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举例¶
求解如下微分线性方程组的通解
\[\left\{ {\begin{array}{lll}
{\frac{{{\rm{d}}{x_1}}}{{{\rm{d}}t}} = 2{x_2} + 2{x_3}}\\
{\frac{{{\rm{d}}{x_2}}}{{{\rm{d}}t}} = 2{x_1} + {x_2} + 2{x_3}}\\
{\frac{{{\rm{d}}{x_3}}}{{{\rm{d}}t}} = 2{x_2} + {x_3}}
\end{array}} \right.
\]
解:
写成矩阵形式 \(\frac{{\rm d}{\bm x}}{{\rm d}t} = {\bm A}{\bm x}\) , 其中
\[{\bm{A}} = \left[ {\begin{array}{ccc}
0&2&2\\
2&1&2\\
0&2&1
\end{array}} \right]
\]
求解矩阵 \({\bm A}\) 的特征值与广义特征向量与 \({\bm P}\)
由 例子(广义特征向量) 知其特征值为 \(\lambda_1 = 4, \lambda_2 = \lambda_3 = -1\) ,对应于 \(\lambda_1\) 的特征向量 \({\bm x}_1 = (5, 6, 4)^T\) , 对应于 \(\lambda_2 = \lambda_3 = -1\) 的广义特征向量 \({\bm x}_2 = (0, 1, -1)^T\) , \({\bm x}_3 = (2, -2, 1)^T\) .
所以取
\[{\bm P} = \left[ {\begin{array}{ccc}
5&0&1\\
6&1& - \\
4&{-1}&{1{\rm{/}}2}
\end{array}1} \right]
\]
由 \({\bm P}^{-1}{\bm A}{\bm P}\) 或者初等因子组求解出Jordan标准形
\[{\bm J} = \left[ {\begin{array}{ccc}
4&{}&{}\\
{}&{ - 1}&1\\
{}&{}&{ - 1}
\end{array}} \right]
\]
求解 \(\frac{{\rm d}{\bm y}}{{\rm d}t} = {\bm J}{\bm y}\) 一般解
\[\left\{ {\begin{array}{lll}
{\frac{{{\rm{d}}{y_1}}}{{{\rm{d}}t}} = 4{y_1}}\\
{\frac{{{\rm{d}}{y_2}}}{{{\rm{d}}t}} = - {y_2} + {y_3}}\\
{\frac{{{\rm{d}}{y_3}}}{{{\rm{d}}t}} = - {y_3}}
\end{array}} \right.\;\; \Rightarrow \left\{ {\begin{array}{lll}
{{y_1} = {c_1}{e^{4t}}}\\
{\frac{{{\rm{d}}{y_2}}}{{{\rm{d}}t}} = - {y_2} + {c_3}{e^{ - t}}}\\
{{y_3} = {c_3}{e^{ - t}}}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{lll}
{{y_1} = {c_1}{e^{4t}}}\\
{{y_2} = {c_2}{e^{ - t}} + {c_3}t{e^{ - t}}}\\
{{y_3} = {c_3}{e^{ - t}}}
\end{array}} \right.
\]
由 \({\bm x} = {\bm P}{\bm y}\) 知
\[{\bm{x}} = {\bm{Py}} = \left[ {\begin{array}{ccc}
5&0&1\\
6&1& - \\
4&{ - 1}&{1{\rm{/}}2}
\end{array}1} \right]\left[ {\begin{array}{ccc}
{{y_1}}\\
{{y_2}}\\
{{y_3}}
\end{array}} \right]
\]
\[\Rightarrow \left\{ {\begin{array}{lll}
{{x_1} = 5{y_1} + {y_3}}\\
{{x_2} = 6{y_1} + {y_2} - {y_3}}\\
{{x_3} = 4{y_1} - {y_2} + {y_3}/2}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{lll}
{{x_1} = 5{c_1}{e^{4t}} + {c_3}{e^{ - t}}}\\
{{x_2} = 6{c_1}{e^{4t}} + {c_2}{e^{ - t}} + {c_3}t{e^{ - t}} - {c_3}{e^{ - t}}}\\
{{x_3} = 4{c_1}{e^{4t}} - {c_2}{e^{ - t}} - {c_3}t{e^{ - t}} + {c_3}{e^{ - t}}/2}
\end{array}} \right.
\]
提示
在Matlab中可以使用 dsolve 函数求解微分方程组, 如对于 \(\frac{{\rm d}\bm x}{{\rm d}t} = -{\bm x}\)
>> syms x(t)
Dx = diff(x);
dsolve(diff(Dx) == -x, Dx(0) == 1)
ans =
sin(t) + C3*cos(t)