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The equilibria of $\dot{x}=(a^2-x)(a-y^2)$, $\dot{y}=x-y$ lie onfor every $a$, and on $x=y=\pm \sqrt{a}$ for $a\ge 0$. They meet in a subcritical pitchfork bifurcation at $a=0$; the positive square-root branch meets the $a^2$ branch and exchanges stability in a transcritical bifurcation at $a=1$.

Forthe equilibrium branches are $x=0$, $x=\pm \sqrt{2\mu }$, and $x=\pm \sqrt{\mu -a}$ wherever the square roots are real. If $a<0$, pitchforks occur at $\mu =a$ and $\mu =0$, and the two nonzero branch pairs meet in two transcritical bifurcations at $\mu =-a$. If $a=0$, all five branches meet in one degenerate fifth-order bifurcation. If $a>0$, separate pitchforks occur at $\mu =0$ and $\mu =a$. A constant perturbation produces imperfect pitchforks and saddle nodes, so none of these unperturbed bifurcations is structurally stable without the reflection symmetry.

Forthe invariant center manifold near $(x,y,k)=(0,0,0)$ is $y=0$, with reduced equation $\dot{x}=x(-k-3x+x^2)$ and hence a transcritical bifurcation. Near $(x,y,k)=(1,2,0)$, put $X=x-1$ and $v=y-2-X$. The extended center manifold beginsand its reduced equation is $\dot{X}=2(X^2-k)+O(3)$, a saddle-node bifurcation.

Chemostat washout means the culture's biomass concentration tends to zero because growth cannot sustain replacement of the cells leaving the vessel. In the ideal Monod equation chemostat with $D>0$, its equilibrium is $(s,x)=(s_{\mathrm{in}},0)$. Its Jacobian matrix has eigenvalues $-D$ and $\mu (s_{\mathrm{in}})-D$: a small inoculum grows when $D<\mu (s_{\mathrm{in}})$, whereas chemostat washout is locally asymptotically stable when $D>\mu (s_{\mathrm{in}})$. For nonnegative initial concentrations, chemostat washout occurs for $D\ge \mu (s_{\mathrm{in}})$; the boundary case also loses biomass asymptotically.

At $D=D_c$ the positive chemostat steady state meets the chemostat washout equilibrium, and one eigenvalue is zero, so linearization alone does not decide the boundary case. On the invariant nutrient-equivalent balance $s+x/Y=s_{\mathrm{in}}$, small positive $x$ obeys $\dot{x}=-[{\mu }^{\prime }(s_{\mathrm{in}})/Y]x^2+O(x^3)$: nutrient consumption lowers the specific growth rate below $D_c$, leading to slow decay rather than a positive biomass concentration. The continued mathematical branches exchange stability in a transcritical bifurcation; the positive-biomass branch becomes physically inadmissible beyond $D_c$.

At $(x,y)=(\mu ,0)$ near $\mu =1$, use $X=x-\mu$, $Y=y$, $\nu =\mu -1$. The stable direction is the $X$ axis and the centre direction is the $Y$ axis. The extended centre manifold has $X=Y^2+O(\nu Y^2,Y^4)$ and reduced flow $\dot{Y}=-\nu Y-2Y^3+\cdots$, a supercritical pitchfork bifurcation in parameter $-\nu$. At $(0,1)$ near $\mu =-1$, use $X=x$, $Y=y-1$, $\nu =\mu +1$. The centre line is $Y=-X/2$, the stable line is $X=0$, and the reduced flow is $\dot{X}=\nu X-2X^2+\cdots$, a transcritical bifurcation. The branch $x=(\mu +1)/2$, $y=\sqrt{(1-\mu )/2}$ joins the two bifurcations and is stable for $-1<\mu <1$.

The equilibria are $0$ and $-\epsilon \pm \sqrt{{\epsilon }^2+\lambda }$. Their linear trace is $\kappa -v_{\ast }^2$ and determinant $\lambda -4\epsilon v_{\ast }-3v_{\ast }^2$. For nonzero $\epsilon$ the symmetry-protected pitchfork bifurcation splits into a transcritical bifurcation at $\lambda =0$ and a saddle-node bifurcation at $\lambda =-{\epsilon }^2$. The energy $H={\dot{v}}^2/2+\lambda v^2/2-2\epsilon v^3/3-v^4/4$ obeys $\dot{H}=(\kappa -v^2){\dot{v}}^2$. For positive $\lambda$, the two outer saddle equilibria have unequal barrier heights, so the symmetric heteroclinic cycle unfolds into separated one-way connections and a homoclinic orbit termination of the central stable limit cycle.

