Existence and Uniqueness Results for a Class of Singular Fractional Boundary Value Problems with the p-Laplacian Operator via the Upper and Lower Solutions Approach

In this paper, we study the existence and uniqueness of positive solutions to a class of multipoint boundary value problems for singular fractional differential equations with the p-Laplacian operator. Here, the nonlinear source term f permits singularity with respect to its time variable t. Some fixed-point theorems such as the Leray-Schauder nonlinear alternative, the Schauder fixed-point theorem, and the Banach contraction mapping principle and the properties of the Gauss hypergeometric function are used to prove our main results. And by employing the upper and lower solutions technique, we derive a new approach to obtain the maximal and minimal solutions to the given problem. Finally, we present some examples to demonstrate our existence and uniqueness results.

Fractional calculus has a history of several hundred years, and many valuable results, which have contributed to the development of mathematical theories and their application to practice, have been created during its historical process (see [1]). Also, fractional differential equations are one of the powerful tools to model and solve scientific and technological problems arising in physics, chemistry, biology, and mechanics, and it has developed more and more in depth (see [2]). In particular, after Leibenson's work dealt with the application of the integer-order differential equation with the p-Laplacian operator to the analysis of turbulent flow in porous media (see [3]), many valuable existence results for this equation have been achieved, and recently, the achievements obtained in this integer-order differential equation are more generalized to the fractional differential equation (see [4][5][6] and the references therein). However, due to the nonlinearity of the p-Laplacian operator, not much has been studied on the solutions to singular fractional differential equations with this operator and many researchers have been paid their attention to those equations. For instance, by using the fixed-point theorem of mixed monotone operators, Jong et al. [7] proved the existence of positive solutions to the boundary value problem (1), in which the nonlinear source term f ðt, xÞ was singular with respect to its space variable x, and proposed a new approach by which the approximate solution of the given problem could be obtained. Unlikely in [7], this paper deals with the boundary value problem (1), in which the function f ðt, xÞ permits singularity with respect to its time variable t.
Many researchers have derived some important results for solutions to boundary value problems of fractional differential equations with singularity with respect to the time variable (see ). In [14], Henderson et al. established the existence and multiplicity of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations with the coupled multipoint boundary conditions where α ∈ ðn − 1, n, β ∈ ðm − 1, m and the functions f , g can be singular at the points t = 0 and/or t = 1. They employed the Guo-Krasnosel'skii fixed-point theorem to prove that their problem has at least one positive solution. And Wu and Zhou [6] used the upper and lower solutions method to study the existence of positive solutions for the fractional-order eigenvalue problem with the p-Laplacian operator where 1 < α ≤ 2, 0 < β, γ ≤ 1 and f can be singular at t = 0, 1 and x = 0 (for more detailed information about the upper and lower solutions method to solve integral and differential equations, see [30]). Taking the previous results together, to our best knowledge, very little is known about the existence and uniqueness of positive solutions of p-Laplacian fractional boundary value problems with singularities with respect to their time variable.
Motivated by the above works, in this paper, we first apply the Leray-Schauder nonlinear alternative to establish the existence of solutions to our problem (1) and then use the Schauder fixed-point theorem and upper and lower control functions to derive the upper and lower solutions method to obtain the maximal and minimal solutions. And we prove the uniqueness of solutions to the given problem by using some useful properties of the Gauss hypergeometric function 2 F 1 ða, b, c, sÞ and the Banach contraction mapping principle.
Throughout this paper, we suppose that

