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Write the vector field asThe equilibrium points of a dynamical system in the closed first quadrant areThe Jacobian matrix is

At $(0,0)$ its eigenvalues are $8,-1$, so this is a saddle equilibrium. Its unstable direction is the positive $x$-axis and its stable direction is the positive $y$-axis; to first order, nearby trajectories satisfy $\dot{x}=8x$, $\dot{y}=-y$.

At $(4,0)$ the eigenvalues are $-8,3$, so it is also a saddle. The $x$-axis is the stable direction, while trajectories entering the quadrant in the $y$ direction move away.

At $(1,\sqrt{3})$,whose eigenvalues areIt is therefore a stable spiral. A point immediately to its right moves upward, so nearby trajectories spiral counterclockwise into the equilibrium. These eigendirections and the inward spiral give the requested local phase-portrait sketches.

The Jacobian matrix isAt $(0,0)$ its eigenvalues are $-k$ and $-1$. It is therefore a stable node for $k>0$, a saddle for $k<0$, and nonhyperbolic at $k=0$, consistent with the proposed transcritical bifurcation.

At $(0,1)$ the matrix iswith eigenvalues $1-k$ and $1$. It is an unstable node for $k<1$, a saddle for $k>1$, and nonhyperbolic at $k=1$, again consistent with a transcritical bifurcation.

Let $X\subseteq {\mathbb{A}}^n$ have ideal $I(X)$, and let $p\in X$. Its Zariski tangent space isIf $X$ is irreducible, $p$ is smooth whenotherwise it is singular. Equivalently, for generators of $I(X)$, the Jacobian has the maximum rank $n-\dim X$ at $p$.

Now let the irreducible affine plane cubic be $X=V(f)\subseteq {\mathbb{A}}^2$. Suppose distinct points $p,q\in X$ were both singular. Parametrize their joining line byand put $h(t)=f(\ell (t))$. Then $\mathrm{deg}h\le 3$. At $t=0$,so zero is a root of multiplicity at least two. The same argument at $t=1$ gives another root of multiplicity at least two. Therefore $h$ has at least four roots counted with multiplicity and must vanish identically. The line through $p,q$ is then contained in $X$, so its linear equation divides $f$, contradicting irreducibility. This proves singular points of an irreducible affine plane cubic are unique when they exist.

For the density statement, embed an irreducible affine variety $X$ of dimension $d$ in ${\mathbb{A}}^n$ and choose generators $f_1,\dots ,f_r$ of its ideal. By dimension from minimum tangent dimension, some point has tangent dimension $d$, so the Jacobian has rank $n-d$ there. Some $(n-d)\times (n-d)$ minor $\Delta$ is consequently nonzero on $X$. The distinguished open setis nonempty, and at each of its points the Jacobian rank is at least $n-d$. Since every tangent space has dimension at least $d$, the rank is also at most $n-d$. Thus every point of $X_{\Delta }$ is smooth. A nonempty open subset of an irreducible topological space is dense, proving density of the smooth locus.

Finally write the smooth irreducible projective hypersurface asfor an irreducible homogeneous polynomial $F$. The closure of ${\pi }^{-1}(X)$ is its affine cone over a projective hypersurfaceAt a nonzero point $y\in Y$, simultaneous vanishing of all partial derivatives of $F$ would make the projective point $[y]\in X$ singular. Since $X$ is smooth, this cannot happen. Hence every nonzero point of $Y$ is smooth, and the origin is its only possible singular point.

Both possibilities occur. If $X$ is a projective hyperplane, then $F$ is linear and $Y$ is an affine linear subspace, hence smooth even at the origin. For a singular cone, take the smooth conicIts affine cone isAll first derivatives vanish at the origin, so the vertex is singular, while the preceding argument shows that it is the only singular point.

At an equilibrium point of a dynamical system, the second equation givesThe choice $x=0$ is impossible because then $\dot{x}=r>0$. Hence $y=s/x$, and substitution into the first equation givesThus the unique fixed point is

Choosewhich satisfies $0<\alpha <r$. LetWe check the direction of the vector field on its four edges.

On $x=\alpha$,On $y=0$,On $y=s/\alpha$, where $x\ge \alpha$,Finally, throughout the system,On the sloping edge $x+y=r+s/\alpha$, the bound $y\le s/\alpha$ implies $x\ge r$, and hence $\dot{x}+\dot{y}\le 0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact trapping region and then remain there. This is the Brusselator trapping region.

