Dynamic and static stability of a drop attached to an inhomogeneous plane wall

Abstract

This article concerns the stability of a drop on a wall for which the contact angle, $\theta _\mathrm{{{w}}}$, varies from place to place. Such a wall may allow unstable equilibria of the drop, i.e. ones for which small perturbations to equilibrium grow, making the equilibrium unrealisable in practice. This will be referred to as dynamic instability and is one of the two versions of instability considered. The other arises from consideration of potential energy, which is the sum of surface (liquid/gas, liquid/solid and solid/gas) components and the gravitational potential energy. Equilibria are extrema of the potential energy with respect to variations of drop geometry which preserve its volume. An equilibrium is said to be statically stable if it is a local minimum of the potential energy for volume-preserving perturbations of the drop. The relationship between static and dynamic stability is the main subject of this paper. The liquid flow is governed by the incompressible Navier–Stokes equations. To allow for the moving contact line, a Navier slip condition with slip length $\lambda$ is used at the wall, as is a prescribed contact angle, $\theta _\mathrm{{{w}}} =\theta _\mathrm{{{w}}} ({x,y})$, at the contact line, where x, y are Cartesian coordinates on the wall. The perturbation is assumed small, allowing linearisation of the governing equations and, in the usual manner of stability analysis, complex modes having the time dependency $e^\mathrm{{{st}}}$ are introduced. This leads to an eigenvalue problem with eigenvalue s, the sign of whose real part determines dynamic stability/instability. A quite different eigenvalue problem, which describes static stability/instability is also derived. It is shown that, despite this difference, the conditions for dynamic and static instability are in fact the same. This conclusion is far from evident a priori but should be good news for interested numerical analysts because determination of static stability is much less numerically costly than a dynamic stability study, whereas it is the latter which gives a true determination of stability.

Appendix A: Mathematical details

1.1 A.1 Derivation of (2.12)

The normal unit vector, ${\mathbf{n}}$, is originally only defined on the perturbed interface, $d=\eta$, but can be extended using

$$\begin{aligned} {\mathbf{n}}=\frac{\nabla F}{\left| {\nabla F} \right| }. \end{aligned}$$

(A.1)

The interface curvature follows from

$$\begin{aligned} \kappa =\nabla {\mathbf{.n}} \end{aligned}$$

(A.2)

evaluated at the interface. Since $F=d-\eta$, $\left| {\nabla d} \right| =1$ and $\nabla d{\mathbf{.}}\nabla \eta =0$,

$$\begin{aligned} \left| {\nabla F} \right| ^{2}=1+\left| {\nabla \eta } \right| ^{2}. \end{aligned}$$

(A.3)

The second term on the right-hand side being of second order, it is neglected, hence ${\mathbf{n}}=\nabla F$ according to (A.1). Thus, (A.2) gives

$$\begin{aligned} \kappa= & {} \nabla ^{2}F =\nabla ^{2}d-\nabla ^{2}\eta , \end{aligned}$$

(A.4)

correct to first order when evaluated at the interface.

For the equilibrium, we set $\eta =0$ in (A.4) and ${\mathbf{u}}=0$ in (2.5) to derive

$$\begin{aligned} \left. {p_{e} } \right| _{d=0} =p_{a} +\left. {\nabla ^{2}d} \right| _{d=0} \end{aligned}$$

(A.5)

for the equilibrium pressure, $p_{e}$, at the equilibrium interface, $d=0$. Making the small displacement $\eta {\mathbf{n}}$ to arrive at the perturbed interface and given that $p_{e} -{\mathbf{G.x}}$ is constant,

$$\begin{aligned} \left. {p_{e} } \right| _{d=\eta } =p_{a} +\left. {\nabla ^{2}d} \right| _{d=0} +\eta \, {\mathbf{G.n}} =p_{a} +\left. {\nabla ^{2}d} \right| _{d=\eta } +\gamma \eta \end{aligned}$$

(A.6)

correct to first order, where

$$\begin{aligned} \gamma ={\mathbf{n.}}\left( {{\mathbf{G}}-\nabla \left( {\nabla ^{2}d} \right) } \right) . \end{aligned}$$

(A.7)

On the other hand, applying (2.5) and (A.4) to the perturbed interface

$$\begin{aligned} \text {Oh }{\mathbf{n.}}\left( {\nabla {\mathbf{u}}+\left( {\nabla {\mathbf{u}}} \right) ^{\mathrm{{T}}}} \right)= & {} {\mathbf{n}}\left( {\left. p \right| _{d=\eta } -p_{a} -\left. {\nabla ^{2}d} \right| _{d=\eta } +\nabla ^{2}\eta } \right) \nonumber \\= & {} {\mathbf{n}}\left( {\left. {{p}'} \right| _{d=\eta } +\left. {p_{e} } \right| _{d=\eta } -p_{a} -\left. {\nabla ^{2}d} \right| _{d=\eta } +\nabla ^{2}\eta } \right) . \end{aligned}$$

(A.8)

Using (A.6),

$$\begin{aligned} \text {Oh }{\mathbf{n.}}\left( {\nabla {\mathbf{u}}+\left( {\nabla {\mathbf{u}}} \right) ^{\mathrm{{T}}}} \right) ={\mathbf{n}}\left( {\nabla ^{2}\eta +\gamma \eta +{p}'} \right) . \end{aligned}$$

(A.9)

Fig. 6

The region $\Sigma$ and its boundary $\partial \Sigma$ in the $q^{\alpha }$ plane

It remains to express the term $\nabla ^{2}\eta$ in terms of $\eta \left( {q^{\alpha }} \right)$. To this end, let $\Sigma$ be a region in the $q^{\alpha }$ plane representing part of the equilibrium interface and V the volume in physical space for which $q^{\alpha }\in \Sigma$ and $0<d<\delta$, where $\delta$ is an infinitesimal constant (see Figs. 6 and 7). The divergence theorem gives

$$\begin{aligned} \int _V {\nabla ^{2}\eta \,\mathrm{{d}}v} =\int _{\partial V} {\frac{\partial \eta }{\partial n}\mathrm{{d}}S} . \end{aligned}$$

(A.10)

The normal vectors of the parts of $\partial V$ with constant d are directed parallel to $\nabla d$. Since $\nabla d{\mathbf{.}}\nabla \eta =0$, $\partial \eta /\partial n=0$. Thus, these parts of $\partial V$ do not contribute to (A.10). The remainder of $\partial V$ gives

$$\begin{aligned} \delta \oint _{\partial S} {{\mathbf{N.}}\nabla \eta \, \mathrm{{d}}s} \end{aligned}$$

(A.11)

for the right-hand side of (A.10), where $\partial S$ is the curve on the equilibrium interface corresponding to the boundary $\partial \Sigma$ in the $q^{\alpha }$ plane, $\mathrm{{d}}s$ is elementary arc length along $\partial S$ and ${\mathbf{N}}$ is the unit vector, tangential to the equilibrium interface and orthogonal to $\partial S$, which is directed outwards from S (see Fig. 7). Thus, (A.10) implies

$$\begin{aligned} \int _S {\nabla ^{2}\eta \, \mathrm{{d}}S} =\oint _{\partial S} {{\mathbf{N.}}\nabla \eta \, \mathrm{{d}}s} , \end{aligned}$$

(A.12)

where S corresponds to $\Sigma$.

Fig. 7

The volume V, surface S, boundary, $\partial S$, of S, and unit vector ${\mathbf{N}}$. The latter is tangential to the equilibrium interface, orthogonal to $\partial S$ and directed outwards from S

From here until the end of this section, we restrict attention to the equilibrium interface. A point on the interface has position vector ${\mathbf{x}}\left( {q^{\alpha }} \right)$, of which derivatives, $\partial {\mathbf{x}}/\partial q^{1}$ and $\partial {\mathbf{x}}/\partial q^{2}$, are tangential to the interface and yield the metric tensor via

$$\begin{aligned} g_{\alpha \beta } =\frac{\partial {\mathbf{x}}}{\partial q^{\alpha }}{\mathbf{.}}\frac{\partial {\mathbf{x}}}{\partial q^{\beta }}. \end{aligned}$$

(A.13)

The components of this tensor form a symmetric, positive-definite matrix. Since ${\mathbf{N}}$ is tangential and $\partial {\mathbf{x}}/\partial q^{1}$, $\partial {\mathbf{x}}/\partial q^{2}$ span the space of such vectors,

$$\begin{aligned} {\mathbf{N}}=\mu ^{\alpha }\frac{\partial {\mathbf{x}}}{\partial q^{\alpha }}. \end{aligned}$$

(A.14)

It is convenient to define $\nu _{\alpha } =g_{\alpha \beta } \mu ^{\beta }$, hence $\mu ^{\alpha }=g^{\alpha \beta }\nu _{\beta }$, where $g^{\alpha \beta }$ is the inverse of the matrix $g_{\alpha \beta }$. Thus,

$$\begin{aligned} {\mathbf{N}}=\nu _{\alpha } g^{\alpha \beta }\frac{\partial {\mathbf{x}}}{\partial q^{\beta }}. \end{aligned}$$

(A.15)

${\mathbf{N.N}}=1$, (A.13), (A.15) and the definition of $g^{\alpha \beta }$ imply

$$\begin{aligned} g^{\alpha \beta }\nu _{\alpha } \nu _{\beta } =1. \end{aligned}$$

(A.16)

An infinitesimal displacement $\mathrm{{d}}q^{\alpha }$ in the $q^{\alpha }$ plane produces the displacement $\mathrm{{d}}{\mathbf{x}}=\mathrm{{d}}q^{\alpha }\partial {\mathbf{x}}/\partial q^{\alpha }$ on $S_{i}$, hence

$$\begin{aligned} \mathrm{{d}}q^{\alpha }=\mathrm{{d}}{\mathbf{x.}}\nabla q^{\alpha }=\mathrm{{d}}q^{\beta }\frac{\partial {\mathbf{x}}}{\partial q_{\beta } }{\mathbf{.}}\nabla q^{\alpha }. \end{aligned}$$

(A.17)

Since (A.17) holds for any choice of $\mathrm{{d}}q^{\alpha }$,

$$\begin{aligned} \frac{\partial {\mathbf{x}}}{\partial q^{\alpha }}{\mathbf{.}}\nabla q^{\beta }=\delta _{\alpha }^{\beta } , \end{aligned}$$

(A.18)

where $\delta _{\alpha }^{\beta }$ is the Kronecker delta. Writing

$$\begin{aligned} \nabla \eta =\frac{\partial \eta }{\partial q^{\alpha }}\nabla q^{\alpha }, \end{aligned}$$

