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## Complex Eigenvalues for Linear Systems

Consider the following linear system: $\overrightarrow{x}'=A(t)\overrightarrow{x}, \;\;\;\;(55)$ $$A(t)$$ is a real-valued continuous function of $$t$$.

Theorem. If $$\overrightarrow{x}=\overrightarrow{u}(t)+i\overrightarrow{v}(t)$$ is a solution of (55) for real-valued functions $$\overrightarrow{u}(t)$$ and $$\overrightarrow{v}(t)$$, then so are $$\overrightarrow{u}(t)$$ and $$\overrightarrow{v}(t)$$.

\begin{align*} (\overrightarrow{u}+i\overrightarrow{v})'&= A(\overrightarrow{u}+i\overrightarrow{v})\\ \overrightarrow{u}'+i\overrightarrow{v}'&= A\overrightarrow{u}+iA\overrightarrow{v}. \end{align*} Comparing real and imaginary parts of both sides, we get $$\overrightarrow{u}'=A\overrightarrow{u}$$ and $$\overrightarrow{v}'=A\overrightarrow{v}$$.

Theorem. If $$A$$ has complex eigenvalue $$\lambda=r+i\theta$$ with corresponding eigenvector $$\overrightarrow{v}$$, then two linearly independent solutions of (55) are $\overrightarrow{x_1}=e^{rt}[\cos(\theta t)Re \overrightarrow{v}-\sin(\theta t)Im \overrightarrow{v}]$ and $\overrightarrow{x_2}=e^{rt}[\cos(\theta t)Im \overrightarrow{v}+\sin(\theta t)Re \overrightarrow{v}].$

We have verified before that $$\overrightarrow{x}=e^{(r+i\theta)t}\overrightarrow{v}$$ is a solution. \begin{align*} \overrightarrow{x}&=e^{rt}[\cos(\theta t)+i\sin(\theta t)][Re \overrightarrow{v}+iIm\overrightarrow{v}]\\ &=e^{rt}[\cos(\theta t)Re \overrightarrow{v}-\sin(\theta t)Im \overrightarrow{v}]+ie^{rt}[\cos(\theta t)Im \overrightarrow{v}+\sin(\theta t)Re \overrightarrow{v}]. \end{align*} The rest follows from the preceding theorem and Wronskian.

Example. $$\overrightarrow{x}'=\left[\begin{array}{rr} 1&-2\\5&3\end{array} \right] \overrightarrow{x}$$.

1. Find the general solution.

2. Draw the phase portrait.

Solution. (a) $$\lambda=2\pm 3i$$ and $$\overrightarrow{v}= \left[\begin{array}{l}-1+3i\\5\end{array} \right]$$. A solution is \begin{align*} \overrightarrow{x}&=e^{2t}[\cos(3t)+i\sin(3t)] \left(\left[\begin{array}{r}-1\\5\end{array} \right] +i\left[\begin{array}{r}3\\0\end{array} \right]\right)\\ &=e^{2t}\left(\cos(3t)\left[\begin{array}{r}-1\\5\end{array} \right]-\sin(3t)\left[\begin{array}{r}3\\0\end{array} \right] \right)+ ie^{2t}\left(\cos(3t)\left[\begin{array}{r}3\\0\end{array} \right]+\sin(3t)\left[\begin{array}{r}-1\\5\end{array} \right]\right) \end{align*} So the general solution is \begin{align*} \overrightarrow{x}&=c_1 \overrightarrow{x_1} +c_2\overrightarrow{x_2}\\ &=c_1e^{2t}\left(\cos(3t)\left[\begin{array}{r}-1\\5\end{array} \right]-\sin(3t)\left[\begin{array}{r}3\\0\end{array} \right] \right)+ c_2e^{2t}\left(\cos(3t)\left[\begin{array}{r}3\\0\end{array} \right]+\sin(3t)\left[\begin{array}{r}-1\\5\end{array} \right]\right)\\ &=c_1e^{2t}\left[\begin{array}{l}-\cos(3t)-3\sin(3t)\\5\cos(3t)\end{array} \right]+ c_2e^{2t}\left[\begin{array}{r}3\cos(3t)- \sin(3t)\\5\sin(3t)\end{array} \right].\\ \end{align*} (b) To draw the spiral for $$\overrightarrow{x_1}$$, get a point on it by plugging $$t=0$$ and then find the tangent vector $$\overrightarrow{x_1}'(0)$$ at the point $$\overrightarrow{x_1}(0)$$. Do the same for $$\overrightarrow{x_2}$$. The origin is an unstable spiral point. Note that the origin would have been an asymptotically stable spiral point if eigenvalues had negative real part.

