《Dichotomies in Stability Theory》W. A. Coppel
p.2
证明:
\(\dot{x}=\{\sin \log (t+1)+\cos \log(t+1)-\alpha\}x\)
的基本解\(x(t)=e^{(t+1)\sin\log(t+1)-\alpha t}\).
当\(\alpha>1\)是渐近稳定、指数稳定的。但当\(\alpha<\sqrt{2}\) 不是一致稳定的。
解：
\(
|x(t)| \leq e \cdot e^{-(\alpha-1) t}\le e, \quad \alpha-1>0,
\)
因此是渐近稳定，且指数稳定的。但当\(\alpha<\sqrt{2}\)是不是一致稳定的：如果是一致稳定的，那么存在 \(k>0\), s.t.:
\(
\left|x(t) x^{-1}(s)\right| \leq k,
\)
因为
\[\begin{aligned}
& x(t) x^{-1}(s)
=e^{(t+1) \sin \log (t+1)-\alpha t+
\alpha s-(s+1)\sin\log (s+1)}
\end{aligned}
\]
于是, 令 \(\bar{k}=\ln k\),
\[\quad(t+1) \sin \log (t+1)-(s+1) \sin \log (s+1)-\alpha(t-s) \leq \bar{k}.
\]
\[\begin{aligned}
& \text {利用微分中值定理} \\
\end{aligned}
\]
\[\begin{aligned}
\bar{k}+\alpha(t-s)
&\ge(t-s) \cdot[(t+1) \sin \log (t+1)]_{t=\xi}^{\prime} \\
& =(t-s) \cdot[\sin \log (\xi+1)+\cos \log (\xi+1)] \\
& =(t-s)\cdot\sqrt{2} \sin \left[\log (\xi+1)+\frac{\pi}{4}\right] \text {. } \\
\end{aligned}
\]
\[\Rightarrow
\bar{k}+\alpha(t-s) \ge
\sqrt{2}(t-s), \ \text{当$\alpha<\sqrt{2}$时，我们不能保证此不等式成立}.
\]
p.4
\(A\) 有\(n\)个互异的特征值\(\lambda_1,\cdots,\lambda_n\), \(p(\lambda)\)是过 \(\{(\lambda_i,e^{t\lambda_i})\}_{i=1}^{n}\) 由插值确定的唯一次数 \(<n\) 的多项式, 证明\(P(A)=e^{tA}\).
证明： 设
\(
A=Q{\rm diag} (\lambda_1,\cdots,\lambda_n)Q^{-1},
\)
既然\(P(\lambda)\)是多项式, 那么
\[P(A)=Q{\rm diag}(P(\lambda_1),\cdots, P(\lambda_n))Q^{-1}=Q{\rm diag}(e^{\lambda_1 t},\cdots, e^{\lambda_n t})Q^{-1}
\]
再计算\(e^{tA}\),
\[e^{tA}=e^{tQ{\rm diag} (\lambda_1,\cdots,\lambda_n)Q^{-1}}=Qe^{t{\rm diag} (\lambda_1,\cdots,\lambda_n)}Q^{-1}=Q{\rm diag}(e^{\lambda_1 t},\cdots, e^{\lambda_n t})Q^{-1}
\]
这就证明了
\(
P(A)=e^{tA}.
