Abstract and Applied Analysis

Volume 2008 (2008), Article ID 829701, 35 pages

http://dx.doi.org/10.1155/2008/829701

## Generalized Solutions of Functional Differential Inclusions

^{1}Center for Integrative Genetics (CIGENE), Norwegian University of Life Sciences, 1432 Aas, , Norway^{2}Department of Mathematical Sciences and Technology, Norwegian University of Life Sciences, Aas 1432, Norway^{3}Department of Algebra and Geometry, Tambov State University, Tambov 392000, Russia^{4}Department of Higher Mathematics, Faculty of Electronics and Computer Sciences, Moscow State Forest University, Moscow 141005, Russia

Received 12 March 2007; Revised 4 July 2007; Accepted 12 September 2007

Academic Editor: Yong Zhou

Copyright © 2008 Anna Machina et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We consider the initial value problem for a functional differential inclusion with a Volterra multivalued mapping that is not necessarily decomposable in . The concept of the decomposable hull of a set is introduced. Using this concept, we define a generalized solution of such a problem and study its properties. We have proven that standard results on local existence and continuation of a generalized solution remain true. The question on the estimation of a generalized solution with respect to a given absolutely continuous function is studied. The density principle is proven for the generalized solutions. Asymptotic properties of the set of generalized approximate solutions are studied.

#### 1. Introduction

During the last years, mathematicians have been intensively studying (see [1, 2]) perturbed inclusions that are generated by the algebraic sum of the values of two multivalued mappings, one of which is decomposable. Many types of differential inclusions can be represented in this form (ordinary differential, functional differential, etc.). In the above-mentioned papers, the authors investigated the solvability problem for such inclusions. Estimates for the solutions were obtained similar to the estimates, which had been obtained by Filippov for ordinary differential inclusions (see [3, 4]). The concept of quasisolutions is introduced and studied. The density principle and the “bang-bang” principle are proven. In papers [5–8], the perturbed inclusions with internal and external perturbations are considered, and the conjecture that “small” internal and external perturbations can significantly change the solution set of the perturbed inclusion is proven. Let us remark that, in the cited papers, the proofs of the obtained results essentially depend on the assumption that the multivalued mapping, which generates the algebraic sum of the values, is decomposable. Therefore, these studies once again confirm V. M. Tikhomirov's conjecture that decomposability is a specific feature of the space and plays the same role as the concept of convexity in Banach spaces. The decomposability is implicitly used in many fields of mathematics: optimization theory, differential inclusions theory, and so forth. If a multivalued mapping is not necessarily decomposable, then the methods known for multivalued mappings cannot even be applied to the solvability problem of the perturbed inclusion. Furthermore, in this case, the equality between the set of quasisolutions of the perturbed inclusion and the solution set of the perturbed inclusion with the decomposable hull of the right-hand side fails. This equality for the ordinary differential inclusions was proven by Ważewski (see [9]). The point is that, in this case, the closure (in the weak topology of ) of the set of the values of this multivalued mapping does not coincide with the closed convex hull of this set. As a result, we have that fundamental properties of the solution sets (the density principle and “bang-bang” principle) do not hold any more (see [3, 10–13]). The situation cannot be improved even if the mapping in question is continuous.

In this paper, we consider the initial value problem for a functional differential inclusion with a multivalued mapping. We assume that this mapping is not necessarily decomposable. Some mathematical models can naturally be described by such an inclusion. For instance, so do certain mathematical models of sophisticated multicomponent systems of automatic control (see [14]), where, due to the failure of some devices, objects are controlled by different control laws (different right-hand sides) with the diverse sets of the control admissible values. This means that the object's control law consists of a set of the controlling subsystems. These subsystems may be linear as well as nonlinear. For example, this occurs in the control theory of the hybrid systems (see [15–20]). Due to the failure of a device, the control object switches from one control law to another. The control of an object must be guaranteed in spite of the fact that failures (switchings) may take place any time. Therefore, the mathematical model should treat all available trajectories (states) corresponding to all switchings. The generalized solutions of the inclusion make up the set of all such trajectories. The concept of a generalized solution should be then introduced and its properties should be studied.

