Existence of nonoscillatory solutions to second-order nonlinear neutral difference equations

Research Article

Existence of nonoscillatory solutions to second-order nonlinear neutral difference equations

Yazhou Tian^a,b, Yuanli Cai^a, Tongxing Li^b,∗

aSchool of Electronic and Information Engineering, Xi’an Jiaotong University, Xi’an, Shaanxi 710049, P. R. China.

bQingdao Technological University, Feixian, Shandong 273400, P. R. China.

Communicated by Martin Bohner

Abstract

We study a class of second-order neutral delay difference equations with positive and negative coefficients

∆(rn(∆(xn+pxn−m))) +pnf(xn−k)−qng(xn−l) = 0, n=n0, n0+ 1, . . . ,

wherep ∈R, m, k, l, n₀ ∈N,p_n, q_n, r_n∈R⁺, f, g ∈C(R, R) withxf(x) >0 and xg(x)>0 (x 6= 0). Some sufficient conditions for the existence of a nonoscillatory solution of the studied equation expressed in terms of P∞

Rnpn < ∞ and P∞

Rnqn < ∞ are obtained, where Rn = Pn s=n0

1

rs, n ≥ n0. 2015 All rightsc reserved.

Keywords: Nonoscillatory solution, neutral delay difference equation, second-order, positive and negative coefficients.

2010 MSC: 34B15, 34B25.

1. Introduction

This paper is concerned with a second-order neutral delay difference equation with positive and negative coefficients

∆(r_n(∆(x_n+pxn−m))) +p_nf(xn−k)−q_ng(xn−l) = 0, n=n₀, n₀+ 1, . . . , (1.1) where ∆ stands for the forward difference operator, ∆xn =xn+1−xn,p∈R, m, k, l, n0 ∈N, pn, qn, rn∈ R⁺, f, g ∈ C(R, R), xf(x) > 0, and xg(x) > 0 for all x 6= 0. Throughout, we suppose that the following assumptions are satisfied.

∗Corresponding author

Email addresses: [email protected](Yazhou Tian),[email protected](Yuanli Cai),[email protected] (Tongxing Li)

Received 2014-12-23

(H₁) f and g satisfy local Lipchitz conditions, Lipchitz constants are denoted by L_f(A) and L_g(A), where Ais the domain that f and g are defined;

(H₂) R_n=Pn s=n0

1

rs, n≥n₀,P∞

R_sp_s<∞, and P∞

R_sq_s<∞.

In recent years, there has been an increasing interest in studying the oscillatory and nonoscillatory be- havior of various classes of differential, difference, and dynamic equations; see, for instance, the monographs [1, 2], papers [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], and the references cited therein. Candan [3] investigated a higher-order nonlinear neutral differential equation

h

r(t)(x(t) +P(t)x(t−τ))⁽ⁿ⁻¹⁾i0

+ (−1)ⁿ[Q₁(t)g₁(x(t−σ₁))−Q₂(t)g₂(x(t−σ₂))−f(t)] = 0, (1.2) where t ≥t0, n ≥ 2 is an integer, r ∈ C([t0,∞), R⁺), P, f ∈C([t0,∞), R), Qi ∈ C([t0,∞), R⁺), i= 1,2, and gi ∈ C(R, R), i = 1,2, satisfy the local Lipschitz condition with xgi(x) > 0, i = 1,2 for x 6= 0.

Using the Banach contraction principle, the author obtained some sufficient conditions for the existence of nonoscillatory solutions to (1.2). Cheng [6] studied the existence of nonoscillatory solution of a second-order linear neutral difference equation

∆²(x_n+pxn−m) +p_nxn−k−q_nxn−l= 0, n=n₀, n₀+ 1, . . . , (1.3) where p ∈R,m, k, l, n0 ∈ N, pn, qn ∈R⁺, and some other special cases of equation (1.3) were considered by Li et al. [9] and Zhang and Zhou [13]. In particular, Cheng [6] established the following result.

Theorem 1.1 (See [6, Theorem 1]). Suppose that p6=−1,P∞

sps <∞, and P∞

sqs <∞. Then equation (1.3)has a nonoscillatory solution.

To the best of our knowledge, there are few results for second-order nonlinear difference equations with positive and negative coefficients. Motivated by the ideas exploited in [3, 6], we obtain the global results (with respect to p), which are some sufficient conditions for the existence of a nonoscillatory solution of (1.1) forp6=−1. The results obtained extend those reported in [6]. An example is considered to illustrate the possible applications.

2. Main results

Theorem 2.1. Assume that p6=−1 and conditions(H1) and (H2) are satisfied. Then (1.1) has a bounded nonoscillatory solution.

