text:
\section{Introduction}
Landau-Lifshitz-Gilbert equation describes physical properties of micromagnetic at temperatures below the critical temperature. The equation is as follows:
\begin{equation}\label{1.1}
\frac{\partial m}{\partial t}=\lambda_1m\times H_{eff}-\lambda_2m\times(m\times H_{eff})
\end{equation}
where $\times$ denotes the vector cross product in $\mathbb{R}^3$ and $H_{eff}$ is effective field while $\lambda_1$ and $\lambda_2$ are real constants.

However, at high temperature, the model must be replaced by following Landau-Lifshitz-Bloch equation(LLB)
\begin{equation}\label{1.2}
\frac{\partial u}{\partial t}=\gamma u\times H_{eff}+L_1\frac{1}{|u|^2}(u\cdot H_{eff})u-L_2\frac{1}{|u|^2}u\times(u\times H_{eff})
\end{equation}
where $\gamma$, $L_1$, $L_2$ are real numbers and $\gamma>0$. $H_{eff}$ is given by
\[H_{eff}=\Delta u-\frac{1}{\chi||}\Big(1+\frac{3T}{5(T-T_c)}|u|^2\Big)u.\]
where $T>T_c>0$ and $\chi||>0$.

Now let us recall some previous results about LLB. In \cite{L}, Le consider the case that $L_1=L_2=:\kappa_1>0$. At that time, he rewrites (\ref{1.2}) as
\begin{eqnarray}\label{1.31}
\frac{\partial u}{\partial t}=\kappa_1\Delta u+\gamma u\times\Delta u-\kappa_2(1+\mu|u|^2)u
\end{eqnarray}
with $\kappa_2:=\frac{\kappa_1}{\chi||}$ and $\mu:=\frac{3T}{5(T-T_c)}$ and assume that $\kappa_2$, $\gamma$, $\mu$ is positive. Le has proven that above equation with Neumann boundary value conditions has global weak solution(the weak solution here is different from ordinary one). Inspired by Le, in \cite{J} Jia introduces following equation
\begin{equation}\label{1.3}
\left\{
\begin{array}{llll}
\partial_t u=\kappa_1\Delta u+\gamma\nabla F(u)\times\Delta u-\kappa_2(1+\mu\cdot F(u))\nabla F(u)\,\,\,\,\s\mbox{in}\,\,\,\,\Omega\times(0,\infty)\\
\frac{\partial u}{\partial\nu}=0\,\,\,\,\s\mbox{on}\,\,\,\,\partial\Omega\times(0,\infty)\\
u(\cdot ,0)=u_0\,\,\,\,\s\mbox{in}\,\,\,\,\Omega
\end{array}
\right.
\end{equation}
where $\Omega$ is a regular bounded domain of $\mathbb{R}^d(d\leqslant3)$, $\nu$ is outer normal direction of $\partial\Omega$ and $F\in C^3(\mathbb{R}^3)$ is a known function. He calls it Generalized Landau-Lifshitz-Bloch equation(GLLB) and gets that (\ref{1.3}) admits a local strong solution provided $u_0\in W^{2,2}(\Omega,\mathbb{R}^3)$ and $\frac{\partial u_0}{\partial\nu}=0$. In \cite{GLZ}, Guo, Li and Zeng consider the coming LLB equation with initial condition
\begin{eqnarray} \label{eq:6}
\left\{ \begin{aligned}
         &u_t=\Delta u+u\times\Delta u-\lambda(1+\mu|u|^2)u\,\,\,\,\s\mbox{in}\,\,\,\,\mathbb{R}^d\times(0,T)\\
                  &u(,0)=u_0\,\,\,\,\s\mbox{in}\,\,\,\,\mathbb{R}^d,
                          \end{aligned} \right.
                          \end{eqnarray}
where the constant $\lambda,\mu>0$. They prove the existence of smooth solutions of (\ref{eq:6}) in $\mathbb{R}^2$ or $\mathbb{R}^3$. And a small initial value condition should be added in the latter case.\\

In this paper, we would like to introduce a equation similar with (\ref{eq:6}) on Riemannian manifold. Before getting to this, we should make some preparation.

Let $\pi:(E,h,D)\longrightarrow(M,g,\nabla)$ denote a smooth vector bundle over an $m-$dimensional smooth closed Riemannian manifold $(M,g,\nabla)$ with $rank(E)=3$. $g$ means Riemannian metric of $M$ and $\nabla$ is its Levi-Civita connection. $h$ and $D$ are respectively metric and connection of $E$ such that $Dh=0$. Sometimes we also write $h$ as $\langle\cdot,\cdot\rangle$.
\subsection{$k-$times continuously differentiable section}Suppose $\Gamma(E)$ is the set of all sections in $E$. Under arbitrary local frame $\{e_{\alpha}:1\leqslant\alpha\leqslant3\}$, a section $s\in\Gamma(E)$ can be written in the form of $s=s^{\alpha}\cdot e_{\alpha}$. If $s^{\alpha}$ is $k-$times continuously differentiable, then we say $s$ is $k-$times continuously differentiable. Since $E$ is smooth, $k-$times continuous differentiability is independent of the choice of local frame. Define
$$\Gamma^k(E):=\{s\in\Gamma(E):\mbox{$s$ is $k-$times continuously differentiable}\}.$$
\subsection{Orientable vector bundle}$E$ is called orientable if there exists an $\omega\in E^*\wedge E^*\wedge E^*$ such that $\omega$ is continuous and for all $p\in M$, $\omega(p)\not=0$, where $E^*$ is dual bundle of $E$.

Suppose $\{e_1,e_2,e_3\}$ is a frame of $E$. It is called adapted to the orientation $\omega$ if
$$\omega(e_1,e_2,e_3)>0.$$
From now on, we always assume that $E$ is orientable unless otherwise stated.
\subsection{Cross product on orientable vector bundle}Suppose $\omega$ is an orientation of $E$. $\{e_{\alpha}:1\leqslant\alpha\leqslant3\}$ is a local frame of $E$ which is adapted to $\omega$. For any $f_1,f_2\in\Gamma(E)$, we assume that $f_1:=f_1^{\alpha}\cdot e_{\alpha}$, $f_2:=f_2^{\alpha}\cdot e_{\alpha}$. Their cross product $\times$ is defined as follow
$$(f_1\times f_2)(p):=f_1(p)\times f_2(p),$$
where
\begin{eqnarray*}
f_1(p)\times f_2(p):&=&\big(f_1^2(p)\cdot f_2^3(p)-f_2^2(p)\cdot f_1^3(p)\big)\cdot e_1(p)\\
&+&\big(f_2^1(p)\cdot f_1^3(p)-f_1^1(p)\cdot f_2^3(p)\big)\cdot e_2(p)\\
&+&\big(f_1^1(p)\cdot f_2^2(p)-f_2^1(p)\cdot f_1^2(p)\big)\cdot e_3(p).
\end{eqnarray*}
It is not hard to verify that $f_1(p)\times f_2(p)$ does not depend upon the choice of local frames which are adapted to $\omega$.
\subsection{Laplace operator on vector bundle}
Define a functional $Energy$ on $\Gamma^{2}(E)$ which is given in the form of
$$Energy(X):=\frac{1}{2}\int_M|DX|^2\,dM.$$
It is not hard to see that the Euler-Lagrange equation of $Energy$ is
$$\Delta X:=g^{ij}\cdot(D^2X)\Big(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\Big)=0,$$
where $g_{ij}:=g\Big(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\Big)$ and $(g^{ij})$ is the inverse matrix of $(g_{ij})$. Then we say that $\Delta$ is the Laplace operator on vector bundle $E$.
\subsection{sections depending on time}
A section depending on time is a map
$$V:I\longrightarrow\Gamma(E),$$
where $I$ is an interval of $\mathbb{R}$. Under arbitrary local frame $\{e_{\alpha}:1\leqslant\alpha\leqslant3\}$, $V(t,x)$ can be written as $V(t,x):=V^{\alpha}(t,x)\cdot e_{\alpha}(x)$. If $V^{\alpha}$ is $k-$times continuously differentiable with respect to $t$, we say $V$ is $k-$times continuously differentiable with respect to $t$ and use the symbol $C^k(I,\Gamma(E))$ to denote all such $V$. Since $E$ is smooth, differentiability with respect to time is independent of the choice of local frame. Moreover, we define
$$(\partial_t^kV)(t,x):=(\partial_t^kV^{\alpha})(t,x)\cdot e_{\alpha}(x).$$
\subsection{Sobolev space on vector bundle} Equip $\Gamma^k(E)$ with a norm $||\cdot||_{H^{k,p}}(p\geqslant1)$ which is defined as follow
$$||s||^p_{H^{k,p}}:=\sum\limits_{i=0}^k\int_M|D^is|^p\,dM.$$
The Sobolev space $H^{k,p}(E)$ is the completion of $\Gamma^k(E)$ with respect to the norm $||\cdot||_{H^{k,p}}$. For convenience, we also denote $H^{k,2}$ by $H^k$ and $||\cdot||_{H^{0,p}}$ by $||\cdot||_p$.
\\

Having the above preparation, we will give the definition of Landau-Lifshitz-Bloch equation(LLB) on Riemannian manifold.

For any $T>0,\lambda>0$ and $\mu>0$, let us consider a section depending on time $V\in C^1([0,T],\Gamma^{2}(E))$. LLB is just the following equation
\begin{eqnarray}\label{eq:1}
\left\{ \begin{aligned}
         &\partial_tV=\Delta V+V\times\Delta V-\lambda\cdot(1+\mu\cdot|V|^2)V \,\,\,\,\s\,\,\,\,\s\mbox{in}\,\,\,\,(0,T]\times M\\
                  &V(0,\cdot)=V_0
                          \end{aligned} \right.
\end{eqnarray}
\\

Our main results are as follow:
\begin{thm}\label{thm4}
Let $\pi:(E,h,D)\longrightarrow(M,g,\nabla)$ denote a smooth vector bundle over an $m-$dimensional smooth closed Riemannian manifold $(M,g,\nabla)$ with $rank(E)=3$ and $Dh=0$. $E$ is orientable. Given $l\geqslant m_0+1$(Here $m_0:=[\frac{m}{2}]+3$ and $[q]$ is the integral part of $q$) and $V_0\in H^l(E)$, there is a $T^*=T^*(||V_0||_{H^{m_0}})>0$ and a unique solution $V$ of (\ref{eq:1}) satisfying that for any $0\leqslant j\leqslant[\frac{l}{\hat{m}}]$($\hat{m}:=\max\{2,[\frac{m}{2}]+1\}$) and $\alpha\leqslant l-\hat{m}j$,
\begin{eqnarray}\label{6}
\partial_t^jD^{\alpha}V\in L^{\infty}([0,T^*],L^2(E)).
\end{eqnarray}
Furthermore, if $V_0\in\Gamma^{\infty}(E)$, then $V\in C^{\infty}([0,T^*],\Gamma^{\infty}(E))$.
\end{thm}

\begin{thm}\label{thm5}
Let $\pi:(E,h,D)\longrightarrow(M,g,\nabla)$ denote a smooth vector bundle over an $m-$dimensional smooth closed Riemannian manifold $(M,g,\nabla)$ with $rank(E)=3$, $m\geqslant3$ and $Dh=0$. $E$ is orientable. For any $T>0$ and $N\geqslant m_0+1$, there exists a $\hat{B}_N>0$ such that for all $V_0\in H^N(E)$ with $||V_0||_{\infty}\leqslant\hat{B}_N$, there is a unique solution of (\ref{eq:1}) satisfying
\begin{eqnarray}\label{7}
\partial_t^jD^{\alpha}V\in L^{\infty}([0,T],L^2(E))\,\,\,\,\forall0\leqslant j\leqslant\Big[\frac{N}{\hat{m}}\Big]\,\,\,\,\forall\alpha\leqslant N-\hat{m}j
\end{eqnarray}
and
\begin{eqnarray}\label{9}
\partial_t^iD^{\beta}V\in L^{2}([0,T],L^2(E))\,\,\,\,\forall0\leqslant i\leqslant\Big[\frac{N+1}{\hat{m}+1}\Big]\,\,\,\,\forall\beta\leqslant N+1-(\hat{m}+1)i.
\end{eqnarray}
Furthermore, if $V_0\in\Gamma^{\infty}(E)$, then $V\in C^{\infty}([0,T],\Gamma^{\infty}(E))$.
\end{thm}

\begin{thm}\label{thm9}
Let $\pi:(E,h,D)\longrightarrow(M,g,\nabla)$ denote a smooth vector bundle over an $2-$dimensional smooth closed Riemannian manifold $(M,g,\nabla)$ with $rank(E)=3$ and $Dh=0$. $E$ is orientable. For any $T>0$,      $N\geqslant5$ and $V_0\in H^5(E)$, there is a unique solution of (\ref{eq:1}) satisfying
\begin{eqnarray*}
\partial_t^jD^{\alpha}V\in L^{\infty}([0,T],L^2(E))\,\,\,\,\forall0\leqslant j\leqslant\Big[\frac{N}{2}\Big]\,\,\,\,\forall\alpha\leqslant N-2j
\end{eqnarray*}
and
\begin{eqnarray*}
\partial_t^iD^{\beta}V\in L^{2}([0,T],L^2(E))\,\,\,\,\forall0\leqslant i\leqslant\Big[\frac{N+1}{3}\Big]\,\,\,\,\forall\beta\leqslant N+1-3i.
\end{eqnarray*}
Furthermore, if $V_0\in\Gamma^{\infty}(E)$, then $V\in C^{\infty}([0,T],\Gamma^{\infty}(E))$.
\end{thm}
\section{Notation and Preliminaries}
In the paper, we appoint that the same indices appearing twice means summing it. And $Q_1\lesssim Q_2$ implies there is a universal constant $C$ such that $Q_1\leqslant C\cdot Q_2$.
\subsection{Riemannian curvature tensor on vector bundle}Using the connection $D$ on $E$, we can define a tensor $R^E$ called Riemannian curvature tensor. For any $X,Y\in TM$ and $s\in\Gamma^2(E)$,
$$R^E(X,Y)s:=D_XD_Ys-D_YD_Xs-D_{[X,Y]}s.$$
Let $R^M$ be the Riemannian curvature tensor of $M$. Being going to represent $R^M$ and $R^E$ in local frame, we appoint $\frac{\partial}{\partial x^i}$ as $\partial_i$. Then,
$$R^M(\partial_i,\partial_j)\partial_r:=(R^M)^h_{ijr}\cdot\partial_h\,\,\,\,\s
\mbox{and}
\,\,\,\,\s R^E(\partial_i,\partial_j)e_{\beta}:=(R^E)^{\alpha}_{ij\beta}\cdot e_{\alpha}.$$
Now we give two tensors
$$\mathcal{R}^M:=(\mathcal{R}^M)_{ijkl}\cdot dx^i\otimes dx^j\otimes dx^k\otimes dx^l$$
and
$$\mathcal{R}^E:=(\mathcal{R}^E)^{\alpha\beta}_{ij}\cdot dx^i\otimes dx^j\otimes e_{\alpha}\otimes e_{\beta},$$
where
$$(\mathcal{R}^M)_{ijkl}:=(R^M)^h_{ijk}\cdot g_{hl}\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s(\mathcal{R}^E)^{\alpha\beta}_{ij}:=(R^E)^{\alpha}_{ij\theta}\cdot h^{\theta\beta}.$$
$(h_{\alpha\beta})$ is the metric matrix of $h$ and $(h^{\theta\beta})$ is its inverse matrix.
\subsection{Cross product of tensors}
We also want to introduce cross product between two tensors. Given $S\in\Gamma(T^*M^{\otimes k}\otimes E)$ and $T\in\Gamma(T^*M^{\otimes l}\otimes E)$, let us define
$$S\times T:=(S_{i_1\cdots i_k}\times T_{j_1\cdots j_l})\otimes dx^{i_1}\otimes\cdots\otimes dx^{i_k}\otimes dx^{j_1}\otimes\cdots\otimes dx^{j_l},$$
where
$$S_{i_1\cdots i_k}:=S(\partial_{i_1},\cdots,\partial_{i_k})\,\,\,\,\s\,\,\,\,\s \mbox{and}\,\,\,\,\s\,\,\,\,\s T_{j_1\cdots j_l}:=T(\partial_{j_1},\cdots,\partial_{j_l}).$$
It is easy to check
\begin{eqnarray}\label{15}
|S\times T|\leqslant|S|\cdot|T|
\end{eqnarray}
\subsection{Properties of cross product}
\begin{thm}\label{thm3}
For any $f_1,f_2\in\Gamma^1(E)$, we have
\begin{eqnarray}\label{1}
D(f_1\times f_2)=(Df_1)\times f_2+f_1\times(Df_2).
\end{eqnarray}
\end{thm}
\textbf{Proof.} Take any $p\in M$. Then there exists a neighbourhood $U$ and a positive number $\delta$ such that the following map
$$\exp_p: N_{\delta}\triangleq\{\hat{v}\in T_pM:||\hat{v}||<\delta\}\longrightarrow U$$
is a diffeomorphism. Take $v\in T_pM$ such that $||v||=1$. Define $\gamma_v(t):=\exp_p(tv)$, where $t\in[0,\delta)$. Now take arbitrary orthonormal basis $\{e_{p\alpha}:1\leqslant\alpha\leqslant3\}$ in $E_p$ which is adapted to $\omega$ and let it move parallelly along $\gamma_v$ to get $$\{e_{\alpha}(t,v):t\in[0,\delta),1\leqslant\alpha\leqslant3\}.$$ Clearly,
$$w(t):=\omega(e_1(t,v),e_2(t,v),e_3(t,v))>0,\,\,\,\,\s\,\,\,\,\s\forall t\in[0,\delta)$$
since $w$ is a continuous function with respect to $t$. In the next, let $v$ range all the direction in $T_pM$ to obtain $$\{e_{\alpha}(t,v):t\in[0,\delta),v\in T_pM,||v||=1,1\leqslant\alpha\leqslant3\}.$$
It is a orthonormal frame on $U$ which is adapted to $\omega$ and
\begin{eqnarray}\label{2}
(De_{\alpha})(p)=0.
\end{eqnarray}

