On a cyclic system of m difference equations having exponential terms

On a cyclic system of m difference equations having exponential terms

Garyfalos Papaschinopoulos

, Nikos Psarros

and Kyriakos B. Papadopoulos

1School of Engineering, Democritus University of Thrace, Xanthi, 67100, Greece

2American University of the Middle East, Kuwait

Received 8 January 2015, appeared 15 February 2015 Communicated by Stevo Stevi´c

Abstract. In this paper we study the asymptotic behavior of the positive solutions of a cyclic system of the followingmdifference equations:

x_n+1⁽ⁱ⁾ =a_ix⁽ⁱ⁺¹⁾n +b_ix⁽ⁱ⁾_n−1e^−xn(i+1), i=1, 2, . . . ,m−1, x_n+1^(m) =amx⁽¹⁾n +bmx^(m)_n−1e^−x(n1),

wheren=0, 1, . . ., anda_i, b_i,i=1, 2, . . . ,mare positive constants and the initial values x⁽ⁱ⁾₋₁,x⁽ⁱ⁾₀ ,i=1, 2, . . . ,mare positive numbers.

Keywords: difference equations, boundedness, persistence, asymptotic behavior.

2010 Mathematics Subject Classification: 39A10.

1 Introduction

In [14] the authors obtained results concerning the global behavior of the positive solutions for the difference equation:

x_n+1= axn+bx_n−1e^−xn, n=0, 1, . . . (1.1) where a, b are positive constants and the initial values x₋₁, x₀ are positive numbers. This equation can be considered as a biological model, since it arises from models studying the amount of litter in a perennial grassland.

In addition, in [23] the authors studied analogous results for the system of difference equations:

x_n+1= ay_n+bx_n−1e^−yn, y_n+1= cx_n+dy_n−1e^−xn (1.2) where a, b, c, d are positive constants and the initial values x−1, x₀, y−1, y₀ are also positive numbers. For a = cand b= d the system is symmetric, so it is a close to symmetric system.

BCorresponding author. Email: gpapas@env.duth.gr

Studying symmetric and close to symmetric systems of difference equations is an area of a considerable recent interest (see, for example, [5,7,8,10,11,22–24,31–35,38–40]).

In this paper we obtain results concerning the behavior of the positive solutions for the following cyclic system of difference equations:

x⁽_nⁱ₊⁾₁ =a_ix⁽_nⁱ⁺¹⁾+b_ix⁽_nⁱ₋⁾₁e^−x(ni+1), i=1, 2, . . . ,m−_1, x⁽_n^m₊⁾₁ =a_mx⁽_n¹⁾+b_mx⁽_n^m₋⁾₁e^−x(n1),

(1.3)

n = 0, 1, . . ., and a_i, b_i, i = 1, 2, . . . ,m are positive constants and the initial values x⁽₋ⁱ⁾₁, x⁽₀ⁱ⁾, i = 1, 2, . . . ,m are positive numbers. More precisely, we study the existence of the unique nonnegative equilibrium of (1.3). In addition, we investigate the boundedness and the per- sistence of the positive solutions of system (1.3). Finally, we investigate the convergence of the positive solutions of (1.3) to the unique nonnegative equilibrium. We note that if a₁= a2 = · · ·= am = a,b₁ =b2 = · · ·= bm = band(x¹_n,x²_n, . . . ,x^m_n)is a solution of (1.3) and x⁽₋¹₁⁾ = x⁽₋²₁⁾ = · · · = x⁽₋^m₁⁾, x⁽₀¹⁾ = x₀⁽²⁾ = · · · = x⁽₀^m), then it is obvious that x⁽n¹⁾ = x⁽n²⁾ = · · · = x⁽_n^m) = x_n, n = 0, 1, . . . , and so x_n is a solution of (1.1). Moreover, if m= 2 then system (1.3) reduces to system (1.2). Studying cyclic systems of difference equations has attracted some attention recently (see [16,36,37] and the related references therein).

