Abstract

Theorems on the unique solvability and nonnegativity of solutions to the characteristic initial value problem 𝑢(1,1)(𝑡,𝑥)=0(𝑢)(𝑡,𝑥)+1(𝑢(1,0))(𝑡,𝑥)+2(𝑢(0,1))(𝑡,𝑥)+𝑞(𝑡,𝑥),𝑢(𝑡,𝑐)=𝛼(𝑡) for 𝑡[𝑎,𝑏],𝑢(𝑎,𝑥)=𝛽(𝑥)for𝑥[𝑐,𝑑] given on the rectangle [𝑎,𝑏]×[𝑐,𝑑] are established, where the linear operators 0, 1, 2 map suitable function spaces into the space of essentially bounded functions. General results are applied to the hyperbolic equations with essentially bounded coefficients and argument deviations.

1. Introduction

On the rectangle 𝒟=[𝑎,𝑏]×[𝑐,𝑑], we consider the linear partial functional-differential equation𝑢(1,1)(𝑡,𝑥)=0(𝑢)(𝑡,𝑥)+1𝑢(1,0)(𝑡,𝑥)+2𝑢(0,1)(𝑡,𝑥)+𝑞(𝑡,𝑥),(1.1) where 𝑢(1,0) and 𝑢(0,1) (resp., 𝑢(1,1)) denote the first-order (resp., the second-order mixed) partial derivatives. The operators 0, 1, and 2 are supposed to be linear and acting from suitable function spaces (see Section 3) to the space of Lebesgue measurable and essentially bounded functions. By a solution to (1.1), we mean a function 𝑢𝒟 absolutely continuous in the sense of Carathéodory possessing some additional properties (namely, inclusions (2.20)) which satisfies equality (1.1) almost everywhere on 𝒟.

Three main initial value problems for the hyperbolic equations are studied in the literature—Darboux, Cauchy, and Goursat problems. In this paper, we consider the Darboux problem in which case the values of a solution 𝑢 to (1.1) are prescribed on both characteristics 𝑡=𝑎 and 𝑥=𝑐, that is, the initial conditions are[][].𝑢(𝑡,𝑐)=𝛼(𝑡)for𝑡𝑎,𝑏,𝑢(𝑎,𝑥)=𝛽(𝑥)for𝑥𝑐,𝑑(1.2) Properties of the initial functions 𝛼 and 𝛽 will be specified in Section 3. It is worth to remember here that various initial and boundary value problems for the hyperbolic equation 𝑢𝑡𝑥=𝑓𝑡,𝑥,𝑢,𝑢𝑡,𝑢𝑥(1.3) with continuous as well as discontinuous right-hand sides but without argument deviations have been studied in detail (see, e.g., [113] and references therein). As for the hyperbolic functional-differential equations, we can mention for example the works [1416] (see also references cited therein) but, as far as the authors know, there is still a broad field for further investigation. We have made the first steps in the papers [17, 18] where the Darboux problem for (1.1) with 1=0 and 2=0 is considered.

2. Notation and Definitions

The following notation is used throughout the paper.(i), , and are the sets of all natural, rational, and real numbers, respectively, +=[0,+[.(ii)𝒟=[𝑎,𝑏]×[𝑐,𝑑], where <𝑎<𝑏<+ and <𝑐<𝑑<+.(iii)The first-order partial derivatives of the function 𝑢𝒟 at the point (𝑡,𝑥)𝒟 are denoted by 𝑢(1,0)(𝑡,𝑥) (or 𝑢𝑡(𝑡,𝑥)) and 𝑢(0,1)(𝑡,𝑥) (or 𝑢𝑥(𝑡,𝑥)). The second-order mixed partial derivatives of the function 𝑢𝒟 at the point (𝑡,𝑥)𝒟 are denoted by 𝑢𝑡𝑥(𝑡,𝑥) and 𝑢𝑥𝑡(𝑡,𝑥) whereas we use 𝑢(1,1)(𝑡,𝑥) if 𝑢𝑡𝑥(𝑡,𝑥)=𝑢𝑥𝑡(𝑡,𝑥).(iv)𝐶(𝒟;) is the Banach space of continuous functions 𝑢𝒟 equipped with the norm 𝑢𝐶||𝑢||=max(𝑡,𝑥)(𝑡,𝑥)𝒟.(2.1)(v)𝐶([𝛼,𝛽];), where <𝛼<𝛽<+, is the linear space of continuous functions 𝑣[𝛼,𝛽].(vi)𝐴𝐶([𝛼,𝛽];), where <𝛼<𝛽<+, is the linear space of absolutely continuous functions 𝑣[𝛼,𝛽].(vii)𝐿(𝒟;) is the Banach space of Lebesgue measurable and essentially bounded functions 𝑝𝒟 equipped with the norm 𝑝𝐿||𝑝||=esssup(𝑡,𝑥)(𝑡,𝑥)𝒟.(2.2)(viii)𝐿(𝒟;+)={𝑝𝐿(𝒟;)𝑝(𝑡,𝑥)0fora.e.(𝑡,𝑥)𝒟}.(ix)For any 𝑧1,𝑧2𝐿(𝒟;), we put 𝑧2𝑧1𝑧2(𝑡,𝑥)𝑧1𝑧(𝑡,𝑥)0fora.e.(𝑡,𝑥)𝒟,2𝑧1𝑧2(𝑡,𝑥)𝑧1(𝑡,𝑥)𝜀fora.e.(𝑡,𝑥)𝒟withsome𝜀>0.(2.3)(x)𝐿([𝛼,𝛽];), where <𝛼<𝛽<+, is the linear space of Lebesgue measurable and essentially bounded functions 𝑓[𝛼,𝛽].(xi)meas 𝐴 denotes the Lebesgue measure of the set 𝐴𝑅𝑚, 𝑚=1,2.(xii)If 𝑋, 𝑌 are Banach spaces and 𝑇𝑋𝑌 is a linear bounded operator then 𝑇 denotes the norm of the operator 𝑇, that is, 𝑇=sup𝑇(𝑧)𝑌𝑧𝑋,𝑧𝑋1.(2.4)

Two subsections below contain a number of definitions used in the sequel.

2.1. Spaces 𝑍[1](𝒟;), 𝑍[2](𝒟;), and Set 𝐶(𝒟;)

Motivated by [19, Section 2], the authors introduce the following assertions and definitions.

Lemma 2.1 (see [19, Section 1, Lemma 1]). Let the function 𝑢𝒟 be such that [][],[][].𝑢(,𝑥)𝑎,𝑏iscontinuousfora.e.𝑥𝑐,𝑑𝑢(𝑡,)𝑐,𝑑ismeasurableforall𝑡𝑎,𝑏(2.5) Then the function max{|𝑢(𝑡,)|𝑡[𝑎,𝑏]}[𝑐,𝑑] is measurable.

Notation 1. 𝑍[1](𝒟;) denotes the linear space of all functions 𝑢𝒟 satisfying conditions (2.5), and ||𝑢||[][]esssupmax(𝑡,𝑥)𝑡𝑎,𝑏𝑥𝑐,𝑑<+.(2.6) If one identifies functions 𝑢1, 𝑢2 from 𝑍[1](𝒟;) such that 𝑢1(,𝑥)𝑢2(,𝑥) for a.e. 𝑥[𝑐,𝑑] then 𝑢𝑍[1]||𝑢||[][]=esssupmax(𝑡,𝑥)𝑡𝑎,𝑏𝑥𝑐,𝑑(2.7) defines a norm in the space 𝑍[1](𝒟;).
Analogously, we introduce the space 𝑍[2](𝒟;) of functions which are “measurable in the first variable and continuous in the second one” and define the norm 𝑍[2] there.

The proof of the following proposition is similar to those presented in [19, Section 2, Lemma 1]. For the sake of completeness we prove the proposition here in detail.

Proposition 2.2. 𝑍[1](𝒟;) and 𝑍[2](𝒟;) are Banach spaces.

