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On the solutions of second order generalized difference equations

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Abstract

In this article, the authors discuss 2 ( ) and c 0 ( ) solutions of the second order generalized difference equation

Δ 2 u(k)+f ( k , u ( k ) ) =0,k[a,),a>0

and we prove the condition for non existence of non-trivial solution where Δ u(k)=u(k+)u(k) for >0. Further we present some formulae and examples to find the values of finite and infinite series in number theory as application of Δ .

MSC:39A12, 39A70, 47B39, 39B60.

1 Introduction

Difference equations usually describe the evolution of some certain phenomena over time and are also important in describing dynamics for fundamentally discrete system, see [1]. For example, in the numerical integration, the standard approach is to use the difference equations. Similarly, the population dynamics have discrete generations; the size of the (k+1)st generation u(k+1) is a function of the k th generation u(k). This can be expressed as difference equation of the form

u(k+1)=f ( u ( k ) ) ,

see for example [2]. Further, the concept of difference equations with many examples in applications such as asymptotic behavior of solutions of difference equations were studied extensively by Elaydi [3] where the analytic and geometric approaches were also combined in order to studying difference equations. Further, in [3], both classical and modern treatment of the difference equations were presented in excellent form. For related results on difference equations, see [48]. In the present article, we study 2 ( ) and c 0 ( ) solutions of the following second order generalized difference equation

Δ 2 u(k)+f ( k , u ( k ) ) =0,k[a,),a>0,
(1)

where Δ u(k)=u(k+)u(k) for >0. We provide some related definitions and development for the present article.

The basic theory of difference equations is based on the operator Δ defined as

Δu(k)=u(k+1)u(k),kN,
(2)

where N={0,1,2,3,,}. Even though many authors [14] have suggested the definition of Δ as

Δu(k)=u(k+)u(k),kN,R{0}
(3)

and there are several research took place on this line. By defining Δ and its inverse Δ 1 , many interesting results and applications in number theory as well as in fluid dynamics can be obtained. By extending the study for sequences of complex numbers and to be real, some new qualitative properties like rotatory, expanding, shrinking, spiral and weblike structures were studied for the solutions of difference equations involving Δ . For similar results, we refer to [913].

In particular, the 2 and c 0 solutions of second order difference equations of (1) when =1, were discussed in [8]. In this article, we discuss 2 ( ) and c 0 ( ) solutions for the second order generalized difference Equation (1) and present some applications of Δ in the finite and infinite series of number theory. Throughout this article, we use the following notation:

  1. (i)

    [k] denotes the integer part of k,

  2. (ii)

    N={0,1,2,3,}, N(a)={a,a+1,a+2,},

  3. (iii)

    N (j)={j,j+,j+2,} and R is the set of all real numbers.

2 Preliminaries

In this section, we present some of the preliminary definitions and related results which will be useful for future discussion. The following three definitions held in [9].

Definition 2.1 Let u:[0,)C and (0,) then, the generalized difference operator Δ is defined as

Δ u(k)=u(k+)u(k).
(4)

Similarly, the generalized difference operator of the r th kind is defined as

Δ r = Δ ( Δ r 1 ) ifr2.
(5)

Definition 2.2 For arbitrary x,yR the h-factorial function is defined by

x h ( y ) = h y Γ ( x h + 1 ) Γ ( x h + 1 y ) ,
(6)

where Γ is the Euler gamma function. Note that when x=k, h=, y=nN(1) Definition 2.2 coincides with Definition 2.1.

Definition 2.3 Let u(k), k[0,) be a real or complex valued function and (0,). Then, the inverse of Δ denoted by Δ 1 and defined as follows.

If Δ v(k)=u(k),thenv(k)= Δ 1 u(k)+ c j ,
(7)

where c j is a constant for all k N (j), j=k k .

Definition 2.4 The generalized polynomial factorial for >0 is defined as

k ( n ) =k(k)(k2) ( k ( n 1 ) ) .
(8)

Lemma 2.5 If>0andn N (1)then,

Δ 1 k ( n ) = 1 ( n + 1 ) ( k ) ( n + 1 ) + c j
(9)

for allk N (j), j=k k and c j is constant.

