# Positive Solutions to Boundary Value Problems of Nonlinear Fractional Differential Equations

Authors: Yige Zhao; Shurong Sun; Zhenlai Han; Qiuping Li
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We study the existence of positive solutions for the boundary value problem of nonlinear fractional differential equationsD0+αu(t)+λf(u(t))=0, 0<t<1, u(0)=u(1)=u′(0)=0, where 2<α≤3 is a real number, D0+α is the Riemann-Liouville fractional derivative, λ is a positive parameter, and f:(0,+∞)→(0,+∞) is continuous. By the properties of the Green function and Guo-Krasnosel'skii fixed point theorem on cones, the eigenvalue intervals of the nonlinear fractional differential equation boundary value problem are considered, some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for the boundary value problem are established. As an application, some examples are presented to illustrate the main results.


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications; see [1–4]. It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations on terms of special functions.Recently, there are some papers dealing with the existence of solutions (or positive solutions) of nonlinear initial fractional differential equations by the use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, Adomian decomposition method, etc.); see [5–11]. In fact, there has the same requirements for boundary conditions. However, there exist some papers considered the boundary value problems of fractional differential equations; see [12–19].Yu and Jiang [19] examined the existence of positive solutions for the following problem: (1.1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=u′(0)=0,
where 2<α≤3 is a real number, f∈C([0,1]×[0,+∞),(0,+∞)), and D0+α is the Riemann-Liouville fractional differentiation. By using the properties of the Green function, they obtained some existence criteria for one or two positive solutions for singular and nonsingular boundary value problems by means of the Krasnosel'skii fixed point theorem and a mixed monotone method.To the best of our knowledge, there is very little known about the existence of positive solutions for the following problem:(1.2)D0+αu(t)+λf(u(t))=0,0<t<1,u(0)=u(1)=u′(0)=0,
where 2<α≤3 is a real number, D0+α is the Riemann-Liouville fractional derivative, λ is a positive parameter and f:(0,+∞)→(0,+∞) is continuous.On one hand, the boundary value problem in [19] is the particular case of problem (1.2) as the case of λ=1. On the other hand, as Yu and Jiang discussed in [19], we also give some existence results by the fixed point theorem on a cone in this paper. Moreover, the purpose of this paper is to derive a λ-interval such that, for any λ lying in this interval, the problem (1.2) has existence and multiplicity on positive solutions.In this paper, by analogy with boundary value problems for differential equations of integer order, we firstly give the corresponding Green function named by fractional Green's function and some properties of the Green function. Consequently, the problem (1.2) is reduced to an equivalent Fredholm integral equation. Finally, by the properties of the Green function and Guo-Krasnosel'skii fixed point theorem on cones, the eigenvalue intervals of the nonlinear fractional differential equation boundary value problem are considered, some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for the boundary value problem are established. As an application, some examples are presented to illustrate the main results.


