Positive solutions for fractional differential equation at resonance under integral boundary conditions

Youyu Wang; Yue Huang; Xianfei Li

doi:10.1515/dema-2022-0026

Article Open Access

Positive solutions for fractional differential equation at resonance under integral boundary conditions

Youyu Wang , Yue Huang and Xianfei Li

Published/Copyright: July 6, 2022

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From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

By using the theory of fixed point index and spectral theory of linear operators, we study the existence of positive solutions for Riemann-Liouville fractional differential equations at resonance. Our approach will provide some new ideas for the study of this kind of problem.

Keywords: fractional differential equation; positive solution; resonance; fixed point index; Green’s function

MSC 2010: 34B10; 34B15

1 Introduction

In this article, we investigate the resonance boundary value problems for the Riemann-Liouville fractional differential equation

(1.1) ( D a + α u ) ( t ) + f ( t , u ( t ) ) = 0 , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s ,

where η ∈ ( a , b ) , α ( b − a ) α − 1 = δ ( η − a ) α , D a + α denotes the standard Riemann-Liouville derivative of order α , f : [ a , b ] × [ 0 , + ∞ ) → R is continuous.

The problem (1.1) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem

(1.2) ( D a + α u ) ( t ) = 0 , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s ,

has nontrivial solutions. Clearly, the resonant condition is α ( b − a ) α − 1 = δ ( η − a ) α .

In the last few decades, there have been numerous results in the study of the existence of positive solutions for boundary value problems involving fractional derivatives. Many authors focused on the study of existence of positive solutions to nonresonant fractional boundary value problems. It is well known that the existence of positive solutions to nonlinear fractional boundary value problems is very difficult when the resonant case is considered.

Recently, there are some papers dealing with the existence of solutions of fractional boundary value problem at resonance; the approach mainly includes Leray-Schauder fixed point theorem, coincidence degree theory, Krasnosel’skii fixed point theorem, and Leggett-Williams fixed point theorem, we refer the readers to [1,2, 3,4,5, 6,7,8, 9,10,11, 12,13] and references therein. For example, in [12], Wang and Liu studied the nonlocal fractional differential equation

(1.3) ( D 0 + α u ) ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = η u ( ξ ) ,

where 1 < α < 2 , 0 < ξ < 1 , η ξ α − 1 = 1 . The existence and uniqueness of positive solutions are obtained by means of the fixed point index theory and iterative techniques.

Wang [13] obtained necessary conditions for the existence of positive solutions to the following fractional boundary value problems at resonance

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

where 1 < α ≤ 2 , η i > 0 , 0 < ξ 1 < … < ξ m − 2 < 1 , ∑ i = 1 m − 2 η i ξ i α − 1 = 1 , f : [ 0 , 1 ] × [ 0 , + ∞ ) × R → R is continuous. The method is based on fixed point index theory and iterative technique.

As far as we know, there is no article to discuss the existence of positive solutions to fractional boundary value problems with integral boundary conditions in the resonant case. Motivated by the works [12,13], in this article, we will investigate the existence of positive solutions to resonant problem (1.1). Compared with the previous articles, this article includes the following features. First, the operator theory is used in the process of solving the problem. Second, the interval in the problem is extended from [ 0 , 1 ] to the general interval [ a , b ] .

The rest of this article is organized as follows. In Section 2, we give some fundamental concepts and lemmas. In Section 3, the existence and uniqueness results of positive solutions are obtained. In Section 4, we give some sufficient conditions for the existence of positive solutions. In Section 5, we give an example to illustrate the application of our main results.

2 Basic definitions and preliminaries

In this section, we will recall some of the necessary definitions and results that will be used in the main results.

Definition 2.1

[14] Let α ≥ 0 and f be a real function defined on [ a , b ] . The Riemann-Liouville fractional integral of order α is defined by ( I a + 0 f ) ≡ f and

( I a + α f ) ( t ) = 1 Γ ( α ) ∫ a t ( t − s ) α − 1 f ( s ) d s , α > 0 , t ∈ [ a , b ] .

Definition 2.2

[14] The Riemann-Liouville fractional derivative of order α ≥ 0 is defined by ( D a + 0 f ) ≡ f and

( D a + α f ) ( t ) = ( D m I a + m − α f ) ( t ) = 1 Γ ( m − α ) d d t m ∫ a t ( t − s ) m − α − 1 f ( s ) d s ,

for α > 0 , where m is the smallest integer greater or equal to α .

Lemma 2.3

[14] If α ≥ 0 and β > 0 , then

I a + α ( t − a ) β − 1 = Γ ( β ) Γ ( β + α ) ( t − a ) β + α − 1 .

Lemma 2.4

[14] Let α > 0 and n be a positive integer such that n − 1 < α ≤ n , L ( a , b ) denotes the space of Lebesgue integrable and measurable real functions on ( a , b ) . If u ∈ C ( a , b ) ∩ L ( a , b ) , then

I a + α D a + α u ( t ) = u ( t ) + c 1 ( t − a ) α − 1 + c 2 ( t − a ) α − 2 + ⋯ + c n ( t − a ) α − n ,

for some c i ∈ R , i = 1 , 2 , … , n .

