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Generalized integral inequalities for discontinuous functions with two independent variables and their applications

Abstract

This paper investigates integral inequalities for discontinuous functions with two independent variables involving two nonlinear terms. We do not require that ω(u) is in the class or the class ȷ in Gallo and Piccirllo’s paper (Nonlinear Stud. 19:115-126, 2012). My main results can be applied to generalize Borysenko and Iovane’s results (Nonlinear Anal., Theory Methods Appl. 66:2190-2230, 2007) and to give results similar to Gallo-Piccirllo’s. Examples to show the bounds of solutions of a partial differential equation with impulsive terms are also given, which cannot be estimated by Gallo and Piccirllo’s results.

MSC:26D15, 26D20.

1 Introduction

Integral inequalities and their various linear and nonlinear generalizations involving continuous or discontinuous functions play very important roles in investigating different qualitative characteristics of solutions for differential equations, partial differential equations and impulsive differential equations such as existence, uniqueness, continuation, boundedness, continuous dependence of parameters, stability, and attraction. The literature on inequalities for continuous functions and their applications is vast (see [111]). In the one-dimensional case, all the main results in the theory of integral inequalities for continuous functions are almost based on the solvability of Chaplygin’s problem [6] for the integral inequality

u(x)φ(x)+ x 0 x Γ ( x , s , u ( s ) ) ds.
(1.1)

Recently, more attention has been paid to generalizations of Gronwall-Bellman’s results for discontinuous functions and their applications (see [1227]). One of the important things is that Samoilenko and Perestyuk [26] studied the following inequality:

u(x)c+ x 0 x f(s)u(s)ds+ x 0 < x i < x β i u( x i 0)
(1.2)

for the nonnegative piecewise continuous function u(x), where c, β i are nonnegative constants, f(x) is a positive function, and x i are the first kind discontinuity points of the function u(x). Then Borysenko [14] investigated integral inequalities with two independent variables,

u ( x , y ) a ( x , y ) + x 0 x y 0 y τ ( s , t ) u ( s , t ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u ( x i 0 , y i 0 ) .
(1.3)

Here u(x,y) is an unknown nonnegative continuous function with the exception of the points ( x i , y i ) where there is a finite jump: u( x i 0, y i 0)u( x i +0, y i +0), i=1,2, .

In 2007, Borysenko and Iovane [16] considered the following inequalities:

u ( x , y ) a ( x , y ) + x 0 x y o y τ ( s , t ) u ( s , t ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) , m > 0 ,
(1.4)
u ( x , y ) a ( x , y ) + x 0 x y o y τ ( s , t ) u m ( s , t ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) , m > 0 ,
(1.5)
u ( x , y ) a ( x , y ) + q ( x , y ) x 0 x y o y τ ( s , t ) u m ( σ ( s ) , σ ( t ) ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) , m > 0 .
(1.6)

Later, Gallo and Piccirllo [24] studied the following inequalities:

u ( x , y ) a ( x , y ) + q ( x , y ) x 0 x y o y τ ( s , t ) ω ( u ( σ ( s ) , τ ( t ) ) ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) , m > 0 .
(1.7)

In this paper, motivated by the work above, we will establish the following much more general integral inequality:

u ( x , y ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t + g ( x , y ) ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) , m > 0 ,
(1.8)

with two independent variables involving two nonlinear terms ω 1 (u) and ω 2 (u) where we do not restrict ω 1 and ω 2 to the class or the class ȷ. Moreover, f n (x,y,s,t) (n=1,2) has a more general form. We also show that many integral inequalities for discontinuous functions such as (1.3)-(1.6) can be reduced to the form of (1.8). Finally, our main result is applied to an estimation of the bounds of the solutions of a partial differential equation with impulsive terms.

2 Main results

Let

Ω= i , j 1 Ω i j , Ω i j = { ( x , y ) : x i 1 x < x i , y j 1 y < y j } ,

for i,j=1,2, , x 0 >0 and y 0 >0, and let D 1 z(x,y) denote the first-order partial derivative of z(x,y) with respect to x and k = 0 1 u k (x,y)=0.

