Jia et al. Boundary Value Problems (2016) 2016:104 DOI 10.1186/s13661-016-0614-7
RESEARCH
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Existence of positive solutions for fractional differential equation with integral boundary conditions on the half-line Mei Jia* , Haibin Zhang and Qiang Chen *
Correspondence:
[email protected] College of Science, University of Shanghai for Science and Technology, Shanghai, 200093, China
Abstract This paper considers the existence of positive solutions for fractional-order nonlinear differential equation with integral boundary conditions on the half-infinite interval. By using the fixed point theorem in a cone, sufficient conditions for the existence of at least one or at least two positive solutions of a boundary value problem are established. These theorems also reveal the properties of solutions on the half-line. MSC: 34B18; 34B10; 26A33 Keywords: Caputo derivative; integral boundary conditions; half-line; positive solutions; fixed point theorems
1 Introduction Boundary value problems are often studies in the areas of applied mathematics and physics. With the development of technology, applications of boundary value problems on the infinite interval attract increasing attention; see [–] and the references therein. Recently, fractional differential equations have also aroused great interest; see [–]. At the same time, the existence of positive solutions for nonlinear fractional differential equation boundary value problems have been widely studied by many authors; see [–] and the references therein. In [], the authors, using fixed point theorems in a cone, established the existence of one positive solution and three positive solutions for the following second-order nonlinear boundary value problems with integral boundary conditions on an infinite interval: ⎧ ⎪ ⎪ ⎨ p(t) (p(t)u (t)) + f (t, u(t)) = ,
t ∈ (, +∞), +∞ a u() – b limt→+ p(t)u (t) = g (u(s))ψ(s) ds, ⎪ ⎪ +∞ ⎩ a limt→+∞ u(t) + b limt→+∞ p(t)u (t) = g (u(s))ψ(s) ds,
where f ∈ C((, +∞) × [, +∞) × R, [, +∞)), f may be singular at t = , g , g : [, +∞) → +∞ [, +∞) and ψ : [, +∞) → (, +∞) are continuous, ψ(s) ds < +∞, p ∈ C[, +∞) ∩ +∞ ds < +∞, a + a > , and bi > for i = , . C (, +∞) with p(t) > on (, +∞) and p(s) Iterative schemes for approximating the solutions of a nonlinear fractional boundary value problem on the half-line were presented in []. The authors, based on the monotone © 2016 Jia et al. 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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iterative technique, obtained the existence of positive solutions of the following fractional boundary value problem: ⎧ ⎨Dα u(t) + q(t)f (t, u(t)) = , t ∈ (, +∞), + ⎩u() = , limt→+∞ Dα– u(t) = λ > , +
where < α < , and Dα+ is the standard Riemann-Liouville fractional derivative. For an overview of the literature on differential equations boundary value problems, see [–] and the references therein. Motivated by all the works mentioned, we study the following fractional boundary value problem on the half-line: ⎧ C α ⎪ ⎪ ⎨ p(t) (p(t) D+ u(t)) + f (t, u(t)) = ,
t ∈ (, +∞),
+∞ a u() – b limt→+ p(t) Dα+ u(t) = g (u(s))ψ (s) ds, ⎪ ⎪ +∞ ⎩ a limt→+∞ u(t) + b limt→+∞ p(t) C Dα+ u(t) = g (u(s))ψ (s) ds, C
(.)
where C Dα is the Caputo fractional derivative of order α ∈ (, ), p ∈ C ([, +∞), (, +∞)), f : (, +∞) × [, +∞) → [, +∞) is a continuous function and may be singular at t = ; ai > , bi > , gi ∈ C([, +∞), [, +∞)), and ψi ∈ L ([, +∞)) is nonnegative for i = , . We assume that the following conditions are satisfied: t α– (H) limt→+∞ (t–s) ds < +∞, ab > M, where p(s) sup M= (α) t∈[,+∞)
t
(t – s)α– ds. p(s)
(.)
(H) There exist functions h ∈ C([, +∞), [, +∞)) and v ∈ C((, +∞), (, +∞)) such that f (t, u) ≤ v(t)h(u),
t ∈ (, +∞);
+∞
p(s)v(s) ds < +∞.
