# On Weakly Singular Versions of Discrete Nonlinear Inequalities and Applications

Authors: Kelong Cheng; Chunxiang Guo; Qingke Zeng
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Some new weakly singular versions of discrete nonlinear inequalities are established, which generalize some existing weakly singular inequalities and can be used in the analysis of nonlinear Volterra type difference equations with weakly singular kernels. A few applications to the upper bound and the uniqueness of solutions of nonlinear difference equations are also involved.


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## Body

## 1. Introduction

Recently, along with the development of the theory of integral inequalities and difference equations, many authors have researched some discrete versions of Gronwall-Bellman type inequalities [1–5]. Starting from the basic form,
(1)u(n)≤a(n)+∑s=0n-1‍f(s)u(s)
discussed originally by Pachpatte in [4], various such new inequalities have been established, which can be used as a powerful tool in the analysis of certain classes of finite difference equations. Among these results, discrete weakly singular integral inequalities also play an important role in the study of the behavior and numerical solutions for singular integral equations [6, 7] and the theory for parabolic equations [8–10]. For example, Dixon and McKee [7] investigated the convergence of discretization methods for the Volterra integral and integrodifferential equations using the following inequality:
(2)xi≤ψi+Mh1-α∑j=0i-1‍xj(i-j)α,i=1,2,…,N,xi≤ψi+Mh1-α∑j=0i-1‍xj(i-j)α,n>0,Nh=T,
and Beesack [6] also discussed the inequality,
(3)xi≤ψi+Mh1+σ-αβ∑j=0i-1‍jσxj(iβ-jβ)α,
for the second kind Abel-Volterra singular integral equations. Henry [9] presented a linear inequality
(4)xn≤an+∑k=0n-1‍(tn-tk)β-1τkbkxk,
to investigate some qualitative properties for a parabolic equation. In particular, to avoid the shortcoming of analysis, Medved [11–13] used a new method to discuss some nonlinear weakly singular integral inequalities and difference inequalities. Following Medved’s work, Ma and Yang [14] improved his method to discuss a more general nonlinear weakly singular integral inequality,
(5)u(t)≤a(t)+b(t)∫0t‍(tσ-sσ)μ-1sτ-1g(s)u(s)ds+c(t)∫0t‍(tα-sα)β-1sγ-1g(s)w(u(s))ds,
and a nonlinear difference inequality [15],
(6)xnα≤an+∑k=0n-1‍(tn-tk)β-1τkbkxkλ.
As for other new weakly singular inequalities, recent work can be found, for example, in [16–25] and references therein.In this paper, we investigate some new nonlinear discrete weakly singular inequalities(7)xn≤an+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk),(8)xn≤an+∑k=0n-1‍(tnσ-tkσ)μ-1tkλ-1τkgkxk+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk).
Compared to the existing result, our result is more concise and can be used to obtain pointwise explicit bounds on solutions of a class of more general weakly singular difference equations of Volterra type. Finally, to illustrate the usefulness of the result, we give some applications to Volterra type difference equation with weakly singular kernels.For convenience, before giving our main results, we first cite some useful lemmas here.Lemma 1 (discrete Jensen inequality, see [15]).
LetA1,A2,…,An be nonnegative real numbers and r>1 a real number. Then
(9)(A1+A2+⋯+An)r≤nr-1(A1r+A2r+⋯+Anr).Lemma 2 (discrete Hölder inequality, see [15]).
Letai,bi(i=1,2,…,n) be nonnegative real numbers, and p,q positive numbers such that 1/p+1/q=1 (or p=1,q=∞). Then
(10)∑i=1n‍aibi≤(∑i=1n‍aip)1/p(∑i=1n‍biq)1/q.Lemma 3 (see [14]).
Letα,β,γ, and p be positive constants. Then
(11)∫0t‍(tα-sα)p(β-1)sp(γ-1)ds=tθαB[p(γ-1)+1α,p(β-1)+1],t∈R+,
where B[ξ,η]=∫01‍sξ-1(1-s)η-1ds  (Reξ>0,Reη>0) is the well-known B-function and θ=p[α(β-1)+γ-1]+1.In what follows, denoteR to be the set of real numbers. Let 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.


