7.8
- Page ID
- 154476
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)7.5/7.8: Suppose the solution to \(\mathrm{x}^{\prime}=A \mathrm{x}\) is
\[ \notag
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]=c_1\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right] e^{r_1 t}+c_2\left[\begin{array}{l}
w_1 \\
w_2
\end{array}\right] e^{r_2 t}
\]
where \(\mathbf{v}, \mathbf{w}\) are linearly independent.
Then
\[ \notag
\frac{x_2(t)}{x_1(t)}=\frac{c_1 v_2 e^{r_1 t}+c_2 w_2 e^{r_2 t}}{c_1 v_1 e^{r_1 t}+c_2 w_1 e^{r_2 t}}=\frac{c_1 v_2+c_2 w_2 e^{\left(r_2-r_1\right) t}}{c_1 v_1+c_2 w_1 e^{\left(r_2-r_1\right) t}}
\]
If \(c_2=0\), then \(\frac{x_2(t)}{x_1(t)}=\frac{v_2}{v_1}\)
If \(c_1=0\), then \(\frac{x_2(t)}{x_1(t)}=\frac{w_2}{w_1}\)
Suppose \(r_1>r_2\) and \(c_1 c_2 \neq 0\),
Then \(\lim _{t \rightarrow \infty} \frac{x_2(t)}{x_1(t)}=\lim _{t \rightarrow \infty} \frac{c_1 v_2+c_2 w_2 e^{\left(r_2-r_1\right) t}}{c_1 v_1+c_2 w_1 e^{\left(r_2-r_1\right) t}}=\frac{v_2}{v_1}\)
Similarly \(\lim _{t \rightarrow-\infty} \frac{x_2(t)}{x_1(t)}=\lim _{t \rightarrow-\infty} \frac{c_1 v_2 e^{\left(r_1-r_2\right) t}+c_2 w_2}{c_1 v_1 e^{\left(r_1-r_2\right) t}+c_2 w_1}=\frac{w_2}{w_1}\)
\(r_1>0>r_2 \quad r_1>r_2>0 \quad 0>r_1>r_2\)
Repeated root case with 2 linearly independent eigenvectors:
If \(r_1=r_2\), then
\[ \notag
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]=e^{r_1 t}\left(c_1\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right]+c_2\left[\begin{array}{l}
w_1 \\
w_2
\end{array}\right]\right)=e^{r_1 t}\left[\begin{array}{l}
c_1 v_1+c_2 w_1 \\
c_1 v_2+c_2 w_2
\end{array}\right]
\]
Then \(\frac{x_2(t)}{x_1(t)}=\frac{e^{r_1 t}\left(c_1 v_2+c_2 w_2\right)}{e^{r_1 t}\left(c_1 v_1+c_2 w_1\right)}=\frac{c_1 v_2+c_2 w_2}{c_1 v_1+c_2 w_1}\)
Repeated root case, \(r_1=r_2\), with only 1 linearly independent eigenvector, \(\mathbf{v}\)
One solution: \(\mathbf{x}=\mathbf{v} e^{r_1 t}\)
Need 2nd solution, Guess \(\mathbf{x}=t \mathbf{v} e^{r_1 t}+\mathbf{w} e^{r_1 t}\)
Then \(\mathbf{x}^{\prime}=r_1 t \mathbf{v} e^{r_1 t}+\mathbf{v} e^{r_1 t}+r_1 \mathbf{w} e^{r_1 t}\)
Plug into \(\mathbf{x}^{\prime}=A \mathbf{x}: \quad r_1 t \mathbf{v} e^{r_1 t}+\mathbf{v} e^{r_1 t}+r_1 \mathbf{w} e^{r_1 t}=A\left(t \mathbf{v} e^{r_1 t}+\mathbf{w} e^{r_1 t}\right)\)
\[ \notag
\begin{aligned}
r_1 t \mathbf{v} e^{r_1 t}+\mathbf{v} e^{r_1 t}+r_1 \mathbf{w} e^{r_1 t} & =t e^{r_1 t} A \mathbf{v}+e^{r_1 t} A \mathbf{w} \\
r_1 t \mathbf{v} e^{r_1 t}+\mathbf{v} e^{r_1 t}+r_1 \mathbf{w} e^{r_1 t} & =t e^{r_1 t} r_1 \mathbf{v}+e^{r_1 t} A \mathbf{w} \\