## 2. Preliminaries

For the convenience of the reader, we give some background materials from fractional calculus theory to facilitate analysis of problem (1.2). These materials can be found in the recent literature; see [19–21].Definition 2.1 (see [20]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function f:(0,+∞)→ℝ is given by
(2.1)D0+αf(t)=1Γ(n-α)(ddt)(n)∫0tf(s)(t-s)α-n+1ds,
where n=[α]+1, [α] denotes the integer part of number α, provided that the right side is pointwise defined on (0,+∞).Definition 2.2 (see [20]).
The Riemann-Liouville fractional integral of orderα>0 of a function f:(0,+∞)→ℝ is given by
(2.2)I0+αf(t)=1Γ(α)∫0t(t-s)α-1f(s)ds,
provided that the right side is pointwise defined on (0,+∞).From the definition of the Riemann-Liouville derivative, we can obtain the following statement.Lemma 2.3 (see [20]).
Letα>0. If we assume u∈C(0,1)∩L(0,1), then the fractional differential equation
(2.3)D0+αu(t)=0
has u(t)=c1tα-1+c2tα-2+⋯+cNtα-N, ci∈ℝ, i=1,2,…,N, as unique solutions, where N is the smallest integer greater than or equal to α.Lemma 2.4 (see [20]).
Assume thatu∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then
(2.4)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cNtα-N,
for some ci∈ℝ, i=1,2,…,N, where N is the smallest integer greater than or equal to α.In the following, we present the Green function of fractional differential equation boundary value problem.Lemma 2.5 (see [19]).
Leth∈C[0,1] and 2<α≤3. The unique solution of problem
(2.5)D0+αu(t)+h(t)=0,0<t<1,u(0)=u(1)=u′(0)=0
is
(2.6)u(t)=∫01G(t,s)h(s)ds,
where
(2.7)G(t,s)={tα-1(1-s)α-1-(t-s)α-1Γ(α),0≤s≤t≤1,tα-1(1-s)α-1Γ(α),0≤t≤s≤1.
Here G(t,s)is called the Green function of boundary value problem (2.5).The following properties of the Green function play important roles in this paper.Lemma 2.6 (see [19]).
The functionG(t,s) defined by (2.7) satisfies the following conditions: (1)
G(t,s)=G(1-s,1-t), for t,s∈(0,1);(2)
tα-1(1-t)s(1-s)α-1≤Γ(α)G(t,s)≤(α-1)s(1-s)α-1, for t,s∈(0,1);(3)
G(t,s)>0, for t,s∈(0,1);(4)
tα-1(1-t)s(1-s)α-1≤Γ(α)G(t,s)≤(α-1)(1-t)tα-1, for t,s∈(0,1).The following lemma is fundamental in the proofs of our main results.Lemma 2.7 (see [21]).
LetX be a Banach space, and let P⊂X be a cone in X. Assume Ω1,Ω2 are open subsets of X with 0∈Ω1⊂Ω¯1⊂Ω2, and let S:P→P be a completely continuous operator such that, either (A1)
∥Sw∥≤∥w∥, w∈P∩∂Ω1, ∥Sw∥≥∥w∥, w∈P∩∂Ω2 or(A2)
∥Sw∥≥∥w∥, w∈P∩∂Ω1, ∥Sw∥≤∥w∥, w∈P∩∂Ω2.Then S has a fixed point in P∩(Ω¯2∖Ω1).For convenience, we setq(t)=tα-1(1-t),k(s)=s(1-s)α-1; then (2.8)q(t)k(s)≤Γ(α)G(t,s)≤(α-1)k(s).