## 2. Some New Nonlinear Weakly Singular Difference Inequalities

Lemma 4.
Suppose thatω(u)∈C(R+,R+) is nondecreasing with w(u)>0 for u>0. Let an,cn be nonnegative and nondecreasing in n. If yn is nonnegative such that
(12)yn≤an+cn∑k=0n-1‍bkω(yk),n∈N,
then
(13)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
(14)M=sup⁡⁡{i:Ω(ai)+ci∑k=0i-1‍bk∈Dom⁡(Ω-1)}.Assume that(S1)
α∈(0,1],β∈(0,1) and 1>(p(γ-1)+1)/α≥p(β-1)+1>0 such that 1/p+α(β-1)+γ-1≥0;(S2)
an,bn are nonnegative functions for n∈N, respectively;(S3)
ω(u)∈C(R+,R+) is nondecreasing and ω(0)=0.Lemma 5.
Suppose that[α,β,γ] satisfies assumption (S1); then for sufficiently small τk, one has
(15)∑k=0n-1‍(tnα-tkα)pi(β-1)tkpi(γ-1)τk≤∫0tn‍(tnα-sα)pi(β-1)spi(γ-1)ds=tnθiαB[pi(γ-1)+1α,pi(β-1)+1].Proof.
Consider theB-function in (15). Consider
(16)B[p(γ-1)+1α,p(β-1)+1]=∫01‍(1-s)p(β-1)+1-1s(p(γ-1)+1)/α-1ds:=∫01‍(1-s)n2-1sn1-1ds,
and denote f(s):=(1-s)n2-1sn1-1 for s∈(0,1), where n1=(p(γ-1)+1)/α and n2=p(β-1)+1. If n2=n1, then f(s) is symmetric about s=1/2. According to assumption (S1),
(17)1>n1=p(γ-1)+1α>p(β-1)+1=n2>0;
that is,
(18)0>n1-1>n2-1>-1.
On the other hand, the zero-point of f′(s) can be obtained as follows:
(19)s0=n1-1n1+n2-2<12.
Therefore, the function f(s) is decreasing on the interval (0,s0] while increasing sharply on the interval [s0,1). Consequently, for some given sufficiently small τk, by the properties of the left-rectangle integral formula, we have
(20)∑k=0n-1‍(1α-tkα)n2-1tkn1-1τk≤∫01‍(1-sα)n2-1sn1-1ds=B[p(γ-1)+1α,p(β-1)+1],
where 0=t0<t1<⋯<tn=1. For the general interval (0,tn], we can easily obtain the corresponding result (15) by the similar method and omit the details here.
Denotea~n=max⁡0≤k≤n,k∈Nak and τ=max⁡0≤k≤n-1,k∈Nτk, where τk is the variable time step. Next, we first discuss inequality (7) and obtain the following result.Theorem 6.
Under assumptions(S1), (S2), and (S3), if xn is nonnegative such that (7), then
(21)xn≤[Ω-1(∑k=0n-1‍bkqΩ(2q-1a~nq)+2q-1τ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p∑k=0n-1‍bkq)(tnθα)q/p]1/q,
for 0≤n≤N1, where Ω(u)=∫u0u‍(1/ωq(s1/q))ds, u≥u0, Ω-1 is the inverse function of Ω, θ=p[α(β-1)+γ-1]+1, and N1 is the largest integer number such that
(22)Ω(2q-1a~nq)+2q-1τ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p×∑k=0n-1‍bkq∈Dom⁡(Ω-1).Proof.
Sincea~n=max⁡0≤k≤n,k∈Nak, according to assumption (S2), a~n is nonnegative and nondecreasing, and a~n≥an. From (7), we have
(23)xn≤a~n+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk).
Due to assumption (S1), take suitable indices p,q such that 1/p+1/q=1. An application of Lemma 2 yields
(24)xn≤a~n+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τk1/pτk1/qbkω(xk)≤a~n+τ1/q∑k=0n-1‍(tnα-tkα)β-1tkγ-1τk1/pbkω(xk)≤a~n+τ1/q[∑k=0n-1‍(tnα-tkα)p(β-1)tkp(γ-1)τk]1/p×[∑k=0n-1‍bkqωq(xk)]1/q.
By Lemma 1, it follows from the inequality above that
(25)xnq≤2q-1a~nq+2q-1τ[∑k=0n-1‍(tnα-tkα)p(β-1)tkp(γ-1)τk]q/p×[∑k=0n-1‍bkqωq(xk)].
Considering
(26)∑k=0n-1‍(tnα-tkα)p(β-1)tkp(γ-1)τk≤∫0tn‍(tnα-sα)p(β-1)sp(γ-1)ds=tnθαB[p(γ-1)+1α,p(β-1)+1],
in which we apply Lemma 5, we have
(27)xnq≤2q-1a~nq+2q-1τ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p×[∑k=0n-1‍bkqωq(xk)].
Letvn=xnq, An=2q-1a~nq, and Cn=2q-1τ(tnθ/α)q/p(B[(p(γ-1)+1)/α,p(β-1)+1])q/p. Obviously, An,Cn are nondecreasing for n∈N and ωq(vk1/q) satisfies assumption (S3). Equation (27) can be rewritten as
(28)vn≤An+Cn(∑k=0n-1‍bkqωq(vk1/q)),
which is similar to inequality (12). Using Lemma 4 to (28), we have
(29)vn≤[Ω-1(Ω(An)+Cn∑k=0n-1‍bkq)],
for 0≤n≤N1, where Ω(u)=∫u0u‍(1/ωq(s1/q))ds, u≥u0, Ω-1 is the inverse function of Ω, and N1 is the largest integer number such that
(30)Ω(An)+Cn∑k=0n-1‍bkq∈Dom⁡(Ω-1).
Therefore, by xn=vn1/q, (21) holds for 0≤n≤N1.Remark 7.
Whenα=1 and γ=1, the inequality was discussed by Medved [12] which is the special case of our result. Moreover, 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 8.
Under assumptions (S1) and (S2), let ν>0, μ>0(ν>μ). If xn is nonnegative such that
(31)xnν≤an+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkxkμ,
then
