# Some New Nonlinear Weakly Singular Inequalities and Applications to Volterra-Type Difference Equation

Authors: Kelong Zheng; Wenqiang Feng; Chunxiang Guo
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Some new nonlinear weakly singular difference inequalities are discussed, which generalize some known weakly singular inequalities and can be used in the analysis of nonlinear Volterra-type difference equations with weakly singular kernel. An application to the upper bound of solutions of a nonlinear difference equation is also presented.


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## Body

## 1. Introduction

The discrete version of the well-known Gronwall-Bellman inequality is an important tool in the development of the theory of difference equations as well as the analysis of the numerical schemes of differential equations. A great deal of interest has been given to these inequalities, and many results on their generalizations have been found; for example, see [1–4]. Among them, one of the fundamental cases is Pachpatte’s result [3] for the difference inequality:
(1)u(n)≤a(n)+∑s=0n-1‍f(s)u(s).In particular, due to the study of the behavior and numerical solutions for the singular integral equations, some discrete weakly singular integral inequalities also have drawn more and more attention [5–7]. Dixon and McKee [8] investigated the convergence of discretization methods for the Volterra integral and integrodifferential equations, by using the following inequality:
(2)xi≤ψi+Mh1-α∑j=0i-1‍xj(i-j)α,i=1,2,…,N,hhhhhn>0,Nh=T.Henry [9] presented a linear integral inequality with weakly kernel:
(3)x(t)≤a(t)+∫0t‍(t-s)β-1b(s)x(s)ds
to investigate some qualitative properties for a parabolic equation. The corresponding discrete version was discussed by Slodic˘ka [10]. But he studied the case τk=τ, that is, the case of constant differences. Furthermore, the first formulation of the inequality with a nonlinearity and τk nonconstant was studied in [6], in which the general nonlinear discrete case as follows:
(4)xn≤an+∑k=0n-1‍(tn-tk)β-1τkbkω(xk)
was considered. However, his results are based on the so-called “(q) condition”: (1) ω satisfies e-qt[ω(u)]q≤R(t)ω(e-qt)uq; (2) there exists c>0 such that ane-τtn≤c. Recently, a new nonlinear difference inequality:
(5)xnα≤an+∑k=0n-1‍(tn-tk)β-1τkbkxkλ
was discussed by Yang et al. [11]. For other new weakly singular inequalities, lots of work can be found, for example, in [12–22] and references therein.In this paper, we investigate the new nonlinear weakly singular inequality:(6)xn≤an+∑k=0n-1‍(tn-tk)β-1τkbkω(xk),
where 0<β≤1, t0=0, τk=tk+1-tk, supk∈ℕτk=τ, and limt→∞tk=∞. Compared to the existing result, our result does not need the so-called “(q) condition” proposed in [6] and can be used to obtain pointwise explicit bounds on solutions for a class of more general weakly singular inequalities of Volterra type. Finally, we also present an application to Volterra-type difference equation with weakly singular kernel.


## 2. Preliminaries

Letℝ be the set of real numbers, ℝ+=(0,∞), and ℕ={0,1,2,…}. 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.Lemma 1 (Discrete Jensen inequality, [11]).
LetA1,A2,…,An be nonnegative real numbers, and let r>1 be a real number. Then,
(7)(A1+A2+⋯+An)r≤nr-1(A1r+A2r+⋯+Anr).Lemma 2 (Discrete Hölder inequality, [11]).
Letai,bi(i=1,2,…,n) be nonnegative real numbers, and let p,q be positive numbers such that (1/p)+(1/q)=1 (or p=1,q=∞). Then,
(8)∑i=1n‍aibi≤(∑i=1n‍aip)1/p(∑i=1n‍biq)1/q.
Furthermore, takep=q=2; then, one gets the discrete Cauchy-Schwarz inequality.Lemma 3.
Suppose thatω(u)∈C(ℝ+,ℝ+) is nondecreasing. Let an,cn be nonnegative and nondecreasing in n. If yn is nonnegative such that
(9)yn≤an+cn∑k=0n-1‍bkω(yk),n∈ℕ.
Then,
(10)yn≤Ω-1[Ω(an)+cn∑k=0n-1‍bk],0≤n≤M,
where Ω(v)=∫v0v‍(1/ω(s))ds,v≥v0, Ω-1 is the inverse function of Ω, and M is defined by
(11)M=sup{i:Ω(ai)+ci∑k=0i-1‍bk∈Dom(Ω-1)}.


