- Email: [email protected]
On a class of third-order nonlinear difference equations
Applied Mathematics and Computation 213 (2009) 479–483
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
On a class of third-order nonlinear difference equations Bratislav Iricˇanin a, Stevo Stevic´ b,* a b
Faculty of Electrical Engineering, Bulevar Kralja Aleksandra 73, 11000 Beograd, Serbia Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia
a r t i c l e
i n f o
a b s t r a c t This paper studies the boundedness character of the positive solutions of the difference equation
Keywords: Positive solution Difference equation Boundedness
xnþ1 ¼ A þ
xpn q xn1 xrn2
;
n 2 N0 ;
where the parameters A; p; q and r are positive numbers. Ó 2009 Elsevier Inc. All rights reserved.
1. Introduction Recently there has been a great interest in studying nonlinear and rational difference equations c.f. [1–50] and the references therein. One of the reasons for this is a necessity for some techniques which can be used in investigating equations arising in mathematical models describing real life situations in population biology, economics, probability theory, genetics, psychology, sociology etc (see, e.g., [6,7,15,16,25,28–32,34,39] and the references therein). In this paper we investigate the boundedness character of positive solutions of the following difference equation
xnþ1 ¼ A þ
xpn q xn1 xrn2
;
n 2 N0 ;
ð1Þ
where the parameters A; p; q and r are positive numbers. Eq. (1) is a natural extension of the equation treated in [45], where the case r ¼ 0 was considered. The case q ¼ 0 will be treated in a separate paper. Note that the case p ¼ 0 is trivial, since from the obvious inequality xn > A; n 2 N, and (1) it follows that
xnþ1 6 A þ
1 ; Aqþr
n ¼ 3; 4; . . . ;
from which the boundedness of all positive solutions of Eq. (1) in this case, directly follows. Hence, of some interest is the case when all parameters A; p, q and r are positive. Closely related equations to Eq. (1) have been investigated, for example, in [1–5,8,17–19,37,46,48,49]. 2. Main results In this section we prove our main results in this paper. The first two results give some sufficient conditions which guarantee the existence of positive unbounded solutions of Eq. (1). * Corresponding author. E-mail addresses: [email protected] (B. Iricˇanin), [email protected] (S. Stevic´). 0096-3003/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2009.03.039
B. Iricˇanin, S. Stevic´ / Applied Mathematics and Computation 213 (2009) 479–483
480
Theorem 1. Assume that p2 P 3q and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q 6 0; and p P 3 or p P ð3 þ qÞ=2, (where at least one of two inequalities, in both combinations, is strict), or (b) p > 1 þ q þ r. Then Eq. (1) has positive unbounded solutions.
Proof. First, note that for every solution of Eq. (1) the following inequality holds
xnþ1 P
xpn q xn1 xrn2
;
n 2 N0 :
ð2Þ
Let yn ¼ ln xn . Taking the logarithm it follows that
ynþ1 pyn þ qyn1 þ ryn2 P 0;
n 2 N0 :
ð3Þ
Let
PðkÞ ¼ k3 pk2 þ qk þ r ¼ 0:
ð4Þ
Then, we have
P0 ðkÞ ¼ 3k2 2pk þ q; from which itfollows that the polynomial PðkÞ has a local maximum at the point k1 ¼ ðp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mum at k2 ¼ p þ p2 3q =3, moreover, we have that
Pðk2 Þ ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 3qÞ=3, and a local mini-
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q 27
ð5Þ
and 0 < k1 < k2 < p. The case k1 ¼ k2 is excluded since if p2 ¼ 3q, then the first condition in (a) is impossible. If conditions in (a) hold, then from (5), it is clear that the first condition in (a) is equivalent with Pðk2 Þ 6 0. On the other hand, it is easy to see that any of the following two conditions p P 3 or p P ð3 þ qÞ=2 is equivalent with k2 P 1. Note also that according to the assumption, at least one of the inequalities Pðk2 Þ 6 0 and p P 3, or Pðk2 Þ 6 0 and p P ð3 þ qÞ=2 is strict. From this and since limk!þ1 PðkÞ ¼ þ1, it follows that there is a k3 > 1 such that Pðk3 Þ ¼ 0. On the other hand, if (b) holds, then Pð1Þ < 0 and since limk!þ1 PðkÞ ¼ þ1, it again follows that there is a k3 > 1 such that Pðk3 Þ ¼ 0. Let
