ASYMPTOTIC BEHAVIOUR OF SECOND-ORDER DIFFERENCE EQUATIONS

ANZIAM J. 462004, 57 70 ASYMPTOTIC BEHAVIOUR OF SECOND-ORDER DIFFERENCE EQUATIONS STEVO STEVIĆ Received 9 December, 200; revised 9 September, 2003 Abstract In this paper we prove several growth theorems for second-order difference equations.. Introduction In this paper we study second-order nonlinear difference equations of the form.c x / = d n f.x n / + g n ; n N;. where x n is the desired solution, and c n, d n and g n are given real sequences. In Section 2 we give and cite several auxiliary results which we shall apply in the sections which follow. In Section 3 we study the second-order linear difference equation c n x n+ b n x n + c x = 0; n N;.2 where x n is the desired solution, and b n and c n are given real sequences. We investigate the asymptotic behaviour of the solutions of that equation under some conditions. We were motivated by [2] and [8]. This equation models, for example, the amplitude of oscillation of the weights on a discretely weighted vibrating string [2, p. 5 7]. A presentation of the results on similar problems for second-order differential equations can be found in [4]. In Section 4 we study the asymptotic behaviour of the second-order nonlinear difference equation.. Mathematical Institute of Serbian Academy of Science, Knez Mihailova 35/I, 000 Beograd, Serbia; e-mail: sstevic@ptt.yu and sstevo@matf.bg.ac.yu. c Australian Mathematical Society 2004, Serial-fee code 446-8735/04 57

58 Stevo Stević [2] 2. Auxiliary results For an investigation into the asymptotic behaviour of the solution x n, we need a few auxiliary lemmas. The first one is a discrete variant of the Bellman-Gronwall lemma. The continuous case of this lemma can be found in [3, 4] and [0]. Applications and further generalisations of this lemma can be found, for example, in [7,, 2, 3, 6, 7, 8, 20]. LEMMA 2. [6, p.2]. If x n ; b n ; c n 0; and then x n a n + b n c i x i ; n N; x n a n + b n a i c i e j=i+ b j c j ; n N: COROLLARY 2.2. If x n ; c n 0, c is a positive constant, and then x n c + c i x i ; n N; x n c exp c i ; n N: PROOF. By Lemma 2. we have x n c + c c i e j=i+ c j ; n N: Applying the well-known inequality x e x, x 0, we obtain as desired. x n c +.e ci /e j=i+ c j = c + e j=i c j e j=i+ c j = c exp The following lemma was proved in [20]. c i ;

[3] Asymptotic behaviour of second-order difference equations 59 LEMMA 2.3. If x n ; c n 0, c is a positive constant, p [0; / and x n c + c i x p i ; n N; then =. p/ x n c p +. p/ c i ; n N: The following lemma is a variant of the discrete version of Bihari s inequality [5]. This lemma generalises a discrete inequality in [], see also [6, p. 4]. LEMMA 2.4. Assume that x n, a n, b n and c n are positive sequences, and that a n and b n satisfy the conditions and a n+ a n M; b n+ b n M; n N; 2. x n a n + b n c i g.x i /; n N; 2.2 where the real function g.x/ is continuous, nondecreasing and g.x/ x, forx > 0. Then x n G G.a / + M ln a nb n + b i+ c i ; n ; n 0 ; 2.3 a b where G.u/ = u.ds=g.s// and " { } n 0 = sup j G.a / + M ln a j jb j + b i+ c i G.R + / : a b PROOF. Let R n = b n c ig.x i /, s n = n c ig.x i / and v n = R n + a n. We can write 2.2 in the following form: x n a n + R n, n N.Fromthatweget 0 v n+ v n = b n+.s + c n g.x n // b n s + a n+ a n =.b n+ b n /s + b n+ c n g.x n / + a n+ a n = b n+ b n R n + b n+ c n g.x n / + a n+ a n : 2.4 b n By the mean value theorem we have G.v n+ / G.v n / =.v n+ v n /=g. n /; 2.5

