Modelling of hydrophobic surfaces by the

Stokes problem with the stick-slip boundary

conditions

S. Fialová1, J. Haslinger2,3, R. Ku£era2, F. Pochylý1, and V. átek2

1Brno University of Technology, Technická 2896/2, 616 69 Brno, CZ2IT4Innovations, VB-TU Ostrava, 17. listopadu 15/2172, 708 33

Ostrava-Poruba, CZ3IG CAS, Studentská 1768, 708 00 Ostrava-Poruba, CZ

January 13, 2016

Abstract

Unlike the Navier boundary condition, the present paper deals withthe case when the slip may occur only when the shear stress attainscertain bound which is given à-priori. The discrete velocity-pressuremodel is derived using P1-bubble/P1 elements. To release the imper-meability condition and to regularize the non-smooth term characteriz-ing the stick-slip behavior in the algebraic formulation, two additionalvectors of Lagrange multipliers were introduced. The resulting min-imization problem in terms of the dual variables (the pressure, thenormal and shear stress) is solved by the interior point type method.

1 Introduction

The biggest part of hydraulic losses arises from uid ow near the surface anddepends on the uid velocity gradient in the normal direction to the surface.The velocity gradient is subject to liquid adhesion. If the surface is perfectlywetted by the liquid, it is possible to characterize the sticking of the liquidby the Dirichlet boundary condition prescribing zero velocity vector alongthe surface. This condition is fullled by the majority of known compounds,e.g. metals-water, water-glass, metals-oil, most plastics-water, etc. Reducedadhesion can be achieved by reducing the gradient of the liquid velocity,

1

thereby reducing hydraulic losses. This phenomena is usually modelled bythe following boundary condition, which expresses the slip of the liquid ateach point of the surface by

σt = −κut, (1)

where ut, σt denote the tangential components of the velocity and the shearstress along the surface, respectively. Further, κ is a positive adhesive func-tion which generally depends on the spatial variable and its knowledge isassumed [5, 6]. If κ = const. > 0 on the surface then (1) is the classicalNavier condition with the adhesive coecient κ.

In this paper we use another type of the slip condition, namely the stick-slip boundary conditions. We assume that a nonnegative slip bound functiong is prescribed along the surface and we require fulllment of the followingrelations at each point of the surface: (i) |σt| ≤ g; (ii) if |σt| < g, thenut = 0; (iii) if |σt| = g, then the slip may occur and its sign is opposite to thesign of σt. It is worth noticing that the partition of the surface between thestick and slip part is one of unknowns of the problem. The condition (iii)yields the existence of κs ≥ 0 such that σt = −κsut. In contrast to (1), itsvalue is known only after solving the problem and depends on the computedtangential velocity component:

κs =g

|ut|, ut 6= 0. (2)

It is worth noticing that if g is chosen in such a way that the slip occurs onthe whole surface then from (2) and (iii) it follows that σt = −κsut. Thuswe obtain (1) with κ = κs.

The conditions analogous to our stick-slip boundary conditions are well-known in contact problems of solid mechanics. They represent the Trescafriction law on common interfaces of solid bodies [7, 1]. As the algebraicproblems arising from nite element approximations are formally the same,the algorithms, which are highly ecient for contact problems could serve asnatural tools also for solving this type of ow problems. Numerical experi-ments in [10] proved that the algorithm based on an interior point method [11]is suitable for this kind of problems, too. In this paper we test this modelfor physically realistic problems.

The paper is organized as follows: Section 2 presents the classical andweak formulation of the Stokes problem with the stick-slip boundary condi-tions. The algebraic formulation of this problem and its dual form in termsof the pressure, shear and normal stress are derived in Section 3. In Sec-tion 4 the path-following interior point method is described. Three modelexamples, namely ow between parallel plates, ow initialized by rotating

2

concentric and eccentric circles are computed. The results of rst two exam-ples are compared with the ones obtained by ANSYS Fluent [16] and alsowith the analytic solutions [3].