Assume that $f_1=f_1^{\alpha}\cdot e_{\alpha}$ and $f_2=f_2^{\beta}\cdot e_{\beta}$. Then, (\ref{2}) yields
$$Df_1(p)=df_1^{\alpha}(p)\otimes e_{\alpha}(p)\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s Df_2(p)=df_2^{\beta}(p)\otimes e_{\beta}(p).$$
Recalling the definition of cross product, we have
\begin{eqnarray*}
f_1\times f_2:=\big(f_1^2\cdot f_2^3-f_2^2\cdot f_1^3\big)\cdot e_1+\big(f_2^1\cdot f_1^3-f_1^1\cdot f_2^3\big)\cdot e_2+\big(f_1^1\cdot f_2^2-f_2^1\cdot f_1^2\big)\cdot e_3.
\end{eqnarray*}
Therefore, since of (\ref{2}), one can get
\begin{eqnarray}\label{3}
[D(f_1\times f_2)](p):&=&\big[df_1^2(p)\cdot f_2^3(p)+f_1^2(p)\cdot df_2^3(p)-df_2^2(p)\cdot f_1^3(p)-f_2^2(p)\cdot df_1^3(p)\big]\otimes e_1(p)\nonumber\\
&+&\big[df_2^1(p)\cdot f_1^3(p)+f_2^1(p)\cdot df_1^3(p)-df_1^1(p)\cdot f_2^3(p)-f_1^1(p)\cdot df_2^3(p)\big]\otimes e_2(p)\\
&+&\big[df_1^1(p)\cdot f_2^2(p)+f_1^1(p)\cdot df_2^2(p)-df_2^1(p)\cdot f_1^2(p)-f_2^1(p)\cdot df_1^2(p)\big]\otimes e_3(p),\nonumber
\end{eqnarray}
\begin{eqnarray}\label{4}
[f_1\times(Df_2)](p)&=&f_1(p)\times(Df_2)(p)\nonumber\\
&=&\big[f_1^2(p)\cdot df_2^3(p)-df_2^2(p)\cdot f_1^3(p)\big]\otimes e_1(p)\nonumber\\
&+&\big[df_2^1(p)\cdot f_1^3(p)-f_1^1(p)\cdot df_2^3(p)\big]\otimes e_2(p)\\
&+&\big[f_1^1(p)\cdot df_2^2(p)-df_2^1(p)\cdot f_1^2(p)\big]\otimes e_3(p),\nonumber
\end{eqnarray}
and
\begin{eqnarray}\label{5}
[(Df_1)\times f_2](p)&=&(Df_1)(p)\times f_2(p)\nonumber\\
&=&\big[df_1^2(p)\cdot f_2^3(p)-f_2^2(p)\cdot df_1^3(p)\big]\otimes e_1(p)\nonumber\\
&+&\big[f_2^1(p)\cdot df_1^3(p)-df_1^1(p)\cdot f_2^3(p)\big]\otimes e_2(p)\\
&+&\big[df_1^1(p)\cdot f_2^2(p)-f_2^1(p)\cdot df_1^2(p)\big]\otimes e_3(p).\nonumber
\end{eqnarray}
This theorem follows easily from combining (\ref{3}) with (\ref{4}) and (\ref{5}).
$\hfill\Box$\\

Because of (\ref{1}), it is easy to verify that
\begin{eqnarray}\label{14}
D(S\times T)=(DS)\times T+S\times(DT),
\end{eqnarray}
provided $S\in\Gamma^1(T^*M^{\otimes k}\otimes E)$ and $T\in\Gamma^1(T^*M^{\otimes l}\otimes E)$.
\subsection{Hamilton's notation}Suppose $k,l,p,q\in\mathbb{N}$, $S\in T^*M^{\otimes k}\otimes E^{\otimes p}$ and $T\in T^*M^{\otimes l}\otimes E^{\otimes q}$, where
$$E^{\otimes p}:=\underbrace{E\otimes\cdots\otimes E}\limits_{\mbox{$p-$times}}.$$
we will write $S\ast T$, following Hamilton \cite{Hamiltion}, to denote a tensor formed by contraction on some indices of $S\otimes T$ using the coefficients $g^{ij}$ or $h_{\alpha\beta}$.
\begin{thm}\label{thm1}
$$|S\ast T|\leqslant|S|\cdot|T|$$
\end{thm}
\textbf{Proof.} We will get the above formula in an orthonormal basis of $M$ and an orthonormal basis of $E$.
\begin{eqnarray*}
|S\ast T|^2&=&\sum\limits_{\substack{\mbox{free}\\ \mbox{indices}}}\Bigg(\sum\limits_{\substack{\mbox{contracted}\\ \mbox{indices}}}S^{\alpha_1\cdots\alpha_p}_{i_1\cdots i_k}\cdot T^{\beta_1\cdots\beta_q}_{j_1\cdots j_l}\Bigg)^2\\
&\leqslant&\sum\limits_{\substack{\mbox{free}\\ \mbox{indices}}}\Bigg[\sum\limits_{\substack{\mbox{contracted}\\ \mbox{indices}}}\Big(S^{\alpha_1\cdots\alpha_p}_{i_1\cdots i_k}\Big)^2\Bigg]\cdot \Bigg[\sum\limits_{\substack{\mbox{contracted}\\ \mbox{indices}}}\Big(T^{\beta_1\cdots\beta_q}_{j_1\cdots j_l}\Big)^2\Bigg]\\
&\leqslant&\Bigg[\sum\limits_{\substack{\mbox{free}\\ \mbox{indices}}}\sum\limits_{\substack{\mbox{contracted}\\ \mbox{indices}}}\Big(S^{\alpha_1\cdots\alpha_p}_{i_1\cdots i_k}\Big)^2\Bigg]\cdot \Bigg[\sum\limits_{\substack{\mbox{free}\\ \mbox{indices}}}\sum\limits_{\substack{\mbox{contracted}\\ \mbox{indices}}}\Big(T^{\beta_1\cdots\beta_q}_{j_1\cdots j_l}\Big)^2\Bigg]\\
&=&|S|^2\cdot|T|^2
\end{eqnarray*}
$\hfill\Box$\\
Because we do not specifically illustrate which indices are contracted, we have to appoint that
$$S_1\ast T_1-S_2\ast T_2:=S_1\ast T_1+S_2\ast T_2.$$

We will use the symbol $\mathfrak{q}_s(T_1,\cdots,T_r)$ for a polynomial in the tensors $T_1,\cdots,T_r$ and their iterated covariant derivatives with the $\ast$ product like
$$\mathfrak{q}_s(T_1,\cdots,T_r):=\sum\limits_{j_1+\cdots+j_r=s}c_{j_1\cdots j_r}\cdot D^{j_1}T_1\ast\cdots\ast D^{j_r}T_r,$$
where for $1\leqslant i\leqslant r$, $T_i\in\Gamma^{j_i}(T^*M^{\otimes t_i}\otimes E^{\otimes q_i})$ and $c_{j_1\cdots j_r}$ are some universal constants.
\subsection{Ricci identity}
Given $s\in\Gamma^2(T^*M^{\otimes k}\otimes E)$, it is obvious to see that $s$ can be written as follow
$$s:=s^{\alpha}_{i_1\cdots i_k}\cdot dx^{i_1}\otimes\cdots\otimes dx^{i_k}\otimes e_{\alpha}.$$
We denote $Ds$ in the form of components
$$Ds:=s^{\alpha}_{i_1\cdots i_k,p}\cdot dx^{i_1}\otimes\cdots\otimes dx^{i_k}\otimes dx^p\otimes e_{\alpha}.$$
At some time, we also employ the coming convention
\begin{eqnarray}\label{10}
Ds:=s_{i_1\cdots i_k,p}\cdot dx^{i_1}\otimes\cdots\otimes dx^{i_k}\otimes dx^p.
\end{eqnarray}
Thanks to the above agreement, Ricci identity is conveniently represented in the next theorem.
\begin{thm}\label{thm2}
\begin{eqnarray*}
&&s^{\alpha}_{i_1\cdots i_k,pq}-s^{\alpha}_{i_1\cdots i_k,qp}\\
&=&\sum\limits_{l=1}^ks^{\alpha}_{i_1\cdots i_{l-1}hi_{l+1}\cdots i_k}\cdot(R^M)^h_{pqi_l}-s^{\beta}_{i_1\cdots i_k}\cdot(R^E)^{\alpha}_{pq\beta}\\
&=&k\cdot s\ast\mathcal{R}^M+s\ast\mathcal{R}^E.
\end{eqnarray*}
\end{thm}
\textbf{Proof.}
The proof is straightforward if one takes normal coordinates. So we omit it.
$\hfill\Box$\\

Given $V\in\Gamma^{k+1}(E)$ and $S\in\Gamma^{k+1}(T^*M\otimes E)$, by Theorem \ref{thm2} and induction, the following formulas are easy.\\
\textbf{Formula 1.}\\
There exist $a_{ij}\in\mathbb{Z}$ and $b_{rl}\in\mathbb{Z}$ such that
\begin{eqnarray}\label{11}
V_{,pi_1\cdots i_k}-V_{,i_1\cdots i_kp}&=&\sum\limits_{i+j=k-1}a_{ij}\cdot D^iV\ast D^j\mathcal{R}^E+\sum\limits_{r+l=k-2}b_{rl}\cdot D^{r+1}V\ast\nabla^l\mathcal{R}^M\nonumber\\
&=&\mathfrak{q}_{k-1}(V,\mathcal{R}^E)+\mathfrak{q}_{k-2}(DV,\mathcal{R}^M)
\end{eqnarray}
\textbf{Formula 2.}\\
There exist $a_{ij}\in\mathbb{Z}$ and $b_{rl}\in\mathbb{Z}$ such that
\begin{eqnarray}\label{12}
S_{p,qi_1\cdots i_k}-S_{p,i_1\cdots i_kq}&=&\sum\limits_{i+j=k-1}a_{ij}\cdot D^iS\ast D^j\mathcal{R}^E+\sum\limits_{r+l=k-1}b_{rl}\cdot D^{r}S\ast\nabla^l\mathcal{R}^M\nonumber\\
&=&\mathfrak{q}_{k-1}(S,\mathcal{R}^E)+\mathfrak{q}_{k-1}(S,\mathcal{R}^M)
\end{eqnarray}
\subsection{Interpolation for sections}
We shall prove Gagliardo-Nirenberg inequality of sections on vector bundle.
\begin{thm}\label{thm6}
$(M,g)$ is a $m-$dimensional smooth closed Riemannian manifold. $(E,h,D)$ is a smooth vector bundle over $M$ with $Dh=0$. $rank(E)$ may not be 3 and $E$ may not be orientable. Let $T$ be a smooth section of $E$. Given $s\in\mathbb{R}^+$ and $j\in\mathbb{Z}^+$, we will have
\begin{eqnarray}\label{29}
||D^jT||_{\frac{2s}{l}}\leqslant C(m,s,k,j)\cdot||D^kT||^{\frac{j}{k}}_{\frac{2s}{l+k-j}}\cdot||T||^{1-\frac{j}{k}}_{\frac{2s}{l-j}},
\end{eqnarray}
provided $k\in[j,\infty)\cap\mathbb{Z}$, $l\in[1,s]\cap[j,s+j+1-k]\cap\mathbb{Z}$.
\end{thm}
\textbf{Proof.} Apply induction for $j$.\\
\textbf{Step 1:} When $j=1$, (\ref{29}) is equivalent to
\begin{eqnarray}\label{30}
||DT||_{\frac{2s}{l}}\leqslant C(m,s,k)\cdot||D^kT||^{\frac{1}{k}}_{\frac{2s}{l+k-1}}\cdot||T||^{1-\frac{1}{k}}_{\frac{2s}{l-1}},
\end{eqnarray}
for all $l\in[1,s]\cap[1,s+2-k]\cap\mathbb{Z}$. In order to show (\ref{30}), we use induction for $k$.

When $k=1$, (\ref{30}) holds obviously.

When $k=2$, by 12.1 Theorem of \cite{Hamiltion} we know (\ref{30}) holds.

Assume that for $2\leqslant\hat{k}\leqslant k$, we obtain
\begin{eqnarray*}
||DT||_{\frac{2s}{l}}\leqslant C_1(m,s,\hat{k})\cdot||D^{\hat{k}}T||^{\frac{1}{\hat{k}}}_{\frac{2s}{l+\hat{k}-1}}\cdot||T||^{1-\frac{1}{\hat{k}}}_{\frac{2s}{l-1}},
\end{eqnarray*}
provided $l\in[1,s]\cap[1,s+2-\hat{k}]\cap\mathbb{Z}$.

