Proof. For any \(\phi\in C^{\infty}_c(\Omega)\) after integrating by parts we have \[\label{eq2466} \langle b\cdot\nabla u, \phi\rangle=-\int_{\Omega} b\cdot(\nabla\phi )u\,dx.\tag{11}\] Therefore we can apply Hölder inequality and get, in case \(q\ge n\), \[\int_{\Omega}|b\cdot\nabla u\phi |~dx\le \|b\|_{n}\|\phi\|_{\frac{nq'}{n-q'}}\|u\|_{1,q}\le C \|b\|_{n}\|\phi\|_{W^{1,q'}}\|u\|_{W^{1,q}},\] and in case \(1\le q<n\), \[\int_{\Omega}|b\cdot\nabla\phi u|~dx\le \|b\|_{n}\|\phi\|_{W^{1,q'}}\|u\|_{\frac{nq}{n-q}}\le C \|b\|_{n}\|\phi\|_{W^{1,q'}}\|u\|_{W^{1,q}}.\] The above shows 9 .

To prove the second estimate 10 , we approximate \(b\) by \(b_k\in C^{\infty}(\Omega)\) with \(\mathop{\rm div}\nolimits b_k=0\) and \(\|b-b_k\|_{n}\rightarrow0\) as \(k\to\infty\). Then \[\begin{align} \|b\cdot\nabla u\|_{W^{-1,q}(\Omega)} &\le \|(b_k- b)\cdot\nabla u\|_{W^{-1,q}(\Omega)} +\|b_k\cdot\nabla u\|_{W^{-1,q}(\Omega)}\\ &\le C \|b_k- b\|_n \|u\|_{1,q} +C\|b_k\|_{L^{\infty}(\Omega)}\|u\|_{q} \\ &\le \varepsilon\|u\|_{1,q}+C(\varepsilon,p,b,\Omega)\|u\|_{q}. \end{align}\] We have used 9 for the second inequality, and chosen \(k\) so that \(C\|b_k- b\|_n \le \varepsilon\). ◻

Proof. We use a similar argument to Lemma 3. For any \(\phi\in C^{\infty}_c(\Omega)\) we still have 11 . By Hölder and Sobolev inequalities in Lorentz spaces (Lemmas 1 and 2), in case \(q\ge n\) we have \[\begin{align} \int\limits_{\Omega}|b\cdot\nabla u\phi |~dx &\le C\|b\|_{L^{n,\infty}(\Omega)}\|\phi\|_{L^{\frac{nq'}{n-q'},q'}(\Omega)}\|u\|_{1,q} \\ &\le C\|b\|_{L^{n,\infty}(\Omega)}\|\phi\|_{W^{1,q'}(\Omega)}\|u\|_{W^{1,q}(\Omega)}. \end{align}\] The case \(1\le q< n\) is done similarly: \[\begin{align} \int\limits_{\Omega}|b\cdot\nabla\phi u|~dx &\le C\|b\|_{L^{n,\infty}(\Omega)}\|\phi\|_{W^{1,q'}(\Omega)}\|u\|_{L^{\frac{nq}{n-q},q}(\Omega)}\\ &\le C\|b\|_{L^{n,\infty}(\Omega)}\|\phi\|_{W^{1,q'}(\Omega)}\|u\|_{W^{1,q}(\Omega)} . \qedhere \end{align}\] ◻

Proof. We may assume \(\lambda=1\). Define bilinear form \(\mathcal{B}: W^{1,2}_{0,\sigma}(\Omega)\times W^{1,2}_{0,\sigma}(\Omega)\rightarrow \Bbb R\) by \[\mathcal{B}(u,v)=\int_{\Omega}\nabla u:\nabla v+(b\cdot\nabla) u \cdot v\, dx.\] The integral \(\int (b\cdot\nabla) u \cdot v\, dx\) is defined by 8 with \(q=2\) (which uses Lemmas 1 and 2 and that \(\Omega\) is bounded and Lipschitz). This bounds enables us to justify by approximation that \[\label{eq3464} \int_\Omega(b\cdot \nabla) u\cdot u = \int_\Omega b\cdot \nabla(\frac{1}{2}|u|^2) = 0,\tag{15}\] using the condition \(\mathop{\rm div}\nolimits b=0\). Thus we get the coercivity of this bilinear form, indeed, \[\mathcal{B}(u,u)=\|\nabla u\|_{L^2(\Omega)}^2, \quad \forall u\in W_{0,\sigma}^{1,2}(\Omega).\] Therefore, by Lax-Milgram theorem, for any \(f \in W^{-1,2}(\Omega)^n\), there exists a unique \(u\in W_{0,\sigma}^{1,2}(\Omega)\) such that \[\mathcal{B}(u,v)=\langle f,v\rangle,\quad \forall v\in W_{0,\sigma}^{1,2}(\Omega).\] This is the weak form of 13 and hence \(u\) is the unique weak solution. By taking \(v=u\), we get \(\left \| u \right \|_{W^{1,2}_0(\Omega)} \le \left \| f \right \|_{W^{-1,2}(\Omega)}\). Lastly, we apply Lemma 5 with \(q=2\) (and using that \(\Omega\) is bounded and Lipschitz) to get an associated pressure \(\pi\in L^2_0(\Omega)\) that satisfies the bound. ◻

