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Mathematics LibreTexts

10.6: Constant Coefficient Homogeneous Systems III

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    9458
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    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 \(\PageIndex{2}\)).

    clipboard_e1fb0a31d1ff1c3e7f0493671743c2916.png
    Figure \(\PageIndex{1}\): Shadow trajectories traversed clockwise.
    clipboard_e34070211f88f614240debf9da80ddd84.png
    Figure \(\PageIndex{2}\): Shadow trajectories traversed counterclockwise

    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 \(\PageIndex{6}\).

    clipboard_e868b4df1e27e9063c5afa2c8702c10da.png
    Figure \(\PageIndex{3}\): \(\alpha >0\); shadow trajectory spiraling outward.
    clipboard_ee60c14e857e8ad3f4fdace20c6d728dc.png
    Figure \(\PageIndex{4}\): \(\alpha >0\); shadow trajectory spiraling outward.
    clipboard_e554d4cf027ed78c945842edd49930e84.png
    Figure \(\PageIndex{5}\): \(\alpha <0\); shadow trajectory spiraling inward.
    clipboard_eddae2c394ff34110da5908f8425dd5b9.png
    Figure \(\PageIndex{6}\): \(\alpha <0\); shadow trajectory spiraling inward.