Mixed Problem with an Integral Two-Space-Variables Condition for a Class of Hyperbolic Equations

Abstract

This paper is devoted to the proof of the existence and uniqueness of the classical solution of mixed problems which combine Neumann condition and integral two-space-variables condition for a class of hyperbolic equations. The proof is based on a priori estimate “energy inequality” and the density of the range of the operator generated by the problem considered.

1. Introduction

The integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow, and population dynamics.

Cannon was the first who drew attention to these problems with an integral one-space-variable condition [1], and their importance has been pointed out by Samarskii [2]. The existence and uniqueness of the classical solution of mixed problem combining a Dirichlet and integral condition for the equation of heat demonstrated by cannon [1] using the potential method.

Always using the potential method, Kamynin established in [3] the existence and uniqueness of the classical solution of a similar problem with a more general representation.

Subsequently, more works related to these problems with an integral one-space-variable have been published, among them, we cite the work of Benouar and Yurchuk [4], Cannon and Van Der Hoek [5, 6], Cannon-Esteva-Van Der Hoek [7], Ionkin [8], Jumarhon and McKee [9], Kartynnik [10], Lin [11], Shi [12] and Yurchuk [13]. In these works, mixed problems related to one-dimensional parabolic equations of second order combining a local condition and an integral condition was discussed. Also, by referring to the articles of Bouziani [14–16] and Bouziani and Benouar [17–19], the authors have studied mixed problems with integral conditions for some partial differential equations, specially hyperbolic equation with integral condition which has been investigated in Bouziani [20].

The present paper is devoted to the study of problems with a boundary integral two-space-variables condition for second-order hyperbolic equation.

2. Setting of the Problem

In the rectangle Ω=(0,1)×(0,𝑇), with 𝑇<∞, we consider the hyperbolic equation: 𝜕ℒ𝑢=²𝑢𝜕𝑡²−𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢(),𝜕𝑥=𝑓𝑥,𝑡 (1) where the coefficient 𝑎(𝑥,𝑡) is a real-valued function belonging to 𝐶¹(Ω) such that 0<𝑐₀≤𝑎(𝑥,𝑡)≤𝑐₁,𝜕𝑎(𝑥,𝑡)𝜕𝑡≤𝑐₂. (2) in the rest of the paper, 𝑘, 𝑐_𝑖, 𝑖=1,…,12, denote strictly positive constants.

we adjoin to (1) the initial conditions ℓ₁()(𝑥)(),ℓ𝑢=𝑢𝑥,0=𝜑,𝑥∈0,1₂𝑢=𝜕𝑢()(𝑥)(),𝜕𝑡𝑥,0=𝜓,𝑥∈0,1 (3) the Neumann condition 𝜕𝑢()𝜕𝑥0,𝑡=0, (4) and the integral condition ^𝛼₀𝑢(𝑥,𝑡)𝑑𝑥+¹_𝛽𝑢()(),𝑥,𝑡𝑑𝑥&=0,𝛼>0,𝛽>0,𝛼+𝛽=1,𝑡∈0,𝑇 (5) where 𝜑 and 𝜓 are known functions.

We will assume that the function 𝜑 and 𝑓 satisfy a compatibility conditions with (5), that is, ^𝛼₀𝜑(𝑥)𝑑𝑥+¹_𝛽𝜑(𝑥)𝑑𝑥=0,_𝛼0𝑓(𝑥,𝑡)𝑑𝑥+¹_𝛽𝑓(𝑥,𝑡)𝑑𝑥=0. (6)(7)

The presence of integral terms in boundary conditions can, in general, greatly complicate the application of standard functional or numerical techniques specially the integral two-space-variables condition. In order to avoid this difficulty, we introduce a technique to transfer this problem to another classically less complicated one which does not contain integral conditions. For that, we establish the following lemma.