60 Stevo Stević [4] for some n.v n ;v n+ /.From2.4 and 2.5 we obtain G.v n+ / G.v n / = bn+ b n R n + b n+ c n g.x n / + a n+ a n g. n / b n+ b n b n since g.a n / g.v n / g. n /. Summing 2.6 from to n, we obtain G.v n / G.a / + b n + b n+ c n + a n+ a n ; 2.6 g.a n / b i+ b i + b i From the conditions of the theorem we get G.v n / G.a / + M b i+ b i + b i+ b i+ c i + b i+ c i + M a i+ a i : g.a i / a i+ a i a i+ : Since every positive nondecreasing sequence y n satisfies the following inequality: we obtain y i+ y i y i+ yn y dt t = ln y n y ; G.v n / G.a / + M ln a nb n + b i+ c i ; a b andso2.3 follows. LEMMA 2.5 [4, p.28]. Let v n > 0 and assume that the series + u n and + v n converge. Then + u n lim = c lim u i=n i n + v + n n + v = c: i=n i 3. The linear equation case We are now in a position to formulate and to prove the main results in the case of a linear equation. In what follows we exclude the trivial solution from consideration. Observe that the difference equation x n+ 2x n + x = 0 has a general solution in the form x n = an + b for some a; b R. In the following theorem we give

[5] Asymptotic behaviour of second-order difference equations 6 one sufficient condition, such that the difference equation.2 has solutions which approach those of x n+ 2x n + x = 0. An equivalent result was proved in [6, p. 377]. We present here a different proof which follows the lines of the proof given in [4] in the continuous case. The proof essentially appears in [9, Theorem 7.7] but contains a gap. Hence we present here a correct proof. THEOREM 3.. Consider.2 where + i. c i + 2 b i /<+. Then the general solution is asymptotic to an + b as n,wherea or b maybezero,but not both simultaneously. PROOF. Without loss of generality we may suppose c n > 0, n N {0}. Let us write.2 in the following form:.c x / = d n x n : 3. It is clear that d n = b n c n c.lety n = x n+ x n. Then from 3.wehave c n y n c y = d n x n ; n N: 3.2 Summing 3.2 from to n, we obtain x n x = c 0 y 0 + c Now, summing 3.3 from to n,weget d i x i : 3.3 x n = x 0 + c 0 y 0 n c i + n c i i d j x j : j= By the condition of the theorem, c n asn. Therefore lim n n n and the sequences. =c n / and.=n/ n M > 0. c i = =c i are bounded, for example, by

62 Stevo Stević [6] It follows that x n n x 0 n + c 0 x x 0 M + M n = x 0 n + c 0 x x 0 M + M n i n d j x j j= n.n i/ d i x i x 0 + c 0 x x 0 M + M i d i x i i. x 0 + c 0 x x 0 M/ exp M i d i. x 0 + c 0 x x 0 M/ exp M c 0 +3M = M < ; where in the third inequality we applied Corollary 2.2. From 3.4 we obtain d i x i M i d i M + i. c i + b i 2 / 3.4 i d i < : 3.5 By 3.3, we can conclude that there exists lim n.x n x / = a. If this limit is not zero, we have x n an as n. In particular, x n = 0 for sufficiently large n. To ensure that lim n.x n x / is not zero, we may choose x and x 0 such that + c 0 x x 0 M i d i > 0: Further, we shall use the fact that i c j x j z n = x n C + = x n C + x x 2 c c j+ x j+2 j= c i x i x i+ is another solution, linearly independent of x n ; see, for example, [5, p. 60]. Therefore + z n = x n c i x i x i+ i=n