2 Classical and weak formulation of the prob-

lem

Let Ω be a bounded domain in R2 with a suciently smooth boundary ∂Ωthat is split into three disjoint parts: ∂Ω = γD ∪ γN ∪ γC . We consider themodel of a viscous incompressible Newtonian uid modelled by the Stokessystem with the Dirichlet and Neumann boundary conditions on γD andγN , respectively, and with the impermeability and the stick-slip boundaryconditions prescribed on γC :

−η∆u+∇p = f in Ω,∇ · u = 0 in Ω,

u = uD on γD,σ = σN on γN ,un = 0 on γC ,|σt| ≤ g on γC ,

|σt(x)| < g(x) ⇒ ut(x) = 0 x ∈ γC ,|σt(x)| = g(x) ⇒ ∃κs := κs(x) ≥ 0 : σt(x) = −κsut(x) x ∈ γC ,

(3)

where σ = η dudn−pn. Here, u = (u1, u2) is the ow velocity, p is the pressure,

f = (f1, f2) represents volume forces acting on the uid, η > 0 is the dynamicviscosity, and uD, σN are given Dirichlet and Neumann boundary data,respectively. Further n = (n1, n2), t = (−n2, n1) is the unit outer normaland tangential vector to ∂Ω, respectively, while un = u · n, ut = u · t isthe normal and tangential component of u along γC , respectively. Finallyσt = σ · t is the shear stress and g ≥ 0 is a given slip bound function on γC .Note that κs is not known à-priori. Its value can be computed by (2) onlyhaving the solution of (3) at our disposal. We will assume that γD 6= ∅ andγC 6= ∅.

Next we present the weak velocity-pressure formulation of (3). For thesake of simplicity we will suppose here that uD = 0. We introduce thefollowing notation:

V (Ω) = v ∈(H1(Ω)

)2: v = 0 on γD, vn = 0 on γC

anda(v,w) = η

∫Ω

∇v : ∇w dx, b(v, q) = −∫

Ω

q∇ · v dx,

3

l(v) =

∫Ω

f · v dx+

∫γN

σN · v ds, j(v) =

∫γC

g|vt| ds,

where ∇v : ∇w = ∇v1 · ∇w1 + ∇v2 · ∇w2, v = (v1, v2), w = (w1, w2)and g is a bounded, nonnegative function on γC . Finally, H1(Ω) stands forthe Sobolev space of functions which are together with their rst derivativessquare integrable in Ω (see [12]).

The velocity-pressure formulation of (3) reads as follows:

Find (u, p) ∈ V (Ω)× L2(Ω) such that

a(u,v − u) + b(v − u, p) + j(v)− j(u) ≥ l(v − u) ∀v ∈ V (Ω),

b(u, q) = 0 ∀q ∈ L2(Ω).

(4)

This problem has a unique solution provided that γN is nonempty. In theopposite case the pressure is determined up to a constant. These resultshave been established in [2, 4]. To discretize (4) we use mixed nite ele-ments satisfying the inf-sup stability condition [4]. In particular, we use theP1-bubble/P1 elements [8], for which we observed a stable behavior of theLagrange multipliers representing the shear stress [10].

3 Algebraic formulation of the problem

The nite element approximation of (4) leads to the following algebraic prob-lem:

Find (u,p) ∈ Rnu × Rnp such that

u>A(v − u) + (v − u)>B>p + g>(|Tv| − |Tu|) ≥ l>(v − u) ∀v ∈ Rnu ,

q>Bu = 0 ∀q ∈ Rnp ,

Nu = 0,

(5)

where u, p is the vector of the nodal values of the velocity u and the pressurep, respectively, A ∈ Rnu×nu is a symmetric and positive denite stinessmatrix, B ∈ Rnp×nu , T,N ∈ Rnc×nu are full row-rank matrices, l ∈ Rnu ,g ∈ Rnc

+ are the load vector and the vector of the nodal values of g onγC , respectively. Further |x| = (|x1|, . . . , |xnc |)> for x ∈ Rnc ; np is thetotal number of the nodes of a used triangulation contained in Ω, nc is thenumber of the nodes lying on γC\γD, and nu is the dimension of the solutioncomponent representing the velocity u.

