On a model for a sliding droplet:Well-posedness and stability of translating circular solutions

TL;DR

This paper analyzes a mathematical model for a viscous droplet sliding down an inclined plane, establishing conditions for well-posedness and demonstrating the stability of circular translating solutions under certain slope and boundary conditions.

Contribution

It provides a rigorous analysis of the well-posedness and stability of the model, identifying critical slopes and conditions for the stability of circular solutions.

Findings

Well-posedness holds for slopes below a critical value.

Translating circular solutions are asymptotically stable under small inclines.

Loss of well-posedness occurs when the linearization is no longer parabolic.

Abstract

In this paper the model for a highly viscous droplet sliding down an inclined plane is analyzed. It is shown that, provided the slope is not too steep, the corresponding moving boundary problem possesses classical solutions. Well-posedness is lost when the relevant linearization ceases to be parabolic. This occurs above a critical incline which depends on the shape of the initial wetted region as well as on the liquid's mass. It is also shown that translating circular solutions are asymptotically stable if the kinematic boundary condition is given by an affine functionand provided the incline is small.

Equations

$$-\Delta u$$

$$u$$

$${\int }_{\Omega (t)}udx$$

$$V$$

$$\Omega (0)$$

$$-\Delta u_0$$

$$u_0$$

$${\int }_{{\Omega }_{\ast }}{u}_{0}dx$$

$$F(-{\partial }_{\nu }{u}_0)$$

$$u_0(r,\theta ):=\frac{\mu r}{8}(R_0^2-r^2)\cos \theta +\frac{2\mathcal{V}}{\pi R_0^4}(R_0^2-r^2),r\in [0,R_0],\theta \in [0,2\pi ],$$

$$F\left(\frac{\mu R_0^2}{4}\cos \theta +\frac{4\mathcal{V}}{\pi R_0^3}\right)=v_0\cos \theta ,\theta \in [0,2\pi ].$$

$$F(q)=aq-b,q\in [-v_0,v_0],a,b>0,$$

$$R_0={\left(\frac{4\mathcal{V}a}{\pi b}\right)}^{1/3},v_0=\frac{\mu a}{4}{\left(\frac{4\mathcal{V}a}{\pi b}\right)}^{2/3},$$

$$\sin \chi \le \frac{16\mathcal{V}}{\pi \mathrm{Bo}{R}_0^5}.$$

$$F\in C^4(\mathbb{R},\mathbb{R}),F^{\prime }>0.$$

$$\begin{split}\operatorname{BUC}^{2+\alpha}(O):=\big{\{}u:O\rightarrow\mathbb{R}\,\big{|}\,&\partial^{\beta}u\text{ is uniformly continuous and bounded for }|\beta|\leq 2\\ &\text{ and }[\partial^{\beta}u]_{\alpha}<\infty\text{ for }|\beta|=2\big{\}},\end{split}$$

$$\Vert u{\Vert }_{2,\alpha }:=\Vert u{\Vert }_{2,\infty }+\underset{|\beta |=2}{\max }[{\partial }^{\beta }u{]}_{\alpha },u\in {\mathrm{BUC}}^{2+\alpha }(O),$$

$$\Vert u{\Vert }_{2,\infty }=\sum _{|\gamma |\le 2}\Vert {\partial }^{\gamma }u{\Vert }_{\infty }\text{ and }[u{]}_{\alpha }:=\underset{x\ne y}{\sup }\frac{|u(x)-u(y)|}{|x-y{|}^{\alpha }}.$$

$$(u,\lambda )\in bu{c}^{2+\alpha }(\Omega )\times \mathbb{R}$$

$$\lambda =\frac{\mathcal{V}}{{\int }_{\Omega }{u}^{(1)}dx}-\mu \frac{{\int }_{\Omega }{u}^{(x^1)}dx}{{\int }_{\Omega }{u}^{(1)}dx}.$$

$$u(\lambda )=\mu u^{(x^1)}+\lambda u^{(1)},$$

$$\mu {\int }_{\Omega }{u}^{(x^1)}dx+\lambda {\int }_{\Omega }{u}^{(1)}dx=\mathcal{V},$$

$$T_{r_0}({\Gamma }_0):=\{x\in {\mathbb{R}}^2||d(x,{\Gamma }_0)|<r_0\}$$

$$\Phi :{\Gamma }_0\times (-r_0,r_0)⟶{\mathbb{R}}^2,(y,r)\mapsto y+r{\nu }_0(y),$$

$${\gamma }_0^{\prime }(s)={\tau }_0({\gamma }_0(s))\text{ and }{d}_s{\nu }_0({\gamma }_0(s))={\kappa }_0({\gamma }_0(s)){\tau }_0({\gamma }_0(s)),$$

$$d_y\Phi (y,r)={\tau }_0(y)+r{\kappa }_0(y){\tau }_0(y)\text{ and }{d}_r\Phi (y,r)={\nu }_0(y).$$

$$\Vert {\kappa }_0{\Vert }_{\infty }<\infty ,$$

$${\mathbb{B}}_{{\Gamma }_0}(y,2r_0)\times (-2r_0,2r_0)\cap {\Gamma }_0={\mathbb{B}}_{{\Gamma }_0}(y,2r_0),$$

$$\bigcup _{|r|<r_0}{\Gamma }_r$$

$$\tilde{\tilde{\nu}}(x):=\nu_{0}\bigl{(}y(x)\bigr{)},\quad x\in T_{\tilde{r}_{0}}(\Gamma_{0}),$$

$$0\leq\eta\leq 1,\quad\eta\big{|}_{[-\tilde{r}_{0}/2,\tilde{r}_{0}/2]}\equiv 1,\quad\text{ and }\ \eta\big{|}_{(-3\tilde{r}_{0}/4,3\tilde{r}_{0}/4)^{\mathsf{c}}}\equiv 0,$$

$$\tilde{\nu}(x):=\begin{cases}\tilde{\tilde{\nu}}(x)\eta\bigl{(}r(x)\bigr{)},&x\in T_{\tilde{r}_{0}}(\Gamma_{0}),\\ 0,&x\notin T_{\tilde{r}_{0}}(\Gamma_{0}).\end{cases}$$

$${\nu }^{\delta }:={\psi }_{\delta }\ast \tilde{\nu }$$

$$\nu^{\delta}\longrightarrow\nu_{0}\circ y\ \text{ in }\ buc^{1+\alpha}\bigl{(}T_{\tilde{r}_{0}/2}(\Gamma_{0})\bigr{)}$$

Full text

1. Introduction

Of interest is the analysis of a model for the motion under gravity of a highly viscous droplet on an inclined homogeneous substrate as depicted below.

