# A Generalized Nonlinear Gronwall-Bellman Inequality with Maxima in Two Variables

Authors: Yong Yan
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

This paper deals with a generalized form of nonlinear retarded Gronwall-Bellman type integral inequality in which the maximum of the unknown function of two variables is involved. This form includes both a nonconstant
term outside the integrals and more than one distinct nonlinear integrals. Requiring neither monotonicity nor separability of given functions, we apply a technique of monotonization to estimate the unknown function. Our result can be used to weaken conditions for some known results. We apply our result to a boundary value problem of a partial differential equation with maxima for uniqueness.


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## Body

## 1. Introduction

The Gronwall-Bellman inequality [1, 2] plays an important role in the study of existence, uniqueness, boundedness, stability, invariant manifolds, and other qualitative properties of solutions of differential equations and integral equations. There can be found a lot of its generalizations in various cases from literatures (see, e.g., [3–18]). In 1956, Bihari [3] discussed the integral inequality
(1)u(t)≤c+∫0tf(s)ω(u(s))ds,t≥0,
where c>0 is a constant, f is a continuous and nonnegative function, and ω is a continuous and nondecreasing positive function. Replacing t by a function b(t) in (1), Lipovan [4] investigated the retarded integral inequality
(2)u(t)≤c+∫t0t‍f(s)ω(u(s))ds+∫b(t0)b(t)‍g(s)ω(u(s))ds,t0≤t<t1.
Their results were further generalized by Agarwal et al. [5] to the inequality
(3)u(t)≤a(t)+∑i=1n‍∫bi(t0)bi(t)‍fi(t,s)ωi(u(s))ds,t0≤t<t1,
where the constant c is replaced with a function a(t), bi’s are continuously differentiable and nondecreasing functions, and ωi’s are continuous and nondecreasing positive functions such that
(4)ω1∝ω2∝⋯∝ωn,
that is, each ratio ωi+1/ωi is also nondecreasing on A⊆ℝ∖{0}, called in [6] that ωi+1 is stronger nondecreasing than ωi. On the basis of this work, Wang [7] considered the inequality of two variables
(5)up(x,y)≤a(x,y)+∑i=1n‍∫bi(x0)bi(x)‍∫ci(y0)ci(y)‍fi(x,y,s,t)ωi(u(s,t))dtds,
where the functions a, fi, and ωi are not required to be monotone, and those ωi’s are not required to be stronger monotone than the one after the next as shown in (4). This inequality belongs to both the case of multivariables, to which great attentions [7–11] have been paid, and to the case that the left-hand side is a composition of the unknown function with a known function, in which Ou-Iang's idea [19] was applied [11–14]. He applied a technique of monotonization to construct a sequence of functions, made each function possess stronger monotonization than the previous one, and gave an estimate for the unknown function u.On the other aspect, many problems in the control theory can be modeled in the form of differential equations with the maxima of the unknown function [20–22]. In connection with the development of the theory of differential equations with maxima (see, e.g., [20, 21, 23]) and partial differential equations with maxima [24, 25], a new type of integral inequalities with maxima is required, respectively. There have been given some results for integral inequalities containing the maxima of the unknown function [23, 26–28]. Concretely, in 2012, Bohner et al. [26] discussed the following system of integral inequalities:
(6)φ(u(t))≤a(t)+∑i=1m‍∫αi(t0)αi(t)‍fi(s)up(s)ωi(u(s))ds+∑j=m+1m+n‍∫αj(t0)αj(t)‍fj(s)up(s)×ωj(maxξ∈[s-h,s]g(u(ξ)))ds,u(t)≤ψ(t),
where a, fi’s, ωi’s, φ, and ψ are nonnegative continuous functions and αi’s are nonnegative continuously differentiable and nondecreasing functions. They required that a(t)≥1, φ is C1 on ℝ+:=[0,+∞) and increasing such that φ(tx)≥tφ(x) for 0≤t≤1, and ωi satisfies the following: (i) ωi∈C1(ℝ+,ℝ+) is an increasing function, and (ii) ωi(tx)≥tωi(x) for all 0≤t≤1 and x>0. Bainov and Hristova [23] considered the following system:
(7)u(x,y)≤a(x,y)+∫x0x‍∫y0y‍f(s,t)up(s,t)dtds+∫α(x0)α(x)‍∫y0y‍h(s,t)(maxξ∈[s-h,s]up(ξ,t))dtds,u(x,y)≤ψ(x,y),
where a(x,y) is nonnegative and nondecreasing in both of its arguments, f, h, and ψ are continuous and nonnegative functions, and p∈(0,1].In this paper, we consider the system of integral inequalities as follows:(8)φ(u(x,y))≤a(x,y)+∑i=1n‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍fi(x,y,s,t)ωi(u(s,t))dtds+∑j=n+1n+m‍∫αj(x0)αj(x)‍∫βj(y0)βj(y)‍fj(x,y,s,t)×ωj(maxξ∈[s-h,s]g(u(ξ,t)))dtds,(x,y)∈[x0,x1)×[y0,y1),u(x,y)≤ψ(x,y),(x,y)∈[α*(x0)-h,x0]×[y0,y1),
where a, fi’s, ωi’s, and g are continuous and nonnegative functions, αi’s and βi’s are nonnegative continuously differentiable and nondecreasing functions, and α*(x0):=min1≤i≤m+nαi(x0). As required in previous works [27–29], we suppose that 0≤αi(t)≤t, 0≤βi(t)≤t, h>0 is constant. In this paper, we require neither monotonicity of a, ωi's, fi's, and g nor a(x,y)≥1. We monotonize those ωi’s to make a sequence of functions in which each one possesses stronger monotonicity than the previous one so as to give an estimation for the unknown function. We can use our result to discuss inequalities (6) and (7), giving the stronger results under weaker conditions. We finally apply the obtained result to a boundary value problem of a partial differential equation with maxima for uniqueness.


