从离散到连续

上一讲解 $u_{k+1}=Au_k$，解里出现 $\lambda^k$。这一讲解

$$\frac{du}{dt} = Au$$

常数系数线性微分方程的解是指数函数，所以答案里出现 $e^{\lambda t}$。两套理论完全平行：幂换成指数，特征值仍然决定一切。

时间轴

一个完整的二维例子

设

$$A=\begin{bmatrix}-1&2\\1&-2\end{bmatrix}, \qquad u(0)=\begin{bmatrix}1\\0\end{bmatrix}$$

先求特征值

矩阵奇异——第二列是第一列的 $-2$ 倍——所以 $\lambda_1=0$。迹为 $-3$，因此 $\lambda_2=-3$。

也可以用特征多项式验证：

$$\det(A-\lambda I) = \begin{vmatrix}-1-\lambda&2\\1&-2-\lambda\end{vmatrix} = \lambda^2+3\lambda = \lambda(\lambda+3)$$

再求特征向量

$\lambda=0$：零空间满足 $-u_1+2u_2=0$，取

$$x_1=\begin{bmatrix}2\\1\end{bmatrix}$$

$\lambda=-3$：解 $(A+3I)x=0$，其中

$$A+3I=\begin{bmatrix}2&2\\1&1\end{bmatrix}$$

零空间由

$$x_2=\begin{bmatrix}1\\-1\end{bmatrix}$$

张成（取 $[-1,1]$ 也可以）。

写出通解

两个纯指数解是 $e^{0t}x_1$ 与 $e^{-3t}x_2$，通解为它们的线性组合：

$$u(t)=c_1 e^{0t}\begin{bmatrix}2\\1\end{bmatrix} + c_2 e^{-3t}\begin{bmatrix}1\\-1\end{bmatrix}$$

为什么它满足方程？对第一项求导得 $\lambda_1 e^{\lambda_1 t}x_1$，而 $Au$ 中对应项是 $e^{\lambda_1 t}Ax_1=\lambda_1 e^{\lambda_1 t}x_1$，两边相等。第二项同理。

用初值定系数

在 $t=0$：

$$c_1\begin{bmatrix}2\\1\end{bmatrix} + c_2\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}$$

即

$$2c_1+c_2=1, \qquad c_1-c_2=0$$

所以 $c_1=c_2=\frac13$，解为

$$u(t)=\frac13\begin{bmatrix}2\\1\end{bmatrix} + \frac13 e^{-3t}\begin{bmatrix}1\\-1\end{bmatrix}$$

读懂这个解

• 第一项不随时间变化，是稳态。
• 第二项以 $e^{-3t}$ 衰减，最终消失。
• 当 $t\to\infty$，$u\to\left[\frac23,\frac13\right]^T$。初始的 $[1,0]$ 有一部分「流」到了第二个分量。

这正是特征值在开头就预告的行为：一个零特征值给出稳态，一个负特征值给出衰减。

稳定、稳态与发散

为什么看实部而不是模

若 $\lambda=-3+6i$，则

$$e^{\lambda t}=e^{-3t}e^{6i t}$$

$e^{6it}=\cos 6t+i\sin 6t$ 的模恒为 1，只在单位圆上转。决定增长或衰减的只有实部 $-3$。虚部只带来振荡。

所以：

• 差分方程的稳定区域是复平面上的单位圆内（$|\lambda|<1$）。
• 微分方程的稳定区域是左半平面（$\operatorname{Re}\lambda<0$）。

