# On the Existence of Solutions for Impulsive Fractional Differential Equations

Authors: Yongliang Guan; Zengqin Zhao; Xiuli Lin
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

We are concerned with a type of impulsive fractional differential equations attached with integral boundary conditions and get the existence of at least one positive solution via global bifurcation techniques.


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## Body

## 1. Introduction

Fractional differential equations have been extensively studied in recent years (see, for instance, [1–7] and their references). In addition, since Rabinowitz established unilateral global bifurcation theorems, there have been many researches in global bifurcation theory and it has been applied to obtain the existence and multiplicity for solutions of differential equations (see, for instance, [8–16] and their references). However, the previous researches seldom involve both global bifurcation techniques and fractional differential equations. In [16], the following problem was studied. (1)D0+αut+rft,ut=0,t∈0,1,tn-αun-2tt=0=tn-αun-3tt=0⋯=tn-αutt=0=u1=0,where the fractional difference was of Riemann-Liouville type. Under suitable conditions, the existence of at least one positive solution and one negative solution was got. We would like to conduct further research on the above problem. For instance, in practical applications, f may rely not only on u(t) but also on u′(t), which will give rise to additional difficulties for the study. Moreover, what if the boundary value conditions are nonlocal rather than local? Can we add impulsive terms into the system? As a reply to above questions, we will tackle the following impulsive Caputo fractional differential equations attached with integral boundary value conditions in this paper.(2)D0+αcut+aft,ut,u′t=0,t∈0,1,u′0=0,u1=∫01ktutdt,Δuξ=∫0ξgtutdt,Δu′ξ=∫0ξhtutdt,where f∈C([0,1]×R×R,R) satisfies f(t,x,y)≥0 when x≥0 and y≤0; k∈C0,1,[0,+∞]; ξ∈(0,1); g,h∈C([0,ξ],(-∞,0]); (3)Δuξ≔limt→ξ+⁡ut-limt→ξ-⁡ut;Δu′ξ≔limt→ξ+⁡u′t-limt→ξ-⁡u′t.We set 1<α≤2 throughout this paper.Through global bifurcation techniques, we get the existence of at least one positive solution of (2) (see Theorem 14). Moreover, for the sake of convenience of use, we give a corollary of Theorem 14 (see Theorem 15).The rest of this paper is organized as follows. In Section2, we will present some preliminary knowledge and some conditions for (2) that we need. We will prove some properties of several functions in Section 3 as a preparation of Section 4. In Section 4, we will present our main results and prove them. In Section 5, we present an example as an application of the main results. Finally, an appendix is given to prove a formula which will be used in the proof of the main results.


