A Mixed Problem with an Integral Two-Space-Variables Condition for Parabolic Equation with The Bessel Operator

Abstract

We study a mixed problem with an integral two-space-variables condition for parabolic equation with the Bessel operator. The existence and uniqueness of the solution in functional weighted Sobolev space are proved. 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 importance of boundary value problems with integral boundary condition has been pointed out by Samarskiĭ [1] and problems with integral conditions for parabolic equations were treated by Kamynin [2], Ionkin [3], Yurchuk [4], Benouar and Yurchuk [5], Bouziani [6], Bouziani and Benouar [7, 8], and Mesloub and Bouziani [9]. Other parabolic problems also arise in plasma physics by Samarskiĭ [1], heat conduction by Cannon [10], Ionkin [3], dynamics of ground waters by Nakhushev [11], Vodakhova [12], Kartynnik [13], and Lin [14]. Regular case of this problem is studied in [15]. The problem where the equation contains an operator of the form (𝜕/𝜕𝑥)(𝑎(𝑥,𝑡)(𝜕𝑢/𝜕𝑥)), instead of Bessel operator, is treated in [16]. Similar problems for second-order parabolic equations are investegated by the potential method in [17]. Later problems with an integral two-space-variables condition for parabolic equations were treated by Marhoune [18], Marhoune and Lakhal [19]. Motivated by this, we study a mixed problem with an integral two-space-variables condition for parabolic equation with the Bessel operator.

2. Setting of the Problem

In the rectangular domain Ω=(0,1)×(0,𝑇), with 𝑇<∞, we consider the following equation: ℒ𝑢=𝜕𝑢−𝜕𝜕𝑡²𝑢𝜕𝑥²+1𝑥𝜕𝑢(),𝜕𝑥=𝑓𝑥,𝑡 (1) with the initial data ℓ𝑢=𝑢(𝑥,0)=𝜑(𝑥),𝑥∈(0,1), (2) Neumann boundary condition 𝜕𝑢()𝜕𝑥1,𝑡=0, (3) and the integral condition 𝑥𝑢(^𝛼₀𝑥,𝑡)𝑑𝑥+¹_𝛽()(),𝑥𝑢𝑥,𝑡𝑑𝑥=0,𝛽>𝛼>0,𝛼+𝛽=1,𝑡∈0,𝑇 (4) where 𝜑 is a known function.

We shall assume that the functions 𝜑 and 𝑓 satisfy the compatibility condition with (4), that is, 𝑥𝜑(^𝛼₀𝑥)𝑑𝑥+¹_𝛽𝑥𝜑(𝑥)𝑑𝑥=0,_𝛼0𝑥𝑓(𝑥,𝑡)𝑑𝑥+¹_𝛽𝑥𝑓(𝑥,𝑡)𝑑𝑥=0. (5)(6) The presence of integral terms in boundary condition can, in general, greatly complicate the application of standard functional or numerical techniques, specially the integral two-space-variables condition. Then to avoid this difficulty, we introduce a technique for transfering this problem to another classically less complicated one that does not contain integral conditions. For that, we establish the following lemma.

Lemma 1. Problem (1)–(4) is equivalent to the following problem (PR): ℒ𝑢=𝜕𝑢−𝜕𝜕𝑡²𝑢𝜕𝑥²+1𝑥𝜕𝑢(),()(𝑥)(),𝛼𝜕𝑥=𝑓𝑥,𝑡ℓ𝑢=𝑢𝑥,0=𝜑,𝑥∈0,1𝜕𝑢()𝜕𝑥𝛼,𝑡=𝛽𝜕𝑢,𝜕𝑥𝛽,𝑡𝜕𝑢()𝜕𝑥1,𝑡=0. (PR)

