MSC 2000. 76D08, 76D45, 76M55.
Keywords. viscous flow, thin film approximation, surface tension gradient.

Many viscous flow problems of practical importance involve small scale flows with free surface whose effects contribute significantly to the dynamics through superficial forces. One prototype problem that has received much attention is that of the draining of viscous layers down an inclined plane driven simultaneously by a surface tension gradient.

We consider the steady behavior of such a trickle of viscous liquid (which we take to be supplied at a prescribed volume flux) when the surface tension gradient is constant. We are particularly interested in the study of Marangoni effect. More specific, we take into account a non-zero tangential (shear) stress and use a lubrication (thin-film) approximation. This approximation linearizes the Navier-Stokes system and enables us to obtain the velocity field, the freesurface velocity, the pressure and the free-surface profile in closed form.

Consider the flow of a uniform thin liquid layer (rivulet) down an inclined solid plane driven simultaneously by a surface tension gradient. This gradient acts on the free surface (the upper of the rivulet) and due to viscous forces can drag the fluid up or down on the incline. Suppose a Newtonian fluid, of constant density $\rho$$\rho$rho\rho$\rho$ and viscosity $\mu$$\mu$mu\mu$\mu$, which undergoes a steady rectilinear flow in the form of a filament, down a plane inclined at an angle $\alpha$$\alpha$alpha\alpha$\alpha$ to the horizontal.

With respect to the Cartesian coordinates system Oxyz as indicated in Fig.1, the velocity of this locally unidirectional flow will be of the form $\overrightarrow{u}=u(y,z)\overrightarrow{i}$$\overrightarrow{u}=u(y,z)\overrightarrow{i}$vec(u)=u(y,z) vec(i)\vec{u}= u(y, z) \vec{i}$\overrightarrow{u}=u(y,z)\overrightarrow{i}$ and the Navier-Stokes equations reduce to

where the subscripts denote partial differentiation.

In the thin-film (lubrication) theory, (see, for example, Acheson [1], Ockendon and Ockendon [11], or Gheorghiu [9]) these equations reduce to

and are to be integrated subject to the boundary conditions

Here $z=H(y)$$z=H(y)$z=H(y)z=H(y)$z=H(y)$ is the known equation of the transverse profile of the substrate (the bottom line) and $z=h(y)$$z=h(y)$z=h(y)z=h(y)$z=h(y)$ is the unknown equation of the transverse profile of the free surface. When the substrate is an inclined plane $H(y)$$H(y)$H(y)H(y)$H(y)$ becomes identically zero.

We impose the following contact conditions:

Here again $z=h(y)$$z=h(y)$z=h(y)z=h(y)$z=h(y)$ is the free-surface profile, $p$$p$pp$p$ is the pressure in the liquid, $p_a$$p_a$p_(a)p_{a}$p_a$ is the atmospheric pressure, $\overrightarrow{g}$$\overrightarrow{g}$vec(g)\vec{g}$\overrightarrow{g}$ is the gravitational acceleration, $\overline{\gamma }$$\overline{\gamma }$bar(gamma)\bar{\gamma}$\overline{\gamma }$ is the reference value of surface tension, $\beta$$\beta$beta\beta$\beta$ is the contact angle at the three-phase contact line, $2a$$2a$2a2 a$2a$ is the width of the layer (trickle), $h_m$$h_m$h_(m)h_{m}$h_m$ is the maximum depth of the liquid, and $\tau$$\tau$tau\tau$\tau$ is the constant shear stress acting on $z=h(y)$$z=h(y)$z=h(y)z=h(y)$z=h(y)$.

