Nonlinear Discrete Inequality in Two Variables with ... - Springer Link

Acta Mathematicae Applicatae Sinica, English Series Vol. 30, No. 2 (2014) 389–400 DOI: 10.1007/s10255-014-0287-x http://www.ApplMath.com.cn & www.SpringerLink.com

Acta Mathemacae Applicatae Sinica, English Series The Editorial Office of AMAS & Springer-Verlag Berlin Heidelberg 2014

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application Hong WANG1 , Ke-long ZHENG1 , Chun-xiang GUO2,† 1 School

of Science, Southwest University of Science and Technology, Mianyang 621010, China

(E-mail: [email protected]) 2 School of Business, Sichuan University, Chengdu 610064, China (E-mail: [email protected])

Abstract

Delay discrete integral inequalities with n nonlinear terms in two variables are discussed, which

generalize some existing results and can be used as powerful tools in the analysis of certain partial difference equations. An application example is also given to show boundedness of solutions of a difference equation. Keywords

discrete inequality; nonlinear; delay; boundedness

2000 MR Subject Classification

1

26D15; 26D20

Introduction

Gronwall-Bellman type integral inequalities play a dominant role in the study of quantitative properties of solutions of differential equations and integral equations. An enormous amount of effort has been devoted to the discovery of new types of inequalities and their applications in various areas (see [1,2,5,7,8]) and the references given therein). Recently, more attention has been paid to the discrete versions of such integral inequalities (for example, [3,4]) and delay inequalities (see [11–13,15,18]). In this paper, we investigate more general nonlinear delay sum-difference inequality: u(m, n) ≤ a(m, n) +

(n−1) k αi (m−1)   βi 

fi (m, n, s, t)wi (u(s, t)),

m, n ∈ N0

(1.1)

i=1 s=αi (0) t=βi (0)

which has k nonlinear terms, where N0 = {0, 1, 2, · · ·}. Our result can generalize some existing results (see [6,8,14]) and be used more effectively to study the boundedness and uniqueness of solutions of certain partial difference equations. Moreover, at the end of this paper, an example is included to illustrate the usefulness of our result.

2

Main Results

In what follows, R denotes the set of real numbers and R+ = (0, ∞). C(X, Y ) denotes the collection of continuous functions from the set X to the set Y . As usual, the empty sum is taken to be 0. We assume that (I1 ) a(m, n) is nonnegative for m, n ∈ N0 and a(0, 0) > 0; Manuscript received August 4, 2010. Revised February 14, 2012. Supported by Doctoral Program Research Foundation of Southwest University of Science and Technology (No.11zx7129) and Fundamental Research Funds for the Central Universities (No. skqy201324). † Corresponding author.

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application

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(I2 ) αi (n), βi (n) are nondecreasing for n ∈ N0 , αi (n), βi (n) ∈ N0 and αi (n) ≤ n, βi (n) ≤ n; (I3 ) fi (m, n, s, t) (i = 1, · · · , k) is nonnegative for m, n, s, t ∈ N0 ; (I4 ) wi (u) ∈ C(R+ , R+ ) (i = 1, · · · , k) is a nondecreasing function. They satisfy the relationship w1 ∝ w2 ∝ · · · ∝ wk , where wi ∝ wi+1 means that wwi+1 is nondecreasing on i (0, ∞). Define a(m, n) = (J1 ) 

max

0≤τ ≤m, 0≤η≤n, τ,η∈N0

(J2 ) fi (m, n, s, t) =

a(τ, η);

max

0≤τ ≤m,0≤η≤n,τ,η∈N0

fi (τ, η, s, t);

(J3 ) Δ1 u(m, n) = u(m + 1, n) − u(m, n) and Δ3 r(m, n, s, t) = r(m, n, s + 1, t) − r(m, n, s, t); u , where ui > 0 is a given constant. From assumption (I4 ), (J4 ) for u ≥ ui , Wi (u) = ui wdz i (z) Wi is strictly increasing so its inverse Wi−1 is well defined, continuous and increasing in its corresponding domain. From (J1 ) and (J2 ) it is clear that clearly  a(m, n) and fi (m, n, s, t) (i = 1, · · · , k) are nonnegative and nondecreasing in m and n, and  a(m, n) ≥ a(m, n) and fi (m, n, s, t) ≥ fi (m, n, s, t). Let αi (−1) = βi (−1) = −1. a(m, n) is nondecreasing in n and u(m, n) Theorem 2.1 Under assumptions (I1 )–(I4 ), if Δ1  is a nonnegative function for m, n ∈ N0 satisfying (1.1), then αk (m−1) βk (n−1) m−1     u(m, n) ≤ Wk−1 Wk ( a(0, n)) + fk (m, n, s, t) + s=αk (0) t=βk (0)