Preliminaries
For the convenience of the readers, we will give some necessary definitions and lemmas here. The Riemann-Liouville fractional integral and the Riemann-Liouville fractional derivative of order α > 0 of a function f : ð0, ∞Þ ⟶ R are given by where n = ½α + 1, provided that the right-hand sides are pointwise defined on ð0, ∞Þ (see [1]).
Lemma 2 (see [32]). (Schauder fixed-point theorem). Let D be a nonempty, closed, bounded, and convex subset of a Banach space X, and suppose A : D ⟶ D is a compact operator.
Then, A has a fixed point.
Lemma 3 (see [33]). Let X be a Banach space with D a closed and convex subset of X. Assume U is a relatively open subset of D, with 0 ∈ U, and A : U ⟶ D is a compact map. Then, either, Take the limit t ′ ⟶ 0 + to obtain This implies lim t→0+ KðtÞ = 0 = Kð0Þ.
If t ∈ ðδ, 1 − δ, we can get In a similar way above, the first term of the right side in Equation (17) can be evaluated as And for the second term of the right side in Equation (17), it can be easily seen that Combining these two inequalities above, we can find For the case t ∈ ð1 − δ, 1, if t < 1, then we have Similarly to that given above, we obtain By simple calculation, we can get So for any t ∈ ð1 − δ, 1Þ, it holds that If t = 1, the definition of an improper integral also implies Kð1Þ = lim t→1− KðtÞ. Taking the limit t ⟶ 1 − on both sides of the inequality (11), we can obtain Therefore, it follows directly from the inequalities (7) and (10), (11), (12), (13) that K is well-defined on ½0, 1. Since z 3 Journal of Function Spaces ∈ Cð0, 1Þ, it is obvious that K is continuous on ð0, 1Þ. Combining this with the continuity of K at t = 0, 1, we can prove that K is a continuous function on ½0, 1.

Lemma 5.
Let z : ð0, 1Þ ⟶ ½0, +∞Þ be a continuous function such that satisfies (6). Then, the boundary value problem has a unique solution which is given by where in which where Remark 6. In Lemma 5, a function v ∈ C½0, 1 with a fractional derivative of order β that belongs to Cð0, 1Þ (i.e., D β 0+ u ∈ Cð0, 1Þ) is said to be a solution of the boundary value problem (14) if it satisfies the fractional differential equation and the boundary conditions of (12).
Proof. As we can see in the proof of Lemma 4, we have So, we can get This implies z ∈ Lð0, 1Þ. Also, Lemma 4 asserts that Since z ∈ Cð0, 1Þ ∩ Lð0, 1Þ and I β 0+ z ∈ C½0, 1, it follows from Lemma 1 that a solution of the boundary value problem (14), For the rest of the proof, it is easy to see that doing as in the proof of Lemma 4 in [12] leads to a conclusion of this lemma.
then the function Hðt, sÞ in Lemma 5 satisfies the following conditions: As we can see in the proof of Lemma 4 in [4], it holds that Lemma 8 (see [34]). Let y ∈ C½0, 1. Then, the boundary value problem has a unique solution which is given by Journal of Function Spaces in which where Lemma 9 (see [34]).
then the function Gðt, sÞ in Lemma 8 satisfies the following conditions: The following useful properties of φ p ð⋅Þ which will be used later can be found in [5]:

Main Results
In this section, we will prove the existence and uniqueness of positive solutions for the boundary value problem (1) and derive the upper and lower solutions method by using some fixed-point theorems.
3.1. The Existence Results for Problem (1) is called a solution of problem (1) if it satisfies the fractional differential equation and the boundary conditions of (1).
The following hypothesis concerned with the function f ðt, xÞ, which permits singularity with respect to time variable, will be used in this article.
Lemma 11. Assume that the hypothesis (H1) holds. Then, the function x ∈ P is a solution of the problem (1) if and only if x is a solution of the integral equation in P, where q is a real number such that ð1/pÞ + ð1/qÞ = 1.
Proof. Suppose that x ∈ P is a solution of the problem (1).
Conversely, suppose that x ∈ P is a solution of the integral Equation (46). Put as follows: So, we can get and it follows that D α we can see This yields D β 0+ ðφ p ðD α 0+ xÞÞ ∈ Cð0, 1Þ. Since the boundary value problems (14) and (16) have unique solutions by Lemma 5 and Lemma 8, we can find that the function x satisfies the fractional differential equation and the boundary conditions of the problem (1). This completes the proof.
Define the operator T on P as follows: Then, the function x ∈ P is a solution of the integral Equation (46) if and only if the operator T has a fixed point x in P.

