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Article

Multiple and Nonexistence of Positive Solutions for a Class of Fractional Differential Equations with p-Laplacian Operator

1
School of Mathematical Sciences, Qufu Normal University, Jining 273165, China
2
Department of Mathematics and Statistics, Missouri S&T, Rolla, MO 65409-0020, USA
*
Author to whom correspondence should be addressed.
Mathematics 2024, 12(23), 3869; https://doi.org/10.3390/math12233869
Submission received: 3 October 2024 / Revised: 18 November 2024 / Accepted: 6 December 2024 / Published: 9 December 2024
(This article belongs to the Special Issue Recent Research on Fractional Calculus: Theory and Applications)

Abstract

:
Research about multiple positive solutions for fractional differential equations is very important. Based on some outstanding results reported in this field, this paper continue the focus on this topic. By using the properties of the Green function and generalized Avery–Henderson fixed point theorem, we derive three positive solutions of a class of fractional differential equations with a p-Laplacian operator. We also study the nonexistence of positive solutions to the eigenvalue problem of the equation. Three examples are given to illustrate our main result.

1. Introduction

In recent years, the application and research of fractional calculus equations in the field of science and engineering have shown a thriving trend. The fractional order viscoelastic theory provides a novel and accurate mathematical tool for the study of the mechanical properties of new materials. The emergence of fractional-order models has also injected new vitality into the study of non-Newtonian fluid mechanics. The introduction of fractional calculus provides suitable mathematical tools for complex diffusion problems such as seepage in geotechnical engineering, and migration of nuclear substances and pollutants in the bottom layer. This has led to an increasing number of authors studying fractional differential equations. In [1,2,3,4,5,6,7,8,9,10,11,12,13,14], the authors studied p-Laplacian fractional differential equations. In [15,16,17], the authors studied fractional differential equations on infinite intervals.
In [1], the authors studied the problem
D 0 + α ( ϕ p ( D 0 + β u ( t ) ) ) = h ( t , u ( t ) ) , 0 t 1 , u ( 0 ) = 0 , u ( 1 ) = a D 0 + γ u ( ξ ) , D 0 + β u ( 0 ) = 0 , D 0 + β u ( 1 ) = b D 0 + β u ( η ) ,
where ϕ p ( v ) = | v | p 2 v , p > 1 . The authors derived two positive solutions from the above problem by applying the monotone iterative method.
In [2], the authors studied the problem
D 0 + α ( φ p ( D 0 + α x ( t ) ) ) = f ( t , x ( t ) , D 0 + α x ( t ) ) , 0 t 1 , x ( i ) ( 0 ) = 0 , [ φ p ( D 0 + α x ) ] i ( 0 ) = 0 , i = 0 , 1 , 2 , , n 2 , [ D 0 + β x ( t ) ] t = 1 = 0 , 0 < β α 1 , [ D 0 + β ( φ p ( D 0 + α x ( t ) ) ) ] t = 1 = 0 ,
where φ p ( v ) = | v | p 2 v , p > 1 . By employing the Legget–Williams fixed point theorem, the authors found three positive solutions for the above problem.
In [3], the author investigated the fractional differential problem
D 0 + γ [ ϕ p ( c D 0 + δ v ( s ) ) ] + g ( s , v ( s ) ) = 0 , s [ 0 , 1 ] , ϕ p ( c D 0 + δ v ( 0 ) ) = [ ϕ p ( c D 0 + δ v ( 0 ) ) ] = ϕ p ( c D 0 + δ v ( 1 ) ) = 0 , v ( 0 ) = v ( 1 ) = 0 , l 1 v ( 0 ) + l 2 v ( 0 ) = 0 1 h ( s ) v ( s ) d s ,
where 2 < δ 3 , 2 < γ 3 and 5 < δ + γ 6 , l 1 , l 2 0 , ; D 0 + γ is the Riemann–Liouville fractional derivative, D 0 + δ c is the Caputo fractional derivative; h : 0 , 1 0 , with h L 1 0 , 1 C ( 0 , 1 ) , g C ( 0 , 1 × 0 , , 0 , ) ; ϕ p ( v ) = | v | p 2 v , p > 1 . The author of [3] gives two positive solutions of (1) by using the Avery–Henderson fixed point theorem.
Based on the work mentioned above, in this work, we study the FBVP (1) and eigenvalue problem (19). Compared with [1,2], the FBVP (1) studies the integral BVP. In [1,2], they just study the Riemann–Liouville fractional derivative. Moreover, in [2], the derivative of order α inside the p-Laplacian operator is the same as that outside the p-Laplacian operator but in our paper, the order of the fractional derivative inside the p-Laplacian can be different from the one outside the p-Laplacian operator, and we investigate two different types of fractional derivatives. Compared with [1,2,3], we investigate the nonexistence of positive solutions.
Now, we want to find three positive solutions of the FBVP (1) and the nonexistence of the positive solution of the FBVP (1). We do not use the Legget–Williams fixed point theorem (from [2]) anymore. In our paper, we use the generalized Avery–Henderson fixed point theorem (from [15]) to find three positive solutions of the FBVP (1). Moreover, the nonexistence of a positive solution in terms of different eigenvalue intervals is also obtained.
In this article, we assume that the following conditions always hold:
(H1)
0 1 h ( s ) d s > 0 , 0 1 s h ( s ) d s > 0 , l 2 > l 1 > 0 1 h ( s ) d s ;
(H2)
g ( s , 0 ) 0 for all s [ 0 , 1 ] .

