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

Authors: Alexander Domoshnitsky; Alexander Lomtatidze; Abraham Maghakyan; Jiří Šremr
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Theorems on the unique solvability and nonnegativity of solutions to the characteristic initial value problemu1,1(t,x)=ℓ0(u)(t,x)+ℓ1(u1,0)(t,x)+ℓ2(u0,1)(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 ℓ0, ℓ1, ℓ2 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.


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## Body

## 1. Introduction

On the rectangle𝒟=[a,b]×[c,d], we consider the linear partial functional-differential equation(1.1)u(1,1)(t,x)=l0(u)(t,x)+l1(u(1,0))(t,x)+l2(u(0,1))(t,x)+q(t,x),
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 ℓ0, ℓ1, and ℓ2 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 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 solutionu to (1.1) are prescribed on both characteristics t=a and x=c, that is, the initial conditions are(1.2)u(t,c)=α(t)fort∈[a,b],u(a,x)=β(x)forx∈[c,d].
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 (1.3)utx=f(t,x,u,ut,ux)
with continuous as well as discontinuous right-hand sides but without argument deviations have been studied in detail (see, e.g., [1–13] and references therein). As for the hyperbolic functional-differential equations, we can mention for example the works [14–16] (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 ℓ1=0 and ℓ2=0 is considered.


## 2. 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,+∞[.(ii)
𝒟=[a,b]×[c,d], where -∞<a<b<+∞ and -∞<c<d<+∞.(iii)
The first-order partial derivatives of the functionu:𝒟→ℝ 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 u:𝒟→ℝ 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;ℝ) is the Banach space of continuous functions u:𝒟→ℝ equipped with the norm(2.1)‖u‖C=max⁡{|u(t,x)|:(t,x)∈D}.(v)
C([α,β];ℝ), where -∞<α<β<+∞, is the linear space of continuous functions v:[α,β]→ℝ.(vi)
AC([α,β];ℝ), where -∞<α<β<+∞, is the linear space of absolutely continuous functions v:[α,β]→ℝ.(vii)
L∞(𝒟;ℝ) is the Banach space of Lebesgue measurable and essentially bounded functions p:𝒟→ℝ equipped with the norm(2.2)‖p‖L∞=esssup⁡{|p(t,x)|:(t,x)∈D}.(viii)
L∞(𝒟;ℝ+)={p∈L∞(𝒟;ℝ):p(t,x)≥0fora.e.(t,x)∈𝒟}.(ix)
For anyz1,z2∈L∞(𝒟;ℝ), we put(2.3)z2≥z1⟺z2(t,x)-z1(t,x)≥0fora.e.(t,x)∈D,z2≫z1⟺z2(t,x)-z1(t,x)≥ɛfora.e.(t,x)∈Dwithsomeɛ>0.(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⊂Rm, m=1,2.(xii)
IfX, Y are Banach spaces and T:X→Y is a linear bounded operator then ∥T∥ denotes the norm of the operator T, that is,(2.4)∥T∥=sup⁡{∥T(z)∥Y:z∈X,∥z∥X≤1}.Two subsections below contain a number of definitions used in the sequel.


### 2.1. SpacesZ[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 functionu:𝒟→ℝ be such that
(2.5)u(⋅,x):[a,b]⟶Riscontinuousfora.e.x∈[c,d],u(t,⋅):[c,d]⟶Rismeasurableforallt∈[a,b].
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
(2.6)esssup⁡{max⁡{|u(t,x)|:t∈[a,b]}:x∈[c,d]}<+∞.
If one identifies functions u1, u2 from Z[1](𝒟;ℝ) such that u1(·,x)≡u2(·,x) for a.e. x∈[c,d] then
(2.7)‖u‖Z[1]=esssup⁡{max⁡{|u(t,x)|:t∈[a,b]}:x∈[c,d]}
defines a norm in the space Z[1](𝒟;ℝ).
Analogously, we introduce the spaceZ[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 spaceZ[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 {ɛi}i=1+∞ with ∑i=1+∞ɛi<+∞ there exists an increasing sequence {ki}i=1+∞ such that(2.8)esssup⁡{max⁡{|un(t,x)-uk(t,x)|:t∈[a,b]}:x∈[c,d]}<ɛi,
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], meas Ei=d-c, such that
(2.9)max⁡{|vi+1(t,x)-vi(t,x)|:t∈[a,b]}<ɛiforx∈Ei,i∈N.
Put E=∩i=1+∞Ei. Then, clearly, we have meas E=d-c and
(2.10)max⁡{|vn(t,x)-vk(t,x)|:t∈[a,b]}≤∑m=kn-1max⁡{|vm+1(t,x)-vm(t,x)|:t∈[a,b]}≤∑m=k+∞ɛmforx∈E,n>k.
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
(2.11)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→+∞∑m=in-1[vm(t,x)-vm+1(t,x)]
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
(2.12)‖uk-u‖Z[1]≤ɛi+∑m=i+∞ɛmfork≥ki,i∈N.
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 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;ℝ) stands for the set of functions u:𝒟→ℝ admitting the integral representation
(2.13)u(t,x)=z+∫atf(s)ds+∫cxg(η)dη+∫at∫cxh(s,η)dηdsfor(t,x)∈D,
where z∈ℝ, f∈L∞([a,b];ℝ), g∈L∞([c,d];ℝ), and h∈L∞(𝒟;ℝ).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]).
Leth:𝒟→ℝ be a Lebesgue integrable function and
(2.14)v(t,x)=∫at∫cxh(s,η)dηdsfor(t,x)∈D.
Then(1)
there exists a setE⊆[a,b] such that measE=b-a and(2.15)v(1,0)(t,x)=∫cxh(t,η)dηfort∈E,x∈[c,d],(2)
there exists a setF⊆[c,d] such that measF=d-c and(2.16)v(0,1)(t,x)=∫ath(s,x)dsfort∈[a,b]andx∈F,(3)
there exists a setG⊆𝒟 such that measG=(b-a)(d-c) and(2.17)v(1,1)(t,x)=h(t,x)for(t,x)∈G.Remark 2.4.
Ifu∈C*(𝒟;ℝ), that is, the function u admits integral representation (2.13), then by using Lemma 2.3 we get
(2.18)u(1,0)(t,x)=f(t)+∫cxh(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],u(0,1)(t,x)=g(x)+∫ath(s,x)dsforallt∈[a,b]anda.e.x∈[c,d],u(1,1)(t,x)=h(t,x)fora.e.(t,x)∈D.
Consequently, for any u∈C*(𝒟;ℝ), we have
(2.19)u(1,0)∈Z[2](D;R),u(0,1)∈Z[1](D;R),u(1,1)∈L∞(D;R).Remark 2.5.
It follows from Remark2.4 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
(2.20)u(1,0)(⋅,c)∈L∞([a,b];R),u(0,1)(a,⋅)∈L∞([c,d];R),u(1,1)∈L∞(D;R).


