 Solution

Write $c=(1-a-b)/a$. The admissible equilibrium points are $O=(0,0)$, $E=(1-a,0)$ when $a\le 1$, and $P=(b,c)$ when $a+b\le 1$. The Jacobian matrix isAt $O$ the eigenvalues are $(1-a)/a$ and $-1$; at $E$ they are $-(1-a)/a$ and $(1-a-b)/b$. At a positive coexistence equilibrium point,It is therefore asymptotically stable. It is a node for $b^2>4a(1-a-b)$, a focus for the reverse inequality, and has repeated real eigenvalues at equality. The node–focus change does not alter the attracting equilibrium point's topology.

There are three open physical parameter regions:

The axes are invariant. On $x=0$, $y$ decays exponentially; on $y=0$, prey obey the logistic differential equation. Positivity follows because each right-hand side has its population as a factor. Also $x$ is bounded by logistic comparison, while $W=x+by$ satisfies $\dot{W}=(1-a)x/a-x^2/a-by$, so all forward trajectories are bounded. The Dulac function $1/(xy)$ has divergence $-1/(ay)<0$ in the interior, excluding periodic orbits by the Bendixson-Dulac criterion.

In the extinction region, $\dot{x}\le -(a-1)x/a$, and then $\dot{y}/y=x/b-1$ is eventually negative. In the prey-only region, the upper bound $\mathrm{limsup}x\le 1-a<b$ implies predator decay and convergence of interior prey to $1-a$. In the coexistence region the Lyapunov functionhas $\dot{\mathcal{V}}=-(x-b{)}^2/a\le 0$. Its positive-quadrant sublevel sets are compact, and the only invariant subset with $x=b$ also has $y=c$. The LaSalle invariance principle proves convergence of every interior trajectory to $P$. These facts justify the complete qualitative phase portraits, not just their local arrows.

At $a=1$, $O$ and $E$ meet in a transcritical bifurcation; positive prey decay algebraically. At $a+b=1$ with $a<1$, $E$ and $P$ exchange stability. On its centre manifold, $x=b-ay+O(y^2)$ gives $\dot{y}=-a{y}^2/b+O(y^3)$, so the physically positive side is still attracting. There is no interior Hopf bifurcation since the coexistence trace never vanishes.