(A.19)

and using (A.15) and (A.18),

$$\begin{aligned} {\mathbf{N.}}\nabla \eta =\nu _{\alpha } g^{\alpha \beta }\frac{\partial \eta }{\partial q^{\beta }}. \end{aligned}$$

(A.20)

Let $\mathrm{{d}}q^{\alpha }$ represent an infinitesimal displacement along the curve $\partial \Sigma$, in the sense, anticlockwise, indicated by the arrow in Fig. 6. The corresponding displacement, $\mathrm{{d}}{\mathbf{x}}=\mathrm{{d}}q^{\alpha }\partial {\mathbf{x}}/\partial q^{\alpha }$, along $\partial S$ is perpendicular to ${\mathbf{N}}$, hence, using (A.13), (A.15) and the definition of $g^{\alpha \beta }$,

$$\begin{aligned} \nu _{\alpha } \mathrm{{d}}\,q^{\alpha }=0. \end{aligned}$$

(A.21)

This result shows that $\nu _{\alpha }$ provides a normal vector to $\partial \Sigma$ in the $q^{\alpha }$ plane, as indicated in Fig. 6, and implies

$$\begin{aligned} \mathrm{{d}}q^{1}=-\tau \nu _{2} ,\quad \mathrm{{d}}q^{2}=\tau \nu _{1} , \end{aligned}$$

(A.22)

where $\tau \ne 0$ is infinitesimal. Thus, the vector $b_{\alpha } =\tau \nu _{\alpha }$ has components $b_{1} =\mathrm{{d}}q^{2}$ and $b_{2} =-\mathrm{{d}}q^{1}$. Given that the displacement along $\partial \Sigma$ is anticlockwise, $b_{\alpha }$, which is normal to $\partial \Sigma$, is directed outwards from $\Sigma$. Next, consider the infinitesimal displacement $b_{\alpha }$ in the $q^{\alpha }$ plane. This produces $d{\mathbf{x}}=b_{\alpha } \partial {\mathbf{x}}/\partial q^{\alpha }$ in physical space. Equation (A.15) and $b_{\alpha } =\tau \nu _{\alpha }$ give

$$\begin{aligned} {\mathbf{N}}=\tau ^{-1}b_{\alpha } g^{\alpha \beta }\frac{\partial {\mathbf{x}}}{\partial q^{\beta }}, \end{aligned}$$

(A.23)

hence

$$\begin{aligned} \mathrm{{d}}{\mathbf{x.N}}=\tau ^{-1}b_{\alpha } b_{\gamma } g^{\alpha \beta }\frac{\partial {\mathbf{x}}}{\partial q^{\beta }}{\mathbf{.}}\frac{\partial {\mathbf{x}}}{\partial q^{\gamma }} =\tau ^{-1}b_{\alpha } b_{\gamma } g^{\alpha \beta }g_{\beta \gamma } =\tau ^{-1}b_{\alpha } b_{\gamma } \delta _{\gamma }^{\alpha } =\tau ^{-1}b_{\alpha } b_{\alpha } . \end{aligned}$$

(A.24)

Because $b_{\alpha }$ is directed outwards from $\Sigma$, $\mathrm{{d}}{\mathbf{x}}$ takes us from $\partial S$ to a location on $S_{i}$ just outside S. By definition, ${\mathbf{N}}$ is a normal vector directed outwards from S. Thus, $\mathrm{{d}}{\mathbf{x.N}}>0$, hence, since $b_{\alpha } b_{\alpha } >0$, (A.24) gives $\tau >0$. Given that $b_{\alpha }$ is directed outwards from $\Sigma$, $\nu _{\alpha } =\tau ^{-1}b_{\alpha }$ is an outwardly directed normal vector to $\partial \Sigma$, as indicated in Fig. 6.

The components of $g^{\alpha \beta }$ are

$$\begin{aligned} g^{11}=g^{-1}g_{22} ,\quad g^{22}=g^{-1}g_{11} ,\quad g^{12}=g^{21}=-g^{-1}g_{12} , \end{aligned}$$

(A.25)

where g is the determinant of the matrix $g_{\alpha \beta }$, which is positive because $g_{\alpha \beta }$ is positive definite. (A.16), (A.22), (A.25) and $\tau >0$ imply

$$\begin{aligned} \tau =\left( {\frac{g_{\alpha \beta }\, \mathrm{{d}}q^{\alpha }\,\mathrm{{d}}q^{\beta }}{g}} \right) ^{1/2}. \end{aligned}$$

(A.26)

Since $\mathrm{{d}}s=\left( {g_{\alpha \beta }\,\mathrm{{d}}q^{\alpha \, }\mathrm{{d}}q^{\beta }} \right) ^{1/2}$, (A.20), (A.22) and (A.26) give

(A.27)

Using Green’s theorem,

$$\begin{aligned} \oint _{\partial S} {{\mathbf{N.}}\nabla \eta \, \mathrm{{d}}s} =\int _\Sigma {\frac{\partial }{\partial q^{\alpha }}\left( {\sqrt{g} g^{\alpha \beta }\frac{\partial \eta }{\partial q^{\beta }}} \right) \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}} . \end{aligned}$$

(A.28)

Consider a rectangular area element, $\mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}$, in the $q^{\alpha }$ plane. This corresponds to a small parallelogram on S with sides ${\mathbf{a}}_{1} \,\mathrm{{d}}q^{1}$ and ${\mathbf{a}}_{2}\, \mathrm{{d}}q^{2}$, where ${\mathbf{a}}_{\alpha } =\partial {\mathbf{x}}/\partial q^{\alpha }$. The area of this parallelogram is $\mathrm{{d}}S=\left| {{\mathbf{a}}_{1} \times {\mathbf{a}}_{2} } \right| \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}=\left( {\left| {{\mathbf{a}}_{1} } \right| ^{2}\left| {{\mathbf{a}}_{2} } \right| ^{2}-\left( {{\mathbf{a}}_{1} {\mathbf{.a}}_{2} } \right) ^{2}} \right) ^{1/2}\mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}$ according to the Lagrange identity. Employing (A.13), $\left| {{\mathbf{a}}_{1} } \right| ^{2}\left| {{\mathbf{a}}_{2} } \right| ^{2}-\left( {{\mathbf{a}}_{1} {\mathbf{.a}}_{2} } \right) ^{2}=g_{11} g_{22} -g_{12}^{2} =g$, hence $\mathrm{{d}}S=\sqrt{g} \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}$, which is the relation between elementary areas in physical space and the $q^{\alpha }$ plane. Using this result and (A.28), (A.12) yields

$$\begin{aligned} \int _\Sigma {\left( {\sqrt{g} \nabla ^{2}\eta -\frac{\partial }{\partial q^{\alpha }}\left( {\sqrt{g} g^{\alpha \beta }\frac{\partial \eta }{\partial q^{\beta }}} \right) } \right) \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}} =0. \end{aligned}$$

(A.29)

Finally, letting $\Sigma$ shrink down to approach a point, $\nabla ^{2}\eta =\Delta \eta$, hence (A.9) gives (2.12).

1.2 A.2 Derivation of (2.14)

The condition that the liquid/gas interface meets the wall at angle $\theta _\mathrm{{{w}}} \left( {x,y} \right)$ is

$$\begin{aligned} n_{z} =\cos \theta _\mathrm{{{w}}}, \end{aligned}$$

(A.30)

at the contact line, where $n_{z}$ is the z-component of ${\mathbf{n}}$. As we saw earlier, ${\mathbf{n}}=\nabla d-\nabla \eta$ correct to first order in the perturbation. Thus,

$$\begin{aligned} n_{z} =\frac{\partial d}{\partial z}-\frac{\partial \eta }{\partial z}. \end{aligned}$$

(A.31)

Let $\left( {x,y} \right)$ be a point on the equilibrium contact line and $\bar{{\theta }}=\theta _\mathrm{{{w}}} \left( {x,y} \right)$ the associated contact angle (see Fig. 8). Setting $\eta =0$ in (A.31), (A.30) gives

$$\begin{aligned} \frac{\partial d}{\partial z}\left( {x,y,z=0} \right) =\cos \bar{{\theta }}. \end{aligned}$$

(A.32)

Since $d\left( {x,y,z=0} \right) =0$, the point $\left( {x+\mathrm{{d}}x,y+\mathrm{{d}}y} \right)$ lies on the perturbed contact line, $d\left( {x,y,z=0} \right) =\eta$, provided

$$\begin{aligned} \frac{\partial d}{\partial x}\mathrm{{d}}x+\frac{\partial d}{\partial y}\mathrm{{d}}y=\eta . \end{aligned}$$

(A.33)

Applying (A.30) and (A.31) at $\left( {x+\mathrm{{d}}x,y+\mathrm{{d}}y} \right)$,

$$\begin{aligned} \frac{\partial d}{\partial z}\left( {x+\mathrm{{d}}x,y+\mathrm{{d}}y,z=0} \right) -\frac{\partial \eta }{\partial z}=\cos \theta _\mathrm{{{w}}} \left( {x+\mathrm{{d}}x,y+\mathrm{{d}}y} \right) . \end{aligned}$$

(A.34)

Fig. 8

The equilibrium interface close to its contact line, which intersects the plane of the figure at right angles. ${\mathbf{N}}$ and $\varvec{\mathcal {N}}$ are unit vectors normal to the contact line and directed outwards from the drop. The former is tangential to $S_{i}$, the latter to the wall

Subtracting (A.32) and recalling that $\bar{{\theta }}=\theta _\mathrm{{{w}}} \left( {x,y} \right)$,

$$\begin{aligned} \frac{\partial ^{2}d}{\partial x\partial z}\mathrm{{d}}x+\frac{\partial ^{2}d}{\partial y\partial z}\mathrm{{d}}y-\frac{\partial \eta }{\partial z}=-\sin \bar{{\theta }}\left( {\frac{\partial \theta _\mathrm{{{w}}} }{\partial x}\mathrm{{d}}x+\frac{\partial \theta _\mathrm{{{w}}} }{\partial y}\mathrm{{d}}y} \right) . \end{aligned}$$

(A.35)

