# Rigorous scaling laws for internally heated convection at infinite Prandtl
number

Ali Arslan Corresponding author<EMAIL_ADDRESS>Dept. of
Aeronautics, Imperial College London, London, SW7 2AZ, UK Giovanni Fantuzzi
Dept. of Aeronautics, Imperial College London, London, SW7 2AZ, UK John Craske
Dept. of Civil and Environmental Engineering, Imperial College London, SW7
2AZ, UK  Andrew Wynn  Dept. of Aeronautics, Imperial College London, London,
SW7 2AZ, UK

###### Abstract

New bounds are proven on the mean vertical convective heat transport,
$\smash{\overline{\left\langle wT\right\rangle}}$, for uniform internally
heated (IH) convection in the limit of infinite Prandtl number. For fluid in a
horizontally-periodic layer between isothermal boundaries, we show that
$\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}-cR^{-2}$, where $R$ is a nondimensional
‘flux’ Rayleigh number quantifying the strength of internal heating and
$c=216$. Then, $\mbox{$\smash{\overline{\left\langle wT\right\rangle}}$}=0$
corresponds to vertical heat transport by conduction alone, while
$\mbox{$\smash{\overline{\left\langle wT\right\rangle}}$}>0$ represents the
enhancement of vertical heat transport upwards due to convective motion. If,
instead, the lower boundary is a thermal insulator, then we obtain
$\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}-cR^{-4}$, with $c\approx 0.0107$. This
result implies that the Nusselt number Nu, defined as the ratio of the total-
to-conductive heat transport, satisfies
$\mbox{{Nu}}\lesssim\mbox{{R}}^{4}$.Both bounds are obtained by combining the
background method with a minimum principle for the fluid’s temperature and
with Hardy–Rellich inequalities to exploit the link between the vertical
velocity and temperature. In both cases, power-law dependence on R improves
the previously best-known bounds, which, although valid at both infinite and
finite Prandtl numbers, approach the uniform bound exponentially with R.

## 1 Introduction

Convective flows driven by internal sources of heat have attracted renewed
interest in recent years [1, 2, 3, 4, 5, 6]. Such flows are commonly
encountered in geophysics, where atmospheric convection [7] and mantle
convection [8, 9] are typical examples. They also exhibit unique features not
seen in boundary-driven Rayleigh–Bénard convection: for instance, it has
recently been observed experimentally that internally heated (IH) convection
can transport heat more efficiently than Rayleigh-Bénard convection [10, 11].
Nevertheless, the former remains much less studied. In particular, it remains
a largely open challenge to rigorously predict how key statistical properties
such as the mean vertical heat flux depend on the heating strength and on the
fluid’s Prandtl number Pr, defined as the ratio between the fluid’s kinematic
viscosity $\nu$ and its thermal diffusivity $\kappa$.

One source of difficulty for mathematical studies of IH convection is that the
mean thermal dissipation, the mean viscous dissipation, and the mean vertical
convective heat flux cannot all be related to each other via a priori
relationships. This is in contrast with Rayleigh-Bénard convection, where such
relationships enable one to rigorously bound the convective heat transfer
through variational analysis of the mean thermal dissipation [12]. Applying
the same strategy to IH flows yields bounds on the mean temperature of the
fluid [13, 14, 15, 16] but not the convective heat flux. Recently, this
variational strategy was extended by taking into account a minimum principle
for the temperature, leading to bounds on the mean convective heat flux that
approach a constant exponentially fast as the heating strength is increased
[2, 5]. Here, we demonstrate that these bounds can be improved to algebraic
powers when the Prandtl number is taken to be infinite.

Using standard non-dimensional variables [15], we consider a fluid in a
horizontally periodic domain
$\Omega=\mathbbm{T}_{[0,L_{x}]}\times\mathbbm{T}_{[0,L_{y}]}\times[0,1]$, the
motion of which is governed by the infinite Prandtl number Boussinessq
equations

$\displaystyle\nabla\cdot\boldsymbol{u}$ $\displaystyle=0\,,$ (1.1a)
$\displaystyle\nabla p$
$\displaystyle=\Delta\boldsymbol{u}+\mbox{{R}}\,T\boldsymbol{\hat{\mathbf{z}}}\,,$
(1.1b) $\displaystyle\partial_{t}T+\boldsymbol{u}\cdot\nabla T$
$\displaystyle=\Delta T+1.$ (1.1c)

Here, $\boldsymbol{u}=(u,v,w)$ is the fluid velocity in cartesian components,
$p$ is the pressure, $T$ is the temperature, and the unit forcing in (1.1c)
represents the non-dimensional internal heating rate. The flow is controlled
by a ‘flux’ Rayleigh number that measures the destabilising effect of the
heating compared to the stabilising effects of diffusion,

$\mbox{{R}}:=\frac{g\alpha Qd^{5}}{\rho c_{p}\nu\kappa^{2}}.$ (1.2)

Here $g$ is the acceleration of gravity, $\rho$ is the density, $c_{p}$ is the
specific heat capacity, $\alpha$ is the thermal expansion coefficient and $Q$
is the heating rate per unit volume.

We consider two separate configurations that differ in the choice of boundary
conditions at the top ($z=1$) and bottom ($z=0$) of the domain. In the first
configuration, referred to as IH1 and sketched in Figure 1(a), the velocity
satisfies no-slip conditions and the temperature of both vertical boundaries
is held at a constant value, which can be taken as zero without loss of
generality. Hence, we enforce

$\text{IH1:}\qquad\boldsymbol{u}|_{z\in\\{0,1\\}}=\boldsymbol{0},\qquad
T|_{z=0}=0,\qquad T|_{z=1}=0.$ (1.3a) In the second configuration, illustrated
by Figure 1(b), the bottom plate is replaced by a perfect thermal insulator,
giving
$\text{IH3:}\qquad\boldsymbol{u}|_{z\in\\{0,1\\}}=\boldsymbol{0},\qquad\partial_{z}T|_{z=0}=0,\qquad
T|_{z=1}=0.$ (1.3b)

We seek bounds on the mean vertical convective heat transport,
$\smash{\overline{\left\langle wT\right\rangle}}$, a quantity that is directly
proportional to the viscous dissipation for both the IH1 and IH3
configurations. Throughout this paper overbars denote infinite time averages,
while angled brackets denote volume averages:

$\displaystyle\left\langle f\right\rangle$
$\displaystyle=\frac{1}{L_{x}L_{y}}\int^{L_{x}}_{0}\int^{L_{y}}_{0}\int^{1}_{0}f(x,y,z,t)\textrm{d}z\,\textrm{d}y\,\textrm{d}x,$
(1.4a) $\displaystyle\overline{\left\langle f\right\rangle}$
$\displaystyle=\limsup_{\tau\rightarrow\infty}\frac{1}{\tau}\int^{\tau}_{0}\left\langle
f\right\rangle\textrm{d}t.$ (1.4b)

For either set of boundary conditions considered in this paper, it is known
that $0\leq\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}$ uniformly in R and Pr [17]. The zero lower
bound is saturated by the (possibly unstable) state in which the flow does not
move and heat is transported vertically by conduction alone. The upper bound
of $\frac{1}{2}$, instead, takes on a different meaning depending on the
thermal boundary conditions. As will become apparent from (1.6) below, for the
IH1 configuration a flow with $\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}=\frac{1}{2}$ would see all heat escape the domain through
the upper boundary.

$z=0$$z=1$$\partial_{z}T=0$$T=0$IH
$z$$T$$z$$x$$y$$z=0$$z=1$$T=0$$T=0$$z$$T$IH(a)(b) Figure 1: IH convection with
(a) isothermal boundaries (1.3a), and (b) insulating lower boundary (1.3b). In
both panels, IH represents the uniform unit internal heat generation. Red
lines denote the conductive temperature profiles ( ) and indicative mean
temperature profiles in the turbulent regime ( ).

This, however, cannot be achieved at any finite value of R. Precisely, our
first main results reveal that the mean vertical heat flux
$\smash{\overline{\left\langle wT\right\rangle}}$ is strictly smaller than
$\frac{1}{2}$ by an amount that cannot decrease faster than quadratically as R
is raised.

######  Theorem 1 (Isothermal boundaries, $\mbox{{Pr}}=\infty$).

Suppose that $\boldsymbol{u}=(u,v,w)$ and $T$ solve (1.1) subject to the no-
slip isothermal boundary conditions (1.3a). There exists a constant $c>0$ such
that, for all sufficiently large $R>0$,

$\overline{\left\langle wT\right\rangle}\leq\frac{1}{2}-c\,R^{-2}.$ (1.5)

######  Remark 1.

It is shown in section 2.4 that (1.5) holds with $c=216$ for any $R>1892$.

The bound on $\smash{\overline{\left\langle wT\right\rangle}}$ can be given a
clear physical interpretation as a measure of the asymmetry of the heat
transport due to heating. Indeed, upon computing $\overline{\left\langle
z\cdot\eqref{nondim_energy}\right\rangle}$ one can show that the average
nondimensional heat fluxes through the top and bottom boundaries, denoted by
$\mathcal{F}_{T}$ and $\mathcal{F}_{B}$ respectively, can be expressed as

$\displaystyle\mathcal{F}_{T}$
$\displaystyle:=-\partial_{z}\overline{\left\langle
T\right\rangle}_{h}|_{z=1}=\frac{1}{2}+\overline{\left\langle
wT\right\rangle},$ (1.6a) $\displaystyle\mathcal{F}_{B}$
$\displaystyle:=\phantom{-}\partial_{z}\overline{\left\langle
T\right\rangle}_{h}|_{z=0}=\frac{1}{2}-\overline{\left\langle
wT\right\rangle},$ (1.6b)

where $\left\langle\cdot\right\rangle_{h}$ denotes a spatial average over the
horizontal directions alone. Thus, our upper bound on
$\smash{\overline{\left\langle wT\right\rangle}}$ immediately implies bounds
on $\mathcal{F}_{T}$ and $\mathcal{F}_{B}$.

######  Corollary 1.1.

For all sufficiently large $R$,

$\mathcal{F}_{T}\leq 1-c\,R^{-2}\qquad\textrm{and}\qquad\mathcal{F}_{B}\geq
c\,R^{-2}.$ (1.7)

The results discussed so far apply to the IH1 configuration, where the two
horizontal domain boundaries are isothermal. Similar results hold also for the
IH3 case, where the bottom boundary is perfectly insulating. In this case,
however, the deviation of $\smash{\overline{\left\langle wT\right\rangle}}$
from $\frac{1}{2}$ may decay as fast as $\mbox{{R}}^{-4}$.

######  Theorem 2 (Insulating bottom & isothermal top, $\mbox{{Pr}}=\infty$).

Suppose that $\boldsymbol{u}=(u,v,w)$ and $T$ solve (1.1) subject to the
boundary conditions (1.3b). There exists a constant $c>0$ such that, for all
sufficiently large $\mbox{{R}}>0$,

$\overline{\left\langle wT\right\rangle}\leq\frac{1}{2}-c\,\mbox{{R}}^{-4}.$
(1.8)

######  Remark 2.

It is shown in section 3.1 that (1.8) holds with $c\approx 0.0107$ for all
$R>2961$.

Since the IH3 boundary conditions imply that the conductive heat flux is
positive, the effects of convection on the enhancement of heat transport in
the system can be described using a Nusselt number, Nu. This is defined as the
ratio of the mean total heat flux to the mean conductive heat flux, and can be
expressed in terms of $\smash{\overline{\left\langle wT\right\rangle}}$ as

$\mbox{{Nu}}=\left(1-2\overline{\left\langle wT\right\rangle}\right)^{-1}.$
(1.9)

Thus, our upper bound on $\smash{\overline{\left\langle wT\right\rangle}}$ can
be transformed into an upper bound on Nu.

######  Corollary 2.1.

For all sufficiently large R, $\mbox{{Nu}}\leq cR^{4}$.

The proofs of Theorems 1 and 1.8 rely on two key ingredients. The first is a
variational problem giving an upper bound on $\smash{\overline{\left\langle
wT\right\rangle}}$. This variational problem is derived by enforcing a minimum
principle for the fluid’s temperature within the classical “background method”
[12, 18, 19], which for simplicity we formulate using the language of a more
general framework for bounding infinite-time averages [20, 21, 22, 23] (see
[21, 24] for further discussion of the link between the two approaches). Using
this minimum principle is essential to obtain bounds on
$\smash{\overline{\left\langle wT\right\rangle}}$ that asymptote to
$\frac{1}{2}$ from below. This has already been shown for the IH1
configuration at finite Pr: for this case, without the miniumum principle one
obtains only $\mbox{$\smash{\overline{\left\langle wT\right\rangle}}$}\lesssim
R^{1/5}$ [2], while with it one can prove that
$\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}-O(R^{1/5}\exp{(-R^{3/5})})$ uniformly in Pr
[5]. A similar (but not identical) exponentially-varying bound of
$\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}-O(R^{-1/5}\exp{(-R^{3/5}))}$ uniformly in
Pr was also obtained for the IH3 configuration [5].

The second key ingredient in our proofs are estimates of Hardy–Rellich type,
obtained by observing that the reduced momentum equation (1.1b) determines the
vertical velocity field as a function of the temperature field. Specifically,
taking the vertical component of the double curl of (1.1b) gives

$\Delta^{2}w=-R\,\Delta_{h}T,$ (1.10)

where $\Delta_{h}:=\partial^{2}_{x}+\partial^{2}_{y}$, is the horizontal
Laplacian. Using the no-slip boundary conditions with the incompressibility
condition (1.1a), the vertical velocity $w$ satisfies

$w|_{z=0}=\partial_{z}w|_{z=0}=w|_{z=1}=\partial_{z}w|_{z=1}=0.$ (1.11)

Equation (1.10) was exploited in Rayleigh-Bénard convection to improve the
scaling of upper bounds on Nu [25]. This was achieved by using (1.10) to
derive inequalities of the Hardy–Rellich type (see Lemma 4 below) that help
the construction of a background field with a logarithmically-varying stable
stratification in the bulk [25, 26]. Here, we use the same inequalities to
construct (different) background fields suited to IH convection, which will
enable us to bound $\smash{\overline{\left\langle wT\right\rangle}}$ in the
infinite Pr limit.

## 2 Bounds for the IH1 configuration

We first consider the IH1 configuration, where the top and bottom plate are
held at zero temperature. In section 2.1, we show that
$\smash{\overline{\left\langle wT\right\rangle}}$ can be bounded from above by
constructing suitably constrained functions of the vertical coordinate $z$.
Section 2.2 describes parametric ansätze for such functions, while section 2.3
establishes auxiliary results that simplify the verification of the
constraints and the evaluation of the bound. We then prove Theorem 1 in
section 2.4 by prescribing R-dependent values of the free parameters in our
ansätze.

