坐标变换的寻找方法

kuzb

$$p_{1}{}'=-\mu _{1}p_{1}-v_{1}q_{1}+\gamma _{11}q_{1}\left ( p_{1}^{2}+q_{1}^{2} \right )+\gamma _{12}q_{1}\left ( p_{2}^{2}+q_{2}^{2} \right )-\delta _{1}\left [ 2p_{1}q_{1}p_{2}-q_{2}\left ( p_{1}^{2}-q_{1}^{2}\right )\right ]\\
q_{1}{}'=-\mu _{1}q_{1}+v_{1}p_{1}-\gamma _{11}p_{1}\left ( p_{1}^{2}+q_{1}^{2} \right )-\gamma _{12}p_{1}\left ( p_{2}^{2}+q_{2}^{2} \right )-\delta _{1}\left [ 2p_{1}q_{1}q_{2}+p_{2}\left ( p_{1}^{2}-q_{1}^{2}\right )\right ]\\
p_{2}{}'=-\mu _{2}p_{2}-v_{2}q_{2}+\gamma _{21}q_{2}\left ( p_{1}^{2}+q_{1}^{2} \right )+\gamma _{22}q_{2}\left ( p_{2}^{2}+q_{2}^{2} \right )+\delta _{2}\left ( 3p_{1}^{2} q_{1}-q_{1}^{3}\right )-s_{22}Fq_{2}\\
q_{2}{}'=-\mu _{2}q_{2}+v_{2}p_{2}-\gamma _{21}p_{2}\left ( p_{1}^{2}+q_{1}^{2} \right )-\gamma _{22}p_{2}\left ( p_{2}^{2}+q_{2}^{2} \right )+\delta _{2}\left ( 3p_{1}q_{1}^{2}-p_{1}^{3}\right )-s_{22}Fp_{2}$$
$$\left ( p_{1},q_{1} ,p_{2},q_{2}\right )\\
\Leftrightarrow \left ( -p_{1},-q_{1} ,-p_{2},-q_{2}\right )\\
\Leftrightarrow \left ( \frac{1}{2} \left ( p_{1} -\sqrt{3}q_{1}\right ),\frac{1}{2}\left ( \sqrt{3}p_{1} +q_{1}\right ),-p_{2},-q_{2}\right )\\
\Leftrightarrow \left ( -\frac{1}{2} \left ( p_{1} -\sqrt{3}q_{1}\right ),-\frac{1}{2}\left ( \sqrt{3}p_{1} +q_{1}\right ),p_{2},q_{2}\right )\\
\Leftrightarrow \left ( -\frac{1}{2} \left ( p_{1} +\sqrt{3}q_{1}\right ),\frac{1}{2}\left ( \sqrt{3}p_{1} -q_{1}\right ),p_{2},q_{2}\right )\\
\Leftrightarrow \left ( \frac{1}{2} \left ( p_{1} +\sqrt{3}q_{1}\right ),-\frac{1}{2}\left ( \sqrt{3}p_{1} -q_{1}\right ),-p_{2},-q_{2}\right )$$
请问，后面那6个坐标变换是怎么找到的（该坐标变换使得上面的系统方程保持不变）

Accompany

从哪找来的

kuzb

Accompany 发表于 2016-12-21 10:55
从哪找来的

帮朋友问的

Agoni

这是数值分析里的吧

kuzb

Agoni 发表于 2016-12-22 08:45
这是数值分析里的吧

不是

sovereign

好长  没看明白