∈0,T,x,y∈ℝ.Choosingu~∈Eα satisfies u~α=1, and we obtain
(26)φwru~=r22∫0TD0tαu~t2+DtTαu~t2dt−∫0TFt,ru~t,D0tαwtdt≤r221+1cosπα2−c1rμ∫0Tu~μdt+c2T,which implies that φru~⟶−∞ as r⟶∞. Hence, we obtain that there exists a β>0 independent of u1 and w such that φwu≤0 for all uα>β.The above discussions combined with the Mountain Pass theorem show that (2) has at least one nontrivial solution uw which can be characterized as
(27)φw′uw=0,φwuw=infmaxg∈Γu∈g0,1φwu,where Γ=g∈C0,1,Eαg0=0,g1=βu~.In order to obtain the existence of solutions for (1), we need the following Lipschitz condition.H3 There exist L1,L2>0 such that
(28)ft,x1,y−ft,x2,y≤L1x1−x2,∀t∈0,T,x1,x2∈0,r1,y∈ℝ,ft,x,y1−ft,x,y2≤L2y1−y2,∀t∈0,T,x∈0,r1,y1,y2∈ℝ,where r1=c2Tα−1/2/Γα2α−11/2; C2 will be determined later.Theorem 10.
Letγ=L2TαΓα+1/1+cosπα2Γα+12−L1T2α, suppose (H0)–(H3) hold, 2R/1+cosπα2<1, and 0<γ<1, then problem (1) has a nontrivial solution.Proof.
Forn=1,2,⋯ and u1=0, we construct a sequence un, where un is the solution of the following problem:
(29)DtTαD0tαunt+D0tαDtTαunt=Ft,unt,D0tαun−1t,a.e.t∈0,T,un0=unT=0.Now, we assumeun−1α≤C2; by the mathematical induction, we will prove that unα≤C2. It follows from un satisfying (29) that
(30)φun−1un=infmaxg∈Γu∈g0,1φun−1u≤maxr∈0,βφun−1ru~≤maxr∈0,βr221+1cosπα2−c1rμ∫0Tu~μdt+c2T.Let(31)hr=r221+1cosπα2−c1rμ∫0Tu~μdt+c2T,then hr can achieve its maximum at r~=cosπα2+1/cosπα2c1μ∫0Tu1μdt1/μ−2, so φun−1un≤hr~. By H1 and H0, we have
(32)hr~≥φun−1un=12∫0T0Dtαunt2+tDTαunt2dt−∫0TFt,unt,0Dtαun−1tdt≥121+cosπα2unα2−∫t∈0,Tunt≤RFt,unt,0Dtαun−1tdt≥121+cosπα2unα2∫t∈0,Tunt≤R∫0untft,s,0Dtαun−1tdsdt≥121+cosπα2unα2−∫t∈0,Tunt≤R∫0Rasbtdsdt−R∫0t0Dtαun−1t2dt≥121+cosπα2unα2−∫0Tbtdt∫0Rasds−Run−1α2.Then,(33)unα2≤2R1+cosπα2un−1α2+2hr~+∫0Tbtdt∫0Rasds1+cosπα2≤2R1+cosπα2n−1u1α2+2hr~+∫0Tbtdt∫0Rasds1+cosπα2∑k=0n−22R1+cosπα2k=2R1+cosπα2n−1u1α2+21+cosπα2hr~+∫0Tbtdt∫0Rasds1+cosπα21+cosπα2−2R.Thus, we can chooseC2=21+cosπα2hr~+∫0Tbtdt∫0Rasds/1+cosπα21+cosπα2−2R.Sinceφ′un+1−φ′unun+1−un=0 and ∥un∥α≤C2, then ∥un∥∞≤r1, where r1 is given by H3. From H3, we have
(34)∫0TD0tαun+1t−unt2+tDTαun+1t−unt2dt=∫0Tft,un+1t,D0tαunt−ft,unt,D0tαun−1tun+1t−untdt=∫0Tft,un+1tD0tαunt−ft,unt,D0tαuntun+1t−untdt+∫0Tft,unt,D0tαunt−ft,unt,D0tαun−1tun+1t−untdt≤L1un+1−unL22+L2∫0T0Dtαunt−0Dtαun−1tun+1t−untdt≤L1un+1−unL22+L2un−un−1αun+1−unL2.Combining the above estimates with (8) and (5), we obtain
