2.1 Governing equations

We represent the system in non-dimensional variables using $a^{\prime }$ , $\unicode[STIX]{x1D705}U_{max}^{\prime }$ , $\unicode[STIX]{x1D707}^{\prime }\unicode[STIX]{x1D705}U_{max}^{\prime }/a^{\prime }$ and $a^{\prime }|E_{\infty }^{\prime }|$ as the characteristic scales for length, velocity, pressure and electrostatic potential, respectively. Here, $\unicode[STIX]{x1D705}$ is the size ratio of the particle radius $a^{\prime }$ to the channel width $l^{\prime }$ , and $\unicode[STIX]{x1D707}^{\prime }$ denotes the viscosity of the electrolyte. If no external concentration gradient is applied, the electrostatic potential ( $\unicode[STIX]{x1D719}$ ) in the ‘electro-neutral’ bulk region (zero charge density) is governed by the Laplace equation

(2.1) $$\begin{eqnarray}\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D719}=0.\end{eqnarray}$$

Since the effect of streaming potential is neglected in this work, a no-flux condition for the electrostatic potential is imposed at the surface of the particle

(2.2) $$\begin{eqnarray}\boldsymbol{e}_{r}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}=0\quad \text{at }r=1.\end{eqnarray}$$

The electric field remains parallel to the walls (located at $z=s$ and $z=1-s$ , where $s\equiv d^{\prime }/l^{\prime }$ ), yielding

(2.3) $$\begin{eqnarray}\boldsymbol{e}_{z}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}=0\quad \text{at walls}.\end{eqnarray}$$

Far away from the particle, the electric potential distribution approaches its undisturbed state ( $\unicode[STIX]{x1D719}_{\infty }\sim -x$ )

(2.4) $$\begin{eqnarray}\unicode[STIX]{x1D719}\rightarrow \unicode[STIX]{x1D719}_{\infty }\quad \text{as }r\rightarrow \infty .\end{eqnarray}$$

The hydrodynamics of the current problem consists of three time scales: (i) viscous time scale ( $t_{visc}^{\prime }\sim a^{\prime 2}/\unicode[STIX]{x1D708}^{\prime }$ ), (ii) convective time scale of the flow ( $t_{conv}^{\prime }\sim a^{\prime }/\unicode[STIX]{x1D705}U_{max}^{\prime }$ ) and (iii) migration (or geometric) time scale ( $t_{mig}^{\prime }\sim a^{\prime }/U_{mig}^{\prime }$ ). Since the time dependence enters the system through the migration time scale, temporal variations can be assumed to be negligible if the migration time scale is much larger than the convective time scale ( $t_{mig}^{\prime }\gg t_{conv}^{\prime }$ ) (Becker, McKinley & Stone Reference Becker, McKinley and Stone1996). Since the migration occurs due to weak inertia, the migration velocity is much smaller than the characteristic velocity ( $U_{mig}^{\prime }\ll \unicode[STIX]{x1D705}U_{max}^{\prime }$ ), which suggests that we can neglect the temporal variations. Therefore, the velocity and pressure distribution in the electroneutral bulk are governed by continuity and quasi-steady Navier–Stokes equation. The term quasi-steady implies that the dynamics depends only on the instantaneous geometric configuration (i.e. particle position between the walls), provided $U_{mig}^{\prime }\ll \unicode[STIX]{x1D705}U_{max}^{\prime }$ . (A detailed derivation is described in the supplementary material S1.) In the frame of reference of the moving particle, the flow is governed by

(2.5a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{U}=0,\quad \unicode[STIX]{x1D6FB}^{2}\boldsymbol{U}-\unicode[STIX]{x1D735}P=\mathit{Re}_{p}(\boldsymbol{U}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{U}).\end{eqnarray}$$

Here, $Re_{p}$ is particle Reynolds number defined as

(2.6) $$\begin{eqnarray}\mathit{Re}_{p}=\frac{\unicode[STIX]{x1D70C}^{\prime }U_{max}^{\prime }\unicode[STIX]{x1D705}a^{\prime }}{\unicode[STIX]{x1D707}^{\prime }}.\end{eqnarray}$$

The sphere moves with a – yet to be determined – translational and rotational velocity: $\boldsymbol{U}_{s}$ and $\unicode[STIX]{x1D734}_{s}$ , respectively. Since the domain is infinite in the $y$ -direction, we can consider: $\boldsymbol{U}_{s}=U_{sx}\boldsymbol{e}_{x}+U_{sz}\boldsymbol{e}_{z}$ and $\unicode[STIX]{x1D734}_{s}=\unicode[STIX]{x1D6FA}_{sy}\boldsymbol{e}_{y}$ (i.e. the particle rotates in the direction of the background flow vorticity). To account for the electrokinetic effects inside the double layer, we employ a macroscale description i.e. the slip-boundary condition derived by Smoluchowski (Reference Smoluchowski1903). We obtain the following boundary condition at the particle surface:

(2.7) $$\begin{eqnarray}\boldsymbol{U}=\unicode[STIX]{x1D734}_{s}\times \boldsymbol{r}+HaZ_{p}\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}\quad \text{at }r=1.\end{eqnarray}$$

Here, $Z_{p}$ is the dimensionless particle surface zeta potential ( $\unicode[STIX]{x1D701}_{p}^{\prime }/|E_{\infty }^{\prime }|a^{\prime }$ ). $Ha$ is a Hartmann-type number: $Ha=\unicode[STIX]{x1D700}^{\prime }E_{\infty }^{\prime 2}a^{\prime }/4\unicode[STIX]{x03C0}\unicode[STIX]{x1D707}^{\prime }\unicode[STIX]{x1D705}U_{max}^{\prime }$ , which denotes the ratio of electrical energy density to shear stress. Here $\unicode[STIX]{x1D700}^{\prime }$ is the electrical permittivity. In this work, based on the experimental conditions (Kim & Yoo Reference Kim and Yoo2009a ; Cevheri & Yoda Reference Cevheri and Yoda2014; Li & Xuan Reference Li and Xuan2018), we address the $Ha\sim O(1)$ and $Z_{p}\sim O(1)$ regime ( $E_{\infty }^{\prime }\approx O(10^{3}){-}O(10^{4})~\text{V}~\text{m}^{-1}$ , $\unicode[STIX]{x1D701}_{p}^{\prime }=50~\text{mV}$ , $a^{\prime }=10~\unicode[STIX]{x03BC}\text{m}$ , $\unicode[STIX]{x1D707}=10^{-3}~\text{Pa}~\text{s}$ , $U_{max}^{\prime }=2.25~\text{mm}~\text{s}^{-1}$ , $\unicode[STIX]{x1D705}=10^{-2}$ , $\unicode[STIX]{x1D700}^{\prime }=80\unicode[STIX]{x1D700}_{vacuum}^{\prime }$ ).

At walls ( $z=s$ and $z=1-s$ ), the flow satisfies the Smoluchowski slip condition

(2.8) $$\begin{eqnarray}\boldsymbol{U}=HaZ_{w}\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}-\boldsymbol{U}_{s}\quad \text{at walls.}\end{eqnarray}$$

Here, $Z_{w}$ is the dimensionless wall surface zeta potential ( $\unicode[STIX]{x1D701}_{w}^{\prime }/|E_{\infty }^{\prime }|a^{\prime }$ ). The flow remains undisturbed far away from the particle and satisfies

(2.9) $$\begin{eqnarray}\boldsymbol{U}\rightarrow \boldsymbol{V}_{\infty }\quad \text{as }r\rightarrow \infty .\end{eqnarray}$$

Here, $\boldsymbol{V}_{\infty }$ denotes the undisturbed background flow (Poiseuille flow combined with electro-osmosis) in the frame of reference of the particle

(2.10) $$\begin{eqnarray}\boldsymbol{V}_{\infty }=(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}z+\unicode[STIX]{x1D6FE}z^{2})\boldsymbol{e}_{x}+HaZ_{w}\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{\infty }-\boldsymbol{U}_{s}.\end{eqnarray}$$

The constants $\unicode[STIX]{x1D6FC}$ , $\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D6FE}$ are

(2.11a-c ) $$\begin{eqnarray}\unicode[STIX]{x1D6FC}=4U_{max}s(1-s),\quad \unicode[STIX]{x1D6FD}=4U_{max}\unicode[STIX]{x1D705}(1-2s),\quad \unicode[STIX]{x1D6FE}=-4U_{max}\unicode[STIX]{x1D705}^{2},\end{eqnarray}$$

where $\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D6FE}$ represent the shear and curvature of the background flow.

