DOI:10.12677/AAM.2018.71009,PDF,HTML,XML,被引量下载: 1,779浏览: 3,946
作者:李 良：云南民族大学，数学与计算机科学院，云南 昆明
关键词:薛定谔映射；k等变解；狄里克雷能量；Schrodinger Map Problem；k-Equivariant Solution；Dirichlet Energy

文章引用：李良. 薛定谔映射问题的k等变解的性质[J]. 应用数学进展, 2018, 7(1): 74-79. https://doi.org/10.12677/AAM.2018.71009

1. 引言

薛定谔映射如下：

$\{\begin{array}{l}{\partial }_{t}u=u\wedge \Delta u\\ {u|}_{t=0}=u_0\in H_0^1,\left(t,x\right)\in R\times R^2,u\left(t,x\right)\in S^2\end{array}$(1)

该方程是由Landau和Lifshitz在1935年研究铁磁流体的色散理论时在 [1] 提出来的，用来描述磁化运动。其应用广泛且意义重大，就如NS方程在流体力学中的地位一样，它是铁磁体材料的基本方程。从数学上讲，它是一个强退化的拟线性抛物型方程，与调和热流、薛定谔方程等著名方程都有密切的联系。其精确解在文献 [2] 中已经构造出来。包括方程的爆破问题在文献 [3] [4] 也给出了许多显式的解。方程目前的研究主要是小能量附近的等变爆破解。研究等变解是因为方程本身的结构所决定的，本文也主要研究的是等变解的一些性质，最后根据文献 [5] 求得解的最小的狄里克雷能量。

2. 一些定义

定义1：方程(1)的如下形式的解称为k等变解： $u\left(t,x\right)={\text{e}}^{k\theta R}|\begin{array}{c}{u}_1\left(t,r\right)\\ u_2\left(t,r\right)\\ u_3\left(t,r\right)\end{array},R=\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right),k\in Z$

其中 $\left(r,\theta \right)$是 $R^2$中的极坐标。

定义2：解的拓扑度为：

定义3：解的狄利克雷能量为： $E\left(u\left(t\right)\right)={\displaystyle \underset{R^2}{\int }{\left|\nabla u\left(t,x\right)\right|}^2\text{d}x}$

例如： $V={\text{e}}^{k\theta R}\left(\begin{array}{c}\frac{2r^k}{1+r^{2k}}\\ 0\\ \frac{r^{2k}-1}{r^{2k}+1}\end{array}\right)$就是一个等变解。根据定义就可以验证。

3. 本文的主要结果

定理1：方程的解保持狄里克雷能量不变，即：

$E\left(u\left(t\right)\right)={\displaystyle \underset{R^2}{\int }{\left|\nabla u\left(t,x\right)\right|}^2\text{d}x}=E(\; u\; o\; )$

定理2：等变解V的拓扑度为k.

定理3：等变解V的狄里克雷能量为 $8\left|k\right|\text{π}$

4. 定理的证明

4.1. 定理1的证明

方程两边点乘 $\Delta u$再积分：

$\Rightarrow \Delta u\cdot {\partial }_{t}u=\Delta u\cdot \left(u\wedge \Delta u\right)=0$

$\Rightarrow {\displaystyle \underset{R^2}{\int }\Delta u\cdot {\partial }_{t}u\text{d}x}=-\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle \underset{R^2}{\int }{\left|\nabla u\right|}^2\text{d}x}0=0$

$\Rightarrow {\displaystyle \underset{R^2}{\int }{\left|\nabla u\right|}^2\text{d}x}=C$

$\Rightarrow E\left(u\left(t\right)\right)=E(\; u\; o\; )$

4.2. 定理2的证明

先用数学归纳法可求得 $R^k=\left(\begin{array}{ccc}\cos \frac{k\text{π}}{2}& -\sin \frac{k\text{π}}{2}& 0\\ \sin \frac{k\text{π}}{2}& \cos \frac{k\text{π}}{2}& 0\\ 0& 0& 0\end{array}\right)$

