Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 14E (Dynamical Systems)

Choose $A=B=\sigma /r$ in the proposed Lyapunov function. The cubic terms cancel and direct differentiation givesFor $0<r<1$ this is negative definite, while $V$ is positive definite and radially unbounded. Finite-dimensional equivalence of these positive quadratic forms gives $\dot{V}\le -cV$ for some $c>0$. Thus $V(t)\le V(0)e^{-ct}$, every solution remains bounded and exists for all forward times, and every solution tends to zero. Therefore the origin is globally asymptotically stable, indeed exponentially stable.

The centre manifold theorem says that near an equilibrium of a sufficiently smooth finite-dimensional vector field, there is a locally invariant centre manifold tangent to the generalized eigenspace of eigenvalues with zero real part. It can be represented as a graph over that space, with graph value and derivative zero at the equilibrium; the dynamics on it determine the local centre-direction behaviour. For parameter-dependent systems the parameters may be adjoined as variables with zero derivative, producing an extended centre manifold.

At $r=1$, the linearization has the $x,y$ block $\left(\begin{array}{cc}-\sigma & \sigma \\ 1& -1\end{array}\right)$ and the independent eigenvalue $-b$. Its other eigenvalues are zero and $-(\sigma +1)$. Hence the equilibrium is nonhyperbolic, the stable subspace has dimension two, and the non-extended centre subspace has dimension one. A centre eigenvector is $(1,1,0)$.

With $\sigma =1$, substitute $x=(u+v)/2$, $y=(u-v)/2$, $r=1+\mu$. The equations becomeThe extended centre variables are $u,\mu$. The centre manifold theorem gives smooth graph functions $V,Z$ with $V(0,0)=Z(0,0)=0$ and all first derivatives zero. Their invariance equations are $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$, after substitution of the graph.

Count $\mu$ as order $u^2$. Then $\dot{u}=O(u^3)$ on the graph, so the left side of the $Z$ equation contributes only at higher order. Its quadratic terms give $0=u^2/4-bZ$, hence $Z=u^2/(4b)+O(u^3)$. The cubic terms of the $V$ equation similarly give $0=-2V-\mu u/2+uZ/2$. ThusTerms purely in $\mu$ vanish because the origin remains an equilibrium for every parameter; these expansions describe the weighted scaling $\mu =O(u^2)$, not an unrestricted one-variable Taylor series. Substitution gives the reduced equationIts cubic coefficient is negative. For $\mu <0$ the origin is attracting in the centre direction; for $\mu >0$ it loses stability and two attracting branches emerge with $u\sim \pm 2\sqrt{b\mu }$. Therefore the bifurcation is a supercritical pitchfork. These branches agree with the exact Lorenz equilibria $x=y=\pm \sqrt{b\mu }$, $z=\mu$.