2 The model

We model the system using the Boussinesq equations and make all variables non-dimensional using $h$ , $h/\unicode[STIX]{x1D705}$ , and $h\unicode[STIX]{x1D6FD}$ , respectively, as the length, time and temperature scales (Otero et al. Reference Otero, Wittenberg, Worthing and Doering2002). The non-dimensional velocity $\boldsymbol{u}(x,y,z,t)$ , pressure $p(x,y,z,t)$ , and perturbations $\unicode[STIX]{x1D703}(x,y,z,t)$ from the conductive temperature profile $T_{c}=-z$ then satisfy (Otero et al. Reference Otero, Wittenberg, Worthing and Doering2002)

where $\boldsymbol{e}_{z}$ is the unit vector in the $z$ direction and $R=\unicode[STIX]{x1D6FC}g\unicode[STIX]{x1D6FD}h^{4}/(\unicode[STIX]{x1D708}\unicode[STIX]{x1D705})$ is the Rayleigh number based on the imposed boundary heat flux. Note that $\mathit{R}$ is related to the Rayleigh number based on the (unknown) mean temperature drop, $\mathit{Ra}$ , by $\mathit{R}=\mathit{Ra}\,\mathit{Nu}$ (Otero et al. Reference Otero, Wittenberg, Worthing and Doering2002). The domain is periodic in the horizontal ( $x$ , $y$ ) directions and the vertical BCs are

Since the average vertical heat flux across the layer is fixed to 1 in non-dimensional units, convection reduces the mean temperature difference between the top and bottom plates and hence the mean conductive heat flux $\overline{\langle -\unicode[STIX]{x2202}_{z}T\rangle }=\overline{1-\langle \unicode[STIX]{x2202}_{z}\unicode[STIX]{x1D703}\rangle }$ (here and throughout this work overlines denote averages over infinite time, while angle brackets denote volume averages). The Nusselt number – the ratio of the average vertical heat flux and the mean conductive flux – is then given by (see also Otero et al. Reference Otero, Wittenberg, Worthing and Doering2002)

3 Upper bound formulation

When $\mathit{R}<120$ , conduction is globally asymptotically stable and $\mathit{Nu}=1$ (Chapman & Proctor Reference Chapman and Proctor1980; Goluskin Reference Goluskin2016). For $\mathit{R}>120$ , convection sets in (Hurle, Jakeman & Pike Reference Hurle, Jakeman and Pike1967) and we look for a positive lower bound $L$ on $\overline{1-\langle \unicode[STIX]{x2202}_{z}\unicode[STIX]{x1D703}\rangle }$ , implying $\mathit{Nu}\leqslant 1/L$ . To find $L$ we use the background method (Doering & Constantin Reference Doering and Constantin1994; Constantin & Doering Reference Constantin and Doering1995; Doering & Constantin Reference Doering and Constantin1996) but we formulate it in the language of the auxiliary functional method (Chernyshenko et al. Reference Chernyshenko, Goulart, Huang and Papachristodoulou2014; Chernyshenko Reference Chernyshenko2017) because of its conceptual simplicity: it relies on one simple inequality, rather than a seemingly ad hoc manipulation of the governing equations.

The analysis starts with the observation that any uniformly bounded and differentiable time-dependent functional ${\mathcal{V}}(t)={\mathcal{V}}\{\unicode[STIX]{x1D703}(\cdot ,t),\boldsymbol{u}(\boldsymbol{\cdot },t)\}$ satisfies $\overline{\text{d}{\mathcal{V}}/\text{d}t}=0$ . Consequently, to prove that $\overline{1-\langle \unicode[STIX]{x2202}_{z}\unicode[STIX]{x1D703}\rangle }\geqslant L$ it suffices to show that at any instant in time

Using the ideas outlined by Chernyshenko (Reference Chernyshenko2017), it can be shown that constructing a background temperature field in the ‘classical’ background method analysis is equivalent to finding constants $a$ , $b$ and $L$ and a function $\unicode[STIX]{x1D711}(z)$ such that (3.1) holds for

We assume that $\boldsymbol{u}$ and $\unicode[STIX]{x1D703}$ are sufficiently regular in time to ensure differentiability of this functional, while uniform boundedness can be proved using estimates similar to those presented in this paper (we do not give a full proof in this work due to space limitations, but outline the argument in appendix A).

