$$+k_{3}\widehat{W}_{1}(\widehat{v'_{1}\partial_{v_{3}}f^{0}}
-\widehat{v_{3}\partial_{v'_{1}}f^{0}})(\eta_{1}+\eta_{k1},\eta_{2}+\eta_{k2},k_{3}t+\eta_{3})$$ $$+(k_{1}\widehat{W}_{2}-k_{2}\widehat{W}_{1})(\widehat{v'_{2}\partial_{v_{1}}f^{0}}
-\widehat{v_{1}\partial_{v'_{2}}f^{0}})(\eta_{1}+\eta_{k1},\eta_{2}+\eta_{k2},k_{3}t+\eta_{3})]$$ $$\leq C\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{1}+\eta_{k1,t-\tau}|}e^{-2\pi\lambda_{0}|\eta_{2}+\eta_{k2,t-\tau}|}e^{-2\pi\lambda'_{0}|\eta_{k1,\tau}|}
e^{-2\pi\lambda'_{0}|\eta_{k2,\tau}|}e^{-2\pi\lambda'_{0}|k_{3}|\tau}\cdot|k_{3}(t-\tau)+\eta_{3}|e^{-2\pi\lambda_{0}|k_{3}(t-\tau)+\eta_{3}|}d\tau$$ $$\leq C\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{1}+\eta_{k1,t-\tau}|}e^{-2\pi\lambda_{0}|\eta_{2}+\eta_{k2,t-\tau}|}e^{-2\pi\lambda'_{0}|\eta_{k1,\tau}|}
e^{-2\pi\lambda'_{0}|\eta_{k2,\tau}|}e^{-2\pi\lambda'_{0}|k_{3}|\tau} e^{-2\pi\frac{(\lambda_{0}+\lambda'_{0})}{2}|k_{3}(t-\tau)+\eta_{3}|}d\tau$$ $$\leq C\bigg(\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{1}+\eta_{k1,t-\tau}|}e^{-2\pi\lambda'_{0}|\eta_{k1,\tau}|}d\tau\bigg)^{\frac{1}{3}}
\bigg(\int^{t}_{0}e^{-2\pi\lambda'_{0}|\eta_{k2,\tau}|}e^{-2\pi\lambda_{0}|\eta_{2}+\eta_{k2,t-\tau}|}d\tau\bigg)^{\frac{1}{3}}
 e^{-2\pi\lambda''_{0}|k_{3}t+\eta_{3}|}$$ $$\begin{aligned}
&=C\bigg(\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{1}+\frac{1}{\Omega}\{k_{2}-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega(t-\tau)+\varphi_{1}]\}|}
e^{-2\pi\lambda'_{0}|\frac{1}{\Omega}\{k_{2}-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega\tau+\varphi_{1}]\}|}d\tau\bigg)^{\frac{1}{3}}\notag\\
&\bigg(\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{2}+\frac{1}{\Omega}[-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega(t-\tau)-\varphi_{2}]-k_{1}]|}
e^{-2\pi\lambda'_{0}|\frac{1}{\Omega}[-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega\tau-\varphi_{2}]-k_{1}]|}d\tau\bigg)^{\frac{1}{3}}
e^{-2\pi\lambda''_{0}|k_{3}t+\eta_{3}|},\notag\\
$$ where $\lambda''_{0}=\lambda'_{0}-\frac{1}{2}(\lambda_{0}-\lambda'_{0}),\tan\varphi_{1}=\frac{k_{2}}{k_{1}},\tan\varphi_{2}=\frac{k_{1}}{k_{2}}.$ $$\begin{aligned}
&VI=-\frac{q}{m}\hat{\rho}(t,k)\ast_{t}\partial_{t}\bigg(\hat{W}_{1}\widehat{\partial_{v'_{1}}f^{0}}+\hat{W}_{2}
\widehat{\partial_{v'_{2}}f^{0}}\bigg)(\eta_{1}+\eta_{k1},\eta_{2}+\eta_{k2},k_{3}t+\eta_{3})\notag\\
&\leq C\bigg(\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{1}+\frac{1}{\Omega}\{k_{2}-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega(t-\tau)+\varphi_{1}]\}|}
e^{-2\pi\lambda'_{0}|\frac{1}{\Omega}\{k_{2}-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega\tau+\varphi_{1}]\}|}d\tau\bigg)^{\frac{1}{3}}\notag\\
&\bigg(\int^{t}_{0}e^{-2\pi\lambda_{0}|\eta_{2}+\frac{1}{\Omega}[-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega(t-\tau)-\varphi_{2}]-k_{1}]|}
e^{-2\pi\lambda'_{0}|\frac{1}{\Omega}[-\sqrt{k^{2}_{1}+k^{2}_{2}}\sin[\Omega\tau-\varphi_{2}]-k_{1}]|}d\tau\bigg)^{\frac{1}{3}}
e^{-2\pi\lambda''_{0}|k_{3}t+\eta_{3}|}.\notag\\
$$ From (1.23)-(1.26) and Corollary 1.2, the results of Theorem 1.1 are obvious.

$$$$ $$$$

Notation and Hybrid analytic norm
=================================

In order to prove our result in the nonlinear case, we have to introduce the hybrid analytic norm that is one of the cornerstones of our analysis, because they will connect well to both estimates in $ x$ on the force field and uniform estimates in $ v. $ First, let us recall that the free transport equation in a constant magnetic field $$\begin{aligned}
\partial_{t}f+v\cdot\nabla_{x}f+\frac{q}{m}(v\times B_{0})\cdot\nabla_{v}f=0\end{aligned}$$ has a strong mixing property: any solution of (1.1) converges weakly in large time to a spatially homogeneous distribution equal to the space-averaging of the initial datum. Let us sketch the proof.

If $ f $ solves (2.1) in $\mathbb{T}^{3}\times\mathbb{R}^{3},$ with initial datum $f_{in}=f(\tau,\cdot),$ then $$\begin{aligned}
f(t,x'_{t},v'_{t})=f_{in}(x'_{\tau}-v'\cdot(\frac{\sin\Omega t}{\Omega},\frac{\sin\Omega t}{\Omega},t),v'_{\tau}),\end{aligned}$$

We give a description on the motion of charged particles in the following way: $$\begin{aligned}
\frac{dx'}{dt}=v',\quad \frac{dv'}{dt}=\frac{q}{m}v'\times B_{0},\end{aligned}$$ where $B_{0}=B_{0}\hat{z}.$

We assume that $x'(t,x,v)=x=(x_{1},x_{2},x_{3}),v'(t,x,v)=v=(v_{x},v_{y},v_{3})=(v_{\perp}\cos\theta,v_{\perp}\sin\theta,v_{3})$ at $t=\tau,$ then the solution of Eq.(2.1) 