# Estimation of Unknown Functions of Iterative Difference Inequalities  and Their Applications

Authors: Ricai Luo; Wu-Sheng Wang; Honglei Xu
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We establish some new nonlinear retarded finite difference inequalities. The results that we propose here can be used as tools in the theory of certain new classes of finite difference equations in various difference situations. We also give applications of the inequalities to show the usefulness of our results.


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## Body

## 1. Introduction

An integral inequality that provides an explicit bound to the unknown function furnishes a handy tool to investigate qualitative properties of solutions of differential and integral equations. One of the best known and widely used inequalities in the study of nonlinear differential equations is Gronwall-Bellman inequality [1, 2], which can be stated as follows. If u and f are nonnegative continuous functions on an interval [a,b] satisfying
(1)u(t)≤c+∫at‍f(s)u(s)ds,t∈[a,b],
for some constant c≥0, then
(2)u(t)≤cexp⁡(∫at‍f(s)ds),t∈[a,b].
Being an important tool in the study of qualitative properties of solutions of differential equations and integral equations, various generalizations of Gronwall inequalities [1, 2] and their applications have attracted great interests of many mathematicians [3–5]. Some recent works can be found in [6–12] and the references therein. Along with the development of the theory of integral inequalities and the theory of difference equations, more and more attentions are paid to discrete versions of Gronwall-type inequalities; see [13–36] and the references cited therein.Sugiyama [13] established the most precise and complete discrete analogue of the Gronwall inequality [1] in the following form. Let u(n) and f(n) be nonnegative functions defined for n∈N, and suppose that f(n)≥0 for every n∈N. If
(3)u(n)<u0+∑s=n0n-1‍f(s)u(s),n∈N,
where N is the set of points n0+k (k=0,1,2,…), n0≥0 is a given integer, and u0 is a nonnegative constant, then
(4)u(n)<u0∏s=n0n-1‍[1+f(s)],n∈N.
Pachpatte [15] established some generalized discrete analogue of the Gronwall inequality in the following form. Let m(s) be a positive and monotone nondecreasing function on N, and let a(s),b(s) be nonnegative functions on N. If u(n) satisfies
(5)u(n)≤m(s)+∑s=n0n-1‍a(s)(u(s)+∑τ=n0s-1‍b(τ)u(τ)),∀n∈N,
then
(6)u(n)≤P(n)m(s),∀n∈N,
where
(7)P(n)=1+∑s=n0n-1‍a(s)∏τ=n0s-1‍[1+a(τ)+b(τ)],∀n∈N.Lemma 1 (see [16]).
Suppose thatu0 is a nonnegative constant and u(n), a(n), b(n), c(n), and d(n) are nonnegative functions defined on N, 1+a(n)-b(n)≥0 for all n∈N. If u(n) satisfies the inequality
(8)u(n)≤u0+∑s=n0n-1‍a(s)u(s)+∑s=n0n-1‍b(s)(∑τ=n0s-1‍c(τ)u(τ)(∑t=n0τ-1‍d(t)u(t))),kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk∀n∈N,
then
(9)u(n)≤u0∏s=n0n-1‍[1+a(s)-b(s)]+∑s=n0n-1‍b(s)R(s)∏t=s+1n-1‍[1+a(t)-b(t)],llllllllllllllllllllllllllllllllllllllllllllllllllki∀n∈N,
where
(10)R(n)≤u0∏s=n0n-1[1+a(s)+b(s)+c(s)Q(s)]1+u0∑s=n0n-1c(s)∏t=n0s[1+a(t)+b(t)+c(t)Q(t)],kkkkkkkkkkkkkkkkkklkkkkkkkkkkkkkkkkkkkkkkkki∀n∈N,
in which
(11)Q(n)≤u0∏s=n0n-1[1+a(s)+b(s)+d(s)]1-u0∑s=n0n-1c(s)∏t=n0s[1+a(t)+b(t)+d(t)],hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh∀n∈N,
and ∑s=n0n-1c(s)∏t=n0s[1+a(t)+b(t)+d(t)]<u0-1 for all n∈N.Lemma 2 (see [14, 18]).