s=0

 Δ3 rk (m, n, s, n) −1 φk (Wk−1 (rk (0, 0, s, 0))) (2.1)

for 0 ≤ m ≤ M1 , 0 ≤ n ≤ N1 , where rk (m, n, s, t) is determined recursively by r1 (m, n, s, t) = a(s, t), αi (s−1) βi (t−1)

ri+1 (m, n, s, t) =Wi (r1 (m, n, 0, t)) +





fi (m, n, τ, η)

τ =αi (0) η=βi (0)

+

s−1 

Δ3 ri (m, n, τ, t) , −1 φ (W i−1 (ri (0, 0, τ, 0))) τ =0 i

(2.2)

i = 1, · · · , k − 1,

i (u) φi (u) = wwi−1 (u) , φ1 (u) = w1 (u), W0 = I (Identity), and M1 and N1 are positive integers satisfying

αi (M1 −1) βi (N1 −1)

Wi ( a(0, N1 )) +  ≤

∞

ui

dz , wi (z)





s=αi (0)

t=βi (0)

i = 1, · · · , k.

fi (M1 , N1 , s, t) +

M 1 −1  s=0

Δ3 ri (M1 , N1 , s, N1 ) −1 φi (Wi−1 (ri (0, 0, s, 0))) (2.3)

∞ Remark 2.1. If wi (i = 1, · · · , k) satisfies ui wdz = ∞, then we may choose N1 = ∞ and i (z) M1 = ∞. As explained in [5], different choices of ui in Wi do not affect our results.

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Remark2.2. (1) Take αi (m − 1) = m − 1 and βi (n − 1) = n − 1 for i = 1, · · · , k, (1.1) can be reduced to the form: u(m, n) ≤ a(m, n) +

k m−1   n−1 

fi (m, n, s, t)wi (u(s, t))

i=1 s=0 t=0

which was discussed in [6]. (2) If α1 (m − 1) = m − 1, β1 (n − 1) = n − 1 and w1 (u) = u, (1.1) for k = 1 becomes the inequality in [14]. If a(m, n) = c, f1 (m, n, s, t) = b(s, t) and k = 1, then the inequality in [14] can be obtained. So these cases mentioned here are special cases of our result. Before giving our proof, we cite the following lemma. Lemma 2.1[6] . For i = 1, · · · , k, Δ3 ri (m, n, s, t) is nonnegative and nondecreasing in m, n and t, and ri (m, n, s, t) is nonnegative and nondecreasing in its arguments.

3

Proof the Main Result

Proof of Theorem 2.1. Fix any arbitrary positive integers M and N with M ≤ M1 , N ≤ N1 . By the definition of functions  a, fi and r1 , we have an auxiliary inequality: u(m, n) ≤ r1 (M, N, m, n) +

(n−1) k αi (m−1)   βi 

fi (M, N, s, t)wi (u(s, t))

(3.1)

i=1 s=αi (0) t=βi (0)

for 0 ≤ m ≤ M, 0 ≤ n ≤ N . Claim that u(m, n) in (3.1) satisfies αk (m−1) βk (n−1)    fk (M, N, s, t) u(m, n) ≤Wk−1 Wk (r1 (M, N, 0, n)) + s=αk (0) t=βk (0)

+

m−1  s=0

 Δ3 rk (M, N, s, n) −1 φk (Wk−1 (rk (0, 0, s, 0)))

(3.2)

for 0 ≤ m ≤ min{M, M2 } and 0 ≤ n ≤ min{N, N2 }, where M2 and N2 are positive integers satisfying αi (M2 −1) βi (N2 −1)





s=αi (0)

t=βi (0)

