In this paper, we consider the following fractional differential equation:

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = ∑ i = 1 m − 2 η i D 0 + β u ( ξ i ) ,

where 1 < α ≤ 2 , 0<β<α−1, 0<ξ1<⋯<ξm−2<1 with ∑i=1m−2ηiξiα−β−1<1, D0+α is the standard Riemann-Liouville derivative, f ∈ C ( ( 0 , 1 ) × ( 0 , + ∞ ) → [ 0 , + ∞ ) ) may be singular at t=0, t=1, and u=0. In this paper, by a positive solution to (1.1), we mean a function u ∈ C [ 0 , 1 ] which satisfies D 0 + α u ∈ L ( 0 , 1 ) , positive on ( 0 , 1 ] and satisfies (1.1).

Recently, many results were obtained dealing with the existence of solutions for nonlinear fractional differential equations by using the techniques of nonlinear analysis; see 1234567891011121314151617181920212223 and references therein. The multi-point boundary value problems (BVP for short) have provoked a great deal of attention, for example 13141516171819. In 10, the authors discussed some positive properties of the Green function for Direchlet-type BVP of nonlinear fractional differential equation

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0 ,

where 1 < α < 2 , D0+α is the standard Riemann-Liouville derivative, f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) → [ 0 , + ∞ ) ) . By using the Krasnosel’skii fixed point theorem, the existence of positive solutions were obtained under suitable conditions on f.

In 14, the authors investigated the existence and multiplicity of positive solutions by using some fixed point theorems for the fractional differential equation

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = a D 0 + β u ( ξ ) ,

where 1 < α ≤ 2 , 0 ≤ β ≤ 1 , 0 < ξ < 1 , 0 ≤ a ≤ 1 with a ξ α − β − 2 < 1 − β , 0 ≤ α − β − 1 , f : [ 0 , 1 ] × [ 0 , + ∞ ) → [ 0 , + ∞ ) satisfied Carathéodory type conditions.

In 2021, the authors considered the fractional differential equation given by

{ D 0 + α u ( t ) + f ( t , u , u ′ , … , u ( n − 2 ) ) = 0 , 0 < t < 1 , n − 1 < α ≤ n , n ≥ 2 , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , u ( n − 2 ) ( 1 ) = 0 .

In order to obtain the existence of positive solutions of (1.4), they considered the following fractional differential equation:

{ D 0 + α − n + 2 v ( t ) + f ( t , I 0 + n − 2 v ( t ) , I 0 + n − 3 v ( t ) , … , I 0 + 1 v ( t ) , v ( t ) ) = 0 , 0 < t < 1 , v ( 0 ) = v ( 1 ) = 0 .

In 20, f = q ( t ) ( g + h ) , and g, h have different monotone properties. By using the fixed point theorem for the mixed monotone operator, Zhang obtained (1.4) and had a unique positive solution u ( t ) = I 0 + n − 2 v ( t ) with v ∈ Q = : { x ( t ) : 1 M t α − n + 1 ≤ x ( t ) ≤ M t α − n + 1 } . But the results are not true since v ( t ) is a positive solution of (1.5), and v ( 1 ) = 0 . What causes it lies in the unsuitable using of properties of the Green function.

In 21, f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × R n − 2 → [ 0 , + ∞ ) ) , f ( t , y 1 , y 2 , … , y n − 1 ) is increasing for y i ≥ 0 , i = 1 , 2 , … , n − 1 . By using the positive properties of the Green function obtained in [ 10 ] and fixed point theory for the u 0 concave operator, the authors obtained the uniqueness of a positive solution for the BVP (1.4).

Motivated by the works mentioned above, in this paper we aim to establish the existence and uniqueness of a positive solution to the BVP (1.1). Our work presented in this paper has the following features. Firstly, the BVP (1.1) possesses singularity, that is, f may be singular at t=0, t=1, and u=0. Secondly, we impose weaker positivity conditions on the nonlocal boundary term, that is, some of the coefficients η i can be negative. Thirdly, the unique positive solution can be approximated by an iterative scheme.