Lemma 12.
If the hypothesis (H1) holds, then Proof. The conclusion of this lemma easily follows from Lemma 4, so we omit the details.

Lemma 13.
If the hypothesis (H1) holds, then the operator T : P ⟶ P is completely continuous.
Proof. We first prove that the operator T is continuous on P.
For this, choose any x ∈ P and any sequence fx n g ∞ n=1 ⊂ P convergent to x. Then, there exists L 1 > 0 such that ∀n, kx n k ≤ L 1 . From the continuity of the function F, we can put From the definition of the function F, we can see that for any 0 < t < 1, Since x n ∈ C½0, 1, we have A simple calculation provides that and therefore, by using the Lebesgue dominated convergence theorem, we know Journal of Function Spaces Combining this with H ∈ Cð½0, 1 × ½0, 1Þ, we can obtain that for any s ∈ ½0, 1, Since G ∈ Cð½0, 1 × ½0, 1Þ, we can get that for any t ∈ ½0, 1, This gives Next, we show that for any bounded set Ω ⊂ P, TðΩÞ is relatively compact. To do this, by the Arzela-Ascoli theorem, it is sufficient to prove that TðΩÞ is uniformly bounded and equicontinuous. Denote Ω = fu ∈ P : kuk ≤ L 2 g. From Lemma 7 and Lemma 9, we see that for any x ∈ Ω and any t ∈ ½0, 1, Put M 2 ≔ max 0≤t≤1 0≤x≤L 2 jFðt, xÞj. Since kxk ≤ L 2 , we have Therefore, we know This means that TðΩÞ is uniformly bounded.
By using the inequality (20), we can get that for any t 1 , t 2 ∈ ½0, 1, Combining this inequality with the uniform continuity of Gðt, sÞ on ½0, 1 × ½0, 1, it holds that for any ε > 0, there exists δ > 0 such that This implies that TðΩÞ is equicontinuous. The proof is completed.
Put M = ½ðα − γÞAΓðαÞ p−1 BΓðβÞ and list more hypotheses to be used in this paper.
Proof. Putting U R = fu ∈ Pjkuk < Rg, it is obvious that U R = B R . By using Lemma 13, we can know that the operator T : U R ⟶ P is completely continuous. Assume that there exist a point x ∈ ∂U R and a number λ ∈ ð0, 1Þ, with x = λTx. By the hypothesis (H2), we see that for any t ∈ ½0, 1, 7 Journal of Function Spaces Using Lemma 7 and Lemma 9, we have Combining these two inequalities, we obtain Now taking ðp − 1Þth power on both sides of the above inequality, since kxk = R, ðp − 1Þðq − 1Þ = 1, and g is a nondecreasing function, we get This is a contradiction to the hypothesis (H3). Therefore, by Lemma 3, the operator T has at least one fixed point in B R . The proof is completed.

Derivation of the Upper and Lower Solutions Method.
We define the upper control function Kðt, xÞ and the lower control function kðt, xÞ on ð0, 1Þ × ½0, R as follows: Then, we know that the functions Kðt, xÞ and kðt, xÞ are nondecreasing in x and for any ðt, xÞ ∈ ð0, 1Þ × ½0, R, Definition 15. The functions x ∈ B R and x ∈ B R are said to be a upper solution and a lower solution of the integral Equation (46) in B R , respectively, if Proof. It is obvious that S is a nonempty, closed, bounded, and convex subset of the Banach space E and the operator T : B R ⟶ P is completely continuous. In a similar way to the proof of the Theorem 14, for any x ∈ B R , we can get By the hypothesis (H3), we have That is, TðB R Þ ⊂ B R . Now we prove TðSÞ ⊂ S. In fact, since the functions Kðt, xÞ and kðt, xÞ are nondecreasing in x, by using the definition of upper and lower solution and the inequality (21), we obtain Therefore, Lemma 2 assures that the problem (1) has at least one solution in S.
Proof. From the inequality (23), similar to the proof of Theorem 14, we can see that for any t ∈ ½0, 1, By using the estimation (22), we know that for any ðt, xÞ ∈ ð0, 1Þ × ½0, R, Since for any t ∈ ½0, 1, 0 ≤ ωðtÞ ≤ ωðtÞ ≤ R, we have In other words, it holds that for any t ∈ ð0, 1Þ, Using the inequalites (24), we obtain ω t ð Þ ≥ This implies that the functions ωðtÞ, ωðtÞ are upper and lower solutions of the integral Equation (46) in B R , respectively. Therefore, by Lemma 16, the problem (1) has at least one solution in S * .

