Phase Plane — Question 4

Question 4 – Solution

Strategy. Compute radial and angular rates separately; multiplication of a field can change timing without changing its curves.

Step 1: Verify the common invariant circles. For $F$, $(x^2+y^2)'=2x(-y)+2yx=0$. Multiplication by $1+x^2+y^2$ preserves this cancellation for $G$, and changing the sign preserves it for $H$. A nonzero initial radius stays fixed and each field is nonzero on that circle.

Step 2: Compute orientation from angular velocity. For $R^2=x^2+y^2>0$, $$\theta'=\frac{xy'-yx'}{R^2}.$$ Substitution gives $\boxed{\theta'_F=1,\quad\theta'_G=1+R^2, \quad\theta'_H=-1}$. Thus $F,G$ traverse circles counterclockwise, while $H$ traverses them clockwise.

Step 3: Calculate periods and the given trajectories. A full revolution requires angular change of magnitude $2\pi$, so $$\boxed{T_F=T_H=2\pi,\qquad T_G=\frac{2\pi}{1+R^2}.}$$ At radius $2$, the requested solutions are $X_F=(2\cos t,2\sin t)$ and $X_G=(2\cos 5t,2\sin 5t)$. They start at the same state and cover the same circle at different rates. The zero equilibrium is not assigned a least positive revolution period.

Step 4: Distinguish geometry, direction and speed. All three fields have the same nonzero orbit sets and the same equilibrium. Arrows separate $H$ from $F,G$; the time parametrization separates $F$ from $G$. A strictly positive multiplier preserves velocity direction wherever the field is nonzero, so it cannot reverse an orbit. A negative multiplier does reverse it. The plotted circle is shared by $F,G$; its arrows show their common orientation, and the caption records their different periods at radius $2$.