We consider a functional differential inclusion with a Volterra-Tikhonov type (in the sequel simply Volterra type) multivalued mapping and we prove that for such an inclusion, the theorem on existence and continuation of a local generalized solution holds true. This justifies one of the requirements, which were formulated in the monograph of Filippov [4] for generalized solutions of differential equations with discontinuous right-hand sides. In the present paper, it is also proven that in the regular case, that is, when a multivalued mapping is decomposable, a generalized solution coincides with an ordinary solution. At the same time, the concept of a generalized solution discussed in the present paper does not satisfy all the requirements that are usually put on generalized (in the sense of the monograph [4]) solutions of differential equations with discontinuous right-hand sides. For instance, the limit of generalized (in the sense of the present paper) solutions is not necessarily a generalized solution itself. The reason for that is that a multivalued mapping that determines a generalized solution (the definition is given below) may not be closed in the weak topology of as this mapping is not necessarily convex-valued.

#### 2. Preliminaries

We start with the notation and some definitions. Let be a normed space with the norm . Let be the closed ball in the space with the center at and of radius ; if then . Let . Then is the closure of , is the convex hull of ; , is the set of all extreme points of ; . Let . Let if and .

Let be the distance from the point to the set in the space ; let be the Hausdorff semideviation of the set from the set ; let be the Hausdorff distance between the subsets and of .

We denote by (resp., ) the set of all nonempty compact subsets of (resp., the set of all nonempty, bounded, closed in the space , and relatively compact in the weak topology on the space subsets of ). Let be the set of all nonempty bounded subsets of

Let be a system of subsets of (a subset of ). We denote by the set of all nonempty convex subsets of belonging to the system (the set of all nonempty convex subsets of belonging to ).

Let be the space of all -dimensional column vectors with the norm . We denote by (resp., ) the space of continuous (resp., absolutely continuous) functions with norm (resp., . Let be a measurable set (—the Lebesgue measure). We denote by the space of all functions such that is integrable (if ) and the space of all measurable, essentially bounded (if ) functions with the norms respectively.

Let . The set is called
integrally bounded if there exists a function such that for each and almost all The set is said to be *decomposable* if for each and every
measurable set the inclusion holds, where is the
characteristic function of the set . We denote by (resp., ) the set of
all nonempty, closed, and integrally bounded (resp., nonempty, bounded,
closed, and decomposable) subsets of the space .

Let be a measurable mapping. Then by definition, . By (resp., ), denote the cone of all nonnegative functions of the space (resp., ).

Let be a mapping between two partially ordered sets and (the partial order of both sets is denoted by ). The mapping is isotonic if whenever

In this paper, the expression “measurability of a single-valued function” is always used in the sense of Lebesgue measurability and “measurability of a multivalued function” in the sense of [21]. Let be a space with finite positive measure and let be a multivalued mapping from to A set () of measurable mappings from to is said to approximate the multivalued mapping if the set is measurable for any , and the set belongs to the closure of its intersection with the set for almost all A multivalued mapping from to is called measurable if there exists a countable set of measurable mappings from to that approximates the mapping

Further, let us introduce the main characteristic properties of a set that is decomposable.

Lemma 2.1. *
Let Then there
exists a function such that for each function and almost all *

*Proof. *Let , be a sequence
of functions such that

Let us show that there exists a sequence of functions , such that the
equality (2.2) holds and for almost all

Indeed, let and , where Since we see that the
sequence , has the
following properties: for almost all , the inequalities (2.3) hold and for each Hence, from
this property and equality (2.2), it follows that the sequence , satisfies
(2.2). Further, we consider a measurable function defined by Since the set is bounded, we
see, using Fatou's lemma (see [22]), that Moreover, by
the definition of the function and due to
(2.2), for every
measurable set Now, let us
show that the function defined by
(2.4) satisfies the assumptions of the lemma. Indeed, if the contrary is true,
then there exist a function and a
measurable set () such that for each This implies
that which
contradicts (2.5). This completes the proof.

Lemma 2.2. *
Let and , be a sequence
that is dense in
Further, let a
measurable set be defined by ** Then *

*Proof. *Since and the
sequence , is dense in we have, due to
the closedness of the set the relation Let us prove
that Let For each , , put which are measurable sets. For , let , and for , let Then if By the
definition of the mapping for each , we have Let , be a sequence
of measurable functions such that Since the set is
decomposable, we see that for each . Moreover, from Lemma 2.1 and the definition of the
set , it follows that for the functions , we have the
estimates where satisfies the
assertions of Lemma 2.1. From (2.8) and (2.10), it follows that in as Since the set is closed, we
have that . Hence Thus

Lemma 2.3. *Let
measurable sets ,
be integrally
bounded, then if and only if for almost all *

*Proof. *First
of all, it is evident that if for almost all , then .

