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Periodicity of solutions of nonhomogeneous linear difference equations
Advances in Difference Equations volume 2012, Article number: 195 (2012)
Abstract
Firstly, sufficient conditions for nonexistence of an ωperiodic solution of the equation x(n+1)+{a}_{0}(n)x(n)=b(n) are presented. Then, sufficient conditions under which every solution of the above equation is asymptotically ωperiodic are given. Next, the results obtained for the firstorder difference equation are generalized for the higherorder nonhomogeneous linear difference equation
Finally, the periodic and asymptotically periodic solutions of this equation are investigated. Many examples illustrate the results given.
MSC:39A11, 39A10.
1 Introduction
We consider a class of korder linear difference equations of the form
where {a}_{0}(n)\ne 0, {a}_{k}(n)\ne 0 for each n\in \mathbf{N}.
For the reader’s convenience, we note that the background for difference equations theory can be found, e.g., in the wellknown 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 firstorder linear difference equations. In [6], the same authors studied periodic oscillation of solutions of nonhomogeneous higherorder difference equations. Popenda and Schmeidel (see [7]) considered the linear difference equation {c}_{n}^{r}{y}_{n+r}+\cdots +{c}_{n}^{1}{y}_{n+1}+{c}_{n}^{0}{y}_{n}={d}_{n} 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}={\sum}_{i=0}^{r}{a}_{n}^{(i)}{x}_{n+i} 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 thirdorder difference equations. In [11], asymptotic behavior of solutions of a linear secondorder difference equation was studied by Trench.
For convenience, we adopt the notation for sequences b=(b(n)) and {a}_{i}=({a}_{i}(n)), where i=0,1,2,\dots ,k. Throughout this paper, we assume that {\sum}_{n=k}^{l}a(n)=0 and {\prod}_{n=k}^{l}a(n)=1 for l<k.
We begin with the following basic wellknown definition.
Definition 1 The sequence y:\mathbf{N}\to \mathbf{R} is called ωperiodic if y(n+\omega )=y(n) for all n\in \mathbf{N}. The sequence y is called asymptotically ωperiodic if there exist two sequences u,v:\mathbf{N}\to \mathbf{R} such that u is ωperiodic, {lim}_{n\to \mathrm{\infty}}v(n)=0, and y(n)=u(n)+v(n) for all n\in \mathbf{N}.
It is clear that every constant function is 1periodic.
If a sequence {a}_{0} is {\omega}_{1}periodic and b is {\omega}_{2}periodic in (1), then throughout this paper, ω is the least common multiple of {\omega}_{1} and {\omega}_{2} (\omega =lcm({\omega}_{1},{\omega}_{1})).
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
is given by
Here, sequences {a}_{0}\equiv 1 and b\equiv 0 are 1periodic, but there are 4periodic solutions.
2 Firstorder difference equations
Periodicity of solutions of firstorder 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 in equation (1) and {a}_{1}\equiv 1. Hence, equation (1) takes the following form:
If {a}_{0}\equiv 1, b\equiv 0, then the general solution of (2) is a constant function, then it is 1periodic.
If {a}_{0}\equiv 1, b\not\equiv 0, then the general solution of (2) is
where c is an arbitrary constant. From (3) we see that a necessary and sufficient condition for the existence of ωperiodic solutions of (2) is b being an ωperiodic sequence such that
The general solution of the associated homogeneous equation of (2) is
If {a}_{0}(n)\ne 0 for any n\in \mathbf{N}, then the necessary and sufficient condition for the existence of a nontrivial ωperiodic solution of the homogeneous equation is that {a}_{0} is an ωperiodic sequence and
If these conditions are satisfied, then all the solutions of the homogeneous equation are ωperiodic. We also note that if {a}_{0}(n)=0, for some n\in \mathbf{N}, then x\equiv 0 for large enough n, and this solution is eventually a 1periodic solution.
From (2) we see that if {a}_{0} is ωperiodic, then the necessary condition for the existence of an ωperiodic solution is ωperiodicity of the sequence b.
Example 2 Consider the equation
Sequences {a}_{0}(n)=2+{(1)}^{n} and b(n)=5+{(1)}^{n} are 2periodic. The solution x(n)=2+{(1)}^{n+1} of the above equation is 2periodic, too. Notice that there are not 2periodic solutions of the associated homogeneous equation.
The following example shows us that in the case {a}_{0} is ωperiodic, ωperiodicity of the sequence b is not sufficient for the existence of an ωperiodic solution of (2).
Example 3 Take in (2)
Sequences {a}_{0} and b are 2periodic sequences. The general solution of the above equation
is not a periodic sequence.
Theorem 1 Let {a}_{0} and b be ωperiodic in (2). The following statements then hold true:

(i)
If
{(1)}^{\omega}\prod _{i=0}^{\omega 1}{a}_{0}(i)\ne 1,(4)
then (2) has an ωperiodic solution with the initial condition

(ii)
If
{(1)}^{\omega}\prod _{i=0}^{\omega 1}{a}_{0}(i)=1,\phantom{\rule{2em}{0ex}}\sum _{j=0}^{\omega 1}({(1)}^{\omega j1}\prod _{i=j+1}^{\omega 1}{a}_{0}(i))b(j)=0,
then every solution of (2) is ωperiodic.

(iii)
If
{(1)}^{\omega}\prod _{i=0}^{\omega 1}{a}_{0}(i)=1,\phantom{\rule{2em}{0ex}}\sum _{j=0}^{\omega 1}({(1)}^{\omega j1}\prod _{i=j+1}^{\omega 1}{a}_{0}(i))b(j)\ne 0,
then there is no ωperiodic solution of (2).
Proof The solution of equation (2) is given by
From the above, the result follows immediately. □
Assume that condition (4) holds. It follows from (i) that equation (2) has a unique ωperiodic solution if and only if the homogeneous equation
has not any nontrivial ωperiodic solution.
In [5] Agarwal and Popenda proved that if {a}_{0} is not periodic, then equation (2) can have at most one periodic solution.
Example 4 The equation
has a unique periodic solution x(n)={(1)}^{n}. Here, the general solution
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 ωperiodic solutions (each of them differs on the subsequence ({a}_{0}(3n1))).
Example 5 Let {a}_{0}=(3,1,a,3,1,a,\dots )={(3,1,a)}_{3}, b={(2,4,2)}_{3}. It is easy to check that the sequence x={(2,4,0)}_{3} is a 3periodic 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}^{\ast} be an ωperiodic sequence which fulfills condition (4) and p\in \{0,1,2,\dots ,\omega 1\}. We define the set {S}_{p} as follows:
Theorem 2 Assume that in equation (2) sequences {a}_{0} and b are ωperiodic and condition (4) holds.
If
and
then every equation of the form
has the same ωperiodic solution x as equation (2) independently on {a}_{0}^{\ast}(p) term.
Proof Let x and {x}^{\ast} 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}^{\ast}(0). Because {a}_{0}(i)={a}_{0}^{\ast}(i) for i=0,1,2,\dots ,p1, we get x(i)={x}^{\ast}(i) for i=0,1,2,\dots ,p1. From (6), (8), and x(0)={x}^{\ast}(0), we have x(p)={x}^{\ast}(p)=0. So, x(i)={x}^{\ast}(i) for i=p,p+1,\dots ,\omega 1. By ωperiodicity of x and {x}^{\ast}, x={x}^{\ast}. □
Now, we turn our attention to asymptotical periodicity of the solutions of (2).
Assume that {a}_{0} is ωperiodic, b(n)=c(n)+d(n), where c is ωperiodic and {lim}_{n\to \mathrm{\infty}}d(n)=0. Let y be a solution of the equation
and z be a solution of the equation
Hence, x=y+z is a solution of
Set {a}_{0}(n)={\alpha}_{0}(n). Multiplying both sides of equation (10) by {\prod}_{i=0}^{n\omega [\frac{n}{\omega}]}\frac{1}{{\alpha}_{0}(i)}, we obtain
Summing the above equality from j=0 to n1, we obtain
and
Hence,
Assuming
and letting n\to \mathrm{\infty}, the right side of the above equality tends to some constant c=z(0)+S, then the left one does too. Utilizing littleo notation, we obtain
Hence,
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} be ωperiodic and b(n)=c(n)+d(n), where c is ωperiodic and the series
converges, then there exists an asymptotically ωperiodic solution of equation (2). Moreover, if conditions
hold, then every solution of equation (2) is asymptotically ωperiodic.
The following three examples illustrate the result presented in Theorem 3:

in the first one sequence {a}_{0} is constant;

in the second sequence {a}_{0} is 3periodic;

in the third sequence {a}_{0} is not periodic.
Example 6 Consider the equation
The assumptions of Theorem 2 hold (c(n)=2{(1)}^{n}, d(n)=0). The general solution of the equation x(n)=c+{(1)}^{n+1} is an asymptotically 2periodic sequence.
Example 7 Assume that
in (2) and b(n)=c(n)+d(n), where
Furthermore,
Hence,
All the assumptions of Theorem 2 are satisfied. Therefore, all the solutions of equation (2) are asymptotically 3periodic. This can be easily seen from the general solution of the considered equation, which is given below.
Example 8 Let us put {a}_{0}(n)=\frac{1}{n+1}, b(3n2)=\frac{6n3}{3n1}, b(3n1)=\frac{9n2}{3n}, b(3n)=\frac{3n2}{3n+1} in (2). Hence, the general solution of the associated homogeneous equation
tends to zero. The 3periodic solution of (2) is x=({(1,2,3)}_{3}). Therefore, every solution of (2) is asymptotically 3periodic.
3 Some results for higherorder 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}\in \{0,1,2,\dots ,k\} such that {sup}_{n\in \mathbf{N}}{a}_{{i}_{0}}(n)=\mathrm{\infty} and {sup}_{n\in \mathbf{N}}{a}_{j}(n)<\mathrm{\infty} for j\ne {i}_{0}, j\in \{0,1,2,\dots ,k\}. Let the sequence b be bounded, too. Then equation (1) has not any asymptotically periodic solution x:\mathbf{N}\to \mathbf{R}\setminus \{0\}.
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}\in \{0,1,2,\dots ,k\} such that the sequence ({a}_{{i}_{0}}(n)) is unbounded. Therefore,
is also unbounded. Hence,
is unbounded, while b is bounded. This contradiction completes the proof. □
The sufficient conditions for the existence of an asymptotically ωperiodic solution of equation (1) are given in the following theorem.
Theorem 5 Assume that {a}_{0}:\mathbf{N}\to \mathbf{R}\setminus \{0\}, the condition
holds for each i\in \{1,2,\dots ,k\} and the sequence (\frac{b(n)}{{a}_{0}(n)}) is asymptotically ωperiodic. Then there exists an asymptotically ωperiodic solution of equation (1).
Proof From the periodicity of the sequence (\frac{b(n)}{{a}_{0}(n)}), there exists a positive constant C such that
From condition (11), for any \epsilon >0, there exists a positive integer N such that
Set c=C+\epsilon. We define the sequence α as follows:
Let {B}_{N} be the Banach space of all real bounded sequences x defined for n\ge N, with usually ‘sup’ norm. Set
It is not difficult to prove that S is a nonempty, closed, convex, and compact subset of {B}_{N}. For example, to show that the set S is convex, let us take sequences x,y\in S and a real constant \beta \in [0,1]. Thus, multiplying (12) by β, we obtain
Analogously, for the sequence y, we have
Summing the above inequalities, we get
It means that the set S is convex.
Let us define a mapping T:S\to {B}_{N} as follows:
We show that T(S)\subset S. Indeed, if x\in S, then x(i)<c for large n>N, and
We see that T is continuous. Hence, by the Schauder fixed point theorem, there exists x\in S such that x(n)=(Tx)(n) for n\ge N, so
hence it is a solution of equation (1). Because x\in S, then x is an asymptotically periodic sequence. This completes the proof. □
Example 9 Consider the equation
By Theorem 5, we get that there exists an asymptotically periodic solution of the above equation. In fact, the asymptotically 2periodic sequence
is such a solution.
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Janglajew, K., Schmeidel, E. Periodicity of solutions of nonhomogeneous linear difference equations. Adv Differ Equ 2012, 195 (2012). https://doi.org/10.1186/168718472012195
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DOI: https://doi.org/10.1186/168718472012195