$K=1$时显然成立，若 $K=n\left(n>1\right)$时成立，则

$\begin{array}{c}{R}^{n+1}=\left(\begin{array}{ccc}\cos \frac{n\text{π}}{2}& -\sin \frac{n\text{π}}{2}& 0\\ \sin \frac{n\text{π}}{2}& \cos \frac{n\text{π}}{2}& 0\\ 0& 0& 0\end{array}\right)\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}-\sin \frac{n\text{π}}{2}& -\cos \frac{n\text{π}}{2}& 0\\ \cos \frac{n\text{π}}{2}& -\sin \frac{n\text{π}}{2}& 0\\ 0& 0& 0\end{array}\right)\\ =\left(\begin{array}{ccc}\cos \frac{\left(n+1\right)\text{π}}{2}& -\sin \frac{\left(n+1\right)\text{π}}{2}& 0\\ \sin \frac{\left(n+1\right)\text{π}}{2}& \cos \frac{\left(n+1\right)\text{π}}{2}& 0\\ 0& 0& 0\end{array}\right)\end{array}$

再计算 ${\text{e}}^{k\theta R}$：

${\text{e}}^{k\theta R}={\displaystyle \underset{h=0}{\overset{\infty }{\sum }}\frac{1}{n!}}{\left(k\theta R\right)}^n={\displaystyle \underset{h=0}{\overset{\infty }{\sum }}\frac{1}{n!}}{\left(k\theta \right)}^n\left(\begin{array}{ccc}\cos \frac{n\text{π}}{2}& -\sin \frac{n\text{π}}{2}& 0\\ \sin \frac{n\text{π}}{2}& \cos \frac{n\text{π}}{2}& 0\\ 0& 0& 0\end{array}\right)$

分别计算得： ${\displaystyle \underset{h=0}{\overset{\infty }{\sum }}\frac{1}{n!}}{\left(k\theta \right)}^n\cos \frac{n\text{π}}{2}={\displaystyle \underset{h=0}{\overset{\infty }{\sum }}\frac{1}{\left(2n\right)!}}{\left(k\theta \right)}^{2n}{\left(-1\right)}^n=\cos k\theta$

同理可得： ${\text{e}}^{k\theta R}=\left(\begin{array}{ccc}\cos k\theta & -\sin k\theta & 0\\ \sin k\theta & \cos k\theta & 0\\ 0& 0& 1\end{array}\right)$

记： $A={\text{e}}^{k\theta R}=\left(\begin{array}{ccc}\cos k\theta & -\sin k\theta & 0\\ \sin k\theta & \cos k\theta & 0\\ 0& 0& 1\end{array}\right)$，， $V=AQ$

$\begin{array}{c}\mathrm{deg}V=\frac{1}{4\text{π}}{\displaystyle \underset{R^2}{\int }\left|\begin{array}{ccc}{V}_{x1}& V& V_{x_2}\end{array}\right|\text{d}x}=\frac{1}{4\text{π}}{\displaystyle \underset{R^2}{\int }\left|\begin{array}{ccc}{\left(AQ\right)}_{x_1}& AQ& {\left(AQ\right)}_{x_2}\end{array}\right|\text{d}x}\\ =\frac{1}{4\text{π}}{\displaystyle \underset{R^2}{\int }\left|\begin{array}{ccc}{A}_{\theta }{\theta }_{x_1}Q+A{Q}_r{R}_{x_1}& AQ& A_{\theta }{\theta }_{x_2}Q+A{Q}_r{R}_{x_2}\end{array}\right|\text{d}x}\end{array}$

$\begin{array}{l}\left|\begin{array}{ccc}{A}_{\theta }{\theta }_{x_1}Q+A{Q}_r{R}_{x_1}& AQ& A_{\theta }{\theta }_{x_2}Q+A{Q}_r{R}_{x_2}\end{array}\right|\\ =\left|\begin{array}{ccc}-\frac{\sin \theta }{r}{A}_{\theta }Q+A{Q}_r\cos \theta & AQ& \frac{\cos \theta }{r}{A}_{\theta }Q+A{Q}_r\sin \theta \end{array}\right|\\ =\left|\begin{array}{ccc}-\frac{\sin \theta }{r}{A}_{\theta }Q& AQ& \frac{\cos \theta }{r}{A}_{\theta }Q+A{Q}_r\sin \theta \end{array}\right|+\left|\begin{array}{ccc}A{Q}_r\cos \theta & AQ& \frac{\cos \theta }{r}{A}_{\theta }Q+A{Q}_r\sin \theta \end{array}\right|\\ =\left|\begin{array}{ccc}-\frac{\sin \theta }{r}{A}_{\theta }Q& AQ& A{Q}_r\sin \theta \end{array}\right|+\left|\begin{array}{ccc}A{Q}_r\cos \theta & AQ& \frac{\cos \theta }{r}{A}_{\theta }Q\end{array}\right|\\ =\frac{{\left(\sin \theta \right)}^2}{r}\left|\begin{array}{ccc}-A_{\theta }Q& AQ& A{Q}_r\end{array}\right|+\frac{{\left(\sin \theta \right)}^2}{r}\left|\begin{array}{ccc}A{Q}_r& AQ& A_{\theta }Q\end{array}\right|=\frac{1}{r}\left|\begin{array}{ccc}A{Q}_r& AQ& A_{\theta }Q\end{array}\right|\end{array}$