Theorem 18.
Assume that the function f ðt, xÞ is nondecreasing in x and the hypotheses (H1)-(H3) are satisfied. Then, the problem (1) has a maximal solution x max ðtÞ and a minimal solution x min ðtÞ in S * and the following estimation holds: Proof. Put x 0 ≔ ω, x 0 ≔ ω. Theorem 17 shows that the functions x 0 , x 0 are upper and lower solutions of the integral Equation (46). Now, we construct the iterative sequences f x n g, fx n g as follows: Since the function f ðt, xÞ is nondecreasing in x, we know that if Considering that the functions x 0 , x 0 are upper and lower solutions of the integral Equation (46), we also have So, it is obvious that for any n ∈ N, This implies that for the sequences f x n g, fx n g, passing to the limit as n ⟶ ∞, the limits exist. And denoting we can get Taking the limit n ⟶ ∞ on both sides of the inequality (25), by the Lebesgue dominated convergence theorem, we obtain Therefore, it holds that the functions x min , x max are solutions to the problem (1).
Then, the following estimation holds Proof. To estimate the function Ð 1 0 Hðs, τÞτ −σ 0 ð1 − τÞ −σ 1 dτ by using (15), we should calculate the fractional integral of order β of the function s −σ 0 ð1 − sÞ −σ 1 . By the properties (i), (iv) of the Gauss hypergeometric function, we have So, we obtain Journal of Function Spaces By putting dðsÞ = s 1−σ 0 χðsÞ, we can see The function dðsÞ is an increasing function on ½0, 1. In fact, it is obvious that dð0Þ = 0 and dðsÞ ≥ 0 for any s ∈ ½0, 1, and by using the property (iii) of the Gauss hypergeometric function, we can get that for any s > 0, The proof is completed.
From the property (ii) of the Gauss hypergeometric function, we can calculate χð1Þ as (H4). There exists L > 0 such that for any t ∈ ½0, 1 and any x 1 , x 2 ∈ ½0, R, Lemma 22. Assume that the hypotheses (H1)-(H4) are satisfied. Then, the followings hold for any x, y ∈ B R : (ii) If p > 2, then Then, the sequences f x n g ∞ n=0 , fx n g ∞ n=0 are convergent to the maximal and minimal solutions of the problem (30), respectively.
Putting L = 1/20 and Fðt, xÞ = ððln ð2 + xÞÞ/10Þ, it holds that for any t ∈ ½0, 1 and any x 1 , x 2 ≥ 0, This assures that the hypothesis (H4) holds. A simple calculation gives Checking the condition (i) of Theorem 23, we obtain Finally, we can find that the boundary value problem (30) has a unique solution and the iterative sequences (32) are convergent to the unique solution of the given problem.

Conclusion
In this paper, we have studied the existence and uniqueness results and upper and lower solution methods for multipoint boundary value problems for singular fractional differential 13 Journal of Function Spaces equations with the p-Laplacian operator. For this purpose, we used some fixed-point theorems such as the Leray-Schauder nonlinear alternative and the Schauder fixed-point theorem to prove the existence of positive solutions to the problem (1). And by employing the upper and lower solutions technique, we derived a new approach to obtain the maximal and minimal solutions to the given problem. After that, some useful properties of the Gauss hypergeometric function were applied to the establishment of our uniqueness results. For the application of this work, two expressive examples were illustrated.

Data Availability
No data were used to support this study.

Conflicts of Interest
There is no competing interest among the authors regarding the publication of the article.