91Ó°ÊÓ

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}$where A is a real $\mathbf{2}\mathbf{×}\mathbf{2}$matrix with complex eigenvalues $\mathit{p}\mathbf{Â\pm }\mathit{i}\mathit{q}$(and $\mathit{q}\mathbf{≠}\mathbf{0}$).

Consider an eigenvectorrole="math" localid="1659877431239" $\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{+}\mathit{i}\stackrel{\mathbf{→}}{\mathbf{w}}$with eigenvalue $\mathit{p}\mathbf{Â\pm }\mathit{i}\mathit{q}$.

Then$\stackrel{\mathbf{→}}{\mathbf{x}}\left(t\right)\mathbf{=}{\mathit{e}}^{\mathbf{p}\mathbf{t}}\mathit{S}\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]{\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$ where $\mathit{S}\mathbf{=}\left[\stackrel{→}{w} \stackrel{→}{v}\right]$. Recall that ${\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$ is the coordinate vector of ${\stackrel{\mathbf{→}}{\mathbf{x}}}_{\mathbf{0}}$ with respect to basis $\stackrel{\mathbf{→}}{\mathbf{w}}\mathbf{,}\stackrel{\mathbf{→}}{\mathbf{v}}$.

Consider the given system as follows.

$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\stackrel{→}{x}$

No, to find the eigenvalues of the coefficient matrix as follows.

$$\begin{gathered} \left[\begin{array}{cc}0-\mathrm{λ}& 4\\ -9& 0-\mathrm{λ}\end{array}\right]=0 \\ \left(0-\mathrm{λ}\right)\left(0-\mathrm{λ}\right)-\left(-9\right)\left(4\right)=0 \\ {\mathrm{λ}}^2+36=0 \\ {\mathrm{λ}}^2=-36 \end{gathered}$$

Simplify further as follows

$$\begin{gathered} {\mathrm{λ}}^2=-36 \\ \mathrm{λ}=Â\pm 6i \end{gathered}$$

Therefore, the eigenvalues are${\mathrm{λ}}_1=6i$and${\mathrm{λ}}_2=-6i$

To find a formula for trajectory as follows.

$$\begin{gathered} E_{6i}=ker\left[\begin{array}{cc}0-6i& 4\\ -9& 0-6i\end{array}\right] \\ =span\left(\left[\begin{array}{c}2\\ 0\end{array}\right]+i\left[\begin{array}{c}0\\ 3\end{array}\right]\right) \end{gathered}$$

Therefore, the eigenvectors are as follows.

$$\begin{gathered} \stackrel{→}{v}=\left[\begin{array}{c}2\\ 0\end{array}\right] \\ \stackrel{→}{w}=\left[\begin{array}{c}0\\ 3\end{array}\right] \end{gathered}$$

Then by the theorem, the general solution for the system $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\stackrel{→}{x}$is as follows.

$\stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\stackrel{→}{x}}_0$where $S=\left[\stackrel{→}{w}  \stackrel{→}{v}\right]$

Similarly, the values of p, q and S are as follows.

$$\begin{gathered} p=0 \\ q=6 \\ S=\left[\begin{array}{cc}0& 2\\ 3& 0\end{array}\right] \end{gathered}$$

Substitute the value 0 for p and 6 for q and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for S in $\stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\stackrel{→}{x}}_0$ as follows.

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\stackrel{→}{x}}_0 \\ \stackrel{→}{x}\left(t\right)=e^{0t}\left[\begin{array}{cc}0& 2\\ 3& 0\end{array}\right]\left[\begin{array}{cc}cos\left(6t\right)& -sin\left(6t\right)\\ sin\left(6t\right)& cos\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=\left[\begin{array}{cc}0\left(cos\left(6t\right)\right)+2\left(sin\left(6t\right)\right)& 0\left(-sin\left(6t\right)\right)+2\left(cos\left(6t\right)\right)\\ 3\left(cos\left(6t\right)\right)+0\left(sin\left(6t\right)\right)& 3\left(-sin\left(6t\right)\right)+0\left(cos\left(6t\right)\right)\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=\left[\begin{array}{cc}2sin\left(6t\right)& 2cos\left(6t\right)\\ 3cos\left(6t\right)& -3sin\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \end{gathered}$$

Hence, the solution for the system$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\stackrel{→}{x}$is$\stackrel{→}{x}\left(t\right)=\left[\begin{array}{cc}2sin\left(6t\right)& 2cos\left(6t\right)\\ 3cos\left(6t\right)& -3sin\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

where x and y are arbitrary constants.