\section{Introduction}\label{intro}
We consider the electroconvection model 
\begin{equation} \label{intro1}
\partial_t q + u \cdot \nabla q + \l q = 0,
\end{equation}
\begin{equation}  \label{intro2}
\partial_t u + u \cdot \nabla u + \nabla p - \Delta u = -qRq,
\end{equation} 
\begin{equation} \label{intro3}
\nabla \cdot u = 0
\end{equation}
in $\mathbb R^2$ describing the evolution of a surface charge density $q$ in a two-dimensional incompressible fluid flowing with a velocity $u$ and a pressure $p$. Here $\l = (-\Delta)^{\frac{1}{2}}$ is the square root of the two-dimensional Laplacian, and $R = \nabla \l^{-1}$ is the two-dimensional Riesz transform. The initial data 
\begin{equation} 
q(\cdot, 0) = q_0
\label{qinit}
\end{equation}
and
\begin{equation}
u(\cdot, 0) = u_0
\label {uinit}
\end{equation}
are assumed to be regular enough and have good decay properties.
The model is motivated by physical and numerical studies of electroconvection
\cite{DDMB,DMT,DDMT}. The nonlocal aspect of the evolution of the charge density
and the nonlocal forcing on the Navier-Stokes equations in the model are due to the fact that the fluid and charges are confined to a thin two dimensional film. The global well-posedness of the system in bounded domains was obtained in \cite{ceiv} using commutator estimates and nonlocal nonlinear analysis.
In \cite{AI}, we investigated the long time dynamics of the model in two dimensions, with periodic boundary conditions and with applied voltage. When the fluid is forced by time-independent smooth mean zero body forces, we proved that the model \eqref{intro1}--\eqref{qinit} has a finite dimensional global attractor. In the absence of body forces, the charge density $q$ converges exponentially in time to a unique limit due to the applied voltage, and the velocity $u$ converges exponentially in time to zero. The rate of exponential decay depends  on the periodic boundary conditions

In this paper, we consider the time asymptotic behavior of solutions of \eqref{intro1}--\eqref{qinit} in ${\mathbb R}^2$, and adapt the Fourier splitting method \cite{S,S1} of Schonbek to the present system. The method was initially used in \cite{S} to prove decay of Leray weak solutions \cite{L} of Navier-Stokes equations and to further decay studies for Navier-Stokes equations \cite{BN,KM,S1,SW,W1} and 
many other partial differential equations (see for instance \cite{BS,CW,DL,NS,Z,ZL}). Different approaches were employed as well to investigate the time decay \cite{OT} and space-time decay  \cite{AGSS, KU1, KU,T} of higher-order derivatives of solutions to Navier-Stokes equations.   








The electroconvection model \eqref{intro1}--\eqref{qinit} couples Navier-Stokes equations to a scalar equation for a surface charge density $q$, evolving via advection by $u$ and diffusion by $\Lambda $.  We obtain in Theorem~\ref{decay} of section \ref{s2} the long time $L^2$ decay of the type
\[
\|q\|_{L^2} = O(t^{-1})
\]
and
\[
\|u\|_{L^2}  = O(t^{-\frac{1}{2}}).
\]
This rate of decay is sharp for the linear uncoupled system if the initial data have non vanishing finite $L^1$ norms, because functions of the form  $Q(t) = e^{-t\l^\alpha}q_0$  obey
\[
\lim_{t\to\infty} t^{\frac{n}{\alpha}}\|Q(t)\|_{L^2({\mathbb R}^n)}^2 = C_{n,\alpha} \left(\int_{{\mathbb R}^n} q_0 dx\right)^2
\]
for any $\alpha>0$ and $n\ge 1$.
The fact that such a decay is sharp for the nonlinear evolution as well is a consequence of Theorem~\ref{nonlin} of section \ref{s3} where we prove that $u- U$  with $U(t) = e^{t\Delta}u_0$ and $q- Q$ with $Q(t) = e^{-t\l}q_0$ decay faster in $L^2$ than $u$ and $q$, respectively. Similar results were proved for critical SQG in \cite{CW}. The critical SQG velocity $u= R^{\perp}q$ decays in $L^2$ like $q$, that is at the rate $t^{-1}$, which helps lower the size of the nonlinear term $u\cdot\nabla q$ in that equation. In our case, the velocity has slower decay in $L^2$ due to the Navier-Stokes equation, namely of the order $t^{-\frac{1}{2}}$, and the nonlinear term is larger. The influence of the charge density $q$ is felt by the Navier-Stokes velocity via the electric force $-qRq$. In order to obtain a key fast enough decay at low wave numbers for the difference 
$v= u- U$, we need to control a moment of $q$,  $\int_{{\mathbb R}^2}|x|^2|q(x,t)|^2dx = M^2(t)$, in view of the inequality
\[
|\widehat{qRq}(\xi)| \le C |\xi| \|q\|_{L^2}M(t)
\]
(see Lemma \ref{lem3}).  We prove that 
\[
M(t) = O(\sqrt{\log t})
\]
for long time, by analyzing the evolution of the quantity $a(x)q(x,t)$ with $a(x)=\sqrt{|x|^2 +1}$. This analysis uses the boundedness of the commutator between $\Lambda$ and multiplication by $a(x)$, which we establish in Lemma \ref{lem1}. In addition, in order to achieve the necessary sharp $L^2$ bounds we have to obtain bounds for the decay of higher norms of both $u$ and $q$.  For instance,  $H^1$
 norms of $q$ are of the order 
\[
\|\nabla q\|_{L^2} = O(t^{-1}).
\]
These bounds are obtained by somewhat involved nonlinear and nonlocal analysis, and they are no longer sharp compared to the generic $t^{-2}$ linear behavior.

The paper is organized as follows. In section \ref{s2}, we study the asymptotic behavior of solutions to the electroconvection model \eqref{intro1}--\eqref{qinit}: we prove that the $L^2$ norm of the surface charge density $q$ decays in time to zero with a rate of order $t^{-1}$ whereas the velocity $u$ decays in time to zero with a rate of order  $t^{-\frac{1}{2}}$. We also investigate the rate of decay of their higher-order derivatives, and we obtain decaying-in-time bounds in H\"older spaces $C^{0, \frac{1}{2}}$. In section \ref{s3}, we prove that the differences $q-Q$ and $u - U$ decay to zero in $L^2$ faster than $q$ and $u$, with rates of order $t^{-1- \frac{3}{4}}$ and $t^{-\frac{3}{4}}$, respectively. In the Appendix, we present results on the existence and uniqueness of solutions to \eqref{intro1}--\eqref{qinit}, based on the Banach fixed point theorem, the Aubin-Lions lemma and commutator estimates. 
     




\section{Long Time Behavior of Solutions} \label{s2}

In this section, we consider the long-time behavior of solutions of the electroconvection model described by \eqref{intro1}--\eqref{qinit}. We show that the charge density $q$ and the velocity $u$ converge to $0$ in the $H^2$ norm, and we investigate the rate of convergence.  

For a function $f \in L^1({\mathbb R}^2)$, we denote its Fourier transform by 
\begin{equation} 
\widehat{f} (\xi) = \int_{{\mathbb R}^2} f(x) e^{- i \xi \cdot x } dx.
\end{equation}

\begin{thm} \label{decay} Let $u_0 \in H^1 \cap L^1$ be divergence-free and $q_0 \in L^4 \cap L^1$. There exist positive constants $\Gamma_0$ and $\Gamma_0'$ depending only on the initial data and some universal constants such that the unique global-in-time solution $(q,u)$ of \eqref{intro1}--\eqref{qinit} obeys
\begin{equation} \label{ltime1}
\|q(t)\|_{L^2}^2 \le \frac{\Gamma_0}{(t + 1)^{2}}
\end{equation} and 
\begin{equation} \label{ltime2}
\|u(t)\|_{L^2}^2 \le \frac{\Gamma_0'}{t+1}
\end{equation}
for all $t \ge 0$. 
\end{thm}

\textbf{Proof:} The proof is divided into several steps.

\textit{\textbf{Step 1} (Basic energy estimates}).
We take the $L^2$ inner product of equation \eqref{intro1} with $\l^{-1} q$ and the $L^2$ inner product of equation \eqref{intro2} with $u$. Then we add the resulting energy equalities. Integrating by parts, we have the cancellations 
\begin{equation} 
(u \cdot \nabla u, u)_{L^2} = (\nabla p, u)_{L^2} = 0
\end{equation} and 
\begin{align}
(u \cdot \nabla q, \l^{-1} q)_{L^2} + (qRq, u)_{L^2} 
&= - (u \cdot \nabla \l^{-1} q, q)_{L^2} + (qRq, u)_{L^2} \nonumber
\\&= - (u \cdot Rq, q)_{L^2} + (qRq, u)_{L^2}
= 0
\end{align} due to the divergence-free condition \eqref{intro3}.
Thus, we obtain 
\begin{equation} \label{weaknormb}
\frac{1}{2} \frac{d}{dt} \left(\|\l^{-\frac{1}{2}}q\|_{L^2}^2 + \|u\|_{L^2}^2  \right)
+ \|q\|_{L^2}^2 + \|\nabla u\|_{L^2}^2
= 0.
\end{equation} We integrate in time from $0$ to $t$ and we take the supremum over all positive times $t \ge 0$. We get 
\begin{equation}  \label{step11}
\sup\limits_{t \ge 0} \left\{\|\l^{-\frac{1}{2}} q(t)\|_{L^2}^2 + \|u(t)\|_{L^2}^2
+ \int_{0}^{t} 2\left(\|q(s)\|_{L^2}^2 + \|\nabla u(s)\|_{L^2}^2 \right) ds  \right\}
= \|\l^{-\frac{1}{2}} q_0\|_{L^2}^2 + \|u_0\|_{L^2}^2
\end{equation} ending the proof of Step 1.

\textit{\textbf{Step 2} (Pointwise bounds for the Fourier transform of the charge density $q$).}
The Fourier transform of $q$ evolves according to 
\begin{equation} 
\partial_t \widehat{q} (\xi, t) + (\widehat{u \cdot \nabla q})(\xi, t) + \widehat{\l q} (\xi, t) = 0.
\end{equation} The fractional Laplacian $\l$ is a Fourier multiplier with symbol $|\xi|$, hence 
\begin{equation} 
\partial_t \widehat{q} + |\xi| \widehat{q} = - \widehat{u \cdot \nabla q}.
\end{equation} We estimate the Fourier transform of the nonlinear term 
\begin{equation} 
|\widehat{u \cdot \nabla q}| = |\widehat{\nabla \cdot (uq)}| \le C|\xi| \|u\|_{L^2} \|q\|_{L^2}
\end{equation} using the divergence-free condition \eqref{intro3}, the boundedness of the Fourier transform of a function by its $L^1$ norm, and the Cauchy-Schwarz inequality. 
This yields the differential inequality 
\begin{equation} 
\partial_t \widehat{q} + |\xi| \widehat{q}  \le C|\xi| \|u\|_{L^2} \|q\|_{L^2}.
\end{equation} We multiply both sides by the integrating factor $e^{|\xi|t}$ and integrate in time from $0$ to $t$. We obtain the bound
\begin{equation} 
|\widehat{q} (\xi, t)| \le |\widehat{q}_0 (\xi)| + C|\xi| \int_{0}^{t} \|u(s)\|_{L^2}\|q(s)\|_{L^2} ds. 
\end{equation} As a consequence of Step 1 and the Cauchy-Schwarz inequality, we get the pointwise bound
\begin{equation} \label{EQ1}
|\widehat{q} (\xi, t)| \le \|q_0\|_{L^1} + C_0|\xi|\sqrt{t} 
\end{equation} where $C_0$ is a time-independent constant depending only on $\|u_0\|_{L^2}$ and $\|\l^{-\frac{1}{2}}q_0\|_{L^2}$. This finishes the proof of Step 2.
 