The functional ${\mathcal{S}}\{\unicode[STIX]{x1D703}(\cdot ,t),\boldsymbol{u}(\boldsymbol{\cdot },t)\}$ corresponding to (3.2) can be expressed in terms of $\boldsymbol{u}$ and $\unicode[STIX]{x1D703}$ using (2.1a )–(2.1c ). Integrating the volume average $\langle \boldsymbol{u}\boldsymbol{\cdot }(2.1a)\rangle$ by parts using incompressibility and the BCs yields

Averaging $\unicode[STIX]{x1D703}\times$  (2.1c ) and $\unicode[STIX]{x1D711}\times$  (2.1c ) in a similar way gives

Combining expressions (3.3)–(3.5) and rearranging we find

To prove that (3.1) holds at all times, we make one key further simplification: we drop the equations of motion and choose $a$ , $b$ , $L$ and $\unicode[STIX]{x1D711}(z)$ such that ${\mathcal{S}}\{\unicode[STIX]{x1D703},\boldsymbol{u}\}\geqslant 0$ for any time-independent fields $\unicode[STIX]{x1D703}=\unicode[STIX]{x1D703}(x,y,z)$ and $\boldsymbol{u}=\boldsymbol{u}(x,y,z)$ that satisfy (2.1b ) and the BCs. Hereafter, we also assume that $a,b>0$ to ensure that ${\mathcal{S}}\{\unicode[STIX]{x1D703},\boldsymbol{u}\}$ is bounded below.

Incompressibility can be incorporated explicitly in (3.6) upon substitution of the horizontal Fourier expansions

where $\boldsymbol{x}=(x,y)$ is the horizontal position vector and $\boldsymbol{k}=(k_{x},k_{y})$ is the wavevector. The $z$ -dependent Fourier amplitudes $\boldsymbol{u}_{\boldsymbol{k}}$ , $\unicode[STIX]{x1D703}_{\boldsymbol{k}}$ satisfy the same vertical BCs as the full fields in (2.2). Using the Fourier-transformed incompressibility constraint one can show that (Doering & Constantin Reference Doering and Constantin1996; Otero et al. Reference Otero, Wittenberg, Worthing and Doering2002)

with

In these equations and in the following we write $k^{2}=k_{x}^{2}+k_{y}^{2}$ , $\Vert \boldsymbol{\cdot }\Vert$ denotes the standard Lebesgue ${\mathcal{L}}^{2}$ norm on the interval $(0,1)$ , and $\tilde{w}_{\boldsymbol{k}}$ is the complex conjugate of $w_{\boldsymbol{k}}$ .

The right-hand side of (3.8) is clearly non-negative if ${\mathcal{S}}_{0}\geqslant 0$ and ${\mathcal{S}}_{\boldsymbol{k}}\geqslant 0$ for all wavevectors $\boldsymbol{k}\neq (0,0)$ . (A standard argument based on the consideration of fields $\unicode[STIX]{x1D703}$ and $\boldsymbol{u}$ with a single Fourier mode shows that these conditions are also necessary, so enforcing the positivity of each ${\mathcal{S}}_{\boldsymbol{k}}$ individually does not introduce conservativeness. However, necessity is not required to proceed with our argument so we omit the details for brevity.) In particular, given $a$ , $b$ and $\unicode[STIX]{x1D711}$ the largest value of $L$ for which ${\mathcal{S}}_{0}\geqslant 0$ is found upon completing the square (in the ${\mathcal{L}}^{2}$ norm sense) in (3.9a ), so we set