Definition 2.5

The Mittag-Leffler function is defined by:

E α ( z ) ≔ ∑ k = 0 ∞ z k Γ ( α k + 1 ) , α > 0 .

The two-parameter Mittag-Leffler function is defined by:

E α , β ( z ) ≔ ∑ k = 0 ∞ z k Γ ( α k + β ) , α > 0 .

From [15], we have [ x ν − 1 E τ , ν ( λ x τ ) ] ( m ) = x ν − 1 − m E τ , ν − m ( λ x τ ) . For brevity, set

F c ( t ) = t c − 1 E α , c ( λ t α ) for c ≥ 0 and t ≥ 0 ,

so, we have F c ( m ) ( t ) = F c − m ( t ) for m = 1 , 2 , 3 , … . It can be concluded that

F α ( t ) = t α − 1 E α , α ( λ t α ) = t α − 1 ∑ k = 0 ∞ λ k Γ ( α k + α ) t k α , F α ′ ( t ) = F α − 1 ( t ) = t α − 2 E α , α − 1 ( λ t α ) = t α − 2 ∑ k = 0 ∞ λ k Γ ( α k + α − 1 ) t k α , F α ″ ( t ) = F α − 2 ( t ) = t α − 3 E α , α − 2 ( λ t α ) = t α − 3 ∑ k = 0 ∞ λ k Γ ( α k + α − 2 ) t k α .

In this article, we use the following notations:

G ( t , s ) = 1 F α ( b − a ) F α ( t − a ) F α ( b − s ) − F α ( b − a ) F α ( t − s ) , a ≤ s ≤ t ≤ b , F α ( t − a ) F α ( b − s ) , a ≤ t ≤ s ≤ b , K ( t , s ) = G ( t , s ) + δ F α ( t − a ) F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ∫ a η G ( t , s ) d t , p ( t ) = α − 2 Γ ( α − 1 ) + ∑ k = 1 ∞ ( b − a ) k α Γ ( α k + α − 2 ) t k .

Now, we list some properties of the function p which shall be useful. It is easy to verify that p ′ ( t ) > 0 on ( 0 , + ∞ ) , and p ( 0 ) = α − 2 Γ ( α − 1 ) < 0 , lim t → + ∞ p ( t ) = + ∞ ; therefore, there exists a unique λ ∗ > 0 such that p ( λ ∗ ) = 0 . We here list the assumptions to be used throughout the article.

( H 1 ) λ ∈ ( 0 , λ ∗ ] is a constant.

( H 2 ) f : [ a , b ] × [ 0 , + ∞ ) → R is continuous and f ( t , x ) + λ x ≥ 0 .

Lemma 2.6

[16] For any λ ∈ R , α > 0 , we have the following results.

For any r ∈ C ( [ a , b ] , R ) , series ∑ k = 0 ∞ λ k I a + k α r ( t ) is convergent and the sum is
∑ k = 0 ∞ λ k I a + k α r ( t ) = r ( t ) + λ ∫ a t ( t − s ) α − 1 E α , α [ λ ( t − s ) α ] r ( s ) d s .
The operator I − λ I a + α : C ( [ a , b ] , R ) → C ( [ a , b ] , R ) is invertible and:
( I − λ I a + α ) − 1 r ( t ) = ∑ k = 0 ∞ λ k I a + k α r ( t ) .

Lemma 2.7

Let σ ∈ L [ a , b ] . Then the boundary value problem

(2.1) − ( D a + α u ) ( t ) + λ u ( t ) = σ ( t ) , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s ,

has the unique solution

u ( t ) = ∫ a b K ( t , s ) σ ( s ) d s .

Proof

Applying I a + α to both sides of equation (2.1), we have:

− I a + α ( D a + α u ) ( t ) + λ I a + α u ( t ) = I a + α σ ( t ) .

From Lemma 2.4,

− u ( t ) + c 0 ( t − a ) α − 2 + c 1 ( t − a ) α − 1 + λ I a + α u ( t ) = I a + α σ ( t ) .

Since u ( a ) = 0 , we obtain immediately that c 0 = 0 , thus

− u ( t ) + c 1 ( t − a ) α − 1 + λ I a + α u ( t ) = I a + α σ ( t ) , ( ( I − λ I a + α ) u ) ( t ) = c 1 ( t − a ) α − 1 − I a + α σ ( t ) .

by Lemmas 2.3 and 2.6, we have

u ( t ) = ( I − λ I a + α ) − 1 ( c 1 ( t − a ) α − 1 ) − ( I − λ I a + α ) − 1 I a + α σ ( t ) = c 1 ∑ k = 0 ∞ λ k I a + k α ( t − a ) α − 1 − ∑ k = 0 ∞ λ k I a + k α I a + α σ ( t ) = c 1 ∑ k = 0 ∞ λ k Γ ( α ) Γ ( k α + α ) ( t − a ) k α + α − 1 − ∑ k = 0 ∞ λ k I a + k α + α σ ( t ) = c 1 Γ ( α ) ( t − a ) α − 1 ∑ k = 0 ∞ λ k ( t − a ) k α Γ ( k α + α ) − ∫ a t ∑ k = 0 ∞ λ k ( t − s ) k α + α − 1 Γ ( k α + α ) σ ( s ) d s = c 1 Γ ( α ) ( t − a ) α − 1 E α , α [ λ ( t − a ) α ] − ∫ a t ( t − s ) α − 1 E α , α [ λ ( t − s ) α ] σ ( s ) d s = c 1 Γ ( α ) F α ( t − a ) − ∫ a t F α ( t − s ) σ ( s ) d s .