To simplify the notation, we introduce two sets of temperature fields that
encode the thermal boundary conditions and the pointwise nonnegativity
constraint implied by the minimum principle:

$\displaystyle\mathcal{H}_{n}:=\\{T\in H^{1}(\Omega):\;T\text{ is horizontally
periodic \& \eqref{bc_T_IH1}}\veebar\eqref{bc_T_IH3}\\},$ (2.1a)
$\displaystyle\mathcal{H}_{+}:=\\{T\in\mathcal{H}_{n}:T(\boldsymbol{x})\geq
0\text{ a.e. }\boldsymbol{x}\in\Omega\\},$ (2.1b)

where $n=$ 1 or 3, depending on the boundary conditions of IH1 and IH3
respectively. In section 2, $T$ belongs to $\mathcal{H}_{1}$.

### 2.1 Bounding framework

To bound $\smash{\overline{\left\langle wT\right\rangle}}$ we employ the
auxiliary function method [20, 24]. The method relies on the observation that
the time derivative of any bounded functional $\mathcal{V}\\{T(t)\\}$ along
solutions of the Boussinesq equations (1.1) averages to zero over infinite
time, so

$\overline{\left\langle wT\right\rangle}=\overline{\left\langle
wT\right\rangle+\tfrac{\rm d}{{\rm d}t}\mathcal{V}\\{T(t)\\}}.$ (2.2)

If $\mathcal{V}$ is chosen such that the quantity being averaged on the right-
hand side is bounded above pointwise in time, then this pointwise bound is
also an upper bound on $\smash{\overline{\left\langle wT\right\rangle}}$.

Following analysis at finite Prandtl number [2], we restrict our attention to
quadratic functionals taking the form

$\mathcal{V}\\{T\\}=\left\langle\tfrac{\beta}{2}\left|T\right|^{2}-[\tau(z)+z-1]T\right\rangle,$
(2.3)

which are parametrized by a positive constant $\beta\in\mathbb{R}_{+}$ and a
piecewise-differentiable function $\tau:[0,1]\to\mathbb{R}$ with square-
integrable derivative. We require $\tau$ to satisfy

$\tau(0)=1,\qquad\tau(1)=0,$ (2.4)

so the coefficient multiplying $T$ in (2.3) vanishes at $z=0$ and $z=1$. This
choice enables us to integrate by parts without picking up boundary terms when
calculating $\tfrac{\rm d}{{\rm d}t}\mathcal{V}\\{T(t)\\}$. To ensure that the
resulting expression $\left\langle wT\right\rangle+\tfrac{\rm d}{{\rm
d}t}\mathcal{V}\\{T\\}$ can be bounded from above poinwise in time, we also
require that the pair $(\beta,\tau)$ satisfies a condition called the
_spectral constraint_.

######  Definition 1 (Spectral constraint).

The pair $(\beta,\tau)$ is said to satisfy the _spectral constraint_ if

$\left\langle\beta\left|\nabla T\right|^{2}+\tau^{\prime}wT\right\rangle\geq
0\qquad T\in\mathcal{H}_{n},$ (2.5)

where $w=-R\Delta^{-2}\Delta_{h}T$ solves (1.10) with the boundary conditions
1.11.

If the spectral constraint is satisfied, then it is possible to bound
$\smash{\overline{\left\langle wT\right\rangle}}$ from above in terms of
$\tau$, $\beta$, and another suitably constrained function
$\lambda:[0,1]\to\mathbb{R}$.

######  Proposition 1 (Bounding framework, IH1).

Suppose that the pair $(\beta,\tau)$ satisfies the spectral constraint and the
boundary conditions in (2.4). Further, let $\lambda\in L^{2}(0,1)$ be a
nondecreasing function such that $\left\langle\lambda\right\rangle=-1$. Then,

$\overline{\left\langle
wT\right\rangle}\leq\tfrac{1}{2}+\left\langle\tfrac{1}{4\beta}\left|\tau^{\prime}-\lambda-\beta\left(z-\tfrac{1}{2}\right)\right|^{2}-\tau\right\rangle=:U(\tau,\lambda,\beta)$

###### Proof.

A standard calculation using integration by parts, the incompressibility
condition (1.1a), and the boundary conditions on $\boldsymbol{u}$ and $T$
yields

$\displaystyle\overline{\left\langle wT\right\rangle}$
$\displaystyle=\overline{\left\langle wT\right\rangle+\tfrac{\rm d}{{\rm
d}t}\mathcal{V}\\{T(t)\\}}$
$\displaystyle=\tfrac{1}{2}+\overline{\left\langle-\beta\left|\nabla
T\right|^{2}-\tau^{\prime}wT+(\tau^{\prime}-\beta
z)\partial_{z}T-\tau\right\rangle}.$ (2.6)

The infinite-time average on the right-hand side is bounded above by the
largest value of the argument over the global attractor of the infinite-
Prandtl-number Boussinesq equations (1.1). The minimum principle for the
temperature field implies that the global attractor is contained in the set
$\mathcal{H}_{+}$ defined in (2.1b). Consequently, we can estimate

$\overline{\left\langle
wT\right\rangle}\leq\tfrac{1}{2}+\sup_{\begin{subarray}{c}T\in\mathcal{H}_{+}\\\
w=-R\Delta^{-2}\Delta_{h}T\end{subarray}}\left\langle-\beta\left|\nabla
T\right|^{2}-\tau^{\prime}wT+(\tau^{\prime}-\beta
z)\partial_{z}T-\tau\right\rangle.$ (2.7)

This upper bound is finite if and only if unless the pair $(\beta,\tau)$
satisfies the spectral constraint (cf. definition 1), in which case the
supremum over $T$ can be evaluated using a technical convex duality argument
detailed in appendix A. The result of the argument is that the bound in 2.7 is
equivalent to

$\overline{\left\langle
wT\right\rangle}\leq\tfrac{1}{2}+\inf_{\begin{subarray}{c}\lambda\in
L^{2}(0,1)\\\ \lambda\text{ \rm nondecreasing}\\\
\langle\lambda\rangle=-1\end{subarray}}\left\langle\tfrac{1}{4\beta}\left|\tau^{\prime}-\lambda-\beta\left(z-\tfrac{1}{2}\right)\right|^{2}-\tau\right\rangle.$
(2.8)

This inequality clearly implies the upper bound on
$\smash{\overline{\left\langle wT\right\rangle}}$ stated in the proposition,
which is therefore proven. ∎

######  Remark 3.

The function $\lambda$ that arises when deriving 2.8 from 2.7 can be viewed as
a Lagrange multiplier enforcing the pointwise nonnegativity of temperature
fields in the set $\mathcal{H}_{+}$. Further details regarding this
interpretation (with slightly different notation) are given in [2, §4.4].

######  Remark 4.

The best upper bound on $\smash{\overline{\left\langle wT\right\rangle}}$
provable with our approach is found upon minimizing the expression
$U(\tau,\lambda,\beta)$ over all choices of $\tau$, $\lambda$ and $\beta$ that
satisfy the conditions of Proposition 1. This is hard to do analytically, but
can be done computationally using a variety of numerical schemes (see [24] and
references therein). We leave such computations to future work and focus on
proving Theorem 1 by constructing suboptimal $\tau$, $\lambda$ and $\beta$
analytically.

### 2.2 Ansätze

To prove the upper bound on $\smash{\overline{\left\langle wT\right\rangle}}$,
we seek $\beta>0$, $\tau(z)$, and $\lambda(z)$ that satisfy the conditions of
Proposition 1 and make the quantity $U(\beta,\tau,\lambda)$ as small as
possible. To simplify this task, we restrict $\tau$ to take the form

$\tau(z):=\begin{dcases}1-\frac{z}{\delta},&0\leq z\leq\delta\\\
A\,\ln\\!{\left(\frac{z(1-\delta)}{\delta(1-z)}\right)},&\delta\leq z\leq
1-\varepsilon\\\
A\,\ln\\!\left(\frac{(1-\varepsilon)(1-\delta)}{\varepsilon\delta}\right)\,\left(\frac{1-z}{\varepsilon}\right),&1-\varepsilon\leq
z\leq 1\end{dcases}$ (2.9)

and $\lambda$ to be given by

$\lambda(z):=\begin{dcases}-\frac{1}{\delta},&0\leq z\leq\delta,\\\
0,&\delta\leq z\leq 1.\end{dcases}$ (2.10)

These piecewise-defined functions, sketched in Figure 2, are fully specified
by the bottom boundary layer width $\delta\in(0,\frac{1}{2})$, the top
boundary layer width $\varepsilon\in(0,\frac{1}{2})$, and the parameter $A>0$
that determines the amplitude of $\tau$ in the bulk of the layer.

$z$$\tau(z)$$1$$1$$\delta$$1-\varepsilon$$z$$\lambda(z)$$1$$\delta$$-1/\delta$
Figure 2: Sketches of the functions $\tau(z)$ in (2.9) and $\lambda(z)$ in
(2.10) used to prove Theorem 1.

We also fix

$\beta:={\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}}{\left\langle|z-\tfrac{1}{2}|^{2}\right\rangle^{-\frac{1}{2}}}=2\sqrt{3}\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}.$
(2.11)

This choice is motivated by the desire to minimize the right-hand side of the
inequality

$\frac{1}{4\beta}\left\langle|\tau^{\prime}-\lambda-\beta\left(z-\tfrac{1}{2}\right)|^{2}\right\rangle\leq\frac{\beta}{2}\left\langle|z-\tfrac{1}{2}|^{2}\right\rangle+\frac{1}{2\beta}\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle,$
(2.12)

which is used later in Lemma 2 by estimating from above the value of the bound
$U(\beta,\tau,\lambda)$ for our choices of $\tau$ and $\lambda$.

For any choice of the parameters $\delta$, $\varepsilon$, and $A$, the
function $\tau$ satisfies the boundary conditions in (2.4), while $\lambda$ is
nondecreasing and satisfies the normalization condition
$\left\langle\lambda\right\rangle=-1$. To establish Theorem 1 using
Proposition 1, we only need to specify parameter values such that
$U(\beta,\tau,\lambda)\leq\frac{1}{2}-O(\mbox{{R}}^{-2})$ while ensuring that
the pair $(\beta,\tau)$ satisfies the spectral constraint. For the purposes of
simplifying the algebra in what follows, we shall fix

$A(\delta)=\frac{3\sqrt{15}}{20}\delta^{\frac{3}{2}}$ (2.13)

from the outset. As explained in remark 7 below, this choice arises when
insisting that the upper estimate on $U(\beta,\tau,\lambda)$ derived in Lemma
2 be strictly less than $\frac{1}{2}$ for some values of $\delta$ and
$\varepsilon$, at least when all other constraints on these parameters are
ignored.

### 2.3 Preliminary estimates

We now derive a series of auxiliary results that make it simpler to specify
the boundary layer widths $\delta$ and $\varepsilon$. The first result gives
estimates on the value of $\beta$ in (2.11).

######  Lemma 1 (Estimates on $\beta$).

Let $\tau(z)$, $\lambda(z)$ and $\beta$ be given by (2.9), (2.10), and (2.11)
with $A$ specified by (2.13). Suppose that the boundary layer widths $\delta$
and $\varepsilon$ satisfy

$\delta\leq\frac{1}{6},\qquad\varepsilon\leq\frac{1}{3},\qquad\delta\ln^{2}{\left(\frac{1}{\delta^{2}}\right)}\leq\varepsilon.$
(2.14a,b,c)

Then,

$\frac{9}{2\sqrt{5}}\delta\leq\beta\leq\frac{9}{2}\delta.$ (2.15)

######  Remark 5.

Condition (2.14a,b,cc) is key to prove the auxiliary results of this section.
The other two restrictions, instead, are introduced to more easily keep track
of constants in our estimates, which is necessary to obtain an explicit
prefactor for the $O(\mbox{{R}}^{-2})$ term. We have not attempted to optimize
this prefactor.

######  Remark 6.

Condition (2.14a,b,cc) implies that $\delta\leq\varepsilon$. We will use this
fact often in the proofs of this section.

###### Proof of Lemma 1.

It suffices to estimate
$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}$ from
above and below. For a lower bound, substitute our choices for $\tau$ and
$\lambda$ to estimate

$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle=\int^{1}_{\delta}|\tau^{\prime}(z)|^{2}\,{\rm
d}z\geq\int^{1-\varepsilon}_{\delta}|\tau^{\prime}(z)|^{2}\,{\rm
d}z=A^{2}\int^{1-\varepsilon}_{\delta}\left|\frac{1}{z}+\frac{1}{1-z}\right|^{2}\,{\rm
d}z.$

Expanding the square, dropping the nonnegative term
$\smash{\frac{2}{z(1-z)}}$, and recalling that $\varepsilon\leq\frac{1}{3}$
and $\delta\leq\frac{1}{6}<\frac{1}{3}$ by assumption, we can further estimate

$\displaystyle\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle\geq
A^{2}\left(\int^{2/3}_{\delta}\frac{1}{z^{2}}\,{\rm
d}z+\int^{2/3}_{1/3}\frac{1}{(1-z)^{2}}\,{\rm
d}z\right)=\frac{A^{2}}{\delta}.$ (2.16)

Taking the square root of both sides and substituting for the value of $A$
from (2.13) gives
$\smash{\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{1/2}}\geq(3\sqrt{15}/20)\,\delta$,
which combined with (2.11) proves the lower bound on $\beta$ stated in (2.15).