P1 ðkÞ ¼
PðkÞ ¼ k2 þ ak þ b k k3
and
un ¼ yn þ ayn1 þ byn2 ;
n 2 N0 ;
(this means that un ¼ LP1 ðyn Þ where LP1 is the linear operator obtained by the polynomial P1 ðkÞ). Then inequality (3) can be written in the following form
unþ1 k3 un P 0:
ð6Þ
Choose y2 ; y1 and y0 such that u0 > 0. For example, it is enough to choose y2 ; y1 and y0 such that
y0 > jajjy1 j þ jbjjy2 j: From this and (6) we have that
un P kn3 u0 :
ð7Þ
Since u0 > 0 and k3 > 1, by letting n ! 1 in (7) we obtain that un ! þ1 as n ! 1, from which it follows that yn is unbounded. Indeed, if yn were bounded then un would be also bounded, which would be a contradiction. Note also that yn P ln A; n 2 N, that is, the sequence yn is bounded from below. Since yn is unbounded then there is a subsequence ðynk Þk2N such that ynk ! þ1 as k ! 1. From this we have that xnk ¼ eynk ! þ1 as k ! 1. Hence the sequence xn ¼ eyn is unbounded too, finishing the proof of the theorem. The next theorem concerns the case when Pðk2 Þ ¼ 0 and k2 ¼ 1, where P is the polynomial in (4). Note that Pð1Þ ¼ 0 is equivalent with p ¼ 1 þ q þ r, while from k2 ¼ 1, it follows that p ¼ ðq þ 3Þ=2 and p 2 ð0; 3. Hence r ¼ ð1 qÞ=2 and consequently q 2 ð0; 1Þ. This explains the constraints posed in the next theorem.
B. Iricˇanin, S. Stevic´ / Applied Mathematics and Computation 213 (2009) 479–483
481
Theorem 2. Assume that p ¼ ðq þ 3Þ=2 and r ¼ ð1 qÞ=2. Then Eq. (1) has positive unbounded solutions. Proof. Since in this case Pð1Þ ¼ P0 ð1Þ ¼ 0, the polynomial P does not have any zero on the interval ð1; 1Þ. Hence, we have to use a method different from that in Theorem 1. Let x0 > maxfx1 ; x2 g. Note that Eq. (1) in this case is qþ3
xnþ1 ¼ A þ
xn2 1q
ð8Þ
:
2 xqn1 xn2
From (8), for every positive solution of the equation, we have that
q 1q 2 xnþ1 A xn xn ¼ þ ; xn xn xn1 xn2
n 2 N0 ;
ð9Þ
n 2 N0 :
ð10Þ
from which with n ¼ 0, it follows that
q 1q x1 x0 x0 2 > > 1: x0 x1 x2 From (9) and by induction, we obtain
xnþ1 > xn
xn xn1
q
xn xn2
1q 2 > 1;
Hence the sequence ðxn Þ1 n¼1 is strictly increasing. Assume that xn is bounded. If it were, then it would exist a finite positive limit limn!1 xn , say l. By letting n ! 1 in (8) we obtain l ¼ A þ l, which is a contradiction. Hence, all the solutions of Eq. (8) with x0 > maxfx1 ; x2 g are unbounded, from which the theorem follows. Remark 1. Note that the condition p ¼ ðq þ 3Þ=2 appearing in Theorem 2, imply p2 P 3q. Now we consider the cases when p2 < 3q or Pðk2 Þ > 0. Theorem 3. Assume that p2 < 3q or
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q > 0:
ð11Þ
Then all positive solutions of Eq. (1) are bounded. Proof. From Eq. (1), we have that
xnþ1
0 1p 0 1p q pp x xpn x A n n1 A ¼Aþ@ q A ; ¼Aþ q r ¼Aþ@ q þ r þq r r p p p xn1 xn2 xn2 xrn3 xpn1 xn2 xpn1 xn2
n 2 N:
ð12Þ
If p2 6 q, then from (12) it follows that
xnþ1 6 A þ
A q
r
Apþp
þ
1
p
rþq
A p þqþrp
;
n P 4;
from which the boundedness follows in this case. Now assume that p2 > q. From (12), and by continuing with iterations, we have that
0
xnþ1
1pqp 1p x n1 B A C ¼Aþ@ q þ@ r A r ðpþqÞ=ðpqpÞ r=ðpqpÞ p xpn1 xn2 xn2 xn3 0 0 A
0
1 !pak 1ppq p A x B C nk ¼Aþ@ q þ@ r A A q q þ þ r akþ1 bkþ1 ðpþqÞ=ðppÞ r=ðppÞ p xnk1 xnk2 xpn1 xn2 xn2 xn3 0 !pak !pqp 1p pakþ1 x A A A ; ¼Aþ@ q þ þ a þ b nk1 r bkþ1 kþ1 kþ1 þq r p xnk2 xnk2 xnk3 xnk1 xpn2 xn3 A
ð13Þ
for every n P k þ 1, where the sequences ak and bk are defined by
akþ1 ¼
bk þ q ; p ak
bkþ1 ¼
r ; p ak
a1 ¼ q=p; b1 ¼ r=p:
ð14Þ
B. Iricˇanin, S. Stevic´ / Applied Mathematics and Computation 213 (2009) 479–483
482
Since p2 > q, we have
q r q r q ¼ a1 < a2 ¼ þq p and b1 ¼ < r p ¼ b2 : p p p p p
ð15Þ
From (14) and (15) it is easy to see that the sequences ak and bk are strictly increasing, as far as ak < p. From (14), we have that
akþ1 ¼
r þ qðp ak1 Þ ; ðp ak Þðp ak1 Þ
k 2 N:
Hence, if ak < p for every k 2 N we have that there is finite limit limk!1 ak ¼ x 2 ð0; p, and that x is a solution of the equation.
