- Open Access
Third order problems with nonlocal conditions of integral type
© Boucherif et al.; licensee Springer 2014
- Received: 23 January 2014
- Accepted: 23 May 2014
- Published: 10 September 2014
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.
- third order differential equations
- integral boundary conditions
- a priori bound on solutions
- fixed point theorems
- lower and upper solutions
- iterative technique
Let satisfy for every . We denote by the set of all such that for every .
It is clear that if and , then .
Let satisfy for every . Let denote the set of all functions such that and .
It is clear that and imply that .
The operators and are continuous and bounded.
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.
Letbe continuous, bounded and satisfy the following condition:
(H ϕ ): for allsuch thatand.
- (ii)We show that there exists a positive constant , independent of , such that any possible solution of (6) satisfies(10)(11)(12)(13)Also, and (12) implyLet . Then any possible solution of (6) satisfies (10).
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 ) 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. □
We should emphasize that, unlike Theorem 6 in , our Theorem 1 gives the uniqueness of the solution and this is essentially utilized in the proof of our Theorem 2 below.
To state and prove our second main result we introduce the following assumptions.
there exists such that any solution of (1), with , satisfies , for all ;
for such that , , and ;
for all , , , and .
(A h ): are continuous and nondecreasing with respect to both arguments.
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.
- 1.The sequence is well defined. Indeed, for any and any we have and . It follows that the function , defined byis 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 .
For each the functions satisfy and the sequence is uniformly bounded.
Claim 1. There exists depending only on , , , , and such that and .
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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