Proof. Let \(\phi= f+\Delta u-\lambda b\cdot\nabla u\). It is a distribution in \(W^{-1,q}(\Omega)^n\) by Lemmas 3 and 4, and it vanishes on \(W^{1,q'}_{0,\sigma}(\Omega)\) by the weak form (similar to 4 ) of 13 . Hence \(\phi\) is given by a pressure gradient, \(\phi=\nabla\pi\) with \(\pi \in L^q_0(\Omega)\) by Lemma 5, and \((u,\pi)\) is a solution pair of 13 . By Theorem 8, \[\|u\|_{W^{1,q}(\Omega)} + \| \pi \|_{L^q(\Omega)} \le C_1\|f-\lambda b\cdot\nabla u\|_{W^{-1,q}(\Omega)}\le C_1\|f\|_{W^{-1,q}(\Omega)}+C_1\|b\cdot \nabla u\|_{W^{-1,q}(\Omega)}.\] For the drift term, if \(b \in L^n(\Omega)\), we can apply Lemma 3 and get that \[\|b\cdot \nabla u\|_{W^{-1,q}(\Omega)}\le \varepsilon \|u\|_{1,q;\Omega}+C \|u\|_{q,\Omega}.\] Taking \(\varepsilon= (4C_1)^{-1}\) and using \(\|u\|_q \le \delta\|u\|_{1,q}+ C(\delta) \|u\|_2\), we get \[\label{3465} \|u\|_{W^{1,q}(\Omega)}+ \| \pi \|_{L^q(\Omega)}\le C_1\|f\|_{-1,q}+\frac{1}{4}\|u\|_{1,q}+C\|u\|_{q}\le C_1\|f\|_{-1,q}+\frac{1}{2}\|u\|_{1,q}+C\|u\|_{2}.\tag{16}\] If \(b \in L^{n,\infty}(\Omega)\), we apply Lemma 4 and assume \(\left \| b \right \|_{n,\infty} \le\varepsilon/C_q\). We still get 16 .

As \(q>2\), we have \(u \in W^{1462}(\Omega)\) and \(\|u\|_{W^{1,2}(\Omega)}\le C(n,q,\Omega)\|f\|_{-1,q}\) by Lemma 6. Using this uniform bound in \(W^{1,2}(\Omega)\) we get from 16 that \[\|u\|_{W^{1,q}(\Omega)} + \| \pi \|_{L^q(\Omega)} \le C\|f\|_{-1,q}. \qedhere\] ◻

Proof of Proposition 3 and Theorem 4 when \(q \ge 2\). By Lemmas 3 and 4, the linear differential operators \[L_\lambda(u,\pi) = -\Delta u+\lambda b\cdot\nabla u + \nabla\pi,\quad 0 \le \lambda\le 1,\] are bounded linear maps from \(\mathfrak X_q\) to \(\mathfrak Y_q\), where \[\label{XqYq} \mathfrak X_q:= W^{1,q}_{0,\sigma}(\Omega) \times L^q_0(\Omega) ,\quad \mathfrak Y_q:=W^{-1,q}(\Omega)^n.\tag{17}\] By Lemma 7, they satisfy \(\left \| x \right \|_{\mathfrak X_q} \le C \left \| L_\lambda x \right \|_{\mathfrak Y_q}\) for all \(\lambda\in [0,1]\) and \(x \in \mathfrak X\), with a uniform constant \(C\). As \(L_0: \mathfrak X_q \to \mathfrak Y_q\) is onto by Theorem 8, the map \(L_1\) is also onto by the method of continuity [13]. This means that for every \(f =\mathop{\rm div}\nolimits\mathbb{G}\in \mathfrak Y_q\) there is a solution \((u,\pi)\in \mathfrak X_q\) of 1 , and shows Proposition 3 and Theorem 4 when \(q \ge 2\). ◻

Proof. The proof for \(q<2\) is by duality. Let \(u\) be a given \(q\)-weak solution of 13 with source \(\mathbb{G}\in L^{q}(\Omega)^{n\times n}\). Take any \(\mathbb{F}\in L^{q'}(\Omega)^{n\times n}\). By Proposition 3 and Theorem 4 when \(q \ge 2\), there is solution pair \((v,\pi)\in \mathfrak X_{q'}\) (see 17 ) to the following equation \[\label{eq3467} -\Delta v-b\cdot\nabla v+\nabla \pi =\mathop{\rm div}\nolimits\mathbb{F},\tag{18}\] with \(\left \| \nabla v \right \|_{q'} \le C \left \| \mathbb{F} \right \|_{q'}\). We can use \(u\) as a test function for 18 and get that \[\int_{\Omega}\nabla v:\nabla u-b\cdot\nabla v \cdot u=-\int_{\Omega} \mathbb{F}:\nabla u.\] For the left hand side we can apply equation 13 for \(u\) with test function \(v\) and get that \[\begin{align} \int_{\Omega}\mathbb{F}:\nabla u&= \int_{\Omega}\mathbb{G}:\nabla v \\ &\le \| \mathbb{G}\|_q \| \nabla v\|_{q'} \le \| \mathbb{G}\|_q C\| \mathbb{F}\|_{q'}. \end{align}\] Since \(\mathbb{F}\in L^{q'}\) is arbitrary, this implies a-priori bound of \(u\): \(\|\nabla u\|_{q}\le C\|\mathbb{G}\|_{q}\). ◻