## 2. Main Result

Consider system (8) of integral inequalities with x0<x1 and y0<y1 in ℝ+:=[0,∞). Let Λ:=[x0,x1)×[y0,y1), Ω:=[α*(x0)-h,x0]×[y0,y1). Suppose that(H1)
αi:[x0,x1)→ℝ+(i=1,2,…,m+n) and βi:[y0,y1)→[y0,y1), i=1,2,…,m+n, are nondecreasing such that αi(x)≤x on [x0,x1),  βi(y)≤y on [y0,y1) and βi(y0)=y0;(H2)
allfi’s (i=1,2,…,m+n) are continuous and nonnegative functions on Λ×[α*(x0),x1)×[y0,y1);(H3)
g,φ:ℝ+→ℝ+ and ψ:[α*(x0)-h,x1)→ℝ+ are continuous, and φ is strictly increasing such that limt→∞φ(t)=+∞;(H4)
allωi’s (i=1,2,…,m+n) are continuous on ℝ+ and positive on (0,+∞);(H5)
a(x,y) is a continuous and nonnegative function on Λ.For thoseωi’s given in (H4), define ω~i(t), i=1,…,m+n, inductively by
(9)ω~1(t):=maxτ∈[0,t]{ω1(τ)},ω~i+1(t):=maxτ∈[0,t]{ωi+1(τ)ω~i(τ)+ϵi}ω~i(t),
for i=1,2,…,m-1 and
(10)ω~m+1(t):=maxτ∈[0,t]{ω^m+1(maxs∈[0,τ]{g(s)})ω~m(τ)+ϵm}ω~m(t),ω~j+1(t):=maxτ∈[0,t]{ω^j+1(maxs∈[0,τ]{g(s)})ω~j(τ)+ϵj}ω~j(t),
for j=m+1,…,m+n, where ω^j(t):=maxτ∈[0,t]{ωj(τ)} for j=m+1,…,m+n, ϵi:=ε1 if ω~i(0)=0 or :=0 if ω~i(0)≠0 for i=1,2,…,m+n-1, and ε1>0 be a given very small constant.Theorem 1.
Suppose that(H1)–(H5) hold, maxs∈[α*(x0)-h,x0]ψ(s,y)≤φ-1(a(x0,y)) for all y∈[y0,y1) and u∈C(Ω,ℝ+) satisfies the system (8) of integral inequalities. Then,
(11)u(x,y)≤φ-1(Wm+n-1(Ωm+n(x,y))),
for all (x,y)∈[x0,X1)×[y0,Y1), where
(12)Ωi(x,y):=Wi(ri(x,y))+∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍max(ι,ξ)∈[x0,x]×[y0,y]fi(ι,ξ,s,t)dtds,Wi-1 is the inverse of the function
(13)Wi(u):=∫uiu‍dsω~i(φ-1(s)),u≥ui,i=1,2,…,m+n,ui>0 is a given constant, ω~i is defined just before the theorem, and ri(x,y) is defined recursively by
(14)r1(x,y)=max(ι,ξ)∈[x0,x]×[y0,y]a(ι,ξ),ri+1(x,y)=Wi-1(Ωi(x,y)),
for i=1,2,…,m+n-1, and X1∈[x0,x1),Y1∈[y0,y1) are chosen such that
(15)Ωi(X1,Y1)≤∫ui∞‍dsω~i(φ-1(s)),
for i=1,2,…,m+n.For the special choice thatn=m=1, ω1(s)=sp, ω2(s)=s, f1(x,y,s,t)=f(s,t), f2(x,y,s,t)=h(s,t), α1(s)=s, α2(s)=α(s), g(s)=sp, and β1(s)=β2(s)=s, where α is a nonnegative continuously differentiable and nondecreasing function, Theorem 1 gives an estimate for the unknown u in the system (7). we require neither the monotonicity of a nor the monotonicity of ωi. Obviously, Lemma 2 and Theorem 1 are applicable to more general forms than Corollary 2.3.4 in [23]. Even if ωi(s) is enlarged to max1≤i≤m+nωi(s) such that (8) is changed into the form of (2.1) in [29], where m=n=1, our theorem gives a better estimate. For example, the system of inequalities