\chi$$e_{1}$$u(t,x)

The droplet is characterized by the wetted region $\Omega(t)=[u(t,\cdot)>0]$ on the substrate and a height field $u(t,x)$ measured in direction normal to the plane of motion at points $x\in\Omega(t)$, as depicted above. It follows that $u(t,x)=0$ for $x\in\Gamma(t):=\partial\Omega(t)$. The system reads

[TABLE]

where the parameter $\mu\geq 0$ is related to the inclination $\chi\geq 0$ via $\mu=\mathrm{Bo}\sin\chi$, where $\mathrm{Bo}$ is the Bond number representing the relative magnitude of gravity and viscous forces. The unknown $\lambda$ can be thought of as a time-dependent Lagrange multiplier for the third condition enforcing conservation of the total volume $\mathcal{V}>0$ (actually, it is a constant resulting from integrating the original fourth-order equation for $u$, see [2]). The vector $\nu$ is the outward unit normal to the boundary $\Gamma(t)$ of the wetted region and $V$ is the speed in normal direction of the same boundary. Coordinates $x=(x^{1},x^{2})\in\mathbb{R}^{2}$ are used to describe points on the inclined plane, where $x^{1}$ is the coordinate in the direction of motion. Observe that the location of the origin of the coordinate system is irrelevant as any translation in the direction of $e_{1}$ is absorbed by the Lagrange multiplier $\lambda$ and the problem is invariant under translations in the direction of $e_{2}$. Equation (1.4) corresponds to a kinematic boundary condition relating the normal velocity to the (small) dynamic contact angle through an empirical law described by the function $F$. The dimensionless equations (1.1)–(1.5) for an inclined plane were derived in [2] under the assumption that a lubrication approximation is applicable and the Navier-Stokes equations thus greatly simplify. Actually, the model for a droplet on a horizontal plane with $\chi=0$ and hence $\mu=0$ was introduced some time ago in [6]. For numerical experiments and numerical schemes in this case we refer to [10] and [5], respectively. For this case local and global existence results for generalized weak solutions are to be found in [12, 13]. Moreover, short time existence of classical solutions was proved in [3] while in [7] circles were identified as the only equilibria and shown to be locally asymptotically stable.

The situation for an inclined plane with $\chi>0$ (and hence $\mu>0$) considered herein is somewhat different. Indeed, in [2] it was shown that for an affine111This means that the normal velocity is proportional to the difference between equilibrium and dynamic contact angle for the fluid-substrate system. Such a form was derived in [2] as the linearization for small contact angles of the “Cox-Voinov law”, a simple particular choice of the many available laws for $F$, see [15, 2]. function $F$ there is a critical inclination of the substrate below which a translating circular solution to (1.1)–(1.5) exists moving at a constant speed, while such a solution ceases to exist if the incline is increased any further. More precisely, consider a droplet sliding down the substrate at a constant speed $v_{0}>0$, so that the wetted region is of the form $\Omega(t)=\Omega_{*}+tv_{0}e_{1}$ with normal velocity $V=v_{0}e_{1}\cdot\nu$, where $e_{1}$ is the unit vector in $x^{1}$-direction and $u(t,x)=u_{0}(x-tv_{0}e_{1})$. Then it is readily seen that (1.1)–(1.4) is equivalent to

[TABLE]

for the unknowns $u_{0}$, $\lambda$, and $\Omega_{*}$ with $\Gamma_{*}=\partial\Omega_{*}$. Using polar coordinates, set

[TABLE]

which is easily checked to be the (unique) solution to (1.6)–(1.8) on the disk $\Omega_{*}=R_{0}\mathbb{B}$. Here $\mathbb{B}=\mathbb{B}_{\mathbb{R}^{2}}$ denotes the two-dimensional unit ball centered at the origin. Moreover, (1.9) becomes

[TABLE]

If $\mu=0$, that is, if the incline vanishes, then (1.11) implies that $v_{0}$ must vanish and $F$ must posses a zero at $4\mathcal{V}/\pi R_{0}^{3}$. In particular, no translating solution can exist on a disk if $\mu=0$. If $\mu>0$, then (1.11) implies that translating solutions only exist provided that $F$ is affine, i.e.

[TABLE]

as in [2]. Therefore, if $F$ is affine, then there are a unique radius and a unique velocity, given by

[TABLE]

such that (1.11) holds. While the velocity is of order $\mu$, the radius is independent of the incline. Consequently, if $\mu>0$, then (1.12) is a sufficient, but also necessary condition for a disk to be a translating geometry solution of (1.1)–(1.5), that is, for the existence a (unique) solution $u_{0}$ to problem (1.6)–(1.9) on a disk $\Omega_{*}=R_{0}\mathbb{B}$ which is then given by (1.10) with uniquely determined $R_{0}$ and $v_{0}$ in dependence of $a$ and $b$. However, to guarantee the positivity of $u_{0}$ in $\Omega_{*}$ we need $16\mathcal{V}\geq\mu\pi R_{0}^{5}$ as is seen by taking $\theta=\pi$ in (1.10). Writing $\mu=\mathrm{Bo}\sin\chi$ we derive the physical restriction on the maximal inclination of the substrate as

[TABLE]

Therefore, circular solutions to (1.1)-(1.4) only exist if either $\mu=0$ and $F$ has a zero or if $\mu>0$ is sufficiently small and $F$ is an affine function. It is an interesting question to determine whether non-circular translating solutions exist for general functions $F$. Observe that this cannot be the case for $\mu=0$ by Serrin’s Rigidity Theorem [17]. For $\mu>0$, however, the assumptions of Serrin’s theorem are not met since the right-hand sides of (1.6) and (1.9) are no longer constant. Uniqueness of (non-circular) translating solutions is therefore not clear.

In this paper we establish the local well-posedness of (1.1)–(1.5) for general initial droplet geometries $\partial\Omega_{0}$ and for general laws $F$ for inclines $\chi$ smaller than a positive critical value, that is, we do not impose any structural conditions on $F$ except that

[TABLE]

In contrast to the often used Hanzawa transformation (e.g. see [4, 3, 7] and the references therein) we shall use to this end a description of the unknown curve $\Gamma(t)$ by means of coordinates induced by a smooth flow transversal to the curve $\Gamma_{0}=\partial\Omega_{0}$ (see Section 2). This approach to moving boundary problems was first introduced in [8] in a more general context and the analysis performed in this paper therefore provides a concrete demonstration of the benefits that it offers. In particular, it yields a significant simplification of the analysis required. In short, problem (1.1)–(1.5) is reduced to a single nonlocal, nonlinear evolution equation and yields a relatively simple and insightful formula for its linearization (see Theorem 3.7). The latter, not only, is the basis for using maximal regularity results to obtain local well-posedness (see Theorem 3.8), but also for the characterization of the critical incline beyond which the model ceases to be parabolic in nature and becomes ill-posed. Finally, we also investigate the stability of the sliding circular droplet, the existence of which is ensured by (1.12). We show that, for small inclines and when starting out with an initial droplet geometry sufficiently close to the circle of radius $R_{0}$, the droplet asymptotically becomes circular of radius $R_{0}$ sliding down the plane with constant speed $v_{0}$ (see Theorem 4.5).

