#### Abstract

By using the extension of Mawhin’s continuation theorem due to Ge, we consider boundary value problems for fractional -Laplacian equation. A new result on the existence of solutions for the fractional boundary value problem is obtained, which generalizes and enriches some known results to some extent from the literature.

#### 1. Introduction and Preliminaries

Recently, fractional differential equations have played an important role in many fields such as physics, electrical circuits, and control theory (see ). Many scholars have paid more attention to boundary value problems for fractional differential equations (see ).

By using a fixed point theorem on a cone, Agarwal et al. (see ) considered a two-point boundary value problem at nonresonance given by where , are real numbers, , and is the Riemann-Liouville fractional derivative.

By using the coincidence degree theory, Bai (see ) considered the following -point fractional boundary value problems: where is a real number, , are given constants such that , and are the Riemann-Liouville differentiation and integration.

The turbulent flow in a porous medium is a fundamental mechanics problem. For studying this type of problems, Leibenson (see ) introduced the -Laplacian equation as follows: where , . Obviously, is invertible and its inverse operator is , where is a constant such that .

In the past few decades, many important results relative to (3) with certain boundary value conditions have been obtained. We refer the reader to  and the references cited therein. However, to the best of our knowledge, there are relatively few results on boundary value problems for fractional -Laplacian equations.

Motivated by the work above, in this paper, we investigate the existence of solutions for boundary value problem (BVP for short) of fractional -Laplacian equation with the following form: where , , is Caputo fractional derivative, and is continuous.

BVP (4) happens to be at resonance in the sense that its associated linear homogeneous boundary value problem has a nontrivial solution , where .

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory, which can be found, for instance, in .

Definition 1. The Riemann-Liouville fractional integral operator of order of a function is given by provided that the right side integral is pointwise defined on .

Definition 2. The Caputo fractional derivative of order of a continuous function is given by where is the smallest integer greater than or equal to , provided that the right side integral is pointwise defined on .

Lemma 3. Assume that , . Then where , , and here is the smallest integer greater than or equal to .
Now, one briefly recalls some notations and an abstract existence result, which can be found in .

Definition 4. Let and be two Banach spaces with norms and , respectively. A continuous operator is said to be quasilinear if(i)Im  is a closed subset of ,(ii)Ker  is linearly homeomorphic to , .

Definition 5. Let be a real Banach space and let . The operator is said to be a projector provided that , for and . The operator is said to be a semiprojector provided .

Definition 6 (see ). Let and let be the complement space of in , and then . On the other hand, suppose that is a subspace of and is the complement space of in so that . Let be a projector, let be a semiprojector, and let be an open and bounded set with origin , where is the origin of a linear space.
Suppose that , , is a continuous operator. Denote by . Let . is said to be -compact in if there is with dim = dim and an operator continuous and compact such that, for ,

Lemma 7 (see  Ge-Mawhin’s continuation theorem). Let and be two Banach spaces with norms and , respectively. is an open and bounded nonempty set. Suppose that is a quasilinear operator and is -compact in . In addition, if(C1), ,(C2), for ,(C3), where , then the equation has at least one solution in .

In this paper, we take with the norm and with the norm . By means of the linear functional analysis theory, we can prove that is a Banach space.

Define the operator by where Define the operator by Then BVP (4) is equivalent to the operator equation. Consider

#### 2. Main Result

We will always assume that the nonlinearity will be retained:(H1)there exist nonnegative functions such that (H2)there exists a constant such thateither or Moreover, we will always assume that is continuous and

Now, we begin with some lemmas below.

Lemma 8. Let be defined by (16), and then and is a quasilinear operator

Proof. By Lemma 3, has solution: which satisfies Combining with the boundary value condition and , we can get that (24) holds.
If , then there exists a function such that . Based on Lemma 3, we have From condition , one has . By the condition , we obtain that Thus, we get (25).
Then we have dim Ker  and closed. Therefore, is a quasilinear operator.

Lemma 9. Let be an open and bounded set; then is -compact in .

Proof. Define the continuous projector and the semiprojector : where and .
Obviously, and . It follows from that . By a simple calculation, we can get . Then we get
By the definition of , we can get Let , where , . It follows from and that . Then, we have Thus
Let be an open and bounded set with . For each , we can get . Thus, . Take any in the type . Since , we can get . So (10) holds. It is easy to verify (11).
Furthermore, define by
By the continuity of , it is easy to get that is continuous on . Moreover, for all , there exists a constant such that , so we can easily obtain that is uniformly bounded. By the Arzelà-Ascoli theorem, we just need to prove that is equicontinuous. Furthermore, for , , we have
By , we have Since is uniformly continuous on , so is equicontinuous. Similarly, we can get that is equicontinuous, and considering that is uniformly continuous on , we get that is also equicontinuous. So we can obtain that is compact.
For each , we have . Thus, which together with yields that It is easy to verify that is the zero operator. So (12) holds.
On the other hand, consider So (13) holds. Then we get that is -compact in . The proof is complete.

Lemma 10. Suppose that , hold; then the set is bounded.

Proof. Take ; then , , and . By (25), we have Then, by the integral mean value theorem, there exists a constant such that . So, from , we get . By , we get Take ; we have Then we have So we get That is, By and , we get So, from , we have which together with and (47) yields that In view of (23), we can see that there exists a constant such that Thus, from (47), we get
Combining (51) with (52), we have Therefore, is bounded. The proof is complete.

Lemma 11. Suppose that holds; then the set is bounded.

Proof. For , we have , and . Then we get which together with implies . Thus, we have Hence, is bounded. The proof is complete.

Lemma 12. Suppose that the first part of holds; then the set is bounded.

Proof. For , we have , , and If , then because of the first part of . If , we can also obtain . Otherwise, if , in view of the first part of , one has which contradicts (58). Therefore, is bounded. The proof is complete.

Remark 13. If the second part of holds, then the set is bounded.

Theorem 14. Let be continuous. Suppose that , hold. Then BVP (4) has at least one solution.

Proof. Set . It follows from Lemmas 8 and 9 that is a quasilinear operator and is -compact on . By Lemmas 10 and 11, we get that the following two conditions are satisfied:(C1),(C2), for .Take According to Lemma 12 (or Remark 13), we know that for . Therefore So the condition of Lemma 7 is satisfied. By Lemma 7, we can get that has at least one solution in . Therefore BVP (4) has at least one solution. The proof is complete.

#### 3. Example

In this section, we will give an example to illustrate our main result.

Example 1. Consider the following BVP: Corresponding to BVP (4), we get that , , , and Choose , , , . By a simple calculation, we can get that and Obviously, BVP (63) satisfies all conditions of Theorem 14. Hence, it has at least one solution.

#### 4. Conclusions

In this paper, the boundary value problem for -Laplacian equation at resonance is investigated. In view of the boundary value problem (4) is equivalent to the operator equation (19); we only need to find a fixed point of the operator equation (19). Firstly, we established the sufficient conditions of existence of boundary value problem for -Laplacian equation. Then, by using the extension of Mawhin’s continuation theorem due to Ge, we got the fixed point of operator equation (19).

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This research was supported by the Fundamental Research Funds for the Central Universities (2013XK03) and the National Natural Science Foundation of China (11271364).