FOURIER PROBLEM FOR WEAKLY NONLINEAR EVOLUTION INCLUSIONS WITH FUNCTIONALS

The Fourier problem or, in other words, the problem without initial conditions for evolution equations and inclusions arise in modeling different nonstationary processes in nature, that started a long time ago and initial conditions do not affect on them in the actual time moment. Thus, we can assume that the initial time is −∞, while 0 is the final time, and initial conditions can be replaced with the behaviour of the solution as time variable turns to −∞. The Fourier problem for evolution variational inequalities (inclusions) with functionals is considered in this paper. The conditions for existence and uniqueness of weak solutions of the problem are set. Also the estimates of weak solutions are obtained.


Introduction
In this paper we consider problem without initial conditions, or, in other words, the Fourier problem for evolution variational inequalities (inclusions) with functionals. Let us introduce an example of the problem being studied here.
Then problem (1.1),(1.2) can be rewritten as following: find a function u ∈ L 2 loc (S; V ) such that u ∈ L 2 loc (S; H), condition (1.2) holds, and, for a.e. t ∈ S, u(t) ∈ K and (u (t) + Au(t) + B(t, u(t)), v − u(t)) ≥ (f (t), v − u(t)) ∀ v ∈ K. (1. 3) Here f ∈ L 2 loc (S; H) is a given function. We remark that variational inequality (1.3) can be written as a subdifferential inclusion. For this purpose we put I K (v) := 0 if v ∈ K, and I K (v) := +∞ if v ∈ V \ K, and also It is easy to verify that the functional Φ : V → R ∪ {+∞} is convex and semilower-continuous. By the known results (see, e.g., [22, p. 83]) it follows that the problem of finding a solution of variational inequality (1.3) can be written as such subdifferential inclusion: to find a function u ∈ L 2 loc (S; V ) such that u ∈ L 2 loc (S; H), condition (1.2) holds and, for a.e. t ∈ S, u(t) ∈ D(∂Φ) and The aim of this paper is to investigate problems for inclusions of type (1.4). Problem without initial conditions or, in other words, the Fourier problem for evolution equations and inclusions arise in modeling different nonstationary processes in nature, that started a long time ago and initial conditions do not affect on them in the actual time moment. Thus, we can assume that the initial time is −∞, while 0 is the final time, and initial conditions can be replaced with the behaviour of the solution as time variable turns to −∞. Such problem appear in modeling in many fields of science such as ecology, economics, physics, cybernetics, etc. The research of the problem without initial conditions for the evolution equations and variational inequalities were conducted in the monographs [16,18,22], the papers [3, 6-8, 13, 15, 17, 19, 21], and others. In particular, R.E. Showalter in the paper [21] proved the existence of a unique solution u ∈ e 2ω· H 1 (S; H), where H is a Hilbert space, of the problem without initial condition for ω + µ > 0 and f ∈ e 2ω· H 1 (S; H), in case when A : H → 2 H is maximal monotone operator such that 0 ∈ A(0). Moreover, if A = ∂ϕ, where ϕ : H → R∪{+∞} is proper, convex and lower-semi-continuous functional such that ϕ(0) = 0 = inf {ϕ(v) : v ∈ H}, then this problem is uniquely solvable for each µ > 0, f ∈ L 2 (S; H) and ω = 0.
As is well known the uniqueness of the solutions of problem without initial conditions for linear parabolic equations and variational inequalities is possible only under some restrictions on the behavior of solutions as time variable terns to −∞. For the first time it was strictly justified by A.N. Tikhonov [23] in the case of heat equation. However, as it was shown by M.M. Bokalo [3], problem without initial conditions for some nonlinear parabolic equations has a unique solution in the class of functions without behavior restriction as time variable terns to −∞. Similar result for evolutionary variational inequalities were also obtained in the paper [4].
Note that in inclusion (1.4) the unknown function can enter both in the differential part and in functional part. Previously, the Fourier problem for evolution integro-differential equations were studied in [5,9,10] (see also references therein). Let us note that problems without initial conditions for variational inequalities or inclusions with functionals have not been considered in the literature, and this serves as one of the motivations for the study of such problems.
The outline of this paper is as follows. In Section 2, we give notation, definitions of needed function spaces and auxiliary results. In Section 3, we formulate the problem and main result. We prove the main result in Section 4.