Proof. Let 𝑢(𝑥,𝑡) be a solution of (1)–(4), we prove that 𝜕𝑢()=𝜕𝑥𝛼,𝑡𝜕𝑢.𝜕𝑥𝛽,𝑡 (7) So, multiplying (1) by 𝑥 and integrating with respect to 𝑥 over (0,𝛼) and (𝛽,1) and taking into account (4) and (6), we obtain 𝑥𝜕𝑢𝜕𝑥^𝑥=𝛼_𝑥=0𝑑𝑥+𝑥𝜕𝑢𝜕𝑥^𝑥=1_𝑥=𝛽𝛼𝑑𝑥=0,thisimplies𝜕𝑢()+𝜕𝑥𝛼,𝑡𝜕𝑢()𝜕𝑥1,𝑡−𝛽𝜕𝑢𝜕𝑥𝛽,𝑡=0. (8) Then, from (3), we obtain 𝛼𝜕𝑢()𝜕𝑥𝛼,𝑡=𝛽𝜕𝑢.𝜕𝑥𝛽,𝑡 (9) Let now 𝑢(𝑥,𝑡) be a solution of (PR), we are bound to prove that 𝑥𝑢(^𝛼₀𝑥,𝑡)𝑑𝑥+¹_𝛽𝑥𝑢(𝑥,𝑡)𝑑𝑥=0. (10) So, multiplying (1) by 𝑥 and integrating with respect to 𝑥 over (0,𝛼) and (𝛽,1) and taking into account 𝛼𝜕𝑢()𝜕𝑥𝛼,𝑡=𝛽𝜕𝑢,𝜕𝑥𝛽,𝑡𝜕𝑢()𝜕𝑥1,𝑡=0, (11) we obtain 𝜕𝜕𝑡^𝛼₀𝑥𝑢(𝑥,𝑡)𝑑𝑥=^𝛼₀()𝜕𝑥𝑓𝑥,𝑡𝑑𝑥,𝜕𝑡¹_𝛽𝑥𝑢(𝑥,𝑡)𝑑𝑥=¹_𝛽𝑥𝑓(𝑥,𝑡)𝑑𝑥, (12) combining the two preceding equations, and from (6) we get 𝑥𝑢(^𝛼₀𝑥,𝑡)𝑑𝑥+¹_𝛽𝑥𝑢(𝑥,𝑡)𝑑𝑥=0. (13)

3. A Priori Estimate

The method used here is one of the most efficient functional analysis methods in solving partial differential equations with integral conditions, the so-called a priori estimate method or the energy-integral method. This method is essentially based on the construction of multiplicators for each specific given problem, which provides the a priori estimate from which it is possible to establish the solvability of the posed problem. More precisely, the proof is based on an energy inequality and the density of the range of the operator generated by the abstract formulation of the stated problem. But here we use the energy inequality method for the equivalent problem (PR) given in Lemma 1; so to investigate the posed problem, we introduce the needed function spaces.

We introduce function spaces needed in our investigation. We denote by 𝐿²_𝜌(Ω) the weighted Lebesgue space that consists of all measurable functions 𝑢 equipped with the finite norm ‖𝑢‖_𝐿²_𝜌(Ω)=_Ω𝜌(𝑥)|𝑢(𝑥,⋅)|²𝑑𝑥^1/2, (14) where 𝜌(𝑥)=(𝑥²+𝑥). If 𝜌(𝑥)=1, 𝐿²_𝜌(Ω) are identified with the standard spaces 𝐿²(Ω).

In this paper, we prove the existence and the uniqueness for solution of the problem (1)–(4) as a solution of the operator equation 𝐿𝑢=ℱ, (15) where 𝐿=(ℒ,ℓ), with domain of definition 𝐵 consisting of functions 𝑢 ∈𝐿²_𝜌(Ω) such that 𝜕𝑢/𝜕𝑡,  𝜕𝑢/𝜕𝑥,  𝜕²𝑢/𝜕𝑥²,  𝜕²𝑢/𝜕𝑡𝜕𝑥∈𝐿²_𝜌(Ω), and 𝑢 satisfy condition (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=_Ω𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+sup_{0≤𝜏≤𝑇}¹₀𝑥²()+𝑥𝜕𝑢𝑥,𝜏𝜕𝑥²𝑑𝑥, (16) and 𝐹 is the Hilbert space consisting of all elements ℱ=(𝑓,𝜑) for which the norm ‖ℱ‖²_𝐹=_Ω𝑓²𝑑𝑥𝑑𝑡+¹₀𝑥²+𝑥𝜕𝜑𝜕𝑥²𝑑𝑥 (17) is finite.

Theorem 2. For any function 𝑢∈𝐵, one has the estimate ‖𝑢‖_𝐵≤𝑐‖𝐿𝑢‖_𝐹, (18) where 𝑐 is a positive constant independent of 𝑢.

