SOLVABILITY OF NONLINEAR DIRICHLET PROBLEM FOR A CLASS OF DEGENERATE ELLIPTIC EQUATIONS

We prove an existence result for solution to a class of nonlinear degenerate elliptic equation associated with a class of partial 
differential operators of the form L u ( x ) = ∑ i , j = 1 n D j ( a i j ( x ) D i u ( x ) ) , with D j = ∂ / ∂ x j , where a i j : Ω → ℝ are functionssatisfying suitable hypotheses.


Introduction
In this paper, we prove the existence of solution in D(A) ⊆ H 0 (Ω) for the following nonlinear Dirichlet problem: −Lu(x) + g u(x) ω(x) = f 0 (x) − n j=1 D j f j (x) on Ω, u(x) = 0 on ∂Ω, (1.1) where L is an elliptic operator in divergence form D j a i j (x)D i u(x) , with D j = ∂ ∂x j (1.2) and the coefficients a i j are measurable, real-valued functions whose coefficient matrix (a i j (x)) is symmetric and satisfies the degenerate ellipticity condition for all ξ ∈ R n and almost every x ∈ Ω ⊂ R n a bounded open set with piecewise smooth boundary (i.e., ∂Ω ∈ C 0,1 ), and ω and v two weight functions (i.e., locally integrable nonnegative functions).
Theorem 1.1.Suppose that the following assumptions are satisfied.
(H1) Dual pairs.Let the dual pairs {X, X + } and {Y ,Y + } be given, where X, X + , Y , and Y + are Banach spaces with corresponding bilinear forms •, • X and •, • Y and the continuous embeddings Y ⊆ X and The dual pairs are compatible, that is, Moreover, the Banach spaces X and Y are separable and X is reflexive.(H2) Operator A. Let the operator A : D(A) ⊆ X → Y + be given, and let K be a bounded closed convex set in X containing the zero point as an interior point and K∩Y ⊆ D(A).
(H3) Local coerciveness.There exists a number α ≥ 0 such that Av,v Y ≥ α for all v ∈ Y ∩ ∂K, where ∂K denotes the boundary of K in the Banach space X.
(H4) Continuity.For each finite-dimensional subspace Y 0 of the Banach space Y , the mapping u → Au,v Y is continuous on K∩Y 0 for all v ∈ Y 0 .
(H5) Generalized condition (M).Let {u n } be a sequence in Y ∩K and let T ∈ X + .Then, from it follows that Au = T. (H6) Quasiboundedness.Let {u n } be a sequence in Y ∩K .Then, from (1.6) and Au n , u n Y ≤ C u X for all n, it follows that the sequence We will apply this theorem to a sufficiently large ball K in the Banach spaces X = H 0 (Ω), X + = (H 0 (Ω)) * , and Y + = Y * .
We make the following basic assumption on the weights ω and v.

The weighted Sobolev inequality (WSI).
Let Ω be an open bounded set in R n .There is an index q = 2σ, σ > 1, such that for every ball B and every f ∈ Lip 0 (B) (i.e, f ∈ Lip(B) whose support is contained in the interior of B), where C B,ω,v is called the Sobolev constant and (1.10) For instance, the WSI holds if ω and v are as in [6, Chapter X, Theorem 4.8], or if ω and v are as in [1,Theorem 1.5].
The following theorem will be proved in Section 3.
Theorem 1.2.Let L be the operator (1.2) and satisfy (1.3).Suppose that the following assumptions are satisfied: where

