The Dirichlet problem for elliptic operators having a BMO anti-symmetric part

Abstract

The present paper establishes the first result on the absolute continuity of elliptic measure with respect to the Lebesgue measure for a divergence form elliptic operator with non-smooth coefficients that have a ${{\,\mathrm{BMO}\,}}$ anti-symmetric part. In particular, the coefficients are not necessarily bounded. We prove that the Dirichlet problem for elliptic equation $\mathrm{div}(A\nabla u)=0$ in the upper half-space $(x,t)\in {\mathbb {R}}^{n+1}_+$ is uniquely solvable when $n\ge 2$ and the boundary data is in $L^p({\mathbb {R}}^n,dx)$ for some $p\in (1,\infty )$. This result is equivalent to saying that the elliptic measure associated to L belongs to the $A_\infty$ class with respect to the Lebesgue measure dx, a quantitative version of absolute continuity.

A Appendix: Weak solution of parabolic equations

Lemma 15

Suppose $u,v\in L^2\left( (0,T),W^{1,2}({\mathbb {R}}^n)\right)$ with $\partial _tu, \partial _tv \in L^2\left( (0,T),\widetilde{W}^{-1,2}({\mathbb {R}}^n)\right)$. Then

• (i) $u\in C\left( [0,T], L^2({\mathbb {R}}^n)\right)$;

• (ii) The mapping $t\mapsto \left\| u(\cdot ,t)\right\| _{L^2({\mathbb {R}}^n)}$ is absolutely continuous, with

  $$\begin{aligned}\frac{d}{dt}\left\| u(\cdot ,t)\right\| _{L^2({\mathbb {R}}^n)}^2=2\mathfrak {R}\langle \partial _tu(\cdot ,t),u(\cdot ,t)\rangle _{{\widetilde{W}}^{-1,2},W^{1,2}} \quad \text {for a.e. }t\in [0,T].\end{aligned}$$

  As a consequence,

  $$\begin{aligned} \frac{d}{dt}\left( u(\cdot ,t),v(\cdot ,t)\right) _{L^2({\mathbb {R}}^n)}=\langle \partial _tu(\cdot ,t),v(\cdot ,t)\rangle _{{\widetilde{W}}^{-1,2},W^{1,2}}+\overline{\langle {\partial _tv(\cdot ,t),u(\cdot ,t)\rangle }}_{\widetilde{W}^{-1,2},W^{1,2}}\,\text {a.e.} \end{aligned}$$

For the proof see, e.g., [8], Section 5.9.2, Theorem 3.

Suppose that $A=A(x)=A^s(x)+A^a(x)$ is a real, $n\times n$ matrix, with $A^s$ being symmetric, elliptic with constant $\lambda _0>0$, $\left\| A^s\right\| _{L^{\infty }({\mathbb {R}}^n)}\le \lambda _0^{-1}$, and $A^a$ being anti-symmetric and $\left\| A^a\right\| _{{{\,\mathrm{BMO}\,}}({\mathbb {R}}^n)}\le \Lambda _0$.

Proposition 13

For any $u_0\in L^2({\mathbb {R}}^n)$, the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu-{{\,\mathrm{div}\,}}(A\nabla u)=0 \quad \text {in }{\mathbb {R}}^n\times (0,\infty ),\\ u(x,0)=u_0(x), \end{array}\right. } \end{aligned}$$

(A.1)

has a unique weak solution $u(x,t)=e^{-tL}(u_0)(x)$. Here, ${{\,\mathrm{div}\,}}={{\,\mathrm{div}\,}}_x$ and $\nabla =\nabla _x$.

Proof

Existence.