Consider (1.8) and assume that

(H1) a(x,y) is defined on Ω and a( x 0 , y 0 )0; β i is a nonnegative constant for any positive integer i;

(H2) f n (x,y,s,t) (n=1,2) are continuous and nonnegative functions on Ω×Ω and satisfy a certain condition: f n (x,y,s,t)=0 (n=1,2) if (s,t) Ω i j , ij for arbitrary i,j=1,2, ;

(H3) ω 1 (u) and ω 2 (u) are continuous and nonnegative functions on [0,) and are positive on (0,) such that ω 2 ( u ) ω 1 ( u ) is nondecreasing;

(H4) g(x,y) is continuous and nonnegative on Ω;

(H5) u(x,y) is nonnegative and continuous on Ω with the exception of the points ( x i , y i ) where there is a finite jump: u( x i 0, y i 0)u( x i +0, y i +0), i=1,2, . Here ( x i , y i )<( x i + 1 , y i + 1 ) if x i < x i + 1 , y i < y i + 1 , i=0,1,2, , and lim i x i =, lim i y i =;

(H6) b n (x) and c n (y) (n=1,2) are continuously differentiable and nondecreasing such that x 0 b n (x)x on [ x 0 ,) and y 0 c n (y)y on [ y 0 ,).

Let W j (u)= u ˜ j u d z ω j ( z ) for u u ˜ j and j=1,2 where u ˜ j is a given positive constant. Clearly, W j is strictly increasing so its inverse W j 1 is well defined, continuous, and increasing in its corresponding domain.

Theorem 2.1 Suppose that (H k ) (k=1,,6) hold and u(x,y) satisfies (1.8) for a positive constant m. If we let u i (x,y)=u(x,y) for (x,y) Ω i i , then the estimate of u(x,y) is recursively given, for (x,y) Ω i i , i=1,2, , by

u i ( x , y ) W 2 1 { W 2 W 1 1 [ W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ] + b 2 ( x i 1 ) b 2 ( x ) c 2 ( y i 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t } ,
(2.1)

where

r 1 ( x , y ) = max x 0 ξ x , y 0 η y | a ( ξ , η ) | , f ˜ n ( x , y , s , t ) = max x 0 ξ x , y 0 η y f n ( ξ , η , s , t ) , r i ( x , y ) = r 1 ( x , y ) + k = 1 i 1 n = 1 2 b n ( x k 1 ) b n ( x k ) c n ( y k 1 ) c n ( y k ) f n ( x , y , s , t ) ω n ( u k ( s , t ) ) d s d t + g ( x , y ) k = 1 i 1 β k u k m ( x k 0 , y k 0 ) ,
(2.2)

provided that

W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t u ˜ 1 d z ω 1 ( z ) , W 2 W 1 1 [ W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ] + b 2 ( x i 1 ) b 2 ( x ) c 2 ( y i 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t u ˜ 2 d z ω 2 ( z ) .
(2.3)

The proof is given in Section 3.

Remark 2.1 If ω j satisfies u ˜ j u d z ω j ( z ) = for j=1,2, then i in Theorem 2.1 can be any nonzero integer. Reference [4] pointed out that different choices of u ˜ j in W j do not affect our results for j=1,2. If a(x,y)0, then define W 1 (0)=0 and (2.1) is still true.

Remark 2.2 If a(x,y) is nondecreasing, Theorem 2.1 generalizes many known results. For example:

  1. (1)

    If we take f 1 (x,y,s,t)=τ(s,t), f 2 (x,y,s,t)=0, ω 1 (u)=u, m=1, b 1 (x)=x, c 1 (y)=y and g(x,y)=1, then (1.8) reduces to (1.3). It is easy to check that W 1 (u)=ln u u ˜ 1 and W 1 1 (u)= u ˜ 1 e u . From Theorem 2.1, we know that for (x,y) Ω i i

    u i (x,y) r i (x,y) e x i 1 x y i 1 y τ ( s , t ) d s d t

with

r i (x,y)=a(x,y)+ k = 1 i 1 x k 1 x k y k 1 y k τ(s,t) u k (s,t)dsdt+ k = 1 i 1 β k u k ( x k 0, y k 0).

