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The phase plane system is $\dot{x}=y$, $\dot{y}=-cy-\sin x$. Its equilibrium points are $(k\pi ,0)$ for every integer $k$. At such a point the Jacobian matrix and characteristic equation areFor even $k$, the equilibrium point is a stable focus for $0<c<2$, with eigenvalues $-c/2\pm i\sqrt{1-c^2/4}$; it is a stable node for $c>2$, with two distinct negative real eigenvalues. The excluded value $c=2$ gives the repeated critical case.

For odd $k$, the eigenvalues are $(-c\pm \sqrt{c^2+4})/2$. One is positive and one negative, so each is a saddle equilibrium for every $c>0$. These classifications follow from the linearization of a dynamical system, since all the relevant eigenvalues have nonzero real part for $c>0$. The even equilibria remain hyperbolic at $c=2$ too, although their linearization then has a repeated eigenvalue; that critical case is excluded from the requested classification.

The equilibrium equations are $x(-1+x^2+\beta y^2)=y(-1+\beta x^2+y^2)=0$. For $\beta \ne 1$ the fixed points are the origin, four axis points $(\pm 1,0),(0,\pm 1)$, and four diagonal points $(\pm s,\pm s)$ with independent signs and $s=(1+\beta {)}^{-1/2}$. The hypothesis $\beta >-1$ ensures that $s$ is real. The Jacobian matrix has diagonal entries $-1+3x^2+\beta y^2$, $-1+\beta x^2+3y^2$ and off-diagonal entries $2\beta xy$.

At the origin both eigenvalues are $-1$, so it is a stable node. At each axis point the eigenvalues are $2,\beta -1$. At each diagonal point they are $2,2(1-\beta )/(1+\beta )$, along the radial and transverse diagonal directions respectively. ThusThe axes and diagonals are invariant lines. For $\beta =1/2$ the axis saddle points have stable directions transverse to their axes and unstable directions along them; the diagonal points are unstable nodes. For $\beta =2$ these roles switch: the diagonal saddle points have stable transverse directions and unstable radial directions, and the axis points are unstable nodes. Their stable manifolds separate trajectories attracted to the origin from outward-escaping trajectories. Symmetry gives the other quadrants.

For $\beta =1$, the entire unit circle consists of equilibria. In polar coordinates $\dot{r}=r(r^2-1)$ and $\dot{\theta }=0$: every ray inside the circle moves toward the origin, and every ray outside moves outward. The circle has neutral tangential and unstable radial directions; its points are not isolated saddle points or unstable nodes.

All three phase portraits are shown below. The potential $\Phi =-(x^2+y^2)/2+(x^4+y^4)/4+\beta x^2{y}^2/2$ satisfies $(\dot{x},\dot{y})=\nabla \Phi$ and $\dot{\Phi }=|\nabla \Phi {|}^2$, ruling out nonconstant periodic orbits. Blue curves show the saddle stable separatrices, integrated backwards from their local stable directions. The flow arrows and invariant lines, together with the equilibrium classification, specify the topology of the sketches.

At an equilibrium point, $x^2-1=0$ and $kxy=0$. If $k\ne 0$, the only equilibria areTheir Jacobian matrix isAt $(1,0)$ its eigenvalues are $2,k$, and at $(-1,0)$ they are $-2,-k$. Hence for $k>0$, $(1,0)$ is an unstable node and $(-1,0)$ is a stable node; for $k<0$, both points are saddles.

If $k=0$, every point on each line $x=1$ and $x=-1$ is an equilibrium. The line $x=1$ is transversely repelling because the nonzero eigenvalue is $2$, while $x=-1$ is transversely attracting because the nonzero eigenvalue is $-2$; the tangential eigenvalue is zero on both nonisolated equilibrium lines.

For $\nu =\mu +1>0$ small, $(1,0)$ has two negative eigenvalues and is a stable node. The two nearby equilibria satisfy $u=\pm \sqrt{\nu }+O(\nu )$ and are saddles. Their stable and unstable separatrices form the local phase portrait; trajectories in the wedge between the stable manifolds approach $(1,0)$, while trajectories crossing an unstable direction leave the neighbourhood.

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.

The phase-plane equations are $a^{\prime }=b$, $b^{\prime }=-cb-f(a)$. Nonnegative equilibria require $b=0$ and $f(a)=0$, hence only $\boxed{(0,0),(\eta ,0)}$. At the origin $f^{\prime }(0)={\eta }^2$, so the linear eigenvalues solveFor $0<c<2\eta$ the roots are complex with negative real part, giving a stable focus. For $c>2\eta$ both are negative real numbers, giving a stable node. Equality is the repeated-root threshold and is not covered by either strict classification.

Let $S(x,y,z)=(-x,-y,z)$. In the first two components of the vector field, replacing $x,y$ by their negatives changes the sign of the whole component; in the third, the product $xy$ is unchanged. Thus $F(Sv)=SF(v)$, establishing the symmetry, with solutions mapped to solutions by $S$.

At an equilibrium point, positivity of $a$ makes $y=x$. The remaining equations give $x(2c-a-z)=0$ and $z=x^2/b$. Hencewhere the two nonzero equilibrium points exist when $0<a<2c$. They merge with the origin at $a_p=2c$, in a pitchfork bifurcation.

