 Solution

Assume first $\beta >0$ and $d\ge 0$. The nonnegative equilibrium points areand, when both coordinates are nonnegative,The Jacobian matrix at $O$ has eigenvalues $1,1-d$, so $O$ cannot attract an initially positive uninfected population. At $U$ the eigenvalues are $-1,\beta -d$. At $I$, for $d<1$, they are $d(1+\beta )-\beta$ and $-(1-d)$. At an interior $C$, the equilibrium equations reduce the Jacobian matrix toThus the coexistence equilibrium is asymptotically stable precisely when $\beta /(1+\beta )<d<\beta$.

For positive initial populations these local results identify the global outcome. The relation $\dot{N}=N(1-N)-dy$ bounds all trajectories. With the Dulac function $1/(xy)$, the divergence of the rescaled vector field is $-1/y-1/x<0$, excluding interior periodic orbits. Boundary saddles cannot attract an interior trajectory except at the merging thresholds. The resulting long-term outcomes areAt $d=\beta /(1+\beta )$, $C$ merges into $I$; at $d=\beta$, it merges into $U$. The merged equilibria still attract interior trajectories, but convergence along the zero-eigenvalue direction is algebraic. For example at $d=\beta$, put $x=1+u$, $y=v$. The local centre manifold has $u=-(1+\beta )v+O(v^2)$, and $\dot{v}=-{\beta }^2{v}^2+O(v^3)$. At the other threshold, using $y=(1-d)+v$, $x=u$, gives $v=(\beta -1)u+O(u^2)$ and $\dot{u}=-{\beta }^2{u}^2+O(u^3)$.

The invariant axes also matter: $y=0$ leads to $U$ whenever $x(0)>0$; $x=0$ leads to $I$ for $d<1$ and to extinction for $d\ge 1$. At $d=1$, $I$ coincides with $O$. If $\beta =0$, disease declines for $d>0$ and any initially positive $x$ tends to $U$. If also $d=0$, every point $x+y=1$ is an equilibrium and the initial population proportions are preserved.