Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 54 / 2

The spatial mean is conserved: averaging every term in the partial differential equation over the periodic cell gives ${\partial }_t⟨T⟩=0$. Adding a constant to $T$ also preserves the equation. We therefore fix the mean, conveniently at zero. This qualification is essential: without it the constant Fourier mode contributes an additional neutral direction, so the centre subspace would have five real dimensions rather than four.

On a Fourier mode $e^{i(mx+ny)}$, the linearized linear operator has eigenvalueOn the fixed-mean subspace, the first eigenvalues to reach zero are those with $m^2+n^2=1$. At $\mu =1$ these are the modes $e^{\pm ix},e^{\pm iy}$; every other nonconstant Fourier mode is strictly damped. The reality condition leaves two complex amplitudes, hence a four-real-dimensional centre manifold on the fixed-mean subspace. The trivial state loses linear stability as $\mu$ increases through one.

Use slow time $\tau ={\epsilon }^{2}t$, and write $T=\epsilon W+{\epsilon }^3{T}_3+\cdots$, whereThere is no quadratic nonlinearity, so no forced second-order correction is needed. At order ${\epsilon }^3$, with $\mu =1+{\epsilon }^2\nu$, projection onto the critical Fourier modes givesThe correction $T_3$ does not contribute to the critical projection because the linear operator vanishes on those modes. To find the nonlinear coefficient of $e^{ix}$, split the divergence term intoThe first term contributes $-3|A{|}^{2}A$. In the second, only the zero-$y$ Fourier coefficient of $W_y^2$ contributes; it is $2|B{|}^2$, and differentiating $W_x$ once more gives $-2|B{|}^{2}A$. The third term has zero $y$-average, and the fourth has no $x$-dependent critical component. Interchanging $x,y$ gives the other amplitude equation. Thus the conserved-mean convection amplitude equations areTo establish the selected planform, put $A=r{e}^{i\theta }$, $B=s{e}^{i\psi }$. The phases are constant and the magnitudes obeyFor $\nu >0$, the square state has $r=s=\sqrt{\nu /5}$. Its magnitude Jacobian matrix is $-(2\nu /5)\left(\begin{array}{cc}3& 2\\ 2& 3\end{array}\right)$, with eigenvalues $-2\nu$ and $-2\nu /5$. Its two neutral phases correspond to translation symmetry. The pure roll has $r^2=\nu /3,s=0$; an infinitesimal $B$ disturbance grows at rate $\nu -2\nu /3=\nu /3>0$. The other pure roll is equivalent, while the trivial state is unstable for $\nu >0$. Consequently squares, with $|A{|}^2=|B{|}^2=\nu /5$, are the stable small-amplitude planform, modulo translations. For $\nu <0$ the trivial state is stable.

Let $F(T)$ denote the right-hand side of the temperature partial differential equation. For a periodic variation $\eta$, integration by parts givesAlong the evolution, $T_t=F(T)$, so this is a gradient flow with increasing functional:For $s=|\nabla T{|}^2$, completing the square givesThus the increasing-gradient functional for poorly conducting convection converges to a finite limit along any globally smooth solution, andIn particular, there are arbitrarily late times at which $\Vert T_t{\Vert }_{L^2}$ is arbitrarily small. A nonstationary time-periodic solution is impossible: it would have a strictly positive increase of $V$ over a temporal period. More generally, if the orbit is precompact in a topology strong enough for the equation, its limiting states are stationary. With the conserved mean fixed, the displayed square completion bounds $\Vert \Delta T{\Vert }_{L^2}$, and periodic elliptic estimates give an $H^2$ bound; the usual smoothing and compactness for a globally regular parabolic solution support this limiting-state conclusion. The inequalities alone do not prove convergence to one uniquely specified equilibrium, and stationary families related by translation symmetry can remain.

Now take a smooth, nonconstant periodic roll $T_0(x)$. Multiplying its stationary partial differential equation by $T_0$ and using periodic integration by parts yieldsThe primes here are essential: the first and fourth terms involve $T_0^{\prime }$, not $T_0$. Write $a=T_0^{\prime }$, $b=T_0^{\prime \prime }$, $M_2=⟨a^2⟩$, $M_4=⟨a^4⟩$, and $N_2=⟨b^2⟩$. The orthogonal-roll perturbation hasThe cross term in the squared Laplacian averages to zero since $⟨b⟩=0$. Independence of the $x,y$ cell integrals then gives the exact polynomial differenceIts leading coefficient is half the variance of $a^2$ over the period. That variance is strictly positive: a nonconstant smooth periodic function has a derivative which vanishes somewhere and is nonzero somewhere, so $a^2$ is not constant. Hence arbitrarily small transverse perturbations increase $V$ above the roll value.

One can turn this energy comparison into linear instability. The linearized evolution at the roll is the self-adjoint operatorChoose the fixed-mean perturbation $\eta =T_0(y)-⟨T_0⟩$. Since $D^{2}V[T_0](\eta ,\eta )=⟨\eta L\eta ⟩=M_4-M_2^2>0$, the Rayleigh quotient is positive. The periodic self-adjoint fourth-order operator therefore has a positive eigenvalue. This proves the transverse energy instability of nonconstant temperature rolls: every nonconstant roll is unstable wherever it exists.

The existence qualification matters. Constants are also stationary and are stable on the fixed-mean subspace for $0<\mu <1$. Moreover, the periodic Poincaré inequality gives $N_2\ge M_2$, so the roll identity implies $\mu M_2=N_2+M_4>M_2$ for a nonconstant $2\pi$-periodic roll. Such rolls require $\mu >1$ on this specified cell; the conclusion for all positive $\mu$ is conditional on a nonconstant roll being available.