A transcritical bifurcation need not already have its persistent branch at coordinate zero. The smooth parameter-dependent shift $u=y-\mu$ turns the displayed crossing of $y=\pm \mu$ into the standard form with parameter $-2\mu$. The crossing branches exchange dynamical stability. A mere nonsmooth relabeling of stable and unstable branches as $\pm |\mu |$ would conceal this smooth exchange.

At $\mu =-1/4$, the two branches meet at $(1/2,1/4)$. The eigenvalues there are $0$ and $1/2-\lambda$. For $\lambda \ne 1/2$, the nonzero quadratic term and transverse stability identify a saddle-node bifurcation. If $\lambda >1/2$, it creates a sink equilibrium and a saddle equilibrium as $\mu$ increases; if $\lambda <1/2$, it creates a source equilibrium and a saddle equilibrium. At $\lambda =1/2$, the zero eigenvalue is double and the ordinary saddle-node description needs the two-parameter unfolding.

At $\mu =0$, the origin and lower branch cross. Since $s_{-}=-\mu +O({\mu }^2)$, the branches cross transversely in the $(\mu ,u)$ diagram. The origin changes from a sink equilibrium to a saddle equilibrium as $\mu$ increases through zero; the lower branch changes from a positive-$u$ saddle equilibrium to a negative-$u$ sink equilibrium. This is a transcritical bifurcation. A local centre-manifold calculation also gives $\dot{u}=\mu u+u^2+\cdots$, because $v=O(u^2)$ and the transverse eigenvalue is $-\lambda <0$.

The figure uses a representative $\lambda >1/2$; the upper branch's additional Hopf change is indicated. For $\lambda <1/2$ the upper branch is unstable everywhere on its existence interval. The exact trace criterion in part (a) specifies stability for every fixed positive $\lambda$.

Put $a=a_1>0$, so $a_2=4a/5$ and $a_3=8a/5$. The second equilibrium equation factors as $B({\lambda }_2+a_{3}A)=0$. The resulting equilibria areThe mixed pair exists strictly when ${\lambda }_2({\lambda }_1-5{\lambda }_2/8)>0$; equality describes its merger with an axial equilibrium. At $O$ the eigenvalues of the Jacobian matrix are $({\lambda }_1,{\lambda }_2)$, and at $P$ they are $(-{\lambda }_1,{\lambda }_2-8{\lambda }_1/5)$. At a nonzero mixed equilibrium,Thus both mixed equilibria are saddle equilibria. The axial equilibria are nodes when their two real eigenvalues have the same sign and saddles when the signs differ. In particular,There are exactly three local bifurcation lines away from their common codimension-two intersection. On ${\lambda }_1=0$, $O$ and $P$ meet in a transcritical bifurcation, visible in $\dot{A}={\lambda }_{1}A+a{A}^2$ on the invariant axis $B=0$. On ${\lambda }_2=0$, the mixed pair meets $O$ in a pitchfork bifurcation. Eliminating $A$ near $O$, for fixed nonzero ${\lambda }_1$, gives $\dot{B}={\lambda }_{2}B-(a_2{a}_3/{\lambda }_1)B^3+\cdots$: this pitchfork is supercritical for ${\lambda }_1>0$ and subcritical for ${\lambda }_1<0$. On ${\lambda }_2=8{\lambda }_1/5$, the mixed pair meets $P$. Writing $A=-{\lambda }_1/a+u$ and $\eta ={\lambda }_2-8{\lambda }_1/5$, elimination gives $\dot{B}=\eta B+(a_2{a}_3/{\lambda }_1)B^3+\cdots$. This pitchfork is subcritical for ${\lambda }_1>0$ and supercritical for ${\lambda }_1<0$. A supercritical pitchfork here need not produce a stable mixed state: its other eigenvalue can already be positive. The negative mixed Jacobian determinant settles the full stability question. There is no Hopf bifurcation in this quadratic truncation.