## 2. Preliminary

Firstly, we introduce several spaces that this paper needs.In this paper, we set(4)PC0,1=g:gtare  continuous  on0,ξ,ξ,1;there  existslimt→ξ+⁡gt as well as (5)PC10,1=g:g,g′∈PC0,1.Secondly, we introduce some knowledge of fractional integral and fractional derivative.Definition 1 (see [3]).
The fractional integral of orderβ>0 for a function g is defined as (6)I0βgt=1Γβ∫0tt-sβ-1gsds,t≥0.Definition 2 (see [3]).
The Riemann-Liouville derivative of fractional orderα for a function g is defined as (7)D0αrlgt=d2dt2I02-αgt,t>0.
Forg∈Lp(a,b)(1≤p≤+∞), it is well known that D0αrlI0αg(t)=g(t) holds almost everywhere on [a,b]. What is more, D0αrlI0αg(t)=g(t) holds on (a,b) if g∈C(a,b).Definition 3 (see [3]).
The Caputo derivative of fractional orderα for a function is defined as (8)D0αcgt=D0αrlgt-g0-g′0t,t∈0,1.
Clearly, (8) is equivalent to (9)D0αcgt=D0αrlgt-g0Γ1-αt-α-g′0Γ2-αt1-α,t∈0,1.Moreover, for g∈C1[0,1] and ϕ∈L[0,1], it is well known that (10)D0αcgt=ϕt⟹gt=I0αϕt+C0+C1t.However, (10) does not necessarily hold if g does not belong to C1[0,1]. Since u(t) and u′(t) considered in (2) are not continuous at t=ξ, we need some modification on Definition 3. So we generalize Definition 3 in the following way, which will be applied to (2).Definition 4.
The Caputo derivative of fractional orderα for a function g∈PC1[0,1] is defined as (11)D0αcgt=D0αrlg-gξt,t∈0,1,where (12)gξt≔g0+g′0t,t∈0,ξ,g0+gξ+-gξ-+g′0t+g′ξ+t-ξ-g′ξ-t-ξ,t∈ξ,1.With the above definition, we have a conclusion similar to (10), which will be represented as (13).Lemma 5.
Forg∈PC1[0,1] and ϕ∈L[0,1], if D0αcg(t)=ϕ(t) for t∈(0,1), there must be (13)gt=I0αϕt+C1+C2t,t∈0,ξ,I0αϕt+C3+C4t,t∈ξ,1.Proof.
Suppose that(14)D0αcgt=ϕt,t∈0,1;that is (15)D0αrlg-gξt=ϕt,t∈0,1.We know that (16)gt-gξt=I0αϕt+d1tα-1+d2tα-2,t∈0,ξ,I0αϕt+d1tα-1+d2tα-2+d3t-ξα-1+d4t-ξα-2,t∈ξ,1.Then (17)g′t-gξ′t=I0α-1ϕt+d1α-1tα-2+d2α-2tα-3,t∈0,ξ,I0α-1ϕt+d1α-1tα-2+d2α-2tα-3+d3α-1t-ξα-2+d4α-2t-ξα-3,t∈ξ,1.Since g,gξ∈PC1[0,1], we know that g-gξ∈PC1[0,1]. Then by (17) we know that (18)d1=d2=d3=d4=0. So (13) turns to be (19)gt-gξt=I0αϕt,t∈0,1;that is (20)gt=I0αϕt+gξt,t∈0,1.Choosing suitable C1,C2,C3,C4, we finish the proof.If we set(21)Ct=a+bt,t∈0,ξ;c+dt,t∈ξ,1,where a,b,c, and d are constant real numbers, then we will have the following conclusion.Lemma 6.
One has(22)D0αcCt=0,t∈0,1.Proof.
It is easy to see thatC(t)-Cξ(t)=0 for t∈[0,1] and then it is clear that (23)D0αcCt=0,t∈0,1.Lemma 7.
Forp(t)∈PC[0,1], u∈X is a solution of (24)D0αcut+pt=0,t∈0,1,u′0=0,u1=∫01ktutdt,Δuξ=∫0ξgtutdt,Δu′ξ=∫0ξhtutdtif and only if(25a)ut=-1Γα∫0tt-sα-1psds+1Γα∫011-sα-1psds-∫0ξgsusds-1-ξ∫0ξhsusds+∫01ksusds,t∈0,ξ;(25b)ut=-1Γα∫0tt-sα-1psds+1Γα∫011-sα-1psds-1-t∫0ξhsusds+∫01ksusds,t∈ξ,1.Proof.
Ifu∈X is a solution of (24), by Lemma 5 we know that (26)ut=-1Γα∫0tt-sα-1psds+c1+c2t,t∈0,ξ;-1Γα∫0tt-sα-1psds+c3+c4t,t∈ξ,1.
Sinceu′(0)=0, Δu(ξ)=∫0ξ‍g(t)u(t)dt, Δu′(ξ)=∫0ξ‍h(t)u(t)dt, and u(1)=∫01‍k(t)u(t)dt, we know that (27)c2=0;c1+∫0ξgtutdt=c3+c4ξ;c2+∫0ξhtutdt=c4;-1Γα∫011-sα-1psds+c3+c4=∫01ktutdt.Then (28)c1=1Γα∫011-sα-1psds-∫0ξgtutdt-1-ξ∫0ξhtutdt+∫01ktutdt;c2=0;c3=1Γα∫011-sα-1psds-∫0ξhtutdt+∫01ktutdt;c4=∫0ξhtutdt;that is, u satisfies (25a) and (25b).
