Question

Find the real solution of the system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left[\begin{array}{cc}0& -3\\ 3& 0\end{array}\right]\overrightarrow{\mathbf{x}}$

Short answer

The solution is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\cos \left(3t\right)& -\sin \left(3t\right)\\ \sin \left(3t\right)& \cos \left(3t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

Step-by-step solution

definition of the theorem.

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$where A is a real $\mathbf{2}\mathbf{\times }\mathbf{2}$matrix with complex eigenvalues role="math" localid="1659881702011" $\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$(and $\mathit{q}\mathbf{\ne }\mathbf{0}$).

Consider an eigenvector $\overrightarrow{\mathbf{v}}\mathbf{+}\mathit{i}\overrightarrow{\mathbf{w}}$with eigenvalue $\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$.

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

To find the eigenvalues

Consider the given system as follows.

$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& -3\\ 3& 0\end{array}\right]\overrightarrow{x}$

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

$$\begin{gathered} \left|\begin{array}{cc}0-\lambda & -3\\ 3& 0-\lambda \end{array}\right|=0 \\ \left(0-\lambda \right)\left(0-\lambda \right)-\left(3\right)\left(-3\right)=0 \\ {\lambda }^2+9=0 \\ {\lambda }^2=-9 \end{gathered}$$

Simplify further as follows

$$\begin{gathered} {\lambda }^2=-9 \\ \lambda =\pm 3i \end{gathered}$$

Therefore, the eigenvalues are$\lambda =3iand\lambda =-3i$

To find the eigenvectors

To find a formula for trajectory as follows.

$$\begin{gathered} E_{3i}=ker\left[\begin{array}{cc}0-3i& -3\\ 3& 0-3i\end{array}\right] \\ =span\left(\left[\begin{array}{c}0\\ 1\end{array}\right]+i\left[\begin{array}{c}1\\ 0\end{array}\right]\right) \end{gathered}$$

Therefore, the eigenvectors are as follows.

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

Then by the theorem, the general solution for the system $\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& -3\\ 3& 0\end{array}\right]\overrightarrow{x}$is as follows.

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

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

$$\begin{gathered} p=0 \\ q=3 \end{gathered}$$

$S=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Substitute the value 0 for p and 3 for q and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for S in $\overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0$as follows.

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

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

where x and y are arbitrary constants.