For the upper bound on $\beta$, recall that condition (2.14a,b,cc) implies
$\delta\leq\varepsilon$, so $1-\varepsilon\leq 1-\delta\leq 1$. Then,

$\displaystyle\frac{\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle}{A^{2}}$
$\displaystyle=$
$\displaystyle\int^{1-\varepsilon}_{\delta}\left|\frac{1}{z}+\frac{1}{1-z}\right|^{2}\textrm{d}z+\frac{1}{\varepsilon}\ln^{2}{\left(\frac{(1-\varepsilon)(1-\delta)}{\varepsilon\delta}\right)}$
$\displaystyle\leq$
$\displaystyle\int^{1-\delta}_{\delta}\left|\frac{1}{z}+\frac{1}{1-z}\right|^{2}\textrm{d}z+\frac{1}{\varepsilon}\ln^{2}\left(\frac{1}{\delta^{2}}\right).$

Using the inequality $(a+b)^{2}\leq 2a^{2}+2b^{2}$ we can further estimate

$\displaystyle\frac{\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle}{A^{2}}$
$\displaystyle\leq$ $\displaystyle
2\left(\int^{1-\delta}_{\delta}\frac{1}{z^{2}}\,{\rm
d}z+\int^{1-\delta}_{\delta}\frac{1}{(1-z)^{2}}\,{\rm
d}z\right)+\frac{1}{\varepsilon}\ln^{2}\left(\frac{1}{\delta^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{4}{\delta}\left(\frac{1-2\delta}{1-\delta}\right)+\frac{1}{\varepsilon}\ln^{2}\left(\frac{1}{\delta^{2}}\right).$

Finally, we observe that $\frac{1-2\delta}{1-\delta}\leq 1$ for all
$\delta\geq 0$ and apply (2.14a,b,cc) to arrive at

$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle\leq\frac{5A^{2}}{\delta}.$
(2.17)

Substituting our choice of $A$ from (2.13) and taking a square root gives
$\smash{\langle|\tau^{\prime}-\lambda|^{2}\rangle^{1/2}}\leq(3\sqrt{3}/4)\delta$,
which combined with (2.11) yields the upper bound on $\beta$ in (2.15). ∎

Our second auxiliary result estimates the upper bound $U(\beta,\tau,\lambda)$
on $\smash{\overline{\left\langle wT\right\rangle}}$ from Proposition 1 in
terms of the bottom boundary layer width $\delta$ alone.

######  Lemma 2 (Estimates on $U(\beta,\tau,\lambda)$).

Let $\tau(z)$, $\lambda(z)$ and $\beta$ be given by (2.9), (2.10), and (2.11)
with $A$ specified by (2.13). Suppose the boundary layer widths $\delta$ and
$\varepsilon$ satisfy the conditions of Lemma 1. Then,

$U(\beta,\tau,\lambda)\leq\frac{1}{2}-\frac{\delta}{8}.$ (2.18)

###### Proof.

Using inequality (2.12), our choice of $\beta$ from (2.11), and the upper
bound on $\langle|\tau^{\prime}-\lambda|^{2}\rangle$ from (2.17) yields

$\displaystyle
U(\beta,\tau,\lambda)\leq\frac{1}{2}+\frac{\sqrt{3}}{6}\,\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}-\left\langle\tau\right\rangle\leq\frac{1}{2}+\frac{\sqrt{15}A}{6}\,\delta^{-\frac{1}{2}}-\left\langle\tau\right\rangle.$

Moreover, since we have chosen $\tau$ to be a non-negative function we can
estimate $\left\langle\tau\right\rangle=\int_{0}^{1}\tau(z)\,{\rm
d}z\geq\int_{0}^{\delta}\tau(z)\,{\rm d}z=\frac{1}{2}\delta$ to obtain

$U(\beta,\tau,\lambda)\leq\frac{1}{2}+\frac{\sqrt{15}A}{6}\,\delta^{-\frac{1}{2}}-\frac{1}{2}\delta.$
(2.19)

Substituting our choice of $A$ from (2.13) into this inequality gives (2.18).
∎

######  Remark 7.

The right-hand side of (2.19) can be strictly smaller than $\frac{1}{2}$ only
if $A\lesssim\delta^{3/2}$. It is this observation that dictates the choice of
$A$ in (2.13). For any fixed value of $R$, one should choose
$A\sim\delta^{3/2}$ with a (possibly $R$-dependent) prefactor that optimises
the balance between the positive and negative terms, subject to constraints on
$A$, $\delta$ and all other parameters that ensure the spectral constraint. To
simplify our proof, however, we choose to fix this prefactor a priori
irrespective of $R$.

Our final auxiliary result gives sufficient conditions on $\delta$ and
$\varepsilon$ that ensure the spectral constraint (cf. definition 1) is
satisfied.

######  Lemma 3 (Sufficient conditions for the spectral constraint).

Let $\tau(z)$, $\lambda(z)$ and $\beta$ be given by (2.9), (2.10), and (2.11)
with $A$ specified by (2.13). Suppose the boundary layer widths $\delta$ and
$\varepsilon$ satisfy the conditions of Lemma 1. Suppose further that

$\delta\leq(24\sqrt{3})^{2}\,R^{-2}\qquad\text{and}\qquad\varepsilon^{3}\delta^{\frac{1}{2}}\,\ln^{2}\\!\left(\frac{1}{\delta^{2}}\right)\leq
8\sqrt{3}R^{-1}.$ (2.20a,b)

Then, the pair $(\beta,\tau)$ satisfies the spectral constraint.

The proof of this result relies on Hardy–Rellich inequalities established in
Ref. [14], which extract a positive term from the a priori indefinite term
$\langle\tau^{\prime}wT\rangle$.

######  Lemma 4 (Hardy–Rellich inequalities [14]).

Let $T,w:\Omega\to\mathbb{R}$ be horizontally periodic functions such that
$\Delta^{2}w=-R\Delta_{h}T$ subject to velocity boundary conditions (1.11).
Then,

$\displaystyle\left\langle\frac{wT}{z}\right\rangle$
$\displaystyle\geq\frac{4}{R}\left\langle\frac{w^{2}}{z^{3}}\right\rangle$ and
$\displaystyle\left\langle\frac{wT}{1-z}\right\rangle$
$\displaystyle\geq\frac{4}{R}\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle.$
(2.21a,b)

###### Proof of Lemma 3.

Let $\mathbbm{1}_{(a,b)}$ denote the indicator function of the interval
$(a,b)$. Define the functions

$\displaystyle
f(z):=\left[\frac{1}{\delta}+\frac{A}{z(1-z)}\right]\mathbbm{1}_{(0,\delta)}(z),$
(2.22a) $\displaystyle
g(z):=\left[\frac{1}{\varepsilon}\ln\\!{\left(\frac{(1-\varepsilon)(1-\delta)}{\varepsilon\delta}\right)}+\frac{1}{z(1-z)}\right]\mathbbm{1}_{(1-\varepsilon,1)}(z).$
(2.22b)

Given our choice of $\tau$ from (2.9), we can rewrite the spectral constraint
as

$0\leq\left\langle\beta|\nabla
T|^{2}+\tau^{\prime}wT\right\rangle=\mathcal{F}\\{T\\}+\mathcal{G}\\{T\\},$

where

$\displaystyle\mathcal{F}\\{T\\}:=\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+A\left\langle\frac{wT}{z}\right\rangle-\left\langle
f(z)wT\right\rangle,$ (2.23a)
$\displaystyle\mathcal{G}\\{T\\}:=\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+A\left\langle\frac{wT}{1-z}\right\rangle-A\left\langle
g(z)wT\right\rangle.$ (2.23b)

Here, $w$ is determined as a function of $T$ by solving (1.10) subject to the
boundary conditions in (1.11). We shall prove that $\mathcal{F}$ and
$\mathcal{G}$ are individually non-negative.

First, let us consider $\mathcal{F}$. The Hardy–Rellich inequality (2.21a)
gives

$\mathcal{F}\\{T\\}\geq\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+\frac{4A}{R}\left\langle\frac{w^{2}}{z^{3}}\right\rangle-\left\langle
f(z)wT\right\rangle.$ (2.24)

Next, we estimate $\langle f(z)wT\rangle$. Since $T$ vanishes at $z=0$ by
virtue of the thermal boundary conditions in (1.3a), we can use the
fundamental theorem of calculus and the Cauchy-Schwarz inequality to estimate
$|T(\cdot,z)|\leq\smash{\sqrt{z}(\int^{1}_{0}|\nabla
T|^{2}\textrm{d}z)^{1/2}}$. Squaring both sides and taking the horizontal
average of which gives $\left\langle T^{2}\right\rangle_{h}\leq
z\left\langle|\nabla T|^{2}\right\rangle$. Then, use of the Cauchy–Schwarz
inequality, substitution for $\left\langle T^{2}\right\rangle_{h}$ and Youngs
inequality gives

$\displaystyle\left\langle
f(z)wT\right\rangle\leq\int^{1}_{0}|f(z)|\left\langle|wT|\right\rangle_{h}\textrm{d}z$
$\displaystyle\leq\int^{1}_{0}|f(z)|\left\langle
w^{2}\right\rangle^{1/2}_{h}\left\langle
T^{2}\right\rangle^{1/2}_{h}\textrm{d}z$ $\displaystyle\leq\left\langle|\nabla
T|^{2}\right\rangle^{\frac{1}{2}}\int^{1}_{0}f(z)z^{2}\left\langle\frac{w^{2}}{z^{3}}\right\rangle^{\frac{1}{2}}_{h}\textrm{d}z$
$\displaystyle\leq\left\langle|\nabla
T|^{2}\right\rangle^{\frac{1}{2}}\left\langle
f(z)^{2}z^{4}\right\rangle^{\frac{1}{2}}\left\langle\frac{w^{2}}{z^{3}}\right\rangle^{\frac{1}{2}}$
$\displaystyle\leq\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+\frac{1}{2\beta}\left\langle
f(z)^{2}z^{4}\right\rangle\left\langle\frac{w^{2}}{z^{3}}\right\rangle.$

Upon using the lower bound on $\beta$ from (2.15) to estimate the last term
from above we obtain

$\left\langle f(z)wT\right\rangle\leq\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+\frac{\sqrt{5}}{9\delta}\left\langle
f(z)^{2}z^{4}\right\rangle\left\langle\frac{w^{2}}{z^{3}}\right\rangle.$
(2.25)

This can be substituted into (2.24) along with our choice of $A$ from (2.13)
to find

$\mathcal{F}\\{T\\}\geq\frac{\sqrt{5}}{9}\delta^{\frac{3}{2}}\left(\frac{27\sqrt{3}}{5R}\,-\frac{\left\langle
f(z)^{2}z^{4}\right\rangle}{\delta^{\frac{5}{2}}}\right)\left\langle\frac{w^{2}}{z^{3}}\right\rangle.$
(2.26)

To conclude, we show that the term in parentheses is nonnegative when $\delta$
satisfies $\delta\leq\frac{1}{6}$ and (2.20a,ba). To do this, we observe that
the function $f(z)$ in (2.22a) is nonnegative function and that it is nonzero
only if $z\leq\delta$. We can therefore bound it from above on the interval
$(0,\delta)$ using the estimates
$\frac{1}{1-z}\leq\frac{1}{1-\delta}\leq\frac{6}{5}$ and, consequently, obtain

$\frac{\left\langle
f(z)^{2}z^{4}\right\rangle}{\delta^{\frac{5}{2}}}\leq\frac{1}{\delta^{\frac{5}{2}}}\int^{\delta}_{0}\left(\frac{1}{\delta}+\frac{6A}{5z}\right)^{2}z^{4}\,{\rm
d}z=\bigg{(}\frac{1}{5}+\frac{9\sqrt{15}}{100}\delta^{\frac{3}{2}}+\frac{81}{500}\delta^{3}\bigg{)}\delta^{\frac{1}{2}}.$

Using the assumption that $\delta\leq\frac{1}{6}$ to estimate the expression
in parentheses by its value at $\delta=\frac{1}{6}$, followed by an
application of assumption (2.20a,ba) to estimate the remaining $\delta^{1/2}$
term in terms of R we arrive at the desired inequality

$\frac{\left\langle
f(z)^{2}z^{4}\right\rangle}{\delta^{\frac{5}{2}}}\leq\frac{9\delta^{\frac{1}{2}}}{40}\leq\frac{27\sqrt{3}}{5\mbox{{R}}}.$

Analogous arguments show that $\mathcal{G}\\{T\\}$ is nonnegative. Using the
Hardy–Rellich inequality (2.21b) we have

$\mathcal{G}\\{T\\}\geq\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+\frac{4A}{R}\left\langle\frac{|w|^{2}}{(1-z)^{3}}\right\rangle-A\left\langle
g(z)wT\right\rangle.$ (2.27)

To estimate the last term, we use (in order) the inequality $\left\langle
T^{2}\right\rangle_{h}\leq\smash{(1-z)\left\langle|\nabla
T|^{2}\right\rangle}$, the Cauchy–Schwarz inequality, and Young’s inequality:

$\displaystyle\left\langle
g(z)wT\right\rangle\leq\int^{1}_{0}|g(z)|\left\langle|wT|\right\rangle_{h}\textrm{d}z$
$\displaystyle\leq\int^{1}_{0}|g(z)|\left\langle
w^{2}\right\rangle^{1/2}_{h}\left\langle
T^{2}\right\rangle^{1/2}_{h}\textrm{d}z$ $\displaystyle\leq\left\langle|\nabla
T|^{2}\right\rangle^{\frac{1}{2}}\int^{1}_{0}g(z)(1-z)^{2}\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle_{h}^{\frac{1}{2}}\textrm{d}z$
$\displaystyle\leq\left\langle|\nabla
T|^{2}\right\rangle^{\frac{1}{2}}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle^{\frac{1}{2}}\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle^{\frac{1}{2}}$
$\displaystyle\leq\frac{\beta}{2A}\left\langle|\nabla
T|^{2}\right\rangle+\frac{A}{2\beta}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle.$

Using the lower bound on $\beta$ from (2.15) gives

$\left\langle g(z)wT\right\rangle\leq\frac{\beta}{2A}\left\langle|\nabla
T|^{2}\right\rangle+\frac{\sqrt{5}A}{9\delta}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle,$
(2.28)

which can be substituted into (2.27) along with the value of $A$ from (2.13)
to obtain

$\mathcal{G}\\{T\\}\geq\frac{3\delta^{\frac{3}{2}}}{16\sqrt{5}}\left(\frac{16\sqrt{3}}{\mbox{{R}}}-\delta^{\frac{1}{2}}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle\right)\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle.$
(2.29)

To conclude the argument we show that the term in parentheses is non-negative.
To demonstrate this, we first estimate $g(z)$ on the interval
$(1-\varepsilon,1)$ from above using the assumption that
$\varepsilon\leq\frac{1}{3}$, so $\frac{2}{3}\leq z\leq 1$ and
$\frac{1}{z(1-z)}\leq\frac{3}{2(1-z)}$. Thus,

$g(z)\leq\frac{1}{\varepsilon}\ln{\left(\frac{1}{\delta^{2}}\right)}+\frac{3}{2(1-z)}\qquad\forall
z\in(1-\varepsilon,1).$

Then, we use the assumptions $\delta\leq\frac{1}{6}$ and
$\delta\leq\varepsilon$ (cf. remark 6) to observe that
$\ln{(\frac{1}{\varepsilon\delta})}\leq\ln{(\frac{1}{\delta^{2}})}\leq\ln^{2}{(\frac{1}{\delta^{2}})}$
and $\frac{3}{4}\leq\frac{3}{4}\ln^{2}{(\frac{1}{\delta^{2}})}$. Combining
these estimates with the upper bound on $g$ derived above gives

$\displaystyle\delta^{\frac{1}{2}}\left\langle g(z)^{2}(1-z)^{4}\right\rangle$
$\displaystyle\leq\delta^{\frac{1}{2}}\int^{1}_{1-\varepsilon}\left(\frac{1}{\varepsilon}\ln{\left(\frac{1}{\delta^{2}}\right)}+\frac{3}{2(1-z)}\right)^{2}(1-z)^{4}\,\textrm{d}z$
$\displaystyle=\delta^{\frac{1}{2}}\varepsilon^{3}\left(\frac{1}{5}\ln^{2}{\left(\frac{1}{\delta^{2}}\right)}+\frac{3}{4}\ln{\left(\frac{1}{\delta^{2}}\right)}+\frac{3}{4}\right)$
$\displaystyle\leq
2\delta^{\frac{1}{2}}\varepsilon^{3}\ln^{2}{\left(\frac{1}{\delta^{2}}\right)}$
$\displaystyle(\text{{by}
\hyperref@@ii[e:delta_eps_cons]{(\ref*{e:delta_eps_cons}b)}})\qquad$
$\displaystyle\leq\frac{16\sqrt{3}}{R},$

as desired. This concludes the proof of Lemma 3. ∎

### 2.4 Proof of Theorem 1

It is now straightforward to prove the upper bound on
$\smash{\overline{\left\langle wT\right\rangle}}$ by specifying boundary layer
widths $\delta$ and $\varepsilon$ that satisfy the conditions of Lemmas 1,
LABEL:, 2, LABEL: and 3.

