2 Second-order linear coefficient delayed equation

Linear even order homogenous difference equation with delay in coefficient

Jan Jekl

Masaryk University, Kotláˇrská 2, Brno, CZ–611 37, Czech Republic Received 31 January 2020, appeared 1 July 2020

Communicated by Stevo Stevi´c

Abstract. We use many classical results known for the self-adjoint second-order linear equation and extend them for a three-term even order linear equation with a delay ap- plied to coefficients. We derive several conditions concerning the oscillation and the ex- istence of positive solutions. Our equation for a choice of parameter is disconjugate, and for a different choice can have positive and oscillatory solutions at the same time.

However, it is still, in a sense, disconjugate if we use a weaker definition of oscillation.

Keywords: coefficient delayed equations, separately disconjugate, oscillation theory, minimal solution, difference equation.

2020 Mathematics Subject Classification: 39A06, 39A21, 39A22, 47B36, 47B39.

1 Introduction

This paper is divided into two parts. In the first part, we analyse the linear second-order homogeneous difference equation with a delay in a coefficient

a_n−ky_n−1+b_ny_n+a_ny_n+1 =0, n∈_Z. (1.1) Equations with a delay in termy_n−1are usually considered. Nevertheless, we did not find a situation where the considered delay is in the coefficient an. This may be because Eq. (1.1) fork=1 is often discussed together with its self-adjoint form4(p_n4y_n) +q_ny_n+1=0.

Properties of this special case were discussed many times. Some necessary and sufficient conditions for the equation to be oscillatory were derived in [6,8,10,19,20,22,29] and for a matrix case in [7]. Properties of eventually positive solutions were observed in [28]. Minimal solutions of the special case were discussed in [14]. Recessive solutions and their connection to oscillation were discussed in [27], for a matrix case in [3], and for nonoscillatory symplectic systems in [33]. Notion of generalized zero was developed in [15] and the Sturm comparison theorem on Z together with the existence of a recessive solutions was discussed in [2,5].

Many classic results about this special case can be found in [21]. Boundedness and growth of the special case were investigated in [30,31]. Generalization of the special case were considered

BEmail: jekl@mail.muni.cz

for example in [24–26,32]. If we consider a continuous case, criteria for oscillation can be found, for example, in [11], and the existence of a principal solution of a 2n-order self-adjoint equation was recently discussed in [34]. Some ideas about how to extend the results for the fourth-order equation can be found in [9].

In Section 2, we would like to extend the results from [14], where the special case is also considered. The results from [14] were already extended in [12,13,17] and for the time scales in [18], but there was used the symmetrical case fork=1. Arbitrary choice ofk∈_Zwill lead to the generalization of some already known results.

We derive equivalent conditions for which the equation has a positive solution, and later through the deriving of a suitable version of the Sturm comparison theorem, we will get criteria of disconjugacy for Eq. (1.1). These results will be used in Section 3 as a tool, as well.

In Section 3 we analyse the linear even order homogeneous difference equation with a de- lay in a coefficient

a_n−kHyn+bn+Hyn+H+an+Hyn+2H =0, n∈_Z, (1.2) which is a generalization of Eq. (1.1). Fork = 0 we get a equation discussed in [16]. We can assume that results obtained in Section 2 can be extended for Eq. (1.2) in the similar way as in [16].

We derive conditions under which Eq. (1.2) can or cannot have positive or eventually positive solutions. We also discuss a situation when Eq. (1.2) has recessive and dominant solutions. Among others, we use a combination of ideas as were established in [19,27]. We find that Eq. (1.2) can have both positive and sign-changing solutions. A situation where an equation has oscillatory and nonoscillatory solutions at the same time was discussed for example in [1]. The same situation can appear in our equation, but we use a weaker version of oscillation to avoid this situation.

2 Second-order linear coefficient delayed equation

Let real valued sequences a_n, b_n satisfy a_n < 0, b_n > 0, for everyn ∈ Z. In the first part we study the equation

a_n−ky_n−1+b_ny_n+a_ny_n+1 =0, k ∈_Z. (2.1) If we consider a solutiony_nof Eq. (2.1), then we have a solution x_n= (−1)ⁿy_nof the equa- tion

a_n−kx_n−1+d_nx_n+a_nx_n+1=0,

where sequenced_n <0 for everyn. In a similar sense if we consider the equation c_n−px_n−1+b_nx_n+c_n+lx_n+1 =0,

where c_n < 0 for every n. Then we can take a_n = c_n+l and this will result in Eq. (2.1) for k=−l−p.

