Gradient Weighted Norm Inequalities for Linear Elliptic Equations with Discontinuous Coefficients

Abstract

Local and global weighted norm estimates involving Muckenhoupt weights are obtained for gradient of solutions to linear elliptic Dirichlet boundary value problems in divergence form over a Lipschitz domain $\Omega$. The gradient estimates are obtained in weighted Lebesgue and Lorentz spaces, which also yield estimates in Lorentz–Morrey spaces as well as Hölder continuity of solutions. The significance of the work lies on its applicability to very weak solutions (that belong to $W^{1,p}_{0}(\Omega )$ for some $p>1$ but not necessarily in $W^{1,2}_{0}(\Omega )$) to inhomogeneous equations with coefficients that may have discontinuities but have a small mean oscillation. The domain is assumed to have a Lipschitz boundary with small Lipschitz constant and as such allows corners. The approach implemented makes use of localized sharp maximal function estimates as well as known regularity estimates for very weak solutions to the associated homogeneous equations. The estimates are optimal in the sense that they coincide with classical weighted gradient estimates in the event the coefficients are continuous and the domain has smooth boundary.

Appendix: Very Weak Solutions in \(W^{1, p}\), \(p > 1\), are Finite Energy Solutions

In this appendix we would like to demonstrate the validity of Brezis’s result [5, Theorem A1.1] up to the boundary. Brezis’s result [5, Lemma A.1] says that very weak solutions of homogeneous linear equation with continuous coefficients that are in $W^{1, p}$ for some $p > 1$ are in fact in $W^{1, q}_{loc}$ for any $q>1$. In this appendix we will show that in fact the statement will remain true for boundary value problems even with coefficients with small BMO and posed over half balls, having a zero boundary condition on the flat part of the boundary. The proof strictly follows the argument used in the proof of [5, Lemma A.1], with natural modification to fit our setting. The main tool we use is the following lemma which is actually the main result of [29]. The result is stated in its general form, to include what are called ‘quasiconvex domains’, see [29, Definition 3.2]. For our purpose we simply note that polygonal convex domains, sector of balls, and ball segments (such as half balls) are all quasiconvex domains.

Lemma A.1

[29, Theorem 1.1] Let $1< p < \infty$ and $\mathbf{f} \in L^{p}(\Omega ;\mathbb {R}^{n})$. Suppose that $\mathbb {A}$ is a symmetric, uniformly elliptic matrix with constants of ellipticity $\lambda$ and $\Lambda$. Then there exists $\delta =\delta (n, p, \lambda , \Lambda ) > 0$ such that whenever $\mathbb {A}$ is $(\delta , \mathfrak {K})$-BMO and $\Omega$ is a $(\delta , \sigma , \mathfrak {K})$-quasiconvex bounded domain, the Dirichlet problem

$$\begin{aligned} \left\{ \begin{aligned} \text {div}\,\mathbb {A}(x)\nabla u(x)&= \text {div}\,\mathbf{f} \,\, \text {in } \Omega , \\ u&= 0\quad \text {on } \partial \Omega , \end{aligned} \right. \end{aligned}$$

has a unique solution $u\in W^{1, p}_{0}(\Omega )$. Moreover, there exists $C=C(\lambda , \Lambda , n, p, \sigma , \mathfrak {K}, \Omega ) > 0$ such that

$$\begin{aligned} \Vert \nabla u\Vert _{L^{p}(\Omega )} \le C \Vert \mathbf{f}\Vert _{L^{p}}. \end{aligned}$$

The main theorem we would like to prove is the following. We note that we have used the result with constant coefficients in the form given in Lemma 6.2.

Theorem A.2

Let $1< r < \infty$, $R > 0$, and $\mathfrak {K} > 0$. Suppose that $\mathbb {A}$ is a symmetric, uniformly elliptic matrix with constants of ellipticity $\lambda$ and $\Lambda$. Suppose also that $u_{*} \in W^{1, r}(B_{R}^{+}(0))$ such that $u_{*}=0$ on $B_{R}(0) \cap \partial \mathbb {R}^{n}_{+}$. Then there exists $\delta =\delta (n, r, \lambda , \Lambda ) > 0$ such that if $\mathbb {A}$ is $(\delta , \mathfrak {K})$-BMO and v solves

$$\begin{aligned} \left\{ \begin{aligned} \text {div} \, {\mathbb {A}}(x)\nabla v&= 0\quad \text {in} \,\,B_{R}^{+}(0),\\ v - u_{*}&\in W^{1, r}_{0}(B_{R}^{+}(0)), \end{aligned}\right. \end{aligned}$$

