2 Three-dimensional rough channel flow

Imagine a channel of average height $h$ , periodic extent $hL_{x}$ in the (streamwise) direction of an applied pressure gradient and periodic spanwise extent $hL_{y}$ . If $\unicode[STIX]{x1D708}$ is the kinematic viscosity, then introduce $\unicode[STIX]{x1D708}/h$ as the unit of speed, $h$ as the unit of length and $h^{2}/\unicode[STIX]{x1D708}$ as the unit of time. The governing Navier–Stokes equations become

(2.1) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}+\unicode[STIX]{x1D735}p-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}=Gr\hat{\boldsymbol{x}}, & \displaystyle\end{eqnarray}$$

(2.2) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0, & \displaystyle\end{eqnarray}$$

where $Gr:=h^{3}{\mathcal{G}}/\unicode[STIX]{x1D708}^{2}$ , the Grashof number, is the non-dimensionalised applied pressure gradient ${\mathcal{G}}$ driving the flow. For convenience, we refer to the cross-channel $z$ -direction as the ‘vertical’ direction and the $x$ and $y$ directions as ‘horizontal’ and imagine rough channel boundaries at $z=f(x,y)$ (the lower boundary) and $z=g(x,y)$ (the upper boundary). The roughness functions $f$ and $g$ will be assumed at least $C^{3}$ (3 times differentiable), periodic over ${\mathcal{A}}:=[0,L_{x}]\times [0,L_{y}]$ such that

(2.3a,b ) $$\begin{eqnarray}\iint _{{\mathcal{A}}}f\,\text{d}x\,\text{d}y=0\quad \text{ and }\quad \frac{1}{A}\iint _{{\mathcal{A}}}g\,\text{d}x\,\text{d}y=1\end{eqnarray}$$

(to preserve the average channel height as 1 in non-dimensional units and non-dimensional volume as $A:=L_{x}L_{y}$ ) and to be such that the channel is never blocked at any point (i.e. $f<g$ for any $(x,y)\in {\mathcal{A}}$ ). In the following we work with a one-dimensional (1-D) family of interior surfaces

(2.4) $$\begin{eqnarray}{\mathcal{S}}(\unicode[STIX]{x1D706}):=\{(x,y,z)|z=F(x,y,\unicode[STIX]{x1D706}):=(1-\unicode[STIX]{x1D706})f(x,y)+\unicode[STIX]{x1D706}g(x,y)\text{ for }(x,y)\in {\mathcal{A}}\}\end{eqnarray}$$

with $\unicode[STIX]{x1D706}\in [0,1]$ which smoothly interpolate between the two rough boundaries: $\unicode[STIX]{x1D706}=0$ giving the lower boundary and $\unicode[STIX]{x1D706}=1$ the upper boundary: see figure 1. Let ${\mathcal{C}}(\unicode[STIX]{x1D706},y^{\ast })$ be the line formed from the intersection of ${\mathcal{S}}(\unicode[STIX]{x1D706})$ with the $y=y^{\ast }$ plane. Let ${\mathcal{V}}(\unicode[STIX]{x1D706})$ be the volume enclosed by ${\mathcal{S}}(\unicode[STIX]{x1D706})$ , ${\mathcal{S}}(0)$ (the lower boundary) and the planes $x=0$ , $x=L_{x}$ , $y=0$ and $y=L_{y}$ , and let $\unicode[STIX]{x2202}{\mathcal{V}}(\unicode[STIX]{x1D706})$ be the boundary of ${\mathcal{V}}(\unicode[STIX]{x1D706})$ . The flow conditions at the edges of each surface are periodic so that the flow is invariant under the transformations $x\rightarrow x+L_{x}$ and $y\rightarrow y+L_{y}$ . A long-time average is defined as

(2.5) $$\begin{eqnarray}\langle (\cdot )\rangle :=\lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}(\cdot )\,\text{d}t.\end{eqnarray}$$

Figure 1.

The flow geometry. The top boundary is given by $z=g(x,y)$ and the bottom by $z=f(x,y)$ . The intermediate surface shown in outline is parametrised by $\unicode[STIX]{x1D706}$ and defined by $z=F(x,y,\unicode[STIX]{x1D706}):=(1-\unicode[STIX]{x1D706})f(x,y)+\unicode[STIX]{x1D706}g(x,y)$ .

We start the ‘boundary layer’ bounding analysis (Otto & Seis Reference Otto and Seis2011; Seis Reference Seis2015) by taking the line integral of (2.1) along ${\mathcal{C}}(\unicode[STIX]{x1D706},y)$ (in the direction of increasing $x$ ) followed by an integration over $y$ ,

(2.6) $$\begin{eqnarray}\int _{0}^{L_{y}}\int _{{\mathcal{C}}(\unicode[STIX]{x1D706},y)}\hat{\boldsymbol{s}}\boldsymbol{\cdot }\biggl[\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}+\unicode[STIX]{x1D735}p-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}\biggr]\,\text{d}s\,\text{d}y=Gr\int _{0}^{L_{y}}\int _{{\mathcal{C}}(\unicode[STIX]{x1D706},y)}\hat{\boldsymbol{s}}\boldsymbol{\cdot }\hat{\boldsymbol{x}}\,\text{d}s\,\text{d}y,\end{eqnarray}$$

where $s$ is the arc-length along ${\mathcal{C}}(\unicode[STIX]{x1D706},y)$ and $\hat{\boldsymbol{s}}:=(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})/\sqrt{1+F_{x}^{2}}$ is the unit tangent vector (subscripts indicate partial derivatives so that $F_{x}=\unicode[STIX]{x2202}F/\unicode[STIX]{x2202}x$ with $y$ and $\unicode[STIX]{x1D706}$ held fixed). Crucially, this procedure kills the pressure term as