二阶矩阵的快速判据

对 $2\times2$ 矩阵，两个特征值实部都为负，等价于

$$\operatorname{tr}A<0 \quad \text{且} \quad \det A>0$$

迹是特征值之和，行列式是乘积。两个实部为负的数，和必为负、积必为正；复共轭情形同理。

单独一个条件不够。反例：

$$\begin{bmatrix}-2&0\\0&1\end{bmatrix}$$

迹为 $-1<0$，但特征值是 $-2$ 与 $1$，仍然发散。所以行列式条件不能省。

用 S 和 Λ 解耦

把 $u=Sv$ 代入 $\frac{du}{dt}=Au$：

$$S\frac{dv}{dt}=ASv$$

左乘 $S^{-1}$：

$$\frac{dv}{dt}=S^{-1}AS\,v=\Lambda v$$

系统变成 $n$ 个独立方程：

$$\frac{dv_i}{dt}=\lambda_i v_i$$

每个解是 $v_i(t)=e^{\lambda_i t}v_i(0)$，即

$$v(t)=e^{\Lambda t}v(0)$$

回到原变量：

$$u(t)=S e^{\Lambda t}S^{-1}u(0)$$

这解释了为什么 $S$ 的列必须是特征向量：换到特征向量基下，系统没有耦合。

矩阵指数

定义

把标量指数函数的幂级数搬到矩阵上：

$$e^{At}=I+At+\frac{(At)^2}{2!}+\frac{(At)^3}{3!}+\cdots$$

这个级数对任何矩阵、任何 $t$ 都收敛，因为分母的阶乘增长得足够快。

与对角化的关系

把 $A=S\Lambda S^{-1}$ 代入每一项：

$$(At)^k=S(\Lambda t)^k S^{-1}$$

级数逐项提出 $S$ 与 $S^{-1}$，中间剩下

$$I+\Lambda t+\frac{(\Lambda t)^2}{2!}+\cdots=e^{\Lambda t}$$

所以

$$e^{At}=S e^{\Lambda t}S^{-1}$$

对角矩阵的指数只需对每个对角元取指数：

$$e^{\Lambda t}= \begin{bmatrix} e^{\lambda_1 t} & & \\ & \ddots & \\ & & e^{\lambda_n t} \end{bmatrix}$$

前提

这个推导要求 $A$ 可对角化。特征向量不足的矩阵没有 $S^{-1}$，需要改用 Jordan 形式或其他方法。课堂上明确把这一类留到后面处理。

把高阶方程化成一阶系统

考虑二阶方程

$$y''+by'+ky=0$$

令

$$u=\begin{bmatrix}y'\\y\end{bmatrix}$$

则

$$u'=\begin{bmatrix}y''\\y'\end{bmatrix} = \begin{bmatrix}-b&-k\\1&0\end{bmatrix} \begin{bmatrix}y'\\y\end{bmatrix}$$

第一行来自原方程，第二行是恒等式 $y'=y'$。特征值满足

$$\lambda^2+b\lambda+k=0$$

与原方程的特征方程一致。这个方法可以推广到任意阶：把最高阶导数放第一行，其余行放移位单位矩阵。

核心回顾

• $\frac{du}{dt}=Au$ 的解是特征向量乘 $e^{\lambda t}$ 的线性组合。
• 通解的系数由初值 $Sc=u(0)$ 决定。
• 稳定看实部：全负趋于零，有零且其余为负趋于稳态，有正则发散。
• $2\times2$ 稳定判据：迹为负且行列式为正。
• 换基到特征向量后系统解耦，得到 $u=Se^{\Lambda t}S^{-1}u(0)$。
• 矩阵指数由幂级数定义，$e^{At}=Se^{\Lambda t}S^{-1}$，要求可对角化。

常见误区

1. 用特征值的模判断连续系统的稳定性。要看实部；模对应的是离散系统。
2. 忽略虚部。虚部不改变增长趋势，但会产生振荡，解不是单调的。
3. 只检查迹为负。还必须检查行列式为正。
4. 把矩阵指数当成逐元素取指数。$e^{At}$ 不是对每个元素求 $e^{a_{ij}t}$。
5. 忘记初始条件的作用。特征值决定行为类型，初值决定各分量占多少。

理解检验

1. 为什么 $\lambda=0$ 对应稳态而不是衰减？

参考答案：对应分量是 $e^{0\cdot t}x=e^{0}x=x$，即 $e^{\lambda t}=1$ 对一切 $t$ 成立，大小和方向都不变。它落在零空间里（$Ax=0$），所以这个模式被「保存」下来，形成不随时间变化的常数项。只有当 $\operatorname{Re}\lambda<0$ 时指数才衰减；$\lambda=0$ 恰好处在稳定与发散的边界上。

2. 若特征值是 $-1\pm 4i$，解会发散吗？会振荡吗？

参考答案：不会发散，会振荡。对复数特征值，

$$e^{(-1+4i)t}=e^{-t}e^{4it}=e^{-t}(\cos 4t+i\sin 4t),$$

模为 $e^{-t}$，实部 $-1<0$ 使它随时间衰减到零；虚部 $\pm4$ 只是让解以角频率 $4$ 旋转、产生振荡，不改变增长趋势。所以解是衰减振荡（螺旋收敛）。