There is a natural relation of Eq. (2.1) to the infinite matrix operator, whose truncations for

n≤ p,n,p∈ Z, are the matrices

dn,p =



















b_n a_n 0 . . . 0

a_n−k+1 b_n+1 a_n+1 . .. ... 0 a_n−k+2 b_n+2 . .. ...

... 0 . .. . .. a_p−1

0 . . . a_p−k b_p



















and we denote their determinants by D_n,p =_det(d_n,p). Note that fork=_{1 is} d_n,psymmetrical.

For simplification of formulas, we takeD_i+1,i = 1 and D_i+j,i = 0 for anyi∈ _Zand j>1, as well as∏ⁱ_i⁻¹x_i =1. Moreover, we will use recurrence relations

D_n,p=b_nD_n+1,p−a_n−k+1a_nD_n+2,p, (2.2) D_n,p=b_pD_n,p−1−a_p−ka_p−1D_n,p−2, (2.3) forn≤ p.

Lemma 2.1. Let n< p and real vectorsX= (x_n, . . . ,x_p)^T,B= (y, 0 . . . , 0,z)^T, then the equation d_n,pX=B,

implies

x_hD_n,p =yD_h+1,p

∏

(−a_j−k) +zD_n,h−1

p−1

∏

(−a_j), (2.4)

where n≤h≤ p.

Proof. The proof follows from the Cramer’s rule. Signs at−a_jand−a_j−k follow from compar- ing the sign and number of terms in a given product.

Lemma 2.2. Let

D_i,j >0, for i≤ j, (2.5)

and let x¹_n, x²_n be two solutions of Eq.(2.1), which satisfy x¹_m = x²_m for some m ∈ Z. If also x¹_h > x²_h (respectively x¹_h =x²_h)for some h>m, then it holds that x¹_j > x²_j (respectively x¹_j = x²_j)for all j> m.

Proof. Obviously, two solutionsx¹_n,x²_nof Eq. (2.1) have to also satisfy Lemma 2.1where y= −a_m−k+1x¹_m =−a_m−k+1x²_m,

z¹ =−a_h−1x¹_h >−a_h−1x²_h= z². Where for i∈ {1, 2}we haveXⁱ = xⁱ_m+1, . . . ,xⁱ_h−1T

andBⁱ = (y, 0 . . . , 0,zⁱ)^T. Together with (2.5), we obtain from (2.4) that

x¹_jD_m+1,h−1 =yD_j+1,h−1

∏

(−a_i−k) +z¹D_m+1,j−1 h−2

∏

(−a_i)

>yD_j+1,h−1

∏

(−a_i−k) +z²D_m+1,j−1 h−2

∏

(−a_i) =x²_jD_m+1,h−1,

holds for alln<j<hand thusx¹_j >x²_j. Takingx¹_j <x²_j for somej> hleads to a contradiction withx¹_h >x²_hin the same manner. Therefore,x¹_j >x²_j for allj>m. The case ofx¹_h= x²_hfollows analogously.

Similarly, we get a version of Lemma 2.2 for someh < mand all j< m. It means that if two solutions of Eq. (2.1) are equal at two points, then they are equal everywhere.

Lemma 2.3. Assume(2.5), then for any h< p it holds that 1

b_h < ^Dh+1,p

D_h,p < ^bh−1 a_h−ka_h−1

, (2.6)

and the sequence x_p = ^D_D^h+1,p

h,p is increasing for any h where h< p.

Proof. Because of (2.2) we get

D_h,p =b_hD_h+1,p−a_h−k+1a_hD_h+2,p< b_hD_h+1,p, which implies the left inequality of (2.6). Further, we compute

0< D_h−1,p =b_h−1D_h,p−a_h−ka_h−1D_h+1,p, a_h−ka_h−1D_h+_1,p <b_h−1D_h,p,

D_h+1,p

D_h,p < ^bh−1 a_h−ka_h−1

, which implies the right inequality in (2.6).

