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Research | Open | Published:

Asymptotically polynomial solutions of difference equations

Abstract

Asymptotic properties of solutions of a difference equation of the form

Δ m x n = a n f(n, x σ ( n ) )+ b n

are studied. We present sufficient conditions under which, for any polynomial φ(n) of degree at most m1 and for any real s0, there exists a solution x of the above equation such that x n =φ(n)+o( n s ). We give also sufficient conditions under which, for given real sm1, all solutions x of the equation satisfy the condition x n =φ(n)+o( n s ) for some polynomial φ(n) of degree at most m1.

MSC:39A10.

1 Introduction

Let , , denote the set of positive integers, the set of all integers and the set of real numbers, respectively. For p,kZ, let N(p)={p,p+1,}, N(p,k)={p,p+1,,k}.

Let mN(1). In this paper we consider the difference equation of the form

Δ m x n = a n f ( n , x σ ( n ) ) + b n , a n , b n R , f : N × R R , σ : N Z , lim σ ( n ) = .
(E)

By a solution of (E) we mean a sequence x:NR satisfying (E) for all large n. We denote the space of all sequences x:NR by SQ. We denote the Banach space of all bounded sequences xSQ with the norm x=sup{| x n |:nN} by BS. If x, y in SQ, then xy denotes the sequence defined by pointwise multiplication xy(n)= x n y n . Moreover, |x| denotes the sequence defined by |x|(n)=| x n | for every n. We use the symbols ‘big O’ and ‘small o’ in the usual sense, but for aSQ, we also regard o(a) and O(a) as subspaces of SQ. More precisely, let o(1)={xSQ:x is convergent to zero}, O(1)={xSQ:x is bounded}, and for aSQ, let

o(a)=ao(1)= { a x : x o ( 1 ) } ,O(a)=aO(1)= { a x : x O ( 1 ) } .

For mN(1), let Pol(m) denote the space Ker Δ m + 1 , i.e., the space of all polynomial sequences of degree at most m. Now we can define asymptotically polynomial sequences. We say a sequence xSQ is asymptotically polynomial of degree at most m if

xPol(m)+o ( n s )

for some s(,m]. Note that if t,sR, then the condition o( n s )o( n t ) is equivalent to st. Hence Pol(m)+o( n s )Pol(m)+o( n t ) if st. In particular if s<0, then Pol(m)+o( n s )Pol(m)+o(1) and the inclusion is proper. Note also that if kN(0,m) and s(,k], then xPol(m)+o( n s ) if and only if

x n = c m n m + c m 1 n m 1 ++ c k n k +o ( n s )

for some fixed constants c m ,, c k . For kN(1), we use the factorial notation

n ( k ) =n(n1)(nk+1)with  n ( 0 ) =1.

The integer part of real number t is denoted by t.

The purpose of this paper is to study the asymptotic behavior of solutions of equation (E). In Section 2 we present some preliminary results. Main results appear in Sections 3 and 4. We establish sufficient conditions under which, for some natural k and for any φPol(m1) such that φσ=O( n k ), there exists a solution x of (E) such that x=φ+o( n s ). We also give sufficient conditions under which all solutions are asymptotically polynomial. The proofs of main theorems are based on the Schauder fixed point theorem (Theorem 3.1) and on the discrete Bihari-type inequality (Theorem 4.1).

Asymptotically polynomial solutions appear in the theory of both differential and difference equations. Especially in the theory of second-order equations, the so-called asymptotically linear solutions, i.e., asymptotically polynomial solutions of degree at most one, are considered. Asymptotically linear solutions of differential equations are considered, for example, in papers [111]. A historical survey of this topic can be found in [12]. The asymptotic linearity of a solution x, called in some papers ‘property (L)’, usually means (passing over some additional properties of a derivative) one of the following two conditions:

x(t)=at+b+o(1)orx(t)=at+o(t).

In [9] and [10] the condition of the form x(t)=at+o( t d ) for some d(0,1) is also considered.

Asymptotically polynomial solutions of differential equations of higher order appear, for example, in papers [1317]. In [14], Naito presented necessary and sufficient conditions under which some neutral differential equation of order m possesses a solution x such that

lim t x ( t ) t k =a,for some fixed kN(1,m1) and aR.

