Contact Problem For Thin Biphasic Cartilage Layers_ Perturbation Solution

CONTACT PROBLEM FOR THIN BIPHASICCARTILAGE LAYERS: PERTURBATION SOLUTION

by I. I. ARGATOV and G. S. MISHURIS†

(Institute of Mathematics and Physics, Aberystwyth University,Ceredigion SY23 3BZ)

[Received 12 November 2010. Revised 28 April 2011. Accepted 19 May 2011]

Summary

A three-dimensional unilateral contact problem for articular cartilage layers is considered inthe framework of the biphasic cartilage model. The articular cartilages bonded to subchondralbones are modelled as biphasic materials consisting of a solid phase and a fluid phase. It isassumed that the subchondral bones are rigid and shaped close to an elliptic paraboloid. Theobtained analytical solution is valid over long time periods and can be used for increasingloading conditions.

1. Introduction

Biomechanical contact problems involving transmission of forces across biological joints are ofconsiderable practical importance in surgery. Many solutions to the axisymmetric problem of con-tact interaction of articular cartilage surfaces in joints are available. Ateshian et al. (1) obtained anasymptotic solution for the contact problem of two identical biphasic cartilage layers attached totwo rigid impermeable spherical bones of equal radii modelled as elliptic paraboloids. Wu et al. (2)extended this solution to a more general model by combining the assumption of the kinetic rela-tionship from classical contact mechanics (3) with the joint contact model for the contact of twobiphasic cartilage (1). An improved solution for the contact of two biphasic cartilage layers whichcan be used for dynamic loading was obtained by Wu et al. (4, 5). These solutions have been widelyused as theoretical background in modelling the articular contact mechanics. Recently, the analysisdeveloped in (2) was extended (6, 7) by formulating the refined contact condition which takes intoaccount the tangential displacements at the contact region.

When studying contact problems for real joint geometries, a numerical analysis, such as the finiteelement method (FEM), is necessary (8) since exact analytical solutions can be only obtained fora simple enough problem configurations. Moreover, FEM allows the investigation of much morecomplicated regimes and material properties (9) where obtaining any analytical solution is hope-less. Such analytical solutions have been obtained for two-dimensional (10), or axisymmetric andsimple geometries (11 to 13). On the other hand, analytical solutions for more complex geometriesconsidered as benchmarks are of extreme importance for researchers as they allow verifying numeri-cal computations made using commercial software that involves various complicated biomechanicalmodels. Moreover, exact formulae allow also analysing various trends in solution behaviours and

†⟨[email protected]⟩Q. Jl Mech. Appl. Math, Vol. 64. No. 3 c⃝ The author 2011. Published by Oxford University Press;

all rights reserved. For Permissions, please email: [email protected] Access publication 28 June 2011. doi:10.1093/qjmam/hbr008

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298 I. I. ARGATOV AND G. S. MISHURIS

determine its sensitivity with respect to material and geometrical parameters without heavy andtime-consuming numerical computations. These arguments show a strong need in evaluation of anexact solutions for the biomechanical problem exhibiting more complex geometry. Such solutionhas been recently obtained (14, 15) for the contact problem for biphasic cartilage layers attached torigid bones shaped like elliptic paraboloids.

In this study, this solution is generalized for a more general three-dimensional (3D) contact. Themethod developed in (16) is used to obtain general relationships between the integral characteristicsof the contact problem. The exact closed-form solution of the elliptic contact problem for biphasiccartilage layers from (14, 15) is used as the basis for constructing the perturbation solution in thegeneral case. In dealing with the perturbed unilateral contact problem, we employ a previouslydeveloped asymptotic technique (17 to 19).