Lemma 1. Problem (1)–(5) is equivalent to the following problem (𝑃𝑅): ()=𝜕𝑃𝑅ℒ𝑢=²𝑢𝜕𝑡²−𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢(),𝜕𝑥=𝑓𝑥,𝑡0<𝑐₀≤𝑎(𝑥,𝑡)≤𝑐₁,𝜕𝑎(𝑥,𝑡)𝜕𝑡≤𝑐₂,ℓ₁()(𝑥)(),ℓ𝑢=𝑢𝑥,0=𝜑,𝑥∈0,1₂𝑢=𝜕𝑢()(𝑥)(),𝜕𝑡𝑥,0=𝜓,𝑥∈0,1𝜕𝑢()𝜕𝑥0,𝑡=0,𝜕𝑢()𝜕𝑥𝛼,𝑡=0,𝜕𝑢𝜕𝑥𝛽,𝑡=0,𝜕𝑢()𝜕𝑥1,𝑡=0. (8)

Proof. Let 𝑢(𝑥,𝑡) be a solution of (1)–(5), we prove that 𝜕𝑢()𝜕𝑥0,𝑡=0,𝜕𝑢()𝜕𝑥𝛼,𝑡=0,𝜕𝑢𝜕𝑥𝛽,𝑡=0,𝜕𝑢()𝜕𝑥1,𝑡=0. (9)
So, by integrating (1) with respect to 𝑥 over (0,𝛼) and (𝛽,1) and taking into account (6) and (7), we obtain ^𝛼₀𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝑑𝑥+¹_𝛽𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝑑𝑥=0, (10) and we get 𝑎(𝛼,𝑡)𝜕𝑢()()𝜕𝑥𝛼,𝑡−𝑎0,𝑡𝜕𝑢()()𝜕𝑥0,𝑡+𝑎1,𝑡𝜕𝑢()𝜕𝑥1,𝑡−𝑎𝛽,𝑡𝜕𝑢𝜕𝑥𝛽,𝑡=0. (11) The integral condition (5) has nothing to do with the coefficient 𝑎(𝑥,𝑡). Then, (11) imposes 𝜕𝑢()=𝜕𝑥𝛼,𝑡𝜕𝑢()=𝜕𝑥0,𝑡𝜕𝑢()=𝜕𝑥1,𝑡𝜕𝑢𝜕𝑥𝛽,𝑡=0. (12)
Let now 𝑢(𝑥,𝑡) be a solution of (𝑃𝑅), then we are bound to prove that ^𝛼₀𝑢(𝑥,𝑡)𝑑𝑥+¹_𝛽𝑢(𝑥,𝑡)𝑑𝑥=0. (13)
So, by integrating (1) with respect to 𝑥 over (0,𝛼) and taking into account that 𝜕𝑢()𝜕𝑥0,𝑡=0,𝜕𝑢()𝜕𝑥𝛼,𝑡=0, (14) we obtain 𝜕²𝜕𝑡^2𝛼₀𝑢(𝑥,𝑡)𝑑𝑥=^𝛼₀𝑓(𝑥,𝑡)𝑑𝑥, (15) and by integrating (1) with respect to 𝑥 over (𝛽,1) and taking in consideration 𝜕𝑢𝜕𝑥𝛽,𝑡=0,𝜕𝑢()𝜕𝑥1,𝑡=0, (16) we obtain 𝜕²𝜕𝑡²¹_𝛽𝑢(𝑥,𝑡)𝑑𝑥=¹_𝛽𝑓(𝑥,𝑡)𝑑𝑥. (17) By combining the two preceding (15) and (17) and taking into account (7), we get ^𝛼₀𝑢(𝑥,𝑡)𝑑𝑥+¹_𝛽𝑢(𝑥,𝑡)𝑑𝑥=0. (18)

