# Positive Solution for the Nonlinear Hadamard Type Fractional Differential Equation withp-Laplacian

Authors: Ya-ling Li; Shi-you Lin
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We study the following nonlinear fractional differential equation involving thep-Laplacian operator
DβφpDαut=ft,ut, 1<t<e, u1=u′1=u′e=0, Dαu1=Dαue=0, where the continuous function f:1,e×0,+∞→[0,+∞), 2<α≤3, 1<β≤2. Dα denotes the standard Hadamard fractional derivative of the order α, the constant p>1, and the p-Laplacian operator φps=sp-2s. We show some results about the existence and the uniqueness of the positive solution by using fixed point theorems and the properties of Green's function and the p-Laplacian operator.


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## Body

## 1. Introduction

Fractional differential equations have attracted more and more attention for their useful applications in various fields, such as economics, science, and engineering; see [1–4]. In the last few decades, much attention has been focused on the study of the existence and uniqueness of solutions for boundary value problems of Riemann-Liouville type or Caputo type fractional differential equations; see [5–19]. There are few papers devoted to the research of the p-Laplacian fractional differential equations; see [20–25].By the use of the fixed point theorem on cones, Chai in [20] obtained the existence and multiplicity of positive solutions for a class of boundary value problem of fractional differential equation with p-Laplacian operator:
(1)D0+β(ϕp(D0+αu(t)))=f(t,u(t)),0<t<1,u(0)=0,u(1)+σD0+γu(1)=0,D0+αu(0)=0,
where 1<α≤2, 0<β≤1, 0<γ≤1, 0≤α-γ-1, σ is a positive constant number, and D0+α,D0+β,D0+γ are the standard Riemann-Liouville derivatives. ϕp(s)=|s|p-2s, p>1, ϕp-1=ϕq, 1/p+1/q=1.Han et al. in [22] studied the following boundary value problem of nonlinear fractional differential equation with p-Laplacian operator:
(2)D0+β(ϕp(D0+αu(t)))+a(t)f(u)=0,0<t<1,u(0)=γu(ξ)+λ,ϕp(D0+αu(0))=(ϕp(D0+αu(1)))′=(ϕp(D0+αu(1)))′′=0,
where 0<α≤1, 2<β≤3, 0≤γ<1, 0≤ξ≤1, λ>0 is a parameter, and D0+α,D0+β are the standard Caputo fractional derivatives. By the properties of Green function and Schauder fixed point theorem, several existence and nonexistence results and the uniqueness of positive solutions are acquired.Liu et al. in [23] investigated the solvability of a fractional differential equation model involving the p-Laplacian operator with boundary value conditions as follows:
(3)(φp(cDαx(t)))′=f(t,x(t)),t∈(0,1),x(0)=r0x(1),x′(0)=r1x′(1),x(i)=0,i=2,3,…,[α]-1,
where 1<α∈ℝ, r0, r1≠1, and cDα is the standard Caputo derivative. By the means of the Banach contraction mapping principle, they obtained the existence and uniqueness of a solution for the model.Lu et al. in [24] considered the following fractional boundary value problem with p-Laplacian operator:
(4)D0+β(ϕp(D0+αu(t)))=f(t,u(t)),0<t<1,u(0)=u′(0)=u′(1)=0,D0+αu(0)=D0+βu(1)=0,
where 2<α≤3, 1<β≤2, and D0+α,D0+β are the standard Riemann-Liouville fractional derivatives. By the properties of Green’s function, the Guo-Krasnosel’skii fixed point theorem, the Leggett-Williams fixed point theorem, and the upper and lower solutions method, some new results on the existence of positive solutions are gained.Motivated by the mentioned papers, we will consider the Hadamard fractional boundary value withp-Laplacian operator as below:
(5)Dβ(φp(Dαu(t)))=f(t,u(t)),1<t<e,u(1)=u′(1)=u′(e)=0,Dαu(1)=Dαu(e)=0,
where 2<α≤3, 1<β≤2, φp(s)=|s|p-2s, and f:[1,e]×[0,+∞)→[0,+∞) is a positive continuous function. Evidently, for any p>1, φp-1=φq, here 1/p+1/q=1. Here Dα is the standard Hadamard fractional derivative of order α which is described as follows.Definition 1 (see [1, Page 111]).