Proof. Multiplying (1) by the following 𝑀𝑢: 𝑥𝑀𝑢=²−𝑥𝜕𝑢+𝑥𝜕𝑥²+𝑥𝜕𝑢𝜕𝑡,0≤𝑥≤𝛼,−𝑥𝜕𝑢+𝑥𝜕𝑥²+𝑥𝜕𝑢𝑥𝜕𝑡,𝛼≤𝑥≤𝛽,₂−𝑥𝜕𝑢+𝑥𝜕𝑥²+𝑥𝜕𝑢𝜕𝑡,𝛽≤𝑥≤1, (19) and integrating over Ω^𝜏, where Ω^𝜏=(0,1)×(0,𝜏), ^Ω=^𝜏∫ℒ𝑢⋅𝑀𝑢𝑑𝑥𝑑𝑡_Ω𝜏𝛼𝑥²−𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡+∫_Ω𝜏𝛼(1−𝑥)𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡+∫_Ω𝜏𝛼𝑥²+𝑥𝜕𝑢𝜕𝑡²−∫𝑑𝑥𝑑𝑡_Ω𝜏𝛼𝑥−𝑥²𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥²𝑑𝑥𝑑𝑡−∫_Ω𝜏𝛼()𝜕𝑥+1𝑥𝜕𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=∫_Ω𝜏𝛼𝑓⋅𝑥²−𝑥𝜕𝑢+𝑥𝜕𝑥₂+𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡,Ω^𝜏_𝛼,∫_{Ω𝜏𝛼,𝛽}𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡−∫_{Ω𝜏𝛼,𝛽}𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡+∫_{Ω𝜏𝛼,𝛽}𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡+∫_{Ω𝜏𝛼,𝛽}𝑥𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥²𝑑𝑥𝑑𝑡−∫_{Ω𝜏𝛼,𝛽}()𝜕𝑥+1𝑥𝜕𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=∫_{Ω𝜏𝛼,𝛽}𝑓⋅−𝑥𝜕𝑢+𝑥𝜕𝑥₂+𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡,Ω^𝜏_𝛼,𝛽,∫_Ω𝜏𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+∫_Ω𝜏𝛽𝑥²−𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡+∫_Ω𝜏𝛽(1−𝑥)𝜕𝑢𝜕𝑥²+∫𝑑𝑥𝑑𝑡_Ω𝜏𝛽𝑥−𝑥²𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥²−∫𝑑𝑥𝑑𝑡_Ω𝜏𝛽()𝜕𝑥+1𝑥𝜕𝑥𝜕𝑢𝜕𝑥𝜕𝑢=∫𝜕𝑡𝑑𝑥𝑑𝑡_Ω𝜏𝛽𝑓⋅𝑥²−𝑥𝜕𝑢+𝑥𝜕𝑥₂+𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡,Ω^𝜏_𝛽. (20) Integration by parts, we