Note that the condition lim t x(t)/ t k =a can be written in the form x(t)=a t k +o( t k ). In [17], Hasanbulli and Rogovchenko obtained sufficient conditions under which every nonoscillatory solution x of some neutral differential equation of order m has the property

lim t x ( t ) t m 1 =aR, i.e. x(t)=a t m 1 +o ( t m 1 ) .

In [15], Philos, Purnaras and Tsamatos presented sufficient conditions under which, for given mN(1) and kN(0,m1), every solution of the equation x ( m ) (t)=f(t,x(t)) fulfills the condition x(t)= c 0 + c 1 t++ c k t k +o(1). Moreover, they obtained sufficient conditions under which, for every polynomial function φ of degree at most m1, there exists a solution x of this equation such that x(t)=φ(t)+o(1).

Asymptotically linear solutions of difference equations are studied, for example, in papers [1822]. Asymptotic linearity, similarly as in the continuous case, usually means one of the following two conditions:

x n =an+b+o(1)orx(n)=an+o(n).

Asymptotically polynomial solutions of difference equations of higher order appear, for example, in papers [2330]. In [25], Popenda and Drozdowicz presented necessary and sufficient conditions under which the equation Δ m x n = a n f( x n ) has a convergent solution (i.e., a solution that is asymptotically polynomial of degree zero). In [26], Zafer obtained sufficient conditions under which the equation Δ m x n =F(n, x g ( n ) )+ b n has a solution x such that lim n x n / n ( m 1 ) =aR, i.e., x n =a n ( m 1 ) +o( n ( m 1 ) ). It is easy to see that the last condition is equivalent to x n =a n m 1 +o( n m 1 ). In [28] sufficient conditions under which, for any φPol(m1), there exists a solution x of the equation Δ m x n = a n f( x n )+ b n such that x n =φ(n)+o(1) are presented. In [30] sufficient conditions under which every solution x of the equation Δ m x n = a n F(n, x g ( n ) )+ b n has the property x n =φ(n)+o(1) for some φPol(m1) are presented. Moreover, sufficient conditions under which, for every φPol(m1) there exists a solution x of this equation such that x n =φ(n)+o(1), are presented.

2 Asymptotically polynomial sequences

In this section we obtain some technical results which will be used in the next sections. The solutions of ‘the simplest’ difference equation Δ m x n =0 are the polynomial sequences. In Theorem 2.1, which is the main result of this section, we show that if aSQ is sufficiently ‘small’ then the solutions of the ‘equation’ Δ m x n =O(a) are asymptotically polynomial sequences. This result will be used in the proofs of our main Theorems 3.1 and 4.1. Lemmas 2.1, 2.2 and 2.3 are used in the proof of Theorem 2.1, Lemma 2.4 is used in the proof of Theorem 2.2 and Lemma 2.5 is used to justify an important example (see Remark 4.1 and Example 4.2).

Lemma 2.1 Assume that s(1,), mN(1) and Δ m x n =o( n s ). Then x n =o( n s + m ).

Proof Induction on m. Let m=1. Using de l’Hospital theorem, we obtain

lim n Δ n s + 1 n s = lim n ( n + 1 ) s + 1 n s + 1 n 1 n s + 1 = lim n ( 1 + n 1 ) s + 1 1 n 1 = lim n ( s + 1 ) ( 1 + n 1 ) s ( n 2 ) n 2 = s + 1 .

So, by the assumption Δx=o( n s ), we obtain

lim Δ x n Δ n s + 1 =lim Δ x n n s n s Δ n s + 1 =lim Δ x n n s lim n s Δ n s + 1 = 0 s + 1 =0.

Since s>1, the sequence ( n s + 1 ) is increasing to infinity. By the Stolz-Cesaro theorem, we obtain x n =o( n s + 1 ). Hence the assertion is true for m=1. Assume that it is true for some m1, and let Δ m + 1 x n =o( n s ). Then Δ m Δ x n =o( n s ), and by an inductive hypothesis, we get Δ x n =o( n s + m ). Hence, by the first part of the proof, we obtain x n =o( n s + m + 1 ). The proof is complete. □

For kN(0), nN(1), let

s n k = ( n + k 1 k ) = n ( n + 1 ) ( n + k 1 ) k ! = ( n + k 1 ) ( k ) k ! .

Lemma 2.2 Assume that mN(1), and let the series n m 1 x n be absolutely convergent. Then there exists exactly one sequence z such that z=o(1) and Δ m z= ( 1 ) m x. The sequence z is defined by

z n = i 1 = n i 2 = i 1 i m = i m 1 x i m = j = 0 s j + 1 m 1 x n + j = k = n s k n + 1 m 1 x k .

