[T050201] 求方程组 \(\begin{cases}x'+y'=y+z\\ y'+z'=z+x\\ z'+x'=x+y\end{cases}\) 的通解.
解 将方程组写成矩阵形式:
\[\begin{aligned}
&\begin{pmatrix}
1&1&0\\0&1&1\\1&0&1
\end{pmatrix}
\begin{pmatrix}
x'\\y'\\z'
\end{pmatrix}=
\begin{pmatrix}
0&1&1\\1&0&1\\1&1&0
\end{pmatrix}
\begin{pmatrix}
x\\y\\z
\end{pmatrix}\\ \\
\Longrightarrow
&\begin{pmatrix}
x'\\y'\\z'
\end{pmatrix}=\begin{pmatrix}
1&1&0\\0&1&1\\1&0&1
\end{pmatrix}^{-1}\begin{pmatrix}
0&1&1\\1&0&1\\1&1&0
\end{pmatrix}\begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
0&1&0\\0&0&1\\1&0&0
\end{pmatrix}\begin{pmatrix}
x\\y\\z
\end{pmatrix}
:=A\begin{pmatrix}
x\\y\\z
\end{pmatrix}
\end{aligned}
\]
则 \(|\lambda E-A|=\begin{vmatrix}\lambda&-1&0\\0&\lambda&-1\\-1&0&\lambda\end{vmatrix}=\lambda^3-1=0\Longrightarrow \lambda_1=1,\lambda_2=-\frac{1}{2}+\frac{\sqrt3}{2}i,\lambda_3=-\frac{1}{2}-\frac{\sqrt3}{2}i\).
对于特征值 \(\lambda_1=1\), 有
\[A-\lambda_1E=
\begin{pmatrix}
-1&1&0\\0&-1&1\\1&0&-1
\end{pmatrix}\xrightarrow{初等行变换}
\begin{pmatrix}
1&0&-1\\0&1&-1\\0&0&0
\end{pmatrix}
\]
于是可取特征向量 \(v_1=(1,1,1)^T\), 从而常系数齐次线性方程组有一个解 \(X_1(t)=e^tv_1\).
对于特征值 \(\lambda_2=-\frac{1}{2}+\frac{\sqrt3}{2}i\), 有
\[A-\lambda_2E=
\begin{pmatrix}
\frac12-\frac{\sqrt{3}}{2}i&1&0\\0&\frac12-\frac{\sqrt3}{2}i&1\\1&0&\frac12-\frac{\sqrt{3}}{2}i
\end{pmatrix}
\xrightarrow{初等行变换}
\begin{pmatrix}
1&0&\frac12-\frac{\sqrt3}{2}i\\0&1&\frac12+\frac{\sqrt3}{2}i\\0&0&0
\end{pmatrix}
\]
于是可取特征向量 \(v_2=\left(-\frac12-\frac{\sqrt3}{2}i,-\frac12+\frac{\sqrt3}{2}i,1\right)^T\). 注意到
\[\begin{aligned}
e^{(-\frac{1}{2}+\frac{\sqrt3}{2}i)t}v_2&=e^{-\frac12t}
\begin{pmatrix}
(-\frac{1}{2}+\frac{\sqrt3}{2}i)e^{\frac{\sqrt3}{2}it}\\
(-\frac{1}{2}-\frac{\sqrt3}{2}i)e^{\frac{\sqrt3}{2}it}\\
e^{\frac{\sqrt3}{2}it}
\end{pmatrix}\\
&=e^{-\frac12t}
\left\{
\begin{pmatrix}
-\frac12\cos\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\
-\frac12\cos\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\
\cos\frac{\sqrt3}{2}t
\end{pmatrix}+i
\begin{pmatrix}
-\frac12\sin\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\
-\frac12\sin\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\
\sin\frac{\sqrt3}{2}t
\end{pmatrix}
\right\}
\end{aligned}
\]
于是另两个线性无关的解为
\[X_2(t)=e^{-\frac12t}\begin{pmatrix}
-\frac12\cos\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\
-\frac12\cos\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\
\cos\frac{\sqrt3}{2}t
\end{pmatrix},\quad
X_3(t)=e^{-\frac12t}\begin{pmatrix}
-\frac12\sin\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\
-\frac12\sin\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\
\sin\frac{\sqrt3}{2}t
\end{pmatrix}
\]
注意到 \(X_1(t),X_2(t),X_3(t)\) 是常系数齐次线性方程组的三个线性无关解, 故通解为
\[\begin{aligned}
X(t)&=c_1X_1(t)+c_2X_2(t)+c_3X_3(t)\\
&=c_1e^t\begin{pmatrix}
1\\1\\1
\end{pmatrix}+
c_2e^{-\frac12t}\begin{pmatrix}
-\frac12\cos\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\
-\frac12\cos\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\
\cos\frac{\sqrt3}{2}t
\end{pmatrix}+
c_3e^{-\frac12t}\begin{pmatrix}
-\frac12\sin\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\
-\frac12\sin\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\
\sin\frac{\sqrt3}{2}t
\end{pmatrix}
\end{aligned}
\]
其中 \(c_1,c_2,c_3\) 为任意常数. #
[T050202] 求常系数非齐次线性方程组的通解:
\[X'=\begin{pmatrix}
-1&2\\-2&3
\end{pmatrix}X+\begin{pmatrix}
1\\0
\end{pmatrix}.
