Asymptotic properties of solutions of difference equation of the form ∆mxn=anϕn(xσ(n)) +bn are studied

Tomus 46 (2010), 1–11

ASYMPTOTIC PROPERTIES OF SOLUTIONS OF NONAUTONOMOUS DIFFERENCE EQUATIONS

Janusz Migda

Abstract. Asymptotic properties of solutions of difference equation of the form

∆^mxn=anϕn(xσ(n)) +bn

are studied. Conditions under which every (every bounded) solution of the equation ∆^myn =bn is asymptotically equivalent to some solution of the above equation are obtained. Moreover, the conditions under which every polynomial sequence of degree less thanmis asymptotically equivalent to some solution of the equation and every solution is asymptotically polynomial are obtained. The consequences of the existence of asymptotically polynomial solution are also studied.

1. Introduction

Let N, N(0),Z,R denote the set of positive integers, the set of nonnegative integers, the set of all integers and the set of real numbers, respectively.

Letm∈N. In this paper we consider the difference equation of the form (E) ∆^mx_n =a_nϕ_n(x_σ(n)) +b_n

n∈N , an, bn ∈R , ϕn:R→R , σ:N →Z , limσ(n) =∞. By a solution of (E) we mean a sequence x:N →Rsatisfying (E) for all large n.

Letn₀= min{n∈N : σ(k)≥1 for allk≥n}. If (E) is satisfied for alln≥n₀we say thatxis a full solution of (E).

Let (X, d), (Y, ρ) be metric spaces, and let Φ be a family of mapsϕ:X →Y. Φ is said to be equicontinuous at a point p∈ X if for every ε > 0 there exists δ >0 such that ifd(x, p)< δ thenρ ϕ(x), ϕ(p)

< ε for allϕ∈Φ. We say that Φ is equicontinuous if it is equicontinuous at every pointp∈X. If for anyε >0 there existsδ >0 such thatρ ϕ(x1), ϕ(x2)

< εfor any pairx1, x2∈X such that d(x1, x2)< δand allϕ∈Φ, then Φ is said to be uniformly equicontinuous. Φ is said to be locally bounded if for any point p∈X there exist a neighborhood U of p inX and a constant M > 0 such that |ϕ(t)| ≤ M for all t ∈U, ϕ∈ Φ. If

|ϕ(t)| ≤ M for all t ∈X, ϕ ∈Φ then we say that Φ is bounded. If ψ: X →R, U ⊆X thenψ|U denotes the restriction ofψ. We say that sequencesx, y are

2000Mathematics Subject Classification: primary 39A10.

Key words and phrases: difference equation, asymptotic behavior, asymptotically polynomial solution.

Received August 27, 2008. Editor O. Došlý.

asymptotically equivalent ifx_n−y_n =o(1). For a given sequencexof real numbers, by Px_n we denote the series whose partial sums are: x₁,x₁+x₂,x₁+x₂+x₃ and so on. We often use the algebraic notation Axto denote A(x) ifAis a linear operator.

Recently, there has been a great interest in the study of asymptotic and oscillatory behavior of solutions of higher order difference equations, see for example [2] [3], [7]–[12]and the references cited therein. The purpose of this paper is to study the asymptotic behavior of solutions of equation (E). Using the Schauder’s fixed point theorem and some technical results based on the properties of iterated rest operator we show that if the seriesP

n^m−1an is absolutely convergent and the family{ϕn}is equicontinuous and locally bounded (uniformly equicontinuous and bounded) then every bounded solution (every solution) of the equation ∆^myn=bnis asymptotically equivalent to some solution of (E). Moreover, if the seriesP

n^m−1bn

is also absolutely convergent then every polynomial sequence of degree less thanmis asymptotically equivalent to some solution of (E). Similar problem for autonomous difference equation was considered in [4]. We also show that if the seriesPn^m−1a_n and Pn^m−1b_n are absolutely convergent and the family{ϕn} is bounded then every solution of (E) is asymptotically polynomial. On the other hand, under some additional assumptions, we show that if there exists asymptotically polynomial solution of (E) then the seriesPn^m−1a_n andPn^m−1b_nare absolutely convergent.

