We study a boundary value problem with multivariables integral type condition for a class of parabolic equations. We prove the existence, uniqueness, and continuous dependence of the solution upon the data in the functional wieghted Sobolev spaces. Results are obtained by using a functional analysis method based on two-sided a priori estimates and on the density of the range of the linear operator generated by the considered problem.

1. Introduction

Certain problems of modern physics and technology can be effectively described in terms of nonlocal problems with integral conditions for partial differential equations.These nonlocal conditions arise mainly when the data on the boundary cannot be measured directly. Motivated by this, we consider in the rectangular domain Ω=(0,1)×(0,𝑇), the following nonclassical boundary value problem of finding a solution 𝑢(𝑥,𝑡) such that ℒ𝑢=ğœ•ğ‘¢ğœ•ğœ•ğ‘¡âˆ’ğ‘Ž(𝑡)2𝑢𝜕𝑥2=𝑓(𝑥,𝑡),(1.1) where the function ğ‘Ž(𝑡) and its derivative are bounded on the interval [0,𝑇]∶

0<𝑐0â‰¤ğ‘Ž(𝑡)≤𝑐1,0<𝑐2â‰¤ğ‘‘ğ‘Ž(𝑡)𝑑𝑡≤𝑐3,(1.2)𝑙𝑢=𝑢(𝑥,0)=𝜑(𝑥),𝑥∈(0,1),(1.3)𝑢(0,𝑡)=𝑢(𝛽,𝑡)=𝑢(𝛾,𝑡)=𝑢(1,𝑡),𝑡∈(0,𝑇),(1.4)𝛼0𝑢(𝑥,𝑡)𝑑𝑥+2𝛾𝛽𝑢(𝑥,𝑡)𝑑𝑥+1𝛿𝑢(𝑥,𝑡)𝑑𝑥=0,0<𝛼<𝛽<𝛾<𝛿<1,𝛼=1−𝛿=𝛾−𝛽,𝑡∈(0,𝑇).(1.5) Here, we assume that the known function 𝜑 satisfies the conditions given in (1.4) and (1.5), that is,

𝜑(0)=𝜑(𝛽)=𝜑(𝛾)=𝜑(1),𝛼0𝜑(𝑥)𝑑𝑥+2𝛾𝛽𝜑(𝑥)𝑑𝑥+1𝛿𝜑(𝑥)𝑑𝑥=0.(1.6) When considering the classical solution of the problem (1.1)–(1.5), along with (1.5), there should be the fulfilled conditions:


Mathematical modelling of different phenomena leads to problems with nonlocal or integral boundary conditions. Such a condition occurs in the case when one measures an averaged value of some parameter inside the domain. This amounts to the specification of the energy or mass contained in a portion of the conductor or porous medium as a function of time. This problems arise in plasma physics, heat conduction, biology and demography, as well as modelling of technological process, see, for example, [1–5]. Boundary-value problems for parabolic equations with integral boundary condition are investigated by Batten [6], Bouziani and Benouar [7], Cannon [8, 9], Cannon, Perez Esteva and van der Hoek [10], Ionkin [11], Kamynin [12], Shi and Shillor [13], Shi [4], Marhoune and Bouzit [14], Marhoune and Hameida [15], Yurchuk [16], and many references therein. The problem with one-variable (resp., two-variable) boundary integral type condition is studied in [5] and by Marhoune and Latrous [17] (resp., in Marhoune [2]).

Mention that in the cited paper [16], the author proved the existence, uniqueness, and continuous dependence of a stronge solution in weighted Sobolev spaces to the problem

𝜕𝑢=ğœ•ğœ•ğ‘¡î‚€ğœ•ğ‘¥ğ‘Ž(𝑥,𝑡)𝜕𝑢𝜕𝑥+𝑓(𝑥,𝑡),(1.8) under the following conditions:

𝑢𝑢(𝑥,0)=0,0≤𝑥≤1,(0,𝑡)=0,0<𝑡≤𝑇,10𝑢(𝑥,𝑡)𝑑𝑥=0.(1.9) This last integral condition in the form

10𝑢(𝑥,𝑡)𝑑𝑥=𝑚(𝑡),0<𝑡≤𝑇,(1.10) arises, for example, in biochemistry in which 𝑚 is a constant, and in this case is known as the conservation of protein [18]. Further, in [5], the author studied a similar problem with the weak integral condition