If (25a) and (25b) hold, by Lemma 6 we know that (29)D0αcut=-D0αcI0αpt.Since p(t)∈PC[0,1], we know that I0αp∈C[0,1] are continuous on [0,1] as well as I0αp(0)=0 and I0αp′(0)=0, so (29) turns to be (30)D0αcut=-D0αcI0αpt=-D0αrlI0αpt=-pt.All the other conditions of (24) can be verified by direct computations and we omit it here.Thirdly, we introduce some knowledge of order cone.Definition 8.
LetX be a real Banach space and let K be a subset of X. Then K is called an order cone if(i)
K is closed, nonempty, and K≠θ;(ii)
a,b∈R,a,b≥0,x,y∈K⇒ax+by∈K;(iii)
x∈K and -x∈K⇒x=θ.
On this basis,u∈K is denoted by u≥θ, while u>θ means that u∈K and u≠θ. Moreover, K is called to be solid if int(K)≠ϕ; that is, K has interior points. u≫θ means that u is an interior point of K.In this paper, we set(31)X=u:u∈PC10,1,limt→0+⁡u′ttα-1exist.Then X is a Banach space endowed with the norm (32)u=supt∈0,1⁡ut+supt∈0,1⁡u′ttα-1.We set the cone(33)K≔u∈X:ut≥0,u′t≤0fort∈0,1.Clearly K is a solid cone in X. By the way, for any u>θ, it is obvious that u(0)>0 and u′(0)=0, which will be useful lately in this paper.Fourthly, we introduce two lemmas which are very important in this paper.Lemma 9 (see [17, Corollary  15.12]).
We set(34)S+=μ,u∈R×X:μ,uis  a  solution  ofu=μLu+Nuwithμ>0,u>θ.If(H1) and(H2) are satisfied, then (r(L)-1,θ) is a bifurcation point of u=μ(Lu+Nu) and S+¯ contains an unbounded solution component C+(r(L)-1) which passes through (r(L)-1,θ).
If additionally(H3) is satisfied, then (μ,u)∈C+(r(L)-1) and μ≠r(L)-1 always imply that μ>0 and u>θ.The conditions that Lemma9 needs are stated as below.(H1)
The operatorsL,N:X→X are compact on the real Banach space X. L+N is positive: that is, (L+N)u≥θ when u≥θ. L is linear and Nu/u→0 as u→0. Moreover, X has an order cone K with X=K-K.(H2)
The spectral radiusr(L) of L is positive.(H3)
L is strongly positive: that is, Lu≫θ for u>θ.Lemma 10 (see [18, Theorem  19.3]).
LetX be a Banach space and let K⊂X be a solid cone. L:X→X is linear, compact, and strongly positive. Then we have the following:(a)
r(L)>0, r(L) is a simple eigenvalue with an eigenvector ν≫θ and there is no other eigenvalue with positive eigenvector.(b)
Fory>θ,λ≤r(L), the equation λu-Lu=y has no solution in K.(c)
LetS:X→X be a linear operator. If Sx-Lx≥θ on K, then r(s)≥r(L), while r(s)>r(L) if Sx-Lx≫θ for x>θ.At last, we present here some conditions that we need in this paper.(C1)
At least one of the following two conditions is satisfied:
(1)
k(s) does not identically vanish on any subinterval of [0,1].(2)
k(0)>0.(C2)
There exista0,a∞∈C0,1,[0,+∞) and b0,b∞∈C([0,1],(-∞,0]) such that (35)limx,y→0fs,x,y-a0sx-b0syx+y=0,limx≥0,y≤0;x+y→∞fs,x,y-a∞sx-b∞syx+y=0for all s∈[0,1] uniformly. What is more, ai(0)>0, i=0,∞.(C3)
There exists a functionφ∈C0,1,[0,+∞) such that (36)fs,x,y≥φsxfor all (s,x,y)∈[0,1]×[0,+∞)×(-∞,0]. In addition, φ(0)>0.


