# The Positive Properties of Green’s Function for Fractional Differential Equations and Its Applications

Authors: Fuquan Jiang; Xiaojie Xu; Zhongwei Cao
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider the properties of Green’s function for the nonlinear fractional differential equation boundary value problem:D0+αu(t)+f(t,u(t))+e(t)=0,0<t<1,u(0)=u'(0)=⋯=u(n-2)(0)=0,u(1)=βu(η), where n-1<α≤n,n≥3,0<β≤1,0≤η≤1, D0+α is the standard Riemann-Liouville derivative. Here our nonlinearity f may be singular at u=0. As applications of Green’s function, we give some multiple positive solutions for singular boundary value problems by means of Schauder fixed-point theorem.


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. This is due to the intensive development of the theory of fractional calculus itself as well as its applications. Apart from diverse areas of mathematics, fractional differential equations arise in rheology, dynamical processes in selfsimilar and porous structures, fluid flows, electrical networks, viscoelasticity, chemical physics, and many other branches of science. For details, see [1–10].It should be noted that most of the 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 and multiplicity of solution to the nonlinear fractional differential equations boundary value problems, see [11–17].Bai [14] investigated the existence and uniqueness of positive solutions for a nonlocal boundary value problem of fractional differential equation
(1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=0,u(1)=βu(η),
by contraction map principle and fixed-point index theory, where 1<α≤2, 0<βηα-1<1, 0<η<1, D0+α is the standard Riemann-Liouville derivative. The function f is continuous on [0,1]×[0,∞).Li et al. [17] investigated the the existence and multiplicity results of positive solutions for the nonlinear differential equation of fractional order
(2)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=0,D0+βu(1)=aD0+βu(ξ),
by using some fixed-point theorems, where 1<α≤2, 0<β≤1, 0≤a≤1, ξ∈(0,1), aξα-β-2≤1-β, 0≤α-β-1, D0+α is the standard Riemann-Liouville derivative.Xu and Fei [18] considered the properties of Green’s function for the nonlinear fractional differential equation boundary value problem
(3)D0+αu(t)+f(t,u(t))+e(t)=0,0<t<1,u(0)=0,D0+βu(1)=aD0+βu(ξ),
where 1<α≤2, 0<β≤1, 0≤a≤1, 0<ξ<1, α-β-1≥0, D0+α is the standard Riemann-Liouville derivative. Here the nonlinearity f may be singular at u=0. As applications of Green’s function, they give some existence of positive solutions for singular boundary value problems by means of Schauder fixed-point theorem. Here they consider the case: γ*=0, γ*≥0, γ*≤0.In this paper, we consider the singular boundary value problem(4)D0+αu(t)+f(t,u(t))+e(t)=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=βu(η),
where n-1<α≤n, n≥3, 0<β, η<1 is a real constant, D0+α is the standard Riemann-Liouville fractional derivative. We will deduce a property of Green’s function. The result we establish in Section 2 can be stated as follow.Theorem 1.
The FunctionG(t,s) defined by (12) is continuous and satisfies
(5)Mtα-1s(1-s)α-1Γ(α)(1-βηα-1)≤G(t,s)≤tα-1(1-s)α-1Γ(α)(1-βηα-1),fort,s∈[0,1],
where 0<M=min{1-βηα-1,βηα-2(1-η),βηα-1}<1.In this paper, we give some existence of positive solutions for singular boundary value problems by means of Schauder fixed-point theorem for the case:γ*=0, γ*≥0, γ*≤0, γ*<0<γ*.The paper is organized as follows. In Section2, we state some known results and give a property of Green’s function. In Section 3, using Schauder fixed-point theorem, the existence of positive solutions to singular problems are obtained.


## 2. Background Materials

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory.Definition 2 (see [7]).
The Riemann-Liouville fractional integral of orderα>0 of a function y:(0,∞)→R is given by
(6)I0+αy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds,
provided the right side is pointwise defined on (0,∞).Definition 3 (see [7]).
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function y:(0,∞)→R is given by
(7)D0+αy(t)=1Γ(n-α)(ddt)n∫0t‍y(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,∞).From the definition of Riemann-Liouville’s derivative, we can obtain the statement.Lemma 4 (see [7]).
Letα>0, if one assumes that u∈C(0,1)∩L(0,1), then the fractional differential equation
(8)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 5 (see [7]).
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,
(9)I0+αD0+αu(t)=u(t)+C1tα-1+C2tα-2+⋯+CNtα-N,
for some Ci∈R, i=1,2,…,N, N is the smallest integer greater than or equal to α.Lemma 6.