3. A Priori Estimate and Its Consequences

In this paper, we prove the existence and the uniqueness of the solution of the problem (1)–(5) and of the operator equation 𝐿𝑢=ℱ, (19) where 𝐿=(ℒ,ℓ₁,ℓ₂) with domain of difinition 𝐵 consisting of functions 𝑢∈𝐿²(Ω) such that 𝜕𝑢/𝜕𝑡, 𝜕𝑢/𝜕𝑥, 𝜕²𝑢/𝜕𝑥²∈𝐿²(Ω), and 𝑢 satisfies conditions (3) and (4); the operator 𝐿 is considered from 𝐵 to 𝐹, where 𝐵 is the Banach space consisting of all functions 𝑢(𝑥,𝑡) having a finite norm ‖𝑢‖²_𝐵=¹₀𝜕𝑢𝜕𝑡²𝑑𝑥+¹₀𝜕𝑢𝜕𝑥²𝑑𝑥, (20) and 𝐹 is the Hilbert space consisting of all elements ℱ=(𝑓,𝜑,𝜓) for which the norm ‖ℱ‖²_𝐹=¹₀𝑓²𝑑𝑥𝑑𝑡+¹₀𝜓²(𝑥)𝑑𝑥+¹₀𝜕𝜑𝜕𝑥²𝑑𝑥 (21) is finite.

Theorem 2. For any function 𝑢∈𝐵, we have the inequality ‖𝑢‖_𝐵≤𝑐‖𝐿𝑢‖_𝐹, (22) where 𝑐 is a positive constant independent of 𝑢.