2 Preliminaries In this section, we present some useful definitions and the related theorems. Definition . (See [, ]) Let α > . For a function u : (, +∞) → R, the RiemannLiouville fractional integral operator of order α of u is defined by α I+ u(t) =
(α)
t
(t – s)α– u(s) ds,
provided that the integral exists. Definition . (See [, ]) The Caputo derivative of order α for a function u : (, +∞) → R is given by C
Dα+ u(t) =
(n – α)
t
u(n) (s) ds, (t – s)δ+–n
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provided that the right side is pointwise defined on (, +∞), where n = [α] + and n – < α < n. If α = n, then C Dα+ u(t) = u(n) (t). Lemma . (See []) Let α > . Then the differential equation C
Dα+ h(t) =
has solutions h(t) = c + c t + c t + · · · + cn– t n– ,
ci ∈ R, i = , , , . . . , n – ,
where n is the smallest integer greater than or equal to α. Lemma . If (H) holds and y ∈ C((, +∞), [, +∞)) with fractional boundary value problem
+∞
p(s)y(s) ds < +∞, then the
⎧ C α ⎪ ⎪ ⎨ p(t) (p(t) D+ u(t)) + y(t) = ,
t ∈ (, +∞), +∞ a u() – b limt→+ p(t) C Dα+ u(t) = g (u(s))ψ (s) ds, ⎪ ⎪ +∞ ⎩ a limt→+∞ u(t) + b limt→+∞ p(t) C Dα+ u(t) = g (u(s))ψ (s) ds
(.)
has a unique solution
+∞
+∞
G(t, s)p(s)y(s) ds + F (t)
u(t) =
g u(s) ψ (s) ds
+∞
+ F (t)
g u(s) ψ (s) ds,
where ⎧ t (t–r)α– τ –r)α– ⎪ dr)(b + a limτ →+∞ s (τ(α)p(r) dr), (b + a (α)p(r) ⎪ ⎨ s (t–r)α– τ (τ –r)α– G(t, s) = (b + a (α)p(r) dr)(b + a limτ →+∞ s (α)p(r) dr) ρ⎪ ⎪ t (t–r)α– ⎩ – b a s (α)p(r) dr, τ (τ – r)α– ρ = a b + a b + a a lim dr, τ →+∞ (α)p(r)
τ t (τ – r)α– (t – r)α– F (t) = b + a lim dr – a dr , τ →+∞ (α)p(r) ρ (α)p(r)
≤ t ≤ s < +∞, ≤ s < t < +∞,
and F (t) =
t (t – r)α– b + a dr . ρ (α)p(r)
Proof It is well known that the fractional differential equation in (.) is equivalent to the integral equation p(t) C Dα+ u(t) = lim+ p(t) C Dα+ u(t) – t→
t
p(s)y(s) ds.
(.)
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Hence,
t α α lim+ p(t) C Dα+ u(t) – I+ u(t) = I+ p(s)y(s) ds + c p(t) t→ p(t) t t α– (t – s) (t – s)α– s ds lim+ p(t) C Dα+ u(t) – = p(s)y(s) dr ds + c , t→ (α)p(s) (α)p(s) where c ∈ R is any constant. It follows from (.) and (.) that lim p(t) C Dα+ u(t) = lim+ p(t) C Dα+ u(t) –
t→+∞
+∞
p(s)y(s) ds
t→
and lim u(t) = lim+ p(t)
t→+∞
C
t→
Dα+ u(t) lim t→+∞
– lim
t→+∞
t
α–
(t – s) (α)p(s)
t
(t – s)α– ds (α)p(s)
s
p(s)y(s) dr ds + c .
By the boundary conditions in (.) we have t (t – s)α– s a b lim c = p(r)y(r) dr ds t→+∞ (α)p(s) ρ +∞ +∞ p(s)y(s) ds + b g u(s) ψ (s) ds + b b
+ a lim
t
t→+∞
α–
(t – s) ds + b (α)p(s)
+∞
g u(s) ψ (s) ds
and lim+ p(t) C Dα+ u(t) =
t→
t (t – s)α– s a a lim p(r)y(r) dr ds t→+∞ (α)p(s) ρ +∞ p(s)y(s) ds + a b
+∞
+ a
g u(s) ψ (s) ds – a
+∞
g u(s) ψ (s) ds .
