Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 359 / 1

The sobolev embedding theorem in three dimensions and the periodic elliptic estimate for the Stokes operator giveThe last step uses the absence of the zero Fourier mode: on mean-zero periodic fields, the Poincaré inequality makes the homogeneous $H^2$ norm controlled by $\Vert \Delta v{\Vert }_2=|Av|$.

Because $v$ is a divergence-free vector field, the product rule givesThe integral of a divergence over the periodic domain vanishes. Hence integration by parts yieldsThis is the skew-symmetry of incompressible transport.

Set $\chi =\eta$ in part (ii). The integral is equal to its own negative, soEquivalently, incompressible advection does not change the scalar's quadratic energy.

Both $P_m$ and the Leray-Helmholtz projection are orthogonal projections, hence contractions in $L^2$. Taking the $H$ norm of the first Galerkin equation givesTherefore

The first equation determines $u_m$ linearly from ${\theta }_m$:Substitution into the second equation gives an ordinary differential equation on the finite-dimensional space ${\tilde{H}}_m$. Its right-hand side is polynomial, and therefore locally Lipschitz continuous. The Picard-Lindelöf theorem supplies a unique local solution. The energy estimate in part (iii) bounds ${\theta }_m$ on every finite time interval, so the finite-dimensional continuation criterion rules out finite-time escape. The solution is consequently unique on every interval $[0,T]$.

Take the $L^2$ inner product of the temperature equation with ${\theta }_m$. The spectral projection disappears against ${\theta }_m\in {\tilde{H}}_m$, and part (a)(iii) cancels transport. ThusThe periodic Poincaré inequality, part (i), and the Cauchy-Schwarz inequality implyThe Gronwall inequality therefore gives, for $0\le t\le T$,This defines a bound $K_0(T)$ independent of $m$, and part (i) then givesIntegrating the energy identity and using the same bound on its right-hand side gives

It remains to estimate the time derivative. For $\phi \in H_{\mathrm{per}}^1$, the Fourier projection ${\Pi }_m$ is a contraction in $H^1$, and the skew identity from part (a) givesThe sobolev embedding theorem and the periodic elliptic estimate for the Stokes operator bound $\Vert u_m{\Vert }_{\infty }$ by $C|A{u}_m|$. Moreover,The already obtained bounds therefore implywith $K_0^{\prime }(T)$ independent of $m$.

The uniform bounds and the Banach-Alaoglu theorem provide a subsequence, not relabelled, and a function $\theta$ such thatandSince $H_{\mathrm{per}}^1$ embeds compactly into $L_{\mathrm{per}}^2$, the Aubin-Lions lemma strengthens the first convergence toThe weak continuity from evolution-space bounds gives a representativeTesting against fixed spatial modes and using ${\theta }_m(0)={\Pi }_m{\theta }_0$ shows that this representative satisfies $\theta (0)={\theta }_0$ weakly.

DefineThe first equation and the bounded inverse of the Stokes operator show that $u\in L^{\infty }(0,T;D(A))$ and thatin $L^{\infty }(0,T;H)$. In fact, the strong convergence from part (i) and convergence of the spectral projections implyIn three dimensions $D(A)\subset H^2\hookrightarrow L^{\infty }$, so $u_m{\theta }_m\to u\theta$ in $L^1(0,T;L^2)$. Because $\nabla \cdot u_m=0$,and the nonlinear term consequently converges in distributions and in the required weak $L^2(0,T;H_{\mathrm{per}}^{-1})$ sense. The linear terms pass by weak convergence, while ${\Pi }_m$ tends strongly to the identity. Hencein $L^2(0,T;H_{\mathrm{per}}^{-1})$. Together with part (i), this proves existence of a global weak solution of the Rayleigh-Bénard convection system on every finite interval.

Let $(u_1,{\theta }_1)$ and $(u_2,{\theta }_2)$ be two weak solutions with the same initial data, and set $U=u_1-u_2$ and $\Theta ={\theta }_1-{\theta }_2$. The diagnostic Stokes equation givesSubtracting the temperature equations yieldsPair this equation with $\Theta$. The term transported by $u_1$ vanishes by the skew-symmetry of incompressible transport, while the other nonlinear term satisfiesThe forcing difference is at most $C\Vert \Theta {\Vert }_2^2$. ConsequentlyThe coefficient is integrable on $[0,T]$ because ${\theta }_2\in L^2(0,T;H^1)$. Since $\Theta (0)=0$, the Gronwall inequality gives $\Theta =0$, and the Stokes equation then gives $U=0$. The weak solution is unique.

Part (c)(i) already gives weak continuity of $t\mapsto \theta (t)$ in $L^2$. In the assumed energy equality, the dissipation integralis absolutely continuous. The forcing integrand is in $L^1(0,T)$ because $u\in L^{\infty }(0,T;L^2)$ and $\theta \in L^{\infty }(0,T;L^2)$. The equality therefore makes $t\mapsto \Vert \theta (t){\Vert }_2^2$ continuous.

Whenever $t_n\to t$, weak continuity gives $\theta (t_n)\rightharpoonup \theta (t)$ and the energy equality gives convergence of their norms. The Radon-Riesz theorem, or directly the strong continuity from weak continuity and an energy equality, now gives $\theta (t_n)\to \theta (t)$ in $L^2$. Thus