obtain−_Ω𝜏𝛼𝑥−𝑥²𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥²=𝑑𝑥𝑑𝑡_𝜏0𝛼2−𝛼²2𝜕𝑢()𝜕𝑥𝛼,𝑡²𝑑𝑡−_Ω𝜏𝛼12−𝑥𝜕𝑢𝜕𝑥²−𝑑𝑥𝑑𝑡,_Ω𝜏𝛼()𝜕𝑥+1𝑥𝜕𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=−^𝜏₀𝛼²+𝛼𝜕𝑢()𝜕𝑥𝛼,𝑡𝜕𝑢()+𝜕𝑡𝛼,𝑡𝑑𝑡_Ω𝜏𝛼𝑥𝜕𝑢𝜕𝑥𝜕𝑢1𝜕𝑡𝑑𝑥𝑑𝑡+2^𝛼₀𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²−1𝑑𝑥2^𝛼₀𝑥²+𝑥𝜕𝜑𝜕𝑥²𝑑𝑥,_{Ω𝜏𝛼,𝛽}𝑥𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥²=𝛽𝑑𝑥𝑑𝑡2^𝜏₀𝜕𝑢𝜕𝑥𝛽,𝑡²𝛼𝑑𝑡−2^𝜏₀𝜕𝑢()𝜕𝑥𝛼,𝑡²−1𝑑𝑡2_Ω^𝜏_𝛼,𝛽𝜕𝑢𝜕𝑥²−𝑑𝑥𝑑𝑡,_{Ω𝜏𝛼,𝛽}()𝜕𝑥+1𝑥𝜕𝑥𝜕𝑢𝜕𝑥𝜕𝑢𝜕𝑡𝑑𝑥𝑑𝑡=−^𝜏₀𝛽²+𝛽𝜕𝑢𝜕𝑥𝛽,𝑡𝜕𝑢+𝜕𝑡𝛽,𝑡𝑑𝑡_𝜏0𝛼²+𝛼𝜕𝑢()𝜕𝑥𝛼,𝑡𝜕𝑢()+𝜕𝑡𝛼,𝑡𝑑𝑡_Ω𝛼,𝛽𝑥𝜕𝑢𝜕𝑥𝜕𝑢+1𝜕𝑡𝑑𝑥𝑑𝑡2^𝛽_𝛼𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²−1𝑑𝑥2^𝛽_𝛼𝑥²+𝑥𝜕𝜑𝜕𝑥²𝑑𝑥,_Ω𝜏𝛽𝑥−𝑥²𝜕𝑢𝜕𝜕𝑥²𝑢𝜕𝑥²=𝑑𝑥𝑑𝑡_𝜏0𝛽²2−𝛽2𝜕𝑢𝜕𝑥𝛽,𝑡²𝑑𝑡−_Ω𝜏𝛽12−𝑥𝜕𝑢𝜕𝑥²−𝑑𝑥𝑑𝑡,_Ω𝜏𝛽()𝜕𝑥+1𝑥𝜕𝑥𝜕𝑢𝜕𝑥𝜕𝑢=𝜕𝑡𝑑𝑥𝑑𝑡_𝜏0𝛽²+𝛽𝜕𝑢𝜕𝑥𝛽,𝑡𝜕𝑢+𝜕𝑡𝛽,𝑡𝑑𝑡_Ω𝜏𝛽𝑥𝜕𝑢𝜕𝑥𝜕𝑢+1𝜕𝑡𝑑𝑥𝑑𝑡2¹_𝛽𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²−1𝑑𝑥2¹_𝛽𝑥²+𝑥𝜕𝜑𝜕𝑥²𝑑𝑥, (21) and by using the Cauchy's 𝜀-inequality, we get _Ω𝜏𝛼𝑥²𝜕𝑢𝜕𝑥𝜕𝑢≤𝜀𝜕𝑡𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛼𝑥²𝜕𝑢𝜕𝑥²1𝑑𝑥𝑑𝑡+2𝜀_Ω^𝜏_𝛼𝑥²𝜕𝑢𝜕𝑡²≤1𝑑𝑥𝑑𝑡4_Ω^𝜏_𝛼𝜕𝑢𝜕𝑥²+1𝑑𝑥𝑑𝑡4𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑥²+𝛼𝑑𝑥𝑑𝑡₂+𝛼2_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡,_Ω𝜏𝛼𝑓⋅𝑥²−𝑥𝜕𝑢+𝑥𝜕𝑥²+𝑥𝜕𝑢≤𝛼𝜕𝑡𝑑𝑥𝑑𝑡₂+𝛼²_Ω𝜏𝛼𝑓²1𝑑𝑥𝑑𝑡+4_Ω^𝜏_𝛼𝜕𝑢𝜕𝑥²+1𝑑𝑥𝑑𝑡4𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑥²+1𝑑𝑥𝑑𝑡2𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡,_{Ω𝜏𝛼,𝛽}𝑓⋅−𝑥𝜕𝑢+𝑥𝜕𝑥²+𝑥𝜕𝑢≤𝛽𝜕𝑡𝑑𝑥𝑑𝑡𝛽−𝛼²−𝛼²+𝛽−𝛼2+12×_{Ω𝜏𝛼,𝛽}𝑓²1𝑑𝑥𝑑𝑡+4_Ω^𝜏_𝛼,𝛽𝜕𝑢𝜕𝑥²+1𝑑𝑥𝑑𝑡4𝛽²−𝛼²+×𝛽−𝛼_{Ω𝜏𝛼,𝛽}𝑥²+𝑥𝜕𝑢𝜕𝑥²+1𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛼,𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