# Multiple Positive Solutions for Nonlinear Fractional Differential Equations with Integral Boundary Value Conditions and a Parameter

Authors: Juan Mao; Daliang Zhao
Journal: Journal of Function Spaces
 (2019)

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## Abstract

By using the theory of fixed-point index on cone for differentiable operators, spectral radii of some related linear integral operators, and properties of Green’s function, the existence of multiple positive solutions to a nonlinear fractional differential system with integral boundary value conditions and a parameter is established. At last, some examples are also provided to illustrate the validity of our main results.


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## Body

## 1. Introduction

In recent years, the interest in the study of fractional differential equations has been growing rapidly since it has many applications in biology, mechanics, electrochemistry, and dynamical processes in self-similar structures, etc.; see [1–3], for instance. Fractional differential equations also serve as an excellent tool for the description of hereditary properties of various materials and processes. Fractional-order models have proved to be more accurate than integer order models and have more applications. In consequence, many meaningful results in these fields have been obtained. For details, see [4–14] and references therein.In addition, several new types of fractional derivatives are also investigated in some excellent papers. In [15], Yang studied a class of fractional derivatives of constant and variable orders for the first time. In [16], the authors introduced a new fractional derivative without singular kernel, and this was an extension of the Riemann-Liouville fractional derivative with singular kernel and had some important applications in the modeling of the fractional-order heat flow.Meanwhile, there are a lot of papers (see [3, 17–28], for instance) dealing with the existence of positive solutions of nonlinear fractional differential equations by use of fixed-point theorems, monotone iterative technique, and upper and lower solution method. For example, in [25], by using the method of upper and lower solutions and Schauder fixed theorem, Vong investigated the positive solutions for the following nonlocal fractional BVP:(1)D0+αCut+ft,ut=0,0<t<1,u′0=···=un-10=0,u1=∫01usdμs,where n≥2,α∈(n-1,n), and μ(s) is a function of bounded variation. f may be singular at t=1.In [26], The authors studied a class of higher order fractional BVP. In view of monotone iterative method, they got an existence and uniqueness result of positive solution.Goodrich [8] dealt with a problem similar to the system in [26] but with local conditions. By deriving properties of the Green’s function and by means of Guo-Krasnoselskii’s fixed-point theorem, the author established some existence results of at least one positive solution provided that f(t,x) satisfies some growth conditions.However, compared with the methods of fixed-point theorems, Leray-Schauder theory, and monotone iterative technique, there is little work investigating positive solution of fractional integral boundary value problem by use of the differentiable operator method. On the other hand, the results about such system with a parameter are relatively scarce.Motivated by all the above-mentioned works, we aim to establish some existence criteria of multiple positive solutions for the following nonlinear fractional BVP with integral boundary conditions and a parameter:(2)D0+αut+ft,ut=0,t∈0,1,uj0=0,0≤j≤3,j≠1,u′1=λ∫0ξusds,where 3<α≤4, 0≤λ<α, 0<ξ≤1, 0≤λξ2<2, D0+α is the Caputo fractional derivative, and f:[0,1]×[0,+∞)→[0,+∞) is continuous. It is worth mentioning that there is no paper studying the multiple positive solutions for the fractional differential systems by using the differentiable operator method. Besides, to the best of our knowledge, no contribution exists investigating the existence of positive solutions for BVP (2). It should be noted that ξ∈(0,1] is more general than those in [23, 25, 26], in which ξ equals one.Contributions of this article are as follows. Firstly, we derive the corresponding Green’s function for fractional BVP (2) and argue its important properties which will play an important role in our proof. Also, this is the key to establish an appropriate cone. Secondly, by using the differential operator method, spectrum theory, and the properties of Green’s function, we firstly investigate some existence results of multiple positive solutions for the considered fractional system. The methods used in the present paper are different from all the above-mentioned works.This paper is organized in the following way. In Section2, we give some necessary preliminaries and lemmas. Section 3 establishes sufficient conditions for the existence of multiple positive solutions to system (2). In Section 4, some examples are given to highlight the value of the new contributions.


## 2. Preliminaries and Some Lemmas

We now recall some results about fractional calculus in common use. These materials can be found in the recent literature; see [2, 3, 29, 30].Definition 1.
The Riemann-Liouville fractional integral of orderα∈(3,4] of a continuous function u:(0,+∞)→R is given by(3)I0+αut=1Γα∫0tt-sα-1usds,provided that the right side integral is pointwise defined on (0,+∞).Definition 2.
