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\title[Dirichlet problems with skew-symmetric drift terms] {Dirichlet problems with skew-symmetric drift terms}
\alttitle{Probl\`{e}mes de Dirichlet avec des termes de drift asym\'{e}triques}

\author{\firstname{Lucio} \lastname{Boccardo}}
\address{Istituto Lombardo \& Sapienza Universit\`{a} di Roma, Italy}
\email{boccardo@mat.uniroma1.it}

\author{\firstname{Juan} \lastname{Casado-Diaz}}
\address {Departamento de Ecuaciones Diferenciales y An\'alisis Num\'erico, Universidad de Sevilla, Spain}
\email{jcasadod@us.es}

\author{\firstname{Luigi} \lastname{Orsina}\IsCorresp}
\address{Dipartimento di Matematica, Sapienza Universit\`{a} di Roma, Italy}
\email{orsina@mat.uniroma1.it}

\keywords{\kwd{Singular drift}
\kwd{Dirichlet problems}
\kwd{nonlinear test functions}}

\begin{abstract}
We prove existence of finite energy solutions for a linear Dirichlet problem with a drift and a convection term of the form $A\,E(x)\nabla u + \mathrm{div}(u\,E(x))$, with $A > 0$ and $E$ in $(\elle{r})^{N}$. The result is obtained using a nonlinear function of $u$ as test function, in order to ``cancel'' this term.
\end{abstract}

\begin{altabstract}
Nous prouvons l'existence de solutions d'\'{e}nergie finie pour un probl\`{e}me de Dirichlet lin\'{e}aire avec un terme de la forme $A\,E(x)\nabla u + \mathrm{div}(u\,E(x))$, o\`{u} $A > 0$ et $E$ est dans $(\elle{r})^{N}$. Le r\'{e}sultat est obtenu en utilisant une fonction non lin\'{e}aire de $u$ comme fonction test, afin d'``annuler'' ce terme.
\end{altabstract}


\dateposted{2024-05-02}
\begin{document}
\maketitle

\section{Introduction}

In~\cite{brca} and~\cite{BrGe} is studied the Dirichlet problem
\[
u\in\w:\;-\Delta u + E(x)\cdot \nabla u +\mathrm{div} (u\, E(x)) = f(x),
\]
where
\[
\text{$\Omega$ is a bounded domain of $\RR^{N}$, and the drift $E$ belongs to $(\elle2)^N$},
\]
and
\[
f(x)\in \elle{\frac{2N}{N+2}}.
\]
Remark that if $u$ belongs to $\w$, then the terms $E(x)\cdot\nabla u$ and $\mathrm{div}(u\,E(x))$ just belong to $L^1(\Om)$ and $W^{-1,\frac{N}{N-1}}(\Om)$ (or $W^{-1,p}(\Om)$, $\forall\, p<2$ if $N=2$) respectively. The key point for the existence of finite energy solutions is that the map
\[
P:u\to E(x)\cdot\nabla u+\mathrm{div}(u\,E(x))
\]
is skew-symmetric and thus, it satisfies the ``cancellation'' property
\begin{equation}\label{canc}
\langle Pu,u\rangle=0,
\end{equation}
in the sense of distributions, if $u$ is smooth enough.

The present paper deals with a more general framework, given by
\begin{equation}\label{dir}
u\in\wq:\;-\mathrm{div}(M(x)\du) +A\, E(x)\cdot \nabla u +\mathrm{div} (u\, E(x))=f(x),
\end{equation}
with
\begin{equation}\label{a}
A>0,
\end{equation}
$M :\Omega \to \RR^{N^2}$ a measurable matrix such that there exist $\alpha,\,\beta>0$, satisfying
\begin{equation}\label{alfa}
\alpha |\xi|^2\leq M(x)\xi\cdot\xi,\qquad |M(x)|\leq \beta, \quad \text{a.e. in $\Omega$, }\ \forall \xi \in \RR^{N},
\end{equation}
\begin{equation}\label{defir}
E\in(L^r(\Om))^N:\;\;
\begin{cases}
r= 2 & \text{if }A\geq 1,\\
r=\frac{N(A+1)}{A(N-1)+1} & \text{if }0<A<1,\,N\geq 3,\\
2<r<1+\frac{1}{A} & \text{if }0<A<1,\,N=2,
\end{cases}
\end{equation}
and
\begin{equation}\label{fm}
f\in \elle{m},\quad m>1\ \text{ if }N=2,\quad m=\frac{N(A+1)}{N+2A}\ \text{ if }N\geq 3.
\end{equation}
Note that $r\geq2$, so that $r'\leq2$; and also $r'>1$.

