Explicit bounds for second-order difference equations and a solution to a question of Stević

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1 J. Math. Anal. Appl Explicit bounds for second-order difference equations and a solution to a question of Stević Kenneth S. Berenhaut,1, Eva G. Goedhart Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109, USA Received 24 June 2004 Available online 19 November 2004 Submitted by J. Henderson Abstract This note gives explicit, applicable bounds for solutions of a wide class of second-order difference equations with nonconstant coefficients. Among the applications is an affirmative answer to a recent question of Stević Elsevier Inc. All rights reserved. Keywords: Explicit bounds; Applicable bounds; Second-order linear difference equations; Growth rates; Nonconstant coefficients 1. Introduction This paper studies explicit, applicable growth rates for second-order difference equations. In particular, we will consider equations of the form b i = 2 + gi 1 b i hi 1 b i 2, 1 for i 2, and provide sharp inequalities for {b i } in terms of the sequences {gi} and {hi}, and the initial values b 0 and b 1. Solutions of difference equations of the form in 1 * Corresponding author. addresses: berenhks@wfu.edu K.S. Berenhaut, goedeg3@wfu.edu E.G. Goedhart. 1 The author acknowledges financial support from a Sterge Faculty Fellowship X/$ see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 2 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl have been studied by many authors cf. [2 4,6 13]. Often the study has focused on the understanding of oscillatory or asymptotic behavior. Our main theorem Theorem 2 implies the following result which partially answers a question of Stević[8]. Theorem 1. Suppose that c 0 and {b i } satisfies 1, with gi = c/i 2. Then, for n 0, b n b 0 + b 1 n k c, 2 χ c where k c = 1 + 4c and { 1, if 0 c 2 or c 6, χ c = 2 kc ck c 22 kc c, if 2 <c<6. Note that c = k c k c 1. Hence, by Lemma 2a, in Section 3, for 2 c 6i.e., 2 k c 3, 2 k c 1 χ c 1, k c k c 1 and by Lemma 2c, 3 χ c 1 + k c k c In [8], Stević proved that b n = On c+1 and correctly conjectured that b n = On k c. The main theorem here is the following. Theorem 2. Suppose B and g are positive functions satisfying the second-order differential equation B x = gxbx, B x exists on x 1, {b i } is a solution to the difference equation in 1, with h1> 1, and 4 7 hn gn, for all n 2. Let V2 = 0 and 8 Vn= 1 6 for n 3, where n 1 min i 1<η i <i<ζ i <i+1 { B ζ i B η i } + Hn, 9

3 Hn= K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl hi<0 hibi In addition, suppose there exist positive constants, c 0 and c 1, satisfying { } B2 B1 + Vn 0 <c 0 min,b1 B0, h1 { } B2 B1 + Vn 0 <c 1 min,b1 B0, g1 for all n 3, then b0 b n + b 1 Bn, 13 c 0 c 1 for n 0. Note that from the conditions in Theorem 2, it follows that B is increasing. If B x is nondecreasing in x and h 0, then Vn 0, for all n, and we obtain the following corollary. Corollary 1. Suppose B and g are positive functions satisfying 7, B x is nondecreasing on x 1, and {b i } is a solution to the difference equation in 1 with h 0. In addition, suppose that there exist positive constants, c 0 and c 1, satisfying 0 <c 0 min { B2 B1, B1 B0 }, 14 { } B2 B1 0 <c 1 min,b1 B0, g1 then 13 is satisfied for all n 0. Similarly, if B x is nonincreasing in x and h 0, we have Vn= 1 n 1 { min B ζ i B η i } 6 i 1<η i <i<ζ i <i+1 = 1 6 n 1 B i + 1 B i 1 = 1 6 B n + B n 1 B 2 B 1, 16 for n 3, and Theorem 2 leads to the following corollary. Corollary 2. Suppose B and g are positive functions satisfying 7, B x is nonincreasing on x 1, with { 1 inf B n + B n 1 B 2 B 1 } C, 17 n 1 6

4 4 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl Table 1 Constants c 0 and c 1 for some pairs g, B Bx gx B x c 0 c 1 x 5 20/x xi1 2 x 1/x x lnx + 1 x + 2/x lnx xe x 1 + 2/x x 5/2 3.75/x for some C, and {b i } is a solution to the difference equation in 1, with h 0. In addition, suppose that there exist positive constants, c 0 and c 1, satisfying 0 <c 0 min { B2 B1 + C,B1 B0 }, { } B2 B1 + C 0 <c 1 min,b1 B0, g1 then 13 is satisfied for all n 0. Note that if B x 0, for all x, then we may take C = 1 6 B 2 + B 1 in 17. Example. In [8], Stević alsoprovedthatifgi = 1/i for i 1, then b n = One n.as noted in Table 1, for that particular g, we actually have b n = O ni 1 2 n, 19 where I k z denotes the modified Bessel function of the first kind cf. [1]. To see how the two bounds compare, note that 1/n ni1 2 n = lim n As shown in Table 1, a more appropriate g for the bound ne n is given by gi = 1 + 2/i. Table 1 gives several noteworthy examples of pairs g, B with associated constants c 0 and c 1. The remainder of the paper proceeds as follows. Section 2 comprises a proof of Theorem 2, while Section 3 includes a proof of Theorem 1 which uses Corollaries 1 and Proof of the main result In this section we prove Theorem 2. We will employ the following elementary lemma cf. Mitrinović [5, p. 362] which follows directly from Taylor s theorem. Lemma 1. Suppose f is defined over the interval n 1,n+1.Iff x exists for n 1 x n + 1, then