## 3. Property ofH,L0,L∞,(I-G)-1L0 and (I-G)-1L∞

In this paper, we set(37)Hut=-1Γα∫0tt-sα-1fs,us,u′sds+1Γα∫011-sα-1fs,us,u′sds,t∈0,1;Gut=-∫0ξgsusds-1-ξ∫0ξhsusds+∫01ksusds,t∈0,ξ;-1-t∫0ξhsusds+∫01ksusds,t∈ξ,1;L0ut=-1Γα∫0tt-sα-1a0sus+b0su′sds+1Γα∫011-sα-1a0sus+b0su′sds,t∈0,1;L∞ut=-1Γα∫0tt-sα-1a1sus+b∞su′sds+1Γα∫011-sα-1a∞sus+b∞su′sds,t∈0,1.Lemma 11.
H : X → X is positive and compact.Proof.
Firstly, we will prove thatHu∈X for all u∈X. With all the other properties of X easily verified, we only show here that (Hu)′(t)/tα-1 converges, while t→0+.
We know that(38)Hu′t=-1Γα-1∫0tt-sα-2fs,us,u′sds.
So,(39)limt→0+⁡Hu′ttα-1=-1Γα-1limt→0+⁡∫0tt-sα-2fs,us,u′sdstα-1=-limt→0+⁡fτ,uτ,u′τΓα-1∫0tt-sα-2dstα-1,τ∈0,t=-limt→0+⁡fτ,uτ,u′τΓα,τ∈0,t=-f0,u0,0Γα.
Secondly, we show thatH is positive.
In fact, foru≥θ, observing the expression of Hu, it is easy to see that (Hu)(t)≥0 and (Hu)′(t)≤0 for t∈[0,1].
Thirdly, we will prove thatH is compact. To this end, we only need to show that (Hu)(t):u∈D and (Hu)′(t)/tα-1:u∈D are uniformly bounded and equicontinuous on (0,1] for any bounded subset of X named D. We only prove the later one since the proof of another one is easier.
SinceD is bounded, we can choose a constant number M>0 such that u≤M for all u∈D, which implies that ut≤M and u′(t)/tα-1≤M and hence u′t≤M. Then there exists a constant number M¯ such that f(s,u(s),u′(s))≤M¯ for all s∈[0,1],u∈D. So we have (40)Hu′ttα-1=1Γα-1∫0tt-sα-2fs,us,u′stα-1ds≤M¯Γα-1∫0tt-sα-2tα-1ds=M¯Γα,which means that (Hu)′(t)/tα-1:u∈D is uniformly bounded on (0,1].
Next, we will prove that(Hu)′(t)/tα-1:u∈D is equicontinuous on (0,1]. We know that (41)Hu′t1t1α-1-Hu′t2t1α-2=1Γα-1∫0t1t1-sα-2fs,us,u′sdst1α-1-∫0t2t2-sα-2fs,us,u′sdst2α-1=1Γα-1∫011-rα-2ft1r,ut1r,u′t1rdr-∫011-rα-2ft2r,ut2r,u′t2rdr≤1Γα-1∫011-rα-2ft1r,ut1r,u′t1r-ft2r,ut2r,u′t2rdr;t1r-t2r≤t1-t2;ut1r-ut2r=u′t∗rt1-t2r≤Mt1-t2,t∗∈t1,t2;u′t1r-u′t2r≤u′t1rt1rα-1t1rα-1+u′t2rt2rα-1t2rα-1≤Mt1α-1+t2α-1.
By the uniform continuity off on [0,1]×[-M,M]×[-M,M] and (41), we know that, for any ε>0, there exists δ1>0 such that t1,t2∈(0,δ1] implies (42)Hu′t1t1α-1-Hu′t2t1α-2≤ε.
Moreover, whenδ1/2≤t1<t2≤1, we have that (43)Hu′t2t2α-1-Hu′t1t1α-1=1Γα-1∫0t2t2-sα-2fs,us,u′sdst2α-1-∫0t1t1-sα-2fs,us,u′sdst1α-1=1Γα-1∫t1t2t2-sα-2fs,us,u′sdst2α-1+∫0t1t2-sα-2t2α-1-t1-sα-2t1α-1fs,us,u′sds≤M¯Γα-1∫t1t2t2-sα-2dst2α-1+∫0t1t1-sα-2t1α-1-t2-sα-2t2α-1ds=2M¯t2-t1α-1Γαt2α-1≤2αM¯Γαδ1α-1t2-t1α-1. So (44)t2-t1<Γαδ1α-1ε2αM¯1/α-1will  lead  toHu′t2t2α-1-Hu′t1t1α-1≤ε,t1,t2∈δ12,1.
We setδ=min⁡δ1/2,Γ(α)δ1α-1ε/2αM¯1/α-1; by (42) and (44), we know that (Hu)′(t2)/t2α-1-(Hu)′(t1)/t1α-1≤ε for all u∈D and t1,t2∈(0,1] satisfy t2-t1<δ.Lemma 12.
L 0 , L ∞ : X → X are linear, compact, and positive.Proof.
Similar to Lemma11, we can verify that L0,L∞:X→X are compact and positive. Moreover, they are obviously linear operators.Lemma 13.
If(C1) holds and G<1, then (I-G)-1L0 and (I-G)-1L∞ are linear, compact, and strongly positive.Proof.
It is obvious thatG:X→X is a positive linear operator. So if G<1, I-G will have a positive linear bounded inverse operator: (45)I-G-1=I+G+G2+⋯. Since L0 and L∞ are compact, we know that (I-G)-1L0 and (I-G)-1L∞ are compact. In the next, we will only prove that (I-G)-1L0 is strongly positive since the proof for another one is similar.
For any givenx∈X, x>θ, if we set (46)y=I-G-1L0x,we will know that y≥L0x>θ (noticing the expression of L0, we can see that L0 will turn any positive element of X to be a positive element of X). By (46), we know that (47)I-Gy=L0x and hence (48)y=Gy+L0x.Since y>θ, we know that y′(t)≤0 for 0∈[0,1], so by (C1) and (48), we will know that (49)yt≥y1=Gy1+L0x1=∫01ksysds>0.
On the other hand, fort∈(0,1], by (C2) and u(0)>0, we will know that (50)y′t=Gy′t+L0x′t≤L0x′t=-1Γα-1∫0tt-sα-2a0sus+b0su′sds≤-1Γα-1∫0tt-sα-2a0susds<0.What is more, similar to (39), we have (51)limt→0+⁡y′ttα-1=-1Γα-1limt→0+⁡1tα-1∫0tt-sα-2a0sus+b0su′sds=-a00u0Γα<0.
By (50) and (51) and the piecewise continuity of y′(t) on [0,1], we know that there must exist a positive number τ such that (52)supt∈0,1⁡y′ttα-1<-τ.
Equations (49) and (52) exactly mean that y≫θ.