Theαth Hadamard fractional order derivative of a function u:[1,+∞)↦ℝ is defined by
(6)Dαu(t)=1Γ(n-α)(tddt)n∫1t‍(logts)n-α-1u(s)sds,
where α>0, n=[α]+1, and [α] denotes the largest integer which is less than or equal to α. Correspondingly, the αth Hadamard fractional order integral of u:[1,+∞)↦ℝ is defined by
(7)Iαu(t)=1Γ(α)∫1t‍(logts)α-1u(s)sds,
where Γ is the gamma function.To the best of our knowledge, there are few contributions to the Hadamard type withp-Laplacian operator; we fill the gap in this paper. In fact, we will discuss the existence and the uniqueness of the positive solutions of (5). The structure of this paper goes on as follows. In Section 2, we will introduce some basic lemmas that will be used. In Section 3, we first give some existence results including Theorems 10 and 11, Corollary 12, and Theorem 13. Then, we will prove Theorems 14 and 15 which reveal the uniqueness of the solution. In Section 4, we give two examples to illustrate our results.


## 2. Preliminary Results

In this section, we will first recall the following preliminary facts that will be used in our main results.Lemma 2 (see [1, Theorem 2.3]).
Letα>0, n=[α]+1; then
(8)IαDαu(t)=u(t)+∑i=1nci(logt)α-i,
where ci, i=1,2,…,n, are some constants in ℝ.The following lemma is the Schauder fixed point theorem which is well known; see Theorem 2.10 in [22].Lemma 3.
IfU is a nonempty closed, bounded, and convex subset of a Banach space X and T:U→U is completely continuous, then T has a fixed point in U.Lemma 4 (see [24, Lemma 2.7]).
LetX be a Banach space, let P⊆X be a cone, and let Ω1, Ω2 be two bounded open balls of E centered at the origin with Ω-1⊂Ω2. Suppose that T:P∩(Ω-2∖Ω1)→P is a completely continuous operator such that either(i)
∥Tx∥≤∥x∥, x∈P∩∂Ω1 and ∥Tx∥≥∥x∥, x∈P∩∂Ω2, or(ii)
∥Tx∥≥∥x∥, x∈P∩∂Ω1 and ∥Tx∥≤∥x∥, x∈P∩∂Ω2,holds. Then T has a fixed point in P∩(Ω-2∖Ω1).The following conclusion is the nonlinear alternative of Leray-Schauder type; see Lemma 2.6 in [10].Lemma 5.
LetX be a Banach space with C⊆X being closed and convex. Assume that U is a relatively open subset of C with 0∈U and A:U¯→C is a continuous, compact map. Then either (1)
A has a fixed point in U¯, or(2)
there existsu∈∂U and λ∈(0,1), with u=λAu.Next, we give several lemmas which will be applied in the proofs of our main results.Lemma 6.
Letu(t) be the solution of the problem (5); then it can be described as below:
(9)u(t)=∫1eG(t,s)φq(∫1e‍H(s,τ)f(τ,u(τ))dττ)dss=Y(t),
where
(10)G(t,s)={(logt)α-1(1-logs)α-2-(log(t/s))α-1Γ(α),s≤t,(logt)α-1(1-logs)α-2Γ(α),t≤s,H(t,s)={(logt)β-1(1-logs)β-1-(log(t/s))β-1Γ(β),s≤t,(logt)β-1(1-logs)β-1Γ(β),t≤s.Proof.
Puttingy(t)=f(t,u(t)), we have Dβ(φp(Dαu(t)))=y(t). By Lemma 2 and the fact that 1<β≤2,
(11)φp(Dαu(t))=Iβy(t)+c1(logt)β-1+c2(logt)β-2.
The boundary value hypotheses give Dαu(1)=Dαu(e)=0. So we can get that
(12)c2=0,c1=-1Γ(β)∫1e‍(1-logτ)β-1y(τ)dττ.
Therefore,
(13)φp(Dαu(t))=Iβy(t)-(logt)β-1Γ(β)∫1e(1-logτ)β-1y(τ)dττ=1Γ(β)∫1t(logtτ)β-1y(τ)dττ-(logt)β-1Γ(β)∫1t(1-logτ)β-1y(τ)dττ-(logt)β-1Γ(β)∫te(1-logτ)β-1y(τ)dττ=-∫1t‍(logt)β-1(1-logτ)β-1-(logt-logτ)β-1Γ(β)×y(τ)dττ-∫te(logt)β-1(1-logτ)β-1Γ(β)y(τ)dττ=-∫1eH(t,τ)y(τ)dττ.