(16)u(x,y)≤3+4∫0x‍∫0y‍tsu(s,t)+1dtds+4∫0x‍∫0y‍ts(maxξ∈[s-h,s]u(ξ,t)+1)dtds,(x,y)∈[0,x1)×[0,y1),u(x,y)≤x+3,(x,y)∈[-h,0]×[0,y1),
implies that
(17)u(x,y)≤3+4∫0x‍∫0y‍ts(u(s,t)+1)dtds+4∫0x‍∫0y‍ts(maxξ∈[s-h,s]u(ξ,t)+1)dtds,(x,y)∈[0,x1)×[0,y1),u(x,y)≤x+3,(x,y)∈[-h,0]×[0,y1),
by enlarging s+1 to s+1. Applying Theorem 1, we obtain
(18)u(x,y)≤(x2y2+4)24exy2,(x,y)∈[0,x1)×[0,y1).
On the other hand, Theorem 2.2 of [29] gives from (17) that
(19)u(t)≤4e(x2y2+xy2),(x,y)∈[0,x1)×[0,y1).
Clearly, (18) is sharper than (19) for large x and y.In order to prove Theorem1, we need the following lemma.Lemma 2.
Suppose that(C1)
αi:[x0,x1)→ℝ+(i=1,2,…,m+n) and βi:[y0,y1)→ℝ+(i=1,2,…,m+n) are nondecreasing such that αi(x)≤x on [α*(x0),x1) and βi(y)≤y on [y0,y1) and βi(y0)=y0;(C2)
ψ∈C([α*(x0)-h,x1)×[y0,y1),ℝ+), bi∈C([α*(x0),x1)×[y0,y1),ℝ+) for i=1,2,…,m+n;(C3)
allhi’s (i=1,2,…,m+n) are continuous and nondecreasing on ℝ+ and positive on (0,+∞) such that h1∝h2∝…∝hm+n;(C4)
a(x,y) is continuously differentiable in x and y, nonnegative on [α*(x0),x1)×[y0,y1), and maxs∈[α*(x0)-h,x1]ψ(s,y)≤a(x0,y) for all y∈[y0,y1).
Ifu∈C([α*(x0)-h,x1)×[y0,y1),ℝ+) satisfies the system of inequalities as follows:
(20)u(x,y)≤a(x,y)+∑i=1m‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍bi(s,t)hi(u(s,t))dtds+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫βj(y0)βj(y)‍hj(s,t)hj(maxξ∈[s-h,s]u(ξ,t))dtds,&+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫βj(y0)βj(y)‍hj(s,t)hj(maxξ∈[s-h,s]ux)(x,y)∈Λ,u(x,y)≤ψ(x,y),(x,y)∈Ω,
then
(21)u(x,y)≤Hm+n-1({∫αm+n(x0)αm+n(x)‍}Hm+n(γm+n(x,y))≤Hm+n-15+∫αm+n(x0)αm+n(x)‍∫βm+n(y0)βm+n(y)‍hm+n(s,t)dtds),
for all (x,y)∈[x0,X*)×[y0,Y*), where Hi-1 is the inverse of the function
(22)Hi(u):=∫uiu‍dxωi(x),u≥ui,i=1,2,…,m+n,ui>0 is a given constant, and γi(x,y) is defined recursively by
(23)γ1(x,y):=a(x0,y0)+∫x0x‍|ax(t,y)|dt+∫y0y‍|ay(x,s)|ds,γi+1(x,y):=Hi-1(Hi(γi(x,y))+∫αi(t0)αi(t)‍∫βi(t0)βi(t)‍bi(s,t)dtds)
for i=1,2,…,m+n-1, and x0≤X*<x1,y0≤Y*<y1 are chosen such that
(24)Hi(γi(X*,Y*))+∫αi(x0)αi(X*)‍∫βi(y0)βi(Y*)‍bi(s,t)dtds≤∫ui∞‍dshi(s),
for i=1,2,…,m+n.Proof.
