薛定谔映射问题的k等变解的性质 The Nature for k-Equivariant Solutions to the Schrodinger Map Problem

doi:10.12677/AAM.2018.71009

Advances in Applied Mathematics
Vol.07 No.01(2018), Article ID:23550,6 pages
10.12677/AAM.2018.71009

The Nature for k-Equivariant Solutions to the Schrodinger Map Problem

Liang Li

●How to Cite this Article

Math and Computer School, Yunnan National University, Kunming Yunnan

Received: Dec. 19^th, 2017; accepted: Jan. 18^th, 2018; published: Jan. 25^th, 2018

ABSTRACT

K-equivariant solutions have a relationship with its homotopy index k in the Schrodinger map problem. We can explain it by topological degree. We give an example to verify it without certification. Then we proved that the equation keeps the Dirichlet energy constant. At last we calculate its minimum value.

Keywords:Schrodinger Map Problem, k-Equivariant Solution, Dirichlet Energy

薛定谔映射问题的k等变解的性质

李良

云南民族大学，数学与计算机科学院，云南昆明

收稿日期：2017年12月19日；录用日期：2018年1月18日；发布日期：2018年1月25日

摘要

薛定谔映射问题中的等变解的同伦指数k与解本身是有一定的关系的，可以通过定义的拓扑度来表达。举了一个例子验证了一下，没有证明。但证明了方程保证了解的狄里克雷能量是保持不变的，并计算出了随着初值变化，解的能量的最小值。

关键词 :薛定谔映射，k等变解，狄里克雷能量

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言

薛定谔映射如下：

${\begin{cases} \partial_{t} u = u \land Δ u \\ {u |}_{t = 0} = u_{0} \in H_{0}^{1}, (t, x) \in R \times R^{2}, u (t, x) \in S^{2} \end{cases}$ (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： $u (t, x) = e^{k θ R} | \begin{matrix} u_{1} (t, r) \\ u_{2} (t, r) \\ u_{3} (t, r) \end{matrix}, R = (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}), k \in Z$

其中 $(r, θ)$ 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： $E (u (t)) = \int_{R^{2}} {| \nabla u (t, x) |}^{2} d x$

例如： $V = e^{k θ R} (\begin{matrix} \frac{2 r^{k}}{1 + r^{2 k}} \\ 0 \\ \frac{r^{2 k} - 1}{r^{2 k} + 1} \end{matrix})$ 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：

$E (u (t)) = \int_{R^{2}} {| \nabla u (t, x) |}^{2} d x = E ( u o )$

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 $8 | k | π$

4. 定理的证明

4.1. 定理1的证明

方程两边点乘 $Δ u$ 再积分：

$\Rightarrow Δ u \cdot \partial_{t} u = Δ u \cdot (u \land Δ u) = 0$

$\Rightarrow \int_{R^{2}} Δ u \cdot \partial_{t} u d x = - \frac{1}{2} \frac{d}{d t} \int_{R^{2}} {| \nabla u |}^{2} d x 0 = 0$

$\Rightarrow \int_{R^{2}} {| \nabla u |}^{2} d x = C$

$\Rightarrow E (u (t)) = E ( u o )$

4.2. 定理2的证明

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

$K = 1$ 时显然成立，若 $K = n (n > 1)$ 时成立，则

$\begin{matrix} R^{n + 1} = (\begin{matrix} \cos \frac{n π}{2} & - \sin \frac{n π}{2} & 0 \\ \sin \frac{n π}{2} & \cos \frac{n π}{2} & 0 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = (\begin{matrix} - \sin \frac{n π}{2} & - \cos \frac{n π}{2} & 0 \\ \cos \frac{n π}{2} & - \sin \frac{n π}{2} & 0 \\ 0 & 0 & 0 \end{matrix}) \\ = (\begin{matrix} \cos \frac{(n + 1) π}{2} & - \sin \frac{(n + 1) π}{2} & 0 \\ \sin \frac{(n + 1) π}{2} & \cos \frac{(n + 1) π}{2} & 0 \\ 0 & 0 & 0 \end{matrix}) \end{matrix}$