The results obtained here generalize the results of [1, 6, 7] and some of those contained in [5]. The results obtained here in Theorems 1–4 are analogous to those obtained in [4] for autonomous equations.

2. Preliminary lemmas

In this section we introduce a rest operator and establish some useful properties of their iterations. These results will be used in the proofs of the main theorems in Section 3.

By SQ we denote the space of all sequences x:N →R. If x ∈SQ then |x|

denote the sequence defined by|x|(n) =|xn|for everyn∈N. The Banach space of all bounded sequences x∈SQwith the norm kxk= sup{|xn| : n∈N} we denote by BS. For k ∈ N, Pol(k) denotes the space of all polynomial sequences (with real coefficients) of degree ≤k. We identify every sequenceβ ∈Pol(k) with the corresponding polynomial.

LetS(0) ={x∈SQ : limx_n= 0},

S(1) ={x∈SQ : the series X

x_n is convergent}. Ifx∈S(1), we may define the sequence r(x) by the formula

r(x)(n) =

∞

X

j=n

x_j.

Obviously r(x) ∈ S(0) and the mapping r:S(1) → S(0), which we call the rest operator, is linear. Let S(2) = {x ∈ S(1) : r(x) ∈ S(1)}. Then S(2) is a linear subspace of S(1) and we may define the operator r²:S(2) → S(0) by

r²(x) =r(r(x)). Ifk∈N then, by induction, we define the spaceS(k+ 1) and the operator r^k+1:S(k+ 1)→S(0) by

S(k+ 1) ={x∈S(k) : r^k(x)∈S(1)}, r^k+1(x) =r r^k(x) .

Ifx∈S(k),n∈N then the valuer^k(x)(n) we denote also byr^k_n(x) or simplyr^k_nx.

Moreover, ifx∈SQ, n∈N then we define r_n⁰x=xn. Note that S(k)⊂S(k−1)⊂ · · · ⊂S(1)⊂S(0)

are linear subspaces ofS(0) andr^k: S(k)→S(0) is a linear operator.

For n ∈N we define the numbers s⁰_n = 1, s¹_n =s⁰₁+s⁰₂+· · ·+s⁰_n = n. Ifk, n∈N then, by induction onk, we define numbers

s^k+1_n =s^k₁+s^k₂+. . .+s^k_n.

Lemma 1. If k∈N,|x| ∈S(k) thenx∈S(k)and|r^kx| ≤r^k|x|.

Proof. Induction on k. The casek= 1 is obvious. Assume the assertion is true for some k ≥ 1 and |x| ∈ S(k+ 1). Then r^k|x| ∈ S(1). Moreover by inductive assumption,x∈S(k) and|r^kx| ≤r^k|x|. Hence, by comparison test of convergence of the series,r^kx∈S(1). Therefore,x∈S(k+ 1). Moreover |r^k+1x|=|r(r^kx)| ≤

r(|r^kx|)≤r(r^k|x|) =r^k+1|x|.

Lemma 2. Assume x∈ SQ, k ∈N. Then |x| ∈S(k) if and only if the series Ps^k−1_i x_i is absolutely convergent. If |x| ∈S(k),n∈N then

r^k_n|x|=s^k−1₁ |xn|+s^k−1₂ |xn+1|+s^k−1₃ |xn+2|+. . . .

Proof. We prove this by induction on k. The casek= 1 is obvious. Assume the assertion is true for some k≥1. If|x| ∈S(k+ 1),n∈N then

r^k+1_n |x|=r_n^k|x|+r_n+1^k |x|+r_n+2^k |x|+r^k_n+3|x|+. . .

=s^k−1₁ |xn|+s^k−1₂ |xn+1|+s^k−1₃ |xn+2|+s^k−1₄ |xn+3|+. . . +s^k−1₁ |xn+1|+s^k−1₂ |xn+2|+s^k−1₃ |xn+3|+. . .

+s^k−1₁ |xn+2|+s^k−1₂ |xn+3|+. . . +s^k−1₁ |xn+3|+. . .

=s^k−1₁ |xn|+ (s^k−1₁ +s^k−1₂ )|xn+1|+ (s^k−1₁ +s^k−1₂ +s^k−1₃ )|xn+2|+. . .

