\section{Introduction}
In this paper we study the Cauchy problem for the Landau-Lifshitz-Gilbert (LLG) equation
\begin{align}\label{eq:Lanlif}
\partial_t s=as\times \Delta s-\varepsilon s\times (s\times \Delta s), \quad s(x,0)=s_0(x),
\end{align}
where $s(x,t):{\mathbb R}^n\times {\mathbb R}\to \mathbb{S}^2\subset {\mathbb R}^3$,  $\times$
denotes the wedge product in ${\mathbb R}^3$, $a\in {\mathbb R}$ and $\varepsilon>0$ is the
Gilbert damping parameter. The equation \eqref{eq:Lanlif} is one of the equations of  ferromagnetic spin chain, which
was proposed by Landau-Lifshitz \cite{LanLif} in studying the
dispersive theory of magnetisation of ferromagnets. Later on, such
equations were also found in the condensed matter physics.  The LLG equation has been studied extensively,  see
\cite{Lak,GuoDing} for an introduction on the equation.

Formally, if $a=0$, then  \eqref{eq:Lanlif} reduces to the heat flow
equations for harmonic maps
\begin{align}\label{eq:Harmap}
\partial_t s=-\varepsilon s\times (s\times \Delta s), \quad s(x,0)=s_0(x),
\end{align}
and if $\varepsilon=0$, then \eqref{eq:Lanlif} reduces to the Schr\"odinger
maps
\begin{align}\label{eq:Schmap}
\partial_t s=as\times \Delta s, \quad s(x,0)=s_0(x).
\end{align}
Both special cases have been objects of intense research. The purpose of this paper is to study the inviscid limit of \eqref{eq:Lanlif}, namely, to prove rigorously that the solutions of \eqref{eq:Lanlif} converges to the solutions of \eqref{eq:Schmap} as $\varepsilon\to 0$ under optimal conditions on the initial data.

The inviscid limit is an important topic in mathematical physics, and has been studied in various settings, e.g. for hyperbolic-dissipative equations such as Navier-Stokes equation to Euler equation (see \cite{HK} and references therein), for dispersive-dissipative equations such as KdV-Burgers equation to KdV equation (see \cite{Guo2}) and Ginzburg-Landau equation to Schr\"odinger equations (see \cite{Wang02, HW}). The LLG equation \eqref{eq:Lanlif} is an equation with both dispersive and dissipative effects. This can be seen from the stereographic projection transform. It was known that (see
\cite{LakNak}) let
\begin{align}
u=\mathbb{P}(s)=\frac{s_1+is_2}{1+s_3},
\end{align}
where $s=(s_1,s_2,s_3)$ is a solution to \eqref{eq:Lanlif}, then $u$ solves the following complex derivative Ginzburg-Landau type equation
\begin{align}\label{eq:dGL}
\begin{split}
(i\partial_t+\Delta-i\varepsilon\Delta)u=&\frac{2a\bar
u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2-\frac{2i\varepsilon\bar
u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2\\
u(x,0)=&u_0.
\end{split}
\end{align}
On the other hand, the projection transform has an inverse
\begin{align}
\mathbb{P}^{-1}(u)=\brk{\frac{u+\bar u}{1+|u|^2},\frac{-i(u-\bar u)}{1+|u|^2},\frac{1-|u|^2}{1+|u|^2}}.
\end{align}
Therefore, \eqref{eq:Lanlif} is equivalent to \eqref{eq:dGL} assuming $\mathbb P$ and $\mathbb P^{-1}$ is well-defined, and we
will focus on \eqref{eq:dGL}.  The previous works \cite{IK,IK2,Bej,Bej2,BIK} on the Schr\"odinger maps ($\varepsilon=0$) were also based on this transform.  Note that \eqref{eq:dGL} is invariant
under the following scaling transform: for $\lambda>0$
\[u(x,t)\to u(\lambda x,\lambda^2 t), \quad u_0(x)\to u_0(\lambda x).\]
Thus the critical Besov space  is $\dot
B_{2,1}^{n/2}$ in the sense of scaling.

To study the inviscid limit, the crucial task is to obtain uniform well-posedness with respect to the inviscid parameter. Energy method was used in \cite{HK}. For dispersive-dissipative equations, one needs to exploit the dispersive effect uniformly. Strichartz estimates and energy estimates were used in \cite{Wang02} for Ginzburg-Landau equations, and Bourgain space was used in \cite{Guo2} for KdV-Burgers equations. In \cite{WW, HWG} the inviscid limit for the derivative Ginzburg-Landau equations were studied by using the Strichartz estimates, local smoothing estimates and maximal function estimates.  However, these results requires high regularities when applied to equation \eqref{eq:dGL}.
In this paper we will use Bourgain-type space and exploit the null structure that are inspired by the latest development for the Schr\"odinger maps ($\varepsilon=0$) (see \cite{BIKT, BIK,Bej,Bej2,IK,IK2,Guo}) to study \eqref{eq:dGL} with small initial data in the critical Besov space. In
\cite{Bej2} and \cite{IK2} it was proved independently that global
well-posedness for \eqref{eq:Schmap} holds for small data in the critical
Besov space. We will extend their results to \eqref{eq:Lanlif} uniformly with respect to $\varepsilon$. We exploit the Bourgain space in a different way from both \cite{Bej2} and \cite{IK2}. One of the novelties is the use of $X^{0,1}$-structure that results in many simplifications even for the Schr\"odinger maps. The presence of dissipative term brings many technical difficulties, e.g. the lack of symmetry in time and incompatibility with $X^{s,b}$ structure. We need to overcome these difficulties when extending the linear estimates for the Schr\"odinger equation to the Schr\"odinger-dissipative equation uniformly with respect to $\varepsilon$.

By scaling we may assume $a=\pm 1$. From now on, we assume $a=1$ since
the other case $a=-1$ is similar. For $Q\in \mathbb{S}^{2}$, the space
$\dot{B}^{s}_Q$ is defined by
\[\dot{B}^{s}_Q=\dot{B}^{s}_Q({\mathbb R}^n;\mathbb{S}^{2})=\{f: {\mathbb R}^n\to {\mathbb R}^3; f-Q\in \dot{B}^{s}_{2,1},\, |f(x)|\equiv 1 \mbox{ a.e. in } {\mathbb R}^n\},\]
where $\dot{B}^{s}_{2,1}$ is the standard Besov space. It was known
the critical space is $\dot{B}^{n/2}_Q$. The main result of this
paper is

\begin{thm}\label{thm:Schmap}
Assume $n\geq 3$. The LLG equation \eqref{eq:Lanlif} is
globally well-posed for small data $s_0\in
\dot{B}^{n/2}_Q({\mathbb R}^n;\mathbb{S}^{2})$, $Q\in \mathbb{S}^{2}$ in a uniform way
with respect to $\varepsilon\in (0,1]$. Moreover, for any $T>0$, the solution
converges to that of Schr\"odinger map \eqref{eq:Schmap} in
$C([-T,T]:\dot{B}^{n/2}_Q)$ as $\varepsilon \to 0$.
\end{thm}

As we consider the inviscid limit in the strongest topology (same space as the initial data), no convergence rate is expected. This can be seen from linear solutions for \eqref{eq:dGL}. However, if assuming initial data has higher regularity, one can have convergence rate $O(\varepsilon T)$ (see \eqref{eq:limitHigh} below).

\section{Definitions and Notations}

For $x, y\in {\mathbb R}$, $x{\lesssim} y$ means that there exists a constant $C$
such that $x\leq Cy$, and $x\sim y$ means that $x{\lesssim} y$ and $y{\lesssim}
x$. We use ${\mathcal{F}}(f)$, $\hat{f}$ to denote the space-time Fourier
transform of $f$, and ${\mathcal{F}}_{x_i,t}f$ to denote the Fourier transform
with respect to $x_i,t$.

Let $\eta: {\mathbb R}\to [0, 1]$ be an even, non-negative, radially
decreasing smooth function such that: a) $\eta$ is compactly
supported in $\{\xi:|\xi|\leq 8/5\}$; b) $\eta\equiv 1$ for
$|\xi|\leq 5/4$. For $k\in {\mathbb Z}$ let
$\chi_k(\xi)=\eta(\xi/2^k)-\eta(\xi/2^{k-1})$, $\chi_{\leq
k}(\xi)=\eta(\xi/2^k)$, $\widetilde
\chi_k(\xi)=\sum_{l=-9n}^{9n}\chi_{k+l}(\xi)$, and then define the
Littlewood-Paley projectors $P_k, P_{\leq k}, P_{\geq k}$ on
$L^2({\mathbb R}^n)$ by
\[\widehat{P_ku}(\xi)=\chi_k(|\xi|)\widehat{u}(\xi),\quad \widehat{P_{\leq
k}u}(\xi)=\chi_{\leq k}(|\xi|)\widehat{u}(\xi),\] and $P_{\geq
k}=I-P_{\leq k-1}$, $P_{[k_1,k_2]}=\sum_{j=k_1}^{k_2}P_j$. We also
define $\wt P_k u=\mathcal{F}^{-1} \widetilde\chi_k(|\xi|)\widehat{u}(\xi)$