2. Preliminaries and Lemmas

The definitions for fractional-order integration and differentiation can be found in [18].
Lemma 1
([3]). The FBVP (1) is tantamount to the equation
v ( s ) = c 0 + c 1 s + 1 Γ ( δ ) 0 s ( s r ) δ 1 Ψ ( r ) d r ,
where
Ψ ( r ) = ϕ q 0 1 Θ ( r , ζ ) g ( ζ , v ( ζ ) ) d ζ ,
Θ ( r , τ ) = 1 Γ ( γ ) ( r r ζ ) γ 1 ( r ζ ) γ 1 , 0 ζ r 1 , ( r r ζ ) γ 1 , 0 r ζ 1 ,
c 1 = 1 Γ ( δ 1 ) 0 1 ( 1 r ) δ 2 Ψ ( r ) d r ,
c 0 = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r ,
ϕ q ( r ) in (3) is the inverse function of ϕ p ( r ) , i . e . , ϕ q ( r ) = | r | q 2 r , 1 p + 1 q = 1 , in (6)
A 1 = l 1 0 1 h ( s ) d s , A 2 = l 2 0 1 s h ( s ) d s ,
and from ( H 1 ) , we know A 2 > A 1 .
Lemma 2
([19]).  Θ ( r , ζ ) in (4) satisfies
(i) 
Θ ( r , ζ ) 0 ;
(ii) 
r γ 1 ( 1 r ) ζ ( 1 ζ ) γ 1 Γ ( γ ) Θ ( r , ζ ) ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) for r , ζ 0 , 1 .
Let E = C 0 , 1 with v = max s [ 0 , 1 ] v ( s ) , define the operator A : E E by
A v ( s ) = c 0 + c 1 s + 1 Γ ( δ ) 0 s ( s r ) δ 1 Ψ ( r ) d r ,
where Ψ ( r ) , c 1 , and c 0 are defined by (3), (5), and (6), respectively. Therefore, v is a solution of the FBVP (1) tantamount to the statement that v is a fixed point of A.
Remark 1.
In [3], the author proved that A v ( s ) is nonincreasing and nonnegative, where A is defined by (8).
Theorem 1.
(Generalized Avery–Henderson fixed point theorem [15]). Let E be a Banach space and K be a cone of E, ψ , σ , ϑ : K R + are three nonnegative, continuous functionals; additionally ψ , σ : K R + are increasing, such that for some constants D > 0 , 0 < k < a , ψ ( θ ) < k , σ ( v ) ϑ ( v ) ψ ( v ) and v D σ ( v ) for all v K ( σ , a ) ¯ = v K | σ ( v ) < a ¯ = v K | σ ( v ) a . Moreover, suppose there exist a completely continuous operator A : K ( σ , a ) ¯ K and a constant c > 0 with 0 < k < c < a such that ϑ ( λ v ) λ ϑ ( v ) for λ 0 , 1 and v K ( ϑ , r ) , and
(i) 
σ ( A v ) < a for all v K ( σ , a ) ,
(ii) 
ϑ ( A v ) > c for all v K ( ϑ , c ) ,
(iii) 
ψ ( A v ) < k for all v K ( ψ , k ) .
Then, A has at least three fixed points v 1 , v 2 , v 3 K ( σ , a ) ¯ such that
0 ψ ( v 1 ) < k < ψ ( v 2 ) , ϑ ( v 2 ) < c < ϑ ( v 3 ) , σ ( v 3 ) < a .
Remark 2.
In Theorem 1, ψ ( θ ) < k K ( ψ , k ) . The reasons are as follows: if ψ ( θ ) < k , then θ K ( ψ , k ) , so K ( ψ , k ) ; however, since K ( ψ , k ) is nonempty, we have ψ ( v 0 ) < k , v 0 K , and ψ is an increasing functional, we obtain ψ ( θ ) ψ ( v 0 ) < k .
Remark 3.
If we assume A θ θ in Theorem 1, then A has at least three positive fixed points v 1 , v 2 , v 3 K ( σ , a ) ¯ such that
0 < ψ ( v 1 ) < k < ψ ( v 2 ) , ϑ ( v 2 ) < r < ϑ ( v 3 ) , σ ( v 3 ) < a .
Now, we introduce a cone
K = v : v E , min s [ 0 , 1 ] v ( s ) 1 l v ,
where
l = A 2 A 2 A 1 > 1 ,
A 1 , A 2 are defined by (7).
Lemma 3.
A : K K is a completely continuous operator.
Proof. 
From Lemma 2, we know Θ is continuous, and g and ϕ q are also continuous, and then from (8), we obtain that A is a continuous operator.
Firstly, we prove A : K K . For all v K , from Remark 1, (8) and (9), we have
min s [ 0 , 1 ] | A v ( s ) | = A v ( 1 ) = c 0 + c 1 + 1 Γ ( δ ) 0 1 ( 1 r ) δ 1 Ψ ( r ) d r = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s +   ( A 2 A 1 ) ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r ( l A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s +   ( A 2 A 1 ) ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r = l 1 ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s + A 2 ( A 1 Γ ( δ 1 ) ) 1 × 0 1 ( 1 r ) δ 2 Ψ ( r ) d r = l 1 A v ( 0 ) = l 1 max s [ 0 , 1 ] A v ( s ) = l 1 A v .
Therefore, A : K K .
Secondly, we choose a bounded set Ω K . Then we have a positive number D such that v D , v Ω , which implies that | v ( s ) | D , s [ 0 , 1 ] . From the continuity of g, there exists N 1 > 0 satisfying 0 g ( s , v ) N 1 for ( s , v ) [ 0 , 1 ] × [ 0 , D ] . Then from (3) and Lemma 2, we have
Ψ ( r ) = ϕ q 0 1 Θ ( r , ζ ) g ( ζ , v ( ζ ) ) d ζ ϕ q 0 1 ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) N 1 d ζ = N 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 : = N 2 ,
where B ( 2 , γ ) = 0 1 ζ ( 1 ζ ) γ 1 d ζ . Thus, from ( H 1 ) , Remark 1, (8) and (10), we have
A v = max s [ 0 , 1 ] | A v ( s ) | = A v ( 0 ) = c 0 = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r N 2 ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 d r d s + A 2 N 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 d r < .
Therefore, A ( Ω ) is uniformly bounded.
Finally, we prove A ( Ω ) is equicontinuous. For all v Ω , s 1 , s 2 [ 0 , 1 ] , from (8) and (10), we have
| A v ( s 1 ) A v ( s 2 ) | = c 1 ( s 1 s 2 ) + ( Γ ( δ ) ) 1 0 s 1 ( s 1 r ) δ 1 Ψ ( r ) d r ( Γ ( δ ) ) 1 0 s 2 ( s 2 r ) δ 1 Ψ ( r ) d r c 1 ( s 1 s 2 ) + ( Γ ( δ ) ) 1 0 s 1 ( s 1 r ) δ 1 Ψ ( r ) d r ( Γ ( δ ) ) 1 0 s 2 ( s 2 r ) δ 1 Ψ ( r ) d r c 1 | s 1 s 2 | + N 2 ( Γ ( δ ) ) 1 0 s 1 ( s 1 r ) δ 1 d r 0 s 2 ( s 2 r ) δ 1 d r = c 1 | s 1 s 2 | + N 2 ( Γ ( δ + 1 ) ) 1 | s 1 δ s 2 δ | .
If s 1 s 2 , then | A v ( s 1 ) A v ( s 2 ) | 0 . Therefore, A is equicontinuous.
From the Arzelà–Ascoli theorem, we know that A : K K is a completely continuous operator. □

