# Some Generalized Gronwall-Bellman Type Impulsive Integral Inequalities and Their Applications

Authors: Yuzhen Mi
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

This paper investigates some generalized Gronwall-Bellman type impulsive integral inequalities containing integration on
infinite intervals. Some new results are obtained, which generalize some existing conclusions. Our result is also applied to study a boundary
value problem of differential equations with impulsive terms.


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## Body

## 1. Introduction

It is well known that Gronwall-Bellman type integral inequalities involving functions of one and more than one independent variables play important roles in the study of existence, uniqueness, boundedness, stability, invariant manifolds, and other qualitative properties of solutions of the theory of differential and integral equations. A lot of contributions to its generalization have been archived by many researchers (see [1–14]). Pachpatte [15] especially studied the following inequality:
(1)u(x)≤a(x)+∫x∞‍b(s)u(s)ds
containing integration on infinite integral and used it in the study of terminal value problems for Gronwall-Bellman type differential equations. Then, Cheung and Ma [16] generalized it into two independent variables with a nonlinear term:
(2)u(x,y)≤a(x,y)+c(x,y)∫x∞‍∫y∞‍d(s,t)ω(u(s,t))dsdt.Along the development of the theory of impulsive differential systems, more and more attention is paid to generalizations of Gronwall-Bellman’s results for discontinuous functions (that is, impulsive integral inequalities) and their applications (see [17–25]). Among them, one of the important things is that Samoilenko and Perestyuk [17] considered
(3)u(x)≤c+∫x0x‍f(s)u(s)ds+∑x0<xi<x‍βiu(xi-0)
about the nonnegative piecewise continuous function u(x) where c, βi are nonnegative constants, f(x) is a positive function, and xi are the first kind discontinuity points of the function u(x). Then Borysenko [18] investigated integral inequalities with two independent variables:
(4)u(x,y)≤a(x,y)+∫x0x‍∫y0y‍τ(s,t)u(s,t)dsdt+∑(x0,y0)<(xi,yi)<(x,y)‍βiu(xi-0,yi-0).
Here u(x,y) is an unknown nonnegative continuous function with the exception of the points (xi,yi) where there is a finite jump u(xi-0,yi-0)≠u(xi+0,yi+0) for i=1,2,…. In 2013, Zheng [25] considered the following delay integral inequalities containing integration on infinite intervals:
(5)u(x)≤c+∫x∞‍f1(x,s)u(τ(s))ds+∫x∞‍f2(x,s)ω(u(τ(s)))ds+∑x<xi<∞‍βiu(xi-0),(6)u(x,y)≤c+∫x∞‍∫y∞‍f1(s,t)u(σ(s),τ(t))dsdt+∫x∞‍∫y∞‍f2(s,t)ω(u(σ(s),τ(t)))dsdt+∑x<xi<∞,y<yi<∞‍βiu(xi-0,yi-0)
with one general nonlinear term ω(u). They assumed that ω∈℘ where the class ℘ consists of all nonnegative, nondecreasing, and continuous functions ω(u) on [0,∞) such that ω(0)=0 and ω(αu)≤ω(α)ω(u) for all α>0 and u≥0. Actually, when we study behaviors of solutions of differential equations with impulsive terms, ω may not satisfy the following condition: ω∈℘. For example, ω(u)=eu does not belong to the class ℘ for any α>1 and large u>0. Thus, it is very interesting to avoid such conditions. Our main aim here, motivated by the work above, is to discuss the following much more general integral inequality:
(7)u(x)≤a(x)+∑k=12‍∫x∞‍fk(x,s)ωk(u(σk(s)))ds+∑x<xi<∞‍βium(xi-0),m>0,(8)u(x,y)≤a(x,y)+∑k=12‍∫x∞‍∫y∞‍fk(x,y,s,t)×ωk(u(σk(s),τk(t)))dsdt+∑x<xi<∞,y<yi<∞‍βium(xi-0,yi-0),m>0
with two nonlinear terms ω1(u) and ω2(u) where we do not restrict ω1 and ω2 to the class ℘. Moreover, our main results are applied to estimate the bounds of solutions of differential equations with impulsive terms.


## 2. Main Results

In what follows,R denotes the set of real numbers, R+=[0,∞), and D1z(x,y) denotes the first-order partial derivative of z(x,y) with respect to x.Consider (7) and assume that(H1)
fk(x,s)(k=1,2) is a continuous and nonnegative function for x,s∈R+ and is bounded in x∈R+ for each fixed s∈R+;(H2)
ω1(u) and ω2(u) are continuous and nonnegative functions on [0,∞) and positive on (0,∞) such that ω2(u)/ω1(u) is nondecreasing;(H3)
u(x) is a nonnegative and continuous function defined on R+ with the first kind of discontinuities at the points xi where i=1,2,…,n and 0<x0<x1<x2<⋯<xn<xn+1=∞;(H4)
a(x) is a continuous and bounded function for x∈R+ and a(∞)≠0; βi is a nonnegative constant for any positive integer i;(H5)
σ1(x) and σ2(x) are continuous and nonnegative functions on R+ such that σk(x)≥x and σk(x)≤xi for x∈[xi-1,xi), i=1,2,…,n+1, and k=1,2.LetWj(u)=∫u~ju‍(dz/ωj(z)) for u≥u~j and j=1,2 where u~j is a given positive constant. Clearly, Wj is strictly increasing so its inverse Wj-1 is well defined, continuous, and increasing in its corresponding domain.Theorem 1.