It is easy to show that (5) is equivalent to the following minimizationproblem:

Find u ∈ V such that J (u) ≤ J (v) ∀v ∈ V, (6)

4

where J (v) = 12v>Av− v>l + g>|Tv| and V = v ∈ Rnu : Nv = 0, Bv =

0. To release the discrete impermeability condition Nv = 0 and to reg-ularize the last non-dierentiable slip term in J , we introduce two alge-braic Lagrange multipliers λn and λt, respectively, and dene the LagrangianL : Rnu ×Λ 7→ R by

L(v,λ) =1

2v>Av − v>l + λ>Cv,

where Λ = λt ∈ Rnc : |λt| ≤ g × Rnc+np , λ = (λ>t ,λ>n ,p

>)> ∈ Λ, andC = (T>,N>,B>)>. The minimization problem (6) is equivalent to thefollowing saddle-point formulation:

Find (u, λ) ∈ Rnu ×Λ such that

L(u,λ) ≤ L(u, λ) ≤ L(v, λ) ∀(v,λ) ∈ Rnu ×Λ.

(7)

From the second inequality in (7) we see that

u = A−1(l−C>λ) . (8)

Inserting (8) into the rst inequality in (7) we get the dual problem in termsof λ only:

Find λ ∈ Λ such that S(λ) ≤ S(λ) ∀λ ∈ Λ (9)

with S(λ) = 12λ>Fλ−λ>d, where F = CA−1C> is symmetric, positive def-

inite and d = CA−1l. The dual formulation (9) is the minimization of thestrictly quadratic function S subject to a small number (nc) of constrainedunknowns versus a large number (nc+np = nc+O(n2

c)) of the unconstrainedones. The optimization algorithm appropriate for problems with this struc-ture is described in the next section.

4 Path-following interior point algorithm

The Lagrangian associated with (9) is dened by

L(λ,µ) = S(λ) + µ>1 (−λt − g) + µ>2 (λt − g),

where µ = (µ>1 ,µ>2 )> ∈ R2nc is the Lagrange multiplier releasing two sided

constraint appearing in the denition of Λ. Let z := −∇µL(λ,µ) be thenew variable used in the function G : R6nc+np 7→ R6nc+np dened by

G(w) := (∇λL(λ,µ)>, (∇µL(λ,µ) + z)>, e>MZ)>,

5

where w = (λ>,µ>, z>)> ∈ R6nc+np , M = diag(µ), Z = diag(z), ande ∈ R2nc is the vector whose all components are equal to 1. The solution λto (9) is the rst component of the vector w = (λ>, µ>, z>)> which satises

G(w) = 0, µ ≥ 0, z ≥ 0, (10)

since (10) is equivalent to the respective Karush-Khun-Tucker conditions.To derive the path-following algorithm, we replace (10) by the following

perturbed problem:

G(w) = (0>,0>, τe>)>, µ > 0, z > 0, (11)

where τ ∈ R+. Solutions wτ to (11) dene a curve C(τ) in R6nc+np called thecentral path. This curve approaches w when τ tends to zero. We combinethe damped Newton method for solving (11) with an appropriate change ofτ which guarantees that the iterations belong to a neighbourhood N (c1, c2)of C(τ) dened by

N (c1, c2) = w = (λ>,µ>, z>)> ∈ R6nc+np : µizi ≥ c1ϑ, i = 1, . . . , 2nc,

µ ≥ 0, z ≥ 0, ‖∇λL(λ,µ)‖ ≤ c2ϑ, ‖∇µL(λ,µ) + z‖ ≤ c2ϑ,

where c1 ∈ (0, 1], c2 ≥ 0, and ϑ := ϑ(w) = µ>z/(2nc). In the k-th iteration,we modify τ := τ (k) by the product of ϑ(k) = ϑ(w(k)) with the centering pa-rameter c(k) chosen as in [13]. To get the monotonically decreasing sequenceϑ(k), the algorithm uses also the Armijo-type condition (13). By J(w) in(12), we denote the Jacobi matrix of G at w. The bounds on the parametersmentioned in the initialization section follow from the convergence analysispresented in [11].