At the origin the Jacobian matrix and its eigenvalues areAt $a=2c$ these are $0,-c,-b$. To see the local pitchfork bifurcation dynamics more explicitly put $\delta =2c-a$, $y=x+v$, and use $z=x^2/b+\cdots$ on the center manifold. The equations becomeAt $\delta =0$, invariance at cubic order gives $v=-x^3/(bc)+\cdots$, hence the reduced equation isIt has the supercritical pitchfork bifurcation form as $\delta$ increases, with the two branches $x\simeq \pm \sqrt{b\delta }$ attracting near onset while the origin becomes unstable.

For $0<a<2c$, the determinant of the planar block is $a(a-2c)<0$, so the origin has one positive and two negative eigenvalues and is a saddle equilibrium. At $a=c$, the planar eigenvalues are explicitly $\pm c$. For $a>2c$, the planar trace $c-a$ is negative and its determinant positive, so the origin is asymptotically stable. To distinguish its type putFor $2c<a<a_{\ast }$ both planar eigenvalues are real negative, giving a stable node. At $a=a_{\ast }$ they coincide at $-c/\sqrt{3}$; the planar block is not a scalar matrix, so it is a defective stable node. For $a>a_{\ast }$ they are nonreal with negative real part, giving a stable spiral in the planar block together with the stable third direction. This refines the requested node-versus-saddle classification.

The origin cannot have a Hopf bifurcation: a planar imaginary pair would require trace zero, namely $a=c$, but the determinant there is $-c^2<0$, so the pair is real and of opposite signs.

For either nonzero equilibrium point put $s=\pm \sqrt{b(2c-a)}$. The Jacobian matrix isIts characteristic polynomial isA nonzero imaginary root of an equation $i\omega$ requires, by separating real and imaginary parts, ${\omega }^2=B$ and $AB=C$. Equivalentlyso the candidate Hopf bifurcation values areAt such an admissible value the factorization is $p(\lambda )=(\lambda +A)({\lambda }^2+bc)$, giving the pair $\pm i\sqrt{bc}$ and a negative third eigenvalue $-A$, since $A=C/B>0$ on the nonzero branches.

The mathematical model parameter restrictions matter. A real candidate requires $b/c\le 17/8$, and the nonzero equilibrium points also require $0<a_h<2c$. The upper-sign value is admissible for $0<b/c<17/8$; the lower-sign value additionally requires $b/c>1$. For fixed $b,c$, differentiation of the characteristic equation at the imaginary root of an equation gives the eigenvalue crossing speedThus the transverse eigenvalue crossing condition holds except at the coincident-root case $b/c=17/8$, $a_h=3c/4$. At a transverse eigenvalue crossing, a generic nonzero first Lyapunov coefficient gives a Hopf bifurcation. The displayed radical is the spectral threshold; it does not justify an unconditional generic-Hopf claim for all positive mathematical model parameters, nor does it by itself exclude a degenerate Hopf bifurcation point. For example $b>17c/8$ yields no real candidate at all. The cubic Routh-Hurwitz stability criterion gives local Lyapunov stability on the side with $A>0$ and $AB-C>0$, where

For the modified equation, the planar block at the origin is $\left(\begin{array}{cc}-a& a\\ c& c\end{array}\right)$, with determinant $-2ac<0$ for the stipulated positive mathematical model parameters. Thus the origin is always a saddle equilibrium for nonzero $a,c$ in that mathematical model parameter range. The nonzero equilibrium points now satisfy $y=x$, $z=2c$ and $x^2=2bc$, so they areFor $b,c>0$ these branches stay a positive distance from the origin and do not merge with it as $a$ varies. The origin also has no zero eigenvalue for $a,c>0$. Therefore the pitchfork bifurcation is absent in the modified system.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

The equilibria satisfy $y=0$ and $x-x^3=0$, giving $(0,0)$ and $(\pm 1,0)$. At the origin,Their product is $-1$, so the origin is a saddle equilibrium for every real $\Omega$.

At either nonzero equilibrium,The determinant is positive and the trace is $-(\Omega +1)$. Thus both equilibria are attracting for $\Omega >-1$, repelling for $\Omega <-1$, and have purely imaginary eigenvalues $\pm i\sqrt{2}$ at $\Omega =-1$. More precisely they are stable foci for $-1<\Omega <-1+2\sqrt{2}$ and stable nodes for $\Omega >-1+2\sqrt{2}$; they are unstable foci for $-1-2\sqrt{2}<\Omega <-1$ and unstable nodes for $\Omega <-1-2\sqrt{2}$. At the two discriminant-zero endpoints they are repeated-eigenvalue degenerate nodes, with a nondiagonalizable Jacobian. At $\Omega =-1$, linearization alone does not determine nonlinear stability or prove a nonlinear centre.

A hyperbolic equilibrium point is a sink when all its eigenvalues have negative real parts. Nearby trajectories converge exponentially to it. In a planar system this includes both stable nodes and stable foci. The condition is stronger than neutral stability: a centre equilibrium is not a sink.