Let $\mathbb{S}^{n-1}$ be the unit sphere in ${\mathbb R}^n$. For $\ve e\in
\mathbb{S}^{n-1}$, define $\widehat{P_{k,\ve
e}u}(\xi)=\widetilde\chi_{k}(|\xi\cdot \ve
e|)\chi_k(|\xi|)\widehat{u}(\xi)$. Since for $|\xi|\sim 2^k$ we have
\[\sum_{l=-5n}^{5n}\chi_{k+l}(\xi_1)+\cdots+\sum_{l=-5n}^{5n}\chi_{k+l}(\xi_n)\sim
1,\] then let
\[\beta_k^j(\xi)=\frac{\sum_{l=-5n}^{5n}\chi_{k+l}(\xi_j)}{\sum_{j=1}^n\sum_{l=-5n}^{5n}\chi_{k+l}(\xi_j)}\cdot \sum_{l=-1}^1\chi_{k+l}(|\xi|), \quad j=1,\cdots, n.\]
Define the operator $\Theta_k^j$ on $L^2({\mathbb R}^n)$ by
$\widehat{\Theta_k^jf}(\xi)=\beta_k^j(\xi)\hat f(\xi)$, $1\leq j\leq n$. Let $\ve
e_1=(1,0,\cdots, 0), \cdots, \ve e_n=(0,\cdots, 0,1)$. Then we have
\begin{align}\label{eq:Pkdec}
P_k=\sum_{j=1}^nP_{k,\ve e_j}\Theta_k^j.
\end{align}
For any $k\in {\mathbb Z}$, we define the modulation projectors $Q_k, Q_{\leq k}, Q_{\geq k}$ on $L^2({\mathbb R}^n\times {\mathbb R})$ by
\[\widehat{Q_ku}(\xi,\tau)=\chi_k(\tau+|\xi|^2)\widehat{u}(\xi,\tau),\quad \widehat{Q_{\leq
k}u}(\xi,\tau)=\chi_{\leq k}(\tau+|\xi|^2)\widehat{u}(\xi,\tau),\]
and $Q_{\geq k}=I-Q_{\leq k-1}$,
$Q_{[k_1,k_2]}=\sum_{j=k_1}^{k_2}Q_j$.

For any $\ve e\in \mathbb{S}^{n-1}$, we can decompose ${\mathbb R}^n=\lambda \ve e
\oplus H_{\ve e}$, where $H_{\ve e}$ is the hyperplane with normal vector
$\ve e$, endowed with the induced measure. For $1\leq p, q< \infty$,
we define $L_{\ve e}^{p,q}$ the anisotropic Lebesgue space by
\[\norm{f}_{L_{\ve e}^{p,q}}=\brk{\int_{\mathbb R}\brk{\int_{H_{\ve e}\times {\mathbb R}} |f(\lambda \ve e+y,t)|^qdydt}^{p/q}d\lambda}^{1/p}\]
with the usual definition if $p=\infty$ or $q=\infty$. We write
$L_{\ve e_j}^{p,q}=L_{x_j}^pL_{\bar{x_j},t}^q$. We use
$\dot{B}^s_{p,q}$ to denote the homogeneous Besov spaces on ${\mathbb R}^n$
which is the completion of the Schwartz functions under the norm
\[\norm{f}_{\dot{B}^s_{p,q}}=(\sum_{k\in {\mathbb Z}}2^{qsk}\norm{P_kf}_{L^p}^q)^{1/q}.\]

To exploit the null-structure we also need the Bourgain-type space associated to the Schr\"odinger equation.
In this paper we use the modulation-homogeneous version as in \cite{Bej2, Guo}. We define $X^{0,b,q}$ to be the completion of the space of Schwartz functions with the norm
\begin{align}\label{eq:Xbq}
\norm{f}_{X^{0,b,q}}=(\sum_{k\in {\mathbb Z}}2^{kbq}\norm{Q_k
f}_{L^2_{t,x}}^q)^{1/q}.
\end{align}
If $q=2$ we simply write $X^{0,b}=X^{0,b,2}$. By the Plancherel's
equality we have
$\norm{f}_{X^{0,1}}=\norm{(i\partial_t+\Delta)f}_{L^2_{t,x}}$. Since
$X^{0,b,q}$  is not closed under conjugation, we also define the
space $\bar X^{0,b,q}$ by the norm $\norm{f}_{\bar
X^{0,b,q}}=\norm{\bar f}_{X^{0,b,q}}$, and similarly write $\bar
X^{0,b}=\bar X^{0,b,2}$. It's easy to see that $X^{0,b,q}$ function
is unique modulo solutions of the homogeneous Schr\"odinger
equation. For a more detailed description of the $X^{0,b,p}$ spaces
we refer the readers to \cite{Tataru98} and \cite{Tao01}. We use
$X_+^{0,b,p}$ to denote the space restricted to the interval
$[0,\infty)$:
\[\norm{f}_{X_+^{0,b,p}}=\inf_{\tilde f: \tilde f=f \mbox{ on } t\in [0,\infty)}\norm{\tilde f}_{X^{0,b,p}}.\]
In particular, we have
\begin{align}\label{eq:X01ext}
\norm{f}_{X_+^{0,1}}\sim \norm{\tilde f}_{X^{0,1}}
\end{align}
where $\tilde f=f(t,x)1_{t\geq 0}+f(-t,x)1_{t<0}$.

Let $L=\partial_t-i\Delta$ and $\bar
L=\partial_t+i\Delta$. We define the main
dyadic function space. If $f(x,t)\in L^2({\mathbb R}^n\times {\mathbb R}_+)$ has
spatial frequency localized in $\{|\xi|\sim 2^k\}$, define
\begin{align*}
\norm{f}_{F_{k}}=&\norm{f}_{X_+^{0,1/2,\infty}}+\norm{f}_{L_t^\infty
L_x^2}+\norm{f}_{L_t^2 L_x^\frac{2n}{n-2}}\\
&+2^{-(n-1)k/2}\sup_{\ve{e}\in
\mathbb{S}^{n-1}}\norm{f}_{L_{\ve{e}}^{2,\infty}}+2^{k/2}\sup_{|j-k|\leq
20}\sup_{\ve{e}\in
\mathbb{S}^{n-1}}\norm{P_{j,\ve e} f}_{L_{\ve{e}}^{\infty,2}},\\
\norm{f}_{Y_{k}}=&\norm{f}_{L_t^\infty L_x^2}+\norm{f}_{L_t^2
L_x^\frac{2n}{n-2}}+2^{-(n-1)k/2}\sup_{\ve{e}\in
\mathbb{S}^{n-1}}\norm{f}_{L_{\ve{e}}^{2,\infty}}\\
&+2^{-k}\inf_{f=f_1+f_2}(\norm{L f_1}_{L_{t,x}^2}+\norm{\bar L f_2}_{L_{t,x}^2}),\\
\norm{f}_{Z_k}=&2^{-k}\norm{L f}_{L^2_{t,x}}\\
\norm{f}_{N_{k}}=&\inf_{f=f_1+f_2+f_3}(\norm{f_1}_{L_t^1L_x^2}+2^{-k/2}\sup_{\ve{e}\in
\mathbb{S}^{n-1}}\norm{f_2}_{L_{\ve{e}}^{1,2}}+\norm{f_3}_{X_+^{0,-1/2,1}})+2^{-k}\norm{f}_{L^2_{t,x}}.
\end{align*}
Then we define the space $F^s, Y^s, Z^s, N^{s}$ with the following norm
\begin{align*}
\norm{u}_{F^{s}}=\sum_{k\in {\mathbb Z}}2^{ks}\norm{P_k u}_{F_{k}},\quad& \norm{u}_{Y^{s}}=\sum_{k\in {\mathbb Z}}2^{ks}\norm{P_k
u}_{Y_{k}},\\
\norm{u}_{Z^{s}}=\sum_{k\in {\mathbb Z}}2^{ks}\norm{P_k u}_{Z_{k}},\quad&
\norm{u}_{N^{s}}=\sum_{k\in {\mathbb Z}}2^{ks}\norm{P_k
u}_{N_{k}}.
\end{align*}
Obviously, $F_k\cap Z_k\subset Y_k$, and thus $F^s\cap Z^s\subset Y^s$.
In the end of this section, we present a standard extension lemma (See Lemma 5.4 in \cite{book}) giving the relation between $X^{s,b}$ and other space-time norm.
\begin{lem}\label{lem:ext}
Let $k\in {\mathbb Z}$ and $B$ be a space-time norm satisfying with some $C(k)$
\[\norm{e^{it_0}e^{it\Delta}P_kf}_B\leq C(k)\norm{P_kf}_2\]
for any $t_0\in {\mathbb R}$ and $f\in L^2$.  Then
\[\norm{P_ku}_B{\lesssim} C(k) \norm{P_ku}_{X^{0,1/2,1}}.\]
\end{lem}


\section{Uniform linear estimates}
In this section we prove some uniform linear estimates for the equation \eqref{eq:dGL}  with respect to the dissipative parameter. First we recall the known linear estimates for the Schr\"odinger equation, see \cite{KT} and \cite{IK2}.