Determination of $\boldsymbol{U}_{s}$ and $\unicode[STIX]{x1D734}_{s}$ . The translational and rotational velocities of the sphere are evaluated by acknowledging that the freely suspended neutral particle (charged surface plus counter-ions) is force-free and torque-free (see § A.1 for details). Therefore, in the formulation above, the overall (hydrodynamic and electric) force ( $\boldsymbol{F}_{H}+\boldsymbol{F}_{M}$ ) and torque ( $\boldsymbol{L}_{H}+\boldsymbol{L}_{M}$ ) on the particle must be zero,

(2.12) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{F}=\boldsymbol{F}_{H}+\boldsymbol{F}_{M}=\int _{S_{p}}\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D72E}_{H}\,\text{d}S+4\unicode[STIX]{x03C0}Ha\int _{S_{p}}\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D72E}_{M}\,\text{d}S=\mathbf{0}, & \displaystyle\end{eqnarray}$$

(2.13) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{L}=\boldsymbol{L}_{H}+\boldsymbol{L}_{M}=\int _{S_{p}}\boldsymbol{n}\times (\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D72E}_{H})\,\text{d}S+4\unicode[STIX]{x03C0}Ha\int _{S_{p}}\boldsymbol{n}\times (\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D72E}_{M})\,\text{d}S=\mathbf{0}. & \displaystyle\end{eqnarray}$$

The above force and torque have been normalized with $\unicode[STIX]{x1D707}^{\prime }a^{\prime }\unicode[STIX]{x1D705}U_{max}^{\prime }$ and $\unicode[STIX]{x1D707}^{\prime }a^{\prime 2}\unicode[STIX]{x1D705}U_{max}^{\prime }$ , respectively. Here, $\unicode[STIX]{x1D72E}_{H}$ is the dimensionless Newtonian stress tensor which contributes to hydrodynamic force; $\unicode[STIX]{x1D72E}_{M}$ is the dimensionless Maxwell stress tensor which contributes to the electric force

(2.14a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D72E}_{H}=-P\unicode[STIX]{x1D644}+\unicode[STIX]{x1D735}\boldsymbol{U}+(\unicode[STIX]{x1D735}\boldsymbol{U})^{\text{T}}\quad \text{and}\quad \unicode[STIX]{x1D72E}_{M}=\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}-{\textstyle \frac{1}{2}}(\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D719})\unicode[STIX]{x1D644}.\end{eqnarray}$$

Here, $\unicode[STIX]{x1D644}$ is the second-order identity tensor and the superscript T denotes transpose.

2.2 Equations governing the disturbance

The solution to (2.1)–(2.5), (2.7)–(2.9), (2.12) and (2.13) is sought in terms of disturbance variables: electrostatic potential ( $\unicode[STIX]{x1D713}=\unicode[STIX]{x1D719}-\unicode[STIX]{x1D719}_{\infty }$ ), velocity ( $\boldsymbol{v}=\boldsymbol{U}-\boldsymbol{V}_{\infty }$ ) and pressure ( $p=P-P_{\infty }$ ). These represent the deviation of actual fields from the undisturbed fields. The disturbance electrostatic potential is governed by

(2.15) $$\begin{eqnarray}\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D713}=0,\end{eqnarray}$$

and is subject to the following boundary conditions:

(2.16) $$\begin{eqnarray}\displaystyle & \boldsymbol{e}_{r}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}=-\boldsymbol{e}_{r}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{\infty }\quad \text{at }r=1, & \displaystyle\end{eqnarray}$$

(2.17) $$\begin{eqnarray}\displaystyle & \boldsymbol{e}_{z}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}=0\quad \text{at walls}, & \displaystyle\end{eqnarray}$$

(2.18) $$\begin{eqnarray}\displaystyle & \unicode[STIX]{x1D713}\rightarrow 0\quad \text{as }r\rightarrow \infty . & \displaystyle\end{eqnarray}$$

The disturbance velocity is governed by

(2.19a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}=0,\quad \unicode[STIX]{x1D6FB}^{2}\boldsymbol{v}-\unicode[STIX]{x1D735}p=Re_{p}(\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}+\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{V}_{\infty }+\boldsymbol{V}_{\infty }\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}),\end{eqnarray}$$

and is subject to the following boundary conditions:

(2.20) $$\begin{eqnarray}\displaystyle & \boldsymbol{v}=\unicode[STIX]{x1D734}_{s}\times \boldsymbol{r}+HaZ_{p}\unicode[STIX]{x1D735}(\unicode[STIX]{x1D713}+\unicode[STIX]{x1D719}_{\infty })-\boldsymbol{V}_{\infty }\quad \text{at }r=1, & \displaystyle\end{eqnarray}$$

(2.21) $$\begin{eqnarray}\displaystyle & \boldsymbol{v}=HaZ_{w}\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}\quad \text{at walls,} & \displaystyle\end{eqnarray}$$

(2.22) $$\begin{eqnarray}\displaystyle & \boldsymbol{v}\rightarrow \mathbf{0}\quad \text{as }r\rightarrow \infty . & \displaystyle\end{eqnarray}$$

The disturbance hydrodynamic and Maxwell stresses are defined as

(2.23a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D748}_{H}=-p\unicode[STIX]{x1D644}+\unicode[STIX]{x1D735}\boldsymbol{v}+(\unicode[STIX]{x1D735}\boldsymbol{v})^{\text{T}}\quad \unicode[STIX]{x1D748}_{M}=\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}-{\textstyle \frac{1}{2}}(\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713})\unicode[STIX]{x1D644}.\end{eqnarray}$$

Since only the disturbance stresses must contribute to force and torque on the particle, we write

(2.24) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{F}=\boldsymbol{F}_{H}+\boldsymbol{F}_{M}=\int _{S_{p}}\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D748}_{H}\,\text{d}S+4\unicode[STIX]{x03C0}Ha\int _{S_{p}}\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D748}_{M}\,\text{d}S=\mathbf{0}, & \displaystyle\end{eqnarray}$$

(2.25) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{L}=\boldsymbol{L}_{H}+\boldsymbol{L}_{M}=\int _{S_{p}}\boldsymbol{n}\times (\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D748}_{H})\,\text{d}S+4\unicode[STIX]{x03C0}Ha\int _{S_{p}}\boldsymbol{n}\times (\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D748}_{M})\,\text{d}S=\mathbf{0}. & \displaystyle\end{eqnarray}$$

2.3 Regular perturbation expansion

The primary objective of this work is to obtain the lift associated with inertia for an asymptotically small $Re_{p}$ . Towards this, we perform a regular perturbation expansion in $Re_{p}$ . This expansion is valid only when the inertia persists as a perturbation throughout the domain. Therefore, the wall particle distance must be shorter than the length scale where the inertial force balances viscous force (Guazzelli & Morris Reference Guazzelli and Morris2011). The order of magnitude of the ratio of inertial to viscous forces shows that the length scale associated with the bulk shear ( $a^{\prime }/\sqrt{Re_{p}}$ ) is much shorter than that based on the electrokinetic slip ( $a^{\prime }/HaZ_{p}Re_{p}$ ). Therefore, based on the inertial length scale of bulk shear, we arrive at the following constraint for the entire domain to be dominated by viscous forces

(2.26) $$\begin{eqnarray}O(Re_{p})\ll \unicode[STIX]{x1D705}^{2}.\end{eqnarray}$$

This allows us to seek the solution for disturbance velocity, pressure, translational and rotational velocities as a perturbation expansion in $Re_{p}$

(2.27) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\boldsymbol{v}=\boldsymbol{v}^{(0)}+Re_{p}\boldsymbol{v}^{(1)}+\cdots \quad p=p^{(0)}+Re_{p}p^{(1)}+\cdots \,,\\ \boldsymbol{U}_{s}=\boldsymbol{U}_{s}^{(0)}+Re_{p}\boldsymbol{U}_{s}^{(1)}+\cdots \quad \unicode[STIX]{x1D734}_{s}=\unicode[STIX]{x1D734}_{s}^{(0)}+Re_{p}\unicode[STIX]{x1D734}_{s}^{(1)}+\cdots \,.\end{array}\right\}\end{eqnarray}$$

We substitute the above expansion into the equations governing hydrodynamics (2.19)–(2.22). We obtain the $O(1)$ problem as

(2.28) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}^{(0)}\, & =\, & 0,\\ \unicode[STIX]{x1D6FB}^{2}\boldsymbol{v}^{(0)}-\unicode[STIX]{x1D735}p^{(0)}\, & =\, & \mathbf{0},\\ \boldsymbol{v}^{(0)}\, & =\, & \unicode[STIX]{x1D734}_{s}^{(0)}\times \boldsymbol{r}+HaZ_{p}\unicode[STIX]{x1D735}(\unicode[STIX]{x1D713}+\unicode[STIX]{x1D719}_{\infty })-\boldsymbol{V}_{\infty }^{(0)}\quad \text{at }r=1,\\ \boldsymbol{v}^{(0)}\, & =\, & HaZ_{w}\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}\quad \text{at walls},\\ \boldsymbol{v}^{(0)}\, & \rightarrow \, & \mathbf{0}\quad \text{as }r\rightarrow \infty ,\end{array}\right\}\end{eqnarray}$$

and the $O(Re_{p})$ problem as:

