Abstract
We consider the following third-order boundary value problem with advanced arguments and Stieltjes integral boundary conditions: and , where , , is continuous, for , and for . Under some suitable conditions, by applying a fixed point theorem due to Avery and Peterson, we obtain the existence of multiple positive solutions to the above problem. An example is also included to illustrate the main results obtained.
1. Introduction
Third-order differential equations arise from a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity driven flows, and so on [1].
Recently, third-order boundary value problems (BVPs for short) have received much attention from many authors; see [2–20] and the references therein. However, it is necessary to point out that all the unknown functions in the above-mentioned papers do not depend on advanced arguments.
In 2012, Jankowski [21] studied the existence of multiple positive solutions to the BVPwhere the unknown function depended on an advanced argument satisfying the following condition:
was continuous and for .
denoted a linear functional on given byinvolving a Stieltjes integral with a suitable function of bounded variation. The measure could be a signed one. The situation with a signed measure was first discussed in [22, 23] for second-order differential equations; it was also discussed in [24, 25] for second-order impulsive differential equations.
Among the boundary conditions in (1), only was related to and a Stieltjes integral. When was also related to and the Stieltjes integral, the authors in [26, 27] obtained the existence and multiplicity of positive solutions to the BVPwhere was defined as in (2) and the condition was imposed on the advanced argument . The main tools used were the Guo-Krasnoselskii fixed point theorem [28, 29] and a fixed point theorem due to Avery and Peterson [30].
In this paper, we are concerned with the following third-order BVP with advanced arguments and Stieltjes integral boundary conditions:wherehere and are suitable functions of bounded variation. It is important to indicate that it is not assumed that is positive to all positive . Throughout this paper, we always assume that , , is continuous, and the advanced argument satisfies the following condition:
is continuous, for and for .
In order to obtain our main results, we need the following concepts and fixed point theorem [30].
Let be a real Banach space and let be a cone in .
A map is said to be a nonnegative continuous convex functional on if is continuous and for all and .
Similarly, a map is said to be a nonnegative continuous concave functional on if is continuous and for all and .
Let and be nonnegative continuous convex functionals on , let be a nonnegative continuous concave functional on , and let be a nonnegative continuous functional on . For positive numbers , we define the following sets:
Theorem 1 (Avery and Peterson fixed point theorem). Let be a real Banach space and let be a cone in . Let and be nonnegative continuous convex functionals on , let be a nonnegative continuous concave functional on , and let be a nonnegative continuous functional on satisfying for , such that, for some positive numbers and ,for all . Suppose is completely continuous and there exist positive numbers with , such that (C1) and for ; (C2) for with ; (C3) and for with .Then has at least three fixed points , such that
2. Main Results
For convenience, we denote
In the remainder of this paper, we always assume that , , and , and for , the following conditions are fulfilled: Then, and .
Lemma 2 (see [21]). One has , .
Lemma 3. For any , the BVPhas the unique solution
Proof. By integrating the differential equation in (82) three times from to and using the boundary condition , we know thatAnd so,In view of (13), (14), and the boundary conditions and , we have Therefore,Substituting (16) into (15), we get
Let be equipped with the maximum norm. Then is a Banach space. If we let wherethen is a cone in . Now, we define operators and on bywhere
Lemma 4. , .
Proof. Let . Then it is easy to know that which shows that is concave down on . In view of we haveSo, , .
Now, we prove that . To do it we consider two cases.
Case 1. Let . Then and there exists such that .
If , then So, which together withimplies thatthat is,If , then So,On the other hand, it follows from and (27) that which together with (31) implies that that is,Case 2. Let . Then and there exists such that .
If , then So,At the same time, since we have which together withimplies thatIn view of (37) and (41), we have that is,If , then So, which together with (40) implies that that is,It follows from (29), (35), (43), and (47) that Finally, we need to show that . Sincewe haveTherefore, . Similarly, we may prove that .
Lemma 5. and have the same fixed points in .
Proof. On the one hand, if is a fixed point of , that is, , thenwhich shows thatSo,which indicates that is a fixed point of .
On the other hand, if is a fixed point of , that is, , then which shows that So,which indicates that is a fixed point of .
Lemma 6. is completely continuous.
Proof. First, by Lemma 4, we know that .
Next, we show that is compact.
Let be a bounded set. Then there exists such that for any . Since and are functions of bounded variation, there exists such that for any partition . Let Then for any , we have which shows that is uniformly bounded.
On the other hand, for any , since is uniformly continuous on , there exists such that, for any with , LetThen for any , with , we have which shows that is equicontinuous. It follows from Arzela-Ascoli theorem that is relatively compact. Thus, we have shown that is a compact operator.
Finally, we prove that is continuous.
Assume that and . Then there exists such that and . For any , since is uniformly continuous on , there exists such that, for any with ,At the same time, since , there exists positive integer such that, for any ,It follows from (63) and (64) that, for any ,which indicates that is continuous.
Therefore, is completely continuous. Similarly, we can prove that is also completely continuous.
For convenience, we denote Let
Theorem 7. Suppose that there exist positive constants , , and with such that the following conditions are fulfilled:, ,, ,, .Then the BVP (4) has at least three positive solutions , , satisfying and
Proof. For , we defineThen it is easy to know that is a nonnegative continuous concave functional on and , and are nonnegative continuous convex functionals on . In order to apply Theorem 1 to prove our main results, we use the operator and take .
First, we assert that .
In fact, if , then , which together with implies that It follows from (70) thatThis shows that .
Next, we claim that and for .
Indeed, the constant function . Moreover, if , then , which together with for implies that , . In view of , we have Since is concave down on , we have So, which together with implies thatTherefore, it follows from (72) and (76) thatThirdly, we assert that for with .
To see this, we suppose and . ThenFinally, we prove that and for with .
Indeed, it follows from that . Moreover, if and , then , , which together with implies that In view of (79), we have as required.
To sum up, all the hypotheses of Theorem 1 are satisfied. Hence, the BVP (4) has at least three positive solutions , , satisfying and
3. An Example
Example 1. Consider the following BVP:whereSince and , a simple calculation shows thatAt the same time, in view of and , we getIf we choose , and , then all the conditions of Theorem 7 are fulfilled. Therefore, it follows from Theorem 7 that the BVP (82) has at least three positive solutions , , satisfying and
Competing Interests
The authors declare that they have no competing interests.