In order to establish this first result, we reduce (1.1) to a first-order equation by means of the substitution
This transforms (1.1) to
Theorem 2.1.
Let
and
in (1.1). Then one has the following.
-
(i)
There are infinitely many solutions,
, such that for each, one of its subsequences,
,
, converges to zero and the other to infinity.
-
(ii)
There exist solutions,
, which
-
(a)
converge to zero if
;
-
(b)
diverge to infinity if
;
-
(c)
are constant if
.
Proof.
Starting with (2.1), let the function
be defined as
. Note that for
,
is a decreasing function since
. Also note that
and
. Hence
has a unique positive fixed point
.
We next compute the expression
and simplify, it including canceling the common factor
from the numerator and denominator, thereby obtaining the following:
where
Note that since
,
and
. Thus the numerator of
has one and only one sign change. Therefore, by Descartes' rule of signs, the numerator of
has exactly one positive root,
.
In addition, we see that
and so, given that
is the only positive root of the numerator of
, we have
for
. Thus, since
and
is continuous, we must have
for
. Therefore,
We consider two cases depending on the initial value
for (2.1).
Case 1 (
).
Using induction and the fact that
is a decreasing function so that
is an increasing function, we have
Thus,
Since
is the only positive fixed point of
, then we must have
and 
Case 2 (
).
The argument is similar to that in Case 1 in showing
and
In both cases, the solution,
, of (2.1) is divided into even and odd subsequences,
and
, where one subsequence converges monotonically to zero and the other to infinity.
We now go back to (1.1) by inferring the behavior of
from
. To do this we first consider
. Without loss of generality, we will assume that
and so
and
.
Next, observe that
From this and our assumption with
, we have
Hence, for
, there exists
such that
for all
. We then have
and by induction, for
,
This, in turn, implies that
The argument is similar in showing that
since
Hence, result (i) is true.
Now consider
. Then
for all
, and so
for all
. Induction then gives us
for all
. We thus have one of the following:
-
(1)
If
(
), then 
-
(2)
If
(
), then 
-
(3)
If
(
), then
is a constant solution 
Thus the result (ii) is true and this completes the proof.
For the next couple of results we rewrite (1.1) in the form
Note that if either
and
, or
and
, then
satisfies the following properties:
-
(P1)
, with
undefined when
.
-
(P2)
-
(P3)
if
.
-
If
we consider the addition restriction that
and
, we also obtain
-
(P4)
if
, then
, or
.
Lemma 2.2.
Let
be a positive solution of (1.1) with
and
. Then there exist
and
such that the following statements are true:
-
(1)
as
,
-
(2)
as
,
-
(3)
, and
and
are undefined; or if either
or
is not zero, then
is a solution of (1.1).
-
(4)
.
Proof.
Statements 1 and 2 follow from the fact that
by properties (P2) and (P3). Statement 3 follows from the fact that either
, and so
and
are undefined by property (P1); or
and
where Statements 1 and 2 and the continuity of
(Property (P1) hold. Finally, Statement 4 follows immediately from Statement 3 and Property (P4).
In the first three results, we characterize the convergence of the odd and even subsequences of solutions of (1.1).
Theorem 2.3.
Let
and
in (1.1). Then for each positive solution,
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number.
Proof.
Consider (1.1) with
,
, and
. Then it follows from Lemma 2.2 that for each positive solution of (1.1),
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number.
Theorem 2.4.
Let
and
in (1.1). Then for each positive solution
, one of the subsequences,
,
, diverges to infinity and the other to a positive number or diverges to infinity.
Proof.
Consider (1.1) with
and
. Using the transformation
convert (1.1) to the equation
Then
, and so it follows from Lemma 2.2 that for each positive solution of (2.16),
, one of the subsequences,
,
, converges to zero and the other to a nonnegative number. Hence, for each positive solution of (1.1),
, one of the subsequences,
,
, diverges to infinity and the other to a positive number or diverges to infinity.
In the following results, we show the existence of monotonic solutions for (1.1). As with Theorem 2.1 we use the substitution 
Theorem 2.5.
Let
and
in (1.1). Then there are positive initial values for which the corresponding solutions,
, decrease monotonically to zero.
Proof.
Note that an equilibrium equation for (2.1) satisfies,
Set
Given Descartes' rule of signs, we have that there exists a unique positive equilibrium,
, where
and
Recall that
and let
for all
. Then
for all
. It follows from induction that
for all
. Since
,
, with
, decreases monotonically to zero.
Theorem 2.6.
Let
and
in (1.1). Then there are positive initial values for which the corresponding solution,
, increases monotonically to infinity.
Proof.
As in the previous proof, an equilibrium equation for (2.1) satisfies (2.17). Setting
we obtain from Descartes' rule of signs, a unique positive equilibrium,
, where
and
Recall that
and let
for all
. Then
for all
. It follows from induction that
for all
. Since
,
, with
, increases monotonically to infinity.