Li et al. Advances in Difference Equations (2017) 2017:126 DOI 10.1186/s13662-017-1185-3
RESEARCH
Open Access
The positive solutions of infinite-point boundary value problem of fractional differential equations on the infinite interval Xiaochen Li, Xiping Liu* , Mei Jia and Luchao Zhang *
Correspondence:
[email protected] College of Science, University of Shanghai for Science and Technology, Shanghai, 200093, China
Abstract In this paper, we consider a class of infinite-point boundary value problems of fractional differential equations on the infinite interval [0, +∞) with a disturbance parameter. By using the method of upper and lower solutions, fixed point index theory and some fixed point theorems, the existence, multiplicity and nonexistence for the positive solution of the boundary value problem are obtained, respectively. The impact of the disturbance parameters on the existence of positive solutions is also given. Finally, some examples are presented to illustrate the wide range of potential applications of our main results. Keywords: fractional differential equations; infinite-point boundary value problem; half line; positive solution; L1 -Carathéodory conditions
1 Introduction In this paper, we are concerned with a class of infinite-point boundary value problems of fractional differential equations on the infinite interval with a disturbance parameter λ as follows: ⎧ δ ⎪ ⎪ ⎨D+ u(t) + q(t)f (t, u(t)) = ,
t ∈ (, +∞),
u() = Dδ– + u(+∞) = , ⎪ ⎪ ⎩ δ– δ– D+ u() = ∞ i= g(ξi )D+ u(ξi ) + λ,
(.)
where Dδ+ is the standard Riemann-Liouville fractional derivative, < δ < . < ξ < + ξ < · · · < ξi < · · · < +∞, i = , , . . . , g(ξi ) ≥ and ∞ i= g(ξi ) is convergent. R = [, +∞), f : R+ × R+ → R+ is an L -Carathéodory function, the disturbance parameter λ ∈ R+ . δ– Dδ– + u(+∞) := limt→+∞ D+ u(t) exists.
In recent years, the theory of fractional differential equations has been widely used in various fields, such as physics, mechanics, chemistry, engineering, etc., see [–]. Meanwhile, the study of boundary value problems of fractional differential equations has gained plenty of meaningful results and has been growing rapidly, see [–]. © The Author(s) 2017. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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In [], the authors studied the existence and nonexistence of the positive solutions for the fractional differential equation with two disturbance parameters ⎧ ⎨Dδ + u(t) + f (t, u(t)) = ,
t ∈ (, ),
⎩limt→+ t –δ u(t) = a,
u() = b,
where < δ < , disturbance parameters a ≥ , b ≥ . Under certain conditions, the authors studied the impact of the disturbance parameters a and b on the existence of positive solutions. As an important part of fractional differential equations, the boundary value problems on infinite intervals have also been extensively researched, see [–]. In [], Liang and Zhang studied the following m-point boundary value problem of fractional differential equations on the infinite interval: ⎧ ⎨Dδ + u(t) + a(t)f (u(t)) = ,
⎩u() = u () = ,
t ∈ (, +∞), m–
Dδ– + u(+∞) =
i=
γi u(ξi ),
δ– < where < δ ≤ , < ξ < ξ < · · · < ξm– < +∞, γi ≥ , i = , , . . . , m – and m– i= γi ξi (δ). By using the Leggett-Williams fixed point theorem, the existence of three positive solutions for the boundary value problem on the infinite interval was obtained. In [], authors investigated the integral boundary value problem of fractional differential equations on infinite intervals with two disturbance parameters ⎧ ⎪ Dδ+ u(t) + f (t, u(t), Dδ– ⎪ + u(t)) = , ⎪ ⎪ ⎪ ⎨u() = , τ ⎪ ⎪ Dδ– u(∞) = g (s)u(s) ds + a, + ⎪ ⎪ ⎪ τ ⎩ δ– D+ u() = g (s)u(s) ds + b,
t ∈ (, +∞),
where < δ ≤ , f satisfies the L -Caratheódory conditions, g , g ∈ L ([, +∞)) are nonnegative, disturbance parameters a, b ∈ [, +∞). The purpose of this paper is to investigate the existence of positive solutions for the infinite-point boundary value problem of fractional differential equations on the half-line (.). Moreover, the impact of the disturbance parameters on the existence and nonexistence of positive solutions is established. Finally, some examples are presented to illustrate the main results.
2 Preliminaries For the convenience of the readers, we present here some basic definitions and lemmas, which are used throughout this paper. The function f : R+ × R+ → R is called an L -Carathéodory function if () for each u ∈ R+ , t → f (t, u) is measurable on t ∈ R+ ; () for a.e. t ∈ R+ , u → f (t, u) is continuous on u ∈ R+ ; () for each r > , there exists ϕr ∈ L (R+ ) with ϕr (t) ≥ on t ∈ R+ such that
f t, + t δ– u ≤ ϕr (t),
for all |u| ≤ r, and a.e. t ∈ R+ .
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Throughout this paper, we always assume that the following hypotheses hold: +∞ (H) q ∈ L (R+ ) is nonnegative and q(s)ϕr (s) ds < +∞ for any r > ; (H) f (t, u) is monotone increasing with respect to u ∈ R+ for each t ∈ R+ . Definition . (see []) Let p > . The Riemann-Liouville fractional integral of order p of a function u : R+ → R is given by p I+ u(t) =
(p)
t
(t – s)p– u(s) ds,
provided the integral exists. Definition . (see []) Let p > . The Riemann-Liouville fractional derivative of order p of a function u : R+ → R is given by p
n–p
D+ u(t) = Dn I+ u(t) =
n t u(s) d ds, (n – p) dt (t – s)p–n+
where n is the smallest integer greater than or equal to p, provided the right-hand side is pointwise defined on R+ . Denote
|u(t)| < +∞ , E = u ∈ C R+ : sup δ– t∈R+ + t endowed with the norm u = supt∈R+
|u(t)| , +t δ–
then E is a Banach space.