Figure 3: Variation with R of the allowed values for the bottom boundary layer
width $\varepsilon$ (shaded region), determined by condition (2.14a,b,cc) in
Lemma 1 and condition (2.20a,bb) in Lemma 3 when the bottom boundary layer
width $\delta$ ( ) is chosen as in (2.30). Also shown are the uniform upper
bounds $\delta\leq\frac{1}{6}$ ( ) and $\varepsilon\leq\frac{1}{3}$ ( )
imposed on these variables. A black vertical line marks the Rayleigh number
$\mbox{{R}}_{0}\simeq 1891.35$ above which all constraints on $\delta$ and
$\varepsilon$ are satisfied.

Since the estimate for the resulting upper bound obtained in Lemma 2 is
minimized when $\delta$ is as large as possible, we choose the largest value
consistent with (2.20a,ba),

$\delta=(24\sqrt{3})^{2}\,R^{-2}.$ (2.30)

With this choice of $\delta$, conditions (2.14a,b,cc) and (2.20a,bb) require
$\varepsilon$ to satisfy

$\frac{27\,648}{R^{2}}\,\ln^{2}{\left(\frac{R}{24\sqrt{3}}\right)}\leq\varepsilon\leq\left(\frac{1}{48}\right)^{\frac{1}{3}}\ln^{-\frac{2}{3}}{\left(\frac{R}{24\sqrt{3}}\right)},$
(2.31)

which is possible for $\mbox{{R}}>\mbox{{R}}_{0}\simeq 1891.35$ (cf. Figure
3). For $\mbox{{R}}>\mbox{{R}}_{0}$, any choice of $\varepsilon$ in this range
is feasible. The optimal value could be determined at the expense of more
complicated algebra either by optimizing the full bound
$U(\beta,\tau,\lambda)$, or by deriving better $\varepsilon$-dependent
estimates for it. However, we expect that any $\varepsilon$-dependent terms
will contribute only higher-order corrections to our bound on
$\overline{\left\langle wT\right\rangle}$.

To conclude the proof of Theorem 1, there remains to verify that our choice of
$\delta$ is no larger than $\frac{1}{6}$ and that any $\varepsilon$ satisfying
(2.31) is no larger than $\frac{1}{3}$. It is easily checked that both
conditions hold when $\mbox{{R}}\geq\mbox{{R}}_{0}$ (see Figure 3 for an
illustration). For all such values of $R$, therefore, Lemma 2 and our choice
of $\delta$ yield the upper bound $\overline{\left\langle wT\right\rangle}\leq
U(\beta,\tau,\lambda)\leq\frac{1}{2}-cR^{-2}$ with $c=216$.

## 3 Bounds for the IH3 configuration

We now move on to studying IH convection in the IH3 configuration, where the
top boundary is maintained at constant (zero) temperature and the bottom
boundary is insulating. First, in section 3.1, we derive a bounding framework
for $\smash{\overline{\left\langle wT\right\rangle}}$ following steps similar
to those used for the IH1 case (cf. section 2.1). In section 3.2 we present
ansätze for $\tau$ and $\lambda$, with which we obtain crucial estimates in
section 3.3 , which give the bound in section 3.4. Throughout this section,
$T$ belongs to $\mathcal{H}_{3}$. Observe that this changes the set of
temperature fields over which the spectral constraint in definition 1 is
imposed. The notation $\mathcal{H}_{+}$ still denotes the subset of
temperature fields in $\mathcal{H}_{3}$ that are nonnegative pointwise almost
everywhere.

### 3.1 Bounding framework

Upper bounds on $\smash{\overline{\left\langle wT\right\rangle}}$ for the IH3
configuration can be derived using a quadratic auxiliary function
$\mathcal{V}\\{T\\}$ similar to that used for the IH1 case. Precisely, we
still take $\mathcal{V}$ to be defined as in (2.3), where the positive
constant $\beta$ and the piecewise-differentiable square-integrable function
$\tau(z)$ are tunable parameters. However, this time we impose only the
boundary condition

$\tau(1)=0.$ (3.1)

These changes result in the following family of parametrized upper bounds on
$\overline{\left\langle wT\right\rangle}$.

######  Proposition 2 (Bounding framework, IH3).

Suppose that the pair $(\beta,\tau)$ satisfies the spectral constraint (cf.
definition 1) and the boundary condition in (3.1). Further, let $\lambda\in
L^{2}(0,1)$ be a nondecreasing function such that $\lambda(0)=-1$. Then,

$\overline{\left\langle
wT\right\rangle}\leq\tfrac{1}{2}+\left\langle\tfrac{1}{4\beta}\left|\tau^{\prime}-\lambda-\beta
z\right|^{2}-\tau\right\rangle=:U(\tau,\lambda,\beta)$

###### Proof.

Proceeding as in the proof of Proposition 1 shows that

$\overline{\left\langle
wT\right\rangle}\leq\tfrac{1}{2}+\sup_{\begin{subarray}{c}T\in\mathcal{H}_{+}\\\
w=-R\Delta^{-2}\Delta_{h}T\end{subarray}}\left\langle-b\left|\nabla
T\right|^{2}-\tau^{\prime}wT+\left(\tau-\beta
z+1\right)\partial_{z}T-\tau\right\rangle.$ (3.2)

The supremum on the right-hand side can be evaluated using the convex duality
argument summarized in appendix B, leading to the equivalent inequality

$\overline{\left\langle
wT\right\rangle}\leq\tfrac{1}{2}+\inf_{\begin{subarray}{c}\lambda\in
L^{2}(0,1)\\\ \lambda\text{ \rm nondecreasing}\\\
\lambda\geq-1\end{subarray}}\left\langle\tfrac{1}{4\beta}\left|\tau^{\prime}-\lambda-\beta
z\right|^{2}-\tau\right\rangle.$ (3.3)

This clearly implies the upper bound stated in the proposition. ∎

### 3.2 Ansätze

The procedure for the proof of an upper bound on $\overline{\left\langle
wT\right\rangle}$ is the same as that employed for isothermal boundaries. We
construct $\beta>0$, $\tau(z)$ and $\lambda(z)$ that satisfy the conditions of
Proposition 2, while trying to minimize the corresponding bound
$U(\beta,\tau,\lambda)$. Due to the Neumann boundary condition on $T$ at
$z=0$, we can no longer employ the Poincaré estimates used in section 2.3 to
control the sign-indefinite term in the spectral constraint at the bottom
boundary. Instead we modify $\tau(z)$ in $(\delta,\frac{1}{2})$ to increase
slower than logarithmically in $z$ and use results established in [26]. The
function $\tau(z)$ is hence chosen to have the form

$\tau(z):=\begin{dcases}\delta-z,&0\leq z\leq\delta,\\\
\frac{A}{1-\alpha}\left(z^{1-\alpha}-\delta^{1-\alpha}\right)-A\ln{\left(\frac{1-z}{1-\delta}\right)},&\delta\leq
z\leq 1-\varepsilon,\\\
\frac{A\,B}{\varepsilon}\left(1-z\right),&1-\varepsilon\leq z\leq
1,\end{dcases}$ (3.4)

where

$\displaystyle
B=B(\epsilon,\delta,\alpha):=\frac{(1-\varepsilon)^{1-\alpha}-\delta^{1-\alpha}}{1-\alpha}+\ln{\left(\frac{1-\delta}{\varepsilon}\right)}.$
(3.5)

On the other hand, the Lagrange multiplier $\lambda(z)$ is still chosen to be

$\lambda(z):=\begin{dcases}-1,&0\leq z\leq\delta,\\\ 0,&\delta\leq z\leq
1.\end{dcases}$ (3.6)

These piecewise functions, sketched in Figure 4, are fully specified by the
bottom and top boundary layer widths $\delta,\varepsilon\in(0,\frac{1}{2})$,
the constant $A>0$, and the exponent $\alpha\in(0,1)$ driving the behaviour of
$\tau(z)$ in the bulk.

For $\beta$ we take

$\beta:=\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}\left\langle
z^{2}\right\rangle^{-\frac{1}{2}}=\sqrt{3}\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}.$
(3.7)

This choice is motivated by minimizing the right hand side of the estimate

$\frac{1}{4\beta}\left\langle|\tau^{\prime}-\lambda-\beta
z|^{2}\right\rangle\leq\frac{\beta}{2}\left\langle
z^{2}\right\rangle+\frac{1}{2\beta}\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle=\frac{1}{\sqrt{3}}\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}=\frac{\beta}{3},$
(3.8)

which is used in Lemma 6 below to estimate the value of the bound
$U(\beta,\tau,\lambda)$ from above when $\tau$ and $\lambda$ are define by
(3.4) and (3.6) respectively.

For any choice of the parameters $\delta$, $\varepsilon$, $A$, and $\alpha$,
the function $\tau$ satisfies the boundary conditions in (3.1), while
$\lambda$ is nondecreasing and satisfies the condition $\lambda(0)=-1$. Thus,
to establish 1.8 using Proposition 2 we need only specify parameter values
such that $U(\beta,\tau,\lambda)\leq\frac{1}{2}-O(R^{-4})$ while ensuring that
$(\beta,\tau)$ satisfy the spectral constraint. For the purposes of
simplifying the algebra in what follows, we shall fix

$A(\delta,\alpha)=\frac{2\sqrt{3}}{9}\sqrt{2\alpha-1}\,\delta^{\alpha+\frac{3}{2}}$
(3.9)

from the outset. This choice arises when insisting that the upper estimate on
$U(\beta,\tau,\lambda)$ derived in Lemma 6 below should be strictly less than
$\frac{1}{2}$ for suitable choices of $\delta$ and $\varepsilon$, at least
when all other constraints on these parameters are ignored.

$z$$\tau(z)$$1$$\delta$$\delta$$1-\varepsilon$$z$$\lambda(z)$$1$$\delta$$-1$
Figure 4: Sketch of the functions $\tau(z)$ (3.4) and $\lambda(z)$ in (3.6)
used to prove 1.8.

### 3.3 Preliminary Estimates

We now derive auxiliary results that simplify the choice of the exponent
$\alpha$ and of the boundary widths $\delta$ and $\varepsilon$. The first
gives estimates on the value of $\beta$ in (3.7).

######  Lemma 5 (Estimates on $\beta$).

Let $\tau(z)$, $\lambda(z)$ and $\beta$ be given by (3.4), (3.6) and (3.7)
with $A$ specified in (3.9). Suppose that $\alpha\in(\frac{1}{2},1)$ and the
boundary layer widths $\delta$ and $\varepsilon$ satisfy

$\delta\leq\frac{1}{3}\left(\frac{1}{2}\right)^{\frac{1}{2\alpha-1}},\qquad\varepsilon\leq\frac{1}{3},\qquad(2\alpha-1)\,\delta^{2\alpha-1}B(\epsilon,\delta,\alpha)^{2}\leq\varepsilon.$
(3.10a,b,c)

Then

$\frac{\sqrt{2}}{3}\,\delta^{2}\leq\beta\leq\frac{4}{3}\,\delta^{2}.$ (3.11)

######  Remark 8.

Condition (3.10a) and the bounds on $\alpha$ imposed in the Lemma imply that
$0\leq\delta\leq\frac{1}{6}$. These uniform bounds will be used repeatedly in
the following proofs.

###### Proof of Lemma 5.

It suffices to estimate
$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}$ from
above and below. For a lower bound, we can substitute our choices of $\tau$
and $\lambda$ and then estimate

$\displaystyle\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle=\int^{1}_{\delta}|\tau^{\prime}(z)|^{2}\textrm{d}z\geq\int^{1-\varepsilon}_{\delta}|\tau^{\prime}(z)|^{2}\textrm{d}z=A^{2}\int^{1-\varepsilon}_{\delta}\left|\frac{1}{z^{\alpha}}+\frac{1}{1-z}\right|^{2}\textrm{d}z.$

Dropping the positive term $1/(1-z)$ from the integrand and integrating the
rest gives

$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle\geq
A^{2}\left(\frac{\delta^{1-2\alpha}-(1-\varepsilon)^{1-2\alpha}}{2\alpha-1}\right).$

For every $\alpha\in(\frac{1}{2},1)$, the second term inside the parentheses
can be estimated upon observing that constraints (3.10a-b) imply

$(1-\varepsilon)^{1-2\alpha}\leq\left(\frac{2}{3}\right)^{1-2\alpha}\leq\frac{1}{2^{2\alpha}}\delta^{1-2\alpha}\leq\frac{1}{2}\delta^{1-2\alpha}.$
(3.12)

Thus, we obtain

$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle\geq
A^{2}\frac{\frac{1}{2}\delta^{1-2\alpha}}{2\alpha-1}.$ (3.13)

Taking the square root of (3.13) and using (3.9) gives
$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{\frac{1}{2}}\geq\frac{\sqrt{6}}{9}\,\delta^{2}$,
which combined with (3.7) proves the lower bound on $\beta$ stated in (3.11).