Hence

r 1 ( x , y ) = a ( x , y ) , u 1 ( x , y ) = a ( x , y ) e x 0 x y 0 y τ ( s , t ) d s d t , r 2 ( x , y ) = a ( x , y ) ( 1 + β 1 ) e x 0 x 1 y 0 y 1 τ ( s , t ) d s d t , u 2 ( x , y ) = a ( x , y ) ( 1 + β 1 ) e x 0 x y 0 y τ ( s , t ) d s d t .

After recursive calculations, we have

u(x,y)a(x,y) Π ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) (1+ β i ) e x 0 x y 0 y τ ( s , t ) d s d t ,

which is the same as the expression in [14];

  1. (2)

    If we take f 1 (x,y,s,t)=τ(s,t), f 2 (x,y,s,t)=0, ω 1 (u)=u, m>0, b 1 (x)=x, c 1 (y)=y and g(x,y)=1, then (1.8) reduces to (1.4) and Theorem 2.1 becomes Theorem 2.1 in [16];

  2. (3)

    If we take f 1 (x,y,s,t)=τ(s,t), f 2 (x,y,s,t)=0, ω 1 (u)= u m , m>0, b 1 (x)=x, c 1 (y)=y and g(x,y)=1, then (1.8) reduces to (1.5) and Theorem 2.1 becomes Theorem 2.2 in [16];

  3. (4)

    If σ (t)>0 on [ t 0 ,) where t 0 =min{ x 0 , y 0 }, then (1.6) can be rewritten as

    u ( x , y ) a ( x , y ) + q ( x , y ) σ ( x 0 ) σ ( x ) σ ( y o ) σ ( y ) τ ( σ 1 ( s ) , σ 1 ( t ) ) σ ( σ 1 ( s ) ) σ ( σ 1 ( t ) ) u m ( s , t ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) .
    (2.4)

If we let f 1 (x,y,s,t)=q(x,y) τ ( σ 1 ( s ) , σ 1 ( t ) ) σ ( σ 1 ( s ) ) σ ( σ 1 ( t ) ) , f 2 (x,y,s,t)=0, and ω 1 (u)= u m , the above inequality is the same as (1.8).

Consider the inequality

φ ( u ( x , y ) ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t + g ( x , y ) ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i ψ ( u ( x i 0 , y i 0 ) ) ,
(2.5)

which looks much more complicated than (1.8).

Corollary 2.1 Suppose that (H k ) (k=1,,6) hold, ψ(u) is positive on (0,), φ(u) is positive and strictly increasing on (0,) and u(x,y) satisfies (2.5). If we let u i (x,y)=u(x,y) for (x,y) Ω i i , then the estimate of u(x,y) is recursively given, for (x,y) Ω i i , i=1,2, , by

u i ( x , y ) φ 1 { W 2 1 [ W 2 W 1 1 ( W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ) + b 2 ( x i 1 ) b 2 ( x ) c 2 ( y i 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t ] } ,
(2.6)

where W j (u)= u ˜ j u d z ω j ( φ 1 ( z ) ) , r 1 (x,y) and f ˜ n (x,y,s,t) are given in Theorem  2.1, r i (x,y) is defined as follows:

r i ( x , y ) = r 1 ( x , y ) + k = 1 i 1 n = 1 2 b n ( x k 1 ) b n ( x k ) c n ( y k 1 ) c n ( y k ) f n ( x , y , s , t ) ω n ( u k ( s , t ) ) d s d t + g ( x , y ) k = 1 i 1 β k ψ ( u k ( x k 0 , y k 0 ) ) .

Proof Let φ(u(x,y))=h(x,y). Since the function φ is strictly increasing on [0,), its inverse φ 1 is well defined. Equation (2.5) becomes

h ( x , y ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω n ( φ 1 ( h ( s , t ) ) ) d s d t + g ( x , y ) ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i ψ ( φ 1 ( h ( x i 0 , y i 0 ) ) ) .
(2.7)

Let ω ˜ n = ω n φ 1 and ψ ˜ =ψ φ 1 . Equation (2.7) becomes

h ( x , y ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω ˜ n ( h ( s , t ) ) d s d t + g ( x , y ) ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i ψ ˜ ( h ( x i 0 , y i 0 ) ) .
(2.8)