(A.1)

then for any $0<\tau < 1$, one has $v\in W^{1, 2}(B_{\tau R}^{+}(0))$ along with the estimate

$$\begin{aligned} \Vert v\Vert _{W^{1, 2}(B_{\tau R}^{+}(0))} \le C_{\tau } \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

The constant $C_{\tau }$ depends only $\lambda , \Lambda , \tau ,$ r, R, n and $\mathfrak {K}$.

Proof

There is nothing to prove if $r \ge 2$. So we assume that $1< r <2$. Suppose that $\mathbf{g} \in C_{c}^{\infty }(B_{R}^{+}(0))$ such that

$$\begin{aligned} \Vert \mathbf{g}\Vert _{L^{s'}(B_{R}^{+}(0))} \le 1, \end{aligned}$$

where $\frac{n}{n-1}<s\le 2$ to be determined and $1/s + 1/s' = 1$.

We now apply Lemma A.1 to obtain $\delta > 0$ such that a unique solution $w\in W^{1, 2}_{0}(B_{R}^{+}(0)) \cap W_{0}^{1, s'}(B_{R}^{+}(0))$ to $\text {div}\, {\mathbb {A}} \nabla w = \text {div}\, \mathbf{g}$ in $B_{R}^{+}(0)$ exists such that

$$\begin{aligned} \quad \Vert w\Vert _{W^{1, s'}(B_{R}^{+}(0))} \le C \Vert \mathbf{g}\Vert _{L^{s'}(B_{R}^{+}(0))} \le C, \end{aligned}$$

(A.2)

provided $\mathbb {A}$ is $(\delta , \mathfrak {K})$-BMO. Note also that by choosing $\delta$ even smaller, we can have $w\in W^{1, r'}(B_{R}^{+}(0))$ where $r'$ is the Hölder conjugate exponent of r. By definition of w, we have that

$$\begin{aligned} \int _{B_{R}^{+}(0)} {\mathbb {A}}\nabla w \cdot \nabla \phi dx = \int _{B_{R}^{+}(0))} \mathbf{g}\cdot \nabla \phi dx, \quad \quad \forall \phi \in C_{c}^{1}(B_{R}^{+}(0)). \end{aligned}$$

Moreover, by density the above equation is valid for all $\phi \in W^{1, r}_{0}(B_{R}^{+}(0))$. Now for any fixed $\zeta \in C_{c}^{\infty }(B_{R}(0))$, we can take $\phi (x) = \zeta (x) v(x)$ as a test function in the above, since by assumption $v\in W^{1, r}(B_{R}^{+}(0))$ and $v = 0$ on $B_{R}(0) \cap \partial \mathbb {R}^{n}_{+}$, and therefore the product $\phi (x) = \zeta (x) v(x) \in W^{1, r}_{0}(B_{R}^{+}(0))$. We then have the following:

$$\begin{aligned} \int _{B_{R}^{+}(0))} {\mathbb {A}} \nabla w\cdot (\zeta \nabla v + v \nabla \zeta ) dx = \int _{B_{R}^{+}(0))} \mathbf{g}\cdot (\zeta \nabla v + v \nabla \zeta )dx. \end{aligned}$$

(A.3)

Again, since $v\in W^{1, r}(B_{R}^{+}(0))$ solves Eq. (A.1), it follows from the definition of v as a solution and density

$$\begin{aligned} \int _{B_{R}^{+}(0)}{\mathbb {A}}\nabla v \cdot \nabla \psi dx =0,\quad \quad \forall \psi \in W_{0}^{1,r'}(B_{R}^{+}(0)). \end{aligned}$$

Now take $\psi = \zeta w \in W_{0}^{1, r'}(B_{R}^{+}(0))$ as a test function in the above to get that

$$\begin{aligned} \int _{B_{R}^{+}(0)}{\mathbb {A}}\nabla v \cdot (w\nabla \zeta + \zeta \nabla w) dx =0. \end{aligned}$$

(A.4)