We will try to maximise this expression over $a$ , $b$ and $\unicode[STIX]{x1D711}$ subject to the non-negativity of the functional ${\mathcal{S}}_{\boldsymbol{k}}$ in (3.9b ) for all wavevectors $\boldsymbol{k}\neq (0,0)$ . Note that ${\mathcal{S}}_{\boldsymbol{k}}$ and the right-hand side of (3.10) reduce, respectively, to the quadratic form and the bound obtained by Otero et al. (Reference Otero, Wittenberg, Worthing and Doering2002) using the ‘classical’ background method analysis if we let $a=b-1$ and identify $[\unicode[STIX]{x1D711}^{\prime }(z)-2b+1]/(2b)$ with the derivative of the background temperature field. We also remark that our analysis appears more general because the choice $a=1-b$ is unjustified at this stage, but its optimality (at least within the context of our proof) will be demonstrated below.

4 An explicit bound

Let $\unicode[STIX]{x1D6FF}\leqslant 1/2$ and consider the piecewise-linear profile $\unicode[STIX]{x1D711}(z)$ shown in figure 1, whose derivative is

To show that $a$ , $b$ and $\unicode[STIX]{x1D6FF}$ can be chosen to make the quadratic form ${\mathcal{S}}_{\boldsymbol{k}}\{\unicode[STIX]{x1D703}_{\boldsymbol{k}},w_{\boldsymbol{k}}\}$ in (3.9b ) positive semidefinite we rewrite $\unicode[STIX]{x1D703}_{\boldsymbol{k}}$ and $w_{\boldsymbol{k}}$ as the sum of functions that are symmetric and antisymmetric with respect to $z=1/2$ . In other words, we decompose

with

(The subscripts $+$ and $-$ denote, respectively, the symmetric and antisymmetric parts.) Since $\unicode[STIX]{x1D711}^{\prime }(z)$ is symmetric with respect to $z=1/2$ by construction we obtain

i.e. we can split the quadratic form ${\mathcal{S}}_{\boldsymbol{k}}\{\unicode[STIX]{x1D703}_{\boldsymbol{k}},w_{\boldsymbol{k}}\}$ into its symmetric and antisymmetric components also. Symmetric and antisymmetric Fourier amplitudes $\unicode[STIX]{x1D703}_{\boldsymbol{k}}$ and $w_{\boldsymbol{k}}$ – for which one term on the right-hand side of (4.4) vanishes – are also admissible, so ${\mathcal{S}}_{\boldsymbol{k}}$ is non-negative if and only if it is so for arguments that are either symmetric or antisymmetric with respect to $z=1/2$ (and, of course, satisfy the correct BCs). As before, the ‘only if’ statement is not needed to proceed but guarantees that no conservativeness is introduced.

Figure 1.

Sketch of the piecewise-linear function $\unicode[STIX]{x1D711}(z)$ .

The decomposition into symmetric and antisymmetric components is the essential ingredient of our proof. In fact, contrary to the case of no-slip boundaries considered by Otero et al. (Reference Otero, Wittenberg, Worthing and Doering2002), the free-slip and fixed-flux BCs cannot be used to control the indefinite term in ${\mathcal{S}}_{\boldsymbol{k}}\{\unicode[STIX]{x1D703}_{\boldsymbol{k}},w_{\boldsymbol{k}}\}$ via the usual elementary functional-analytic estimates. However, $\unicode[STIX]{x1D703}_{\pm }$ and $w_{\pm }$ (or the appropriate derivatives) are known not only at the boundaries, but also on the symmetry plane. In particular, for small $\unicode[STIX]{x1D6FF}$ the indefinite term in (3.9b ) can be controlled without recourse to the free-slip, fixed-flux BCs using