(32)xn≤[2q-1a~n(ν-μ)/ν+ν-μν2q-1τ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p∑k=0n-1‍bkq(tnθα)q/p]1/(ν-μ)q,
for n≥0.Proof.
Letyn=xnν; then xn=yn1/ν or xnμ=ynμ/ν. From (31) we have
(33)yn≤an+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkykμ/ν.
Denote ω(yk)=ykμ/ν. Clearly, ω satisfies assumption (S3). With the definition of Ω in Theorem 6, letting u0=0, we have
(34)Ω(u)=∫0u‍dssμ/ν=νν-μu(ν-μ)/ν,(35)Ω-1(u)=(ν-μνu)ν/(ν-μ),Dom⁡(Ω-1)=[0,∞).
Substituting (34) and (35) into (29), we get
(36)yn≤[2q-1a~n(ν-μ)/ν+ν-μν2q-1τ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p∑k=0n-1‍bkq(tnθα)q/p]ν/(ν-μ)q.
In view of xn=yn1/ν, we can obtain (32).Remark 9.
In [15], Yang et al. investigated inequality (6). Clearly, let α=1 and γ=1 in (31), and we can get the same formula.Remark 10.
Letν=2 and μ=1; we can get the interesting Henry’s version of the Ou-Iang-Pachpatte type difference inequality [26]. Thus, our results are a more general discrete analogue for such inequality.Remark 11.
Ma and Pečarić discussed the continuous case of (2.15) in [27] and here we present the discrete version of their result. Furthermore, the result in [27] is established for the cases when the ordered parameter group [α,β,γ] obeys distribution I or II (for details, see [27]) which makes the application of inequality more inconvenient. Clearly, our result is based on the concise assumption to overcome this weakness.Corollary 12.
Under assumptions (S1) and (S2), if xn is nonnegative such that
(37)xn≤an+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkxk,
then
(38)xn≤2(q-1)/qa~n×exp⁡⁡(2q-1qτ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p∑k=0n-1‍bkq2q-1qτ(tnθα)q/p),
for n≥0.Proof.
In (7), ω(u)=u also satisfies assumption (S3). Thus, we have
(39)Ω(u)=∫u0u‍dss=ln⁡uu0,Ω-1(u)=u0exp⁡⁡(u),Dom⁡(Ω-1)=[0,∞).
Similar to the computation in Corollary 8, estimate (38) holds.
Now, we discuss inequality (8)
(40)xn≤an+∑k=0n-1‍(tnσ-tkσ)μ-1tkλ-1τkgkxk+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk).
Since there are two different parameter groups [σ,μ,λ] and [α,β,γ], assumption (S1) is revised as follows: (S4)
σ∈(0,1],μ∈(0,1), 1>(p(λ-1)+1)/σ≥p(μ-1)+1>0 and α∈(0,1],β∈(0,1), 1>(p(γ-1)+1)/α≥p(β-1)+1>0 such that 1/p+σ(μ-1)+λ-1≥0 and 1/p+α(β-1)+γ-1≥0(p>1).Theorem 13.
Under assumptions (S3) and (S4), suppose that gk,bk are nonnegative for n∈N. If xn is nonnegative such that (8), then
(41)xn≤[Ω-1(∑k=0n-1‍bkqΩ(2q-1a~nqQnq)+2q-1Qnqτ(tnθ2α)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p×∑k=0n-1‍bkq)]1/q,
for 0≤n≤N2, where Ω and Ω-1 are defined as in Theorem 6,
(42)Qn=2(q-1)/q×exp⁡⁡(2q-1qτ(tnθ1σ)q/p×(B[p(λ-1)+1σ,p(μ-1)+1])q/p∑k=0n-1‍gkq(tnθ1σ)q/p),θ1=p[σ(μ-1)+λ-1]+1,θ2=p[α(β-1)+γ-1]+1,
and N2 is the largest integer number such that
(43)Ω(2q-1a~nqQnq)+2q-1Qnqτ(tnθ2α)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p×∑k=0n-1‍bkq∈Dom⁡(Ω-1).Proof.
By the definition ofa~n, we have
(44)xn≤a~n+∑k=0n-1‍(tnσ-tkσ)μ-1tkλ-1τkgkxk+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk).
Let
(45)Pn=a~n+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk),
which yields directly
(46)xn≤Pn+∑k=0n-1‍(tnσ-tkσ)μ-1tkλ-1τkgkxk.
Using Corollary 12, from the inequality above, we get
(47)xn≤2(q-1)/qPn×exp⁡⁡(2q-1qτ(tnθ1σ)q/p×(B[p(λ-1)+1σ,p(μ-1)+1])q/p∑k=0n-1‍gkq2q-1qτ(tnθ1σ)q/p),
where θ1=p[σ(μ-1)+λ-1]+1. Letting
(48)Qn=2(q-1)/qexp⁡⁡(2q-1qτ(tnθ1σ)q/p×(B[p(λ-1)+1σ,p(μ-1)+1])q/p×∑k=0n-1‍gkq(tnθ1σ)q/p),
from (47), we get
(49)un≤PnQn≤a~nQn+Qn∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkω(xk).
Clearly, inequality (49) is similar to (7). According to Theorem 6, we obtain
(50)xn≤[(tnθ2α)q/pΩ-1((tnθ2α)q/pΩ(2q-1a~nqQnq)+2q-1Qnqτ(tnθ2α)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p×∑k=0n-1‍bkq(tnθ2α)q/p)]1/q,
for 0≤n≤N2.Remark 14.
Our result for inequality (8) is also the discrete analogue of inequality (5). In fact, with the different choice of the parameter groups [α,β,γ] and [σ,μ,λ] in [14], the complicate results must be presented by four cases, respectively. Apparently, compared to their results, our result is quite simple.