Preliminaries
Set S := (−∞, 0]. Let V and H be separable Hilbert spaces with the scalar products (·, ·) V , (·, ·) and norms · , | · |, respectively. Suppose that V ⊂ H with dense, continuous and compact injection, i.e., the closure of V in H coincides with H, and there exists a constant λ > 0 such that and for every sequence Let V and H be the dual spaces to V and H, respectively. We suppose (after appropriate identification of functionals), that the space H is a subspace of V . Identifying the spaces H and H by the Riesz-Fréchet representation theorem, we obtain dense and continuous embeddings Note that in this case g, v V = (g, v) for every v ∈ V, g ∈ H, where ·, · V is the scalar product for the duality [V , V ]. Therefore, further we can use the notation (·, ·) instead of ·, · V .
We introduce some spaces of functions and distributions. Let X be an arbitrary Hilbert space with the scalar product (·, ·) X and the norm · X . By C(S; X) we mean the linear space of continuous functions defined on S with values in X. We say that w m −→ m→∞ w in C(S; X) if for each t 1 , t 2 ∈ S, t 1 < t 2 , we have max Denote by L 2 loc (S; X) the linear space of measurable functions defined on S with values in X, whose restrictions to any segment [t 1 , t 2 ] ⊂ S belong to the space L 2 (t 1 , t 2 ; X). We say that a sequence {w m } is bounded (respectively, strongly, weakly or * -weakly convergent to w) in L 2 loc (S; X), if for each t 1 , t 2 ∈ S, t 1 < t 2 , the sequence of restrictions of {w m } on the segment [t 1 , t 2 ] is bounded (respectively, strongly, weakly or * -weakly convergent to the restriction of w on this segment) in L 2 (t 1 , t 2 ; X).
Let ν ∈ R. Put by definition This space is a Hilbert space with the scalar product and the corresponding norm Also we introduce the space By D (−∞, 0; V ) we mean the space of continuous linear functionals on D(−∞, 0) with values in V w (hereafter D(−∞, 0) is space of test functions, that is, the space of infinitely differentiable on (−∞, 0) functions with compact supports, equipped with the corresponding topology, and V w is the linear space V equipped with weak topology). It is easy to see (using (2.2)), that spaces L 2 loc (S; V ), In this paper we use the following well-known facts.
Lemma 2.1 (Cauchy-Schwarz inequality [14, p. 158]). Suppose that t 1 , t 2 ∈ R, t 1 < t 2 , and X is a Hilbert space with the scalar product (·, ·) X . Then, for v, w ∈ L 2 (t 1 , t 2 ; X), we have (w(·), v(·)) X ∈ L 1 t 1 , t 2 and Note that, we understand embedding (2.5) as follows: if a sequence {w m } is bounded in the space L q (t 1 , t 2 ; W) and the sequence {w m } is bounded in the space and strongly in L q (t 1 , t 2 ; L).
. Now, according to the diagonal process we select the desired subsequence as {w m j,j } ∞ j=1 , and we define the function w as follows: for each k ∈ N we take w(t) := w k (t) for t ∈ (−k, −k + 1].

Statement of the problem and main result
which satisfies the conditions: i.e., the functional Φ is convex, i.e., the functional Φ is lower semicontinuous.
Recall that the subdifferential of functional Φ is a mapping ∂Φ : V → 2 V , defined as follows We identify the subdifferential ∂Φ with its graph, assuming that [20,Theorem A] proves that the subdifferential ∂Φ is a maximal monotone operator, that is, and for every element Let, for each t ∈ S, B(t, ·) : H → H be an operator which satisfies the condition: (B) for any v ∈ H the mapping B(·, v) : S → S is measurable, and there exists a constant L ≥ 0 such that following inequality holds for a.e. t ∈ S, and for all v 1 , v 2 ∈ H; in addition, B(t, 0) = 0 for a.e. t ∈ S.
Remark 3.1. From the condition (B) it follows that for a.e. t ∈ S, and for every v ∈ H the following estimate is valid: Let us consider the evolutionary variational inequality where f : S → V is a given measurable function and u : S → V is an unknown function.
The solution of variational inequality (3.3) is a function u : S → V that satisfies the following conditions: 3) there exists a function g ∈ L 2 loc (S; V ) such that, for a.e. t ∈ S, g(t) ∈ ∂Φ u(t) and For variational inequality (3.3) consider the problem: find its solution which satisfies the condition lim where γ ∈ R is given. The problem of finding a solution of variational inequality (3.3) (for given Φ,B,f ) satisfying the condition (3.4) for given γ, is called the Fourier problem or, in other words, the problem without initial conditions for the evolution variational inequality (3.3). This problem, in short, be called the problem P (Φ, B, f, γ), and the function u is called its solution.
Additionally, assume that the following conditions hold: Now we shall formulate the main result.

