Complex Eigenvalues — Question 3

Question 3 – Solution

Strategy. Use coordinates that balance the oscillatory coupling, then compare the adapted quadratic size with Euclidean distance.

Step 1: Solve the damped rotation. The characteristic polynomial is $(\lambda+1)^2+4$, so the rates are $-1\pm 2i$. Direct substitution verifies $$\boxed{x=4e^{-t}\sin 2t,\qquad y=e^{-t}\cos 2t.}$$ For any nonzero state, $xy'-yx'=-x^2/2-8y^2<0$; rotation is clockwise. The transformed coordinates $(x,4y)$ rotate uniformly with decay.

Step 2: Exhibit temporary growth. Initially the norm is one, whereas $\boxed{\|X(\pi/4)\|=4e^{-\pi/4}>1}$. More generally, $$(\|X\|^2)'=-2x^2+15xy-2y^2,$$ which equals $11>0$ at $(1,1)$. Its value at the given initial state is $-2$. Hence even this trajectory can first shrink and subsequently exceed its initial distance. The negative real parts guarantee eventual decay, not a sign for this derivative at every state.

Step 3: Find a decreasing quadratic size. Differentiating $x^2+c y^2$ cancels the mixed term when $c=16$. Then $\boxed{H=x^2+16y^2,\quad H'=-2H}$, so $H(t)=H(0)e^{-2t}$. Since $\|X\|^2\le H$, every initial state tends to zero. Together with the strict rotation sign this identifies a stable spiral, even though the drawn orbit can move outside the unit circle.

Step 4: Give a permanent-entry bound. For the specified data, $H(0)=16$ and $\|X(t)\|\le 4e^{-t}$. Thus every $t\ge\boxed{\ln(4/\rho)}$ satisfies $\|X(t)\|\le\rho$. This bound controls all later times. It need not be the first entry: $H$ overestimates the ordinary squared norm except where $y=0$. Transient contraction and expansion can occur before the guaranteed time; no claim of a sharp first-entry time is made.