f ðxÞ ¼ xðx pÞ2 r qðp xÞ ¼ 0: Let x ¼ p y , then y is a solution of the equation
y3 py2 þ qy þ r ¼ 0; which is equivalent with PðyÞ ¼ 0. 0 Recall that P 0 ðyÞ ¼ 3y2 2py þ q. If p2 < 3q, then the roots k1;2 of the polynomial P are complex numbers. If (11) holds, then Pðk2 Þ > 0. Since Pð0Þ ¼ r > 0 it follows that PðyÞ does not have positive roots, which is a contradiction. This implies that there is the least k0 2 N such that ak0 1 < p and ak0 P p. From this and (13) with k ¼ k0 1, it follows that
0 B xnþ1 ¼ A þ @
6Aþ
0
A
0
pak
xnk00
1pak
A A þ @ þ @ a þ b þq bk k0 k0 0 r xnk0 xnk x x xn1 xn2 nk nk 1 1 2 0 0 0 q p
r p
0 1
1pqp 1p C A A
pak 1 pqp !p 0 1 1 þ þ þ : qþr Aak0 þbk0 1 Aak0 þbk0 þqpþr A p 1 1
The last expression is an upper bound for the sequence xn , as desired. The next two theorems concern the case when p 2 ð0; 1. Theorem 4. Assume that p 2 ð0; 1Þ. Then every positive solution of Eq. (1) is bounded. Proof. Since xn > A; n 2 N, we have that
xnþ1 6 A þ
xpn ; Arþq
ð16Þ
for n ¼ 3; 4; . . ., where ðxn Þ is an arbitrary positive solution of (1). Let yn be the solution of the difference equation
ynþ1 ¼ A þ
ypn ; Arþq
n ¼ 3; 4; . . . ;
ð17Þ
such that y3 ¼ x3 . Since the function
f ðxÞ ¼ A þ
xp Arþq
is increasing, by induction it follows that xn 6 yn , for n P 3. Since the function f is also concave (we use here the condition p 2 ð0; 1Þ) it follows that there is a unique fixed point x of the equation f ðxÞ ¼ x, and the function f satisfies the condition
ðf ðxÞ xÞðx x Þ < 0;
x 2 ð0; 1Þ n fx g:
ð18Þ
By (18) we see that if y3 2 ð0; x the sequence ðyn ÞnP3 is nondecreasing and bounded above by x and if y0 P x , it is nonincreasing and bounded below by x . Hence the sequence ðyn Þ is bounded and consequently the sequence ðxn Þ is bounded too. Theorem 5. Assume that p ¼ 1 and A > 1. Then every positive solution of Eq. (1) is bounded. Proof. First note that if ðxn Þ is an arbitrary positive solution of (1) then inequality (16) holds, with p ¼ 1. Let ðyn Þ be the solution of Eq. (17) with p ¼ 1, such that y3 ¼ x3 . As in the proof of Theorem 4 we have that xn 6 yn , for n P 3.
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483
This equation is a first order linear difference equation with constant coefficients, which can be solved explicitly, and proved that
lim yn ¼
n!1
Arþqþ1 ; Arþq 1
from which the boundedness of ðyn Þ follows and consequently the boundedness of ðxn Þ, as claimed.