²,−_Ω𝜏𝛽𝑥²𝜕𝑢𝜕𝑥𝜕𝑢≤𝜀𝜕𝑡𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛽𝑥𝜕𝑢𝜕𝑥²1𝑑𝑥𝑑𝑡+2𝜀_Ω^𝜏_𝛽𝑥𝜕𝑢𝜕𝑡²≤𝜀𝑑𝑥𝑑𝑡1−𝛽2_Ω^𝜏_𝛽𝜕𝑢𝜕𝑥²+1𝑑𝑥𝑑𝑡2𝜀_Ω^𝜏_𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²≤1𝑑𝑥𝑑𝑡4_Ω^𝜏_𝛽𝜕𝑢𝜕𝑥²+𝑑𝑥𝑑𝑡1−𝛽_Ω𝜏𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡,_Ω𝜏𝛽𝑓⋅𝑥²−𝑥𝜕𝑢+𝑥𝜕𝑥²+𝑥𝜕𝑢≤𝜕𝑡𝑑𝑥𝑑𝑡1−𝛽²+1−𝛽²_Ω𝜏𝛽𝑓²+14_Ω^𝜏_𝛽𝜕𝑢𝜕𝑥²+141−𝛽²+1−𝛽^Ω𝜏_𝛽𝑥²+𝑥𝜕𝑢𝜕𝑥²+121−𝛽²+1−𝛽^Ω𝜏_𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡². (22) Substituting (21)–(22) into ∫_Ω𝜏ℒ𝑢⋅𝑀𝑢𝑑𝑥𝑑𝑡, we obtain 𝛼1−²+𝛼2−12𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑡²+1𝑑𝑥𝑑𝑡2^𝛼₀𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²𝑑𝑥_𝜏0𝛼2−𝛼²2𝜕𝑢()𝜕𝑥𝛼,𝑡²−𝑑𝑡_𝜏0𝛼²+𝛼𝜕𝑢()𝜕𝑥𝛼,𝑡𝜕𝑢()≤𝛼𝜕𝑡𝛼,𝑡𝑑𝑡₂+𝛼²_Ω𝜏𝛼𝑓²+1𝑑𝑥𝑑𝑡2^𝛼₀𝑥²+𝑥𝜕𝜑𝜕𝑥²+1𝑑𝑥2𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡,Ω^𝜏_𝛼,12_Ω^𝜏_𝛼,𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²+12^𝛽_𝛼𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²−𝑑𝑥_𝜏0𝛽²+𝛽𝜕𝑢𝜕𝑥𝛽,𝑡𝜕𝑢+𝜕𝑡𝛽,𝑡𝑑𝑡_𝜏0𝛼2−𝛼²2𝜕𝑢()𝜕𝑥𝛼,𝑡²𝛽𝑑𝑡2^𝜏₀𝜕𝑢𝜕𝑥𝛽,𝑡²𝛼𝑑𝑡−2^𝜏₀𝜕𝑢()𝜕𝑥𝛼,𝑡²≤𝑑𝑡𝛽−𝛼²𝛽+𝛼+12+12_Ω^𝜏_𝛼,𝛽𝑓²+1𝑑𝑥𝑑𝑡2^𝛽_𝛼𝑥²+𝑥𝜕𝜑𝜕𝑥²+1𝑑𝑥4𝛽²−𝛼²+×𝛽−𝛼_{Ω𝜏𝛼,𝛽}𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡,Ω^𝜏_𝛼,𝛽,1𝛽−21−𝛽²+1−𝛽^Ω𝜏_𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²+1𝑑𝑥𝑑𝑡2¹_𝛽𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²𝑑𝑥_𝜏0𝛽²2−𝛽2𝜕𝑢𝜕𝑥𝛽,𝑡²+𝑑𝑡_𝜏0𝛽²+𝛽𝜕𝑢𝜕𝑥𝛽,𝑡𝜕𝑢≤𝜕𝑡𝛽,𝑡𝑑𝑡1−𝛽²+1−𝛽²_Ω𝜏𝛽𝑓²+1𝑑𝑥𝑑𝑡2¹_𝛽𝑥²+𝑥𝜕𝜑𝜕𝑥²+1𝑑𝑥41−𝛽²+×1−𝛽_Ω𝜏𝛽𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡,Ω^𝜏_𝛽. (23)(24)(25) Now, by