Positive solutions for a system of fractional differential equations with p-Laplacian operator and multi-point boundary conditions

>We investigate the existence and nonexistence of positive solutions for a system of nonlinear Riemann–Liouville fractional differential equations with parameters and p-Laplacian operator subject to multi-point boundary conditions, which contain fractional derivatives. The proof of our main existence results is based on the Guo–Krasnosel'skii fixed-point theorem.

The paper is organized as follows.In Section 2, we investigate two nonlocal boundary value problems for fractional differential equations with p-Laplacian, and we present some properties of the associated Green functions.Section 3 contains the main existence theorems for the positive solutions with respect to a cone for our problem (S)-(BC) based on the Guo-Krasnosel'skii fixed-point theorem (see [4]).In Section 4, we study the nonexistence of positive solutions of (S)-(BC), and in Section 5, an example is given to support our results.In Appendix we prove a relation between the supremum limits of two functions, which is used in the proof of the second existence result.

R. Luca
Next, we consider the nonlinear fractional differential equation with the boundary conditions where We denote by and by G 2 , g 3 , g 4 the following Green functions: In a similar manner as above, we obtain the following result.

Existence of positive solutions
In this section, we present sufficient conditions on the functions f , g and intervals for the parameters λ, µ such that positive solutions with respect to a cone for our problem (S)-(BC) exist.
We present now the assumptions that we will use in the sequel.
, we introduce the following extreme limits: By using Lemmas 1 and 2 (relations ( 12) and ( 18)) a solution of the nonlinear system of integral equations is solution of problem (S)-(BC).
Nonlinear Anal.Model.Control, 23(5):771-801 We consider the Banach space X = C[0, 1] with the supremum norm • and the Banach space Y = X × X with the norm (u, v) Y = u + v .We define the cones and We define now the operators and Proof.Let (u, v) ∈ P be an arbitrary element.Because and problem ( 13)-( 14) for k(t) = µg(t, u(t), v(t)), t ∈ [0, 1], respectively, then we obtain and so Therefore, for all t ∈ [0, 1], we conclude that )) ∈ P , and then Q(P ) ⊂ P .By the continuity of the functions f , g, G 1 , G 2 and the Ascoli-Arzela theorem we can show that Q 1 and Q 2 are completely continuous operators (compact operators, that is they map bounded sets into relatively compact sets, and continuous), and then Q is a completely continuous operator.
where J 1 and J 2 are defined in Lemma 3. First, for where Theorem 1. Assume that (H1) and (H2) hold, Proof.We consider the above cone P ⊂ Y and the operators Q 1 , Q 2 , and Q.Because the proofs of the above cases are similar, in what follows, we will prove some representative cases.
(i) We have and .
By using (H2) and the definitions of f s 0 and g s 0 , we deduce that there exists We define the set and by Lemma 3 we obtain Therefore, we have In a similar manner, we conclude Next, by the definitions of Nonlinear Anal.Model.Control, 23(5):771-801