Proof of Proposition 3 and Theorem 4 when \(q< 2\). It is by the same proof for the case \(q>2\), but we use the a priori estimate Lemma 8 instead of Lemma 7. ◻

Proof. (a) Let \(x=(x',x_n)\) be any point in \(\Omega_R\), and \(y=(y',y_n)\) be any point in \(B\). We want to show \((1-t)y + tx \in \Omega_R\) for any \(t \in (0,1)\). We have \((1-t)y + tx \in B_R\) as \(B_R\) is convex. It remains to show \((1-t)y_n + tx_n > \gamma((1-t)y '+ tx' )\). Indeed, \[\begin{align} &(1-t)y_n + tx_n - \gamma((1-t)y '+ tx' )\\ & >(1-t)(1-2\varepsilon) R + t\gamma(x') - [\gamma(x') + L (1-t)|y'-x'|] \\ &\ge (1-t)\left \{ (1-2\varepsilon)R -LR - L(1+\varepsilon)R \right \}, \end{align}\] which is nonnegative if \(\varepsilon\le \frac{1-2L}{2+L}\).

(b) We may assume \(R=1\) by rescaling \(\tilde{f}(x) = f(Rx)\) and \(\tilde{v}(x) = R^{-1}v(Rx)\), noting that 19 and the Lipschitz constant \(L\) are scaling invariant. Let \(\Phi\in C^\infty_c(\mathbb{R}^n)\) be supported in \(|x|\le \frac{1}{2}\) with \(\int_{\mathbb{R}^n} \Phi=1\). Let \(\varepsilon\) and \(B\) be as in Part (a), and \(\phi(x) =\varepsilon^{-n} \Phi(\varepsilon^{-1}(x', x_n - 1+\varepsilon))\), which is supported in \(B\). By Part (a), \(\Omega_R\) is star-shaped with respect to any point in the ball \(B\). We use the Bogovskii formula for \(f \in C^\infty_c(\Omega_R)\) with \(\int_\Omega f=0\): \[\label{eq44623} v(x)=\int_{\Omega} N(x,y)f(y)\,dy, \quad N(x,y)=\frac{x-y}{|x-y|^n} \int_{|x-y|}^\infty \phi\left ( y+r \frac{x-y}{|x-y|} \right )r^{n-1} dr.\tag{20}\] By the properties of the Bogovskii formula (see [1]), we have \(v \in W^{1,q}_0(\Omega_R)\), \(\mathop{\rm div}\nolimits v= f\) and \(\left \| \nabla v \right \|_{L^q} \le C_1 \left \| f \right \|_{L^q}\). Note that formula 20 does not depend on the choice of \(\gamma(x')\), hence nor does the constant \(C_1\). It shows 19 . ◻

Proof. By [1], there exists a unique pair \((v,\pi) \in W^{1,2}_0(\Omega_0)^n\times L^2_0(\Omega_0)\) satisfying \[-\Delta v+\nabla\pi =\mathop{\rm div}\nolimits\mathbb{F},\quad \mathop{\rm div}\nolimits v=0,\quad v|_{\partial\Omega_0}=0,\quad\int_{\Omega_0}\pi=0.\] We take \(v\) as a test function to get that \[\label{vL2bound} \|\nabla v\|_{L^{2}(\Omega_0)}\le \|\mathbb{F}\|_{L^2(\Omega_0)}.\tag{23}\] Here the constant is equal to one and does not depend on the domain. Next, since \(\int_{\Omega_0}\pi=0\), we can apply the Bogovskii map to construct a function \(\phi_0\in W^{1,2}_0(\Omega_0)^n\) \[\label{phi950} \mathop{\rm div}\nolimits\phi_0=\pi, \quad\tag{24}\] with the following \(L^2\) estimate \[\label{phibound} \|\nabla\phi_0\|_{L^2(\Omega_0)}\le c\|\pi\|_{L^2(\Omega_0)},\tag{25}\] where \(c=c(n,L)\) is independent of \(r\) or \(\gamma(x')\). Indeed, in the internal case it follows from the property of the Bogovski map in a ball, and in the boundary case it follows from Lemma 9. We now use \(\phi_0\) as a test function in 21 and get that \[\int_{\Omega_0}\nabla v:\nabla\phi_0-\pi\mathop{\rm div}\nolimits\phi_0~dx=-\int_{\Omega_0}\mathbb{F}: \nabla\phi_0~dx.\] From 24 and 25 we get that \[\begin{align} \|\pi\|_{L^2(\Omega_0)}^2&\le \|\nabla\phi_0\|_{L^2(\Omega_0)}(\|\mathbb{F}\|_{L^2(\Omega_0)}+\|\nabla v\|_{L^2(\Omega_0)}) \\&\le c\|\pi\|_{L^2(\Omega_0)}(\|\mathbb{F}\|_{L^2(\Omega_0)}+\|\nabla v\|_{L^2(\Omega_0)}). \end{align}\] Therefore from 23 we get the pressure estimate \[\|\pi\|_{2,\Omega_0}\le c_1 \|\mathbb{F}\|_{L^2(\Omega_0)},\] with \(c_1=2c(n,L)\). ◻