Lemma 2.6 ([13] Product formula)

Letu(k)andv(k)be any two functions. Then

Δ { u ( k ) v ( k ) } = u ( k + ) Δ v ( k ) + v ( k ) Δ u ( k ) = v ( k + ) Δ u ( k ) + u ( k ) Δ v ( k ) , k N ( a ) .
(10)

Lemma 2.7 ([12])

Let>0, nN(2), k(,)and k ( n ) 0. Then,

Δ 1 1 k ( n ) = 1 ( n 1 ) ( k ) ( n 1 ) + c j .
(11)

Definition 2.8 A function u(k), k[a,) is said to be in the space 2 ( ) , if

γ = 0 |u(a+j+γ) | 2 <for allj[0,).
(12)

If lim r |u(a+j+r)|=0, for all 0j< then u(k) is said to be in the space c 0 ( ) .

Lemma 2.9 ([9] Summation formula of finite series)

If real valued functionu(k)is defined for allk[0,), then

Δ 1 u(k)= r = 1 k u(kr)+ c j ,
(13)

where c j is a constant for allk N (j), j=k k . Since[0,)= 0 j < N (j), each complex number c j , (0j<) is called an initial value ofk N (j). Usually, each initial value c j is taken from any one of the valuesu(j), u(j+), u(j+2), etc.

Lemma 2.10 (Summation formula of infinite series)

If lim k u(k)=0and>0, then

Δ 1 u(k)= r = 0 u(k+r).
(14)

Proof Assume z(k)= r = 0 u(k+r). Then,

Δ z(k)= r = 0 u(k++r) r = 0 u(k+r)=u(k).

Now, the proof follows from lim k u(k)=0 and Definition 2.3. □

Theorem 2.11 If lim k u(k)=0and>0, then

Δ 2 u(k)= r 1 = 0 r 2 = 0 u(k+ r 1 + r 2 ).
(15)

Proof The proof follows by taking Δ 1 on (14). □

Corollary 2.12 Letk[,)and(0,). Then

Δ 1 1 k ( k ) = 1 ( k )

and hence

r = 0 1 ( k + r ) ( k + r ) = 1 ( k ) .
(16)

Proof The proof follows from Equation (14) and c j =0 as k. □

The following example illustrates Corollary 2.12.

Example 2.13 Taking =0.8, k=1 in (16), we obtain

1 1 × 0.2 + 1 1.8 × 1 + 1 2.6 × 1.8 += 1 0.8 × 0.2 .

The following example shows that 1 k ( n ) c 0 ( ) and 2 ( ) .

Example 2.14 Assume nN(2) and k[n,). Let u(k)= 1 k ( n ) . By Lemmas 2.7 and 2.10, we obtain

1 ( n 1 ) k ( n 1 ) = r = 0 1 ( k + r ) ( n ) .

Since c j =0 as k. Replacing k by a+j, we get

r = 0 1 ( a + j + r ) ( n ) = 1 ( n 1 ) ( a + j ) ( n 1 ) ,foran.
(17)

Since

| 1 ( a + j + r ) ( n ) | 2 < 1 ( a + j + r ) ( n ) ,

for an thus Equation (17) yields

r = 0 |u(a+j+r) | 2 < r = 0 1 ( a + j + r ) ( n ) = 1 ( n 1 ) ( a + j ) ( n 1 ) <.

By Definition 2.8, the function 1 k ( n ) 2 ( ) . Since

lim r 1 ( a + j + r ) ( n ) =0, 1 k ( n ) c 0 ( ) .

Now taking a=n then u(k) is an 2 ( ) space function.

3 Main results

In this section, we present the condition for non existence of non-trivial solution of (1).

Lemma 3.1 Leta2andk[a,). Then

1 k < 4 ( k + + k ) ( k + k ) .

Proof We have

4 ( k + + k ) ( k + k ) = 4 ( k + k ) ( k k ) 2 = 4 2 k k [ ( 1 + k ) 1 2 1 ] [ 1 ( 1 k ) 1 2 ] = 4 k 2 [ 1 + 1 2 k 1 2 ! 1 4 ( k ) 2 + 1 3 ! 1 4 3 2 ( k ) 3 1 4 ! 1 4 3 2 5 2 ( k ) 4 + ] × [ 1 ( 1 1 2 k 1 2 ! 1 4 ( k ) 2 1 3 ! 1 4 3 2 ( k ) 3 1 4 ! 1 4 3 2 5 2 ( k ) 4 ) ] .