For the readers interested in the research area we leave the following research problem. Research problem. Investigate the boundedness of Eq. (1) for the case when
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q 6 0; 1 6 p < q þ r þ 1 and p 6 minf3; ð3 þ qÞ=2g; excluding those subcases which are covered by the cases in Theorems 1–5. References [1] K. Berenhaut, J. Foley, S. Stevic´, The global attractivity of the rational difference equation yn ¼ 1 þ ðynk =ynm Þ, Proc. Am. Math. Soc. 135 (2007) 1133– 1140. [2] K. Berenhaut, J. Foley, S. Stevic´, The global attractivity of the rational difference equation yn ¼ A þ ðynk =ynm Þp , Proc. Am. Math. Soc. 136 (2008) 103– 110. [3] K.S. Berenhaut, S. Stevic´, A note on positive nonoscillatory solutions of the difference equation xnþ1 ¼ a þ ðxpnk =xpn Þ, J. Differ. Eqs. Appl. 12 (5) (2006) 495–499. [4] K. Berenhaut, S. Stevic´, The behaviour of the positive solutions of the difference equation xn ¼ A þ ðxn2 =xn1 Þp , J. Differ. Eqs. Appl. 12 (9) (2006) 909– 918. P [5] K. Berenhaut, S. Stevic´, The difference equation xnþ1 ¼ a þ xnk = k1 i¼0 c i xni has solutions converging to zero, J. Math. Anal. Appl. 326 (2007) 1466– 1471. [6] L. Berezansky, E. Braverman, On impulsive Beverton–Holt difference equations and their applications, J. Differ. Eqs. Appl. 10 (9) (2004) 851–868. [7] L. Berezansky, E. Braverman, Sufficient Conditions for the global stability of nonautonomous higher order difference equations, J. Differ. Eqs. Appl. 11 (9) (2005) 785–798. [8] L. Berg, On the asymptotics of nonlinear difference equations, Z. Anal. Anwendungen 21 (4) (2002) 1061–1074. [9] L. Berg, Inclusion theorems for non-linear difference equations with applications, J. Differ. Eqs. Appl. 10 (4) (2004) 399–408. [10] L. Berg, Oscillating solutions of rational difference equations, Rostock. Math. Kolloq. 58 (2004) 31–35. [11] L. Berg, Nonlinear difference equations with periodic solutions, Rostock. Math. Kolloq. 61 (2005) 12–19. [12] L. Berg, On the asymptotics of difference equation xn3 ¼ xn ð1 þ xn1 xn2 Þ, J. Differ. Eqs. Appl. 14 (1) (2008) 105–108. [13] L. Berg, S. Stevic´, Periodicity of some classes of holomorphic difference equations, J. Differ. Eqs. Appl. 12 (8) (2006) 827–835. [14] L. Berg, S. Stevic´, Linear difference equations mod 2 with applications to nonlinear difference equations, J. Differ. Eqs. Appl. 14 (7) (2008) 693–704. [15] M. De la Sen, About the properties of a modified generalized Beverton–Holt equation in ecology models, Discrete Dyn. Nat. Soc. 2008 (2008) 23. Article ID 592950. [16] M. De la Sen, S. Alonso-Quesada, A control theory point of view on Beverton–Holt equation in population dynamics and some of its generalizations, Appl. Math. Comput. 199 (2) (2008) 464–481. [17] R. DeVault, C. Kent, W. Kosmala, On the recursive sequence xnþ1 ¼ p þ ðxnk =xn Þ, J. Differ. Eqs. Appl. 9 (8) (2003) 721–730. [18] J. Feuer, On the behaviour of solutions of xnþ1 ¼ p þ ðxn1 =xn Þ, Appl. Anal. 83 (6) (2004) 599–606. [19] L. Gutnik, S. Stevic´, On the behaviour of the solutions of a second-order difference equation 2007 (2007) 14. Article ID 27562. [20] L.H. Hu, W.T. Li, S. Stevic´, Global asymptotic stability of a second order rational difference equation, J. Differ. Eqs. Appl. 14 (8) (2008) 779–797. [21] B. Iricˇanin, A global convergence result for a higher-order difference equation, Discrete Dyn. Nat. Soc. 2007 (2007) 7. Article ID 91292. [22] G.L. Karakostas, Convergence of a difference equation via the full limiting sequences method, Differ. Eqs. Dyn. Syst. 1 (4) (1993) 289–294. [23] G.L. Karakostas, Asymptotic 2-periodic difference equations with diagonally self-invertible responses, J. Differ. Eqs. Appl. 6 (2000) 329–335. [24] G.L. Karakostas, Asymptotic behavior of the solutions of the difference equation xnþ1 ¼ x2n f ðxn1 Þ, J. Differ. Eqs. Appl. 9 (6) (2003) 599–602. [25] A.D. Mishkis, On some problems of the theory of differential equations with deviating argument, Uspekhi Mat. Nauk 32:2 (194) (1977) 173–202. [26] A.Y. Ozban, On the system of rational difference equations xn ¼ a=yn3 ; yn ¼ byn3 =xnq ynq , Appl. Math. Comput. 