\documentclass[12pt]{article}
\usepackage{latexsym}
\input psfig.sty
%\usepackage{maple2e}
% \def\emptyline{\vspace{12pt}}
%\DefineParaStyle{Maple Output}
%\DefineParaStyle{Maple Plot}
%\DefineCharStyle{2D Math}
%\DefineCharStyle{2D Output}
\input spec_frac
\begin{document}
\title{Solutions, 316-XXIII}
\maketitle
23( 4/15) Complex eigenvalues\\
          9.6[1,2*,5*,6,13(a*,b,c,d)]\\
CAUTION: there may be errors!!!

\section{Problem 9.6.1}
Find a general solution of the system ${\bf x}^\prime = {\bf A} {\bf x}$ for
the matrix
\[ {\bf A} = \left( \begin{array}{rr}
2 & -4 \\ 2 & -2 \end{array} \right) \ .\]
{\bf Solution:}\\
Form the characteristic polynomial:
\begin{eqnarray*}
 P(r) & := & \det \left({\bf A}-r{\bf I}\right) =
\det \left( \begin{array}{rr}
2-r & -4 \\ 2 & -2-r \end{array} \right) \\
& =& (r+2)(r-2)+8 = r^2+4 = (r+2i)(r-2i) = 0
\end{eqnarray*}
so the problem has complex eigenvalues/vectors. We work with the
eigenvalue $r_1 = +2i$; the results for $r_2 = -2i$ are found by
{\em complex conjugation} of the answer (i.e. change the sign of $i$
everywhere). We have
\begin{eqnarray*}
 \left( \begin{array}{rr}
-2-2i & -4 \\ 2 & -2-2i \end{array} \right)
\left( \begin{array}{r} x \\ y \end{array} \right) & = &
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \\
\left( \begin{array}{r} x \\ y \end{array} \right)  & = &
\left( \begin{array}{r} 2 \\ -1-i \end{array} \right) = 
  \left( \begin{array}{r} 2 \\ -1 \end{array} \right) + 
i \left( \begin{array}{r} 0 \\ -1 \end{array} \right) 
\end{eqnarray*}
We know from the theory that when the matrix has a complex eigenvalue
$r=\alpha \pm i \beta$ with eigenvector ${\bf u} = {\bf a} \pm i {\bf b}$
then the two linearly independent solutions to 
${\bf x}^\prime = {\bf Ax}$ are
${\bf x}_1 = e^{\alpha t} \cos{\beta t} {\bf a} 
           - e^{\alpha t} \sin{\beta t} {\bf b}$,
and
${\bf x}_2 = e^{\alpha t}\sin{\beta t}{\bf a} 
+ e^{\alpha t}\cos{\beta t}{\bf b}$.
Here this gives, with $r = 0 + 2i$:
\[
{\bf x}_1 = \cos{2t} \left( \begin{array}{r} 2 \\ -1 \end{array} \right)
          - \sin{2t} \left( \begin{array}{r} 0 \\ -1 \end{array} \right) \]
and 
\[
{\bf x}_2 = \sin{2t} \left( \begin{array}{r} 2 \\ -1 \end{array} \right)
          + \cos{2t} \left( \begin{array}{r} 0 \\ -1 \end{array} \right) \]
so that the general solution is given by
\[
{\bf x}_{gen}(t) = C_1 \left(\begin{array}{rr}
2 \cos{2t} \\ - \cos{2t} + \sin{2t} \end{array} \right) +
C_2 \left(\begin{array}{rr} 2 \sin{2t} \\  -\sin{2t} - \cos{2t}
\end{array} \right) \ . \]
\newpage
\section{Problem 9.6.2}
Find a general solution of the system ${\bf x}^\prime = {\bf A} {\bf x}$ for
the matrix
\[ {\bf A} = \left( \begin{array}{rr}
-2 & -2 \\ 4 &  2 \end{array} \right) \ .\]