2. Reformulation

System (1.1)–(1.5) can be reduced to a nonlocal geometric evolution for the unknown closed curve $\Gamma(t)$ enclosing the simply connected domain $\Omega(t)$. The derivation of a suitable description of this evolution is the purpose of this section.

Working in a classical regularity context, we consider domains $\Omega$ with boundary $\partial\Omega$ of class $buc^{2+\alpha}$ with $\alpha\in(0,1)$, i.e. domains the boundary of which are locally the graph of functions belonging to the little Hölder space $buc^{2+\alpha}$. Recall that, for an open subset $O$ of $\mathbb{R}^{n}$, the space $\operatorname{BUC}^{2+\alpha}(O)$, defined by

[TABLE]

is a Banach space with respect to the norm

[TABLE]

for

[TABLE]

Then $buc^{2+\alpha}(O)$ is the closure of $\operatorname{BUC}^{2+\alpha+\epsilon}(O)$ in $\operatorname{BUC}^{2+\alpha}(O)$ for any $\epsilon>0$. Given any smooth closed curve $\Gamma\in buc^{2+\alpha}$, the space $buc^{2+\alpha}(\Gamma)$ can be defined in the standard way via local charts and a partition of unity.

In the rest of the paper we use the notation $x\cdot y$ and $(x|y)$ interchangeably for the scalar product of vectors $x,y\in\mathbb{R}^{2}$ to enhance the readability of some formulæ.

For a fixed domain $\Omega$ with boundary $\Gamma\in buc^{2+\alpha}$ we first solve the sub-problem (1.1)–(1.3) to obtain a solution \bigl{(}u,\lambda\bigr{)}. We will use the notation $u^{(f)}$ to indicate the solution of $-\Delta u=f$ in $\Omega$ which vanishes on the boundary $\Gamma=\partial\Omega$.

Proposition 2.1**.**

For any domain $\Omega$ bounded by a closed curve $\Gamma\in buc^{2+\alpha}$ and any $\mu\geq 0$, problem (1.1)–(1.3) has a unique solution

[TABLE]

with

[TABLE]

There are $0<\mu_{0}\leq\mu_{1}$ depending on $\Omega$ and ${\mathcal{V}}$ such that $u>0$ in $\Omega$ and $-\partial_{\nu}u>0$ on $\Gamma$ if $\mu\in[0,\mu_{0})$ and $-\partial_{\nu}u<0$ on $\Gamma$ if $\mu\geq\mu_{1}$.

Proof.

Since the right-hand side of (1.1) is smooth, classical theory for elliptic boundary value problems ensures that (1.1)–(1.2) has, for any fixed $\lambda$, a unique solution $u(\lambda)\in buc^{2+\alpha}(\Omega)$ and

[TABLE]

by linearity of the equation. The unknown parameter $\lambda$ is then determined by (1.3) from

[TABLE]

which yields formula (3.7). Next note that we can always fix the origin in such a way that $x^{1}$ is negative throughout the domain $\Omega$. Thus, since $u^{(1)}>0$ in $\Omega$ by the maximum principle, (3.7) implies that there are $\mu_{1}\geq\mu_{0}>0$ such that $\lambda+\mu x^{1}\geq 0$ in $\Omega$ for $\mu\in(0,\mu_{0})$ while $\lambda+\mu x^{1}\leq 0$ in $\Omega$ for $\mu\geq\mu_{1}$. Consequently, if $\mu\in[0,\mu_{0})$, then $u>0$ in $\Omega$ by the maximum principle and $\partial_{\nu}u<0$ on $\Gamma$ by Hopf’s Lemma, while if $\mu\geq\mu_{1}$, then $-\partial_{\nu}u<0$ on $\Gamma$. ∎

With $u$ in hand, equations (1.4)–(1.5) amount to a nonlocal (geometric) evolution for the closed curve $\Gamma(t)$. In order to obtain an evolution equation for it, it is necessary to gain a local understanding of the manifold $\mathcal{M}^{2+\alpha}$ of closed curves in $\mathbb{R}^{2}$ of class $buc^{2+\alpha}$. A convenient local parametrization about a given fixed initial curve $\Gamma_{0}$ is particularly useful. Let $\tau_{0}$, $\nu_{0}$ be the unit tangent and normal vectors for $\Gamma_{0}$, the latter pointing out of the domain enclosed by $\Gamma_{0}$.

Lemma 2.2** (Tubular Neighborhood).**

Given $\Gamma_{0}\in\mathcal{M}^{2+\alpha}$, there is $r_{0}>0$ such that

[TABLE]

is an open neighborhood of $\Gamma_{0}$ diffeomorphic to $\Gamma_{0}\times(-r_{0},r_{0})$. The notation $d(\cdot,\Gamma_{0})$ is used for the signed distance function to $\Gamma_{0}$.

Proof.

Define the map

[TABLE]

and recall that, if $\gamma_{0}=\gamma_{0}(s)$ is an arc length parametrization of $\Gamma_{0}$ about the point $y\in\Gamma_{0}$, then

[TABLE]

where $\kappa_{0}$ denotes the curvature of $\Gamma_{0}$. Hence, computing in the above coordinates,

[TABLE]

The assumption on $\Gamma_{0}$ implies that

[TABLE]

and thus that $d\Phi(y,r)$ is invertible at least as long as $1+r\kappa_{0}(y)>0$, since its columns are orthogonal. This can be ensured by choosing $r_{0}<1/\|\kappa_{0}\|_{\infty}$ and the inverse function theorem yields local invertibility of $\Phi$. Assuming without loss of generality that $r_{0}$ is also so small that

[TABLE]

for each $y\in\Gamma_{0}$, it follows that $x\in T_{r_{0}}(\Gamma_{0})$ has a unique representation by $\Phi$, yielding global injectivity. ∎

Remark 2.3**.**

The map $\Phi$ yields a foliation of $T_{r_{0}}(\Gamma_{0})$ by $buc^{1+\alpha}$ curves since it uses $\nu_{0}\in buc^{1+\alpha}(\Gamma_{0})$. For technical reasons this regularity is not sufficient.

