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Hussein A.H. Salem [1] investigated the existence of pseudo-solutions for the nonlinear m-point boundary value problem of the fractional case,. Dα. 0+x(t) + q(t)f.
Xu and Zhang Boundary Value Problems (2018) 2018:34 https://doi.org/10.1186/s13661-018-0944-8

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Multiple positive solutions to singular positone and semipositone m-point boundary value problems of nonlinear fractional differential equations Xiaojie Xu* and Huina Zhang *

Correspondence: [email protected] College of Science, China University of Petroleum Huadong, Qingdao, China

Abstract In this paper, we consider the properties of Green’s function for the nonlinear fractional differential equation boundary value problem Dα0+ u(t) + f (t, u(t)) = 0, u(0) = 0,

u(1) =

0 < t < 1,

m–2 

βi u(ηi ),

i=1

where 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < ηm–2 < 1,  m–2 α –1 < 1, Dα is the standard Riemann–Liouville derivative. Here our 0+ i=1 βi ηi nonlinearity f may be singular at u = 0. As an application of Green’s function, we give some multiple positive solutions for singular positone and semipositone boundary value problems by means of the Leray–Schauder nonlinear alternative and a fixed point theorem on cones. MSC: 34B15 Keywords: Multiple positive solutions; Singular fractional differential equation; Semipositone; m-point boundary value problem

1 Introduction In this paper, we consider the existence and multiplicity of positive solutions of the nonlinear fractional differential equation semipositone boundary value problem:   Dα0+ u(t) + f t, u(t) = 0, u(0) = 0,

u(1) =

0 < t < 1,

m–2  βi u(ηi ),

(1.1)

i=1

 α–1 where 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < ηm–2 < 1, m–2 < 1, i=1 βi ηi α D0+ is the standard Riemann–Liouville derivative. Here our nonlinearity f may be singular © The Author(s) 2018. 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.

Xu and Zhang Boundary Value Problems (2018) 2018:34

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at u = 0. The nonlinear fractional differential equation for the multi-point boundary value problem has been studied extensively. For details, see [1–14] and the references therein. For m = 3, Bai [15] investigated the existence and uniqueness of positive solutions for a nonlocal boundary value problem of the fractional differential equation   Dα0+ u(t) + f t, u(t) = 0, u(0) = 0,

0 < t < 1, (1.2)

u(1) = βu(η),

via the contraction map principle and fixed point index theory, where 1 < α ≤ 2, 0 < βηα–1 < 1, 0 < η < 1, Dα0+ is the standard Riemann–Liouville derivative. The function f is continuous on [0, 1] × [0, ∞). Wang, Xiang and Liu [16] investigated the existence and uniqueness of a positive solution to nonzero three-point boundary values problem for a coupled system of fractional differential equations. Ahmad and Nieto [17] considered the three point boundary value problems of the fractional order differential equation. By using some fixed point theorems, they obtained the existence and multiplicity result of positive solution to this problem. They considered the case when f has no singularities. Xu, Jiang et al. [18] deduced some new properties of Green’s function of (1.2). By using some fixed point theorems, they obtained the existence, uniqueness and multiplicity of positive solutions to singular positone and semipositone problems. Hussein A.H. Salem [1] investigated the existence of pseudo-solutions for the nonlinear m-point boundary value problem of the fractional case,   Dα0+ x(t) + q(t)f t, x(t) = 0, 

x(0) = x (0) = · · · = x

(n–1)

a.e. on [0, 1], α ∈ (n – 1, n], n ≥ 2,

= 0,

x(1) =

m–2 

(1.3) ζi x(ηi ),

i=1

where 0 < η1 < η2 < · · · < ηm–2 < 1, ζi > 0 with

m–2 i=1

ζi ηiα–1 < 1. It is assumed that q is a

real-valued continuous function and f is a nonlinear Pettis integrable function. However, no paper to date has discussed the multiplicity for the semipositone singular problem. This paper attempts to fill this gap in the literature, and as a corollary, we give a result for singular positone problems.

2 Background materials For convenience of the reader, we present here the necessary definitions from fractional calculus theory. Definition 2.1 ([19]) The Riemann–Liouville fractional integral of order α > 0 of a function y : (0, ∞) → R is given by α y(t) = I0+

1 (α)



t

(t – s)α–1 y(s) ds, 0

provided the right side is pointwise defined on (0, ∞).

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Definition 2.2 ([19]) The Riemann–Liouville fractional derivative of order α > 0 of a continuous function y : (0, ∞) → R is given by Dα0+ y(t) =

 n  t d 1 y(s) ds, α–n+1 (n – α) dt 0 (t – s)

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, one has the following results. Lemma 2.1 ([19]) Assume that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then α I0+ Dα0+ u(t) = u(t) + C1 t α–1 + C2 t α–2 + · · · + CN t α–N ,

for some Ci ∈ R, i = 1, 2, . . . , N , where N is the smallest integer greater than or equal to α. Lemma 2.2 ([20]) Let X be a Banach space, and let P ⊂ X be a cone in X. Assume 1 , 2 are open subsets of X with 0 ∈ 1 ⊂ 1 ⊂ 2 , and let S : P → P be a completely continuous operator such that either 1. Sw ≤ w , w ∈ P ∩ ∂1 , Sw ≥ w , w ∈ P ∩ ∂2 , or 2. Sw ≥ w , w ∈ P ∩ ∂1 , Sw ≤ w w ∈ P ∩ ∂2 . Then S has a fixed point in P ∩ (2 \1 ).

