Research Article | Open Access
Positive Solutions for a Mixed-Order Three-Point Boundary Value Problem for -Laplacian
The author investigates the existence and multiplicity of positive solutions for boundary value problem of fractional differential equation with -Laplacian operator. The main tool is fixed point index theory and Leggett-Williams fixed point theorem.
In this paper, we are interested in the existence of single and multiple positive solutions to nonlinear mixed-order three-point boundary value problem for -Laplacian. Consider the following: where , , is the Caputo's derivative, , and with . We assume the following conditions throughout.(H1).(H2) is nonnegative and on any subinterval of .
The equation with a -Laplacian operator arises in the modeling of different physical and natural phenomena, non-Newtonian mechanics, nonlinear elasticity and glaciology, combustion theory, population biology, nonlinear flow laws, and so on. Liang et al. in  used the fixed point theorem of Avery and Henderson to show the existence of at least two positive solutions. Zhao et al.  studied the existence of at least three positive solutions by using Leggett-Williams fixed point theorem. Chai  obtain results for the existence of at least one nonnegative solution and two positive solutions by using fixed point theorem on cone. Su et al.  studied the existence of one and two positive solutions by using the fixed point index theory. Su  studied the existence of one and two positive solution by using the method of defining operator by the reverse function of Green function and the fixed point index theory. Tang et al.  studied the existence of positive solutions of fractional differential equation with -laplacian by using the coincidence degree theory.
The organization of this paper is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our results. In Section 3, we discuss the existence of at least one positive solution for (1) and (2). In Section 4, we discuss the existence of multiple positive solutions for (1) and (2). Finally, we give some examples to illustrate our results in Section 5.
Definition 1. Let be a real Banach space. A nonempty closed convex set is called cone if(1), then ,(2), then .
Definition 2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Remark 3. By the positive solution of (1) and (2) we understand a function which is positive on and satisfies the differential equation (1) and the boundary conditions (2).
We will consider the Banach space equipped with standard norm: The proof of existence of solution is based upon an application of the following theorem.
Lemma 5 (see ). Let with , . If and , then holds on .
Proof. Integrating both sides of (1) on , we have
From Lemma 5, we have
From (2), and .
Now, consider the following: by the boundary value condition , we have so Splitting the second integral in two parts of the form we have ; thus, therefore, This completes the proof.
Lemma 7. Let be fixed. The kernel, , satisfies the following properties.(1) for all ,(2) for all .
Proof. As and , we have
thus, . Note then, is increasing as a function of ; therefore,
For , we have where (a) If , on the other hand, Since and(i), ,(ii), (iii);
thus, we have From (20), we obtain It follows from (20), (21), and (23), that item in the proof hold.
(b) If , It follows from (24) that item in the proof holds.
Define the cone by and the operator by
Lemma 10. is completely continuous and .
Proof. By Lemma 8, . In view of the assumption of nonnegativeness and continuity of functions with and , we conclude that is continuous.
Let be bounded; that is, there exists such that for all .
Then from and from Lemmas 6 and 7, we have Hence, is bounded.
On the other hand, let , with ; Then The continuity of implies that the right-side of the above inequality tends to zero if . Therefore, is completely continuous by Arzela-Ascoli theorem.
We introduce the notation where . Consider the following:
3. Single Solutions
In what follows, the number .
Proof. Without loss of generality, we suppose that . For any , we have
We define two open subsets of ,
For , by (33), we have
For , by , (27), and Lemma 7, we have
Then by Theorem 4,
On the other hand, as , we have thus, by , (27), Lemma 7, and (29), we have Therefore, Then by Theorem 4, By (38) and (42): Then, has a fixed point , is positive solution of problem (1)-(2), and .
Proof. By , for , there exists a suitably small positive number , as and , such that
Let and , then, by (44), condition holds.
By , for , there exists a sufficiently large such that Thus, when , one has Let . Then, by (46), condition holds.
Hence, from Theorem 11 the desired result hold.
Proof. By , for , there exists a sufficiently small , such that
Thus, when , we have
Let . Then by (48), condition holds.
By , for , there exist a sufficiently large such that We consider the following two cases.
(a) Suppose that is unbounded, then we know from that there is a such that Since , then from (49) and (50) we have Letting , we have Thus, holds.
(b) Suppose that is bounded, say In this case, taking sufficiently large , then letting , we have Thus, holds.
Hence, from Theorem 11 the desired result holds.
4. Multiple Solutions
To show the existence of multiple solutions we will use the Leggett-Williams fixed point theorem . To this end define the following subsets of a cone as
Definition 14. A map is said to be a nonnegative continuous concave functional on a cone of a real Banach space if is continuous and
for all and .
Theorem 15 (see ). Suppose is completely continuous and suppose that there exists a concave positive functional on such that for . Suppose that there exist constants such that(B1) and if ,(B2) if ,(B3) for with .
Then, has at least three fixed points , , and such that , and with .
Proof. By Lemma 10, is completely continuous.
Let It is obvious that is a nonnegative continuous concave functional on with , for . Now we will show that the conditions of Theorem 15 are satisfied. For , then . For by (27), (29), and , one has This implies . By the same method, if , then we can get and therefore is satisfied. Next, we assert that and for all . In fact, the constant function On the other hand, for , we have Thus, in view of (27), Lemma 7, and , one has as required and therefore is satisfied.
Finally, we assert that if with then . To see this, suppose that and , then it follows from Lemma 10 that Thus, is satisfied.
Therefore, by the conclusion of Theorem 15, the operator has at least three fixed points. This implies that (1)-(2) has at least three solutions.
Example 1. Consider the boundary value problem with -Laplacian:
where , , , , , , , , and , .
Then Next, Then, for we have , so condition holds.
Now, Consider Then, for and we have , so condition holds. Therefore, by Corollary 12, (63) has at least one positive solution.
Example 2. Consider the boundary value problem with -Laplacian:
where , , , , , , , , and .
Let By Example 1, we have and .
Choosing , , and , then Then the conditions are satisfied. Therefore, it follows from Theorem 16 that (67) has at least three positive solutions , , and such that
Francisco J. Torres was partially supported by DIUDA 221231, Universidad de Atacama.
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Copyright © 2013 Francisco J. Torres. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.