{\bf Solution:}\\
Form the characteristic polynomial:
\begin{eqnarray*}
 P(r) & := & \det \left({\bf A}-r{\bf I}\right) =
\det \left( \begin{array}{rr}
-2-r & -2 \\ 4 & 2-r \end{array} \right) \\
& =& (r+2)(r-2)+8 = r^2+4 = (r+2i)(r-2i) = 0
\end{eqnarray*}
so the problem has complex eigenvalues/vectors. We work with the
eigenvalue $r_1 = +2i$; the results for $r_2 = -2i$ are found by
{\em complex conjugation} of the answer (i.e. change the sign of $i$
everywhere). We have
\begin{eqnarray*}
 \left( \begin{array}{rr}
-2-2i & -2 \\ 4 & 2-2i \end{array} \right)
\left( \begin{array}{r} x \\ y \end{array} \right) & = &
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \\
\left( \begin{array}{r} x \\ y \end{array} \right)  & = &
\left( \begin{array}{r} 1 \\ -1-i \end{array} \right) = 
  \left( \begin{array}{r} 1 \\ -1 \end{array} \right) + 
i \left( \begin{array}{r} 0 \\ -1 \end{array} \right) 
\end{eqnarray*}
We know from the theory that when the matrix has a complex eigenvalue
$r=\alpha \pm i \beta$ with eigenvector ${\bf u} = {\bf a} \pm i {\bf b}$
then the two linearly independent solutions to
${\bf x}^\prime = {\bf Ax}$ are
${\bf x}_1 = e^{\alpha t} \cos{\beta t} {\bf a} 
           - e^{\alpha t} \sin{\beta t} {\bf b}$,
and
${\bf x}_2 = e^{\alpha t}\sin{\beta t}{\bf a} 
+ e^{\alpha t}\cos{\beta t}{\bf b}$.
Here this gives, with $r = 0 + 2i$:
\[
{\bf x}_1 = \cos{2t} \left( \begin{array}{r} 1 \\ -1 \end{array} \right)
          - \sin{2t} \left( \begin{array}{r} 0 \\ -1 \end{array} \right) \]
and
\[
{\bf x}_2 = \sin{2t} \left( \begin{array}{r} 1 \\ -1 \end{array} \right)
          + \cos{2t} \left( \begin{array}{r} 0 \\ -1 \end{array} \right) \]
so that the general solution is given by
\[
{\bf x}_{gen}(t) = C_1 \left(\begin{array}{rr}
 \cos{2t} \\ - \cos{2t} + \sin{2t} \end{array} \right) +
C_2 \left(\begin{array}{rr}  \sin{2t} \\  -\sin{2t} - \cos{2t}
\end{array} \right) \ . \]

\newpage
\section{Problem 9.6.5}
Find a fundamental matrix for the system ${\bf x}^\prime = {\bf A} {\bf x}$
for the matrix
\[ {\bf A} = \left( \begin{array}{rr}
-1 & -2 \\ 8 & -1 \end{array} \right) \ .\]

{\bf Solution:}\\
Form the characteristic polynomial:
\begin{eqnarray*}
 P(r) & := & \det \left({\bf A}-r{\bf I}\right) =
\det \left( \begin{array}{rr}
-1-r & -2 \\ 8 & -1-r \end{array} \right) \\
& =& (r+1)^2+16 = (r-1-4i)(r-1+4i) = 0
\end{eqnarray*}
so the problem has complex eigenvalues/vectors. We work with the
eigenvalue $r_1 = -1+4i$; the results for $r_2 = -1-4i$ are found by
{\em complex conjugation} of the answer (i.e. change the sign of $i$
everywhere). We have
\begin{eqnarray*}
 \left( \begin{array}{rr}
-4i & -2 \\ 8 & -4i \end{array} \right)
\left( \begin{array}{r} x \\ y \end{array} \right) & = &
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \\
\left( \begin{array}{r} x \\ y \end{array} \right)  & = &
\left( \begin{array}{r} 1 \\ -2i \end{array} \right) = 
  \left( \begin{array}{r} 1 \\ 0 \end{array} \right) + 
i \left( \begin{array}{r} 0 \\ -2 \end{array} \right) 
\end{eqnarray*}
We know from the theory that when the matrix has a complex eigenvalue
$r=\alpha \pm i \beta$ with eigenvector ${\bf u} = {\bf a} \pm i {\bf b}$
then the two linearly independent solutions to
${\bf x}^\prime = {\bf Ax}$ are
${\bf x}_1 = e^{\alpha t} \cos{\beta t} {\bf a} 
           - e^{\alpha t} \sin{\beta t} {\bf b}$,
and
${\bf x}_2 = e^{\alpha t}\sin{\beta t}{\bf a} 
+ e^{\alpha t}\cos{\beta t}{\bf b}$.
Here this gives, with $r = -1 + 4i$:
\[
{\bf x}_1 = \cos{2t} \left( \begin{array}{r} 1 \\  0 \end{array} \right)
          - \sin{2t} \left( \begin{array}{r} 0 \\ -2 \end{array} \right) \]
and
\[
{\bf x}_2 = \sin{2t} \left( \begin{array}{r} 1 \\  0 \end{array} \right)
          + \cos{2t} \left( \begin{array}{r} 0 \\ -2 \end{array} \right) \]
so that the Fundamental Matrix is given by
\[
{\bf \Phi}(t) = \left(\begin{array}{rr}
 \cos{2t}  &     \sin{2t}  \\
2 \sin{2t} &  - 2\cos{2t}
\end{array} \right) \ . \]
\newpage
\section{Problem 9.6.13a}
Find the solution to the IVP:
\[ {\bf x}^\prime = \left( \begin{array}{rr}
-3 & -1 \\ 2 & -1 \end{array} \right) {\bf x} \ ,\ 
{\bf x}(0) = \left( \begin{array}{r} -1 \\ 0 \end{array} \right) {\bf x} \ .\]
{\bf Solution:}\\
Form the characteristic polynomial:
\begin{eqnarray*}
P(r) & := & \det \left({\bf A}-r{\bf I}\right) = 
\det \left( \begin{array}{ll}
-3-r & -1 \\ 2 & -1-r \end{array} \right) \\
& =& (r+1)(r+3)+2 = r^2+4r+5 = (r+2)^2+1 = 0
\end{eqnarray*}
so the problem has complex eigenvalues/vectors. We work with the
eigenvalue $r_1 = -2+i$; the eigenvector is found from 
\begin{eqnarray*}
 \left( \begin{array}{ll}
-3+2-i & -1 \\ 2 & -1+2-i \end{array} \right)
\left( \begin{array}{l} x \\ y \end{array} \right) & = &
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \\
\left( \begin{array}{r} x \\ y \end{array} \right)  & = &
\left( \begin{array}{l} 1 \\ -1-i \end{array} \right) = 
\left( \begin{array}{r} 1 \\ -1 \end{array} \right) + 
i \left( \begin{array}{r} 0 \\ -1 \end{array} \right) 
\end{eqnarray*}