Givenh∈C(0,1) the problem
(10)D0+αu(t)+h(t)=0,0<t<1,n-1<α≤n,n≥3,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=βu(η),0<β,η<1
is equivalent to
(11)u(t)=∫01‍G(t,s)h(s)ds,
where
(12)G(t,s)={((t-s)α-1(1-βηα-1)[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1))×((1-βηα-1)Γ(α))-1,0≤s≤t≤1,s≤η,([t(1-s)]α-1-(t-s)α-1(1-βηα-1))×((1-βηα-1)Γ(α))-1,0<η≤s≤t≤1,([t(1-s)]α-1-βtα-1(η-s)α-1)×((1-βηα-1)Γ(α))-1,0≤t≤s≤η<1,([t(1-s)]α-1)×((1-βηα-1)Γ(α))-1,0≤t≤s≤1,η≤s.Proof.
We can apply Lemma5 to reduce (10) to an equivalent integral equation
(13)u(t)=-I0+αh(t)+C1tα-1+C2tα-2+⋯+Cntα-n,
for some C1,C2,…,Cn∈R. Consequently, the general solution of (10) is
(14)u(t)=-1Γ(α)∫0t‍(t-s)α-1h(s)ds+C1tα-1+C2tα-2+⋯+Cntα-n.
By u(0)=u′(0)=⋯=u(n-2)(0)=0,  this is C2=C3=⋯=Cn=0.
On the other hand,u(1)=βu(η) combining with
(15)u(1)=-1Γ(α)∫01‍(1-s)α-1h(s)ds+C1,u(η)=-1Γ(α)∫0η‍(η-s)α-1h(s)ds+C1ηα-1
yields
(16)C1=11-βηα-1∫01‍(1-s)α-1Γ(α)h(s)ds-β1-βηα-1∫0η‍(η-s)α-1Γ(α)h(s)ds.
Therefor, the unique solution of problem (10) is
(17)u(t)=-∫0t‍(t-s)α-1Γ(α)h(s)ds+11-βηα-1∫01‍(t(1-s))α-1Γ(α)h(s)ds-β1-βηα-1∫0η‍(t(η-s))α-1Γ(α)h(s)ds.
For t≤η, we have
(18)u(t)=-∫0t‍(t-s)α-1Γ(α)h(s)ds+11-βηα-1[(∫0t‍+∫tη‍+∫η1‍)(t(1-s))α-1Γ(α)h(s)ds]-β1-βηα-1[(∫0t‍+∫tη‍)(t(η-s))α-1Γ(α)h(s)ds]=∫0t(((1-βηα-1)Γ(α))-1([t(1-s)]α-1-β[t(η-s)]α-1-(t-s)α-1(1-βηα-1))×((1-βηα-1)Γ(α))-1)h(s)ds+∫tη‍[t(1-s)]α-1-β[t(η-s)]α-1(1-βηα-1)Γ(α)h(s)ds+∫η1‍[t(1-s)]α-1(1-βηα-1)Γ(α)h(s)ds=∫01‍G(t,s)h(s)ds.
For t≥η, we have
(19)u(t)=-[(∫0η‍+∫ηt‍)(t-s)α-1Γ(α)h(s)ds]-β1-βηα-1∫0η‍(t(η-s))α-1Γ(α)h(s)ds+11-βηα-1[(∫0η‍+∫ηt‍+∫t1‍)(t(1-s))α-1Γ(α)h(s)ds]=∫0η(((1-βηα-1)Γ(α))-1((t-s)α-1(1-βηα-1)[t(1-s)]α-1-β[t(η-s)]α-1-(t-s)α-1(1-βηα-1))×((1-βηα-1)Γ(α))-1)h(s)ds+∫ηt‍[t(1-s)]α-1-(1-βηα-1)(t-s)α-1(1-βηα-1)Γ(α)h(s)ds+∫t1‍[t(1-s)]α-1(1-βηα-1)Γ(α)h(s)ds=∫01‍G(t,s)h(s)ds.
The proof is complete.Proof of Theorem1.
It is easy to prove thatG(t,s) is continuous on [0,1]×[0,1], here we omit it. In the following, we consider (1-βηα-1)Γ(α)G(t,s). When 0≤s<t≤1, s<η, let
(20)g1(t)=[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1).
We have
(21)g1(t)=[t(1-s)]α-1-βtα-1(η-s)α-1-(t-s)α-1(1-βηα-1)=[t(1-s)]α-1-βηα-1tα-1(1-sη)α-1-tα-1(1-st)α-1(1-βηα-1)≥[t(1-s)]α-1-βηα-1tα-1(1-sη)α-1-tα-1(1-s)α-1(1-βηα-1)=tα-1βηα-1[(1-s)α-1-(1-sη)α-1]=tα-1βηα-1[(1-s)α-1-(1-sη)α-2(1-sη)]≥tα-1βηα-1(1-s)α-2[1-s-(1-sη)]=tα-1βηα-1(1-s)α-2s1-ηη≥βηα-1tα-1(1-s)α-1s1-ηη≥Mtα-1(1-s)α-1s.