3 Semipositone and positone singular problem In this section, we consider the existence and multiplicity of positive solutions of the nonlinear fractional differential equation semipositone and positone boundary value problem   Dα0+ u(t) + f t, u(t) = 0, u(0) = 0,

u(1) =

m–2 

0 < t < 1, (3.1) βi u(ηi ),

i=1

 α–1 where 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < ηm–2 < 1, m–2 < 1, i=1 βi ηi α D0+ is the standard Riemann–Liouville derivative. Here our nonlinearity f may be singular at u = 0. Lemma 3.1 Given y ∈ C(0, 1) the problem Dα0+ u(t) + y(t) = 0, u(0) = 0,

u(1) =

0 < t < 1, m–2  i=1

is equivalent to  u(t) =

1

G(t, s)y(s) ds, 0

(3.2) βi u(ηi ),

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where G(t, s) = G1 (t, s) + G2 (t, s),

G1 (t, s) =

⎧ [t(1–s)]α–1 –β1 t α–1 (η1 –s)α–1 –(t–s)α–1 (1–β1 η1α–1 ) ⎪ ⎪ , 0 ≤ s ≤ t ≤ 1, s ≤ η1 , ⎪ (1–β1 ηα–1 )(α) ⎪ ⎪ ⎪ [t(1–s)]α–1 –(t–s)α–1 (1–β1 1 η1α–1 ) ⎪ ⎨ , 0 < η1 ≤ s ≤ t ≤ 1, (1–β ηα–1 )(α) 1 1

[t(1–s)]α–1 –β1 t α–1 (η1 –s)α–1 ⎪ ⎪ , ⎪ (1–β1 η1α–1 )(α) ⎪ ⎪ ⎪ α–1 ⎪ ⎩ [t(1–s)] , α–1 (1–β1 η1

G2 (t, s) = H(s)t

α–1

0 ≤ t ≤ s ≤ 1, η1 ≤ s.

)(α)

 m–2  K1 K i Ki H(s) = M0 qi (s) + pi (s) , 1 – K1 1 – K1 i=1

,

where Ki = βi ηiα–1 , M0 =

(3.3)

0 ≤ t ≤ s ≤ η1 < 1,

(α)(1–

1 m–2

α–1 ) i=1 βi ηi

(3.4)

,

  s α–1 I[0≤s≤ηi ] , qi (s) = (1 – s)α–1 – 1 – ηi     s α–1 s α–1 pi (s) = 1 – I[0≤s≤ηi ] – 1 – I[0≤s≤η1 ] , ηi η1 ⎧ ⎨1, s ∈ [0, η ], i I[0≤s≤ηi ] = ⎩0, s ∈/ [0, ηi ]. Proof By Lemma 2.1, the solution of (3.2) can be written as  u(t) = C1 t α–1 + C2 t α–2 – 0

t

(t – s)α–1 y(s) ds. (α)

From u(0) = 0, we know that C2 = 0.  On the other hand, together with u(1) = m–2 i=1 βi u(ηi ), we have C1 =

(α)(1 –

1 m–2 i=1

 βi ηiα–1 )

1

(1 – s)

α–1

y(s) ds –

0

m–2 



(ηi – s)

βi

i=1



ηi

α–1

y(s) ds .

0

Therefore, the unique solution of fractional boundary value problem (3.2) is t α–1 m–2



1 α–1

m–2 



ηi

 α–1

(1 – s) y(s) ds – βi (ηi – s) y(s) ds (α)(1 – i=1 βi ηiα–1 ) 0 0 i=1  t (t – s)α–1 y(s) ds – (α) 0

 1   η1 t α–1 α–1 α–1 = (1 – s) y(s) ds – β (η – s) y(s) ds 1 1 (α)(1 – β1 η1α–1 ) 0 0  t (t – s)α–1 y(s) ds – (α) 0 

 1  η1 t α–1 α–1 α–1 – (1 – s) y(s) ds – β (η – s) y(s) ds 1 1 (α)(1 – β1 η1α–1 ) 0 0

u(t) =

+

(α)(1 –

t α–1 m–2 i=1

βi ηiα–1 )

Xu and Zhang Boundary Value Problems (2018) 2018:34



1

×

(1 – s)α–1 y(s) ds – 0

Page 5 of 18

m–2  i=1

 βi

ηi

 (ηi – s)α–1 y(s) ds

0

= u1 (t) + u2 (t),

(3.5)

where

 1   η1 t α–1 α–1 α–1 (1 – s) y(s) ds – β (η – s) y(s) ds 1 1 (α)(1 – β1 η1α–1 ) 0 0  t α–1 (t – s) y(s) ds, – (α) 0

 1   η1 t α–1 α–1 α–1 (1 – s) y(s) ds – β1 (η1 – s) y(s) ds u2 (t) = – (α)(1 – β1 η1α–1 ) 0 0  1 t α–1 + (1 – s)α–1 y(s) ds m–2 α–1 (α)(1 – i=1 βi ηi ) 0   ηi m–2  α–1 – βi (ηi – s) y(s) ds . u1 (t) =

(3.6)

(3.7)