Let and let , be a countable
set, which is dense in and which
approximates (see [21]).
Thus for each and by the
definition of the set we have that for almost all Since the
sequence , approximates
the map it follows from
the previous inclusion that for almost all

Corollary 2.4. *
Let and let , be measurable
sets such that Then for almost all *

*Remark 2.5. *If then a
measurable set , that satisfies uniquely
determines the set

#### 3. Decomposable Hull of a Set in the Space of Integrable Functions

We introduce the concept of the decomposable hull of a set in the space We consider a multivalued mapping that is not necessarily decomposable. For such a mapping, we construct its decomposable hull and investigate topological properties of this hull.

*Definition 3.1. *Let be a nonempty
subset of By , we denote the set of all finite combinations of elements , where the
disjoint measurable subsets , of the segment are such that

Lemma 3.2. *
The set
is decomposable
for any nonempty set *

*Proof. *Let Let also be a measurable
set. Without loss of generality, it can be assumed that where , , and the
measurable disjoint sets , , are such that , (if the number
of summands in (3.2) is not the same, we may use arbitrary functions multiplied
by the characteristic functions of the empty sets). Further, from the equality it follows that Hence, the set is
decomposable.

*Remark 3.3. *Note that even if a set is bounded, the
set is not
necessarily bounded. For example, let us check that Indeed, let and , be measurable
sets with the following properties: if , , for each , the inequality holds. Then , and Therefore, and
consequently, the equality (3.4) holds.

*Remark 3.4. *From (3.4), it follows that if a set is relatively
compact in the weak topology of then the set does not
necessarily possess this property.

*Remark 3.5. *Note
that if a set is convex in then this set
is not necessarily decomposable. The ball is an example
of such a set.

*Remark 3.6. *If a set is integrally
bounded, then by Lemma 2.2, for the set , there exists a measurable and integrally bounded
mapping such that

Lemma 3.7. *If a
set is
decomposable, then *

*Proof. *Evidently, . We claim that . The proof is made by induction over . By the
definition of the switching convexity, any expression (3.1) including two
elements and two
measurable sets belongs to .

Suppose now that for , the combination of the form (3.1) belongs to . Let and let be disjoint
measurable sets such that . Let By the inductive assumption, and therefore . Since we have that Hence . This concludes the proof.

Corollary 3.8. *If then the set is the minimal
set which is decomposable and which contains *

*Proof. *Consider any set which is
decomposable and which satisfies Then, by Lemma
3.7, we have

Lemma 3.9. *If a
set is convex, then
so is the set *

*Proof. *Let be given by the
formula (3.2). It follows from the convexity of the set and the
equality that for any Thus, the set is convex.

Similar to the definition of the convex hull in a
normed space, the set will, in the
sequel, be called the *decomposable hull of the set **in the space of
integrable functions*, or simply the *decomposable
hull of the set *. Likewise, is addressed as
the *closed decomposable hull of the set *.

*Remark 3.10. *If then the closed
decomposable hull of the set (the set ) can be
constructed as described in Remark 3.6. To do it, one needs a measurable and
integrally bounded (see Remark 3.6)
mapping that satisfies
(3.7). Note that finding this mapping is easier than
constructing the set . At the same time, when one studies the metrical
relations between the sets and their
decomposable hulls (see Lemma 3.12), it is more convenient to use Definition
3.1.

Lemma 3.11. *Let and let a set
be
decomposable. Then for any disjoint measurable sets such that , one has *

*Proof. *Indeed,
let and satisfy It follows from
this estimate that This yields

Further, let us show that the opposite inequality is
valid. Let , where is the set of
of all mappings from restricted to , and suppose
that the functions , satisfy Since the set is
decomposable, it follows that the map defined by belongs to the set By (3.15), we
have This implies that Comparing (3.14)
and (3.18), we obtain (3.12).