### 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 spaceC(𝒟;ℝ).Definition 2.6.
A linear operatorℓ:C(𝒟;ℝ)→L∞(𝒟;ℝ) is said to be positive if the relation
(2.21)l(u)(t,x)≥0fora.e.(t,x)∈D
holds whenever the function u∈C(𝒟;ℝ) is such that
(2.22)u(t,x)≥0for(t,x)∈D.Example 2.7.
For anyv∈C(𝒟;ℝ), we put
(2.23)l0(v)(t,x)=p0(t,x)v(τ0(t,x),μ0(t,x))fora.e.(t,x)∈D,
where p0∈L∞(𝒟;ℝ) and τ0:𝒟→[a,b], μ0:𝒟→[c,d] are measurable functions. Then the operator ℓ0:C(𝒟;ℝ)→L∞(𝒟;ℝ) is linear and bounded. Moreover, ℓ0 is positive if and only if p0(t,x)≥0 for a.e. (t,x)∈𝒟.Definition 2.8.
A linear operatorℓ:C(𝒟;ℝ)→L∞(𝒟;ℝ) is called (a,c)-Volterra operator if, for any (t0,x0)∈𝒟 and u∈C(𝒟;ℝ) such that
(2.24)u(t,x)=0for(t,x)∈[a,t0]×[c,x0],
we have
(2.25)l(u)(t,x)=0fora.e.(t,x)∈[a,t0]×[c,x0].Remark 2.9.
It can be shown by using Lemma5.8 stated below that the operator ℓ0 given by formula (2.23) is an (a,c)-Volterra one if and only if
(2.26)|p0(t,x)|(τ0(t,x)-t)≤0fora.e.(t,x)∈D,|p0(t,x)|(μ0(t,x)-x)≤0fora.e.(t,x)∈D.Now we introduce analogous notions for linear operators defined on the spacesZ[1](𝒟;ℝ) and Z[2](𝒟;ℝ).Definition 2.10.
We say that a linear operatorℓ:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) (resp., ℓ:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ)) is positive if relation (2.21) is satisfied for every function u∈Z[1](𝒟;ℝ) (resp., u∈Z[2](𝒟;ℝ)) such that
(2.27)u(t,x)≥0fort∈[a,b]anda.e.x∈[c,d](resp.,u(t,x)≥0fora.e.t∈[a,b]andallx∈[c,d]).Example 2.11.
For anyv∈Z[2](𝒟;ℝ) (resp., v∈Z[1](𝒟;ℝ)), we put
(2.28)l1(v)(t,x)=p1(t,x)v(t,μ1(t,x))fora.e.(t,x)∈D,
respectively,
(2.29)l2(v)(t,x)=p2(t,x)v(τ2(t,x),x)fora.e.(t,x)∈D,
where p1,p2∈L∞(𝒟;ℝ) and μ1:𝒟→[c,d], τ2:𝒟→[a,b] are measurable functions. Then the operators ℓ1:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ) and ℓ2:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) are linear and bounded. Moreover, ℓ1 (resp., ℓ2) 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 operatorℓ:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) (resp., ℓ:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ)) is an a-Volterra operator (resp., a c-Volterra operator) if, for any t0∈[a,b] (resp., x0∈[c,d]) and u∈Z[1](𝒟;ℝ) (resp., u∈Z[2](𝒟;ℝ)) such that
(2.30)u(t,x)=0fort∈[a,t0]anda.e.x∈[c,d](resp.,u(t,x)=0fora.e.t∈[a,b]andallx∈[c,x0]),
we have
(2.31)l(u)(t,x)=0fora.e.(t,x)∈[a,t0]×[c,d](resp.,l(u)(t,x)=0fora.e.(t,x)∈[a,b]×[c,x0]).Remark 2.13.
One can show by using Lemma5.9 (resp., Lemma 5.10) stated below that the operator ℓ1 (resp., ℓ2) given by formula (2.28) (resp., (2.29)) is a c-Volterra one (resp., an a-Volterra one) if and only if
(2.32)|p1(t,x)|(μ1(t,x)-x)≤0fora.e.(t,x)∈D,
respectively,
(2.33)|p2(t,x)|(τ2(t,x)-t)≤0fora.e.(t,x)∈D.


## 2.1. SpacesZ[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 functionu:𝒟→ℝ be such that
(2.5)u(⋅,x):[a,b]⟶Riscontinuousfora.e.x∈[c,d],u(t,⋅):[c,d]⟶Rismeasurableforallt∈[a,b].
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
(2.6)esssup⁡{max⁡{|u(t,x)|:t∈[a,b]}:x∈[c,d]}<+∞.
If one identifies functions u1, u2 from Z[1](𝒟;ℝ) such that u1(·,x)≡u2(·,x) for a.e. x∈[c,d] then
(2.7)‖u‖Z[1]=esssup⁡{max⁡{|u(t,x)|:t∈[a,b]}:x∈[c,d]}
defines a norm in the space Z[1](𝒟;ℝ).
Analogously, we introduce the spaceZ[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 spaceZ[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 {ɛi}i=1+∞ with ∑i=1+∞ɛi<+∞ there exists an increasing sequence {ki}i=1+∞ such that(2.8)esssup⁡{max⁡{|un(t,x)-uk(t,x)|:t∈[a,b]}:x∈[c,d]}<ɛi,
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], meas Ei=d-c, such that
(2.9)max⁡{|vi+1(t,x)-vi(t,x)|:t∈[a,b]}<ɛiforx∈Ei,i∈N.
Put E=∩i=1+∞Ei. Then, clearly, we have meas E=d-c and
(2.10)max⁡{|vn(t,x)-vk(t,x)|:t∈[a,b]}≤∑m=kn-1max⁡{|vm+1(t,x)-vm(t,x)|:t∈[a,b]}≤∑m=k+∞ɛmforx∈E,n>k.
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
(2.11)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→+∞∑m=in-1[vm(t,x)-vm+1(t,x)]
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
(2.12)‖uk-u‖Z[1]≤ɛi+∑m=i+∞ɛmfork≥ki,i∈N.
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 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;ℝ) stands for the set of functions u:𝒟→ℝ admitting the integral representation
(2.13)u(t,x)=z+∫atf(s)ds+∫cxg(η)dη+∫at∫cxh(s,η)dηdsfor(t,x)∈D,
where z∈ℝ, f∈L∞([a,b];ℝ), g∈L∞([c,d];ℝ), and h∈L∞(𝒟;ℝ).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]).
Leth:𝒟→ℝ be a Lebesgue integrable function and
(2.14)v(t,x)=∫at∫cxh(s,η)dηdsfor(t,x)∈D.
Then(1)
there exists a setE⊆[a,b] such that measE=b-a and(2.15)v(1,0)(t,x)=∫cxh(t,η)dηfort∈E,x∈[c,d],(2)
there exists a setF⊆[c,d] such that measF=d-c and(2.16)v(0,1)(t,x)=∫ath(s,x)dsfort∈[a,b]andx∈F,(3)
there exists a setG⊆𝒟 such that measG=(b-a)(d-c) and(2.17)v(1,1)(t,x)=h(t,x)for(t,x)∈G.Remark 2.4.
Ifu∈C*(𝒟;ℝ), that is, the function u admits integral representation (2.13), then by using Lemma 2.3 we get
(2.18)u(1,0)(t,x)=f(t)+∫cxh(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],u(0,1)(t,x)=g(x)+∫ath(s,x)dsforallt∈[a,b]anda.e.x∈[c,d],u(1,1)(t,x)=h(t,x)fora.e.(t,x)∈D.
Consequently, for any u∈C*(𝒟;ℝ), we have
(2.19)u(1,0)∈Z[2](D;R),u(0,1)∈Z[1](D;R),u(1,1)∈L∞(D;R).Remark 2.5.
It follows from Remark2.4 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
(2.20)u(1,0)(⋅,c)∈L∞([a,b];R),u(0,1)(a,⋅)∈L∞([c,d];R),u(1,1)∈L∞(D;R).


## 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 spaceC(𝒟;ℝ).Definition 2.6.
A linear operatorℓ:C(𝒟;ℝ)→L∞(𝒟;ℝ) is said to be positive if the relation
(2.21)l(u)(t,x)≥0fora.e.(t,x)∈D
holds whenever the function u∈C(𝒟;ℝ) is such that
(2.22)u(t,x)≥0for(t,x)∈D.Example 2.7.
For anyv∈C(𝒟;ℝ), we put
(2.23)l0(v)(t,x)=p0(t,x)v(τ0(t,x),μ0(t,x))fora.e.(t,x)∈D,
where p0∈L∞(𝒟;ℝ) and τ0:𝒟→[a,b], μ0:𝒟→[c,d] are measurable functions. Then the operator ℓ0:C(𝒟;ℝ)→L∞(𝒟;ℝ) is linear and bounded. Moreover, ℓ0 is positive if and only if p0(t,x)≥0 for a.e. (t,x)∈𝒟.Definition 2.8.
A linear operatorℓ:C(𝒟;ℝ)→L∞(𝒟;ℝ) is called (a,c)-Volterra operator if, for any (t0,x0)∈𝒟 and u∈C(𝒟;ℝ) such that
(2.24)u(t,x)=0for(t,x)∈[a,t0]×[c,x0],
we have
(2.25)l(u)(t,x)=0fora.e.(t,x)∈[a,t0]×[c,x0].Remark 2.9.
It can be shown by using Lemma5.8 stated below that the operator ℓ0 given by formula (2.23) is an (a,c)-Volterra one if and only if
(2.26)|p0(t,x)|(τ0(t,x)-t)≤0fora.e.(t,x)∈D,|p0(t,x)|(μ0(t,x)-x)≤0fora.e.(t,x)∈D.Now we introduce analogous notions for linear operators defined on the spacesZ[1](𝒟;ℝ) and Z[2](𝒟;ℝ).Definition 2.10.
We say that a linear operatorℓ:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) (resp., ℓ:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ)) is positive if relation (2.21) is satisfied for every function u∈Z[1](𝒟;ℝ) (resp., u∈Z[2](𝒟;ℝ)) such that
(2.27)u(t,x)≥0fort∈[a,b]anda.e.x∈[c,d](resp.,u(t,x)≥0fora.e.t∈[a,b]andallx∈[c,d]).Example 2.11.
For anyv∈Z[2](𝒟;ℝ) (resp., v∈Z[1](𝒟;ℝ)), we put
(2.28)l1(v)(t,x)=p1(t,x)v(t,μ1(t,x))fora.e.(t,x)∈D,
respectively,
(2.29)l2(v)(t,x)=p2(t,x)v(τ2(t,x),x)fora.e.(t,x)∈D,
where p1,p2∈L∞(𝒟;ℝ) and μ1:𝒟→[c,d], τ2:𝒟→[a,b] are measurable functions. Then the operators ℓ1:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ) and ℓ2:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) are linear and bounded. Moreover, ℓ1 (resp., ℓ2) 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 operatorℓ:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) (resp., ℓ:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ)) is an a-Volterra operator (resp., a c-Volterra operator) if, for any t0∈[a,b] (resp., x0∈[c,d]) and u∈Z[1](𝒟;ℝ) (resp., u∈Z[2](𝒟;ℝ)) such that
(2.30)u(t,x)=0fort∈[a,t0]anda.e.x∈[c,d](resp.,u(t,x)=0fora.e.t∈[a,b]andallx∈[c,x0]),
we have
(2.31)l(u)(t,x)=0fora.e.(t,x)∈[a,t0]×[c,d](resp.,l(u)(t,x)=0fora.e.(t,x)∈[a,b]×[c,x0]).Remark 2.13.
One can show by using Lemma5.9 (resp., Lemma 5.10) stated below that the operator ℓ1 (resp., ℓ2) given by formula (2.28) (resp., (2.29)) is a c-Volterra one (resp., an a-Volterra one) if and only if
(2.32)|p1(t,x)|(μ1(t,x)-x)≤0fora.e.(t,x)∈D,
respectively,
(2.33)|p2(t,x)|(τ2(t,x)-t)≤0fora.e.(t,x)∈D.