using the conditions (PR) in (23), (24), and (25), respectively, we obtain 𝛼1−²+𝛼2−12𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑡²+1𝑑𝑥𝑑𝑡2^𝛼₀𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²≤𝛼𝑑𝑥₂+𝛼²_Ω𝜏𝛼𝑓²1𝑑𝑥𝑑𝑡+2^𝛼₀𝑥²+𝑥𝜕𝜑𝜕𝑥²+1𝑑𝑥2𝛼²+𝛼_Ω^𝜏_𝛼𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡,Ω^𝜏_𝛼,12_Ω^𝜏_𝛼,𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²+12^𝛽_𝛼𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²≤𝑑𝑥𝛽−𝛼²𝛽+𝛼+12+12_Ω^𝜏_𝛼,𝛽𝑓²+1𝑑𝑥𝑑𝑡2^𝛽_𝛼𝑥²+𝑥𝜕𝜑𝜕𝑥²+1𝑑𝑥4𝛽²−𝛼²+×𝛽−𝛼_{Ω𝜏𝛼,𝛽}𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡,Ω^𝜏_𝛼,𝛽,1𝛽−21−𝛽²+1−𝛽^Ω𝜏_𝛽𝑥²+𝑥𝜕𝑢𝜕𝑡²+1𝑑𝑥𝑑𝑡2¹_𝛽𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²≤𝑑𝑥1−𝛽²+1−𝛽²_Ω𝜏𝛽𝑓²+1𝑑𝑥𝑑𝑡2¹_𝛽𝑥²+𝑥𝜕𝜑𝜕𝑥²+1𝑑𝑥41−𝛽²+1−𝛽×_Ω𝜏𝛽𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡.Ω^𝜏_𝛽. (26)(27)(28) Or in Ω^𝜏, by combining (26), (27), and (28), we get 𝛼min1−²+𝛼2−12𝛼²,1+𝛼2,1𝛽−21−𝛽²+×1−𝛽_Ω𝜏𝑥²+𝑥𝜕𝑢𝜕𝑡²+1𝑑𝑥𝑑𝑡2¹₀𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²𝛼𝑑𝑥≤max²+𝛼²,𝛽−𝛼²𝛽+𝛼+12+12,1−𝛽²+1−𝛽²×_Ω𝜏𝑓²1𝑑𝑥𝑑𝑡+2¹₀𝑥²+𝑥𝜕𝜑𝜕𝑥²1𝑑𝑥+max2𝛼²,1+𝛼4𝛽²−𝛼²+,1𝛽−𝛼41−𝛽²+×1−𝛽_Ω𝜏𝑥²+𝑥𝜕𝑢𝜕𝑥²𝑑𝑥𝑑𝑡. (29)
Using Lemma 1 in [20], one has _Ω𝑥²+𝑥𝜕𝑢𝜕𝑡²𝑑𝑥𝑑𝑡+¹₀𝑥²()+𝑥𝜕𝑢⋅,𝜏𝜕𝑥²𝑑𝑥≤𝑘_Ω𝑓²𝑑𝑥𝑑𝑡+¹₀𝑥²+𝑥𝜕𝜑𝜕𝑥²𝑑𝑥, (30) where 𝑘=max𝛼²+𝛼²,𝛽−𝛼²𝛽+𝛼+12+12,1−𝛽²+1−𝛽²×𝛼min1−²+𝛼2−12𝛼²,1+𝛼2,1𝛽−21−𝛽²+1−𝛽⁻¹×𝑒^{(max(1/2(𝛼2+𝛼),1/4((𝛽2−𝛼2)+(𝛽−𝛼)),1/4((1−𝛽)2+(1−𝛽)))𝑇).} (31)
The right-hand side of (30) 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=(𝑘)^1/2.