2. Formulation of the contact problem

We consider a frictionless contact between two thin linear biphasic cartilage layers firmly attached torigid bones shaped like elliptic paraboloids. Introducing the Cartesian coordinate system (x1, x2, z),we write the equations of the two cartilage surfaces (n = 1, 2) in the form z = (−1)n8(n)

ε (x), wherex = (x1, x2) (see Fig. 1). We assume that the two cartilage-bone systems occupy convex domainsz 6 −8(1)

ε (x) and z > 8(2)ε (x), whereas in the undeformed state, they are in contact with the plane

z = 0 at a single point chosen as the coordinate origin. In the particular case of bones shaped likeslightly irregular elliptic paraboloids, we have

8(n)ε (x) = x2

1

2R(n)1

+ x22

2R(n)2

+ εφn(x), (2.1)

where R(n)1 and R(n)

2 are the curvature radii of the nth bone middle surface at its apex, ε is a smallpositive dimensionless parameter and the function εφn(x) describes the deviation of the nth bonesurface from the elliptic paraboloid shape (n = 1, 2).

Fig. 1 Schematic diagram of the contact of articular cartilage surfaces 1 and 2 under load F(t). The dashedlines denote the surfaces’ profiles in the undeformed state; w1(x, t) and w2(x, t) are local deformations of thecontacting surfaces 1 and 2, respectively; δε(t) is the relative approach between the subchondral bones coveredwith cartilages. The difference between the sum w1(x, t) + w2(x, t) and δε(t) is due to the initial gap betweenthe cartilage surfaces

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CONTACT PROBLEM FOR THIN BIPHASIC CARTILAGE LAYERS 299

We denote the vertical approach of the bones by −δε(t). Then, the linearized unilateral contactcondition that the boundary points of the cartilage layers do not penetrate one into another can bewritten as follows:

δε(t) − w1(x, t) − w2(x, t) 6 8(1)ε (x) + 8(2)

ε (x). (2.2)

An asymptotic solution obtained in (1) for the vertical displacement of the boundary points ofa biphasic cartilage layer, wn(x, t), in the axisymmetric problem of acting contact pressure on itssurface can be generalized for the 3D case as follows:

wn(x, t) = h3n

3µsn

{1x Pε(x, t) + 3µsnkn

h2n

∫ t

01x Pε(x, τ )dτ

}. (2.3)

Here, µsn is the shear modulus of the solid phase of the cartilage tissue (n = 1, 2), h1 and h2 are thethicknesses of the cartilage layers, k1 and k2 are the cartilage permeabilities, Pε(x, t) is the contactpressure, 1x = ∂2/∂x2

1 + ∂2/∂x22 is the Laplace operator.

The equality in relation (2.2) determines the contact region ωε(t). In other words, the followingequation holds within the contact area:

w1(x, t) + w2(x, t) = δε(t) − 8ε(x), x ∈ ωε(t). (2.4)

Here, we introduced the notation 8ε(x) = 8(1)ε (x) + 8(2)

ε (x). In the case (2.1), we have

8ε(x) = 80(x) + εφ(x), (2.5)

where φ(x) = φ1(x) + φ2(x),

80(x) = x21

2R1+ x2

22R2

, (2.6)

and the parameters R1 and R2 are determined by the formulas

1R1

= 1

R(1)1

+ 1

R(2)1

,1R2

= 1

R(1)2

+ 1

R(2)2

.

Substituting the expressions for the displacements w1(x, t) and w2(x, t) given by (2.3) into (2.4),we obtain the contact condition in the following form (we assume that x ∈ ωε(t)):

1x Pε(x, t) + χ

∫ t

01x Pε(x, τ )dτ = m(8ε(x) − δε(t)). (2.7)

Here, we introduced the notation

χ = 3µs1k1

h21

+ 3µs2k2

h22

, m =(

h31

3µs1+ h3

23µs2

)−1

. (2.8)

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Equation (2.7) will be used to find the contact pressure density Pε(x, t). The contour Ɣε(t) of thecontact area ωε(t) is determined from the condition that the contact pressure is positive and vanishesat the contour of the contact area:

Pε(x, t) > 0, x ∈ ωε(t); Pε(x, t) = 0, x ∈ Ɣε(t). (2.9)

Moreover, in the case of contact problem for a biphasic cartilage layer in which the contact pressureis carried primarily by the fluid phase, it is additionally assumed a smooth transition of the surfacenormal stresses from the contact region x ∈ ωε(t) to the outside region x ∈ ωε(t) (1). Thus, weimpose the following boundary condition:

∂ Pε

∂n(x, t) = 0, x ∈ Ɣε(t). (2.10)

Here, ∂/∂n is the normal derivative directed outward from ωε(t). Note that for determining theinitial contact radius in the case of axisymmetric contact, a new approach has been suggested in (20)which does not make use of the zero pressure gradient condition (2.10) and is based on the analyticalsolutions of the axisymmetric contact problem for an incompressible single-phasic elastic layer (21).While the contact conditions (2.9) are widely used for describing unilateral contact between elasticbodies, the pressure gradient boundary condition (2.10) appears rarely (e.g. in contact between thinincompressible elastic layers (22)). In (23), an FEM analysis of the contact interaction betweentwo biphasic layers has been performed which shows that the boundary condition (2.10) is quitereasonable.

We assume that the density Pε(x, t) is defined on the entire plane such that

Pε(x, t) = 0, x ∈ ωε(t). (2.11)

Finally, from the physical point of view, the contact pressure under a blunt punch with a smoothsurface should satisfy the regularity condition, i.e. in the case (2.5), the function Pε(x, t) is assumedto be analytical in the domain ωε(t).

The equilibrium equation for the whole system is∫∫ωε(t)

Pε(x, t)dx = F(t), (2.12)

where F(t) denotes the external load.For non-decreasing loads when d F(t)/dt > 0, the contact zone should increase. Thus, we assume

that the following monotonicity condition holds:

ωε(t1) ⊂ ωε(t2), t1 6 t2. (2.13)

The aim of this study is to derive an asymptotic solution for the 3D contact problem for biphasiccartilage layers formulated in (2.7) under the monotonicity condition (2.13). Notice that in theparticular case of axisymmetric contact zone, the limit contact problem under consideration (ε = 0)coincides with that studied in detail in (1, 4).

3. Solution to the problem: general relationships

Here, we consider the problem in its general formulation transforming the problem to a set ofequations more suitable for further analysis.

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3.1 Equation for the displacement parameter δε(t)

Integrating (2.7) over the contact domain ωε(t), we get∫∫ωε(t)

1Pε(y, t)dy + χ

∫∫ωε(t)

∫ t

01Pε(y, τ )dy dτ = m

∫∫ωε(t)

(8ε(y) − δε(t))dy. (3.1)

Here, we used the notation y = (y1, y2) and dy = dy1 dy2.In view of (2.11) and (2.13), we have ωε(τ ) ⊂ ωε(t) and Pε(y, τ ) ≡ 0 for y ∈ ωε(τ ). Therefore,

the second integral on the left-hand side of (3.1) takes the form∫∫ωε(t)

∫ t

01Pε(y, τ ) dy dτ =

∫ t

0

∫∫ωε(τ )

1Pε(y, τ )dy dτ. (3.2)

Note that the density Pε(x, t) is a smooth function of x1 and x2 on the entire plane.Using the second Green’s formula∫∫

ω

(u(y)1v(y) − v(y)1u(y))dy =∫

Ɣ

(u(y)

∂v

∂n(y) − v(y)

∂u∂n

(y)

)ds, (3.3)

where ds is the element of the arc length, we obtain∫∫ωε(τ )

1Pε(y, τ )dy =∫

Ɣε(τ )

∂ Pε

∂n(y, τ )ds. (3.4)

Thus, taking into account (3.2) and (3.4), we rewrite (3.1) as follows:∫Ɣε(t)

∂ Pε

∂n(y, t)ds + χ

∫ t

0

∫Ɣε(τ )

∂ Pε

∂n(y, τ )ds dτ = m

∫∫ωε(t)

8ε(y)dy − m Aε(t)δε(t). (3.5)

Here, Aε(t) is the area of ωε(t) given by the integral

Aε(t) =∫∫

ωε(t)dy. (3.6)

Finally, in view of the boundary condition (2.10), from (3.5) it follows that

δε(t) = 1Aε(t)

∫∫ωε(t)

8ε(y)dy. (3.7)

Equation (3.7) connects the unknown displacement parameter δε(t) with the integral characteristicof the contact domain ωε(t). In the case of the axisymmetric problem (24), it coincides with theresults obtained in (1, 4).

3.2 Equation for the integral characteristics of the contact domain

Substituting the functions u(x) = Pε(x, t) and v(x) = (x21 + x2

2)/4 into Green’s formula (3.3) andtaking into account the boundary conditions (2.9) and (2.10), we obtain the relation

14

∫∫ωε(t)

|y|21Pε(y, t)dy =∫∫

ωε(t)Pε(y, t)dy. (3.8)

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Using (3.8), we can evaluate the contact load (2.12). Indeed, multiplying the both sides of (2.7)by (x2

1 + x22)/4 and then integrating over the contact domain ωε(t), we obtain∫∫

ω(t)Pε(y, t)dy + χ

∫ t

0

∫∫ωε(τ )

Pε(y, τ )dy dτ = m4

∫∫ωε(t)

|y|2(8ε(y) − δε(t))dy. (3.9)

Taking the notation (2.12) into account, we rewrite (3.9) as follows:

F(t) + χ

∫ t

0F(τ )dτ = m

4

∫∫ωε(t)

|y|28ε(y)dy − δε(t)m4

∫∫ωε(t)

|y|2 dy. (3.10)

Excluding δε(t) from (3.10) by means of (3.7), we obtain

F(t) + χ

∫ t

0F(τ )dτ = m

4

∫∫ωε(t)

(|y|2 − Jε(t)

Aε(t)

)8ε(y)dy. (3.11)

Here, Jε(t) is the polar moment of inertia of ωε(t) given by the integral

Jε(t) =∫∫

ωε(t)|y|2 dy. (3.12)

Equation (3.11) connects the integral characteristic of the unknown contact domain ωε(t) andthe known contact load F(t). In the case of the axisymmetric problem, it coincides with the resultsobtained in (4).

3.3 Equation for the pressure

Let us rewrite (2.7) in the form

1pε(x, t) = m(8ε(x) − δε(t)), x ∈ ωε(t), (3.13)

where we introduced the notation

pε(x, t) = KPε(x, t) (3.14)

and we have defined the operator on the right-hand side of (3.14) by

Ky(t) = y(t) + χ

∫ t

0y(τ )dτ. (3.15)

The inverse operator to K denoted by K−1 is defined by the formula

K−1Y (t) = Y (t) − χ

∫ t

0Y (τ )e−χ(t−τ)dτ. (3.16)

Finally, in view of the boundary conditions (2.9) and (2.10), the function p(x, t) must satisfy thefollowing boundary conditions:

pε(x, t) = 0, x ∈ Ɣε(t), (3.17)

∂pε

∂n(x, t) = 0, x ∈ Ɣε(t). (3.18)

Thus, (3.11) allows us to find a domain ωε(t). Then, from (3.6) and (3.7), one can determine thedisplacement parameter δε(t). Finally, solving the problem (3.13)–(3.18), we obtain the completesolution to the problem. However, this is a complicated nonlinear problem in the general case.

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Section 4 will deal with the problem for the limiting case ε = 0 when the function 80(x) representsan ellipse.

4. Limiting case problem ε = 0

4.1 Elliptical contact domain

The solution to this problem was first presented in (14) with more details in (15). Here, we adopt itin the form necessary for use in the further asymptotic analysis to construct a more general solutionto the problem with the slightly perturbed boundary of an arbitrary shape.

In this case, the right-hand side of (3.13) takes the form m(80(x)−δ0(t)). This suggests assumingthat the domain ω0(t) is elliptical and so we put

p0(x, t) = q0(t)

(1 − x2

1a2(t)

− x22

b2(t)

)2

. (4.1)

In other words, the contour Ɣ0(t) is an ellipse with the semi-axes a(t) and b(t). It is not hard tocheck that p0 satisfies the boundary conditions (3.17) and (3.18) exactly.

Substituting (4.1) into (3.13), we obtain after some algebra the following system of algebraicequations:

δ0(t) = 4q0(t)m

(1

a2(t)+ 1

b2(t)

), (4.2)

12R1

= 4q0(t)ma2(t)

(3

a2(t)+ 1

b2(t)

),

12R2

= 4q0(t)mb2(t)

(1

a2(t)+ 3

b2(t)

). (4.3)

The form of the ellipse Ɣ0(t) can be characterized by its aspect ratio s defined as follows:

s = b(t)/a(t). (4.4)

From (4.3), it immediately follows that

R2

R1= s2(3s2 + 1)

3 + s2 . (4.5)

Equation (4.5) can be reduced to a quadratic equation for s2. In this way, one can obtain

s2 =√(

R1 − R2

6R1

)2

+ R2

R1− (R1 − R2)

6R1. (4.6)

Further, (3.7) takes the form

δ0(t) = 18

(1R1

+ s2

R2

)a2(t), (4.7)

Excluding the quantity δ0(t) from (4.2) and (4.7), we obtain

q0(t) = m32

s2

(s2 + 1)

(1R1

+ s2

R2

)a4(t). (4.8)

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Finally, (3.11) becomes

KF(t) = mπ

384

(3s − s3

R1+ 3s5 − s3

R2

)a6(t). (4.9)

This allows us to determine the major semi-axis a(t) of the contact domain ω0(t):

a(t) =[

mπ

384

(3s − s3

R1+ 3s5 − s3

R2

)]−1/6

(KF(t))1/6. (4.10)

Now, (4.7) and (4.8) determine δ0(t) and q0(t), respectively. Again, in the case of the axisym-metric problem s = 1, (4.10) coincides with the corresponding result obtained in (4).

4.2 Pressure in case of elliptical contact

In view of (3.15) and (4.1), we obtain the following operator equation for the contact pressuredensity P0(x, t):

KP0(x, t) = q0(t)

(1 − x2

1a2(t)

− x22

b2(t)

)2

, x ∈ ω0(t). (4.11)

Taking the relation (3.15), a solution of (4.11) can be represented as

P0(x, t) = K−1

(

1 − x21

a2(t)− x2

2b2(t)

)2

q0(t)H

(1 − x2

1a2(τ )

− x22

b2(τ )

) , (4.12)

where H(x) is the Heaviside step function: H(x) = 1 for x > 0 and H(x) = 0 for x 6 0.In view of the notation (3.16), this can be written as follows:

P0(x, t) =(

1 − x21

a2(t)− x2

2b2(t)

)2

q0(t)

− χ

∫ t

0

(1 − x2

1a2(τ )

− x22

b2(τ )

)2

H

(1 − x2

1a2(τ )

− x22

b2(τ )

)q0(τ )e−χ(t−τ) dτ. (4.13)

It is clear that if the point belongs to the initial contact zone x ∈ ω0(0), i.e.

1 − x21

a2(0)− x2

2b2(0)

> 0,

then (4.13) simplifies to

P0(x, t) =(

1 − x21

a2(t)− x2

2b2(t)

)2

q0(t) − χ

∫ t

0

(1 − x2

1a2(τ )

− x22

b2(τ )

)2

q0(τ )e−χ(t−τ)dτ.

