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Advances in Difference Equations

, 2012:195 | Cite as

Periodicity of solutions of nonhomogeneous linear difference equations

  • Klara Janglajew
  • Ewa SchmeidelEmail author
Open Access
Research
Part of the following topical collections:
  1. Progress in Functional Differential and Difference Equations

Abstract

Firstly, sufficient conditions for nonexistence of an ω-periodic solution of the equation x ( n + 1 ) + a 0 ( n ) x ( n ) = b ( n ) Open image in new window are presented. Then, sufficient conditions under which every solution of the above equation is asymptotically ω-periodic are given. Next, the results obtained for the first-order difference equation are generalized for the higher-order nonhomogeneous linear difference equation

i = 0 k a i ( n ) x ( n + i ) = b ( n ) . Open image in new window

Finally, the periodic and asymptotically periodic solutions of this equation are investigated. Many examples illustrate the results given.

MSC:39A11, 39A10.

Keywords

nonhomogeneous linear difference equation asymptotically periodic solution 

1 Introduction

We consider a class of k-order linear difference equations of the form
i = 0 k a i ( n ) x ( n + i ) = b ( n ) , n N = { 0 , 1 , 2 , } , Open image in new window
(1)

where a 0 ( n ) 0 Open image in new window, a k ( n ) 0 Open image in new window for each n N Open image in new window.

For the reader’s convenience, we note that the background for difference equations theory can be found, e.g., in the well-known monograph by Agarwal [1] as well as in those by Elaydi [2], Kelley and Peterson [3] or Kocić and Ladas [4].

The investigation of linear difference equations attracted the attention of many mathematicians. Agarwal and Popenda, in [5], set together various basic statements on the periodicity of the solutions of first-order linear difference equations. In [6], the same authors studied periodic oscillation of solutions of nonhomogeneous higher-order difference equations. Popenda and Schmeidel (see [7]) considered the linear difference equation c n r y n + r + + c n 1 y n + 1 + c n 0 y n = d n Open image in new window and presented sufficient conditions for the existence of an asymptotically constant solution of the above equation. In [8], the conditions which guarantee that the linear difference equation x n + 1 a n x n = i = 0 r a n ( i ) x n + i Open image in new window possesses an asymptotically periodic solution were given by the same authors. In [9], Popenda and Schmeidel studied the linear difference equation, where one of the coefficients is periodic or constant and the others asymptotically approach zero, and obtained sufficient conditions for the existence of asymptotically periodic solutions. Smith (see [10]) investigated oscillatory and asymptotic behavior of solutions of linear third-order difference equations. In [11], asymptotic behavior of solutions of a linear second-order difference equation was studied by Trench.

For convenience, we adopt the notation for sequences b = ( b ( n ) ) Open image in new window and a i = ( a i ( n ) ) Open image in new window, where i = 0 , 1 , 2 , , k Open image in new window. Throughout this paper, we assume that n = k l a ( n ) = 0 Open image in new window and n = k l a ( n ) = 1 Open image in new window for l < k Open image in new window.

We begin with the following basic well-known definition.

Definition 1 The sequence y : N R Open image in new window is called ω-periodic if y ( n + ω ) = y ( n ) Open image in new window for all n N Open image in new window. The sequence y is called asymptotically ω-periodic if there exist two sequences u , v : N R Open image in new window such that u is ω-periodic, lim n v ( n ) = 0 Open image in new window, and y ( n ) = u ( n ) + v ( n ) Open image in new window for all n N Open image in new window.

It is clear that every constant function is 1-periodic.

If a sequence a 0 Open image in new window is ω 1 Open image in new window-periodic and b is ω 2 Open image in new window-periodic in (1), then throughout this paper, ω is the least common multiple of ω 1 Open image in new window and ω 2 Open image in new window ( ω = lcm ( ω 1 , ω 1 ) Open image in new window).

In the paper, we are looking for the periodic solutions of (1) with the period less than or equal to ω. We are not interested in the solutions of (1) with the period greater than ω, but such solutions can exist.

