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 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 $\stackrel{\mathbf{→}}{\mathbf{v}}\mathbf{+}\mathit{i}\stackrel{\mathbf{→}}{\mathbf{w}}$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 ${\stackrel{\mathit{→}}{\mathit{x}}}_{\mathit{0}}$with respect to basis $\stackrel{\mathbf{→}}{\mathbf{w}}\mathbf{,}\stackrel{\mathbf{→}}{\mathbf{v}}$.

Consider the given system as follows.

$\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}-11& 15\\ -6& 7\end{array}\right]\stackrel{→}{x}$

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

$$\begin{gathered} \left[\begin{array}{cc}11-\mathrm{λ}& 15\\ -6& 7-\mathrm{λ}\end{array}\right]=0 \\ \left(-11-\mathrm{λ}\right)\left(7-\mathrm{λ}\right)-\left(-6\right)\left(15\right)=0 \\ \left({\mathrm{λ}}^2+4\mathrm{λ}-77\right)+90=0 \\ {\mathrm{λ}}^2+4\mathrm{λ}+13=0 \end{gathered}$$

Simplify further as follows

$$\begin{gathered} \mathrm{λ}=\frac{-4Â\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(13\right)}}{2\left(1\right)}\left\{Qa=1,b=4,c=13\right\} \\ \mathrm{λ}=\frac{-4Â\pm \sqrt{16-52}}{2} \\ \mathrm{λ}=\frac{-4Â\pm \sqrt{-36}}{2} \\ \mathrm{λ}=\frac{-4Â\pm i6}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{λ}=\frac{-4Â\pm i6}{2} \\ \mathrm{λ}=-2Â\pm i3 \\ {\mathrm{λ}}_1=-2+3i \\ {\mathrm{λ}}_2=-2-3i \end{gathered}$$

Therefore, the eigenvalues ${\mathrm{λ}}_1=-2+3i$are and ${\mathrm{λ}}_2=-2-3i$.

To find a formula for trajectory as follows.

$$\begin{gathered} E_{-2+3i}=ker\left[\begin{array}{cc}-11-\left(-2+3i\right)& 15\\ -6& 7-\left(-2+3i\right)\end{array}\right] \\ =ker\left[\begin{array}{cc}-9-3i& 15\\ -6& 9-3i\end{array}\right] \\ =span\left(\left[\begin{array}{c}5\\ 3\end{array}\right]+i\left[\begin{array}{c}0\\ 1\end{array}\right]\right) \end{gathered}$$

Therefore, the eigenvectors are as follows.

$$\begin{gathered} \stackrel{→}{v}=\left[\begin{array}{c}5\\ 3\end{array}\right] \\ \stackrel{→}{w}=\left[\begin{array}{c}0\\ 1\end{array}\right] \end{gathered}$$

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.

$$\begin{gathered} p=-2 \\ q=3 \\ S=\left[\begin{array}{cc}0& 5\\ 1& 3\end{array}\right] \end{gathered}$$

Substitute the value -2 for p and 3 for q and $\left[\begin{array}{cc}0& 5\\ 1& 3\end{array}\right]$for S in

role="math" localid="1659870301019" $\stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\stackrel{→}{x}}_0$

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\stackrel{→}{x}}_0 \\ \stackrel{→}{x}\left(t\right)=e^{-2t}\left[\begin{array}{cc}0& 5\\ 1& 3\end{array}\right]\left[\begin{array}{cc}cos\left(3t\right)& -sin\left(3t\right)\\ sin\left(3t\right)& cos\left(3t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=e^{-2t}\left[\begin{array}{cc}0\left(cos\left(3t\right)\right)+5\left(sin\left(3t\right)\right)& 0\left(-sin\left(3t\right)\right)+5\left(cos\left(3t\right)\right)\\ 1\left(cos\left(3t\right)\right)+3\left(sin\left(3t\right)\right)& 1\left(-sin\left(3t\right)\right)+3\left(cos\left(3t\right)\right)\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=e^{-2t}\left[\begin{array}{cc}5sin\left(3t\right)& 5cos\left(3t\right)\\ cos\left(3t\right)+3\left(sin\left(3t\right)\right)& -sin\left(3t\right)+3\left(cos\left(3t\right)\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \end{gathered}$$

Hence, the solution for the system $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}-11& 15\\ -6& 7\end{array}\right]\stackrel{→}{x}$is $\stackrel{→}{x}\left(t\right)=e^{-2t}\left[\begin{array}{cc}5sin\left(3t\right)& 5cos\left(3t\right)\\ cos\left(3t\right)+3\left(sin\left(3t\right)\right)& -sin\left(3t\right)+3\left(cos\left(3t\right)\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$, where x and y are arbitrary constants.