(4.14)

If the point x lies outside of the initial contact zone x /∈ ω0(0), i.e.

1 − x21

a2(0)− x2

2b2(0)

6 0,

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then (4.13) can be rewritten as

P0(x, t) =(

1 − x21

a2(t)− x2

2b2(t)

)2

q0(t) − χ

∫ t

t∗(x)

(1 − x2

1a2(τ )

− x22

b2(τ )

)2

q0(τ )e−χ(t−τ)dτ,

(4.15)

where t∗(x) is the time when the contour of the contact zone first reaches the point x. The quantityt∗(x) is determined by the equation a2(t∗) = x2

1 + (x2/s)2 or in accordance with (4.7) by thefollowing one:

F(t∗) + χ

∫ t∗

0F(τ ) dτ = mπ

384

(3s − s3

R1+ 3s5 − s3

R2

)(x2

1 + x22

s2

)3

. (4.16)

In the case of a stepwise loading, we have F(t) = F0, and (4.16) admits the following closed-form solution:

t∗(x) = mπ

384χ F0

(3s − s3

R1+ 3s5 − s3

R2

)(x21 + x2

2s2

)3

− a6(0)

, (4.17)

where a(0) is the initial value of the major semi-axis of contact domain, given by

a6(0) = 384mπ

(3s − s3

R1+ 3s5 − s3

R2

)−1

F0. (4.18)

Finally, using the Heaviside function and (4.18), we can rewrite (4.17) as

t∗(x) = 1χ

((x2

1 + s−2x22)3

a6(0)− 1

)H((x2

1 + s−2x22)3 − a6(0)). (4.19)

Thus, in the case of a stepwise loading, formula (4.15), where quantity t∗(x) is determined by(4.19), represents the sought for solution of (2.7) in the case of the gap between the contactingsurfaces shaped as the elliptic paraboloid (2.5). Note that in the case of the axisymmetric problem,the derived expression for the contact pressure coincides with the result obtained previously in (7).

5. Slightly perturbed contact problem

5.1 Perturbation of the contact domain

Now, we consider the punch defined by the function 8ε given in a general form (2.5) with smallε > 0. The solution corresponding to the limiting case ε = 0 was defined above and denotedby p0(x, t) and δ0(t) with the contact domain ω0(t) bounded by an ellipse Ɣ0(t). Moreover, thefollowing relations hold true:

1p0(x, t) = m(80(x) − δ0(t)), x ∈ ω0(t), (5.1)

p0(x, t) = 0,∂p0

∂n(x, t) = 0, x ∈ Ɣ0(t). (5.2)

In accordance with (3.7), the displacement parameter δ0(t) is given by

δ0(t) = 1A0(t)

∫∫ω0(t)

80(x) dx, (5.3)

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where A0(t) is the area of ω0(t). Moreover, in view of (3.11), we will have

KF(t) = m4

∫∫ω0(t)

A0(x, t)80(x) dx, (5.4)

where, with J0(t) being the polar moment of inertia of ω0(t),

A0(x, t) = |x|2 − J0(t)A0(t)

. (5.5)

Now, we represent the solution to the perturbed auxiliary contact problem (3.13), (3.17), (3.18)as follows:

pε(x, t) = p0(x, t) + εp1(x, t) + O(ε2), (5.6)

δε(t) = δ0(t) + εδ1(t) + O(ε2). (5.7)

We assume that the contact load F(t) is specified, while the contact approach δε(t) is unknowna priori.

Substituting the expansions (5.6) and (5.7) into (3.13), we arrive at the equation

1p1(x, t) = m(φ(x) − δ1(t)

), x ∈ ω0(t). (5.8)

Let us assume that the unknown boundary Ɣε(t) of the contact domain ωε(t) (see Fig. 2) isdescribed by the equation

n = hε(σ, t), s ∈ Ɣ0(t). (5.9)

Here, σ is the arc length along Ɣ0(t), n is the (signed) distance measured along the outward (withrespect to the domain ω0(t)) normal to the curve Ɣ0(t). The function hε(σ, t) should be determinedby considering the boundary conditions for (5.8).