Definitions and basic results
Let ω be a locally integrable nonnegative function in R n and assume that 0 < ω < ∞ almost everywhere.We say that ω belongs to the Muckenhoupt class 208 Solvability for a class of nonlinear Dirichlet problem for all balls B ⊂ R n , where | • | denotes the n-dimensional Lebesgue measure in R n .If 1 < q ≤ p, then A q ⊂ A p (see [4,5] for more information about A p -weights).The weight ω satisfies the doubling condition if ω(2B) ≤ Cω(B), for all balls B ⊂ R n , where ω(B) = B ω(x)dx and 2B denotes the ball with the same center as B which is twice as large.If ω ∈ A p , then ω is doubling (see [5,Corollary 15.7]).We say that the pair of weights (v,ω) satisfies the condition A p (1 < p < ∞ and (v,ω) ∈ A p ) if and only if there is a constant C 2 such that 1 for every ball B⊂R n .
Given a measurable subset Ω of R n , we will denote by L p (Ω,ω), 1 ≤ p < ∞, the Banach space of all measurable functions f defined on Ω for which We will denote by W k,p (Ω,ω), the weighted Sobolev spaces, the set of all functions u ∈ L p (Ω,ω) such that the weak derivatives D α u ∈ L p (Ω,ω), 1 ≤ |α| ≤ k.The norm in the space W k,p (Ω,ω) is defined by When k = 1 and p = 2, the spaces W 1,2 (Ω,ω) and W 1,2 0 (Ω,ω) are Hilbert spaces.We will denote by H 0 (Ω) the closure of C ∞ 0 ( Ω) with respect to the norm where Ꮽ(x) = [a i j (x)] (the coefficient matrix) and the symbol ∇ indicates the gradient.
We introduce the following definition of (weak) solutions for problem (1.1).

Definition 2.4. A function u
Remark 2.5.Using that p > 4, we have that v ∈ A 2 ⊂ A p/2 and
Step 5. Global coerciveness of operator A. Using the condition (1.3) and hypothesis (ii), we obtain Step 6. Generalized condition (M).Let T ∈ (H 0 (Ω)) * and let {u n } be a sequence in Y with We want to show that this implies that Au = T. Using that the operator A 1 is linear and continuous, we obtain Because of (3.16), it is sufficient to prove that u ∈ D(A) and Therefore, it is sufficient to show that Using the same argument in Step 3, we obtain Therefore, it is sufficient to show that

Albo Carlos Cavalheiro 213
The continuity of g implies that g(u n (x))u n (x)ω(x) → g(u(x))u(x)ω(x) for almost all x ∈ Ω.Therefore, by Fatou lemma, we have that is, u ∈ D(A).Now we want to show that g(u n (x)) → g(u(x)) in L 1 (Ω,ω).Let a > 0 be fixed.For each x ∈ Ω, we have either Hence, for all ε > 0, we have if a is sufficiently large and ω(X) is sufficiently small.Therefore, for all ε > 0, there exists with ω(X) < δ.Thus, the Vitali convergence theorem tells us that (3.22) holds.
Step 7. Quasiboundedness of the operator A. Let {u n } be a sequence in Y with u n u in H 0 (Ω) and suppose that for all u,v ∈ D(A).Therefore, if u,v ∈ D(A) and Au = Av = T, we obtain that u = v.
Au n ,u n Y ≤ C u n H0(Ω) , ∀n.(3.33)We want to show that the sequence {Au n } is bounded in Y * .In fact, the boundedness of {u n } in H 0 (Ω) implies that lim n→∞ Au n ,u n Y ≤ C. (3.34) Suppose by contradiction that the sequence {Au n } is unbounded in Y * .Then there exists a subsequence, again denoted by {u n }, such that Au n Y * −→ ∞ as n −→ ∞.(3.35)By the same arguments as in Step 6, we obtain that Au n ,ϕ Y −→ Au,ϕ Y as n −→ ∞, ∀ϕ ∈ Y. (3.36)The uniform boundedness principle tells us that the sequence {Au n } is bounded (which is a contradiction with (3.35)).Therefore, by Theorem 1.1, the equation Au = T, with T ∈ (H 0 (Ω)) * , has a solution u ∈ D(A) ⊆ H 0 (Ω), and it is the solution for problem (1.1).(II) Uniqueness.If the function g : R → R is monotone increasing, we have that (g(a) − g(b))(a − b) ≥ 0, for all a,b ∈ R. Then Au − Av,u − v Y = Ω Ꮽ∇(u − v),∇(u − v) dx that is, the equation Au = T has a solution u for each T ∈ X + .
.22)Note that it is sufficient to prove (3.22) for a subsequence of {u n }.If (v,ω) ∈ A 2 and ω ≤ v, then ω ∈ A 2 (see Remark 2.1).By Lemma 2.3, This implies u n →u in L 2 (Ω,ω).Using again that ω ∈ A 2 , we have u n → u in L 1 (Ω).Thus, there exists a subsequence, again denoted by {u n }, such that u n (x) → u(x) for almost all x ∈ Ω.The continuity of g implies that g(u n (x)) → g(u(x)) for almost all x ∈ Ω.Moreover, since u n u in H 0 (Ω), it follows that with C independent of n.