This article is part of the series Singular Boundary Value Problems for Ordinary Differential Equations.

Open Access Research Article

Homoclinic Solutions of Singular Nonautonomous Second-Order Differential Equations

Irena Rachůnková* and Jan Tomeček

Author Affiliations

Department of Mathematical Analysis and Applications of Mathematics, Faculty of Science, Palacký University, 17 listopadu 12, 771 46 Olomouc, Czech Republic

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Boundary Value Problems 2009, 2009:959636  doi:10.1155/2009/959636


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2009/1/959636


Received:27 April 2009
Revisions received:1 September 2009
Accepted:15 September 2009
Published:13 October 2009

© 2009 The Author(s)

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.

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 .

1. Introduction

Having a positive parameter we consider the problem

(11)

(12)

under the following basic assumptions for and

(13)

(14)

(15)

(16)

(17)

(18)

Then problem (1.1), (1.2) generalizes some models arising in hydrodynamics or in the nonlinear field theory (see [15]). However (1.1) is singular at because .

Definition 1.1.

If , thena 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, 46]. 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,

(19)

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

(110)

(111)

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

(112)

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

(113)

(114)

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

(115)

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

(116)

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.

For each there exists a maximal such that problem (1.1), (1.10) has a unique solution on and

(21)

Further, for each there exists such that

(22)

Proof.

Let be a solution of problem (1.1), (1.10) on . By (1.1), we have

(23)

and multiplying by and integrating between and , we get

(24)

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

(25)

A function is a solution of problem (1.1), (1.2) on if and only if it is a fixed point of the operator . Using the Lipschitz property of we can prove that the operator is contractive for each sufficiently small and from the Banach Fixed Point Theorem we conclude that there exists exactly one solution of problem (1.1), (1.2) on . This solution has the form

(26)

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

(27)

Put

(28)

Then

(29)

and, by "per partes" integration we derive . Multiplying (2.7) by and integrating it over , we get

(210)

Estimates (2.2) follow from (2.7)–(2.10) for

(211)

Remark 2.2.

The proof of Lemma 2.1 yields that if , then .

Let us put

(212)

and consider an auxiliary equation

(213)

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]).

For each , and each , there exists such that for any ,

(214)

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

Proof.

Choose , , . Let be the Lipschitz constant for on . By (2.6) for , , , ,

(215)

From the Gronwall inequality, we get

(216)

Similarly, by (2.6), (2.9), and (2.16),

(217)

If we choose such that

(218)

we get (2.14).

Remark 2.4.

Choose and , and consider the initial conditions

(219)

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.

Let be a solution of problem (1.1), (1.10). Assume that there exists such that

(220)

Then for and

(221)

Proof.

By (1.13) and (2.20), on and thus and are positive on . Consequently, there exists . Further, by (1.1),

(222)

and, by multiplication and integration over ,

(223)

Therefore,

(224)

and hence exists. Since is bounded on , we get

(225)

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.

Let be a solution of problem (1.1), (1.10). Assume that there exist and such that

(226)

Then for all and (2.21) holds.

Proof.

Since fulfils (2.26), we can find a maximal such that for and consequently for . By (4.23) and (2.26), on and thus and are negative on . So, is positive and decreasing on which yields (otherwise, we get , contrary to (2.26)). Consequently there exists . By multiplication and integration (2.22) over , we obtain

(227)

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.

A solution of problem (1.1), (1.10) (or of problem (2.13), (1.10)) on is calleddamped if

(31)

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.

Let be a damped solution of problem (1.1), (1.10). By Remark 2.2, we have in Lemma 2.1 and hence

(32)

Step 1.

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

Step 2.

Assume that is the first zero of on . Then, due to Remark 2.4, . Let for . By virtue of (1.4), for and thus is decreasing. Let be positive on . Then is also decreasing, is increasing and , due to (3.1). Consequently, . Letting in (2.22), we get , which is impossible because is bounded below. Therefore there are and satisfying (2.26) and, by Lemma 2.6, either fulfils (2.21) or has the second zero with . So is positive on and has just one local maximum in . Moreover, putting and in (2.23), we have

(33)

and hence

(34)

Step 3.

Let have no other zeros. Then for . Assume that is negative on . Then, due to (2.1), . Putting in (2.23) and letting , we obtain

(35)

Therefore, exists and, since is bounded, we deduce that

(36)

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

(37)

which together with (3.4) yields

(38)

Step 4.

Assume that has its third zero . Then we prove as in Step 2 that has just one negative minimum in and (3.8) is valid. Further, as in Step 2, we deduce that either fulfils (2.21) or has the fourth zero , is positive on with just one local maximum on , and . This together with (3.8) yields

(39)

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.

A solution of problem (1.1), (1.10) fulfils the condition

(310)

if and only if fulfils the condition

(311)

Proof.