In the second part of the proof, we will proceed by induction. First, we assume p =h+1 and we get

D_h+1,h+2 D_h,h+2

− ^Dh+1,h+1 D_h,h+1

= ^{Dh,h+1Dh+1,h+2}−D_h+1,h+1D_h,h+2 D_h,h+2D_h,h+1

= ^{ah−k+1ah−k+2ahah+1} D_h,h+2D_h,h+1

>0.

Next, again by (2.2), we get D_h,p D_h+1,p

− ^Dh,p+1 D_h+1,p+1

=a_h−k+1a_h

D_h+2,p+1

D_h+1,p+1

− ^Dh+2,p D_h+1,p

>_0, by the induction assumption, which together with (2.5) results in

D_h,p D_h+1,p

> ^Dh,p+1

D_h+1,p+1

, D_h+1,p

D_h,p < ^Dh+1,p+1 D_h,p+1

.

Therefore, the sequence is increasing and the proof is complete.

Similarly, using (2.3), we get forn< hthat 1

b_h < ^Dn,h−1

D_n,h < ^bh+1 a_h−k+1a_h, and the sequencex_n= ^D_D^n,h−1

n,h is decreasing for anyhwhichn< h.

Now, thanks to Lemma2.3, we can define the sequences c⁺_n = lim

p→_∞

D_n+1,p

D_n,p , c⁻_n = lim

p→−_∞

D_p,n−1 D_p,n , and

u(j,n) =











1, j=n,

∏^j_h⁻₌¹_n(−a_h)c⁻_h, n< j,

∏ⁿ_h⁻=¹j(−a_h−k+1)c⁺_h+1, n> j.

Notice that by Lemma2.3together with a_i <0 for every i, we get thatu(j,n)>0 for any j,n.

Definition 2.4. We say that a solutionunof Eq. (2.1) is minimal on[j+1,∞)∩_Zif any linearly independent solution v_n of Eq. (2.1) such thatu_j = v_j satisfies u_k < v_k for every k ≥ j+1.

The minimal solution on(−_∞,j−1]∩_Zis defined analogously.

Lemma 2.5. Assume(2.5), thenα_n=u(j,n)is a positive minimal solution of Eq.(2.1)on the interval [j+_1,∞)∩_Zand also on the interval(−_∞,j−₁]∩_Z.

Proof. Using Lemma2.1with y=−a_j−k+1 andz=0 we obtain that

v_n(j,p) =











1, n =j,

∏ⁿ_h⁻=¹j(−a_h−k+1)^D_D^n+1,p

j+1,p, j+1≤n≤ p,

0, n = p+1,

is a solution on the interval[j+1,p]∩Z. Moreover, it holds thatu(j,n) =lim_p→_∞v_n(j,p)and soα_n= u(j,n)is a solution on the interval[j+_1,∞)∩_{Z, where}α_j = u(j,j) =_1.

Next, we assume that there is a positive solution vn such that v_j = α_j and which is also linearly independent on α_n. Then we know that v_p+1 > v_p+1(j,p) =0 and v_j = v_j(j,p) =1, for every p. Therefore, due to Lemma 2.2, we know that v_n > v_n(j,p) _{for all} p. Because αn = limp→_∞vn(j,p), we get that vn ≥ αn. But vn is linearly independent and, again by Lemma2.2, this inequality must hold strictly, i.e. v_n >α_n.

Similarly, we get thatα_n =u(j,n)is a solution on interval(−_∞,j−1]∩_Zusing function

vn(j,m) =











1, n= j,

∏^j_h⁻=¹n(−a_h)^D_D^m,n−1

m,j−1, m≤n≤ j−1,

0, n=m−1.

Further, we will use the following notation. We define

u⁺_n =











1, n=0,

u(_0,n)_, n∈_N, u(n, 0)⁻¹, −n ∈_N,

and u⁻_n =











1, n=0,

u(n, 0)⁻¹_, n∈_N, u(0,n), −n∈ _N.

Lemma 2.6. Assume(2.5), then u^±_n are positive solutions of Eq.(2.1)onZ.

Proof. From Lemma 2.5 we know, thatu⁺_n is a solution onN. Moreover, for arbitrary n,B ∈ N∪ {0},n<B, it holds

u(−B, 0) =u(−B,−n)u(−n, 0), and so

u⁺_−n= ¹

u(−n, 0) = ^u(−B,−n) u(−B, 0) ^.

