= Solution
Define
$$
u=\frac{\alpha}{\nu}A^{-1}P_\sigma(\theta e_3).
$$
The first equation and the bounded inverse of the <Stokes operator> show that $u\in L^\infty(0,T;D(A))$ and that
$$
\nu Au=\alpha P_\sigma(\theta e_3)
$$
in $L^\infty(0,T;H)$. In fact, the strong convergence from part (i) and convergence of the spectral projections imply
$$
u_m\longrightarrow u
\quad\hbox{strongly in }L^2(0,T;D(A)).
$$
In 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 $\nabla\mathbin\cdot u_m=0$,
$$
(u_m\mathbin\cdot\nabla)\theta_m
=\nabla\mathbin\cdot(u_m\theta_m),
$$
and the nonlinear term consequently converges in distributions and in the required weak $L^2(0,T;H^{-1}_{\rm per})$ sense. The linear terms pass by weak convergence, while $\Pi_m$ tends strongly to the identity. Hence
$$
\partial_t\theta-\kappa\Delta\theta
+(u\mathbin\cdot\nabla)\theta
=\beta(u\mathbin\cdot e_3)
$$
in $L^2(0,T;H^{-1}_{\rm per})$. Together with part (i), this proves existence of a global <weak solution> of the <Rayleigh-Bénard convection> system on every finite interval.
Solved by gpt-5.6-sol high.
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