Proof. For \(x\in \overline{\Omega}\) and \(R>0\), denote \(\Omega_{x,R}=\Omega\cap B(x,R)\). Because \(\partial\Omega\) is Lipschitz and compact, there is a finite number of points \(x_i\in \partial\Omega\), \(i=1,\ldots,N\), and a radius \(R_0\in (0,1]\), such that \(\partial\Omega\) is contained in \(\cup_{i=1}^N \Omega_{x_i,R_0}\), and \[\Omega_{x_i,2R_0}= \{ (x',x_n) \in B_{2R_0}(0) \subset \mathbb{R}^n,\;x_n>\gamma_i(x')\},\] after suitable coordinate rotation and translation, where \(\gamma_i(x')\) is a Lipschitz function defined for \(x'\in B_{2R_0}'(0) \subset \mathbb{R}^{n-1}\) with Lipschitz constant \(L\), and \(\gamma_i(0)=0\), for \(i=1,\ldots,N\).

To prove 27 , we first consider the interior case and take \(B_{\rho}=B_{\rho}(x_0)\subset\Omega\). Let \(\bar{u}=u-k\) where \(k\) is an arbitrary real constant that we will choose later. By Remark 10, we can decompose \(\pi - (\pi)_{B_\rho}=\pi_1+\pi_2+\pi_3\) in \(B_{\rho}\), where \(\pi_1,\pi_2,\pi_3\) are given by \[\nabla\pi_1=\mathcal{W}_{2,B_{\rho}}(\Delta \bar{u}),\quad \nabla\pi_2=\mathcal{W}_{2,B_{\rho}}(-\mathop{\rm div}\nolimits(b\otimes \bar{u})),\quad \nabla\pi_3=\mathcal{W}_{2,B_{\rho}}(\mathop{\rm div}\nolimits\mathbb{G}).\] By Theorem 9, we have the following bound for some global constant \(c_1\) \[\label{wolfbound} \|\pi_1\|_{2,B_{\rho}}\le c_1\|\nabla \bar{u}\|_{2,B_{\rho}},\quad\|\pi_2\|_{2,B_{\rho}}\le c_1 \|b\bar{u}\|_{2,B_{\rho}},\quad \|\pi_3\|_{2,B_{\rho}}\le c_1\|\mathbb{G}\|_{2,B_{\rho}}.\tag{28}\] We may replace \((u,\pi)\) by \((\bar u,\pi-(\pi)_{B_\rho})\) in 1 and multiply the equation by \(\eta^4 \bar{u}\) where \(\eta\) is a smooth cutoff function on \(B_{\rho}\), with \(\eta=1\) in \(B_r\), \(0<r<\rho\). Integrating, we get \[\begin{align} \label{3461} \int_{B_{\rho}} |\nabla (\eta^2\bar{u})|^2 &\le \int_{B_{\rho}} |\bar{u}|^2|\nabla \eta^2|^2+\int_{B_{\rho}} |\bar{u}|^2|b||\nabla \eta|\eta^3\\ &\quad +\int_{B_{\rho}} (|\pi_1|+|\pi_2|+|\pi_3|)|\bar{u}||\nabla\eta|\eta^3+\langle\mathop{\rm div}\nolimits\mathbb{G}, \bar{u}\eta^4\rangle_{B_{\rho}}. \end{align}\tag{29}\] Next we estimate each of the terms on the right hand side separately: \[\begin{align} \int_{B_{\rho}} |\bar{u}|^2|b||\nabla \eta|\eta^3 &\le \frac{c}{\rho-r} \|b\|_{L^{n,\infty}}\|\bar{u}\eta\|_{2,B_{\rho}}\|\bar{u}\eta^2\|_{L^{\frac{2n}{n-2},2}(B_{\rho})} \\ &\le \frac{c}{(\rho-r)^2} \|b\|_{L^{n,\infty}(B_{\rho})}^2\|\bar{u}\eta\|^2_{2,B_{\rho}}+\frac{1}{16}\|\bar{u}\eta^2\|_{W^{1,2}(B_{\rho})}^2. \end{align}\] Pressure terms are estimated in the following way \[\begin{align} \int_{B_{\rho}}|\pi_1||\bar{u}||\nabla\eta|\eta^3 &\le \frac{c}{\rho-r}\|\pi_1\|_{2,B_{\rho}}\|\eta^2 \bar{u}\|_{2,B_{\rho}} \\ &\le \frac{c}{\rho-r}\|\nabla \bar{u}\|_{2,B_{\rho}}\|\eta^2 \bar{u}\|_{2,B_{\rho}}\le \frac{1}{16}\|\nabla \bar{u}\|_{2,B_{\rho}}^2+\frac{c}{(\rho-r)^2}\|\bar{u}\eta^2\|^2_{2,B_{\rho}}, \end{align}\] \[\begin{align} \int_{B_{\rho}}|\pi_2||\bar{u}||\nabla\eta|\eta^3 &\le \|\pi_2\|_{2,B_{\rho}}\|\eta^2\bar{u}\|_{2,B_{\rho}}\le c\|\bar{u}\|_{2,B_{\rho}}\|b\|_{L^{n,\infty}(B_{\rho})}\|\bar{u}\|_{L^{\frac{2n}{n-2},2}(B_{\rho})} \\ &\le C(b)\|\bar{u}\|_{2,B_{\rho}}^2+\frac{1}{16}\|\bar{u}\|^2_{W^{1,2}(B_{\rho})}, \end{align}\] \[\begin{align} \int_{B_{\rho}}|\pi_3||\bar{u}||\nabla\eta|\eta^3 &\le \frac{c}{\rho-r}\|\pi_3\|_{2,B_{\rho}}\|\eta^2 \bar{u}\|_{2,B_{\rho}}\le \frac{c}{\rho-r}\|\mathbb{G}\|_{2,B_{\rho}}\|\eta^2 \bar{u}\|_{2,B_{\rho}} \\ &\le \frac{1}{4} \|\mathbb{G}\|_{2,B_{\rho}}^2+\frac{c}{(\rho-r)^2}\|\eta^2 \bar{u}\|_{2,B_{\rho}}^2. \end{align}\] Next we estimate the term from the right hand side of 1 , \[\begin{align} \langle\mathop{\rm div}\nolimits\mathbb{G}, \bar{u}\eta^4\rangle_{B_{\rho}} &=-\int_{B_{\rho}} \mathbb{G}:\nabla(\bar{u}\eta^4)=-\int_{B_{\rho}} \mathbb{G}: \nabla(\eta^2\bar{u})\eta^2-2\int_{B_{\rho}} \mathbb{G}:\bar{u}\nabla\eta\eta^3 \\ &\le \frac{1}{16}\|\nabla(\eta^2\bar{u})\|^2_{2,B_{\rho}}+\frac{C}{(\rho-r)^2}\|\eta^2 \bar{u}\|_{2,B_{\rho}}^2+ \frac{1}{4}\|\mathbb{G}\|_{2,B_{\rho}}^2. \end{align}\] Combining these estimates with 29 we get the following (with \(c(b)\) depending on \(b\) only through \(\left \| b \right \|_{L^{n,\infty}(\Omega)}\)) \[\int |\nabla (\eta^2\bar{u})|^2\le \frac{1}{4}\|\bar{u}\|^2_{W^{1,2}(B_{\rho})}+\frac{c(b)}{(\rho-r)^2}\|\bar{u}\|_{2,B_{\rho}}^2+\|\mathbb{G}\|_{2,B_{\rho}}^2.\] Finally, we get the inequality for all \(0<r<\rho\le R_0 \le 1\), \[\int_{B_r} |\nabla \bar{u}|^2\le \frac{1}{4}\|\nabla \bar{u}\|^2_{2,B_{\rho}}+\frac{c(b)}{(\rho-r)^2}\|\bar{u}\|_{2,B_{\rho}}^2+\|\mathbb{G}\|_{2,B_{\rho}}^2.\] Here we apply Lemma 3.1 from Giaquinta [40] to remove the first term on the right side: \[\int_{B_r} |\nabla \bar{u}|^2\le c \left ( \frac{c(b)}{(\rho-r)^2}\|\bar{u}\|_{2,B_{\rho}}^2+\|\mathbb{G}\|_{2,B_{\rho}}^2 \right ).\] By taking \(\rho=2r \le R_0\), we get the following inequality \[\int_{B_r} |\nabla \bar{u}|^2\le \frac{c(b)}{r^2}\|\bar{u}\|_{2,B_{2r}}^2+c\|\mathbb{G}\|_{2,B_{2r}}^2.\] Since \(\bar{u}=u-k\) and constant \(k\) was arbitrary we can choose \(k=(u)_{B_{2r}}\) and apply Poincaré inequality to get \[\label{54612} \frac{1}{|B_r|}\int_{B_r} |\nabla \bar{u}|^2 \le c(b) \Big(\frac{1}{|B_{2r}|}\int_{B_{2r}}|\nabla u|^{\frac{2n}{n+2}}\Big)^{\frac{n+2}{n}}+\frac{c}{|B_{2r}|}\|\mathbb{G}\|_{2,B_{2r}}^2.\tag{30}\]