188 (1) (2007) 833–837. [27] S. Ozen, I. Ozturk, F. Bozkurt, On the recursive sequence ynþ1 ¼ ða þ yn1 Þ=ðb þ yn Þ þ yn1 =yn , Appl. Math. Comput. 188 (1) (2007) 180–188. [28] E.C. Pielou, An introduction to mathematical ecology, Wiley Interscience, 1969. [29] E.C. Pielou, Population and community ecology, Gordon and Breach, 1974. [30] E.P. Popov, Automatic Regulation and Control, Nauka, Moscow, Russia, 1966, Russian. [31] S. Stevic´, Behaviour of the positive solutions of the generalized Beddington–Holt equation, Panamer. Math. J. 10 (4) (2000) 77–85. [32] S. Stevic´, Asymptotic behaviour of a sequence defined by iteration with applications, Colloq. Math. 93 (2) (2002) 267–276. [33] S. Stevic´, A global convergence results with applications to periodic solutions, Indian J. Pure Appl. Math. 33 (1) (2002) 45–53. [34] S. Stevic´, Asymptotic behaviour of a nonlinear difference equation, Q IndianJ. Pure Appl. Math. 34 (12) (2003) 1681–1687. Q 2ðkþ1Þ [35] S. Stevic´, On the recursive sequence xnþ1 ¼ A= ki¼0 xni þ 1= j¼kþ2 xnj , Taiwanese J. Math. 7 (2) (2003) 249–259. [36] S. Stevic´, A note on periodic character of a difference equation, J. Differ. Eqs. Appl. 10 (10) (2004) 929–932. [37] S. Stevic´, On the recursive sequence xnþ1 ¼ a þ ðxpn1 =xpn Þ, J. Appl. Math. Comput. 18 (1-2) (2005) 229–234. [38] S. Stevic´, On the recursive sequence xnþ1 ¼ ða þ bxnk Þ=f ðxn ; . . . ; xnkþ1 Þ, Taiwanese J. Math. 9 (4) (2005) 583–593. [39] S. Stevic´, A short proof of the Cushing–Henson conjecture, Discrete Dyn. Nat. Soc. 2006 (2006) 5. Article ID 37264. [40] S. Stevic´, Asymptotic behaviour of a class of nonlinear difference equations, Discrete Dyn. Nat. Soc. 2006 (2006) 10. Article ID 47156. [41] S. Stevic´, Global stability and asymptotics of some classes of rational difference equations, J. Math. Anal. Appl. 316 (2006) 60–68. [42] S. Stevic´, On positive solutions of a ðk þ 1Þth order difference equation, Appl. Math. Lett. 19 (5) (2006) 427–431. [43] S. Stevic´, Asymptotics of some classes of higher order difference equations, Discrete Dyn. Nat. Soc. 2007 (2007) 20. Article ID 56813. [44] S. Stevic´, Existence of nontrivial solutions of a rational difference equation, Appl. Math. Lett. 20 (2007) 28–31. ðxpn =xrn1 Þ, Discrete Dyn. Nat. Soc. 2007 (2007) 9. Article ID 40963. [45] S. Stevic´, On the recursive sequence xnþ1 ¼ A þ P Pm k [46] S. Stevic´, On the recursive sequence xn ¼ 1 þ i¼1 ai xnpi = j¼1 bj xnqj , Discrete Dyn. Nat. Soc. 2007 (2007) 7. Article ID 39404. [47] S. Stevic´, Boundedness and global stability of a higher-order difference equation, J. Differ. Eqs. Appl. 14 (10–11) (2008) 1035–1044. [48] S. Stevic´, On the difference equation xnþ1 ¼ a þ ðxn1 =xn Þ, Comput. Math. Appl. 56 (5) (2008) 1159–1171. [49] S. Stevic´, Boundedness character of a class of difference equations, Nonlin. Anal. TMA 70 (2009) 839–848. [50] Y. Yang, X. Yang, On the difference equation xnþ1 ¼ ðpxns þ xnt Þ=ðqxns þ xnt Þ, Appl. Math. Comput. 203 (2) (2008) 903–907.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
On a class of third-order nonlinear difference equations Bratislav Iricˇanin a, Stevo Stevic´ b,* a b
Faculty of Electrical Engineering, Bulevar Kralja Aleksandra 73, 11000 Beograd, Serbia Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia
a r t i c l e
i n f o
a b s t r a c t This paper studies the boundedness character of the positive solutions of the difference equation
Keywords: Positive solution Difference equation Boundedness
xnþ1 ¼ A þ
xpn q xn1 xrn2
;
n 2 N0 ;
where the parameters A; p; q and r are positive numbers. Ó 2009 Elsevier Inc. All rights reserved.
