2.1 Normalisation and unit cell

Using $h$ as a length unit we employ the Cartesian coordinates $(x,y,z)$ with the $x$ -axis running perpendicular to the grooves, anti-parallel to the applied gradient, and the liquid domain lying within $0<y<2$ . The geometry is accordingly periodic in the $x$ -direction, the dimensionless semi-period being $\unicode[STIX]{x1D6EC}=l/h$ . The plane $x=0$ is chosen as that passing between the centres of an opposing pair of solid slats. (Note the different choice of origin in Yariv Reference Yariv2017.) The dimensionless geometry is portrayed in figure 1(b). It is convenient to label the slats such that the plane passing through the centre of the $n$ th slat (with $n\in \mathbb{Z}$ ) is $x=2n\unicode[STIX]{x1D6EC}$ .

We employ $Gh$ as a temperature unit for defining the dimensionless temperature $t$ , a function of $x$ and $y$ . Since the temperature is defined to within an arbitrary additive constant, we may assume – with no loss of generality – that $t$ is an odd function of $x$ . In particular, the externally imposed gradient is represented by the linear macroscopic variation,

(2.1) $$\begin{eqnarray}t=-x,\end{eqnarray}$$

which applies ‘deep’ within the solid substrate. Assuming long slats, the temperature at the end of the $n$ th slat (and in particular at the associated solid–liquid interface) is uniform, namely

(2.2) $$\begin{eqnarray}t\equiv -2n\unicode[STIX]{x1D6EC}.\end{eqnarray}$$

Since the liquid motion is driven by Marangoni stresses at the surfaces, we employ $-G\unicode[STIX]{x1D70E}_{T}$ as a stress unit (positive in the common case of negative $\unicode[STIX]{x1D70E}_{T}$ ). With the intrinsic length scale in the flow direction being $l$ , it is natural to use $-G\unicode[STIX]{x1D70E}_{T}l/\unicode[STIX]{x1D707}$ as the velocity unit. The two-dimensional velocity field is $\boldsymbol{u}=\hat{\boldsymbol{e}}_{x}u+\hat{\boldsymbol{e}}_{y}v$ , with $u$ and $v$ being functions of $x$ and $y$ .

Owing to the obvious periodicity of $\boldsymbol{u}$ , it is sufficient to consider the liquid domain restricted to the unit cell $|x|<\unicode[STIX]{x1D6EC}$ . Given the symmetry about the mid-plane $y=1$ , the domain of interest is further restricted to the ‘semi-cell’ $0<y<1$ .

2.2 Thermal problem

Neglecting heat advection, the temperature is governed by Laplace’s equation,

(2.3) $$\begin{eqnarray}\unicode[STIX]{x1D6FB}^{2}t=0\quad \text{for }-\unicode[STIX]{x1D6EC}<x<\unicode[STIX]{x1D6EC},\quad 0<y<1.\end{eqnarray}$$

At $y=0$ it satisfies (i) the temperature-continuity condition at the solid–liquid interface, which, upon making use of (2.2) reads

(2.4) $$\begin{eqnarray}t=0\quad \text{for }|x|<\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC};\end{eqnarray}$$

and (ii) the no-flux condition at the liquid–air interface,

(2.5) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}t}{\unicode[STIX]{x2202}y}=0\quad \text{for }\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<|x|<\unicode[STIX]{x1D6EC}.\end{eqnarray}$$

At $y=1$ it satisfies the symmetry condition

(2.6) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}t}{\unicode[STIX]{x2202}y}=0\quad \text{at }y=1,\end{eqnarray}$$

while at the lateral ends of the semi-cell it satisfies

(2.7) $$\begin{eqnarray}t=\mp \unicode[STIX]{x1D6EC}\quad \text{at }x=\pm \unicode[STIX]{x1D6EC}\quad \text{for }0<y<1,\end{eqnarray}$$

representing the imposed macroscopic gradient.

It is convenient to define the periodic function $\unicode[STIX]{x1D703}$ via the relation

(2.8) $$\begin{eqnarray}t(x,y)=-x+\unicode[STIX]{x1D703}(x,y).\end{eqnarray}$$

With $t$ being an odd function of $x$ , so must be $\unicode[STIX]{x1D703}$ . The ‘excess temperature’ $\unicode[STIX]{x1D703}$ satisfies the same boundary-value problem as $t$ , with two exceptions: the homogeneous condition (2.4) is replaced by the inhomogeneous condition,

(2.9) $$\begin{eqnarray}\unicode[STIX]{x1D703}=x\quad \text{for }|x|<\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC},\end{eqnarray}$$

while the jump condition (2.7) becomes

(2.10) $$\begin{eqnarray}\unicode[STIX]{x1D703}=0\quad \text{at }x=\pm \unicode[STIX]{x1D6EC}\quad \text{for }0<y<1.\end{eqnarray}$$

With advection being neglected, the thermal problem is uncoupled from the flow problem which we consider next.

2.3 Flow problem

The flow is governed by the continuity equation,

(2.11) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}x}+\frac{\unicode[STIX]{x2202}v}{\unicode[STIX]{x2202}y}=0,\end{eqnarray}$$

and the Stokes equations,

(2.12a,b ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}x}=\unicode[STIX]{x1D6FB}^{2}u,\quad \frac{\unicode[STIX]{x2202}p}{\unicode[STIX]{x2202}y}=\unicode[STIX]{x1D6FB}^{2}v,\end{eqnarray}$$

in which $p$ is the pressure field. At $y=0$ the flow satisfies impermeability,

(2.13) $$\begin{eqnarray}v=0.\end{eqnarray}$$

In addition, it satisfies there a no-slip condition at the edge of the solid slat,