Suppose that(H1)–(H5) hold and u(x) satisfies (7) for a positive constant m. If one lets ui-1(x)=u(x) for x∈[xi-1,xi),  i=1,2,…,n+1, then the estimate of u(x) is recursively given by
(9)ui-1(x)≤W2-1[W2∘W1-1(W1(ri-1(x))+∫xxif~1‍(x,s)ds)+∫xxi‍f~2(x,s)ds(W1(ri-1(x))+∫xxi‍f1~(x,s)ds)]
for x∈[xi-1,xi) and i=1,2,…,n+1, where
(10)rn(x)=sup⁡x≤τ<∞|a(τ)|,f~k(x,s)=sup⁡x≤τ<∞fk(τ,s),k=1,2,ri-1(x)=rn(x)+∑j=in‍∑k=12‍∫xjxj+1‍fk(x,s)ωk(uj(σ(s)))ds+∑j=in‍βjujm(xj-0),i=1,2,…,n,
provided that
(11)W1(ri-1(x))+∫xxi‍f~1(x,s)ds≤∫u~1∞‍dzω1(z),W2∘W1-1(W1(ri-1(x))+∫xxi‍f~1(x,s)ds)+∫xxi‍f~2(x,s)ds≤∫u~2∞‍dzω2(z).Proof.
From the assumptions, we know thatrn(x) and f~k(x,s)(k=1,2) are well defined. Moreover, rn(x) is nonnegative and nonincreasing in x and f~k(x,s) is nonnegative and nonincreasing in x and satisfies a(x)≤rn(x), fk(x,s)≤f~k(x,s).
Case  1. If x∈[xn,∞) (in fact, xn+1=∞), from the definition of σk, we have σk(x)∈[xn,∞)(k=1,2). According to (7) and (10) we get
(12)u(x)≤rn(x)+∑k=12‍∫x∞‍f~k(x,s)ωk(u(σk(s)))ds.
Take any fixed T∈[xn,∞), and we investigate the following inequality:
(13)u(x)≤rn(T)+∑k=12‍∫x∞‍f~k(T,s)ωk(u(σk(s)))ds
for x∈[T,∞). Let
(14)z(x)=∑k=12‍∫x∞‍f~k(T,s)ωk(u(σk(s)))ds
and let z(∞)=0. Hence, u(x)≤rn(T)+z(x). Clearly, z(x) is a nonnegative, nonincreasing, and differentiable function for x∈[T,∞). The assumption a(∞)≠0 yields that rn(T)+z(x)>0. Thus
(15)z′(x)ω1(rn(T)+z(x))=-f~1(T,x)ω1(u(σ1(x)))-f~2(T,x)ω2(u(σ2(x)))ω1(rn(T)+z(x))≥(-f~1(T,x)ω1(rn(T)+z(σ1(x)))-f~2(T,x)ω2(rn(T)+z(σ2(x))))×(ω1(rn(T)+z(x)))-1≥-f~1(T,x)ω1(rn(T)+z(x))ω1(rn(T)+z(x))-f~2(T,x)ω2(rn(T)+z(x))ω1(rn(T)+z(x))≥-f~1(T,x)-f~2(T,x)ω2(rn(T)+z(x))ω1(rn(T)+z(x)).
Integrating both sides of the above inequality from x to ∞, we obtain
(16)W1(rn(T))-W1(rn(T)+z(x))≥-∫x∞‍f~1(T,s)ds-∫x∞‍f~2(T,s)ϕ(rn(T)+z(s))ds
for x∈[T,∞), where ϕ(x)=ω2(x)/ω1(x), so
(17)W1(rn(T)+z(x))≤W1(rn(T))+∫x∞‍f~1(T,s)ds+∫x∞‍f~2(T,s)ϕ(rn(T)+z(s))ds
or, equivalently,
(18)ξ(x)≤W1(rn(T))+∫x∞‍f~1(T,s)ds+∫x∞‍f~2(T,s)ϕ(W1-1(ξ(s)))ds≜z1(x),
where
(19)ξ(x)=W1(rn(T)+z(x)).
It is easy to check that ξ(x)≤z1(x), z1(∞)=W1(rn(T)) and z1(x) is differentiable, positive, and nonincreasing on [T,∞). Since ϕ(W1-1(u)) is nondecreasing, from the assumption (H2), we have
(20)z1′(x)ϕ(W1-1(z1(x)))=-f~1(T,x)ϕ(W1-1(z1(x)))-f~2(T,x)ϕ(W1-1(ξ(x)))ϕ(W1-1(z1(x)))≥-f~1(T,x)ϕ[W1-1(W1(rn(T))+∫x∞‍f~1(T,s)ds)]-f~2(T,x).
Note that
(21)∫x∞‍z1′(s)ϕ(W1-1(z1(s)))ds=∫x∞‍ω1(W1-1(z1(s)))z1′(s)ω2(W1-1(z1(s)))ds=W2∘W1-1(z1(∞))-W2∘W1-1(z1(x))=W2(rn(T))-W2∘W1-1(z1(x)).
