Complex Eigenvalues — Question 2

Question 2 – Solution

Strategy. A linear coordinate change can turn a circle into an ellipse while preserving elapsed time and period.

Step 1: Solve the oscillatory IVP. The characteristic polynomial is $\lambda^2+4$, giving $\pm 2i$. From $x''=-4x$ and $x'(0)=0$, $$\boxed{x=2\cos 2t,\qquad y=-\sin 2t.}$$ Differentiation verifies both first-order equations and the initial data.

Step 2: Recover the invariant ellipse. The derivative of $H=x^2+4y^2$ is $8xy-8xy=0$. Thus the orbit is $\boxed{x^2+4y^2=4}$, with intercepts $(\pm 2,0)$ and $(0,\pm 1)$. The parametrization covers the whole ellipse. Its unequal semiaxes must remain unequal in an equal-scale drawing; it is not a circle in the original Euclidean coordinates.

Step 3: Determine period and orientation. The least positive period is $\pi$. At $(2,0)$ the velocity is $(0,-2)$, so traversal is clockwise. The coordinates $u=x$, $v=2y$ satisfy $u'=2v$, $v'=-2u$ and $u^2+v^2=4$. In that transformed plane the angular velocity is constantly $-2$. The invertible transformation preserves the least period.

Step 4: Compute the actual angular speed. In the original plane, $$\theta'=-\frac{x^2+4y^2}{x^2+y^2}=-\frac 4{x^2+y^2}.$$ On the ellipse, $1\le x^2+y^2\le 4$, so $\boxed{-4\le\theta'\le-1}$. It equals $-1$ at the horizontal intercepts and $-4$ at the vertical ones. Thus imaginary eigenvalues determine a temporal frequency, but do not force a Euclidean circle or constant Euclidean angular speed. The average signed angular velocity over a complete traversal is nevertheless $-2$.