Solvable product-type system of difference equations whose associated polynomial is of the fourth order

Solvable product-type system of difference equations whose associated polynomial is of the fourth order

Stevo Stevi´c

Mathematical Institute of the Serbian Academy of Sciences Knez Mihailova 36/III, 11000 Beograd, Serbia

Operator Theory and Applications Research Group, Department of Mathematics King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia

Received 12 January 2017, appeared 14 March 2017 Communicated by Jeff R. L. Webb

Abstract. The solvability problem for the following system of difference equations zn+1=αz^a_nw^b_n, wn+1=βw^c_n−1z^d_n−2, n∈N0,

where a,b,c,d ∈ Z, α,β ∈ C\ {0}, z−2,z−1,z0,w−1,w0 ∈ C\ {0}, is solved. In the main case whenbd 6= 0, a polynomial of the fourth order is associated to the system, and its solutions are represented in terms of the parameters, through the roots of the polynomial in all possible cases (the roots are given in terms of parameters a,b,c,d).

This is also the first paper which successfully deals with the associated polynomial (to a product-type system) of the fourth order in detail, which is the main achievement of the paper.

Keywords: system of difference equations, product-type system, solvable in closed form, polynomial of fourth order.

2010 Mathematics Subject Classification: 39A20, 39A45.

1 Introduction

Concrete nonlinear difference equations and systems is a research field of some recent interest (see, e.g., [2,4,9–15,17–45]). Among the systems, symmetric and related ones have attracted at- tention of some experts, especially after the publication of several papers by Papaschinopoulos and Schinas almost twenty years ago (see, e.g., [2,4,9–14,17,18,21,23–28,30,31,34–40,42–45]).

On the other hand, solvability of the equations and systems has re-attracted some recent attention (see, e.g., [2,15,21–36,38–45]). Some of them are solved by the method of transfor- mation (see, e.g., [15,21,22,24,38–41] and the references therein). For somewhat more complex methods see [33] and [34]. An interesting related method has been recently applied to partial difference equations in [29] and [32]. Books [1,5–8] contain many classical methods for solving difference equations and systems.

BEmail: sstevic@ptt.rs

If initial values and coefficients of product-type systems are positive they can be solved by transforming them to the linear ones with constant coefficients, by using the logarithm. If the initial values and coefficients are not positive the method is not of a special use. Therefore, the solvability of product-type systems with non-positive initial values and coefficients is a problem of interest. We started studying the problem in [36], where we showed the solvability of the system

z_n+1 = ^w

an

z^b_n−1, w_n+1= ^z

cn

w^d_n−1, n∈_N0, (1.1)

for a,b,c,d ∈ _{Z and} z₋₁,z₀,w₋₁,w₀ ∈ _C\ {₀}, and gave many results on the long-term behavior of solutions to (1.1) by using the obtained closed form formulas. Product-type equa- tions appeared also in the study of the difference equation in [33], as its special cases. The max-type system in [23] is solved by reducing it to a product-type one. They also appeared indirectly in the study of some max-type and related difference equations and systems, as their boundary cases (see, e.g., [19,20,37]). The study was continued in [42], in [30] where a three-dimensional system was considered, in [28] where it was noticed for the first time that some coefficients can be added to a product-type system so that the solvability is preserved, and later in [31,35,43–45] where various new details and methods are presented.

This paper continues investigating the solvability problem, by studying the following product-type system

z_n+1=αz^a_nw^b_n, w_n+1= βw^c_n−1z^d_n−2, n ∈_N0, (1.2) wherea,b,c,d∈_Z,α,β∈_Candz−2,z−1,z0,w−1,w0∈ _C.

Clearly, the domain of undefinable solutions [24] to system (1.2) is a subset of

U ={(z−2,z₋₁,z₀,w₋₁,w₀)∈_C⁵ _:z−2 =_{0 or}z₋₁ =_{0 or}z₀=_{0 or}w₋₁ =_{0 or}w₀ =₀}_. Thus, from now on we will assume thatz−2,z−1,z₀,w−1,w₀ ∈ _C\ {0}. Since the casesα = 0 and β = 0 are trivial or produce solutions which are not well-defined we will also assume thatαβ6=0.

