- Research Article
- Open Access

# Homoclinic Solutions of Singular Nonautonomous Second-Order Differential Equations

- Irena Rachůnková
^{1}Email author and - Jan Tomeček
^{1}

**2009**:959636

https://doi.org/10.1155/2009/959636

© I. Rachůnková and J. Tomeček. 2009

**Received:**27 April 2009**Accepted:**15 September 2009**Published:**13 October 2009

## Abstract

This paper investigates the singular differential equation , having a singularity at . The existence of a strictly increasing solution (a homoclinic solution) satisfying , is proved provided that has two zeros and a linear behaviour near .

## Keywords

- Differential Equation
- Functional Equation
- Local Maximum
- Point Theorem
- Lipschitz Constant

## 1. Introduction

Having a positive parameter we consider the problem

under the following basic assumptions for and

Then problem (1.1), (1.2) generalizes some models arising in hydrodynamics or in the nonlinear field theory (see [1–5]). However (1.1) is singular at because .

Definition 1.1.

If
, then*a solution* of (1.1) on
is a function
satisfying (1.1) on
. If
is a solution of (1.1) on
for each
, then
is a solution of (1.1) on
.

Definition 1.2.

Let
be a solution of (1.1) on
. If
moreover fulfils conditions (1.2), it is called *a solution of problem*(1.1), (1.2).

Clearly, the constant function is a solution of problem (1.1), (1.2). An important question is the existence of a strictly increasing solution of (1.1), (1.2) because if such a solution exists, many important physical properties of corresponding models can be obtained. Note that if we extend the function in (1.1) from the half–line onto (as an even function), then any solution of (1.1), (1.2) has the same limit as and . Therefore we will use the following definition.

Definition 1.3.

A strictly increasing solution of problem (1.1), (1.2) is called a *homoclinic solution.*

Numerical investigation of problem (1.1), (1.2), where and , , can be found in [1, 4–6]. Problem (1.1), (1.2) can be also transformed onto a problem about the existence of a positive solution on the half-line. For , and for , , such transformed problem was solved by variational methods in [7, 8], respectively. Some additional assumptions imposed on were needed there. Related problems were solved, for example, in [9, 10].

Here, we deal directly with problem (1.1), (1.2) and continue our earlier considerations of papers [11, 12], where we looked for additional conditions which together with (1.3)–(1.8) would guarantee the existence of a homoclinic solution.

Let us characterize some results reached in [11, 12] in more details. Both these papers assume (1.3)–(1.8). In [11] we study the case that has at least three zeros . More precisely, the conditions,

are moreover assumed. Then there exist , and a solution of (1.1) on such that

We call such solution *an escape solution*. The main result of [11] is that (under (1.3)–(1.8), (1.9)) the set of solutions of (1.1), (1.10) for
consists of escape solutions and of oscillatory solutions (having values in
) and of at least one homoclinic solution. In [12] we omit assumptions (1.9) and prove that assumptions (1.3)–(1.8) are sufficient for the existence of an escape solution and also for the existence of a homoclinic solution provided the
fulfils

If (1.12) is not valid, then the existence of both an escape solution and a homoclinic solution is proved in [12], provided that satisfies moreover

Assumption (1.13) characterizes the case that
has just two zeros
and
in the interval
. Further, we see that if (1.14) holds, then
is either bounded on
or
is unbounded earlier and has *a sublinear* behaviour near
.

This paper also deals with the case that
satisfies (1.13) and is unbounded above on
. In contrast to [12], here we prove the existence of a homoclinic solution for
having *a linear* behaviour near
. The proof is based on a full description of the set of all solutions of problem (1.1), (1.10) for
and on the existence of an escape solutions in this set.

Finally, we want to mention the paper [13], where the problem

is investigated under the assumptions that is continuous, it has three distinct zeros and satisfies the sign conditions similar to those in [11, (3.4)]. In [13], an approach quite different from [11, 12] is used. In particular, by means of properties of the associated vector field together with the Kneser's property of the cross sections of the solutions' funnel, the authors provide conditions which guarantee the existence of a strictly increasing solution of (1.15). The authors apply this general result to problem

and get a strictly increasing solution of (1.16) for a sufficiently small . This corresponds to the results of [11], where may be arbitrary.

## 2. Initial Value Problem

In this section, under the assumptions (1.3)–(1.8) and (1.13) we prove some basic properties of solutions of the initial value problem (1.1), (1.10), where .

Lemma 2.1.

Proof.