The Caputo fractional derivative of orderα∈(3,4] of a continuous function u:(0,∞)→R is given by(4)D0+αCut=1Γ4-α∫0tu4st-sα-3ds,provided that the right side integral is pointwise defined on [0,∞).Lemma 3.
Let3<α≤4. Then(5)I0+αD0+αCut=ut+c0+c1t+c2t2+c3t3, for some ci∈R,i=0,1,2,3.LetE=C[0,1] be endowed with the norm u=max{ut:t∈[0,1]} for each u∈E. Clearly, (E,·) is a Banach space.Lemma 4.
Suppose functionx∈C[0,1]. Then the unique solution of the following system(6)D0+αut+xt=0,t∈0,1,uj0=0,0≤j≤3,j≠1,u′1=λ∫0ξusds,where 3<α≤4, 0≤λ<α, 0<ξ≤1, 0≤λξ2<2, is given by(7)ut=∫01Gt,sxsds,where G(t,s) is the Green’s function of system (6) given by(8)Gt,s=α-1t1-sα-2-λ/αξ-sαt-1-λξ2/2t-sα-11-λξ2/2Γα,0≤s≤minξ,t,α-1t1-sα-2-1-λξ2/2t-sα-11-λξ2/2Γα,0≤ξ≤s≤t≤1,α-1t1-sα-2-λ/αξ-sαt1-λξ2/2Γα,0≤t≤s≤ξ≤1,α-1t1-sα-21-λξ2/2Γα,maxξ,t≤s≤1.Proof.
From Lemma3, it follows that (9)ut=-I0+αxt+c0+c1t+c2t2+c3t3=-∫0tt-sα-1Γαxsds+c0+c1t+c2t2+c3t3,where ci∈R,i=0,1,2,3.
So, one has(10)u′t=-∫0tt-sα-2Γα-1xsds+c1+2c2t+3c3t2,u′′t=-∫0tt-sα-3Γα-2xsds+2c2+6c3t,u′′′t=-∫0tt-sα-4Γα-3xsds+6c3. From uj(0)=0,0≤j≤3,j≠1,u′(1)=λ∫0ξu(s)ds, we get (11)c0=c2=c3=0,c1=∫011-sα-2Γα-1xsds+λ∫0ξusds. So we have,(12)u′1=-∫011-sα-2Γα-1xsds+∫011-sα-2Γα-1xsds+λ∫0ξusds,(13)ut=-∫0tt-sα-1Γαxsds+t∫011-sα-2Γα-1xsds+λ∫0ξusds.By direct integration to (13), it is easy to get(14)∫0ξusds=-1Γα∫0ξ∫0xx-sα-1xsdsdx+∫0ξsds∫011-sα-2Γα-1xsds+λ∫0ξusds=-1Γα∫0ξξ-αααxsds+ξ22∫011-sα-2Γα-1xsds+λ∫0ξusds.Through the conditions u′(1)=λ∫0ξu(s)ds, (12), and (14), one has(15)-∫011-sα-2Γα-1xsds+∫011-sα-2Γα-1xsds+λ∫0ξusds=-λΓα∫0ξξ-αααxsds+λξ22∫011-sα-2Γα-1xsds+λ∫0ξusds.Then, we obtain(16)∫011-sα-2Γα-1xsds+λ∫0ξusds=∫011-sα-21-λξ2/2Γα-1xsds-∫0ξλξ-sαα1-λξ2/2Γαxsds.Therefore, we can get the following solution of (6) by (13),(17)ut=-∫0tt-sα-1Γαxsds+∫011-sα-2t1-λξ2/2Γα-1xsds-∫0ξλ/αξ-sαt1-λξ2/2Γαxsds.
In the following, we divide the proof into two cases.
Case  1 ( t ≤ ξ ) (18) u t = - ∫ 0 t t - s α - 1 Γ α x s d s + ∫ 0 t + ∫ t ξ + ∫ ξ 1 ⁡ 1 - s α - 2 t 1 - λ ξ 2 / 2 Γ α - 1 x s d s - ∫ 0 t + ∫ t ξ ⁡ λ / α ξ - s α t 1 - λ ξ 2 / 2 Γ α x s d s = ∫ 0 t α - 1 t 1 - s α - 2 - λ / α ξ - s α t - 1 - λ ξ 2 / 2 t - s α - 1 1 - λ ξ 2 / 2 Γ α x s d s + ∫ t ξ α - 1 t 1 - s α - 2 - λ / α ξ - s α t 1 - λ ξ 2 / 2 Γ α x s d s + ∫ ξ 1 α - 1 t 1 - s α - 2 1 - λ ξ 2 / 2 Γ α x s d s = ∫ 0 1 G t , s x s d s .