## 3. Main Results

In this section, we establish the existence of positive solutions for boundary value problem (1.2).Let Banach spaceE=C[0,1] be endowed with the norm ∥u∥=max0≤t≤1|u(t)|. Define the cone P⊂E by (3.1)P={u∈E:u(t)≥q(t)α-1‖u‖,t∈[0,1]}.Suppose thatu is a solution of boundary value problem (1.2). Then (3.2)u(t)=λ∫01G(t,s)f(u(s))ds,t∈[0,1].We define an operatorAλ:P→E as follows: (3.3)(Aλu)(t)=λ∫01G(t,s)f(u(s))ds,t∈[0,1].By Lemma2.6, we have (3.4)‖Aλu‖≤λΓ(α)∫01(α-1)k(s)f(u(s))ds,(Aλu)(t)≥λΓ(α)∫01q(t)k(s)f(u(s))ds≥q(t)α-1‖Aλu‖.
Thus, Aλ(P)⊂P.Then we have the following lemma.Lemma 3.1.
Aλ:P→P is completely continuous.Proof.
The operatorAλ:P→P is continuous in view of continuity of G(t,s) and f(u(t)). By means of the Arzela-Ascoli theorem, Aλ:P→P is completely continuous.
For convenience, we denote(3.5)F0=lim⁡u→0+sup⁡f(u)u,F∞=lim⁡u→+∞sup⁡f(u)u,f0=lim⁡u→0+inf⁡f(u)u,f∞=lim⁡u→+∞inf⁡f(u)u,C1=1Γ(α)∫01(α-1)k(s)ds,C2=1Γ(α)∫011(α-1)q(s)k(s)ds,C3=1Γ(α)∫011(α-1)k(s)ds.Theorem 3.2.
If there existsl∈(0,1) such that q(l)f∞C2>F0C1 holds, then for each
(3.6)λ∈((q(l)f∞C2)-1,(F0C1)-1),
the boundary value problem (1.2) has at least one positive solution. Here we impose (q(l)f∞C2)-1=0 if f∞=+∞ and (F0C1)-1=+∞ if F0=0.Proof.
Letλ satisfy (3.6) and ɛ>0 be such that
(3.7)(q(l)(f∞-ɛ)C2)-1≤λ≤((F0+ɛ)C1)-1.
By the definition of F0, we see that there exists r1>0 such that
(3.8)f(u)≤(F0+ɛ)u,for0<u≤r1.
So if u∈P with ∥u∥=r1, then by (3.7) and (3.8), we have
(3.9)‖Aλu‖≤λΓ(α)∫01(α-1)k(s)f(u(s))ds≤λΓ(α)∫01(α-1)k(s)(F0+ɛ)r1ds=λ(F0+ɛ)r1C1≤r1=‖u‖.
Hence, if we choose Ω1={u∈E:∥u∥<r1}, then
(3.10)‖Aλu‖≤‖u‖,foru∈P∩∂Ω1.
Letr3>0 be such that(3.11)f(u)≥(f∞-ɛ)u,foru≥r3.
If u∈P with ∥u∥=r2=max{2r1,r3}, then by (3.7) and (3.11), we have
(3.12)‖Aλu‖≥Aλu(l)=λ∫01G(l,s)f(u(s))ds≥λΓ(α)∫01q(l)k(s)f(u(s))ds≥λΓ(α)∫01q(l)k(s)(f∞-ɛ)u(s)ds≥λΓ(α)∫01q(l)α-1q(s)k(s)(f∞-ɛ)‖u‖ds=λq(l)C2(f∞-ɛ)‖u‖≥‖u‖.
Thus, if we set Ω2={u∈E:∥u∥<r2}, then
(3.13)‖Aλu‖≥‖u‖,foru∈P∩∂Ω2.
Now, from (3.10), (3.13), and Lemma 2.7, we guarantee that Aλ has a fixed-point u∈P∩(Ω¯2∖Ω1) with r1≤∥u∥≤r2, and clearly u is a positive solution of (1.2). The proof is complete.Theorem 3.3.
If there existsl∈(0,1) such that q(l)C2f0>F∞C1 holds, then for each
(3.14)λ∈((q(l)f0C2)-1,(F∞C1)-1),
the boundary value problem (1.2) has at least one positive solution. Here we impose (q(l)f0C2)-1=0 if f0=+∞ and (F∞C1)-1=+∞ if F∞=0.Proof.
Letλ satisfy (3.14) and ɛ>0 be such that
(3.15)(q(l)(f0-ɛ)C2)-1≤λ≤((F∞+ɛ)C1)-1.
From the definition of f0, we see that there exists r1>0 such that
(3.16)f(u)≥(f0-ɛ)u,for0<u≤r1.
Further, if u∈P with ∥u∥=r1, then similar to the second part of Theorem 3.2, we can obtain that ∥Aλu∥≥∥u∥. Thus, if we choose Ω1={u∈E:∥u∥<r1}, then
(3.17)‖Aλu‖≥‖u‖,foru∈P∩∂Ω2.
Next, we may chooseR1>0 such that(3.18)f(u)≤(F∞+ɛ)u,foru≥R1.
We consider two cases.Case 1.
Supposef is bounded. Then there exists some M>0, such that
(3.19)f(u)≤M,foru∈(0,+∞).
We define r3=max{2r1,λMC1}, and u∈P with ∥u∥=r3, then
(3.20)‖Aλu‖≤λΓ(α)∫01(α-1)k(s)f(u(s))ds≤λMΓ(α)∫01(α-1)k(s)ds≤λMC1≤r3≤‖u‖.
Hence,
(3.21)‖Aλu‖≤‖u‖,foru∈Pr3={u∈P:‖u‖≤r3}.Case 2.
Supposef is unbounded. Then there exists some r4>max{2r1,R1}, such that
(3.22)f(u)≤f(r4),for0<u≤r4.
Let u∈P with ∥u∥=r4. Then by (3.15) and (3.18), we have
(3.23)‖Aλu‖≤λΓ(α)∫01(α-1)k(s)f(u(s))ds≤λΓ(α)∫01(α-1)k(s)(F∞+ɛ)‖u‖ds≤λC1(F∞+ɛ)‖u‖≤‖u‖.
Thus, (3.21) is also true.In both Cases 1 and 2, if we set Ω2={u∈E:∥u∥<r2=max{r3,r4}}, then
(3.24)‖Aλu‖≤‖u‖,foru∈P∩∂Ω2.
Now that we obtain (3.17) and (3.24), it follows from Lemma 2.7 that Aλ has a fixed-point u∈P∩(Ω¯2∖Ω1) with r1≤∥u∥≤r2. It is clear u is a positive solution of (1.2). The proof is complete.Theorem 3.4.
