= Solution
Use the layer depth $d$, the thermal diffusion time $d^2/\kappa$, velocity $\kappa/d$, and the imposed temperature difference $\Delta T$ as scales. Let $\theta$ be the temperature departure from the conductive profile, $\mathbf u$ the velocity, $w=\mathbf u\cdot\hat{\mathbf z}$, and let pressure absorb hydrostatic terms. The dimensionless <Boussinesq equations> for <Rayleigh-Bénard convection> are
$$
\nabla\cdot\mathbf u=0,\qquad \frac1\sigma(\partial_t\mathbf u+\mathbf u\cdot\nabla\mathbf u)=-\nabla p+R\theta\hat{\mathbf z}+\nabla^2\mathbf u,\qquad \partial_t\theta+\mathbf u\cdot\nabla\theta=w+\nabla^2\theta.
$$
The <Prandtl number> and <Rayleigh number> are $\sigma=\nu/\kappa$ and $R=g\alpha_T\Delta T d^3/(\nu\kappa)$, with <kinematic viscosity> $\nu$, <thermal diffusivity> $\kappa$ and <coefficient of thermal expansion> $\alpha_T$. At $z=0,1$, the <stress-free boundary condition> and <perfectly conducting thermal boundary condition> give $w=0$, $\partial_z u_x=\partial_z u_y=0$, and $\theta=0$.
For the fundamental vertical <normal mode>, take $w=W\sin(\pi z)e^{i\mathbf k\cdot\mathbf x+\lambda t}$ and $\theta=\vartheta\sin(\pi z)e^{i\mathbf k\cdot\mathbf x+\lambda t}$. Here $k=|\mathbf k|>0$ is the horizontal <wavenumber>. With $s=k^2+\pi^2$, projection onto <incompressible flow> eliminates pressure and gives
$$
(\lambda/\sigma+s)sW=Rk^2\vartheta,\qquad (\lambda+s)\vartheta=W.
$$
Thus the <stress-free convection growth-rate polynomial> is $(\lambda+\sigma s)(\lambda+s)=\sigma Rk^2/s$.
For the infinite-<Prandtl number> limit, the finite thermal growth rate is
$$
\boxed{\lambda_\infty(k)=\frac{Rk^2}{(k^2+\pi^2)^2}-(k^2+\pi^2).}
$$
The other, viscously damped root tends to $-\sigma s$. For $\sigma=1$, both roots are explicit:
$$
\boxed{\lambda_\pm(k)=-s\pm\sqrt{\frac{Rk^2}{s}}.}
$$
The plus root determines instability. A higher vertical harmonic replaces $\pi^2$ by $n^2\pi^2$; the fundamental is the most favorable one.
For the large-$R$ maximum in the infinite-<Prandtl number> limit, put $x=k^2$ and $p=\pi^2$. Differentiating gives $d\lambda_\infty/dx=R(p-x)/(p+x)^3-1$. The unique maximum for sufficiently large $R$ therefore satisfies $R(p-x)=(p+x)^3$, whence
$$
\boxed{k_{\rm max}^2=p-\frac{8p^3}{R}+O(R^{-2})\simeq\pi^2,\qquad \lambda_{\rm max}=\frac{R}{4\pi^2}-2\pi^2+O(R^{-1})\simeq\frac{R}{4\pi^2}.}
$$
This is <fastest growth in infinite-Prandtl convection>. The large-$R$ expansion is taken after the infinite-<Prandtl number> limit; at finite $\sigma$, neglecting inertia additionally requires $|\lambda|\ll\sigma s$. It selects the fastest growing mode far above onset, rather than the neutral-curve minimum $k^2=\pi^2/2$.
Back to article page