(2.29) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}^{(1)}\, & =\, & 0,\\ \unicode[STIX]{x1D6FB}^{2}\boldsymbol{v}^{(1)}-\unicode[STIX]{x1D735}p^{(1)}\, & =\, & \boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{V}_{\infty }^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{V}_{\infty }^{(0)},\\ \boldsymbol{v}^{(1)}\, & =\, & \boldsymbol{U}_{s}^{(1)}+\unicode[STIX]{x1D734}_{s}^{(1)}\times \boldsymbol{r}\quad \text{at }r=1,\\ \boldsymbol{v}^{(1)}\, & =\, & \mathbf{0}\quad \text{at walls},\\ \boldsymbol{v}^{(1)}\, & \rightarrow \, & \mathbf{0}\quad \text{as }r\rightarrow \infty .\end{array}\right\}\end{eqnarray}$$

We do not perform a similar expansion for the electrostatic potential because the problem governing electrostatics is decoupled from the flow field.

Owing to (2.27), the hydrodynamic force and torque can be expanded as

(2.30a,b ) $$\begin{eqnarray}\boldsymbol{F}_{H}=\boldsymbol{F}_{H}^{(0)}+Re_{p}\boldsymbol{F}_{H}^{(1)}\quad \text{and}\quad \boldsymbol{L}_{H}=\boldsymbol{L}_{H}^{(0)}+Re_{p}\boldsymbol{L}_{H}^{(1)}.\end{eqnarray}$$

Here, $\boldsymbol{F}_{H}^{(0)}$ represents the hydrodynamic force in the Stokes regime and therefore acts in the $x$ -direction alone. As the flow is two-dimensional, the hydrodynamic force due to inertia ( $\boldsymbol{F}_{H}^{(1)}$ ) is of the following form: $\{F_{x}^{(1)},0,F_{z}^{(1)}\}$ . The cross-stream inertial migration arises from $F_{z}^{(1)}$ , whose evaluation is the primary objective of this work. One way to obtain this is by solving (2.28) and (2.29), and evaluating $Re_{p}\,\boldsymbol{e}_{z}\boldsymbol{\cdot }\int _{S_{p}}\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x1D748}_{H}^{(1)}\,\text{d}S$ . Alternatively, an application of the Lorentz reciprocal theorem allows us to determine the inertial lift force ( $F_{I}\equiv Re_{p}F_{z}^{(1)}$ ) associated with $\boldsymbol{v}^{(1)}$ , without solving for (2.29). In the next section, we will elaborate on this approach which was proposed originally by Ho & Leal (Reference Ho and Leal1974).

2.4 Lift force

The reciprocal theorem relates the properties of an unknown Stokes flow ( $\unicode[STIX]{x1D748},\boldsymbol{v}$ ) to a known test flow field ( $\unicode[STIX]{x1D748}^{t},\boldsymbol{u}^{t}$ ), provided both fields correspond to the same geometry. The test field is taken to be that generated by a sphere moving in the positive $z$ -direction with unit velocity $\boldsymbol{e}_{z}$ in a quiescent fluid between the walls. The test field is governed by

(2.31) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}^{t}\, & =\, & 0,\quad \unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}^{t}-\unicode[STIX]{x1D735}p^{t}=\mathbf{0},\\ \boldsymbol{u}^{t}\, & =\, & \boldsymbol{e}_{z}\quad \text{at }r=1,\\ \boldsymbol{u}^{t}\, & =\, & \mathbf{0}\quad \text{at the walls,}\\ \boldsymbol{u}^{t}\, & \rightarrow \, & \mathbf{0}\quad \text{at r}\rightarrow \infty .\end{array}\right\}\end{eqnarray}$$

The generalized reciprocal theorem (Kim & Karrila Reference Kim and Karrila2013) states

(2.32) $$\begin{eqnarray}\int _{S_{p}}\boldsymbol{v}\boldsymbol{\cdot }(\unicode[STIX]{x1D748}^{t}\boldsymbol{\cdot }\boldsymbol{n}^{t})\,\text{d}S+\int _{V_{f}}\boldsymbol{v}\boldsymbol{\cdot }(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D748}^{t})\,\text{d}V=\int _{S_{p}}\boldsymbol{u}^{t}\boldsymbol{\cdot }(\unicode[STIX]{x1D748}\boldsymbol{\cdot }\boldsymbol{n})\,\text{d}S+\int _{V_{f}}\boldsymbol{u}^{t}\boldsymbol{\cdot }(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D748})\,\text{d}V.\end{eqnarray}$$

Here $\unicode[STIX]{x1D748}^{t}$ and $\unicode[STIX]{x1D748}$ are the stress tensors corresponding to the test Stokes field and unknown Stokes field. The Navier–Stokes equations (2.29) and (2.31) are expressed in terms of stress tensors as

(2.33a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D748}_{H}^{(1)}=\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{V}_{\infty }^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{V}_{\infty }^{(0)},\quad \text{and}\quad \unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D748}^{t}=\mathbf{0}.\end{eqnarray}$$

Substituting $\unicode[STIX]{x1D748}_{H}^{(1)}$ as $\unicode[STIX]{x1D748}$ in (2.32) and using the boundary conditions expressed in (2.29), we obtain

(2.34) $$\begin{eqnarray}\displaystyle & & \displaystyle {\boldsymbol{U}_{s}}^{(1)}\boldsymbol{\cdot }\int _{S_{p}}(\unicode[STIX]{x1D748}^{t}\boldsymbol{\cdot }\boldsymbol{n}^{t})\,\text{d}S+\int _{S_{p}}({\unicode[STIX]{x1D734}_{s}}^{(1)}\times \boldsymbol{r})\boldsymbol{\cdot }(\unicode[STIX]{x1D748}^{t}\boldsymbol{\cdot }\boldsymbol{n}^{t})\nonumber\\ \displaystyle & & \displaystyle \quad =\boldsymbol{e}_{z}\boldsymbol{\cdot }\int _{S_{p}}(\unicode[STIX]{x1D748}_{H}^{(1)}\boldsymbol{\cdot }\boldsymbol{n})\,\text{d}S+\int _{V_{f}}\boldsymbol{u}^{t}\boldsymbol{\cdot }(\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{V}_{\infty }^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{V}_{\infty }^{(0)})\,\text{d}V.\qquad \quad\end{eqnarray}$$

The surface integral in the first term on the left-hand side denotes the hydrodynamic force on the sphere due to the $z$ -direction motion ${\approx}-6\unicode[STIX]{x03C0}(1+O(\unicode[STIX]{x1D705}))\boldsymbol{e}_{z}$ (with first-order wall correction). The second term can be shown to be zero by invoking symmetry arguments. Also, since the particle is freely suspended and neutrally buoyant, the first term on the right-hand side is zero. This yields the following expression for the migration velocity of the particle due to inertia:

(2.35) $$\begin{eqnarray}U_{sz}^{(1)}=\frac{-1}{6\unicode[STIX]{x03C0}(1+O(\unicode[STIX]{x1D705}))}\int _{V_{f}}\boldsymbol{u}^{t}\boldsymbol{\cdot }(\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{V}_{\infty }^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{V}_{\infty }^{(0)})\,\text{d}V.\end{eqnarray}$$

Following Ho & Leal (Reference Ho and Leal1974), at the present order of approximation (for migration of $O(Re_{p})$ ), we alternatively prescribe $U_{sz}\equiv 0$ in (2.34). In other words, the particle is not allowed to migrate across streamlines. (Since the particle is free to move in the flow direction, at $O(Re_{p}^{0})$ the problem remains force-free in $x$ -direction and torque-free in $y$ -direction (Ho & Leal Reference Ho and Leal1974).) This implies that a finite inertial force $\boldsymbol{F}_{I}$ acts on the particle

(2.36) $$\begin{eqnarray}\displaystyle F_{I} & \equiv & \displaystyle Re_{p}\left(\boldsymbol{e}_{z}\boldsymbol{\cdot }\int _{S_{p}}(\unicode[STIX]{x1D748}_{H}^{(1)}\boldsymbol{\cdot }\boldsymbol{n})\,\text{d}S\right)\nonumber\\ \displaystyle & = & \displaystyle -Re_{p}\int _{V_{f}}\boldsymbol{u}^{t}\boldsymbol{\cdot }(\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{V}_{\infty }^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{v}^{(0)}+\boldsymbol{v}^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{V}_{\infty }^{(0)})\,\text{d}V.\end{eqnarray}$$

We have now obtained the inertia-induced lift force as a volume integral which can be evaluated by solving (2.28) (equations governing the creeping flow $\boldsymbol{v}^{(0)}$ ), equations (2.15)–(2.18) (equations governing the electrostatic potential $\unicode[STIX]{x1D713}$ ) and (2.31) (equations governing the test field). In the next section, we evaluate the disturbance fields and the test field.