Notice the fact thatφp-1=φq, 1/p+1/q=1; we have
(14)Dαu(t)+φq(∫1eH(t,τ)f(τ,u(τ))dττ)=0.
Putting x(t)=φq(∫1e‍H(t,τ)f(τ,u(τ))(dτ/τ)), it follows from Lemma 2 that
(15)u(t)=-Iαx(t)+C1(logt)α-1+C2(logt)α-2+C3(logt)α-3.
This, combined with the fact that u(1)=u′(1)=u′(e)=0, yields
(16)C3=0,C2=0,C1=1Γ(α)∫1e(1-logs)α-2x(s)dss.
Thus,
(17)u(t)=∫1eG(t,s)x(s)dss=∫1eG(t,s)φq(∫1e‍H(s,τ)f(τ,u(τ))dττ)dss,
where
(18)G(t,s)={(logt)α-1(1-logs)α-2-(log(t/s))α-1Γ(α),s≤t,(logt)α-1(1-logs)α-2Γ(α),t≤s,H(t,s)={(logt)β-1(1-logs)β-1-(log(t/s))β-1Γ(β),s≤t,(logt)β-1(1-logs)β-1Γ(β),t≤s.
This completes the proof of Lemma6.Lemma 7.
Suppose that2<α≤3, 1<β≤2. Then the functions G(t,s) and H(t,s) defined in (10) have the following properties: (1)
G(t,s), H(t,s) are continuous on [1,e]×[1,e];(2)
for anyt,s∈[1,e], G(t,s)≥0, H(t,s)≥0;(3)
for anyt,s∈[1,e], G(t,s)≤G(e,s), H(t,s)≤H(s,s);(4)
there exist two positive functionsγ1,γ2∈C(1,e) such that
(19)mine1/4≤t≤e3/4G(t,s)≥γ1(s)max1≤t≤eG(t,s)=γ1(s)G(e,s),fors∈(1,e),(20)mine1/4≤t≤e3/4H(t,s)≥γ2(s)max1≤t≤eH(t,s)=γ2(s)H(s,s),fors∈(1,e).Proof.
(1) and (2) are evident from the expression ofG(t,s) and H(t,s). Since, for any fixed number s∈[1,e], G(t,s) is an increasing function on [1,e] and H(t,s) is a decreasing function on [s,e] and increasing on [1,s], we get (3). To prove (4), suppose that
(21)γ1(s)={(1/4)α-1(1-logs)α-2-(1/4-logs)α-1(1-logs)α-2-(1-logs)α-1,s∈(1,e1/4],(1/4)α-1(1-logs)α-2(1-logs)α-2-(1-logs)α-1,s∈[e1/4,e),
and put
(22)p1(t,s)=(logt)α-1(1-logs)α-2-(log(t/s))α-1Γ(α),p2(t,s)=(logt)α-1(1-logs)α-2Γ(α).
The monotonicity of G(t,s) gives
(23)mine1/4≤t≤e3/4G(t,s)hhhhhh={p1(e1/4,s),s∈(1,e1/4],p2(e1/4,s),s∈[e1/4,e),hhhhhh={1Γ(α)[(14)α-1(1-logs)α-2-(14-logs)α-1],s∈(1,e1/4],1Γ(α)(14)α-1(1-logs)α-2,s∈[e1/4,e),max1≤t≤eG(t,s)=G(e,s)hhhhhhhhh=1Γ(α)[(1-logs)α-2-(1-logs)α-1],hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhs∈(1,e).
Which implies that (19) holds.
Similarly, by writing(24)q1(t,s)=(logt)β-1(1-logs)β-1-(log(t/s))β-1Γ(β),q2(t,s)=(logt)β-1(1-logs)β-1Γ(β)
and applying the monotonicity of H(t,s), we have
(25)mine1/4≤t≤e3/4H(t,s)hhhhhh={q1(e3/4,s),s∈(1,e1/4],min{q1(e3/4,s),q2(e1/4,s)},s∈[e1/4,e3/4],q2(e1/4,s),s∈[e3/4,e),hhhhhh={q1(e3/4,s),s∈(1,r],<?layout cmd="vspace" value="15"?>q2(e1/4,s),s∈[r,e),hhhhhh={1Γ(β)[(34-logs)β-1(34)β-1(1-logs)β-1-(34-logs)β-1],s∈(1,r],1Γ(β)(14)β-1(1-logs)β-1,s∈[r,e),max1≤t≤eH(t,s)=H(s,s)hhhhhhhhh=1Γ(β)[logs(1-logs)]β-1,hhhhhhhhhhhhhhhhhhhhhhhhs∈(1,e),
where e1/4<r<e3/4 is the unique solution of the equation
(26)[34(1-logs)]β-1-(34-logs)β-1=[14(1-logs)]β-1.