0

i=1

When t ≤ η1 , we have

 t  η1  1   (t – s)α–1 t α–1 α–1 (1 – s) y(s) ds y(s) ds + + + u1 (t) = – (α) (α)(1 – β1 η1α–1 ) t 0 0 η1  t  η1  t α–1 + β1 (η1 – s)α–1 y(s) ds – (α)(1 – β1 η1α–1 ) 0 t  t [t(1 – s)]α–1 – β1 (η1 – s)α–1 t α–1 – (t – s)α–1 (1 – β1 η1α–1 ) y(s) ds = (α)(1 – β1 η1α–1 ) 0  1  η1 [t(1 – s)]α–1 – β1 (η1 – s)α–1 t α–1 [t(1 – s)]α–1 y(s) ds + y(s) ds + α–1 α–1 (α)(1 – β1 η1 ) t η1 (α)(1 – β1 η1 )  1 = G1 (t, s)y(s) ds. 

t

0

When t ≥ η1 , we have 

 t

η1

(t – s)α–1 y(s) ds (α) 0 η1

 η1  t  1   t α–1 α–1 (1 – s) y(s) ds + + + (α)(1 – β1 η1α–1 ) 0 η1 t  η1 t α–1 β1 (η1 – s)α–1 y(s) ds – (α)(1 – β1 η1α–1 ) 0  η1 [t(1 – s)]α–1 – β1 (η1 – s)α–1 t α–1 – (t – s)α–1 (1 – β1 η1α–1 ) y(s) ds = (α)(1 – β1 η1α–1 ) 0  1  t [t(1 – s)]α–1 – (t – s)α–1 (1 – β1 η1α–1 ) [t(1 – s)]α–1 y(s) ds + y(s) ds + α–1 α–1 (α)(1 – β1 η1 ) η1 t (α)(1 – β1 η1 )  1 = G1 (t, s)y(s) ds,

u1 (t) = –

0

+

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where G1 (t, s) is defined by (3.3). t α–1 (α)(1 – β1 η1α–1 )

u2 (t) = – +



(α)(1 – m–2  i=2

=

=

=

=

=



t α–1 m–2 i=1



1

(ηi – s)

α–1

η1

(1 – s)α–1 y(s) ds – β1 0

 (η1 – s)α–1 y(s) ds

0



βi ηiα–1 )

ηi

βi





1

(1 – s)

α–1

η1

y(s) ds – β1

0

(η1 – s)α–1 y(s) ds

0



y(s) ds

0

t α–1 m–2

m–2

βi ηiα–1 1 – β1 η1α–1



1

i=2

(1 – s)α–1 y(s) ds (α)(1 – i=1 βi ηiα–1 ) 0 m–2   η1 βi ηiα–1 – i=2 α–1 β1 (η1 – s)α–1 y(s) ds 1 – β1 η1 0  ηi m–2 α–1  t β (ηi – s)α–1 y(s) ds – m–2 i (α)(1 – i=1 βi ηiα–1 ) i=2 0  m–2 α–1  1 t α–1 i=2 βi ηi (1 – s)α–1 y(s) ds m–2 (α)(1 – i=1 βi ηiα–1 ) 1 – β1 η1α–1 0 m–2  η1 βi ηiα–1 – i=2 α–1 β1 (η1 – s)α–1 y(s) ds 1 – β1 η1 0   ηi m–2 1 – β1 η1α–1  βi (ηi – s)α–1 y(s) ds – 1 – β1 η1α–1 i=2 0  m–2 α–1  1 t α–1 i=2 βi ηi (1 – s)α–1 y(s) ds m–2 (α)(1 – i=1 βi ηiα–1 ) 1 – β1 η1α–1 0 m–2   η1  α–1 s α–1 i=2 βi ηi α–1 1– – β1 η1 y(s) ds η1 1 – β1 η1α–1 0   ηi  m–2  s α–1 1 α–1 1 – β η y(s) ds – i i ηi 1 – β1 η1α–1 i=2 0     m–2 s α–1 β1 η1α–1  α–1 ηi 1– + βi ηi y(s) ds ηi 1 – β1 η1α–1 i=2 0

m–2 α–1  1 t α–1 i=2 βi ηi (1 – s)α–1 y(s) ds  α–1 (α)(1 – m–2 ) 1 – β1 η1α–1 0 i=1 βi ηi    ηi  s α–1 1– y(s) ds – ηi 0      η1  α–1  ηi  β1 η1α–1 m–2 s α–1 s α–1 i=2 βi ηi 1 – 1 – y(s) ds – y(s) ds + ηi η1 1 – β1 η1α–1 0 0  1 G2 (t, s)y(s) ds, 0

where G2 (t, s) is defined by (3.4). The proof is complete.



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Lemma 3.2 ([18]) The Green’s function G1 (t, s) defined by Lemma 3.1 has the following properties: (1) G(t, s) ≤ G(s, s) ≤

sα–1 (1 – s)α–1 (α)(1 – β1 η1α–1 )

for t, s ∈ [0, 1], β1 η1α–2 ≤ 1;

(3.8)

(2) t α–1 (1 – s)α–1 Mt α–1 s(1 – s)α–1 ≤ G(t, s) ≤ (α)(1 – β1 η1α–1 ) (α)(1 – β1 η1α–1 )

for t, s ∈ [0, 1];

(3.9)

(3) Mt α–1 s(1 – s)α–1 s(1 – s)α–1 t α–2 (1 + β1 η1α–2 ) ≤ G(t, s) ≤ α–1 (α)(1 – β1 η1 ) (α)(1 – β1 η1α–1 ) for t, s ∈ (0, 1];

(3.10)

(4) Mts(1 – s)α–1 s(1 – s)α–1 (1 + β1 η1α–2 ) ≤ t 2–α G(t, s) ≤ α–1 (α)(1 – β1 η1 ) (α)(1 – β1 η1α–1 ) for t, s ∈ [0, 1],