\begin{lem}\label{lem:linearStr}
Assume $n\geq 3$. For any $k\in {\mathbb Z}$ we have
\begin{align}
\norm{e^{it\Delta}P_kf}_{L_t^2 L_x^\frac{2n}{n-2}\cap L_t^\infty L_x^2}{\lesssim}& \norm{P_kf}_2,\\
\sup_{\ve{e}\in
\mathbb{S}^{n-1}}\norm{e^{it\Delta}P_kf}_{L_{\ve{e}}^{2,\infty}}{\lesssim}& 2^{\frac{(n-1)k}{2}}\norm{P_kf}_2,\\
\sup_{\ve{e}\in
\mathbb{S}^{n-1}}\norm{e^{it\Delta}P_{k,\ve e}f}_{L_{\ve{e}}^{\infty,2}}{\lesssim}& 2^{-\frac{k}{2}}\norm{P_kf}_2.
\end{align}
\end{lem}

\begin{lem}\label{lem:linearXsb}
Assume $n\geq 3$, $u,F$ solves the equation: for $\varepsilon>0$
\[u_t-i\Delta u-\varepsilon \Delta u=F(x,t), \quad u(x,0)=u_0.\]
Then for $b\in [1,\infty]$
\begin{align}
\norm{P_ku}_{X_+^{0,1/2,b}}{\lesssim}&
\norm{P_k u_0}_{L^2}+\norm{P_kF}_{X_+^{0,-1/2,b}},\\
\norm{P_ku}_{Z_k}{\lesssim}&\varepsilon^{1/2}\norm{P_ku_0}_{L^2}+
2^{-k}\norm{P_kF}_{L_{t,x}^2},
\end{align}
where the implicit constant is independent of $\varepsilon$.
\end{lem}
\begin{proof}
First we show the second inequality.  We have
\begin{align}
u=e^{it\Delta+\varepsilon t\Delta}u_0+\int_0^te^{i(t-s)\Delta+\varepsilon (t-s)\Delta}F(s)ds
\end{align}
and thus
\begin{align*}
\norm{P_ku}_{Z_k}{\lesssim}& \varepsilon 2^k\norm{P_ku}_{L^2_{t,x}}+
2^{-k}\norm{P_kF}_{L_{t,x}^2}\\
{\lesssim}&\varepsilon^{1/2}\norm{P_ku_0}_{L^2}+2^{-k}\norm{P_kF}_{L_{t,x}^2}.
\end{align*}
Now we show the first inequality. We only prove the case $b=\infty$ since the other cases are similar.
First we assume $F=0$. Then $u=e^{it\Delta+\varepsilon t\Delta}u_0$. Let
$\tilde u=e^{it\Delta+\varepsilon |t|\Delta}u_0$, then $\tilde u$ is an
extension of $u$. Then
\begin{align*}
\norm{P_ku}_{X_+^{0,1/2,\infty}}{\lesssim}
&\sup_j 2^{j/2}\norm{\mathcal{F}_t(e^{-it|\xi|^2-\varepsilon|t|\cdot |\xi|^2})(\tau)\hat u_0(\xi)\chi_k(\xi)\chi_j(\tau+|\xi|^2)}_{L_{\tau,\xi}^2}\\
{\lesssim}&\sup_j
2^{j/2}\norm{(\varepsilon|\xi|^2)^{-1}\bigg(1+\frac{|\tau+|\xi|^2|^2}{(\varepsilon|\xi|^2)^2}\bigg)^{-1}\hat
u_0(\xi)\chi_k(\xi)\chi_j(\tau+|\xi|^2)}_{L_{\tau,\xi}^2}\\
{\lesssim}& \norm{P_ku_0}_2.
\end{align*}
Next we assume $u_0=0$. Fix an extension $\tilde F$ of $F$ such that
\[\norm{\tilde F}_{X^{0,-1/2,\infty}}\leq
2\norm{F}_{X_+^{0,-1/2,\infty}}.\] Then define $\tilde
u=\mathcal{F}^{-1}\frac{1}{\tau+|\xi|^2+i\varepsilon |\xi|^2}\mathcal{F} \tilde F$. We see
$\tilde u$ is an extension of $u$ and then
\begin{align*}
\norm{P_ku}_{X_+^{0,1/2,\infty}}{\lesssim}& \norm{\tilde
u}_{X^{0,1/2,\infty}}\\
{\lesssim}&\sup_j 2^{j/2}\norm{\chi_k(\xi)\chi_j(\tau+|\xi|^2)\frac{1}{\tau+|\xi|^2+i\varepsilon |\xi|^2}\mathcal{F} \tilde F}_{L_{\tau,\xi}^2}\\
{\lesssim}&\norm{\tilde F}_{X^{0,-1/2,\infty}}{\lesssim}
\norm{F}_{X_+^{0,-1/2,\infty}}.
\end{align*}
Thus we complete the proof.
\end{proof}

\begin{lem}\label{lem:max}
Let $n\geq 3$, $k\in {\mathbb Z},\varepsilon\geq 0$. Assume $u,F$ solves the equation
\[u_t-i\Delta u-\varepsilon \Delta u=F(x,t), \quad u(x,0)=u_0.\]
Then for any $\ve e\in \mathbb{S}^{n-1}$ we have
\begin{align}
\norm{P_{k}u}_{L_{\ve e}^{2,\infty}}{\lesssim}&
2^{k(n-1)/2}\norm{u_0}_{L^2}+2^{k(n-2)/2}\sup_{\ve e\in
\mathbb{S}^{n-1}}\norm{F}_{L_{\ve e}^{1,2}},\\
\norm{P_{k,\ve e}u}_{L_{\ve e}^{\infty,2}}{\lesssim}&
2^{-k/2}\norm{u_0}_{L^2}+2^{-k}\sup_{\ve e\in
\mathbb{S}^{n-1}}\norm{F}_{L_{\ve e}^{1,2}},
\end{align}
where the implicit constant is independent of $\varepsilon$.
\end{lem}

\begin{proof}
If $\varepsilon=0$, then the inequalities were proved in \cite{IK2}. By the
scaling and rotational invariance, we may assume $k=0$ and $\ve
e=(1,0,\cdots,0)$.  Then the second inequality follows from Proposition 2.5,
2.7 in \cite{HWG}.  We prove the first inequality by the following two steps.

{\bf Step 1:} $F=0$.

From the fact
\begin{align*}
|e^{\varepsilon t\Delta}f(\cdot,t)(x)|\leq& (\varepsilon t)^{-n/2}\int
e^{-\frac{|x-y|^2}{2\varepsilon t}}|f(y,t)|dy\\
{\lesssim}& (\varepsilon t)^{-n/2}\int e^{-\frac{|x-y|^2}{2\varepsilon
t}}\norm{f(y,t)}_{L_t^\infty}dy
\end{align*}
we get
\begin{align*}
\norm{e^{it\Delta+\varepsilon
t\Delta}P_0u_0}_{L_{x_1}^2L_{\bar{x}_1,t}^{\infty}} {\lesssim}
\norm{e^{it\Delta}P_0u_0}_{L_{x_1}^2L_{\bar{x}_1,t}^{\infty}}{\lesssim}
\norm{u_0}_2.
\end{align*}

{\bf Step 2:} $u_0=0$.

Decompose $P_0u=U_1+\cdots U_n$ such that ${\mathcal{F}}_xU_i$ is supported in
$\{|\xi|\sim 1: |\xi_i|\sim 1\}\times {\mathbb R}$. Thus it suffices to show
\begin{align}
\norm{U_i}_{L_{x_1}^2L_{\bar{x},t}^{\infty}}{\lesssim} \norm{F}_{L_{\ve
e_i}^{1,2}}.
\end{align}
We only show the estimate for $U_1$. We still write $u=U_1$. We
assume ${\mathcal{F}}_xF$ is supported in $\{|\xi|\sim 1: \xi_1\sim 1\}\times
{\mathbb R}$. We have
\begin{align*}
u(t,x)=&\int_{{\mathbb R}^{n+1}}\frac{e^{it\tau}e^{ix\xi}}{\tau+|\xi|^2+i\varepsilon
|\xi|^2}\widehat{F}(\xi,\tau)d\xi
d\tau\\
=&\int_{{\mathbb R}^{n+1}}\frac{e^{it\tau}e^{ix\xi}}{\tau+|\xi|^2+i\varepsilon
|\xi|^2}\widehat{F}(\xi,\tau)(1_{\{-\tau-|\bar \xi|^2\sim
1\}^c}+1_{-\tau-|\bar\xi|^2\sim 1, |\tau+|\xi|^2|{\lesssim} \varepsilon}\\
&+1_{-\tau-|\bar\xi|^2\sim 1, |\tau+|\xi|^2|\gg \varepsilon})d\xi d\tau\\
:=&u_1+u_2+u_3.
\end{align*}
For $u_1$, we simply use the Plancherel equality and get
\[\norm{\Delta u_1}_{L^2}+\norm{\partial_t u_1}_2\leq \norm{F}_2,\]
and thus by Sobolev embedding and Bernstein's inequality we obtain
the desired estimate. For $u_2$, using the Lemma \ref{lem:ext}, Lemma \ref{lem:linearStr} and
Bernstein's inequality we get
\begin{align*}
\norm{u_2}_{L_{x_1}^2L_{\bar x_1,t}^{\infty}}{\lesssim} \norm{u_2}_{\dot
X^{0,1/2,1}}{\lesssim} \varepsilon^{-1/2}\norm{\widehat F}_{L^2}{\lesssim}
\norm{F}_{L_{\ve e_1}^{1,2}}.
\end{align*}
Now we estimate $u_3$. Since $|\tau+|\xi|^2|\gg \varepsilon$, we have
\begin{align*}
\frac{1}{\tau+|\xi|^2+i\varepsilon
|\xi|^2}=\frac{1}{\tau+|\xi|^2}+\sum_{k=1}^\infty \frac{(-i\varepsilon
|\xi|^2)^k}{(\tau+|\xi|^2)^{k+1}}.
\end{align*}
Moreover, let $s=(-\tau-|\bar \xi|^2)^{1/2}$. Then
$\tau+|\xi|^2=-(s-\xi_1)(s+\xi_1)$, and thus we get $|s-\xi_1|\gg \varepsilon,
|s+\xi_1|\sim 1$
\[(\tau+|\xi|^2)^{-1}=-\frac{1}{2s}(\frac{1}{s-\xi_1}+\frac{1}{s+\xi_1})=-\frac{1}{2s(s-\xi_1)}(1+\frac{s-\xi_1}{s+\xi_1}).\]
Hence
\begin{align*}
\frac{1}{\tau+|\xi|^2+i\varepsilon
|\xi|^2}=&\frac{1}{\tau+|\xi|^2}+\sum_{k=1}^\infty \frac{(-i\varepsilon
|\xi|^2)^k}{(2s(\xi_1-s))^{k+1}}\\
&+\sum_{k=1}^\infty \frac{(-i\varepsilon
|\xi|^2)^k}{(2s(\xi_1-s))^{k+1}}[(1+\frac{s-\xi_1}{s+\xi_1})^{k+1}-1]\\
:=&a_1(\xi,\tau)+a_2(\xi,\tau)+a_3(\xi,\tau).
\end{align*}
Inserting this identity into the expression of $u_3$, then we have
$u_3=u_3^1+u_3^2+u_3^3$, where
\[u_3^j=\int_{{\mathbb R}^{n+1}}{e^{it\tau}e^{ix\xi}}a_j(\xi,\tau)\widehat{F}(\xi,\tau)1_{-\tau-|\bar\xi|^2\sim 1, |\tau+|\xi|^2|\gg \varepsilon}d\xi d\tau, \quad j=1,2,3.\]
For $u_3^1$, this corresponds to the case $\varepsilon=0$ which is proved in
\cite{IK2}. For $u_3^3$, we can control it similarly as $u_1$, since
\[|a_3(\xi,\tau)|{\lesssim} \sum_{k=1}^\infty \frac{\varepsilon^k
|\xi|^{2k}}{(2|s(s-\xi_1)|)^{k+1}}(k+1)\frac{|s-\xi_1|}{|s+\xi_1|}{\lesssim}
1.\]