Difference equations and systems of difference equations containing exponential terms have numerous potential applications in biology. A large number of papers dealing with such or related equations have been published. See for example, [2,14,18,20–23,26,29,41] and the references cited therein. We also note that since difference equations have many applications in applied sciences, there is a quite rich bibliography concerning theory and applications of difference equations (see, for example, [1–41] and the references cited therein).

2 Existence and uniqueness of a nonnegative equilibrium for (1.3).

In this section we study the existence and the uniqueness of the positive equilibrium of (1.3).

Proposition 2.1. The following statements are true.

I. Suppose that

a_i,b_i ∈(0, 1), a_i+b_i >1, i=1, 2, . . . ,m. (2.1) Then system(1.3)has a unique positive equilibrium(x¯₁, ¯x₂, . . . , ¯x_m).

II. Consider that a_i,b_i, i =1, 2, . . . ,m are positive constants such that:

a_i,b_i ∈ (0, 1), a_i+b_i <1. (2.2) Then, the zero equilibrium(0, 0, . . . , 0)is the unique nonnegative equilibrium of system(1.3).

Proof. I. We consider the functionsh_i: R⁺ →_R⁺,R⁺= (_0,∞)_, h_i(x) = ^aix

1−b_ie^−x, i=_{1, . . . ,}m.

Then we define

∏

h_j =h_j◦h_j+1◦ · · · ◦h_k, k≥ j,

∏

h_j = I,

where I is the identity function. If we set x_m+1 = x₁ we consider the system of algebraic equations

x_i =h_i(x_i+1), i=1, 2, . . . ,m. (2.3)

From (2.3) for aj∈ {1, 2, . . . ,m}we get

x_j = h_j(x_j+1) =h_j◦h_j+1(x_j+2) =

∏

h_j◦

j−1 s

∏

h_s(x_j) =h_j◦

∏

h_j◦

j−1

∏

h_s(x_j)

= a_j

∏

h_s◦

j−1

∏

h_s(x_j) 1−b_je^{−∏mk=j+1hk◦∏j}

−1

k=1h_k(x_j) =

a_jh_j+1◦

∏

h_s◦

j−1

∏

h_s(x_j) 1−b_je^{−∏mk=j+1hk◦∏j}

−1 k=1h_k(x_j)

=

a_ja_j+1

∏

hs◦

j−1

∏

hs(x_j)

1−b_je^{−∏mk=j+1hk◦∏jk−=11hk(xj)}

1−b_j+₁e^{−∏mk=j+2hk◦∏kj−=11hk(xj)} ...

=

m−1

∏

ashm◦

j−1

∏

hs(_xj)

m−1

∏

1−b_se^{−∏mk=j+1hk◦∏kj−=11hk(xj)}

=

∏

a_s

j−1

∏

h_s(x_j)

m−1

∏

1−bse^{−∏mk=j+1hk◦∏j}

−1

k=1h_k(x_j)

1−bme^{−∏kj−=11hk(xj)}

=

∏

ash₁◦

j−1 s

∏

hs(x_j)

m−1

∏

1−b_se^{−∏mk=j+1hk◦∏kj−=11hk(xj)}

1−b_me^{−∏kj−=11hk(xj)}

=

a₁

∏

as j−1 s

∏

hs(x_j)

m−1

∏

1−b_se^{−∏mk=j+1hk◦∏kj−=11hk(xj)}

1−b_me^{−∏kj−=11hk(xj)}

1−b₁e^{−∏kj−=12hk(xj)}

=

a₁

∏

a_s

j−1

∏

h_s(x_j)

∏

1−b_se^{−∏mk=j+1hk◦∏jk−=11hk(xj)}

1−b₁e^{−∏jk−=12hk(xj)}

=

a₁

∏

a_sh₂◦

j−1

∏

h_s(x_j)

∏

1−b_se^{−∏mk=j+1hk◦∏jk−=11hk(xj)}

1−b₁e^{−∏jk−=12hk(xj)}

=

a₁a2

∏

as j−1

∏

hs(x_j)

∏

1−bse^{−∏mk=j+1hk◦∏}

j−1

k=1hk(xj)

1−b₁e^{−∏jk−=12hk(xj)}

1−b2e^{−∏jk−=13hk(xj)} ...