The boundary condition u ( b ) = δ ∫ a η u ( s ) d s yields

c 1 Γ ( α ) F α ( b − a ) − ∫ a b F α ( b − s ) σ ( s ) d s = δ ∫ a η u ( s ) d s ,

we obtain

c 1 = 1 Γ ( α ) F α ( b − a ) ∫ a b F α ( b − s ) σ ( s ) d s + δ Γ ( α ) F α ( b − a ) ∫ a η u ( s ) d s ,

therefore,

u ( t ) = c 1 Γ ( α ) F α ( t − a ) − ∫ a t F α ( t − s ) σ ( s ) d s = F α ( t − a ) F α ( b − a ) ∫ a b F α ( b − s ) σ ( s ) d s − ∫ a t F α ( t − s ) σ ( s ) d s + δ F α ( t − a ) F α ( b − a ) ∫ a η u ( s ) d s = ∫ a b G ( t , s ) σ ( s ) d s + δ F α ( t − a ) F α ( b − a ) ∫ a η u ( s ) d s .

Integrating both sides of the aforementioned equation from a to η , we obtain

∫ a η u ( t ) d t = ∫ a η ∫ a b G ( t , s ) σ ( s ) d s d t + δ F α ( b − a ) ∫ a η F α ( t − a ) d t ⋅ ∫ a η u ( s ) d s ;

note the fact that

F α ( b − a ) − δ ∫ a η F α ( t − a ) d t = ∑ k = 0 ∞ λ k Γ ( α k + α ) ( b − a ) k α + α − 1 − δ ∑ k = 0 ∞ λ k Γ ( α k + α ) ( η − a ) k α + α k α + α = ( b − a ) α − 1 ∑ k = 0 ∞ λ k Γ ( α k + α ) ( b − a ) k α − ( η − a ) k α k + 1 > 0 ,

so,

∫ a η u ( s ) d s = F α ( b − a ) F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ∫ a η ∫ a b G ( t , s ) σ ( s ) d s d t ,

therefore,

u ( t ) = ∫ a b G ( t , s ) σ ( s ) d s + δ F α ( t − a ) F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ∫ a η ∫ a b G ( t , s ) σ ( s ) d s d t = ∫ a b G ( t , s ) σ ( s ) d s + δ F α ( t − a ) F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ∫ a b σ ( s ) ∫ a η G ( t , s ) d t d s = ∫ a b G ( t , s ) + δ F α ( t − a ) F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ∫ a η G ( t , s ) d t σ ( s ) d s = ∫ a b K ( t , s ) σ ( s ) d s .

□

Lemma 2.8

Suppose that ( H 1 ) holds. The function K ( t , s ) has the following properties:

G ( t , s ) > 0 , for t , s ∈ ( a , b ) ;
ω 2 ( s ) t − a b − a α − 1 ≤ K ( t , s ) ≤ ω 1 ( s ) t − a b − a α − 1 , for t , s ∈ [ a , b ] , where
ω 1 = F α ( b − s ) + δ F α ( b − a ) ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t , ω 2 = δ ( b − a ) α − 1 ∫ a η G ( t , s ) d t Γ ( α ) [ F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ] .

Proof

(1). It is easy to verify the following results, for a < t ≤ b ,

F α ′ ( t − a ) = ( t − a ) α − 2 ∑ k = 0 ∞ λ k Γ ( α k + α − 1 ) ( t − a ) k α > 0 , F α ″ ( t − a ) = ( t − a ) α − 3 α − 2 Γ ( α − 1 ) + ∑ k = 1 ∞ λ k ( t − a ) k α Γ ( α k + α − 2 ) = ( t − a ) α − 3 p λ ( t − a ) α ( b − a ) α < ( t − a ) α − 3 p ( λ ) ≤ ( t − a ) α − 3 p ( λ ∗ ) = 0 ,

so, F α ′ ( t ) > 0 , F α ″ ( t ) < 0 for t > 0 . When a < t ≤ s < b , noticing F α ( 0 ) = 0 , and the monotonicity of F α ( t ) , it is clear that F α ( t − a ) F α ( b − s ) > 0 . When a < s ≤ t < b , we have

∂ ∂ s [ F α ( t − a ) F α ( b − s ) − F α ( b − a ) F α ( t − s ) ] = F α ( b − a ) F α ′ ( t − s ) − F α ( t − a ) F α ′ ( b − s ) ≥ F α ( b − a ) F α ′ ( b − s ) − F α ( t − a ) F α ′ ( b − s ) = F α ′ ( b − s ) [ F α ( b − a ) − F α ( t − a ) ] .