## 3. Statement of Problem

On the rectangle𝒟, we consider the linear nonhomogeneous Darboux problem (1.1), (1.2) in which ℓ0:C(𝒟;ℝ)→L∞(𝒟;ℝ), ℓ1:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ), and ℓ2:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) are linear bounded operators, q∈L∞(𝒟;ℝ), and α∈AC([a,b];ℝ), β∈AC([c,d];ℝ) are such that α′∈L∞([a,b];ℝ),β′∈L∞([c,d];ℝ), 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.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(3.1)utx=p0(t,x)u+p1(t,x)ut+p2(t,x)ux+q(t,x),
where p0,p1,p2,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 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, Theorems4.1 and 4.4) will be illustrated on the hyperbolic differential equation with argument deviations(3.2)u(1,1)(t,x)=p0(t,x)u(τ0(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),
in which coefficients p0,p1,p2,q∈L∞(𝒟;ℝ) and argument deviations τ0,τ2:𝒟→[a,b], μ0,μ1:𝒟→[c,d] are measurable functions. We obtain this equation from (1.1) if the operators ℓ0, ℓ1, and ℓ2 are defined by formulas (2.23), (2.28), and (2.29), respectively. Let us also mention that in the case, where(3.3)τk(t,x)=t,μj(t,x)=xfora.e.(t,x)∈D(k=0,2,j=0,1),
equation (3.2) takes form (3.1).