(2.14) $$\begin{eqnarray}u=0\quad \text{for }|x|<\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC},\end{eqnarray}$$

as well as the Marangoni condition at the liquid–air interfaces,

(2.15) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}=\unicode[STIX]{x1D6EC}^{-1}\frac{\unicode[STIX]{x2202}t}{\unicode[STIX]{x2202}x}=\unicode[STIX]{x1D6EC}^{-1}\left(\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}{\unicode[STIX]{x2202}x}-1\right)\quad \text{for }\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<|x|<\unicode[STIX]{x1D6EC}.\end{eqnarray}$$

The symmetry about $y=1$ implies the relations

(2.16) $$\begin{eqnarray}v=\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}=0\quad \text{at }y=1.\end{eqnarray}$$

Noting that $u$ is an even function of $x$ , while $v$ is an odd one, the periodicity conditions adopt the form

(2.17) $$\begin{eqnarray}v=\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}x}=0\quad \text{at }x=\pm \unicode[STIX]{x1D6EC}\quad \text{for }0<y<1.\end{eqnarray}$$

The preceding equations are supplemented by the condition of pressure periodicity,

(2.18) $$\begin{eqnarray}p(x=\unicode[STIX]{x1D6EC},y)=p(x=-\unicode[STIX]{x1D6EC},y)\quad \text{for }0<y<1,\end{eqnarray}$$

representing the absence of a macroscopic pressure gradient. (Note that in the longitudinal problem this absence is reflected in the longitudinal velocity being governed by Laplace’s equation.) The Stokes equation (2.12b ) and the symmetry conditions (2.17) imply that $p$ is uniform at $x=\pm \unicode[STIX]{x1D6EC}$ ; since it is defined to within an additive constant, we may set it to zero at $x=0$ with no loss of generality. We conclude that $p$ is an odd function of $x$ , whereby (2.18) becomes

(2.19) $$\begin{eqnarray}p(x=\unicode[STIX]{x1D6EC})=0\quad \text{for }0<y<1.\end{eqnarray}$$

The preceding structure of two consecutive linear problems is reminiscent of classical thermocapillary problems involving drops and bubbles (Young, Goldstein & Block Reference Young, Goldstein and Block1959; Subramanian Reference Subramanian1981). With a vanishing Marangoni number, the dimensionless problem is governed by two geometric parameters, namely the semi-period $\unicode[STIX]{x1D6EC}=l/h$ and the solid fraction $\unicode[STIX]{x1D719}$ . Our interest is not in the detailed solution, but rather in the mean velocity,

(2.20) $$\begin{eqnarray}\bar{u}=\int _{0}^{1}u(x,y)\,\text{d}y,\end{eqnarray}$$

which coincides with (half of) the volumetric flux through the channel (per unit length in the $z$ -direction), This quantity, which is necessarily independent of $x$ , can only depend upon $\unicode[STIX]{x1D6EC}$ and $\unicode[STIX]{x1D719}$ .

2.4 Integral force balance

Integration of the $x$ -momentum balance (2.12a ) using the two-dimensional variants of the divergence and gradient theorems yields, upon making use of the pertinent symmetries about $x=0$ and the homogeneous conditions (2.16)–(2.18),

(2.21) $$\begin{eqnarray}\int _{-\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}\left.\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right|_{y=0}\,\text{d}x+2\int _{\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D6EC}}\left.\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right|_{y=0}\,\text{d}x=0.\end{eqnarray}$$

This relation may be interpreted as an integral force balance (per unit length in the $z$ -direction) in the $x$ -direction, where the first and second terms on the left-hand side represent the viscous forces which respectively act on the solid and free-surface portions of the compound boundary, and the nil right-hand side represents the absence of a net hydrodynamic force on the unit cell – a consequence of the absence of external forces. (Recall that the Marangoni forces originate in surface-tension gradients and should accordingly be considered as internal forces.)

Use of (2.15) together with (2.4) and (2.7) thus reveals that

(2.22) $$\begin{eqnarray}\int _{\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D6EC}}\left.\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right|_{y=0}\,\text{d}x=-1.\end{eqnarray}$$

Substitution into (2.21) then furnishes the integral relation

(2.23) $$\begin{eqnarray}\int _{-\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}\left.\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}\right|_{y=0}\,\text{d}x=2.\end{eqnarray}$$

Since this relation has been derived from the governing differential equations, it does not provide any independent information. Nonetheless, it comes in handy in resolving the small solid-fraction limit, addressed in § 6, where it is used for both the prediction of asymptotic scaling and the subsequent analysis.

3.1 An analogy with longitudinal shear-driven flow

We begin by embedding the harmonic field $t(x,y)$ in a complex potential, say

(3.1) $$\begin{eqnarray}t(x,y)+\text{i}\unicode[STIX]{x1D714}(x,y),\end{eqnarray}$$

wherein $\unicode[STIX]{x1D714}$ , the harmonic conjugate of $t$ , is related to $t$ via the Cauchy–Riemann conditions,

(3.2a,b ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}t}{\unicode[STIX]{x2202}x}=\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}y},\quad \frac{\unicode[STIX]{x2202}t}{\unicode[STIX]{x2202}y}=-\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}x}.\end{eqnarray}$$

Making use of these conditions and the boundary conditions (2.4)–(2.7) satisfied by $t$ we note that the harmonic function $\unicode[STIX]{x1D714}$ satisfies: (i) the mixed conditions at $y=0$ ,

(3.3a,b ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}y}=0\quad \text{for }|x|<\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC},\quad \unicode[STIX]{x1D714}=0\quad \text{for }\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<|x|<\unicode[STIX]{x1D6EC};\end{eqnarray}$$

(ii) the homogeneous Neumann conditions at the periodic boundaries,

(3.4) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}x}=0\quad \text{at }x=\pm \unicode[STIX]{x1D6EC};\end{eqnarray}$$

and (iii) the Dirichlet condition,

(3.5) $$\begin{eqnarray}\unicode[STIX]{x1D714}=\text{const.}\quad \text{at }y=1.\end{eqnarray}$$