Hence, setting
(27)γ2(s)={[(3/4)(1-logs)]β-1-(3/4-logs)β-1[logs(1-logs)]β-1,s∈(1,r],1(4logs)β-1,s∈[r,e),
we obtain (20). This completes the proof of Lemma 7.LetX=C[1,e], ∥u∥=max1≤t≤e|u(t)|. we define the cone P={u∈X∣u(t)≥0} and the bounded closed set U={u∈X∣0≤u(t)≤K}.The operatorT:X→X is defined as the following form:
(28)Tu(t)=Y(t)=∫1eG(t,s)φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss.
Evidently, the solutions of boundary value problem (5) are the corresponding fixed points of the operator T.Lemma 8.
Suppose thatT:P→X is an operator as above; then T:P→P is completely continuous.Proof.
It is easy to see thatT:P→P is continuous. Let Ω⊂P be a bounded set; then there is a positive constant A>0 such that ∥u∥≤A for any u∈Ω. Write B=max1≤t≤e,0≤u≤Af(t,u(t))+1. For any u∈Ω, we have
(29)|Tu(t)|=∫1eG(t,s)φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss≤Bq-1∫1eG(e,s)dssφq(∫1eH(s,τ)f(τ,u(τ)))dττ<+∞,
which shows that TΩ is uniformly bounded.
Next, the continuity ofG(t,s) implies that, for any ε>0, there exists a constant δ such that, for any t1,t2∈[1,e], if |t1-t2|<δ, then
(30)|G(t1,s)-G(t2,s)|<εBq-1φq(∫1eH(τ,τ)(dτ/τ)).
Therefore, for any u∈Ω,
(31)|Tu(t1)-Tu(t2)|=|∫1e(G(t1,s)-G(t2,s))0000000×φq(∫1e‍H(s,τ)f(τ,u(τ))dττ)dss|≤Bq-1∫1e|G(t1,s)-G(t2,s)|×φq(∫1eH(τ,τ)dττ)dss<ε.
That is, TΩ is equicontinuity. By the means of Arzela-Ascoli theorem [26], we have that T:P→P is completely continuous. This completes the proof of Lemma 8.In the final part of this section, we list the following basic properties of thep-Laplacian operator.Lemma 9.
(1) If1<p<2, xy>0, and |x|,|y|≥m>0, then
(32)|φp(x)-φp(y)|≤(p-1)mp-2|x-y|.
(2) Ifp>2,|x|,|y|≤M, then
(33)|φp(x)-φp(y)|≤(p-1)Mp-2|x-y|.


## 3. Proofs of the Main Results

In this section, first, we consider the existence of the solutions of problem (5).Theorem 10.
If[max1≤t≤e,0≤u≤Kf(t,u)]q-1∫1eG(e,s)(ds/s)φq(∫1eH(τ,τ)(dτ/τ))≤K, then the boundary value problem (5) has at least one positive solution.Proof.
For anyu∈U, by the assumption as above and the nonnegativeness of G(t,s), H(t,s), and f(t,u), we have
(34)0≤Tu(t)=∫1eG(t,s)φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss≤[max1≤t≤e,0≤u≤Kf(t,u)]q-1×∫1eG(e,s)dssφq(∫1eH(τ,τ)dττ)≤K.
Therefore, T is a mapping from U to U. This, combined with the continuity of G(t,s), H(t,s), and f(t,u), implies that T:U→U is continuous.
LetΩ⊂U be a bounded set; then there exists a positive constant A such that ∥u∥≤A for any u∈Ω. So we have, for any u∈Ω,
(35)|Tu(t)|=∫1eG(t,s)φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss≤[max1≤t≤e,0≤u≤Af(t,u)]q-1×∫1eG(e,s)dssφq(∫1eH(τ,τ)dττ)≤[max1≤t≤e,0≤u≤Kf(t,u)]q-1×∫1eG(e,s)dssφq(∫1eH(τ,τ)dττ)≤K≤+∞.