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, in the same way as in Theorem 2.1, for (x,y) Ω i i , i=1,2, , we can obtain the estimate of u(x,y),

u i ( x , y ) φ 1 { W 2 1 [ W 2 W 1 1 ( W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ) + b 2 ( x i 1 ) b 2 ( x ) c 2 ( y i 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t ] } .
(2.9)

This completes the proof of Corollary 2.1. □

If φ(u)= u λ where λ>0 is a constant, we can study the inequality

u λ ( x , y ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t + g ( x , y ) ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i ψ ( u ( x i 0 , y i 0 ) ) .
(2.10)

According to Corollary 2.1, we have the following result.

Corollary 2.2 Suppose that (H k ) (k=1,,6) hold, ψ(u)>0, and u(x,y) satisfies (2.10). If we let u i (x,y)=u(x,y) for (x,y) Ω i i , then the estimate of u(x,y) is recursively given, for (x,y) Ω i i , i=1,2, , by

u i ( x , y ) { W 2 1 [ W 2 W 1 1 ( W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ) + b 2 ( x i 1 ) b 2 ( x ) c 2 ( y i 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t ] } 1 λ ,
(2.11)

where W j (u)= u ˜ j u d z ω ( z 1 λ ) , r 1 (x,y), r i (x,y) and f ˜ n (x,y,s,t) are given in Corollary  2.1.

3 Proof of Theorem 2.1

Obviously, for any (x,y)Ω, r 1 (x,y) is positive and nondecreasing with respect to x and y, f ˜ n (x,y,s,t) (n=1,2) is nonnegative and nondecreasing with respect to x and y for each fixed s and t. They satisfy a(x,y) r 1 (x,y) and f n (x,y,s,t) f ˜ n (x,y,s,t).

We first consider (x,y) Ω 11 ={(x,y): x 0 x< x 1 , y 0 y< y 1 } and have from (1.8)

u ( x , y ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t r 1 ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f ˜ n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t .
(3.1)

Take any fixed x ˜ ( x 0 , x 1 ), y ˜ ( y 0 , y 1 ), and for arbitrary x[ x 0 , x ˜ ], y[ y 0 , y ˜ ] we have

u(x,y) r 1 ( x ˜ , y ˜ )+ n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f ˜ n ( x ˜ , y ˜ ,s,t) ω n ( u ( s , t ) ) dsdt.
(3.2)

Let

z(x,y)= r 1 ( x ˜ , y ˜ )+ n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f ˜ n ( x ˜ , y ˜ ,s,t) ω n ( u ( s , t ) ) dsdt
(3.3)

and z( x 0 ,y)= r 1 ( x ˜ , y ˜ ). Hence, u(x,y)z(x,y). Clearly, z(x,y) is a nonnegative, nondecreasing and differentiable function for x[ x 0 , x ˜ ] and y[ y 0 , y ˜ ]. Moreover, b n (x) (or c n (y)) is differentiable and nondecreasing in x[ x 0 , x ˜ ] (or y[ y 0 , y ˜ ]) for n=1,2. Thus, b n (x)0 (or c n (y)0) for x[ x 0 , x ˜ ] (or y[ y 0 , y ˜ ]). Since r 1 ( x ˜ , y ˜ )>0 and ω 1 (z(x,y))>0, we have

D 1 z ( x , y ) ω 1 ( z ( x , y ) ) c 1 ( y o ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) ω 1 ( u ( b 1 ( x ) , t ) ) d t ω 1 ( z ( x , y ) ) + c 2 ( y o ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) ω 2 ( u ( b 2 ( x ) , t ) ) d t ω 1 ( z ( x , y ) ) c 1 ( y o ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) ω 1 ( z ( b 1 ( x ) , t ) ) d t ω 1 ( z ( x , y ) ) + c 2 ( y o ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) ω 2 ( z ( b 2 ( x ) , t ) ) d t ω 1 ( z ( x , y ) ) c 1 ( y o ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) ω 1 ( z ( x , t ) ) d t ω 1 ( z ( x , y ) ) + c 2 ( y o ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) ω 2 ( z ( b 2 ( x ) , t ) ) d t ω 1 ( z ( x , y ) ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) d t + c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) ω 2 ( z ( b 2 ( x ) , t ) ) ω 1 ( z ( b 2 ( x ) , t ) ) d t .
(3.4)

Integrating both sides of the above inequality from x 0 to x, we obtain

W 1 ( z ( x , y ) ) W 1 ( z ( x 0 , y ) ) x 0 x c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( s ) , t ) b 1 ( s ) d s d t + x 0 x c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( s ) , t ) b 2 ( s ) ω 2 ( z ( b 2 ( s ) , t ) ) ω 1 ( z ( b 2 ( s ) , t ) ) d s d t .