To prove the upper bound on $\beta$, we start by using the inequality
$(a+b)^{2}\leq 2a^{2}+2b^{2}$, evaluating exactly the integral of
$\tau^{\prime}-\lambda$, and dropping the negative terms to get

$\displaystyle\frac{\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle}{A^{2}}$
$\displaystyle=\int^{1-\varepsilon}_{\delta}\left|\frac{1}{z^{\alpha}}+\frac{1}{1-z}\right|^{2}\textrm{d}z+\frac{B^{2}}{\varepsilon}$
$\displaystyle\leq\int^{1-\varepsilon}_{\delta}\frac{2}{z^{2\alpha}}+\frac{2}{(1-z)^{2}}\textrm{d}z+\frac{B^{2}}{\varepsilon}$
(since $\alpha>\frac{1}{2}$)
$\displaystyle\leq\frac{2\delta^{1-2\alpha}}{2\alpha-1}+\frac{2}{\varepsilon}+\frac{B^{2}}{\varepsilon}.$
(3.14)

Using assumption (3.10c), the second and final term in (3.14) can be estimated
from above to arrive at

$\frac{\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle}{A^{2}}\leq\frac{\delta^{1-2\alpha}}{2\alpha-1}\left(2+\frac{2}{B^{2}}+1\right).$
(3.15)

Next, we observe that for all $\varepsilon\in(0,\frac{1}{3})$,
$\delta\in(0,\frac{1}{6})$, and $\alpha\in(\frac{1}{2},1)$ we can estimate

$B(\varepsilon,\delta,\alpha)\geq
B\left(\frac{1}{3},\frac{1}{6},\frac{1}{2}\right)=\frac{\sqrt{6}}{3}+\ln{\left(\frac{5}{2}\right)}>\sqrt{2},$

so $B^{-2}\leq 1/2$. Using this estimate in (3.15), taking a square root, and
substituting in the value of $A$ given in (3.9) leads to the inequality
$\left\langle|\tau^{\prime}-\lambda|^{2}\right\rangle^{1/2}\leq(4\sqrt{3}/9)\,\delta^{2}$.
Combining this with (3.7) yields the upper bound on $\beta$ stated in (3.11)
and concludes the proof of Lemma 5. ∎

The second auxiliary result of this section estimates the upper bound
$U(\beta,\tau,\lambda)$ on $\overline{\left\langle wT\right\rangle}$ given by
Proposition 2 using only the bottom boundary layer width $\delta$.

######  Lemma 6 (Estimates on $U(\beta,\tau,\lambda)$).

Let $\tau(z)$, $\lambda(z)$ and $\beta$ be specified by (3.4), (3.6) and (3.7)
with $A$ given by (3.9). Suppose $\alpha$ and the boundary layer widths
$\delta$, $\varepsilon$ satisfy the conditions of Lemma 5. Then,

$U(\beta,\tau,\lambda)\leq\frac{1}{2}-\frac{\delta^{2}}{18}.$ (3.16)

###### Proof.

Inequality (3.8) and the upper bound on $\beta$ from (3.11) give

$U(\beta,\tau,\lambda)\leq\frac{1}{2}+\frac{\beta}{3}-\left\langle\tau\right\rangle\leq\frac{1}{2}+\frac{4}{9}\delta^{2}-\left\langle\tau\right\rangle.$
(3.17)

The result follows upon observing that $\tau$ is non-negative, so
$\left\langle\tau\right\rangle=\int^{1}_{0}\tau(z)\textrm{d}z\geq\int^{\delta}_{0}\tau(z)\textrm{d}z=\frac{1}{2}\delta^{2}$.
∎

######  Remark 9.

The right hand side of (3.17) can be strictly smaller than $\frac{1}{2}$ when
$\delta^{2}$ is small only if $A\lesssim\delta^{\alpha+3/2}$. This observation
dictates the choice of $A$ in (3.9). For any fixed value of $R$, one should
choose $A\sim\delta^{\alpha+3/2}$ with a (possibly $R$-dependent) prefactor
that optimises the balance between the positive and negative terms, subject to
constraints on $A$, $\delta$ and all other parameters that ensure the spectral
constraint. To simplify our proof, however, we choose to fix this prefactor a
priori irrespective of $R$.

Our final auxiliary result gives the sufficient conditions on $\delta$ and
$\varepsilon$ that ensure the spectral constraint (cf. definition 1) is
satisfied.

######  Lemma 7 (Sufficient conditions for spectral constraint).

Let $\tau(z)$, $\lambda(z)$ and $\beta$ be specified by (3.4), (3.6) and (3.7)
with $A$ given by (3.9). Suppose that $\alpha$ and the boundary layer widths
satisfy the conditions of Lemma 5. Further, let

$h(\alpha)=2(2\alpha-1)(1-\alpha^{2}).$ (3.18)

and suppose that

$\displaystyle\delta\leq\varepsilon,$ (3.19a) $\displaystyle\delta\leq
h(\alpha)\,R^{-2},$ (3.19b)
$\displaystyle\frac{\sqrt{2\alpha-1}}{(1-\alpha)^{2}}\,\delta^{\alpha-\frac{1}{2}}\varepsilon^{3}\ln^{2}{(\delta^{-1})}\leq\frac{4}{3}\sqrt{6}R^{-1}.$
(3.19c)

Then, the pair $(\beta,\tau)$ satisfies the spectral constraint.

Unlike the analogous result obtained in section 2.3, Lemma 7 cannot be proven
using only the Hardy–Rellich inequalities stated in Lemma 4. The lack of a
fixed boundary temperature at $z=0$, makes it impossible to gain sufficient
control on the contribution of the bottom boundary layer to the quadratic form
in 2.5. This difficulty can be overcome using the following result, obtained
as a particular case of a more general analysis by Whitehead and Wittenberg
[26, Eqs. (59) & (77)], which upon setting (in their notation)
$\nu_{1}=\frac{\alpha}{2}-\nu_{2}$ and
$\nu_{2}=\frac{1}{2}\left(-1+\sqrt{2(1-\alpha^{2})}\right)$.

######  Lemma 8 (Adapted from [26]).

Fix $\alpha\in(\frac{1}{2},1)$ and $\mu,R>0$. Suppose
$\varphi(z):[0,1]\rightarrow\mathbb{R}$, be a non-negative function,
satisfying

$\left\langle\varphi(z)z^{1+\alpha/2}\right\rangle^{2}\leq\frac{\mu}{R}\frac{\sqrt{2}(3-\alpha)(2+\alpha)\sqrt{1-\alpha^{2}}}{3+\alpha}.$
(3.20)

Then, for every $w$ and $T$ solving (1.10) subject to the boundary conditions
(1.11),

$\left\langle\frac{1}{2}|\nabla
T|^{2}+\mu\frac{wT}{z^{\alpha}}-\varphi(z)wT\right\rangle\geq 0.$ (3.21)

We are now ready to prove Lemma 7.

###### Proof of Lemma 7.

Let $\mathbbm{1}_{(a,b)}$ denote the indicator function of the interval
$(a,b)$ and define the functions

$\displaystyle
f(z):=\left[1+\frac{A(\delta,\alpha)}{z^{\alpha}}+\frac{A(\delta,\alpha)}{1-z}\right]\mathbbm{1}_{(0,\delta)}(z),$
(3.22a) $\displaystyle
g(z):=\left[\frac{B(\varepsilon,\delta,\alpha)}{\varepsilon}+\frac{1}{z^{\alpha}}+\frac{1}{1-z}\right]\mathbbm{1}_{(1-\varepsilon,1)}(z).$
(3.22b)

Given our choice of $\tau$, we can rewrite the spectral constraint as

$0\leq\left\langle\beta|\nabla
T|^{2}+\tau^{\prime}wT\right\rangle=\mathcal{F}\\{T\\}+\mathcal{G}\\{T\\},$

where

$\displaystyle\mathcal{F}\\{T\\}:=\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+A\left\langle\frac{wT}{z^{\alpha}}\right\rangle\,-\left\langle
f(z)\,wT\right\rangle,$ (3.23a)
$\displaystyle\mathcal{G}\\{T\\}:=\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+A\left\langle\frac{wT}{1-z}\right\rangle-\,A\left\langle
g(z)\,wT\right\rangle.$ (3.23b)

Observe that $\mathcal{F}\\{T\\}$ and $\mathcal{G}\\{T\\}$ are functionals of
the temperature field only because $w$ is determined as a function of $T$ by
solving (1.10) subject to the boundary conditions in (1.11). We shall prove
that $\mathcal{F}\\{T\\}$ and $\mathcal{G}\\{T\\}$ are individually non-
negative for all temperatures $T$ from the space $\mathcal{H}_{3}$, which is
sufficient for the spectral constraint to hold.

To prove that $\mathcal{F}\\{T\\}\geq 0$, we apply Lemma 8 with
$\varphi(z)=f(z)/\beta$ and $\mu=A/\beta$, where $f$ is given by (3.22a) and
$A$ given by (3.9). We therefore need to check that

$\left\langle f(z)z^{1+\frac{\alpha}{2}}\right\rangle^{2}\leq\frac{\beta
A}{R}\frac{\sqrt{2}(3-\alpha)(2+\alpha)\sqrt{1-\alpha^{2}}}{3+\alpha}.$ (3.24)

To verify this inequality, we first bound from above the weighted integral on
the left-hand side. By assumption we have $0\leq z\leq\delta\leq\frac{1}{6}$
and $\alpha\in(\frac{1}{2},1)$, from which we obtain
$\frac{1}{1-z}\leq\frac{1}{z^{\alpha}}$. Using this estimate and the
definition of $A$ from (3.9) we can therefore estimate

$\left\langle
f(z)z^{1+\frac{\alpha}{2}}\right\rangle\leq\int^{\delta}_{0}z^{1+\frac{\alpha}{2}}+2Az^{1-\frac{\alpha}{2}}\textrm{d}z=\left[\frac{2}{4+\alpha}+\frac{8\sqrt{3}}{9}\frac{\sqrt{2\alpha-1}}{(4-\alpha)}\delta^{\frac{3}{2}}\right]\delta^{2+\frac{\alpha}{2}}.$

Using again that $0\leq\delta\leq\frac{1}{6}$ and $\alpha\in(\frac{1}{2},1)$,
the bracketed expression can be bounded from above to obtain

$\left\langle
f(z)z^{1+\frac{\alpha}{2}}\right\rangle\leq\left[\frac{4}{9}+\frac{8\sqrt{3}}{9}\cdot\frac{1}{3}\cdot\left(\frac{1}{6}\right)^{\frac{3}{2}}\right]\delta^{2+\frac{\alpha}{2}}<\frac{1}{2}\delta^{2+\frac{\alpha}{2}}.$
(3.25)

Next, we bound from below the right-hand-side of (3.24). Using the lower bound
on $\beta$ from Lemma 5, the definition (3.9) of $A$, and the fact that
$\alpha\in(\frac{1}{2},1)$, we have

$\displaystyle\frac{\beta A\sqrt{2}(3-\alpha)(2+\alpha)}{(3+\alpha)}$
$\displaystyle\geq\frac{4\sqrt{3}}{27}\delta^{\alpha+\frac{7}{2}}\sqrt{2\alpha-1}\cdot\frac{(3-\alpha)(2+\alpha)}{(3+\alpha)}$
$\displaystyle\geq\frac{5}{9\sqrt{3}}\delta^{\alpha+\frac{7}{2}}\sqrt{2\alpha-1}.$
(3.26)

Combining (3.25) and (3.26), we conclude that (3.24) holds if

$\delta^{\frac{1}{2}}\leq\frac{20}{9\sqrt{3}}\frac{\sqrt{(2\alpha-1)(1-\alpha^{2})}}{R}\leq\sqrt{2}\frac{\sqrt{(2\alpha-1)(1-\alpha^{2})}}{R},$

which is true because $\delta$ satisfies (3.19a) by assumption. This proves
that $\mathcal{F}\\{T\\}\geq 0$, as desired.

We now prove that $\mathcal{G}\\{T\\}$ is also nonnegative. This can be done
following the same steps used in section 2.3. The Hardy–Rellich inequality
(2.21b) gives

$\mathcal{G}\\{T\\}\geq\frac{\beta}{2}\left\langle|\nabla
T|^{2}\right\rangle+\frac{4A}{R}\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle-A\left\langle
g(z)wT\right\rangle.$ (3.27)

To estimate the last term, as before we use the inequality $\left\langle
T^{2}\right\rangle_{h}\leq\smash{(1-z)\left\langle|\nabla
T|^{2}\right\rangle}$, the Cauchy–Schwarz inequality, and Young’s inequality:

$\displaystyle\left\langle
g(z)wT\right\rangle\leq\frac{\beta}{2A}\left\langle|\nabla
T|^{2}\right\rangle+\frac{A}{2\beta}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle.$

Using the lower bound on $\beta$ from (3.11) gives

$\left\langle g(z)wT\right\rangle\leq\frac{\beta}{2A}\left\langle|\nabla
T|^{2}\right\rangle+\frac{3\sqrt{2}\,A}{4\delta^{2}}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle\left\langle\frac{w^{2}}{(1-z)^{3}}\right\rangle,$

which can be substituted into (3.27) along with the value of $A$ from (3.9) to
obtain

$\mathcal{G}\\{T\\}\geq\frac{\sqrt{6(2\alpha-1)}\,A}{6}\left(\frac{4\sqrt{6}}{\sqrt{2\alpha-1}\,R}-\delta^{\alpha-\frac{1}{2}}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle\right)\left\langle\frac{|w|^{2}}{(1-z)^{3}}\right\rangle.$
(3.28)

To conclude the argument we need to show that term in the parentheses is non-
negative. To demonstrate this, we first estimate from above the function
$g(z)$ given in (3.22b) on the interval $(1-\varepsilon,1)$. Our assumption
that $\delta\leq\varepsilon$ implies that $1-\varepsilon\leq 1-\delta\leq 1$
and $\ln(\frac{1}{\varepsilon})\leq\ln(\frac{1}{\delta})$. Thus, for all
$\delta\leq\frac{1}{6}$ and $\alpha\in(\frac{1}{2},1)$ the first term in
$g(z)$ can be bounded as

$\displaystyle\frac{B}{\varepsilon}\leq\frac{1}{\varepsilon}\left(\frac{1}{1-\alpha}+\ln{\left(\frac{1}{\delta}\right)}\right)$
$\displaystyle\leq\frac{2\ln{\left(\frac{1}{\delta}\right)}}{\varepsilon(1-\alpha)}.$
(3.29)