Since the domain of L (denoted by D(L)) is dense in $W^{1,2}({\mathbb {R}}^n)$, and thus dense in $L^2({\mathbb {R}}^n)$, we can find a sequence $\left\{ u_{0,\epsilon }\right\} \subset D(L)$ such that $u_{0,\epsilon }$ converges to $u_0$ in $L^2({\mathbb {R}}^n)$. Denote $u_\epsilon (x,t):=e^{-tL}(u_{0,\epsilon })(x)$. Then by semigroup theory,

$$\begin{aligned} \partial _tu_\epsilon +Lu_\epsilon =0 \quad \text {in } L^2({\mathbb {R}}^n) \quad \forall \, t\ge 0. \end{aligned}$$

(A.2)

For any $0<\tau <T$, and any $\varphi \in L^2\left( (0,T),W^{1,2}({\mathbb {R}}^n)\right)$, with $\partial _t\varphi \in L^2\left( (0,T),\widetilde{W}^{-1,2}({\mathbb {R}}^n)\right)$, (A.2) implies

$$\begin{aligned} \int _{\tau }^{T}\left( \partial _tu_\epsilon ,\varphi \right) _{L^2}dt+\int _{\tau }^T\left( Lu_\epsilon ,\varphi \right) _{L^2}dt=0. \end{aligned}$$

(A.3)

Since $\partial _tu_\epsilon \in L^2_{{{\,\mathrm{loc\ }\,}}}\left( (0,\infty ),L^2({\mathbb {R}}^n)\right)$ (see [13] Theorem 4.9), and by Lemma 15 (ii), (A.3) can be written as

$$\begin{aligned}&\int _{{\mathbb {R}}^n}u_\epsilon (x,T)\overline{\varphi (x,T)}dx+\int _{\tau }^T\int _{{\mathbb {R}}^n}A\nabla u_\epsilon \cdot \overline{\nabla \varphi }dxdt\nonumber \\&\quad =\int _{{\mathbb {R}}^n}u_\epsilon (x,\tau )\overline{\varphi (x,\tau )}dx+\int _{\tau }^T\overline{\langle \partial _t\varphi ,u_\epsilon \rangle }_{\widetilde{W}^{-1,2},W^{1,2}}. \end{aligned}$$

(A.4)

Notice that $u_\epsilon \rightarrow u$ in $C((\tau ,T),W^{1,2}({\mathbb {R}}^n))$ (see [13] Theorem 4.9), and so letting $\epsilon \rightarrow 0^+$ we get

$$\begin{aligned}&\int _{{\mathbb {R}}^n}u(x,T)\overline{\varphi (x,T)}dx+\int _{\tau }^T\int _{{\mathbb {R}}^n}A\nabla u\cdot \overline{\nabla \varphi }dxdt\nonumber \\&\quad =\int _{{\mathbb {R}}^n}u(x,\tau )\overline{\varphi (x,\tau )}dx+\int _{\tau }^T\overline{\langle \partial _t\varphi ,u\rangle }_{\widetilde{W}^{-1,2},W^{1,2}}. \end{aligned}$$

(A.5)

Letting $\varphi =u_\epsilon$ in (A.4) and applying Lemma 15 (ii) again, one obtains

$$\begin{aligned} \int _{{\mathbb {R}}^n}\left| u_\epsilon (x,T)\right| ^2dx+2\mathfrak {R}\int _{\tau }^T\int _{{\mathbb {R}}^n}A\nabla u_\epsilon \cdot \overline{\nabla u_\epsilon }dxdt =\int _{{\mathbb {R}}^n}\left| u_\epsilon (x,\tau )\right| ^2dx. \end{aligned}$$

By ellipticity and the definition of $u_\epsilon$, we have

$$\begin{aligned}&2\lambda _0\int _{\tau }^T\int _{{\mathbb {R}}^n}\left| \nabla u_\epsilon \right| ^2dxdt\le \left\| e^{-\tau L}(u_{0,\epsilon })\right\| _{L^2({\mathbb {R}}^n)}^2\\&\quad \le 2\left\| e^{-\tau L}(u_{0,\epsilon }-u_0)\right\| _{L^2({\mathbb {R}}^n)}^2+2\left\| e^{-\tau L}(u_{0})\right\| _{L^2({\mathbb {R}}^n)}^2. \end{aligned}$$