## 4. Main Results and the Proof

Theorem 14.
Suppose that(C1)–(C3) hold and G<1; then there must exist at least one positive solution of (2) if 1/r((I-G)-1L∞)<a<1/r((I-G)-1L0) or 1/r((I-G)-1L∞)>a>1/r((I-G)-1L0).Proof.
The proof is divided into three parts.
InPart 1, we consider the auxiliary equation (see (53)) whose solutions of the kind μ=1 will be the solutions of (2) and we transform it into a functional operator equation (see (57)).
InPart 2, we will verify that (I-G)-1L0, (I-G)-1N0 satisfy all the conditions required to apply Lemmas 9 and 10, where N0 will be defined inPart 1.
InPart 3, we apply Lemmas 9 and 10 to get the existence of at least one positive solution of (2).
Part 1. We consider the following problem. (53)D0+αcut+μaft,ut,u′t=0,t∈0,1,u′0=0,u1=∫01ktutdt,Δuξ=∫0ξgtutdt,Δu′ξ=∫0ξhtutdt.
We call(μ,u)∈R×X to be a solution of (53) if it satisfies (53). It is clear that any solution of (53) of the form (1,u) yields a solution u of (2).
Due to Lemma7, (μ,u)∈R×X is a solution of (53) if and only if (54)u=μaHu+Gu;that is, (55)u=μaI-G-1Hu.If we set (56)N0u=Hu-L0u, then u is a solution of (53) if and only if (57)u=μaI-G-1L0u+I-G-1N0u.
Part 2. By Lemmas 11–13, we have confirmed that (I-G)-1H,(I-G)-1L0,(I-G)-1N0 are compact, (I-G)-1H is positive, and (I-G)-1L0 is strongly positive. Now we only need to verify that (58)lim∥u∥→0⁡I-G-1N0uu=0.∀ε>0, by (C2), there must exist a number δ>0 such that x<δ and y<δ imply (59)fs,x,y-a0sx-b0syx+y<Γα+12ε.Then, if u<δ, there will be (60)N0utu=-1Γα∫0tt-sα-1fs,us,u′s-a0sus-b0su′suds+1Γα∫011-sα-1fs,us,u′s-a0sus-b0su′suds≤1Γα∫0tt-sα-1fs,us,u′s-a0sus-b0su′suds+1Γα∫011-sα-1fs,us,u′s-a0sus-b0su′suds≤1Γα∫0tt-sα-1fs,us,u′s-a0sus-b0su′sus+u′sds+1Γα∫011-sα-1fs,us,u′s-a0sus-b0su′sus+u′sds≤αε2∫0tt-sα-1ds+αε2∫011-sα-1ds=εtα2+ε2≤ε.
Moreover, we know that(61)N0u′ttα-1u=1Γα-1∫0tt-sα-2fs,us,u′s-a0sus-b0su′stα-1uds≤1Γα-1∫0tt-sα-2fs,us,u′s-a0sus-b0su′stα-1uds≤1Γα-1∫0tt-sα-2tα-1fs,us,u′s-a0sus-b0su′sus+u′sds≤1Γα-1Γα+12ε∫0tt-sα-2tα-1ds=1Γα-1Γα+12ε1α-1=αε2≤ε.
By (60) and (61), we will know that (62)limu→0⁡N0uu=0.Equation (62) together with the boundedness of (I-G)-1 will lead to (63)limu→0⁡I-G-1N0uu⟶0.
Part 3. Applying Lemmas 9 and 10, we can draw a conclusion as below.
For (57), from (1/ar((I-G)-1L0),θ) there emanates an unbounded continua of positive solutions C+⊂D¯+, where (64)D+=μ,u∈R×X:u=μaI-G-1Huwithμ>0,u>θ.Furthermore, (μ,u)∈C+ and μ≠1/ar((I-G)-1L0) always imply that u>θ.