3. Multiplicity

Theorem 2.
Assume 0 < k < c < a and g satisfies the following conditions:
(C1)
g ( s , v ) < D 1 , ( s , v ) [ 0 , 1 ] × [ 0 , l a ] ,
(C2)
g ( s , v ) > D 2 , ( s , v ) [ 0 , 1 ] × [ c , l c ] ,
(C3)
g ( s , v ) < D 3 , ( s , v ) [ 0 , 1 ] × [ 0 , k ] ,
(C4)
g ( s , 0 ) > 0 , s [ 0 , 1 ] ,
where
D 1 = Γ ( γ 1 ) B ( 2 , γ ) a L 1 p 1 , L 1 = 2 l 1 A 1 Γ ( δ + 1 ) + l 2 A 1 Γ ( δ ) ,
D 2 = Γ ( γ ) B ( 2 , γ ) c L 2 p 1 ,
L 2 = ( l 2 l 1 ) B ( γ q γ q + 2 , δ + q 2 ) A 1 Γ ( δ 1 ) + B ( γ q γ q + 2 , δ + q 1 ) Γ ( δ ) ,
D 3 = Γ ( γ 1 ) B ( 2 , γ ) k L 3 p 1 , L 3 = 0 1 h ( s ) d s A 1 Γ ( δ + 1 ) + A 2 A 1 Γ ( δ ) .
Then the FBVP (1) has at least three positive solutions v 1 , v 2 and v 3 such that
0 < ψ ( v 1 ) < k < ψ ( v 2 ) , ϑ ( v 2 ) < c < ϑ ( v 3 ) , σ ( v 3 ) < a .
Proof. 
Step 1. Let us verify that all the assumptions, including (i), (ii), and (iii) of Theorem 1 are satisfied. Now we give three functionals on K,
ψ ( v ) = max s [ 0 , 1 ] | v ( s ) | , ϑ ( v ) = σ ( v ) = min s [ 0 , 1 ] | v ( s ) | .
Then, ψ , σ , ϑ : K R + are three nonnegative, continuous functionals; additionally, ψ , σ : K R + are increasing. Moreover, we have ψ ( 0 ) = 0 < k , σ ( v ) ϑ ( v ) ψ ( v ) . For any v K ( σ , a ) ¯ , we obtain min s [ 0 , 1 ] v ( s ) l 1 v , that is, σ ( v ) l 1 v , therefore, we obtain v l σ ( v ) . For any v K ( ϑ , r ) , 0 < λ 1 , we have
ϑ ( λ v ) = min s [ 0 , 1 ] | λ v ( s ) | = λ min s [ 0 , 1 ] | v ( s ) | = λ ϑ ( v ) .
Step 2. We prove that (i) of Theorem 1 holds. Let v K ( σ , a ) , that is, σ ( v ) = min s [ 0 , 1 ] | v ( s ) | = a . Thus, v l σ ( v ) = l a . Then 0 v ( s ) l a for s [ 0 , 1 ] and v K ( σ , a ) . By means of ( C 1 ) and Lemma 2, we have
Ψ ( r ) = ϕ q 0 1 Θ ( r , ζ ) g ( ζ , v ( ζ ) ) d ζ < 0 1 ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) D 1 d ζ q 1 = D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 .
From Remark 1, ( H 1 ) , (7), (8), (11), (14), we obtain
σ ( A v ) = min s [ 0 , 1 ] | A v ( s ) | = A v ( 1 ) = c 0 + c 1 + 1 Γ ( δ ) 0 1 ( 1 r ) δ 1 Ψ ( r ) d r = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r   ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r 0 1 h ( s ) d s ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d s r   + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r l 1 ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r   + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r = ( l 1 + A 1 ) ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r 2 l 1 ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r   + l 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r < 2 l 1 ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 d r   + l 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 d r = 2 l 1 ( A 1 Γ ( δ + 1 ) ) 1 D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 + l 2 ( A 1 Γ ( δ ) ) 1 D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 = 2 l 1 ( A 1 Γ ( δ + 1 ) ) 1 + l 2 ( A 1 Γ ( δ ) ) 1 D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 = L 1 D 1 B ( 2 , γ ) Γ ( γ 1 ) q 1 = a .
Step 3. We prove that (ii) of Theorem 1 holds. Let v K ( ϑ , c ) , that is, ϑ ( v ) = min s [ 0 , 1 ] v ( s ) = c . Consquently, v l min s [ 0 , 1 ] v ( s ) = l ϑ ( v ) = l c . Thus, r v ( s ) l c for s [ 0 , 1 ] and v K ( ϑ , c ) . By means of ( C 2 ) and Lemma 2, we obtain
Ψ ( r ) = ϕ q 0 1 Θ ( r , ζ ) g ( ζ , v ( ζ ) ) d ζ > ϕ q D 2 0 1 r γ 1 ( 1 r ) ζ ( 1 ζ ) γ 1 Γ ( γ ) d ζ = D 2 r γ 1 ( 1 r ) B ( 2 , γ ) Γ ( γ ) q 1 .
In view of ( H 1 ) and (7), we have