Using Lemma 2.5 we obtain that u⁺_n is a solution on interval [−B+1,∞)∩_{Z. Because} B is arbitrary, we have that u⁺_n is a solution on Z. The second part involving u⁻_n is done in the similar way.

Theorem 2.7. Condition(2.5)holds if and only if there is a positive solution of Eq.(2.1).

Proof. The sufficiency of (2.5) comes directly from Lemma2.6. For the second part, we assume the existence of a positive solution un. Then, using Lemma2.1 for arbitrary n, n < _{p, with} y=−a_n−ku_n−1,z=−apu_p+1, we get from (2.4) that

u_nD_n,p=−a_n−ku_n−1D_n+1,p−a_pu_p+1 p−1

∏

(−a_j)_.

If we put p = n+1, then because D_n+1,n+1 = b_n+1 > 0 we obtain that the right-hand side is positive which implies the positivity of D_n,n+1 > 0. Next, by induction we obtain that if Dn+1,p > 0, then also Dn,p > 0 through the same procedure. Therefore, the condition (2.5) is satisfied.

We emphasize that for k = _{1 is} d_n,p symmetrical, thus condition (2.5) gives the positive definiteness of alldn,p. Now we recall the definitions of generalized zero and disconjugacy.

Definition 2.8. Solutiony_nhas a generalized zero at n₀if y_n0 =0 ory_n0−1y_n0 <0.

Definition 2.9. The given difference equation is disconjugate on an intervalI if every nontriv- ial solution has at most one generalized zero on I.

Lemma 2.10. Let Eq.(2.1)be disconjugate on interval[a,b]then the boundary value problem a_n−ky_n−1+b_ny_n+a_ny_n+1 =0,

yn₁ = A, yn₂ =B,

where a≤n₁ <n2 ≤b and A,B∈R, has an unique solution.

Proof. General solution of Eq. (2.1) is

y_n=Cz¹_n+Dz²_n,

for some linearly independentz¹_nandz²_n. The boundary conditions result in the system Cz¹_n1+Dz²_n1 = A,

Cz¹_n2+Dz²_n2 =B.

We see that the boundary value problem has a solution whenever det

z¹_n1 z²_n1 z¹_n2 z²_n2

6=_0.

Now assume that this determinant is equal to zero. Then there would exist constantsC,D∈_R such that

Cz¹_n1+Dz²_n1 =0, Cz¹_n2+Dz²_n2 =0.

Thus, yn₁ =yn2 =0. This contradicts that Eq. (2.1) is disconjugate.

Theorem 2.11. Let Eq.(2.1)be disconjugate onZ, then(2.5)holds.

Proof. We will show that D_i,i+k−1 > 0 by induction onk ∈ _N for arbitraryi. Because b_i > 0 we have that D_i,i >0.

Lety_n be a solution of

a_n−ky_n−1+bnyn+any_n+₁=0, y_i−1=_0, y_i+k+1=_1,

and assume thatD_i,i+k−1>0. By Lemma2.10, we know that suchynexists and it must satisfy system

d_i,i+ky=_b,

wherey= (y_i, . . . ,y_i+k)^T,b= (0, . . . , 0,−a_i+k). Now, using Lemma2.1 we get that y_i+kD_i,i+k = −a_i+kD_i,i+k−1.

By disconjugacy we know thaty_i+_k >0 and together with the assumptionD_i,i+_k−1>0 we see that D_i,i+k >0, as well.

Corollary 2.12. Let Eq.(2.1)be disconjugate onZ, then there exists a positive solution of Eq.(2.1).

Proof. This is a direct consequence of Theorem2.7.

The natural question is whether the converse statement is valid as well. We will solve this problem by formulating an appropriate version of Sturm’s comparison theorem. Nevertheless, it can be solved using Theorem2.7and Lemma2.6 together withu^±_n being minimal solutions as well. Note that we have two separate situations whereu⁺_n =u⁻_n andu⁺_n 6=u⁻_n.

Lemma 2.13. If ynis a nontrivial solution of Eq.(2.1)such that yn₀ =0, then y_n0−1y_n0+1<0.

Proof. If y_n is a nontrivial solution and y_n0 = 0 for some n₀ ∈ _Z, then y_n0−1 6= 0 6= y_n0+1. The rest follows from ynbeing a solution of Eq. (2.1).

Lemma 2.14. Assume(2.5). If a nontrivial solution y_n of Eq. (2.1) has two generalized zeros at n₁ and n2, then any other linearly independent solution has a generalized zero in[n₁,n2].

Proof. Without loss of generality assume that there are not other generalized zeros of y_n on (n₁,n₂). Now by contradiction, we assume thaty_n >_{0 on}(n₁,n₂)and that there is a linearly independent solution zn such that zn > 0 on [n₁,n₂] and z_n1−1 ≥ 0, i.e. it does not have a generalized zero on [n₁,n₂]. We consider some n₀ from (n₁,n₂) and we can find K ∈ _R such that Kz_n0 = y_n0. Because y_n2 ≤ 0 and it has to hold that y_n1 = _{0 or} y_n1−1 < _{0 we can} use Lemma 2.2 to get that Kzn > yn. Moreover, un = Kzn−yn is also a solution of Eq. (2.1) and u_n0 = 0, u_n > 0 for n 6= n₀. Finally, u_n0−1u_n0+1 > 0 gives us a contradiction with Lemma2.13.

Theorem 2.15. Eq.(2.1)is disconjugate onZif and only if it has a positive solution onZ.

Proof. We already have the first part from Corollary 2.12. Next, assume that Eq. (2.1) has a positive solution. By Theorem 2.7 we know, that (2.5) holds and so does Lemma 2.14.

However, because we have a positive solution, then by Lemma 2.14, we know that there cannot be a solution with more than one generalized zero.

3 Even order linear coefficient delayed equation

In this section we will focus on the equation

a_n−kHy_n+b_n+Hy_n+H+a_n+Hy_n+2H =0, (3.1) forn∈ Z, with the parameters H∈_N,k∈_Z.

Lemma 3.1. If a_i < 0for every i and there is a subsequence b_nl such that b_nl ≤ 0 for n_l → _∞ then Eq.(3.1) cannot have an eventually positive solution (i.e. a solution y_n, where y_n > 0for all n ≥ N, for some N∈_Z).

Proof. Suppose that there exist an eventually positive solutionyn. It implies a_nl−k·Hyn_l +bn_l+Hyn_l+H+an_l+Hyn_l+2H <0,

forn_l →_∞. This is a contradiction withy_nbeing a solution of Eq. (3.1).

Similar statement holds even if n_l → −_∞ and y_n > 0 for all n ≤ N for some N ∈ _Z.

Because of this, we will again assume thata_j <0,b_j >0 for everyj.

Theorem 3.2. The following statements are true.

1. Let H be an even number, then Eq.(3.1)has a solution ynif and only if it has a solution(−1)ⁿ_yn. 2. Let H be an odd number, then Eq.(3.1) cannot have a solution (−1)ⁿpn where pn > 0 for all

|n| ≥ N and some N∈_N.

Proof. For the first part, it suffices to usez_n = (−1)ⁿy_n in Eq. (3.1) and the rest follows from H being even.

To prove the second part, we suppose that Eq. (3.1) has a solution(−1)ⁿp_n. Then we have that

a_n−k·Hp_n+b_n+H(−1)^Hp_n+H+a_n+Hp_n+2H =0.

For|n|sufficiently large, the terms are negative, hence the left-hand side cannot be equal zero and such a solution cannot exist.

Corollary 3.3. Let H be an even number, then Eq.(3.1)has at least on solution, which is not eventually positive.

Proof. Assume that all solutions of Eq. (3.1) are eventually positive. Then there is a solution yn, which is positive for n greater than some N. However, because H is an even number, then (−1)ⁿy_n is also a solution of Eq. (3.1) and is not eventually positive. Thus we arrive to a contradiction.

We obtain further generalization if we let p^k_nbe real sequences and consider a linear equa-

tion m

k

∑

p^k_ny_n+2k =_{0. (3.2)}

Then Eq. (3.2) has a solution, which is not eventually positive.