The rest of the paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. We also develop some new positive properties of the Green function. In Section 3, we discuss the existence and uniqueness of a positive solution of the BVP (1.1), we also give an example to demonstrate the application of our theoretical results.

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in recent literature.

Definition 2.1 The fractional integral of order α > 0 of a function u : ( 0 , + ∞ ) → R is given by

I 0 + α u ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 u ( s ) d s

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

Definition 2.2 The fractional derivative of order α>0 of a continuous function u:(0,+∞)→R is given by

D 0 + α u ( t ) = 1 Γ ( n − α ) ( d d t ) n ∫ 0 t ( t − s ) n − α − 1 u ( s ) d s ,

where n = [ α ] + 1 , [ α ] denotes the integral part of the number α, provided the right-hand side is pointwisely defined on (0,+∞).

Definition 2.3 By u ∈ L ( 0 , 1 ) , we mean ∫ 0 1 | u ( t ) | d t < ∞ .

Lemma 2.1 (3)

Let α>0. Then the following equality holds for u∈L(0,1), D0+αu∈L(0,1),

I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α − 1 + c 2 t α − 2 + ⋅ ⋅ ⋅ + c n t α − n ,

where c i ∈ R , i = 1 , 2 , … , n , n − 1 < α ≤ n .

Set

where

q ( s ) = p ( s ) − p ( 0 ) Γ ( α ) p ( 0 ) ( 1 − s ) α − β − 1 , p ( 0 ) = 1 − ∑ i = 1 m − 2 η i ξ i α − β − 1 .

For the convenience in presentation, we here list the assumption to be used throughout the paper.

( H 1 ) p ( 0 ) > 0 , q ( s ) ≥ 0 on [ 0 , 1 ] .

Remark 2.1 If η i = 0 (i = 1 , … , m − 2 ), we have p ( 0 ) = 1 and q ( s ) ≡ 0 . If η i ≥ 0 (i = 1 , … , m − 2 ) and ∑i=1m−2ηiξiα−β−1<1, we have p ( 0 ) > 0 and q(s)≥0 on [0,1].

Lemma 2.2 (14)

Assume that g ∈ L ( 0 , 1 ) and α > 1 ≥ β ≥ 0 . Then

D 0 + β ∫ 0 t ( t − s ) α − 1 g ( s ) d s = Γ ( α ) Γ ( α − β ) ∫ 0 t ( t − s ) α − β − 1 g ( s ) d s .

Lemma 2.3 Assume (H1) holds, and y ∈ L ( 0 , 1 ) . Then the unique solution of the problem

{ D 0 + α u ( t ) + y ( t ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = ∑ i = 1 m − 2 η i D 0 + β u ( ξ i ) ,

is

u ( t ) = ∫ 0 1 G ( t , s ) y ( s ) d s ,

where G ( t , s ) is called the Green function of BVP (2.5).

Proof From Lemma 2.1, we have the solution of (2.5) given by

u ( t ) = − I 0 + α y ( t ) + c 1 t α − 1 + c 2 t α − 2 .

Consequently,

u ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + c 1 t α − 1 + c 2 t α − 2 .

From u ( 0 ) = 0 , we have c 2 = 0 .

By Lemma 2.2, we have

D 0 + β u ( t ) = − 1 Γ ( α − β ) ∫ 0 t ( t − s ) α − β − 1 y ( s ) d s + c 1 Γ ( α ) Γ ( α − β ) t α − β − 1 .

Therefore,

D 0 + β u ( 1 ) = − 1 Γ ( α − β ) ∫ 0 1 ( 1 − s ) α − β − 1 y ( s ) d s + c 1 Γ ( α ) Γ ( α − β ) ,

and

D 0 + β u ( ξ i ) = − 1 Γ ( α − β ) ∫ 0 ξ i ( ξ i − s ) α − β − 1 y ( s ) d s + c 1 Γ ( α ) Γ ( α − β ) ξ i α − β − 1 .