Case  2 ( t ≥ ξ ) (19) u t = - ∫ 0 ξ + ∫ ξ t t - s α - 1 Γ α x s d s + ∫ 0 ξ + ∫ ξ t + ∫ t 1 ⁡ 1 - s α - 2 t 1 - λ ξ 2 / 2 Γ α - 1 x s d s - ∫ 0 ξ λ / α ξ - s α t 1 - λ ξ 2 / 2 Γ α x s d s = ∫ 0 ξ α - 1 t 1 - s α - 2 - λ / α ξ - s α t - 1 - λ ξ 2 / 2 t - s α - 1 1 - λ ξ 2 / 2 Γ α x s d s + ∫ ξ t α - 1 t 1 - s α - 2 - 1 - λ ξ 2 / 2 t - s α - 1 1 - λ ξ 2 / 2 Γ α x s d s + ∫ t 1 α - 1 t 1 - s α - 2 1 - λ ξ 2 / 2 Γ α x s d s = ∫ 0 1 G t , s x s d s .The conclusion of this lemma followsNext, we give some important properties about Green’s functionG(t,s) belonging to system (6).Lemma 5.
The Green’s functionG(t,s) of system (6) satisfies the following inequalities:
( i )   G ( t , s ) ≤ α - 1 / ( 1 - λ ξ 2 / 2 ) Γ ( α ) t ( 1 - s ) α - 2 ,  ∀ t , s ∈ [ 0,1 ];
( i i )   G ( t , s ) ≤ α - 1 / ( 1 - λ ξ 2 / 2 ) Γ ( α ) ( 1 - s ) α - 2 ,  ∀ t , s ∈ [ 0,1 ];
( i i i )   G ( t , s ) ≥ α - 1 - λ / α - ( 1 - λ ξ 2 / 2 ) / ( 1 - λ ξ 2 / 2 ) Γ ( α ) t ( 1 - s ) α - 2 ,  ∀ t , s ∈ ( 0,1 );
( i v )   G ( t , s ) > 0 ,  ∀ t , s ∈ ( 0,1 ).Proof.
We denoteQ(s)=α-1-λ/α-(1-λξ2/2)/(1-λξ2/2)Γ(α)(1-s)α-2 and take H(t,s)=G(t,s)/Q(s).
For0≤s≤min{ξ,t}<1, one has(20)H1,s=α-11-sα-2-λ/αξ-sα-1-λξ2/21-sα-1α-1-λ/α-1-λξ2/21-sα-2>α-11-sα-2-λ/α1-sα-2-1-λξ2/21-sα-2α-1-λ/α-1-λξ2/21-sα-2=1,Hs,s=α-1s1-sα-2-λ/αξ-sαsα-1-λ/α-1-λξ2/21-sα-2>α-1-λ/α1-sα-2sα-1-λ/α-1-λξ2/21-sα-2>s,and(21)∂2H∂t2t,s=-1-λξ2/2α-1α-2t-sα-3α-1-λ/α-1-λξ2/21-sα-2≤0.This means that H(·,s) is concave on [s,1]. So we get H(t,s)≥t.
For0≤ξ≤s≤t<1, we have(22)H1,s=α-11-sα-2-1-λξ2/21-sα-1α-1-λ/α-1-λξ2/21-sα-2>α-1-1-λξ2/21-sα-2α-1-λ/α-1-λξ2/21-sα-2>1,Hs,s=α-1s1-sα-2α-1-λ/α-1-λξ2/21-sα-2>s, and(23)∂2H∂t2t,s=-1-λξ2/2α-1α-2t-sα-3α-1-λ/α-1-λξ2/21-sα-2≤0. These three facts can also indicate H(t,s)≥t.
For0≤t≤s<ξ≤1, we have H(0,s)=0, and(24)Hs,s=α-11-sα-2-λ/αξ-sαsα-1-λ/α-1-λξ2/21-sα-2>α-1-λ/α1-sα-2sα-1-λ/α-1-λξ2/21-sα-2>s,∂H∂tt,s=α-11-sα-2-λ/αξ-sαα-1-λ/α-1-λξ2/21-sα-2>α-1-λ/α1-sα-2α-1-λ/α-1-λξ2/21-sα-2>1.So, we can get that H(t,s)≥t.