=s^k₁|xn|+s^k₂|xn+1|+s^k₃|xn+2|+s^k₄|xn+3|+. . . . Hence

(1) r^k_n|x|+r^k_n+1|x|+r^k_n+2|x|+. . .=s^k₁|xn|+s^k₂|xn+1|+s^k₃|xn+2|+. . . . Forn= 1 we obtain the convergence of the seriesP∞

i=1s^k_ixi. Conversely, assume the series P∞

i=1s^k_ixi is absolutely convergent. Since 0≤s^k−1_i ≤s^k_i for alli∈N, the series P∞

i=1s^k−1_i x_i is absolutely convergent. Hence, by inductive assumption,

|x| ∈ S(k). Let n = 1. By (1), the series P∞

i=1r_i^k|x| is convergent. Therefore r^k|x| ∈S(1). Hence|x| ∈S(k+ 1). The proof is complete.

Lemma 3. Assume x∈SQ, k∈ N(0). Then |x| ∈S(k+ 1) if and only if the series Pn^kx_n is absolutely convergent.

Proof. First we show thats^k_n≤n^k for all n∈N. It is obvious ifk= 0 ork= 1.

Assumes^k_n≤n^k for somek≥1 and anyn∈N. Thens^k+1_n =s^k₁+s^k₂+. . .+s^k_n≤ n^k+n^k+. . .+n^k =nn^k =n^k+1.

Using the known equalityPn i=1

k+i−1 k

= ^k+n_k+1

it is easy to show that s^k_n=

k+n−1 k

= n(n+ 1). . .(n+k−1)

k! .

Hence, n^k ≤ n(n+ 1). . .(n+k−1) = k!s^k_n. Since s^k_n ≤ n^k ≤ k!s^k_n, absolute convergence of the seriesPn^kx_n is equivalent to the absolute convergence of the seriesPs^k_nx_n. The assertion follows now from Lemma 2.

Lemma 4. AssumeM >0,k∈N,a, b∈SQ,|b| ≤M, and the seriesP

n^k−1|an| is convergent. Then ba,|a| ∈S(k)and|r^k(ba)| ≤M r^k|a|.

Proof. By Lemma 3,|a|,|ba| ∈S(k). Hence, by Lemma 1,ba∈S(k) and|r^k(ba)| ≤ r^k|ba|. By Lemma 2,r^k|ba| ≤r^k(M|a|) =M r^k|a|. Hence|r^k(ba)| ≤M r^k|a|.

Lemma 5. If k∈N andx∈S(k), then ∆^kr^kx= (−1)^kx.

Proof. We prove this by induction onk. Ifk= 1, then

∆rx(n) =rx(n+ 1)−rx(n) =

∞

X

k=n+1

xk−

∞

X

k=n

xk =−xn.

Hence ∆rx=−x. Assume the assertion is true for somek≥1 and letx∈S(k+ 1).

Sincer^kx∈S(1), we have ∆r(r^kx) =−r^kx. Hence,

∆^k+1r^k+1x= ∆^k∆rr^kx= ∆^k(−r^kx) = (−1)∆^kr^kx= (−1)^k+1x .

Lemma 6. Let xbe a sequence convergent toc∈R and letk∈N. Then

∆^kx∈S(k), r^k∆^kx= (−1)^k(x−c).

Proof. Letk= 1. Then ∆x₁+∆x₂+. . .+∆x_n=x_n+1−x1. Hence the seriesP∆x_n is convergent i.e., ∆x∈S(1). Moreover ∆x_n+ ∆x_n+1+. . .+ ∆x_p=x_p+1−x_n. Hence r_n¹∆x=c−x_n. Therefore the assertion is true fork= 1. Assume it is true for some k≥1. Since the sequence ∆xis convergent to zero we have

∆^k+1x= ∆^k∆x∈S(k), r^k∆^k+1x=r^k∆^k∆x= (−1)^k∆x .

Since ∆x ∈ S(1), it follows that (−1)^k∆x ∈ S(1). Hence, r^k∆^k+1x ∈ S(1).

Therefore ∆^k+1x∈S(k+ 1). Moreover

r^k+1∆^k+1x=rr^k∆^k∆x=r((−1)^k∆x) = (−1)^kr∆x

= (−1)^k(−1)(x−c) = (−1)^k+1(x−c).

The proof is complete.

Lemma 7. Let k ∈ N and let W = {x ∈ SQ : x_n = 0 for n ≥ k}. Then

∆^m(W) =W.

Proof. It is easy to see that W is a finite dimensional linear subspace of SQ and ∆^m(W)⊆W. Since W∩Ker ∆^m =W ∩Pol(m−1) = 0, the linear opera- tor ∆^m|W:W →W is monomorphic. Hence dim(∆^m(W)) = dimW. Therefore

∆^m(W) =W.

Remark 1. It is easy to see that ifx,z∈SQ,p∈N, andzn=xn for alln≥p then ∆^kz_n= ∆^kx_n forn≥p. Analogously, by easy induction onk, one can show that ifx∈S(k),z∈SQ, andz_n =x_n forn≥pthenz∈S(k) andr^k_nz=r^k_nxfor n≥p.

Lemma 8 ([6]). If X andY are metric spaces,X is compact, andΦis equiconti- nuous family of maps ϕ:X →Y, thenΦis uniformly equicontinuous.

Lemma 9 ([6]). If X, Y are metric spaces, X is compact, and Φ is a locally bounded family of mapsϕ:X→Y, thenΦis bounded.

3. Main results

Theorem 1. Assume the seriesPn^m−1an is absolutely convergent,yis a bounded solution of the equation∆^myn=bn, and Y is the set of values of the sequence y.

If there exists a neighbourhoodU of the closureY such that the family {ϕn|U} is locally bounded and equicontinuous, then there exists a solutionxof (E)such that x=y+o(1).

Proof. Since the setY is bounded, the closureY is compact. Hence, there exists an open set V such that V is compact andY ⊆V ⊆V ⊆ U. Using Lemma 8 and Lemma 9 one can show that the family{ϕn|V}is bounded and uniformly equicontinuous. SinceY is compact, so there exists a numberc >0 such that if

s∈Y , t∈R , |s−t| ≤c

then t∈V. There existsM >0 such that|ϕn(t)| ≤M for allt∈V and alln∈N. Letρ=r^m|a|. By definition ofr^m,ρ=o(1). Choosep∈N such thatM ρn ≤c for any n≥p. Let

T ={x∈BS : xn= 0 for n < p and |xn| ≤M ρn for n≥p}.

ObviouslyT is a convex and closed subset ofBS. Choose anε >0. Then there existsm∈N such thatM ρn< εfor anyn≥m. Forn= 1, . . . , mletGndenote a finiteε-net for the interval [−M ρn, M ρn] and let

G={x∈T : xn∈Gn for n≤m and xn= 0 for n > m}.

ThenGis a finiteε-net forT. HenceT is a complete and totally bounded metric space and so,T is compact. HenceT is a convex and compact subset of the Banach spaceBS. Let

S ={x∈SQ : x_n=y_n for n < p and |x_n−y_n| ≤M ρ_n for n≥p}

and letF:T →S be a map given byF(x)(n) =x_n+y_n. The formulad(x, z) = sup_n∈N|xn−z_n|defines a metric onS such thatF is an isometry ofT ontoS. By

Schauder’s fixed point theorem, every continuous mapB:T →T has a fixed point.

Since the spaceS is homeomorphic toT, every continuous mapA:S→S has a fixed point too.

Let x∈S. Then|x_i−y_i| ≤c for anyi∈N. Hencex_i∈V for alli∈N. Therefore

|ϕ_n(x_i)| ≤M for alln∈N and alli∈N. Hence, ifx∈S then|ϕ_n(x_σ(n))| ≤M for alln≥n₀. Forx∈SQlet ¯xbe defined by

¯ xn =

(0 for n < n0

a_nϕ_n(x_σ(n)) for n≥n₀.

Ifx∈S then|¯x| ≤M|a|. Hence, by Lemma 3,|¯x| ∈S(m) for allx∈S. By Lemma 1, ¯x ∈ S(m) for all x ∈ S. Since limσ(n) = ∞, there exists p1 ≥ p such that σ(n)≥pfor alln≥p1. Forx∈S we define the sequenceA(x) by

A(x)(n) =

(yn for n < p1

yn+ (−1)^mr_n^mx¯ for n≥p1. Then|A(x)−y| ≤ |r^mx| ≤¯ r^m|¯x|. Hence, by Lemma 2,

|A(x)−y| ≤r^m(M|a|) =M ρ . ThereforeA(x)∈S for allx∈S.

Letε >0. Since the family{ϕn|V}is uniformly continuous, there existsδ >0 such that if t, s∈V and|t−s|< δthen|ϕ_n(t)−ϕ_n(s)|< εfor anyn∈N. Let x, z ∈ S, kx−zk < δ. Then x_i, z_i ∈ V and |x_i−z_i| < δ for all i ∈ N. Hence

|ϕ_n(x_i)−ϕ_n(z_i)|< εfor alln∈N and alli∈N. Hence|¯x−z| ≤¯ ε|a|. Therefore kA(x)−A(z)k= sup

n≥p1

|r_n^mx¯−r^m_nz|¯

= sup

n≥p1

|r_n^m(¯x−z)| ≤¯ r^m_p1|¯x−¯z| ≤εr^m_p1|a|=ερp₁.

Hence, the mapping A: S →S is continuous. Therefore there existsx∈S such thatA(x) =x. Thenxn=yn+ (−1)^mr^m_nx¯for any n≥p1. Hence, using Lemma 5 we obtain

∆^mxn= ∆^myn+ ∆^m (−1)^mr^m_nx¯

=bn+ ¯xn=anϕn(x_σ(n)) +bn

for n ≥ p₁. Moreover, since r^mx¯ = o(1), we have x = y+o(1). The proof is

complete.

Corollary 1. If the series Pn^m−1a_n is absolutely convergent, and the family {ϕn}is equicontinuous and locally bounded, then for any bounded solutiony of the equation ∆^my_n=b_n there exists a solution xof(E)such thatx=y+o(1).

Proof. TakeU =R in Theorem 1.

Corollary 2. If the series P

n^m−1an is absolutely convergent, and the family {ϕn} is equicontinuous and bounded, then for any bounded full solutiony of the equation ∆^my=b there exists a full solution xof (E)such thatx=y+o(1).

Proof. ChooseM >0 such that|ϕn(t)| ≤M for alln∈N,t∈R. In the proof of Theorem 1 we can chooseV andc such that c > M ρ₁. Then we can take p= 1,

p₁=n₀.

The next Corollary generalizes Theorem 1 of [7].

Corollary 3. If the seriesPn^m−1an,Pn^m−1bn are absolutely convergent, and the family {ϕn} is equicontinuous and locally bounded, then for anyc∈R there exists a solution x of (E) such that limxn = c. If moreover, the family {ϕn} is bounded, then for every c ∈ R there exists a full solution x of (E) such that limxn=c.

Proof. Let c ∈ R, u = (−1)^mr^mb, y = c+u. Then u = o(1) and ∆^my =

∆^mc+ ∆^mu= ∆^mu=b. Hence y is a bounded solution of the equation ∆^my=b.

By Corollary 1 there exists a solution xof (E) such that x = y+o(1). Hence x=c+u+o(1) =c+o(1). This means that limxn =c. The second assertion

follows from Corollary 2.

Remark 2. Assume f, g: R → R are continuous functions and (αn), (βn) are bounded sequences of real numbers. Then the family {ϕn} defined by ϕn = α_nf+β_ngis equicontinuous and locally bounded. Moreover iff anggare uniformly continuous then the family {ϕn}is uniformly equicontinuous.

Remark 3. Assume f0, f1, . . . , f_p−1: R → R are continuous functions. Then the family {ϕn} defined by ϕjp+k = fk for k = 0,1, . . . , p−1, j = 0,1, . . . is equicontinuous and locally bounded.

Example 1. Assume c₀, c₁, c₂ ∈R. Let b₀ = 9(c₀−c₂), b₁ = 9(c₁−c₀), b₂ = 9(c₀−c₂) and let (c_n), (b_n) be defined by

c0, c1, c2, c0, c1, c2, . . . b0, b1, b2, b0, b1, b2, . . .

respectively. It is easy to see that for everyα∈Rthe sequence (α+cn) is a bounded solution of the equation ∆⁵yn=bn. Hence, by Remark 2 and by Corrollary 1, for every α∈Rthere exists a solutionxof the equation

∆⁵xn= 1 n⁶

1 + 1

n

e^xn+ sinnπ

3

x²_n +bn

such that limx3n =α+c₀, limx3n+1=α+c₁, limx3n+2=α+c₂. Theorem 2. If the seriesP

n^m−1an is absolutely convergent, and the family{ϕn} is bounded and uniformly equicontinuous, then for every full solution y of the equation ∆^myn=bn there exists a full solutionxof (E)such thatx=y+o(1).

Proof. Let ybe a full solution of the equation ∆^my_n=b_n. ChooseM >0 such that |ϕn(t)| ≤M for everyn∈N and everyt∈R. Letρ=r^m|a|,

T ={x∈BS : |x| ≤M ρ}, S={x∈SQ : |x−y| ≤M ρ}.

Let x∈S, u∈SQ,u_n =a_nϕ_n(x_σ(n)) forn≥n₀,A(x) =y+ (−1)^mr^mu. As in the proof of Theorem 1 one can show that there exists x∈S such that A(x) =x.

Then x = y+ (−1)^mr^mu = x+o(1) and ∆^mx = ∆^my +u = b+u. Hence

∆^mx_n =b_n+a_nϕ_n(x_σ(n)) forn≥n₀. Thereforexis a full solution of (E).

The next Corollary generalizes Theorem 2 of [7].

Corollary 4. If the family{ϕn}is bounded and uniformly equicontinuous, and the seriesP

n^m−1an,P

n^m−1bn are absolutely convergent, then for every polynomial β ∈Pol(m−1) there exists a full solutionxof (E)such thatx=β+o(1).

Proof. Let β ∈ Pol(m−1), u = (−1)^mr^mb, y = β +u. Then u = o(1) and

∆^my = ∆^mβ + ∆^mu = ∆^mu = b. Hence y is a full solution of the equation

∆^my=b. By Theorem 2 there exists a full solutionxof (E) such thatx=y+o(1).

Hence x=β+u+o(1) =β+o(1).

Theorem 3. Assumeλ∈R, and the family{ϕn|[λ,∞)} is bounded and uniformly equicontinuous. If the series Pn^m−1a_n,Pn^m−1b_n are absolutely convergent, then for every polynomialβ ∈Pol(m−1)such thatlimβ(n) =∞there exists a solution xof (E) such thatx=β+o(1).

Proof. Assumeβ ∈Pol(m−1), limβ(n) =∞. ChooseM ≥1 such that|ϕn(t)| ≤ M for all t∈[λ,∞) and alln∈N. Letρ=r^m(|a|+|b|). Thenρ=o(1). Choose p∈N such thatβ(n)≥λ+M ρ1 forn≥p. Let

T ={x∈BS : xn= 0 for n < p and |xn| ≤M ρn for n≥p}.

S={x∈SQ : xn =β(n) for n < p and |xn−β(n)| ≤M ρn for n≥p}. Ifx∈S,n≥pthenxn≥β(n)−M ρn ≥λ+M ρ1−M ρn≥λ. Choosep1≥psuch thatσ(n)≥pfor anyn≥p1. Thenxσ(n)≥λfor alln≥p1. Hence|ϕn(xσ(n))| ≤M forn≥p1. Forx∈SQwe define ¯xby

x¯_n=

(0 for n < n₀

anϕn(x_σ(n)) +bn for n≥n0.

Ifx∈S then|¯x| ≤M|a|+|b| ≤M(|a|+|b|). Hence, by Lemma 3,|¯x| ∈S(m) for allx∈S. By Lemma 1, ¯x∈S(m) for all x∈S. Forx∈S let

A(x)(n) =

(β(n) for n < p₁ β(n) + (−1)^mr^m_nx¯ for n≥p1. Then|A(x)−β| ≤ |r^m¯x| ≤r^m|¯x|. Hence, by Lemma 2,

|A(x)−β| ≤r^m(M(|a|+|b|)) =M ρ .

As in the proof of Theorem 1 one can show that there exists x ∈ S such that A(x) =x. Thenxn=β(n) + (−1)^mr^m_nx¯forn≥p1. Hencex=β+o(1) and

∆^mxn= ∆^mβ(n) + ∆^m((−1)^mr_n^mx) = 0 + ¯¯ xn=anϕn(x_σ(n)) +bn

forn≥p₁. The proof is complete.

The proof of the following theorem is analogous to that of Theorem 3 and hence it is omitted.

Theorem 4. Assume there exists λ ∈ R such that the family {ϕn|(−∞, λ]} is bounded and uniformly equicontinuous. If the series P

n^m−1an,P

n^m−1bn are absolutely convergent, then for every polynomialβ ∈Pol(m−1)such thatlimβ(n) =

−∞there exists a solution xof(E)such that x=β+o(1).

The following corollary is an immediate consequence of Theorems 3 and 4.

Corollary 5. Assume the series P

n^m−1an, P

n^m−1bn are absolutely convergent, and there existsλ∈Rsuch that the family{ϕn|(−∞,−λ]∪[λ,∞)}is bounded and uniformly equicontinuous. Then for every nonconstant polynomialβ ∈Pol(m−1) there exists a solution xof(E)such that x=β+o(1).

Theorem 5. Assume the seriesP

n^m−1an,P

n^m−1bn are absolutely convergent and the family {ϕn} is bounded. Then for every solution xof (E) there exist a polynomial β∈Pol(m−1)such thatx=β+o(1). If, moreover,{ϕn} is uniformly equicontinuous, then for every polynomialβ∈Pol(m−1)there exists a full solution xof (E) such thatx=β+o(1).

Proof. Let xbe a solution of (E). Forn≥n0 letun=anϕn(x_σ(n)). The series Pn^m−1an is absolutely convergent and there exists a constant M such that

|ϕn(x_σ(n))| ≤M forn≥n0. Hence the seriesP

n^m−1un is absolutely convergent.

Therefore the series P

n^m−1(un +bn) is also absolutely convergent. Let z = (−1)^mr^m(u+b). Then, by definition ofr^m,z =o(1) and, by Lemma 5, ∆^mz= u+b. Hence there exists k ∈ N such that ∆^mzn = ∆^mxn for all n ≥ k. Let v = ∆^mx−∆^mz. Then vn = 0 forn≥k. By Lemma 7, there exists a sequence w∈SQsuch that ∆^mw=v andwn= 0 forn≥k. Then 0 = ∆^mx−∆^mz−v=

∆^mx−∆^mz−∆^mw= ∆^m(x−z−w). Hencex−z−w=β for someβ ∈Pol(m−1).

Thenx=β+z+w=β+o(1). The second assertion follows from Corollary 4.

The next theorem generalizes Theorem 1 of [1].

Theorem 6. AssumeU is a neighborhood of some c∈R,ε≥0, the sequences (a_n),(b_n)are nonoscillatory and one of the following conditions holds

(a) anbn≥0, for largen,ϕn(t)≥εfor largenand t∈U, (b) anbn≤0, for largen,ϕn(t)≤ −εfor largenandt∈U.

If there exists a solutionxof (E)such that limx_n =c, then the seriesPn^m−1b_n is absolutely convergent. If moreover ε >0, then the seriesPn^m−1a_n is absolutely convergent.

Proof. Assume the condition (a) is satisfied and an ≥0 for all largen. The proof in other cases is similar and will be omitted. Letxbe a solution of (E) such that limx_n=c. Thenx=c+z for somez=o(1). Letu= ∆^mx. Then, by Lemma 6, u= ∆^m(c+z) = ∆^mz∈S(m). There exists p∈N such that

u_n =a_nϕ_n(x_σ(n)) +b_n, ϕ_n(x_σ(n))≥ε , a_n≥0, b_n≥0 forn≥p. Leth=|u| −u. Thenh_n= 0 forn≥p. Hence, by Lemma 3,h∈S(m).

Therefore |u| = u+h ∈ S(m). Hence by Lemma 3 the series Pn^m−1|u_n| is convergent. Since|b_n| ≤ |u_n|for largen, the seriesPn^m−1|b_n|is convergent too.