When $\hat{k}=k+1$, pick any $l\in[1,s]\cap[1,s+2-(k+1)]\cap\mathbb{Z}$. Clearly,
$$l+1\in[1,s]\cap[1,s+2-k]\cap\mathbb{Z},$$
since $k\geqslant2$. Using induction hypothesis, we get
\begin{eqnarray}\label{31}
||D^2T||_{\frac{2s}{l+1}}\leqslant C_2(m,s,k)\cdot||D^k(DT)||^{\frac{1}{k}}_{\frac{2s}{l+k}}\cdot||DT||^{1-\frac{1}{k}}_{\frac{2s}{l}}.
\end{eqnarray}
Because $1\leqslant l\leqslant s+2-(k+1)<s$, using induction hypothesis for $k=2$ gives
\begin{eqnarray}\label{32}
||DT||_{\frac{2s}{l}}\leqslant C(m,s)\cdot||D^2T||^{\frac{1}{2}}_{\frac{2s}{l+1}}\cdot||T||^{\frac{1}{2}}_{\frac{2s}{l-1}}.
\end{eqnarray}
Combing (\ref{31}) with (\ref{32}) yields
$$||DT||_{\frac{2s}{l}}\leqslant C_3(m,s,k)\cdot||D^{k+1}T||^{\frac{1}{2k}}_{\frac{2s}{l+k}}\cdot||DT||^{\frac{1}{2}(1-\frac{1}{k})}_{\frac{2s}{l}}\cdot||T||^{\frac{1}{2}}_{\frac{2s}{l-1}},$$
which implies
$$||DT||_{\frac{2s}{l}}\leqslant C(m,s,k+1)\cdot||D^{k+1}T||^{\frac{1}{k+1}}_{\frac{2s}{l+k}}\cdot||T||^{1-\frac{1}{k+1}}_{\frac{2s}{l-1}}.$$
\textbf{Step 2:} Suppose that for all the indices not greater than $j$, (\ref{29}) is true. Now we consider $j+1$. At this moment, we take any $k\in[j+1,\infty)\cap\mathbb{Z}$ and any $l\in[1,s]\cap[j+1,s+j+2-k]\cap\mathbb{Z}$. It is easy to deduce that
$$k-1\in[j,\infty)\cap\mathbb{Z}\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s l\in[1,s]\cap[j,s+j+1-(k-1)]\cap\mathbb{Z}.$$

Using induction hypothesis leads to
\begin{eqnarray}\label{33}
||D^j(DT)||_{\frac{2s}{l}}\leqslant C_1(m,s,k,j)\cdot||D^{k-1}(DT)||^{\frac{j}{k-1}}_{\frac{2s}{l+k-1-j}}\cdot||DT||^{1-\frac{j}{k-1}}_{\frac{2s}{l-j}}.
\end{eqnarray}
Since $l-j\in[1,s]\cap[1,s+2-k]\cap\mathbb{Z}$, by Step 1 we have
\begin{eqnarray}\label{34}
||DT||_{\frac{2s}{l-j}}\leqslant C(m,s,k)\cdot||D^kT||^{\frac{1}{k}}_{\frac{2s}{l-j+k-1}}\cdot||T||^{1-\frac{1}{k}}_{\frac{2s}{l-j-1}}.
\end{eqnarray}
Combining (\ref{33}) with (\ref{34}) gives
\begin{eqnarray*}
||D^{j+1}T||_{\frac{2s}{l}}&\leqslant& C(m,s,k,j+1)\cdot||D^kT||^{\frac{j}{k-1}}_{\frac{2s}{l+k-1-j}}\cdot||D^kT||^{\frac{1}{k}(1-\frac{j}{k-1})}_{\frac{2s}{l+k-j-1}}\cdot||T||^{(1-\frac{1}{k})(1-\frac{j}{k-1})}_{\frac{2s}{l-j-1}}\\
&=&C(m,s,k,j+1)\cdot||D^kT||^{\frac{j+1}{k}}_{\frac{2s}{l+k-1-j}}\cdot||T||^{1-\frac{j+1}{k}}_{\frac{2s}{l-j-1}}.
\end{eqnarray*}
This completes the proof.$\hfill\Box$\\
\begin{thm}\label{thm7}
$(M,g)$ is a $m-$dimensional smooth closed Riemannian manifold. $(E,h,D)$ is a smooth vector bundle over $M$ with $Dh=0$. $rank(E)$ may not be 3 and $E$ may not be orientable. Let $T$ be a smooth section of $E$. If $r,q\geqslant2$, then there is a universal constant $C=C(m,r,q,j,k)$ such that
\begin{eqnarray}\label{35}
||D^jT||_p\leqslant C\cdot||D^kT||_r^{\frac{j}{k}}\cdot||T||_q^{1-\frac{j}{k}},
\end{eqnarray}
provided
$$1\leqslant j\leqslant k\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s\frac{k}{p}=\frac{j}{r}+\frac{k-j}{q}.$$
\end{thm}
\textbf{Proof.} We consider 3 cases.\\
\textbf{Case 1:} When $2\leqslant r<q\leqslant\infty$, there exist $s$ and $l$ such that
$$q=\frac{2s}{l-j}\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s r=\frac{2s}{l+k-j}.$$
Since
$$\frac{k}{p}=\frac{j}{r}+\frac{k-j}{q},$$
we have $p=\frac{2s}{l}$. From Theorem \ref{thm6} it follows that
$$
||D^jT||_{\frac{2s}{l}}\leqslant C(m,s,k,j)\cdot||D^kT||^{\frac{j}{k}}_{\frac{2s}{l+k-j}}\cdot||T||^{1-\frac{j}{k}}_{\frac{2s}{l-j}},
$$
which means
$$
||D^jT||_p\leqslant C(m,r,q,j,k)\cdot||D^kT||_r^{\frac{j}{k}}\cdot||T||_q^{1-\frac{j}{k}}.
$$
\textbf{Case 2:} When $2\leqslant q<r\leqslant\infty$, the proof is similar.\\
\textbf{Case 3:} When $2\leqslant q=r$, clearly we have $p=q=r$. From 12.1 Theorem in \cite{Hamiltion} it follows that
$$
||DT||_p\leqslant C(m,p)\cdot||D^2T||^{\frac{1}{2}}_p\cdot||T||_p^{\frac{1}{2}},
$$
which implies
$$
||D^jT||_p\leqslant C(m,p)\cdot||D^{j+1}T||^{\frac{1}{2}}_p\cdot||D^{j-1}T||_p^{\frac{1}{2}}.
$$
Let $f(j):=||D^jT||_p$. It is easy to check that $f$ meets the condition of 12.5 Corollary in \cite{Hamiltion}. Then we conclude this theorem.$\hfill\Box$\\
\section{Proof of Theorem \ref{thm4}}
Given any $T>0$, define an operator
$$P:C^1([0,T],\Gamma^2(E))\longrightarrow C([0,T],\Gamma(E)),$$
here
$$P(V):=\partial_tV-\Delta V-V\times\Delta V-\lambda(1+\mu|V|^2)V.$$
It is not difficult to check that the leading coefficient of the linearised operator of $P$ meets Legendre-Hadamard condition. By Main Theorem 1 in page 3 of \cite{Baker} we know (\ref{eq:1}) admits a unique local smooth solution $V$ provided $V_0\in\Gamma^{\infty}(E)$.

In the sequel, we would like to know the lower bound of maximal existence time $T_{\max}$ of the above smooth solution. Our strategy is to deduce a Gronwall inequality. That is to say, we shall control $\frac{d}{dt}||V(t)||^2_{H^l}$. Before getting to this, it is important to obtain an upper bound of $||V(t)||_{\infty}$.

Taking inner product with $|V|^{p-2}V(p>2)$ in (\ref{eq:1}), and integrating the result over $M$, we get
\begin{eqnarray*}
\int_M|V|^{p-2}\langle V,\partial_tV\rangle\,dM&=&\int_M|V|^{p-2}\langle V,\Delta V\rangle\,dM-\lambda\int_M(1+\mu|V|^2)|V|^p\,dM\\
&\leqslant&-\int_M|V|^{p-2}\cdot|DV|^2\,dM-(p-2)\int_M|V|^{p-4}\cdot|\langle V,DV\rangle|^2\,dM\leqslant0.
\end{eqnarray*}
The left hand side of the above inequality is $\frac{1}{p}\frac{d}{dt}||V(t)||_p^p$, so this inequality means
\begin{eqnarray*}
||V(t)||_p\leqslant||V_0||_p\,\,\,\,\s\,\,\,\,\s\forall t\in[0,T_{\max}).
\end{eqnarray*}
Taking the limit $p\rightarrow\infty$ leads to
\begin{eqnarray}\label{8}
||V(t)||_{\infty}\leqslant||V_0||_{\infty}\,\,\,\,\s\,\,\,\,\s\forall t\in[0,T_{\max}).
\end{eqnarray}

Given $k\geqslant1$, recalling our appointment (\ref{10}), we have the next identity
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\int_M|D^kV|^2\,dM&=&\int_Mg^{pq}g^{i_1j_1}\cdots g^{i_kj_k}\langle V_{,j_1\cdots j_k},V_{,pqi_1\cdots i_k}\rangle\,dM\\
&&+\int_Mg^{pq}g^{i_1j_1}\cdots g^{i_kj_k}\langle V_{,j_1\cdots j_k},(V\times V_{,p})_{,qi_1\cdots i_k}\rangle\,dM\\
&&-\lambda\cdot\mu\int_Mg^{i_1j_1}\cdots g^{i_kj_k}\langle V_{,j_1\cdots j_k},(|V|^2V)_{,i_1\cdots i_k}\rangle\,dM\\
&&-\lambda\int_M|D^kV|^2\,dM
\end{eqnarray*}
Applying (\ref{11}) and (\ref{12}) to exchange the order of derivatives yields
\begin{eqnarray}\label{13}
\frac{1}{2}\frac{d}{dt}\int_M|D^kV|^2\,dM&=&-\int_M|D^{k+1}V|^2\,dM+\int_MD^kV\ast\mathfrak{q}_k(V,\mathcal{R}^E)\,dM\\
&&+\int_MD^kV\ast\mathfrak{q}_{k-1}(DV,\mathcal{R}^M)\,dM\nonumber\\
&&-\int_Mg^{pq}g^{i_1j_1}\cdots g^{i_kj_k}\langle V_{,qj_1\cdots j_k},(V\times V_{,p})_{,i_1\cdots i_k}\rangle\,dM\nonumber\\
&&-\int_MD^kV\ast\mathfrak{q}_{k-1}(V\times DV,\mathcal{R}^M)\,dM\nonumber\\
&&-\int_MD^kV\ast\mathfrak{q}_{k-1}(V\times DV,\mathcal{R}^E)\,dM\nonumber\\
&&-\lambda\int_M|D^kV|^2\,dM-\lambda\mu\int_M|V|^2\cdot|D^kV|^2\,dM\nonumber\\
&&-\lambda\mu\sum\limits_{i+j=k-1}b_{ij}\cdot\int_MD^kV\ast D^{i+1}(|V|^2)\ast D^jV\,dM\nonumber\\
&&+\int_M\mathfrak{q}_k(V,\mathcal{R}^E)\ast D^{k-1}(V\times DV)\,dM\nonumber\\
&&+\int_M\mathfrak{q}_{k-1}(DV,\mathcal{R}^M)\ast D^{k-1}(V\times DV)\,dM,\nonumber
\end{eqnarray}
here $b_{ij}\in\mathbb{Z}^+$ are some universal constants. Note that
\begin{eqnarray*}
g^{pq}g^{i_1j_1}\cdots g^{i_kj_k}\langle V_{,qj_1\cdots j_k},(V\times V_{,p})_{,i_1\cdots i_k}\rangle=\sum\limits_{i+j=k-1}a_{ij}\cdot D^{k+1}V\ast(D^{i+1}V\times D^{j+1}V)
\end{eqnarray*}
where $a_{ij}\in\mathbb{Z}^+$ are some universal constants. Taking norms on the right hand side of (\ref{13}) leads to
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}\int_M|D^kV|^2\,dM&\leqslant&-\int_M|D^{k+1}V|^2\,dM+\sum\limits_{i=0}^kC_i\int_M|D^kV|\cdot|D^iV|\,dM\\
&&+\sum\limits_{i+j=k-1}a_{ij}\int_M|D^{k+1}V|\cdot|D^{i+1}V|\cdot|D^{j+1}V|\,dM\\
&&+\sum\limits_{i=0}^{k-1}\bar{C}_i\int_M|D^kV|\cdot|D^i(V\times DV)|\,dM\\
&&+\lambda\mu\sum\limits_{i+j=k-1}b_{ij}\int_M|D^kV|\cdot|D^{i+1}(|V|^2)|\cdot|D^jV|\,dM\\
&&+\sum\limits_{i=0}^kC_i\int_M|D^iV|\cdot|D^{k-1}(V\times DV)|\,dM,
\end{eqnarray*}
where $C_i$ and $\bar{C}_i$ depend upon $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations. Applying (\ref{15}) and (\ref{14}) yields
\begin{eqnarray}\label{16}
\frac{1}{2}\frac{d}{dt}\int_M|D^kV|^2\,dM&\leqslant&-\int_M|D^{k+1}V|^2\,dM+\sum\limits_{i+j=k-1}a_{ij}\int_M|D^{k+1}V|\cdot|D^{i+1}V|\cdot|D^{j+1}V|\,dM\nonumber\\
&&+\tilde{C}_k\Big\{||D^kV||_2\cdot||V||_{H^k}+\sum\limits_{0\leqslant r+q\leqslant k-1}\int_M|D^kV|\cdot|D^rV|\cdot|D^{q+1}V|\,dM\nonumber\\
&&+\sum\limits_{r+q+j=k}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\\
&&+\sum\limits_{i=0}^k\sum\limits_{r+q=k-1}\int_M|D^iV|\cdot|D^rV|\cdot|D^{q+1}V|\,dM\Big\},\nonumber
\end{eqnarray}
where $\tilde{C}_k$ depends upon $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations.
\begin{lem}\label{lem1}
There is a $C'_{m_0}>0$ depending on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations such that, for any $t\in[0,T_{\max})$, we have
\begin{eqnarray*}
\frac{d}{dt}||V(t)||^2_{H^{m_0}}\leqslant C'_{m_0}\cdot(1+||V_0||^2_{H^{m_0}})\cdot\{||V(t)||^2_{H^{m_0}}+||V(t)||^4_{H^{m_0}}\}.
\end{eqnarray*}
\end{lem}
\textbf{Proof.} Given $1\leqslant k\leqslant m_0$, we consider
\begin{eqnarray*}
&&\sum\limits_{r+q+j=k}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\\
&=&\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}=k}}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\\
&&+\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}\leqslant k-1}}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM.
\end{eqnarray*}
Clearly,
\begin{eqnarray*}
\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}=k}}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\lesssim||D^kV||_2^2\cdot||V||^2_{\infty}\leqslant||D^kV||_2^2\cdot||V_0||^2_{\infty}.
\end{eqnarray*}
And we want to derive the following
\begin{eqnarray*}
&&\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}\leqslant k-1}}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\\
&\leqslant&\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}\leqslant k-1}}||D^kV||_2\cdot||D^rV||_{p_r}\cdot||D^qV||_{p_q}\cdot||D^jV||_{p_j}.
\end{eqnarray*}
where $p_r$, $p_q$ and $p_j$, belonging to $[1,\infty]$, will be determined later and satisfy
\begin{eqnarray}\label{20}
\frac{1}{p_r}+\frac{1}{p_q}+\frac{1}{p_j}=\frac{1}{2}.
\end{eqnarray}
And then we employ Theorem 2.1 due to \cite{DW1} to obtain
\begin{eqnarray*}
||D^rV||_{p_r}\lesssim||V||^{a_r}_{H^{m_0}}\cdot||V||_2^{1-a_r}\leqslant||V||_{H^{m_0}},
\end{eqnarray*}
\begin{eqnarray*}
||D^qV||_{p_q}\lesssim||V||^{a_q}_{H^{m_0}}\cdot||V||_2^{1-a_q}\leqslant||V||_{H^{m_0}},
\end{eqnarray*}
and
\begin{eqnarray*}
||D^jV||_{p_j}\lesssim||V||^{a_j}_{H^{m_0}}\cdot||V||_2^{1-a_j}\leqslant||V||_{H^{m_0}}.
\end{eqnarray*}
We hope $p_r$, $p_q$ and $p_j$ meet the next conditions:\\
\textbf{Condition 1.}
\begin{eqnarray*}
\frac{1}{p_r}=\frac{r}{m}+\frac{1}{2}-a_r\cdot\frac{m_0}{m}\,\,\,\,\s\,\,\,\,\s\mbox{with}\,\,\,\,\s\,\,\,\,\s a_r\in\Big[\frac{r}{m_0},1\Big),
\end{eqnarray*}
which is equivalent to
\begin{eqnarray}\label{17}
\frac{1}{p_r}\in\Big(\frac{r-m_0}{m}+\frac{1}{2},\frac{1}{2}\Big].
\end{eqnarray}
\textbf{Condition 2.}
\begin{eqnarray*}
\frac{1}{p_q}=\frac{q}{m}+\frac{1}{2}-a_q\cdot\frac{m_0}{m}\,\,\,\,\s\,\,\,\,\s\mbox{with}\,\,\,\,\s\,\,\,\,\s a_q\in\Big[\frac{q}{m_0},1\Big),
\end{eqnarray*}
which is equivalent to
\begin{eqnarray}\label{18}
\frac{1}{p_q}\in\Big(\frac{q-m_0}{m}+\frac{1}{2},\frac{1}{2}\Big].
\end{eqnarray}
\textbf{Condition 3.}
\begin{eqnarray*}
\frac{1}{p_j}=\frac{j}{m}+\frac{1}{2}-a_j\cdot\frac{m_0}{m}\,\,\,\,\s\,\,\,\,\s\mbox{with}\,\,\,\,\s\,\,\,\,\s a_j\in\Big[\frac{j}{m_0},1\Big),
\end{eqnarray*}
which is equivalent to
\begin{eqnarray}\label{19}
\frac{1}{p_j}\in\Big(\frac{j-m_0}{m}+\frac{1}{2},\frac{1}{2}\Big].
\end{eqnarray}
We claim there exist $p_r$, $p_q$ and $p_j$ which are in $[1,\infty]$ and satisfy (\ref{20}), (\ref{17}), (\ref{18}) and (\ref{19}). Obviously, that this claim holds is equivalent to
\begin{eqnarray}\label{21}
\Big(\frac{r-m_0}{m}+\frac{1}{2}\Big)+\Big(\frac{q-m_0}{m}+\frac{1}{2}\Big)+\Big(\frac{j-m_0}{m}+\frac{1}{2}\Big)<\frac{1}{2}\Longleftrightarrow k<3m_0-m.
\end{eqnarray}
Since $k\leqslant m_0$ and $m_0>\frac{m}{2}$, (\ref{21}) is true. In other words,
\begin{eqnarray*}
\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}\leqslant k-1}}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\lesssim||D^kV||_2\cdot||V||^3_{H^{m_0}}
\end{eqnarray*}
In conclusion,
\begin{eqnarray}\label{23}
&&\sum\limits_{r+q+j=k}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\\
&\lesssim&||D^kV||_2\cdot||V||^3_{H^{m_0}}+||D^kV||_2^2\cdot||V_0||^2_{\infty}.\nonumber
\end{eqnarray}
For the other terms of (\ref{16}), using the same methods, we get similar estimations:\\
\textbf{Estimation 1.}
\begin{eqnarray*}
&&\sum\limits_{i+j=k-1}a_{ij}\int_M|D^{k+1}V|\cdot|D^{i+1}V|\cdot|D^{j+1}V|\,dM\\
&\lesssim&||D^{k+1}V||_2\cdot||D^kV||_2\cdot||DV||_{\infty}+\sum\limits_{\substack{i+j=k-1\\ \max\{i,j\}\leqslant k-2}}a_{ij}\cdot||D^{k+1}V||_2\cdot||D^{i+1}V||_{p_i}\cdot||D^{j+1}V||_{p_j}\\
&\lesssim&||D^{k+1}V||_2\cdot||D^kV||_2\cdot||DV||_{\infty}+||D^{k+1}V||_2\cdot||V||^2_{H^{m_0}}\\
&\lesssim&||D^{k+1}V||_2\cdot||D^kV||_2\cdot||V||_{H^{m_0}}+||D^{k+1}V||_2\cdot||V||^2_{H^{m_0}},
\end{eqnarray*}
\textbf{Estimation 2.}
\begin{eqnarray*}
\sum\limits_{0\leqslant r+q\leqslant k-1}\int_M|D^kV|\cdot|D^rV|\cdot|D^{q+1}V|\,dM\lesssim||V_0||_{\infty}\cdot||D^kV||^2_2+||D^kV||_2\cdot||V||^2_{H^{m_0}},
\end{eqnarray*}
\textbf{Estimation 3.}
\begin{eqnarray*}
\sum\limits_{i=0}^k\sum\limits_{r+q=k-1}\int_M|D^iV|\cdot|D^rV|\cdot|D^{q+1}V|\,dM\lesssim||V||_{H^k}\cdot||D^kV||_2\cdot||V_0||_{\infty}+||V||_{H^k}\cdot||V||^2_{H^{m_0}}.
\end{eqnarray*}
Summing $k$ from $0$ to $m_0$ gives
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}||V||^2_{H^{m_0}}&\leqslant&-||V||^2_{H^{m_0+1}}+L_{m_0}\cdot||V||_{H^{m_0+1}}\cdot||V||^2_{H^{m_0}}\\
&&+\tilde{L}_{m_0}\cdot\Big\{||V||^2_{H^{m_0}}+||V_0||_{\infty}\cdot||V||^2_{H^{m_0}}\\
&&+||V||^3_{H^{m_0}}+||V_0||^2_{\infty}\cdot||V||^2_{H^{m_0}}+||V||^4_{H^{m_0}}\Big\},
\end{eqnarray*}
where $L_{m_0}$ is universal and $\tilde{L}_{m_0}$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations. Then the result follows easily from Young's inequality and Sobolev embedding
$$||V_0||_{\infty}\lesssim||V_0||_{H^{m_0}}.$$
This completes the proof.
$\hfill\Box$\\

