Complex Eigenvalues — Question 5

Question 5 – Solution

Strategy. An identity valid for all initial states determines quadratic coefficients; the remaining sign is geometric orientation.

Step 1: Compare polynomial coefficients. Differentiation gives $2ax^2+2(b+4c)xy+8dy^2=-2x^2-8y^2$. Every point can be an initial state, so coefficients must agree: $$\boxed{a=d=-1,\qquad b=-4c.}$$

Step 2: Use the spectral determinant. The eigenvalues require trace $-2$ and determinant $10$. The trace is already correct, while $\det A=1-bc=1+4c^2=10$ gives $c=\pm 3/2$. Thus precisely two matrices remain: $$\begin{pmatrix}-1&6\\-3/2&-1\end{pmatrix},\qquad \begin{pmatrix}-1&-6\\3/2&-1\end{pmatrix}.$$ Both have the required quadratic decay and the required eigenvalues.

Step 3: Use orientation and solve. At $(2,0)$, $y'=2c<0$, so select $c=-3/2$, $b=6$. The IVP is $$\boxed{x=2e^{-t}\cos 3t,\qquad y=-e^{-t}\sin 3t.}$$ It satisfies the selected equations, initial data and $x^2+4y^2=4e^{-2t}$. Moreover $xy'-yx'=-3x^2/2-6y^2<0$ off zero, confirming clockwise rotation. The shrinking ellipses are size contours, not individual orbits.

Step 4: Separate timing, contraction and sign. The coordinates $(x,2y)$ make one clockwise revolution in $2\pi/3$. An invertible positive diagonal rescaling preserves a full turn, so this is also the time to return to the initial positive ray. The whole state then equals $e^{-2\pi/3}X(0)$; this is not a periodic return. Without the direction observation, the second matrix would remain equally possible, with counterclockwise rotation and the same contraction factor. The unordered conjugate eigenvalues do not determine orientation.