Lemma 3.12. *If and there
exists a function such that ** for any
measurable set , then ** for any
measurable set *

*Proof. *Let be a measurable
set, . Let and , Suppose also
that the functions and disjoint
measurable sets , such that satisfy the
equality

Further, by , we denote the
restrictions of these functions to and put ,

From (3.21) and Lemma 3.11, it follows that

From (3.19), we obtain that for each

Therefore, (3.22) and (3.23) imply Since (3.24)
holds for any it follows from
(3.24) that (3.20) holds as well.

*Remark 3.13. *Note that the function (see (3.19))
provides a uniform with respect to measurable sets estimate for
the Hausdorff semideviation of the set from the set

*Remark 3.14. *The
inequality (3.20) holds true even if the set is replaced
with its closure ,

We say that a multivalued mapping is *integrally bounded on a set * if the image is integrally
bounded.

Let We introduce an operator by the formula

Note that even if a mapping is continuous, the mapping given by (3.25) may be discontinuous. To illustrate this, let us consider an example.

*Example 3.15. *We define an
integrable function by

We also define a multivalued mapping by the formula

Note that for any , but at the same time, for any

Using Lemma 3.12, we obtain the following continuity conditions for the operator given by (3.25).

*Definition 3.16. *Let One says that a
mapping is symmetric on
the set if for any One says that a
mapping is continuous
in the second variable at a point belonging to
the diagonal of if for any
sequence such that as it holds that One says that a
mapping is continuous
in the second variable on the diagonal of if is continuous
in the second variable at each point of this diagonal. Continuity in the fist
variable is defined similarly.

*Definition 3.17. *Let Suppose also
that for any One says that a
mapping has property on the set if it is
continuous in the second variable on the diagonal of it has property on the set if it is
continuous in the first variable on the diagonal of it has property on the set if it is
continuous on the diagonal of and symmetric
on the set

Theorem 3.18. *
Let
Suppose also
that for a mapping there exists a
mapping such that ** for any and any
measurable set Then for the
mapping given by
(3.25), the inequality (3.30), where is satisfied as
well as for any and any
measurable set *

Corollary 3.19. *
If
the mapping in Theorem 3.18
has property (resp., , ) on the set
then the
operator given by (3.25)
is Hausdorff lower semicontinuous (resp., Hausdorff upper
semicontinuous, Hausdorff continuous) on the set *

We say that the mapping satisfying the
inequality (3.30) for any measurable set is a *majorant* mapping for on the set

Let a mapping , be measurable as a composite function for every Let also be integrally bounded for every bounded set Consider a mapping given by where the mapping , is the Nemytskii operator generated by the mapping , For the operator given by (3.31), the majorant mapping can be defined as

It follows from Theorem 3.18 that the operator given by (3.32) is also a majorant mapping for the mapping given by (3.25), where If the mapping , is Hausdorff lower semicontinuous (resp., Hausdorff upper semicontinuous and Hausdorff continuous) in the second variable, then by Corollary 3.19, the mapping given by (3.25) is Hausdorff lower semicontinuous (resp., Hausdorff upper semicontinuous and Hausdorff continuous).

*Definition 3.20. *One says that a
multivalued mapping *has Property * (resp., and ) if for this
mapping there exists a majorant mapping satisfying
Property (resp., and ).

#### 4. Basic Properties of Generalized Solutions of Functional Differential Inclusions

Using decomposable hulls, we introduce in this section the concept of a generalized solution of a functional differential inclusion with a right-hand side which is not necessarily decomposable. Using, as mentioned in Section 3, basic topological properties of a mapping given by (3.25), we study the properties of a generalized solution of the initial value problem.

Consider the initial value problem for the functional differential inclusion where the mapping satisfies the following condition: for every bounded set , the image is integrally bounded. Note that the right-hand side of the inclusion (4.1) is not necessarily decomposable. Note also that in (4.1) is not treated as a derivative at a point but as an element of (see [10, 23–25]). When we study such a problem, there may appear some difficulties described in the introduction. In this connection, we will introduce the concept of a generalized solution of the problem (4.1) and study the properties of this solution. Using the Nemytskii operator, which is decomposable, the initial value problem for a classical differential inclusion, that is, one without delay (see [10, 23–25]), can be reduced to (4.1).

*Definition 4.1. *An absolutely
continuous function is called a
generalized solution of the problem (4.1) if

Note that from Lemma 3.7, it follows that if the set (see(4.1)) is decomposable, then a generalized solution of the problem (4.1) coincides with a classical solution.

