2. Problem set-up

We consider the flow of a Newtonian fluid of density $\rho$, viscosity $\nu$, thermal diffusivity $\kappa$, and specific heat capacity $c_p$, driven by buoyancy forces resulting from internal heating. The fluid is confined between two horizontal no-slip plates with a gap of width $d$, and the heat is produced at a constant volumetric rate of $H^\ast$ (with units $\mathrm {W\ m}^{-3}=\mathrm {kg\ ms}^{-3}$). We consider the two configurations sketched in figure 1, one where both plates are kept at constant temperature $T_0^\ast$ (IH1) and one where the top plate is kept at a constant temperature $T_0^\ast$ while the bottom plate is insulating (IH3). We assume that the fluid properties are a weak function of the temperature, and use the Naiver–Stokes equations under the Boussinesq approximation to model the problem. Various justifications have been put forward for the Boussinesq approximation; see, for example, Spiegel & Veronis (Reference Spiegel and Veronis1960) and Rajagopal, Ruzicka & Srinivasa (Reference Rajagopal, Ruzicka and Srinivasa1996). In their non-dimensional form, the governing equations are

(2.1a)$$\begin{gather} \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{u} = 0, \end{gather}$$

(2.1b)$$\begin{gather}\partial_t \boldsymbol{u} + \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \boldsymbol{u}+ \boldsymbol{\nabla} p = Pr\,\nabla^2 \boldsymbol{u} + Pr\,R T \boldsymbol{e}_z, \end{gather}$$

(2.1c)$$\begin{gather}\partial_t T + \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} T = \boldsymbol{\nabla}^2 T + 1, \end{gather}$$

where we have used the following non-dimensionalization for the variables:

(2.2a–e)\begin{equation} \boldsymbol{x} = \frac{\boldsymbol{x}^\ast}{d}, \quad t = \frac{t^\ast}{d^2/ \kappa}, \quad \boldsymbol{u} = \frac{\boldsymbol{u}^\ast}{\kappa/d}, \quad p = \frac{p^\ast{-} p_0}{\rho \kappa^2/d^2}, \quad T = \frac{T^\ast{-} T_0^\ast}{d^2 H^\ast{/} (\kappa \rho c_p)}. \end{equation}

Here, $\boldsymbol {x}$, $t$, $\boldsymbol {u}$, $p$ and $T$ denote the non-dimensional position, time, velocity, pressure and temperature, respectively, whereas $p_0$ is the dimensional hydrostatic ambient pressure. The quantities with a star superscript are dimensional. The non-dimensional governing parameters of the flow are the Prandtl number and the Rayleigh number, given by

(2.3a,b)\begin{equation} Pr = \frac{\nu}{\kappa} \quad\text{and}\quad R = \frac{g \alpha d^5 H^\ast}{\rho c_p \nu \kappa^2}, \end{equation}

where $\alpha$ is the coefficient of thermal expansion. Our choice of non-dimensionalization implies that the heating source appears as a unit force in (2.1c).

We use the Cartesian coordinates $\boldsymbol {x} = (x, y, z)$ and place the origin of the coordinate system at the bottom plate. The $z$ direction points vertically upwards, and the $x$ and $y$ directions are horizontal. In this coordinate system, we write the velocity vector as $\boldsymbol {u} = (u, v, w)$, where $u$, $v$ and $w$ are the velocity components in the $x$, $y$ and $z$ directions, respectively. In this coordinate system, the boundary conditions at the top and bottom plates for velocity and temperature can be written as

(2.4a)$$\begin{gather} \boldsymbol{u}(x, y, 0, t) = \boldsymbol{u}(x, y, 1, t) = \boldsymbol{0}, \end{gather}$$

(2.4b)$$\begin{gather}T(x, y, 0, t) = T(x, y, 1, t) = 0 \quad \text{for IH1,} \end{gather}$$

(2.4c)$$\begin{gather}\partial_z T(x, y, 0, t) = T(x, y, 1, t) = 0 \quad \text{for IH3.} \end{gather}$$

We further assume that the fluid layer is periodic in the horizontal directions $x$ and $y$ with lengths $L_x$ and $L_y$, meaning that the domain of interest is $\varOmega = \mathbb {T}_{[0, L_x]} \times \mathbb {T}_{[0, L_y]} \times [0, 1]$.

Throughout the paper, spatial averages, long-time horizontal averages and long-time volume averages will be denoted, respectively, by

(2.5a)$$\begin{gather} {\int\hskip -1,05em -\,}_{\varOmega} [{\cdot} ]\,{\rm d}\boldsymbol{x} = \frac{1}{L_x L_y} \int_{0}^{1} \int_{0}^{L_y} \int_{0}^{L_x} [{\cdot} ] \, \textrm{d}\kern0.06em x \, \textrm{d} y \, \textrm{d}z, \end{gather}$$

(2.5b)$$\begin{gather}\overline{[{\cdot} ]} = \lim_{\tau \to \infty} \frac{1}{\tau L_x L_y} \int_{0}^\tau \int_{0}^{L_y} \int_{0}^{L_x} [{\cdot} ] \, \textrm{d}\kern0.06em x \, \textrm{d} y \, \textrm{d}t, \end{gather}$$

(2.5c)$$\begin{gather}\langle[{\cdot} ]\rangle = \lim_{\tau \to \infty} \frac{1}{\tau} \int_{0}^\tau {\int\hskip -1,05em -\,}_{\varOmega} [{\cdot} ] \, \textrm{d}\boldsymbol{x} \, \textrm{d}t. \end{gather}$$

3. The auxiliary functional method

A bound on the mean vertical heat flux can be derived using the auxiliary function method. The formulation of the method given here is very similar to the one given by Arslan et al. (Reference Arslan, Fantuzzi, Craske and Wynn2021b) for isothermal boundaries, but we repeat it to make the paper self-contained and highlight the changes required when the lower boundary is insulating.

