Open AccessArticle Existence and Uniqueness of Solutions to Four-Point Impulsive Fractional Differential Equations with p-Laplacian Operator by Limin ChuLimin Chu SciProfiles Scilit Preprints.org Google Scholar 1,2, Weimin HuWeimin Hu SciProfiles Scilit Preprints.org Google Scholar 1,3, Youhui SuYouhui Su SciProfiles Scilit Preprints.org Google Scholar 2,* and Yongzhen YunYongzhen Yun SciProfiles Scilit Preprints.org Google Scholar 2 1 School of Mathematics and Statistics, Yili Normal University, Yining 835000, China 2 School of Mathematics and Statistics, Xuzhou University of Technology, Xuzhou 221018, China 3 Institute of Applied Mathematics, Yili Normal University, Yining 835000, China * Author to whom correspondence should be addressed. Mathematics 2022, 10(11), 1852; https://doi.org/10.3390/math10111852 Submission received: 26 April 2022 / Revised: 10 May 2022 / Accepted: 23 May 2022 / Published: 28 May 2022 (This article belongs to the Special Issue Fractional Dynamical Systems and Its Applications in Science and Engineering) Download keyboard_arrow_down Download PDF Download PDF with Cover Download XML Download Epub Versions Notes Abstract: In this paper, by using fixed-point theorems, the existence and uniqueness of positive solutions to a class of four-point impulsive fractional differential equations with p-Laplacian operators are studied. In addition, three examples are given to justify the conclusion. The interest of this paper is to study impulsive fractional differential equations with p-Laplacian operators. Keywords: impulsive fractional differential equations; p-Laplacian operator; boundary value problem; existence; uniqueness MSC: 34B15; 34B27; 34B37 1. IntroductionIt is well known that fractional differential equations describe a number of phenomena in natural life, especially in describing natural, physical and chemical phenomena more accurately and generally [1,2,3,4,5]. In the past few years, fractional differential equations described problems in greater depth and specificity, especially in the study of the liver, mumps virus, glucose and other problems [6,7,8,9].Impulsive differential equations, which offer a depiction of surveyed evolving processes, were called mathematical instruments for the comprehension of several actual problems in science [10,11,12,13]. However, the theory of impulsive equations is more expressive than the corresponding theory of equations without impulse impacts since straightforward impulsive differential equations emerged for several fresh phenomena, such as rhythmical beating and the merging of solutions [14,15,16,17]. For the theory and applications of impulsive differential equations, we refer to the references [18,19,20,21].In recent years, fractional differential equations with p-Laplacian operators are becoming significant, they can be used to depict a number of proliferate phenomena, which have been comprehensively applied in the fields of fluid mechanics, material memory and chemistry [22,23,24,25,26,27].For convenience, among this paper, we mark φ p ( u ) as a p-Laplacian operator, i.e., φ p ( u ) = u p − 2 u , ( φ p ) − 1 = φ q , where p > 1 , 1 / p + 1 / q = 1 . In [24], Li et al. discussed the existence of positive solutions to four-point boundary value problems, c D 0 + β ( φ p ( c D 0 + α u ( t ) ) ) + f ( t , u ( t ) ) = 0 , t ∈ ( 0 , 1 ) , ( c D 0 + α u ( 0 ) ) ( i ) = 0 , ( i = 1 , 2 , ⋯ , n − 1 ) , ( φ p ( c D 0 + α u ( 0 ) ) ) ( j ) = 0 , ( j = 1 , 2 , ⋯ , m − 1 ) , u ( 0 ) − a c D 0 + α u ( ξ ) = 0 , u ( 1 ) + b c D 0 + α u ( η ) = 0 , (1) where φ p is p-Laplacian operator, c D 0 + α u ( t ) and c D 0 + β u ( t ) are Caputo fractional derivatives, 0 < n − 1 < α ≤ n , 0 < m − 1 < β ≤ m , m + n − 1 < α + β < m + n , 0 < ξ < η < 1 . By using the monotone iterative technique, they obtained the existence of positive solutions to the fractional differential equation.In [20], Tian et al. discussed the existence of solutions to a class of three-point impulse boundary value problems. c D q u ( t ) = f ( t , u ( t ) ) , 0 < t < 1 , t ≠ t k , k = 1 , 2 , ⋯ , p , Δ u ( t k ) = I k ( u ( t k ) ) , Δ u ′ ( t k ) = I k ¯ ( u ( t k ) ) , k = 1 , 2 , ⋯ , p , u ( 0 ) + u ′ ( 0 ) = 0 , u ( 1 ) + u ′ ( ξ ) = 0 , (2) where c D q is a Caputo fractional derivative, and q ∈ R , 1 < q ≤ 2 , f : [ 0 , 1 ] × R → R is a continuous function. By using a fixed-point theorem, the existence of solutions was found.In [24], the authors considered a four-point fractional differential Equation ( 1 ) with p-Laplacian operators; however, the boundary value conditions of this differential equations were impulsive. As we all know, the impulsive differential equations with p-Laplacian not only contain general dynamic equations but also have profound engineering physical significance—for instance, a p-Laplacian equation is needed to study the turbulence of non-Newtonian fluida, gas in porous media and the spontaneous combustion theory of chemically active gas.Thus, in this paper, we studied the existence of solutions to four-point impulsive fractional differential equations with p-Laplacian operators. The studying of the existence of solutions to impulsive fractional differential equations with p-Laplacian is extremely important. In this paper, we discuss the existence and uniqueness of positive solutions to four-point impulsive fractional differential equations given by ( c D 0 + β ( φ p ( c D 0 + α u ( t ) ) ) ) ′ = λ f ( t , u ( t ) , u ′ ( t ) ) , 2 < α , β ≤ 3 , t ∈ J ′ , Δ u ( t k ) = I k ( u ( t k ) ) , Δ u ′ ( t k ) = I k ′ ( u ( t k ) ) , Δ u ″ ( t k ) = I k ″ ( u ( t k ) ) , k = 1 , 2 , 3 , ⋯ , m , a u ( 0 ) − b u ′ ( ξ ) = 0 , c u ( 1 ) + d u ′ ( η ) = 0 , ( c D 0 + α u ( t ) ) ( j ) = 0 , ( j = 0 , 1 , 2 ) , u ″ ( 0 ) = 0 , (3) where D 0 + α c u ( t ) and D 0 + β c u ( t ) are Caputo fractional differentials, λ is a parameter, φ p is a p-Laplacian operator, and φ p ( s ) = | s | p − 2 s , p > 1 , ( φ p ) − 1 ( s ) = φ q ( s ) , 1 p + 1 q = 1 , f ∈ C ( J × R × R , R ) , I k , I k ′ , I k ″ ∈ C ( R , R ) , a , b , c , d ∈ [ 0 , + ∞ ) , a c + b c + d a = 0 , J = [ 0 , 1 ] , 0 = t 0 < t 1 < ⋯ < t k < ⋯ < t m < t m + 1 = 1 , J ′ = J ∖ { t 1 , t 2 , ⋯ , t m } , c i ∈ R , ( i = 1 , 2 , ⋯ , 9 ) , 0 < ξ < η < 1 , ξ ≠ t k , η ≠ t k ( k = 1 , 2 , ⋯ , m ) , Δ u ( t k ) = u ( t k + ) − u ( t k − ) , Δ u ′ ( t k ) = u ′ ( t k + ) − u ′ ( t k − ) , Δ u ″ ( t k ) = u ″ ( t k + ) − u ″ ( t k − ) , where u ( t k + ) and u ( t k − ) indicate the right and the left limits of u ( t ) at t = t k ( k = 1 , 2 , ⋯ , m ) , respectively, u ′ ( t k ) and u ″ ( t k ) have a kinsmanship implication for u ′ ( t ) and u ″ ( t ) . The objective of this paper is to establish the existence and uniqueness of positive solutions to a class of four-point impulsive fractional differential equations (Equation (3)) by using a fixed-point theorem and contraction mapping principle. In addition, three examples are given to justify the conclusion. The main features of this paper are as follows. First, the nonlinear term involved the fractional order derivative. Next, the studied Equation (3) was used to find the first derivative on the outside of c D 0 + β ( φ p ( c D 0 + α u ( t ) ) ) .Compared with [20,24], our nonlinear terms are more general. Finally, our main results are optimal. This paper contributes to the further development of impulsive fractional differential equations. The rest of the paper is as follows: In Section 2, some basic definitions and related lemmas are given. The existence of positive solutions and the uniqueness of positive solutions for the four-point impulsive fractional differential Equation (3) with p-Laplacian operators under certain assumptions are proven in Section 3. Three examples are given in the Section 4. 2. PreliminariesIn this section, we present some basic definitions, related lemmas and Green functions.Let J 0 = [ 0 , t 1 ] , J 1 = ( t 1 , t 2 ] , ⋯ , J m − 1 = ( t m − 1 , t m ] , J m = ( t m , 1 ] , we define the spaces as follows. Let P C ( J , R ) = { u : J → R ∣ u ∈ C ( J k ) , k = 1 , 2 , ⋯ , m } , P C 1 ( J , R ) = { u : J → R ∣ u ∈ C 1 ( J k ) , k = 1 , 2 , ⋯ , m } , P C 2 ( J , R ) = { u : J → R ∣ u ∈ C 2 ( J k ) , k = 1 , 2 , ⋯ , m } , and endowed with norm ∥ u ( t ) ∥ = sup t ∈ J | u ( t ) | , ∥ u ( t ) ∥ P C 1 = max { ∥ u ( t ) ∥ , ∥ u ′ ( t ) ∥ } , ∥ u ( t ) ∥ P C 2 = max { ∥ u ( t ) ∥ , ∥ u ′ ( t ) ∥ , ∥ u ″ ( t ) ∥ } . where u ( t k + ) indicate the right limit of u ( t ) at t = t k ( k = 1 , 2 , ⋯ , m ) , u ′ ( t k ) and u ″ ( t k ) have a kinsmanship implication for u ′ ( t ) and u ″ ( t ) . Clearly, P C ( J , R ) , P C 1 ( J , R ) , P C 2 ( J , R ) is a complete linear norm space. Thus, P C ( J , R ) , P C 1 ( J , R ) , P C 2 ( J , R ) are Banach spaces.Definition 1 ([18]). Let f ( t ) ∈ ( 0 , 1 ) . The Caputo fractional derivative is defined as D 0 + α c f ( t ) = 1 Γ ( n − α ) ∫ 0 t ( t − s ) n − α − 1 f ( n ) ( s ) d s , n = [ α ] + 1 , where [ α ] indicates the integer part of real number α.Definition 2 ([18]). Let f ( t ) ∈ ( 0 , 1 ) . The Riemann–Liouville fractional integral of order is defined as I 0 + α f ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s , α > 0 , provided the integral exists.Lemma 1 ([18]). For α > 0 , u ∈ C [ 0 , 1 ] , the general solution of the fractional differential equation D 0 + α c ( u ( t ) ) = 0 is stated by u ( t ) = C 0 + C 1 + ⋯ + C N − 1 t N − 1 , C i ∈ R ( i = 1 , 2 , ⋯ , N ) , N = [ α ] + 1 , where [ α ] indicates the integer part of real number α.Lemma 2 ([18]). Let u ( t ) ∈ C [ 0 , 1 ] ∩ L [ 0 , 1 ] , α > 0 and D 0 + α c ( u ( t ) ) ∈ C [ 0 , 1 ] ∩ L [ 0 , 1 ] , I 0 + α D 0 + α c u ( t ) = u ( t ) + C 0 + C 1 + ⋯ + C N − 1 t N − 1 , (4) C i ∈ R ( i = 1 , 2 , ⋯ , N ) , N = [ α ] + 1 . Lemma 3 ([18]). [Contraction Mapping Principle] ( X , P ) is a complete distance space, and T is a compressed map from ( X , P ) to itself, and thus T has a unique fixed point on X.Lemma 4 ([18]). [Arzela–Ascoli Theorem] Let M : ζ ∩ ( Ω 2 ¯ ∖ Ω 1 ) → ζ ; we say M is a compact operator if it is uniformly bounded and equicontinuous.Lemma 5 ([18]). Let E be a Banach space. Assume that T : E → E is a completely continuous operator and that the set V = { u ∈ E ∣ u = μ T u , 0 < μ < 1 } is bounded, then T has a fixed point in E.Lemma 6 ([18]). Let E be a Banach space. Assume that Ω is an open bounded subset of E with θ ∈ Ω and let T : Ω ¯ → E be a completely continuous operator such that ‖ T u ‖ ≤ ‖ u ‖ , ∀ u ∈ ∂ Ω , then T has a fixed point in Ω ¯ , where θ is a fixed point.Lemma 7. Let y ∈ C [ 0 , 1 ] , ξ , η ∈ ( t l , t l + 1 ) , l is a nonnegative integer, 0 ≤ l ≤ m . then the solution of the fractional differential equations D 0 + β c ( φ p D 0 + α c u ( t ) ) ) ) ′ = λ y ( t ) , 2 < α , β ≤ 3 , t ∈ J ′ , Δ u ( t k ) = I k ( u ( t k ) ) , Δ u ′ ( t k ) = I k ′ ( u ( t k ) ) , Δ u ″ ( t k ) = I k ″ ( u ( t k ) ) , k = 1 , 2 , 3 , ⋯ , m , a u ( 0 ) − b u ′ ( ξ ) = 0 , c u ( 1 ) + d u ′ ( η ) = 0 , D 0 + α c u ( t ) ) ( j ) = 0 , ( j = 0 , 1 , 2 ) , u ″ ( 0 ) = 0 , (5) has the solution u ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 4 + c 5 t , t ∈ J 0 , ∫ t k t ( t − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t k − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ( t − t k ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ( t − t k ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k I i ( u ( t i ) ) + ∑ i = 1 k − 1 ( t k − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 k − 1 ( t k − t i ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 k ( t − t k ) I i ′ ( u ( t i ) ) + ∑ i = 1 k ( t − t k ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) I i ″ ( u ( t i ) ) + c 4 + c 5 t , t ∈ J k , k = 1 , 2 , ⋯ , m , (6) where c 4 = 1 a c + a d + b c b c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) I i ″ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) 2 2 I i ″ ( u ( t i ) ) + b d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( η − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) − b ( c + d ) ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( ξ − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) , c 5 = 1 a c + a d + b c a c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) I i ″ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) 2 2 I i ″ ( u ( t i ) ) + a d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( η − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) + b c ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( ξ − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) . Proof. Let u be a solution of (5). By (4), we find φ p ( D 0 + α c u ( t ) ) = λ Γ ( β ) ∫ 0 t ( t − s ) β − 1 ∫ 0 s y ( r ) d r d s + c 1 + c 2 t + c 3 t 2 , according to conditions ( D 0 + α c u ( t ) ) ( 0 ) = 0 , ( j = 0 , 1 , 2 ) , we find c 1 = c 2 = c 3 = 0 , and c D 0 + α u ( t ) = φ q λ Γ ( β ) ∫ 0 t ( t − s ) β − 1 ∫ 0 s y ( r ) d r d s . In addition, u ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 