## 4. Main Results

At first, we put(4.1)Ak(z)=lk(φk(z))forz∈L∞(D;R),k=0,1,2,
where (4.2)φ0(z)(t,x)=∫at∫cxz(s,η)dηdsfor(t,x)∈D,φ1(z)(t,x)=∫cxz(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],φ2(z)(t,x)=∫atz(s,x)dsfort∈[a,b],a.e.x∈[c,d].Clearly,φ0:L∞(𝒟;ℝ)→C(𝒟;ℝ), φ1:L∞(𝒟;ℝ)→Z[2](𝒟;ℝ), φ2:L∞(𝒟;ℝ)→Z[1](𝒟;ℝ) and thus the operators A0, A1, A2 mapping the space L∞(𝒟;ℝ) into itself are linear and bounded.Theorem 4.1.
LetA=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 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).Theorem4.1 implies the following.Corollary 4.2.
If the inequality(4.3)(b-a)(d-c)‖l0‖+(d-c)‖l1‖+(b-a)‖l2‖<1
holds then problem (1.1), (1.2) is uniquely solvable for arbitrary 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).Remark 4.3.
On the rectangle[a,b]×[c,d], we consider the equation
(4.4)u(1,1)(t,x)=p0u(b,d)+p1u(1,0)(t,d)+p2u(0,1)(b,x)
subjected to the initial conditions
(4.5)u(t,c)=0fort∈[a,b],u(a,x)=0forx∈[c,d],
where
(4.6)p0=m0(b-a)(d-c),p1=m1d-c,p2=m2b-a.
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)≡(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).If the operatorsℓ0, ℓ1, and ℓ2 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 operatorsℓ0, ℓ1, ℓ2 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 functionz0∈L∞(𝒟;ℝ+) such that z0≫A(z0).(2)
The spectral radius of the operatorA is less than one.(3)
Problem (1.1), (1.2) is uniquely solvable for arbitrary 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).If, in addition, the initial functionsα,βand the forcing term qare such that
(4.7)α(a)≥0,α′(t)≥0,β′(x)≥0,q(t,x)≥0fora.e.(t,x)∈D,
then the solution uto problem (1.1), (1.2) satisfies
(4.8)u(t,x)≥0for(t,x)∈D,u(1,0)(t,x)≥0fora.e.t∈[a,b]and allx∈[c,d],u(0,1)(t,x)≥0fort∈[a,b]anda.e.x∈[c,d].(4)
There exists a functionγ∈C*(𝒟;ℝ) such that(4.9)γ(1,1)≫l0(γ)+l1(γ(1,0))+l2(γ(0,1)),(4.10)γ(a,c)≥0,(4.11)γ(1,0)(t,c)≥0fora.e.t∈[a,b],γ(0,1)(a,x)≥0fora.e.x∈[c,d],(4.12)γ(1,1)(t,x)≥0fora.e.(t,x)∈D.For Volterra-type operatorsℓ0, ℓ1, and ℓ2, we derive from the previous theorem the following.Corollary 4.5.
Letℓ0, ℓ1, and ℓ2 be positive (a,c)-Volterra, c-Volterra, and a-Volterra operators, respectively, such that the inequalities
(4.13)l1(y)(t,x)≤y(t)l1(1)(t,x)fora.e.(t,x)∈D(here,ℓ1(y) means ℓ1(y¯) in which  y¯(t,x)=y(t) for a.e.t∈[a,b] and all x∈[c,d])  and
(4.14)l2(z)(t,x)≤z(x)l2(1)(t,x)fora.e.(t,x)∈D(byℓ2(z) we mean ℓ2(z¯), where z¯(t,x)=z(x)forallt∈[a,b] and a.e.  x∈[c,d]) hold for every y∈L∞([a,b];ℝ) and z∈L∞([c,d];ℝ).
Then problem (1.1), (1.2) is uniquely solvable for arbitrary 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). 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, whereℓ1=0 and ℓ2=0 (see [18]), we can introduce the following.Definition 4.6.
Letℓ0:C(𝒟;ℝ)→L∞(𝒟;ℝ), ℓ1:Z[2](𝒟;ℝ)→L∞(𝒟;ℝ), and ℓ2:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ). We say that the triplet (ℓ0,ℓ1,ℓ2) belongs to the set 𝒮ac′ if the implication
(4.15)u∈C*(D;R),u(1,1)(t,x)≥l0(u)(t,x)+l1(u(1,0))(t,x)+l2(u(0,1))(t,x),fora.e.(t,x)∈D,u(a,c)≥0,u(1,0)(t,c)≥0fora.e.t∈[a,b],u(0,1)(a,x)≥0fora.e.x∈[c,d],⟹usatisfies(4.8)
holds.Remark 4.7.
If(ℓ0,ℓ1,ℓ2)∈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 (ℓ0,ℓ1,ℓ2)∈𝒮ac′ is fulfilled.Theorem4.4 immediately yields the following.Corollary 4.8.
If one of assertions (1)–(4) stated in Theorem4.4 holds then (ℓ0,ℓ1,ℓ2)∈𝒮ac′.Remark 4.9.
The inclusion(ℓ0,ℓ1,ℓ2)∈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,q∈L∞(𝒟;ℝ) and argument deviations τ0,τ2:𝒟→[a,b], μ0,μ1:𝒟→[c,d] are measurable functions.As a consequence of Corollary4.2 we obtain the following.Corollary 4.10.
If the inequality(4.16)(b-a)(d-c)‖p0‖L∞+(d-c)‖p1‖L∞+(b-a)‖p2‖L∞<1
holds, then problem (3.2), (1.2) is uniquely solvable for arbitrary 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).If the coefficientsp0, 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.
Letp0,p1,p2∈L∞(𝒟;ℝ+) and
(4.17)esssup⁡{p0(t,x)(τ0(t,x)-a)(μ0(t,x)-c)+p1(t,x)(μ1(t,x)-c)+p2(t,x)(τ2(t,x)-a):(t,x)∈D}<1.
Then problem (3.2), (1.2) is uniquely solvable for arbitrary 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). 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, Corollary4.5 implies the following.Corollary 4.12.
Letp0,p1,p2∈L∞(𝒟;ℝ+) 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 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). 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 (4.18)τk(t,x)-t,μj(t,x)-x(k=0,2,j=0,1)
are small enough, that is, if (3.2) with deviating arguments is “close” to (3.1).Corollary 4.13.
Letp0,p1,p2∈L∞(𝒟;ℝ+),pk≢0(k=1,2), and
(4.19)esssup⁡{∫tτ0(t,x)∫cμ0(t,x)p0(s,η)dηds+∫at∫xμ0(t,x)p0(s,η)dηds+2‖p2‖L∞(τ0(t,x)-t)+2‖p1‖L∞(μ0(t,x)-x):(t,x)∈D∫at}≤ωe(1+ln⁡1ω),esssup⁡⁡{∫at∫xμ1(t,x)p0(s,η)dηds+2‖p1‖L∞(μ1(t,x)-x):(t,x)∈D}<ωe,esssup⁡⁡{∫tτ2(t,x)∫cxp0(s,η)dηds+2‖p2‖L∞(τ2(t,x)-t):(t,x)∈D}<ωe,where
(4.20)ω=2min⁡⁡{‖p1‖L∞,‖p2‖L∞}‖p0‖L∞max⁡⁡{b-a,d-c}+2min⁡⁡{‖p1‖L∞,‖p2‖L∞}.
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 operatorsℓ0, ℓ1, and ℓ2, the operators A0, A1, and A2 are defined by relations (4.1), (4.2). Moreover, having 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), we put
(5.1)y=l0(-α(a)+α+β)+l1(α′)+l2(β′)+q(byℓ0(-α(a)+α+β) the authors understand ℓ0(σ) in which σ(t,x)=-α(a)+α(t)+β(x) for (t,x)∈𝒟. Similarly, ℓ1(α′)(resp.,ℓ2(β′)) means ℓ1(α0)(resp.,ℓ2(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, y∈L∞(𝒟;ℝ).Lemma 5.1.
Ifu is a solution to problem (1.1), (1.2) then u(1,1) is a solution to the equation
(5.2)z=(A0+A1+A2)(z)+y
in the space L∞(𝒟;ℝ), 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, ifz is a solution to (5.2) in the space L∞(𝒟;ℝ) with the operators A0, A1, and A2 defined by relations (4.1), (4.2) and the function y given by formula (5.1), then(5.3)u(t,x)=-α(a)+α(t)+β(x)+∫at∫cxz(s,η)dηdsfor(t,x)∈D
is a solution to problem (1.1), (1.2).Proof.
Ifu is a solution to problem (1.1), (1.2) then, by virtue of Remark 2.4, we get u(1,1)∈L∞(𝒟;ℝ),
(5.4)u(t,x)=-α(a)+α(t)+β(x)+∫at∫cxu(1,1)(s,η)dηdsfor(t,x)∈D,u(1,0)(t,x)=α′(t)+∫cxu(1,1)(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],u(0,1)(t,x)=β′(x)+∫atu(1,1)(s,x)dsforallt∈[a,b]anda.e.x∈[c,d].
Consequently, (1.1) yields that
(5.5)u(1,1)=(A0+A1+A2)(u(1,1))+y,