Consider an ODE
\begin{eqnarray}\label{eq:2}
\left\{ \begin{aligned}
        \frac{df}{dt}&=C'_{m_0}\cdot(1+||V_0||^2_{H^{m_0}})\cdot(f+f^2)\\
                  f(0)&=||V_0||^2_{H^{m_0}}.
                          \end{aligned} \right.
                          \end{eqnarray}
Solving the above equation to get an expression of $f$, we know that the maximal existence time of the solution $f(\cdot,||V_0||_{H^{m_0}})$ to (\ref{eq:2}) is not smaller than
$$T^*:=\frac{1}{C'_{m_0}\cdot(1+||V_0||^2_{H^{m_0}})}\log\Big(\frac{1+2||V_0||^2_{H^{m_0}}}{2||V_0||^2_{H^{m_0}}}\Big).$$
And $f(t,||V_0||_{H^{m_0}})$ is monotone increasing with respect to $t$. In other words, for all $t\in[0,T^*]$,
$$f(t,||V_0||_{H^{m_0}})\leqslant f(T^*,||V_0||_{H^{m_0}})=1+2||V_0||^2_{H^{m_0}}.$$
By comparison principle of ODE, we know that for any $t\in[0,\min\{T_{\max},T^*\})$,
$$||V(t)||_{H^{m_0}}\leqslant\sqrt{1+2||V_0||^2_{H^{m_0}}}.$$

In the sequel, we focus on the case that $k$ is sufficiently big.
\begin{lem}\label{lem2}
When $k\geqslant m_0+1$, there is a $Q_k>0$ depending on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations such that, for any $t\in[0,T_{\max})$, we have
\begin{eqnarray}\label{22}
&&\frac{d}{dt}||V(t)||^2_{H^k}+||V(t)||^2_{H^{k+1}}\\
&\leqslant&Q_k\cdot\Big[||V(t)||^2_{H^k}\cdot||V(t)||^2_{H^{m_0}}+||V(t)||^4_{H^{k-1}}+||V(t)||^2_{H^k}\cdot(1+||V_0||_{\infty}+||V_0||^2_{\infty})\nonumber\\
&&+||V(t)||^2_{H^{k-1}}\cdot(||V(t)||^2_{H^{m_0}}+||V(t)||^4_{H^{m_0}})+||V(t)||^6_{H^{k-1}}\Big].\nonumber
\end{eqnarray}
\end{lem}
\textbf{Proof.} Firstly, let us calculate one term of (\ref{16}). Applying the same method of (\ref{23}), one can see easily that there are $p_i$ belonging to $[1,\infty]$ such that the following inequalities hold
\begin{eqnarray}\label{24}
&&\sum\limits_{i+j=k-1}a_{ij}\int_M|D^{k+1}V|\cdot|D^{i+1}V|\cdot|D^{j+1}V|\,dM\\
&\lesssim&||D^{k+1}V||_2\cdot||D^kV||_2\cdot||DV||_{\infty}+||D^{k+1}V||_2\cdot||D^{k-1}V||_2\cdot||D^2V||_{\infty}\nonumber\\
&&+\sum\limits_{\substack{i+j=k-1\\ \max\{i,j\}\leqslant k-3}}a_{ij}\cdot||D^{k+1}V||_2\cdot||D^{i+1}V||_{p_i}\cdot||D^{j+1}V||_{p_j}\nonumber\\
&\lesssim&||D^{k+1}V||_2\cdot||D^kV||_2\cdot||V||_{H^{m_0}}+||D^{k+1}V||_2\cdot||D^{k-1}V||_2\cdot||V||_{H^{m_0}}\nonumber\\
&&+||D^{k+1}V||_2\cdot||V||^2_{H^{k-1}}.\nonumber
\end{eqnarray}
By the same procedure, we get the next estimations:\\
\textbf{Estimation 4.}
\begin{eqnarray}\label{25}
&&\sum\limits_{0\leqslant r+q\leqslant k-1}\int_M|D^kV|\cdot|D^rV|\cdot|D^{q+1}V|\,dM\\
&\lesssim&||D^kV||_2^2\cdot||V_0||_{\infty}+||D^kV||_2\cdot||D^{k-1}V||_2\cdot||V||_{H^{m_0}}\nonumber\\
&&+||D^kV||_2\cdot||D^{k-2}V||_2\cdot||V||_{H^{m_0}}+||D^kV||_2\cdot||V||^2_{H^{k-1}}.\nonumber
\end{eqnarray}
\textbf{Estimation 5.}
\begin{eqnarray}\label{26}
&&\sum\limits_{r+q+j=k}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\\
&\lesssim&\int_M|D^kV|^2\cdot|V|^2\,dM+\int_M|D^kV|\cdot|D^{k-1}V|\cdot|DV|\cdot|V|\,dM\nonumber\\
&&+\sum\limits_{\substack{r+q+j=k\\ \max\{r,q,j\}\leqslant k-2}}\int_M|D^kV|\cdot|D^rV|\cdot|D^qV|\cdot|D^jV|\,dM\nonumber\\
&\lesssim&||D^kV||_2^2\cdot||V||^2_{\infty}+||D^kV||_2\cdot||D^{k-1}V||_2\cdot||DV||_{\infty}\cdot||V||_{\infty}+||D^kV||_2\cdot||V||^3_{H^{k-1}}\nonumber\\
&\lesssim&||D^kV||_2^2\cdot||V_0||^2_{\infty}+||D^kV||_2\cdot||D^{k-1}V||_2\cdot||V||^2_{H^{m_0}}+||D^kV||_2\cdot||V||^3_{H^{k-1}}.\nonumber
\end{eqnarray}
\textbf{Estimation 6.}
\begin{eqnarray}\label{27}
&&\sum\limits_{i=0}^k\sum\limits_{r+q=k-1}\int_M|D^iV|\cdot|D^rV|\cdot|D^{q+1}V|\,dM\\
&\lesssim&||V||_{H^k}\cdot||V||_{\infty}\cdot||D^kV||_2+||V||_{H^k}\cdot||D^{k-1}V||_2\cdot||DV||_{\infty}\nonumber\\
&&+||V||_{H^k}\cdot||D^{k-2}V||_2\cdot||D^2V||_{\infty}+||V||_{H^k}\cdot||V||^2_{H^{k-1}}\nonumber\\
&\lesssim&||V||_{H^k}\cdot||V_0||_{\infty}\cdot||D^kV||_2+||V||_{H^k}\cdot||D^{k-1}V||_2\cdot||V||_{H^{m_0}}\nonumber\\
&&+||V||_{H^k}\cdot||D^{k-2}V||_2\cdot||V||_{H^{m_0}}+||V||_{H^k}\cdot||V||^2_{H^{k-1}}\nonumber
\end{eqnarray}

Substituting (\ref{24}), (\ref{25}), (\ref{26}) and (\ref{27}) into (\ref{16}) and then summing $k$ lead to
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}||V||^2_{H^k}&\leqslant&-||V||^2_{H^{k+1}}+\hat{Q}_k\cdot(||V||_{H^{k+1}}\cdot||V||_{H^{k}}\cdot||V||_{H^{m_0}}+||V||_{H^{k+1}}\cdot||V||^2_{H^{k-1}})\\
&&+\tilde{Q}_k\cdot(||V||^2_{H^{k}}+||V||^2_{H^{k}}\cdot||V_0||_{\infty}+||V||_{H^{k}}\cdot||V||_{H^{k-1}}\cdot||V||_{H^{m_0}}\\
&&+||V||_{H^{k}}\cdot||V||^2_{H^{k-1}}+||V||^2_{H^{k}}\cdot||V_0||^2_{\infty}+||V||_{H^{k}}\cdot||V||_{H^{k-1}}\cdot||V||^2_{H^{m_0}}\\
&&+||V||_{H^{k}}\cdot||V||^3_{H^{k-1}}),
\end{eqnarray*}
where $\tilde{Q}_k>0$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations. Using Young's inequality, we conclude this theorem.
$\hfill\Box$\\

Note that (\ref{22}) is linear for $||V||^2_{H^k}$. It is now clear that inductively using (\ref{22}) one can show the existence of $N_k=N(||V_0||_{H^k},Q_k,Q_{k-1},\cdots,Q_{m_0+1})$ for any $k\geqslant m_0+1$ such that
$$||V(t)||_{H^k}\leqslant N_k\,\,\,\,\s\,\,\,\,\s\forall t\in[0,\min\{T_{\max},T^*\}),$$
which implies
$$T_{\max}\geqslant T^*.$$\\

Now we return to prove Theorem \ref{thm4}. Define
$$h(x):=\frac{1}{C'_{m_0}\cdot(1+x)}\log\Big(\frac{1+2x}{2x}\Big)$$
and we observe that it is a monotone decreasing function. Given $l\geqslant m_0+1$ and $V_0\in H^l(E)$, there are $V_{i0}\in\Gamma^{\infty}(E)$ such that as $i\rightarrow\infty$,
$$V_{i0}\longrightarrow V_0\,\,\,\,\s\,\,\,\,\s\mbox{strongly in $H^l(E)$}.$$
By the above discussion we know there exist $$T^*_i\geqslant h(||V_{i0}||^2_{H^{m_0}})>0\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s V_i\in C^{\infty}([0,T^*_i),\Gamma^{\infty}(E))$$ such that
\begin{eqnarray} \label{eq:3}
\left\{ \begin{aligned}
         &\partial_tV_i=\Delta V_i+V_i\times\Delta V_i-\lambda\cdot(1+\mu|V_i|^2)V_i \\
                 &V_i(0,\cdot)=V_{i0},
                          \end{aligned} \right.
                          \end{eqnarray}
here $T^*_i$ is the maximal existence time of $V_i$. Obviously, when $i$ is enough large,
$$||V_{i0}||^2_{H^{m_0}}\leqslant||V_{0}||^2_{H^{m_0}}+1\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s||V_{i0}||_{H^{l}}\leqslant||V_{0}||_{H^{l}}+1,$$
which imply
$$T^*_i\geqslant h(||V_{0}||^2_{H^{m_0}}+1):=2\tilde{\delta}>0$$
and
$$||V_i(t)||_{H^l}\leqslant N(||V_{0}||_{H^{l}}+1,Q_k,Q_{k-1},\cdots,Q_{m_0+1})\,\,\,\,\s\,\,\,\,\s\forall t\in[0,\tilde{\delta}].$$
Then $V_i$ is a bounded sequence in $L^{\infty}([0,\tilde{\delta}],H^l(E))$. It is not hard to verify that $\partial_tV_i$ is a bounded sequence in $L^{\infty}([0,\tilde{\delta}],L^2(E))$. So there exists a $V\in L^{\infty}([0,\tilde{\delta}],H^{l}(E))$ and a subsequence which is still denoted by $\{V_i\}$ such that
$$V_i\rightharpoonup V\,\,\,\,\s\,\,\,\,\s\mbox{weakly $*$ in $L^{\infty}([0,\tilde{\delta}],H^{l}(E))$}.$$
By Aubin-Lions lemma, one can find a subsequence still denoted by $\{V_i\}$ such that
$$V_i\longrightarrow V\,\,\,\,\s\,\,\,\,\s\mbox{strongly in $L^{\infty}([0,\tilde{\delta}],H^{l-1}(E))$}.$$
Because $l-1\geqslant m_0$, $H^{l-1}(E)$ can be embedded into $\Gamma^2(E)$. In other words, $V$ is a solution to (\ref{eq:1}). Using LLB to transform time derivatives into spatial derivatives gives that for all $0\leqslant j\leqslant\big[\frac{l}{\hat{m}}\big]$ and all $\alpha\leqslant l-\hat{m}j$, we have
\begin{eqnarray}\label{28}
\partial^j_tD^{\alpha}V\in L^{\infty}([0,\tilde{\delta}],L^2(E)).
\end{eqnarray}
\begin{rem}
The proof of (\ref{28}) is easy if one employs induction for $j$.
\end{rem}
At last, since $l\geqslant[\frac{m}{2}]+4$, by the same method of Theorem 3 in \cite{GLZ} it is not difficult to know that the solution of (\ref{eq:1}) with initial data $V_0\in H^l(E)$ is unique. This completes the proof.
$\hfill\Box$\\
\section{Proof of Theorem \ref{thm5}}
Now we focus on global existence of LLB. Suppose that $V$ is the local smooth solution of (\ref{eq:1}). Our trick is to deduce a uniform estimation for $||V||_{H^k}$. To this goal, firstly we should get a linear Gronwall inequality.