[7] Asymptotic behaviour of second-order difference equations 63 is another solution, linearly independent of x n. It is well-defined since x n = 0for sufficiently large n and x n an as n. By Lemma 2.5, we obtain + =.c i=n i x i x i+ / lim z x n n = lim n n n lim n =n = a lim n =.c n x n x n+ / =.n.n + // = a : Thus the solution az n is asymptotic to as n, and therefore every solution of our difference equation is asymptotic to an + b as n. If y n is an arbitrary solution of.2 and if lim n + y n is finite, then y n = cz n, n N for some c R. Thus if lim n + y n = 0 we obtain c = 0, that is, y n is a trivial solution. In the other cases lim n + y n = and so a = 0. Before formulating the following result we would like to point out that recently W. Trench investigated principal and nonprincipal solutions of the nonoscillatory equation 3. in[9]. Let us investigate what happens in the case of + i d i =+. The simplest case is when d n = =n Þ, Þ.0; 2] and c n = for all n N. THEOREM 3.2. Consider the equation x n+ 2x n + x = d n x n ; 3.6 where d i = c=i Þ, i N, c R, Þ.0; 2]. Then for every solution of 3.6 the asymptotic formula x n = O n c +,forþ = 2, 2 x n = O ne c n2 Þ =.2 Þ/,forÞ.0; 2/ holds. PROOF. Let y n = x n+ x n : As in Theorem 3. we have x n x = y = y 0 + c i x i: 3.7 Þ Now, summing 3.7 from to n, and by a simple calculation we obtain It follows that x n = x 0 + n.x x 0 / + c.n i/ i x i: 3.8 Þ x n x 0 +n x x 0 +n c i x i Þ

64 Stevo Stević [8] and further x n n x 0 n + x x 0 + c i Þ x i x i x 0 + x x 0 + c : i Þ i Applying the discrete Bellman-Gronwall lemma, we obtain x n n. x 0 + x x 0 / exp c i Þ : We have Thus we have i Þ + n dt ; for Þ.; 2]; t Þ dt ; for Þ.0; ]: t Þ i i Þ + ln.n /; From all of the above, the result follows. +.n /2 Þ ; for Þ.; 2/; 2 Þ n t Þ dt = n2 Þ ; for Þ.0; ]: 2 Þ EXAMPLE. Consider the difference equation x n+ 2x n + x = 2 n 2 x n; n : This equation is derived from 3.6 by putting c = 2. Its solution is x n = n 2. EXAMPLE 2. Consider the difference equation x n+ 2x n + x = 6 n 2 x n; n : This equation is derived from 3.6 by putting c = 6. Its solution is x n = n 3.

[9] Asymptotic behaviour of second-order difference equations 65 QUESTION. These two examples motivate us to conjecture that in Theorem 3.2 x n = O n.+ 4 c +/=2 holds. Is it really so and for what c R does this formula hold? These examples show that for a fixed Þ the growth of the solution of 3.6 really depends on the parameter c. THEOREM 3.3. There exists a sequence d n such that lim n + d n = 0, + i d i = + and for some solutions of 3.6, n k x n holds for every k N. PROOF. Consider the equation We know that x n+ 2x n + x = n Þ x n; Þ.0; 2/: x n = x 0 + n.x x 0 / +.n i/ i x i: 3.9 Þ Let x 0 = 0andx =. It is easy to see that in that case x n 0 for every n N. Thus we have x n n, forn N. Applying this in 3.9 we obtain Since x n n + i þ.n i/ i i = n + n Þ dt t þ i Þ n þ ; for þ = þ i Þ 2 : we have that there is a c > 0 such that x n n + c n 3 Þ c n 3 Þ ; 3.0 for all n N. Hence n þ x n,forþ<3 Þ. Applying 3.0 in3.9 we obtain x n n + c.n i/ i i 3 Þ = n + nc Þ i c 2Þ 3 i ; 2Þ 4 from which it follows that there is a c 2 > 0 such that x n n + c 2 n 5 2Þ, for all n N. Repeating the previous procedure and by induction we obtain that for every k N, there is a constant c k > 0 such that x n n + c k n 2k+ kþ c k n.2 Þ/k+ ; for every n N. From this and since Þ.0; 2/ the result follows.

66 Stevo Stević [0] 4. The nonlinear equation case In this section we shall study the asymptotic behaviour of the second-order nonlinear difference equation.. THEOREM 4.. Consider. where a c n Ž>0, n n 0 ; b g n is an arbitrary real sequence; c d n is a real sequence such that + i d i < + ; d f is a real function such that f.x/ L x Þ, x R,forsomeL > 0 and some Þ [0; ]. Then the following asymptotic formula holds: x n = O n + n.n i/ g i as n + : PROOF. Let y n = c n.x n+ x n /. Then from. wehave As in Theorem 3. we can obtain x n = x 0 + c 0 y 0 y n y = d n f.x n / + g n ; n N: 4. n c i + n c i i.d j f.x j / + g j / : By conditions a, d and some simple calculations, we obtain x n x 0 +n c 0 x x 0 M + M.n i/ g i +nm d i f.x i / x 0 +n c 0 x x 0 M + M.n i/ g i +nml d i x i Þ ; where M is an upper bound for the sequence. =c n /. Let A n =.n i/ g i. By the well-known inequality x Þ + x, x R, Þ [0; ],wehave for some c. j= x n + c. + n + A n / + cn d i. x i +/;

[] Asymptotic behaviour of second-order difference equations 67 By Lemma 2. and condition c, we obtain x n + c. + n + A n / + c 2 n. + i + A i / d i e c where c = c 2 e c + i di. From 4.2 we obtain j=i+ j d j c. + n + A n / + c n. + i + A i / d i ; 4.2 x n + c.=n + / + c. + i/ d i ; 4.3 n + na n since A i is nondecreasing. From 4.3, the result follows. REMARK. If Þ>, then the theorem does not hold. EXAMPLE 3. Consider the difference equation x n+ 2x n + x = 2 n x Þ ; 2Þ n n ; Þ>: This equation satisfies all conditions of Theorem 4. except Þ [0; ]. For this equation x n = n 2 is a solution, but x n is not O.n/. Also lim n.x n+ x n / is not finite. EXAMPLE 4. Consider the difference equation x n+ 2x n + x = 6 n x Þ ; 3Þ n n ; Þ>: For this equation x n = n 3 is a solution, but x n is not O.n/. THEOREM 4.2. Consider. where a c n Ž>0, n n 0 ; b g n is a real sequence such that + g i < + ; c d n is a real sequence such that + i Þ d i < +,forsomeþ [0; /; d f is a real function such that f.x/ L x Þ, x R,forsomeL > 0. Then for every solution x n of., x n = O.n/ as n + and the following limit is finite: lim n + c.x n x /:

68 Stevo Stević [2] PROOF. As in the proof of Theorem 4. we have x n x 0 +n c 0 x x 0 M + M.n i/ g i +nml d i x i Þ c + n + n g i + cn d i x i Þ ; for some c > 0. Since + g i < +,wehave x n n c + c d i x i Þ c + c i Þ xi Þ d i : i For Þ [0; /, bylemma2.3 we get =. Þ/ x n n c Þ +. Þ/c i Þ d i =. Þ/ + c Þ +. Þ/c i Þ d i < + ; thus the first part of the theorem follows. From the above we know that there exists M > 0 such that x n Mn, for every n N. Summing 4. from n + ton + p, we obtain Hence y n+p y n = y n+p y n n+p i=n+ n+p i=n+ n+p i=n+ g i + g i + n+p i=n+ n+p g i +M Þ d i f.x i /: i=n+ n+p d i x i Þ i=n+ i Þ d i : By the conditions of the theorem and Cauchy s criteria we obtain the result. REMARK 2. Theorem4.2 is a generalisation of the main result in [8]. The result also holds in the case Þ = see, for example, [, Problem 6.24.40]. Using Corollary 2.2 instead of Lemma 2.3 in the above proof we can prove the theorem in this case. REMARK 3. Example 3 shows that we cannot allow that + i Þ d i =+,for some Þ [0; /. Indeed, in that case d n = 2=n 2Þ and + i Þ d i = + 2=i Þ = +, ifþ [0; /. On the other hand x n = n 2 is a solution such that x n = O.n/.

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