(3.11)

where 0 < M = min{1 – β1 η1α–1 , β1 η1α–2 (1 – η1 )(α – 1), β1 η1α–1 } < 1. Theorem 3.1 The Green’s function G(t, s) defined by Lemma 3.1 has the following properties: (1) H(s) ≥ 0,

(3.12)

(2)

Mt

s(1 – s)α–1 + H(s) (α)(1 – β1 η1α–1 )

≤ t 2–α G(t, s) ≤



1 + β1 η1α–2





s(1 – s)α–1 H(s) + (α)(1 – β1 η1α–1 )

 for t, s ∈ [0, 1],

(3.13)

for t, s ∈ [0, 1],

(3.14)

(3)  s(1 – s)α–1 + H(s) (α)(1 – β1 η1α–1 ) 

(1 – s)α–1 α–1 H(s) + ≤ G(t, s) ≤ t (α)(1 – β1 η1α–1 )

Mt α–1

where 0 < M = min{1 – β1 η1α–1 , β1 η1α–2 (1 – η1 )(α – 1), β1 η1α–1 } < 1.

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Proof From (3.4), we know that Ki = βi ηiα–1 > 0, M0 =

(α)(1–

> 0, and pi (s) ≥

1 m–2

α–1 ) i=1 βi ηi

0, qi (s) ≥ 0 since 0 < η1 < η2 < · · · < ηm–2 < 1. Thus, H(s) ≥ 0. From Lemma 3.2 and the definition of G2 (t, s), it is easy to check that (2)–(3) hold. Thus the proof of Theorem 3.1 is complete.  For convenience, in this article, we let ω =

M , G∗ (t, s) = t 2–α G(t, s). 1+β1 η1α–2

Lemma 3.3 Suppose e ∈ C[0, 1], e(t) > 0, for t ∈ (0, 1), 1 < α ≤ 2, and γ (t) is the unique solution of Dα0+ u(t) + e(t) = 0, u(0) = 0,

u(1) =

0 < t < 1, m–2 

βi u(ηi ),

i=1

where 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < ηm–2 < 1, then there exists a constant C0 such that 0 ≤ φ(t) := t 2–α γ (t) ≤ ωC0 t, here C0 =

1+β1 η1α–2 M

1 0

[H(s) +

m–2 i=1

βi ηiα–1 < 1,

0 ≤ t ≤ 1,

(1–s)α–1 ]e(s) ds. (α)(1–β1 η1α–1 )

Proof In fact, from Lemma 3.1, we have 

1

γ (t) =

G(t, s)e(s) ds. 0

Thus, from Theorem 3.1, we have  φ(t) = t

2–α

1

 G(t, s)e(s) ds ≤

0

1

t

2–α α–1

0

t

H(s) +

 (1 – s)α–1 e(s) ds = ωC0 t.  (α)(1 – β1 η1α–1 )

The above lemma together with Lemma 2.2 and the Leray–Schauder alternative principle establishes our main result. Theorem 3.2 Suppose the following conditions are satisfied: (H1 ) f : (0, 1] × (0, ∞) → R is continuous and there exists a function e(t) ∈ C[0, 1], e(t) > 0 for t ∈ (0, 1), with f (t, u) + e(t) ≥ 0 for (t, u) ∈ (0, 1] × (0, ∞); (H2 ) f ∗ (t, u) = f (t, u) + e(t), and f ∗ (t, t α–2 u) ≤ q(t)[g(u) + h(u)] on (0, 1] × (0, ∞)with g > 0 continuous and nonincreasing on (0, ∞), h ≥ 0continuous on [0, ∞) and hg nondecreasing on (0, ∞), q ∈ L1 [0, 1], q > 0 on (0, 1); 1 (H3 ) a0 = 0 q(s)g(s) ds < +∞; (H4 ) ∃K0 with g(ab) ≤ K0 g(a)g(b), ∀a > 0, b > 0; (H5 ) ∃r > C0 with r g(ω(r – C0 )){1 +

h(r) } g(r)

> K0 b0 ,

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where  b0 = 0

1

1 + β1 η1α–2



H(s) +

 s(1 – s)α–1 q(s)g(s) ds; (α)(1 – β1 η1α–1 )

(3.15)

(H6 ) there exists 0 < a < 12 (choose and fix it) and a continuous, nonincreasing function g1 : (0, ∞) → (0, ∞), and a continuous function h1 : [0, ∞) → (0, ∞) with hg11 nondecreasing on (0, ∞) and with f ∗ (t, t α–2 u) ≥ q1 (t)[g1 (u) + h1 (u)] for (t, u) ∈ [a, 1] × (0, ∞), q1 (t) ∈ C([0, 1], [0, ∞)); (H7 ) ∃R > r with Rg1 (εaωR) ≤ g1 (R)g1 (εωaR) + g1 (R)h1 (εωaR)



1

G∗ (σ , s)q1 (s) ds

here ε > 0 is any constant (choose and fix it) so that 1 – R > r > C0 ) and 0 ≤ σ ≤ 1 is such that 

1

q1 (s)G∗ (σ , s) ds = sup



1

(3.16)

a C0 R

≥ ε (note ε exists since

q1 (s)G∗ (t, s) ds;

t∈[0,1] a

a

(H8 ) for each L > 0, there exists a function ϕL ∈ C[0, 1], ϕL > 0 for t ∈ (0, 1) such that f ∗ (t, t α–2 u) ≥ ϕL (t) for (t, u) ∈ (0, 1) × (0, L], and ϕr (t) > e(t), t ∈ (0, 1), where r is as in H5 . Then (3.1) has at least two solution u1 , u2 with u1 (t) > 0, u2 (t) > 0 for t ∈ (0, 1). Proof To show (3.1) has two nonnegative solutions we will look at the boundary value problem   Dα0+ u(t) + f ∗ t, u(t) – γ (t) = 0, u(0) = 0,

u(1) =

m–2 

0 < t < 1, (3.17)