r_1 t \mathbf{v}+\mathbf{v}+r_1 \mathbf{w} & =t r_1 \mathbf{v}+A \mathbf{w} \\
\mathbf{v}+r_1 \mathbf{w} & =A \mathbf{w}
\end{aligned}
\]
Thus \(\mathbf{v}=A \mathbf{w}-r_1 \mathbf{w}=\)
Hence \(\mathbf{v}=\)
If \(r_1\) is an eigenvalue with eigenvector \(\mathbf{v}\), then \(\mathbf{w}\) is a generalized eigenvector corresponding to eigenvalue \(r_1\), if \(\mathbf{w} \neq 0\) and
\[\notag
\left(A-r_1 I\right) \mathbf{w}=\mathbf{v}
\]
General solution:
\[ \notag
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]=c_1\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right] e^{r_1 t}+c_2\left(t\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right]+\left[\begin{array}{l}
w_1 \\
w_2
\end{array}\right]\right) e^{r_1 t}
\]
or equivalently,
\[ \notag
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]=e^{r_1 t}\left(c_1\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right]+c_2\left(t\left[\begin{array}{l}
v_1 \\
v_2
\end{array}\right]+\left[\begin{array}{l}
w_1 \\
w_2
\end{array}\right]\right)\right)
\]
Then
\[ \notag
\frac{x_2(t)}{x_1(t)}=\frac{e^{r_1 t}\left(c_1 v_2+c_2\left(t v_2+w_2\right)\right)}{e^{r_1 t}\left(c_1 v_1+c_2\left(t v_1+w_1\right)\right)}=\frac{c_1 v_2+c_2 w_2+c_2 t v_2}{c_1 v_1+c_2 w_1+c_2 t v_1}
\]
If \(c_2=0\), then \(\frac{x_2(t)}{x_1(t)}=\frac{v_2}{v_1}\). No other constant slopes, but \(\lim _{t \rightarrow \pm \infty} \frac{x_2(t)}{x_1(t)}=\frac{v_2}{v_1}\)
Solve \(\mathbf{x}^{\prime}=\left[\begin{array}{ll}3 & 1 \\ 0 & 3\end{array}\right] \mathbf{x}\)
Repeated root eigenvalue \(r=3\)
- Answer
-
To find eigenvector corresponding to eigenvalue 3 , solve \((A-3 I) \mathbf{x}=\mathbf{0}\) :
\[
A-3 I=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]
\]Thus \(x_1\) is free and \(x_2=0\). Thue \(\mathbf{v}=\left[\begin{array}{l}1 \\ 0\end{array}\right]\) is an eigenvector corresponding to eigenvalue 3.
To find generalized eigenvector \(\mathrm{w}\) corresponding to eigenvalue 3 and eigenvector \(\mathrm{v}\),
\[
\text { solve }(A-3 I) \mathbf{w}=\mathbf{v}=\left[\begin{array}{l}
1 \\
0
\end{array}\right]
\]Solve \(\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{l}w_1 \\ w_2\end{array}\right]=\left[\begin{array}{l}1 \\ 0\end{array}\right]\)
Thus can choose \(\left[\begin{array}{l}w_1 \\ w_2\end{array}\right]=\left[\begin{array}{l}0 \\ 1\end{array}\right]\)
Hence general solution is \(\mathbf{x}=c_1\left[\begin{array}{l}1 \\ 0\end{array}\right] e^{3 t}+c_2\left(t\left[\begin{array}{l}1 \\ 0\end{array}\right]+\left[\begin{array}{l}0 \\ 1\end{array}\right]\right) e^{3 t}.\)