Integrating both sides of (20) from x to ∞, we obtain
(22)W2(rn(T))-W2∘W1-1(z1(x))=∫x∞‍z1′(s)ϕ(W1-1(z1(s)))ds≥-∫x∞‍f~1(T,s)ϕ[W1-1(W1(rn(T))+∫s∞‍f~1(T,τ)dτ)]ds-∫x∞‍f~2(T,s)ds≥-W2∘W1-1(W1(rn(T))+∫x∞‍f~1(T,s)ds)+W2(rn(T))-∫x∞‍f~2(T,s)ds.
Thus,
(23)W2∘W1-1(z1(x))≤W2∘W1-1(W1(rn(T))+∫x∞‍f~1(T,s)ds)+∫x∞‍f~2(T,s)ds.
We have by (11)
(24)u(x)≤z(x)+rn(T)≤W1-1(ξ(x))≤W1-1(z1(x))≤W2-1[W2∘W1-1(W1(rn(T))+∫x∞‍f~1(T,s)ds)+∫x∞‍f~2(T,s)ds].
Since the inequality above is true for any x∈[T,∞), we obtain
(25)u(T)≤W2-1[W2∘W1-1(W1(rn(T))+∫T∞‍f~1(T,s)ds)+∫T∞‍f~2(T,s)ds].
Replacing T by x and ∞ by xn+1 yields
(26)u(x)≤W2-1[W2∘W1-1(W1(rn(x))+∫xxn+1‍f~1(x,s)ds)+∫xxn+1‍f~2(x,s)ds].
This means that (9) is true for x∈[xn,∞) if we replace u(x) with un(x).
Case  2. If x∈[xn-1,xn), (7) becomes
(27)u(x)≤rn(x)+∑k=12‍∫xnxn+1‍fk(x,s)ωk(un(σk(s)))ds+βnunm(xn-0)+∑k=12‍∫xxn‍fk(x,s)ωk(u(σk(s)))ds≤rn-1(x)+∑k=12‍∫xxn‍fk(x,s)ωk(u(σk(s)))ds,
where the definition of rn-1(x) is given in (10), which is similar to (12). Then, we obtain
(28)u(x)≤W2-1[W2∘W1-1(W1(rn-1(x))+∫xxn‍f~1(x,s)ds)+∫xxn‍f~2(x,s)ds].
This implies that (9) is true for x∈[xn-1,xn) if we replace u(x) by un-1(x).
Case  3. If (7) is true for x∈[xi,xi+1), that is,
(29)ui(x)≤W2-1[W2∘W1-1(W1(ri(x))+∫xxi+1‍f~1(x,s)ds)+∫xxi+1‍f~2(x,s)ds],
then, for x∈[xi-1,xi), (7) becomes
(30)u(x)≤rn(x)+∑j=in‍∑k=12‍∫xjxj+1‍fk(x,s)ωk(uj(σk(s)))ds+∑j=in‍βjujm(xj-0)+∑k=12‍∫xxi‍fk(x,s)ωk(u(σk(s)))ds≤ri-1(x)+∑k=12‍∫xxi‍fk(x,s)ωk(u(σk(s)))ds,
where we use the fact that the estimate of u(x) is already known for x∈[xj,xj+1),j=i,i+1,…,n. By assumption (29), again (30) is the same as (27) if we replace rn-1(x) by ri-1(x) and xn by xi. Thus, by (28), we have
(31)u(x)≤W2-1[W2∘W1-1(W1(ri-1(x))+∫xxi‍f~1(x,s)ds)+∫xxi‍f~2(x,s)ds].
This completes the proof of Theorem 1 by mathematical induction.Remark 2.
Zheng [25] investigated (5) which is the special case of (7). His results are under the assumptions that a(x)=c, f1(x,s), f2(x,s) are decreasing in s for every fixed s and ω∈℘. In our result, these assumptions are avoided.Consider the inequality(32)φ(u(x))≤a(x)+∑k=12‍∫x∞‍fk(x,s)ωk(u(σk(s)))ds+∑x<xi<∞‍βiψ(u(xi-0)),
which looks much more complicated than (7).Corollary 3.
In addition to the assumptions(H1)–(H5), suppose that ψ(u) is positive on (0,∞), φ(u) is positive and strictly increasing on (0,∞), and u(x) satisfies (32). If one lets ui-1(x)=u(x) for x∈[xi-1,xi), i=1,2,…,n+1, then the estimate of u(x) is recursively given by
(33)ui-1(x)≤φ-1{W2-1[∫xxi‍f1~(x,s)dsW2∘W1-1×(W1(ri-1(x))+∫xxi‍f~1(x,s)ds)+∫xxi‍f~2(x,s)ds]},
where Wj(u)=∫u~ju‍(dz/ωj(φ-1(z))), rn(x), and f~k(x,s) are given in Theorem 1 and ri-1(x) is defined as follows:
(34)ri-1(x)=rn(x)+∑j=in‍∑k=12‍∫xjxj+1‍fk(x,s)ωk(uj(σk(s)))ds+∑j=in‍βjψ(uj(xj-0)),i=1,2,…,n,
provided that
(35)W1(ri-1(x))+∫xxi‍f~1(x,s)ds≤∫u~1∞‍dzω1(z),W2∘W1-1(W1(ri-1(x))+∫xxi‍f~1(x,s)ds)+∫xxi‍f~2(x,s)ds≤∫u~2∞‍dzω2(z).Proof.
