Abstract
We discuss the existence of solutions of nonlinear third order ordinary differential equations with integral boundary conditions. We provide sufficient conditions on the nonlinearity and the functions appearing in the boundary conditions that guarantee the existence of at least one solution to our problem. We rely on the method of lower and upper solutions to generate an iterative technique, which is not necessarily monotone.
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Introduction
The purpose of this paper is to establish the existence of solutions for a class of nonlinear third order ordinary differential equations with integral boundary conditions. More specifically, we consider the following problem:
where , are continuous functions, and is a nonnegative real number. Several papers have been devoted to the study of third order differential equations with two-point and three-point boundary conditions. See [1]–[3], and [4] for references. Problems with integral boundary conditions have been used in the description of many phenomena in the applied sciences. We refer the interested reader to [5] and the references therein. Very few papers have dealt with nonlocal conditions for third order differential equations. We can mention [6], [7] and [8]. For higher order differential equations with functional boundary conditions the interested reader can consult [9]. In this work we use the method of lower and upper solutions to generate a sequence of modified nonlinear problems, each having a unique solution; in this way, we obtain a sequence of functions, which is uniformly bounded together with their first and second order derivatives. We then extract a subsequence converging uniformly to a solution of our original problem (1)-(4). Contrary to many works in the literature, we develop an iterative technique, which is not necessary monotone. We should point out that our approach is totally different from that of [9].
Preliminaries
Let denote the real interval . is the Banach space of real-valued continuous functions on , equipped with the norm , for . Let denote the set of all real-valued functions which are three times continuously differentiable on . We define the norm of by
Let . Then is a Banach space.
Definition 1
A solution of problem (1)-(4) is a function that satisfies (1) for every and the conditions (3) and (4).
Definition 2
Let satisfy for every . We denote by the set of all such that for every .
It is clear that if and , then .
Definition 3
Let satisfy for every . Let denote the set of all functions such that and .
Remark 1
It is clear that and imply that .
Definition 4
Let satisfy for every . Define the operator by
and the operator by
Remark 2
The operators and are continuous and bounded.
Main results
In this section we state and prove our main results. The first result is of independent interest and plays a key role in the proof of our second result.
Theorem 1
Letbe continuous, bounded and satisfy the following condition:
(H ϕ ): for allsuch thatand.
Then for any, the boundary value problem
has a unique solution.
Proof
Uniqueness. Suppose that problem (5) has two solutions and in . Put . Then . We show that . Suppose this is not true. Then either or . We consider the case . From the condition at it follows that . Since is nonnegative we have , which implies that is increasing to the right of . Since there must exist such that . Then
The differential equation in (5) and (H ϕ ) imply
This is a contradiction. Similarly, if we consider the case we will arrive at a contradiction. Hence . Now, we have a function continuous on with . Then there exists such that
Proceeding as before we show that . So that for all . This shows that for all . Since it follows that for all , which shows the uniqueness of the solution.
Existence. For consider the family of problems
For problem (6) has only the trivial solution. Thus, we consider the case .
-
(i)
is a solution of (6) if and only if it satisfies, for all ,
(7)Indeed, it is clear that the differential equation in (6) implies
(8)Then
(9)It follows that
But , so that
and consequently
-
(ii)
We show that there exists a positive constant , independent of , such that any possible solution of (6) satisfies
(10)The boundedness of implies that there exists such that for all , so that . Then
(11)and
(12)Combining relations (9), (11), and (12) we see that
(13)Since it follows that
Also, and (12) imply
-
(iii)
Define an operator by = the right-hand side of (7). Let . Then it is easily seen that is uniformly bounded and equicontinuous. Ascoli-Arzela theorem implies that the operator is compact. Moreover, the set of all solutions of the equation is bounded (see (10)). It follows from Schaefer theorem (see [10]) that has at least one solution. Thus, (6) has at least one solution for , which is, in fact, unique from the previous step. Thus, is a solution of (5). This completes the proof of the theorem. □
Remark 3
We should emphasize that, unlike Theorem 6 in [9], our Theorem 1 gives the uniqueness of the solution and this is essentially utilized in the proof of our Theorem 2 below.