where the first equation was used to find the components of the eigenvector.
Following problem (1), we find an independent set as (with $r = -2 + i$):
\[
{\bf x}_1 = e^{-2t}\cos{t} \left( \begin{array}{r} 1 \\ -1 \end{array} \right)
          - e^{-2t}\sin{t} \left( \begin{array}{r} 0 \\ -1 \end{array} \right) 
 = 
e^{-2t}\left( \begin{array}{r} \cos{t}
\\   \sin{t} -\cos{t} \end{array} \right) 
\]
and 
\[
{\bf x}_2 = e^{-2t}\sin{t} \left( \begin{array}{r} 1 \\ -1 \end{array} \right)
          + e^{-2t}\cos{t} \left( \begin{array}{r} 0 \\ -1 \end{array} \right) 
 = 
e^{-2t}\left( \begin{array}{r} \sin{t}
\\ -\cos{t} - \sin{t} \end{array} \right) 
\]
so that the general solution is given by
\[
{\bf x}_{gen}(t) = C_1 e^{-2t}\left(\begin{array}{r}
\cos{t}\\ \sin{t}- \cos{t}  \end{array} \right) +
C_2 e^{-2t}\left(\begin{array}{r} \sin{t} \\  -\cos{t}-\sin{t}
\end{array} \right) \ . \]
A fundamental matrix for the system is then
\[
\Phi(t) = 
e^{-2t}\left(\begin{array}{rr} 
\cos{t} & \sin{t} \\
\sin{t}- \cos{t} &  - \cos{t}-\sin{t} \end{array} \right) \ ,\]
and the general solution is given in terms of the Fundamental matrix by the
equivalent form
\[ 
{\bf x}_{gen}(t)=\Phi(t)\left(\begin{array}{r}C_1\\C_2\end{array}\right)
\]
Solving for the IC:
\begin{eqnarray*}
\Phi((0))\left(\begin{array}{r}C_1\\C_2\end{array}\right) & = & 
\left( \begin{array}{r} -1 \\ 0 \end{array} \right) \\
\left(\begin{array}{rr} 
 1  &   0 \\
-1  &  -1 \end{array} \right) 
\left(\begin{array}{r}C_1\\C_2\end{array}\right)  & = &
\left( \begin{array}{r} -1 \\ 0 \end{array} \right) \\
\left(\begin{array}{r}C_1\\C_2\end{array}\right)  & = &
\frac{1}{-1}\left(\begin{array}{rr} 
-1 &  0 \\
 1 &  1 \end{array} \right) \left( \begin{array}{r} -1 \\ 0 \end{array} \right)
= \left( \begin{array}{r} -1 \\ 1 \end{array} \right)
\end{eqnarray*}