We see that, in some cases, the studied equation cannot have a positive solution. Later we show that there is an equation that has positive and sign-changing solutions at the same time, which is a case that fork =0 cannot occur. For this reason, it is more useful to focus on the situation when the equation has a positive solution. Nevertheless, we start by reminding us of the lemma, which can be found in [21].

Lemma 3.4. Let us consider the equation

∑

p^k_nu_n+k =0, (3.3)

where p^k_n, k ∈ {0, . . . ,m}, are real sequences, for some m ∈ _{N. If Eq.}(3.3) has a solution u_n, then Eq.(3.3)has another solution in the form vnun, where vnsolves the equation

m−1 k

∑

∑

pⁱ_nu_n+i

!

4v_n+k =0. (3.4)

Proof. We expand the sum ∑^mk=₀p^k_nv_n+ku_n+k by Abel’s summation formula and use the fact that u_nis a solution of Eq. (3.3) to obtain Eq. (3.4).

Assume that we have a solution un of Eq. (3.1) and using Lemma 3.4 we obtain other solution asv_nu_n, where v_nsolves

a_n−k·Hun H−1

j

∑

4v_n+j+ (a_n−k·Hun+b_n+Hu_n+H)

H−1

∑

4v_n+H+j =0.

Using the substitutionz_n=v_n+H−v_nwe get usingu_nbeing a solution of Eq. (3.1) that 0= a_n−kHunzn+ (a_n−k·Hun+b_n+Hu_n+H)z_n+H =a_n−kHunzn−a_n+Hu_n+2Hz_n+H. (3.5) Wheneveru_n6=0 for alln, then the solution of Eq. (3.5) is

zn= ^D∏

−k−1 j=1 a_n+jH

u_nu_n+H∏⁰_j=−ka_n+jH

,

for someD∈R. Finally, we can use the fact thatz_n=v_n+H−v_n. Hence, v_n=−

∑

g=0

z_n+gH, (3.6)

v_n=

∑

z_n−gH.

Definition 3.5. We say that a solutionun of Eq. (3.1) is minimal on[µ,∞)∩_Zif any linearly independent solutionv_nof Eq. (3.1) withu_µ =v_µ, . . . ,u_µ+H−1 = v_µ+H−1satisfiesv_n> u_n, for everyn≥µ+H.

Theorem 3.6. Let Eq.(3.1)have a positive solution u_nonZ, which is minimal on an interval[l,∞), where l∈Z. Then for everyµ∈_Zit holds

∑

∏^g_j=⁻g^k+⁻1¹(−a_µ+jH)

uµ+gHu_µ+(g+1)H∏^g_j=g−k(−aµ+jH) = _∞. (3.7) Proof. Assume that for someµ∈_Zthe sum in (3.7) is finite. Sinceu_nis a positive solution, by (3.6) we know that also

wn=







un∑^∞g=0

∏^g_j=⁻g^k+⁻1¹(−a_n+jH)

un+gHu_n+(g+1)H∏^gj=g−k(−an+jH), n≡µ(mod H),

u_n, n6≡µ(mod H),

is a positive solution.

Next, we introduce

w^∗_n= ^wn wµ

u_µ, when n≡µ(mod H).

Therefore,w^∗_nis also a solution where values ofw^∗_nandu_nare equal forHconsecutive indices aroundµ. Because the sum in (3.7) is finite, we get

lim inf

n→_∞

w^∗_n u_n = ^uµ

w_µ lim

n→_∞

∑

∏^g_j=⁻g^k+⁻1¹(−a_n+jH)

u_n+gHu_n+(g+1)H∏^g_j=g−k(−a_n+jH) =0.

It means that from some N > l we have w^∗_N < u_N which is a contradiction with u_n being a minimal solution on[l,∞).

Through similar means as were used in [27], we can deduce the following statements. But first, we have to define a generalization of Casoratian as

ω_n,µ=det

u_µ+nH v_µ+nH

u_µ+(n+1)H v_µ+(n+1)H

.

Lemma 3.7. Let u_n,v_nbe two solutions of Eq.(3.1), thenω_n,µsatisfies for allµ∈_Zthe equation ω_n+1,µ = −a_µ+(n−k)H

−a_µ+(n+1)Hω_n,µ. Proof. Becauseun,vn are solutions of (3.1) we have

ωn,µ =det −^a_a^µ+(n+1)H

µ+(n−k)Hu_µ+(n+2)H −^a_a^µ+(n+1)H

µ+(n−k)Hv_µ+(n+2)H u_µ+(n+1)H v_µ+(n+1)H

!