Substituting them into (.), we get
t (t – s)α– (t – s)α– s p(r)y(r) dr ds ds a a lim t→+∞ (α)p(s) (α)p(s) +∞ + a b p(s)y(s) ds
u(t) = ρ
t
t +∞ (t – s)α– s a b lim p(r)y(r) dr ds + b b p(s)y(s) ds + t→+∞ (α)p(s) ρ t α– s (t – s) p(r)y(r) dr ds – (α)p(s) +∞ +∞ g u(s) ψ (s) ds + F (t) g u(s) ψ (s) ds + F (t)
(.)
Jia et al. Boundary Value Problems (2016) 2016:104
t t (t – s)α– (t – r)α– ds a a lim dr p(s)y(s) ds t→+∞ (α)p(s) s (α)p(r)
+∞ p(s)y(s) ds + a b
= ρ
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t
t t +∞ (t – r)α– a b lim dr p(s)y(s) ds + b b p(s)y(s) ds + t→+∞ ρ s (α)p(r) t t α– (t – r) – drp(s)y(s) ds s (α)p(r) +∞ +∞ g u(s) ψ (s) ds + F (t) g u(s) ψ (s) ds. + F (t) b , a
By (.), M <
t
p(s)y(s) s
and
+∞
p(s)y(s) ds < +∞, we have
(t – r)α– dr ≤ Mp(s)y(s) ∈ L [, +∞) . (α)p(r)
Hence,
t
lim
t→+∞
t
p(s)y(s) s
(t – r)α– dr ds = (α)p(r)
+∞
p(s)y(s) lim
t→+∞ s
t
(t – r)α– dr ds. (α)p(r)
Therefore, the unique solution of the fractional boundary value problem (.) is
+∞
+∞
G(t, s)p(s)y(s) ds + F (t)
u(t) =
g u(s) ψ (s) ds
+∞
+ F (t)
g u(s) ψ (s) ds.
For convenience, we denote GM =
(b + a M)(b + a M) , a b + a b
FM =
b + a M , a b + a b
Fm =
Gm =
b (b – a M) , a b + a b + a a M
b – a M , a b + a b + a a M
b + a M b , , Fm = a b + a b a b + a b + a a M b b Gm Fm Fm , γ = min , , , G= GM FM FM ρ
τ b (τ – r)α– b + a lim dr . F = , F = τ →+∞ (α)p(r) ρ ρ FM =
Lemma . If (H) holds, then G(t, s), F (t), and F (t) defined in Lemma . satisfy () G(t, s) is a continuous function and G(t, s) > for (t, s) ∈ [, +∞) × [, +∞); () F (t), F (t) are continuous functions, and F (t), F (t) ≥ for t ∈ [, +∞); () Gm ≤ G(t, s) ≤ GM Fim ≤ Fi (t) ≤ FiM
for (t, s) ∈ [, +∞) × [, +∞), for t ∈ [, +∞), i = , ;
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() there exist constants < l < l < +∞ such that G(t, s) ≥ γ GM Fi (t) ≥ γ FiM
for (t, s) ∈ [l , l ] × [, +∞), for t ∈ [l , l ] and i = , ;
() for any s ∈ [, +∞), limt→+∞ G(t, s) = G < +∞, limt→+∞ F (t) = F < +∞, limt→+∞ F (t) = F < +∞. Proof () For ≤ t ≤ s, it is easy to see that G(t, s) > . For ≤ s < t, by (H) and (.) we have
t b a b (t – r)α– G(t, s) ≥ b b – b a dr ≥ – M > . ρ ρ a s (α)p(r) Hence, G(t, s) > for (t, s) ∈ [, +∞) × [, +∞). It is easy to see that G(t, s) is a continuous function. () It follows from (.) and (H) that () holds. () By (H), for t, s ∈ [, +∞), we have
t (t – r)α– b a b G(t, s) ≥ b b – b a dr ≥ – M = Gm ρ ρ a s (α)p(r) and
t τ (t – r)α– (τ – r)α– b + a dr b + a lim dr ≤ GM . G(t, s) ≤ τ →+∞ s (α)p(r) ρ (α)p(r) It is easy to see that Fim ≤ Fi (t) ≤ FiM for t ∈ [, +∞), i = , . () By () there exist constants < l < l < +∞ such that G(t, s) ≥ Gm =
Gm · GM ≥ γ GM GM
for (t, s) ∈ [l , l ] × [, +∞).
Similarly, we have Fi (t) ≥ γ FiM
for t ∈ [l , l ] and i = , .