Taking the displacement $\left( {\mathrm{{d}}x,\mathrm{{d}}y} \right)$ perpendicular to the equilibrium contact line, it lies in the direction of the vector $\left( {\frac{\partial d}{\partial x},\frac{\partial d}{\partial y}} \right)$ and (A.33) implies

$$\begin{aligned} \left( {\mathrm{{d}}x,\mathrm{{d}}y} \right) =\eta \left( {\left( {\frac{\partial d}{\partial x}} \right) ^{2}+\left( {\frac{\partial d}{\partial y}} \right) ^{2}} \right) ^{-1}\left( {\frac{\partial d}{\partial x},\frac{\partial d}{\partial y}} \right) . \end{aligned}$$

(A.36)

With this displacement,

$$\begin{aligned} \frac{\partial ^{2}d}{\partial x\partial z}\mathrm{{d}}x+\frac{\partial ^{2}d}{\partial y\partial z}\mathrm{{d}}y=\frac{1}{2}\eta \left( {\left( {\frac{\partial d}{\partial x}} \right) ^{2}+\left( {\frac{\partial d}{\partial y}} \right) ^{2}} \right) ^{-1}\frac{\partial }{\partial z}\left( {\left( {\frac{\partial d}{\partial x}} \right) ^{2}+\left( {\frac{\partial d}{\partial y}} \right) ^{2}} \right) , \end{aligned}$$

(A.37)

and

$$\begin{aligned} \frac{\partial \theta _\mathrm{{{w}}} }{\partial x}\mathrm{{d}}x+\frac{\partial \theta _\mathrm{{{w}}} }{\partial y}\mathrm{{d}}y=\eta \left( {\left( {\frac{\partial d}{\partial x}} \right) ^{2}+\left( {\frac{\partial d}{\partial y}} \right) ^{2}} \right) ^{-1/2}{\varvec{\mathcal {N}}}.\nabla \theta _\mathrm{{{w}}} , \end{aligned}$$

(A.38)

where

$$\begin{aligned} \varvec{\mathcal {N}}=\left( {\left( {\frac{\partial d}{\partial x}} \right) ^{2}+\left( {\frac{\partial d}{\partial y}} \right) ^{2}} \right) ^{-1/2}\left( {\frac{\partial d}{\partial x},\frac{\partial d}{\partial y}} \right) , \end{aligned}$$

(A.39)

is a unit vector, normal to the equilibrium contact line and tangential to the wall, which is directed outwards from the wetted region (see Fig. 8). Thus, $\varvec{\mathcal {N}}.\nabla \theta _\mathrm{{{w}}}$ is the normal derivative of $\theta _\mathrm{{{w}}} \left( {x,y} \right)$ at the contact line. Since $\left| {\nabla d} \right| =1$,

$$\begin{aligned} \left( {\frac{\partial d}{\partial x}} \right) ^{2}+\left( {\frac{\partial d}{\partial y}} \right) ^{2}=1-\left( {\frac{\partial d}{\partial z}} \right) ^{2}, \end{aligned}$$

(A.40)

hence (A.37) and (A.38) give

$$\begin{aligned}&\frac{\partial ^{2}d}{\partial x\partial z}\mathrm{{d}}x+\frac{\partial ^{2}d}{\partial y\partial z}\mathrm{{d}}y=-\eta \left( {1-\left( {\frac{\partial d}{\partial z}} \right) ^{2}} \right) ^{-1}\frac{\partial d}{\partial z}\frac{\partial ^{2}d}{\partial z^{2}}, \end{aligned}$$

(A.41)

$$\begin{aligned}&\frac{\partial \theta _\mathrm{{{w}}} }{\partial x}\mathrm{{d}}x+\frac{\partial \theta _\mathrm{{{w}}} }{\partial y}\mathrm{{d}}y=\eta \left( {1-\left( {\frac{\partial d}{\partial z}} \right) ^{2}} \right) ^{-1/2}{\varvec{\mathcal {N}}}.\nabla \theta _\mathrm{{{w}}} . \end{aligned}$$

(A.42)

Using (A.32) and (A.35),

$$\begin{aligned} \frac{\partial \eta }{\partial z}=\eta \left( {\varvec{\mathcal {N}}}.\nabla \theta _\mathrm{{{w}}} -\frac{\cos \bar{{\theta }}}{\sin ^{2}\bar{{\theta }}}\frac{\partial ^{2}d}{\partial z^{2}} \right) . \end{aligned}$$

(A.43)

Let ${\mathbf{N}}$ be the unit vector shown in Fig. 8, which is tangential to the equilibrium interface and normal to the contact line. Keeping $\left( {x,y} \right)$ constant, $\mathrm{{d}}\eta =\mathrm{{d}}z\,\partial \eta /\partial z$ is the change in $\eta$ for the increment $\mathrm{{d}}z$. $\nabla {{d}}. \nabla \eta =0$ implies that the component of $\nabla \eta$ normal to the interface is zero. Thus, only the component of displacement, $-\mathrm{{d}}z{\mathbf{N}}\sin \bar{{\theta }}$, parallel to the interface produces a change in $\eta$, hence $d\eta =-\mathrm{{d}}z\sin \bar{{\theta }}{\mathbf{N.}}\nabla \eta$. It follows that $\partial \eta /\partial z=-\sin \bar{{\theta }}{\mathbf{N.}}\nabla \eta$ so (A.43) gives

$$\begin{aligned} {\mathbf{N.}}\nabla \eta =J\eta \end{aligned}$$

(A.44)

as the contact-line condition, where

$$\begin{aligned} J=\frac{1}{\sin \bar{{\theta }}}\left( {\frac{\cos \bar{{\theta }}}{\sin ^{2}\bar{{\theta }}}\frac{\partial ^{2}d}{\partial z^{2}}-{\varvec{\mathcal {N}}}.\nabla \theta _\mathrm{{{w}}} } \right) . \end{aligned}$$

(A.45)

Taking the surface S, used in Sect. A.1 and shown in Fig. 7, to be the entire equilibrium interface, $\partial S$ is the contact line and ${\mathbf{N}}$ is the vector defined above. Using (A.20), (A.44) yields (2.14). Note that, as stated following (2.14), its left-hand side is the derivative, ${\mathbf{N.}}\nabla \eta$, of $\eta$, taken tangential to $S_{i}$, normal to the contact line and outwards from $S_{i}$.

1.3 A.3 A frequently used identity

The following identity will often be used in subsequent sections:

$$\begin{aligned} \int _{S_{i} } {f_{1} \Delta f_{2} \,\mathrm{{d}}S} =\oint _C {f_{1} g^{\alpha \beta }\nu _{\alpha } \frac{\partial f_{2} }{\partial q^{\beta }}\mathrm{{d}}s} -\int _{S_{i} } {g^{\alpha \beta }\frac{\partial f_{1} }{\partial q^{\alpha }}\frac{\partial f_{2} }{\partial q^{\beta }}\mathrm{{d}}S} , \end{aligned}$$

(A.46)

where $f_{1}$ and $f_{2}$ are any functions defined on the equilibrium liquid/gas interface $S_{i}$ and C is the equilibrium contact line. This identity can be derived as follows. Using $\mathrm{{d}}S=\sqrt{g} \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}$ and (2.13),

$$\begin{aligned} \int _{S_{i} } {f_{1} \Delta f_{2} \mathrm{{d}}S} =\int _{\Sigma _{i} } {f_{1} \frac{\partial }{\partial q^{\alpha }}\left( {\sqrt{g} g^{\alpha \beta }\frac{\partial f_{2} }{\partial q^{\beta }}} \right) \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}} , \end{aligned}$$

(A.47)

where $\Sigma _{i}$ represents the entire equilibrium interface in the $q^{\alpha }$ plane. Writing

$$\begin{aligned} f_{1} \frac{\partial }{\partial q^{\alpha }}\left( {\sqrt{g} g^{\alpha \beta }\frac{\partial f_{2} }{\partial q^{\beta }}} \right) =\frac{\partial }{\partial q^{\alpha }}\left( {f_{1} \sqrt{g} g^{\alpha \beta }\frac{\partial f_{2} }{\partial q^{\beta }}} \right) -\sqrt{g} g^{\alpha \beta }\frac{\partial f_{1} }{\partial q^{\alpha }}\frac{\partial f_{2} }{\partial q^{\beta }}, \end{aligned}$$

(A.48)

the first term can be treated using the two-dimensional divergence theorem in the $q^{\alpha }$ plane:

$$\begin{aligned} \int _{\Sigma _{i} } {\frac{\partial }{\partial q^{\alpha }}\left( {f_{1} \sqrt{g} g^{\alpha \beta }\frac{\partial f_{2} }{\partial q^{\beta }}} \right) \mathrm{{d}}q^{1}\,\mathrm{{d}}q^{2}} =\oint _{\partial \Sigma _{i} } {f_{1} \sqrt{g} g^{\alpha \beta }\frac{\partial f_{2} }{\partial q^{\beta }}\frac{\nu _{\alpha } }{\sqrt{\nu _{\delta } \nu _{\delta } } }\mathrm{{d}}s_{q} } , \end{aligned}$$

(A.49)

where $\mathrm{{d}}s_{q}$ is elementary arc length on $\partial \Sigma _{i}$. Transforming to physical space, $\partial \Sigma _{i}$ becomes the equilibrium contact line, C, and it can be shown (using (A.22), (A.26) and $\mathrm{{d}}s=\left( {g_{\alpha \beta }\, \mathrm{{d}}q^{\alpha }\,\mathrm{{d}}q^{\beta }} \right) ^{1/2})$ that $\sqrt{g} \mathrm{{d}}s_{q} /\sqrt{\nu _{\delta } \nu _{\delta } } =\mathrm{{d}}s$, where $\mathrm{{d}}s$ is elementary arc length on C. Thus,

$$\begin{aligned} \oint _{\partial \Sigma _{i} } {f_{1} \sqrt{g} g^{\alpha \beta }\frac{\partial f_{2} }{\partial q^{\beta }}\frac{\nu _{\alpha } }{\sqrt{\nu _{\delta } \nu _{\delta } } }\mathrm{{d}}s_{q} } =\oint _C {f_{1} g^{\alpha \beta }\nu _{\alpha } \frac{\partial f_{2} }{\partial q^{\beta }}\mathrm{{d}}s} . \end{aligned}$$

(A.50)

Again using $dS=\sqrt{g} \mathrm{{d}}q^{1}\mathrm{{d}}q^{2}$,

$$\begin{aligned} \int _{\Sigma _{i} } {\sqrt{g} g^{\alpha \beta }\frac{\partial f_{1} }{\partial q^{\alpha }}\frac{\partial f_{2} }{\partial q^{\beta }}\mathrm{{d}}q^{1}\mathrm{{d}}q^{2}} =\int _{S_{i} } {g^{\alpha \beta }\frac{\partial f_{1} }{\partial q^{\alpha }}\frac{\partial f_{2} }{\partial q^{\beta }}\mathrm{{d}}S}. \end{aligned}$$

(A.51)

Equations (A.47)–(A.51) give (A.46).