\textit{\textbf{Step 3} (Decaying bound for the $L^2$ norm of the charge density).}
The $L^2$ norm of $q$ evolves according to 
\begin{equation} \label{step22}
\frac{1}{2} \frac{d}{dt} \|q\|_{L^2}^2 + \|\l^{\frac{1}{2}} q\|_{L^2}^2 = 0.
\end{equation} In view of Parseval's identity and the fact that $\l^{\frac{1}{2}}$ is a Fourier multiplier with symbol $|\xi|^{\frac{1}{2}}$, we have 
\begin{equation} 
\|\l^{\frac{1}{2}} q\|_{L^2}^2 = \|\widehat{\l^{\frac{1}{2}}q}\|_{L^2}^2 = \int_{\mathbb R^2} |\xi| |\widehat{q}(\xi,t)|^2 d\xi. 
\end{equation} We bound the dissipation from below 
\begin{equation} 
\int_{\mathbb R^2} |\xi| |\widehat{q}(\xi,t)|^2 d\xi 
\ge \int_{|\xi| > \rho(t)} |\xi| |\widehat{q}(\xi,t)|^2 d\xi 
\end{equation} where $\rho(t)$ is the function defined on $[0, \infty)$ by 
\begin{equation} 
\rho(t) = \frac{r}{2(t+1)}
\end{equation} for some positive constant $r$ to be determined later. We note that 
\begin{align} 
\int_{|\xi| > \rho (t)} |\xi| |\widehat{q}(\xi,t)|^2 d\xi 
&\ge \rho(t) \int_{|\xi| > \rho(t)} |\widehat{q}(\xi, t)|^2 d\xi \nonumber
\\&= \rho(t) \int_{\mathbb R^2} |\widehat{q}(\xi, t)|^2 d\xi - \rho(t) \int_{|\xi| \le \rho(t)} |\widehat{q}(\xi, t)|^2 d\xi \nonumber
\\&= \rho(t) \|q\|_{L^2}^2 - \rho(t) \int_{|\xi| \le \rho(t)} |\widehat{q}(\xi, t)|^2 d\xi
\end{align} where we used Parseval's identity. 
Consequently, we obtain the energy inequality
\begin{equation} 
\frac{d}{dt} \|q\|_{L^2}^2 + 2\rho(t) \|q\|_{L^2}^2 \le 2\rho(t) \int_{|\xi| \le \rho(t)} |\widehat{q}(\xi, t)|^2 d\xi.
\end{equation}
By the pointwise bound \eqref{EQ1} and Fubini's theorem for spherical coordinates, we estimate
\begin{align}
\int_{|\xi| \le \rho(t)} |\widehat{q}(\xi, t)|^2 d\xi
&\le \int_{|\xi| \le \rho(t)} \left(\|q_0\|_{L^1} + C_0|\xi|\sqrt{t} \right)^2 d\xi
= C\int_{0}^{\rho(t)} r\left(\|q_0\|_{L^1} + C_0 r\sqrt{t} \right)^2 dr \nonumber
\\&\le C\int_{0}^{\rho(t)} r\left(\|q_0\|_{L^1}^2 + C_0^2 r^2 t  \right) dr
\le \Gamma_1 \left(\rho(t)^2 + t\rho(t)^4 \right) 
\end{align} where $\Gamma_1$ depends only on the initial data. 
We obtain 
\begin{equation} 
\frac{d}{dt} \|q\|_{L^2}^2 + 2\rho(t) \|q\|_{L^2}^2 \le 2\Gamma_1 (\rho(t)^3 + t\rho(t)^5) 
\end{equation} for all $t \ge 0$. We multiply both sides of the inequality by the integrating factor 
\begin{equation} 
e^{2\int_{0}^{t} \rho(s) ds} = e^{r \int_{0}^{t} \frac{1}{s+1} ds} = e^{r \ln(t+1)} = (t+1)^{r}
\end{equation} and then we integrate in time from $0$ to $t$. We get
\begin{align} 
\|q(t)\|_{L^2}^2 
&\le \frac{\|q_0\|_{L^2}^2}{(t+1)^{r}} + \frac{\Gamma_2}{(t+1)^{r}} \int_{0}^{t} \left(\frac{1}{(s+1)^3} + \frac{1}{(s+1)^4}\right) (s+1)^{r} ds \nonumber
\\&\le \frac{\|q_0\|_{L^2}^2}{(t+1)^r} + \frac{\Gamma_2}{(t+1)^{r}} \left(\frac{(t+1)^{r-2}}{r-2} - \frac{1}{r-2} + \frac{(t+1)^{r-3}}{r-3} - \frac{1}{r-3} \right) \nonumber
\\&\le \frac{\|q_0\|_{L^2}^2}{(t+1)^{r}} + \frac{\Gamma_2}{(r-2)(t+1)^{2}} + \frac{\Gamma_2}{(r-3)(t+1)^{3}} \label{step33}
\end{align} for any $r > 3$. Here $\Gamma_2$ depends on $r$ and the initial data. We choose $r=4$ and we obtain the bound
\begin{equation} \label{EQ3}
\|q\|_{L^2}^2 \le \frac{\Gamma_3}{(t+1)^2}
\end{equation} where $\Gamma_3$ is a positive constant depending only on the initial data. This completes the proof of \eqref{ltime1} and Step~3. 

\textit{\textbf{Step 4} (Pointwise bounds for the Fourier transform of the velocity $u$).}
Applying the Leray Projector $\mathbb P$ to equation \eqref{intro2}, we have 
\begin{equation} 
\partial_t u + \mathbb P(u \cdot \nabla u) - \Delta u = -\mathbb P(qRq),
\end{equation} where we used the incompressibility condition \eqref{intro3} and the fact that $\mathbb P$ and $-\Delta$ are Fourier multipliers so they commute. 
Hence the Fourier transform of $u$ obeys 
\begin{equation} 
\partial_t \widehat{u} + \widehat{ \mathbb P(u \cdot \nabla u)} - \widehat{\Delta u} = -\widehat{\mathbb P(qRq)}.
\end{equation} 
We estimate 
\begin{equation} 
|\widehat{ \mathbb P(u \cdot \nabla u)}(\xi, t)| \le C|\xi||\widehat{u}(\xi,t)|^2 \le C|\xi|\|u(t)\|_{L^2}^2 
\end{equation} and 
\begin{equation} 
|\widehat{\mathbb P(qRq)}(\xi,t)| \le C\|(qRq)(t)\|_{L^1} \le C\|q(t)\|_{L^2}^2 
\end{equation} in view of the boundedness of the Riesz transforms on $L^2({\mathbb R}^2)$.
We obtain 
\begin{equation} 
\partial_t \widehat{u} + |\xi|^2 \widehat{u} \le C|\xi|\|u\|_{L^2}^2 + C\|q\|_{L^2}^2
\end{equation} and hence
\begin{equation} 
|\widehat{u}(\xi,t)| \le \|u_0\|_{L^1} + C|\xi| \int_{0}^{t} \|u(s)\|_{L^2}^2 ds + C\int_{0}^{t}\|q(s)\|_{L^2}^2 ds
\end{equation} for all $\xi \in {\mathbb R}^2$ and $t \ge 0$. 
In view of the bound \eqref{step11}, we get 
\begin{equation} \label{EQ2}
|\widehat{u}(\xi,t)| \le \Gamma_4  + C|\xi| \int_{0}^{t} \|u(s)\|_{L^2}^2 ds 
\end{equation} where $\Gamma_4$ is a positive constant depending only on the initial data. This completes the proof of Step~4.

\textit{\textbf{Step 5} (Decaying bounds for the $L^4$ norm of $q$).} The evolution of the $L^4$ norm of $q$ is described by the energy equality
\begin{equation} 
\frac{1}{4} \frac{d}{dt} \|q\|_{L^4}^4 + \int_{\mathbb R^2} q^3 \l q dx = 0.
\end{equation} In view of the C\'ordoba-C\'ordoba inequality \cite{CC}, the dissipation is bounded from below 
\begin{equation} 
\int_{\mathbb R^2} q^3 \l q dx 
\ge \frac{1}{2} \|\l^{\frac{1}{2}}(q^2)\|_{L^2}^2 
\end{equation} and thus
\begin{equation} 
\int_{\mathbb R^2} q^3 \l q dx  \ge c\|q\|_{L^8}^4 
\end{equation} due to Gagliardo-Nirenberg inequalities. Using interpolation inequalities in $L^p$ spaces and the uniform boundedness of the $L^2$ norm of the charge density $q$ by $\|q_0\|_{L^2}$, we have the bound
\begin{equation} 
\|q\|_{L^4} \le \|q\|_{L^2}^{\frac{1}{3}}  \|q\|_{L^8}^{\frac{2}{3}}\le \|q_0\|_{L^2}^{\frac{1}{3}} \|q\|_{L^8}^{\frac{2}{3}}
\end{equation} from which we conclude that
\begin{equation} 
\int_{\mathbb R^2} q^3 \l q dx \ge C \|q_0\|_{L^2}^{-2} \|q\|_{L^4}^6
\end{equation} and hence
\begin{equation} 
\frac{d}{dt} \|q\|_{L^4}^4 + \frac{C}{\|q_0\|_{L^2}^2} \|q\|_{L^4}^6 \le 0.
\end{equation} Letting $y = \|q\|_{L^4}^4$, we obtain the Bernouilli ordinary differential inequality 
\begin{equation} 
\frac{dy}{dt}  + \frac{C}{\|q_0\|_{L^2}^2} y^{\frac{3}{2}} \le 0.
\end{equation} We apply a change of variable given by $u = y^{-\frac{1}{2}}$ and we get
\begin{equation} 
\frac{-2}{u^3} \frac{du}{dt}  + \frac{C}{\|q_0\|_{L^2}^2} \frac{1}{u^3} \le 0
\end{equation} so
\begin{equation} 
\frac{du}{dt} \ge \frac{C}{2\|q_0\|_{L^2}^2}.
\end{equation} Integrating in time from $0$ to $t$, we arrive at the bound
\begin{equation} 
\|q\|_{L^4}^{-2} \ge \|q_0\|_{L^4}^{-2} + \frac{C}{\|q_0\|_{L^2}^2}t \ge \Gamma_5(1+t) 
\end{equation} where $\Gamma_5$ is a constant depending on the initial data. Consequently, we obtain 
\begin{equation} \label{Bernouilli}
\|q\|_{L^4} \le \frac{1}{\sqrt{\Gamma_5}} \frac{1}{(1+t)^{\frac{1}{2}}}
\end{equation} for all $t \ge 0$.

\textit{\textbf{Step 6} (Decaying bound for the $L^2$ norm of the velocity).}
The $L^2$ norm of the velocity evolves according to 
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2 = -\int_{{\mathbb R}^2} qRq \cdot u dx.
\end{equation}
In view of H\"older's inequality, the boundedness of the Riesz transforms on $L^4({\mathbb R}^2)$, and Ladyzhenskaya's interpolation inequality, we bound
\begin{equation} 
\left|\int_{{\mathbb R}^2} qRq \cdot u \right| \le C\|q\|_{L^2}\|q\|_{L^4}\|u\|_{L^2}^{\frac{1}{2}}\|\nabla u\|_{L^2}^{\frac{1}{2}}
\end{equation} yielding 
\begin{equation} 
 \frac{d}{dt} \|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2 \le C\|q\|_{L^2}^{\frac{4}{3}} \|q\|_{L^4}^{\frac{4}{3}} \|u\|_{L^2}^{\frac{2}{3}}.
\end{equation} By Parseval's identity, we have
\begin{equation} 
\frac{d}{dt} \|u\|_{L^2}^2 + \int_{\mathbb R^2} |\xi|^2 |\widehat{u}(\xi,t)|^2 d\xi \le C\|q\|_{L^2}^{\frac{4}{3}} \|q\|_{L^4}^{\frac{4}{3}} \|u\|_{L^2}^{\frac{2}{3}}.
\end{equation} For a positive function $\rho_1(t)$ continuous on $[0, \infty)$, we have  
\begin{align}
\int_{\mathbb R^2} |\xi|^2 |\widehat{u}(\xi,t)|^2 d\xi
&\ge \int_{|\xi| > \rho_1(t)} |\xi|^2 |\widehat{u}(\xi,t)|^2 d\xi
\ge \rho_1(t)^2 \int_{|\xi| > \rho_1(t)} |\widehat{u}(\xi,t)|^2 d\xi \nonumber
\\&\ge \rho_1(t)^2 \left(\int_{\mathbb R^2} |\widehat{u}(\xi,t)|^2 d\xi - \int_{|\xi| \le \rho_1(t)}  |\widehat{u}(\xi,t)|^2 d\xi \right) \nonumber
\\&= \rho_1(t)^2 \|u\|_{L^2}^2 - \rho_1(t)^2 \int_{|\xi| \le \rho_1(t)}  |\widehat{u}(\xi,t)|^2 d\xi.
\end{align} 
Consequently, we obtain the energy inequality 
\begin{equation} 
\frac{d}{dt} \|u\|_{L^2}^2 + \rho_1(t)^2 \|u\|_{L^2}^2 \le C\|q\|_{L^2}^{\frac{4}{3}} \|q\|_{L^4}^{\frac{4}{3}} \|u\|_{L^2}^{\frac{2}{3}} + \rho_1(t)^2\int_{|\xi| \le \rho_1(t)}  |\widehat{u}(\xi,t)|^2 d\xi.
\end{equation}
Using \eqref{EQ2}, we have 
\begin{align} 
&\int_{|\xi| \le \rho_1(t)}  |\widehat{u}(\xi,t)|^2 d\xi
\le C\int_{0}^{\rho_1(t)} r \left(\Gamma_4^2 + C r^2 \left\{\int_{0}^{t} \|u(s)\|_{L^2}^2 ds \right\}^2 \right) dr \nonumber
\\&\le \Gamma_6 \rho_1(t)^2 +  C\rho_1(t)^4 \left(\int_{0}^{t} \|u(s)\|_{L^2}^2 ds  \right)^2 
\end{align} and thus
\begin{align} 
\frac{d}{dt} \|u\|_{L^2}^2 + \rho_1(t)^2 \|u\|_{L^2}^2 \le  \Gamma_6 \rho_1(t)^4 + C\rho_1(t)^6  \left(\int_{0}^{t} \|u(s)\|_{L^2}^2 ds \right)^2 + C\|q\|_{L^2}^{\frac{4}{3}} \|q\|_{L^4}^{\frac{4}{3}} \|u\|_{L^2}^{\frac{2}{3}}.
\end{align}
Multiplying by the integrating factor $e^{\int_{0}^{t} \rho_1(s)^2 ds}$, and integrating in time from $0$ to $t$, we obtain
\begin{align} 
\|u(t)\|_{L^2}^2 
&\le \frac{\|u_0\|_{L^2}^2}{e^{\int_{0}^{t} \rho_1(s)^2 ds}}
+ \frac{\Gamma_{6}}{e^{\int_{0}^{t} \rho_1(s)^2 ds}} \int_{0}^{t} e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau} \rho_1(s)^4 ds  \nonumber
\\&+ \frac{C}{e^{\int_{0}^{t} \rho_1(s)^2 ds}} \int_{0}^{t} \left(e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau}\rho_1(s)^6\right) \left(\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \right)^2 ds  \nonumber
\\&+ \frac{C}{e^{\int_{0}^{t} \rho_1(s)^2 ds}} \int_{0}^{t} \|q\|_{L^2}^{\frac{4}{3}} \|q\|_{L^4}^{\frac{4}{3}} \|u\|_{L^2}^{\frac{2}{3}} e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau} ds.  
\end{align} 
In view of
\eqref{step11}, \eqref{EQ3} and \eqref{Bernouilli}, we estimate 
\begin{equation} 
 \int_{0}^{t} \|q\|_{L^2}^{\frac{4}{3}} \|q\|_{L^4}^{\frac{4}{3}} \|u\|_{L^2}^{\frac{2}{3}} e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau} ds
\le \Gamma_7 \int_{0}^{t} \frac{e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau}}{(s+1)^{2}} ds 
\end{equation} for any $t \ge 0$, and so
\begin{align} \label{abstractbound}
\|u(t)\|_{L^2}^2 
&\le \frac{\|u_0\|_{L^2}^2}{e^{\int_{0}^{t} \rho_1(s)^2 ds}}
+ \frac{\Gamma_{6}}{e^{\int_{0}^{t} \rho_1(s)^2 ds}} \int_{0}^{t} e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau} \rho_1(s)^4 ds  \nonumber
\\&+ \frac{C}{e^{\int_{0}^{t} \rho_1(s)^2 ds}} \int_{0}^{t} \left(e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau}\rho_1(s)^6\right) \left(\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \right)^2 ds   \nonumber
\\&+  \frac{\Gamma_7}{e^{\int_{0}^{t} \rho_1(s)^2 ds}} \int_{0}^{t} \frac{e^{\int_{0}^{s} \rho_1(\tau)^2 d\tau}}{(s+1)^{2}} ds 
\end{align} for any $t \ge 0$.

