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# 10.6: Constant Coefficient Homogeneous Systems III

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ 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} \nonumber$

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

$A({\bf u}-i{\bf v})=(\alpha-i\beta)({\bf u}-i{\bf v}), \nonumber$

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 will not 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.

Theorem $$\PageIndex{1}$$

Let $$A$$ be an $$n\times n$$ matrix with real entries$$.$$ Let $$\lambda=\alpha+i\beta$$ ($$\beta\ne0$$) 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

${\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \quad \text{and} \quad {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t), \nonumber$

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 $${\bf y}'=A{\bf y}$$.

Proof

A function of the form Equation \ref{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 Equation \ref{eq:10.6.1} and collecting the real and imaginary parts of the result yields

$A({\bf u}+i{\bf v})=(\alpha{\bf u}-\beta{\bf v})+i(\alpha{\bf v}+\beta{\bf u}). \nonumber$

Equating real and imaginary parts on the two sides of this equation yields

$\begin{array}{rcl} A{\bf u}&=&\alpha{\bf u}-\beta{\bf v}\\ A{\bf v}&=&\alpha{\bf v}+\beta{\bf u}. \end{array}\nonumber$

We leave it to you (Exercise 10.6.25) to show from this that $${\bf u}$$ and $${\bf v}$$ are both nonzero. Substituting from these equations into Equation \ref{eq:10.6.4} yields

\begin{aligned} 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{aligned}\nonumber

This is true if

$\begin{array}{rcr} f_1'&=&\alpha f_1+\beta f_2\phantom{,}\\ f_2'&=&-\beta f_1+\alpha f_2, \end{array} \quad \text{or equivalently} \quad \begin{array}{rcr} f_1'-\alpha f_1&=&\phantom{-}\beta f_2\phantom{.}\\ f_2'-\alpha f_2&=&-\beta f_1. \end{array}\nonumber$

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

$\begin{array}{rcr} g_1'&=&\beta g_2\phantom{.}\\ g_2'&=&-\beta g_1, \end{array}\nonumber$

which implies that

$g_1''=\beta g_2'=-\beta^2 g_1, \nonumber$

so

$g_1''+\beta^2 g_1=0. \nonumber$

The general solution of this equation is

$g_1=c_1\cos\beta t+c_2\sin\beta t. \nonumber$

Moreover, since $$g_2=g_1'/\beta$$,

$g_2=-c_1\sin\beta t+c_2\cos\beta t. \nonumber$

Multiplying $$g_1$$ and $$g_2$$ by $$e^{\alpha t}$$ shows that

\begin{aligned} 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{aligned}\nonumber

Substituting these into Equation \ref{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 Equation \ref{eq:10.6.3} (Exercise 10.6.26), and that they are linearly independent (Exercise 10.6.27).

Example $$\PageIndex{1}$$

Find the general solution of

$\label{eq:10.6.6} {\bf y}'=\left[\begin{array}{cc}{4}&{-5}\\{5}&{-2}\end{array}\right]{\bf y}.$

Solution

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.6.6} is

$\left|\begin{array}{cc} 4-\lambda&-5\\ 5&-2-\lambda \end{array}\right|=(\lambda-1)^2+16. \nonumber$

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

$\left[\begin{array}{cccr} 3-4i&-5&\vdots&0\\ 5&-3-4i&\vdots&0 \end{array}\right], \nonumber$

which is row equivalent to

$\left[\begin{array}{cccr} 1&-{3+4i\over5}&\vdots&0\\ 0&0&\vdots&0 \end{array}\right]. \nonumber$

Therefore $$x_1=(3+4i)x_2/5$$. Taking $$x_2=5$$ yields $$x_1=3+4i$$, so

${\bf x}=\left[\begin{array}{c}3+4i\\5\end{array}\right] \nonumber$

is an eigenvector. The real and imaginary parts of

$e^t(\cos4t+i\sin4t)\left[\begin{array}{c}3+4i\\5\end{array}\right] \nonumber$

are

${\bf y}_1=e^t\left[\begin{array}{c}3\cos4t-4\sin 4t\\5\cos4t\end{array}\right]\quad\text{ and }\quad {\bf y}_2=e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin 4t\end{array}\right], \nonumber$

which are linearly independent solutions of Equation \ref{eq:10.6.6}. The general solution of Equation \ref{eq:10.6.6} is