3.1 Electrostatic potential

When the particle is not too close to the wall ( $\unicode[STIX]{x1D705}/s\ll 1$ ), the solution can be sought in terms of successive reflections (Brenner Reference Brenner1962). The first reflection represents the disturbance due to a particle in an unbounded domain. The second reflection corrects the previous disturbance by imposing the boundary condition at the walls and ignoring the particle. Subsequently, each reflection imposes the boundary condition alternatively at the particle and walls. Every successive pair of reflections increases the accuracy by $O(\unicode[STIX]{x1D705})$ (Happel & Brenner Reference Happel and Brenner2012). This iterative process is performed until a desired accuracy is attained. We seek the disturbance potential as a sum of reflections

(3.1) $$\begin{eqnarray}\unicode[STIX]{x1D713}=\unicode[STIX]{x1D713}_{1}+\unicode[STIX]{x1D713}_{2}+\cdots \,.\end{eqnarray}$$

Here, $\unicode[STIX]{x1D713}_{i}$ represents the $i$ th reflection. Substituting this in (2.15)–(2.18), we obtain for the first reflection

(3.2) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D713}_{1}\, & =\, & 0,\\ \boldsymbol{e}_{r}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{1}\, & =\, & -\boldsymbol{e}_{r}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{\infty }\quad \text{at }r=1,\\ \unicode[STIX]{x1D713}_{1}\, & \rightarrow \, & 0\quad \text{as }r\rightarrow \infty ,\end{array}\right\}\end{eqnarray}$$

and for the second reflection

(3.3) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D713}_{2}\, & =\, & 0,\\ \boldsymbol{e}_{z}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{2}\, & =\, & -\boldsymbol{e}_{z}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{1}\quad \text{at walls.}\end{array}\right\}\end{eqnarray}$$

Solution to $\unicode[STIX]{x1D713}_{1}$ . Equation (3.2) suggests that $\unicode[STIX]{x1D713}_{1}$ is a harmonic function which decays as $r\rightarrow \infty$ . The boundary condition suggests that it must be linear in the ‘driving force’: $\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{\infty }(=-\boldsymbol{e}_{x})$ . Therefore, the solution is sought in terms of spherical solid harmonics (Guazzelli & Morris Reference Guazzelli and Morris2011) as

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D713}_{1}=\frac{1}{2}\frac{\boldsymbol{r}}{r^{3}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{\infty }=-\frac{1}{2}\frac{x}{r^{3}}.\end{eqnarray}$$

Solution to $\unicode[STIX]{x1D713}_{2}$ . To evaluate $\unicode[STIX]{x1D713}_{2}$ , we adopt the approach devised by Faxén (Reference Faxén1922) in the context of bounded viscous flows. Using this, the disturbance around the particle is expressed in an integral form which satisfies the boundary conditions at the walls. The first reflection is characterized by particle scale ( $a^{\prime }$ ). The second reflection $\unicode[STIX]{x1D713}_{2}$ is characterized by a length scale ( $l^{\prime }$ ) which is $O(1/\unicode[STIX]{x1D705})$ , because the walls are remotely located. Therefore, the coordinates for second reflection are stretched, and are termed as ‘outer’ coordinates. These outer coordinates (denoted by capital letters) are defined as

(3.5a-d ) $$\begin{eqnarray}R=r\unicode[STIX]{x1D705}\quad X=x\unicode[STIX]{x1D705}\quad Y=y\unicode[STIX]{x1D705}\quad Z=z\unicode[STIX]{x1D705}.\end{eqnarray}$$

Using Faxén’s transformations (see § A.2), we obtain the second reflection

(3.6) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D713}}_{2}=\frac{\unicode[STIX]{x1D705}^{2}}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{+\infty }\int _{-\infty }^{+\infty }\text{e}^{\text{i}\unicode[STIX]{x1D6E9}}(\text{e}^{(-\unicode[STIX]{x1D706}Z/2)}b_{2}+\text{e}^{(+\unicode[STIX]{x1D706}Z/2)}b_{3})\left(\frac{\text{i}\unicode[STIX]{x1D709}}{\unicode[STIX]{x1D706}}\right)\,\text{d}\unicode[STIX]{x1D709}\,\text{d}\unicode[STIX]{x1D702}.\end{eqnarray}$$

Here $b_{2}$ and $b_{3}$ are

(3.7a,b ) $$\begin{eqnarray}b_{2}=\frac{1}{8}\frac{(1+\text{e}^{(1-s)\unicode[STIX]{x1D706}})}{-1+\text{e}^{\unicode[STIX]{x1D706}}},\quad b_{3}=\frac{1}{8}\frac{(1+\text{e}^{s\unicode[STIX]{x1D706}})}{-1+\text{e}^{\unicode[STIX]{x1D706}}}.\end{eqnarray}$$

The next reflection ( $\unicode[STIX]{x1D713}_{3}$ ) is $O(\unicode[STIX]{x1D705}^{3})$ . As we restrict $\unicode[STIX]{x1D705}\ll 1$ , the first two reflections capture the leading-order contribution to the disturbance potential.

Solution of $\boldsymbol{F}_{M}$ and $\boldsymbol{L}_{M}$ . Before determining the inertial lift $F_{I}$ through the evaluation of velocity disturbances in the creeping flow limit, we find the electrical force and torque on the particle caused by Maxwell stress. These are used later in the calculation of $\boldsymbol{U}_{s}^{(0)}$ and $\unicode[STIX]{x1D734}_{s}^{(0)}$ (see § 3.3).

We substitute (3.4) and (3.6) in both (2.24) and (2.25). Upon simplification (see § A.3) we obtain

(3.8) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{F}_{M}=4\unicode[STIX]{x03C0}Ha\left(\frac{3\unicode[STIX]{x03C0}}{16}\unicode[STIX]{x1D705}^{4}(\mathfrak{Z}(4,s)-\mathfrak{Z}(4,1-s))\right)\boldsymbol{e}_{z}+O(\unicode[STIX]{x1D705}^{7}), & \displaystyle\end{eqnarray}$$

(3.9) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{L}_{M}=\mathbf{0}. & \displaystyle\end{eqnarray}$$

Here, $\mathfrak{Z}$ is the generalized Riemann zeta function (Abramowitz & Stegen Reference Abramowitz and Stegen1972). This leading-order electrical force acts only in the $z$ -direction and always away from the walls (i.e. positive below the channel centreline and vice versa). As the particle approaches walls ( $s\rightarrow 0$ or 1), equation (3.8) asymptotically matches with the single wall expression derived by Yariv (Reference Yariv2006). We also find that the Maxwell stresses exert no torque on the particle, at the present level of approximation.

Having obtained the leading-order potential distribution, we next evaluate the velocity disturbance.

3.2 Velocity disturbance

We seek the disturbance ( $\boldsymbol{v}^{(0)},p^{(0)}$ ) as successive reflections

(3.10a,b ) $$\begin{eqnarray}\boldsymbol{v}^{(0)}=\boldsymbol{v}_{1}^{(0)}+\boldsymbol{v}_{2}^{(0)}+\boldsymbol{v}_{3}^{(0)}+\cdots \,,\quad p^{(0)}=p_{1}^{(0)}+p_{2}^{(0)}+p_{3}^{(0)}+\cdots \,.\end{eqnarray}$$

Upon substituting the above expansion in (2.28), we obtain the following set of problems:

(3.11) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}_{1}^{(0)}\, & =\, & 0,\quad \unicode[STIX]{x1D6FB}^{2}\boldsymbol{v}_{1}^{(0)}-\unicode[STIX]{x1D735}p_{1}^{(0)}=\mathbf{0},\\ \boldsymbol{v}_{1}^{(0)}\, & =\, & U_{sx}^{(0)}\boldsymbol{e}_{x}+\unicode[STIX]{x1D6FA}_{sy}^{(0)}\boldsymbol{e}_{y}\times \boldsymbol{r}+HaZ_{p}(\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{1}+\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{2})\\ \, & \, & -(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}z+\unicode[STIX]{x1D6FE}z^{2}+Ha(Z_{p}-Z_{w}))\boldsymbol{e}_{x}\quad \text{at }r=1,\\ \boldsymbol{v}_{1}^{(0)}\, & \rightarrow \, & \mathbf{0}\quad \text{at }r\rightarrow \infty .\end{array}\right\}\end{eqnarray}$$