Proof. Multiplying the (1) by the following 𝑀𝑢: 𝑀𝑢=𝑀𝑢₁=2(1−𝑥)𝜕𝑢𝜕𝑡,0≤𝑥≤𝛼,𝑀𝑢₂=2𝜕𝑢𝜕𝑡,𝛼≤𝑥≤𝛽,𝑀𝑢₃=2(2−𝑥)𝜕𝑢𝜕𝑡,𝛽≤𝑥≤1 (23) and integrating over Ω^𝜏, where Ω^𝜏=(0,1)×(0,𝜏),(1) by integrating over Ω^𝜏_𝛼=Ω_𝛼=(0,𝛼)×(0,𝜏). Consequently, _Ω𝛼ℒ𝑢⋅𝑀𝑢₁𝑑𝑥𝑑𝑡=2_Ω𝛼(1−𝑥)𝜕𝑢𝜕𝜕𝑡²𝑢𝜕𝑡²𝑑𝑥𝑑𝑡−2_Ω𝛼(1−𝑥)𝜕𝑢𝜕𝜕𝑡𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝑑𝑥𝑑𝑡=2_Ω𝛼(1−𝑥)𝜕𝑢𝜕𝑡𝑓𝑑𝑥𝑑𝑡. (24)
Employing integration by parts in (24), and taking into account the boundary conditions in (𝑃𝑅), we obtain 2_Ω𝛼(1−𝑥)𝜕𝑢𝜕𝜕𝑡²𝑢𝜕𝑡²𝑑𝑥𝑑𝑡=^𝛼₀(1−𝑥)𝜕𝑢𝜕𝑡²𝑑𝑥−^𝛼₀(1−𝑥)𝜓²(𝑥)𝑑𝑥,−2_Ω𝛼(1−𝑥)𝜕𝑢𝜕𝜕𝑡𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝑑𝑥𝑑𝑡=−^𝜏₀2(1−𝑥)𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑡𝜕𝑢𝜕𝑥^𝑥=𝛼_𝑥=0𝑑𝑡+2_Ω𝛼(1−𝑥)𝑎(𝑥,𝑡)𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡−2_Ω𝛼𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=−^𝛼₀(1−𝑥)𝑎(𝑥,0)𝜕𝜑𝜕𝑥²+𝑑𝑥_𝛼0(1−𝑥)𝑎(𝑥,𝜏)𝜕𝑢𝜕𝑥²−𝑑𝑥_Ω𝛼()()1−𝑥𝜕𝑎𝑥,𝑡𝜕𝑡𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡−2_Ω𝛼𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡. (25)
Using the Cauchy’s inequality, we get 2_Ω𝛼(1−𝑥)𝜕𝑢𝜕𝑡𝑓𝑑𝑥𝑑𝑡≤_Ω𝛼𝑓²𝑑𝑥𝑑𝑡+_Ω𝛼𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡. (26)
Substituting (25) and (26) into (24), we obtain ^𝛼₀(1−𝑥)𝜕𝑢𝜕𝑡²𝑑𝑥+^𝛼₀(1−𝑥)𝑎(𝑥,𝜏)𝜕𝑢𝜕𝑥²≤𝑑𝑥_Ω𝛼𝑓²𝑑𝑥𝑑𝑡+_Ω𝛼𝜕𝑢𝜕𝑡²+𝑑𝑥𝑑𝑡_𝛼0(1−𝑥)𝑎(𝑥,0)𝜕𝜑𝜕𝑥²𝑑𝑥+^𝛼₀(1−𝑥)𝜓²(𝑥)+𝑑𝑥_Ω𝛼()()1−𝑥𝜕𝑎𝑥,𝑡𝜕𝑡𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡+2_Ω𝛼𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡. (27)
Then, using the Cauchy’s inequality and according to conditions (2), we get 𝛽^𝛼₀𝜕𝑢𝜕𝑡²𝑑𝑥+𝛽𝑐₀^𝛼₀𝜕𝑢𝜕𝑥²≤𝑑𝑥_Ω𝛼𝑓²𝑑𝑥𝑑𝑡+𝑐₁+1_Ω𝛼𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+^𝛼₀𝜓²(𝑥)+𝑐𝑑𝑥₁+𝑐_2Ω𝛼𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡+𝑐₁^𝛼₀𝜕𝜑𝜕𝑥²𝑑𝑥, (28) so, we obtain ^𝛼₀𝜕𝑢𝜕𝑡²𝑑𝑥+^𝛼₀𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_3Ω𝛼𝑓²𝑑𝑥𝑑𝑡+^𝛼₀𝜓²(𝑥)+𝑑𝑥_𝛼0𝜕𝜑𝜕𝑥²𝑑𝑥+_Ω𝛼𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+𝑐_4Ω𝛼𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡, (29) where 𝑐₃=𝑐₁+1min𝛽,𝛽𝑐₀,𝑐₄=𝑐₁+𝑐₂min𝛽,𝛽𝑐₀. (30)
