Volume 2011, Issue 1 356356
Research Article
Open Access

L-Solutions for Some Nonlinear Degenerate Elliptic Equations

Albo Carlos Cavalheiro

Corresponding Author

Albo Carlos Cavalheiro

Department of Mathematics, State University of Londrina, 86051-990 Londrina, PR, Brazil uel.br

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First published: 15 December 2011
Citations: 2
Academic Editor: Toka Diagana

Abstract

We are interested in the existence of solutions for Dirichlet problem associated to the degenerate quasilinear elliptic equations f(x),   on  Ω in the setting of the weighted Sobolev spaces .

1. Introduction

In this paper we prove the existence of (weak) solutions in the weighted Sobolev spaces for the homogeneous Dirichlet problem:
( P )
where L is the partial differential operator:
(1.1)
where Ω is a bounded open set in N (N ≥ 2), ω1 and ω2 are two weight functions, and the functions 𝒜 : Ω × × NN, g : Ω × , and H : Ω × × N are Carathéodory functions.

By a weight, we will mean a locally integrable function ω on N such that ω(x) > 0 for a.e. xN. Every weight ωi (i = 1,  2) gives rise to a measure on the measurable subsets on N through integration. This measure will be denoted by μi. Thus, μi(E) = ∫Eωi(x)dx (i = 1,2) for measurable sets EN.

In general, the Sobolev spaces Wk,p(Ω) without weights occur as spaces of solutions for elliptic and parabolic partial differential equations. For degenerate partial differential equations, that is, equations with various types of singularities in the coefficients, it is natural to look for solutions in weighted Sobolev spaces (see [14]).

A class of weights, which is particularly well understood, is the class of Ap-weights (or Muckenhoupt class) that was introduced by Muckenhoupt (see [5]). These classes have found many useful applications in harmonic analysis (see [6, 7]). Another reason for studying Ap-weights is the fact that powers of distance to submanifolds of N often belong to Ap (see [8]). There are, in fact, many interesting examples of weights (see [4] for p-admissible weights).

Equations like (1.1) have been studied by many authors in the nondegenerate case (i.e., with ω(x) ≡ 1) (see, e.g., [9] and the references therein). The degenerate case with different conditions has been studied by many authors. In [2] Drábek et al. proved that under certain condition, the Dirichlet problem associated with the equation −div (a(x, u, ∇u)) = h, has at least one solution , and in [1] the author proved the existence of solution when the nonlinear term H(x, η, ξ) is equal to zero.

Firstly, we prove an L estimate for the bounded solutions of (P): we assume that f/ω1Lq(Ω, ω1), with r2/(r2 − 1) < q < (where r2 > 1 as in Theorem 2.5), and we prove that any that solves (P) satisfies , where C depends only on the data, that is, Ω, N, p, q, α1, α2, C0, C1 and . After that, we prove the existence of solution for problem (P) if f/ω1    Lq(Ω, ω1), with pr2/(r2 − 1) < q < .

Note that, in the proof of our main result, many ideas have been adapted from [911].

The following theorem will be proved in Section 3.

Theorem 1.1. Let ω1 and ω2 be Ap-weights, 1 < p < , with ω1ω2. Suppose the following.

  • (H1)

    x𝒜(x, η, ξ) is measurable in Ω for all (η, ξ) ∈ × N: (η, ξ) ↦ 𝒜(x, η, ξ) is continuous in × N for almost all x ∈ Ω.

  • (H2)

    [𝒜(x, η, ξ) − 𝒜(x, η, ξ)]·(ξξ)   >   0, whenever ξ, ξN, ξξ.

  • (H3)

    𝒜(x, η, ξ) · ξα1 | ξ|p, with 1 < p < , where α1 > 0.

  • (H4)

    |𝒜(x, η, ξ) |     K2(x) + h1(x) | η|p/p + h2(x) | ξ|p/p, where K2, h1, and h2 are positive functions, with h1 and h2L(Ω), and K2  Lp(Ω, ω2) (1/p + 1/p = 1).

  • (H5)

    x  g(x, η) is measurable in Ω for all η     :   η  g(x, η) is continuous in for almost all x    Ω.

  • (H6)

    |g(x, η)| ≤ K1(x) + h3(x) | η|p/p, where K1 and h3 are positive functions, with h3L(Ω) and K1Lp(Ω, ω1).

  • (H7)

    g(x, η)  ηα0 | η|p, for all η, where α0 > 0.

  • (H8)

    xH(x, η, ξ) is measurable in Ω for all (η, ξ) ∈ × N: (η, ξ) ↦ H(x, η, ξ) is continuous in × N for almost all x ∈ Ω.

