New Explicit Bounds on Gamidov Type Integral Inequalities for Functions in Two Variables and Their Applications

Abstract

Some linear and nonlinear Gamidov type integral inequalities in two variables are established, which can give the explicit bounds on the solutions to a class of Volterra-Fredholm integral equations. Some examples of application are presented to show boundedness and uniqueness of solutions of a Volterra-Fredholm type integral equation.

1. Introduction

Integral inequalities that give explicit bounds on unknown functions provide a very useful and important device in the study of many qualitative as well as quantitative properties of solutions of differential and integral equations. During the past few years, an enormous amount of effort has been devoted to the discovery of new types of inequalities and their applications in various branches of ordinary and partial differential and integral equations; see [1–14] and the references given therein.

In particular, Gamidov [6], while studying the boundary value problem for higher order differential equations, initiated the study of obtaining explicit upper bounds on the integral inequalities of the forms 𝑢(𝑡)≤𝑐+^𝑡_𝛼𝑓(𝑠)𝑢(𝑠)𝑑𝑠+^𝛽_𝛼𝑔(𝑠)𝑢(𝑠)𝑑𝑠, (1) for 𝑡∈[𝛼,𝛽], under some suitable conditions on the functions involved in (1). Later, such inequalities (usually called Gamidov type inequalities) are also discussed by Bainov and Simeonov [15] and Pachpatte [16, 17]. In [16], Pachpatte established more general Gamidov inequalities as follows: 𝑢(𝑡)≤𝑐+^𝑡_𝛼𝑎(𝑡,𝑠)𝑢(𝑠)𝑑𝑠+^𝛽_𝛼𝑏(𝑡,𝑠)𝑢(𝑠)𝑑𝑠 (2) and also generalized it to a general version in two independent variables [18] as follows: 𝑢𝑥,𝑦≤𝑐+^𝑥_𝑥0^𝑦_𝑦0𝑎𝑥,𝑦,𝑠,𝑡𝑢(𝑠)𝑑𝑠𝑑𝑡+^𝑀_𝑥0^𝑁_𝑦0𝑏𝑥,𝑦,𝑠,𝑡𝑢(𝑠)𝑑𝑠𝑑𝑡. (3) Based on the results of inequalities (3), the retarded cases with power nonlinearity of inequalities (3) are studied by Ma and Pečarić [19]. As a matter of fact, the results of inequality (2) are not right. In the proof of Theorem 1 in [16] about inequalities (2), by definition 𝑧(𝛼)=𝑐+∫^𝛽_𝛼𝑏(𝑡,𝑠)𝑢(𝑠)𝑑𝑠 and inequalities 𝑢(𝑡)≤𝑧(𝛼)exp(∫^𝑡_𝛼𝑎(𝑡,𝜎)𝑑𝜎), they obtained 𝑧(𝛼)≤𝑐/(1−𝑝(𝑡)) (for details, see [16, page 355]). However, we can see easily that 𝑧(𝛼) is a function in 𝑡, while it was treated as a constant in [16].

In this paper, we discuss some linear and nonlinear Gamidov type integral inequalities and obtain new explicit bounds on these inequalities. Moreover, we also give some examples to show boundedness and uniqueness of solutions of a Volterra-Fredholm type integral equation.

2. Linear Gamidov Type Integral Inequalities

In what follows, 𝑅 denotes the set of real numbers, 𝑅₊=[0,∞), 𝑅₀=(0,∞), 𝐼₁=[0,𝑀], and 𝐼₂=[0,𝑁] are given subsets of 𝑅. Let Δ=𝐼₁×𝐼₂. 𝐶(𝑈,𝑉) denotes the collection of continuous functions from 𝑈 to 𝑉.

Lemma 1. Assume 𝑢(𝑥,𝑦), 𝑎(𝑥,𝑦), 𝑐(𝑥,𝑦), 𝑔(𝑥,𝑦)∈𝐶(Δ,𝑅₊) and 𝑢𝑥,𝑦≤𝑎𝑥,𝑦+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡, (4) for (𝑥,𝑦)∈Δ. If ∫^𝑀₀∫^𝑁₀𝑐(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡<1, then the following explicit estimate 𝑢+𝑐∫𝑥,𝑦≤𝑎𝑥,𝑦𝑥,𝑦^𝑀₀∫^𝑁₀𝑎(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡1−∫^𝑀₀∫^𝑁₀𝑐(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡 (5) holds for (𝑥,𝑦)∈Δ.

