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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.

The nullclines are $y=\mu x$ and $y=x^2/[\nu (1+x^2)]$. At a nonzero fixed point, cancellation of $x$ yields $\mu \nu =x/(1+x^2)$. The latter function attains its maximum $1/2$ at $x=1$. HenceFor $0<\mu <{\mu }_c$, $x_{-}<1<x_{+}$ and $x_{-}{x}_{+}=1$.

The Jacobian matrix isAt $O$ the distinct eigenvalues are $-\mu ,-\nu$, so the origin is a stable node. At a nonzero fixed point its trace is $-(\mu +\nu )<0$ andThe discriminant is $(\mu -\nu {)}^2+8x/(1+x^2{)}^2>0$. Thus $P_{-}$ is a saddle and $P_{+}$ is a stable node, never a focus.

Above the straight nullcline $\dot{x}>0$; below it $\dot{x}<0$. Below the curved nullcline $\dot{y}>0$; above it $\dot{y}<0$. At the saddle, an eigenvector can be chosen as $(1,\mu +\lambda )$: the stable eigenvalue gives negative slope, and the unstable eigenvalue gives positive slope. The stable manifold is the basin boundary. The two branches of the unstable manifold lead respectively to the origin and the positive stable node, as illustrated for representative parameters.

The nonnegative quadrant is forward invariant. Since $\dot{y}\le 1-\nu y$, $y$ stays bounded, and then $\dot{x}=-\mu x+y$ also bounds $x$. Moreover the divergence is the negative constant $-(\mu +\nu )$, excluding periodic or homoclinic loops by the Bendixson-Dulac criterion. The Poincaré-Bendixson theorem then leaves convergence to a fixed point. Initial data on the saddle's stable manifold approach the saddle; data in the two open basins approach extinction at $O$ or persistence at $P_{+}$, respectively.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.

Use the complete system in the PDF: the TeX omits $\dot{y}=y(4x-1)/8$. Both coordinate axes are invariant, and the equilibrium points are $(0,0)$, $(1,0)$ and $(1/4,3/4)$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1,-1/8$, and at $(1,0)$ they are $-1,3/8$: both points are saddle equilibria. At the coexistence point,Thus the interior point is a stable spiral. The nontrivial nullclines are $x+y=1$ and $x=1/4$: $x$ increases below the first and decreases above it, while $y$ increases to the right of the second and decreases to the left. The rotation near coexistence is counterclockwise.

A local linearization by itself does not prove that every interior solution has the same limit. For this logistic predator-prey model, let $x_{\ast }=1/4,y_{\ast }=3/4$ and use the Lyapunov functionThe inequality $u-1-\log u\ge 0$ makes $V$ nonnegative, with equality only at coexistence. Its sublevel sets are compact inside the positive quadrant, because $V$ diverges both at the axes and at infinity. Direct differentiation cancels the predator-prey cross terms and givesOn $\{\dot{V}=0\}$ one has $x=x_{\ast }$, and remaining there requires $\dot{x}=-x_{\ast }(y-y_{\ast })=0$. The largest invariant subset is therefore the coexistence point. The LaSalle invariance principle provesOn $x=0$, $y$ decays as $e^{-t/8}$ toward the origin. On $y=0$, the logistic differential equation takes every $x(0)>0$ to one. The origin itself stays fixed. These boundary trajectories are the exceptions to the interior convergence statement.

The non-axis nullclines are $u_1+A{u}_2=1$ and $B{u}_1+u_2=1$. For $A>1$, $B<1$, the first has intercepts $(1,0),(0,1/A)$ and the second has $(1/B,0),(0,1)$. Their axis intercept order rules out a positive intersection. Below the first line $u_1$ increases and above it decreases; below the second $u_2$ increases and above it decreases. Thus $S_2$ is a saddle point and $S_3=(0,1)$ is stable.

For strictly positive initial populations, both coordinates stay positive and are bounded above by their individual logistic equations. There are no interior periodic orbits: the Dulac function $1/(u_1{u}_2)$ makes the divergence of the multiplied vector field $-1/u_2-\rho /u_1<0$, so the Bendixson-Dulac theorem applies. The stable manifold of $S_2$ is the invariant red-only axis, and $S_1$ is repelling. The phase portrait, or the Poincaré-Bendixson theorem with these boundary directions, therefore givesThe qualification $u_2(0)>0$ matters: a grey population identically absent cannot drive extinction. In dimensional variables, positive grey presence leads to $N_1\to 0$, $N_2\to K_2$ in this no-hunting case.