Wi (r1 (M, N, 0, N2 )) +  ≤

∞

ui

fi (M, N, s, t) +

M 2 −1  s=0

Δ3 ri (M, N, s, N2 ) −1 φi (Wi−1 (ri (0, 0, s, 0)))

dz , i = 1, · · · , k. wi (z)

(3.3)

Notice that we may choose M1 ≤ M2 and N1 ≤ N2 . In fact, by Lemma 2.1 and the definition (J2 ), ri (M, N, m, n), Δ3 ri (M, N, m, n) and fi (M, N, m, n) are nondecreasing in M and N . Thus, M2 and N2 satisfying (3.3) get smaller as M and N are chosen larger. Moreover, M2 and N2 satisfy the same (2.3) as M1 and N1 for M = M1 and N = N1 . Next, we use the mathematical induction to prove our result. (I). k = 1. α1 (m−1)  β1 (n−1)   Let z(m, n) = f1 (M, N, s, t)w1 (u(s, t)). Obviously, z(0, n) = 0 and z(m, n) s=α1 (0) t=β1 (0)

is nonnegative and nondecreasing in each variable. It follows from (3.1) that u(m, n) ≤ r1 (M, N, m, n) + z(m, n),

(3.4)

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application

392

for 0 ≤ m ≤ M, 0 ≤ n ≤ N . By assumptions (I2 ) and (I3 ), definition (J3 ) and Lemma 2.1, we get β1 (n−1)



Δ1 z(m, n) =z(m + 1, n) − z(m, n) =

f1 (M, N, α1 (m), t)w1 (u(α1 (m), t))

t=β1 (0) β1 (n−1)



≤

f1 (M, N, α1 (m), t)w1 (r1 (M, N, α1 (m), t) + z(α1 (m), t))

t=β1 (0) β1 (n−1)



≤w1 (r1 (M, N, m, n) + z(m, n))

f1 (M, N, α1 (m), t),

(3.5)

t=β1 (0)

a(m, n) > 0, we have where we used the fact that w1 is nondecreasing. Since r1 (M, N, m, n) =  Δ1 z(m, n) + Δ3 r1 (M, N, m, n) w1 (z(m, n) + r1 (M, N, m, n)) β1 (n−1)



≤

f1 (M, N, α1 (m), t) +

t=β1 (0) β1 (n−1)



≤

t=β1 (0)

Δ3 r1 (M, N, m, n) w1 (z(m, n) + r1 (M, N, m, n))

Δ3 r1 (M, N, m, n) . f1 (M, N, α1 (m), t) + w1 (r1 (0, 0, m, 0))

(3.6)

Then 

z(m+1,n)+r1 (M,N,m+1,n)

z(m,n)+r1 (M,N,m,n)

 ≤

z(m+1,n)+r1 (M,N,m+1,n)

z(m,n)+r1 (M,N,m,n)

dτ w1 (τ ) dτ w1 (z(m, n) + r1 (M, N, m, n))

Δ1 z(m, n) + Δ3 r1 (M, N, m, n) ≤ w1 (z(m, n) + r1 (M, N, m, n)) β1 (n−1)



≤

t=β1 (0)

Δ3 r1 (M, N, m, n) . f1 (M, N, α1 (m), t) + w1 (r1 (0, 0, m, 0))

(3.7)

So 

z(m,n)+r1 (M,N,m,n)

z(0,n)+r1 (M,N,0,n)

≤

(n−1) m−1  β1

m−1   z(s+1,n)+r1 (M,N,s+1,n) dτ dτ = w1 (τ ) w1 (τ ) s=0 z(s,n)+r1 (M,N,s,n)

f1 (M, N, α1 (s), t) +

s=0 t=β1 (0) α1 (m−1) β1 (n−1)

=





s=α1 (0) t=β1 (0)

m−1  s=0

f1 (M, N, s, t) +

m−1  s=0

Δ3 r1 (M, N, s, n) w1 (r1 (0, 0, s, 0))

Δ3 r1 (M, N, s, n) . w1 (r1 (0, 0, s, 0))

(3.8)

H. Wang, K.L. ZHENG, C.X. GUO

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The definition of W1 and z(0, n) = 0 implies α1 (m−1) β1 (n−1)



W1 (z(m, n) + r1 (M, N, m, n)) ≤ W1 (r1 (M, N, 0, n)) +



f1 (M, N, s, t)

s=α1 (0) t=β1 (0)

+

m−1  s=0

Δ3 r1 (M, N, s, n) , w1 (r1 (0, 0, s, 0))

0 ≤ m ≤ M, 0 ≤ n ≤ N.

