在本节中,我们将给出系统光滑解的全局存在唯一性(1.1)。根据经典双曲线方法,存在一个有限时间\(T_{0}\)这样系统(1.1)区间内局部位置是否良好\([0,T_{0}]\)在里面\(H^{s}\)具有\(s>2\)因此,在区间内建立先验估计就足够了\([0,T]\)对于给定的\(T>T_{0}\).我们将证明定理1.1采用精细的非线性能量法。此外,在本文中,我们使用C来表示可能因线而异的正常数。
2.1
\(L^{2}\)估计\((u,B)\)
引理2.1
假设定理中的条件1.1持有,那么它就支持了
$$\bigl\Vert u(t)\bigr\Vert_{L^{2}}^{2{+\bigl\ Vert B(t)\ bigr\Vert_{L_{2}{2}+2\int_{0}^{t}\bigl Vert\textit{\pounds}^{\frac{1}{2{}u(\tau)\biger\Vert_L^{2]}^{2\,d\tau=\Vert u{0}\Vert_{L^{2}}^{2{+\VertB_{0}\Vert_{L^}2}^{2}$$
(2.1)
证明
获取的内部产品(1.1)1具有u个和(1.1)2具有B类,添加结果并使用以下取消标识:
$$\int_{R^{2}}(B\cdot\nabla)B\cdot u\,dx+\int_[R^{2]}(B\cdot\nabla)u\cdot B\,dx=0$$
(2.2)
我们有
$$\frac{1}{2}\frac}{d}{dt}\bigl(\bigl\Vertu(t)\bigr\Vert_{L^{2}}^{2{+\bigl\ VertB(t)\ bigr\Vert_{L ^{2neneneep}^{2}\bigr)+\bigle\Vert\text{\pounds}^{\frac#1}{2{}}u\bigr\ Vert_L^{2{2}}^2}=0$$
(2.3)
多亏了普朗彻定理
$$\begin{aligned}\int _{R^{2}}\text{\bounds}u\cdot u\,dx=&\int _{R^{2}}\widehat{\text{\bounds}u(\xi)}\widehat{u(\xi)}\,d\xi=\int _{R^{2}}}\vert\xi\vert^{2} 克(xi)\bigl\vert\widehat{u(\xi)}\bigr\vert^{2}\,d\xi\end{aligned}$$
(2.4)
$$\begin{aligned}=&\int_{R^{2}}\bigl\vert\widehat{\text{\pounds}^{\frac{1}{2}u(\xi)}\bigr\vert^{2{\,d\xi=\bigl\ vert\text{\ponds}^}\frac}1}{2\}u\bigr\ vert_2}},\end{alinged}$$
(2.5)
然后对上述不等式进行积分,得到
$$\bigl\Vert u(t)\bigr\Vert_{L^{2}}^{2{+\bigl\ Vert B(t)\ bigr\Vert_{L_{2}{2}+2\int_{0}^{t}\bigl Vert\text{\pounds}^{\frac{1}{2{}u(\tau)\biger\Vert_{L^2}}}^2},d\tau=\Vert u{0}\垂直_{L^{2}}^{2{+\垂直B_{0}\垂直_(L^{2)}^{2}$$
(2.6)
□
2.2
\(H^{1}\)估计\((u,B)\)
引理2.2
假设定理中的条件1.1持有,那么存在一个正常数 C类 依赖于 T型,\(u{0}\) 和 \(B_{0}\),这样的话
$$\bigl\Vert\omega(t)\bigr\Vert_{L^{2}}^{2{+\bigl \Vert j(t)\ bigr\Vert_{L_{2}{2}+\int_{0}^{t}\bigl\ Vert\textit{\pounds}^{\frac{1}{2{}\omega\bigr\ Vert_{L2}^{2\,d\tau\leq C(t,u{0},B_{0})$$
(2.7)
证明
首先,我们通过应用∇×到MHD方程(1.1),其中涡度\(ω=nabla\times u=\partial_{x{1}}u_{2}-\部分{x{2}}u{1}\)和电流\(j=nabla\times B=partial_{x_{1}}B_{2}-\部分{x{2}}B_{1}\),对应的涡度方程:
$$\begin{aligned}\partial _{t}\omega+(u\cdot\nabla)\omega+\nu\text{\bounds}\omega=(B\cdot\nabla)j,\end{aligned}$$
(2.8)
$$\begin{aligned}\partial _{t}j+(u\cdot\nabla)j=(B\cdot\nabla\omega)+t(\nabla u,\nabla B),\end{alinged}$$
(2.9)
哪里\(T(nabla u,nabla B)=2\partial _{x}B_{1}.