𝛼0𝑢(𝑥,𝑡)𝑑𝑥=0,0<𝛼<1.(1.11) The same problem with the new integral condition

𝛼0𝑢(𝑥,𝑡)𝑑𝑥+1𝛽𝑢(𝑥,𝑡)𝑑𝑥=0,𝛼+𝛽=1,(1.12) was investigated in [2]. The present paper is an extension in the same direction. By constructing a suitable multiplicator, we will try to establish existence and uniqueness of solution of problem (1.1)–(1.5). Note that the multivariables integral type condition (1.5) is considerably much weaker and better than that used in [2]. In fact, some physical problems have motivated specialists to consider nonlocal integral condition (1.5), which tells us the integral total effect of the solution 𝑢 over several independent portions [0,𝛼], [𝛽,𝛾], and [𝛿,1] of interval 𝐼=(0,1) at certain time 𝑡 that give this effect over the entire or part of this interval.

We associate with (1.1)–(1.5) the operator 𝐿=(ℒ,𝑙), defined from 𝐸 into 𝐹, where 𝐸 is the Banach space of functions 𝑢∈𝐿2(Ω), satisfying (1.4) and (1.5), with the finite norm ‖𝑢‖2𝐸=Ω𝑝(𝑥)3|||𝜕𝑢|||𝜕𝑡2+||||𝜕2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡+sup0≤𝑡≤𝑇10𝑝(𝑥)2|||𝜕𝑢|||𝜕𝑥2𝑑𝑥+sup0≤𝑡≤𝑇𝛼0|𝑢|2𝑑𝑥+sup0≤𝑡≤𝑇𝛾𝛽|𝑢|2𝑑𝑥+sup0≤𝑡≤𝑇1𝛿|𝑢|2𝑑𝑥,(1.13) and 𝐹 is the Hilbert space of vector-valued functions ℱ=(𝑓,𝜑) obtained by completion of the space 𝐿2(Ω)×𝑊22(0,1) with respect to the norm

‖ℱ‖2𝐹=‖(𝑓,𝜑)‖2𝐹=Ω𝑝(𝑥)3||𝑓||2𝑑𝑥𝑑𝑡+10𝑝(𝑥)2|||𝑑𝜑|||𝑑𝑥2𝑑𝑥+𝛼0||𝜑||2𝑑𝑥+𝛾𝛽||𝜑||2𝑑𝑥+1𝛿||𝜑||2𝑑𝑥,(1.14) where

âŽ§âŽªâŽªâŽ¨âŽªâŽªâŽ©ğ‘¥ğ‘(𝑥)=2]],,𝑥∈0,𝛼(𝛾−𝛽)2[]∪[],,𝑥∈𝛼,𝛽𝛾,𝛿(𝛾−𝑥)2+(𝑥−𝛽)2[],,𝑥∈𝛽,𝛾(1−𝑥)2[[.,𝑥∈𝛿,1(1.15) Using the energy inequalities method proposed in [16], we establish two-sided a priori estimates. Then, we prove that the operator 𝐿 is a linear homeomorphism between the spaces 𝐸 and 𝐹.

2. Two-Sided A Priori Estimates

Theorem 2.1. For any function 𝑢∈𝐸, one has the a priori estimate ‖𝐿𝑢‖2𝐹≤𝑐4‖𝑢‖2𝐸,(2.1) where the constant 𝑐4 is independent of 𝑢. In fact, 𝑐4=2max(1,𝑐21).

Proof. Using (1.1) and initial condition (1.3), we obtain Ω𝑝(𝑥)3||||ℒ𝑢2𝑑𝑥𝑑𝑡≤2Ω𝑝(𝑥)3|||𝜕𝑢|||𝜕𝑡2+𝑐21||||𝜕2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡,10𝑝(𝑥)2|||𝑑𝜑|||𝑑𝑥2𝑑𝑥≤sup0≤𝑡≤𝑇10𝑝(𝑥)2|||𝜕𝑢|||𝜕𝑥2𝑑𝑥,𝛼0||𝜑||2𝑑𝑥≤sup0≤𝑡≤𝑇𝛼0|𝑢|2𝑑𝑥,𝛾𝛽||𝜑||2𝑑𝑥≤sup0≤𝑡≤𝑇𝛾𝛽|𝑢|2𝑑𝑥,1𝛿||𝜑||2𝑑𝑥≤sup0≤𝑡≤𝑇1𝛿|𝑢|2𝑑𝑥.(2.2) Combining the inequalities in (2.2), we obtain (2.1) for 𝑢∈𝐸.