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, letz be a solution to (5.2) in the space L∞(𝒟;ℝ) 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,(5.6)u(t,x)=α(a)+∫atα′(s)ds+∫cxβ′(η)dη+∫at∫cxz(s,η)dηdsfor(t,x)∈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
(5.7)u(1,0)(t,x)=α′(t)+∫cxz(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],u(0,1)(t,x)=β′(x)+∫atz(s,x)dsforallt∈[a,b]anda.e.x∈[c,d],u(1,1)(t,x)=z(t,x)fora.e.(t,x)∈D.
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 setK 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)
ifx∈K and -x∈K 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 coneK⊆X is solid if its interior IntK is nonempty.Remark 5.5.
The presence of a coneK in a Banach space X allows one to introduce a natural partial ordering there. More precisely, two elements x1,x2∈X are said to be in the relation x2≥Kx1 if and only if they satisfy the inclusion x2-x1∈K.  If, moreover, K is a solid cone then we write x2≫Kx1 if and only if x2-x1∈IntK.Definition 5.6.
A coneK⊆X is said to be normal if there is a constant N≥0 such that, for every x,y∈X with the property  0≤Kx≤Ky, the relation ∥x∥X≤N∥y∥X holds.The proof of the main part of Theorem4.4 is based on the following result.Lemma 5.7 (see [24, Theorem 5.6]).
LetK 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 x0∈IntK such that δx0-A(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 Corollary4.5.Lemma 5.8.
Letℓ0:C(𝒟;ℝ)→L∞(𝒟;ℝ) be a positive (a,c)-Volterra operator. Then, for any function γ∈C(𝒟;ℝ) satisfying
(5.8)γ(t1,x1)≤γ(t2,x2)fora≤t1≤t2≤b,c≤x1≤x2≤d,
one has
(5.9)l0(γ)(t,x)≤l0(1)(t,x)γ(t,x)fora.e.(t,x)∈D.Proof.
We first show that, for any(t,x)∈]a,b]×]c,d], we have
(5.10)l0(γ)(s,η)≤l0(1)(s,η)γ(t,x)fora.e.(s,η)∈[a,t]×[c,x].
Indeed, let (t,x)∈]a,b]×]c,d] be arbitrary but fixed. Put
(5.11)γ0(s,η)=γ(min⁡{s,t},min⁡{η,x})for(s,η)∈D.
Then, clearly γ0∈C(𝒟;ℝ),
(5.12)γ0(s,η)≤γ(t,x)for(s,η)∈D,γ0(s,η)=γ(s,η)for(s,η)∈[a,t]×[c,x].
Since the operator ℓ0 is positive, we obtain
(5.13)l0(γ0)(s,η)≤l0(γ(t,x))(s,η)=γ(t,x)l0(1)(s,η)fora.e.(s,η)∈D.
On the other hand, the operator ℓ0 is supposed to be an (a,c)-Volterra one which guarantees the equality
(5.14)l0(γ0)(s,η)=l0(γ)(s,η)fora.e.(s,η)∈[a,t]×[c,x],
and thus the desired relation (5.10) holds.
Now we put(5.15)u(t,x)=∫at∫cxl0(γ)(s,η)dηds,v(t,x)=∫at∫cxl0(1)(s,η)dηdsfor(t,x)∈D.
It follows from Lemma 2.3 that there exists a set E1⊆]a,b], meas E1=b-a, such that
(5.16)u(1,0)(t,x)=∫cxl0(γ)(t,η)dηfort∈E1,x∈[c,d],v(1,0)(t,x)=∫cxl0(1)(t,η)dηfort∈E1,x∈[c,d],
and, moreover, there is a set E⊆E1×]c,d], meas E=(b-a)(d-c), with the properties
(5.17)u(1,1)(t,x)=l0(γ)(t,x),v(1,1)(t,x)=l0(1)(t,x)for(t,x)∈E.
Let(t,x)∈E be arbitrary but fixed. Then, relation (5.10) yields
(5.18)1hk∫t-ht∫x-kxl0(γ)(s,η)dηds≤γ(t,x)hk∫t-ht∫x-kxl0(1)(s,η)dηds
for h∈]0,t-a] and k∈]0,x-c], whence we get
(5.19)1k[u(t,x)-u(t-h,x)h-u(t,x-k)-u(t-h,x-k)h]≤γ(t,x)k[v(t,x)-v(t-h,x)h-v(t,x-k)-v(t-h,x-k)h],forh∈]0,t-a],k∈]0,x-c].
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
(5.20)1k[u(1,0)(t,x)-u(1,0)(t,x-k)]≤γ(t,x)k[v(1,0)(t,x)-v(1,0)(t,x-k)],
for k∈]0,x-c]. Now, letting k→0+ in the previous relation and using equalities (5.17) give
(5.21)l0(γ)(t,x)=u(1,1)(t,x)≤γ(t,x)v(1,1)(t,x)=γ(t,x)l0(1)(t,x).
That is, the desired inequality (5.9) holds because (t,x)∈E was arbitrary.Lemma 5.9.
Letℓ2:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) be a positive a-Volterra operator such that inequality (4.14) holds for every z∈L∞([c,d];ℝ). Then, for any function γ∈Z[1](𝒟;ℝ) with the property
(5.22)γ(t1,x)≤γ(t2,x)fora≤t1≤t2≤banda.e.x∈[c,d],
one has
(5.23)l2(γ)(t,x)≤l2(1)(t,x)γ(t,x)fora.e.(t,x)∈D.Proof.
LetE1⊆[c,d], meas E1=d-c, be a set such that, for any x∈E1, we have γ(·,x)∈C([a,b];ℝ) and
(5.24)γ(t1,x)≤γ(t2,x)fora≤t1≤t2≤b.
We first show that the relation
(5.25)l2(γ)(s,x)≤l2(1)(s,x)γ(t,x)fora.e.(s,x)∈[a,t]×[c,d]
holds for every t∈]a,b]. Indeed, let t∈]a,b] be arbitrary but fixed. Put
(5.26)γ0(s,x)=γ(min⁡{s,t},x)fors∈[a,b],x∈E1,γ1(x)=γ(t,x)forx∈E1.
Then, clearly, γ0∈Z[1](𝒟;ℝ),γ1∈L∞([c,d];ℝ),
(5.27)γ0(s,x)≤γ1(x)fors∈[a,b],x∈E1,γ0(s,x)=γ(s,x)fors∈[a,t],x∈E1.
Since the operator ℓ2 is positive and satisfies condition (4.14), we obtain
(5.28)l2(γ0)(s,x)≤l2(γ1)(s,x)≤γ1(x)l2(1)(s,x)fora.e.(s,x)∈D(by ℓ2(γ1) the authors mean ℓ2(γ¯1), where γ¯1(s,x)=γ1(x) for all s∈[a,b] and x∈E1). On the other hand, the operator ℓ2 is supposed to be an a-Volterra one which guarantees the equality
(5.29)l2(γ0)(s,x)=l2(γ)(s,x)fora.e.(s,x)∈[a,t]×[c,d],
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=d-c such that
(5.30)l2(γ)(s,x)≤l2(1)(s,x)γ(t,x)forx∈At,s∈Bt(x),
where, for each x∈At, we have Bt(x)⊆[a,b] with meas Bt(x)=t-a.
PutE2=∩t∈CAt, where C=]a,b]∩ℚ. Clearly, meas E2=d-c because the set C is countable. Moreover, relation (5.30) yields that(5.31)∫t0tl2(γ)(s,x)ds≤γ(t,x)∫t0tl2(1)(s,x)dsforx∈E2,t∈C,t0∈[a,t].
Let t∈]a,b], t0∈[a,t[, and x∈E1∩E2 be arbitrary but fixed. Then there exists a sequence {tn}n=1+∞⊂[t0,b]∩ℚ such that tn→t as n→+∞. It follows from relation (5.31) that
(5.32)∫t0tnl2(γ)(s,x)ds≤γ(tn,x)∫t0tnl2(1)(s,x)dsforn∈N,
whence we get
(5.33)∫t0tl2(γ)(s,x)ds≤γ(t,x)∫t0tl2(1)(s,x)ds.
Consequently, we have proved that
(5.34)∫t0tl2(γ)(s,x)ds≤γ(t,x)∫t0tl2(1)(s,x)ds,forx∈E1∩E2,a≤t0<t≤b.
Now we put(5.35)u(t,x)=∫atl2(γ)(s,x)ds,v(t,x)=∫atl2(1)(s,x)dsfort∈[a,b],x∈E1∩E2.
Lemma 2.3 guarantees that there exists a set E⊆]a,b]×(E1∩E2) such that meas E=(b-a)(d-c) and
(5.36)u(1,0)(t,x)=l2(γ)(t,x),v(1,0)(t,x)=l2(1)(t,x)for(t,x)∈E.
Let (t,x)∈E be arbitrary but fixed. We choose a sequence {hn}n=1+∞ of numbers from the interval ]0,t-a] such that
(5.37)lim⁡n→+∞hn=0.
Then relation (5.34) yields that
(5.38)1hn∫t-hntl2(γ)(s,x)ds≤γ(t,x)hn∫t-hntl2(1)(s,x)dsforn∈N.
Letting n→+∞ in the previous relation and using equalities (5.36) give
(5.39)l2(γ)(t,x)=u(1,0)(t,x)≤γ(t,x)v(1,0)(t,x)=γ(t,x)l2(1)(t,x).
Consequently, desired inequality (5.23) holds because (t,x)∈E was arbitrary.Lemma 5.10.
Letℓ1: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
(5.40)γ(t,x1)≤γ(t,x2)fora.e.t∈[a,b]andallc≤x1≤x2≤d,
one has
(5.41)l1(γ)(t,x)≤l1(1)(t,x)γ(t,x)fora.e.(t,x)∈D.Proof.
Lemma can be proven analogously as Lemma5.9 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 Section4.Proof of Theorem4.1.
Since the spectral radius of the linear bounded operatorA: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 Corollary4.2.