In order to obtain the sharp decaying bound for the velocity $u$, we need the following three sub-steps: 

\textit{\textbf{Step 6.1} (Logarithmic decaying bound for the $L^2$ norm of the velocity).} We take $\rho_1(t) = (e + t)^{-\frac{1}{2}} \left[\ln (e + t)\right]^{-\frac{1}{2}}$. In this case, the integrating factor is given by 
\begin{equation} 
e^{\int_{0}^{t} \rho_1(s)^2 ds} = e^{\int_{0}^{t} \frac{1}{(e+s)\ln(e+s)} ds} = e^{\ln \left[\ln (e+t) \right]} = \ln (e+t)
\end{equation} and so \eqref{abstractbound} becomes
\begin{align}
&\|u(t)\|_{L^2}^2 
\le \frac{\|u_0\|_{L^2}^2}{\ln (e+t)}
+ \frac{\Gamma_{6}}{\ln (e +t)} \int_{0}^{t} \frac{1}{(e+s)^2 \ln (e+s)} ds  \nonumber
\\&+ \frac{C \|u_0\|_{L^2}^2}{\ln (e +t)} \int_{0}^{t} \frac{s^2}{(e+s)^3 \left[\ln (e+s)\right]^2} ds  
+  \frac{\Gamma_7}{\ln (e+t)} \int_{0}^{t} \frac{\ln (e+s)}{(s+1)^{2}} ds. 
\end{align} in view of the uniform boundedness of $\|u\|_{L^2}$ by $\|u_0\|_{L^2}$. We note that 
\begin{equation} 
\int_{0}^{t} \frac{s^2}{(e+s)^3 \left[\ln (e+s)\right]^2} ds 
\le \int_{0}^{t} \frac{1}{(e+s) \left[\ln (e+s) \right]^2} ds
= 1 - \frac{1}{\ln (e+t)} \le 1
\end{equation} for any $t \ge 0$. Therefore, 
\begin{equation} \label{EQ4}
\|u(t)\|_{L^2}^2 \le \frac{\Gamma_8}{\ln (e+t)}
\end{equation} for all $t \ge 0$, where $\Gamma_8$ is a constant depending only on the initial data. 

\textit{\textbf{Step 6.2} (Almost sharp decaying bound for the $L^2$ norm of the velocity).} In order to improve the logarithmic decay \eqref{EQ4}, we take $\rho_1(t) = r^{\frac{1}{2}} (t+1)^{-\frac{1}{2}}$ for some $r$ to be chosen later. In this case, the integrating factor is given by
\begin{equation} 
e^{\int_{0}^{t} \rho_1(s)^2 ds} = e^{r \int_{0}^{t} \frac{1}{(s+1)} ds} = e^{r \ln (t+1)} = (t+1)^r
\end{equation} and so \eqref{abstractbound} becomes
\begin{equation} \label{abstractbound1}
\|u(t)\|_{L^2}^2 
\le \frac{\|u_0\|_{L^2}^2}{(t+1)^r}
+ \frac{\Gamma_{9}}{(t+1)^r} \int_{0}^{t} \frac{(s+1)^r}{(s+1)^2} ds  \nonumber
+ \frac{C}{(t+1)^r} \int_{0}^{t} \frac{(s+1)^r}{(s+1)^3} \left(\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \right)^2 ds   
\end{equation} for all $t \ge 0$. Here $\Gamma_9$ is a constant depending only on the initial data and $r$.
We have 
\begin{equation} 
\frac{\Gamma_{9}}{(t+1)^{r}} \int_{0}^{t} \frac{(s+1)^{r}}{(s+1)^2} ds
= \frac{\Gamma_{9}}{(r-1)(t+1)^{r}} \left((t+1)^{r-1} - 1 \right)
\le \frac{\Gamma_{9}}{(r-1)(t+1)} 
\end{equation} for any $r > 1$. 
Moreover, applying the Cauchy-Schwarz inequality in the time variable yields
\begin{equation} 
\left(\int_{0}^{s} \|u(\tau)\|_{L^2}^2 d\tau \right)^2 \le s \int_{0}^{s} \|u(\tau)\|_{L^2}^4 d\tau, 
\end{equation} so that 
\begin{align} 
\frac{C}{(t+1)^r} \int_{0}^{t} \frac{(s+1)^r}{(s+1)^3} \left(\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \right)^2 ds
&\le \frac{C}{(t+1)^r} \left(\int_{0}^{t} (s+1)^{r-2} ds\right) \left(\int_{0}^{t} \|u(s)\|_{L^2}^4 ds \right) \nonumber
\\&\le \frac{C}{(r-1)(t+1)} \left(\int_{0}^{t} \|u(s)\|_{L^2}^4 ds \right)
\end{align} for any $r > 1$. Taking $r =2$ and using \eqref{EQ4} give 
\begin{equation} 
\|u(t)\|_{L^2}^2 \le \frac{\Gamma_{10}}{t+1} + \frac{\Gamma_{10}}{t+1} \int_{0}^{t} \frac{\|u(s)\|_{L^2}^2}{\ln (e+s)} ds
\end{equation} and so
\begin{equation}  
(t+1)\|u(t)\|_{L^2}^2 \le \Gamma_{10} + C'\Gamma_{10} \int_{0}^{t} \frac{(s+1)\|u(s)\|_{L^2}^2}{(s+e) \ln (e+s)} ds
\end{equation} for any $t \ge 0$. By Gronwall's inequality, we obtain 
\begin{align} 
&(t+1) \|u(t)\|_{L^2}^2 
\le \Gamma_{10} + C'\Gamma_{10}^2 \int_{0}^{t} \frac{e^{\int_{s}^{t} \frac{1}{(e + \tau ) \ln (e + \tau)} d\tau}}{(e + s) \ln (e+s)} ds \nonumber
\\&=  \Gamma_{10} + C'\Gamma_{10}^2 \int_{0}^{t} \frac{\ln (e+t)}{(e + s) \left[\ln (e+s)\right]^2} ds
\le \Gamma_{10} + C'\Gamma_{10}^2 \ln (e + t).
\end{align}  Therefore, 
\begin{equation} 
\|u(t)\|_{L^2}^2 \le \frac{\Gamma_{11} \ln (t + e)}{t + 1}
\end{equation} for any $t \ge 0$, where $\Gamma_{11}$ is a constant depending only on the initial data. 

\textit{\textbf{Step 6.3} (Sharp decaying bound for the $L^2$ norm of the velocity).} Finally, we prove \eqref{ltime2}. We take $\rho_1(t) = \sqrt{2} (t+1)^{-\frac{1}{2}}$ as in the previous sub-step, and we obtain the bound  
\begin{equation} \label{abstractbound1}
\|u(t)\|_{L^2}^2 
\le \frac{\|u_0\|_{L^2}^2}{(t+1)^2}
+ \frac{\Gamma_{12}}{t+1}
+ \frac{C}{(t+1)^2} \int_{0}^{t} \frac{1}{s+1} \left(\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \right)^2 ds   
\end{equation} for all $t \ge 0$. We note that 
\begin{equation}
\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \le \Gamma_{13} \int_{0}^{s} \frac{\ln (\tau + e)}{\tau + 1}  d\tau
\le C\Gamma_{13}\int_{0}^{s} \frac{\ln (\tau + e)}{\tau + e}  d\tau 
\le \Gamma_{14} \left[\ln (s + e)\right]^2
\end{equation} and so
\begin{equation} 
\int_{0}^{t} \frac{1}{s+1} \left(\int_{0}^{s} \|u(\tau)\|_{L^2} ^2 d\tau \right)^2 ds 
\le \Gamma_{15} \int_{0}^{t} \frac{1}{\sqrt{s+1}} ds 
\end{equation} for all $t \ge 0$. Therefore,
\begin{equation} 
\|u(t)\|_{L^2}^2 \le \frac{\Gamma_{16}}{t+1}
\end{equation} for all $t \ge 0$, where $\Gamma_{16}$ is a positive constant depending only on the initial data. This ends the proof of Theorem \ref{decay}. 


Now we study the rate of convergence of the gradients of the charge density and the velocity. 

\begin{thm} \label{decay2} Let $u_0 \in H^1 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$. Let $q_0 \in H^1 \cap L^1$ such that $\nabla q_0 \in L^1$. There exist positive constants $K_0$ and $K_0'$ depending only on the initial data and some universal constants such that the unique global-in-time solution $(q,u)$ of \eqref{intro1}--\eqref{qinit} obeys 
\begin{equation} \label{lt1}
\|\nabla u(t)\|_{L^2}^2 \le \frac{K_0}{t+1}
\end{equation} and
\begin{equation} \label{lt2}
\|\nabla q(t)\|_{L^2}^2 \le \frac{K_0'}{(t + 1)^{2}}
\end{equation} 
for all $t \ge 0$. 
\end{thm}

\textbf{Proof:} The proof is divided into 5 steps.


\textit{\textbf{Step 1} (Pointwise bounds for the Fourier transform of $\nabla u$).} The Fourier transform of the gradient of $u$ satisfies 
\begin{equation} 
\partial_t \widehat{\nabla u} + \widehat{\nabla \mathbb P(u \cdot \nabla u)} - \widehat{\nabla \Delta u} = - \widehat{\nabla \mathbb P(qRq)}
\end{equation} yielding the differential inequality
\begin{equation} 
\partial_t \widehat{\nabla u} + |\xi|^2 \widehat{\nabla u} \le |\xi|^2 \|u\|_{L^2}^2 + |\xi| \|q\|_{L^2}^2.
\end{equation} Integrating in time from $0$ to $t$, and using \eqref{step11}, we obtain
\begin{equation} \label{decay2st1}
|\widehat{\nabla u} (\xi,t)| \le \|\nabla u_0\|_{L^1} + K_1|\xi| + K_2 |\xi|^2 t
\end{equation} where $K_1$ and $K_2$ are positive constants depending only on the initial data. 

