Research Article | Open Access

Francisco J. Torres, "Positive Solutions for a Mixed-Order Three-Point Boundary Value Problem for -Laplacian", *Abstract and Applied Analysis*, vol. 2013, Article ID 912576, 8 pages, 2013. https://doi.org/10.1155/2013/912576

# Positive Solutions for a Mixed-Order Three-Point Boundary Value Problem for -Laplacian

**Academic Editor:**Jaume Giné

#### Abstract

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.

#### 1. Introduction

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 [1] used the fixed point theorem of Avery and Henderson to show the existence of at least two positive solutions. Zhao et al. [2] studied the existence of at least three positive solutions by using Leggett-Williams fixed point theorem. Chai [3] 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. [4] studied the existence of one and two positive solutions by using the fixed point index theory. Su [5] 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. [6] studied the existence of positive solutions of fractional differential equation with -laplacian by using the coincidence degree theory.

Motivated by the above works, we obtain some sufficient conditions for the existence of at least one and three positive solutions for (1) and (2).

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.

#### 2. Preliminaries

*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.

Theorem 4 (see [7, 8]). *Let be a Banach space and let be a cone of . For , define and assume that is a completely continuous operator such that for all . *(1)*If for all , then .*(2)*If for all , then .*

Lemma 5 (see [9]). *Let with , . If and , then
**
holds on .*

Lemma 6. *The three-point boundary value problem (1)-(2) has a unique solution
**
where
*

*Proof. *Integrating both sides of (1) on , we have
So
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.

Lemma 8 (see [10]). *The unique solution of (1), (2) is nonnegative and satisfies
*

Define the cone by and the operator by

*Remark 9. *By Lemma 6, the problem (1)-(2) has a positive solution if and only if is a fixed point of .

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 .

Let

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 .

Theorem 11. *Suppose that conditions (H1) and (H2) hold. Assume that also satisfy * *, , ,* *, , ,**
where and . Then (1)-(2) has at least one positive solution such that .*

*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
Therefore,
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 .

Corollary 12. *Suppose that conditions (H1) and (H2) hold. Assume that also satisfy * * ,* * .**Then (1)-(2) has at least one positive solution such that .*

*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.

Corollary 13. *Suppose that conditions (H1) and (H2) hold. Assume that also satisfy * * ,* * .**Then (1)-(2) has at least one positive solution such that .*

*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 [11]. 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 [11]). *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 .*

Theorem 16. *Suppose that there exist , , with such that**(C1)**, ,**(C2)**, ,**(C3)**, .**
Then (1)-(2) has at least three positive solutions.*

*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.

#### 5. Examples

*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

#### Acknowledgment

Francisco J. Torres was partially supported by DIUDA 221231, Universidad de Atacama.

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#### Copyright

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.