An application of (A.46) is the following. Let $\tilde{{\eta }}$ be a function, possibly complex, on $S_{i}$ such that (3.13) holds on C. It follows from (3.16) and (A.46) that

$$\begin{aligned} \int _{S_{i} } {\tilde{{\eta }}\left( {\Delta \tilde{{\eta }}^{*}+\gamma \tilde{{\eta }}^{*}} \right) \mathrm{{d}}S} =-\tilde{{E}}_{\mathrm{{s}}} \left[ {\tilde{{\eta }}} \right] , \end{aligned}$$

(A.52)

where * denotes complex conjugation.

1.4 A.4 Derivation of the energy Eq. (2.16)

Writing ${\varvec{\upsigma }}=-{p}'{\mathbf{I}}+2\,\text {Oh }{\mathbf{e}}$, where ${\mathbf{I}}$ is the identity tensor and ${\mathbf{e}}=\left( {\nabla {\mathbf{u}}+\left( {\nabla {\mathbf{u}}} \right) ^{\mathrm{{T}}}} \right) /2$ the strain-rate tensor, ${\varvec{\upsigma }}$ is the perturbation of the stress tensor. Equations (2.2) and (2.9) give

$$\begin{aligned} \frac{\partial {\mathbf{u}}}{\partial t}=\nabla .{\varvec{\upsigma }}, \end{aligned}$$

(A.53)

while (2.12) implies

$$\begin{aligned} {\varvec{\upsigma }} .\mathbf{n} ={\mathbf{n}}\left( {\Delta \eta +\gamma \eta } \right) \end{aligned}$$

(A.54)

on the interface. Employing (2.2) and the definitions of ${\varvec{\upsigma }}$ and ${\mathbf{e}}$, the identity ${\mathbf{u.}}\left( {\nabla .{\varvec{\upsigma }}} \right) =\nabla {\mathbf{.}}\left( {{\mathbf{u}}.{\varvec{\upsigma }}} \right) -{\varvec{\upsigma }} :\nabla {\mathbf{u}}$ yields ${\mathbf{u.}}\left( {\nabla .{\varvec{\upsigma }}} \right) =\nabla .\left( {{\mathbf{u}}.{\varvec{\upsigma }}} \right) -2\,\text {Oh }{\mathbf{e:e}}$. Using this result, scalar multiplying (A.53) by ${\mathbf{u}}$ and integrating over the drop, D, the divergence theorem leads to

$$\begin{aligned} \frac{1}{2}\frac{\mathrm{{d}}}{\mathrm{{d}}t}\int _D {\left| {{\mathbf{u}}} \right| ^{2}\mathrm{{d}}v} =\int _{\partial D} {{\mathbf{u}}.{\varvec{\upsigma }} .{\mathbf{n}}\, \mathrm{{d}}S} -2\,\text {Oh}\int _D {{\mathbf{e:e}}\, \mathrm{{d}}v} . \end{aligned}$$

(A.55)

Equation (2.3) and the definitions of ${\varvec{\upsigma }}$ and ${\mathbf{e}}$ give

$$\begin{aligned} \int _{S_\mathrm{{{w}}} } {{\mathbf{u}}.{\varvec{\upsigma }} .{\mathbf{n}}\,\mathrm{{d}}S}= & {} 2\,\text {Oh}\int _{S_\mathrm{{{w}}} } {{\mathbf{u.e.n}} \, \mathrm{{d}}S} \nonumber \\= & {} -2\,\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {u_{x} e_{xz} +u_{y} e_{yz} } \right) \mathrm{{d}}S} \nonumber \\= & {} -\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {u_{x} \frac{\partial u_{x} }{\partial z}+u_{y} \frac{\partial u_{y} }{\partial z}} \right) \mathrm{{d}}S} \nonumber \\= & {} -\lambda \text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {\left( {\frac{\partial u_{x} }{\partial z}} \right) ^{2}+\left( {\frac{\partial u_{y} }{\partial z}} \right) ^{2}} \right) \mathrm{{d}}S} \end{aligned}$$

(A.56)

as the wall contribution to the surface integral in (A.55). Note that $S_\mathrm{{{w}}}$ is the equilibrium wetted area of the wall.

The interfacial contribution to (A.55) follows from (2.11) and (A.54) as

$$\begin{aligned} \int _{S_{i} } {{\mathbf{u}}.{\varvec{\upsigma }} .{\mathbf{n}}\,\mathrm{{d}}S} =\int _{S_{i} } {\frac{\partial \eta }{\partial t}\left( {\Delta \eta +\gamma \eta } \right) \mathrm{{d}}S} . \end{aligned}$$

(A.57)

Applying (A.46),

$$\begin{aligned} \int _{S_{i} } {\frac{\partial \eta }{\partial t}\Delta \eta \,\mathrm{{d}}S} =\oint _C {\frac{\partial \eta }{\partial t}g^{\alpha \beta }\nu _{\alpha } \frac{\partial \eta }{\partial q^{\beta }}\mathrm{{d}}s} -\int _{S_{i} } {g^{\alpha \beta }\frac{\partial ^{2}\eta }{\partial q^{\alpha }\partial t}\frac{\partial \eta }{\partial q^{\beta }}\mathrm{{d}}S} . \end{aligned}$$

(A.58)

Using (2.14),

$$\begin{aligned} \oint _C {\frac{\partial \eta }{\partial t}g^{\alpha \beta }\nu _{\alpha } \frac{\partial \eta }{\partial q^{\beta }}\mathrm{{d}}s} =\oint _C {J\frac{\partial \eta }{\partial t}\eta \,\mathrm{{d}}s} =\frac{1}{2}\frac{\mathrm{{d}}}{\mathrm{{d}}t}\oint _C {J\eta ^{2}\,\mathrm{{d}}s} . \end{aligned}$$

(A.59)

Symmetry of $g^{\alpha \beta }$ implies

$$\begin{aligned} \int _{S_{i} } {g^{\alpha \beta }\frac{\partial ^{2}\eta }{\partial q^{\alpha }\partial t}\frac{\partial \eta }{\partial q^{\beta }}\mathrm{{d}}S} =\frac{1}{2}\frac{\mathrm{{d}}}{\mathrm{{d}}t}\int _{S_{i} } {g^{\alpha \beta }\frac{\partial \eta }{\partial q^{\alpha }}\frac{\partial \eta }{\partial q^{\beta }}\mathrm{{d}}S} . \end{aligned}$$

(A.60)

Finally,

$$\begin{aligned} \int _{S_{i} } {\frac{\partial \eta }{\partial t}\gamma \eta \, \mathrm{{d}}S} =\frac{1}{2}\frac{\mathrm{{d}}}{\mathrm{{d}}t}\int _{S_{i} } {\gamma \eta ^{2} \mathrm{{d}}S} . \end{aligned}$$

(A.61)

Equations (A.57)–(A.61) give

$$\begin{aligned} \int _{S_{i} } {{\mathbf{u}}.{\varvec{\upsigma }} .{\mathbf{n}}\,\mathrm{{d}}S} =-\frac{\mathrm{{d}}E_\mathrm{{{s}}} }{\mathrm{{d}}t} \end{aligned}$$

(A.62)

as the interfacial contribution to (A.55), where $E_\mathrm{{{s}}}$ is given by (2.18). Combining (A.55), (A.56) and (A.62) yields (2.16).

1.5 A.5 Derivation of (3.2) and (3.3)

Using Lagrange multipliers for the constraints (3.1), we look for extrema of

$$\begin{aligned} Q\left[ \omega \right] =\frac{1}{2}\left( {\int _{S_{i} } {\left( {g^{\alpha \beta }\frac{\partial \omega }{\partial q^{\alpha }}\frac{\partial \omega }{\partial q^{\beta }}-\left( {\gamma +\zeta } \right) \omega ^{2}-2\phi \omega } \right) \mathrm{{d}}S} -\oint _C {J\omega ^{2}\mathrm{{d}}s} } \right) . \end{aligned}$$

(A.63)

Symmetry of $g^{\alpha \beta }$ implies

$$\begin{aligned} \delta Q=\int _{S_{i} } {\left( {g^{\alpha \beta }\frac{\partial \delta \omega }{\partial q^{\alpha }}\frac{\partial \omega }{\partial q^{\beta }}-\delta \omega \left( {\left( {\gamma +\zeta } \right) \omega +\phi } \right) } \right) \mathrm{{d}}S} -\oint _C {J\omega \delta \omega \, \mathrm{{d}}s}, \end{aligned}$$

(A.64)

for the variation of $Q\left[ \omega \right]$ due to the infinitesimal variation, $\delta \omega$, of $\omega$. Using (A.46),

$$\begin{aligned} \delta Q=-\int _{S_{i} } {\delta \omega \left( {\Delta \omega +\left( {\gamma +\zeta } \right) \omega +\phi } \right) \mathrm{{d}}S} +\oint _C {\delta \omega \left( {\nu _{\alpha } g^{\alpha \beta }\frac{\partial \omega }{\partial q^{\beta }}-J\omega } \right) \mathrm{{d}}s} . \end{aligned}$$

(A.65)

The condition for an extremum is $\delta Q=0$ for any $\delta \omega$. Thus, we obtain (3.2) and (3.3). The constraints (3.1) also need to be imposed. Of these, the first contributes to the eigenvalue problem, whereas the second provides a normalisation condition.

1.6 A.6 Derivation of (3.4) and (3.6)

The first of Eq. (3.1), together with (3.2) and (3.3), gives

$$\begin{aligned}&\int _{S_{i} } {\omega _{n}\quad \mathrm{{d}}S} =0 \end{aligned}$$

(A.66)

$$\begin{aligned}&\Delta \omega _{n} +\left( {\gamma +\zeta _{n} } \right) \omega _{n} +\phi _{n} =0 \end{aligned}$$

(A.67)

on $S_{i}$ and

$$\begin{aligned} \nu _{\alpha } g^{\alpha \beta }\frac{\partial \omega _{n} }{\partial q^{\beta }}=J\omega _{n} \end{aligned}$$

(A.68)

on C. Multiplying (A.67) by $\omega _{m}$, integrating over $S_{i}$ and using (A.66) with n replaced by m,

$$\begin{aligned} \int _{S_{i} } {\omega _{m} \left( {\Delta \omega _{n} +\left( {\gamma +\zeta _{n} } \right) \omega _{n} } \right) \mathrm{{d}}S} =0. \end{aligned}$$

(A.69)

Using (A.46),

$$\begin{aligned} \int _{S_{i} } {\left( {\left( {\gamma +\zeta _{n} } \right) \omega _{n} \omega _{m} -g^{\alpha \beta }\frac{\partial \omega _{n} }{\partial q^{\alpha }}\frac{\partial \omega _{m} }{\partial q^{\beta }}} \right) \mathrm{{d}}S} +\oint _C {J\omega _{n} \omega _{m}\, \mathrm{{d}}s} =0. \end{aligned}$$

(A.70)

Permuting n and m and subtracting, symmetry of $g^{\alpha \beta }$ yields

$$\begin{aligned} \left( {\zeta _{n} -\zeta _{m} } \right) \int _{S_{i} } {\omega _{n} \omega _{m}\, \mathrm{{d}}S} =0. \end{aligned}$$

(A.71)

If $\zeta _{n} \ne \zeta _{m}$, (A.71) gives orthogonality of $\omega _{n}$ and $\omega _{m}$. In the case of a degenerate eigenvalue, the associated eigenfunctions can be orthogonalised, so (3.4) holds for all $n\ne m$. It also applies when $n=m$, thanks to the normalisation resulting from the second equation of (3.1).

Using (3.5) in (2.18),

$$\begin{aligned} E_\mathrm{{{s}}} \left[ \eta \right] =\frac{1}{2}\sum \limits _{n,m} {E_{nm} c_{n} c_{m} }, \end{aligned}$$

(A.72)

where

$$\begin{aligned} E_{nm} =\int _{S_{i} } {\left( {g^{\alpha \beta }\frac{\partial \omega _{n} }{\partial q^{\alpha }}\frac{\partial \omega _{m} }{\partial q^{\beta }}-\gamma \omega _{n} \omega _{m} } \right) \mathrm{{d}}S} -\oint _C {J\omega _{n} \omega _{m}\,\mathrm{{d}}s} . \end{aligned}$$

(A.73)

Employing (3.4) and (A.70),

$$\begin{aligned} E_{nm} =\zeta _{n} \int _{S_{i} } {\omega _{n} \omega _{m\,} \mathrm{{d}}S} =\zeta _{n} \delta _{nm} , \end{aligned}$$

(A.74)

hence (A.72) gives (3.6).

1.7 A.7 Derivation of (3.15)

Let ${{\tilde{\varvec{\upsigma }}}}=-\tilde{{p}}{\mathbf{I}}+2\,\text {Oh }{{\tilde{\mathbf{e}}}}$, where ${{\tilde{\mathbf{e}}}}=\left( {\nabla {{\tilde{\mathbf{u}}}}+\left( {\nabla {{\tilde{\mathbf{u}}}}} \right) ^{\mathrm{{T}}}} \right) /2$. (3.8) and (3.9) imply

$$\begin{aligned} s{{\tilde{\mathbf{u}}}}=\nabla .{\varvec{\tilde{{\sigma }}}}, \end{aligned}$$

(A.75)

while (3.12) gives

$$\begin{aligned} {{\tilde{\varvec{\upsigma }}.n}}={\mathbf{n}}\left( {\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}} \right) \end{aligned}$$

(A.76)

on the interface. Taking the complex conjugate of (A.75), scalar multiplying by ${{\tilde{\mathbf{u}}}}$ and integrating over the drop D, ${{\tilde{\mathbf{u}}.}}\left( {\nabla {{.\tilde{\varvec{\upsigma }}}}^{*}} \right) =\nabla {\mathbf{.}}\left( {{{\tilde{\mathbf{u}}.\tilde{\varvec{\upsigma }}}}^{*}} \right) -2\,\text {Oh }{{\tilde{\mathbf{e}}:\tilde{{\mathbf{e}}}}}^{*}$ and the divergence theorem lead to

$$\begin{aligned} s^{*}\int _D {\left| {{{\tilde{\mathbf{u}}}}} \right| ^{2}\mathrm{{d}}v} =\int _{\partial D} {{{\tilde{\mathbf{u}}.\tilde{\varvec{\upsigma }}}}^{*}{\mathbf{.n}}\mathrm{{d}}S} -2\,\text {Oh}\int _D {{{\tilde{\mathbf{e}}:\tilde{{\mathbf{e}}}}}^{*}\mathrm{{d}}v} . \end{aligned}$$

(A.77)

Equation (3.10) and the definitions of ${{\tilde{\varvec{\upsigma }}}}$ and ${{\tilde{\mathbf{e}}}}$ give

$$\begin{aligned} \int _{S_\mathrm{{{w}}} } {{{\tilde{\mathbf{u}}.\tilde{\varvec{\upsigma }}}}^{*}{\mathbf{.n}}\, \mathrm{{d}}S}= & {} 2\,\text {Oh}\int _{S_\mathrm{{{w}}} } {{{\tilde{\mathbf{u}}.\tilde{{\mathbf{e}}}}}^{*}{\mathbf{.n}}\mathrm{{d}}S} \nonumber \\= & {} -2\,\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {\tilde{{u}}_{x} \tilde{{e}}_{xz}^{*} +\tilde{{u}}_{y} \tilde{{e}}_{yz}^{*} } \right) \mathrm{{d}}S} \nonumber \\= & {} -\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {\tilde{{u}}_{x} \frac{\partial \tilde{{u}}_{x}^{*} }{\partial z}+\tilde{{u}}_{y} \frac{\partial \tilde{{u}}_{y}^{*} }{\partial z}} \right) \mathrm{{d}}S} \nonumber \\= & {} -\lambda \text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {\left| {\frac{\partial \tilde{{u}}_{x} }{\partial z}} \right| ^{2}+\left| {\frac{\partial \tilde{{u}}_{y} }{\partial z}} \right| ^{2}} \right) \mathrm{{d}}S} \end{aligned}$$

(A.78)

as the wall contribution to the surface integral in (A.77).

Using (3.11), (A.52) and (A.76),

$$\begin{aligned} \int _{S_{i} } {{{\tilde{\mathbf{u}}.\tilde{\varvec{\upsigma }}}}^{*}{\mathbf{.n}}\quad \mathrm{{d}}S} =-s\tilde{{E}}_{\mathrm{{s}}} \left[ {\tilde{{\eta }}} \right] \end{aligned}$$

(A.79)

for the interfacial contribution to (A.77). Equation (3.15) follows from (A.77)–(A.79).

1.8 A.8 Nonexistence of $s\ne 0$ modes such that $s\rightarrow 0$ as $\text {Oh}\rightarrow {\text {Oh}}_{c} >0$

In order to derive a contradiction, suppose an $s\ne 0$ mode which approaches $s=0$ as $\text {Oh}\rightarrow \text {Oh}_{c} >0$. The limit would be an $s=0$ mode. Unless $\zeta =0$ is an eigenvalue there are no such modes and we already have a contradiction. The case in which $\zeta =0$ is an eigenvalue is more complicated and is treated below.

Let $\omega _{n}$ be one of the $\zeta =0$ eigenfunctions introduced in Sect. 3.1. Thus, $\omega _{n}$, which is real and corresponds to an $s=0$ mode, satisfies

$$\begin{aligned}&\int _{S_{i} } {\omega _{n}\, \mathrm{{d}}S} =0, \end{aligned}$$

(A.80)

$$\begin{aligned}&\Delta \omega _{n} +\gamma \omega _{n} +\phi _{n} =0, \end{aligned}$$

(A.81)

with constant $\phi _{n}$, and

$$\begin{aligned} \nu _{\alpha } g^{\alpha \beta }\frac{\partial \omega _{n} }{\partial q^{\beta }}=J\omega _{n} \end{aligned}$$

(A.82)

on the contact line. Define ${\mathbf{v}}_{n} \left( {{\mathbf{x}}} \right)$ and $\chi _{n} \left( {{\mathbf{x}}} \right)$ via

$$\begin{aligned}&\int _D {\chi _{n}\,\mathrm{{d}}v} =0, \end{aligned}$$

(A.83)

$$\begin{aligned}&\text {Oh}_{c} \nabla ^{2}{\mathbf{v}}_{n} =\nabla \chi _{n}, \end{aligned}$$

(A.84)

$$\begin{aligned}&\nabla {\mathbf{.v}}_{n} =0, \end{aligned}$$

(A.85)

$$\begin{aligned}&\lambda \frac{\partial v_{nx} }{\partial z}=v_{nx} ,\quad \lambda \frac{\partial v_{ny} }{\partial z}=v_{ny},\quad v_{nz} =0,\quad z=0, \end{aligned}$$

(A.86)

and

$$\begin{aligned}&{\mathbf{v}}_{n} {\mathbf{.n}}=\omega _{n} , \end{aligned}$$

(A.87)

$$\begin{aligned}&{\mathbf{n.}}\left( {\nabla {\mathbf{v}}_{n} +\left( {\nabla {\mathbf{v}}_{n} } \right) ^{\mathrm{{T}}}} \right) {\mathbf{.t}}=0 \end{aligned}$$

(A.88)

on $S_{i}$, where ${\mathbf{t}}$ is any tangent vector to $S_{i}$. Taking two independent choices for ${\mathbf{t}}$, (A.88) gives two boundary conditions. Equations (A.83)–(A.88) can be interpreted as follows. Equations (A.84) and (A.85) mean that ${\mathbf{v}}_{n} \left( {{\mathbf{x}}} \right)$ and $\chi _{n} \left( {{\mathbf{x}}} \right)$ are the velocity and pressure of a steady, incompressible Stokes flow within D. This flow is subject to the Navier conditions (A.86) on the wall and Eqs. (A.87), (A.88) on $S_{i}$. Equation (A.87) specifies the normal component of velocity as $\omega _{n}$, while (A.88) means that the tangential components of the surface force are zero. Equation (A.80) is required for a solution. This follows from integration of (A.85) over D, use of the divergence theorem, $v_{nz} =0$ on the wall and (A.87) on $S_{i}$. Equation (A.83) makes the solution for $\chi _{n} \left( {{\mathbf{x}}} \right)$, which would otherwise be only determined up to an additive constant, unique.