Since each positive term is greater than the consecutive negative term in the first expression, we find

4 k 2 [ 1 2 k 1 2 ! 1 4 ( k ) 2 ] × [ 1 2 k + 1 2 ! 1 4 ( k ) 2 + 1 3 ! 1 4 3 2 ( k ) 3 + 1 4 ! 1 4 3 2 5 2 ( k ) 4 + ] = 4 2 [ 2 2 1 4 k ] [ 1 2 k + 1 2 ! 1 4 ( k ) 2 + 1 3 ! 1 4 3 2 ( k ) 3 + 1 4 ! 1 4 3 2 5 2 ( k ) 4 + ] = 4 2 2 [ 1 2 k + 1 2 ! 1 4 ( k ) 2 + 1 3 ! 1 4 3 2 ( k ) 3 + 1 4 ! 1 4 3 2 5 2 ( k ) 4 + ] 4 2 2 1 4 k [ 1 2 k + 1 2 ! 1 4 ( k ) 2 + 1 3 ! 1 4 3 2 ( k ) 3 + 1 4 ! 1 4 3 2 5 2 ( k ) 4 + ] = 1 k + 2 [ 1 2 ! 1 4 ( k ) 2 + 1 3 ! 1 4 3 2 ( k ) 3 + 1 4 ! 1 4 3 2 5 2 ( k ) 4 + ] 2 [ 1 2 ! 1 4 ( k ) 2 + 1 2 ! 1 4 1 4 ( k ) 3 + 1 3 ! 1 4 1 4 1 4 ( k ) 4 + ] = 1 k + 2 4 [ 1 3 ! ( 3 2 3 4 ) ( k ) 3 + 1 4 ! 3 2 ( 5 2 4 4 ) ( k ) 4 + 1 5 ! 3 2 5 2 ( 7 2 5 4 ) ( k ) 5 + 1 6 ! 3 2 5 2 7 2 ( 9 2 6 4 ) ( k ) 6 + ] > 1 k ,

since the second term is positive. □

Lemma 3.2 Leta2andk[a,). Then

k + k k k + + k <1.
(18)

Proof From the Binomial theorem for rational index, we find

k + k k k + + k = ( 1 + k ) 1 2 k 2 [ ( k + ) 1 2 ( k ) 1 2 ] = 1 + 1 2 k 1 2 ! 1 2 1 2 ( k ) 2 + 1 3 ! 1 2 1 2 3 2 ( k ) 3 k 2 [ 1 + 1 2 k 1 2 ! 1 2 1 2 ( k ) 2 + 1 3 ! 1 2 1 2 3 2 ( k ) 3 ( 1 1 2 k 1 2 ! 1 2 1 2 ( k ) 2 1 3 ! 1 2 1 2 3 2 ( k ) 3 ) ] = 1 + 1 2 k 1 2 ! 1 2 1 2 ( k ) 2 + 1 3 ! 1 2 1 2 3 2 ( k ) 3 k 2 [ k + 1 3 ! 1 2 1 2 3 2 ( k ) 3 + ] .

Since each negative terms is greater than the next consecutive positive term and k2, we get

k + k k k + + k =1+ 1 2 k 1 2 = 1 2 + 1 2 k <1.

 □

Lemma 3.3 Leta2. If

Δ z(k)α(k)+β(k)z(k)
(19)

and k <β< 2 k 2 for all k[a,) then

Δ ( z ( k ) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 ) α(k) r = 0 k a ( 1 + β ( j + a + r ) ) 1 ,
(20)

wherej=ka k a .

Proof From the inequality (19) and 1+β(k)>0 for all k[a,), we find,

z ( k + ) 1 + β ( k ) z(k) α ( k ) 1 + β ( k )

which yields,

z ( k + ) 1 + β ( k ) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 z ( k ) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 α ( k ) 1 + β ( k ) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 .

Now (20) follows by taking r= k a and j+a+ k a =k. □

The following theorem shows the nonexistence of solutions of (3).

Theorem 3.4 For all(k,u)[a,)×R, let the functionf(k,u)be defined and

|f(k,u)| 2 2 k 2 |u|.
(21)

Then, ifu(k) 2 ( ) is a solution of (3), there exists a real k 1 a (a2) such thatu(k)=0for allk[ k 1 ,).