We are going to prove the existence of a distributional solution $u$ of~\eqref{dir} in the Sobolev space $W_0^{1,r'}(\Om)$, with $r'=\frac{r}{r-1}$, the H\"older conjugate exponent of $r$. In particular, $u$ is in $W_0^{1,2}(\Om)$ if $A\geq 1$.

The proof of the result is based on the use of a nonlinear function of $u$ as test function in~\eqref{dir}, in such way that a cancellation property similar to~\eqref{canc} still holds.

We complete this introduction with some references about the existence of solutions for linear elliptic equations with a first order term whose coefficients have poor summability.

For a measurable matrix function $M :\Omega \to \RR^{N^2}$, which satisfies~\eqref{alfa},
\begin{equation}\label{conbEF}
E,F\in
\begin{cases}
(L^N(\Om))^{N} &\text{if }N>2,\\
(L^p(\Om))^{N},\ p>2 & \text{if }N=2,
\end{cases}
\qquad a\in
\begin{cases}
L^\frac{N}{2}(\Om) &\text{if }N>2,\\
L^q(\Om),\ q>1 & \text{if }N=2,
\end{cases}
\end{equation}
such that
\begin{equation}\label{hippos}
-\mathrm{div}(E(x))+a(x)\geq 0\ \text{ in }\Om,
\end{equation}
it has been proved in~\cite{st, GiTr} that the weak maximum principle holds for the equation
\[
{\mathcal L}u=f\ \text{ in }\Om,
\]
with
\begin{equation}\label{defopL}
{\mathcal L}u =-\mathrm{div}\big(M(x)\nabla u - u\,E(x)) + F(x)\cdot\nabla u + a(x)\,u.
\end{equation}
Thus, the corresponding Dirichlet problem has at most one solution in $W^{1,2}_0(\Om)$, for every $f\in W^{-1,2}(\Om)$. By the Fredholm theory, for every $f\in W^{-1,2}(\Om)$ there exists a unique solution $u$ for problems
\begin{equation}\label{problla}
u\in W^{1,2}_0(\Om),\quad {\mathcal L}(u)=f\ \text{ in }\Om,\qquad\quad u\in W^{1,2}_0(\Om),\quad {\mathcal L}^\ast(u)=f\ \text{ in }\Om,
\end{equation}
where
\[
{\mathcal L}^\ast(u)=-\mathrm{div}\,\big(M(x)^t\nabla u + u\,F(x)) - E(x)\cdot\nabla u + a(x)\,u,
\]
is the adjoint operator of ${\mathcal L}.$ We emphasize that these problems are not coercive.

If $E$ or $F$ do not satisfy~\eqref{conbEF}, the operators $u\mapsto -\mathrm{div}(u\,E(x))$ or $u\mapsto F(x)\cdot\nabla u$ do not apply $W^{1,2}_0(\Om)$ into $W^{-1,2}(\Om)$, but some existence results have still been proved.

