A well-posed model for dynamic contact angles

A well-posed model for dynamic contact angles Ben Schweizer

February 27, 1998

Abstract

We consider uid systems with a free boundary and with a point of contact of the free boundary with a solid wall. We contribute to the discussion on the conditions for the dynamic contact angle and wellposedness of the equations. An energy equality suggests a constant angle. With the help of symmetric extensions we prove a resolvent estimate in the case of a 90 degree contact angle. A technique developed by Renardy can be applied and yields an existence result for the nonlinear problem.

1 Introduction We analyze from a mathematical point of view the instationary problem of a

uid that has both a free boundary ( uid-air interface) and a xed boundary ( uid-solid interface). The contact line or, in two dimensions, the point of contact is de ned to be the point where the two interfaces meet. The tangent planes intersect under the contact angle. In 1805 Young analyzed the static contact angle and claimed that the contact angle is a constant, depending only on the materials. One counterintuitive consequence is that there are no stationary solutions for liquid drops on a tilted plane [2], [3]. A further question concerns the dynamic contact angle. Using the usual

uid equations and the well accepted equations on the free surface together with a no-slip boundary condition along the wall, one arrives at the unphysical result that the forces are not integrable [4], [8]. Starting in the seventies several alternative models were discussed. One was to replace the no-slip boundary condition along the wall by a slip condition; one assumes that the tangential stress is proportional to the tangential speed [4], [7], [12]. We mention here another interesting approach by Huh and Mason [6]. Their model includes a narrow zone in which no tangential forces are exerted between wall and uid. It seems that meanwhile the slip condition found broad acceptance. 1

2

Ben Schweizer

As Hocking describes in [5], a discussion focused on the following question: Can one assume that the dynamic contact angle is a constant, depending only on the materials, or does the contact angle have a dependence on the speed of the point of contact? While for instance Dussan and Chow [2] argue in favor of a speed-dependence, Hocking points out that there might very well be a discrepancy between the actual (microscopic) and the apparent contact angle, justifying the more economical model of a constant angle. We remark that in the instationary case a constant angle does not lead to the above mentioned diculties in the tilted plane problem. The purpose of this paper is two-fold: We start from the elementary energy-decay equality and consider two dierent implicit time-step schemes. Their analysis emphasizes the one-to-one correspondence between dissipation of energy along the line of contact and the equation for the dynamic contact angle. In particular: The assumption of vanishing energy dissipation along the line of contact suggests a constant contact angle in the dynamic equations. We furthermore study the system of equations with a constant contact angle in the two-dimensional case. Contradictory equations in the point of contact imply limitations for the regularity of solutions. In general the velocity v will not be of class C 1 (  ). We consider a function space of slightly less regularity: In the case of a rectangular geometry we can estimate v 2 H s ( ); 1 < s < 2 for the stationary equations. The proof uses symmetric extensions of the variables. The regularity turns out to be sucient to demonstrate well-posedness of the nonlinear, time dependent free boundary problem. The treatment of the nonlinearity is involved. Without higher regularity properties we can not apply a direct method as in Beale [1] and later works. Instead we use the technique developed by Renardy in [9]. He considers a free surface problem with in ow and out ow boundary conditions where the boundary conditions x the point of contact at the in ow boundary and do not involve the contact angle. Despite of this dierence there is an important analogy: The velocity has estimates only in H s ; s < s < 2. Renardy's treatment of the nonlinearity can be applied and yields an existence result for small data.

