Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 59 / 2

Use the Galerkin method: substitute the retained trigonometric modes and project onto them with the $L^2$ inner product over the rectangle. Each mode satisfies the printed boundary conditions. Write $X=\alpha x$, $Z=\pi z$, and abbreviate the four mode prefactors by $C=2\sqrt{2}\beta /\alpha$, $D=2\sqrt{2}/\beta$, $V_1=2\sqrt{2}\sigma \pi R_{\Omega }/(\alpha \beta )$ and $V_2=\sigma R_{\Omega }/\alpha$. The Jacobian determinant in the advection terms is $J(f,g)=f_x{g}_z-f_z{g}_x$.

The stream function is a single Laplacian eigenfunction, so ${\nabla }^2\psi =-{\beta }^2\psi$ and $J(\psi ,{\nabla }^2\psi )=0$ exactly. For the temperature modes, direct multiplication givesFor the transverse-velocity modes, the retained products areFor example, $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$ and $\sin X\cos 2X=(\sin 3X-\sin X)/2$ supply the signs of the two retained $ac$ and $ae$ couplings. The third harmonics are orthogonal to the retained modes and are discarded by this Galerkin method; they are not claimed to vanish in the full partial differential equations.

Before changing time, the five coefficient equations areDividing by ${\beta }^2$ gives the stated five-mode rotating-convection truncation, withThe squared rotation parameter is $q=r_{\Omega }^2\ge 0$. The printed $R_{\Omega }$ is proportional to rotation rate, so it is a square-root-type rotation parameter rather than the more usual Taylor number proportional to squared rotation rate; the algebra uses the coefficient as printed. At zero rotation the physical transverse velocity modes vanish; the normalized $d,e$ equations can still be understood through their continuous zero-rotation limit.

Let $q=r_{\Omega }^2$ and $h=\varpi$. The $c,e$ eigenvalues at the conductive equilibrium point are $-h$ and $-(4-h)\sigma$, both negative since $0<h<4$ and $\sigma >0$. The coupled $(a,b,d)$ Jacobian matrix has characteristic polynomialwhereA simple zero eigenvalue occurs on $\boxed{r_P=1+q}$. The reflection $(a,b,d)\mapsto (-a,-b,-d)$, with $c,e$ fixed, makes its generic steady-state bifurcation a pitchfork bifurcation.

For a nonzero imaginary pair, substitute $s=i\omega$ into $P$: the real and imaginary parts require $A_3=A_1{A}_2$ and ${\omega }^2=A_2>0$. HenceWith $0<\sigma <1$, this Hopf bifurcation curve is present only for $q>q_{TB}=(1+\sigma )/(1-\sigma )$. At its intersection with the pitchfork bifurcation curve,and $P(s)=s^2(s+1+2\sigma )$. The zero eigenvalue has geometric multiplicity one: its eigenvector has $b=a$, $d=a/\sigma$. Thus this is a reflection-symmetric Bogdanov–Takens bifurcation, rather than two independent zero modes.

The Routh-Hurwitz stability criterion gives the stable trivial state below $r_P$ for $q<q_{TB}$ and below $r_H$ for $q>q_{TB}$. The rest of the line $r_P$, beyond $q_{TB}$, is a stationary bifurcation of an already unstable trivial state. Likewise the extension of $r_H$ below $q_{TB}$ is only an algebraic line, since its putative frequency is not real. The middle panel of the bifurcation diagram in part 1(b) distinguishes the actual curves from this extension.

To determine the nonlinear pitchfork bifurcation, solve the four slaved steady equations exactly. SetThe nonzero equilibrium points satisfyThusIf $k\le 1$, the coefficient of $u$ is positive for every physical $q\ge 0$: there is no change of pitchfork direction at a nonnegative rotation parameter. If $k>1$, the change occurs atBelow $q_D$ the nonzero branch lies on $r>r_P$; above $q_D$ it initially lies on $r<r_P$. At $q_D$, $r-r_D=k{u}^2+O(u^3)$, so $|a|$ grows like $(r-r_D{)}^{1/4}$ rather than the ordinary square-root scaling.

The rational equilibrium curve at a degenerate pitchfork also gives the nearby saddle-node bifurcations of the two reflection-related nonzero branches. For $q>q_D$,andThe two saddle-node bifurcations occur on one parameter curve by reflection symmetry; its separation from the pitchfork bifurcation curve is quadratic in $q-q_D$.

The stability statement needs a further condition. At $r_P$, the other coupled eigenvalues have polynomial $s^2+(1+2\sigma )s+\sigma [(1+\sigma )-(1-\sigma )q]$. They are stable only for $q<q_{TB}$. If $q_D<q_{TB}$, a one-dimensional centre manifold gives the local form $a^{\prime }=\chi a[\Delta -[1-q(k-1)]a^2-k{a}^4+\cdots ]$, where $\Delta =r-r_P$ and $\chi =\sigma /[(1+\sigma )-(1-\sigma )q_D]>0$. The quintic term stabilizes the large branch. The interval $r_{SN}<r<r_P$ then has a stable conductive state and two stable finite-amplitude states, with the two small unstable branches separating their basins of attraction. This is the usual subcritical convection hysteresis. If $q_D>q_{TB}$, the branch still reverses its direction, but the trivial state already has an additional unstable mode; this equilibrium-curve calculation alone does not imply a stable bistable wedge. At equality a two-dimensional centre manifold is essential.

A codimension-three degeneracy is possible when the double-zero eigenvalue condition and the vanishing cubic restoring coefficient of the pitchfork bifurcation coincide. Equating $q_D$ and $q_{TB}$ givesEquivalently $L=\sqrt{2{\sigma }^2/(1+\sigma )}$. For every $0<\sigma <1$, this is a physically admissible aspect ratio and $0<h<4$. The other two parameters are $q=(1+\sigma )/(1-\sigma )$ and $r=2/(1-\sigma )$.

At this point the fifth-order restoring term is nonzero: the steady relation is $r-r_{TB}=k{a}^4+O(a^6)$ with $k>0$. It is therefore a degenerate reflection-symmetric Bogdanov–Takens bifurcation with three independent conditions: zero determinant, zero frequency, and zero cubic restoring coefficient. Varying the aspect ratio supplies the third parameter, in addition to $r,q$. At a fixed generic aspect ratio it is not a codimension-two event available by varying only $r,q$.

It is not a triple-zero eigenvalue: the three-mode characteristic polynomial still has the noncritical factor $s+1+2\sigma$, while the $c,e$ eigenvalues remain strictly negative. The extra degeneracy is nonlinear. A fifth-order, two-dimensional normal form is needed for its complete unfolding; the one-dimensional quintic pitchfork bifurcation used away from $q_{TB}$ cannot be substituted at the double-zero point.