Letw(n,r) be a real-valued function defined for n∈N, 0≤r<∞ and monotone nondecreasing with respect to r for any fixed n∈N. Let u(n) be a real-valued function defined for n∈N such that
(12)Δu(n)≤w(n,u(n)),∀n∈N.
Let r(n) be a solution of
(13)Δr(n)=w(n,r(n)),r(0)=r0,∀n∈N,
such that u(0)≤r(0). Then
(14)u(n)≤r(n),∀n∈N.Pachpatte [18, 19] also established some difference inequalities of product form as follows. Let u,a,b be nonnegative functions defined on N and let c be a nonnegative constant. Let w(n,r) be a nonnegative function defined for n∈N, 0≤r<∞ and monotone nondecreasing with respect to r for any fixed n∈N. If u(n) satisfies
(15)u2(n)≤c2+2∑s=n0n-1‍u(s)[a(s)(u(s)+∑t=n0s-1‍b(t)u(t))kkkkkkkkkkkkkkkk[a(s)(u(s)+∑t=n0s-1‍b(t)u(t))+w(s,u(s))],∀n∈N,
then
(16)u(n)≤P(n)r(n),∀n∈N,
where P(n) is defined by (7), and r(n) is a solution of
(17)Δr(n)=w(n,P(n)r(n)),r(0)=c,∀n∈N.
Let u,a,b be nonnegative functions defined for n∈N and let c be a nonnegative constant. Let w(n,r) be a nonnegative function defined for n∈N, 0≤r<∞ and monotone nondecreasing with respect to r for any fixed n∈N. If u(n) satisfies
(18)u2(n)≤c2+∑s=n0n-1‍a(s)(u(s+1)+u(s))kkkkkkk×[(u(s)+∑τ=n0s-1‍b(τ)u(τ))kkkkkkkkkkkk∑τ=n0s-1‍+w(s,u(s))],∀n∈N,
then
(19)u(n)≤P(n)r(n),∀n∈N,
where P(n) is defined by (7), and r(n) is a solution of the difference equation
(20)Δr(n)=a(n)w(n,P(n)r(n)),r(0)=c,∀n∈N.Motivated by the results given in [16, 18, 19], in this paper, we discuss new nonlinear finite difference inequalities:
(21)u2(n)≤c2+∑s=n0n-1‍f(s)(u(s+1)+u(s))kkkkkk×[(u(s)+∑t=n0s-1‍g(t)u(t)kkkkkkkkkkkkkkkk×∑τ=n0t-1‍h(τ)u(τ))hhhhhkkkk(u(s)+∑t=n0s-1‍g(t)u(t)∑τ=n0t-1‍h(τ)u(τ))+w(s,u(s))],∀n∈N.
Our inequalities can be used as tools in the study of certain classes of finite difference equations. We also present some immediate applications to show the importance of our results to study the various problems in the theory of finite difference equations.


## 2. Main Results

Throughout this paper, letR=(-∞,+∞), R+=[0,+∞). Let N∶={n0,n0+1,n0+2,…} and NT∶={n0,n0+1,n0+2,…,T},T∈N. For function u(n), n∈N, we define the operator Δ by Δu(n)=u(n+1)-u(n). Obviously, the linear difference equation Δu(n)=f(n) with the initial condition u(n0)=0 has the solution u(n)=∑s=n0n-1f(s). For convenience, in the sequel we complementarily define that ∑s=n0n0-1f(s)=0 and ∏s=n0n0-1f(s)=1.Theorem 3.
