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The existence of symmetric positive solutions for a seconder-order difference equation with sum form boundary conditions
Advances in Difference Equations volume 2014, Article number: 237 (2014)
In this paper, we consider the existence of positive solutions for a second-order discrete boundary value problem subject to the boundary conditions: , , where , for , is symmetric on , is symmetric on , is continuous, for all , and is nonnegative and symmetric on . By the fixed point theorem and the Hölder inequality, we study the existence of symmetric positive solutions for the above difference equation with sum form boundary conditions.
A class of boundary value problems (BVPs) with integral boundary conditions arise in thermal conduction problems, semiconductor problems, and hydrodynamic problems [1–3]. Recently, such problems have been investigated by many authors [4–10]. The equation , , describes many phenomena in the fields of gas dynamics, nuclear physics, chemically reacting systems and atomic structures [11–15]. In , Feng considered the following differential equation BVP with integral boundary conditions:
Applying the fixed point index theorem and the Hölder inequality, the author studied the existence of symmetric positive solutions for BVP (1.1)-(1.3).
Motivated by the above works, we will study the following BVP with sum form boundary conditions:
Throughout this paper, the following conditions are assumed:
(A1) , is symmetric on , and there exists such that on , for , and is symmetric on , h is nonnegative, symmetric on , and , where , is continuous and is symmetric on for all .
Remark 1 The conditions that g and h are symmetric on the different sets, which can guarantee the symmetry of associated kernel function for BVP (1.4)-(1.6). The kernel functions are then used to obtain the existence of symmetric positive solutions for BVP (1.4)-(1.6) by constructing a suitable operator.
In order to study the existence of symmetric positive solutions of problem (1.4)-(1.6), we need the following lemmas.
Lemma 1.1 
Let P be a cone of the real Banach space E and Ω be a bounded open subset of E and . Assume is a completely continuous operator and satisfies , , . Then .
Lemma 1.2 
Suppose is a completely continuous operator, and satisfies
, , .
Lemma 1.3 (Hölder)
Suppose is a real-valued column, let
where p, q satisfy the condition , which are called conjugate exponents, and for . If , then
which can be denoted as
Let . It is well known that E is a real Banach space with the norm defined by . Let K be a cone of E,
In our main results, we will use the following lemmas.
Lemma 2.1 Assume that (A1) holds. Then for any , the BVP
has a unique solution u given by
and , .
Proof From the properties of the difference operator, it is easy to see that
then we have
From the above equalities, we can obtain
Let , then
It follows that
By the boundary conditions, we get
where is defined by (2.5). Multiplying the above equation with , and summing from 1 to , we can get
One deduces that
where is defined by (2.4). The proof is complete. □
From the above work, we can prove that and have the following properties.
Proposition 2.1 If (A1) holds, then we have
where , , .
Proof It is clear that (2.6) holds. Now we prove (2.7) holds.
If , then , from (2.5) and (A1) we get
Similarly, we can prove that , . So we have , for . From (2.4) and (A1), we have
So, (2.7) is established. Next we prove (2.8) holds. In fact, for , if , then
Similarly, we can prove that , for . Therefore . For , we can get
On the other hand, from (2.5), we have
So, by (2.4), for , we can obtain
The proof is completed. □
Remark 2 The symmetry of on can guarantee that is symmetric for , and the symmetry of on can guarantee that is symmetric for .
Next, we can construct a cone in E by
where . Then we define an operator
It can be observed that u is a solution of problem (1.4)-(1.6) if and only if u is a fixed point of operator T.
We can get the following lemma from Lemma 2.1.
Lemma 2.2 Suppose (A1) holds. If u is a solution of the equation
then u is a solution of BVP (1.4)-(1.6).
Lemma 2.3 Assume (A1) holds. Then and is completely continuous.
Proof For , from (2.9), we obtain . By Proposition 2.1, it is to see that , for . Using the fact that w, u, are symmetric on , we have
then Tu is symmetric on for . And from (2.8) we can see
Similarly, by (2.8) we obtain
Thus, and . It is clear that is completely continuous. □
Remark 3 The symmetry of the kernel function for can guarantee that Tu is symmetric on for .
3 Main results
In this section, we will establish that problem (1.4)-(1.6) has at least one positive solution with Lemma 1.1 and Lemma 1.2. We need consider the following situations: , , . Next, we will prove a theorem for . At first, we define
where β denotes 0 or ∞, and
Theorem 3.1 Assume that conditions (A1) hold. In addition, suppose that
(A2) , and , or
(A3) , and
are satisfied. Then problem (1.4)-(1.6) has at least one symmetric positive solution.
Proof We only consider (A2) case, (A3) is similar to (A2). If , then there exist , such that and for all , we have
For all , from Lemma 1.3 we obtain
So , for , . From Lemma 1.1, we can get . Next, we prove it satisfies Lemma 1.2. Because , there exist , such that
Let , then , and
Now we prove that , , . If not, then there exist and such that ; thus we have
i.e., , which is a contradiction. In addition, because , so , from Lemma 1.2 we have . On the other hand, from the above work with the additivity of the fixed point index, we get
So, T has at least one fixed point on . Then it follows that problem (1.4)-(1.6) has a symmetric positive solution . The proof is complete. □
Remark 4 From the proof of Theorem 3.1, we can establish that problem (1.4)-(1.6) has another nonnegative solution , .
The following corollary deals with the case .
Corollary 3.1 Suppose that (A1), (A2) hold. Then problem (1.4)-(1.6) has at least one symmetric positive solution.
Proof It is similar to the proof of Theorem 3.1. Let replace and repeat the argument of Theorem 3.1. □
Finally, we consider the case of .
Corollary 3.2 Assume that (A1), (A2) hold. Then problem (1.4)-(1.6) has at least one symmetric positive solution.
Proof It is similar to the proof of Theorem 3.1. For all , we have
So , , . By Lemma 1.1, we can get . This together with in the proof of Theorem 3.1 completes the proof. □
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The authors express their sincere thanks to the referees for the careful and details reading of the manuscript and very helpful suggestions. The project was supported by the Natural Science Foundation of China (11371120), the Natural Science Foundation of Hebei Province (A2013208147) and the Education Department of Hebei Province Science and Technology Research Project (Z2014095).
The authors declare that they have no competing interests.
All authors contributed equally to the manuscript and read and approved the final draft.
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Guo, Y., Ji, Y. & Lv, X. The existence of symmetric positive solutions for a seconder-order difference equation with sum form boundary conditions. Adv Differ Equ 2014, 237 (2014). https://doi.org/10.1186/1687-1847-2014-237
- difference equation
- sum form boundary conditions
- symmetric positive solutions