By (\ref{16}) and H\"older inequality, we have
\begin{eqnarray}\label{37}
\frac{1}{2}\frac{d}{dt}||D^kV||_2^2&\leqslant&-||D^{k+1}V||_2^2+\sum\limits_{i+j=k-1}a_{ij}\cdot||D^{k+1}V||_2\cdot\Big|\Big||D^{i+1}V|\cdot|D^{j+1}V|\Big|\Big|_2\nonumber\\
&&+\tilde{C}_k\cdot\Big\{||V||^2_{H^k}+||D^kV||_2\sum\limits_{0\leqslant r+q\leqslant k-1}\Big|\Big||D^rV|\cdot|D^{q+1}V|\Big|\Big|_2\\
&&||D^kV||_2\sum\limits_{r+q+j=k}\Big|\Big||D^rV|\cdot|D^qV|\cdot|D^jV|\Big|\Big|_2\nonumber\\
&&+||V||_{H^k}\sum\limits_{r+q=k-1}\Big|\Big||D^rV|\cdot|D^{q+1}V|\Big|\Big|_2\Big\}.\nonumber
\end{eqnarray}

For the second term on the right hand side of (\ref{37}),
$$
\Big|\Big||D^{i+1}V|\cdot|D^{j+1}V|\Big|\Big|_2\leqslant||D^{i+1}V||_{\frac{2k+2}{i+1}}\cdot||D^{k-i}V||_{\frac{2k+2}{k-i}},
$$
since $i+j=k-1$. Theorem \ref{thm6} implies
$$
||D^{i+1}V||_{\frac{2k+2}{i+1}}\lesssim||D^{k+1}V||_2^{\frac{i+1}{k+1}}\cdot||V||_{\infty}^{\frac{k-i}{k+1}}
$$
and
$$
||D^{k-i}V||_{\frac{2k+2}{k-i}}\lesssim||D^{k+1}V||_2^{\frac{k-i}{k+1}}\cdot||V||_{\infty}^{\frac{i+1}{k+1}}.
$$
So
\begin{eqnarray}\label{38}
&&\sum\limits_{i+j=k-1}a_{ij}\cdot||D^{k+1}V||_2\cdot\Big|\Big||D^{i+1}V|\cdot|D^{j+1}V|\Big|\Big|_2\\
&\leqslant&B_k\cdot||V||_{\infty}\cdot||D^{k+1}V||_2^2\leqslant B_k\cdot||V_0||_{\infty}\cdot||D^{k+1}V||_2^2,\nonumber
\end{eqnarray}
where $B_k$ is a universal constant. By the same way, we will get
\begin{eqnarray}\label{39}
&&\sum\limits_{0\leqslant r+q\leqslant k-1}\Big|\Big||D^rV|\cdot|D^{q+1}V|\Big|\Big|_2\leqslant\sum\limits_{0\leqslant r+q\leqslant k-1}||D^rV||_{\frac{2r+2q+2}{r}}\cdot||D^{q+1}V||_{\frac{2r+2q+2}{q+1}}\\
&\lesssim&\sum\limits_{0\leqslant r+q\leqslant k-1}\Big[||D^{r+q+1}V||_2^{\frac{r}{r+q+1}}\cdot||V||^{\frac{q+1}{r+q+1}}_{\infty}\Big]\cdot\Big[||D^{r+q+1}V||_2^{\frac{q+1}{r+q+1}}\cdot||V||^{\frac{r}{r+q+1}}_{\infty}\Big]\nonumber\\
&=&\sum\limits_{0\leqslant r+q\leqslant k-1}||D^{r+q+1}V||_2\cdot||V||_{\infty}\lesssim||V||_{H^k}\cdot||V_0||_{\infty}\nonumber
\end{eqnarray}
and
\begin{eqnarray}\label{40}
\sum\limits_{r+q=k-1}\Big|\Big||D^rV|\cdot|D^{q+1}V|\Big|\Big|_2\lesssim||D^kV||_{2}\cdot||V_0||_{\infty}.
\end{eqnarray}
Moreover, Theorem \ref{thm6} yields
\begin{eqnarray}\label{41}
&&\sum\limits_{r+q+j=k}\Big|\Big||D^rV|\cdot|D^qV|\cdot|D^jV|\Big|\Big|_2\leqslant\sum\limits_{r+q+j=k}||D^rV||_{\frac{2k}{r}}\cdot||D^qV||_{\frac{2k}{q}}\cdot||D^jV||_{\frac{2k}{j}}\\
&\lesssim&\sum\limits_{r+q+j=k}\Big(||D^kV||^{\frac{r}{k}}_2\cdot||V||^{\frac{q+j}{k}}_{\infty}\Big)\cdot\Big(||D^kV||^{\frac{q}{k}}_2\cdot||V||^{\frac{r+j}{k}}_{\infty}\Big)\cdot\Big(||D^kV||^{\frac{j}{k}}_2\cdot||V||^{\frac{q+r}{k}}_{\infty}\Big)\nonumber\\
&\lesssim&||D^kV||_2\cdot||V||^2_{\infty}\leqslant||D^kV||_2\cdot||V_0||^2_{\infty}.\nonumber
\end{eqnarray}
Substituting (\ref{38}), (\ref{39}), (\ref{40}) and (\ref{41}) into (\ref{37}) leads to
\begin{eqnarray}\label{42}
&&\frac{1}{2}\frac{d}{dt}||D^kV||_2^2+(1-B_k\cdot||V_0||_{\infty})\cdot||D^{k+1}V||_2^2\\
&\leqslant&G_k\cdot\{||V||^2_{H^k}+||D^kV||_2\cdot||V||_{H^k}\cdot||V_0||_{\infty}+||D^kV||_2^2\cdot||V_0||^2_{\infty}\}\nonumber\\
&\leqslant&G_k\cdot(1+||V_0||_{\infty}+||V_0||^2_{\infty})\cdot||V||^2_{H^k},\nonumber
\end{eqnarray}
where $G_k$ depends upon $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant differentiations. In the sequel, using Gronwall inequality gives the following theorem.
\begin{thm}\label{thm8}
Given $N\in\mathbb{N}$, there exists an $\tilde{B}_N>0$ such that if $||V_0||_{\infty}\leqslant\tilde{B}_N$, we will obtain
\begin{eqnarray}\label{36}
||D^kV(t)||_2^2+\int^t_0||D^{k+1}V(s)||_2^2\,ds\leqslant C_k(||V_0||_{H^k},\tilde{B}_N,t),
\end{eqnarray}
provided $0\leqslant k\leqslant N$ and $t\in[0,T_{\max})$. Here $C_k(x,y,t)$ is monotone increasing with respect to $x$ and $t$.
\end{thm}
\textbf{Proof.} Employ induction for $N$.

In the case $N=0$, let $\tilde{B}_0:=1$. Taking inner product with $V$ in (\ref{eq:1}) and then integrating the result over $M$, we get
$$
\frac{1}{2}\frac{d}{dt}||V(t)||_2^2+||DV(t)||_2^2+\lambda\int_M(1+\mu|V(t)|^2)\cdot|V(t)|^2\,dM=0
$$
which is equivalent to
\begin{eqnarray}\label{*}
&&||V(t)||_2^2+2\int_0^t||DV(s)||_2^2\,ds\\
&&+2\lambda\int_0^tds\int_M(1+\mu|V(s)|^2)\cdot|V(s)|^2\,dM=||V_0||_2^2.\nonumber
\end{eqnarray}

Assume that for all the indices not larger than $N$, (\ref{36}) holds. Now we consider $N+1$.

Take $\tilde{B}_{N+1}:=\min\{\tilde{B}_N,\frac{1}{2B_{N+1}}\}$. If $||V_0||_{\infty}\leqslant\tilde{B}_{N+1}$, (\ref{42}) gives
\begin{eqnarray}\label{43}
&&\frac{1}{2}\frac{d}{dt}||D^{N+1}V||_2^2+\frac{1}{2}||D^{N+2}V||_2^2\\
&\leqslant& G_{N+1}\cdot(1+\tilde{B}_{N+1}+\tilde{B}^2_{N+1})\cdot\Big\{||D^{N+1}V||_2^2+\sum\limits_{k=0}^NC_k(||V_0||_{H^k},\tilde{B}_N,t)\Big\}.\nonumber
\end{eqnarray}
Then this theorem follows easily from Gronwall inequality. This completes the proof.$\hfill\Box$\\

Now we return to prove Theorem \ref{thm5}. Given $T>0$ and $N\geqslant m_0+1=[\frac{m}{2}]+4$, we take any $V_0\in H^N(E)$ with $||V_0||_{\infty}\leqslant\frac{1}{2}\tilde{B}_N:=\hat{B}_N$(This $\tilde{B}_N$ is from Theorem \ref{thm8}). Then there are $V_{0i}\in\Gamma^{\infty}(E)$ converging to $V_0$ strongly in $H^N(E)$.

Suppose $V_i$ satisfies
\begin{eqnarray}\label{eq:4}
\left\{ \begin{aligned}
         &\partial_tV_i=\Delta V_i+V_i\times\Delta V_i-\lambda\cdot(1+\mu\cdot|V_i|^2)V_i \\
                  &V_i(0,\cdot)=V_{0i}
                          \end{aligned} \right.
\end{eqnarray}
and its maximal existence time is $T^*_i$. As $i$ is large enough, we have $$||V_{0i}||_{\infty}\leqslant2||V_0||_{\infty}\leqslant\tilde{B}_N,\,\,\,\,\s\,\,\,\,\s||V_{0i}||_{H^{m_0}}\leqslant2||V_0||_{H^{m_0}}$$
and $||V_{0i}||_{H^{N}}\leqslant2||V_0||_{H^{N}}$.

If $T^*_i<T$, then by Theorem \ref{thm8},
\begin{eqnarray*}
||V_i(t)||_{H^{m_0}}^2+\int^t_0||DV_i(s)||_{H^{m_0}}^2\,ds\leqslant C_{m_0}(||V_{0i}||_{H^{m_0}},\tilde{B}_N,t)\leqslant C_{m_0}(2||V_{0}||_{H^{m_0}},\tilde{B}_N,T),
\end{eqnarray*}
provided $t\in[0,T^*_i)$. Review that in the proof of Theorem \ref{thm4} we have defined a monotone decreasing function
$$
h(x):=\frac{1}{C'_{m_0}\cdot(1+x)}\log\Big(\frac{1+2x}{2x}\Big).
$$
So for arbitrary $t\in[0,T^*_i)$, it is obvious to see
$$h(||V_i(t)||^2_{H^{m_0}})\geqslant h\big(C_{m_0}(2||V_{0}||_{H^{m_0}},\tilde{B}_N,T)\big):=\delta_0>0.$$
Now we bring in a new system
\begin{eqnarray}\label{eq:5}
\left\{ \begin{aligned}
         &\partial_t\hat{V}_i=\Delta \hat{V}_i+\hat{V}_i\times\Delta \hat{V}_i-\lambda\cdot(1+\mu\cdot|\hat{V}_i|^2)\hat{V}_i \,\,\,\,\s\mbox{in}\,\,\,\,\s \Big(T^*_i-\frac{\delta_0}{2},\infty\Big)\times M\\
                  &\hat{V}_i\Big(T^*_i-\frac{\delta_0}{2},\cdot\Big)=V_i\Big(T^*_i-\frac{\delta_0}{2},\cdot\Big)
                          \end{aligned} \right.
\end{eqnarray}
The maximal existence time of $\hat{V}_i$ is not smaller than $h(||V_i(T^*_i-\frac{\delta_0}{2})||^2_{H^{m_0}})$ which is not smaller than $\delta_0$. By the uniqueness we know that for any $t\in[T^*_i-\frac{\delta_0}{2},T^*_i)$, $\hat{V}_i(t)=V_i(t)$. It means that $V_i$ can be extended to $[0,T^*_i+\frac{\delta_0}{2})$. Because $T^*_i$ is maximal, we get a contradiction. So $V_i\in C^{\infty}([0,T],\Gamma^{\infty}(E))$ and for all $t\in[0,T]$,
$$
||V_i(t)||_{H^N}^2+\int^t_0||DV_i(s)||_{H^N}^2\,ds\leqslant C_N(2||V_0||_{H^N},\tilde{B}_N,T).
$$

By the same method we prove local well-posedness one can know there is a
$$V\in L^{\infty}([0,T],H^N(E))\cap L^2([0,T],H^{N+1}(E))$$
such that $V_i$ converges to $V$ strongly in $L^{\infty}([0,T],H^{N-1}(E))$(in the sense of picking subsequence). It means $V$ is a solution of LLB.\\

At last, we claim (\ref{7}) and (\ref{9}) are true. Since (\ref{7}) is easy, we only prove (\ref{9}).\\
\textbf{Proof.} Employ induction for $i$.

When $i=0$, (\ref{9}) holds.

Suppose that for all the indices not bigger than $i$, (\ref{9}) is true.

Now we consider $i+1$. Choose any $\beta\in[0,N+1-(\hat{m}+1)(i+1)]\cap\mathbb{Z}$. Applying $\partial_t^iD^{\beta}$ to both sides of (\ref{eq:1}), we get
\begin{eqnarray*}
\partial_t^{i+1}D^{\beta}V=\partial_t^iD^{\beta}\Delta V-\partial_t^iD^{\beta}(V\times\Delta V)-\lambda\cdot\partial_t^iD^{\beta}\big[(1+\mu|V|^2)V\big],
\end{eqnarray*}
which implies
\begin{eqnarray}\label{44}
&&\int_0^{T}||\partial_t^{i+1}D^{\beta}V(s)||_2^2\,ds\\
&\lesssim&\int_0^{T}||\partial_t^{i}D^{\beta+2}V(s)||_2^2\,ds+\int_0^{T}||\partial_t^{i}D^{\beta}V(s)||_2^2\,ds\nonumber\\
&&+\sum\limits_{i',\beta'}\int_0^{T}\Big|\Big||\partial_t^{i'}D^{\beta'}V(s)|\cdot|\partial_t^{i-i'}D^{\beta-\beta'+2}V(s)|\Big|\Big|^2_2\,ds\nonumber\\
&&+\sum\limits_{\substack{i_1+i_2+i_3=i\\ \beta_1+\beta_2+\beta_3=\beta}}\int_0^{T}\Big|\Big||\partial_t^{i_1}D^{\beta_1}V(s)|\cdot|\partial_t^{i_2}D^{\beta_2}V(s)|\cdot|\partial_t^{i_3}D^{\beta_3}V(s)|\Big|\Big|^2_2\,ds.\nonumber
\end{eqnarray}
Because $\hat{m}:=\max\{2,[\frac{m}{2}]+1\}$,
$$i'\leqslant i\leqslant\Big[\frac{N+1}{\hat{m}+1}\Big]\leqslant\Big[\frac{N}{\hat{m}}\Big]\,\,\,\,\s\,\,\,\,\s\mbox{and}\,\,\,\,\s\,\,\,\,\s\beta'\leqslant\beta\leqslant N+1-(\hat{m}+1)(i+1)\leqslant N-\hat{m}\cdot i'-\hat{m},$$
(\ref{7}) yields
$$||\partial_t^{i'}D^{\beta'}V(t)||_{\infty}\leqslant||\partial_t^{i'}D^{\beta'}V(t)||_{H^{\hat{m}}}<\infty\,\,\,\,\s\,\,\,\,\s\forall t\in[0,T].$$
And since
$$\hat{m}(i-i')+(\beta-\beta'+2)\leqslant\hat{m}\cdot i+\beta+2\leqslant\hat{m}\cdot i+N+1-(\hat{m}+1)(i+1)+2=N-i-\hat{m}+2\leqslant N-i\leqslant N,$$
by (\ref{7}) we have
$$
\int_0^{T}||\partial_t^{i-i'}D^{\beta-\beta'+2}V(s)||_2^2\,ds\leqslant\sup\limits_{t\in[0,T]}\{||\partial_t^{i-i'}D^{\beta-\beta'+2}V(t)||^2_2\}\cdot T<\infty.
$$
So
\begin{eqnarray*}
&&\int_0^{T}\Big|\Big||\partial_t^{i'}D^{\beta'}V(s)|\cdot|\partial_t^{i-i'}D^{\beta-\beta'+2}V(s)|\Big|\Big|^2_2\,ds\\
&\leqslant&\sup\limits_{t\in[0,T]}\{||\partial_t^{i'}D^{\beta'}V(t)||^2_{\infty}\}\cdot\int_0^{T}||\partial_t^{i-i'}D^{\beta-\beta'+2}V(s)||_2^2\,ds<\infty
\end{eqnarray*}
For the other terms on the right hand side of (\ref{44}), using similar method we know all of them are strictly smaller than $\infty$.