Proof. We only prove the assertion for the space 𝑍[1](𝒟;), the assertion of the lemma concerning the space 𝑍[2](𝒟;) can be proven analogously by exchanging the roles of the variables 𝑡 and 𝑥.
Let {𝑢𝑘}+𝑘=1 be an arbitrary Cauchy sequence in 𝑍[1](𝒟;). For a decreasing sequence of positive numbers {𝜀𝑖}+𝑖=1 with +𝑖=1𝜀𝑖<+ there exists an increasing sequence {𝑘𝑖}+𝑖=1 such that||𝑢esssupmax𝑛(𝑡,𝑥)𝑢𝑘||[][](𝑡,𝑥)𝑡𝑎,𝑏𝑥𝑐,𝑑<𝜀𝑖,(2.8) for every 𝑛,𝑘𝑘𝑖, 𝑖. Let 𝑣𝑖=𝑢𝑘𝑖(𝑖=1,2,). Then, for any 𝑖, there is a set 𝐸𝑖[𝑐,𝑑], meas 𝐸𝑖=𝑑𝑐, such that ||𝑣max𝑖+1(𝑡,𝑥)𝑣𝑖||[](𝑡,𝑥)𝑡𝑎,𝑏<𝜀𝑖for𝑥𝐸𝑖,𝑖.(2.9) Put 𝐸=+𝑖=1𝐸𝑖. Then, clearly, we have meas 𝐸=𝑑𝑐 and ||𝑣max𝑛(𝑡,𝑥)𝑣𝑘||[](𝑡,𝑥)𝑡𝑎,𝑏𝑛1𝑚=𝑘||𝑣max𝑚+1(𝑡,𝑥)𝑣𝑚||[](𝑡,𝑥)𝑡𝑎,𝑏+𝑚=𝑘𝜀𝑚for𝑥𝐸,𝑛>𝑘.(2.10) Consequently, for any fixed 𝑥𝐸, the sequence {𝑣𝑖(,𝑥)}+𝑖=1 converges uniformly on [𝑎,𝑏], say to 𝑢(,𝑥). Hence, {𝑣𝑖(𝑡,)}+𝑖=1 converges point-wise on 𝐸 to 𝑢(𝑡,) for every fixed 𝑡[𝑎,𝑏]. Therefore, the function 𝑢 satisfies conditions (2.5). Since 𝑢𝑘(𝑡,𝑥)𝑢(𝑡,𝑥)=𝑢𝑘(𝑡,𝑥)𝑢𝑘𝑖(𝑡,𝑥)+lim𝑛+𝑣𝑖(𝑡,𝑥)𝑣𝑛(𝑡,𝑥)=𝑢𝑘(𝑡,𝑥)𝑢𝑘𝑖(𝑡,𝑥)+lim𝑛+𝑛1𝑚=𝑖𝑣𝑚(𝑡,𝑥)𝑣𝑚+1(𝑡,𝑥)(2.11) holds for 𝑖,𝑘, all 𝑡[𝑎,𝑏] and a.e. 𝑥[𝑐,𝑑], in view of (2.8) and (2.9), we obtain 𝑢𝑘𝑢𝑍[1]𝜀𝑖++𝑚=𝑖𝜀𝑚for𝑘𝑘𝑖,𝑖.(2.12) Hence, 𝑢𝑍[1](𝒟;) and 𝑢𝑛𝑢 in 𝑍[1](𝒟;), that is, the space 𝑍[1](𝒟;) is complete.

For the investigation of hyperbolic differential equations with discontinuous right-hand side, the concept of a Carathéodory solution is usually used (see, e.g., [7, 10, 20, 21]), that is, solutions are considered in the class of absolutely continuous functions. One possible definition of absolute continuity of functions of two variables was given by Carathéodory in his monograph [22]. It is also known that such functions admit a certain integral representation. Following the concept mentioned, we introduce the following.

Notation 2. 𝐶(𝒟;) stands for the set of functions 𝑢𝒟 admitting the integral representation 𝑢(𝑡,𝑥)=𝑧+𝑡𝑎𝑓(𝑠)d𝑠+𝑥𝑐𝑔(𝜂)d𝜂+𝑡𝑎𝑥𝑐(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟,(2.13) where 𝑧, 𝑓𝐿([𝑎,𝑏];), 𝑔𝐿([𝑐,𝑑];), and 𝐿(𝒟;).

The next lemma on differentiating of an indefinite double integral plays a crucial role in our investigation.

Lemma 2.3 (see [23, Proposition 3.5]). Let 𝒟 be a Lebesgue integrable function and 𝑣(𝑡,𝑥)=𝑡𝑎𝑥𝑐(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟.(2.14) Then (1)there exists a set 𝐸[𝑎,𝑏] such that meas𝐸=𝑏𝑎 and 𝑣(1,0)(𝑡,𝑥)=𝑥𝑐[],(𝑡,𝜂)d𝜂for𝑡𝐸,𝑥𝑐,𝑑(2.15)(2)there exists a set 𝐹[𝑐,𝑑] such that meas𝐹=𝑑𝑐 and 𝑣(0,1)(𝑡,𝑥)=𝑡𝑎[](𝑠,𝑥)d𝑠for𝑡𝑎,𝑏and𝑥𝐹,(2.16)(3)there exists a set 𝐺𝒟 such that meas𝐺=(𝑏𝑎)(𝑑𝑐) and 𝑣(1,1)(𝑡,𝑥)=(𝑡,𝑥)for(𝑡,𝑥)𝐺.(2.17)

Remark 2.4. If 𝑢𝐶(𝒟;), that is, the function 𝑢 admits integral representation (2.13), then by using Lemma 2.3 we get 𝑢(1,0)(𝑡,𝑥)=𝑓(𝑡)+𝑥𝑐[][],𝑢(𝑡,𝜂)d𝜂fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)(𝑡,𝑥)=𝑔(𝑥)+𝑡𝑎[][],𝑢(𝑠,𝑥)d𝑠forall𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(1,1)(𝑡,𝑥)=(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟.(2.18) Consequently, for any 𝑢𝐶(𝒟;), we have 𝑢(1,0)𝑍[2](𝒟;),𝑢(0,1)𝑍[1](𝒟;),𝑢(1,1)𝐿(𝒟;).(2.19)

Remark 2.5. It follows from Remark 2.4 and [22, Satz 1, page 654] that 𝑢𝐶(𝒟;) if and only if 𝑢𝒟 is absolutely continuous in the sense of Carathéodory with the properties 𝑢(1,0)(,𝑐)𝐿([]𝑎,𝑏;),𝑢(0,1)(𝑎,)𝐿([]𝑐,𝑑;),𝑢(1,1)𝐿(𝒟;).(2.20)

2.2. Positive and Volterra-Type Operators

We recall here some definitions from the theory of linear operators. We start with the operators acting on the space 𝐶(𝒟;).

Definition 2.6. A linear operator 𝐶(𝒟;)𝐿(𝒟;) is said to be positive if the relation (𝑢)(𝑡,𝑥)0fora.e.(𝑡,𝑥)𝒟(2.21) holds whenever the function 𝑢𝐶(𝒟;) is such that 𝑢(𝑡,𝑥)0for(𝑡,𝑥)𝒟.(2.22)

Example 2.7. For any 𝑣𝐶(𝒟;), we put 0(𝑣)(𝑡,𝑥)=𝑝0𝜏(𝑡,𝑥)𝑣0(𝑡,𝑥),𝜇0(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟,(2.23) where 𝑝0𝐿(𝒟;) and 𝜏0𝒟[𝑎,𝑏], 𝜇0𝒟[𝑐,𝑑] are measurable functions. Then the operator 0𝐶(𝒟;)𝐿(𝒟;) is linear and bounded. Moreover, 0 is positive if and only if 𝑝0(𝑡,𝑥)0 for a.e. (𝑡,𝑥)𝒟.

Definition 2.8. A linear operator 𝐶(𝒟;)𝐿(𝒟;) is called (𝑎,𝑐)-Volterra operator if, for any (𝑡0,𝑥0)𝒟 and 𝑢𝐶(𝒟;) such that 𝑢(𝑡,𝑥)=0for(𝑡,𝑥)𝑎,𝑡0×𝑐,𝑥0,(2.24) we have (𝑢)(𝑡,𝑥)=0fora.e.(𝑡,𝑥)𝑎,𝑡0×𝑐,𝑥0.(2.25)

Remark 2.9. It can be shown by using Lemma 5.8 stated below that the operator 0 given by formula (2.23) is an (𝑎,𝑐)-Volterra one if and only if ||𝑝0||𝜏(𝑡,𝑥)0||𝑝(𝑡,𝑥)𝑡0fora.e.(𝑡,𝑥)𝒟,0(||𝜇𝑡,𝑥)0(𝑡,𝑥)𝑥0fora.e.(𝑡,𝑥)𝒟.(2.26)

Now we introduce analogous notions for linear operators defined on the spaces 𝑍[1](𝒟;) and 𝑍[2](𝒟;).

Definition 2.10. We say that a linear operator 𝑍[1](𝒟;)𝐿(𝒟;) (resp., 𝑍[2](𝒟;)𝐿(𝒟;)) is positive if relation (2.21) is satisfied for every function 𝑢𝑍[1](𝒟;) (resp., 𝑢𝑍[2](𝒟;)) such that [][][][]𝑢(𝑡,𝑥)0for𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(resp.,𝑢(𝑡,𝑥)0fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑).(2.27)

Example 2.11. For any 𝑣𝑍[2](𝒟;) (resp., 𝑣𝑍[1](𝒟;)), we put 1(𝑣)(𝑡,𝑥)=𝑝1(𝑡,𝑥)𝑣𝑡,𝜇1(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟,(2.28) respectively, 2(𝑣)(𝑡,𝑥)=𝑝2𝜏(𝑡,𝑥)𝑣2(𝑡,𝑥),𝑥fora.e.(𝑡,𝑥)𝒟,(2.29) where 𝑝1,𝑝2𝐿(𝒟;) and 𝜇1𝒟[𝑐,𝑑], 𝜏2𝒟[𝑎,𝑏] are measurable functions. Then the operators 1𝑍[2](𝒟;)𝐿(𝒟;) and 2𝑍[1](𝒟;)𝐿(𝒟;) are linear and bounded. Moreover, 1 (resp., 2) is positive if and only if 𝑝1(𝑡,𝑥)0 (resp., 𝑝2(𝑡,𝑥)0) for a.e. (𝑡,𝑥)𝒟.