Proof. The proof of Theorem 2.1 will be divided into five cases, depending on the five different ranges of the parameter p. Let l^∞_n0 be the Banach space which is composed of all bounded real sequences x ={x_n}^∞_n=n

0

with the norm||x||= sup_n≥n0|x_n|.

Case 1. p = 1. By (H₁) and (H₂), one can choose an n∗ ≥n₀ + max{m, k, l} sufficiently large such that, for all n≥n∗,

∞

X

u=n

(R_u−Rn−1)p_u ≤ 1 α,

∞

X

u=n

(R_u−Rn−1)q_u ≤ 1 β,

∞

X

u=n

(Ru−Rn−1)(pu+qu)<min 1

L,1 α + 1

β

,

(2.1)

whereα= max1≤x≤3{f(x)},β = max1≤x≤3{g(x)}, and L= max{L_f([1,3]), Lg([1,3])}.

We define a bounded, closed, and convex subset S inl_n^∞₀ by

S={x={x_n} ∈l^∞_n0 : 1≤x_n≤3, n≥n₀}.

Consider the operatorT :S→l^∞_n0 defined by

(T x)_n=









 2−

∞

X

j=1

n+2jm

X

s=n+(2j−1)m

1 r_s

∞

X

u=s

(puf(xu−k)−qug(xu−l))

!

, n≥n∗,

(T x)_n∗, n₀≤n≤n∗.

Clearly,T xn is a real sequence. It is not difficult to show that T is a continuous mapping on S. For every x={x_n} ∈S andn≥n∗, we obtain

(T x)n≤2 +

∞

X

j=1





n+2jm

X

s=n+(2j−1)m

1 rs

∞

X

u=s

qug(xu−l) +

n+(2j−1)m

X

s=n+(2j−2)m

1 rs

∞

X

u=s

qug(xu−l)





= 2 +

∞

X

s=n

1 r_s

∞

X

u=s

q_ug(xu−l) = 2 +

∞

X

u=n u

X

s=n

1

r_sq_ug(xu−l)

= 2 +

∞

X

u=n

(Ru−Rn−1)qug(xu−l)≤2 +β

∞

X

u=n

(Ru−Rn−1)qu ≤3.

On the other hand, we have (T x)_n≥2−

∞

X

j=1





n+2jm

X

s=n+(2j−1)m

1 r_s

∞

X

u=s

p_uf(xu−k) +

n+(2j−1)m

X

s=n+(2j−2)m

1 r_s

∞

X

u=s

p_uf(xu−k)





= 2−

∞

X

s=n

1 r_s

∞

X

u=s

puf(xu−k) = 2−

∞

X

u=n u

X

s=n

1

r_spuf(xu−k)

= 2−

∞

X

u=n

(R_u−Rn−1)p_uf(x_u−l)≥2−α

∞

X

u=n

(R_u−Rn−1)p_u≥1.

Thus, we conclude that T S⊆S.

Next, we prove thatT is a contraction mapping onS. As a matter of fact, for everyx, y∈S andn≥n∗, we get

|T x_n−T y_n| ≤

∞

X

j=1

n+2jm

X

s=n+(2j−1)m

1 rs

∞

X

u=s

(p_u|f(xu−k)−f(yu−k)|+q_u|g(x_u−l)−g(yu−l)|)

!

≤L||x−y||

∞

X

s=n

1 r_s

∞

X

u=s

(pu+qu) =L||x−y||

∞

X

u=n u

X

s=n

1

r_s(pu+qu)

=L||x−y||

∞

X

u=n

(Ru−Rn−1)(pu+qu) =p0||x−y||, which implies that

||T x−T y|| ≤p0||x−y||, wherep0 =LP∞

u=n(Ru−Rn−1)(pu+qu). Using (2.1), we havep0 <1, and thusT is a contraction mapping.

Consequently,T has a unique fixedx such that (T x)_n=x_n, that is,

x_n=









 2−

∞

X

j=1

n+2jm

X

s=n+(2j−1)m

1 r_s

∞

X

u=s

[puf(xu−k)−qug(xu−l)], n≥n∗, (T x)_n∗, n₀ ≤n≤n∗.

Furthermore, we have

xn+xn−m = 4−

∞

X

j=1





n+2jm

X

s=n+(2j−1)m

+

n+(2j−1)m

X

s=n+(2j−2)m



 1 r_s

∞

X

u=s

(puf(xu−k)−qug(xu−l))

= 4−

∞

X

s=n

1 rs

∞

X

u=s

(p_uf(xu−k)−q_ug(xu−l)).