() By (H) and < α < , for any s ∈ [, +∞), we can show that
s τ (t – r)α– (τ – r)α– lim dr b + a lim dr b + a τ →+∞ s (α)p(r) ρ t→+∞ (α)p(r) t (t – r)α– – b a dr s (α)p(r)
s t (t – r)α– (t – r)α– b + a dr b + a lim dr lim = t→+∞ s (α)p(r) ρ t→+∞ (α)p(r) t (t – r)α– dr – b a lim t→+∞ s (α)p(r)
lim G(t, s) =
t→+∞
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t t (t – r)α– (t – r)α– b b + a lim dr – b a lim dr = t→+∞ s (α)p(r) t→+∞ s (α)p(r) ρ = G. It is obvious that limt→+∞ F (t) = F < +∞ and limt→+∞ F (t) = F < +∞.
Let E = u ∈ C[, +∞) : lim u(t) < +∞
(.)
t→+∞
be a Banach space with the norm u = supt∈[,+∞) |u(t)|, and P = u ∈ E : u(t) ≥ , t ∈ [, +∞), inf u(t) ≥ γ u t∈[l ,l ]
be a cone in E. For r > , we denote ∂Kr = u ∈ P : u = r , Kr = u ∈ P : u < r , Sr := sup h(u) : ≤ u ≤ r , Sr := sup g (u) : ≤ u ≤ r , and Sr := sup g (u) : ≤ u ≤ r . It follows from (H) that Sr , Sr , and Sr < +∞. We define the operator T : P → E by
+∞
Tu(t) =
G(t, s)p(s)f s, u(s) ds + F (t)
+∞
g u(s) ψ (s) ds
+∞
+ F (t)
g u(s) ψ (s) ds,
t ∈ [, +∞).
We can easily get the following Lemma . from Lemma .. Lemma . If u ∈ P, then the boundary value problem (.) is equivalent to the integral equation u(t) =
+∞
G(t, s)p(s)f s, u(s) ds + F (t)
+ F (t)
+∞
g u(s) ψ (s) ds
+∞
g u(s) ψ (s) ds,
t ∈ [, +∞).
Lemma . (See [, ]) Let E be defined by (.), and ⊂ E. Then is relatively compact in E if the following conditions hold: (a) is uniformly bounded in E; (b) the functions belonging to M are equicontinuous on any compact interval of [, +∞);
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(c) the functions from are equiconvergent, that is, for any given ε > , there exists T(ε) > such that |f (t) – f (+∞)| < ε for any t > T(ε) and f ∈ . Lemma . If (H) and (H) hold, then T : P → P is completely continuous. Proof We divide the proof into three steps. Step : We show that T : P → P is well defined. For u ∈ P, there exists a constant r > such that u ≤ r . By (H) and Lemma ., for t, s ∈ [, +∞), we have G(t, s)p(s)f s, u(s) ≤ G(t, s)p(s)v(s)h(u) ≤ GM p(s)v(s)Sr . Since G(t, s), F (t), F (t) are continuous with respect to t, by using the Lebesgue dominated convergence theorem, for t ∈ [, +∞), we have
G(t, s)p(s)f s, u(s) ds + lim F (t)
+∞
lim Tu(t) = lim
t→t
t→t
t→t
+∞
+ lim F (t) t→t
+∞
=
+∞
g u(s) ψ (s) ds
g u(s) ψ (s) ds
G(t , s)p(s)f s, u(s) ds + F (t )
+∞
g u(s) ψ (s) ds
+∞
+ F (t )
g u(s) ψ (s) ds.
So, Tu ∈ C[, +∞), and we get
+∞
lim Tu(t) =
t→+∞
Gp(s)f s, u(s) ds + F
+∞
g u(s) ψ (s) ds
g u(s) ψ (s) ds < +∞.
+∞
+ F
It is obvious that Tu(t) ≥ , t ∈ [, +∞), by Lemma .. Moreover,
+∞
inf Tu(t) ≥ inf
t∈[l ,l ]
t∈[l ,l ]
G(t, s)p(s)f s, u(s) ds + inf F (t) t∈[l ,l ]
+∞
+ inf F (t) t∈[l ,l ]
+∞
g u(s) ψ (s) ds
g u(s) ψ (s) ds.
