= Solution
The four dimensional equations express <mass conservation>, <conservation of momentum>[momentum balance] with <pressure>, <gravity> and <viscous stress>, heat advection and conduction, and a linear thermal-expansion <equation of state>. For heating from below, suitable plate conditions are $\mathbf u=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf u=0$, $T=T_0$ at $z=d$. <No-slip boundary condition>[No slip] and <thermal insulation> on the sidewall provide one suitable lateral completion; prescribed lateral <temperatures> would be another specified thermal setup.
Let $\kappa=k/(\rho_0c_p)$ be <thermal diffusivity> and $\nu=\mu/\rho_0$ <kinematic viscosity>. Use $d$, $d^2/\kappa$, $\kappa/d$ and $\Delta T$ as length, time, <velocity> and <temperature> scales. Subtract the static reference <pressure> $p_H$ with $\nabla p_H=-\rho_0g\mathbf k$ and scale its remainder by $\mu\kappa/d^2$. Then
$$
\boxed{B=\alpha\Delta T,\qquad Pr=\nu/\kappa,\qquad Ra=\frac{g\alpha\Delta T d^3}{\nu\kappa}.}
$$
<Density> becomes $\rho_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales give
$$
-BT_t+\nabla\cdot[(1-BT)\mathbf u]=0,\qquad
Pr^{-1}(1-BT)(\mathbf u_t+\mathbf u\cdot\nabla\mathbf u)=-\nabla p+Ra\,T\mathbf k+\Delta\mathbf u,
$$
$$
(1-BT)(T_t+\mathbf u\cdot\nabla T)=\Delta T.
$$
In the Boussinesq small-<density>-contrast limit retain <buoyancy> while neglecting the other $B$ factors. At infinite <Prandtl number>, after any fast initial adjustment, the equations are
$$
\boxed{\nabla\cdot\mathbf u=0,\qquad0=-\nabla p+Ra\,T\mathbf k+\Delta\mathbf u,\qquad T_t+\mathbf u\cdot\nabla T=\Delta T.}
$$
The <velocity> is instantaneously constrained by a viscous elliptic balance; the <temperature> still evolves.
Take the <divergence> of the <Poisson equation>[vector Poisson equation]. Derivatives commute, and <divergence> of a <curl> is zero, so $\Delta(\nabla\cdot\mathbf A)=0$. Its boundary value is zero, hence the <maximum principle> gives $\nabla\cdot\mathbf A=0$ throughout $D$. The vector identity now gives $\operatorname{curl}\operatorname{curl}\mathbf A=\operatorname{curl}\mathbf u$, or
$$
\operatorname{curl}(\operatorname{curl}\mathbf A-\mathbf u)=0.
$$
The layer domain is <simply connected> because its horizontal section is <simply connected>. For a <curl>-free field on such a domain, define $\chi(x)$ as its <line integral> from a fixed base point to $x$. Contractibility of loops and <Stokes theorem> make the <integral> <path independent>, and differentiation gives its <gradient> equal to the field. Thus $\operatorname{curl}\mathbf A-\mathbf u=\nabla\chi$.
<Incompressibility> then gives $\Delta\chi=0$, while the prescribed normal <curl> gives $\partial_n\chi=0$. <Green's first identity>[Green's identity] yields
$$
\int_D|\nabla\chi|^2\,dV=\int_{\partial D}\chi\partial_n\chi\,dS-\int_D\chi\Delta\chi\,dV=0.
$$
Hence $\chi$ is constant on the connected domain and $\boxed{\operatorname{curl}\mathbf A=\mathbf u}$. The extra condition $\mathbf A\cdot\mathbf n=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the <divergence> and normal-<curl> conditions. On a multiply connected domain a <curl>-free field need not be a globally single-valued <gradient>.
Finally $\operatorname{curl}(T\mathbf k)=(T_y,-T_x,0)$. Taking the <curl> of the infinite-Prandtl <conservation of momentum>[momentum balance] and using $\operatorname{curl}\mathbf u=-\Delta\mathbf A$ gives
$$
0=Ra(T_y\mathbf i-T_x\mathbf j)-\Delta^2\mathbf A,\qquad
\boxed{\nabla^4\mathbf A+Ra(T_x\mathbf j-T_y\mathbf i)=0.}
$$
The derivation is an instance of <velocity reconstruction from a mixed-boundary vector Poisson problem>.
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