Variational Analysis, Optimization, and Fixed Point Theory 2014View this Special Issue
The Existence of Positive Solutions for a Fourth-Order Difference Equation with Sum Form Boundary Conditions
We consider the fourth-order difference equation: , subject to the boundary conditions: , , , where and for , is continuous. is nonnegative ; is nonnegative for . Using fixed point theorem of cone expansion and compression of norm type and Hölder’s inequality, various existence, multiplicity, and nonexistence results of positive solutions for above problem are derived, which extends and improves some known recent results.
Boundary value problems (BVPs) for ordinary differential equations arise in different areas of applied mathematics and so on. The existence of solutions for second order and higher order nonlocal boundary value problems has been studied by several authors; for example, see [1–11] and the references therein. Many authors have also discussed the existence of positive solutions for higher order difference equation BVPs [12, 13], by using fixed point theorem of cone expansion and compression of norm type, sufficient conditions for the existence of positive solutions for fourth-order and third-order difference equation BVPs are established, respectively. Recently, there has been much attention on the existence of positive solutions for the fourth-order differential equations with integral boundary conditions [14–17]. In , Zhang and Ge considered the differential equation BVP: where, is symmetric on , for some , and it is symmetric on the interval,is continuous, andfor all, andare nonnegative, symmetric on. The authors made use of fixed point theorem of cone expansion and compression of norm type and Hölder inequality to prove the existence of positive solutions for the above problem.
Motivated by the above works, we intend to study the existence and nonexistence of positive solutions of the following fourth-order difference BVP with sum form boundary conditions:
Throughout this paper, we make the following assumptions: is symmetric onand; there is asuch thatfor;is continuous;is nonnegative onand,is nonnegative onand, where,;,.
In order to establish the existence of positive solutions of the problem (2), we need the following definitions, theorem, and lemma.
Definition 1. A functionis said to be a solution of problem (2) ifsatisfying BVP (2).
Definition 2 (see ). Letbe a real Banach space over . A nonempty closed setis said to be a cone provided that(i)for alland alland(ii)implies.
Every coneinduces an ordering ingiven byif and only if.
Theorem 3 (see ). Letbe a cone in a real Banach space. Assume thatare bounded open sets inwith. If is completely continuous such that either(i),and,, or(ii),and,, thenhas at least one fixed point in.
Lemma 4 (Hölder). Suppose thatis a real-valued column; let , satisfy the conditionwhich are called conjugate exponent, andfor. If, then which can be recorded as
Let; is a real Banach space with the normdefined by
Letbe a cone of, and where.
In our main results, we will use the following lemmas and properties.
Lemma 5. Suppose thatandhold and; then, for all, the BVP has unique solutiongiven by where where,.
Proposition 6. Assume that; then
Proposition 7. Assume that; then where
Proposition 8. Suppose that; then where
Proof. From Lemma 5 and Proposition 7, we have
On the other hand, using, we get
Lemma 9. Suppose that, for all; the BVP has unique solutiongiven by where
Proof. From the properties of difference operator, we can get
then we have
It can imply that
Let; we have That is thus,
We can get
From the boundary conditions, we have thus, Because we have Thus, we get where
Multiplying the above equation with, summing them fromto, we have So, we can get
Proposition 10. Assume that; then
Proposition 11. For, one has where.
Proposition 12. If, then, for all, one has where, .
We construct a cone onby where Obviously,is a closed convex cone of.
Define an operatoras Let Then we can obtain the following properties.
Proposition 13. Ifandhold, then
Proposition 14. Ifandhold, then, for all , one has
Lemma 15. Suppose thathold; ifis a solution of the equation thenis a solution of the BVP (2).
Lemma 16. Assume thathold; thenandis completely continuous.
Proof. From above works, for all, we have
That is to sayfor.
On the other hand, for, we have Similarly, we can get So,and. It is easy to see thatis completely continuous.
3. The Existence of One Positive Solution
In this part, we apply Theorem 3 and Lemma 4 to prove the existence of one positive solution for BVP (2). We need consider the following cases for:,, and.
Let wheredenotesor, and
Remark 17. If we only consider the case, then we can take
Firstly, the following theorem deals with the case .
Theorem 18. Suppose thathold. If there exist two constants, withsuch thatfor allandfor all; orfor allandfor all, then BVP (2) has at least one positive solution.
Proof. We only consider condition. For, from the definition ofwe obtain that
So, for, we have, which implies that. Thus, for, fromwe get
that is,implies that
On the other hand, for, we have, which implies that; therefore for, fromwe have that is,implies that
From the above works, we apply (i) of Theorem 3 to yield thathas a fixed point, and. Thus, it follows that BVP (2) has at least one positive solution.
The following theorem deals with the case.
Theorem 19. Suppose thathold and or holds. Then BVP (2) has at least one positive solution.
Proof. Letreplace and repeat the argument of Theorem 18.
Finally we consider the case.
Theorem 20. Suppose thathold and or holds. Then BVP (2) has at least one positive solution.
Proof. Similar to the proof of Theorem 18, for, we have So, for, we have. And from the proof of Theorem 18,for. Thus Theorem 20 is proved.
Theorem 21. Assume thathold. If one of the following conditions is satisfiedand(particularly,and);and (particularly,and), then BVP (2) has at least one positive solution.
Proof. We only consider the case. The proof of caseis similar to case. Considering, there existssuch thatfor,, wheresatisfies. Then, for,, fromwe have
that is,implies that.
Next, considering, then there existssuch that wheresatisfies; assume that then Choosing.
Thus, for, we obtain