# A Class of  Volterra-Fredholm  Type Weakly Singular Difference Inequalities with Power Functions and Their Applications

Authors: Yange Huang; Wu-Sheng Wang; Yong Huang
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We discuss a class of Volterra-Fredholm type difference inequalities with weakly singular. The upper bounds of the embedded unknown functions are estimated explicitly by analysis techniques. An application of the obtained inequalities to the estimation of Volterra-Fredholm type difference equations is given.


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## Body

## 1. Introduction

Being an important tool in the study of existence, uniqueness, boundedness, stability, invariant manifolds, and other 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 [6–28].In 1981, Henry [12] discussed the following linear singular integral inequality:
(1)u(t)≤a+b∫0t‍(t-s)β-1u(s)ds.
In 2007, Ye et al. [18] discussed linear singular integral inequality
(2)u(t)≤a(t)+b(t)∫0t‍(t-s)β-1u(s)ds.
In 2014, Cheng et al. [28] discussed the following inequalities:
(3)um(t)≤a(t)+b(t)∫0t‍f(s)un(s)ds+c(t)∫0T‍g(s)ur(s)ds,um(t)≤a(t)+b(t)∫0t‍(tα1-sα1)β1-1sγ1-1f(s)un(s)ds+c(t)∫0T‍(Tα2-sα2)β2-1sγ2-1g(s)ur(s)ds.On the other hand, difference inequalities which give explicit bounds on unknown functions provide a very useful and important tool in the study of many qualitative as well as quantitative properties of solutions of nonlinear difference equations. More attentions are paid to some discrete versions of Gronwall-Bellman type inequalities (such as [29–50]).In 2002, Pachpatte [36] discussed the following difference inequality:
(4)u(n)≤c+∑s=αn-1‍f(n,s)u(s)ds+∑s=αβ‍g(n,s)u(s),yyyyyyyyyyyyyyyyyyyyyn∈N∩[α,β].
In 2010, Ma [45] discussed the following difference inequality with two variables:
(5)ui(m,n)≤a(m,n)+∑s=m0m-1‍∑t=n0n-1‍f(s,t)uj(s,t)+∑s=m0M-1‍∑t=n0N-1‍g(s,t)ur(s,t).
In 2014, Huang at el. [50] discussed the following linear singular difference inequality:
(6)u(n)≤a(n)+b(n)∑s=0n-1‍(tn-ts)β-1τsw1(u(s))×[u(s)+h(s)+∑σ=0s-1‍(ts-tσ)β-1τσw2(u(σ))].Motivated by the results given in [6, 11, 28, 36, 45, 49, 50], in this paper, we discuss the following inequalities:
(7)u(m,n)≤a(m,n)+b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t)u(s,t)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t),(8)ui(m,n)≤a(m,n)+b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t)uj(s,t)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)ur(s,t),(9)ui(n)≤a(n)+b(n)∑s=0n-1‍(tn-ts)β-1tsγ-1τsf(s)uj(s)+c(n)∑s=0N-1‍(tN-ts)β-1tsγ-1τsg(s)ur(s).