Thus,

W 1 ( z ( x , y ) ) W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t + b 2 ( x 0 ) b 2 ( x ) c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , s , t ) ϕ ( z ( s , t ) ) d s d t

for x 0 x x ˜ and y 0 y y ˜ , where ϕ(u)= ω 2 ( u ) ω 1 ( u ) , or equivalently

ξ ( x , y ) W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t + b 2 ( x 0 ) b 2 ( x ) c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , s , t ) ϕ ( W 1 1 ( ξ ( s , t ) ) ) d s d t z 1 ( x , y ) ,
(3.5)

where

ξ(x,y)= W 1 ( z ( x , y ) ) .

It is easy to check that ξ(x,y) z 1 (x,y), z 1 ( x 0 ,y)= W 1 ( r 1 ( x ˜ , y ˜ )) and z 1 (x,y) is differentiable, positive and nondecreasing on ( x 0 , x ˜ ] and ( y 0 , y ˜ ]. Since ϕ( W 1 1 (u)) is nondecreasing from the assumption (H3), we have

D 1 z 1 ( x , y ) ϕ ( W 1 1 ( z 1 ( x , y ) ) ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) d t ϕ ( W 1 1 ( z 1 ( x , y ) ) ) + c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) ϕ ( W 1 1 ( ξ ( b 2 ( x ) , t ) ) ) d t ϕ ( W 1 1 ( z 1 ( x , y ) ) ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) d t ϕ [ W 1 1 ( W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t ) ] + c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) ϕ ( W 1 1 ( z 1 ( x , t ) ) ) d t ϕ ( W 1 1 ( z 1 ( x , y ) ) ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( x ) , t ) b 1 ( x ) d t ϕ [ W 1 1 ( W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t ) ] + c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( x ) , t ) b 2 ( x ) d t .
(3.6)

Note that

x 0 x D 1 z 1 ( s , y ) ϕ ( W 1 1 ( z 1 ( s , y ) ) ) d s = x 0 x D 1 z 1 ( s , y ) ω 1 ( W 1 1 ( z 1 ( s , y ) ) ) ω 2 ( W 1 1 ( z 1 ( s , y ) ) ) d s = W 1 1 ( z 1 ( x 0 , y ) ) W 1 1 ( z 1 ( x , y ) ) d u ω 2 ( u ) = W 2 W 1 1 ( z 1 ( x , y ) ) W 2 W 1 1 ( z 1 ( x 0 , y ) ) = W 2 W 1 1 ( z 1 ( x , y ) ) W 2 W 1 1 ( W 1 ( r 1 ( x ˜ , y ˜ ) ) ) = W 2 W 1 1 ( z 1 ( x , y ) ) W 2 ( r 1 ( x ˜ , y ˜ ) ) .

Integrating both sides of the inequality (3.6) from x 0 to x, we obtain

W 2 W 1 1 ( z 1 ( x , y ) ) W 2 ( r 1 ( x ˜ , y ˜ ) ) = x 0 x D 1 z 1 ( s , y ) ϕ ( W 1 1 ( z 1 ( s , y ) ) ) d s x 0 x c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , b 1 ( s ) , t ) b 1 ( s ) d t ϕ [ W 1 1 ( W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( s ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , τ , t ) d τ d t ) ] d s + x 0 x c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , b 2 ( s ) , t ) b 2 ( s ) d s d t W 2 W 1 1 [ W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t ] W 2 ( r 1 ( x ˜ , y ˜ ) ) + b 2 ( x 0 ) b 2 ( x ) c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , s , t ) d s d t .