It remains to control $u_3^2$. Let $G_k(x_1,\bar
\xi,\tau)={\mathcal{F}}_{x_1}^{-1}1_{-\tau-|\bar \xi|^2\sim 1}1_{|\xi|\sim
1}|\xi|^{2k}\widehat{F}$. Note that \[\norm{G_k}_{L_{\ve
e_1}^{1,2}}{\lesssim} c^k\norm{F}_{L_{\ve e_1}^{1,2}}.\] Then
\begin{align*}
u_3^2=&\sum_{k=1}^\infty (-i\varepsilon)^k\int_{\mathbb R} \int_{{\mathbb R}^{n+1}}
\frac{e^{it\tau}e^{ix\xi}1_{|s-\xi_1|\gg \varepsilon}}{(2s(\xi_1-s))^{k+1}}[e^{-iy_1\xi_1}G_k(y_1,\bar \xi,\tau)]d\xi d\tau dy_1\\
=&\sum_{k=1}^\infty (-i\varepsilon)^k\int_{\mathbb R} T^k_{y_1}(G(y_1,\cdot))(t,x)
dy_1
\end{align*}
where \[T^k_{y_1}(f)(t,x)=\int_{{\mathbb R}^{n+1}}
\frac{e^{it\tau}e^{ix\xi}1_{|s-\xi_1|\gg
\varepsilon}}{(2s(\xi_1-s))^{k+1}}[e^{-iy_1\xi_1}f(y_1,\bar \xi,\tau)]d\xi
d\tau.\] We have
\begin{align*}
T^k_{y_1}(f)(t,x)=&\int_{{\mathbb R}^{n}} \int_{\mathbb R}
\frac{e^{i(x_1-y_1)\xi_1}1_{|s-\xi_1|\gg \varepsilon}}{(\xi_1-s)^{k+1}}
d\xi_1\cdot (2s)^{-k-1}{e^{it\tau}e^{i\bar x\cdot
\bar\xi}}[f(y_1,\bar \xi,\tau)]d\bar\xi d\tau\\
=&\int_{\mathbb R} \frac{e^{i(x_1-y_1)\xi_1}1_{|\xi_1|\gg \varepsilon}}{(\xi_1)^{k+1}}
d\xi_1\cdot \int_{{\mathbb R}^{n}} e^{i(x_1-y_1)s}
(2s)^{-k-1}{e^{it\tau}e^{i\bar x\cdot \bar\xi}}[f(y_1,\bar
\xi,\tau)]d\bar\xi d\tau.
\end{align*}
Then we get
\[|T^k_{y_1}(f)(t,x)|{\lesssim} M^{-k}\varepsilon^{-k}\bigg|\int_{{\mathbb R}^{n}} e^{i(x_1-y_1)s}\cdot
(2s)^{-k-1}{e^{it\tau}e^{i\bar x\cdot \bar\xi}}[f(y_1,\bar
\xi,\tau)]d\bar\xi d\tau\bigg|.\] Making a change of variable
$\eta_1=s=\sqrt{-\tau-|\bar\xi|^2}$, $d\tau=-2\eta_1 d\eta_1$, we
obtain
\begin{align*}
&\int_{{\mathbb R}^{n}} e^{i(x_1-y_1)s}\cdot (2s)^{-k-1}{e^{it\tau}e^{i\bar
x\cdot \bar\xi}}[f(y_1,\bar \xi,\tau)]d\bar\xi d\tau\\
=&\int_{{\mathbb R}^{n}} e^{i(x_1-y_1)\eta_1}\cdot
(2\eta_1)^{-k}{e^{it(\eta_1^2+|\bar \xi|^2)}e^{i\bar x\cdot
\bar\xi}}[f(y_1,\bar \xi,\eta_1^2+|\bar \xi|^2)]d\bar\xi d\tau.
\end{align*}
Thus, by the linear estimate (see Lemma \ref{lem:linearStr}) we get
\begin{align*}
\norm{T^k_{y_1}(f)}_{_{L_{x_1}^2L_{\bar x,t}^{\infty}}}{\lesssim}
M^{-k}\varepsilon^{-k} \norm{f}_2,
\end{align*}
which suffices to give the estimate for $u_3^2$. We complete the
proof of the lemma.
\end{proof}

\begin{lem} Let $n\geq 3$, $k\in {\mathbb Z},\varepsilon\geq 0$. Assume $u,F$ solves the equation
\[u_t-i\Delta u-\varepsilon \Delta u=F(x,t), \quad u(x,0)=u_0.\]
Then for any $\ve e\in \mathbb{S}^{n-1}$ we have
\begin{align}
\norm{P_ku}_{L_t^\infty L_x^2\cap L_t^2L_x^{\frac{2n}{n-2}}}{\lesssim}
\norm{u_0}_{L^2}+\norm{F}_{N_k},
\end{align}
where the implicit constant is independent of $\varepsilon$.
\end{lem}
\begin{proof}
Since we have for $t>0$
\begin{align*}
\norm{e^{it\Delta+\varepsilon t\Delta}u_0}_{L_x^\infty}{\lesssim}&
t^{-n/2}\norm{u_0}_{L_x^1}\\
\norm{e^{it\Delta+\varepsilon t\Delta}u_0}_{L_x^2}{\lesssim}& \norm{u_0}_{L_x^2}
\end{align*}
where the implicit constant is independent of $\varepsilon$, then by the
abstract framework of Keel-Tao \cite{KT} we get the Strichartz
estimates
\begin{align*}
\norm{P_ku}_{L_t^\infty L_x^2\cap L_t^2L_x^{\frac{2n}{n-2}}}{\lesssim}
\norm{u_0}_{L^2}+\norm{F}_{L_t^1L_x^2},
\end{align*}
with the implicit constant independent of $\varepsilon$.

By the same argument as in Step 2 of the proof of Lemma
\ref{lem:max}, we get
\begin{align*}
\norm{P_ku}_{L_t^\infty L_x^2\cap L_t^2L_x^{\frac{2n}{n-2}}}{\lesssim}
\norm{u_0}_{L^2}+2^{-k/2}\sup_{\ve e\in \mathbb{S}^{n-1}}\norm{F}_{L_{\ve
e}^{1,2}}.
\end{align*}
On the other hand, by Lemma \ref{lem:ext}, Lemma \ref{lem:linearStr} and Lemma \ref{lem:linearXsb}, we get
\begin{align*}
\norm{P_ku}_{L_t^\infty L_x^2\cap L_t^2L_x^{\frac{2n}{n-2}}}{\lesssim}
\norm{P_ku}_{X_+^{0,1/2,1}}{\lesssim}
\norm{u_0}_{L^2}+\norm{F}_{X_+^{0,-1/2,1}}.
\end{align*}
Thus we complete the proof.
\end{proof}


Gathering the above lemmas, we can get the following linear estimates:

\begin{lem}[Linear estimates]\label{lem:linear}
Assume $n\geq 3$, $u,F$ solves the equation: for $\varepsilon>0$
\[u_t-i\Delta u-\varepsilon \Delta u=F(x,t), \quad u(x,0)=u_0.\]
Then for $s\in {\mathbb R}$
\begin{align}
\norm{u}_{F^s\cap Z^s}{\lesssim}&
\norm{u_0}_{\dot B^s_{2,1}}+\norm{F}_{N^s},
\end{align}
where the implicit constant is independent of $\varepsilon$.
\end{lem}




\section{Nonlinear estimates}

In this section we prove some nonlinear estimates.
The nonlinear term in the Landau-Lifshitz equation is
\[G(u)=\frac{\bar u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2.\]
By Taylor's expansion, if $\norm{u}_\infty <1$ we have
\[G(u)=\sum_{k=0}^\infty \bar u(-1)^k|u|^{2k}\sum_{j=1}^n(\partial_{x_j}u)^2.\]
So we will need to do multilinear estimates.

\begin{lem}
(1) If $j\geq 2k-100$ and $X$ is a space-time translation invariant
Banach space, then $Q_{\leq j}P_k$ is bounded on $X$ with bound
independent of $j,k$.