Proof. Obviously, ∫^𝑀₀∫^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡 is a constant. Letting ∫^𝑀₀∫^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡=𝑘, from (4), we have 𝑢𝑥,𝑦≤𝑎𝑥,𝑦+𝑐𝑥,𝑦𝑘. (6) Since 𝑔(𝑥,𝑦) is positive, then 𝑔𝑥,𝑦𝑢𝑥,𝑦≤𝑎𝑥,𝑦𝑔𝑥,𝑦+𝑐𝑥,𝑦𝑔𝑥,𝑦𝑘. (7) Integrating for both sides of (7) on Δ, we have 𝑘=^𝑀₀^𝑁₀𝑔()𝑢()≤𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡_𝑀0𝑁0𝑎(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑘^𝑀₀^𝑁₀𝑐(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡. (8) Consequently, it follows from the inequality above that ∫𝑘≤^𝑀0∫^𝑁₀𝑎(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡1−∫^𝑀₀∫^𝑁₀𝑐()𝑔(),𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡 (9) where ∫^𝑀₀∫^𝑁₀𝑐(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡<1. Substituting the inequality above into (6), we get the explicit estimate (5) for 𝑢(𝑥,𝑦).

Theorem 2. Assume 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦), 𝑓(𝑥,𝑦), 𝑔(𝑥,𝑦)∈𝐶(Δ,𝑅₊), and 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦) are nondecreasing in 𝑥 and 𝑦. If 𝑢(𝑥,𝑦)∈𝐶(Δ,𝑅₊) satisfies 𝑢𝑥,𝑦≤𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡, (10) then, under the condition that ^𝑀₀^𝑁₀𝑐(𝑠,𝑡)𝑔(𝑠,𝑡)exp𝑏(𝑠,𝑡)^𝑠₀^𝑡₀𝑓(𝑢,𝑣)𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡<1, (11) the following explicit estimate 𝑢≤𝑎×𝑥,𝑦𝑥,𝑦+𝑐𝑥,𝑦_𝑀0𝑁0𝑎()𝑔()𝑏()𝑠,𝑡𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝑓()×𝑢,𝑣𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀𝑐()𝑔()𝑏()𝑠,𝑡𝑠,𝑡×exp𝑠,𝑡×^𝑠₀^𝑡₀𝑓(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹×exp𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑑𝑠𝑑𝑡 (12) holds for (𝑥,𝑦)∈Δ.