In the main case when bd 6= 0, a polynomial of the fourth order is associated to the system, and its solutions are represented in terms of the parameters, through the roots of the polynomial in all the cases (the roots are given in terms of parametersa,b,c,d), which is the main achievement of the paper. This is the first paper which deals with the associated polynomial (to a product-type system) of the fourth order in detail. An associated polynomial of the fourth order appears yet in [42], but almost without any analysis of its roots and their influence on the solutions to the system therein.

In this paper, we will use the following standard convention∑^mi=ka_i =0, whenm< k.

2 Auxiliary results

In this section we quote two auxiliary results which are used in the proofs of the main results.

The first one is the following lemma which is well-known (see, e.g., [6,8]). For a proof of a more general result see [35].

Lemma 2.1. Let i∈_N0and

s⁽_nⁱ⁾(z) =1+2ⁱz+3ⁱz²+· · ·+nⁱzⁿ⁻¹, n∈_N (2.1) where z∈_C.

Then

s⁽_n⁰⁾(z) = ¹−zⁿ

1−z , (2.2)

s⁽_n¹⁾(z) = ¹−(n+₁)zⁿ+nzⁿ⁺¹

(1−z)^{2 , (2.3)}

for every z∈ _C\ {1}and n ∈_N.

The following lemma is also known, and can be proved, for example, by using the Lagrange interpolation polynomial or the calculus of residue (see, for example, [6] and [42]).

Lemma 2.2. Assume thatλ_j, j=1,k,are pairwise different zeros of the polynomial P(z) =α_kz^k+α_k−1z^k−1+· · ·+α₁z+α₀,

withα_kα₀ 6=0.

Then

∑

λ^l_j P⁰(λ_j) =0 for l =0,k−2, and

∑

λ^k_j⁻¹ P⁰(λ_j) = ¹

α_k.

3 Main results

The main results in this paper are formulated and proved in this section.

3.1 Solvability of system (1.2)

The first result concerns the solvability problem of system (1.2).

Theorem 3.1. Assume that a,b,c,d∈ _Z,α,β∈_C\ {0}and z−2,z−1,z₀,w−1,w₀ ∈_C\ {0}. Then system(1.2)is solvable in closed form.

Proof. Case b=0. In this case system (1.2) becomes

z_n+1 =αz^a_n, w_n+1= βw^c_n−1z^d_n−2, n∈_N0. (3.1) From the first equation in (3.1) we get

zn= α^∑

n−1 j=0 a^j

z^a₀ⁿ, n∈_N, (3.2)

from which it follows that

z_n=α

1−an

1−a z₀^an, n∈_{N, (3.3)}

whena 6=1, and

zn= αⁿz₀, n∈_N, (3.4)

whena =1.

Employing (3.2) in the second equation in (3.1), we get w_n =βα^d∑

n−4 j=0a^j

z^da₀ ⁿ⁻³w^c_n−2, (3.5)

forn≥4, from which it follows that w_2n+i = βα^d∑

2n+i−4 j=0 a^j

z^da₀ ²ⁿ⁺ⁱ⁻³w^c_2n+i−2, (3.6) forn≥2 andi=0, 1.

Assume that for somek ∈_Nit has been proved that w_2n+i =β^∑

k−1 j=0c^j

α^d∑

k−1

j=0c^j∑²ⁿl=⁻0^2j+i−4a^l

z^d∑

k−1

j=0c^ja^{2n−2j+i−3}

0 w^c₂^k_(n−k)+i, (3.7) forn≥ k+_{1 and}i=_{0, 1.}

By using (3.6) withnreplaced byn−k and inserting into (3.7) we get w_2n+i = β^∑

k−1 j=0c^j

α^d∑

k−1

j=0c^j∑²ⁿl=⁻0^2j+i−4a^l

z^d∑

k−1

j=0c^ja^{2n−2j+i−3} 0

βα^d∑

2(n−k)+i−4

j=0 a^j

z^da₀ ^{2(n−k)+i−3}w^c_{2(n−k−1)+i}c^k

= β^∑

kj=0c^j

α^d∑

kj=0c^j∑²ⁿl=⁻0^2j+i−4a^lz^d∑

kj=0c^ja^{2n−2j+i−3}

0 w^c₂^k₍⁺_n¹_−k−1)+i, (3.8) forn≥ k+2 andi=0, 1, from which along with (3.6) and the method of induction it follows that (3.7) holds for everyn≥k+1 andi=0, 1.

