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Positive Solutions for Higher Order Nonlocal Fractional Differential Equation with Integral Boundary Conditions
In this paper, by using the spectral analysis of the relevant linear operator and Gelfand’s formula, some properties of the first eigenvalue of a fractional differential equation were obtained; combining fixed point index theorem, sufficient conditions for the existence of positive solutions are established. An example is given to demonstrate the application of our main results.
Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, control theory, signal and image processing, and aerodynamics (see [1, 2]). In the last decade, boundary value problems of fractional calculus have received a great attention, based on various analytic techniques, and a variety of results concerning the existence of solutions can be found in the literature [3–19]. For example, Zhang  studied the following singular fractional boundary value problem (FBVP): where , is the standard Riemann-Liouville derivative, the nonlinear term , and and have different monotone properties. By using a mixed monotone method, a unique positive solution is obtained.
Integral boundary conditions arise in thermal conduction problems , semiconductor problems , and hydrodynamic problems . Recently, the existence and multiplicity of positive solutions for nonlinear ordinary differential equations with integral boundary value problems have received a great deal of attention. To identify a few, see [23–27] and the references therein for integer order integral boundary value problems and [28–33] for fractional order integral boundary value problems. In , authors using monotone iterative technique investigated the existence and uniqueness of the positive solutions of higher-order nonlocal fractional differential equations of the typewhere By means of Schauders fixed point theorem, FBVP (2) are also studied in . In , the authors investigated problems (2) with replaced by , and the existence and multiplicity of positive solutions are obtained by means of the fixed point index theory in cones.
Inspired by the work of the above papers, the aim of this paper is to establish the existence of positive solutions for the following nonlinear fractional differential equation involving Stieltjes integrals conditions:where is the Riemann-Liouville fractional derivative, , is a linear functional on given by a Stieltjes integral with representing a suitable function of bounded variation, and can be a signed measure. is a continuous function, the nonlinearity may be singular at , and Here and .
Our work presented in this paper has the following features. First of all, we discuss the boundary value problem with the Stieltjes integral boundary conditions appearing in the boundary conditions of FBVP (2) as more general which was not considered in the literature , covering two-point, multipoint, and nonlocal problems as special cases. The second new feature is that the nonlinearity is allowed to depend on higher order derivatives of unknown function up to order which was not considered in the literature [28–30]. Thirdly, we allow that the nonlinearity may be singular at and Therefore, our study improves and extends the previous results to some degree in the relevant literature [17, 28–30].
The rest of this paper is organized as follows. In Section 2, we present some lemmas that are used to prove our main results. In Section 3, the existence of positive solutions is established under some sufficient conditions. In Section 4, an example is given to demonstrate the application of our theoretical results.
2. Basic Definitions and Preliminaries
Let , and then is a Banach space with the norm , for any . Let be a cone in and construct a subcone of as follows: For any , let , , and .
Lemma 1 (see ). Let . Then the boundary value problem has a unique solution
Lemma 2. Suppose that for and , and then the functions and have the following properties: (1) and are nonnegative and continuous for (2) satisfies (3) satisfies where
Throughout this paper, we adopt the following assumptions: (H0) is a function of bounded variation, for and (H1) is a continuous function and, for any ,
In order to overcome the difficulty due to the dependance of on derivatives, we consider the following modified problem:where
Define a nonlinear operator and a linear operator as follows:Observe that problem (14) has solutions if the operator equation has fixed points.
Lemma 4 (Krein-Rutmann ). Let be a continuous linear operator, be a total cone, and . If there exist and a positive constant such that , then the spectral radius and has a positive eigenfunction corresponding to its first eigenvalue .
Lemma 5 (Gelfand’s formula ). For a bounded linear operator and the operator norm , the spectral radius of satisfies
Lemma 6. Assume that holds. Then defined by (16) is a completely continuous linear operator, and the spectral radius ; moreover has a positive eigenfunction corresponding to its first eigenvalue .
Proof. For any , by Lemma 2, we can obtainOn the other hand, from Lemma 2, we also haveThen (18) and (19) yield that Consequently, . And, from the uniform continuity of , , we have that is a completely continuous linear operator.
Next we will show that has the first eigenvalue by using Krein-Rutmann’s theorem. In fact, by Lemma 2, there exists such that . Thus there exists such that and for all . Choose such that and for all . Then, for any , we have So there exists such that for . From Lemma 4, we know that the spectral radius and has a positive eigenfunction corresponding to its first eigenvalue ; i.e., The proof is completed.
Lemma 7. Assume that hold; then is completely continuous.
Proof. First, we prove . In fact, for any , Lemma 2 implies that Hence, On the other hand, from Lemma 2, we also have Therefore, .
Next, for any , we assert thatwhich implies that is well defined. In fact, it follows from that there exists a natural number such thatFor any , as , we haveandwhere is the beta function. So, for any , (27) and (28) yield thatFrom (26) and (29), we havewhere Thus (25) is true which implies that is uniformly bounded on any bounded set.
Now we will show that is continuous. Let and For any , by , there exists a natural number such thatSince is uniformly continuous on , we have that holds uniformly on . It follows from the Lebesgue control convergence theorem that Thus, for the above , there exists a natural number such that, for , we haveIt follows from (32) and (35) that when , Hence, is continuous.
For any bounded set , we show that is equicontinuous. In fact, by , for any , there exists a natural number such that Let Since is uniformly continuous on , for the above and a fixed , there exists such that for , . Hence, for , , and , we have which show that is equicontinuous. According to the Ascoli-Arzela theorem, is completely continuous. The proof is completed.
Lemma 8 (see ). Let be a cone in Banach space . Suppose that is a completely continuous operator.(i)If there exists such that for any and , then .(ii)If for any and , then .
3. Main Results
Proof. By (41), there exists such thatLet be the positive eigenfunction of corresponding to , and thus . For any , by virtue of (27) and (28), that isfrom (43) and (44), we haveWe may suppose that has no fixed point on (otherwise, the proof is finished). Now we show thatIf, otherwise, there exist and such that , obviously, and Let , and then and . Since , we have Therefore, by (45), we havewhich contradicts the definition of . Hence (46) is true and it follows from Lemma 8 thatNow we choose a constant and define a linear operator , and then is a bounded linear operator and . Moreover , so the spectral radius of is and also has the first eigenvalue By Gelfand’s formula, we knowLet ; by (49), there exists a sufficiently large natural number such that implies that For any , definewhere is the identity operator. Clearly, is also the norm of .
On the other hand, it follows from (42) that there exists such thatLet and then, by (25), we know that . For the sake of convenience, let and choose In the following we will prove thatIf, otherwise, there exist and such that . It follows from and that there exists such that . Thus let , and then we have and . Now set and then, for any , we have and . It follows from (51) and Lemma 2 thatNoticing that is a bounded linear operator, and from (56), we haveThen (57) yields that which leads toSince and , from (50), we have