Let for some . Then (2.4) yields , which is not possible, because is decreasing on . Therefore for .

Let . Consider the Banach space (with the maximum norm) and an operator defined by

for . Hence, can be extended onto each interval where is bounded. So, we can put .

Let . Then there exists such that for . So, (2.6) yields

Remark 2.2.

The proof of Lemma 2.1 yields that if , then .

Let us put

Similarly as in the proof of Lemma 2.1 we deduce that problem (2.13), (1.10) has a unique solution on . Moreover the following lemma is true.

Lemma 2.3 ([12]).

Here is a solution of problem (2.13), (1.10) with , .

Proof.

we get (2.14).

Remark 2.4.

Arguing as in the proof of Lemma 2.1, we get that problem (2.13), (2.19) has a unique solution on . In particular, for and , the unique solution of problem (2.13), (2.19) (and also of problem (1.1), (2.19)) is and , respectively.

Lemma 2.5.

Proof.

By (1.3), (1.8), and (2.22), exists and, since is bounded on , we get . Hence, letting in (2.22), we obtain . Therefore, and (2.21) is proved.

Lemma 2.6.

Then for all and (2.21) holds.

Proof.

By similar argument as in the proof of Lemma 2.5 we get that and . Therefore (2.21) is proved.

## 3. Damped Solutions

In this section, under assumptions (1.3)–(1.8) and (1.13) we describe a set of all damped solutions which are defined in the following way.

Definition 3.1.

Remark 3.2.

We see, by (2.12), that is a damped solution of problem (1.1), (1.10) if and only if is a damped solution of problem (2.13), (1.10). Therefore, we can borrow the arguments of [12] in the proofs of this section.

Theorem 3.3.

If is a damped solution of problem (1.1), (1.10), then has a finite number of isolated zeros and satisfies (2.21); or is oscillatory (it has an unbounded set of isolated zeros).

Proof.

Step 1.

If has no zero in , then for and, by Lemma 2.5, fulfils (2.21).

Step 2.

Step 3.

Letting in (2.22), we get , which contradicts the fact that is bounded above. Therefore, cannot be negative on the whole interval and there exists such that . Moreover, according to (3.2), .

Then, Lemma 2.5 yields that fulfils (2.21). Since is positive on , has just one minimum on . Moreover, putting and in (2.23), we have

Step 4.

If has no other zeros, we deduce as in Step 3 that has just one negative minimum in , and fulfils (2.21).

Step 5.

If has other zeros, we use the previous arguments and get that either has a finite number of zeros and then fulfils (2.21) or is oscillatory.

Remark 3.4.

According to the proof of Theorem 3.3, we see that if is oscillatory, it has just one positive local maximum between the first and the second zero, then just one negative local minimum between the second and the third zero, and so on. By (3.8), (3.9), (1.4)–(1.6) and (1.13), these maxima are decreasing (minima are increasing) for increasing.

Lemma 3.5.

Proof.

Due to (1.4), we see that is strictly decreasing for as long as . Thus, there are two possibilities. If for all , then from Lemma 2.6 we get (2.21), which contradicts (3.10). If there exists such that , then in view Remark 2.4 we have . Using the arguments of Steps 3–5 of the proof of Theorem 3.3, we get that is damped, contrary to (3.10). Therefore, such cannot exist and on . Consequently, . So, fulfils (3.11). The inverse implication is evident.

Remark 3.6.

According to Definition 1.3 and Lemma 3.5, is a homoclinic solution of problem (1.1), (1.10) if and only if is a homoclinic solution of problem (2.13), (1.10).

Theorem 3.7 (on damped solutions).

Let satisfy (1.5) and (1.6). Assume that is a solution of problem (1.1), (1.10) with . Then is damped.

Proof.

Therefore, on , and letting , we get . This together with (3.16) contradicts (3.13). We have proved that is damped.

Theorem 3.8.

Let be the set of all such that corresponding solutions of problem (1.1), (1.10) are damped. Then is open in .

Proof.

- (a)
Let be oscillatory. Then its first local maximum belongs to . Lemma 2.3 guarantees that if is sufficiently close to , the corresponding solution of (2.13), (1.10) has also its first local maximum in . This means that there exist and such that satisfies (2.26). Now, we can continue as in the proof of Theorem 3.3 using the arguments of Steps 2–5 and Remark 3.2 to get that is damped.

- (b)

Consequently, integrating (2.13) over and using (3.19)–(3.23), we get for

Assume that there is such that , . Then, since if and , we get and for , contrary to (3.25). Hence we get that fulfils (3.1).