(35)1+cosπα2−L1T2αΓα+12un+1−unα≤L2TαΓα+1un−un−1α,that is un+1−unα≤γun−un−1α. Since 0<γ<1, the above inequality implies that un is a Cauchy sequence in Eα. Thus, there is a u such that un converges strongly to u in Eα.Now, we show thatu≡0. By the proof of Lemma 9, we know φun≥σ.Ifφun⟶φu, thenφu>σ, and since φ0=0, we obtain u≡0. In order to show φun⟶φu, we only need to show ∫0TFt,un,D0tαun−1tdt⟶∫0TFt,u,D0tαutdt. In fact,
(36)∫0TFt,un,D0tαun−1tdt−∫0TFt,u,D0tαutdt=∫0T∫0unft,x,D0tαun−1tdt−∫0T∫0uft,x,D0tαutdt=∫0T∫uunft,x,D0tαun−1t−ft,x,D0tαutdt≤L2∫0TD0tαun−1t−D0tαutun−udt≤L2D0tαun−1t−D0tαutL2un−uL2≤L2TαΓα+1un−uα2⟶0.Next, we show that for anyv∈Eα,
(37)∫0TD0tαutD0tαvt+DtTαutDtTαvtdt=∫0Tft,ut,D0tαutvtdt.It remains only to show(38)∫0Tft,unt,D0tαun−1tvtdt⟶∫0Tft,ut,D0tαutvtdt.Note that(39)∫0Tft,unt,D0tαun−1t−ft,ut,D0tαutvtdt=∫0Tft,unt,D0tαun−1t−ft,ut,D0tαun−1tvtdt+∫0Tft,ut,D0tαun−1t−ft,ut,D0tαutvtdt≤L1un−uL2vL2+L2un−1−uαvL2≤Cun−uα+un−1−uαvα,where C is a constant. Thus, we obtain a nontrivial solution of problem (1).


## 4. Existence Result for (3)

In Section3, condition H3 implies that the nonlinearity is linear growth about a fractional derivative of solutions; this section will consider the fractional boundary value problem (3) in which the nonlinearity is quadratic growth about the fractional derivative of solutions.Assume thatg satisfies the following conditions:H1′ There are constants μ>2 and R>0 such that, forx≥R,0<µGx≤gxx, where Gx=∫0xgsds.H2′ limx→0gx/x=0.Similar to Section3, since (3) is not variational, given w∈Eα, we consider the following problem which is independent on the fractional derivative of the solution:
(40)DtTαD0tαut+D0tαDtTαut=atgut+λ0Dtαwt2,a.e.t∈0,T,u0=uT=0.Then, the corresponding functionalϕ:Eα⟶Eα is given by
(41)ϕwu=12∫0TD0tαut2+DtTαut2dt−∫0TatGutdt−λ∫0TD0tαwt2utdt,ϕw′uv=∫0TD0tαtD0tαvt+DtTαutDtTαvtdt−∫0Tatgutvtdt−λ∫0TD0tαwt2vtdt,where v∈Eα.H1′ implies that there are constants c1,c2>0 such that
(42)Gx≥c1xµ−c2.ByH2′, there is a δ>0 such that x≤δ implies
(43)gx≤Γα22α−11+cosπα24∫0TatdtT2a−1x2.For convenience of our statement, let us give some notations and denote(44)a=μ−21+cosπα22μ,b=λμ−1Tα−1/2Γα2α−11/2μ,d=μ−1Γα2α−11/2δ1+cosπα232Tα−1/2μ,where μ is given in H1′ and δ is given in (43). Assuming a2>4bd, let
(45)ε1=a−a2−4bd2b,ε2=a+a2−4bd2b.For a fixed functionν∈Eα with v=1, we denote
(46)mt=1+1cosπα2r2−c1rμ∫0Tatvμdt+rΓα2α−11/2δ1+cosπα24Tα−1/2+c2∫0Tat,(47)M=maxr∈0,∞mt,M1=M+∫0Tatdtmaxx≤RGx−1μgx.Forε∈ε1,ε2, denote
(48)R1=R1ε=M1εaε−bε2−d,R2=R2ε=R1Tα−1/2Γα2α−11/2.We also need the following Lipschitz condition:H3′There exists LR2>0 such that gx1−gx2≤LRx1−x2 for 0<x1,x2<R2. Then we have the following result.Theorem 11.