*Example 4.2. *Consider an ordinary differential equation, Its solution is the function

We assume that the parameter may take two
values: 1 or 2. Then the trajectories of such a system are described by the
differential inclusion where is a
multivalued function with the values from the set Note that that is, the
set in the right-hand side of the inclusion is decomposable. In this case, a
generalized solution of the inclusion coincides with a classical solution.

The latter differential inclusion describes the model
that is controlled by the differential equation either with the parameter value or with the
parameter value In this model,
switchings from one law (equation) to another may take place any time.

In the limit case, all possible solutions fill
entirely the set of all points between the graphs of the functions and

*Example 4.3. * Consider a
simple pendulum. It consists of a mass hanging from a
string of length and fixed at a
pivot point . When displaced to an initial angle and released, the
pendulum will swing back and forth with periodic motion. The equation of motion
for the pendulum is given by where is the
angular displacement at the moment , , is the
acceleration of gravity, and is the length
of the string.

If the amplitude of angular displacement is small
enough that the small angle approximation holds true, then the equation of
motion reduces to the equation of simple harmonic motion Let us now assume that the length of the string may change,
that is, it may take an value from a finite set In this case,
the equation of simple harmonic motion transforms to the differential inclusion
with a multivalued mapping where

We assume that switching from one length (equation) to
another may take place any time. Then the generalized solutions of the
inclusion treat all available trajectories (states) corresponding to all
switchings.

*Definition 4.4. *An operator is called a
Volterra-Tikhonov (or simply a Volterra) operator (see [26]) if the equality on , implies where is the set of
all functions from restricted to

In what follows, we assume that the operator (the right-hand side of the inclusion (4.1)) is a Volterra operator. This implies that the operator given by (3.25) is also a Volterra operator.

Let Let us determine the continuous mapping by

*Definition 4.5. *One says that an
absolutely continuous function is a
generalized solution of the problem (4.1) on the interval , if satisfies and where the
continuous mapping is given by (4.8).

A function which is absolutely continuous on any interval , is called a generalized solution of the problem (4.1) on the interval if for each the restriction of to is a generalized solution of the problem (4.1) on the interval

A generalized solution of the problem (4.1) on the interval is said to be nonextendable if there is no generalized solution of the problem (4.1) on any larger interval (here, if and if ) such that for each

In Theorems 4.6–4.12 below, we assume that the mapping has Property Due to Corollary 3.19, the mapping given by (3.25) is lower semicontinuous. Due to [27, 28], the mapping admits a continuous selection. Therefore, the following propositions on local solutions of the problem (4.1) are straightforward.

Theorem 4.6. *There exists
such that a
generalized solution of the problem (4.1) is defined on the interval . *

Theorem 4.7. *
A
generalized solution of the problem
(4.1) admits a continuation if and only if *

Theorem 4.8. *If
is a
generalized solution of the problem (4.1) on the interval ,
then there
exists a nonextendable solution
of the problem
(4.1) defined on the interval ,
or on the
entire interval such that for each *

Let be the set of all generalized solutions of the problem (4.1) on the interval

We say that generalized solutions of the problem (4.1) *admit a uniform a priori estimate* if there exists a number such that for
every , there is no generalized solution satisfying

Theorems 4.6–4.8 yield the following result.

Theorem 4.9. *Let
the generalized solutions of the problem (4.1) admit a uniform a priori
estimate. Then for any
and there
exists a number such that for any , *

*Definition 4.10. *One says that a
mapping has Property if there exists
an isotonic continuous operator satisfying the
following conditions:(i)for any function and any
measurable set , one has where the
continuous mapping is given by (ii)the local solutions of the problem admit a uniform
a priori estimate.

Lemma 4.11. *Suppose that a multivalued mapping has Property Then so does
the mapping given by
(3.25). *

*Proof. *It
suffices to show that for any
function and any
measurable set Indeed, let a
function be as in (3.1).
By (4.9), for each Hence, we have
that for the function , the estimate is satisfied as well. This gives the inequality (4.12).
The proof is complete.