βi u(ηi ),

i=1

 α–1 < 1, where 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < ηm–2 < 1, m–2 i=1 βi ηi and γ is as in Lemma 3.1. We will show, using Lemma 2.2 and the Leray–Schauder alternative principle, that there exist two solutions, u1 , u2 to (3.17) with u1 (t) > γ (t), u2 (t) > γ (t) for t ∈ (0, 1). If this is true then u¯ i (t) = ui (t) – γ (t), 0 ≤ t ≤ 1 are nonnegative solutions (positive on (0,1)) of (3.1), i = 1, 2, since Dα0+ u¯ i (t)

  = Dα0+ ui (t) – Dα0+ γ (t) = –f ∗ t, ui (t) – γ (t) + e(t)     = – f t, ui (t) – γ (t) + e(t) + e(t)   = –f t, u¯ i (t) , 0 < t < 1.

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As a result, we will concentrate our study on (3.17). Suppose that u is a solution of (3.17). Then 

1

  G(t, s)f ∗ s, u(s) – γ (s) ds

1

  t α–2 G∗ (t, s)f ∗ s, u(s) – γ (s) ds,

u(t) = 

0

=

0 ≤ t ≤ 1.

(3.18)

0

Let y(t) := t 2–α u(t), then from (3.18) we have 

1

  G∗ (t, s)f ∗ s, sα–2 y(s) – γ (s) ds

1

   G∗ (t, s)f ∗ s, sα–2 y(s) – φ(s) ds,

y(t) = 

0

=

(3.19)

0

where γ and φ are as in Lemma 3.2. Let E = C[0, 1] be endowed with maximum norm, u = max0≤t≤1 |u(t)|, and define the cone K ⊂ E by   K = u ∈ E|u(t) ≥ ωt u , and let   1 = u ∈ E; u < r ,

  2 = u ∈ E; u < R .

¯ 2 \ 1 ) → E be defined by Next let A : K ∩ ( 

1

(Ay)(t) =

   G∗ (t, s)f ∗ s, sα–2 y(s) – φ(s) ds,

0 ≤ t ≤ 1.

(3.20)

0

¯ 2 \ 1 ) then r ≤ y ≤ R First we show A is well defined. To see this notice if y ∈ K ∩ ( and y(t) ≥ ωt y ≥ ωtr, 0 ≤ t ≤ 1. Also notice for t ∈ (0, 1) that Lemma 3.1 implies y(t) – φ(t) ≥ ωtr – ωtC0 = ωt(r – C0 ),

t ∈ [0, 1],

and for t ∈ (0, 1), from (H2 ) we have       f ∗ t, t α–2 y(t) – φ(t) = f t, t α–2 y(t) – φ(t) + e(t)      ≤ q(t) g y(t) – φ(t) + h y(t) – φ(t)     h(y(t) – φ(t)) = q(t)g y(t) – φ(t) 1 + g(y(t) – φ(t))     h(R) ≤ q(t)g ωt(r – C0 ) 1 + g(R)     h(R) . ≤ K0 q(t)g(t)g ω(r – C0 ) 1 + g(R) ¯ 2 \ 1 ) → E is well defined. These inequalities with (H3 ) guarantee that A : K ∩ (

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¯ 2 \ 1 ), then we have If y ∈ K ∩ ( ⎧ 1 s(1–s)α–1 ∗ α–2 ⎪ (y(s) – φ(s))) ds,

Ay ≤ 0 (1 + β1 η1α–2 [H(s) + (α)(1–β ⎪ α–1 ]f (s, s ⎪ 1 η1 ) ⎨ 1 s(1–s)α–1 (Ay)(t) ≥ 0 Mt[H(s) + (α)(1–β ηα–1 ) ]f ∗ (s, sα–2 (y(s) – φ(s))) ds ⎪ 1 1 ⎪ ⎪ ⎩ ≥ ωt Ay , t ∈ [0, 1], ¯ 2 \ 1 ) → K . i.e., Ay ∈ K so A : K ∩ ( ¯ 2 \ 1 ) → K Similarly reasoning as in the proof of Theorem 3.1 [18] shows that A : K ∩ ( is continuous and completely continuous. We now show

Ay ≤ y

for K ∩ ∂1 .

(3.21)

To see this, let y ∈ K ∩ ∂1 . Then y = r and y(t) ≥ ωtr for t ∈ [0, 1]. Now for t ∈ (0, 1) (as above) y(t) – φ(t) ≥ ωt(r – C0 ) > 0, then we have  (Ay)(t) = 0

1

   G∗ (t, s)f ∗ s, sα–2 y(s) – φ(s) ds

 1 + β1 η1α–2 H(s) +

 s(1 – s)α–1 (α)(1 – β1 η1α–1 ) 0      × q(s) g y(s) – φ(s) + h y(s) – φ(s) ds 

   1     s(1 – s)α–1 h(r) q(s)g ωs(r – C 1 + β1 η1α–2 H(s) + ds ) 1 + ≤ 0 g(r) (α)(1 – β1 η1α–1 ) 0

  1   s(1 – s)α–1 1 + β1 η1α–2 H(s) + ≤ (α)(1 – β1 η1α–1 ) 0     h(r) ds. × q(s)K0 g ω(r – C0 ) g(s) 1 + g(r) 



1

This together with (H5 ) yields

Ay < r = y , so (3.21) is satisfied. Next we show

Ay ≥ y

for K ∩ ∂2 .