(2.7) $$\begin{eqnarray}\int _{{\mathcal{C}}(\unicode[STIX]{x1D706},y)}\hat{\boldsymbol{s}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}p\,\text{d}s=p(L_{x},y,F(L_{x},y,\unicode[STIX]{x1D706}),t)-p(0,y,F(0,y,\unicode[STIX]{x1D706}),t)=0\end{eqnarray}$$

by periodicity in $x$ . With this, and converting the line integral to an integration over $x$ , we get

(2.8) $$\begin{eqnarray}A\,Gr=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\,\boldsymbol{\cdot }\,\boldsymbol{u}\,\text{d}x\,\text{d}y+\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }[\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]\,\text{d}x\,\text{d}y.\end{eqnarray}$$

The first term on the right-hand side can be dropped after long-time averages, assuming that the kinetic energy remains bounded in time, to leave

(2.9) $$\begin{eqnarray}Gr=\left\langle \frac{1}{A}\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }[\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]\,\text{d}x\,\text{d}y\,\right\rangle ,\end{eqnarray}$$

which is an identity for any $\unicode[STIX]{x1D706}\in [0,1]$ . We now generate volume integrals from this expression which can be related to the long-time-averaged energy dissipation rate per unit mass (in units of $\unicode[STIX]{x1D708}^{3}/h^{4}$ )

(2.10) $$\begin{eqnarray}\unicode[STIX]{x1D700}:=\left\langle \frac{1}{A}\int \!\!\!\int \!\!\!\int \!\!|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}\,\text{d}V\right\rangle :=\left\langle \frac{1}{A}\int _{0}^{L_{x}}\int _{0}^{L_{y}}\int _{z=f(x,y)}^{z=g(x,y)}|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}\,\text{d}z\,\text{d}y\,\text{d}x\right\rangle .\end{eqnarray}$$

To do this, (2.9) is integrated over $\unicode[STIX]{x1D706}\in [0,\unicode[STIX]{x1D6EC}]$ to give

(2.11) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}Gr=\left\langle \frac{1}{A}\int _{0}^{\unicode[STIX]{x1D6EC}}\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }[\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]\,\text{d}x\,\text{d}y\,\text{d}\unicode[STIX]{x1D706}\right\rangle .\end{eqnarray}$$

This can be converted into a volume integral by converting the integral over $\unicode[STIX]{x1D706}$ to one over $z$ with the addition of a (Jacobian) scaling factor $\unicode[STIX]{x2202}z/\unicode[STIX]{x2202}\unicode[STIX]{x1D706}|_{x,y}=\,F_{\unicode[STIX]{x1D706}}(x,y,\unicode[STIX]{x1D706})$ :

(2.12) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}Gr=\left\langle \frac{1}{A}\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\boldsymbol{a}\boldsymbol{\cdot }[\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]\,\text{d}x\,\text{d}y\,\text{d}z\right\rangle \quad \forall \unicode[STIX]{x1D6EC}\in [0,1],\end{eqnarray}$$

where

(2.13) $$\begin{eqnarray}\boldsymbol{a}:=\frac{\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}}}{F_{\unicode[STIX]{x1D706}}}=\frac{\hat{\boldsymbol{x}}+[f_{x}+(z-f)[\ln (g-f)]_{x}]\hat{\boldsymbol{z}}}{(g-f)}\end{eqnarray}$$

eliminating $\unicode[STIX]{x1D706}$ in favour of $z$ (recall $g=g(x,y)$ and $f=f(x,y)$ ). The key now is to lift spatial derivatives off $\boldsymbol{u}$ and onto $\boldsymbol{a}$ by judicious use of the divergence theorem. This leads to

(2.14) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EC}Gr & = & \displaystyle \frac{1}{A}\left\langle \oint _{\unicode[STIX]{x2202}{\mathcal{V}}(\unicode[STIX]{x1D6EC})}(\boldsymbol{a}\boldsymbol{\cdot }\boldsymbol{u})\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{n}}+\boldsymbol{u}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}-\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}\,\text{d}S\right.\nonumber\\ \displaystyle & & \displaystyle \left.-\,\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}\,\text{d}V\,\right\rangle \quad \forall \unicode[STIX]{x1D6EC}\in [0,1]\end{eqnarray}$$

(note the last term requires the roughness functions $f$ and $g$ to be at least $C^{3}$ ). Due to periodicity over $(x,y)\in [0,L_{x}]\times [0,L_{y}]$ , the only parts of $\unicode[STIX]{x2202}{\mathcal{V}}(\unicode[STIX]{x1D6EC})$ which need to be considered are the lower boundary ${\mathcal{S}}(0)$ where $\boldsymbol{u}=\mathbf{0}$ and the interior surface ${\mathcal{S}}(\unicode[STIX]{x1D6EC})$ . Hence, in fact