When 0<η≤s<t≤1, let
(22)g2(t)=[t(1-s)]α-1-(t-s)α-1(1-βηα-1).
We have
(23)g2(t)=[t(1-s)]α-1-(t-s)α-1(1-βηα-1)=[t(1-s)]α-1-tα-1(1-st)α-1(1-βηα-1)≥[t(1-s)]α-1-tα-1(1-s)α-1(1-βηα-1)=βηα-1(t-ts)α-1≥Mtα-1(1-s)α-1s.
When 0≤t≤s≤η<1, let g3(t)=[t(1-s)]α-1-βtα-1(η-s)α-1, we have
(24)g3(t)=[t(1-s)]α-1-βtα-1(η-s)α-1=[t(1-s)]α-1-βηα-1tα-1(1-sη)α-1≥[t(1-s)]α-1-βηα-1tα-1(1-s)α-1=(1-βηα-1)tα-1(1-s)α-1≥Mtα-1(1-s)α-1s.
When 0≤t≤s≤1, η≤s, we have
(25)(1-βηα-1)Γ(α)G(t,s)=[t(1-s)]α-1≥Mtα-1(1-s)α-1s.
It is easy to see thatG(t,s)≤tα-1(1-s)α-1/Γ(α)(1-βηα-1). Thus, the proof is complete.Let us fix some notations to be used in the following. “For a.e.” means “for almost every”. Givena∈L1(0,1), we write a≻0 if a≥0 for a.e. t∈[0,1], and it is positive in a set of positive measure, we write f∈Car((0,1)×(0,+∞),(0,+∞)) if f:(0,1)×(0,+∞)→(0,+∞) is a L1-caratheodory function, that is, the map x↦f(t,x) is continuous for a.e. t∈(0,1), and the map t↦f(t,x) is measurable for all x∈(0,+∞).Let us define(26)γ*=supt∈[0,1]∫01‍G(t,s)tα-1e(s)ds,γ*=inft∈[0,1]∫01‍G(t,s)tα-1e(s)ds.
Then,
(27)tα-1γ*≤∫01‍G(t,s)e(s)ds≤tα-1γ*.


## 3. Main Results

In this section, we establish the existence of positive solutions for equation(28)D0+αu(t)+f(t,u(t))+e(t)=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=βu(η),
where n-1<α≤n,  n≥3,  0<β≤1,  0≤η≤1, f∈Car((0,1)×(0,+∞),(0,+∞)), e(t)∈L1[0,1], D0+α is the standard Riemann-Liouville derivative. The following is the first main result in this section.Theorem 7.
Suppose that the following conditions are satisfied.(H1)
For eachL>0, there exists a function ϕL≻0 such that f(t,tα-1x)≥ϕL(t) for a.e. t∈(0,1), all x∈(0,L].(H2)
There existg(x), h(x), and k(t)≻0, such that
(29)0≤f(t,x)≤k(t){g(x)+h(x)}fora.e.t∈(0,1),allx∈(0,∞),
here
(30)g:(0,+∞)⟶[0,+∞)iscontinuousandnonincreasing,h:[0,+∞)⟶[0,+∞)iscontinuous,andhgisnondecreasing.(H3)
There exist two positive constantsR>r>0 such that
(31)R>ΦR1+γ*≥r>0,∫01‍k(s)g(rsα-1)ds<+∞,R≥(1+h(R)g(R))∫01‍(1-s)α-1Γ(α)(1-βηα-1)k(s)g(rsα-1)ds+γ*,
and here
(32)ΦR1=∫01‍Ms(1-s)α-1Γ(α)(1-βηα-1)ϕR(s)ds.Then, (28) has at least one positive solution.Proof.
LetE=(C[0,1],∥·∥), and Ω is a closed convex set defined as
(33)Ω={x∈C[0,1]:tα-1r≤x(t)≤tα-1R∀t∈[0,1]},
here E=C[0,1] is the Banach space of continuous functions defined on [0,1] with the norm
(34)∥x∥:=maxt∈[0,1]|x(t)|,
and R>r>0 are positive constants to be given below.
Now, we define an operatorT:Ω→E by
(35)(Tx)(t):=∫01‍G(t,s)[f(s,x(s))+e(s)]ds.
Then, (28) is equivalent to the fixed-point problem
(36)x=Tx.
Let R be the positive constant satisfying (H3) and
(37)ΦR1+γ*≥r.
Then, we have R>r>0. Now, we prove T(Ω)⊂Ω.