Lemma 2.4** (Generalized Tubular Neighborhood).**

Given $\Gamma_{0}\in\mathcal{M}^{2+\alpha}$, there is $r_{0}>0$ and curves $\Gamma_{r}\in\mathcal{M}^{2+\alpha}$ for $r\in(-r_{0},r_{0})$ such that

[TABLE]

is an open neighborhood of $\Gamma_{0}$ that is diffeomorphic to $\Gamma_{0}\times(-r_{0},r_{0}).$

Proof.

By the preceding lemma, there exists $\tilde{r}_{0}$ such that, for each $x\in T_{\tilde{r}_{0}}(\Gamma_{0})$, there is a unique (y,r)=\bigl{(}y(x),r(x)\bigr{)}\in\Gamma_{0}\times(-\tilde{r}_{0},\tilde{r}_{0}) with $x=y+r\nu_{0}(y)$. In $T_{\tilde{r}_{0}}(\Gamma_{0})$ define the field

[TABLE]

take a smooth cutoff function $\eta:\mathbb{R}\to\mathbb{R}$ with

[TABLE]

and set

[TABLE]

Then $\tilde{\nu}\in buc^{1+\alpha}(\mathbb{R}^{2},\mathbb{R}^{2})$ is a global vector field. Now take a compactly supported smooth mollifier $\psi_{\delta}$ and define

[TABLE]

componentwise. It follows that $\nu^{\delta}\in\operatorname{BUC}^{\infty}(\mathbb{R}^{2},\mathbb{R}^{2})$ and that

[TABLE]

as $\delta\to 0$. In particular it holds that

[TABLE]

where $c(\delta)\to 0$ as $\delta\to 0$. The vector field is therefore uniformly transversal to $\Gamma_{0}$ for $\delta>0$ fixed small enough. Let then $\varphi^{\delta}=\varphi^{\delta}(y,r)$ be the flow generated by the ODE

[TABLE]

It is easily seen that there is $r_{0}>0$ such that $\varphi^{\delta}$ is defined on $\Gamma_{0}\times(-r_{0},r_{0})$ with

[TABLE]

and standard ODE arguments yield that

[TABLE]

is a diffeomorphism of class $buc^{2+\alpha}$. ∎

In the following we use the notation introduced in the proof of Lemma 2.4 for a fixed $\Gamma_{0}\in\mathcal{M}^{2+\alpha}$, that is, having $\delta>0$ chosen small enough we let $\varphi^{\delta}:\Gamma_{0}\times(-r_{0},r_{0})\rightarrow\mathbb{R}^{2}$ denote the flow induced by the vector field $\nu^{\delta}{\in\operatorname{BUC}^{\infty}(\mathbb{R}^{2},\mathbb{R}^{2})}$ transversal to $\Gamma_{0}$. It is convenient to define

[TABLE]

for $(y,r)\in\Gamma_{0}\times(-r_{0},r_{0})$, hence

[TABLE]

Note that $\nu^{\delta}_{0}=\nu^{\delta}$ since $\varphi^{\delta}_{0}=\operatorname{id}$. The previous lemma provides coordinates $(y,r)$ for a neighborhood of $\Gamma_{0}$, which can be denoted by $T^{\nu^{\delta}}_{r_{0}}(\Gamma_{0})$ since it is constructed based on $\nu^{\delta}$. Explicitly this means

[TABLE]

We show next that curves close to $\Gamma_{0}$ can be conveniently parametrized in these coordinates.

Lemma 2.5**.**

Let $\Gamma_{0}$ and $\Gamma\in\mathcal{M}^{2+\alpha}$ be close in the $\operatorname{BUC}^{1}$-topology, that is, let they satisfy

[TABLE]

where $d_{H}$ denotes the Haussdorff distance between compact sets. Then there is a unique function $\rho\in buc^{2+\alpha}(\Gamma_{0})$ such that

[TABLE]

Proof.

We refer to [8, Lemma 2.6] for a complete proof. ∎

3. The Equation for $\Gamma(t)$ and its Linearization

We now focus on equations (1.4)–(1.5) for a given simply connected domain $\Omega^{0}$ of class $buc^{2+\alpha}$ with $\alpha\in(0,1)$, that is, $\Gamma_{0}:=\partial{\Omega^{0}}\in\mathcal{M}^{2+\alpha}$. Using the corresponding coordinates introduced in (2.4) with $\delta>0$ small enough, it is possible to reduce the evolution for $\Gamma(t)$ to one for the function $\rho(t,\cdot):\Gamma_{0}\to\mathbb{R}$ given in Lemma 2.5 through

[TABLE]

Denote the unit tangent and normal vectors to $\Gamma_{\rho(t)}$ at the point $\varphi^{\delta}_{\rho(t,y)}(y)$ by $\tau_{\rho(t)}(y)$ and $\nu_{\rho(t)}(y)$, respectively. Then (2.3) implies

[TABLE]

with superscript dot indicating a derivative with respect to time. Since the normal velocity $V_{\Gamma_{\rho(t)}}$ of $\Gamma_{\rho(t)}$ at a point $\varphi^{\delta}_{\rho(t,y)}(y)$ is given by the component of the projection of the tangent vector to the curve $t\mapsto\varphi^{\delta}_{\rho(t,y)}(y)$ onto the unit normal vector $\nu_{\rho(t)}(y)$ at that point, it holds that

[TABLE]

To keep notation simple we often omit the time variable. Therefore, (1.4)–(1.5) is equivalent to the evolution equation for $\rho$,

[TABLE]

where $(u_{\rho},\lambda_{\rho})$ denotes the solution of (1.1)–(1.3) from Proposition 2.1 in $\Omega_{\rho}$, the domain bounded by $\Gamma_{\rho}$, and a given fixed $\mu\geq 0$ (note that $\rho\equiv 0$ gives the initial domain, that is, $\Omega_{0}=\Omega^{0}$). This is a nonlinear, nonlocal equation for $\rho:[0,\infty)\times\Gamma_{0}\to\mathbb{R}$. Notice that, for $\delta<<1$ and $|\rho|<<1$, the factor of $\dot{\rho}$ in the expression for $V_{\Gamma_{\rho}}$ satisfies

[TABLE]

In order to obtain local well-posedness for (3.1) using maximal regularity techniques, its linearization at the initial datum $\rho_{0}\equiv 0$ is computed. For this we first note

Proposition 3.1**.**

There exists an open zero-neighborhood $\mathcal{O}$ in $buc^{2+\alpha}(\Gamma_{0})$ such that

[TABLE]

Proof.