(2.15) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EC}Gr & = & \displaystyle \frac{1}{A}\left\langle \int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}(\boldsymbol{a}\boldsymbol{\cdot }\boldsymbol{u})\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{n}}+\boldsymbol{u}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}-\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}\,\text{d}S\right.\nonumber\\ \displaystyle & & \displaystyle -\,\int _{{\mathcal{S}}(0)}\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}\,\text{d}S\nonumber\\ \displaystyle & & \displaystyle -\left.\,\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}\,\text{d}V\,\right\rangle \quad \forall \unicode[STIX]{x1D6EC}\in [0,1].\end{eqnarray}$$

Now the strategy is to separately average (2.15) over $\unicode[STIX]{x1D6EC}\in [0,\ell ]$ and over $\unicode[STIX]{x1D6EC}\in [1-\ell ,1]$ and then compute the difference in order to eliminate the $\unicode[STIX]{x1D6EC}$ -independent ${\mathcal{S}}(0)$ boundary term. Subtracting $1/\ell \int _{0}^{\ell }$ (2.15) $\text{d}\unicode[STIX]{x1D6EC}$ from $1/\ell \int _{1-\ell }^{1}$ (2.15) $\text{d}\unicode[STIX]{x1D6EC}$ gives

(2.16) $$\begin{eqnarray}\displaystyle (1-\ell )Gr & = & \displaystyle \frac{1}{A}\left\langle \frac{1}{\ell }\int _{1-\ell }^{1}\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}(\boldsymbol{a}\boldsymbol{\cdot }\boldsymbol{u})\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{n}}+\boldsymbol{u}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}-\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right.\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{\ell }\int _{0}^{\ell }\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}(\boldsymbol{a}\boldsymbol{\cdot }\boldsymbol{u})\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{n}}+\boldsymbol{u}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}-\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{\ell }\int _{1-\ell }^{1}\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\,\,\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{1}{\ell }\int _{0}^{\ell }\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle .\end{eqnarray}$$

This is the generalisation of Seis’s (Reference Seis2015) expression (4.10) which can be recovered by setting $\boldsymbol{a}=\hat{\boldsymbol{x}}$ , taking ${\mathcal{S}}(\unicode[STIX]{x1D6EC})$ as the horizontal plane $z=\unicode[STIX]{x1D6EC}$ over $[0,L_{x}]\times [0,L_{y}]$ and $\hat{\boldsymbol{n}}=\hat{\boldsymbol{z}}$ . Notably, there are new integrated volume terms because the linear momentum directed along ${\mathcal{C}}(\unicode[STIX]{x1D706},y)$ is not conserved but these turn out to be subdominant in what follows. From here, the exercise is to bound the right-hand side of (2.16) in terms of the energy dissipation rate $\unicode[STIX]{x1D700}$ and then to optimise over $\ell \in [0,1/2]$ to produce the best bound on $Gr$ as in Seis (Reference Seis2015). Since our focus is on establishing the scaling exponent of how $Gr$ scales with $\unicode[STIX]{x1D700}$ rather than the secondary issue of producing the best estimate for the numerical prefactor, we proceed by employing straightforward conservative estimates. This has the advantage of securing our key result quickly and relatively clearly but more careful estimates could lower the (numerical) bound on $Gr$ but not, we contend, the scaling exponent. Firstly,

(2.17) $$\begin{eqnarray}\displaystyle Gr & {\leqslant} & \displaystyle \frac{1}{A\ell (1-\ell )}\left\{\!\left\langle \int _{1-\ell }^{1}\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{a}\boldsymbol{\cdot }\boldsymbol{u}||\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{n}}|\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle +\left\langle \int _{1-\ell }^{1}\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}\!|\boldsymbol{u}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}|\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle \right.\nonumber\\ \displaystyle & & \displaystyle +\left\langle \int _{1-\ell }^{1}\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}|\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle +\left\langle \int _{0}^{\ell }\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{a}\boldsymbol{\cdot }\boldsymbol{u}||\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{n}}|\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle ^{(1)}\nonumber\\ \displaystyle & & \displaystyle +\left\langle \int _{0}^{\ell }\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{u}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}|\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle ^{(2)}+\left\langle \int _{0}^{\ell }\int _{{\mathcal{S}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{a}\boldsymbol{\cdot }(\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{u}|\,\text{d}S\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle ^{(3)}\nonumber\\ \displaystyle & & \displaystyle +\left\langle \int _{1-\ell }^{1}\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}|\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle +\left\langle \int _{1-\ell }^{1}\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}|\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle \nonumber\\ \displaystyle & & \displaystyle +\left.\left\langle \int _{0}^{\ell }\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735})\boldsymbol{a}|\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle ^{(4)}+\left\langle \int _{0}^{\ell }\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}|\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}|\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle ^{(5)}\right\}\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(the superscripts in parentheses are just labels). Each of the integrals in (2.17) need to be bounded by a function of $\unicode[STIX]{x1D700}$ . We focus on integrals concentrated at the lower boundary and treat first the integral, $I_{1}$ , labelled $^{(1)}$ in (2.17). Firstly converting the surface integral into one over ${\mathcal{A}}$

(2.18) $$\begin{eqnarray}I_{1}\leqslant \left\langle \int _{0}^{\ell }\iint _{{\mathcal{A}}}|\boldsymbol{a}|\sqrt{1+F_{x}^{2}+F_{y}^{2}}|\boldsymbol{u}|^{2}\,\text{d}x\,\text{d}y\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle .\end{eqnarray}$$