Definition 2.12. We say that a linear operator 𝑍[1](𝒟;)𝐿(𝒟;) (resp., 𝑍[2](𝒟;)𝐿(𝒟;)) is an 𝑎-Volterra operator (resp., a 𝑐-Volterra operator) if, for any 𝑡0[𝑎,𝑏] (resp., 𝑥0[𝑐,𝑑]) and 𝑢𝑍[1](𝒟;) (resp., 𝑢𝑍[2](𝒟;)) such that 𝑢(𝑡,𝑥)=0for𝑡𝑎,𝑡0[][]anda.e.𝑥𝑐,𝑑resp.,𝑢(𝑡,𝑥)=0fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑥0,(2.30) we have (𝑢)(𝑡,𝑥)=0fora.e.(𝑡,𝑥)𝑎,𝑡0×[][]×𝑐,𝑑resp.,(𝑢)(𝑡,𝑥)=0fora.e.(𝑡,𝑥)𝑎,𝑏𝑐,𝑥0.(2.31)

Remark 2.13. One can show by using Lemma 5.9 (resp., Lemma 5.10) stated below that the operator 1 (resp., 2) given by formula (2.28) (resp., (2.29)) is a 𝑐-Volterra one (resp., an 𝑎-Volterra one) if and only if ||𝑝1||𝜇(𝑡,𝑥)1(𝑡,𝑥)𝑥0fora.e.(𝑡,𝑥)𝒟,(2.32) respectively, ||𝑝2||𝜏(𝑡,𝑥)2(𝑡,𝑥)𝑡0fora.e.(𝑡,𝑥)𝒟.(2.33)

3. Statement of Problem

On the rectangle 𝒟, we consider the linear nonhomogeneous Darboux problem (1.1), (1.2) in which 0𝐶(𝒟;)𝐿(𝒟;), 1𝑍[2](𝒟;)𝐿(𝒟;), and 2𝑍[1](𝒟;)𝐿(𝒟;) are linear bounded operators, 𝑞𝐿(𝒟;), and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) are such that 𝛼𝐿([𝑎,𝑏];),𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐). By a solution to problem (1.1), (1.2), we mean a function 𝑢𝐶(𝒟;) possessing property (1.2) and satisfying equality (1.1) almost everywhere on 𝒟. Let us mention that, in view of Remark 2.4, the definition of a solution to the problem considered is meaningful.

We are interested in question on the unique solvability of problem (1.1), (1.2), and nonnegativity of its solutions. Clearly, the second-order hyperbolic differential equation𝑢𝑡𝑥=𝑝0(𝑡,𝑥)𝑢+𝑝1(𝑡,𝑥)𝑢𝑡+𝑝2(𝑡,𝑥)𝑢𝑥+𝑞(𝑡,𝑥),(3.1) where 𝑝0,𝑝1,𝑝2,𝑞𝐿(𝒟;), is a particular case of (1.1). It follows from the results due to Deimling (see [20, 21]) that, among others, problem (3.1), (1.2) has a unique solution without any additional assumptions imposed on the coefficients 𝑝0, 𝑝1, and 𝑝2. We would like to get solvability conditions for general problem (1.1), (1.2) which conform to those well known for (3.1), (1.2).

The main results (namely, Theorems 4.1 and 4.4) will be illustrated on the hyperbolic differential equation with argument deviations𝑢(1,1)(𝑡,𝑥)=𝑝0𝜏(𝑡,𝑥)𝑢0(𝑡,𝑥),𝜇0(𝑡,𝑥)+𝑝1(𝑡,𝑥)𝑢(1,0)𝑡,𝜇1(𝑡,𝑥)+𝑝2(𝑡,𝑥)𝑢(0,1)𝜏2(𝑡,𝑥),𝑥+𝑞(𝑡,𝑥),(3.2) in which coefficients 𝑝0,𝑝1,𝑝2,𝑞𝐿(𝒟;) and argument deviations 𝜏0,𝜏2𝒟[𝑎,𝑏], 𝜇0,𝜇1𝒟[𝑐,𝑑] are measurable functions. We obtain this equation from (1.1) if the operators 0, 1, and 2 are defined by formulas (2.23), (2.28), and (2.29), respectively. Let us also mention that in the case, where𝜏𝑘(𝑡,𝑥)=𝑡,𝜇𝑗(𝑡,𝑥)=𝑥fora.e.(𝑡,𝑥)𝒟(𝑘=0,2,𝑗=0,1),(3.3) equation (3.2) takes form (3.1).

4. Main Results

At first, we put𝐴𝑘(𝑧)=𝑘𝜑𝑘(𝑧)for𝑧𝐿(𝒟;),𝑘=0,1,2,(4.1) where 𝜑0(𝑧)(𝑡,𝑥)=𝑡𝑎𝑥𝑐𝜑𝑧(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟,1(𝑧)(𝑡,𝑥)=𝑥𝑐[][],𝜑𝑧(𝑡,𝜂)d𝜂fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑2(𝑧)(𝑡,𝑥)=𝑡𝑎[][].𝑧(𝑠,𝑥)d𝑠for𝑡𝑎,𝑏,a.e.𝑥𝑐,𝑑(4.2)

Clearly, 𝜑0𝐿(𝒟;)𝐶(𝒟;), 𝜑1𝐿(𝒟;)𝑍[2](𝒟;), 𝜑2𝐿(𝒟;)𝑍[1](𝒟;) and thus the operators 𝐴0, 𝐴1, 𝐴2 mapping the space 𝐿(𝒟;) into itself are linear and bounded.

Theorem 4.1. Let 𝐴=𝐴0+𝐴1+𝐴2, where the operators 𝐴0, 𝐴1, 𝐴2 are defined by relations (4.1), (4.2). If the spectral radius of the operator 𝐴 is less than one then problem (1.1), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];),𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐).

Theorem 4.1 implies the following.

Corollary 4.2. If the inequality (𝑏𝑎)(𝑑𝑐)0+(𝑑𝑐)1+(𝑏𝑎)2<1(4.3) holds then problem (1.1), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];),𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐).

Remark 4.3. On the rectangle [𝑎,𝑏]×[𝑐,𝑑], we consider the equation 𝑢(1,1)(𝑡,𝑥)=𝑝0𝑢(𝑏,𝑑)+𝑝1𝑢(1,0)(𝑡,𝑑)+𝑝2𝑢(0,1)(𝑏,𝑥)(4.4) subjected to the initial conditions [][],𝑢(𝑡,𝑐)=0for𝑡𝑎,𝑏,𝑢(𝑎,𝑥)=0for𝑥𝑐,𝑑(4.5) where 𝑝0=𝑚0(𝑏𝑎)(𝑑𝑐),𝑝1=𝑚1𝑑𝑐,𝑝2=𝑚2.𝑏𝑎(4.6) Clearly (4.4) is a particular case of (1.1). If 𝑚0+𝑚1+𝑚2=1, then problem (4.4), (4.5) has the trivial solution 𝑢(𝑡,𝑥)0 and the nontrivial solution 𝑢(𝑡,𝑥)(𝑡𝑎)(𝑥𝑐). It justifies that the strict inequality (4.3) in the previous corollary is essential and cannot be replaced by the nonstrict one. On the other hand, it is worth to mention that the inequality indicated is very restrictive and thus it is far from being optimal for a wide class of equations (1.1).

If the operators 0, 1, and 2 on the right-hand side of (1.1) are positive then we can estimate the spectral radius of the operator 𝐴 by using the well-known results due to Krasnosel'skij and we thus obtain the following.

Theorem 4.4. Let the operators 0, 1, 2 be positive and 𝐴=𝐴0+𝐴1+𝐴2, where the operators 𝐴0, 𝐴1, 𝐴2 are defined by relations (4.1), (4.2). Then the following four assertions are equivalent. (1)There exists a function 𝑧0𝐿(𝒟;+) such that 𝑧0𝐴(𝑧0).(2)The spectral radius of the operator 𝐴 is less than one.(3)Problem (1.1), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];),𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];), 𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐).If, in addition, the initial functions 𝛼,𝛽and the forcing term 𝑞are such that 𝛼(𝑎)0,𝛼(𝑡)0,𝛽(𝑥)0,𝑞(𝑡,𝑥)0fora.e.(𝑡,𝑥)𝒟,(4.7) then the solution 𝑢to problem (1.1), (1.2) satisfies 𝑢𝑢(𝑡,𝑥)0for(𝑡,𝑥)𝒟,(1,0)[][],𝑢(𝑡,𝑥)0fora.e.𝑡a,𝑏andall𝑥𝑐,𝑑(0,1)[][].(𝑡,𝑥)0for𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(4.8)(4)There exists a function 𝛾𝐶(𝒟;) such that𝛾(1,1)0(𝛾)+1𝛾(1,0)+2𝛾(0,1),𝛾(4.9)𝛾(𝑎,𝑐)0,(4.10)(1,0)[](𝑡,𝑐)0fora.e.𝑡𝑎,𝑏,𝛾(0,1)[]𝛾(𝑎,𝑥)0fora.e.𝑥𝑐,𝑑,(4.11)(1,1)(𝑡,𝑥)0fora.e.(𝑡,𝑥)𝒟.(4.12)

For Volterra-type operators 0, 1, and 2, we derive from the previous theorem the following.