2 Model equations and energy decay We denote the independent variables as (x; y), the velocity eld by v = (v1 ; v2 ) and the pressure by p, de ned on the time dependent domain =

t = f(x; y)j0 < x < L; ?1 < y < ht (x)g. The rigid walls are  = 1 [ 2

On the dynamic contact angle

3

with 1 := f(0; y)j ? 1 < y < h(0)g and 2 := f(L; y)j ? 1 < y < h(L)g. The rigid bottom is given by ?0 = fy = ?1g and the free surface by ? = ?t = graph(ht : [0; L] ! IR) with exterior normal n and tangent vector  . We introduce the strain tensor S and the stress tensor T as   Sv := 12 rv + (rv)T ; T (v; p) := ?p id + 2Sv : On ? we denote normal speed and normal strain by

vn := v
n; and Svn := n
Sv
n: The mean curvature H (h) of the boundary ? is given by

H (h) := ?@x

!

@xh p : 1 + j@xhj2

(2.1)

With these notations the evolutionary equations are the following.

@tv + (v
r)v ? 
v + rp = 0 in ; div v = 0 in ; p ? 2Svn = H (h) + f on ?; 
Sv
n = 0 on ?; @t h ? v2 + (@xh)v1 = 0 on ?; v = 0 on ?0 ; v1 = 0 on i ; i = 0; 1; @xv2 ? i v2 = 0 on i ; i = 0; 1; some equation in the points of contact:

(2.2) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8) (2.9) (2.10)

These equations are generally accepted to be a good model for the free boundary problem. We left three of the equations ambiguous:

 The force f in equation (2.4) is to be chosen. Usually it is set to zero.  With the parameter = 0 = ? 1 2 [0; 1] we can select as boundary condition the no-slip condition ( = 1), perfect slip ( = 0) and a linear slip condition.

 An equation (2.10) must be speci ed.

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Ben Schweizer

We want to analyze equations (2.2){(2.10) on the basis of energy considerations. In the context of static contactRangles one introduces the following energies: The kinetic energy Ekin = 12 jvj2 , the potential energy of the free boundary E? = j?j corresponding to surface tension, and the energy of a wall-liquid interface, E = jj. For curves j:j denotes the length, and  are material constants. The static contact angle s is calculated by minimizing the potential energy of the system and is given by

sin(s ? =2) = :

(2.11)

We give the following remark on equation (2.10). Remark 2.1 (The number of conditions) Consider as equation (2.10) one of the following:

h(0) = 0 = h(L)

or

h0 (0) = const = ?h0 (L):

In both cases the stationary equations corresponding to (2.2){(2.9) have a unique solution for small force f . Observe that in the case of a no{slip condition on  the rst choice of boundary conditions is necessary in order to have smooth solutions. Proof: Compare [13] and [12]. Idea: Solve the Stokes equations (2.2) with boundary conditions (2.5) and (2.6). Evaluating the left-hand-side of (2.4) and solving for h is possible if some suitable boundary conditions for h are imposed. Without specifying equation (2.10) we can perform the following calculation for the energy decay. Lemma 2.2 Let (v; h) be a solution of the evolutionary equations (2.2){ (2.9). De ne the contact angles 0 and 1 by

tan(0 ? =2) = h0 (0); Then the total energy

Etot = 12

satis es the decay equation

Z

tan(1 ? =2) = ?h0 (L): Z



jvj2 + j?j + jj Z

(2.12) (2.13)

Z

@tEtot = ?2 jSv j2 ?  jv2 j2 ? v f

 ? +v2 (0; h(0)) [ ? sin(0 ? =2)] +v2 (L; h(L)) [ ? sin(1 ? =2)] :

(2.14)

On the dynamic contact angle

5

Proof: The proof is a direct calculation: Z Z 1 @t Etot = hv; @t vi + 2 jvj2 v
n ?