In fact, for eachx∈Ω and for all t∈(0,1), by (H1) and (H3)
(38)(Tx)(t)≥∫01‍tα-1Ms(1-s)α-1Γ(α)(1-βηα-1)ϕR(s)ds+∫01‍G(t,s)e(s)ds≥tα-1[ΦR1+γ*]≥tα-1r.
On the other hand, by conditions (H2) and (H3), we have
(39)(Tx)(t)≤∫01‍tα-1(1-s)α-β-1Γ(α)(1-aξα-β-1)k(s)×[g(sα-1x(s)sα-1)+h(sα-1x(s)sα-1)]ds+∫01‍G(t,s)e(s)ds≤tα-1[∫01‍(1-s)α-1Γ(α)(1-βηα-1)k(s)g(sα-1x(s)sα-1)×{1+h(sα-1(x(s)/sα-1))g(sα-1(x(s)/sα-1))}ds+γ*]≤tα-1×[(1+h(R)g(R))∫01‍(1-s)α-1Γ(α)(1-βηα-1)k(s)g(rsα-1)ds+γ*]≤tα-1R.
In conclusion, T(Ω)⊂Ω.
Finally, it is standard thatT:Ω→Ω is a continuous and completely continuous operator. By a direct application of Schauder’s fixed-point theorem, (28) has at least one positive solution x(t)∈C[0,1], the proof is finished.Case 1 (γ*=0).
As an application of Theorem7, we consider the case γ*=0. The following corollary is a direct result of Theorem 7 with r=ΦR1.Corollary 8.
Suppose thatf(t,x) satisfies conditions (H1)-(H2). Furthermore, assume the following.(H3*)
There exists a positive constantR>0 such that
(40)R>ΦR1>0,∫01‍k(s)g[(ΦR1)sα-1]ds<+∞,R≥(1+h(R)g(R))×∫01‍(1-s)α-1Γ(α)(1-βηα-1)k(s)g[(ΦR1)sα-1]ds+γ*,
and here
(41)ΦR1=∫01‍Ms(1-s)α-1Γ(α)(1-βηα-1)ϕR(s)ds.Then, (28) has at least one positive solution.From now on, let us define(42)β1=∫01‍Ms1-λ(α-1)(1-s)α-1Γ(α)(1-βηα-1)k(s)ds,β2=∫01‍s-λ(α-1)(1-s)α-1Γ(α)(1-βηα-1)k(s)ds.Example 9.
Suppose that the nonlinearity in (28) is
(43)f(t,x)=k(t)xλ,
where k≻0, 0<λ<1 and
(44)ω(λ):=∫01‍k(s)s-λ(α-1)ds<+∞.
If γ*=0, then (28) has at least one positive solution.Proof.
We will apply Corollary8. To this end, we take
(45)ϕL(t)=k(t)(tα-1L)λ,g(x)=1xλ,h(x)≡0,
then (H1) and (H2) are satisfied since ω(λ)<+∞, and the existence condition (H3*) becomes
(46)(Rλβ1)λβ2+γ*≤R,R>β1Rλ,ω(λ)<+∞,
for some R>0. Since 0<λ<1, we can choose R>0 large enough such that (46) is satisfied, and the proof is finished.Example 10.
Suppose that the nonlinearity in (28) is
(47)f(t,x)=k(s)(x-λ+μxν),
where 0<λ<1, ν≥0 and μ≥0 is a nonnegative parameter. For each e(t) with γ*=0,ω(λ)<+∞,(i)
ifλ+ν<1-λ2, then (28) has at least one positive solution for each μ≥0.(ii)
Ifλ+ν≥1-λ2, then (28) has at least one positive solution for each 0≤μ<μ1, where μ1 is some positive constant.Proof.
We will apply Corollary8. To this end, we take
(48)ϕL(t)=k(t)(tα-1L)λ,g(x)=x-λ,h(x)=μxν.
Then, (H1)-(H2) are satisfied since ω(λ)<+∞. Now, the existence condition (H3*) becomes ω(λ)<+∞, and
(49)μ≤R1-λ2β1λ-γ*β1λR-λ2-β2β2Rλ+ν,
for some R>0 with R1+λ>β1. So, (28) has at least one positive solution for
(50)0<μ<μ1=supR>β11/1+λR1-λ2β1λ-γ*β1λR-λ2-β2β2Rλ+ν.
Note that μ1=∞ if λ+ν<1-λ2 and if λ+ν≥1-λ2, set
(51)l(R):=R1-λ2β1λ-γ*β1λR-λ2-β2β2Rλ+ν,
then, we have
(52)l′(R)=1Rλ+ν+1β2[(1-λ2-λ-ν)R1-λ2β1λ+(λ2+λ+ν)γ*β1λR-λ2+(λ+ν)β2].