\)
p3. 一个非自治方程的基本解矩阵的计算
证明
\[\begin{equation}
\dot{x}=A(t)x
\end{equation}
\]
的基本解矩阵为
\[X(t)=
\begin{pmatrix}
e^t(\cos t+1/2\sin t) & e^{-3t}(\cos t-1/2\sin t)\\
e^t(\sin t-1/2\cos t) & e^{-3t}(\sin t+1/2\cos t)
\end{pmatrix},
\]
\[A(t)=U^{-1}(t) A_{0} U(t),~~~
U(t)=\begin{pmatrix}
\cos t &\sin t\\
-\sin t & \cos t
\end{pmatrix},
~~~~~~
A_0=\begin{pmatrix}
-1&-5\\
0 & -1
\end{pmatrix}.\]
把\(A(t)\)带入(1)中\(U(t)\dot{x}=A_0 U(t)x\)
展开
\[\begin{pmatrix}
\cos t \dot{x}_1+\sin t \dot{x}_2\\
-\sin t \dot{x}_1+\cos t\dot{x}_2
\end{pmatrix}
=A_0
\begin{pmatrix}
\cos t x_1+\sin t x_2\\
-\sin t x_1+\cos t x_2
\end{pmatrix}
=:A_0 y(t)
\]
即\(y(t)=U(t)x(t).\)注意到
\[\dot{y}(t)=\begin{pmatrix}
\cos t \dot{x}_1+\sin t \dot{x}_2\\
-\sin t \dot{x}_1+\cos t\dot{x}_2
\end{pmatrix}+
\begin{pmatrix}
-\sin t x_1+\cos t x_2\\
-\cos t x_1-\sin t x_2
\end{pmatrix}=A_0 y(t)+
\begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix}y(t)
\]
即
\[\dot{y}=\left(A_0+\begin{pmatrix}0 & 1\\-1 & 0
\end{pmatrix}\right)y=:By\]
我们可以计算出其基本解矩阵为\(Y(t)=e^{Bt}\), 进而\(X(t)=U^{-1}(t)Y(t).\)
具体计算:
\[B=\begin{pmatrix}-1 & -4\\-1 & -1\end{pmatrix}
\]
有两个特征值: \(\lambda_1=1\) 和 \(\lambda_2=-3\), 对应的特征向量分别为\(\xi_1=(-2,1)^T\) 和 \(\xi_2=(-2,-1)^T\). 于是
\[\begin{pmatrix}-2 &-2\\1 &-1\end{pmatrix}^{-1}B
\begin{pmatrix}-2 & -2\\1 &-1\end{pmatrix}=
\begin{pmatrix}1 & 0\\0 &-3\end{pmatrix}\]
进而
\[\begin{aligned}e^{Bt}&=
\begin{pmatrix}-2 & -2\\1 &-1\end{pmatrix}
\begin{pmatrix} e^t& 0\\ 0 &e^{-3t}\end{pmatrix}
\begin{pmatrix}-2 & -2\\1 &-1\end{pmatrix}^{-1}
=
\begin{pmatrix}-2 & -2\\1 &-1\end{pmatrix}
\begin{pmatrix} e^t& 0\\ 0 &e^{-3t}\end{pmatrix}
(\frac{1}{4})
\begin{pmatrix}-1 & 2\\-1 &-2\end{pmatrix}
\\
&=\begin{pmatrix}-1 & -1\\1/2 &-1/2\end{pmatrix}
\begin{pmatrix} e^t& 0\\ 0 &e^{-3t}\end{pmatrix}
\begin{pmatrix}-1/2 &1\\-1/2&-1\end{pmatrix}
=
\begin{pmatrix}1 & 1\\-1/2 &1/2\end{pmatrix}
\begin{pmatrix} e^t& 0\\ 0 &e^{-3t}\end{pmatrix}
\begin{pmatrix}1/2 &-1\\1/2&1\end{pmatrix}
\\
&=\begin{pmatrix}
e^t& e^{-3t}\\ -(1/2)e^t &(1/2)e^{-3t}\end{pmatrix}
\begin{pmatrix}1/2 &-1\\1/2&1\end{pmatrix}
\end{aligned}
\]
(这里纯粹是为了凑出书上的结果) 把
\[Y(t)=\begin{pmatrix} e^t& e^{-3t}\\ -(1/2)e^t &(1/2)e^{-3t}\end{pmatrix}
\]
取作\(\dot{y}=By\)的基本解矩阵, 那么 \(\dot{x}=A(t)x\)的基本解矩阵为
\[\begin{aligned}X(t)&=U(t)^{-1}Y(t)=
\begin{pmatrix}\cos t &-\sin t\\
\sin t & \cos t
\end{pmatrix}
\begin{pmatrix} e^t& e^{-3t}\\ -(1/2)e^t &(1/2)e^{-3t}\end{pmatrix}\\
&=\begin{pmatrix}
e^t\left (\cos t +\frac{1}{2}\sin t\right) & e^{-3t}\left(\cos t-\frac{1}{2}\sin t\right)\\
e^t\left(\sin t - \frac{1}{2}\cos t\right) &e^{-3t}\left(\sin t+\frac{1}{2}\cos t\right)\end{pmatrix}
\end{aligned}.
\]
p6. Proof of Proposition 5
Proposition 5
Let \(A(t)\) be \(n\times n\) matrix function defined on an interval \(J=[0,+\infty)\) such that
\(\textbf{(i)}\) every eigenvalue of \(A(t)\) has real part \(\le \alpha\) for all \(t\in J\),
\(\textbf{(ii)}\) \(|A(t)|\le M\) for all \(t\in J\).
