# Some Results on Fractional m-Point Boundary Value Problems

Authors: Huijuan Zhu; Baozhi Han; Jun Shen
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we will apply some fixed-point theorems to discuss the existence of solutions for fractional m-point boundary value problemsD0+qu″t=htfut,t∈0,1,1<q≤2,u′0=u″0=u1=0,u″1−∑i=1m−2αiu‴ξi=0. In addition, we also present Lyapunov’s inequality and Ulam-Hyers stability results for the given m-point boundary value problems.


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## Body

## 1. Introduction

Mathematical models due to fractional differential equations can describe the natural phenomenon in physics, population dynamics, chemical technology, biotechnology, aerodynamics, electrodynamics of complex medium, polymer rheology, and control of dynamical systems (see [1–4]). Due to the nonlocal characteristics and the rapid development of the theory of fractional operators, some authors have investigated different aspects of fractional differential equations including existence of solutions, Lyapunov’s inequality, and Hyers-Ulam stability for fractional differential equations by different mathematical techniques. For example, first, many authors have discussed the existence of nontrivial solutions of fractional differential equations in nonsingular case as well as singular case. Usually, the proof is based on either the method of upper and lower solutions, fixed-point theorems, alternative principle of Leray-Schauder, topological degree theory, or critical point theory. We refer the readers to [5–20]. Second, Lyapunov, during his study of general theory of stability of motion in 1892, introduced the stability criterion for second-order differential equations, which yielded a counter inequality be called Lyapunov inequality (see [21, 22]). Since then, we can find considerable modifications of Lyapunov-type inequality of differential equations, such as linear differential-algebraic equations, fractional differential equations, extreme Pucci equations, and dynamic equations, which are applied to study the stability and disconjugacy or oscillatory criterion for the mentioned problems, and we refer the readers to [23–32]. Finally, the stability of functional equations was originally raised by Hyers in 1941 (see [33, 34]). Thereafter, the stability properties of all kinds of equations have attracted the attention of many mathematicians. To see more details on the Ulam-Hyers stability and Ulam-Hyers-Rassias of differential equations, we refer the readers to [35–38].Inspired by the references, this paper is mainly concerned with the existence, Lyapunov’s inequality, and Ulam-Hyers stability results for the m-point boundary value problems.(1)D0+qu″t=htfut,t∈0,1,1<q≤2,u′0=u″0=u1=0,u″1−∑i=1m−2αiu‴ξi=0,where αi, ξi, h, and f satisfy the following assumptions:(H1)∑i=1m−2αiξiq−2>1/q−1 and ∑i=2m−2αi≤1−ξm−2q−1/q−1ξm−2q−21−ξm−2q−1/q−1ξm−2q−2(H2)h:0,1⟶ℝ is Lebesgue integral(H3)f:ℝ⟶ℝ is continuousFor these goals, we first convert problem (1) into an integral equation via Green function. Furthermore, we study the properties and estimates of the Green function. Then, on the basis of these properties, we apply some fixed-point theorems to establish some existence results of problem (1) under some suitable conditions. In addition, the Lyapunov inequality and Hyers-Ulam stability of the proposed problem are also considered.


