Open Access

Strictly Increasing Solutions of Nonautonomous Difference Equations Arising in Hydrodynamics

Advances in Difference Equations20102010:714891

DOI: 10.1155/2010/714891

Received: 19 December 2009

Accepted: 10 March 2010

Published: 16 March 2010

Abstract

The paper provides conditions sufficient for the existence of strictly increasing solutions of the second-order nonautonomous difference equation , , where is a parameter and is Lipschitz continuous and has three real zeros . In particular we prove that for each sufficiently small there exists a solution such that is increasing, and . The problem is motivated by some models arising in hydrodynamics.

1. Formulation of Problem

We will investigate the following second-order non-autonomous difference equation

(1.1)

where is supposed to fulfil

(1.2)
(1.3)
(1.4)

Let us note that means that for each there exists such that for all . A simple example of a function satisfying (1.2)–(1.4) is , where is a positive constant.

A sequence which satisfies (1.1) is called a solution of (1.1). For each values there exists a unique solution of (1.1) satisfying the initial conditions

(1.5)

Then is called a solution of problem (1.1), (1.5).

In [1] we have shown that (1.1) is a discretization of differential equations which generalize some models arising in hydrodynamics or in the nonlinear field theory; see [26]. Increasing solutions of (1.1), (1.5) with has a fundamental role in these models. Therefore, in [1], we have described the set of all solutions of problem (1.1), (1.6), where

(1.6)

In this paper, using [1], we will prove that for each sufficiently small there exists at least one such that the corresponding solution of problem (1.1), (1.6) fulfils

(1.7)

Note that an autonomous case of (1.1) was studied in [7]. We would like to point out that recently there has been a huge interest in studying the existence of monotonous and nontrivial solutions of nonlinear difference equations. For papers during last three years see, for example, [822]. A lot of other interesting references can be found therein.

2. Four Types of Solutions

Here we present some results of [1] which we need in next sections. In particular, we will use the following definitions and lemmas.

Definition 2.1.

Let be a solution of problem (1.1), (1.6) such that
(2.1)

Then is called a damped solution.

Definition 2.2.

Let be a solution of problem (1.1), (1.6) which fulfils
(2.2)

Then is called a homoclinic solution.

Definition 2.3.

Let be a solution of problem (1.1), (1.6). Assume that there exists , such that is increasing and
(2.3)

Then is called an escape solution.

Definition 2.4.

Let be a solution of problem (1.1), (1.6). Assume that there exists , , such that is increasing and
(2.4)

Then is called a non-monotonous solution.

Lemma 2.5 (see [1] (on four types of solutions)).

Let be a solution of problem (1.1), (1.6). Then is just one of the following four types:
  1. (I)

    is an escape solution;

     
  2. (II)

    is a homoclinic solution;

     
  3. (III)

    is a damped solution;

     
  4. (IV)

    is a non-monotonous solution.

     

Lemma 2.6 (see [1] (estimates of solutions)).

Let be a solution of problem (1.1), (1.6). Then there exists a maximal satisfying
(2.5)
Further, if , then moreover
(2.6)
(2.7)
for if , and for if , where
(2.8)

In [1] we have proved that the set consisting of damped and non-monotonous solutions of problem (1.1), (1.6) is nonempty for each sufficiently small . This is contained in the next lemma.

Lemma 2.7 (see [1] (on the existence of non-monotonous or damped solutions)).

Let , where is defined by (1.4). There exists such that if , then the corresponding solution of problem (1.1), (1.6) is non-monotonous or damped.

In Section 4 of this paper we prove that also the set of escape solutions of problem (1.1), (1.6) is nonempty for each sufficiently small . Note that in our next paper [23] we prove this assertion for the set of homoclinic solutions.

3. Properties of Solutions

Now, we provide other properties of solutions important in the investigation of escape solutions.

Lemma 3.1.

Let be an escape solution of problem (1.1), (1.6). Then is increasing.

Proof.

Due to (1.1), fulfils
(3.1)
According to Definition 2.3 there exists , such that is increasing and (2.3) holds. By (1.3) we get . Consequently, by (3.1) and (2.3), and . Similarly and
(3.2)

This yields that is increasing.

Lemma 3.2.

Assume that for . Choose an arbitrary . Let and let and be a solution of problem (1.1), (1.6) with and , respectively. Let be the Lipschitz constant for on . Then
(3.3)
(3.4)

where , .