Example 1 The general solution of
x ( n + 4 ) x ( n ) = 0 Open image in new window
is given by
x ( n ) = c + c ( 1 ) n + C cos ( n π 2 ) + C sin ( n π 2 ) . Open image in new window

Here, sequences a 0 1 Open image in new window and b 0 Open image in new window are 1-periodic, but there are 4-periodic solutions.

2 First-order difference equations

Periodicity of solutions of first-order linear nonhomogeneous difference equations was considered by Agarwal and Popenda in [5]. The authors contemplate the class of equations which have the same periodic solutions.

Let k = 1 Open image in new window in equation (1) and a 1 1 Open image in new window. Hence, equation (1) takes the following form:
x ( n + 1 ) + a 0 ( n ) x ( n ) = b ( n ) . Open image in new window
(2)

If a 0 1 Open image in new window, b 0 Open image in new window, then the general solution of (2) is a constant function, then it is 1-periodic.

If a 0 1 Open image in new window, b 0 Open image in new window, then the general solution of (2) is
x ( n ) = c + j = 0 n 1 b ( j ) , Open image in new window
(3)
where c is an arbitrary constant. From (3) we see that a necessary and sufficient condition for the existence of ω-periodic solutions of (2) is b being an ω-periodic sequence such that
j = 0 ω 1 b ( j ) = 0 . Open image in new window
The general solution of the associated homogeneous equation of (2) is
x ( n ) = c ( 1 ) n i = 0 n 1 a 0 ( i ) . Open image in new window
If a 0 ( n ) 0 Open image in new window for any n N Open image in new window, then the necessary and sufficient condition for the existence of a nontrivial ω-periodic solution of the homogeneous equation is that a 0 Open image in new window is an ω-periodic sequence and
( 1 ) ω i = 0 ω 1 a 0 ( i ) = 1 . Open image in new window

If these conditions are satisfied, then all the solutions of the homogeneous equation are ω-periodic. We also note that if a 0 ( n ) = 0 Open image in new window, for some n N Open image in new window, then x 0 Open image in new window for large enough n, and this solution is eventually a 1-periodic solution.

From (2) we see that if a 0 Open image in new window is ω-periodic, then the necessary condition for the existence of an ω-periodic solution is ω-periodicity of the sequence b.

Example 2 Consider the equation
x ( n + 1 ) + ( 2 + ( 1 ) n ) x ( n ) = 5 + ( 1 ) n . Open image in new window

Sequences a 0 ( n ) = 2 + ( 1 ) n Open image in new window and b ( n ) = 5 + ( 1 ) n Open image in new window are 2-periodic. The solution x ( n ) = 2 + ( 1 ) n + 1 Open image in new window of the above equation is 2-periodic, too. Notice that there are not 2-periodic solutions of the associated homogeneous equation.

The following example shows us that in the case a 0 Open image in new window is ω-periodic, ω-periodicity of the sequence b is not sufficient for the existence of an ω-periodic solution of (2).

Example 3 Take in (2)
a 0 ( 2 n 1 ) = 2 , a 0 ( 2 n ) = 1 2 , b ( 2 n 1 ) = 2 , b ( 2 n ) = 1 . Open image in new window
Sequences a 0 Open image in new window and b are 2-periodic sequences. The general solution of the above equation
{ x ( 2 n 1 ) = c 3 + 2 n , x ( 2 n ) = 2 c + 4 4 n , Open image in new window

is not a periodic sequence.