SupposeH1′–H3′ hold, a2>4bd; if there existsε∈ε1,ε2, such that
(49)λR12≤Γα22α−1δ1+cosπα28T2α−1,(50)2λR1Tα−1/2Γα2α−11/2<1+cosπα2−T2α−1LR2∫0TatdtΓα22α−1.
Then, (3) has a nontrivial weak solution.Proof.
We first verify that for a givenw∈Eα with wα≤R1, (40) has at least a nontrivial weak solution.
In order to use the Mountain Pass theorem, we first show thatϕ satisfies the (PS) condition. Let un⊂Eα, ϕwun is bounded, and ϕw′un⟶0, we show that un is bounded.
In fact,(51)μϕωun−ϕω′unun=μ2−1∫0T0Dtαunt2+tDTαunt2dt+∫0Tatgunun−μGundt+λ1−μ∫0T0Dtαωt2untdt≥μ2−11+cosπα2unα2+λ1−μωα2un∝+∫t∈0,Tunt≤Ratuntgunt−μGuntdt≥μ2−11+cosπα2unα2+λR121−μTα−1/2Γα2α−11/2unα+∫t∈0,Tunt≤Ratuntgunt−μGuntdt,which implies that un is bounded, and similar to the last part of Lemma 8, we get that ϕwu satisfies the (PS) condition.
Letρ=Γα2α−11/2δ/Ta−1/2 and choose u∈Eα with u=ρ, then u∞≤δ. From (43), (6), and (49), we have
(52)ϕwu=12∫0TD0tαut2+DtTαut2dt−∫0TatGutdt−λ∫0TD0tαwt2utdt≥121+cosπα2uα2−Γα22α−11+cosπα24T2α−1u∞2−λwα2u∞≥141+cosπα2uα2−λTα−1/2R12Γα2α−11/2uα=141+cosπα2ρ−λTα−1/2R12Γα2α−11/2ρ≥181+cosπα2ρ2.
Then, we obtain that there existsβ>0 such that for uα=ρ, ϕwu≥β uniformly for w∈Eα with wα≤R1.
Letv∈Eα with ν=1, then
(53)ϕwrv≤r221+1cosπα2−c1rμ∫0Tatvμdt+λr∫0TD0tαwt2vdt+c2∫0Tatdt≤r221+1cosπα2−c1rμ∫0Tatvμdt+rλR12Tα−1/2Γα2α−11/2vα+c2∫0Tatdt⟶−∞asr⟶∞,which implies there exists r0>ρ such that ϕwr0ν<0. Then, from the Mountain Pass theorem, we get that ϕw has a nontrivial weak solution uw which can be characterized as
(54)ϕw′uw=0,ϕwuw=infmaxη∈Γu∈η0,1ϕwu,where Γ=η∈C0,1,Eαη0=0,η1=r0v.
Letu1=0, we can obtain that ϕu1 has a nontrivial critical point u2. For n=1,2,⋯, we construct a sequence un, where un is the critical point of ϕun−1. Now, we assume that un−1α≤R1; by the mathematical induction, we will prove that unα≤R1. In fact, by (54), we have
(55)ϕun−1un≤maxr∈0,r0ϕun−1rv≤1+1cosπα2r2−∫0Tatc1rμvμ−c2dt+λr∫0TD0tαun−1t2vtdt≤1+1cosπα2r2−c1rμ∫0Tatvμdt+λrTα−1/2Γα2α−11/2un−1α2+c2∫0Tatdt≤1+1cosπα2r2−c1rμ∫0Tatvμdt+λrTα−1/2R12Γα2α−11/2+c2∫0Tatdt≤1+1cosπα2r2−c1rμ∫0Tatvμdt+rΓα2α−11/2δ1+cosπα24Tα−1/2+c2∫0Tatdt≤M,where M is given in (47).
