Complex Eigenvalues — Question 9

Question 9 – Solution

Strategy. Decouple the rotating plane from the real exponential axis, and distinguish convergence of a state from convergence of its direction.

Step 1: Separate the blocks. The eigenvalues are $-1\pm 2i$ and $1$. The complete solution is $$\boxed{\begin{aligned} x&=e^{-t}(p\cos 2t-q\sin 2t),\\ y&=e^{-t}(p\sin 2t+q\cos 2t),\\ z&=r e^t. \end{aligned}}$$ The planar rotation is invertible at every time, so the arbitrary three initial components are represented without loss.

Step 2: Classify boundedness in both time directions. Exactly $r=0$ gives forward boundedness. For backward boundedness, $x^2+y^2=(p^2+q^2)e^{-2t}$ forces $p=q=0$; the remaining $z$ component then tends to zero backward. The intersection of this axis with the forward-bounded plane is zero, so $\boxed{X\equiv 0}$ is the only solution bounded on the entire real line. Rotation cannot cancel the planar radius.

Step 3: Find the direction with an unstable component. If $r\ne 0$, the ratio of planar radius to $|z|$ is $\sqrt{p^2+q^2}\,e^{-2t}/|r|\to 0$. Consequently $\boxed{X/\|X\|\to(0,0,\operatorname{sgn}r)^T}$. The state itself grows without bound; its direction can still converge because the real growing mode dominates both decaying components.

Step 4: Test direction on the stable plane. If $r=0$ and $s=\sqrt{p^2+q^2}>0$, then $X\to 0$, but its normalized planar state is $R(2t)(p,q)^T/s$. At $t_n=n\pi$ it equals $(p,q)^T/s$; at $u_n=n\pi+\pi/2$ it equals its negative. Both sequences tend to infinity and the two unit vectors are distinct, so no directional limit exists. At $n\pi+\pi/4$ the direction is perpendicular to its value at $n\pi$, so even the unoriented line of approach has no limit. A convergent state need not behave like a nonzero real dominant eigenmode.