Then using the fundamental theorem of calculus since $\boldsymbol{u}(x,y,f(x,y))=\mathbf{0}$ (i.e. on ${\mathcal{S}}(0)$ ) and Cauchy–Schwarz gives

(2.19) $$\begin{eqnarray}\displaystyle I_{1} & {\leqslant} & \displaystyle \left\langle \int _{0}^{\ell }\iint _{{\mathcal{A}}}|\boldsymbol{a}|\sqrt{1+F_{x}^{2}+F_{y}^{2}}\,\biggl|\int _{f(x,y)}^{z}\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}\overline{z}}\,\text{d}\overline{z}\biggr|^{2}\,\text{d}x\,\text{d}y\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \left\langle \int _{0}^{\ell }\iint _{{\mathcal{A}}}|\boldsymbol{a}|\sqrt{1+F_{x}^{2}+F_{y}^{2}}\biggl(\int _{f(x,y)}^{z}1^{2}\,\text{d}\overline{z}\int _{f(x,y)}^{z}\biggl|\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}\overline{z}}\biggr|^{2}\,\text{d}\overline{z}\,\biggr)\,\text{d}x\,\text{d}y,\text{d}\unicode[STIX]{x1D6EC}\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \left\langle \int _{0}^{\ell }\iint _{{\mathcal{A}}}\,|\boldsymbol{a}|\sqrt{1+F_{x}^{2}+F_{y}^{2}}\,\unicode[STIX]{x1D6EC}F_{\unicode[STIX]{x1D706}}\int _{f(x,y)}^{z}\biggl|\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}\overline{z}}\biggr|^{2}\,\text{d}\overline{z}\,\text{d}x\,\text{d}y\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \max _{x,y,\unicode[STIX]{x1D706}\in {\mathcal{V}}(\ell )}\biggl\{|\boldsymbol{a}|F_{\unicode[STIX]{x1D706}}\sqrt{1+F_{x}^{2}+F_{y}^{2}}\biggr\}\int _{0}^{\ell }\unicode[STIX]{x1D6EC}d\unicode[STIX]{x1D6EC}\left\langle \int \!\!\!\int \!\!\!\int \!\!|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}\,\text{d}V\,\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \max _{x,y,\unicode[STIX]{x1D706}\in {\mathcal{V}}(\ell )}\biggl\{\sqrt{1+F_{x}^{2}}\sqrt{1+F_{x}^{2}+F_{y}^{2}}\biggr\}\times \frac{1}{2}\ell ^{2}A\unicode[STIX]{x1D700}.\end{eqnarray}$$

Integral $I_{4}$ (labelled $^{(4)}$ in (2.17) ) is treated similarly albeit with an extra integration in $\unicode[STIX]{x1D6EC}$ giving rise to $\ell ^{3}$ in the estimate:

(2.20) $$\begin{eqnarray}\displaystyle I_{4} & {\leqslant} & \displaystyle \left\langle \int _{0}^{\ell }\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}|\unicode[STIX]{x1D735}\boldsymbol{a}||\boldsymbol{u}|^{2}\,\text{d}V\,\text{d}\unicode[STIX]{x1D6EC}\,\right\rangle \leqslant \left\langle \int _{0}^{\ell }\int _{0}^{\unicode[STIX]{x1D6EC}}\iint _{{\mathcal{A}}}F_{\unicode[STIX]{x1D706}}|\unicode[STIX]{x1D735}\boldsymbol{a}||\boldsymbol{u}|^{2}\,\text{d}x\,\text{d}y\,\text{d}\unicode[STIX]{x1D706}\,\text{d}\unicode[STIX]{x1D6EC}\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \max _{x,y,\unicode[STIX]{x1D706}\in {\mathcal{V}}(\ell )}\{F_{\unicode[STIX]{x1D706}}^{2}|\unicode[STIX]{x1D735}\boldsymbol{a}|\}\int _{0}^{\ell }\int _{0}^{\unicode[STIX]{x1D6EC}}\unicode[STIX]{x1D706}\,\text{d}\unicode[STIX]{x1D706}\,\text{d}\unicode[STIX]{x1D6EC}\left\langle \int \!\!\!\int \!\!\!\int \!\!|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}\,\text{d}V\,\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \max _{x,y,\unicode[STIX]{x1D706}\in {\mathcal{V}}(\ell )}\{F_{\unicode[STIX]{x1D706}}^{2}|\unicode[STIX]{x1D735}\boldsymbol{a}|\}\times \frac{1}{6}\ell ^{3}A\unicode[STIX]{x1D700}.\end{eqnarray}$$

The remaining integrals, $I_{2}$ , $I_{3}$ and $I_{5}$ , are treated as follows:

(2.21) $$\begin{eqnarray}\displaystyle I_{2} & {\leqslant} & \displaystyle \left\langle \iint _{{\mathcal{A}}}\int _{f(x,y)}^{F(x,y,\ell )}\frac{|\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{a}|\sqrt{1+F_{x}^{2}+F_{y}^{2}}}{F_{\unicode[STIX]{x1D706}}}|\boldsymbol{u}|\,\text{d}z\,\text{d}x\,\text{d}y\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}\frac{|\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{a}|^{2}(1+F_{x}^{2}+F_{y}^{2})}{F_{\unicode[STIX]{x1D706}}^{2}}\,\text{d}V\biggr]^{1/2}\biggl[\left\langle \int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}|\boldsymbol{u}|^{2}\,\,\text{d}V\right\rangle \biggr]^{1/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}\frac{|\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{a}|^{2}(1+F_{x}^{2}+F_{y}^{2})}{F_{\unicode[STIX]{x1D706}}^{2}}\,\text{d}V\biggr]^{1/2}\times \Bigl[\max _{(x,y)\in {\mathcal{A}}}F_{\unicode[STIX]{x1D706}}^{2}\Bigr]^{1/2}\times \sqrt{\frac{1}{2}\ell ^{2}A\unicode[STIX]{x1D700}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\frac{1}{A}\iint _{{\mathcal{A}}}\max _{\unicode[STIX]{x1D6EC}\in [0,\ell ]}\{|\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{a}|^{2}(1+F_{x}^{2}+F_{y}^{2})\}\frac{\text{d}x\,\text{d}y}{F_{\unicode[STIX]{x1D706}}}\biggr]^{1/2}\times \Bigl[\max _{(x,y)\in {\mathcal{A}}}F_{\unicode[STIX]{x1D706}}^{2}\Bigr]^{1/2}\times A\sqrt{\frac{1}{2}\ell ^{3}\unicode[STIX]{x1D700}},\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $\hat{\boldsymbol{n}}=(-F_{x}\hat{\boldsymbol{x}}-F_{y}\hat{\boldsymbol{y}}+\hat{\boldsymbol{z}})/\sqrt{1+F_{x}^{2}+F_{y}^{2}}$ and the last two lines follow using the same type of estimates as in (2.19) and noting that $F_{\unicode[STIX]{x1D706}}=g(x,y)-f(x,y)$ is independent of $\unicode[STIX]{x1D706}$ and strictly positive. Respectively

(2.22) $$\begin{eqnarray}\displaystyle I_{3} & {\leqslant} & \displaystyle \left\langle \iint _{{\mathcal{A}}}\int _{f(x,y)}^{F(x,y,\ell )}\frac{|\boldsymbol{a}|\sqrt{1+F_{x}^{2}+F_{y}^{2}}}{F_{\unicode[STIX]{x1D706}}}|\unicode[STIX]{x1D735}\boldsymbol{u}|\,\text{d}z\,\text{d}x\,\text{d}y\,\right\rangle \nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}\frac{|\boldsymbol{a}|^{2}(1+F_{x}^{2}+F_{y}^{2})}{F_{\unicode[STIX]{x1D706}}^{2}}\,\text{d}V\biggr]^{1/2}\times \sqrt{A\unicode[STIX]{x1D700}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\iint _{{\mathcal{A}}}\int _{0}^{\ell }\frac{(1+F_{x}^{2})(1+F_{x}^{2}+F_{y}^{2})}{F_{\unicode[STIX]{x1D706}}^{3}}\,\text{d}\unicode[STIX]{x1D6EC}\,\text{d}x\,\text{d}y\biggr]^{1/2}\times \sqrt{A\unicode[STIX]{x1D700}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\frac{1}{A}\iint _{{\mathcal{A}}}\max _{\unicode[STIX]{x1D6EC}\in [0,\ell ]}\{(1+F_{x}^{2})(1+F_{x}^{2}+F_{y}^{2})\}\,\,\frac{\text{d}x\,\text{d}y}{F_{\unicode[STIX]{x1D706}}^{3}}\biggr]^{1/2}\times A\sqrt{\ell \unicode[STIX]{x1D700}}\end{eqnarray}$$

and

(2.23) $$\begin{eqnarray}\displaystyle I_{5} & {\leqslant} & \displaystyle \int _{0}^{\ell }\biggl[\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}|\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}|^{2}\,\text{d}V\biggr]^{1/2}\biggl[\left\langle \int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}|\boldsymbol{u}|^{2}\,\text{d}V\right\rangle \biggr]^{1/2}\,\text{d}\unicode[STIX]{x1D6EC}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\frac{1}{A}\int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\ell )}|\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}|^{2}\,\text{d}V\biggr]^{1/2}\times \Bigl[\max _{(x,y)\in {\mathcal{A}}}F_{\unicode[STIX]{x1D706}}^{2}\Bigr]^{1/2}\times \ell ^{2}A\sqrt{\frac{1}{2}\unicode[STIX]{x1D700}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl[\frac{1}{A}\iint _{{\mathcal{A}}}F_{\unicode[STIX]{x1D706}}\max _{\unicode[STIX]{x1D6EC}\in [0,\ell ]}~|\unicode[STIX]{x1D6FB}^{2}\boldsymbol{a}|^{2}\,\text{d}x\,\text{d}y\biggr]^{1/2}\times \Bigl[\max _{(x,y)\in {\mathcal{A}}}F_{\unicode[STIX]{x1D706}}^{2}\Bigr]^{1/2}\times A\sqrt{\frac{1}{2}\ell ^{5}\unicode[STIX]{x1D700}}.\end{eqnarray}$$

The estimates for the corresponding integrals centred at the top boundary are exactly analogous, and grouping the contributions together for each term, we get a simplified ‘inequality’ version of (2.16)