Theorem 1 Let a 0 Open image in new window and b be ω-periodic in (2). The following statements then hold true:
  1. (i)
    If
    ( 1 ) ω i = 0 ω 1 a 0 ( i ) 1 , Open image in new window
    (4)
     
then (2) has an ω-periodic solution with the initial condition
x ( 0 ) = ( 1 ( 1 ) ω i = 0 ω 1 a 0 ( i ) ) 1 [ j = 0 ω 1 ( ( 1 ) ω j 1 i = j + 1 ω 1 a 0 ( i ) ) b ( j ) ] . Open image in new window
(5)
  1. (ii)
    If
    ( 1 ) ω i = 0 ω 1 a 0 ( i ) = 1 , j = 0 ω 1 ( ( 1 ) ω j 1 i = j + 1 ω 1 a 0 ( i ) ) b ( j ) = 0 , Open image in new window
     
then every solution of (2) is ω-periodic.
  1. (iii)
    If
    ( 1 ) ω i = 0 ω 1 a 0 ( i ) = 1 , j = 0 ω 1 ( ( 1 ) ω j 1 i = j + 1 ω 1 a 0 ( i ) ) b ( j ) 0 , Open image in new window
     

then there is no ω-periodic solution of (2).

Proof The solution of equation (2) is given by
x ( n ) = ( ( 1 ) n i = 0 n 1 a 0 ( i ) ) x ( 0 ) + j = 0 n 1 ( ( 1 ) n j 1 i = j + 1 n 1 a 0 ( i ) ) b ( j ) . Open image in new window
(6)

From the above, the result follows immediately. □

Assume that condition (4) holds. It follows from (i) that equation (2) has a unique ω-periodic solution if and only if the homogeneous equation
x ( n + 1 ) + a 0 ( n ) x ( n ) = 0 Open image in new window

has not any nontrivial ω-periodic solution.

In [5] Agarwal and Popenda proved that if a 0 Open image in new window is not periodic, then equation (2) can have at most one periodic solution.

Example 4 The equation
x ( n + 1 ) + 1 2 n x ( n ) = ( 1 ) n + 1 ( 1 1 2 n ) Open image in new window
has a unique periodic solution x ( n ) = ( 1 ) n Open image in new window. Here, the general solution
x ( n ) = c ( 1 ) n 1 2 n ( n 1 ) 2 Open image in new window

of the associated homogeneous equation has not any nontrivial periodic solution.

The following example shows us that there exists a class of equations (2) which have the same ω-periodic solutions (each of them differs on the subsequence ( a 0 ( 3 n 1 ) ) Open image in new window).

Example 5 Let a 0 = ( 3 , 1 , a , 3 , 1 , a , ) = ( 3 , 1 , a ) 3 Open image in new window, b = ( 2 , 4 , 2 ) 3 Open image in new window. It is easy to check that the sequence x = ( 2 , 4 , 0 ) 3 Open image in new window is a 3-periodic solution of (2) independently of the values taken for a.

This leads to the problem of defining the class of equations which have the same periodic solutions.

Let a 0 Open image in new window be an ω-periodic sequence which fulfills condition (4) and p { 0 , 1 , 2 , , ω 1 } Open image in new window. We define the set S p Open image in new window as follows:
S p = { a 0 = { a 0 ( n ) } : a 0 ( i ) = a 0 ( i )  for  i p , a 0 ( i ) a 0 ( i )  for  i = p } . Open image in new window

Theorem 2 Assume that in equation (2) sequences a 0 Open image in new window and b are ω-periodic and condition (4) holds.

then every equation of the form
x ( n + 1 ) + a 0 ( n ) x ( n ) = b ( n ) , a 0 S p , Open image in new window
(9)

has the same ω-periodic solution x as equation (2) independently on a 0 ( p ) Open image in new window term.

Proof Let x and x Open image in new window be the solutions of equations (2) and (9) respectively. The assumptions of Theorem 1 hold for equations (2) and (9), then by (5) and (7), we get that x ( 0 ) = x ( 0 ) Open image in new window. Because a 0 ( i ) = a 0 ( i ) Open image in new window for i = 0 , 1 , 2 , , p 1 Open image in new window, we get x ( i ) = x ( i ) Open image in new window for i = 0 , 1 , 2 , , p 1 Open image in new window. From (6), (8), and x ( 0 ) = x ( 0 ) Open image in new window, we have x ( p ) = x ( p ) = 0 Open image in new window. So, x ( i ) = x ( i ) Open image in new window for i = p , p + 1 , , ω 1 Open image in new window. By ω-periodicity of x and x Open image in new window, x = x Open image in new window. □

Now, we turn our attention to asymptotical periodicity of the solutions of (2).