Letφ(u(x))=h(x). Since the function φ is strictly increasing on [0,∞), its inverse φ-1 is well defined. And (32) becomes
(36)h(x)≤a(x)+∑k=12‍∫x∞‍fk(x,s)ωk(φ-1(h(σk(s))))ds+∑x<xi<∞‍βiψ(φ-1(h(xi-0))).
Let ω~k=ωk∘φ-1 and ψ~=ψ∘φ-1; (36) becomes
(37)h(x)≤a(x)+∑k=12‍∫x∞‍fk(x,s)ω~k(h(σk(s)))ds+∑x<xi<∞‍βiψ~(h(xi-0)).
It is easy to see that ψ~(u)>0, ω~1(u) and ω~2(u) are continuous and nonnegative functions on [0,∞), and ω~2(u)/ω~1(u) is nondecreasing on (0,∞). Even though ψ~(u) is much more general, using the same way in Theorem 1, for x∈[xi-1,xi),i=1,2,…,n+1, we can obtain the estimate of u(x):
(38)ui-1(x)≤φ-1{W2-1[W2∘W1-1(W1(ri-1(x))+∫xxi‍f~1(x,s)ds)+∫xxi‍f~2(x,s)ds(W1(ri-1(x))+∫xxi‍f1~(x,s)ds)]}.
This completes the proof of Corollary 3.Ifφ(u)=uλ where λ>0 is a constant, we can study the inequality
(39)uλ(x)≤a(x)+∑k=12‍∫x∞‍fk(x,s)ωk(u(σk(s)))ds+∑x<xi<∞‍βiψ(u(xi-0)).
According to Corollary 3, we have the following result.Corollary 4.
In addition to the assumptions(H1)–(H5), suppose that ψ(u) is positive on (0,∞) and u(x) satisfies (39). If one lets ui-1(x)=u(x) for x∈[xi-1,xi),i=1,2,…,n+1, then the estimate of u(x) is recursively given by
(40)ui-1(x)≤{W2-1[W2∘W1-1(W1(ri-1(x))+∫xxi‍f~1(x,s)ds)+∫xxi‍f~2(x,s)ds(W1(ri-1(x))+∫xxi‍f1~(x,s)ds)]}1/λ,
where Wj(u)=∫u~ju‍(dz/ω(z1/λ)), rn(x), ri-1(x), and f~k(x,s) are given in Corollary 3.Let(41)Ω=∪i,j≥1Ωij,Ωij={(x,y):xi-1≤x<xi,yj-1≤y<yj},
for ,j=1,2,…,n+1, 0<x0<x1<x2<⋯<xn<xn+1=∞,  and  0<y0<y1<y2<⋯<yn<yn+1=∞.Consider (8) and assume that(C1)
fk(x,y,s,t)(k=1,2) is continuous and nonnegative on Ω×Ω and bounded in (x,y)∈Ω for each fixed (s,t)∈Ω and satisfies fk(x,y,s,t)=0(k=1,2) if (s,t)∈Ωij, i≠j for arbitrary i,j=1,2,…,n+1;(C2)
ω1(u) and ω2(u) are continuous and nonnegative functions on [0,∞) and are positive on (0,∞) such that ω2(u)/ω1(u) is nondecreasing;(C3)
u(x,y) is nonnegative and continuous on Ω with the exception of the points (xi,yi) where there is a finite jump: u(xi-0,yi-0)≠u(xi+0,yi+0),i=1,2,…,n;(C4)
a(x,y) is continuous and bounded for (x,y)∈Ω and a(∞,∞)≠0; βi is a nonnegative constant for any positive integer i;(C5)
σk(x) and τk(y)(k=1,2) are continuous and nonnegative such that σk(x)≥x and σk(x)≤xi for x∈[xi-1,xi),i=1,2,…,n+1, and τk(y)≥y and τk(y)≤yi for y∈[yi-1,yi),i=1,2,…,n+1.Theorem 5.
Suppose that(C1)–(C5) hold and u(x,y) satisfies (8) for a positive constant m. If one lets ui(x,y)=u(x,y) for (x,y)∈Ωii,i=1,2,…,n, then the estimate of u(x,y) is recursively given by
(42)ui-1(x,y)≤W2-1[(W1(ri-1(x,y))+∫xxi‍∫yyi‍f1~(x,y,s,t)dsdt)W2∘W1-1×(W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt)+∫xxi‍∫yyi‍f~2(x,y,s,t)dsdt(W1(ri-1(x,y))+∫xxi‍∫yyi‍f1~(x,y,s,t)dsdt)],
for (x,y)∈Ωii,i=1,2,…,n+1, where
(43)rn(x,y)=sup⁡⁡x≤ξ<∞sup⁡⁡y≤η<∞|a(ξ,η)|,f~k(x,y,s,t)=sup⁡⁡x≤ξ<∞sup⁡⁡y≤η<∞fk(ξ,η,s,t),ri-1(x,y)=rn(x,y)+∑j=in‍∑k=12‍∫xjxj+1‍∫yjyj+1‍fk(x,y,s,t)ωk(uj(s,t))dsdt+∑j=in‍βjujm(xj-0,yj-0),i=1,2,…,n,
provided that
(44)W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt≤∫u~1∞‍dzω1(z),W2∘W1-1[W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt]+∫xxi‍∫yyi‍f~2(x,y,s,t)dsdt≤∫u~2∞‍dzω2(z).Proof.
