Linear Hyperbolic Functional-Differential Equations with Essentially Bounded Right-Hand Side

and Applied Analysis 3 vii L∞ D;R is the Banach space of Lebesgue measurable and essentially bounded functions p : D → R equipped with the norm ∥p∥L∞ ess sup {∣∣p t, x ∣∣ : t, x ∈ D}. 2.2 viii L∞ D;R {p ∈ L∞ D;R : p t, x ≥ 0 for a.e. t, x ∈ D}. ix For any z1, z2 ∈ L∞ D;R , we put z2 ≥ z1 ⇐⇒ z2 t, x − z1 t, x ≥ 0 for a.e. t, x ∈ D, z2 z1 ⇐⇒ z2 t, x − z1 t, x ≥ ε for a.e. t, x ∈ D with some ε > 0. 2.3 x L∞ α, β ;R , where −∞ < α < β < ∞, is the linear space of Lebesgue measurable and essentially bounded functions f : α, β → R. xi measA denotes the Lebesgue measure of the set A ⊂ R, m 1, 2. xii If X, Y are Banach spaces and T : X → Y is a linear bounded operator then ‖T‖ denotes the norm of the operator T , that is, ‖T‖ sup{‖T z ‖Y : z ∈ X, ‖z‖X ≤ 1}. 2.4 Two subsections below contain a number of definitions used in the sequel. 2.1. Spaces Z 1 D;R , Z 2 D;R , and Set C∗ D;R 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 u : D → R be such that u ·, x : a, b −→ R is continuous for a.e. x ∈ c, d , u t, · : c, d −→ R is measurable for all t ∈ a, b . 2.5 Then the function max{|u t, · | : t ∈ a, b } : c, d → R is measurable. Notation 1. Z 1 D;R denotes the linear space of all functions u : D → R satisfying conditions 2.5 , and ess sup{max{|u t, x | : t ∈ a, b } : x ∈ c, d } < ∞. 2.6 4 Abstract and Applied Analysis If one identifies functions u1, u2 from Z 1 D;R such that u1 ·, x ≡ u2 ·, x for a.e. x ∈ c, d then ‖u‖Z 1 ess sup{max{|u t, x | : t ∈ a, b } : x ∈ c, d } 2.7 defines a norm in the space Z 1 D;R . Analogously, we introduce the space Z 2 D;R 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;R and Z 2 D;R are Banach spaces. Proof. We only prove the assertion for the space Z 1 D;R , the assertion of the lemma concerning the space Z 2 D;R 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;R . For a decreasing sequence of positive numbers {εi} ∞ i 1 with ∑ ∞ i 1 εi < ∞ there exists an increasing sequence {ki} ∞ i 1 such that ess sup{max{|un t, x − uk t, x | : t ∈ a, b } : x ∈ c, d } < εi, 2.8 for every n, k ≥ ki, i ∈ N. Let vi uki i 1, 2, . . . . Then, for any i ∈ N, there is a set Ei ⊆ c, d , measEi d − c, such that max{|vi 1 t, x − vi t, x | : t ∈ a, b } < εi for x ∈ Ei, i ∈ N. 2.9 Put E ∩ ∞ i 1Ei. Then, clearly, we have measE d − c and max{|vn t, x − vk t, x | : t ∈ a, b } ≤ n−1 ∑ m k max{|vm 1 t, x − vm t, x | : t ∈ a, b } ≤ ∞ ∑ m k εm for x ∈ E, n > k. 2.10 Consequently, for any fixed x ∈ E, 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 lim n→ ∞ vi t, x − vn t, x uk t, x − uki t, x lim n→ ∞ n−1 ∑ m i vm t, x − vm 1 t, x 2.11 Abstract and Applied Analysis 5 holds for i, k ∈ N, all t ∈ a, b and a.e. x ∈ c, d , in view of 2.8 and 2.9 , we obtain ‖uk − u‖Z 1 ≤ εi ∞ ∑and Applied Analysis 5 holds for i, k ∈ N, all t ∈ a, b and a.e. x ∈ c, d , in view of 2.8 and 2.9 , we obtain ‖uk − u‖Z 1 ≤ εi ∞ ∑ m i εm for k ≥ ki, i ∈ N. 2.12 Hence, u ∈ Z 1 D;R and un → u in Z 1 D;R , that is, the space Z 1 D;R is complete. For the investigation of hyperbolic differential equations with discontinuous righthand 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;R stands for the set of functions u : D → R admitting the integral representation