While $\unicode[STIX]{x1D714}$ is defined to within an arbitrary additive constant, that freedom has already been exploited in setting the constant to zero in the Dirichlet condition (3.3b ). It follows that the value of the constant appearing in (3.5) cannot be set arbitrarily, implying in turn that an additional constraint is required. To obtain it we note that conditions (3.4), while reflecting the uniform temperature values at $x=\pm \unicode[STIX]{x1D6EC}$ , do not express the corresponding temperature-jump magnitude, as implied by (2.7). Rewriting (2.7) as

(3.6) $$\begin{eqnarray}\int _{-\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D6EC}}\frac{\unicode[STIX]{x2202}t}{\unicode[STIX]{x2202}x}\,\text{d}x=-2\unicode[STIX]{x1D6EC},\end{eqnarray}$$

where the integration may be carried out at any value of $y\in (0,1)$ , and making use of (3.2a ), we obtain the requisite constraint

(3.7) $$\begin{eqnarray}\int _{-\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D6EC}}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}y}\,\text{d}x=-2\unicode[STIX]{x1D6EC}.\end{eqnarray}$$

It is now readily observed that conditions (3.3)–(3.5) are identical to the boundary conditions governing the unidirectional longitudinal velocity between an ‘interchanged’ boundary, on which the liquid–air interface is at $|x|<\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}$ and the solid slats are at $\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<x<\unicode[STIX]{x1D6EC}$ , and a solid boundary which moves rigidly in the $z$ -direction. Constraint (3.7), which represents a force per unit area of unit magnitude which acts on the latter boundary in the negative $z$ -direction, uniquely determines that problem. Since that unidirectional field is also governed by Laplace’s equation, we have a perfect analogy. (The spirit of this analogy is reminiscent of a similar observation due to Hasimoto (Reference Hasimoto1958).)

The solution to the problem governing $\unicode[STIX]{x1D714}$ is derived in § 8 of Philip (Reference Philip1972a ), where the complex potential is provided in terms of the Jacobian elliptic functions. The associated force per unit area is provided in § 3.5 of Philip (Reference Philip1972b ).

3.2 An integral representation of $\bar{u}$

Consider now the flow problem, as specified by (2.11)–(2.17) and (2.19). The only inhomogeneous equation in that problem is the Marangoni condition (2.15), which is conveniently written

(3.8) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u}{\unicode[STIX]{x2202}y}=\unicode[STIX]{x1D6EC}^{-1}\frac{\text{d}t(x,0)}{\text{d}x}\quad \text{for }\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<|x|<\unicode[STIX]{x1D6EC}.\end{eqnarray}$$

The problem linearity then implies that the flow variables are linear functionals of (the $x$ -derivative of) $t(x,0)$ , as evaluated for $\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<|x|<\unicode[STIX]{x1D6EC}$ . It then follows from (2.20) that the same is true for the flux $\bar{u}$ , which is the quantity of interest.

Making use of Lorentz reciprocity (Happel & Brenner Reference Happel and Brenner1965), the requisite functional representation may be obtained without the need to solve the flow problem. To this end, we employ two velocity fields, namely the present field $\boldsymbol{u}=\hat{\boldsymbol{e}}_{x}u+\hat{\boldsymbol{e}}_{y}v$ and an auxiliary field $\tilde{\boldsymbol{u}}=\hat{\boldsymbol{e}}_{x}\tilde{u} +\hat{\boldsymbol{e}}_{y}\tilde{v}$ . Denoting the respective stress fields by $\unicode[STIX]{x1D748}$ and $\tilde{\unicode[STIX]{x1D748}}$ , the Lorentz reciprocal theorem reads

(3.9) $$\begin{eqnarray}\oint \text{d}a\,\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D748}\boldsymbol{\cdot }\tilde{\boldsymbol{u}}=\oint \text{d}a\,\hat{\boldsymbol{n}}\boldsymbol{\cdot }\tilde{\unicode[STIX]{x1D748}}\boldsymbol{\cdot }\boldsymbol{u},\end{eqnarray}$$

wherein $\text{d}a$ is a differential area element, normalised by $h^{2}$ , $\hat{\boldsymbol{n}}$ is an outward-pointing unit vector, and the integration is carried over a rectangular box, constructed by translating the semi-cell ( $-\unicode[STIX]{x1D6EC}<x<\unicode[STIX]{x1D6EC}$ , $0<y<1$ ) by a unit length in the $z$ -direction.

We take $\tilde{\boldsymbol{u}}$ as a pressure-driven flow which would exist in a channel which is bounded by two superhydrophobic walls, one at $y=0$ and one at $y=2$ . It is governed by a problem of the form (2.11)–(2.18), with the Marangoni condition (2.15) being replaced by the shear-free condition

(3.10) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\tilde{u} }{\unicode[STIX]{x2202}y}=0\quad \text{for }\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}<|x|<\unicode[STIX]{x1D6EC}\end{eqnarray}$$

and condition (2.18) replaced by

(3.11) $$\begin{eqnarray}\tilde{p}(x=-\unicode[STIX]{x1D6EC},y)-\tilde{p}(x=\unicode[STIX]{x1D6EC},y)=2\unicode[STIX]{x1D6EC}\quad \text{for }0<y<1.\end{eqnarray}$$

With the fields $\boldsymbol{u}$ and $\tilde{\boldsymbol{u}}$ depending only upon $x$ and $y$ , the contributions to (3.9) from the ‘front’ and ‘back’ sides, parallel to the $xy$ -plane, trivially vanish. Given the periodicity conditions, the contributions from the ‘side walls’ $x=\pm \unicode[STIX]{x1D6EC}$ to the left-hand side of (3.9) also vanish. On the other hand, the pressure-jump condition (3.11) implies that the side walls make a non-zero contribution to the right-hand side of (3.9); making use of definition (2.20), that contribution is readily seen to be given by $2\unicode[STIX]{x1D6EC}\bar{u}$ .