By virtue of Lemma  7.1 in [21] and by using it twice, we find ^𝛼₀𝜕𝑢𝜕𝑡²𝑑𝑥+^𝛼₀𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_5Ω𝛼𝑓²𝑑𝑥𝑑𝑡+^𝛼₀𝜓²(𝑥)𝑑𝑥+^𝛼₀𝜕𝜑𝜕𝑥²𝑑𝑥, (31) where 𝑐₅=𝑐₃exp𝑐₄𝑇exp𝑐₃𝑇exp𝑐₄𝑇. (32)(2) by integrating over Ω^𝜏_𝛼,𝛽=Ω_𝛼,𝛽=(𝛼,𝛽)×(0,𝜏). Consequently, _Ω𝛼,𝛽ℒ𝑢⋅𝑀𝑢₂𝑑𝑥𝑑𝑡=2_Ω𝛼,𝛽𝜕²𝑢𝜕𝑡²𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡−2_Ω𝛼,𝛽𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=2_Ω𝛼,𝛽𝑓𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡. (33)
Employing integration by parts in (33) and taking into account the boundary conditions in (𝑃𝑅), we obtain 2_Ω𝛼,𝛽𝜕²𝑢𝜕𝑡²𝜕𝑢=𝜕𝑡𝑑𝑥𝑑𝑡_𝛽𝛼𝜕𝑢𝜕𝑡²𝑑𝑥−^𝛽_𝛼𝜓²(𝑥)𝑑𝑥,−2_Ω𝛼,𝛽𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=−^𝜏₀2𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑡𝜕𝑢𝜕𝑥^𝑥=𝛽_𝑥=𝛼𝑑𝑡+2_Ω𝛼,𝛽𝑎(𝑥,𝑡)𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡=−^𝛽_𝛼𝑎(𝑥,0)𝜕𝜑𝜕𝑥²𝑑𝑥+^𝛽_𝛼𝑎(𝑥,𝜏)𝜕𝑢𝜕𝑥²−𝑑𝑥_Ω𝛼,𝛽𝜕𝑎(𝑥,𝑡)𝜕𝑡𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡. (34)
By virtue of the Cauchy’s inequality, we obtain _Ω𝛼,𝛽𝑓𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡≤2_Ω𝛼,𝛽𝑓²1𝑑𝑥𝑑𝑡+2_Ω𝛼,𝛽𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡. (35)
Substituting (34) and (35) into (33) and according to conditions (2), we get ^𝛽_𝛼𝜕𝑢𝜕𝑡²𝑑𝑥+^𝛽_𝛼𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_6Ω𝛼,𝛽𝑓²𝑑𝑥𝑑𝑡+^𝛽_𝛼𝜓²(𝑥)+𝑑𝑥_𝛽𝛼𝜕𝜑𝜕𝑥²𝑑𝑥+_Ω𝛼,𝛽𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+𝑐_7Ω𝛼,𝛽𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡, (36) where 𝑐₆=𝑐₂min1,𝑐₀,𝑐₇=max2,𝑐₁min1,𝑐₀. (37)
By virtue of Lemma  7.1 in [21] and by using it twice, we find ^𝛽_𝛼𝜕𝑢𝜕𝑡²𝑑𝑥+^𝛽_𝛼𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_8Ω𝛼,𝛽𝑓²𝑑𝑥𝑑𝑡+^𝛽_𝛼𝜓²(𝑥)𝑑𝑥+^𝛽_𝛼𝜕𝜑𝜕𝑥²𝑑𝑥, (38) where 𝑐₈=𝑐₆exp𝑐₇𝑇exp𝑐₆𝑇exp𝑐₇𝑇. (39)(3) by integrating over Ω^𝜏_𝛽=Ω_𝛽=(𝛽,1)×(0,𝜏). Consequently, _Ω𝛽ℒ𝑢⋅𝑀𝑢₃𝑑𝑥𝑑𝑡=2_Ω𝛽(2−𝑥)𝜕𝑢𝜕𝜕𝑡²𝑢𝜕𝑡²𝑑𝑥𝑑𝑡−2_Ω𝛽(2−𝑥)𝜕𝑢𝜕𝜕𝑡𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝑑𝑥𝑑𝑡=2_Ω𝛽(2−𝑥)𝜕𝑢𝜕𝑡𝑓𝑑𝑥𝑑𝑡. (40)