Definition . We say that u = u(t) is a solution of boundary value problem (.), if u ∈ E, Dδ+ u ∈ L (R+ ) and satisfies (.). Moreover, if u(t) ≥ , t ∈ R+ , we say that u is a positive solution of boundary value problem (.). Lemma . Suppose h ∈ L (R+ ) and < δ < , then the following boundary value problem ⎧ δ ⎪ ⎪D+ u(t) + h(t) = , ⎨ ⎪ ⎪ ⎩
t ∈ (, +∞),
u() = Dδ– + u(+∞) = , ∞ δ– Dδ– i= g(ξi )D+ u(ξi ) + λ + u() =
(.)
has a unique solution u(t) =
+∞
G(t, s)h(s) ds +
λt δ– , (δ – )
(.)
where G(t, s) = G (t, s) + G (t, s), ⎧ ⎨t δ– – (t – s)δ– , G (t, s) = (δ) ⎩t δ– ,
(.) ≤ s ≤ t < +∞, ≤ t < s < +∞,
(.)
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∞
G (t, s) =
g(ξi )χ(ξi , s)t δ– , (δ – ) i=
(.)
and the characteristic function χ is defined by
χ(ξi , s) =
⎧ ⎨,
s ≥ ξi , ⎩, s < ξi ,
i = , , . . . .
Proof Since Dδ+ u(t) + h(t) = , we have
u(t) = – (δ)
t
(t – s)δ– h(s) ds + c t δ– + c t δ– + c t δ– ,
(.)
where ci ∈ R, i = , , . Since u() = , then c = and Dδ– + u(t) = –
t
Dδ– + u(t) = –
h(s) ds + c (δ),
t
(t – s)h(s) ds + c (δ)t + c (δ – ).
By the boundary conditions, we can get ⎧ ⎨– +∞ h(s) ds + c (δ) = , ⎩c (δ – ) = ∞ g(ξi ) +∞ h(s) ds + λ, i= ξi and c = So
(δ)
+∞
u(t) = –
h(s) ds, c =
(δ)
(δ–)
+∞ λ i= g(ξi ) ξi h(s) ds + (δ–) .
∞
t
(t – s)δ– h(s) ds + ∞
t δ– g(ξi ) (δ – ) i=
(δ)
+∞
t δ– h(s) ds
+∞
λt δ– (δ – ) ξi +∞ ∞ g(ξi )t δ– +∞ λt δ– G (t, s)h(s) ds + i= χ(ξi , s)h(s) ds + = (δ – ) (δ – ) +∞ +∞ δ– λt G (t, s)h(s) ds + G (t, s)h(s) ds + = (δ – ) +∞ λt δ– = . G(t, s)h(s) ds + (δ – ) +
h(s) ds +
On the other hand, if u satisfies (.), we can easily show that u satisfies (.) and Dδ+ u ∈ L (R+ ). The proof is completed.
Lemma . Let G(t, s), G (t, s) defined by (.) and (.) satisfy the following properties: () G(t, s), G (t, s) ≥ for any (t, s) ∈ R+ × R+ ; + + () G (t, s) and G+t (t,s) δ– are continuous on (t, s) ∈ R × R ; ∞ G(t,s) L + () ≤ +t δ– ≤ (δ) for any t, s ∈ R , where the constant L = + (δ – ) i= g(ξi );
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() for a constant k > , we have min
G (t, s) G (t, s) sup ≥ + t δ– k ( + k δ– ) t∈R+ + t δ–
min
G(t, s) G(t, s) sup ≥ . + t δ– k ( + k δ– ) t∈R+ + t δ–
k ≤t≤k
and
k ≤t≤k
Proof By (.) and (.), the definitions of G(t, s) and G (t, s), the results () and () can be easily obtained. () According to the definition of G(t, s), we have ∞
G(t, s) G (t, s) G (t, s) L + . = + ≤ g(ξi ) = δ– δ– δ– +t +t +t (δ) (δ – ) i= (δ) () The first inequality can be found in []. t δ– k δ– We can easily show that min ≤t≤k { +t δ– } = +k δ– for k > . Thus k
∞
G (t, s) t δ– = g(ξ )χ(ξ , s) min i i + t δ– (δ – ) i= + t δ– k ≤t≤k k ≤t≤k min
∞
=
k δ– g(ξi )χ(ξi , s) (δ – ) i= + k δ–
>
g(ξi )χ(ξi , s) (δ – ) i= + k δ–
>
t δ– g(ξi )χ(ξi , s) sup δ– δ– + k (δ – ) i= t∈R+ + t
=
G (t, s) sup . + k δ– t∈R+ + t δ–
∞
∞
So min
k ≤t≤k
G(t, s) G (t, s) G (t, s) ≥ min + min δ– δ– +t +t + t δ– k ≤t≤k k ≤t≤k ≥ ≥
G (t, s) G (t, s) sup + sup k ( + k δ– ) t∈R+ + t δ– + k δ– t∈R+ + t δ– k ( + k δ– )
Let P = u ∈ E : u(t) ≥ , t ∈ R+ . Then P ⊂ E is a cone.