Letβ>0 be a constant, x,y positive functions defined on N, x a monotone increasing function, and y a monotone decreasing function. Let ϕ be a nonnegative function defined on N such that
(22)x(n+1)y(n+1)-x(n)y(n)≤ϕ(n)[x(n)y(n+1)]β,hhhhhhhhhhhoik∀n∈N.(i)
Supposeβ>1. If 1+(1-β)[x(n0)y(n0)]β-1∑s=n0n-1ϕ(s)>0, then
(23)x(n)≤x(n0)y(n0)y(n){1+(1-β)[x(n0)y(n0)]β-1∑s=n0n-1ϕ(s)}1/(β-1),kkkkkkkkkkkkkkkkkkkkkkkkkkkhkkkkkkkkik∀n∈N.(ii)
Suppose0<β<1. Then
(24)x(n)≤y-1(n){[x(n0)y(n0)]1-β∑s=n0n-1‍kkkkkkkk+∑s=n0n-1‍(1-β)ϕ(s)}1/(1-β),∀n∈N.Proof.
(i) We apply mean value theorem for differentiation to the function(25)F(z)=z1-β(1-β),z>0,
and then there exists ξ between x(n)y(n) and x(n+1)y(n+1) such that
(26)[x(n+1)y(n+1)]1-β-[x(n)y(n)]1-β=(1-β)ξ-β[x(n+1)y(n+1)-x(n)y(n)].
Because x(n) is monotone increasing and y(n) is monotone decreasing and -β<0, we see that [x(n)y(n+1)]-β≥[x(n+1)y(n+1)]-β and [x(n)y(n+1)]-β≥[x(n)y(n)]-β. So for all values of ξ between x(n)y(n) and x(n+1)y(n+1) we have
(27)[x(n)y(n+1)]-β≥ξ-β.
From (22) and (27), we have
(28)ξ-β[x(n+1)y(n+1)-x(n)y(n)]≤[x(n)y(n+1)]-βϕ(n)[x(n)y(n+1)]β=ϕ(n).
Since 1-β<0, from (26) and (28) we have
(29)[x(n+1)y(n+1)]1-β-[x(n)y(n)]1-β≥(1-β)ϕ(n).
Taking n=s in (29) and summing up over s from n0 to n-1, we obtain
(30)[x(n)y(n)]1-β≥[x(n0)y(n0)]1-β+∑s=n0n-1‍(1-β)ϕ(s).
From (30), we obtain our required estimation (23).
(ii) Now by following the same steps as in the proof of (i) before (29) we have
(31)[x(n+1)y(n+1)]1-β-[x(n)y(n)]1-β≤(1-β)ϕ(n),
because 1-β>0. Taking n=s in (31) and summing up over s from n0 to n-1, we obtain
(32)[x(n)y(n)]1-β≤[x(n0)y(n0)]1-β+∑s=n0n-1‍(1-β)ϕ(s).
From (32), we obtain our required estimation (24).Theorem 4.
Letm(s) be a positive and monotone nondecreasing function defined on N and f(s),g(s),h(s) nonnegative functions defined on N. If u(n) satisfies
(33)u(n)≤m(n)+∑s=n0n-1‍f(s)u(s)kkkkkkkk+∑s=n0n-1‍f(s)(∑t=n0s-1‍g(t)u(t)kkkkkkkkkkkkkkkkkkkk×(∑τ=n0t-1‍h(τ)u(τ))),kkkkkkkkkkkkkkkikkkkkkkkkkkkkikkkkkk∀n∈N,
then
(34)u(n)≤m(n)+∑s=n0n-1‍f(s)U(s),∀n∈N,
where
(35)U(n)≤m(n)∏s=n0n-1[1+f(s)+g(s)V(s)]1+m(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)V(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
in which
(36)V(n)=m(n)∏s=n0n-1[1+f(s)+h(s)]1-m(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
and ∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)]<m-1(n) for all n∈N.Proof.
FixT∈N, where T is chosen arbitrarily, since m(t) is a nonnegative and monotone nondecreasing function, from (33), we have
(37)u(n)≤m(T)+∑s=n0n-1‍f(s)u(s)kkkkkkk+∑s=n0n-1‍f(s)(∑t=n0s-1‍g(t)u(t)kkkkkkkkkkkkkkkkkkk×(∑τ=n0t-1‍h(τ)u(τ))),kkkkkkkkkkkkkkkkkkkkkkkkkkkkkk∀n∈NT.