=

x_j

∏

a_s

∏

1−bse^{−∏mk=s+1hk◦∏jk−=11hk(xj)j−1}

∏

1−bse^{−∏kj−=1s+1hk(xj)} .

We consider the function

F_j(x) =

∏

a_s

∏

1−bse^{−∏mk=s+1hk◦∏kj−=11hk(x)j−1}

∏

1−bse^{−∏jk−=1s+1hk(x)}

−1. (2.4)

Sinceh_k(0) =0 fork=1, 2, . . . ,m, from (2.4) we can prove that

F_j(0) =

∏

a_s

∏

(1−bs)

−1. (2.5)

Then from (2.1) we have that F_j(0) > 0. Moreover, since lim_x→_∞h_k(x) = _∞, k = 1, 2 . . . ,m, from (2.1) and (2.4) we get

xlim→_∞F_j(x) =

∏

a_s−1<0.

Therefore there exists an ¯x_j, j = 1, 2, . . . ,m such that F_j(x¯_j) = 0, j = 1, 2, . . . ,m. So, (x¯₁, ¯x2, . . . , ¯xm) is a positive equilibrium for (1.3). Moreover since h⁰_k(x) = a_k¹⁻₍₁^b₋^ke_b^−x(x+1)

ke^−x)² then

from (2.1) ande^−x(x+1)<1 we geth⁰_k(x)>0, k=1, 2, . . . ,m. Therefore for allk=1, 2, . . . ,m the functions h_k are increasing. Hence for all j = 1, 2, . . . ,m the functions F_j are decreasing.

This implies that (x¯₁, ¯x2, . . . , ¯xm)is the unique positive equilibrium for (1.3). This completes the proof of statement I.

II. From (2.2) and (2.5) we have F_j(0)<0 for all j=1, 2, . . . ,m. Since for all j=1, 2, . . . ,m F_j are decreasing functions it is obvious that the zero equilibrium is the only nonnegative equilibrium. This completes the proof of the proposition.

3 Asymptotic behavior of the positive solutions of (1.3)

In this section we study the boundedness and persistence of the positive solutions of (1.3) and the convergence of the positive solutions of (1.3) to the unique nonnegative equilibrium.

Proposition 3.1.

I. Suppose that

a_i,b_i ∈(_{0, 1})_, i=1, 2, . . . ,m. (3.1) Then every positive solution of (1.3)is bounded.

II. Consider that a_i,b_i, i = 1, 2, . . . ,m are positive constants such that (2.1) holds. Then, every positive solution of (1.3)persists.

Proof. I. Let x⁽_n¹⁾,x⁽_n²⁾, . . . ,x⁽_n^m)

be an arbitrary solution of (1.3) and M =max

x⁽_jⁱ⁾, ln 1

1−a_i

, i=1, 2, . . . ,m, j=−1, 0

. Then arguing as in Lemma 1 of [14] and Theorem 3.1 of [22] we can prove that

x⁽_nⁱ⁾≤ M, i=1, 2, . . . ,m, n=1, 2, . . . and so the solution x⁽_n¹⁾,x_n⁽²⁾, . . . ,x⁽_n^m)

is bounded.