Suppose there existl∈(0,1), r2>r1>0 such that q(l)>(α-1)r1/r2, and f satisfy
(3.25)min⁡(q(l)/(α-1))r1≤u≤r1f(u)≥r1λ(α-1)q(l)C3,max⁡0≤u≤r2f(u)≤r2λC1.
Then the boundary value problem (1.2) has a positive solution u∈P with r1≤∥u∥≤r2.Proof.
ChooseΩ1={u∈E:∥u∥<r1}; then for u∈P∩∂Ω1, we have
(3.26)‖Aλu‖≥Aλu(l)=λ∫01G(l,s)f(u(s))ds≥λΓ(α)∫01q(l)k(s)f(u(s))ds≥λΓ(α)∫01q(l)k(s)min⁡(q(l)/(α-1))r1≤u≤r1f(u(s))ds≥λ(α-1)q(l)C3r1λ(α-1)q(l)C3=r1=‖u‖.
On the other hand, chooseΩ2={u∈E:∥u∥<r2}, then for u∈P∩∂Ω2, we have (3.27)‖Aλu‖≤λΓ(α)∫01(α-1)k(s)f(u(s))ds≤λΓ(α)∫01(α-1)k(s)max⁡0≤u≤r2f(u(s))ds≤λC1r2λC1=r2=‖u‖.
Thus, by Lemma 2.7, the boundary value problem (1.2) has a positive solution u∈P with r1≤∥u∥≤r2. The proof is complete.For the reminder of the paper, we will need the following condition.(H)
(minu∈[(q(l)/(α-1))r,r]f(u))/r>0, where l∈(0,1).Denote(3.28)λ1=sup⁡r>0rC1max⁡0≤u≤r⁡f(u),(3.29)λ2=inf⁡r>0rC3min⁡(q(l)/(α-1))r≤u≤r⁡f(u).In view of the continuity off(u) and (H), we have 0<λ1≤+∞ and 0≤λ2<+∞.Theorem 3.5.
Assume(H) holds. If f0=+∞ and f∞=+∞, then the boundary value problem (1.2) has at least two positive solutions for each λ∈(0,λ1).Proof.
Define(3.30)a(r)=rC1max⁡0≤u≤rf(u).
By the continuity of f(u), f0=+∞ and f∞=+∞, we have that a(r):(0,+∞)→(0,+∞) is continuous and
(3.31)lim⁡r→0a(r)=lim⁡r→+∞a(r)=0.
By (3.28), there exists r0∈(0,+∞), such that
(3.32)a(r0)=sup⁡r>0a(r)=λ1;
then for λ∈(0,λ1), there exist constants c1,c2(0<c1<r0<c2<+∞) with
(3.33)a(c1)=a(c2)=λ.
Thus,
(3.34)f(u)≤c1λC1,foru∈[0,c1],(3.35)f(u)≤c2λC1,foru∈[0,c2].
On the other hand, applying the conditionsf0=+∞ and f∞=+∞, there exist constants d1, d2(0<d1<c1<r0<c2<d2<+∞) with(3.36)f(u)u≥1q2(l)λC3,foru∈(0,d1)∪(q(l)α-1d2,+∞).
Then
(3.37)min⁡(q(l)/(α-1))d1≤u≤d1f(u)≥d1λ(α-1)q(l)C3,(3.38)min⁡(q(l)/(α-1))d2≤u≤d2f(u)≥d2λ(α-1)q(l)C3.
By (3.34) and (3.37), (3.35) and (3.38), combining with Theorem 3.4 and Lemma 2.7, we can complete the proof.Corollary 3.6.
Assume(H) holds. If f0=+∞ or f∞=+∞, then the boundary value problem (1.2) has at least one positive solution for each λ∈(0,λ1).Theorem 3.7.
Assume(H) holds. If f0=0 and f∞=0, then for each λ∈(λ2,+∞), the boundary value problem (1.2) has at least two positive solutions.Proof.
Define(3.39)b(r)=rC3min⁡(q(l)/(α-1))r≤u≤r⁡f(u).
By the continuity of f(u), f0=0 and f∞=0, we easily see that b(r):(0,+∞)→(0,+∞) is continuous and
(3.40)lim⁡r→0b(r)=lim⁡r→+∞b(r)=+∞.
By (3.29), there exists r0∈(0,+∞), such that
(3.41)b(r0)=inf⁡r>0b(r)=λ2.
For λ∈(λ2,+∞), there exist constants d1, d2(0<d1<r0<d2<+∞) with
(3.42)b(d1)=b(d2)=λ.
Therefore,
(3.43)f(u)≥d1λ(α-1)q(l)C3,foru∈[q(l)α-1d1,d1],f(u)≥d2λ(α-1)q(l)C3,foru∈[q(l)α-1d2,d2].
On the other hand, usingf0=0, we know that there exists a constant c1(0<c1<d1) with(3.44)f(u)u≤1λC1,foru∈(0,c1),(3.45)max⁡0≤u≤c1f(u)≤c1λC1.
In view of f∞=0, there exists a constant c2∈(d2,+∞) such that
(3.46)f(u)u≤1λC1,foru∈(c2,+∞).
Let(3.47)M=max⁡0≤u≤c2f(u),c2≥λC1M.
It is easily seen that
(3.48)max⁡0≤u≤c2f(u)≤c2λC1.
By (3.45) and (3.48), combining with Theorem 3.4 and Lemma 2.7, the proof is complete.Corollary 3.8.
Assume(H) holds. If f0=0 or f∞=0, then for each λ∈(λ2,+∞), the boundary value problem (1.2) has at least one positive solution.By the above theorems, we can obtain the following results.Corollary 3.9.
Assume(H) holds. If f0=+∞, f∞=d, or f∞=+∞, f0=d, then for any λ∈(0,(dC1)-1), the boundary value problem (1.2) has at least one positive solution.Corollary 3.10.
Assume(H) holds. If f0=0,f∞=d, or if f∞=0, f0=d, then for any λ∈((q(l)dC2)-1,+∞), the boundary value problem (1.2) has at least one positive solution.Remark 3.11.
For the integer derivative caseα=3, Theorems 3.2–3.7 also hold; we can find the corresponding existence results in [22].