Theorem 2.2. For any function 𝑢∈𝐸, one has the a priori estimate ‖𝑢‖2𝐸≤𝑐5‖𝐿𝑢‖2𝐹,(2.3) with the constant 𝑐5=exp(𝑐𝑇)max49,2𝑐1min13/32,𝑐0,𝑐20,/2(2.4) and 𝑐 is such that 𝑐𝑐0≥𝑐3.(2.5)

Before proving this theorem, we need the following lemma.

Lemma 2.3 (see [19]). For 𝑢∈𝐸, one has î€œğ‘ğ‘Ž||||𝑏𝑥𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2𝑑𝑥≤4ğ‘ğ‘Ž(ğ‘¥âˆ’ğ‘Ž)2|||𝜕𝑢|||𝜕𝑡2𝑑𝑥,ğ‘ğ‘Ž||||î€œğ‘¥ğ‘Žğœ•ğ‘¢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2𝑑𝑥≤4ğ‘ğ‘Ž(𝑏−𝑥)2|||𝜕𝑢|||𝜕𝑡2𝑑𝑥.(2.6)

Proof of Theorem 2.2. Define âŽ§âŽªâŽªâŽªâŽªâŽªâŽ¨âŽªâŽªâŽªâŽªâŽªâŽ©ğ‘¥ğ‘€ğ‘¢=2𝜕𝑢𝜕𝑡+2𝑥𝛼𝑥𝜕𝑢(𝜉,𝑡)]],𝜕𝑡𝑑𝜉,𝑥∈0,𝛼(𝛾−𝛽)2𝜕𝑢[]∪[],𝜕𝑡,𝑥∈𝛼,𝛽𝛾,𝛿(𝛾−𝑥)2𝜕𝑢𝜕𝑡+(𝑥−𝛽)2𝜕𝑢𝜕𝑡+2(𝛾−𝑥)𝑥𝛽𝜕𝑢(𝜉,𝑡)𝜕𝑡𝑑𝜉+2(𝑥−𝛽)𝛾𝑥𝜕𝑢(𝜉,𝑡)[],𝜕𝑡𝑑𝜉,𝑥∈𝛽,𝛾(1−𝑥)2𝜕𝑢𝜕𝑡+2(1−𝑥)𝑥𝛿𝜕𝑢(𝜉,𝑡)[[.𝜕𝑡𝑑𝜉,𝑥∈𝛿,1(2.7) We consider for 𝑢∈𝐸 the quadratic formula Re𝜏010exp(−𝑐𝑡)ℒ𝑢𝑀𝑢𝑑𝑥𝑑𝑡,(2.8) with the constant 𝑐 satisfying (2.5), obtained by multiplying (1.1) by exp(−𝑐𝑡)𝑀𝑢, by integrating over Ω𝜏, where Ω𝜏=(0,1)×(0,𝜏), with 0≤𝜏≤𝑇, and by taking the real part. Integrating by parts in (2.8) by report to 𝑥 with the use of boundary conditions (1.4) and (1.5), we obtain Re𝜏010exp(−𝑐𝑡)ℒ𝑢=𝑀𝑢𝑑𝑥𝑑𝑡𝜏010|||𝑝(𝑥)exp(−𝑐𝑡)𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡+𝜏0𝛼0||||exp(−𝑐𝑡)𝛼𝑥𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2+𝑑𝑥𝑑𝑡𝜏0𝛾𝛽||||exp(−𝑐𝑡)𝑥𝛽𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2+||||𝛾𝑥𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2+𝑑𝑥𝑑𝑡𝜏01𝛿||||exp(−𝑐𝑡)𝑥𝛿𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2𝑑𝑥𝑑𝑡+Re𝜏010exp(−𝑐𝑡)𝑝(𝑥)ğ‘Žğœ•ğ‘¢ğœ•ğœ•ğ‘¥2𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡+2Re𝜏0𝛼0𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡+4Re𝜏0𝛾𝛽𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡+2Re𝜏01𝛿𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢ğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡.(2.9) On the other hand, by using the elementary inequalities we get Re𝜏010exp(−𝑐𝑡)ℒ𝑢≥𝑀𝑢𝑑𝑥𝑑𝑡𝜏010𝑝|||(𝑥)exp(−𝑐𝑡)𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡+Re𝜏010exp(−𝑐𝑡)𝑝(𝑥)ğ‘Žğœ•ğ‘¢ğœ•ğœ•ğ‘¥2𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡+2Re𝜏0𝛼0𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡+4Re𝜏0𝛾𝛽𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡+2Re𝜏01𝛿𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢ğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡.(2.10) Again, integrating by parts the second, third, fourth, and fifth terms of the right-hand side of the inequality (2.10) by report to 𝑡 and taking into account the initial condition (1.3) and (2.5) gives Re𝜏010exp(−𝑐𝑡)𝑝(𝑥)ğ‘Žğœ•ğ‘¢ğœ•ğœ•ğ‘¥2𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡≥10exp(−𝑐𝜏)2|||𝑝(𝑥)ğ‘Ž(𝜏)𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2−1𝑑𝑥210𝑝|||(𝑥)ğ‘Ž(0)𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥,Re𝜏0𝛼0𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡â‰¥ğ›¼0exp(−𝑐𝜏)2||||ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥−𝛼0ğ‘Ž(0)2||||𝑙𝑢2𝑑𝑥;Re𝜏0𝛾𝛽𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡â‰¥ğ›¾ğ›½exp(−𝑐𝜏)2||||ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2î€œğ‘‘ğ‘¥âˆ’ğ›¾ğ›½ğ‘Ž(0)2||||𝑙𝑢2𝑑𝑥,Re𝜏01𝛿𝜕exp(−𝑐𝑡)ğ‘Žğ‘¢ğ‘¢î€œğœ•ğ‘¡ğ‘‘ğ‘¥ğ‘‘ğ‘¡â‰¥1𝛿exp(−𝑐𝜏)2||||ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥−1ğ›¿ğ‘Ž(0)2||||𝑙𝑢2𝑑𝑥.(2.11) Using (2.11) in (2.10), we get Re𝜏010exp(−𝑐𝑡)ℒ𝑢𝑀𝑢𝑑𝑥𝑑𝑡+𝛼0||||ğ‘Ž(0)𝑙𝑢2𝑑𝑥+2𝛾𝛽||||ğ‘Ž(0)𝑙𝑢2+𝑑𝑥1𝛿||||ğ‘Ž(0)𝑙𝑢21𝑑𝑥+210|||𝑝(𝑥)ğ‘Ž(0)𝑑𝑙𝑢|||𝑑𝑥2≥𝑑𝑥𝜏010|||𝑝(𝑥)exp(−𝑐𝑡)𝜕𝑢|||𝜕𝑡2+1𝑑𝑥𝑑𝑡210|||exp(−𝑐𝜏)𝑝(𝑥)ğ‘Ž(𝜏)𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2𝑑𝑥+𝛼0||||exp(−𝑐𝜏)ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥+2𝛾𝛽||||exp(−𝑐𝜏)ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥+1𝛿||||exp(−𝑐𝜏)ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥.(2.12) By using the 𝜀-inequalities on the first integral in the left-hand side of (2.12) and Lemma 2.3, we obtain 1532𝜏010|||𝑝(𝑥)exp(−𝑐𝑡)𝜕𝑢|||𝜕𝑡21𝑑𝑥𝑑𝑡+210|||exp(−𝑐𝜏)𝑝(𝑥)ğ‘Ž(𝜏)𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2+𝑑𝑥𝛼0||||exp(−𝑐𝜏)ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥+2𝛾𝛽||||exp(−𝑐𝜏)ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2+𝑑𝑥1𝛿||||exp(−𝑐𝜏)ğ‘Ž(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥≤16𝜏010||||𝑝(𝑥)exp(−𝑐𝑡)ℒ𝑢2+𝑑𝑥𝑑𝑡𝛼0||||ğ‘Ž(0)𝑙𝑢21𝑑𝑥+210|||𝑝(𝑥)ğ‘Ž(0)𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥+2𝛾𝛽||||ğ‘Ž(0)𝑙𝑢2𝑑𝑥+1𝛿||||ğ‘Ž(0)𝑙𝑢2𝑑𝑥.(2.13) Now, from (1.1), we have 𝑐206𝜏010||||𝜕𝑝(𝑥)exp(−𝑐𝑡)2𝑢𝜕𝑥2||||2≤𝑑𝑥𝑑𝑡𝜏010𝑝(𝑥)3||||exp(−𝑐𝑡)ℒ𝑢2𝑑𝑥𝑑𝑡+𝜏010𝑝(𝑥)3|||exp(−𝑐𝑡)𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡.(2.14) Combining inequalities (2.13) and (2.14), we get exp(−𝑐𝑇)1332𝜏010𝑝(𝑥)3|||𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡+𝑐010𝑝(𝑥)2|||𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2𝑑𝑥+𝑐0𝛼0||||𝑢(𝑥,𝜏)2𝑑𝑥+2𝑐0𝛾𝛽||||𝑢(𝑥,𝜏)2𝑑𝑥+𝑐01𝛿||||𝑢(𝑥,𝜏)2𝑐𝑑𝑥+206𝜏010𝑝(𝑥)3||||𝜕2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡≤49Ω𝑝(𝑥)3||||ℒ𝑢2𝑑𝑥𝑑𝑡+𝑐110𝑝(𝑥)2|||𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥+𝑐1𝛼0||||𝑙𝑢2𝑑𝑥+2𝑐1𝛾𝛽||||𝑙𝑢2𝑑𝑥+𝑐11𝛿||||𝑙𝑢2𝑑𝑥.(2.15) As the right-hand side of (2.15) is independent of 𝜏, by replacing the left-hand side by its upper bound with respect to 𝜏 in the interval [0,𝑇], we obtain the desired inequality.

