On Volterra-type integral equations in noncompact metric space
© Zahariev et al.; licensee Springer. 2014
Received: 10 December 2013
Accepted: 20 June 2014
Published: 22 July 2014
In the present work, we consider one of the possible generalizations of linear and nonlinear Volterra integral equations of the second kind in the case when the independent variable belongs to an arbitrary noncompact metric space. Sufficient conditions are obtained for the existence of solutions of Volterra-type integral equations in the nonhomogeneous case. Some applications of the obtained results to the integral inequalities are given.
MSC:35Q99, 35R35, 65M12, 65M70.
Integral equations and inequalities have been of considerable significance in mathematics and have held a central place in the attention of mathematicians during the last few decades. In the past few years, integral equations have proved to be of tremendous use in several applied fields, such as population dynamics, spread of epidemics, automatic control theory, network theory and the dynamics of nuclear reactors. The main application of the integral inequalities is that they provide an explicit bounds of the unknown functions, which are a very useful and important device in the study of many qualitative as well as quantitative properties of solutions of nonlinear differential equations. In the theory of integral inequalities, an enormous amount of effort has been devoted to the polishing of classical approaches to proving the inequalities. The techniques of these proofs, in general based on the classical mathematical analysis, lead up to virtuosity and significantly depend on the number of independent variables and the geometry of the domain of integration. The mathematical literature provides a good deal of information about the integral equations and inequalities and an excellent amount of results may be found in the monographs [1–4] and the comprehensive list of references therein. We can find several different directions and approaches to this field of study in [5, 6] and . All monographs mentioned above are proposed because the authors have an essential contribution to the theory presented in their books.
The aim of this paper is to develop the approach introduced for the linear case in the work  for the nonlinear one. The main idea introduced in  is simple - establishing conditions under which the unique solution of equation (2.1) is an upper bound of all solutions of inequality (2.2). A similar idea was used in  too. In view of the applications, it is important to study also in which cases, as in the one-dimensional case, the exponential function is a sharp estimation for the Gronwall-type multidimensional inequalities. In the case when the domains of integration are the Cartesian product of bounded and closed intervals this is true (see ), but some results and examples given in  indicate that the answer of this problem generally speaking is negative, even in the linear case. Moreover, the type of the sharp estimation function will depend on the geometry of the domains of integration in the high dimensional cases.
The paper is organized as follows. In Section 2 we introduce an abstract nonlinear analogue of the Volterra equations and its corresponding inequality. In Section 3 we discuss the possibility to replace without loss of generality one compact domain of integration with another one which is close to it in the measure and metric sense, but have better properties. Section 4 is devoted to the study of Volterra equation (2.1) introduced below. Section 5 includes the results concerning the integral inequality (2.2), and in Section 6 some applications of our results obtained in the previous sections are given as well as examples illustrating the applications are presented.
We note that different kinds of results used in the conceptual plan of our approach introduced in  are received in [11–14] and some interesting ideas in the atomic case are developed in  and .
Let Ω be a complete metric space with metric ρ, let denote the σ-algebra of the Borel subsets of Ω, and let be a nontrivial σ-finite Borel measure. We will denote by the open balls with a center point , and radius . If G is an arbitrary subset of Ω, then ∂G denotes the boundary of G and denotes the ε-neighborhood of G.
Let B be a real Banach space with the norm , and let be a cone in B. Then we can introduce a partial ordering in B associated with the cone V, i.e., when . We shall write to indicate that , but . Denote by , is an arbitrary compact subset, the Banach space of all continuous maps with the norm , by the Banach space of all bounded continuous maps with the norm and by the linear topological space of all continuous maps ( if for each we have ).
the symmetric difference of the sets G and H.
Definition 2.1 A set is called an atom for the measure μ if , and for each with one has .
Partially, if for some point we have that , then we say that this point is an atom for the measure μ; otherwise, we say that the point is nonatomic.
where the operator and , .
For every point , the set is compact.
For each and every , there exists such that, for each with , we have that .
For every , the inclusion holds for each .
There exists such that .
For every map for which conditions (A) hold, we will define .