4 + c 5 t + c 6 t 2 , (7) u ′ ( t ) = 1 Γ ( α − 1 ) ∫ 0 t ( t − s ) α − 2 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 5 + 2 c 6 t , u ″ ( t ) = 1 Γ ( α − 2 ) ∫ 0 t ( t − s ) α − 3 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + 2 c 6 . If t ∈ J 1 , u ( t ) = 1 Γ ( α ) ∫ t 1 t ( t − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 7 + c 8 ( t − t 1 ) + c 9 ( t − t 1 ) 2 , u ′ ( t ) = 1 Γ ( α − 1 ) ∫ t 1 t ( t − s ) α − 2 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 8 + 2 c 9 ( t − t 1 ) , u ″ ( t ) = 1 Γ ( α − 2 ) ∫ t 1 t ( t − s ) α − 3 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + 2 c 9 . Thus u ( t 1 − ) = 1 Γ ( α ) ∫ 0 t 1 ( t 1 − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 4 + c 5 t 1 + c 6 t 1 2 , u ( t 1 + ) = e 1 , u ′ ( t 1 − ) = 1 Γ ( α − 1 ) ∫ 0 t 1 ( t 1 − s ) α − 2 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 5 + 2 c 6 t 1 , u ′ ( t 1 + ) = e 2 , u ″ ( t 1 − ) = 1 Γ ( α − 2 ) ∫ 0 t 1 ( t 1 − s ) α − 3 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + 2 c 6 , u ″ ( t 1 + ) = 2 e 3 . Due to Δ u ( t 1 ) = u ( t 1 + ) − u ( t 1 − ) = I 1 ( u ( t 1 ) ) , Δ u ′ ( t 1 ) = u ′ ( t 1 + ) − u ′ ( t 1 − ) = I 1 ′ ( u ( t 1 ) ) , Δ u ″ ( t 1 ) = u ″ ( t 1 + ) − u ″ ( t 1 − ) = I 1 ″ ( u ( t 1 ) ) , we find c 7 = 1 Γ ( α ) ∫ 0 t 1 ( t 1 − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 4 + c 5 t 1 + c 6 t 1 2 + I 1 ( u ( t 1 ) ) , c 8 = 1 Γ ( α − 1 ) ∫ 0 t 1 ( t 1 − s ) α − 2 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + c 5 + 2 c 6 t 1 + I 1 ′ ( u ( t 1 ) ) , 2 c 9 = 1 Γ ( α − 2 ) ∫ 0 t 1 ( t 1 − s ) α − 3 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + 2 c 6 + I 1 ″ ( u ( t 1 ) ) . Consequently, u ( t ) = 1 Γ ( α ) ∫ t 1 t ( t − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + 1 Γ ( α ) ∫ 0 t 1 ( t 1 − s ) α − 1 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + I 1 ( u ( t 1 ) ) + t − t 1 Γ ( α − 1 ) ∫ 0 t 1 ( t 1 − s ) α − 2 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ( t − t 1 ) I 1 ′ ( u ( t 1 ) ) + ( t − t 1 ) 2 2 Γ ( α − 2 ) ∫ 0 t 1 ( t 1 − s ) α − 3 φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ( t − t 1 ) 2 2 I 1 ″ ( u ( t 1 ) ) + c 4 + c 5 t + c 6 t 2 , t ∈ J 1 . Thus, t ∈ J k , k = 1 , 2 , ⋯ , m , and we get u ( t ) = ∫ t k t ( t − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t k − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ( t − t k ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ( t − t k ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k I i ( u ( t i ) ) + ∑ i = 1 k − 1 ( t k − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 k − 1 ( t k − t i ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 k ( t − t k ) I i ′ ( u ( t i ) ) + ∑ i = 1 k ( t − t k ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) I i ″ ( u ( t i ) ) + c 4 + c 5 t + c 6 t 2 . (8) Due to a u ( 0 ) − b u ′ ( ξ ) = 0 , c u ( 1 ) + d u ′ ( η ) = 0 , u ″ ( 0 ) = 0 , we find c 4 = 1 a c + a d + b c b c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) I i ″ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) 2 2 I i ″ ( u ( t i ) ) + b d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( η − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) − b ( c + d ) ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( ξ − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) , c 5 = 1 a c + a d + b c a c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) I i ″ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) 2 2 I i ″ ( u ( t i ) ) + a d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( η − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) + b c ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( ξ − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) and c 6 = 0 . Substituting the value of c 4 , c 5 , c 6 in (7) and (8), we can find (6). This completes the proof. □ 3. Main ResultsIn this section, the existence of positive solutions for the four-point impulsive fractional differential Equation (3) with p-Laplacian operators is proven.We set σ 1 = a d + 2 a c + 2 b c a d + a c + b c , σ 2 = a d + 2 b c + 2 b d a d + a c + b c , σ 3 = a c + 2 b c + 2 a d . Define the operator T : P C ( J , R ) → P C ( J , R ) as ( T u ) ( t ) = ∫ t k t ( t − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t k − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k ( t − t k ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) d r ) d τ ) d s + ∑ i = 1 k ( t − t k ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 k I i ( u ( t i ) ) + ∑ i = 1 k − 1 ( t k − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 k − 1 ( t k − t i ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 k ( t − t k ) I i ′ ( u ( t i ) ) + ∑ i = 1 k ( t − t k ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) I i ″ ( u ( t i ) ) + g 1 + g 2 t . where g 1 = 1 a c + a d + b c b c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) I i ″ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) 2 2 I i ″ ( u ( t i ) ) + b d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( η − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) − b ( c + d ) ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( ξ − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) and g 2 = 1 a c + a d + b c a c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) I i ″ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) I i ′ ( u ( t i ) ) + ∑ i = 1 m ( 1 − t m ) 2 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 ( t m − t i ) 2 2 I i ″ ( u ( t i ) ) + a d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( η − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) + b c ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ y ( r ) d r d τ d s + ∑ i = 1 l I i ′ ( u ( t i ) ) + ∑ i = 1 l ( ξ − t l ) I i ″ ( u ( t i ) ) + ∑ i = 1 l − 1 ( t l − t i ) I i ″ ( u ( t i ) ) . Using Lemma 7 with y ( t ) = f ( t , u ( t ) , u ′ ( t ) ) , problem (3) reduces to a fixed-point problem u = T u . Thus, fractional differential Equation (3) has a solution if and only if the operator T has a fixed point.Theorem 1. The operator T : P C ( J , R ) → P C ( J , R ) is completely continuous.Proof. It is clear that T is continuous by the continuity of f , I k , I k ′ a n d I k ″ . Let Ω ∈ P C ( J , R ) be bounded. Afterward, there exist positive constants M i > 0 ( i = 1 , 2 , 3 , 4 ) , such that ∣ ( f ( t , u ( t ) , u ′ ( t ) ) ) ∣ ≤ φ p ( M 1 ) , ∣ I k ( u ) ∣ ≤ M 2 , ∣ I k ′ ( u ) ∣ ≤ M 3 , ∣ I k ″ ( u ) ∣ ≤ M 4 , ∀ u ∈ Ω . Thus, Ω ∈ P C ( J , R ) , set A = | φ q ( 1 Γ ( β + 1 ) ) | , λ = 1 , we find | g 1 | ≤ 1 a c + a d + b c b c A M 1 ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m − 1 1 2 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m 1 2 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m M 2 + ∑ i = 1 m − 1 M 3 + ∑ i = 1 m − 1 M 4 2 + ∑ i = 1 m M 3 + ∑ i = 1 m M 4 2 + b d A M 1 ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m M 3 + ∑ i = 1 m − 1 M 4 + ∑ i = 1 m M 4 − b ( c + d ) A M 1 ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m M 3 + ∑ i = 1 m − 1 M 4 + ∑ i = 1 m M 4 ≤ 1 a c + a d + b c b c ( 1 + m ) A M 1 Γ ( α + 1 ) + ( 2 b ( c + d ) m + b ( 2 d + c m ) ) A M 1 Γ ( α ) + ( 4 b ( c + d ) m − b ( 5 2 c + 2 d ) ) A M 1 Γ ( α − 1 ) + b c m M 2 + ( b c ( 2 m − 1 ) + b ( c + 2 d ) m ) M 3 + ( b c ( 2 m − 3 2 ) + b ( c + 2 d ) ( 2 m − 1 ) M 4 ) , and | g 2 | ≤ 1 a c + a d + b c a c ( 1 + m ) A M 1 Γ ( α + 1 ) + ( a ( 2 c + d ) m + b c ( 1 + m ) − a ( c − d ) ) A M 1 Γ ( α ) + ( a c ( 2 m − 3 2 ) + ( a d + b c ) ( 2 m − 1 ) ) A M 1 Γ ( α − 1 ) + a c m M 2 + ( a c ( 2 m − 1 ) + ( a d + b c ) m ) M 3 + ( a c ( 2 m − 3 2 ) + ( a d + b c ) ( 2 m − 1 ) ) M 4 . Therefore, | ( T u ) ( t ) | ≤ A M 1 ∫ t k t ( t − s ) α − 1 Γ ( α ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m − 1 1 2 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m 1 2 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 I i ′ ( u ( t i ) ) + ∑ i = 1 m − 1 1 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m I i ′ ( u ( t i ) ) + ∑ i = 1 m 1 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 I i ″ ( u ( t i ) ) + | g 1 | + | g 2 | ≤ σ 1 ( 1 + m ) A M 1 Γ ( α + 1 ) + σ 1 m M 2 + ( σ 1 ( 2 m − 1 ) + σ 2 ( 1 + m ) ) A M 1 Γ ( α ) + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) A M 1 Γ ( α − 1 ) + ( σ 1 ( 2 m − 1 ) + σ 2 m ) M 3 + ( σ 1 ( 2 m − 3 2 ) + σ 1 ( 2 m − 1 ) ) M 4 : = M . Since t ∈ [ 0 , 1 ] , there exists a positive constant M , such that ‖ T u ‖ ≤ M , which implies that the operator T is uniformly bounded.However, for any t ∈ J k , 0 ≤ k ≤ m , we find | ( T u ) ′ ( t ) | ≤ A M 1 ∫ t k t ( t − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m − 1 A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m A M 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m I i ′ ( u ( t i ) ) + ∑ i = 1 m − 1 I i ″ ( u ( t i ) ) + ∑ i = 1 m I i ″ ( u ( t i ) ) + | g 2 | ≤ 1 a c + a d + b c a c A M 1 ( m + 1 ) Γ ( α + 1 ) + a c A M 1 ( 2 m − 1 ) + A M 1 σ 3 ( 1 + m ) Γ ( α ) + a c A M 1 ( 2 m − 3 2 ) + A M 1 σ 3 ( 2 m − 1 ) Γ ( α − 1 ) + a c m M 2 + ( a c ( 2 m − 1 ) + m σ 3 ) M 3 + ( a c ( 2 m − 3 2 ) + σ 3 ( 2 m − 1 ) ) M 4 : = M ¯ . Thus, for t 1 , t 2 ∈ J k , t 1 < t 2 , 0 ≤ k ≤ m , we have ∣ ( T u ) ( t 2 ) − ( T u ) ( t 1 ) ∣ ≤ ∫ t 1 t 2 ∣ ( T u ) ′ ( s ) ∣ d s ≤ M ¯ ( t 2 − t 1 ) , which implies that T is equicontinuous on all J k , k = 1 , 2 , ⋯ , m . Therefore, by Lemma 4, the operator T is completely continuous. □Theorem 2. Assume that there exist positive constants H i ( i = 1 , 2 , 3 , 4 ) , such that ∣ ( f ( t , u , u ′ ) ∣ ≤ φ p ( H 1 ) , ∣ I k ( u ) ∣ ≤ H 2 , ∣ I k ′ ( u ) ∣ ≤ H 3 , ∣ I k ″ ( u ) ∣ ≤ H 4 , the fractional differential Equation (3) has a solution.Proof. First, we prove that B = { u ∈ P C ( J , R ) ∣ u = λ T u , 0 < λ < 1 } is bounded. Let u ∈ B , and then u ∈ λ T u , 0 < λ < 1 . For any t ∈ J , we find | u ( t ) | = λ | ( T u ) ( t ) | ≤ A H 1 ∫ t k t ( t − s ) α − 1 Γ ( α ) d s + ∑ i = 1 m A H 1 ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) d s + ∑ i = 1 m − 1 A H 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m − 1 1 2 A H 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m A H 1 ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) d s + ∑ i = 1 m 1 2 A H 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m − 1 A H 1 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) d s + ∑ i = 1 m I i ( u ( t i ) ) + ∑ i = 1 m − 1 I i ′ ( u ( t i ) ) + ∑ i = 1 m − 1 1 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m I i ′ ( u ( t i ) ) + ∑ i = 1 m 1 2 I i ″ ( u ( t i ) ) + ∑ i = 1 m − 1 I i ″ ( u ( t i ) ) + | g 1 | + | g 2 | ≤ σ 1 ( 1 + m ) A H 1 Γ ( α + 1 ) + σ 1 m H 2 + ( σ 1 ( 2 m − 1 ) + σ 2 ( 1 + m ) ) A H 1 Γ ( α ) + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) A H 1 Γ ( α − 1 ) + ( σ 1 ( 2 m − 1 ) + σ 2 m ) H 3 + ( σ 1 ( 2 m − 3 2 ) + σ 1 ( 2 m − 1 ) ) H 4 : = H , which illustrates that ∥ u ∥ ≤ H . For any t ∈ J , the set B is bounded. Hence, by Lemma 5, the fractional differential Equation (3) has a solution. □Theorem 3. Let lim u → 0 f ( t , u , u ′ ) φ p ( u ) = 0 , lim u → 0 I k ( u ) u = 0 , lim u → 0 I k ′ ( u ) u = 0 , lim u → 0 I k ″ ( u ) u = 0 , and sup σ 1 ( m + 1 ) A Γ ( α + 1 ) δ 1 + ( σ 1 ( 2 m − 1 ) + σ 2 ( m + 1 ) ) A Γ ( α ) δ 1 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) A Γ ( α − 1 ) δ 1 + σ 1 m δ 2 + ( σ 1 ( 2 m − 1 ) + σ 2 m ) δ 3 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) δ 4 ≤ 1 . for some δ i > 0 ( i = 1 , 2 , 3 , 4 ) , then the fractional differential Equation (3) has a solution.Proof. Since lim u → 0 f ( t , u , u ′ ) φ p ( u ) = 0 , lim u → 0 I k ( u ) u = 0 , lim u → 0 I k ′ ( u ) u = 0 and lim u → 0 I k ″ ( u ) u = 0 ; therefore, there exists a constant s > 0 , such that | f ( t , u , u ′ ) | ≤ φ q ( δ 1 ) | φ p ( u ) | , | I k ( u ) | ≤ δ 2 | u | , | I k ′ ( u ) | ≤ δ 3 | u | and | I k ″ ( u ) | ≤ δ 4 | u | , for 0 < | u | < s , sup σ 1 ( m + 1 ) A Γ ( α + 1 ) δ 1 + ( σ 1 ( 2 m − 1 ) + σ 2 ( m + 1 ) ) A Γ ( α ) δ 1 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) A Γ ( α − 1 ) δ 1 + σ 1 m δ 2 + ( σ 1 ( 2 m − 1 ) + σ 2 m ) δ 3 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) δ 4 ≤ 1 . Hence, it follows that ‖ T u ‖ ≤ ‖ u ‖ , u ∈ ∂ Ω . Thus, by Lemma 6, the operator T has one fixed point. Thus, the fractional differential Equation (3) has a solution u ∈ Ω ¯ . □Theorem 4. There exist positive constants L i > 0 ( i = 1 , 2 , 3 ) such that | φ q ( f ( t , u , u ′ ) ) − φ q ( f ( t , v , v ′ ) ) | ≤ L 1 | u − v | , | I k ( u ) − I k ( v ) | ≤ L 2 | u − v | , | I k ′ ( u ) − I k ′ ( v ) | ≤ L 3 | u − v | , | I k ″ ( u ) − I k ″ ( v ) | ≤ L 4 | u − v | , for t ∈ J , u , v ∈ C ( [ 0 , 1 ] , R ) and k = 1 , 2 , ⋯ , m . If K = max t ∈ J σ 1 ( m + 1 ) A Γ ( α + 1 ) L 1 + ( σ 1 ( 2 m − 1 ) + σ 2 ( m + 1 ) ) A Γ ( α ) L 1 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) A Γ ( α − 1 ) L 1 + σ 1 m L 2 + ( σ 1 ( 2 m − 1 ) + σ 2 