For a map , the difference equation
is always well defined whatever the initial points are, even though the are subsets of , rather than points.
A point is said to be an equilibrium for the map h if . The equilibrium μ is said to be stable if, for each neighborhood V of μ in , there is a neighborhood W of in such that for all n, whenever .
The equilibrium μ is said to be an attractor in a neighborhood V of μ in , if for all n and in , whenever for .
Definition 1 The point μ is said to be an unconditional equilibrium of h (respectively, unconditional stable equilibrium, unconditional attractor in V) if it is an equilibrium (respectively, stable equilibrium, attractor in V) of for all .
Definition 2 We define the equilibria, stable equilibria, attractors, unconditional equilibria, unconditional stable equilibria and unconditional attractors of a continuous function to be those of .
3.1 Sufficient condition for unconditional convergence
After giving Definitions 1 and 2 we are going to prove a result guaranteeing that a general second order difference equation as in (4) has an unconditional stable attractor.
Let and consider in the sequel a continuous function , satisfying the following conditions:
(H1) , whenever and .
(H2) There exists such that
The functions and are defined in the obvious way. Notice that is the limit of a monotone increasing sequence of continuous functions, thus it is lower-semicontinuous, likewise is an upper-semicontinuous function. Remember that we denote both φ and by φ.
The next lemma, which we prove at the end of this section, shows that if (H1) and (H2) holds we can get some information about the behavior and properties of φ and .
Lemma 2 Let , where , be a continuous function satisfying (H1) and (H2). Then the function in (H2) is unique and it is a continuous nonincreasing map, thus it has a unique fixed point . Furthermore,
for all and for all .
If , and , then .
for all .
is decreasing in .
We are in conditions of presenting and proving our main result.
Theorem 1 Let , where , be a continuous function satisfying (H1) and (H2). If and , then is an unconditional stable attractor of φ in .
Proof of Theorem 1 Consider and denote
for some . Notice that for all n, as a consequence of (i) in Lemma 2.
We are going to prove first that is a stable equilibrium of . By (iii) in Lemma 2, as
we see that is an equilibrium.
Let . Because of the continuity of , there is such that
As and , we have
If , then
By replacing a, b by , in Lemma 2(i), we see that
whenever for , thus is an unconditional stable equilibrium of φ.
Now, if we see that
we are done with the whole proof. Indeed, for each accumulation point of , one would have
because of the continuity of . As , this implies .
Therefore, as a consequence of Lemma 1, it suffices to find an increasing sequence of natural numbers such that
Here, and are defined as in Lemma 1, with and ,
Let , so that
Having in mind that satisfies (3), we find from as follows. Denote
and momentarily assume and in such a way that
for all .
As a consequence, the nonincreasing sequence is bounded below by . It cannot be the case that , because in such a case there is a subsequence converging to and such that converges to a point .
By applying (H2), we see that
Therefore, and there exists such that
Analogously, we see that there exists such that
Proof of Lemma 2
(i): It suffices to prove the first assertion, because is a closed set and, by definition,
for all . Assume now that . We consider the following three possible situations. If , it is obvious that .On the other hand, if and , then
Finally, if and , then
(ii): Suppose . Since , then or . In any event, it cannot be the case that
which contradicts hypothesis (H2), thus .
(iii): Since , it is worth considering the following three cases for each : first, , and then (after probing continuity, monotonicity and statement (iv)), we proceed with the case , and finally with , .Case and : Since when and when , we see that
because of the continuity of .
If , then , or , thus
then there exist two sequences with
and, by (ii), one has . Since , this would imply for all , which is impossible.
and because of the continuity of , we have
Analogously, it can be seen that .
Suppose for all n. Then
which implies , eventually for all n.Since , we reach a contradiction. Therefore,
Analogously, we see that when . □