sup t∈R+
G(t, s) . + t δ–
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For u ∈ P, let (Tu)(t) =
+∞
λt δ– . G(t, s)q(s)f s, u(s) ds + (δ – )
Then T : P → E. Lemma . (see []) Let V = {u ∈ E : u ≤ l}, l > , V = {v = +tuδ– : u ∈ V }. If V is equicontinuous on any compact interval of R+ and equiconvergent at infinity, then V is relatively compact on E. Lemma . Assume (H) holds, then T : P → P is completely continuous. u(t) Proof For u ∈ P ⊂ E, since supt∈R+ +t δ– < +∞, there exists a constant l > such that +∞ Tu(t) L λ u ≤ l. Then +tδ– < (δ) q(s)ϕl (s) ds + (δ–) < +∞, and Tu(t) is continuous with re+ spect to t ∈ R . So Tu ∈ E and T : P → E is well defined. Since G, f , q are nonnegative, then Tu(t) ≥ , which implies Tu ∈ P for any u ∈ P. un (t) u(t) (i) Let {un } ⊂ P, u ∈ P such that un – u → as n → +∞, that is, +t δ– → +t δ– . Then there exists a constant r > such that un ≤ r, u ≤ r. Since f satisfies the L Carathéodory conditions for a.e. s ∈ R+ , then
f s, un (s) – f s, u(s) → ,
as n → +∞,
and
f s, un (s) – f s, u(s) = f s, + sδ– un (s) – f s, + sδ– u(s) ≤ ϕr (s). + sδ– + sδ– By the Lebesgue dominated convergence theorem,
+∞
q(s)f s, un (s) – f s, u(s) ds → ,
as n → +∞.
Therefore,
|Tun (t) – Tu(t)| +∞ G(t, s) = q(s) f s, un (s) – f s, u(s) ds + t δ– + t δ– +∞
G(t, s) q(s)f s, un (s) – f s, u(s) ds ≤ δ– +t +∞
L q(s)f s, un (s) – f s, u(s) ds → , ≤ (δ)
as n → +∞.
Hence, Tun – Tu → as n → +∞, which implies that T is a continuous operator. Let B ⊂ P be a nonempty bounded closed subset. There exists a constant lB > such that u ≤ lB for all u ∈ B, and there exists ϕlB ∈ L (R+ ) such that
u(s)
≤ ϕlB (s). f s, u(s) = f s, + sδ– + sδ–
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For any u ∈ B, Tu(t) + t δ– =
+∞
L ≤ (δ)
G(t, s) t δ– λ q(s)f s, u(s) ds + + t δ– (δ – ) + t δ– +∞
q(s)ϕlB (s) ds +
λ < +∞. (δ – )
So, Tu < +∞. Then T(B) is bounded and T is uniformly bounded. (ii) For any T > , let I = [, T ] be a compact interval. Because G+t (t,s) δ– is continuous for δ–
δ–
t t (t, s) ∈ I × I, +t δ– , +t δ– are continuous for t ∈ I, then they are uniformly continuous. So, for any ε > , there exists a constant < δ < ε such that
G (t , s ) G (t , s ) < ε , – + t δ– δ– (δ) + t δ– δ– t t tδ– tδ– – – < ε, + t δ– + t δ– < ε + t δ– + t δ– for all t , t , s , s ∈ I and |t – t | < δ , |s – s | < δ . From the definition of G (t, s), for s > t, δ– G (t , s) G (t , s) tδ– ≤ t – – + t δ– + t δ– + t δ– < (δ) ε. δ– (δ) + t Similarly, we can get δ– ∞ ∞ t G (t , s) G (t , s) tδ– ε ≤ – g(ξ ) – g(ξi ). < i + t δ– + tδ– (δ – ) i= + tδ– + tδ– (δ – ) i= Then, for each u ∈ B, Tu(t ) Tu(t ) – + t δ– + t δ– +∞ G (t , s) G (t , s)
q(s)f s, u(s) ds ≤ – + t δ– δ– + t +∞ G (t , s) G (t , s)
tδ– λ tδ– q(s)f s, u(s) ds + + – – + t δ– (δ – ) + tδ– + tδ– + tδ– T G (t , s) G (t , s)
q(s)f s, u(s) ds – ≤ + t δ– δ– + t +∞ G (t , s) G (t , s)
+ + t δ– – + t δ– q(s) f s, u(s) ds T +∞ ∞
λ ε ε g(ξi ) q(s)f s, u(s) ds + + (δ – ) i= (δ – ) ≤
ε (δ)
T
q(s)ϕlB (s) ds + ∞
+
ε g(ξi ) (δ – ) i=
ε (δ)
+∞ T
q(s)ϕlB (s) ds
+∞
q(s)ϕlB (s) ds +
λ ε (δ – )
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∞
≤ q(s)ϕlB (s) ds + g(ξi ) (δ – ) i=
+∞ λ L ε. q(s)ϕlB (s) ds + = (δ) (δ – ) (δ)
Therefore,
Tu(t) +t δ–
+∞
+∞
λ ε q(s)ϕlB (s) ds + (δ – )
is equicontinuous on I. δ–
t (iii) We prove that T : P → P is equiconvergent at t = +∞. Since limt→+∞ +t δ– = G (t,s) G (t,s) t δ– and limt→+∞ +tδ– = , then limt→+∞ +tδ– = and limt→+∞ +tδ– = . Therefore G(t,s) limt→+∞ +t δ– = . For any u ∈ B, we have
+∞
q(s)f s, u(s) ds ≤
+∞
q(s)ϕlB (s) ds < +∞
and Tu(t) = lim lim δ– t→+∞ + t t→+∞
+∞
t δ– λ G(t, s) = < +∞. q(s)f s, u(s) ds + + t δ– (δ – ) + t δ–
Hence, T(B) is equiconvergent at infinity. Consequently, in view of Lemma ., T(B) is relatively compact, thus T is a compact operator. So T is completely continuous and the proof is finished. Lemma . If boundary value problem (.) has a positive solution u, then for t ∈ R+ , min
k ≤t≤k
u(t) u . ≥ δ– +t k ( + k δ– )
Proof By Lemma ., we have u(t) =
+∞
G(t, s)q(s)f s, u(s) ds +
λ t δ– . (δ – )
Then, according to () of Lemma ., for t ∈ R+ , u(t) = min ≤t≤k + t δ– k k ≤t≤k
+∞
min
λ t δ– G(t, s) q(s)f s, u(s) ds + + t δ– (δ – ) + t δ–
λ G(t, s) t δ– min q(s)f s, u(s) ds + + t δ– (δ – ) k ≤t≤k + t δ– k ≤t≤k +∞
G(t, s) k δ– λ ≥ sup q(s)f s, u(s) ds + k ( + k δ– ) t∈R+ + t δ– (δ – ) + k δ– +∞
λ G(t, s) ≥ q(s)f s, u(s) ds + sup δ– δ– k ( + k ) t∈R+ +t (δ – ) + k δ– +∞
λ t δ– G(t, s) ≥ q(s)f s, u(s) ds + sup k ( + k δ– ) t∈R+ + t δ– (δ – ) + t δ–
≥ min
= The proof is finished.