3. Solvability of the Problem

From estimates (2.1) and (2.3), it follows that the operator 𝐿∶𝐸→𝐹 is continuous and its range is closed in 𝐹. Therefore, the inverse operator 𝐿−1 exists and is continuous from the closed subspace 𝑅(𝐿) onto 𝐸, which means that 𝐿 is an homeomorphism from 𝐸 onto 𝑅(𝐿). To obtain the uniqueness of solution, it remains to show that 𝑅(𝐿)=𝐹. The proof is based on the following lemma.

Lemma 3.1. Let 𝐷0(𝐿)={𝑢∈𝐸∶𝑙𝑢=0}.(3.1) If for 𝑢∈𝐷0(𝐿) and some 𝑤∈𝐿2(Ω), one has î€œÎ©ğ‘ž(𝑥)ℒ𝑢𝑤𝑑𝑥𝑑𝑡=0,(3.2) where ⎧⎪⎨⎪⎩]],[],[[,ğ‘ž(𝑥)=𝑥,𝑥∈0,𝛼𝛾−𝛽,𝑥∈𝛼,𝛿1−𝑥,𝑥∈𝛿,1(3.3) then 𝑤=0.

Proof. From (3.2) we have î€œÎ©ğ‘ž(𝑥)𝜕𝑢𝜕𝑡𝑤𝑑𝑥𝑑𝑡=Î©ğœ•ğ‘ž(𝑥)ğ‘Ž(𝑡)2𝑢𝜕𝑥2𝑤𝑑𝑥𝑑𝑡.(3.4) Now, for given 𝑤(𝑥,𝑡), we introduce the function âŽ§âŽªâŽªâŽ¨âŽªâŽªâŽ©î€œğ‘£(𝑥,𝑡)=𝑤(𝑥,𝑡)−𝛼𝑥𝑤(𝜉,𝑡)𝜉]],𝑤[],𝑑𝜉,𝑥∈0,𝛼(𝑥,𝑡),𝑥∈𝛼,𝛿𝑤(𝑥,𝑡)−𝑥𝛿𝑤(𝜉,𝑡)𝜉[[.𝑑𝜉,𝑥∈𝛿,1(3.5) Integrating by parts with respect to 𝜉, we obtain âŽ§âŽªâŽªâŽªâŽ¨âŽªâŽªâŽªâŽ©î€œğ‘ž(𝑥)𝑤=𝑥𝑣+𝛼𝑥]],[]∪[],𝑣(𝜉,𝑡)𝑑𝜉,𝑥∈0,𝛼(𝛾−𝛽)𝑣,𝑥∈𝛼,𝛽𝛾,𝛿(𝛾−𝛽)𝑣+𝛾𝛽[],𝑣(𝜉,𝑡)𝑑𝜉,𝑥∈𝛽,𝛾(1−𝑥)𝑣+𝑥𝛿[[,𝑣(𝜉,𝑡)𝑑𝜉,𝑥∈𝛿,1(3.6) which implies that 𝛼0𝑣(𝜉,𝑡)𝑑𝜉+2𝛾𝛽𝑣(𝜉,𝑡)𝑑𝜉+1𝛿𝑣(𝜉,𝑡)𝑑𝜉=0.(3.7) Then, from (3.4), we obtain −Ω𝜕𝑢𝜕𝑡𝑁𝑣𝑑𝑥𝑑𝑡=Ω𝐴(𝑡)𝑢𝑣𝑑𝑥𝑑𝑡,(3.8) where 𝜕𝑁𝑣=ğ‘ž(𝑥)𝑣,𝐴(𝑡)𝑢=âˆ’î‚€ğœ•ğ‘¥ğ‘ž(𝑥)ğ‘Ž(𝑡)𝜕𝑢.𝜕𝑥(3.9) If we introduce the smoothing operators with respect to 𝑡 [16], ℑ𝜀−1=(𝐼+𝜀𝜕/𝜕𝑡)−1 and (ℑ𝜀−1)∗, then these operators provide the solutions of the respective problems: 𝜀𝑑𝑔𝜀(𝑡)𝑑𝑡+𝑔𝜀(𝑡)=𝑔(𝑡),𝑔𝜀(𝑡)𝑡=0=0,(3.10)−𝜀𝑑𝑔∗𝜀(𝑡)𝑑𝑡+𝑔∗𝜀(𝑡)=𝑔(𝑡),𝑔∗𝜀(𝑡)𝑡=𝑇=0,(3.11) and also have the following properties: for any 𝑔∈𝐿2(0,𝑇), the functions 𝑔𝜀=(ℑ𝜀−1)𝑔 and 𝑔∗𝜀=(ℑ𝜀−1)∗𝑔 are in 𝑊12(0,𝑇) such that 𝑔𝜀(𝑡)𝑡=0=0 and 𝑔∗𝜀(𝑡)𝑡=𝑇=0. Morever, ℑ𝜀−1 commutes with 𝜕/𝜕𝑡, so ∫𝑇0|𝑔𝜀−𝑔|2𝑑𝑡→0 and ∫𝑇0|𝑔∗𝜀−𝑔|2𝑑𝑡→0 for 𝜀→0.
Putting ∫𝑢=𝑡0exp(𝑐𝜏)𝑣∗𝜀(𝑥,𝜏)𝑑𝜏 in (3.8), where the constant 𝑐 satisfies 𝑐𝑐0−𝑐3−𝜀𝑐23/𝑐0≥0, and using (3.11), we obtain −Ωexp(𝑐𝑡)𝑣∗𝜀𝑁𝑣𝑑𝑥𝑑𝑡=Ω𝜕𝐴(𝑡)𝑢exp(−𝑐𝑡)𝑢𝜕𝑡𝑑𝑥𝑑𝑡−𝜀Ω𝜕𝐴(𝑡)𝑢𝑣∗𝜀𝜕𝑡𝑑𝑥𝑑𝑡.(3.12)
Integrating by parts each term in the right-hand side of (3.12) and taking the real parts yield 2ReΩ𝜕𝐴(𝑡)𝑢exp(−𝑐𝑡)𝑢𝜕𝑡𝑑𝑥𝑑𝑡=10|||ğ‘Ž(𝑇)ğ‘ž(𝑥)exp(−𝑐𝑇)𝜕𝑢(𝑥,𝑇)|||𝜕𝑡2+î€œğ‘‘ğ‘¥Î©î‚µğ‘ž(𝑥)exp(−𝑐𝑡)ğ‘ğ‘Ž(𝑡)âˆ’ğ‘‘ğ‘Ž(𝑡)|||𝑑𝑡𝜕𝑢|||𝜕𝑥2𝑑𝑥𝑑𝑡;(3.13)Re−𝜀Ω𝜕𝐴(𝑡)𝑢𝑣∗𝜀𝜀𝜕𝑡𝑑𝑥𝑑𝑡=ğ‘…ğ‘’Î©ğ‘‘ğ‘Ž(𝑡)ğ‘‘ğ‘¡ğ‘ž(𝑥)𝜕𝑢𝜕𝜕𝑥𝑣∗𝜀𝜕𝑥𝑑𝑥𝑑𝑡+ğœ€Î©ğœ•ğ‘Ž(𝑡)exp(𝑐𝑡)ğ‘ž(𝑥)𝑣∗𝜀𝜕𝑥𝑑𝑥𝑑𝑡.(3.14) Using 𝜀-inequalities, we obtain Re−𝜀Ω𝜕𝐴(𝑡)𝑢𝑣∗𝜀𝜕𝑡𝑑𝑥𝑑𝑡≥−𝜀𝑐232𝑐0Ω|||ğ‘ž(𝑥)exp(−𝑐𝑡)𝜕𝑢|||𝜕𝑥2𝑑𝑥𝑑𝑡.(3.15) Combining (3.13) and (3.15), we get ReΩexp(𝑐𝑡)ğ‘£âˆ—ğœ€î‚¶î€œğ‘ğ‘£ğ‘‘ğ‘¥ğ‘‘ğ‘¡â‰¤âˆ’Î©îƒ©ğ‘ž(𝑥)exp(−𝑐𝑡)𝑐𝑐0−𝑐3−𝜀𝑐23𝑐0|||𝜕𝑢|||𝜕𝑥2𝑑𝑥𝑑𝑡≤0.(3.16) From (3.16), we deduce that ReΩexp(𝑐𝑡)𝑣∗𝜀𝑁𝑣𝑑𝑥𝑑𝑡≤0.(3.17) Then, for 𝜀→0, we obtain ReΩexp(𝑐𝑡)𝑣𝑁𝑣𝑑𝑥𝑑𝑡=Î©ğ‘ž(𝑥)exp(𝑐𝑡)|𝑣|2𝑑𝑥𝑑𝑡≤0.(3.18) We conclude that 𝑣=0, hence 𝑤=0, which ends the proof of the the lemma.