This completes the proof.$\hfill\Box$\\
\section{Proof of Theorem \ref{thm9}}
In this section, we need some formulas. Their proofs are tedious. So we only list the results.

\textbf{Formula 3.} Suppose that $V\in\Gamma^2(E)$. Then we will obtain
\begin{eqnarray*}
||\Delta V||_2^2&=&||D^2V||_2^2+2\int_M\langle DV,DV\ast\mathcal{R}^E\rangle\,dM\\
&&+\int_M\langle DV,V\ast D\mathcal{R}^E\rangle\,dM+\int_M\langle DV,DV\ast\mathcal{R}^M\rangle\,dM.
\end{eqnarray*}
\begin{rem}
Formula 3 easily implies
\begin{eqnarray}\label{5.4}
||D^2V||_2^2\leqslant||\Delta V||_2^2+\eta\cdot(||DV||_2^2+||V||_2^2),
\end{eqnarray}
where $\eta$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives.
\end{rem}
\textbf{Formula 4.} Given $V\in\Gamma^3(E)$,
\begin{eqnarray*}
||\Delta DV||_2^2&=&||D^3V||_2^2+3\int_M\langle D^2V,D^2V\ast\mathcal{R}^M\rangle\,dM+2\int_M\langle D^2V,D^2V\ast\mathcal{R}^E\rangle\,dM\\
&&+\int_M\langle D^2V,DV\ast\nabla\mathcal{R}^M\rangle\,dM+\int_M\langle D^2V,DV\ast D\mathcal{R}^E\rangle\,dM.
\end{eqnarray*}
\begin{rem}
From Formula 4 it follows that
\begin{eqnarray}\label{45}
||D^3V||_2^2\leqslant||\Delta DV||_2^2+\eta_2\cdot(||D^2V||_2^2+||DV||_2^2),
\end{eqnarray}
where $\eta_2$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives. Since by (\ref{11}) we have
\begin{eqnarray*}
\Delta DV=D\Delta V+\mathfrak{q}_1(V,\mathcal{R}^E)+\mathfrak{q}_0(DV,\mathcal{R}^M),
\end{eqnarray*}
integration by parts and H\"older's inequality yield
\begin{eqnarray}\label{47}
||\Delta DV||_2^2\leqslant||D\Delta V||_2^2+\eta_3\cdot(||D^2V||_2^2+||DV||_2^2+||V||_2^2),
\end{eqnarray}
where $\eta_3$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives. Substituting (\ref{47}) into (\ref{45}) gives
\begin{eqnarray}\label{5.5}
||D^3V||_2^2\leqslant||D\Delta V||_2^2+(\eta_2+\eta_3)\cdot(||D^2V||_2^2+||DV||_2^2+||V||_2^2).
\end{eqnarray}
\end{rem}
\textbf{Formula 5.} If $V\in\Gamma^4(E)$, then
\begin{eqnarray*}
||D^4V||_2^2&=&||\Delta^2V||_2^2+\int_M\langle\mathfrak{q}_3(V,\mathcal{R}^E),D^3V\rangle\,dM+\int_M\langle\mathfrak{q}_2(DV,\mathcal{R}^M),D^3V\rangle\,dM\\
&&+\int_M\langle\mathfrak{q}_1(DV,\mathcal{R}^E),\mathfrak{q}_1(DV,\mathcal{R}^E)\rangle\,dM+\int_M\langle\mathfrak{q}_1(DV,\mathcal{R}^E),\mathfrak{q}_1(DV,\mathcal{R}^M)\rangle\,dM\\
&&+\int_M\langle\mathfrak{q}_1(DV,\mathcal{R}^M),\mathfrak{q}_1(DV,\mathcal{R}^M)\rangle\,dM.
\end{eqnarray*}
\begin{rem}
Formula 5, H\"older's inequality and Young's inequality lead to
\begin{eqnarray}\label{5.6}
||D^4V||_2^2&\leqslant&||\Delta^2V||_2^2+\eta_5\cdot(||D^3V||_2^2+||D^2V||_2^2+||DV||_2^2+||V||_2^2)\nonumber\\
&\leqslant&||\Delta^2V||_2^2+\eta_6\cdot(||D\Delta V||_2^2+||\Delta V||_2^2+||DV||_2^2+||V||_2^2),
\end{eqnarray}
where we have used (\ref{5.4}), (\ref{5.5}) and $\eta_5$, $\eta_6$ depend on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives.
\end{rem}
\textbf{Formula 6.} Assume that $V\in\Gamma^4(E)$. Then we get
\begin{eqnarray}\label{48}
||D^2\Delta V||_2^2&=&||\Delta^2V||_2^2+\int_M\langle\mathfrak{q}_1(\Delta V,\mathcal{R}^E),D\Delta V\rangle\,dM+\int_M\langle\mathfrak{q}_0(D\Delta V,\mathcal{R}^M),D\Delta V\rangle\,dM\nonumber\\
&\lesssim&||\Delta^2V||_2^2+\eta_8\cdot(||D\Delta V||_2^2+||\Delta V||_2^2),
\end{eqnarray}
where we have applied H\"older's inequality, Young's inequality and $\eta_8$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives.\\

Now let us go on to prove Theorem \ref{thm9}. Suppose that $V\in C^{\infty}\big([0,T^*),\Gamma^{\infty}(E)\big)$ is the unique local smooth solution of (\ref{eq:1}), where $T^*$ is its maximal existence time. First of all, we shall estimate $||DV(t)||_{\infty}$ for all $t\in[0,T^*)$. By Sobolev embedding it is easy to see that we only need to get a uniform upper bound of $||DV(t)||_{H^2}$(Note that in this section $m=2$). Combining (\ref{5.4}) and (\ref{5.5}) one can know that we only need to estimate
$$||DV(t)||_2^2+||\Delta V(t)||_2^2+||D\Delta V(t)||_2^2.$$

Using the same method of (2.2) in \cite{GLZ} we can get
\begin{eqnarray}\label{5.3}
||DV(t)||_2^2+\int_0^t||\Delta V(s)||_2^2\,ds\leqslant\lambda^2\cdot(1+\mu||V_0||^2_{\infty})^2\cdot||V_0||_2^2\cdot t+||DV_0||_2^2.
\end{eqnarray}
\\

For $||\Delta V(t)||_2^2$, our trick is to deduce a Gronwall's inequality. (\ref{eq:1}) yields
\begin{eqnarray}\label{49}
&&\frac{1}{2}\frac{d}{dt}||\Delta V(t)||_2^2+\int_M|D\Delta V(t)|^2\,dM+\lambda\cdot||\Delta V(t)||_2^2\\
&=&-\int_M[DV(t)\times\Delta V(t)]\ast D\Delta V(t)\,dM-\lambda\mu\cdot\int_M\langle\Delta[|V(t)|^2\cdot V(t)],\Delta V(t)\rangle\,dM\nonumber\\
&\leqslant&\int_M|DV(t)|\cdot|\Delta V(t)|\cdot|D\Delta V(t)|\,dM+C\cdot||V(t)||^2_{\infty}\cdot(||\Delta V(t)||^2_2+||DV(t)||_4^2)\nonumber\\
&\leqslant&||DV(t)||_4\cdot||\Delta V(t)||_4\cdot||D\Delta V(t)||_2+C\cdot||V(t)||^2_{\infty}\cdot(||\Delta V(t)||^2_2+||DV(t)||_4^2),\nonumber
\end{eqnarray}
here $C$ is a universal constant. Theorem 2.1 of \cite{DW1}, (\ref{5.4}) and (\ref{5.5}) give
\begin{eqnarray}\label{5.1}
||DV(t)||_4&\lesssim&||DV(t)||^{\frac{1}{4}}_{H^2}\cdot||DV(t)||_2^{\frac{3}{4}}\nonumber\\
&\leqslant&\eta_4\cdot(||D\Delta V(t)||_2^{\frac{1}{4}}+||\Delta V(t)||_2^{\frac{1}{4}}+||DV(t)||_2^{\frac{1}{4}}+||V(t)||_2^{\frac{1}{4}})\cdot||DV(t)||_2^{\frac{3}{4}}
\end{eqnarray}
and
\begin{eqnarray}\label{5.2}
||\Delta V(t)||_4&\lesssim&||\Delta V(t)||^{\frac{1}{2}}_{H^1}\cdot||\Delta V(t)||_2^{\frac{1}{2}}\nonumber\\
&\lesssim&(||D\Delta V(t)||_2^{\frac{1}{2}}+||\Delta V(t)||_2^{\frac{1}{2}})\cdot||\Delta V(t)||_2^{\frac{1}{2}},
\end{eqnarray}
where $\eta_4$ depends upon $\eta_2$, $\eta_3$ and $\eta$. Thus we derive
\begin{eqnarray*}
&&||DV(t)||_4\cdot||\Delta V(t)||_4\cdot||D\Delta V(t)||_2\\
&\lesssim&\eta_4\cdot||DV(t)||_2^{\frac{3}{4}}\cdot||\Delta V(t)||_2^{\frac{1}{2}}\cdot||D\Delta V(t)||_2^{\frac{7}{4}}\\
&&+\eta_4\cdot||DV(t)||_2^{\frac{3}{4}}\cdot||\Delta V(t)||_2^{\frac{3}{4}}\cdot||D\Delta V(t)||_2^{\frac{3}{2}}\\
&&+\eta_4\cdot||DV(t)||_2\cdot||\Delta V(t)||_2^{\frac{1}{2}}\cdot||D\Delta V(t)||_2^{\frac{3}{2}}\\
&&+\eta_4\cdot||DV(t)||_2^{\frac{3}{4}}\cdot||\Delta V(t)||_2\cdot||D\Delta V(t)||_2^{\frac{5}{4}}\\
&&+\eta_4\cdot||DV(t)||_2^{\frac{3}{4}}\cdot||\Delta V(t)||_2^{\frac{5}{4}}\cdot||D\Delta V(t)||_2\\
&&+\eta_4\cdot||DV(t)||_2\cdot||\Delta V(t)||_2\cdot||D\Delta V(t)||_2\\
&&+\eta_4\cdot||V(t)||_2^{\frac{1}{4}}\cdot||DV(t)||^{\frac{3}{4}}_2\cdot||\Delta V(t)||_2^{\frac{1}{2}}\cdot||D\Delta V(t)||_2^{\frac{3}{2}}\\
&&+\eta_4\cdot||V(t)||_2^{\frac{1}{4}}\cdot||DV(t)||^{\frac{3}{4}}_2\cdot||\Delta V(t)||_2\cdot||D\Delta V(t)||_2\\
&\leqslant&\eta_4\gamma_1\cdot(||\Delta V(t)||_2^{\frac{1}{2}}\cdot||D\Delta V(t)||_2^{\frac{7}{4}}+||\Delta V(t)||_2^{\frac{3}{4}}\cdot||D\Delta V(t)||_2^{\frac{3}{2}}\\
&&+||\Delta V(t)||_2^{\frac{1}{2}}\cdot||D\Delta V(t)||_2^{\frac{3}{2}}+||\Delta V(t)||_2\cdot||D\Delta V(t)||_2^{\frac{5}{4}}\\
&&+||\Delta V(t)||_2^{\frac{5}{4}}\cdot||D\Delta V(t)||_2+||\Delta V(t)||_2\cdot||D\Delta V(t)||_2),
\end{eqnarray*}
where $\gamma_1$ relies on $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$ and $t$. From Young's inequality it follows that
\begin{eqnarray}\label{50}
&&||DV(t)||_4\cdot||\Delta V(t)||_4\cdot||D\Delta V(t)||_2\\
&\leqslant&\gamma_2\cdot(||\Delta V(t)||_2^4+1)+\frac{1}{4}\cdot||D\Delta V(t)||_2^2,\nonumber
\end{eqnarray}
where $\gamma_2$ is dependent of $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$ and $t$.
By the same way, we have
\begin{eqnarray}\label{51}
&&||V(t)||^2_{\infty}\cdot(||\Delta V(t)||_2^2+||DV(t)||_4^2)\\
&\lesssim&||V(t)||^2_{\infty}\cdot||\Delta V(t)||_2^2\nonumber\\
&&+\eta_4^2\cdot||V_0||^2_{\infty}\cdot(||D\Delta V(t)||_2^{\frac{1}{2}}+||\Delta V(t)||_2^{\frac{1}{2}}+||DV(t)||_2^{\frac{1}{2}}+||V(t)||_2^{\frac{1}{2}})\cdot||DV(t)||_2^{\frac{3}{2}}\nonumber\\
&\leqslant&\gamma_3\cdot(||\Delta V(t)||_2^2+1)+\frac{1}{4}||D\Delta V(t)||_2^2,\nonumber
\end{eqnarray}
where $\gamma_3$ is dependent of $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$ and $t$. Substituting (\ref{50}) and (\ref{51}) into (\ref{49}) and Young's inequality lead to
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}||\Delta V(t)||_2^2+\frac{1}{2}||D\Delta V(t)||_2^2+\lambda||\Delta V(t)||_2^2\leqslant\gamma_4\cdot(||\Delta V(t)||_2^4+1),
\end{eqnarray*}
here $\gamma_4$ relies on $\gamma_2$ and $\gamma_3$. The generalized Gronwall's inequality says that if
$$\frac{df}{dt}\leqslant C\cdot f\cdot g+C,$$
then
$$f\leqslant C\cdot\exp\Big(\int_0^tg(s)\,ds\Big)+C.$$
So if we replace $f$ and $g$ by $||\Delta V(t)||_2^2$ and note that (\ref{5.3}) implies the boundedness of $\int_0^tg(s)\,ds$, then
\begin{eqnarray}\label{56}
||\Delta V(t)||_2^2\leqslant\gamma_5
\end{eqnarray}
which implies
\begin{eqnarray}\label{Eq:1}
\int_0^t||D\Delta V(s)||_2^2\,ds\leqslant2\gamma_4\cdot(\gamma_5^2+1)\cdot t,
\end{eqnarray}
where $\gamma_5$ is dependent of $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$, $||\Delta V_0||_2$ and $t$.\\