In~\cite{boc1} and~\cite{boc2}, are considered the cases $F=0$ and $E=0$ respectively, and a Stampacchia--Cald\'eron--Zygmund theory for finite or infinite energy solutions (depending on the summability of $f(x)$) is proved. For example, if $E\in (L^2(\Om))^N$, $F=0$, $f\in L^1(\Om)$, it is proved the existence of a solution for the problem
\begin{equation}\label{pbDg}
{\mathcal L}u=f\ \text{ in }\Om,\quad u=0\ \text{ on }\partial\Om,
\end{equation}
in a very weak sense: the solution $u$ is such that $\log(1+|u|)$ belongs to $\w$ and is an ``entropy solution'' of the equation (for the theory of entropy, or renormalized, solutions, see e.g.~\cite{BBGGPV}, \cite{BocGal}, \cite{DMMuOrPr}) in the sense that:
\[
\into \Big[M(x)\nabla u\cdot\nabla T_{k}(u-\varphi) - u\,E(x)\cdot\nabla T_k(u-\varphi)\Big]\leq \int f(x)\,T_{k}(u-\varphi),
\]
for every $\varphi$ in $\w \cap \elle{\infty}$, and for every $k > 0$. Here
\begin{equation}\label{Tk}
T_k(s)=
\begin{cases}
-k &\text{if }s<-k,\\
s &\text{if }-k\leq s\leq k,\\
k &\text{if }s>k,
\end{cases}
\end{equation}
is the usual truncation function at levels $\pm k$. In~\cite{boc3}, adding a zero order term greater than a positive constant, it has been proved that the above function $u$ is also in $L^1(\Om)$. A duality argument then shows that problem~\eqref{pbDg} has a solution in $W^{1,2}_0(\Om)\cap L^\infty(\Om)$ if $E = 0$, $F$ is in $(L^2(\Om))^N$ and $a$ is greater than a positive constant.


\section{Proof of the main result}

We devote this section to the proof of an existence result for problem~\eqref{dir}. It will be based on the introduction of an approximate problem and then the use of a nonlinear test function, which will provide the estimates needed to pass to the limit.

\begin{theo}\label{tepr}
Assume that $M:\Om\to\RR^{N^2}$ is a measurable matrix which satisfies~\eqref{alfa}, that $A>0$, and that $E$ belongs to $(L^r(\Om))^N$, with $r$ defined by~\eqref{defir}. Then, for every $f\in L^m(\Om)$, with $m$ given by~\eqref{fm}, there exists a distributional solution $u$ of~\eqref{dir} in $W_0^{1,r'}(\Om)$. Moreover, $u$ satisfies
\begin{equation}\label{acpuh1}
|u|^\frac{A-1}{2}u\in \w,\quad u\in
\begin{cases}
L^\frac{N(A+1)}{N-2}(\Om) &\text{if }N\geq 3,\\
L^p(\Om),\ \forall p\in [1,\infty) &\text{if }N=2.
\end{cases}
\end{equation}
\end{theo}

\begin{rema}
We point out a regularizing effect of the problem: for $A\geq 1$, $u$ belongs to $\w$, even if the term $A\, E(x)\cdot \nabla u$ only belongs to $L^1(\Om)$.
\end{rema}

\begin{rema}
Defining for $\mu,\lambda>0$, $\hat E(x)=\lambda E(x)$ and $A=\mu\lambda$, we deduce from Theorem~\ref{tepr} the existence of a distributional solution for problem
\[
u\in W^{1,r'}_0(\Om):\;-\mathrm{div}(M(x)\du) +\lambda\, E(x)\cdot \nabla u +\mu\,\mathrm{div} (u\, E(x))=f(x).
\]
\end{rema}

