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4.6: Constant Coefficient Homogeneous Systems III

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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

We now consider the system ${\bf y}'=A{\bf y}$, where $A$ has a complex
eigenvalue $\lambda=\alpha+i\beta$ with $\beta\ne0$. We continue to
assume that $A$ has real entries, so the characteristic polynomial of
$A$ has real coefficients. This implies that
$\overline\lambda=\alpha-i\beta$ is also an eigenvalue of $A$.

An eigenvector ${\bf x}$ of $A$ associated with
$\lambda=\alpha+i\beta$ will have complex entries, so we'll write
$${\bf x}={\bf u}+i{\bf v}$$
where ${\bf u}$ and ${\bf v}$ have real entries; that is, ${\bf u}$
and ${\bf v}$ are the real and imaginary parts of ${\bf x}$. Since
$A{\bf x}=\lambda {\bf x}$,
\label{eq:10.6.1}
A({\bf u}+i{\bf v})=(\alpha+i\beta)({\bf u}+i{\bf v}).

Taking complex conjugates here and recalling that $A$ has real entries
yields
$$A({\bf u}-i{\bf v})=(\alpha-i\beta)({\bf u}-i{\bf v}),$$
which shows that ${\bf x}={\bf u}-i{\bf v}$ is an eigenvector
associated with $\overline\lambda=\alpha-i\beta$. The complex
conjugate eigenvalues $\lambda$ and $\overline\lambda$ can be
separately associated with linearly independent solutions ${\bf y}'=A{\bf y}$;     however, we won't pursue this approach, since solutions
obtained in this way turn out to be complex--valued. Instead, we'll
obtain solutions of ${\bf y}'=A{\bf y}$ in the form
\label{eq:10.6.2}
{\bf y}=f_1{\bf u}+f_2{\bf v}

where $f_1$ and $f_2$ are real--valued scalar functions.
The next theorem shows how to do this.

\begin{theorem}\color{blue} \label{thmtype:10.6.1}
Let $A$ be an $n\times n$ matrix with real entries$.$ Let
$\lambda=\alpha+i\beta$ {\rm(}$\beta\ne0${\rm}) be a complex
eigenvalue of $A$ and let ${\bf x}={\bf u}+i{\bf v}$ be an associated
eigenvector$,$ where ${\bf u}$ and ${\bf v}$ have real components$.$ Then
${\bf u}$ and ${\bf v}$ are both nonzero and
$${\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \mbox{\quad and\quad} {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),$$
which are the real and imaginary parts of
\label{eq:10.6.3}
e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}),

are linearly independent solutions of  ${\bf y}'=A{\bf y}$.
\end{theorem}

\proof
A function of the form \eqref{eq:10.6.2} is a solution of ${\bf y}'=A{\bf y}$ if and only if
\label{eq:10.6.4}
f_1'{\bf u}+f_2'{\bf
v}=f_1A{\bf u}+f_2A{\bf v}.