In the sequel, we are going to estimate $||D\Delta V(t)||_2$ for all $t\in[0,T^*)$. (\ref{eq:1}) gives
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}||D\Delta V(t)||_2^2&=&-||\Delta^2V(t)||_2^2-\int_M\langle\Delta^2V(t),\Delta[V(t)\times\Delta V(t)]\rangle\,dM\\
&&+\lambda\int_M\langle\Delta[(1+\mu|V(t)|^2)\cdot V(t)],\Delta^2 V(t)\rangle\,dM\\
&=&-||\Delta^2V(t)||_2^2-2\int_M\Delta^2V(t)\ast[DV(t)\times D\Delta V(t)]\,dM\\
&&+\lambda\int_M\langle\Delta[(1+\mu|V(t)|^2)\cdot V(t)],\Delta^2 V(t)\rangle\,dM\\
&=&-||\Delta^2V(t)||_2^2-2\int_M\Delta^2V(t)\ast[DV(t)\times D\Delta V(t)]\,dM\\
&&+\lambda\mu\int_M\langle\Delta\big[|V(t)|^2\cdot V(t)\big],\Delta^2 V(t)\rangle\,dM-\lambda||D\Delta V(t)||_2^2.
\end{eqnarray*}
On the other hand, H\"older inequality yields
\begin{eqnarray}\label{52}
&&\Big|\int_M\Delta^2V(t)\ast[DV(t)\times D\Delta V(t)]\,dM\Big|\\
&\leqslant&||DV(t)||_{\frac{16}{5}}\cdot||D\Delta V(t)||_{\frac{16}{3}}\cdot||\Delta^2V(t)||_2\nonumber
\end{eqnarray}
and
\begin{eqnarray*}
&&\Big|\int_M\langle\Delta\big[|V(t)|^2\cdot V(t)\big],\Delta^2 V(t)\rangle\,dM\Big|\\
&\lesssim&||V(t)||^2_{\infty}\cdot||\Delta^2V(t)||_2\cdot||\Delta V(t)||_2+||V(t)||_{\infty}\cdot||\Delta^2V(t)||_2\cdot||DV(t)||_4^2.
\end{eqnarray*}
By Sobolev Embedding, we have
\begin{eqnarray}\label{53}
||DV(t)||_{\frac{16}{5}}\lesssim||DV(t)||^{\frac{1}{8}}_{H^3}\cdot||DV(t)||^{\frac{7}{8}}_2.
\end{eqnarray}
Combining (\ref{5.4}), (\ref{5.5}) and (\ref{5.6}) we arrive at
\begin{eqnarray}\label{54}
||DV(t)||^{\frac{1}{8}}_{H^3}&\leqslant&\eta_7\cdot(||\Delta^2V(t)||_2^{\frac{1}{8}}+||D\Delta V(t)||_2^{\frac{1}{8}}\\
&&+||\Delta V(t)||_2^{\frac{1}{8}}+||DV(t)||_2^{\frac{1}{8}}+||V(t)||_2^{\frac{1}{8}}),\nonumber
\end{eqnarray}
where $\eta_7$ is dependent of $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives. Moreover,
\begin{eqnarray*}
||D\Delta V(t)||_{\frac{16}{3}}\lesssim||D\Delta V(t)||^{\frac{5}{8}}_{H^1}\cdot||D\Delta V(t)||_2^{\frac{3}{8}}
\end{eqnarray*}
and
\begin{eqnarray*}
||D\Delta V(t)||_{H^1}\lesssim||D^2\Delta V(t)||_2+||D\Delta V(t)||_2.
\end{eqnarray*}
By (\ref{48}) we are led to
\begin{eqnarray*}
||D\Delta V(t)||_{H^1}\lesssim||\Delta^2V(t)||_2+(\sqrt{\eta_8}+1)\cdot||D\Delta V(t)||_2+\sqrt{\eta_8}\cdot||\Delta V(t)||_2,
\end{eqnarray*}
which implies
\begin{eqnarray}\label{55}
||D\Delta V(t)||_{\frac{16}{3}}&\lesssim&\big[||\Delta^2V(t)||^{\frac{5}{8}}_2+(\eta_8^{\frac{5}{16}}+1)\cdot||D\Delta V(t)||_2^{\frac{5}{8}}\\
&&+\eta_8^{\frac{5}{16}}\cdot||\Delta V(t)||_2^{\frac{5}{8}}\big]\cdot||D\Delta V(t)||_2^{\frac{3}{8}}.\nonumber
\end{eqnarray}
Furthermore, substituting (\ref{53}), (\ref{54}) and (\ref{55}) into (\ref{52}) we arrive at
\begin{eqnarray}\label{5.7}
&&\Big|\int_M\Delta^2V(t)\ast[DV(t)\times D\Delta V(t)]\,dM\Big|\\
&\lesssim&\eta_7\cdot(||\Delta^2V(t)||_2^{\frac{1}{8}}+||D\Delta V(t)||_2^{\frac{1}{8}}+||\Delta V(t)||_2^{\frac{1}{8}}+||DV(t)||_2^{\frac{1}{8}}+||V(t)||_2^{\frac{1}{8}})\nonumber\\
&&\cdot\big[||\Delta^2V(t)||^{\frac{5}{8}}_2+(\eta_8^{\frac{5}{16}}+1)\cdot||D\Delta V(t)||_2^{\frac{5}{8}}+\eta_8^{\frac{5}{16}}\cdot||\Delta V(t)||_2^{\frac{5}{8}}\big]\nonumber\\
&&\cdot||D\Delta V(t)||_2^{\frac{3}{8}}\cdot||\Delta^2V(t)||_2\cdot||DV(t)||_2^{\frac{7}{8}}.\nonumber
\end{eqnarray}
Substituting the upper bounds of $||V(t)||_2$, $||DV(t)||_2$ and $||\Delta V(t)||_2$ into (\ref{5.7}) leads to
\begin{eqnarray*}
&&\Big|\int_M\Delta^2V(t)\ast[DV(t)\times D\Delta V(t)]\,dM\Big|\\
&\leqslant&\eta_9\cdot(||\Delta^2V(t)||_2^{\frac{1}{8}}+||D\Delta V(t)||_2^{\frac{1}{8}}+1)\\
&&\cdot(||\Delta^2V(t)||_2^{\frac{5}{8}}+||D\Delta V(t)||_2^{\frac{5}{8}}+1)\cdot||D\Delta V(t)||_2^{\frac{3}{8}}\cdot||\Delta^2V(t)||_2\\
&\lesssim&\eta_9\cdot(||\Delta^2V(t)||_2^{\frac{3}{4}}+||D\Delta V(t)||_2^{\frac{3}{4}}+1)\cdot||D\Delta V(t)||_2^{\frac{3}{8}}\cdot||\Delta^2V(t)||_2\\
&\leqslant&\eta_9\cdot||\Delta^2V(t)||_2^{\frac{7}{4}}\cdot||D\Delta V(t)||_2^{\frac{3}{8}}+\eta_9\cdot||\Delta^2V(t)||_2\cdot||D\Delta V(t)||_2^{\frac{9}{8}}\\
&&+\eta_9\cdot||\Delta^2V(t)||_2\cdot||D\Delta V(t)||_2^{\frac{3}{8}}\\
&\lesssim&\eta_9\cdot\Big(\varepsilon\cdot||\Delta^2V(t)||_2^2+\frac{1}{\varepsilon}\cdot||D\Delta V(t)||_2^3\Big)+\eta_9\cdot\Big(\varepsilon\cdot||\Delta^2V(t)||_2^2+\frac{1}{\varepsilon}\cdot||D\Delta V(t)||_2^{\frac{9}{4}}\Big)\\
&&+\eta_9\cdot\Big(\varepsilon\cdot||\Delta^2V(t)||_2^2+\frac{1}{\varepsilon}\cdot||D\Delta V(t)||_2^{\frac{3}{4}}\Big)\\
&\lesssim&\eta_9\cdot\Big(\varepsilon\cdot||\Delta^2V(t)||_2^2+\frac{1}{\varepsilon}\cdot||D\Delta V(t)||_2^4+\frac{1}{\varepsilon}\Big),
\end{eqnarray*}
where $\eta_9$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$, $||\Delta V_0||_2$ and $t$.

Moreover, there is a universal constant $\kappa_1$ such that
\begin{eqnarray*}
&&\Big|\int_M\langle\Delta\big[|V(t)|^2\cdot V(t)\big],\Delta^2 V(t)\rangle\,dM\Big|\\
&\leqslant&\frac{1}{4}||\Delta^2V(t)||_2^2+\kappa_1\cdot||V(t)||^4_{\infty}\cdot||\Delta V(t)||_2^2+\kappa_1\cdot||V(t)||^2_{\infty}\cdot||DV(t)||_4^4.
\end{eqnarray*}
Recalling (\ref{5.1}) and (\ref{56}) we obtain
\begin{eqnarray}\label{5.8}
&&\Big|\int_M\langle\Delta\big[|V(t)|^2\cdot V(t)\big],\Delta^2 V(t)\rangle\,dM\Big|\leqslant\frac{1}{4}||\Delta^2V(t)||_2^2+\kappa_1\cdot||V_0||^4_{\infty}\cdot\gamma_5\\
&&+\kappa_1\cdot||V_0||^2_{\infty}\cdot\eta_4\cdot(||D\Delta V(t)||_2+||\Delta V(t)||_2+||DV(t)||_2+||V(t)||_2)\cdot||DV(t)||_2^3.\nonumber
\end{eqnarray}
Substituting the upper bounds of $||\Delta V(t)||_2$, $||DV(t)||_2$ and $||V(t)||_2$ into (\ref{5.8}) gives
\begin{eqnarray*}
&&\Big|\int_M\langle\Delta\big[|V(t)|^2\cdot V(t)\big],\Delta^2 V(t)\rangle\,dM\Big|\\
&\leqslant&\frac{1}{4}||\Delta^2V(t)||_2^2+\kappa_2\cdot(||D\Delta V(t)||_2+1)\\
&\leqslant&\frac{1}{4}||\Delta^2V(t)||_2^2+\frac{\kappa_2}{2}\cdot(||D\Delta V(t)||^2_2+3),
\end{eqnarray*}
where $\kappa_2$ relies on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$, $||\Delta V_0||_2$ and $t$.

In conclusion,
\begin{eqnarray*}
&&\frac{1}{2}\frac{d}{dt}||D\Delta V(t)||_2^2+||\Delta^2V(t)||_2^2+\lambda\cdot||D\Delta V(t)||_2^2\\
&\leqslant&\eta'_9\cdot\Big(\varepsilon\cdot||\Delta^2V(t)||_2^2+\frac{1}{\varepsilon}\cdot||D\Delta V(t)||_2^4+\frac{1}{\varepsilon}\Big)+\frac{1}{4}||\Delta^2V(t)||_2^2+\frac{\kappa_2}{2}\cdot(||D\Delta V(t)||^2_2+3),
\end{eqnarray*}
where $\eta'_9$ depends on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$, $||\Delta V_0||_2$ and $t$. Let $\varepsilon$ be small enough. From Young's inequality it follows that
\begin{eqnarray*}
\frac{1}{2}\frac{d}{dt}||D\Delta V(t)||_2^2+\frac{1}{2}||\Delta^2V(t)||_2^2+\lambda\cdot||D\Delta V(t)||_2^2\leqslant\eta_{10}\cdot(1+||D\Delta V(t)||_2^4),
\end{eqnarray*}
where $\eta_{10}$ is dependent of $\eta_9$ and $\kappa_2$. Since of (\ref{Eq:1}), the generalized Gronwall's inequality implies
$$||D\Delta V(t)||_2^2\leqslant\gamma_6,$$
here $\gamma_6$ relies on $||D\Delta V_0||_2$, $\eta_{10}$, $\gamma_4$, $\gamma_5$ and $t$. Substituting the upper bounds of $||\Delta V(t)||^2_2$ and $||D\Delta V(t)||_2^2$ into (\ref{5.4}) and (\ref{5.5}) yields
$$||DV(t)||_{H^2}\leqslant\gamma_7$$
which implies
$$||DV(t)||_{\infty}\lesssim\gamma_7,$$
where $\gamma_7$ relies on $\mathcal{R}^M$, $\mathcal{R}^E$ and their covariant derivatives, $||V_0||_{\infty}, ||V_0||_2, ||DV_0||_2$, $||\Delta V_0||_2$, $||D\Delta V_0||_2$ and $t$.\\

Now we return to prove Theorem \ref{thm9}. Reviewing (\ref{37}), we know that the key is to estimate
$$\sum\limits_{i+j=k-1}a_{ij}\cdot||D^{k+1}V||_2\cdot\Big|\Big||D^{i+1}V|\cdot|D^{j+1}V|\Big|\Big|_2.$$
H\"older inequality yields
$$\Big|\Big||D^{i+1}V|\cdot|D^{j+1}V|\Big|\Big|_2\leqslant||D^{i+1}V||_{\frac{2k-2}{i}}\cdot||D^{j+1}V||_{\frac{2k-2}{k-1-i}}.$$
From Theorem \ref{thm6}, it follows that
$$||D^{i+1}V||_{\frac{2k-2}{i}}\lesssim||D^kV||_2^{\frac{i}{k-1}}\cdot||DV||_{\infty}^{\frac{k-1-i}{k-1}}$$
and
$$||D^{j+1}V||_{\frac{2k-2}{k-1-i}}\lesssim||D^kV||_2^{\frac{k-1-i}{k-1}}\cdot||DV||^{\frac{i}{k-1}}_{\infty}.$$
So one can get
\begin{eqnarray}\label{5.9}
\Big|\Big||D^{i+1}V|\cdot|D^{j+1}V|\Big|\Big|_2\lesssim||D^kV||_2\cdot||DV||_{\infty}.
\end{eqnarray}
Substituting (\ref{5.9}), (\ref{39}), (\ref{40}) and (\ref{41}) into (\ref{37}) we obtain
\begin{eqnarray*}
&&\frac{1}{2}\frac{d}{dt}||D^kV||_2^2+||D^{k+1}V||_2^2\\
&\leqslant&\kappa_3\cdot||D^{k+1}V||_2\cdot||D^kV||_2\cdot||DV||_{\infty}+G_k\cdot(1+||V_0||_{\infty}+||V_0||^2_{\infty})\cdot||V||^2_{H^k}\\
&\leqslant&\frac{1}{2}\cdot||D^{k+1}V||_2^2+\frac{\kappa_3^2}{2}\cdot||D^kV||_2^2\cdot||DV||^2_{\infty}+G_k\cdot(1+||V_0||_{\infty}+||V_0||^2_{\infty})\cdot||V||^2_{H^k},
\end{eqnarray*}
where $\kappa_3$ is a universal constant. Note the fact
$$||DV(t)||_{\infty}\lesssim\gamma_7\,\,\,\,\s\,\,\,\,\s\forall\,\,\,\, t\in[0,T^*).$$
Summing $k$ from 0 to $N$ and applying Gronwall's inequality we are led to
$$
||V(t)||^2_{H^N}+\int_0^t||V(s)||^2_{H^{N+1}}\,ds\leqslant C_N(||V_0||_{H^N},t,\gamma_7,||V_0||_{\infty},\kappa_3,G_0,\cdots,G_N).
$$
The remaining part of the proof of Theorem \ref{thm9} is as the same as that of Theorem \ref{thm5}. So we omit it. This completes the proof.$\hfill\Box$\\

{

redpajama_set_name: RedPajamaArXiv


text:
Jason Day withdraws from Arnold Palmer Invitational
Written By Luke Sheehan
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The Australian 26-year-old picked up the injury on his way to winning the World Match Play Championship in February, and was sidelined from the WGC-Cadillac Championship at Doral.
Day is not the only top-10 player to bypass the PGA Tour event at the Bay Hill Club and Lodge in Orlando, Florida, with each of Phil Mickelson, Rory McIlroy, Dustin Johnson and Sergio Garcia also absent.
The Queenslander's withdrawal means he will have been without a strokeplay event for over two months heading to Augusta, a place Day has finished second and third at in his past three visits.
Top-three Tiger Woods, Adam Scott and Henrik Stenson will be teeing off, though, with the former still carrying his mysterious back problem.
Woods is expected to compete this week at an event he has won eight times and is the two-time defending champion of, before an 18-day hiatus to rest up for the Masters at Augusta.
Scott, the Masters champion, has performed consistently in his few appearances on the tour this year, with the Australian producing three finishes inside the top-12, while he was tied 25th at his most recent outing at Doral.
Sweden's Stenson bounced back from missing the cut at the Honda Classic in February, to finish tied for 16th at Doral in a return to form for the world number three.
Champion in 2011, Scotland's Martin Laird will be in action, having had a month between tournaments since finishing tied for 65th at the Northern Trust Open.
Other former winners in the field include Vijay Singh (2007) and Rod Pampling (2006).
In-form players set to contend at Bay Hill include Patrick Reed, Jamie Donaldson and Bubba Watson, with that group occupying the top-three placings at Doral, while John Senden will look to back up his drought-breaking title at the Valspar Championship as the Australian tees off on Thursday among the 120-strong field.
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Running in boots? You're my hero! That's awesome…no one has any excuse not to get a workout in! I did squats, calf raises and wall push ups in the airplane bathroom yesterday, haha. Not the best situation but it was better than nothing!
Something is always better than nothing, right? I want there to be small gyms in airports so I can get my workout on while I wait to fly! How can we make THAT happen?!
Oh, it's my dream to do this. I think the problem the gyms have is that they can't afford the rent in the airports (space is limited) and the ones on this updated site require that you leave the airport and take a shuttle: http://www.airportgyms.com/airport-gyms.php So totally not ideal when you just want to get a quick run in. Any investors out there want to go in on this with me? Anyone? Bueller?
#TBOMGBecky Still love this workout.


Join hundreds of party people Friday Night for the kick off to this year's Lobsterfest. The "crawl" will begin immediately after sunset and continue until Midnight!! Participants will receive an official Lobsterfest '19 T-Shirt*, souvenir Corona bottle opener/lanyard and fantastic drink specials at all participating bars. The cost is just $20 and a portion of the proceeds benefit the Key West High School's Scholarship Fund.
*T-Shirts now available in sizes Small thru XXL.
Pick up your T-Shirt, Lanyard and instructions from the Rick's Ship to Shore Store. 202 Duval Street. Packages are available to pick up August 8th & 9th noon to 7:30pm.
Can't Make it to Lobsterfest?