Next we consider the boundary case, firstly for \(\Omega_{\rho}=\Omega\cap B_{\rho}(x_0)\) with \(x_0\) being one of \(x_i\in\partial{\Omega}\), \(i=1,\ldots,N\), \(0<\rho<R_0/6\). We do similar pressure decomposition \(\pi-(\pi)_{\Omega_\rho}=\pi_1+\pi_2+\pi_3\) where \[\nabla\pi_1=\mathcal{W}_{2,\Omega_{\rho}}(\Delta u),\quad \nabla\pi_2=\mathcal{W}_{2,\Omega_{\rho}}(-\mathop{\rm div}\nolimits(b\otimes u)),\quad \nabla\pi_3=\mathcal{W}_{2,\Omega_{\rho}}(\mathop{\rm div}\nolimits\mathbb{G}).\] By Theorem 9, the local pressure projection \(\mathcal{W}_{2,\Omega_{\rho}}\) is well defined, and we have the bound 22 with constant \(c_1\) independent of \(\rho\). Take \(r\le \rho/2\) and \(\eta\) is a cutoff function on \(B_{\rho}(x_0)\) such that \(\eta=1\) on \(B_r(x_0)\). We extend \(\mathbb{G},b,u\) as well as all parts of the pressure \(\pi_k,~k=1,2,3\) with \(0\) to the region \(B_{\rho}(x_0)\setminus\Omega\). Since \(u\) satisfies the boundary condition \(u|_{\partial\Omega}=0\), the function \(\eta^4u\) is still an admissible test function for 1 with \(\pi\) replaced by \(\pi-(\pi)_{\Omega_\rho}\). Therefore we get similar to 29 that \[\begin{align} \int_{B_{\rho}} |\nabla (\eta^2u)|^2 &\le \int_{B_{\rho}} |u|^2|\nabla \eta^2|^2+\int_{B_{\rho}} |u|^2|b||\nabla \eta|\eta^3\\ &\quad +\int_{B_{\rho}} (|\pi_1|+|\pi_2|+|\pi_3|)|u||\nabla\eta|\eta^3-\int_{B_{\rho}} \mathbb{G}:\nabla(u\eta^4). \end{align}\] Due to our zero extension of pressure we have the following estimates \[\begin{align} \|\pi_1\|_{2,B_{\rho}}= \|\pi_1\|_{2,\Omega_{\rho}}\le c\|\nabla u\|_{2,\Omega_{\rho}}=c\|\nabla u\|_{2,B_{\rho}} \\ \|\pi_2\|_{2,B_{\rho}}= \|\pi_1\|_{2,\Omega_{\rho}}\le c\|bu\|_{2,\Omega_{\rho}}=c\|b u\|_{2,B_{\rho}} \\ \|\pi_3\|_{2,B_{\rho}}= \|\pi_1\|_{2,\Omega_{\rho}}\le c\|\mathbb{G}\|_{2,\Omega_{\rho}}=c\|\mathbb{G}\|_{2,B_{\rho}}. \end{align}\] Therefore, we can apply similar approach as in the internal case and get \[\begin{align} \label{54616} \frac{1}{|B_r|}\int_{B_r} |\nabla u|^2 &\le \frac{c(b)}{r^2}\frac{1}{|B_{2r}|}\|u\|_{2,B_{2r}}^2+\frac{c}{|B_{2r}|}\|\mathbb{G}\|_{2,B_{2r}}^2 \\ &\le c(b) \Big(\frac{1}{|B_{2r}|}\int_{B_{2r}}|\nabla u|^{\frac{2n}{n+2}}\Big)^{\frac{n+2}{n}}+\frac{c}{|B_{2r}|}\|\mathbb{G}\|_{2,B_{2r}}^2. \end{align}\tag{31}\] To get the second inequality we have used Poincare inequality, unlike the internal case, using \(u=0\) on the set \(\Omega^c\cap B_{\rho}(x_0)\) of non zero measure, with the ratio \(\frac{|\Omega^c\cap B_{\rho}(x_0)|}{|B_{\rho}(x_0)|}>c\) bounded from below by a constant \(c=c(n,L)\) independent of \(\rho\).

Lastly, for arbitrary ball \(B_{2r}(x_0)\) with \(B_{2r}(x_0)\cap\partial\Omega\neq\emptyset\), there exists \(y_0\in B_{2r}(x_0)\cap\partial\Omega\). Then \[B_r(x_0)\subset B_{3r}(y_0)\subset B_{6r}(y_0)\subset B_{8r}(x_0).\] Using 31 for \(B_{3r}(y_0)\) we get \[\begin{align} \label{54617} \frac{1}{|B_r|}\int_{B_r(x_0)} |\nabla u|^2 &\le \frac{C}{|B_{3r}|}\int_{B_{3r}(y_0)} |\nabla u|^2 \\ &\le c(b) \Big(\frac{1}{|B_{6r}|}\int_{B_{6r}(y_0)}|\nabla u|^{\frac{2n}{n+2}}\Big)^{\frac{n+2}{n}}+\frac{C}{|B_{6r}|}\|\mathbb{G}\|_{2,B_{6r}(y_0)}^2 \\ & \le c(b) \Big(\frac{1}{|B_{8r}|}\int_{B_{8r}(x_0)}|\nabla u|^{\frac{2n}{n+2}}\Big)^{\frac{n+2}{n}}+\frac{C}{|B_{8r}|}\|\mathbb{G}\|_{2,B_{8r}(x_0)}^2. \end{align}\tag{32}\]

Combining the internal estimate 30 and the boundary estimate 32 , we conclude \[\label{54619} \frac{1}{|B_r|}\int_{B_r(x_0)} |\nabla u|^2 \le c(b) \Big(\frac{1}{|B_{8r}|}\int_{B_{8r}(x_0)}|\nabla u|^{\frac{2n}{n+2}}\Big)^{\frac{n+2}{n}}+\frac{C}{|B_{8r}|}\|\mathbb{G}\|_{2,B_{8r}(x_0)}^2,\tag{33}\] for any \(x_0\) in a big cube containing \(\overline{\Omega}\), and any \(r\le R_0/6\). We can now apply Proposition 1.1 of Giaquinta [40] (also see [24]) to get that \(\nabla u \in L^p_\mathrm{loc}\) for \(p \in (2,p_0)\) for some \(p_0 >2\), and \[\label{0619} \left ( \fint_{B_r(x_0)} |\nabla u|^p \right )^{\frac{1}{p}} \le c_1\left ( \fint_{B_{8r}(x_0)} |\nabla u|^2 \right )^{\frac{1}{2}} + c_1\left ( \fint_{B_{8r}(x_0)} |\mathbb{G}|^p \right )^{\frac{1}{p}},\tag{34}\] for all \(x_0 \in \overline{\Omega}\) and \(r\le R_0/6\), with \(p_0\) and \(c_1\) depending only on \(n\), \(c(b)\), and \(R_0\). Summing 34 over a finite cover of \(\overline{\Omega}\) of balls of radius \(R_0/6\), and using the a priori \(W^{1,2}\)-estimate in Lemma 6, we get 27 . ◻

Proof of Theorem 5. First we consider case \(q>2\). Let \(p_0(\Omega,\|b\|_{L^{n,\infty}(\Omega)})>2\) be the constant in Theorem 11 for given \(\Omega\) and \(b\). We may assume \(p_0<n\) (so that we can use 8 for \(q \le p_0\)). Take any \(q\in (2,p_0)\) and \(\mathbb{G}\in L^q(\Omega)^{n\times n}\). By Lemma 6 we know that there exists a weak solution pair \(u\in W^{1,2}(\Omega)\) and \(\pi\in L^2_0(\Omega)\) of 1 . Next we apply Theorem 11 and get that \[\|u\|_{W^{1,q}(\Omega)}\le C(b)\|\mathbb{G}\|_{L^q(\Omega)}.\] This bound also implies uniqueness of the solutions. Since \(u\in W^{1,q}\), \(b\in L^{n,\infty}\), and \(\mathop{\rm div}\nolimits b=0\), we get that \(\mathbb{G}+b\otimes u\in L^q(\Omega)\) by 8 , and we can apply Lemma 5 to get that the pressure \(\pi\) is indeed in \(L^q_0(\Omega)\) and \[\|\pi\|_{L^q(\Omega)}\le C(b)\|\mathbb{G}\|_{L^q(\Omega)}.\] This proves the case \(q>2\).

For case \(q<2\) we will use a duality argument. First, we will prove a priori estimate in \(W^{1,q}\) for \(q\in(p_0',2)\). Suppose \(u\) is a \(q\)-weak solution of 1 with right hand side \(\mathop{\rm div}\nolimits\mathbb{G}\). For any \(\mathbb{G}' \in L^{q'}(\Omega)^{n \times n}\), \(2<q'<p_0\), let \(v\) be the unique \(q'\)-weak solution of the following system \[\label{drifteq95test} -\Delta v -b\cdot\nabla v+\nabla \pi=\mathop{\rm div}\nolimits\mathbb{G}',\quad \mathop{\rm div}\nolimits v=0, \quad v|_{\partial\Omega}=0,\tag{35}\] Therefore, we have an a-priori bound \[\|v\|_{W^{1,q'}(\Omega)}\le C(b)\|\mathbb{G}'\|_{L^{q'}(\Omega)},\] guaranteed by the first part of this theorem. Next we use \(u\) as a test function in equation 1 for function \(v\) and integrate by part to get that \[\begin{align} \int_{\Omega}\mathbb{G}' :\nabla u\,dx&=-\int_{\Omega}(\nabla v+bv):\nabla u \, dx=-\int_{\Omega}(\nabla v-bu):\nabla v \, dx \\ &=\int_{\Omega} \mathbb{G}:\nabla v\,dx\le \|\mathbb{G}\|_{L^{q}(\Omega)}\|\nabla v\|_{L^{q'}(\Omega)}\le C(b) \|\mathbb{G}\|_{L^{q}(\Omega)}\|\mathbb{G}'\|_{L^{q'}(\Omega)}. \end{align}\] Since the matrix function \(\mathbb{G}'\in L^{q'}(\Omega)\) is arbitrary we get that \[\label{smallparpriori} \|u\|_{W^{1,q}(\Omega)}\le C(b) \|\mathbb{G}\|_{L^{q}(\Omega)}.\tag{36}\]