Carrying out the multiplication indicated on the  right side
of \eqref{eq:10.6.1} and collecting the  real and imaginary parts of the
result yields
$$A({\bf u}+i{\bf v})=(\alpha{\bf u}-\beta{\bf v})+i(\alpha{\bf v}+\beta{\bf u}).$$
Equating real and imaginary parts on the two sides of this equation yields
$$\begin{array}{rcl} A{\bf u}&=&\alpha{\bf u}-\beta{\bf v}\\ A{\bf v}&=&\alpha{\bf v}+\beta{\bf u}. \end{array}$$
We leave it to you (Exercise~\ref{exer:10.6.25}) to show from this that
${\bf u}$ and
${\bf v}$ are both nonzero.
Substituting from these equations into \eqref{eq:10.6.4} yields
\begin{eqnarray*}
f_1'{\bf u}+f_2'{\bf v}
&=&f_1(\alpha{\bf u}-\beta{\bf v})+f_2(\alpha{\bf v}+\beta{\bf u})\\
&=&(\alpha f_1+\beta f_2){\bf u}+(-\beta f_1+\alpha f_2){\bf v}.
\end{eqnarray*}
This is true if
$$\begin{array}{rcr} f_1'&=&\alpha f_1+\beta f_2\phantom{,}\\ f_2'&=&-\beta f_1+\alpha f_2, \end{array} \mbox{\quad or, equivalently,\quad} \begin{array}{rcr} f_1'-\alpha f_1&=&\phantom{-}\beta f_2\phantom{.}\\ f_2'-\alpha f_2&=&-\beta f_1. \end{array}$$
If we let $f_1=g_1e^{\alpha t}$ and $f_2=g_2e^{\alpha t}$, where
$g_1$ and $g_2$ are to be determined, then the last two equations
become
$$\begin{array}{rcr} g_1'&=&\beta g_2\phantom{.}\\ g_2'&=&-\beta g_1, \end{array}$$
which implies that
$$g_1''=\beta g_2'=-\beta^2 g_1,$$
so
$$g_1''+\beta^2 g_1=0.$$
The general solution of this equation is
$$g_1=c_1\cos\beta t+c_2\sin\beta t.$$
Moreover, since $g_2=g_1'/\beta$,
$$g_2=-c_1\sin\beta t+c_2\cos\beta t.$$
Multiplying  $g_1$  and $g_2$  by $e^{\alpha t}$ shows that
\begin{eqnarray*}
f_1&=&e^{\alpha t}(\phantom{-}c_1\cos\beta t+c_2\sin\beta t ),\\
f_2&=&e^{\alpha t}(-c_1\sin\beta t+c_2\cos\beta t).
\end{eqnarray*}
Substituting these into \eqref{eq:10.6.2} shows that
\label{eq:10.6.5}
\begin{array}{rcl}
{\bf y}&=&e^{\alpha t}\left[(c_1\cos\beta t+c_2\sin\beta t){\bf u}
+(-c_1\sin\beta t+c_2\cos\beta t){\bf v}\right]\\
&=&c_1e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)
+c_2e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)
\end{array}

is a solution of ${\bf y}'=A{\bf y}$ for any choice of the constants
$c_1$ and $c_2$. In particular, by first taking $c_1=1$ and $c_2=0$
and then taking $c_1=0$ and $c_2=1$, we see that ${\bf y}_1$ and ${\bf y}_2$ are solutions of ${\bf y}'=A{\bf y}$. We leave it to you to
verify that they are, respectively, the real and imaginary parts of
\eqref{eq:10.6.3} (Exercise~\ref{exer:10.6.26}), and that they are linearly
independent (Exercise~\ref{exer:10.6.27}).

\begin{example}\label{example:10.6.1} \rm
Find the general solution of
\label{eq:10.6.6}
{\bf y}'=\twobytwo4{-5}5{-2}{\bf y}.

\end{example}

\solution The characteristic polynomial of the coefficient matrix $A$
in \eqref{eq:10.6.6} is
$$\left|\begin{array}{cc} 4-\lambda&-5\\ 5&-2-\lambda \end{array}\right|=(\lambda-1)^2+16.$$
Hence, $\lambda=1+4i$ is an eigenvalue of $A$. The associated
eigenvectors satisfy $\left(A-\left(1+4i\right)I\right){\bf x}={\bf 0}$. The augmented matrix of this system is
$$\left[\begin{array}{cccr} 3-4i&-5&\vdots&0\\ 5&-3-4i&\vdots&0 \end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{cccr} 1&-{3+4i\over5}&\vdots&0\\ 0&0&\vdots&0 \end{array}\right].$$
Therefore $x_1=(3+4i)x_2/5$. Taking $x_2=5$ yields $x_1=3+4i$, so
$${\bf x}=\left[\begin{array}{c}3+4i\\5\end{array}\right]$$
is an eigenvector.  The real and imaginary parts of
$$e^t(\cos4t+i\sin4t)\left[\begin{array}{c}3+4i\\5\end{array}\right]$$
are
$${\bf y}_1=e^t\left[\begin{array}{c}3\cos4t-4\sin 4t\\5\cos4t\end{array}\right]\quad\mbox{ and }\quad {\bf y}_2=e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin 4t\end{array}\right],$$
which are linearly independent solutions of  \eqref{eq:10.6.6}.
The general solution of \eqref{eq:10.6.6} is
$${\bf y}= c_1e^t\left[\begin{array}{c}3\cos4t-4\sin 4t\\5\cos4t\end{array}\right]+ c_2e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin 4t\end{array}\right].$$

