In order to establish this first result, we reduce (1.1) to a firstorder 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.