Moreover,

| z n | k = n k m 1 | x k | for any nN(1).

Proof The first assertion is an immediate consequence of Lemma 4 in [31]. The second assertion follows from the inequality s k n + 1 m 1 < k m 1 . □

Lemma 2.3 Assume that u is a positive and nondecreasing sequence, mN(1) and

n = 1 n m 1 u n | a n |<.

Then there exists a sequence wo( u 1 ) such that Δ m w=a.

Proof Since u is positive and nondecreasing, we have n = 1 n m 1 | a n |<. By Lemma 2.2, there exist sequences x,zo(1) such that Δ m x= ( 1 ) m a and Δ m z= ( 1 ) m |ua|. Moreover, using Lemma 2.2, we obtain

| u n x n | = | s 1 m 1 u n a n + s 2 m 1 u n a n + 1 + | s 1 m 1 u n | a n | + s 2 m 1 u n | a n + 1 | + s 1 m 1 u n | a n | + s 2 m 1 u n + 1 | a n + 1 | + s 3 m 1 u n + 2 | a n + 2 | + = z n = o ( 1 ) .

Hence x=o( u 1 ), and we can take w= ( 1 ) m x. □

Theorem 2.1 Assume that mN(1), s(,m1], Δ m xO(a), and

n = 1 n m 1 s | a n |<.

Then xPol(m1)+o( n s ).

Proof Assume that s0. Since Δ m xO(a), we have n m 1 s | Δ m x n |<. Let u n = n s . By Lemma 2.3, there exists a sequence w=o( n s ) such that Δ m w= Δ m x. Then

xwKer Δ m =Pol(m1)andx=xw+wPol(m1)+o ( n s ) .

Let s(0,m1]. Choose kN(1,m1) such that k1<sk. Then

n = 1 n m k 1 n k s | Δ m x|<,

and by Lemma 2.3 there exists w=o( n s k ) such that Δ m k w= Δ m x. Choose zSQ such that Δ k z=w. Since sk>1, so by Lemma 2.1, z=o( n s ). Moreover,

Δ m z= Δ m k Δ k z= Δ m k w= Δ m x.

Hence x=xz+zPol(m1)+o( n s ). The proof is complete. □

Lemma 2.4 Let sR and nN(2). Then

Δ n s = k = 1 ( s k ) n s k .

Proof If |x|<1, then ( 1 + x ) s = k = 0 ( s k ) x k . Hence

n s Δ n s = n s ( ( n + 1 ) s n s ) = ( 1 + 1 n ) s 1= k = 0 ( s k ) n k 1= k = 1 ( s k ) n k ,

and we obtain the result. □

Theorem 2.2 If mN(1) and sR, then

Δ m n s = s ( m ) n s m +o ( n s m ) =O ( n s m ) .
(1)

Proof First we show, by induction on m, that there exists a sequence c such that

Δ m n s = s ( m ) n s m + k = m + 1 c k n s k .
(2)

For m=1 this assertion follows from Lemma 2.4. Assume it is true for some m1. By Lemma 2.1 and by an inductive hypothesis, we obtain sequences c , b, b , b , c and c such that

Δ m + 1 n s = Δ Δ m n s = Δ ( s ( m ) n s m + k = m + 1 c k n s k ) = s ( m ) Δ n s m + k = m + 1 c k Δ n s k = s ( m ) ( ( s m ) n s m 1 + k = 2 b k n s m k ) + k = m + 1 c k ( ( s k ) n s k 1 + j = 2 b j n s k j ) = s ( m + 1 ) n s m 1 + k = 2 b k n s m k + k = m + 1 c k n s k 1 = s ( m + 1 ) n s m 1 + k = m + 2 b k m n s k + k = m + 2 c k 1 n s k = s ( m + 1 ) n s m 1 + k = m + 2 c k n s k .

Hence we obtain (2). The assertion (1) is an easy consequence of (2). □

Lemma 2.5 Assume that s,aR and mN(1). The following conditions are equivalent:

(1) ( n a ) Pol(m)+o ( n s ) ,(2)a<soraN(0,m).

Proof If a<s, then ( n a )o( n s )Pol(m)+o( n s ). If aN(0,m), then

( n a ) Pol(m)Pol(m)+o ( n s ) .