Let the function l(R) possess a maximum at R0, then
(53)(λ2+λ+ν-1)R01-λ2β1λ=(λ2+λ+νλ)γ*β1λR0-λ2+(λ+ν)β2,
so we have
(54)(λ+ν-1+λ2)β1λR0≥(λ+ν)β2R0λ2,
it is easy to find that R0>(β2/β1λ)1/(1-λ2) since λ+ν≥1-λ2, and 0<λ<1. Finally, it would remain to prove R0>β11/1+λ. This is easily verified through elementary computations since β1≤β2. We have the desired results (i) and (ii).Case 2 (γ*>0).
The next result explores the case whenγ*>0. In this case r=γ*.Corollary 11.
Suppose thatf(t,x) satisfies (H2). Furthermore, assume the following.(H4)
There existsR>0, ∫01‍k(s)g(γ*sα-1)ds<+∞, such that
(55)(1+h(R)g(R))∫01‍(1-s)α-1Γ(α)(1-βηα-1)k(s)g(γ*sα-1)ds+γ*≤R.If γ*>0, then (28) has at least one positive solution.Example 12.
Suppose that the nonlinearity in (28) be (43) with k≻0, λ>0. If γ*>0, ω(λ)<+∞, then (28) has at least one positive solution.Proof.
We will apply Corollary11. Take k(t), g(x), and h(x) as the same in the proof of Example 9. Then, (H2) is satisfied, and the existence condition (H4) is satisfied if we take R>0 with R≥β2/γ*λ+γ*, and ω(λ)<+∞.Example 13.
Let the nonlinearity in (28) be (47) with λ>0 and ν≥0. For each e(t) with γ*>0, ω(λ)<+∞,(i)
ifλ+ν<1, then (28) has at least one positive solution for each μ≥0.(ii)
Ifλ+ν≥1, then (28) has at least one positive solution for each 0≤μ<μ2, where μ2 is some positive constant.Proof.
We will apply Corollary11. To this end, we take g(x), h(x), and k(t) as the same in the proof of Example 10, then (H2) is satisfied, and the existence condition (H4) becomes ω(λ)<+∞,
(56)μ≤Rγ*λ-γ*γ*λ-β2β2Rλ+ν,
for some R>0. So, (28) has at least one positive solution for
(57)0<μ<μ2=supR>0Rγ*λ-γ*γ*λ-β2β2Rλ+ν.
Note that μ2=∞ if λ+ν<1 and μ2=γ*λ/β2 if λ+ν=1, and if λ+ν>1 set
(58)l(R):=Rγ*λ-γ*γ*λ-β2β2Rλ+ν.
The function l(R) possesses a maximum at
(59)R0:=(λ+ν)(γ*γ*λ+β2)(λ+ν-1)γ*λ,
then μ2=l(R0). We have the desired results (i) and (ii).Case 3 (γ*≤0).
The next result considers the caseγ*≤0.Corollary 14.
Suppose thatf(t,x) satisfies (H1)-(H2). Furthermore, assume the following.(H5)
There exist two positive constantsR>r>0 such that
(60)R>ΦR1+γ*≥r>0,∫01‍k(s)g(rsα-1)ds<+∞,R≥(1+h(R)g(R))∫01‍(1-s)α-1Γ(α)(1-βηα-1)k(s)g(rsα-1)ds,
here
(61)ΦR1=∫01‍Ms(1-s)α-1Γ(α)(1-βηα-1)ϕR(s)ds.Then, (28) has at least one positive solution.Example 15.
Suppose that the nonlinearity in (28) be (43) with k≻0, λ>0. If γ*≤0, ω(λ)<+∞,
(62)γ*≥[β1λ2β2λ]1/(1-λ2)(1-1λ2),
then (28) has at least one positive solution.Proof.
We will apply Corollary14. Take k(t), g(x) as the same in the proof of Example 9. Then, (H2) is satisfied, and the existence condition (H5) is satisfied if we take R>r>0 with
(63)β1Rλ+γ*≥r,R≥β2rλ,
and ω(λ)<+∞. If we fix R=β2/rλ, then the first inequality holds if r satisfies
(64)β1β2λrλ2+γ*≥r,
or equivalently
(65)γ*≥l(r):=r-β1β2λrλ2.
The function l(r) possesses a minimum at
(66)r0:=[β1β2λλ2]1/(1-λ2).
Taking r=r0, then the first inequality in (63) holds if γ*≥l(r0), which is just condition (62). The second inequality holds directly from the choice of R, so it remains to prove that R=β2/rλ>r0. This is easily verified through elementary computations.Example 16.
Let the nonlinearity in (28) be (47) with k≻0,λ>0 and ν≥0. If γ*≤0,ω(λ)<+∞,
(67)γ*≥m0λ[β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ-β1],
here m0 is the unique solution of the equation
(68)β21/λ(m1-λ2+μm1-λ-ν-λ2)1/λ-1×[m1-λ2+μ(1-λ-ν)m1-λ-ν-λ2]=λ2β1.
