2.1 The background method

The background method is to construct the functional

(2.2) $$\begin{eqnarray}{\mathcal{L}}[\boldsymbol{u},\unicode[STIX]{x1D742}]:=\left\langle \int _{0}^{1}\overline{|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}}\,\text{d}z\right\rangle -\unicode[STIX]{x1D6FC}\left\langle \int _{0}^{1}\overline{\unicode[STIX]{x1D742}\boldsymbol{\cdot }(\boldsymbol{N}\boldsymbol{S})}\,\text{d}z\right\rangle ,\end{eqnarray}$$

where $\unicode[STIX]{x1D6FC}$ is a balance (scalar) parameter (usually ‘a’ in past work), $\unicode[STIX]{x1D742}(\boldsymbol{x},t)$ is a Lagrange multiplier field and

(2.3) $$\begin{eqnarray}\overline{(\cdot )}:=\frac{1}{L_{x}L_{y}}\int _{0}^{L_{x}}\int _{0}^{L_{y}}(\cdot )\,\text{d}y\,\text{d}x\end{eqnarray}$$

is an average over the rectangle $(x,y)\in [0,L_{x}]\times [0,L_{y}]$ . The key step is to restrict the difference between $\boldsymbol{u}$ and $\unicode[STIX]{x1D742}$ by defining a background field $\unicode[STIX]{x1D719}(z)$ such that

(2.4) $$\begin{eqnarray}\boldsymbol{u}(\boldsymbol{x},t)=\unicode[STIX]{x1D719}(z)\hat{\boldsymbol{x}}+\unicode[STIX]{x1D742}(\boldsymbol{x},t),\end{eqnarray}$$

where $\unicode[STIX]{x1D719}$ carries the mass flux of the flow but vanishes at the boundaries. In particular, if the energy dissipation rate is sought in terms of the mean flow $U$ rather than the imposed pressure gradient ( $Gr$ ) then

(2.5) $$\begin{eqnarray}U:=\int _{0}^{1}\overline{\boldsymbol{u}\boldsymbol{\cdot }\hat{\boldsymbol{x}}}\,\text{d}z=\int _{0}^{1}\unicode[STIX]{x1D719}\,\text{d}z\end{eqnarray}$$

(see (2.29), Kerswell Reference Kerswell2016). Rewriting ${\mathcal{L}}$ in terms of $\unicode[STIX]{x1D742}$ and $\unicode[STIX]{x1D719}$ , and then integrating by parts, the boundedness of the kinetic energy and the fact that both $\unicode[STIX]{x1D719}$ and $\unicode[STIX]{x1D742}$ vanish on $z=0$ and $1$ leads to the simplified expression

(2.6) $$\begin{eqnarray}{\mathcal{L}}[\unicode[STIX]{x1D742},\unicode[STIX]{x1D719}]=\int _{0}^{1}\unicode[STIX]{x1D719}^{\prime 2}\,\text{d}z-\langle {\mathcal{G}}(\unicode[STIX]{x1D742};\unicode[STIX]{x1D719},\unicode[STIX]{x1D6FC})\rangle ,\end{eqnarray}$$

where

(2.7) $$\begin{eqnarray}{\mathcal{G}}:=\int _{0}^{1}(\unicode[STIX]{x1D6FC}-1)\overline{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D742}|^{2}}+\overline{\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D708}_{1}\unicode[STIX]{x1D708}_{3}\unicode[STIX]{x1D719}^{\prime }}-(\unicode[STIX]{x1D6FC}-2)\overline{\unicode[STIX]{x1D708}_{1}\unicode[STIX]{x1D719}^{\prime \prime }}\,\text{d}z.\end{eqnarray}$$

Then the background method (Doering & Constantin Reference Doering and Constantin1994) is the observation that