Proof Since u(k) is a solution of (3) and belong to 2 ( ) , we have r = 0 |u(a+j+r) | 2 < which yields lim k u(k)=0 and hence

lim k Δ u(k)= lim k Δ 2 u(k)=0.
(22)

By using Equations (3) and (22), and applying Δ 1 on Equation (3) with Lemma 2.10, we obtain

Δ u(k)= r = 0 f ( k + r , u ( k + r ) ) .
(23)

Now by applying again Δ 1 on both sides, and by Theorem 2.10, we get

u(k)= r = 0 s = 0 f ( k + r + s , u ( k + r + s ) )
(24)

which yields

u(k)= r = 0 (r+1)f ( k + r , u ( k + r ) ) ,k[a,).
(25)

Therefore, from (21), we obtain

|u(k)| 2 2 v(k),
(26)

where

v(k)= r = 0 (r+1) ( k + r ) 2 |u(k+r)|,for allk[a,).
(27)

Obviously v(k)0 for all k[a,) and lim k v(k)=0.

If v(k+j)=0, j[0,), for some k= k 1 a, then (r+1) ( k + j + r ) 2 u(k+j+r)=0 for all r=0,1,2, . Hence u(k)=0 for all k k 1 . In this case, the proof is complete.

Now, we suppose that v(k)>0 for all k[a,), from (27), we have

Δ v(k)= r = 0 ( k + r ) 2 |u(k+r)|

and

Δ 2 v(k)= k 2 |u(k)|.

From (26), we have

Δ 2 v(k) 2 2 k 2 v(k)for allk[a,).
(28)

From (27), a2, r + 1 k + r 1 , by Schwartz’s inequality, we obtain

v(k) 1 r = 0 ( k + r ) 1 |u(k+r)| 1 ( r = 0 ( k + r ) 2 ) 1 2 ( r = 0 | u ( k + r ) | 2 ) 1 2 .

By using Corollary 2.12, we get

v(k) 3 2 1 k ( r = 0 | u ( k + r ) | 2 ) 1 2 .

If w(k)= 3 2 k v(k), then

w(k) ( r = 0 u ( k + r ) 2 ) 1 2 ,for allk[a,).
(29)

Hence we have

w(k)0andw(k)>0,k[a,).
(30)

By applying Lemma 2.6 to Equation (29) twice, we obtain

Δ 2 w(k)= 3 2 ( k + Δ 2 v ( k ) + 2 Δ v ( k ) Δ k + v ( k ) Δ 2 k ) .
(31)

Again from Lemma 2.6 and Equation (29), we obtain

Δ v(k)= 3 2 ( 1 k Δ w ( k ) + w ( k ) Δ 1 k ) .
(32)

From (31), (32) and by Lemma 2.6, we find that

Δ ( 1 k Δ w ( k ) ) = 1 k Δ 2 w ( k ) ( k ( k ) ) Δ w ( k ) = 3 2 k { k + Δ 2 v ( k ) + 2 Δ v ( k ) Δ k + v ( k ) Δ 2 k } ( k ( k ) ) Δ w ( k ) = 3 2 k { k + Δ 2 v ( k ) + 2 3 2 [ 1 k Δ w ( k ) + 2 w ( k ) k Δ 1 k ] Δ k + 3 2 k v ( k ) Δ 2 k } ( k ( k ) ) Δ w ( k ) = 3 2 ( k + k ) Δ 2 v ( k ) + 2 k 3 2 k v ( k ) Δ 1 k Δ k + 3 2 k v ( k ) Δ 2 k + 2 k k Δ w ( k ) Δ k k ( k ) Δ w ( k ) 3 2 ( 2 k + 2 k 3 ) v ( k ) + 2 3 2 k k v ( k ) Δ k Δ 1 k + 3 2 k v ( k ) Δ 2 k + ( 2 ( k ) k k Δ k k ) 1 k Δ w ( k )

which in view of (28), (30) gives

Δ z(k)α(k)+β(k)z(k),
(33)

where

z(k)= 1 k Δ w(k),
(34)
α(k)= 3 2 ( 2 k + 2 k 3 + 2 k k Δ k Δ 1 k + 1 k Δ 2 k ) v(k)
(35)

and

β(k)= ( 2 ( k ) k k ) Δ k k .
(36)