To estimate the other terms in $g(z)$, we observe that the assumptions
$\varepsilon\leq\frac{1}{3}$ and $\alpha\in(\frac{1}{2},1)$ imply that
$\frac{1}{z^{\alpha}}\leq\frac{1}{2(1-\alpha)(1-z)}$ and
$\frac{1}{1-z}\leq\frac{1}{2(1-\alpha)(1-z)}$. Consequently, we arrive at

$g(z)\leq\frac{2\ln{(\frac{1}{\delta})}}{\varepsilon(1-\alpha)}+\frac{1}{(1-\alpha)(1-z)}.$
(3.30)

Finally, using (3.30) and evaluating the integral in the parentheses of (3.28)
with the fact that $\ln(\frac{1}{\delta})\leq\ln^{2}{(\frac{1}{\delta})}$ and
$\frac{1}{3}\leq\frac{1}{3}\ln{(\frac{1}{\delta})}$ gives

$\displaystyle\delta^{\alpha-\frac{1}{2}}\left\langle
g(z)^{2}(1-z)^{4}\right\rangle$ $\displaystyle\leq$
$\displaystyle\delta^{\alpha-\frac{1}{2}}\int^{1}_{1-\varepsilon}\left(\frac{2\ln{\left(\frac{1}{\delta}\right)}}{\varepsilon(1-\alpha)}+\frac{1}{(1-\alpha)(1-z)}\right)^{2}(1-z)^{4}\,\textrm{d}z$
$\displaystyle=$
$\displaystyle\frac{\delta^{\alpha-\frac{1}{2}}\varepsilon^{3}}{(1-\alpha)^{2}}\left[\frac{4}{5}\ln^{2}{\left(\frac{1}{\delta}\right)}+\ln{\left(\frac{1}{\delta}\right)}+\frac{1}{3}\right]$
$\displaystyle\leq$
$\displaystyle\frac{3\,\delta^{\alpha-\frac{1}{2}}\varepsilon^{3}\ln^{2}{\left(\frac{1}{\delta}\right)}}{(1-\alpha)^{2}}$
$\displaystyle{(\text{by
\hyperref@@ii[e:delta_eps_lemma]{(\ref*{e:delta_eps_lemma}b)}})}\qquad$
$\displaystyle\leq$ $\displaystyle\frac{4\sqrt{6}}{\sqrt{2\alpha-1}\,R}.$
(3.31)

This concludes the proof of Lemma 7.

∎

### 3.4 Proof of 1.8

To prove 1.8, we only need to specify $R$-dependent values for $\alpha$ and
for the boundary layer widths $\delta$ and $\varepsilon$ that satisfy the
conditions of Lemmas 5, LABEL:, 6, LABEL: and 7.

(a)(b)

Figure 5: (a) Variation with R of the allowed values for the bottom boundary
layer width $\delta$ (3.32) ( ) and the feasible region of $\varepsilon$
(3.34) (shaded region). Also shown are uniform upper bounds of
$\delta\leq\frac{1}{6}$ ( ), and $\varepsilon\leq\frac{1}{3}$ ( ) imposed on
the variables. A black vertical line marks the Rayleigh number,
$\mbox{{R}}_{0}\approx 2960.89$ above which all constraints on are satisfied.
(b) Plot of the function $h(\alpha)$ (3.18) ( ). Shown also is the optimal
$\alpha^{*}=(1+\sqrt{13})/6$ ( ).

Motivated by the desire to minimize the upper bound on $U(\beta,\tau,\lambda)$
stated in Lemma 6, we choose

$\delta=h(\alpha^{*})R^{-2}$ (3.32)

where $\alpha^{*}=(1+\sqrt{13})/6$ is the unique maximizer of $h(\alpha)$ on
the interval $(\frac{1}{2},1)$ (see Figure 5(b)). With these choices,
conditions (3.10c) and (3.19b) require $\varepsilon$ to satisfy

$\displaystyle
c_{0}R^{-4\alpha^{*}+2}B(\delta,\varepsilon,\alpha^{*})^{2}\leq\varepsilon\leq
c_{2}R^{-\frac{2-2\alpha^{*}}{3}}\ln^{-\frac{2}{3}}{(\delta^{-1})},$ (3.33)

where $c_{0}$, $c_{1}$ and $c_{2}$ are non-negative constants independent of
R. Using the upper bound on $B$ from (3.29), it suffices to find $\varepsilon$
such that

$\displaystyle\frac{4c_{0}}{(1-\alpha^{*})^{2}}R^{-4\alpha^{*}+2}\ln^{2}{\left(\frac{R^{2}}{h(\alpha^{*})}\right)}\leq\varepsilon\leq
c_{2}R^{-\frac{2-2\alpha^{*}}{3}}\ln^{-\frac{2}{3}}\left(\frac{R^{2}}{h(\alpha^{*})}\right).$
(3.34)

Figure 5 shows that suitable values of $\varepsilon$ exist when
$\mbox{{R}}\geq\mbox{{R}}_{0}\approx 2960.89$. One can also check that for all
such values of R and any $\varepsilon$ in the range given by 3.34 one has
$\delta\leq\frac{1}{6}$, $\varepsilon\leq\frac{1}{3}$, and
$\delta\leq\varepsilon$. We have therefore verified all conditions of Lemmas
5, LABEL:, 6, LABEL: and 7.

To conclude the proof of 1.8, we simply substitute our choice of $\delta$ from
(3.32) into Lemma 6 to find the upper bound $\overline{\left\langle
wT\right\rangle}\leq U(\beta,\tau,\lambda)\leq\frac{1}{2}-cR^{-4}$, where
$c=\frac{h^{2}(\alpha^{*})}{18}\simeq 0.0107$.

######  Remark 10.

The top boundary layer width $\varepsilon$ is not uniquely determined in our
construction. Its optimal value could be obtained by considering more refined
estimates on $U(\beta,\tau,\lambda)$ than Lemma 6, but we expect such
estimates to provide only higher-order corrections to the eventual bound on
$\smash{\overline{\left\langle wT\right\rangle}}$.

## 4 Conclusions

We have proven upper bounds on the mean vertical convective heat transport
$\smash{\overline{\left\langle wT\right\rangle}}$ for two configurations of
infinite-Prandtl-number convection driven by uniform internal heating between
no-slip boundaries. In the first case, where both boundaries are held at a
constant temperature, we find $\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}-O(\mbox{{R}}^{-2})$ for all sufficiently
large R (cf. Theorem 1). This result implies that the outward heat fluxes
through the top and bottom are bounded by $\mathcal{F}_{T}\leq
1-O(\mbox{{R}}^{-2})$ and $\mathcal{F}_{B}\geq O(\mbox{{R}}^{-2})$,
respectively. In the second configuration, where the top boundary remains
isothermal but the bottom one is insulating (no-flux condition), we find
$\mbox{$\smash{\overline{\left\langle
wT\right\rangle}}$}\leq\frac{1}{2}-O(\mbox{{R}}^{-4})$ (cf. 1.8). In this
case, we conclude from (1.9) that the Nusselt number is bounded above by
$\mbox{{Nu}}\leq O(\mbox{{R}}^{4})$. Explicit suboptimal values for the
prefactors in the Rayleigh-dependent terms were also obtained (cf. remarks 1
and 2).

All of these results were derived using the background method, which we
formulated as a search over quadratic auxiliary functionals of the form (2.3)
and augmented using a minimum principle for the fluid’s temperature. Similar
to previous works on infinite-Prandtl-number Rayleigh–Bénard convection, the
background temperature fields used vary linearly in thin boundary layers, and
increase either logarithmically (IH1 configuration) or as a power law (IH3
configuration) in the bulk of the fluid layer. This bulk behaviour enables us
to use Hardy–Rellich inequalities from [14] (Lemma 4) and an integral estimate
from [26] (Lemma 8) that were originally developed in the context of
Rayleigh–Bénard convection. In contrast to the latter, however, our background
fields lack symmetry in the vertical direction, which reflects the lack of
vertical symmetry of IH convection problems.

In our choice of background fields, allowing the bottom boundary layer width
$\delta$ to be smaller than the top boundary layer width $\varepsilon$ is
essential to prove Theorems 1 and 1.8. For the IH1 configuration, forcing
$\delta=\varepsilon$ worsens the R-dependent correction to $\frac{1}{2}$ in
Theorem 1 to $O(\mbox{{R}}^{-2}\ln^{-2}{(\mbox{{R}})})$. For the IH3
configuration, instead, no upper bound on $\smash{\overline{\left\langle
wT\right\rangle}}$ that asymptotes to $\frac{1}{2}$ from below as R increases
can be obtained with our method of proof if $\delta=\varepsilon$. This
boundary layer asymmetry contrasts the construction of background fields for
IH convection at finite Pr [5], where taking $\delta\neq\varepsilon$ appears
to bring no qualitative improvement to the exponentially-varying upper bounds
on $\smash{\overline{\left\langle wT\right\rangle}}$ . We also stress that the
a priori uniform limits on the allowed values of $\delta$ and $\varepsilon$
imposed throughout sections 2 and 3 have been chosen with the only goal of
simplifying the algebra in our proofs. Varying these limits affects the
prefactors of the R-dependent terms in Theorems 1 and 1.8, as well as the
range of R values for which they hold. Both could be optimized further if
desired.

One crucial difference between our constructions for the IH1 and IH3
configurations is the leading-order behaviour of the background temperature
fields—or, more precisely, of the function $\tau(z)$—as the bottom boundary
layer edge is approached from the bulk region. For the IH1 configuration, it
suffices for $\tau$ to have the same logarithmic behaviour as the background
temperature fields used to study Rayleigh–Bénard convection [25]. For the IH3
configuration, however, this choice does not work due to the loss of control
on the temperature of the bottom boundary, and we are instead forced to take
$\tau(z)\sim z^{1-\alpha}$ with $\alpha\in(0,1)$. This modification was
already used in the context of Rayleigh–Bénard convection between imperfectly
conducting boundaries [26], where the optimal exponent $\alpha$ depended
logarithmically on the Rayleigh number. Within our proof, instead, the optimal
$\alpha$ is a constant. Whether this difference is due to our choice of
estimates or the inherent differences between Rayleigh-Bénard and IH
convection remains an open question.

More generally, we do not know whether the upper bounds on
$\smash{\overline{\left\langle wT\right\rangle}}$ stated in Theorems 1 and 1.8
are qualitatively sharp. To check if the $O(\mbox{{R}}^{-2})$ and
$O(\mbox{{R}}^{-4})$ corrections to the asymptotic value of $\frac{1}{2}$ are
optimal within our bounding framework, one could employ a variation of the
computational approach taken in [2] and optimize the tunable parameters
$\tau$, $\beta$, and $\lambda$ in full (see also [24] and references therein
for more details on the numerical optimization of bounds). A more interesting
but also more challenging problem is to identify which convective flows
maximize $\smash{\overline{\left\langle wT\right\rangle}}$ and the
corresponding optimal scaling of this quantity with R. Considerable insight in
this direction can be gained through (i) direct numerical simulations, which
to the best of our knowledge are currently lacking; (ii) the calculation of
steady but unstable solution of the Boussinesq equations (1.1) that, as
recently observed in the context of Rayleigh–Bénard convection [27, 28], may
transport heat more efficiently than turbulence; and (iii) the explicit design
of optimally-cooling flows [29, 30, 6]. Finally, it would be interesting to
investigate if more sophisticated PDE analysis techniques used for
Rayleigh–Bénard convection [31] can be extended to IH convection to
interpolate between the algebraic bounds on $\smash{\overline{\left\langle
wT\right\rangle}}$ proved in this paper for infinite-Pr fluids with the
finite-Pr exponential bounds obtained in [5].

Acknowledgements A.A. acknowledges funding by the EPSRC Centre for Doctoral
Training in Fluid Dynamics across Scales (award number EP/L016230/1). G.F. was
supported by an Imperial College Research Fellowship and would like to thank
the Isaac Newton Institute for Mathematical Sciences, Cambridge, for support
and hospitality during the programme “Mathematical aspects of turbulence:
where do we stand?” (EPSRC grant number EP/R014604/1) where work on this paper
was undertaken.

##  Appendix A Convex duality for the IH1 configuration

The equivalence between (2.7) and (2.8) follow from a relatively standard
convex duality argument. It will be enough to show that

$\sup_{T\in\mathcal{H}_{+}}-\Phi\\{T\\}=\inf_{\begin{subarray}{c}\lambda\in
L^{2}(0,1)\\\ \lambda\text{ \rm nondecreasing}\\\
\langle\lambda\rangle=-1\end{subarray}}\tfrac{1}{4\beta}\left\langle\left|\tau^{\prime}-\lambda-\beta\left(z-\tfrac{1}{2}\right)\right|^{2}\right\rangle,$
(A.1)

where

$\Phi\\{T\\}:=\left\langle\beta\left|\nabla
T\right|^{2}-R\tau^{\prime}(\Delta^{-2}\Delta_{h}T)T-(\tau^{\prime}-\beta
z)\partial_{z}T\right\rangle.$

To establish this identity, we start by rewriting the maximization on the
left-hand side as a minimization problem for the Legendre transform of $\Phi$.
Recall that the Legendre transform of a functional
$\Lambda:\mathcal{H}\to\mathbb{R}$ is the functional

$\Lambda^{*}\\{\mu\\}:=\sup_{T\in\mathcal{H}}\left(\mu\\{T\\}-\Lambda\\{T\\}\right),$

which acts on the dual space $\mathcal{H}^{*}$ of bounded linear functionals
$T\mapsto\mu\\{T\\}$ on $\mathcal{H}$. We shall write
$\mathcal{H}^{*}_{+}\subset\mathcal{H}^{*}$ for the subset of nonnegative
bounded linear functionals on $\mathcal{H}$, meaning that
$\mu\in\mathcal{H}^{*}_{+}$ if and only if $\mu\in\mathcal{H}^{*}$ and
$\mu\\{T\\}\geq 0$ for all $T\in\mathcal{H}_{+}$.