(2) For any $j,k$, $Q_{\leq j}P_{k,\ve e}$ is bounded on $L_{\ve
e}^{p, 2}$ and $Q_{\leq j}$ is bounded on $L_t^pL_x^2$ for $1\leq p
\leq \infty$, with bound independent of $j,k$.
\end{lem}

\begin{proof}
See the proof of Lemma 5.4 in \cite{Guo}.
\end{proof}

\begin{lem}\label{lem:L2est}
Assume $n\geq 3$, $k_1,k_2,k_3\in {\mathbb Z}$. Then
\begin{align}
\norm{P_{k_1}uP_{k_2}v}_{L_{t,x}^{2}}{\lesssim}&
2^{(n-1)k_1/2}2^{-k_2/2}\norm{P_{k_1}u}_{Y_{k_1}+F_{k_1}}\norm{P_{k_2}v}_{F_{k_2}},\\
\norm{P_{k_3}(P_{k_1}uP_{k_2}v)}_{L_{t,x}^{2}}{\lesssim}&
2^{\frac{(n-2)\min(k_1,k_2,k_3)}{2}}\norm{P_{k_1}u}_{Y_{k_1}}\norm{P_{k_2}v}_{Y_{k_2}}.
\end{align}
\end{lem}
\begin{proof}
For the first inequality, we have
\begin{align*}
\norm{P_{k_1}uP_{k_2} v}_{L_{t,x}^{2}}
{\lesssim}& \sum_{j=1}^n\norm{P_{k_1} uP_{k_2,\ve e_j}\Theta_{k_2}^jv}_{L_{t,x}^{2}}{\lesssim} \sum_{j=1}^n\norm{P_{k_1} u}_{L_{\ve
e_j}^{2,\infty}}\norm{P_{k_2,\ve e_j}v}_{L_{\ve e_j}^{\infty,2}}\\
{\lesssim}&2^{(n-1)k_1/2}2^{-k_2/2}\norm{P_{k_1}u}_{Y_{k_1}+F_{k_1}}\norm{P_{k_2}v}_{F_{k_2}}.
\end{align*}
For the second inequality, if $k_3\leq \min(k_1,k_2)$, then
\begin{align*}
\norm{P_{k_3}(P_{k_1}uP_{k_2}v)}_{L_{t,x}^{2}}{\lesssim}&2^{k_3(n-2)/2}\norm{P_{k_1}uP_{k_2}v}_{L_{t}^{2}L_x^{\frac{2n}{2n-2}}}\\
{\lesssim}&2^{k_3(n-2)/2}\norm{P_{k_1}u}_{L_{t}^{2}L_x^{\frac{2n}{n-2}}}\norm{P_{k_2}v}_{L_{t}^\infty L_x^2}.
\end{align*}
If $k_1\leq \min(k_2,k_3)$, then
\begin{align*}
\norm{P_{k_3}(P_{k_1}uP_{k_2}v)}_{L_{t,x}^{2}}
{\lesssim}&\norm{P_{k_1}u}_{L_{t}^{2}L_x^\infty}\norm{P_{k_2}v}_{L_{t}^\infty L_x^2}\\
{\lesssim}&2^{k_1(n-2)/2}\norm{P_{k_1}u}_{L_{t}^{2}L_x^{\frac{2n}{n-2}}}\norm{P_{k_2}v}_{L_{t}^\infty L_x^2}.
\end{align*}
If $k_2\leq \min(k_1,k_3)$, the proof is identical to the above case.
\end{proof}


\begin{lem}[Algebra properties]\label{lem:alg}
If $s\geq n/2$, then we have
\begin{align*}
\norm{uv}_{Y^{s}}{\lesssim}&
\norm{u}_{Y^{s}}\norm{v}_{Y^{n/2}}+\norm{u}_{Y^{n/2}}\norm{v}_{Y^{s}}.
\end{align*}
\end{lem}
\begin{proof}
We only show the case $s=n/2$. By the embedding $\dot
B^{n/2}_{2,1}\subset L_x^\infty$ we get
\[\norm{u}_{L^\infty_{x,t}}\leq \norm{u}_{L^\infty_{t}\dot
B_{2,1}^{n/2}}{\lesssim} \norm{u}_{Y^{n/2}}.\] The Lebesgue component can
be easily handled by para-product decomposition and H\"older's
inequality. Now we deal with $X^{s,b}$-type space.  By \eqref{eq:X01ext} it suffices to
show
\begin{align}
\sum_k2^{nk/2}2^{-k}\norm{P_k(fg)}_{X^{0,1}+\bar X^{0,1}}{\lesssim}&
\norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}},\label{eq:algeXsb}
\end{align}
We have
\begin{align*}
&\sum_k2^{nk/2}2^{-k}\norm{P_k(fg)}_{X^{0,1}+\bar X^{0,1}}\\
{\lesssim}& \sum_{k_i}2^{nk_3/2}2^{-k_3}\norm{P_{k_3}(P_{k_1}fP_{k_2}g)}_{X^{0,1}+\bar X^{0,1}}\\
{\lesssim}& (\sum_{k_i:k_1\leq k_2}+\sum_{k_i:k_1>
k_2})2^{k_3n/2}2^{-k_3}\norm{P_{k_3}(P_{k_1}fP_{k_2}g)}_{X^{0,1}+\bar X^{0,1}}\\
:=&I+II.
\end{align*}
By symmetry, we only need to estimate the term $I$.

Assume $P_{k_1}f=P_{k_1}f_1+P_{k_1}f_2$,
$P_{k_2}g=P_{k_2}g_1+P_{k_2}g_2$ such that
\begin{align*}
\norm{P_{k_1}f_1}_{X^{0,1}}+\norm{P_{k_1}f_2}_{\bar X^{0,1}}{\lesssim}& \norm{P_{k_1}f}_{X^{0,1}+\bar X^{0,1}},\\
\norm{P_{k_2}g_1}_{X^{0,1}}+\norm{P_{k_2}g_2}_{\bar X^{0,1}}{\lesssim}& \norm{P_{k_2}g}_{X^{0,1}+\bar X^{0,1}}.
\end{align*}
Then we have
\begin{align*}
I{\lesssim}& \sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}\norm{P_{k_3}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}}\\
&+\sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}\norm{P_{k_3}(P_{k_1}f_jP_{k_2}g_2)}_{\bar X^{0,1}}\\
:=&I_1+I_2.
\end{align*}
We only estimate the term $I_1$ since the term $I_2$ can be
estimated in a similar way. We have
\begin{align}\label{eq:I1}
I_1{\lesssim}&\sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}\norm{P_{k_3}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}}.
\end{align}
First we assume $k_3\leq k_1+5$ in the summation of \eqref{eq:I1}. We
have
\begin{align*}
I_1{\lesssim}& \sum_{k_i:k_1\leq k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}(\norm{P_{k_3}Q_{\leq k_1+k_2+9}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}}\\
&\qquad+\norm{P_{k_3}Q_{\geq k_1+k_2+10}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}})\\
:=&I_{11}+I_{12}.
\end{align*}
For the term $I_{11}$ we have
\begin{align*}
I_{11}{\lesssim} &\sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{k_3n/2}2^{-k_3}2^{k_1+k_2}\norm{P_{k_1}f_j}_{L_t^\infty
L_{x}^n}\norm{P_{k_2}g_1}_{L_t^2L_x^{\frac{2n}{n-2}}}\\
{\lesssim}& \sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{k_3n/2}2^{-k_3}2^{k_2}2^{k_1n/2}\norm{P_{k_1}f_j}_{L_t^\infty
L_{x}^2}\norm{P_{k_2}g_1}_{L_t^2L_x^{\frac{2n}{n-2}}}\\
{\lesssim}&\norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}}.
\end{align*}
For the term $I_{12}$, we need to exploit the nonlinear interactions as in \cite{Guo}. We have
\begin{align*}
&{\mathcal{F}} P_{k_3}Q_{\geq k_1+k_2+10}(P_{k_1}f_jP_{k_2}g_1)\\
&=\chi_{k_3}(\xi_3)\chi_{\geq k_1+k_2+10}(\tau_3+|\xi_3|^2)
\int_{\xi_3=\xi_1+\xi_2,\tau_3=\tau_1+\tau_2}\chi_{k_1}(\xi_1)\widehat{f_j}(\tau_1,\xi_1)\chi_{k_2}(\xi_2)\widehat{g_1}(\tau_2,\xi_2).
\end{align*}
We assume $j=1$ since $j=2$ is similar.  On the plane $\{\xi_3=\xi_1+\xi_2,\tau_3=\tau_1+\tau_2\}$ we have
\begin{align}
\tau_3+|\xi_3|^2=\tau_1+|\xi_1|^2+\tau_2+|\xi_2|^2-H(\xi_1,\xi_2)
\end{align}
where $H$ is the resonance function in the product
$P_{k_3}(P_{k_1}f_jP_{k_2}g_1)$
\begin{align}
H(\xi_1,\xi_2)=|\xi_1|^2+|\xi_2|^2-|\xi_1+\xi_2|^2.
\end{align}
Since $|H|{\lesssim} 2^{k_1+k_2}$, then
one of $P_{k_1}f_j$, $P_{k_2}g_1$ has modulation larger than the
output modulation, namely
\[\max(|\tau_1+|\xi_1|^2|,|\tau_2+|\xi_2|^2|){\gtrsim} |\tau_3+|\xi_3|^2|.\]
If $P_{k_1}f_j$ has larger modulation, then
\begin{align*}
I_{12}{\lesssim}& \sum_{k_i:k_1\leq k_2}2^{nk_3/2}2^{-k_3}\norm{2^{j_3}\norm{P_{k_3}Q_{j_3}(P_{k_1}f_jP_{k_2}g_1)}_{L^2_{t,x}}}_{l^2_{j_3\geq k_1+k_2+10}}\\
{\lesssim}&
\sum_{k_i:k_1\leq k_2}\sum_{j=1}^22^{nk_3/2}2^{nk_3/2}2^{-k_3}(\sum_{j_3\geq k_1+k_2+10}2^{2j_3}\norm{Q_{\geq j_3}P_{k_1}f_j}^2_{L_{t,x}^2})^{1/2}\norm{P_{k_2}g_1}_{L^\infty_tL_x^2}\\
{\lesssim}&\sum_{k_i:k_1\leq
k_2}2^{nk_3}2^{-k_3}(\norm{P_{k_1}f_1}_{X^{0,1}}+\norm{P_{k_1}f_2}_{\bar X^{0,1}})\norm{P_{k_2}g_1}_{Y_{k_2}}\\
{\lesssim}&
\norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}}.
\end{align*}
If $P_{k_2}g_1$ has larger modulation, then
\begin{align*}
I_{12}{\lesssim}& \sum_{k_i:k_1\leq k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}\norm{P_{k_1}f_j}_{L_{t,x}^\infty}(\sum_{j_3\geq k_1+k_2}2^{2j_3}\norm{P_{k_2}Q_{\geq j_3}g_1}^2_{L^2_tL_x^2})^{1/2}\\
{\lesssim}&\sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}2^{nk_1/2}\norm{P_{k_1}f_j}_{Y_{k_1}}\norm{P_{k_2}g_1}_{X^{0,1}}\\
{\lesssim}&
\norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}}.
\end{align*}