Formax{ξ,t}≤s<1, we obtain H(0,s)=0, and(25)Hs,s=α-11-sα-2sα-1-λ/α-1-λξ2/21-sα-2>s,∂H∂tt,s=α-11-sα-2α-1-λ/α-1-λξ2/21-sα-2>1.Hence, we also have H(t,s)≥t.
Consequently, one gets the validity of(iii) and (iv). The establishment of (i) and (ii) is obvious.Lemma 6 (see [29]).
LetX be a Banach space, P be a cone in X, and Ω(P) be a bounded open subset in P. Suppose that A:Ω(P)¯→P is a completely continuous operator. Then the following results hold:
(i) if there existsu0∈P∖{θ} such that u≠Au+λu0, for any u∈∂Ω(P),λ≥0, then i(A,Ω(P),P)=0;
(ii) ifθ∈Ω(P), Au≠λu, for any u∈∂Ω(P),λ≥1, then i(A,Ω(P),P)=1.Lemma 7 (see [29]).
LetP be a cone in a Banach space E, A:P→P be completely continuous, and Aθ=θ. Suppose that A is differentiable at θ along P and 1 is not an eigenvalue of A+′(θ) corresponding to a positive eigenvector. Moreover, if A+′(θ) has no positive eigenvectors corresponding to an eigenvalue greater than one, then there exists r0>0 such that i(A,Pr,P)=1, for 0<r≤r0, where Pr={x∈P:u<r}.Lemma 8 (see [29]).
LetP be a cone in a Banach space E and A:P→P be completely continuous. Suppose that A is differentiable at ∞ along P and 1 is not an eigenvalue of A+′(∞) corresponding to a positive eigenvector. Moreover, if A+′(∞) has no positive eigenvectors corresponding to an eigenvalue greater than one, then there exists R0>0 such that i(A,PR,P)=1, for R≥R0, where PR={x∈P:u<R}.Define a coneP by(26)P=u∈E:ut≥Mtu,t∈0,1, where M=α-1-λ/α-(1-λξ2/2)/α-1. Let Pr={u∈P:u<r}(r>0).Define integral operatorsL and T by(27)Lut=∫01Gt,susds,u∈E;Tut=∫01Gt,sfs,usds,u∈P.Lemma 9 (see [14]).
L : P → P is completely continuous and the spectral radius r(L)>0.By a similar process in the proof of Lemma9, we have the following lemma.Lemma 10.
Letf∈C([0,1]×[0,+∞),[0,+∞)), and then operator T maps cone P into P and T is completely continuous.


## 3. Main Results

For convenience in the following discussion, we present the following notations and assumptions:(28)f∞=liminfu→+∞mint∈0,1ft,uu,f0=liminfu→0+mint∈0,1ft,uu,f∞=limsupu→+∞maxt∈0,1ft,uu,f0=limsupu→0+maxt∈0,1ft,uu.(H1)
f(t,0)=0,fu∈C([0,1]×[0,+∞)), and fu(t,0)>0, ∀t∈[0,1].(H2)
∫01(1-t)α-2fu(t,0)dt<(1-λξ2/2)Γ(α)/α-1.(H3)
There existsϕ∈C(0,1,0,+∞),ϕ≢0, such that limu→+∞f(t,u)/u=ϕ(t) uniformly holds with respect to t on [0,1], and ∫01(1-t)α-2ϕ(t)dt<(1-λξ2/2)Γ(α)/α-1.Adopting regular approaches similar to that in [14], we give the following two lemmas.Lemma 11.
Suppose that (H1) and (H2) hold. ThusT is differentiable at θ along P,Tθ=θ, and (29)T+′θu=∫01Gt,sfus,0usds,u∈P.In addition, T+′(θ) has no positive eigenvectors corresponding to an eigenvalue greater than or equal to one.Proof.
Obviously,Tθ=θ by f(t,0)=0. From mean value theorem, it follows that(30)ft,u=ft,u-ft,0=fut,u0u,for some u0∈(0,u). Notice fu∈C([0,1]×[0,+∞)), and we can obtain that for any ε>0, there exists a δ>0 such that 0<u0<u<δ and(31)ft,u-fut,0u≤ft,u-fut,u0u+fut,u0-fut,0u=fut,u0-fut,0u<1-λξ22Γαεu.So for any u∈P,u<δ, noticing (31) and Lemma 5, we get(32)Tut-∫01Gt,sfus,0usds≤∫01Gt,sfs,u-fus,0ds≤1-λξ22Γαεu∫01Gt,sds≤εu,which means that(33)T+′θu=∫01Gt,sfus,0usds,u∈P.