Employing integration by parts in (40) and taking into account the boundary conditions in (𝑃𝑅), we obtain 2_Ω𝛽(2−𝑥)𝜕𝑢𝜕𝜕𝑡²𝑢𝜕𝑡²=𝑑𝑥𝑑𝑡_1𝛽(2−𝑥)𝜕𝑢𝜕𝑡²𝑑𝑥−¹_𝛽(2−𝑥)𝜓²(𝑥)𝑑𝑥,−2_Ω𝛽(2−𝑥)𝜕𝑢𝜕𝜕𝑡𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥𝑑𝑥𝑑𝑡=−2^𝜏₀(2−𝑥)𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑡𝜕𝑢𝜕𝑥^𝑥=𝛼_𝑥=0𝑑𝑡+2_Ω𝛽(2−𝑥)𝑎(𝑥,𝑡)𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡−2_Ω𝛽𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=−¹_𝛽(2−𝑥)𝑎(𝑥,0)𝜕𝜑𝜕𝑥²+𝑑𝑥_1𝛽(2−𝑥)𝑎(𝑥,𝜏)𝜕𝑢𝜕𝑥²−𝑑𝑥_Ω𝛽()()2−𝑥𝜕𝑎𝑥,𝑡𝜕𝑡𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡−2_Ω𝛽𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡. (41)
Using the Cauchy’s inequality, we get 2_Ω𝛼(2−𝑥)𝜕𝑢𝜕𝑡𝑓𝑑𝑥𝑑𝑡≤2_Ω𝛼𝑓²𝑑𝑥𝑑𝑡+2_Ω𝛼𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡. (42)
Substituting (41) and (42) into (40), we obtain ¹_𝛽(2−𝑥)𝜕𝑢𝜕𝑡²𝑑𝑥+¹_𝛽(2−𝑥)𝑎(𝑥,𝜏)𝜕𝑢𝜕𝑥²𝑑𝑥≤2_Ω𝛽𝑓²𝑑𝑥𝑑𝑡+2_Ω𝛽𝜕𝑢𝜕𝑡²+𝑑𝑥𝑑𝑡_1𝛽(2−𝑥)𝑎(𝑥,0)𝜕𝜑𝜕𝑥²𝑑𝑥+¹_𝛽(2−𝑥)𝜓²(𝑥)+𝑑𝑥_Ω𝛽()()2−𝑥𝜕𝑎𝑥,𝑡𝜕𝑡𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡+2_Ω𝛽𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡. (43)
Then, using the Cauchy’s 𝜀-inequality and according to conditions (2), we get ¹_𝛽𝜕𝑢𝜕𝑡²𝑑𝑥+𝑐₀¹_𝛽𝜕𝑢𝜕𝑥²𝑑𝑥≤2_Ω𝛽𝑓²𝑑𝑥𝑑𝑡+𝑐₁+2_Ω𝛽𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+2¹_𝛽𝜓²(𝑥)𝑑𝑥+𝑐₁+2𝑐_2Ω𝛽𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡+2𝑐₁¹_𝛽𝜕𝜑𝜕𝑥²𝑑𝑥, (44) so, we obtain ¹_𝛽𝜕𝑢𝜕𝑡²𝑑𝑥+¹_𝛽𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_9Ω𝛽𝑓²𝑑𝑥𝑑𝑡+¹_𝛽𝜓²(𝑥)𝑑𝑥+¹_𝛽𝜕𝜑𝜕𝑥²𝑑𝑥+_Ω𝛽𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+𝑐_10Ω𝛽𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡, (45) where 𝑐₉=2𝑐₁+2min1,𝑐₀,𝑐₁₀=𝑐₁+2𝑐₂min1,𝑐₀. (46)
By virtue of Lemma  7.1 in [21] and by using it twice, we find ¹_𝛽𝜕𝑢𝜕𝑡²𝑑𝑥+¹_𝛽𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_11Ω𝛽𝑓²𝑑𝑥𝑑𝑡+¹_𝛽𝜓²(𝑥)𝑑𝑥+¹_𝛽𝜕𝜑𝜕𝑥²𝑑𝑥, (47) where 𝑐₁₁=𝑐₉exp𝑐₁₀𝑇exp𝑐₉𝑇exp𝑐₁₀𝑇. (48)
Combining inequalities (31), (38), and (47), we obtain ¹₀𝜕𝑢𝜕𝑡²𝑑𝑥+¹₀𝜕𝑢𝜕𝑥²𝑑𝑥≤𝑐_12Ω𝑓²𝑑𝑥𝑑𝑡+¹₀𝜓²(𝑥)𝑑𝑥+¹₀𝜕𝜑𝜕𝑥²𝑑𝑥, (49) where 𝑐₁₂=max𝑐₅,𝑐₈,𝑐₁₁. (50)
The right-hand side of (49) is independent of 𝜏; hence replacing the left-hand side by its upper bound with respect to 𝜏 from 0 to 𝑇, we obtain the desired inequality, where 𝑐=(𝑐₁₂)^1/2.

Corollary 3. A solution of the problem (1)–(5) is unique if it exists and depends continuously on ℱ∈𝐹.

4. Solvability of the Problem

To show the existence of solutions, we prove that 𝑅(𝐿) is dense in 𝐹 for all 𝑢∈𝐵 and for arbitrary ℱ=(𝑓,𝜙,𝜓)∈𝐹.

Theorem 4. Suppose the conditions of Theorem 2 are satisfied. Then, the problem (1)–(5) admits a unique solution 𝑢=𝐿⁻¹ℱ.

Proof. First, we prove that 𝑅(𝐿) is dense in 𝐹 for the special case where 𝐷(𝐿)≡𝐵 is reduced to 𝐷₀(𝐿), where 𝐷₀(𝐿)={𝑢,𝑢∈𝐷(𝐿)∶ℓ₁𝑢=0,ℓ₂𝑢=0}.

Proposition 5. Let the conditions of Theorem 4 be satisfied. If for 𝜔∈𝐿²(Ω) and for all 𝑢∈𝐷₀(𝐿), we have _Ωℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡=0 (51) then, 𝜔 vanishes almost everywhere in Ω.

Proof. The scalar product of 𝐹 is defined by (𝐿𝑢,𝜔)_𝐹=_Ωℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡+¹₀𝜕ℓ₁𝑢𝜕𝑥𝜕𝜔₀+𝜕𝑥𝑑𝑥₁₀ℓ₂𝑢𝜔₁𝑑𝑥. (52) the equality (51) can be written as follows:_Ω𝜕²𝑢𝜕𝑡²⋅𝜔𝑑𝑥𝑑𝑡=_Ω𝜕𝑎()𝜕𝑥𝑥,𝑡𝜕𝑢𝜕𝑥⋅𝜔𝑑𝑥𝑑𝑡. (53)
If we put 𝑢=ℑ_𝑡𝑒^𝑘𝑡𝑧=^𝑡₀𝑒^𝑘𝜏𝑧(𝑥,𝜏)𝑑𝜏, (54) where 𝑘 is a constant such that 𝑘𝑐₀−𝑐₂≥0 and 𝑧,𝜕𝑧/𝜕𝑥,(𝜕/𝜕𝑥)(𝑎(𝜕ℑ_𝑡(𝑒^𝑘𝑡𝑧)/𝜕𝑥))∈𝐿²(Ω), then, 𝑢 satisfies the boundary conditions in (𝑃𝑅). As a result of (53), we obtain the equality _Ω𝜕𝑒^𝑘𝑡𝑧𝜕𝑡𝜔𝑑𝑥𝑑𝑡=_Ω𝜕𝑎𝜕𝑥𝜕ℑ_𝑡𝑒^𝜏𝑧𝜕𝑥⋅𝜔𝑑𝑥𝑑𝑡. (55)
The left-hand side of (55) shows that the mapping 𝐿²(Ω)∋𝑧→^𝑇₀𝜕𝑎𝜕𝑥𝜕ℑ_𝑡𝑒^𝜏𝑧𝜕𝑥⋅𝜔𝑑𝑡 (56) is a continuous linear functional of 𝑧. From the right-hand side of (55), It is true if the function 𝜔 has the following properties: 𝜕𝜔𝜕𝑥,ℑ^⋆_𝑡(𝑎𝜔),ℑ^⋆_𝑡𝑎𝜕𝜔,𝜕𝜕𝑥ℑ𝜕𝑥^⋆𝑡(),𝜕𝑎𝜔ℑ𝜕𝑥^⋆𝑡𝑎𝜕𝜔𝜕𝑥∈𝐿²(Ω),𝜔()()𝜔()𝑥,0=0,𝜔0,𝑡=0,1,𝑡=0;ℑ^⋆_𝑡(𝑎𝜔)_𝑥=1=0. (57)
In terms of the given function 𝜔 and from the equality (55), we give the function 𝜔 in terms of 𝑧 as follows: 𝜔(𝑥,𝑡)=𝑧(𝑥,𝑡), (58) and 𝑧 satisfies the same conditions of the function 𝜔.
Replacing 𝜔 in (55) by its representation (58), we obtain _Ω𝜕𝑒^𝑘𝑡𝑧=𝜕𝑡𝑧𝑑𝑥𝑑𝑡_Ω𝜕𝑎𝜕∫𝜕𝑥^𝑡0𝑒^𝑘𝜏𝑧(𝑥,𝜏)𝑑𝜏𝜕𝑥𝑧𝑑𝑥𝑑𝑡. (59)
Integrating by parts the right-hand side of (59) with respect to 𝑥 and 𝑡 and by taking the conditions of the function 𝑧 yields _Ω𝜕𝑒^𝑘𝑡𝑧=𝜕𝑡𝑧𝑑𝑥𝑑𝑡_𝑇0𝑎(𝑥,𝑡)𝜕ℑ_𝑡𝑒^𝑘𝜏𝑧(𝑧)𝜕𝑥^𝑥=1_𝑥=0𝑑𝑡=−_Ω𝑎()𝜕ℑ𝑥,𝑡^𝑡𝑒^𝑘𝜏𝑧𝜕𝑥𝜕𝑧1𝜕𝑥𝑑𝑥𝑑𝑡=−2¹₀𝑒^−𝑘𝑡𝑎(𝑥,𝑡)𝜕ℑ_𝑡𝑒^𝑘𝜏𝑧𝜕𝑥^2𝑡=𝑇_𝑡=0−1𝑑𝑥2_Ω𝑒^−𝑘𝑡()−()𝑘𝑎𝑥,𝑡𝜕𝑎𝑥,𝑡×𝜕𝑡𝜕ℑ_𝑡𝑒^𝑘𝜏𝑧𝜕𝑥²𝑑𝑥𝑑𝑡. (60)
According to condition (2), we obtain _Ω𝜕𝑒^𝑘𝑡𝑧1𝜕𝑡𝑧𝑑𝑥𝑑𝑡≤−2𝑘𝑐₀−𝑐_2Ω𝑒^−𝑘𝑡𝜕ℑ_𝑡𝑒^𝑘𝜏𝑧𝜕𝑥²𝑑𝑥𝑑𝑡≤0. (61)
Integrating by parts the left-hand side of (61) with respect to 𝑡, and by taking the conditions of the function 𝑧, we obtain _Ω𝜕𝑒^𝑘𝑡𝑧=𝜕𝑡𝑧𝑑𝑥𝑑𝑡_Ω𝑒^𝑘𝑡𝜕𝑧𝜕𝑡𝑧+𝑘𝑒^𝑘𝑡𝑧²=1𝑑𝑥𝑑𝑡2¹₀𝑒^𝑘𝑡𝑧(𝑥,𝑇)²+3𝑑𝑥2_Ω𝑘𝑒^𝑘𝑡𝑧²𝑑𝑥𝑑𝑡. (62)
Then, by (61) and (62) we obtain 12¹₀𝑒^𝑘𝑡(𝑧(𝑥,𝑇))²3𝑑𝑥+2_Ω𝑘𝑒^𝑘𝑡𝑧²𝑑𝑥𝑑𝑡≤0. (63) And thus 𝑧=0 in Ω, hence 𝜔=0 in Ω. This proves Proposition 5.

