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.

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
.

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

Proof.

For equality (2.13) see Lemma 4.6 in [

8]. Let us prove (2.15). Using the per partes integration, we get for

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

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

Lemma 2.7.

and assume that the sequence

is bounded above. Then there exists

such that

Proof.

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

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

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

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.

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

because

for

. Further, there exists

satisfying

Therefore, according to (2.12),

Then inequalities (2.15) and (2.66) imply

Step 2 (estimate of
from below).

Since

is a solution of (1.1) on

, we have

where

, are such that

Integrating (2.70) over

, we get

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

,

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),

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

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.

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.

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.