doi:10.1155/2009/985161
Research Article
Convergence Results on a Second-Order Rational
Difference Equation with Quadratic Terms
D. M. Chan, C. M. Kent, and N. L. Ortiz-Robinson
Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Harris Hall, 1015 Floyd Avenue, P.O. Box 842014, Richmond, VA 23284-2014, USA
Correspondence should be addressed to C. M. Kent,[email protected]
Received 6 March 2009; Accepted 20 June 2009
Recommended by Martin J. Bohner
We investigate the global behavior of the second-order difference equationxn1 xn−1αxn
βxn−1/AxnBxn−1, where initial conditions and all coefficients are positive. We find conditions onA, B, α, βunder which the even and odd subsequences of a positive solution converge, one to zero and the other to a nonnegative number; as well as conditions where one of the subsequences diverges to infinity and the other either converges to a positive number or diverges to infinity. We also find initial conditions where the solution monotonically converges to zero and where it diverges to infinity.
Copyrightq2009 D. M. Chan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction and Preliminaries
There are a number of studies published on second-order rational difference equationssee, e.g.,1–9 . We investigate the global behavior of the second-order difference equation
xn1xn−1 αx
nβxn−1
AxnBxn−1
, 1.1
where the numerator is quadratic and the denominator is linear with A, B, α, β ∈ 0,∞. Under various hypotheses on the parameters, we establish the existence of different behaviors of even and odd subsequences of solutions of1.1. Our results are summarized below.
iLetα < Aandβ > B, then we have the following.
aThere are infinitely many solutions, {xn}∞n−1, such that for each, one of its
subsequences, {x2n}∞n0,{x2n−1}∞n0, converges to zero and the other diverges
bThere exist solutions,{xn}∞n0, which 1converge to zero ifAB > αβ;
2diverge to infinity ifAB < αβ;
3are constant ifABαβ.
iLet α A and β > B. Then for each positive solution {xn}∞n−1, one of the
subsequences,{x2n}∞n0,{x2n−1}∞n0, diverges to infinity and the other to a positive
number that can be arbitrarily large depending on initial values. Further there, are positive initial values for which the corresponding solution,{xn}∞n−1, increases
monotonically to infinity.
iiLet α < A and β B. Then for each positive solution {xn}∞n−1, one of the
subsequences,{x2n}∞n0,{x2n−1}∞n0, converges to zero and the other to a nonnegative
number. Further, there are positive initial values for which the corresponding solution,{xn}∞n−1, decreases monotonically to zero.
We note that the following results address and solve the first five conjectures posed by Sedaghat in10 .
2. Results
In order to establish this first result, we reduce1.1to a first-order equation by means of the substitutionrnxn/xn−1.This transforms1.1to
rn1
αrnβ
Ar2 nBrn
. 2.1
Theorem 2.1. Letα < Aandβ > Bin1.1. Then one has the following.
iThere are infinitely many solutions,{xn}∞n−1, such that for each, one of its subsequences, {x2n}∞n0,{x2n−1}∞n0, converges to zero and the other to infinity.
iiThere exist solutions,{xn}∞n−1, which
aconverge to zero ifAB > αβ; bdiverge to infinity ifAB < αβ;
care constant ifABαβ.
Proof. Starting with2.1, let the functiong : 0,∞ → 0,∞ be defined asgr αr
β/Ar2Br. Note that forr ∈0,∞,gris a decreasing function sincegr −Aαr2
2AβrBβ/Ar2Br2 <0. Also note that lim
r→0gr−r ∞and limr→∞gr−r −∞. Henceghas a unique positive fixed pointr.
We next compute the expression g2r− r and simplify, it including canceling the
common factor Ar Br from the numerator and denominator, thereby obtaining the following:
g2r−r a4r4a3r3a2r2a1r
b3r3b2r2b1rb0
where
and only one sign change. Therefore, by Descartes’ rule of signs, the numerator ofg2r−r0
has exactly one positive root,r.
We consider two cases depending on the initial valuer0for2.1.
Hence, for 0< < β/B−1, there existsN≥0 such that
Thus the resultiiis true and this completes the proof.
For the next couple of results we rewrite1.1in the form
Note that if eitherα ≤ Aand β < B, or α < Aand β ≤ B, thenf satisfies the following properties:
P1f ∈C0,∞2− {0,0},0,∞ , withfu, vundefined whenuv0.