## 3. Applications

In this section, we apply our results to discuss the upper bound and the uniqueness of solutions of a Volterra type difference equation with certain weakly singular kernels.Example 15.
Consider the following inequality:(51)xn≤12+∑k=0n-1‍(tn-tk)-1/3tk-1/4τkxk+∑k=0n-1‍(tn-tk)-1/3tk-1/3τkxk.
Obviously, (51) is the special case of inequality (8), and we get
(52)an=12,σ=1,μ=23,λ=34,α=1,β=23,γ=23,gk=1,bk=1.
Next, we discuss the choice of parameter p. By assumption (S4), from the conditions 1>(p(λ-1)+1)/σ≥p(μ-1)+1>0 and 1/p+σ(μ-1)+λ-1≥0, we have 1<p<12/7. From the conditions 1>(p(γ-1)+1)/α≥p(β-1)+1>0 and 1/p+α(β-1)+γ-1≥0(p>1), we have 1<p<3/2. Thus, we can take p=4/3; then q=4, q/p=3. According to Theorem 13, we obtain
(53)a~n=12,θ1=29,θ2=19,B[p(λ-1)+1σ,p(μ-1)+1]=B[23,59],B[p(γ-1)+1α,p(β-1)+1]=B[59,59],q-1q=34,2q-1q=2;Qn=23/4exp⁡(2τ(tn2/9)3(B[23,59])3n),Ω(u)=∫0u‍dss=2u,Ω-1(u)=u24,2q-1a~nqQnq=4exp⁡⁡(8τtn1/5B3[23,59]n).
Substituting the results above into (41), we can get the upper bound of xn and omit the details for its complicated formula.Example 16.
Consider the linear weakly singular difference equation(54)xn=an+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkxk,(55)yn=cn+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbkyk,
where |an-cn|<ϵ, ϵ is an arbitrary positive number, and the parameter group [α,β,γ] satisfies assumption (S1). From (54) and (55), we get
(56)|xn-yn|≤|an-cn|+∑k=0n-1‍(tnα-tkα)β-1tkγ-1τkbk|xk-yk|,
which is the form of inequality (37). Applying Corollary 12, we have
(57)|xn-yn|≤2(q-1)/qϵ×exp⁡⁡(2q-1qτ(tnθα)q/p×(B[p(γ-1)+1α,p(β-1)+1])q/p×∑k=0n-1‍bkq(tnθα)q/p),
for n∈N. If an=cn, let ϵ→0 and we obtain the uniqueness of the solution of (54).



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