(2.24) $$\begin{eqnarray}Gr\leqslant \frac{1}{\ell (1-\ell )}\{(B_{1}\ell ^{2}+B_{4}\ell ^{3})\unicode[STIX]{x1D700}+(B_{2}\ell +B_{3}+B_{5}\ell ^{2})\sqrt{\ell \unicode[STIX]{x1D700}}\}.\end{eqnarray}$$

where the coefficients $B_{i}$ represent the O(1) numerical factors (in the sense of $Gr\rightarrow \infty$ ) of the $i$ th integral bound (summed for both boundaries) when a factor of $A$ is factored out and the dominant $\ell$ and $\unicode[STIX]{x1D700}$ behaviour (which both are not $O(1)$ ) is separated off. The idea now is to minimise the right-hand side over the choice of $\ell \in (0,1/2)$ in the turbulent limit $Gr\rightarrow \infty$ . Here, $\unicode[STIX]{x1D700}\rightarrow \infty$ (naive scalings suggest $Gr\lesssim \unicode[STIX]{x1D700}\lesssim Gr^{2}$  ) and the optimal $\ell \rightarrow 0$ so, working to leading order,

(2.25) $$\begin{eqnarray}Gr\leqslant B_{1}\ell \unicode[STIX]{x1D700}+B_{3}\sqrt{\unicode[STIX]{x1D700}/\ell }+\text{h.o.t.}\end{eqnarray}$$

and now it is clear that all the new integral contributions due to the roughness are subdominant. Instead the roughness manifests itself as adjusted numerical coefficients for the integrals which arise in the smooth situation (Seis Reference Seis2015). The minimising $\ell$ is $(B_{3}/2B_{1})^{2/3}\unicode[STIX]{x1D700}^{-1/3}$ and

(2.26a,b ) $$\begin{eqnarray}Gr\leqslant \frac{3}{2}(2B_{1}B_{3}^{2})^{1/3}\times \unicode[STIX]{x1D700}^{2/3}\quad \text{or}\quad \frac{2\sqrt{3}}{9B_{3}\sqrt{B_{1}}}\times Gr^{3/2}\leqslant \unicode[STIX]{x1D700},\end{eqnarray}$$

where

(2.27) $$\begin{eqnarray}\displaystyle & \displaystyle B_{1}=\mathop{\sum }_{\unicode[STIX]{x1D706}=0}^{1}\max _{(x,y)\in {\mathcal{A}}}\biggl\{\frac{1}{2}\sqrt{1+F_{x}^{2}}\sqrt{1+F_{x}^{2}+F_{y}^{2}}\biggr\}, & \displaystyle\end{eqnarray}$$

(2.28) $$\begin{eqnarray}\displaystyle & \displaystyle B_{3}=\mathop{\sum }_{\unicode[STIX]{x1D706}=0}^{1}\biggl[\frac{1}{A}\iint _{{\mathcal{A}}}\frac{(1+F_{x}^{2})(1+F_{x}^{2}+F_{y}^{2})}{[g(x,y)-f(x,y)]^{3}}\,\text{d}x\,\text{d}y\biggr]^{1/2} & \displaystyle\end{eqnarray}$$

(since $\ell \ll 1$ , it is sufficient to leading order to replace $\max _{\unicode[STIX]{x1D6EC}\in [0,\ell ]}$ by the value at $\unicode[STIX]{x1D6EC}=0$ and similarly for the upper boundary). The lower bound in (2.26) indicates that turbulence decreases dissipation for a given applied pressure gradient as per the smooth wall case (Constantin & Doering Reference Constantin and Doering1995; Seis Reference Seis2015). However, as there, rewriting the result in terms of the a priori unknown mean flow recovers the familiar upper bound situation. Taking $\langle \int \!\int \!\int \boldsymbol{u}\boldsymbol{\cdot }(2.1)\,\text{d}V\rangle$ connects the mean flow $U$ with the applied pressure gradient and ensuing dissipation rate,

(2.29) $$\begin{eqnarray}U:=\left\langle \frac{1}{A}\int \!\!\!\int \!\!\!\int \!\!\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{x}}\,\text{d}V\right\rangle =\unicode[STIX]{x1D700}/Gr.\end{eqnarray}$$

This can be used to eliminate the pressure gradient ( $Gr$ ) to get

(2.30a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D700}\leqslant {\textstyle \frac{27}{4}}B_{1}B_{3}^{2}U^{3}\quad \text{or}\quad \unicode[STIX]{x1D700}\leqslant {\textstyle \frac{27}{4}}B_{1}B_{3}^{2}\quad (\text{in units of }U^{3}/h)\end{eqnarray}$$

so that the bound predicts that the energy dissipation rate becomes independent of the viscosity as the viscosity goes to zero – Kolmogorov scaling – just as in the smooth wall calculation (Constantin & Doering Reference Constantin and Doering1995; Seis Reference Seis2015, and Plasting & Kerswell Reference Plasting and Kerswell2005 for pipe flow).