Corollary 4.5. Let 0, 1, and 2 be positive (𝑎,𝑐)-Volterra, 𝑐-Volterra, and 𝑎-Volterra operators, respectively, such that the inequalities 1(𝑦)(𝑡,𝑥)𝑦(𝑡)1(1)(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟(4.13)(here,1(𝑦) means 1(𝑦) in which 𝑦(𝑡,𝑥)=𝑦(𝑡) for 𝑎.𝑒.𝑡[𝑎,𝑏] and all 𝑥[𝑐,𝑑]) and 2(𝑧)(𝑡,𝑥)𝑧(𝑥)2(1)(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟(4.14)(by2(𝑧) we mean 2(𝑧), where 𝑧(𝑡,𝑥)=𝑧(𝑥)forall𝑡[𝑎,𝑏] and a.e. 𝑥[𝑐,𝑑]) hold for every 𝑦𝐿([𝑎,𝑏];) and 𝑧𝐿([𝑐,𝑑];).
Then problem (1.1), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];),𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐). If, in addition, the initial functions 𝛼, 𝛽 and the forcing term 𝑞 are such that relations (4.7) hold, then the solution 𝑢 to problem (1.1), (1.2) satisfies inequalities (4.8).

Following our previous results concerning the case, where 1=0 and 2=0 (see [18]), we can introduce the following.

Definition 4.6. Let 0𝐶(𝒟;)𝐿(𝒟;), 1𝑍[2](𝒟;)𝐿(𝒟;), and 2𝑍[1](𝒟;)𝐿(𝒟;). We say that the triplet (0,1,2) belongs to the set 𝒮𝑎𝑐 if the implication 𝑢𝐶𝑢(𝒟;),(1,1)(𝑡,𝑥)0(𝑢)(𝑡,𝑥)+1𝑢(1,0)(𝑡,𝑥)+2𝑢(0,1)𝑢(𝑡,𝑥),fora.e.(𝑡,𝑥)𝒟,𝑢(𝑎,𝑐)0,(1,0)[],𝑢(𝑡,𝑐)0fora.e.𝑡𝑎,𝑏(0,1)[],(𝑎,𝑥)0fora.e.𝑥𝑐,𝑑𝑢satises(4.8)(4.15) holds.

Remark 4.7. If (0,1,2)𝒮𝑎𝑐, we usually say that a certain theorem on differential inequalities holds for (1.1). It should be noted here that there is another terminology which says that a certain maximum principle holds for (1.1) if the inclusion (0,1,2)𝒮𝑎𝑐 is fulfilled.

Theorem 4.4 immediately yields the following.

Corollary 4.8. If one of assertions (1)–(4) stated in Theorem 4.4 holds then (0,1,2)𝒮ac.

Remark 4.9. The inclusion (0,1,2)𝒮𝑎𝑐 ensures that every solution 𝑢 to problem (1.1), (1.2) with (4.7) satisfies relations (4.8). However, we do not know whether this inclusion also guarantees the unique solvability of problem (1.1), (1.2) for arbitrary 𝑞, 𝛼, and 𝛽. Consequently, we cannot reverse the assertion of the previous corollary.
The reason lays in the question whether the Fredholm alternative holds for problem (1.1), (1.2) or not. In fact, we are not able to prove compactness of the operator 𝐴 appearing in Theorem 4.4 which plays a crucial role in the proofs of the Fredholm alternative for problem (1.1), (1.2) as well as a continuous dependence of its solutions on the initial data and parameters.

Now we apply general results to (3.2) with argument deviations in which coefficients 𝑝0,𝑝1,𝑝2,𝑞𝐿(𝒟;) and argument deviations 𝜏0,𝜏2𝒟[𝑎,𝑏], 𝜇0,𝜇1𝒟[𝑐,𝑑] are measurable functions.

As a consequence of Corollary 4.2 we obtain the following.

Corollary 4.10. If the inequality 𝑝(b𝑎)(𝑑𝑐)0𝐿𝑝+(𝑑𝑐)1𝐿𝑝+(𝑏𝑎)2𝐿<1(4.16) holds, then problem (3.2), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];), 𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐).

If the coefficients 𝑝0, 𝑝1, 𝑝2 in the previous corollary are non-negative then the assertion of the corollary follows also from implication (4)(3) of Theorem 4.4. More precisely, the following statement holds.

Corollary 4.11. Let 𝑝0,𝑝1,𝑝2𝐿(𝒟;+) and 𝑝esssup0𝜏(𝑡,𝑥)0𝜇(𝑡,𝑥)𝑎0(𝑡,𝑥)𝑐+𝑝1𝜇(𝑡,𝑥)1(𝑡,𝑥)𝑐+𝑝2(𝜏𝑡,𝑥)2(𝑡,𝑥)𝑎(𝑡,𝑥)𝒟<1.(4.17) Then problem (3.2), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];), 𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐). If, in addition, the initial functions 𝛼, 𝛽, and the forcing term 𝑞 are such that relations (4.7) hold, then the solution 𝑢 to problem (3.2), (1.2) satisfies inequalities (4.8).

Finally, Corollary 4.5 implies the following.

Corollary 4.12. Let 𝑝0,𝑝1,𝑝2𝐿(𝒟;+) and argument deviations 𝜏0, 𝜇0, 𝜇1, and 𝜏2 satisfy inequalities (2.26), (2.32), and (2.33). Then problem (3.2), (1.2) is uniquely solvable for arbitrary 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];), 𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐). If, in addition, the initial functions 𝛼, 𝛽 and the forcing term 𝑞 are such that relations (4.7) hold, then the solution 𝑢 to problem (3.2), (1.2) satisfies inequalities (4.8).

The assumptions of the previous corollary require, in fact, that (3.2) is delayed in all its deviating arguments. Observe that in the case, where (3.3) holds, the inequalities (2.26), (2.32), and (2.33) are satisfied trivially and Corollary 4.12 thus conform to the results well known for (3.1). The following statements show that the assertion of Corollary 4.12 remains true if the deviations 𝜏0, 𝜇0, 𝜇1, and 𝜏2 are not necessarily delays but the differences 𝜏𝑘(𝑡,𝑥)𝑡,𝜇𝑗(𝑡,𝑥)𝑥(𝑘=0,2,𝑗=0,1)(4.18) are small enough, that is, if (3.2) with deviating arguments is “close” to (3.1).

Corollary 4.13. Let 𝑝0,𝑝1,𝑝2𝐿(𝒟;+),𝑝𝑘0(𝑘=1,2), and esssup𝜏0𝑡(𝑡,𝑥)𝜇0𝑐(𝑡,𝑥)𝑝0(𝑠,𝜂)d𝜂d𝑠+𝑡𝑎𝜇0𝑥(𝑡,𝑥)𝑝0𝑝(𝑠,𝜂)d𝜂d𝑠+22𝐿𝜏0𝑝(𝑡,𝑥)𝑡+21𝐿𝜇0𝜔(𝑡,𝑥)𝑥(𝑡,𝑥)𝒟e11+ln𝜔,esssup𝑡𝑎𝜇1𝑥(𝑡,𝑥)𝑝0𝑝(𝑠,𝜂)d𝜂d𝑠+21𝐿𝜇1<𝜔(𝑡,𝑥)𝑥(𝑡,𝑥)𝒟e,esssup𝜏2𝑡(𝑡,𝑥)𝑥𝑐𝑝0𝑝(𝑠,𝜂)d𝜂d𝑠+22𝐿𝜏2<𝜔(𝑡,𝑥)𝑡(𝑡,𝑥)𝒟e,(4.19)where 𝑝𝜔=2min1𝐿,𝑝2𝐿𝑝0𝐿𝑝max{𝑏𝑎,𝑑𝑐}+2min1𝐿,𝑝2𝐿.(4.20) Then the assertion of Corollary 4.12 holds.

5. Auxiliary Statements and Proofs of Main Results

The proofs use several auxiliary statements given in the next subsection.

5.1. Auxiliary Statements

Remember that, for given operators 0, 1, and 2, the operators 𝐴0, 𝐴1, and 𝐴2 are defined by relations (4.1), (4.2). Moreover, having 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];), 𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];), 𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐), we put 𝑦=0(𝛼(𝑎)+𝛼+𝛽)+1𝛼+2𝛽+𝑞(5.1)(by0(𝛼(𝑎)+𝛼+𝛽) the authors understand 0(𝜎) in which 𝜎(𝑡,𝑥)=𝛼(𝑎)+𝛼(𝑡)+𝛽(𝑥) for (𝑡,𝑥)𝒟. Similarly, 1(𝛼)(resp.,2(𝛽)) means 1(𝛼0)(resp.,2(𝛽0)), where 𝛼0(𝑡,𝑥)=𝛼(𝑡) for a.e.𝑡[𝑎,𝑏] and all 𝑥[𝑐,𝑑](resp.,𝛽0(𝑡,𝑥)=𝛽(𝑥) for all 𝑡[𝑎,𝑏]anda.e.𝑥[𝑐,𝑑])). Clearly, 𝑦𝐿(𝒟;).