Therefore, TΩ is uniformly bounded.
SinceG(t,s) is continuous, for any ε>0, there exists a constant δ>0 satisfying that, for any t1,t2∈[1,e] and |t1-t2|<δ,
(36)|G(t1,s)-G(t2,s)|<ε[max1≤t≤e,0≤u≤Af(t,u)+1]q-1φq(∫1eH(τ,τ)(dτ/τ)).
Then, for any u∈Ω,
(37)|Tu(t1)-Tu(t2)|≤∫1e|G(t1,s)-G(t2,s)|×φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss≤[max1≤t≤e,0≤u≤Af(t,u)+1]q-1×φq(∫1eH(τ,τ)dττ)×∫1e|G(t1,s)-G(t2,s)|dss<ε,
which shows that TΩ is equicontinuous. By Arzela-Ascoli theorem [26], T:U→U is a completely continuous operator. It follows from Lemma 3 that T has a fixed point u in U. That is, problem (5) has at least one positive solution. This completes the proof of Theorem 10.Let us denote(38)Λ1=(∫e1/4e3/4γ1(s)G(e,s)dss×φq(∫e1/4e3/4γ2(τ)H(τ,τ)dττ))-1,Λ2=(∫1eG(e,s)dssφq(∫1eH(τ,τ)dττ))-1.Theorem 11.
Suppose thatf(t,u):[1,e]×[0,+∞) is a continuous function and there exist two constants r2>r1>0 satisfying that (i)
f(t,u)≥φp(Λ1r1), for (t,u)∈[1,e]×[0,r1];(ii)
f(t,u)≤φp(Λ2r2), for (t,u)∈[1,e]×[0,r2]. Then the boundary value problem (5) has at least one positive solution u which satisfies that r1≤∥u∥≤r2.Proof.
LetΩ1={u∈P∣∥u∥<r1}. For any u∈∂Ω1, we have 0≤u(t)≤r1 for t∈[1,e]. By the assumption (i), for any t∈[e1/4,e3/4],
(39)|Tu(t)|=∫1e‍G(t,s)×φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss≥Λ1r1∫1eG(t,s)φq(∫1eH(s,τ)dττ)dss≥Λ1r1∫e1/4e3/4G(t,s)φq(∫e1/4e3/4H(s,τ)dττ)dss≥Λ1r1∫e1/4e3/4‍γ1(s)G(e,s)×φq(∫e1/4e3/4γ2(τ)H(τ,τ)dττ)dss=r1=∥u∥.
Hence,
(40)∥Tu∥≥∥u∥,foru∈∂Ω1.
Similarly, let Ω2:={u∈P∣∥u∥<r2}. For any u∈∂Ω2, we get 0≤u(t)≤r2, t∈[1,e]. It follows from (ii) that for any t∈[1,e],
(41)|Tu(t)|=∫1e‍G(t,s)φq(∫1e‍H(s,τ)f(τ,u(τ))dττ)dss≤Λ2r2∫1e‍G(t,s)φq(∫1e‍H(s,τ)dττ)dss≤Λ2r2∫1e‍G(e,s)φq(∫1e‍H(τ,τ)dττ)dss=r2=∥u∥.
Therefore,
(42)∥Tu∥≤∥u∥,foru∈∂Ω2.
By Lemmas 4 and 8, T has a fixed point in Ω¯2∖Ω1. Therefore, the boundary value problem (5) has one positive solution in Ω¯2∖Ω1. This completes the proof of Theorem 11.Corollary 12.
Suppose thatf(t,u):[1,e]×[0,+∞) is a continuous function and there exist two constants r2>r1>0 satisfying that (i)
f(t,u)≤φp(Λ2r1), for (t,u)∈[1,e]×[0,r1];(ii)
f(t,u)≥φp(Λ1r2), for (t,u)∈[1,e]×[0,r2]. Then the boundary value problem (5) has at least one positive solution u which satisfies that r1≤∥u∥≤r2.Proof.
The proof of Corollary12 is similar to the one of Theorem 11. So we omit the detail.Theorem 13.