Consider a mode with $s\ne 0$ and let ${\mathbf{v}}=s^{-1}{{\tilde{\mathbf{u}}}}$, $\chi =s^{-1}\left( {\tilde{{p}}-\tilde{{p}}_{0} } \right)$, where the constant $\tilde{{p}}_{0}$ is such that

$$\begin{aligned} \int _D {\chi \,\mathrm{{d}}v} =0. \end{aligned}$$

(A.89)

Equations (3.8)–(3.14) give

$$\begin{aligned}&s{\mathbf{v}}=-\nabla \chi +\text {Oh}\nabla ^{2}{\mathbf{v}}, \end{aligned}$$

(A.90)

$$\begin{aligned}&\nabla {\mathbf{.v}}=0 \end{aligned}$$

(A.91)

with

$$\begin{aligned} \lambda \frac{\partial v_{x} }{\partial z}=v_{x} ,\quad \lambda \frac{\partial v_{y} }{\partial z}=v_{y},\quad v_{z} =0 \end{aligned}$$

(A.92)

on the wall,

$$\begin{aligned}&\tilde{{\eta }}={\mathbf{v.n}}, \end{aligned}$$

(A.93)

$$\begin{aligned}&s\,{\mathrm{Oh}}\,{\mathbf{n.}}\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{\mathrm{{T}}}} \right) ={\mathbf{n}}\left( {\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}+\tilde{{p}}_{0} +s\chi } \right) \end{aligned}$$

(A.94)

on the equilibrium interface,

$$\begin{aligned} \nu _{\alpha } g^{\alpha \beta }\frac{\partial \tilde{{\eta }}}{\partial q^{\beta }}=J\tilde{{\eta }} \end{aligned}$$

(A.95)

at the equilibrium contact line and

$$\begin{aligned} \int _{S_{i} } {\tilde{{\eta }}\quad \mathrm{{d}}S} =0. \end{aligned}$$

(A.96)

The normal and tangential components of (A.94) give

$$\begin{aligned} s\,{\mathrm{Oh}}\,{\mathbf{n.}}\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{T}} \right) {\mathbf{.n}}=\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}+\tilde{{p}}_{0} +s\chi , \end{aligned}$$

(A.97)

and

$$\begin{aligned} {\mathbf{n.}}\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{\mathrm{{T}}}} \right) {\mathbf{.t}}=0, \end{aligned}$$

(A.98)

where ${\mathbf{t}}$ is any tangent vector to $S_{i}$.

Using (A.91), (A.90) can be rewritten as

$$\begin{aligned} s{\mathbf{v}}=\nabla {{.\hat{\varvec{\upsigma }}}}, \end{aligned}$$

(A.99)

where ${{\hat{\varvec{\upsigma }}}}=-\chi {\mathbf{I}}+2\,\text {Oh }{ \hat{{\mathbf{e}}}}$ and ${\hat{{\mathbf{e}}}}=\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{\mathrm{{T}}}} \right) /2$. Equation (A.94) gives

$$\begin{aligned} s{{\hat{\varvec{\upsigma }}.n}}={\mathbf{n}}\left( {\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}+\tilde{{p}}_{0} } \right) \end{aligned}$$

(A.100)

on $S_{i}$. Scalar multiplying (A.99) by ${\mathbf{v}}_{n}$ and integrating over D,

$$\begin{aligned} s\int _D {{\mathbf{v}}_{n} {\mathbf{.v}}\,d \mathrm{{d}}v} =\int _D {{\mathbf{v}}_{n} {\mathbf{.}}\left( {\nabla {{.\hat{\varvec{\upsigma }}}}} \right) \mathrm{{d}}v} . \end{aligned}$$

(A.101)

Using ${\mathbf{v}}_{n} {\mathbf{.}}\left( {\nabla {{.\hat{\varvec{\upsigma }}}}} \right) =\nabla {\mathbf{.}}\left( {{\mathbf{v}}_{n} {{.\hat{\varvec{\upsigma }}}}} \right) -2\,\text {Oh }{{\hat{\mathbf{e}}}}_{n} {:\hat{{\mathbf{e}}}}$, where ${{\hat{\mathbf{e}}}}_{n} =\left( {\nabla {\mathbf{v}}_{n} +\left( {\nabla {\mathbf{v}}_{n} } \right) ^{\mathrm{{T}}}} \right) /2$, and the divergence theorem,

$$\begin{aligned} s\int _D {{\mathbf{v}}_{n} {\mathbf{.v}}\,\mathrm{{d}}v} =\int _{\partial D} {{\mathbf{v}}_{n} {{.\hat{\varvec{\upsigma }}.n}}\,\mathrm{{d}}S} -2\,\text {Oh}\int _D {{\hat{{\mathbf{e}}}}_{n} {:\hat{{\mathbf{e}}}}\,\mathrm{{d}}v} . \end{aligned}$$

(A.102)

The contribution of $S_\mathrm{{{w}}}$ to the surface integral can be evaluated using ${{\hat{\varvec{\upsigma }}}}=-\chi {\mathbf{I}}+2\text { Oh }{\hat{{\mathbf{e}}}}$, ${\hat{{\mathbf{e}}}}=\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{T}} \right) /2$ and $v_{z} =v_{nz} =0$. Thus,

$$\begin{aligned} \int _{S_\mathrm{{{w}}} } {{\mathbf{v}}_{n} {{.\hat{\varvec{\upsigma }}.n}}\mathrm{{d}}S} =-\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {v_{nx} \frac{\partial v_{x} }{\partial z}+v_{ny} \frac{\partial v_{y} }{\partial z}} \right) \mathrm{{d}}S} . \end{aligned}$$

(A.103)

Equations (A.80), (A.87) and (A.100) imply

$$\begin{aligned} \int _{S_{i} } {{\mathbf{v}}_{n} {{.\hat{\varvec{\upsigma }}.n}}\mathrm{{d}}S} =s^{-1}\int _{S_{i} } {\omega _{n} \left( {\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}} \right) \mathrm{{d}}S} . \end{aligned}$$

(A.104)

Using (A.46), (A.82), (A.95) and symmetry of $g^{\alpha \beta }$,

$$\begin{aligned} \int _{S_{i} } {\omega _{n} \Delta \tilde{{\eta }}\, \mathrm{{d}}S} =\oint _C {J\omega _{n} \tilde{{\eta }}\, \mathrm{{d}}s} -\int _{S_{i} } {g^{\alpha \beta }\frac{\partial \omega _{n} }{\partial q^{\alpha }}\frac{\partial \tilde{{\eta }}}{\partial q^{\beta }}\, \mathrm{{d}}S} =\int _{S_{i} } {\tilde{{\eta }}\Delta \omega _{n}\, \mathrm{{d}}S} , \end{aligned}$$

(A.105)

hence

$$\begin{aligned} \int _{S_{i} } {{\mathbf{v}}_{n} {{.\hat{\varvec{\upsigma }}.n}}\,\mathrm{{d}}S} =s^{-1}\int _{S_{i} } {\tilde{{\eta }}\left( {\Delta \omega _{n} +\gamma \omega _{n} } \right) \mathrm{{d}}S} , \end{aligned}$$

(A.106)

which is zero according to (A.81) and (A.96). Thus, (A.102) and (A.103) give

$$\begin{aligned} s\int _D {{\mathbf{v}}_{n} {\mathbf{.v}}\mathrm{{d}}v} =-2\,\text {Oh }\int _D {{\hat{{\mathbf{e}}}}_{n} {:\hat{{\mathbf{e}}}}\;\mathrm{{d}}v} -\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {v_{nx} \frac{\partial v_{x} }{\partial z}+v_{ny} \frac{\partial v_{y} }{\partial z}} \right) \mathrm{{d}}S} \end{aligned}$$

(A.107)

for any $n\in N$, where N denotes the set of n for which $\zeta _{n} =0$.

Suppose the given $s\ne 0$ mode approaches $s=0$ as $\text {Oh}\rightarrow \text {Oh}_{c}$ and is normalised using

$$\begin{aligned} \int _{S_{i} } {\left| {\tilde{{\eta }}} \right| ^{2}\mathrm{{d}}S} =1. \end{aligned}$$

(A.108)

In the limit $\text {Oh}\rightarrow \text {Oh}_{c}$, $s\rightarrow 0$, $\tilde{{\eta }}\rightarrow \tilde{{\eta }}_{c}$, ${\mathbf{v}}\rightarrow {\mathbf{v}}_{c}$, $\chi \rightarrow \chi _{c}$ and $\tilde{{p}}_{0} \rightarrow \tilde{{p}}_{0c}$, where, according to (A.95)–(A.97), $\tilde{{\eta }}_{c}$ and $\tilde{{p}}_{0c}$ satisfy the $s=0$ problem, (3.13), (3.14) and (3.21). Equations (A.89)–(A.93) and (A.98) give

$$\begin{aligned}&\int _D {\chi _{c}\, \mathrm{{d}}v} =0, \end{aligned}$$

(A.109)

$$\begin{aligned}&\text {Oh}_{c} \nabla ^{2}{\mathbf{v}}_{c} =\nabla \chi _{c} , \end{aligned}$$

(A.110)

$$\begin{aligned}&\nabla {\mathbf{.v}}_{c} =0, \end{aligned}$$

(A.111)

$$\begin{aligned}&\lambda \frac{\partial v_{cx} }{\partial z}=v_{cx} ,\quad \lambda \frac{\partial v_{cy} }{\partial z}=v_{cy} ,\quad v_{cz} =0,\quad z=0, \end{aligned}$$

(A.112)

and

$$\begin{aligned}&{\mathbf{v}}_{c} {\mathbf{.n}}=\tilde{{\eta }}_{c} , \end{aligned}$$

(A.113)

$$\begin{aligned}&{\mathbf{n.}}\left( {\nabla {\mathbf{v}}_{c} +\left( {\nabla {\mathbf{v}}_{c} } \right) ^{T}} \right) {\mathbf{.t}}=0 \end{aligned}$$

(A.114)

on $S_{i}$. Given $\tilde{{\eta }}_{c}$, (A.109)–(A.114) determine ${\mathbf{v}}_{c}$ and $\chi _{c}$. Using (A.86), (A.107) has the limiting form

$$\begin{aligned} 2\int _D {{\hat{{\mathbf{e}}}}_{n} {:\hat{{\mathbf{e}}}}_{c} \mathrm{{d}}v} +\lambda \int _{S_\mathrm{{{w}}} } {\left( {\frac{\partial v_{nx} }{\partial z}\frac{\partial v_{cx} }{\partial z}+\frac{\partial v_{ny} }{\partial z}\frac{\partial v_{cy} }{\partial z}} \right) \mathrm{{d}}S} =0, \end{aligned}$$

(A.115)

for $n\in N$, where ${\hat{{\mathbf{e}}}}_{c} =\left( {\nabla {\mathbf{v}}_{c} +\left( {\nabla {\mathbf{v}}_{c} } \right) ^{\mathrm{{T}}}} \right) /2$.