(3.12) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}_{2}^{(0)}\, & =\, & 0,\quad \unicode[STIX]{x1D6FB}^{2}\boldsymbol{v}_{2}^{(0)}-\unicode[STIX]{x1D735}p_{2}^{(0)}=\mathbf{0},\\ \boldsymbol{v}_{2}^{(0)}\, & =\, & HaZ_{w}(\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{1}+\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}_{2})-\boldsymbol{v}_{1}^{(0)}\quad \text{at the walls,}\\ \boldsymbol{v}_{2}^{(0)}\, & \rightarrow \, & \mathbf{0}\quad \text{at }r\rightarrow \infty .\end{array}\right\}\end{eqnarray}$$

(3.13) $$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}_{3}^{(0)}\, & =\, & 0,\quad \unicode[STIX]{x1D6FB}^{2}\boldsymbol{v}_{3}^{(0)}-\unicode[STIX]{x1D735}p_{3}^{(0)}=\mathbf{0},\\ \boldsymbol{v}_{3}^{(0)}\, & =\, & -\boldsymbol{v}_{2}^{(0)}\quad \text{at }r=1,\\ \boldsymbol{v}_{3}^{(0)}\, & \rightarrow \, & \mathbf{0}\quad \text{at }r\rightarrow \infty .\end{array}\right\}\end{eqnarray}$$

Solution of $\boldsymbol{v}_{1}^{(0)}$ . The odd reflections are found using Lamb’s general method (Lamb Reference Lamb1975). We obtain the following leading terms for $\boldsymbol{v}_{1}^{(0)}$ :

(3.14) $$\begin{eqnarray}\boldsymbol{v}_{1}^{(0)}=A_{1}\left(\boldsymbol{e}_{x}+\frac{x\boldsymbol{r}}{r^{2}}\right)\frac{1}{r}+B_{1}\left(-\boldsymbol{e}_{x}+\frac{3x\boldsymbol{r}}{r^{2}}\right)\frac{1}{r^{3}}+C_{1}\left(\frac{z\boldsymbol{e}_{x}}{r^{3}}-\frac{x\boldsymbol{e}_{z}}{r^{3}}\right)+D_{1}\left(\frac{zx\boldsymbol{r}}{r^{5}}\right).\end{eqnarray}$$

The coefficients $A_{1}$ , $B_{1}$ , $C_{1}$ and $D_{1}$ are associated with the Stokeslet, source-dipole, rotlet (generated due to relative rotation of the particle) and stresslet flow disturbances, respectively. These are

(3.15) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}A_{1}={\displaystyle \frac{3}{4}}\left(U_{sx}^{(0)}-\unicode[STIX]{x1D6FC}-{\displaystyle \frac{\unicode[STIX]{x1D6FE}}{3}}-Ha(Z_{p}-Z_{w})+\unicode[STIX]{x1D705}^{3}HaZ_{p}\hspace{2.22198pt}\mathfrak{s}_{x}(\tilde{\unicode[STIX]{x1D713}}_{2})\right),\\ B_{1}={\displaystyle \frac{1}{2}}(HaZ_{p})-{\displaystyle \frac{1}{4}}\left(U_{sx}^{(0)}-\unicode[STIX]{x1D6FC}-{\displaystyle \frac{3\unicode[STIX]{x1D6FE}}{5}}-Ha(Z_{p}-Z_{w})+\unicode[STIX]{x1D705}^{3}HaZ_{p}\hspace{2.22198pt}\mathfrak{s}_{x}(\tilde{\unicode[STIX]{x1D713}}_{2})\right),\\ C_{1}=\unicode[STIX]{x1D6FA}_{sy}^{(0)}-\unicode[STIX]{x1D6FD}/2,\quad \text{and}\quad D_{1}=-5\unicode[STIX]{x1D6FD}/2.\end{array}\right\}\end{eqnarray}$$

Here, $\boldsymbol{\mathfrak{s}}(\tilde{\unicode[STIX]{x1D713}}_{2})$ is a vector denoting the contribution to slip at the particle surface due to the wall correction of the disturbance potential. This correction enters the solution at $O(\unicode[STIX]{x1D705}^{3})$ .

(3.16) $$\begin{eqnarray}\boldsymbol{\mathfrak{s}}(\tilde{\unicode[STIX]{x1D713}}_{2})=\int _{0}^{\infty }-{\textstyle \frac{1}{4}}(b_{2}+b_{3})\unicode[STIX]{x1D706}^{2}\,\text{d}\unicode[STIX]{x1D706}\boldsymbol{e}_{x}.\end{eqnarray}$$

Solution to $\boldsymbol{v}_{2}^{(0)}$ . Similar to § 3.1, we use Faxén transformation technique to obtain the solution to $\tilde{\boldsymbol{v}}_{2}^{(0)}=\{\tilde{u} _{2}^{(0)},\tilde{v}_{2}^{(0)},\tilde{w}_{2}^{(0)}\}$ as

(3.17) $$\begin{eqnarray}\tilde{u} _{2}^{(0)}=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{+\infty }\int _{-\infty }^{+\infty }\text{e}^{\text{i}\unicode[STIX]{x1D6E9}}\left(\begin{array}{@{}c@{}}\text{e}^{(-\unicode[STIX]{x1D706}Z/2)}\left(\ell _{4}+{\displaystyle \frac{\unicode[STIX]{x1D709}^{2}}{\unicode[STIX]{x1D706}^{2}}}\left(\ell _{5}+{\displaystyle \frac{\unicode[STIX]{x1D706}Z}{2}}\ell _{6}-{\displaystyle \frac{HaZ_{w}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D706}}{2}}b_{2}\right)\right)\\ +\,\text{e}^{(+\unicode[STIX]{x1D706}Z/2)}\left(\ell _{7}+{\displaystyle \frac{\unicode[STIX]{x1D709}^{2}}{\unicode[STIX]{x1D706}^{2}}}\left(\ell _{8}-{\displaystyle \frac{\unicode[STIX]{x1D706}Z}{2}}\ell _{9}-{\displaystyle \frac{HaZ_{w}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D706}}{2}}b_{3}\right)\right)\end{array}\right)\text{d}\unicode[STIX]{x1D709}\,\text{d}\unicode[STIX]{x1D702},\end{eqnarray}$$

(3.18) $$\begin{eqnarray}\tilde{v}_{2}^{(0)}=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{+\infty }\int _{-\infty }^{+\infty }\text{e}^{\text{i}\unicode[STIX]{x1D6E9}}\left(\begin{array}{@{}c@{}}\text{e}^{(-\unicode[STIX]{x1D706}Z/2)}\left(\ell _{5}+{\displaystyle \frac{\unicode[STIX]{x1D706}Z}{2}}\ell _{6}-{\displaystyle \frac{HaZ_{w}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D706}}{2}}b_{2}\right)\\ +\,\text{e}^{(+\unicode[STIX]{x1D706}Z/2)}\left(\ell _{8}-{\displaystyle \frac{\unicode[STIX]{x1D706}Z}{2}}\ell _{9}-{\displaystyle \frac{HaZ_{w}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D706}}{2}}b_{3}\right)\end{array}\right)\left(\frac{\unicode[STIX]{x1D709}\unicode[STIX]{x1D702}}{\unicode[STIX]{x1D706}^{2}}\right)\text{d}\unicode[STIX]{x1D709}\,\text{d}\unicode[STIX]{x1D702},\end{eqnarray}$$

(3.19) $$\begin{eqnarray}\displaystyle \tilde{w}_{2}^{(0)} & = & \displaystyle \frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{+\infty }\int _{-\infty }^{+\infty }\text{e}^{\text{i}\unicode[STIX]{x1D6E9}}\left(\begin{array}{@{}c@{}}\text{e}^{(-\unicode[STIX]{x1D706}Z/2)}\left(\ell _{4}+\ell _{5}+\left(1+{\displaystyle \frac{\unicode[STIX]{x1D706}Z}{2}}\right)\ell _{6}-{\displaystyle \frac{HaZ_{w}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D706}Z}{2}}b_{2}\right)\\ -\,\text{e}^{(+\unicode[STIX]{x1D706}Z/2)}\left(\ell _{7}+\ell _{8}+\left(1-{\displaystyle \frac{\unicode[STIX]{x1D706}Z}{2}}\right)\ell _{9}-{\displaystyle \frac{HaZ_{w}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D706}Z}{2}}b_{3}\right)\end{array}\right)\nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{\text{i}\unicode[STIX]{x1D709}}{\unicode[STIX]{x1D706}}\right)\text{d}\unicode[STIX]{x1D709}\,\text{d}\unicode[STIX]{x1D702}.\end{eqnarray}$$

Here, the terms $\ell _{4},\ell _{5},\ldots ,\ell _{9}$ are functions of the Fourier variable $\unicode[STIX]{x1D706}$ and the coefficients ( $A_{1}$ , $B_{1}$ , $C_{1}$ and $D_{1}$ ), defined in (3.15). Details of the derivation are skipped for brevity and can be found in the supplementary material.