取中第一个方程的内积(2.8)与ω,中的第二个方程(2.8)与j个将它们相加,并使用不可压缩条件和以下事实
$$\int_{R^{2}}(B\cdot\nablaj)\omega\,dx+\int_[R^{2]}(B\cdot\nabla\omega)j\,dx=0$$
(2.10)
那么我们很容易获得
$$\frac{1}{2}\frac}{d}{dt}\bigl(\Vert\omega\Vert_{L{^{2}}}^{2{+\bigl\Vertj(t)\bigr\Vert_{L^{2neneneep}^{2}\bigr)+\bigr\ Vert\text{\pounds}^{1}{2}{}\omega\bigr\Vert_{L^}}}^{2{2}=\ int_{R^{2}}t(nabla u,nabla B)j\,dx$$
(2.11)
对于任意函数
$$\begin{aligned}\Vert\nabla u\Vert_{L^{2}}^{2{=&C\bigl\Vert\xi\widehat{u(\xi)}\bigr\Vert_2}^{2}\\=&C\int_{R^{2{}\Vert\xi\Vert^2}\bigl\ Vert\widehat{u(\ xi){,d\xi\\=&C\ int_{xi\Vert\leq 1}\Vert\xi\ Vert^{2}\bigl\Vert\widehat{u(\xi})\bigr\Vert^}\,d\xi+C\int_{\vert\xi\vert\geq1}\vert\xi\vert^{2}\bigl\vert\widehat{u(\xi})\bigr\vert_{2}\,d\xi\\=&C\int_}\vert_xi\vert\leq1}\ widehat{u(\ xi}^{2} 米(\xi)\bigl\vert\widehat{u(\xi})\bigr\vert^{2}\,d\xi\\leq&C\int_{R^{2{}\bigl\ vert\wide hat{u(\xi)}\bigr\vert_{2},d\xi+C\int__{vert\xi\vert\geq1}\frac{1}{m(1)}\vert\xi\ vert^{2} 米(\xi)\bigl\vert\widehat{u(\xi})\bigr\vert^{2}\,d\xi\\leq&C_{1}\vert u\vert_{L^{2{}^{2neneneep+C_{2}\bigl\ vert\text{\pounds}^{\frac{1}{2}}u\bigr\vert_L^{2]}^{2],\end{aligned}$$
(2.12)
我们最终获得
$$\Vert\nabla u\Vert_{L^{2}}\leq C_{1}\Vert u\Vert_{L^}}+C_{2}\bigl\Vert\text{\pounds}^{\frac{1}{2}u\bigr\Vert_2}}$$
(2.13)
利用估算\(\|\nabla u\|_{L^{2}}\)英寸(2.13)以及插值不等式,我们得到
$$开始{aligned}和\int_{R^{2}}T(\nabla u,\nablaB)j\,dx\\&\quad\leq C\Vert\nabla u\Vert_{L^{2{}}\Vert\natbla B\Vert_ ^{2}\\&\quad\leq C\Vert\omega\Vert_{L^{2{}}\Vert j\Vert_}L^{2]}\Vert\nabla j\Vert_{L^}}\\&\quad\leq C\Vert\omega\Vert_{L^{2}}\Vert\nabla B\Vert_}L^{2]}\Verd\nabla j\Vert_ r \Vert_{L^{2}}\bigr)\bigl(\Vert j\Vert_\\&\quad\leq\frac{1}{2}\bigl\Vert\text{\pounds}^{\frac}1}{2]}\omega\bigr\Vert_{L^{2}}^{2{+C\bigl(1+\VertB\Vert_{L^}}^2}+bigl\Vert\text{\ponds}^}^{1}}{2{B\bigr\ Vert_{2}{}^{2}\biger)\bigl(\Vert\omega\Vert_{L^{2}}^{2{+\Vertj\Vert_}L^{2]}^{2\biger)。\结束{对齐}$$
(2.14)
因此,结合(2.14)和(2.11)这意味着
$$开始{对齐}和\frac{d}{dt}\bigl \quad\leq C\bigl(1+\Vert B\Vert_{L^{2}}^{2{+\bigl\Vert\text{\pounds}^{\frac{1}{2}{B\bigr\Vert_{L^}}^}{2{bigr)\bigl(\Vert\omega\Vert_{L^{2}}^{2{+\Vert j\Vert_}L^{2]}^{2\biger)。\结束{对齐}$$
(2.15)
Gronwall引理为我们提供了
$$\bigl\Vert\omega(t)\bigr\Vert_{L^{2}}^{2{+\bigl \Vert j(t)\ bigr\Vert_{L_{2}{2}+\int_{0}^{t}\bigl\ Vert\text{\pounds}^{\frac{1}{2{}\omega\bigr\ Vert_{2}^{2},d\tau\leq C(t,u_{0},B_{0})$$