Abbreviate $A={\alpha }_{12}$, $B={\alpha }_{21}$. The coordinate factors immediately give the three boundary equilibria $S_1,S_2,S_3$. The Jacobian matrix isAt $S_1$, the eigenvalues are $1,\rho \gamma >0$, so it is a repelling equilibrium. At the other two points, triangularity givesTherefore $S_2$ is asymptotically stable for $\gamma <B$ and a saddle point for $\gamma >B$; $S_3$ is asymptotically stable for $\gamma >1/A$ and a saddle point for $\gamma <1/A$. Equality gives a zero eigenvalue, so linearization alone is inconclusive there.

For later use, the coexistence point is physically positive exactly when $B<\gamma <1/A$ if $AB<1$, or $1/A<\gamma <B$ if $AB>1$. Its stability is as supplied in the question. If $AB=1$ the displayed formula for $S_4$ does not apply: when also $\gamma =B=1/A$, the coincident non-axis nullclines give a whole segment of equilibria.

TODO: biological interpretations and organism-specific propagation are user-deferred. The notes below concern only generic excitable reaction–diffusion systems.

A front solution joins different asymptotic states in space. In a reaction–diffusion system, local nonlinear dynamics select possible states while diffusion couples neighboring points. A travelling wave with profile $u(x,t)=U(\xi )$, $\xi =x-ct$, converts a partial differential equation into a boundary value problem for the shape and speed. A pulse instead returns to the same resting state on both sides and can contain leading and trailing fronts.

A useful nondimensional cubic fast-slow reaction-diffusion model, in a convention of the FitzHugh-Nagumo model, iswhere $D>0$, $\epsilon \ll 1$, and $\beta >0$. Here $u$ is a fast activation variable, $w$ a slow negative-feedback variable and $I$ a generic drive; no physiological variable is assigned. Without spatial coupling, nullclines are $w=f(u)+I$ and $w=u/\beta$. At $I=0$, the resting state $(0,0)$ has Jacobian matrixIt has linear stability. With appropriate parameter choices a sufficiently large perturbation makes a fast excursion to the other outer branch of the cubic, followed by slow feedback and return. Thus a stable rest state can be excitable without being linearly unstable. Other parameters or applied drive can yield multiple equilibria or oscillations; these behaviours should not be asserted for every parameter set.

The fast-subsystem front can be obtained explicitly by freezing $w=I=0$. For the bistable cubic reaction-diffusion equation, $u=0,1$ are stable and $u=\alpha$ is unstable. The decreasing logistic front of a bistable cubic equation isTo check it, $U^{\prime }=-U(1-U)/\sqrt{2D}$ and substitution into $D{U}^{\prime \prime }+c{U}^{\prime }+f(U)=0$ cancels every term. For $\alpha <1/2$, the high state advances to the right; for $\alpha >1/2$, the direction reverses. At $\alpha =1/2$ the front is stationary. More generally, multiplying the travelling-wave equation by $U^{\prime }$ and integrating yieldsThis links propagation direction to the bias between the two stable states and recovers Maxwell balance for a scalar reaction-diffusion front at zero speed. It is a bistable selected-speed problem, rather than a universal Fisher-type minimum-speed formula.

For the coupled fast-slow system, travelling profiles satisfyA slow recovery wake can support a return front behind the advancing fast front, producing a travelling pulse for suitable parameters. Its width reflects the slow recovery time and propagation speed, while each sharp edge has the faster diffusion–reaction scale. The frozen-recovery speed above is a controlled subsystem result, not an exact speed for every coupled pulse. A definite pulse speed requires parameters and connecting boundary condition; none are specified by this brief-note question.

Consider the original analytic initial value problemHere $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact solutions of a differential equation, with no measured data or empirical parameter fit. For $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear solution of a differential equation uses only the interval containing $0$ where $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ is locally Lipschitz in $y$ everywhere, the Picard-Lindelöf theorem ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.

For $a=1/2$, the exact solution of a differential equation is $y=e^{-x}/(2-x)$ on $(-\infty ,2)$. It first decreases, reaches a local minimum $y(1)=e^{-1}$, then increases without bound. At the endpoint,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, rather than from a singular coefficient in the first-order linear differential equation. The same algebraic expression for $y$ is negative on $(2,\infty )$, but that is a separate solution of a differential equation on a disjoint interval; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty )$, giving a distinct decaying branch whose finite blowup is instead in the past.

The direction field also explains the positive turning point. Its nullclines are $y=0$ and $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. The moving nullcline $y=e^{-x}$ is not a solution of a differential equation: its derivative is $-e^{-x}$, whereas the prescribed slope there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its local minimum.