3.1. Linear stability

The Rayleigh number up to which the flow is linearly stable is identified by analysing the evolution of perturbations from the linearised system of (2.1). In the non-rotating case, for selected thermal boundary conditions, the marginally stable states are stationary (Davis Reference Davis1969; Herron Reference Herron2003). When the flow is subject to rotation, the condition for steady rolls, as opposed to oscillatory-in-time rolls, at the onset of convection is unknown. For comparison, in the case of RBC, the onset modes are steady provided $Pr \geq 0.68$, such that assuming a sufficiently large $Pr$ fluid removes the question of oscillatory motion in the analysis we carry out here. The precise structure of the motion at the onset should not affect the asymptotic behaviour of $R_L$ for a small Ekman number. It is noteworthy that in the case of the rotating internally heated fluid sphere, the first convective modes are always unsteady (Roberts Reference Roberts1968; Busse Reference Busse1970; Jones, Soward & Mussa Reference Jones, Soward and Mussa2000) and recent evidence from numerical simulations confirms the existence subcritical convection (Guervilly & Cardin Reference Guervilly and Cardin2016; Kaplan et al. Reference Kaplan, Schaeffer, Vidal and Cardin2017).

Taking the set-up as described in § 2, we start by decomposing the temperature field into perturbations from the conductive profile $T_c(z)$,

(3.3)\begin{equation} T(\boldsymbol{x},t) = \vartheta(\boldsymbol{x},t) + T_c(z), \end{equation}

to obtain the temperature perturbation, $\vartheta$ equation from (2.1c) and boundary conditions of

(3.4a)$$\begin{gather} \partial_t \vartheta + \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla} \vartheta = \nabla^2 \vartheta + T_c' w, \end{gather}$$

(3.4b)$$\begin{gather}\vartheta|_{z=\{0,1\}} = 0, \end{gather}$$

where primes denote derivative with respect to $z$. Then, we look at the marginally stable stationary states of the linearised system of (2.1) by considering the $z$ component of the double curl of the momentum equation and the vertical component of the vorticity equation; we thus have

(3.5a)$$\begin{gather} \nabla^4 w = E^{{-}1}\partial_z \zeta - R \nabla_H^2 \vartheta , \end{gather}$$

(3.5b)$$\begin{gather}\nabla^2 \zeta ={-} E^{{-}1} \partial_z w, \end{gather}$$

(3.5c)$$\begin{gather}\nabla^2 \vartheta = T_c' w. \end{gather}$$

Given horizontal periodicity, we take a Fourier series expansion in the horizontal ($x$ and $y$) directions of the form,

(3.6)\begin{equation} \begin{bmatrix} \vartheta(x,y,z)\\ \boldsymbol{u}(x,y,z) \\ \zeta(x,y,z) \end{bmatrix}= \sum_{\boldsymbol{k}} \begin{bmatrix} \vartheta_{\boldsymbol{k}}(z)\\ \boldsymbol{u}_{\boldsymbol{k}}(z) \\ \zeta_{\boldsymbol{k}}(z) \end{bmatrix} \textrm{e}^{{\rm i}(k_x x + k_y y)}, \end{equation}

where the sum is over wavevectors $\boldsymbol {k} = (k_x,k_y)$ with magnitude $k = \sqrt {k_x^2 + k_y^2}$ and variables with subscripts $\boldsymbol {k}$ are functions of $z$ only. Then, substituting (3.6) into (3.5) gives

(3.7a)$$\begin{gather} (D^2 - k^2)^2 w_{\boldsymbol{k}} = E^{{-}1} \zeta'_{\boldsymbol{k}} + R k^2 \vartheta_{\boldsymbol{k}} , \end{gather}$$

(3.7b)$$\begin{gather}(D^2 - k^2) \zeta_{\boldsymbol{k}} ={-} E^{{-}1} w'_{\boldsymbol{k}}, \end{gather}$$

(3.7c)$$\begin{gather}(D^2 - k^2) \vartheta_{\boldsymbol{k}} = T_c' w_{\boldsymbol{k}}, \end{gather}$$

where $D^2 = {\textrm {d}^2}/{\textrm {d}z^2}$. For no-slip or stress-free boundary conditions, $\zeta _{\boldsymbol {k}}$ and $\vartheta _{\boldsymbol {k}}$ can be eliminated from (3.7) to give

(3.8)\begin{equation} (D^2 - k^2)^3 w_{\boldsymbol{k}} + E^{{-}2} w''_{\boldsymbol{k}} = Rk^2 T_c' w_{\boldsymbol{k}}. \end{equation}

Here, (3.8) is an eigenvalue problem and can be solved numerically by fixing $E$ and finding the $R$ at which the first eigenvalue changes sign. The equivalent problem for RBC is well documented (Chandrasekhar Reference Chandrasekhar1961), however unlike RBC, the ODE in (3.8) admits solutions which are hypergeometric functions because $T_c$ in (3.1) is a non-constant function. As such, even in the case of stress-free boundary conditions where $w_{\boldsymbol {k}} = w''_{\boldsymbol {k}} = w''''_{\boldsymbol {k}} = 0$ at both boundaries, (3.8) becomes complicated to solve. Instead, we consider the asymptotic regime of small $E$ where a simplified form of (3.8) gives the desired asymptotic relation between $R_L$ and $E$. Following the argument first presented in Chandrasekhar (Reference Chandrasekhar1961), we posit that for $R$ close to $R_L$, the wavenumber $k$ tends to infinity so that we retain only terms in $E$, $R$ and the highest power in $k$ and (3.8) becomes

(3.9)\begin{equation} E^{{-}2} w''_{\boldsymbol{k}} = k^2(R(\tfrac12 - z ) + k^4) w_{\boldsymbol{k}}. \end{equation}

Since (3.9) is of second order, we require only two boundary conditions. However, in the simplest set of boundary conditions of stress-free boundaries where $w_{\boldsymbol {k}}$ and $w_{\boldsymbol {k}}''$ are zero, the problem is over-determined with four boundary conditions. It suffices to take $w_{\boldsymbol {k}}(0)=w''_{\boldsymbol {k}}(0) = 0$, such that we can make the ansätz that

(3.10)\begin{equation} w_{\boldsymbol{k}} = Ai(n -mz) - \frac{Ai(n)}{Bi(n)} Bi(n - m z), \end{equation}

where $Ai(z)$ and $Bi(z)$ are the Airy functions of the first and second kind. Substituting (3.10) back into (3.9) gives

(3.11)\begin{equation} E^{{-}2}m^2(n-m z) = \tfrac12 k^2 R + k^6 - k^2 R z, \end{equation}

from which we require the choice that $m = (k^2 R E^2)^{1/3}$. Substituting for $m$ and rearranging, we obtain

(3.12)\begin{equation} -\tfrac12 E^{2/3} k^{2/3} R + n R^{2/3} -E^{2/3} k^{ 14/3 } = 0. \end{equation}

Given that (3.12) is a cubic equation in $R^{1/3}$, by application of the cubic formula, we find the real root to be

(3.13) \begin{align} R_L^{1/3} &= \frac{2 n}{3 E^{ 2/3 } k^{2/3}} + \frac{1}{3E^{2/3} k^{2/3}} \left({-}8n^3 + 27 E^2 k^6 + 27 E^2 k^3 \sqrt{k^{6} - \frac{16n^3}{27 E^2}}\, \right)^{1/3} \nonumber\\ &\quad + \frac{4n^2}{3 E^{2/3} k^{2/3} } \left({-}8n^3 + 27 E^2 k^6 + 27 E^2 k^3 \sqrt{k^{6} - \frac{16n^3}{27 E^2}}\, \right)^{{-}1/3}. \end{align}

Then, we want to find the smallest possible $R_L$ in (3.13) by finding the minimising $k$ by solving $\partial _k R_L^{1/3} = 0$ and substituting back into (3.13) to obtain $R_L$ as a function of only $E$. In figure 3 we plot $R_L$ from (3.13) against $k$ for a wide range of $E$ ($10^{-37}$ to $10^{-13}$), highlighting the minimum in $k$ found. The inset in figure 3 demonstrates that the $k_m$ varies as $E^{-1/3}$. Given figure 3, and noting that in the asymptotic limit of small $E$ we assume large wavenumbers $k$, it is natural in (3.13) to take the minimising wavenumber to be

(3.14)\begin{equation} k_m^6 = \frac{16n^3}{27E^2}, \end{equation}

such that the terms in the cube roots are real and positive. Then, substituting $k_m$ back into (3.13) and the minimal $R_L(n,E)$ is achieved with $n=1$, such that (3.13) simplifies to

(3.15)\begin{equation} R_L = \frac{(1+2^{2/3}+2^{4/3})^3}{9\cdot 2^{1/3}} E^{{-}4/3} \sim E^{{-}4/3}. \end{equation}

The asymptotic scaling is equivalent to rotating RBC, unsurprising given the equivalence of the momentum equations, albeit with different prefactors, and highlights the inhibiting effect of rotation on the Rayleigh number for the system to become linearly unstable. In the rest of the paper, we use (3.15) to arrive at heuristic scaling arguments and later contextualise the bounds proven.