## 2. Preliminaries

Before beginning the main results, we state some classic and modified definitions and lemmas from fractional calculus.Definition 1 [4].
The fractional integral of orderq>0 of a function u:0,+∞⟶R is given by
(2)I0+qut=1Γq∫0tt−sq−1usds,provided the right-hand side is pointwise defined on 0,+∞.Definition 2 [4].
The fractional derivative of orderq>0 of a continuous function u:0,+∞⟶R is given by
(3)D0+qut=1Γn−qddtn∫0tust−sq−n−1ds,where n=q+1, provided that the right-hand side is pointwise defined on 0,+∞.Definition 3 [21].
Assume thatq>0, then
(4)I0+qD0+qut=ut+∑i=1nCitq−i,for some Ci∈R,i=1,2,⋯,n, where n is the smallest integer greater than or equal to q.Lemma 4.
Assume that (H1) holds. Then, for anyyt∈L10,1, the boundary value problem
(5)D0+qu″t=yt,t∈0,1,1<q≤2,u′0=u″0=u1=0,u″1−∑i=1m−2αiu‴ξi=0,has a unique solution ut=∫01Gt,sysds. Let p=1−q−1∑i=1m−2αiξiq−2<0,, and we have
(i)
fors≤t, s≤ξ1(6)Gt,s=1Γq+2t−sq+1−1−sq+1+1−tq+1−1−tq+11−1−sq−1p(ii)
fort≤s≤ξ1(7)Gt,s=1Γq+2−1−sq+1+1−tq+1−1−tq+11−1−sq−1p(iii)
fors≤t, ξj≤s≤ξj+1, j=1,2,⋯,m−3(8)Gt,s=1Γq+21−tq+1p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2+t−sq+1−1−sq+1(iv)
fort≤s, ξj≤s≤ξj+1, j=1,2,⋯,m−3(9)Gt,s=1Γq+21+tq+1p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2−1−sq+1(v)
forξm−2≤s≤t(10)Gt,s=1Γq+2t−sq+1−1−sq+1+1−tq+1p1−sq−1(vi)
forξm−2≤s, t≤s(11)Gt,s=1Γq+2−1−sq+1+1−tq+1p1−sq−1Proof.
From Definition3, it follows that
(12)u″t=I0+qyt−C1tq−1−C2tq−2.
Sinceu″0=0, it is clear that C2=0. Then,
(13)u″t=I0+qyt−C1tq−1=1Γq∫0tt−sq−1ysds−C1tq−1.
On one hand, taking the derivative ofu′′t, we can get
(14)u‴t=q−1Γq∫0tt−sq−2ysds−q−1C1tq−2.
On the other hand, combining the boundary conditionsu1=u′0=0, we have
(15)u′t=1Γq∫0t∫0ττ−sq−1ysdτ−C1qtq=1Γq+1∫0tt−sqysds−C1qtq.
Furthermore, we have(16)ut=−1Γq+1∫t1∫0ττ−sqysdsdτ+C1q∫t1τqdτ=−1Γq+2∫0t1−sq+1ysds−∫0tt−sq+1ysds+C11−tq+1qq+1.
According to these above expressions, we have(17)u″1=1Γq∫011−sq−1ysds−C1,∑i=1m−2αiu‴ξi=q−1Γq∑i=1m−2αi∫0ξiξi−sq−2ysds−q−1C1∑i=1m−2αiξiq−2.
Then, fromu″1−∑i=1m−2αiu‴ξ1=0, it follows that
(18)C1=1pΓq∫011−sq−1ysds−q−1∑i=1m−2αi∫0ξiξi−sq−2ysds,which yields
(19)ut=1Γq+2∫0tt−sq+1ysds−∫011−sq+1ysds+1−tq+1p∫011−sq−1ysds−q−1∑i=1m−2αi∫0ξiξi−sq−2ysds.
