Abstract

This paper investigates the existence, multiplicity, nonexistence, and uniqueness of positive solutions to a kind of two-point boundary value problem for nonlinear fractional differential equations with -Laplacian operator. By using fixed point techniques combining with partially ordered structure of Banach space, we establish some criteria for existence and uniqueness of positive solution of fractional differential equations with -Laplacian operator in terms of different value of parameter. In particular, the dependence of positive solution on the parameter was obtained. Finally, several illustrative examples are given to support the obtained new results. The study of illustrative examples shows that the obtained results are applicable.

1. Introduction

In this paper, we consider boundary value problem for a fractional differential equation with -Laplacian operator where is a constant and is a parameter. , is the -Laplacian operator; that is, . denotes the Caputo fractional derivative.

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various fields of sciences and engineering such as physics, chemistry, aerodynamics, electrodynamics of complex medium, electrical circuits, and biology (see [17] and their references). Differential equations with -Laplacian arise naturally in non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow laws. Since the -Laplacian operator and fractional calculus arise from so many applied fields, the fractional -Laplacian differential equations are worth studying. Recently, there have appeared a very large number of papers which are devoted to the existence of solutions of boundary value problems and initial value problems for the fractional differential equations (see [812]), and the existence of solutions of boundary value problems for the fractional -Laplacian differential equations has just begun in recent years (see [1322]). On the other hand, there are few papers that consider the eigenvalue intervals of fractional boundary value problems (see [23, 24]). In [23], the author discussed the following system: By the use of appropriate conditions with respect to and , the author proved that the above problem has at least one or two positive solutions for some , where . But almost all the results which the author obtained depend on both and ; the case depends on one of and ; the author only discussed and ; the case or has not been discussed. On the other hand, there exist several results on the existence of one solution to fractional -Laplacian boundary value problems (BVPs); there are, to the best of our knowledge, relatively few results on the nonexistence and the uniqueness of positive solutions to fractional -Laplacian differential equation with parameter.

Motivated by the above questions, in this paper, we will establish several sufficient conditions for the existence of positive solutions of (1) by using fixed point theorem and fixed point index theory.

The work is organized in the following fashion. In Section 2, we provide some necessary background. In particular, we will introduce some lemmas and definitions associated with fixed point index theory. The main results will be stated and proved in Section 3. Two examples are given in Section 4.

2. Preliminaries

In this section, we introduce definitions and preliminary facts which are used throughout this paper.

Definition 1 (see [1]). The Riemann-Liouville fractional integral of order of a function is given by provided that the right side is point-wise defined on and is the Gamma function.

Definition 2 (see [1]). The Caputo fractional derivative of order of a continuous function is given by where , provided that the right side is point-wise defined on .

Lemma 3 (see [1]). Let , and assume that , and then the fractional differential equation has unique solutions:

Lemma 4 (see [1]). Let . Then, where .

Lemma 5 (see [25]). Let be a cone in a real Banach space , and let be a bounded open set of . Assume that operator is completely continuous. If there exists a such that then .

Lemma 6 (see [26]). Let be a Banach space and a cone, and and are open set with , and let be completely continuous operator such that either(i), and or(ii), and holds. Then has a fixed point in .

3. Main Results

In this section, we present some new results on the existence, multiplicity, nonexistence, and the uniqueness of positive solution of problem (1) and dependence of the positive solution on the parameter .

Let ; then is a real Banach space with the norm defined by .

Lemma 7. Let and ; then the solution of the problem is given by where and .

Proof. It is easy to see by integration of (8) that By the boundary condition , we can easily get that , and we obtain that that is By Lemmas 3 and 4, we get that Using the boundary condition , , we get and Thus, Direct differentiation of (9) implies By differentiation of (17), we get
Thus the solution of problem (8) is nonincreasing and concave on , and On the other hand, as we assume that , we see that Therefore, and is concave for . So for every , Therefore, Since , we have that is, The Lemma is proved.

We construct a cone in by where is defined in Lemma 7. It is easy to see is a closed convex cone of .

Define by

It is clear that the fixed points of the operator are the solutions of the boundary value problems (1).

We make the following hypotheses:(H1) is continuous;(H2) is continuous and not identical zero on any closed subinterval of with ;(H3); (H4), where uniformly for ;(H5), where uniformly for ;(H6) for any and .

Lemma 8. Assume that (H1)–(H3) hold. Then is completely continuous.

Proof. First, we show that is continuous. It is easy to see . For and , as , by the continuous of , we get , as . This implies that So we have Denote ; then
Hence, is continuous.
Second, we show that is compact.
For , by condition (H1), . Denote .
Then that is, maps bounded sets into bounded sets in .
For , , By mean value theorem, we obtain that Thus This shows that , as , so is equicontinuous. Therefore, the operator is completely continuous by the Arzela-Ascoli theorem.

Now for convenience we introduce the following notations. Let

Theorem 9. Assume that (H1)–(H5) hold.(i)If , then there exists such that, for every , problem (1) has a positive solution satisfying associated with where and are two positive finite numbers.(ii)If , then there exists such that, for every , problem (1) has a positive solution satisfying for any where and are two positive finite numbers.(iii)If , then there exists such that, for any , problem (1) has a positive solution satisfying for any where is a positive finite number.(iv)If , then there exists such that, for every , problem (1) has a positive solution satisfying for any where is a positive finite number.(v)If there exists such that , then problem (1) has positive solution satisfying for any where is a positive finite number.