We note that the pointwise estimate \eqref{decay2st1} is not the sharpest. Indeed, one can use the decaying bounds for the $L^2$ norms of the velocity and the surface charge density derived in Theorem \ref{decay} instead of using \eqref{step11}, and this will result in a bound whose growth in $t$ is slower when $t \ge 1$. However, this will not improve the decay in the upcoming step, so we disregard this observation. 

\textit{\textbf{Step 2} (Decaying bound for the $L^2$ norm of $\nabla u$).}
We take the $L^2$ inner product of equation \eqref{intro2} with $-\Delta u$ and we get
\begin{equation} \label{gradu}
\frac{1}{2} \frac{d}{dt} \|\nabla u\|_{L^2}^2 + \|\Delta u\|_{L^2}^2 =  \int_{\mathbb R^2} qRq \cdot \Delta u.
\end{equation} The nonlinear term $(u \cdot \nabla u, \Delta u)_{L^2}$ vanishes due to the fact that the matrix $M^tM^2$ has a zero  trace where $M$ is the two-by-two traceless matrix whose entries are given by $M_{ij} = \frac{\partial u_i}{\partial x_j}$ and $M^t$ is its transpose.  
In view of H\"older's inequality with exponents 4,4,2, the boundedness of the Riesz transforms on $L^4({\mathbb R}^2)$ and Young's inequality, we obtain 
\begin{equation} \label{difna}
\frac{d}{dt} \|\nabla u\|_{L^2}^2 + \|\Delta u\|_{L^2}^2 \le C\|q\|_{L^4}^4.
\end{equation}
Using the $L^4$ estimate \eqref{Bernouilli}, we have 
\begin{equation}  \label{intl4}
\|q\|_{L^4}^4 \le K_3 (1+t)^{-2}
\end{equation} where $K_3$ depends on the initial data.
We note that the initial charge density is assumed to be in $H^1$ and so it belongs to $L^4$ due to the Sobolev embedding of $H^1(\mathbb R^2)$ into $L^4(\mathbb R^2)$. 
Going back to \eqref{difna}, we have 
\begin{equation} 
\frac{d}{dt} \|\nabla u\|_{L^2}^2 + \|\Delta u\|_{L^2}^2 \le CK_3(t+1)^{-2}.
\end{equation} For $t\in [0, \infty)$, we let 
\begin{equation} \label{ro2}
\rho_2(t) = r^{\frac{1}{2}}(t+1)^{-\frac{1}{2}}
\end{equation} for some $r>0$ to be chosen later. By Parseval's identity, we get
\begin{equation} 
\frac{d}{dt} \|\nabla u\|_{L^2}^2 + \rho_2(t)^2 \|\nabla u\|_{L^2}^2 
\le CK_3 (1+t)^{-2} + \rho_2(t)^2 \int_{|\xi| \le \rho_2(t)} |\widehat{\nabla u} (\xi, t)|^2 d\xi. 
\end{equation}
In view of the pointwise bound \eqref{decay2st1}, we have
\begin{align} 
\frac{d}{dt} \|\nabla u\|_{L^2}^2 + \rho_2(t)^2 \|\nabla u\|_{L^2}^2 
&\le K_4(t+1)^{-2} + K_5 \rho_2(t)^2 \left[\rho_2(t)^2 + \rho_2(t)^4 + \rho_2(t)^6t^2\right] 
\end{align} for $t \ge 0$.
We multiply by the integrating factor $(t+1)^{r}$ and then we integrate in time from $0$ to $t$. We obtain
\begin{align} 
\|\nabla u\|_{L^2}^2 
&\le \frac{\|\nabla u_0\|_{L^2}^2}{(t+1)^{r}} 
+ \frac{K_4}{(t+1)^{r}} \int_{0}^{t} \frac{(s+1)^{r}}{(s+1)^{2}} ds \nonumber
\\&+ \frac{K_6}{(t+1)^{r}} \int_{0}^{t} (1+s)^{r}\left(\frac{1}{(1+s)^2} + \frac{1}{(1+s)^3} \right) ds 
\end{align} where $K_6$ depends only on the initial data. We can take any $r > 2$ and we obtain the bound \eqref{lt1}.

\textit{\textbf{Step 3} (Bounds for $\int_{0}^{t} (s+1)^{\gamma}\|\Delta u(s)\|_{L^2}^2 ds$ where $\gamma\ne 1$ is a real number).} 
Let $\gamma \ne 1$. The differential inequality \eqref{difna} yields 
\begin{equation} 
\frac{d}{dt} ((t+1)^{\gamma} \|\nabla u\|_{L^2}^2) - \gamma (t+1)^{\gamma -1} \|\nabla u\|_{L^2}^2 + (t+1)^{\gamma} \|\Delta u\|_{L^2}^2 \le C(t+1)^{\gamma} \|q\|_{L^4}^4
\end{equation} for all $t \ge 0$. Integrating in time from $0$ to $t$ and using \eqref{Bernouilli} and \eqref{lt1}, we obtain
\begin{align} \label{disb}
\int_{0}^{t} (s+1)^{\gamma} \|\Delta u\|_{L^2}^2 ds 
&\le \|\nabla u_0\|_{L^2}^2 + (\gamma K_0 + K_7) \int_{0}^{t} (s+1)^{\gamma - 2} ds  \nonumber
\\&= \|\nabla u_0\|_{L^2}^2 + \frac{\gamma K_0 + K_7}{\gamma - 1} \left[(t+1)^{\gamma -1} - 1 \right] 
\end{align} for some positive constant $K_7$ depending on $\|q_0\|_{L^2}$ and $\|q_0\|_{L^4}$. 


\textit{\textbf{Step 4} (Pointwise bounds for the Fourier transform of $\nabla q$).}
The Fourier transform of the gradient of $q$ satisfies
\begin{equation}  
\partial_t \widehat{\nabla q} + \widehat{\nabla (u \cdot \nabla q)} + \widehat{\nabla \l q} = 0,
\end{equation} hence
\begin{equation} 
\partial_t \widehat{\nabla q} + |\xi| \widehat{\nabla q} \le |\xi|^2 \|u\|_{L^2}\|q\|_{L^2}.
\end{equation}
Using \eqref{step11}, we obtain the pointwise bound
\begin{equation} \label{naqf}
|\widehat{\nabla q} (\xi, t)| \le \|\nabla q_0\|_{L^1} + K_8 |\xi|^2 \sqrt{t} 
\end{equation} for all $\xi \in \mathbb R^2$ and $t \ge 0$. Here $K_8 $ depends only on the initial data. 

\textit{\textbf{Step 5} (Decaying bound for the $L^2$ norm of $\nabla q$).}
The $L^2$ norm of the gradient of $q$ evolves according to the energy equality
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|\nabla q\|_{L^2}^2 + \|\l^{\frac{3}{2}}q\|_{L^2}^2 = (u \cdot \nabla q, \Delta q)_{L^2}.
\end{equation}
In view of the Ladyzhenskaya interpolation inequality 
\begin{equation} 
\|\nabla u\|_{L^4} \le C\|\nabla u\|_{L^2}^{\frac{1}{2}} \|\Delta u\|_{L^2}^{\frac{1}{2}}
\end{equation} and the interpolation inequality \cite{AI}
\begin{equation} 
\|\nabla q\|_{L^{\frac{8}{3}}}^2 \le C\|q\|_{L^4}^{\frac{1}{2}} \|\l^{\frac{3}{2}}q\|_{L^2}^{\frac{3}{2}},
\end{equation} we estimate the nonlinear term
\begin{equation} 
|(u \cdot \nabla q, \Delta q)_{L^2}| \le \|\nabla u\|_{L^4} \|\nabla q\|_{L^{\frac{8}{3}}}^2 
\le C\|\nabla u\|_{L^2}^{\frac{1}{2}} \|\Delta u\|_{L^2}^{\frac{1}{2}} \|q\|_{L^4}^{\frac{1}{2}} \|\l^{\frac{3}{2}}q\|_{L^2}^{\frac{3}{2}}.
\end{equation}
Applying Young's inequality, we obtain
\begin{equation} \label{thtwo} 
\frac{d}{dt} \|\nabla q\|_{L^2}^2 + \|\l^{\frac{3}{2}} q\|_{L^2}^2 \le C\|\nabla u\|_{L^2}^2\|\Delta u\|_{L^2}^2\|q\|_{L^4}^2.
\end{equation}
In view of \eqref{lt1} and \eqref{intl4}, we have 
\begin{equation} 
\frac{d}{dt} \|\nabla q\|_{L^2}^2 + \|\l^{\frac{1}{2}} \nabla q\|_{L^2}^2 \le \frac{K_{9}}{(t+1)^{2}} \|\Delta u\|_{L^2}^2
\end{equation} for all $t \ge 0$. 	Here $K_{9}$ depends only the initial data. Letting 
\begin{equation} 
\rho_3(t)= r(t+1)^{-1},
\end{equation} we split the dissipation term,
\begin{equation} 
\|\l^{\frac{3}{2}} q\|_{L^2}^2  \ge \rho_3(t) \|\nabla q\|_{L^2}^2 - \rho_3(t) \int_{|\xi| \le \rho_3(t)} |\widehat{\nabla q} (\xi, t)|^2 d\xi
\end{equation} yielding the differential inequality
\begin{equation} 
\frac{d}{dt} \|\nabla q\|_{L^2}^2 + \rho_3(t) \|\nabla q\|_{L^2}^2 \le \frac{K_{9}}{(t+1)^{2}} \|\Delta u\|_{L^2}^2 + \rho_3(t) \int_{|\xi| \le \rho_3(t)} |\widehat{\nabla q} (\xi, t)|^2 d\xi
\end{equation}
In view of the pointwise bound for $\widehat{\nabla q}$ given by \eqref{naqf}, we have 
\begin{equation} 
\int_{|\xi| \le \rho_3(t)} |\widehat{\nabla q} (\xi, t)|^2 d\xi
\le K_{10} \left(\rho_3(t)^2 + \rho_3(t)^6 t \right)
\end{equation} and so
\begin{equation} 
\frac{d}{dt} \|\nabla q\|_{L^2}^2 + \rho_3(t) \|\nabla q\|_{L^2}^2 \le \frac{K_{9}}{(t+1)^{2}} \|\Delta u\|_{L^2}^2 + K_{10}\left(\rho_3(t)^3 + \rho_3(t)^7t \right). 
\end{equation}
We multiply both sides by $(t+1)^{r}$ and we integrate in time from $0$ to $t$. We obtain
\begin{align} \label{grq1}
\|\nabla q(t)\|_{L^2}^2 &\le \frac{\|\nabla q_0\|_{L^2}^2}{(t+1)^{r }}
+ \frac{K_{11}}{(t+1)^{r}} \int_{0}^{t} (s+1)^{r-2}\|\Delta u(s)\|_{L^2}^2 ds\\
&+ \frac{K_{12}}{(t+1)^{r}} \int_{0}^{t} \left[(s+1)^{r-3} - (s+1)^{r-6}\right] ds.
\end{align}
In view of \eqref{disb} applied with $\gamma = r-2$, we have
\begin{equation} 
\int_{0}^{t} (s+1)^{r-2}\|\Delta u(s)\|_{L^2}^2 ds
\le  \|\nabla u_0\|_{L^2}^2 + \frac{(r -2)K_0 + K_7}{r - 3} \left[(t+1)^{r-3} - 1 \right] 
\end{equation} for any $r \ne 3$, and so
\begin{equation} \label{grq2}
\frac{K_{11}}{(t+1)^{r}} \int_{0}^{t} (s+1)^{r-2}\|\Delta u(s)\|_{L^2}^2 ds
\le \frac{K_{13}}{(t+1)^3}
\end{equation} for any $r > 3$. Here $K_{13}$ depends on the initial data and $r$. Putting \eqref{grq1} and \eqref{grq2} together and choosing $r = 6$ give the desired decay \eqref{lt2}. This completes the proof of Theorem \ref{decay2}.