To prove this, recall that for any symmetric or antisymmetric quantity $q(z)$

Symmetry, equation (4.1), and the identity $\left|\unicode[STIX]{x1D703}_{\pm }\tilde{w}_{\pm }\right|=\left|\unicode[STIX]{x1D703}_{\pm }w_{\pm }\right|$ ( $\tilde{w}_{\pm }$ is the complex conjugate of $w_{\pm }$ ) yield

with $M:=(1+a+b)/b$ . Since $w_{\pm }(0)=0$ the product $\unicode[STIX]{x1D703}_{\pm }w_{\pm }$ vanishes at $z=0$ and for any $z\leqslant \unicode[STIX]{x1D6FF}\leqslant 1/2$ the fundamental theorem of calculus implies

Using the fact that $w_{\pm }(0)=0$ once again, the fundamental theorem of calculus for $\unicode[STIX]{x1D709}\leqslant 1/2$ , the Cauchy–Schwarz inequality, and (4.6) we also obtain

Furthermore, the conditions in (4.5) imply that the product $\unicode[STIX]{x1D703}_{\pm }w_{\pm }^{\prime }$ vanishes at the symmetry plane, so similar estimates as above yield

Upon inserting (4.9) and (4.10) into (4.8), applying the Cauchy–Schwarz inequality, and using (4.6) we arrive at

Substituting this estimate into (4.7) and integrating gives an estimate for the indefinite term in (3.9b ), and after dropping the term $ak^{2}\Vert w_{\pm }\Vert ^{2}/(b\mathit{R})$ we conclude that

Recalling the definition of $M$ and that a quadratic form $\unicode[STIX]{x1D6FC}u^{2}+\unicode[STIX]{x1D6FD}uv+\unicode[STIX]{x1D6FE}v^{2}$ is positive semidefinite if $\unicode[STIX]{x1D6FD}^{2}\leqslant 4\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FE}$ , the right-hand side of (4.12) is non-negative if we set

Having chosen $\unicode[STIX]{x1D6FF}$ to ensure the non-negativity of ${\mathcal{S}}_{\boldsymbol{k}}$ , all that is left to do is optimise the eventual bound $\mathit{Nu}\leqslant L^{-1}$ over $a$ and $b$ as a function of $\mathit{R}$ . Substituting (4.1) into (3.10) for our choice of $\unicode[STIX]{x1D6FF}$ yields

In order to maximise this expression with respect to $a,b>0$ we set the partial derivatives $\unicode[STIX]{x2202}L/\unicode[STIX]{x2202}a$ and $\unicode[STIX]{x2202}L/\unicode[STIX]{x2202}b$ to zero. After some rearrangement it can be verified that

A few lines of simple algebra show that setting to zero the second factor in (4.15b ) leads to a solution with negative $a$ or $b$ , so we must choose $b=1+a$ where $a>0$ satisfies

(No positive roots exist if $R\leqslant 4A^{4}\approx 43.92$ , but we are only interested in $\mathit{R}\geqslant 120$ because conduction is globally asymptotically stable otherwise. It can also be checked that this stationary point is a maximum; the algebra is straightforward but lengthy and uninteresting, so we do not report it for brevity.) In particular, when $\mathit{R}$ tends to infinity (4.16) admits an asymptotic solution of the form $a=a_{1}\mathit{R}^{-1/3}+O(\mathit{R}^{-2/3})$ . Substituting this expansion into (4.16) and solving for the leading-order terms gives $a_{1}=A^{4/3}/4$ . We then set $b=1+a$ and $a=a_{1}\mathit{R}^{-1/3}$ in (4.14), simplify, and estimate

Note that this bound is sharp as $\mathit{R}\rightarrow \infty$ . Consequently,

Recalling that $\mathit{R}=\mathit{Nu}\,\mathit{Ra}$ (Otero et al. Reference Otero, Wittenberg, Worthing and Doering2002) we can also express this bound in terms of the Rayleigh number $\mathit{Ra}$ based on the average temperature drop across the layer: