For each
define
by
Then
with
By (A3), it follows that
Now let us consider the auxiliary family of the equations
Lemma 3.1 (see [1, Proposition
]).
Let (A1), (A2) hold. If
is a nontrivial solution of (3.4), (3.5), then
for some
.
Let
be such that
Note that
Let us consider
as a bifurcation problem from the trivial solution 
Equation (3.8) can be converted to the equivalent equation
Further we note that
for
near
in 
The results of Rabinowitz [8] for (3.8) can be stated as follows. For each integer
, there exists a continuum
of solutions of (3.8) joining
to infinity in
Moreover, 
Proof of Theorem 1.5.
Let us verify that
satisfies all of the conditions of Lemma 2.3.
Since
condition
in Lemma 2.3 is satisfied with
. Obviously
and accordingly,
holds.
can be deduced directly from the Arzela-Ascoli Theorem and the definition of
. Therefore, the superior limit of
, contains an unbounded connected component
with
.
From the condition (A2), applying Lemma 2.2 with
in [10], we can show that the initial value problem
has a unique solution on
for every
and
. Therefore, any nontrivial solution
of (1.1), (1.2) has only simple zeros in
and
. Meanwhile, (A1) implies that
[1, proposition 4.1]. Since
, we conclude that
. Moreover,
by (1.1) and (1.2).
We divide the proof into three cases.
Case 1 (
).
In this case, we show that
.
Assume on the contrary that
then there exists a sequence
such that
for some positive constant
depending not on
. From Lemma 2.6, we have
Set
Then
Now, choosing a subsequence and relabelling if necessary, it follows that there exists
with
such that
Since
, we can show that
The proof is similar to that of the step 1 of Theorem
in [7]; we omit it. So, we obtain
and subsequently,
for
. This contradicts (3.16). Therefore
Case 2 (
).
In this case, we can show easily that
joins
with
by using the same method used to prove Theorem
in [2].
Case 3 (
).
In this case, we show that
joins
with
.
Let
be such that
If
is bounded, say,
, for some
depending not on
, then we may assume that
Taking subsequences again if necessary, we still denote
such that
. If
, all the following proofs are similar.
Let
denote the zeros of
in
. Then, after taking a subsequence if necessary,
. Clearly,
. Set
. We can choose at least one subinterval
which is of length at least
for some
. Then, for this
if
is large enough. Put
.
Obviously, for the above given
and
have the same sign on
for all
. Without loss of generality, we assume that
Moreover, we have
Combining this with the fact
and using the relation
we deduce that
must change its sign on
if
is large enough. This is a contradiction. Hence
is unbounded. From Lemma 2.6, we have that
Note that
satisfies the autonomous equation
We see that
consists of a sequence of positive and negative bumps, together with a truncated bump at the right end of the interval
with the following properties (ignoring the truncated bump) (see, [1]):
(i)all the positive (resp., negative) bumps have the same shape (the shapes of the positive and negative bumps may be different);
(ii)each bump contains a single zero of
, and there is exactly one zero of
between consecutive zeros of
;
(iii)all the positive (negative) bumps attain the same maximum (minimum) value.
Armed with this information on the shape of
it is easy to show that for the above given
is an unbounded sequence. That is
Since
is concave on
, for any
small enough,
This together with (3.31) implies that there exist constants
with
, such that
Hence, we have
Now, we show that
.
Suppose on the contrary that, choosing a subsequence and relabeling if necessary,
for some constant
. This implies that
From (3.28) we obtain that
must change its sign on
if
is large enough. This is a contradiction. Therefore
.
Proof of Theorem 1.6.
and
are immediate consequence of Theorem 1.5
and
, respectively.
To prove
, we rewrite (1.1), (1.2) to
By Lemma 2.5, for every
and
,
where 
Let
be such that
Then for
and
,
This means that
By Lemma 2.6 and Theorem 1.5, it follows that
is also an unbounded component joining
and
in
. Thus, (3.40) implies that for
(1.1), (1.2) has at least two solutions in
.