\begin{example}\label{example:10.6.2} \rm
Find  the general solution of
\label{eq:10.6.7}
{\bf y}'=\twobytwo{-14}{39}{-6}{16}{\bf y}.

\end{example}

\solution The characteristic polynomial of the coefficient matrix $A$
in \eqref{eq:10.6.7} is
$$\left|\begin{array}{cc}-14-\lambda&39\\-6&16-\lambda \end{array}\right|=(\lambda-1)^2+9.$$
Hence, $\lambda=1+3i$ is an eigenvalue of $A$. The associated
eigenvectors satisfy $\left(A-(1+3i)I\right){\bf x}={\bf 0}$. The
augmented augmented matrix of this system is
$$\left[\begin{array}{cccr}-15-3i&39&\vdots&0\\ -6&15-3i&\vdots&0 \end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{cccr} 1&{-5+i\over2}&\vdots&0\\ 0&0&\vdots&0 \end{array}\right].$$
Therefore $x_1=(5-i)/2$. Taking $x_2=2$ yields $x_1=5-i$, so
$${\bf x}=\left[\begin{array}{c}5-i\\2\end{array}\right]$$
is an eigenvector. The real and imaginary parts of
$$e^t(\cos3t+i\sin3t)\left[\begin{array}{c}5-i\\2\end{array}\right]$$
are
$${\bf y}_1=e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos 3t\end{array}\right]\quad\mbox{ and }\quad {\bf y}_2=e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin 3t\end{array}\right],$$
which are  linearly independent solutions of  \eqref{eq:10.6.7}.
The general solution of \eqref{eq:10.6.7} is
$${\bf y}=c_1e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos 3t\end{array}\right]+ c_2e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin 3t\end{array}\right].$$

\begin{example}\label{example:10.6.3}  \rm
Find the general solution of
\label{eq:10.6.8}
{\bf y}'=\threebythree{-5}54{-8}76100{\bf y}.

\end{example}

\solution  The characteristic
polynomial of the  coefficient matrix $A$ in  \eqref{eq:10.6.8} is
$$\left|\begin{array}{ccc}-5-\lambda&5&4\\-8&7-\lambda& 6\\ \phantom{-}1 &0&-\lambda\end{array}\right|=-(\lambda-2)(\lambda^2+1).$$
Hence, the eigenvalues of $A$ are $\lambda_1=2$, $\lambda_2=i$, and
$\lambda_3=-i$.
The augmented matrix of $(A-2I){\bf x=0}$ is
$$\left[\begin{array}{rrrcr}-7&5&4&\vdots&0\\-8& 5&6&\vdots&0\\ 1&0&-2&\vdots&0 \end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&-2& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].$$
Therefore $x_1=x_2=2x_3$.  Taking $x_3=1$ yields
$${\bf x}_1=\threecol221,$$
so
$${\bf y}_1=\threecol221e^{2t}$$
is a solution of  \eqref{eq:10.6.8}.