Obviously, for any(x,y)∈Ω,  rn(x,y) is positive and nonincreasing with respect to x and y; f~k(x,y,s,t)(k=1,2) is nonnegative and nonincreasing with respect to x and y for each fixed s and t. They satisfy a(x,y)≤rn(x,y) and fk(x,y,s,t)≤f~k(x,y,s,t).
Case  1. If (x,y)∈Ωn+1,n+1={(x,y):xn≤x<xn+1,yn≤y<yn+1}, we have from (8)
(45)u(x,y)≤rn(x,y)+∑k=12‍∫x∞‍∫y∞‍f~k(x,y,s,t)ωk×(u(σk(s),τk(t)))dsdt.
Take any fixed x~∈[xn,∞), y~∈[yn,∞), and for arbitrary x∈[x~,∞), y∈[y~,∞), we get
(46)u(x,y)≤rn(x~,y~)+∑k=12‍∫x∞‍∫y∞‍f~k(x~,y~,s,t)ωk×(u(σk(s),τk(t)))dsdt.
Let
(47)z(x,y)=rn(x~,y~)+∑k=12‍∫x∞‍∫y∞‍f~k(x~,y~,s,t)ωk×(u(σk(s),τk(t)))dsdt
and let z(∞,y)=rn(x~,y~). Hence, u(x,y)≤z(x,y). Clearly, z(x,y) is a nonnegative, nonincreasing, and differentiable function for x∈[x~,∞) and y∈[y~,∞). Since a(∞,∞)≠0 and ω1(z(x,y))>0, we have
(48)D1z(x,y)ω1(z(x,y))=-∫y∞‍f~1(x~,y~,x,t)ω1(u(σ1(x),τ1(t)))dtω1(z(x,y))-∫y∞‍f~2(x~,y~,x,t)ω2(u(σ2(x),τ2(t)))dtω1(z(x,y))≥-∫y∞‍f~1(x~,y~,x,t)ω1(z(σ1(x),τ1(t)))dtω1(z(x,y))-∫y∞‍f~2(x~,y~,x,t)ω2(z(σ2(x),τ2(t)))dtω1(z(x,y))≥-∫y∞‍f~1(x~,y~,x,t)ω1(z(x,t))dtω1(z(x,y))-∫y∞‍f~2(x~,y~,x,t)ω2(z(x,t))dtω1(z(x,y))≥-∫y∞‍f~1(x~,y~,x,t)dt-∫y∞‍f~2(x~,y~,x,t)ω2(z(x,t))ω1(z(x,t))dt.
Integrating both sides of the above inequality from x to ∞, we obtain
(49)W1(z(∞,y))-W1(z(x,y))≥-∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt-∫x∞‍∫y∞‍f~2(x~,y~,s,t)ω2(z(s,t))ω1(z(s,t))dsdt.
Thus,
(50)W1(z(x,y))≤W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt+∫x∞‍∫y∞‍f~2(x~,y~,s,t)ϕ(z(s,t))dsdt
for x~≤x<∞ and y~≤y<∞, where ϕ(u)=ω2(u)/ω1(u), or equivalently
(51)ξ(x,y)≤W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt+∫x∞‍∫y∞‍f~2(x~,y~,s,t)ϕ(W1-1(ξ(s,t)))dsdt≜z1(x,y),
where
(52)ξ(x,y)=W1(z(x,y)).
It is easy to check that ξ(x,y)≤z1(x,y), z1(∞,y)=W1(rn(x~,y~)), and z1(x,y) is differentiable, positive, and nonincreasing on [x~,∞) and [y~,∞). Since ϕ(W1-1(u)) is nondecreasing, from assumption (C2), we have
(53)D1z1(x,y)ϕ(W1-1(z1(x,y)))=-∫y∞‍f~1(x~,y~,x,t)dtϕ(W1-1(z1(x,y)))-∫y∞‍f~2(x~,y~,x,t)ϕ(W1-1(ξ(x,t)))dtϕ(W1-1(z1(x,y)))≥-∫y∞‍f~1(x~,y~,x,t)dtϕ[W1-1(W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt)]-∫y∞‍f~2(x~,y~,x,t)ϕ(W1-1(z1(x,t)))dtϕ(W1-1(z1(x,y)))≥-∫y∞‍f~1(x~,y~,x,t)dtϕ[W1-1(W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt)]-∫y∞‍f~2(x~,y~,x,t)dt.
Note that
(54)∫x∞‍D1z1(s,y)ϕ(W1-1(z1(s,y)))ds=∫x∞‍D1z1(s,y)ω1(W1-1(z1(s,y)))ω2(W1-1(z1(s,y)))ds=∫W1-1(z1(x,y))W1-1(z1(∞,y))‍duω2(u)=W2∘W1-1(z1(∞,y))-W2∘W1-1(z1(x,y))=W2∘W1-1(W1(rn(x~,y~)))-W2∘W1-1(z1(x,y))=W2(rn(x~,y~))-W2∘W1-1(z1(x,y)).