  • (H9)

    |H(x, η, ξ)| ≤   C0 + C1 | ξ|p, where C0 and C1 are positive constants.

  • (H10)

    f/ω1Lq(Ω, ω1), with r2/(r2 − 1) < q < (where r2 > 1 as in Theorem 2.5) and ω2/ω1Lq(Ω, ω1).

Let be a solution of problem (P). Then there exists a constant C > 0, which depends only on Ω, n, p, α1, α0, C0, C1 and , such that .

The main result of this article is given in the next theorem, which is proved in Section 4.

Theorem 1.2. Assume that (H1)–(H9) hold true and suppose that

  • (H11)

    f/ω1Lq(Ω, ω1), with pr2/(r2 − 1) < q < ;

  • (H12)

    H(x, η, ξ)  η ≥ 0, for all η.

Then there exists at least one solution of the problem P.

Theorem 1.2 will be proved by approximating problem (P) with the following problems:
( P m )
where Hm(x, η, ξ) = H(x, η, ξ)/1 + (1/m)|H(x, η, ε)|, for m. Note that |Hm | ≤|H| and that |Hm | ≤ m.

2. Definitions and Basic Results

Let ω be a locally integrable nonnegative function in N and assume that 0 < ω(x) < almost everywhere. We say that ω belongs to the Muckenhoupt class Ap, 1 < p < , or that ω is an Ap-weight, if there is a constant C = Cp,ω such that
(2.1)
for all balls BN, where |·| denotes the N-dimensional Lebesgue measure in N. If 1 < q    p, then AqAp (see [4, 7, 12] or [13] for more information about Ap-weights). The weight ω satisfies the doubling condition if μ(2B)    Cμ(B), for all balls B    n, where μ(B) = ∫Bω(x)dx and 2B denotes the ball with the same center as B which is twice as large. If ωAp, then ω is doubling (see Corollary  15.7 in [4]).

As an example of Ap-weight, the function ω(x) = |x|α, xN, is in Ap if and only if −N < α < N(p − 1) (see Corollary  4.4, Chapter IX in [7]). If φ    BMO(N), then ω(x) = eαφ(x)A2 for some α > 0 (see [6]).

Definition 2.1. Let ω be a weight, and let Ω  N be open. For 0 < p < , we define Lp(Ω, ω) as the set of measurable functions f on Ω such that

(2.2)

Remark 2.2. If ωAp, 1 < p < , then since ω−1/(p−1) is locally integrable, we have for every open set Ω (see Remark  1.2.4 in [13]). It thus makes sense to talk about weak derivatives of functions in Lp(Ω, ω).

Definition 2.3. Let Ω ⊂ N be open, 1 < p < , and let ω1 and ω2 be Ap-weights, 1 < p < . We define the weighted Sobolev space W1,p(Ω, ω1, ω2) as the set of functions u  Lp(Ω, ω1) with weak derivatives DjuLp(Ω, ω2), for j = 1, …, N. The norm of u in W1,p(Ω, ω1, ω2) is given by

(2.3)
The space is the closure of with respect to the norm . The dual space of is the space
(2.4)

Remark 2.4. (a) If ωAp, 1 < p < , then C(Ω) is dense in W1,p(Ω, ω) = W1,p(Ω, ω, ω) (see Corollary  2.16 in [13]).

(b) If ω1ω2, then

(2.5)

In this paper we use the following four results.

Theorem 2.5 (The Weighted Sobolev Inequality). Let Ω be an open bounded set in N(N  2) and ω2Ap(1 < p < ). There exist constants CΩ and δ positive such that for all and all r2 satisfying 1 ≤ r2N/(N − 1) + δ,

(2.6)
where p* = pr2.

Proof. See Theorem  1.3 in [3].

The following lemma is due to Stampacchia (see [14], Lemma  4.1).

Lemma 2.6. Let α, β, C, and k0 be real positive numbers, where β > 1.

Let φ : ++ be a decreasing function such that

(2.7)
for all l > k    k0. Then φ(k0 + d) = 0, where .

Lemma 2.7. If ωAp, then (|E|/|B|) pCp,ω(μ(E)/μ(B)), whenever B is a ball in N and E is a measurable subset of B.

Proof. See Theorem  15.5 Strong doubling of Ap-weights in [4].

By Lemma 2.7, if μ(E) = 0, then |E | = 0.