Choosingk =n−1 in (3.7) we obtain w_2n= β^∑

n−2 j=0 c^j

α^d∑

n−2

j=0 c^j∑²ⁿi=⁻0^2j−4aⁱ

z^d∑

n−2

j=0 c^ja^2n−2j−3

0 w^c₂ⁿ⁻¹, (3.9)

and

w_2n+1= β^∑

n−2 j=0c^j

α^d∑

n−2

j=0 c^j∑²ⁿ_i=⁻0^2j−3aⁱ

z^d∑

n−2

j=0 c^ja^2n−2j−2

0 w^c₃ⁿ⁻¹, (3.10)

forn≥2.

From the second equation in (3.1), withn=0, 1, 2, we have

w₁= βw^c₋₁z^d₋₂, w₂ =βw^c₀z^d₋₁, w₃ =βw₁^cz^d₀ =β^1+cw₋^c2₁z^cd₋₂z^d₀. (3.11) Then by using (3.11) into (3.9) and (3.10) we have

w_2n= β^∑

n−2 j=0c^j

α^d∑

n−2

j=0 c^j∑²ⁿi=⁻0^2j−4aⁱ

z^d∑

n−2

j=0 c^ja^2n−2j−3

0 (βw^c₀z^d₋₁)^cn−1

= β^∑

n−1 j=0c^j

α^d∑

n−2

j=0 c^j∑²ⁿi=⁻0^2j−4aⁱ

w₀^cnz^dc₋₁ⁿ⁻¹z^d∑

n−2

j=0 c^ja^2n−2j−3

0 (3.12)

and

w_2n+1 =β^∑

n−2 j=0 c^j

α^d∑

n−2

j=0c^j∑²ⁿi=⁻0^2j−3aⁱ

z^d∑

n−2

j=₀c^ja^2n−2j−2

0 (β^1+cw^c₋²₁z^cd₋₂z^d₀)^cn−1

=β^∑

nj=0c^j

α^d∑

n−2

j=0 c^j∑²ⁿi=⁻0^2j−3aⁱ

w^c₋ⁿ⁺₁¹z^dc₋₂ⁿz^d∑

n−1

j=0c^ja^2n−2j−2

0 , (3.13)

forn≥2.