Z Lq 1 + jh0 (x)j2dx + @t h(0) + @t h(L) + @t =

0

Z



Z

hv; ?(v
r)v + 
v ? rpi + hv; (v
r)vi

0 0 + ph (x)@th0 (x)2 dx + v2(0; h(0)) + v2 (L; h(L)) 1 + jhZ(x)j 0 Z Z = ?2 jSv j2 ? v
(pI ? 2Sv )
n dS ?  v2@x v2 dy

?  Z Z L 0 + v2 @xv2 dy + p h (x0) 2 @x (v2 ? h0 v1)dx 1 + jh (x)j  0 +v2(0; h(0)) + v2(L; h(L)) Z Z Z L = ?2 jSv j2 ?  jv2 j2 ? (v2 ? h0 v1 ) H (h)dx 0

 Z Z L 0 ? vn f + p1 +h j(hx0)(x)j2 @x(v2 ? h0 v1 )dx 0 ? +v2Z(0; h(0)) + v 2 (L; h(L))Z Z = ?2 jSv j2 ?  jv2 j2 ? vn f

 ? # " 0 (0) h +v2(0; h(0))  ? p 1 + jh0 (0)j2 " # 0 (L) h +v2(L; h(L))  + p : 1 + jh0 (L)j2 Z L

0

1

3 The constant contact angle EquationR (2.14) contains several terms with immediate physicalR interpretation: 2 jSv j2 is the energy dissipation due to inner friction,   jv2 j2 is the loss of energy due to friction along the wall. If energy is dissipated only by viscosity and by friction along the wall, then for f = 0 (2.14) implies that the quantity [ ? sin( ? =2)] v2 vanishes in the points of contact. We will analyze the following equations that represent an implicit time step for equations (2.2){(2.9): (~v; h~ ) and (v; h) are the solutions for time

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Ben Schweizer

t and t +
t respectively. The equations are linear, the time derivative is replaced by a dierence quotient with
t = ?1 . The equations are stated

on the xed domain ~ with the upper boundary ?~ and lateral boundaries ~ i prescribed by h~ .

v + (~v
r)v ? 
v + rp = v~ in ~ ; (3.1) ~ div v = 0 in ; (3.2) n ~ (3.3) p ? 2Sv + @xG(h) = f on ?; ~ 
Sv
n = 0 on ?; (3.4) 0 h ? v2 + h~ v1 = h~ on ?~ ; (3.5) v = 0 on ?0 ; (3.6) v1 = 0 on ~ i ; (3.7) ~ @xv2 ? iv2 = 0 on i ; (3.8) q with n = (?h~ 0 ; 1)=s;  = (1; h~ 0 )=s; s := 1 + jh~ 0 j2 : We assume that s and v~ are bounded and div (~v) = 0. We used the function ~ ~ G(h) := q @x h + q@x (h ? h)3 1 + jh~ 0 j2 1 + jh~ 0 j2 which is the linear approximation of p

h0

= sin(tan?1 (h0 )): 0 2 1 + jh j

Testing equation (3.1) with v and equation (3.5) with @x G(h) yields the following discrete analogue of lemma 2.2. Lemma 3.1 Let (v; h) be a solution of (3.1){(3.8). Then there holds Z hv; (v ? v~)i + ~ v~n 12 jvj2 + (h ? h~ )(0) + (h ? h~ )(L) ? Z L Z Z + G(h)@x (h ? h~ ) dx = ?2 ~ jSv j2 ?  ~ jv2 j2 0

Z L





f (x)(v2(x; h~ (x)) ? h~ (x)v1 (x; h~ (x))) dx 0 +v2 (0; ~h(0)) [ ? G(h)(0)] + v2(L; h~ (L)) [ + G(h)(L)] :

?

0

We will use the following assumptions to derive approximate equations.