All that remain are the contributions from the ‘bottom’ ( $y=0$ ) and ‘top’ ( $y=1$ ) sides. Making use of the boundary conditions at $y=0$ and $y=1$ it is readily seen that these sides do not contribute to the right-hand side of (3.9). When considering the left-hand side of (3.9) we similarly find that the top side does not contribute, but that the bottom side does contribute. Making use of (3.8) and noting the vanishing of $\tilde{u}$ at the solid edge, the latter contribution is

(3.12) $$\begin{eqnarray}-2\unicode[STIX]{x1D6EC}^{-1}\int _{\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D6EC}}\frac{\text{d}t(x,0)}{\text{d}x}\tilde{u} (x,0)\,\text{d}x.\end{eqnarray}$$

We conclude that

(3.13) $$\begin{eqnarray}\bar{u}=-\unicode[STIX]{x1D6EC}^{-2}\int _{\unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}}^{\unicode[STIX]{x1D6EC}}\frac{\text{d}t(x,0)}{\text{d}x}\tilde{u} (x,0)\,\text{d}x.\end{eqnarray}$$

To evaluate $\bar{u}$ using representation (3.13) it is required to have both $t(x,0)$ and $\tilde{u} (x,0)$ available. As explained in § 3.1, the field $t$ may be obtained using the complex potential obtained in § 8 of Philip (Reference Philip1972a ). We are not aware of any closed-form representation of the field $\tilde{\boldsymbol{u}}$ , but semi-numerical solutions have been obtained using Fourier-series expansions (Teo & Khoo Reference Teo and Khoo2009). In principle, one may substitute these solutions into (3.13) to evaluate $\bar{u}$ for any numerical values of $\unicode[STIX]{x1D6EC}$ and $\unicode[STIX]{x1D719}$ .

In what follows, we supplement (3.13) with asymptotic approximations for $\bar{u}$ , obtained in three different geometric limits.

4.1 Thermal problem

Consider first the thermal problem in the near-boundary region. Since $t$ varies by an $O(\unicode[STIX]{x1D6EC})$ amount between $X=\pm 1$ , we postulate the asymptotic expansion

(4.4) $$\begin{eqnarray}t=\unicode[STIX]{x1D6EC}T(X,Y)+\cdots \,.\end{eqnarray}$$

Note that $T$ represents non-dimensionalisation by $Gl$ . This rescaled temperature satisfies Laplace’s equation in liquid domain,

(4.5) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}X^{2}}+\frac{\unicode[STIX]{x2202}^{2}T}{\unicode[STIX]{x2202}Y^{2}}=0\quad \text{in }{\mathcal{D}}.\end{eqnarray}$$

At $Y=0$ it satisfies the homogeneous Dirichlet condition at the solid–liquid interface,

(4.6) $$\begin{eqnarray}T=0\quad \text{for }|X|<\unicode[STIX]{x1D719},\end{eqnarray}$$

together with the no-flux condition at the liquid–gas interfaces,

(4.7) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}Y}=0\quad \text{for }\unicode[STIX]{x1D719}<|X|<1.\end{eqnarray}$$

As $Y\rightarrow \infty$ it satisfies

(4.8) $$\begin{eqnarray}T\rightarrow -X\quad \text{as }Y\rightarrow \infty ,\end{eqnarray}$$

while at $X=\pm 1$ is satisfies the jump conditions

(4.9) $$\begin{eqnarray}T=\mp 1\quad \text{at }X=\pm 1.\end{eqnarray}$$

4.2 Flow problem

It follows from (2.15) and (4.4) that $u$ is $O(1)$ in the near-boundary region. The continuity equation (2.11) then implies that so must be $v$ , while the Stokes equations (2.12) suggest $O(\unicode[STIX]{x1D6EC}^{-1})$ pressure variations. We accordingly employ the expansions,

(4.10a-c ) $$\begin{eqnarray}u=U(X,Y)+\cdots \,,\quad v=V(X,Y)+\cdots \,,\quad p=\unicode[STIX]{x1D6EC}^{-1}P(X,Y)+\cdots \,.\end{eqnarray}$$

The leading-order flow is governed by the continuity and Stokes equations,

(4.11a-c ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}X}+\frac{\unicode[STIX]{x2202}V}{\unicode[STIX]{x2202}Y}=0,\quad \frac{\unicode[STIX]{x2202}P}{\unicode[STIX]{x2202}X}=\frac{\unicode[STIX]{x2202}^{2}U}{\unicode[STIX]{x2202}X^{2}}+\frac{\unicode[STIX]{x2202}^{2}U}{\unicode[STIX]{x2202}Y^{2}},\quad \frac{\unicode[STIX]{x2202}P}{\unicode[STIX]{x2202}Y}=\frac{\unicode[STIX]{x2202}^{2}V}{\unicode[STIX]{x2202}X^{2}}+\frac{\unicode[STIX]{x2202}^{2}V}{\unicode[STIX]{x2202}Y^{2}}.\end{eqnarray}$$

At $Y=0$ it satisfies the impermeability condition,

(4.12) $$\begin{eqnarray}V=0\quad \text{at }Y=0;\end{eqnarray}$$

in addition, it satisfies a no-slip condition at the edge of the solid slats,

(4.13) $$\begin{eqnarray}U=0\quad \text{for }|X|<\unicode[STIX]{x1D719},\end{eqnarray}$$

as well as the Marangoni condition at the liquid–air interfaces,

(4.14) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}Y}=\frac{\unicode[STIX]{x2202}T}{\unicode[STIX]{x2202}X}\quad \text{for }\unicode[STIX]{x1D719}<|X|<1.\end{eqnarray}$$

At $X=\pm 1$ it satisfies the periodicity conditions,

(4.15) $$\begin{eqnarray}V=\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}X}=0\quad \text{at }X=\pm 1.\end{eqnarray}$$

Finally, it satisfies the pressure-periodicity conditions,

(4.16) $$\begin{eqnarray}P(X=1)=P(X=-1)\quad \text{for }Y>0,\end{eqnarray}$$

and the condition

(4.17) $$\begin{eqnarray}\lim _{Y\rightarrow \infty }\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}Y}=0,\end{eqnarray}$$

reflecting the necessary matching with a uniform flow.