## 4. Nonexistence

In this section, we give some sufficient conditions for the nonexistence of positive solution to the problem (1.2).Theorem 4.1.
Assume(H) holds. If F0<+∞ and F∞<∞, then there exists a λ0>0 such that for all 0<λ<λ0, the boundary value problem (1.2) has no positive solution.Proof.
SinceF0<+∞ and F∞<+∞, there exist positive numbers m1,m2,r1, and r2, such that r1<r2 and
(4.1)f(u)≤m1u,foru∈[0,r1],f(u)≤m2u,foru∈[r2,+∞).
Let m=max{m1,m2,maxr1≤u≤r2{f(u)/u}}. Then we have
(4.2)f(u)≤mu,foru∈[0,+∞).
Assume v(t) is a positive solution of (1.2). We will show that this leads to a contradiction for 0<λ<λ0:=(mC1)-1. Since Aλv(t)=v(t) for t∈[0,1],
(4.3)‖v‖=‖Aλv‖≤λΓ(α)∫01(α-1)k(s)f(v(s))ds≤mλΓ(α)‖v‖∫01(α-1)k(s)ds<‖v‖,
which is a contradiction. Therefore, (1.2) has no positive solution. The proof is complete.Theorem 4.2.
Assume(H) holds. If f0>0 and f∞>0, then there exists a λ0>0 such that for all λ>λ0, the boundary value problem (1.2) has no positive solution.Proof.
Byf0>0 and f∞>0, we know that there exist positive numbers n1,n2,r1, and r2, such that r1<r2 and
(4.4)f(u)≥n1u,foru∈[0,r1],f(u)≥n2u,foru∈[r2,+∞).
Let n=min{n1,n2,minr1≤u≤r2{f(u)/u}}>0. Then we get
(4.5)f(u)≥nu,foru∈[0,+∞).
Assume v(t) is a positive solution of (1.2). We will show that this leads to a contradiction for λ>λ0:=(q(l)nC2)-1. Since Aλv(t)=v(t) for t∈[0,1],
(4.6)‖v‖=‖Aλv‖≥λΓ(α)∫01q(l)k(s)f(v(s))ds>‖v‖,
which is a contradiction. Thus, (1.2) has no positive solution. The proof is complete.