## 2. Difference Inequalities with Two Variables

Throughout this paper, letN0:={0,1,2,…}, N:={1,2,…}, and ΩX,Y={(m,n):m0≤m≤X,n0≤n≤Y,m,n,X,Y∈N}. For a function z(m,n), its first-order difference is defined by Δ1z(m,n)=z(m+1,n)-z(m,n). Obviously, the linear difference equation Δz(n)=b(n) with the initial condition z(n0)=0 has the solution z(n)=∑s=n0n-1‍b(s). For convenience, in the sequel, we complementarily define that ∑s=n0n0-1‍b(s)=0.Lemma 1.
Assume thatu(m,n), a(m,n), c(m,n), and g(m,n) are nonnegative functions on ΩM,N={(m,n):m0≤m≤M,n0≤n≤N,m,n,M,N∈N}. If ∑s=m0m-1‍∑t=n0n-1‍g(s,t)c(s,t)<1 and u(m,n) satisfies the following difference inequality:
(10)u(m,n)≤a(m,n)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t),ddddddddddddddddddd∀(m,n)∈ΩM,N,
then
(11)u(m,n)≤a(m,n)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)a(s,t)1-∑s=m0M-1‍∑t=n0N-1‍g(s,t)c(s,t),fffffffffffffffffffffffffffffffffffffffffddd∀(m,n)∈ΩM,N.Proof.
Since∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t) is a constant. Let ∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t)=K. From (10), we have
(12)u(m,n)≤a(m,n)+c(m,n)K,∀(m,n)∈ΩM,N.
Since g(m,n) is nonnegative, we have
(13)g(m,n)u(m,n)≤g(m,n)a(m,n)+c(m,n)g(m,n)K.
Let s=m and t=n in (13) and substituting s=m0,m1,m2,…,M-1 and t=n0,n1,n2,…,N-1, successively, we obtain
(14)K=∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t)≤∑s=m0M-1‍∑t=n0N-1‍g(s,t)a(s,t)+∑s=m0M-1‍∑t=n0N-1‍g(s,t)c(s,t)K.
From (14), we have
(15)K≤∑s=m0M-1‍∑t=n0N-1‍g(s,t)a(s,t)1-∑s=m0M-1‍∑t=n0N-1‍g(s,t)c(s,t),
where ∑s=m0m-1‍∑t=n0n-1‍g(s,t)c(s,t)<1. Substituting inequality (15) into (13), we get the explicit estimation (11) for u(m,n).Theorem 2.
Assume thatu(m,n), a(m,n), b(m,n), c(m,n), f(m,n), and g(m,n) are nonnegative functions on ΩM,N and a(m,n), b(m,n), and c(m,n) are nondecreasing in both m and n. If
(16)∑s=m0M-1‍∑t=n0N-1‍g(s,t)c(s,t)×exp⁡(b(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ))<1,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∀(m,n)∈ΩM,N,
and u(m,n) satisfies the difference inequality (7), then
(17)u(m,n)≤exp⁡(b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t))×[×exp⁡(b(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ)))-1)a(m,n)+c(m,n)TTT×(∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ)))-1exp⁡(b(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ))(exp⁡(b(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ))∑s=m0M-1‍∑t=n0N-1‍g(s,t)a(s,t)GGGGGIIggg×exp⁡(b(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ)))IIIIIIIIIdd×(×exp⁡(b(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ)))-1)1-∑s=m0M-1‍∑t=n0N-1‍g(s,t)c(s,t)iiiiiiiiiiiiiiiiiiiiiid×exp⁡(∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ)b(s,t)ttttttttttttttttttttttttti×∑τ=m0s-1‍∑ξ=n0t-1‍f(τ,ξ)))-1)],
for all (m,n)∈ΩM,N.Proof.
Fixing any arbitrary(X,Y)∈ΩM,N, from (7), we have
(18)u(m,n)≤a(X,Y)+b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t)u(s,t)+c(X,Y)∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t),