Algorithm PF: Given c1 ∈ (0, 1], c2 ≥ 1, 0 < cmin ≤ cmax ≤ 1/2, ω ∈ (0, 1),and ε ≥ 0. Let w(0) ∈ N (c1, c2) and set k := 0.

(i). Choose c(k) ∈ [cmin, cmax];

(ii). SolveJ(w(k))∆w(k+1) = −G(w(k)) + (0>,0>, c(k)ϑ(k)e>)>; (12)

(iii). Set w(k+1) = w(k) +α(k)∆w(k+1) with the largest α(k) ∈ (0, 1] satisfyingw(k+1) ∈ N (c1, c2) and

ϑ(k+1) ≤ (1− α(k)ω(1− c(k)))ϑ(k); (13)

(iv). Return w = w(k+1), if err (k) := ‖w(k+1) −w(k)‖/‖w(k+1)‖ ≤ ε, else setk := k + 1 and go to step (i).

6

The computational eciency depends on a method used for solving theinner linear systems (12). The Jacobi matrix is non-symmetric and indenitewith the following block structure:

J(w(k)) =

F J12 0J>12 0 I0 Z M

, J12 =

(−I I

0 0

).

Eliminating the 2nd and 3rd unknown in ∆w(k+1), we get the reduced linearsystem for ∆λ(k+1) with the Schur complement:

JSC = F + M1Z−11 + M2Z

−12 ,

where Z = diag(Z1,Z2) and M = diag(M1,M2). As µ(k) > 0, z(k) > 0,the matrix JSC is symmetric, positive denite and the reduced linear systemcan be solved by the conjugate gradient method. In order to guarantee itsconvergence, we use the preconditioner:

PSC = D + M1Z−11 + M2Z

−12 ,

where D = diag(F). All eigenvalues of the preconditioned matrix P−1SCJSC

belong to an interval which does not depend on the iteration and the spectralcondition number is bounded by (see [11]):

cond(P−1SCJSC) ≤ cond(D)cond(F).

In computations we approximate D so that A−1 in F is replaced by diag(A)−1.The conjugate gradient method in the k-th step of Algorithm PF is initial-ized and terminated adaptively. The initial iteration is taken as the computedresult of the previous iteration and the (inner) iterations are terminated, ifthe relative residuum is less than the stopping tolerance given by

tol (k) = minrtol × err (k−1), cfact × tol (k−1),

where 0 < rtol < 1, 0 < cfact < 1, err (−1) = 1, and tol (−1) = rtol/cfact .

5 Examples of uid slippage ow

In this section we present numerical results of three physically realistic prob-lems [6, 14, 15] governed by the Stokes system with the stick-slip boundaryconditions. To justify this ow model we compare some characteristic quan-tities computed using the solutions of the rst two examples with the ones

7

resulting from the Stokes problem with the Navier boundary condition (1)and also with known analytic solutions. The second and the third examplesare motivated by a ow inside a concentric and eccentric bearing, respec-tively.

All our codes are implemented in Matlab 2013b [17]. The computationswere performed by ANSELM supercomputer at IT4I VB-TU Ostrava. Weuse Algorithm PF with c1 = 0.01, c2 = 109, cmin = 10−12, cmax = 0.5,ω = 0.01, ε = 10−4, rtol = 0.5, cfact = 0.9. It turns out that these values areoptimal as follows from numerical tests in [11].

5.1 Flow between parallel plates

Let Ω = (−0.5, 0.5) × (−0.0075, 0.0075) (in meters). The boundary ∂Ω issplit into the following parts: γD = −0.5×(−0.0075, 0.0075), γN = 0.5×(−0.0075, 0.0075), and γC = γC,1∪ γC,2, γC,1 = (−0.5, 0.5)×0.0075, γC,2 =(−0.5, 0.5) × −0.0075; see Figure 1. Further the liquid density ρ = 998.2[kg·m−3], the kinematical viscosity ν = 1.00481·10−6 [m2·s−1], the dynamicalviscosity η = ρν = 0.001003 [Pa ·s], the volume forces f = 0 [N ·m−3], theinow velocity uD = (0.1, 0) [m ·s−1], and the outow stress σN = 0 [Pa].The computations are carried out with nu = 2000, np = 1025, nc = 80.