Since $v$ vanishes outside the near-boundary region, the following decay condition applies,

(4.18) $$\begin{eqnarray}\lim _{Y\rightarrow \infty }V=0.\end{eqnarray}$$

On the other hand, (4.1) implies

(4.19) $$\begin{eqnarray}\lim _{Y\rightarrow \infty }U=u_{\infty }.\end{eqnarray}$$

Finally, note that in terms of the near-boundary variables, the integral balance (2.23) reads

(4.20) $$\begin{eqnarray}\int _{-\unicode[STIX]{x1D719}}^{\unicode[STIX]{x1D719}}\left.\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}Y}\right|_{Y=0}\,\text{d}X=2.\end{eqnarray}$$

4.3 Lorentz’s reciprocity

The only inhomogeneous equation in the problem (4.11)–(4.17) governing the near-boundary flow is the Marangoni condition (4.14), which is conveniently written

(4.21) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}U}{\unicode[STIX]{x2202}Y}=\frac{\text{d}T(X,0)}{\text{d}X}\quad \text{for }\unicode[STIX]{x1D719}<|X|<1.\end{eqnarray}$$

The problem linearity then implies that the flow variables are linear functionals of (the $X$ -derivative of) $T(X,0)$ . It then follows from (4.19) that the same is true for $u_{\infty }$ , which is the quantity of interest.

In principle, one may obtain the functional representation of $u_{\infty }$ by degenerating representation (3.13), obtained using Lorentz’s reciprocity, in the deep-channel limit. However, since we do not have available a closed-form expression for $\tilde{\boldsymbol{u}}$ in the above representation, we prefer to employ Lorentz’s reciprocity directly in the near-boundary problem.

As in the derivation of the general representation (3.13), we employ two velocity fields, namely the present field $\boldsymbol{U}=\hat{\boldsymbol{e}}_{x}U+\hat{\boldsymbol{e}}_{y}V$ and an auxiliary field $\tilde{\boldsymbol{U}}=\hat{\boldsymbol{e}}_{x}\tilde{U} +\hat{\boldsymbol{e}}_{y}\tilde{V}$ . Denoting the respective stress fields by $\unicode[STIX]{x1D72E}$ and $\tilde{\unicode[STIX]{x1D72E}}$ , the Lorentz reciprocal theorem reads

(4.22) $$\begin{eqnarray}\oint \text{d}A\,\hat{\boldsymbol{n}}\boldsymbol{\cdot }\unicode[STIX]{x1D72E}\boldsymbol{\cdot }\tilde{\boldsymbol{U}}=\oint \text{d}A\,\hat{\boldsymbol{n}}\boldsymbol{\cdot }\tilde{\unicode[STIX]{x1D72E}}\boldsymbol{\cdot }\boldsymbol{U},\end{eqnarray}$$

wherein $\text{d}A$ is a differential area element, normalised by $l^{2}$ , the integration is carried over a closed surface whose interior lies entirely in the liquid domain, and $\hat{\boldsymbol{n}}$ is an outward-pointing unit vector. We choose the integration surface as the boundary of a rectangular box, constructed by translating the rectangle $\{(X,Y)|-1<X<1,0<Y<Y_{\infty }\}$ a unit length in the $z$ -direction. In the above, $Y_{\infty }>0$ is arbitrary.

We take $\tilde{\boldsymbol{U}}$ as the velocity field due to applied shear, in the absence of any thermocapillary effects. It is governed by a problem of the form (4.11)–(4.17), with the Marangoni condition (4.14) being replaced by the shear-free condition

(4.23) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\tilde{U} }{\unicode[STIX]{x2202}Y}=0\quad \text{for }\unicode[STIX]{x1D719}<|X|<1\end{eqnarray}$$

and the matching condition (4.17) being replaced by the imposed-shear condition

(4.24) $$\begin{eqnarray}\tilde{U} \sim Y\quad \text{as }Y\rightarrow \infty .\end{eqnarray}$$

This large- $Y$ linear variation represents the leading-order term in a large- $Y$ asymptotic expansion. In particular, this term is followed by a constant term, namely the ‘slip length’ $\unicode[STIX]{x1D6FD}$ :

(4.25) $$\begin{eqnarray}\lim _{Y\rightarrow \infty }\{\tilde{U} -Y\}=\unicode[STIX]{x1D6FD}.\end{eqnarray}$$

With the chosen fields $\boldsymbol{U}$ and $\tilde{\boldsymbol{U}}$ , which depend only upon $X$ and $Y$ , the contribution from the ‘front’ and ‘back’ sides, parallel to the $XY$ -plane, trivially vanishes. Given the periodicity conditions, the contributions from the ‘side walls’ $X=\pm 1$ also vanish. All that remain are the contributions from the ‘bottom’ ( $Y=0$ ) and ‘top’ ( $Y=Y_{\infty }$ ) sides. Forming the limit $Y_{\infty }\rightarrow \infty$ we consider first the left-hand side of (4.22). Making use of (4.17) and (4.24) we find that the top side does not contribute. The only remaining contribution is from the bottom side, which, given the vanishing of $\tilde{U}$ at the solid edge, is

(4.26) $$\begin{eqnarray}-2\int _{\unicode[STIX]{x1D719}}^{1}\frac{\text{d}T(X,0)}{\text{d}X}\tilde{U} (X,0)\,\text{d}X.\end{eqnarray}$$

Similarly, considering the right-hand side of (4.22) in the same limit we find from (4.13) and (4.23) that the bottom side does not contribute. The only remaining contribution is from the top side; upon making use of (4.19) and (4.24), we find that contribution to be $2u_{\infty }$ . We conclude that

(4.27) $$\begin{eqnarray}u_{\infty }=-\int _{\unicode[STIX]{x1D719}}^{1}\frac{\text{d}T(X,0)}{\text{d}X}\tilde{U} (X,0)\,\text{d}X.\end{eqnarray}$$

4.4 Complex potential

Representation (4.27) constitutes the transverse counterpart of the original representation of Baier et al. (Reference Baier, Steffes and Hardt2010), derived in the longitudinal problem. In that problem the longitudinal temperature gradient is uniform, so the associated quadrature may be readily evaluated. This is not the case in the present problem, where evaluation of the quadrature appearing in (4.27) requires detailed knowledge of $T(X,0)$ . This obstacle has led Baier et al. (Reference Baier, Steffes and Hardt2010) to solve the transverse problem using numerical simulations.