The augmented  matrix of $(A-iI){\bf x=0}$ is
$$\left[\begin{array}{ccrccc}-5-i&5&4&\vdots&0\\-8& 7-i&6&\vdots&0\\ \phantom{-}1&0&-i&\vdots&0 \end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{ccccc} 1&0&-i&\vdots&0\\ 0&1&1-i& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].$$
Therefore $x_1=ix_3$ and $x_2=-(1-i)x_3$. Taking $x_3=1$ yields
the eigenvector
$${\bf x}_2=\left[\begin{array}{c} i\\-1+i\\ 1\end{array} \right].$$
The real and imaginary parts of
$$(\cos t+i\sin t)\left[\begin{array}{c}i\\-1+i\\1\end{array}\right]$$
are
$${\bf y}_2= \left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right] \quad\mbox{ and }\quad {\bf y}_3=\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin t\end{array}\right],$$
which are solutions of  \eqref{eq:10.6.8}.
Since the  Wronskian of $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$
at $t=0$  is
$$\left|\begin{array}{rrr} 2&0&1\\2&-1&1\\1&1&0\end{array}\right|=1,$$
$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is a
fundamental set of solutions of \eqref{eq:10.6.8}. The general solution of
\eqref{eq:10.6.8} is
$${\bf y}=c_1 \threecol221e^{2t} +c_2\left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right] +c_3\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin t\end{array}\right].$$

\begin{example}\label{example:10.6.4}  \rm
Find the general solution of
\label{eq:10.6.9}
{\bf y}'=\threebythree1{-1}{-2}1321{-1}2{\bf y}.

\end{example}

\solution The characteristic
polynomial of the  coefficient matrix $A$ in  \eqref{eq:10.6.9} is
$$\left|\begin{array}{ccc} 1-\lambda&-1&-2\\ 1&3-\lambda& \phantom{-}2\\ 1 &-1&2-\lambda\end{array}\right|= -(\lambda-2)\left((\lambda-2)^2+4\right).$$
Hence, the eigenvalues of $A$ are $\lambda_1=2$, $\lambda_2=2+2i$,
and $\lambda_3=2-2i$.
The augmented matrix of $(A-2I){\bf x=0}$ is
$$\left[\begin{array}{rrrcr}-1&-1&-2&\vdots&0\\1& 1&2&\vdots&0\\ 1&-1&0&\vdots&0 \end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1&0&1&\vdots&0\\ 0&1&1& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].$$
Therefore $x_1=x_2=-x_3$.  Taking $x_3=1$ yields
$${\bf x}_1=\threecol{-1}{-1}1,$$
so
$${\bf y}_1=\threecol{-1}{-1}1e^{2t}$$
is a solution of  \eqref{eq:10.6.9}.

The augmented  matrix of $\left(A-(2+2i)I\right){\bf x=0}$ is
$$\left[\begin{array}{ccrcc}-1-2i&-1&-2&\vdots&0\\ 1& 1-2i&\phantom{-}2&\vdots&0\\ 1&-1&-2i&\vdots&0 \end{array}\right],$$
which is row equivalent to
$$\left[\begin{array}{rrrcr} 1&0&-i&\vdots&0\\ 0&1&i& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right].$$
Therefore $x_1=ix_3$ and $x_2=-ix_3$. Taking $x_3=1$ yields
the eigenvector
$${\bf x}_2=\threecol i{-i}1$$
The real and imaginary parts of
$$e^{2t}(\cos2t+i\sin2t)\threecol i{-i}1$$
are
$${\bf y}_2=e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos 2t\end{array}\right]\quad\mbox{ and }\quad {\bf y}_2=e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin 2t\end{array}\right],$$
which are solutions of  \eqref{eq:10.6.9}.
Since  the Wronskian of $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$
at $t=0$ is
$$\left|\begin{array}{rrr} -1&0&1\\-1&0&-1\\1&1&0\end{array}\right|=-2,$$
$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is a fundamental set of solutions
of \eqref{eq:10.6.9}.  The general solution of \eqref{eq:10.6.9} is
$${\bf y}=c_1\threecol{-1}{-1}1e^{2t}+ c_2e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos 2t\end{array}\right]+ c_3e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin 2t\end{array}\right].$$