## 3. Main Results

Assume that(A1) an,bn are nonnegative functions for n∈ℕ, respectively;(A2) ω(u)∈C(ℝ+,ℝ+) is nondecreasing and ω(0)=0.Definea~n=max0≤k≤n,k∈ℕak and τ=max0≤k≤n-1,k∈ℕτk, where τk is the variable time step.Theorem 4.
Under assumptions (A1) and (A2), if xn is nonnegative such that (6), then(1)
for0<β≤1/2, letting p=1+β and q=(1+β)/β, one has
(12)xn≤[Ω-1(∑k=0n-1‍Ω(2q-1a~nq)+2q-1τ1-(q/p)β2×Kq/p(β)eqτtn∑k=0n-1‍e-qτtkbkq)]1/q,
for 0≤n≤N1, where Ω(u)=∫u0u‍(1/ωq(s1/q))ds,u≥u0≥0, Ω-1 is the inverse function of Ω,
(13)K(β)=(1+β)-β2Γ(β2),
and N1 is the largest integer number such that
(14)Ω(2q-1a~nq)+2q-1τ1-(q/p)β2Kq/p(β)eqτtn×∑k=0n-1‍e-qτtkbkq∈Dom(Ω-1);(2)
for1/2<β≤1, letting p=2 and q=2, one has
(15)xn≤[Ω-1(Ω(2a~n2)+B(β)τ2-2βe2τtn∑k=0n-1‍e-2τtkbk2)]1/2,
for 0≤n≤N2, where Ω(u)=∫u0u‍(1/ω2(s1/2))ds,u≥u0≥0,
(16)B(β)=41-βΓ(2β-1),β>12,
and N2 is the largest integer number such that
(17)Ω(2a~n2)+B(β)τ2-2βe2τtn∑k=0n-1‍e-2τtkbk2∈Dom(Ω-1).Proof.
By definition ofa~n and assumption (A1), a~n is nonnegative and nondecreasing and a~n≥an. It follows from (6) that
(18)xn≤a~n+∑k=0n-1‍(tn-tk)β-1τkbkω(xk).(1)
If0<β≤1/2, using Lemma 2 with the indices p=1+β,q=(1+β)/β for (18), we get
(19)xn≤a~n+∑k=0n-1‍(tn-tk)β-1τk1/pτk1/qeτtke-τtkbkω(xk)≤a~n+τ1/q∑k=0n-1‍(tn-tk)β-1τk1/peτtke-τtkbkω(xk)≤a~n+τ1/q[∑k=0n-1‍(tn-tk)p(β-1)τkepτtk]1/p×[∑k=0n-1‍e-qτtkbkqωq(xk)]1/q.
By Lemma 1, the inequality above yields
(20)xnq≤2q-1a~nq+2q-1τ[∑k=0n-1‍(tn-tk)p(β-1)τkepτtk]q/p×[∑k=0n-1‍e-qτtkbkqωq(xk)].
Consider that
(21)∑k=0n-1‍(tn-tk)p(β-1)τkepτtk≤∫0tn‍(tn-s)p(β-1)epτsds=epτtn∫0tn‍ηp(β-1)e-pτηdη,=epτtn(pτ)1+p(β-1)∫0pτtn‍σp(β-1)e-σdσ≤K(β)τ-β2epτtn,
where K(β)=(1+β)-β2Γ(β2) and Γ(z)=∫0∞‍uz-1e-udu,(Rez>0) is the well-known G-function. Thus, we have
(22)xnq≤2q-1a~nq+2q-1τ1-(q/p)β2×Kq/p(β)eqτtn∑k=0n-1‍e-qτtkbkqωq(xk).
Let vn=xnq, An=2q-1a~nq, and Cn=2q-1τ1-(q/p)β2Kq/p(β)eqτtn. Obviously, An,Cn are nondecreasing for n∈ℕ and ωq(vk1/q) satisfies the assumption (A2). Equation (22) can be rewritten as
(23)vn≤An+Cn∑k=0n-1‍e-qτtkbkqωq(vk1/q),
which is similar to inequality (9). Using Lemma 3, from (23), we have
(24)vn≤Ω-1[(Ω(An)+Cn∑k=0n-1‍e-qτtkbkq)],
for 0≤n≤N1, where N1 is the largest integer number such that
(25)Ω(An)+Cn∑k=0n-1‍e-qτtkbkq∈Dom(Ω-1).
Therefore, by xn=vn1/q, (12) holds for 0≤n≤N1.(2)
If1/2<β≤1, applying Cauchy-Schwarz inequality for (18), that is, p=q=2, we get
(26)xn≤a~n+∑k=0n-1‍(tn-tk)β-1τk1/2τk1/2eτtke-τtkbkω(xk)≤a~n+τ1/2∑k=0n-1‍(tn-tk)β-1τk1/2eτtke-τtkbkω(xk)≤a~n+τ1/2[∑k=0n-1‍(tn-tk)2(β-1)τke2τtk]1/2×[∑k=0n-1‍e-2τtkbk2ω2(xk)]1/2.
By Lemma 1, the inequality above yields
(27)xn2≤2a~n2+2τ[∑k=0n-1‍(tn-tk)2(β-1)τke2τtk]×[∑k=0n-1‍e-2τtkbk2ω2(xk)].
Because
(28)∑k=0n-1‍(tn-tk)2(β-1)τke2τtk≤∫0tn‍(tn-s)2(β-1)e2τsds=e2τtn(2τ)2β-1∫02τtn‍σ2(β-1)e-σdσ≤12B(β)τ1-2βe2τtn,
where B(β)=41-βΓ(2β-1),β>1/2, it follows from (27) that
(29)xn2≤2a~n2+B(β)τ2-2βe2τtn[∑k=0n-1‍e-2τtkbk2ω2(xk)].
Let vn=xn2, An=2a~n2, and Cn=B(β)τ2-2βe2τtn. Similarly, An,Cn also are nondecreasing for n∈ℕ and ω2(vk1/2) also satisfies the assumption (A2). Equation (29) can be rewritten as