We return to the proof of Theorem 4. We have already noted that it is sufficient to prove that the set 𝑅(𝐿) dense in 𝐹.

Suppose that for some 𝑊=(𝜔,𝜔₀,𝜔₁)∈𝑅(𝐿)^⟂ and for all 𝑢∈𝐷(𝐿)≡𝐵 it holds (𝐿𝑢,𝜔)_𝐹=_Ω+ℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡₁₀𝜕ℓ₁𝑢𝜕𝑥𝜕𝜔₀+𝜕𝑥𝑑𝑥₁₀ℓ₂𝑢𝜔₁𝑑𝑥=0. (64)

Then, we must prove that 𝑊=0. Putting 𝑢∈𝐷₀(𝐿) in (64), we have _Ωℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡=0,𝑢∈𝐷₀(𝐿). (65)

Hence, Proposition 5 implies that 𝜔=0. Thus, (64) takes the form ¹₀𝜕ℓ₁𝑢𝜕𝑥𝜕𝜔₀+𝜕𝑥𝑑𝑥₁₀ℓ₂𝑢𝜔₁𝑑𝑥=0,𝑢∈𝐷(𝐿). (66)

Since the ranges of the trace operators ℓ₁ and ℓ₂ are independents and the ranges of values ℓ₁ and ℓ₂ are everywhere dense in the Hilbert space 𝐹 with the norm ¹₀𝜕ℓ₁𝑢𝜕𝑥²+ℓ₂𝑢²𝑑𝑥^1/2, (67) then the equality (66) implies that 𝜔₀=0, 𝜔₁=0 (we recall satisfies a compatibility conditions). Hence 𝑊=0 implies (𝑅(𝐿)=𝐹). Therefore, the proof of Theorem 4 is complete.