Figure 1: Geometry of the channel Ω with diameter d = 0.015 [m] and lengthL = 1 [m].

The outow velocity prole for the Stokes system with the Navier condi-tion (1) and κ = const. > 0 prescribed on the walls of a channel is given by(see [3]):

u1 =∂p

∂x1

(x2

2

2η− r

κ− r2

2η) on γN , (14)

8

where u1 is the rst component of the velocity u. The value of κ [Pa·s·m−1]can be computed from (14):

κ = − ∂p

∂x1

(xc)r

u1(xc), (15)

where xc := γC,1∩γN . To compare the model with the Navier and the stick-slip conditions we introduce a new quantity κnum(xc) which will be computedby (15) but using the solution to (3) this time.

Now we solve the problem (3) for several suciently small constant thresh-olds g chosen in such a way that the slip occurs along the whole γC . Fromthe found solution we compute κs at xc by (2) and compare with κnum at xcfrom (15). Both values practically coincide as seen from Figure 2(a). Notethat if g is constant on the whole γC then due to the symmetry of the settingalso |ut| is constant there and consequently κs is constant. Figure 2(b) showsthat the computed ut(xc) depends linearly on g for small values of g andhence on σt which conrms (1). The main benet of the stick-slip conditionsis automatic switching between the stick/slip mode for g large enough asapparent from Figure 2(b).

g0 0.005 0.01 0.015 0.02 0.025 0.03

κ(x

c)

0

0.2

0.4

0.6

0.8

1

1.2

κs

κnum

g0 0.01 0.02 0.03 0.04 0.05 0.06

|ut(x

c)|

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

(a) (b)

Figure 2: (a) Comparison of κnum and κs at xc; (b) dependence of |ut(xc)|on the slip bound g.

The computed outow velocity proles on γC of solutions to (3) for severalvalues of g are depicted in Figure 3.

Now we solve the Stokes system with the Navier condition (1) prescribedon γC by ANSYS Fluent (based on the nite volume method) for dierentadhesive coecients κ. The computed values of |σt(xc)| and ut(xc) are sum-marized in Table 1. In the next step we solve problem (3) with the constant

9

−8 −6 −4 −2 0 2 4 6 8

x 10−3

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

γN

u 1

κs = ’noslip’ / g=1

κs = 1 / g=0.029

κs = 0.5 / g=0.0225

κs = 0.1 / g=0.008

−8 −6 −4 −2 0 2 4 6 8

x 10−3

−0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

γN

u 1

κs = 0.69 / g

1=0.029

κs = 0.34 / g

1=0.0225

κs = 0.066 / g

1=0.008

κs = 0 / g

1=0

(a) (b)

Figure 3: Outow velocity proles. (a) the same value of g on both parts ofγC ; (b) dierent values of g := g1 on γC,1 and xed value g = 1 on γC,2 leadto the sticking eect on γC,2. The values κs := κs(xc) are computed by (2),and (15).

slip bounds g := |σt(xc)| with |σt(xc)| taken from Table 1. Using the re-spective numerical solutions of (3), the values of κs(xc), κnum(xc) computedby (2), and (15), respectively and ut(xc) are shown in the columns of Table 2.Comparing Table 1 and 2 we see a very good agreement of the results.

κ |σt(xc)| ut(xc)1 0.029822 0.0298220.8 0.027836 0.0347940.5 0.023190 0.0463800.1 0.008203 0.0820280.01 0.000979 0.0979170.001 0.000099 0.099788

Table 1: Stokes problem with the Navier condition (1) for dierent κ (byANSYS Fluent).

