= Solution
{c}
The <nullclines> are
$$
\boxed{\dot x=0:\ y=\beta;\qquad \dot y=0:\ x=-1,0,1.}
$$
On $y=\beta$, away from the <equilibrium points>, the <vector field> is vertical. Its crossing direction is downwards for $x<-1$, upwards for $-1<x<0$, downwards for $0<x<1$, and upwards for $x>1$, because $\dot y=x(x^2-1)$. On each of $x=-1,0,1$, the <vector field> is horizontal: it points right when $y>\beta$ and left when $y<\beta$.
Throughout each complementary region, $\dot x$ has the sign of $y-\beta$, and $\dot y$ has the four signs just listed. These signs orient the <level sets> from part (ii). In particular, the inner <periodic orbits> are clockwise. The saddle level is $h=1/4$, whose <separatrices> satisfy
$$
\boxed{(y-\beta)^2=\frac{(x^2-1)^2}{2}.}
$$
For $-1<x<1$, the upper arc $y-\beta=(1-x^2)/\sqrt2$ goes from $(-1,\beta)$ to $(1,\beta)$; the lower arc goes back from $(1,\beta)$ to $(-1,\beta)$. These two <heteroclinic orbits> form a closed boundary around the family of <periodic orbits>. The remaining branches of the same saddle level extend outwards. The sketch shows the surrounding nonclosed trajectories as well as the <nullclines> and <equilibrium points>.
\Image[/past-exam-of-the-differential-equations-1-course-of-the-university-of-oxford/2021-phase-plane.png]
{title=The phase plane in coordinates x and y minus beta, showing clockwise periodic orbits and saddle connections}
Back to article page