Integrating both sides of (53) from x to ∞, we obtain
(55)W2(rn(x~,y~))-W2∘W1-1(z1(x,y))=∫x∞‍D1z1(s,y)ϕ(W1-1(z1(s,y)))ds≥-∫x∞‍∫y∞‍f~1(x~,y~,s,t)dtϕ[W1-1(W1(rn(x~,y~))+∫s∞‍∫y∞‍f~1(x~,y~,τ,t)dτdt)]ds-∫x∞‍∫y∞‍f~2(x~,y~,s,t)dsdt≥-W2∘W1-1[W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt]+W2(rn(x~,y~))-∫x∞‍∫y∞‍f~2(x~,y~,s,t)dsdt.
Thus,
(56)W2∘W1-1(z1(x,y))≤W2∘W1-1[W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt]+∫x∞‍∫y∞‍f~2(x~,y~,s,t)dsdt.
Hence,
(57)u(x,y)≤z(x,y)≤W1-1(ξ(x,y))≤W1-1(z1(x,y))≤W2-1[W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)W2∘W1-1×(W1(rn(x~,y~))+∫x∞‍∫y∞‍f~1(x~,y~,s,t)dsdt)+∫x∞‍∫y∞‍f~2(x~,y~,s,t)dsdt].
Since the above inequality is true for any x∈[x~,∞), y∈[y~,∞), we obtain
(58)u(x~,y~)≤W2-1[∫x~∞‍∫y~∞‍f~1(x~,y~,s,t)dsdtW2∘W1-1×(W1(rn(x~,y~))+∫x~∞‍∫y~∞‍f~1(x~,y~,s,t)dsdt)+∫x~∞‍∫y~∞‍f~2(x~,y~,s,t)dsdt].
Replacing x~, y~, and ∞ by x, y, and xn+1, respectively, yields
(59)u(x,y)≤W2-1[W1(rn(x,y))+∫xxn+1‍∫yyn+1‍W2∘W1-1×(W1(rn(x,y))+∫xxn+1‍∫yyn+1‍f~1(x,y,s,t)dsdt)+∫xxn+1‍∫yyn+1‍f~2(x,y,s,t)dsdt].
This means that (42) is true for (x,y)∈Ωn+1,n+1 and i=n if we replace u(x,y) with un(x,y).
Case  2. If (x,y)∈Ωn,n={(x,y):xn-1≤x<xn,yn-1≤y<yn}, (8) becomes
(60)u(x,y)≤rn(x,y)+∑k=12‍∫xnxn+1‍∫ynyn+1‍fk(x,y,s,t)ωk×(un(σk(s),τk(t)))dsdt+βnunm(xn-0,yn-0)+∑k=12‍∫xxn‍∫yyn‍fk(x,y,s,t)ωk×(u(σk(s),τk(t)))dsdt≤rn-1(x,y)+∑k=12‍∫xxn‍∫yyn‍fk(x,y,s,t)ωk×(u(σk(s),τk(t)))dsdt,
where the definition of rn-1(x,y) is given in (43). Note that the estimate of un(x,y) is known. Clearly, (60) is the same as (45) if we replace rn(x,y) and (∞,∞) by rn-1(x,y) and (xn,yn). Thus, by (59), we have
(61)u(x,y)≤W2-1[W1(rn-1(x,y))+∫xxn‍∫yyn‍W2∘W1-1×(W1(rn-1(x,y))+∫xxn‍∫yyn‍f~1(x,y,s,t)dsdt)+∫xxn‍∫yyn‍f~2(x,y,s,t)dsdt].
This implies that (42) is true for (x,y)∈Ωn,n and i=n-1 if we replace u(x,y) by un-1(x,y).
Case  3. Assume that (42) is true for (x,y)∈Ωi+1,i+1={(x,y):xi≤x<xi+1,yi≤y<yi+1}. Then for (x,y)∈Ωi,i={(x,y):xi-1≤x<xi,yi-1≤y<yi}, (8) becomes
(62)u(x,y)≤rn(x,y)+∑j=in‍∑k=12‍∫xixi+1‍∫yiyi+1‍fk(x,y,s,t)ωk×(uj(σk(s),τk(t)))dsdt+∑j=in‍βjujm(xj-0,yj-0)+∑k=12‍∫xxi‍∫yyi‍fk(x,y,s,t)ωk×(u(σk(s),τk(t)))dsdt≤ri-1(x,y)+∑k=12‍∫xxi‍∫yyi‍fk(x,y,s,t)ωk×(u(σk(s),τk(t)))dsdt,
where we use the fact that the estimate of u(x,y) is already known for (x,y)∈Ωjj,  j=i,…,n. Again, (62) is the same as (60) if we replace rn-1(x,y) and (xn,yn) by ri-1(x,y) and (xi,yi). Thus, by (61), we have
(63)u(x,y)≤W2-1[W1(ri-1(x,y))+∫xxi‍W2∘W1-1×(W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt)+∫xxi‍∫yyi‍f~2(x,y,s,t)dsdt].
This yields that (42) is true for (x,y)∈Ωi,i if we replace u(x,y) by ui-1(x,y). By mathematical induction, we know that (42) holds for (x,y)∈Ωi,i for any nonnegative integer i. This completes the proof of Theorem 5.Remark 6.