Along ${\lambda }_2={\lambda }_1-\Delta$, the axial curves are $A=0$ and $A=-{\lambda }_1/a$. The mixed pair has the same plotted amplitude for its two signs of $B$:Consequently the mixed curves exist outside the interval whose endpoints are $\Delta$ and $-5\Delta /3$. The stable portions of the axial curves areFor $\Delta >0$, stability passes directly from $O$ to $P$ at the transcritical bifurcation ${\lambda }_1=0$; the two pitchfork bifurcations lie at $-5\Delta /3$ and $\Delta$. For $\Delta <0$, $O$ loses stability at ${\lambda }_1=\Delta$ and $P$ becomes stable only at ${\lambda }_1=-5\Delta /3$. Between these points none of the equilibria is stable. This local statement does not prescribe a bounded attractor of the quadratic truncation. All mixed portions are unstable.

Write $h={\mu }^2+\epsilon$. The entire unfolded diagram is obtained from $g(x)=h$, with the same dynamical stability sign $-x(x+2)$ as before. A saddle-node bifurcation at $x=0$ requires $h=0$, while a saddle-node bifurcation at $x=-2$ requires $h=4/3$.

For $0<\epsilon <4/3$, the crossing separates into a positive stable branch and a negative unstable branch, with no saddle-node bifurcation near $\mu =0$. Three equilibria exist for $|\mu |<\sqrt{4/3-\epsilon }$, and the two negative branches annihilate at the outer saddle-node bifurcations $\mu =\pm \sqrt{4/3-\epsilon }$. Outside that interval only the positive stable equilibrium remains. If $\epsilon >4/3$, the diagram has just that one branch everywhere; equality is the limiting saddle-node bifurcation contact at $\mu =0,x=-2$.

For $\epsilon <0$, let $d=-\epsilon >0$. Two inner saddle-node bifurcations occur at $\mu =\pm \sqrt{d},x=0$, and two outer saddle-node bifurcations at $\mu =\pm \sqrt{4/3+d},x=-2$. For $|\mu |<\sqrt{d}$ there is a single stable equilibrium below $-3$; for $\sqrt{d}<|\mu |<\sqrt{4/3+d}$ there are three equilibria with stable–unstable–stable ordering; outside the outer saddle-node bifurcations only the positive stable equilibrium remains. The preceding figure shows representative small unfoldings. In particular this unrestricted perturbation destroys the exact transcritical bifurcation crossing.

Set $g(x)=x^2+x^3/3$. Equilibria satisfy ${\mu }^2=g(x)$, and their scalar linear eigenvalue isIt vanishes only at $x=0$ or $x=-2$. Substitution in the equilibrium relation gives the three nonhyperbolic parameter-state pointsThe branches with $x<-2$ and $x>0$ are stable, while $-2<x<0$ is unstable. For $0<|\mu |<2/\sqrt{3}$ there are three equilibria; beyond the two outer saddle-node bifurcations there is just the positive stable one. At $\mu =0$ there is also the distinct stable equilibrium $x=-3$.

Near the crossing, use the smooth coordinate $y=x\sqrt{1+x/3}$, whose derivative is positive near zero. The equation becomes $\dot{y}=p(y)({\mu }^2-y^2)$ with $p>0$. A positive time rescaling removes $p$, and the parameter-dependent state shift $u=y-\mu$ givesThus the crossing at $\mu =0$ is a transcritical bifurcation, with parameter $-2\mu$. The two smooth branches $y=\pm \mu$ exchange dynamical stability; labeling the stable branch by $|\mu |$ instead would obscure that exchange. The parameter-dependent coordinate shift in a transcritical bifurcation explains why the original coordinate need not display a branch identically equal to zero.

Put $w=\dot{v}$. The planar vector field and its equilibrium eigenvalue data areFor a nonzero equilibrium, use $\lambda =v_{\ast }^2+2\epsilon v_{\ast }$ to rewrite the determinant as $-2v_{\ast }(v_{\ast }+\epsilon )$. These formulas determine the local bifurcations of the imperfect soft Duffing-van der Pol oscillator directly.

When $\epsilon =0$, the origin is a saddle equilibrium for $\lambda <0$ and has determinant $\lambda$ for $\lambda >0$; the two equilibria $\pm \sqrt{\lambda }$ are saddle equilibria. They meet the origin in a symmetry-protected pitchfork bifurcation at $\lambda =0$ for $\kappa \ne 0$. At $\kappa =0,\lambda >0$ the origin undergoes a supercritical Hopf bifurcation. To determine criticality, take $v=r\cos (\sqrt{\lambda }t)+\cdots$: averaging $\dot{H}=(\kappa -v^2)w^2$ gives $\dot{r}=\kappa r/2-r^3/8+\cdots$, so the stable small limit cycle has $r^2=4\kappa +\cdots$.