\boxit{Geometric Properties of Solutions when  $n=2$}

\noindent
We'll now  consider the geometric properties of solutions of a
$2\times2$ constant coefficient system
\label{eq:10.6.10}
\twocol{y_1'}{y_2'}=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}
\end{array}\right]\twocol{y_1}{y_2}

under the assumptions of this section; that is, when the matrix
$$A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right]$$
has a complex eigenvalue $\lambda=\alpha+i\beta$
{\rm(}$\beta\ne0${\rm}) and ${\bf x}={\bf u}+i{\bf v}$ is an
associated eigenvector, where ${\bf u}$ and ${\bf v}$ have real
components. To describe the trajectories accurately it's necessary to
introduce a new rectangular coordinate system in the $y_1$-$y_2$
plane. This raises a point that hasn't come up before: It is always
possible to choose ${\bf x}$ so that $({\bf u},{\bf v})=0$. A special
effort is required to do this, since not every eigenvector has this
property. However, if we know an eigenvector that doesn't, we can
multiply it by a suitable complex constant to obtain one that does. To
see this, note that if ${\bf x}$ is a $\lambda$-eigenvector of $A$ and
$k$ is an arbitrary real number, then
$${\bf x}_1=(1+ik){\bf x}=(1+ik)({\bf u}+i{\bf v}) =({\bf u}-k{\bf v})+i({\bf v}+k{\bf u})$$
is also a $\lambda$-eigenvector of $A$, since
$$A{\bf x}_1= A((1+ik){\bf x})=(1+ik)A{\bf x}=(1+ik)\lambda{\bf x}= \lambda((1+ik){\bf x})=\lambda{\bf x}_1.$$
The real and imaginary parts of ${\bf x}_1$ are
\label{eq:10.6.11}
{\bf u}_1={\bf u}-k{\bf v} \mbox{\quad and \quad}
{\bf v}_1={\bf v}+k{\bf u},

so
$$({\bf u}_1,{\bf v}_1)=({\bf u}-k{\bf v},{\bf v}+k{\bf u}) =-\left[({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k -({\bf u},{\bf v})\right].$$
Therefore $({\bf u}_1,{\bf v}_1)=0$ if
\label{eq:10.6.12}
({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf
v})=0.

If $({\bf u},{\bf v})\ne0$ we can use the quadratic formula to find
two real values of $k$ such that $({\bf u}_1,{\bf v}_1)=0$
(Exercise~\ref{exer:10.6.28}).

\begin{example}\label{example:10.6.5} \rm  In Example~\ref{example:10.6.1}
we found the eigenvector
$${\bf x}=\ctwocol{3+4i}5=\twocol35+i\twocol40$$
for the matrix of the system \eqref{eq:10.6.6}. Here ${\bf u}=\dst{\twocol35}$
and ${\bf v}=\twocol40$ are not orthogonal, since $({\bf u},{\bf v})=12$. Since $\|{\bf v}\|^2-\|{\bf u}\|^2=-18$, \eqref{eq:10.6.12} is
equivalent to
$$2k^2-3k-2=0.$$
The zeros of this equation are $k_1=2$ and $k_2=-1/2$. Letting $k=2$
in \eqref{eq:10.6.11} yields
$${\bf u}_1={\bf u}-2{\bf v}=\twocol{-5}{\phantom{-}5}\mbox{\quad and \quad} {\bf v}_1={\bf v}+2{\bf u}=\twocol{10}{10},$$
and $({\bf u}_1,{\bf v}_1)=0$.
Letting $k=-1/2$
in \eqref{eq:10.6.11} yields
$${\bf u}_1={\bf u}+{{\bf v}\over2}=\twocol{5}5\mbox{\quad and \quad} {\bf v}_1={\bf v}-{{\bf u}\over2}={1\over2}\twocol{-5}{\phantom{-}5},$$
and again $({\bf u}_1,{\bf v}_1)=0$. \bbox
\end{example}

(The numbers don't always work out as nicely as in this example.
You'll need a calculator or computer to do
Exercises~\ref{exer:10.6.29}-\ref{exer:10.6.40}.)