A 2 A 1 = l 2 0 1 s h ( s ) d s l 1 0 1 h ( s ) d s = l 2 l 1 + 0 1 h ( s ) d s 0 1 s h ( s ) d s l 2 l 1 > 0 .
From Remark 1, (8), (12), (15), and (16), we obtain
ϑ ( A v ) = min s [ 0 , 1 ] | A v ( s ) | = A v ( 1 ) = c 0 + c 1 + 1 Γ ( δ ) 0 1 ( 1 r ) δ 1 Ψ ( r ) d r = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r   ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r   + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r = ( A 2 A 1 ) ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r ( l 2 l 1 ) ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r > ( l 2 l 1 ) ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 D 2 r γ 1 ( 1 r ) B ( 2 , γ ) Γ ( γ ) q 1 d r   + ( Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 D 2 r γ 1 ( 1 s ) B ( 2 , γ ) Γ ( γ ) q 1 d r = ( l 2 l 1 ) ( A 1 Γ ( δ 1 ) ) 1 D 2 B ( 2 , γ ) Γ ( γ ) q 1 0 1 ( 1 r ) δ 2 ( r γ 1 ( 1 r ) ) q 1 d r   + ( Γ ( δ ) ) 1 D 2 B ( 2 , γ ) Γ ( δ ) q 1 0 1 ( 1 r ) δ 1 ( r γ 1 ( 1 r ) ) q 1 d r = ( l 2 l 1 ) ( A 1 Γ ( δ 1 ) ) 1 D 2 B ( 2 , γ ) Γ ( γ ) q 1 B ( γ q γ q + 2 , δ + q 2 )   + ( Γ ( δ ) ) 1 D 2 B ( 2 , γ ) Γ ( γ ) q 1 B ( γ q γ q + 2 , δ + q 1 ) = ( l 2 l 1 ) B ( γ q γ q + 2 , δ + q 2 ) A 1 Γ ( δ 1 ) + B ( γ q γ q + 2 , δ + q 1 ) Γ ( δ )   × D 2 B ( 2 , γ ) Γ ( γ ) q 1 = L 2 D 2 B ( 2 , γ ) Γ ( γ ) q 1 = c .
Step 4. We prove that (iii) of Theorem 1 holds. Let v K ( ψ , h ) , that is, ψ ( v ) = max s [ 0 , 1 ] v ( s ) = k . Thus, 0 v ( s ) k , for s [ 0 , 1 ] and v K ( ψ , k ) . By means of ( C 3 ) and Lemma 2, we obtain
Ψ ( r ) = ϕ q 0 1 Θ ( r , ζ ) g ( ζ , v ( ζ ) ) d ζ < ϕ q D 3 0 1 ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) d ζ = D 3 B ( 2 , γ ) Γ ( γ 1 ) q 1 .
From Remark 1, (8), (13), and (17), we obtain
ψ ( A v ) = max s [ 0 , 1 ] | A v ( s ) | = | A v ( 0 ) | = c 0 = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ ( r ) d r d s   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r 0 1 h ( s ) d s ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ ( r ) d r   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ ( r ) d r < 0 1 h ( s ) d s ( A 1 Γ ( δ + 1 ) ) 1 D 3 B ( 2 , γ ) Γ ( γ 1 ) q 1   + A 2 ( A 1 Γ ( δ ) ) 1 D 3 B ( 2 , γ ) Γ ( γ 1 ) q 1 = 0 1 h ( s ) d s ( A 1 Γ ( δ + 1 ) ) 1 + A 2 ( A 1 Γ ( δ ) ) 1 D 3 B ( 2 , γ ) Γ ( γ 1 ) q 1 = L 3 D 3 B ( 2 , γ ) Γ ( γ 1 ) q 1 = k .
Step 5. We verify that A θ θ holds. In view of ( H 2 ) , we know that g ( s , 0 ) 0 on [ 0 , 1 ] , and g ( s , 0 ) is continuous on [ 0 , 1 ] . Due to Lemma 2, we can see Θ ( r , ζ ) is continuous on [ 0 , 1 ] × [ 0 , 1 ] and Θ ( r , ζ ) 0 on [ 0 , 1 ] × [ 0 , 1 ] . From ( C 4 ) , we let D 4 = min s [ 0 , 1 ] g ( s , 0 ) > 0 , and then
Ψ 0 ( r ) = ϕ q 0 1 Θ ( r , ζ ) g ( ζ , 0 ) d ζ ϕ q D 4 0 1 Θ ( r , ζ ) d ζ > 0 .
From (8), (16), and (18), we obtain
A θ = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 t ( s r ) δ 1 Ψ 0 ( r ) d r d s   + A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ 0 ( r ) d r   s ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ 0 ( r ) d r + ( Γ ( δ ) ) 1 0 s ( s r ) δ 1 Ψ 0 ( r ) d r
A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ 0 ( r ) d r ( Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ 0 ( r ) d r = ( A 2 A 1 ) ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ 0 ( r ) d r > 0 .
We can see that A θ θ holds.
Conclusion. Therefore, in view of Theorem 1 and Remark 3, we see that the FBVP (1) has at least three positive solutions v 1 , v 2 , v 3 , and
0 < ψ ( v 1 ) < k < ψ ( v 2 ) , ϑ ( v 2 ) < c < ϑ ( v 3 ) , σ ( v 3 ) < a .
The proof is complete. □