Now we establish decaying bounds for higher order derivatives. We need the following proposition.

\begin{prop} \label{threetwo} Let $u_0 \in H^1 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$. Let $q_0 \in H^1 \cap L^1$ such that $\nabla q_0 \in L^1$. Let $\beta > 3$. There exist a positive universal constant $C$ and positive constants $c_1$, $c_2$, and $c_3$ depending only on the initial data such that the solution $q$ of \eqref{intro1}--\eqref{qinit} obeys 
\begin{equation}  \label{tht2}
\int_{0}^{t} (s+1)^{\beta} \|\l^{\frac{3}{2}}q(s)\|_{L^2}^2 ds 
\le \|\nabla q_0\|_{L^2}^2 + C\|\nabla u_0\|_{L^2}^2 + C\frac{(\beta - 2)c_1 + c_2}{\beta - 3} (t+1)^{\beta - 3}
+ \frac{\beta c_3}{\beta - 2} (t+1)^{\beta - 2}
\end{equation} for all $t \ge 0$. 
\end{prop}

\textbf{Proof:} In view of the differential inequality \eqref{thtwo}, we have
\begin{equation} 
\frac{d}{dt} (t+1)^{\beta} \|\nabla q\|_{L^2}^2 -\beta(t+1)^{\beta -1} \|\nabla q\|_{L^2}^2 + (t+1)^{\beta} \|\l^{\frac{3}{2}}q\|_{L^2}^2 \le C(t+1)^{\beta} \|\nabla u\|_{L^2}^2 \|\Delta u\|_{L^2}^2 \|q\|_{L^4}^2.
\end{equation} Integrating in time from $0$ to $t$, using the bounds \eqref{Bernouilli} and \eqref{lt1} and applying \eqref{disb} with $\gamma = \beta -2$, we obtain \eqref{tht2}.


\begin{thm} \label{decay3} Let $u_0 \in H^2 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$ and $\Delta u_0 \in L^1$. Let $q_0 \in H^2 \cap L^1$ such that $\nabla q_0 \in L^1$ and $\Delta q_0 \in L^1$. There exist positive constants $M_0$ and $M_0'$ depending only on the initial data and some universal constants such that the unique global-in-time solution $(q,u)$ of \eqref{intro1}--\eqref{qinit} obeys 
\begin{equation} \label{deltau1}
\|\Delta u(t)\|_{L^2}^2 \le \frac{M_0}{t+1}
\end{equation} and
\begin{equation} \label{deltaq2}
\|\Delta q(t)\|_{L^2}^2 \le \frac{M_0'}{(t + 1)^2}
\end{equation} 
for all $t \ge 0$. 
\end{thm}

\textbf{Proof:} The Fourier transform of $\Delta u$ obeys
\begin{equation} 
\partial_t \widehat{\Delta u} + |\xi|^2 \widehat{\Delta u}
\le C|\xi|^3 \|u\|_{L^2}^2 + C|\xi|^2 \|q\|_{L^2}^2.
\end{equation} Consequently, it satisfies the pointwise bound 
\begin{equation} \label{ptb}
|\widehat{\Delta u} (\xi, t)| \le \|\Delta u_0\|_{L^1} + M_1 |\xi|^3 t + M_2 |\xi|^2
\end{equation} for all $\xi \in \mathbb R^2$ and all $t \ge 0$. Here $M_1$ and $M_2$ are positive constants depending only on the initial data. 
The $L^2$ norm of $\Delta u$ evolves according to the energy equality 
\begin{equation} \label{deltaeq1}
\frac{1}{2} \frac{d}{dt} \|\Delta u\|_{L^2}^2 + \|\nabla \Delta u\|_{L^2}^2
= - \int_{{\mathbb R}^2} \Delta (qRq) \cdot \Delta u dx 
- \int_{{\mathbb R}^2} \Delta (u \cdot \nabla u) \cdot \Delta u dx.
\end{equation} Integrating by parts, using \eqref{intro3}, and applying Ladyzhenskaya's interpolation inequality, we estimate the second term on the right hand side in \eqref{deltaeq1} as 
\begin{equation} 
\left|\int_{{\mathbb R}^2} \Delta (u \cdot \nabla u) \cdot \Delta u dx \right|
\le C\|\nabla u\|_{L^2}\|\Delta u\|_{L^2}\|\nabla \Delta u\|_{L^2}.
\end{equation} In view of the boundedness of the Riesz transforms on $L^4$ and the continuous embedding of $\dot{H}^{\frac{1}{2}}$ in $L^4$, we obtain for the first term on the right hand side in \eqref{deltaeq1} 
\begin{equation} \label{deltaeq2}
\left|\int_{{\mathbb R}^2} \Delta (qRq) \cdot \Delta u dx  \right|
\le C\|q\|_{L^4}\|\l^{\frac{3}{2}}q\|_{L^2}\|\nabla \Delta u\|_{L^2}.
\end{equation}
From \eqref{deltaeq1}--\eqref{deltaeq2} and using Young's inequality, we obtain the energy inequality 
\begin{equation} 
\frac{d}{dt} \|\Delta u\|_{L^2}^2 + \|\nabla \Delta u\|_{L^2}^2 
\le C\|q\|_{L^4}^2 \|\l^{\frac{3}{2}}q\|_{L^2}^2 + C\|\nabla u\|_{L^2}^2 \|\Delta u\|_{L^2}^2.
\end{equation}
In view of Parseval's identity, we have 
\begin{equation} 
\frac{d}{dt} \|\Delta u\|_{L^2}^2 + \rho_2(t)^2 \|\Delta u\|_{L^2}^2
\le C\|q\|_{L^4}^2 \|\l^{\frac{3}{2}}q\|_{L^2}^2 
+ C\|\nabla u\|_{L^2}^2 \|\Delta u\|_{L^2}^2 
+ \rho_2(t)^2 \int_{|\xi| \le \rho_2(t)} |\widehat{\Delta u}(\xi, t)|^2 d\xi
\end{equation}
where $\rho_2$ is the function defined by \eqref{ro2}. 
The decay bounds \eqref{intl4} and \eqref{lt1} together with the pointwise bound for the Fourier transform of $\Delta u$ given by \eqref{ptb} yield 
\begin{align} 
\frac{d}{dt} \|\Delta u\|_{L^2}^2 + \rho_2(t)^2 \|\Delta u\|_{L^2}^2 
\le \frac{M_3}{t+1} \|\l^{\frac{3}{2}}q\|_{L^2}^2 
+ \frac{M_4}{t+1} \|\Delta u\|_{L^2}^2 
+ M_5 \rho_2(t)^2 \left[\rho_2(t)^2 + \rho_2(t)^8 t^2 + \rho_2(t)^6 \right]
\end{align} for all $t \ge 0$, where $M_3, M_4$ and $M_5$ are positive constants depending only on the initial data. Multiplying by the integrating factor and integrating in time from $0$ to $t$, we obtain
\begin{align} 
&\|\Delta u\|_{L^2}^2 
\le \frac{\|\Delta u_0\|_{L^2}^2}{(t+1)^{r}}
+ \frac{M_3}{(t+1)^{r}} \int_{0}^{t} (s+1)^{r - 1}\|\l^{\frac{3}{2}}q\|_{L^2}^2 ds \nonumber
\\&+ \frac{M_4}{(t+1)^{r}} \int_{0}^{t} (s+1)^{r-1} \|\Delta u\|_{L^2}^2 ds 
+ \frac{M_5}{(t+1)^{r }} \int_{0}^{t} \left[(s+1)^{r-2} + (s+1)^{r-3} +(s+1)^{r-4}\right] ds.
\end{align}
We choose $r = 5$. In view of the bound \eqref{disb} applied with $\gamma = r-1$ and Proposition \ref{threetwo} applied with $\beta = r-1$, we obtain \eqref{deltau1}. 

The Fourier transform of the Laplacian of $q$ satisfies
\begin{equation} 
\partial_t \widehat{\Delta q} + |\xi| \widehat{\Delta q} \le |\xi|^3 \|u\|_{L^2}\|q\|_{L^2}
\end{equation} and hence
\begin{equation} \label{Deltaqf}
|\widehat{\Delta q} (\xi, t)| \le \|\Delta q_0\|_{L^1} + M_8 |\xi|^3 \sqrt{t} 
\end{equation} for all $\xi \in \mathbb R^2$ and $t \ge 0$. Here $M_8 $ depends only on the initial data. 
Now, we establish decaying estimate for $\|\Delta q\|_{L^2}^2$ which evolves according to 
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|\Delta q\|_{L^2}^2 + \|\l^{\frac{5}{2}}q\|_{L^2}^2
= 2\int_{{\mathbb R}^2} (\nabla u \cdot \nabla (\nabla q)) \Delta q dx
+ \int_{{\mathbb R}^2} (\Delta u \cdot \nabla q) \Delta q dx.
\end{equation} In view of the Gagliardo-Nirenberg interpolation inequality
\begin{equation} 
\|\Delta q\|_{L^2} \le C\|\l^{\frac{5}{2}} q\|_{L^2}^{\frac{4}{5}}\|q\|_{L^2}^{\frac{1}{5}},
\end{equation} the Sobolev embedding inequality
\begin{equation} 
\|\Delta q\|_{L^4} \le C\|\l^{\frac{5}{2}}q\|_{L^2},
\end{equation} and the bound
\begin{equation} 
\|\nabla \nabla q\|_{L^2} = \|\nabla \l^{-1} \nabla \l^{-1} \Delta q\|_{L^2} \le C\|\Delta q\|_{L^2}
\end{equation} that follows from the boundedness of the Riesz transforms on $L^2$, we obtain 
\begin{align} 
\frac{1}{2} \frac{d}{dt} \|\Delta q\|_{L^2}^2 + \|\l^{\frac{5}{2}}q\|_{L^2}^2 
&\le C\|\nabla u\|_{L^4} \|\Delta q\|_{L^2} \|\Delta q\|_{L^4} + C\|\Delta u\|_{L^2}\|\nabla q\|_{L^4} \|\Delta q\|_{L^4} \nonumber
\\&\le C\|\nabla u\|_{L^2}^{\frac{1}{2}}\|\Delta u\|_{L^2}^{\frac{1}{2}} \|\l^{\frac{5}{2}} q\|_{L^2}^{\frac{9}{5}} \|q\|_{L^2}^{\frac{1}{5}} + C\|\Delta u\|_{L^2} \|\l^{\frac{3}{2}}q\|_{L^2}\|\Delta q\|_{L^4} \nonumber
\\&\le \frac{1}{2} \|\l^{\frac{5}{2}}q\|_{L^2}^2 
+ C\|\nabla u\|_{L^2}^5 \|\Delta u\|_{L^2}^5\|q\|_{L^2}^2 
+ C\|\Delta u\|_{L^2}^{2} \|\l^{\frac{3}{2}}q \|_{L^2}^2.
\end{align} Consequently, 
\begin{align} 
&\frac{d}{dt} \|\Delta q\|_{L^2}^2 + \rho_2(t)^2 \|\Delta q\|_{L^2}^2 
\le  C\|\nabla u\|_{L^2}^5 \|\Delta u\|_{L^2}^5\|q\|_{L^2}^2 
\\&\quad\quad+ C\|\Delta u\|_{L^2}^{2} \|\l^{\frac{3}{2}}q \|_{L^2}^2 \nonumber
+ \rho_2(t)^2 \int_{|\xi| \le \rho_2(t)^2} |\widehat{\Delta q} (\xi,t)|^2 d\xi
\end{align} where $\rho_2$ is defined by \eqref{ro2}.
In view of the estimates \eqref{ltime1}, \eqref{lt1} and \eqref{deltau1}, Proposition \ref{threetwo} applied with $\beta = r-1$, and the pointwise bound for the Fourier transform of $\Delta q$ given by \eqref{Deltaqf}, we obtain \eqref{deltaq2}. This ends the proof of Theorem \ref{decay3}.
 
Let $C^{0,\frac{1}{2}}$ be the space of bounded $1/2$-H\"older continuous functions on $\mathbb{R}^2$ with 
\begin{equation} 
\|f\|_{C^{0,\frac{1}{2}}} = \|f\|_{L^{\infty}} + \sup\limits_{x, y \in \mathbb{R}^2, x 
\neq y} \frac{|f(x) - f(y)|}{|x-y|^{\frac{1}{2}}}.
\end{equation}
In view of the continuous Sobolev embedding of $W^{1,4}$ into $C^{0, \frac{1}{2}}$, the Ladyzhenskaya interpolation inequality, and Theorems \ref{decay}, \ref{decay2}, and \ref{decay3}, we obtain the following statement.

\begin{cor}
Let $u_0 \in H^2 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$ and $\Delta u_0 \in L^1$. Let $q_0 \in H^2 \cap L^1$ such that $\nabla q_0 \in L^1$ and $\Delta q_0 \in L^1$. There exist positive constants $A_0$ and $A_0'$ depending only on the initial data and some universal constants such that the unique global-in-time solution $(q,u)$ of \eqref{intro1}--\eqref{qinit} obeys 
\begin{equation} \label{hold1}
\|u(t)\|_{C^{0,\frac{1}{2}}}^2 \le \frac{A_0}{t+1}
\end{equation} and
\begin{equation} \label{hold2}
\|q(t)\|_{C^{0, \frac{1}{2}}}^2 \le \frac{A_0'}{(t + 1)^{2}}
\end{equation} 
for all $t \ge 0$. 
\end{cor}