The potential energy of the conservative part and the exact energy balance areAt zero imperfection the stable limit cycle grows to a heteroclinic cycle connecting the two outer saddle equilibria. Near the double-zero point its leading connection curve can be calculated, rather than inferred just from symmetry. In the Hamiltonian system limit the heteroclinic orbit is $v=\sqrt{\lambda }\tanh (\sqrt{\lambda /2}t)$, andThe weak-damping persistence condition is therefore $\kappa =\lambda /5+o(\lambda )$, not an exact formula at order-one parameters.

Now take $\epsilon >0$; negative imperfection reflects the phase portrait. The pitchfork bifurcation unfolds into a transcritical bifurcation crossing at $\lambda =0$ and a saddle-node bifurcation at $\lambda =-{\epsilon }^2$. The latter occurs at $v=-\epsilon$. Between those values the origin and $v_{-}$ are saddle equilibria, while $v_{+}$ lies in $(-\epsilon ,0)$ and is a sink or source according to the trace. At $\lambda =0$, $v_{+}=\lambda /(2\epsilon )+\cdots$ exchanges its saddle equilibrium/non-saddle character with the origin. The double-zero points are now at $(\lambda ,\kappa )=(0,0)$ and $(-{\epsilon }^2,{\epsilon }^2)$.

The original Hopf bifurcation line $\kappa =0,\lambda >0$ remains, still supercritical. A second Hopf bifurcation curve occurs on the intermediate nonzero equilibrium:It is important not to assign the same criticality to its entire length. Write $s=\sqrt{{\epsilon }^2+\lambda }$, so $q=-\epsilon +s$. Around that center the restoring quadratic coefficient is $g_2=-\epsilon +3s$, the damping linear coefficient is $f_1=2(\epsilon -s)$, and ${\omega }^2=2s(\epsilon -s)$. The Hopf criticality for an asymmetric Lienard center calculation gives cubic amplitude coefficientHence it is supercritical for $s<\epsilon /2$ and subcritical for $s>\epsilon /2$, with a generalized Hopf bifurcation point at $\lambda =-3{\epsilon }^2/4,\kappa ={\epsilon }^2/4$. Higher-order terms are needed exactly there. For small imperfection the leading quintic coefficient is positive: writing $q_1=v-q$ and using the coordinate that normalizes the potential energy, $y=q_1\sqrt{1-a{q}_1-b{q}_1^2}$, with $a=2g_2/(3{\omega }^2)$ and $b=1/(2{\omega }^2)$, the coefficient of $y^4$ in $(f_1{q}_1-q_1^2)d{q}_1/dy$ at vanishing cubic coefficient is $10b/3>0$. Weak-damping averaging therefore supplies the nearby saddle-node bifurcation of stable and unstable limit cycles in the generalized Hopf bifurcation unfolding. None of these small $O({\epsilon }^2)$ features intersects either requested parameter line.

Globally, the two saddle equilibrium barriers are no longer equal. Their leading energy difference is $H(v_{+},0)-H(v_{-},0)=-4\epsilon {\lambda }^{3/2}/3+\cdots$. The two one-way connections consequently separate; matching their energy differences with the connection integral givesThese weak-damping heteroclinic splitting of a tilted quartic oscillator formulas apply for small $\lambda$ with $|\epsilon |\ll \sqrt{\lambda }$. Between the separated connection curves, the central stable limit cycle terminates in a homoclinic orbit to the lower-barrier saddle equilibrium, $v_{+}$ for positive imperfection. The simultaneous symmetric heteroclinic-cycle destruction is therefore replaced by two basin-changing one-way connections and a distinct homoclinic orbit loss of the limit cycle. Its period diverges at that loss.

Along $\kappa +\lambda =1$, decrease $\lambda$ from above one. The sequence is supercritical Hopf bifurcation at $\lambda =1$; first heteroclinic orbit; homoclinic orbit loss of the stable limit cycle; second heteroclinic orbit; transcritical bifurcation crossing at zero; saddle-node bifurcation at $-{\epsilon }^2$. The connection locations are global, not obtained by treating the Melnikov approximation as exact. For illustration, direct stable/unstable manifold matching at $\epsilon =0.04$ places them at approximately $\lambda =0.882,0.858,0.783$, respectively. The diagram below gives the qualitative ordering without prescribing those example values for arbitrary imperfection.