In § 3.1 we have identified an analogy between the harmonic conjugate of the temperature field and a complementary pressure-driven problem. In principle, one may obtain $T$ by using that analogy and degenerating it to the deep-channel limit. Given the complexity of the expression provided by Philip (Reference Philip1972a ), however, we elect here to identify a direct analogy in context of the near-boundary problem. That analogy makes use of a complementary shear-driven problem.

We begin by embedding the harmonic field $T(X,Y)$ in a complex potential, say

(4.28) $$\begin{eqnarray}G(Z)=T(X,Y)+\text{i}\unicode[STIX]{x1D6FA}(X,Y),\end{eqnarray}$$

wherein $Z=X+\text{i}Y$ and $\unicode[STIX]{x1D6FA}$ is the harmonic conjugate of $T$ (cf. (3.1)). Making use of the Cauchy–Riemann conditions and the boundary conditions (4.6)–(4.9) satisfied by $T$ we note that the harmonic function $\unicode[STIX]{x1D6FA}$ satisfies: (i) the mixed conditions at $Y=0$ ,

(4.29a,b ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FA}}{\unicode[STIX]{x2202}Y}=0\quad \text{for }|X|<\unicode[STIX]{x1D719},\quad \unicode[STIX]{x1D6FA}=D\quad \text{for }\unicode[STIX]{x1D719}<|X|<1,\end{eqnarray}$$

wherein $D$ is a constant; (ii) the homogeneous Neumann conditions at the periodic boundaries,

(4.30) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FA}}{\unicode[STIX]{x2202}X}=0\quad \text{at }X=\pm 1;\end{eqnarray}$$

and (iii) the far-field condition,

(4.31) $$\begin{eqnarray}\unicode[STIX]{x1D6FA}\sim -Y\quad \text{as }Y\rightarrow \infty .\end{eqnarray}$$

Since $\unicode[STIX]{x1D6FA}$ is defined to within an arbitrary additive constant, we may set $D=0$ ; this does not affect the remaining conditions. It is now readily observed that the above problem, governing the temperature field due to transverse gradient, is analogous to the problem of shear-driven longitudinal velocity over an ‘interchanged’ boundary on which the liquid–air interface is at $|X|<\unicode[STIX]{x1D719}$ while the solid slats are at $\unicode[STIX]{x1D719}<|X|<1$ .

The solution to that problem is provided in § 5 of Philip (Reference Philip1972a ), where the velocity is represented as the imaginary part of a complex potential; in the present notation, this potential is

(4.32) $$\begin{eqnarray}-\frac{2}{\unicode[STIX]{x03C0}}\cos ^{-1}\frac{\cos (\unicode[STIX]{x03C0}Z/2)}{\cos (\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}/2)},\end{eqnarray}$$

for which the inverse cosine is defined such that $X$ and the real part of the potential increase together from $-1$ to $1$ . (Note that the problem was also solved by Crowdy Reference Crowdy2011 using a method which avoids the branch points associated with Philip’s solution.) Expression (4.32) accordingly provides $G(Z)$ , up to a possible addition of a real constant. Inspection reveals that no such addition is necessary. Considering (4.27), we actually need the temperature at the free-surface portions of $Y=0$ . Forming the real part of (4.32) we obtain

(4.33) $$\begin{eqnarray}T(X,0)=-\frac{2\operatorname{sgn}\{X\}}{\unicode[STIX]{x03C0}}\cos ^{-1}\frac{\cos (\unicode[STIX]{x03C0}X/2)}{\cos (\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}/2)}\quad \text{for }\unicode[STIX]{x1D719}<|X|<1.\end{eqnarray}$$

This expression varies from $0$ at $|X|=\unicode[STIX]{x1D719}$ to $\pm 1$ at $X=\pm 1$ , as required.

4.5 Calculation of $u_{\infty }$ : approximate and exact solutions

With $T(X,0)$ determined, all that is required for the calculation of $u_{\infty }$ using (4.27) is $\tilde{U} (X,0)$ . The field $\tilde{\boldsymbol{U}}$ , as specified after (4.22), was calculated in § 11 of Philip (Reference Philip1972a ). In particular, Philip (Reference Philip1972a ) provides the velocity at the liquid–air interfaces,

(4.34) $$\begin{eqnarray}\tilde{U} (X,0)=\frac{1}{\unicode[STIX]{x03C0}}\cosh ^{-1}\frac{\sin (\unicode[STIX]{x03C0}X/2)}{\sin (\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}/2)}\quad \text{for }\unicode[STIX]{x1D719}<|X|<1.\end{eqnarray}$$

(Note the error in (11.2) of Philip (Reference Philip1972a ), where $a$ should be replaced by $a^{2}$ .) Substitution of (4.33)–(4.34) into (4.27) yields the requisite integral representation of $u_{\infty }$ ,

(4.35) $$\begin{eqnarray}u_{\infty }=\frac{1}{\unicode[STIX]{x03C0}}\int _{\unicode[STIX]{x1D719}}^{1}\frac{\sin (\unicode[STIX]{x03C0}X/2)}{\sqrt{\cos ^{2}(\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}/2)-\cos ^{2}(\unicode[STIX]{x03C0}X/2)}}\cosh ^{-1}\frac{\sin (\unicode[STIX]{x03C0}X/2)}{\sin (\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}/2)}\,\text{d}X.\end{eqnarray}$$

The quadrature (4.35) may be evaluated numerically for any value of $\unicode[STIX]{x1D719}$ . The resulting variation of $u_{\infty }$ with $\unicode[STIX]{x1D719}$ is portrayed in figure 2.

