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Continuous dynamical system with eigenvalues$\mathit{p}\mathbf{卤}\mathit{i}\mathit{q}$

Consider the linear system$\frac{\mathbf{d}\stackrel{\mathbf{鈫�}}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\stackrel{\mathbf{鈫�}}{\mathbf{x}}$where$\mathit{A}$is a real$\mathbf{2}\mathbf{脳}\mathbf{2}$matrix with complex eigenvalues$\mathit{p}\mathbf{卤}\mathit{i}\mathit{q}$(and$\mathit{q}\mathbf{鈮�}\mathbf{0}$).

Consider an eigenvector$\stackrel{\mathbf{鈫�}}{\mathbf{v}}\mathbf{+}\mathit{i}\stackrel{\mathbf{鈫�}}{\mathbf{w}}$with eigenvalue$\mathit{p}\mathbf{卤}\mathit{i}\mathit{q}$.

Then$\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{p}\mathbf{t}}\mathit{S}\mathbf{\left[}\begin{array}{cc}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}& \mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}\\ \mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}& \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{q}\mathbf{t}\mathbf{\right)}\end{array}\mathbf{\right]}{\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$where$\mathit{S}\mathbf{=}\mathbf{\left[}\stackrel{\mathbf{鈫�}}{\mathbf{w}}\mathbf{\text{鈥�}}\stackrel{\mathbf{鈫�}}{\mathbf{v}}\mathbf{\right]}$. Recall that${\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$is the coordinate vector of${\stackrel{\mathbf{鈫�}}{\mathbf{x}}}_{\mathbf{0}}$with respect to basis$\stackrel{\mathbf{鈫�}}{\mathbf{w}}\mathbf{,}\stackrel{\mathbf{鈫�}}{\mathbf{v}}$.

Consider a$2脳2$matrix$A$with eigenvalues$卤蟺i$. Let$\stackrel{鈫�}{v}+i\stackrel{鈫�}{w}$be an eigenvector of$A$with eigenvalue$蟺i$.

Now, consider the system as follows.

$\frac{d\stackrel{鈫�}{x}}{dt}=A\stackrel{鈫�}{x}$with the initial condition${\stackrel{鈫�}{x}}_0=\stackrel{鈫�}{w}$.

If$A$has eigenvalue$p卤iq(q鈮�0)$and$\stackrel{鈫�}{v}+i\stackrel{鈫�}{w}$is the eigenvector of$p卤iq$, then the solution as follows.

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

Substitute the value$0$for$t$in$\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}$as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺\left(0\right)\right)& -sin\left(蟺\left(0\right)\right)\\ sin\left(蟺\left(0\right)\right)& cos\left(蟺\left(0\right)\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(0\right)& -sin\left(0\right)\\ sin\left(0\right)& cos\left(0\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\text{鈥夆赌�}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(0\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{l}\stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(0\right)=\stackrel{鈫�}{w}\end{array}$

$\begin{array}{l}\stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(0\right)=\stackrel{鈫�}{w}\end{array}$

Substitute the value$\frac{1}{2}$for$t$in$\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}$as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺\left(\frac{1}{2}\right)\right)& -sin\left(蟺\left(\frac{1}{2}\right)\right)\\ sin\left(蟺\left(\frac{1}{2}\right)\right)& cos\left(蟺\left(\frac{1}{2}\right)\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(\frac{蟺}{2}\right)& -sin\left(\frac{蟺}{2}\right)\\ sin\left(\frac{蟺}{2}\right)& cos\left(\frac{蟺}{2}\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\text{鈥夆赌�}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]{\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]}^{-1}\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\left[\begin{array}{cc}{v}_1& -w_1\\ v_2& -w_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]\end{array}$

Simplify further as follows.

$\stackrel{鈫�}{x}\left(\frac{1}{2}\right)=\stackrel{鈫�}{v}$

Substitute the value$1$for$t$in$\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}$as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(1\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺\left(1\right)\right)& -sin\left(蟺\left(1\right)\right)\\ sin\left(蟺\left(1\right)\right)& cos\left(蟺\left(1\right)\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(1\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺\right)& -sin\left(蟺\right)\\ sin\left(蟺\right)& cos\left(蟺\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& 鈭�1\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(1\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\text{鈥夆赌�}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}-w_1& -v_1\\ -w_2& -v_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(1\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}-w_1& -v_1\\ -w_2& -v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{l}\stackrel{鈫�}{x}\left(1\right)=\left[\begin{array}{cc}-w_1& -v_1\\ -w_2& -v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(1\right)=\left[\begin{array}{c}-w_1\\ -w_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(1\right)=-\stackrel{鈫�}{w}\end{array}$

Substitute the value$2$for$t$in$\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}cos\left(蟺t\right)& -sin\left(蟺t\right)\\ sin\left(蟺t\right)& cos\left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}$as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(t\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}\cos \left(蟺t\right)& -\sin \left(蟺t\right)\\ \sin \left(蟺t\right)& \cos \left(蟺t\right)\end{array}\right]S^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(2\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}\cos \left(蟺\left(2\right)\right)& -\sin \left(蟺\left(2\right)\right)\\ \sin \left(蟺\left(2\right)\right)& \cos \left(蟺\left(2\right)\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ \stackrel{鈫�}{x}\left(2\right)=\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}\cos \left(2蟺\right)& -\sin \left(2蟺\right)\\ \sin \left(2蟺\right)& \cos \left(2蟺\right)\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\right]}^{-1}\stackrel{鈫�}{w}\end{array}$

Simplify further as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(2\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{cc}\stackrel{鈫�}{w}& \stackrel{鈫�}{v}\end{array}\text{鈥夆赌�}\right]}^{-1}\stackrel{鈫�}{w}\\ =\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{v}_2& -v_1\\ -w_2& w_1\end{array}\right]\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ =\frac{1}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}{v}_2{w}_1-v_1{w}_2\\ -w_2{w}_1+w_1{w}_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(2\right)=\frac{v_2{w}_1-v_1{w}_2}{w_1{v}_2-w_2{v}_1}\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Simplify further as follows.

$\begin{array}{l}\stackrel{鈫�}{x}\left(2\right)=\left[\begin{array}{cc}{w}_1& v_1\\ w_2& v_2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(2\right)=\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]\\ \stackrel{鈫�}{x}\left(2\right)=\stackrel{鈫�}{w}\end{array}$

The vectors $\stackrel{鈫�}{x}\left(0\right),\stackrel{鈫�}{x}\left(\frac{1}{2}\right),\stackrel{鈫�}{x}\left(1\right)$and $\stackrel{鈫�}{x}\left(2\right)$is pointed in the figure 1 as follows.

Thus, the vectors $\stackrel{鈫�}{x}\left(0\right),\stackrel{鈫�}{x}\left(\frac{1}{2}\right),\stackrel{鈫�}{x}\left(1\right)$and$\stackrel{鈫�}{x}\left(2\right)$are shown in figure 1.