We consider $\beta$$\beta$beta\beta$\beta$ to be a prescribed constant such that $\beta <\pi /2$$\beta <\pi /2$beta < pi//2\beta<\pi / 2$\beta <\pi /2$. A constant value of $\beta$$\beta$beta\beta$\beta$ means that any contact angle "hysteresis" is ignored. We appreciate this hypothesis as reasonable for these rivulets. On the physical nature of the shear stress $\tau$$\tau$tau\tau$\tau$ we have to make the following important remark. The origin of $\tau$$\tau$tau\tau$\tau$ could be very diverse. We are mainly interested when this stress comes from a local variation of surface tension, such that $\tau =\frac{\partial \gamma }{\partial x}$$\tau =\frac{\partial \gamma }{\partial x}$tau=(del gamma)/(del x)\tau=\frac{\partial \gamma}{\partial x}$\tau =\frac{\partial \gamma }{\partial x}$. Actually, $\gamma$$\gamma$gamma\gamma$\gamma$ may vary somewhat along $x$$x$xx$x$-axis and then the flow will not be truly unidirectional. However, the above approach may still be approximately correct if $\gamma$$\gamma$gamma\gamma$\gamma$ varies only slowly. A more sophisticated model that takes into account reactions, fluxes of surfactants on the free surface, surface diffusion and convection, change of metrics, etc., is available for example in [14]. To consider such models is beyond our computational capabilities. Consequently, we simply solve the differential system (2)-(5), using the strategy from Wilson and Duffy [15] and finally plug in $\frac{\partial \gamma }{\partial x}$$\frac{\partial \gamma }{\partial x}$(del gamma)/(del x)\frac{\partial \gamma}{\partial x}$\frac{\partial \gamma }{\partial x}$ for the shear stress $\tau$$\tau$tau\tau$\tau$.

About the boundary condition (4) we observe that we have taken the surface curbature to be $h^{\prime \prime }$$h^{\prime \prime }$h^('')h^{\prime \prime}$h^{\prime \prime }$. The boundary condition (4) makes the difference between our study and those of Allen and Biggin [2], Duffy and Moffatt [7], Towel and Rothfeld[13], or Wilson and Duffy [15]. It means that the Marangoni effect is taken into account. Some mathematical aspects (existence, uniqueness, etc.) of this effect are addressed in Amick [3], Amick and Fraenkel [4] and Wagner [14].

With respect to the angle to the horizontal, we consider the following three cases:
i). $0<\alpha <\frac{\pi }{2}$$0<\alpha <\frac{\pi }{2}$0 < alpha < (pi)/(2)0<\alpha<\frac{\pi}{2}$0<\alpha <\frac{\pi }{2}$,
ii) $\alpha =\frac{\pi }{2}$$\alpha =\frac{\pi }{2}$alpha=(pi)/(2)\alpha=\frac{\pi}{2}$\alpha =\frac{\pi }{2}$,
iii) $\frac{\pi }{2}<\alpha <\pi$$\frac{\pi }{2}<\alpha <\pi$(pi)/(2) < alpha < pi\frac{\pi}{2}<\alpha<\pi$\frac{\pi }{2}<\alpha <\pi$.

Then the solution read as follows: the velocity

the free-surface velocity, $u_s:=u(y,h)$$u_s:=u(y,h)$u_(s):=u(y,h)u_{s}:=u(y, h)$u_s:=u(y,h)$,

and the pressure is
(8) $p(z)=p_a-\rho \cdot g\cdot z\cdot \cos \alpha +\tan \beta \sqrt{\rho \cdot g\cdot \overline{\gamma }\cdot |\cos \alpha |}\cdot \{\begin{array}{ll}\coth B,& \alpha \in (0,\frac{\pi }{2})\\ B^{-1},& \alpha =\frac{\pi }{2}\\ \cot B,& \alpha \in (\frac{\pi }{2},\pi ).\end{array}$$p(z)=p_a-\rho \cdot g\cdot z\cdot \cos \alpha +\tan \beta \sqrt{\rho \cdot g\cdot \overline{\gamma }\cdot |\cos \alpha |}\cdot \left\{\begin{array}{l}\coth B,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \in \left(0,\frac{\pi }{2}\right)\\ B^{-1},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha =\frac{\pi }{2}\\ \cot B,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \in \left(\frac{\pi }{2},\pi \right).\end{array}\right.$p(z)=p_(a)-rho*g*z*cos alpha+tan betasqrt(rho*g* bar(gamma)*|cos alpha|)*{[coth B",",alpha in(0,(pi)/(2))],[B^(-1)",",alpha=(pi)/(2)],[cot B",",alpha in((pi)/(2),pi).]:}p(z)=p_{a}-\rho \cdot g \cdot z \cdot \cos \alpha+\tan \beta \sqrt{\rho \cdot g \cdot \bar{\gamma} \cdot|\cos \alpha|} \cdot \begin{cases}\operatorname{coth} B, & \alpha \in\left(0, \frac{\pi}{2}\right) \\ B^{-1}, & \alpha=\frac{\pi}{2} \\ \cot B, & \alpha \in\left(\frac{\pi}{2}, \pi\right) .\end{cases}$p(z)=p_a-\rho \cdot g\cdot z\cdot \cos \alpha +\tan \beta \sqrt{\rho \cdot g\cdot \overline{\gamma }\cdot |\cos \alpha |}\cdot \{\begin{array}{ll}\coth B,& \alpha \in (0,\frac{\pi }{2})\\ B^{-1},& \alpha =\frac{\pi }{2}\\ \cot B,& \alpha \in (\frac{\pi }{2},\pi ).\end{array}$

