Weighted Calderón-Zygmund Estimates for Weak Solutions of Quasi-Linear Degenerate Elliptic Equations

Abstract

This paper studies regularity estimates in Sobolev spaces for weak solutions of a class of degenerate and singular quasi-linear elliptic problems of the form div[A(x,u,∇u)] = div[F] with non-homogeneous Dirichlet boundary conditions over bounded non-smooth domains. The coefficients A could be be singular, degenerate or both in x in the sense that they behave like some weight function μ, which is in the A₂ class of Muckenhoupt weights. Global and interior weighted W^1,p(Ω,ω)-regularity estimates are established for weak solutions of these equations with some other weight function ω. The results obtained are even new for the case μ = 1 because of the dependence on the solution u of A. In case of linear equations, our W^1,p-regularity estimates can be viewed as the Sobolev’s counterpart of the Hölder’s regularity estimates established by B. Fabes, C. E. Kenig, and R. P. Serapioni.

Appendix : Proof of Lemma 4

Proof

We follow the method used in [6, 27] with some modifications fitting to our setting. For each x ∈ C, let us define

$$\phi(\rho) = \frac{\omega(C\cap B_{\rho}(x))}{\omega(B_{\rho}(x))}. $$

By the Lebesgue Differentiation Theorem, for almost every x ∈ C, ϕ is continuous and $\phi (0) = \displaystyle {\lim _{\rho \rightarrow 0^{+}} \phi (\rho )} = 1$. Moreover, by (i), ϕ(r₀) < 𝜖. Therefore, for almost every x ∈ C, there is 0 < ρ_x < r₀ such that

$$ \begin{array}{ll} \omega(C \cap B_{\rho_{x}}(x)) & = \epsilon \omega (B_{\rho_{x}}(x)), \quad \text{and} \\ \omega(C\cap B_{\rho}(x)) & < \epsilon \omega (B_{\rho}(x)), \quad \rho > \rho_{x}. \end{array} $$

(A.1)

Now, observe that the family of balls $\{B_{\rho _{x}}(x)\}_{x \in C}$ covers C. Therefore, by Vitali’s Covering Lemma, there exists a countable {x_k}_k∈ℕ in C such that the balss {B_ρk(x_k)}_k∈ℕ with $\rho _{k} = \rho _{x_{k}}$ are disjoint and

$$C \subset \displaystyle{\cup_{k = 1}^{\infty} B_{5\rho_{k}}(x_{k})}. $$

From this, Eq. A.1 and Lemma 1, we infer that

$$\begin{array}{@{}rcl@{}} \omega(C) &\leq& \omega\left( C\cap \left( \cup_{k = 1}^{\infty} B_{5\rho_{k}}(x_{k}) \right) \right) \leq \sum\limits_{k = 1}^{N} \omega\left( C\cap B_{5\rho_{k}}(x_{k}) \right) \\ && < \epsilon \sum\limits_{k = 1}^{N}\omega (B_{5\rho_{k}}(x_{k})) \leq \epsilon M 5^{np}\sum\limits_{k = 1}^{N} \omega(B_{\rho_{k}}(x_{k})). \end{array} $$

Observe that by (i), and since $B_{\rho _{k}}(x_{k})$ are all disjoint,

$$\left( \cup_{k = 1}^{n} B_{\rho_{k}}(x_{k}) \right) \cap {\Omega}_{R}(y_{0}) \subset D. $$

We claim that

$$ \omega(B_{\rho}(x)) \leq \max M\left\{A^{-1}, 4^{n} \right\}^{q} \omega(B_{\rho}(x) \cap {\Omega}_{R}(y_{0})), \quad \forall x \in {\Omega}_{R}(y_{0}), \ \rho \in (0, r_{0}). $$

(A.2)

From this claim, it follows that

$$\omega(C) \leq \epsilon^{\prime} \sum\limits_{k = 1}^{n} \omega\left( B_{\rho_{k}}(x_{k}) \cap {\Omega}_{R}(y_{0})\right) = \epsilon^{\prime} \omega \left( \left( \cup_{k = 1}^{n} B_{\rho_{k}}(x_{k}) \right) \cap {\Omega}_{R}(y_{0})\right) \leq \epsilon^{\prime} \omega(D). $$

It now remains to prove Eq. A.2. It follows from Lemma 1 that

$$\frac{\omega(B_{\rho}(x))}{\omega(B_{\rho}(x) \cap {\Omega}_{R}(y_{0}))} \leq M \left( \frac{|B_{\rho}(x)|}{|B_{\rho}(x) \cap {\Omega}_{R}(y_{0})|} \right)^{q}. $$

Hence, it suffices to prove

$$ \sup_{x \in {\Omega}_{R}(y_{0})}\sup_{0 <\rho < r_{0}} \frac{|B_{\rho}(x)|}{|B_{\rho}(x) \cap {\Omega}_{R}(y_{0})|} \leq \left\{ A^{-1}, 4^{n} \right\}. $$

(A.3)

We fix x ∈Ω_R(y₀) and ρ ∈ (0,r₀). Observe that if B_ρ(x) ⊂Ω_R(y₀). Then Eq. A.3 is trivial. Hence, we only need to consider the case B_ρ(x) ∩ ∂Ω_R(y₀)≠∅. We divide this situation into three subcases.

Case 1

If $B_{\rho }(x) \cap \left (\partial B_{R}(y_{0}) \cap \overline {\Omega } \right ) \not =\emptyset$ and $B_{\rho }(x) \cap \partial {\Omega } \cap \overline {B}_{R}(y_{0}) = \emptyset$. Then, it follows that Ω_R(y₀) ∩ B_ρ(x) = B_R(y₀) ∩ B_ρ(x). From this, a simple calculation shows

$$\frac{|B_{\rho}(x)|}{|B_{\rho}(x) \cap {\Omega}_{R}(y_{0})|} = \frac{|B_{\rho}(x)|}{|B_{\rho}(x) \cap B_{R}(y_{0})|} \leq 4^{n}. $$

Case 2

If $B_{\rho }(x) \cap \partial {\Omega } \cap \overline {B}_{R}(y_{0}) \not = \emptyset$ and $B_{\rho }(x) \cap \partial B_{R}(y_{0}) \cap \overline {\Omega } =\emptyset$. In this case, by the proof of [27, Lemma 3.8], it follows that

$$\frac{|B_{\rho}(x)|}{|B_{\rho}(x) \cap {\Omega}_{R}(y_{0})|} \leq \left( \frac{2}{1-4\delta} \right)^{n} \leq 4^{n}. $$

Case 3

If $B_{\rho }(x) \cap \partial {\Omega } \cap \overline {B}_{R}(y_{0}) \not = \emptyset$ and $B_{\rho }(x) \cap \partial B_{R}(y_{0}) \cap \overline {\Omega } \not =\emptyset$. Then, by the definition of type (A,r₀) domain, we see that

$$|{\Omega}_{R}(y_{0}) \cap B_{\rho}(x)| \geq A|B_{\rho}(x)|. $$

Therefore,

$$\frac{|B_{\rho}(x)|}{|{\Omega}_{R}(y_{0}) \cap B_{\rho}(x)|} \leq A^{-1}. $$

This completes the proof.

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