for all (m,n)∈ΩX,Y, where a(m,n), b(m,n), and c(m,n) are nondecreasing in both m and n.
Define a functionz(m,n) by the right side of (18); that is,
(19)z(m,n):=a(X,Y)+b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t)u(s,t)+c(X,Y)∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t),
for all (m,n)∈ΩX,Y. Obviously, we have
(20)u(m,n)≤z(m,n),∀(m,n)∈ΩX,Y,(21)z(m0,n)=a(X,Y)+c(X,Y)∑s=m0M-1∑t=n0N-1‍g(s,t)u(s,t).
Using the difference formula Δ1z(m,n)=z(m+1,n)-z(m,n) and relation (20), from (21), we have
(22)Δ1z(m,n)=b(X,Y)∑t=n0n-1‍f(m,t)u(m,t)≤b(X,Y)∑t=n0n-1‍f(m,t)z(m,t)≤b(X,Y)z(m,n)∑t=n0n-1‍f(m,t),
where we have used the monotonicity of z in n. From (22), we observe that
(23)Δ1z(m,n)z(m,n)≤b(X,Y)∑t=n0n-1‍f(m,t),∀(m,n)∈ΩX,Y.
On the other hand, by the mean-value theorem for integrals, for arbitrarily given integers m, n with (m+1,n), (m,n)∈ΩX,Y, there exists ξ in the open interval (z(m,n),z(m,n+1)) such that
(24)ln⁡⁡z(m+1,n)-ln⁡⁡z(m,n)=∫z(m,n)z(m+1,n)‍dss=Δ1z(m,n)ξ≤Δ1z(m,n)z(m,n).
From (23) and (24), we have
(25)ln⁡⁡z(m+1,n)-ln⁡⁡z(m,n)≤b(X,Y)∑t=n0n-1‍f(m,t),&(m+1,n)-ln⁡z(m,n)(m+1,n)-ln⁡z(m,n)(m+1,n)-ln⁡z(m,n)(m+1,n)-ln⁡z(m,n)iiiiiiiiii∀(m,n)∈ΩX,Y.
Let s=m and t=n in (25), and substituting s=m0,m1,m2,…,m-1 and t=n0,n1,n2,…,n-1, successively, we obtain
(26)ln⁡⁡z(m,n)-ln⁡⁡z(m0,n)≤b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t),dddddddd∀(m,n)∈ΩX,Y.
It implies that
(27)z(m,n)≤z(m0,n)exp⁡(b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t)),∀(m,n)∈ΩX,Y.
Using (20) and (21), from (27), we have
(28)u(m,n)≤(a(X,Y)+c(X,Y)∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t))×exp⁡(b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t))=a(X,Y)exp⁡(b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t))+c(X,Y)exp⁡(b(X,Y)∑s=m0m-1‍∑t=n0n-1‍f(s,t))×∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t),
for all (m,n)∈ΩX,Y. Taking m=X and n=Y in (28), we have
(29)u(X,Y)≤a(X,Y)exp⁡(b(X,Y)∑s=m0X-1‍∑t=n0Y-1‍f(s,t))+c(X,Y)exp⁡(b(X,Y)∑s=m0X-1‍∑t=n0Y-1‍f(s,t))×∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t).
Since X, Y are chosen arbitrarily, we replace X and Y in (29) with m and n, respectively, and obtain that
(30)u(m,n)≤a(m,n)exp⁡(b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t))+c(m,n)exp⁡(b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t))×∑s=m0M-1‍∑t=n0N-1‍g(s,t)u(s,t),
for all (m,n)∈ΩM,N. Applying the result of Lemma 1 to inequality (30), we obtain desired estimation (17).Lemma 3 (see [39]).
Leta≥0, i≥j≥0, and i≠0. Then,
(31)aj/i≤jiK(i-j)/ia+i-jiKj/i,∀K>0.Theorem 4.
Assume thatu(m,n), a(m,n), b(m,n), c(m,n), f(m,n), and g(m,n) are defined as in Theorem 2 and that i≥j>0 and i≥r>0. If
(32)∑s=m0M-1‍∑t=n0N-1‍G(s,t)C(s,t)exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ))<1,fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff∀(m,n)∈ΩM,N,
and u(m,n) satisfies difference inequality (8), then
