反对称矩阵$A$ 微分方程$u'(t)=A\bf u$

hbghlyj

6.3 #14.a Introduction to Linear Algebra: Strang
反对称矩阵$A^T = −A$
微分方程
$$\frac{d \mathbf{u}}{d t}=\left[\begin{array}{rrr}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right] \mathbf{u}$$
即\begin{array}{l}u_{1}^{\prime}=c u_{2}-b u_{3} \\ u_{2}^{\prime}=a u_{3}-c u_{1} \\ u_{3}^{\prime}=b u_{1}-a u_{2}\end{array}这道题证明了$\|𝐮(t)\|^2=\|𝐮(0)\|^2$, 接下来如何解这个微分方程$𝐮'(t)=A\bf u$

hbghlyj

下面假定$a^2+b^2+c^2=1$

Output:\begin{align*}u_{1}\left(t\right) &= a^{2} u_{1}\left(0\right) + a b u_{2}\left(0\right) + c a u_{3}\left(0\right) - {\left(a^{2} u_{1}\left(0\right) + a b u_{2}\left(0\right) + c a u_{3}\left(0\right) - u_{1}\left(0\right)\right)} \cos\left(t\right)+{\left(c u_{2}\left(0\right)-b u_{3}\left(0\right)\right)} \sin\left(t\right)\\ u_{2}\left(t\right) &= a b u_{1}\left(0\right) + b^{2} u_{2}\left(0\right) + c b u_{3}\left(0\right) - {\left(a b u_{1}\left(0\right) + b^{2} u_{2}\left(0\right) + c b u_{3}\left(0\right) - u_{2}\left(0\right)\right)} \cos\left(t\right) + {\left(a u_{3}\left(0\right) - c u_{1}\left(0\right)\right)} \sin\left(t\right)\\ u_{3}\left(t\right) &= ca u_{1}\left(0\right) + bc u_{2}\left(0\right)+c^{2} u_{3}\left(0\right) - \left(ca u_{1}\left(0\right) + bc u_{2}\left(0\right)+c^{2} u_{3}\left(0\right)-u_{3}\left(0\right)\right) \cos\left(t\right) + {\left(b u_{1}\left(0\right) - a u_{2}\left(0\right)\right)} \sin\left(t\right)\end{align*}

Czhang271828

一般用 Jordan 分解做. 这里也可以直接观察到\[[au_1+bu_2+cu_3]'=0,\]然后换元.