3. 用迹与行列式判断 $\begin{bmatrix}-1&2\\1&-2\end{bmatrix}$ 是否稳定。

参考答案：$\operatorname{tr}A=-1+(-2)=-3<0$，$\det A=(-1)(-2)-2\cdot1=2-2=0$。稳定判据要求 $\operatorname{tr}<0$ 且 $\det>0$；这里 $\det=0$，判据不满足，所以它不是渐近稳定的。直接求特征值：特征多项式为 $\lambda^2+3\lambda=\lambda(\lambda+3)$，$\lambda_1=0$、$\lambda_2=-3$。零特征值给出稳态，负特征值给出衰减，因此解有界并收敛到非零的稳态向量（正文中为 $\left[\frac23,\frac13\right]^T$），而不是收敛到零。

4. 写出 $y''+3y'+2y=0$ 对应的一阶系统与特征方程。

参考答案：取 $u=\begin{bmatrix}y'\\y\end{bmatrix}$，则

$$u'=\begin{bmatrix}y''\\y'\end{bmatrix} = \begin{bmatrix}-3&-2\\1&0\end{bmatrix} \begin{bmatrix}y'\\y\end{bmatrix}.$$

特征方程为

$$\det\begin{bmatrix}-3-\lambda&-2\\1&-\lambda\end{bmatrix} =\lambda^2+3\lambda+2=(\lambda+1)(\lambda+2)=0,$$

所以 $\lambda_1=-1$、$\lambda_2=-2$。两个特征值都为负，解单调衰减到零。

5. 推导 $e^{At}=Se^{\Lambda t}S^{-1}$ 时，哪一步用到了 $S$ 可逆？

参考答案：第一步就把 $A$ 写成 $A=S\Lambda S^{-1}$，这需要 $S^{-1}$ 存在；等价地，需要 $A$ 有 $n$ 个线性无关的特征向量。代入幂级数后，$(At)^k=S(\Lambda t)^kS^{-1}$，相邻项之间的 $S^{-1}S$ 才能消掉，级数才能整理成 $S\left(\sum(\Lambda t)^k/k!\right)S^{-1}$。若 $A$ 不可对角化，就没有这样的 $S$，必须改用 Jordan 形式或别的方法。

与前后课程的关系

上一讲是离散时间，这一讲是连续时间。接下来课程转向对称矩阵：它们的特征值必为实数、特征向量必正交，因此一定可以对角化，而且分解是正交的——这解决了本章遗留的「坏情况」中的一大类。

From discrete to continuous

The previous lecture solved $u_{k+1}=Au_k$, with $\lambda^k$ in the answer. This one solves

$$\frac{du}{dt} = Au$$

Solutions of constant-coefficient linear equations are exponentials, so $e^{\lambda t}$ appears. The two theories run in parallel: powers become exponentials, and eigenvalues still decide the behavior.

Timeline

A complete two-dimensional example

Let

$$A=\begin{bmatrix}-1&2\\1&-2\end{bmatrix}, \qquad u(0)=\begin{bmatrix}1\\0\end{bmatrix}$$

Eigenvalues first

The matrix is singular — the second column is $-2$ times the first — so $\lambda_1=0$. The trace is $-3$, so $\lambda_2=-3$.

The characteristic polynomial confirms it:

$$\det(A-\lambda I) = \begin{vmatrix}-1-\lambda&2\\1&-2-\lambda\end{vmatrix} = \lambda^2+3\lambda = \lambda(\lambda+3)$$

Then eigenvectors

For $\lambda=0$, the null space satisfies $-u_1+2u_2=0$, so take

$$x_1=\begin{bmatrix}2\\1\end{bmatrix}$$

For $\lambda=-3$, solve $(A+3I)x=0$ with

$$A+3I=\begin{bmatrix}2&2\\1&1\end{bmatrix}$$

whose null space is spanned by

$$x_2=\begin{bmatrix}1\\-1\end{bmatrix}$$

(using $[-1,1]$ is equally fine).

The general solution

The two pure exponential solutions are $e^{0t}x_1$ and $e^{-3t}x_2$, and the general solution is their combination:

$$u(t)=c_1 e^{0t}\begin{bmatrix}2\\1\end{bmatrix} + c_2 e^{-3t}\begin{bmatrix}1\\-1\end{bmatrix}$$

Why does it solve the equation? Differentiate the first term: $\lambda_1 e^{\lambda_1 t}x_1$. On the other side, $Au$ contains $e^{\lambda_1 t}Ax_1=\lambda_1 e^{\lambda_1 t}x_1$. The two sides match; the second term works the same way.