Proof.

By (3.1) we have
(3.5)
Summing it for , we get by (1.6)
(3.6)
Summing it again for , we get
(3.7)
and similarly
(3.8)
From this and by using summation by parts we easily obtain
(3.9)
By the discrete analogue of the Gronwall-Bellman inequality (see, e.g., [24, Lemma ]), we get
(3.10)

which yields (3.3).

By (3.6) and (3.3) we have for , ,
(3.11)

4. Existence of Escape Solutions

Lemma 4.1.

Assume that and . Let be a solution of problem (1.1), (1.6) with , . For choose a maximal such that for if is finite, and for if , and is increasing if . Then there exists such that for any there exists a unique , , such that
(4.1)

Moreover, if the sequence is unbounded, then there exists such that the solution of problem (1.1), (1.6) with is an escape solution.

Proof.

Choose such that
(4.2)

For denote by a solution of problem (1.1), (1.6) with . The existence of is guaranteed by Lemma 2.6. By Lemma 2.5, is just one of the types (I)–(IV), and if , then the monotonicity of yields a unique , , satisfying (4.1).

For , consider the sequence and assume that it is unbounded. Then we have
(4.3)
(otherwise we take a subsequence.) Assume on the contrary that for any , is not an escape solution. Choose . If is damped, then by Definition 2.1, we have and
(4.4)
If is homoclinic, then by Definition 2.2, we have and
(4.5)
If is non-monotonous, then by Definition 2.4, we have and
(4.6)
To summarize if is not an escape solution, then by (4.4), (4.5), and (4.6), we have
(4.7)
Since , there exists satisfying
(4.8)
Consider (3.5) with . By dividing it by , multiplying such obtained equality by and summing in from 1 to we get
(4.9)
Denote
(4.10)
Then we get
(4.11)
Let us put and to (4.11) and subtract. By (4.7) and (4.8) we get
(4.12)
Let us put and to (4.10) and subtract. We get
(4.13)
Choose and such that
(4.14)
Let . Then (4.6) holds. Since , and , (3.1) yields
(4.15)
and hence
(4.16)
Clearly, if , then by (4.4) and (4.5), inequality (4.16) holds, as well. By (1.2), is integrable on . So, having in mind (4.1), we can find such that if
(4.17)
then
(4.18)
Therefore, due to (1.3) and (4.7),
(4.19)
Let be such that
(4.20)

If , then (2.7) implies (4.17) and hence (4.19) holds.

Now, let us put and choose . Then, (4.2), (4.14), (4.20), and (4.13)–(4.19) yield
(4.21)
Finally, (4.12) and (4.21) imply
(4.22)

Letting , we obtain, by (4.3), that , contrary to (4.17). Therefore an escape solution of problem (1.1), (1.6) with must exist.

Now, we are in a position to prove the next main result.

Theorem 4.2 (On the existence of escape solutions).

There exists such that for any the initial value problem (1.1), (1.6) has an escape solution for some .

Proof.

We have the following steps.

Step 1.

Let us define
(4.23)
and consider an auxiliary equation
(4.24)
Let be the constant of Lemma 4.1 for problem (4.24), (1.6). Choose , and let be the Lipschitz constant for on . Consider a sequence such that . Then, for each there exists such that
(4.25)
Let for . Then the sequence is the unique solution of problem (4.24), (1.6) with . Let be a solution of problem (4.24), (1.6) with , , and let be the sequence corresponding to by Lemma 4.1. We prove that is unbounded. According to Lemma 3.2, for each ,
(4.26)
Consequently, (4.25) and (4.26) give
(4.27)
and hence
(4.28)
Therefore
(4.29)

which yields that is unbounded. By Lemma 4.1, the auxiliary initial value problem (4.24), (1.6) has an escape solution for some . Denote this solution by .

Step 2.

By Definition 2.3, there exists such that
(4.30)

Now, consider the solution of our original problem (1.1), (1.6) with . Due to (4.23), for . Using (4.30) and Definition 2.3, we get that is an escape solution of problem (1.1), (1.6).

Declarations

Acknowledgments

The paper was supported by the Council of Czech Government MSM 6198959214. The authors thank the referees for valuable comments.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Palacký University

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Copyright

© L. Rachůnek and I. Rachůnková. 2010

This article is published under license to BioMed Central Ltd. 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.