By Lemma . () we have
+∞
inf Tu(t) ≥ γ sup
t∈[l ,l ]
+∞
G(t, s)p(s)f s, u(s) ds + F (t)
t∈[,+∞)
+∞
+ F (t)
g u(s) ψ (s) ds
≥ γ Tu . Hence, T : P → P is well defined. Step : We can verify that T : P → P is continuous.
g u(s) ψ (s) ds
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Let un , u ∈ P and un – u → as n → +∞. Then there exists a constant r > such that un , u ≤ r . We have ≤ GM p(s)f s, un (s) – f s, u(s) + FM g un (s) – g u(s) ψ (s) + FM g un (s) – g u(s) ψ (s) ≤ GM p(s)v(s) h un (s) + h u(s) + FM g un (s) + g u(s) ψ (s) + FM g un (s) + g u(s) ψ (s) ≤ GM Sr p(s)v(s) + FM Sr ψ (s) + FM Sr ψ (s) ∈ L [, +∞) and, for s ∈ [, +∞), f s, un (s) – f s, u(s) → as n → +∞, gi un (s) – gi u(s) → as n → +∞, i = , . Then, by the Lebesgue dominated convergence theorem we have
GM p(s)f s, un (s) – f s, u(s) ds
+∞
Tun – Tu ≤
+∞
g un (s) – g u(s) ψ (s) ds
+ FM
+∞
g un (s) – g u(s) ψ (s) ds
+ FM
→
as n → +∞.
Therefore, T : P → P is a continuous operator. Step : We can show that T : P → P is relatively compact. Let be a bounded subset of P. Then there exists a constant r > such that u ≤ r for each u ∈ . By Lemma . and (H) we have sup
Tu =
t∈[,+∞)
+∞
+∞
+ F (t) ≤
+∞
G(t, s)p(s)f s, u(s) ds + F (t)
g u(s) ψ (s) ds
+∞
g u(s) ψ (s) ds
+∞
+ FM
g u(s) ψ (s) ds
≤ GM Sr
+∞
g u(s) ψ (s) ds
GM p(s)v(s)h u(s) ds + FM
+∞
p(s)v(s) ds + FM Sr
< +∞. So, T() is uniformly bounded.
+∞
ψ (s) ds + FM Sr
+∞
ψ (s) ds
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For any T ∈ (, +∞), since G(t, s), F (t), and F (t) are continuous, we have that G is uniformly continuous on [, T] × [, T] and F and F are uniformly continuous on [, T]. This implies that, for any ε > , there exists δ > such that, when t , t ∈ [, T], whenever |t – t | < δ and s ∈ [, T], we have G(t , s) – G(t , s) < ε, Fi (t ) – Fi (t ) < ε, i = , . Therefore, for t , t ∈ [, T], whenever |t – t | < δ and u ∈ , we can show that Tu(t ) – Tu(t ) +∞ G(t , s) – G(t , s)p(s)f s, u(s) ds ≤
+ F (t ) – F (t )
+∞
g u(s) ψ (s) ds
+∞
g u(s) ψ (s) ds
+ F (t ) – F (t )
< ε Sr
+∞
p(s)v(s) ds + Sr
+∞
ψ (s) ds + Sr
+∞
ψ (s) ds .
Hence, T() is locally equicontinuous on [, +∞). Since +∞ +∞ Gp(s)f s, u(s) ds + F g u(s) ψ (s) ds Tu(+∞) =
+∞
+ F
g u(s) ψ (s) ds.
By Lemma . we conclude that Tu(t) – Tu(+∞) +∞ G(t, s) – Gp(s)f s, u(s) ds + F (t) – F ≤
+ F (t) – F
+∞
g u(s) ψ (s) ds
+∞
g u(s) ψ (s) ds
→ as t → +∞. Hence, T() is equiconvergent at infinity. By Lemma . we obtain that T : P → P is completely continuous.
Lemma . (See []) Let E be a Banach space, P ⊆ E be a cone, and , be two bounded open subsets of E with θ ∈ ⊂ ⊂ . Suppose that T : P ∩ ( \ ) → P is a completely continuous operator such that either (i) Tx ≤ x , x ∈ P ∩ ∂ and Tx ≥ x , x ∈ P ∩ ∂ , or (ii) Tx ≥ x , x ∈ P ∩ ∂ , and Tx ≤ x , x ∈ P ∩ ∂ , holds. Then the operator T has at least one fixed point in P ∩ ( \ ).