It is easy to show that the norms of the linear bounded operatorsA0,A1,A2:L∞(𝒟;ℝ)→L∞(𝒟;ℝ) defined by relations (4.1), (4.2) satisfy the estimates
(5.42)‖A0‖≤(b-a)(d-c)‖l0‖,‖A1‖≤(d-c)‖l1‖,‖A2‖≤(b-a)‖l2‖.
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 Theorem4.4.
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∈IntK 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 ℓ0, ℓ1, and ℓ2 are supposed to be positive. Therefore, the assumptions z0∈K and z0≫A(z0) yield that z0≫0 as well.
Applying Lemma5.7 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
(5.43)z=y+A(y)+A2(y)+⋯.
Consequently, in view of Lemma 5.1, problem (1.1), (1.2) has a unique solution u 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).
Assume, in addition, that the initial functionsα, 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
(5.44)u(t,x)=α(a)+∫atα′(s)ds+∫cxβ′(η)dη+∫at∫cxz(s,η)dηdsfor(t,x)∈D,
in which z 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
(5.45)γ(1,1)(t,x)=l0(γ)(t,x)+l1(γ(1,0))(t,x)+l2(γ(0,1))(t,x)+1,(5.46)γ(t,c)=0fort∈[a,b],γ(a,x)=0forx∈[c,d]
has a unique solution γ and the function γ∈C*(𝒟;ℝ) satisfies the inequalities
(5.47)γ(t,x)≥0for(t,x)∈D,γ(1,0)(t,x)≥0fora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)≥0fort∈[a,b]anda.e.x∈[c,d].
Hence, conditions (4.9)–(4.11) are fulfilled. Moreover, inequality (4.12) follows from (5.45) because the operators ℓ0, ℓ1, and ℓ2 are positive and γ satisfies relations (5.47).
(4)⇒(1): assume that there exists a function γ∈C*(𝒟;ℝ) satisfying inequalities (4.9)–(4.12). Then, by virtue of Remark 2.4, we get γ(1,1)∈L∞(𝒟;ℝ+),
(5.48)γ(t,x)≥∫at∫cxγ(1,1)(s,η)dηdsfor(t,x)∈D,γ(1,0)(t,x)≥∫cxγ(1,1)(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)≥∫atγ(1,1)(s,x)ds,forallt∈[a,b],a.e.x∈[c,d].
Consequently,  assertion (1) of the theorem holds with z0=γ(1,1) because the operators ℓ0, ℓ1, and ℓ2 are positive.Proof of Corollary4.5.
We put(5.49)γ(t,x)=e∫at∫cx(1+l0(1)(s,η))dηds+2g2(t-a)+2g1(x-c)for(t,x)∈D,
where g1=∥ℓ1(1)∥L∞ and g2=∥ℓ2(1)∥L∞. It can be verified that γ∈C*(𝒟;ℝ) and satisfies the inequality
(5.50)γ(t,x)≥1for(t,x)∈D.
In view of Lemma 2.3, from (5.49) we get
(5.51)γ(1,0)(t,x)=(∫cx(1+l0(1)(t,η))dη+2g2)γ(t,x),fora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)=(∫at(1+l0(1)(s,x))ds+2g1)γ(t,x),fort∈[a,b]anda.e.x∈[c,d].
Therefore, inequalities (4.10) and (4.11) are fulfilled and
(5.52)γ(1,0)(t,x)≥0,γ(0,1)(t,x)≥0fora.e.(t,x)∈D,
because the operator ℓ0 is positive. Moreover, by using Lemma 2.3 and inequalities (5.50) and (5.52), it follows from equalities (5.51)   that
(5.53)γ(1,1)(t,x)=(1+l0(1)(t,x))γ(t,x)+(∫cx(1+l0(1)(t,η))dη+2g2)γ(0,1)(t,x)=(1+l0(1)(t,x))γ(t,x)+12(∫at(1+l0(1)(s,x))ds+2g1)γ(1,0)(t,x)+12(∫cx(1+l0(1)(t,η))dη+2g2)γ(0,1)(t,x)≥1+l0(1)(t,x)γ(t,x)+g1γ(1,0)(t,x)+g2γ(0,1)(t,x)≥l0(1)(t,x)γ(t,x)+l1(1)(t,x)γ(1,0)(t,x)+l2(1)(t,x)γ(0,1)(t,x)+1,
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 inequalities(5.54)γ(t1,x1)≤γ(t2,x2)fora≤t1≤t2≤b,c≤x1≤x2≤d,γ(1,0)(t,x1)≤γ(1,0)(t,x2)fora.e.t∈[a,b]andallc≤x1≤x2≤d,γ(0,1)(t1,x)≤γ(0,1)(t2,x)fora≤t1≤t2≤banda.e.x∈[c,d].
Therefore, by using Lemmas 5.8–5.10, from inequality (5.53) we get
(5.55)γ(1,1)(t,x)≥l0(γ)(t,x)+l1(γ(1,0))(t,x)+l2(γ(0,1))(t,x)+1fora.e.(t,x)∈D.
that is, relation (4.9) holds.
Consequently, we have found a functionγ satisfying conditions (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.Proof of Corollary4.10.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0, ℓ1, and ℓ2 are defined by formulas (2.23), (2.28), and (2.29), respectively (see Examples 2.7 and 2.11). Since we have
(5.56)‖lk‖=‖pk‖L∞fork=0,1,2,
the validity of the corollary follows immediately from Corollary 4.2.Proof of Corollary4.11.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0,ℓ1, and ℓ2 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 ɛ>0 such that
(5.57)1≥p0(t,x)(τ0(t,x)-a)(μ0(t,x)-c)+p1(t,x)(μ1(t,x)-c)+p2(t,x)(τ2(t,x)-a)+ɛfora.e.(t,x)∈D.
Therefore, the function γ defined by the relation
(5.58)γ(t,x)=(t-a)(x-c)for(t,x)∈D
satisfies inequalities (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.Proof of Corollary4.12.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0, ℓ1, and ℓ2 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 ℓ0, ℓ1 and ℓ2 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 Corollary4.5.Proof of Corollary4.13.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0, ℓ1, and ℓ2 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).
Letp̃k=∥pk∥L∞(k=0,1,2) and y=e/ω. According to (4.19) and (4.20), there exist numbers ɛ>0 and δ∈]0,1[ such that inequalities
(5.59)∫tτ0(t,x)∫cμ0(t,x)(ɛ+p0(s,η))dηds+∫at∫xμ0(t,x)(ɛ+p0(s,η))dηds+2p̃2(τ0(t,x)-t)+2p̃1(μ0(t,x)-x)≤1yln⁡(y+yδey[∫ab∫cd(ɛ+p0(s,η))dηds+2p̃2(b-a)+2p̃1(d-c)]-δ),∫at∫xμ1(t,x)(ɛ+p0(s,η))dηds+2p̃1(μ1(t,x)-x)≤1yln⁡2p̃2y(ɛ+p̃0)(d-c)+2p̃2,∫tτ2(t,x)∫cx(ɛ+p0(s,η))dηds+2p̃2(τ2(t,x)-t)≤1yln⁡2p̃1y(ɛ+p̃0)(b-a)+2p̃1,
hold for a.e. (t,x)∈𝒟. Now we put
(5.60)γ(t,x)=eyz(t,x)-δfor(t,x)∈D,
where
(5.61)z(t,x)=∫at∫cx(ɛ+p0(s,η))dηds+2p̃2(t-a)+2p̃1(x-c)for(t,x)∈D.
It can be verified that γ∈C*(𝒟;ℝ) and satisfies
(5.62)γ(t,x)≥1-δ>0for(t,x)∈D.
In view of Lemma 2.3, from (5.62) we get
(5.63)γ(1,0)(t,x)=y(∫cx(ɛ+p0(t,η))dη+2p̃2)(γ(t,x)+δ),fora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)=y(∫at(ɛ+p0(s,x))ds+2p̃1)(γ(t,x)+δ),fort∈[a,b]anda.e.x∈[c,d].
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
(5.64)γ(1,1)(t,x)=y(ɛ+p0(t,x))(γ(t,x)+δ)+y(∫cx(ɛ+p0(t,η))dη+2p̃2)γ(0,1)(t,x)=yɛ(γ(t,x)+δ)+yp0(t,x)(γ(t,x)+δ)+y2(∫at(ɛ+p0(s,x))ds+2p̃1)γ(1,0)(t,x)+y2(∫cx(ɛ+p0(t,η))dη+2p̃2)γ(0,1)(t,x)≥yɛ+p0(t,x)y(γ(t,x)+δ)+yp̃1γ(1,0)(t,x)+yp̃2γ(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)+yɛ,
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
(5.65)z(τ0(t,x),μ0(t,x))-z(t,x)≤1yln⁡(y+yδeyz(τ0(t,x),μ0(t,x))-δ),z(t,μ1(t,x))-z(t,x)≤1yln⁡2p̃2y∫cμ1(t,x)(ɛ+p0(t,η))dη+2p̃2,z(τ2(t,x),x)-z(t,x)≤1yln⁡2p̃1y∫aτ2(t,x)(ɛ+p0(s,x))ds+2p̃1,
for a.e. (t,x)∈𝒟 and thus, by using relations (5.60), (5.63), we get
(5.66)y(γ(t,x)+δ)≥γ(τ0(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)
for a.e. (t,x)∈𝒟. Consequently, it follows from (5.64) that inequality (4.9) holds.
We have constructed a functionγ satisfying conditions (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.