Since ( 2 ( k ) k k ) Δ k >0, from ( 1 + k ) 1 2 <1+ 1 2 k , we obtain

k <β(k)< 2 k 2 ,wherek[a,).
(37)

Further, since

k k Δ k Δ 1 k = ( k + k ) ( k k ) = 2 ( k + + k ) ( k + k )

and

From Lemmas 3.1 and 3.2

γ ( k ) < 3 2 k k ( 2 k + 2 k 4 ( k + + k ) ( k + k ) + 2 2 + k ( k k + ) ( k + + k ) ( k + k ) ) v ( k ) = 2 3 2 k k ( k + + k ) ( k + k ) ( 2 k + k 2 k k + + k 2 ) v ( k ) = 2 7 2 k k ( k + + k ) ( k + k ) ( k + k k k + + k 1 ) v ( k ) .
(38)

By Lemma 3.2, we find γ(k)<0 for all k[a,). Thus from Lemma 3.3 and γ(k)<0, we find

Δ ( z ( k ) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 ) <0,for allk[a+,).

That is,

z(k) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1

is decreasing by steps.

If

z(k) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 >0

for all k[a+,), then z(k)>0. From (34) we find Δ w(k)>0 and hence w(k) is increasing by steps, but this contradicts (30).

If there exists a real Ka+ such that

z(K) r = 0 K a 1 ( 1 + β ( j + a + r ) ) 1 = p j <0

for all 0j<, then

z(k) r = 0 k a 1 ( 1 + β ( j + a + r ) ) 1 < p j

for all k[K,), that is,

z(k)< p j r = 0 k a 1 ( 1 + β ( j + a + r ) ) .

However from (37), since 1+β(k)>(k)/k>0 and j=k k a , it follows that z(k)< p j (j+a)/(k), and hence from (34), we find Δ w(k)< p j (j+a). Further, since

w(k)0,kK+2 1 (kK)1

we get w(k+)<w(k)+ p j (j+a) which yields w(k)<w(k)+ p j (j+a) and hence we get

w(k)<w(K+)+ 1 p j (j+a)(kK)

for all k[K+2,), since

kK+2kK2, 1 (kK)1.

But this implies that w(k), and again we get a contradiction to (30).

Thus combining the above arguments, we conclude that our assumption v(k)>0 for all k[a,) is not correct, and this completes the proof. □

Theorem 3.5 For all(k,u)[0,)×R, let the functionf(k,u)be defined and

|f(k,u)| q k q |u|,q> 5 2 .
(39)

Then, ifu(k)is a solution of (3) c 0 ( ) , there exists an integer k 1 a (a4) such thatu(k)=0for allk[ k 1 ,).

Proof Let u(k) be a solution of (3) such that lim k |u(k)|=0. Then,

lim k Δ u(k)= lim k Δ 2 u(k)=0

for all >0. Thus, for this solution also the relation (24) holds. Further, since there exists a constant c j >0 such that |u(k)| c j for all k[a,), where 0j=k k <, we find that

r = 0 ( r + 1 ) | f ( ( k + r ) , u ( k + r ) ) | r = 0 ( r + k q ( k + r ) q | u ( k + r ) | ) = r = 0 ( k + r ) 1 q q 1 | u ( k + r ) | c j q 1 r = 0 ( k + r ) 1 q where j = k k = c j q 1 [ k 1 q + r = 1 ( k + r ) 1 q ] = c j q 1 [ k 1 q + 1 q r = 1 ( k + r ) 1 q ] = c j q 1 [ k 1 q + 1 q [ ( k ) 2 q 2 q + r ] k ] = c j q 1 [ k 1 q + ( k 2 q ( q 2 ) ) ] < ,

for all k[ k 1 ,). Therefore, this solution also has the representation (24).

Now as in Theorem 3.4, we define

v ¯ (k)= r = 0 (r+1) ( k + r ) q |u(k+r)|= r = 0 q (r+1) ( k + r ) q |u(k+r)|.

Since q> 5 2 , we find

v ¯ (k) q r = 0 (r+1) ( k + r ) 2 |u(k+r)|= 2 q r = 0 (r+1) ( k + r ) 2 |u(k+r)|

then it follows that

v ¯ (k) 2 q ( 3 2 k ) { r = 0 | u ( k + r ) | 2 } 1 2 .