Lemma 2.8. Let ω1 and ω2 be Ap-weights, 1 < p < , ω1ω2, and a sequence {un}, satisfies the following:

  • (i)

    unu in and μ2-a.e. in Ω;

  • (ii)

    Ω𝒜(x, un, ∇un) − 𝒜(x, un, ∇u), ∇(unu)〉  ω2  dx → 0 with n  .

Then unu in .

Proof. The proof of this lemma follows the line of Lemma  5 in [10].

Definition 2.9. We say that is a (weak) solution of problem (P) if

(2.8)
for all .

3. Proof of Theorem 1.1

Set λ = (C1/α1) + 1 and define for k > 0 the functions ϕC1() and GkW1,() by
(3.1)
If is a solution of problem (P), define the set A(k) = {x  Ω:|u(x)| > k}. We will use the test functions v(x) = ϕ(Gk(u(x))). Since and Ω is a bounded set in N, then and
(3.2)
where χA(k) is the characteristic function of the set A(k) (by sign (u(x)) we mean the function equal to +1 for u(x) > 0 and to −1 for u(x) < 0).

Since , we have that .

Using the function v in (2.8) we obtain
(3.3)
We have the following estimates.
  • (i)

    By (H3) we obtain

(3.4)
  • (ii)

    By (H7) we obtain

(3.5)
  • (iii)

    Using (H9) we obtain

(3.6)
And we also have
(3.7)
Hence in (3.3) we obtain
(3.8)
Since λ = (C1/α1) + 1, we have for s  0
(3.9)
Hence in (3.8) we obtain
(3.10)
Using (3.9) and k<|u(x)| if x    A(k), we obtain
(3.11)
Let us define the function ψk by ψk(x) = ϕ((|u(x)| − k) +/p). We have that and
(3.12)
We have the following.
  • (a)

    For all s    0, eλs − 1    (eλs/p − 1) p.

  • (b)

    There exists, a constant C2 > 0 (C2 = C2(λ, p)) such that for all s    1

(3.13)
This implies the following.
  • (I1)

    a.e. on Ω.

  • (I2)

    If x ∈ A(k + 1), then

(3.14)
Combining (I1) and (I2) with (3.11) and (3.12) we obtain
(3.15)
Define the function h = |f | + C0ω2. Since f/ω1Lq(Ω, ω1) and ω2/ω1  Lq(Ω, ω1), we have that h/ω1Lq(Ω, ω1). Hence
(3.16)
If xA(k) − A(k + 1), we have k<|u | < k + 1. Hence
(3.17)
and we obtain
(3.18)
By Theorem 2.5, (3.15), and (3.18) we have
(3.19)
Therefore, there exist positive constants C3 and C4 (depending only on Ω, α1, p, λ, C2, and CΩ) such that
(3.20)
Since r2/(r2 − 1) < q, then q < r2 and p < p  q < p*. For 0 < θ < 1 such that 1/p  q = (θ/p) + (1 − θ)/p*, using an interpolation inequality, Young’s inequality (with 0 < γ < ), and Hölder’s inequality with exponents q and q we thus obtain
(3.21)
Hence in (3.20) we obtain
(3.22)
Now, we can choose γ in order to have (1 − θ)  γ1/(1−θ) = C3/2 and k0 such that
(3.23)
We obtain, from (3.22), that for every k    k0 it results
(3.24)
where . Hence for all k    k0 we have
(3.25)
Let us now take l > k    k0. Then we have
(3.26)
Therefore for all l > k    k0 we obtain (by (3.25) and (3.26))
(3.27)
that is, .
Let φ(k) = μ1(A(k)). Since β = p*/p  q > 1, by Lemma 2.6 there exists a constant C7 > 0 such that
(3.28)

Using Lemma 2.7 we have |A(k)| = 0 for all k    C7. Therefore any solution u of problem (P) satisfies the estimate .

4. Proof of Theorem 1.2

Step 1. Let us define for m the approximation

(4.1)
We have that |Hm(x, η, ξ)|≤|H(x, η, ξ)|, |Hm(x, η, ξ)| < m, and Hm(x, η, ξ) satisfies the conditions (H9) and (H12). We consider the approximate problem
( P m )
We say that is (weak) solution of problem (Pm) if
(4.2)
for all . We will prove that there exists at least one solution um of the problem (Pm). For we define
(4.3)
Then is a (weak) solution of problem (Pm) if
(4.4)
Let a(u, v, φ) = B(u, v, φ) + Bm(u, φ).
  • (i)

    Using (H4) we obtain

(4.5)
  
  • (ii)

    Using (H6) and |Hm(x, η, ξ)| ≤ m, we obtain

(4.6)
   Hence,
(4.7)
For each we have that a(u, v.) is linear and continuous. Hence, there exists a linear and continuous operator
(4.8)
such that 〈A(u, v), φ〉 = a(u, v, φ). We set
(4.9)
The operator is semimonotone; that is, by similar arguments as in the proof of Theorem  2 in [11] we have the following.
  • (i)

    A(u, u) − A(u, v), uv〉≥0 for all .