There is yet another verication of the computed results. If a couple(u, p) satises the Stokes system (3)1,2 in the channel Ω with the right handside f = 0 together with the impermeability condition (3)5 on γC then dueto the special geometry of Ω , the following identity holds :

10

g κs(xc) by (2) κnum(xc) by (15) ut(xc)0.029822 1.091641 1.097414 0.0273180.027836 0.865347 0.869922 0.0321660.023190 0.533203 0.536026 0.0434920.008203 0.102527 0.103072 0.0800060.000979 0.010030 0.010082 0.0976140.000099 0.001000 0.001006 0.099752

Table 2: Values of κs(xc), κnum(xc) and ut(xc) for dierent g.

pin − pout :=1

d(

∫γD

p dx2 −∫γN

p dx2) =1

d(

∫γC,1

σt dx1 −∫γC,2

σt dx1) (16)

making use of the Green theorem.The left-hand side in (16) represents the hydraulic losses. In particular,

(16) holds for solutions (u, p) to (3) whose right hand side f = 0. Let usobserve that the velocity uD = (uD,1, 0) prescribed on γD does not appearexplicitly but only implicitly through σt and p in (16). Suppose that we solve(3) with uD := u∗D xed and g := g∗, where g∗ is chosen in such a waythat the slip occurs along the whole γC . Then necessarily σt = g∗ on γC,1and σt = −g∗ on γC,2 as follows from the stick-slip boundary conditions.Consequently, the right hand side of (16) depends solely on g∗ . Thereforethe hydraulic losses are the same for any velocity uD used in (3)3 whichguarantees that the respective shear stress σt still satises σt = g∗ on γC,1and σt = −g∗ on γC,2. In Table 3 we check this property for dierent uD,1and xed g∗ = 9.79165 · 10−4 on γC . One can observe the satisfaction of (16)under the level of the terminating tolerance ε = 10−4.

uD,1 κs(xc) by (2) κnum(xc) by (15) (pin − pout)− 2g∗L/d0.01 0.129467 0.129830 5 · 10−6

0.03 0.035521 0.035617 5 · 10−6

0.05 0.020588 0.020646 6 · 10−6

0.07 0.014492 0.014532 5 · 10−6

0.09 0.111820 0.011213 5 · 10−6

Table 3: Verication of (16) for dierent uD,1.

It is evident that the slip bound g (and the adhesive coecient κ) signif-icantly inuences the hydraulic losses.One can expect that wall features willfundamentally aect more sophisticated turbulence models.

11

Figure 4 shows the dependence of the hydraulic losses on the adhesivecoecient κnum which is equal to κs dened by (2) as mentioned above.From the practical point of view, surfaces with κnum < 2 are desired.

0 1 2 3 40

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

κnum

hydr

aulic

loss

es

Figure 4: Dependence of the hydraulic losses on the adhesive coecient κnum.

5.2 Flow initialized by a rotating concentric inner cylin-

der

Let Ω = (x1, x2) ∈ R2 : R21 < x2

1 + x22 < R2

2, where R1 = 0.02 [m]and R2 = 0.0203 [m]. The decomposition of the boundary ∂Ω is as follows:γD = (x1, x2) ∈ R2 : x2

1+x22 = R2

1, γC = (x1, x2) ∈ R2 : x21+x2

2 = R22, and

γN = ∅. The problem (3) is solved for f = 0, η = 0.001003, and uD = ω1R1t,where t is the unit vector tangential to γD and ω1 = 2πfq = 62.8319 [rad·s−1]is the angular velocity of rotating cylinder γD with the frequency fq = 10 [Hz];see Figure 5(a). Assuming γD as perfectly wetted, the nonzero velocity uDsimulates the initialization of owing by a rotation of the concentric innercylinder. We assume the outer cylinder γC partially wettable that is modelledby dierent g (and κ). Our computations are performed with nu = 55826,np = 31203, nc = 3317. A MATLAB nite element mesh generator was usedfor the triangulation of Ω, see [9].