Hence,(56)M≥ϕun−1un=12∫0TD0tαunt2+DtTαunt2dt−∫0TatGuntdt−λ∫0TD0tαun−1t2untdt≥12∫0TD0tαunt2+DtTαunt2dt−1μ∫0Tatguntuntdt−λ∫0TD0tαun−1t2untdt−∫0Tatdtmaxx≤RGx−1μgx=12−1μ∫0TD0tαunt2+DtTαunt2dt+1μϕun−1′unun−λ1−1μ∫0TD0tαun−1t2untdt−∫0Tatdtmaxx≤RGx−1μgx≥12−1μ1+cosπα2unα2−λ1−1μ∫0TD0tαun−1t2untdt−∫0Tatdtmaxx≤RGx−1μgx.
So,(57)12−1μ1+cosπα2unα2≤M+λ1−1μ∫0TD0tαun−1t2untdt+∫0Tatdtmaxx≤RGx−1μgx≤M1+1−1μλTα−1/2Γα2α−11/2un−1α2unα≤M1+λμ−1Tα−1/2Γα2α−11/2μεunα2+R124εun−1α2.
By (49), we obtain
(58)μ−21+cosπα22μ−λεμ−1Tα−1/2Γα2α−11/2μunα2≤λR12μ−1Tα−1/24εΓα2α−11/2μun−1α2+M1≤μ−1Γα2α−11/2δ1+cosπα232Tα−1/2μεun−1α2+M1.
That is(59)a−bεunα2≤dεun−1α2+M1.
Whenε∈ε1,ε2, where ε1 and ε2 are given in (45), we have
(60)dε<a−bε.
Then,(61)unα2≤d/εa−bεun−1α2+M1a−bε≤d/εa−bεn−1u1α2+M1a−bε∑k=0n−2d/εa−bεk≤u1α2+M1εaε−bε2−d.
Therefore, the above argument implies thatunα≤R1.
Finally, we show thatun is convergent to a nontrivial solution of (3).
Sinceunα≤R1, we haveun∞≤R2, then by H3′, we obtain
(62)1+cosπα2un+1−unα2≤∫0TD0tαun+1−unt2+DtTαun+1−unt2dt=ϕun′un+1−ϕun−1′unun+1−unt+∫0Tatfun−fun+1un+1−undt+λ∫0TD0tαunt2−D0tαun−12un+1−untdt≤LR2un+1−un∞2∫0Tatdt+λ∫0TD0tαuntD0tαun−un−1dtun+1−un∞+λ∫0TD0tαun−1tD0tαun−un−1dtun+1−un∞≤LR2un+1−un∞2∫0Tatdt+λunαun−un−1αun+1−un∞+λun−1αun−un−1αun+1−un∞≤T2α−1LR2∫0TatdtΓα22α−1un+1−unα2+2λR1Tα−1/2Γα2α−11/2un−un−1αun+1−unα.
So(63)1+cosπα2−T2α−1LR2∫0TatdtΓα22α−1un+1−unα≤2λR1Tα−1/2Γα2α−11/2un−un−1α.
From (50), we know that un is a Cauchy sequence in Eα and u is a weak solution of (3). Since ϕun−1un>β>0 for n=1,2,⋯ and β does not depend on n, we obtain that u is a nontrivial weak solution of (3).Corollary 12.
SupposeH1′-H3′ hold; if the right-hand side of (50) is greater than 0, then there exists a constant λ0>0, such that (1) has a solution when λ∈0,λ0.



---

*Source: 7139795-2020-05-13.xml*