Thus,

W 2 W 1 1 ( z 1 ( x , y ) ) W 2 W 1 1 [ W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t ] + b 2 ( x 0 ) b 2 ( x ) c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , s , t ) d s d t .

Hence

u ( x , y ) z ( x , y ) W 1 1 ( ξ ( x , y ) ) W 1 1 ( z 1 ( x , y ) ) W 2 1 [ W 2 W 1 1 ( W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t ) + b 2 ( x 0 ) b 2 ( x ) c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x ˜ , y ˜ , s , t ) d s d t ] .

Since the above inequality is true for any x[ x 0 , x ˜ ], y[ y 0 , y ˜ ], we obtain

u ( x ˜ , y ˜ ) W 2 1 [ W 2 W 1 1 ( W 1 ( r 1 ( x ˜ , y ˜ ) ) + b 1 ( x 0 ) b 1 ( x ˜ ) c 1 ( y 0 ) c 1 ( y ˜ ) f ˜ 1 ( x ˜ , y ˜ , s , t ) d s d t ) + b 2 ( x 0 ) b 2 ( x ˜ ) c 2 ( y 0 ) c 2 ( y ˜ ) f ˜ 2 ( x ˜ , y ˜ , s , t ) d s d t ] .
(3.7)

Replacing x ˜ by x and y ˜ by y yields

u ( x , y ) W 2 1 [ W 2 W 1 1 ( W 1 ( r 1 ( x , y ) ) + b 1 ( x 0 ) b 1 ( x ) c 1 ( y 0 ) c 1 ( y ) f ˜ 1 ( x , y , s , t ) d s d t ) + b 2 ( x 0 ) b 2 ( x ) c 2 ( y 0 ) c 2 ( y ) f ˜ 2 ( x , y , s , t ) d s d t ] .
(3.8)

This means that (2.1) is true for (x,y) Ω 11 and i=1 if replace u(x,y) with u 1 (x,y).

For i=2 and (x,y) Ω 22 ={(x,y): x 1 x< x 2 , y 1 y< y 2 }, (1.8) becomes

u ( x , y ) r 1 ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x 1 ) c n ( y o ) c n ( y 1 ) f n ( x , y , s , t ) ω n ( u 1 ( s , t ) ) d s d t + g ( x , y ) β 1 u 1 m ( x 1 0 , y 1 0 ) + n = 1 2 b n ( x 1 ) b n ( x ) c n ( y 1 ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t r 2 ( x , y ) + n = 1 2 b n ( x 1 ) b n ( x ) c n ( y 1 ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t ,
(3.9)

where the definition of r 2 (x,y) is given in (2.2). Note that the estimate of u 1 (x,y) is known. Clearly, (3.9) is the same as (3.1) if replace r 1 (x,y) and ( x 0 , y 0 ) by r 2 (x,y) and ( x 1 , y 1 ). Thus, by (3.8) we have

u ( x , y ) W 2 1 [ W 2 W 1 1 ( W 1 ( r 2 ( x , y ) ) + b 1 ( x 1 ) b 1 ( x ) c 1 ( y 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ) + b 2 ( x 1 ) b 2 ( x ) c 2 ( y 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t ] .

This implies that (2.1) is true for (x,y) Ω 22 and i=2 if replace u(x,y) by u 2 (x,y).

Assume that (2.1) is true for (x,y) Ω i i ={(x,y): x i 1 x< x i , y i 1 y< y i }, i.e.,

u i ( x , y ) W 2 1 [ W 2 W 1 1 ( W 1 ( r i ( x , y ) ) + b 1 ( x i 1 ) b 1 ( x ) c 1 ( y i 1 ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ) + b 2 ( x i 1 ) b 2 ( x ) c 2 ( y i 1 ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t ] .
(3.10)