Now, we are ready to show thatT+′(θ) has no positive eigenvectors corresponding to an eigenvalue greater than or equal to one. Otherwise, there exist u∗∈P∖{θ} and λ∗≥1 such that T+′(θ)u∗=λ∗u∗. Then(34)u∗t≤λ∗u∗t=∫01Gt,sfus,0u∗sds≤α-11-λξ2/2Γα∫011-sα-2fus,0u∗sds. Notice fu(t,0)>0, and it follows(35)1-tα-2fut,0u∗t≤α-11-λξ2/2Γα1-tα-2fut,0∫011-sα-2fus,0u∗sds.By direct integration to (35) on [0,1], we obtain(36)∫011-tα-2fut,0u∗tdt≤α-11-λξ2/2Γα∫011-tα-2fut,0dt∫011-sα-2fus,0u∗sds.Notice ∫01(1-t)α-2fu(t,0)u∗(t)dt>0, from which we infer that(37)α-11-λξ2/2Γα∫011-tα-2fut,0dt≥1,which contradicts (H2). The proof is thus completed.Lemma 12.
Suppose that (H3) holds. ThusT is differentiable at ∞ along P and (38)T+′∞u=∫01Gt,sϕsusds,u∈P.In addition, T+′(∞) has no positive eigenvectors corresponding to an eigenvalue greater than or equal to one.Proof.
From (H3), it follows that(39)ft,u-ϕtu<C+εu,∀u≥0,t∈0,1,where ε>0, and C is a constant corresponding to ε.
Thus, for anyu∈P,t∈[0,1], it has(40)Tut-∫01Gt,sϕsusds≤∫01Gt,sfs,us-ϕsusds≤C+εu∫01Gt,sds≤C+εu1-λξ2/2Γα, which implies that(41)Tut-∫01Gt,sϕsusdsu≤C1-λξ2/2Γαu+ε1-λξ2/2Γα<21-λξ2/2Γαε, when u>C/ε. Therefore, T+′(∞)u=∫01G(t,s)ϕ(s)u(s)ds,u∈P.
Suppose thatT+′(∞) has positive eigenvectors corresponding to an eigenvalue greater than or equal to one. Then we can choose a function u∗∈P∖{θ} and a constant λ∗≥1 satisfying T+′(θ)u∗=λ∗u∗, and thus(42)u∗t≤λ∗u∗t=∫01Gt,sϕsu∗sds≤α-11-λξ2/2Γα∫011-sα-2ϕsu∗sds.By means of ϕ∈C([0,1],[0,+∞)),ϕ≢0, one has(43)1-tα-2ϕtu∗t≤α-11-λξ2/2Γα1-tα-2ϕt∫011-sα-2ϕsu∗sds.Integrating (43) with respect to t on [0,1] gives(44)∫011-tα-2ϕtu∗tdt≤α-11-λξ2/2Γα∫011-tα-2ϕtdt∫011-sα-2ϕsu∗sds.From the condition ∫01(1-t)α-2ϕ(t)u∗(t)dt>0, it follows that(45)α-11-λξ2/2Γα∫011-tα-2ϕtdt≥1.Notice (H3); this is a contradict. Hence, Lemma 12 holds.The following theorems are main results in this paper.Theorem 13.
Suppose that (H1), (H2), andf∞>r-1(L) hold. Then BVP (2) has at least one positive solution.Proof.
From Lemmas9, 11, and 7, it follows that(46)iT,Pr0,P=1,where r0>0 is a suitable constant.
Let(47)Lnut=∫1/n1Gt,susds,n>1.
Standard arguments can show thatLn:P→P is completely continuous and r(Ln)>0. Let λn=r-1(Ln),λ1=r-1(L), and limn→∞λn=λ0. The proof of λ0=λ1 follows by arguing as in [14], so we omit the proof. Hence, there exist N0 and ε0 such that λN0<λ1+ε0; that is (48)rLN0>1r-1L+ε0.According to Krein-Rutman theorem, we can obtain some ψN0∈E∖{θ} with ψN0≥0 such that(49)ψN0t=r-1LN0∫1/N01Gt,sψN0sds.Then, for all t,τ∈[0,1], one has(50)ψN0t≥Mtr-1LN0∫1/N01Gτ,sψN0sds=MtψN0τ. Thus, ψN0(t)≥MtψN0, which means ψN0(t)∈P∖{θ}.