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Microfeathering: Everything You Need to Know About the New Brow-Filling Technique
Join Allure Beauty BoxFLASH SALE! Sign up today and get over 40% off your first box with code BEAUTYBFF at checkout!JOIN NOW
Brows of a Feather
By Devon Abelman
Now that everyone and their mother knows what microblading is, a more minimalist take on the brow trend is becoming popular. (I've lost count of the amount of times even the most random people in my life have asked me if they should get their brows microbladed.) Microfeathering is like the no-makeup makeup version of the semipermanent brow tattoos, and celebrities like Mandy Moore are visiting Kristie Streicher at STRIIIKE, a beauty studio based in Beverly Hills, Cali., to get it done.
The biggest difference between microblading and microfeathering is surface area. Microblading usually involves tattooing over your whole brow area to create fullness and shape. People with alopecia or super-thin brows often get it done so they don't have to constantly draw on their brows. Microfeathering, on the other hand, is for those with naturally full brows. The name comes from the look Streicher created called the Feathered Brow. (Not to be confused with feather brows, of course.) She helps her clients have the most lush, bold but natural-looking brows in the game. For six to eight weeks, they must avoid all forms of hair removal on their brow area, including tweezing, waxing, and threading. Then, Streicher performs her magic. If there are any naturally sparse areas, that's where microfeathering comes in.
"My main goal with this technique is to really mimic the natural hair and its growth patterns for the perfect Feathered Brow, which is soft, natural and beautiful," Streicher tells Allure. "Microfeathering is typically used on areas of the brow that are surrounded with natural hair. Gaps and sparsely covered areas can be filled in, resulting in the most natural look possible." Once microfeathering is done, Streicher says you shouldn't be able to tell the difference between real eyebrow hairs and microfeather strokes. The technique is that seamless. Plus, the ink used is closely matched to the eyebrows's true color, not to the client's actual hair color.
Amy Jean, a brow expert, uses a similar technique at her salons in Australia. She calls it the "feather touch brow tattoo" and involves "less strokes, less color intensity and finer blades," she tells Allure. Its popularity is rising, as of late, due to health concerns. "The cosmetic tattoo industry is seeing a quick shift away from typical microblading due to novice operators literally scarring clients' skin from cutting too deep into the brow line," Amy Jean explains. "People are turning to more experienced technicians who know how to maneuver the blades along the skin in a more delicate fashion."And she's really not kidding. Earlier this month, we reported on a woman whose brows literally fell off after getting an infection from microblading.
As for what to expect when getting your brows microfeathered, the technician creates "precise 'hair-like' incisions that deposit dye in the top layer of the brow skin," Streicher explains. If your first thought was "ouch," don't worry, the area is numbed first with a topical cream. Also, it doesn't use needles or a tattoo gun. Instead, Streicher uses a fine, precise blade. "Only state of the art pigments are used, which makes the fade out process merely invisible," she adds. "The hair-like strokes created by microfeathering gradually fades over the course of a year, leaving no residual pigment behind in the skin."
You can expect to shell out about $1,000 for microfeathering and spend about an hour and a half getting it done, Streicher says. She typically completes the whole process in two appointments. The second usually is scheduled for six to eight weeks after the first one. "During this appointment, the healing process is evaluated and additional hair strokes will be added and perfected," Streicher says. Once you're done, all you have to worry about for eight months to a year is plucking away strays. For anyone who's into that low-maintenance, bushy brow look, microfeathering seems like a pretty awesome option, if you ask me. As someone who never considered themselves to be a microblading candidate because I have naturally full brows, I'd consider this technique to fill in the pesky sparse spots at the front of them that I fill in every day.
Brow Exfoliation Is the Part of Your Skin-Care Routine You're Forgetting
This Woman's Eyebrows Fell Off After She Got Microblading
I Tried Brow Wigs After Trichotillomania Led Me to Pull Out My Eyebrows
Now, learn about 100 years of brow history:
Follow Devon Abelman on Twitter and Instagram.
Devon Abelman is a former digital beauty editor for Allure. A reporter with nearly a decade of experience covering the intricacies and intersectionality of beauty, culture, and K-pop, Devon previously served as an editor at FamilyFun and held roles at Parents and Country Living. In 2014, she graduated from... Read more
Former Beauty Editor, Allure.com
Keywordsmicrobladingeyebrows


text:
Foreign Currency Deposit Accounts
Personal Foreign Currency Accounts
An individual including a minor who is a Sri Lankan national.
An individual of Sri Lankan origin including a minor who is a resident outside Sri Lanka.
A non-national resident in Sri Lanka.
A non-national either on temporary visit to Sri Lanka or intending to visit Sri Lanka.
An administrator or executor of the estates of a deceased person, who maintained
A Personal Foreign Currency Account (PFCA) with that authorized dealer or restricted dealer until the completion of the administration of the deceased person's estate.
Accounts may be opened in the form of a Savings Account , Fixed Deposit Account or current Accounts (Without cheque drawing facility) in any following Designated Foreign Currency.
Sterling Pounds
Japanese Yens
Swedish Kroners
Danish Kroners
Norwegian Kroners
Chinese Yuan (Renminbi)
Required documents and information
The following information and documents shall be obtained prior to opening a PFCA, as applicable;
Information / Documents
i An individual including a Minor who is a Sri Lankan national.
Full name, nationality, address and contact details.
A copy of National Identity Card or Passport (photo and detailed page).
A photocopy of Birth Certificate, if a Minor.
Account Opening Application.
FATCA Declaration.
Any other documents stipulated by the Bank from time to time.
ii An individual of Sri Lankan origin including a minor who is a resident outside Sri Lanka
Full name, nationality, foreign address and contact details.
A copy of the Passport (photo and detailed page).
Date of Departure from Sri Lanka.
Citizenship or nature of visa granted by a foreign Government.
iii A non-national resident in Sri Lanka
Address of workplace in Sri Lanka and Designation, if any.
A copy of the employment contract and salary particulars.
A copy of the Passport (photo and detailed page) and visa granted by the Dept. of Immigration & Emigration.
iv A non-national either on temporary visit to Sri Lanka or intending to visit Sri Lanka
v An Administrator or Executor of the estates of a deceased person, who maintained a Personal Foreign Currency Account (PFCA)
A copy of Letter of Administration or Probate issued by the Courts.
Debit Card issued free of charge
Credit Card facilities are available against the funds in the Accounts.
Monthly Interest.
Inward Remittances credited to Accounts free of charge
Outward Remittances
Instant loans up to 90% of the FC A/C balances.
Internet banking, Mobile banking and Voice banking facilities
Ability to transfer funds within the Bank or different Banks.
Free insurance cover * (Minimum balance of USD 250/- should be maintained for 6 months continuously prior to the incident )
Accident insurance cover up to Rs. 1,000,000/- for death /permanent disability
Medical cover up to Rs.100,000/-
Free baggage insurance cover up to Rs. 50,000/-
* Applicable according to the eligibility and the a/c holder and and subject to other Terms and Conditions.
Permitted Credits
Remittances in foreign exchange received from outside Sri Lanka in favor of the account holder through the banking system.
Unutilized foreign currency obtained by the account holder by debiting the Personal Foreign Currency Account for travel purpose as permitted by directions issued by the Central Bank.
Foreign currency brought into Sri Lanka by the account holder upon declaration [i.e., where such foreign currency notes exceeds or equals to USD 15,000 or its equivalent in other foreign currency on a declaration to Customs Department or where such foreign currency notes are less than USD 15,000 or its equivalent in other foreign currency, on an appropriate declaration to the authorized dealer or restricted dealer].
Transfers from Personal Foreign Currency Accounts (PFCAs), Business Foreign Currency Accounts (BFCAs) or accounts maintained in the Offshore Banking Unit, irrespective of the account holder.
Where the account holder is a resident outside Sri Lanka or a non-national resident in Sri Lanka, transfers from an Inward Investment Account (IIA) of the same account holder.
Transfers from Capital Transactions Rupee Account (CTRA) of the same account holder as permitted by directions issued by the Central Bank.
Transfers from Diplomatic Foreign Currency Accounts (DFAs) of any other person.
Capital, capital gains and other receipts arising from investments made in Sri Lanka in foreign currency, if the original transfer had been made by debiting the Personal Foreign Currency Account (PFCA) [or from the Non-Resident Foreign Currency Account (NRFC), Resident Foreign Currency Account (RFC) or Resident Non-National Foreign Currency Account (RNNFC) of the account holder, prior to this order come into effect].
Capital and capital gains received from foreign investments into an Outward Investment Account (OIA), if the original investment had been made by debiting the Non-Resident Foreign Currency Account (NRFC), Resident Foreign Currency Account (RFC) or Resident Non-National Foreign Currency Account (RNNFC) of the account holder, prior to this order come into effect.
Unutilized balance remaining in the Foreign Travel Card of the same account holder, up to the amount the Foreign Travel Card had been loaded from the Personal Foreign Currency Account (PFCA).
Where the account holder is a non-national employee resident in Sri Lanka, transfers from employers' Sri Lanka Rupee account as monthly salary, employment benefits and other related benefits.
Where the account holder is a non-national employee resident in Sri Lanka, amounts in Sri Lanka Rupees authorized by the Central Bank for remittance abroad, converted at the rate of exchange obtained on the day of credit.
Transactions or Transfers from any other category of accounts that may be permitted to maintain by the Central Bank from time to time, to the extent such transactions or transfers have been permitted by directions issued by the Central Bank.
Remittances originating outside Sri Lanka from the account holder in terms of section 8(3) of the Act
Interest earned in foreign currency on the funds held in the account.
Permitted Debits
Any outward remittances made outside Sri Lanka.
Disbursements in Sri Lanka in Sri Lanka Rupees.
Transfers to Personal Foreign Currency Accounts or accounts maintained in the Offshore Banking Units, irrespective of the account holder.
Where the account holder is a resident outside Sri Lanka or a non-national resident in Sri Lanka transfers to an Inward Investment Account (IIA) of the same account holder.
Transfer of funds for uploading a Foreign Travel Card.
Transfers in respect of any investment permitted to be made in Sri Lanka in foreign currency utilizing funds in the Personal Foreign Currency Account (PFCA).
Withdrawal in foreign currency notes up to USD 10,000 (or equivalent in any other foreign currency) for travel purpose as permitted by directions issued by the Central Bank.
Where the account holder is a non-national resident outside Sri Lanka who is on temporary visit to Sri Lanka, withdrawals in foreign currency.
Where the account holder is a non-national employee resident in Sri Lanka, transfers to Business Foreign Currency Accounts (BFCAs) of the employer in respect of over payment of salaries and any other payment due to such employer.
Transfers to employers' Sri Lanka Rupee account in the event of an over payment to a non-national employee resident in Sri Lanka, as per the sub paragraph (k) under the "Permitted Credits".
Payments to residents in foreign exchange in respect of transactions or categories of transactions, to the extent such transactions have been permitted by directions issued by the Central Bank.
Transactions or Transfers to any other category of accounts that may be permitted to maintain by the Central Bank from time to time, to the extent such transactions or transfers have been permitted by directions issued by the Central Bank.
Transfer of funds to a Foreign Currency Loan Account (FCLA) or an LKR Loan Account (LLA) of the account holder.
For more information please visit the nearest branch of People's Bank or contact us
Manager - Foreign Currency Department , People's Bank,
Overseas Customer Services
Contact No :+94 112332745
General Nos : 0112332746/2334278/2437621/2446409
email :nrfc@peoplesbank.lk
Special benefits:


Currently working with two small boutique recruiting firms, The LRJ Group, specializing in Life Sciences and Ascend IT staffing, specializing in placing IT professionals. As a recruiter, I see thousands of resumes every year. I read just as many, if not more, of your online profiles on sites like LinkedIn, Indeed, Dice and others. We live in the digital age, where we define ourselves by what we post online on our social media accounts. Defining ourselves in our resume is the topic of today's discussion.
That "the average time spent by recruitment managers for analyzing a resume is 5 to 7 seconds?" by Mr. Oompfh, Interesting Facts about Interviews, Job-seeking and Resumes!, from 7/1/2015. Even if they spent 30 seconds, is my resume formatted in a way that is informative and enticing to the reader?
Is there a particular length a resume should be? The best answer I have heard so far, I found in an article by Kim Isaacs, Monster Resume Expert.
I know, a "little vague" but brilliant. After all, is that not the goal of our resume?
Mainly, I see two different types of resumes. The "cut and dry" resume, telling you nothing at all. And, the "verbose" resume, telling your entire life's story from childhood, to present day, without skipping a beat. On a personal level, I favor a balance between the two. One of the many articles I have read suggested writing the verbose and then, as I would say "surgically" editing it! Yes, we need the basics, where you worked and the "from-until" dates. Go beyond that and, in a few brief bullet points, describe your position, in short, descriptive terms.
Did you know that having an unprofessional email address or photo on your resume increase the chances that it will be passed over without further thought! Take 10 seconds, go to Google or Yahoo and create a new "personalized and professional" email address!
Yes, you can keep your funny, witty or flirtatious emails for your personal emails (although, you should know that 93% of the recruiters say that they are likely to look at a candidate's social media behavior). I can't tell you how many people use a business email address, from their previous employer, on their resume! That email I just sent to you with new and exciting job possibilities, just went to your old boss and, might I add, he is not the one looking for a new job!
This should not be your resumes final destination!
The simple answer is, your resume should be updated whenever information that is pertinent to your getting your dream job changes. If you have changed your phone number, your email addresses, if you have moved, started a new job, and well you get the idea.
Use spell check, grammar check, proofread, let someone else proofread!
I use a chrome app called Grammarly to proof all of my documents. You can start with the free version and upgrade if you like. But, the free version will help you immensely. I have the app running in the background as I work during the day. It is amazing how many spelling and punctuation errors I find in resumes every day. Being the gracious sort that I am, will correct any spacing, spelling or punctuation errors I find before I submit your resume to the hiring managers.
Most hiring managers are not so kind, and he only "position" your resume will find will be crumpled up in the trash or the shape of a paper airplane on the floor (if they miss they basket)! The free version of Grammarly does not correct grammatical errors, but it could make the difference.
No, this is not a paid advertisement.
The bottom line is that we are not all writers. There are hundreds of articles on how to write the perfect resume. Google the topic and you will find hundreds, if not thousands of articles on the subject. At the end of the day, the only thing that matters is getting your dream job. So do your research and if you are feeling a bit overwhelmed, admit to yourself that resume writing is not your strong suite!
Resume writers are out there, and it is their job to make you look good!
Tweak your social media presence.
Here is a little tip that could help out as well, make sure your social media sites and digital resumes match the written resumes you present to the hiring manager. I have lost candidates, that were qualified for positions because their LinkedIn accounts said one thing and their resumes another!
In the end, there are companies who sole purpose is to make you succeed. It's that simple, it is!
Remember if you are going to leave a digital imprint, make sure it's a lasting one!
Looking for a new career? Be sure to check out our Open Positions!


text:
Q: How to convert Iron Python dictionary to Python dictionary I need to convert a received IronPython dictionary variable to a regular Python dictionary. From what I understand the variable is so-called:
'System.Collections.Generic.Dictionary`2[System.String,System.String]'

If I use:
for each in IronDictionary:
    print type(each)
    print each

I am getting:
type: '<class 'System.Collections.Generic.0, Culture=neutral, PublicKeyToken=b77a5c561934e089]]'>'
'[MultiProcess, False]' # this line is a result of `print each` command

So in an essence each is just like a regular Python list...
It would be great if I could convert this entire Iron dictionary to a regular Python one. If it is not possible, I wouldn't mind to iterate using for each in IronDict: and converting every Iron each into a regular Python list or a Python string maybe?... How to accomplish this correctly?

A: I can only suspect what you are asking for.
Assuming your dictionary uses string as key and value, if you would like to create python dictionary from clr one, try:
pythondict = dict(clrdict)

And reverse direction:
from System.Collections.Generic import Dictionary
clrdict = Dictionary[str,str](pythondict)

You may also just try to use it as-is without converting.
Can you show why do you need a conversion?
EDIT:
This is working example:
from System.Collections.Generic import Dictionary
clrdict = Dictionary[str,str]()
clrdict.Add('k1','v1')
clrdict.Add('k2','v2')
pythondict=dict(clrdict)

print clrdict
print type(clrdict)
print pythondict
print type(pythondict)

Which produces:
Dictionary[str, str]({'k1' : 'v1', 'k2' : 'v2'})
<type 'Dictionary[str, str]'>
{'k2': 'v2', 'k1': 'v1'}
<type 'dict'>

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