\begin{proof}[Proof of Theorem~\ref{tepr}] Let $n$ in $\mathbb{N}$: the starting point is the nonlinear Dirichlet problem: $\un\in\w:$
\begin{multline}\label{dirn}
\io M(x)\nabla u_n\cdot\nabla v - \io\frac{u_n}{1+\frac1n|\un|}
\frac{E_n(x)}{1+\frac1n|\nabla\un|}\cdot \nabla v\\
+A\io \frac{1}{1+\frac1n|\un|}\frac{E_n(x)}{1+\frac1n|\nabla\un|}\cdot \nabla u_n\,v =\io f_n(x)\, v,\qquad \forall v\in\w,
\end{multline}
where
\[
E_n(x)=\frac{E(x)}{1+\frac1n|E(x)|},\quad f_n(x)=\frac{f(x)}{1+\frac1n|f(x)|}.
\]
The existence of $\un$ is a consequence of the use of the Schauder fixed point theorem (see also~\cite{boc0}), since all the terms are bounded. Note that $\un$ is the solution of the Dirichlet problem
\[
\un \in \w:\ -\mathrm{div}(M(x)\,\nabla\un) = -\mathrm{div}(G_{n}(x)) + g_{n}(x)\,,
\]
where
\[
G_{n}(x) = \frac{u_n}{1+\frac1n|\un|}
\frac{E_n(x)}{1+\frac1n|\nabla\un|}\,,
\quad g_{n}(x) = f_{n}(x) -A\,\frac{E_n(x)}{1+\frac1n|\un|}\cdot\frac{\nabla\un}{1+\frac1n|\nabla\un|} \,.
\]
Since
\[
|G_{n}(x)| \leq n^{2}\,,
\qquad |g_{n}(x)| \leq n^{2}\,,
\]
from Lax--Milgram theorem and from a result by Stampacchia (see~\cite{st}, Th\'{e}or\`{e}me~4.1), it follows that there exists $C>0$ such that
\[
\norma{\un}{\w} + \norma{\un}{L^\infty(\Om)}
\leq C\,n^2.
\]
Since every $\un$ is a bounded function, it is possible to use a nonlinear composition of $\un$ as test function in~\eqref{dirn}.

In the following, we recall that the Sobolev exponent $2^\ast$ is $2N/(N-2)$, if $N > 2$. For $N = 2$, we define $2^\ast$ as a positive number bigger than 2 to be chosen later.

\begin{proof}[Step 1]
In this step, we assume $ A\geq1$ and we will prove the existence of finite energy solutions.

We use $|\un|^{A-1}\un$ as test function in~\eqref{dirn}, and we have
\begin{multline}\label{30}
A\io M(x)\nabla u_n\cdot\nabla \un \, |\un|^{A-1} - A \io\frac{u_n}{1+\frac1n|\un|}\frac{E_n(x)}{1+\frac1n|\nabla\un|}\cdot \nabla u_n\,|\un|^{A-1}\\
+A\io \frac{\un}{1+\frac1n|\un|}\frac{E_n(x)}{1+\frac1n|\nabla\un|}\cdot \nabla u_n\,|\un|^{A-1} =\io f_n(x)\,|\un|^{A-1}\un.
\end{multline}
Thus, after cancellation of equal terms, we have that
\[
A\io M(x)\nabla u_n\cdot\nabla \un\,|\un|^{A-1} =\io f_n(x)\,|\un|^{A-1}\un,
\]
which, by~\eqref{alfa} and Sobolev's inequality, implies that
\begin{equation}\label{sabato}
\frac{4A\alpha}{(A+1)^2}\Biggl[\io|\un|^{\frac{2^*}{2}(A+1)}\Biggr]^\frac2{2^*}
\leq
\mathcal{S}\,A\,\alpha \io|\nabla u_n|^2 |\un|^{A-1}
\leq \mathcal{S}\,\norma{f}{L^m(\Om)}\Biggl[\io |\un|^{A\,m'}\Biggr]^\frac1{m'}.
\end{equation}
If $N > 3$, the condition $\frac{2^\ast}{2}(A+1)=A\,m'$ holds if $m = \frac{N\,(A+1)}{N + 2A}$ (which is our assumption); thus, from~\eqref{sabato} we obtain that
\begin{equation}\label{stimaleb}
\frac{4A\alpha}{(A+1)^2}\|u_n\|_{L^{\frac{2^\ast}{2}(A+1)}(\Om)}
\leq
\mathcal{S}\,\norma{f}{L^m(\Om)}.
\end{equation}
If $N = 2$, taking into account that $2^\ast$ can be chosen arbitrarily large, we deduce that $m$ can be chosen any number bigger than one in order to have again~\eqref{stimaleb}.