Since $\tilde{{\eta }}_{c}$ and $\tilde{{p}}_{0c}$ satisfy the $s=0$ problem, (3.13), (3.14) and (3.21), $\tilde{{\eta }}_{c}$ can be expressed as a linear combination of the $\zeta =0$ eigenfunctions of (3.1), (3.2) and (3.3) , i.e.

$$\begin{aligned} \tilde{{\eta }}_{c} =\sum \limits _{n\in N} {c_{n} \omega _{n} } , \end{aligned}$$

(A.116)

where the coefficients $c_{n}$ may be complex and, using (3.4) and (A.108),

$$\begin{aligned} \sum \limits _{n\in N} {\left| {c_{n} } \right| ^{2}} =1. \end{aligned}$$

(A.117)

Comparing (A.83)–(A.88) with (A.109)–(A.114), we see that

$$\begin{aligned} {\mathbf{v}}_{c} =\sum \limits _{n\in N} {c_{n} {\mathbf{v}}_{n} } ,\quad \chi _{c} =\sum \limits _{n\in N} {c_{n} \chi _{n} } . \end{aligned}$$

(A.118)

Using the first of Eqs. (A.118) in (A.115),

$$\begin{aligned} \sum \limits _{m\in N} {A_{nm} c_{m} } =0, \quad n\in N, \end{aligned}$$

(A.119)

where

$$\begin{aligned} A_{nm} =2\int _D {{\hat{{\mathbf{e}}}}_{n} {:\hat{{\mathbf{e}}}}_{m} \;\mathrm{{d}}v} +\lambda \int _{S_\mathrm{{{w}}} } {\left( {\frac{\partial v_{nx} }{\partial z}\frac{\partial v_{mx} }{\partial z}+\frac{\partial v_{ny} }{\partial z}\frac{\partial v_{my} }{\partial z}} \right) \mathrm{{d}}S} \end{aligned}$$

(A.120)

is a real, square, symmetric matrix defined for $n,\;m\in N$. Let $\bar{{c}}_{n}$ be real and ${\bar{{\mathbf{v}}}}=\sum \nolimits _{n\in N} {\bar{{c}}_{n} {\mathbf{v}}_{n} }$, then

$$\begin{aligned} \sum \limits _{n,m\in N} {A_{nm} \bar{{c}}_{n} \bar{{c}}_{m} } =2\int _D {\bar{{\mathbf{e}}}}:{\bar{{\mathbf{e}}}}\;{\mathrm{{d}}v} +\lambda \int _{S_\mathrm{{{w}}} } {\left( {\left( {\frac{\partial \bar{{v}}_{x} }{\partial z}} \right) ^{2}+\left( {\frac{\partial \bar{{v}}_{y} }{\partial z}} \right) ^{2}} \right) \mathrm{{d}}S}, \end{aligned}$$

(A.121)

where ${\bar{{\mathbf{e}}}}=\left( {\nabla {\bar{{\mathbf{v}}}}+\left( {\nabla {\bar{{\mathbf{v}}}}} \right) ^{\mathrm{{T}}}} \right) /2$. Equation (A.121) is obviously positive or zero. If it were zero, then ${\bar{{\mathbf{e}}}}=0$, hence ${\bar{{\mathbf{v}}}}$ is a combination of a translation and a rotation. On the other hand, (A.86) implies

$$\begin{aligned} \lambda \frac{\partial \bar{{v}}_{x} }{\partial z}=\bar{{v}}_{x} ,\,\, \lambda \frac{\partial \bar{{v}}_{y} }{\partial z}=\bar{{v}}_{y} ,\,\, \bar{{v}}_{z} =0,\, z=0, \end{aligned}$$

(A.122)

thus, ${\bar{{\mathbf{v}}}}=0$, hence $\bar{{c}}_{n} =0$. We conclude that $A_{nm}$ is positive definite. It follows from (A.119) that $c_{n} =0$, which is incompatible with (A.117). This contradiction means $s\ne 0$ modes cannot approach $s=0$ as $\text {Oh}\rightarrow \text {Oh}_{c} >0$.

1.9 A.9 Some properties of inviscid modes

Given (3.30), $\nabla {\mathbf{.}}\left( {\psi \nabla \psi ^{*}} \right) -\left| {\nabla \psi } \right| ^{2}=\psi \nabla ^{2}\psi ^{*}=0$. Integrating over D, the divergence theorem, (3.31) and (3.32) give

$$\begin{aligned} \int _{S_{i} } {\psi \tilde{{\eta }}^{*}\mathrm{{d}}S} =-\int _D {\left| {\nabla \psi } \right| ^{2}\mathrm{{d}}v} . \end{aligned}$$

(A.123)

On the other hand, (3.14) and (3.33) implies

$$\begin{aligned} \sigma \int _{S_{i} } {\psi \tilde{{\eta }}^{*}\mathrm{{d}}S} =-\int _{S_{i} } {\tilde{{\eta }}^{*}\left( {\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}} \right) \mathrm{{d}}S} . \end{aligned}$$

(A.124)

Using (3.13) and (A.46),

$$\begin{aligned} \int _{S_{i} } {\tilde{{\eta }}^{*}\Delta \tilde{{\eta }}\, \mathrm{{d}}S} =\oint _C {J\left| {\tilde{{\eta }}} \right| ^{2}\mathrm{{d}}s} -\int _{S_{i} } {g^{\alpha \beta }\frac{\partial \tilde{{\eta }}^{*}}{\partial q^{\alpha }}\frac{\partial \tilde{{\eta }}}{\partial q^{\beta }}\mathrm{{d}}S} . \end{aligned}$$

(A.125)

Equations (A.123)–(A.125) yield

$$\begin{aligned} \sigma \int _D {\left| {\nabla \psi } \right| ^{2}\mathrm{{d}}S} =\oint _C {J\left| {\tilde{{\eta }}} \right| ^{2}\,\mathrm{{d}}s} -\int _{S_{i} } {\left( {g^{\alpha \beta }\frac{\partial \tilde{{\eta }}^{*}}{\partial q^{\alpha }}\frac{\partial \tilde{{\eta }}}{\partial q^{\beta }}-\gamma \left| {\tilde{{\eta }}} \right| ^{2}} \right) \mathrm{{d}}S} . \end{aligned}$$

(A.126)

The integral on the left-hand side cannot be zero, otherwise $\psi$ is constant and $\tilde{{\eta }}=0$ from (3.32). (3.23) implies $\psi =0$, hence $\tilde{{p}}_{0} =0$ from (3.33). Thus, all unknowns would be zero, which is not allowed for an eigenvalue problem. Symmetry of $g^{\alpha \beta }$ makes the right-hand side of (A.126) real. We conclude that $\sigma$ is real. For each eigenvalue $\sigma _{k}$, $\psi _{k}$, $\tilde{{\eta }}_{k}$ and $\tilde{{p}}_{0k}$ are chosen real from here on. Because the integral on the left-hand side of (A.126) is positive, the $\psi _{k}$ can be normalised such that (3.34) holds when $k=l$.

Equation (3.30) implies $\nabla {\mathbf{.}}\left( {\psi _{k} \nabla \psi _{l} } \right) -\nabla \psi _{k} {\mathbf{.}}\nabla \psi _{l} =\psi _{k} \nabla ^{2}\psi _{l} =0$. Integrating over D, the divergence theorem, (3.31) and (3.32) give

$$\begin{aligned} \int _{S_{i} } {\psi _{k} \tilde{{\eta }}_{l}\, \mathrm{{d}}S} =-\int _D {\nabla \psi _{k} {\mathbf{.}}\nabla \psi _{l}\, \mathrm{{d}}v} , \end{aligned}$$

(A.127)

while (3.14) and (3.33) imply

$$\begin{aligned} \sigma _{k} \int _{S_{i} } {\psi _{k} \tilde{{\eta }}_{l}\, \mathrm{{d}}S} =-\int _{S_{i} } {\tilde{{\eta }}_{l} \left( {\Delta \tilde{{\eta }}_{k} +\gamma \tilde{{\eta }}_{k} } \right) \mathrm{{d}}S} . \end{aligned}$$

(A.128)

Using (3.13), (A.46) and symmetry of $g^{\alpha \beta }$,

$$\begin{aligned} \int _{S_{i} } {\tilde{{\eta }}_{l} \Delta \tilde{{\eta }}_{k}\, \mathrm{{d}}S} =\oint _C {J\tilde{{\eta }}_{k} \tilde{{\eta }}_{l} \mathrm{{d}}s} -\int _{S_{i} } {g^{\alpha \beta }\frac{\partial \tilde{{\eta }}_{k} }{\partial q^{\alpha }}\frac{\partial \tilde{{\eta }}_{l} }{\partial q^{\beta }}\, \mathrm{{d}}S} . \end{aligned}$$

(A.129)

Equations (A.127)–(A.129) yield

$$\begin{aligned} \sigma _{k} \int _D {\nabla \psi _{k} {\mathbf{.}}\nabla \psi _{l}\, \mathrm{{d}}S} =\oint _C {J\tilde{{\eta }}_{k} \tilde{{\eta }}_{l}\, \mathrm{{d}}s} -\int _{S_{i} } {\left( {g^{\alpha \beta }\frac{\partial \tilde{{\eta }}_{k} }{\partial q^{\alpha }}\frac{\partial \tilde{{\eta }}_{l} }{\partial q^{\beta }}-\gamma \tilde{{\eta }}_{k} \tilde{{\eta }}_{l} } \right) \mathrm{{d}}S} . \end{aligned}$$

(A.130)

Thus,

$$\begin{aligned} \sigma _{k} \int _D {\nabla \psi _{k} {\mathbf{.}}\nabla \psi _{l}\, \mathrm{{d}}S} =-\tilde{{E}}_{kl} , \end{aligned}$$

(A.131)

where the matrix $\tilde{{E}}_{kl}$ is given by (3.35) and is symmetric. Using symmetry of $\tilde{{E}}_{kl}$, permutation of k, l and subtraction gives

$$\begin{aligned} \left( {\sigma _{k} -\sigma _{l} } \right) \int _D {\nabla \psi _{k} {\mathbf{.}}\nabla \psi _{l}\,\mathrm{{d}}S} =0. \end{aligned}$$

(A.132)

When $\sigma _{k} \ne \sigma _{l}$, (A.132) implies

$$\begin{aligned} \int _D {\nabla \psi _{k} {\mathbf{.}}\nabla \psi _{l}\,\mathrm{{d}}S} =0. \end{aligned}$$

(A.133)

If $\sigma _{k} =\sigma _{l}$ is a degenerate eigenvalue, its eigenfunctions can be orthogonalised such that (A.133) applies for $k\ne l$. Thus, (A.133) holds for all $k\ne l$. Given the normalisation referred to above, we obtain (3.34). (3.36) follows from (3.34) and (A.131).