Henceforth, we'll assume that $({\bf u},{\bf v})=0$. Let ${\bf U}$ and
${\bf V}$ be unit vectors in the directions of ${\bf u}$ and ${\bf v}$, respectively; that is, ${\bf U}={\bf u}/\|{\bf u}\|$ and ${\bf V}={\bf v}/\|{\bf v}\|$. The new rectangular coordinate system will
have the same origin as the $y_1$-$y_2$ system. The coordinates of a
point in this system will be denoted by $(z_1,z_2)$, where $z_1$ and
$z_2$ are the displacements in the directions of ${\bf U}$ and ${\bf V}$, respectively.

From \eqref{eq:10.6.5}, the solutions of \eqref{eq:10.6.10} are given by
\label{eq:10.6.13} {\bf y}=e^{\alpha t}\left[(c_1\cos\beta
t+c_2\sin\beta t){\bf u} +(-c_1\sin\beta t+c_2\cos\beta t){\bf
v}\right]. For convenience, let's call the curve
traversed by $e^{-\alpha t}{\bf y}(t)$ a {\color{blue}\it shadow trajectory\/} of
\eqref{eq:10.6.10}. Multiplying \eqref{eq:10.6.13} by $e^{-\alpha t}$ yields
$$e^{-\alpha t}{\bf y}(t)=z_1(t){\bf U}+z_2(t){\bf V},$$
where
\begin{eqnarray*}
z_1(t)&=&\|{\bf u}\|(c_1\cos\beta t+c_2\sin\beta
t)\\ z_2(t)&=&\|{\bf v}\|(-c_1\sin\beta t+c_2\cos\beta t).
\end{eqnarray*}
Therefore
$${(z_1(t))^2\over\|{\bf u}\|^2}+{(z_2(t))^2\over\|{\bf v}\|^2} =c_1^2+c_2^2$$
(verify!), which means that the shadow trajectories of \eqref{eq:10.6.10}
are ellipses centered at the origin, with axes of symmetry parallel to
${\bf U}$ and ${\bf V}$. Since
$$z_1'={\beta\|{\bf u}\|\over\|{\bf v}\|} z_2\mbox{\quad and \quad} z_2'=-{\beta\|{\bf v}\|\over\|{\bf u}\|} z_1,$$
the vector from the
origin to a point on the shadow ellipse rotates in the same direction
that ${\bf V}$ would have to be rotated by $\pi/2$ radians to bring it
into coincidence with ${\bf U}$ (Figures~\ref{figure:10.6.1} and
\ref{figure:10.6.2}).

\enlargethispage{3in}
\begin{figure}[H]
\color{blue}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.65}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100601}}
\caption{ Shadow trajectories traversed clockwise}
\label{figure:10.6.1}
\end{minipage}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.65}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100602}}
\caption{ Shadow trajectories traversed counterclockwise}
\label{figure:10.6.2}
\end{minipage}
\end{figure}

If $\alpha=0$, then any trajectory of \eqref{eq:10.6.10} is a shadow
trajectory of \eqref{eq:10.6.10};  therefore, if $\lambda$ is purely
imaginary, then the trajectories of \eqref{eq:10.6.10} are ellipses
traversed
periodically as indicated in Figures~\ref{figure:10.6.1} and
\ref{figure:10.6.2}.

If $\alpha>0$, then
$$\lim_{t\to\infty}\|{\bf y}(t)\|=\infty\mbox{\quad and \quad} \lim_{t\to-\infty}{\bf y}(t)=0,$$
so the trajectory spirals away from the origin as $t$ varies from
$-\infty$ to $\infty$. The direction of the spiral depends upon the
relative orientation of ${\bf U}$ and ${\bf V}$, as shown in
Figures~\ref{figure:10.6.3} and \ref{figure:10.6.4}.