Therefore,

∆(r_n(∆(x_n+xn−m))) +p_nf(xn−k)−q_ng(xn−l) = 0, and x_n is obviously a positive solution of (1.1). This completes the proof of Case 1.

Case 2. p ∈ (0,1). By virtue of conditions (H1) and (H2), we can choose an n1 ≥ n0 + max{m, k, l}

sufficiently large such that

∞

X

s=n

Rsps≤ p−(1−N1) α₁ ,

∞

X

s=n

Rsqs≤ 1−p−pN1−M1

β1

,

∞

X

s=n

R_s(p_s+q_s)< 1−p L₁

(2.2)

hold for all n ≥ n₁, where N₁ ≥ M₁ > 0, 1−N₁ < p < (1−M₁)/(1 +N₁), α₁ = max_M1≤x≤N1{f(x)}, β₁ = max_M1≤x≤N₁{g(x)}, andL₁= max{L_f([M₁, N₁]), L_g([M₁, N₁])}. Set

A₁ ={x={x_n} ∈l_n^∞₀ :M₁≤x_n≤N₁, n≥n₀}.

Define an operatorT :A1 →l_n^∞₀ by

(T x)_n=



















1−p−pxn−m+Rn−1

∞

X

s=n−1

(psf(xs−k)−qsg(xs−l))

+

n−2

X

s=n1

Rs(psf(xs−k)−qsg(xs−l)), n≥n1, (T x)_n1, n₀ ≤n≤n₁.

For everyx∈A₁ andn≥n₁, we have

(T x)_n≤1−p+α₁Rn−1

∞

X

s=n−1

p_s+α₁

n−2

X

s=n1

R_sp_s

≤1−p+α1

∞

X

s=n1

Rsps ≤N1.

Furthermore, we get

(T x)n≥1−p−pN1−Rn−1

∞

X

s=n−1

qsg(xs−l)−

n−2

X

s=n1

Rsqsg(xs−l)

≥1−p−pN₁−β₁

∞

X

s=n1

R_sq_s≥M₁,

and henceT A₁ ⊆A₁.

Now, for x, y∈A1 and n≥n1, we obtain

|T x_n−T yn| ≤p|x_n−m−yn−m|+Rn−1

∞

X

s=n−1

ps|f(xs−k)−f(ys−k)|

+Rn−1

∞

X

s=n−1

q_s|g(x_s−l)−g(ys−l)|+

n−2

X

s=n1

R_sp_s|f(xs−k)−f(ys−k)|

+

n−2

X

s=n1

R_sq_s|g(x_s−l)−g(ys−l)|

≤p||x−y||+L₁||x−y||

∞

X

s=n1

R_s(p_s+q_s)

= ˆq1||x−y||, where ˆq₁ =p+L₁P∞

s=n1R_s(p_s+q_s)<1 due to (2.2). This immediately yields

||T x−T y|| ≤qˆ₁||x−y||,

and so T is a contraction mapping. Consequently, T has a unique fixed x, which is obviously a positive solution of (1.1). This completes the proof of Case 2.

Case 3. p∈(1,∞). From (H₁) and (H₂), one can choose an n₂≥n₀+ max{m, k, l}sufficiently large such

that _∞

X

s=n

R_sp_s≤ 1−p(1−N₂)

α₂ ,

∞

X

s=n

Rsqs≤ (1−M2)p−(1 +N2)

β₂ ,

∞

X

s=n

Rs(ps+qs)< p−1 L2

(2.3)

hold for all n ≥n2, where N2 ≥ M2 > 0, (1−M2)p > 1 +N2, p(1−N2) < 1, α2 = maxM2≤x≤N₂{f(x)}, β2 = maxM2≤x≤N2{g(x)}, andL2= max{L_f([M2, N2]), Lg([M2, N2])}. Set

A2 ={x={x_n} ∈l_n^∞₀ :M2≤xn≤N2, n≥n0}.

Define an operatorT :A₂ →l_n^∞₀ as

(T x)n=

















 1−1

p −1

px_n+m+1

pRn+m−1

∞

X

s=n+m−1

(p_sf(xs−k)−q_sg(xs−l))

+ 1 p

n+m−2

X

s=n2

R_s(p_sf(xs−k)−q_sg(xs−l)), n≥n₂, (T x)n2, n0 ≤n≤n2.

For everyx∈A2 andn≥n2, we get (T x)n≤1− 1

p+1

pα2Rn+m−1

∞

X

s=n+m−1

ps+1 pα2

n+m−2

X

s=n2

Rsps

≤1− 1 p+1

pα₂

∞

X

s=n2

R_sp_s ≤N₂.

