Phase Plane — Question 6

Question 6 – Solution

Strategy. Separate the radial attraction from the angular motion; approaching a circle does not require stopping on it.

Step 1: Find the radial and angular equations. For a nonzero state, the equilibrium coefficient matrix has determinant $(1-r^2)^2+1>0$, so it cannot annihilate that state. The only equilibrium is the origin. Direct differentiation gives $$\boxed{r'=r(1-r^2),\qquad\theta'=1\quad(r>0).}$$ The formulas follow from $rr'=xx'+yy'$ and $r^2\theta'=xy'-yx'$.

Step 2: Identify the periodic orbit and radial directions. At $r=1$, the radius is constant and the angle increases at unit rate. This circle is a counterclockwise periodic orbit of least period $2\pi$. For $0<r<1$, $r'>0$; for $r>1$, $r'<0$. Both sides move radially toward the circle rather than away from it.

Step 3: Prove the forward radial limit. The squared radius satisfies $u'=2u(1-u)$, with solution $$\boxed{u(t)=\frac 1{1+(r_0^{-2}-1)e^{-2t}}\quad(t\ge 0).}$$ The denominator is positive for every $t\ge 0$, and direct differentiation and evaluation at zero verify the formula. It tends to $1$, hence $r\to 1$. With $\theta=\theta_0+t$, this also supplies a forward-global solution. At $r_0=0$ the solution remains at the equilibrium; the reciprocal formula is not used there.

Step 4: Rule out convergence to a single point. Along times $t=2\pi n$ the states tend to $(\cos\theta_0,\sin\theta_0)$. Along $t=2\pi n+\pi$ they tend to its negative. These are distinct, so no nonzero solution has a single limiting point. Its distance to the unit circle, $|r-1|$, tends to zero while it continues rotating. The figure shows the unit orbit and sample inward/outward spirals; all arrows represent increasing time.