(2.8) $$\begin{eqnarray}{\mathcal{L}}\leqslant \int _{0}^{1}\unicode[STIX]{x1D719}^{\prime 2}\,\text{d}z-\min _{\unicode[STIX]{x1D742}}{\mathcal{G}}(\unicode[STIX]{x1D742};\unicode[STIX]{x1D719},\unicode[STIX]{x1D6FC}),\end{eqnarray}$$

where only steady fields now need to be considered. The important point is $\min {\mathcal{G}}$ only exists for $\unicode[STIX]{x1D6FC}>1$ and $\unicode[STIX]{x1D719}$ which satisfy the spectral constraint (Doering & Constantin Reference Doering and Constantin1994). The best bound is found by then minimising the whole right-hand side over the (convex) set of such $\unicode[STIX]{x1D719}$ and $\unicode[STIX]{x1D6FC}$ .

2.2 The Otto–Seis ‘boundary layer’ method

The starting point for the Otto–Seis ‘boundary layer’ approach is again (2.2) and the same decomposition $\boldsymbol{u}=\unicode[STIX]{x1D719}(z)\hat{\boldsymbol{x}}+\unicode[STIX]{x1D742}$ is used. The key difference now is that ${\mathcal{L}}$ is re-expressed in terms of $\boldsymbol{u}$ and $\unicode[STIX]{x1D719}$ rather than $\unicode[STIX]{x1D742}$ and $\unicode[STIX]{x1D719}$ . So

(2.9) $$\begin{eqnarray}\displaystyle {\mathcal{L}}[\boldsymbol{u},\unicode[STIX]{x1D719}] & = & \displaystyle \left\langle \int _{0}^{1}\overline{|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}}\,\text{d}z\right\rangle -\unicode[STIX]{x1D6FC}\left\langle \int _{0}^{1}\overline{(\boldsymbol{u}-\unicode[STIX]{x1D719}(z)\hat{\boldsymbol{x}})\boldsymbol{\cdot }(\boldsymbol{N}\boldsymbol{S})}\,\text{d}z\right\rangle \nonumber\\ \displaystyle & = & \displaystyle \left\langle \int _{0}^{1}\overline{|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}}\,\text{d}z\right\rangle -\unicode[STIX]{x1D6FC}\left\langle \frac{\text{d}}{\text{d}t}\int _{0}^{1}\overline{\frac{1}{2}\boldsymbol{u}^{2}}\,\text{d}z+\int _{0}^{1}\overline{|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}}\,\text{d}z-GrU\right\rangle \nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D6FC}\left\langle \int _{0}^{1}\unicode[STIX]{x1D719}^{\prime }(\overline{uw}-\overline{u}_{z})\,\text{d}z\right\rangle -\unicode[STIX]{x1D6FC}GrU\end{eqnarray}$$

(2.10) $$\begin{eqnarray}\displaystyle & = & \displaystyle (1-\unicode[STIX]{x1D6FC})\left\langle \int _{0}^{1}\overline{|\unicode[STIX]{x1D735}\boldsymbol{u}|^{2}}\,\text{d}z\right\rangle -\unicode[STIX]{x1D6FC}\left\langle \int _{0}^{1}\unicode[STIX]{x1D719}^{\prime }(\overline{uw}-\overline{u}_{z})\,\text{d}z\right\rangle .\end{eqnarray}$$

At this point, the Euler–Lagrange equations

(2.11) $$\begin{eqnarray}\frac{\unicode[STIX]{x1D6FF}{\mathcal{L}}}{\unicode[STIX]{x1D6FF}\boldsymbol{u}}=\mathbf{0}\quad \text{and}\quad \frac{\unicode[STIX]{x1D6FF}{\mathcal{L}}}{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}}=\frac{\unicode[STIX]{x1D6FF}{\mathcal{L}}}{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6FC}}=0\end{eqnarray}$$

contain the background method bound as a solution but there is no means to identify it as such (i.e. appreciate that the associated value of ${\mathcal{L}}$ is a bound on the dissipation rate). Instead, the Otto–Seis approach appears to be to select a simplifying value of $\unicode[STIX]{x1D6FC}=1$ so that the right-hand side reduces to

