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# Superlinear Singular Problems on the Half Line

*Boundary Value Problems*
**volume 2010**, Article number: 429813 (2010)

## Abstract

The paper studies the singular differential equation , which has a singularity at . Here the existence of strictly increasing solutions satisfying is proved under the assumption that has two zeros 0 and and a superlinear behaviour near . The problem generalizes some models arising in hydrodynamics or in the nonlinear field theory.

## 1. Introduction

Let us consider the problem

where is a positive real parameter.

Definition 1.1.

Let . A function satisfying (1.1) on is called *a solution of ( 1.1 ) on*.

Definition 1.2.

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

Definition 1.3.

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

In this paper we are interested in the existence of strictly increasing solutions and, in particular, of homoclinic solutions. In what follows we assume

Under assumptions (1.3)–(1.8) problem (1.1), (1.2) generalizes some models arising in hydrodynamics or in the nonlinear field theory (see [1–5]). If a homoclinic solution exists, many important 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 . This is a motivation for Definition 1.3. Equation (1.1) is singular at because . In [6, 7] we have proved that assumptions (1.3)–(1.8) are sufficient for the existence of strictly increasing solutions and homoclinic solutions provided

Here we assume that (1.9) is not valid. Then

and the papers [6, 8] provide existence theorems for problem (1.1), (1.2) if has a sublinear or linear behaviour near . The case that has a superlinear behaviour near is studied in this paper. To this aim we consider the initial conditions

where , and introduce the following definition.

Definition 1.4.

Let and let be a solution of (1.1) on satisfying (1.11). Then is called *a solution of problem ( 1.1 ), ( 1.11 ) on*. If moreover fulfils

then is called *an escape solution of problem ( 1.1 ), ( 1.11 )*.

We have proved in [6, 8] that for sublinear or linear the existence of a homoclinic solution follows from the existence of an escape solution of problem (1.1), (1.11). Therefore our first task here is to prove that at least one escape solution of (1.1), (1.11) exists, provided (1.3)–(1.8), (1.10), and

hold, and has a superlinear behaviour near . This is done in Section 2. Using the results of Section 2 "Theorem 2.10", and of [6, Theroms 13, 14 and 20] we get the existence of a homoclinic solution in Section 3.

Note that by Definitions 1.3 and 1.4 just the values of a solution which are less than are important for a decision whether the solution is homoclinic or escape one. Therefore condition (1.13) can be assumed without any loss of generality.

Close problems about the existence of positive solutions have been studied in [9–11].

## 2. Escape Solutions

In this section we assume that (1.3)–(1.8), (1.10), and (1.13) hold. We will need some lemmas.

Lemma 2.1 (see [6, Lemma 3]).

For each , problem (1.1), (1.11) has a unique solution on such that

In what follows by a solution of (1.1), (1.11) we mean a solution on .

Remark 2.2 (see [6, Remark 4]).

Choose and , and consider the initial conditions

Problem (1.1), (2.2) has a unique solution on . In particular, for and , we get and , respectively. Clearly, for , and are solutions of (1.1) on the whole interval .

Lemma 2.3.

Let and let be a solution of problem (1.1), (1.11) which is not an escape solution. Let us denote

Then holds and is increasing on . If , then and

Proof.

The inequality yields . By (1.1) and (1.10), we get on and hence is increasing on . As , one has on and consequently on . Therefore .

Let . Then is the first zero of and . Remark 2.2 yields that is not possible. This implies that . As is strictly increasing on and is not an escape solution, we have on . Thus on and hence is decreasing on . This gives (2.4).

Lemma 2.4.

Let and let be a solution of problem (1.1), (1.11) which is not an escape solution. Assume that is given by Lemma 2.3. Then

Proof.

From (1.1), we have

and, by multiplication and integration over ,

?(1) Assume that . The definition of yields on . Since is not an escape solution, it is bounded above and there exists

Therefore the following integral is bounded and, since it is increasing, it has a limit

So, by (2.7), exists. By virtue of (2.8), we get

If , then by (1.4), (1.10) and (2.6) we get , which contradicts (2.10). Hence, . In particular, if is defined as in Lemma 2.3, then

? (2) Assume that . Then the continuity of gives and of Lemma 2.3 fulfils . We deduce that on as in the proof of Lemma 2.3. Remark 2.2 yields that if , then neither nor can occur. Therefore .

Denote

Lemma 2.5.

Let and let be a solution of problem (1.1), (1.11). Further assume maximal such that and for . Then

For , let us denote

Then

Proof.

For equality (2.13) see Lemma 4.6 in [8]. Let us prove (2.15). Using the per partes integration, we get for

where

By multiplication and integration of (1.1) we obtain

and by the per partes integration,

To compute , we use (1.1) and get

By the per partes integration we derive

We have proved that (2.15) is valid.

Lemma 2.6.

Let , and let be solutions of problem (1.1), (1.11) with , . Let us denote

Then for each there exists a unique satisfying

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

Proof.

Choose . The monotonicity and continuity of in give a unique . If is unbounded we argue as in the proof of Lemma 4.8 in [8].

Let and let , , and be sequences from Lemma 2.6. Assume that for any , is not an escape solution of problem (1.1), (1.11). Lemma 2.6 implies that

We can assume that that either there exists such that

or

Otherwise we take a subsequence. Some additional properties of are given in the next two lemmas.

Lemma 2.7.

Denote

and assume that the sequence is bounded above. Then there exists such that

Proof.

By Lemma 2.4 we have

Step 1 (sequence is bounded).

Assume on the contrary that is unbounded. We may write

(otherwise we take a subsequence). Equality (2.13) yields for and ,

Using (1.4), (1.6), (1.10), and the fact that for , we get

Consequently, inequality in (2.31) leads to

for . Therefore

We will consider two cases.

Case 1.

If (2.25) holds, then (2.34) gives for

By (2.30), for each sufficiently large , we get

Putting it to (2.35), we have , contrary to (2.29).

Case 2.

If (2.26) holds, then (2.34) gives for

Due to (2.30), we have

for each sufficiently large . Putting it to (2.37), we get for . Integrating it over , we obtain . Equation (1.1) and condition (1.13) yield for , and so , contrary to (2.29).

We have proved that there exists such that

Step 2 (estimate for ).

Choose . By (2.32) we get

This together with (2.31) and (2.39) imply

According to (2.27) and Lemma 2.3 we see that is the first zero of . Since the sequence is bounded above, there exists such that , . Then (1.8) and (2.41) give

Put

Then, by virtue of (2.4), inequality (2.28) is valid.

Lemma 2.8.

Consider and satisfying (2.23) and (2.24). Let , be given by (2.27). Assume that

Then there exists such that

Proof.

Assume on the contrary that

By Lemma 2.3, is increasing on , . Therefore

and therefore there exists such that

Moreover (2.23), (2.24), (2.27), (2.44), and the monotonicity of and yield

Integrating the last inequality over , we obtain , so , a contradiction.

Lemma 2.9.

Let real sequences , , be given and assume that

Let and

(for we assume ) be such that

Assume that is given by (2.14) with . Then

Proof.

By (2.50), . Condition (2.52) yields that there exists such that

Therefore

Hence

where . Consequently,

where , because is less than the critical value . We have proved (2.53).

Now we are ready to prove the following main result of this paper.

Theorem 2.10.

Assume that

for some . Further, let and be such that (2.51) and (2.52) are valid. Then there exists such that the corresponding solution of problem (1.1), (1.11) is an escape solution.

Proof.

Assumption (2.51) implies , and hence we can choose and define by (2.14). According to (1.4), (1.10), and (2.56), there exists such that for . Consequently, we can find such that

Let , , , be sequences defined in Lemma 2.6. Moreover, let

Assume that for any , is not an escape solution of problem (1.1), (1.11). By Lemma 2.4 we have

Condition (2.60) gives such that

Choose an arbitrary . We will construct a contradiction.

Step 1 (inequality for ).

Since is increasing on , (2.62) gives a unique satisfying

By (2.59) we have

because for . Further, there exists satisfying

Therefore, according to (2.12),

Let us put

Then inequalities (2.15) and (2.66) imply

Step 2 (estimate of from below).

Since is a solution of (1.1) on , we have

Therefore

where , are such that

Integrating (2.70) over , we get

Hence

By (2.52), (2.60) and , we deduce that

Since for , we get

Due to (2.58), there exists such that . Then and . Hence for each there exists such that, for ,

Consequently

Having in mind (2.75), we can choose in (2.62) such that for all the inequality holds. Hence (2.72) and the first inequality in (2.77) yield

Put . Then , and

On the other hand, by (2.76),

By (2.79), this yields

Step 3 (estimate of ).

The inequality gives . Hence there exists such that

Having in mind (2.76), we choose to this and then, by the second inequality in (2.77), we obtain

Therefore

By Lemmas 2.7 and 2.8 there exists such that

Here , , if (2.25) holds and , , if (2.26) holds. In addition there exists such that , . (Note that if in Lemma 2.8 is not bounded but does not fulfil (2.44), we work with a proper subsequence fulfilling (2.44).) By virtue of (2.84) and (2.85) we get

Inequalities (2.84) and (2.86) yield

Step 4 (final contradictions).