This follows from (1.14) and the fact that the flow $\varphi_{r}^{\delta}=\varphi^{\delta}(\cdot,r)$ is smooth with respect $r$ which implies that the solution $(u_{\rho},\lambda_{\rho})$ of (1.1)–(1.3) depends smoothly on $\rho$ since $\Gamma_{\rho}=\varphi^{\delta}_{\rho}(\Gamma_{0})$. We refer to [8, Theorem 3.6] for a more detailed and explicit calculation of this derivative which yields the desired smoothness. ∎

We next verify that

[TABLE]

is the generator of an analytic semigroup. Observe that

[TABLE]

for $h\in buc^{2+\alpha}(\Gamma_{0})$ since $\nu^{\delta}_{0}=\nu^{\delta}$. It then remains to compute the derivatives in the curly brackets. This is where the choice of coordinate system from Lemma 2.4 delivers its benefits yielding a particularly insightful representation.

Lemma 3.2**.**

It holds that

[TABLE]

for $h\in buc^{2+\alpha}(\Gamma_{0})$, where ′ denotes differentiation with respect to arc length.

Proof.

It follows from $\tau_{\rho}\cdot\nu_{\rho}\equiv 0$ and $\nu_{\rho}\cdot\nu_{\rho}\equiv 1$ that

[TABLE]

This implies that

[TABLE]

It is therefore enough to compute $\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}\tau_{\epsilon h}$. Now, since \gamma_{\rho}(s):=\varphi^{\delta}\Bigl{(}\gamma_{0}(s),\rho\bigl{(}\gamma_{0}(s)\bigr{)}\Bigr{)} is a parametrization of $\Gamma_{\rho}$ whenever $\gamma_{0}$ is an arc length parametrization of $\Gamma_{0}$, one has from (2.3) that

[TABLE]

is a tangent vector to $\Gamma_{\epsilon h}$ and thus that $\tau_{\epsilon h}={\tilde{\tau}_{\epsilon h}}/{|\tilde{\tau}_{\epsilon h}|}$. The latter yields

[TABLE]

Then, using (2.3) again,

[TABLE]

Combining (3.3)–(3.5) and noticing $\tilde{\tau}_{0}=d_{y}\varphi^{\delta}(\cdot,0)[\tau_{0}]=\tau_{0}$ since $\varphi^{\delta}_{0}=\operatorname{id}_{\Gamma_{0}}$, we get

[TABLE]

and the claim follows from $\nu^{\delta}_{0}=\nu^{\delta}$. ∎

Lemma 3.3**.**

It holds that

[TABLE]

for $h\in buc^{2+\alpha}(\Gamma_{0})$.

Proof.

Owing to the previous lemma it only remains to compute

[TABLE]

and to note that

[TABLE]

∎

Remark 3.4**.**

Notice that, when $\delta\simeq 0$ one has that

[TABLE]

uniformly on $\Gamma_{0}$. If $\delta$ can be set equal to zero, then

[TABLE]

Lemma 3.5**.**

Given $h\in buc^{2+\alpha}(\Gamma_{0})$ and $\epsilon>0$ small, let $u_{\epsilon h}$ solve the boundary value problem

[TABLE]

for $f\in\operatorname{C}^{\infty}(\mathbb{R}^{n})$. Then

[TABLE]

where the Dirichlet-to-Neumann operator $DtN_{\Gamma}$ is the operator that yields the normal derivative $\partial_{\nu_{\Gamma}}u_{g}$ (Neumann datum) of the harmonic function $u_{g}$ with Dirichlet datum $g$, that is

[TABLE]

and where $u_{0}$ is the solution corresponding to the boundary value problem in $\Omega_{0}$.

Proof.

Assume first that $h\leq 0$, hence $\Omega_{\epsilon h}\subset\Omega_{0}$. We look for $u_{\epsilon h}$ in the form

[TABLE]

Then $v_{\epsilon h}$ satisfies

[TABLE]

and

[TABLE]

It is known that the mapping

[TABLE]

is a smooth local section of the corresponding bundle. Indicating with a superscript $*$ the pull-back operation it follows that (see [8, Section 3] for more details)

[TABLE]

since the first term after the first equality sign vanishes in view of the homogeneous Dirichlet condition satisfied by $u_{0}$ and the third in view of Lemma 3.2 and of the boundary condition again. It remains to show that the result remains valid for any $h$ without the restriction that $h\leq 0$. To that end, define $\Gamma_{r_{0}}=\varphi^{\delta}(\Gamma_{0},r_{0})$ for $r_{0}>0$ small enough and replace the solution $u_{0}$ by the solution $u_{r_{0}}$ in the above argument. At the end of the calculation, formula (3) is obtained with all terms after the first equality sign non-vanishing. Letting $r_{0}$ tend to zero makes them vanish and delivers the claim. For more details we refer to the proof of [8, Theorem 3.7]. ∎

Lemma 3.6**.**

Given $h\in buc^{2+\alpha}(\Gamma_{0})$ it holds that

[TABLE]

where

[TABLE]

Proof.

Recall from Proposition 2.1 that

[TABLE]

with $\lambda_{\epsilon h}$ given as in the statement. Lemma 3.5 implies that

[TABLE]

It follows that

[TABLE]

∎

Combining the results of (3.2), Lemma 3.3, and Lemma 3.6 the linearization of $G$ at zero is seen to be given by the expression

[TABLE]

for $h\in buc^{2+\alpha}(\Gamma_{0})$, where $(u_{0},\lambda_{0})$ is the solution to (1.1)–(1.3) in $\Omega_{0}$ from Proposition 2.1. From this formula we derive the following generation result.

Theorem 3.7**.**

Suppose (1.14) and let $\Gamma_{0}\in\mathcal{M}^{2+\alpha}$. Then

[TABLE]

for $\mu\in[0,\mu_{0})$, where $\mu_{0}>0$ is small enough depending on $\Omega_{0}$ and ${\mathcal{V}}$. In other words, $DG(0)$ generates an analytic $\operatorname{C}^{0}$-semigroup on $buc^{1+\alpha}(\Gamma_{0})$ for $\mu<\mu_{0}$ and its domain of definition coincides with $buc^{2+\alpha}(\Gamma_{0})$. There is $\mu_{1}\geq\mu_{0}$ such that

[TABLE]

for $\mu\geq\mu_{1}$, which makes the evolution equation linearly ill-posed.

Proof.