Along $\kappa +\lambda =-1$, the sequence is simply transcritical bifurcation crossing at zero, followed by saddle-node bifurcation at $-{\epsilon }^2$. Before the latter, $\kappa <0$ in the relevant region, so $H$ decreases strictly on nonstationary trajectories and there is no periodic orbit or saddle equilibrium loop. After it only a saddle equilibrium remains, which cannot be enclosed by a periodic orbit by the Poincare-index obstruction to a periodic orbit. The origin does not undergo a Hopf bifurcation at $\lambda =-1$, since its determinant there is negative.

Write $c=(1-a-b)/a$. The admissible equilibrium points are $O=(0,0)$, $E=(1-a,0)$ when $a\le 1$, and $P=(b,c)$ when $a+b\le 1$. The Jacobian matrix isAt $O$ the eigenvalues are $(1-a)/a$ and $-1$; at $E$ they are $-(1-a)/a$ and $(1-a-b)/b$. At a positive coexistence equilibrium point,It is therefore asymptotically stable. It is a node for $b^2>4a(1-a-b)$, a focus for the reverse inequality, and has repeated real eigenvalues at equality. The node–focus change does not alter the attracting equilibrium point's topology.

There are three open physical parameter regions:

The axes are invariant. On $x=0$, $y$ decays exponentially; on $y=0$, prey obey the logistic differential equation. Positivity follows because each right-hand side has its population as a factor. Also $x$ is bounded by logistic comparison, while $W=x+by$ satisfies $\dot{W}=(1-a)x/a-x^2/a-by$, so all forward trajectories are bounded. The Dulac function $1/(xy)$ has divergence $-1/(ay)<0$ in the interior, excluding periodic orbits by the Bendixson-Dulac criterion.

In the extinction region, $\dot{x}\le -(a-1)x/a$, and then $\dot{y}/y=x/b-1$ is eventually negative. In the prey-only region, the upper bound $\mathrm{limsup}x\le 1-a<b$ implies predator decay and convergence of interior prey to $1-a$. In the coexistence region the Lyapunov functionhas $\dot{\mathcal{V}}=-(x-b{)}^2/a\le 0$. Its positive-quadrant sublevel sets are compact, and the only invariant subset with $x=b$ also has $y=c$. The LaSalle invariance principle proves convergence of every interior trajectory to $P$. These facts justify the complete qualitative phase portraits, not just their local arrows.

At $a=1$, $O$ and $E$ meet in a transcritical bifurcation; positive prey decay algebraically. At $a+b=1$ with $a<1$, $E$ and $P$ exchange stability. On its centre manifold, $x=b-ay+O(y^2)$ gives $\dot{y}=-a{y}^2/b+O(y^3)$, so the physically positive side is still attracting. There is no interior Hopf bifurcation since the coexistence trace never vanishes.

The quadratic discrete predator-prey map equals the identity plus the vector field of part (a), so its admissible fixed points are the same $O,E,P$. Their Jacobian matrices are the identity plus the corresponding continuous-time Jacobian matrices, but map stability requires eigenvalues inside the unit circle rather than the left half-plane.

At $O$ the multipliers are $1/a,0$: it is a sink for $a>1$ and a saddle for $a<1$. At $E$ they are $2-1/a$ and $(1-a)/b$. ThereforeFor $a<1/3$, the prey multiplier is below $-1$: $E$ is a saddle when $b>1-a$ and a repeller when $b<1-a$. For $1/3<a<1$, $b<1-a$ makes it a saddle. The $+1$ crossings at $a=1$ and $a+b=1$ are the two transcritical bifurcations already encountered. The crossing at $a=1/3$ is a prey period-doubling bifurcation.

At positive coexistence,For clarity, derive the relevant Schur stability criterion directly: setting $z=(1+s)/(1-s)$ in $z^2-Tz+D$ givesThe open unit disk maps to $\mathrm{Re}s<0$. A real quadratic has both roots there precisely when all three displayed coefficients have the same positive sign. Since $1-T+D=c>0$, the coexistence fixed point is a sink precisely whenThis region exists only for $a>1/9$. In the coexistence triangle $a+b<1$, put $K=1+T+D=(3a+1-3b)/a$. If $K<0$, one root lies below $-1$ and the other in $(-1,1)$, so $P$ is a saddle. If $K>0$ but $D>1$, it is a repeller; if $K>0,D<1$, it is a sink. These signs give the complete local fixed-point partition, including real and complex multipliers.