4. Nonexistence

Next, we study nonexistence results of positive solutions for the eigenvalue problem of (1)
D 0 + γ [ ϕ p ( c D 0 + δ v ( s ) ) ] + λ g ( s , v ( s ) ) = 0 , s [ 0 , 1 ] , ϕ p ( c D 0 + δ v ( 0 ) ) = [ ϕ p ( c D 0 + δ v ( 0 ) ) ] = ϕ p ( c D 0 + δ v ( 1 ) ) = 0 , v ( 0 ) = v ( 1 ) = 0 , l 1 v ( 0 ) + l 2 v ( 0 ) = 0 1 h ( s ) v ( s ) d s .
From Lemma 1, we know that the BVP (19) is tantamount to the equation
v ( s ) = c 0 + c 1 s + 1 Γ ( δ ) 0 s ( s r ) δ 1 Ψ λ ( r ) d r = : A λ v ( s ) ,
where
Ψ λ ( r ) = ϕ q 0 1 Θ ( r , ζ ) λ g ( ζ , v ( ζ ) ) d ζ ,
Θ ( r , ζ ) , c 1 and c 0 are defined by (4), (5), and (6). Moreover, from Lemma 3, we know that A λ is a completely continuous operator.
For convenience, we provide the following notations:
g 0 = lim v 0 + sup max s [ 0 , 1 ] g ( s , v ) ϕ p ( v ) , g = lim v sup max s [ 0 , 1 ] g ( s , v ) ϕ p ( v ) ,
g 0 = lim v 0 + inf min s [ 0 , 1 ] g ( s , v ) ϕ p ( v ) , g = lim v inf min s [ 0 , 1 ] g ( s , v ) ϕ p ( v ) .
Theorem 3.
If g 0 < and g < , then there exists λ > 0 such that for all 0 < λ < λ , the BVP (19) has no positive solution.
Proof. 
From g 0 < and g < , there exist positive constants K 1 , K 2 , l 1 and l 2 such that l 1 < l 2 and
g ( s , v ) K 1 ϕ p ( v ) , ( s , v ) [ 0 , 1 ] × [ 0 , l 1 ] ; g ( s , v ) K 2 ϕ p ( v ) , ( s , v ) [ 0 , 1 ] × [ l 2 , ) .
Let K = max K 1 , K 2 , max ( s , v ) [ 0 , 1 ] × [ l 1 , l 2 ] g ( s , v ) ϕ p ( v ) . Then we have
g ( s , v ) K ϕ p ( v ) , ( s , v ) [ 0 , 1 ] × [ 0 , ) .
Set
λ = Γ ( γ 1 ) K B ( 2 , γ ) ϕ p 1 0 1 h ( s ) d s ( A 1 Γ ( δ + 1 ) ) 1 + A 2 ( A 1 Γ ( δ ) ) 1 .
Assume there exists λ 1 ( 0 , λ ) such that BVP (19) has a positive solution v 1 ( s ) , i.e., A λ 1 v 1 ( s ) = v 1 ( s ) , s [ 0 , 1 ] . From (21) and Lemma 2, we have
Ψ λ 1 ( r ) = ϕ q 0 1 Θ ( r , ζ ) λ 1 g ( ζ , v 1 ( ζ ) ) d ζ ϕ q 0 1 ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) λ 1 g ( ζ , v 1 ( ζ ) ) d ζ .
From Remark 1, (6), (7), (20), (22), (23), and (24), we obtain
v 1 = A λ 1 v 1 = max s [ 0 , 1 ] | A λ 1 v 1 ( s ) | = | A λ 1 v 1 ( 0 ) | = c 0 = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ λ 1 ( r ) d r d s +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ λ 1 ( r ) d r 0 1 h ( s ) d s ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 Ψ λ 1 ( r ) d r +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ λ 1 ( r ) d r 0 1 h ( s ) d s ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 ϕ q 0 1 ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) λ 1 g ( ζ , v 1 ( ζ ) ) d ζ d r +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 ϕ q 0 1 ζ ( 1 ζ ) γ 1 Γ ( γ 1 ) λ 1 g ( ζ , v 1 ( ζ ) ) d ζ d r 0 1 h ( s ) d s ( A 1 Γ ( δ ) ) 1 0 1 ( 1 r ) δ 1 ϕ q B ( 2 , γ ) Γ ( γ 1 ) λ 1 K ϕ p ( v 1 ) d r +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 ϕ q B ( 2 , γ ) Γ ( γ 1 ) λ 1 K ϕ p ( v 1 ) d r = 0 1 h ( s ) d s ( A 1 Γ ( δ + 1 ) ) 1 + A 2 ( A 1 Γ ( δ ) ) 1 ϕ q B ( 2 , γ ) Γ ( γ 1 ) λ 1 K ϕ p ( v 1 ) < 0 1 h ( s ) d s ( A 1 Γ ( δ + 1 ) ) 1 + A 2 ( A 1 Γ ( δ ) ) 1 ϕ q B ( 2 , γ ) Γ ( γ 1 ) λ K ϕ p ( v 1 ) =   v 1 ,