1. Introduction Recently there has been a great interest in studying nonlinear and rational difference equations c.f. [1–50] and the references therein. One of the reasons for this is a necessity for some techniques which can be used in investigating equations arising in mathematical models describing real life situations in population biology, economics, probability theory, genetics, psychology, sociology etc (see, e.g., [6,7,15,16,25,28–32,34,39] and the references therein). In this paper we investigate the boundedness character of positive solutions of the following difference equation
xnþ1 ¼ A þ
xpn q xn1 xrn2
;
n 2 N0 ;
ð1Þ
where the parameters A; p; q and r are positive numbers. Eq. (1) is a natural extension of the equation treated in [45], where the case r ¼ 0 was considered. The case q ¼ 0 will be treated in a separate paper. Note that the case p ¼ 0 is trivial, since from the obvious inequality xn > A; n 2 N, and (1) it follows that
xnþ1 6 A þ
1 ; Aqþr
n ¼ 3; 4; . . . ;
from which the boundedness of all positive solutions of Eq. (1) in this case, directly follows. Hence, of some interest is the case when all parameters A; p, q and r are positive. Closely related equations to Eq. (1) have been investigated, for example, in [1–5,8,17–19,37,46,48,49]. 2. Main results In this section we prove our main results in this paper. The first two results give some sufficient conditions which guarantee the existence of positive unbounded solutions of Eq. (1). * Corresponding author. E-mail addresses: [email protected] (B. Iricˇanin), [email protected] (S. Stevic´). 0096-3003/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2009.03.039
B. Iricˇanin, S. Stevic´ / Applied Mathematics and Computation 213 (2009) 479–483
480
Theorem 1. Assume that p2 P 3q and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q 6 0; and p P 3 or p P ð3 þ qÞ=2, (where at least one of two inequalities, in both combinations, is strict), or (b) p > 1 þ q þ r. Then Eq. (1) has positive unbounded solutions.
Proof. First, note that for every solution of Eq. (1) the following inequality holds
xnþ1 P
xpn q xn1 xrn2
;
n 2 N0 :
ð2Þ
Let yn ¼ ln xn . Taking the logarithm it follows that
ynþ1 pyn þ qyn1 þ ryn2 P 0;
n 2 N0 :
ð3Þ
Let
PðkÞ ¼ k3 pk2 þ qk þ r ¼ 0:
ð4Þ
Then, we have
P0 ðkÞ ¼ 3k2 2pk þ q; from which itfollows that the polynomial PðkÞ has a local maximum at the point k1 ¼ ðp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mum at k2 ¼ p þ p2 3q =3, moreover, we have that
Pðk2 Þ ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 3qÞ=3, and a local mini-
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q 27
ð5Þ
and 0 < k1 < k2 < p. The case k1 ¼ k2 is excluded since if p2 ¼ 3q, then the first condition in (a) is impossible. If conditions in (a) hold, then from (5), it is clear that the first condition in (a) is equivalent with Pðk2 Þ 6 0. On the other hand, it is easy to see that any of the following two conditions p P 3 or p P ð3 þ qÞ=2 is equivalent with k2 P 1. Note also that according to the assumption, at least one of the inequalities Pðk2 Þ 6 0 and p P 3, or Pðk2 Þ 6 0 and p P ð3 þ qÞ=2 is strict. From this and since limk!þ1 PðkÞ ¼ þ1, it follows that there is a k3 > 1 such that Pðk3 Þ ¼ 0. On the other hand, if (b) holds, then Pð1Þ < 0 and since limk!þ1 PðkÞ ¼ þ1, it again follows that there is a k3 > 1 such that Pðk3 Þ ¼ 0. Let
P1 ðkÞ ¼
PðkÞ ¼ k2 þ ak þ b k k3
and
un ¼ yn þ ayn1 þ byn2 ;
n 2 N0 ;
(this means that un ¼ LP1 ðyn Þ where LP1 is the linear operator obtained by the polynomial P1 ðkÞ). Then inequality (3) can be written in the following form
unþ1 k3 un P 0:
ð6Þ