The Jacobian matrix at the unique fixed point isIts determinant and trace areWhen $s-1>r^2$, the trace is positive, so linear stability of a planar equilibrium shows that the fixed point is a repeller. It lies in the interior of $R$.

Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the Poincare-Bendixson theorem gives a periodic orbit. HenceThis is the Brusselator periodic-orbit criterion.

At a nontrivial equilibrium $I^{\ast }>0$, the equation $I^{\prime }=0$ givesThe equation $R^{\prime }=0$ gives $R^{\ast }=\nu I^{\ast }/f$, and conservation of population then yieldsThese values are positive precisely in the epidemic regime $\beta N>\nu$.

Eliminate $R=N-S-I$. The two-dimensional system isSet $a=\beta I^{\ast }=f(\beta N-\nu )/(f+\nu )>0$. The Jacobian matrix at the endemic equilibrium isso its eigenvalues satisfyTheir sum is $-(f+a)<0$ and their product is $a(f+\nu )>0$, proving local asymptotic stability.

The discriminant isWriting $r=\beta N-\nu >0$, we have $a=fr/(f+\nu )$. As $f\to 0^{+}$,whereas as $f\to \infty$,Hence sufficiently slow immunity loss gives a stable focus: $I(t)$ approaches $I^{\ast }$ through damped oscillations, corresponding to successively smaller epidemic waves. Sufficiently rapid immunity loss gives a stable node: small disturbances are sums of two decaying real modes and show no forced oscillation. These conclusions and the equilibrium formulas are collected in the Endemic equilibrium of the SIR model with waning immunity.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

Write the reaction terms asA nonzero homogeneous equilibrium satisfies $v_{\ast }=\beta u_{\ast }$ and $1+u_{\ast }-\gamma v_{\ast }=0$, henceFor physically positive populations it exists exactly when

The reaction Jacobian matrix at this equilibrium isIts trace and determinant areThe equilibrium is therefore stable to spatially uniform perturbations when

For a spatial Fourier mode of wavenumber $k$, put $q=k^2$. The linearized reaction-diffusion system has matrixIts trace is smaller than $\mathrm{tr}J$, whileThe two-species diffusion-driven instability criterion says that this upward-opening quadratic becomes negative for some $q>0$ precisely whenCombining all conditions, a Turing instability may occur in the region

At onset the discriminant vanishes, and the double root isThe threshold relation givesso the critical wavenumber isIf $d=1$, uniform stability requires $\beta >1$ whereas diffusion-driven instability requires $d>\beta$. These inequalities are incompatible, so the Turing region vanishes.

The fixed point branches and their stability are as follows.

which is a negative-definite matrix, so this branch is stable.

Thus the $x$-versus-$\mu$ bifurcation diagram has the stable $x=0$ branch up to $\mu =1$, a stable branch $x=\sqrt{\mu -1}$ joining the bifurcation points $(1,0)$ and $(2,1)$, and the branch $x=\sqrt{\mu /2}$ changing from unstable to stable at $\mu =2$. The origin branch also lies at $x=0$ but remains unstable throughout.

SetThe scalar conservation law is $u_t+a(t)f^{\prime }(u)u_x=0$. Its characteristic curve issuing from $\xi$ satisfiesThe Jacobian of the one-dimensional characteristic map isBefore characteristic crossing, differentiation with respect to $\xi$ givesBecause $u_0$ has compact support, $m$ is continuous and vanishes outside a compact set. It therefore attains its minimumby the hypothesis. Since $a(t)>\epsilon$, the function $A$ is strictly increasing and tends to infinity. There is consequently a unique first time $T_{\ast }>0$ satisfyingAt a minimizer of $m$, the numerator $u_0^{\prime }$ is nonzero because $f^{\prime \prime }$ is nonzero, while the denominator tends to zero as $t\uparrow T_{\ast }$. Hence the classical solution has gradient blow-up:

A perturbation of a renormalization-group fixed point is relevant if it grows as the coarse-graining length increases. For a continuous flow $d\delta g/d\ell =M\delta g$, an eigenvector of the linearized matrix with eigenvalue $\sigma$ grows as $e^{\sigma \ell }$; a positive real part is relevant, a negative real part is irrelevant and a zero real part requires nonlinear analysis. For a discrete transformation with length factor $b>1$, its Jacobian matrix multiplier $\lambda$ instead gives $\delta g_n={\lambda }^n\delta g_0$: $|\lambda |>1$ is relevant, $|\lambda |<1$ is irrelevant and $|\lambda |=1$ requires nonlinear analysis. For a positive multiplier the scaling exponent is $y=\log \lambda /\log b$. Thus a positive discrete eigenvalue smaller than one is an irrelevant operator direction, while the sign criterion belongs to the continuous generator.