(3.9)

By (3.3) it is easy to check that the right side of (3.9) is in the domain of W1−1 for all 0 ≤ m ≤ M and 0 ≤ n ≤ N . Thus by the monotonicity of W1−1 , we obtain u(m, n) ≤z(m, n) + r1 (M, N, m, n) ≤W1−1

α1 (m−1) β1 (n−1)    W1 (r1 (M, N, 0, n)) + f1 (M, N, s, t) s=α1 (0) t=β1 (0)

+

Δ3 r1 (M, N, s, n)  w1 (r1 (0, 0, s, 0))

m−1  s=0

(3.10)

for 0 ≤ m ≤ M and 0 ≤ n ≤ N , that is, (3.2) is true for k = 1. (II). Assume that (3.2) is true for k. Consider u(m, n) ≤ r1 (M, N, m, n) +

(n−1) k+1  αi (m−1)  βi 

fi (M, N, s, t)wi (u(s, t))

(3.11)

i=1 s=αi (0) t=βi (0)

for 0 ≤ m ≤ M and 0 ≤ n ≤ N . Let z(m, n) =

(n−1) k+1  αi (m−1)  βi 

fi (M, N, s, t)wi (u(s, t)).

i=1 s=αi (0) t=βi (0)

Obviously, z(0, n) = 0 holds and z(m, n) is nonnegative and nondecreasing in m and n. (3.11) can be rewritten as u(m, n) ≤ r1 (M, N, m, n) + z(m, n) for 0 ≤ m ≤ M , 0 ≤ n ≤ N . Moreover, we have Δ1 z(m, n) =

(n−1) k+1  βi 

fi (M, N, αi (m), t)wi (u(αi (m), t))

i=1 t=βi (0)

≤

(n−1) k+1  βi 

fi (M, N, αi (m), t)wi (r1 (M, N, αi (m), t) + z(αi (m), t)). (3.12)

i=1 t=βi (0)

Since wi is nondecreasing and r1 (M, N, m, n) > 0, by Lemma 2.1 we have Δ1 z(m, n) + Δ3 r1 (M, N, m, n) w1 (z(m, n) + r1 (M, N, m, n)) k+1  βi (n−1) 

≤

i=1 t=βi (0)

fi (M, N, αi (m), t)wi (z(αi (m), t) + r1 (M, N, αi (m), t)) w1 (z(m, n) + r1 (M, N, m, n))

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application

394

+

Δ3 r1 (M, N, m, n) w1 (r1 (M, N, m, n))

β1 (n−1)

≤



Δ3 r1 (M, N, m, n) f1 (M, N, α1 (m), t) + w1 (r1 (0, 0, m, 0))

t=β1 (0)

+

(n−1) k+1  βi  i=2 t=βi (0)

β1 (n−1)

≤



Δ3 r1 (M, N, m, n) f1 (M, N, α1 (m), t) + w1 (r1 (0, 0, m, 0))

t=β1 (0)

+

wi (z(αi (m), t) + r1 (M, N, αi (m), t)) fi (M, N, αi (m), t) w1 (z(αi (m), t) + r1 (M, N, αi (m), t))

(n−1) k βi+1  

fi+1 (M, N, αi+1 (m), t)υi+1 (z(αi+1 (m), t)

i=1 t=βi+1 (0)

+ r1 (M, N, αi+1 (m), t))

(3.13)

for 0 ≤ m ≤ M , 0 ≤ n ≤ N , where υi+1 (u) =

wi+1 (u) w1 (u)

(3.14)

for i = 1, · · · , k. Notice that   ≤

z(m+1,n)+r1 (M,N,m+1,n) z(m,n)+r1 (M,N,m,n) z(m+1,n)+r1 (M,N,m+1,n) z(m,n)+r1 (M,N,m,n)

dτ w1 (τ ) dτ w1 (z(m, n) + r1 (M, N, m, n))

Δ1 z(m, n) + Δ3 r1 (M, N, m, n) ≤ w1 (r1 (M, N, m, n) + z(m, n)) β1 (n−1)



≤

t=β1 (0)

+

Δ3 r1 (M, N, m, n) f1 (M, N, α1 (m), t) + w1 (r1 (0, 0, m, 0))

(n−1) k βi+1  

fi+1 (M, N, αi+1 (m), t)υi+1 (z(αi+1 (m), t)

i=1 t=βi+1 (0)

+ r1 (M, N, αi+1 (m), t)).

(3.15)

Therefore, 

z(m,n)+r1 (M,N,m,n) z(0,n)+r1 (M,N,0,n)

α1 (m−1) β1 (n−1)



≤



dτ w1 (τ )

f1 (M, N, s, t) +

s=α1 (0) t=β1 (0)

+

(n−1) k m−1   βi+1

m−1  s=0

Δ3 r1 (M, N, s, n) w1 (r1 (0, 0, s, 0))

fi+1 (M, N, αi+1 (s), t)υi+1 (z(αi+1 (s), t)

i=1 s=0 t=βi+1 (0)

+ r1 (M, N, αi+1 (s), t)),

(3.16)

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395

that is, W1 (z(m, n) + r1 (M, N, m, n)) α1 (m−1) β1 (n−1)



≤W1 (r1 (M, N, 0, n)) +



f1 (M, N, s, t)

s=α1 (0) t=β1 (0)

+

m−1  s=0

+

Δ3 r1 (M, N, s, n) w1 (r1 (0, 0, s, 0))

(m−1) βi+1 (n−1) k αi+1  

fi+1 (M, N, s, t)υi+1 (z(s, t) + r1 (M, N, s, t)),

(3.17)

i=1 s=αi+1 (0) t=βi+1 (0)

or equivalently Ξ(m, n) ≤ θ1 (M, N, m, n) +

(m−1) βi+1 (n−1) k αi+1  

fi+1 (M, N, s, t)υi+1 (W1−1 (Ξ(s, t))) (3.18)

i=1 s=αi+1 (0) t=βi+1 (0)

for 0 ≤ m ≤ M and 0 ≤ n ≤ N , which is the same as (3.1) for k, where Ξ(m, n) = W1 (z(m, n) + r1 (M, N, m, n)) α1 (m−1) β1 (n−1)

θ1 (M, N, m, n) = W1 (r1 (M, N, 0, n)) +





f1 (M, N, s, t)

s=α1 (0) t=β1 (0)

+

m−1  s=0

Δ3 r1 (M, N, s, n) . w1 (r1 (m0 , n0 , s, n0 ))

(3.19)

From the assumption (I4 ), each υi+1 (W1−1 ), i = 1, · · · , k, is continuous and nondecreasing on [0, ∞) and is positive on (0, ∞) since W1−1 is continuous and nondecreasing on [0, ∞). Moreover, υ2 (W1−1 ) ∝ υ3 (W1−1 ) ∝ · · · ∝ υk+1 (W1−1 ). By the inductive assumption, we have Ξ(m, n)

≤Υ−1 k+1

αk+1 (m−1) βk+1 (n−1)    Υk+1 (θ1 (M, N, 0, n)) + fk+1 (M, N, s, t) s=αk+1 (0) t=βk+1 (0)

+

m−1  s=0

 Δ3 ck (M, N, s, n) ϕk+1 (Υ−1 k (θk (0, 0, s, 0)))

(3.20)

for 0 ≤ m ≤ min{M, M3 } and 0 ≤ n ≤ min{N, N3 }, where  u dz , Υi+1 (u) = −1  ui+1 υi+1 (W1 (z))

(3.21)

u > 0, Φ1 = I (Identity), u i+1 = W1 (ui+1 ), Υ−1 i+1 is the inverse of Υi+1 , ϕi+1 (u) =

wi+1 (W1−1 (u)) υi+1 (W1−1 (u)) = , υi (W1−1 (u)) wi (W1−1 (u))

i = 1, · · · , k,

αi+1 (m−1) βi+1 (n−1)



θi+1 (M, N, m, n) = Υi+1 (θ1 (M, N, 0, n)) +



fi+1 (M, N, s, t)

s=αi+1 (0) t=βi+1 (0)

+

m−1  s=0

Δ3 θi (M, N, s, n) , ϕi+1 (Υ−1 i (θi (0, 0, s, 0)))

i = 1, · · · , k − 1,

(3.22)

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application

396

and M3 and N3 are positive integers satisfying αi+1 (M3 −1) βi+1 (N3 −1)

Υi+1 (θ1 (M, N, 0, N3 )) +

+  ≤

M 3 −1 



s=αi+1 (0)

t=βi+1 (0)

fi+1 (M, N, s, t)

Δ3 θi (M, N, s, N3 ) ϕi+1 (Υ−1 i (θi (0, 0, s, 0)))

s=0 W1 (∞)

 ui+1



dz , υi+1 (W1−1 (z))

i = 1, · · · , k.

(3.23)

Note that  u dz w1 (W1−1 (z))dz = Υi (u) = −1 −1  ui υi (W1 (z)) W1 (ui ) wi (W1 (z))  W1−1 (u) dz = Wi ◦ W1−1 , = i = 2, · · · , k + 1 w i (z) ui 

u

(3.24)

and ϕi+1 (Υ−1 i (u)) = =

wi+1 (W1−1 (W1 (Wi−1 (u)))) wi+1 (W1−1 (Υ−1 i (u))) = wi (W1−1 (Υ−1 wi (W1−1 (W1 (Wi−1 (u)))) i (u)))

wi+1 (Wi−1 (u)) = φi+1 (Wi−1 (u)), wi (Wi−1 (u))

i = 1, · · · , k.

(3.25)

Thus, using the fact that θ1 (M, N, 0, n) = W1 (r1 (M, N, 0, n)), and by (3.20) we have u(m, n) ≤r1 (M, N, m, n) + z(m, n) = W1−1 (Ξ(m, n)) αk+1 (m−1) βk+1 (n−1)    −1 Wk+1 (W1−1 (θ1 (M, N, 0, n))) + ≤Wk+1 fk+1 (M, N, s, t) s=αk+1 (0) t=βk+1 (0)

+

m−1  s=0

−1 ≤Wk+1

Δ3 θk (M, N, s, n) φk+1 (Wk−1 (θk (0, 0, s, 0)))



αk+1 (m−1) βk+1 (n−1)    Wk+1 (r1 (M, N, 0, n)) + fk+1 (M, N, s, t) s=αk+1 (0) t=βk+1 (0)

+

m−1  s=0

 Δ3 θk (M, N, s, n) φk+1 (Wk−1 (θk (0, 0, s, 0)))

(3.26)

for 0 ≤ m ≤ min{M, M3 } and 0 ≤ n ≤ min{N, N3 }. In the following, we prove that θi (M, N, m, n) = ri+1 (M, N, m, n) by induction again. It is clear that θ1 (M, N, m, n) = r2 (M, N, m, n) for i = 1. Suppose that θl (M, N, m, n) = rl+1 (M, N, m, n)

(3.27)

for i = l. We have αl+1 (m−1) βl+1 (n−1)

θl+1 (M, N, m, n) =Υl+1 (θ1 (M, N, 0, n)) +





s=αl+1 (0) t=βl+1 (0)

fl+1 (M, N, s, t)

H. Wang, K.L. ZHENG, C.X. GUO

+

397

n−1  s=0

Δ3 θl (M, N, s, n) ϕl+1 (Υ−1 l (θl (0, 0, s, 0))) αl+1 (m−1) βl+1 (n−1)

=Wl+1 (r1 (M, N, 0, n)) +





fl+1 (M, N, s, t)

s=αl+1 (0) t=βl+1 (0)

+

m−1  s=0

Δ3 rl+1 (M, N, s, n) φl+1 (Wl−1 (rl+1 (0, 0, s, 0)))

=rl+2 (M, N, m, n),

(3.28)

where θ1 (M, N, 0, n) = W1 (r1 (M, N, 0, n)) is applied. It implies that it is true for i = l + 1. Thus, θi (M, N, m, n) = ri+1 (M, N, m, n) for i = 1, · · · , k. Thus, (3.23) becomes αi+1 (M3 −1) βi+1 (N3 −1)

Wi+1 (r1 (M, N, 0, N3 )) +





s=αi+1 (0)

t=βi+1 (0)

fi+1 (M, N, s, t)

M 3 −1 

Δ3 ri+1 (M, N, s, N3 ) −1 φ i+1 (Wi (ri+1 (0, 0, s, 0))) s=0  W1 (∞)  ∞  W1 (∞) dz w1 (W1−1 (z)) dz = dz = ≤ −1 −1 w υ (W (z)) w (W (z)) i+1 (z)  ui+1  ui+1 ui+1 i+1 i+1 1 1 +

(3.29)

for i = 1, · · · , k. It implies that we may choose M3 = M2 and N3 = N2 . Thus, (3.26) becomes αk+1 (m−1) βk+1 (n−1)    −1 Wk+1 (r1 (M, N, 0, n)) + fk+1 (M, N, s, t) u(m, n) ≤Wk+1 s=αk+1 (0) t=βk+1 (0)

+

m−1  s=0

 Δ3 rk+1 (M, N, s, n) −1 φk+1 (Wk (rk+1 (0, 0, s, 0)))

(3.30)

for 0 ≤ m ≤ M and 0 ≤ n ≤ N . This shows that (3.2) is true for k + 1. Thus, the claim is proved. Now we prove (2.1). Replacing m and n by M and N in (3.2), respectively, we have u(M, N )

≤Wk−1

αk (m−1) βk (n−1)    Wk (r1 (M, N, 0, N )) + fk (M, N, s, t) s=αk (0) t=βk (0)

+

M−1  s=0

Δ3 rk (M, N, s, N )  . −1 φk (Wk−1 (rk (0, 0, s, 0)))

(3.31)

Since (3.2) is true for any M ≤ M1 and N ≤ N1 , we replace M and N by m and n to get αk (m−1) βk (n−1)    fk (m, n, s, t) u(m, n) ≤Wk−1 Wk (r1 (m, n, 0, n)) + s=αk (0) t=βk (0)

+

m−1  s=0

 Δ3 rk (m, n, s, n) . −1 φk (Wk−1 (rk (0, 0, s, 0)))

(3.32)

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application

398

This is exactly (2.1) since r1 (m, n, 0, t) =  a(0, t). This proves Theorem 2.1.

2

Remark 3.1. If we exchange the order of the two sum symbols for s and t, (1.1) also holds. Therefore, with a suitable modification in the proof, another form of our result can be obtained as follows: αk (m−1) βk (n−1)    u(m, n) ≤Wk−1 Wk ( a(m, 0)) + fk (m, n, s, t) s=αk (0) t=βk (0)

+

n−1  t=0

 Δ4 rk (m, n, m, t) , −1 φk (Wk−1 (rk (0, 0, 0, t)))

1 , 0 ≤ n ≤ N 1 , 0≤m≤M (3.33)

where rk (m, n, s, t) is determined by a(s, t), r1 (m, n, s, t) =  αi (m−1) βi (n−1)



ri+1 (m, n, s, t) = Wi (r1 (m, n, s, 0)) +



fi (m, n, τ, η)

s=αi (0) t=βi (0)

+

t−1 

Δ4 ri (m, n, s, η) , −1 φ (W i−1 (ri (0, 0, 0, η))) η=0 i

i = 1, · · · , k − 1,

Δ4 ri (m, n, s, t) = ri (m, n, s, t + 1) − ri (m, n, s, t),

i = 1, · · · , k,

(3.34)

1 and N 1 are positive integers satisfying M 1 , 0)) + a(M Wi (  ≤

∞

ui

dz , wi (z)

1 −1) βi (N 1 −1) αi (M 



s=αi (0)

t=βi (0)

1 , N 1 , s, t) + fi (M

 N 1 −1 t=0

1 , N 1 , M 1 , t) Δ4 ri (M −1 φi (Wi−1 (ri (0, 0, 0, t)))

i = 1, · · · , k

(3.35)

and other functions are defined as in Theorem 2.1. Here, the condition in Theorem 2.1 that Δ1  a(m, n) is nondecreasing in n is replaced by the condition that Δ2  a(m, n) =  a(m, n + 1) −  a(m, n) is nondecreasing in m.

4

Application

In this section, we apply our theorem to study the boundedness of solutions of a nonlinear difference equation α1 (m−1) β1 (n−1)

v(m, n) =c(m, n) +





F (m, n, s, t, v(s, t))

s=α1 (0) t=β1 (0) α2 (m−1) β2 (n−1)

+





G(m, n, s, t, v(s, t))

(4.1)

s=α2 (0) t=β2 (0)

for m, n ∈ N0 , where v : N20 → R is an unknown function, c maps from N20 to R, and F and G map from N40 × R to R. Theorem 4.1. conditions

Suppose that c(0, 0) = 0 and the functions F and G in (4.1) satisfy the

|F (m, n, s, t, y)| ≤ f1 (m, n, s, t) |y|,

|G(m, n, s, t, y)| ≤ f2 (m, n, s, t)|y|,

(4.2)

H. Wang, K.L. ZHENG, C.X. GUO

399

where f1 , f2 : N40 → [0, ∞). If v(m, n) is a solution of (4.1) on N20 , then |v(m, n)| ≤ a(0, n) exp

(n−1)  α2 (m−1)  β2

f2 (m, n, s, t)

s=α2 (0) t=β2 (0) β1 (n−1) 

+

m−1 

t=β1 (0)

√1a(s,n) f1 (m, n, α1 (s), t) + Δ a(s,n) 

(4.3)

h(s)

s=0

where  a(m, n) =

max

0≤τ ≤m,0≤η≤n,τ,η∈N0

f1 (m, n, s, t) = f2 (m, n, s, t) = h(s) =

Proof.

|c(τ, η)|,

max

f1 (τ, η, s, t),

max

f2 (τ, η, s, t),

0≤τ ≤m,0≤η≤n,τ,η∈N0 0≤τ ≤m,0≤η≤n,τ,η∈N0

s−1

a(τ, 0) 1  Δ1 

 a(0, 0) + . 2 τ =0  a(τ, 0)

Using (4.1) and (4.2), the solution u(m, n) satisfies α1 (m−1) β1 (n−1)

u(m, n) ≤a(m, n) +





f1 (m, n, s, t)w1 (u(s, t))

s=α1 (0) t=β1 (0) α2 (m−1) β2 (n−1)



+



m, n ∈ N0 ,

f2 (m, n, s, t)w2 (u(s, t)),

(4.4)

s=α2 (0) t=β2 (0)

where u(m, n) = |v(m, n)|,

a(m, n) = |c(m, n)|,

w1 (u) =

√ u,

w2 (u) = u.

Clearly,  a(m, n) > 0 for all m, n ∈ N0 since c(0, 0) = 0. For positive constants u1 , u2 , we have  u √ √ dz u √ = 2( u − u1 ), W1−1 (u) = ( + u1 )2 , W1 (u) = w (z) 2 1 u  u1 dz u W2 (u) = = ln , W2−1 (u) = u2 exp(u), w (z) u 2 2 u2 r1 (m, n, s, t) =  a(s, t) > 0,

r1 (m, n, 0, t) =  a(0, t),

α1 (s−1) β1 (t−1) s−1   

√ Δ1  a(τ, t) 

a(0, t) − u1 ) + f1 (m, n, τ, η) + r2 (m, n, s, t) = 2(  ,  a(τ, 0) τ =0 τ =α (0) η=β (0) 1

1

β1 (t−1)

Δ3 r2 (m, n, s, t) =



η=β1

a(s, t) Δ1  f1 (m, n, α1 (s), η) +

,  a (s, 0) (0)

h(s) = φ2 (W1−1 (r2 (0, 0, s, 0))) =

φ2 (u) =

s−1

1  Δ1  a(τ, 0)

 a(0, 0) + . 2 τ =0  a(τ, 0)

w2 (u) √ = u, w1 (u)

400

Nonlinear Discrete Inequality in Two Variables with Delay and Its Application

It is obvious that w1 and w2 satisfy (I4 ). Applying Theorem 2.1, we have u(m, n) ≤ a(0, n) exp

 α2 (m−1) (n−1)  β2

f2 (m, n, s, t)

s=α2 (0) t=β2 (0) β1 (n−1) 

+

m−1  s=0

which implies (4.3).

√1a(s,n) f1 (m, n, α1 (s), η) + Δ a(s,0) η=β1 (0) h(s)

.

(4.5) 2

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