Introduction
On the rectangle D a, b × c, d , we consider the linear partial functional-differential equation u 1,1 t, x 0 u t, x 1 u 1,0 t, x 2 u 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 u : D → R absolutely

Notation and Definitions
The following notation is used throughout the paper.
i N, Q, and R are the sets of all natural, rational, and real numbers, respectively, R 0, ∞ .iii The first-order partial derivatives of the function u : D → R at the point t, x ∈ D are denoted by u 1,0 t, x or u t t, x and u 0,1 t, x or u x t, x .The second-order mixed partial derivatives of the function u : D → R at the point t, x ∈ D are denoted by u tx t, x and u xt t, x whereas we use u
xi meas A denotes the Lebesgue measure of the set A ⊂ R m , m 1, 2.
xii If X, Y are Banach spaces and T : X → Y is a linear bounded operator then T denotes the norm of the operator T , that is, Two subsections below contain a number of definitions used in the sequel.

2.6
If one identifies functions u 1 , u 2 from Z 1 D; R such that u 1 •, x ≡ u 2 •, x for a.e.x ∈ c, d then defines a norm in the space Z 1 D; R .Analogously, we introduce the space Z 2 D; R 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.Proof.We only prove the assertion for the space Z 1 D; R , the assertion of the lemma concerning the space Z 2 D; R can be proven analogously by exchanging the roles of the variables t and x.
Let {u k } ∞ k 1 be an arbitrary Cauchy sequence in Z 1 D; R .For a decreasing sequence of positive numbers

2.10
Consequently, for any fixed x ∈ E, the sequence {v i •, x } ∞ i 1 converges uniformly on a, b , say to u •, x .Hence, {v i 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 holds for i, k ∈ N, all t ∈ a, b and a.e.x ∈ c, d , in view of 2.8 and 2.9 , we obtain For the investigation of hyperbolic differential equations with discontinuous righthand 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.where The next lemma on differentiating of an indefinite double integral plays a crucial role in our investigation.

2.18
Consequently, for any u ∈ C * D; R , we have

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; R .

2.26
Now we introduce analogous notions for linear operators defined on the spaces Z 1 D; R and Z 2 D; R .Definition 2.10.We say that a linear operator : Definition 2.12.We say that a linear operator :

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

Statement of Problem
On the rectangle D, we consider the linear nonhomogeneous Darboux problem 1.1 , 1.2 in which 0 : 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 equation u tx p 0 t, x u p 1 t, x u t p 2 t, x u x q t, x , 3.1 where p 0 , p 1 , p 2 , q ∈ L ∞ D; R , 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 p 0 , p 1 , and p 2 .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 deviations

Main Results
At first, we put where

4.2
Clearly, ϕ 0 : and thus the operators A 0 , A 1 , A 2 mapping the space L ∞ D; R into itself are linear and bounded.

Theorem 4.1. Let
where the operators A 0 , A 1 , A 2 are defined by relations 4.1 ,  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 A 0 A 1 A 2 , where the operators A 0 , A 1 , A 2 are defined by relations 4.1 , 4.2 .Then the following four assertions are equivalent.

. If the spectral radius of the operator A is less than one then problem
2 The spectral radius of the operator A is less than one.
If, in addition, the initial functions α, β and the forcing term q are such that α a ≥ 0, α t ≥ 0, β x ≥ 0, q t, x ≥ 0 for a.e.  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.
We say that the triplet 0 , 1 , 2 belongs to the set S ac if the implication Remark 4.9.The inclusion 0 , 1 , 2 ∈ S ac 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 p 0 , p 1 , p 2 , q ∈ L ∞ D; R and argument deviations τ 0 , τ 2 : D → a, b , μ 0 , μ 1 : D → c, d are measurable functions.
As a consequence of Corollary 4.2 we obtain the following.
4.17  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 τ 0 , μ 0 , μ 1 , and τ 2 are not necessarily delays but the differences where Then the assertion of Corollary 4.12 holds.