+∞
u . k ( + k δ– )
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By Lemma ., we can easily show that the following lemma holds. Lemma . Assume u ∈ E, Dδ+ u ∈ L (R+ ). Then boundary value problem (.) has a positive solution if and only if the operator T has a fixed point in P. Lemma . (see []) Let P be a cone of the Banach space E, ⊂ E be a bounded open set and θ ∈ . Suppose T : P ∩ → P is a completely continuous operator. If u = μTu for any u ∈ P ∩ ∂ and μ ∈ [, ], then i(T, P ∩ , P) = .
3 Comparison principle Definition . Let α ∈ E, Dδ+ α ∈ L (R+ ), we say that α = α(t) is a lower solution of boundary value problem (.) if α satisfies ⎧ δ ⎪ ⎪–D+ α(t) ≤ q(t)f (t, α(t)), ⎨ ⎪ ⎪ ⎩
t ∈ (, +∞),
Dδ– + α(+∞) = , ∞ δ– Dδ– i= g(ξi )D+ u(ξi ) + λ. + α() ≤ α() = ,
Let β ∈ E, Dδ+ β ∈ L (R+ ), we say β = β(t) is an upper solution of boundary value problem (.) if β satisfies ⎧ δ ⎪ ⎪ ⎨–D+ β(t) ≥ q(t)f (t, β(t)), t ∈ (, +∞), β() = , Dδ– + β(+∞) = , ⎪ ⎪ ⎩ δ– ∞ D+ β() ≥ i= g(ξi )Dδ– + u(ξi ) + λ. Lemma . If u ∈ E, Dδ+ u ∈ L (R+ ) and satisfies ⎧ ⎨Dδ + u(t) ≤ ,
t ∈ (, +∞),
⎩u() = Dδ– + u(+∞) = ,
Dδ– + u() ≥
∞
δ– i= g(ξi )D+ u(ξi ),
then u(t) ≥ for t ∈ R+ . ∞ δ– Proof Let –Dδ+ u(t) = y(t) ≥ for a.e. t ∈ R+ , and Dδ– i= g(ξi )D+ u(ξi ) + λ, then + u() = λ ≥ . According to Lemma ., we know that the boundary value problem ⎧ δ ⎪ ⎪ ⎨–D+ u(t) = y(t), ⎪ ⎪ ⎩
t ∈ (, +∞),
u() = Dδ– + u(+∞) = , δ– δ– D+ u() = ∞ i= g(ξi )D+ u(ξi ) + λ
has a unique solution u(t) =
+∞
G(t, s)y(s) ds +
λ t δ– . (δ – )
From Lemma ., we can obtain that u(t) ≥ for t ∈ R+ .
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Theorem . Suppose (H) and (H) hold if boundary value problem (.) has a nonnegative lower solution α and an upper solution β satisfies α(t) ≤ β(t) for t ∈ R+ . Then boundary value problem (.) has at least one positive solution u that satisfies α(t) ≤ u(t) ≤ β(t) for t ∈ R+ . Proof Let ⎧ ⎪ ⎪ ⎨f (t, β(t)), u > β(t), F(t, u) = f (t, u), α(t) ≤ u ≤ β(t), ⎪ ⎪ ⎩ f (t, α(t)), u < α(t). Since f is an L -Carathéodory function, then F is an L -Carathéodory function, too. By Lemma ., the boundary value problem ⎧ δ ⎪ ⎪ ⎨D+ u(t) + q(t)F(t, u(t)) = ,
t ∈ (, +∞),
u() = Dδ– + u(+∞) = , ⎪ ⎪ ⎩ δ– δ– D+ u() = ∞ i= g(ξi )D+ u(ξi ) + λ
(.)
is equivalent to the integral equation
+∞
u(t) =
G(t, s)q(s)F s, u(s) ds +
λ t δ– . (δ – )
Define the operator Q : P → P by (Qu)(t) =
+∞
G(t, s)q(s)F s, u(s) ds +
λ t δ– . (δ – )
For any u ∈ P and t ∈ R+ , by (H), we can get
β(t) ≤ F t, u(t) ≤ f t, β(t) = f t, + t δ– ≤ ϕ β (t). + t δ– +∞ L Let = {u ∈ P : u ≤ R}, where the constant R = (δ) q(s)ϕ β (s) ds + Obviously, is a closed and convex set. Then, for any u ∈ ,
λ . (δ–)
|Qu(t)| +∞ G(t, s) t δ– λ = q(s)F s, u(s) ds + + t δ– + t δ– (δ – ) + t δ– +∞ L λ ≤ = R. q(s)ϕ β (s) ds + (δ) (δ – ) That is, Qu ≤ R, which implies Q : → . We can easily show that Q is completely continuous since its proof is similar to Lemma .. According to the Schauder fixed point theorem, we know that Q has at least one fixed point u. By Lemma ., boundary value problem (.) has a positive solution u. Next, we prove α(t) ≤ u(t) ≤ β(t) for t ∈ R+ , and u = u(t) is a solution of boundary value problem (.).