Finally, the solution to the IVP is:
\begin{eqnarray*}
{\bf x}(t) & = & \Phi(t)\left(\begin{array}{r}-1 \\ 
1 \end{array} \right) \\
& = & \left({\bf x}_2 (t)- {\bf x}_1 (t) \right) \\
& = & e^{t}
\left( \begin{array}{r} -\cos{t} + \sin{t} \\-2\sin{t} \end{array} \right) \ . 
\end{eqnarray*}
\newpage
\section{Problem 9.6.13d modified}
Find the solution to the IVP:
\[ {\bf x}^\prime = \left( \begin{array}{rr}
3 & -1 \\ 5 & -1 \end{array} \right) {\bf x} \ ,\ 
{\bf x}(\pi/2) = \left( \begin{array}{r} 0 \\ 1 \end{array} \right) \ .\]
{\bf Solution:}\\
Form the characteristic polynomial:
\begin{eqnarray*}
P(r) & := & \det \left({\bf A}-r{\bf I}\right) = 
\det \left( \begin{array}{ll}
3-r & -1 \\ 5 & -1-r \end{array} \right) \\
& =& (r+1)(r-3)+3 = r^2-2r+2 = (r-1)^2+1 = 0
\end{eqnarray*}
so the problem has complex eigenvalues/vectors. We work with the
eigenvalue $r_1 = 1+i$; the eigenvector is found from 
\begin{eqnarray*}
 \left( \begin{array}{ll}
3-1-i & -1 \\ 5 & -1-1-i \end{array} \right)
\left( \begin{array}{l} x \\ y \end{array} \right) & = &
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \\
\left( \begin{array}{r} x \\ y \end{array} \right)  & = &
\left( \begin{array}{l} 2+i \\ 5 \end{array} \right) = 
\left( \begin{array}{r} 2 \\ 5 \end{array} \right) + 
i \left( \begin{array}{r} 1 \\ 0 \end{array} \right) 
\end{eqnarray*}

where the second equation was used to find the components of the eigenvector.
Following problem (1), we find an independent set as (with $r = 1 + i$):
\[
{\bf x}_1 = e^{t}\cos{t} \left( \begin{array}{r} 2 \\ 5 \end{array} \right)
          - e^{t}\sin{t} \left( \begin{array}{r} 1 \\ 0 \end{array} \right) 
 = 
e^{t}\left( \begin{array}{l} 2\cos{t}-\sin{t} 
\\ 5 \cos{t} \end{array} \right) 
\]
and 
\[
{\bf x}_2 = e^{t}\sin{t} \left( \begin{array}{r} 2 \\ 5 \end{array} \right)
          + e^{t}\cos{t}\left( \begin{array}{r} 1 \\ 0 \end{array} \right) 
 = 
e^{t}\left( \begin{array}{r} \cos{t} + 2\sin{t}
\\ 5 \sin{t} \end{array} \right) 
\]
so that the general solution is given by
\[
{\bf x}_{gen}(t) = C_1 e^{t}\left(\begin{array}{ll}
2\cos{t}-\sin{t} \\ 5 \cos{t}  \end{array} \right) +
C_2 e^{t}\left(\begin{array}{rr} \cos{t}+2\sin{t} \\  5\sin{t}
\end{array} \right) \ . \]
A fundamental matrix for the system is then
\[
\Phi(t) = 
e^{t}\left(\begin{array}{lr} 
2\cos{t}-\sin{t} & \cos{t}+2\sin{t} \\
 5 \cos{t} &  5\sin{t} \end{array} \right) \ ,\]
and the general solution is given in terms of the Fundamental matrix by the
equivalent form
\[ 
{\bf x}_{gen}(t)=\Phi(t)\left(\begin{array}{r}C_1\\C_2\end{array}\right)
\]
Solving for the IC:
\begin{eqnarray*}
\Phi((\pi/2))\left(\begin{array}{r}C_1\\C_2\end{array}\right) & = & 
\left( \begin{array}{r} 0 \\ 1 \end{array} \right) \\
e^{\pi/2}\left(\begin{array}{rr} 
-1 & 2 \\
 0     &  5 \end{array} \right) 
\left(\begin{array}{r}C_1\\C_2\end{array}\right)  & = &
\left( \begin{array}{r} 0 \\ 1 \end{array} \right) \\
\left(\begin{array}{r}C_1\\C_2\end{array}\right)  & = &
\frac{e^{-\pi/2}}{-5}\left(\begin{array}{rr} 
5 & -2 \\
0 & -1 \end{array} \right) \left( \begin{array}{r} 0 \\ 1 \end{array} \right)
=\frac{e^{-\pi/2}}{5} \left( \begin{array}{r} 2 \\ 1 \end{array} \right)
\end{eqnarray*}

Finally, the solution to the IVP is:
\begin{eqnarray*}
{\bf x}(t) & = & \frac{e^{-\pi/2}}{5} \Phi(t)\left(\begin{array}{r}2 \\ 
1 \end{array} \right) \\
& = & \frac{e^{-\pi/2}}{5} \left(2 {\bf x}_1 (t)+ {\bf x}_2 (t) \right) \\
& = & e^{t-\pi/2}
\left( \begin{array}{r} \cos{t} \\ \sin{t}+ 2\cos{t} \end{array} \right) \ . 
\end{eqnarray*}
\end{document}