Comparing equations (A.3) and (A.4) we obtain that

$$\begin{aligned} \begin{aligned} \int _{B_{R}^{+}(0)} \zeta \nabla v \cdot \mathbf{g} dx =&-\int _{B_{R}^{+}(0)} w\left( {\mathbb {A}} \nabla v \cdot \nabla \zeta \right) dx + \int _{B_{R}^{+}(0)}v \left( {\mathbb {A}}\nabla w \cdot \nabla \zeta \right) dx \\&- \int _{B_{R}^{+}(0)} v \mathbf{g}\cdot \nabla \zeta dx\\ =&I_{1} + I_{2} + I_{3} \end{aligned} \end{aligned}$$

Since $r<2\le n$, and by the Sobolev embedding,

$$\begin{aligned} \Vert v\Vert _{L^{r^{*}} (B_{R}^{+}(0))} \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))},\quad \text {where } \quad \frac{1}{r^{*}} = \frac{1}{r} - \frac{1}{n}. \end{aligned}$$

We estimate $I_{1}, I_{2}$ and $I_{3}$, by choosing $s\in (n/(n-1), 2]$ in the following way: if $r^{*} \le 2$, then choose $s = r^{*}$, if $r^{*}> 2$, choose $s=2$.

1.1 Case 1

$r^{*} > 2$: In this case choose $s=2$, ($s' = 2$ ). Then we have $w\in W^{1, 2}(B_{R}^{+}(0))$ and from (A.2) we have

$$\begin{aligned} \Vert w\Vert _{W^{1, 2}(B_{R}^{+}(0))} \le C. \end{aligned}$$

Now when $n\ge 3$, the assumption $r^{*} > 2$ is equivalent to $r' < 2^{*}$; when $n = 2$, Sobolev imbedding implies that $W^{1, 2}(B_{R}^{+}(0)) \hookrightarrow L^{q}(B_{R}^{+}(0))$ for any $q\in [2, \infty )$. Combining the two we find that

$$\begin{aligned} \Vert w\Vert _{L^{r'}(B_{R}^{+}(0))} \le C. \end{aligned}$$

With this at hand, we can now estimate $|I_{i}|$, $i= 1, 2, 3$. To that end,

$$\begin{aligned} |I_{1}|\le & {} \int _{B_{R}^{+}(0)} |w||{\mathbb {A}} \nabla v||\nabla \zeta | dx \le C \Vert \nabla \zeta \Vert _{L^{\infty }} \Vert w\Vert _{L^{r'}(B_{R}^{+}(0)) } \Vert \nabla v\Vert _{L^{r}(B_{R}^{+}(0))} \\\le & {} C \Vert \nabla v\Vert _{L^{r}(B_{R}^{+}(0))}. \end{aligned}$$

We also notice from Sobolev embedding that

$$\begin{aligned} \Vert v\Vert _{L^{2}(B_{R}^{+}(0))} \le C \Vert v\Vert _{L^{r^{*} } (B_{R}^{+}(0))} \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

As a consequence,

$$\begin{aligned} |I_{2}|\le & {} \int _{B_{R}^{+}(0))} |v||{\mathbb {A}}\nabla w| |\nabla \zeta |dx \le C \Vert \nabla \zeta \Vert _{L^{\infty }} \Vert v\Vert _{L^{2}(B_{R}^{+}(0))} \Vert \nabla w\Vert _{L^{2}(B_{R}^{+}(0))} \\\le & {} C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}, \end{aligned}$$

and

$$\begin{aligned} |I_{3}| \le C \Vert \nabla \zeta \Vert _{L^{\infty }}\Vert \mathbf{g}\Vert _{L^{2}}\Vert v\Vert _{L^{2}} \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

Combining the estimates for $|I_{i}|$, we get that

$$\begin{aligned} \left| \int _{B_{R}^{+} (0)} \zeta \nabla v \cdot \mathbf{g} dx\right| \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}, \quad \text {provided}~ \Vert \mathbf{g}\Vert _{L^{2}} \le 1. \end{aligned}$$

In particular, by choosing the cut off function $\zeta$ appropriately, for a given $\tau > 0$, $v\in W^{1,2}(B_{\tau R}^{+}(0))$ and

$$\begin{aligned} \Vert v\Vert _{W^{1,2} (B_{\tau R}^{+}(0))} \le C_{\tau } \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

1.2 Case 2

$r^{*} \le 2$: In this case, take $s = r^{*}$. We then have that

$$\begin{aligned} \frac{1}{r'} = 1- \frac{1}{r} = \frac{1}{r^{*}} + \frac{1}{s'} - \frac{1}{r} = \frac{1}{s'} - \frac{1}{n}. \end{aligned}$$

