 Solution

Use the approximate reaction-rate equations from part (a), with axes $u={\mu }_1$ and $v={\mu }_2$. In every case the positive quadrant is forward invariant, since the drift on either zero-concentration axis points inward. The nullclines specify the arrow signs: right if $\lambda >u(\beta +Cv)$, left if the inequality is reversed; up if $\lambda >v(\beta +Cu)$, down otherwise.

For $C=0$ and $\beta >0$, the nullclines are the vertical line $u=\lambda /\beta$ and the horizontal line $v=\lambda /\beta$. The two components relax independently:Every nonstationary trajectory is a straight ray towards the unique stable equilibrium $(\lambda /\beta ,\lambda /\beta )$, whose Jacobian matrix is $-\beta I$.

For $\beta =0$ and $C>0$, both derivatives equal $\lambda -Cuv$, so the two nullclines coincide onEvery point of this hyperbola is an equilibrium. The difference $u-v=d$ is conserved, and trajectories therefore run along straight lines of slope one. Below the hyperbola they point up and right; above it they point down and left. Each such line converges to its own intersection with the hyperbola,At this point the Jacobian matrix is $-C\left(\begin{array}{cc}{v}_{\ast }& u_{\ast }\\ v_{\ast }& u_{\ast }\end{array}\right)$, with eigenvalues $0$ and $-C(u_{\ast }+v_{\ast })$. Thus the equilibrium curve attracts transversely but has no restoring drift along its tangent. Its points are not individually asymptotically stable, because arbitrarily close initial states on different constant-difference lines converge to different equilibria.

For $\beta >0$ and $C>0$, the nullclines are the decreasing curves $u=\lambda /(\beta +Cv)$ and $v=\lambda /(\beta +Cu)$, reflected in the diagonal. At an intersection, subtracting the two stationary equations gives $u=v=\mu$. Hence there is exactly one intersection, whereThe Jacobian matrix at this point isIts symmetric eigenvector $(1,1)$ has eigenvalue $-(\beta +2C\mu )$, and the antisymmetric eigenvector $(1,-1)$ has eigenvalue $-\beta$. Both are negative, so the symmetric equilibrium is a stable node. All trajectories are bounded because each component above $\lambda /\beta$ decreases. The divergence $-2\beta -C(u+v)<0$ excludes periodic orbits by the Bendixson-Dulac criterion, and the unique interior equilibrium attracts the positive-quadrant trajectories. As $\beta$ becomes small, the drift restoring the difference becomes slow, so paths first approach the near-hyperbolic nullclines and then drift towards the symmetric intersection. These sketches show the three cases printed in the PDF; the middle case is $\beta =0$, not the different cases introduced by the converted TeX.