By D 0 + β u ( 1 ) = ∑ i = 1 m − 2 η i D 0 + β u ( ξ i ) , we have

c 1 = ∫ 0 1 ( 1 − s ) α − 1 y ( s ) d s − ∑ i = 1 m − 2 η i ∫ 0 ξ i ( ξ i − s ) α − β − 1 y ( s ) d s Γ ( α ) p ( 0 ) = ∫ 0 1 ( 1 − s ) α − β − 1 p ( s ) y ( s ) d s Γ ( α ) p ( 0 ) .

Therefore, the solution of (2.5) is

u ( t ) = c 1 t α − 1 − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s = ∫ 0 1 G ( t , s ) y ( s ) d s .

 □

Lemma 2.4 The function G 0 ( t , s ) has the following properties:

(1) G 0 ( t , s ) > 0 , for t , s ∈ ( 0 , 1 ) ;

(2) Γ ( α ) G 0 ( t , s ) ≤ t α − 1 , for t , s ∈ [ 0 , 1 ] ;

(3) β t α − 1 h ( s ) ≤ Γ ( α ) G 0 ( t , s ) ≤ h ( s ) t α − 2 , for t,s∈(0,1), where

h ( s ) = s ( 1 − s ) α − β − 1 .

Proof It is obvious that (1), (2) hold. In the following, we will prove (3).

(i) When 0 < s ≤ t < 1 , noticing 0 < β < α − 1 ≤ 1 , we have

Therefore,

t α − 2 s ( 1 − s ) α − β − 1 − ( t α − 1 ( 1 − s ) α − β − 1 − ( t − s ) α − 1 ) ≥ t α − 2 s − t α − 1 + ( t − s ) α − 1 = − t α − 2 ( t − s ) + ( t − s ) α − 1 ≥ 0 ,

which implies

Γ ( α ) G 0 ( t , s ) ≤ h ( s ) t α − 2 .

On the other hand, we have

∂ ∂ s { β s + ( 1 − s ) β } ≤ 0 , s ∈ [ 0 , 1 ) .

Therefore, β s + ( 1 − s ) β ≤ 1 , which implies

[ 1 − ( 1 − s ) β ] ≥ β s .

Then

Γ ( α ) G 0 ( t , s ) = t α − 1 ( 1 − s ) α − β − 1 − ( t − s ) α − 1 ≥ t α − 1 ( 1 − s ) α − β − 1 − ( t − s ) β ( t − t s ) α − β − 1 = [ 1 − ( 1 − s t ) β ] t α − 1 ( 1 − s ) α − β − 1 ≥ [ 1 − ( 1 − s ) β ] t α − 1 ( 1 − s ) α − β − 1 ≥ β t α − 1 h ( s ) .

(ii) When 0 < t ≤ s < 1 , we have

Γ ( α ) G 0 ( t , s ) = t α − 1 ( 1 − s ) α − β − 1 = t α − 2 t ( 1 − s ) α − β − 1 ≤ t α − 2 s ( 1 − s ) α − β − 1 = h ( s ) t α − 2 .

On the other hand, clearly we have

Γ ( α ) G 0 ( t , s ) = t α − 1 ( 1 − s ) α − β − 1 ≥ β t α − 1 h ( s ) .

(2.8)-(2.11) implies (3) holds. □

By Lemma 2.4 we have the following results.

Lemma 2.5 Assume (H1) holds, then the Green function defined by (2.3) satisfies:

(1) G ( t , s ) > 0 , ∀ t , s ∈ ( 0 , 1 ) ;

(2) G ( t , s ) ≤ t α − 1 ( 1 Γ ( α ) + q ( s ) ) , ∀ t , s ∈ [ 0 , 1 ] ;

(3) β t α − 1 Φ ( s ) ≤ G ( t , s ) ≤ t α − 2 Φ ( s ) , ∀t,s∈(0,1), where

Φ ( s ) = h ( s ) Γ ( α ) + q ( s ) .

