Repeated Eigenvalues — Question 10

Question 10 – Solution

Strategy. A repeated stable eigenvalue can have a two-dimensional eigenspace or a defective chain; the limiting direction detects that difference.

Step 1: Determine the eigenspaces. The eigenvalues are $-1$ (multiplicity two) and $2$. For $-1$, $z=0$ and $\varepsilon y=0$. Thus its eigenspace is the whole plane $z=0$ when $\varepsilon=0$, but only the $x$-axis when $\varepsilon>0$. The $2$ eigenspace is always $\operatorname{span}(1,0,1)^T$. There is a full eigenbasis exactly when $\boxed{\varepsilon=0}$; this diagonalizable three-dimensional matrix is not a scalar matrix.

Step 2: Solve the coupled components. First $z=re^{2t}$ and $y=qe^{-t}$. Then $(e^t x)'=\varepsilon q+3re^{3t}$, giving $$\boxed{X=e^{-t}\begin{pmatrix}p-r+\varepsilon qt\\q\\0\end{pmatrix} +re^{2t}\begin{pmatrix}1\\0\\1\end{pmatrix}.}$$ This satisfies every component and recovers $(p,q,r)^T$ at zero.

Step 3: Classify the stable-plane directions. Forward boundedness holds exactly when $r=0$. On that plane all states decay. For $\varepsilon=0$, the nonzero direction stays constantly $(p,q,0)^T/\sqrt{p^2+q^2}$. For $\varepsilon>0$ and $q\ne 0$, the normalized direction tends to $(\operatorname{sgn}q,0,0)^T$. If $q=0$, $p\ne 0$, it is constantly $(\operatorname{sgn}p,0,0)^T$. The zero solution has no normalized direction.

Step 4: Compare the two orders of limits. For the specified initial state, the normalized vector is $(\varepsilon t,1,0)^T/\sqrt{\varepsilon^2t^2+1}$. Taking $t\to\infty$ first with $\varepsilon>0$, then $\varepsilon\downarrow 0$, gives $\boxed{(1,0,0)^T}$. Reversing the operations gives $\boxed{(0,1,0)^T}$. At every fixed finite time the solution depends continuously on $\varepsilon$, but long-time direction is not uniform near zero. The eigenvalues stay fixed while the eigenspace dimension changes.