Assume that as and aN(0,m). Obviously, ( n a )o( n s ). Hence the assumption ( n a )Pol(m)+o( n s ) implies the existence of a nonzero constant c such that n a =c n k +o( n k )+o( n s ) for some kN(0,m). If k>a, then dividing by n k , we obtain o(1)=c+o(1), which is impossible. If k<a, then dividing by n a , we have 1=o(1), which is impossible too. Hence, in this case, ( n a )Pol(m)+o( n s ). The proof is complete. □

3 Asymptotically polynomial solutions

In this section we consider the first issue of the Abstract. In Theorem 3.2 we establish sufficient conditions under which, for any φPol(m1) and for any real s0, there exists a solution x of (E) such that x=φ+o( n s ). However, the main result of the section is more general Theorem 3.1 in which we establish sufficient conditions under which, for some natural k and for any φPol(m1) such that φσ=O( n k ), there exists a solution x of (E) such that x=φ+o( n s ). In the second part of the section (Theorems 3.3, 3.4 and Corollary 3.1), we present some consequences of Theorem 3.1.

Theorem 3.1 generalizes Theorem 2 of [28]. The method of the proof of Theorem 3.1 shows certain similarities to the method used in the continuous case in the proof of Theorem 1 of [15].

In this section we regard N×R as a metric subspace of the plane R 2 . Moreover, we assume a,bSQ, mN(1), s(,0] and

n = 1 n m 1 s | a n |<, n = 1 n m 1 s | b n |<.
(3)

Theorem 3.1 Assume that kN(0), g:[0,)[0,) is continuous, f is continuous and

| f ( n , t ) | g ( | t | n k ) for (n,t)N×R.
(4)

Then for any φPol(m1) such that φσ=O( n k ), there exists a solution x of (E) such that

x=φ+o ( n s ) .

Proof Let x be a solution of equation (E), and let φPol(m1), φ(σ(n))=O( n k ). Let us denote y n = x n φ(n). Then Δ m y n = Δ m x n and equation (E) takes the form

Hence, we have to prove that equation (E*) has a solution y such that y=o( n s ). Let n 0 =min{nN:σ(k)1 for all kn}. For ySQ, we define y ¯ and y ˜ by

y ¯ n = { y σ ( n ) + φ σ ( n ) for  n n 0 , 0 for  n < n 0 , y ˜ n = a n f(n, y ¯ n )+ b n .

Let T={ySQ:y1}. Since φ(σ(n))=O( n k ), there exists a constant L such that ( y ¯ n / n k )L for yT. Hence, by the continuity of g, there exists a constant M1 such that g(| y ¯ n |/ n k )M for every yT and every n. Moreover,

| y ˜ n |M| a n |+| b n |M ( | a n | + | b n | )
(5)

for every yT and every n. Let

d n =M j = n s j n + 1 m 1 ( | a j | + | b j | ) .
(6)

Choose p such that p n 0 and d n 1 for np. Let

S= { z SQ : | z n | d n  for  n p  and  z n = 0  for  n < p } .

Then ST and S is a convex subset of the Banach space BS. Moreover, as in the proof of Theorem 1 in [30], one can show that S is compact. If yT, then | y ˜ |M(|a|+|b|). Hence, by (3) we have n = 1 n m 1 | y ˜ n |<. For yT, let

H(y)(n)= { ( 1 ) m j = n s j n + 1 m 1 y ˜ j for  n p , 0 for  n < p .

Then, for yS and np, by (5) and (6), we get

| H ( y ) ( n ) | =| j = n s j n + 1 m 1 y ˜ j | j = n s j n + 1 m 1 | y ˜ j | d n .

Hence H(S)S. Let ε>0. Choose ε 1 >0 and ε 2 >0 such that ε 1 + ε 2 =ε. There exists qp such that 2 d n < ε 2 for nq. Choose α>0 such that

α j = p q s j n + 1 m 1 | a j |< ε 1 .

Let yS, and let j{p,,q}. By the continuity of f at a point (j, y ¯ j ), there exists δ j >0 such that | y ¯ j s|< δ j implies |f(j, y ¯ j )f(j,s)|<α. Let δ=min{ δ p , δ p + 1 ,, δ q }, and let zS, yz<δ. If j{p,,q}, then | y ¯ j z ¯ j |=| y σ ( j ) z σ ( j ) |<δ. Hence

| y ˜ j z ˜ j |= | a j f ( j , y ¯ j ) a j f ( j , z ¯ j ) | <| a j |α

and

j = p q s j n + 1 m 1 | y ˜ j z ˜ j |< j = p q s j n + 1 m 1 | a j |α< ε 1 .