Figure 3.

Plot of $R_L$ against $k$ as given by (3.13) for Ekman numbers ranging from $10^{-37}$ (red) to $10^{-13}$ (yellow). The blue dots represent the minimum $R_L$ in $k$, and the inset demonstrates that the minimum wavenumber $k_m$ varies as $E^{-1/3}$ with a compensated plot.

4.1. The auxiliary functional method

Here, we outline the main steps in constructing the variational problem to obtain an upper bound on $\overline {{\langle {wT} \rangle }}$. See previous works for a detailed derivation (Arslan et al. Reference Arslan, Fantuzzi, Craske and Wynn2021b, Reference Arslan, Fantuzzi, Craske and Wynn2023). To prove an upper bound on $\smash {\overline {{\langle {wT} \rangle }}}$, we employ the auxiliary function method (Chernyshenko et al. Reference Chernyshenko, Goulart, Huang and Papachristodoulou2014; Fantuzzi et al. Reference Fantuzzi, Arslan and Wynn2022). The method relies on the observation that the time derivative of any bounded and differentiable functional ${\mathcal {V}}\{T(t)\}$ along solutions of the Boussinesq equations (2.1) averages to zero over infinite time, so that

(4.1)\begin{equation} \overline{{\langle {wT} \rangle}} = \overline{\langle wT \rangle + \frac{\rm d}{ {\rm d} t}{\mathcal{V}}\{T(t)\} }. \end{equation}

Two key simplifications follow. The first is that we can estimate (4.1) by the pointwise-in-time maximum along the solutions of the governing equations, and this value is estimated by the maximum it can take over all velocity and temperature fields that are periodic in $x$ and $y$, satisfying incompressibility (2.1a), the boundary conditions (2.3) and the maximum principle Lemma 2.1.

We restrict our attention to quadratic functionals taking the form

(4.2)\begin{equation} {\mathcal{V}}\{T\}: = \left\langle \frac{\beta}{2} |T|^2 - [\tau(z)+z-1] T \right\rangle, \end{equation}

that are parametrised by a positive constant $\beta > 0$, referred to as the balance parameter and a piecewise-differentiable function $\tau :[0,1] \to \mathbb {R}$ with a square-integrable derivative that we call the background temperature field. Here $\tau (z)$ satisfies

(4.3a,b)\begin{equation} \tau(0)=1, \quad \tau(1)=0. \end{equation}

Introducing a constant, $U$, and rearranging, (4.1) can be written as

(4.4)\begin{equation} \overline{{\langle {wT} \rangle}} \leq U - U + \langle wT \rangle + \frac{\textrm{d}}{\textrm{d}t}{\mathcal{V}}\{T\} \leq U, \end{equation}

where the final inequality holds given that, $U - \langle wT \rangle - ({\textrm {d}}/{\textrm {d}t}){\mathcal {V}}\{T\} \geq 0$, where we can substitute for the Lie derivative of ${\mathcal {V}}\{T\}$ by using (2.1c). However, the minimum principle, Lemma 2.1, is necessary to obtain an $R$-dependent bound on $\overline {{\langle {wT} \rangle }}$ that approaches $\tfrac 12$ from below as $R$ increases. The condition is enforced with a Lagrange multiplier, $\lambda (z)$, so that the problem statement after computations as outlined in previous work (Arslan et al. Reference Arslan, Fantuzzi, Craske and Wynn2021b, Reference Arslan, Fantuzzi, Craske and Wynn2023) becomes

(4.5)\begin{equation} \overline{{\langle {wT} \rangle}} \leq \inf_{U,\beta,\tau(z),\lambda(z)} \{U \,|\, {\mathcal{S}}\{\boldsymbol{u},T\} \geq 0 \enspace \forall(\boldsymbol{u},T) \in {\mathcal{H}}_+ \}, \end{equation}

where

(4.6)\begin{equation} {\mathcal{H}}_+= \{(\boldsymbol{u},T)\,| \,\text{horizontally periodic},\boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{u}=0, (2.3a), (2.4a), T(\boldsymbol{x})\geq 0\ \text{a.e.}\ \boldsymbol{x}\in V \}, \end{equation}

provided $\lambda (z)$ is a non-decreasing function, and

(4.7)\begin{equation} {\mathcal{S}}\{\boldsymbol{u},T\} := \langle \beta|\boldsymbol{\nabla} T|^2 + \tau'(z) wT + (\beta z - \tau'(z) + \lambda(z))\partial_z T + \tau(z)+ U \rangle - \tfrac12. \end{equation}

Ensuring the positivity of the quadratic terms in (4.7) is referred to as the spectral constraint and is defined as

(4.8)\begin{equation} \langle \beta|\boldsymbol{\nabla} T|^2 + \tau'(z) wT \rangle \geq 0, \end{equation}

where $w$ and $T$ are related by (2.4) and subject to the boundary conditions (2.3). As has been established previously (Arslan et al. Reference Arslan, Fantuzzi, Craske and Wynn2023), provided the spectral constraint is satisfied, then the non-negativity of ${\mathcal {S}}\{\boldsymbol {u},T\}$ is ensured when $U$ is given by

(4.9)\begin{equation} \overline{{\langle {wT} \rangle}} \leq U :=\frac12 + \inf_{\substack{ \lambda \in L^{2}(0,1)\\ \lambda\ \mbox{non-decreasing} \\ \langle \lambda \rangle={-}1}} \left\langle \frac{1}{4\beta} \left|\beta\left(z-\frac12\right) - \tau'(z)+\lambda(z) \right|^2 - \tau(z) \right\rangle. \end{equation}

4.2. Preliminaries

To establish a bound on $\overline {{\langle {wT} \rangle }}$, we state the main choices used in the proof that minimise $U(\beta,\tau,\lambda )$ as defined in (4.9). We make the following choice of background temperature field,

(4.10)\begin{equation} \tau(z):=\begin{cases} 1- \dfrac{1-a}{\delta}z, & 0 \leq z \leq \delta, \\ a, & \delta \leq z \leq 1- \varepsilon, \\ \dfrac{a(1-\varepsilon)}{\varepsilon} \left(\dfrac{1-z}{z}\right), & 1-\varepsilon \leq z \leq 1, \end{cases} \end{equation}

and set $\lambda (z)$ to be

(4.11)\begin{equation} \lambda(z):=\begin{cases} -\dfrac{1-a}{\delta}, & 0 \leq z \leq \delta, \\ -\dfrac{a}{1-\delta}, & \delta \leq z \leq 1. \end{cases} \end{equation}

The piecewise functions $\tau (z)$ and $\lambda (z)$ are quantified by the boundary layer widths $\delta \in (0,\tfrac 13)$ and $\varepsilon \in (0,\tfrac 13)$, where $\delta \leq \varepsilon$, and parameter $a>0$ that determines the value of $\tau (z)$ in the bulk. See figure 5 for a sketch of the functions.

Figure 5.

