Complex Eigenvalues — Question 7

Question 7 – Solution

Strategy. The state components have different phases: displacement zeros and velocity zeros mark different events on the spiral.

Step 1: Solve the state IVP. The characteristic polynomial is $\lambda^2+2\lambda+5$, giving $-1\pm 2i$. The initial conditions yield $$\boxed{x=e^{-t}(\cos 2t+\tfrac 12\sin 2t),\qquad y=-\tfrac 52e^{-t}\sin 2t.}$$ Differentiation gives $x'=y$ and $y'=-5x-2y$, including $y'(0)=-5$.

Step 2: Locate the first displacement crossing. Write $\alpha=\arctan 2\in(0,\pi/2)$. The first positive root of $\cos 2t+\tfrac 12\sin 2t=0$ occurs at $$\boxed{t_0=(\pi-\alpha)/2,\qquad y(t_0)=-\sqrt 5e^{-t_0}.}$$ Indeed $\sin(\pi-\alpha)=2/\sqrt 5$ and $\cos(\pi-\alpha)=-1/\sqrt 5$. The negative velocity shows a crossing from positive to negative displacement, not a turning point.

Step 3: Separate the turning events. Velocity vanishes exactly at $t=n\pi/2$, for integers $n$. The first positive turning time is $\pi/2$, with displacement $-e^{-\pi/2}$. There $x''=y'=5e^{-\pi/2}>0$, so it is a minimum. Successive extrema have $x(n\pi/2)=(-1)^n e^{-n\pi/2}$; their absolute values have ratio $\boxed{e^{-\pi/2}}$ for consecutive forward extrema. Successive positive maxima are one full oscillation apart and have ratio $e^{-\pi}$.

Step 4: Interpret dissipation at a zero derivative. The energy derivative is $E'=yy'+5xx'=-2y^2\le 0$. At zero, $y=0$ so $E'=0$, but the acceleration is $-5$ and the state immediately changes. On any nontrivial time interval this solution has $y\ne 0$ somewhere, so its energy strictly drops over that interval. A nonconstant periodic solution would instead require zero total energy loss over a period, forcing $y\equiv 0$ and then $x\equiv 0$, a contradiction. The phase arrows distinguish the crossing and turning points.