The free-surface profile $z=h(y)$$z=h(y)$z=h(y)z=h(y)$z=h(y)$ is given by

where $B$$B$BB$B$ is the Bond number for the flow, $B\ne 0$$B\ne 0$B!=0B \neq 0$B\ne 0$ for $\alpha \ne \pi /2$$\alpha \ne \pi /2$alpha!=pi//2\alpha \neq \pi / 2$\alpha \ne \pi /2$,

The maximum depth $h_m$$h_m$h_(m)h_{m}$h_m$ of the liquid, $h_m:=h(0)$$h_m:=h(0)$h_(m):=h(0)h_{m}:=h(0)$h_m:=h(0)$ satisfies

The scales of $h_m$$h_m$h_(m)h_{m}$h_m$ and $a$$a$aa$a$ in eqs. (10) and (11) differ essentially by the small factor $\tan \beta$$\tan \beta$tan beta\tan \beta$\tan \beta$ (and indeed in case ii) $h_m/a=\frac{1}{2}\tan \beta$$h_m/a=\frac{1}{2}\tan \beta$h_(m)//a=(1)/(2)tan betah_{m} / a=\frac{1}{2} \tan \beta$h_m/a=\frac{1}{2}\tan \beta$ ). This reflects the fact that the depth of the layer is much less than its width. The solution is physically sensible only for $h(y)\ge 0$$h(y)\ge 0$h(y) >= 0h(y) \geq 0$h(y)\ge 0$.

The volume flux of liquid running down the plane is

It is convenient to introduce appropriate nondimensional variables defined by: $x^{\ast }=\frac{x}{a},y^{\ast }=\frac{y}{a},z^{\ast }=\frac{z}{a}{h}^{\ast }=\frac{h}{a},{\gamma }^{\ast }=\frac{\gamma }{\overline{\gamma }},u^{\ast }=\frac{u}{U},p^{\ast }=\frac{p}{P}$$x^{\ast }=\frac{x}{a},y^{\ast }=\frac{y}{a},z^{\ast }=\frac{z}{a}{h}^{\ast }=\frac{h}{a},{\gamma }^{\ast }=\frac{\gamma }{\overline{\gamma }},u^{\ast }=\frac{u}{U},p^{\ast }=\frac{p}{P}$x^(**)=(x)/(a),y^(**)=(y)/(a),z^(**)=(z)/(a)h^(**)=(h)/(a),gamma^(**)=(gamma)/(( bar(gamma))),u^(**)=(u)/(U),p^(**)=(p)/(P)x^{*}=\frac{x}{a}, y^{*}=\frac{y}{a}, z^{*}=\frac{z}{a} h^{*}=\frac{h}{a}, \gamma^{*}=\frac{\gamma}{\bar{\gamma}}, u^{*}=\frac{u}{U}, p^{*}=\frac{p}{P}$x^{\ast }=\frac{x}{a},y^{\ast }=\frac{y}{a},z^{\ast }=\frac{z}{a}{h}^{\ast }=\frac{h}{a},{\gamma }^{\ast }=\frac{\gamma }{\overline{\gamma }},u^{\ast }=\frac{u}{U},p^{\ast }=\frac{p}{P}$ (where $P=\rho ga$$P=\rho ga$P=rho gaP=\rho g a$P=\rho ga$ is the reference pressure), $Q^{\ast }=\frac{Q}{\overline{Q}}$$Q^{\ast }=\frac{Q}{\overline{Q}}$Q^(**)=(Q)/(( bar(Q)))Q^{*}=\frac{Q}{\bar{Q}}$Q^{\ast }=\frac{Q}{\overline{Q}}$, where $\overline{Q}=U{a}^2$$\overline{Q}=U{a}^2$bar(Q)=Ua^(2)\bar{Q}=U a^{2}$\overline{Q}=U{a}^2$.