\section{Decomposition of the solution} \label{s3}

In this section, we decompose the charge density $q$ and the velocity $u$ solutions of \eqref{intro1}--\eqref{qinit} in the sum of solutions $Q$ and $U$ of the linear equations 
\begin{equation} \label{linear}
\partial_t Q + \l Q = 0
\end{equation} and
\begin{equation} \label{linear2}
\partial_t U - \Delta U = 0
\end{equation} with initial datum $Q(0) = q_0$ and $U(0) =u_0$ and remainders. We study the decays of the remainders  $q-Q$ and $u - U$ in $L^2$ and we show that they are faster than the decays of the $L^2$ norms of $q$ and $u$ respectively.
The solutions of \eqref{linear} and \eqref{linear2} are given explicitly by 
\begin{equation} 
Q(t) = \int_{\mathbb R^2} K_t^1(x-w) q_0(w) dw 
\end{equation} 
and 
\begin{equation} 
U(t) = \int_{\mathbb R^2} K_t^2(x-w) u_0(w) dw
\end{equation}
where $K_t^s$ is the kernel defined by its Fourier transform 
\begin{equation} 
\mathcal{F} (K_t^s)(\xi) = e^{-|\xi|^st}.
\end{equation}

The following proposition describes the decay of $\nabla Q$ and $\nabla U$ in $L^{\infty}$.

\begin{prop} Suppose $q_0 \in L^1$ such that $\int_{{\mathbb R}^2} |\xi| |\widehat{q_0}(\xi)| d\xi < \infty $ and $u_0 \in L^1$ such that $\int_{{\mathbb R}^2} |\xi| |\widehat{u_0}(\xi)| d\xi < \infty $. Then there exist positive constants $R_0$ and $R_0'$ depending only on the initial data such that the solutions $Q$ and $U$ of the linear equations \eqref{linear} and \eqref{linear2} satisfy
\begin{equation} \label{linearinf}
\|\nabla Q(t)\|_{L^{\infty}} \le \frac{R_0}{(t+1)^3}
\end{equation} and 
\begin{equation} \label{linearinf2}
\|\nabla U(t)\|_{L^{\infty}} \le \frac{R_0'}{(t+1)^{\frac{3}{2}}}
\end{equation} for all $t \ge 0$. 
\end{prop}

\textbf{Proof:} In view of Parseval's identity and the translation property of the Fourier transform, we have
\begin{equation} 
\nabla Q (x) = \nabla K_t^1 * q_0 (x) = \int_{\mathbb R^2} \nabla K_t^1 (x-w) q_0(w) dw = \int_{{\mathbb R}^2} e^{-2\pi i x \cdot \xi} \widehat{\nabla K_t^1}(\xi) \widehat{q_0}(\xi) d\xi.
\end{equation} On one hand,
\begin{equation} 
\|\nabla Q\|_{L^{\infty}} \le C\|\widehat{q_0}\|_{L^{\infty}} \int_{{\mathbb R}^2} |\xi| |\widehat{K_t^1}(\xi)| d\xi
\le C\|q_0\|_{L^1} \int_{0}^{\infty} r^2 e^{-rt} dr \le C\|q_0\|_{L^1} t^{-3}
\end{equation} and so
\begin{equation} 
t^3 \|\nabla Q\|_{L^{\infty}} \le C\|q_0\|_{L^1}. 
\end{equation} On the other hand, 
\begin{equation} 
\|\nabla Q\|_{L^{\infty}} \le C \int_{{\mathbb R}^2} |\xi| |\widehat{K_t^1}(\xi)||\widehat{q_0}(\xi)| d\xi
\le C \int_{{\mathbb R}^2} |\xi| |\widehat{q_0}(\xi)| d\xi
\end{equation} for all $t \ge 0$. Hence 
\begin{equation} 
(1+t)^3 \|\nabla Q\|_{L^{\infty}} \le 4(1+t^3)\|\nabla Q\|_{L^{\infty}} \le C\left(\|q_0\|_{L^1} + \int_{{\mathbb R}^2} |\xi| |\widehat{q_0}(\xi)| d\xi \right) 
\end{equation} for all $t \ge 0$, yielding \eqref{linearinf}. Similarly, we have 
\begin{equation} 
(1+t)^{\frac{3}{2}} \|\nabla U\|_{L^{\infty}} \le C(1+t^{\frac{3}{2}})\|\nabla U\|_{L^{\infty}} \le C\left(\|u_0\|_{L^1} + \int_{{\mathbb R}^2} |\xi| |\widehat{u_0}(\xi)| d\xi \right) 
\end{equation} for all $t \ge 0$, yielding \eqref{linearinf2}.

\begin{rem} The assumptions $\int_{{\mathbb R}^2} |\xi| |\widehat{q_0}(\xi)| d\xi < \infty $ and $\int_{{\mathbb R}^2} |\xi| |\widehat{u_0}(\xi)| d\xi < \infty $ are required to obtain the uniform-in-time boundedness of the $L^{\infty}$ norms $\nabla Q$ and $\nabla U$ for small times $t \in (0,1)$. This imposed regularity can be dropped since we are interested in studying the long-time behavior of solutions.  
\end{rem}

Next, we consider the pointwise behavior of the Fourier transforms of the differences $q-Q$ and $u- U$. We need first the following lemmas.


\begin{lem} \label{lem1} For $f \in L^2({\mathbb R}^2)$ and $x \in {\mathbb R}^2$, we let 
\begin{equation} 
Tf(x) = \lim\limits_{\epsilon \to 0} \int_{|x-y| > \epsilon} \frac{\sqrt{|y|^2 +1} - \sqrt{|x|^2 +1}}{|x-y|^{3}} f(y)dy.
\end{equation} There exists a universal constant $C>0$ (independent of $f$) such that 
\begin{equation} 
\|Tf\|_{L^2} \le C\|f\|_{L^2}.
\end{equation}
\end{lem}

\textbf{Proof:} We write
\begin{equation} 
Tf(x) = \lim\limits_{\epsilon \to 0} \int_{|x-y| > \epsilon} (a(y) - a(x))k(x-y) f(y)dy.
\end{equation} where $a(x)$ is the function defined on ${\mathbb R}^2$ by
\begin{equation} 
a(x) = \sqrt{|x|^2 + 1}
\end{equation} and $k(x)$ is the function defined on ${\mathbb R}^2 \setminus \left\{0\right\}$ by 
\begin{equation} 
k(x) = \frac{1}{|x|^3}.
\end{equation} We note that $k$ is homogeneous of degree $-3$. Moreover, the gradient of $a$ is given by 
\begin{equation} 
\nabla a(x) = \left(\frac{x_1}{\sqrt{|x|^2 +1}}, \frac{x_2}{\sqrt{|x|^2 + 1}} \right)
\end{equation} and satisfies $\|\nabla a\|_{L^{\infty}} \le 1$. Therefore, $T$ is a well-defined operator and bounded on $L^2$ (see page 435 in Section 2 of \cite{C}).

Using Lemma \ref{lem1}, we study the evolution of $(\sqrt{|x|^2 + 1})q(x)$ in $L^2(\mathbb R^2)$.

\begin{lem} \label{lem2} Let $u_0 \in H^1 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$. Let $q_0 \in H^1 \cap L^1$ such that $\nabla q_0 \in L^1$. Furthermore, suppose that $\int_{{\mathbb R}^2} |x|^2q_0(x)^2 dx < \infty$. Then there exists a positive constant $R_1 > 0$ depending only on the initial data such that 
\begin{equation} \label{moment}
\|(\sqrt{|\cdot|^2 + 1}) q(\cdot,t)\|_{L^2} \le R_1 \ln(t+1) + \|(\sqrt{|\cdot|^2 + 1}) q_0 (\cdot)\|_{L^2} 
\end{equation} holds for all $t \ge 0$.
\end{lem}

\textbf{Proof:} Let $a(x) = \sqrt{|x|^2 + 1}$. The evolution of $aq$ is described by 
\begin{equation} 
\partial_t (aq) + au \cdot \nabla q + a\l q = 0.
\end{equation} Multiplying by $aq$ and integrating in the space variable over ${\mathbb R}^2$, we obtain
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|aq\|_{L^2}^2 + \int_{{\mathbb R}^2} (a \l q) a q 
= - \int_{{\mathbb R}^2} (a u \cdot \nabla q) a q .
\end{equation} The cancellation 
\begin{equation} 
\int_{{\mathbb R}^2} (u \cdot\nabla (aq)) aq  = 0 
\end{equation} holds due to \eqref{intro3}, so we can rewrite the nonlinear term as 
\begin{equation} 
-\int_{{\mathbb R}^2} (a u \cdot \nabla q) a q
= \int_{{\mathbb R}^2} (u \cdot \nabla a) q^2 a.
\end{equation} By H\"older's inequality, Ladyzhenskaya's interpolation inequality, and the decaying bounds for the $L^2$ norms of $q, u, \nabla u$ and $\nabla q$ given by \eqref{ltime1}, \eqref{ltime2}, \eqref{lt1} and \eqref{lt2}, respectively,  we estimate 
\begin{align} 
&\left|\int_{{\mathbb R}^2} (u \cdot \nabla a) q^2 a \right|
\le \|\nabla a\|_{L^{\infty}} \|q\|_{L^4} \|u\|_{L^4} \|aq\|_{L^2} \nonumber
\\&\quad\quad\le C\|q\|_{L^2}^{\frac{1}{2}} \|\nabla q\|_{L^2}^{\frac{1}{2}} \|u\|_{L^2}^{\frac{1}{2}}\|\nabla u\|_{L^2}^{\frac{1}{2}}\|aq\|_{L^2}
\le R_2 (t+1)^{-\frac{3}{2}} \|aq\|_{L^2}
\end{align} for some constant $R_2$ depending only on the initial data. 
Now we write the linear term as the sum
\begin{align} 
\int_{{\mathbb R}^2} (a \l q) a q
&= \int_{{\mathbb R}^2} aq \l (aq)
+ \int_{{\mathbb R}^2} (aq) \left[a\l q - \l (aq) \right] \nonumber
\\&= \|\l^{\frac{1}{2}} (aq)\|_{L^2}^2 + \int_{{\mathbb R}^2} (aq) \left[a\l q - \l (aq) \right] .
\end{align} By the Cauchy-Schwarz inequality, we bound
\begin{equation} 
\left|\int_{{\mathbb R}^2} (aq) \left[a\l q - \l (aq) \right]\right| \le \|aq\|_{L^2} \|a\l q - \l(aq)\|_{L^2}.
\end{equation} The pointwise formula for the fractional Laplacian of order 1 yields
\begin{align} 
(a \l q - \l (aq))(x) 
&= C\int_{{\mathbb R}^2} \left[\frac{a(x)q(x) - a(x) q(y)}{|x-y|^3} - \frac{a(x) q(x) - a(y) q(y)}{|x-y|^3} \right] dy \nonumber
\\&= C\int_{{\mathbb R}^2} \frac{a(y) - a(x)}{|x-y|^3} q(y) dy
\end{align} where $C$ is  positive universal constant. As a consequence of Lemma \ref{lem1} and \eqref{ltime1}, we obtain
\begin{equation} 
\|a\l q - \l(aq)\|_{L^2} \le C\|q\|_{L^2}\le  C(t+1)^{-1}.
\end{equation}
Therefore, the $L^2$ norm of $aq$ obeys the energy inequality
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|aq\|_{L^2}^2 + \|\l^{\frac{1}{2}} (aq)\|_{L^2} \le \left[R_2(t+1)^{-\frac{3}{2}} + C(t+1)^{-1}\right] \|aq\|_{L^2}
\end{equation} so
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|aq\|_{L^2}^2 \le R_3(t+1)^{-1} \|aq\|_{L^2}
\end{equation} for some positive constant $R_3$ depending only on the initial data. Dividing both sides of the inequality by $\|aq\|_{L^2}$, we get 
\begin{equation} 
\frac{d}{dt} \|aq\|_{L^2} \le R_3 (t+1)^{-1}.
\end{equation} Integrating in time from $0$ to $t$, we obtain \eqref{moment}.

The following lemma is needed to obtain a growth in $|\xi|$ for the Fourier transform of $\mathbb P(qRq)$.

\begin{lem} \label{lem3} Let $f \in L^2({\mathbb R}^2)$ such that $\int_{{\mathbb R}^2} |x|^2f(x)^2 dx < \infty$. Then 
\begin{equation} \label{lemma}
|\widehat{\mathbb P(fRf)} (\xi)| \le C|\xi| \|f\|_{L^2} \left(\int_{{\mathbb R}^2} |x|^2 |f(x)|^2 dx\right)^{\frac{1}{2}}.
\end{equation} where $\mathbb P$ is the Leray projector and $R = (R_1, R_2)$ is the Riesz transform vector on $\mathbb R^2$.
\end{lem}

\textbf{Proof:} The Leray projector is a Fourier multiplier with a symbol denoted by $m(\xi)$. We have  
\begin{equation} 
\widehat{\mathbb P(fRf)}(\xi) = m (\xi) \widehat{fRf} (\xi)
\end{equation} for all $\xi \in {\mathbb R}^2$.
We note that $m(\xi)$ is bounded uniformly in $\xi$. Now, the Fourier transform of $fRf$ at $\xi$ is given by 
\begin{equation} 
\widehat{fRf} (\xi) = \int_{{\mathbb R}^2} f(x)Rf(x) e^{- i \xi \cdot x} dx
\end{equation} for $\xi \in {\mathbb R}^2$. Since the Riesz transform is antisymmetric, we have 
\begin{equation} 
\int_{{\mathbb R}^2} f(x)Rf(x) dx = 0
\end{equation} and so we can write $\widehat{fRf}$ at $\xi$ as 
\begin{equation}  
\widehat{fRf}(\xi)= \int_{{\mathbb R}^2} f(x)Rf(x) \left(e^{-i \xi \cdot x} -1 \right) dx.
\end{equation} Using the identity
\begin{equation} 
|e^{-i \xi \cdot x} - 1| \le  |\xi| |x|
\end{equation} that holds for all $x, \xi \in {\mathbb R}^2$, we estimate
\begin{equation} 
|\widehat{fRf}(\xi)| \le  |\xi| \int_{{\mathbb R}^2} |x| |f(x)| |Rf(x)| dx
\le |\xi| \|Rf\|_{L^2} \left(\int_{{\mathbb R}^2} |x|^2 |f(x)|^2 dx\right)^{\frac{1}{2}}
\end{equation} in view of the Cauchy-Schwarz inequality. This gives the pointwise estimate \eqref{lemma}. 