Due to the setting of the problem, the velocity eld u solving the Stokessystem with the Navier condition on γC is rotationally symmetric, i.e. itcan be expressed in the polar coordinates with the radial function u := u(r)

12

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

gκ

κs

κnum

(a) (b)

Figure 5: (a) Rotating cylinder γD and the stick-slip boundary conditions onγC ; (b) comparison of κs and κnum on γC .

which depends only on the radius r. The velocity prole of this ow in Ω isgiven by

u(r) =ω1R

21

R22 +R2

1(2η/(κR2)− 1)

(R2

2

r− r +

2ηr

κR2

), r ∈ 〈R1, R2〉. (17)

The tangential velocity ut(x), x ∈ γC is equal to u(R2), i.e.

u(R2) =ω1R

21

R22 +R2

1(2η/(κR2)− 1)

(2η

κ

). (18)

Let us mention that ut and σt are constant on γC . The constant value of κon γC can be expressed from (18) as

κ = 2ηR21

ω1 − u(R2)/R2

(R22 −R2

1)u(R2). (19)

The presentation of the results has the same structure as in the previousexample. First, problem (3) is computed for small values of g = const. whichguarantee that the slip occurs on γC . Clearly, ut and σt are again constanton γC and so κs is. In Figure 5(b) we compare κs computed by (2) withκnum computed by (19) but using the solution to (3) for dierent values ofg. We see that κs and κnum are the same. The dependence of ut on g,κnum is depicted in Figure 6(a), and (b), respectively. We see that ut is a

13

linear function of g which conrms (1) while the graph of ut as a function ofκnum coincides with the graph of the right hand side of (18) considered as afunction of κ.

0 0.5 1 1.5 2 2.5 3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

g

u t

0 1 2 3 4 5 6 7 8 9

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

κnum

u t

(a) (b)

Figure 6: Dependence of ut on (a) the slip bound g, (b) the adhesive coe-cient κnum.

Then we compute by ANSYS Fluent the Stokes system with the Naviercondition (1) for dierent κ (Table 4-left). The rst row (no slip) of this tableis obtained from the solution with the no slip boundary condition ut = 0 onγC . The resulting |σt| denes the slip bound g which will be used now in (3).Table 4-right now collects the results: κs computed by (2), the tangentialvelocity ut and the tangential velocity ut computed by (18) but using κs inplace of κ.

5.3 Flow initialized by a rotating eccentric inner cylin-

der

The previous two examples were used to compare solutions of the Stokessystem with the Navier boundary condition on one hand and the stick-slipconditions on the other hand. In the next example we chose data to obtainall three modes: complete stick or slip on γC and simultaneous stick on onepart and slip on another part of γC .

We modify the domain from the previous example by shifting the innercylinder. Let Ω = (x1, x2) ∈ R2 : x2

1 + x22 < R2

2 \ ω, where ω = (x1, x2) ∈R2 : (x1 − e)2 + x2

2 < R21 with e = 0.00015 [m], γD = ∂ω and γC = ∂Ω \ γD;

14

Navier by Fluent Stick-slip by MATLABκ |σt| u(R2) g κs by (2) ut ut by (18)

no slip 4.1695 0 4.1695 2.58e4 1.73e-4 1.62e-42 1.5784 0.7892 1.5784 2.0212 0.7926 0.78821 0.9745 0.9751 0.9745 1.0120 0.9773 0.97400.8 0.8180 1.0226 0.8180 0.8099 1.0250 1.02200.5 0.5523 1.1046 0.5523 0.5066 1.1064 1.10440.1 0.1236 1.2362 0.1236 0.1387 1.2375 1.23710.01 0.0127 1.2691 0.0127 0.0101 1.2715 1.27150.001 0.0011 1.2731 0.0011 0.0010 1.2749 1.2750

Table 4: left - Stokes problem with the Navier condition for dierent κ (solvedby ANSYS Fluent); right - the adhesive coecient κs and ut on γC for dif-ferent g (except the rst row) solved by MATLAB.

see Figure 7(a). The remaining data of the problem are the same as inSubsection 5.2. Our computations are performed with nu = 30300, np =17541, nc = 2408. A MATLAB nite element mesh generator was used forthe triangulation of Ω, see [9]. A part of the zoomed mesh is in Figure 7(b).

x1

-0.0204 -0.0203 -0.0202 -0.0201 -0.02 -0.0199 -0.0198 -0.0197

x 2

×10-4

0

1

2

3

4

5

(a) (b)

Figure 7: Example: (a) eccentric rotating cylinder; (b) zoom of the mesh.

First we solve problem (3) with the slip bound g = 0.18094 when theslip occurs on the whole γC . In Figure 8(a),(b) we see the distribution of κscomputed by (2) and ut, respectively along γC .