The coexistence period-doubling bifurcation lies at $b=a+1/3$; its segment adjoining the sink region has $1/9<a<1/3$. The other multiplier there is $2-1/(3a)$ and lies strictly inside the unit circle on this segment. This coexistence flip is subcritical on that segment. To calculate its direction, normalize the $-1$ eigenvector by $q_f=(1,(2-b/a)/b{)}^T$ and the left eigenvector by $p_f{q}_f=1$. The symmetric quadratic derivative $\mathcal{B}$ is given explicitly in part (d). Eliminating the quadratic term from the one-dimensional centre manifold map gives the effective cubic flip coefficientFor a scalar reduced map $u^{\prime }=(-1+\delta )u+c_f{u}^3+\cdots$, its second iterate satisfies $u^{\prime \prime }-u=-2\delta u-2c_f{u}^3+\cdots$. Thus $u^2=-\delta /c_f>0$ on the stable-fixed-point side $\delta >0$, and the two-cycle multiplier is $1+4\delta +\cdots >1$. An unstable coexistence two-cycle meets the stable fixed point as $b$ increases to the flip curve. The other multiplier remains inside the unit circle. The complex unit-modulus crossing lies at $a+2b=1$, with $1/9<a<1$. The two curves meet at $(a,b)=(1/9,4/9)$ with a double multiplier $-1$. The prey flip and predator invasion curves meet at $(1/3,2/3)$ with multipliers $-1,+1$.

The sketches mark the sink regions and all these local thresholds. They are a local atlas, not a claim that the map has only these global attractors. Indeed the full nonnegative quadrant is not invariant: $x^{\prime }=x(1-x-ay)/a$ becomes negative when $x+ay>1$. On the prey axis the logistic map already supplies new dynamics. Its nontrivial two-cycle, obtained by factoring $F^2(x)-x$ after removing the fixed-point factors, isIts longitudinal multiplier is $4+2/a-1/a^2$, so it is stable on the axis for $1/(1+\sqrt{6})<a<1/3$. The transverse multiplier is $x_{+}{x}_{-}/b^2=a(1+a)/b^2$, making the cycle stable in the plane when additionally $b>\sqrt{a(1+a)}$. Further flips and more complicated logistic dynamics show why the differential equation's three global regions cannot be transferred unchanged to its unit-step map.

At a fixed point $y=x$, and the first equation factors. The branches areThe three branches meet at $a=0$, and the positive square-root branch meets the quadratic branch at $a=1$. The Jacobian matrix at a general fixed point isOn $x=a^2$, its trace is $-1$ and determinant $a-a^4=a(1-a^3)$: the branch is attracting for $0<a<1$ and a saddle for $a<0$ or $a>1$. On $x=\pm \sqrt{a}$, the eigenvalues are $-1$ and $2a(1\mp a^{3/2})$. The negative branch is always a saddle for $a>0$; the positive branch is a saddle for $0<a<1$ and attracting for $a>1$. At the bifurcation values the zero eigenvalue requires the nonlinear analysis below.

For $a=1+\mu$, $u=x-1$, $v=y-x$, the equations areThe extended centre manifold has $v=h(u,\mu )$ with no linear terms. Since $h_u\dot{u}$ has degree at least three, its invariance equation to degree two is $0=-h_2-(2{\mu }^2-5\mu u+2u^2)$. ThusThe reduced quadratic factors as $(u-2\mu )(2u-\mu )$. Its two roots have reduced eigenvalues $3\mu$ and $-3\mu$, respectively, and the transverse eigenvalue remains negative. They exchange stability at $\mu =0$, a transcritical bifurcation. The exact branches have $u=2\mu +{\mu }^2$ and $u=\mu /2-{\mu }^2/8+\cdots$, consistent with this second-order evolution equation.

Near $a=0$, the slow equation has leading form $\dot{x}=x^3-ax$, so this is a subcritical pitchfork bifurcation with the parameter orientation displayed. For $a<0$ there is one nearby saddle, whose unstable direction is close to $y=x$ and stable direction close to vertical. For $a>0$ the central branch is attracting and the outer two equilibria near $\pm \sqrt{a}$ are saddles; their separatrices distinguish attraction toward the central equilibrium from departure outwards. The actual central location is $(a^2,a^2)$, not exactly the origin away from $a=0$.