Moreover, by (5) and (6), we have

j = q s j n + 1 m 1 | y ˜ j z ˜ j | j = q s j n + 1 m 1 | y ˜ j |+ j = q s j n + 1 m 1 | z ˜ j | d q + d q < ε 2 .

Hence

H ( y ) H ( z ) = sup n p | H ( y ) ( n ) H ( z ) ( n ) | = sup n p | j = n s j n + 1 m 1 ( y ˜ j z ˜ j ) | j = p s j n + 1 m 1 | y ˜ j z ˜ j | j = p q s j n + 1 m 1 | y ˜ j z ˜ j | + j = q s j n + 1 m 1 | y ˜ j z ˜ j | < ε 1 + ε 2 = ε .

This shows that the map H is continuous. By the Schauder fixed point theorem, there exists a sequence yS such that H(y)=y. Then y n = ( 1 ) m j = n s j n + 1 m 1 y ˜ j for np. By Lemma 2.2 we obtain

Δ m y n = y ˜ n = a n f(n, y ¯ n )+ b n = a n f(n, y σ ( n ) + φ σ ( n ) )+ b n

for np. Hence y is a solution of (E*). Moreover, by (5) we have

| Δ m y n | =| y ˜ n |M ( | a n | + | b n | ) .

Hence Δ m y=O(|a|+|b|), and by Theorem 2.1 we obtain yPol(m1)+o( n s ). Moreover, s0 and y=o(1). Hence y=o( n s ). The proof is complete. □

Remark 3.1 Let k 1 denote the greatest natural number such that for every polynomial φPol( k 1 ), there exists a solution x of (E) such that x=φ+o(1). Note that if in Theorem 3.1 σ(n)=np for some fixed integer p, then the condition φ(σ(n))=O( n k ) takes the form φ(n)=O( n k ). Hence k 1 =min(k,m1). But if the sequence σ is of another form, then k 1 may be greater than k. In the following example, we have k=1 and k 1 =2.

Example 3.1 Let m=3, s=0, k=1, σ(n)= n , b n =0, g(x)=x ,

f(n,x)= x n , a n = 6 n σ ( n ) ( σ ( n ) + 1 ) ( n + 3 ) ( 3 ) .

Then the conditions of Theorem 3.1 are satisfied and equation (E) takes the form

Δ 3 x n = 6 n ( n + 1 ) ( n + 3 ) ( 3 ) x n .
(7)

Note, that for a polynomial φ, the condition φ( n )=O(n) is equivalent to the condition φPol(2). Thus, by Theorem 3.1, for any φPol(2), equation (7) has a solution ( x n ) such that x n = φ n +o(1). One such solution is x n = n 2 + 1 n .

Theorem 3.2 Assume that σ(n)=O(n), g:[0,)[0,) is continuous, f is continuous and

| f ( n , t ) | g ( | t | n m 1 ) for (n,t)N×R.

Then, for any φPol(m1), there exists a solution x of (E) such that x=φ+o( n s ).

Proof Let φPol(m1). Choose M, K such that |φ(n)/ n m 1 |M and σ(n)/nK. Then

| φ ( σ ( n ) ) n m 1 |=| φ ( σ ( n ) ) σ ( n ) m 1 σ ( n ) m 1 n m 1 |M K m 1 .

Hence φσ=O( n m 1 ). Now the assertion follows from Theorem 3.1.

We say that a function f:N×RR is locally equibounded if for every tR, there exists a neighborhood U of t such that f is bounded on N×U. Obviously, every bounded function f:N×RR is locally equibounded. □

Example 3.2 Let f 1 (n,t)=t and f 2 (n,t)=n. Then f 1 is continuous, unbounded and locally equibounded, f 2 is continuous but not locally equibounded.

Example 3.3 Let g:RR be continuous, and let f(n,t)=g(t). Then f is continuous and locally equibounded. If g 0 ,, g p 1 :RR are continuous and f(n,t)= g n mod p (t), then f is continuous and locally equibounded.

Example 3.4 Assume that g,h:RR are continuous, α,βSQ are bounded, and let f(n,t)= α n g(t)+ β n h(t). Then f is continuous and locally equibounded.