which is a contradiction. Therefore, the BVP (19) has no positive solution for λ ( 0 , λ ) . □
Theorem 4.
If g ( s , v ) > 0 , ( s , v ) [ 0 , 1 ] × ( 0 , ) , g 0 > 0 and g > 0 , then there exists λ > 0 such that for all λ > λ , the BVP (19) has no positive solution.
Proof. 
From g 0 > 0 , and g > 0 , there exist positive constants k 1 , k 2 , l 3 and l 4 such that l 3 < l 4 and
g ( s , v ) k 1 ϕ p ( v ) , ( s , v ) [ 0 , 1 ] × [ 0 , l 3 ] ; g ( s , v ) k 2 ϕ p ( v ) , ( s , v ) [ 0 , 1 ] × [ l 4 , ) .
Let k = min k 1 , k 2 , min ( s , v ) [ 0 , 1 ] × [ l 3 , l 4 ] g ( s , v ) ϕ p ( v ) > 0 . Then we have
g ( s , v ) k ϕ p ( v ) , ( s , v ) [ 0 , 1 ] × [ 0 , ) .
Set
λ = Γ ( γ ) B ( 2 , γ ) k ϕ p l 1 A 1 Γ ( δ ) 0 1 h ( s ) 0 s ( s r ) δ 1 r γ 1 ( 1 r ) q 1 d r d s + A 2 A 1 Γ ( δ 1 ) B ( γ q γ q + 2 , δ + q 2 ) 1 .
Assume there exists λ 2 > λ such that (19) has a positive solution v 2 ( s ) , i.e., A λ 2 v 2 ( s ) = v 2 ( s ) , s [ 0 , 1 ] . From Lemma 2 and (21), we have
Ψ λ 2 ( r ) = ϕ q 0 1 Θ ( r , ζ ) λ 2 g ( ζ , v 2 ( ζ ) ) d ζ ϕ q 0 1 r γ 1 ( 1 r ) ζ ( 1 ζ ) γ 1 Γ ( γ ) λ 2 g ( ζ , v 2 ( ζ ) ) d ζ .
From Remark 1, (6), (7), (20), (25), (26), and (27), we obtain
v 2   =   A λ 2 v 2   =   max s [ 0 , 1 ] | A λ 2 v 2 ( s ) | =   | A λ 2 v 2 ( 0 ) | = c 0 = ( A 1 Γ ( δ ) ) 1 0 1 h ( s ) 0 s ( s r ) δ 1 Ψ λ 2 ( r ) d r d s +   A 2 ( A 1 Γ ( δ 1 ) ) 1 0 1 ( 1 r ) δ 2 Ψ λ 2 ( r ) d r 1 A 1 Γ ( δ ) 0 1 h ( s ) 0 s ( s r ) δ 1 ϕ q 0 1 r γ 1 ( 1 r ) ζ ( 1 ζ ) γ 1 Γ ( γ ) λ 2 g ( ζ , v 2 ( ζ ) ) d ζ d r d s +   A 2 A 1 Γ ( δ 1 ) 0 1 ( 1 r ) δ 2 ϕ q 0 1 r γ 1 ( 1 r ) ζ ( 1 ζ ) γ 1 Γ ( γ ) λ 2 g ( ζ , v 2 ( ζ ) ) d ζ d r 1 A 1 Γ ( δ ) 0 1 h ( s ) 0 s ( s r ) δ 1 ϕ q r γ 1 ( 1 r ) B ( 2 , γ ) Γ ( γ ) λ 2 k ϕ p ( v 2 ) d r d s +   A 2 A 1 Γ ( δ 1 ) 0 1 ( 1 r ) δ 2 ϕ q r γ 1 ( 1 r ) B ( 2 , γ ) Γ ( γ ) λ 2 k ϕ p ( v 2 ) d r 1 A 1 Γ ( δ ) ϕ q B ( 2 , γ ) Γ ( γ ) λ 2 k ϕ p v 2 l 0 1 h ( s ) 0 s ( s r ) δ 1 r γ 1 ( 1 r ) q 1 d r d s + A 2 A 1 Γ ( δ 1 ) ϕ q B ( 2 , γ ) Γ ( γ ) λ 2 k ϕ p v 2 l 0 1 ( 1 r ) δ 2 r γ 1 ( 1 r ) q 1 d r = ϕ q B ( 2 , γ ) Γ ( γ ) λ 2 k ϕ p v 2 l 1 A 1 Γ ( δ ) 0 1 h ( s ) 0 s ( s r ) δ 1 r γ 1 ( 1 r ) q 1 d r d s + A 2 A 1 Γ ( δ 1 ) B ( γ q γ q + 2 , δ + q 2 ) > ϕ q B ( 2 , γ ) Γ ( γ ) λ k ϕ p v 2 l 1 A 1 Γ ( δ ) 0 1 h ( s ) 0 s ( s r ) δ 1 r γ 1 ( 1 r ) q 1 d r d s + A 2 A 1 Γ ( δ 1 ) B ( γ q γ q + 2 , δ + q 2 ) =   v 2 ,
which is a contradiction. Therefore, the BVP (19) has no positive solution for λ > λ . □