For our second main result we introduce the notion of lower and upper solutions of problem (1), (2), (3), (4).
Definition 5
-
(a)
We say that is a lower solution of problem (1), (3), (4) if
-
(b)
We say that is an upper solution of problem (1), (3), (4) if
To state and prove our second main result we introduce the following assumptions.
(A f ): is continuous and satisfies
-
(1)
there exists such that any solution of (1), with , satisfies , for all ;
-
(2)
for such that , , and ;
-
(3)
for all , , , and .
(A h ): are continuous and nondecreasing with respect to both arguments.
Remark 4
There are several sufficient conditions that imply (A f )(1). See for instance [6], Lemma 1], [3], Lemma 1].
Theorem 2
Letbe, respectively, a lower and an upper solution of problem (1), (3), (4) such thaton. Assume that the conditions (A f ) and (A h ) are satisfied for the pair, whereis a given lower solution andis a given upper solution. Then problem (1), (3), (4) has at least one solution.
Proof
We modify problem (1), (2), (3), (4) as follows. We define two functions by
and
where is the constant from condition (A f )(1). Consider the modified problem
Define a sequence of functions in as follows. Let
where , and for ,
We shall show that the sequence of modified problems (17) is such that each problem has a unique solution, which is uniformly bounded, together with its first and second order derivatives. Then we rely on Bolzano-Weierstrass theorem to extract a uniformly convergent subsequence, whose limit is the solution of our original problem.
-
1.
The sequence is well defined. Indeed, for any and any we have and . It follows that the function , defined by
is continuous and bounded for all and . Moreover, condition (A f )(2) shows that satisfies condition (H ϕ ) in Theorem 1. It follows from this theorem that (17) has a unique solution , for each .
-
2.
For each the functions satisfy and the sequence is uniformly bounded.
It is clear that for . Since it follows that , so that . On the other hand, we have ; so that . It follows that , and consequently . Also, since then . Suppose, by induction, that we have and there exists such that . Let
Claim 1. There exists depending only on , , , , and such that and .
To prove the claim we start with
which leads to
The boundary conditions imply
On the other hand
It is readily seen that
Since (by the induction hypothesis) it follows that
It follows from (18) and (19) that
Claim 2. . Since it suffices to show that , i.e.. We, first, prove that . For this purpose, set , for . We show that for all . Suppose by contradiction, that there exists such that . Since is continuous, it follows that there exists such that . Hence we have , and . Thus,
But and . Therefore, and . Hence,
It follows that
Since
we infer that
The above inequality is not possible by (A f )(3). Now, if , then , , and . It follows that
Since
we get
The monotonicity of leads to
This is not possible by the properties of the lower solution . Finally, . Indeed,
by definition of the lower solution . Similarly we show that . Thus, we have shown that , which implies that . Therefore, we have proved that the sequences , , and are uniformly bounded on the interval . Bolzano-Weierstrass theorem implies that we can extract subsequences , and that are uniformly convergent on . Using the diagonalization process, if necessary, we shall assume that , and . To complete the proof of our second main result we prove that , and is the desired solution of our original problem. Since , it follows from the uniform convergence of the two subsequences that , and this equality implies that and . Also, we have , which implies that , from which we readily get . It is clear that
The differential equation , for , implies that . The continuity of and the uniform convergence of the respective subsequences imply that , so that
The definition of and (21) show that
Similarly we can show that
and
We see that is a solution of (1), (2), (3), (4). Moreover, . This completes the proof of our main result. □
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Acknowledgements
A Boucherif is grateful to King Fahd University of Petroleum and Minerals for its constant support. The authors are grateful to the anonymous referees and the handling editor for comments that led to the improvement of the presentation of the manuscript.
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Boucherif, A., Bouguima, S.M., Benbouziane, Z. et al. Third order problems with nonlocal conditions of integral type. Bound Value Probl 2014, 137 (2014). https://doi.org/10.1186/s13661-014-0137-z
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DOI: https://doi.org/10.1186/s13661-014-0137-z