From (23), we see that γ1(x,y) is nondecreasing on Λ, a(x,y)≤γ1(x,y), and maxs∈[α*(x0)-h,x1]ψ(s,y)≤a(x0,y)≤γ1(x0,y) for y∈[y0,y1). It implies from (20) that
(25)u(x,y)≤γ1(x,y)+∑i=1m‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍bi(s,t)hi(u(s,t))dtds+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫βj(y0)βj(y)‍bj(s,t)×hj(maxξ∈[s-h,s]u(ξ,t))dtds
for all (x,y)∈Λ. Let(26)z(x,y):={γ1(x,y)+∑i=1m∫αi(x0)αi(x)∫βi(y0)βi(y)bi(s,t)hi(u(s,t))dtds+∑j=m+1m+n∫αj(x0)αj(x)∫βi(y0)βi(y)bj(s,t)hj(maxξ∈[s-h,s]u(ξ,t))dtds,(x,y)∈Λ,γ1(x0,y),(x,y)∈Ω.Clearly, z(x,y) is nondecreasing in x. Then, we have
(27)u(x,y)≤z(x,y),(x,y)∈[α*(x0)-h,ξ)×[y0,η),maxξ∈[s-h,s]u(ξ,y)≤maxξ∈[s-h,s]z(ξ,y)=z(s,y),(s,y)∈[α*(x0),x1)×[y0,y1).
From (25), (27), and (28) and the definition of z(x,y) on Λ, we get
(28)z(x,y)≤γ1(x,y)+∑i=1m‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍bi(s,t)hi(u(s,t))dtds+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫βi(y0)βi(y)‍bj(s,t)hj(z(s,t))dtds.
Applying Theorem  1 of [7] to the case that fi(x,y,s,t)=bi(s,t), a(x,y)=γ1(x,y), p=1, and ωi(t)=hi(t), i=1,2,…,m+n, we obtain (21) from (28). This completes the proof.Proof of Theorem1.
First of all, we monotonize some given functionsfi, ωi, g, and a in the system (8) of integral inequalities. Let
(29)g~(t):=maxτ∈[0,t]{g(τ)},t≥0,a~(x,y):=max(τ,ξ)∈[x0,x]×[y0,y]{a(τ,ξ)},{a(τ,ξ)}(x,y)∈[x0,x1)×[y0,y1).
From (13), we see that the function Wi is strictly increasing, and therefore its inverse Wi-1 is well defined, continuous, and increasing in its domain. The sequence {ω~i(t)}, defined by ωi(s), consists of nondecreasing nonnegative functions on ℝ+ and satisfies
(30)ωi(t)≤ω~i(t),i=1,2,…,m,ωi(t)≤ω^i(t),i=m+1,…,m+n,ω^i(g~(t))≤ω~i(t),i=m+1,…,m+n.
Moreover,
(31)ω~i∝ω~i+1,i=1,2,…,m+n,
because the ratios ω~i+1/ω~i, i=1,2,…,m+n, are all nondecreasing. Furthermore, let
(32)f~i(x,y,s,t):=max(ι,ξ)∈[x0,x]×[y0,y]fi(ι,ξ,s,t),
which is nondecreasing in x and y for each fixed t and s and satisfies f~i(x,y,s,t)≥fi(x,y,s,t)≥0 for all i=1,2,…,m+n. The monotonicity of g~ implies that
(33)maxξ∈[s-h,s]g(u(ξ,y))≤maxξ∈[s-h,s]g~(u(ξ,y))≤g~(maxξ∈[s-h,s]u(ξ,y))
for (s,y)∈[α*(x0)-h,x1)×[y0,y1). From (8) and the definition of f~i(x,y,s,t), we obtain
(34)φ(u(x,y))≤a~(x,y)+∑i=1n‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍f~i(x,y,s,t)×ωi(u(s,t))dtds+∑j=n+1n+m‍∫αj(x0)αj(x)‍∫βj(y0)βj(y)‍f~j(x,y,s,t)×ωj(maxξ∈[s-h,s]g(u(ξ,t)))dtds,(x,y)∈Λ,u(x,y)≤ψ(x,y),(x,y)∈Ω.
Concerning (34), we consider the auxiliary system of inequalities
(35)φ(u(x,y))≤a~(X,Y)+∑i=1n‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍f~i(X,Y,s,t)×ωi(u(s,t))dtds+∑j=1m‍∫αj(x0)αj(x)‍∫βi(y0)βi(y)‍f~j(X,Y,s,t)×ωj(maxξ∈[s-h,s]g(u(ξ,t)))dtds,(x,y)∈[x0,X)×[y0,Y),u(x,y)≤ψ(x,y),(x,y)∈Ω,
where x0≤X≤X1 and y0≤Y≤Y1 are chosen arbitrarily, and claim
(36)u(x,y)≤φ-1(Wm+n-1(Ω~i(X,Y,x,y))),
for all x0≤x≤min{X,X2}, y0≤y≤min{Y,Y2}, where
(37)Ω~i(X,Y,x,y):=Wi(r~i(X,Y,x,y))+∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍f~(X,Y,s,t)dtds,i=1,2,…,m+n, r~i(X,Y,x,y) is defined inductively by
(38)r~1(X,Y,x,y):=a~(X,Y),r~i+1(X,Y,x,y):=Wi-1(Wi(Ω~i(X,Y,x,y))),
for i=1,2,…,m+n-1, and X2,Y2 are chosen such that
(39)Ω~i(X,Y,X2,Y2)≤∫ui∞‍dsω~i(φ-1(s)),
for i=1,2,…,m+n.