(3.22)

To see this let y ∈ K ∩ ∂2 , then y = R and y(t) ≥ ωtR for t ∈ [0, 1]. Also for t ∈ [0, 1] we have   C0 ≥ εωtR. y(t) – φ(t) ≥ ωt(R – C0 ) ≥ ωtR 1 – R

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Now with σ as in the statement of Theorem 3.1, we have 

1

  G∗ (σ , s)f ∗ (s, sα–2 y(s) – φ(s) ds

1

    G∗ (σ , s)q1 (s) g1 y(s) – φ(s) + h1 (y(s) – φ(s) ds

(Ay)(σ ) = 0

 ≥

a

    h1 (y(s) – φ(s)) ds G∗ (σ , s)q1 (s)g1 y(s) – φ(s) 1 + g1 (y(s) – φ(s)) a    1 h1 (εωaR) ∗ G (σ , s)q1 (s) 1 + ≥ g1 (R) ds. g1 (εωaR) a 

1

=

This together with (3.16) yields (Ay)(σ ) ≥ R = y . Thus Ay ≥ y , (3.22) holds. ¯ 2 \ 1 ), i.e. r ≤ y1 ≤ R and Now Lemma 2.2 implies A has a fixed point y1 ∈ K ∩ ( y1 (t) ≥ ωtr for t ∈ [0, 1]. Thus y1 (t) is a solution of (3.19) with y1 (t) > φ(t) for t ∈ (0, 1) and t α–2 y1 (t) is a solution of (3.17) for t ∈ [0, 1]. Therefore, t α–2 y1 (t) – γ (t) is a positive solution of (3.1). The existence of another solution is proved using the Leray–Schauder alternative principle, together with a truncation technique. Since (H5 ) holds, we can choose n0 ∈ {1, 2, . . .} such that n10 < r – C0 and   h(r) 1 b0 + < r. K0 g ω(r – C0 ) 1 + g(r) n0 



(3.23)

Let N0 = {n0 , n0 + 1, . . .}. Consider the family of equations 

1

(Tn y)(t) = 0

   1 G∗ (t, s)fn∗ s, sα–2 y(s) – φ(s) ds + , n

(3.24)

where n ∈ N0 and  ∗

fn

⎧ ⎨f ∗ (s, sα–2 u), u ≥ 1 ,  n s, sα–2 u = ⎩f ∗ (s, sα–2 1 ), u ≤ 1 . n n

Similarly reasoning as in the proof of Theorem 3.1 shows that Tn is well defined and maps E into K . Moreover, Tn is continuous and completely continuous. We consider y = λTn y + (1 – λ)

1 n

i.e.  y(t) = λ 0

1

   1 G∗ (t, s)fn∗ s, sα–2 y(s) – φ(s) ds + , n

(3.25)

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where λ ∈ [0, 1]. We claim that any fixed point y of (3.25) for any λ ∈ [0, 1] must satisfy

x = r. Otherwise, assume that y is a fixed point of (3.25) for some λ ∈ [0, 1] such that

y = r. Note that 1 y(t) – = λ n



1

0



   G∗ (t, s)fn∗ s, sα–2 y(s) – φ(s) ds

1

≤λ 0

 1 + β1 η1α–2 H(s) +

    s(1 – s)α–1 fn∗ s, sα–2 y(s) – φ(s) ds, α–1 (α)(1 – β1 η1 )

then we have   y – 

 

 1      1 s(1 – s)α–1 α–2 ≤λ fn∗ s, sα–2 y(s) – φ(s) ds. 1 + β H(s) + η 1 1  α–1 n (α)(1 – β1 η1 ) 0

On the other hand, we have 1 y(t) – = λ n



1

0

≥ λMt

   G∗ (t, s)fn∗ s, sα–2 y(s) – φ(s) ds  1 H(s) + 0

  ≥ ωt  y – By the choice of n0 ,   y(t) ≥ ωt  y –

1 n

 1 . n



1 n0

    s(1 – s)α–1 fn∗ s, sα–2 y(s) – φ(s) ds α–1 (α)(1 – β1 η1 )

< r – C0 . Hence, for all t ∈ [0, 1], we have

   1  + 1 ≥ ωt y – 1 + 1 ≥ ωtr + (1 – ωt) 1 . n n n n n

Therefore,   1 1 1 1 + > y(t) – φ(t) ≥ ωtr + (1 – ωt) – ωtC0 ≥ ωt r – C0 – n n n n and 1 1 y(t) – φ(t) ≥ ωtr + (1 – ωt) – ωtC0 ≥ ωt(r – C0 ) + [1 – ωt] > ωt(r – C0 ). n n Thus we have from condition (H2 ), for all t ∈ [0, 1], 

1

y(t) = λ 0



   1 G∗ (t, s)fn∗ s, sα–2 y(s) – φ(s) ds + n

   1 G∗ (t, s)f ∗ s, sα–2 y(s) – φ(s) ds + n 0 

 1   s(1 – s)α–1 1 + β1 η1α–2 H(s) + ≤ (α)(1 – β1 η1α–1 ) 0 1



     1 × q(s) g y(s) – φ(s) + h y(s) – φ(s) ds + n

    1 α–1     s(1 – s) h(r) 1 1 + β1 η1α–2 H(s) + ) 1 + q(s)g ωs(r – C ds + ≤ 0 α–1 g(r) n (α)(1 – β1 η1 ) 0