If $\alpha<0$,  then
$$\lim_{t\to-\infty}\|{\bf y}(t)\|=\infty\mbox{\quad and \quad} \lim_{t\to\infty}{\bf y}(t)=0,$$
\enlargethispage{2in}
so the trajectory spirals toward the origin as $t$ varies from
$-\infty$ to $\infty$. Again, the direction of the spiral depends upon
the relative orientation of ${\bf U}$ and ${\bf V}$, as shown in
Figures~\ref{figure:10.6.5} and \ref{figure:10.6.6}.
\newpage

\begin{figure}[H]
\color{blue}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.65}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010603}}
\caption{ $\alpha>0$;   shadow trajectory spiraling outward}
\label{figure:10.6.3}
\end{minipage}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.65}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010604}}
\caption{ $\alpha>0$;   shadow trajectory spiraling outward}
\label{figure:10.6.4}
\end{minipage}
\end{figure}

\begin{figure}[H]
\color{blue}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.65}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010605}}
\caption{ $\alpha<0$;   shadow trajectory spiraling inward}
\label{figure:10.6.5}
\end{minipage}
\begin{minipage}[b]{0.5\linewidth}
\centering
\scalebox{.65}{
\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig100606}}
\caption{ $\alpha<0$;   shadow trajectory spiraling inward}
\label{figure:10.6.6}
\end{minipage}
\end{figure}

\exercises
In Exercises~\ref{exer:10.6.1}--\ref{exer:10.6.16}  find the general
solution.

\begin{exerciselist}
\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.1}  $\dst{{\bf y}'=\twobytwo{-1}2{-5}5{\bf y} }$&
\item\label{exer:10.6.2}  $\dst{{\bf y}'=\twobytwo{-11}4{-26}9{\bf y} }$
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.3}  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[29]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.5}  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[30]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.7} $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[31]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.9}  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[32]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.11}  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[33]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.13}  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[34]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.15}  $\dst{{\bf y}'=\threebythree60{-3}{-3}331{-2}6{\bf y}'}$&
\item\label{exer:10.6.16}  $\dst{{\bf y}'=\threebythree12{-2}02{-1}100{\bf y}'}$
\end{tabular}

\exercisetext{In Exercises~\ref{exer:10.6.17}--\ref{exer:10.6.24}
solve the initial value problem.}

\item\label{exer:10.6.17}
$\dst{{\bf y}'=\twobytwo4{-6}3{-2}{\bf y},\quad {\bf y}(0)=\twocol52}$

\item\label{exer:10.6.18}
$\dst{{\bf y}'=\twobytwo7{15}{-3}1{\bf y},\quad {\bf y}(0)=\twocol51}$

\item\label{exer:10.6.19}
$\dst{{\bf y}'=\twobytwo7{-15}3{-5}{\bf y},\quad {\bf y}(0)=\twocol{17}7}$

\item\label{exer:10.6.20}
$\dst{{\bf y}'={1\over6}\twobytwo4{-2}52{\bf y},\quad {\bf y}(0)=\twocol1{-1}}$

\item\label{exer:10.6.21}
$\dst{{\bf y}'=\threebythree52{-1}{-3}22132{\bf y},\quad {\bf y}(0)=\threecol406}$

\item\label{exer:10.6.22}
$\dst{{\bf y}'=\threebythree4408{10}{-20}23{-2}{\bf y},\quad {\bf y}(0)=\threecol865}$

\item\label{exer:10.6.23}
$\dst{{\bf y}'=\threebythree1{15}{-15}{-6}{18}{-22}{-3}{11}{-15}{\bf y},\quad {\bf y}(0)=\threecol{15}{17}{10}}$

\item\label{exer:10.6.24}
$\dst{{\bf y}'=\threebythree4{-4}4{-10}3{15}2{-3}1{\bf y},\quad {\bf y}(0)=\threecol{16}{14}6}$

\item\label{exer:10.6.25}
Suppose an $n\times n$ matrix $A$ with
real entries has a complex eigenvalue $\lambda=\alpha+i\beta$
($\beta\ne0$) with associated eigenvector ${\bf x}={\bf u}+i{\bf v}$,
where ${\bf u}$ and ${\bf v}$ have real components. Show that ${\bf u}$ and ${\bf v}$ are both nonzero.