Then for any \(\varepsilon>0\) there exits \(\delta=\delta(M,\varepsilon)>0\) such that if \(\textbf{(iii)}\)
\[|A(t_1)-A(t_2)|\le \delta|t_1-t_2|,~~~\forall t_1,t_2\in J,
\]
the fundamental matrix \(X_t(t)\) the equation \(\dot{x}=A(t)x\) satisfies the inequality
\[|X(t)X^{-1}(s)|\le K_\varepsilon e^{(\alpha+\varepsilon)(t-s)},~~t\ge s
\]
where \(K_\varepsilon=\max\{(\frac{4M}{\varepsilon})^{n-1},1\}\).
Before proving Proposition 5, we need to recall Propositions 3 and 4:
Proposition 3. Let \(A\) be an \(n\times n\) matrix such that \(\textbf{(i)}\) every eigenvalue of \(A\) has real part \(\le \alpha\) and \(\textbf{(ii)}\) \(|A|\le M\) then
\[|e^{tA}|\le \max\left\{\left(\frac{2M}{\varepsilon}\right)^{n-1},1\right\}e^{(\alpha+\varepsilon)t},~~~\forall t\ge 0,\forall \varepsilon>0.
\]
Proposition 4. Let \(A(t)\) be an \(n\times n\) matrix function defined on an interval \(J\) such that \(\textbf{(i)}\)
\[|A(t_1)-A(t_2)|\le \delta|t_1-t_2|,~~\forall t_1,t_2\in J
\]
and \(\textbf{(ii)}\) \(|e^{\tau A(t)}|\le K e^{\alpha \tau}\) for all \(\tau\ge 0\) and all \(t\in J\)
then the fundamental matrix \(X(t)\) of the equation \(\dot{x}=A(t)x\) satisfies the inequality
\[|X(t)X^{-1}(s)|\le K e^{\beta(t-s)}, ~~~\forall t \in J,~~~t\ge s,
\]
where \(\beta=\alpha+(\delta K \log K)^{1/2}\).
我们证明命题5:
固定\(t\in J\), 命题5的条件\(\textbf{(i)-(ii)}\)满足命题3的条件\(\textbf{(i)-(ii)}\), 因此利用命题3我们得到, 在命题3中取\(\varepsilon=\varepsilon/2\),
\[|e^{\tau A(t)}|\le \max\left\{\left(\frac{4M}{\varepsilon}
\right)^{n-1},1\right\}e^{(\alpha+\varepsilon/2)\tau}=:\tilde{K}e^{\tilde{\alpha}\tau},~~~\tau\ge 0,~~~t\in J.
\]
它结合命题5的条件\(\textbf{(iii)}\)验证了命题4的条件，利用命题4我们得到了这个方程\(\dot{x}=A(t)x\)的基本解矩阵满足
\[|X(t)X^{-1}(s)|\le \tilde{K} e^{\tilde{\beta}(t-s)}.
\]
其中\(\tilde{\beta}=\tilde{\alpha}+(\delta \tilde{K}\log \tilde{K})^{1/2}\). 最后为了保证
\[\tilde{\beta}=\tilde{\alpha}+(\delta \tilde{K}\log \tilde{K})^{1/2}=\alpha+\varepsilon/2+(\delta \tilde{K}\log \tilde{K})^{1/2}\le \alpha+\varepsilon
\]
我们需要 \(\delta\le \delta(M,\varepsilon),\) 其中
\[\delta(M,\varepsilon)=\frac{\varepsilon^2}{4\tilde{K}\log \tilde{K}}=\frac{\varepsilon^2}{4\max\left\{\left(\frac{4M}{\varepsilon}
\right)^{n-1},1\right\}\log \left(\max\left\{\left(\frac{4M}{\varepsilon}
\right)^{n-1},1\right\}\right)}.
\]
命题5下面一句话：当\(\alpha<0\)时, 取\(\varepsilon>0\) 充分小使得\(\alpha+\varepsilon<0\), 然后确定了\(\delta(M,\varepsilon)\). 再让\(\delta<\delta(M,\varepsilon)\), 那么当
\[|A(t_2)-A(t_1)|\le \delta|t_2-t_1|
\]
时, 方程\(\dot{x}=A(t)x\)的基本解矩阵都满足
\[|X(t)X(s)^{-1}|\le Ke^{\beta(t-s)}, ~~t\ge s,~~K>0,~~~\beta<0.
\]
因此一致渐近稳定.