Sketches of the functions $\tau (z)$ in (4.10) and $\lambda (z)$ in (4.11) used to prove (1), where $\delta$ is the boundary layer width at the bottom, $\varepsilon$ the boundary layer width at the top of the domain and $a$ is (4.12).

We further fix

(4.12)\begin{equation} a = \tfrac12 \delta \varepsilon^{1/2}, \end{equation}

and

(4.13)\begin{equation} \beta = \frac{\langle |\tau'(z)-\lambda(z)|^2 \rangle^{1/2}}{\langle |z-\tfrac12|^2 \rangle^{1/2}}. \end{equation}

In the following subsections, we prove bounds for different regimes of the Ekman number. We achieve this by using different estimates on the spectral constraint (4.8). However, the expression for the upper bound on $\overline {{\langle {wT} \rangle }}$ in (4.9) remains the same. Therefore, first, we use our choices of $\tau (z)$ in (4.10) and $\lambda (z)$ in (4.11) to estimate (4.9).

4.3. Estimating the upper bound

Starting with (4.9) an application of the triangle inequality and the choice of $\beta$ in (4.13) gives

(4.14)\begin{equation} U \leq \frac12 + \frac{1}{\sqrt{12}} \langle |\tau'(z) - \lambda(z)|^2 \rangle^{1/2} - \langle \tau \rangle. \end{equation}

Then, evaluating the sign positive integral with $\tau (z)$ from (4.10) and $\lambda (z)$ in (4.11), gives

(4.15)\begin{align} \langle |\tau'(z)-\lambda(z) |^2 \rangle &= \int^{1}_\delta |\tau'(z)-\lambda(z)|^2\, \textrm{d}z\nonumber\\ & = \frac{a^2}{(1-\delta)^2}(1-\varepsilon-\delta) + \frac{a^2}{\varepsilon^2} \int^{1}_{1-\varepsilon} \left(\frac{1-\varepsilon}{z^{2}} - \frac{\varepsilon}{1-\delta}\right)^2 \, \textrm{d}z. \end{align}

We will require an upper and lower bound on (4.15). Starting with a lower bound, given that $\varepsilon \leq \frac 13$ and $\delta \leq \frac 13$, with $z$ in the range $(1-\varepsilon,1)$, we make the suboptimal but simple estimate that

(4.16)\begin{equation} \left(\frac{1-\varepsilon}{z^{2}} - \frac{\varepsilon}{1-\delta}\right)^2 \geq \frac19, \end{equation}

such that we get

(4.17)\begin{equation} \langle |\tau'(z)-\lambda(z)|^2 \rangle \geq \frac{a^2}{\varepsilon^2} \int^{1}_{1-\varepsilon} \left(\frac{1-\varepsilon}{z^{2}} - \frac{\varepsilon}{1-\delta}\right)^2 \textrm{d}z\geq \frac{a^2}{9\varepsilon}. \end{equation}

For an upper bound on (4.15), given that $\varepsilon$ and $\delta$ are positive, bounded above by $\frac 13$ and that $\delta \leq \varepsilon$, we use the estimate $(1-\varepsilon -\delta )(1-\delta )^{-2}\leq \frac 34 \varepsilon ^{-1}$ and $z^{-2}\leq (1-\varepsilon )^{-2}$, to obtain

(4.18)\begin{align} \langle |\tau'(z) - \lambda(z)|^2 \rangle & \leq \frac{3a^2}{4\varepsilon} + \frac{a^2}{\varepsilon^2} \int^{1}_{1-\varepsilon} \left(\frac{1-\varepsilon}{z^{2}} - \frac{\varepsilon}{1-\delta}\right)^2 \textrm{d}z \nonumber\\ &\leq \frac{3a^2}{4\varepsilon} + \frac{a^2}{\varepsilon^2} \int^{1}_{1-\varepsilon} \frac{1}{(1-\varepsilon)^2} + \frac{\varepsilon^2}{(1-\delta)^2} \,\textrm{d}z \nonumber\\ &\leq \frac{3a^2}{4\varepsilon} + \frac{a^2}{\varepsilon} \frac{1+\varepsilon^2}{(1-\varepsilon)^2} \leq \frac{4a^2}{\varepsilon}. \end{align}

Moving on to the integral of $\tau (z)$ in (4.9), we have that

(4.19)\begin{equation} \int^{1}_0 \tau(z)\, \textrm{d}z = \frac12 \delta(1-a) - \frac{a}{\varepsilon}(1-\varepsilon)\ln(1-\varepsilon). \end{equation}

Substituting (4.19) and (4.18) back into (4.14), taking $a$ as given by (4.12) and $\varepsilon,\delta \leq \frac 13$ such that $\delta ^2\varepsilon ^{1/2} \leq \frac {\sqrt {3}}{9} \delta$, gives

(4.20)\begin{equation} U \leq \tfrac12 - n \delta - \tfrac{1}{3} \delta \varepsilon^{{-}1/2} |\ln{(1-\varepsilon)}| , \end{equation}

where $n = {(18-7\sqrt {3})}/{36}$.

4.4. Large Ekman numbers

To obtain bounds for large $E$ in this subsection, we use Lemma 2.2. The estimates in Lemma 2.2 are pointwise estimates of the vertical velocity in Fourier space. Therefore, we exploit the horizontal periodicity of $\boldsymbol {u}$ and $T$ and take a Fourier decomposition of $w$ and $T$ in the spectral constraint (4.8). Taking that

(4.21)\begin{equation} \begin{bmatrix} T(x,y,z)\\ \boldsymbol{u}(x,y,z) \end{bmatrix}= \sum_{\boldsymbol{k}} \begin{bmatrix} T_{\boldsymbol{k}}(z)\\ \boldsymbol{u}_{\boldsymbol{k}}(z) \end{bmatrix} \textrm{e}^{{\rm i}(k_x x + k_y y)}, \end{equation}

where the sum is over non-zero wavevectors $\boldsymbol {k}=(k_x,k_y)$ for the horizontal periods $L_x$ and $L_y$ and magnitude of each wavevector is $k = \sqrt {k_{\smash {x}}^2 + k_{\smash {y}}^2}$. Inserting the Fourier expansions (4.21) into (4.8) gives

(4.22)\begin{equation} \int^{1}_0 \beta |{T}'_{\boldsymbol{k}} |^2 + \beta k^2 |{T}_{\boldsymbol{k}} |^2 + \tau'(z) |{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}| \,\textrm{d}z \geq 0, \end{equation}

where the complex conjugate relations of ${w}_{\boldsymbol {k}} = {w}^{*}_{\boldsymbol {k}}$ holds, and $w_{\boldsymbol {k}}$ and $T_{\boldsymbol {k}}$ are subject to the boundary conditions

(4.23a)$$\begin{gather} {w}_{\boldsymbol{k}}(0) = {w}_{\boldsymbol{k}}(1) = {w}'_{\boldsymbol{k}}(0) = {w}'_{\boldsymbol{k}}(1) =0, \end{gather}$$

(4.23b)$$\begin{gather}{T}_{\boldsymbol{k}}(0) = {T}_{\boldsymbol{k}}(1) = 0. \end{gather}$$

Based on the boundary conditions, we infer the following two estimates. Given (4.23a) applying the fundamental theorem of calculus and Hölders inequality gives

(4.24)\begin{equation} |{w}_{\boldsymbol{k}}| = \int^{z}_{0} \int^{\sigma}_{0} |\partial^2_\eta {w}_{\boldsymbol{k}}(\eta)| \,\textrm{d}\eta \,\textrm{d}\sigma \leq \frac12 z^2 \lVert {w}''_{\boldsymbol{k}} \rVert_{\infty}, \end{equation}

and for ${T}_{\boldsymbol {k}}$, the fundamental theorem of calculus and the Cauchy–Schwarz inequality gives

(4.25)\begin{equation} |{T}_{\boldsymbol{k}}| = \int^{z}_{0} |\partial_\eta {T}_{\boldsymbol{k}}(\eta)|\, \textrm{d}\eta \leq \sqrt{z} \lVert {T}'_{\boldsymbol{k}} \rVert_{2}. \end{equation}