Corollary 3. A solution of the problem (1)–(4) is unique if it exists and depends continuously on ℱ∈𝐹.

4. Existence of Solution

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

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

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

Proposition 5. Let the conditions of Theorem 4 be satisfied. if for 𝜔∈𝐿²(Ω) and for all 𝑢∈𝐷₀(𝐿), one has _Ωℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡=0, (32) then 𝜔 vanishes almost everywhere in Ω.

Proof. The scalar product of 𝐹 is defined by (𝐿𝑢,𝜔)_𝐹=_Ω+ℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡₁₀𝑥²+𝑥𝜕ℓ𝑢𝜕𝑥𝜕𝜔₀𝜕𝑥𝑑𝑥. (33) the equality (32) can be written as follows: _Ω𝜕𝑢𝜕𝑡⋅𝜔𝑑𝑥𝑑𝑡=_Ω1𝑥𝜕𝑥𝜕𝑥𝜕𝑢𝜕𝑥⋅𝜔𝑑𝑥𝑑𝑡, (34) if we put 𝑢=ℑ_𝑡𝑒^𝑘𝑡𝑧=^𝑡₀𝑒^𝑘𝜏𝑧(𝑥,𝜏)𝑑𝜏, (35) where 𝑘 is a strictly positive constant and 𝑧,𝜕𝑧/𝜕𝑥,(𝜕/𝜕𝑥)(𝜕ℑ_𝑡(𝑒^𝑘𝑡𝑧)/𝜕𝑥)∈𝐿²(Ω), then, 𝑢 satisfies the boundary conditions in (PR). As a result of (34), we obtain the equality _Ω𝑒^𝑘𝑡𝑧𝜔𝑑𝑥𝑑𝑡=_Ω1𝑥𝜕𝑥𝜕𝑥𝜕ℑ_𝑡𝑒^𝑘𝑡𝑧𝜕𝑥⋅𝜔𝑑𝑥𝑑𝑡. (36) The left-hand side of (36) shows that the mapping 𝐿²(Ω)∋𝑧⟶_Ω1𝑥𝜕𝑥𝜕𝑥𝜕ℑ_𝑡𝑒^𝑘𝑡𝑧𝜕𝑥⋅𝜔𝑑𝑥𝑑𝑡 (37) is a continuous linear functional of 𝑧. From the right-hand side of (36) there follows that is true if the function 𝜔 has the following properties: 1𝑥𝜕𝜔,𝜕𝑥𝜕ℑ^⋆_𝑡𝜔,1𝜕𝑥𝑥𝜕ℑ^⋆_𝑡𝜔,𝜕𝜕𝑥²ℑ^⋆_𝑡𝜔𝜕𝑥²,ℑ^⋆_𝑡𝜔𝑥²∈𝐿²(Ω),𝜕𝜔()=𝜕𝑥𝛼,0𝜕𝜔=𝜕𝑥𝛽,0𝜕𝜔()𝜕𝑥1,0=0,𝜕𝜔()𝜕𝑥1,𝑡=0. (38) In terms of the given function 𝜔 and from the equality (36), we give the function 𝜔 in terms of 𝑧 as follows: −𝑥𝜔=𝑥−²2𝜕𝑧𝜕𝑥,0≤𝑥≤𝛼,−𝑥𝜕𝑧−𝑥𝜕𝑥,𝛼≤𝑥≤𝛽,𝑥−²2𝜕𝑧𝜕𝑥,𝛽≤𝑥≤1, (39) and 𝑧 satisfies the same conditions of the function 𝑢 in (PR): 𝛼𝜕𝑧()𝜕𝑥𝛼,𝑡=𝛽𝜕𝑧,𝜕𝑥𝛽,𝑡𝜕𝑧()()𝜕𝑥1,𝑡=0,𝑧𝛼,𝑡=𝑧𝛽,𝑡=0. (40)(41) Replacing 𝜔 in (36) by its representation (39) and integrating by parts each term of (36) and by taking the conditions (40) and (41), we obtain _Ω𝜏𝑒^𝑘𝑡=−1𝑧𝜔𝑑𝑥𝑑𝑡2∫^𝜏₀𝛼𝛼−²2ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧(𝛼,𝑡)𝜕𝑥²−1𝑑𝑥𝑑𝑡2∫_Ω𝜏𝛼(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²Ω𝑑𝑥𝑑𝑡,_𝜏𝛼=()×(),−𝛽0,𝛼0,𝜏2∫^𝜏₀ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝛽,𝑡𝜕𝑥²+𝛼𝑑𝑥𝑑𝑡2∫^𝜏₀ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧(𝛼,𝑡)𝜕𝑥²−1𝑑𝑥𝑑𝑡2∫_{Ω𝜏𝛼,𝛽}ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²Ω𝑑𝑥𝑑𝑡,_{𝜏𝛼,𝛽}=×(),1𝛼,𝛽0,𝜏2∫^𝜏₀𝛽𝛽−²2ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝛽,𝑡𝜕𝑥²−1𝑑𝑥𝑑𝑡2∫_Ω𝜏𝛽(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²Ω𝑑𝑥𝑑𝑡,_𝜏𝛽=𝛽,1×(0,𝜏). (42)