Notice that we may takeX2=X1 and Y2=Y1. In fact, the monotonicity that r~i(X,Y,x,y) and f~i(X,Y,x,y) are both nondecreasing in X and Y for fixed x, y. Furthermore, it is easy to check that r~i+1(X,Y,X,Y)=ri(X,Y), for i=1,2,…,m+n. If X2, Y2 are replaced with X1, Y1, respectively, on the left side of (39), we get from (15) that
(40)Ω~i(X,Y,X1,Y1)=Wi(r~i(X,Y,X1,Y1))+∫αi(x0)αi(X1)‍∫βi(y0)βi(Y1)‍f~(X,Y,s,t)dtds≤Wi(r~i(X1,Y1,X1,Y1))+∫αi(x0)αi(X1)‍∫βi(y0)βi(Y1)‍f~(X1,Y1,s,t)dtds=ri(X1,Y1)+∫αi(x0)αi(X1)‍∫βi(y0)βi(Y1)‍f~(X1,Y1,s,t)dtds=Ωi(X1,Y1)≤∫ui∞‍dsω~i(φ-1(s)).
Thus, it means that we can take X2=X1, Y2=Y1.
Now, we prove (36) by induction. From (33), (35), and the definitions of g~(t), ω~i(t), and ω^i(t), we obtain
(41)φ(u(x,y))≤a~(X,Y)+∑i=1m‍∫αi(x0)αi(x)‍∫y0y‍f~i(X,Y,s,t)ωi(u(s,t))+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫y0y‍f~j(X,Y,s,t)×ω^j(g~(maxξ∈[s-h,s]u(ξ,t)))dtds≤a~(X,Y)+∑i=1m‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍f~i(X,Y,s,t)ω~i(u(s))ds+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫βj(y0)βj(y)‍f~j(X,Y,s,t)×ω~j(maxξ∈[s-h,s]u(ξ))dtds,
for all (x,y)∈[x0,X)×[y0,Y), where x0≤X≤X1 and y0≤Y≤Y1 are chosen arbitrarily. Since maxs∈[α*(x0)-h,x0]ψ(s,y)≤φ-1(a(x0,y)) and a(x0,y)≤a~(x0,y)≤a~(X,Y), we have maxs∈[α*(x0)-h,x0]ψ(s,y)≤φ-1(a~(X,Y)). Define a function z(x,y):[α*(x0)-h,X)×[y0,Y)→ℝ+ by(42)z(x,y)={a~(X,Y)+∑i=1m∫αi(x0)αi(x)∫βi(y0)βi(y)f~i(X,Y,s,t)ω~i(u(s,t))dtds+∑j=m+1m+n∫αj(x0)αj(x)∫βi(y0)βi(y)f~j(s,t)ω~j(maxξ∈[s-h,s]u(ξ,t))dtds,(x,y)∈[x0,X)×[y0,Y),a~(X,Y),(x,y)∈[α*(x0)-h,x0]×[y0,Y).Clearly, z(x,y) is nondecreasing in x. By (41) and the definition of z(x,y), we have
(43)u(x,y)≤φ-1(z(x,y)),(x,y)∈[α*(x0)-h,X)×[y0,Y).
Then noting that z(x,y) is nondecreasing and φ(t) is strictly increasing, from (43), we obtain
(44)maxξ∈[s-h,s]u(ξ,y)≤maxξ∈[s-h,s]φ-1(z(ξ,y))≤maxξ∈[s-h,s]φ-1(z(s,y))≤φ-1(maxξ∈[s-h,s]z(ξ,y)),(s,y)∈[α*(x0),X)×[y0,Y).
It follows from (43), (44), and the definition of z(x,y) that
(45)z(x,y)≤a~(X,Y)+∑i=1m‍∫αi(x0)αi(x)‍∫βi(y0)βi(y)‍f~i(X,Y,s,t)×ω~i(φ-1(z(s,t)))dtds+∑j=m+1m+n‍∫αj(x0)αj(x)‍∫βi(y0)βi(y)‍f~i(X,Y,s,t)×ω~j(φ-1(maxξ∈[s-h,s]z(ξ,t)))dtds,(x,y)∈[x0,X)×[y0,Y],z(x,y)≤a~(X,Y),(x,y)∈[α*(x0)-h,x0]×[y0,Y].