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3 The existence of positive solutions For convenience, we give the following notation: hϕ = lim sup u→ϕ
giϕ = lim sup u→ϕ
h(u) , u
fϕ = lim inf inf
gi (u) , u
gi,ϕ = lim inf
u→ϕ t∈[l ,l ]
u→ϕ
f (t, u) , u
gi (u) , u
where ϕ = or +∞, and i = , . We denote A = max GM
+∞
B = γ · min Gm
l
p(s) ds, Fm l
+∞
p(s)v(s) ds, FM
+∞
ψ (s) ds, FM
l
l
ψ (s) ds, Fm l
ψ (s) ds ,
ψ (s) ds .
l
Theorem . Suppose that (H) and (H) hold. If A h + g + g < < B(f+∞ + g,+∞ + g,+∞ ), then the boundary value problem (.) has at least one positive solution. Proof Since A(h + g + g ) < , then there exists a constant r > such that, for u ≤ r , we have
ε u, h(u) ≤ h +
ε gi (u) ≤ gi + u,
i = , ,
(.)
where ε satisfies A(h + g + g + ε ) ≤ . Therefore, for any t ∈ [, +∞), u ∈ ∂Kr , we can get Tu(t) =
+∞
G(t, s)p(s)f s, u(s) ds + F (t)
+∞
+ F (t)
g u(s) ψ (s) ds
+∞
g u(s) ψ (s) ds
+∞
ε ε u(s) ds + FM u(s)ψ (s) ds g + GM p(s)v(s) h + ≤ +∞
ε u(s)ψ (s) ds g + + FM ≤ A h + g + g + ε u
+∞
≤ u . On the other hand, since B(f+∞ + g,+∞ + g,+∞ ) > , there exist constants r > and Mi , i = , , , with f+∞ > M > , g,+∞ > M > , g,+∞ > M > such that, for t ∈ [l , l ],
ε f (t, u) ≥ M – u,
ε gi (u) ≥ Mi – u,
where ε satisfies B(M + M + M – ε ) ≥ .
i = , ,
(.)
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Let r = max{r , γr }. In view of the definition of P, inf u(t) ≥ γ u
for u ∈ P.
t∈[l ,l ]
(.)
According to Lemma ., for u ∈ ∂Kr , we have sup
Tu =
t∈[,+∞)
+F (t) ≥
l
G(t, s)p(s)f s, u(s) ds + F (t)
+∞
l
+ Fm
+∞
g u(s) ψ (s) ds
+∞ g u(s) ψ (s) ds
Gm p(s)f s, u(s) ds + Fm
l
g u(s) ψ (s) ds
l l
g u(s) ψ (s) ds.
l
By (.) and (.) we have
l
ε ε Tu ≥ u(s) ds + Fm u(s)ψ (s) ds M – Gm p(s) M – l l l
ε M – u(s)ψ (s) ds + Fm l
l
≥ B(M + M + M – ε ) u ≥ u . Therefore, by (i) of Lemma . and Lemma ., the boundary value problem (.) has at least one positive solution u ∈ K r \ Kr . Remark . It follows from the proof of Theorem . that the boundary value problem (.) has at least one positive solution u ∈ P if one of the conditions f+∞ = +∞, g,+∞ = +∞, and g,+∞ = +∞ holds. Theorem . Suppose that (H) and (H) hold. If A h+∞ + g+∞ + g+∞ < < B(f + g, + g, ), then the boundary value problem (.) has at least one positive solution. Proof It follows from B(f + g, + g, ) > that there exist constants r > , Mi , i = , , , with f > M > , g, > M > , g, > M > such that, for t ∈ [l , l ] and < u ≤ r , we have
ε u, f (t, u) ≥ M –
ε gi (u) ≥ Mi – u,
where i = , , and ε satisfies B(M + M + M – ε ) ≥ .
(.)
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Thus, for any t ∈ [, +∞) and u ∈ ∂Kr , we have inft∈[l ,l ] u(t) ≥ γ u and sup
Tu =
t∈[,+∞)
+∞
+∞
+ F (t)
l
+∞
g u(s) ψ (s) ds
g u(s) ψ (s) ds
Gm p(s)f s, u(s) ds + Fm
l
≥
G(t, s)p(s)f s, u(s) ds + F (t)
l
g u(s) ψ (s) ds
l
g u(s) ψ (s) ds.