## 5. Examples

In this section, we will present some examples to illustrate the main results.Example 5.1.
Consider the boundary value problem(5.1)D0+5/2u(t)+λua=0,0<t<1,a>1,u(0)=u(1)=u′(0)=0.
Sinceα=5/2, we have(5.2)C1=1Γ(α)∫01(α-1)k(s)ds=1Γ(5/2)∫0132s(1-s)3/2ds=0.1290,C2=1Γ(α)∫011(α-1)q(s)k(s)ds=1Γ(5/2)∫0123s5/2(1-s)5/2ds=0.0077.
Let f(u)=ua,a>1. Then we have F0=0, f∞=+∞. Choose l=1/2. Then q(1/2)=2/8=0.1768. So q(l)C2f∞>F0C1 holds. Thus, by Theorem 3.2, the boundary value problem (5.1) has a positive solution for each λ∈(0,+∞).Example 5.2.
Discuss the boundary value problem(5.3)D0+5/2u(t)+λub=0,0<t<1,0<b<1,u(0)=u(1)=u′(0)=0.
Sinceα=5/2, we have C1=0.1290 and C2=0.0077. Let f(u)=ub,0<b<1. Then we have F∞=0,f0=+∞. Choose l=1/2. Then q(1/2)=2/8=0.1768. So q(l)C2f0>F∞C1 holds. Thus, by Theorem 3.3, the boundary value problem (5.3) has a positive solution for each λ∈(0,+∞).Example 5.3.
Consider the boundary value problem(5.4)D0+5/2u(t)+λ(200u2+u)(2+sin⁡u)u+1=0,0<t<1,a>1,u(0)=u(1)=u′(0)=0.
Sinceα=5/2, we have C1=0.129 and C2=0.0077. Let f(u)=(200u2+u)(2+sinu)/(u+1). Then we have F0=f0=2,F∞=600, f∞=200, and 2u<f(u)<600u.
(i)
Choosel=1/2. Then q(1/2)=2/8=0.1768. So q(l)C2f∞>F0C1 holds. Thus, by Theorem 3.2, the boundary value problem (5.4) has a positive solution for each λ∈(3.6937,3.8759).(ii)
By Theorem4.1, the boundary value problem (5.4) has no positive solution for all λ∈(0,0.0129).(iii)
By Theorem4.2, the boundary value problem (5.4) has no positive solution for all λ∈(369.369,+∞).Example 5.4.
Consider the boundary value problem(5.5)D0+5/2u(t)+λ(u2+u)(2+sin⁡u)150u+1=0,0<t<1,a>1,u(0)=u(1)=u′(0)=0.
Sinceα=5/2, we have C1=0.129 and C2=0.0077. Let f(u)=(u2+u)(2+sinu)/(150u+1). Then we have F0=f0=2, F∞=1/50, f∞=1/150, and u/150<f(u)<2u.
(i)
Choosel=1/2. Then q(1/2)=2/8=0.1768. So q(l)C2f0>F∞C1 holds. Thus, by Theorem 3.3, the boundary value problem (5.5) has a positive solution for each λ∈(369.369,387.5968).(ii)
By Theorem4.1, the boundary value problem (5.5) has no positive solution for all λ∈(0,3.8759).(iii)
By Theorem4.2, the boundary value problem (5.5) has no positive solution for all λ∈(110810.6911,+∞).



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*Source: 390543-2010-12-26.xml*