## 4. Escape Solutions

During the whole section, we assume (1.3)–(1.8) and (1.13). We prove that problem (1.1), (1.10) has at least one escape solution. According to Section 1 and Remark 2.2, we work with the following definitions.

Definition 4.1.

Definition 4.2.

Remark 4.3.

If is an escape solution of problem (2.13), (1.10), then is an escape solution of problem (1.1), (1.10) on some interval .

Theorem 4.4 (on three types of solutions.).

Let be a solution of problem (1.1), (1.10). Then is just one of the following three types

(I) is damped;

(II) is homoclinic;

(III) is escape.

Proof.

By Definition 3.1, is damped if and only if (3.1) holds. By Lemma 3.5 and Definition 1.3, is homoclinic if and only if (3.10) holds. Let be neither damped nor homoclinic. Then there exists such that is bounded on , , . So, has its first zero and on . Assume that there exist such that and . Then, by Lemma 2.6, either fulfils (2.21) or has its second zero and, arguing as in Steps 2–5 of the proof of Theorem 3.3, we deduce that is a damped solution. This contradiction implies that on . Therefore, by Definition 4.1, is an escape solution.

Theorem 4.5.

Let be the set of all such that the corresponding solutions of (1.1), (1.10) are escape solutions. The set is open in .

Proof.

Let and be a solution of problem (1.1), (1.10) with . So, fulfils (4.1) for some . Let be a solution of problem (2.13), (1.10) with . Then on and is increasing on . There exists and such that . Let be a solution of problem (2.13), (1.10) for some . Lemma 2.3 yields such that if , then . Therefore, is an escape solution of problem (2.13), (1.10). By Remark 4.3, is also an escape solution of problem (1.1), (1.10) on some interval .

To prove that the set of Theorem 4.5 is nonempty we will need the following two lemmas.

Lemma 4.6.

Proof.

Step 1.

So, is an increasing solution of problem (1.1), (1.10) on and for . Therefore the nonempty interval exists.

Step 2.

This relation together with (4.6) implies (4.3).

Remark 4.7.

Consider a solution of Lemma 4.6. If is an escape solution, then . Assume that is not an escape solution. Then both possibilities and can occur. Let . By Theorem 4.4 and Lemma 2.5, , . Let . We write , . Using Lemmas 3.5 and 2.5 and Theorem 4.4, we obtain and either or .

Lemma 4.8.

Let and let . Then for each

(i)there exists a solution of problem (1.1), (1.10) with ,

(ii)there exists such that is the maximal interval on which the solution is increasing and its values in this interval are contained in ,

(iii)there exists satisfying .

If the sequence is unbounded, then there exists such that is an escape solution.

Proof.

which contradicts (4.20). Therefore, at least one escape solution of (1.1), (1.10) with must exist.

Theorem 4.9 (on escape solution).

Then there exists such that the corresponding solution of problem (1.1), (1.10) is an escape solution.

Proof.

By (4.24) we can find such that for . We assume that for any , is not an escape solution and we construct a contradiction.

Step 1.

Step 2.

So, by virtue of (4.37), there exists such that for .

Step 3.

In order to get a contradiction, we distinguish two cases.

Case 1.

Therefore , contrary to (4.25).

Case 2.

which yields, by (4.26), and also , contrary to (4.25). These contradictions obtained in both cases imply that there exists such that is an escape solution.

## 5. Homoclinic Solution

The following theorem provides the existence of a homoclinic solution under the assumption that the function in (1.1) has a linear behaviour near . According to Definition 1.2, a homoclinic solution is a strictly increasing solution of problem (1.1), (1.2).

Theorem 5.1 (On homoclinic solution.).

Let the assumptions of Theorem 4.9 be satisfied. Then there exists such that the corresponding solution of problem (1.1), (1.10) is a homoclinic solution.

Proof.

For denote by the corresponding solution of problem (1.1), (1.10). Let and be the set of all such that is a damped solution and an escape solution, respectively. By Theorems 3.7, 3.8, 4.5, and 4.9, the sets and are nonempty and open in . Therefore, the set is nonempty. Choose . Then, by Theorem 4.4, is a homoclinic solution. Moreover, due to Theorem 3.7, .

Example 5.2.

where is a negative constant, satisfies the conditions (1.3)–(1.6), (1.13), and (4.23).

## Declarations

### Acknowledgments

The authors thank the referee for valuable comments. This work was supported by the Council of Czech Government MSM 6198959214.

## Authors’ Affiliations

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