\item\label{exer:10.6.26}
Verify that
$${\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \mbox{\quad and\quad} {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),$$
are the real and imaginary parts of
$$e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}).$$

\item\label{exer:10.6.27}
Show that if the vectors ${\bf u}$ and ${\bf v}$ are not both ${\bf 0}$ and $\beta\ne0$ then the vector functions
$${\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)\quad \mbox{ and }\quad {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)$$
are linearly independent on every interval. \hint{There are
two
cases to consider: {\rm(i)} $\{{\bf u},{\bf v}\}$ linearly
independent, and {\rm(ii)}
$\{{\bf u},{\bf v}\}$ linearly dependent. In either case, exploit the
the linear independence of $\{\cos\beta t,\sin\beta t\}$
on every interval.}

\item\label{exer:10.6.28}
Suppose ${\bf u}=\dst{\twocol{u_1}{u_2}}$ and ${\bf v}=\dst{\twocol{v_1}{v_2}}$ are not orthogonal; that is, $({\bf u},{\bf v})\ne0$.
\begin{alist}
\item % (a)
Show that  the quadratic equation
$$({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0$$
has a positive root $k_1$ and a negative root $k_2=-1/k_1$.
\item % (b)
Let ${\bf u}_1^{(1)}={\bf u}-k_1{\bf v}$, ${\bf v}_1^{(1)}={\bf v}+k_1{\bf u}$, ${\bf u}_1^{(2)}={\bf u}-k_2{\bf v}$, and ${\bf v}_1^{(2)}={\bf v}+k_2{\bf u}$, so that $({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0$, from the discussion
given above. Show that
$${\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1} \mbox{\quad and \quad} {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}.$$
\item % (c)
Let ${\bf U}_1$, ${\bf V}_1$, ${\bf U}_2$, and ${\bf V}_2$ be unit
vectors in the directions of ${\bf u}_1^{(1)}$, ${\bf v}_1^{(1)}$,
${\bf u}_1^{(2)}$, and ${\bf v}_1^{(2)}$, respectively. Conclude from
\part{a} that ${\bf U}_2={\bf V}_1$ and ${\bf V}_2=-{\bf U}_1$, and
that therefore the counterclockwise angles from ${\bf U}_1$ to ${\bf V}_1$ and from ${\bf U}_2$ to ${\bf V}_2$ are both $\pi/2$ or both
$-\pi/2$.
\end{alist}

\exercisetext{In Exercises~\ref{exer:10.6.29}-\ref{exer:10.6.32} find vectors
${\bf U}$ and ${\bf V}$ parallel to the axes of symmetry of the
trajectories, and plot some typical trajectories.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.29} \CGex\, $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[50]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.31} \CGex\,  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[51]/span, line 1, column 1 $
\end{tabular}

\exercisetext{In Exercises~\ref{exer:10.6.33}-\ref{exer:10.6.40} find vectors
${\bf U}$ and ${\bf V}$ parallel to the axes of symmetry of the shadow
trajectories, and plot a typical trajectory.}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.33} \CGex\,  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[53]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.35} \CGex\,  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[54]/span, line 1, column 1 $
\end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.37} \CGex\,  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[55]/span, line 1, column 1 $
\end{tabular}

\enlargethispage{1in}
\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}
\item\label{exer:10.6.39} \CGex\,  $\dst ParseError: invalid DekiScript (click for details) Callstack: at (Courses/Mount_Royal_University/MATH_3200:_Mathematical_Methods/4:_Linear_Systems_of_Ordinary_Differential_Equations_(LSODE)/4.6:_Constant_Coefficient_Homogeneous_Systems_III), /content/body/p[56]/span, line 1, column 1 $
\end{tabular}

\end{exerciselist}