Auxiliary Statements and Proofs of Main Results
The proofs use several auxiliary statements given in the next subsection.

Auxiliary Statements
Remember that, for given operators 0 , 1 , and 2 , the operators A 0 , A 1 , and A 2 are defined by relations 4.1 , 4.2 .Moreover, having q

5.4
Consequently, 1.1 yields that where the operators A 0 , A 1 , and A 2 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; R with the operators A 0 , A 1 , and A 2 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.

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 y ∈ K for all x, y ∈ K, ii λx ∈ K for all x ∈ K and an arbitrary λ ≥ 0, iii if x ∈ K and −x ∈ K then x 0.
Remark 5.3.In the original terminology introduced by Kreȋn 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 K ⊆ X is solid if its interior Int K 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 x 1 , x 2 ∈ X are said to be in the relation x 2 ≥ K x 1 if and only if they satisfy the inclusion x 2 − x 1 ∈ K. If, moreover, K is a solid cone then we write x 2 K x 1 if and only if x 2 − x 1 ∈ Int K. Definition 5.6.A cone K ⊆ X is said to be normal if there is a constant N ≥ 0 such that, for every x, y ∈ X with the property 0 ≤ K x ≤ K y, the relation x X ≤ N y X 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 A : X → X leave invariant the cone K, that is, A K ⊆ K.If there exists a constant δ > 0 and an element x 0 ∈ Int K such that δx 0 − A x 0 ∈ Int K, 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.

5.13
On the other hand, the operator 0 is supposed to be an a, c -Volterra one which guarantees the equality 0 γ 0 s, η 0 γ s, η for a.e.s, η ∈ a, t × c, x ,

5.15
It follows from Lemma 2.3 that there exists a set

5.16
Abstract and Applied Analysis 17 and, moreover, there is a set 5.17 Let t, x ∈ E be arbitrary but fixed.Then, relation 5.10 yields for h ∈ 0, t − a and k ∈ 0, x − c , whence we get

5.19
For any k ∈ 0, x − c fixed, we pass to the limit h → 0 in the latter inequality and thus, in view of equalities 5.16 , we get for k ∈ 0, x − c .Now, letting k → 0 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.

5.24
We first show that the relation 2 γ s, x ≤ 2 1 s, x γ t, x for a.e.s, x ∈ a, t × c, d 5.25 holds for every t ∈ a, b .Indeed, let t ∈ a, b be arbitrary but fixed.Put

5.27
Since the operator 2 is positive and satisfies condition 4.14 , we obtain Proof.Lemma can be proven analogously as Lemma 5.9 by exchanging the roles of the variables t and x.

Proofs of Main Results
Now we are in a position to prove the main results stated in Section 4.

5.42
Consequently, the assumption 4.3 guarantees that the spectral radius of the operator A 0 A 1 A 2 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; R .It is not difficult to verify that K forms a normal and solid cone in the Banach space L ∞ D; R .Moreover, a function z ∈ L ∞ D; R satisfies the relation z 0 if and only if the inclusion z ∈ Int K 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 z 0 ∈ K and z 0 A z 0 yield that z 0 0 as well.Applying Lemma 5.7 with X L ∞ D; R 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 y ∈ L ∞ D; R , 5.2 has a unique solution z and, moreover, this solution admits the series representation Consequently, in view of Lemma 5.1, problem 1.1 , 1.2 has a unique solution u for every

5.48
Consequently, assertion 1 of the theorem holds with z 0 γ 1,1 because the operators 0 , 1 , and 2 are positive.We have constructed a function γ satisfying conditions 4.9 -4.12 and thus the assertion of the corollary follows from Theorem 4.4.