Moreover, usingf∞>r-1(L), we can choose a constant R>0 such that(51)ft,u>r-1L+ε0u,∀u≥R,t∈0,1.Choose R0>max{r0,N0/MR}. Thus for any u∈∂PR0, it follows(52)ut≥Mtu=MtR0>R,t∈1N0,1.
If we can choose a functionu2∈∂PR0 and a constant μ0>0 such that(53)u2t-Tu2t=μ0ψN0t,t∈0,1,then we know u2(t)≥μ0ψN0(t),t∈[0,1].
Denote(54)μ∗=supμ∣u2t≥μψN0t,t∈0,1.Obviously, 0<μ0≤μ∗<+∞ and u2(t)≥μ∗ψN0(t),t∈[0,1]. Noticing (48), (49), (51), and (53), we obtain(55)u2t=Tu2t+μ0ψN0t=∫01Gt,sfs,u2sds+μ0ψN0t≥r-1L+ε0∫1/N01Gt,su2sds+μ0ψN0t≥r-1L+ε0μ∗∫1/N01Gt,sψN0sds+μ0ψN0t=r-1L+ε0μ∗rLN0ψN0t+μ0ψN0t>μ∗+μ0ψN0t,t∈0,1. This is in contradiction with the definition of μ∗. Then we have(56)ut-Tut≠μψN0t,∀u∈∂PR0,μ≥0,t∈0,1. Using Lemma 6 yields that i(T,PR0,P)=0. This together with (46) gives(57)iT,PR0∖P¯r0,P=iT,PR0,P-iT,Pr0,P=-1. Therefore, system (2) has at least one positive solution.Theorem 14.
Suppose that (H3) andf0>r-1(L) hold. Then BVP (2) has at least one positive solution.Proof.
Sincef0>r-1(L), we know that there exists a positive constant r0 such that(58)ft,u>r-1Lu,∀u∈0,r0,t∈0,1.By a similar way in Theorem 13, there exists a function ψ(t)∈P∖{θ} such that(59)ψt=r-1L∫01Gt,sψsds.Next, we are ready to show(60)ut-Tut≠μψt,∀u∈∂Pr0,μ≥0,t∈0,1.In fact, if not, then there exist u1∈∂Pr0 and μ1>0 such that(61)u1t-Tu1t=μ1ψt,t∈0,1,which indicates u1(t)≥μ1ψ(t),t∈[0,1].
Denote(62)μ∗=supμ∣u1t≥μψt,t∈0,1. Obviously, 0<μ1≤μ∗<+∞ and u1(t)≥μ∗ψ(t),for anyt∈[0,1]. This together with (58), (59), and (61) gives(63)u1t=Tu1t+μ1ψt=∫01Gt,sfs,u1sds+μ1ψt≥r-1L∫01Gt,su1sds+μ1ψt≥r-1Lμ∗∫01Gt,sψsds+μ1ψt=μ∗+μ1ψt,t∈0,1. By means of the definition of μ∗, we get a contradiction. Thus, from Lemma 6, one gets(64)iT,Pr0,P=0.One the other hand, by Lemmas 9, 12, and 8, we can choose a constant R0>r0 such that i(T,PR0,P)=1. From this and (64), we get(65)iT,PR0∖P¯r0,P=iT,PR0,P-iT,Pr0,P=1.Consequently, BVP (2) has at least one positive solution.Theorem 15.
Suppose that (H1), (H2), andf∞<r-1(L) hold. In addition, suppose there exists r>0 such that(66)ft,u>m0r,∀u∈M2r,r,t∈0,1,where m0=maxt∈[0,1]∫1/21G(t,s)ds-1. Then BVP (2) has at least two positive solutions.Proof.
By Lemmas9, 11, and 7, we can choose a constant r0∈(0,r) such that(67)iT,Pr0,P=1.
Letψ(t)∈P∖{θ} be the eigenfunction of L corresponding to r(L). For u∈∂Pr, one has u(t)≥tMu≥M/2r, for any t∈[1/2,1]. Thus, it follows(68)Tut=∫01Gt,sfs,usds≥∫1/21Gt,sfs,usds>∫1/21Gt,sds·m0r,which yields Tu>maxt∈[0,1]⁡∫1/21G(t,s)ds·m0r=r. Then, for u∈∂Pr,μ>0, we can obtain Tu+μψ≥Tu>r. This means u≠Tu+μψ, for u∈∂Pr,μ>0. By Lemma 6, we get(69)iT,Pr,P=0.