再计算 $e^{k θ R}$ ：

$e^{k θ R} = \sum_{h = 0}^{\infty} \frac{1}{n!} {(k θ R)}^{n} = \sum_{h = 0}^{\infty} \frac{1}{n!} {(k θ)}^{n} (\begin{matrix} \cos \frac{n π}{2} & - \sin \frac{n π}{2} & 0 \\ \sin \frac{n π}{2} & \cos \frac{n π}{2} & 0 \\ 0 & 0 & 0 \end{matrix})$

分别计算得： $\sum_{h = 0}^{\infty} \frac{1}{n!} {(k θ)}^{n} \cos \frac{n π}{2} = \sum_{h = 0}^{\infty} \frac{1}{(2 n)!} {(k θ)}^{2 n} {(- 1)}^{n} = \cos k θ$

同理可得： $e^{k θ R} = (\begin{matrix} \cos k θ & - \sin k θ & 0 \\ \sin k θ & \cos k θ & 0 \\ 0 & 0 & 1 \end{matrix})$

记： $A = e^{k θ R} = (\begin{matrix} \cos k θ & - \sin k θ & 0 \\ \sin k θ & \cos k θ & 0 \\ 0 & 0 & 1 \end{matrix})$ ，， $V = A Q$

$\begin{matrix} \deg V = \frac{1}{4 π} \int_{R^{2}} | \begin{matrix} V_{x 1} & V & V_{x_{2}} \end{matrix} | d x = \frac{1}{4 π} \int_{R^{2}} | \begin{matrix} {(A Q)}_{x_{1}} & A Q & {(A Q)}_{x_{2}} \end{matrix} | d x \\ = \frac{1}{4 π} \int_{R^{2}} | \begin{matrix} A_{θ} θ_{x_{1}} Q + A Q_{r} R_{x_{1}} & A Q & A_{θ} θ_{x_{2}} Q + A Q_{r} R_{x_{2}} \end{matrix} | d x \end{matrix}$

$\begin{array}{l} | \begin{matrix} A_{θ} θ_{x_{1}} Q + A Q_{r} R_{x_{1}} & A Q & A_{θ} θ_{x_{2}} Q + A Q_{r} R_{x_{2}} \end{matrix} | \\ = | \begin{matrix} - \frac{\sin θ}{r} A_{θ} Q + A Q_{r} \cos θ & A Q & \frac{\cos θ}{r} A_{θ} Q + A Q_{r} \sin θ \end{matrix} | \\ = | \begin{matrix} - \frac{\sin θ}{r} A_{θ} Q & A Q & \frac{\cos θ}{r} A_{θ} Q + A Q_{r} \sin θ \end{matrix} | + | \begin{matrix} A Q_{r} \cos θ & A Q & \frac{\cos θ}{r} A_{θ} Q + A Q_{r} \sin θ \end{matrix} | \\ = | \begin{matrix} - \frac{\sin θ}{r} A_{θ} Q & A Q & A Q_{r} \sin θ \end{matrix} | + | \begin{matrix} A Q_{r} \cos θ & A Q & \frac{\cos θ}{r} A_{θ} Q \end{matrix} | \\ = \frac{{(\sin θ)}^{2}}{r} | \begin{matrix} - A_{θ} Q & A Q & A Q_{r} \end{matrix} | + \frac{{(\sin θ)}^{2}}{r} | \begin{matrix} A Q_{r} & A Q & A_{θ} Q \end{matrix} | = \frac{1}{r} | \begin{matrix} A Q_{r} & A Q & A_{θ} Q \end{matrix} | \end{array}$

$\deg V = \frac{1}{4π} \int_{R^{2}} \frac{1}{r} | \begin{matrix} A Q_{r} & A Q & A_{θ} Q \end{matrix} | d x$