Ifs≤t, s≤ξ1, we have
(20)Gt,s=1Γq+2t−sq+1−1−sq+1+1−tq+1p1−sq−1−q−1∑i=1m−2αiξi−sq−2=1Γq+2t−sq+1−1−sq+1+1−tq+1−1−tq+1p1−1−sq−1.☐In the similar way, we also can get the expression ofGt,s on other intervals.Lemma 5.
Assume that (H1) holds. Then,Gt,s satisfies the following properties:
(I)
Sign ofGt,s(i)
Gt,s≥0,for0≤s≤ξ1(ii)
Gt,s≤0,forξ1<s≤1(II)
The range ofGt,s(1)
For0≤s≤ξ1(21)0≤Gt,s<1Γq+21−1−sq+1−1−1−sq−1p(2)
Forξj≤s≤ξj+1, j=1,2,⋯,m−3(22)1Γq+21p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2−1−sq+1≤Gt,s≤0(3)
Forξm−2≤s≤1(23)1Γq+21−sq−1p−1−sq+1≤Gt,s≤0Proof.
For0≤s≤ξ1, by the definition of Gt,s, it is clear that Gt,s is continuous and derivativable with respect to t at 0,1. On one hand, if s≤t≤1, we have
(24)∂G∂t=1Γq+1t−sq−tq+tq1−1−sq−1p≤0.
On the other hand, if0≤t<s, we have
(25)∂G∂t=tqΓq+1−1+1−1−sq−1p≤0.
Then,Gt,s is nonincreasing on t, which yields that
(26)minGt,s:t∈0,1=G1,s=0,maxGt,s:t∈0,1=G0,s=1Γq+21−1−sq+1−1−1−sq−1p.
So for0≤s<ξ1, 0≤t≤1, it concludes that
(27)0≤Gt,s<1Γq+21−1−sq+1−1−1−sq−1p.
Forξj≤s≤ξj+1j=1,2,⋯,m−3, we have
(28)minGt,s:t∈0,1=G0,s=1Γq+21p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2−1−sq+1,maxGt,s:t∈0,1=G1,s=0.
Forξm−2≤s≤1, we have
(29)minGt,s:t∈0,1=G0,s=1Γq+21−sq−1p−1−sq+1,maxGt,s:t∈0,1=G1,s=0.☐Let(30)G1=maxs∈0,ξ11Γq+21−1−sq+1−1−1−sq−1p=1Γq+21−1−ξ1q+1−1−1−ξ1q−1p,G2=maxs∈ξm−2,11Γq+21−sq−1p−1−sq+1=1Γq+21−ξm−2q+1−1−ξm−2q−1p,G3=max1≤j≤m−3Gj3,where
(31)Gj3=maxs∈ξj,ξj+11Γq+21p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2−1−sq+1.From Lemma5, it is clear that Gt,s≤G¯, where G¯=maxG1,G2,G3.Lemma 6.
Assume that (H1) holds andξ1>1−1/21/q−1. Then, G¯=G1.Proof.
Let(32)Gjs=1Γq+21p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2−1−sq+1,s∈ξj,ξj+1,j=1,2,⋯,m−3.
From (H1) andξ1>1−1/21/q−1, we can verify that
(33)G1−Gjs=1Γq+21−1−ξ1q+1−1p1−1−ξ1q−1−1−sq+1+1p1−sq−1−q−1∑i=j+1m−2αiξi−sq−2≥1Γq+21−1−ξ1q+1−1−ξjq+1−1p1−1−ξ1q−1−1−ξ1q−1−q−1p∑i=j+1m−2αiξi−ξjq−2≥1Γq+21−21−ξ1q+1−1p1−21−ξ1q−1−q−1p∑i=j+1m−2αiξi−ξjq−2≥1Γq+21−122/q−1−q−1p∑i=j+1m−2αiξi−ξj>0.
Also, we can verify that(34)G1−G2=1Γq+21−1−ξ1q+1−1p1−1−ξ1q−1−1−ξm−2q+1+1p1−ξm−2q−1≥1Γq+21−21−ξ1q+1−1p1−21−ξ1q−1≥1Γq+21−122/q−1>0.
So, it concludes thatG1>G2,G1>G3, namely, G¯=G1.☐