In order to demonstrate the basic condition of monotonicity, let h(t):=φ-1(t), which is clearly a continuous and nondecreasing function on ℝ+. Thus, each ω~i(h(t)) is continuous and nondecreasing on ℝ+ and satisfies ω~i(h(t))>0 for t>0. Moreover, since ω~i(t)∝ω~i+1(t), ω~i+1(h(t))/ω~i(h(t)) is also continuous and nondecreasing on ℝ+ and positive on (0,+∞), implying that ω~i(h(t))∝ω~i+1(h(t)), for i=1,2,…,m+n-1. By Lemma 2 and (45),
(46)z(x,y)≤Wm+n-1({∫αm+n(x0)αm+n(x)‍}W(r~m+n(X,Y,x,y))+∫αm+n(x0)αm+n(x)‍∫βm+n(y0)βm+n(y)‍f~m+n(X,Y,s,t)dtds),
for x0≤x<X2 and y0≤y<Y2. It follows from (43) and (46) that
(47)u(x,y)≤φ-1({+∫αm+n(x0)αm+n(x)∫βm+n(y0)βm+n(y)f~m+n(X,Y,s,t)dtds))}Wm+n-1({∫βm+n(y0)βm+n(y)‍}W(r~m+n(X,Y,x,y))+∫αm+n(x0)αm+n(x)∫βm+n(y0)βm+n(y)f~m+n(X,Y,s,t)dtds)),
for x0≤x<X2 and y0≤y<Y2. This proves the claimed (36).
Takingx=X,y=Y, X2=X1, and Y2=Y1 in (36), we have
(48)u(X,Y)≤φ-1({+∫αm+n(x0)αm+n(X)‍∫βm+n(y0)βm+n(Y)‍f~m+n(X,Y,s,t)dtds))}Wm+n-1({∫βm+n(y0)βm+n(y)‍}Wm+n(r~m+n(X,Y,X,Y))+∫αm+n(x0)αm+n(X)∫βm+n(y0)βm+n(Y)f~m+n(X,Y,s,t)dtds)),
for all x0≤X<X1, y0≤Y<Y1. It is easy to verify  r~m+n(X,Y,X,Y)=rm+n(X,Y). Thus, (48) can be written as
(49)u(X,Y)≤φ-1({+∫αm+n(x0)αm+n(X)‍∫βm+n(y0)βm+n(Y)‍f~m+n(X,Y,s,t)dtds))}Wm+n-1({∫βm+n(y0)βm+n(Y)‍}Wm+n(rm+n(X,Y))+∫αm+n(x0)αm+n(X)∫βm+n(y0)βm+n(Y)f~m+n(X,Y,s,t)dtds)).
Since X,Y are arbitrary, replacing X and Y with x and y, respectively, we get
(50)u(x,y)≤φ-1({+∫αm+n(x0)αm+n(x)‍∫βm+n(y0)βm+n(y)‍f~m+n(x,y,s,t)dtds))}Wm+n-1({∫βm+n(y0)βm+n(y)‍}Wm+n(rm+n(x,y))+∫αm+n(x0)αm+n(x)∫βm+n(y0)βm+n(y)f~m+n(x,y,s,t)dtds)),
for all (x,y)∈[x0,X1)×[y0,Y1). This completes the proof.


## 3. Applications

In this section, we apply our result to prove the boundedness of solutions for a differential equation with the maxima.Consider a system of partial differential equations with maxima(51)∂2z(x,y)∂x∂y=F(x,y,z(x,y),maxs∈[α(x),β(x)]ω(z(s,y))),(x,y)∈[x0,x1)×[y0,y1)z(x,y)=ψ(x,y),(x,y)∈[β(x0)-h,x0]×[y0,y1),z(x,y0)=f(x),z(x0,y)=g(y),x≥x0,y≥y0,
where F∈C([x0,x1)×[y0,y1)×ℝ2,ℝ), ω∈C([0,∞),ℝ+), α,β∈C1([x0,x1),ℝ+) are nondecreasing such that α(x)≤x, β(x)≤x, and 0<β(x)-α(x)≤h (h is a positive constant) for x∈[x0,x1], ψ∈C([β(x0)-h,x0]×[y0,y1), and f∈C([x0,x1),ℝ), g∈C([y0,y1),ℝ) satisfy that f(x0)=g(y0) and g(y)=ψ(x0,y), for all y∈[y0,y1).Equation (51) is more general than the equation considered in Section 2.4 of [23]. The following result gives an estimate for its solutions.Corollary 3.