Next we solve problem (3) with the constant slip bound g = 1.60313 onγC (a part of the circle of radius g is in red in Figure 9(a)). The distribution

15

0 500 1000 1500 2000 25000

0.2

0.4

0.6

0.8

1

1.2

1.4

clockwise from A

κ s

γC

0 500 1000 1500 2000 25000

0.5

1

1.5

2

clockwise from A

u t

γC

(a) (b)

Figure 8: Distribution of: (a) adhesive coecient κs (b) tangential velocityut on γC for g = 0.18094.

of |σt| is in green and γC itself in blue. One can see that there is a small partof γC (on the left) where |σt| is below g and accordingly to (3)7 the tangentialcomponent ut should be equal to zero there. This is conrmed by Figure 9(b)which depicts the distribution of ut on γC .

|σt|

γC

g

clockwise from A0 500 1000 1500 2000 2500

ut

0

0.5

1

1.5

γC

(a) (b)

Figure 9: Distribution of (a) shear stress |σt| (b) tangential velocity ut on γCfor g = 1.60313.

Finally, we solve (3) with g = 50. In this case |σt| is strictly less than gon the whole γC and so ut = 0 on γC , see Figure 10. Let us note that in thiscase the solution is the same as the one to the Stokes system with the no-slipboundary condition u = 0 on γC .

One can observe the reverse ow in a part of the domain which is aconsequence of the continuity equation for incompressible ows, see Fig-

16

ure 11(g = 50). This eect is caused by a relatively high velocity gradient inthe wider part of the domain.

g

|σt|

γC

Figure 10: The distribution of |σt| on γC for g = 50.

Velocity elds in the part of Ω highlighted in Figure 7(a) and the dis-tribution of the pressure on γC are shown in Figure 11 and 12, respectively.

x1

-0.0204 -0.0203 -0.0202 -0.0201 -0.02 -0.0199 -0.0198 -0.0197

x 2

×10-4

0

1

2

3

4

5

6

7

γC γ

D

x1

-0.0204 -0.0203 -0.0202 -0.0201 -0.02 -0.0199 -0.0198 -0.0197

x 2

×10-4

0

1

2

3

4

5

6

7

γD

γC

x1

-0.0204 -0.0203 -0.0202 -0.0201 -0.02 -0.0199 -0.0198 -0.0197

x 2

×10-4

0

1

2

3

4

5

6

7

γD

γC

g = 0.18094 g = 1.60313 g = 50

Figure 11: Zoomed velocity elds u.

17

clockwise from A0 500 1000 1500 2000 2500

p

0

200

400

600

800

1000

1200

γC clockwise from A

0 500 1000 1500 2000 2500

p

0

100

200

300

400

500

600

700

800

γC clockwise from A

0 500 1000 1500 2000 2500

p

0

500

1000

1500

2000

2500

γC

g = 0.18094 g = 1.60313 g = 50

Figure 12: Distribution of the pressure p on γC .

6 Conclusions and future plans

The paper is devoted to numerical solution of the Stokes system with stick-slip boundary conditions prescribed on a part of the boundary. Unlike theclassical Navier condition, this time the slip may occur only when the shearstress attains a threshold g given à-priori. The resulting numerical modelleads to minimization of a strictly convex quadratic function subject to asmall number of simple constraints. The minimization itself was carriedout by the interior point method. In order to validate this ow model, theobtained numerical results of two model examples for small values of g arecompared with the ones computed by ANSYS Fluent and also with the knownanalytic solutions. The results are practically identical.

As the next step of our research we plan to extend the stick-slip boundaryconditions to the case when the threshold g depends on the solution itself(this seems to be more practical for the real hydrophobic materials) , moreprecisely g := g(|ut|) and to the stick-slip conditions of the Coulomb type.

Acknowledgement

This work has been supported by the IT4IXS - IT4Innovations Excellence inScience project (LQ1602) (JH,RK,VS). Grant Agency of the Czech Republicwithin the project and GA101/15-06621S is gratefully acknowledged for thesupport of this work (SF,FP).

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