Hence we define

w ¯ ( k ) = q 1 2 k v ¯ ( k ) , z ¯ ( k ) = 1 k Δ w ¯ ( k ) , γ ¯ ( k ) = q 1 2 ( q k + 2 k q + 1 + 2 k k Δ k Δ 1 k + 1 k Δ 2 k ) v ¯ ( k ) , β ¯ ( k ) = ( 2 ( k ) k k ) Δ k k ,

and applying similar arguments as in the previous theorem one can see that there exists a positive integer k 1 such that u(k)=0 for all k[ k 1 ,). □

In the next we present some formulae and examples to find the values of finite and infinite series in number theory as application of Δ . First of all we need the following theorem.

Theorem 3.6 Letk[,)and(0,). Then

r = 1 [ k ] + s ( k r + 2 ) 2 3 2 r ( k r + 4 ) ( 2 ) ( k r + ) ( k r + ) = c j k 1 ( k + 3 ) k ( k ) ,
(40)

wheres=1fork N (), s=0fork N ()and each c j is a constant for allk N (j), j=k[ k ]. In particular c j is obtained from (40) by substitutingk=+j. Further

r = 0 ( k + r ) 3 3 r ( ( k + r ) 2 2 2 ) ( 2 ) ( k + r + ) ( k + r + ) = 1 ( ( k ) 2 2 2 ) k ( k ) .
(41)

Proof By Definition 2.1, we find

Δ 1 ( ( k + 2 ) 2 3 2 ) k ( k + 4 ) ( 2 ) ( k + ) ( k + ) = c j k ( k + 3 ) k ( k )

and (40) follows by Lemma 2.9 and

( k ( [ k ] + s ) + 2 ) 2 3 2 ( k ( [ k ] + s ) + 4 ) ( 2 ) ( k ( [ k ] + s ) + ) ( k ( [ k ] + s ) + ) 0.

 □

The following example illustrates Theorem 3.6.

Example 3.7 By taking =1.7, k=2 and j=0.3 in (40), we get c j = 85 81 and hence (40) becomes

r = 1 [ k ] ( k 1.7 r + 2 ( 1.7 ) ) 2 3 ( 1.7 ) 2 1.7 r ( k 1.7 r + 4 ( 1.7 ) ) 1.7 ( 2 ) ( k 1.7 r + 1.7 ) 1.7 ( k 1.7 r + 1.7 1.7 ) = 85 81 ( 1.7 ) k 1.7 1 ( k + 3 ( 1.7 ) ) k 1.7 ( k 1.7 ) , k = 2 , 3.7 , 5.4 ,

Example 3.8 Taking =3.5 in (41), we obtain

r = 0 ( k + 3.5 r ) 3 3.5 3 3.5 r ( ( k + 3.5 r ) 2 2 ( 3.5 ) 2 ) ( 2 ) ( k + 3.5 r + 3.5 ) 3.5 ( k + 3.5 r + 3.5 3.5 ) = 1 ( ( k 3.5 ) 2 2 ( 3.5 ) 2 ) k 3.5 ( k 3.5 ) .

In particular, when k=9, above series becomes

9 3 3.5 3 ( 9 2 2 ( 3.5 ) 2 ) 3.5 ( 2 ) 12.5 3.5 ( 4 ) + 12.5 3 3.5 3 3.5 ( 12.5 2 2 ( 3.5 ) 2 ) 3.5 ( 2 ) 16 3.5 ( 5 ) + 16 3 3.5 3 3.5 2 ( 16 2 2 ( 3.5 ) 2 ) 3.5 ( 2 ) 19.5 3.5 ( 6 ) + = 1 ( 5.5 2 2 ( 3.5 ) 2 ) 9 3.5 ( 3 ) .

4 Concluding remarks

In the difference equations there are several interesting development, see for example, [46], and [816]. Recently, in [7], the fractional h-difference equations was studied. In the present work we study the 2 ( ) and c 0 ( ) solutions of the second order generalized difference equation

Δ 2 u(k)+f ( k , u ( k ) ) =0,k[a,),a>0

and we prove the condition for non existence of non-trivial solution.

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Acknowledgements

The authors would also like to thank the referee(s) for valuable remarks and suggestions on the previous version of the manuscript.

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Correspondence to Adem Kılıçman.

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Keywords

  • generalized difference equation
  • generalized difference operator