Let $\mathcal {V}\{\boldsymbol {u}, T\}$ be a functional that is uniformly bounded in time along solutions $\boldsymbol {u}(t)$ and $T(t)$ of the governing equations (2.1a–c). Further, let $\mathcal {L}\{\boldsymbol {u}, T\}$ be the Lie derivative of $\mathcal {V}\{\boldsymbol {u}, T\}$, meaning a functional such that

(3.1)\begin{equation} \mathcal{L}\{\boldsymbol{u}(t), T(t)\} = \frac{{\rm d}}{{\rm d} t} \mathcal{V}\{\boldsymbol{u}(t), T(t)\} ,\end{equation}

when $\boldsymbol {u}(t)$ and $T(t)$ solve the governing equations. Then a simple calculation shows that the long-time average of $\mathcal {L}\{\boldsymbol {u}(t), T(t)\}$ vanishes, and given any constant $B$, we can rewrite the mean vertical heat flux as

(3.2)\begin{align} \langle w T \rangle & = \lim_{\tau \to \infty} \frac{1}{\tau} \int_{0}^{\tau} \left[ {\int\hskip -1,05em -\,}_{\varOmega} w T \,{\rm d}\boldsymbol{x} + \mathcal{L}\{\boldsymbol{u}(t), T(t)\} \right] {\rm d}t, \nonumber\\ & = B + \lim_{\tau \to \infty} \frac{1}{\tau} \int_{0}^{\tau} \left[ {\int\hskip -1,05em -\,}_{\varOmega} w T \,{\rm d}\boldsymbol{x} + \mathcal{L}\{\boldsymbol{u}(t), T(t)\} - B \right] {\rm d}t. \end{align}

If the functional $\mathcal {V}$ can be chosen such that

(3.3)\begin{equation} \mathcal{S}^\ast\{\boldsymbol{u}, T\} := {\int\hskip -1,05em -\,}_{\varOmega} w T \,{\rm d}\boldsymbol{x} + \mathcal{L} \{\boldsymbol{u}, T\} - B \leq 0, \end{equation}

for any solution of the governing equations, then it follows that $\langle w T \rangle \leq B$. Of course, it is intractable to impose (3.3) only over the set of solutions of the governing equation, because they are not known explicitly. However, to obtain a (possibly conservative) bound, it suffices to enforce the stronger condition that (3.3) holds for all pairs of divergence-free velocity fields $\boldsymbol {u}$ and temperature fields $T$ that satisfy the boundary conditions (2.4a–c).

Following Arslan et al. (Reference Arslan, Fantuzzi, Craske and Wynn2021b), we choose the functional $\mathcal {V}$ to be

(3.4)\begin{equation} \mathcal{V} \{\boldsymbol{u}, T\} = {\int\hskip -1,05em -\,}_{\varOmega} \left[\frac{a}{2\,Pr\,R} |\boldsymbol{u}|^2 + \frac{b}{2} |T|^2 - (\psi(z) + z - 1) T \right] {\rm d} \boldsymbol{x}, \end{equation}

where the function $\psi (z)$ and the non-negative scalars $a$ and $b$ are to be optimized to obtain the best possible bound. The profile $[\psi (z)+z-1]/b$ corresponds exactly to the background temperature field. Differentiating this functional in time along solutions of the governing equations, followed by standard integrations by parts using the divergence-free and boundary conditions, yields an expression for $\mathcal {L}\{\boldsymbol {u},T\}$ that can be substituted into (3.3) to obtain

(3.5)\begin{align} \mathcal{S}^\ast\{\boldsymbol{u}, T\} &= {\int\hskip -1,05em -\,}_{\varOmega} \left[ \frac{a}{R}\,|\boldsymbol{\nabla} \boldsymbol{u}|^2 + b\,|\boldsymbol{\nabla} T|^2 - (a - \psi^\prime) w T + (bz - \psi^\prime)\,\frac{\partial T}{\partial z} + \psi \right] {\rm d} \boldsymbol{x} \nonumber\\ &\quad + T(0) - T(1) + \psi(1) \left. \overline{\frac{\partial T}{\partial z}}\right|_{z = 1} - (\psi(0) - 1) \left. \overline{\frac{\partial T}{\partial z}} \right|_{z = 0} + B - \frac{1}{2} \geq 0. \end{align}

This inequality needs to be satisfied for all $\boldsymbol {u}$ and $T$ satisfying (2.1a), (2.4a), and either (2.4b) for IH1 or (2.4c) for IH3.

A crucial improvement to the best upper bound $B$ implied by (3.5) can be achieved by imposing the minimum principle, which says that $T\geq 0$ at all times if it is so initially, and that any negative component decays exponentially quickly (Arslan et al. Reference Arslan, Fantuzzi, Craske and Wynn2021b). We may therefore restrict attention to non-negative temperature fields, thereby relaxing inequality (3.5). As explained by Arslan et al. (Reference Arslan, Fantuzzi, Craske and Wynn2021b), the constraint can be enforced with the help of a non-decreasing Lagrange multiplier function $q(z)$ by adding the term

(3.6)\begin{equation} {\int\hskip -1,05em -\,}_{\varOmega} q^\prime(z)\,T\, {\rm d} \boldsymbol{x}\end{equation}

to the right-hand side of (3.5). Integrating by parts and rearranging leads to the weaker constraint

(3.7)\begin{equation} \mathcal{S}\{\boldsymbol{u}, T\} := \mathcal{S}^\ast\{\boldsymbol{u}, T\} + {\int\hskip -1,05em -\,}_{\varOmega} q(z)\,\frac{\partial T}{\partial z}\, {\rm d}\boldsymbol{x} + q(0)\,T(0) - q(1)\,T(1) \geq 0, \end{equation}

and the best upper bound on $\langle wT \rangle$ implied by this inequality is

(3.8)\begin{align} \langle w T \rangle \leq \inf_{B,\psi(z),q(z),a,b}\ \Big\lbrace B: &\quad q(z) \text{ non-decreasing},\nonumber\\& \quad\mathcal{S}\{\boldsymbol{u}, T\} \geq 0\ \forall\,(\boldsymbol{u},T)\text{ satisfying (2.1a) and (2.4)}\Big\rbrace.\end{align}

Moreover, since no derivatives of the Lagrange multiplier $q(z)$ appear in inequality (3.7), one can perform the optimization over non-decreasing Lagrange multipliers that are not necessarily differentiable everywhere and may even be discontinuous. A rigorous justification of this statement is given by Arslan et al. (Reference Arslan, Fantuzzi, Craske and Wynn2021b).

To prove an explicit rigorous bound on $\langle wT \rangle$, it is convenient to replace inequality (3.7) with a stronger condition that is more amenable to analytical treatment. To achieve this, we introduce the following Fourier series decomposition of the variables in the $x$ and $y$ directions:

(3.9)\begin{equation} \begin{bmatrix} \boldsymbol{u}(\boldsymbol{x}) \\ T(\boldsymbol{x}) \end{bmatrix} = \sum_{\boldsymbol{k} \in K} \begin{bmatrix} \hat{\boldsymbol{u}}_{\boldsymbol{k}}(z) \\ \hat{T}_{\boldsymbol{k}}(z) \end{bmatrix} \, {\rm e}^{{\rm i} k_x x + {\rm i} k_y y},\end{equation}

where

(3.10)\begin{equation} K \equiv \left\{ (k_x, k_y) = \left. \left(\frac{2 m {\rm \pi}}{L_x}, \frac{2 n {\rm \pi}}{L_y}\right) \right| (m, n) \in \mathbb{Z}^2 \right\}. \end{equation}