That is, the Sobolev conjugate of $s'$ is $r'$ and that $s' = \frac{r^{*}}{r^{*} - 1} < n$. As a consequence, by (A.2) and Sobolev embedding again we have

$$\begin{aligned} \Vert w\Vert _{L^{r'} (B_{R}^{+}(0))} \le C \Vert w\Vert _{W^{1, s'}(B_{R}^{+}(0))} \le C. \end{aligned}$$

We again use this to estimate $|I_{i}|,$ $i=1, 2, 3$. We begin with $I_{1}$:

$$\begin{aligned} |I_{1}|\le & {} \int _{B_{R}^{+}(0)} |w||{\mathbb {A}} \nabla v||\nabla \zeta | dx \le C \Vert \nabla \zeta \Vert _{L^{\infty }} \Vert w\Vert _{L^{r'}(B_{R}^{+}(0)) } \Vert \nabla v\Vert _{L^{r}(B_{R}^{+}(0))} \\\le & {} C \Vert \nabla v\Vert _{L^{r}(B_{R}^{+}(0))}. \end{aligned}$$

Next, since $s = r^{*}$, we have that $\frac{1}{r*} + \frac{1}{s'} = 1$, and applying Hölder’s inequality with exponents $r^{*}$ and $s'$ we obtain that

$$\begin{aligned} |I_{2}|\le & {} \int _{B_{R}^{+}(0))} |v||{\mathbb {A}}\nabla w| |\nabla \zeta |dx \le C \Vert \nabla \zeta \Vert _{L^{\infty }} \Vert v\Vert _{L^{r^{*}}(B_{R}^{+}(0))} \Vert \nabla w\Vert _{L^{s'}(B_{R}^{+}(0))}\\\le & {} C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0) }. \end{aligned}$$

Finally, by Sobolev embedding,

$$\begin{aligned} |I_{3}| \le C \Vert \nabla \zeta \Vert _{L^{\infty }}\Vert \mathbf{g}\Vert _{L^{s^{'}}}\Vert v\Vert _{L^{r*}} \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

Combining the above inequalities, we obtain that there exists a constant C, that depends on $\zeta$, such that

$$\begin{aligned} \left| \int _{B_{R}^{+}(0)}\zeta \nabla v \cdot \mathbf{g} dx \right| \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

As this holds for all $\mathbf{g} \in C_c^\infty (B_{R}^{+}(0))$ such that $\Vert \mathbf{g} \Vert _{L^{s'}} \le 1$, by duality we get

$$\begin{aligned} \left( \int _{B_{R}^{+}(0)}|\zeta \nabla v|^{r^*} dx \right) ^{\frac{1}{r^*}} \le C \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

Now, given $\epsilon \in (0,1)$, we may choose $\zeta \in C_{c}^{\infty }(B_{R}(0))$ such that $0\le \zeta \le 1$, $\zeta = 1$ on $B_{\epsilon R}(0)$. For this choice of $\zeta$ we have in particular that $v\in W^{1, r^{*}} (B_{\epsilon R}^{+}(0))$ and

$$\begin{aligned} \Vert v\Vert _{W^{1, r^{*} } (B_{\epsilon R}^{+}(0)) } \le C_{\epsilon } \Vert v\Vert _{W^{1, r}(B_{R}^{+}(0))}. \end{aligned}$$

It is easy to see that $r < r^{*}$. If $r^{*} =2$, then we are done. Otherwise, applying the argument of Case 2 yields $r^{**} > r^{*}$ that $v \in W^{1, r^{**} }(B_{(\epsilon /2) R}^{+}(0))$, and then $v \in W^{1, r^{***} }(B_{(\epsilon /3)R}^{+}(0))$ and so on, until $r^{**\cdots *}$ reaches the first value bigger than 2, at which point we apply Case 1 to obtain that $v \in W^{1, 2}(B_{(\epsilon /m)R}^{+}(0) )$ for some positive integer $m = m(r)$. In particular, taking $\tau =\varepsilon /m$ gives the desired result. We emphasize that for the argument to work we need to verify that the solution w belongs to $W^{1, r^{**\cdots *}}_{0} (B_{R}^{+}(0))$, and for that we must choose $\delta$ sufficiently small. $\square$