## 5.1. Auxiliary Statements

Remember that, for given operatorsℓ0, ℓ1, and ℓ2, the operators A0, A1, and A2 are defined by relations (4.1), (4.2). Moreover, having 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), we put
(5.1)y=l0(-α(a)+α+β)+l1(α′)+l2(β′)+q(byℓ0(-α(a)+α+β) the authors understand ℓ0(σ) in which σ(t,x)=-α(a)+α(t)+β(x) for (t,x)∈𝒟. Similarly, ℓ1(α′)(resp.,ℓ2(β′)) means ℓ1(α0)(resp.,ℓ2(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, y∈L∞(𝒟;ℝ).Lemma 5.1.
Ifu is a solution to problem (1.1), (1.2) then u(1,1) is a solution to the equation
(5.2)z=(A0+A1+A2)(z)+y
in the space L∞(𝒟;ℝ), 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, ifz is a solution to (5.2) in the space L∞(𝒟;ℝ) with the operators A0, A1, and A2 defined by relations (4.1), (4.2) and the function y given by formula (5.1), then(5.3)u(t,x)=-α(a)+α(t)+β(x)+∫at∫cxz(s,η)dηdsfor(t,x)∈D
is a solution to problem (1.1), (1.2).Proof.
Ifu is a solution to problem (1.1), (1.2) then, by virtue of Remark 2.4, we get u(1,1)∈L∞(𝒟;ℝ),
(5.4)u(t,x)=-α(a)+α(t)+β(x)+∫at∫cxu(1,1)(s,η)dηdsfor(t,x)∈D,u(1,0)(t,x)=α′(t)+∫cxu(1,1)(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],u(0,1)(t,x)=β′(x)+∫atu(1,1)(s,x)dsforallt∈[a,b]anda.e.x∈[c,d].
Consequently, (1.1) yields that
(5.5)u(1,1)=(A0+A1+A2)(u(1,1))+y,
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, letz be a solution to (5.2) in the space L∞(𝒟;ℝ) 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,(5.6)u(t,x)=α(a)+∫atα′(s)ds+∫cxβ′(η)dη+∫at∫cxz(s,η)dηdsfor(t,x)∈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
(5.7)u(1,0)(t,x)=α′(t)+∫cxz(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],u(0,1)(t,x)=β′(x)+∫atz(s,x)dsforallt∈[a,b]anda.e.x∈[c,d],u(1,1)(t,x)=z(t,x)fora.e.(t,x)∈D.
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 setK 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)
ifx∈K and -x∈K 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 coneK⊆X is solid if its interior IntK is nonempty.Remark 5.5.
The presence of a coneK in a Banach space X allows one to introduce a natural partial ordering there. More precisely, two elements x1,x2∈X are said to be in the relation x2≥Kx1 if and only if they satisfy the inclusion x2-x1∈K.  If, moreover, K is a solid cone then we write x2≫Kx1 if and only if x2-x1∈IntK.Definition 5.6.
A coneK⊆X is said to be normal if there is a constant N≥0 such that, for every x,y∈X with the property  0≤Kx≤Ky, the relation ∥x∥X≤N∥y∥X holds.The proof of the main part of Theorem4.4 is based on the following result.Lemma 5.7 (see [24, Theorem 5.6]).
LetK 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 x0∈IntK such that δx0-A(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 Corollary4.5.Lemma 5.8.
Letℓ0:C(𝒟;ℝ)→L∞(𝒟;ℝ) be a positive (a,c)-Volterra operator. Then, for any function γ∈C(𝒟;ℝ) satisfying
(5.8)γ(t1,x1)≤γ(t2,x2)fora≤t1≤t2≤b,c≤x1≤x2≤d,
one has
(5.9)l0(γ)(t,x)≤l0(1)(t,x)γ(t,x)fora.e.(t,x)∈D.Proof.
We first show that, for any(t,x)∈]a,b]×]c,d], we have
(5.10)l0(γ)(s,η)≤l0(1)(s,η)γ(t,x)fora.e.(s,η)∈[a,t]×[c,x].
Indeed, let (t,x)∈]a,b]×]c,d] be arbitrary but fixed. Put
(5.11)γ0(s,η)=γ(min⁡{s,t},min⁡{η,x})for(s,η)∈D.
Then, clearly γ0∈C(𝒟;ℝ),
(5.12)γ0(s,η)≤γ(t,x)for(s,η)∈D,γ0(s,η)=γ(s,η)for(s,η)∈[a,t]×[c,x].
Since the operator ℓ0 is positive, we obtain
(5.13)l0(γ0)(s,η)≤l0(γ(t,x))(s,η)=γ(t,x)l0(1)(s,η)fora.e.(s,η)∈D.
On the other hand, the operator ℓ0 is supposed to be an (a,c)-Volterra one which guarantees the equality
(5.14)l0(γ0)(s,η)=l0(γ)(s,η)fora.e.(s,η)∈[a,t]×[c,x],
and thus the desired relation (5.10) holds.
Now we put(5.15)u(t,x)=∫at∫cxl0(γ)(s,η)dηds,v(t,x)=∫at∫cxl0(1)(s,η)dηdsfor(t,x)∈D.
It follows from Lemma 2.3 that there exists a set E1⊆]a,b], meas E1=b-a, such that
(5.16)u(1,0)(t,x)=∫cxl0(γ)(t,η)dηfort∈E1,x∈[c,d],v(1,0)(t,x)=∫cxl0(1)(t,η)dηfort∈E1,x∈[c,d],
and, moreover, there is a set E⊆E1×]c,d], meas E=(b-a)(d-c), with the properties
(5.17)u(1,1)(t,x)=l0(γ)(t,x),v(1,1)(t,x)=l0(1)(t,x)for(t,x)∈E.
Let(t,x)∈E be arbitrary but fixed. Then, relation (5.10) yields
(5.18)1hk∫t-ht∫x-kxl0(γ)(s,η)dηds≤γ(t,x)hk∫t-ht∫x-kxl0(1)(s,η)dηds
for h∈]0,t-a] and k∈]0,x-c], whence we get
(5.19)1k[u(t,x)-u(t-h,x)h-u(t,x-k)-u(t-h,x-k)h]≤γ(t,x)k[v(t,x)-v(t-h,x)h-v(t,x-k)-v(t-h,x-k)h],forh∈]0,t-a],k∈]0,x-c].
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
(5.20)1k[u(1,0)(t,x)-u(1,0)(t,x-k)]≤γ(t,x)k[v(1,0)(t,x)-v(1,0)(t,x-k)],
for k∈]0,x-c]. Now, letting k→0+ in the previous relation and using equalities (5.17) give
(5.21)l0(γ)(t,x)=u(1,1)(t,x)≤γ(t,x)v(1,1)(t,x)=γ(t,x)l0(1)(t,x).
That is, the desired inequality (5.9) holds because (t,x)∈E was arbitrary.Lemma 5.9.
Letℓ2:Z[1](𝒟;ℝ)→L∞(𝒟;ℝ) be a positive a-Volterra operator such that inequality (4.14) holds for every z∈L∞([c,d];ℝ). Then, for any function γ∈Z[1](𝒟;ℝ) with the property
(5.22)γ(t1,x)≤γ(t2,x)fora≤t1≤t2≤banda.e.x∈[c,d],
one has
(5.23)l2(γ)(t,x)≤l2(1)(t,x)γ(t,x)fora.e.(t,x)∈D.Proof.
LetE1⊆[c,d], meas E1=d-c, be a set such that, for any x∈E1, we have γ(·,x)∈C([a,b];ℝ) and
(5.24)γ(t1,x)≤γ(t2,x)fora≤t1≤t2≤b.
We first show that the relation
(5.25)l2(γ)(s,x)≤l2(1)(s,x)γ(t,x)fora.e.(s,x)∈[a,t]×[c,d]
holds for every t∈]a,b]. Indeed, let t∈]a,b] be arbitrary but fixed. Put
(5.26)γ0(s,x)=γ(min⁡{s,t},x)fors∈[a,b],x∈E1,γ1(x)=γ(t,x)forx∈E1.
Then, clearly, γ0∈Z[1](𝒟;ℝ),γ1∈L∞([c,d];ℝ),
(5.27)γ0(s,x)≤γ1(x)fors∈[a,b],x∈E1,γ0(s,x)=γ(s,x)fors∈[a,t],x∈E1.
Since the operator ℓ2 is positive and satisfies condition (4.14), we obtain
(5.28)l2(γ0)(s,x)≤l2(γ1)(s,x)≤γ1(x)l2(1)(s,x)fora.e.(s,x)∈D(by ℓ2(γ1) the authors mean ℓ2(γ¯1), where γ¯1(s,x)=γ1(x) for all s∈[a,b] and x∈E1). On the other hand, the operator ℓ2 is supposed to be an a-Volterra one which guarantees the equality
(5.29)l2(γ0)(s,x)=l2(γ)(s,x)fora.e.(s,x)∈[a,t]×[c,d],
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=d-c such that
(5.30)l2(γ)(s,x)≤l2(1)(s,x)γ(t,x)forx∈At,s∈Bt(x),
where, for each x∈At, we have Bt(x)⊆[a,b] with meas Bt(x)=t-a.
PutE2=∩t∈CAt, where C=]a,b]∩ℚ. Clearly, meas E2=d-c because the set C is countable. Moreover, relation (5.30) yields that(5.31)∫t0tl2(γ)(s,x)ds≤γ(t,x)∫t0tl2(1)(s,x)dsforx∈E2,t∈C,t0∈[a,t].
Let t∈]a,b], t0∈[a,t[, and x∈E1∩E2 be arbitrary but fixed. Then there exists a sequence {tn}n=1+∞⊂[t0,b]∩ℚ such that tn→t as n→+∞. It follows from relation (5.31) that
(5.32)∫t0tnl2(γ)(s,x)ds≤γ(tn,x)∫t0tnl2(1)(s,x)dsforn∈N,
whence we get
(5.33)∫t0tl2(γ)(s,x)ds≤γ(t,x)∫t0tl2(1)(s,x)ds.
Consequently, we have proved that
(5.34)∫t0tl2(γ)(s,x)ds≤γ(t,x)∫t0tl2(1)(s,x)ds,forx∈E1∩E2,a≤t0<t≤b.
Now we put(5.35)u(t,x)=∫atl2(γ)(s,x)ds,v(t,x)=∫atl2(1)(s,x)dsfort∈[a,b],x∈E1∩E2.
Lemma 2.3 guarantees that there exists a set E⊆]a,b]×(E1∩E2) such that meas E=(b-a)(d-c) and
(5.36)u(1,0)(t,x)=l2(γ)(t,x),v(1,0)(t,x)=l2(1)(t,x)for(t,x)∈E.
Let (t,x)∈E be arbitrary but fixed. We choose a sequence {hn}n=1+∞ of numbers from the interval ]0,t-a] such that
(5.37)lim⁡n→+∞hn=0.
Then relation (5.34) yields that
(5.38)1hn∫t-hntl2(γ)(s,x)ds≤γ(t,x)hn∫t-hntl2(1)(s,x)dsforn∈N.
