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

Definition 4.2.

A solution

of problem (2.13), (1.10) is called

*an escape solution*, if there exists

such that

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

Proof.

Step 1.

We show that the interval

is nonempty. Since

and

satisfies (1.3), (1.13), we can find

such that

Integrating (1.1) over

we obtain

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

Using the "per partes" integration, we get for

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

(otherwise, we take a subsequence). Assume on the contrary that for any

,

is not an escape solution. Choose

. Then, by Remark 4.7,

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

satisfying

By (i) and (ii),

satisfies

Integrating it over

we get

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

Integrating (4.14) over

and using (4.10), we obtain

Conditions (1.8) and (4.8) yield

, which implies

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

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

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

and, by Lemma 4.8, the sequence

is bounded. Therefore there exists

such that

Choose an arbitrary

. According to Lemma 4.6,

satisfies equality (4.3), that is

Since

and

is increasing on

, there exists a unique

such that

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

By virtue of (1.6) and (1.13), we see that

is decreasing on

, which yields

Since

and

, the monotonicity of

yields

for

, and consequently

Therefore (4.27) and (4.32) give

Step 2.

We prove that the sequence

is bounded below by some positive number. Since

is a solution of (1.1) on

, we have

where

satisfies

and

. Having in mind (1.8), we see that

is increasing and

for

. Consequently

Integrating (4.36) over

, we obtain

which, due to (4.39), yields

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

Step 3.

We construct a contradiction. Putting

in (4.34), we have

Due to (4.23),

. Therefore,

, and consequently, by (4.24),

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

Case 1.

Let

, that is, we can find

,

,

, such that

Then, by (4.43), for each sufficiently large

, we get

Putting it to (4.42), we have

Therefore
, contrary to (4.25).

Case 2.

Let

. We may assume

(otherwise we take a subsequence). Then there exists

,

, such that

Due to (4.43), for each sufficiently large

, we get

Putting it to (4.42), we have

Therefore,

for

. Integrating it over

, we obtain

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.