Subcase a6=16=c, c 6= a². In this case we have w_2n= β^∑

n−1 j=0c^j

α^d∑

n−2

j=0 c^j∑²ⁿi=⁻0^2j−4aⁱ

w^c₀ⁿz^d∑

n−2

j=₀ c^ja^2n−2j−3 0 z^dc₋₁ⁿ⁻¹

= β

1−cn 1−cα^d∑

n−2

j=0 c^j1−a2n₁₋⁻_a^2j−3

w^c₀ⁿz^da

a2n−2−cn−1 a2−c

0 z^dc₋₁ⁿ⁻¹

= β

1−cn 1−cα

d 1−a 1−cn−1

1−c −^{a(a2n−2−cn−1)}

a2−c

w^c₀ⁿz^da

a2n−2−cn−1 a2−c

0 z^dc₋₁ⁿ⁻¹

= β

1−cn 1−cα

d(a2−c+(a+c)(1−a)cn−1−(1−c)a2n−1) (1−a)(1−c)(a2−c) w₀^cnz^da

a2n−2−cn−1 a2−c

0 z^dc₋₁ⁿ⁻¹, (3.14) forn≥2, and

w_2n+1 =β^∑

nj=0c^j

α^d∑

n−2

j=0c^j∑²ⁿi=⁻0^2j−3aⁱ

w^c₋ⁿ₁⁺¹z^d∑

n−1

j=0c^ja^2n−2j−2

0 z^dc₋₂ⁿ

=β

1−cn+1 1−c α^d∑

n−2

j=0c^j1−a2n₁₋⁻_a^2j−2

w^c₋ⁿ₁⁺¹z^d

a2n−cn a2−c

0 z^dc₋₂ⁿ

=β

1−cn+1 1−c α

d

1−a 1−cn−1

1−c −^{a2(a2n−2−cn−1)}

a2−c

w^c₋ⁿ₁⁺¹z^d

a2n−cn a2−c

0 z^dc₋₂ⁿ

=β

1−cn+1 1−c α

d(a2−c+(₁−a2)cn−(1−c)a2n)

(1−a)(1−c)(a2−c) w^c₋ⁿ⁺₁¹z^d

a2n−cn a2−c

0 z^dc₋₂ⁿ, (3.15)

forn∈_N.

Subcase a²6=16=c, c= a². In this case we have w_2n= β^∑

n−1 j=0a^2j

α^d∑

n−2

j=0 a^2j∑²ⁿi=⁻0^2j−4aⁱ

w^a₀²ⁿz^d∑

n−2

j=0a^2ja^2n−2j−3 0 z^da₋₁²ⁿ⁻²

= β

1−a2n 1−a2 α^d∑

n−2

j=0 a^2j1−a2n₁₋⁻_a^2j−3

w^a₀²ⁿz₀^{d(n−1)a2n−3}z^da₋₁²ⁿ⁻²

= β

1−a2n 1−a2

α

1−da 1−a2n−2

1−a2 −(n−1)a²ⁿ⁻³

w^a₀²ⁿz^d₀^{(n−1)a2n−3}z^da₋₁²ⁿ⁻²

= β

1−a2n 1−a2 α

d(1−(n−1)a2n−3−a2n−2+(n−1)a2n−1)

(a−1)²(a+1) w^a₀²ⁿz^d₀^{(n−1)a2n−3}z^da₋₁²ⁿ⁻² (3.16) forn≥2, and

w_2n+₁= β^∑

n j=0a^2j

α^d∑

n−2

j=0 a^2j∑²ⁿi=⁻0^2j−3aⁱ

w^a₋²ⁿ₁⁺²z^d∑

n−1

j=0a^2ja^2n−2j−2 0 z^da₋₂²ⁿ

= β

1−a2n+₂ 1−a2

α^d∑

n−2

j=0 a^2j1−a2n₁₋⁻_a^2j−2

w^a₋²ⁿ₁⁺²z^dna₀ ²ⁿ⁻²z^da₋₂²ⁿ

= β

1−a2n+2 1−a2

α

d

1−a 1−a2n−2

1−a2 −(n−1)a²ⁿ⁻²

w^a₋²ⁿ₁⁺²z^dna₀ ²ⁿ⁻²z^da₋₂²ⁿ

= β

1−a2n+₂ 1−a2

α

d(1−na2n−2+(n−1)a2n)

(a−1)2(a+1) w₋^a2n₁⁺²z^dna₀ ²ⁿ⁻²z^da₋₂²ⁿ, (3.17) forn≥2.

Subcase a²6=₁= c.In this case we have w_2n =β^∑

n−1 j=0 1

α^d∑

n−2

j=0∑²ⁿi=⁻0^2j−4aⁱ

w₀z^d∑

n−2 j=0 a^2n−2j−3

0 z^d₋₁

=βⁿα^d∑

n−2

j=0 1−a2n−2j−3 1−a w₀z^da

a2n−2−1 a2−1

0 z^d₋₁

=βⁿα

d

1−a n−1−^{a(a2n−2−1)}

a2−1

w₀z^da

a2n−2−1 a2−1

0 z^d₋₁

=βⁿα

d(a2n−1+(n−1)(1−a2)−a) (a−1)2(a+1) w₀z^da

a2n−2−1 a2−1

0 z^d₋₁, (3.18)