II. Let

r=minn

x⁽_jⁱ⁾, z_i i=1, 2, . . . ,m, j=−1, 0o wherez_i =ln(₁₋^bi_a

i). Arguing as in the proof of Proposition 3.1 of [23] we have the following:

If fori=2, 3, . . . ,m

x⁽₀ⁱ⁾ ≤z_i−1, then

x⁽₁ⁱ⁻¹⁾≥minn

x⁽₀ⁱ⁾,x₋⁽ⁱ⁻₁¹⁾o . In addition, if

x⁽₀ⁱ⁾ >z_i−1, x⁽₋ⁱ⁻₁¹⁾ ≤z_i−1, we take

x₁⁽ⁱ⁻¹⁾ >x⁽₋ⁱ⁻₁¹⁾. Finally, if

x⁽₀ⁱ⁾ >z_i−1, x⁽₋ⁱ⁻₁¹⁾ >z_i−1, we get

x⁽₁ⁱ⁻¹⁾ >z_i−1. So, we have that

x⁽₁ⁱ⁻¹⁾≥r.

We consider now the case

x⁽₀¹⁾ ≤zm, then

x₁^(m) ≥minn

x⁽₀¹⁾,x⁽₋^m₁⁾o .

Furthermore, if

x₀⁽¹⁾>zm, x⁽₋^m₁⁾≤zm, we take

x⁽₁^m)> x⁽₋^m₁⁾. Finally, if

x₀⁽¹⁾>zm, x⁽₋^m₁⁾>zm, we get

x₁^(m) >z_m. So, we have that

x⁽₁^m) ≥r.

Arguing as above and using the method of induction we can prove that:

x⁽_nⁱ⁾ ≥r, n=1, 2, . . . , i=1, 2 . . . ,m.

This completes the proof of the proposition.

In the following proposition we study the convergence of the positive equilibrium of (1.3) to the unique positive equilibrium.

Proposition 3.2. Consider system(1.3)such that relations(2.1)hold. Suppose also that there exists a v∈ {1, 2, . . . ,m}such that

a_j ≤ a_v, b_j ≤

∏

a_s, j=1, 2, . . . ,m. (3.2) Then every positive solution of (1.3)tends to unique positive equilibrium.

Proof. Let x⁽n¹⁾,x⁽n²⁾, . . . ,x⁽n^m)

be an arbitrary solution of (1.3). From Proposition 3.1 there exists numbersl_i,L_i,i=1, 2, . . . ,m, 0<l_i < L_i <_∞such that

lim inf

n→_∞ x⁽_nⁱ⁾=l_i, lim sup

n→_∞

x_n⁽ⁱ⁾ =L_i, i=1, 2, . . . ,m. (3.3) Moreover, since relations (2.1) hold, from Proposition2.1 System (1.3) has a unique positive equilibrium(x¯₁, ¯x₂, . . . , ¯x_m).

First of all we prove that

l_i ≤ L_i ≤ x¯_i, i=1, 2, . . . ,m. (3.4) From (3.3) for everyethere exists an₀ such that for everyn≥n₀

l_i−e≤x⁽_nⁱ⁾≤ L_i+e, i=1, 2, . . . ,m. (3.5) So, relations (1.3) and (3.5) imply that fori=1, 2, . . . ,m−1 andn=0, 1, . . . ,

x⁽_nⁱ₊⁾₁= a_ix⁽_nⁱ⁺¹⁾+b_ix⁽_nⁱ₋⁾₁e^−x(ni+1) ≤a_ix⁽_nⁱ⁺¹⁾+b_i(L_i+e)e^−x(ni+1). (3.6) We setg_Li+e(x) =a_ix+b_i(L_i+e)e^−x. Sinceg⁰⁰_L

i+e(x) =b_i(L_i+e)e^−x >0 we have

g_Li+e(x)≤_max{g_Li+e(l_i+1−e)_, g_Li+e(L_i+1+e)}, x∈[l_i+1−_e, L_i+1+e]_{. (3.7)}