3 Oscillatory flow across topography: the tidal problem

As a simple application of the above result, we now consider the ocean tidal problem of oscillatory flow across bottom topography (and a free top surface). This situation is modelled by assuming an oscillatory pressure gradient driving the flow directly i.e. ${\mathcal{G}}\sim \unicode[STIX]{x1D714}^{\ast }U^{\ast }$ where the tidal frequency $\unicode[STIX]{x1D714}^{\ast }=2\unicode[STIX]{x03C0}/12~\text{hours}=1.4\times 10^{-4}~\text{s}^{-1}$ and $U^{\ast }\approx 0.1~\text{m}~\text{s}^{-1}$ . We consider a symmetrised problem where the top boundary is defined by $g(x,y)=1-f(x,y)$ so that the real free surface position on average is the (symmetric) midplane of the extended domain and $h/2$ is then the mean ocean depth. By symmetry, the maximum dissipation in the extended domain is exactly double the maximum dissipation in the ‘ocean’ domain. Further restricting the flow in the extended domain to have zero vertical motion and zero stress across $z=1/2$ (the horizontal free-surface approximation) can only reduce the maximum dissipation possible (by restricting the competitor space) so halving the dissipation bound from the unrestricted, extended domain should be an upper bound on the dissipation in a flat surface ocean ( $z\leqslant 1/2$ ).

Taking a typical ocean depth $h/2=1000~\text{m}$ , the Grashof number $Gr=1.12\times 10^{17}\gg 1$ and the non-dimensionalised tidal frequency

(3.1) $$\begin{eqnarray}1\,\ll \unicode[STIX]{x1D714}:=\frac{h^{2}\unicode[STIX]{x1D714}^{\ast }}{\unicode[STIX]{x1D708}}=\sqrt{\frac{h{\mathcal{G}}}{U^{2}}}\sqrt{Gr}\,\ll Gr,\end{eqnarray}$$

where $\unicode[STIX]{x1D6F6}:=h{\mathcal{G}}/U^{2}$ is a second non-dimensional number independent of $\unicode[STIX]{x1D708}$ which is here $O(1)$ . This makes it clear that in the limit $\unicode[STIX]{x1D708}\rightarrow 0$ (with everything else kept fixed), $Gr\rightarrow \infty$ with $\unicode[STIX]{x1D714}\sim \sqrt{Gr}$ . The governing equation for the problem is then

(3.2) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\boldsymbol{u}}{\unicode[STIX]{x2202}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}+\unicode[STIX]{x1D735}p-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}=Gr\,\unicode[STIX]{x1D6E4}(t)\hat{\boldsymbol{x}},\end{eqnarray}$$

the flow is assumed incompressible and a planetary rotation $\unicode[STIX]{x1D734}$ is also included for completeness ( $|\unicode[STIX]{x1D734}|=\unicode[STIX]{x1D714}/2$ ). The oscillatory function $\unicode[STIX]{x1D6E4}(t)$ is defined such that

(3.3) $$\begin{eqnarray}\max _{t\in [0,2\unicode[STIX]{x03C0}/\unicode[STIX]{x1D714}]}|\unicode[STIX]{x1D6E4}(t)|=1\end{eqnarray}$$

but is otherwise left unspecified for clarity. As before, the analysis starts by taking the line integral of (3.2) along ${\mathcal{C}}(\unicode[STIX]{x1D706},y)$ and integrating over $y$ as above in § 2 but now also multiplying by $\unicode[STIX]{x1D6E4}(t)$ to rectify the forcing pressure gradient. This leads to an extended version of (2.8)

(3.4) $$\begin{eqnarray}\displaystyle AGr\unicode[STIX]{x1D6E4}(t)^{2} & = & \displaystyle \unicode[STIX]{x1D6E4}(t)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }\boldsymbol{u}\,\text{d}x\,\text{d}y\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6E4}(t)\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }[2\unicode[STIX]{x1D734}\times \boldsymbol{u}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]\,\text{d}x\,\text{d}y.\end{eqnarray}$$

Now long-time averaging leads to

(3.5) $$\begin{eqnarray}\displaystyle AGr\langle \unicode[STIX]{x1D6E4}(t)^{2}\rangle & = & \displaystyle -\left\langle \unicode[STIX]{x1D6E4}_{t}\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }\boldsymbol{u}\,\text{d}x\,\text{d}y\right\rangle \nonumber\\ \displaystyle & & \displaystyle +\,\left\langle \unicode[STIX]{x1D6E4}(t)\iint _{{\mathcal{A}}}(\hat{\boldsymbol{x}}+F_{x}\hat{\boldsymbol{z}})\boldsymbol{\cdot }[2\unicode[STIX]{x1D734}\times \boldsymbol{u}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]\,\text{d}x\,\text{d}y\right\rangle ,\end{eqnarray}$$

where integration by parts in time has been used to transfer the time derivative onto  $\unicode[STIX]{x1D6E4}$ . Integrating over $\unicode[STIX]{x1D706}\in [0,\unicode[STIX]{x1D6EC}]$ and converting to a volume integral over ${\mathcal{V}}(\unicode[STIX]{x1D6EC})$ , gives

(3.6) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}Gr\langle \unicode[STIX]{x1D6E4}(t)^{2}\rangle =\frac{1}{A}\left\langle \int \!\!\!\int \!\!\!\int _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\unicode[STIX]{x1D6E4}(t)\boldsymbol{a}\boldsymbol{\cdot }[\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}-\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}]+[\unicode[STIX]{x1D6E4}\boldsymbol{a}\times 2\unicode[STIX]{x1D734}-\unicode[STIX]{x1D6E4}_{t}\boldsymbol{a}]\boldsymbol{\cdot }\boldsymbol{u}\,\text{d}x\,\text{d}y\,\text{d}z\right\rangle\end{eqnarray}$$

for all $\unicode[STIX]{x1D6EC}\in [0,1]$ . The first term on the right-hand side is as before, albeit with an extra factor of $\unicode[STIX]{x1D6E4}(t)$ . Since this is bounded in modulus by 1, the subsequent estimates are unchanged. The new second term on the right-hand side is of the form of $I_{5}$ and can be similarly estimated as adding a new term