l
+ Fm l
It follows from (.) that
l
ε ε M – u(s) ds + Fm u(s)ψ (s) ds Gm p(s) M – Tu ≥ l l l
ε + Fm M – u(s)ψ (s) ds l ≥ B M + M + M – ε u
l
≥ u . On the other hand, since A(h+∞ + g+∞ + g+∞ ) < , there exists a constant r > such that, for u ≥ r , we have
ε +∞ u, h(u) ≤ h +
ε +∞ gi (u) ≤ gi + u
for i = , ,
where ε with A(h+∞ + g+∞ + g+∞ + ε ) < . Let +∞ +∞ +∞ GM Sr p(s)v(s) ds + FM Sr ψ (s) ds + FM Sr ψ (s) ds r > max r , r , . – A(h+∞ + g+∞ + g+∞ + ε ) For any t ∈ [, +∞), u ∈ ∂Kr , we denote D = t ∈ [, +∞) : u(t) ≥ r , u ∈ ∂Kr , D = t ∈ [, +∞) : ≤ u(t) ≤ r u ∈ ∂Kr . We have Tu(t) =
+∞
G(t, s)p(s)f s, u(s) ds + F (t)
+∞
+ F (t)
D
+ F (t) D
g u(s) ψ (s) ds
g u(s) ψ (s) ds
g u(s) ψ (s) ds D
+∞
g u(s) ψ (s) ds
GM p(s)v(s)h u(s) ds + F (t)
≤
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GM p(s)v(s)h u(s) ds + F (t)
+ D
g u(s) ψ (s) ds
D
g u(s) ψ (s) ds
+ F (t) D
≤ A h+∞ + g+∞ + g+∞ + ε u + GM Sr + FM Sr
+∞
ψ (s) ds + FM Sr
+∞
p(s)v(s) ds
+∞
ψ (s) ds
≤ u . Hence, by using (ii) of Lemma . and Lemma ., the boundary value problem (.) has at least one positive solution u ∈ K r \ Kr . Remark . It follows from the proof of Theorem . that the boundary value problem (.) has one positive solution u ∈ P if at least one of the conditions f = +∞, g, = +∞, and g, = +∞ holds. Theorem . Suppose that (H) and (H) hold. If () B(f + g, + g, ) > , B(f+∞ + g,+∞ + g,+∞ ) > and () there exists a constant c > such that max{Sc , Sc , Sc } := S∗ < A– c, then the boundary value problem (.) has at least two positive solutions. Proof Since B(f + g, + g, ) > , similarly to the proof of Theorem ., there exists a constant < r < c with Tu ≥ u ,
u = r.
Since B(f+∞ + g,+∞ + g,+∞ ) > , there also exists a constant R > c such that Tu ≥ u ,
u = R.
On the other hand, by condition (), for any u ∈ ∂Kc , sup
Tu =
t∈[,+∞)
+∞
+∞
+ F (t) ≤
+∞
+∞
g u(s) ψ (s) ds
+∞
g u(s) ψ (s) ds
+∞
+ FM
g u(s) ψ (s) ds
GM p(s)v(s)h u(s) ds + FM
g u(s) ψ (s) ds
≤ GM Sc
+∞
≤ S GM ∗
< c.
G(t, s)p(s)f s, u(s) ds + F (t)
p(s)v(s) ds + FM Sc
+∞
+∞
p(s)v(s) ds + FM
ψ (s) ds + FM Sc
+∞
ψ (s) ds + FM
+∞
ψ (s) ds
+∞
ψ (s) ds
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Namely, Tu < c = u ,
u ∈ ∂Kc .
According to Lemma ., the boundary value problem (.) has at least two positive solu tions u , u with < u < c < u . Remark . It follows from the proof of Theorem . that the boundary value problem (.) has at least two positive solutions u ∈ P if one of the conditions f = +∞, g, = +∞, g, = +∞, f+∞ = +∞, g,+∞ = +∞, and g,+∞ = +∞ holds. Theorem . Suppose that (H) and (H) hold. If () A(h + g + g ) < , A(h+∞ + g+∞ + g+∞ ) < , and () there exists a constant C > such that, for any t ∈ [l , l ] and u ∈ [γ C, C], we have min f (t, u), g (u), g (u) > γ B– C, then the boundary value problem (.) has at least two positive solutions. Proof The proof is similar to that of Theorem .. It is easy to get the two positive solutions u , u with < u < C < u .