For (x,y) Ω i + 1 , i + 1 ={(x,y): x i x< x i + 1 , y i y< y i + 1 }, (1.8) becomes

u ( x , y ) a ( x , y ) + n = 1 2 b n ( x 0 ) b n ( x ) c n ( y o ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t + g ( x , y ) ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i u m ( x i 0 , y i 0 ) r 1 ( x , y ) + k = 1 i n = 1 2 b n ( x k 1 ) b n ( x k ) c n ( y k 1 ) c n ( y k ) f n ( x , y , s , t ) ω n ( u k ( s , t ) ) d s d t + g ( x , y ) k = 1 i β k u k m ( x k 0 , y k 0 ) + n = 1 2 b n ( x i ) b n ( x ) c n ( y i ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t r i + 1 ( x , y ) + n = 1 2 b n ( x i ) b n ( x ) c n ( y i ) c n ( y ) f n ( x , y , s , t ) ω n ( u ( s , t ) ) d s d t ,
(3.11)

where we use the fact that the estimate of u(x,y) is already known for (x,y) Ω i i (i=1,2,). Again, (3.11) is the same as (3.1) if replace r 1 (x,y) and ( x 0 , y 0 ) by r i + 1 (x,y) and ( x i , y i ). Thus, by (3.8) we have

u ( x , y ) W 2 1 [ W 2 W 1 1 ( W 1 ( r i + 1 ( x , y ) ) + b 1 ( x i ) b 1 ( x ) c 1 ( y i ) c 1 ( y ) f 1 ˜ ( x , y , s , t ) d s d t ) + b 2 ( x i ) b 2 ( x ) c 2 ( y i ) c 2 ( y ) f 2 ˜ ( x , y , s , t ) d s d t ] .

This shows that (2.1) is true for (x,y) Ω i + 1 , i + 1 if replace u(x,y) by u i + 1 (x,y). By induction, we know that (2.1) holds for (x,y) Ω i + 1 , i + 1 for any nonnegative integer i. This completes the proof of Theorem 2.1.

4 Applications

Consider the following partial differential equation with an impulsive term:

{ 2 v ( x , y ) x y = H ( x , y , v ( x , y ) ) , ( x , y ) Ω i i , x x i , y y i , v | x = x i , y = y i = I i ( v ) , v ( x , y 0 ) = ϕ 1 ( x ) , v ( x 0 , y ) = ϕ 2 ( y ) , ϕ 1 ( x 0 ) = ϕ 2 ( y 0 ) 0 ,
(4.1)

where vR, HR, I i R, and i=1,2, .

Assume that

(C1) |H(x,y,v(x,y))| h 1 (x,y) e | v ( x , y ) | + h 2 (x,y) e 2 | v ( x , y ) | where h 1 , h 2 are nonnegative and continuous on Ω, h 1 (x,y)=0, h 2 (x,y)=0 for (x,y) Ω i j , ij, i,j=1,2, ;

(C2) | I i (v)| β i | v | m where β i and m are nonnegative constants.

Corollary 4.1 Suppose that (C1) and (C2) hold. If we let v i (x,y)=v(x,y) for (x,y) Ω i i , then the solution of system (4.1) has an estimate for (x,y) Ω i i

| v i ( x , y ) | 1 2 ln [ ( e r i ( x , y ) x i 1 x y i 1 y h 1 ( s , t ) d s d t ) 2 x i 1 x y i 1 y h 2 ( s , t ) d s d t ] ,
(4.2)

where

r 1 ( x , y ) = max x 0 ξ x , y 0 η y | ϕ 1 ( ξ ) + ϕ 2 ( η ) ϕ 1 ( x 0 ) | > 0 , r i ( x , y ) = r 1 ( x , y ) + k = 1 i 1 x k 1 x k y k 1 y k h 1 ( s , t ) e | v k ( s , t ) | d s d t + k = 1 i 1 x k 1 x k y k 1 y k h 2 ( s , t ) e 2 | v k ( s , t ) | d s d t + k = 1 i 1 β k | v k ( x k 0 , y k 0 ) | m , ( e r i ( x , y ) x i 1 x y i 1 y h 1 ( s , t ) d s d t ) 2 x i 1 x y i 1 y h 2 ( s , t ) d s d t > 0 .