Theorem 3.2. The range 𝑅(𝐿) of 𝐿 coincides with 𝐹.

Proof. Since 𝐹 is a Hilbert space, we have 𝑅(𝐿)=𝐹 if and only if the relation Ω𝑝(𝑥)3ℒ𝑢𝑓𝑑𝑥𝑑𝑡+10𝑝(𝑥)2𝑑𝑙𝑢𝑑𝑑𝑥𝜑𝑑𝑥𝑑𝑥+𝛼0𝑙𝑢𝜑𝑑𝑥+𝛾𝛽𝑙𝑢𝜑𝑑𝑥+1𝛿𝑙𝑢𝜑𝑑𝑥=0,(3.19) for arbitrary 𝑢∈𝐸 and (𝑓,𝜑)∈𝐹, implies that 𝑓=0 and 𝜑=0.
Putting 𝑢∈𝐷0(𝐿) in (3.19), we conclude from Lemma 3.1 that 𝜃𝑓=0, where ⎧⎪⎨⎪⎩]],[],[[,𝜃𝑓=𝑥𝑓,𝑥∈0,𝛼(𝛾−𝛽)𝑓,𝑥∈𝛼,𝛿(1−𝑥)𝑓,𝑥∈𝛿,1(3.20) then 𝑓=0.
Taking 𝑢∈𝐸 in (3.19) yields 10𝑝(𝑥)2𝑑𝑙𝑢𝑑𝑑𝑥𝜑𝑑𝑥𝑑𝑥+𝛼0𝑙𝑢𝜑𝑑𝑥+𝛾𝛽𝑙𝑢𝜑𝑑𝑥+1𝛿𝑙𝑢𝜑𝑑𝑥=0.(3.21) The range of the operator 𝑙 is everywhere dense in Hilbert space with the norm 10𝑝(𝑥)2|||𝑑𝜑|||𝑑𝑥2𝑑𝑥+𝛼0||𝜑||2𝑑𝑥+𝛾𝛽||𝜑||2𝑑𝑥+1𝛿||𝜑||2𝑑𝑥1/2,(3.22) hence, 𝜑=0.


The authors would like to thank the referee for helpful suggestions and comments.