Theorem 3.3 If cR, U is a neighborhood of c and the function f|N×U is bounded and continuous, then there exists a solution x of (E) such that x=c+o( n s ). Moreover, if f is continuous and locally equibounded, then for any cR, there exists a solution x of (E) such that x=c+o( n s ).

Proof Choose α,βR such that α<c<β and [α,β]U. Let

f 1 :N×RR, f 1 (n,t)= { f ( n , α ) for  t α , f ( n , t ) for  t ( α , β ) , f ( n , β ) for  t β .

Then f 1 is continuous and bounded. Choose M>0 such that | f 1 (n,t)|M for any (n,t). Let g:[0,)[0,), g(t)=M, and let km1. Then

| f 1 ( n , t ) | M=g ( | t | / n k )

for any (n,t)N×R. Hence, by Theorem 3.1, there exists a solution x of the equation

Δ m x n = a n f 1 (n, x σ ( n ) )+ b n

such that x=c+o( n s ). Since σ(n), x σ ( n ) (α,β) for large n. Hence f 1 (n, x σ ( n ) )=f(n, x σ ( n ) ) for large n. Therefore x is a solution of (E). The second assertion is an easy consequence of the first one. □

Theorem 3.4 Assume that σ(n)=O( n p ) for some p>0, λ(0,), φPol(m1) and that at least one of the following conditions is satisfied:

  1. (1)

    f is continuous and bounded on N×[λ,) and φ(n),

  2. (2)

    f is continuous and bounded on N×(,λ] and φ(n),

  3. (3)

    f is continuous and bounded on N×((,λ][λ,)) and φ is nonconstant.

Then there exists a solution x of (E) such that x=φ+o( n s ).

Proof

Assume (1) is satisfied. Let

f 1 :N×RR, f 1 (n,t)= { f ( n , λ ) for  t λ , f ( n , t ) for  t > λ .

Then f 1 is continuous and bounded. Choose M>0 such that | f 1 (n,t)|M for any (n,t). Let g:[0,)[0,), g(t)=M, and let k(m1)p. Then φσ=O( n k ) and

| f 1 ( n , t ) | M=g ( | t | / n k )

for any (n,t)N×R. Then, by Theorem 3.1, there exists a solution x of the equation

Δ m x n = a n f 1 (n, x σ ( n ) )+ b n

such that x=φ+o( n s ). Moreover, x σ ( n ) >λ for large n. Therefore x is a solution of (E). The assertion (1) is proved. (2) is analogous to (1) and (3) is a consequence of (1) and (2). □

Corollary 3.1 Assume that σ(n)=O( n p ) for some p>0, cR, λ(0,), φPol(m1), g:RR and that at least one of the following conditions is satisfied:

  1. (1)

    g is continuous on some neighborhood of c and φ=c,

  2. (2)

    g is continuous and bounded on [λ,) and φ(n),

  3. (3)

    g is continuous and bounded on (,λ] and φ(n),

  4. (4)

    g is continuous and bounded on (,λ][λ,) and φ is nonconstant.

Then there exists a solution x of the equation

Δ m x n = a n g( x σ ( n ) )+ b n
(E1)

such that x=φ+o( n s ).

Proof This is a consequence of Theorems 3.3 and 3.4. □

Example 3.5 Assume that ( 1 ) m a n >0, b n =0, s=0, cR and

g(t)= { 1 for  t c , 0 for  t > c .

By Corollary 3.1, for every nonconstant polynomial φPol(m1), there exists a solution x of (E1) such that x=φ+o(1). Moreover, by Corollary 3.1, for every real c 1 c, there exists a solution x of (E1) such that x= c 1 +o(1). We will show that a solution of (E1), which is convergent to c, does not exist. Assume that m is even. Then a n >0. Let x be a solution of (E1) such that lim x n =c. Then

Δ Δ m 1 x n = Δ m x= a n g( x σ ( n ) )0

for large n. Moreover, Δ m 1 x n = Δ m 1 (c+o(1))=o(1). Hence Δ m 1 x n 0 for large n. Thus Δ Δ m 2 x n 0 for large n and so on. After (m1)-steps, we obtain Δ x n 0 for large n. Choose p such that Δ x n 0 for np. Then x n c for np. If x m =c for some mp, then x n =c for every nm and Δ m x n =0 for nm. On the other hand, if σ(n)m, we obtain Δ m x n = a n g( x σ ( n ) )= a n g(c)= a n >0, which is impossible. Hence x n >c for np. Choose p 1 such that σ(n)p for every n p 1 . Then g( x σ ( n ) )=0 for n p 1 . Hence Δ m x n =0 for n p 1 , and there exists a polynomial sequence φPol(m1) such that x n =φ(n) for n p 1 . But the only polynomial φPol(m1) which satisfies the condition φ(n)=c+o(1) is the constant polynomial φ(n)=c. Hence we again obtain x n =c for large n, which is impossible. Similarly, if m is odd, one can show that a solution of (E1), which is convergent to c, does not exist.