Remark 2.2 Generally speaking it is not necessary for the set to include the point x. A simple example when this is true is and .
3 Domains of integration
It is well known that even for finite dimensional metric spaces with Lebesgue measure two compact subsets in them can be very close in the measure sense (partially, having equal measures) and at the same time very far in the metric sense (having very different diameters). From this fact it follows that using as domains of integration arbitrary compact sets cannot be convenient enough in lots of cases. It is easy to see that generally this is true even for the compact sets belonging to the set M. The aim of this section is to find some other set , whose elements have better properties. Since all these sets will be used as domains of integration, the best result will be if for every element we can find an element such that , and both elements are close in the metric sense too.
Let be an arbitrary set.
Definition 3.1 The nonatomic point will be called essential for the set G if for each we have .
Definition 3.2 The nonatomic point will be called unessential for the set G if there exists such that .
Let us consider the points .
Lemma 3.3 For every point which is unessential for G, there exists such that the open ball does not include essential points for G.
which is impossible. □
Corollary 3.4 Every isolated point is either unessential for G, or an atom for the measure μ.
Then either and is an atom, or and then is an unessential point. □
Corollary 3.5 The nonatomic point is essential for G if for each in the open ball , there exists at least one point z, which is essential for G.
Proof Let for each the open ball include at least one point z, essential for G and put . Then and therefore . □
Remark 3.6 It is easy to see that if G includes internal points, then all internal points of G are essential for the set G. Every external point of G is either nonessential for the set G, or an atom for the measure μ. Moreover, if the set G includes at least one essential for G point, then we have .
Definition 3.7 ()
The sets will be called μ-equivalent if .
Let be two arbitrary sets. If the set includes at least one atom, then the sets G and H cannot be μ-equivalent.
Definition 3.8 The set is called μ-dense if each is an essential for G point.
Everywhere in our exposition below, we will assume that the measure μ is nonatomic.
Let G be an arbitrary closed subset of Ω. Denote by the set of all points , which are essential for G and denote by the set .
The set is either empty or μ-dense and a closed subset of Ω.
The set is either empty or and every point is unessential for G.
Proof 1. If G is closed and , then obviously . Let and be an arbitrary essential point. Then from the definition of the set it follows that is μ-dense. Let be an arbitrary convergent sequence, and . Then, for each in the neighborhood , there exists at least one essential point of G. Then Corollary 3.5 implies that is an essential point of G. Thus we can conclude that the set is a closed subset of G.
2. If , then from point 1, Lemma 3.3 and Corollary 3.4 it follows that and all points are unessential for G. □
Corollary 3.10 Let G be an arbitrary closed subset of Ω. Then if G includes at least one essential point, we have that .
Proof Let be an arbitrary essential point. Then, for each , we have and therefore . □
Lemma 3.11 Let G be a compact subset of Ω, and . Then the set is a nonempty compact set, and the sets G and are μ-equivalent.
Proof Suppose that . Then , and according to Lemma 3.9, for , there exists such that . Since G is a compact subset of Ω and is an open cover of G, then there exist points and positive numbers such that , which is impossible. Therefore , and according to Lemma 3.9, the set is a closed subset of G, and we can conclude that is compact.
where are open subsets of Ω. Since G is a compact set, then there exist points and positive numbers such that , and for each i we have . Therefore, and the sets G and are μ-equivalent. □
The next example illustrates the fact that the statement of Lemma 3.11 can be not true when the measure μ is atomic and the set G is only closed, but not compact.
Obviously, , , and therefore the sets and G are not μ-equivalent.
Lemma 3.12 Let, for the map , conditions (A1) and (A2) hold. Then, for every for which , there exists such that for each with , we have .
Proof Suppose that the statement of Lemma 3.12 is not true. Then there exist a strictly decreasing sequence from positive numbers with and a sequence , with for each . Since , then from condition (A2) it follows that . On the other hand, since for each , then we have that , which is impossible. □
Corollary 3.13 Let the conditions of Lemma 3.12 hold. Then, for each , there exists such that for each with , we have .
For every map , we define the associated map for each by the following relation .
Lemma 3.14 ()
Let, for the map , conditions (A) hold. Then, for the map , conditions (A) hold too.
Remark 3.15 From Lemma 3.14 and Lemma 3.11 it follows that for every map for which conditions (A) hold, without loss of generality, we can use for every as a domain of integration instead of the set its μ-equivalent set which is μ-dense.
Let the map be arbitrary and denote by KerM the set .
Conditions (A) hold.
Ω is a connected metric space and .
Proof (i) Let us consider the set and define . Condition (A4) implies that , and let us assume that too.
First we will prove that is a closed subset of Ω. Let be an arbitrary point. Then there exists a sequence such that . Since , then there exist a number and a constant such that for each . For an arbitrary number from condition (A2) it follows that there exists a number such that for each we have . Then and therefore . Thus we proved that is a closed subset of Ω. It is easy to see that for each from condition (A2) it follows that . Then is also a closed set and therefore , where , which is impossible. Thus we can conclude that .
and the inequality holds, which is impossible. Then, for every point , we have that either or .
Define and . We will prove that G is closed. Since , let be an arbitrary convergent sequence with boundary , i.e., . Then we have , and there exists integer such that for each the inequality holds.
holds, which is impossible. Therefore , i.e., .
Since and Ω is connected, then H cannot be a closed set. There exist a point and a sequence such that , and therefore . There exists integer such that for each we have . Since the sequence , then we have that and therefore . Then the following estimation holds, which is impossible. □
Remark 3.17 The assertion of Lemma 3.16 was proved in  with the additional assumption that all sets are μ-dense.
We will say that condition (AM) is fulfilled if the following condition holds:
(AM) For each , the set is μ-dense.
Before considering the next important theorem, we draw attention to that from the viewpoint of the applications the case when is meaningless.
Conditions (A) and (AM) hold.
Ω is a connected metric space and .
Then, for every for which , there exist a point and a sequence such that .
Proof Let be an arbitrary point, for which . In virtue of Lemma 3.16, there exists at least one connected component of , such that . For the closure of , we have that , then is compact. Since is compact, then if we assume that , then there exists such that and , which is impossible.
Define and . In a similar way as in the proof of Lemma 3.16, we can prove that G is closed, and too. Moreover, for each , we have . Then there exist a point and a sequence such that and therefore . In the same way as in the proof of Lemma 3.16, we conclude that this is impossible.
Then we can conclude that , and therefore there exists at least a point . Then, since is connected, there exists such that . □
Theorem 3.19 Let conditions (A) and (AM) hold. Then, for every point and each , there exists such that for we have that and .
holds. Taking into account that is compact, there exists a convergent subsequence such that , where . Since , then from the last inequality above and condition (A2) the relation follows. Therefore is an unessential point of , which is impossible. Then there exists such that for every the inclusion holds.
Let be an arbitrary fixed point. Then we have and therefore . □
Remark 3.20 For the sets which are μ-dense, Theorem 3.19 illustrates the important fact that if these sets are close in the measure sense, they are close in the metric sense too.
Definition 3.21 We say that the set is M-star if for every the inclusion holds.
Remark 3.22 It is easy to see that condition (A3) implies that for each the set is an M-star set. Moreover, the union and the intersection of an arbitrary family of M-star sets are the M-star set. The set KerM is an M-star set too.
4 Main results
For every M-star compact set , the operator Q is continuous in the set .
- (S2)For each and every , there exist numbers and such that for every functions for which , the following inequality holds:
- (S3)For every and arbitrary , there exists a constant such that for every two functions , the following inequality holds:
- (S4)For each M-star continuum , there exists a number such that for every two functions and for each , the following inequality holds:
Conditions (A), (AM), (S1) and (S2) hold.
Ω is a connected metric space and .
Then the operator K defined by equality (4.1) maps continuously for each .
hold, then such an extension will be called a middle extension.
Lemma 4.3 Let conditions (A) and (AM) be fulfilled. Then, for each , every function has at least one middle extension.
hold for each , then is a middle extension of the function . □
Definition 4.4 We say that equation (2.1) has a local solution in some M-star set for some if there exist a point and a function for which and f satisfies equation (2.1) for each . If and f satisfies equation (2.1) for each , then we say that f is a solution of (2.1) in .