Then (28) has at least one positive solution.Proof.
We will apply Corollary14. To this end, we take g(x),  h(x), and  k(t) as the same in the proof of Example 10, then (H2) is satisfied, and the existence condition (H5) is satisfied if we take R>r>0 with
(69)γ*≥r-β1R-λ,(1+μRλ+ν)β2r-λ≤R,
and ω(λ)<+∞. If we fix R=(1+μRλ+ν)β2r-λ, then the first inequality holds if R satisfies
(70)γ*≥[(1+μRλ+ν)β2R]1/λ-β1R-λ.
Let m=1/R, then
(71)γ*≥mλ[β21/λ(m1-λ2+μm1-λ-ν-λ2)1/λ-β1]=:F(m).
Then, we have(72)F′(m)=λmλ-1[(β2(m1-λ2+μm1-λ-ν-λ2))1/λ-β1]+mλ[1λ(β2(m1-λ2+μm1-λ-ν-λ2))1/λ-1×((1-λ2)β2m-λ2+μ(1-λ-ν-λ2)β2m-λ-ν-λ2m-λ2)1λ(β2(m1-λ2+μm1-λ-ν-λ2))1/λ-1]=1λmλ-1{λ2[(β2(m1-λ2+μm1-λ-ν-λ2))1/λ-β1]+β21/λ[(m1-λ2+μm1-λ-ν-λ2)1/λ-1×((1-λ2)m1-λ2+μ(1-λ-ν-λ2)m1-λ-ν-λ2m1-λ2)(m1-λ2+μm1-λ-ν-λ2)1/λ-1](β2(m1-λ2+μm1-λ-ν-λ2))1/λ}=1λmλ-1{β21/λ(m1-λ2+μm1-λ-ν-λ2)1/λ-1×[((1-λ2)m1-λ2+μ(1-λ-ν-λ2)m1-λ-ν-λ2)λ2(m1-λ2+μm1-λ-ν-λ2)+((1-λ2)m1-λ2+μ(1-λ-ν-λ2)m1-λ-ν-λ2)]-λ2β1(m1-λ2+μm1-λ-ν-λ2)1/λ-1}=1λmλ-1{β21/λ(m1-λ2+μm1-λ-ν-λ2)1/λ-1×[m1-λ2+μ(1-λ-ν)m1-λ-ν-λ2]-λ2β1(m1-λ2+μm1-λ-ν-λ2)1/λ-1}.
LetF′(m)=0, then we have
(73)1λmλ-1{β21/λ(m1-λ2+μm1-λ-ν-λ2)1/λ-1×[m1-λ2+μ(1-λ-ν)m1-λ-ν-λ2]-λ2β1(m1-λ2+μm1-λ-ν-λ2)1/λ-1}=0.
Now, let us defineΦ(m) by
(74)Φ(m)=:β21/λ(m1-λ2+μm1-λ-ν-λ2)1/λ-1×[m1-λ2+μ(1-λ-ν)m1-λ-ν-λ2].
It is easy to see thatΦ(m) is a nondecreasing function for m∈[0,+∞) and Φ(m)→+∞, as m→+∞. Thus, Φ(m)=λ2β1 has a unique solution m0 such that
(75)β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ-1×[m01-λ2+μ(1-λ-ν)m01-λ-ν-λ2]=λ2β1,
and F(m0)=infm>0F(m).
So, it remains to prove thatR>r=[(1+μRλ+ν)β2/R]1/λ, that is,
(76)m0λ+1β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ<1.
In fact, by (75), we have
(77)λ2β1≥(β2m01-λ2)1/λ-1(β2m01-λ2)=(β2m01-λ2)1/λ,
that is,
(78)m0≤(λ2β1β21/λ)λ/(1-λ2).
Also we have(79)λ2β1≥β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ-1×[(1-λ-ν)m01-λ2+μ(1-λ-ν)m01-λ-ν-λ2]=(1-λ-ν)β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ,
that is,
(80)β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ≤λ2β11-λ-ν.
Thus, we have(81)m0λ+1β21/λ(m01-λ2+μm01-λ-ν-λ2)1/λ<((λ2β1β21/λ)λ/(1-λ2))λ+1λ2β11-λ-ν=(λ2β1β21/λ)λ/(1-λ)λ2β11-λ-ν=λ2λ/(1-λ)(β1β2)1/(1-λ)λ21-λ-ν<1
since 0<λ,ν<1, and 1-λ-ν-λ2>0.
We have the desired results.Case 4 (γ*<0<γ*).Example 17.