At the leading order, it suffices to represent the velocity disturbance field using the first two reflections. However, the solution is incomplete because the translational and angular velocities are unknown (see (3.15)). These can be found by imposing force-free and torque-free conditions on the particle at $O(Re^{0})$ : $\boldsymbol{F}_{H}^{(0)}+\boldsymbol{F}_{M}=\mathbf{0}$ and $\boldsymbol{L}_{H}^{(0)}+\boldsymbol{L}_{M}=\mathbf{0}$ . Since the electric force (3.8) acts only along the $z$ -axis and the electric torque (3.9) is zero, $U_{s\,x}^{(0)}$ and $\unicode[STIX]{x1D6FA}_{s\,y}^{(0)}$ are found by hydrodynamic force and torque balance in $x$ and $y$ directions, respectively.

3.3 Evaluation of $\boldsymbol{U}_{s}^{(0)}$ and $\unicode[STIX]{x1D734}_{s}^{(0)}$

For a particle–wall system, Happel & Brenner (Reference Happel and Brenner2012, p. 239) derived the hydrodynamic force and torque on a spherical particle in terms of successive reflections

(3.20a,b ) $$\begin{eqnarray}\boldsymbol{F}_{H}^{(0)}=\boldsymbol{F}_{H\,1}^{(0)}+\boldsymbol{F}_{H\,3}^{(0)}+\cdots \quad \text{and}\quad \boldsymbol{L}_{H}^{(0)}=\boldsymbol{L}_{H\,1}^{(0)}+\boldsymbol{L}_{H\,3}^{(0)}+\cdots \,.\end{eqnarray}$$

Since the particle is absent in the formulation of even-numbered reflections, the hydrodynamic drag and torque due to even fields vanishes. Thus, the contribution to force and torque arises only from the odd-numbered reflections.

At the current order of reflections of the velocity disturbance field ( $\boldsymbol{v}^{(0)}=\boldsymbol{v}_{1}^{(0)}+\boldsymbol{v}_{2}^{(0)}$ ), imposing force-free and torque-free conditions provides translational and rotational velocities without the wall correction. $\boldsymbol{F}_{H\,3}^{(0)}$ incorporates the first-order wall effects into $U_{s\,x}^{(0)}$ and $\unicode[STIX]{x1D6FA}_{s\,y}^{(0)}$ , which requires $\boldsymbol{v}_{3}^{(0)}$ . Section A.4 describes the determination of $\boldsymbol{v}_{3}^{(0)}$ using Lamb’s general solution (Lamb Reference Lamb1975).

The force and torque on a spherical particle can be expressed through the coefficients of Stokeslet and rotlet disturbances, respectively. Following Kim & Karrila (Reference Kim and Karrila2013, p. 88), we write (3.20) as

(3.21a,b ) $$\begin{eqnarray}F_{H\,x}^{(0)}=-4\unicode[STIX]{x03C0}(A_{1}+A_{3}+\cdots )\quad \text{and}\quad L_{Hy}^{(0)}=-8\unicode[STIX]{x03C0}(C_{1}+C_{3}+\cdots ).\end{eqnarray}$$

The coefficients $A_{1},C_{1}$ and $A_{3},C_{3}$ are associated with Lamb’s solution of the first reflection of the velocity disturbance (3.14) and third reflection (see (A 17)), respectively. $A_{3}$ and $C_{3}$ can be expressed in terms of $A_{1}$ , $B_{1}$ , $C_{1}$ and $D_{1}$ as

(3.22) $$\begin{eqnarray}\displaystyle & A_{3}=A_{1}(\unicode[STIX]{x1D705}W_{A})+B_{1}(\unicode[STIX]{x1D705}^{3}W_{B})+C_{1}(\unicode[STIX]{x1D705}^{2}W_{C})+D_{1}(\unicode[STIX]{x1D705}^{2}W_{D})-{\textstyle \frac{3}{4}}\unicode[STIX]{x1D705}^{3}HaZ_{w}\hspace{2.22198pt}\mathfrak{s}_{x}(\tilde{\unicode[STIX]{x1D713}}_{2}), & \displaystyle\end{eqnarray}$$

(3.23) $$\begin{eqnarray}\displaystyle & C_{3}=A_{1}(\unicode[STIX]{x1D705}^{2}{\mathcal{X}}_{A})+B_{1}(\unicode[STIX]{x1D705}^{3}{\mathcal{X}}_{B})+C_{1}(\unicode[STIX]{x1D705}^{4}{\mathcal{X}}_{C}). & \displaystyle\end{eqnarray}$$

Here, $\unicode[STIX]{x1D705}W_{A},\unicode[STIX]{x1D705}^{3}W_{B},\unicode[STIX]{x1D705}^{2}W_{C}$ and $\unicode[STIX]{x1D705}^{2}W_{D}$ represent the wall corrections to hydrodynamic drag due to the reflection of Stokeslet, source-dipole, rotlet and stresslet disturbances, respectively. Similarly, $\unicode[STIX]{x1D705}^{2}{\mathcal{X}}_{A}$ , $\unicode[STIX]{x1D705}^{3}{\mathcal{X}}_{B}$ and $\unicode[STIX]{x1D705}^{4}{\mathcal{X}}_{C}$ represent the wall correction to hydrodynamic torque. These corrections are integrals over Fourier variable $\unicode[STIX]{x1D706}$ and are a function of distance from the walls.

Substitution of (3.22) and (3.23) into (3.21) results in a system of two equations for: $U_{s\,x}^{(0)}$ and $\unicode[STIX]{x1D6FA}_{s\,y}^{(0)}$ . Imposing the hydrodynamic force and torque to be zero and upon expanding the coefficients $A_{1}$ , $B_{1}$ , $C_{1}$ and $D_{1}$ , we obtain

(3.24) $$\begin{eqnarray}U_{s\,x}^{(0)}\approx \left\{\begin{array}{@{}l@{}}[(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}/3)(1+\unicode[STIX]{x1D705}W_{A})-10\unicode[STIX]{x1D705}^{2}\unicode[STIX]{x1D6FD}W_{D}/9]/(1+\unicode[STIX]{x1D705}W_{A})\quad \\ \quad +\,Ha(Z_{p}-Z_{w})\left(1-{\displaystyle \frac{2}{3}}{\displaystyle \frac{W_{B}\unicode[STIX]{x1D705}^{3}}{(1+\unicode[STIX]{x1D705}W_{A})}}\right)-\unicode[STIX]{x1D705}^{3}\hspace{2.22198pt}\mathfrak{s}_{x}(\tilde{\unicode[STIX]{x1D713}}_{2})Ha\left(Z_{p}-{\displaystyle \frac{Z_{w}}{(1+\unicode[STIX]{x1D705}W_{A})}}\right),\quad \end{array}\right.\end{eqnarray}$$

(3.25) $$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{s\,y}^{(0)}\approx \frac{\unicode[STIX]{x1D6FD}(1+\unicode[STIX]{x1D705}W_{A})-10{\mathcal{X}}_{D}\unicode[STIX]{x1D705}^{3}\unicode[STIX]{x1D6FD}/3}{2(1+\unicode[STIX]{x1D705}W_{A})}-Ha(Z_{p}-Z_{w})\frac{{\mathcal{X}}_{B}\unicode[STIX]{x1D705}^{4}}{2(1+{\mathcal{X}}_{C}\unicode[STIX]{x1D705}^{3})}.\end{eqnarray}$$

Validation of the above two velocities with the literature can be found in appendix B.

With (3.14), (3.17)–(3.19), (3.24) and (3.25), we have obtained the solution to $\boldsymbol{v}^{(0)}$ . We use a similar procedure to evaluate the test field $\boldsymbol{u}^{t}$ governed by (2.31) (the details are described in § A.5). The calculation of the inertial lift force using the disturbance and test field is discussed next.

