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Some consequences of an existence result by Kiguradze and Partsvania for singular Dirichlet problems
Boundary Value Problems volume 2014, Article number: 160 (2014)
A sharp theorem by Kiguradze and Partsvania ensures the existence of extremal solutions between given lower and upper solutions for singular Dirichlet problems. This paper has a twofold purpose: first, we present a new sufficient condition for one of Kiguradze and Partsvania’s assumptions, and we illustrate its applicability in the study of a new family of examples; second, we combine Kiguradze and Partsvania’s theorem with Heikkilä’s iterative technique to obtain a new result on the existence of extremal solutions for a more general class of discontinuous and singular functional boundary value problems. In particular, our framework includes classical equations with delay (or advance), singularities with respect to the independent variable, and implicit functional boundary conditions.
MSC: 34A12, 34A36.
1 Introduction and first results
We are going to review the results in a paper by Kiguradze and Partsvania  and then we are going to use them in the proof of a new more general existence result of extremal solutions for functional and singular second-order problems.
Our revision of the results in  is not merely a reproduction, as it includes some contributions of our own. Specifically, Proposition 1.1 provides us with a new sufficient condition for a technical assumption in , and we will use it in the analysis of a new family of examples.
Let us start by recalling the main results in . Consider the Dirichlet problem
where , (), and satisfies
for all the mapping is measurable;
for almost all (a.a.) the mapping is continuous;
for every the function defined almost everywhere by
is integrable on for every sufficiently small .
Let denote the set of all real valued functions which are absolutely continuous over each compact subinterval of .
A lower (upper) solution of (1.1) is a function such that
and admits the representation
where and is a nondecreasing (nonincreasing) function whose first derivative is equal to zero almost everywhere on ;
holds almost everywhere on ;
the limits and exist and
We say that is a solution of (1.1) if it is both a lower and an upper solution, and we are interested in solutions of (1.1) which are extremal in the following sense.
We say that a solution of (1.1) is the greatest (smallest) solution of (1.1) in if and () on for any other solution .
When both the smallest and the greatest solutions of (1.1) in exist we call them the extremal solutions of (1.1) in .
The main contribution in  concerns the existence of extremal solutions of (1.1) in the presence of lower and upper solutions and . Kiguradze and Partsvania proved their existence result for nonlinearities which belong to the class defined as follows: a function belongs to if there exist two constants, , and a function such that is continuous in and for any , and continuous , an arbitrary solution of the differential equation
such that for admits the estimate
The main result in  reads as follows.
, Theorem 1]
Letbe a function satisfying (i), (ii), and (iii) and assume thatandare lower and upper solutions of (1.1) such thatfor all. If, then (1.1) has extremal solutions in the set
Notice that the condition implies, moreover, that the solutions provided by the previous theorem satisfy the estimate
The condition ‘’ is hard to check in practice. For that reason, the authors included some sufficient hypotheses which imply it. One of them is a one-sided Nagumo condition, in the spirit of . The following is a slight modification of Corollary 11 in .
Assume that there exist, which are, respectively, a lower and an upper solution of problem (1.1) such thatfor alland that the following conditions hold:
(H1):For eachand allthe mappingis measurable.
(H2):For almost all, alland allthe functionsandare continuous.
(H3):For everythe function
is integrable onfor every sufficiently small.
(H4):There exist, , , andsuch that for a.a. , alland allwe have
and for a.a. , alland allwe have
there exists a continuous function such that and for all we have
Then problem (1.1) has the extremal solutions in. Moreover, ifis a solution of (1.1) such that, thenfor all.
We only have to show that and then the conclusion follows from Theorem 1.1. To do this, let , be such that , , and is an arbitrary solution of the differential equation
We claim that for all . Indeed, as
there exists such that .
Reasoning by contradiction, assume that for some we have and, without loss of generality, that for all or . Then
so, in case , we have
and if , then
The previous two inequalities contradict (1.5), thus proving that is possible for no .
One can prove in an analogous way that for all . □
Finding a function to check (1.5) with is not easy, but the following proposition quite simplifies that task.
Let, , , , and.
If the functions
are well defined and integrable in, then condition (1.5) holds with
The assumptions guarantee that is continuous and integrable.
Now let be fixed. If , then
On the other hand, the identity yields ; hence
The proof is similar if . □
We close this section with a new example of the applicability of Kiguradze and Partsvania’s results. Notice how our Proposition 1.1 simplifies the verification of condition (H4) in Corollary 1.1.
For every and every the singular Dirichlet problem
has the extremal solutions between the lower solution and the upper solution .
Conditions (H1) through (H3) in Corollary 1.1 are obviously satisfied. The hardest part concerns showing that (H4) is satisfied with, for instance, the following choice of the functions and the constants: let us define for all , for all , , , and, according to (1.4), .