On the dynamic contact angle

7

(A1) Energy is dissipated by viscous forces in the interior of the liquid and by friction along the wall. No energy is dissipated in the point of contact. (A2) Equations with a singular force are approximated as follows: The force is distributed over an "-neighborhood such that the approximate solutions are smooth. To satisfy (A1) we have a choice. We can set f = f0 = 0 and prescribe a xed contact angle by

G(h)(0) =  = ? G(h)(L):

(3.9)

Let (v0 ; h0 ) be a solution to equations (3.1) { (3.8) together with (3.9). From now on we assume 2 (0; 1), that is, a slip condition on i . Remark 3.2 In general we can not expect the solutions (v0; h0 ) to be smooth: Assuming that @xh0 and @x v0 are continuous up to the boundary we can calculate for h~ = 0,  = 0:

G(h)(0) = 0; therefore

v20 (0; h0 (0) = @x v20(0; h0 (0)) = @xh0 (0) = 0: We nd the unphysical conclusion that the point of contact does not move. Alternatively we can use (A2) to de ne C 1 (  ) approximate solutions (v" ; h" ). There is no loss of energy in the point of contact in x = 0 if either v2(0) = 0 or f (x) = [ ? G(h)(0)] 0 (x)e2; (3.10) where 0 is the Dirac distribution in 0 in the x-variable. The rst possibility implies a constant contact angle h0 (0) = h~ 0 (0) by (3.8). Therefore, following assumption (A2), we will analyze a force f" (x; h~ (x)) := [ ? G (h" )(0)] "(x); RL (3.11) supp(")  [0; "); 0 " = 1:

To estimate products with f" we provide inequality (3.12) for the left point of contact. Using the primitive " : (0; ") ! IR, @x" = ", " (0) = ?1 we calculate for a function g : (0; ") ! IR Z " "



Z "

 (x)g(x)dx ? g(0) = " (x)@xg(x)dx 0  k@xgkL (0;")  p"kgkH (0;L=2): 0

1

1



(3.12)

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Ben Schweizer

We consider a solution (v" ; h" ) of (3.1) { (3.8) that is of class C 1 up to the boundary with right hand side f " . Assume h~ 0 (0) = h~ 0 (L) = 0. Due to the regularity we do not have to specify any additional condition in the point of contact. There holds

@xh" (0) = @xv2" (0) = v2" (0) = h" (0):

(3.13)

This equation guarantees uniqueness of (v" ; h" ). The linear equations admit the following a priori estimate. Lemma 3.3 Let (v" ; h" ) be a C 1 (  )-solution of (3.1){(3.8) with force f = f" (x) as in (3.11) and with analogous construction in x = L. Assume that h~ 0 (0) = ~h0 (L) = 0. Then for   0 (; v~) and "  "0 ( ; ; ) there holds an estimate kh" kH (0;L) + j@xh" (0)j + j@x h" (L)j  C0 (~v; h~ ) (3.14) with C0 independent of " and . 1

Proof: We use lemma 3.1. We collect positive terms on the left hand side and estimate linear terms with Young's inequality. Two terms remain to be estimated: Z 1 C1 " 2 " j2   kv " k2 v ~ j v ~ ) +  kv kL ( ~ ) : 2 ~ n H (

? For small  and large  we can absorb this sum in the left hand side. We use (3.12) with g(x) := [ ? G(h" )(0)] (v2" j?~ ? h~ 0 v1" j?~ )(x) to calculate 1



?

Z L=2

0

2



f"(x)(v2" j?~ ? h~ 0 v1" j?~ ) dx + [ ? G(h" )(0)] v2" (0)

 (jj + j@xh" (0)j)p"k(h" ? h~ )kH (0;L=2) p = (jj + jh" (0)j) "kh" ? h~ kH (0;L=2)  Cp"(1 + kh" k2H (0;L=2)): 1

1

1

For small " the last term can be absorbed in the left hand side and we conclude

kh" k2H (0;L)  const: Using again j@x h" (0)j = j h" (0)j  ckh" kH (0;L) we proved the lemma. 1

1

In the following proposition we compare the two constructions of a time step.