On the other hand, chooseε0 such that 0<(r-1(L)-ε0)L<1. Then by f∞<r-1(L), there exists R>0 such that (70)ft,u≤r-1L-ε0u,u≥R,t∈0,1.Let m1=max(t,u)∈[0,1]×[0,R]⁡f(t,u). Thus, we get (71)ft,u≤r-1L-ε0u+m1,∀u≥0,t∈0,1.Choose R0>{m1/(1-(r-1(L)-ε0)L)(1-λξ2/2)Γ(α),r}. Now, we are ready to show(72)Tut≠μut,∀u∈∂PR0,μ≥1,t∈0,1.Otherwise, suppose u1∈∂PR0 and μ1≥1 such that Tu1(t)=μ1u1(t). Then,(73)u1t≤μ1u1t=Tu1t=∫01Gt,sfs,u1sds≤∫01Gt,sr-1L-ε0u1s+m1ds=r-1L-ε0Lu1t+m1∫01Gt,sdsand, noticing that 0<(r-1(L)-ε0)L<1, we get that the inverse operator of I-(r-1(L)-ε0)L exists and equals(74)I-r-1L-ε0L-1=∑n=0∞r-1L-ε0nLn,from which one has (I-(r-1(L)-ε0)L)-1(P)⊆P. This together with (73) gives (75)u1t≤I-r-1L-ε0L-1m1∫01Gt,sds.In addition, by (I-(r-1(L)-ε0)L)-1≤1/1-(r-1(L)-ε0)L and Lemma 5, we have(76)R0=u1≤I-r-1L-ε0L-1m11-λξ2/2Γα≤m11-r-1L-ε0L1-λξ2/2Γα<R0,which is a contradiction. By Lemma 6, we get i(T,PR0,P)=1. This together with (67) and (69) guarantees that(77)iT,Pr∖P¯r0,P=-1,iT,PR0∖P¯r,P=1.Therefore, BVP (2) has at least two positive solutions.Theorem 16.
Suppose that (H3) andf0<r-1(L) hold. In addition, suppose there exists r>0 such that(78)ft,u>M0r,∀u∈0,r,t∈0,1, where M0=maxt∈[0,1]⁡∫01G(t,s)ds-1. Then BVP (2) has at least two positive solutions.Proof.
Letψ(t)∈P∖{θ} be the eigenfunction of L corresponding to r(L). For u∈∂Pr, it yields(79)Tut=∫01Gt,sfs,usds>∫01Gt,sds·M0r,which means Tu>maxt∈[0,1]⁡∫01G(t,s)ds·M0r=r. Thus, we can easily get u≠Tu+μψ, for u∈∂Pr,μ>0. This together with Lemma 6 gives(80)iT,Pr,P=0.
Byf0<r-1(L), there exists a constant r0∈(0,r) such that (81)ft,u≤r-1L-ε0u,u≤r0,t∈0,1,where ε0 satisfies 0<(r-1(L)-ε0)L<1.
If there existu1∈∂Pr0 and μ1≥1 such that Tu1(t)=μ1u1(t), we have(82)u1t≤μ1u1t=Tu1t=∫01Gt,sfs,u1sds≤∫01Gt,sr-1L-ε0u1sds=r-1L-ε0Lu1t.Since 0<(r-1(L)-ε0)L<1, we know that I-(r-1(L)-ε0)L is reversible and(83)I-r-1L-ε0L-1=∑n=0∞r-1L-ε0nLn. By (82), one has (84)u1t≤I-r-1L-ε0L-10=0. Noticing u1∈∂Pr0, we get a contradiction. That is to say,(85)Tut≠μut,∀u∈∂Pr0,μ≥1,t∈0,1. By Lemma 6, we obtain(86)iT,Pr0,P=1.
On the other hand, in view of Lemmas9, 12, and 8, we can choose a constant R0>r such that i(T,PR0,P)=1. Combining this with (86) and (80), we easily get(87)iT,PR0∖P¯r,P=1,iT,Pr∖P¯r0,P=-1.Hence, BVP (2) has at least two positive solutions.Theorem 17.