Moreover, from (3.5), (3.6) and (3.7) it follows that x⁽_nⁱ₊⁾₁≤ gL_i+e(x⁽_nⁱ⁺¹⁾)≤max

gL_i+e(l_i+1−e), gL_i+e(L_i+1+e) , which implies that

L_i ≤max

g_Li+e(l_i+1−e)_, g_Li+e(L_i+1+e) _. Then fore→0 it holds

L_i ≤_max{g_Li(l_i+₁)_, g_Li(L_i+₁)}, i=1, 2, . . . ,m−_{1. (3.8)} We claim that

L_i ≤g_Li(L_i+1). (3.9)

Sinceg⁰_Li(x) =a_i−b_iL_ie^−x, forx≥ln(^bi_a^Li

i )we haveg⁰_Li(x)>0. So, from (3.8) forl_i+1 ≥ln(^bi_a^Li

i )

we have that (3.9) is true. Forl_i+₁ <ln(^bi_a^Li

i )we take

g_Li(l_i+1)≤ g_Li(0) =b_iL_i <L_i. (3.10) Then using (3.8) and (3.10), for l_i+₁ <_ln(^bi_a^Li

i )relation (3.9) is satisfied. Hence, our claim (3.9) is true. Then using (3.9) we take

L_i ≤a_iL_i+1+b_iL_ie^−Li+1, i=1, 2. . . . ,m−1 and so we get

L_i ≤ ^aiLi+1

1−b_ie^−Li+1, i=1, 2. . . . ,m−1. (3.11) Similarly we can prove that

Lm ≤ ^amL1

1−bme^−L1. (3.12)

Therefore since the functions h_i, i = 1, 2, . . . ,m defined in the proof of Proposition 2.1 are increasing, then from relations (3.11), (3.12) and arguing as in Proposition 2.1 we can prove that

F_i(L_i)≥0, i=1, 2, . . . ,m. (3.13) Then from (3.13) and since the functionsF_i,i=1, 2, . . . ,mare decreasing and from Proposition 2.1 F(x¯_i) =0,i=1, 2, . . . ,mwe take relations (3.4).

We prove now that

¯

x_j ≤l_j ≤L_j, j=1, 2, . . . ,m. (3.14) Since(x¯₁, ¯x₂, . . . , ¯xm)is the unique positive equilibrium of (1.3) we have that ¯x_i, i=1, 2, . . . ,m satisfy system (2.3). Then by setting ¯x_m+1= x¯₁we have

¯

x_j+1 =ln

b_jx¯_j

¯

x_j−a_jx¯_j+1

, j=1, 2, . . . ,m.

Then sinceb_j <1 we have,

¯

x_j+1 ≤ ^bjx¯j

¯

x_j−a_jx¯_j+₁

−1≤ ^ajx¯j+1

¯

x_j−a_jx¯_j+₁

, j=1, 2, . . . ,m, and so

¯

x_j−a_jx¯_j+1 ≤a_j, j=1, 2, . . . ,m−_1, x¯_m−a_mx¯₁ ≤a_m. (3.15)

From (3.15) for a j∈ {1, 2, . . . ,m}it holds

¯

x_j−a_jx¯_j+1≤ a_j, x¯_j+1−a_j+1x¯_j+2≤ a_j+1, . . . , x¯_m−a_mx¯₁≤ a_m,

¯

x₁−a₁x¯₂≤ a₁, x¯₂−a₂x¯₃≤ a₂, . . . , x¯_j−1−a_j−1x¯_j ≤ a_j−1. (3.16) From the first two relations of (3.16) we get

¯

x_j−a_ja_j+1x¯_j+2≤ a_j+a_ja_j+1

and working similarly to (3.16) we can prove that

¯ x_j ≤

∑

^s

∏

a_i +

∏

a_s

j−1 r

∑

^r

w

∏

a_w

!