(33)u(m,n)≤{×exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))-1)]a(m,n)+exp⁡(B(m,n)∑s=m0m-1‍∑t=n0n-1‍F(s,t))ii×[×exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))-1A(m,n)+C(m,n)IIIid×(exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))(exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))∑s=m0M-1‍∑t=n0N-1‍G(s,t)A(s,t)I0000000IIIiooid×exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))IIIIIIIIIiiiii×(1-∑s=m0M-1‍∑t=n0N-1‍G(s,t)C(s,t)iiiiiiiiiiiiidiiiiii000d×exp⁡(∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)B(s,t)yyyyyyyyyyydyyyypyd0d×∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))-1)]}1/i,
for all (m,n)∈ΩM,N, where
(34)A(m,n)≔b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t)(jiK1(i-j)/ia(s,t)+i-jiK1j/i)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)(riK2(i-r)/ia(s,t)dddddddddddddddddddddddd+i-riK2r/i),(35)B(m,n):=jb(m,n)i,C(m,n):=rc(m,n)i,(36)F(m,n):=f(s,t)K1(i-j)/i,G(m,n):=g(s,t)K2(i-r)/i,
and K1,K2 are arbitrary constants.Proof.
Define a functionv(m,n) by
(37)v(m,n)=b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t)uj(s,t)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)ur(s,t),
for all (m,n)∈ΩM,N. Then, from (8), we have
(38)u(m,n)≤(a(m,n)+v(m,n))1/i,∀(m,n)∈ΩM,N.
Applying Lemma 3 to (38), we obtain
(39)uj(m,n)≤(a(m,n)+v(m,n))j/i≤jiK1(i-j)/i(a(m,n)+v(m,n))+i-jiK1j/i,ur(m,n)≤(a(m,n)+v(m,n))r/i≤riK2(i-r)/i(a(m,n)+v(m,n))+i-riK2r/i,
for all (m,n)∈ΩM,N. Substituting (39) into (37), we obtain
(40)v(m,n)≤b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t)∑s=m0m-1‍∑t=n0n-1‍∑s=m0m-1‍∑t=n0n-1‍f(s,t)×(jiK1(i-j)/i(a(s,t)+v(s,t))uuuuuuuuuuuuuuuui+i-jiK1j/i)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)iiiiiiiiiiiiiiiiiiiiiiggggii×(riK2(i-r)/i(a(s,t)+v(s,t))gggggggggggggooohuuu+i-riK2r/i)=b(m,n)∑s=m0m-1‍∑t=n0n-1‍f(s,t)(jiK1(i-j)/ia(s,t)hhhhhhhhhhhhhhhooooohhhh+i-jiK1j/i)+c(m,n)∑s=m0M-1‍∑t=n0N-1‍g(s,t)∑s=m0M-1‍∑t=n0N-1‍∑s=m0M-1‍∑t=n0N-1‍×(riK2(i-r)/ia(s,t)+i-riK2r/i)+jb(m,n)i∑s=m0m-1‍∑t=n0n-1‍f(s,t)K1(i-j)/iv(s,t)+rc(m,n)i∑s=m0M-1‍∑t=n0N-1‍g(s,t)K2(i-r)/iv(s,t)=A(m,n)+B(m,n)∑s=m0m-1‍∑t=n0n-1‍F(s,t)v(s,t)+C(m,n)∑s=m0M-1‍∑t=n0N-1‍G(s,t)v(s,t),
for all (m,n)∈ΩM,N, where A, B, C and F, G are defined by (34), (35), and (36), respectively. Since a(m,n), b(m,n), and c(m,n) are nonnegative and nondecreasing in both m and n and by (34), (35), and (36), A(m,n), B(m,n), and C(m,n) are also nonnegative and nondecreasing in both m and n. Using Theorem 2, from (40), we obtain
(41)v(m,n)≤exp⁡(B(m,n)∑s=m0m-1‍∑t=n0n-1‍F(s,t))×[exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))-1A(m,n)+C(m,n)d×(exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))-1(exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ))∑s=m0M-1‍∑t=n0N-1‍G(s,t)A(s,t)iiiiid×exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))d×(exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ))1-∑s=m0M-1‍∑t=n0N-1‍G(s,t)C(s,t)d×exp⁡(B(s,t)∑τ=m0s-1‍∑ξ=n0t-1‍F(τ,ξ)))-1)],
for all (m,n)∈ΩM,N. Substituting (41) into (38), we get our required estimation (33) of unknown function in (8).