Suppose that (H1) and (H2) hold. In addition, suppose that there existr1,r2>0 with 0<r1<r2 such that(88)ft,u>m0r1,∀u∈Mr12,r1,t∈0,1,ft,u<M0r2,∀u∈0,r2,t∈0,1,hold, where m0,M0 are the same as in Theorems 15 and 16. Then BVP (2) has at least two positive solutions.Proof.
Similar to the proof of Theorem15, we can choose 0<r0<r2 such that(89)iT,Pr1,P=0,iT,Pr0,P=1.Moreover, for u∈∂Pr2, we get(90)Tut=∫01Gt,sfs,usds<∫01Gt,sds·M0r2,which gives Tu<maxt∈[0,1]⁡∫01G(t,s)ds·M0r2=r2. This implies that Tu≠μu for all u∈∂Pr2,μ≥1. By Lemma 6, we have i(T,Pr2,P)=1. This together with (89) guarantees(91)iT,Pr2∖P¯r1,P=1,iT,Pr1∖P¯r0,P=-1. Therefore, BVP (2) has at least two positive solutions.


## 4. Examples

Example 1.
Consider the following system:(92)D0+13/4ut+1+t3ueu=0,t∈0,1,u0=u′′0=u′′′0=0,u′1=12∫03/4usds.Then BVP (92) has at least one positive solution.Proof.
BVP (92) can be regarded as a BVP of the form (2), where(93)ft,u=1+t3ueu,and α=13/4,λ=1/2,ξ=3/4,0≤λξ2≈0.28<2. Obviously, fu=1+t/3(eu+ueu), and f(t,0)=0, fu(t,0)=1/3(1+t)>0,t∈[0,1]. By direct calculation, we have (94)∫011-tα-2fut,0dt=∫011-tα-21+t3dt<23<0.97≈1-λξ2/2Γαα-1,and f∞=+∞. Thus, (H1) and (H2) hold. By Theorem 13, BVP (92) has at least one positive solution.Example 2.
Consider the following system of fractional differential equations:(95)D0+7/2ut+arctan1+t2sin2u+tet/213u+3+t2=0,t∈0,1,u0=u′′0=u′′′0=0,u′1=3∫01/6usds.Then BVP (95) has at least one positive solution.Proof.
(95) is a BVP of type (2), where(96)ft,u=arctan⁡1+t2sin2⁡u+tet/213u+3+t2,and α=7/2,λ=3,ξ=1/6,0≤λξ2≈0.08<2. By direct calculation, we get that limu→+∞⁡f(t,u)/u=1/3tet/2=ϕ(t) uniformly holds with respect to t on [0,1], ∫01(1-t)α-2ϕ(t)dt<2/3(e1/2-1)<1.28≈(1-λξ2/2)Γ(α)/α-1, and f0=+∞.
Clearly,f(t,u) satisfies conditions (H3). Hence, our conclusion follows from Theorem 14.Example 3.
Consider the following system of fractional differential equations:(97)D0+13/4ut+η2+e1+ηuet/2+3600sin2⁡uln⁡3+t=0,t∈0,1,u0=u′′0=u′′′0=0,u′1=52∫01/5usds.Here η is a positive constant which satisfies 0<η<r-1(L). Then BVP (97) has at least two positive solutions.Proof.
We will apply Theorem15. Corresponding to the problem (2), we take(98)ft,u=η2+e1+ηuet/2+3600sin2⁡uln⁡3+t, and α=13/4,λ=5/2,ξ=1/5,0≤λξ2=0.1<2.
Obviously,fu=η/(2+e)(1+η)et/2+3600sin⁡2uln⁡(3+t)∈C([0,1]×[0,+∞)), and f(t,0)=0, fu(t,0)=η/(2+e)(1+η)et/2>0,t∈[0,1]. Thus,(99)∫011-tα-2fut,0dt<η2+e1+ηe<e2+e<1.075≈1-λξ2/2Γαα-1,and f∞<η/(2+e)(1+η)e<η/1+η<η<r-1(L). This means (H1) and (H2) hold.
In addition, by Lemma5, we get m0=maxt∈[0,1]⁡∫1/21G(t,s)ds-1≤48.9. By direct calculation, we have M=α-1-λ/α-(1-λξ2/2)/α-1≈0.236,sin2M/2≈0.014. Taking r=1, we obtain f(t,u)>3600ln⁡3+tsin2⁡M/2>50.4>m0r, for any u∈[0.0118,1]=[M/2r,r].
By Theorem15, system (97) has at least two positive solutions.



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*Source: 2787569-2019-05-02.xml*