(2.16)
□
2.3
\(L_{t}^{infty}L^{inffy}\)估计u个
引理2.3
假设定理中的条件1.1持有,那么它就支持了
$$\Vertu\Vert_{L_{t}^{\infty}L^{\infty}}\leq C(t,u_{0},B_{0{),\quad\textit{表示[0,t]中的任意}$$
(2.17)
证明
根据经典双曲线理论,系统(1.1)在区间中局部适定\([0,T_{0}]\)在里面\(H^{s}\)具有\(s>2\),我们有估计
$$\Vert u\Vert_{L^{infty}_{T_{0}}H^{s}}+\Vert B\Vert_{L_{T_0}}^{inffy}H_{s}{leq C(u_{0},B_{0{)$$
(2.18)
因此,根据嵌入定理,
$$\Vertu\Vert_{L^{infty}_{T_{0}}L^{infty}}\leq C(u_{0},B_{0{)$$
(2.19)
因此,建立一个先验估计就足够了(2.19)对有效\(t>t_{0}\).我们重写了系统中的第一个方程(1.1)作为
$$\部分_{t} u个+\nu\text{\pounds}u=\nabla\cdot(B\otimes B)-\nabla\ cdot(u\otimesu)-\nab la p$$
(2.20)
根据线性非齐次方程的解(2.20)可以通过显式给定
$$u(t)=K(t)\ast u_{0}+\int_{0}^{t}K(t-\tau)\ast\bigl[\nabla\cdot(B\otimes B)-\nabla \cdot$$
(2.21)
拿\(L^{\infty}\)利用空间变量范数和Young不等式,我们得到
$$\begin{aligned}\bigl\Vert u(t)\bigr\Vert _{L^{infty}}=&\bigl\ Vert K(t)\ ast u_{0}\bigr\ Vert _}+\int_{0{0}^{t}\bigle\Vert\nabla K(t-\tau)\ ast\bigl[(B\otimes B)-(u\otimesu)\ bigr](\tau}\,d\tau\\&{}+\int_{0}^{t}\bigl\Vert-div\bigl(K(t-\tau)\bigr)p\bigr\Vert_{L^{infty}}\,2\tau\end{aligned}$$
(2.22)
$$\开始{aligned}\leq&C\bigl\Vert K(t)\bigr\Vert_{L^{1}}\Vert u_{0}\Vert_{L^}\infty}}+C\int_{0{t}\bigl\ Vert\nabla K(t-\tau)\biger\Vert_2}}\bigl\Vert\bigl[(B\otimes B)\\&{}-(u\otimesB)\bigra](\tau)\bigr\Vert_{L^{2}}\,d\tau+\int_{0}^{t}\bigl\Vert-div\bigl,d\tau\\leq&C\bigl\Vert K(t)\bigr\Vert_{L^{1}}\Vert u_{0}\Vert_{L^}\infty}}+C\int_{0{0}^{t}\bigl\ Vert\nabla K(t-\tau)\bigr\Vert_{L ^{2}}\bigr\ Vert B(\tau bigr\Vert_{L^{4}}\,d\tau\\leq&C\bigl\Vert K(t)\bigr\Vert_{L_{1}}\Vert u_{0}\Vert_{L^}\infty}}+C\bigl\ Vert\nabla K(\tau)\bigr\Vert_{L_{t}^{1}L^{2}}\bigl\Vert u(\tau \Vert_{L^{\infty}}+C(t,u_{0},B_{0{)\\leq&C\bigl(1+t^{3}\bigr)\VertB_{0}\Vert_{L^}\infty}}+C(t,u,u,B_}0}),\end{aligned}$$
(2.23)
我们在哪里使用了不等式\(K(t)\|_{L^{1}}\leq C(1+t^{-1}+t^}),这是引理2.5中证明的算子的性质[14],因此我们得到
$$\Vertu\Vert_{L_{t}^{\infty}L^{\infty}}\leq C(t,u_{0},B_{0{),\quad\text{表示}t>t_{0}$$
(2.24)
组合(2.19)和(2.24),我们有
$$\Vertu\Vert_{L_{t}^{\infty}L^{\infty}}\leq C(t,u_{0},B_{0{),\quad\text{表示[0,t]中的任意}$$
(2.25)
□
2.4
\(升^{2}_{t} L(左)^{infty}\)估计ω
引理2.4
假设定理中的条件1.1持有,那么它就支持了
$$\int_{0}^{t}\bigl\Vert\omega(\tau)\bigr\Vert_{L^{infty}}^{2}\,d\tau\leq C(t,u_{0{,B_{0neneneep)$$
