Well posedness of a nonlinear mixed problem for a parabolic equation with integral condition

The aim of this work is to prove the well posedness of some posed linear and nonlinear mixed problems with integral conditions. First, an a priori estimate is established for the associated linear problem and the density of the operator range generated by the considered problem is proved by using the functional analysis method. Subsequently, by applying an iterative process based on the obtained results for the linear problem, the existence, uniqueness of the weak solution of the nonlinear problems is established.


Introduction and statement of the problem
Some problems related to physical and technical issues can be effectively described in terms of nonlocal problems with integral conditions in partial differential equations. These nonlocal conditions arise mainly when the values on the boundary cannot be measured directly, while their average values are known. The problem of parabolic equation with integral condition is stated as follows: Let us consider the rectangular domain Q = ]0, 1[ × ]0, T[, then the problem is to find a solution σ (x, t) of the following nonclassical boundary value problem: with the initial condition lσ = σ (x, 0) = ϕ(x), for x ∈ [0, 1], (1.2) and the Dirichlet boundary condition σ (0, t) = 0, for t ∈ [0, T], (1.3) and the nonlocal condition α 0 σ (x, t) dx + 1 β σ (x, t) dx = 0, 0 ≤ α ≤ β < 1 ∀t ∈ [0, T]. (1.4) In addition, we assume that the function a(x, t) and its derivatives satisfy the conditions (1.5) where the functions g(x, t, σ , ∂σ ∂x ), ϕ(x) are given, and we assume that the following matching conditions are satisfied: We also assume that there exists a positive constant d such that g x, t, σ 1 , for all (x, t) ∈ Q. This type of problem can be found in various physic problems such as heat conduction [1][2][3][4], plasma physics [5], thermoelasticity [6], electrochemistry [7], chemical diffusion [8] and underground water flow [9][10][11]. Several research papers such as found in [1-4, 7, 12-18] have studied and solved the parabolic equation by combining the integral condition with Dirichlet condition or Newmann condition, or with purely integral conditions, using various methods. For hyperbolic equations, the unicity and existence of the solution have been studies in [13,[19][20][21][22] and the mixed-type equations in [23][24][25][26][27]. The elliptic equations were considered in [28,29] and [30].
The linear problem associated to the problem stated in (1.1)-(1.4), for α = β = 0, has been studied in [18] and for β = 1 in [16]. Meanwhile in [31] the solved problem is for the case α + β = 1. It is worth mentioning that in [32] the author studied the same case where ∂ ∂x (a ∂σ ∂x ) was replaced by the Bessel operator. In the present paper the motivation is to study and find a solution to the stated problem without imposing any conditions on the constants α and β in the interval [0, 1]. In addition, the nonlinear problem of the parabolic equation with integral condition defined on two parts of the boundary is solved.
First, an a priori estimate is established for the associated linear problem and the density of the operator range generated by the considered problem is proved by using the functional analysis method. Subsequently, by applying an iterative process based on the obtained results for the linear problem, the existence and uniqueness of the weak solution of the nonlinear problems is established.
The rest of the paper is organized as follows. In Sect. 2, the associated linear problem is stated. Section 3 deals with the proof of the uniqueness of the solution using an a priori estimate. Section 4 gives the solvability of the considered linear problem. Finally, in Sect. 5, on the basis of the obtained results in Sects. 3 and 4, and on the use of an iterative process, we prove the existence and uniqueness of the solution of the nonlinear problem.

Statement of the associated linear problem
In this section we introduce the linear problem and the different function spaces needed to investigate the mixed nonlocal problem given by the equation. The operator L is an operator acting on E into F, where E is the Banach space of functions u ∈ L 2 (Q), with a finite norm Then we show that the operator L has a closure L and later on, in Sect. 3, we establish an energy inequality of the following type (see Theorem 3.1): Since the points of the graph of the operator L are limits of sequences of points of the graph of L, we can extend the a priori estimate (2.4) to be applied to strong solutions by taking the limits, that is, we have the inequality (2.5) From this inequality, we deduce the uniqueness of a strong solution, if it exists, and that the range of the operator L coincides with the closure of the range of L.
By virtue of the uniqueness of the limit in D (Q), we conclude that f = 0. According to (2.7), we also conclude that The following a priori estimate gives the uniqueness of the solution of the posed linear problem.

An energy inequality and its application
In this section, the uniqueness of the solution will be proved using an energy inequality method.
Theorem 3.1 There exists a positive constant K , such that for each function u ∈ D(L) we have and λ, k and δ are a positives scalar parameters such that Substituting Mu by its expression in the first term in the right-hand side of (3.3), we obtain Integrating by parts the second term in the right-hand side of the last equality of (3.4) with respect to x, using the fact that ∂u ∂t = 1 k xe δ(x-1) ∂g ∂x , then integrating by parts with respect to x, we obtain using this equality The last term in the previous equality becomes Similarly integrating by parts the last term of (3.4) with respect to x, we obtain From (3.7) and (3.6), equality (3.4) becomes Similarly, substituting Mu by its expression in the last term in the right-hand side of (3.3), integrating by parts with respect to x, using the Dirichlet condition (1.3) and the integral Integrating by parts the first two terms with respect to t in (3.9), using the condition (1.2) we have then from the above equalities and equalities (3.8) and (3.9), (3.3) becomes Using the Young inequality in the last four terms in the left-hand side of (3.10), and using the facts that We choose ε 1 = 8, ε 2 = 2, ε 3 = δ k , and ε 4 = 2 and c > 0 such that therefore by combining the previous inequalities with (3.10), we get the following expression: where Substituting Mu by its expression in the first term in the right-hand side of (3.12), we obtain each term in the right-hand side of (3.13), can be, respectively, controlled by and Re Q s The combination of the previous inequalities with (3.12) yields where  This last inequality implies the following corollaries.
then from Theorem 3.1" we deduce that u E ≤ 0, which implies that u 1 = u 2 .

Corollary 3.2 The range R(L) of L is closed in F and R(L) = R(L).
Proof First, we prove that R(L) is closed. Let T ∈ R(L), then there exists a sequence , we deduce that the convergence of LU n in F implies the convergence of U n in E, say U n − → n→∞ U, in E. Since L is closed, (U n ) is a sequence in D(L) and

Solvability of the linear problem
In order to prove the solvability of problem (2.1)-(1.4) it is sufficient to show that R(L) is dense in F. The proof is based on the following lemma. where

then w vanishes almost everywhere in .
Proof Equality (4.1), can be written as follows: We introduce the smoothing operators J -1 ε = (Iε ∂ ∂t ) -1 and (J -1 ε ) * = (I + ε ∂ ∂t ) -1 from L 2 (0, T) into the space H 1 (0, T) with respect to t, then these operators provide the solution of the problems: We also have the following properties: If g ∈ D(L), then J -1 g ∈ D(L) and we have ⎧ ⎨ ⎩ lim J -1 gg L 2 (0,T) = 0, for ε → 0, Substituting the function u in (4.2) by the smoothing function u ε and using the relation Since the operator A(t) has a continuous inverse in L 2 (0, 1) defined by where the functions C 1 (t) satisfy the following expression: , the function K(x) is given by x -1, (β, 1).
Then we have α 0 A -1 (t)u dx + 1 β A -1 (t)u dx = 0, hence, the function J -1 u = u ε can be represented in the form Consequently, equality (4.3) becomes u ∂ρ * ∂t dx dt = A(t)uh dx dt, (4.4) where The left-hand side of (4.4) is a continuous linear functional of u, hence the function h has the derivatives ∂h ∂x , ∂ 2 h ∂x 2 ∈ L 2 (Q) and the following conditions are satisfied: For a sufficiently small and the operator then the function ρ(x) can be expressed as follows: .
Taking u ∈ D(L) in (4.6) yields Since the two terms in the previous equality vanish independently and since the range of the trace operator is everywhere dense in Hilbert space with the norm

Study of the nonlinear problem
This section is devoted to the proof of the existence, uniqueness of the solution of the problem (1.1)-(1.4).
If the solution of problem (1.1)-(1.4) exists, it can be expressed in the form θ = w + U, where U is a solution of the homogeneous problem and w is a solution of the problem We shall prove that the problem (5.5)-(5.8) has a weak solution by using an approximation process and passing to the limit. Assume that v and w ∈ C 1 (Q), and the following conditions are satisfied: Taking the scalar product in L 2 (Q) of Eq. (5.5) and the integrodifferential operator by taking the real part, we obtain Nv dx dt. (5.11) Substituting the expression of Nv in the first integral of the right-hand side of (5.11), integrating by parts with respect to t, using the condition (5.10), we get Substituting the expression of Nv in the second integral of the right hind-side of (5.11), integrating by parts with respect to x, using the condition (5.10), we get Insertion of (5.12), (5.13) into (5.11) yields where obtained by integrating by parts the right-hand side of (5.11) with respect to x. Definition 5.1 By a weak solution of problem (5.5)-(5.8) we mean a function w ∈ L 2 (0, T : V 1,0 (0, 1)) satisfying the identity (5.14) and the integral condition (5.8).