EE263 Homework 5

课程主页：https://see.stanford.edu/Course/EE263

这次回顾EE263作业5。

10.2

(a)

特征值：

$$\det(\lambda I-A)=\lambda^2+w^2=0\Rightarrow \lambda =\pm iw$$

resolvent：

$$\begin{aligned} (sI-A)^{-1} &=\left[\begin{array}{cc}{s} & {-\omega} \\ {\omega} & {s}\end{array}\right]\\ &=\left[ \begin{array}{cc}{\frac s{s^2+w^2}} & \frac \omega{s^2+w^2} \\ -\frac \omega{s^2+w^2} & {\frac s{s^2+w^2}}\end{array}\right] \end{aligned}$$

状态转移矩阵：

$$\begin{aligned} \Phi(t)&=\mathcal{L}^{-1}\left((s I-A)^{-1}\right)\\ &=\left[\begin{array}{rr}{\cos wt} & {\sin wt} \\ {-\sin wt} & {\cos wt}\end{array}\right] \end{aligned}$$

所以

$$x(t)=\left[\begin{array}{rr}{\cos wt} & {\sin wt} \\ {-\sin wt} & {\cos wt}\end{array}\right]x(0)$$

(b)略过

(c)因为$\left[\begin{array}{rr}{\cos wt} & {\sin wt} \\ {-\sin wt} & {\cos wt}\end{array}\right]$是正交矩阵，所以结论成立。

(d)

$$\begin{aligned} \frac d{dt}\| x(t)\|^2 &=\frac d{dt}\left( x(t)^Tx(t) \right)\\ &=2\dot x(t)^T x(t)\\ &= x(t)^T \left[\begin{array}{cc}{0} & {-\omega} \\ {\omega} & {0}\end{array}\right]x(t)\\ &=0 \end{aligned}$$

所以

$$\dot x(t)^T x(t) =0$$

结论成立。

10.3

(a)

$$\begin{aligned} e^{A}e^{B} &=\left(\sum_{i=0}^{\infty} \frac {A^i}{i!} \right) \left(\sum_{j=0}^{\infty} \frac {B^j}{j!} \right)\\ &=\sum_{k=0}^{\infty} \sum_{i+j=k}\frac 1 {i! j!} A^iB^j\\ &=\sum_{k=0}^{\infty}\frac 1{k!}\sum_{i+j=k}\frac {k!} {i! j!} A^iB^j\\ &=\sum_{k=0}^{\infty}\frac 1{k!}(A+B)^k & 由AB=BA\\ &=e^{(A+B)} \end{aligned}$$

(b)

$$\begin{aligned} \frac{d}{d t} e^{A t} &=\frac{d}{d t}\left(\sum_{i=0}^{\infty} \frac {(At)^i}{i!} \right)\\ &=\sum_{i=1}^{\infty} \frac {A^it^{i-1}}{(i-1)!}\\ &=A\sum_{i=1}^{\infty} \frac {(At)^{i-1}}{(i-1)!}\\ &=Ae^{At}\\ &=\left(\sum_{i=1}^{\infty} \frac {(At)^{i-1}}{(i-1)!} \right) A\\ &=e^{At} A \end{aligned}$$

10.4

(a)因为

$$A^2=\left[\begin{array}{ll}{-1} & {1} \\ {-1} & {1}\end{array}\right]\left[\begin{array}{ll}{-1} & {1} \\ {-1} & {1}\end{array}\right]=0$$

所以

$$\begin{aligned} e^{tA} &=\sum_{i=0}^{\infty} \frac {(tA)^i}{i!} \\ &=\left[\begin{matrix} 1& 0\\ 0&1 \end{matrix}\right] +\left[\begin{array}{ll}{-1} & {1} \\ {-1} & {1}\end{array}\right]t\\ &=\left[\begin{array}{ll}1-t & {t} \\ {-t} & {1+t}\end{array}\right] \end{aligned}$$