(2.26)
证明
我们写出了(2.8)作为
$$\partial_{t}\omega+\nu\text{\pounds}\omega=(B\cdot\nabla)j-(u\cdot\nabla)\omega=\nabla\cdot(B\otimes j)-\nabla/cdot(u\otimes\omega)$$
(2.27)
根据线性非齐次方程的解(2.27)可以显式地写为
$$\omega=K(t)\ast\omega _{0}+\int _{0}^{t} K(K)(t-\tau)\ast\bigl[\nabla\cdot(B\otimes j)-\nabla \cdot,(u\otimes\omega)\bigr](\tau,)\,d\tau$$
(2.28)
拿\(L^{\infty}\)利用Young不等式,我们得到了空间变量的范数
$$\begin{aligned}\Vert\omega\Vert_{L^{infty}}=&\bigl\Vert K(t)\ast\omega_{0}\bigr\Vert_}+\int_{0{0}^{t}\bigl\ Vert\nabla K(t-\tau)\ast\ bigl[(B\otimes j)-(u\otimesj)\bigr](\tau d\tau\\leq&C\bigl\VertK(t)\bigr\Vert_{L^{2}}\Vert\omega_{0}\Vert_{L^}}+C\int_{0{t}\bigl\Vert\nabla K(t-\tau)\bigr\Vert_{L^{2}}。\结束{对齐}$$
(2.29)
为了获得\(L_{t}^{2} L(左)^{\infty}\)估计ω,我们采取\(L^{2}\)时间变量范数和卷积Young不等式及其估计(2.25)以获得
$$\开始{aligned}\Vert\omega\Vert_{L_{t}^{2}}{L^{infty}}}\leq&C\bigl\Vert K(t)\bigr\Vert_}^{2} L(左)^{2} }\Vert\omega_{0}\Vert_{L^{2}}+C\bigl\Vert\nabla K(t)\bigr\Vert_{L_{t}^{1} L(左)^{2} {\bigl(\Vert B\otimes j\Vert _{L_{t}^{2} L(左)^{2} }+\bigl\Vert(u\otimes j)\bigr\Vert_{L_{t}^{2} L(左)^{2} }\bigr)\\leq&C\bigl\Vert K(t)\bigr\Vert_{L_{t}^{2} L(左)^{2} }\Vert\omega_{0}\Vert_{L^{2}}+C\bigl\Vert\nabla K(t)\bigr\Vert_{L_{t}^{1} L(左)^{2} }\bigl(\Vert B\Vert_{L_{t}^{2} L(左)^{\infty}}\垂直j\垂直_{L_{t}^{\inffy}L^{2}}\\&{}+\Vertu\Vert_{L_}^{4} L(左)^{4} }\垂直j\垂直_{L_{t}^{4} L(左)^{4} }\bigr)\\leq&C(T,u_{0},B_{0{),\end{aligned}$$
(2.30)
其中,由于线性非均匀方程,我们使用了以下已知估计:
$$\int_{0}^{t}\bigl\Vert K(\tau)\bigr\Vert_{L^{2}}^{2{,d\tau\leq C(t,u_{0{,B_{0}),\qquad\int_{0}^{t}\bigle\Vert K(\t au)\ bigr\Vert_{L^}\infty}}\,d\tau\leqC(t、u_{0},B_{0})$$
(2.31)
另一方面,我们还利用了以下事实
$$\垂直u\垂直_{L_{t}^{4} L(左)^{4} }\leq C\Vertu\Vert_{L_{t}^{infty}L^{2}}^{\frac{1}{2}{\Vert\omega\Vert_}L{t}^2}L^{1}}\leqC(t,u_{0},B_{0{)$$
(2.32)
和
$$\开始{aligned}\Vert\omega\Vert_{L_{t}^{4} L(左)^{4} }\leq&C\Vert\omega\Vert_{L_{t}^{infty}L^{2}}^{\frac{1}{2}{\Vert\nabla\omega\Vert_{L{t}^2}}^{\frac{1{2}}\\leq&C\ Vert\omega \Vert_{L{t}^{2{}}{2}}\bigl}\大)\\leq&C(T,u_{0},B_{0{)。\结束{对齐}$$
(2.33)
因此,有人
$$\int_{0}^{t}\bigl\Vert\omega(\tau)\bigr\Vert_{L^{infty}}^{2}\,d\tau\leq C(t,u_{0{,B_{0neneneep)$$
(2.34)
□
2.5对\(int _{0}^{t}\|\nabla j(\tau)\|_{L ^{infty}}\,d \tau)
引理2.5
假设定理中的条件1.1持有,那么它就支持了
$$\begin{aligned}&\bigl\Vert\omega(t)\bigr\Vert_{L^{infty}}\leq C(t,u_{0},B_{0{),\end{alinged}$$
(2.35)
$$\begin{aligned}和\Vert\nabla u\Vert_{L^{infty}}\leq C\Vert\natbla u\ Vert_{L^{2}}^{\frac{1}{2}{\Vert_nabla\omega\Vert_ L^}{\infty{}},\end{alinged}$$
(2.36)
$$\begin{aligned}&\int _{0}^{t}\Vert\nabla u\Vert _{L^{\infty}}^{2}\,d\tau\leq C(t,u_{0},B_{0})。\结束{对齐}$$
(2.37)
证明
由(2.28),我们可以检查一下
$$\omega=K(t)\ast\omega_{0}+\int_{0{^{t} K(K)(t-\tau)\ast\bigl[\nabla\cdot(B\otimes j)-\nabla \cdot,(u\otimes\omega)\bigr](\tau,)\,d\tau$$
(2.38)
接受手术∇然后取\(L^{\infty}\)根据空间变量的范数,并使用Young不等式,我们得到
$$\begin{aligned}和\Vert\nabla\omega\Vert_{L_{\infty}}\\和\quad\leq\bigl\Vert_nabla K \bigr\Vert_{L^{\infty}}\,d\tau\\&\quad\leq C\bigl\Vert\nabla K(t)\bigr\ Vert_{L ^{2}}\Vert\omega_{0}\Vert_{L^{2}}+C\int_{0{t}\bigl\Vert\nabla^{2} K(K)(t-\tau)\bigr\Vert_{L^{1}}\bigl(\bigl\Vert(B\otimesj)(\tau。\结束{对齐}$$
(2.39)
然后,我们通过取\(L^{1}\)时间变量范数与卷积Young不等式的应用
$$\开始{aligned}&\bigl\Vert\nabla\omega(t)\bigr\Vert_{L_{t}^{1} L(左)^{\infty}}\\&\quad\leq C\bigl\Vert\nabla K(t)\bigr\Vert_{L_{t}^{1} L(左)^{2} }\Vert\omega_{0}\Vert_{L^{2}}+C\bigl\Vert\nabla^{2} K(K)(t) \bigr\Vert_{L_{t}^{1} L(左)^{1} }\bigl(\Vert B\otimes j\Vert_{L_{t}^{1}L^{\infty}}+\Vert u\otimes\omega\Vert_{L_}t}^}1}L ^{\infty}{}\bigr)\\&\quad\leq C\bigl\Vert\nabla K(t)\bigr\Vert_ L_{t}^{1} L(左)^{2} }\Vert\omega_{0}\Vert_{L^{2}}+C\bigl\Vert\nabla^{2} K(K)(t) \bigr\版本_{L_{t}^{1} L(左)^{1} }\bigl(\Vert Bj\Vert_{L_{t}^{1} L(左)^{\infty}}+\垂直u\垂直_{L_{t}^{2} L(左)^{\infty}}\Vert\omega\Vert_{L_{t}^{2} L(左)^{\infty}}\bigr)\\&\quad\leq C\bigl\Vert\nabla K(t)\bigr\Vert_{L_{t}^{1} L(左)^{2} }\Vert\omega_{0}\Vert_{L^{2}}+C\bigl\Vert\nabla^{2} K(K)(t) \bigr\Vert_{L_{t}^{1} L(左)^{1} }\\&\qquad\bigl(\Vert B\Vert_{L_{t}^{\infty}L^{\infty}}\Vert j\Vert_}L_{t}^{1} L(左)^{\infty}}+\Vertu\Vert_{L_{t}^{\inffy}L^{2}}^{\压裂{1}{2}{\Vert\omega\Vert_{L_}^{1} L(左)^{\infty}}^{\frac{1}{2}}\Vert\omega\Vert_{L_{t}^{2} L(左)^{\fity}}\bigr)。\结束{对齐}$$
(2.40)
根据之前的估计\(K{L_{t}^{1} L(左)^{2}}\),很容易证明\(K\|{L_{t}^{1}L^{2}}\|j{0}\|{L^{2]})非递减函数是否满足
$$开始{aligned}和\bigl\Vert\nabla K(t)\bigr\Vert_{L_{t}^{1}}L^{2}}\Vert\omega_{0}\Vert_ L^{2]}\leq C(t,u_{0neneneep,B_{0{),\end{aligned}$$
(2.41)
$$\开始{aligned}&\bigl\Vert\nabla^{2} K(K)(t) \bigr\Vert_{{L_{t}^{1}}L^{1{}}\VertB\Vert_}{t}^{infty}L^{inffy}}\leq C(t,u_{0},B_{0{)。\结束{对齐}$$