1−

∏

a_i

. (3.17)

Then from (3.2) and (3.17) we get

¯

x_j ≤ ^av 1−av

, j=1, 2, . . . ,m. (3.18)

From (1.3) and (3.5) for ane>0 there exists an₀such that for n≥n₀ we get

x⁽_n^j₊⁾₁≥ a_jx⁽_n^j+1)+b_j(l_j−e)e^−x(nj+1) (3.19) wherej∈ {1, 2, . . . ,m−1}. We consider the function

k_lj−e(y) =a_jy+b_j(l_j−e)e^−y. Then sincek⁰_l

j−e(y) =a_j−b_j(l_j−e)e^−ywe have thatk_lj−eis increasing fory≥ln(b_j(l_j−e)/a_j). We claim that

l_j+₁−e

∏

a_s >_ln

b_j(l_j−e) a_j

. (3.20)

From (3.4) and (3.18) we get

l_j ≤ x¯_j ≤ ^av 1−av

and so _a^lj_v −1≤ l_j. Therefore since av <1 for ane>0 we get l_j−e

av

−₁< ^lj−eav

av

−₁≤l_j−_{e. (3.21)}

Moreover, from (1.3) we take

l_j−1≥ a_j−1l_j, l_j−2 ≥a_j−2l_j−1, . . . , l₁ ≥a₁l2, lm ≥ aml₁, l_m−1≥ a_m−1l_m, l_m−2 ≥a_m−2l_m−1, . . . , l_j+₁ ≥a_j+₁l_j+₂.

Then we have l_j ≤ ^lj−1

a_j−1

≤ ^lj−2 a_j−1a_j−2

≤ · · · ≤ ^l1

j−1

∏

as

≤ ^lm am

j−1

∏

as

≤ ^lm−1 a_m−1am

j−1

∏

as

≤ · · · ≤ ^l_m^j+1

s=

∏

a_s

. (3.22)

Then, from (3.21) and (3.22) we have, l_j−e

a_v −1≤ ^l_m^j+1

s=

∏

a_s

−e

and so

l_j−e av

∏

a_s−

∏

a_s≤l_j+₁−e

∏

a_s. Then from (2.1) and (3.2) it follows that

b_j

a_j(l_j−e)−₁<l_j+1−e

∏

a_s. (3.23)

Therefore, from (3.23) and since lnx ≤ x−1 our claim (3.20) is true. Moreover, there exists a n₁such that for n≥n₁

x⁽_n^j+1) ≥l_j+₁−e

∏

a_s. (3.24)

Since k_lj−e is an increasing function for y ≥ ln(b_j(l_j−e)/a_j), then from (3.20) and (3.24) we take

k_lj−e(x⁽_n^j+1))≥k_lj−e l_j+1−e

∏

as

! . Then from (3.19) it follows that

x⁽_n^j₊⁾₁≥ a_j l_j+1−e

∏

a_s

!

+b_j(l_j−e)e^{−(lj+1−e∏ms=1,s6=jas)} and so

l_j ≥a_j l_j+1−e

∏

as

!

+b_j(l_j−e)e^{−(lj+1−e∏ms=1,s6=jas)}. (3.25) For e→0 to (3.25) we have

l_j ≥ a_jl_j+1+b_jl_je^−lj+1, j=1, 2, . . . ,m−1. (3.26) Similarly we can prove that

l_m ≥a_ml₁+b_ml_me^−l1. (3.27) From (3.26) and (3.27) we can prove that

F_j(_lj)≤0, j=1, 2, . . . ,m. (3.28) But since from Proposition2.1 F_j(x¯_j) =0, j=1, 2, . . . ,mwe get

F_j(l_j)≤ F_j(x¯_j), j=1, 2, . . . ,m.

Since F_j, j=1, 2, . . . ,mare decreasing functions we take (3.14). Then from (3.4) and (3.14) we have ¯x_j =l_j = L_j, j=1, 2, . . . ,m. This completes the proof of the proposition.

In the last proposition we study the convergence of the positive solutions of (1.3) to the zero equilibrium.