As a consequence of lemmas \ref{lem2} and \ref{lem3}, we obtain the following statement.

\begin{prop} Let $u_0 \in H^1 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$ and $q_0 \in H^1 \cap L^1$ such that $\nabla q_0 \in L^1$. Furthermore, suppose that $\int_{{\mathbb R}^2} |x|^2q_0(x)^2 dx < \infty$. Let $(q,u)$ be the solution of \eqref{intro1}--\eqref{qinit}. Let $\zeta = q - Q$ and $v = u - U$. Then there exist positive constants $R_4$, $R_5$ and $R_6$ depending only on the initial data such that the Fourier transforms of $\zeta$ and $v$ satisfy the pointwise bounds
\begin{equation} \label{fz}
|\widehat{\zeta} (\xi,t)| \le R_4 |\xi|
\end{equation} and 
\begin{equation} \label{fv}
|\widehat{v}(\xi, t)| \le R_5 |\xi| \ln (t+1) + R_6 |\xi|\ln^2(t+1)
\end{equation}
for all $\xi \in \mathbb R^2$ and $t \ge 0$.
\end{prop}

\textbf{Proof:} The Fourier transform of $\zeta$ obeys
\begin{equation} 
\partial_t \widehat{\zeta} + |\xi| \widehat{\zeta} = - \widehat{u \cdot \nabla q} \le |\xi| \|u\|_{L^2}\|q\|_{L^2}.
\end{equation} Consequently, 
\begin{equation} 
|\widehat{\zeta}(\xi, t)| \le \int_{0}^{t} |\xi| \|u\|_{L^2} \|q\|_{L^2} \le R_4 |\xi|
\end{equation} in view of the decaying bounds \eqref{ltime1} and \eqref{ltime2}.
The Fourier transform of $v$ evolves according to 
\begin{equation} 
\partial_t \widehat{v} + |\xi|^2 \widehat{v} = - \widehat{\mathbb P(u \cdot \nabla u)} - \widehat{\mathbb P(qRq)}.
\end{equation} Thus
\begin{equation} 
|\widehat{v}(\xi,t)| \le C|\xi| \int_{0}^{t} \|u\|_{L^2}^2 ds + C |\xi|\int_{0}^{t} \|q\|_{L^2} \left(\int_{{\mathbb R}^2}
|x|^2q(x)^2 dx \right)^{\frac{1}{2}} ds
\end{equation} by Lemma \ref{lem3}. In view of Lemma \ref{lem2} and the decaying estimates \eqref{ltime1} and \eqref{ltime2}, we obtain \eqref{fv}.
 

\begin{thm}\label{nonlin} Let $u_0 \in H^1 \cap L^1$ be divergence-free such that $\nabla u_0 \in L^1$. Let $q_0 \in H^1 \cap L^1$ such that $\nabla q_0 \in L^1$. Furthermore, suppose that $\int_{{\mathbb R}^2} |x|^2q_0(x)^2 dx < \infty$, $\int_{{\mathbb R}^2} |\xi| |\widehat{q_0}(\xi)| d\xi < \infty $, and $\int_{{\mathbb R}^2} |\xi| |\widehat{u_0}(\xi)| d\xi < \infty $. Let $(q, u)$ be the solution of \eqref{intro1}--\eqref{qinit}. Then there exist positive constants $R_7$ and $R_8$ depending only on the initial data such that the differences $q - Q$ and $u-U$ satisfy 
\begin{equation} \label{improved}
\|q(t) - Q(t)\|_{L^2}^2 \le \frac{R_7}{(t+1)^{2 + \frac{3}{2}}}
\end{equation} and 
\begin{equation} \label{improved2}
\|u(t) - U(t)\|_{L^2}^2 \le \frac{R_8}{(t+1)^{1+ \frac{1}{2}}}
\end{equation}
for all $t \ge 0$.
\end{thm}


\textbf{Proof:} Let $\zeta = q - Q$ and $v = u - U$. We have
\begin{equation} \label{EQ6}
\partial_t \zeta + \l \zeta = - u \cdot \nabla q.
\end{equation} Taking the $L^2$ inner product of equation \eqref{EQ6} with $\zeta$ and using \eqref{intro3}, we obtain 
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|\zeta\|_{L^2}^2 + \|\l^{\frac{1}{2}} \zeta\|_{L^2}^2 
= \int_{{\mathbb R}^2} (u \cdot \nabla q)Q dx
\le \|u\|_{L^2}\|q\|_{L^2}\|\nabla Q\|_{L^{\infty}}.
\end{equation} As a consequence of Theorem \ref{decay} and the bound \eqref{linearinf}, we obtain the energy inequality
\begin{equation} 
\frac{d}{dt} \|\zeta\|_{L^2}^2 + \|\l^{\frac{1}{2}} \zeta\|_{L^2}^2
\le \frac{R_9}{(t+1)^{4 + \frac{1}{2}}}
\end{equation} where $R_9$ is a positive constant depending only on the initial data. 
For a fixed $r$, we let $\rho(t) = r(t+1)^{-1}$. Then
\begin{equation} 
\frac{d}{dt} \|\zeta\|_{L^2}^2 + \rho(t) \|\zeta\|_{L^2}^2
\le \frac{R_9}{(t+1)^{4 + \frac{1}{2}}} 
+ \rho(t) \int_{|\xi| \le \rho(t)} |\widehat{\zeta}(\xi, t)|^2 d\xi.
\end{equation}
Using \eqref{fz}, we estimate
\begin{equation}
\int_{|\xi| \le \rho(t)} |\widehat{\zeta}(\xi, t)|^2 d\xi
\le R_{10} \rho(t)^4
\end{equation} and we obtain 
\begin{equation} 
\frac{d}{dt} \|\zeta\|_{L^2}^2 + \rho(t) \|\zeta\|_{L^2}^2
\le  \frac{R_{9}}{(t+1)^{4 + \frac{1}{2} }}  + R_{10} \rho(t)^5.
\end{equation} 
Multiplying by the factor $(s+1)^r$, integrating in the time variable $s$ from $0$ to $t$, and choosing any $r > 4$, we obtain the desired bound \eqref{improved}.  
Now, $v$ obeys 
\begin{equation} 
\partial_t v - \Delta v = - u \cdot \nabla u - qRq - \nabla p.
\end{equation} Taking the $L^2$ inner product of this latter equation with $v$ and using the fact that $v$ is divergence-free, we  get the energy equation 
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|v\|_{L^2} + \|\nabla v\|_{L^2}^2 = \int_{{\mathbb R}^2} (u \cdot \nabla u) \cdot U dx
- \int_{{\mathbb R}^2} (qRq) \cdot v dx.
\end{equation} We estimate
\begin{equation} 
\int_{{\mathbb R}^2} (u \cdot \nabla u) \cdot U dx \le \|u\|_{L^2}^2 \|\nabla U\|_{L^{\infty}} \le \frac{R_{11}}{(t+1)^{1+ \frac{3}{2}}}
\end{equation} in view of the bounds \eqref{ltime2} and \eqref{linearinf2}, and 
\begin{equation} 
\int_{{\mathbb R}^2} (qRq) \cdot v dx \le C\|q\|_{L^4}^2\|v\|_{L^{2}} 
\le C\|q\|_{L^2} \|\nabla q\|_{L^2} \|v\|_{L^2}
\le \frac{R_{12}}{(t+1)^{1+\frac{3}{2}}}
\end{equation} in view of the decaying estimate \eqref{ltime1}, \eqref{lt2} and \eqref{hold1}. This yields the energy inequality
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|v\|_{L^2}^2 + \rho(t) \|v\|_{L^2}^2
\le \frac{R_{13}}{(t+1)^{1+\frac{3}{2}}} + \rho(t) \int_{|\xi| \le \sqrt{\rho(t)}} |\widehat{v}(\xi,t)|^2 d\xi
\end{equation} where $\rho(t) = r(t+1)^{-1}$. Using the pointwise bound for the Fourier transform of $v$ given by \eqref{fv}, we have
\begin{equation} 
 \int_{|\xi| \le \sqrt{\rho(t)}} |\widehat{v}(\xi,t)|^2 d\xi 
\le R_{14}  \left[\ln^2(t+1) + \ln^4(t+1) \right] \rho(t)^2
\le R_{15} \sqrt{t+1} \rho(t)^2, 
\end{equation} hence
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|v\|_{L^2}^2 + \rho(t) \|v\|_{L^2}^2
\le \frac{R_{13}}{(t+1)^{1+\frac{3}{2}}} + R_{15} \sqrt{t+1} \rho(t)^3.
\end{equation} We multiply both sides by $(s+1)^r$, we integrate from $0$ to $t$, we choose any $r > 3/2$, and we obtain \eqref{improved2}.

\section{Appendix: Existence and Uniqueness of Solutions}

In this appendix, we prove the existence of weak and strong solutions for the electroconvection model \eqref{intro1}--\eqref{qinit}. 

\begin{defi}
A solution $(q,u)$ of \eqref{intro1}--\eqref{qinit} is said to be a weak solution on $[0,T]$ if it solves \eqref{intro1}--\eqref{qinit} in the sense of distributions, $u$ is divergence-free in the sense of distributions, 
\begin{equation}
u \in L^{\infty}(0,T;L^2) \cap L^2(0,T; H^1)
\end{equation} and 
\begin{equation}
q \in L^{\infty}(0,T; L^2) \cap L^2(0,T; H^{1/2}).
\end{equation}
\end{defi}

\begin{thm} \label{weak} Let $u_0\in L^2$ be divergence-free, let $q_0\in L^2$. Let $T > 0$ be arbitrary. There exists a weak solution $(q,u)$ of the system \eqref{intro1}--\eqref{qinit} on $[0,T]$.
\end{thm}