= (−1) −^aµ+(n+1)H a_µ+(n−k)H

!

ω_n+1,µ = −a_µ+(n+1)H

−a_µ+(n−k)Hω_n+1,µ.

Hence, we can compute for some D∈ _Rthat ω_n,µ= ^D

∏ⁿj=n−k −a_µ+jH

n−k j=

∏

−a_µ+(j−1)H .

Note that if for some ω_n,µ is D < 0, we get by swapping values of u_n and v_n on the set {µ+jH|j∈_Z}_thatu_nandv_n are still solutions of Eq. (3.1) andD>_0.

Theorem 3.8. If Eq.(3.1)has two independent eventually positive solutions, then there are two inde- pendent eventually positive solutions un, vnfor whichlimn→_∞ ^u_vnⁿ =0. Moreover, for arbitraryµ∈_Z sufficiently large

∑

∏ⁿ_j=⁻n^k+2(−a_µ+(j−1)H)

u_µ+nHu_µ+(n+1)H∏ⁿj=n−k(−a_µ+jH) =_∞, (3.8)

∑

∏ⁿ_j=⁻n^k+2(−a_µ+(j−1)H)

vµ+nHv_µ+(n+1)H∏ⁿj=n−k(−a_µ+jH) <_∞. (3.9) Proof. We can expect thatun,vnare linearly independent, eventually positive and also that in ωn,µ isD<0, for all µ. Considering µsufficiently large we have

4

u_µ+nH

v_µ+nH

= ^{uµ+nHvµ+(n+1)H}−u_µ+(n+1)Hvµ+nH

v_µ+nHv_µ+(n+1)H

= ^D

v_µ+nHv_µ+(n+1)H∏ⁿj=n−k −a_µ+jH

n−k j=

∏

−a_µ+(j−1)H

. (3.10) Hence, (3.10) is negative, therefore ^u_v^µ+nH

µ+nH is strictly decreasing inn, but ^u_v^µ+nH

µ+nH is also positive and thus bounded from below. We have that limn→_∞ ^u_v^µ+nH

µ+nH = L_µ ≥0. In case that for someµis Lµ >0, we replaceu_nbyu_n−Lµv_n, forn∈ {µ+jH|j∈_Z}. Hence, u_n will still be a solution and we get that lim_n→_∞ ^u_vⁿ_n =0.

Moreover, by summing equality (3.10) we obtain D

n−1 g

∑

1

v_µ+gHv_µ+(g+1)H∏^g_j=g−k −a_µ+jH

g−k j=

∏

−a_µ+(j−1)H

= ^uµ+nH vµ+nH

−^uµ+kH v_µ+kH

,

n→_∞

−−−→D

∑

1

vµ+gHv_µ+(g+1)H∏^g_j=_g−k −a_µ+jH

g−k j=

∏

−a_µ+(j−1)H

=−^uµ+kH v_µ+kH

,

which confirms the validity of (3.9). Using the unboundedness of ^v_u^µ+nH

µ+nH, we get (3.8).

Corollary 3.9. Let for someµbe

∑

∏ⁿ_j=⁻n^k+2(−a_µ+(j−1)H)

∏ⁿj=n−k(−a_µ+jH) =_∞,

and every solution of Eq.(3.1)be eventually bounded, then Eq.(3.1)has at most one linearly indepen- dent eventually positive solution.

Proof. Suppose that Eq. (3.1) has two such solutions. Then from Theorem3.8 there has to be a solution v_n such that 0 < v_n < M for n sufficiently large and some M. Moreover, for ν sufficiently large and satisfyingν≡µ(mod H)we get from (3.9) that

∞>

∑

∏ⁿ_j=⁻n^k+2(−a_ν+(j−1)H)

v_ν+nHv_ν+(n+1)H∏ⁿ_j=n−k(−a_ν+jH) > ¹ M²

∑

∏ⁿ_j=⁻n^k+2(−a_ν+(j−1)H)

∏ⁿ_j=n−k(−a_ν+jH) ^. Which is a contradiction.