To prove existence we use approximation approach. For \(\mathbb{G}\in L^q(\Omega)\), choose a sequence \(\mathbb{G}_n\in L^2(\Omega)\) such that \(\mathbb{G}_n\rightarrow\mathbb{G}\) in \(L^q(\Omega)\). There exists a sequence of weak \(W^{1,2}\) solutions \(u_n\) of 1 with source term \(\mathop{\rm div}\nolimits\mathbb{G}_n\) instead of \(\mathop{\rm div}\nolimits\mathbb{G}\) and they satisfy 36 , \[\|u_n\|_{W^{1,q}(\Omega)} \le C \|\mathbb{G}_n\|_{L^{q}(\Omega)} \le C\|\mathbb{G}\|_{L^{q}(\Omega)} .\] Therefore there exists \(u\in W^{1,q}(\Omega)\) and a subsequence \(u_n\rightharpoonup u\) in \(W^{1,q}(\Omega)\) weakly, with \(\|u\|_{W^{1,q}(\Omega)} \le C\|\mathbb{G}\|_{L^{q}(\Omega)}\). The last step is passing to the limit and constructing pressure. For any divergence free test function \(\zeta\in C^{\infty}_{c,\sigma}(\Omega)\), we have the weak form 4 for \(u_n\), \[\int_{\Omega} (\nabla u_n - b \otimes u_n): \nabla\zeta = -\int_\Omega\mathbb{G}_n : \nabla\zeta.\] We can pass to the limit \(n\rightarrow\infty\) and get that \[\int_{\Omega} (\nabla u - b \otimes u): \nabla\zeta = -\int_\Omega\mathbb{G}: \nabla\zeta.\] Hence \(u\) is a \(q\)-weak solution of 1 with right side \(\mathop{\rm div}\nolimits\mathbb{G}\) satisfying the weak form 4 , uniqueness follows from the a priori estimate, similarly to the case \(q>2\). Lastly we apply Lemma 5 to prove that there exists pressure \(\pi\in L^q_0(\Omega)\) and \[\|\pi\|_{L^q(\Omega)}\le C(b) \|\mathbb{G}\|_{L^{q}(\Omega)}.\] This finishes the proof of Theorem 5. ◻

Proof of Proposition 6. We follow the same steps of the proof of Theorem 5. Let \(p_0>2\) be the constant from our a priori estimate 40 . Take any \(q\in (2,p_0)\) and \(G\in L^q(\Omega)\). From \(L^2\) theory (Lemma 10) we know that there exists a weak solution \(v\in W^{1,2}(\Omega)\) of 7 . Next we apply 40 and get that \[\|v\|_{W^{1,q}(\Omega)}\le C(b)\|G\|_{L^q(\Omega)}.\] This finishes the proof of (a).

For case (b) we will use a duality argument. First, we will prove an a priori estimate in \(W^{1,q}\) for \(q\in(p_0',2)\). Suppose \(u\) is a \(q\)-weak solution of 6 with right hand side \(F\). For any \(G \in L^{q'}(\Omega)^{n}\), \(2<q'<p_0\), let \(v\) be the unique \(q'\)-weak solution of the dual system 7 with right hand side \(G\) and \[\|v\|_{W^{1,q'}(\Omega)}\le C(b)\|G\|_{L^{q'}(\Omega)},\] guaranteed by the first part of this theorem. We use \(u\) as a test function in equation 7 and use 6 to get that \[\begin{align} \int_{\Omega}G\cdot\nabla u\,dx&=-\int_{\Omega}(\nabla v\cdot\nabla u-(b\nabla v) u )\, dx \\ &=\int_{\Omega} F\cdot\nabla v\,dx\le \|F\|_{L^{q}(\Omega)}\|\nabla v\|_{L^{q'}(\Omega)}\le C(b) \|F\|_{L^{q}(\Omega)}\|G\|_{L^{q'}(\Omega)}. \end{align}\] Since function \(G\in L^{q'}(\Omega)\) is arbitrary we get that \[\label{scalarapri} \|u\|_{W^{1,q}(\Omega)}\le C(b) \|F\|_{L^{q}(\Omega)}.\tag{41}\]

Finally, we prove existence using the approximation approach: For \(F\in L^q(\Omega)\), choose a sequence \(F_n\in L^2(\Omega)\) such that \(F_n\rightarrow F\) in \(L^q(\Omega)\). By Lemma 10, there exists a sequence of weak \(W^{1,2}\) solutions \(u_n\) of 6 with right side \(F_n\) and they satisfy 41 , \[\|u_n\|_{W^{1,q}(\Omega)} \le C \|F_n\|_{L^{q}(\Omega)} \le C\|F\|_{L^{q}(\Omega)} .\] Therefore there exists \(u\in W^{1,q}(\Omega)\) and a subsequence \(u_n\rightharpoonup u\) in \(W^{1,q}(\Omega)\) weakly, with \(\|u\|_{W^{1,q}(\Omega)} \le C\|F\|_{L^{q}(\Omega)}\). The last step is passing to the limit. For any smooth test function \(\zeta\in C^{\infty}_{c}(\Omega)\), we have the weak form of 6 for \(u_n\), \[\int_{\Omega} (\nabla u_n - b u_n)\cdot \nabla\zeta = -\int_\Omega F_n : \nabla\zeta.\] We can pass to the limit \(n\rightarrow\infty\) to show that \(u\) satisfies the weak form of 6 . This gives (b). ◻