${\bf y}= c_1e^t\left[\begin{array}{c}3\cos4t-4\sin 4t\\5\cos4t\end{array}\right]+ c_2e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin 4t\end{array}\right]. \nonumber$

Example $$\PageIndex{2}$$

Find the general solution of

$\label{eq:10.6.7} {\bf y}'=\left[\begin{array}{cc}{-14}&{39}\\{-6}&{16}\end{array}\right]{\bf y}.$

Solution

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.6.7} is

$\left|\begin{array}{cc}-14-\lambda&39\\-6&16-\lambda \end{array}\right|=(\lambda-1)^2+9. \nonumber$

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

$\left[\begin{array}{cccr}-15-3i&39&\vdots&0\\ -6&15-3i&\vdots&0 \end{array}\right], \nonumber$

which is row equivalent to

$\left[\begin{array}{cccr} 1&{-5+i\over2}&\vdots&0\\ 0&0&\vdots&0 \end{array}\right]. \nonumber$

Therefore $$x_1=(5-i)/2$$. Taking $$x_2=2$$ yields $$x_1=5-i$$, so

${\bf x}=\left[\begin{array}{c}5-i\\2\end{array}\right] \nonumber$

is an eigenvector. The real and imaginary parts of

$e^t(\cos3t+i\sin3t)\left[\begin{array}{c}5-i\\2\end{array}\right] \nonumber$

are

${\bf y}_1=e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos 3t\end{array}\right]\quad\text{ and }\quad {\bf y}_2=e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin 3t\end{array}\right], \nonumber$

which are linearly independent solutions of Equation \ref{eq:10.6.7}. The general solution of Equation \ref{eq:10.6.7} is

${\bf y}=c_1e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos 3t\end{array}\right]+ c_2e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin 3t\end{array}\right]. \nonumber$

Example $$\PageIndex{3}$$

Find the general solution of

$\label{eq:10.6.8} {\bf y}'=\left[\begin{array}{ccc}{-5}&{5}&{4}\\{-8}&{7}&{6}\\{1}&{0}&{0}\end{array}\right]{\bf y}.$

Solution

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.6.8} is

$\left|\begin{array}{ccc}-5-\lambda&5&4\\-8&7-\lambda& 6\\ \phantom{-}1 &0&-\lambda\end{array}\right|=-(\lambda-2)(\lambda^2+1). \nonumber$

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

$\left[\begin{array}{rrrcr}-7&5&4&\vdots&0\\-8& 5&6&\vdots&0\\ 1&0&-2&\vdots&0 \end{array}\right], \nonumber$

which is row equivalent to

$\left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&-2& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber$

Therefore $$x_1=x_2=2x_3$$. Taking $$x_3=1$$ yields

${\bf x}_1=\threecol221, \nonumber$

so

${\bf y}_1=\threecol221e^{2t} \nonumber$

is a solution of Equation \ref{eq:10.6.8}.

The augmented matrix of $$(A-iI){\bf x=0}$$ is

$\left[\begin{array}{ccrccc}-5-i&5&4&\vdots&0\\-8& 7-i&6&\vdots&0\\ \phantom{-}1&0&-i&\vdots&0 \end{array}\right], \nonumber$

which is row equivalent to

$\left[\begin{array}{ccccc} 1&0&-i&\vdots&0\\ 0&1&1-i& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber$

Therefore $$x_1=ix_3$$ and $$x_2=-(1-i)x_3$$. Taking $$x_3=1$$ yields the eigenvector

${\bf x}_2=\left[\begin{array}{c} i\\-1+i\\ 1\end{array} \right]. \nonumber$

The real and imaginary parts of

$(\cos t+i\sin t)\left[\begin{array}{c}i\\-1+i\\1\end{array}\right] \nonumber$

are

${\bf y}_2= \left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right] \quad\text{ and }\quad {\bf y}_3=\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin t\end{array}\right], \nonumber$

which are solutions of Equation \ref{eq:10.6.8}. Since the Wronskian of $$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ at $$t=0$$ is