To verify the existence of at least one positive solution of (2), we only need to show that C+ crosses the hyperplane {1}×X in R×X. To this end, it will be enough to show that C+ joins (1/ar((I-G)-1L0),θ) to (1/ar((I-G)-1L∞),+∞).
Suppose that(μn,un)∈C+ satisfy μn+un→∞. To begin with, we will show that {μn} is bounded. Defining Lφ as (65)Lφut=-1Γα∫0tt-sα-1φsusds+1Γα∫011-sα-1φsusds,t∈0,1, similar to Lemma 13, we can verify that (I-G)-1Lφ:K→K is linear, compact, and strongly positive. Then, due to Lemma 10(b), we see that there is a contradiction for sufficiently large n in the following inequality: (66)un=μnaI-G-1Hun≥μnaI-G-1Lφun. Then, {μn} is bounded and hence limn→∞⁡un→∞.
We choose a subsequence of{μn} converging to μ∗. Without loss of generality, we relabel the subsequence to be {μn} just for convenience.
Now we set(67)N∞u=Hu-L∞u; then (68)vn=unun=μnaI-G-1L∞unun+μnaI-G-1N∞unun=μnaI-G-1L∞vn+μnaI-G-1N∞unun.Since (I-G)-1L∞ is compact, we can find a subsequence of {(I-G)-1L∞vn} named {(I-G)-1L∞vnk} converging to v0∈X. Moreover, we know that (69)limu∈K,u→∞⁡I-G-1N∞uu=θsee  the  proof  in  the  appendix, so (70)limk→∞⁡vnk=μ∗av0≔v∗. It is obvious that v∗>θ and v∗=1. Let k→∞ in vnk=μnka(I-G)-1L∞vnk+μnka(I-G)-1N∞unk/unk; we know that (71)v∗=μ∗aI-G-1L∞v∗.
Since(I-G)-1L∞:X→X is linear, compact, and strongly positive, by Lemma 10 and (71) we know that (72)μ∗=1arI-G-1L∞, which implies that C+ joins (1/ar((I-G)-1L0),θ) to (1/ar((I-G)-1L∞),+∞).As a corollary of Theorem14, we have the following conclusion.Theorem 15.
Suppose that(C1)–(C3) hold. If (73)-∫0ξgsds-1-ξ∫0ξhsds+∫01ksds<1,-∫0ξhsds<ξα-1,then there must exist at least one positive solution of (2) when 1/r((I-G)-1L∞)<a<1/r(I-G)-1L0 or 1/r((I-G)-1L∞)>a>1/r((I-G)-1L0).Proof.
With all the other conditions of Theorem14 satisfied, we only need to verify that G<1. For any u∈X and t∈[0,ξ], we have (74)Gut=-∫0ξgsusds-1-ξ∫0ξhsusds+∫01ksusds≤-∫0ξgsusds-1-ξ∫0ξhsusds+∫01ksusds≤-∫0ξgsds-1-ξ∫0ξhsds+∫01ksdsu.
For anyu∈X and t∈(ξ,1], we have (75)Gut=-1-t∫0ξhsusds+∫01ksusds≤-1-ξ∫0ξhsusds+∫01ksusds≤-1-ξ∫0ξhsds+∫01ksdsu.
For anyu∈X and t∈[0,ξ], we have (76)Gu′t=0.For any u∈X and t∈(ξ,1], we have (77)Gu′ttα-1=∫0ξhsusdstα-1≤-∫0ξhsusdstα-1<-∫0ξhsdsξα-1u.By (74)–(77), we know that G<1.


