Example 6.3.4.
In Example 6.3.2, we had solutions
\begin{equation*}
\mathbf x_1(t) = e^{2t} \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix},
\mathbf x_2(t) = e^{t} \begin{pmatrix} 5 \\ -7 \\ 13 \end{pmatrix}, \quad \text{and} \quad
\mathbf x_3(t) = e^{-t} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}.
\end{equation*}
The straight line through the origin and the point \((1, -1, 2)\) is a stable line. That is, for any initial condition \({\mathbf x}(0) = (x_0, y_0, z_0)\) lying on this line, our solution will tend toward the origin as \(t \to \infty\text{.}\) On the other hand, the plane spanned by \((2, -3, 5)\) and \((5, -7, 13)\) is unstable plane. Solutions on this plane move away from the origin as \(t \to \infty\text{.}\) Of course, \((0, 0, 0)\) is an equlibrium solution for our system. We say that the origin is a saddle in this example (Figure 6.3.5).
a stable line and an unstable plane in three dimensions with a transformation to another unstable plane and stable line