Choose y2 ; y1 and y0 such that u0 > 0. For example, it is enough to choose y2 ; y1 and y0 such that
y0 > jajjy1 j þ jbjjy2 j: From this and (6) we have that
un P kn3 u0 :
ð7Þ
Since u0 > 0 and k3 > 1, by letting n ! 1 in (7) we obtain that un ! þ1 as n ! 1, from which it follows that yn is unbounded. Indeed, if yn were bounded then un would be also bounded, which would be a contradiction. Note also that yn P ln A; n 2 N, that is, the sequence yn is bounded from below. Since yn is unbounded then there is a subsequence ðynk Þk2N such that ynk ! þ1 as k ! 1. From this we have that xnk ¼ eynk ! þ1 as k ! 1. Hence the sequence xn ¼ eyn is unbounded too, finishing the proof of the theorem. The next theorem concerns the case when Pðk2 Þ ¼ 0 and k2 ¼ 1, where P is the polynomial in (4). Note that Pð1Þ ¼ 0 is equivalent with p ¼ 1 þ q þ r, while from k2 ¼ 1, it follows that p ¼ ðq þ 3Þ=2 and p 2 ð0; 3. Hence r ¼ ð1 qÞ=2 and consequently q 2 ð0; 1Þ. This explains the constraints posed in the next theorem.
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Theorem 2. Assume that p ¼ ðq þ 3Þ=2 and r ¼ ð1 qÞ=2. Then Eq. (1) has positive unbounded solutions. Proof. Since in this case Pð1Þ ¼ P0 ð1Þ ¼ 0, the polynomial P does not have any zero on the interval ð1; 1Þ. Hence, we have to use a method different from that in Theorem 1. Let x0 > maxfx1 ; x2 g. Note that Eq. (1) in this case is qþ3
xnþ1 ¼ A þ
xn2 1q
ð8Þ
:
2 xqn1 xn2
From (8), for every positive solution of the equation, we have that
q 1q 2 xnþ1 A xn xn ¼ þ ; xn xn xn1 xn2
n 2 N0 ;
ð9Þ
n 2 N0 :
ð10Þ
from which with n ¼ 0, it follows that
q 1q x1 x0 x0 2 > > 1: x0 x1 x2 From (9) and by induction, we obtain
xnþ1 > xn
xn xn1
q
xn xn2
1q 2 > 1;
Hence the sequence ðxn Þ1 n¼1 is strictly increasing. Assume that xn is bounded. If it were, then it would exist a finite positive limit limn!1 xn , say l. By letting n ! 1 in (8) we obtain l ¼ A þ l, which is a contradiction. Hence, all the solutions of Eq. (8) with x0 > maxfx1 ; x2 g are unbounded, from which the theorem follows. Remark 1. Note that the condition p ¼ ðq þ 3Þ=2 appearing in Theorem 2, imply p2 P 3q. Now we consider the cases when p2 < 3q or Pðk2 Þ > 0. Theorem 3. Assume that p2 < 3q or
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 27r þ 9pq 2p3 ð2p2 6qÞ p2 3q > 0:
ð11Þ
Then all positive solutions of Eq. (1) are bounded. Proof. From Eq. (1), we have that
xnþ1
0 1p 0 1p q pp x xpn x A n n1 A ¼Aþ@ q A ; ¼Aþ q r ¼Aþ@ q þ r þq r r p p p xn1 xn2 xn2 xrn3 xpn1 xn2 xpn1 xn2
n 2 N:
ð12Þ
If p2 6 q, then from (12) it follows that
xnþ1 6 A þ
A q
r
Apþp
þ
1
p
rþq
A p þqþrp
;
n P 4;
from which the boundedness follows in this case. Now assume that p2 > q. From (12), and by continuing with iterations, we have that
0
xnþ1
1pqp 1p x n1 B A C ¼Aþ@ q þ@ r A r ðpþqÞ=ðpqpÞ r=ðpqpÞ p xpn1 xn2 xn2 xn3 0 0 A
0
1 !pak 1ppq p A x B C nk ¼Aþ@ q þ@ r A A q q þ þ r akþ1 bkþ1 ðpþqÞ=ðppÞ r=ðppÞ p xnk1 xnk2 xpn1 xn2 xn2 xn3 0 !pak !pqp 1p pakþ1 x A A A ; ¼Aþ@ q þ þ a þ b nk1 r bkþ1 kþ1 kþ1 þq r p xnk2 xnk2 xnk3 xnk1 xpn2 xn3 A
ð13Þ
for every n P k þ 1, where the sequences ak and bk are defined by
akþ1 ¼
bk þ q ; p ak
bkþ1 ¼
r ; p ak
a1 ¼ q=p; b1 ¼ r=p:
ð14Þ
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482
Since p2 > q, we have
q r q r q ¼ a1 < a2 ¼ þq p and b1 ¼ < r p ¼ b2 : p p p p p
ð15Þ
From (14) and (15) it is easy to see that the sequences ak and bk are strictly increasing, as far as ak < p. From (14), we have that
akþ1 ¼
r þ qðp ak1 Þ ; ðp ak Þðp ak1 Þ
k 2 N:
Hence, if ak < p for every k 2 N we have that there is finite limit limk!1 ak ¼ x 2 ð0; p, and that x is a solution of the equation.
