Phase Plane — Question 3

Question 3 – Solution

Strategy. Read component signs from the two defining expressions, then distinguish instantaneous tangency from remaining on a curve.

Step 1: Locate the nullclines and their intersection. The $x$-nullcline is $y=x^2$, and the $y$-nullcline is $x=1$. Their sole intersection is the equilibrium $\boxed{(1,1)}$.

Step 2: Determine the regional directions. Above the parabola $x'>0$; below it $x'<0$. To the left of $x=1$, $y'>0$; to its right, $y'<0$. Combining these independent signs gives right/up, left/up, right/down or left/down wherever the corresponding inequalities hold. The requested velocities are $F(0,0)=(0,1)$, $F(0,1)=(1,1)$ and $F(1,0)=(-1,0)$.

Step 3: Check nullcline tangency correctly. On $y=x^2$, away from $(1,1)$, the nonzero velocity is vertical. The parabola has finite-slope tangent $(1,2x)$, so this velocity is not tangent to that parabola. On $x=1$, away from the equilibrium, the velocity is horizontal and not tangent to the vertical line. Thus neither complete nullcline is invariant. Each is crossed by typical solutions; the equilibrium is the exception where both velocities vanish.

Step 4: Examine departure from the origin. Differentiating the equations gives $x''=y'-2xx'$ and $y''=-x'$. At the initial state, $$\boxed{x'=0,\quad y'=1,\quad x''=1,\quad y''=0.}$$ Hence $x(t)=t^2/2+o(t^2)$ and $y(t)=t+o(t^2)$ as $t\to 0$. For small positive $t$, $y-x^2=t+o(t)>0$, so the solution leaves the nullcline and acquires positive horizontal velocity. A vertical tangent at one instant does not mean that $x$ stays constant. The figure shows nullclines and local velocity arrows, not guessed solution curves.