Define a function z(n) by the right-hand side of (37). Then z(n) is a positive and monotone nondecreasing function defined on N. We have
(38)z(n0)=m(T),u(n)≤z(n),∀n∈N.
Using the definitions of the operator Δ and z, we obtain
(39)Δz(n)=f(n){u(n)+[∑t=n0n-1‍g(t)u(t)(∑τ=n0t-1‍h(τ)u(τ))]}≤f(n){z(n)+[∑t=n0n-1‍g(t)z(t)(∑τ=n0t-1‍h(τ)z(τ))]},kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk∀n∈NT.
Let
(40)z1(n)=z(n)+[∑t=n0n-1‍g(t)z(t)(∑τ=n0t-1‍h(τ)z(τ))],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈NT.
Then
(41)z1(n0)=z(n0)=m(T),z(n)≤z1(n),∀n∈NT.
It follows that
(42)Δz1(n)=Δz(n)+g(n)z(n)(∑τ=n0n-1‍h(τ)z(τ))≤f(n)z1(n)+g(n)z1(n)(∑τ=n0n-1‍h(τ)z1(τ)),kkkkkkkkkkkkkkkkkkkkkkkkkkkkoi∀n∈NT.
Adding g(n)z12(n) to both sides of the above inequality we have
(43)Δz1(n)+g(n)z12(n)≤f(n)z1(n)+g(n)z1(n)×[z1(n)+(∑τ=n0n-1‍h(τ)z1(τ))],kkkkkkkkkkkkkkkkkkkk∀n∈NT.
Put
(44)z2(n)=z1(n)+∑τ=n0n-1‍h(τ)z1(τ),∀n∈NT,
and then z1(n)≤z2(n), z2(n0)=z1(n0)=m(T) and
(45)Δz2(n)=Δz1(n)+h(n)z1(n)≤f(n)z2(n)+g(n)z22(n)+h(n)z2(n),lllllllkkkkkkkkkkkkkkkkkkkkkk∀n∈NT.
We see that the inequality
(46)z2(n+1)-(1+f(n)+h(n))z2(n)≤g(n)z22(n),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈NT.
Define a function
(47)P1(n)=∏s=n0n-1‍(1+f(s)+h(s))-1,∀n∈NT.
Multiplying by P1(n+1) to both sides of (46) we obtain
(48)z2(n+1)P1(n+1)-z2(n)P1(n)≤P1-1(n+1)g(n)[z2(n)P1(n+1)]2,jjjjjjjjkkkkkkkkkkkkkkkkkikkoi∀n∈NT.
Let x(n)=z2(n), y(n)=P1(n), ϕ(n)=P1-1(n+1)g(n), and β=2. Because z2(n) is monotone increasing, P1(n) is monotone decreasing and 2>0; applying Theorem 3 to (48) we obtain
(49)z2(n)≤z2(n0)P1-1(n)1-z2(n0)∑s=n0n-1g(s)P1-1(s+1)=m(T)∏s=n0n-1[1+f(s)+h(s)]1-m(T)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],ujjjjjjjjuuuuuuuuuuuuuuuuuuuuuuuuu∀n∈NT,
where P1(n0)=1, z2(n0)=m(T) are used. Define a function V~ of the right-hand side of (49). Substituting (49) in (43) we obtain
(50)z1(n+1)-(1+f(n)+g(n)V~(n))z1(n)≤-g(n)z12(n),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈NT.
Performing the same derivation as in (46)–(49), we obtain from (50) that
(51)z1(n)≤m(T)∏s=n0n-1[1+f(s)+g(s)V~(s)]1+m(T)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)V~(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈NT.
Define a function U~ of the right-hand side of (51). Substituting (51) in (39) we obtain
(52)Δz(n)=f(n)U~(n),∀n∈NT.
Using (38), from (52) it follows that
(53)u(n)≤m(T)+∑s=n0n-1‍f(s)U~(s),∀n∈NT.
Since T∈N is arbitrary, from (53), we get the required estimate (35).Theorem 5.
Letu,f,g,h be nonnegative functions defined for n∈N and c a nonnegative constant. Let w(n,r) be a real-valued function defined for n∈N, 0≤r<∞, and monotone nondecreasing with respect to r for any fixed n∈N. If u(n) satisfies (21), then
(54)u(n)≤v(n)+∑s=n0n-1‍f(s)W1(s),∀n∈N,
where
(55)W1(n)≤v(n)∏s=n0n-1[1+f(s)+g(s)W2(s)]1+v(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)W2(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
in which
(56)W2(n)=v(n)∏s=n0n-1[1+f(s)+h(s)]1-v(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
and v(n) is a solution of the difference equation
(57)Δr(n)=f(n)w(n,r(n)+∑s=n0n-1‍f(s)W3(s)),r(0)=c,∀n∈N,
where
(58)W3(n)≤r(n)∏s=n0n-1[1+f(s)+g(s)W4(s)]1+r(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)W4(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
in which
(59)W4(n)=r(n)∏s=n0n-1[1+f(s)+h(s)]1-r(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
and ∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)]<v-1(n) for all n∈N.Proof.
We first assume thatc>0 and define a function z(n) by the right-hand side of (21). Then z(n) is a positive and monotone nondecreasing function defined on N. We have
(60)z(0)=c2,u(n)≤z(n),∀n∈N.
Using the definitions of the operator Δ and z, we obtain
(61)Δz(n)=f(n)(u(n+1)+u(n))×[(u(n)+∑t=n0n-1‍g(t)u(t)∑τ=n0t-1‍h(τ)u(τ))kkkkkkk∑t=n0n-1‍+w(n,u(n))]≤f(n)(z(n+1)+z(n))×[(z(n)+∑t=n0n-1‍g(t)z(t)kkkkkkk×∑τ=n0t-1‍h(τ)z(τ))+w(n,z(n))],kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkki∀n∈N.
From (61) it follows that the inequality
(62)Δ(z(n))=Δz(n)z(n+1)+z(n)≤f(n)[(z(n)+∑t=n0n-1‍g(t)z(t)kkkkkkkkkkkkkkkk×∑τ=n0t-1‍h(τ)z(τ))kkkkkk∑t=n0n-1‍ki+w(n,z(n))]
holds for all n∈N. Setting n=s in (62) and substituting s=n0,1,2,…,n-1, successively, we get
(63)z(n)≤c+∑s=n0n-1‍f(s)kkkkk×[((z(s))+∑t=n0s-1‍g(t)z(t)kkkkkkkkkkkkkkkkkkkk×∑τ=n0t-1‍h(τ)z(τ))kkkkkkkkkk∑τ=n0t-1‍+w(s,z(s))],kkkkkkkkkkkkkkkkkkkkkkkkkkikkk∀n∈N.
Define a function z1(n) by
(64)z1(n)=c+∑s=n0n-1‍f(s)w(s,z(s)),∀n∈N.
Then z1(n)=c and
(65)Δz1(n)=f(n)w(n,z(n)),∀n∈N.
Using (64), the inequality (63) can be written as
(66)z(n)≤z1(n)+∑s=n0n-1‍f(s)(z(s)+∑t=n0s-1‍g(t)z(t)jjjjjjjhjhkkkkkkkkkkkk×∑τ=n0t-1‍h(τ)z(τ)),ujjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjkkkkkkkkkkkkk∀n∈N.
Since z1(n) is positive and monotone nondecreasing for n∈N, f(s),g(s),h(s) satisfy the conditions in Theorem 4. Now an application of Theorem 4 to (66) yields
(67)z(n)≤z1(n)+∑s=n0n-1‍f(s)W~1(s),∀n∈N,
where
(68)W~1(n)≤z1(n)∏s=n0n-1[1+f(s)+g(s)W~2(s)]1+z1(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)W~2(t)],jjjjjjjjjkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk∀n∈N,
in which
(69)W~2(n)=z1(n)∏s=n0n-1[1+f(s)+h(s)]1-z1(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N.
Since w(n,r) is monotone nondecreasing with respect to r for any fixed n∈N, from (65) and (67), we have
(70)Δz1(n)≤f(n)w(n,z1(n)+∑s=n0n-1‍f(s)W~1(s)),∀n∈N.