Now we assume $k_3\geq k_1+6$ in the summation of \eqref{eq:I1} and thus $|k_2-k_3|\leq 4$.  We have
\begin{align*}
I_1{\lesssim}& \sum_{k_i:k_1\leq k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}(\norm{P_{k_3}Q_{\leq k_1+k_2+9}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}}\\
&\qquad+\norm{P_{k_3}Q_{\geq k_1+k_2+10}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}})\\
:=&\tilde I_{11}+\tilde I_{12}.
\end{align*}
By Lemma \ref{lem:L2est} we get
\begin{align*}
\tilde I_{11}{\lesssim}& \sum_{k_i:k_1\leq k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}(\norm{P_{k_3}Q_{\leq k_1+k_2+9}(P_{k_1}f_jP_{k_2}g_1)}_{X^{0,1}}\\
{\lesssim}&\sum_{k_i:k_1\leq
k_2}2^{nk_3/2}2^{k_1}2^{(n-2)k_1/2}\norm{P_{k_1}f_j}_{Y_{k_1}}\norm{P_{k_2}g_1}_{Y_{k_2}}{\lesssim}
\norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}}.
\end{align*}
For the term $\tilde I_{12}$, similarly as the term $I_{12}$,
one of $P_{k_1}f_j$, $P_{k_2}g_1$ has modulation larger than the
output modulation. If $P_{k_1}f_j$ has larger modulation, then
\begin{align*}
\tilde I_{12}{\lesssim}& \sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}(\sum_{j_3}2^{2j_3}\norm{P_{k_1}Q_{\geq
j_3}f_j}^2_{L_t^2L_x^\infty})^{1/2} \norm{P_{k_2}g_1}_{L_t^\infty
L_x^2}\\
{\lesssim}&\norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}}.
\end{align*}
If $P_{k_2}g_1$ has larger modulation, then
\begin{align*}
\tilde I_{12}{\lesssim}& \sum_{k_i:k_1\leq
k_2}\sum_{j=1}^22^{nk_3/2}2^{-k_3}\norm{P_{k_1}f_j}_{L_t^\infty L_x^\infty}
(\sum_{j_3}2^{2j_3}\norm{P_{k_2}Q_{\geq j_3}g_1}^2_{L_t^2
L_x^2})^{1/2}\\
{\lesssim}& \norm{f}_{Y^{n/2}}\norm{g}_{Y^{n/2}}.
\end{align*}
Thus, we complete the proof.
\end{proof}

\begin{lem}\label{lem:L2non}
We have
\begin{align}\label{eq:nonL2est}
\sum_{k_j}2^{k_3(n-2)/2}\norm{P_{k_3}[u\sum_{i=1}^n (\partial_{x_i}P_{k_1}v
\partial_{x_i}P_{k_2}w)]}_{L_{t,x}^2}{\lesssim}
\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align}
\end{lem}
\begin{proof}
We have
\begin{align*}
\mbox{LHS of \eqref{eq:nonL2est}}{\lesssim}&
\sum_{k_j}2^{k_3(n-2)/2}\norm{P_{k_3}[P_{\geq
k_3-10}u\sum_{i=1}^n
(\partial_{x_i}P_{k_1}v \partial_{x_i}P_{k_2}w)]}_{L_{t,x}^2}\\
&+ \sum_{k_j}2^{k_3(n-2)/2}\norm{P_{k_3}[P_{\leq k_3-10}u\sum_{i=1}^n
(\partial_{x_i}P_{k_1}v \partial_{x_i}P_{k_2}w)]}_{L_{t,x}^2}\\
:=&I+II.
\end{align*}
By symmetry, we may assume $k_1\leq k_2$ in the above summation.
For the term $II$, since $n\geq 3$, then we have
\begin{align*}
II{\lesssim}& \norm{u}_{Y^{n/2}} \sum_{k_j}2^{k_3(n-2)/2}\norm{\tilde
P_{k_3}\sum_{i=1}^n (\partial_{x_i}P_{k_1}v
\partial_{x_i}P_{k_2}w)]}_{L_{t,x}^2}\\
{\lesssim}&\norm{u}_{Y^{n/2}} \sum_{k_1,k_3\leq k_2+5}2^{k_3(n-2)/2}2^{k_1+k_2}2^{[(n-1)k_1-k_2]/2}\norm{P_{k_1}v}_{F_{k_1}}\norm{P_{k_2}w}_{F_{k_2}}\\
{\lesssim}& \norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
For the term $I$, if $k_3\leq k_2+20$, then we get from Lemma \ref{lem:L2est} that
\begin{align*}
I{\lesssim}& \sum_{k_j}2^{nk_3/2}2^{k_3(n-2)/2}\norm{P_{\geq k_3-10}u}_{L_t^\infty
L_x^2}\norm{\sum_{i=1}^n (\partial_{x_i}P_{k_1}v
\partial_{x_i}P_{k_2}w)]}_{L_{t,x}^2}\\
{\lesssim}&\sum_{k_j}2^{nk_3/2}2^{k_3(n-2)/2}2^{(n+1)k_1/2}2^{k_2/2}\norm{P_{\geq
k_3-10}u}_{L_t^\infty L_x^2}\norm{P_{k_1}v}_{F_{k_1}}\norm{P_{k_2}w}_{F_{k_2}}\\
{\lesssim}& \norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
If $k_3\geq k_2+20$, then $u$ has frequency $\sim 2^{k_3}$, and thus we get
\begin{align*}
I{\lesssim}& \sum_{k_j}2^{k_3(n-2)/2}\norm{P_{k_3}u}_{L_t^\infty
L_x^2}\norm{\sum_{i=1}^n (\partial_{x_i}P_{k_1}v
\partial_{x_i}P_{k_2}w)]}_{L_{t}^2L_x^\infty}\\
{\lesssim}& \sum_{k_j}2^{k_3(n-2)/2}\norm{P_{k_3}u}_{L_t^\infty
L_x^2}2^{nk_2/2}\norm{\sum_{i=1}^n (\partial_{x_i}P_{k_1}v
\partial_{x_i}P_{k_2}w)]}_{L_{t}^2L_x^2}\\
{\lesssim}&\sum_{k_j}2^{k_3(n-2)/2}2^{k_1+k_2}2^{(n-1)k_1/2}2^{(n-1)k_2/2}\norm{P_{k_3}u}_{L_t^\infty L_x^2}\norm{P_{k_1}v}_{F_{k_1}}\norm{P_{k_2}w}_{F_{k_2}}\\
{\lesssim}& \norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
Therefore we complete the proof.
\end{proof}


\begin{lem}\label{lem:Nnon}
We have
\begin{align}
\norm{u\sum_{i=1}^n(\partial_{x_i}v \partial_{x_i}w)}_{N^{n/2}}{\lesssim}&
\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}\cap Z^{n/2}}\norm{w}_{F^{n/2}\cap Z^{n/2}}.
\end{align}
\end{lem}
\begin{proof}
By the definition of $N^{n/2}$, the $L^2$ component was handled by the previous lemma. We only need to control
\begin{align}\label{eq:tripf1}
\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\sum_{i=1}^n(P_{k_2}\partial_{x_i}v
\partial_{x_i}P_{k_3}w)]}_{N_{k_4}}.
\end{align}
By symmetry we may assume $k_2\leq k_3$ in the above summation.
If in the above summation we assume $k_4\leq k_1+40$, then
\begin{align*}
\eqref{eq:tripf1}{\lesssim}& \sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}uP_{k_2}\partial_{x_i}v
\partial_{x_i}P_{k_3}w]}_{L_t^1L_x^{2}}\\
{\lesssim}& \sum_{k_i}2^{k_1n/2}\norm{P_{k_1}u}_{L_t^\infty L_x^2}2^{k_2}\norm{P_{k_2}v}_{L_t^2L_x^{\infty}}2^{k_3}\norm{P_{k_3}w}_{L_t^2L_x^{\infty}}\\
{\lesssim}&\sum_{k_i}2^{k_1n/2}\norm{P_{k_1}u}_{L_t^\infty L_x^2}2^{k_2n/2}\norm{P_{k_2}v}_{L_{t}^{2}L_x^{\frac{2n}{n-2}}}2^{k_3n/2}\norm{P_{k_3}w}_{L_{t}^{2}L_x^{\frac{2n}{n-2}}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
Thus from now on we assume $k_4\geq k_1+40$ in the summation of \eqref{eq:tripf1}. We bound the summation case by case.