5. Three Examples

We provide Example 1 to illustrate the effectiveness of the results of Theorem 2.
Example 1.
Let δ = 2.2 , γ = 2.9 , h ( s ) = s , l 1 = 2 , l 2 = 3 , p = q = 2 . Then we consider
D 0 + 2.9 [ ϕ p ( C D 0 + 2.2 v ( s ) ) ] + g ( s , v ( s ) ) = 0 , s [ 0 , 1 ] , ϕ p ( C D 0 + 2.2 v ( 0 ) ) = [ ϕ p ( C D 0 + 2.2 v ( 0 ) ) ] = ϕ p ( C D 0 + 2.2 v ( 1 ) ) = 0 , v ( 0 ) = v ( 1 ) = 0 , 2 v ( 0 ) + 3 v ( 0 ) = 0 1 h ( s ) v ( s ) d s ,
where
g ( s , v ) = 3 , ( s , v ) [ 0 , 1 ] × [ 0 , 0.5 ] , 3 + 690 ( v 0.5 ) , ( s , v ) [ 0 , 1 ] × [ 0.5 , 64 35 ] , 6438 7 , ( s , v ) [ 0 , 1 ] × [ 64 35 , + ) .
First, we verify that ( H 1 ) , ( H 2 ) hold. We let h ( s ) = s L 1 [ 0 , 1 ] C ( 0 , 1 ) . Then 0 1 s d s = 1 2 > 0 and 0 1 s 2 d t = 1 3 > 0 . Let l 2 = 3 > 2 = l 1 , and l 1 = 2 > 0 1 s d s = 1 2 . Then ( H 1 ) holds. From (29), we can see that g ( s , v ) is continuous on [ 0 , 1 ] × [ 0 , ) and g ( s , 0 ) = 3 > 0 for s [ 0 , 1 ] . Then ( H 2 ) holds.
Now, we prove that the conclusion of Theorem 2 holds. Let k = 0.5 , c = 0.8 , a = 320 . From (7), (9), (11), (12), and (13), we have
A 1 = 2 0 1 s d s = 3 2 , A 2 = 3 0 1 s 2 d s = 8 3 , l = A 2 A 2 A 1 = 16 7 , l d = 5120 7 , l c = 64 35 , L 1 = 4 A 1 Γ ( 3.2 ) + 3 A 1 Γ ( 2.2 ) 2.915258 , D 1 = Γ ( 1.9 ) B ( 2 , 2.9 ) 320 L 1 2 1 1196.304668 , L 2 = ( 3 2 ) B ( 2.9 , 2.2 ) A 1 Γ ( 1.2 ) + B ( 2.9 , 3.2 ) Γ ( 2.2 ) 0.080717 , D 2 = Γ ( 2.9 ) B ( 2 , 2.9 ) 0.8 L 2 2 1 205.232533 , L 3 = 0 1 s d s A 1 Γ ( 3.2 ) + A 2 A 1 Γ ( 2.2 ) 1.750987 , D 3 = Γ ( 1.9 ) B ( 2 , 2.9 ) 0.5 L 3 2 1 3.112117 ,
and
max ( s , v ) [ 0 , 1 ] × [ 0 , 5120 7 ] g ( s , v ) = 6438 7 < D 1 1196.304668 ,
min ( s , v ) [ 0 , 1 ] × [ 0.8 , 64 35 ] g ( s , v ) = 3 + 690 ( 0.8 0.5 ) = 210 > D 2 205.232533 ,
max ( s , v ) [ 0 , 1 ] × [ 0 , 0.5 ] g ( s , v ) = 3 < D 3 3.112117 , min s [ 0 , 1 ] g ( s , 0 ) = 3 > 0 .
Then from Theorem 2, we know the FBVP (28) has at least three positive solutions satisfying
0 < ψ ( v 1 ) < 0.5 < ψ ( v 2 ) , ϑ ( v 2 ) < 0.8 < ϑ ( v 3 ) , σ ( v 3 ) < 320 .
We provide Example 2 to illustrate the effectiveness of the results of Theorem 3.
Example 2.
Let δ = 2.2 , γ = 2.6 , h ( s ) = s , l 1 = 2 , l 2 = 3 , p = 3 / 2 , q = 3 . Then we consider
D 0 + 2.6 [ ϕ p ( C D 0 + 2.2 v ( s ) ) ] + λ g ( s , v ( s ) ) = 0 , s [ 0 , 1 ] , ϕ p ( C D 0 + 2.2 v ( 0 ) ) = [ ϕ p ( C D 0 + 2.2 v ( 0 ) ) ] = ϕ p ( C D 0 + 2.2 v ( 1 ) ) = 0 , v ( 0 ) = v ( 1 ) = 0 , 2 v ( 0 ) + 3 v ( 0 ) = 0 1 s v ( s ) d s ,
where
g ( s , v ) = s v 1 2 , ( s , v ) [ 0 , 1 ] × [ 0 , 1 ] , s v 1 2 + 10 3 ( v 1 2 1 ) , ( s , v ) [ 0 , 1 ] × [ 1 , ) .
Obviously, g : [ 0 , 1 ] × [ 0 , ) [ 0 , ) is continuous and
g 0 = 1 < , g = lim v sup max s [ 0 , 1 ] s v 1 2 + 10 3 ( v 1 2 1 ) v 3 2 1 = 1001 < .
We choose K = 1001 . Then g ( s , v ) < 1001 ϕ p ( v ) = 1001 v 1 2 , ( s , v ) [ 0 , 1 ] × [ 0 , ) , and
A 1 = 2 0 1 s d s = 3 2 , A 2 = 3 0 1 s 2 d s = 8 3 , λ = Γ ( 2.6 1 ) 1001 B ( 2 , 2.6 ) ϕ p 1 0 1 s d s ( A 1 Γ ( 2.2 + 1 ) ) 1 + A 2 ( A 1 Γ ( 2.2 ) ) 1 0.006063 .
From Theorem 3, BVP (30) has no positive solution for λ ( 0 , 0.006063 ) .
We provide Example 3 to illustrate the effectiveness of the results of Theorem 4.
Example 3.
Let δ = 2.2 , γ = 2.6 , h ( s ) = s , l 1 = 2 , l 2 = 3 , p = 2 , q = 2 . Then we consider
D 0 + 2.6 [ ϕ p ( C D 0 + 2.2 v ( s ) ) ] + λ g ( s , v ( s ) ) = 0 , s [ 0 , 1 ] , ϕ p ( C D 0 + 2.2 v ( 0 ) ) = [ ϕ p ( C D 0 + 2.2 v ( 0 ) ) ] = ϕ p ( C D 0 + 2.2 v ( 1 ) ) = 0 , v ( 0 ) = v ( 1 ) = 0 , 2 v ( 0 ) + 3 v ( 0 ) = 0 1 s v ( s ) d s ,
where
g ( s , v ) = e s v , ( s , v ) [ 0 , 1 ] × ( 0 , ) .
Obviously, g : [ 0 , 1 ] × ( 0 , ) ( 0 , ) is continuous, and
g 0 = lim v 0 + inf min s [ 0 , 1 ] e s v v 2 1 = 1 > 0 , g = lim v inf min s [ 0 , 1 ] e s v v 2 1 = 1 > 0 .
We choose k = 0.5 . Then g ( s , v ) = e s v > 0.5 v , ( s , v ) [ 0 , 1 ] × ( 0 , ) , and
A 1 = 2 0 1 s d s = 3 2 , A 2 = 3 0 1 s 2 d s = 8 3 , l = A 2 A 2 A 1 = 16 7 ,
λ = Γ ( 2.6 ) B ( 2 , 2.6 ) k ϕ p l 1 A 1 Γ ( 2.2 ) 0 1 s 0 s ( s r ) 2.2 1 r 2.6 1 ( 1 r ) 2 1 d r d s + A 2 A 1 Γ ( 2.2 1 ) B ( 2.6 × 2 2.6 2 + 2 , 2.2 + 2 2 ) 1 32.224568 .
From Theorem 4, BVP (32) has no positive solution for λ > 32.224568 .

