Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 78 / 1

Put $y=\dot{x}$. The planar autonomous differential equation isIts equilibria have $y=0$, $x^2=p$. There are none for $p<0$, one at $p=0$, and two $x_{\pm }=\pm \sqrt{p}$ for $p>0$. At an equilibrium $x_{\ast }$ the stability matrix isWriting $s=\sqrt{p}>0$, $x_{+}=s$ is always a saddle equilibrium. At $x_{-}=-s$, the determinant is $2s>0$ and the trace is $s-q$: it is attracting for $q>s$ and repelling for $q<s$.

At $p=0$ and $q\ne 0$, the eigenvalues are $0,-q$; the nonzero quadratic term and parameter unfolding give a saddle-node bifurcation. At $p>0$, the trace crosses zero at $q=s$, with eigenvalues $\pm i\sqrt{2s}$, giving a Hopf bifurcation. It is nondegenerate: around $x=-s$ the equation is $\ddot{\xi }+(q-s)\dot{\xi }+2s\xi -{\xi }^2+\xi \dot{\xi }=0$. At $q=s$, a small oscillatory expansion of leading amplitude $r$ has $⟨\xi {\dot{\xi }}^2⟩=r^4/8+O(r^5)$, so the first nonlinear damping is positive and nonzero. For example, its second-order oscillatory expansion is $\xi =r\cos (\Omega t)+r^2/(2{\Omega }^2)-r^2\cos (2\Omega t)/(6{\Omega }^2)-r^2\sin (2\Omega t)/(6\Omega )$ with ${\Omega }^2=2s$. Thus the small attracting limit cycle is on the $q<s$ side. At $(p,q)=(0,0)$ the matrix has a double zero eigenvalue and a nontrivial Jordan block, the Bogdanov–Takens bifurcation organizing the two curves. The local bifurcation curves areA change between node and focus away from these curves is not itself a local bifurcation.

For the requested global exclusion, suppose $x(t)$ were periodic with period $T$. Integrate the original equation over a period. The integrals of $\ddot{x}$, $q\dot{x}$, and $x\dot{x}=\frac{1}{2}(d/dt)x^2$ vanish. ThereforeThe left side is nonnegative and the right side is negative when $p<0$. This contradiction proves there are no periodic orbits for $p<0$, independently of $q$.

With the stated time rescaling, $\dot{x}={\epsilon }^3{u}^{\prime }$ and $\ddot{x}={\epsilon }^4{u}^{\prime \prime }$, where primes refer to $\tilde{t}$. Dividing by ${\epsilon }^4$ givesAt $\epsilon =0$, put $v=u^{\prime }$ and define the HamiltonianIts orbital derivative is $H^{\prime }=v(u^2-\kappa )+(\kappa -u^2)v=0$, so this is a conserved quantity. The Hamiltonian system has $u^{\prime }=H_v=v$, $v^{\prime }=-H_u=u^2-\kappa$.

For $\kappa >0$, set $s=\sqrt{\kappa }$. The potential $U(u)=\kappa u-u^3/3$ has a minimum at $-s$ and a maximum at $s$. The corresponding center equilibrium and saddle equilibrium have energies $H_c=-2s^3/3$ and $H_s=2s^3/3$. Levels $H_c<H<H_s$ contain closed ovals around $(-s,0)$, as well as separate unbounded components on the right. At $H=H_s$, the center's ovals limit to a homoclinic orbit to the saddle, turning at $u=-2s$. Its equation isThe same energy also has external unbounded saddle separatrices for $u>s$. Levels above $H_s$ pass over the potential barrier and are open. Along the upper halves $u^{\prime }=v>0$, so the closed loops are traversed clockwise. If $\kappa =0$, the two critical points coalesce into a degenerate equilibrium; if $\kappa <0$, the potential is strictly decreasing and there is no center or homoclinic loop.

For small $\epsilon$, the conserved quantity from part (b) instead obeys $H^{\prime }=-\epsilon (\lambda +u)v^2$. The Melnikov energy-balance method evaluates the first-order net change along the unperturbed homoclinic orbit:Using the homoclinic curve $v^2=\frac{2}{3}(s-u{)}^2(u+2s)$, these are integrals of $vdu$ on its two halves. Their weighted mean is $I_1/I_0=-5s/7$. Thus the first-order splitting vanishes at $\lambda =5s/7$. Since its derivative with respect to $\lambda$ is $-I_0\ne 0$, the implicit function theorem gives a nearby curve where the stable and unstable saddle separatrices reconnect. This is the homoclinic balance for a quadratic-force oscillator and therefore a global, rather than local, bifurcation.

Returning to the original parameters, $q={\epsilon }^2\lambda$ and $\sqrt{p}={\epsilon }^2\sqrt{\kappa }$. With fixed positive $\kappa$ and $\epsilon \to 0$, the homoclinic bifurcation curve has leading formThe energy integral locates the leading curve; the perturbed orbit itself is no longer a constant-$H$ curve.

With $y=\dot{x}$, the feedback system isFor $p=0$, $(x,y,z)=(0,0,0)$ is an equilibrium. Its stability matrix and characteristic polynomial areThus there is always a zero eigenvalue, with eigenvector $(1,0,0)$, so it is a nonhyperbolic equilibrium. If $q\ne -1$, the other two eigenvalues are off the imaginary axis: for $q>-1$ their real parts are negative, while for $q<-1$ their product is negative. At $q=-1$, all three eigenvalues are zero, explaining the exclusion in the next part.

Adjoin $\dot{p}=0$ and use an extended centre manifold with local coordinates $(x,p)$. Count $p$ as order $x^2$, as appropriate to a steady-state fold, and writeThe centre-manifold invariance equation differentiates these graphs along $\dot{x}=y$. The derivatives $y_{x}y$ and $z_{x}y$ first contribute at weighted order three. Comparing the coefficients of $p$ and $x^2$ in the right sides therefore givesSince $q+1\ne 0$, their solution isHence the reduced normal form isBoth the quadratic and unfolding coefficients are nonzero, so the steady-state bifurcation is a saddle-node, with equilibria on the $p>0$ side. The exact steady equations confirm this: $y=0$, $z=x^2$, and $p=z+z^3=x^2+x^6$, giving two nearby equilibria for $p>0$ and none for $p<0$.

For $q>-1$ the transverse modes are attracting; the negative-$x$ branch is locally attracting and the positive-$x$ branch has one unstable direction. For $q<-1$, one transverse eigenvalue is already positive, so neither full equilibrium is asymptotically stable, although the reduced fold is still present with reversed flow orientation. The parameter $q=-1$ requires a higher-dimensional degeneracy analysis and is outside the stated assumption.