{\bf Case 1:} $k_2\leq k_1+20$

In this case we have $k_4\geq k_2+20$ and hence $|k_4-k_3|\leq 5$. By Lemma \ref{lem:L2est} we
get
\begin{align*}
\eqref{eq:tripf1}{\lesssim}&
\sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_4}[P_{k_1}uP_{k_2}\partial_{x_i}
v
\partial_{x_i}P_{k_3}w]}_{L_{\ve{e}}^{1,2}}\\
{\lesssim}& \sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_1}uP_{k_3}\partial_{x_i}w}_{L_{x,t}^{2}}\norm{P_{k_2}\partial_{x_i}v}_{L_{\ve{e}}^{2,\infty}}\\
{\lesssim}& \sum_{k_i}2^{k_4n/2}2^{-k_4/2}2^{(n-1)k_1/2}2^{-k_3/2}2^{(n-1)k_2/2}\norm{P_{k_1}u}_{Y_{k_1}}\norm{P_{k_3}\partial_{x_i}w}_{F_{k_3}}\norm{P_{k_2}\partial_{x_i}v}_{F_{k_2}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}

{\bf Case 2:} $k_2\geq k_1+21$

In this case we have $k_4\leq k_3+40$. Let $g=\sum_{i=1}^n(P_{k_2}\partial_{x_i}v\cdot
P_{k_3}\partial_{x_i}w)$. Then we have
\begin{align*}
\eqref{eq:tripf1}{\lesssim}&\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}uQ_{\leq k_2+k_3}g]}_{N_{k_4}}+\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}uQ_{\geq k_2+k_3}g]}_{N_{k_4}}\\
:=&I+II.
\end{align*}
First we estimate the term $II$.  We have
\begin{align*}
II{\lesssim} &\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}Q_{\geq k_2+k_3-10}u\cdot Q_{\geq k_2+k_3}g]}_{N_{k_4}}\\
&+\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}Q_{\leq k_2+k_3-10}u\cdot Q_{\geq k_2+k_3}g]}_{N_{k_4}}\\
:=&II_1+II_2.
\end{align*}
For the term $II_1$ we have
\begin{align*}
II_1{\lesssim} &\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}Q_{\geq k_2+k_3-10}u\cdot Q_{\geq k_2+k_3}g]}_{L_t^1L_x^2}\\
{\lesssim} &\sum_{k_i}2^{k_4n/2}\norm{P_{k_1}Q_{\geq k_2+k_3-10}u}_{L_t^2L_x^\infty}\norm{Q_{\geq k_2+k_3}g]}_{L_t^2L_x^2}\\
{\lesssim} &\sum_{k_i}2^{k_4n/2}2^{k_1n/2}\norm{P_{k_1}Q_{\geq k_2+k_3-10}u}_{L_t^2L_x^2}\norm{Q_{\geq k_2+k_3}g]}_{L_t^2L_x^2}\\
{\lesssim} &\sum_{k_i}2^{k_4n/2}2^{k_1n/2}2^{-(k_2+k_3)}\norm{P_{k_1}Q_{\geq 2k_1+10}u}_{X^{0,1}}\\
&\cdot 2^{[(n-1)k_2-k_3]/2}2^{k_2+k_3}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{F_{k_3}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
For the term $II_2$, since $k_4\geq k_1+40$, then we may assume $g$
has frequency of size $2^{k_4}$.  The resonance function in the
product $P_{k_1}u\cdot P_{k_4}g$ is of size ${\lesssim} 2^{k_1+k_4}$.
 Thus the output modulation is of size ${\gtrsim} 2^{k_2+k_3}$.  Then we get
\begin{align*}
II_2{\lesssim} &\sum_{k_i}2^{k_4n/2}2^{-(k_2+k_3)/2}\norm{P_{k_4}[P_{k_1}Q_{\leq k_2+k_3-10}u\cdot Q_{\geq k_2+k_3}g]}_{L^2_{t,x}}\\
{\lesssim} &\sum_{k_i}2^{k_4n/2}2^{-(k_2+k_3)/2}2^{k_1n/2}\norm{P_{k_1}u}_{L_t^\infty L_x^2}\cdot \norm{g}_{L^2_{t,x}}\\
{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{-(k_2+k_3)/2}2^{k_1n/2}2^{[(n-1)k_2-k_3]/2}2^{k_2+k_3} \norm{P_{k_1}u}_{Y_{k_1}}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{F_{k_3}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}

Now we estimate the term $I$.  We have
\begin{align*}
I{\lesssim} &\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_2+k_3}\sum_{i=1}^n(P_{k_2}\partial_{x_i}Q_{\geq k_2+k_3+40}v\cdot
P_{k_3}\partial_{x_i}w)]}_{N_{k_4}}\\
&+\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_2+k_3}\sum_{i=1}^n(P_{k_2}\partial_{x_i}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}\partial_{x_i}w)]}_{N_{k_4}}\\
:=&I_1+I_2.
\end{align*}
For the term $I_1$, since the resonance function in the product
$P_{k_2}v\cdot P_{k_3}w$ is of size ${\lesssim} 2^{k_2+k_3}$, then we may
assume $P_{k_3}w$ has modulation of size ${\gtrsim} 2^{k_2+k_3}$.  Then
we get
\begin{align*}
I_1{\lesssim}& \sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_4}[P_{k_1}u\cdot
Q_{\leq k_2+k_3}\sum_{i=1}^n(P_{k_2}\partial_{x_i}Q_{\geq
k_2+k_3+40}v\cdot
P_{k_3}\partial_{x_i}Q_{\geq k_2+k_3-5}w)]}_{L^{1,2}_{\ve e}}\\
{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_1}u}_{L^\infty_{t,x}}2^{k_2+k_3}\norm{P_{k_2}v}_{L^{2,\infty}_{\ve e}}\norm{P_{k_3}Q_{\geq k_2+k_3-5}w}_{L^2_{t,x}}\\
{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{-k_4/2}2^{(k_2+k_3)/2}2^{(n-1)k_2/2}2^{k_1n/2}\norm{P_{k_1}u}_{Y_{k_1}}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{F_{k_3}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
Finally, we estimate the term $I_2$.  For this term, we need to use
the null structure observed by Bejenaru \cite{Bej}.  We can rewrite
\begin{align}
-2\nabla u\cdot \nabla v=(i\partial_t+\Delta)u\cdot v+u\cdot
(i\partial_t+\Delta)v-(i\partial_t+\Delta)(u\cdot v).
\end{align}
Then we have
\begin{align*}
I_2=&\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_2+k_3}(P_{k_2}LQ_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{N_{k_4}}\\
&+\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_2+k_3}(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}Lw)]}_{N_{k_4}}\\
&+\sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_2+k_3}L(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{N_{k_4}}\\
:=&I_{21}+I_{22}+I_{23}.
\end{align*}
For the term $I_{21}$, we have
\begin{align*}
I_{21}{\lesssim}& \sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_2+k_3}(P_{k_2}L Q_{\leq k_2+k_3+39}v\cdot P_{k_3}w)]}_{L_t^{1}L^2_{x}}\\
{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{k_1n/2}\norm{P_{k_1}u}_{L_t^\infty
L_x^2}2^{k_2(n-2)/2}\norm{P_{k_2}L v}_{L^2_{t,x}}\norm{P_{k_3}w}_{L_t^{2}L_x^{\frac{2n}{n-2}}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{Z^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
For the term $I_{22}$, we may assume  $w$ has modulation ${\lesssim} 2^{k_2+k_3}$. Then we get
\begin{align*}
I_{22}{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_4}[P_{k_1}u\cdot
Q_{\leq k_2+k_3}(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}Q_{\leq k_2+k_3+100}Lw)]}_{L^{1,2}_{\ve e}}\\
{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{-k_4/2}2^{k_1n/2}\norm{P_{k_1}u}_{L_t^\infty
L_x^2}\norm{P_{k_2}v}_{L^{2,\infty}_{\ve e}}\norm{
P_{k_3}Q_{\leq k_2+k_3+100}Lw)]}_{L^{2}_{t,x}}\\
{\lesssim}&\sum_{k_i}2^{k_4n/2}2^{-k_4/2}2^{nk_1/2}2^{(n-1)k_2/2}2^{(k_2+k_3)/2}\norm{P_{k_1}u}_{Y_{k_1}}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{X^{0,1/2,\infty}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
Next we estimate the term $I_{23}$. We have
\begin{align*}
I_{23}{\lesssim}& \sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot
Q_{[k_1+k_4+100, k_2+k_3]}L(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{N_{k_4}} \\
&+ \sum_{k_i}2^{k_4n/2}\norm{P_{k_4}[P_{k_1}u\cdot Q_{\leq
k_1+k_4+99}L(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{N_{k_4}} \\
:=&I_{231}+I_{232}.
\end{align*}
For the term $I_{232}$ we have
\begin{align*}
I_{232}{\lesssim}& \sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_4}[P_{k_1}u\cdot
Q_{\leq k_1+k_4+99}L(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{L^{1,2}_{\ve e}} \\
{\lesssim} & \sum_{k_i}2^{k_4n/2}2^{-k_4/2}\norm{P_{k_1}u}_{L^{2,\infty}_{\ve e}}2^{k_1+k_4}2^{[(n-1)k_2-k_3]/2}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{F_{k_3}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
For the term $I_{231}$ we have
\begin{align*}
I_{231}{\lesssim}&
\sum_{k_i}\sum_{j_2=k_1+k_4+100}^{k_2+k_3}2^{k_4n/2}\norm{P_{k_4}Q_{\leq
j_2-10}[P_{k_1}u\cdot Q_{j_2}L(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{N_{k_4}}\\
&+\sum_{k_i}\sum_{j_2=k_1+k_4+100}^{k_2+k_3}2^{k_4n/2}\norm{P_{k_4}Q_{\geq
j_2-9}[P_{k_1}u\cdot Q_{j_2}L(P_{k_2}Q_{\leq k_2+k_3+39}v\cdot
P_{k_3}w)]}_{N_{k_4}}\\
:=&I_{2311}+I_{2312}.
\end{align*}
For the term $I_{2312}$ we have
\begin{align*}
I_{2312}{\lesssim}&
\sum_{k_i}\sum_{j_2=k_1+k_4+100}^{k_2+k_3}\sum_{j_3\geq
k_2-9}2^{k_4n/2}2^{-j_3/2}\\
&\cdot\norm{P_{k_4}Q_{j_3}[P_{k_1}u\cdot Q_{j_2}L(P_{k_2}Q_{\leq
k_2+k_3+39}v\cdot
P_{k_3}w)]}_{L^2_{t,x}}\\
{\lesssim}& \sum_{k_i}2^{k_4n/2}2^{k_1n/2}2^{(k_2+k_3)/2}2^{[(n-1)k_2-k_3]/2}\norm{P_{k_1}u}_{Y_{k_1}}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{F_{k_3}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
For the term $I_{2311}$ we have
\begin{align*}
I_{2311}{\lesssim}&
\sum_{k_i}\sum_{j_2=k_1+k_4+100}^{k_2+k_3}2^{k_4n/2}\norm{P_{k_4}Q_{\leq
j_2-10}[P_{k_1}\tilde Q_{j_2}u\cdot Q_{j_2}L(P_{k_2}Q_{\leq
k_2+k_3+39}v\cdot
P_{k_3}w)]}_{L_t^1L_x^2}\\
{\lesssim}&\sum_{k_i}\sum_{j_2=k_1+k_4+100}^{k_2+k_3}2^{k_4n/2}2^{k_1n/2}\norm{P_{k_1}\tilde Q_{j_2}u}_{L^2_{t,x}}2^{j_2}2^{[(n-1)k_2-k_3]/2}\norm{P_{k_2}v}_{F_{k_2}}\norm{P_{k_3}w}_{F_{k_3}}\\
{\lesssim}&\norm{u}_{Y^{n/2}}\norm{v}_{F^{n/2}}\norm{w}_{F^{n/2}}.
\end{align*}
Therefore, we complete the proof.
\end{proof}