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Let v(t) = u(t) – α(t). According to (H), we get
Dδ+ v(t) = Dδ+ u(t) – Dδ+ α(t) = –q(t)F t, u(t) – Dδ+ α(t)
≤ –q(t)F t, u(t) + q(t)f t, α(t) ≤ , v() = u() – α() = , δ– δ– Dδ– + v(+∞) = D+ u(+∞) – D+ α(+∞) = ,
and δ– δ– Dδ– + v() = D+ u() – D+ α() ∞
≥
=
g(ξi )Dδ– + u(ξi ) + λ –
∞
g(ξi )Dδ– + α(ξi ) – λ
i=
i=
∞
∞
u(ξ g(ξi )Dδ– ) – α(ξ ) = g(ξi )Dδ– + i i + v(ξi ).
i=
i=
From Lemma ., we have v(t) ≥ for t ∈ R+ , which implies that u(t) ≥ α(t) for t ∈ R+ . Similarly, we can show that u(t) ≤ β(t) for t ∈ R+ . Therefore, each solution u of boundary value problem (.) satisfies α(t) ≤ u(t) ≤ β(t) for t ∈ R+ . That is, F(t, u(t)) = f (t, u(t)), and u is a positive solution of boundary value problem (.).
4 The properties of positive solutions Theorem . Assume (H) and (H) hold. () If there exists a constant λ = λ ≥ such that boundary value problem (.) has a positive solution u = u(t), then for each ≤ λ ≤ λ, boundary value problem (.) has λt δ– a positive solution u and (δ–) ≤ u(t) ≤ u(t) for t ∈ R+ . () If there exists a constant λ = λ ≥ such that boundary value problem (.) does not have positive solutions, then for each λ ≥ λ, boundary value problem (.) does not have positive solutions. Proof () Since u = u(t) is a positive solution of boundary value problem (.) with λ = λ, then by Lemma ., u(t) =
+∞
λt δ– . G(t, s)q(s)f s, u(s) ds + (δ – )
Therefore, we can obtain that for any ≤ λ ≤ λ, u(t) ≥ λt δ–
λt δ– (δ–)
≥
λt δ– , (δ–)
t ∈ R+ .
We take α = (δ–) and β = u, obviously, α ≤ β. We can easily show that α is a lower solution and β is an upper solution of boundary value problem (.), respectively. Then, according to Theorem ., we can obtain that for any ≤ λ ≤ λ, boundary value λt δ– ≤ u(t) ≤ u(t) for t ∈ R+ . problem (.) has a positive solution u and (δ–) () Assume that there exists a constant λ ≥ λ such that boundary value problem (.) has a positive solution. In view of () in this theorem, for λ = λ ≤ λ , boundary value problem (.) has a positive solution, which is a contradiction.
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Denote f = lim sup sup u→+ t∈R+
ρ =
f (t, ( + t δ– )u) , u
(δ) +∞ , δ – + L q(s) ds
ρ =
f (t, ( + t δ– )u) , u→+∞ t∈[ ,k] u k
f∞ = lim inf inf
k ( + k δ– ) (δ) . k q(s) ds k
Theorem . Suppose (H) holds, if f < ρ , then there exists a constant λ∗ ≥ such that boundary value problem (.) with λ = λ∗ has at least one positive solution. Proof Because f < ρ , there exists a constant r > such that f (t, ( + t δ– )u) < ρ u ≤ ρ r for any t ∈ R+ and u ∈ (, r ]. Set B = {u ∈ P : u ≤ r } and ≤ λ∗ ≤ ρ r . Then, for any u ∈ B,
u(s)
u(s) ≤ ρ r f s, u(s) = f s, + sδ– ≤ ρ δ– +s + sδ– and
t δ– λ∗ G(t, s) q(s)f s, u(s) ds + + t δ– (δ – ) + t δ– +∞ Lρ r λ∗ ≤ q(s) ds + (δ) (δ – ) Lρ r +∞ ρ r ≤ q(s) ds + (δ) (δ – ) +∞
ρ r = L q(s) ds + δ – (δ) = r .
Tu(t) = + t δ–
+∞
So T(B) ⊂ B. According to the Schauder fixed point theorem, we can obtain that T has at least one fixed point on B. By Lemma ., boundary value problem (.) has at least one positive solution. Theorem . Suppose (H) holds, if f∞ > ρ , then there exist large enough positive constants λˆ such that boundary value problem (.) with λ = λˆ has no positive solution. Proof Assume that there exists a constant λˆ > and λˆ is large enough, boundary value ˆ problem (.) with λ = λˆ has a positive solution uˆ = u(t). By f∞ > ρ , we know that
f t, + t δ– u > ρ u, for t ∈ [ k , k] and u ≥ r , where constant r > is large enough. Take λˆ > (δ – )( + k δ– )r . By Lemma ., ˆ = u(t)
+∞
ˆ G(t, s)q(s)f s, u(s) ds +
λˆ t δ– , (δ – )
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then ˆ k ) u( + ( k )δ–
≥
( k )δ– k λˆ λˆ λˆ = > > r , (δ – ) + ( k )δ– (δ – ) + k δ– (δ – ) + k δ–
ˆ > r . Then, for s ∈ [ k , k], and u
u(s)
ˆ ˆ u(s) δ– ˆ . f s, u(s) = f s, + s > ρ δ– +s + sδ– According to Lemma ., we have min ≤t≤k k
ˆ k ) u( + ( k )δ–
+ ( k )
≥
ˆ q(s)f s, u(s) ds + δ–
G( k , s)
+∞
=
G ( k , s)
k
+ ( k )δ–
k
ˆ u(t) +t δ–
k
ˆ q(s)f s, u(s) ds +
( k )δ–
≥
ρ (δ)
>
ˆ ρ u k ( + k δ– ) (δ)
k
+ ( k )δ–
≥
ˆ u . k (+k δ– )
Then
( k )δ– λˆ (δ – ) + ( k )δ–
k λˆ (δ – ) + k δ–
ˆ u(s) λˆ q(s) ds + δ– +s (δ – ) + k δ– k q(s) ds + r k
ˆ + r . = u ˆ > u ˆ + r , which is a contradiction. Thus, there exist large enough constants That is, u ˆλ > such that boundary value problem (.) with λ = λˆ has no positive solution. Theorem . Let I ⊂ [, +∞) be a bounded set. Suppose (H) holds, f∞ > ρ , then for each λ ∈ I, there exists a constant τ > such that u ≤ τ , where u = u(t) is a solution of boundary value problem (.). Proof Because I is a bounded set, then for each λ ∈ I, there exists a constant σ > such that ≤ λ ≤ σ . Since f∞ > ρ , there exists a constant r > such that f (t, ( + t δ– )u) > ρ u for any t ∈ [ k , k] and u ≥ r. Let τ = k ( + k δ– )r. Assume that boundary value problem (.) has a solution u = u(t) that satisfies u > τ . Then min
k ≤t≤k
u(t) u > τ = r. ≥ δ– δ– +t k ( + k ) k ( + k δ– )
By Lemmas . and ., we have u( k ) + ( k )δ–
+∞
=
+ ( k )
≥
k k
q(s)f s, u(s) ds + δ–
G( k , s)
G ( k , s) + ( k )δ–
q(s)f s, u(s) ds
( k )δ– λ (δ – ) + ( k )δ–
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k
( k )δ–
≥
ρ (δ)
>
ρ u k ( + k δ– ) (δ)
k
+ ( k )δ–
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u(s) q(s) ds + sδ– k q(s) ds = u , k
which is a contradiction. So, for all solutions of boundary value problem (.), u = u(t) satisfies that u ≤ τ .