P2f ∈C0,∞×0,∞,0,∞
P3fu, v< vifu, v∈0,∞.
If we consider the addition restriction thatα < AandβB, we also obtain
P4iffu, v v, thenu0, v >0, oru >0, v0.
Lemma 2.2. Let{xn}∞n−1 be a positive solution of 1.1withα < Aand β B. Then there exist
Lo≥0andLe≥0such that the following statements are true:
1x2n−1↓Loasn → ∞, 2x2n↓Leasn → ∞,
3LoLe 0, andfLo, LeandfLe, Loare undefined; or if eitherLoorLeis not zero, thenLo, Le, Lo, Le, . . .is a solution of1.1.
4Lo·Le0.
Proof. Statements 1 and 2 follow from the fact that
0< x2n1fx2n, x2n−1< x2n−1, 0< x2n2fx2n1, x2n< x2n 2.14
by propertiesP2andP3. Statement 3 follows from the fact that eitherLoLe0, and so
fLo, LeandfLe, Loare undefined by propertyP1; orLo/Leand
Lonlim
→ ∞x2n1nlim→ ∞fx2n, x2n−1 fLe, Lo
Lenlim
→ ∞x2n2nlim→ ∞fx2n1, x2n fLo, Le,
2.15
where Statements 1 and 2 and the continuity off PropertyP1hold. Finally, Statement 4 follows immediately from Statement 3 and PropertyP4.
In the first three results, we characterize the convergence of the odd and even subsequences of solutions of1.1.
Theorem 2.3. Letα < Aandβ Bin1.1. Then for each positive solution,{xn}∞n−1, one of the subsequences,{x2n}∞n0,{x2n−1}∞n0, converges to zero and the other to a nonnegative number.
Proof. Consider1.1withα < A,βB, andfu, v vαuβv/AuBv. Then it follows fromLemma 2.2that for each positive solution of1.1,{xn}∞n−1, one of the subsequences, {x2n}∞n0,{x2n−1}∞n0, converges to zero and the other to a nonnegative number.
Proof. Consider1.1withαAandβ > B. Using the transformationyn1/xn,convert1.1
to the equation
yn1yn−1 By
nAyn−1
βynαyn−1
. 2.16
Thenfu, v vAvBu/αvβu, and so it follows fromLemma 2.2that for each positive solution of2.16,{yn}∞n−1, one of the subsequences,{y2n}∞n0,{y2n−1}∞n0, converges to zero
and the other to a nonnegative number. Hence, for each positive solution of1.1,{xn}∞n−1,
one of the subsequences,{x2n}∞n0,{x2n−1}∞n0, 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 for1.1. As withTheorem 2.1we use the substitutionrnxn/xn−1.
Theorem 2.5. Letα < Aand β Bin 1.1. Then there are positive initial values for which the corresponding solutions,{xn}∞n−1, decrease monotonically to zero.
Proof. Note that an equilibrium equation for2.1satisfies,
Ar3Br2−αr−β0. 2.17
Setpr Ar3Br2−αr−β.Given Descartes’ rule of signs, we have that there exists a unique
positive equilibrium,r < 1, wherep0 < 0 andp1 > 0.Recall thatrn xn/xn−1,and let
rnrfor alln≥0. Thenxn/xn−1 rfor alln≥0. It follows from induction thatxnrn1x−1
for alln≥0. Sincer <1,{xn}∞n−1, withx0rx−1, decreases monotonically to zero.
Theorem 2.6. Letα Aand β > Bin 1.1. Then there are positive initial values for which the corresponding solution,{xn}∞n−1, increases monotonically to infinity.
Proof. As in the previous proof, an equilibrium equation for 2.1 satisfies 2.17. Setting
pr Ar3Br2−αr−β,we obtain from Descartes’ rule of signs, a unique positive equilibrium,
r >1, wherep0<0 and limr→ ∞pr>0.Recall thatrnxn/xn−1,and letrnrfor alln≥0.
Thenxn/xn−1rfor alln≥0. It follows from induction thatxnrn1x−1for alln≥0. Since
r >1,{xn}∞n−1, withx0rx−1, increases monotonically to infinity.
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