5 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl fn+ 1 2fn+ fn 1 = f n f ζ f η, 21 for some n 1 <η<n<ζ <n Proof of Theorem 2. Suppose B and g satisfy the assumptions of the theorem and define and ε i = Bi b i i 0, 23 ε i 1 = ε i ε i 1 i First, set b 0 = c 0 and b 1 = 0. We will show that in this case, {b i } i>0 and {b i+1 b i } i>0 are nonnegative sequences, i.e., b i+1 b i 0, for i 1. Note that b 1 = 0, and since h1> 1, we have that b 2 = 1 + h1c 0 > 0. Thus assume 25 holds for 1 i M 2, for some M 3. By 1, the induction hypothesis, and 8, we have b M b M 1 = 1 + gm 1 b M hm 1 b M 2 Thus 25 holds for all i 1. Now, by 23 and 11, we have gm 1b M 1 hm 1b M 2 gm 1b M 2 hm 1b M 2 25 = gm 1 hm 1 b M ε 0 = ε 1 ε 0 = B1 b 1 B0 b 0 = B1 B0 c Also, employing 1 and 11, ε 1 = B2 b 2 B1 b 1 = B2 B1 1 + g1 b h1 b 0 = B2 B1 1 + h1 c For ε 2, we have, for some ζ 2 and η 2 satisfying 1 <η 2 < 2 <ζ 2 < 3, ε 2 = B3 B2 1 + g2 b h2 b 1 = B3 2B2 + B1 g2b2 + g2 B2 b 2 + B2 b 2 B1 b1 + h2b1 = 1 6 B ζ 2 B η 2 + B 2 g2b2 + g2ε 2 + ε 1 + h2b 1 = 1 6 B ζ 2 B η 2 + g2ε 2 + ε 1 + h2b 1, 29

6 6 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl where we have used 1, Lemma 1, and 7. Repeating the process in 29, successively, to rewrite ε 3, ε 4,..., ε n 1, we obtain ε n 1 = ε n ε n 1 = 1 n 1 B ζ i B η i n 1 n 1 + giε i + ε 1 + hib i 1 6 n 1 1 B ζ i B η i + giε i + ε hi<0 n 1 hib i 1 by 25 n 1 = Vn+ giε i + ε 1 hi<0 hiε i 1 = Vn+ B2 B1 1 + h1 n 1 c 0 + giε i + hi<0 n 1 hi ε i 1 giε i + hi<0 hi ε i 1, 30 where ζ i and η i satisfy i 1 <η i <i<ζ i <i+ 1, for i {2, 3,...,n 1}. The inequality in 30 follows from 11. Now, ε 0 = B0 b 0 = B0 + c 0 > 0, and hence from 27 and 28, we obtain ε 2 ε 1 ε 0 > Thus, assume ε i 0, for 0 i N 1, for some N 3. Then, by 30, the induction hypothesis, and the fact that g is positive, we have N 1 ε N = ε N 1 + ε N 1 giε i + hi<0 and the induction is complete. Combining this with 25 gives hi εi 1 0, 32 b n Bn, 33 for all n 0. A similar argument also holds when, in place of the sequence {b i }, we consider the solution {bi } of 1 with starting values b0 = 0 and b 1 = c We then have

7 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl bn Bn, 35 for all n 1. To complete the proof, note that for the solution {b i }, with arbitrary starting values b 0 and b 1,wehave b b i = 0 b i + b 1 b i c 0 c b 0 b i 1 c + b 1 b i 0 c b 0 1 c Bi + b 1 0 c Bi 1 b 0 = + b 1 Bi. 36 c 0 c 1 This completes the proof of Theorem A question of Stević In this section, we employ Corollaries 1 and 2 to prove Theorem 1. First we require the following technical lemma. Lemma 2. Define, for x 1, the functions z, r 1, and r 2 via zx def = 1 6 xx 1x 2, r 1x def = 2 x 2 xx 1, and r 2 x def = 2 x 4 zx 2 x = 2 x 3 8 zx 4 + zx. We then have the following inequalities: a r 1 x 0 for 2 x 3; b r 1 x 0 for 1 x 2, x 3; c r 2 x 0 for 2 x 3. Proof. Note that r 1 x = 2x ln 2 2 2, 37 and hence r 1 x 0for1 x x 0 and r 1 x 0forx x 0, where def x 0 = ln 2 ln ln 2 Since 1/4 < ln 2 2 < 1/2, the definition in 38 leads to 2 <x 0 < 3 in fact, x The inequalities in a and b now follow via concavity considerations, upon noting that r 1 1 = r 1 2 = r 1 3 = 0. For c, suppose 2 x 3, and note that for such x, 0 zx x We then have

8 8 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl zx r 2x = 2 x 2 2 = 2 x zx 8 zx 1 + zx 4 2 x zx 4 2 x zx = 2 x zx 4 1 zx 1 + zx zx zx The two inequalities in 40 follow from 39. Expanding about x = 2, and employing 39, this gives 2 8 zx r 2x 1 + x 2 ln x 2 28 = ln 2 12 x 2 0, and part c follows. We are now in a position to prove Theorem 1. Proof of Theorem 1. Set h 0, and note that for c 0i.e.,k c 1, B and g defined by Bx = x k c and gx = c satisfy x 2 c B x = k c k c 1x kc 2 = cx kc 2 = x 2 x k c = gxbx, 42 and hence 7 is satisfied. Now, for x 1, { B 4 x = k c k c 1k c 2k c 3x k c 4 0, if 2 <c<6, 43 0, otherwise, thus B x is nonincreasing on x 1, when 2 <c<6i.e.,2<k c < 3 and nondecreasing when 0 c 2orc 6. Note that B2 = 2 k c, B1 = 1, B0 = 0, and g1 = c. Case 1. c 6or0 c 2i.e.,k c 3or1 k c 2. Here, min{b2 B1, B1 B0}=1 and by Lemma 2b, { } { B2 B1 2 k c } 1 min,b1 B0 = min 1 + g1 1 + c, 1 = Applying Corollary 1 with c 0 = c 1 = 1 gives 2. Case 2. 2 <c<6i.e.,2<k c < 3. Here, B x = k c k c 1k c 2x k c 3, 45 thus B x 0asx tends to infinity. Taking C = 1 6 B 2 + B 1, wehave

9 where K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl min { B2 B1 + C,B1 B0 } { = min B2 B1 6 1 B 2 + B 1 }, 1 = min { Jc,1 } = 1, 46 Jc def = 2 k c k ck c 1k c 2 2 k c The last equality in 46 follows by Lemma 2c. In fact Jc 3. Also, since Jc 2 k c 1, by Lemma 2a we have { } { } B2 B1 + C Jc min,b1 B0 = min 1 + g1 1 + c, 1 = Jc 1 + c. 48 Applying Corollary 2, with c 0 = 1 and c 1 = Jc/1 + c = χ c, gives 2. Remark. After completion of this manuscript, Prof. Stević kindly shared with us a preliminary draft of a short note [9] also confirming his conjecture in [8]. The result therein is of a purely asymptotic nature, whereas here we are interested in explicit and applicable bounds. Since the question provided some of our original motivation and the bound in this case is quite simple and informative, we have chosen to leave our handling of his question among our examples. The interested reader is encouraged to seek out [9] for a different perspective on the particular case when gi = c/i 2 for some c 0, all i 1, and h 0 in 1. Acknowledgment We are very thankful to a referee for comments and insights that substantially improved this manuscript. References [1] M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions, Dover, New York, [2] V.B. Demidovič, Conditions for boundedness of the solutions of second-order linear difference equations with periodic coefficients, Vestnik Moskov. Univ. Ser. I Mat. Meh [3] E. Kurpinar, On the boundedness of solutions of second-order difference equation with monotonical coefficients, J. Fac. Sci. Ege Univ. Fen. Fak. Derg. Ser. A B [4] E. Kurpinar, G.Sh. Guseinov, The boundedness of solutions of second-order difference equations, Indian J. Math [5] D.S. Mitrinović, Analytic Inequalities, in cooperation with P.M. Vasić, Grundlehren Math. Wiss., Band 1965, Springer-Verlag, New York, [6] F.W.J. Olver, Error bounds for linear recurrence relations, Math. Comp [7] S. Stević, Growth theorems for homogeneous second-order difference equations, ANZIAM J [8] S. Stević, Asymptotic behavior of second-order difference equations, ANZIAM J [9] S. Stević, Asymptotic behavior of a second-order difference equation, preprint, [10] W.F. Trench, Asymptotic behavior of solutions of a linear second-order difference equation, in: Asymptotic Methods in Analysis and Combinatorics, J. Comput. Appl. Math

10 10 K.S. Berenhaut, E.G. Goedhart / J. Math. Anal. Appl [11] W.F. Trench, Linear perturbations of a nonoscillatory second-order difference equation, J. Math. Anal. Appl [12] W.F. Trench, Linear perturbations of a nonoscillatory second-order difference equation, II, J. Math. Anal. Appl [13] R. Wong, H. Li, Asymptotic expansions for second-order linear difference equations, in: Asymptotic Methods in Analysis and Combinatorics, J. Comput. Appl. Math