(2.42)
因此,它由(2.40)那个
$$\开始{aligned}&\int_{0}^{t}\bigl\Vert\nabla\omega(\tau)\bigr\Vert_{L^{infty}}\,d\tau\\&\quad\leq C\bigl\ Vert\nabra K(t)\biger\Vert_{L_{t}^{1} L(左)^{2} }\Vert\omega_{0}\Vert_{L^{2}}+\bigl(C\bigl\Vert\nabla^{2{K(t)\bigr\Vert_{L_{t}^{1} L(左)^{1} }+\Vertu\Vert_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}{\Vert\omega\Vert_}L_{t1}^{2{L^{\infty}}\\&\qquad{}+C\VertB\Vert_{L_}{t}^{\infty}L^{{\inffy}}\bigl\Vert\nabla^{2neneneep K(t)\bigr\Vert_{L_{t}^{1}L^{1{}\bigr)\int_{0}^{t}\bigl\Vert\omega(\tau)\bigr\Vert_{L^{\infty}}\,d\tau。\结束{对齐}$$
(2.43)
将涡度方程两侧乘以\(|\omega|^{p-2}\omega\)和集成\(R^{2}\),我们获得
$$开始{对齐}\frac{1}{p}\frac{d}{dt}\bigl\Vert\omega(t)\bigr\Vert_{L^{p}}^{p{=&\int_{R^{2}}(B\cdot\nabla j)\omega\Vert\omega\Vert^{p-2}\,dx-\nu\int_R^{2{}\text{\pounds}\omega\ Vert\omega \Vert^p-2}\omega,dx\\leq&\Vert B\Vert_{L^{infty}}\Vert\nabla j\Vert_}L^{p}}\Vert\omega\Vert_ L^{p}}^{p-1},\结束{对齐}$$
(2.44)
在这里我们使用了几乎拉普拉斯速度耗散的非负性,那么我们有
$$\begin{aligned}\frac{d}{dt}\bigl\Vert\omega(t)\bigr\Vert_{L^{p}}\leq\Vert B\Vert_}L^{infty}}\Vert\nabla j\Vert_ L^}}。\结束{对齐}$$
(2.45)
随着时间的推移,我们已经
$$\begin{aligned}\bigl\Vert\omega(t)\bigr\Vert_{L^{p}}\leq\bigl\ Vert_omega(0)\biger\Vert_{L^}}+\int_{0}^{t}\bigle\Vert B(\tau)\bigr\Vert_{L^{\infty}}\bigl\Vert\nabla j(\tau)\bigra\Vert_L^{p{}}\,d\tau。\结束{对齐}$$
(2.46)
出租\(p\rightarrow\infty\),这个产量
$$\begin{aligned}\bigl\Vert\omega(t)\bigr\Vert_{L^{infty}}\leq\bigl\ Vert_omega(0)\biger\Vert_}L^{infty}}+\int_{0}^{t}\bigr\ Vert B迪托。\结束{对齐}$$
(2.47)
根据之前的估计,我们得出
$$\begin{aligned}\bigl\Vert\omega(t)\bigr\Vert_{L^{infty}}\leq\bigl\ Vert_omega(0)\birgr\Vert_{L1^{inffy}}+C(t)\int_{0}^{t}\bigle\Vert\nabla j(\tau)\biger\Vert_{L^}\,d\tau。\结束{对齐}$$
(2.48)
现在,假设
$$\开始{对齐}H(t)=H_{1}(t)+H_{2}(t)\int_{0}^{t}\bigl\Vert\omega(\tau)\bigr\Vert_{L^{infty}}\,d\tau,\end{aligned}$$
(2.49)
它是从(2.43)那个
$$\开始{aligned}\int_{0}^{t}\bigl\Vert\nabla\omega(\tau)\bigr\Vert_{L^{infty}}\,d\tau\leq H(t)。\结束{对齐}$$
(2.50)
由于(2.48),我们很容易获得
$$\开始{对齐}\frac{d}{dt}H(t)=&H_{2}(t)\bigl\Vert\omega(t)\bigr\Vert_{L^{infty}}\\leq&H_2}(t)\ biggl bigr\Vert_{L^{\infty}}\,d\tau\biggr)\\leq&H_{2}(t)\bigl(\bigl\Vert\omega(0)\bigr\Vert_{L_{\inffy}}+C(t)H(t)\biger)。\结束{对齐}$$