On the dynamic contact angle

9

Proposition 3.4 Let (v" ; h" ) with uniformly bounded contact angles as in

lemma 3.3 and let (v0; h0 ) be the solution of (3.1){ (3.8) with force f = f0 = 0 and xed contact angle as in (3.9). Then for   0(; v~) there holds in L2( ~ )  H 1 (0; L):

lim (v" ; h" ) = (v0; h0 )

"!0

(3.15)

Proof: We collect the equations for the dierence v = v" ? v0 with pressure distribution p = p" ? p0 and h = h" ? h0 . They are (3.1){(3.8) with ~ = 0, q v h~ = 0, f" as in (3.11) and G replaced by G (h) = s?3 @xh with s = 1 + jh~ 0 j2 . We multiply equation (3.1) with v and (3.5) with @x G (h) to derive the estimate Z

 ~ jvj2 + 

Z

Z L

0

s?3j@x hj2 + 2

Z L

Z

Z

jSv j2 +  ~ jv2 j2 ~ 



 = ? 12 ~ v~n jvj2 ? f" (x) v2 (x; h~ (x)) ? h~ 0 v1 (x; h~ (x)) (3.16) 0 ? + G (h)(L)h(L) ? G (h)(0)h(0): 

The rst term on the right hand side is treated as in lemma 3.3. We use 3.12 and insert g = [ ? G(h" )(0)] (v2 ? ~h0 v1)j?~ : Due to G(h0 ) =  this function satis es

g(0) = ? G (h)(0)h(0): We arrive at

p f (x)h(x)dx + G (h)(0)h(0)  "khkH

Z " "

0



1

(0;L) j ? G(h

" )(0)j

and an analogous bound in x = L. Using lemma 3.3 the right hand side of (3.16) can be bounded by

p

C2 "khkH (0;L)  khk2H 1

C2  (0;L) + 4 "; 2

1

for any  > 0. Choosing  small and subtracting the rst term on both sides of (3.16) we proved the proposition. We read this proposition as follows: We can either prescribe in every time step the static contact angle or approximate every time step with smooth solutions that compensate the loss of energy in the point of contact. The

10

Ben Schweizer

two constructions lead to the same results. In the remainder of this article we use as equation (2.10)

h0 (0) = ?h0 (L) = tan (s ? =2) with s as in (2.11). Existence theorems for free boundary uid systems are usually proved by showing estimates for the resolvent in high Sobolev norms. Remark 3.2 gives limitations for the maximal regularity for the equations with a constant contact angle. In the two-dimensional case we can not expect the velocity eld to have a regularity higher than H 2 ( ). H 2 turns out to be critical.

4 Regularity of steady solutions The aim of this section is to derive regularity results for solutions of the nonlinear steady equations. We have to restrict ourselves to the rectangular domain, the case s = =2 corresponding to  = 0. In this case the estimates can be derived with the help of symmetric extensions of the solution. In the standard way ([1], [9]) we transform the equations onto a xed domain with independent variables (X1 ; X2 ) = (X; Y ), in our case the rectangle (0; L)  (?1; 0). The transformation is given by 



(x1 ; x2 ) = (x; y) = X; (1 + Y )(1 + h~ (X; Y )) ? 1 :

(4.1)

Here ~h satis es h~ (X; 0) = h(X ) and is given by h~ = Eh where E is a linear extension operator such that h~ has maximal regularity. We can assume that   @x = ~h satis es @X h~ = 0 on i . We introduce the Jacobian J = det @X 1 + h~ + (1 + Y )@Y h~ and calculate i

j

!

@Xi = 1 @xj J

J 0 ?(1 + Y ) @X h~ 1 :

(4.2)

Observe that nondiagonal entries vanish identically on i . One introduces new dependent variables v~, p~ and T~ by

@xi v~ (X; Y; t); vi (x; y; t) = J1 @X j

p~(X; Y; t) = p(x; y; t);

j

T~(X; Y; t) = T (x; y; t):

(4.3) (4.4)

On the dynamic contact angle

11

We next state equations (2.2){(2.10) in the new variables. 1 @xi @ v~ +  (~v; h~ ; @ h~ )

J @Xj

t j

i

t

!

!