Lemma 5.1. If 𝑢 is a solution to problem (1.1), (1.2) then 𝑢(1,1) is a solution to the equation 𝐴𝑧=0+𝐴1+𝐴2(𝑧)+𝑦(5.2) in the space 𝐿(𝒟;), where the operators 𝐴0, 𝐴1, and 𝐴2 are defined by relations (4.1), (4.2) and the function 𝑦 is given by formula (5.1).
Conversely, if 𝑧 is a solution to (5.2) in the space 𝐿(𝒟;) with the operators 𝐴0, 𝐴1, and 𝐴2 defined by relations (4.1), (4.2) and the function 𝑦 given by formula (5.1), then 𝑢(𝑡,𝑥)=𝛼(𝑎)+𝛼(𝑡)+𝛽(𝑥)+𝑡𝑎𝑥𝑐𝑧(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟(5.3) is a solution to problem (1.1), (1.2).

Proof. If 𝑢 is a solution to problem (1.1), (1.2) then, by virtue of Remark 2.4, we get 𝑢(1,1)𝐿(𝒟;), 𝑢(𝑡,𝑥)=𝛼(𝑎)+𝛼(𝑡)+𝛽(𝑥)+𝑡𝑎𝑥𝑐𝑢(1,1)(𝑢𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟,(1,0)(𝑡,𝑥)=𝛼(𝑡)+𝑥𝑐𝑢(1,1)[][],𝑢(𝑡,𝜂)d𝜂fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)(𝑡,𝑥)=𝛽(𝑥)+𝑡𝑎𝑢(1,1)[][].(𝑠,𝑥)d𝑠forall𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(5.4) Consequently, (1.1) yields that 𝑢(1,1)=𝐴0+𝐴1+𝐴2𝑢(1,1)+𝑦,(5.5) where the operators 𝐴0, 𝐴1, and 𝐴2 are defined by relations (4.1), (4.2) and the function 𝑦 is given by formula (5.1).
Conversely, let 𝑧 be a solution to (5.2) in the space 𝐿(𝒟;) with the operators 𝐴0, 𝐴1, and 𝐴2 defined by relations (4.1), (4.2) and the function 𝑦 given by formula (5.1). Moreover, let the function 𝑢 be defined by relation (5.3), that is,𝑢(𝑡,𝑥)=𝛼(𝑎)+𝑡𝑎𝛼(𝑠)d𝑠+𝑥𝑐𝛽(𝜂)d𝜂+𝑡𝑎𝑥𝑐𝑧(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟.(5.6) Then the function 𝑢 belongs to the set 𝐶(𝒟;) and verifies initial conditions (1.2). Furthermore, by using Lemma 2.3, we get 𝑢(1,0)(𝑡,𝑥)=𝛼(𝑡)+𝑥𝑐[][],𝑢𝑧(𝑡,𝜂)d𝜂fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)(𝑡,𝑥)=𝛽(𝑥)+𝑡𝑎[][],𝑢𝑧(𝑠,𝑥)d𝑠forall𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(1,1)(𝑡,𝑥)=𝑧(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟.(5.7) Consequently, (5.2) implies that 𝑢 is also a solution to (1.1).

Now we recall some definitions from the theory of linear operators leaving invariant a cone in a Banach space (see, e.g., [24, 25] and references therein).

Definition 5.2. A nonempty closed set 𝐾 in a Banach space 𝑋 is called a cone if the following conditions are satisfied: (i)𝑥+𝑦𝐾 for all 𝑥,𝑦𝐾,(ii)𝜆𝑥𝐾 for all 𝑥𝐾 and an arbitrary 𝜆0,(iii)if 𝑥𝐾 and 𝑥𝐾 then 𝑥=0.

Remark 5.3. In the original terminology introduced by Kreĭn and Rutman [25], a set 𝐾 satisfying conditions (i) and (ii) of Definition 5.2 is called a linear semigroup.

Definition 5.4. We say that a cone 𝐾𝑋 is solid if its interior Int𝐾 is nonempty.

Remark 5.5. The presence of a cone 𝐾 in a Banach space 𝑋 allows one to introduce a natural partial ordering there. More precisely, two elements 𝑥1,𝑥2𝑋 are said to be in the relation 𝑥2𝐾𝑥1 if and only if they satisfy the inclusion 𝑥2𝑥1𝐾. If, moreover, 𝐾 is a solid cone then we write 𝑥2𝐾𝑥1 if and only if 𝑥2𝑥1Int𝐾.

Definition 5.6. A cone 𝐾𝑋 is said to be normal if there is a constant 𝑁0 such that, for every 𝑥,𝑦𝑋 with the property 0𝐾𝑥𝐾𝑦, the relation 𝑥𝑋𝑁𝑦𝑋 holds.

The proof of the main part of Theorem 4.4 is based on the following result.

Lemma 5.7 (see [24, Theorem 5.6]). Let 𝐾 be a normal and solid cone in a Banach space 𝑋 and the operator 𝐴𝑋𝑋 leave invariant the cone 𝐾, that is, 𝐴(𝐾)𝐾. If there exists a constant 𝛿>0 and an element 𝑥0Int𝐾 such that 𝛿𝑥0𝐴(𝑥0)Int𝐾, then the spectral radius of the operator 𝐴 is less than 𝛿.

Finally, we establish three lemmas dealing with Volterra type operators which we need to prove Corollary 4.5.

Lemma 5.8. Let 0𝐶(𝒟;)𝐿(𝒟;) be a positive (𝑎,𝑐)-Volterra operator. Then, for any function 𝛾𝐶(𝒟;) satisfying 𝛾𝑡1,𝑥1𝑡𝛾2,𝑥2for𝑎𝑡1𝑡2𝑏,𝑐𝑥1𝑥2𝑑,(5.8) one has 0(𝛾)(𝑡,𝑥)0(1)(𝑡,𝑥)𝛾(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟.(5.9)

Proof. We first show that, for any (𝑡,𝑥)]𝑎,𝑏]×]𝑐,𝑑], we have 0(𝛾)(𝑠,𝜂)0[]×[](1)(𝑠,𝜂)𝛾(𝑡,𝑥)fora.e.(𝑠,𝜂)𝑎,𝑡𝑐,𝑥.(5.10) Indeed, let (𝑡,𝑥)]𝑎,𝑏]×]𝑐,𝑑] be arbitrary but fixed. Put 𝛾0(𝑠,𝜂)=𝛾(min{𝑠,𝑡},min{𝜂,𝑥})for(𝑠,𝜂)𝒟.(5.11) Then, clearly 𝛾0𝐶(𝒟;), 𝛾0𝛾(𝑠,𝜂)𝛾(𝑡,𝑥)for(𝑠,𝜂)𝒟,0[]×[].(𝑠,𝜂)=𝛾(𝑠,𝜂)for(𝑠,𝜂)𝑎,𝑡𝑐,𝑥(5.12) Since the operator 0 is positive, we obtain 0𝛾0(𝑠,𝜂)0(𝛾(𝑡,𝑥))(𝑠,𝜂)=𝛾(𝑡,𝑥)0(1)(𝑠,𝜂)fora.e.(𝑠,𝜂)𝒟.(5.13) On the other hand, the operator 0 is supposed to be an (𝑎,𝑐)-Volterra one which guarantees the equality 0𝛾0(𝑠,𝜂)=0[]×[](𝛾)(𝑠,𝜂)fora.e.(𝑠,𝜂)𝑎,𝑡𝑐,𝑥,(5.14) and thus the desired relation (5.10) holds.
Now we put 𝑢(𝑡,𝑥)=𝑡𝑎𝑥𝑐0(𝛾)(𝑠,𝜂)d𝜂d𝑠,𝑣(𝑡,𝑥)=𝑡𝑎𝑥𝑐0(1)(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟.(5.15) It follows from Lemma 2.3 that there exists a set 𝐸1]𝑎,𝑏], meas 𝐸1=𝑏𝑎, such that 𝑢(1,0)(𝑡,𝑥)=𝑥𝑐0(𝛾)(𝑡,𝜂)d𝜂for𝑡𝐸1[],𝑣,𝑥𝑐,𝑑(1,0)(𝑡,𝑥)=𝑥𝑐0(1)(𝑡,𝜂)d𝜂for𝑡𝐸1[],,𝑥𝑐,𝑑(5.16) and, moreover, there is a set 𝐸𝐸1×]𝑐,𝑑], meas 𝐸=(𝑏𝑎)(𝑑𝑐), with the properties 𝑢(1,1)(𝑡,𝑥)=0(𝛾)(𝑡,𝑥),𝑣(1,1)(𝑡,𝑥)=0(1)(𝑡,𝑥)for(𝑡,𝑥)𝐸.(5.17)
Let (𝑡,𝑥)𝐸 be arbitrary but fixed. Then, relation (5.10) yields 1𝑘𝑡𝑡𝑥𝑥𝑘0(𝛾)(𝑠,𝜂)d𝜂d𝑠𝛾(𝑡,𝑥)𝑘𝑡𝑡𝑥𝑥𝑘0(1)(𝑠,𝜂)d𝜂d𝑠(5.18) for ]0,𝑡𝑎] and 𝑘]0,𝑥𝑐], whence we get 1𝑘𝑢(𝑡,𝑥)𝑢(𝑡,𝑥)𝑢(𝑡,𝑥𝑘)𝑢(𝑡,𝑥𝑘)𝛾(𝑡,𝑥)𝑘𝑣(𝑡,𝑥)𝑣(𝑡,𝑥)𝑣(𝑡,𝑥𝑘)𝑣(𝑡,𝑥𝑘),]]]].for0,𝑡𝑎,𝑘0,𝑥𝑐(5.19) For any 𝑘]0,𝑥𝑐] fixed, we pass to the limit 0+ in the latter inequality and thus, in view of equalities (5.16), we get 1𝑘𝑢(1,0)(𝑡,𝑥)𝑢(1,0)(𝑡,𝑥𝑘)𝛾(𝑡,𝑥)𝑘𝑣(1,0)(𝑡,𝑥)𝑣(1,0)(𝑡,𝑥𝑘),(5.20) for 𝑘]0,𝑥𝑐]. Now, letting 𝑘0+ in the previous relation and using equalities (5.17) give 0(𝛾)(𝑡,𝑥)=𝑢(1,1)(𝑡,𝑥)𝛾(𝑡,𝑥)𝑣(1,1)(𝑡,𝑥)=𝛾(𝑡,𝑥)0(1)(𝑡,𝑥).(5.21) That is, the desired inequality (5.9) holds because (𝑡,𝑥)𝐸 was arbitrary.