m ) L 3 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) L 4 < 1 , then the fractional differential Equation (3) has a unique solution.Proof. For u , v ∈ P C ( J , R ) , we obtain | T u ( t ) − T v ( t ) | = ∫ t k t ( t − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 k ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 k − 1 ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 k − 1 ( t k − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 k ( t − t k ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 k ( t − t k ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 k | I i ( u ( t i ) ) − I i ( v ( t i ) ) | + ∑ i = 1 k − 1 ( t k − t i ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 k − 1 ( t k − t i ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 k ( t − t k ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 k ( t − t k ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 k − 1 ( t − t k ) ( t k − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + 1 a c + a d + b c { b c ( ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 m | I i ( u ( t i ) ) − I i ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) + b d ( ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( η − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) − b ( c + d ) ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( ξ − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) } + t 1 a c + a d + b c a c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 m | I i ( u ( t i ) ) − I i ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) + a d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( η − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) + b c ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( ξ − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) } ) ≤ A ∫ t k t ( t − s ) α − 1 Γ ( α ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m A ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m − 1 A ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m − 1 1 2 A ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m A ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m 1 2 A ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m − 1 A ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( f ( t , u ( s ) , u ′ ( s ) ) − f ( t , v ( s ) , v ′ ( s ) ) ) | d s + ∑ i = 1 m | I i ( u ( t i ) ) − I i ( v ( t i ) ) | + ∑ i = 1 m − 1 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m − 1 | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m − 1 1 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m 1 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + 1 a c + a d + b c b c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m | I i ( u ( t i ) ) − I i ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) + b d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( η − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) − b ( c + d ) ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( ξ − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) } + t 1 a c + a d + b c a c ∫ t m 1 ( 1 − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ∫ t i − 1 t i ( t i − s ) α − 1 Γ ( α ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( t m − t i ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m ( 1 − t m ) 2 2 ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 m − 1 ( 1 − t m ) ( t m − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 m | I i ( u ( t i ) ) − I i ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) ( t m − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 m ( 1 − t m ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 m − 1 ( t m − t i ) 2 2 | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) + a d ∫ t l η ( η − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ( η − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( η − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) + b c ∫ t l ξ ( ξ − s ) α − 2 Γ ( α − 1 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l ∫ t i − 1 t i ( t i − s ) α − 2 Γ ( α − 1 ) | φ q λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r d τ | d s + ∑ i = 1 l ( ξ − t l ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l − 1 ( t l − t i ) ∫ t i − 1 t i ( t i − s ) α − 3 Γ ( α − 2 ) | φ q ( λ Γ ( β ) ∫ 0 s ( s − τ ) β − 1 ( ∫ 0 τ f ( r , u ( r ) , u ′ ( r ) ) − f ( r , v ( r ) , v ′ ( r ) ) d r ) d τ ) | d s + ∑ i = 1 l | I i ′ ( u ( t i ) ) − I i ′ ( v ( t i ) ) | + ∑ i = 1 l ( ξ − t l ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | + ∑ i = 1 l − 1 ( t l − t i ) | I i ″ ( u ( t i ) ) − I i ″ ( v ( t i ) ) | ) } ) ≤ σ 1 ( m + 1 ) A Γ ( α + 1 ) L 1 + ( σ 1 ( 2 m − 1 ) + σ 2 ( m + 1 ) ) A Γ ( α ) L 1 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) A Γ ( α − 1 ) L 1 + σ 1 m L 2 + ( σ 1 ( 2 m − 1 ) + σ 2 m ) L 3 + ( σ 1 ( 2 m − 3 2 ) + σ 2 ( 2 m − 1 ) ) L 4 ∥ u − v ∥ . Consequently, we find ∥ T u − T v ∥ ≤ K ∥ u − v ∥ . Hence, T is a contraction. Accordingly, the conjecture of the theorem follows by Lemma 3. This completes the proof. □ 4. ExampleIn this section, three examples are given to prove our conclusions.Example 1. Consider the following impulsive fractional differential equation ( D 0 + β c ( φ p D 0 + α c u ( t ) ) ) ) ′ = c o s u ( t ) 3 + u 3 ( t ) , 1 < α , β ≤ 2 , 0 < t < 1 , t ≠ 2 3 , Δ u ( 2 3 ) = e − u 2 ( 2 3 ) + 5 s i n ( 1 + e u ( 2 3 ) ) , Δ u ′ ( 2 3 ) = 7 + 3 u 2 ( 2 3 ) 2 + 5 u 2 ( 2 3 ) , Δ u ″ ( 2 3 ) = e 2 3 c o s u ( 2 3 ) 5 + u 3 ( 2 3 ) , u ( 0 ) − 2 u ′ ( 1 5 ) = 0 , 3 u ( 1 ) + 5 u ′ ( 1 7 ) = 0 , D 0 + α c u ( 0 ) = 0 , u ″ ( 0 ) = 0 , (9) where m = 1 a n d λ = 1 . Clearly, H 1 = 1 2 , H 2 = 6 , H 3 = 3 5 and H 4 = 4 5 , and the condition of Theorem 2 holds. Hence, by Theorem 2, the fractional differential Equation (11) has at least one solution.Example 2. Consider the following impulsive fractional differential equation ( D 0 + β c ( φ p D 0 + α c u ( t ) ) ) ) ′ = t 2 u 5 ( t ) + u 3 ( t ) a r c t a n 4 u ( t ) , 1 < α , β ≤ 2 , 0 < t < 1 , t ≠ 1 5 , Δ u ( 1 5 ) = e u 2 ( 1 5 ) u 2 ( 1 5 ) , Δ u ′ ( 1 5 ) = 2 u 5 ( 1 5 ) 3 + 2 u 3 ( 1 5 ) , Δ u ″ ( 1 5 ) = u 2 ( 1 5 ) + s i n ( 1 − c o s u ( 1 5 ) ) , 4 u ( 0 ) − 3 u ′ ( 2 7 ) = 0 , 2 u ( 1 ) + 5 u ′ ( 4 5 ) = 0 , D 0 + α c u ( 0 ) = 0 , u ″ ( 0 ) = 0 , (10) where m = 1 a n d λ = 1 . lim u → 0 f ( t , u , u ′ ) φ p ( u ) = lim u → 0 t 2 u 5 ( t ) + u 3 ( t ) a r c t a n 4 u ( t ) u | u | = 0 , lim u → 0 e u ( t ) u 2 ( t ) u = 0 , lim u → 0 2 u 5 ( t ) 3 + 2 u 3 ( t ) u = 0 , lim u → 0 u 2 ( t ) + s i n ( 1 − c o s u ( t ) ) u = 0 . Clearly, the condition of Theorem 3 holds, and we can find that the fractional differential Equation (10) has a solution.Example 3. Consider the following impulsive fractional differential equation ( D 0 + β c ( φ p ( c D 0 + α u ( t ) ) ) ) ′ = s i n u ( t ) 5 + u 2 ( t ) , 1 < α , β ≤ 2 , 0 < t < 1 , t ≠ 3 7 , Δ u ( 3 7 ) = e − u 2 ( 3 7 ) + 3 s i n ( 1 + e u ( 3 7 ) ) , Δ u ′ ( 3 7 ) = 5 + 3 u 2 ( 3 7 ) 6 + 3 u 2 ( 3 7 ) , Δ u ″ ( 3 7 ) = e 3 7 c o s u ( 2 3 ) 5 + u 3 ( 3 7 ) , 2 u ( 0 ) − u ′ ( 1 5 ) = 0 , u ( 1 ) + 3 u ′ ( 1 7 ) = 0 , D 0 + α c u ( 0 ) = 0 , u ″ ( 0 ) = 0 , (11) where m = 1 a n d λ = 1 . Clearly, L 1 = 1 3 , L 2 = 6 , L 3 = 4 5 , L 4 = 3 5 , and we gain the condition of Theorem 4. Hence, by Theorem 4, the fractional differential Equation (11) has a unique solution. 5. ConclusionsIn this paper, by using fixed-point theorems, the existence and uniqueness of positive solutions to a class of four-point impulsive fractional differential equations with p-Laplacian operators are studied. Three examples are provided at the end to sustain and validate the potentiality of all our acquired epilogs. 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Mathematics 2022, 10, 1852. https://doi.org/10.3390/math10111852 AMA Style Chu L, Hu W, Su Y, Yun Y. Existence and Uniqueness of Solutions to Four-Point Impulsive Fractional Differential Equations with p-Laplacian Operator. Mathematics. 2022; 10(11):1852. https://doi.org/10.3390/math10111852 Chicago/Turabian Style Chu, Limin, Weimin Hu, Youhui Su, and Yongzhen Yun. 2022. "Existence and Uniqueness of Solutions to Four-Point Impulsive Fractional Differential Equations with p-Laplacian Operator" Mathematics 10, no. 11: 1852. https://doi.org/10.3390/math10111852 APA Style Chu, L., Hu, W., Su, Y., & Yun, Y. (2022). Existence and Uniqueness of Solutions to Four-Point Impulsive Fractional Differential Equations with p-Laplacian Operator. Mathematics, 10(11), 1852. https://doi.org/10.3390/math10111852 Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. 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