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Question

For solutions of x=Ax\overrightarrow{\textbf{x}}'=\textbf{A}\overrightarrow{\textbf{x}} with A skew-symmetric, the length of vector x\overrightarrow{\textbf{x}} is constant. What does this mean in terms of phase-plane trajectories? Use IDE's Matrix Element Input tool, or an open-ended graphic DE solver, to verify this for the following systems, and explain the role of the parameter k. (a)

x=[0110]x\overrightarrow{\textbf{x}}'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\overrightarrow{\textbf{x}}

(b)

x=[0kk0]x,k>0\overrightarrow{\textbf{x}}'=\begin{bmatrix}0&k\\-k&0\end{bmatrix}\overrightarrow{\textbf{x}},k>0

.

Solution

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(a)(a) We have the system

x=[0110]x\overrightarrow{\textbf{x}}'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\overrightarrow{\textbf{x}}

or

{x=yy=x\begin{align*}\begin{cases}x'=y\\y'=-x\end{cases}\end{align*}

Trajectories of this system are circles centered around the origin and the length of the vector x\overrightarrow{\textbf{x}} is a constant. We can see trajectories on following photo:

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