Next, we substitute (4.10) into the sign indefinite term in (4.22) to obtain

(4.26)\begin{equation} \int^{1}_0 \tau'(z) |{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}| \,\textrm{d}z ={-}\frac{1-a}{\delta} \int^{\delta}_0 |{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}| \,\textrm{d}z - \frac{a(1-\varepsilon)}{\varepsilon} \int^{1}_{1-\varepsilon} z^{{-}2} |{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}| \,\textrm{d}z. \end{equation}

As Lemma 2.2 contains two estimates for different $\boldsymbol {k}$, we will split the sign indefinite term in half. Then, given that $a\leq 1$, use of (4.24), the Cauchy–Schwarz inequality, (2.7) and (2.8) from Lemma 2.2 gives that

(4.27)\begin{align} \frac{1-a}{ \delta } \int^{\delta}_0 |{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}} |\, \textrm{d}z &\leq \frac{1}{2\delta} \int^{\delta}_0 z^{5/2} \,\textrm{d}z \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty \nonumber\\ &\leq \frac{ \delta^{5/2}}{14} \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty + \frac{\delta^{5/2}}{14} \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty \nonumber\\ &\leq \frac{c_1}{14} \delta^{5/2} R(1+ \tfrac14 E^{{-}2})^{1/4} \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {T}_{\boldsymbol{k}} \rVert_2 \nonumber\\ & \quad + \frac{ c_2}{14} \delta^{5/2} \lVert {T}'_{\boldsymbol{k}} \rVert_2 (R \sqrt{k} \lVert {T}_{\boldsymbol{k}} \rVert_2 + R E^{{-}1} \lVert {T}_{\boldsymbol{k}} \rVert_2 ). \end{align}

Taking the term of order $\sqrt {k}$ in (4.27), we estimate further by noting that from (4.25) we have a standard Poincaré inequality of

(4.28)\begin{equation} \lVert {T}_{\boldsymbol{k}} \rVert_2 \leq \frac{1}{\sqrt{2}}\lVert {T}'_{\boldsymbol{k}} \rVert_2, \end{equation}

such that the use of Young's inequality twice gives

(4.29)\begin{align} \frac{1}{14} \delta^{5/2} R \sqrt{k} c_2 \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {T}_{\boldsymbol{k}} \rVert_2 &\leq \frac{\beta}{2} k \lVert {T}_{\boldsymbol{k}} \rVert_2 \lVert {T}'_{\boldsymbol{k}} \rVert_2 + \frac{c_2^2}{392 \beta} \delta^5 R^2 \lVert {T}_{\boldsymbol{k}} \rVert_2 \lVert {T}'_{\boldsymbol{k}} \rVert_2 \nonumber\\ &\leq \frac{\beta}{2}k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 + \frac{\beta}{8}\lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{c_2^2 }{392\sqrt{2}\beta} \delta^5 R^2 \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

Then, substituting (4.29) into (4.27), the integral at the lower boundary becomes

(4.30)\begin{align} &\frac{1-a}{\delta}\int^{\delta}_0|{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z \leq \frac{\beta}{2}k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 + \frac{\beta}{8}\lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{ c_2^2 }{392 \sqrt{2}\beta}\delta^5 R^2 \lVert {T}'_{\boldsymbol{k}} \rVert_2^2 \nonumber\\ &\quad +\frac{ c_1}{14\sqrt{2}} \delta^{5/2} R (1+ \tfrac14 E^{{-}2})^{1/4} \lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{ c_2}{14\sqrt{2}} \delta^{5/2} R E^{{-}1} \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

We realise that for a sufficiently small Ekman number, the term of order $E^{-1}$ is larger than $(1+E^{-2})^{1/4}$ such that if we make the estimate

(4.31)\begin{equation} c_1(1+\tfrac14 E^{{-}2})^{1/4} \leq c_2E^{{-}1}, \end{equation}

we get a quadratic form in terms of $E^2$, that places an upper bound on $E$ of

(4.32)\begin{equation} E\leq E_m = \frac{1}{2\sqrt{2}}({-}1 + \sqrt{1 + 64 (c_2^4 / c_1^4) })^{1/2} = 41.4487. \end{equation}

Now, (4.30) becomes

(4.33)\begin{align} \frac{1-a}{\delta} \int^{\delta}_0|{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}} | \,\textrm{d}z &\leq \frac{\beta}{2}k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 + \frac{\beta}{8}\lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{ c_2^2 }{392 \sqrt{2}\beta}\delta^5 R^2 \lVert {T}'_{\boldsymbol{k}} \rVert_2^2\nonumber\\ &\quad +\frac{c_2}{7\sqrt{2}} \delta^{5/2} R E^{{-}1} \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

Returning to the integral at the upper boundary in (4.26), we apply the same procedure, where (4.24) and (4.25) are instead

(4.34a,b)\begin{equation} |{w}_{\boldsymbol{k}}| \leq \tfrac12(1-z)^2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty, \quad |{T}_{\boldsymbol{k}}| \leq \sqrt{1-z} \lVert {T}'_{\boldsymbol{k}} \rVert_2. \end{equation}

Given $\varepsilon \leq \frac 13$ we use that $z^{-2} \leq (1-\varepsilon )^{-2}$ to get

(4.35)\begin{equation} \frac{a(1-\varepsilon)}{\varepsilon}\int^{1}_{1-\varepsilon}z^{{-}2} |{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z \leq \frac{a}{\varepsilon(1-\varepsilon)}\int^{1}_{1-\varepsilon} |{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z \leq \frac{3a}{2\varepsilon}\int^{1}_{1-\varepsilon}|{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z. \end{equation}

By use of Lemma 2.2, along with (4.34), (4.28), Young's inequality and (4.31), we can estimate the integral at the upper boundary to obtain

(4.36)\begin{align} &\frac{a(1-\varepsilon)}{\varepsilon}\int^{1}_{1-\varepsilon}z^{{-}2} |{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z \leq \frac{3a}{2\varepsilon}\int^{1}_{1-\varepsilon}|{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z \leq \frac{\beta}{2}k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 + \frac{\beta}{8}\lVert {T}'_{\boldsymbol{k}} \rVert_2^2 \nonumber\\ &\quad + \frac{9c_2^2}{6272\sqrt{2} \beta} \delta^2 \varepsilon^6 R^2 \lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{3}{28\sqrt{2}} c_2 \delta \varepsilon^3 R E^{{-}1} \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

Substituting (4.33) and (4.36) back into the spectral constraint (4.22) gives

(4.37)\begin{align} \left(\frac{3\beta}{4} - \frac{c_2^2}{392\sqrt{2}} \frac{\delta^5 R^2}{\beta} - \frac{ c_2}{7\sqrt{2}} \delta^{5/2} R E^{{-}1} - \frac{3 c_2}{28\sqrt{2}} \delta \varepsilon^3 R E^{{-}1} -\frac{9 c_2^2}{6272\sqrt{2} } \frac{\delta^2 \varepsilon^6 R^2}{\beta } \right) \lVert T'_{\boldsymbol{k}} \rVert_2^2 \geq 0. \end{align}

The spectral condition is satisfied provided the term in the brackets of (4.37) is non-negative. Note that we have an explicit expression for $\beta$ in (4.13), which in conjunction with the lower bound in (4.17) gives the following lower bound on $\beta$ of

(4.38)\begin{equation} \beta \geq \frac{\sqrt{3} }{3} \delta. \end{equation}

After estimating $\beta$ from below with (4.38) and making the choice

(4.39)\begin{equation} \delta = (\tfrac{9}{16})^{1/3} \varepsilon^2, \end{equation}

the condition for the positivity of (4.37) becomes, after rearranging,

(4.40)\begin{equation} 1 - \frac{c_2^2}{49\sqrt{2}} \delta^3 R^2 - \frac{8 c_2}{7\sqrt{6}} \delta^{3/2} R E^{{-}1} \geq 0. \end{equation}

In (4.40), two possible choices of $\delta =\delta (R,E)$ guarantee the non-negativity of the left-hand side. If the second negative term dominates the first, i.e.