$\mathrm{deg}V=\frac{1}{\text{4π}}{\displaystyle \underset{R^2}{\int }\frac{1}{r}\left|\begin{array}{ccc}A{Q}_r& AQ& A_{\theta }Q\end{array}\right|\text{d}x}$

记： $Q_r={\left(\begin{array}{c}\frac{2r^k}{1+r^{2k}}\\ 0\\ \frac{r^{2k}-1}{r^{2k}+1}\end{array}\right)}_r=\left(\begin{array}{c}{M}_r\\ 0\\ N_r\end{array}\right)$， $A_{\theta }=\left(\begin{array}{ccc}-k\sin k\theta & -k\cos k\theta & 0\\ k\cos k\theta & -k\sin k\theta & 0\\ 0& 0& 0\end{array}\right)$

$\begin{array}{c}\mathrm{deg}V=\frac{1}{4\text{π}}{\displaystyle \underset{0}{\overset{+\infty }{\int }}{\displaystyle \underset{0}{\overset{\text{2π}}{\int }}\left|\begin{array}{ccc}\cos k\theta M_r& \cos k\theta M& -k\sin k\theta M\\ \sin k\theta M_r& \sin k\theta M& k\cos k\theta M\\ N_r& N& 0\end{array}\right|\text{d}r\text{d}\theta }}\\ =\frac{1}{4\text{π}}{\displaystyle \underset{0}{\overset{+\infty }{\int }}{\displaystyle \underset{0}{\overset{\text{2π}}{\int }}\left|\begin{array}{ccc}0& \cos k\theta M& -k\sin k\theta M\\ 0& \sin k\theta M& k\cos k\theta M\\ N_r-\frac{M_r}{M}N& N& 0\end{array}\right|\text{d}r\text{d}\theta }}\\ =\frac{1}{4\text{π}}{\displaystyle \underset{0}{\overset{+\infty }{\int }}{\displaystyle \underset{0}{\overset{\text{2π}}{\int }}\left|\begin{array}{ccc}0& 0& -k\sin k\theta M\\ 0& \frac{1}{\sin k\theta }M& k\cos k\theta M\\ N_r-\frac{M_r}{M}N& N& 0\end{array}\right|\text{d}r\text{d}\theta }}\end{array}$

$\begin{array}{l}=\frac{k}{2}{\displaystyle \underset{0}{\overset{+\infty }{\int }}{M}^2{N}_r-MN{M}_r}\text{d}r=\frac{k}{2}\left({M^{2}N|}_0^{+\infty }-3{\displaystyle \underset{0}{\overset{+\infty }{\int }}MN{M}_r}\text{d}r\right)=-\frac{3k}{2}{\displaystyle \underset{0}{\overset{+\infty }{\int }}MN{M}_r}\text{d}r\\ =-\frac{3k}{4}{\displaystyle \underset{0}{\overset{+\infty }{\int }}{\left(M^2\right)}_{r}N\text{d}r}=\frac{3k}{4}{\displaystyle \underset{0}{\overset{+\infty }{\int }}{M}^2{N}_r}\text{d}r=\frac{3k}{4}{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{4r^{2k}}{{\left(1+r^{2k}\right)}^2}\cdot \frac{4k{r}^{2k-1}}{{\left(1+r^{2k}\right)}^2}\text{d}r}\\ =12k^2{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{r^{4k-1}}{{\left(1+r^{2k}\right)}^4}\text{d}r}=12k^2{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{s^{\frac{4k-1}{2k}}}{{\left(1+s\right)}^4}\cdot \frac{1}{2k}\cdot s^{\frac{1}{2k}-1}}\text{d}s=6k{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{s}{{\left(1+s\right)}^4}\text{d}s}\\ =6k{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{1}{{\left(1+x\right)}^3}-\frac{1}{{\left(1+x\right)}^4}\text{d}x}=6k\times \left({\frac{1}{3{\left(1+x\right)}^3}-\frac{1}{2{\left(1+x\right)}^2}|}_0^{+\infty }\right)=k\end{array}$