Now assumeε >0. Thenanϕn(x_σ(n))≤un forn≥p. Hence 0≤an≤ε⁻¹un for n≥p. Therefore the seriesPn^m−1|an| is convergent.

Theorem 7. Assumeλ∈R,ε≥0,U = (λ,∞),(U = (−λ,−∞)) the sequences (a_n),(b_n)are nonoscillatory and one of the following conditions holds

(a) a_nb_n≥0 for largen, ϕ_n(t)≥ε for largenandt∈U, (b) a_nb_n≤0 for largen, ϕ_n(t)≤ −ε for largen andt∈U. If there exists a solutionxof (E) andβ∈Pol(m−1)such that

x=β+o(1), limx_n=∞ (limx_n=−∞),

then the series Pn^m−1b_n is absolutely convergent. If moreover ε >0 then the series Pn^m−1a_n is absolutely convergent.

Proof. Assume the condition (a) is satisfied,an ≥0 for all largenandU = (λ,∞).

The proof in other cases is similar. Letxbe a solution of (E) such thatx=β+z for some β ∈Pol(m−1) andz =o(1). Letu= ∆^mx. Thenu= ∆^m(β+z) =

∆^mβ+ ∆^mz= ∆^mz∈S(m). The rest of the proof is the same as the second part

of the proof of Theorem 6.

Remark 4. It is easy to see that ifxis a convergent sequence such that ∆^mxn≥0 forn≥pthen (−1)^m+k∆^kxn≥0 fork∈ {1,2, . . . , m} andn≥p.

Example 2. Assume k∈N,m= 2k,an>0 forn∈N, the series P

n^m−1an is convergent,c0∈R,

ϕ(t) =

(1 for t≤c0

0 for t > c₀.

By Corollary 5 for every nonconstant polynomial β ∈Pol(m−1) there exists a solutionxof the equation

(E1) ∆^mxn=anϕ(xn)

such that x = β +o(1). By Theorem 1 for every constant c 6= c0 there exists a solution x of (E1) such that limxn = c. We will show the equation (E1) has no solutions convergent to c0. Assumexis a solution of (E1), limxn =c0. Then

∆^mxn =anϕ(xn)≥0 for large n. Hence, by Remark 4, there exists p∈N such that ∆xn ≤0 for all n≥p. Thenxn ≥c0 for n≥p. If xq =c0 for someq ≥p then xn = c0 for all n ≥ q. Hence ∆^mxn = 0 for n ≥ q. On the other hand

∆^mx_n = a_nϕ(x_n) = a_n > 0 for n ≥ q. This is impossible. Hence x_n > c₀ for all n ≥p. Therefore ∆^mx_n = a_nϕ(x_n) = 0 for n≥ p. Hence xis for large n a polynomial sequence convergent toc₀. It follows,x_n=c₀ for largen. As above it is impossible.

Example 3. Assumean>0 forn∈N, and the seriesP

n³an is convergent. By Corollary 1, for everyc∈R there exists a solutionxof the equation

(E2) ∆⁴xn=anx²_n

such that limxn =c. Let A= 8 inf{a⁻¹_n : n∈N}. We will show that if c ≥A then (E2) has no full solutions convergent to c. Assumexis a full solution of (E2) such that limx_n =c≥A. Then ∆⁴x_n=a_nx²_n for alln∈N. Hence ∆⁴x≥0. By Remark 4, ∆x≤0. Thereforex≥c >0. There existsp∈N such thatA= 8a⁻¹_p .

Thenca_p≥8 and

x_p+4−4x_p+3+ 6x_p+2−4x_p+1+x_p= ∆⁴x_p=a_px²_p,

x_p+4= 4x_p+3−6x_p+2+ 4x_p+1−x_p+a_px²_p> a_px²_p−6x_p+2−x_p. Since ∆x≤0, we havexp+2≤xp. Hence−6xp+2≥ −6xp. Therefore

xp+4> apx²_p−6xp−xp =apx²_p−7xp= (apxp−7)xp.

Sincea_px_p≥a_pc≥8, we obtaina_px_p−7≥1. Hencex_p+4> x_p. This contradicts the fact thatxis nonincreasing.

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Faculty of Mathematics and Computer Science A. Mickiewicz University

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