(30)vn≤An+Cn(∑k=0n-1‍e-2τtkbk2ω2(vk1/2)),
which also is similar to inequality (9). Using Lemma 3, from (30), we have
(31)vn≤[Ω-1(Ω(An)+Cn∑k=0n-1‍e-2τtkbk2)],
for 0≤n≤N2, and N2 is the largest integer number such that
(32)Ω(An)+Cn∑k=0n-1‍e-2τtkbk2∈Dom(Ω-1).
Clearly, by xn=vn1/2, (15) also holds for 0≤n≤N2.Remark 5.
Here, we note that the most significant work in the study of weakly singular inequalities is Medved˘’s method, originally presented in the paper [6] and also applied in the paper [18]. But his result holds under the assumption “ω(u) satisfies the condition (q),” that is, “e-qt[ω(u)]q≤R(t)ω(e-qtuq), where R(t) is a continuous, nonnegative function.” In our result, the condition (q) is eliminated.Corollary 6.
Under assumptions (A1) and (A2), let ν>0, μ>0(ν>μ). If xn is nonnegative such that
(33)xnν≤an+∑k=0n-1‍(tn-tk)β-1τkbkxkμ,
then(1)
if0<β≤1/2, let p=1+β and q=(1+β)/β, and one gets
(34)xn≤[∑k=0n-1‍(2q-1a~nq)(ν-μ)/ν+ν-μν2q-1τ1-(q/p)β2×Kq/p(β)eqτtn∑k=0n-1‍e-qτtkbkq]1/(ν-μ)q
for n≥0, where K(β) is defined as in Theorem 4;(2)
if1/2<β≤1, let p=q=2, and one gets
(35)xn≤[∑k=0n-1‍(2a~n2)(ν-μ)/ν+ν-μνB(β)τ2-2β×e2τtn∑k=0n-1‍e-2τtkbk2]1/2(ν-μ),
for n≥0, where B(β) is defined as in Theorem 4Proof.
Letzn=xnν, then xn=zn1/ν and xnμ=znμ/ν. From (33), we have
(36)zn≤an+∑k=0n-1‍(tn-tk)β-1τkbkzkμ/ν.
Clearly,ω(zk)=zkμ/ν satisfies the assumption (A2). According to the definition of Ω in Theorem 4, for 0<β≤1/2, letting u0=0, we have
(37)Ω(u)=∫u0u‍1ωq(s1/q)ds=∫0u‍dssμ/ν=νν-μu(ν-μ)/ν,(38)Ω-1(u)=(ν-μνu)ν/(ν-μ),Dom(Ω-1)=[0,∞).
It can be seen easily from (38) that N1=∞. Substituting (37) and (38) into (12), we get
(39)zn≤[∑k=0n-1‍(2q-1a~nq)(ν-μ)/ν+ν-μν2q-1τ1-(q/p)β23×Kq/p(β)eqτtn∑k=0n-1‍e-qτtkbkq]ν/(ν-μ)q.
In view ofxn=zn1/ν, we can obtain (34). For the case that 1/2<β≤1, in fact, Ω and Ω-1 are the same as (37) and (38), respectively. So, it follows from (37), (38), and (15) that
(40)xn≤[∑k=0n-1‍(2a~n2)(ν-μ)/ν+ν-μνB(β)τ2-2β×e2τtn∑k=0n-1‍e-2τtkbk2]1/2(ν-μ),
for n>0.Remark 7.
In [11], Yang  et al. investigated inequality (33), under the assumption that an is nondecreasing. Clearly, our result does not need such condition, and we get a more concise formula.Remark 8.
Lettingν=2 and μ=1, we can get the interesting Henry version of the Ou-Iang-Pachpatte-type difference inequality [3]. Thus, our result is a more general discrete analogue for such inequality.Corollary 9.
Under assumptions (A1) and (A2), if xn is nonnegative such that
(41)xn≤an+∑k=0n-1‍(tn-tk)β-1τkbkxk,
then(1)
if0<β≤1/2, let p=1+β and q=(1+β)/β, and one gets
(42)xn≤2(q-1)/qa~nexp(∑k=0n-1‍2(q-1)/qτ1-(q/p)β2Kq/p(β)×eqτtn∑k=0n-1‍e-qτtkbkq),
for n≥0, where K(β) is defined as in Theorem 4;(2)
if1/2<β≤1, let p=q=2, and one gets
(43)xn≤2a~nexp(12B(β)τ2-2βe2τtn∑k=0n-1‍e-2τtkbk2),
for n≥0, where B(β) is defined as in Theorem 4.Proof.
In (41), ω(u)=u also satisfies the assumption (A2). Thus, we have
(44)Ω(u)=∫u0u‍dss=lnuu0,Ω-1(u)=u0exp(u),Dom(Ω-1)=[0,∞).
Similarly to the computation in Corollary6, the estimates (42) and (43) hold, respectively.