## 3. Main Results

### 3.1. Existence Results

Theorem 7.
Assume that (H1)-(H3) hold. In addition, there exists a positive constantL>0 such that
(35)fu−fv≤Lu−v,∀u,v∈ℝ.Then, problem (1) has a unique solution if LG¯hL1<1.Proof.
LetC0,1=xt:xtiscontinuouson0,1 is a Banach space with the norm x=max0≤t≤1xt. From Lemma 4, it is clear that solutions of (1) can be rewritten as fixed points of operator T, which is defined by
(36)Tut=∫01Gt,shsfusds.
Now, we show thatT:Br⟶Br and T is a contraction map, where Br=u∈E:u<r with
(37)r>G¯hL1f01−G¯LhL1.
On one hand, for anyu∈Br, we have
(38)Tut=∫01Gt,shsfusds≤G¯∫01hsfusds≤G¯∫01hsfu−f0+f0ds≤G¯∫01hsLr+f0ds≤G¯hL1Lr+f0≤r,which implies that TBr⊂Br.
On the other hand, for anyu, v∈E, we have
(39)Tu−Tv=∫01Gt,shsfusds≤G¯∫01hsfus−fvsds≤G¯L∫01hsus−vsds≤LG¯hL1ut−vt,which implies that T is a contraction map.☐Therefore, by the Banach contraction mapping principle, it follows that the operatorT has a unique fixed point, which is the unique solution for problem (1).Theorem 8.
Assume that (H1)-(H3) hold. In addition, there exists a positive constantK such that fu≤K for u ∈ R. Then, problem (1) has at least one solution.The proof is based on the following fixed-point theorem.Lemma 9 [39].
LetE be a Banach space, E1 is a closed, convex subset of E, Ω an open subset of E1, and 0∈Ω. Suppose that T:Ω⟶E1 is completely continuous. Then, either
(i)
T has a fixed point in Ω, or(ii)
there areu∈∂Ω (the boundary of Ω in E1) and ρ∈0,1 with u=ρTuProof of Theorem 8.
First, we show that the operatorT is uniformly bounded.
For anyu∈Ω¯δ=u∈C0,1:u≤δ, we have
(40)Tut=∫01Gt,shsfusds≤G¯KhL1,which implies that TΩδ is uniformly bounded.
Second, for0≤s≤ξ1, from Lemma 4, we have
(i)
ifs≤t≤1(41)∂Gt,s∂t=1Γq+1t−sq−tq+tq1−1−sq−1p≤1Γq+1t−sq+tq−tq1−1−sq−1p≤1Γq+12−1p.(ii)
if0≤t<s(42)∂Gt,s∂t=1Γq+1−tq+tq1−1−sq−1p≤1Γq+1tq−tq1−1−sq−1p≤1Γq+11−1p,which implies that ∂Gt,s/∂t is bounded for 0≤s<ξ1, 0≤t≤1. In the similar way, we know that there exists a S>0 such that ∂Gt,s/∂t≤S for 0≤s, t≤1.
Furthermore, fort1,t2∈0,1, we have
(43)Tut2−Tut1=∫01Gt2,shsfusds−∫01Gt1,shsfusds=∫01Gt2,s−Gt1,shsfusds≤SKhL1t2−t1.
Therefore, applying the Arzela-Ascoli theorem [39], we can find that TΩδ is relatively compact.
Third, we claim thatT:Ω¯δ⟶ℝ is continuous. Assume that unn=1∞⊂Ω¯δ, which converges to u0 uniformly on [0,1]. Since Tuntn=1∞ is uniformly bounded and equicontinuous on [0,1], from the Arzela-Ascoli theorem, it follows that there exists a uniformly convergent subsequence in Tuntn=1∞. Let Tunmtm=1∞ be a subsequence which converges to vt uniformly on [0,1]. Observe that
(44)Tunmt=∫01Gt,shsfunmsds.
Furthermore, by Lebesgue’s dominated convergence theorem and lettingm⟶∞, we have
(45)vt=∫01Gt,shsfu0sds,namely, vt=Tu0t. This shows that each subsequence of Tuntn=1∞ uniformly converges to vt. Therefore, the sequence Tuntn=1∞ uniformly converges to Tu0t. This means that T is continuous at u0∈Ωδ. So, T is continuous on Ω¯δ. Thus, T is completely continuous.
Finally, letΩδ=u∈C0,1:u<δ with δ=G¯KhL1+1. If u is a solution of problem (1), then, for ρ∈0,1, u∈∂Ωδ, we have
(46)u=ρTut=ρ∫01Gt,shsfusds≤ρG¯∫01hsfusds≤G¯KhL1,which yields a contradiction. Therefore, by Lemma 9, the operator T has a fixed point in Ωδ.Theorem 10.
Assume that (H1)-(H3) hold. In addition,f satisfies the following assumptions:
(H4) There exists a nondecreasing functionψ:ℝ+⟶ℝ+ such that
(47)fu≤ψu,∀u∈ℝ.
(H5) There exists a constantR>0 such that R/G¯hL1ψR>1. Then, problem (1) has at least one solution.Proof.
Now we show that (ii) of Lemma9 does not hold. If u is a solution of problem (1), then for ρ∈0,1, we obtain
(48)u=ρTut≤ρ∫01Gt,shsfusds≤ρG¯∫01hsfusds≤G¯hL1ψu.
LetBR=u∈C0,1:u<R. From the above inequality and (H5), it yields a contradiction. Therefore, by Lemma 9, the operator T has a fixed point in BR.☐


