10.4: Distinct Complex-Conjugate Eigenvalues
Example: Find the general solution of \(\dot{x}_{1}=-\frac{1}{2} x_{1}+x_{2}, \dot{x}_{2}=-x_{1}-\frac{1}{2} x_{2}\) .
The equations in matrix form are
\[\frac{d}{d t}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{cc} -1 / 2 & 1 \\ -1 & -1 / 2 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \nonumber \]
The ansatz \(\mathrm{x}=v e^{\lambda t}\) leads to the equation
\[\begin{aligned} 0 &=\operatorname{det}(\mathrm{A}-\lambda \mathrm{I}) \\ &=\lambda^{2}+\lambda+\frac{5}{4} \end{aligned} \nonumber \]
Therefore, \(\lambda=-1 / 2 \pm i\) ; and we observe that the eigenvalues occur as a complex conjugate pair. We will denote the two eigenvalues as
\[\lambda=-\frac{1}{2}+i \quad \text { and } \quad \bar{\lambda}=-\frac{1}{2}-i \nonumber \]
Now, if \(\mathrm{A}\) a real matrix, then \(\mathrm{Av}=\lambda_{\mathrm{v}}\) implies \(\mathrm{A} \overline{\mathrm{v}}=\bar{\lambda} \overline{\mathrm{v}}\) , so the eigenvectors also occur as a complex conjugate pair. The eigenvector \(v\) associated with eigenvalue \(\lambda\) satisfies \(-i v_{1}+v_{2}=0\) , and normalizing with \(v_{1}=1\) , we have
\[\mathrm{v}=\left(\begin{array}{l} 1 \\ i \end{array}\right) \nonumber \]
We have therefore determined two independent complex solutions to the ode, that is,
\[\mathrm{v} e^{\lambda t} \text { and } \overline{\mathrm{v}} e^{\bar{\lambda} t}, \nonumber \]
and we can form a linear combination of these two complex solutions to construct two independent real solutions. Namely, if the complex functions \(z(t)\) and \(\bar{z}(t)\) are written as
\[z(t)=\operatorname{Re}\{z(t)\}+i \operatorname{Im}\{z(t)\}, \quad \bar{z}(t)=\operatorname{Re}\{z(t)\}-i \operatorname{Im}\{z(t)\}, \nonumber \]
then two real functions can be constructed from the following linear combinations of \(z\) and \(\bar{z}\) :
\[\frac{z+\bar{z}}{2}=\operatorname{Re}\{z(t)\} \quad \text { and } \quad \frac{z-\bar{z}}{2 i}=\operatorname{Im}\{z(t)\} . \nonumber \]
Thus the two real vector functions that can be constructed from our two complex vector functions are
\[\begin{aligned} \operatorname{Re}\left\{\mathbf{v} e^{\lambda t}\right\} &=\operatorname{Re}\left\{\left(\begin{array}{l} 1 \\ i \end{array}\right) e^{\left(-\frac{1}{2}+i\right) t}\right\} \\ &=e^{-\frac{1}{2} t} \operatorname{Re}\left\{\left(\begin{array}{l} 1 \\ i \end{array}\right)(\cos t+i \sin t)\right\} \\ &=e^{-\frac{1}{2} t}\left(\begin{array}{r} \cos t \\ -\sin t \end{array}\right) ; \end{aligned} \nonumber \]
and
\[\begin{aligned} \operatorname{Im}\left\{\mathbf{v} e^{\lambda t}\right\} &=e^{-\frac{1}{2} t} \operatorname{Im}\left\{\left(\begin{array}{l} 1 \\ i \end{array}\right)(\cos t+i \sin t)\right\} \\ &=e^{-\frac{1}{2} t}\left(\begin{array}{c} \sin t \\ \cos t \end{array}\right) . \end{aligned} \nonumber \]
Taking a linear superposition of these two real solutions yields the general solution to the ode, given by
\[x=e^{-\frac{1}{2} t}\left[A\left(\begin{array}{r} \cos t \\ -\sin t \end{array}\right)+B\left(\begin{array}{l} \sin t \\ \cos t \end{array}\right)\right] \nonumber \]
The corresponding phase portrait is shown in Fig. 10.2. We say the origin is a spiral point . If the real part of the complex eigenvalue is negative, as is the case here, then solutions spiral into the origin. If the real part of the eigenvalue is positive, then solutions spiral out of the origin.
The direction of the spiral-here, it is clockwise-can be determined easily. If we examine the ode with \(x_{1}=0\) and \(x_{2}=1\) , we see that \(\dot{x}_{1}=1\) and \(\dot{x}_{2}=-1 / 2\) . The trajectory at the point \((0,1)\) is moving to the right and downward, and this is possible only if the spiral is clockwise. A counterclockwise trajectory would be moving to the left and downward.