## 4. Application

In this section, we apply our results to discuss the upper bound of solution of a Volterra type difference equation with weakly singular kernel.Consider the following the inequality:(45)xn≤1+∑k=0n-1‍(tn-tk)-1/2τkxk.Obviously, (45) is the special case of inequality (6), then we get
(46)an=1,β=12,ω=u.Thus, we can takep=1+β=3/2 and q=(1+β)/β=3; then, q/p=2. Moreover,
(47)a~n=1,K(β)=(1+β)-β2Γ(β2)=(32)-1/4Γ(14),Ω(u)=∫0u‍dss=2u,Ω-1(u)=u24.According to Theorem4, we obtain
(48)xn≤[∑k=0n-1‍Ω-1(Ω(2q-1a~nq)+2q-1τ1-(q/p)β2)×Kq/p(β)eqτtn∑k=0n-1‍e-qτtkbkq]1/q=[∑k=0n-1‍Ω-1(Ω(4)+4τ1/2(32)-1/2)×Γ2(14)e3τtn∑k=0n-1‍e-3τtkbk3]1/3=[Ω-1(4+436τ1/2Γ2(14)e3τtn∑k=0n-1‍e-3τtkbk3)]1/3=4-1/3(4+436τ1/2Γ2(14)e3τtn∑k=0n-1‍e-3τtkbk3)2/3
for n>0, which indicates that we get the upper bound of xn.



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