(2.12) $$\begin{eqnarray}\unicode[STIX]{x1D700}=-\left\langle \int _{0}^{1}\unicode[STIX]{x1D719}^{\prime }(\overline{uw}-\overline{u}_{z})\,\text{d}z\right\rangle\end{eqnarray}$$

and then to choose a simple trial function

(2.13) $$\begin{eqnarray}\unicode[STIX]{x1D719}(z):=\frac{U}{1-\ell }\times \left\{\begin{array}{@{}ll@{}}(1-z)/\ell & 1-\ell \leqslant z\leqslant 1\\ 1 & \ell \leqslant z\leqslant 1-\ell \\ z/\ell & 0\leqslant z\leqslant \ell \end{array}\right.\end{eqnarray}$$

(designed so that $\int _{0}^{1}\unicode[STIX]{x1D719}\,\text{d}z=U$ with boundary layers of size $\ell$ ). This converts (2.12) into

(2.14) $$\begin{eqnarray}(1-\ell )Gr=\frac{1}{\ell }\int _{1-\ell }^{1}(\overline{uw}-\overline{u}_{z})\,\text{d}z-\frac{1}{\ell }\int _{0}^{\ell }(\overline{uw}-\overline{u}_{z})\,\text{d}z\end{eqnarray}$$

(after using $\unicode[STIX]{x1D700}=UGr$ ) which is (4.10) in Seis (Reference Seis2015) and then the strategy is to bound the terms on the right-hand side using powers of $\unicode[STIX]{x1D700}$ . The fundamental observation is that the derivative of the background field is what appears in the Otto–Seis ‘boundary layer’ method (Chernyshenko Reference Chernyshenko2017).

2.3 Background field for the rough problem

In the rough channel flow problem, the background (vector) field corresponding to the ‘boundary layer’ method as applied in Kerswell (Reference Kerswell2016) is $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D706})\boldsymbol{a}$ (generalised from $\unicode[STIX]{x1D719}(z)\hat{\boldsymbol{x}}$ in the smooth-walled problem) where

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D706}:=\frac{z-f(x,y)}{g(x,y)-f(x,y)},\quad \boldsymbol{a}:=\frac{\hat{\boldsymbol{x}}+F_{x}(x,y,\unicode[STIX]{x1D706})\hat{\boldsymbol{z}}}{g(x,y)-f(x,y)}\nonumber\\ \displaystyle & & \displaystyle \quad \text{and}\quad F(x,y,\unicode[STIX]{x1D706}):=(1-\unicode[STIX]{x1D706})f(x,y)+\unicode[STIX]{x1D706}g(x,y).\nonumber\end{eqnarray}$$

For example

(2.15) $$\begin{eqnarray}\iiint _{{\mathcal{V}}(\unicode[STIX]{x1D6EC})}\unicode[STIX]{x1D719}(\unicode[STIX]{x1D706})\boldsymbol{a}\boldsymbol{\cdot }(\boldsymbol{N}\boldsymbol{S})\,\text{d}V=\int _{0}^{1}\unicode[STIX]{x1D719}(\unicode[STIX]{x1D706})\int _{0}^{L_{y}}\int _{{\mathcal{C}}(\unicode[STIX]{x1D706},y)}\hat{\boldsymbol{s}}\boldsymbol{\cdot }(\boldsymbol{N}\boldsymbol{S})\,\text{d}s\,\text{d}y\,\text{d}\unicode[STIX]{x1D706},\end{eqnarray}$$

where the right-hand side is (2.6) of Kerswell (Reference Kerswell2016) before integrating with a general weight $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D706})$ over $\unicode[STIX]{x1D706}\in [0,1]$ . Taking $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D706})$ again as the piecewise-linear trial function defined in (2.13) allows the shears associated with it to be controllable. However, $\boldsymbol{a}$ varies spatially throughout the domain and so the shears associated with the combination do not vanish at some controlled distance from the boundary. This is what prevents the background method working and also has to break the boundary layer method. Unfortunately, this is a known limitation of the background method with no work-around currently on the horizon.