Assume that a 0 Open image in new window is ω-periodic, b ( n ) = c ( n ) + d ( n ) Open image in new window, where c is ω-periodic and lim n d ( n ) = 0 Open image in new window. Let y be a solution of the equation
y ( n + 1 ) + a 0 ( n ) y ( n ) = c ( n ) Open image in new window
and z be a solution of the equation
z ( n + 1 ) + a 0 ( n ) z ( n ) = d ( n ) . Open image in new window
(10)
Hence, x = y + z Open image in new window is a solution of
x ( n + 1 ) + a 0 ( n ) x ( n ) = c ( n ) + d ( n ) . Open image in new window
Set a 0 ( n ) = α 0 ( n ) Open image in new window. Multiplying both sides of equation (10) by i = 0 n ω [ n ω ] 1 α 0 ( i ) Open image in new window, we obtain
i = 0 n ω [ n ω ] 1 α 0 ( i ) z ( n + 1 ) i = 0 n 1 ω [ n 1 ω ] 1 α 0 ( i ) z ( n ) = i = 0 n ω [ n ω ] 1 α 0 ( i ) d ( n ) . Open image in new window
Summing the above equality from j = 0 Open image in new window to n 1 Open image in new window, we obtain
j = 0 n 1 ( i = 0 j ω [ j ω ] 1 α 0 ( i ) z ( j + 1 ) i = 0 j 1 ω [ j 1 ω ] 1 α 0 ( i ) z ( j ) ) = j = 0 n 1 ( i = 0 j ω [ j ω ] 1 α 0 ( i ) d ( j ) ) , Open image in new window
and
j = 0 n 1 [ Δ ( i = 0 j 1 ω [ j 1 ω ] 1 α 0 ( i ) z ( j ) ) ] = j = 0 n 1 ( i = 0 j ω [ j ω ] 1 α 0 ( i ) d ( j ) ) . Open image in new window
Hence,
( i = 0 n 1 ω [ n 1 ω ] 1 α 0 ( i ) ) z ( n ) = z ( 0 ) i = 0 1 ω [ 1 ω ] 1 α 0 ( i ) + j = 0 n 1 ( i = 0 j ω [ j ω ] 1 α 0 ( i ) ) d ( j ) . Open image in new window
Assuming
j = 0 ( i = 0 j ω [ j ω ] 1 α 0 ( i ) ) d ( j ) + z ( 0 ) i = 0 1 ω [ 1 ω ] 1 α 0 ( i ) = S Open image in new window
and letting n Open image in new window, the right side of the above equality tends to some constant c = z ( 0 ) + S Open image in new window, then the left one does too. Utilizing little-o notation, we obtain
( i = 0 n 1 ω [ n 1 ω ] 1 α 0 ( i ) ) z ( n ) = c + o ( 1 ) . Open image in new window
Hence,
x ( n ) = y ( n ) + c ( i = 0 n 1 ω [ n 1 ω ] ( a 0 ( i ) ) ) + o ( 1 ) ( i = 0 n 1 ω [ n 1 ω ] ( a 0 ( i ) ) ) . Open image in new window

From above, we get sufficient conditions for asymptotical periodicity of the solutions of (2) which are presented in the following theorem.

Theorem 3 Let the sequence a 0 Open image in new window be ω-periodic and b ( n ) = c ( n ) + d ( n ) Open image in new window, where c is ω-periodic and the series
j = 0 ( i = 0 j ω [ j ω ] 1 a 0 ( i ) ) d ( j ) Open image in new window
converges, then there exists an asymptotically ω-periodic solution of equation (2). Moreover, if conditions
( 1 ) ω i = 0 ω 1 a 0 ( i ) = 1 , j = 0 ω 1 ( ( 1 ) ω j i = j + 1 ω 1 a 0 ( i ) ) c ( j ) = 0 Open image in new window

hold, then every solution of equation (2) is asymptotically ω-periodic.