$\left|\begin{array}{rrr} 2&0&1\\2&-1&1\\1&1&0\end{array}\right|=1, \nonumber$

$$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ is a fundamental set of solutions of Equation \ref{eq:10.6.8}. The general solution of Equation \ref{eq:10.6.8} is

${\bf y}=c_1 \threecol221e^{2t} +c_2\left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right] +c_3\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin t\end{array}\right]. \nonumber$

Example $$\PageIndex{4}$$

Find the general solution of

$\label{eq:10.6.9} {\bf y}'=\left[\begin{array}{ccc}{1}&{-1}&{-2}\\{1}&{3}&{2}\\{1}&{-1}&{2}\end{array}\right]{\bf y}.$

Solution

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.6.9} is

$\left|\begin{array}{ccc} 1-\lambda&-1&-2\\ 1&3-\lambda& \phantom{-}2\\ 1 &-1&2-\lambda\end{array}\right|= -(\lambda-2)\left((\lambda-2)^2+4\right). \nonumber$

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

$\left[\begin{array}{rrrcr}-1&-1&-2&\vdots&0\\1& 1&2&\vdots&0\\ 1&-1&0&\vdots&0 \end{array}\right], \nonumber$

which is row equivalent to

$\left[\begin{array}{rrrcr} 1&0&1&\vdots&0\\ 0&1&1& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber$

Therefore $$x_1=x_2=-x_3$$. Taking $$x_3=1$$ yields

${\bf x}_1=\threecol{-1}{-1}1, \nonumber$

so

${\bf y}_1=\threecol{-1}{-1}1e^{2t} \nonumber$

is a solution of Equation \ref{eq:10.6.9}. The augmented matrix of $$\left(A-(2+2i)I\right){\bf x=0}$$ is

$\left[\begin{array}{ccrcc}-1-2i&-1&-2&\vdots&0\\ 1& 1-2i&\phantom{-}2&\vdots&0\\ 1&-1&-2i&\vdots&0 \end{array}\right], \nonumber$

which is row equivalent to

$\left[\begin{array}{rrrcr} 1&0&-i&\vdots&0\\ 0&1&i& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber$

Therefore $$x_1=ix_3$$ and $$x_2=-ix_3$$. Taking $$x_3=1$$ yields the eigenvector

${\bf x}_2=\threecol i{-i}1 \nonumber$

The real and imaginary parts of

$e^{2t}(\cos2t+i\sin2t)\threecol i{-i}1 \nonumber$

are

${\bf y}_2=e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos 2t\end{array}\right]\quad\text{ and }\quad {\bf y}_2=e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin 2t\end{array}\right], \nonumber$

which are solutions of Equation \ref{eq:10.6.9}. Since the Wronskian of $$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ at $$t=0$$ is

$\left|\begin{array}{rrr} -1&0&1\\-1&0&-1\\1&1&0\end{array}\right|=-2, \nonumber$

$$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ is a fundamental set of solutions of Equation \ref{eq:10.6.9}. The general solution of Equation \ref{eq:10.6.9} is

${\bf y}=c_1\threecol{-1}{-1}1e^{2t}+ c_2e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos 2t\end{array}\right]+ c_3e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin 2t\end{array}\right]. \nonumber$

## Geometric Properties of Solutions when $$n=2$$

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

$A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right] \nonumber$

has a complex eigenvalue $$\lambda=\alpha+i\beta$$ ($$\beta\ne0$$) 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 is 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

${\bf x}_1=(1+ik){\bf x}=(1+ik)({\bf u}+i{\bf v}) =({\bf u}-k{\bf v})+i({\bf v}+k{\bf u}) \nonumber$

is also a $$\lambda$$-eigenvector of $$A$$, since

$A{\bf x}_1= A((1+ik){\bf x})=(1+ik)A{\bf x}=(1+ik)\lambda{\bf x}= \lambda((1+ik){\bf x})=\lambda{\bf x}_1. \nonumber$

The real and imaginary parts of $${\bf x}_1$$ are

$\label{eq:10.6.11} {\bf u}_1={\bf u}-k{\bf v} \quad \text{and} \quad {\bf v}_1={\bf v}+k{\bf u},$

so

$({\bf u}_1,{\bf v}_1)=({\bf u}-k{\bf v},{\bf v}+k{\bf u}) =-\left[({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k -({\bf u},{\bf v})\right]. \nonumber$

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 10.6.28).