First observe that for $\partial_{\nu^{\delta}}u_{0}$ we have

[TABLE]

by definition of $u_{0}\in buc^{2+\alpha}(\Omega_{0})$ and

[TABLE]

for $f(x):=\mu x^{1}+\lambda$, where we used for the second equality that $\partial_{\tau_{0}}u_{0}=0$ owing to the Dirichlet boundary condition and for the third equality the fact that $\partial_{\nu_{0}}\nu_{0}=0$ along with

[TABLE]

Consequently, classical theory of boundary value problems implies that

[TABLE]

From this it follows that

[TABLE]

Next notice that the map that associates with a curve $\Gamma$ the corresponding $\lambda$ from Proposition 2.1 is well-defined in a neighborhood of $\Gamma_{0}$ in $\mathcal{M}^{2+\alpha}$. Consequently, its tangential map

[TABLE]

is a linear, rank 1, and, hence, compact operator. Now term $I$ is the most important one and defines an elliptic pseudodifferential operator of order 1 whenever

[TABLE]

The second condition is satisfied by assumption (1.14). The first holds true if $-\partial_{\nu_{0}}u_{0}>0$ on $\Gamma_{0}$, which follows from Proposition 2.1. Indeed, if $\delta$ is small enough, then $\nu^{\delta}\cdot\nu_{0}\simeq 1$ (which also guarantees the validity of the third condition), and therefore we have that

[TABLE]

thanks to the uniform convergence in $buc^{1+\alpha}$ in (2.2). This implies that $I$ is in fact the generator of an analytic $\operatorname{C}^{0}$-semigroup on $buc^{1+\alpha}(\Gamma_{0})$ with domain $buc^{2+\alpha}(\Gamma_{0})$. A complete argument would require a standard localization argument based on the smoothness of the coefficients and a symbol analysis of the corresponding frozen coefficients operator (see e.g. [4, 3] for more details). In this case, the latter has the explicit form $a_{0}|\xi|$ because of the particularly insightful form of the main term of $DG(0)$, which, it is reminded, is itself a consequence of the use of coordinates constructed by means of the flow $\varphi^{\delta}$. As for the remaining terms, they are lower order perturbations (as multiplication operators), like $II-IV$ thanks to regularity of the coefficients (using also (3)), or a small perturbation, as for $V$, thanks to

[TABLE]

Recall for the latter that the set of analytic generators is open in the natural operator topology of \mathcal{L}\bigl{(}buc^{2+\alpha}(\Gamma_{0}),buc^{1+\alpha}(\Gamma_{0})\bigr{)}. The first assertion is therefore proved. As for the second, we note that Proposition 2.1 yields $\mu_{1}\geq\mu_{0}$ such $-\partial_{\nu^{\delta}}u_{0}<0$ on $\Gamma_{0}$ for $\mu\geq\mu_{1}$, and the claim follows by approximation as above. ∎

Existence results based on maximal regularity can now be used to derive the following

Theorem 3.8**.**

Given any $\Gamma_{0}\in buc^{2+\alpha}$, there is a $\mu_{0}>0$ (depending on $\Gamma_{0}$ and ${\mathcal{V}}$) such that, for all $\mu\in[0,\mu_{0})$, system (1.1)–(1.5) is well-posed on some maximal interval $[0,T_{0})$. The solution $(u,\Gamma)$ satisfies

[TABLE]

with

[TABLE]

and

[TABLE]

for $\Omega(t)=\Omega_{\rho(t,\cdot)}$. There is $\mu_{1}\geq\mu_{0}$ such that the system (1.1)–(1.5) is linearly ill-posed for $\mu\geq\mu_{1}$.

Proof.

Since \mathcal{H}\bigl{(}buc^{2+\alpha}(\Gamma_{0}),buc^{1+\alpha}(\Gamma_{0})\bigr{)} is open in \mathcal{L}\bigl{(}buc^{2+\alpha}(\Gamma_{0}),buc^{1+\alpha}(\Gamma_{0})\bigr{)}, it follows from Theorem 3.7 and Proposition 3.1 that we may assume with loss of generality that

[TABLE]

and the existence assertion follows e.g. from [14, Theorem 8.4.1] (or from [8, Thorem 5.6]) and the fact that the little Hölder spaces are stable under continuous interpolation. ∎

Remark 3.9**.**

The above result confirms and quantifies the physical intuition that a solution exists only for small incline angle and ceases to exist for larger angles. Obviously, the determining critical size of the angle depends on $\Gamma_{0}$ and ${\mathcal{V}}$, i.e. the shape of the initial wetted region and the mass of liquid since these determine the sign of $\partial_{\nu_{0}}u_{0}$.

4. Stability Analysis for Translating Circular Solutions

We assume throughout the following that (1.12) is satisfied, that is,

[TABLE]

for some $a,b>0$. Recall that if $R_{0}$ and $v_{0}$ are as in (1.13), then $u_{0}$ defined in (1.10) solves (1.6)-(1.9) on the disk $\Omega_{*}=R_{0}\mathbb{B}$ (and is positive provided $\mu$ is small). We now investigate the asymptotic stability of this translating circular solution for small inclines, i.e. for small $\mu\geq 0$.

4.1. Reformulation

We rewrite problem (1.1)-(1.5) by introducing the translations

[TABLE]

Substituting this into (1.1)–(1.5) and dropping again the tildes everywhere for ease of notation, it is readily seen that (1.1)–(1.5) is equivalent to

[TABLE]

and $(u_{0},\Omega_{*})$ is a stationary solution to (4.1)-(4.5). As in the previous section, we can consider this problem as a single equation for the geometry. Let $\nu_{0}(y)=y/R_{0}$ denote the normal at $y\in\Gamma_{0}:=R_{0}\mathbb{S}^{1}$. Note that in this case, since $\nu_{0}$ is smooth, we can take $\nu^{\delta}=\nu_{0}$ in Section 2, and the flow in (2.3) is simply given by $\varphi^{\delta}(y,r)=(1+r/R_{0})y$. Thus, the evolution of $\Gamma(t)$ is described by the evolution of the function $\rho(t,\cdot):\Gamma_{0}\to\mathbb{R}$ through

[TABLE]

and $\rho$ is governed by

[TABLE]

provided that $\partial\Omega^{0}=\Gamma_{\rho_{0}}$, which is possible for $\Omega^{0}$ sufficiently close to $\Omega_{*}$ (i.e. for $\rho_{0}$ is small enough). Here, $\nu_{\rho}=\nu_{\Gamma_{\rho}}$, and $u_{\rho}$ with corresponding $\lambda_{\rho}$ denotes the solution of (1.1)–(1.3) from Proposition 2.1 in $\Omega_{\rho}$, the domain enclosed by $\Gamma_{\rho}$. Clearly, $\Omega_{0}=\Omega_{*}=R_{0}\mathbb{B}$. Note that

[TABLE]

according to Proposition 3.1 with $\mathcal{O}$ being an open zero-neighborhood in $buc^{2+\alpha}(\Gamma_{0})$, and $\rho=0$ gives rise to the stationary solution $(u_{0},\Omega_{0})$. We shall prove that $\rho=0$ is locally asymptotically stable for equation (4.6) by using the principle of linearized stability.

