Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 60 / 2

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.

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.

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.