Since $\boldsymbol {u}$ and $T$ in (3.9) are real-valued, the Fourier expansion coefficients satisfy $\hat {w}_{\boldsymbol {k}}^\ast = \hat {w}_{-\boldsymbol {k}}$ and $\hat {T}_{\boldsymbol {k}}^\ast = \hat {T}_{-\boldsymbol {k}}$ for all $\boldsymbol {k} \in K$, subject to the boundary conditions

(3.11a)$$\begin{gather} \hat{w}_{\boldsymbol{k}}(0) = \hat{w}_{\boldsymbol{k}}'(0) = \hat{w}_{\boldsymbol{k}}(1) = \hat{w}_{\boldsymbol{k}}'(1)= 0, \end{gather}$$

(3.11b)$$\begin{gather}\hat{T}_{\boldsymbol{k}}(0)=\hat{T}_{\boldsymbol{k}}(1) = 0, \quad \textrm{IH1,} \end{gather}$$

(3.11c)$$\begin{gather}\hat{T}_{\boldsymbol{k}}'(0)=\hat{T}_{\boldsymbol{k}}(1) = 0, \quad \textrm{IH3.} \end{gather}$$

Substituting (3.9) in (3.7), using the incompressibility condition on $\boldsymbol {u}$, applying the inequality of arithmetic and geometric means (AM–GM inequality), and dropping positive terms in $\hat {u}_{\boldsymbol {k}}$ and $\hat {v}_{\boldsymbol {k}}$, we can estimate

(3.12)\begin{equation} \mathcal{S}\{\boldsymbol{u}, T\} \geq \mathcal{S}_{\boldsymbol{0}} \{\hat{T}_{\boldsymbol{0}}\} + \sum_{\boldsymbol{k} \neq \boldsymbol{0}} \mathcal{S}_{\boldsymbol{k}} \{\hat{w}_{\boldsymbol{k}}, \hat{T}_{\boldsymbol{k}}\} , \end{equation}

where

(3.13)\begin{align} \mathcal{S}_{\boldsymbol{0}} \{\hat{T}_{\boldsymbol{0}}\} &:= \int_0^1 \bigl[b\, |\hat{T}_{\boldsymbol{0}}^{\prime}|^2 + (b z - \psi^\prime + q) \hat{T}_{\boldsymbol{0}}^\prime + \psi\bigr]\, {\rm d}z + (q(0) + 1)\, \hat{T}_{\boldsymbol{0}}(0) \nonumber\\ &\quad - (q(1) + 1)\,\hat{T}_{\boldsymbol{0}}(1) + \psi(1)\, \hat{T}_{\boldsymbol{0}}^\prime(1) - (\psi(0) - 1)\,\hat{T}^\prime_{\boldsymbol{0}}(0) + B - \tfrac{1}{2} \end{align}

and

(3.14)\begin{align} \mathcal{S}_{\boldsymbol{k}} \{\hat{w}_{\boldsymbol{k}}, \hat{T}_{\boldsymbol{k}}\} &:= \int_{0}^{1} \left[\frac{a}{R}\left(\frac{1}{k^2}\,|\hat{w}_{\boldsymbol{k}}^{\prime \prime}|^2 + 2 |\hat{w}_{\boldsymbol{k}}^{\prime}|^2 + k^2 |\hat{w}_{\boldsymbol{k}}|^2 \right) \right. \nonumber\\ &\quad \left. \vphantom{\frac{a}{R}}+ b |\hat{T}_{\boldsymbol{k}}^{\prime}|^2 + b k^2\,|\hat{T}_{\boldsymbol{k}}|^2 - (a - \psi^\prime) \hat{w}_{\boldsymbol{k}}\hat{T}_{\boldsymbol{k}}^{{\ast}}\right] {\rm d}z. \end{align}

In the last expression, $k = \sqrt {k_x^2 + k_y^2}$.

To establish inequality (3.7), therefore, it suffices to check the non-negativity of the right-hand side of (3.12). As all the different Fourier modes $\hat {w}_{\boldsymbol {k}}$ and $\hat {T}_{\boldsymbol {k}}$ can be chosen independently, this requires $\mathcal {S}_{\boldsymbol {k}} \{\hat {w}_{\boldsymbol {k}}, \hat {T}_{\boldsymbol {k}}\} + \mathcal {S}_{-\boldsymbol {k}} \{\hat {w}_{-\boldsymbol {k}}, \hat {T}_{-\boldsymbol {k}}\} \geq 0$ for all wavevectors $\boldsymbol {k} \in K$, which in turn holds true if and only if $\mathcal {S}_{\boldsymbol {k}} \{\operatorname {Re}\{\hat {w}_{\boldsymbol {k}}\}, \operatorname {Re}\{\hat {T}_{\boldsymbol {k}}\}\} \geq 0$ and $\mathcal {S}_{\boldsymbol {k}} \{\operatorname {Im}\{\hat {w}_{\boldsymbol {k}}\}, \operatorname {Im}\{\hat {T}_{\boldsymbol {k}}\}\} \geq 0$ for all wavevectors $\boldsymbol {k} \in K$. This, combined with the fact that the real and imaginary parts of $\hat {w}_{\boldsymbol {k}}$ and $\hat {T}_{\boldsymbol {k}}$ can be chosen independently, implies that we may take $\hat {w}_{\boldsymbol {k}}$ and $\hat {T}_{\boldsymbol {k}}$ to be real-valued without loss of generality, and impose

(3.15a)$$\begin{gather} \mathcal{S}_{\boldsymbol{0}} \{\hat{T}_{\boldsymbol{0}}\} \geq 0, \end{gather}$$

(3.15b)$$\begin{gather}\mathcal{S}_{\boldsymbol{k}}\{\hat{w}_{\boldsymbol{k}},\hat{T}_{\boldsymbol{k}}\} \geq 0 \quad \forall \, \boldsymbol{k} \in K, \ \boldsymbol{k}\neq \boldsymbol{0}. \end{gather}$$

From the non-negativity condition on $\mathcal {S}_{\boldsymbol {0}} \{\hat {T}_{\boldsymbol {0}}\}$, it is possible to extract the bound $B$ explicitly. First, the non-negativity of $\mathcal {S}_{\boldsymbol {0}} \{\hat {T}_{\boldsymbol {0}}\}$ requires

(3.16a)$$\begin{gather} \psi(0) = 1, \psi(1) = 0\quad \text{for IH1}, \end{gather}$$

(3.16b)$$\begin{gather}q(0) ={-}1, \psi(1) = 0\quad \text{for IH3}, \end{gather}$$

otherwise it is possible to choose a profile $\hat {T}_{\boldsymbol {0}}(z)$ that is non-zero only near the boundaries and for which $\mathcal {S}_{\boldsymbol {0}} \{\hat {T}_{\boldsymbol {0}}\} \leq 0$. With these simplifications, one can write