The following three examples illustrate the result presented in Theorem 3:
Example 6 Consider the equation
x ( n + 1 ) x ( n ) = 2 ( 1 ) n . Open image in new window

The assumptions of Theorem 2 hold ( c ( n ) = 2 ( 1 ) n Open image in new window, d ( n ) = 0 Open image in new window). The general solution of the equation x ( n ) = c + ( 1 ) n + 1 Open image in new window is an asymptotically 2-periodic sequence.

Example 7 Assume that
a 0 ( 3 n 2 ) = 2 , a 0 ( 3 n 1 ) = 1 2 , a 0 ( 3 n ) = 1 , Open image in new window
in (2) and b ( n ) = c ( n ) + d ( n ) Open image in new window, where
c ( 3 n 2 ) = 2 , c ( 3 n 1 ) = 0 , c ( 3 n ) = 1 , d ( 3 n 2 ) = 8 n , d ( 3 n 1 ) = 5 4 8 n , d ( 3 n ) = 31 32 8 n . Open image in new window
Furthermore,
j = 1 n | i = 1 j 1 a 0 ( i ) d ( j ) | 5 2 j = 1 n 8 j . Open image in new window
Hence,
j = 1 | i = 1 j 1 a 0 ( i ) d ( j ) | < . Open image in new window
All the assumptions of Theorem 2 are satisfied. Therefore, all the solutions of equation (2) are asymptotically 3-periodic. This can be easily seen from the general solution of the considered equation, which is given below.
{ x ( 3 n 2 ) = c + 2 + 1 4 8 n , x ( 3 n 1 ) = 2 c 2 + 1 2 8 n , x ( 3 n ) = c + 1 + 8 n . Open image in new window
Example 8 Let us put a 0 ( n ) = 1 n + 1 Open image in new window, b ( 3 n 2 ) = 6 n 3 3 n 1 Open image in new window, b ( 3 n 1 ) = 9 n 2 3 n Open image in new window, b ( 3 n ) = 3 n 2 3 n + 1 Open image in new window in (2). Hence, the general solution of the associated homogeneous equation
x ( n ) = c ( n 1 ) ! Open image in new window

tends to zero. The 3-periodic solution of (2) is x = ( ( 1 , 2 , 3 ) 3 ) Open image in new window. Therefore, every solution of (2) is asymptotically 3-periodic.

3 Some results for higher-order equations

In this part, we study equation (1). In the following theorem, sufficient conditions under which equation (1) has no asymptotically periodic solution are given.

Theorem 4 Assume that there exists i 0 { 0 , 1 , 2 , , k } Open image in new window such that sup n N | a i 0 ( n ) | = Open image in new window and sup n N | a j ( n ) | < Open image in new window for j i 0 Open image in new window, j { 0 , 1 , 2 , , k } Open image in new window. Let the sequence b be bounded, too. Then equation (1) has not any asymptotically periodic solution x : N R { 0 } Open image in new window.

Proof Suppose to the contrary that (1) has such an asymptotically periodic solution x. It implies that the sequence x is bounded. Choose i 0 { 0 , 1 , 2 , , k } Open image in new window such that the sequence ( a i 0 ( n ) ) Open image in new window is unbounded. Therefore,
a i 0 ( n ) x ( n + i 0 ) Open image in new window
is also unbounded. Hence,
i = 0 k a i ( n ) x ( n + i ) Open image in new window

is unbounded, while b is bounded. This contradiction completes the proof. □

The sufficient conditions for the existence of an asymptotically ω-periodic solution of equation (1) are given in the following theorem.

Theorem 5 Assume that a 0 : N R { 0 } Open image in new window, the condition
lim n a i ( n ) a 0 ( n ) = 0 Open image in new window
(11)

holds for each i { 1 , 2 , , k } Open image in new window and the sequence ( b ( n ) a 0 ( n ) ) Open image in new window is asymptotically ω-periodic. Then there exists an asymptotically ω-periodic solution of equation (1).