Lemma 2.6 Assume (H1) holds, then the function G ∗ ( t , s ) = : t 2 − α G ( t , s ) satisfies:

(1) G ∗ ( t , s ) > 0 , ∀t,s∈(0,1);

(2) G ∗ ( t , s ) ≤ t ( 1 Γ ( α ) + q ( s ) ) , ∀t,s∈[0,1];

(3) β t Φ ( s ) ≤ G ∗ ( t , s ) ≤ Φ ( s ) , ∀ t , s ∈ [ 0 , 1 ] .

For convenience, we list here two more assumptions to be used later:

( H 2 ) f ( t , u ) = g ( t , u , u ) , here g ∈ C ( ( 0 , 1 ) × [ 0 , + ∞ ) × ( 0 , + ∞ ) → [ 0 , + ∞ ) ) , g ( t , u , v ) is nondecreasing on u, nonincreasing on v, and there exists μ ∈ ( 0 , 1 ) such that

g ( t , r u , v r ) ≥ r μ g ( t , u , v ) , ∀ u , v > 0 , r ∈ ( 0 , 1 ) .

( H 3 ) 0 < ∫ 0 1 g ( s , s α − 1 , s α − 1 ) d s < + ∞ .

Remark 2.2 Inequality (2.12) is equivalent to

g ( t , u r , r v ) ≤ r − μ g ( t , u , v ) , ∀ u , v > 0 , r ∈ ( 0 , 1 ) .

Let E = C [ 0 , 1 ] be endowed with the maximum norm ∥ u ∥ = max 0 ≤ t ≤ 1 | u ( t ) | . Define a cone P by

P = { u ∈ E :  there exists  l u > 0  such that  β ∥ u ∥ t ≤ u ( t ) ≤ l u t } .

Let

A ( u , v ) ( t ) = ∫ 0 1 G ∗ ( t , s ) g ( s , s α − 2 u ( s ) , s α − 2 v ( s ) ) d s .

Set Q = P ∖ { θ } , where θ is the zero element of E. We have the following lemma.

Lemma 2.7 Suppose that (H1)-(H3) hold. Then A : Q × Q → Q .

Proof For any u , v ∈ Q , there exists l 1 , l 2 > 0 , such that

β ∥ u ∥ t ≤ u ( t ) ≤ l 1 t , β ∥ v ∥ t ≤ v ( t ) ≤ l 2 t .

By (H2), (H3) and (2) of Lemma 2.6, we get

A ( u , v ) ( t ) = ∫ 0 1 G ∗ ( t , s ) g ( s , s α − 2 u ( s ) , s α − 2 v ( s ) ) d s ≤ t ∫ 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , s α − 2 u ( s ) , s α − 2 v ( s ) ) d s ≤ t ∫ 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , l 1 s α − 1 , β ∥ v ∥ s α − 1 ) d s ≤ t ∫ 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , ( 1 + l 1 ) s α − 1 , β ∥ v ∥ ( 1 + β ) ( 1 + ∥ v ∥ ) s α − 1 ) d s ≤ L 1 t ∫ 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , s α − 1 , s α − 1 ) d s < + ∞ ,

where L 1 = max { ( 1 + l 1 ) μ , ( ( 1 + β ) ( 1 + ∥ v ∥ ) β ∥ v ∥ ) μ } . This implies that A is well defined in Q × Q .

On the other hand, by (3) of Lemma 2.6, we have

Therefore, A ( u , v ) ( t ) ≥ β ∥ A ( u , v ) ∥ t . Combining with (2.16), we have A:Q×Q→Q. □

Remark 2.3 By (H2) and (2.15), A is a mixed monotone operator.

Theorem 3.1 Suppose that (H1)-(H3) hold. Then the BVP (1.1) has a unique positive solution.

Proof For any r ∈ ( 0 , 1 ) , by Remark 2.2, we have

A ( u r , r v ) ≤ r − μ A ( u , v ) , ∀ u , v ∈ Q .

For any w ∈ Q , noticing A ( w , w ) ∈ Q , we can choose r 0 ∈ ( 0 , 1 ) small enough such that

r 0 1 − μ w ≤ A ( w , w ) ≤ r 0 − ( 1 − μ ) w .

Set

u 0 = r 0 w , v 0 = r 0 − 1 w .