1.10 A.10 Nonexistence of $s\ne 0$ modes such that $s\rightarrow 0$ as ${\text {Oh}}\rightarrow 0$

Modes which continue to be affected by viscosity as $\text {Oh}\rightarrow 0$ are discussed in the main text. A mode of this type has $s=O\left( {\text {Oh}} \right)$ and hence approaches $s=0$ as $\text {Oh}\rightarrow 0$, but it is decaying and hence unimportant from a stability point of view. Here, we consider modes of the other type, i.e. those which approach an inviscid limit.

The analysis given here has many similarities with that of Sect. A.8. One significant difference is that ${\mathbf{v}}_{n}$ and $\eta _{n}$ are replaced by ${\mathbf{v}}_{k} =\nabla \psi _{k}$ and $\tilde{{\eta }}_{k}$, where $\psi _{k}$, $\tilde{{\eta }}_{k}$ are $\sigma =0$ eigenfunctions of the inviscid problem (3.13), (3.14), (3.23) and (3.30)–(3.33). As in Sect. A.8, given an $s\ne 0$ mode, let ${\mathbf{v}}=s^{-1}{{\tilde{\mathbf{u}}}}$ and $\chi =s^{-1}\left( {\tilde{{p}}-\tilde{{p}}_{0} } \right)$, where the constant $\tilde{{p}}_{0}$ is determined by (A.89). Equaions (A.90)–(A.107), with ${\mathbf{v}}_{n}$, $\eta _{n}$ and ${\hat{{\mathbf{e}}}}_{n}$ replaced by ${\mathbf{v}}_{k} =\nabla \psi _{k}$, $\tilde{{\eta }}_{k}$ and ${\hat{{\mathbf{e}}}}_{k} =\left( {\nabla {\mathbf{v}}_{k} +\left( {\nabla {\mathbf{v}}_{k} } \right) ^{\mathrm{{T}}}} \right) /2$, follow as before.

Letting $\psi =s^{-1}\chi$, (A.89)–(A.96) give

$$\begin{aligned}&\int _D {\psi \, \mathrm{{d}}v} =0, \end{aligned}$$

(A.134)

$$\begin{aligned}&{\mathbf{v}}=-\nabla \psi +s^{-1}\text {Oh}\nabla ^{2}{\mathbf{v}}, \end{aligned}$$

(A.135)

$$\begin{aligned}&\nabla {\mathbf{.v}}=0 \end{aligned}$$

(A.136)

with

$$\begin{aligned} \lambda \frac{\partial v_{x} }{\partial z}=v_{x} ,\, \lambda \frac{\partial v_{y} }{\partial z}=v_{y} ,\, v_{z} =0, \end{aligned}$$

(A.137)

on the wall,

$$\begin{aligned}&\tilde{{\eta }}={\mathbf{v.n}}, \end{aligned}$$

(A.138)

$$\begin{aligned}&s\,{\mathrm{Oh}}\,{\mathbf{n.}}\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{\mathrm{{T}}}} \right) ={\mathbf{n}}\left( {\Delta \tilde{{\eta }}+\gamma \tilde{{\eta }}+\tilde{{p}}_{0} +s^{2}\psi } \right) \end{aligned}$$

(A.139)

on the equilibrium interface,

$$\begin{aligned} \nu _{\alpha } g^{\alpha \beta }\frac{\partial \tilde{{\eta }}}{\partial q^{\beta }}=J\tilde{{\eta }} \end{aligned}$$

(A.140)

at the equilibrium contact line and

$$\begin{aligned} \int _{S_{i} } {\tilde{{\eta }}\,\mathrm{{d}}S} =0. \end{aligned}$$

(A.141)

Equation (A.107), with ${\mathbf{v}}_{n}$, ${\hat{{\mathbf{e}}}}_{n}$ replaced by ${\mathbf{v}}_{k}$, ${\hat{{\mathbf{e}}}}_{k}$, gives

$$\begin{aligned} \int _D {{\mathbf{v}}_{k} {\mathbf{.v}}\mathrm{{d}}v} =-2s^{-1}\text {Oh}\int _D {{\hat{{\mathbf{e}}}}_{k} {:\hat{{\mathbf{e}}}}\mathrm{{d}}v} -s^{-1}\text {Oh}\int _{S_\mathrm{{{w}}} } {\left( {v_{kx} \frac{\partial v_{x} }{\partial z}+v_{ky} \frac{\partial v_{y} }{\partial z}} \right) \mathrm{{d}}S} , \end{aligned}$$

(A.142)

where ${\hat{{\mathbf{e}}}}=\left( {\nabla {\mathbf{v}}+\left( {\nabla {\mathbf{v}}} \right) ^{\mathrm{{T}}}} \right) /2$. Recalling that ${\mathbf{v}}_{k} =\nabla \psi _{k}$, where $\psi _{k}$ is a $\sigma =0$ inviscid eigenfunction, (A.142) applies for all k for which $\sigma _{k} =0$. We denote the set of those k by K.

Suppose an $s\ne 0$ mode approaches an inviscid limit with $s=0$ as $\text {Oh}\rightarrow 0$. In order that the viscous term in (A.135) be negligible in the limit, $s^{-1}\text {Oh}\rightarrow 0$. Normalising the mode using (A.108), $\tilde{{\eta }}\rightarrow \tilde{{\eta }}_{c}$, ${\mathbf{v}}\rightarrow {\mathbf{v}}_{c}$, $\tilde{{p}}_{0} \rightarrow \tilde{{p}}_{0c}$ and $\psi \rightarrow \psi _{c}$. Equation (A.134) implies

$$\begin{aligned} \int _D {\psi _{c}\,\mathrm{{d}}v} =0. \end{aligned}$$

(A.143)

Equation (A.135) and $s^{-1}\text {Oh}\rightarrow 0$ give

$$\begin{aligned} {\mathbf{v}}_{c} =-\nabla \psi _{c} . \end{aligned}$$

(A.144)

Thus,

$$\begin{aligned} \nabla ^{2}\psi _{c} =0, \end{aligned}$$

(A.145)

according to (A.136). As discussed following Eq. (3.30), the inviscid problem only allows one wall boundary condition, rather than the three expressed by (A.137) for the viscous problem. The first two equations of (A.137) drop out in the inviscid limit, leaving $v_{z} =0$, hence

$$\begin{aligned} \frac{\partial \psi _{c} }{\partial z}=0, \end{aligned}$$

(A.146)

on the wall. Eqations (A.138)–(A.141) yield

$$\begin{aligned}&\frac{\partial \psi _{c} }{\partial n}=-\tilde{{\eta }}_{c} , \end{aligned}$$

(A.147)

$$\begin{aligned}&\Delta \tilde{{\eta }}_{c} +\gamma \tilde{{\eta }}_{c} +\tilde{{p}}_{0c} =0 \end{aligned}$$

(A.148)

on the equilibrium interface,

$$\begin{aligned} \nu _{\alpha } g^{\alpha \beta }\frac{\partial \tilde{{\eta }}_{c} }{\partial q^{\beta }}=J\tilde{{\eta }}_{c} \end{aligned}$$

(A.149)

at the equilibrium contact line and

$$\begin{aligned} \int _{S_{i} } {\tilde{{\eta }}_{c}\,\mathrm{{d}}S} =0. \end{aligned}$$

(A.150)

Equations (A.148)–(A.150) show that $\tilde{{\eta }}_{c}$ and $\tilde{{p}}_{0c}$ satisfy the $s=0$ problem, (3.13), (3.14) and (3.21), while (A.143) and (A.145)–(A.147) correspond to (3.23) and (3.30)–(3.32) and determine $\psi _{c}$ given $\tilde{{\eta }}_{c}$. Since $s^{-1}\text {Oh}\rightarrow 0$, (A.142) implies

$$\begin{aligned} \int _D {{\mathbf{v}}_{k} {\mathbf{.v}}_{c}\,\mathrm{{d}}v} =0 \end{aligned}$$

(A.151)

for all $k\in K$.

That $\tilde{{\eta }}_{c}$ and $\tilde{{p}}_{0c}$ satisfy the $s=0$ problem, (3.13), (3.14) and (3.21), indicates that $\tilde{{\eta }}_{c}$ is a $\sigma =0$ eigenfunction, hence

$$\begin{aligned} \tilde{{\eta }}_{c} =\sum \limits _{k\in K} {\tilde{{c}}_{k} \tilde{{\eta }}_{k} } . \end{aligned}$$

(A.152)

Equations (A.143), (A.145)–(A.147) and the corresponding equations for $\psi _{k}$ and $\tilde{{\eta }}_{k}$ imply

$$\begin{aligned} \psi _{c} =\sum \limits _{k\in K} {\tilde{{c}}_{k} \psi _{k} } \end{aligned}$$

(A.153)

so

$$\begin{aligned} {\mathbf{v}}_{c} =-\sum \limits _{k\in K} {\tilde{{c}}_{k} \nabla \psi _{k} }, \end{aligned}$$

(A.154)

according to (A.144). Scalar multiplying by $\nabla \psi _{l}$, where $l\in K$, integrating over D and using (3.34), ${\mathbf{v}}_{l} =\nabla \psi _{l}$ and (A.151),

$$\begin{aligned} \tilde{{c}}_{l} =-\int _D {{\mathbf{v}}_{c} {\mathbf{.}}\nabla \psi _{l}\, \mathrm{{d}}v} =-\int _D {{\mathbf{v}}_{c} {\mathbf{.v}}_{l}\,\mathrm{{d}}v} =0. \end{aligned}$$

(A.155)

This result means that $\tilde{{\eta }}_{c} =0$, which is incompatible with the normalisation (A.108). Thus, we have a contradiction and conclude that $s\ne 0$ modes which approach an inviscid limit with $s=0$ as $\text {Oh}\rightarrow 0$ do not exist.