Proof. Fixing any arbitrary (𝑋,𝑌)∈Δ, then for (𝑥,𝑦)∈Δ₁=[0,𝑋]×[0,𝑌], from (10), we have 𝑢𝑥,𝑦≤𝑎(𝑋,𝑌)+𝑏(𝑋,𝑌)^𝑥₀^𝑦₀𝑓()𝑢()()𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡+𝑐𝑋,𝑌^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡, (13) where we apply that 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), and 𝑐(𝑥,𝑦) are nondecreasing in 𝑥 and 𝑦.
Define a function 𝑣(𝑥,𝑦), (𝑥,𝑦)∈Δ₁ by the right side of (13). Then, 𝑣(𝑥,𝑦) is positive and nondecreasing in 𝑥 and 𝑦, and 𝑢𝑥,𝑦≤𝑣𝑥,𝑦. (14) Furthermore, we have 𝑣0,𝑦=𝑎(𝑋,𝑌)+𝑐(𝑋,𝑌)^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡,𝜕𝑣()𝜕𝑥𝑥,𝑦=𝑏𝑋,𝑌^𝑦₀𝑓()𝑢()()𝑥,𝑡𝑥,𝑡𝑑𝑡≤𝑏𝑋,𝑌^𝑦₀𝑓()𝑣()𝑏()𝑥,𝑡𝑥,𝑡𝑑𝑡≤𝑣𝑥,𝑦𝑋,𝑌^𝑦₀𝑓(𝑥,𝑡)𝑑𝑡. (15)(16) Since 𝑣(𝑥,𝑦) is nondecreasing in 𝑦, from (16), one gets (𝜕/𝜕𝑥)𝑣𝑥,𝑦𝑣()𝑥,𝑦≤𝑏𝑋,𝑌^𝑦₀𝑓(𝑥,𝑡)𝑑𝑡. (17) Keeping 𝑦 fixed in (17), setting 𝑥=𝑠, and integrating it from 0 to 𝑥, we get 𝑣𝑥,𝑦≤𝑣0,𝑦exp𝑏(𝑋,𝑌)^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑑𝑠𝑑𝑡. (18) It follows from (14) and (15) that 𝑢≤𝑎()()𝑥,𝑦𝑋,𝑌+𝑐𝑋,𝑌^𝑀₀^𝑁₀𝑔()𝑢()𝑏()𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡×exp𝑋,𝑌^𝑥₀^𝑦₀𝑓()()𝑏()𝑠,𝑡𝑑𝑠𝑑𝑡=𝑎𝑋,𝑌exp𝑋,𝑌^𝑥₀^𝑦₀𝑓()()𝑏()𝑠,𝑡𝑑𝑠𝑑𝑡+𝑐𝑋,𝑌exp𝑋,𝑌^𝑥₀^𝑦₀𝑓()×𝑠,𝑡𝑑𝑠𝑑𝑡_𝑀0𝑁0𝑔()𝑢()=𝐴+𝐶𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡𝑥,𝑦;𝑋,𝑌𝑥,𝑦;𝑋,𝑌^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡, (19) where 𝐴𝑥,𝑦;𝑋,𝑌=𝑎(𝑋,𝑌)exp𝑏(𝑋,𝑌)^𝑥₀^𝑦₀𝑓(),𝐶()𝑏()𝑠,𝑡𝑑𝑠𝑑𝑡𝑥,𝑦;𝑋,𝑌=𝑐𝑋,𝑌exp𝑋,𝑌^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑑𝑠𝑑𝑡. (20) Now, a suitable application of Lemma 1 yields 𝑢≤𝐴+𝐶×∫𝑥,𝑦𝑥,𝑦;𝑋,𝑌𝑥,𝑦;𝑋,𝑌_𝑀0∫^𝑁₀𝐴(𝑠,𝑡;𝑋,𝑌)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡1−∫^𝑀₀∫^𝑁₀𝐶()𝑔().𝑠,𝑡;𝑋,𝑌𝑠,𝑡𝑑𝑠𝑑𝑡 (21) Since the inequality above holds for all (𝑥,𝑦)∈Δ₁, taking 𝑥=𝑋 and 𝑦=𝑌 in (21), we have 𝑢()≤𝐴()+𝐶()×∫𝑋,𝑌𝑋,𝑌;𝑋,𝑌𝑋,𝑌;𝑋,𝑌_𝑀0∫^𝑁₀𝐴(𝑠,𝑡;𝑋,𝑌)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡1−∫^𝑀₀∫^𝑁₀𝐶()𝑔()=𝐴()+𝐶()×∫𝑠,𝑡;𝑋,𝑌𝑠,𝑡𝑑𝑠𝑑𝑡𝑋,𝑌𝑋,𝑌_𝑀0∫^𝑁₀𝐴(𝑠,𝑡)𝑔(𝑠,𝑡)𝑑𝑠𝑑𝑡1−∫^𝑀₀∫^𝑁₀𝐶()𝑔(),𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡 (22) for (𝑋,𝑌)∈Δ, where 𝐴(𝑋,𝑌)=𝑎(𝑋,𝑌)exp𝑏(𝑋,𝑌)^𝑋₀^𝑌₀𝑓(),𝐶()()𝑏()𝑠,𝑡𝑑𝑠𝑑𝑡𝑋,𝑌=𝑐𝑋,𝑌exp𝑋,𝑌^𝑋₀^𝑌₀𝑓(𝑠,𝑡)𝑑𝑠𝑑𝑡. (23) Substituting (23) into (22) and simplifying it, we get 𝑢()≤𝑎()()×𝑋,𝑌𝑋,𝑌+𝑐𝑋,𝑌_𝑀0𝑁0𝑎()𝑔()𝑏()𝑠,𝑡𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝑓()×𝑢,𝑣𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀𝑐()𝑔()𝑏()𝑠,𝑡𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝑓(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹×exp𝑏(𝑋,𝑌)^𝑋₀^𝑌₀𝑓(𝑠,𝑡)𝑑𝑠𝑑𝑡. (24) Considering that 𝑋 and 𝑌 are arbitrary, we replace 𝑋 and 𝑌 by 𝑥 and 𝑦, respectively, and get the desired estimate (12). The proof is complete.