4 Illustration Example We consider the following boundary value problem: ⎧ tC ⎪ ⎪ et (e D+ u(t)) + f (t, u(t)) = , t ∈ (, +∞), ⎨ +∞ (u(s)) u() – limt→+ p(t) C Dα+ u(t) = g+s ds, ⎪ ⎪ +∞ ⎩ C α limt→+∞ u(t) + limt→+∞ p(t) D+ u(t) = where f (t, u) =
–u ) e–t (+t)(u+e √ , t
(.) g (u(s)) +s
ds,
a = , a = , b = , b = , p(t) = et , ψ (s) = ψ (s) =
, +s
and
⎧ ⎨u, ≤ u ≤ , g (u) = g (u) = √ ⎩ u – , < u < +∞. It is obvious that f : (, +∞) × [, +∞) → [, +∞) is a continuous function and singular at t = . Let v(t) =
e–t ( + t) , √ t
h(u) =
u + e–u .
Then we have M ≈ ., A = ., h+∞ = , g+∞ = g+∞ = , and A(h+∞ + g+∞ + g+∞ ) = . < . On the other hand, let l = , l = . So we have B ≈ ., f = +∞, gi, = , i = , . Namely, B(f + g, + g, ) = +∞ > . By using Theorem . the boundary value problem (.) has at least one positive solution.
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Competing interests The authors declare that no competing interests exist. Authors’ contributions The authors contributed equally to this paper. All authors read and approved the final manuscript. Acknowledgements This work is supported by the National Natural Science Foundation of China (No. 11171220) and the Hujiang Foundation of China (B14005). Received: 4 December 2015 Accepted: 13 May 2016 References 1. Lian, H, Ge, G: Existence of positive solutions for Sturm-Liouville boundary value problems on the half-line. J. Math. Anal. Appl. 321, 781-792 (2006) 2. Liu, L, Wang, Z, Wu, Y: Multiple positive solutions of the singular boundary value problems for second-order differential equations on the half-line. Nonlinear Anal. 71, 2564-2575 (2009) 3. Yoruk, F, Hamal, NA: Existence results for nonlinear boundary value problems with integral boundary conditions on an infinite interval. Bound. Value Probl. 2012, 127 (2012) 4. Chen, J, Liu, X, Xiao, Y: Existence of multiple positive solutions for second-order differential equation with integral boundary value problems on the half-line. J. Shandong Univ. Nat. Sci. 46, 37-41 (2011) 5. Miller, KB, Ross, B: An Introduction to the Fractional Calculus and Fractional Differential Equations. Wiley, New York (1993) 6. Podlubny, I: Fractional Differential Equations. Academic Press, New York (1999) 7. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam (2006) 8. Diethelm, K: The Analysis of Fractional Differential Equation. Springer, Heidelberg (2010) 9. Jia, M, Liu, X: Multiplicity of solutions for integral boundary value problems of fractional differential equations with upper and lower solutions. Appl. Math. Comput. 232, 313-323 (2014) 10. Jia, M, Liu, X: Three nonnegative solutions for fractional differential equations with integral boundary conditions. Comput. Math. Appl. 62, 1405-1412 (2011) 11. Ge, F, Zhou, H: Existence of solutions for fractional differential equations with three-point boundary conditions at resonance in Rn . Electron. J. Qual. Theory Differ. Equ. 2014, 68 (2014) 12. Zhi, E, Liu, X, Li, F: Nonlocal boundary value problem for fractional differential equations with p-Laplacian. Math. Methods Appl. Sci. 37, 2651-2662 (2014) 13. Liu, X, Lin, L, Fang, H: Existence of positive solutions for nonlocal boundary value problem of fractional differential equations. Cent. Eur. J. Phys. 11, 1423-1432 (2013) 14. Liu, X, Jia, M, Ge, W: Multiple solutions of a p-Laplacian model involving a fractional derivative. Adv. Differ. Equ. 2013, 126 (2013) 15. Xie, W, Luo, Z, Xiao, J: Successive iteration and positive solutions of a fractional boundary value problem on the half-line. Adv. Differ. Equ. 2013, 210 (2013) 16. Zhao, X, Ge, W: Unbounded solutions for a fractional boundary value problems on the infinite interval. Acta Appl. Math. 109, 495-505 (2010) 17. Su, X, Zhang, S: Unbounded solutions to a boundary value problem of fractional order on the half-line. Comput. Math. Appl. 61, 1079-1087 (2011) 18. Agarwal, RP, O’Regan, D: Infinite Interval Problems for Differential, Difference and Integral Equations. Kluwer Academic, Dordrecht (2001) 19. Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, New York (1988)