Complex Eigenvalues — Question 4

Question 4 – Solution

Strategy. The real part sets radial growth and the imaginary part sets turning time; a ray return becomes periodic only when the radius is restored.

Step 1: Read radial and angular evolution. The eigenvalues are $a\pm 3i$. For initial radius $r_0>0$ and angle $\theta_0$, the solution has $$\boxed{r(t)=r_0e^{at},\qquad\theta(t)=\theta_0+3t.}$$ Equivalently $X(t)=e^{at}R(3t)X(0)$, where $R(\phi)=\begin{pmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{pmatrix}$. The only equilibrium is zero because the determinant is $a^2+9>0$.

Step 2: Classify all parameter cases. For $a<0$ the origin is a stable spiral; for $a>0$ it is an unstable spiral. At $a=0$ it is a center: every nonzero state follows a circle with least period $2\pi/3$. For $a\ne 0$, the radius is strictly monotone, so no nonzero solution is periodic. Zero remains constant and has no least positive period.

Step 3: Compute the return map. The first return to the positive ray takes $T=2\pi/3$, independent of $r$ and $a$. Consequently $$\boxed{P_a(r)=e^{2\pi a/3}r.}$$ For $r>0$, the fixed-point equation holds for every radius if $a=0$ and for none if $a\ne 0$. These fixed radii are exactly the nonzero periodic trajectories. Returning to an angle alone is insufficient.

Step 4: Recover damping from one revolution. The measurement gives $e^{2\pi a/3}=1/2$, hence $$\boxed{a=-\frac{3\ln 2}{2\pi},\qquad r_n=r_0\,2^{-n}.}$$ The coefficient $a$ is unique because the real exponential is one-to-one. The same multiplier applies to every radius by linearity; a single nonzero measurement therefore determines the damping rate in this specified family.