(4.41)\begin{equation} \frac{c_2^2}{49\sqrt{2}} \delta^3 R^2 \leq \frac{8 \, c_2}{7\sqrt{6}} \delta^{3/2} R E^{{-}1}, \end{equation}

then (4.40) becomes

(4.42)\begin{equation} \delta \leq \left(\frac{7\sqrt{6}}{16c_2 }\right)^{2/3} R^{{-}2/3}E^{2/3}. \end{equation}

Taking $\delta$ as large as possible in (4.42) and substituting back into (4.40) implies that $E\leq 8 (\sqrt {2}/3)^{1/2}$. In the opposite scenario, where

(4.43)\begin{equation} \frac{8 c_2}{7\sqrt{6}} \delta^{3/2} R E^{{-}1} \leq \frac{c_2^2}{49\sqrt{2}} \delta^3 R^2, \end{equation}

then (4.40) becomes

(4.44)\begin{equation} \delta \leq \left( \frac{49\sqrt{2}}{2c_2^2} \right)^{1/3} R^{{-}2/3}, \end{equation}

which holds for $E\geq 8(\sqrt {2}/3)^{1/2}$. In summary, the spectral condition holds if the condition in (4.40) is satisfied and (4.40) is guaranteed when we take $\delta$ as large as possible in (4.42) and (4.44). As a result, we have that

(4.45) \begin{equation} \delta =\begin{cases} \left(\dfrac{7\sqrt{6}}{16c_2}\right)^{2/3} R^{{-}2/3}E^{2/3}, & E \leq 8(\sqrt{2}/3)^{1/2}, \\ \left( \dfrac{49\sqrt{2}}{2c_2^2} \right)^{1/3} R^{{-}2/3}, & E \geq 8(\sqrt{2}/3)^{1/2}, \end{cases} \end{equation}

and, by (4.39), that

(4.46)\begin{equation} \varepsilon =\begin{cases} \left(\dfrac{7\sqrt{6}}{12c_2}\right)^{1/3} R^{{-}1/3}E^{1/3}, & E \leq 8(\sqrt{2}/3)^{1/2}, \\ \left( \dfrac{392\sqrt{2}}{9c_2^2} \right)^{1/6} R^{{-}1/3}, & E \geq 8(\sqrt{2}/3)^{1/2}. \end{cases} \end{equation}

Therefore, substituting (4.45) and (4.46) back into (4.20), along with the fact that $\overline {{\langle {wT} \rangle }} \leq U$ and (1.1) to obtain

(4.47) \begin{align} {\mathcal{F}}_B \geq \begin{cases} d_1 R^{{-}2/3}E^{2/3} + d_2 R^{{-}1/2}E^{1/2} | \ln (1- d_3 R^{{-}1/3}E^{1/3})|, & E \leq 8(\sqrt{2}/3)^{1/2}, \\ d_4 R^{{-}2/3} + d_5 R^{{-}1/2} | \ln(1-d_6 R^{{-}1/3})|, & 8(\sqrt{2}/3)^{1/2} \leq E \leq E_m, \end{cases} \end{align}

where the constants $d_1$ to $d_6$ are collated in Appendix A. Finally, in § 4.2 we chose that both boundary layer widths are in $(0,\frac 13)$, therefore given (4.45) and (4.46) the bound obtained in (4.47) holds for all $R \geq 0.4715$.

4.5. Small Ekman numbers

Next, we demonstrate a proof of the bound on ${\mathcal {F}}_B$ valid for small $E$ in (1). Here, we use Lemma 2.3 to demonstrate the non-negativity of the spectral constraint (4.8). The estimates used in this subsection do not require estimates in Fourier space.

Starting with the spectral constraint in (4.8), we start by substituting for $\tau (z)$ from (4.10) into the sign-indefinite term and using the estimate $z^{-2} \leq (1-\varepsilon )^{-2}$ at the upper boundary gives

(4.48)\begin{equation} \langle \tau'(z)wT \rangle \geq{-} \frac{1-a}{\delta} \left\langle\int^{\delta}_0 wT \,\textrm{d}z \right\rangle_h - \frac{a}{\varepsilon}(1-\varepsilon)^{{-}1} \Bigg\langle\int^{1}_{1-\varepsilon} wT \,\textrm{d}z \Bigg\rangle_h . \end{equation}

We first consider the integral in (4.48) near $z=0$ and obtain an estimate on $wT$. Since we require a lower bound on the right-hand side of (4.48), we can rearrange the order of integration of the first term on the right-hand side of (4.48) and estimate the integral from above. Given the boundary conditions (2.3), use of the fundamental theorem of calculus, (2.4b) and integration by parts gives

(4.49)\begin{align} \langle |wT| \rangle_h &= \left\langle \left|\int^{z}_0 \partial_s (wT) \,\textrm{d}s\right| \right\rangle_h = \left\langle \left| \int^{z}_0 T\partial_s w + w \partial_s T \,\textrm{d}s \right| \right\rangle_h = \left\langle \left|\int^{z}_0 - E T \nabla^2 \zeta + w \partial_s T \,\textrm{d}s\right| \right\rangle_h \nonumber\\ &= E \langle |\zeta' T| \rangle_h + \left\langle \left|\int^{z}_0 E\boldsymbol{\nabla} \zeta \boldsymbol{\nabla} T + w \partial_s T \,\textrm{d}s \right| \right\rangle_h . \end{align}

Then, given the boundary condition on the velocity and temperature in (2.3), we have that

(4.50a,b)\begin{equation} |w(\cdot,z)| \leq \frac23 z^{3/2}\Bigg(\int^{1}_0 |w''(\cdot,z)|^2 \,\textrm{d}z\Bigg)^{1/2}, \quad \langle T^2 \rangle_h^{1/2} \leq \sqrt{z} \langle |\boldsymbol{\nabla} T|^2 \rangle^{1/2}, \end{equation}

use of which, along with multiple applications of the Cauchy–Schwarz inequality in (4.49), and that for $f\in L^2(0,1)$ we have $\langle |\,f'|^{2} \rangle _h \leq \langle |\boldsymbol {\nabla } f|^2 \rangle _h \leq \langle |\boldsymbol {\nabla } f|^2 \rangle$, gives

(4.51)\begin{align} \langle |wT| \rangle_h &\leq E \langle |\zeta'|^2 \rangle_h^{1/2} \langle |T|^2 \rangle_h^{1/2} + E \langle |\boldsymbol{\nabla}\zeta|^2 \rangle^{1/2}\langle |\boldsymbol{\nabla} T|^2 \rangle^{1/2} +\frac{z^2}{3} \left\langle\int^{z}_0 |\partial_s T|^2 \,\textrm{d}s\right\rangle^{1/2}_h\langle |w''|^2 \rangle^{1/2} \nonumber\\ &\leq \left( (\sqrt{z} +1 )E\langle |\boldsymbol{\nabla} \zeta|^2 \rangle^{1/2} + \frac13 z^2 \langle |\nabla^2 w|^2 \rangle^{1/2}\right) \langle |\boldsymbol{\nabla} T|^2 \rangle^{1/2} . \end{align}

Next, we use Lemma 2.3, to bound both $\langle |\boldsymbol {\nabla } \zeta |^2 \rangle$ and $\langle |\nabla ^2 w|^2 \rangle$ from above and given that $T$ is horizontally periodic with Dirichlet boundary conditions at $z=0$ and $1$ we have the standard Poincaré inequality $\langle |T|^2 \rangle \leq (1/{\rm \pi} ^2) \langle |\boldsymbol {\nabla } T|^2 \rangle$, which gives

(4.52)\begin{equation} \langle |wT| \rangle_h \leq \left(\frac{ER}{\sqrt{2} {\rm \pi}}(1 + \sqrt{z}) + \frac{1}{3{\rm \pi}}z^2R \right) \langle |\boldsymbol{\nabla} T|^2 \rangle. \end{equation}