Suppose thatf(t,u):[1,e]×[0,+∞) is a positive continuous function and there exists a constant r>0 such that
(43)r>[max1≤t≤e,0≤u≤rf(t,u)+1]q-1×∫1eG(e,s)dssφq(∫1eH(τ,τ)dττ).
Then the boundary value problem (5) has at least one positive solution.Proof.
Let(44)E={u∈P:∥u∥<r}.
From Lemma 8, we know T:E¯→P is completely continuous. Assume that there exist u∈E¯, λ∈(0,1) such that u=λTu. Then we have
(45)|u(t)|=|λTu|≤|∫1eG(t,s)φq(∫1eH(s,τ)f(τ,u(τ))dττ)dss|≤[max1≤t≤e,0≤u≤rf(t,u)+1]q-1×∫1eG(e,s)dssφq(∫1e‍H(τ,τ)dττ).
Thus,
(46)∥u∥≤[max1≤t≤e,0≤u≤rf(t,u)+1]q-1×∫1e‍G(e,s)dssφq(∫1e‍H(τ,τ)dττ).
By (43), we can imply that ∥u∥<r, which means that u∉∂E. That is to say, there is no u∈∂E such that u=λTu for some λ∈(0,1). Therefore, by Lemma 5, we conclude that the problem (5) has at least one positive solution. This completes the proof of Theorem 13.Now we turn to the uniqueness of solution for boundary value problem (5).Theorem 14.
Suppose thatp>2. If there exists a nonnegative function g satisfying that (1)
for any(t,u)∈[1,e]×[0,+∞), f(t,u)≥g(t);(2)
N=∫e1/4e3/4‍γ2(τ)H(τ,τ)g(τ)(dτ/τ)>0;(3)
for anyt∈[1,e], u,v∈[0,+∞), |f(t,u)-f(t,v)|≤L|u-v|, where
(47)0<L<(∫1eG(e,s)dss∫1eH(τ,τ)dττ(q-1)Nq-2)-1,then the boundary value problem (5) has a unique solution.Proof.
Assume thatu,v are two positive solutions of problem (5). It is easy to see that
(48)∫1eH(s,τ)f(τ,u(τ))dττ≥∫e1/4e3/4γ2(τ)H(τ,τ)g(τ)dττ=N>0;
then by the fact p>2 (i.e., its dual exponent 1<q<2) and Lemma 9, we have
(49)|u(t)-v(t)|≤∫1eG(t,s)(q-1)Nq-2×|∫1e‍H(s,τ)f(τ,u(τ))dττ-∫1eH(s,τ)f(τ,v(τ))dττ|dss≤∫1eG(t,s)(q-1)Nq-2×∫1eH(s,τ)|f(τ,u(τ))-f(τ,v(τ))|dττdss≤∫1eG(e,s)(q-1)Nq-2×∫1eH(τ,τ)L∥u-v∥dττdss=∫1e‍G(e,s)dss×∫1eH(τ,τ)dττ(q-1)Nq-2L∥u-v∥=δ1∥u-v∥,
where δ1=∫1eG(e,s)(ds/s)∫1eH(τ,τ)(dτ/τ)(q-1)Nq-2L. So we can get
(50)∥u-v∥≤δ1∥u-v∥.
By the third hypothesis, 0<δ1<1, which implies that u(t)=v(t). And this completes the proof of Theorem 14.By using the same way, we can prove the last one of our main uniqueness results.Theorem 15.
Suppose that1<p<2. If there exists a nonnegative function h satisfying that (1)
for any(t,u)∈[1,e]×[0,+∞), f(t,u)≤h(t);(2)
M=∫1e‍H(τ,τ)h(τ)(dτ/τ)>0;(3)
for anyt∈[1,e], u,v∈[0,+∞), |f(t,u)-f(t,v)|≤L|u-v|, where
(51)0<L<(∫1e‍G(e,s)dss∫1e‍H(τ,τ)dττ(q-1)Mq-2)-1,then the boundary value problem (5) has a unique solution.


## 4. Examples

In this section we give several examples to illustrate our main results.Example 16.
Consider the boundary value problem:(52)D3/2(φ5/2(D5/2u(t)))=(1-logt)1/2(1+sin2u),1<t<e,u(1)=u′(1)=u′(e)=0,D5/2u(1)=D5/2u(e)=0.
Then the boundary value problem has a unique positive solution.Proof.
Sinceα=5/2, β=3/2, a straightforward calculation gives
(53)∫1eG(e,s)dss=∫1e(1-logs)α-2-(1-logs)α-1Γ(α)dss=1Γ(α)∫1e(1-logs)α-2dss-1Γ(α)∫1e(1-logs)α-1dss=1Γ(α)∫01‍(1-t)α-2dt-1Γ(α)∫01(1-t)α-1dt=1(α-1)Γ(α+1)=1(3/2)Γ(7/2)=1645π,∫1eH(τ,τ)dττ=∫1e(logτ)β-1(1-logτ)β-1Γ(β)dττ=1Γ(β)∫01tβ-1(1-t)β-1dt=Γ(β)Γ(2β)=Γ(3/2)Γ(3)=π4.
TakingK=1,f(t,u)=(1-logt)1/2(1+sin2u), and p=5/2>2 (its dual exponent q=5/3), we have
(54)max1≤t≤e,0≤u≤1f(t,u)=1+sin21<2,[max1≤t≤e,0≤u≤1f(t,u)](5/3)-1∫1eG(e,s)dssφ5/3(∫1eH(τ,τ)dττ)<22/3·1645π·(π4)2/3≈0.185<1.
By Theorem 10, the boundary value problem (52) has at least one positive solution.
Choosing the nonnegative functiong(t)=(1-logt)1/2, for any (t,u)∈[1,e]×[0,+∞), we gain that f(t,u)=(1-logt)1/2(1+sin2u)≥g(t). Then
(55)N=∫e1/4e3/4γ2(τ)H(τ,τ)g(τ)dττ=∫e1/4r1Γ(3/2){[34(1-logτ)]3/2-100000000000-(34-logτ)3/2-1}·(1-logτ)1/2dττ+∫re3/41Γ(3/2)(14)3/2-1(1-logτ)3/2-1·(1-logτ)1/2dττ≥∫re3/41π(1-logτ)dττ=12π(1-logr)2-132π=12π(1-1+36)2-132π=196π,
where r=e1-(3/6) is the solution of (26) when β=3/2. For any t∈[1,e], u,v∈[0,+∞), taking L=2, we obtain
(56)|f(t,u)-f(t,v)|=(1-logt)1/2|sin2u-sin2v|≤2|u-v|=L|u-v|.
Thus,
(57)(∫1eG(e,s)dss∫1eH(τ,τ)dττ(q-1)Nq-2)-1=(1645π·π4·23N-1/3)-1≥1358(196π)1/3≈3.045>2=L.
By Theorem 14, the boundary value problem (52) has a unique solution.Example 17.
Consider the following nonlinear boundary value problem:(58)D3/2(φ3/2(D5/2u(t)))=log1/2tsin2u,1<t<e,u(1)=u′(1)=u′(e)=0,D5/2u(1)=D5/2u(e)=0.
Then the boundary value problem has a unique positive solution.Proof.
Takingr=1, since α=5/2, β=3/2, p=3/2<2 (its dual exponent q=3) and f(t,u)=log1/2tsin2u, we obtain
(59)max1≤t≤e,0≤u≤1f(t,u)=sin21<1,(60)[max1≤t≤e,0≤u≤1f(t,u)+1]3-1∫1eG(e,s)dssφ3(∫1eH(τ,τ)dττ)<4·1645π·π16≈0.158<1.
By means of Theorem 13, the boundary value problem (58) has at least one positive solution.
Taking the nonnegative functionh(t)=log1/2t, for (t,u)∈[1,e]×[0,+∞), it is easy to obtain f(t,u)=log1/2tsin2u≤h(t) and
(61)M=∫1eH(τ,τ)h(τ)dττ=1Γ(3/2)∫1elogτ(1-logτ)1/2dττ=815π.
Choosing L=2, for any t∈[1,e], u,v∈[0,+∞), we have
(62)|f(t,u)-f(t,v)|=log1/2t|sin2u-sin2v|hhhhhhhhhhhh1≤2|u-v|=L|u-v|,(∫1eG(e,s)dss∫1eH(τ,τ)dττ(q-1)Mq-2)-1hhhhhh=(1645π·π4·2·815π)-1hhhhhh≈18.694>2=L.
From Theorem 15, the boundary value problem (58) has a unique solution.



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*Source: 951643-2013-10-03.xml*