5 Existence, nonexistence and multiplicity of positive solutions Theorem . Suppose that (H), (H) hold, f < ρ and f∞ > ρ , then there exists a constant λ∗ ∈ (, +∞) such that the following results hold: () Boundary value problem (.) has at least one positive solution for λ = and λ = λ∗ ; () Boundary value problem (.) has at least two positive solutions for each λ ∈ (, λ∗ ); () Boundary value problem (.) does not have any positive solutions for each λ ∈ (λ∗ , +∞). Proof Let = λ ∈ [, +∞) : the λ such that boundary value problem (.) has at least one positive solution . Then by Theorem . we know that = ∅. In view of Theorem ., [, λ˜ ] ⊂ if and only if λ˜ ∈ . According to f∞ > ρ and Theorem ., we can show that is a bounded set. Let M = ˜ then M is a bounded set. Therefore, M has a supremum which is denoted by [, λ], ˜ λ∈ ∗ λ = sup M > . Next, we will prove that boundary value problem (.) has at least one positive solution for λ = λ∗ . Since λ∗ = sup M, there exists a sequence {λm } ⊂ M that satisfies λm < λ∗ such that λm → λ∗ as m → +∞. Let um (t) be the solution of boundary value problem (.) with λ = λm . In view of Lemma ., we know that boundary value problem (.) with λ = λm is equivalent to
+∞
um (t) =
G(t, s)q(s)f s, um (s) ds +
λm t δ– , (δ – )
m = , , . . . .
According to Theorem ., there exists a constant τ such that um ≤ τ , which implies that {um (t)} is uniformly bounded. By Lemma ., we can easily show that {um (t)} is equicontinuous. Then we know that {um (t)} has a convergent subsequence, we assume that {um (t)} itself converges uniformly to u on R+ , and u ∈ P. Since f is an L -Carathéodory function, then by the Lebesgue dominated convergence theorem, as m → +∞, we have u(t) =
+∞
G(t, s)q(s)f s, u(s) ds +
λ∗ t δ– . (δ – )
Hence, boundary value problem (.) has a positive solution u for λ = λ∗ . By Theorem ., boundary value problem (.) has at least one positive solution for λ ∈ [, λ∗ ]. And by
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the definition of λ∗ we know that boundary value problem (.) does not have positive solutions for each λ ∈ (λ∗ , +∞). Finally, we prove that boundary value problem (.) has at least two positive solutions for λ ∈ (, λ∗ ). For each λ ∈ (, λ∗ ), there exist λ, λ ∈ M such that < λ < λ < λ. Let u, u be the solutions of boundary value problem (.) for λ = λ, λ = λ, respectively. Then, according to Theorem ., boundary value problem (.) has a positive solution u = u (t) for λ = λ , and u(t) ≤ u (t) ≤ u(t). Let α = u and β = u. We can easily verify that α is a lower solution and β is an upper solution of boundary value problem (.), and α(t) < β(t). Choose λˆ > λ∗ satisfies λ < λ∗ < λˆ . ˆ × P → E by We define K : [λ, λ]
+∞
K(r, u) =
G(t, s)q(s)f s, u(s) ds +
r t δ– . (δ – )
Let ⎧ ⎪ ⎪ ⎨f (t, β(t)), u > β(t), F(t, u) = f (t, u), ⎪ ⎪ ⎩ f (t, α(t)),
α(t) ≤ u ≤ β(t), u < α(t).