Fixing the coefficients

At $t=0$:

$$c_1\begin{bmatrix}2\\1\end{bmatrix} + c_2\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}$$

so

$$2c_1+c_2=1, \qquad c_1-c_2=0$$

Hence $c_1=c_2=\frac13$, and

$$u(t)=\frac13\begin{bmatrix}2\\1\end{bmatrix} + \frac13 e^{-3t}\begin{bmatrix}1\\-1\end{bmatrix}$$

Reading the answer

• The first term never changes: a steady state.
• The second decays as $e^{-3t}$ and disappears.
• As $t\to\infty$, $u\to\left[\frac23,\frac13\right]^T$. Part of the initial $[1,0]$ has flowed into the second component.

The eigenvalues predicted this at the outset: a zero eigenvalue produces the steady state, a negative one produces decay.

Stability, steady state, blow-up

Why the real part, not the modulus

If $\lambda=-3+6i$, then

$$e^{\lambda t}=e^{-3t}e^{6i t}$$

and $e^{6it}=\cos 6t+i\sin 6t$ has modulus 1, circling the unit circle. Only the real part $-3$ decides growth or decay; the imaginary part produces oscillation.

So:

• For difference equations, the stable region is the unit disk ($|\lambda|<1$).
• For differential equations, it is the left half-plane ($\operatorname{Re}\lambda<0$).

A quick test for $2 \times 2$

Both eigenvalues have negative real part exactly when

$$\operatorname{tr}A<0 \quad \text{and} \quad \det A>0$$

The trace is the sum and the determinant is the product. Two numbers with negative real parts have a negative sum and a positive product; the complex-conjugate case behaves the same way.

One condition alone is not enough. Counterexample:

$$\begin{bmatrix}-2&0\\0&1\end{bmatrix}$$

The trace is $-1<0$, but the eigenvalues are $-2$ and $1$, so the system blows up. The determinant condition is essential.

Decoupling with $S$ and $\Lambda$

Substitute $u=Sv$ into $\frac{du}{dt}=Au$:

$$S\frac{dv}{dt}=ASv$$

Left-multiply by $S^{-1}$:

$$\frac{dv}{dt}=S^{-1}AS\,v=\Lambda v$$

The system becomes $n$ independent equations:

$$\frac{dv_i}{dt}=\lambda_i v_i$$

Each solves as $v_i(t)=e^{\lambda_i t}v_i(0)$, so

$$v(t)=e^{\Lambda t}v(0)$$

and in the original variables

$$u(t)=S e^{\Lambda t}S^{-1}u(0)$$

This is why the columns of $S$ must be eigenvectors: in the eigenvector basis the system has no coupling.

The matrix exponential

Definition

Move the scalar power series to matrices:

$$e^{At}=I+At+\frac{(At)^2}{2!}+\frac{(At)^3}{3!}+\cdots$$

The series converges for every matrix and every $t$, because the factorial in the denominator grows fast enough.

Relation to diagonalization

Substitute $A=S\Lambda S^{-1}$ into each term:

$$(At)^k=S(\Lambda t)^k S^{-1}$$

Factor $S$ out of the whole series on the left and $S^{-1}$ on the right; what remains is

$$I+\Lambda t+\frac{(\Lambda t)^2}{2!}+\cdots=e^{\Lambda t}$$

Therefore

$$e^{At}=S e^{\Lambda t}S^{-1}$$

The exponential of a diagonal matrix is just the diagonal of scalar exponentials:

$$e^{\Lambda t}= \begin{bmatrix} e^{\lambda_1 t} & & \\ & \ddots & \\ & & e^{\lambda_n t} \end{bmatrix}$$

The assumption

This derivation requires $A$ to be diagonalizable. A matrix with too few eigenvectors has no $S^{-1}$ and needs Jordan form or another method. The lecture explicitly leaves that case aside.

Turning a higher-order equation into a first-order system

Consider

$$y''+by'+ky=0$$

Let

$$u=\begin{bmatrix}y'\\y\end{bmatrix}$$

Then

$$u'=\begin{bmatrix}y''\\y'\end{bmatrix} = \begin{bmatrix}-b&-k\\1&0\end{bmatrix} \begin{bmatrix}y'\\y\end{bmatrix}$$

The first row comes from the equation; the second is the identity $y'=y'$. The eigenvalues satisfy

$$\lambda^2+b\lambda+k=0$$

which is the original characteristic equation. The trick generalizes to any order: the coefficient row goes first, and shifted identity rows fill the rest.