Example $$\PageIndex{5}$$

In Example 10.6.1 we found the eigenvector

${\bf x}=\ctwocol{3+4i}5=\twocol35+i\twocol40 \nonumber$

for the matrix of the system Equation \ref{eq:10.6.6}. Here $$\bf {u}=\twocol{3}{5}$$ and $${\bf v}=\twocol40$$ are not orthogonal, since $$({\bf u},{\bf v})=12$$. Since $$\|{\bf v}\|^2-\|{\bf u}\|^2=-18$$, Equation \ref{eq:10.6.12} is equivalent to

$2k^2-3k-2=0. \nonumber$

The zeros of this equation are $$k_1=2$$ and $$k_2=-1/2$$. Letting $$k=2$$ in Equation \ref{eq:10.6.11} yields

${\bf u}_1={\bf u}-2{\bf v}=\twocol{-5}{\phantom{-}5} \quad \text{and} \quad {\bf v}_1={\bf v}+2{\bf u}=\twocol{10}{10}, \nonumber$

and $$({\bf u}_1,{\bf v}_1)=0$$. Letting $$k=-1/2$$ in Equation \ref{eq:10.6.11} yields

${\bf u}_1={\bf u}+{{\bf v}\over2}=\twocol{5}5 \quad \text{and} \quad {\bf v}_1={\bf v}-{{\bf u}\over2}={1\over2}\twocol{-5}{\phantom{-}5}, \nonumber$

and again $$({\bf u}_1,{\bf v}_1)=0$$.

(The numbers don’t always work out as nicely as in this example. You’ll need a calculator or computer to do Exercises 10.6.29-10.6.40.) 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 Equation \ref{eq:10.6.5}, the solutions of Equation \ref{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 shadow trajectory of Equation \ref{eq:10.6.10}. Multiplying Equation \ref{eq:10.6.13} by $$e^{-\alpha t}$$ yields

$e^{-\alpha t}{\bf y}(t)=z_1(t){\bf U}+z_2(t){\bf V}, \nonumber$

where

\begin{aligned} 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{aligned}\nonumber

Therefore

${(z_1(t))^2\over\|{\bf u}\|^2}+{(z_2(t))^2\over\|{\bf v}\|^2} =c_1^2+c_2^2 \nonumber$

(verify!), which means that the shadow trajectories of Equation \ref{eq:10.6.10} are ellipses centered at the origin, with axes of symmetry parallel to $${\bf U}$$ and $${\bf V}$$. Since

$z_1'={\beta\|{\bf u}\|\over\|{\bf v}\|} z_2 \quad \text{and} \quad z_2'=-{\beta\|{\bf v}\|\over\|{\bf u}\|} z_1, \nonumber$

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 $$\PageIndex{1}$$ and $$\PageIndex{2}$$).

If $$\alpha=0$$, then any trajectory of Equation \ref{eq:10.6.10} is a shadow trajectory of Equation \ref{eq:10.6.10}; therefore, if $$\lambda$$ is purely imaginary, then the trajectories of Equation \ref{eq:10.6.10} are ellipses traversed periodically as indicated in Figures $$\PageIndex{1}$$ and $$\PageIndex{2}$$. If $$\alpha>0$$, then

$\lim_{t\to\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to-\infty}{\bf y}(t)=0,\ \nonumber$

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 $$\PageIndex{3}$$ and $$\PageIndex{4}$$. If $$\alpha<0$$, then

$\lim_{t\to-\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to\infty}{\bf y}(t)=0, \nonumber$

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 $$\PageIndex{5}$$ and $$\PageIndex{6}$$.