Proof The solution of (4.1) with an initial value is given by

v ( x , y ) = v ( x , y 0 ) + v ( x 0 , y ) v ( x 0 , y 0 ) + x 0 x y 0 y H ( s , t , v ( s , t ) ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) I i ( v ( x i 0 , y i 0 ) ) = ϕ 1 ( x ) + ϕ 2 ( y ) ϕ 1 ( x 0 ) + x 0 x y 0 y H ( s , t , v ( s , t ) ) d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) I i ( v ( x i 0 , y i 0 ) ) ,
(4.3)

which implies

| v ( x , y ) | | ϕ 1 ( x ) + ϕ 2 ( y ) ϕ 1 ( x 0 ) | + x 0 x y 0 y h 1 ( s , t ) e | v ( s , t ) | d s d t + x 0 x y 0 y h 2 ( s , t ) e 2 | v ( s , t ) | d s d t + ( x 0 , y 0 ) < ( x i , y i ) < ( x , y ) β i | v ( x i 0 , y i 0 ) | m .
(4.4)

Let

u ( x , y ) = | v ( x , y ) | , a ( x , y ) = | ϕ 1 ( x ) + ϕ 2 ( y ) ϕ 1 ( x 0 ) | , b 1 ( x ) = b 2 ( x ) = x , c 1 ( y ) = c 2 ( y ) = y , g ( x , y ) = 1 , ω 1 ( u ) = e u , ω 2 ( u ) = e 2 u , f 1 ( x , y , s , t ) = h 1 ( s , t ) , f 2 ( x , y , s , t ) = h 2 ( s , t ) .

Thus, (4.4) is the same as (1.8). It is easy to see that for any positive constants u ˜ 1 and u ˜ 2

r 1 ( x , y ) = max x 0 ξ x , y 0 η y | a ( ξ , η ) | > 0 , f ˜ 1 ( x , y , s , t ) = h 1 ( s , t ) , f ˜ 2 ( x , y , s , t ) = h 2 ( s , t ) , W 1 ( u ) = u ˜ 1 u d z ω 1 ( z ) = u ˜ 1 u e z d z = e u ˜ 1 e u , W 1 1 ( u ) = ln ( e u ˜ 1 u ) , W 2 ( u ) = u ˜ 2 u d z ω 2 ( z ) = u ˜ 2 u e 2 z d z = 1 2 ( e 2 u ˜ 2 e 2 u ) , W 2 1 ( u ) = 1 2 ln ( e 2 u ˜ 2 2 u ) , r i ( x , y ) = r 1 ( x , y ) + k = 1 i 1 x k 1 x k y k 1 y k h 1 ( s , t ) e | v k ( s , t ) | d s d t + k = 1 i 1 x k 1 x k y k 1 y k h 2 ( s , t ) e 2 | v k ( s , t ) | d s d t + k = 1 i 1 β k | v k ( x k 0 , y k 0 ) | m .

Therefore, for any nonnegative i and (x,y) Ω i i

| v i ( x , y ) | 1 2 ln [ ( e r i ( x , y ) x i 1 x y i 1 y h 1 ( s , t ) d s d t ) 2 x i 1 x y i 1 y h 2 ( s , t ) d s d t ]

provided that

( e r i ( x , y ) x i 1 x y i 1 y h 1 ( s , t ) d s d t ) 2 x i 1 x y i 1 y h 2 (s,t)dsdt>0.

 □

Remark 4.1 From (4.4), we know ω 1 (u)= e u . Clearly, ω 1 (2u)= e 2 u ω 1 (2) ω 1 (u)= e 2 e u does not hold for large u>0. Thus, ω 1 (u)= e u does not belong to the class in [24]. Again ω 1 ( u 2 )= e u 2 1 2 ω 1 (u)= 1 2 e u does not hold for large u>0 so ω 1 (u) does not belong to the class ȷ in [24]. Hence, the results in [24] cannot be applied to the inequality (4.1).

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Acknowledgements

The author is grateful to Dr. Shengfu Deng for many helpful conversations during the course of this work. This research was supported by National Natural Science Foundation of China (No. 11371314), Guangdong Natural Science Foundation (No. S2013010015957), and the Project of Department of Education of Guangdong Province, China (No. 2012KJCX0074).

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Mi, Y. Generalized integral inequalities for discontinuous functions with two independent variables and their applications. J Inequal Appl 2014, 524 (2014). https://doi.org/10.1186/1029-242X-2014-524

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Keywords

  • integral inequalities
  • discontinuous functions
  • impulse
  • nonlinear
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