  • (ii)

    For each , the operator v  A(u, v) is hemicontinuous and bounded from to and,

for each the operator u    A(u, v) is hemicontinuous and bounded from to .
  • (iii)

    If unu in and 〈A(un, un) − A(un, u), unu〉→  0, then A(un, u)⇀A(u, v) in as n for all .

  • (iv)

    If , un  u in , and in , then as n.

  • (v)

    The operator is bounded.

Hence the operator is pseudomonotone (see [15]).
  • (vi)

    By (H3), (H7), and (H12) we have

(4.10)
where α = min {α0, α1}. Since p > 1, we have
(4.11)
that is, the operator is coercive. Then, by Theorem  27.B in [15], for each , the equation
(4.12)
has a solution. Therefore, the problem (Pm), has a solution .

Step 2. We will show that umL(Ω) and , where C is independent of m. If is a solution of problem (Pm), we define

(4.13)
We have Diun = Diu if |u(x)|    n. For k > 0, let us define the function ψn(x) = sign (un(x))  max {|un(x)| − k, 0}. We have .

Now consider the function

(4.14)
Since Φ is a Lipschitz function and Φ(0) = 0, then . Moreover, DiΦ(ψn) = Φ(ψn)Diψ and
(4.15)
We also have, for all measurable subset E  Ω,
(4.16)
By applying Vitali’s Convergence Theorem, with ψ = Φ(u), we obtain
(4.17)
Since ω1ω2, we obtain
(4.18)
Hence ψn    ψ  in Lp(Ω, ω1). Since is a solution of problem (Pm) and , we have
(4.19)
Using (H4), (H6), |Hm(x, η, ξ) |     m, (4.17), and (4.18), we obtain in (4.19) as n  
(4.20)
Using φ = ψχA(k) in (4.2) (where A(k) = {x    Ω :    | u(x)| > k}) we obtain
(4.21)
Since
(4.22)
we obtain the following.
  • (i)

    By (H7) we have g(x, η)  η    0 for all η    , and

(4.23)
  • (ii)

    Using (H12) we have Hm(x, η, ξ)  η  0 for all η, and

(4.24)
We have ∇ψ = ∇u in A(k). Using (H3), (i), and (ii) we obtain in (4.21)
(4.25)
By Theorem  2.1.14 in [13] there is a positive constant C such that
(4.26)
Then we obtain
(4.27)
Using (4.27) and Young’s inequality we obtain in (4.25) (for all ε > 0)
(4.28)
where C(ε) = (εp) p/p/p. We can choose ε > 0 so that Cε = α1/2, and there exists a constant C8 such that
(4.29)
Using Sobolev’s inequality (Theorem 2.5) and Hölder’s inequality with exponents q and q we obtain (since q > p (r/(r − 1)) > p)
(4.30)
Let us now take l > k > 0 and observe that A(l) ⊂ A(k). Then, from the previous inequality, it follows that
(4.31)
Hence we obtain
(4.32)
Since (1 − (p/q))(p*/p) > 1, by Lemma 2.6 there exists a constant C10 > 0 such that μ1(A(k)) = 0 for all k    C10, and using Lemma 2.7 we obtain |A(k)| = 0. Therefore if um is a solution of problem (Pm), we have and C10 is independent of m.

Step 3. Since and , then the sequence {um} is relative compact in the strong topology of (by apply the analogous results of [10] and Lemma 2.8). Then, by extracting a subsequence which strongly converges in (i.e., there exists such that in ), we have for all

(4.33)
Therefore is the solution of problem (P).

Example 4.1. Let Ω = {(x, y) ∈ 2 : x2 + y2 < 1}, and consider the weights ω1(x, y) = (x2 + y2) 1/2 and ω2 = (x2 + y2) 1/4 (ω1 and ω2A2), and the functions 𝒜 : Ω × × 22, g : Ω × , and H : Ω × × 2 defined by

(4.34)
where . Let us consider the partial differential operator
(4.35)
and f(x, y) = (x2 + y2) 3/2q  cos (1/(x2 + y2)), with 2r2/(r2 − 1) < q < 6. Therefore, by Theorem 1.2, the problem
( P )
has a solution .

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