Integrating the above inequality from a to s , we obtain

F α ( t − a ) F α ( b − s ) − F α ( b − a ) F α ( t − s ) ≥ ∫ a s F α ′ ( b − τ ) [ F α ( b − a ) − F α ( t − a ) ] d τ = [ F α ( b − a ) − F α ( t − a ) ] [ F α ( b − a ) − F α ( b − s ) ] > 0 ,

so, we obtain G ( t , s ) > 0 for t , s ∈ ( a , b ) .

(2). It is easy to check that

( t − a ) α − 1 Γ ( α ) ≤ F α ( t − a ) = ( t − a ) α − 1 E α , α [ λ ( t − a ) α ] = ( t − a ) α − 1 ∑ k = 0 ∞ λ k Γ ( α k + α ) ( t − a ) k α ≤ ( t − a ) α − 1 ∑ k = 0 ∞ λ k Γ ( α k + α ) ( b − a ) k α = t − a b − a α − 1 F α ( b − a ) ,

so,

0 ≤ G ( t , s ) ≤ F α ( t − a ) F α ( b − s ) F α ( b − a ) ≤ t − a b − a α − 1 F α ( b − s ) ,

K ( t , s ) = G ( t , s ) + δ ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t F α ( t − a ) ≤ t − a b − a α − 1 F α ( b − s ) + δ ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t t − a b − a α − 1 F α ( b − a ) = F α ( b − s ) + δ F α ( b − a ) ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t t − a b − a α − 1 = ω 1 ( s ) t − a b − a α − 1 , K ( t , s ) = G ( t , s ) + δ ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t F α ( t − a ) ≥ δ ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t F α ( t − a ) ≥ ( t − a ) α − 1 Γ ( α ) ⋅ δ ∫ a η G ( t , s ) d t F α ( b − a ) − δ ∫ a η F α ( t − a ) d t = ω 2 ( s ) t − a b − a α − 1 .

This completes the proof.□

Let E = C [ a , b ] be endowed with the maximum norm ‖ u ‖ = max a ≤ t ≤ b ∣ u ( t ) ∣ , θ is the zero element of E , B r = { u ∈ E : ‖ u ‖ < r } . Define a cone P by

P = { u ∈ E : u ( t ) ≥ 0 , t ∈ [ a , b ] } .

Obviously, problem (1.1) is equivalent to

− ( D a + α u ) ( t ) + λ u ( t ) = λ u ( t ) + f ( t , u ( t ) ) , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s .

Let

(2.2) Au ( t ) = ∫ a b K ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) ] d s ,

(2.3) Tu ( t ) = ∫ a b K ( t , s ) u ( s ) d s .

Clearly, by Lemma 2.7, the fixed point of the operator A is a solution of problem (1.1).

Lemma 2.9

The operator A : P → P is completely continuous.

Proof

The operator A transforms P into itself in view of nonnegativeness of K ( t , s ) and f ( s , u ) + λ u ( s ) . Next, we divide the proof into three steps.

Step 1. A u ( t ) is continuous with respect to u ( t ) ∈ P .

Suppose that { u n ( t ) } is a sequence in P , and { u n ( t ) } converges to u ( t ) ∈ P . Because of f ( t , x ) being continuous with respect to x ∈ [ 0 , + ∞ ) , then, for any positive number ε , there exists an integer N . When n > N , we have

(2.4) ‖ f ( s , u n ( s ) ) − f ( s , u ( s ) ) ‖ + λ ∗ ‖ u n ( s ) − u ( s ) ‖ < ε ∫ a b ω 1 ( s ) d s .

It follows from (2.2) and (2.4) that

‖ ( A u n ) ( t ) − ( A u ) ( t ) ‖ = ∫ a b K ( t , s ) [ f ( s , u n ( s ) ) + λ u n ( s ) ] d s − ∫ a b K ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) ] d s = ∫ a b K ( t , s ) [ f ( s , u n ( s ) ) − f ( s , u ( s ) ) ] d s + λ ∫ a b K ( t , s ) [ u n ( s ) − u ( s ) ] d s ≤ ∫ a b K ( t , s ) d s ( ‖ f ( s , u n ( s ) ) − f ( s , u ( s ) ) ‖ + λ ∗ ‖ u n ( s ) − u ( s ) ‖ ) ≤ ∫ a b ω 1 ( s ) d s ( ‖ f ( s , u n ( s ) ) − f ( s , u ( s ) ) ‖ + λ ∗ ‖ u n ( s ) − u ( s ) ‖ ) < ε .

Thus, the operator A is continuous in P .

Step 2. A maps a bounded set in P into a bounded set.

Assume that D ∈ P is a bounded set with ‖ u ( t ) ‖ ≤ M for any u ∈ D . Let L = max 0 ≤ t ≤ 1 , 0 ≤ u ≤ M ∣ f ( t , u ) ∣ + λ ∗ M , then we have

‖ ( A u ) ( t ) ‖ = ∫ a b K ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) ] d s ≤ L ∫ a b K ( t , s ) d s ≤ L ∫ a b ω 1 ( s ) d s .