\]
解 记 \(X'=AX+F\), 于是
\[|\lambda E-A|=\begin{vmatrix}
\lambda+1&-2\\2&\lambda-3
\end{vmatrix}=(\lambda-1)^2\Longrightarrow \lambda=1 \ (二 \ 重)
\]
从而对应的常系数齐次线性方程组有两个解形如 \(X(t)=(R_0+R_1t)e^t\), 其中 \(R_0,R_1\) 满足
\[\begin{cases}
(A-E)R_0=R_1\\
(A-E)^2R_0=0
\end{cases}
\]
注意到
\[(A-E)^2=\begin{pmatrix}
-2&2\\-2&2
\end{pmatrix}^2=
O
\]
故可取 \(R_0=(1,0)^T\) 和 \((0,1)^T\), 从而 \(R_1=(-2,-2)^T\) 和 \((2,2)\), 从而对应的常系数齐次线性方程组有两个线性无关解
\[X_1(t)=\left[
\begin{pmatrix}
1\\0
\end{pmatrix}+
\begin{pmatrix}
-2\\-2
\end{pmatrix}t
\right]e^t=e^t\begin{pmatrix}
1-2t\\-2t
\end{pmatrix}\\ \\
X_2(t)=\left[
\begin{pmatrix}
0\\1
\end{pmatrix}+
\begin{pmatrix}
2\\2
\end{pmatrix}t
\right]e^t=e^t\begin{pmatrix}
2t\\1+2t
\end{pmatrix}
\]
由此可得对应的常系数齐次线性方程组的基解矩阵为 \(\Phi(t)=e^t\begin{pmatrix}
1-2t&2t\\-2t&1+2t
\end{pmatrix}\),\(\Phi^{-1}(t)=e^{-t}\begin{pmatrix}
1+2t&-2t\\2t&1-2t
\end{pmatrix}\), 于是原常系数非齐次线性方程组有特解
\[\begin{aligned}
\bar X(t)&=\Phi(t)\int_0^t\Phi^{-1}(s)F\mathrm{~d}s=e^t\begin{pmatrix}
1-2t&2t\\-2t&1+2t
\end{pmatrix}\begin{pmatrix}
-3e^{-t}-2te^{-t}+3\\-2e^{-t}-2te^{-t}+2
\end{pmatrix}\\
&=e^t\begin{pmatrix}
-3e^{-t}+3-2t\\-2e^{-t}+2-2t
\end{pmatrix}=\begin{pmatrix}
-3+(1-2t)e^t+2e^t\\-2+(-2t)e^t+2e^t
\end{pmatrix}
\end{aligned}
\]
故原方程组的通解
\[\begin{aligned}
X(t)&=c_1X_1(t)+c_2X_2(t)+\bar X(t)\\
&=c_1e^t\begin{pmatrix}
1-2t\\-2t
\end{pmatrix}+c_2e^t\begin{pmatrix}
2t\\1+2t
\end{pmatrix}+\begin{pmatrix}
-3+(1-2t)e^t+2e^t\\-2+(-2t)e^t+2e^t
\end{pmatrix}\\
&=c_1e^t\begin{pmatrix}
1-2t\\-2t
\end{pmatrix}+c_2e^t\begin{pmatrix}
2t\\1+2t
\end{pmatrix}+\begin{pmatrix}
-3+2e^t\\-2+2e^t
\end{pmatrix}\\
&=c_1e^t\begin{pmatrix}
1-2t\\-2t
\end{pmatrix}+c_2e^t\begin{pmatrix}
2t\\1+2t
\end{pmatrix}+\begin{pmatrix}
-3\\-2
\end{pmatrix}. \quad\quad\#
\end{aligned}
\]
