Show that the system \[ \frac d{dt}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix} 0&1&0\\ 2&0&0\\ 0&0&3 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} \] has a solution whose norm tends to \(\infty\) as \(t\to-\infty\) and tends to \(0\) as \(t\to+\infty\).
Solution
Let \[ M = \begin{pmatrix} 0&1&0\\ 2&0&0\\ 0&0&3 \end{pmatrix}. \]
1. The vector\[ v = \begin{pmatrix} 1\\ -\sqrt2\\ 0 \end{pmatrix} \]is an eigenvector of \(M\) with eigenvalue \(-\sqrt2\).
Direct multiplication gives \[ Mv = \begin{pmatrix} -\sqrt2\\ 2\\ 0 \end{pmatrix} = -\sqrt2 \begin{pmatrix} 1\\ -\sqrt2\\ 0 \end{pmatrix} = -\sqrt2 v. \]
2. The function\[ X(t) \coloneqq e^{-\sqrt2 t}v \]is a solution of the system.
Using step 1, \[ X'(t) = -\sqrt2 e^{-\sqrt2 t}v = e^{-\sqrt2 t}Mv = MX(t). \]
3. This solution has the required limiting behavior:\[ \norm{X(t)}\longrightarrow0 \quad(t\to+\infty), \]while\[ \norm{X(t)}\longrightarrow\infty \quad(t\to-\infty). \]
Since \(v\neq0\), \[ \norm{X(t)} = e^{-\sqrt2 t}\norm v. \] The scalar factor tends to \(0\) as \(t\to+\infty\) and to \(\infty\) as \(t\to-\infty\).
4. Q.E.D.
Steps <1>2–<1>3 exhibit the requested solution explicitly.