: [λ, λ] ˆ × P → E by Define an integral operator K u) = K(r,
+∞
G(t, s)q(s)F s, u(s) ds +
r t δ– . (δ – )
are completely continuous for each r ∈ [λ, λ] ˆ according We can easily prove that K and K to Lemma .. In view of Lemma ., u is a positive solution of boundary value problem (.) if and only if u = K(λ, u). By Theorem ., there exists a constant τ such that the fixed point u of K satisfies u ≤ τ for each r ∈ [λ, λˆ ]. Let = u ∈ P : u < τ , α(t) < u(t) < β(t), t ∈ R+ . Obviously, ⊂ P is a nonempty open-bounded subset, then u ∈ . Since F is an L -Carathéodory function, then F(t, u(t)) ≤ ϕ β (t). For any (r, u) ∈ [λ, λˆ ] × K(r,u) P, there exists a constant R > τ > such that +t δ– < R. Let B(θ , R) = {u ∈ E : u < R}. Then ⊂ P ∩ B(θ , R) and u = μK u for u ∈ P ∩ ∂B(θ , R) and any μ ∈ [, ]. Otherwise, if there u , then R = u = μ K u < μR < R, which is exists u ∈ P ∩ ∂B(θ , R) such that u = μK ˆ a contradiction. Hence, according to Lemma ., for each r ∈ [λ, λ],
u), P ∩ B(θ , R), P = . i K(r, ˆ does not have a fixed point on P ∩ (B(θ , R)\), thus for any r ∈ [λ, λ], Since K
(r, u), P ∩ B(θ , R)\ , P = . i K
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= K , then by the excision property of the fixed point index, we can obtain that Since K| for each r ∈ [λ, λˆ ],
(r, u), P ∩ , P i K(r, u), P ∩ , P = i K
(r, u), P ∩ B(θ , R), P – i K (r, u), P ∩ B(θ , R)\ , P =i K = .
(.)
Since λˆ > λ∗ , we know that boundary value problem (.) does not have positive solutions for each λ ∈ (λ∗ , +∞), then K(λˆ , u) = u for any u ∈ P. Hence,
ˆ u), P ∩ B(θ , R), P = . i K(λ,
(.)
Define H : [, ] × P ∩ B(θ , R) → E by
H(μ, u) = K ( – μ)λ + μλˆ , u . Obviously, H is completely continuous. We have H(μ, u) = u for (μ, u) ∈ [, ] × P ∩ ∂B(θ , R). Otherwise, if there exists (μ , u ) ∈ [, ] × P ∩ ∂B(θ , R) such that H(μ , u ) = u , then
K ( – μ )λ + μ λˆ , u = u ,
u ∈ P, u = R.
Therefore, u = u (t) is a solution of boundary value problem (.) with λ = ( – μ )λ + μ λˆ . Then u ≤ τ , which is a contradiction. By (.) and the homotopy invariance of the fixed point index, we have
i K(λ, u), P ∩ B(θ , R), P = i H(, u), P ∩ B(θ , R), P
= i H(, u), P ∩ B(θ , R), P
= i K(λˆ , u), P ∩ B(θ , R), P = .
(.)
According to (.), (.) and by using the additivity property of the fixed point index, we have
i K(λ, u), P ∩ B(θ , R)\, P = –. Therefore, boundary value problem (.) has a solution u ∈ P ∩ B(θ , R)\. Because u ∈ , we have u = u . Hence, boundary value problem (.) has at least two positive solutions for λ ∈ (, λ∗ ). The proof is completed. Definition . (see [, ]) Suppose that E is a Banach space, P ⊂ E is a cone. We say that γ is a nonnegative, continuous, concave functional on P if γ : P → [, +∞) is continuous, and
γ μx + ( – μ)y ≥ μγ (x) + ( – μ)γ (y) for all x, y ∈ P and μ ∈ [, ].
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Set Pc = u ∈ P : u < c ,
P(γ , a, b) = u ∈ P : a ≤ γ (u), u ≤ b .
Lemma . (see []) Suppose that there exist constants < a < b < d ≤ c, T : Pc → Pc is a completely continuous operator, γ : P → [, +∞) is a continuous concave functional, for u ∈ Pc , we have γ (u) ≤ u , and the following conditions hold: (C) {u ∈ P(γ , b, d) | γ (u) > b} = ∅, and for u ∈ P(γ , b, d), γ (Tu) > b; (C) for u ∈ Pa , we have Tu < a; (C) for u ∈ P(γ , b, c) and Tu > d, we have γ (Tu) > b. Then T has at least three fixed points u , u and u satisfying u < a < u ,
γ (u ) < b < γ (u ).
Theorem . Suppose (H) holds. Let < a < b < d ≤ c, λ < aρ , bρ < cρ and suppose that f satisfies the following conditions: (H) f (t, ( + t δ– )u) < aρ , (t, u) ∈ R+ × [, a]; (H) f (t, ( + t δ– )u) > bρ , (t, u) ∈ [ k , k] × [b, c]; (H) f (t, ( + t δ– )u) < cρ , (t, u) ∈ R+ × [, c]. Then boundary value problem (.) has at least three positive solutions u , u and u such that u < a,
b < γ (u ),
a < u
and γ (u ) < b.
Proof Define a nonnegative, continuous, concave functional on E by γ (u) = min
k ≤t≤k
u(t) . + t δ–
Next, we prove that the conditions of Lemma . hold. u(t) For u ∈ Pc , we have u ≤ c, then for t ∈ R+ , ≤ +t δ– ≤ c. According to assumption (H), we get
u(s)
δ– < cρ . f s, u(s) = f s, + s + sδ– Hence,
t δ– λ G(t, s) q(s)f s, u(s) ds + + t δ– (δ – ) + t δ– +∞ L λ ≤ q(s) ds + cρ (δ) (δ – )
+∞ ρ ≤ q(s) ds + δ – c = c. L (δ)
Tu(t) = + t δ–
+∞
Therefore, T : Pc → Pc . According to Lemma ., we have T is completely continuous. Similarly, it follows from assumption (H) that if u ∈ Pa , we have Tu ≤ a. Condition (C) of Lemma . holds.