Then, substituting back into the integral at the boundary, whereby (4.12) we have $1-a\leq 1$, we get that

(4.53)\begin{equation} \frac{1-a}{\delta} \int^{\delta}_0 \langle |wT| \rangle_h \,\textrm{d}z \leq \left( \frac{E R}{\sqrt{2} {\rm \pi}}\left(1+\frac{2}{3} \delta^{1/2}\right) + \frac{1}{9{\rm \pi}}\delta^2 R \right) \langle |\boldsymbol{\nabla} T|^2 \rangle. \end{equation}

The same estimates hold at the upper boundary with $a$ given by (4.12) and $z$ replaced by $1-z$. Then, since $\varepsilon \leq \frac 13$ we can take $1-\varepsilon \leq 1$ to obtain

(4.54)\begin{align} \frac12\delta \varepsilon^{{-}1/2}(1-\varepsilon)^{{-}1} \int^{1}_{1-\varepsilon} \langle |wT| \rangle_h \,\textrm{d}z \leq \frac34 \delta\varepsilon^{1/2}\left( \frac{ER}{\sqrt{2}{\rm \pi}}\left(1+\frac{2}{3} \varepsilon^{1/2}\right) + \frac{1}{9{\rm \pi}}\varepsilon^2 R\right) \langle |\boldsymbol{\nabla} T|^2 \rangle. \end{align}

Substituting (4.53) and (4.54) back into (4.48) and then into (4.8), gives after use of the lower bound on $\beta$ from (4.38) that

(4.55)\begin{align} \Bigg(\frac{\sqrt{3}\delta}{3} - \frac{ER}{\sqrt{2}{\rm \pi}}\left(1 + \frac{2\delta^{1/2}}{3} \right) - \frac{\delta^2 R}{9{\rm \pi}} - \frac{3\delta\varepsilon^{1/2}ER}{4\sqrt{2}{\rm \pi}} \left( 1 + \frac{2\varepsilon^{1/2}}{3} \right) - \frac{\delta\varepsilon^{5/2}R}{12{\rm \pi}} \Bigg)\langle |\boldsymbol{\nabla} T|^2 \rangle\geq 0. \end{align}

Since $\varepsilon \leq \frac 13$ and $\delta \leq \frac 13$, then, we will make estimates, $1+ 2\delta ^{1/2}/3 \leq \sqrt {2}{\rm \pi} /3$ and $1+ 2\varepsilon ^{1/2}/3 \leq \sqrt {2}{\rm \pi} /3$, such that after rearranging the spectral constraint becomes

(4.56)\begin{equation} \sqrt{3}\delta - E R - \frac{\delta^2 R}{3{\rm \pi}} - \frac34 \delta\varepsilon^{1/2} E R- \frac{\delta\varepsilon^{5/2}R}{4{\rm \pi}} \geq 0. \end{equation}

In (4.56), the first negative term does not contain an explicit $\delta$ dependence, and so we will, at the very least, choose that

(4.57)\begin{equation} E R \leq \frac{\sqrt{3}}{2} \delta. \end{equation}

Using (4.57) and further making the choice

(4.58)\begin{equation} \delta = \frac34 \varepsilon^{5/2}, \end{equation}

gives

(4.59)\begin{equation} \frac{\sqrt{3}}{2} - \frac{2}{3{\rm \pi}}\delta R - \left(\frac34\right)^{4/5} \delta^{1/5} E R \geq 0. \end{equation}

Similar to the proof of a bound for large Ekman numbers in § 4.4, we have a constraint that we consider in two cases. If in (4.59), the second negative term dominates the first such that

(4.60)\begin{equation} \frac{2}{3{\rm \pi}} \delta R \leq \left(\frac34\right)^{4/5} \delta^{1/5} E R, \end{equation}

then

(4.61)\begin{equation} \delta \leq \frac{\sqrt{3}}{36} E^{{-}5} R^{{-}5} , \end{equation}

however, if this $\delta$ is to satisfy the spectral constraint then its implication on (4.57) and (4.59) need to be checked. If we take $\delta$ to be as large as possible in (4.61), then, up to constants, we get from (4.57) and (4.59) that

(4.62a,b)\begin{equation} E \gtrsim R^{{-}4/5} \quad \textrm{and} \quad E \lesssim R^{{-}1}, \end{equation}

which leads to a contradiction, and the initial assumption cannot be true. Assuming, instead, that

(4.63)\begin{equation} \left(\frac34\right)^{4/5}\delta^{1/5} E R \leq \frac{2}{3{\rm \pi}} \delta R, \end{equation}

gives in (4.59) that

(4.64)\begin{equation} \delta \leq \frac{3\sqrt{3}{\rm \pi}}{8} R^{{-}1}. \end{equation}

Taking $\delta$ as large as possible in (4.64), the constraints from (4.57) implies that

(4.65)\begin{equation} E \leq \frac{9{\rm \pi}}{16} R^{{-}2}, \end{equation}

Finally, substituting the largest $\delta$ from (4.64) into (4.58) gives

(4.66)\begin{equation} \varepsilon = \left(\frac{\sqrt{3}{\rm \pi}}{2}\right)^{2/5}R^{{-}2/5}. \end{equation}

Then, taking $\delta$ as large as possible in (4.64), (4.66), substituting into (4.20), along with the fact that $\overline {{\langle {wT} \rangle }} \leq U$ and (1.1) we get that

(4.67a,b)\begin{equation} {\mathcal{F}}_B \geq d_7 R^{{-}1} + d_8 R^{{-}4/5}|\ln{(1- d_9 R^{{-}2/5})} |, \quad E \leq \frac{9{\rm \pi}}{16} R^{{-}2}, \end{equation}

where the constants $d_7$ to $d_9$ are collated in Appendix A. Finally, for completeness, we verify that the choice of $\varepsilon,\delta \in (0,\frac 13)$ made in § 4.2 is not restrictive. Since $\delta$ is smaller than $\varepsilon$ as expressed in (4.58), taking $\delta$ as large as possible in (4.64) the bound obtained in (4.67) holds for all $R \geq 6.1216$.

5.1. The auxiliary functional method

By the auxiliary functional method, we obtain an explicit variational problem that we solve to obtain a bound on $\overline {{\langle {T} \rangle }}$. The following derivation, albeit in the language of the classic background field method approach, appears in previous papers (Whitehead & Doering Reference Whitehead and Doering2011a, Reference Whitehead and Doering2012). Here, we only present an outline within the auxiliary functional framework.

Starting with the quadratic auxiliary functional

(5.1)\begin{equation} {\mathcal{V}}\{T\} = \langle -\tfrac{1}{2}|T|^2 + 2\varphi(z) T \rangle, \end{equation}

where $\varphi (z)$ is the background temperature field subject to the boundary conditions

(5.2)\begin{equation} \varphi(0)=\varphi(1)=0. \end{equation}

Then, provided ${\mathcal {V}}\{T(t)\}$ is bounded along solutions of (2.1), the time derivative of the long-time average of ${\mathcal {V}}\{T\}$ is zero, such that we can write

(5.3)\begin{equation} \overline{{\langle {T} \rangle}}=L - \left(L - \overline{\langle T \rangle + \frac{\textrm{d}}{\textrm{d}t}{\mathcal{V}}\{T\}} \right) \geq L, \end{equation}

where the inequality comes from assuming that, $L-\overline { \langle T \rangle + ({\textrm {d}}/{\textrm {d}t}){\mathcal {V}}\{T\} } \leq L-\langle T \rangle - ({\textrm {d}}/{\textrm {d}t}){\mathcal {V}}\{T\} \leq 0$. Noting that we again bound the terms in the long-time integral by the pointwise in time maximum over all periodic $\boldsymbol {u}$ and $T$, where $\boldsymbol {\nabla } \boldsymbol {\cdot } \boldsymbol {u}=0$, subject to the boundary conditions (2.3). Hence, after substituting for the Lie derivative of ${\mathcal {V}}\{T\}$ by use of (2.1c) and rearranging, we have, after appropriate manipulations, the following variational problem