## 3. Difference Inequality with Weakly Singular

For the reader’s convenience, we present some necessary Lemmas.Lemma 5 (discrete Jensen inequality [47]).
LetA1,A2,…,An be nonnegative real numbers, k>1 a real number, and n a natural number. Then,
(42)(A1+A2+⋯+An)k≤nk-1(A1k+A2k+⋯+Ank).Lemma 6 (discrete Hölder inequality [48]).
Letai,bi  (i=1,2,…,n) be nonnegative real numbers and p, q positive numbers such that (1/q)+(1/p)=1. Then,(43)∑i=0n-1‍aibi≤(∑i=0n-1‍aip)1/p(∑i=0n-1‍biq)1/q.Lemma 7 (see [15, 49]).
Lett0=0, τs=ts+1-ts>0, and sup⁡s∈N,0≤s≤n-1{τs,s∈N}=τ. If β∈(0.5,1), γ>1.5-β, and p=1/β, then
(44)∑s=0n-1‍(tn-ts)p(β-1)tsp(γ-1)τs≤tnθB[p(γ-1)+1,p(β-1)+1],
where θ=p(β+γ-2)+1>0 and B(ξ,η):=∫01‍sξ-1(1-s)η-1ds is the well-known B-function.Now, we consider the weakly singular difference inequality (9).Theorem 8.
Lett0=0, τs=ts+1-ts>0, sup⁡s∈N,0≤s≤n-1{τs,s∈N}=τ, β∈(0.5,1), and γ>1.5-β. Assume that i≥j>0, i≥r>0, u(n), a(n), b(n), c(n), f(n), and g(n) are nonnegative functions on N0 and a(n), b(n), and c(n) are nondecreasing. If
(45)∑s=0N-1‍G~(s)C~(s)exp⁡(B~(s)∑τ=0s-1‍F~(τ))<1,n∈N0,n<N,
and u(n) satisfies (9), then
(46)u(n)≤{×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]a(n)+exp⁡(B~(n)∑s=0n-1‍F~(s))×[×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]A~(n)+C~(n)ii×(exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1(∑s=0N-1‍G~(s)A~(s)iiiiiiiiiiiiiiii×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))iiiiiiiiiiiiiiiiiiiii×(1-∑s=0N-1‍G~(s)C~(s)iiiiiiiiiiiiiiiiiiiiiiiiiiii×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]}1/i,ddddddddddddddddddddddddddddddn∈N0,n<N,
where
(47)A~(n):=b~(n)∑s=0n-1‍fq(s)(jiK1(i-j)/ia~(s)+i-jiK1j/i)+c~(n)∑s=n0N-1‍gq(s)(riK2(i-r)/ia~(s)+i-riK2r/i),B~(n):=jb~(n)i,C~(n):=rc~(n)i,F~(n):=fq(n)K1(i-j)/i,G~(n):=gq(n)K2(i-r)/i,a~(n):=3q-1aq(n),b~(n):=3q-1bq(n)τ×(tnθB[p(γ-1)+1,p(β-1)+1])q/p,c~(n):=3q-1cq(n)τ×(tNθB[p(γ-1)+1,p(β-1)+1])q/p,θ:=p(β+γ-2)+1,
and p=1/β, q=1/(1-β), and K1, K2 are arbitrary constants.Proof.
Applying Lemma6 with p=1/β, q=1/(1-β) to (8), we obtain that
(48)ui(n)≤a(n)+b(n)τ(p-1)/p×(∑s=n0n-1‍(tn-ts)p(β-1)tsp(γ-1)τs)1/p×(∑s=n0n-1‍fq(s)uqj(s))1/q+c(n)τ(p-1)/p×(∑s=n0N-1‍(tN-ts)p(β-1)tsp(γ-1)τs)1/p×(∑s=n0n-1‍gq(s)uqr(s))1/q,
for all n∈N0, n<N, where τs<τ is used. Applying Lemma 5 to (48), we have
(49)ui(n)≤a(n)+b(n)τ(p-1)/p×(tnθB[p(γ-1)+1,p(β-1)+1])1/p×(∑s=n0n-1‍fq(s)uqj(s))1/q+c(n)τ(p-1)/p×(tNθB[p(γ-1)+1,p(β-1)+1])1/p×(∑s=n0n-1‍gq(s)uqr(s))1/q,
for all n∈N0, n<N. By discrete Jensen inequality (42) with n=2, k=q, from (49), we obtain that
(50)uqi(n)≤3q-1aq(n)+3q-1bq(n)τq(p-1)/p×(tnθB[p(γ-1)+1,p(β-1)+1])q/p×∑s=n0n-1‍fq(s)uqj(s)+3q-1cq(n)τq(p-1)/p×(tNθB[p(γ-1)+1,p(β-1)+1])q/p×∑s=n0n-1‍gq(s)uqr(s)=3q-1aq(n)+3q-1bq(n)τ×(tnθB[p(γ-1)+1,p(β-1)+1])q/p×∑s=n0n-1‍fq(s)uqj(s)+3q-1cq(n)τ×(tNθB[p(γ-1)+1,p(β-1)+1])q/p×∑s=n0n-1‍gq(s)uqr(s)=a~(n)+b~(n)∑s=n0n-1‍fq(s)uqj(s)+c~(n)∑s=n0n-1‍gq(s)uqr(s),n∈N0,n<N.
Applying Theorem 4 to (50), we have
(51)u(n)≤{exp⁡⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]a(n)+exp⁡(B~(n)∑s=0n-1‍F~(s))×[exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)A~(n)+C~(n)×(exp⁡⁡(B~(s)∑τ=0s-1‍F~(τ)))-1(exp⁡(B~(s)∑τ=0s-1‍F~(τ))exp⁡(B~(s)∑τ=0s-1‍F~(τ))∑s=0N-1‍G~(s)A~(s)hhhhhh×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))×(exp⁡⁡(B~(s)∑τ=0s-1‍F~(τ))1-∑s=0N-1‍G~(s)C~(s)i×exp⁡⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]}1/i,dddddddddddddddddddffddddsn∈N0,n<N.
This is our required estimation (46) of unknown function in (9).