## 5. Example

Example 1.
Consider the following problem.E1D0+3/2cut+a2ut+eu′tsin⁡ut-u′t=0,t∈0,1,u′0=0,u1=∫011-tutdt,Δu12=-∫01/2t2utdt,Δu′12=-∫01/2tutdt. If we set (78)L0ut=-1Γ3/2∫0tt-sα-13us-u′sds+1Γ3/2∫011-sα-13us-u′sds;L∞ut=-1Γ3/2∫0tt-sα-12us-u′sds+1Γ3/2∫011-sα-12us-u′sds;Gut=∫01/2s2usds+12∫01/2susds+∫011-susds,t∈0,12;1-t∫01/2susds+∫011-susds,t∈12,1then there must be at least one positive solution of E1 when 1/r((I-G)-1L∞)>a>1/r(I-G)-1(L0). (We can verify that r((I-G)-1L0)≥r((I-G)-1L∞) by Lemma 10.)Proof.
Letα=3/2, ξ=1/2, f(s,x,y)=2x+eysin⁡x-y, g(s)=-s2, h(s)=-s, k(s)=1-s; then E1 turns to be (2). Now we set a0(s)≡3, b0(s)≡-1, a∞(s)≡2, b∞(s)≡-1, φ(s)≡1. We can verify that (C1)–(C3) hold. Then by Theorem 15 we know that there must be at least one positive solution of E1 when 1/r((I-G)-1L∞)>a>1/r(I-G)-1(L0).



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*Source: 1207456-2017-12-07.xml*