Now as a suitable application of Lemma 2, we obtain
(71)z1(n)≤v(n),∀n∈N,
where v(n) is a solution of (57). Using (60), (67), and (71), we obtain our required estimation (54).
Ifc is nonnegative, we can carry out the above procedure with c+ϵ instead of c where ϵ is an arbitrary small number. Letting ϵ→0, we obtain (54).


## 3. Application to Finite Difference Equations

In this section, we consider the following difference equation:(72)Δx(n)=f(n)×[F(n,x(n),∑t=n0n-1‍g(t)x(t)kgkkkkk×∑τ=n0t-1‍H(t,τ,x(τ)))+K(n,x(n))],jjjjjjjjklklklklklklklklklklklgklklklklklk∀n∈N,(73)x(n0)=x0,
where F, H, K are real-valued functions defined, respectively, on N×R2, N2×R, N×R, f is as defined in Theorem 5, and x0 is a constant. We assume that
(74)|K(n,x(n))|≤w(n,|x(n)|),|H(n,t,x(t))|≤∑t=n0n-1‍h(t)|x(t)|,|F(n,x(n),y(n))|≤|x(n)|+|y(n)|,
where g, h, w are as defined in Theorem 5. Using the definitions of the operator Δ, from (72), we see that the inequality
(75)x(n+1)-x(n)=f(n)[F(n,x(n),∑t=n0n-1‍g(t)x(t)hhhhhhhhhhhh×∑τ=n0t-1‍H(t,τ,x(τ)))+K(n,x(n))]
holds for all n∈N. It follows that
(76)x2(n+1)-x2(n)=f(n)[x(n+1)-x(n)]×[F(n,x(n),∑t=n0n-1‍g(t)x(t)∑τ=n0t-1‍H(t,τ,x(τ)))ffffffff∑t=n0n-1‍+K(n,x(n))],∀n∈N.
From (76), we have
(77)x2(n)=x2(n0)+∑s=n0n-1‍f(s)[x(s+1)-x(s)]kkkkkkkkkk×[F(s,x(s),∑t=n0s-1‍g(t)x(t)kkkkkkkkkkkkkkkkkk×∑τ=n0t-1‍H(t,τ,x(τ)))fffffffkkkkkkkk∑t=n0s-1‍+K(s,x(s))],∀n∈N.
Using the conditions (74), we obtain
(78)|x(n)|2=x2(n0)+∑s=n0n-1‍f(s)[|x(s+1)|-|x(s)|]kkkkkkkkkkk×[|x(s)|+∑t=n0s-1‍g(t)|x(t)|kkkkkkkkkkkkkkkkkkkkkkkk×∑τ=n0t-1‍h(τ)|x(τ)|ggggggkkkkkkkkk∑t=n0s-1‍+w(s,|x(s)|)],∀n∈N.
Now an application of Theorem 5 to (78) yields the estimation of the difference equation (72), that is,
(79)|x(n)|≤v(n)+∑s=n0n-1‍f(s)W5(s),∀n∈N,
where
(80)W5(n)≤v(n)∏s=n0n-1[1+f(s)+g(s)W6(s)]1+v(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)W6(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
in which
(81)W6(n)=v(n)∏s=n0n-1[1+f(s)+h(s)]1-v(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
and v(n) is a solution of the difference equation
(82)Δr(n)=f(n)w(n,r(n)+∑s=n0n-1‍f(s)W7(s)),r(0)=|x0|,∀n∈N,
where
(83)W7(n)≤r(n)∏s=n0n-1[1+f(s)+g(s)W8(s)]1+r(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+g(t)W8(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
in which
(84)W8(n)=r(n)∏s=n0n-1[1+f(s)+h(s)]1-r(n)∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)],iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀n∈N,
and ∑s=n0n-1g(s)∏t=n0s[1+f(t)+h(t)]<v-1(n) for all n∈N.



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*Source: 763761-2014-07-22.xml*