Suppose that the nonlinearity in (28) be (43) with k≻0,λ>0. If γ*<0<γ*,ω(λ)<+∞,
(82)γ*≥m0λ[β21/λm0(1-λ2)/λ-β1(1+γ*m0)-λ],
and here m0 is the unique solution of the equation
(83)β21/λm(1-λ2)/λ(1+γ*m)λ+1=λ2β1,
then (28) has at least one positive solution.Proof.
We will apply Theorem7. Take k(t),g(x) as the same in the proof of Example 9. Then, (H2) is satisfied, and the existence condition (H3) is satisfied if we take R>r>0 with
(84)γ*≥r-β1R-λ,β2r-λ+γ*≤R,
and ω(λ)<+∞. If we fix R=β2/rλ+γ*, then the first inequality holds if R satisfies
(85)γ*≥(β2R-γ*)1/λ-β1R-λ.
Taking m=1/(R-γ*), then for m∈(0,+∞), we have
(86)γ*≥(β2m)1/λ-β1mλ(1+γ*m)λ=mλ[β21/λm(1-λ2)/λ-β1(1+γ*m)-λ]=:F(m).
Then, we have(87)F′(m)=λmλ-1[β21/λm(1-λ2)/λ-β1(1+γ*m)-λ]+mλ[1-λ2λβ21/λm(1-λ2)/λ-1+λβ1γ*(1+γ*m)-λ-1]=1λmλ-1{1-λ2λλ2[β21/λm(1-λ2)/λ-β1(1+γ*m)-λ]+λm[1-λ2λβ21/λm(1-λ2)/λ-1+λβ1γ*(1+γ*m)-λ-11-λ2λ]}=1λmλ-1[β21/λm(1-λ2)/λ-λ2β1(1+γ*m)-λ-1]=1λmλ-1(1+γ*m)-λ-1[β21/λm(1-λ2)/λ(1+γ*m)λ+1-λ2β1].
LetF′(m)=0, then we have
(88)1λmλ-1(1+γ*m)-λ-1[β21/λm(1-λ2)/λ(1+γ*m)λ+1-λ2β1]=0.
Now, let us defineΦ(m) by
(89)Φ(m)=:β21/λm(1-λ2)/λ(1+γ*m)λ+1.
It is easy to see thatΦ(m) is a nondecreasing function for m∈[0,+∞), and Φ(m)→+∞, as m→+∞. Thus, Φ(m)=λ2β1 has a unique solution m0 such that
(90)β21/λm(1-λ2)/λ(1+γ*m)λ+1=λ2β1,
and F(m0)=infm>0F(m).
So, it remains to prove thatR>r=[β2/(R-γ*)]1/λ, that is,
(91)β21/λm01+1/λ≤1+γ*m0.
In fact, by (90), we have
(92)β21/λm0(1-λ2)/λ≤λ2β1,
that is,
(93)m0≤(λ2β1β21/λ)λ/(1-λ2).
Thus, we have(94)β21/λm01+1/λ<β21/λ((λ2β1β21/λ)λ/(1-λ2))(1+λ)/λm01+1/λ=λ2/(1-λ)(β1β2)1/(1-λ)<1<1+γ*m0.
The proof is complete.Example 18.
Let the nonlinearity in (28) be (47) with k≻0,λ>0 and ν≥0. If γ*<0<γ*,ω(λ)<+∞,
(95)γ*≥m0λ[β21/λ(m01-λ2+μ(1+γ*m0)λ+νm01-λ-ν-λ2)1/λ-β1(1+γ*m0)-λ(m01-λ2+μ(1+γ*m0)λ+νm01-λ-ν-λ2)1/λ],
here m0 is the unique solution of the equation
(96)β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1×[m1-λ2(1+γ*m)λ+1+μ(1-λ-ν+γ*m)m1-λ-ν-λ2×(1+γ*m)2λ+ν]=λ2β1.
Then (28) has at least one positive solution.Proof.
We will apply Theorem7. To this end, we take g(x),  h(x), and k(t) as the same in the proof of Example 10, then (H2) is satisfied, and the existence condition (H3) is satisfied if we take R>r>0 with
(97)γ*≥r-β1R-λ,(1+μRλ+ν)β2r-λ+γ*≤R,
and ω(λ)<+∞. If we fix R=(1+μRλ+ν)β2r-λ+γ*, then the first inequality holds if
(98)γ*≥[(1+μRλ+ν)β2R-γ*]1/λ-β1R-λ.
Let m=1/(R-γ*), then
(99)γ*≥mλ[β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-β1(1+γ*)-λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ]=:F(m).