So, we obtain _Ω𝜏𝑒^𝑘𝑡=−1𝑧𝜔𝑑𝑥𝑑𝑡2∫_Ω𝜏𝛼(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²Ω𝑑𝑥𝑑𝑡,_𝜏𝛼=()×(),−10,𝛼0,𝜏2∫_{Ω𝜏𝛼,𝛽}ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²Ω𝑑𝑥𝑑𝑡,_{𝜏𝛼,𝛽}=×(),−1𝛼,𝛽0,𝜏2∫_Ω𝜏𝛽(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²Ω𝑑𝑥𝑑𝑡,_𝜏𝛽=𝛽,1×(0,𝜏). (43)
Now, by combining over Ω^𝜏, we get 12_Ω^𝜏_𝛼(1−𝑥)𝑒^𝑘𝑡𝑧²1𝑑𝑥𝑑𝑡+2_Ω^𝜏_𝛼,𝛽𝑒^𝑘𝑡𝑧²+1𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛽(1−𝑥)𝑒^𝑘𝑡𝑧²=𝛼𝑑𝑥𝑑𝑡₂4^𝜏₀ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧(𝛼,𝑡)𝜕𝑥²−𝛽𝑑𝑥𝑑𝑡₂4^𝜏₀ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝛽,𝑡𝜕𝑥²−1𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛼(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²−1𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛼,𝛽ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²−1𝑑𝑥𝑑𝑡2_Ω^𝜏_𝛽(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²𝑑𝑥𝑑𝑡, (44) by taking into accont the boundary condition (40) of the function 𝑧 yields 12∫_Ω𝜏𝛼(1−𝑥)𝑒^𝑘𝑡𝑧²𝑑𝑥𝑑𝑡+∫_{Ω𝜏𝛼,𝛽}𝑒^𝑘𝑡𝑧²𝑑𝑥𝑑𝑡+∫_Ω𝜏𝛽(1−𝑥)𝑒^𝑘𝑡𝑧²1𝑑𝑥𝑑𝑡=−2∫_Ω𝜏𝛼(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²𝑑𝑥𝑑𝑡+∫_{Ω𝜏𝛼,𝛽}ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²∫𝑑𝑥𝑑𝑡_Ω𝜏𝛽(1−𝑥)ℑ_𝑡𝑒^𝑘𝑡𝜕𝑧𝜕𝑥²𝑑𝑥𝑑𝑡≤0. (45)
Then, we obtain 12_Ω^𝜏_𝛼(1−𝑥)𝑒^𝑘𝑡𝑧²𝑑𝑥𝑑𝑡+_{Ω𝜏𝛼,𝛽}𝑒^𝑘𝑡𝑧²+𝑑𝑥𝑑𝑡_Ω𝜏𝛽(1−𝑥)𝑒^𝑘𝑡𝑧²𝑑𝑥𝑑𝑡≤0. (46) And thus 𝑧=0 in Ω, then 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)∈𝑅(𝐿)^⟂ and for all 𝑢∈𝐷(𝐿)≡𝐵, it holds (𝐿𝑢,𝜔)_𝐹=_Ωℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡+¹₀(𝑥+1)𝜕ℓ𝑢𝜕𝑥𝜕𝜔₀𝜕𝑥𝑑𝑥. (47)

Then we must prove that 𝑊=0. Putting 𝑢∈𝐷₀(𝐿) in (47), we have _Ωℒ𝑢⋅𝜔𝑑𝑥𝑑𝑡=0,𝑢∈𝐷₀(𝐿). (48) Hence Proposition 5 implies that 𝜔=0. Thus (47) takes the form ¹₀(𝑥+1)𝜕ℓ𝑢𝜕𝑥𝜕𝜔₀(𝐿).𝜕𝑥𝑑𝑥=0,𝑢∈𝐷 (49) Since the range of the trace operator ℓ is everywhere dense in the Hilbert space 𝐹 with the norm (𝑥+1)¹₀𝜕ℓ𝑢𝜕𝑥^21/2, (50) the equality (49) implies that 𝜔₀=0. Hence 𝑊=0 implies (𝑅(𝐿)=𝐹). Therefore, the proof of Theorem 4 is complete.