Letting n→+∞ in the previous relation and using equalities (5.36) give
(5.39)l2(γ)(t,x)=u(1,0)(t,x)≤γ(t,x)v(1,0)(t,x)=γ(t,x)l2(1)(t,x).
Consequently, desired inequality (5.23) holds because (t,x)∈E was arbitrary.Lemma 5.10.
Letℓ1: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
(5.40)γ(t,x1)≤γ(t,x2)fora.e.t∈[a,b]andallc≤x1≤x2≤d,
one has
(5.41)l1(γ)(t,x)≤l1(1)(t,x)γ(t,x)fora.e.(t,x)∈D.Proof.
Lemma can be proven analogously as Lemma5.9 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 Section4.Proof of Theorem4.1.
Since the spectral radius of the linear bounded operatorA: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 Corollary4.2.
It is easy to show that the norms of the linear bounded operatorsA0,A1,A2:L∞(𝒟;ℝ)→L∞(𝒟;ℝ) defined by relations (4.1), (4.2) satisfy the estimates
(5.42)‖A0‖≤(b-a)(d-c)‖l0‖,‖A1‖≤(d-c)‖l1‖,‖A2‖≤(b-a)‖l2‖.
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 Theorem4.4.
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∈IntK 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 ℓ0, ℓ1, and ℓ2 are supposed to be positive. Therefore, the assumptions z0∈K and z0≫A(z0) yield that z0≫0 as well.
Applying Lemma5.7 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
(5.43)z=y+A(y)+A2(y)+⋯.
Consequently, in view of Lemma 5.1, problem (1.1), (1.2) has a unique solution u 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).
Assume, in addition, that the initial functionsα, 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
(5.44)u(t,x)=α(a)+∫atα′(s)ds+∫cxβ′(η)dη+∫at∫cxz(s,η)dηdsfor(t,x)∈D,
in which z 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
(5.45)γ(1,1)(t,x)=l0(γ)(t,x)+l1(γ(1,0))(t,x)+l2(γ(0,1))(t,x)+1,(5.46)γ(t,c)=0fort∈[a,b],γ(a,x)=0forx∈[c,d]
has a unique solution γ and the function γ∈C*(𝒟;ℝ) satisfies the inequalities
(5.47)γ(t,x)≥0for(t,x)∈D,γ(1,0)(t,x)≥0fora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)≥0fort∈[a,b]anda.e.x∈[c,d].
Hence, conditions (4.9)–(4.11) are fulfilled. Moreover, inequality (4.12) follows from (5.45) because the operators ℓ0, ℓ1, and ℓ2 are positive and γ satisfies relations (5.47).
(4)⇒(1): assume that there exists a function γ∈C*(𝒟;ℝ) satisfying inequalities (4.9)–(4.12). Then, by virtue of Remark 2.4, we get γ(1,1)∈L∞(𝒟;ℝ+),
(5.48)γ(t,x)≥∫at∫cxγ(1,1)(s,η)dηdsfor(t,x)∈D,γ(1,0)(t,x)≥∫cxγ(1,1)(t,η)dηfora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)≥∫atγ(1,1)(s,x)ds,forallt∈[a,b],a.e.x∈[c,d].
Consequently,  assertion (1) of the theorem holds with z0=γ(1,1) because the operators ℓ0, ℓ1, and ℓ2 are positive.Proof of Corollary4.5.
We put(5.49)γ(t,x)=e∫at∫cx(1+l0(1)(s,η))dηds+2g2(t-a)+2g1(x-c)for(t,x)∈D,
where g1=∥ℓ1(1)∥L∞ and g2=∥ℓ2(1)∥L∞. It can be verified that γ∈C*(𝒟;ℝ) and satisfies the inequality
(5.50)γ(t,x)≥1for(t,x)∈D.
In view of Lemma 2.3, from (5.49) we get
(5.51)γ(1,0)(t,x)=(∫cx(1+l0(1)(t,η))dη+2g2)γ(t,x),fora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)=(∫at(1+l0(1)(s,x))ds+2g1)γ(t,x),fort∈[a,b]anda.e.x∈[c,d].
Therefore, inequalities (4.10) and (4.11) are fulfilled and
(5.52)γ(1,0)(t,x)≥0,γ(0,1)(t,x)≥0fora.e.(t,x)∈D,
because the operator ℓ0 is positive. Moreover, by using Lemma 2.3 and inequalities (5.50) and (5.52), it follows from equalities (5.51)   that
(5.53)γ(1,1)(t,x)=(1+l0(1)(t,x))γ(t,x)+(∫cx(1+l0(1)(t,η))dη+2g2)γ(0,1)(t,x)=(1+l0(1)(t,x))γ(t,x)+12(∫at(1+l0(1)(s,x))ds+2g1)γ(1,0)(t,x)+12(∫cx(1+l0(1)(t,η))dη+2g2)γ(0,1)(t,x)≥1+l0(1)(t,x)γ(t,x)+g1γ(1,0)(t,x)+g2γ(0,1)(t,x)≥l0(1)(t,x)γ(t,x)+l1(1)(t,x)γ(1,0)(t,x)+l2(1)(t,x)γ(0,1)(t,x)+1,
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 inequalities(5.54)γ(t1,x1)≤γ(t2,x2)fora≤t1≤t2≤b,c≤x1≤x2≤d,γ(1,0)(t,x1)≤γ(1,0)(t,x2)fora.e.t∈[a,b]andallc≤x1≤x2≤d,γ(0,1)(t1,x)≤γ(0,1)(t2,x)fora≤t1≤t2≤banda.e.x∈[c,d].
Therefore, by using Lemmas 5.8–5.10, from inequality (5.53) we get
(5.55)γ(1,1)(t,x)≥l0(γ)(t,x)+l1(γ(1,0))(t,x)+l2(γ(0,1))(t,x)+1fora.e.(t,x)∈D.
that is, relation (4.9) holds.
Consequently, we have found a functionγ satisfying conditions (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.Proof of Corollary4.10.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0, ℓ1, and ℓ2 are defined by formulas (2.23), (2.28), and (2.29), respectively (see Examples 2.7 and 2.11). Since we have
(5.56)‖lk‖=‖pk‖L∞fork=0,1,2,
the validity of the corollary follows immediately from Corollary 4.2.Proof of Corollary4.11.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0,ℓ1, and ℓ2 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 ɛ>0 such that
(5.57)1≥p0(t,x)(τ0(t,x)-a)(μ0(t,x)-c)+p1(t,x)(μ1(t,x)-c)+p2(t,x)(τ2(t,x)-a)+ɛfora.e.(t,x)∈D.
Therefore, the function γ defined by the relation
(5.58)γ(t,x)=(t-a)(x-c)for(t,x)∈D
satisfies inequalities (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.Proof of Corollary4.12.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0, ℓ1, and ℓ2 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 ℓ0, ℓ1 and ℓ2 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 Corollary4.5.Proof of Corollary4.13.
It is clear that (3.2) is a particular case of (1.1) in which the operators ℓ0, ℓ1, and ℓ2 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).
Letp̃k=∥pk∥L∞(k=0,1,2) and y=e/ω. According to (4.19) and (4.20), there exist numbers ɛ>0 and δ∈]0,1[ such that inequalities
(5.59)∫tτ0(t,x)∫cμ0(t,x)(ɛ+p0(s,η))dηds+∫at∫xμ0(t,x)(ɛ+p0(s,η))dηds+2p̃2(τ0(t,x)-t)+2p̃1(μ0(t,x)-x)≤1yln⁡(y+yδey[∫ab∫cd(ɛ+p0(s,η))dηds+2p̃2(b-a)+2p̃1(d-c)]-δ),∫at∫xμ1(t,x)(ɛ+p0(s,η))dηds+2p̃1(μ1(t,x)-x)≤1yln⁡2p̃2y(ɛ+p̃0)(d-c)+2p̃2,∫tτ2(t,x)∫cx(ɛ+p0(s,η))dηds+2p̃2(τ2(t,x)-t)≤1yln⁡2p̃1y(ɛ+p̃0)(b-a)+2p̃1,
hold for a.e. (t,x)∈𝒟. Now we put
(5.60)γ(t,x)=eyz(t,x)-δfor(t,x)∈D,
where
(5.61)z(t,x)=∫at∫cx(ɛ+p0(s,η))dηds+2p̃2(t-a)+2p̃1(x-c)for(t,x)∈D.
It can be verified that γ∈C*(𝒟;ℝ) and satisfies
(5.62)γ(t,x)≥1-δ>0for(t,x)∈D.
In view of Lemma 2.3, from (5.62) we get
(5.63)γ(1,0)(t,x)=y(∫cx(ɛ+p0(t,η))dη+2p̃2)(γ(t,x)+δ),fora.e.t∈[a,b]andallx∈[c,d],γ(0,1)(t,x)=y(∫at(ɛ+p0(s,x))ds+2p̃1)(γ(t,x)+δ),fort∈[a,b]anda.e.x∈[c,d].
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
(5.64)γ(1,1)(t,x)=y(ɛ+p0(t,x))(γ(t,x)+δ)+y(∫cx(ɛ+p0(t,η))dη+2p̃2)γ(0,1)(t,x)=yɛ(γ(t,x)+δ)+yp0(t,x)(γ(t,x)+δ)+y2(∫at(ɛ+p0(s,x))ds+2p̃1)γ(1,0)(t,x)+y2(∫cx(ɛ+p0(t,η))dη+2p̃2)γ(0,1)(t,x)≥yɛ+p0(t,x)y(γ(t,x)+δ)+yp̃1γ(1,0)(t,x)+yp̃2γ(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)+yɛ,
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
(5.65)z(τ0(t,x),μ0(t,x))-z(t,x)≤1yln⁡(y+yδeyz(τ0(t,x),μ0(t,x))-δ),z(t,μ1(t,x))-z(t,x)≤1yln⁡2p̃2y∫cμ1(t,x)(ɛ+p0(t,η))dη+2p̃2,z(τ2(t,x),x)-z(t,x)≤1yln⁡2p̃1y∫aτ2(t,x)(ɛ+p0(s,x))ds+2p̃1,
for a.e. (t,x)∈𝒟 and thus, by using relations (5.60), (5.63), we get
(5.66)y(γ(t,x)+δ)≥γ(τ0(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)
for a.e. (t,x)∈𝒟. Consequently, it follows from (5.64) that inequality (4.9) holds.
We have constructed a functionγ satisfying conditions (4.9)–(4.12) and thus the assertion of the corollary follows from Theorem 4.4.



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*Source: 242965-2011-08-18.xml*