Then, we have(100)F′(m)=λmλ-1[β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-β1(1+γ*m)-λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ]+mλ[1λβ21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1×((1-λ2)m-λ2+μ(1+γ*m)λ+ν×(1-λ-ν-λ2)m-λ-ν-λ2+μ(λ+ν)(1+γ*m)λ+ν-1m1-λ-ν-λ2γ*(1-λ2)m-λ2+μ(1+γ*m)λ+ν)+β1λ(1+γ*m)-λ-1γ*β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-11λ]=1λmλ-1{λ2β21/λ[(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-λ2β1(1+γ*m)-λ+β21/λ(m1-λ2+μ(1+γ*m)λ+ν×m1-λ-ν-λ2)1/λ-1×[(1-λ2)m1-λ2+μ(1+γ*m)λ+ν×(1-λ-ν-λ2)m1-λ-ν-λ2+μ(λ+ν)(1+γ*m)λ+ν-1×m1-λ-ν-λ2mγ*](m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ]+λ2β1(1+γ*m)-λ-1mγ*(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ}=1λmλ-1{β21/λ[(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)]1/λ-1×[λ2(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)+(1-λ2)m1-λ2+μ(1+γ*m)λ+ν(1-λ-ν-λ2)m1-λ-ν-λ2+μ(λ+ν)(1+γ*m)λ+ν-1m1-λ-ν-λ2mγ*]×λ2β1(1+γ*m)-λ-1mγ*-λ2β1(1+γ*m)-λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2))1/λ-1}=1λmλ-1(1+γ*m)-λ-1×{β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1×[m1-λ2(1+γ*m)λ+1+μ(1-λ-ν+γ*m)×m1-λ-ν-λ2(1+γ*m)2λ+ν]-λ2β1(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1}.
LetF′(m)=0, then we have
(101)1λmλ-1(1+γ*m)-λ-1×{β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1×[m1-λ2(1+γ*m)λ+1+μ(1-λ-ν+γ*m)m1-λ-ν-λ2×(1+γ*m)2ν+νm1-λ2]-λ2β1(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1}=0.
Now, let us defineΦ(m) by
(102)Φ(m)=:β21/λ(m1-λ2+μ(1+γ*m)λ+νm1-λ-ν-λ2)1/λ-1×[m1-λ2(1+mγ*m)λ+1+μ(1-λ-ν+γ*m)m1-λ-ν-λ2(1+γ*m)2λ+ν].
It is easy to see thatΦ(m) is a nondecreasing function for m∈[0,+∞), and Φ(m)→+∞, as m→+∞. Thus, Φ(m)=λ2β1 has a unique solution m0 such that
(103)β21/λ(m01-λ2+μ(1+γ*m)λ+νm01-λ-ν-λ2)1/λ-1×[m01-λ2(1+γ*m)λ+1+μ(1-λ-ν+γ*m)m01-λ-ν-λ2(1+γ*m0)2λ+ν]=λ2β1,
and F(m0)=infm>0F(m).
So, it remains to prove thatR>r=[(1+μRλ+ν)β2/(R-γ*)]1/λ, that is,
(104)m0λ+1β21/λ(m01-λ2+μ(1+γ*m)λ+νm01-λ-ν-λ2)1/λ<1+γ*m0.
In fact, by (103), we have
(105)λ2β1≥(β2m01-λ2)1/λ-1(β2m01-λ2)=(β2m01-λ2)1/λ,
that is,
(106)m0≤(λ2β1β21/λ)λ/(1-λ2).
Also we have(107)λ2β1≥β21/λ(m01-λ2+μ(1+γ*m)λ+νm01-λ-ν-λ2)1/λ-1×[(1-λ-ν)m01-λ2+μ(1-λ-ν)m01-λ-ν-λ2(1+γ*m)λ+ν]=(1-λ-ν)β21/λ(m01-λ2+μ(1+γ*m)λ+νm01-λ-ν-λ2)1/λ,
that is,
(108)β21/λ(m01-λ2+μ(1+γ*m)λ+νm01-λ-ν-λ2)1/λ≤λ2β11-λ-ν.
Thus, we have(109)m0λ+1β21/λ(m01-λ2+μ(1+γ*m)λ+νm01-λ-ν-λ2)1/λ<((λ2β1β21/λ)λ/(1-λ2))λ+1λ2β11-λ-ν=(λ2β1β21/λ)λ/(1-λ)λ2β11-λ-ν=λ2λ/(1-λ)(β1β2)1/(1-λ)λ21-λ-ν<1<1+γ*m,
since 0<λ,ν<1 and 1-λ-ν-λ2>0.
Thus, the proof is complete.Remark 19.
It is easy to find that analogous results to Examples10, 13, 16, and 18 for the general equation with the nonlinearity in (28) are
(110)f(t,x)=b(t)xλ+μc(t)xν+e(t),
with b,c≻0, but the notation becomes cumbersome. Here we consider the nonlinearity (47) only for the simplicity.



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*Source: 531038-2013-02-24.xml*