Suppose that functionsF and ψ in (51) satisfy
(52)|F(x,y,s,t)|≤h1(x,y)μ1(|s|)+h2(x,y)μ2(|t|),|ψ(x,y)|≤|g(y)|,∀(x,y)∈[β(x0)-h,x0]×[y0,y1),
where hi∈C([x0,x1)×[y0,y1),ℝ+) and μi∈C(ℝ+,(0,∞)), i=1,2. Then, any solution z(x,y) of (51) has the estimate
(53)|z(x,y)|≤Q2-1(Q2(γ(x,y))+∫x0x‍∫y0y‍h2(s,t)dtds),
for all (x,y)∈[x0,X1)×[y0,Y1), where
(54)γ(x,y):=Q1-1(Q1(γ1(x,y))+∫x0x‍∫y0y‍h1(s,t)dtds),γ1(x,y):=max(ξ,η)∈[x0,x]×[y0,y](|f(ξ)+g(η)-f(x0)|),Q2(u):=∫u2u‍ds{maxτ∈[0,s]{μ~2(ω~(τ))/maxτ1∈[0,τ]{μ1(τ1)}}maxτ∈[0,s]{μ1(τ)}},Q1(u):=∫u1u‍dsmaxτ∈[0,s]{μ1(τ)},ω~(t):=maxs∈[0,t]{ω(s)},μ~2(t):=maxs∈[0,t]{μ2(s)},
and X1, Y1 are given as in Theorem 1, and constants u1>0, u2>0 are given arbitrarily.Proof.
From (51), we obtain
(55)z(x,y)=f(x)+g(y)-f(x0)+∫x0x‍∫y0y‍F(s,t,z(s,t),maxξ∈[α(s),β(s)]ω(z(ξ,t)))dtds,(x,y)∈Λ,z(x,y)=ψ(x,y),(x,y)∈(x,y)∈[β(x0)-h,x0]×[y0,y1).
From (52) and (55), we get
(56)|z(x,y)|≤|f(x)+g(y)-f(x0)|+∫x0x‍∫y0y‍h1(s,t)μ1(|z(s,t)|)dtds+∫x0x‍∫y0y‍h2(s,t)μ2(|maxξ∈[α(s),β(s)]ω(z(ξ,t))|)dtds≤|f(x)+g(y)-f(x0)|+∫x0x‍∫y0y‍h1(s,t)μ1(|z(s,t)|)dtds+∫x0x‍∫y0y‍h2(s,t)μ~2(maxξ∈[α(s),β(s)]ω~(|z(ξ,t)|))dtds,(x,y)∈Λ,|z(x,y)|≤|ψ(x,y)|,(x,y)∈[β(x0)-h,x0]×[y0,y1).
Set v(x,y)=|z(x,y)| for (x,y)∈[β(x0)-h,x1)×[y0,y1). Noting that maxξ∈[α(s),β(s)]z(ξ,y)≤maxξ∈[β(s)-h,β(s)]z(ξ,y), from (56), we get
(57)v(x,y)≤γ1(x,y)+∫x0x‍∫y0y‍h1(s,t)μ1(v(s,t))dtds+∫β(x0)β(x)‍∫y0y‍h2(β-1(η),t)(β-1(η))′μ~2×(maxξ∈[η-h,η]ω~(v(ξ,t)))dtdηv(x,y)≤|ψ(x,y)|,(x,y)∈[β(x0)-h,x0]×[y0,y1).
Applying Theorem 1 to specified m=n=1, φ(u)=u, f1(x,y,s,t)=h1(s,t), α1(t)=t, and α2(t)=β(t), βi(t)=t, i=1,2, f2(s,t)=h2(β-1(s),t)(β-1(s))′, and ωi(u)=μi(u),i=1,2, we obtain (53) from (57).Next, we discuss the uniqueness of solutions for system (51).Corollary 4.
Suppose thatg(s)=s and
(58)|F(x,y,s1,t1)-F(x,y,s2,t2)|≤h1(x,y)μ1(|x1-x2|)+h2(x,y)μ2(|y1-y2|),
for all (x,y)∈[x0,x1]×[y0,y1] and all xi,yi∈ℝ(i=1,2), where hi∈C([x0,x1]×[y0,y1],ℝ+) and μi∈C([ℝ,ℝ+) are both nondecreasing such that μi(0)=0, μi(u)>0 for u>0, μ2/μ1 is also nondecreasing, and ∫01‍ds/μi(s)=+∞, i=1,2. Then, system (51) has at most one solution on [x0,x1)×[y0,y1).Proof.
g
(
s
)
=
s. From (51), we get
(59)∂2z(x,y)∂x∂y=F(x,y,z(x,y),maxs∈[α(x),β(x)]z(s,y)),(x,y)∈[x0,x1)×[y0,y1),z(x,y)=ψ(x,y),(x,y)∈[α(x0)-h,x0]×[y0,y1],z(x,y0)=f(x),z(x0,y)=g(y),x≥x0,y≥y0.