Letting $\epsilon \rightarrow 0^+$, $\tau \rightarrow 0^+$, $T\rightarrow \infty$, we obtain $\int _0^{\infty }\int _{{\mathbb {R}}^n}\left| \nabla u\right| ^2dxdt\le \lambda _0^{-1}\left\| u_0\right\| _{L^2}^2<\infty$. This enables us to take limit as $\tau$ go to $0^+$ on both sides of (A.5) and get

$$\begin{aligned}&\int _{{\mathbb {R}}^n}u(x,T)\overline{\varphi (x,T)}dx+\int _0^T\int _{{\mathbb {R}}^n}A\nabla u\cdot \overline{\nabla \varphi }dxdt\\&\quad =\int _{{\mathbb {R}}^n}u(x,0)\overline{\varphi (x,0)}dx+\int _0^T\overline{\langle \partial _t\varphi ,u\rangle }_{\widetilde{W}^{-1,2},W^{1,2}}, \end{aligned}$$

i.e. u(x, t) is a weak solution of (A.1).

Uniqueness.

Let v be a weak solution of (A.1). We first show that $\partial _tv\in L^2\left( (0,T),\widetilde{W}^{-1,2}({\mathbb {R}}^n)\right)$ for any $T\in (0,\infty )$. Define a semilinear functional F on $L^2\left( [0,T],W^{1,2}({\mathbb {R}}^n)\right)$ as follows: for any $\varphi \in L^2\left( [0,T],W^{1,2}({\mathbb {R}}^n)\right)$, let

$$\begin{aligned} \langle F,\varphi \rangle :=\int _0^T\int _{{\mathbb {R}}^n}A\nabla v\cdot \overline{\nabla \varphi }dxdt. \end{aligned}$$

Obviously,

$$\begin{aligned} \left| \langle F,\varphi \rangle \right| \le C\left\| \nabla v\right\| _{L^2\left( [0,T],L^2({\mathbb {R}}^n)\right) }\left\| \nabla \varphi \right\| _{L^2\left( [0,T],L^2({\mathbb {R}}^n)\right) }. \end{aligned}$$

Then by Riesz representation theorem, there exists $w(x,t)\in L^2\left( [0,T],W^{1,2}({\mathbb {R}}^n)\right)$ such that

$$\begin{aligned} \langle F,\varphi \rangle&=\int _0^T\int _{{\mathbb {R}}^n}(\nabla w\cdot \nabla {\overline{\varphi }}+w{\overline{\varphi }})dxdt\\&=\int _0^T\langle -\Delta w(\cdot ,t)+w(\cdot ,t),\varphi \rangle _{{\widetilde{W}}^{-1,2},W^{1,2}}dt, \end{aligned}$$

and

$$\begin{aligned} \left\| -\Delta w+w\right\| _{L^2\left( [0,T],\widetilde{W}^{-1,2}({\mathbb {R}}^n)\right) }\le \left\| w\right\| _{L^2([0,T],W^{1,2}({\mathbb {R}}^n))}\le C\left\| \nabla v\right\| _{L^2\left( [0,T],L^2({\mathbb {R}}^n)\right) }. \end{aligned}$$

Choose $\varphi (x,t)=\Psi (x){\overline{\eta }}(t)$ as a test function in (A.1), where $\Psi \in W^{1,2}({\mathbb {R}}^n)$, $\eta \in C_0^1\left( (0,T)\right)$. Then since v is a weak solution, we have

$$\begin{aligned} \int _0^T\left( v(\cdot ,t),\Psi \right) _{L^2}\eta '(t)dt&=\int _0^T\int _{{\mathbb {R}}^n}A\nabla v\cdot \overline{\nabla \Psi }\eta (t)dxdt\\&=\int _0^T\langle -\Delta w(\cdot ,t)+w(\cdot ,t),\Psi \rangle _{{\widetilde{W}}^{-1,2},W^{1,2}}\eta (t)dt. \end{aligned}$$