Xu and Zhang Boundary Value Problems (2018) 2018:34



1

≤ 0

 1 + β1 η1α–2 H(s) +

Page 14 of 18

   s(1 – s)α–1 K0 q(s)g ω(r – C0 ) α–1 (α)(1 – β1 η1 )

  1 h(r) ds + × g(s) 1 + g(r) n     h(r) 1 b0 + . = K0 g ω(r – C0 ) 1 + g(r) n Therefore,

   h(r) 1 r = y ≤ K0 g ω(r – C0 ) 1 + b0 + . g(r) n 

This is a contradiction to the choice of n0 and the claim is proved. From this claim, the Leray–Schauder alternative principle guarantees that 

1

y(t) = 0

   1 G∗ (t, s)fn∗ s, sα–2 y(s) – φ(s) ds + n

has a fixed point, denoted by yn , in Br = {y ∈ E, y < r}. Next we claim that yn (t) – φ(t) has a uniform positive lower bound, i.e., there exists a constant δ > 0, independent of n ∈ N0 , such that   min yn (t) – φ(t) ≥ δt,

(3.26)

t∈[0,1]

for all n ∈ N0 . Since (H8 ) holds, there exists a continuous function ϕr (t) > 0 such that f ∗ (t, t α–2 u) > ϕr (t) > e(t) for all (t, u) ∈ (0, 1] × (0, r]. Since yn (t) – φ(t) < r, we have 

1

yn (t) – φ(t) = 0

 ≥

1

0

 ≥

0

1

   1 G∗ (t, s)fn∗ s, sα–2 yn (s) – φ(s) ds + – n



1

G∗ (t, s)e(s) ds

0

  G∗ (t, s) ϕr (s) – e(s) ds    s(1 – s)α–1 Mt H(s) + ϕr (s) – e(s) ds := δt. (α)(1 – β1 η1α–1 )

Next note {yn } is equicontinuous on [0, 1] for all n ∈ N0 .

(3.27)

Similarly reasoning as in the proof of Theorem 3.2 [18], we can easily verify it. The facts yn < r and (3.27) show that {yn }n∈N0 is a bounded and equicontinuous family on [0, 1]. Now the Arzela–Ascoli Theorem guarantees that {yn }n∈N0 has a subsequence {ynk }nk ∈N0 , converging uniformly on [0, 1] to a function y ∈ E. From the facts yn < r and (3.26), y satisfies δt ≤ y(t) – φ(t) < r for all t ∈ [0, 1]. Moreover, ynk satisfies the integral equation  ynk =

0

1

   1 G∗ (t, s)fn∗ s, sα–2 ynk (s) – φ(s) ds + . nk

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Letting k → ∞, we arrive at 

1

y=

   G∗ (t, s)f ∗ s, sα–2 y(s) – φ(s) ds.

0

Therefore, y is a positive solution of (3.19) and satisfies 0 < y < r. Therefore, t α–2 y(t) – γ (t) is a positive solution of (3.1). Thus (3.1) has at least two positive solutions u1 = t α–2 y1 (t)–γ (t), u2 (t) = t α–2 y(t)–γ (t).  It is easy to see that if e(t) ≡ 0, we can have a corollary for the singular positone boundary value problems. Here we omit it. Example 3.1 Consider the boundary value problem   Dα0+ u(t) + μ u–a (t) + ub (t) – At (2–α)a = 0, u(0) = 0,

u(1) =

m–2 

(3.28) βi u(ηi ),

i=1 1–a where 0 < a < α – 1 < 1 < b < 2–α , 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < m–2 α–1 ηm–2 < 1, i=1 βi ηi < 1, 0 < A < 1, μ ∈ (0, μ0 ) is such that

ωa μ20 < , (1 – μ0 c0 )a 2 d0

μ0
0, u2 (t) > 0 for t ∈ (0, 1). Proof We will apply Theorem 3.2. To this end we take   f (t, u) = μ u–a + ub – At (2–α)a , then     f t, t α–2 y = μ t a(2–α) y–a + t b(α–2) yb – At (2–α)a . Let q1 (t) = t –a(α–2) , q(t) = t b(α–2) and g(y) = g1 (y) = μy–a , h(y) = h1 (y) = μyb , e(t) = μAt (2–α)a , K0 = 1, ϕL (t) = μt (2–α)a L–a then (H1 )–(H4 ) and (H6 ) are satisfied since 0 < a < α – 1 < 1 < 1–a b < 2–α , 1 < α < 2.

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Also we have 

1

b0 = 0

1 + β1 η1α–2



H(s) +

 s(1 – s)α–1 q(s)g(s) ds (α)(1 – β1 η1α–1 )

(1 + β1 η1α–2 )(1 – a – b(2 – α)) (α + 1 – a – b(2 – α))  

α–a–b(2–α) 1 – a – b(2 – α) 1 – β1 η1α–a–b(2–α) 1 – m–2 i=1 βi ηi × + +  α–1 1 – β1 η1α–1 (1 – β1 η1α–1 )(2 – a – b(2 – α)) 1 – m–2 i=1 βi ηi



:= μd0 and C0 = =

1 + β1 η1α–2 M

 1 H(s) + 0

 (1 – s)α–1 e(s) ds (α)(1 – β1 η1α–1 )

μA(1 + β1 η1α–2 )(1 + a(2 – α)) M(α + 1 + a(2 – α))   α+(2–α)a 1 – β1 η1α+(2–α)a 1 – m–2 1 i=1 βi ηi × + +  α–1 1 – β1 η1α–1 1 – β1 η1α–1 1 – m–2 i=1 βi ηi

:= μc0 . Now the existence conditions (H5 ) and (H8 ) become μ2
C0 ,

and μt (2–α)a r–a > μAt (2–α)a . Thus, (H5 ) and (H8 ) hold with r = 1 since (3.29) and A < 1. Finally, (H7 ) becomes μ≥ 1 a

R1+a q1 (s)G∗ (σ , s) ds[1 + (ωεa)a+b Ra+b ]

.