(3.17)\begin{align} \mathcal{S}_{\boldsymbol{0}} \{\hat{T}_{\boldsymbol{0}}\} &= \int_0^1 \left[ \sqrt{b} \hat{T}_{\boldsymbol{0}}^{\prime} + \frac{(b z - \psi^\prime + q)}{2 \sqrt{b}}\right]^2 {\rm d}z + B - \frac{1}{4 b} \int_0^1 (b z - \psi^\prime + q)^2 \, {\rm d}z \nonumber\\ &\quad + \int_0^1 \psi(z)\, {\rm d}z -\frac{1}{2}. \end{align}

Therefore, $\mathcal {S}_{\boldsymbol {0}} \{\hat {T}_{\boldsymbol {0}}\}$ is non-negative if we choose $B$ to cancel the negative and sign-indefinite terms. After gathering (3.8), (3.9), (3.11), (3.15b) and (3.16), we conclude that

(3.18)\begin{equation} \langle w T \rangle \leq \inf_{a, b, \psi(z), q(z)} \left\{ \frac{1}{2} + \frac{1}{4 b} \int_0^1 (b z - \psi^\prime + q)^2 \, {\rm d}z - \int_0^1 \psi(z)\, {\rm d}z \right\}, \end{equation}

provided that

(3.19a)$$\begin{gather} q(z) \text{ is a non-decreasing function}, \end{gather}$$

(3.19b)$$\begin{gather}\psi(0) = 1, \psi(1) = 0\quad \text{for IH1}, \end{gather}$$

(3.19c)$$\begin{gather}q(0) ={-}1, \psi(1) = 0\quad \text{for IH3}, \end{gather}$$

(3.19d)$$\begin{gather}\mathcal{S}_{\boldsymbol{k}} \{\hat{w}_{\boldsymbol{k}}, \hat{T}_{\boldsymbol{k}}\} \geq 0 \quad \forall\, \hat{w}_{\boldsymbol{k}}, \hat{T}_{\boldsymbol{k}}:{3.11 a-c)}, \ \forall\, \boldsymbol{k}\neq \boldsymbol{0}. \end{gather}$$

Explicit constructions for which the right-hand side of (3.18) is strictly less than $1/2$ at all Rayleigh numbers are given in §§ 4 and 5 for the IH1 and IH3 configurations, respectively. First, however, we summarize our proof strategy to explain the intuition behind our constructions. From (3.18), we see that the competition between the second term (which is always positive) and the third term will decide if $\langle w T \rangle$ can be less than $1/2$ as long as we are able to enforce that $\mathcal {S}_{\boldsymbol {k}} \{\hat {w}_{\boldsymbol {k}}, \hat {T}_{\boldsymbol {k}}\} \geq 0$. For previous studies using the background method, the standard approach has been to choose a profile $\psi (z)$ that is linear in boundary layers near the walls, whereas in the bulk region, $\psi (z)$ is chosen such that the sign-indefinite term in $\mathcal {S}_{\boldsymbol {k}}$ is zero. Unfortunately, in the present case, for a profile of $\psi (z)$ that is linear in the boundary layers, we are unable to show that the magnitude of the second term in (3.18) is smaller than the third term unless we violate the constraint (3.15b). However, if we use a $z^{-1}$ profile in $\psi (z)$ in the outer layer of a two-layer lower boundary layer, then we gain an extra factor of a logarithm in the integral of $\psi$. Such a boundary layer structure, along with the choice $q(z) = \psi '(z)$ in the bottom boundary layer to cancel the otherwise large contribution of this layer to the quadratic term in (3.18), matches the numerically optimal profiles computed by Arslan et al. (Reference Arslan, Fantuzzi, Craske and Wynn2021b, figure 7). This makes it possible to show that sum of the last two terms in (3.18) is negative without violating $\mathcal {S}_{\boldsymbol {k}} \{\hat {w}_{\boldsymbol {k}}, \hat {T}_{\boldsymbol {k}}\} \geq 0$. To establish this non-trivial result, we rely on the following Hardy and Rellich inequalities, proofs of which are provided for completeness in Appendix A.

Lemma 3.1 (Hardy inequality)

Let $f: [0, \infty ) \to \mathbb {R}$ be a function such that $f, f' \in L^2(0, \infty )$ and such that $f(0) = 0$. Then for any $\epsilon > 0$ and any $\alpha \geq 0$,

(3.20)\begin{equation} \int_{0}^{\alpha} \frac{|f|^2}{(z + \epsilon)^2}\, {\rm d}z \leq 4 \int_{0}^{\alpha} |f'|^2 \, {\rm d} z. \end{equation}

Lemma 3.2 (Rellich inequality)

Let $f: [0, \infty ) \to \mathbb {R}$ be a function such that $f, f', f'' \in L^2(0, \infty )$ and such that $f(0) = f'(0) = 0$. Then for any $\epsilon > 0$ and any $\alpha \geq 0$,

(3.21)\begin{equation} \int_{0}^{\alpha} \frac{|f|^2}{(z + \epsilon)^4}\, {\rm d}z \leq \frac{16}{9} \int_{0}^{\alpha} |f''|^2 \, {\rm d} z. \end{equation}

We now present detailed proofs of the main results. Our emphasis is on the steps necessary to obtain an $R$-dependent bound on $\langle w T \rangle$, and we do not attempt to optimize the constants appearing in our estimates.

4. Bound on heat flux in the IH1 configuration

To prove the bound in (1.5a), we start by setting

(4.1a,b) \begin{align} \psi(z) = \begin{cases} 1 - \dfrac{z}{4 \sigma \delta}, & 0 \leq z \leq 2 \sigma \delta, \\ \dfrac{\sigma \delta}{z}, & 2 \sigma \delta \leq z \leq \delta, \\ \sigma + a (z - \delta), & \delta \leq z \leq 1 - \gamma, \\ (1-z)\,\dfrac{\sigma + a(1-\gamma-\delta)}{\gamma}, & 1 - \gamma \leq z \leq 1, \end{cases} \quad q(z) = \begin{cases} - \dfrac{1}{4 \sigma \delta}, & 0 \leq z \leq 2 \sigma \delta, \\[6pt] - \dfrac{\sigma \delta}{z^2}, & 2 \sigma \delta \leq z \leq \delta, \\ 0, & \delta \leq z \leq 1. \end{cases}\end{align}

These functions are sketched in figure 2. In the definition of $\psi$, the parameter $\delta$ denotes the thickness of the boundary layer near the bottom plate. The parameter $\sigma$ is the value of $\psi$ taken at the edge of the lower boundary layer ($z = \delta$). The lower boundary layer itself is divided into two parts, an inner sublayer where $\psi$ is linear, and an outer sublayer where $\psi \sim z^{-1}$. These sublayers meet at an intermediate point ($z=2\sigma \delta$) where both the value and slope of $\psi$ are equal. The inverse-$z$ scaling of $\psi$ in the outer part of the lower boundary layer is one of the key ingredients in proving (1.5a). The linear inner sublayer, instead, is used to satisfy the boundary condition $\psi (0) = 1$ from (3.19b). In the bulk of the layer ($\delta \leq z \leq 1-\gamma$), we have $\psi ^\prime = a$, so the indefinite sign term in (3.14) is zero. Thus we need only to control the indefinite-sign term in the boundary layers. The parameter $\gamma$ is the thickness of the boundary layer near the upper boundary in which the profile of $\psi$ is linear.