Combining all the estimates above we get
\begin{lem}[Nonlinear estimates]\label{lem:nonlinear}
Assume $u\in F^{n/2}\cap Z^{n/2}$ with $\norm{u}_{Y^{n/2}}\ll 1$. Then
\begin{align*}
\normo{\frac{\bar
u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2}_{N^{n/2}}{\lesssim}&
\frac{\norm{u}_{Y^{n/2}}}{1-\norm{u}_{Y^{n/2}}^2}
\norm{u}_{F^{n/2}\cap Z^{n/2}}\norm{u}_{F^{n/2}\cap Z^{n/2}}.
\end{align*}
\end{lem}
\begin{proof}
Since $Y^{n/2}\subset L^\infty$, then
\[\frac{\bar
u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2=\sum_{k=0}^\infty \bar u(-1)^k|u|^{2k}\sum_{j=1}^n(\partial_{x_j}u)^2.\]
The lemma follows from Lemma \ref{lem:Nnon}, Lemma \ref{lem:L2non} and Lemma \ref{lem:alg}.
\end{proof}


\section{The limit behaviour}

In this section we prove Theorem \ref{thm:Schmap}. It is equivalent
to prove

\begin{thm}\label{thm:dGL}
Assume $n\geq 3$, $\varepsilon\in [0,1]$. There exists $0<\delta\ll 1$ such
for any $\phi\in \dot B_{2,1}^{n/2}$ with $\norm{\phi}_{\dot
B_{2,1}^{n/2}}\leq \delta$, there exists a unique global solution
$u_\varepsilon$ to \eqref{eq:dGL} such that
\[\norm{u_\varepsilon}_{F^{n/2}\cap Z^{n/2}}{\lesssim}
\delta,\] where the implicit constant is independent of $\varepsilon$. The
map $\phi\to u_\varepsilon$ is Lipshitz from $B_\delta(\dot B_{2,1}^{n/2})$
to $C({\mathbb R};\dot B_{2,1}^{n/2})$ and the Lipshitz constant is
independent of $\varepsilon$. Moreover, for any $T>0$,
\[\lim_{\varepsilon\to 0+}\norm{u_\varepsilon-u_0}_{C([0,T];\dot B_{2,1}^{n/2})}=0.\]
\end{thm}

For the uniform global well-posedness, we can prove it by standard
Picard iteration argument by using the linear and nonlinear
estimates proved in the previous section.
Indeed, define
\begin{align*}
\Phi_{u_0}(u):=&e^{it\Delta+\varepsilon t\Delta}u_0\\
&-i\int_0^t e^{i(t-s)\Delta+\varepsilon (t-s)\Delta}\bigg[\frac{2\bar
u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2-\frac{2i\varepsilon\bar
u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2\bigg]ds.
\end{align*}
Then using the Lemma \ref{lem:linear} and Lemma \ref{lem:nonlinear}
we can show $\Phi_{u_0}$ is a contraction mapping in the set
\[\{u:\norm{u}_{F^{n/2}\cap Z^{n/2}}\leq C\delta\}\]
if $\norm{u_0}_{\dot B^{n/2}_{2,1}}\leq \delta$ with $\delta>0$ sufficiently small.  Thus we have existence and uniqueness.
Moreover, by standard arguments we immediately have the persistence of regularity, namely if $u_0\in \dot B^{s}_{2,1}$ for some $s>n/2$, then $u\in F^s\cap Z^s$ and
\begin{align}\label{eq:pers}
\norm{u}_{F^s\cap Z^s}{\lesssim} \norm{u_0}_{\dot B^{s}_{2,1}}
\end{align}
uniformly with respect to $\varepsilon \in (0,1]$.

Now we prove the limit
behaviour.  Assume $u_\varepsilon$ is a solution to the Landau-Lifshitz
equation with small initial data $\phi_1\in \dot B^{n/2}_{2,1}$, and
$u$ is a solution to the Schr\"odinger map with small initial data
$\phi_2\in \dot B^{n/2}_{2,1}$. Let $w=u_\varepsilon-u$,
$\phi=\phi_1-\phi_2$, then $w$ solves
\begin{align}\label{eq:dGLw}
(i\partial_t+\Delta)w=&i\varepsilon\Delta u_\varepsilon+\bigg[\frac{2\bar u_\varepsilon}{1+|u_\varepsilon|^2}\sum_{j=1}^n(\partial_{x_j}u_\varepsilon)^2-\frac{2\bar u}{1+|u|^2}\sum_{j=1}^n(\partial_{x_j}u)^2\bigg]\nonumber\\
&-\frac{2i\varepsilon\bar u_\varepsilon}{1+|u_\varepsilon|^2}\sum_{j=1}^n(\partial_{x_j}u_\varepsilon)^2,\\
w(0)=&\phi.\nonumber
\end{align}
First we assume in addition $\phi_1 \in \dot B^{(n+4)/2}_{2,1}$. By
the linear and nonlinear estimates, for any $T>0$ we get
\begin{align*}
\norm{w}_{F^{n/2}\cap Z^{n/2}}{\lesssim} \norm{\phi}_{\dot
B^{n/2}_{2,1}}+\varepsilon T \norm{u_\varepsilon}_{L_t^\infty \dot
B^{(n+4)/2}_{2,1}}+\delta^2\norm{w}_{F^{n/2}\cap
Z^{n/2}}+\varepsilon\norm{u_\varepsilon}^3_{F^{n/2}\cap Z^{n/2}}.
\end{align*}
Then we get by \eqref{eq:pers}
\begin{align}\label{eq:limitHigh}
\norm{w}_{F^{n/2}\cap Z^{n/2}}{\lesssim} &\norm{\phi}_{\dot
B^{n/2}_{2,1}}+\varepsilon T \norm{\phi_1}_{\dot
B^{(n+4)/2}_{2,1}}+\varepsilon\delta^3.
\end{align}
Now we assume $\phi_1=\phi_2=\varphi\in \dot B^{n/2}_{2,1}$ with
small norm. For fixed $T>0$, we need to prove that $\forall\
\eta>0$, there exists $\sigma>0$ such that if $0<\varepsilon<\sigma$ then
\begin{align}\label{eq:limitHs}
\norm{S_T^\varepsilon(\varphi)-S_T(\varphi)}_{C([0,T];\dot
B^{n/2}_{2,1})}<\eta
\end{align}
where $S_T^\varepsilon$ is the solution map corresponding to \eqref{eq:dGLw}
and $S_T=S_T^0$. We denote $\varphi_K=P_{\leq K}\varphi$. Then we
get
\begin{align*}
&\norm{S_T^\varepsilon(\varphi)-S_T(\varphi)}_{C([0,T];\dot
B^{n/2}_{2,1})}\nonumber\\
\leq&\norm{S_T^\varepsilon(\varphi)-S_T^\varepsilon(\varphi_K)}_{C([0,T];\dot
B^{n/2}_{2,1})}\nonumber\\
&+\norm{S_T^\varepsilon(\varphi_K)-S_T(\varphi_K)}_{C([0,T];\dot
B^{n/2}_{2,1})}+\norm{S_T(\varphi_K)-S_T(\varphi)}_{C([0,T];\dot
B^{n/2}_{2,1})}.
\end{align*}
From the uniform global well-posedness and \eqref{eq:limitHigh}, we
get
\begin{align}
\norm{S_T^\epsilon(\varphi)-S_T(\varphi)}_{C([0,T];\dot
B^{n/2}_{2,1})}{\lesssim} \norm{\varphi_K-\varphi}_{\dot B^{n/2}_{2,1}}+\varepsilon
C(T,K, \norm{\varphi}_{\dot B^{n/2}_{2,1}}).
\end{align}
We first fix $K$ large enough, then let $\varepsilon$ go to zero, therefore
\eqref{eq:limitHs} holds.

\subsection*{Acknowledgment}
Z. Guo is supported in part by NNSF of China (No.11371037), and
C. Huang is supported in part by NNSF of China (No. 11201498).