######  Lemma 9.

If the pair $(\beta,\tau)$ satisfies the spectral constraint,
$\sup_{T\in\mathcal{H}_{+}}-\Phi\\{T\\}=\inf_{\mu\in\mathcal{H}^{*}_{+}}\Phi^{*}\\{\mu\\}.$

###### Proof.

Define a functional $\Psi$ from $\mathcal{H}$ into
$\mathbb{R}\cup\\{+\infty\\}$ via

$\Psi\\{T\\}:=\begin{cases}0&T\in\mathcal{H}_{+},\\\
+\infty&\text{otherwise}.\end{cases}$

Its Legendre transform is

$\Psi^{*}\\{\mu\\}=\begin{cases}0&\text{if }-\mu\in\mathcal{H}^{*}_{+},\\\
+\infty&\text{otherwise}.\end{cases}$

We claim that

$\displaystyle\sup_{T\in\mathcal{H}_{+}}-\Phi\\{T\\}$
$\displaystyle=-\inf_{T\in\mathcal{H}_{+}}\Phi\\{T\\}$
$\displaystyle=-\inf_{T\in\mathcal{H}}\left(\Phi\\{T\\}+\Psi\\{T\\}\right)$
$\displaystyle=\inf_{\mu\in\mathcal{H}^{*}}\left(\Phi^{*}\\{-\mu\\}+\Psi^{*}\\{\mu\\}\right)$
$\displaystyle=\inf_{\mu\in\mathcal{H}^{*}_{+}}\Phi^{*}\\{\mu\\}.$

The first, second and fourth equalities are immediate consequences of the
definitions of $\inf$, $\sup$, $\Psi$, and $\Psi^{*}$. The third one, instead,
follows from the Fenchel–Rockafellar minmax theorem when $\mathcal{H}$ is
viewed as a Hilbert space with the inner product $\left\langle\nabla
T_{1}\cdot\nabla T_{2}\right\rangle$ and norm $\langle\left|\nabla
T\right|^{2}\rangle$. To apply this theorem as stated in [32, Theorem 1.12],
we need to verify that the functionals $\Phi$ and $\Psi$ are convex, and that
$\Phi$ is continuous (with respect to the norm on $\mathcal{H}$) at some
$T_{0}\in{\rm dom}(\Phi)\cap{\rm dom}(\Psi)=\mathcal{H}_{+}$.

The convexity of $\Psi$ is obvious, while that of $\Phi$ follows from the
assumption that the pair $(\beta,\tau)$ satisfies the spectral constraint (cf.
definition 1). To see this, write
$\Phi\\{T\\}=\mathcal{Q}\\{T,T\\}-(\tau^{\prime}-\beta z)\partial_{z}T$ where
$\mathcal{Q}$ is the bilinear form

$\mathcal{Q}\\{T_{1},T_{2}\\}:=\left\langle\beta\nabla T_{1}\cdot\nabla
T_{2}-R\tau^{\prime}(\Delta^{-2}\Delta_{h}T_{1})T_{2}\right\rangle,$

and observe that the spectral constraint ensures $\mathcal{Q}\\{T,T\\}\geq 0$
for all $T$ in the linear space $\mathcal{H}$. Thus, for any $\lambda\in[0,1]$
we can set $\bar{\lambda}=1-\lambda$ and estimate

$\displaystyle\Phi\\{\lambda T_{1}+\bar{\lambda}T_{2}\\}$
$\displaystyle=\mathcal{Q}\\{\lambda T_{1}+\bar{\lambda}T_{2},\lambda
T_{1}+\bar{\lambda}T_{2}\\}-(\tau^{\prime}-\beta z)\partial_{z}[\lambda
T_{1}+\bar{\lambda}T_{2}]$
$\displaystyle=\lambda\Phi\\{T_{1},T_{1}\\}+\bar{\lambda}\Phi\\{T_{2},T_{2}\\}-\lambda\bar{\lambda}\mathcal{Q}\\{T_{1}-T_{2},T_{1}-T_{2}\\}$
$\displaystyle\leq\lambda\Phi\\{T_{1},T_{1}\\}+\bar{\lambda}\Phi\\{T_{2},T_{2}\\},$

proving that $\Phi$ is convex. The continuity of $\Phi$ at any
$T_{0}\in\mathcal{H}_{+}$ follows because $\Phi$ is continuous on the whole
space $\mathcal{H}$. Indeed, the terms $\langle\left|\nabla
T\right|^{2}\rangle$ and $\langle(\tau^{\prime}-\beta z)\partial_{z}T\rangle$
in the expression for $\Phi\\{T\\}$ are clearly continuous on $\mathcal{H}$.
To see that the remaining term is also continuous, it is enough to establish
that $T_{k}\to T$ in $\mathcal{H}$ implies
$\varphi_{k}:=\smash{\Delta^{-2}\Delta_{h}T_{k}}\to\smash{\Delta^{-2}\Delta_{h}T}=:\varphi$
in $\mathcal{H}$. This can be shown by combining the Poincaré inequalities
$\langle\varphi^{2}\rangle\lesssim\langle\left|\Delta\varphi\right|^{2}\rangle$
and $\langle T^{2}\rangle\lesssim\langle\left|\nabla T\right|^{2}\rangle$ with
the estimate

$\langle\left|\Delta\varphi\right|^{2}\rangle=\left|\langle\varphi\Delta^{2}\varphi\rangle\right|=\left|\langle\varphi\Delta_{h}T\rangle\right|=\left|\langle\Delta_{h}\varphi
T\rangle\right|\leq\langle\left|\Delta\varphi\right|^{2}\rangle^{\frac{1}{2}}\langle\left|T\right|^{2}\rangle^{\frac{1}{2}},$

which implies $\langle\left|\Delta\varphi\right|^{2}\rangle\leq\langle
T^{2}\rangle$. This concludes the proof of Lemma 9. ∎

Next, we prove that since $\Phi\\{T\\}$ is invariant under horizontal
translations of the temperature field $T$, the minimization of its Legendre
transform $\Phi^{*}$ can be restricted to functionals
$\mu\in\mathcal{H}^{*}_{+}$ that are translation invariant. Specifically, for
any real numbers $r,s$ define the translation map
$\mathscr{S}_{r,s}:\mathcal{H}\to\mathcal{H}$ and its adjoint
$\mathscr{S}_{r,s}^{*}:\mathcal{H}^{*}\to\mathcal{H}^{*}$ via

$\displaystyle\mathscr{S}_{r,s}T$ $\displaystyle:=T(x+r,y+s,z)$
$\displaystyle\mathscr{S}_{r,s}^{*}\mu$
$\displaystyle:=\mu\circ\mathscr{S}_{r,s}.$

The functional $\mu\in\mathcal{H}^{*}$ is translation invariant if
$\mathscr{S}_{r,s}^{*}\mu=\mu$ for all $r$ and $s$.

######  Lemma 10.

Suppose $\Phi:\mathcal{H}\to\mathbb{R}$ satisfies
$\Phi\\{\mathscr{S}_{r,s}T\\}=\Phi\\{T\\}$ for all $r,s\in\mathbb{R}$. Then,

$\inf_{\mu\in\mathcal{H}^{*}}\Phi^{*}\\{\mu\\}=\inf_{\begin{subarray}{c}\mu\in\mathcal{H}^{*}\\\
\mu\text{ \rm transl. inv.}\end{subarray}}\Phi^{*}\\{\mu\\}.$

###### Proof.

It suffices to show that for every $\nu\in\mathcal{H}^{*}$ there exists a
translation-invariant $\mu\in\mathcal{H}^{*}$ such that
$\Phi^{*}\\{\mu\\}=\Phi^{*}\\{\nu\\}$. Since every temperature field
$T\in\mathcal{H}$ is horizontally periodic, such a $\mu$ can be constructed
simply by averaging the functionals $\mathscr{S}_{r,s}^{*}\nu$ over horizontal
translations $r$ and $s$, i.e., by letting

$\mu(T):=\fint_{r,s}\mathscr{S}_{r,s}^{*}\nu\\{T\\}=\fint_{r,s}\nu\\{\mathscr{S}_{r,s}T\\}\qquad\forall
T\in\mathcal{H}.$

To show that $\Phi^{*}\\{\mu\\}=\Phi^{*}\\{\nu\\}$, we establish the
complementary inequalities $\Phi^{*}\\{\mu\\}\leq\Phi^{*}\\{\nu\\}$ and
$\Phi^{*}\\{\mu\\}\geq\Phi^{*}\\{\nu\\}$. For the first one, use the
translation invariance of $\Phi$ and the definition of the Legendre transform
to estimate

$\nu\\{\mathscr{S}_{r,s}T\\}-\Phi\\{T\\}=\nu\\{\mathscr{S}_{r,s}T\\}-\Phi\\{\mathscr{S}_{r,s}T\\}\leq\Phi^{*}\\{\nu\\}$

for all $T\in\mathcal{H}$ and all horizontal shifts $r,s\in\mathbb{R}$.
Averaging over horizontal shifts shows that
$\mu\\{T\\}-\Phi\\{T\\}\leq\Phi^{*}\\{\nu\\}$ for all $T\in\mathcal{H}$, which
implies $\Phi^{*}\\{\mu\\}\leq\Phi^{*}\\{\nu\\}$.

To obtain the reverse inequality observe that, by definition of $\Phi^{*}$,
for any $\varepsilon>0$ there exists $T_{\varepsilon}\in\mathcal{H}$ such that
$\nu\\{T_{\varepsilon}\\}-\Phi\\{T_{\varepsilon}\\}\geq\Phi^{*}\\{\nu\\}-\varepsilon$.
Then, since $\Phi$ is translation invariant,

$\displaystyle\Phi^{*}\\{\nu\\}-\varepsilon$
$\displaystyle\leq\nu\\{T_{\varepsilon}\\}-\Phi\\{T_{\varepsilon}\\}$
$\displaystyle=(\mathscr{S}_{r,s}^{*}\nu)\\{\mathscr{S}_{-r,-s}T_{\varepsilon}\\}-\Phi\\{\mathscr{S}_{-r,-s}T_{\varepsilon}\\}$
$\displaystyle\leq\Phi^{*}\\{\mathscr{S}_{r,s}^{*}\nu\\}.$

Upon averaging this inequality over all horizontal shifts $r,s\in\mathbb{R}$
and applying Jensen’s inequality to $\Phi^{*}$, which is concave because it is
the supremum of linear functions, we find

$\Phi^{*}(\nu)-\varepsilon\leq\fint_{r,s}\Phi^{*}\left(\mathscr{S}_{r,s}^{*}\nu\right)\leq\Phi^{*}\\!\left(\fint_{r,s}\mathscr{S}_{r,s}^{*}\nu\right)=\Phi^{*}(\mu).$

Letting $\varepsilon\to 0$ yields $\Phi^{*}\\{\mu\\}\geq\Phi^{*}\\{\nu\\}$, as
desired. Lemma 10 is therefore proved. ∎

To establish identity (A.1) we now need to show that its right-hand side
coincides with the infimum of $\Phi^{*}$ over translation-invariant
functionals $\mu\in\mathcal{H}^{*}_{+}$. For this, we use a characterization
of such $\mu$ established in [2, Appendix C].

######  Lemma 11.

Let $\mathcal{H}^{*}_{+}$ be the set of positive linear functionals on the
temperature space $\mathcal{H}$ defined in (2.1a). If
$\mu\in\mathcal{H}^{*}_{+}$ is translation invariant, there exists a
nondecreasing function $\lambda\in L^{2}(0,1)$ with
$\left\langle\lambda\right\rangle=-1$ such that
$\mu\\{T\\}=\left\langle-{\lambda}(z)\,\partial_{z}T\right\rangle$.

Thanks to this representation, all that remains to do is to calculate

$\Phi^{*}\\{\mu\\}=\sup_{T\in\mathcal{H}}\left\langle(\tau^{\prime}-\lambda+\beta
z)\partial_{z}T-\beta\left|\nabla
T\right|^{2}+R\tau^{\prime}(\Delta^{-2}\Delta_{h}T)T\right\rangle.$ (A.2)

To solve this maximization problem, let $\eta(z)=\langle T\rangle_{h}(z)$ be
the horizontal mean of $T$ and set $\xi=T-\eta$. Since $\Delta_{h}\eta=0$ and
$\langle\xi\rangle_{h}(z)=0$ by construction, we can therefore substitute
$T=\eta+\xi$ in A.2 and solve the equivalent problem

$\Phi^{*}\\{\mu\\}=\sup_{\begin{subarray}{c}\eta=0\text{ if }z\in\\{0,1\\}\\\
\xi=0\text{ if }z\in\\{0,1\\}\\\
\langle\xi\rangle_{h}=0\end{subarray}}\Big{\\{}\left\langle(\tau^{\prime}-\lambda+\beta
z)\eta^{\prime}-\beta\left|\eta^{\prime}\right|^{2}-\beta\left|\nabla\xi\right|^{2}+R\tau^{\prime}(\Delta^{-2}\Delta_{h}\xi)\xi\right\rangle\\\\[-19.37492pt]
\left\langle(\tau^{\prime}-\lambda+\beta
z)\partial_{z}\xi-2\beta\,\eta^{\prime}\,\partial_{z}\xi+R\tau^{\prime}\eta\,(\Delta^{-2}\Delta_{h}\xi)\right\rangle\Big{\\}}.$
(A.3)

The boundary conditions on $\eta$ and $\xi$ follow from those on $T$. The
three terms on the second line vanish identically because
$\langle\xi\rangle_{h}(z)=0$ at all $z\in[0,1]$. To verify this claim, observe
that

$\left\langle
f(z)\partial_{z}\xi\right\rangle=\int_{0}^{1}f(z)\left\langle\xi\right\rangle_{h}^{\prime}(z){\rm
d}z=0$

for any function $f(z)$ that depends only on the vertical direction.
Similarly, one can show that

$\left\langle R\tau^{\prime}\eta\,(\Delta^{-2}\Delta_{h}\xi)\right\rangle=0$

because the function
$w=-R\smash{\Delta^{-2}\Delta_{h}T}=-R\smash{\Delta^{-2}\Delta_{h}\xi}$ also
has zero horizontal mean. Indeed, taking the horizontal average of (1.10)
yields the ODE $\langle w\rangle_{h}^{\prime\prime}(z)=0$, whose only solution
satisfying the boundary conditions in (1.11) is $\langle w\rangle_{h}(z)=0$.
The minimization in (A.3) therefore simplifies into

$\displaystyle\Phi^{*}\\{\mu\\}=\sup_{\begin{subarray}{c}\eta=0\text{ if
}z\in\\{0,1\\}\\\ \xi=0\text{ if }z\in\\{0,1\\}\\\
\langle\xi\rangle_{h}=0\end{subarray}}\left\langle(\tau^{\prime}-\lambda-\beta
z)\eta^{\prime}-\beta\left|\eta^{\prime}\right|^{2}-\beta\left|\nabla\xi\right|^{2}-R\tau^{\prime}w\xi\right\rangle.$
(A.4)

Since the pair $(\beta,\tau)$ was assumed to satisfy the spectral constraint
(cf. definition 1), the choice $\xi=0$ is optimal. The optimal $\eta$,
instead, satisfies the Euler–Lagrange equation
$2\beta\eta^{\prime\prime}=(\tau^{\prime}-\lambda-\beta z)^{\prime}$. Solving
this equation using the boundary conditions, the constraints $\tau(0)=1$ and
$\tau(1)=0$, and the normalization $\langle\lambda\rangle=-1$ gives
$\eta^{\prime}=\frac{1}{2b}[\tau^{\prime}-\lambda-b(z-\frac{1}{2})]$, which
can be substituted back into (A.4) to give

$\Phi^{*}\\{\mu\\}=\tfrac{1}{4\beta}\langle\left|\tau^{\prime}-\lambda^{\prime}-\beta\left(z-\tfrac{1}{2}\right)\right|^{2}\rangle.$
(A.5)

Minimizing the left-hand side over translation invariant $\mu$ in
$\mathcal{H}^{*}_{+}$ is the same as minimizing the right-hand side over
$\lambda$ satisfying the conditions in Lemma 11, which is exactly the problem
on the right-hand side of (A.1).