Clearly,

u 0 , v 0 ∈ Q , u 0 ≤ v 0 .

Let

u n = A ( u n − 1 , v n − 1 ) , v n = A ( v n − 1 , u n − 1 ) , n = 1 , 2 , … .

It is easy to see that

u 0 ≤ u 1 ≤ ⋯ ≤ u n ≤ ⋯ ≤ v n ≤ ⋯ ≤ v 1 ≤ v 0 .

Noticing

u 1 = A ( r 0 w , r 0 − 1 w ) ≥ r 0 μ A ( w , w ) , v 1 = A ( r 0 − 1 w , r 0 w ) ≤ r 0 − μ A ( w , w ) ,

therefore,

u 1 ≥ r 0 2 μ v 1 .

Suppose that u n ≥ r 0 2 μ n v n , then v n ≤ r 0 − 2 μ n u n , and

u n + 1 = A ( u n , v n ) ≥ A ( r 0 2 μ n v n , r 0 − 2 μ n u n ) ≥ r 0 2 μ n + 1 A ( v n , u n ) .

By induction, we can get

u n ≥ r 0 2 μ n v n , n = 1 , 2 , … .

By (3.4), (3.5), we have

0 ≤ u n + m − u n ≤ v n − u n ≤ ( 1 − r 0 2 μ n ) v n ≤ ( 1 − r 0 2 μ n ) v 0 ,

which implies { u n } is a Cauchy sequence. Similarly, { v n } is a Cauchy sequence. Noticing (3.4), there exist u ∗ , v ∗ ∈ Q , such that {un} converges to u ∗ and {vn} converges to v ∗ . Moreover,

u n ≤ u ∗ ≤ v ∗ ≤ v n , n = 1 , 2 , … .

(3.5) and (3.6) imply that

∥ v ∗ − u ∗ ∥ ≤ ∥ v n − u n ∥ ≤ ( 1 − r 0 μ n ) ∥ v 0 ∥ , n = 1 , 2 , … .

This implies that u ∗ = v ∗ .

By the mixed monotone property of A and (3.6), we have

A ( u ∗ , v ∗ ) ≥ A ( u n , v n ) = u n + 1 , A ( v ∗ , u ∗ ) ≤ A ( v n , u n ) = v n + 1 , n = 1 , 2 , … .

Let n → + ∞ , we get

u ∗ ≤ A ( u ∗ , v ∗ ) = A ( v ∗ , u ∗ ) ≤ v ∗ .

Since u∗=v∗, we have u∗ is a positive fixed point of A.

In the following, we will prove the positive fixed point of A is unique.

Suppose u ≠ u ∗ is a positive fixed point of A. By Lemma 2.6, we can get u ∈ Q . Let

r 1 = sup { r ∈ ( 0 , 1 ) : r u ∗ ≤ u ≤ r − 1 u ∗ } .

Then 0 < r 1 < 1 , and r 1 u ∗ ≤ u ≤ r 1 − 1 u ∗ . Therefore

u = A ( u , u ) ≥ A ( r 1 u ∗ , r 1 − 1 u ∗ ) ≥ r 1 μ A ( u ∗ , u ∗ ) = r 1 μ u ∗ , u = A ( u , u ) ≤ A ( r 1 − 1 u ∗ , r 1 u ∗ ) ≤ r 1 − μ A ( u ∗ , u ∗ ) = r 1 − μ u ∗ .

Thus, r 1 μ u ∗ ≤ u ≤ r 1 − μ u ∗ , which contradicts the definition of r 1 . Consequently, the positive fixed point of A is unique.

It is clear that y ( t ) = t α − 2 u ∗ ( t ) satisfies

y ( t ) = ∫ 0 1 G ( t , s ) g ( s , y ( s ) , y ( s ) ) d s = ∫ 0 1 G ( t , s ) f ( s , y ( s ) ) d s , t ∈ ( 0 , 1 ] .