4.1 Inner integral

The order of magnitude of the velocity fields in terms of size ratio $\unicode[STIX]{x1D705}$ and radial distance $r$ , at the leading order are

(4.3a,b ) $$\begin{eqnarray}\boldsymbol{v}_{1}^{(0)}\sim O(1/r^{3})Ha\,Z_{p}+O(\unicode[STIX]{x1D705}/r^{2}),\quad \boldsymbol{v}_{2}^{(0)}\sim O(\unicode[STIX]{x1D705}^{3})+O(\unicode[STIX]{x1D705}^{3})Ha\,(Z_{p}-Z_{w}),\end{eqnarray}$$

(4.4a,b ) $$\begin{eqnarray}\boldsymbol{u}_{1}^{(t)}\sim O(1/r)+O(1/r^{3}),\quad \boldsymbol{u}_{2}^{(t)}\sim O(\unicode[STIX]{x1D705})+O(\unicode[STIX]{x1D705}^{3}).\end{eqnarray}$$

Substituting the translational velocity (3.24) into the undisturbed background flow ( $\boldsymbol{V}_{\infty }^{(0)}=(\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FD}z+\unicode[STIX]{x1D6FE}z^{2})\boldsymbol{e}_{x}+HaZ_{w}\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{\infty }-\boldsymbol{U}_{s}^{(0)}$ ), we obtain the order of magnitude of $\boldsymbol{V}_{\infty }^{(0)}$ as

(4.5) $$\begin{eqnarray}\boldsymbol{V}_{\infty }^{(0)}\sim O(\unicode[STIX]{x1D705}r)+O(\unicode[STIX]{x1D705}^{2}r^{2})+O(1)HaZ_{p}+\cdots \,.\end{eqnarray}$$

Using (4.3)–(4.5) in (4.1), we obtain the integrand as

(4.6) $$\begin{eqnarray}\displaystyle \boldsymbol{u}^{t}\boldsymbol{\cdot }\boldsymbol{f} & {\sim} & \displaystyle O\left(\frac{(HaZ_{p})^{2}}{r^{5}}+\frac{(HaZ_{p})^{2}}{r^{7}}+\cdots \right)+\unicode[STIX]{x1D705}O\left(\frac{HaZ_{p}}{r^{4}}+\frac{HaZ_{p}}{r^{6}}+\cdots \right)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D705}^{2}O\left(\frac{1}{r^{3}}+\frac{HaZ_{p}}{r^{5}}+\cdots \right)+\unicode[STIX]{x1D705}^{3}O\left(\frac{1}{r^{2}}+\frac{1}{r^{4}}+\cdots \right)+\cdots \,.\end{eqnarray}$$

The volume integral of the first term is identically zero. The leading-order contribution arises from the second $O(\unicode[STIX]{x1D705})$ term

(4.7) $$\begin{eqnarray}\lim _{\unicode[STIX]{x1D6FF}\rightarrow \infty }F_{V1}=-Re_{p}\left(0+\left(\frac{47\unicode[STIX]{x03C0}}{20}HaZ_{p}\right)\unicode[STIX]{x1D6FD}+0+O(\unicode[STIX]{x1D705}^{3})+\cdots \right).\end{eqnarray}$$

In the above expression, the proportionality of $Ha\,Z_{p}$ suggests that the $O(\unicode[STIX]{x1D705})$ contribution (implicit in background shear $\unicode[STIX]{x1D6FD}$ ) arises due to electrophoresis. The contribution due to stresslet disturbances and wall reflections is $O(\unicode[STIX]{x1D705}^{3})$ . For electric field in the same direction as the flow ( $Ha>0$ ), a positive $Z_{p}$ corresponds to a leading particle. Below the centreline, the shear ( $\unicode[STIX]{x1D6FD}$ ) is positive and the overall expression in (4.7) is therefore negative. This corresponds to a lift in the negative $z$ -direction which pushes the leading particle towards the wall. A similar explanation can be used to show that a lagging particle ( $Z_{p}<0$ ) experiences a lift (4.7) towards the centre. These effects were also experimentally observed by Kim & Yoo (Reference Kim and Yoo2009a ,Reference Kim and Yoo b ), Cevheri & Yoda (Reference Cevheri and Yoda2014), Yuan et al. (Reference Yuan, Pan, Zhang, Yan, Zhao, Alici and Li2016), Li & Xuan (Reference Li and Xuan2018) and Yee & Yoda (Reference Yee and Yoda2018). Expression (4.7) also reveals that the wall zeta potential does not affect the lift because its contribution to the disturbance is $O(\unicode[STIX]{x1D705}^{3})$ (see (4.3)).

In the absence of electrophoresis, the inner integral lift contributes at $O(\unicode[STIX]{x1D705}^{3})$ , as also previously reported by Ho & Leal (Reference Ho and Leal1974) and Hood et al. (Reference Hood, Lee and Roper2015). Next we estimate the contribution arising from the outer integral.

4.2 Outer integral

Since the walls lie at distances $O(1/\unicode[STIX]{x1D705})$ , the velocity field expressions are rescaled with respect to $R(=\unicode[STIX]{x1D705}r)$ (similar to (3.5)). The inertial lift force expression in the outer coordinates is

(4.8) $$\begin{eqnarray}\displaystyle F_{V2} & = & \displaystyle -Re_{p}\int _{V_{2}}(\tilde{\boldsymbol{u}}_{1}^{t}+\tilde{\boldsymbol{u}}_{2}^{t})\boldsymbol{\cdot }\left[\begin{array}{@{}c@{}}(\tilde{\boldsymbol{v}}_{1}^{(0)}+\tilde{\boldsymbol{v}}_{2}^{(0)})\boldsymbol{\cdot }\unicode[STIX]{x1D705}\tilde{\unicode[STIX]{x1D735}}(\tilde{\boldsymbol{v}}_{1}^{(0)}+\tilde{\boldsymbol{v}}_{2}^{(0)})\\ +\,\tilde{\boldsymbol{V}}_{\infty }^{(0)}\boldsymbol{\cdot }\unicode[STIX]{x1D705}\tilde{\unicode[STIX]{x1D735}}(\tilde{\boldsymbol{v}}_{1}^{(0)}+\tilde{\boldsymbol{v}}_{2}^{(0)})+(\tilde{\boldsymbol{v}}_{1}^{(0)}+\tilde{\boldsymbol{v}}_{2}^{(0)})\boldsymbol{\cdot }\unicode[STIX]{x1D705}\tilde{\unicode[STIX]{x1D735}}\tilde{\boldsymbol{V}}_{\infty }^{(0)}\end{array}\right]\unicode[STIX]{x1D705}^{-3}\,\text{d}\tilde{V}\nonumber\\ \displaystyle & = & \displaystyle -Re_{p}\,\unicode[STIX]{x1D705}^{-2}\int _{V_{2}}\tilde{\boldsymbol{u}}^{t}\boldsymbol{\cdot }\tilde{\boldsymbol{f}}\,\text{d}\tilde{V}.\end{eqnarray}$$

In the outer coordinate representation, the order of magnitude of the velocity fields in terms of size ratio $\unicode[STIX]{x1D705}$ , at the leading order are

(4.9a,b ) $$\begin{eqnarray}\tilde{\boldsymbol{v}}_{1}^{(0)}\sim O(\unicode[STIX]{x1D705}^{3})+O(\unicode[STIX]{x1D705}^{3})Ha\,Z_{p},\quad \tilde{\boldsymbol{v}}_{2}^{(0)}\sim O(\unicode[STIX]{x1D705}^{3})+O(\unicode[STIX]{x1D705}^{3})Ha\,(Z_{p}-Z_{w}),\end{eqnarray}$$

(4.10a,b ) $$\begin{eqnarray}\tilde{\boldsymbol{u}}_{1}^{(t)}\sim O(\unicode[STIX]{x1D705})+O(\unicode[STIX]{x1D705}^{3}),\quad \tilde{\boldsymbol{u}}_{2}^{(t)}\sim O(\unicode[STIX]{x1D705})+O(\unicode[STIX]{x1D705}^{3}).\end{eqnarray}$$

The order of magnitude of the undisturbed background velocity is

(4.11) $$\begin{eqnarray}\tilde{\boldsymbol{V}}_{\infty }\sim O(1)+O(1)HaZ_{p}+O(\unicode[STIX]{x1D705}^{2})+\cdots \,.\end{eqnarray}$$

Substituting (4.9)–(4.11) in (4.8), we find that the integrand is $O(\unicode[STIX]{x1D705}^{2})$ . Since the contributions from the inner and outer integrals are comparable, it is important that the integration must be carried out from the particle surface to the walls.

4.3 Evaluation of volume integral

The region of integration extends from the surface of the spherical particle to the rectangular walls, which extend to infinity in the $x$ and $y$ directions. A direct numerical approach would be computationally intensive as the evaluation of the lift force involves numerous nested Fourier integrals, apart from integrals in $x$ , $y$ and $z$ coordinates. (Faxén transformation provides the wall reflected fields in terms of Fourier integrals, which makes it cumbersome to evaluate the lift force volume integral, as it consists of 7 nested integrals. Using certain transformations we reduce this number to 5. See supplementary material for further details.) For computational convenience, we resort to a splitting of the region over which the integral is evaluated. We perform the integration in two sub-regions, where the region of integration extends from: (i) the spherical surface to a circumscribing cylinder and (ii) the cylindrical surface to the walls ( $z$ -direction) and infinity ( $x$ and $y$ directions). This simplifies the calculation as; in sub-region (i) we employ a spherical coordinate system and evaluate the integral with the help of trigonometric identities; in sub-region (ii) a cylindrical coordinate system is employed to perform the integration. We evaluate the integral in both the sub-regions and include it in our results, which ensures no loss of accuracy.