4 Approximations of solutions

In this section we consider the second issue of the Abstract. In Theorem 4.1 we establish sufficient conditions under which, for given real sm1, all solutions x of (E) satisfy the condition x=φ+o( n s ) for some φPol(m1). In the second part of the section, we present some consequences of Theorem 4.1. Moreover, in Example 4.2 we show that the assertion of Theorem 4.1 is in some sense optimal.

Theorem 4.1 generalizes Theorem 4 of [26]. The way Theorem 4.1 is proved partially resembles the methods used in the continuous case in the proofs of Theorem 1 in [5] and Theorem 2 in [15].

In this section we assume a,bSQ, mN(1), s(,m1] and

n = 1 n m 1 s | a n |<, n = 1 n m 1 s | b n |<.
(8)

Lemma 4.1 Assume that φ:[0,)[0,) is a continuous and nondecreasing function such that φ(t)>0 for t1 and 1 φ ( s ) 1 ds=. Let c>0 and let a, u be sequences of nonnegative real numbers such that

n = 1 a n <and u n c+ i = 1 n 1 a n φ( u n ) for any n.

Then the sequence u is bounded.

Proof The assertion is an easy consequence of Theorem 1 in [32]. □

The next lemma can be found in [33].

Lemma 4.2 Let ( a n ) be a sequence of real numbers, and let p,mN(1), nN(p+1). Then

i m = p n 1 i m 1 = p i m 1 i 1 = p i 2 1 a i 1 = k = p n 1 ( n k 1 ) m 1 ̲ ( m 1 ) ! a k .

Theorem 4.1 Assume that σ(n)n, g:[0,)[0,) is nondecreasing and continuous, f is continuous, and assume that

| f ( n , t ) | g ( | t | n m 1 ) for all (n,t)N×R, 1 d t g ( t ) =
(9)

and x is a solution of (E). Then xPol(m1)+o( n s ).

Proof Assume that x is a solution of (E). Then, by (9) for large n, we have

| Δ m x n | | a n |g ( | x σ ( n ) | n m 1 ) +| b n |.
(10)

From the identity x n = x 1 + i = 1 n 1 Δ x i , we get | x n || x 1 |+ i = 1 n 1 |Δ x i |. Similarly, |Δ x i ||Δ x 1 |+ j = 1 i 1 | Δ 2 x j |. Hence we have

| x n || x 1 |+ i = 1 n 1 ( | Δ x 1 | + j = 1 i 1 | Δ 2 x j | ) i = 1 n 1 ( | x 1 | + | Δ x 1 | + j = 1 i 1 | Δ 2 x j | ) .

Analogously, | Δ 2 x j || Δ 2 x 1 |+ k = 1 j 1 | Δ 3 x k | and then

| x n | i = 1 n 1 ( | x 1 | + | Δ x 1 | + j = 1 i 1 ( | Δ 2 x 1 | + k = 1 j 1 | Δ 3 x k | ) ) i = 1 n 1 j = 1 i 1 ( | x 1 | + | Δ x 1 | + | Δ 2 x 1 | + k = 1 j 1 | Δ 3 x k | ) ,

and so on. After m1 steps, we get

| x n | i m = 1 n 1 i m 1 = 1 i m 1 i 2 = 1 i 3 1 ( | x 1 | + | Δ x 1 | + + | Δ m x 1 | + i 1 = 1 i 2 1 | Δ m x i 1 | ) .

Let A=| x 1 |+|Δ x 1 |++| Δ m x 1 |. By Lemma 4.2 we obtain

| x n | k = 1 n 1 ( n k 1 ) m 2 ̲ ( m 2 ) ! ( A + j = 1 n 1 | Δ m x j | ) .