Proof. (i) It follows from that there exists , such that
Let ; we show that is required. When , we have we need ; this implies that , as , so .
Let . Then we may assume that if not, then there exists such that , and then (35) already holds for .
Define ; then , and . We now show that In fact, if there exists such that , then (42) implies that . On the other hand ; we may choose ; then . Therefore Consequently, for any , (26) and (44) imply that which implies that , which is a contradiction to the definition of . Thus, (42) holds and, by Lemma 5, the fixed point index On the other hand, by the fact that the fixed point index of constants operator is 1, so where is the zero operator. It follows therefore from (46) and (47) and the homotopy invariance property that there exist and such that , but , which implies that ; it follows that we get ; that is, and then From the proof above, for any , there exists a positive solution associated with .
Thus with .
Next, we show that . In fact, which implies that that is, it follows that we get In conclusion, .
(ii) It follows from that there exist such that Let , the following proof are similar to (i).
(iii) It follows from , there exists , such that Let ; we show that is required. When , we have ; since , then we need , as , so holds.
Let ; we proceed in the same way as in the proof of (i): replacing (42) we may assume that , and replacing (43) we can prove . It follows from Lemma 5 that . Note that ; we can easily show that there exists and such that . Hence (37) holds for .
The proof of Theorem 9(iv) follows by the method similar to Theorem 9(iii); we omit it here.
(v) It follows that ; for any , we have and ; then Let . The following proof is similar to that of (iv). This finished the proof of (v).

Remark 10. In Theorem 9, all the criteria obtained depend on one of and .

Let ; the following theorems give out the multiply, nonexistence, and the dependence of parameter.

Theorem 11. Assume that (H1)–(H5) hold.(i)If and , uniformly for , then there exists such that the problem (1) has two solutions for any .(ii)If and , then there exists such that, for any , problem (1) has no solution.

Proof. (i) It follows from that, for any , there exists , such that
Since , there exists such that where satisfied .
For , we have This implied Let and ; then ; we have and which implies
Next, for , there exist such that where satisfied .
Similar to the above proof, we get
Applying Lemma 7 to (61), (65), and (63) yields that has two fixed points such that and .
(ii) It follows from and that there exists , such that then for , by the continuous of , there exists such that Let ; then Assuming is a positive solution of problem (1), we will show that this leads to a contradiction for , where . It follows from (26) that This implies that , which is a contradiction. This finishes the proof of (ii).

Remark 12. In (i), the condition uniformly for can be replaced with condition (H6). The same methods can be used to prove the results.

Remark 13. From the above proof, the condition is important to keep the problem (1) having at least two positive solutions, if this condition is not satisfied, then there exists small enough such that the problem (1) has no positive solutions for all .

Theorem 14. Assume that (H1)–(H6) hold.(i)If and , then there exists such that problem (1) has at least two solutions for .(ii)If and , then there exists such that problem (1) has no solution for all .

Proof. (i) It follows from that there exists , such that where satisfies .
Then for , we have for and which implies that Next, from , there exists , such that where satisfies Let ; then for we have Similar to the above proof, we can get Let , , and . Then for , we have Hence, This implies that Applying Lemma 7 to (72), (76), and (79) yields that has two fixed points such that and .
(ii) By the proof of (i) and the continuity of , there exists such that .
Let ; since (H6) holds, then and Assuming is a positive solution of problem (1), we will show that this leads to contradiction for all , where . It follows form (26) that This implies that , which is a contradiction. This completes the proof of (ii).

Corollary 15. Assume that (H1)–(H5) holds.(i)If uniformly for , and one of and is satisfied, then there exists such that the problem (1) has at least one positive solution for any .(ii)If one of and holds, then there exists such that problem (1) has at least one positive solutions for any .

Proof. (i) The conclusion is a direct consequence of Theorem 11(i).
(ii) The conclusion is a direct consequence of Theorem 14(i).

Theorem 16. Assume that (H1)–(H5) hold. Then the following conclusions hold.(i)If and , then the problem (1) has a positive solution for all .(ii)If and , then the problem (1) has a positive solution for all .

Proof. We only prove (i); the proof of (ii) is similar, so we omit it here.
Let ; since , there exists such that where satisfied Similar to the proof of Theorem 14, we obtain that From , there exists such that where satisfied We get that Applying Lemma 7 to (84), (87) yields that has a fixed point such that .

Theorem 17. Assume that (H1)–(H6) hold. Then the following conclusions hold.
If and , then there exists such that problem (1) has no positive solution for any and .

Proof. It follows from and that there exist and such that Let ,. By the condition (H6), . Then ; we have where and .
Let and . If problem (1) has positive solution in , then which is a contradiction. This finishes the proof of Theorem 17.

4. Examples

In this section, we give out same examples to illustrate our main results can be used in practice.

Example 1. Consider the following fractional differential equation: where , and .

By simple computation, . Then, it is easy to see all the conditions of Theorem 9(ii) are satisfied; thus ; the problem (91) has at least one positive solution.

Example 2. Consider the boundary value problem where , and : Let ; then so , and . We have,, and . By Theorem 11(ii), for any , the problem has no positive solution.

5. Conclusions

Recently, differential equations with -Laplacian operator were widely discussed by several authors. In this paper, by using fixed point techniques combining with partially ordered structure of Banach space, we obtained the existence, multiply, and the dependence of parameter. These new results we presented can be used in numerical computation and analyze mathematical models of physical phenomena, mechanics, nonlinear dynamics, and many other related fields.

Conflict of Interests

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

Acknowledgments

The authors are very grateful to anonymous referees for their valuable and detailed suggestions and comments to improve the original paper. This work is supported by the Program for New Century Excellent Talents in University (NCET-10-0097).