非齐次矩阵指数方程求解：一阶线性非齐次微分方程组问询

Alright, let's work through solving this first-order linear nonhomogeneous system of ODEs step by step. I'll break it down into clear parts so you can follow along easily.

Problem Statement #

We need to solve:
$$\frac{d}{du} \begin{bmatrix}a\b \end{bmatrix} = \begin{bmatrix}-x& y\ -y&-x\end{bmatrix} \begin{bmatrix}a\ b\end{bmatrix} + \begin{bmatrix}\cos(zu)\ -\sin(zu)\end{bmatrix}$$
with initial condition $\begin{bmatrix}a(0)\b(0) \end{bmatrix} = \begin{bmatrix}0\0 \end{bmatrix}$.

Step 1: Solve the Homogeneous System #

First, let's handle the homogeneous part:
$$\frac{d}{du}\begin{bmatrix}a\b\end{bmatrix} = P\begin{bmatrix}a\b\end{bmatrix}, \quad P = \begin{bmatrix}-x& y\ -y&-x\end{bmatrix}$$

Notice we can rewrite $P$ as $-xI + Q$, where $I$ is the identity matrix and $Q = \begin{bmatrix}0&y\-y&0\end{bmatrix}$. Since $-xI$ and $Q$ commute, the matrix exponential $e^{Pu}$ can be split as:
$$e^{Pu} = e^{-xu}e{Qu}$$

For the anti-symmetric matrix $Q$, we use the Taylor series expansion of the matrix exponential (and the fact that $Q^2 = -y^2I$, $Q^3=-y2Q$, etc.):
$$e^{Qu} = \cos(yu)I + \frac{\sin(yu)}{y}Q$$

Substituting back, we get:
$$e^{Pu} = e^{-xu}\begin{bmatrix}\cos(yu) & \sin(yu)\ -\sin(yu) & \cos(yu)\end{bmatrix}$$
This works even when $y=0$ (since $\sin(yu)/y \to u$ as $y\to0$, but $Q$ becomes zero, so $e^{Qu}=I$ which matches).

Step 2: Variation of Parameters for Nonhomogeneous Solution #

The general solution to a linear nonhomogeneous system $\frac{d\mathbf{u}}{du}=P\mathbf{u}+\mathbf{f}(u)$ is:
$$\mathbf{u}(u) = e^{Pu}\mathbf{C} + e^{Pu}\int_0u e^{-Ps}\mathbf{f}(s)ds$$

Applying our initial condition $\mathbf{u}(0)=\mathbf{0}$: when $u=0$, $e^{P0}=I$, so $\mathbf{C}=\mathbf{0}$. Our solution simplifies to:
$$\begin{bmatrix}a(u)\b(u)\end{bmatrix} = e^{Pu}\int_0u e^{-Ps}\begin{bmatrix}\cos(zs)\ -\sin(zs)\end{bmatrix}ds$$

First compute $e^{-Ps}$ (the inverse of $e^{Ps}$):
$$e^{-Ps} = e^{xs}\begin{bmatrix}\cos(ys) & -\sin(ys)\ \sin(ys) & \cos(ys)\end{bmatrix}$$

Multiply this by the nonhomogeneous term $\mathbf{f}(s)$:
$$e^{-Ps}\mathbf{f}(s) = e^{xs}\begin{bmatrix}\cos(ys)\cos(zs) + \sin(ys)\sin(zs)\ \sin(ys)\cos(zs) - \cos(ys)\sin(zs)\end{bmatrix}$$

Using trigonometric identities ($\cos(A-B)=\cos A\cos B+\sin A\sin B$, $\sin(A-B)=\sin A\cos B-\cos A\sin B$), this simplifies to:
$$e^{-Ps}\mathbf{f}(s) = e^{xs}\begin{bmatrix}\cos((y-z)s)\ \sin((y-z)s)\end{bmatrix}$$

Step 3: Evaluate the Integral #

We need to compute the integral of each component. Use the standard integrals for exponential-trigonometric products:

• $\int e^{ks}\cos(ms)ds = \frac{e^{ks}(k\cos ms + m\sin ms)}{k^2+m2} + C$
• $\int e^{ks}\sin(ms)ds = \frac{e^{ks}(k\sin ms - m\cos ms)}{k^2+m2} + C$

Here, $k=x$ and $m=y-z$. Let $D = x^2 + (y-z)^2$ (the denominator term). Evaluating from $0$ to $u$:

1. First component integral:
   $$\int_0^u e^{xs}\cos((y-z)s)ds = \frac{e^{xu}\left(x\cos((y-z)u) + (y-z)\sin((y-z)u)\right) - x}{D}$$

2. Second component integral:
   $$\int_0^u e^{xs}\sin((y-z)s)ds = \frac{e^{xu}\left(x\sin((y-z)u) - (y-z)\cos((y-z)u)\right) + (y-z)}{D}$$

Step 4: Combine Results to Get the Solution #

Multiply the integral vector by $e^{Pu}$ (which includes the $e^{-xu}$ factor that cancels the $e^{xu}$ in the integral results). After expanding and simplifying using trigonometric identities again, we get the final solution for $y \neq z$:

Solution for $y \neq z$ #

$$
a(u) = \frac{x\left(\cos(zu) - \cos(yu)\right) + (y-z)\left(\sin(yu) - \sin(zu)\right)}{x^2 + (y-z)^2}
$$
$$
b(u) = \frac{x\left(\sin(yu) - \sin(zu)\right) + (y-z)\left(\cos(yu) - \cos(zu)\right)}{x^2 + (y-z)^2}
$$

Special Case: $y = z$ (Resonance) #

When $y=z$, the denominator $D=x^2$, and the nonhomogeneous term matches the frequency of the homogeneous solution (resonance). Recomputing the integral directly (since $m=0$ now):
$$\int_0^u e^{xs}\begin{bmatrix}1\0\end{bmatrix}ds = \begin{bmatrix}\frac{e^{xu}-1}{x}\0\end{bmatrix}$$

Multiplying by $e^{Pu}$ gives:
$$
a(u) = \frac{(1 - e^{-xu})\cos(zu)}{x}
$$
$$
b(u) = \frac{(e^{-xu} - 1)\sin(zu)}{x}
$$

You can verify this matches the limit of the general solution as $y\to z$ using L'Hospital's Rule, but direct computation is simpler here.

内容的提问来源于stack exchange，提问作者nomad609