forn≥2, and

w_2n+1= β^∑

nj=01

α^d∑

n−2

j=0 ∑²ⁿi=⁻0^2j−3aⁱ

w−1z^d∑

n−1 j=₀ a^2n−2j−2

0 z^d₋₂

= βⁿ⁺¹α^d∑

n−2

j=0 1−a2n−2j−2 1−a w−1z^d

a2n−1 a2−1

0 z^d₋₂

= βⁿ⁺¹α

1−da n−1−^{a2(a2n−2−1)}

a2−1

w−1z^d

a2n−1 a2−1

0 z^d₋₂

= βⁿ⁺¹α

d(a2n−na2+n−1) (a−1)2(a+₁) w−1z^d

a2n−1 a2−1

0 z^d₋₂, (3.19)

forn∈ _N.

Subcase a=−1, c=1. In this case we have w_2n=βⁿα^d∑

n−2

j=0∑²ⁿi=⁻0^2j−4(−1)ⁱ

w₀z^d∑

n−2

j=0(−1)^2n−2j−3

0 z^d₋₁

=βⁿα^d∑

n−2 j=0

1−(−1)²ⁿ−2j−3

1−(−1) w₀z^d₀⁽¹⁻ⁿ⁾z^d₋₁

=βⁿα^d(n−1)w₀z^d₀⁽¹⁻ⁿ⁾z^d₋₁, (3.20) forn∈ _{N, and}

w_2n+1 =β^∑

nj=01

α^d∑

n−2

j=0∑²ⁿi=⁻0^2j−3aⁱ

w−1z^d∑

n−1 j=0 a^2n−2j−2

0 z^d₋₂

=βⁿ⁺¹α^d∑

n−2

j=0∑²ⁿi=⁻0^2j−3(−1)ⁱ

w−1z^d∑

n−1

j=0(−1)^2n−2j−2

0 z^d₋₂

=βⁿ⁺¹α^d∑

n−2 j=0

1−(−1)²ⁿ−2j−2

1−(−1) w−1z^dn₀ z^d₋₂

=βⁿ⁺¹w−1z^dn₀ z^d₋₂, (3.21)

forn∈ _N0.

Subcase a=1, c6=1. In this case, by using formula (2.3), we get w2n =β^∑

n−1 j=0 c^j

α^d∑

n−2

j=0c^j∑²ⁿi=⁻0^2j−41

w^c₀ⁿz^d∑

n−2 j=0c^j 0 z^dc₋₁ⁿ⁻¹

=β

1−cn 1−c α^d∑

n−2

j=0(2n−2j−3)c^j

w^c₀ⁿz^d

1−cn−1 1−c

0 z^dc₋₁ⁿ⁻¹

=β

1−cn

1−c α^{d (2n−1)}

1−cn−1

1−c −2^{1−ncn−1+(n−1)cn} (1−c)²

w^c₀ⁿz^d1

−cn−1 1−c

0 z^dc₋₁ⁿ⁻¹

=β

1−cn 1−c α^d

2n−3−(2n−1)c+cn−1+cn (1−c)2 w^c₀ⁿz^d

1−cn−1 1−c

0 z^dc₋₁ⁿ⁻¹, (3.22)

forn≥2, and

w_2n+1 =β^∑

n j=₀c^j

α^d∑

n−2

j=₀c^j∑²ⁿ_i=⁻0^2j−31

w^c₋ⁿ⁺₁¹z^d∑

n−1 j=0c^j 0 z^dc₋₂ⁿ

=β

1−cn+1 1−c α^d∑

n−2

j=0(2n−2j−2)c^j

w^c₋ⁿ₁⁺¹z^d

1−cn 1−c

0 z^dc₋₂ⁿ

=β

1−cn+1 1−c α^{d 2n}

1−cn−1

1−c −2^1−ncn−₍₁¹₋⁺⁽_c)ⁿ₂^−1)cn

w^c₋ⁿ⁺₁¹z^d

1−cn 1−c

0 z^dc₋₂ⁿ

=β

1−cn+1 1−c α

2d(n−1−nc+cn)

(1−c)² w^c₋ⁿ₁⁺¹z^d

1−cn 1−c

0 z^dc₋₂ⁿ, (3.23)

forn∈ _N.