Figure 2.

Dependence upon $\unicode[STIX]{x1D719}$ of the leading-order outer velocity, as determined from quadrature (4.35). The dashed lines depict the small solid-fraction approximation (4.43) and the small gas-fraction approximation (4.39).

The preceding calculation of $u_{\infty }$ may be supplemented by approximate evaluations of the quadrature (4.27). Consider first the limit of small gas fraction, $\unicode[STIX]{x1D6E5}=1-\unicode[STIX]{x1D719}\ll 1$ . Defining the scaled coordinate

(4.36) $$\begin{eqnarray}\hat{x}=\frac{1-X}{\unicode[STIX]{x1D6E5}}\end{eqnarray}$$

the representation (4.27) becomes

(4.37) $$\begin{eqnarray}u_{\infty }=\int _{0}^{1}\tilde{U} (1-\hat{x}\unicode[STIX]{x1D6E5},0)\frac{\text{d}}{\text{d}\hat{x}}T(1-\hat{x}\unicode[STIX]{x1D6E5},0)\,\text{d}\hat{x}.\end{eqnarray}$$

For $\unicode[STIX]{x1D6E5}\ll 1$ and $\hat{x}=O(1)$ we readily obtain from (4.33) and (4.34)

(4.38a,b ) $$\begin{eqnarray}T(1-\hat{x}\unicode[STIX]{x1D6E5},0)\sim -\frac{2}{\unicode[STIX]{x03C0}}\cos ^{-1}\hat{x},\quad \tilde{U} (1-\hat{x}\unicode[STIX]{x1D6E5},0)\sim \unicode[STIX]{x1D6E5}\frac{(1-\hat{x}^{2})^{1/2}}{2}.\end{eqnarray}$$

Substitution into (4.43) yields

(4.39) $$\begin{eqnarray}u_{\infty }\sim \frac{\unicode[STIX]{x1D6E5}}{\unicode[STIX]{x03C0}}\quad \text{for }\unicode[STIX]{x1D6E5}\ll 1.\end{eqnarray}$$

The second limit we consider is that of small solid fractions, $\unicode[STIX]{x1D719}\ll 1$ , for which (4.33) gives $T(X,0)\rightarrow -X$ . Substitution into (4.27) implies that $u_{\infty }$ is given by the asymptotic limit of

(4.40) $$\begin{eqnarray}\int _{\unicode[STIX]{x1D719}}^{1}\tilde{U} (X,0)\,\text{d}X\end{eqnarray}$$

as $\unicode[STIX]{x1D719}\rightarrow 0$ . The representation (4.40) is reminiscent to that encountered in the longitudinal problem, where the temperature variation is linear. Following Baier et al. (Reference Baier, Steffes and Hardt2010), the integral (4.40) may be evaluated using, again, Lorentz reciprocity, with the first flow field being $\tilde{\boldsymbol{U}}-Y\hat{\boldsymbol{e}}_{x}$ and the second one being the simple Couette flow $Y\hat{\boldsymbol{e}}_{x}$ . Noting from (4.25) that the first field approaches the uniform velocity $\unicode[STIX]{x1D6FD}\hat{\boldsymbol{e}}_{x}$ at large $Y$ , with the accompanying stress field decaying there, we find that

(4.41) $$\begin{eqnarray}\int _{\unicode[STIX]{x1D719}}^{1}\tilde{U} (X,0)\,\text{d}X=\unicode[STIX]{x1D6FD}.\end{eqnarray}$$

(Note that this relation holds for all $\unicode[STIX]{x1D719}$ -values.) The value of $\unicode[STIX]{x1D6FD}$ for the transverse flow problem calculated by Philip (Reference Philip1972a ) was determined by Lauga & Stone (Reference Lauga and Stone2003) as

(4.42) $$\begin{eqnarray}\unicode[STIX]{x1D6FD}=-\frac{1}{\unicode[STIX]{x03C0}}\ln \sin \frac{\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}}{2};\end{eqnarray}$$

Substitution into (4.40) of (4.41)–(4.42) gives

(4.43) $$\begin{eqnarray}u_{\infty }\sim \frac{1}{\unicode[STIX]{x03C0}}\ln \frac{2}{\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}}+o(1)\quad \text{for }\unicode[STIX]{x1D719}\ll 1.\end{eqnarray}$$

Approximations (4.39) are (4.43) are portrayed in figure 2. The agreement with the exact solution is evident.

6.1 Leading-order flow problems

The leading-order outer flow is governed by the continuity and Stokes equations. Since at leading order the solid edge appears as a singular point, the shear-stress condition (2.15) applies at the entire $y=0$ boundary (sans one point), namely for $0<|x|<\unicode[STIX]{x1D6EC}$ . With the excess temperature being $\ll \unicode[STIX]{x1D719}$ in the outer region, the Marangoni term in that condition does not affect the leading-order balance, which is accordingly homogeneous,

(6.5) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u^{\ast }}{\unicode[STIX]{x2202}y}=0\quad \text{for }0<|x|<\unicode[STIX]{x1D6EC}.\end{eqnarray}$$

In addition, $u^{\ast }$ also satisfies the appropriate counterparts of (2.17)–(2.18). Since all these homogeneous conditions involve the velocity derivatives, it follows that $u^{\ast }$ possesses a uniform solution. Considering the problem governing $v^{\ast }$ and noticing the homogeneous symmetry condition (2.13) and the comparable Dirichlet condition implied by the impermeability condition (2.16), we conclude that $v^{\ast }\equiv 0$ .