Since b > 1, the right-hand side goes to 0 as R → ∞. Thus, Eq. (3.28) has at least two positive solutions, u1 , u2 with u1 (t) > 0, u2 (t) > 0 for t ∈ (0, 1). 

4 Conclusion Prompted by the application of multi-point boundary value problems to applied mathematics and physics, these problems have provoked a great deal of attention by many authors. In this paper, we considered the properties of Green’s function for the nonlinear

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fractional differential equation boundary value problem   Dα0+ u(t) + f t, u(t) = 0, u(0) = 0,

u(1) =

m–2 

0 < t < 1, βi u(ηi ),

i=1

 α–1 where 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < · · · < ηm–2 < 1, m–2 < 1, i=1 βi ηi α D0+ is the standard Riemann–Liouville derivative. Here our nonlinearity f may be singular at u = 0. Unlike the classical expression, we gave a new expression of the Green’s function and obtained some properties. As an application of Green’s function, we gave some multiple positive solutions for singular positone and semipositone boundary value problems by means of the Leray–Schauder nonlinear alternative, a fixed point theorem on cones. The results show that: (1) G(t, s) = G1 (t, s) + G2 (t, s),

G1 (t, s) =

⎧ [t(1–s)]α–1 –β1 t α–1 (η1 –s)α–1 –(t–s)α–1 (1–β1 η1α–1 ) ⎪ ⎪ , ⎪ (1–β1 η1α–1 )(α) ⎪ ⎪ α–1 –(t–s)α–1 (1–β ηα–1 ) ⎪ [t(1–s)] ⎪ 1 1 ⎨ , α–1 (1–β1 η1 )(α) [t(1–s)]α–1 –β1 t α–1 (η1 –s)α–1 ⎪ ⎪ , ⎪ (1–β1 η1α–1 )(α) ⎪ ⎪ ⎪ α–1 ⎪ [t(1–s)] ⎩ , (1–β1 η1α–1 )(α)

G2 (t, s) = H(s)t α–1 ,

where Ki = βi ηiα–1 , M0 =

H(s) = M0

(α)(1–

1 m–2

0 ≤ s ≤ t ≤ 1, s ≤ η1 , 0 < η1 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ η1 < 1, 0 ≤ t ≤ s ≤ 1, η1 ≤ s,

 m–2  K1 K i Ki qi (s) + pi (s) , 1 – K1 1 – K1 i=1

α–1 ) i=1 βi ηi

,

  s α–1 I[0≤s≤ηi ] , qi (s) = (1 – s)α–1 – 1 – ηi     s α–1 s α–1 pi (s) = 1 – I[0≤s≤ηi ] – 1 – I[0≤s≤η1 ] . ηi η1 (2) The Green’s function G(t, s) has the following properties: (i) H(s) ≥ 0, s(1–s)α–1 (ii) Mt[ (α)(1–β + H(s)] ≤ t 2–α G(t, s) ≤ ηα–1 ) 1 1

(1 + β1 η1α–2 )[H(s) + (iii)

s(1–s)α–1 Mt α–1 [ (α)(1–β α–1 1 η1 )

s(1–s)α–1 ] (α)(1–β1 η1α–1 )

for t, s ∈ [0, 1], (1–s)α–1 ] for (α)(1–β1 η1α–1 ) – 1), β1 η1α–1 } < 1.

+ H(s)] ≤ G(t, s) ≤ t α–1 [H(s) +

t, s ∈ [0, 1],

where 0 < M = min{1 – β1 η1α–1 , β1 η1α–2 (1 – η1 )(α (3) Suppose the conditions (H1 )–(H8 ) hold. Then (3.1) has at least two solutions, u1 , u2 with u1 (t) > 0, u2 (t) > 0 for t ∈ (0, 1). (4) The boundary value problem   Dα0+ u(t) + μ u–a (t) + ub (t) – At (2–α)a = 0, u(0) = 0,

u(1) =

m–2  i=1

βi u(ηi ),

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1–a where 0 < a < α – 1 < 1 < b < 2–α , 1 < α < 2, 0 < βi < 1, i = 1, 2, . . . , m – 2, 0 < η1 < η2 < m–2 α–1 · · · < ηm–2 < 1, i=1 βi ηi < 1, 0 < A < 1, has at least two positive solutions, u1 , u2 with u1 (t) > 0, u2 (t) > 0 for t ∈ (0, 1), μ ∈ (0, μ0 ), μ0 is some constant.

Funding The work was supported by National Natural Science Foundation of China (No. 11571207), Shandong Provincial Natural Science Foundation of China (No. Zr2015AL003), The Fundamental Research Funds for the central Universities (No. 16CX02015A, No. 15CX08011A, No. 17CX02049). Competing interests The authors declare that they have no competing financial interests. Authors’ contributions XX developed the study concept and design, analyzed and drafted the manuscript. HZ corrected some spelling mistakes and inaccuracies of the expressions. All authors read and approved the final manuscript.

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