Lemma 5.9. Let 2𝑍[1](𝒟;)𝐿(𝒟;) be a positive 𝑎-Volterra operator such that inequality (4.14) holds for every 𝑧𝐿([𝑐,𝑑];). Then, for any function 𝛾𝑍[1](𝒟;) with the property 𝛾𝑡1𝑡,𝑥𝛾2,𝑥for𝑎𝑡1𝑡2[]𝑏anda.e.𝑥𝑐,𝑑,(5.22) one has 2(𝛾)(𝑡,𝑥)2(1)(𝑡,𝑥)𝛾(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟.(5.23)

Proof. Let 𝐸1[𝑐,𝑑], meas 𝐸1=𝑑𝑐, be a set such that, for any 𝑥𝐸1, we have 𝛾(,𝑥)𝐶([𝑎,𝑏];) and 𝛾𝑡1𝑡,𝑥𝛾2,𝑥for𝑎𝑡1𝑡2𝑏.(5.24) We first show that the relation 2(𝛾)(𝑠,𝑥)2[]×[](1)(𝑠,𝑥)𝛾(𝑡,𝑥)fora.e.(𝑠,𝑥)𝑎,𝑡𝑐,𝑑(5.25) holds for every 𝑡]𝑎,𝑏]. Indeed, let 𝑡]𝑎,𝑏] be arbitrary but fixed. Put 𝛾0[](𝑠,𝑥)=𝛾(min{𝑠,𝑡},𝑥)for𝑠𝑎,𝑏,𝑥𝐸1,𝛾1(𝑥)=𝛾(𝑡,𝑥)for𝑥𝐸1.(5.26) Then, clearly, 𝛾0𝑍[1](𝒟;),𝛾1𝐿([𝑐,𝑑];), 𝛾0(𝑠,𝑥)𝛾1[](𝑥)for𝑠𝑎,𝑏,𝑥𝐸1,𝛾0[](𝑠,𝑥)=𝛾(𝑠,𝑥)for𝑠𝑎,𝑡,𝑥𝐸1.(5.27) Since the operator 2 is positive and satisfies condition (4.14), we obtain 2𝛾0(𝑠,𝑥)2𝛾1(𝑠,𝑥)𝛾1(𝑥)2(1)(𝑠,𝑥)fora.e.(𝑠,𝑥)𝒟(5.28)(by 2(𝛾1) the authors mean 2(𝛾1), where 𝛾1(𝑠,𝑥)=𝛾1(𝑥) for all 𝑠[𝑎,𝑏] and 𝑥𝐸1). On the other hand, the operator 2 is supposed to be an 𝑎-Volterra one which guarantees the equality 2𝛾0(𝑠,𝑥)=2[]×[](𝛾)(𝑠,𝑥)fora.e.(𝑠,𝑥)𝑎,𝑡𝑐,𝑑,(5.29) and thus desired relation (5.25) holds for every 𝑡]𝑎,𝑏]. It means that, for any 𝑡]𝑎,𝑏], there exists a set 𝐴𝑡[𝑐,𝑑] with meas 𝐴𝑡=𝑑𝑐 such that 2(𝛾)(𝑠,𝑥)2(1)(𝑠,𝑥)𝛾(𝑡,𝑥)for𝑥𝐴𝑡,𝑠𝐵𝑡(𝑥),(5.30) where, for each 𝑥𝐴𝑡, we have 𝐵𝑡(𝑥)[𝑎,𝑏] with meas 𝐵𝑡(𝑥)=𝑡𝑎.
Put 𝐸2=𝑡𝐶𝐴𝑡, where 𝐶=]𝑎,𝑏]. Clearly, meas 𝐸2=𝑑𝑐 because the set 𝐶 is countable. Moreover, relation (5.30) yields that𝑡𝑡02(𝛾)(𝑠,𝑥)d𝑠𝛾(𝑡,𝑥)𝑡𝑡02(1)(𝑠,𝑥)d𝑠for𝑥𝐸2,𝑡𝐶,𝑡0[].𝑎,𝑡(5.31) Let 𝑡]𝑎,𝑏], 𝑡0[𝑎,𝑡[, and 𝑥𝐸1𝐸2 be arbitrary but fixed. Then there exists a sequence {𝑡𝑛}+𝑛=1[𝑡0,𝑏] such that 𝑡𝑛𝑡 as 𝑛+. It follows from relation (5.31) that 𝑡𝑛𝑡02(𝑡𝛾)(𝑠,𝑥)d𝑠𝛾𝑛,𝑥𝑡𝑛𝑡02(1)(𝑠,𝑥)d𝑠for𝑛,(5.32) whence we get 𝑡𝑡02(𝛾)(𝑠,𝑥)d𝑠𝛾(𝑡,𝑥)𝑡𝑡02(1)(𝑠,𝑥)d𝑠.(5.33) Consequently, we have proved that 𝑡𝑡02(𝛾)(𝑠,𝑥)d𝑠𝛾(𝑡,𝑥)𝑡𝑡02(1)(𝑠,𝑥)d𝑠,for𝑥𝐸1𝐸2,𝑎𝑡0<𝑡𝑏.(5.34)
Now we put 𝑢(𝑡,𝑥)=𝑡𝑎2(𝛾)(𝑠,𝑥)d𝑠,𝑣(𝑡,𝑥)=𝑡𝑎2([]1)(𝑠,𝑥)d𝑠for𝑡𝑎,𝑏,𝑥𝐸1𝐸2.(5.35) Lemma 2.3 guarantees that there exists a set 𝐸]𝑎,𝑏]×(𝐸1𝐸2) such that meas 𝐸=(𝑏𝑎)(𝑑𝑐) and 𝑢(1,0)(𝑡,𝑥)=2(𝛾)(𝑡,𝑥),𝑣(1,0)(𝑡,𝑥)=2(1)(𝑡,𝑥)for(𝑡,𝑥)𝐸.(5.36) Let (𝑡,𝑥)𝐸 be arbitrary but fixed. We choose a sequence {𝑛}+𝑛=1 of numbers from the interval ]0,𝑡𝑎] such that lim𝑛+𝑛=0.(5.37) Then relation (5.34) yields that 1𝑛𝑡𝑡𝑛2(𝛾)(𝑠,𝑥)d𝑠𝛾(𝑡,𝑥)𝑛𝑡𝑡𝑛2(1)(𝑠,𝑥)d𝑠for𝑛.(5.38) Letting 𝑛+ in the previous relation and using equalities (5.36) give 2(𝛾)(𝑡,𝑥)=𝑢(1,0)(𝑡,𝑥)𝛾(𝑡,𝑥)𝑣(1,0)(𝑡,𝑥)=𝛾(𝑡,𝑥)2(1)(𝑡,𝑥).(5.39) Consequently, desired inequality (5.23) holds because (𝑡,𝑥)𝐸 was arbitrary.

Lemma 5.10. Let 1𝑍[2](𝒟;)𝐿(𝒟;) be a positive 𝑐-Volterra operator such that inequality (4.13) holds for every 𝑦𝐿([𝑎,𝑏];). Then, for any function 𝛾𝑍[2](𝒟;) with the property 𝛾𝑡,𝑥1𝛾𝑡,𝑥2[]fora.e.𝑡𝑎,𝑏andall𝑐𝑥1𝑥2𝑑,(5.40) one has 1(𝛾)(𝑡,𝑥)1(1)(𝑡,𝑥)𝛾(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟.(5.41)

Proof. Lemma can be proven analogously as Lemma 5.9 by exchanging the roles of the variables 𝑡 and 𝑥.

5.2. Proofs of Main Results

Now we are in a position to prove the main results stated in Section 4.

Proof of Theorem 4.1. Since the spectral radius of the linear bounded operator 𝐴𝐿(𝒟;)𝐿(𝒟;) is less than one, (5.2) has a unique solution for an arbitrary 𝑦𝐿(𝒟;) and thus, in view of Lemma 5.1, problem (1.1), (1.2) has a unique solution for every 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];),𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];),𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐).

Proof of Corollary 4.2. It is easy to show that the norms of the linear bounded operators 𝐴0,𝐴1,𝐴2𝐿(𝒟;)𝐿(𝒟;) defined by relations (4.1), (4.2) satisfy the estimates 𝐴0(𝑏𝑎)(𝑑𝑐)0,𝐴1(𝑑𝑐)1,𝐴2(𝑏𝑎)2.(5.42) Consequently, the assumption (4.3) guarantees that the spectral radius of the operator 𝐴0+𝐴1+𝐴2 is less than one. Hence, the assertion of the corollary follows from Theorem 4.1.

Proof of Theorem 4.4. To prove the theorem, it is sufficient to show the following four implications.
(1)(2): assume that the assertion (1) of the theorem holds. We put 𝐾=𝐿(𝒟;+). It is not difficult to verify that 𝐾 forms a normal and solid cone in the Banach space 𝐿(𝒟;). Moreover, a function 𝑧𝐿(𝒟;) satisfies the relation 𝑧0 if and only if the inclusion 𝑧Int𝐾 holds.
On the other hand, by virtue of (4.1) and (4.2), the operator 𝐴 leaves the cone 𝐾 invariant, that is, 𝐴(𝐾)𝐾 because the operators 0, 1, and 2 are supposed to be positive. Therefore, the assumptions 𝑧0𝐾 and 𝑧0𝐴(𝑧0) yield that 𝑧00 as well.
Applying Lemma 5.7 with 𝑋=𝐿(𝒟;) and 𝛿=1, we obtain the desired assertion (2) of our theorem.
(2)(3): assume that the spectral radius of the operator 𝐴 is less than one. Then, for an arbitrary 𝑦𝐿(𝒟;), (5.2) has a unique solution 𝑧 and, moreover, this solution admits the series representation 𝑧=𝑦+𝐴(𝑦)+𝐴2(𝑦)+.(5.43) Consequently, in view of Lemma 5.1, problem (1.1), (1.2) has a unique solution 𝑢 for every 𝑞𝐿(𝒟;) and 𝛼𝐴𝐶([𝑎,𝑏];),𝛽𝐴𝐶([𝑐,𝑑];) such that 𝛼𝐿([𝑎,𝑏];), 𝛽𝐿([𝑐,𝑑];), and 𝛼(𝑎)=𝛽(𝑐).
Assume, in addition, that the initial functions 𝛼, 𝛽 and the forcing term 𝑞 are such that relations (4.7) hold. The above-used Lemma 5.1 guarantees that the solution 𝑢 to problem (1.1), (1.2) admits the integral representation 𝑢(𝑡,𝑥)=𝛼(𝑎)+𝑡𝑎𝛼(𝑠)d𝑠+𝑥𝑐𝛽(𝜂)d𝜂+𝑡𝑎𝑥𝑐𝑧(𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟,(5.44) in which 𝑧 is given by formula (5.43) with 𝑦 defined by relation (5.1). Clearly, assumption (4.7) yields 𝑦0 and thus the series representation (5.43) ensures that 𝑧0 because the operator 𝐴 leaves invariant the cone 𝐿(𝒟;+). Hence, by virtue of Remark 2.4, desired property (4.8) of the solution 𝑢 follows from integral representation (5.44).
(3)(4): assume that the assertion (3) of the theorem holds. Then, clearly the problem 𝛾(1,1)(𝑡,𝑥)=0(𝛾)(𝑡,𝑥)+1𝛾(1,0)(𝑡,𝑥)+2𝛾(0,1)(𝑡,𝑥)+1,(5.45)[][]𝛾(𝑡,𝑐)=0for𝑡𝑎,𝑏,𝛾(𝑎,𝑥)=0for𝑥𝑐,𝑑(5.46) has a unique solution 𝛾 and the function 𝛾𝐶(𝒟;) satisfies the inequalities 𝛾𝛾(𝑡,𝑥)0for(𝑡,𝑥)𝒟,(1,0)[][],𝛾(𝑡,𝑥)0fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)[][].(𝑡,𝑥)0for𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(5.47) Hence, conditions (4.9)–(4.11) are fulfilled. Moreover, inequality (4.12) follows from (5.45) because the operators 0, 1, and 2 are positive and 𝛾 satisfies relations (5.47).
(4)(1): assume that there exists a function 𝛾𝐶(𝒟;) satisfying inequalities (4.9)–(4.12). Then, by virtue of Remark 2.4, we get 𝛾(1,1)𝐿(𝒟;+), 𝛾(𝑡,𝑥)𝑡𝑎𝑥𝑐𝛾(1,1)(𝛾𝑠,𝜂)d𝜂d𝑠for(𝑡,𝑥)𝒟,(1,0)(𝑡,𝑥)𝑥𝑐𝛾(1,1)[][],𝛾(𝑡,𝜂)d𝜂fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)(𝑡,𝑥)𝑡𝑎𝛾(1,1)[][].(𝑠,𝑥)d𝑠,forall𝑡𝑎,𝑏,a.e.𝑥𝑐,𝑑(5.48) Consequently, assertion (1) of the theorem holds with 𝑧0=𝛾(1,1) because the operators 0, 1, and 2 are positive.

Proof of Corollary 4.5. We put 𝛾(𝑡,𝑥)=e𝑡𝑎𝑥𝑐(1+0(1)(𝑠,𝜂))d𝜂d𝑠+2𝑔2(𝑡𝑎)+2𝑔1(𝑥𝑐)for(𝑡,𝑥)𝒟,(5.49) where 𝑔1=1(1)𝐿 and 𝑔2=2(1)𝐿. It can be verified that 𝛾𝐶(𝒟;) and satisfies the inequality 𝛾(𝑡,𝑥)1for(𝑡,𝑥)𝒟.(5.50) In view of Lemma 2.3, from (5.49) we get 𝛾(1,0)(𝑡,𝑥)=𝑥𝑐1+0(1)(𝑡,𝜂)d𝜂+2𝑔2[][],𝛾𝛾(𝑡,𝑥),fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)(𝑡,𝑥)=𝑡𝑎1+0(1)(𝑠,𝑥)d𝑠+2𝑔1[][].𝛾(𝑡,𝑥),for𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(5.51) Therefore, inequalities (4.10) and (4.11) are fulfilled and 𝛾(1,0)(𝑡,𝑥)0,𝛾(0,1)(𝑡,𝑥)0fora.e.(𝑡,𝑥)𝒟,(5.52) because the operator 0 is positive. Moreover, by using Lemma 2.3 and inequalities (5.50) and (5.52), it follows from equalities (5.51) that 𝛾(1,1)(𝑡,𝑥)=1+0(1)(𝑡,𝑥)𝛾(𝑡,𝑥)+𝑥𝑐1+0(1)(𝑡,𝜂)d𝜂+2𝑔2𝛾(0,1)=(𝑡,𝑥)1+0+1(1)(𝑡,𝑥)𝛾(𝑡,𝑥)2𝑡𝑎1+0(1)(𝑠,𝑥)d𝑠+2𝑔1𝛾(1,0)(+1𝑡,𝑥)2𝑥𝑐1+0(1)(𝑡,𝜂)d𝜂+2𝑔2𝛾(0,1)(𝑡,𝑥)1+0(1)(𝑡,𝑥)𝛾(𝑡,𝑥)+𝑔1𝛾(1,0)(𝑡,𝑥)+𝑔2𝛾(0,1)(𝑡,𝑥)0(1)(𝑡,𝑥)𝛾(𝑡,𝑥)+1(1)(𝑡,𝑥)𝛾(1,0)(𝑡,𝑥)+2(1)(𝑡,𝑥)𝛾(0,1)(𝑡,𝑥)+1,(5.53) for a.e. (𝑡,𝑥)𝒟 and thus inequality (4.12) is satisfied.
On the other hand, equalities (5.49), (5.51) guarantee the validity of the inequalities𝛾𝑡1,𝑥1𝑡𝛾2,𝑥2for𝑎𝑡1𝑡2𝑏,𝑐𝑥1𝑥2𝛾𝑑,(1,0)𝑡,𝑥1𝛾(1,0)𝑡,𝑥2[]fora.e.𝑡𝑎,𝑏andall𝑐𝑥1𝑥2𝛾𝑑,(0,1)𝑡1,𝑥𝛾(0,1)𝑡2,𝑥for𝑎𝑡1𝑡2[].𝑏anda.e.𝑥𝑐,𝑑(5.54) Therefore, by using Lemmas 5.85.10, from inequality (5.53) we get 𝛾(1,1)(𝑡,𝑥)0(𝛾)(𝑡,𝑥)+1𝛾(1,0)(𝑡,𝑥)+2𝛾(0,1)(𝑡,𝑥)+1fora.e.(𝑡,𝑥)𝒟.(5.55) that is, relation (4.9) holds.
Consequently, we have found a function 𝛾 satisfying conditions (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.

Proof of Corollary 4.10. It is clear that (3.2) is a particular case of (1.1) in which the operators 0, 1, and 2 are defined by formulas (2.23), (2.28), and (2.29), respectively (see Examples 2.7 and 2.11). Since we have 𝑘=𝑝𝑘𝐿for𝑘=0,1,2,(5.56) the validity of the corollary follows immediately from Corollary 4.2.

Proof of Corollary 4.11. It is clear that (3.2) is a particular case of (1.1) in which the operators 0,1, and 2 are defined by formulas (2.23), (2.28), and (2.29), respectively, and that the operators indicated are positive (see Examples 2.7 and 2.11). Moreover, in view of inequality (4.17), there exists 𝜀>0 such that 1𝑝0𝜏(𝑡,𝑥)0𝜇(𝑡,𝑥)𝑎0(𝑡,𝑥)𝑐+𝑝1𝜇(𝑡,𝑥)1(𝑡,𝑥)𝑐+𝑝2(𝜏𝑡,𝑥)2(𝑡,𝑥)𝑎+𝜀fora.e.(𝑡,𝑥)𝒟.(5.57) Therefore, the function 𝛾 defined by the relation 𝛾(𝑡,𝑥)=(𝑡𝑎)(𝑥𝑐)for(𝑡,𝑥)𝒟(5.58) satisfies inequalities (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.

Proof of Corollary 4.12. It is clear that (3.2) is a particular case of (1.1) in which the operators 0, 1, and 2 are defined by formulas (2.23), (2.28), and (2.29), respectively, and that conditions (4.13) and (4.14) are fulfilled (see Examples 2.7 and 2.11). Moreover, the operators 0, 1 and 2 are positive and, in view of Remarks 2.9 and 2.13, the operators indicated are, respectively, (𝑎,𝑐)-Volterra, 𝑐-Volterra, and 𝑎-Volterra ones.
Consequently, the validity of the corollary follows from Corollary 4.5.

Proof of Corollary 4.13. It is clear that (3.2) is a particular case of (1.1) in which the operators 0, 1, and 2 are defined by formulas (2.23), (2.28), and (2.29), respectively, and that the operators indicated are positive (see Examples 2.7 and 2.11).
Let ̃𝑝𝑘=𝑝𝑘𝐿(𝑘=0,1,2) and 𝑦=e/𝜔. According to (4.19) and (4.20), there exist numbers 𝜀>0 and 𝛿]0,1[ such that inequalities 𝜏0𝑡(𝑡,𝑥)𝜇0𝑐(𝑡,𝑥)𝜀+𝑝0(𝑠,𝜂)d𝜂d𝑠+𝑡𝑎𝜇0𝑥(𝑡,𝑥)𝜀+𝑝0(𝑠,𝜂)d𝜂d𝑠+2̃𝑝2𝜏0(𝑡,𝑥)𝑡+2̃𝑝1𝜇01(𝑡,𝑥)𝑥𝑦ln𝑦+𝑦𝛿e𝑦[𝑏𝑎𝑑𝑐(𝜀+𝑝0(𝑠,𝜂))d𝜂d𝑠+2̃𝑝2(𝑏𝑎)+2̃𝑝1(𝑑𝑐)],𝛿𝑡𝑎𝜇1𝑥(𝑡,𝑥)𝜀+𝑝0̃p(𝑠,𝜂)d𝜂d𝑠+21𝜇11(𝑡,𝑥)𝑥𝑦ln2̃𝑝2𝑦𝜀+̃𝑝0(𝑑𝑐)+2̃𝑝2,𝜏2𝑡(𝑡,𝑥)𝑥𝑐𝜀+𝑝0(𝑠,𝜂)d𝜂d𝑠+2̃𝑝2𝜏21(𝑡,𝑥)𝑡𝑦ln2̃𝑝1𝑦𝜀+̃𝑝0(𝑏𝑎)+2̃𝑝1,(5.59) hold for a.e. (𝑡,𝑥)𝒟. Now we put 𝛾(𝑡,𝑥)=e𝑦𝑧(𝑡,𝑥)𝛿for(𝑡,𝑥)𝒟,(5.60) where 𝑧(𝑡,𝑥)=𝑡𝑎𝑥𝑐𝜀+𝑝0(𝑠,𝜂)d𝜂d𝑠+2̃𝑝2(𝑡𝑎)+2̃𝑝1(𝑥𝑐)for(𝑡,𝑥)𝒟.(5.61) It can be verified that 𝛾𝐶(𝒟;) and satisfies 𝛾(𝑡,𝑥)1𝛿>0for(𝑡,𝑥)𝒟.(5.62) In view of Lemma 2.3, from (5.62) we get 𝛾(1,0)(𝑡,𝑥)=𝑦𝑥𝑐𝜀+𝑝0(𝑡,𝜂)d𝜂+2̃𝑝2[][],𝛾(𝛾(𝑡,𝑥)+𝛿),fora.e.𝑡𝑎,𝑏andall𝑥𝑐,𝑑(0,1)(𝑡,𝑥)=𝑦𝑡𝑎𝜀+𝑝0(𝑠,𝑥)d𝑠+2̃𝑝1[][].(𝛾(𝑡,𝑥)+𝛿),for𝑡𝑎,𝑏anda.e.𝑥𝑐,𝑑(5.63) Therefore, relations (4.10), (4.11), and (5.52) are fulfilled. Moreover, by using Lemma 2.3 and inequalities (5.52) and (5.62), it follows from equalities (5.63) that 𝛾(1,1)(𝑡,𝑥)=𝑦𝜀+𝑝0(𝑡,𝑥)(𝛾(𝑡,𝑥)+𝛿)+𝑦𝑥𝑐𝜀+𝑝0(𝑡,𝜂)d𝜂+2̃𝑝2𝛾(0,1)(𝑡,𝑥)=𝑦𝜀(𝛾(𝑡,𝑥)+𝛿)+𝑦𝑝0+𝑦(𝑡,𝑥)(𝛾(𝑡,𝑥)+𝛿)2𝑡𝑎𝜀+𝑝0(𝑠,𝑥)d𝑠+2̃𝑝1𝛾(1,0)+𝑦(𝑡,𝑥)2𝑥𝑐𝜀+𝑝0(𝑡,𝜂)d𝜂+2̃𝑝2𝛾(0,1)(𝑡,𝑥)𝑦𝜀+𝑝0(𝑡,𝑥)𝑦(𝛾(𝑡,𝑥)+𝛿)+𝑦̃𝑝1𝛾(1,0)(𝑡,𝑥)+𝑦̃𝑝2𝛾(0,1)(𝑡,𝑥)𝑝0(𝑡,𝑥)𝑦(𝛾(𝑡,𝑥)+𝛿)+𝑝1(𝑡,𝑥)𝑦𝛾(1,0)(𝑡,𝑥)+𝑝2(𝑡,𝑥)𝑦𝛾(0,1)(𝑡,𝑥)+𝑦𝜀,(5.64) for a.e. (𝑡,𝑥)𝒟 and thus inequality (4.12) is satisfied. Now observe that inequalities (5.59) can be rewritten to the forms 𝑧𝜏0(𝑡,𝑥),𝜇01(𝑡,𝑥)𝑧(𝑡,𝑥)𝑦ln𝑦+𝑦𝛿e𝑦𝑧(𝜏0(𝑡,𝑥),𝜇0(𝑡,𝑥)),𝑧𝛿𝑡,𝜇11(𝑡,𝑥)𝑧(𝑡,𝑥)𝑦ln2̃𝑝2𝑦𝜇1𝑐(𝑡,𝑥)𝜀+𝑝0(𝑡,𝜂)d𝜂+2̃𝑝2,𝑧𝜏21(𝑡,𝑥),𝑥𝑧(𝑡,𝑥)𝑦ln2̃𝑝1𝑦𝜏2𝑎(𝑡,𝑥)𝜀+𝑝0(𝑠,𝑥)d𝑠+2̃𝑝1,(5.65) for a.e. (𝑡,𝑥)𝒟 and thus, by using relations (5.60), (5.63), we get 𝑦𝜏(𝛾(𝑡,𝑥)+𝛿)𝛾0(𝑡,𝑥),𝜇0(𝑡,𝑥)fora.e.(𝑡,𝑥)𝒟,𝑦𝛾(1,0)(𝑡,𝑥)𝛾(1,0)𝑡,𝜇1(𝑡,𝑥),𝑦𝛾(0,1)(𝑡,𝑥)𝛾(0,1)𝜏2(𝑡,𝑥),𝑥(5.66) for a.e. (𝑡,𝑥)𝒟. Consequently, it follows from (5.64) that inequality (4.9) holds.
We have constructed a function 𝛾 satisfying conditions (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.

Acknowledgment

For A. Lomtatidze and J. Šremr, the research was supported by the Academy of Sciences of the Czech Republic, Institutional Research Plan no. AV0Z10190503.