f ðxÞ ¼ xðx pÞ2 r qðp xÞ ¼ 0: Let x ¼ p y , then y is a solution of the equation
y3 py2 þ qy þ r ¼ 0; which is equivalent with PðyÞ ¼ 0. 0 Recall that P 0 ðyÞ ¼ 3y2 2py þ q. If p2 < 3q, then the roots k1;2 of the polynomial P are complex numbers. If (11) holds, then Pðk2 Þ > 0. Since Pð0Þ ¼ r > 0 it follows that PðyÞ does not have positive roots, which is a contradiction. This implies that there is the least k0 2 N such that ak0 1 < p and ak0 P p. From this and (13) with k ¼ k0 1, it follows that
0 B xnþ1 ¼ A þ @
6Aþ
0
A
0
pak
xnk00
1pak
A A þ @ þ @ a þ b þq bk k0 k0 0 r xnk0 xnk x x xn1 xn2 nk nk 1 1 2 0 0 0 q p
r p
0 1
1pqp 1p C A A
pak 1 pqp !p 0 1 1 þ þ þ : qþr Aak0 þbk0 1 Aak0 þbk0 þqpþr A p 1 1
The last expression is an upper bound for the sequence xn , as desired. The next two theorems concern the case when p 2 ð0; 1. Theorem 4. Assume that p 2 ð0; 1Þ. Then every positive solution of Eq. (1) is bounded. Proof. Since xn > A; n 2 N, we have that
xnþ1 6 A þ
xpn ; Arþq
ð16Þ
for n ¼ 3; 4; . . ., where ðxn Þ is an arbitrary positive solution of (1). Let yn be the solution of the difference equation
ynþ1 ¼ A þ
ypn ; Arþq
n ¼ 3; 4; . . . ;
ð17Þ
such that y3 ¼ x3 . Since the function
f ðxÞ ¼ A þ
xp Arþq
is increasing, by induction it follows that xn 6 yn , for n P 3. Since the function f is also concave (we use here the condition p 2 ð0; 1Þ) it follows that there is a unique fixed point x of the equation f ðxÞ ¼ x, and the function f satisfies the condition
ðf ðxÞ xÞðx x Þ < 0;
x 2 ð0; 1Þ n fx g:
ð18Þ
By (18) we see that if y3 2 ð0; x the sequence ðyn ÞnP3 is nondecreasing and bounded above by x and if y0 P x , it is nonincreasing and bounded below by x . Hence the sequence ðyn Þ is bounded and consequently the sequence ðxn Þ is bounded too. Theorem 5. Assume that p ¼ 1 and A > 1. Then every positive solution of Eq. (1) is bounded. Proof. First note that if ðxn Þ is an arbitrary positive solution of (1) then inequality (16) holds, with p ¼ 1. Let ðyn Þ be the solution of Eq. (17) with p ¼ 1, such that y3 ¼ x3 . As in the proof of Theorem 4 we have that xn 6 yn , for n P 3.
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483
This equation is a first order linear difference equation with constant coefficients, which can be solved explicitly, and proved that
lim yn ¼
n!1
Arþqþ1 ; Arþq 1
from which the boundedness of ðyn Þ follows and consequently the boundedness of ðxn Þ, as claimed.
For the readers interested in the research area we leave the following research problem. Research problem. Investigate the boundedness of Eq. (1) for the case when
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