##  Appendix B Convex duality for the IH3 configuration

The equivalence between the upper bounds 3.2 and 3.3 for the IH3 configuration
follows from the identity

$\sup_{T\in\mathcal{H}_{+}}-\Phi\\{T\\}=\inf_{\begin{subarray}{c}\lambda\in
L^{2}(0,1)\\\ \lambda\text{ \rm nondecreasing}\\\
\lambda\geq-1\end{subarray}}\left\langle\tfrac{1}{4\beta}\left|\tau^{\prime}-\lambda-\beta
z\right|^{2}\right\rangle,$ (B.1)

where

$\Phi\\{T\\}:=\left\langle\beta\left|\nabla
T\right|^{2}-R\tau^{\prime}(\Delta^{-2}\Delta_{h}T)T-\left(\tau^{\prime}-\beta
z+1\right)\partial_{z}T\right\rangle.$ (B.2)

This identity can be proven using a convex duality argument analogous to that
in appendix A. Indeed, Lemmas 9 and 10 apply to the functional $\Phi$
considered in this section with no changes to their proofs. Consequently,

$\sup_{T\in\mathcal{H}_{+}}-\Phi\\{T\\}=\inf_{\begin{subarray}{c}\mu\in\mathcal{H}^{*}\\\
\mu\text{ \rm transl. inv.}\end{subarray}}\Phi^{*}\\{\mu\\}.$ (B.3)

To calculate the Legendre transform $\Phi^{*}$, however, we must replace Lemma
11 with a different characterization of translation-invariant linear
functionals $\mu\in\mathcal{H}^{*}_{+}$. This is due to the different boundary
conditions imposed on the temperature space $\mathcal{H}$.

######  Lemma 12.

Let $\mathcal{H}^{*}_{+}$ be the set of positive linear functionals on the
temperature space $\mathcal{H}$ defined in (2.1a). If
$\mu\in\mathcal{H}^{*}_{+}$ is translation invariant, there exists a
nondecreasing function $\lambda\in L^{2}(0,1)$ nonnegative almost everywhere
and such that
$\mu\\{T\\}=\left\langle-{\lambda}(z)\,\partial_{z}T\right\rangle$.

###### Proof.

Straightforward modifications to the proof of [2, Lemma 3] reveal that any
translation-invariant $\mu\in\mathcal{H}^{*}$ admits the representation
$\mu\\{T\\}=\left\langle-{\lambda}(z)\,\partial_{z}T\right\rangle$ for some
function $\lambda\in L^{2}(0,1)$. If $\mu$ is also positive, then the argument
in the proof of [2, Lemma 2] shows that $\lambda$ must be nondecreasing. To
see that we must have $\lambda(z_{0})\geq 0$ at almost every $z_{0}\in(0,1)$,
fix $\varepsilon>0$ sufficiently small and consider the temperature field
$T_{\varepsilon}\in\mathcal{H}_{+}$ given by

$T_{\varepsilon}(\boldsymbol{x}):=\begin{cases}2&z\in(0,z_{0}-\varepsilon),\\\
2-\varepsilon^{-1}(z-z_{0}+\varepsilon)&z\in(z_{0}-\varepsilon,z_{0}+\varepsilon),\\\
0&z\in(z_{0}+\varepsilon,1).\end{cases}$

Then, since $\mu$ is a positive functional by assumption,

$0\leq\mu(T_{\varepsilon})=\frac{1}{\varepsilon}\int_{z_{0}-\varepsilon}^{z_{0}+\varepsilon}\lambda(z){\rm
d}z.$

By Lebesgue’s differentiation theorem, the right-hand side tends to
$2\lambda(z_{0})$ for almost all $z_{0}\in(0,1)$ as $\varepsilon\to 0$. Thus,
we must have $\lambda\geq 0$ almost everywhere on $(0,1)$. ∎

To conclude the argument, we need to calculate $\Phi^{*}\\{\mu\\}$ for
translation-invariant $\mu\in\mathcal{H}^{*}_{+}$, which by Lemma 12 is given
by

$\Phi^{*}\\{\mu\\}=\sup_{T\in\mathcal{H}}\left\langle\left(\tau^{\prime}-\lambda-\beta
z+1\right)\partial_{z}T-\beta\left|\nabla
T\right|^{2}+R\tau^{\prime}(\Delta^{-2}\Delta_{h}T)T\right\rangle.$

Since the pair $(\beta,\tau)$ was assumed to satisfy the spectral constraint,
this maximization problem can be restricted to temperature fields that depend
only on the vertical coordinate $z$ (this can be proven by splitting $T$ into
its horizontal mean $\eta$ and a perturbation $\xi$ with zero horizontal mean,
as outlined at the end of appendix A). The optimal value can then be shown to
be

$\Phi^{*}\\{\mu\\}=\tfrac{1}{4\beta}\left\langle\left|\tau^{\prime}-\lambda-\beta
z+1\right|^{2}\right\rangle$

and can be substituted into (B.3) to arrive at

$\sup_{T\in\mathcal{H}_{+}}-\Phi\\{T\\}=\inf_{\begin{subarray}{c}\lambda\in
L^{2}(0,1)\\\ \lambda\text{ \rm nondecreasing}\\\ \lambda\geq
0\end{subarray}}\tfrac{1}{4\beta}\left\langle\left|\tau^{\prime}-\lambda-\beta
z+1\right|^{2}\right\rangle.$

Changing the optimization variable on the right-hand side to
$\tilde{\lambda}=\lambda-1$ yields (B.1).

## References

  * Arslan et al. [2021a] A. Arslan, G. Fantuzzi, J. Craske, and A. Wynn.  Bounds for internally heated convection with fixed boundary heat flux.  _J. Fluid Mech._ , 922:R1, 2021a.  (doi:10.1017/jfm.2021.527).
  * Arslan et al. [2021b] A. Arslan, G. Fantuzzi, J. Craske, and A. Wynn.  Bounds on heat transport for convection driven by internal heating.  _J. Fluid Mech._ , 919:A15, 2021b.  (doi:10.1017/jfm.2021.360).
  * Wang et al. [2020] Q. Wang, D. Lohse, and O. Shishkina.  Scaling in internally heated convection: a unifying theory.  _Geophysical Research Letters_ , 47:e2020GL091198, 2020\.  (doi:10.1029/2020GL091198).
  * Creyssels [2021] M. Creyssels.  Model for thermal convection with uniform volumetric energy sources.  _J. Fluid Mech._ , 919:A13, 2021.  (doi:10.1017/jfm.2021.392).
  * Kumar et al. [2022] A. Kumar, A. Arslan, G. Fantuzzi, J. Craske, and A. Wynn.  Analytical bounds on the heat transport in internally heated convection.  _J. Fluid Mech._ , 938:A26, 2022.  (doi:10.1017/jfm.2022.170).
  * Tobasco [2022] I. Tobasco.  Optimal cooling of an internally heated disc.  _Phil. Trans. Roy. Soc. A_ , 310(1):20210040, 2022\.  (doi:10.1098/rsta.2021.0040).
  * Emanuel [1994] K. A. Emanuel.  _Atmospheric convection_.  Oxford University Press, 1994.
  * Schubert et al. [2001] G. Schubert, D. L. Turcotte, and P. Olson.  _Mantle convection in the Earth and planets_.  Cambridge University Press, 2001.
  * Mulyukova and Bercovici [2020] E. Mulyukova and D. Bercovici.  Mantle convection in terrestrial planets.  _Oxford Research Encyclopedia of Planetary Science_ , 03 2020.  (doi:10.1093/acrefore/9780190647926.013.109).
  * Miquel et al. [2019] B. Miquel, S. Lepot, V. Bouillaut, and B. Gallet.  Convection driven by internal heat sources and sinks: Heat transport beyond the mixing-length or “ultimate” scaling regime.  _Phys. Rev. Fluids_ , 4(12):121501, 2019.  (doi:10.1103/PhysRevFluids.4.121501).
  * Bouillaut et al. [2019] V. Bouillaut, S. Lepot, S. Aumaître, and B. Gallet.  Transition to the ultimate regime in a radiatively driven convection experiment.  _J. Fluid Mech._ , 861, 2019.  (doi:10.1017/jfm.2018.972).
  * Doering and Constantin [1996] C. R. Doering and P. Constantin.  Variational bounds on energy dissipation in incompressible flows. III. Convection.  _Phys. Rev. E_ , 53(6):5957, 1996.  (doi:10.1103/PhysRevE.53.5957).
  * Lu et al. [2004] L. Lu, C. R. Doering, and F. H. Busse.  Bounds on convection driven by internal heating.  _J. Math. Phys._ , 45(7):2967–2986, 2004.  (doi:10.1063/1.1760845).
  * Whitehead and Doering [2011] J. P. Whitehead and C. R. Doering.  Internal heating driven convection at infinite Prandtl number.  _J. Math. Phys._ , 52(9):093101, 2011.  (doi:10.1063/1.3637032).
  * Goluskin [2016] D. Goluskin.  _Internally heated convection and Rayleigh-Bénard convection_.  (doi:10.1007/978-3-319-23941-5).
  * Whitehead and Doering [2012] J. P. Whitehead and C. R. Doering.  Rigid bounds on heat transport by a fluid between slippery boundaries.  _J. Fluid Mech._ , 707:241–259, 2012.  (doi:10.1017/jfm.2012.274).
  * Goluskin and Spiegel [2012] D. Goluskin and E. A. Spiegel.  Convection driven by internal heating.  _Phys. Lett. A_ , 377(1-2):83–92, 2012.  (doi:10.1016/j.physleta.2012.10.037).
  * Doering and Constantin [1994] C. R. Doering and P. Constantin.  Variational bounds on energy dissipation in incompressible flows: shear flow.  _Phys. Rev. E_ , 49(5):4087, 1994.  (doi:10.1103/PhysRevE.49.4087).
  * Constantin and Doering [1995] P. Constantin and C. R. Doering.  Variational bounds on energy dissipation in incompressible flows. II. Channel flow.  _Phys. Rev. E_ , 51(4):3192, 1995.  (doi:10.1103/PhysRevE.51.3192).
  * Chernyshenko et al. [2014] S. I. Chernyshenko, P. J. Goulart, D. Huang, and A. Papachristodoulou.  Polynomial sum of squares in fluid dynamics: a review with a look ahead.  _Phil. Trans. Roy. Soc. A_ , 372(2020):20130350, 2014.  (doi:10.1098/rsta.2013.0350).
  * Chernyshenko [2022] S. I. Chernyshenko.  Relationship between the methods of bounding time averages.  _Phil. Trans. Roy. Soc. A_ , 380(1):20210044., 2022.  (doi:10.1098/rsta.2021.0044).
  * Tobasco et al. [2018] I. Tobasco, D. Goluskin, and C. R. Doering.  Optimal bounds and extremal trajectories for time averages in nonlinear dynamical systems.  _Phys. Lett. A_ , 382(6):382–386, 2018.  (doi:10.1016/j.physleta.2017.12.023).
  * Rosa and Temam [2020] R. Rosa and R. M. Temam.  Optimal minimax bounds for time and ensemble averages of dissipative infinite-dimensional systems with applications to the incompressible Navier–Stokes equations.  arXiv:2010.06730 [math.AP], 2020.
  * Fantuzzi et al. [2022] G. Fantuzzi, A. Arslan, and A. Wynn.  The background method: Theory and computations.  _Phil. Trans. Roy. Soc. A_ , 380(1):20210038, 2022\.  (doi:10.1098/rsta.2021.0038).
  * Doering et al. [2006] C. R. Doering, F. Otto, and M. G. Reznikoff.  Bounds on vertical heat transport for infinite-Prandtl-number Rayleigh–Bénard convection.  _J. Fluid Mech._ , 560:229–241, 2006.  (doi:10.1017/S0022112006000097).
  * Whitehead and Wittenberg [2014] J. P. Whitehead and R. W. Wittenberg.  A rigorous bound on the vertical transport of heat in Rayleigh–Bénard convection at infinite Prandtl number with mixed thermal boundary conditions.  _J. Math. Phys._ , 55(9):093104, 2014.  (doi:10.1063/1.4896223).
  * Wen et al. [2020] B. Wen, D. Goluskin, M. Leduc, G. P. Chini, and C. R. Doering.  Steady Rayleigh-Bénard convection between stress-free boundaries.  _J. Fluid Mech._ , 905:R4(1–13), 2020.  (doi:10.1017/jfm.2020.812).
  * Wen et al. [2022] B. Wen, D. Goluskin, and C. R. Doering.  Steady rayleigh–bénard convection between no-slip boundaries.  _J. Fluid Mech._ , 933:R4, 2022.  (doi:10.1017/jfm.2021.1042).
  * Tobasco and Doering [2017] I. Tobasco and C. R. Doering.  Optimal wall-to-wall transport by incompressible flows.  _Phys. Rev Lett._ , 118(26):264502, 2017.  (doi:10.1103/PhysRevLett.118.264502).
  * Doering and Tobasco [2019] C. R. Doering and I. Tobasco.  On the optimal design of wall-to-wall heat transport.  _Comm. Pure Appl. Math._ , 72(11):2385–2448, 2019\.  (doi:10.1002/cpa.21832).
  * Choffrut et al. [2016] A. Choffrut, C. Nobili, and F. Otto.  Upper bounds on Nusselt number at finite Prandtl number.  _J. Diff. Eqs._ , 260(4):3860–3880, 2016.  (doi:10.1016/j.jde.2015.10.051).
  * Brezis [2010] H. Brezis.  _Functional Analysis, Sobolev Spaces and Partial Differential Equations_.  Springer, 2010.  (doi:10.1007/978-0-387-70914-7).