@ @Xk T~ ? T~ @ @Xk ; = @X (4.5) ij ij @Xk @xj k @xj div(~v) = 0; (4.6) T~ = 2Sv~ ? p~I + (~v; h~ ): (4.7) The nonlinearities i ;  are dierentiable mappings i : H s ( )  H s+1 ( )  H s ( ) ?! H s?1 ( ) and  : H s ( )  H s+1 ( ) ?! H s?1 ( ) for s 2 (1; 2).

Their Frechet derivatives vanish at the origin. The boundary conditions for (~v; h) are

v~ @t h v~1 @X1 T @xj 2j @X2 T @xj ij

= = = =

0 on ?0 = (0; L)  f?1g; v~2 + v~? on ? = (0; L)  f0g; 0 on  = f0; Lg  (?1; 0);

i v~2 on i ;

(4.8) (4.9) (4.10) (4.11)

ni @X2 h on ?; (4.12) (1 + j@X hj2 )(1 + h + @Y ~h) 0 = @X h(0) = @X h(L): (4.13) Observe that we included velocity data v~? in equation (4.9). In the =

condition for the normal stress both sides of the equation coincide with p ni Tij 1 + j@X hj2 J ?1 . In the following we analyze a linear resolvent problem for (V; P; h). is a rectangle and the boundaries ?; ?0 and i are straight lines. It suces to consider G with G12 = G21 = 0 along i due to equation (4.2) andR @X h~ = 0 on i . We assume throughout that the integrability condition 0L v? = 0 holds.

V ? 
V + rP r
V V h ? V2  (@1 V2 + @2 V1) ?P + 2@2V2 ?
x h V1

= = = = = = =

F + div G in ; 0 in ; 0 on ?0 ; v? on ?; ?G12 + 1 on ?; ?G22 + 2 on ?; 0 on i ;

(4.14) (4.15) (4.16) (4.17) (4.18) (4.19) (4.20)

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Ben Schweizer

@1V2 ? i V2 = 0 on i ; (4.21) 0 0 h (0) = 0 = h (L): (4.22) We will decompose V and P as V = u+v and P = p+q. The construction will be such that (u; p; h) have symmetric extensions across 0 . We denote them as (u; p; h ). Observe that these functions are x-periodic in the domain

 := (?L; L)  (?1; 0) with upper boundary ? := (?L; L)  f0g. The

symmetries are as follows. u1 (?x; y) = ?u1 (x; y); u2 (?x; y) = u2 (x; y); p(?x; y) = p(x; y); h(?x) = h(x); and consistently F2; G 11 ; G 22 ; 2 ; v? even and F1; G 12 ; G21 ; 1 odd : In a similar way we denote by (~v; q~) the symmetric extension of (v; q) across ?. The domain is ~ := (0; L)  (?1; 1) and the symmetries are v~1(x; ?y) = v1(x; y); v~2(x; ?y) = ?v2(x; y); q~(x; ?y) = q(x; y):

?

( L; 0)

?

~ (~v; q~)

?~

 (u; p) ? ?1)

( L;

(u + v; p + q) (0;

?1)

(L;

?1)

De nition 4.1 (The equations for (u; p; h)) Let g be a distribution on

(0; L) and g its even symmetric extension to ? . We de ne (u; p; h ) to be the periodic and symmetric solution of u ? 
u + rp = F + div G in  ; r
u = 0 in  ; h = u2 + v? on ? ; (4.23) u = 0 on ? 0 ;  (@1 u2 + @2u1 ) = ?G 12 + 1 on ? ; ?p + 2@2u2 =
x h ? G 22 + 2 + g on ? : We denote by (u; p) the restriction (u; p)j and by h the restriction h jf0 3=2 we require G12 j = G21 j = 0. Then the equations (4.14) { (4.22) have a unique solution (V; P; h) that satis es the regularity

n

kV kH ( ) + kP kH ? ( ) + khkH (0;L)  C kF kHo? ( )0 +kGkH ? ( ) + kkH ? (0;L) + kv?kH ? (0;L) : s

s

s

1

2 s

s+1=2

1

s

3=2

s

1=2

(4.31)