4.3. 定理3的证明

由定理2知：

$V=\left(\begin{array}{c}\frac{2r^k}{1+r^{2k}}\cos \left(k\theta \right)\\ \frac{2r^k}{1+r^{2k}}\sin \left(k\theta \right)\\ \frac{r^{2k}-1}{r^{2k}+1}\end{array}\right)=\left(\begin{array}{c}A\cos \left(k\theta \right)\\ A\sin \left(k\theta \right)\\ B\end{array}\right)$

$\begin{array}{l}\frac{\partial u}{\partial x_1}=\frac{\partial u}{\partial r}\cdot \frac{\partial r}{\partial x_1}+\frac{\partial u}{\partial \theta }\cdot \frac{\partial \theta }{\partial x_1}=\cos \theta \cdot \frac{\partial u}{\partial r}-\frac{\sin \theta }{r}\frac{\partial u}{\partial \theta }\\ \frac{\partial u}{\partial x_2}=\frac{\partial u}{\partial r}\cdot \frac{\partial r}{\partial x_2}+\frac{\partial u}{\partial \theta }\cdot \frac{\partial \theta }{\partial x_2}=\sin \theta \cdot \frac{\partial u}{\partial r}+\frac{\cos \theta }{r}\frac{\partial u}{\partial \theta }\end{array}$

$\begin{array}{c}{\left|\nabla u\right|}^2={\left|\left(\frac{\partial u}{\partial x_1},\frac{\partial u}{\partial x_2},\frac{\partial u}{\partial t}\right)\right|}^2={\left|\frac{\partial u}{\partial x_1}\right|}^2+{\left|\frac{\partial u}{\partial x_2}\right|}^2+{\left|\frac{\partial u}{\partial t}\right|}^2\\ ={\left|\cos \theta \cdot \frac{\partial u}{\partial r}-\frac{\sin \theta }{r}\frac{\partial u}{\partial \theta }\right|}^2+{\left|\sin \theta \cdot \frac{\partial u}{\partial r}+\frac{\cos \theta }{r}\frac{\partial u}{\partial \theta }\right|}^2\\ ={\left|\frac{\partial u}{\partial r}\right|}^2+{\left|\frac{\partial u}{\partial \theta }\cdot \frac{1}{r}\right|}^2\end{array}$

$E\left(V\right)={\displaystyle \underset{R^2}{\int }{\left|\nabla V\right|}^2\text{d}x}={\displaystyle \underset{R^2}{\int }{\left|\frac{\partial V}{\partial r}\right|}^2+{\left|\frac{\partial V}{\partial \theta }\cdot \frac{1}{r}\right|}^2\text{d}x}$

${\left|\frac{\partial V}{\partial r}\right|}^2=A_r^2{\cos }^2\left(k\theta \right)+A_r^2{\sin }^2\left(k\theta \right)+B_r^2=A_r^2+B_r^2$

${\left|\frac{\partial V}{\partial \theta }\cdot \frac{1}{r}\right|}^2=\frac{1}{r^2}\left(A^2{k}^2{\sin }^2\left(k\theta \right)+A^2{k}^2{\cos }^2\left(k\theta \right)+0\right)=\frac{k^2}{r^2}{A}^2$

$E\left(V\right)={\displaystyle \underset{R^2}{\int }{A}_r^2+B_r^2+\frac{k^2}{r^2}{A}^2\text{d}x}={\displaystyle \underset{R^2}{\int }\frac{8k^2{r}^{2k-2}}{{\left(1+r^{2k}\right)}^2}\text{d}x}=16\text{π}{k}^2{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{r^{2k-2}}{{\left(1+r^{2k}\right)}^2}}\cdot r\text{d}r$

时，显然 $E\left(V\right)=0$

$k>0$时，令 $S=r^{2k}$，则 $\text{d}r=\frac{1}{2k}{S}^{\frac{1}{2k}-1}\text{d}s$，带入得： $16\text{π}{k}^2{\displaystyle \underset{0}{\overset{+\infty }{\int }}\frac{1}{2k{\left(1+s\right)}^2}\text{d}s}=8k\text{π}$

$k<0$时，令 $S=r^{2k}$，则 $\text{d}r=\frac{1}{2k}{S}^{\frac{1}{2k}-1}\text{d}s$，带入得： $16\text{π}{k}^2{\displaystyle \underset{+\infty }{\overset{0}{\int }}\frac{1}{2k{\left(1+s\right)}^2}\text{d}s}=-8k\text{π}$

综上： $E\left(V\right)=8\text{π}\left|k\right|$。

参考文献

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