Lemma 2 . 1 5
Z 1 D; R , Z 2 D; R , and Set C * D; R Motivated by 19, Section 2 , the authors introduce the following assertions and definitions.see 19, Section 1, Lemma 1 .Let the function u: D → R be such that u •, x : a, b −→ R is continuous for a.e.x ∈ c, d , u t, • : c, d −→ R is measurable for all t ∈ a, b .2.Then the function max{|u t,• | : t ∈ a, b } : c, d → R is measurable.Notation 1. Z 1 D; R denotes the linear space of all functions u : D → R satisfying conditions 2.5 , and ess sup{max{|u t, x | : t ∈ a, b } : x ∈ c, d } < ∞.

Notation 2 .
C * D; R stands for the set of functions u : D → R admitting the integral representation η dη ds for t, x ∈ D, 2.13 and α a β c .By a solution to problem 1.1 , 1.2 , we mean a function u ∈ C * D; R 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.
η dη for a.e.t ∈ a, b and all x ∈ c, d , ϕ 2 z t, x t a z s, x ds for t ∈ a, b , a.e.x ∈ c, d .

Corollary 4 . 2 .
and α a β c .Theorem 4.1 implies the following.If the inequality b

Corollary 4 . 11 .
and α a β c .If the coefficients p 0 , p 1 , p 2 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.Let p 0 , p 1 , p 2 ∈ L ∞ D; R and

Lemma 5 . 8 .
Let 0 : C D; R → L ∞ D; R be a positive a, c -Volterra operator.Then, for any function γ ∈ C D; R satisfying , η dη ds for t, x ∈ D.

Proof of Theorem 4 . 1 .
Since the spectral radius of the linear bounded operator A : L ∞ D; R → L ∞ D; R is less than one, 5.2 has a unique solution for an arbitrary y ∈ L ∞ D; R and thus, in view of Lemma 5.1, problem 1.1 , 1.2 has a unique solution for every q∈ L ∞ D; R and α ∈ AC a, b ; R , β ∈ AC c, d ; R such that α ∈ L ∞ a, b ; R , β ∈ L ∞ c, d ; R ,and α a β c .Proof of Corollary 4.2.It is easy to show that the norms of the linear bounded operators A 0 , A 1 , A 2 : L ∞ D; R → L ∞ D; R defined by relations 4.1 , 4.2 satisfy the estimates 1 s, η dη ds for t, x ∈ D,γ 1,0 t, x ≥ x c γ1,1 t, η dη for a.e.t ∈ a, b and all x ∈ c, d , γ 0,1 t, x ≥ t a γ 1,1 s, x ds, for all t ∈ a, b , a.e.x ∈ c, d .

19
Remark 2.5.It follows from Remark 2.4 and 22, Satz 1, page 654 that u ∈ C * D; R if and only if u : D → R is absolutely continuous in the sense of Carathéodory with the properties b and a.e.x ∈ c, d is an a-Volterra operator resp., a c-Volterra operator if, for anyt 0 ∈ a, b resp., x 0 ∈ c, d and u ∈ Z 1 D; R resp., u ∈ Z 2 D; R such that e. t ∈ a, b and all x ∈ c, x 0 , 2.30 we have u t, x 0 for a.e.t, x ∈ a, t 0 × c, d resp., u t, x 0 for a.e.t, x ∈ a, b × c, x 0 .
0 , 1 , and 2 are defined by formulas 2.23 , 2.28 , and 2.29 , respectively.Let us also mention that in the case, where d are measurable functions.We obtain this equation from 1.1 if the operators Clearly 4.4 is a particular case of 1.1 .If m 0 m 1 m 2 1, then problem 4.4 , 4.5 has the trivial solution u t, x ≡ 0 and the nontrivial solution u t, x ≡ t − a x − c .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 .
Remark 4.3.On the rectangle a, b × c, d , we consider the equation u 1,1 t, x p 0 u b, d p 1 u 1,0 t, d p 2 u 0,1 b, x For Volterra-type operators 0 , 1 , and 2 , we derive from the previous theorem the following.Let 0 , 1 , and 2 be positive a, c -Volterra, c-Volterra, and a-Volterra operators, respectively, such that the inequalities t, x ∈ D, 4.7 then the solution u to problem 1.1 , 1.2 satisfies u t, x ≥ 0 for t, x ∈ D, u 1,0 t, x ≥ 0 for a.e.t ∈ a, b and all x ∈ c, d , u 0,1 t, x ≥ 0 for t ∈ a, b and a.e.x ∈ c, d .γ 1,0 t, c ≥ 0 for a.e.t ∈ a, b , γ 0,1 a, x ≥ 0 for a.e.x ∈ c, d , 4.11 γ 1,1 t, x ≥ 0 for a.e.t, x ∈ D. 4.12 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 for a.e.t, x ∈ D 4.14 by 2 z we mean 2 z , where z t, x z x for all t ∈ a, b and a.e.x ∈ c, d hold for every y If 0 , 1 , 2 ∈ S ac , 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 ∈ S ac is fulfilled.
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 .Let p 0 , p 1 , p 2 ∈ L ∞ D; R and argument deviations τ 0 , μ 0 , μ 1 , and τ 2 satisfy inequalities 2.26 , 2.32 , and 2.33 .Then problem 3.2 , 1.2 is uniquely solvable for arbitrary q Similarly, 1 α resp., 2 β means 1 α 0 resp., 2 β 0 , where α If u is a solution to problem 1.1 , 1.2 then u 1,1 is a solution to the equation ∞ D; R , where the operators A 0 , A 1 , and A 2 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; R with the operators A 0 , A 1 , and A 2 defined by relations 4.1 , 4.2 and the function y given by formula 5.1 , then 0 t, x α t for a.e.t ∈ a, b and all x ∈ c, d resp., β 0 t, x β x for all t ∈ a, b and a.e.x ∈ c, d .Clearly, y ∈ L ∞ D; R .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; R , u t, x −α a α t β x t a x c u 1,1 s, η dη ds for t, x ∈ D, x c u 1,1 t, η dη for a.e.t ∈ a, b and all x ∈ c, d , u 0,1 t, x β x t a u 1,1 s, x ds for all t ∈ a, b and a.e.x ∈ c, d .
Then the function u belongs to the set C * D; R and verifies initial conditions 1.2 .Furthermore, by using Lemma 2.3, we get On the other hand, the operator 2 is supposed to be an a-Volterra one which guarantees the equality Put E 2 ∩ t∈C A t , where C a, b ∩ Q.Clearly, meas E 2 d − c because the set C is countable.Moreover, relation 5.30 yields that , t 0 ∈ a, t , and x ∈ E 1 ∩ E 2 be arbitrary but fixed.Then there exists a sequence{t n } ∞ n 1 ⊂ t 0 , b ∩ Q such that t n → tas n → ∞.It follows from relation 5.31 that Let 1 : Z 2 D; R → L ∞ D; R be a positive c-Volterra operator such that inequality 4.13 holds for every y ∈ L ∞ a, b ; R .Then, for any function γ ∈ Z 2 D; R with the property γ t, x 1 ≤ γ t, x 2 for a.e.t ∈ a, b and all c ≤ x 1 ≤ x 2 ≤ d, x for a.e.s, x ∈ D 5.28 by 2 γ 1 the authors mean 2 γ 1 , where γ 1 s, x γ 1 x for all s ∈ a, b and x ∈ E 1 .2 γ 0 s, x 2 γ s, x for a.e.s, x ∈ a, t × c, 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 A t ⊆ c, d with meas A t d − c such that 2 γ s, x ≤ 2 1 s, x γ t, x for x ∈ A t , s ∈ B t x , 5.30 where, for each x ∈ A t , we have B t x ⊆ a, b with meas B t x t − a.
is given by formula 5.43 with y defined by relation 5.1 .Clearly, assumption 4.7 yields y ≥ 0 and thus the series representation 5.43 ensures that z ≥ 0 because the operator A leaves invariant the cone L ∞ D; R .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 has a unique solution γ and the function γ ∈ C * D; R satisfies the inequalities γ t, x ≥ 0 for t, x ∈ D, γ 1,0 t, x ≥ 0 for a.e.t ∈ a, b and all x ∈ c, d , γ 0,1 t, x ≥ 0 for t ∈ a, b and a.e.x ∈ c, d .