7)
where C 1 is a positive constant depending on K 1 , K 2 , L and γ only.
Remark 3.3. The problem P(Φ, B, f, γ) can be replaced by the following problem. Let K be a convex and closed set in V , A : V → V be a monotone, bounded and semi-continuous operator such that The problem is to find a function u ∈ W 2,loc (S; V ) satisfying the condition (3.4) and, for a.e. t ∈ S, u(t) ∈ K and
Proof of the Theorem 3.2. We divide the proof into five steps.
Step 1 (auxiliary statements). The following statements will be used in the sequel.
From this, choosing α > 0 such that inequality C 2 /(2α) < 1 holds, we obtain that operator A is a contraction. Hence, we may apply the Banach fixed-point theorem [12,Theorem 5.7] and deduce that there exists a unique function w ∈ M such that Aw = w, i.e., we have proved Lemma 4.3.
Step 2 (solution approximation). We construct a sequence of functions which, in some sense, approximate the solution of the problem P (Φ, B, f, γ).
For each k ∈ N, let f k (t) := f (t) for t ∈ S k := (−k, 0] and let us consider the problem of finding a function u k ∈ C(S k ; H) ∩ H 1 (S k ; H), where H 1 (S k ; H) := w ∈ L 2 (S k ; H) w ∈ L 2 (S k ; H) , such that, for a.e. t ∈ S k , we have u k (t) ∈ D(∂Φ H ) and Inclusion (4.23) means that there exists a function g k ∈ L 2 (S k ; H) such that, for a.e. t ∈ S k , we have g k (t) ∈ ∂Φ H ( u k (t)) and According to the definition of the subdifferential of a functional and the fact that g k (t) ∈ ∂Φ( u k (t)) for a.e. t ∈ S k , we have Φ(0) ≥ Φ( u k (t)) + ( g k (t), 0 − u k (t)) for a.e. t ∈ S k .
Using this and condition (A 4 ) we obtain ( g k (t), u k (t)) ≥ Φ( u k (t)) ≥ K 2 u k (t) 2 for a.e. t ∈ S k . (4.26) Since the left side of this chain of inequalities belongs to L 1 (S k ), then u k belongs to L 2 (S k ; V ). For each k ∈ N we extend functions f k , u k and g k by zero for the entire interval S, and denote these extensions by f k , u k and g k respectively. From the above it follows that, for each k ∈ N, the function u k belongs to L 2 (S; V ), its derivative u k belongs to L 2 (S; H) and, for a.e. t ∈ S, the inclusion g k (t) ∈ ∂Φ H u k (t) and the following equality (see (4.25)) hold In order to show the convergence {u k } ∞ k=1 to the solution of the problem P(Φ, B, f, γ) we need some estimates of the functions u k , k ∈ N.
Let σ 1 , σ 2 ∈ S be arbitrary numbers such that σ 1 < σ 2 , and k ∈ N. Multiplying identity (4.27), for a.e. t ∈ S, by e 2γt u k (t) and integrating from σ 1 to σ 2 , we obtainˆσ From this taking into account (2.3) and using the integration-by-parts formula, we obtain (4.28) Acccording to the definition of u k and (4.26), we obtain Let us estimate the third term on the left-hand side of inequality (4.28). From (3.5) and (4.29) for arbitrary δ ∈ (0, 1) we obtain Now, let us estimate the last item on the left-hand side of inequality (4.28). Using the Cauchy-Shwarz inequality, (3.2) we have ˆσ Using the Cauchy inequality we estimate the right-hand side of (4.28) as followŝ where ε > 0 is arbitrary.
Step 5 (completion of proof ). In order to complete the proof of the theorem it remains only to show that u(t) ∈ D(∂Φ) and g(t) ∈ ∂Φ u(t) for a.e. t ∈ S.
Let k ∈ N be an arbitrary number. Since u k (t) ∈ D(∂Φ H ) and g k (t) ∈ ∂Φ H u k (t) for every t ∈ S \ S k , where S k ⊂ S is a set of measure zero, applying the monotonicity of the subdifferential ∂Φ H , we obtain that for every t ∈ S \ S k the following equality holds (4.58) Let σ ∈ S, h > 0 be arbitrary numbers. We integrate (4.58) on (σ − h; σ): Now according to (4.52) and (4.53) we pass to the limit in (4.59) as k → ∞. As a result we obtain     This inequality and condition (F) imply that u satisfies condition (3.4). Thus Theorem 3.2 is proved.