## 4. Applications

In this section, we apply our results to discuss the boundedness of solutions of an iterative difference equation with a weakly singular kernel.Example 9.
Suppose thatu(n) satisfies the difference equation
(52)x3(n)=a(n)+b(n)∑s=0n-1‍(tn-ts)-0.3ts-0.1τsf(s)x2(s)+c(n)∑s=0N-1‍(tN-ts)-0.3ts-0.1τsg(s)x(s),
where t0=0, τs=ts+1-ts>0, sup⁡s∈N,0≤s≤n-1{τs,s∈N}=τ, u(n), a(n), b(n), c(n), f(n), and g(n) are nonnegative functions on N0, and a(n), b(n), and c(n) are nondecreasing. From (52), we have
(53)|x(n)|3≤a(n)+b(n)∑s=0n-1‍(tn-ts)-0.3ts-0.1τsf(s)|x(s)|2+c(n)∑s=0N-1‍(tN-ts)-0.3ts-0.1τsg(s)|x(s)|.
Let p=10/7, q=10/3, and K1, K2 are arbitrary constants, and
(54)θ:=37,a~(n):=37/3a10/3(n),b~(n):=37/3b10/3(n)τ(tnθB[67,47])7/3,c~(n):=37/3c10/3(n)τ(tNθB[67,47])7/3,A~(n):=b~(n)∑s=0n-1‍f10/3(s)(23K11/3a~(s)+13K12/3)+c~(n)∑s=n0N-1‍g10/3(s)(13K22/3a~(s)+23K21/3),B~(n):=2b~(n)3,C~(n):=c~(n)3,F~(n):=f10/3(n)K11/3,G~(n):=g10/3(n)K22/3.
If
(55)∑s=0N-1‍G~(s)C~(s)exp⁡(B~(s)∑τ=0s-1‍F~(τ))<1,ddddddddddddddddn∈N0,n<N.
Applying Theorem 8 to (53), we obtain the estimation of the solutions of difference equation (52)
(56)|x(n)|≤{exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]a(n)+exp⁡(B~(n)∑s=0n-1‍F~(s))×[exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)A~(n)+C~(n)×(exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1(∑s=0N-1‍G~(s)A~(s)I×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))×(1-∑s=0N-1‍G~(s)C~(s)×exp⁡(B~(s)∑τ=0s-1‍F~(τ)))-1)]}1/i,fffffffffffffffffffffddddddhhfn∈N0,n<N.



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*Source: 826173-2014-08-14.xml*