6. Conclusions

This paper obtains the existence of at least three positive solutions for fractional differential equations with p-Laplacian operators by using the generalized Avery–Henderson fixed point theorem. We also obtained nonexistence of a positive solution for the eigenvalue problem of the equation. Moreover, three examples to demonstrate the effectiveness of the main results were also provided. Compared with some existing results, the derivative type in this paper is complex and one more positive solution was obtained by adding the nature of the conditions to ensure the third solution is not the zero solution.
At the same time, we know that there are still more works to be performed in the future. For example, how to obtain the existence of positive solutions of FBVP (1) with an eigenvalue parameter D 0 + γ [ ϕ p ( c D 0 + δ v ( s ) ) ] + λ g ( s , v ( s ) ) = 0 , or a disturbance parameter l 1 v ( 0 ) + l 2 v ( 0 ) = 0 1 h ( s ) v ( s ) d s + λ , or semipositone problems, or with more generalized fractional derivative.

Author Contributions

Writing—original draft, H.Z.; Writing—review & editing, Z.H. and M.B. All authors have read and agreed to the published version of the manuscript.

Funding

This research is supported by NSFC (11871302), the Fund of the Natural Science of Shandong Province (ZR2014AM034, ZR2021MA097), and colleges and universities of Shandong province science and technology plan projects (J13LI01).

Data Availability Statement

The original contributions presented in the study are included in the article, further inquiries can be directed to the corresponding author.

Acknowledgments

The authors are grateful to the Editor and Referees for their valuable suggestions and comments, which significantly improved the quality of the paper.

Conflicts of Interest

The authors declare no conflicts of interest.

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Zhang, H.; Hao, Z.; Bohner, M. Multiple and Nonexistence of Positive Solutions for a Class of Fractional Differential Equations with p-Laplacian Operator. Mathematics 2024, 12, 3869. https://doi.org/10.3390/math12233869

AMA Style

Zhang H, Hao Z, Bohner M. Multiple and Nonexistence of Positive Solutions for a Class of Fractional Differential Equations with p-Laplacian Operator. Mathematics. 2024; 12(23):3869. https://doi.org/10.3390/math12233869

Chicago/Turabian Style

Zhang, Haoran, Zhaocai Hao, and Martin Bohner. 2024. "Multiple and Nonexistence of Positive Solutions for a Class of Fractional Differential Equations with p-Laplacian Operator" Mathematics 12, no. 23: 3869. https://doi.org/10.3390/math12233869

APA Style

Zhang, H., Hao, Z., & Bohner, M. (2024). Multiple and Nonexistence of Positive Solutions for a Class of Fractional Differential Equations with p-Laplacian Operator. Mathematics, 12(23), 3869. https://doi.org/10.3390/math12233869

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