取$t=1$得到

$$\begin{aligned} e^{A} &=\left[\begin{array}{ll}0 & {1} \\ {-1} & {2}\end{array}\right] \end{aligned}$$

(b)由结论可得

$$\begin{aligned} x(t) &=e^{tA}x(0) \\ &=\left[\begin{array}{ll}1-t & {t} \\ {-t} & {1+t}\end{array}\right] \left[\begin{matrix} 1\\ a \end{matrix}\right] \\ \end{aligned}$$

令$t=1$得到

$$\begin{aligned} x(1) &=\left[\begin{array}{ll}0 & {1} \\ {-1} & {2}\end{array}\right] \left[\begin{matrix} 1\\ a \end{matrix}\right] \\ &=\left[\begin{matrix} a \\ 2a-1 \end{matrix}\right] \end{aligned}$$

由题意可得

$$2a-1=2\Rightarrow a=\frac 3 2$$

因此

$$\begin{aligned} x(t) &=\left[\begin{array}{ll}1-t & {t} \\ {-t} & {1+t}\end{array}\right] \left[\begin{matrix} 1\\ \frac 3 2 \end{matrix}\right] \\ &=\left[\begin{matrix} 1+\frac 12 t\\ \frac 3 2 +\frac 1 2t \end{matrix}\right]\\ x(2) &=\left[\begin{matrix} 2 \\ \frac 5 2 \end{matrix}\right] \end{aligned}$$

补充题

1

求导可得

$$\begin{aligned} \dot x(t) &=a(t) \exp \left(\int_{0}^{t} a(\tau) d \tau\right) x(0)\\ &=a(t)x(t) \end{aligned}$$

作为反例，考虑（参考自解答）

$$A(t)=\left\{\begin{array}{ll}{A_{1}} & {0 \leq t<1} \\ {A_{2}} & {t \geq 1}\end{array}\right.$$

那么

$$A_{1}=\left[\begin{array}{ll}{0} & {1} \\ {0} & {0}\end{array}\right], \qquad A_{2}=\left[\begin{array}{ll}{0} & {0} \\ {1} & {0}\end{array}\right]$$

那么

$$\begin{aligned} x(2) &=\left(\exp A_{2}\right)\left(\exp A_{1}\right) x(0) \\ &=\left[\begin{array}{cc}{1} & {1} \\ {0} & {1}\end{array}\right]\left[\begin{array}{cc}{1} & {0} \\ {1} & {1}\end{array}\right] x(0) \\ &=\left[\begin{array}{cc}{1} & {1} \\ {1} & {2}\end{array}\right] x(0) \end{aligned}$$

但是，在上述公式中

$$\begin{aligned} \int_0^2 A(t) dt &=\int_0^1 A_1 dt +\int_1^2 A_2 dt\\ &=\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]\\ &\triangleq B \end{aligned}$$

不难验证

$$B^{2k}=I,B^{2k+1}= B$$

所以

$$\begin{aligned} \exp(tB) &=\sum_{i=0}^{\infty} \frac{B^i}{i!}\\ &=\sum_{k=0}^{\infty} \frac{B^{2k}}{(2k)!}+ \sum_{k=0}^{\infty} \frac{B^{2k+1}}{(2k+1)!}\\ &=I\sum_{k=0}^{\infty} \frac{1}{(2k)!} +B \sum_{k=0}^{\infty} \frac{1}{(2k+1)!}\\ &= I\frac {e+e^{-1}} 2 +B\frac {e-e^{-1}} 2\\ &=\left[\begin{array}{cc}{1.5431} & {1.1752} \\ {1.1752} & {1.5431}\end{array}\right] \end{aligned}$$

因此

$$x(2)=\left[\begin{array}{cc}{1.5431} & {1.1752} \\ {1.1752} & {1.5431}\end{array}\right]x(0)$$

所以上述事实对高维情形不成立。