We choose to evaluate the volume integral in the outer coordinates, over the entire volume ( $V_{f}$ ). For sub-region (ii), the region of integration can be visualized as a cylinder $(Z_{C}\in [-s,1-s]\text{ and }R_{C}\in [0,\infty ))$ excluding an asymptotically small cylindrical ‘cavity’ (radius $\in [0,\unicode[STIX]{x1D705}]$ , height $\in [-\unicode[STIX]{x1D705},\unicode[STIX]{x1D705}]$ ). The velocities are transformed into cylindrical coordinates $(R_{C},\unicode[STIX]{x1D703}_{C},Z_{C})$ . The lift force volume integral corresponding to sub-region (ii) is

(4.12) $$\begin{eqnarray}F_{I}=-Re_{p}\int _{0}^{2\unicode[STIX]{x03C0}}\unicode[STIX]{x1D705}^{-2}\left\{\begin{array}{@{}c@{}}\displaystyle \int _{0}^{\infty }\int _{-s}^{-\unicode[STIX]{x1D705}}\tilde{\boldsymbol{u}}^{t}\boldsymbol{\cdot }\tilde{\boldsymbol{f}}\hspace{2.22198pt}R_{C}\,\text{d}Z_{C}\,\text{d}R_{C}\\ \displaystyle +\int _{\unicode[STIX]{x1D705}}^{\infty }\int _{-\unicode[STIX]{x1D705}}^{+\unicode[STIX]{x1D705}}\tilde{\boldsymbol{u}}^{t}\boldsymbol{\cdot }\tilde{\boldsymbol{f}}\hspace{2.22198pt}R_{C}\,\text{d}Z_{C}\,\text{d}R_{C}\\ \displaystyle +\int _{0}^{\infty }\int _{\unicode[STIX]{x1D705}}^{1-s}\tilde{\boldsymbol{u}}^{t}\boldsymbol{\cdot }\tilde{\boldsymbol{f}}\hspace{2.22198pt}R_{C}\,\text{d}Z_{C}\,\text{d}R_{C}\end{array}\right\}\,\text{d}\unicode[STIX]{x1D703}_{C}.\end{eqnarray}$$

The integration in the above equation is carried out analytically in the $\unicode[STIX]{x1D703}_{C}$ , $R_{C}$ and $Z_{C}$ directions. The integration over the Fourier parameters is performed numerically using the Gauss–Kronrod quadrature built into Mathematica 11.0. The contribution from sub-region (i) is evaluated analytically (see appendix C) and incorporated in our results.

6.1 Validation with earlier theoretical studies

We first compare the results obtained in our work with those in the literature (Ho & Leal Reference Ho and Leal1974; Vasseur & Cox Reference Vasseur and Cox1976). Figure 3(a) shows the lift force experienced by a neutrally buoyant sphere freely suspended in Couette flow. There exists a single equilibrium position at the centre where the particle experiences no lift. Below the centreline ( $s<0.5$ ) the force acting on the particle is positive, resulting in a lift in the positive $z$ -direction (i.e. towards the centreline). Above the centreline ( $s>0.5$ ) the lift force is in the negative $z$ -direction (i.e. again towards the centreline). Hence, a neutrally buoyant particle in a Couette flow always migrates to the centre of the channel (a stable equilibrium) as depicted by the arrows on the axis.

For Poiseuille flow (figure 3 b), three equilibrium positions ‘a’, ‘b’ and ‘c’ exist ( $s\approx 0.19$ , 0.5 and 0.81). Points ‘a’ and ‘c’ are characterized by a negative slope of the lift force curve and are therefore stable equilibrium positions, while point ‘b’ is unstable (arrows on the axis depict the direction of migration).

Our results (for both Couette and Poiseuille flow) obtained by using Ho and Leal’s formulation are in good agreement with Vasseur & Cox (Reference Vasseur and Cox1976) who used a Green’s function technique to approximate the particle as a point force. Near the walls, lift force deviates significantly from Ho & Leal (Reference Ho and Leal1974). This may have been due to a failure in their numerical convergence near the walls, as pointed out by Vasseur & Cox (Reference Vasseur and Cox1976).

Figure 3. Comparison with the previous theoretical studies (Ho & Leal Reference Ho and Leal1974; Vasseur & Cox Reference Vasseur and Cox1976) for the inertial lift force experienced by a particle, as a function of distance from the bottom wall for: (a) Couette flow and (b) Poiseuille flow.

6.2 Insights into the origins of the inertial lift

The approach adopted in this work allows us to find different contributions to the lift force on a particle suspended in a Poiseuille flow. These constituents are interpreted using an order-of-magnitude analysis.

In the outer coordinates, the order of magnitude of background velocity is

(6.1) $$\begin{eqnarray}\tilde{\boldsymbol{V}}_{\infty }^{(0)}\sim O(1/\unicode[STIX]{x1D705})\unicode[STIX]{x1D6FD}+O(1/\unicode[STIX]{x1D705}^{2})\unicode[STIX]{x1D6FE}+O(1)\unicode[STIX]{x1D6FE},\end{eqnarray}$$

the disturbance velocity (stresslet) is

(6.2) $$\begin{eqnarray}\tilde{\boldsymbol{v}}^{(0)}\sim O(\unicode[STIX]{x1D705}^{2})\unicode[STIX]{x1D6FD}(1+W_{D}),\end{eqnarray}$$

and the test velocity field is

(6.3) $$\begin{eqnarray}\tilde{\boldsymbol{u}}^{t}\sim O(\unicode[STIX]{x1D705})(1+W_{A}^{t}).\end{eqnarray}$$

Here, $W_{D}$ and $W_{A}^{t}$ are the wall correction factors for stresslet and Stokeslet (test field) disturbances, respectively.

Upon substitution of (6.1)–(6.3) into the lift force expression (4.12), we obtain

(6.4) $$\begin{eqnarray}O\left(\frac{F_{I}}{Re_{p}}\right)\sim \int _{V_{f}}[O(1)\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FD}(1+W_{D})+O(1/\unicode[STIX]{x1D705})\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}(1+W_{D})]\,\text{d}\tilde{V}.\end{eqnarray}$$

The first term in the above equation is the stresslet-wall force (known in the literature as the ‘wall lift’ force). It originates from the interaction of the stresslet disturbance with the wall. This results in a pressure asymmetry around the particle. The pressure is larger on the side facing the closer wall (Feng et al. Reference Feng, Hu and Joseph1994). This force always acts away from the wall, even if the shear direction (the sign of $\unicode[STIX]{x1D6FD}$ ) is reversed, as it is proportional to $\unicode[STIX]{x1D6FD}^{2}$ . The second term is the stresslet-curvature force (known in the literature as the ‘shear-gradient lift force’), arising from the interaction of the stresslet disturbance ( $\unicode[STIX]{x1D6FD}(1+W_{D})$ ) with curvature in the background flow $(\unicode[STIX]{x1D6FE})$ . Owing to the proportionality to $\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}$ , this force always acts towards the walls.

Figure 4. Variation of leading constituents of inertial lift force as a function of distance from the bottom wall for Poiseuille flow. Here, stresslet-wall force (SWF) is the ‘wall lift force’ and stresslet-curvature force (SCF) is the ‘shear-gradient lift force’.

Figure 4 shows the variation of the above two components with the lateral position. The stresslet-wall force (SWF) acts to push the particle towards the centreline. Since SWF depends on the background shear ( $\unicode[STIX]{x1D6FD}^{2}$ ), it vanishes at the centre and increases near the walls. On the other hand, the stresslet-curvature force (SCF) has a destabilizing effect at the centreline i.e. a slight perturbation to the particle would push it away from the centre. SCF increases (in magnitude) away from the centreline, and decreases near the stationary walls due to increased viscous resistance.

From (6.4) it can be observed that the direction of SCF is proportional to $\text{sgn}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})$ . If either the shear or the curvature direction is changed, the direction of SCF can reverse. This can be exploited to control the positions at which the particle gets focused. Lee et al. (Reference Lee, Nam, Kim, Di Carlo and Lee2018) recently showed that the curvature of the background flow profile can be controlled by using stratified flows with different viscosities. By varying the viscosity ratio, an external control over the focusing position was experimentally demonstrated.