The following theorem is proved by utilizing the idea of a retract method, which is well known for ordinary differential equations and goes back to Ważewski [13]. In the next theorem, we assume that the function f, except for the indicated conditions, satisfies all the assumptions given in Section 2. Namely, we assume that the function f is bounded and Lipschitz continuous on an open set S and .
Theorem 2 Let . Let , be Δ-differentiable functions on such that for each . If, moreover, every point is the point of strict egress for the set Ω with respect to system (1), then there exists an rd-continuous initial function satisfying
(6)
such that the initial problem
(7)
defines a solution y of (1) on the interval satisfying
(8)
Proof The idea of the proof is the following. By contrary, we assume that a solution y satisfying (8) does not exist. Then we are able to prove (by a construction of a chain of auxiliary mappings) an existence of a retraction of an n-dimensional ball into its boundary. However, it is well known that the boundary of an n-dimensional ball cannot be its retract (see, e.g., [12]) and we get a contradiction.
Without any special comment, throughout the proof, we use the fact that the initial value problem has a unique solution and this solution depends continuously on the initial data (this is guaranteed by Theorem 1). Suppose now that the initial function satisfying (6) generates the solution which does not satisfy (8) for at least one . This means, in general, that for any rd-continuous initial function satisfying the inequality
(9)
there exists , such that, for a corresponding solution of the initial problem
we have
and
Let us define auxiliary mappings , and . Note that in this part of the proof, without loss of generality, we admit the eventuality that the function instead of (9) satisfies weaker restrictions
and .
First, define a mapping . For with ,
with having the value defined above except for the case . In the latter case, we put .
Second, we define a mapping for every . For this we will need a set
with if and
if and , are such that , and . It is clear that . Further,
is obviously convex. We define, moreover, the y-part of the boundary of as
where if and
if , where and are as above.
Now we are ready to consider an auxiliary mapping . Let . Then, due to the convexity of , there exists a unique intersection of the segment connecting the points and with . We denote this point as and define
Note that if , we get the particular case and .
Third, we define a mapping . For this we need a subset of the y-boundary . For , let
Now we are ready to consider an auxiliary continuous mapping ,
defined for as a point , where with
(10)
It is easy to see that the is a retract of and satisfies all the assumptions from Definition 3 (with and ).
We show that the composite mapping
is continuous due to the continuous dependence of the solutions on the initial data, the convexity of sets of the type defined above and the continuity of the mapping . Let be sufficiently small and be the initial functions such that
and
(11)
Further, let
(12)
(13)
Suppose now that (which is equivalent to ). We consider all the possible settings of and characters of .
(I)Point .
First, assume that the point is dense, i.e., . Then, due to the solutions depending continuously on initial data, we have and, consequently,
In this case the mapping is continuous and the composite mapping is continuous as well because
and
Further, is also continuous because of the continuity of mappings and .
Second, if the point is left-scattered and right-dense, i.e., , we can proceed analogously as in the case before. In this case, for fixed δ, either (if ) or (if ). In the alternative , it is obvious that
and thus
Hence the composite mapping is continuous. Further, is also continuous because of the continuity of mappings and . The alternative with can be proved by the same limit process as in the first case, where the point is dense.
Third, let the point be left-dense and right-scattered, i.e., . Then the approach used in previous cases can be modified as follows. Let, as before, (11), (12) and (13) hold. Then, for fixed δ, either (if ) or (if ). The alternative with can be proved by the same limit process as in the first case, where the point is dense. However, the alternative takes into account a possibility that the mapping cannot be continuous. If is valid for , then, due to the convexity of , there exists a unique intersection of the segment connecting the points and with . We denote this point as and, in accordance with the above definition, . We wish to show that if . However, in view of the definition of the mapping and its geometric meaning,
and
Hence the continuity of is proved. Further, is also continuous (for the same reason as before).
Fourth, suppose now that is an isolated point, i.e., . Let, as before, (11), (12) and (13) hold. Then, for sufficiently small δ, we have either (if ) or (if ). Without any special comment, in the alternative , we proceed in the same way as in the second case, where is left-scattered and right-dense. Furthermore, in the alternative , we proceed in the same way as in the third case, where is left-dense and right-scattered.
(II)Point . Then the point is left-scattered and only two cases are possible (either is left-scattered and right-dense or is isolated). In both mentioned cases, we can proceed in the same way. Let, as before, and (11), (12) and (13) hold. Then, for sufficiently small δ, we have only and, of course, . Further,
Due to the convexity of , there exists a unique intersection of the segment connecting the points and with . We denote this point by . In view of the definition of the mapping and its geometric meaning, we can observe that
Hence the composite mapping is continuous. Moreover, is also continuous because of the continuity of mappings and .
We proved that the composite mapping is continuous. Note that we omitted the special case (which is equivalent to ). However, this part can be shown in an analogous and simpler way to the one used above.
Now we are able to finish the proof. We proved that
where , is continuous. Moreover,
is an identity mapping. In this situation, we have proved that there exists a retraction of the set B onto the set A (see Definition 3). In view of the above-mentioned fact, this is impossible. Our assumption is false and there exists initial problem (7) such that the corresponding solution satisfies (8) for every . The theorem is proved. □