= Solution
A direct elimination of <pressure> gives the same result as taking a double <curl> of the momentum equation. <Incompressibility> gives $\nabla^2p=\mathrm{Ra}\,\mathrm{Pr}\,\partial_z\theta$. Apply $\nabla^2$ to the vertical momentum equation and subtract this pressure derivative to obtain
$$
\boxed{\partial_t\nabla^2w=\mathrm{Ra}\,\mathrm{Pr}\,\nabla_h^2\theta+\mathrm{Pr}\nabla^4w,\qquad
\partial_t\theta=w+\nabla^2\theta.}
$$
Here $\nabla_h^2=\partial_x^2+\partial_y^2$ is the horizontal <Laplacian>. At $z=0,1$, impermeability gives $w=0$ and fixed temperature gives $\theta=0$. The <stress-free boundary condition> gives $\partial_zu=\partial_zv=0$, since horizontal derivatives of the boundary value $w=0$ vanish. Differentiating <incompressibility> once in $z$ therefore gives
$$
\boxed{w=\partial_z^2w=\theta=0\quad\text{at }z=0,1.}
$$
For a horizontal <Fourier mode> $X=e^{i(k_xx+k_yy)}$, $\nabla_h^2X=-\lambda^2X$ with real $\lambda=\sqrt{k_x^2+k_y^2}$. More generally the nonnegative spectrum of $-\nabla_h^2$ follows by <integration by parts> for periodic or square-integrable horizontal disturbances. Substituting the separated <normal mode> gives, with $D=d/dz$,
$$
\sigma(D^2-\lambda^2)W=-\mathrm{Ra}\,\mathrm{Pr}\lambda^2\Theta+\mathrm{Pr}(D^2-\lambda^2)^2W,\qquad
(D^2-\lambda^2-\sigma)\Theta=-W.
$$
The <boundary conditions> admit a complete <Fourier sine series> in $z$. For a nonzero horizontal <wavenumber>, take $W=W_n\sin(n\pi z)$, $\Theta=\Theta_n\sin(n\pi z)$, and set $a_n^2=\lambda^2+n^2\pi^2$. The two amplitude equations are
$$
a_n^2(\sigma+\mathrm{Pr}a_n^2)W_n=\mathrm{Ra}\,\mathrm{Pr}\lambda^2\Theta_n,\qquad
(\sigma+a_n^2)\Theta_n=W_n.
$$
Thus \b[the <stress-free convection growth-rate polynomial> is]
$$
\boxed{(\sigma+\mathrm{Pr}a_n^2)(\sigma+a_n^2)=\frac{\mathrm{Ra}\,\mathrm{Pr}\lambda^2}{a_n^2}.}
$$
Its <discriminant> is $(\mathrm{Pr}-1)^2a_n^4+4\mathrm{Ra}\,\mathrm{Pr}\lambda^2/a_n^2$, so both <growth rates> are real for $\mathrm{Ra}>0$. For $\mathrm{Ra}<0$, its two coefficients after the leading term are positive: their sum gives $\sigma_1+\sigma_2=-(1+\mathrm{Pr})a_n^2<0$, and their product is $\mathrm{Pr}(a_n^4-\mathrm{Ra}\lambda^2/a_n^2)>0$. Real roots are both negative; nonreal roots have real part $-(1+\mathrm{Pr})a_n^2/2<0$. \b[Hence stable heating from above damps every mode.] Horizontally uniform thermal modes and uncoupled horizontal viscous modes are also purely damped; impermeability and <incompressibility> exclude a horizontally uniform nonzero $w$.
Because the unstable-side roots are real, the onset is stationary, an <exchange of stabilities in stress-free convection>. Setting $\sigma=0$ gives the <free-slip convection neutral curve>
$$
\mathrm{Ra}_n(\lambda)=\frac{(\lambda^2+n^2\pi^2)^3}{\lambda^2}.
$$
With $x=\lambda^2$, its derivative has the sign of $2x-n^2\pi^2$. Its minimum is at $x=n^2\pi^2/2$, with value $27n^4\pi^4/4$. Minimizing also over $n\geq1$ gives \b[the critical values]
$$
\boxed{\lambda_c=\frac\pi{\sqrt2},\qquad\mathrm{Ra}_c=\frac{27\pi^4}{4}.}
$$
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