Furthermore, we have

(T x)_n≥1−1 p− 1

pN₂− 1

pβ₂Rn+m−1

∞

X

s=n+m−1

q_s−1 pβ₂

n+m−2

X

s=n2

R_sq_s

≥1−1 p− 1

pN2− 1 pβ2

∞

X

s=n2

Rsqs ≥M2,

and thusT A₂ ⊆A₂. SinceA₂ is a bounded, closed, and convex subset of l^∞_n0, we have to prove that T is a contraction mapping onA2 to apply the contraction principle.

Now, for x, y∈A2 and n≥n2, we obtain

|T x_n−T y_n| ≤ 1

p|x_n+m−y_n+m|+1

pRn+m−1

∞

X

s=n+m−1

p_s|f(xs−k)−f(ys−k)|

+1

pRn+m−1

∞

X

s=n+m−1

q_s|g(xs−l)−g(ys−l)|+1 p

n+m−2

X

s=n2

R_sp_s|f(xs−k)−f(ys−k)|

+1 p

n+m−2

X

s=n2

Rsqs|g(x_s−l)−g(ys−l)|

≤ 1

p||x−y||+ 1

pL2||x−y||

∞

X

s=n2

Rs(ps+qs)

= ˆq₂||x−y||, which yields

||T x−T y|| ≤qˆ2||x−y||.

From (2.3), we have ˆq₂ = 1/p(1 +L₂Σ^∞_s=n2R_s(p_s +q_s)) < 1. Therefore, T is a contraction mapping.

Consequently, T has a unique fixed x, which is obviously a positive solution of (1.1). The proof of Case 3 is complete.

Case 4. p∈(−1,0). Combining (H₁) and (H₂), we can choose ann₃ ≥n₀+ max{m, k, l}sufficiently large such that

∞

X

s=n

Rsps≤ (1 +p)N3−(1 +p)

α₃ ,

∞

X

s=n

Rsqs≤ 1 +p−M₃(1 +p) β3

,

∞

X

s=n

R_s(p_s+q_s)< 1 +p L₃

(2.4)

hold for all n ≥ n₃, where M₃ and N₃ are positive constants satisfying 0 < M₃ < 1 < N₃, α₃ = max_M3≤x≤N₃{f(x)},β₃ = max_M3≤x≤N₃{g(x)}, and L₃ = max{L_f([M₃, N₃]), L_g([M₃, N₃])}. Set

A₃ ={x={x_n} ∈l^∞_n0 :M₃ ≤x_n≤N₃, n≥n₀}.

Define an operatorT :A₃ →l_n^∞₀ by

(T x)n=



















1 +p−pxn−m+Rn−1

∞

X

s=n−1

(p_sf(xs−k)−q_sg(xs−l))

+

n−2

X

s=n3

R_s(p_sf(xs−k)−q_sg(xs−l)), n≥n₃, (T x)_n3, n₀ ≤n≤n₃.

For everyx∈Aand n≥n₃, we have

(T x)_n≤1 +p−pN₃+α₃Rn−1

∞

X

s=n−1

p_s+α₃

n−2

X

s=n3

R_sp_s

≤1 +p−pN3+α3

∞

X

s=n3

Rsps≤N3.

Furthermore, we conclude that

(T x)n≥1 +p−pM3−Rn−1

∞

X

s=n−1

qsg(xs−l)−

n−2

X

s=n3

Rsqsg(xs−l)

≥1 +p−pM₃−β₃

∞

X

s=n3

R_sq_s≥M₃,

and thus T A3⊆A3.

Next, we prove thatT is a contraction mapping on A₃. In fact, for everyx, y∈A₃ and n≥n₃, we have

|T x_n−T yn| ≤ −p|x_n−m−yn−m|+Rn−1

∞

X

s=n−1

ps|f(xs−k)−f(ys−k)|

+Rn−1

∞

X

s=n−1

qs|g(x_s−l)−g(ys−l)|+

n−2

X

s=n3

Rsps|f(xs−k)−f(ys−k)|

+

n−2

X

s=n3

R_sq_s|g(x_s−l)−g(ys−l)|

≤ −p||x−y||+L₃||x−y||

∞

X

s=n3

R_s(p_s+q_s)

= ˆq3||x−y||.

This immediately yields

||T x−T y|| ≤qˆ3||x−y||, where ˆq₃ = −p+L₃P∞

s=n3R_s(p_s+q_s) < 1 due to (2.4), which implies that T is a contraction mapping.

Consequently, T has a unique fixedx, which is obviously a positive solution of (1.1). This completes the proof of Case 4.