Proposition 3.3. Consider system (1.3) such that the constants a_i,b_i, i = 1, 2, . . . ,m satisfy(2.2).

Then every positive solution of (1.3)tends to the zero equilibrium(0, 0, . . . , 0).

Proof. Since (2.2) holds, from Proposition 2.1 the only nonnegative equilibrium is the zero equilibrium. Let x⁽_n¹⁾,x⁽_n²⁾, . . . ,x_n^(m)

be an arbitrary solution of (1.3). From (1.3) we take x⁽_nⁱ₊⁾₁ ≤a_ix⁽_nⁱ⁺¹⁾+b_ix⁽_nⁱ₋⁾₁, i=1, 2, . . . ,m−1,

x⁽_n^m₊⁾₁ ≤a_mx⁽_n¹⁾+b_mx⁽_n^m₋⁾₁.

(3.29) We consider the system of difference equations

y⁽_nⁱ₊⁾₁ =a_iy⁽_nⁱ⁺¹⁾+b_iy⁽_nⁱ₋⁾₁, i=1, 2, . . . ,m−1, y⁽_n^m₊⁾₁ =a_my⁽_n¹⁾+b_my⁽_n^m₋⁾₁.

(3.30) Let y⁽_n¹⁾,y⁽_n²⁾, . . . ,y⁽_n^m)

be a solution of (3.29) with initial values y⁽⁻₋₁ⁱ⁾ = x⁽₋ⁱ⁾₁, y⁽₀ⁱ⁾ = x⁽₀ⁱ⁾, i=1, 2, . . . ,m. Then from (3.29) and (3.30), by induction we can easily prove that

x⁽_nⁱ⁾ ≤y⁽_nⁱ⁾, i=1, 2, . . . ,m, n=0, 1, . . . (3.31) We prove that every positive solution of (3.30) tends to the zero equilibrium(0, 0, . . . , 0). Sys- tem (3.30) is equivalent to the system

¯

y_n+1 = Ay¯_n, (3.32)

where ¯y_n = col y⁽_n¹⁾,y⁽_n²⁾, . . . ,y⁽_n^m),y⁽_n¹₋⁾₁,y⁽_n²₋⁾₁, . . . ,y⁽_n^m₋⁾₁

and A is a matrix where in the (i)th line 1 ≤ i ≤ m−1 the only non zero elements are a_i which is the (i+₁)th element and b_i which is the(i+m)th element, in the(m)th line the only non zero elements are am which is the first element andb_m which is the last element, and finally in the(m+j)th line, 1≤ j≤m the unique nonzero element is the(j)th element which is 1.

Let

T=diag 1,e⁻¹,e⁻², . . . ,e^−2m+1 whereeis a positive number such that

a_i+b_i <e^m, i=1, 2, . . . ,m. (3.33) We take the change of variables ¯y_n =Tz¯_nand we get the system

¯

z_n+1 =T⁻¹ATz¯_n, (3.34)

T⁻¹AT is matrix where in the (i)_{th line 1} ≤ i ≤ m−1 the only non zero elements aree⁻¹a_i which is the (i+1)th element and e^−mb_i which is the (i+m)th element, in the (m)th line the only non zero elements aree^m−1a_m which is the first element and e^−mb_m which is the last element, and finally in the(m+j)_{th line, 1}≤ j ≤ mthe unique nonzero element is the (j)_th element which ise^m.

If for a 2m×2m matrixC = (c_ij)we take the norm |C| = sup_0≤i≤2m

∑^2mj=1|c_ij| then we take

|T⁻¹AT|=_maxⁿe⁻¹a₁+e^−mb₁, e⁻¹a₂+e^−mb₂, . . . ,e⁻¹a_m−1+e^−mb_m−1, e^m−1a_m+e^−mb_m,e^m

o

≤max{e^−m(a_i+b_i), 1≤i≤m}.

(3.35)