Thus, we proved that
\begin{equation}\label{domenica}
\text{the sequence $\{\un\}$ is bounded in $\elle{\frac{2^\ast}{2}(A+1)}$.}
\end{equation}
Using this result in~\eqref{sabato} yields that
\begin{equation}\label{domenica2}
\text{the sequence $\{|\un|^\frac{A-1}{2} \un\}$ is bounded in $W^{1,2}_0(\Om)$.}
\end{equation}
We now use $T_1(\un)$ as test function in~\eqref{dirn} and we have (thanks to Young's inequality)
\begin{align*}
&\alpha\io|\nabla T_1(\un)|^2 \\
&\quad \leq \io|\un||E(x)||\nabla T_1(\un)| + A\io |E(x)||\nabla u_n| + \io |f(x)|
\\
& \quad\leq (A+1)\io |E(x)||\nabla T_1(\un)| + A\io |E(x)||\nabla T_1(\un)|+\frac{2A}{A+1}\io |E(x)||\nabla (|\un|^\frac{A-1}{2}u_n)| + \io |f(x)|
\\
&\quad \leq \frac\alpha2 \io|\nabla T_1(\un)|^2 + \io |f(x)| + C\left(\io |E(x)|^2 + \io \big|\nabla (|\un|^\frac{A-1}{2}u_n)\big|^2\right)
\end{align*}
which proves that the sequence $\{T_1(u_n)\}$ is bounded in $\w$. Combined with~\eqref{domenica2}, we get that
\begin{equation}\label{giuseppe}
\text{the sequence $\{\un\}$ is bounded in }W^{1,2}_0(\Om).
\end{equation}
The reflexivity of $\w$ and Rellich theorem then imply the existence of a subsequence $\{u_{n_j}\}$ and a function $u$ in $\w$, with $|u|^\frac{A-1}{2}u$ in $\w$, such that
\begin{equation}\label{conunj}
\left\{
\begin{aligned}
u_{n_j}&\rightharpoonup u &&\text{in }\w \vphantom{\frac11}\\
|u_{n_j}|^\frac{A-1}{2}u_{n_j} &\rightharpoonup |u|^{\frac{A-1}{2}}u &&\text{in }\w,\\
u_{n_j}&\rightarrow u &&\text{in }L^\rho(\Om),\ 1\leq\rho< \frac{2^\ast}{2}(A+1).
\end{aligned}\right.
\end{equation}

Note that the boundedness of $\{\un\}$ in $\w$ implies that $\norma{\frac1n \nabla\un}{(\elle{2})^{N}}\to0$ which implies (up to a subsequence) that $\frac1n \nabla\un(x)\to0$ a.e. in $\Om$. Thus we can pass to the limit in~\eqref{dirn} to prove that $u$ is a weak solution of~\eqref{dir}; that is $u$ belongs to $\w$ and is such that
\begin{equation}\label{dirw}
\io M(x)\nabla u\cdot\nabla v -
\io u\,[E(x) \cdot \nabla v] + A\io [E(x) \cdot \nabla u]\,v =
\io f \, v,\qquad \forall v\in C^1_0(\Omega).
\end{equation}
\let\qed\relax
\end{proof}

\begin{proof}[Step 2]
In this step, we assume $0<A<1$ and we will prove the existence of infinite energy solutions.