4.2. Linearization

We now express the Fréchet derivative $DH(0)$ in terms of Fourier expansions. For this we use polar coordinates $(r,\theta)$ and Cartesian coordinates $x$ interchangeably and observe that, if $h\in buc^{2+\alpha}(\Gamma_{0})$, then the form of $F$ implies that

[TABLE]

Therefore, Lemma 3.2 and Lemma 3.6 entail that

[TABLE]

with $\tau_{0}$ denoting the unit tangent vector to $\Gamma_{0}$. We compute the different terms separately. Let $h\in buc^{2+\alpha}(\Gamma_{0})$ be given with representation

[TABLE]

Then

[TABLE]

solves

[TABLE]

and thus

[TABLE]

Next, since (1.10) implies

[TABLE]

it follows that

[TABLE]

In summary, we derive that

[TABLE]

Similarly, we have from (1.10)

[TABLE]

hence

[TABLE]

Before computing the third term of the right-hand side of (4.8), we focus on the last term. Clearly, \big{(}d_{y}\nu_{0}[\tau_{0}]\big{|}\nu_{0}\big{)}=0 so that

[TABLE]

where the derivative with respect to arc length is $h^{\prime}=\frac{1}{R_{0}}\frac{d}{d\theta}h$. Consequently, since $v_{0}=\mu aR_{0}^{2}/4$ by (1.13),

[TABLE]

As for the third term on the right-hand side of (4.8) recall, formula (3.7) for $\lambda_{\epsilon h}$. Then

[TABLE]

and thus, on using

[TABLE]

according to (1.10), we deduce

[TABLE]

We then invoke the transport theorem to get

[TABLE]

since $u_{0}^{(f)}$ vanishes on $\Gamma_{0}$. Consequently,

[TABLE]

It remains to compute the derivatives for which we proceed as in [7]. Given a smooth function $f$ on $\bar{\Omega}_{\epsilon h}$, let

[TABLE]

be a representation of the solution to the Dirichlet problem

[TABLE]

with Green’s function ${K}_{\epsilon h}$ on $\Omega_{\epsilon h}$. In particular, for the circle $\Omega_{0}=R_{0}\mathbb{B}$ we have

[TABLE]

Then, noticing that ${K}_{\epsilon h}$ vanishes on $\Omega_{\epsilon h}\times\Gamma_{\epsilon h}$, we obtain

[TABLE]

Next observe [9, 16] that

[TABLE]

and that

[TABLE]

is the unique harmonic function in $\Omega_{0}$ with boundary value $g$ on $\Gamma_{0}$. Therefore,

[TABLE]

since ${K}_{0}(\bar{r},\bar{\theta},\bar{r},\phi)$ is symmetric with respect to the angular (and radial) variables. Using this, we obtain

[TABLE]

Integrating this yields similarly

[TABLE]

and finally

[TABLE]

Exactly the same way one computes

[TABLE]

It now follows from (4.12)-(4.14) that

[TABLE]

Consequently, gathering (4.8)-(4.11) and (4.15) and using $\partial_{\nu_{0}}u^{(1)}_{0}=-R_{0}/2$, we obtain the Fourier expansion of $DH(0)$ in the form

[TABLE]

for $h\in buc^{2+\alpha}(\Gamma_{0})$ with $h(\theta)=\sum_{n\in\mathbb{Z}}\hat{h}_{n}e^{in\theta}$, where $\mathbb{Z}^{*}:=\mathbb{Z}\setminus\{0\}$. Note that the matrix is tridiagonal if $\mu>0$. Since $DH(0)$ has a compact resolvent, its spectrum is discrete and contains only eigenvalues. More information is found in the following lemma.

Lemma 4.1**.**

Let $\mu\geq 0$ be sufficiently small. The kernel of $DH(0)$ is two-dimensional and spanned by $\{e^{-i\theta},e^{i\theta}\}$. Moreover, there is $\omega>0$ independent of $\mu$ such that $\sigma(DH(0))\subset[\mathrm{Re}z\leq-\omega]\cup\{0\}$.

Proof.

It readily follows from (4.16) using an induction argument that $h\in buc^{2+\alpha}(\Gamma_{0})$ with $h=\sum_{n\in\mathbb{Z}}\hat{h}_{n}e^{in\theta}$ belongs to the kernel of $DH(0)$ if and only if $\hat{h}_{n}=0$ for $n\in\mathbb{Z}\setminus\{\pm 1\}$. Hence the kernel of $DH(0)$ is spanned by $\{e^{-i\theta},e^{i\theta}\}$. Next,

[TABLE]

with -A\in\mathcal{H}\big{(}buc^{2+\alpha}(\Gamma_{0}),buc^{1+\alpha}(\Gamma_{0})\big{)} (see Theorem 3.7 with $\mu=0$ therein) given by

[TABLE]

and B\in\mathcal{L}\big{(}buc^{2+\alpha}(\Gamma_{0}),buc^{1+\alpha}(\Gamma_{0})\big{)} given by

[TABLE]

Set $\omega:=\frac{\mathcal{V}a}{\pi R_{0}^{4}}$. Since $\sigma(A)=-4\omega\mathbb{N}$, [1, I.Corollary 1.4.3] implies that there are $\mu_{0}>0$ and $\vartheta\in(0,\pi/2)$ such that $\sigma(A+\mu B)\subset\omega+\Sigma_{\vartheta}$ for $\mu\in[0,\mu_{0})$, where $\Sigma_{\vartheta}:=[\arg\left(z\right)\in(\pi-\vartheta,\pi+\vartheta)]$. Since zero is the only eigenvalue of $A$ in $\Sigma_{\vartheta}\cap[\mathrm{Re}\,z>-\omega]$ and since $A+\mu B\rightarrow A$ in the generalized sense of [11, IV.Theorem 2.24] as $\mu\rightarrow 0$, it follows from [11, IV.Theorem 3.16] that zero is the only eigenvalue of $DH(0)=A+\mu B$ in $\Sigma_{\vartheta}\cap[\mathrm{Re}\,z>-\omega]$ if $\mu\geq 0$ is sufficiently small. This proves the assertion. ∎

4.3. Stability Analysis

To analyze now the stability of the zero solution to (4.6) we need to split off the zero eigenvalue of the linearization $DH(0)$. For this it is useful to use a slightly different description of the curves $\Gamma_{\rho}$ as provided by the next lemma.

Lemma 4.2**.**

For each $\rho\in\mathcal{O}$ there is a unique (z,\bar{\rho})\in\mathbb{R}^{2}\times\big{(}\mathrm{ker}(DH(0))\big{)}^{\perp} such that $\Gamma_{\rho}=z+\Gamma_{\bar{\rho}}$. Moreover, $H(\rho)=H(\bar{\rho})$.

Proof.