记： $Q_{r} = {(\begin{matrix} \frac{2 r^{k}}{1 + r^{2 k}} \\ 0 \\ \frac{r^{2 k} - 1}{r^{2 k} + 1} \end{matrix})}_{r} = (\begin{matrix} M_{r} \\ 0 \\ N_{r} \end{matrix})$ ， $A_{θ} = (\begin{matrix} - k \sin k θ & - k \cos k θ & 0 \\ k \cos k θ & - k \sin k θ & 0 \\ 0 & 0 & 0 \end{matrix})$

$\begin{matrix} \deg V = \frac{1}{4 π} \int_{0}^{+ \infty} \int_{0}^{2π} | \begin{matrix} \cos k θ M_{r} & \cos k θ M & - k \sin k θ M \\ \sin k θ M_{r} & \sin k θ M & k \cos k θ M \\ N_{r} & N & 0 \end{matrix} | d r d θ \\ = \frac{1}{4 π} \int_{0}^{+ \infty} \int_{0}^{2π} | \begin{matrix} 0 & \cos k θ M & - k \sin k θ M \\ 0 & \sin k θ M & k \cos k θ M \\ N_{r} - \frac{M_{r}}{M} N & N & 0 \end{matrix} | d r d θ \\ = \frac{1}{4 π} \int_{0}^{+ \infty} \int_{0}^{2π} | \begin{matrix} 0 & 0 & - k \sin k θ M \\ 0 & \frac{1}{\sin k θ} M & k \cos k θ M \\ N_{r} - \frac{M_{r}}{M} N & N & 0 \end{matrix} | d r d θ \end{matrix}$

$\begin{array}{l} = \frac{k}{2} \int_{0}^{+ \infty} M^{2} N_{r} - M N M_{r} d r = \frac{k}{2} ({M^{2} N |}_{0}^{+ \infty} - 3 \int_{0}^{+ \infty} M N M_{r} d r) = - \frac{3 k}{2} \int_{0}^{+ \infty} M N M_{r} d r \\ = - \frac{3 k}{4} \int_{0}^{+ \infty} {(M^{2})}_{r} N d r = \frac{3 k}{4} \int_{0}^{+ \infty} M^{2} N_{r} d r = \frac{3 k}{4} \int_{0}^{+ \infty} \frac{4 r^{2 k}}{{(1 + r^{2 k})}^{2}} \cdot \frac{4 k r^{2 k - 1}}{{(1 + r^{2 k})}^{2}} d r \\ = 12 k^{2} \int_{0}^{+ \infty} \frac{r^{4 k - 1}}{{(1 + r^{2 k})}^{4}} d r = 12 k^{2} \int_{0}^{+ \infty} \frac{s^{\frac{4 k - 1}{2 k}}}{{(1 + s)}^{4}} \cdot \frac{1}{2 k} \cdot s^{\frac{1}{2 k} - 1} d s = 6 k \int_{0}^{+ \infty} \frac{s}{{(1 + s)}^{4}} d s \\ = 6 k \int_{0}^{+ \infty} \frac{1}{{(1 + x)}^{3}} - \frac{1}{{(1 + x)}^{4}} d x = 6 k \times ({\frac{1}{3 {(1 + x)}^{3}} - \frac{1}{2 {(1 + x)}^{2}} |}_{0}^{+ \infty}) = k \end{array}$

4.3. 定理3的证明

由定理2知：

$V = (\begin{matrix} \frac{2 r^{k}}{1 + r^{2 k}} \cos (k θ) \\ \frac{2 r^{k}}{1 + r^{2 k}} \sin (k θ) \\ \frac{r^{2 k} - 1}{r^{2 k} + 1} \end{matrix}) = (\begin{matrix} A \cos (k θ) \\ A \sin (k θ) \\ B \end{matrix})$