Adding positive $\epsilon$ breaks the pitchfork symmetry. The leading equilibrium equation near zero is $x^3-ax+\epsilon =0$: the diagram unfolds into a single branch plus a saddle-node pair, with the latter appearing for positive $a$ of order ${\epsilon }^{2/3}$. More precisely the fold is at $x=(\epsilon /2{)}^{1/3}$, $a=3(\epsilon /2{)}^{2/3}$ at leading order. Near one the reduced equation becomes $2u^2-5\mu u+2{\mu }^2+\epsilon =0$. Its discriminant is $9{\mu }^2-8\epsilon$; the crossing is replaced by two saddle-node endpoints with a no-equilibrium gap for $|\mu |<\sqrt{8\epsilon }/3$ to leading order. Outside that gap, the lower reduced root is stable and the upper one unstable. These local approximations describe the sketch; the exact perturbed curves have higher-order corrections.

Representative normal forms are $\dot{x}=\mu +x^2$ for a saddle-node bifurcation, $\dot{x}=\mu x-x^2$ for a transcritical bifurcation, and $\dot{x}=\mu x-x^3$ for a supercritical pitchfork bifurcation. Replacing the last cubic term by $+x^3$ gives the subcritical pitchfork bifurcation.

Treat $\mu$ as a variable with $\dot{\mu }=0$. The extended center manifold is tangent to $y=0$, so write $y=h(x,\mu )=A{x}^2+Bx\mu +C{\mu }^2+O((|x|+|\mu |{)}^3)$. Its invariance equation is $h_x\dot{x}=\dot{y}$. At quadratic order this readsMatching coefficients gives $A=2$, $B=-4$, $C=4$. Substitution into the first equation therefore givesHere $O(3)$ denotes total degree at least three in $(x,\mu )$. With $T=t$, $X=x-2\mu$ and $\tilde{\mu }=\mu$, this becomes $dX/dT=\tilde{\mu }+X^2+O(3)$. Thus $\alpha =1$, $\beta =2$, $\gamma (\mu )=\mu$, and the bifurcation is a saddle-node. For negative parameter, the negative-$X$ branch is stable and the positive-$X$ branch unstable along the center manifold; the transverse eigenvalue is near $-1$.

The fixed-point equation is $x[a{x}^2+bx+\mu -1]=0$. Thuswhen the discriminant is nonnegative. At zero the multiplier is $F^{\prime }(0)=\mu$, so it is linearly stable for $-1<\mu <1$ and unstable for $|\mu |>1$. At a nonzero fixed point, eliminate $\mu$ to obtain the branch and its multiplier:The branch is stable precisely when $-2<x(b+2ax)<0$. These formulas specify stability throughout the diagram, including any extra flip points outside the region requested.

If $b=0$ and $a>0$, the two nonzero branches $x=\pm \sqrt{(1-\mu )/a}$ exist for $\mu <1$ and have multiplier $3-2\mu >1$. Thus $\mu =1$ is a subcritical pitchfork bifurcation. If $b\ne 0$, the branch through zero has $x\sim (1-\mu )/b$ and $M\sim 2-\mu$; its stability exchanges with zero at $\mu =1$, a transcritical bifurcation. The two nonzero roots meet atwhere $M=1$. The graph of $\mu (x)$ has nonzero second derivative $-2a$, proving a saddle-node bifurcation. At zero and $\mu =-1$, the multiplier is $-1$, indicating a period-doubling bifurcation; generically its nondegeneracy coefficient is $a+b^2$. If $a+b^2=0$, necessarily $b\ne 0$ here, composition at $\mu =-1$ instead gives $F^2(x)-x=4b^4{x}^5+O(x^6)$. For $\delta =\mu +1>0$ small, the leading return equation is $-2\delta x+4b^4{x}^5=0$, producing a two-cycle with leading amplitudes $x\sim \pm [\delta /(2b^4){]}^{1/4}$. Its return multiplier is $1+8\delta +o(\delta )>1$. Thus this special case still has a flip bifurcation, but it is degenerate and subcritical rather than the generic cubic normal form.

For $a,b>0$, the fold lies above $\mu =1$. Between it and the transcritical point, the branch $-b/(2a)<x<0$ is stable provided its multiplier remains above $-1$, while the other branch is unstable. For $a,b<0$, the fold lies below $\mu =1$: immediately above it the branch $x<-b/(2a)$ is stable and the branch between $-b/(2a)$ and zero is unstable; the branch crossing zero becomes stable for $\mu >1$. All these statements are subject to the displayed exact multiplier criterion if the diagram is extended. The sketch uses representative parameter values and stops before additional flips, as stipulated.