Core review

• The solution of $\frac{du}{dt}=Au$ is a combination of eigenvectors times $e^{\lambda t}$.
• The coefficients come from the initial condition via $Sc=u(0)$.
• Stability is about the real part: all negative decays, one zero and the rest negative gives a steady state, any positive blows up.
• The $2\times2$ stability test is negative trace and positive determinant.
• In the eigenvector basis the system decouples, giving $u=Se^{\Lambda t}S^{-1}u(0)$.
• The matrix exponential is defined by a power series, and $e^{At}=Se^{\Lambda t}S^{-1}$ when $A$ diagonalizes.

Common misconceptions

1. Using the modulus for continuous-time stability. Look at the real part; the modulus belongs to discrete time.
2. Ignoring the imaginary part. It does not change growth, but it produces oscillation rather than monotone decay.
3. Checking only a negative trace. The determinant must be positive as well.
4. Treating $e^{At}$ as elementwise exponentiation. It is not $e^{a_{ij}t}$ entry by entry.
5. Forgetting the initial condition. Eigenvalues determine the type of behavior; the initial vector determines how much of each mode is present.

Check your understanding

1. Why does $\lambda=0$ produce a steady state rather than decay?

Reference answer: The corresponding mode is $e^{0\cdot t}x=e^{0}x=x$, so $e^{\lambda t}=1$ for all $t$ and neither magnitude nor direction changes. It lies in the nullspace ($Ax=0$), so the mode is preserved as a constant term. Decay requires $\operatorname{Re}\lambda<0$; $\lambda=0$ sits exactly on the boundary between stability and divergence.

2. If the eigenvalues are $-1\pm4i$, does the solution blow up? Does it oscillate?

Reference answer: It does not blow up; it oscillates. For complex eigenvalues,

$$e^{(-1+4i)t}=e^{-t}e^{4it}=e^{-t}(\cos 4t+i\sin 4t),$$

whose modulus is $e^{-t}$; the real part $-1<0$ drives it to zero. The imaginary part $\pm4$ only rotates the solution at angular frequency $4$, producing oscillation without changing the growth trend. The solution is a decaying spiral.

3. Use trace and determinant to decide whether $\begin{bmatrix}-1&2\\1&-2\end{bmatrix}$ is stable.

Reference answer: $\operatorname{tr}A=-1+(-2)=-3<0$ and $\det A=(-1)(-2)-2\cdot1=2-2=0$. The stability test requires $\operatorname{tr}<0$ and $\det>0$; here $\det=0$, so the test fails and the system is not asymptotically stable. Computing directly: the characteristic polynomial is $\lambda^2+3\lambda=\lambda(\lambda+3)$, so $\lambda_1=0$ and $\lambda_2=-3$. The zero eigenvalue gives a steady state and the negative one decays, so the solution stays bounded and converges to a nonzero steady vector (in the text, $\left[\tfrac23,\tfrac13\right]^T$) rather than to zero.

4. Write the first-order system and characteristic equation for $y''+3y'+2y=0$.

Reference answer: Set $u=\begin{bmatrix}y'\\y\end{bmatrix}$; then

$$u'=\begin{bmatrix}y''\\y'\end{bmatrix} = \begin{bmatrix}-3&-2\\1&0\end{bmatrix} \begin{bmatrix}y'\\y\end{bmatrix}.$$

The characteristic equation is

$$\det\begin{bmatrix}-3-\lambda&-2\\1&-\lambda\end{bmatrix} =\lambda^2+3\lambda+2=(\lambda+1)(\lambda+2)=0,$$

so $\lambda_1=-1$, $\lambda_2=-2$. Both are negative, and the solution decays monotonically to zero.

5. In the derivation of $e^{At}=Se^{\Lambda t}S^{-1}$, which step uses invertibility of $S$?

Reference answer: The very first step writes $A=S\Lambda S^{-1}$, which needs $S^{-1}$ to exist — equivalently, $A$ must have $n$ linearly independent eigenvectors. Substituting into the power series gives $(At)^k=S(\Lambda t)^kS^{-1}$, so the adjacent factors $S^{-1}S$ cancel and the series collapses to $S\left(\sum(\Lambda t)^k/k!\right)S^{-1}$. Without diagonalizability there is no such $S$, and one must use Jordan form or another method.

How this connects

The previous lecture handled discrete time; this one handles continuous time. The course next turns to symmetric matrices: their eigenvalues are real and their eigenvectors orthogonal, so they always diagonalize, and the decomposition is orthogonal. That resolves a large part of the chapter's "bad cases".