Let a continuous operator be given by

Theorem 4.12. *Suppose that a mapping has Property Then the set is nonempty for
any
and there
exists a number such that for any , *

*Proof. *Indeed,
let (). From Lemma 4.11,
it follows that for any , where the function is given by
(4.15). Due to the theorem on integral inequalities for an isotonic operator
(see [29]), this implies that we actually have where is the upper
solution of the problem (4.11). Thus, there is no satisfying the
inequality From this, it
follows that the set of all local generalized solutions of the problem (4.1)
admits a uniform a priori estimate. Applying Theorem 4.9 completes the proof.

Let a linear continuous operator be given by We say that is the *operator of integration.*

Theorem 4.13. *
Let
the set of all local generalized solutions of the problem (4.1) admit a uniform
a priori estimate. Suppose also that has Property Then for any
function and any , there exists a generalized solution of the problem
(4.1) such that ** for any
measurable set **If then the
theorem is also valid for *

*Proof. *Let have Property Then by
Corollary 3.19, the mapping given by (3.25)
is continuous. Therefore (see [30–32]), given a number and a function , there exists a continuous mapping satisfying and for any and any
measurable set It follows from
Theorem 4.9 that for any , and that there exists a number such that for each , Now, we show
that there exists satisfying
(4.18). Consider the problem where the
continuous mapping is given by We denote by the set of all
solutions of the problem (4.20). Let us show that It follows from
the definition of the mapping (see (4.21))
that Let us prove
that Assume the
converse. Then there exists such that Since we have This implies
that there exists a number such that ( is the
restriction of the function to ). By (4.21),
we have This contradicts
to the definition of the number Hence, Consider a
continuous operator given by where the
operator is the operator
of integration defined by (4.17), and is a continuous
selection of the mapping given by
(3.25). The function ia also assumed
to satisfy (4.19). Since the operator is bounded, we
obtain that the image is a relatively
compact subset of Hence, the set is a convex
compact set. Since the operator given by (4.22)
takes the set into itself, we
have, by Schauder theorem, that the mapping has a fixed
point. This fixed point is the solution
of the problem (4.20). It follows from the above equality that this
solution is a
generalized solution of the problem (4.1). Since we see that
(4.19) implies (4.18).

Let us prove the second statement of the theorem. Let Suppose also that has Property Then by Lemma 3.9, Hence for each , there exists a generalized solution of the problem (4.1) such that for any measurable set , the inequality (4.18) is valid for and Since the set is bounded, we see that the sequence is weakly compact in Without loss of generality, it can be assumed that weakly in and in as Let us show that is a generalized solution of the problem (4.1). In other words, we have to prove that Assume that the functions , satisfy (as these functions do exist). It follows from (4.23) that Since the mapping given by (3.25) is continuous, we obtain, by (4.24), that in as Since weakly in as we have that weakly in as Therefore, the convexity of the set implies that (see [21]). Thus, is a generalized solution of the problem (4.1).

Further, let us show that (4.19) holds for the solution and for Since weakly in as we have, by [21], that for each , there exist numbers , , satisfying the following conditions: the sequence tends to in Since for each it follows, due to the choice of the sequence that for each .

Since it follows that letting in the previous inequality, we obtain Finally, note that by the decomposability of the set this equality holds for any measurable set This completes the proof.

Theorems 4.12 and 4.13 yield the following result.

Corollary 4.14. *Suppose that a mapping has
Properties and Then for any function and any , there exists a generalized solution of the problem
(4.1) such that (4.18) holds for any measurable set **If then the
corollary is also valid for *

*Remark 4.15. *
Consider the convex compact set where the
mapping is given by Here, the
operators and are determined
by (3.25) and (4.21), respectively. If a number is such that for any , then due to the
the coincidence of the sets and (see the proof
of Theorem 4.13),

*Definition 4.16. *Given , , , one says that a mapping has Property if there exists
an isotonic and continuous Volterra operator satisfying the
following conditions:(i)(ii)for any functions and any
measurable set , one has where the
continuous mapping is determined
by (4.10);(iii)the set of all local solutions of the problem admits a
uniform a priori estimate.

Given and , the following estimate will be used in the sequel: for each measurable set

Theorem 4.17. *Let
functions and satisfy the
inequality (4.32) for each measurable set
Suppose that a
mapping has Property where , , and is the initial
condition of the problem (4.1). Then for any generalized solution of the
problem (4.1) satisfying ** for any
measurable set , the following conditions are satisfied:
*(1)* for each where the
function is the upper
solution of the problem (4.31) for and and the mapping is given by
(4.15);*(2)* for almost all *