### 3.2. Lyapunov’s Inequality

Theorem 11.
Assume that (H1)-(H3) hold. In addition,fu is a concave function on ℝ. Then, for any nontrivial solution of problem (1), we have
(49)∫01htdt>uG¯maxu∈u∗,u∗fu,where
(50)u∗=mint∈0,1ut,u∗=maxt∈0,1ut.Proof.
Ifut is a nontrivial solution of problem (1), then by Lemma 4, we have
(51)ut=∫01Gt,shsfusds.
Furthermore, by Lemma6, we have
(52)ut≤∫01Gt,shsfusds.
Sincef is continuous and concave, then from Jensen’s inequality, it follows that
(53)ut≤maxt∈0,1∫01Gt,shsfusds≤∫01maxt∈0,1Gt,shsfusds≤G¯htL1∫01hsΛtL1fusds≤G¯maxu∈u∗,u∗fuhL1,namely,
(54)∫01htdt>uG¯maxu∈u∗,u∗fu.☐


### 3.3. Stability Analysis

Definition 12 [34].
Equation (1) is said to be Ulam-Hyers-Rassias stability with respect to Ψ∈C0,1 if there exists a nonzero positive real number μ such that for every ε>0 and each solution v∈C0,1 of the inequality
(55)D0+qv″t−htfvt≤εΨt,t∈0,1,there exists a solution u∈C0,1 of problem (1) such that ut−vt≤μεΨt, t∈0,1.Theorem 13.
Assume that (H1)-(H3) hold. In addition, there exists a positive constantL>0 such that
(56)fu−fv≤Lu−v,∀u,v∈C0,1.Then, problem (1) is Ulam-Hyers-Rassias stability if LGhL1<1.Proof.
Letv∈C0,1 be the solution of the inequality (55); then,
(57)D0+1v″t−htfvt≤εΨt,t∈0,1.
Thus, forε>0, we get
(58)vt−∫01Gt,shsfvsds≤εΨt,t∈0,1.
By Theorem7, problem (1) has a solution ut satisfies
(59)ut=∫01Gt,shsfusds.
Then, fort∈0,1, we have
(60)vt−ut=vt−∫01Gt,shsfusds≤vt−∫01Gt,shsfvsds+∫01Gt,shsfus−fvsds≤εΨt+L∫01Gt,shsus−vsds≤εΨt+LG¯∫01hsus−vsds≤εΨt+LG¯hL1ut−vt,which yields
(61)vt−ut≤εΨt1−LG¯hL1=μεΨt,tε0,1.
Therefore, problem (1) is Ulam-Hyers-Rassias stability.☐


## 3.1. Existence Results

Theorem 7.
Assume that (H1)-(H3) hold. In addition, there exists a positive constantL>0 such that
(35)fu−fv≤Lu−v,∀u,v∈ℝ.Then, problem (1) has a unique solution if LG¯hL1<1.Proof.
LetC0,1=xt:xtiscontinuouson0,1 is a Banach space with the norm x=max0≤t≤1xt. From Lemma 4, it is clear that solutions of (1) can be rewritten as fixed points of operator T, which is defined by
(36)Tut=∫01Gt,shsfusds.
Now, we show thatT:Br⟶Br and T is a contraction map, where Br=u∈E:u<r with
(37)r>G¯hL1f01−G¯LhL1.
On one hand, for anyu∈Br, we have
(38)Tut=∫01Gt,shsfusds≤G¯∫01hsfusds≤G¯∫01hsfu−f0+f0ds≤G¯∫01hsLr+f0ds≤G¯hL1Lr+f0≤r,which implies that TBr⊂Br.
On the other hand, for anyu, v∈E, we have
(39)Tu−Tv=∫01Gt,shsfusds≤G¯∫01hsfus−fvsds≤G¯L∫01hsus−vsds≤LG¯hL1ut−vt,which implies that T is a contraction map.☐Therefore, by the Banach contraction mapping principle, it follows that the operatorT has a unique fixed point, which is the unique solution for problem (1).Theorem 8.
Assume that (H1)-(H3) hold. In addition, there exists a positive constantK such that fu≤K for u ∈ R. Then, problem (1) has at least one solution.The proof is based on the following fixed-point theorem.Lemma 9 [39].
LetE be a Banach space, E1 is a closed, convex subset of E, Ω an open subset of E1, and 0∈Ω. Suppose that T:Ω⟶E1 is completely continuous. Then, either
(i)
T has a fixed point in Ω, or(ii)
there areu∈∂Ω (the boundary of Ω in E1) and ρ∈0,1 with u=ρTuProof of Theorem 8.
First, we show that the operatorT is uniformly bounded.
For anyu∈Ω¯δ=u∈C0,1:u≤δ, we have
(40)Tut=∫01Gt,shsfusds≤G¯KhL1,which implies that TΩδ is uniformly bounded.
Second, for0≤s≤ξ1, from Lemma 4, we have
(i)
ifs≤t≤1(41)∂Gt,s∂t=1Γq+1t−sq−tq+tq1−1−sq−1p≤1Γq+1t−sq+tq−tq1−1−sq−1p≤1Γq+12−1p.(ii)
if0≤t<s(42)∂Gt,s∂t=1Γq+1−tq+tq1−1−sq−1p≤1Γq+1tq−tq1−1−sq−1p≤1Γq+11−1p,which implies that ∂Gt,s/∂t is bounded for 0≤s<ξ1, 0≤t≤1. In the similar way, we know that there exists a S>0 such that ∂Gt,s/∂t≤S for 0≤s, t≤1.
Furthermore, fort1,t2∈0,1, we have
(43)Tut2−Tut1=∫01Gt2,shsfusds−∫01Gt1,shsfusds=∫01Gt2,s−Gt1,shsfusds≤SKhL1t2−t1.
Therefore, applying the Arzela-Ascoli theorem [39], we can find that TΩδ is relatively compact.
Third, we claim thatT:Ω¯δ⟶ℝ is continuous. Assume that unn=1∞⊂Ω¯δ, which converges to u0 uniformly on [0,1]. Since Tuntn=1∞ is uniformly bounded and equicontinuous on [0,1], from the Arzela-Ascoli theorem, it follows that there exists a uniformly convergent subsequence in Tuntn=1∞. Let Tunmtm=1∞ be a subsequence which converges to vt uniformly on [0,1]. Observe that
(44)Tunmt=∫01Gt,shsfunmsds.
Furthermore, by Lebesgue’s dominated convergence theorem and lettingm⟶∞, we have
(45)vt=∫01Gt,shsfu0sds,namely, vt=Tu0t. This shows that each subsequence of Tuntn=1∞ uniformly converges to vt. Therefore, the sequence Tuntn=1∞ uniformly converges to Tu0t. This means that T is continuous at u0∈Ωδ. So, T is continuous on Ω¯δ. Thus, T is completely continuous.
Finally, letΩδ=u∈C0,1:u<δ with δ=G¯KhL1+1. If u is a solution of problem (1), then, for ρ∈0,1, u∈∂Ωδ, we have
(46)u=ρTut=ρ∫01Gt,shsfusds≤ρG¯∫01hsfusds≤G¯KhL1,which yields a contradiction. Therefore, by Lemma 9, the operator T has a fixed point in Ωδ.Theorem 10.
Assume that (H1)-(H3) hold. In addition,f satisfies the following assumptions:
(H4) There exists a nondecreasing functionψ:ℝ+⟶ℝ+ such that
(47)fu≤ψu,∀u∈ℝ.
(H5) There exists a constantR>0 such that R/G¯hL1ψR>1. Then, problem (1) has at least one solution.Proof.
Now we show that (ii) of Lemma9 does not hold. If u is a solution of problem (1), then for ρ∈0,1, we obtain
(48)u=ρTut≤ρ∫01Gt,shsfusds≤ρG¯∫01hsfusds≤G¯hL1ψu.
LetBR=u∈C0,1:u<R. From the above inequality and (H5), it yields a contradiction. Therefore, by Lemma 9, the operator T has a fixed point in BR.☐