Putting (2.81) and (2.87) to (2.68) and using (1.6), (1.10) and , we obtain

First, let us assume that (2.26) holds and , . So, conditions (2.85), and (2.88) yield

Letting we get a contradiction to (2.53).

Finally, let us assume that (2.25) holds and , . Then (2.61), (2.88), and yield

contrary to (2.53).

Remark 2.11.

We assume that in Theorem 2.10. In particular for and , , the function can behave in neighbourhood of as a function for arbitrary .

Now, let (2.58) hold for . Then and therefore

which is the first condition in (1.9). We have proved in [6, 7] that, in this case, assumptions (1.3)–(1.8) are sufficient for the existence of an escape solution.

Example 2.12.

Let and , . Then and

Hence, for condition (2.58) is satisfied. The critical value is equal to 3. By Theorem 2.10, if fulfils (2.52) with , problem (1.1), (1.11) has an escape solution.

Example 2.13.

Let , . Then and

Hence, for condition (2.58) is satisfied. The critical value is equal to 5. By Theorem 2.10, if fulfils (2.52) with , problem (1.1), (1.11) has an escape solution.

## 3. Homoclinic Solutions

Having an escape solution we can deduce the existence of a homoclinic solution by the same arguments as in [6]. For completeness we bring here the main ideas. Remember that our basic assumptions (1.3)–(1.8), (1.10) and (1.13) are fulfilled in this section.

By Lemma 11 in [6], a solution of problem (1.1), (1.11) is homoclinic if and only if

By Theorem 16 in [6], a solution of problem (1.1), (1.11) is an escape solution if and only if

The third type of solutions of problem (1.1), (1.11) is characterized in the next definition.

Definition 3.1.

A solution of problem (1.1), (1.11) is called damped, if

The following properties of damped and escape solutions are important for the existence of homoclinic solutions.

Theorem 3.2 (see [6, Theorem 13] (on damped solutions)).

Let be of (1.5) and (1.6). Assume that is a solution of problem (1.1), (1.11) with . Then is damped.

Theorem 3.3 (see [6, Theorem 14]).

Let be the set of all such that corresponding solutions of problem (1.1), (1.11) are damped. Then is open in .

Theorem 3.4 (see [6, Theorem 20]).

Let be the set of all such that corresponding solutions of problem (1.1), (1.11) are escape ones. Then is open in .

Having these theorems we get the main result of this section.

Theorem 3.5 (On a homoclinic solution).

Assume that the assumptions of Theorem 2.10 are satisfied. Then problem (1.1), (1.2) has a homoclinic solution.

Proof.

By Theorems 3.2 and 3.3, the set is nonempty and open in . By Theorem 3.4, the set is open in . Using Theorem 2.10, we get that is nonempty. Therefore the set is nonempty and if , then the corresponding solution of problem (1.1), (1.11) is neither damped nor an escape solution. Therefore , and by Lemma 11 in [6], such solution is homoclinic.

The proof of Theorem 3.5 implies that if problem (1.1), (1.11) has an escape solution, then it has also a homoclinic solution. Hence the following corollary is true.

Corollary 3.6.

Assume that the assumptions of Theorem 2.10 are satisfied. Let problem (1.1), (1.11) have no homoclinic solution. Then it has no escape solution.

If we assume (2.51) and (2.52), then the growth of at is less than the critical value . This is necessary for the existence of homoclinic solutions of some types of (1.1). See the next example.

Example 3.7.

Let , , . Consider (1.1), where and for and for . Then and satisfy conditions (1.3)–(1.8), (1.10), (1.13), (2.52) and (2.58) with . By Theorem 3.5, if

then problem (1.1), (1.11) has a homoclinic solution. But if

then we have proved in [12] that problem (1.1), (1.11) has no homoclinic solution and consequently no escape solution.

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## Acknowledgment

This paper was supported by the Council of Czech Government MSM 6198959214.

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Rachunková, I., Tomecek, J. Superlinear Singular Problems on the Half Line.
*Bound Value Probl* **2010**, 429813 (2010). https://doi.org/10.1155/2010/429813

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DOI: https://doi.org/10.1155/2010/429813

### Keywords

- Differential Equation
- Field Theory
- Partial Differential Equation
- Unique Solution
- Ordinary Differential Equation