Conditions (A), (AM), (S1) and (S3) hold.
Ω is a connected metric space and .
(the case when is trivial).
Then, for every such that and for every , equation (2.1) has at least one local solution in .
According to Lemma 4.3, for every function and for each , there exists a middle extension such that . Therefore, the set for every .
where , .
hold. Obviously, first of them implies that .
Therefore, the operator maps into and is a contraction. Therefore, equation (2.1) has a solution . Moreover, since , then . □
Remark 4.6 Generally speaking, if the set is not connected, then the local solution of equation (2.1) can be not unique. The next example confirms that the connectivity of the set is essential for the uniqueness of the local solution of equation (2.1).
Example Let , , for , for and for . Then, for each according to Theorem 4.5, there exists such that equation (2.1) has a solution in . Moreover, since for , then from conditions (A) it follows that if we choose small enough, then for . By the Urison lemma, there exists a continuous function such that for and when . Then, obviously, the function is another solution of equation (2.1) in .
The metric space Ω is connected and .
Conditions (A), (S1), (S4) and (AM) hold and for each the sets are connected.
Then, for each , equation (2.1) has exactly one solution .
holds. Using Theorem 2 and Theorem 3 from  we get for each , which contradicts our supposition. □
Corollary 4.8 Let the conditions of Theorem 4.7 be fulfilled, and let the set be connected for any . Then, for each , equation (2.1) has exactly one solution .
Proof It is enough to apply Theorem 4.7 for . □
The metric space Ω is connected and .
Conditions (A), (AM), (S1), and (S4) hold, and for any the set is connected.
For each , there exists an M-star continuum with such that .
For arbitrary , the operator Q is monotonic in with regard to the order induced by the cone .
Then, for each and every , equation (2.1) has a unique solution in .
into itself, and therefore K has at least one fixed point in the interval (see , Chapter 8).
From Theorem 4.7 it follows that for each for which and for any solutions and , we have that for , and therefore the global solution defined with (2.1) for all is unique. Note that the solution is continuous in for each .
It is easy to see that if Ω is a local compact metric space, then the solution is continuous in Ω, i.e., . □
In this section we apply the results obtained in Section 4 to study inequality (2.2).
Definition 5.1 We say that inequality (2.2) has a local solution in some M-star set for some if there exist a point and a function for which and g satisfies inequality (2.2) for each . If and g satisfies inequality (2.2) for each , then we say that g is a solution of (2.2) in .
Ω is a connected metric space and .
For each , the sets are connected.
Conditions (A) hold.
The function for every two fixed elements and is a monotonously increasing function of ν.
holds for each , then for .
Condition (A4) implies that there exists such that . Then, according to condition (A2), for each , we have that . Since for each , then (5.1) implies that and therefore .
Then there exists a set such that and for each . Then, for each , we have that and from condition (A2) it follows that , which contradicts our assumption. Thus we prove that W is a closed set.
It is easy to see that if , then obviously , i.e., W is an M-star set. Since , then for each , , there exists an open ball such that . Then the set is an open cover of and . Since is compact, there exists a finite number of balls such that . Then, according to Theorem 3.19, there exists such that for each , we have that , and therefore we prove that W is an open set.
and therefore for each . □
Remark 5.3 It is easy to see that if we replace the strict inequality (5.1) by a non-strict one, the assertion of the theorem still holds for every .
Conditions of Theorem 4.9 hold.
Condition 4 of Theorem 5.2 holds.
Then, for every solution of inequality (2.2) and for every , the inequality holds, where is the unique solution of equation (2.1).
then we can conclude that (f is the unique solution of equation (2.1)). For other points for which inequality (5.2) is strict, the assertion of Corollary 5.4 follows from Theorem 5.2. □
As an illustration of the results obtained in the previous sections, we will consider integral inequalities with maxima.
for and for . It is simple to verify that all conditions of Corollary 5.4 are fulfilled and therefore we have that for , where is the unique solution of (6.1) and is an arbitrary solution of (6.2).
This paper is partially supported by Plovdiv University NPD grant NI13 FMI-002.
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