A renormalization-group transformation integrates out selected short-distance degrees of freedom and restores chosen cutoff and field-normalization conventions. It defines a map $R_b$ on couplings at a length factor $b$. Its discrete Jacobian matrix multipliers differ from eigenvalues of the generator of a continuous renormalization-group flow; the relation for a scaling direction is $\lambda =b^y$.

A spatially homogeneous equilibrium of a reaction-diffusion system is a constant state $u_{\ast }$ satisfying $f(u_{\ast })=0$. Since its spatial derivatives vanish, its stability to homogeneous perturbations is determined by the Jacobian matrix $Df(u_{\ast })$; diffusion can additionally destabilize nonconstant modes.

For a chosen orientation and phase-reduced steady amplitude $a$ and nonresonant oscillatory amplitude $z$, a schematic normal form is $\dot{a}=r_{s}a-g_s{a}^3-h_{s}a|z{|}^2$, $\dot{z}=(r_o+i{\omega }_0)z-g_o|z{|}^{2}z-h_o{a}^{2}z$. For $r_s,g_s>0$ the pure steady branch exists and is radially attracting; its oscillatory perturbation is damped when $r_o-h_{o,r}{r}_s/g_s<0$, where $h_{o,r}=\mathrm{Re}(h_o)$. For $r_o,g_{o,r}>0$, $g_{o,r}=\mathrm{Re}(g_o)$, the pure oscillatory branch exists and is radially attracting; its steady perturbation is damped when $r_s-h_s{r}_o/g_{o,r}<0$. The transverse inequalities are not existence or full stability tests by themselves. Mixed intensities $u=a^2>0$, $v=|z{|}^2>0$ solve $g_{s}u+h_{s}v=r_s$, $h_{o,r}u+g_{o,r}v=r_o$. Their intensity Jacobian matrix has trace $-2(g_{s}u+g_{o,r}v)$ and determinant $4uv(g_s{g}_{o,r}-h_s{h}_{o,r})$, giving attraction in intensities when the trace is negative and determinant positive. Resonant phases, oppositely travelling modes or zero-frequency limits need additional retained amplitudes; this schematic pair is not a universal complete normal form.

For additive Gaussian white noise and a time-even order parameter, a midpoint discretization of the Onsager–Machlup functional evaluates drift derivatives at the midpoint configurations. Reversing a trajectory visits the same midpoints in reverse order, so the noise-to-path Jacobian determinant is invariant under reversal. It may depend on the trajectory, and must be distinguished from the constant normalization of the noise measure.

For nonconserved order-parameter dynamics with $\dot{p}=-\Gamma \mu [p]+f$, the factor at one time step, apart from a path-independent power of the step size, isHere the fields have first been restricted to a finite spatial grid, and ${\mathcal{H}}_n$ is the Hessian matrix of the free energy.

For mixed conserved and nonconserved order-parameter dynamics, the required noises for a joint configuration and flux trajectory arewhere $\mathsf{D},\mathsf{G}$ represent the divergence and gradient, and $\Delta =\mathsf{DG}$ represents the Laplacian. The Jacobian matrix isTaking its Schur complement gives the joint factorFor periodic boundary conditions, discretizing the two spatial operators compatibly gives $\mathsf{D}=-{\mathsf{G}}^{\mathsf{T}}$, so $\Gamma I-M\Delta$ is the positive relaxation operator when $\Gamma ,M>0$. Reversal leaves ${\mathcal{H}}_n$ unchanged and reverses the sign of the flux. The product of the joint factors is consequently the same for both histories, justifying its cancellation from the joint path probability of an order parameter and its flux. The functional chain rule used in the action ratio holds in the continuum limit of this midpoint convention.

For a real $2\times 2$ Jacobian matrix $J$, both eigenvalue have negative real part exactly whenThe linearization stability theorem then makes a hyperbolic equilibrium locally asymptotically stable.

A real $2\times 2$ Jacobian matrix has both eigenvalues in the open left half-plane exactly when its trace is negative and its determinant is positive. At zero trace the test no longer determines nonlinear stability.

A planar equilibrium is an unstable node when its two Jacobian eigenvalues are real and positive. Every nearby nonstationary trajectory moves away from it in forward time.