Since $\Psi \in W^{1,2}({\mathbb {R}}^n)$ is arbitrary,

$$\begin{aligned} \int _0^Tv(x,t)\eta '(t)dt=\int _0^T(-\Delta w+w)\eta (t)dt\qquad \text {in }\widetilde{W}^{-1,2}({\mathbb {R}}^n), \end{aligned}$$

which gives $\partial _tv=\Delta w-w\in L^2\left( (0,T),\widetilde{W}^{-1,2}({\mathbb {R}}^n)\right)$. Therefore, we can take $\varphi =v$ as a test function in (A.1) and get

$$\begin{aligned} \int _{{\mathbb {R}}^n}\left| v(x,T)\right| ^2+\int _0^T\int _{{\mathbb {R}}^n}A\nabla v\cdot \nabla {\overline{v}}dxdt\!=\!\int _0^T\overline{\langle \partial _tv,v\rangle }_{\widetilde{W}^{\!-\!1,2},W^{1,2}}\!+\!\int _{{\mathbb {R}}^n}\left| v(x,0)\right| ^2dx. \end{aligned}$$

Using this and Lemma 15 (ii), we have

$$\begin{aligned} \int _{{\mathbb {R}}^n}\left| v(x,T)\right| ^2+2\mathfrak {R}\int _0^T\int _{{\mathbb {R}}^n}A\nabla v\cdot \nabla {\overline{v}}dxdt=\int _{{\mathbb {R}}^n}\left| v(x,0)\right| ^2dx. \end{aligned}$$

So we get

$$\begin{aligned} \int _{{\mathbb {R}}^n}\left| v(x,T)\right| ^2+2\lambda _0\int _0^T\int _{{\mathbb {R}}^n}\left| \nabla v\right| ^2dxdt \le \int _{{\mathbb {R}}^n}\left| v(x,0)\right| ^2dx, \end{aligned}$$

which implies that if $v(x,0)=0$ then $v\equiv 0$. $\square$

Remark 3

Let u(x, t) be the weak solution to (A.1). Since the coefficients are independent of t, a standard argument shows that $\partial _tu$ is a weak solution to $\partial _tv-{{\,\mathrm{div}\,}}(A\nabla v)=0$ in ${\mathbb {R}}^n\times (0,\infty )$. That is, for any $T>0$, any $\varphi \in L^2\left( [0,T],W^{1,2}({\mathbb {R}}^n)\right)$ with $\partial _t\varphi \in L^2\left( [0,T],\widetilde{W}^{-1,2}({\mathbb {R}}^n)\right)$ and $\varphi =0$ when $0\le t\le \varepsilon$ for some $0<\varepsilon <T$,

$$\begin{aligned} \int _{{\mathbb {R}}^n}\partial _tu(x,T)\overline{\varphi (x,T)}dx+\int _0^T\int _{{\mathbb {R}}^n}A\nabla (\partial _tu)\cdot \overline{\nabla \varphi }dxdt=\int _0^T\overline{\langle \partial _t\varphi ,\partial _tu\rangle }_{\widetilde{W}^{-1,2},W^{1,2}}dt. \end{aligned}$$

Moreover, since $\partial _t^lu\in L^2_{{{\,\mathrm{loc\ }\,}}}\left( (0,\infty ),L^2({\mathbb {R}}^n)\right)$ and $\partial _t^l\nabla u\in L^2_{{{\,\mathrm{loc\ }\,}}}\left( (0,\infty ),L^2({\mathbb {R}}^n)\right)$ for any $l\in {\mathbb {N}}$, one can show that for any $l\in {\mathbb {N}}$, $\partial _t^l u$ is a weak solution to $\partial _tv-{{\,\mathrm{div}\,}}(A\nabla v)=0$ in ${\mathbb {R}}^n\times (0,\infty )$.