Figure 2.

Sketches of the functions (a) $\psi (z)$ and (b) $q(z)$, from (4.1a,b), used to obtain a bound on the heat flux $\langle w T \rangle$ in the IH1 configuration.

The sole purpose behind the choice of the function $q(z)$ is to ensure $\psi ^\prime - q = 0$ in the lower boundary layer, thereby making the positive contribution from the second term in the bound (3.18) small in this layer. All parameters are taken to satisfy

(4.2)\begin{equation} a, b, \sigma, \delta, \gamma \leq 1,\end{equation}

and this assumption will be implicit in the proof below.

The goal now is to adjust the free parameters $a$, $b$, $\sigma$, $\delta$ and $\gamma$ such that the spectral constraint (3.19d) is satisfied and, at the same time, the bound (3.18) is as small as possible. We begin by estimating from above the second term in the bound (3.18):

(4.3)\begin{align} \frac{1}{4 b} \int_0^1 (b z - \psi^\prime + q)^2 \, \textrm{d}z & \leq \frac{1}{2 b} \int_{0}^{1} b^2 z^2 \, \textrm{d}z + \frac{1}{2 b}\|\psi^\prime(z) - q(z) \|_2^2 \nonumber\\ & = \frac{b}{6} + \frac{1}{2 b}\int_{\delta}^{1} |\psi^\prime(z) - q(z)|^2 \, \textrm{d}z \nonumber\\ & \leq \frac{b}{6} + \frac{1}{b}\int_{\delta}^{1} |\psi^\prime(z)|^2 \, \textrm{d}z + \frac{1}{b}\int_{\delta}^{1} |q(z)|^2 \, \textrm{d}z \nonumber\\ & \leq \frac{b}{6} + \frac{(\sigma + a)^2}{b \gamma} + \frac{a^2}{b}\nonumber\\ &\leq \frac{b}{6} + \frac{2(\sigma + a)^2}{b \gamma}. \end{align}

Next, we estimate from below the last term in the bound (3.18):

(4.4)\begin{align} \int_{0}^{1} \psi\,{\rm d}z & = \frac{3 \sigma \delta}{2} - \sigma \delta \log(2 \sigma) + \frac{(2 \sigma + a (1 - \gamma- \delta))}{2} + \frac{(\sigma + a (1 - \gamma - \delta)) \gamma}{2} \nonumber\\ & \geq{-} \sigma \delta \log(\sigma). \end{align}

Combining (4.3) and (4.4) with (3.18), we obtain

(4.5)\begin{equation} \langle w T \rangle \leq \frac{1}{2} + \frac{b}{6} + \frac{2 (\sigma + a)^2}{b \gamma} + \sigma \delta \log(\sigma). \end{equation}

Assuming that

(4.6a,b)\begin{equation} \frac{b}{6} \leq{-} \frac{1}{4} \sigma \delta \log(\sigma), \quad \frac{2 (\sigma + a)^2}{b \gamma} \leq{-} \frac{1}{4} \sigma \delta \log(\sigma), \end{equation}

which will be the case for the choices of $a$, $b$, $\sigma$, $\delta$, $\gamma$ made below, the right-hand side of (4.5) can be further estimated from above to obtain

(4.7)\begin{equation} \langle w T \rangle \leq \tfrac{1}{2} + \tfrac{1}{2}\sigma \delta \log(\sigma). \end{equation}

We now shift our focus to the constraint (3.19d). Dropping the positive terms proportional to $\vert \hat {w}_{\boldsymbol {k}} \vert ^2$, $\vert \hat {w}_{\boldsymbol {k}}'' \vert ^2$ and $\vert \hat {T}_{\boldsymbol {k}} \vert ^2$, it is enough to verify that

(4.8)\begin{equation} \tilde{\mathcal{S}}(\hat{w}, \hat{T}) := \int_{0}^{1} \left[\frac{2 a}{R}\,|\hat{w}^{\prime}|^2 + b\,|\hat{T}^{\prime}|^2 - (a - \psi^\prime) \hat{w}\hat{T}\right] {\rm d}z \geq 0. \end{equation}

Here, $\hat {w}$ and $\hat {T}$ satisfy the boundary conditions

(4.9a)$$\begin{gather} \hat{w}(0) = \hat{w}^\prime(0) = \hat{T}(0) = 0, \end{gather}$$

(4.9b)$$\begin{gather}\hat{w}(1) = \hat{w}^\prime(1) = \hat{T}(1) = 0, \end{gather}$$

where $\hat {w}^\prime (0) = \hat {w}^\prime (1) =0$ is a result of the no-slip boundary condition and the incompressibility of the flow field. For brevity, we have dropped the $\boldsymbol {k}$ subscript. The positive terms that we have dropped could be retained, at the expense of a more complicated algebra, in order to improve various prefactors in the eventual bounds. Since this is not our primary goal and the functional form of the bound that one obtains does not change, we work with the stronger constraint (4.8) to ease the presentation.

Substituting the expression for $\psi$ from (4.1a,b) into (4.8) gives

(4.10)\begin{align} \tilde{S}(\hat{w}, \hat{T}) &= \int_{0}^{2 \sigma \delta} \left[\frac{2 a}{R}\,|\hat{w}^\prime|^2 + b\,|\hat{T}^\prime|^2 - \left(a + \frac{1}{4 \sigma \delta}\right) \hat{w} \hat{T}\right] {\rm d} z \nonumber\\ &\quad + \int_{2 \sigma \delta}^{\delta} \left[ \frac{2 a}{R}\,|\hat{w}^\prime|^2 + b\,|\hat{T}^\prime|^2 - \left(a + \frac{\sigma \delta}{z^2}\right) \hat{w} \hat{T}\right] {\rm d} z \nonumber\\ &\quad + \int_{1-\gamma}^{1} \left[ \frac{2 a}{R}\,|\hat{w}^\prime|^2 + b\,|\hat{T}^\prime|^2 - \left(\frac{\sigma + a (1 - \delta)}{\gamma}\right) \hat{w} \hat{T}\right] {\rm d} z. \end{align}

Since $\tilde {S}(\hat {w}, \hat {T}) \geq \tilde {S}(|\hat {w}|, |\hat {T}|)$, with equality when $w$ and $T$ are non-negative, we will assume without loss of generality that $\hat {w},\hat {T} \geq 0$. We further observe that if

(4.11)\begin{equation} 8 a \delta \leq \sigma,\end{equation}

then

(4.12) \begin{equation} \left.\begin{gathered} \dfrac{9}{2}\,\dfrac{\sigma \delta}{(z + \sigma \delta)^2} \geq a + \dfrac{1}{4 \sigma \delta} \quad \text{when } 0 \leq z \leq 2 \sigma \delta, \\[6pt] \dfrac{9}{2}\,\dfrac{\sigma \delta}{(z + \sigma \delta)^2} \geq a + \dfrac{\sigma \delta}{z^2} \quad \text{when } 2 \sigma \delta \leq z \leq \delta. \end{gathered}\right\} \end{equation}

Assuming that $8 a \delta \leq \sigma$, therefore, we can combine the first two terms in (4.10) to conclude that

(4.13)\begin{equation} \tilde{S}(\hat{w}, \hat{T}) \geq \tilde{S}_B(\hat{w}, \hat{T}) + \tilde{S}_T(\hat{w}, \hat{T}), \end{equation}

where

(4.14a)$$\begin{gather} \tilde{S}_B(\hat{w}, \hat{T}) = \int_{0}^{\delta} \left[\frac{2 a}{R}\,|\hat{w}^\prime|^2 + b\,|\hat{T}^\prime|^2 - \frac{9}{2}\,\frac{\sigma \delta}{(z + \sigma \delta)^2}\,\hat{w} \hat{T}\right] \textrm{d} z, \end{gather}$$

(4.14b)$$\begin{gather}\tilde{S}_T(\hat{w}, \hat{T}) = \int_{1-\gamma}^{1} \left[ \frac{2 a}{R}\,|\hat{w}^\prime|^2 + b\,|\hat{T}^\prime|^2 - \frac{(\sigma + a)}{\gamma}\,\hat{w} \hat{T}\right] \textrm{d} z. \end{gather}$$

Next, we derive conditions that ensure that $\tilde {S}_B(\hat {w}, \hat {T})$ and $\tilde {S}_T(\hat {w}, \hat {T})$ are individually non-negative, thereby implying the non-negativity of $\tilde {S}(\hat {w}, \hat {T})$.

First, we deal with $\tilde {S}_T(\hat {w}, \hat {T})$. Using the boundary conditions (4.9b), along with the fundamental theorem of calculus and the Cauchy–Schwarz inequality, leads to

(4.15a,b)\begin{equation} |\hat{w}|^2 \leq (1-z) \int_{1-\gamma}^{1}|\hat{w}'|^2 \, \textrm{d} z, \quad |\hat{T}|^2 \leq (1-z) \int_{1-\gamma}^{1}|\hat{T}'|^2 \, \textrm{d} z. \end{equation}

Using (4.15a,b) in the expression (4.14b) of $\tilde {S}_T$, along with the AM–GM inequality, implies that $\tilde {S}_T \geq 0$ if

(4.16)\begin{equation} \gamma (\sigma + a) \leq 4 \sqrt{\frac{2a b}{R}}.\end{equation}

A condition for the non-negativity of $\tilde {S}_B(\hat {w}, \hat {T})$, instead, can be derived using the Hardy inequality given in Lemma 3.1. First, using the AM–GM inequality, we write

(4.17)\begin{align} \tilde{S}_B(\hat{w}, \hat{T}) \geq \int_{0}^{\delta} \left[\frac{2 a}{R}\,|\hat{w}^\prime|^2 + b\,|\hat{T}^\prime|^2 - \frac{9}{4}\,\frac{\sigma \delta \beta}{(z + \sigma \delta)^2} |\,\hat{w}|^2 - \frac{9}{4}\,\frac{\sigma \delta}{(z + \sigma \delta)^2 \beta} |\,\hat{T}|^2 \right] \textrm{d} z, \end{align}

for some constant $\beta > 0$ to be specified later. Then we can apply Lemma 3.1 to estimate

(4.17a,b)\begin{equation} \int_{0}^{\delta} \frac{|\hat{w}|^2}{(z + \sigma \delta)^2}\, {\rm d}z \leq 4 \int_{0}^{\delta} |\hat{w}'|^2 \, {\rm d} z, \quad \int_{0}^{\delta} \frac{|\hat{T}|^2}{(z + \sigma \delta)^2}\, {\rm d}z \leq 4 \int_{0}^{\delta} |\hat{T}'|^2 \, {\rm d} z. \end{equation}

Using (4.17a,b) and (4.17), and choosing

(4.18)\begin{equation} \beta = \sqrt{\frac{2 a}{b R}}, \end{equation}

we conclude that $\tilde {S}_B(\hat {w}, \hat {T})$ is non-negative if

(4.19)\begin{equation} \sigma \delta \leq \frac{1}{9} \sqrt{\frac{2ab}{R}}.\end{equation}

Given (4.16) and (4.19), and the functional forms of (4.6a,b) with respect to the variables, one can show that the bound (4.7) is optimized when $a$ is proportional to $\sigma$ and $\delta$ is proportional to $\gamma$. For simplicity, therefore, we take $a = \sigma$ and $\delta = \gamma$; we expect that different choices affect only the value of various prefactors appearing in the final bound, but not its functional form or the powers of $R$. With these additional simplifications, the constraints (4.16), (4.19) and (4.6a,b) are satisfied if we take

(4.20a)$$\begin{gather} a = \sigma = \exp ({-}2^{{8}/{5}} 3^{{8}/{5}} R^{{3}/{5}}), \end{gather}$$

(4.20b)$$\begin{gather}b = 2^{{7}/{5}} 3^{{6}/{5}} R^{{1}/{5}} \exp ({-}2^{{8}/{5}} 3^{{8}/{5}} R^{{3}/{5}}), \end{gather}$$

(4.20c)$$\begin{gather}\delta = \gamma = 2^{{6}/{5}}3^{-{7}/{5}} R^{-{2}/{5}}. \end{gather}$$

These choices satisfy the inequalities (4.2) and (4.11) assumed in our derivation provided that $R \geq 2^{{21}/{2}} 3^{-{7}/{2}} \approx 30.97$. We therefore conclude from (4.7) that

(4.21)\begin{equation} \langle w T \rangle \leq \tfrac{1}{2} - 2^{{7}/{5}} 3^{{1}/{5}} R^{{1}/{5}} \exp ({-}2^{{8}/{5}} 3^{{8}/{5}} R^{{3}/{5}}) \quad \forall \, R \geq 2^{{21}/{2}} 3^{-{7}/{2}}. \end{equation}

We end this section with two remarks. First, the scaling of the upper boundary layer thickness given by (4.20c) is stronger (i.e. the boundary layer is thinner) than the scalings $\gamma \sim R^{-1/4}$ and $\gamma \sim R^{-1/3}$ implied by classical (Malkus Reference Malkus1954; Priestley Reference Priestley1954) and ultimate (Spiegel Reference Spiegel1963) scaling arguments for Rayleigh–Bénard convection, respectively (for further details, see § 3 in Arslan et al. Reference Arslan, Fantuzzi, Craske and Wynn2021b). Second, if instead of using the Hardy inequality in (4.14) we had used the Cauchy–Schwarz and AM–GM inequalities, as we did in the upper boundary layer, then we would have obtained the condition

(4.22)\begin{equation} - \frac{9}{2} \sigma \delta \left(\frac{1}{1 + \sigma} + \log \left(\frac{\sigma}{1 + \sigma}\right)\right) \leq \frac{1}{2} \sqrt{\frac{2 a b}{R}}, \end{equation}

and therefore $\sigma \delta \log \sigma \lesssim \sqrt {a b /R}$. This is worse than condition (4.19) by a factor of $\log \sigma ^{-1}$, and as a result, no bound on $\langle w T \rangle$ strictly smaller than $1/2$ can be obtained beyond a certain Rayleigh number.

5. Bound on heat flux in IH3

We now prove the bound (1.5b) for the IH3 configuration. Similar to the previous section, the key ingredients of the proof are (i) a profile of $\psi$ proportional to $1/z$ near the bottom boundary, and (ii) the use of a non-standard Rellich inequality.

We start by choosing the functions $\psi (z)$ and $q(z)$:

(5.1a,b) \begin{align} \psi(z) = \begin{cases} 2 \sqrt{\sigma \delta} - z, & 0 \leq z \leq \sqrt{\sigma \delta}, \\ \dfrac{\sigma \delta}{z}, & \sqrt{\sigma \delta} \leq z \leq \delta, \\ \sigma + a (z - \delta), & \delta \leq z \leq 1 - \gamma, \\ (1-z)\,\dfrac{\sigma + a(1-\gamma-\delta)}{\gamma}, & 1 - \gamma \leq z \leq 1, \end{cases} \quad q(z) = \begin{cases} - 1, & 0 \leq z \leq \sqrt{\sigma \delta}, \\ - \dfrac{\sigma \delta}{z^2}, & \sqrt{\sigma \delta} \leq z \leq \delta, \\ 0, & \delta \leq z \leq 1. \end{cases}\end{align}

These choices are sketched in figure 3, and the parameters $\sigma$, $\delta$ and $\gamma$ have the same purpose as in the previous section. The difference between these profiles and those used for the IH1 configuration in § 4 is in the bottom boundary layer ($0 \leq z \leq \delta$). Here, we require $q(0) = -1$ and at the same time want $q - \psi ^\prime = 0$ in the lower boundary. To satisfy these requirements, we take the linear boundary sublayer of $\psi$ near the bottom boundary ($0 \leq z \leq \sqrt {\sigma \delta }$) to have slope equal to $-1$. As before, in the outer part of the bottom boundary layer ($\sqrt {\sigma \delta } \leq z \leq \delta$), $\psi$ behaves like $z^{-1}$ and matches smoothly with the inner part up to the first derivative. At the edge of the bottom boundary layer ($z = \delta$), the value of $\psi$ is $\sigma$. In the proof below, we assume

(5.2)\begin{equation} a, b, \sigma, \delta, \gamma \leq 1.\end{equation}

Figure 3.

Sketches of the functions (a) $\psi (z)$, and (b) $q(z)$, from (5.1a,b), used to obtain a bound on the heat flux $\langle w T \rangle$ in the IH3 configuration.

Estimating the second term in the bound (3.18) from above gives

(5.3)\begin{equation} \frac{1}{4 b} \int_0^1 (b z - \psi^\prime + q)^2 \, {\rm d}z \leq \frac{b}{6} + \frac{2 (\sigma + a)^2}{b \gamma}, \end{equation}

while the last term can be estimated from below as

(5.4)\begin{equation} \int_{0}^{1} \psi\,{\rm d}z \geq{-} \frac{1}{2}\sigma \delta \log \left(\frac{\sigma}{\delta}\right). \end{equation}

Combining (5.3) and (5.4) with (3.18), we obtain

(5.5)\begin{equation} \langle w T \rangle \leq \frac{1}{2} + \frac{b}{6} + \frac{2 (\sigma + a)^2}{b \gamma} + \frac{1}{2}\sigma \delta \log \left(\frac{\sigma}{\delta}\right). \end{equation}

Finally, we assume that

(5.6a,b)\begin{equation} \frac{b}{6} \leq{-} \frac{1}{8} \sigma \delta \log \left(\frac{\sigma}{\delta}\right), \quad \frac{2 (\sigma + a)^2}{b \gamma} \leq{-} \frac{1}{8} \sigma \delta \log \left(\frac{\sigma}{\delta}\right) \end{equation}

(these constraints will be verified later), and estimate the right-hand side of (5.5) to arrive at the simpler bound

(5.7)\begin{equation} \langle w T \rangle \leq \frac{1}{2} + \frac{1}{4}\sigma \delta \log \left(\frac{\sigma}{\delta}\right). \end{equation}

For this bound to be valid, we need to adjust the parameters $a$, $b$, $\delta$, $\gamma$ and $\sigma$ such that the spectral condition (3.19d) is satisfied. Dropping the positive terms proportional to $\vert \hat {w}_{\boldsymbol {k}} \vert ^2$, $\vert \hat {w}_{\boldsymbol {k}}' \vert ^2$ and $\vert \hat {T}_{\boldsymbol {k}}' \vert ^2$, we will verify the stronger inequality

(5.8)\begin{equation} \tilde{\mathcal{S}}(\hat{w}, \hat{T}) := \int_{0}^{1} \left[\frac{a}{Rk^2}\,|\hat{w}^{\prime \prime}|^2 + b k^2\,|\hat{T}|^2 - (a - \psi^\prime) \hat{w}\hat{T}\right] {\rm d}z \geq 0 ,\end{equation}

for all $z$-dependent functions $\hat {w}$ and $\hat {T}$ satisfying the boundary conditions

(5.9a)$$\begin{gather} \hat{w}(0) = \hat{w}^\prime(0) = \hat{T}'(0) = 0, \end{gather}$$

(5.9b)$$\begin{gather}\hat{w}(1) = \hat{w}^\prime(1) = \hat{T}(1) = 0. \end{gather}$$

Again, we have dropped the subscript $\boldsymbol {k}$ to lighten the notation.

Using arguments similar to those used in § 4 and noticing that if

(5.10)\begin{equation} 8 a \delta \leq \sigma,\end{equation}

then

(5.11)$$\begin{gather} \frac{9}{2}\,\frac{\sigma \delta}{(z + \sqrt{\sigma \delta})^2} \geq a + 1 \quad \text{when } 0 \leq z \leq \sqrt{\sigma \delta}, \end{gather}$$

(5.12)$$\begin{gather}\frac{9}{2}\,\frac{\sigma \delta}{(z + \sqrt{\sigma \delta})^2} \geq a + \frac{\sigma \delta}{z^2} \quad \text{when } \sqrt{\sigma \delta} \leq z \leq \delta, \end{gather}$$

we can write

(5.13)\begin{equation} \tilde{S}(\hat{w}, \hat{T}) \geq \tilde{S}_B(\hat{w}, \hat{T}) + \tilde{S}_T(\hat{w}, \hat{T}), \end{equation}

where

(5.14a)$$\begin{gather} \tilde{S}_B(\hat{w}, \hat{T}) = \int_{0}^{\delta} \left[\frac{a}{Rk^2}\,|\hat{w}^{\prime \prime}|^2 + b k^2\,|\hat{T}|^2 - \frac{9}{2} \frac{\sigma \delta}{(z + \sqrt{\sigma \delta})^2}\,\hat{w} \hat{T}\right] \textrm{d} z, \end{gather}$$

(5.14b)$$\begin{gather}\tilde{S}_T(\hat{w}, \hat{T}) = \int_{1-\gamma}^{1} \left[\frac{a}{Rk^2}\,|\hat{w}^{\prime \prime}|^2 + b k^2\,|\hat{T}|^2 - \frac{(\sigma + a)}{\gamma}\,\hat{w} \hat{T}\right] \textrm{d} z. \end{gather}$$

Finding a condition under which $\tilde {S}_T(\hat {w}, \hat {T}) \geq 0$ is straightforward. Using the fundamental theorem of calculus, the boundary conditions on $\hat {w}$ and the Cauchy–Schwarz inequality, we obtain

(5.15)\begin{equation} |\hat{w}|^2 \leq \frac{4(1-z)^{3}}{9} \int_{1-\gamma}^1 |\hat{w}''|^2 \, {\rm d}z. \end{equation}

Then substituting (5.15) in (5.14b) and using the AM–GM inequality shows that $\tilde {S}_T(\hat {w}, \hat {T})$ is non-negative as long as

(5.16)\begin{equation} (\sigma + a) \gamma \leq 6 \sqrt{\frac{ a b}{R}}.\end{equation}

To show that $\tilde {S}_B(\hat {w}, \hat {T})$ is non-negative, instead, we rely on the Rellich inequality stated in Lemma 3.2. First, using the AM–GM inequality we estimate

(5.17)\begin{equation} \tilde{S}_B(\hat{w}, \hat{T}) \geq \int_{0}^{\delta} \left[\frac{a}{Rk^2}\,|\hat{w}^{\prime \prime}|^2 + b k^2\,|\hat{T}|^2 - \frac{9}{4}\,\frac{\sigma \delta \beta}{(z + \sqrt{\sigma \delta})^4}\, |\hat{w}|^2 - \frac{9}{4}\,\frac{\sigma \delta}{\beta}\,|\hat{T}|^2\right] \textrm{d} z , \end{equation}

for a positive constant $\beta$ to be specified below. Next, using Lemma 3.2 we obtain

(5.18)\begin{equation} \int_{0}^{\delta} \frac{|\hat{w}|^2}{(z + \sqrt{\sigma \delta})^4}\, {\rm d}z \leq \frac{16}{9} \int_{0}^{\delta} |\hat{w}''|^2 \, {\rm d} z.\end{equation}

Combining (5.18) and (5.17), and setting

(5.19)\begin{equation} \beta = \frac{3}{4 k^2} \sqrt{\frac{a}{b R}}, \end{equation}

we conclude that $\tilde {S}_B(\hat {w}, \hat {T})$ is non-negative if

(5.20)\begin{equation} \sigma \delta \leq \frac{1}{3} \sqrt{\frac{a b}{R}}.\end{equation}

At this stage, all that remains is to choose values for $a$, $b$, $\delta$, $\gamma$ and $\sigma$ such that (5.6a,b), (5.16) and (5.20) hold, at least for sufficiently large Rayleigh numbers, while minimizing the right-hand side of (5.7). For the same reasons as explained at the end of § 4, we simplify the algebra by choosing $a = \sigma$ and $\delta = \gamma$. Then, optimizing the bound (5.7) subject to (5.16) and (5.20) leads to

(5.21a)$$\begin{gather} a = \sigma = \frac{2^{{4}/{5}}}{3^{{3}/{5}}}\,\frac{1}{R^{{2}/{5}}} \exp ({-}2^{{14}/{5}} 3^{{2}/{5}} R^{{3}/{5}}), \end{gather}$$

(5.21b)$$\begin{gather}b = \frac{2^{{12}/{5}}3^{{1}/{5}}}{R^{{1}/{5}}} \exp ({-}2^{{14}/{5}} 3^{{2}/{5}} R^{{3}/{5}}), \end{gather}$$

(5.21c)$$\begin{gather}\delta = \gamma = \frac{2^{{4}/{5}}}{3^{{3}/{5}}}\,\frac{1}{R^{{2}/{5}}}. \end{gather}$$

These choices satisfy the constraints in (5.6a,b) assumed in our proof for all $R\geq 2^{{19}/{2}} 3^{-{3}/{2}} \approx 139.35$. Thus from (5.7) we obtain

(5.22)\begin{equation} \langle w T \rangle \leq \frac{1}{2} - \frac{2^{{12}/{5}}}{3^{{4}/{5}}}\, \frac{1}{R^{{1}/{5}}} \exp ({-}2^{{14}/{5}} 3^{{2}/{5}} R^{{3}/{5}}) \quad \forall \, R\geq 2^{{19}/{2}} 3^{-{3}/{2}}. \end{equation}

It is interesting to note that only the boundary layer thicknesses $\delta$ and $\gamma$ have the same $O(R^{-2/5})$ scaling as for the IH1 configuration. The parameters $\sigma$, $a$, $b$ and the correction to $1/2$ in the bound (5.22), instead, are all $O(R^{{2}/{5}})$ smaller than their corresponding values for the IH1 case.