First, notice that for all , all and all we have
and this implies (1.2) and (1.3).
Let us use now Proposition 1.1 to show that (1.5) is satisfied too. First, we have to define a function on by
and for all we define
Now we have to check that the following (continuous) functions are also integrable on :
Notice that the relation between and implies that for all , so our problem reduces to proving that is integrable.
For all we have
which is integrable on because .
Similar computations show that is integrable on , and then it is integrable on the whole of because it is continuous. Hence, Proposition 1.1 ensures that condition (1.5) holds with .
2 Singular functional problems
In this section we are concerned with a generalization of problem (1.1) which includes both past and future dependence. To do so, let us fix , define
and consider the problem
where and .
Notice that, in general, solutions of (1.1) are defined on an interval which is greater than and, moreover, the differential equation depends upon the behavior of the solution in that extended interval. This implies, in particular, that we can study equations with deviated arguments of the form
where is measurable. Moreover, classical start-final functions for problems with delay or advance are also included in (2.1): we only have to take
where and represent, respectively, the initial and the final state of the solution. Problem (2.1) also includes multipoint boundary conditions, which have received a lot of attention in the last few years; see , .
We begin by introducing the concepts of lower and upper solutions for problem (2.1). We do this simply by extending Definition 1.1 in the obvious way.
We say that is a lower solution (respectively, an upper solution) of problem (2.1) if it satisfies the following conditions:
and admits the representation
where and is a nondecreasing (respectively, nonincreasing) function whose first derivative is equal to zero almost everywhere on ;
holds almost everywhere on ;
for all we have
and for all we have
Our existence result for (2.1) will be proven by means of Kiguradze and Partsvania’s results described in the previous section along with Heikkilä’s generalized iterative technique. The following result will be essential.
, Theorem 1.2.2]
Letbe a subset of an ordered metric space, a nonempty order interval in, anda nondecreasing mapping. Ifconverges inwheneveris a monotone sequence in, thenhas inthe extremal fixed points, and, which satisfy
We are at last in a position to introduce and prove our main result.
Letandbe, respectively, a lower and an upper solution of problem (2.1) such thaton, put
and assume that the following conditions hold:
(F1):For eachthe mappingsatisfies conditions (i), (ii), and (iii), anduniformly in, in the sense that the functiondoes not depend on.
(F2):For a.a. , alland allthe mappingis nonincreasing on.
(B1):For eachthere existssuch that for anythe inequalityimplies
for alland all.
(B2):There existssuch that for alland allwe have
(B3):For alland all, the mappingis nondecreasing in.
Then problem (2.1) has extremal solutions in.
Consider the set , which becomes an ordered metric space (in fact, a Banach space) when endowed with the supremum norm and the pointwise ordering. In consider the subset and the ordered interval . Now, define an operator as follows: for each , is the greatest solution in of the Dirichlet problem
where for each (respectively, each ), (respectively, ) is the unique fixed point of the contractive mapping (respectively, ).
Claim 1: Operatoris well defined. Fix . First, the definitions of lower and upper solutions and condition (B3) ensure that for any we have
and . Hence has at least one fixed point in , which is unique by virtue of (B2). Hence, the number is well defined for every ; let us prove now that is continuous on . To do so, let us fix and notice that
and therefore is continuous by virtue of (B1).
A similar argument shows that is well defined and continuous too.
Now, condition (F1) guarantees by application of Theorem 1.1 that the extremal solutions in of problem exist and, in particular, is well defined.
Moreover, condition (F1) also ensures that there exists a continuous and integrable function such that for all and all .
Claim 2: Operatoris nondecreasing. Let be such that . Then by condition (F2) we have for a.a. that
Let us prove now that for each . Indeed, condition (B3) yields
which implies that has its fixed point, namely , inside the interval . Hence, on .
A similar argument shows that for all .
Summing up, is a lower solution of problem , whose greatest solution between and is precisely . Therefore on the whole of .
Claim 3: Operatorhas the extremal fixed points. Let be a monotone sequence. As is nondecreasing, the sequence is monotone and bounded and therefore it has a pointwise limit, say .
We are going to prove that tends to uniformly on and that .
First, for , , and every we have
Second, for and we have
Analogously, for and we have
Since the are continuous at and , we deduce from (2.4), (2.5), (2.6), and condition (B1) that is uniformly equicontinuous on . Hence tends to uniformly on and is continuous on .
Moreover, (2.4) implies that is absolutely continuous on , thus proving that .
Now we can apply Lemma 2.1 to ensure that operator has the extremal fixed points in .