(5.4)\begin{equation} \overline{{\langle {T} \rangle}} \geq \inf_{L, \varphi(z)} \{L \,\vert\,{\mathcal{S}}\{\boldsymbol{u},T\} \geq 0 \enspace \forall (\boldsymbol{u},T)\in{\mathcal{H}}_+\}, \end{equation}

where

(5.5)\begin{equation} {\mathcal{S}}\{\boldsymbol{u},T\} := \langle |\boldsymbol{\nabla} T|^2 + \varphi'(z)wT - \varphi'(z)\partial_z T + \varphi(z) - L \rangle. \end{equation}

In this case, the spectral constraint is given by

(5.6)\begin{equation} \langle |\boldsymbol{\nabla} T|^2 + 2\varphi'(z)wT \rangle\geq 0. \end{equation}

Then, by optimising the linear terms in (5.5) the explicit expression for $L$ is given

(5.7)\begin{equation} L := 2\langle \varphi(z) \rangle - \langle |\varphi'(z)|^2 \rangle. \end{equation}

5.2. Preliminaries

To establish a lower bound on $\overline {{\langle {T} \rangle }}$, we take the following class of background fields

(5.8) \begin{equation} \varphi(z):=\begin{cases} \tfrac12(1-\delta)z, & 0 \leq z \leq \delta, \\ \tfrac12\delta(1-\delta), & \delta \leq z \leq 1- \delta,\\ \tfrac12(1-\delta)(1-z), & 1-\delta \leq z \leq 1. \end{cases} \end{equation}

The background field is determined entirely by $\delta \in (0,\frac 13)$, the width of the boundary layers at the top and bottom of the domain, sketched in figure 7. In contrast to § 4, no advantage follows from background fields with boundary layers of different widths.

Figure 7.

Sketches of the functions $\varphi (z)$ in (5.8) used to prove (1), where $\delta$ is the boundary layer widths.

The lower bounds on $\overline {{\langle {T} \rangle }}$ for different Ekman numbers rely on different estimates of the velocity in the spectral constraint (5.6) given the diagnostic (2.4). Given the choice of background profile (5.8), the expression for the lower bound $L=L(\delta )$ is unchanged based on the estimates we use to demonstrate the non-negativity of the spectral constraint for different Ekman numbers. Thus, we now estimate $L$ in (5.7) given (5.8). Substituting for $\varphi (z)$ from (5.8), we get

(5.9)\begin{equation} L = \tfrac12 \delta (1-\delta)^2. \end{equation}

Since $\delta \leq \frac 13$, then the estimate $1-\delta \geq \frac 23$, gives

(5.10)\begin{equation} L \geq \tfrac29 \delta . \end{equation}

5.3. Large Ekman numbers

We can now proceed to establish a lower bound on $\overline {{\langle {T} \rangle }}$ by establishing the conditions under which the spectral constraint of (5.4) is satisfied, given the choice of background temperature field (5.8) and by use of Lemma 2.2. The estimates in Lemma 2.2 apply to (2.4) in Fourier space so we exploit horizontal periodicity and substitute (4.21) into (5.6) such that the spectral constraint in Fourier space is

(5.11)\begin{equation} \int^{1}_0 |{T}'_{\boldsymbol{k}}|^2 + k^2|{T}_{\boldsymbol{k}}|^2 + 2\varphi'(z)\,\textrm{Re}\{ {w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}^*\} \,\textrm{d}z \geq 0. \end{equation}

Substituting for $\varphi (z)$ and using the fact that $\delta \leq \frac 13$ gives $-(1-\delta ) \geq -1$, the sign-indefinite of (5.11) is

(5.12)\begin{align} 2\int^{1}_0 \varphi'(z) \,\textrm{Re}\{ {w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}^* \} \,\textrm{d}z &\geq{-} 2\int^{1}_0 |\varphi'(z) {w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}|\, \textrm{d}z \nonumber\\ &\geq{-} \int^{\delta}_0 | {w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}| \,\textrm{d}z - \int^{1}_{1-\delta} | {w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}| \,\textrm{d}z. \end{align}

Due to the symmetry in the boundary conditions and $\varphi (z)$, we only demonstrate estimates on the integral at the lower boundary since that at the top gives an identical contribution. Using the estimates of (4.24) and (4.25) on $w_{\boldsymbol {k}}$ and $T_{\boldsymbol {k}}$, Lemma 2.2 and Young's inequality, we get

(5.13)\begin{align} \int^{\delta}_0 |{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}} | \,\textrm{d}z &\leq \frac{1}{2} \int^{\delta}_0 z^{5/2} \,\textrm{d}z \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty \nonumber\\ &= \frac{1}{14} \delta^{7/2} \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty + \frac{1}{14} \delta^{7/2} \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {w}''_{\boldsymbol{k}} \rVert_\infty \nonumber\\ &\leq \frac{ \delta^{7/2}}{14} c_1 R(1+ \tfrac14 E^{{-}2})^{1/4} \lVert {T}_{\boldsymbol{k}}' \rVert_2 \lVert {T}_{\boldsymbol{k}} \rVert_2 \nonumber\\ & \quad + \frac{\delta^{7/2}}{14} \lVert {T}_{\boldsymbol{k}}' \rVert_2 c_2 (R \sqrt{k} \lVert {T}_{\boldsymbol{k}} \rVert_2 + R E^{{-}1} \lVert {T}_{\boldsymbol{k}} \rVert_2 ). \end{align}

Taking the term of order $\sqrt {k}$, estimating by use of Young's inequality twice and (4.28) gives

(5.14)\begin{align} \frac{1}{14} \delta^{7/2} R \sqrt{k} c_2 \lVert {T}'_{\boldsymbol{k}} \rVert_2 \lVert {T}_{\boldsymbol{k}} \rVert_2 &\leq \frac{\sqrt{2}}{2} k \lVert {T}_{\boldsymbol{k}} \rVert_2 \lVert {T}'_{\boldsymbol{k}} \rVert_2 + \frac{c_2^2}{392\sqrt{2}} \delta^7 R^2 \lVert {T}_{\boldsymbol{k}} \rVert_2 \lVert {T}'_{\boldsymbol{k}} \rVert_2 \nonumber\\ &\leq \frac{1}{2}k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 + \frac{1}{4}\lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{ c_2^2}{784} \delta^7 R^2 \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

Therefore, the integral from 0 to $\delta$ simplifies to

(5.15)\begin{align} \int^{\delta}_0|{w}_{\boldsymbol{k}}{T}_{\boldsymbol{k}} | \,\textrm{d}z & \leq \frac12 k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 + \frac{1}{4}\lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{ c_2^2}{784} \delta^7 R^2 \lVert {T}'_{\boldsymbol{k}} \rVert_2^2\nonumber\\ &\quad +\frac{\sqrt{2}}{28} c_1 \delta^{7/2} R (1+ \tfrac14 E^{{-}2})^{1/4} \lVert {T}'_{\boldsymbol{k}} \rVert_2^2 + \frac{\sqrt{2}}{28} c_2 \delta^{7/2} R E^{{-}1} \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

By equivalent estimates at the upper boundary, we get the same upper bound such that (5.12) becomes

(5.16)\begin{align} &2\int^{1}_0 \varphi'(z) \,\textrm{Re}\{{w}_{\boldsymbol{k}} {T}_{\boldsymbol{k}}^* \}\,\textrm{d}z \geq{-} k^2 \lVert {T}_{\boldsymbol{k}} \rVert_2^2 - \frac{1}{2} \lVert {T}'_{\boldsymbol{k}}\rVert_2^2 - \left(\frac{c_2^2 }{392} \delta^7 R^2 \right.\nonumber\\ &\quad \left. +\frac{\sqrt{2}}{14} c_1 \delta^{7/2} R (1+ \tfrac14 E^{{-}2})^{1/4} + \frac{\sqrt{2}}{14} c_2 \delta^{7/2} R E^{{-}1} \right) \lVert {T}'_{\boldsymbol{k}} \rVert_2^2. \end{align}