Case 5. p ∈(−∞,−1). From (H₁) and (H2), one can choose an n4 ≥n0+ max{m, k, l} sufficiently large such that

∞

X

s=n

R_sp_s≤ −(p+ 1)(N4−1)

β₄ ,

∞

X

s=n

Rsqs ≤ −(1 +p)(1−M4) α4

,

∞

X

s=n

R_s(p_s+q_s)< −(p+ 1) L4

(2.5)

hold for all n ≥ n4, where M4 and N4 are positive constants satisfying 0 < M4 < 1 < N4, α₄ = max_M4≤x≤N₄{f(x)},β₄ = max_M4≤x≤N₄{g(x)}, and L₄ = max{L_f([M₄, N₄]), L_g([M₄, N₄])}. Set

A₄ ={x={x_n} ∈l^∞_n0 :M₄ ≤x_n≤N₄, n≥n₀}.

Define an operatorT :A₄ →l_n^∞₀ as

(T x)n=

















 1 +1

p −1

px_n+m+1

pRn+m−1

∞

X

s=n+m−1

(p_sf(xs−k)−q_sg(xs−l))

+ 1 p

n+m−2

X

s=n4

R_s(q_sf(xs−k)−q_sg(xs−l)), n≥n₄, (T x)n4, n0 ≤n≤n4.

For everyx∈A4 andn≥n4, we get (T x)n≤1 +1

p− 1

pN4− 1

pβ4Rn+m−1

∞

X

s=n+m−1

qs−1 pβ4

n+m−2

X

s=n4

Rsqs

≤1 +1 p− 1

pN₄− 1 pβ₄

∞

X

s=n4

R_sq_s ≤N₄.

Furthermore, we have

(T x)n≥1 +1 p −1

pM4+1

pα4Rn+m−1

∞

X

s=n+m−1

ps+ 1 pα4

n+m−2

X

s=n4

Rsps

≥1 +1 p −1

pM₄+1 pα₄

∞

X

s=n4

R_sq_s≥M₄,

and so T A4 ⊆ A4. Since A4 is a bounded, closed, and convex subset of l^∞_n0, we have to prove that T is a contraction mapping onA4 to apply the contraction principle.

Now, for x, y∈A₄ and n≥n₄, we have

|T x_n−T yn| ≤ −1

p|x_n+m−yn+m| − 1

pRn+m−1

∞

X

s=n+m−1

ps|f(x_s−k)−f(ys−k)|

−1

pRn+m−1

∞

X

s=n+m−1

qs|g(x_s−l)−g(ys−l)| −1 p

n+m−2

X

s=n4

Rsps|f(xs−k)−f(ys−k)|

−1 p

n+m−2

X

s=n4

Rsqs|g(x_s−l)−g(ys−l)|

≤ −1

p||x−y|| − 1

pL₄||x−y||

∞

X

s=n4

R_s(p_s+q_s)

= ˆq₄||x−y||.

This immediately implies that

||T x−T y|| ≤qˆ₄||x−y||.

By virtue of (2.5), we get ˆq₄ = 1/p(−1−L₄Σ^∞_s=n4R_s(p_s+q_s)) < 1, which proves that T is a contraction mapping. Consequently, T has a unique fixed x, which is obviously a positive solution of (1.1). This completes the proof of Case 5. Therefore, the proof of Theorem 2.1 is complete.

Remark 2.2. One can easily see that Theorem 2.1 includes Theorem 1.1 whenrn= 1 andf(u) =g(u) =u.

3. Applications

Example 3.1. Consider a second-order difference equation

∆(rn∆(xn+xn−1)) +pnxn−2−qnx³_n−2= 0, n= 2,3, . . . , (3.1) wherep= 1,rn= 1/n,f(x) =x,g(x) =x³,

p_n= 2(n−2)

(n+ 2)(n+ 1)n(n−1)(2n−3), and

q_n= 8(n−2)³

(n+ 2)(n+ 1)n(n−1)(2n−3)³. It is easy to verify that

Rn=

n

X

s=2

1 rs

=

n

X

s=2

s= 1

2(n+ 2)(n−1),

∞

X

n=2

Rnpn<∞, and

∞

X

n=2

Rnqn<∞.

Therefore, conditions (H1) and (H2) are satisfied. By Theorem 2.1, equation (3.1) has a bounded nonoscil- latory solution. As a matter of fact, the sequence{x_n}={2 + 1/n} is a nonoscillatory solution of (3.1).

Acknowledgements

This research was supported by the NNSF of China (Grant Nos. 11171178 and 11271225) and the NSF of Shandong Province (ZR2012AL03).

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