Superlinear Singular Problems on the Half Line
© Rachunková and J. Tomecek. 2010
Received: 19 October 2010
Accepted: 7 December 2010
Published: 15 December 2010
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.
where is a positive real parameter.
Let . A function satisfying (1.1) on is called a solution of ( 1.1 ) on .
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 ).
A strictly increasing solution of problem (1.1), (1.2) is called a homoclinic solution.
where , and introduce the following definition.
then is called an escape solution of problem ( 1.1 ), ( 1.11 ).
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.
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]).
In what follows by a solution of (1.1), (1.11) we mean a solution on .
Remark 2.2 (see [6, Remark 4]).
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 .
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).
? (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 .
We have proved that (2.15) is valid.
If the sequence is unbounded, then there exists an escape solution in .
Choose . The monotonicity and continuity of in give a unique . If is unbounded we argue as in the proof of Lemma 4.8 in .
Otherwise we take a subsequence. Some additional properties of are given in the next two lemmas.
Step 1 (sequence is bounded).
We will consider two cases.
Putting it to (2.35), we have , contrary to (2.29).
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).
Step 2 (estimate for ).
Then, by virtue of (2.4), inequality (2.28) is valid.
Integrating the last inequality over , we obtain , so , a contradiction.
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.
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.
Choose an arbitrary . We will construct a contradiction.
Step 1 (inequality for ).
Step 2 (estimate of from below).
Step 3 (estimate of ).
Step 4 (final contradictions).
Letting we get a contradiction to (2.53).
contrary to (2.53).
We assume that in Theorem 2.10. In particular for and , , the function can behave in neighbourhood of as a function for arbitrary .
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.
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 . 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.
The third type of solutions of problem (1.1), (1.11) is characterized in the next definition.
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.
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 , 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.
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.
then we have proved in  that problem (1.1), (1.11) has no homoclinic solution and consequently no escape solution.
This paper was supported by the Council of Czech Government MSM 6198959214.
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