To determine the value of $u^{\ast }$ we consider the inner problem. The leading-order flow in the half-space $y^{\prime }>0$ is governed by the continuity and Stokes equations. In addition, at $y^{\prime }=0$ it satisfies impermeability (cf. (2.13)),

(6.6) $$\begin{eqnarray}v^{\prime }=0,\end{eqnarray}$$

together with the mixed conditions (cf. (2.14)–(2.15))

(6.7a,b ) $$\begin{eqnarray}u^{\prime }=0\quad \text{for }|x^{\prime }|<1,\quad \frac{\unicode[STIX]{x2202}u^{\prime }}{\unicode[STIX]{x2202}y^{\prime }}=0\quad \text{for }|x^{\prime }|>1.\end{eqnarray}$$

Note that the preceding problem is entirely homogeneous and accordingly does not possess a unique solution. The homogeneity is a consequence of the small asymptotic magnitude of the Marangoni term in (2.15).

In principle, one can proceed to the next asymptotic order, where the Marangoni term appears explicitly. The alternative approach adopted here makes use instead of the integral condition (2.23). This condition, which applies here at the inner region, introduces the inhomogeneous dependence upon the thermal forcing. At leading order, it reads

(6.8) $$\begin{eqnarray}\int _{-1}^{1}\left.\frac{\unicode[STIX]{x2202}u^{\prime }}{\unicode[STIX]{x2202}y^{\prime }}\right|_{y^{\prime }=0}\,\text{d}x^{\prime }=2.\end{eqnarray}$$

6.2 Analogy with the flow about a finite line segment

The structure of the continuity and Stokes equations in conjunction with the homogeneous conditions (6.6)–(6.7) allow for the forming of a symmetric extension of the flow problem about $y^{\prime }=0$ , with

(6.9a,b ) $$\begin{eqnarray}u^{\prime }(-y^{\prime })=u^{\prime }(y^{\prime }),\quad v^{\prime }(-y^{\prime })=-v(y^{\prime }).\end{eqnarray}$$

In the extended region the boundary conditions applying above and below the line segment $\{-1<x^{\prime }<1,y^{\prime }=0\}$ are the same as those applying over a stationary solid segment; with that interpretation, the integral constraint (6.8) represents a hydrodynamic force of magnitude $4$ which acts on that segment in the $x$ -direction.

Considering now large distances away from the segment, ${x^{\prime }}^{2}+{y^{\prime }}^{2}\gg 1$ , we retrieve the problem of a two-dimensional point force. In the present dimensionless notation, a concentrated force $\boldsymbol{F}$ (per unit length in the $z$ -direction) which acts on the fluid at the origin results in the velocity field (Pozrikidis Reference Pozrikidis1992)

(6.10) $$\begin{eqnarray}\frac{\boldsymbol{F}}{8\unicode[STIX]{x03C0}}\boldsymbol{\cdot }\left\{-\unicode[STIX]{x1D644}\ln ({x^{\prime }}^{2}+{y^{\prime }}^{2})+2\frac{\boldsymbol{x}^{\prime }\boldsymbol{x}^{\prime }}{|\boldsymbol{x}^{\prime }|^{2}}\right\}\end{eqnarray}$$

wherein $\boldsymbol{x}^{\prime }=x^{\prime }\hat{\boldsymbol{e}}_{x}+y^{\prime }\hat{\boldsymbol{e}}_{y}$ and $\unicode[STIX]{x1D644}$ is the idemfactor. With $\boldsymbol{F}$ being here $-4\hat{\boldsymbol{e}}_{x}$ we obtain, as ${x^{\prime }}^{2}+{y^{\prime }}^{2}\rightarrow \infty$

(6.11a,b ) $$\begin{eqnarray}u^{\prime }\sim \frac{1}{2\unicode[STIX]{x03C0}}\ln ({x^{\prime }}^{2}+{y^{\prime }}^{2})-\frac{{x^{\prime }}^{2}}{\unicode[STIX]{x03C0}({x^{\prime }}^{2}+{y^{\prime }}^{2})}+o(1),\quad v^{\prime }\sim -\frac{x^{\prime }y^{\prime }}{\unicode[STIX]{x03C0}({x^{\prime }}^{2}+{y^{\prime }}^{2})}+o(1).\end{eqnarray}$$

In terms of the outer coordinates, the $x$ -component (6.11a ) becomes

(6.12) $$\begin{eqnarray}\frac{1}{\unicode[STIX]{x03C0}}\left\{\ln \frac{1}{\unicode[STIX]{x1D719}}+\ln \sqrt{x^{2}+y^{2}}-\frac{x^{2}}{\unicode[STIX]{x03C0}(x^{2}+y^{2})}+o(1)\right\}.\end{eqnarray}$$

The need for asymptotic matching implies that the corresponding leading-order outer velocity component should approach (6.12) as $(x,y)\rightarrow (0,0)$ . Given expansion (6.4a ) and recalling that $u^{\ast }$ is uniform we readily conclude that

(6.13) $$\begin{eqnarray}u^{\ast }\equiv 1/\unicode[STIX]{x03C0}.\end{eqnarray}$$

Comparing with definition (2.20) it follows that the mean velocity is

(6.14) $$\begin{eqnarray}\bar{u}\sim \frac{1}{\unicode[STIX]{x03C0}}\ln \frac{1}{\unicode[STIX]{x1D719}}+\cdots \quad \text{for }\unicode[STIX]{x1D719}\ll 1.\end{eqnarray}$$

Note that this is just half of the corresponding result in the longitudinal case: see indeed (3.7) of Yariv (Reference Yariv2018). At leading order, approximation (6.14) agrees with (4.43). On the other hand, it does not agree with (5.17). We conclude that the respective limits of small solid fraction and shallow channels do not commute. This is hardly surprising.

While the calculation of the excess temperature in the inner region is not required for deriving (6.13), its distribution is still of some interest. It is presented in appendix A.