(3.7) $$\begin{eqnarray}\biggl[\frac{1}{A}\iint _{{\mathcal{A}}}F_{\unicode[STIX]{x1D706}}\max _{\unicode[STIX]{x1D6EC}\in [0,\ell ]}|\unicode[STIX]{x1D6E4}\boldsymbol{a}\times 2\unicode[STIX]{x1D734}-\unicode[STIX]{x1D6E4}_{t}\boldsymbol{a}|^{2}\,\text{d}x\,\text{d}y\biggr]^{1/2}\times \Bigl[\max _{(x,y)\in {\mathcal{A}}}F_{\unicode[STIX]{x1D706}}^{2}\Bigr]^{1/2}\times A\sqrt{\frac{1}{2}\ell ^{5}\unicode[STIX]{x1D700}}\end{eqnarray}$$

to (2.24). Since both $\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D6E4}_{t}=O(\unicode[STIX]{x1D714}\unicode[STIX]{x1D6E4})$ are $O(\sqrt{Gr})$ , the leading-order version of (2.24) is potentially modified to

(3.8) $$\begin{eqnarray}Gr\leqslant B_{1}\ell \unicode[STIX]{x1D700}+(B_{3}+B_{5}\ell ^{2}\sqrt{Gr})\sqrt{\unicode[STIX]{x1D700}/\ell }+\text{h.o.t.}\end{eqnarray}$$

The minimiser, however, remains $\ell \sim \unicode[STIX]{x1D700}^{-1/3}$ with the new term $O(Gr^{-1/2})$ smaller than the other two terms. Hence the rotation and oscillation are not important at leading order in the bound on the energy dissipation rate for tidal flow over topography.

Figure 2.

(a) The 2-D ridge tidal problem with the ocean represented by the region $z\leqslant 1/2$ . (b) A cartoon of the Moody diagram for pipe flow (Moody Reference Moody1944):  $C_{f}:=\unicode[STIX]{x1D700}/(U^{3}/h)$ , laminar flow has $C_{f}\sim 1/Re$ , turbulent smooth pipe data suggest $C_{f}\sim 1/(\log Re)^{2}$ at least to $Re=O(10^{8})$ and turbulent rough pipe data have $C_{f}\sim O(1)$ as $Re\rightarrow \infty$ . Three lines are shown for rough pipe flow with the data increasing to higher $C_{f}$ as the representative height of the roughness increases.

3.1 A simple example

Here we consider a 2-D ridge of height $a<1/2$ and horizontal extent $a\unicode[STIX]{x1D6FF}$

(3.9) $$\begin{eqnarray}f(x):=a\text{sech}^{2}\biggl(\frac{x-{\textstyle \frac{1}{2}}L}{a\unicode[STIX]{x1D6FF}}\biggr)-\frac{2a^{2}\unicode[STIX]{x1D6FF}}{L}\text{tanh}\biggl(\frac{L}{2a\unicode[STIX]{x1D6FF}}\biggr)\end{eqnarray}$$

over $x\in [0,L]$ (the second term on the right-hand side ensures the roughness has zero mean over $x$ ): see figure 2. This roughness function is not periodic over $[0,L]$ but in the steep ( $\unicode[STIX]{x1D6FF}\ll 1$ ) isolated ( $a\unicode[STIX]{x1D6FF}\ll L$ ) ridge limit this can be ignored. In this limit, it is straightforward to approximate $B_{1}$ and $B_{3}$ as follows

(3.10) $$\begin{eqnarray}B_{1}=\max _{x\in [0,L]}(1+f_{x}^{2})\approx \frac{16}{27\unicode[STIX]{x1D6FF}^{2}},\quad B_{3}=2\biggl[\frac{1}{L}\int \,\frac{(1+f_{x}^{2})^{2}}{(1-2f)^{3}}\,\text{d}x\biggr]^{1/2}\lesssim \biggl[\frac{2a}{L(1-2a)^{3}\unicode[STIX]{x1D6FF}^{3}}\biggr]^{1/2},\end{eqnarray}$$

(using the fact that $2048/1155<2$ ) so the dissipation rate per unit mass is

(3.11) $$\begin{eqnarray}\unicode[STIX]{x1D700}\leqslant \frac{4aU^{3}}{L(1-2a)^{3}\unicode[STIX]{x1D6FF}^{5}}\quad (\text{in units of }\unicode[STIX]{x1D708}^{3}/h^{4})\end{eqnarray}$$

(including the $1/2$ to reflect the fact that the ocean is strictly in $z\leqslant 1/2$ ). This can be re-expressed using inertial units to give the familiar bound independent of the viscosity

(3.12) $$\begin{eqnarray}\unicode[STIX]{x1D700}\leqslant \frac{4a}{L(1-2a)^{3}\unicode[STIX]{x1D6FF}^{5}}\quad (\text{in units of }U^{3}/h).\end{eqnarray}$$

Since this is the leading term due to the ridge (the dissipation bound for zero topography is $O(\unicode[STIX]{x1D6FF}^{5})$ smaller), this expression can be viewed as giving the enhanced dissipation caused by the ridge per unit length of the domain $[0,L]$ .