This implies that the operator A maps a bounded set into a bounded set in P .

Step 3. A is equicontinuous in P .

It suffices to show that for any u ( t ) ∈ D and any a < t 1 < t 2 < b , A u ( t 1 ) → A u ( t 2 ) as t 1 → t 2 . Since D is bounded, then there exists Q > 0 such that ∣ f ( s , u ) + λ u ∣ ≤ Q , then for any f ∈ D , we have

‖ ( A u ) ( t 2 ) − ( A u ) ( t 1 ) ‖ ≤ ∫ a b ∣ K ( t 2 , s ) − K ( t 1 , s ) ∣ ∣ f ( s , u ( s ) ) + λ u ( s ) ∣ d s ≤ Q ∫ a b ∣ K ( t 2 , s ) − K ( t 1 , s ) ∣ d s ≤ Q ∫ a b ∣ G ( t 2 , s ) − G ( t 1 , s ) ∣ d s + δ Q ∣ F α ( t 2 − a ) − F α ( t 1 − a ) ∣ F α ( b − a ) − δ ∫ a η F α ( t − a ) d t ∫ a b ∫ a η G ( t , s ) d t d s .

By the uniform continuity of G ( t , s ) for any ( t , s ) ∈ [ a , b ] × [ a , b ] and the continuity of F α ( t − a ) on [ a , b ] , we obtain

‖ ( A u ) ( t 2 ) − ( A u ) ( t 1 ) ‖ → 0 ( as t 2 → t 1 ) .

By means of the Arzela-Ascoli theorem, we have that A : P → P is completely continuous. The proof is complete.□

Lemma 2.10

The operator T : P → P is completely continuous.

Proof

The proof is similar to that of Lemma 2.9, so we omit it here.□

Lemma 2.11

[17] Suppose that T : E → E is a completely continuous linear operator and T ( P ) ⊂ P . If there exist ψ ∈ E ⧹ ( − P ) and a constant c > 0 such that c T ψ ≥ ψ , then the spectral radius r ( T ) ≠ 0 and T has a positive eigenfunction φ 1 corresponding to its first eigenvalue λ 1 = ( r ( T ) ) − 1 .

Lemma 2.12

Suppose T is defined by (2.3), then the spectral radius r ( T ) > 0 and T has a positive eigenfunction φ 1 corresponding to its first eigenvalue λ 1 = ( r ( T ) ) − 1 .

Proof

By Lemma 2.8, K ( t , s ) > 0 for all s , t ∈ ( a , b ) . Take [ t 1 , t 2 ] ⊂ ( a , b ) and ψ ∈ C [ a , b ] such that ψ ( t ) ≥ 0 , ∀ t ∈ [ a , b ] ; ψ ( t ) > 0 , ∀ t ∈ ( t 1 , t 2 ) , and ψ ( t ) = 0 , ∀ t ∉ ( t 1 , t 2 ) . Then for t ∈ [ t 1 , t 2 ] ,

T ψ ( t ) = ∫ a b K ( t , s ) ψ ( s ) d s ≥ ∫ t 1 t 2 K ( t , s ) ψ ( s ) d s > 0 .

So, there exists a constant c > 0 such that c T ψ ≥ ψ , ∀ t ∈ [ a , b ] . By Lemma 2.11, we complete the proof.□

Since λ = 0 is the eigenvalue of the linear problems

( D a + α u ) ( t ) + λ u = 0 , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s ,

and ( t − a ) α − 1 is the corresponding eigenfunction, i.e., the solution of system

( D a + α u ) ( t ) = 0 , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s ,

or equivalent

− ( D a + α u ) ( t ) + λ u ( t ) = λ u ( t ) , a < t < b , 1 < α ≤ 2 , u ( a ) = 0 , u ( b ) = δ ∫ a η u ( s ) d s ,

can be written as (by Lemma 2.7)

u ( t ) = ∫ a b K ( t , s ) ( λ u ( s ) ) d s = λ ∫ a b K ( t , s ) u ( s ) d s = λ T u ( t ) .

So, we have the following result.

Lemma 2.13

Suppose that ( H 1 ) holds, then the first eigenvalue of T is λ 1 = λ , and φ 1 ( t ) = ( t − a ) α − 1 is the positive eigenfunction corresponding to λ 1 , i.e., φ 1 = λ T φ 1 .

Lemma 2.14

[18] Let P be a cone in a Banach space E , and Ω be a bounded open set in E. Suppose that A : Ω ¯ ∩ P → P is a completely continuous operator. If there exists u 0 ∈ P with u 0 ≠ θ such that

u − A u ≠ λ u 0 , ∀ λ ≥ 0 , u ∈ ∂ Ω ∩ P ,

then the fixed point index i ( A , Ω ∩ P , P ) = 0 .

Lemma 2.15

[18] Let P be a cone in a Banach space E , and Ω be a bounded open set in E . Suppose that A : Ω ¯ ∩ P → P is a completely continuous operator. If

A u ≠ λ u , ∀ λ ≥ 1 , u ∈ ∂ Ω ∩ P ,

then the fixed point index i ( A , Ω ∩ P , P ) = 1 .

3 The uniqueness result

Theorem 3.1

Assume that (H1), (H2) hold and there exists μ ∈ ( 0 , λ ) such that

∣ f ( t , u ) + λ u − f ( t , v ) − λ v ∣ ≤ μ ∣ u − v ∣ , for t ∈ [ a , b ] , u , v ∈ [ 0 , ∞ ) ,

then (1.1) has a unique nonnegative solution.

Proof

We divided our proof into two parts.

(I). We first prove the operator A has fixed point in P . Set

Q = u ∈ P : ∃ L 1 , L 2 > 0 such that L 2 t − a b − a α − 1 ≤ u ( t ) ≤ L 1 t − a b − a α − 1 .

For any u ∈ P \ { θ } , define

L 1 ( u ) = ∫ a b ω 1 ( s ) u ( s ) d s , L 2 ( u ) = ∫ a b ω 2 ( s ) u ( s ) d s .

By Lemma 2.8, it is easy to check that L 1 ( u ) , L 2 ( u ) > 0 , and

L 2 ( u ) t − a b − a α − 1 ≤ ( T u ) ( t ) ≤ L 1 ( u ) t − a b − a α − 1 ,

this means that

T : P \ { θ } → Q .

For any u 0 ∈ P \ { θ } , define

u n = A ( u n − 1 ) , n = 1 , 2 , …

We may assume that u 1 − u 0 ≠ θ (otherwise the theorem holds). Then L 1 ( ∣ u 1 − u 0 ∣ ) > 0 , and

T ( ∣ u 1 − u 0 ∣ ) ≤ L 1 ( ∣ u 1 − u 0 ∣ ) t − a b − a α − 1 = L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 ( b − a ) α − 1 .

Therefore,

∣ u 2 − u 1 ∣ = ∫ a b K ( t , s ) [ f ( s , u 1 ( s ) ) + λ u 1 ( s ) − f ( s , u 0 ( s ) ) − λ u 0 ( s ) ] d s ≤ ∫ a b K ( t , s ) ∣ f ( s , u 1 ( s ) ) + λ u 1 ( s ) − f ( s , u 0 ( s ) ) − λ u 0 ( s ) ∣ d s ≤ μ ∫ a b K ( t , s ) ∣ u 1 ( s ) − u 0 ( s ) ∣ d s = μ T ( ∣ u 1 − u 0 ∣ ) ≤ μ L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 ( b − a ) α − 1 = μ ( b − a ) α − 1 L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 ,

∣ u 3 − u 2 ∣ = ∫ a b K ( t , s ) [ f ( s , u 2 ( s ) ) + λ u 2 ( s ) − f ( s , u 1 ( s ) ) − λ u 1 ( s ) ] d s ≤ ∫ a b K ( t , s ) ∣ f ( s , u 2 ( s ) ) + λ u 2 ( s ) − f ( s , u 1 ( s ) ) − λ u 1 ( s ) ∣ d s ≤ μ ∫ a b K ( t , s ) ∣ u 2 ( s ) − u 1 ( s ) ∣ d s ≤ μ 2 ( b − a ) α − 1 ∫ a b K ( t , s ) L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 d s = μ 2 ( b − a ) α − 1 L 1 ( ∣ u 1 − u 0 ∣ ) T φ 1 = μ 2 λ ( b − a ) α − 1 L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 , … … … … … …

By induction, we can obtain

∣ u n + 1 − u n ∣ ≤ 1 ( b − a ) α − 1 μ λ n − 1 μ L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 .

Then, for any n , m ∈ N , we have

∣ u n + m − u m ∣ ≤ ∣ u n + m − u n + m − 1 ∣ + ∣ u n + m − 1 − u n + m − 2 ∣ + ⋯ + ∣ u m + 1 − u m ∣ ≤ 1 ( b − a ) α − 1 μ λ n + m − 2 + μ λ n + m − 3 + ⋯ + μ λ m − 1 μ L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 ≤ 1 ( b − a ) α − 1 μ λ m − 1 1 − μ λ μ L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 = μ m L 1 ( ∣ u 1 − u 0 ∣ ) λ m − 2 ( λ − μ ) ⋅ φ 1 ( b − a ) α − 1 .

So,

‖ u n + m − u m ‖ ≤ μ m L 1 ( ∣ u 1 − u 0 ∣ ) λ m − 2 ( λ − μ ) → 0 , m → ∞ ,

this means { u n } is a Cauchy sequence, there is a point u ∗ ∈ P , such that lim n → ∞ u n = u ∗ and

u ∗ = lim n → ∞ u n = lim n → ∞ A ( u n − 1 ) = A ( u ∗ ) ,

u ∗ is a fixed point of A .

(II). We prove the fixed point of A is unique.

Suppose v ∗ ≠ u ∗ be another fixed point of A. Then L 1 ( ∣ u ∗ − v ∗ ∣ ) > 0 and

T ( ∣ u ∗ − v ∗ ∣ ) ≤ L 1 ( ∣ u ∗ − v ∗ ∣ ) t − a b − a α − 1 = L 1 ( ∣ u ∗ − v ∗ ∣ ) φ 1 ( b − a ) α − 1 .