(1) Ifa(x,y) is nonincreasing in each variable x,y∈R+ and we take f1(x,y,s,t)=b(x,y)c(s,t), f2(x,y,s,t)=0, σk(x)=x, τk(y)=y, and u(x,y) being continuous on R+2, then (8) reduces to (2) and Theorem 1 becomes Theorem  2.2 in [16].
(2) Zheng [25] investigated (6) which is the special case of (8). His results are under the assumptions that a(x,y)=c, fk(x,y,s,t)=fk(s,t), and ω∈℘. In our results, these assumptions are avoided.
Consider the inequality(64)φ(u(x,y))≤a(x,y)+∑k=12‍∫x∞‍∫y∞‍fk(x,y,s,t)ωk×(u(σk(s),τk(t)))dsdt+∑x<xi<∞,y<yi<∞‍βiψ(u(xi-0,yi-0)),
which looks much more complicated than (8).Corollary 7.
In addition to the assumptions(C1)–(C5), suppose that ψ(u) is positive on (0,∞), φ(u) is positive and strictly increasing on (0,∞), and u(x,y) satisfies (64) for a positive constant m. If one lets ui(x,y)=u(x,y) for (x,y)∈Ωii, then the estimate of u(x,y) is recursively given by
(65)ui-1(x,y)≤φ-1{W2-1[∫yyi‍W2∘W1-1×(+∫xxi‍∫yyi‍f1~(x,y,s,t)dsdt)W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt)+∫xxi‍∫yyi‍f~2(x,y,s,t)dsdt]},
where Wj(u)=∫u~ju‍(dz/ωj(φ-1(z))),  rn(x,y), and f~k(x,y,s,t) are given in Theorem 5; ri-1(x,y) is defined as follows:
(66)ri-1(x,y)=rn(x,y)+∑j=in‍∑k=12‍∫xjxj+1‍∫yjyj+1‍fk(x,y,s,t)ωk×(uj(s,t))dsdt+∑j=in‍βjψ(uj(xj-0,yj-0)),i=1,2,…,n,
provided that
(67)W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt≤∫u~1∞‍dzω1(z),W2∘W1-1[W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt]+∫xxi‍∫yyi‍f~2(x,y,s,t)dsdt≤∫u~2∞‍dzω2(z).The proof is similar to Corollary3.Ifφ(u)=uλ, where λ>0 is a constant, we can study the inequality
(68)uλ(x,y)≤a(x,y)+∑k=12‍∫x∞‍∫y∞‍fk(x,y,s,t)×ωk(u(σk(s),τk(t)))dsdt+∑x<xi<∞,y<yi<∞‍βiψ(u(xi-0,yi-0)).
According to Corollary 7, we have the following result.Corollary 8.
In addition to the assumptions(C1)–(C5), suppose that ψ(u) is positive on (0,∞) and u(x,y) satisfies (68) for a positive constant m. If one lets ui(x,y)=u(x,y) for (x,y)∈Ωii, then the estimate of u(x,y) is recursively given by
(69)ui-1(x,y)≤{W2-1[∫xxi‍∫yyi‍W2∘W1-1×(∫xxi‍∫yyi‍W1(ri-1(x,y))+∫xxi‍∫yyi‍f~1(x,y,s,t)dsdt)+∫xxi‍∫yyi‍f~2(x,y,s,t)dsdt]}1/λ,
where Wj(u)=∫u~ju‍(dz/ω(z1/λ)),  rn(x,y),  ri-1(x,y), and f~k(x,y,s,t) are given in Corollary 7.


## 3. Applications

Example 9.
Consider the following impulsive differential equation:(70)dgdx=F(x,g),x≠xi,(71)Δg|x=xi=Ii(x),g(∞)=θ≠0,
where g:R→R, F:R2→R, Ii:R→R and i=1,2,…,n, 0<x0<x1<x2<⋯<xn<xn+1=∞. Here, θ is a constant.
Assume that(A1)
|F(x,g)|≤h1(x)e|g|+h2(x)e2|g| where h1 and h2 are nonnegative, bounded, and continuous on R+;(A2)
|Ii(g)|≤βi|g|m where βi and m are nonnegative constants.Theorem 10.
Suppose that(A1) and (A2) hold. If one lets gi-1(x)=g(x) for x∈[xi-1,xi), i=1,2,…,n+1, then the solution of (70) has an estimate for x∈[xi-1,xi):
(72)|gi-1(x)|≤-12ln⁡[(e-ri-1(x)-∫xxi‍h1(s)ds)2-2∫xxi‍h2(s)ds],
where rn(x)=|θ| and
(73)ri-1(x)=rn(x)+∑j=in‍∫xjxj+1‍h1(s)e|gj(s)|ds+∑j=in‍∫xjxj+1‍h2(s)e2|gj(s)|ds+∑j=in‍βj|gj(xj-0)|m,i=1,2,…,n,(e-ri-1(x)-2∫xxi‍h1(s)ds)2-2∫xxi‍h2(s)ds>0.Proof.
Integrating (70) from x to ∞ and using the initial conditions (71), we get
(74)g(x)=θ-∫x∞‍F(s,g)ds-∑x<xi<∞‍Ii(g(xi-0)),
which implies that
(75)|g(x)|≤|θ|+∫x∞‍h1(s)e|g(s)|ds+∫x∞‍h2(s)e2|g(s)|ds+∑x<xi<∞‍βi|g(xi-0)|m.