Proof: We assume that the right hand side in (4.31) is bounded. As in lemma 4.4 we have the energy estimate for kV kH ( ). It implies a bound for i := i V2j 2 H 1=2 (i ). Using lemma 4.3 we conclude that g := n
T (v; q)
nj? 2 H s?3=2 (?) is bounded. Now lemma 4.4 implies inequality (4.31). We state here also Renardy's conclusion concerning the nonlinear stationary equations. It follows directly from theorem 4.5 with the help of the implicit function theorem. Theorem 4.6 Let v? 2 H s?1=2 (0; L) with s 2 (1; 2), v? suciently small and satisfying the integrability condition. Then the equations (4.5) { (4.13) have a steady solution with the regularity v~ 2 H s ( ), p~ 2 H s?1 ( ), h 2 H s+1=2 (0; L). The solution is unique among solutions of suciently small norm. 1

i

5 The time dependent linear and the nonlinear problem In this section we indicate the steps that lead to the existence result for the xed dynamic contact angle. The construction is introduced by Renardy in [9]. For  2 (1=2; 3=2) one de nes the spaces X = fv 2 H ( )jdiv(v) = 0; v(0; Y ) = v(L; Y ) = 0; v(X; ?1) = 0g : (5.1) Let X? be the dual of X. For a matrix-valued function A we consider the following general linear problem for (v; h) 2 X1  H 1 (0; L). R 



R

@v

(A(X; Y; t) @t u +R Sv Du = (F; u) ? 0L @X h(X; t)@X u2 (X; M )dX; @th = v2(:; M ); 0 = h0 (0) = h0 (L):

8u 2 X1 ; (5.2)

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Ben Schweizer

The following two theorems repeat [9], theorem 7 and theorem 9. Cb denotes the space of bounded, uniformly Holder continuous functions and k:kk;l the norms of H k ((0; 1); Xl ) and H k ((0; 1); H l (0; L)). Theorem 5.1 Assume that  > 0 and

A 2 H 1=2+ ((0; 1); H 1+ ( )) \ Cb1+ ([0; 1); L2( ));

(5.3)

A symmetric and close to the identity matrix in the norm of this space. Moreover, assume that F 2 L2((0; 1); X?1 ), v0 2 X0 ; h0 2 H 1 (0; L). Then

there exists a unique solution of (5.2) which satis es

kvk0;1 + kAvk1;?1 + khk0;3=2 + khk1;1=2 (5.4)  C (kF k0;?1 + kv0k0 + kh0k1 ) : Theorem 5.2 Let s 2 (1; 2) and v0 2 X(s+1)=2, v0t 2 X0, and h0 2

H 2 (0; L) be given and assume that their norms are suciently small. Then

there exists a unique solution of (4.5){(4.13) assuming the initial conditions and with regularity ?




(v; h) 2 H1 ((0; 1); X1 ) \ L2((0; 1); H s ( ))  H 2 ((0; 1); H 1=2(0; L)) \ H 1((0; 1); H 3=2 (0; L)) \L2 ((0; 1); H s+1=2(0; L)) :

(5.5)

The proof of both theorems is as in [9]. The only alteration for theorem 5.1 is the following: While in [9] h(0) = 0 is used to insure a Poincare inequality, RL in the contact angle equations one uses instead 0 h dx = 0. The existence of a solution is shown with a Crank-Nicholson scheme. To prove theorem 5.2 one uses theorem 5.1 for time-derivatives of the physical variables. This yields existence and regularity in time. To conclude spatial regularity one uses theorem 4.6. An iteration over the nonlinearity yields theorem 5.2. It provides the global existence for small solutions of the contact angle equations.

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On the dynamic contact angle

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