Finally, we use the estimate in (4.31), which places an upper limit on $E=E_m$ in (4.32), for which the bounds are valid. Then, substituting (5.16) into (5.11) with (4.31), the spectral constraint becomes

(5.17)\begin{equation} \frac{1}{2} - \frac{c_2^2}{392} \delta^7 R^2 - \frac{ \sqrt{2}}{7} c_2 \delta^{7/2} R E^{{-}1} \geq 0. \end{equation}

In (5.17), two possible choices of $\delta =\delta (R,E)$ guarantee the non-negativity of the left-hand side. If

(5.18)\begin{equation} \frac{c_2^2}{392} \delta^7 R^2 \leq \frac{ \sqrt{2}}{7} c_2 \delta^{7/2} R E^{{-}1}, \end{equation}

then from (5.17) we have that

(5.19)\begin{equation} \delta \leq \left( \frac{7\sqrt{2}}{8c_2}\right)^{2/7} R^{{-}2/7} E^{2/7}. \end{equation}

Taking (5.19) as large as possible, substituting back into (5.17) and rearranging, we find that the spectral constraint holds when $E \leq 8$. In the opposite scenario, where

(5.20)\begin{equation} \frac{ \sqrt{2}}{7} c_2 \delta^{7/2} R E^{{-}1} \leq \frac{c_2^2}{392} \delta^7 R^2, \end{equation}

substituted back into (5.17) gives that

(5.21)\begin{equation} \delta \leq \left( \frac{196}{c_2^2} \right)^{1/7} R^{{-}2/7}, \end{equation}

which holds for $E\geq 8$. In summary, we have that

(5.22) \begin{equation} \delta =\begin{cases} \left( \dfrac{7\sqrt{2}}{8c_2}\right)^{2/7} R^{{-}2/7} E^{2/7}, & E\leq 8,\\ \left( \dfrac{196}{c_2^2} \right)^{1/7} R^{{-}2/7}, & E\geq 8, \end{cases} \end{equation}

which we can finally substitute into (5.10) remembering that $\overline {{\langle {T} \rangle }}\geq L$ to obtain that

(5.23)\begin{equation} \overline{{\langle {T} \rangle}} \geq\begin{cases} d_{10} R^{{-}2/7} E^{2/7}, & E\leq 8,\\ d_{11} R^{{-}2/7}, & 8 \leq E \leq E_m, \end{cases} \end{equation}

where $d_{10}$ and $d_{11}$ are given in Appendix A. A final check is on the validity of the bounds given choices made in § 5.2 that $\delta$ is in $(0,\frac 13)$. Given (5.22) the bound obtained in (5.23) holds for all $R \geq 7.1363$ when $E \leq 8$ and $R\geq 10.0922$ when $E \geq 8$.

5.4. Small Ekman numbers

Next, we move on to the proof of the lower bound on $\overline {{\langle {T} \rangle }}$ in (1) valid for a small $E$. Here, we use Lemma 2.3, which does not require estimates in Fourier space. Starting with the spectral constraint (5.6) and substituting for $\varphi (z)$ from (5.8), the sign-indefinite term becomes

(5.24)\begin{equation} 2\langle \varphi'(z) w T \rangle \geq{-}2 \langle |\varphi'(z) wT| \rangle \geq{-}\left\langle \int^{\delta}_0|wT| \,\textrm{d}z \right\rangle_h - \Bigg\langle \int^{1}_{1-\delta}|wT| \,\textrm{d}z \Bigg\rangle_h. \end{equation}

In § 4.5, we established the estimate (4.52) and can substitute directly into the integral at the lower boundary in (5.24) to obtain

(5.25)\begin{equation} \left\langle \int^{\delta}_0|wT| \,\textrm{d}z \right\rangle_h \leq \left[\frac{E\,R}{\sqrt{2}{\rm \pi}}\left(\delta +\frac{ 2\delta^{3/2}}{3}\right) + \frac{1}{9{\rm \pi}}\delta^3 R \right] \langle |\boldsymbol{\nabla} T|^2 \rangle. \end{equation}

The integral at the upper boundary gives an identical estimate such that the spectral constraint of (5.6) becomes

(5.26)\begin{align} &\langle |\boldsymbol{\nabla} T|^2 + 2\varphi'(z)wT \rangle \geq \langle |\boldsymbol{\nabla} T|^2 - 2|\varphi'(z)wT| \rangle \nonumber\\[4pt] &\quad \geq \langle |\boldsymbol{\nabla} T|^2 \rangle - 2\left[\frac{E\,R}{\sqrt{2}{\rm \pi}}\left(\delta +\frac{ 2\delta^{3/2}}{3}\right) + \frac{1}{9{\rm \pi}}\delta^3 R \right] \langle |\boldsymbol{\nabla} T|^2 \rangle \geq 0. \end{align}

Therefore, the spectral constraint becomes the condition

(5.27)\begin{equation} 1 - \frac{2 ER}{\sqrt{2}{\rm \pi}}\left(\delta +\frac{ 2 \delta^{3/2}} {3 }\right) - \frac{1}{9{\rm \pi}}\delta^3 R \geq 0. \end{equation}

Given that $\delta \leq \frac 13$, we use the estimate that $1+2\delta ^{1/2}/3 \leq \frac {\sqrt {2}{\rm \pi} }{3}$ to obtain the condition

(5.28)\begin{equation} 1 - \frac23 E R \delta - \frac{1}{9{\rm \pi}} \delta^3 R \geq 0. \end{equation}

Once more, the condition admits $\delta =\delta (R,E)$ valid for two cases. If we take

(5.29)\begin{equation} \frac23 E R \delta \leq \frac{1}{9{\rm \pi}} \delta^3 R, \end{equation}

then (5.28) is non-negative if

(5.30)\begin{equation} \delta \leq \left( \frac{9{\rm \pi}}{2} \right)^{1/3} R^{{-}1/3}, \end{equation}

where when taking $\delta$ as large as possible the choice holds when $E \leq \frac {3}{4}({2}/{9{\rm \pi} })^{1/3}R^{-2/3}$. In the case where

(5.31)\begin{equation} \frac{1}{9{\rm \pi}} \delta^3 R \leq \frac23 E R \delta, \end{equation}

then

(5.32)\begin{equation} \delta \leq \tfrac{3}{4} R^{{-}1} E^{{-}1}, \end{equation}

for all $E \geq \frac {3}{4}({2}/{9{\rm \pi} })^{1/3}R^{-2/3}$. Therefore, the spectral constraint is satisfied if

(5.33) \begin{equation} \delta =\begin{cases} \left( \dfrac{9{\rm \pi}}{8} \right)^{1/3} R^{{-}1/3}, & E \leq \dfrac{3}{4}\left(\dfrac{2}{9{\rm \pi}}\right)^{1/3}R^{{-}2/3},\\ \dfrac{3}{4} R^{{-}1} E^{{-}1}, & E \geq \dfrac{3}{4}\left(\dfrac{2}{9{\rm \pi}}\right)^{1/3}R^{{-}2/3}. \end{cases} \end{equation}

Substituting into (5.10) and remembering further that $\overline {{\langle {T} \rangle }}\geq L$, gives

(5.34)\begin{equation} \overline{{\langle {T} \rangle}} \geq \begin{cases} d_{12} R^{{-}1/3}, & E \leq \dfrac{3}{4}\left(\dfrac{2}{9{\rm \pi}}\right)^{1/3}R^{{-}2/3}, \\ d_{13} R^{{-}1}E^{{-}1}, & E \geq \dfrac{3}{4}\left(\dfrac{2}{9{\rm \pi}}\right)^{1/3}R^{{-}2/3}, \end{cases} \end{equation}

where $d_{12}$ and $d_{13}$ are in Appendix A. Finally, for completeness, we verify that the choice of $\delta \in (0,\frac 13)$ made in § 5.2 is not restrictive. Given (5.33) the bound obtained in (5.34) holds for all $R \geq 95.4259$ when $E \leq \frac {3}{4}({2}/{9{\rm \pi} })^{1/3}R^{-2/3}$ and for all $R\geq 1$ otherwise.