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Let u∗ (t) = b+c ( + t δ– ), t ∈ R+ . We can show that u∗ ∈ P and u∗ = b+c < c. According ∗ > b. Hence, u ∈ {u ∈ P(γ , b, d) | γ (u) > b} = ∅. to the definition of γ (u), we get γ (u∗ ) = b+c u(t) On the other hand, if u ∈ P(γ , b, d), then b ≤ min ≤t≤k +tδ– and u ≤ d ≤ c. Hence, for t ∈ [ k , k], we have b ≤
u(t) +t δ–
k
≤ c. Then by assumption (H), we get
u(s)
f s, u(s) = f s, + sδ– > bρ . + sδ– Therefore, γ (Tu) = min
k ≤t≤k
Tu(t) + t δ– +∞
= min
k ≤t≤k
≥ ≥ ≥ >
=
t δ– λ G(t, s) q(s)f s, u(s) ds + + t δ– (δ – ) + t δ–
G(t, s) q(s)f s, u(s) ds δ– +t +∞
G(t, s) sup q(s)f s, u(s) ds δ– δ– k ( + k ) t∈R+ + t k
G ( k , s) q(s)f s, u(s) ds k ( + k δ– ) k + ( k )δ– k bρ q(s) ds δ– (δ)k ( + k ) k + k δ– k bρ q(s) ds = b. (δ)k ( + k δ– ) k +∞
min
k ≤t≤k
Hence, for all u ∈ P(γ , b, d), we have γ (Tu) > b, which implies that condition (C) of Lemma . holds. u(t) Finally, for u ∈ P(γ , b, c) and Tu > d, we have b ≤ min ≤t≤k +t δ– and u ≤ c. Hence, k
u(t) b ≤ +t δ– ≤ c for t ∈ [ k , k]. According to assumption (H), we can obtain γ (Tu) > b. Condition (C) of Lemma . holds. By Lemma ., T has at least three fixed points u , u , u such that u < a, b < γ (u ), a < u and γ (u ) < b. These fixed points are positive solutions of (.).
6 Examples To illustrate our main results, we present the following examples. Example . We consider the infinite-point boundary value problem of nonlinear fractional differential equations on the infinite interval ⎧ ⎪ ⎪ D+ u(t) + e–t–t u = , t ∈ (, +∞), ⎪ ⎨ u() = D+ u(+∞) = , ⎪ ⎪ ⎪ ⎩D u() = ∞ D u(i) + λ, +
i= i
+
(.)
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where δ = , q(t) = e–t , f (t, u) = e–t u , it is easy to show that (H) holds. Let g(t) = t , π –t ∈ L (R+ ). We have ξi = i, then ∞ i= g(ξi ) = is convergent. Let ϕr (t) = r ( + t ) e + f (t, ( + t )u) ≤ ϕr (t) for u ≤ r and t ∈ R . Then f is an L -Carathéodory function and +∞ π q(s)ϕr (s) ds ≈ .r < +∞. Let k = , we have L = + ∞ i= g(ξi ) = + , ρ = ( ) L+
≈ ., ρ =
√ ( )(+ ) –s e ds
≈ ..
f (t, ( + t )u) e–t ( + t ) u f = lim sup sup = lim sup sup = < ρ , u u u→+ t∈R+ u→+ t∈R+
and
f (t, ( + t )u) e–t ( + t ) u = lim inf inf = +∞ > ρ . f∞ = lim inf inf u→+∞ t∈[ ,] u→+∞ t∈[ ,] u u
() Choose r = ., let λ < ρ r ≈ .. We can obtain f (t, ( + t )u) < ρ r for u ∈ (, .], t ∈ R+ . By Theorems . and ., boundary value problem (.) has a positive solution for λ ∈ [, .]. () Choose r = ,, let λ > ( )( + )r ≈ ,.. We have f (t, ( + t )u) > ρ r for u ∈ [,, +∞) and t ∈ [ , ]. By Theorems . and ., boundary value problem (.) does not have a positive solution for λ ∈ (,, +∞). () According to Theorem ., we know that there exists a constant λ∗ ∈ (., ,) such that boundary value problem (.) has at least one positive solution for each λ ∈ [, λ∗ ], two positive solutions for any λ ∈ (, λ∗ ), and no positive solutions for λ ∈ (λ∗ , +∞). Example . We consider the boundary value problem ⎧ ⎪ ⎪ D u(t) + e–t f (t, u(t)) = , t ∈ (, +∞), ⎪ ⎨ + u() = D+ u(+∞) = , ⎪ ⎪ ⎪ ⎩D u() = ∞ D u(i) + ., +
(.)
+
i= (i–)!
where ⎧ t u – ⎪ ≤ u ≤ , , ⎪e ⎪ ) (+t ⎪ ⎨ t u ≤ u ≤ , f (t, u) = e– ( + (u – )), ) ⎪ (+t ⎪ ⎪ ⎪e– t ( u + ), ⎩ u ≥ . (+t )
In this case, δ = , q(t) = e–t , g(t) = r(+t )
– t
,ξ (t–)! i
= i. Then
∞
i= g(ξi ) = e
is convergent. Choose
ϕr (t) = ( + )e ∈ L (R ). We have f (t, ( + t )u) ≤ ϕr (t) for u ≤ r and t ∈ R+ . It is +∞ easy to show that f is an L -Carathéodory function and q(s)ϕr (s) ds ≈ , + .r < +∞. Take a = , b = , c = d = and k = . We can get that L = + ∞ i= g(ξi ) = + .e, ρ =
( ) L+δ–
≈ ., ρ =
+
√ ( )(+ ) –s e ds
≈ ., λ = . < aρ ≈ ..
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Then the function f satisfies
< aρ ≈ . for t ∈ R+ , ≤ u ≤ ; f t, + t u <
f t, + t u > ,. > bρ ≈ ,. for ≤ t ≤ , ≤ u ≤ ;
f t, + t u < × < cρ ≈ . × for t ∈ R+ , ≤ u ≤ . Then, by Theorem ., we know that boundary value problem (.) has at least three positive solutions u , u and u such that u < , < γ (u ), < u and γ (u ) < . Competing interests The authors declare that they have no competing interests. Authors’ contributions The authors contributed equally to this paper. All authors read and approved the final manuscript. Acknowledgements The authors sincerely thank the anonymous reviewers and editor for their valuable suggestions, which improved the presentation of this paper. This work is supported by the National Natural Science Foundation of China (Grant No. 11171220) and the Hujiang Foundation of China (B14005).
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