(2.51)
Gronwall引理为我们提供了
$$\开始{aligned}G(t)\leq C(t,u_{0},B_{0{),结束{aligned}$$
(2.52)
这进一步意味着
$$\begin{aligned}\int_{0}^{t}\bigl\Vert\nabla j(\tau)\bigr\Vert_{L^{infty}}\,d\tau\leq C(t,u_{0{,B_{0})。\结束{对齐}$$
(2.53)
梳理(2.53)和(2.48),我们获得
$$\开始{aligned}\bigl\Vert\omega(t)\bigr\Vert_{L^{infty}}\leq C(t,u_{0},B_{0{)。\结束{对齐}$$
(2.54)
根据插值不等式
$$\begin{aligned}\Vert\nabla u\Vert_{L^{infty}}\leq C\Vert\natbla u\ Vert_{L^{2}}^{压裂{1}{2}}\Vert_nabla\omega\Vert_ L^}{2}{结束{aligned}$$
(2.55)
和(2.16),我们获得
$$\开始{aligned}\int_{0}^{t}\Vert\nabla u\Vert_{L^{infty}}^{2}\,d\tau\leq C(t,u_{0{,B_{0neneneep)。\结束{对齐}$$
(2.56)
□
2.6全球的\(H^{s}(s>2)\)估计\((u,B)\)
引理2.6
假设定理中的条件1.1持有,那么它就支持了
$$\bigl\Vert u(t)\bigr\Vert_{H^{s}}+\bigl\ Vert B(t)\ bigr\Vert_{H_{s}{+\int_{0}^{t}\bigl Vert\textit{\pounds}^{\frac{1}{2}}u(t$$
(2.57)
证明
获取的全局界限\((u,B)\)在里面\(H^{s}(s>2)\),我们申请\(\Lambda^{s}\)具有\(\Lambda=(I-\Delta)^{\frac{1}{2}}\)到原始方程u个和B类,然后使用\(L^{2}\)所得方程的内积\(\兰姆达^{s} u个,\兰姆达^{s} B类)\)得到能量不等式。
$$开始{对齐}和\frac{1}{2}\frac}d}{dt}\bigl{2}\\&\quad=-\int_{R^{2}}\bigl[\Lambda^{s},u\cdot\nabla\bigr]u\cdot \Lambda ^{s{u\,dx+\int_R^{2{}}\bigl[\Lambda^},B\cdot\nabla\bigr]B\cdot\Lambda^{s}u\,dx\\&\qquad{}-\int_{R^{2}}\bigl[\Lambda{s},u\cdot\nabla\bigr]B\cdot \Lambada{s}B\,dx+\int_R^{2{}\bigr[\Lambeda^{s},B\cdote\nabla\sbigr]u\cdot \ Lambda{s}B \\&\quad=I_{1}+I_{2}+I{3}+I{4},结束{对齐}$$
(2.58)
哪里\([m,n]\)是标准的换向器符号,即\([a,b]=ab-ba\)利用以下加藤-庞塞不等式[16].
$$\begin{aligned}\bigl\Vert\bigl[\Lambda^{s},f\bigr]g\bigr\Vert_{L^{p}}\leq C\bigl(\Vert\nabla f\Vert_}L^{infty}}\bigl\Vert_Lambda^{s-1}克\bigr\Vert_{L^{p}}+\Vert g\Vert_}L^{\infty}}\bigl\Vert\Lambda^{s} (f)\bigr\Vert_{L^{p}}\bigr),\quad 1<p<\infty。\结束{对齐}$$
(2.59)
然后,我们估计能量项如下:
$$开始{对齐}和I{1}\leq C\Vert\nabla u\Vert_{L^{infty}}\Vert u\Vert_{H^{s}}^{2}较大),\结束{对齐}$$
(2.60)
$$\开始{对齐}和I_{3}+I_{4}\leq C\bigl(\Vert\nabla u\Vert_{L^{infty}}+\Vert\natbla B\Vert_{L^}\infty{}\bigr)\ bigl。\结束{对齐}$$
(2.61)
最后,我们得到
$$\开始{对齐}和\frac{d}{dt}\bigl(\bigl\Vertu(t)\bigr\Vert_{H^{s}}^{2}+\bigl\ VertB(t)\ bigr\Vert_{H2}}^2}\bigr)+\bigle\Vert\text{\pounds}^{1}{2}u\bigr\ Vert_{H^}}}q C\bigl(\Vert\nabla u\Vert_{L^{\infty}}+\Vert\nabla B\Vert_}^{2} \更大)。\结束{对齐}$$
(2.62)
我们需要估计\(\|\nabla u\|_{L^{infty}}\)和\(\|\omega\|_{L^{infty}}\),根据以下带对数校正的Sobolev外推不等式[17].