For a transcritical bifurcation, use $\boxed{x_{n+1}=x_n+\mu x_n-x_n^2}$. The fixed points $x=0$ and $x=\mu$ cross and exchange stability at $\mu =0$.

For a flow, a hyperbolic fixed point has no eigenvalue of its linearization with zero real part; a nonhyperbolic fixed point has at least one such eigenvalue. At $(\mu ,0)$ the eigenvalues are $-\mu ,1-\mu$, while at $(0,\pm 1)$ they are $\mu +1,-2$. Thus nonhyperbolicity occurs at the two stated parameters. The following centre manifold reductions show the associated actual bifurcations.

Near $(1,0)$ put $X=x-\mu$, $Y=y$, $\nu =\mu -1$. The transformed equations areAt $\nu =0$, the non-extended stable subspace is the $X$ axis and the centre subspace is the $Y$ axis. In the extended centre manifold, write $X=h(Y,\nu )$, with $h(0,\nu )=0$. The symmetry $Y\mapsto -Y$ allows an even graph. At quadratic order its invariance equation has left side of degree at least three and right side $-h+Y^2$, so $h=Y^2+O(\nu Y^2,Y^4)$. Substitution givesThis is a supercritical pitchfork bifurcation in parameter $-\nu =1-\mu$. The central branch is stable for $\mu >1$ and unstable for $\mu <1$ near one; the two nonzero branches for $\mu <1$ are stable. At the bifurcation itself the cubic term attracts on the centre manifold.

Near $(0,1)$ put $X=x$, $Y=y-1$, $\nu =\mu +1$. ThenAt $\nu =0$, the stable subspace is $X=0$ and the centre line is $Y=-X/2$. Thus the leading extended centre graph is $Y=-X/2+O(X^2,\nu X)$, and the reduced flow isThis is a transcritical bifurcation: $X=0$ is stable for $\nu <0$, while the branch $X=\nu /2+\cdots$ is stable for $\nu >0$; the opposite branches are unstable. At the critical point the quadratic term is only one-sided attracting.

The exact additional branch in $y>0$ isIt meets $(\mu ,0)$ at $\mu =1$ and $(0,1)$ at $\mu =-1$, displaying the connecting pitchfork and transcritical branches in a quadratic-product flow. Its Jacobian has determinant $1-{\mu }^2$ and trace $(\mu -3)/2$. Hence it is stable for $-1<\mu <1$ and a saddle for $\mu <-1$. The negative-$y$ companion is needed for the pitchfork sketch near $(1,0)$; no separate analysis of $(0,-1)$ is required.

Now $G(x)=x[{\mu }^2-(x^2-a{)}^2]$. The zero point exists for all parameters and has eigenvalue $G^{\prime }(0)={\mu }^2-a^2$, so it is stable for $|\mu |<|a|$ and unstable for $|\mu |>|a|$. At every nonzero fixed point,Both outer branches are stable and both inner branches unstable.

For $a>0$, the nonzero locations are as in part (i). At $(\mu ,x)=(\pm a,0)$ the local equation is $\dot{x}=({\mu }^2-a^2)x+2a{x}^3+O(x^5)$: these are subcritical pitchfork bifurcations, with the two unstable inner branches present on the side where zero is stable. At the bifurcation itself $\dot{x}\sim 2a{x}^3$, so zero is unstable. At $(0,\pm \sqrt{a})$ the inner and outer branches cross and exchange stability in transcritical bifurcations. The positive crossing is semistable, attracting from the right; the negative crossing is semistable, attracting from the left, because the additional factor $x$ reverses the local flow there.

For $a<0$, the outer stable pair exists only for $|\mu |>|a|$. The points $(\pm |a|,0)$ are supercritical pitchfork bifurcations, since the cubic coefficient $2a$ is negative. Zero loses stability as $|\mu |$ increases across $|a|$, and a stable symmetric pair appears. At the threshold zero remains nonlinearly stable, with $\dot{x}\sim 2a{x}^3$. There are no further bifurcations. The second row of the preceding figure gives the diagrams. The plots use $a=\pm 1$; the general locations are obtained from the formulas above.