Since $0<A<1$, we need to modify our test function, and use $[(h+|\un|)^{A}-h^A]\,\mathrm{sign}(\un)$, with $h>0$. Then we have
\begin{multline*}
A\io M(x)\nabla u_n\cdot\nabla \un \,(h+|\un|)^{A-1} - A\io\frac{u_n}{1+\frac1n|\un|} \frac{E_n(x)}{1+\frac1n|\nabla\un|}\cdot \nabla u_n (h+|\un|)^{A-1}
\\
+ A\io \frac{1}{1+\frac1n|\un|}\frac{E_n(x)}{1+\frac1n|\nabla\un|}\cdot \nabla u_n\, [(h+|\un|)^{A}-h^A]\,\mathrm{sign}(\un)
\leq
\io |f| (h+|\un|)^{A}.
\end{multline*}
Since
\[
(h+|\un|)^{A}\mathrm{sign}(\un)=(h+|\un|)^{A-1}\big(h\,\mathrm{sign}(\un)+u_n\big),
\]
the above identity implies that
\begin{align*}
A\io M(x)\nabla u_n\cdot\nabla \un \,(h+|\un|)^{A-1}
+ A\io \frac{1}{1\!+\!\frac1n|\un|}\frac{E_n}{1\!+\!\frac1n|\nabla\un|}\cdot \nabla u_n\, [(h+|\un|)^{A-1}h-h^A]\,\mathrm{sign}(\un)
\\
\leq \io |f(x)| (h+|\un|)^{A}.
\end{align*}
Since $0<A<1$, one has that
\[
(h+|\un|)^{A-1}h\leq h^A,
\]
so that we can use Lebesgue theorem to pass to the limit as $h$ tends to zero in the second term. Using also the monotone convergence theorem in the first one and the Lebesgue theorem in the third one, we get
\begin{equation}\label{san}
A\io \frac{M(x)\nabla u_n\cdot\nabla \un}{|\un|^{1-A}}
\leq\io |f(x)| |\un|^{A}.
\end{equation}
Thus, the inequalities in~\eqref{sabato} still hold. As above, this proves~\eqref{domenica} and~\eqref{domenica2}. For $1\leq r'<2$, H\"older's inequality also gives
\[
\io|\nabla u_n|^{r'}=\io \frac{|\nabla u_n|^{r'}}{|u_n|^\frac{{r'}(1-A)}{2}}|u_n|^\frac{{r'}(1-A)}{2}
\leq
\left(\io \frac{|\nabla u_n|^2}{|u_n|^{1-A}}\right)^\frac{r'}{2}\left(\io |u_n|^\frac{r'(1-A)}{2-r'}\right)^{2-\frac{r'}{2}}.
\]

Since $\frac{{r'}(1-A)}{2-{r'}}=\frac{2^\ast}{2}(1+A)$, in the case $N\geq 3$, from~\eqref{domenica} and~\eqref{domenica2}, it follows that the sequence $\{\un\}$ is bounded in $W^{1,r'}_0(\Om)$. The same result is true in the case $N=2$ if we define $2^\ast>2$ (recall that $1+A<r'<2$ by~\eqref{defir}) by
\[
2^\ast=2\frac{r'(1-A)}{(2-r')(1+A)}.
\]
Thus it is possible to pass to the limit as in the first step to get that $u$ is a solution of~\eqref{dir}.
\end{proof}
\let\qed\relax
\end{proof}

We now prove that, under the assumption that $f(x) \geq 0$, the solution $u$ is not only positive, but cannot be zero in a set of positive measure.

\begin{prop}
If $f(x)\geq0$, then $u(x)\geq0$. Moreover, if $f(x)$ is not identically zero, then $u(x)$ can be zero at most in a set of zero measure.
\end{prop}

\begin{proof}
We give the proof in the case $A \geq 1$; the case $0 < A < 1$ can be proved modifying the test function, as in Step 2 of the proof of Theorem~\ref{tepr}. Choosing $v = |\un|^{A-1}\un^-$ as test function in~\eqref{dirn} we obtain identity~\eqref{30} with $\un^-$ instead of $\un$. Thus, one can cancel two equal terms to obtain that
\[
-A\io M(x)|\un|^{A-1}\nabla u_n^-\cdot\nabla \un^- = A\io M(x)|\un|^{A-1}\nabla u_n\cdot\nabla \un^- =\io f_n(x)\, \un^-|\un|^{A-1}.
\]
Since the right hand side is positive, and the left hand side is negative, one has that $\un^- = 0$, so that $\un \geq 0$. Recalling that $u$ is the limit of the sequence $\{\un\}$, we have proved that $u(x) \geq 0$.

The second statement can be proved exactly as in~\cite[Theorem~3.1 and Theorem~4.1]{bo-icd}.
\end{proof}

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