$\begin{array}{l} \frac{\partial u}{\partial x_{1}} = \frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial x_{1}} + \frac{\partial u}{\partial θ} \cdot \frac{\partial θ}{\partial x_{1}} = \cos θ \cdot \frac{\partial u}{\partial r} - \frac{\sin θ}{r} \frac{\partial u}{\partial θ} \\ \frac{\partial u}{\partial x_{2}} = \frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial x_{2}} + \frac{\partial u}{\partial θ} \cdot \frac{\partial θ}{\partial x_{2}} = \sin θ \cdot \frac{\partial u}{\partial r} + \frac{\cos θ}{r} \frac{\partial u}{\partial θ} \end{array}$

$\begin{matrix} {| \nabla u |}^{2} = {| (\frac{\partial u}{\partial x_{1}}, \frac{\partial u}{\partial x_{2}}, \frac{\partial u}{\partial t}) |}^{2} = {| \frac{\partial u}{\partial x_{1}} |}^{2} + {| \frac{\partial u}{\partial x_{2}} |}^{2} + {| \frac{\partial u}{\partial t} |}^{2} \\ = {| \cos θ \cdot \frac{\partial u}{\partial r} - \frac{\sin θ}{r} \frac{\partial u}{\partial θ} |}^{2} + {| \sin θ \cdot \frac{\partial u}{\partial r} + \frac{\cos θ}{r} \frac{\partial u}{\partial θ} |}^{2} \\ = {| \frac{\partial u}{\partial r} |}^{2} + {| \frac{\partial u}{\partial θ} \cdot \frac{1}{r} |}^{2} \end{matrix}$

$E (V) = \int_{R^{2}} {| \nabla V |}^{2} d x = \int_{R^{2}} {| \frac{\partial V}{\partial r} |}^{2} + {| \frac{\partial V}{\partial θ} \cdot \frac{1}{r} |}^{2} d x$

${| \frac{\partial V}{\partial r} |}^{2} = A_{r}^{2} \cos^{2} (k θ) + A_{r}^{2} \sin^{2} (k θ) + B_{r}^{2} = A_{r}^{2} + B_{r}^{2}$

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

$E (V) = \int_{R^{2}} A_{r}^{2} + B_{r}^{2} + \frac{k^{2}}{r^{2}} A^{2} d x = \int_{R^{2}} \frac{8 k^{2} r^{2 k - 2}}{{(1 + r^{2 k})}^{2}} d x = 16 π k^{2} \int_{0}^{+ \infty} \frac{r^{2 k - 2}}{{(1 + r^{2 k})}^{2}} \cdot r d r$

时，显然 $E (V) = 0$

$k > 0$ 时，令 $S = r^{2 k}$ ，则 $d r = \frac{1}{2 k} S^{\frac{1}{2 k} - 1} d s$ ，带入得： $16 π k^{2} \int_{0}^{+ \infty} \frac{1}{2 k {(1 + s)}^{2}} d s = 8 k π$

$k < 0$ 时，令 $S = r^{2 k}$ ，则 $d r = \frac{1}{2 k} S^{\frac{1}{2 k} - 1} d s$ ，带入得： $16 π k^{2} \int_{+ \infty}^{0} \frac{1}{2 k {(1 + s)}^{2}} d s = - 8 k π$

综上： $E (V) = 8 π | k |$ 。

文章引用

李良. 薛定谔映射问题的k等变解的性质
The Nature for k-Equivariant Solutions to the Schrodinger Map Problem[J]. 应用数学进展, 2018, 07(01): 74-79. http://dx.doi.org/10.12677/AAM.2018.71009

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https://doi.org/10.1063/1.1402955

3. Guo, B.L., Han, Y.Q. and Yang, G.S. (2000) Blow up Problem for Lan-dau-Lifshitz Equations in Two Dimensions. Communications in Nonlinear Science & Numerical Simulation, 5, 43-44.
https://doi.org/10.1016/S1007-5704(00)90023-6

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