Remark 3. If 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), and 𝑐(𝑥,𝑦) are not nondecreasing, our result also holds, since we can replace it by 𝑎(𝑥,𝑦)=max_{0≤𝑠≤𝑥,0≤𝑡≤𝑦}𝑎(𝑠,𝑡), 𝑏̃(𝑥,𝑦)=max_{0≤𝑠≤𝑥,0≤𝑡≤𝑦}𝑏(𝑠,𝑡), and 𝑐(̃𝑥,𝑦)=max_{0≤𝑠≤𝑥,0≤𝑡≤𝑦}𝑐(𝑠,𝑡).

Remark 4. As for inequality (3) discussed by Pachpatte [18] or a more general case as follows: 𝑢𝑥,𝑦≤𝑎𝑥,𝑦+^𝑥₀^𝑦₀𝑓𝑢(𝑠)+𝑥,𝑦,𝑠,𝑡𝑑𝑠𝑑𝑡_𝑀0𝑁0𝑔𝑥,𝑦,𝑠,𝑡𝑢(𝑠)𝑑𝑠𝑑𝑡, (25) there are some technical problems which we do not discuss here. Clearly, inequalities (4) are the special case of inequalities (25); that is, 𝑓(𝑥,𝑦,𝑠,𝑡)=𝑏(𝑥,𝑦)𝑓(𝑠,𝑡) and 𝑔(𝑥,𝑦,𝑠,𝑡)=𝑐(𝑥,𝑦)𝑔(𝑠,𝑡).

3. Nonlinear Gamidov Type Integral Inequalities

In this section, we discuss some integral inequalities with some power nonlinear terms.

Lemma 5 (see [20]). Let 𝑎≥0, 𝑚≥𝑛≥0, and 𝑚≠0. Then 𝑎^𝑛/𝑚≤𝑛𝑚𝐾^{(𝑛−𝑚)/𝑚}𝑎+𝑚−𝑛𝑚𝐾^𝑛/𝑚 (26) for any 𝐾>0.

Theorem 6. Assume that 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦), 𝑓(𝑥,𝑦), and 𝑔(𝑥,𝑦) are defined as in Theorem 2. If 𝑢(𝑥,𝑦)∈𝐶(Δ,𝑅₊) satisfies 𝑢^𝑚𝑥,𝑦≤𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑢^𝑛(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢^𝑟(𝑠,𝑡)𝑑𝑠𝑑𝑡, (27) where 𝑚≥𝑛≥0, 𝑚≥𝑟≥0, and 𝑚, 𝑛, and 𝑟 are constants, then, under the condition that ^𝑀₀^𝑁₀𝐶(𝑠,𝑡)𝐺(𝑠,𝑡)exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡<1, (28) the following explicit estimate 𝑢≤𝑎+𝑥,𝑦𝑥,𝑦𝐴𝑥,𝑦+𝐶×𝑥,𝑦_𝑀0𝑁0𝐴(𝑠,𝑡)𝐺(𝑠,𝑡)×exp𝐵()×𝑠,𝑡_𝑠0𝑡0𝐹()×𝑢,𝑣𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀𝐶(𝑠,𝑡)𝐺(𝑠,𝑡)×exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹×exp𝐵𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡)𝑑𝑠𝑑𝑡^1/𝑚 (29) holds for (𝑥,𝑦)∈Δ, where 𝐴𝑥,𝑦=𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓()×𝑛𝑠,𝑡𝑚𝐾₁^{(𝑛−𝑚)/𝑚}()𝑎()+𝑠,𝑡𝑠,𝑡𝑚−𝑛𝑚𝐾₁^𝑛/𝑚(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔()×𝑟𝑠,𝑡𝑚𝐾₂^{(𝑟−𝑚)/𝑚}()𝑎()+𝑠,𝑡𝑠,𝑡𝑚−𝑟𝑚𝐾₂^𝑟/𝑚(𝑠,𝑡)𝑑𝑠𝑑𝑡,𝐵=𝑛𝑥,𝑦𝑚𝑏𝑥,𝑦,𝐶=𝑛𝑥,𝑦𝑚𝑐𝑥,𝑦,𝐹𝑥,𝑦=𝑓𝑥,𝑦𝐾₁^{(𝑛−𝑚)/𝑚}𝑥,𝑦,𝐺𝑥,𝑦=𝑔𝑥,𝑦𝐾₂^{(𝑟−𝑚)/𝑚}𝑥,𝑦 (30) for any 𝐾_𝑖(𝑥,𝑦)∈𝐶(Δ,𝑅₀)(𝑖=1,2).

Proof. Define a function 𝑤(𝑥,𝑦) by 𝑤𝑥,𝑦=𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑢^𝑛(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢^𝑟(𝑠,𝑡)𝑑𝑠𝑑𝑡, (31) for (𝑥,𝑦)∈Δ. Then, from (27), we have 𝑢^𝑚𝑥,𝑦≤𝑎𝑥,𝑦+𝑤𝑥,𝑦 (32) or 𝑢𝑥,𝑦≤𝑎𝑥,𝑦+𝑤𝑥,𝑦^1/𝑚. (33)
Applying Lemma 5, for any 𝐾_𝑖(𝑥,𝑦)∈𝐶(Δ,𝑅₀)(𝑖=1,2), we get 𝑢^𝑛𝑥,𝑦≤𝑎𝑥,𝑦+𝑤𝑥,𝑦^𝑛/𝑚≤𝑛𝑚𝐾₁^{(𝑛−𝑚)/𝑚}𝑎+𝑥,𝑦𝑥,𝑦+𝑤𝑥,𝑦𝑚−𝑛𝑚𝐾₁^𝑛/𝑚,𝑢𝑥,𝑦_𝑟𝑥,𝑦≤𝑎𝑥,𝑦+𝑤𝑥,𝑦^𝑟/𝑚≤𝑟𝑚𝐾₂^{(𝑟−𝑚)/𝑚}𝑎+𝑥,𝑦𝑥,𝑦+𝑤𝑥,𝑦𝑚−𝑟𝑚𝐾₂^𝑟/𝑚𝑥,𝑦. (34) Substituting (34) into (31), we get 𝑤𝑥,𝑦≤𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓()×𝑛𝑠,𝑡𝑚𝐾₁^{(𝑛−𝑚)/𝑚}(𝑎()(+𝑠,𝑡)(𝑠,𝑡+𝑤𝑠,𝑡))𝑚−𝑛𝑚𝐾₁^𝑛/𝑚(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔()×𝑟𝑠,𝑡𝑚𝐾₂^{(𝑟−𝑚)/𝑚}(𝑎()(+𝑠,𝑡)(𝑠,𝑡+𝑤𝑠,𝑡))𝑚−𝑟𝑚𝐾₂^𝑟/𝑚(𝑠,𝑡)𝑑𝑠𝑑𝑡=𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑛𝑠,𝑡)𝑚𝐾₁^{(𝑛−𝑚)/𝑚}()𝑎()+𝑠,𝑡𝑠,𝑡𝑚−𝑛𝑚𝐾₁^𝑛/𝑚(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑟𝑠,𝑡)𝑚𝐾₂^{(𝑟−𝑚)/𝑚}()𝑎()+𝑠,𝑡𝑠,𝑡𝑚−𝑟𝑚𝐾₂^𝑟/𝑚(+𝑛𝑠,𝑡)𝑑𝑠𝑑𝑡𝑚𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝐾₁^{(𝑛−𝑚)/𝑚}()𝑤()+𝑟𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡𝑚𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝐾₂^{(𝑟−𝑚)/𝑚}()𝑤()=𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡𝐴𝑥,𝑦+𝐵𝑥,𝑦^𝑥₀^𝑦₀𝐹()𝑤()+𝑠,𝑡𝑠,𝑡𝑑𝑠𝑑𝑡𝐶𝑥,𝑦^𝑀₀^𝑁₀𝐺(𝑠,𝑡)𝑤(𝑠,𝑡)𝑑𝑠𝑑𝑡, (35) which is similar to inequality (10), where 𝐴(𝑥,𝑦), 𝐵(𝑥,𝑦), 𝐶(𝑥,𝑦), 𝐹(𝑥,𝑦), and 𝐺(𝑥,𝑦) are defined as in (30). It is easy to see that 𝐴(𝑥,𝑦), 𝐵(𝑥,𝑦), and 𝐶(𝑥,𝑦) are nonnegative, continuous, and nondecreasing for (𝑥,𝑦)∈Δ.
A suitable application of Theorem 2 to (35) yields 𝑤≤𝑥,𝑦𝐴𝑥,𝑦+𝐶×𝑥,𝑦_𝑀0𝑁0𝐴(𝑠,𝑡)𝐺(𝑠,𝑡)×exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹()×𝑢,𝑣𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀𝐶(𝑠,𝑡)𝐺(𝑠,𝑡)×exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹×exp𝐵𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡)𝑑𝑠𝑑𝑡. (36) Substituting (36) into (33), we can get (29).

Remark 7. In [19], Ma and Pečarić discussed the retarded Gamidov type integral inequality: 𝑢^𝑚𝑥,𝑦≤𝑎𝑥,𝑦+₀^𝛼(𝑥)₀^𝛽(𝑦)𝑓𝑥,𝑦,𝑠,𝑡𝑢^𝑛()+𝑠,𝑡𝑑𝑠𝑑𝑡_{0𝛼(𝑀)0𝛽(𝑁)}𝑔𝑥,𝑦,𝑠,𝑡𝑢^𝑟(𝑠,𝑡)𝑑𝑠𝑑𝑡. (37) As mentioned above, the results of inequality (3) obtained by Pachpatte [18] are not right. So the results in [19] are not valid.

When 𝑚=2,  𝑛=𝑟=1 in Theorem 6, an Ou-Iang version of Gamidov type inequality is obtained as follows.

Corollary 8. Assume that 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦), 𝑓(𝑥,𝑦), and 𝑔(𝑥,𝑦) are defined as in Theorem 2. If 𝑢(𝑥,𝑦)∈𝐶(Δ,𝑅₊) satisfies 𝑢²𝑥,𝑦≤𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡, (38) Then, under the condition that ^𝑀₀^𝑁₀12𝑐(𝑠,𝑡)𝐺()1𝑠,𝑡×exp2𝑏(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡<1, (39) the following explicit estimate 𝑢≤𝑎+𝑥,𝑦𝑥,𝑦𝐴+1𝑥,𝑦2𝑐×𝑥,𝑦_𝑀0𝑁0𝐴(𝑠,𝑡)𝐺()1𝑠,𝑡×exp2𝑏(𝑠,𝑡)^𝑠₀^𝑡₀𝐹()×𝑢,𝑣𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀12𝑐(𝑠,𝑡)𝐺()1𝑠,𝑡×exp2𝑏(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹1×exp2𝑏𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡)𝑑𝑠𝑑𝑡^1/2, (40) holds for (𝑥,𝑦)∈Δ, where 𝐴𝑥,𝑦=𝑏𝑥,𝑦^𝑥₀^𝑦₀12𝑓()×𝐾𝑠,𝑡_1−1/2(𝑠,𝑡)𝑎(𝑠,𝑡)+𝐾₁^1/2(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀12𝑔()×𝐾𝑠,𝑡_2−1/2(𝑠,𝑡)𝑎(𝑠,𝑡)+𝐾₂^1/2(𝑠,𝑡)𝑑𝑠𝑑𝑡,𝐹𝑥,𝑦=𝑓𝑥,𝑦𝐾₁^−1/2𝑥,𝑦,𝐺𝑥,𝑦=𝑔𝑥,𝑦𝐾₂^−1/2𝑥,𝑦 (41) for any 𝐾_𝑖(𝑥,𝑦)∈𝐶(Δ,𝑅₀)(𝑖=1,2).

When 𝑚=𝑛=1, 0≤𝑟≤1 in Theorem 6, we can get an interesting result as follows.

Corollary 9. Assume that 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦), 𝑓(𝑥,𝑦), and 𝑔(𝑥,𝑦) are defined as in Theorem 2. If 𝑢(𝑥,𝑦)∈𝐶(Δ,𝑅₊) satisfies 𝑢𝑥,𝑦≤𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑢(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)𝑢^𝑟(𝑠,𝑡)𝑑𝑠𝑑𝑡, (42) then, under the condition that 𝑟𝑐(^𝑀₀^𝑁₀𝑠,𝑡)𝐺()𝑏()𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡<1, (43) the following explicit estimate 𝑢≤𝑎+𝑥,𝑦𝑥,𝑦𝐴×𝑥,𝑦+𝑐𝑥,𝑦_𝑀0𝑁0𝐴(𝑠,𝑡)𝐺()𝑏()𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝐹()×𝑢,𝑣𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀𝑟𝑐(𝑠,𝑡)𝐺()𝑏()𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹×exp𝑏𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡)𝑑𝑠𝑑𝑡^1/2, (44) holds for (𝑥,𝑦)∈Δ, where 𝐴𝑥,𝑦=𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)𝑎(𝑠,𝑡)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔()×𝑠,𝑡𝑟𝐾^𝑟−1()𝑎()+()𝐾𝑠,𝑡𝑠,𝑡1−𝑟^𝑟(𝑠,𝑡)𝑑𝑠𝑑𝑡,𝐺𝑥,𝑦=𝑔𝑥,𝑦𝐾^𝑟−1𝑥,𝑦 (45) for any 𝐾(𝑥,𝑦)∈𝐶(Δ,𝑅₀).

4. Applications

In this section, we present some applications of our results to study boundedness and uniqueness of solutions of a Volterra-Fredholm integral equation in two independent variables.

Considering the following integral equation: 𝑧^𝑚𝑥,𝑦=𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡,𝑧)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝐺(𝑠,𝑡,𝑧)𝑑𝑠𝑑𝑡, (46) for (𝑥,𝑦)∈Δ, where 𝑧(𝑥,𝑦)∈𝐶(Δ,𝑅), 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦)∈𝐶(Δ,𝑅₊), 𝐹(𝑥,𝑦,𝑧), 𝐺(𝑥,𝑦,𝑧)∈𝐶(Δ×𝑅,𝑅), and 𝑚>1 is a constant.

Theorem 10. Assume that the functions 𝐹 and 𝐺 in (46) satisfy the conditions |𝐹(𝑠,𝑡,𝑧)|≤𝑓(𝑠,𝑡)|𝑧|^𝑛,|𝐺(𝑠,𝑡,𝑧)|≤𝑔(𝑠,𝑡)|𝑧|^𝑟, (47) where 𝑓(𝑠,𝑡) and 𝑔(𝑠,𝑡) are defined as in Theorem 2, 𝑛 and 𝑟 satisfying 0<𝑛≤𝑚, 0<𝑟≤𝑚 are constants. If ^𝑀₀^𝑁₀𝐶(𝑠,𝑡)𝐺(𝑠,𝑡)exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡<1, (48) and 𝑧(𝑥,𝑦) is a solution of (46) and (47), then 𝑧≤𝑎+𝑥,𝑦𝑥,𝑦𝐴𝑥,𝑦+𝐶×𝑥,𝑦_𝑀0𝑁0𝐴(𝑠,𝑡)𝐺(𝑠,𝑡)×exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹()×𝑢,𝑣𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡1−^𝑀₀^𝑁₀𝐶(𝑠,𝑡)𝐺(𝑠,𝑡)×exp𝐵(𝑠,𝑡)^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣×𝑑𝑠𝑑𝑡⁻¹×exp𝐵𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡)𝑑𝑠𝑑𝑡^1/𝑚, (49) for (𝑥,𝑦)∈Δ and any 𝐾_𝑖(𝑥,𝑦)∈𝐶(Δ,𝑅₀)(𝑖=1,2), where 𝐴(𝑥,𝑦), 𝐵(𝑥,𝑦), 𝐶(𝑥,𝑦), 𝐹(𝑥,𝑦), and 𝐺(𝑥,𝑦) are defined as in Theorem 6.

Proof. Letting 𝑧(𝑥,𝑦) be a solution of (46), from (47), we have 𝑧𝑥,𝑦^𝑚≤𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)|𝑧(𝑠,𝑡)|^𝑛𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)|𝑧(𝑠,𝑡)|^𝑟𝑑𝑠𝑑𝑡. (50) An application of the inequality given in Theorem 6 to (50) yields (49). The right-hand side of (49) gives us the bound on the solution 𝑧(𝑥,𝑦) of (46) in terms of the unknown functions.
Considering the following integral equation: 𝑧𝑥,𝑦=𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡,𝑧)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝐺(𝑠,𝑡,𝑧)𝑑𝑠𝑑𝑡, (51) for (𝑥,𝑦)∈Δ, where 𝑧(𝑥,𝑦)∈𝐶(Δ,𝑅), 𝑎(𝑥,𝑦), 𝑏(𝑥,𝑦), 𝑐(𝑥,𝑦)∈𝐶(Δ,𝑅₊), 𝐹(𝑥,𝑦,𝑧), 𝐺(𝑥,𝑦,𝑧)∈𝐶(Δ×𝑅,𝑅).

Theorem 11. Assume that the functions 𝐹 and 𝐺 in (46) satisfy the conditions |𝐹()(𝑧(𝑧|,|𝐺()(𝑧(𝑧|𝑠,𝑡,𝑧−𝐹𝑠,𝑡,)|≤𝑓𝑠,𝑡)|𝑧−𝑠,𝑡,𝑧−𝐺𝑠,𝑡,)|≤𝑔𝑠,𝑡)|𝑧−^𝑟, (52) where 𝑓(𝑠,𝑡) and 𝑔(𝑠,𝑡) are defined as in Theorem 2, 0<𝑟≤1 is a constant. If 𝑟𝑐(^𝑀₀^𝑁₀𝑠,𝑡)𝐺()𝑏()𝑠,𝑡×exp𝑠,𝑡^𝑠₀^𝑡₀𝐹(𝑢,𝑣)𝑑𝑢𝑑𝑣𝑑𝑠𝑑𝑡<1, (53) where 𝐹 and 𝐺 are defined as in Corollary 9 and 𝑧(𝑥,𝑦) is a solution of (51) and (52), then (51) has at most one solution.

Proof. Let 𝑧(𝑥,𝑦) and 𝑧(𝑥,𝑦) be two solutions of (51); that is, 𝑧𝑥,𝑦=𝑎𝑥,𝑦+𝑏𝑥,𝑦^𝑥₀^𝑦₀𝐹(𝑠,𝑡,𝑧)𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝐺(𝑠,𝑡,𝑧)𝑑𝑠𝑑𝑡. (54) From (51) and (54), we have 𝑧𝑥,𝑦−𝑧𝑥,𝑦≤𝑏𝑥,𝑦^𝑥₀^𝑦₀|𝐹(𝑠,𝑡,𝑧)−𝐹(𝑠,𝑡,𝑧)|𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀|𝐺(𝑠,𝑡,𝑧)−𝐺(𝑠,𝑡,𝑧)|𝑑𝑠𝑑𝑡≤𝑏𝑥,𝑦^𝑥₀^𝑦₀𝑓(𝑠,𝑡)|𝑧−𝑧|𝑑𝑠𝑑𝑡+𝑐𝑥,𝑦^𝑀₀^𝑁₀𝑔(𝑠,𝑡)|𝑧−𝑧|^𝑟𝑑𝑠𝑑𝑡. (55) According to Corollary 9, we obtain that |𝑧(𝑥,𝑦)−𝑧(𝑥,𝑦)|≤0, which implies 𝑧(𝑥,𝑦)=𝑧(𝑥,𝑦) for (𝑥,𝑦)∈Δ.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are very grateful to the referees for their helpful comments and valuable suggestions. This work is supported by the Doctoral Program Research Funds of Southwest University of Science and Technology (no. 11zx7129) and the Fundamental Research Funds for the Central Universities (no. skqy201324).