Similar to the proof in the part (I), we have

∣ A u ∗ − A v ∗ ∣ = ∫ a b K ( t , s ) [ f ( s , u ∗ ( s ) ) + λ u ∗ ( s ) − f ( s , v ∗ ( s ) ) − λ v ∗ ( s ) ] d s ≤ ∫ a b K ( t , s ) ∣ f ( s , u ∗ ( s ) ) + λ u ∗ ( s ) − f ( s , v ∗ ( s ) ) − λ v ∗ ( s ) ∣ d s ≤ μ ∫ a b K ( t , s ) ∣ u ∗ ( s ) − v ∗ ( s ) ∣ d s = μ T ( ∣ u ∗ − v ∗ ∣ ) ≤ μ L 1 ( ∣ u ∗ − v ∗ ∣ ) φ 1 ( b − a ) α − 1 = μ ( b − a ) α − 1 L 1 ( ∣ u ∗ − v ∗ ∣ ) φ 1 .

By induction, we can obtain

∣ A n u ∗ − A n v ∗ ∣ ≤ 1 ( b − a ) α − 1 μ λ n − 1 μ L 1 ( ∣ u 1 − u 0 ∣ ) φ 1 .

So,

‖ u ∗ − v ∗ ‖ = ‖ A n u ∗ − A n v ∗ ‖ ≤ μ λ n − 1 μ L 1 ( ∣ u 1 − u 0 ∣ ) → 0 , n → ∞ .

Therefore, the fixed point of A is unique.□

4 Existence of positive solutions

Theorem 4.1

Assume that ( H 1 ), ( H 2 ), and the following assumptions hold:

(4.1) liminf u → 0 + min t ∈ [ a , b ] f ( t , u ) u > 0 ,

(4.2) limsup u → + ∞ max t ∈ [ a , b ] f ( t , u ) u < 0 .

Then (1.1) has at least one positive solution.

Proof

It follows from (4.1) that there exists r 1 > 0 such that

(4.3) f ( t , u ) ≥ 0 , ∀ ( t , u ) ∈ [ a , b ] × [ 0 , r 1 ] .

Thus, for any u ∈ ∂ B r 1 ∩ P , we obtain

(4.4) A u ( t ) = ∫ a b K ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) ] d s ≥ λ ∫ a b K ( t , s ) u ( s ) d s = λ T u ( t ) .

Suppose that A has no fixed point on ∂ B r 1 ∩ P (otherwise, the proof is completed). Now we show that

(4.5) u − A u ≠ ν φ 1 , ∀ u ∈ ∂ B r 1 ∩ P , ν ≥ 0 .

If not, then there exist u 1 ∈ ∂ B r 1 ∩ P and ν 0 > 0 such that u 1 − A u 1 = ν 0 φ 1 . Then

u 1 = A u 1 + ν 0 φ 1 ≥ ν 0 φ 1 .

Let

ν ∗ = sup { ν ∣ u 1 ≥ ν φ 1 } .

It is easy to see that ν ∗ ≥ ν 0 and u 1 ≥ ν ∗ φ 1 . Since T ( P ) ⊂ P , we have λ T u 1 ≥ ν ∗ λ T φ 1 = ν ∗ φ 1 . Then

u 1 = A u 1 + ν 0 φ 1 ≥ λ T u 1 + ν 0 φ 1 ≥ ( ν ∗ + ν 0 ) φ 1 ,

which contradicts the definition of ν ∗ . Hence, (4.5) holds and we see from Lemma 2.14 that

(4.6) i ( A , B r 1 ∩ P , P ) = 0 .

On the other hand, it follows from (4.2) that there exist 0 < σ < 1 and r 2 > r 1 such that

(4.7) f ( t , u ) ≤ ( σ − 1 ) λ u , ∀ t ∈ [ a , b ] , u ≥ r 2 .

Let T 1 u = σ λ T u , then T 1 : C [ a , b ] → C [ a , b ] is a bounded linear operator and T 1 ( P ) ⊂ P . Let

(4.8) W = { u ∈ P ∣ u = ν A u , 0 ≤ ν ≤ 1 } .

In the following, we will prove that W is bounded.

For any u ∈ W , set u ˜ ( t ) = min { u ( t ) , r 2 } . Then

f ( t , u ( t ) ) + λ u ( t ) ≤ σ λ u ( t ) + f ( t , u ˜ ( t ) ) + λ u ˜ ( t ) .

Therefore,

u ( t ) = ν A u ( t ) ≤ A u ( t ) ≤ σ λ T u ( t ) + A u ˜ ( t ) ≤ T 1 u ( t ) + M ,

where

M = max ( t , u ) ∈ [ a , b ] × [ 0 , r 2 ] ( f ( t , u ) + λ u ) ∫ a b ω 1 ( s ) d s .

Thus, ( I − T 1 ) u ( t ) ≤ M , t ∈ [ a , b ] . Noticing λ is the first eigenvalue of T and 0 < σ < 1 , we have ( r ( T 1 ) ) − 1 = σ − 1 > 1 . Therefore, the inverse operator ( I − T 1 ) − 1 exists and

( I − T 1 ) − 1 = I + T 1 + T 1 2 + ⋯ + T 1 n + ⋯ .

