91Ó°ÊÓ

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{}$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 eigenvector $\stackrel{\mathit{→}}{\mathit{v}}\mathit{+}\mathit{i}\stackrel{\mathit{→}}{\mathit{w}}$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.

$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\stackrel{→}{x}$

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

$$\begin{gathered} \left[\begin{array}{cc}2-\mathrm{λ}& 4\\ -4& 2-\mathrm{λ}\end{array}\right]=0 \\ \left(2-\mathrm{λ}\right)\left(2-\mathrm{λ}\right)-\left(-4\right)\left(4\right)=0 \\ \left({\mathrm{λ}}^2-4\mathrm{λ}+4\right)+16=0 \\ {\mathrm{λ}}^2-4\mathrm{λ}+20=0 \end{gathered}$$

Simplify further as follows

$$\begin{gathered} \mathrm{λ}=\frac{4Â\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(20\right)}}{2\left(1\right)}     \left\{Qa=1,b=4,c=20\right\} \\ \mathrm{λ}=\frac{4Â\pm \sqrt{16-80}}{2} \\ \mathrm{λ}=\frac{4Â\pm \sqrt{-64}}{2} \\ \mathrm{λ}=\frac{4Â\pm i8}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{λ}=\frac{4Â\pm i8}{2} \\ \mathrm{λ}=2Â\pm i4 \\ {\mathrm{λ}}_1=2+i4 \\ {\mathrm{λ}}_2=2-4i \end{gathered}$$

Therefore, the eigenvalues are ${\mathrm{λ}}_1=2+4iand{\mathrm{λ}}_2=2-4i$

To find a formula for trajectory as follows.

$$\begin{gathered} E_{2+4i}=ker\left[\begin{array}{cc}2-\left(2+4i\right)& 4\\ -4& 2-\left(2-4i\right)\end{array}\right] \\ =ker\left[\begin{array}{cc}-4i& 4\\ -4& -4i\end{array}\right] \\ =span\left(\left[\begin{array}{c}1\\ 0\end{array}\right]+i\left[\begin{array}{c}0\\ 1\end{array}\right]\right) \end{gathered}$$

Therefore, the eigenvectors are as follows.

$$\begin{gathered} \stackrel{→}{v}=\left[\begin{array}{c}1\\ 2\end{array}\right] \\ \stackrel{→}{w}=\left[\begin{array}{c}0\\ 1\end{array}\right] \end{gathered}$$

Then by the theorem, the general solution for the system $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\stackrel{→}{x}$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.

$$\begin{gathered} p=2 \\ q=4 \\ S=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \end{gathered}$$

Substitute the value 2 for p and 4 for q and $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]forSin$

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

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\stackrel{→}{x}}_0 \\ \stackrel{→}{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}cos\left(4t\right)& -sin\left(4t\right)\\ sin\left(4t\right)& cos\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}0\left(cos\left(4t\right)\right)+1\left(sin\left(4t\right)\right)& 0\left(-sin\left(4t\right)\right)+1\left(cos\left(4t\right)\right)\\ 1\left(cos\left(4t\right)\right)+0\left(sin\left(4t\right)\right)& 1\left(-sin\left(4t\right)\right)+0\left(cos\left(4t\right)\right)\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}sin\left(4t\right)& cos\left(4t\right)\\ cos\left(4t\right)& -sin\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \end{gathered}$$

Hence, the solution for the system$\frac{\mathrm{d}\stackrel{→}{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\stackrel{→}{\mathrm{x}}$ is $\stackrel{→}{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}sin\left(4t\right)& cos\left(4t\right)\\ cos\left(4t\right)& -sin\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$, where x and y are arbitrary constants.