Note that there are 3 types of trajectories for the eigenvalue $$\lambda=r\pm i \theta$$ in a system of 2 linear ODEs:

1. $$r < 0$$, the origin is an asymptotically stable spiral point

2. $$r > 0$$, the origin is an unstable spiral point

3. $$r=0$$, the origin is a stable center (trajectories are circle or ellipse)

Example. $$\overrightarrow{x}'=\left[\begin{array}{rr} -1&-2\\1&1\end{array} \right] \overrightarrow{x}$$.

1. Find the general solution.

2. Draw the phase portrait.

Solution. (a) $$\lambda=\pm i$$ and $$\overrightarrow{v}= \left[\begin{array}{l}-1+i\\1\end{array} \right]$$. A solution is \begin{align*} \overrightarrow{x}&=[\cos(t)+i\sin(t)] \left(\left[\begin{array}{r}-1\\1\end{array} \right] +i\left[\begin{array}{r}1\\0\end{array} \right]\right)\\ \end{align*} So the general solution is \begin{align*} \overrightarrow{x}&=c_1 \overrightarrow{x_1} +c_2\overrightarrow{x_2}\\ &=c_1\left(\cos(t)\left[\begin{array}{r}-1\\1\end{array} \right]-\sin(t)\left[\begin{array}{r}1\\0\end{array} \right] \right)+ c_2\left(\cos(t)\left[\begin{array}{r}1\\0\end{array} \right]+\sin(t)\left[\begin{array}{r}-1\\1\end{array} \right]\right)\\ &=c_1 \left[\begin{array}{l}-\cos t-\sin t\\\cos t\end{array} \right]+ c_2\left[\begin{array}{r}\cos t-\sin t\\\sin t\end{array} \right] \end{align*} (b) Let's draw the trajectory for $$\overrightarrow{x}=\overrightarrow{x_1}=\left[\begin{array}{l}-\cos t-\sin t\\\cos t\end{array} \right]$$. It gives the trajectory $x_1=-\cos t-\sin t,\; x_2=\cos t.$ Solving for $$\cos t$$ and $$\sin t$$, we get $$\sin t=-(x_1+x_2),\; \cos t=x_2$$. Using $$\sin^2 t+\cos^2 t=1$$, we get the trajectory which is the ellipse $(x_1+x_2)^2+x_2^2=1$ whose one axis is $$x_1+x_2=0$$. To determine the direction of a trajectory, plot tangent vector at a point, say $$[1, \; 0]^T$$: $\overrightarrow{x}'=\left[\begin{array}{rr} -1&-2\\1&1\end{array} \right] \left[\begin{array}{r}1\\0\end{array} \right]=\left[\begin{array}{r}-1\\1\end{array} \right].$ So trajectories move counterclockwise. Note that if $$A$$ is not given, then plug $$t=\pi/2$$ to get a point $$\overrightarrow{x_1}(\pi/2)$$ and the tangent vector $$\overrightarrow{x_1}'(\pi/2)$$ at the point $$\overrightarrow{x_1}(\pi/2)$$. The origin is a stable center, but not asymptotically stable as trajectories do not move toward the origin as $$t$$ increases.

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