It follows from T 1 ( P ) ⊂ P that ( I − T 1 ) − 1 ( P ) ⊂ P . So we have

u ( t ) ≤ ( I − T 1 ) − 1 M , t ∈ [ a , b ] ,

and W is bounded.

Choose r 3 > max { r 2 , ‖ ( I − T 1 ) − 1 M ‖ } . Then by Lemma 2.15, we have

(4.9) i ( A , B r 3 ∩ P , P ) = 1 .

By (4.6) and (4.9), we have

i ( A , ( B r 3 ⧹ B ¯ r 1 ) ∩ P , P ) = i ( A , B r 3 ∩ P , P ) − i ( A , B r 1 ∩ P , P ) = 1 ,

which implies that A has at least one fixed point on ( B r 3 ⧹ B ¯ r 1 ) ∩ P . This means that problem (1.1) has at least one positive solution. This completes the proof.□

Theorem 4.2

Assume that ( H 1 ), ( H 2 ), and the following assumptions hold:

(4.10) limsup u → 0 + max t ∈ [ a , b ] f ( t , u ) u < 0 ,

and f ( t , 0 ) ≢ 0 on [ a , b ] . Then (1.1) has at least one positive solution.

Proof

It follows from (4.10) that there exists r 4 > 0 such that

(4.11) f ( t , u ) ≤ 0 , ∀ ( t , u ) ∈ [ a , b ] × [ 0 , r 4 ] .

Define T 2 u = λ T u , then r ( T 2 ) = 1 .

Suppose that A has no fixed point on ∂ B r 4 ∩ P (otherwise, the proof is completed). Next we will prove that

(4.12) A u ≠ ν u , ∀ u ∈ ∂ B r 4 ∩ P , ν > 1 .

If not, then there exist u 1 ∈ ∂ B r 4 ∩ P and ν 0 > 1 such that A u 1 = ν 0 u 1 . Then ν 0 u 1 = A u 1 ≤ T 2 u 1 , and

ν 0 n u 1 ≤ T 2 n u 1 , n = 1 , 2 , …

So,

ν 0 n u 1 ≤ T 2 n u 1 ≤ ‖ T 2 n ‖ ‖ u 1 ‖ .

Therefore, r ( T 2 ) = lim n → + ∞ ‖ T 2 n ‖ n ≥ ν 0 > 1 , which contradicts r ( T 2 ) = 1 . Then by Lemma 2.15, we have

(4.13) i ( A , B r 4 ∩ P , P ) = 1 .

Since f ( t , 0 ) ≢ 0 on [ a , b ] and A θ ≠ θ . So (4.13) implies that problem (1.1) has at least one positive solution.□

5 An example

Example 5.1

Consider the following resonant problem:

(5.1) ( D 0 + 3 2 u ) ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 12 ∫ 0 1 4 u ( s ) d s .

Since

p ( t ) = − 1 2 π + ∑ k = 1 ∞ t k Γ 3 2 k − 1 2 = − 1 2 π + t + ∑ k = 2 ∞ t k Γ 3 2 k − 1 2 ≤ − 1 2 π + t + ∑ k = 2 ∞ t k Γ ( k ) = − 1 2 π + t e t ,

and − 1 2 π ≈ − 0.282 , 1 5 e 1 5 ≈ 0.244 , we have p 1 5 < 0 , thus λ ∗ > 1 5 .

Let

f ( t , x ) = 1 4 ( 1 + t ) ( 1 − 2 λ ) x , ( t , x ) ∈ [ 0 , 1 ] × [ 0 , 4 ) , 1 2 ( 1 + t ) ( x − λ x ) , ( t , x ) ∈ [ 0 , 1 ] × [ 4 , + ∞ ) ,

where λ ∈ 0 , 1 5 . It is clear that ( H 1 ) and ( H 2 ) hold. Moreover,

liminf u → 0 + min t ∈ [ 0 , 1 ] f ( t , u ) u = 1 − 2 λ 4 > 0 , limsup u → + ∞ max t ∈ [ 0 , 1 ] f ( t , u ) u = − λ < 0 .

Therefore, the assumptions of Theorem 4.1 are satisfied. Thus, Theorem 4.1 ensures that (5.1) has at least one positive solution.

6 Conclusion

In this article, by using the theory of fixed point index and spectral theory of linear operator, we study the existence of positive solutions for Riemann-Liouville fractional differential equations at resonance. We provide a general method to solve this kind of problem (Lemmas 2.6 and 2.7), our approach will provide some new ideas for the study of other boundary value problems.

Acknowledgements

The authors would like to thank the handling editor and the referees for their helpful comments and suggestions.

Funding information: This work was supported by the Tianjin Natural Science Foundation (grant no. 20JCYBJC00210).
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2022-01-17

Revised: 2022-05-11

Accepted: 2022-06-03

Published Online: 2022-07-06

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2022-0026

Keywords for this article

fractional differential equation; positive solution; resonance; fixed point index; Green’s function

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