Claim 4: The greatest fixed point ofcorresponds with the greatest solution of problem (2.1) in. Let be the greatest fixed point of . As , it is clear that is a solution of problem (2.1). Now, let be another solution of (2.1). In this case, both and solve , and so taking into account that is the greatest of such solutions. Now, condition (2.2) implies that .
The existence of the least solution of (2.1) in follows from a proper redefinition of the operator . □
The next less general version of Theorem 2.1 is easier to use in practice.
Under the conditions of Theorem 2.1, replace (F1) by
:For eachthe functionsatisfies (H1)-(H4) with the same functionin (H4) for all.
Then the conclusion of Theorem 2.1holds.
3 An example of application
We finish this paper with a corollary of our main result which applies to a large family of boundary conditions satisfying stronger conditions that (B1)-(B3). This corollary is illustrated later with a concrete example.
Letsatisfy (F1) (or) and (F2) and assume that the functions, can be written as follows:
Then the following conditions are sufficient to guarantee the conclusion of Theorem 2.1:
():The functionis uniformly continuous in.
():For alland allthe functionis Lipschitzian with Lipschitz constant, and the functionis nondecreasing. Moreover, there exists.
():For allthe functionis nonnegative, differentiable, and
We will show that these conditions imply (B1)-(B3).
To see (B1), we take and apply () to obtain for and that
and now (B1) follows from () and the fact that is bounded in the compact set .
To see (B2) we notice that for , , condition () provides that there exists such that
Then (B2) follows from (3.1).
Finally, condition () follows from the facts that is nondecreasing and is nonnegative. □
Let , , be nonincreasing functions (not necessarily continuous), and consider the following functional problem, which includes both a past and a future dependence:
We claim that for sufficiently small value of problem (3.2) has the extremal solutions between the lower solution and the upper solution . Notice that and are solutions of the differential equation in (3.2) but not solutions of the whole problem.
This problem falls inside the scope of our main result. To see it, simply define the functions
We are going to use Corollary 3.1 to prove that problem (3.2) has the extremal solutions between and provided that
First we have to prove that and are, in fact, lower and upper solutions. To see that is a lower solution, notice that and . Now, to check that is an upper solution, we use
Let us check the assumptions required in Corollary 3.1. For all we have
and then the condition (H4) is satisfied as in Example 1.1 (with the same function for all ), thanks to condition (3.3).
The remaining conditions over are easily verified, and therefore we omit further explanations.
Now for the boundary conditions. First, both and are bounded w.r.t. and nondecreasing w.r.t. , and for and we have
so is Lipschitzian for all with Lipschtitz constant . On the other hand, the function satisfies
and so condition () holds.
Finally, the function is nonnegative for and satisfies
and so condition () holds.
Kiguradze IT, Partsvania N: On minimal and maximal solutions of two-point singular boundary value problems. Mem. Differ. Equ. Math. Phys. 2005, 36: 147-152.
Kiguradze IT: On some singular boundary value problems for nonlinear second order ordinary differential equations. Differ. Uravn. 1968, 4(10):1753-1773. (Russian); Differ. Equ. 4, 901-910 (1968) (English translation)
Kiguradze IT, Staněk S:On periodic boundary value problem for the equation with one-sided growth restrictions on . Nonlinear Anal. 2002, 48: 1065-1075. 10.1016/S0362-546X(00)00235-2
Adje A: Sur et sous-solutions généralisées et problèmes aux limites du second ordre. Bull. Soc. Math. Belg., Sér. B 1990, 42: 347-368.
De Coster C, Habets P: Two-Point Boundary Value Problems: Lower and Upper Solutions. Elsevier, Amsterdam; 2006.
López Pouso R: Some general comparison results for second order ordinary differential equations. J. Math. Anal. Appl. 2000, 252: 710-728. 10.1006/jmaa.2000.7110
Cabada A, O’Regan D, Pouso RL: Second order problems with functional conditions including Sturm-Liouville and multipoint conditions. Math. Nachr. 2008, 281: 1254-1263. 10.1002/mana.200510675
Kiguradze IT, Lomtatidze A, Partsvania N: Some multi-point boundary value problems for second order singular differential equations. Mem. Differ. Equ. Math. Phys. 2012, 56: 133-141.
Heikkilä S, Lakshmikantham V: Monotone Iterative Techniques for Discontinuous Nonlinear Differential Equations. Dekker, New York; 1994.
This work was partially supported by FEDER and Ministerio de Educación y Ciencia, Spain, project MTM2010-15314.
The authors declare they have no competing interest.
The authors worked jointly on this paper and their respective contributions were of similar quality and quantity. All authors read and approved the final manuscript.
About this article
- discontinuous differential equations
- singular differential equations
- functional differential equations
- boundary value problems
- equations with delay
- equations with advance