Since the kernel of $DH(0)$ is spanned by $\{e^{-i\theta},e^{i\theta}\}$ according to Lemma 4.1, the existence of a unique $(z,\bar{\rho})\in\mathbb{R}^{2}\times\left(\mathrm{ker}(DH(0))\right)^{\perp}$ with $\Gamma_{\rho}=z+\Gamma_{\bar{\rho}}$ is shown in [7, Lemma 5.2]. That $H(\rho)=H(\bar{\rho})$ follows as in [7, Lemma 5.1] from the translation invariance of the problem. Indeed, if $\Gamma=\partial\Omega$ and $u_{\Gamma}$ solves

[TABLE]

then $u_{\Gamma_{\bar{\rho}}}=u_{\Gamma_{\rho}}(\cdot+z)$ on $\Omega_{\bar{\rho}}=-z+\Omega_{\rho}$ and $\nu_{\Gamma_{\bar{\rho}}}=\nu_{\Gamma_{\rho}}(\cdot+z)$ on $\Gamma_{\bar{\rho}}=-z+\Gamma_{\rho}$. The definition of $H$ implies the assertion. ∎

We next derive the evolution in the new coordinates $(z,\bar{\rho})=(z(\rho),\bar{\rho}(\rho))$. Let $\nu_{0}=(\cos,\sin)$ and $\tau_{0}=(-\sin,\cos)$ denote the unit normal and unit tangent vector to $\Gamma_{0}$, respectively, and let

[TABLE]

be an arbitrary point on $\Gamma_{\rho(t,\cdot)}=z+\Gamma_{\bar{\rho}(t,\cdot)}$. We often suppress the time variable $t$ in the following. Differentiating with respect to $\phi$ implies that

[TABLE]

Now, since V_{\Gamma_{\rho}}(p)=\big{(}\dot{z}+\dot{\bar{\rho}}(\phi)\nu_{0}(\phi)\big{)}\cdot\nu_{\rho}(p) with dots indicating time derivatives, it follows from the definition of $H$ and $H(\bar{\rho})=H(\rho)$ that

[TABLE]

Therefore, from (4.19) we derive that

[TABLE]

Let $\Pi^{(\pm 1)}\in\mathcal{L}(buc^{1+\alpha}(\Gamma_{0}))$ denote the projection onto the subspace spanned by $e^{\pm i\theta}$, that is,

[TABLE]

and let $\Pi^{\perp}\in\mathcal{L}(buc^{1+\alpha}(\Gamma_{0}))$ denote the projection onto \big{(}\mathrm{ker}(DH(0))\big{)}^{\perp}. We then apply these projections to (4.20) to derive the evolution for $(z,\bar{\rho})$. For a more compact notation we introduce

[TABLE]

and observe that

[TABLE]

is invertible if $\bar{\rho}\in\mathcal{O}$ is small. Further set

[TABLE]

and

[TABLE]

Then we obtain from (4.20):

Proposition 4.3**.**

The evolution for $\rho$ governed by (4.6) is equivalent to the system

[TABLE]

for $(z,\bar{\rho})$.

Next, we introduce

[TABLE]

Then $\mathcal{O}_{\perp}:=\mathcal{O}\cap buc_{\perp}^{2+\alpha}(\Gamma_{0})$ is an open zero-neighborhood in $buc_{\perp}^{2+\alpha}(\Gamma_{0})$ and (4.7) entails that

[TABLE]

Moreover, for the linearization of \Pi^{\perp}H\in C^{2}\big{(}\mathcal{O}_{\perp},buc_{\perp}^{1+\alpha}(\Gamma_{0})\big{)} at zero we have:

Proposition 4.4**.**

Let $\mu\geq 0$ be sufficiently small. Then

[TABLE]

and there is $\omega_{0}>0$ such that

[TABLE]

Proof.

Observe that, owing to (4.17), (4.18),

[TABLE]

where

[TABLE]

and

[TABLE]

for $h\in buc_{\perp}^{2+\alpha}(\Gamma_{0})$ with $h(\theta)=\sum\limits_{n\in\mathbb{Z},\,n\not=\pm 1}\hat{h}_{n}e^{in\theta}.$ Since -A\in\mathcal{H}\big{(}buc^{2+\alpha}(\Gamma_{0}),buc^{1+\alpha}(\Gamma_{0})\big{)} by Theorem 3.7 (with $\mu=0$), one readily deduces from the obvious fact

[TABLE]

that -A_{\perp}\in\mathcal{H}\big{(}buc_{\perp}^{2+\alpha}(\Gamma_{0}),buc_{\perp}^{1+\alpha}(\Gamma_{0})\big{)}. Observing then that

[TABLE]

the assertion follows from the perturbation result [1, I.Proposition 1.4.2]. ∎

Due to Proposition 4.4 and (4.23) we are in a position to apply the principle of linearized stability from [14, Theorem 9.1.2] to the equation (4.22) for $\bar{\rho}$. Thus, there are $r>0$ and $M>0$ such that for any initial value $\bar{\rho}_{0}\in buc_{\perp}^{2+\alpha}(\Gamma_{0})$ with $\|\bar{\rho}_{0}\|_{buc_{\perp}^{2+\alpha}}<r$, the solution $\bar{\rho}$ to (4.22) with $\bar{\rho}(0)=\bar{\rho}_{0}$ exists globally in time and

[TABLE]

Plugging this into (4.21) and observing that the right-hand side of (4.21) is of order O\bigl{(}\|\bar{\rho}\|_{buc_{\perp}^{2+\alpha}}\bigr{)} owing to (4.7), it readily follows that there is $z_{\infty}\in\mathbb{R}^{2}$ such that

[TABLE]

Therefore, using again the original coordinates instead of $(z,\bar{\rho})$ and noticing that $\|\bar{\rho}(\rho)\|_{buc_{\perp}^{2+\alpha}}$ is small if and only if $\|\rho\|_{buc^{2+\alpha}}$ is, we arrive at:

Theorem 4.5**.**

Assume that (1.12) holds. Then, for small incline $\mu>0$, the stationary solution $(u_{0},\Omega_{*})$ to (4.1)-(4.5) from (1.10) is asymptotically exponentially stable. More precisely, if $\mu>0$ is small and given an initial geometry

[TABLE]

with $\|\rho_{0}\|_{buc^{2+\alpha}(\Gamma_{0})}$ sufficiently small, there exists

[TABLE]

and $u_{\rho}$ with

[TABLE]

for $t\geq 0$ such that $(\rho,u_{\rho})$ satisfies (4.6). Moreover,

[TABLE]

with

[TABLE]

satisfying (4.24) and (4.25).

The above theorem shows that, for small inclines and when starting out with a droplet geometry sufficiently close to the disk of radius $R_{0}$, the droplet asymptotically becomes circular of radius $R_{0}$ sliding down the plane with constant speed $v_{0}$.