Using the inequality

( n k 1 ) m 2 ̲ ( m 2 ) ! n m 2 ,

we have | x n | n m 1 (A+ j = 1 n 1 | Δ m x j |). Hence

| x σ ( n ) | n m 1 | x σ ( n ) | σ ( n ) m 1 A + j = 1 σ ( n ) 1 | Δ m x j | A + j = 1 n 1 | Δ m x j | A + j = 1 n 1 ( | a j | g ( | x σ ( j ) | j m 1 ) + | b j | ) B + j = 1 n 1 | a j | g ( | x σ ( j ) | j m 1 ) ,

where B=A+ j = 1 | b j |. By Lemma 4.1, the sequence (| x σ ( n ) |/ n m 1 ) is bounded. Hence, by (10) and the continuity of g, there exists a constant M1 such that

| Δ m x n | | a n |M+| b n |M ( | a n | + | b n | )

for all large n. Hence Δ m x=O(|a|+|b|), and from Theorem 2.1, we obtain xPol(m1)+o( n s ). The proof is complete. □

Corollary 4.1 Assume that σ(n)n, α(0,1], pα(m1) and x is a solution of the equation

Δ m x n = a n | x σ ( n ) | α n p + b n .

Then xPol(m1)+o( n s ).

Proof Let g:[0,)[0,), g(t)= t α and f(n,t)= | t | α / n p . Then

| f ( n , t ) | =f(n,t)= | t | α n p | t | α n α ( m 1 ) = ( | t | n m 1 ) α =g ( | t | n m 1 ) .

Hence the assertion follows from Theorem 4.1. □

Corollary 4.2 Assume that σ(n)n, α(0,1], pα(m1),

n = 1 n m 1 s + p | a n |<, n = 1 n m 1 s | b n |<

and x is a solution of the equation Δ m x n = a n | x σ ( n ) | α + b n . Then xPol(m1)+o( n s ).

Proof If a n = n p a n , then n = 1 n m 1 s | a n |< and x is a solution of the equation

Δ m x n = a n | x σ ( n ) | α n p + b n .

Hence xPol(m1)+o( n s ) by Corollary 4.1. □

Example 4.1 Let m2, s=1,

f(n,x)=0,g(x)=1, a n =0, b n = ( 1 ) m 1 ( m 1 ) ! ( n + m ) m ̲ .

Then the conditions of Theorem 4.1 are satisfied and equation (E) takes the form

Δ m x n = ( 1 ) m 1 ( m 1 ) ! ( n + m ) m ̲ .
(11)

The general solution x of (11) can be written in the form

x n = c m 1 n m 1 + c m 2 n m 2 ++ c 1 n+ c 0 + k = 1 n 1 1 k .

Using the formula k = 1 n 1 k 1 =lnn+o(1)=o(n), we obtain xPol(m1)+o(n). Note that xPol(m1)+o(1).

Remark 4.1 If the assumptions of Theorem 4.1 are satisfied, then every solution of (E) is an element of the space

Pol(m1)+o ( n s ) .

We will show that for every ε>0, there exist sequences a, b and a function f such that the assumptions of Theorem 4.1 are satisfied and equation (E) has a solution x such that

xPol(m1)+o ( n s ε ) .

Example 4.2 Let mN(1), σ(n)=n, ε>0, 0<λ<ε, s(,m1], sλN(0), b n =0, β=sλ, μ=ms1+λ, 0<α<λ/μ, x n = n β ,

g(t)= t α ,f(n,t)=g ( | t | n m 1 ) = ( | t | n m 1 ) α and a n = n α μ Δ m n β .

Since 0<α<1, we have 1 g 1 (t)dt=. By Lemma 2.5,

xPol(m1)+o ( n s ) andxPol(m1)+o ( n s ε ) .

By Theorem 2.2,

n m s 1 a n = n m s 1 n α μ Δ m n β = n m s 1 + α μ O ( n β m ) =O ( n α μ + β s 1 ) .

Moreover, αμ+βs1<λ+βs1=1. Hence n m s 1 a n =O( n δ ) for some δ<1. Therefore n = 1 n m s 1 | a n |<. Moreover,

a n f(n, x n )+ b n = n α μ Δ m n β ( n 1 m n β ) α = n α ( μ + 1 m + β ) Δ m n β = n 0 Δ m n β = Δ m x n .

Hence x is a solution of the equation Δ m x n = a n f(n, x n )+ b n .

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The author would like to thank the referees for their valuable suggestions and comments.

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Correspondence to Janusz Migda.

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Keywords

  • difference equation
  • asymptotically polynomial solution
  • prescribed asymptotic behavior