Proof From the periodicity of the sequence ( b ( n ) a 0 ( n ) ) Open image in new window, there exists a positive constant C such that
| b ( n ) a 0 ( n ) | < C . Open image in new window
From condition (11), for any ε > 0 Open image in new window, there exists a positive integer N such that
i = 1 k | a i ( n ) a 0 ( n ) | < ε for  n N . Open image in new window
Set c = C + ε Open image in new window. We define the sequence α as follows:
α ( n ) = i = 1 k | a i ( n ) a 0 ( n ) | . Open image in new window
Let B N Open image in new window be the Banach space of all real bounded sequences x defined for n N Open image in new window, with usually ‘sup’ norm. Set
S = { x ( n ) B N : b ( n ) a 0 ( n ) c α ( n ) x ( n ) b ( n ) a 0 ( n ) + c α ( n )  for  n N } . Open image in new window
(12)
It is not difficult to prove that S is a nonempty, closed, convex, and compact subset of B N Open image in new window. For example, to show that the set S is convex, let us take sequences x , y S Open image in new window and a real constant β [ 0 , 1 ] Open image in new window. Thus, multiplying (12) by β, we obtain
β b ( n ) a 0 ( n ) c β α ( n ) β x ( n ) β b ( n ) a 0 ( n ) + c β α ( n ) . Open image in new window
Analogously, for the sequence y, we have
( 1 β ) b ( n ) a 0 ( n ) c ( 1 β ) α ( n ) ( 1 β ) y ( n ) ( 1 β ) b ( n ) a 0 ( n ) + c ( 1 β ) α ( n ) . Open image in new window
Summing the above inequalities, we get
b ( n ) a 0 ( n ) c α ( n ) β x ( n ) + ( 1 β ) y ( n ) b ( n ) a 0 ( n ) + c α ( n ) for  n N . Open image in new window

It means that the set S is convex.

Let us define a mapping T : S B N Open image in new window as follows:
( T x ) ( n ) = b ( n ) a 0 ( n ) i = 1 k a i ( n ) a 0 ( n ) x ( n + i ) for  n N . Open image in new window
We show that T ( S ) S Open image in new window. Indeed, if x S Open image in new window, then | x ( i ) | < c Open image in new window for large n > N Open image in new window, and
| ( T x ) ( n ) b ( n ) a 0 ( n ) | c i = 1 k | a i ( n ) a 0 ( n ) | = c α ( n ) . Open image in new window
We see that T is continuous. Hence, by the Schauder fixed point theorem, there exists x S Open image in new window such that x ( n ) = ( T x ) ( n ) Open image in new window for n N Open image in new window, so
x ( n ) = b ( n ) a 0 ( n ) i = 1 k a i ( n ) a 0 ( n ) x ( n + i ) for  n N , Open image in new window

hence it is a solution of equation (1). Because x S Open image in new window, then x is an asymptotically periodic sequence. This completes the proof. □

Example 9 Consider the equation
1 ( n + 2 ) ( n + 3 ) x ( n + 3 ) + 1 n + 2 x ( n + 2 ) + 1 n + 1 x ( n + 1 ) + x ( n ) = ( 1 ) n + ( 1 ) n 2 n 2 + 7 n + 4 n ( n + 2 ) 2 + ( 1 ) n + 1 n 4 + 11 n 3 + 43 n 2 + 69 n + 40 ( n + 1 ) 2 ( n + 2 ) ( n + 3 ) 2 . Open image in new window
By Theorem 5, we get that there exists an asymptotically periodic solution of the above equation. In fact, the asymptotically 2-periodic sequence
x ( n ) = ( 1 ) n + ( 1 ) n 1 n Open image in new window

is such a solution.

Notes

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© Janglajew and Schmeidel; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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Authors and Affiliations

  1. 1.Institute of MathematicsUniversity of BiałystokBiałystokPoland

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