On the other hand, since u ∗ ∈ Q , we have β ∥ u ∗ ∥ t ≤ u ∗ ( t ) ≤ l u ∗ t . Then, β ∥ u ∗ ∥ t α − 1 ≤ y ( t ) ≤ l u ∗ t α − 1 . By Lemma 2.5 and (H2), (H3), we can get g ( t , y ( t ) , y ( t ) ) ∈ L ( 0 , 1 ) . Moreover,

lim t → 0 y ( t ) = lim t → 0 ∫ 0 1 G ( t , s ) g ( s , y ( s ) , y ( s ) ) d s ≤ t α − 1 ∫ 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , y ( s ) , y ( s ) ) d s = 0 .

Lemma 2.3 implies y(t)=tα−2u∗(t) is a positive solution of (1.1).

On the other hand, if y ( t ) is a positive solution of (1.1), then

y ( t ) = ∫ 0 1 G ( t , s ) g ( s , y ( s ) , y ( s ) ) d s .

By Lemma 2.5, we have there exists l1,l2>0 such that

l 2 t α − 1 ≤ y ( t ) ≤ l 1 t α − 1 .

Set u ( t ) = t 2 − α y ( t ) , we have

l 2 t ≤ u ( t ) ≤ l 1 t ,

and

u ( t ) = ∫ 0 1 G ∗ ( t , s ) g ( s , s α − 2 u ( s ) , s α − 2 v ( s ) ) d s ,

which implies u is a positive fixed point of A.

Then y(t)=tα−2u∗(t) is the unique positive solution of the BVP (1.1). □

Remark 3.1 The unique positive solution y of (1.1) can be approximated by the iterative schemes: for any w∈Q, let u0, v 0 be defined as (3.2) and u n = A ( u n − 1 , v n − 1 ) , v n = A ( v n − 1 , u n − 1 ) , n = 1 , 2 , …  , then t α − 2 u n → y .

Example 3.1 (A 4-point BVP with coefficients of both signs)

Consider the following problem:

{ D 0 + 7 4 u ( t ) + f ( t , u ( t ) ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = 0 , D 0 + 1 4 u ( 1 ) = D 0 + 1 4 u ( 1 4 ) − 1 2 D 0 + 1 4 u ( 4 9 )

with

f ( t , x ) = − x 1 2 ln t − x − 1 3 ln ( 1 − t ) + x − 1 3 ln ( 1 + x 1 2 ) .

Then

G 0 ( t , s ) = 1 Γ ( 7 4 ) { t 3 4 ( 1 − s ) 1 2 , 0 ≤ t ≤ s ≤ 1 , t 3 4 ( 1 − s ) 1 2 − ( t − s ) 3 4 , 0 ≤ s ≤ t ≤ 1

and

p ( s ) = { 1 − ( 1 4 − s 1 − s ) 1 2 − 1 2 ( 4 9 − s 1 − s ) 1 2 , 0 ≤ s ≤ 1 4 , 1 − 1 2 ( 4 9 − s 1 − s ) 1 2 , 1 4 < s ≤ 4 9 , 1 , 4 9 < s ≤ 1 .

By direct calculations, we have p ( 0 ) = 5 6 and q(s)≥0, which implies (H1) holds.

Let

g ( t , x , y ) = − x 1 2 ln t − y − 1 3 ln ( 1 − t ) + y − 1 3 ln ( 1 + x 1 2 ) .

Obviously, g ∈ C ( ( 0 , 1 ) × [ 0 , + ∞ ) × ( 0 , + ∞ ) → [ 0 , + ∞ ) ) , g ( t , x , y ) is nondecreasing on x and nonincreasing on y. It is easy to see that

ln ( 1 + r x ) ≥ r ln ( 1 + x ) , ∀ x ≥ 0 , r ∈ ( 0 , 1 ) .

Then

g ( t , r x , y r ) ≥ r 5 6 g ( t , x , y ) , ∀ x , y > 0 , r ∈ ( 0 , 1 ) .

Therefore (H2) holds. It is easy to get that (H3) holds. Therefore, the assumptions of Theorem 3.1 are satisfied. Thus Theorem 3.1 ensures that the BVP (3.8) has a unique positive solution.

The authors declare that they have no competing interests.

The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.