Existence of weak solutions for a nonlinear parabolic equations by Topological degree

We prove the existence of a weak solution for the nonlinear parabolic initial boundary value problem associated to the equation ut − div a(x, t, u,∇u) = f(x, t), by using the Topological degree theory for operators of the form L+ S, where L is a linear densely de ned maximal monotone map and S is a bounded demicontinuous map of class (S+) with respect to the domain of L. We will therefore use the Topological degree theory to study a parabolic equation in the space Lp(0, T ;W 1,p 0 (Ω)) where the exponent p is not necessarily equal to 2 (p ≥ 2).

to its dual H * = L q (0, T ; W −1,q (Ω)) where 1 p + 1 q = 1. The right-hand side f is assumed to belong to L q (Q). Lions prove in [10] that there exists at least a solution u ∈ H of the following nonlinear parabolic Cauchy-Dirichlet problem where −div a(x, t, u, ∇u) is a pseudomonotone, coercive, uniformly elliptic operator acting from H to H * and f is a measurable function on Q that belongs to H * . Afterwards, Boccardo et al. in [6] concerned with certain results of existence and regularity for solutions of (2), according to the summability of the data f . Asfaw in [1] proved the existence of at least one weak solution for problem (1) in L 2 (0, T ; W 1,2 0 (Ω)) using the topological degree theory.
In this paper, we will combine and generalize this works: using the topological degree and proving the existence of at lest one weak solution in the space H (p is not necessarily equal 2 ).
The theory of topological degree has been widely used in the study of nonlinear dierential equations as a very eective tool, often those of elliptical type. For more details about the history of this theory and its use, the reader can refer, for example, to [1,2,3,4,5,8,9,11] and references therein.
The rest of this paper is organized as follows. In Section 2, we state some mathematical preliminaries about the functional framework where we will treat our problem. In Section 3, we introduce some classes of operators and then the associated topological degree. We will prove the main results in Section 4.

Preliminaries
Let Ω ⊆ R N be a bounded open set with smooth boundary. Let p ≥ 2 and q = p p−1 . We denote by L p (Ω) the Banach space of all measurable functions u dened in Ω for which Let W 1,p 0 (Ω) be the closure of C ∞ 0 (Ω) in the Sobolev space equipped with the norm Due to the Poincaré inequality, the norm · 1,p on W 1,p 0 (Ω) is equivalent to the norm · W 1,p 0 (Ω) given by Note that the Sobolev space W 1,p 0 (Ω) is a uniformly convex Banach space and the embedding W 1,p 0 (Ω) → → L p (Ω) is compact (see [12].) Next, we consider the following functional space which is a separable and reexive Banach space endowed with the norm or, by Poincaré inequality, the equivalent norm

Classes of mappings and Topological degree
Let X be a real separable reexive Banach space with dual X * and with continuous pairing . , . and let Ω be a nonempty subset of X. The symbol → ( ) stands for strong (weak) convergence.
We consider a multi-values mapping T from X to 2 X * (i.e., with values subsets of X * ). With each such map, we associate its graph The multi-values mapping T is said to be monotone if for any pair of elements (u 1 , T is said to be maximal monotone if it is monotone and maximal in the sense of graph inclusion among monotone multi-values mappings from X to 2 X * . An equivalent version of the last clause is that for any We recall that a mapping T : Let L be a linear maximal monotone map from D(L) ⊂ X to X * such that D(L) is dense in X. For each open and bounded subset G on X, we consider the following classes of operators: Note that the class H G (class of admissible homotopies) includes all ane homotopies L + (1 − t)S 1 + tS 2 with (L + S i ) ∈ F G , i = 1, 2.
We introduce the topological degree for the class F G due to Berkovits and Mustonen [4].

(Normalization)
L + J is a normalising map, where J is the duality mapping of X into X * , that is, d(L + J, G, h) = 1, whenever h ∈ (L + J)(G ∩ D(L)). Lemma 3.1. Let L + S ∈ F X and h ∈ X * . Assume that there exists R > 0 such that for all u ∈ ∂B R (0) ∩ D(L). Then (L + S)(D(L)) = X * .

Proof.
• Let's show that the operator S is bounded. By using the Hölder's inequality, we have for all u, v ∈ H Thanks to the growth condition (4) we can easily prove that a(x, t, u, ∇u) L q (Ω) is bounded for all u ∈ W 1,p 0 (Ω). Therefore By the continuous embedding H → L 1 (0, T ; W 1,p 0 (Ω)), we concludes that which means that the operator S is bounded.
• To show that S is continuous, let u n → u in H. Then u n → u and ∇u n → ∇u in L p (Q). Hence there exist a subsequence (u k ) of (u n ) and measurable functions α in L p (Q) and β in (L p (Q)) N such that u k → u and ∇u k → ∇u, for a.e. (x, t) ∈ Q and all k ∈ N. Since a satises the Carathéodory condition, we obtain that a(x, t, u k (x, t), ∇u k (x, t)) → a(x, t, u(x, t), ∇u(x, t)) a.e. (x, t) ∈ Q.
It follows from (4) that for a.e. (x, t) ∈ Q and for all k ∈ N. Since the dominated convergence theorem imply that a(x, t, u, ∇u k ) → a(x, t, u, ∇u) in L q (Q).
Thus the entire sequence a(x, t, u n , ∇u n ) converges to a(x, t, u, ∇u) in L q (Q). Then, ∀v ∈ H; Su n , v → Su, v , which implies that the operator S is continuous.
• It remains to show that the operator S is of class (S + ). Let (u n ) n be a sequence in D(S) such that We will prove that u n −→ u in H. (a(x, t, u n , ∇u n ) − a(x, t, u, ∇u)).(∇u n − ∇u) dx dt ≤ 0.
From the Theorem 5.1 of Lions [10] we know that H embeds compactly in L p (Q). So, there is a subsequence still denoted by (u n ) such that u n → u in L p (Q) and a.e in Q.
But not that the hypothesis (5) implies that ξ → a(x, t, η, ξ) is strictly increasing, and by hypothesis (6) and a result of Browder [7], we can get that and so conclude that S is of type (S + ).
Our main result is the following existence theorem: Theorem 4.1. Let f ∈ H * and u 0 ∈ L 2 (Ω). There exists at least one weak solution u ∈ D(L) of problem (1) in the following sense The operator L is generated by ∂/∂t via the relation , v(t) dt, for all u ∈ D(L), v ∈ H.
One can verify, as in [12] that L is a densely dened maximal monotone operator. By the monotonicity of L ( Lu, u ≥ 0 for all u ∈ D(L)) and the condition (6), we get Since the right side of the above inequality approaches ∞ as u H → ∞, then for each h ∈ H * there exists R = R(f ) such that Lu + Su − f, u > 0 for all u ∈ B R (0) ∩ D(L). By applying Lemma 3.1, we conclude that the equation Lu + Su = f is solvable in D(L); that is, (1) admits at least one-weak solution.