The Reynolds number is $\mathrm{Re}=\frac{UL}{\nu }=\frac{UL\rho }{\mu }$$\mathrm{Re}=\frac{UL}{\nu }=\frac{UL\rho }{\mu }$Re=(UL)/(nu)=(UL rho)/(mu)\operatorname{Re}=\frac{U L}{\nu}=\frac{U L \rho}{\mu}$\mathrm{Re}=\frac{UL}{\nu }=\frac{UL\rho }{\mu }$ and the Weber number is $We=\frac{\rho }{\rho L{U}^2}$$We=\frac{\rho }{\rho L{U}^2}$We=(rho)/(rho LU^(2))W e= \frac{\rho}{\rho L U^{2}}$We=\frac{\rho }{\rho L{U}^2}$, where we assume $\nu =\frac{\mu }{\rho },\sigma =\overline{\gamma },L=a$$\nu =\frac{\mu }{\rho },\sigma =\overline{\gamma },L=a$nu=(mu )/(rho),sigma= bar(gamma),L=a\nu=\frac{\mu}{\rho}, \sigma=\bar{\gamma}, L=a$\nu =\frac{\mu }{\rho },\sigma =\overline{\gamma },L=a$ and $U=\frac{\rho g{a}^2\sin \alpha }{\mu }$$U=\frac{\rho g{a}^2\sin \alpha }{\mu }$U=(rho ga^(2)sin alpha)/(mu)U=\frac{\rho g a^{2} \sin \alpha}{\mu}$U=\frac{\rho g{a}^2\sin \alpha }{\mu }$. Consequently, $\mathrm{Re}=\frac{Lg{\rho }^2{a}^2\sin \alpha }{{\mu }^2},We=\frac{\rho {\mu }^2}{{\rho }^{3}L{g}^2{a}^4\sin \alpha }$$\mathrm{Re}=\frac{Lg{\rho }^2{a}^2\sin \alpha }{{\mu }^2},We=\frac{\rho {\mu }^2}{{\rho }^{3}L{g}^2{a}^4\sin \alpha }$Re=(Lgrho^(2)a^(2)sin alpha)/(mu^(2)),We=(rhomu^(2))/(rho^(3)Lg^(2)a^(4)sin alpha)\operatorname{Re}=\frac{L g \rho^{2} a^{2} \sin \alpha}{\mu^{2}}, W e=\frac{\rho \mu^{2}}{\rho^{3} L g^{2} a^{4} \sin \alpha}$\mathrm{Re}=\frac{Lg{\rho }^2{a}^2\sin \alpha }{{\mu }^2},We=\frac{\rho {\mu }^2}{{\rho }^{3}L{g}^2{a}^4\sin \alpha }$ and then $\mathrm{Re}\cdot We=\frac{\rho }{\rho g{a}^2}$$\mathrm{Re}\cdot We=\frac{\rho }{\rho g{a}^2}$Re*We=(rho)/(rho ga^(2))\operatorname{Re} \cdot W e=\frac{\rho}{\rho g a^{2}}$\mathrm{Re}\cdot We=\frac{\rho }{\rho g{a}^2}$. With the Bond number $B{o}_1=\frac{\rho g{L}^2}{\sigma }$$B{o}_1=\frac{\rho g{L}^2}{\sigma }$Bo_(1)=(rho gL^(2))/(sigma)B o_{1}=\frac{\rho g L^{2}}{\sigma}$B{o}_1=\frac{\rho g{L}^2}{\sigma }$, which relates the gravitational forces $g$$g$gg$g$ to the cappilarity, the nondimensional velocity of the fluid is given by

The nondimensional free surface velocity is

and the nondimensional pressure becomes

The nondimensional version for the volume is given by

A) The limit $a\to 0(B\to 0)$$a\to 0(B\to 0)$a rarr0(B rarr0)a \rightarrow 0(B \rightarrow 0)$a\to 0(B\to 0)$

We note that for all $\alpha \in (0,\pi /2)\cup (\pi /2,\pi )$$\alpha \in (0,\pi /2)\cup (\pi /2,\pi )$alpha in(0,pi//2)uu(pi//2,pi)\alpha \in(0, \pi / 2) \cup(\pi / 2, \pi)$\alpha \in (0,\pi /2)\cup (\pi /2,\pi )$,

B). The limit $\beta \to 0$$\beta \to 0$beta rarr0\beta \rightarrow 0$\beta \to 0$

We have respectively

and

In case iii) we have