Assume that (59) has two different solutions u(x,y) and v(x,y). From the equivalent integral equation system (55), we have
(60)|u(x,y)-v(x,y)|≤∫x0x‍∫y0y‍h1(s,t)μ1(|u(s,t)-v(s,t)|)dtds+∫x0x‍∫y0y‍h2(s,t)μ2(|maxξ∈[α(s),β(s)]u(ξ,t)-maxξ∈[α(s),β(s)]v(ξ,t)|)dtds≤∫x0x‍∫y0y‍h1(s,t)μ1(|u(s)-v(s)|)dtds+∫x0x‍∫y0y‍h2(s,t)μ2×(maxξ∈[α(s),β(s)]|u(ξ,t)-v(ξ,t)|)ds,
for all (x,y)∈[x0,x1]×[y0,y1]. The continuity of the function u(x,y) implies that for any fixed points s∈[x0,x] and t∈[y0,y] there exists a point η∈[α(s),β(s)] such that the inequality maxξ∈[α(s),β(s)]u(ξ,t)=u(η,t) holds, and therefore
(61)|maxξ∈[α(s),β(s)]u(ξ,t)-maxξ∈[α(s),β(s)]v(ξ,t)|=|u(η,t)-maxξ∈[α(s),β(s)]v(ξ,t)|≤|u(η,t)-v(η,t)|≤maxξ∈[α(s),β(s)]|u(ξ,t)-v(ξ,t)|.
Hence,
(62)|u(x,y)-v(x,y)|≤∫x0x‍∫y0y‍h1(s,t)μ1(|u(s,y)-v(s,y)|)dtds+∫x0x‍∫y0y‍h2(s,t)μ2(maxξ∈[α(s),β(s)]|u(ξ,t)-v(ξ,t)|)ds.
Let
(63)ϕ(x,y):=|u(x,y)-v(x,y)|,(x,y)∈[β(x0)-h,x1]×[y0,y1].
Because maxξ∈[α(s),β(s)]u(ξ,y)≤maxξ∈[β(s)-h,β(s)]u(ξ,y), from (62), we obtain
(64)ϕ(x,y)≤∫x0x‍∫y0y‍h1(s,t)μ1(ϕ(s,t))dtds+∫β(x0)β(x)‍∫y0y‍h2(β-1(η),t)(β-1(η))′×μ2(maxξ∈[η-h,η]ϕ(ξ,t))dtdη,(x,y)∈[x0,x1)×[y0,y1),ϕ(x,y)≤0,(x,y)∈[β(x0)-h,x0]×[y0,y1).
Applying Theorem 1 to specified m=n=1, φ(u)=u, f1(s,t)=h1(s,t), α1(t)=t, α2(t)=β(t), f2(s,t)=h2(β-1(t),s)(β-1(t))′, g(t)=t, a(x,y)=0, and ωi(t)=μi(t),i=1,2, from (64), we obtain
(65)ϕ(x,y)≤Q^2-1(Q^2(γ-2(s,t))+∫x0x‍∫y0y‍h2(s,t)dtds),
for all (x,y)∈[x0,X1]×[y0,Y1], where
(66)Q^1(u):=∫1u‍dsμ1(s),Q^2(u):=∫1u‍dsμ2(τ),(67)r-1(x,y):=0,(68)r-2(x,y):=Q^1-1(Q^1(r-1(x,y))+∫x0x‍∫y0y‍h1(s,t)dtds).
By the definition of Q^i and properties of μi, noting that ∫01‍ds/μi(s)=+∞(i=1,2), we obtain
(69)limu→0+Q^i(u)=-∞,limu→-∞Q^i-1(u)=0,i=1,2.
Since ∫x0x‍∫y0y‍h1(s,t)dtds is finite on a finite interval, [x0,x1] and [y0,y1], by (67), we obtain
(70)Q^1(r-1(x,y))+∫x0x‍∫y0y‍h1(s,t)dtds=-∞.
Thus, we obtain γ-2(x,y)=0 from (68), (69), and (70) immediately. Similarly, noting that ∫x0x‍∫y0y‍h2(s,t)dtds is finite on finite interval, [x0,x1] and [y0,y1], from (69), we obtain
(71)Q^2(r-2(x,y))+∫x0x‍∫y0y‍h2(s,t)dtds=-∞.
Thus, we conclude from (65), (69), and (71) that |u(x,y)-v(x,y)|≤0, which implies that u(x,y)=v(x,y), for all (x,y)∈[x0,X1)×[y0,Y1), where X1, Y1 are given as in Theorem 1. The uniqueness is proved.



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*Source: 853476-2013-03-20.xml*