R. Luca
We define the set In a similar manner as in the proof of case (i), for any (u, v) ∈ P ∩ ∂Ω 1 , we obtain and so The second part of the proof is the same as the corresponding one from case (i).For Ω 2 defined in case (i) and for any (u, v) ∈ P ∩ ∂Ω 2 , we conclude and then Therefore, we deduce the conclusion of the theorem.
By using (H2) and the definitions of f s 0 and g s 0 we deduce that there exists R 1 > 0 such that We define the set In a similar manner as in the proof of case (i), for any (u, v) ∈ P ∩ ∂Ω 1 , we obtain and so For the second part of the proof, by the definition of f i ∞ there exists R 2 > 0 such that We consider R 2 = max{2R 1 , R 2 /γ} and define Then by Lemma 3 we have Therefore, we deduce the conclusion of the theorem.
Proof.We consider the cone P ⊂ Y and the operators Q 1 , Q 2 , and Q defined at the beginning of this section.Because the proofs of the above cases are similar, in what follows we will prove some representative cases.
(i) We have Nonlinear Anal.Model.Control, 23(5):771-801 R. Luca By using (H2) and the definitions of f i 0 and g i 0 we deduce that there exists R 3 > 0 such that then by Lemma 3 we obtain Therefore, we conclude In a similar manner, we deduce Hence, we get Now we define the functions f * , g Then The functions f * (t, •), g * (t, •) are nondecreasing for every t ∈ [0, 1], and they satisfy the conditions (see Appendix) Therefore, for ε > 0, there exists R 4 > 0 such that for all x R 4 and t ∈ [0, 1], we have By the definitions of f * and g * we conclude Nonlinear Anal.Model.Control, 23(5):771-801 Then for all t ∈ [0, 1], we obtain and so In a similar manner, we deduce and then By using Lemma 4, ( 21), (23), and the Guo-Krasnosel'skii fixed-point theorem we conclude that Q has a fixed point (u, v) The first part of the proof is the same as the corresponding one from case (i).For Ω 3 defined in case (i), for (u, v) ∈ P ∩ ∂Ω 3 , we obtain and so For the second part, we use the same functions f * and g * from case (i), which satisfy in this case the conditions Therefore, for ε > 0 there exists R 4 > 0 such that for all x R 4 and t ∈ [0, 1], we have and so f * (t, x) εx r1−1 and g * (t, x) By the definitions of f * and g * we obtain relations (22).In addition, in a similar manner as in the proof of case (i), we conclude and so Therefore, we deduce the conclusion of the theorem.
By (H2) and the definition of g i 0 we deduce that there exists R 3 > 0 such that We define by Lemma 3 we obtain For the second part of the proof, we consider the functions f * and g * from case (i), which satisfy in this case the conditions Then for ε > 0, there exists R 4 > 0 such that for all x R 4 and t ∈ [0, 1], we have By the definitions of f * and g * we deduce relations (22).In addition, in a similar manner as in the proof of case (i), we conclude and so https://www.mii.vu.lt/NATherefore, we obtain the conclusion of the theorem.
By using (H2) and the definition of f i 0 we deduce that there exists R 3 > 0 such that We denote For the second part of the proof, we consider the functions f * and g * from case (i), which satisfy in this case the conditions Then for ε > 0, there exists R 4 > 0 such that for all x R 4 and t ∈ [0, 1], we have and so f * (t, x) εx r1−1 and g * (t, x) εx r2−1 .We consider R 4 = max{2R 3 , R 4 } and denote By the definitions of f * and g * we obtain relations (22).In addition, in a similar manner as in the proof of case (i), we deduce and so Therefore, we obtain the conclusion of the theorem.
Remark 1.In the proof of Theorem 3, we can also define λ 0 = (α 1 /B) r1−1 /M 1 and ∞, then there exist positive constants M 1 , M 2 such that relation (24) holds, and then we obtain the conclusion of Theorem 3.
then there exists a positive constant λ 0 such that for every λ > λ 0 and µ > 0, the boundary value problem (S)-(BC) has no positive solution.
Let µ > µ 0 and λ > 0. We suppose that (S)-(BC) has a positive solution Then we deduce R. Luca then there exist positive constants λ0 and μ0 such that for every λ > λ0 and µ > μ0 , the boundary value problem (S)-(BC) has no positive solution.
(i) If for c 1 , c 2 with 0 < c 1 < c 2 1, we have f i 0 , f i ∞ > 0 and f (t, u, v) > 0 for all t ∈ [c 1 , c 2 ] and u, v 0 with u + v > 0, then relation (25) holds, and we obtain the conclusion of Theorem 4. (ii) If for c 1 , c 2 with 0 < c 1 < c 2 1, we have g i 0 , g i ∞ > 0 and g(t, u, v) > 0 for all t ∈ [c 1 , c 2 ] and u, v 0 with u + v > 0, then relation (26) holds, and we obtain the conclusion of Theorem 5.
and u, v 0 with u + v > 0, then relation (27) holds, and we obtain the conclusion of Theorem 6.
we deduce the conclusion of the theorem.https://www.mii.vu.lt/NA

Theorem 5 .
which is a contradiction.Therefore, the boundary value problem (S)-(BC) has no positive solution.Assume that (H1) and (H2) hold.If there exist positive numbers c 1 , c 2 with 0

Theorem 6 .
which is a contradiction.Therefore, the boundary value problem (S)-(BC) has no positive solution.Assume that (H1) and (H2) hold.If there exist positive numbers c 1 , c 2 with 0