Assume that fulfils (3.10). Then there exists such that , for . Otherwise , due to Lemma 2.5. Let be such that on , . By Remark 2.4 and (3.10), . Integrating (1.1) over , we get

(312)

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.

Let be a solution of (1.1), (1.10) with . Then, by (1.4)–(1.6),

(313)

Assume on the contrary that is not damped. Then is defined on the interval and or there exists such that , and for . If the latter possibility occurs, (2.22) and (3.13) give by integration

(314)

a contradiction. If , then, by Lemma 3.5, fulfils (3.11). So has a unique zero . Integrating (2.22) over , we get

(315)

and so

(316)

Integrating (2.22) over , we obtain for

(317)

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.

Let and be a solution of (1.1), (1.10) with . So, is damped and is also a solution of (2.13).

(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) Let have at most a finite number of zeros. Then, by Theorem 3.3, fulfils (2.21). Choose . Since fulfils (2.22), we get by integration over

(318)

For we get, by (2.21),

(319)

Therefore, we can find such that

(320)

Let be the constant of Lemma 2.1. Choose . Assume that and is a corresponding solution of problem (2.13), (1.10). Using Lemma 2.1, Lemma 2.3 and the continuity of , we can find such that if , then

(321)

moreover for and

(322)

Therefore, we have

(323)

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

(324)

We get for . Therefore, for and, due to (1.4)–(1.6),

(325)

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.

Let . A solution of problem (1.1), (1.10) on is called an escape solution if

(41)

Definition 4.2.

A solution of problem (2.13), (1.10) is called an escape solution, if there exists such that

(42)

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.

Let . Assume that is a solution of problem (1.1), (1.10) on and is a maximal interval where is increasing and for . Then

(43)

Proof.

Step 1.

We show that the interval is nonempty. Since and satisfies (1.3), (1.13), we can find such that

(44)

Integrating (1.1) over we obtain

(45)

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

Step 2.

By multiplication of (1.1) by and integration over we obtain

(46)

Using the "per partes" integration, we get for

(47)

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.

Similar arugmets can be found in [12]. By Lemma 2.1, the assertion (i) holds. The arguments in Step 1 of the proof of Lemma 4.6 imply (ii). The strict monotonicity of and Remark 4.7 yields a unique . Assume that is unbounded. Then

(48)

(otherwise, we take a subsequence). Assume on the contrary that for any , is not an escape solution. Choose . Then, by Remark 4.7,

(49)

Due to (4.9), (1.2) and (ii) there exists satisfying

(410)

By (i) and (ii), satisfies

(411)

Integrating it over we get

(412)

Put

(413)

Then, by (4.12),

(414)

We see that is decreasing. From (1.4) and (1.6) we get that is increasing on and consequently by (4.9) and (4.13), we have

(415)

Integrating (4.14) over and using (4.10), we obtain

(416)

where

(417)

Further, by (4.15),

(418)

(419)

Conditions (1.8) and (4.8) yield , which implies

(420)

By (4.13) and (4.18),

(421)

and consequently

(422)

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

Theorem 4.9 (on escape solution).

Assume that (1.3)–(1.8) and (1.13) hold and let

(423)

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

Proof.

Let and let , , and be sequences from Lemma 4.8. Moreover, let

(424)

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.

We derive some inequality for . By Remark 4.7, we have

(425)

and, by Lemma 4.8, the sequence is bounded. Therefore there exists such that

(426)

Choose an arbitrary . According to Lemma 4.6, satisfies equality (4.3), that is

(427)

Since and is increasing on , there exists a unique such that

(428)

Having in mind, due to (1.4)–(1.8), that the inequality

(429)

holds, we get

(430)

By virtue of (1.6) and (1.13), we see that is decreasing on , which yields

(431)

Hence,

(432)

Since and , the monotonicity of yields for , and consequently

(433)

Therefore (4.27) and (4.32) give

(434)

Step 2.

We prove that the sequence is bounded below by some positive number. Since is a solution of (1.1) on , we have

(435)

Integrating it, we get

(436)

where satisfies and . Having in mind (1.8), we see that is increasing and for . Consequently

(437)

Integrating (4.36) over , we obtain

(438)

and hence

(439)

By (4.23) we get

(440)

which, due to (4.39), yields

(441)

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

Step 3.

We construct a contradiction. Putting in (4.34), we have

(442)

Due to (4.23), . Therefore, , and consequently, by (4.24),

(443)

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

Case 1.

Let , that is, we can find , , , such that

(444)

Then, by (4.43), for each sufficiently large , we get

(445)

Putting it to (4.42), we have

(446)

Therefore , contrary to (4.25).

Case 2.

Let . We may assume (otherwise we take a subsequence). Then there exists , , such that

(447)

Due to (4.43), for each sufficiently large , we get

(448)

Putting it to (4.42), we have

(449)

Therefore, for . Integrating it over , we obtain

(450)

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.

The function

(51)

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

Acknowledgments

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

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