Let
(76)u(x)=|g(x)|,a(x)=|θ|,σ1(x)=σ2(x)=x,f1(x,s)=h1(s),f2(x,s)=h2(s),ω1(u)=eu,ω2(u)=e2u.
Thus, (75) is the same as (7). It is easy to obtain that for any positive constants u~1 and u~2(77)rn(x)=|θ|,f~1(x,s)=h1(s),f~2(x,s)=h2(s),W1(u)=∫u~1u‍dzω1(z)=∫u~1u‍e-zdz=e-u~1-e-u,W1-1(u)=-ln⁡(e-u~1-u),W2(u)=∫u~2u‍dzω2(z)=∫u~2u‍e-2zdz=12(e-2u~2-e-2u),W2-1(u)=-12ln⁡(e-2u~2-2u),ri-1(x)=rn(x)+∑j=in‍∫xjxj+1‍h1(s)e|gj(s)|ds+∑j=in‍∫xjxj+1‍h2(s)e2|gj(s)|ds+∑j=in‍βj|gj(xj-0)|m.
Therefore, for any nonnegative i and x∈[xi-1,xi)(78)|gi-1(x)|≤-12ln⁡[(e-ri-1(x)-∫xxi‍h1(s)ds)2-2∫xxi‍h2(s)ds],
provided that
(79)(e-ri-1(x)-2∫xxi‍h1(s)ds)2-2∫xxi‍h2(s)ds>0.Example 11.
Consider the following partial differential equation with an impulsive term:(80)∂2v(x,y)∂x∂y=H(x,y,v(x,y)),(x,y)∈Ωii,x≠xi,y≠yi,Δv|x=xi,y=yi=Ii(v),v(x,∞)=ϕ1(x),v(∞,y)=ϕ2(y),ϕ1(∞)=ϕ2(∞)≠0,
where v:R2→R, H:R3→R, Ii:R→R, and i=1,2,…,n+1.
Assume that(B1)
|H(x,y,v(x,y))|≤h1(x,y)e|v(x,y)|+h2(x,y)e2|v(x,y)| where h1,h2 are nonnegative, bounded, and continuous on Ω, h1(x,y)=0, h2(x,y)=0 for (x,y)∈Ωij,  i≠j,  i,j=1,2,…,n+1;(B2)
|Ii(v)|≤βi|v|m where βi and m are nonnegative constants.Theorem 12.
Suppose that(B1) and (B2) hold. If one lets vi(x,y)=v(x,y) for (x,y)∈Ωii, then the solution of system (80) has an estimate for (x,y)∈Ωii:
(81)|vi(x,y)|≤-12ln⁡[(e-ri-1(x,y)-∫xxi‍∫yyi‍h1(s,t)dsdt)2-2∫xxi‍∫yyi‍h2(s,t)dsdt(e-ri-1(x,y)-∫xxi‍∫yyi‍h1(s,t)dsdt)2],
where
(82)rn(x,y)=sup⁡x≤ξ<∞sup⁡y≤η<∞|ϕ1(ξ)+ϕ2(η)-ϕ1(∞)|>0,ri-1(x,y)=rn(x,y)+∑j=in‍∫xjxj+1‍∫yjyj+1‍h1(s,t)e|vj(s,t)|dsdt+∑j=in‍∫xjxj+1‍∫yjyj+1‍h2(s,t)e2|vj(s,t)|dsdt+∑j=in‍βj|vj(xj-0,yj-0)|m,(e-ri-1(x,y)-∫xxi‍∫yyi‍h1(s,t)dsdt)2-2∫xxi‍∫yyi‍h2(s,t)dsdt>0.Proof.
The solution of (80) with an initial value is given by
(83)v(x,y)=v(x,∞)+v(∞,y)-v(∞,∞)+∫x∞‍∫y∞‍H(s,t,v(s,t))dsdt+∑x<xi<∞,y<yi<∞‍Ii(v(xi-0,yi-0))=ϕ1(x)+ϕ2(y)-ϕ1(∞)+∫x∞‍∫y∞‍H(s,t,v(s,t))dsdt+∑x<xi<∞,y<yi<∞‍Ii(v(xi-0,yi-0)),
which implies that
(84)|v(x,y)|≤|ϕ1(x)+ϕ2(y)-ϕ1(∞)|+∫x∞‍∫y∞‍h1(s,t)e|v(s,t)|dsdt+∫x∞‍∫y∞‍h2(s,t)e2|v(s,t)|dsdt+∑x<xi<∞,y<yi<∞‍βi|v(xi-0,yi-0)|m.
Similar to Theorem 10, we can obtain, for (x,y)∈Ωii,
(85)|vi(x,y)|≤-12ln⁡[(e-ri-1(x,y)-∫xxi‍∫yyi‍h1(s,t)dsdt)2-2∫xxi‍∫yyi‍h2(s,t)dsdt(e-ri-1(x,y)-∫xxi‍∫yyi‍h1(s,t)dsdt)2].Remark 13.
From Examples9 and 11, we know that ω1(u)=eu. Clearly, ω1(2u)=e2u≤ω1(2)ω1(u)=e2eu does not hold for large u>0. Thus, ω1(u)=eu does not belong to class ℘ in [25]. Hence, the results in [25] can not be applied to inequality (75).



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*Source: 353210-2014-06-12.xml*