\textbf{Proof.} We briefly sketch the main ideas of the proof. For $0 < \epsilon \le 1$, we consider a viscous approximation of \eqref{intro1}--\eqref{qinit} given by 
\begin{equation} \label{sysv}
\begin{cases}  \partial_t q^{\epsilon} + J_{\epsilon}(u^{\epsilon}  \cdot \nabla q^{\epsilon}) + \Lambda q^{\epsilon} - \epsilon \Delta q^{\epsilon} = 0 
\\ \partial_t u^{\epsilon}  + J_{\epsilon}(u^{\epsilon}  \cdot \nabla u^{\epsilon})   - \Delta u^{\epsilon} + \nabla p^{\epsilon}  = -J_{\epsilon} (q^{\epsilon} Rq^{\epsilon})  
\\\nabla \cdot u^{\epsilon} = 0 
\end{cases} 
\end{equation} with smoothed out initial data, where $J_{\epsilon}$ is a standard mollifier operator, $u^{\epsilon} = J_{\epsilon}u$, $q^{\epsilon}= J_{\epsilon}q$ and $p^{\epsilon} = J_{\epsilon}p$.   
For each $\epsilon > 0$, we consider the map
\begin{equation} 
(q(t), u(t)) \mapsto \Phi_{\epsilon}((q,u))(t) = (e^{\epsilon t \Delta} J_{\epsilon}q_0 - \mathcal{A}^{\epsilon}_t (q^{\epsilon}, u^{\epsilon}), e^{t\Delta}J_{\epsilon}u_0  - \mathcal{B}^{\epsilon}_t (q^{\epsilon}, u^{\epsilon}))
\end{equation} where
\begin{equation} 
\mathcal{A}^{\epsilon}_t (q^{\epsilon}, u^{\epsilon}) = \int_{0}^{t} e^{\epsilon(t-s)\Delta} J_{\epsilon}(u^{\epsilon} \cdot \nabla q^{\epsilon})(s) ds + \int_{0}^{t} e^{\epsilon(t-s)\Delta} \l q^{\epsilon}(s) ds
\end{equation} and 
\begin{equation} 
\mathcal{B}^{\epsilon}_t (q^{\epsilon}, u^{\epsilon}) = \int_{0}^{t} e^{(t-s)\Delta} J_{\epsilon}\mathbb P(u^{\epsilon} \cdot \nabla u^{\epsilon})(s) ds + \int_{0}^{t} e^{(t-s)\Delta}J_{\epsilon} \mathbb P(q^{\epsilon}Rq^{\epsilon})(s) ds.
\end{equation}
There exists a time $T_{\epsilon} = T_{\epsilon} (\epsilon, \|u_0\|_{L^2}, \|q_0\|_{L^2}) > 0$ such that the map $\Phi_{\epsilon}$ is a contraction on the Banach space 
\begin{equation} 
X_T = L^{\infty}(0,T; \bar{B}_{L^2} (2\|q_0\|_{L^2}) \oplus L^{\infty}(0,T; \bar{B}_{L^2_{\sigma}} (2\|u_0\|_{L^2}) 
\end{equation}
where $\bar{B}_{L^2}(r)$ is the closed ball in $L^2$, and $\bar{B}_{L^2_{\sigma}}$ is the closed ball in the space of $L^2$ divergence-free vectors. Consequently, $\Phi_{\epsilon}$ has a fixed point $(q^{\epsilon}, u^{\epsilon}) \in X_{T_{\epsilon}}$ solving \eqref{sysv}. This solution extends to the time interval $[0,T]$, and this can be obtained by  establishing uniform-in-time bounds for $(q^{\epsilon}, u^{\epsilon})$ on $[0,T]$. 
Indeed, we have 
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \left(\|\l^{-\frac{1}{2}}q^{\epsilon} \|_{L^2}^2 + \|u^{\epsilon} \|_{L^2}^2 \right) + \|q^{\epsilon}\|_{L^2}^2 + \|\nabla u^{\epsilon}\|_{L^2}^2 + \epsilon \|\l^{\frac{1}{2}}q^{\epsilon}\|_{L^2}^2 = 0
\end{equation} as shown in \eqref{weaknormb}. Hence the family of mollified velocities $(u^{\epsilon})_{\epsilon}$ is uniformly bounded in $L^{\infty}(0,T; L^2) \cap L^2(0,T; H^1)$. On the other hand, the $L^2$ norm of $q^{\epsilon}$ evolves according to 
\begin{equation} 
\frac{1}{2} \frac{d}{dt} \|q^{\epsilon}\|_{L^2}^2 + \|\l^{\frac{1}{2}} q^{\epsilon}\|_{L^2}^2 
+ \epsilon \|\nabla q^{\epsilon}\|_{L^2}^2 
= 0, 
\end{equation} and so the family of mollified charge densities $(q^{\epsilon})_{\epsilon}$ is uniformly bounded in $L^{\infty}(0,T; L^2) \cap L^2(0,T; H^{\frac{1}{2}})$. The $q^{\epsilon}$ and $u^{\epsilon}$ equations imply that the sequence of time derivatives $(\partial_t q^{\epsilon})_{\epsilon}$ and $(\partial_t u^{\epsilon})_{\epsilon}$ are uniformly bounded in $L^{2}(0,T; H^{-\frac{3}{2}})$ and $L^2(0,T; H^{-1})$ respectively. By the Aubin-Lions lemma, the sequence $((q^{\epsilon}, u^{\epsilon}))_{\epsilon}$ has a subsequence that converges strongly in $L^2(0,T; L^2)$ to a weak solution $(q,u)$ of \eqref{intro1}--\eqref{qinit}. We omit further details.  

\begin{defi} A weak solution $(q,u)$ of \eqref{intro1}--\eqref{qinit} is said to be a strong solution on $[0,T]$ if 
\begin{equation}
u \in L^{\infty}(0,T;H^1) \cap L^2(0,T; H^2)
\end{equation}
and 
\begin{equation} 
q \in L^{\infty}(0,T; L^4) \cap L^2(0,T; H^{1/2}).
\end{equation}
\end{defi}

\begin{thm} \label{strong} Let $u_0\in H^1$ be divergence-free and $q_0\in L^4$.  Let $T>0$ be arbitrary. There exists a unique strong solution $(u,q)$ of the system \eqref{intro1}--\eqref{qinit} on $[0,T]$.
\end{thm}


\textbf{Proof.} We take the $L^2$ inner product of the equation satisfied by $q^{\epsilon}$ in   \eqref{sysv} with $(q^{\epsilon})^3$. In view of the divergence-free condition satisfied by $u^{\epsilon}$, the nonlinear term vanishes, that is
\begin{equation} 
\int_{{\mathbb R}^2} u^{\epsilon} \cdot \nabla q^{\epsilon} (q^{\epsilon})^3 dx = 0. 
\end{equation} By the C\'ordoba-C\'ordoba inequality (\cite{CC}), we have 
\begin{equation} 
\int_{{\mathbb R}^2} (q^{\epsilon})^3 \l q^{\epsilon} dx  \ge 0 
\end{equation} and 
\begin{equation} 
-\int_{{\mathbb R}^2} (q^{\epsilon})^3 \Delta q^{\epsilon} dx \ge 0.
\end{equation}
Consequently, we obtain 
\begin{equation} 
\frac{1}{4} \frac{d}{dt} \|q^{\epsilon}\|_{L^4}^4  \le 0
\end{equation} which yields the boundedness of $q$ in $L^{\infty} (0, T; L^4({\mathbb R}^2))$ by the Banach Alaoglu theorem and the lower semi-continuity of the norm.  
The $L^2$ norm of $\nabla u^{\epsilon}$ obeys the energy inequality 
\begin{equation} 
\frac{d}{dt} \|\nabla u^{\epsilon}\|_{L^2}^2 + \|\Delta u^{\epsilon}\|_{L^2}^2 \le C \|q^{\epsilon}\|_{L^4}^4 
\end{equation} as shown in \eqref{gradu}, yielding the boundedness of $u$ in $L^{\infty}(0,T; H^1) \cap L^2(0,T; H^2)$. 
Now we prove the uniqueness of strong solutions. Suppose $(q_1, u_1)$ and $(q_2, u_2)$ are strong solutions of \eqref{intro1}--\eqref{qinit} with same initial data. Let $q = q_1 - q_2, u = u_1 - u_2$ and $p = p_1 - p_2$. Then $q$ satisfies 
\begin{equation} \label{uniqq}
\partial_t q + \l q = - u_1 \cdot \nabla q - u \cdot \nabla q_2
\end{equation} and $u$ satisfies
\begin{equation} \label{uniqu}
\partial_t u - \Delta u + \nabla p = - qRq_1 - q_2Rq - u_1 \cdot \nabla u - u \cdot \nabla u_2.
\end{equation} 
We take the $L^2$ inner product of \eqref{uniqq} with $\l^{-1}q$ and the $L^2$ inner product of \eqref{uniqu} with $u$. We add the resulting energy equalities. We have a cancellation 
\begin{equation} \label{uinitial}
- \int_{{\mathbb R}^2} (u \cdot \nabla q_2) \l^{-1} q dx - \int_{{\mathbb R}^2} (q_2Rq) \cdot u dx = 0 
\end{equation} obtained from integration by parts. In view of the Ladyzhenskaya's interpolation inequality, we estimate
\begin{equation} 
\left|\int_{{\mathbb R}^2} (qRq_1) \cdot u dx \right|
\le C\|q\|_{L^2}\|q_1\|_{L^4}\|u\|_{L^2}^{\frac{1}{2}} \|\nabla u\|_{L^2}^{\frac{1}{2}}
\le \frac{1}{4} \|\nabla u\|_{L^2}^2 + \frac{1}{4} \|q\|_{L^2}^2 + C\|q_1\|_{L^4}^4 \|u\|_{L^2}^2
\end{equation} and 
\begin{equation} 
\left|\int_{{\mathbb R}^2} (u \cdot \nabla u_2) \cdot u  dx \right|
\le \|u\|_{L^4}^2 \|\nabla u_2\|_{L^2}
\le \frac{1}{4} \|\nabla u\|_{L^2}^2 + C\|\nabla u_2\|_{L^2}^2 \|u\|_{L^2}^2.
\end{equation}
Now we write
\begin{equation} 
\int_{{\mathbb R}^2} (u_1 \cdot \nabla q)  \l^{-1}q dx   
= \int_{{\mathbb R}^2} \left(\l^{-\frac{1}{2}}(u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q \right) \l^{-\frac{1}{2}}q dx 
\end{equation} via integration by parts, and we show below that 
\begin{equation}\label{ufinal}
\left|\int_{{\mathbb R}^2} \left(\l^{-\frac{1}{2}}(u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q \right) \l^{-\frac{1}{2}}q dx  \right|
\le  C\|u_1\|_{H^2} \|q\|_{L^2}\|\l^{-\frac{1}{2}}q\|_{L^2}.
\end{equation}
Putting \eqref{uinitial}-\eqref{ufinal} together, we obtain the energy inequality 
\begin{equation} 
\frac{d}{dt} \left[\|\l^{-\frac{1}{2}}q\|_{L^2}^2 + \|u\|_{L^2}^2 \right]
\le C\left[\| u_1\|_{H^2}^2 + \|\nabla u_2\|_{L^2}^2 + \|q_1\|_{L^4}^4 \right] \left[\|\l^{-\frac{1}{2}}q\|_{L^2}^2 + \|u\|_{L^2}^2 \right]
\end{equation} from which we obtain uniqueness. 
Finally, we show that the estimate \eqref{ufinal} holds by establishing the commutator estimate 
\begin{equation}  \label{commutator}
\|\l^{-\frac{1}{2}}(u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q  \|_{L^2} 
\le C\| u_1\|_{H^2} \|q\|_{L^2}.
\end{equation} 
Indeed, let $w\in L^2({\mathbb R}^2)$. By Parseval's identity, we have 
\begin{equation} 
\int_{{\mathbb R}^2} (\l^{-\frac{1}{2}} (u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q)(x) w (x) dx
=  \int_{{\mathbb R}^2} \mathcal{F}(\l^{-\frac{1}{2}} (u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q)(\xi) \mathcal{F}w (\xi) d\xi.
\end{equation} But 
\begin{equation} 
\mathcal{F} (\l^{-\frac{1}{2}}(u_1 \cdot \nabla q))(\xi) 
= \int_{{\mathbb R}^2} |\xi|^{-\frac{1}{2}} (\xi \cdot \mathcal{F}u_1 (\xi - y)) \mathcal{F}q (y) dy
\end{equation} and 
\begin{equation} 
\mathcal{F} (u_1 \cdot \nabla \l^{-\frac{1}{2}}q) (\xi)
= \int_{{\mathbb R}^2} |y|^{-\frac{1}{2}} (\xi \cdot \mathcal{F}u_1 (\xi - y)) \mathcal{F}q (y) dy.
\end{equation} 
Consequently,
\begin{align}
&\left|\int_{{\mathbb R}^2} (\l^{-\frac{1}{2}} (u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q)(x) w (x) dx\right| \nonumber
\\&\le \int_{{\mathbb R}^2} \int_{{\mathbb R}^2} \min \left\{|\xi|, |y| \right\} \left||\xi|^{-\frac{1}{2}} - |y|^{-\frac{1}{2}}  \right| |\mathcal{F} u_1 (\xi - y) | |\mathcal{F} q(y)| |\mathcal{F} w(\xi) | dy d\xi
\end{align} where we used 
\begin{equation} 
|\xi \cdot \mathcal{F}u_1 (\xi - y)| \le \min \left\{|\xi|, |y| \right\} |\mathcal{F} u_1 (\xi - y)|
\end{equation} which holds due to the fact that the velocity is divergence-free. 
We note that 
\begin{equation} 
\min \left\{|\xi|, |y| \right\} \left||\xi|^{-\frac{1}{2}} - |y|^{-\frac{1}{2}}  \right|
\le \frac{\min \left\{|\xi|, |y| \right\}}{|\xi|^{\frac{1}{2}}|y|^{\frac{1}{2}}} |\xi - y|^{\frac{1}{2}}
\le |\xi - y|^{\frac{1}{2}}
\end{equation} for all $\xi, y \in {\mathbb R}^2$. Therefore, 
\begin{align} 
\left|\int_{{\mathbb R}^2} (\l^{-\frac{1}{2}} (u_1 \cdot \nabla q) - u_1 \cdot \nabla \l^{-\frac{1}{2}}q)(x) w (x) dx\right| 
&\le \||.|^{\frac{1}{2}} \mathcal{F} u_1(.) \|_{L^1} \|q\|_{L^2}\|w\|_{L^2} \nonumber 
\\&\le C\|u_1\|_{H^2} \|q\|_{L^2}\|w\|_{L^2}
\end{align} by H\"older's inequality and Young's convolution inequality. This gives \eqref{commutator} completing the proof of Theorem \ref{strong}.\\

{\bf{Acknowledgment.}} The research of M.I. was partially supported be NSF grant DMS 2204614.\\

{\bf{Data Availability Statement.}} The research does not have any associated data.\\

{\bf{Conflict of Interest.}} The authors declare that they have no conflict of interest.\\