$$\begin{aligned}\Vert\nabla u\Vert_{L^{infty}}\leq C\bigl(1+\Vert u\Vert_{L^}(R^{2})}+\Vert\omega\Vert_}{L^\\infty{}}(R ^{2{)}\bigr。\结束{对齐}$$
(2.63)
根据上述估算,我们有
$$\开始{对齐}和\frac{d}{dt}\bigl(\bigl\Vertu(t)\bigr\Vert_{H^{s}}^{2}+\bigl\ VertB(t)\ bigr\Vert_{H2}}^2}\bigr)+\bigle\Vert\text{\pounds}^{1}{2}u\bigr\ Vert_{H^}}}q C\bigl(1+\Vert\omega\Vert_{L^{infty}}+\Vert_nabla B\Vert_{L^}\infty{}\bigr)ln\bigl(e+\Vert-u\ Vert_{H^{s}}^{2}+\Vert B\Vert_{H^{s}}^{2}\bigr)\\&\qquad\bigl(\Vert u\Vert_{H^}s}}^{2{+\Vert B \Vert_2H^{s}}^}2}\bigr)。\结束{对齐}$$
(2.64)
利用log-Gronwall型不等式和估计(2.34)和(2.41),我们获得
$$开始{aligned}\bigl\Vertu(t)\bigr\Vert_{H^{s}}+\bigl\ VertB(t)\ bigr\Vert_{H2^{s{}+\int_{0}^{t}\bigle\Vert\text{\pounds}^{\frac{1}{2}}u(t。\结束{对齐}$$
(2.65)
组合(2.13)我们获得
$$\begin{aligned}\int _{0}^{t}\bigl\Vert u(t)\bigr\Vert _{H^{s+1}}^{2}\,d\tau\leq C(t)。\结束{对齐}$$
(2.66)
根据前面结果中的六步先验估计,我们给出了定理的证明1.1.我们介绍了\(\varrho_{N} (f)\)由提供
$$\开始{aligned}(\varrho_{N} (f))(x)=N^{2}\int_{R^{2{}\eta\bigl(N(x-y)\bigr)f(y)\,dy,\end{aligned}$$
(2.67)
哪里\(C_{0}^{infty}(R^{2})中的0(|x|)满足\(\nint_{R^{2}}\eta(y)\,dy=1\).我们将制度正规化(1.1)如下:
$$\开始{aligned}\textstyle\begin{cases}{\partial_{t}}u^{N}+P\varrho_{N}((\varrho_{N} u个^{N} \cdot\nabla)\varrho_{N} u个^{N} )+\nu J\text{\pounds}{1}\varrho_{N}u^{N}=P\varrho{N}(_{N} B类^{N} \cdot\nabla)\varrho_{N} B类^{N} ,\\{\partial_{t}}B^{N}+\varrho_{N}(\varrho_{N} u个\cdot\nabla)\varrho_{N} B类=\varrho_{N}((\varrho _{N}B^{N}\cdot\nabla)_{N} u个_{0}(x),B^{N}(x,0)=\rho_{N} B_{0}(x),\end{cases}\displaystyle\end{aligned}$$
哪里P(P)表示Leray投影操作符。对于任何固定\(N>0),使用软化子的性质和在证明全局界时获得的先验估计\(t\in(0,\infty)\),
$$\begin{aligned}和\bigl\Vert u^{N}(t)\bigr\Vert_{H^{s}}+\bigl\ Vert B^{N{}(t)\biger\Vert_{H^}s}}+\int_{0}^{t}\bigl \Vert\text{\pounds}^{{{1}{2}u^{N}(\tau)\bigr\Vert_H^{s2}}},d\ tau\leq C(t),结束{对齐}$$
(2.68)
$$\开始{对齐}&\int_{0}^{t}\bigl\Vertu^{N}(\tau)\bigr\Vert_{H^{s+1}}^{2}\,d\tau\leq C(t),\end{aligned}$$
(2.69)
它们在N个,根据标准的Alaogu定理和紧性参数,我们可以提取子序列\((u^{N_{i}},B^{N_2}})\)并传递到极限\(N\rightarrow\infty\)得到极限函数\((u,B)\)确实是这个问题的全球经典解决方案(1.1)。唯一性也很容易获得。这就完成了定理的证明1.1. □