## 3.2. Lyapunov’s Inequality

Theorem 11.
Assume that (H1)-(H3) hold. In addition,fu is a concave function on ℝ. Then, for any nontrivial solution of problem (1), we have
(49)∫01htdt>uG¯maxu∈u∗,u∗fu,where
(50)u∗=mint∈0,1ut,u∗=maxt∈0,1ut.Proof.
Ifut is a nontrivial solution of problem (1), then by Lemma 4, we have
(51)ut=∫01Gt,shsfusds.
Furthermore, by Lemma6, we have
(52)ut≤∫01Gt,shsfusds.
Sincef is continuous and concave, then from Jensen’s inequality, it follows that
(53)ut≤maxt∈0,1∫01Gt,shsfusds≤∫01maxt∈0,1Gt,shsfusds≤G¯htL1∫01hsΛtL1fusds≤G¯maxu∈u∗,u∗fuhL1,namely,
(54)∫01htdt>uG¯maxu∈u∗,u∗fu.☐


## 3.3. Stability Analysis

Definition 12 [34].
Equation (1) is said to be Ulam-Hyers-Rassias stability with respect to Ψ∈C0,1 if there exists a nonzero positive real number μ such that for every ε>0 and each solution v∈C0,1 of the inequality
(55)D0+qv″t−htfvt≤εΨt,t∈0,1,there exists a solution u∈C0,1 of problem (1) such that ut−vt≤μεΨt, t∈0,1.Theorem 13.
Assume that (H1)-(H3) hold. In addition, there exists a positive constantL>0 such that
(56)fu−fv≤Lu−v,∀u,v∈C0,1.Then, problem (1) is Ulam-Hyers-Rassias stability if LGhL1<1.Proof.
Letv∈C0,1 be the solution of the inequality (55); then,
(57)D0+1v″t−htfvt≤εΨt,t∈0,1.
Thus, forε>0, we get
(58)vt−∫01Gt,shsfvsds≤εΨt,t∈0,1.
By Theorem7, problem (1) has a solution ut satisfies
(59)ut=∫01Gt,shsfusds.
Then, fort∈0,1, we have
(60)vt−ut=vt−∫01Gt,shsfusds≤vt−∫01Gt,shsfvsds+∫01Gt,shsfus−fvsds≤εΨt+L∫01Gt,shsus−vsds≤εΨt+LG¯∫01hsus−vsds≤εΨt+LG¯hL1ut−vt,which yields
(61)vt−ut≤εΨt1−LG¯hL1=μεΨt,tε0,1.
Therefore, problem (1) is Ulam-Hyers-Rassias stability.☐


## 4. Examples

Now we give some examples to illustrate our main results.Example 1.
We consider the following problem:(62)D0+3/2u″t=6tarctanu,t∈0,1,u′0=u″0=u1=0,u″1−2u‴45−18u‴67=0,where ht=12t and fu=1/2arctanu. It is obvious that (H1)-(H3) hold. Via some computations, we have
(63)ξ1=45>1−121/q−1=34,p=1−52−7166,G¯=8105π1−155−5−15−5/2−35/166≈0.170285.
Since(64)f′u=12arctanu′=121+u2≤12=L,the function f satisfies the condition
(65)fu−fv≤Lu−v,∀u,v∈C0,1.
Furthermore, we can verify thatGLhL1≈0.510855<1. Therefore, by Theorem 7 and Theorem 13, problem (62) has a solution ut, which is Ulam-Hyers-Rassias stability.Example 2.
Let us consider the following problem:(66)D0+3/2u″t=6t2arctanu1/2+sinu,t∈0,1,u′0=u″0=u1=0,u″1−2u‴45−18u‴67=0,where ht=6t and
(67)fu=2arctanu1/2+sinu≤2u1/2+u=ψu.It is obvious that (H1)-(H4) hold. By computations of Example 2, we have
(68)G¯≈0.170285.Furthermore, forR>4.362954, the inequality R/G¯hL1ψR>1 holds, Therefore, by Theorem 10, problem (66) has at least one solution.



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