In view of (5.6) and (5.7), we put

hε(σ, t) = εh(σ, t), (5.10)

Fig. 2 Schematic representation of the contact domain ωε(t) with the boundary Ɣε(t) and the limit domainω0(t) with the boundary Ɣ0(t)

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where the function h(σ, t) does not depend on ε.Applying the perturbation technique (see, e.g. (26)), we will have

pε|Ɣ = pε|Ɣ0 + hε

∂pε

∂n

∣∣∣∣Ɣ0

+ O(ε2). (5.11)

Further, taking into account the formula

n = (1 − κ(σ, t)hε(σ, t))n0 − h′ε(σ, t)t0√

(1 − κ(σ, t)hε(σ, t))2 + h′2ε (σ, t)

, (5.12)

where n = n(ε) is the unit outward normal vector to the curve Ɣε(t), while t0 and n0 are the unittangential and outward normal vectors to the curve Ɣ0(t), κ(σ, t) is the curvature of Ɣ0(t), and theprime denotes the derivative with respect to σ , we obtain

∂pε

∂n

∣∣∣∣Ɣ

= ∂pε

∂n

∣∣∣∣Ɣ0

− h′ε

∂pε

∂σ

∣∣∣∣Ɣ0

+ O(ε2). (5.13)

Thus, substituting the expansion (5.6) into (5.11), (5.13) and taking into account the boundaryconditions (5.2) for the function p0(x, t), we derive the following boundary conditions at the unper-turbed contact boundary:

p1(x, t) = 0, x ∈ Ɣ0(t), (5.14)

∂p1

∂n(x, t) = −h(σ, t)

∂2 p0

∂n2 (x, t), x ∈ Ɣ0(t). (5.15)

Here, in accordance with (5.1), we have

∂2 p0

∂n2 (x, t) = m(80(x) − δ0(t)), x ∈ Ɣ0(t). (5.16)

Note that the right-hand side of (5.16) is strictly positive. This can be checked by using the explicitformula obtained in section 4 or proved using the maximum principle for harmonic functions.

Finally, (3.7) and (3.11) yield

A0(t)δ1(t) =∫∫

ω0(t)φ(x)dx +

∫Ɣ0(t)

80(x)h(σ, t) dsx − A1(t)δ0(t) (5.17)

and

0 =∫∫

ω0(t)A0(x, t)φ(x)dx − δ0(t)

(J1(t) − J0(t)

A0(t)A1(t)

)

+∫

Ɣ0(t)A0(x, t)80(x)h(σ, t)dσx . (5.18)

Here, A1(t) and J1(t) are the first-order perturbation coefficients of Aε(t) and Jε(t), respectively,given by

A1(t) =∫

Ɣ0(t)h(σ, t)dσx , (5.19)

J1(t) =∫

Ɣ0(t)|x|2h(σ, t)dσx . (5.20)

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Equations (5.8), (5.14), (5.15), (5.17) and (5.18) constitute the first-order perturbation problem.By using (5.16) and (5.19), it is not hard to check that (5.17) coincides with the solvability condition(see (3.4)) of the boundary-value problem (5.8), (5.14) and (5.15).

5.2 Determination of the boundary of the contact domain

First, let us represent the solution to (5.8) in the form

p1(x, t) = m(

p(0)1 (x, t) + p(1)

1 (x, t)), (5.21)

where

p(0)1 (x, t) = Y (0)

φ (x) − δ1(t)Y(0)1 (x), (5.22)

and

Y (0)φ (x) = 1

2π

∫∫ω0(t)

φ(y) ln |x − y| dy, Y (0)1 (x) = 1

2π

∫∫ω0(t)

ln |x − y| dy. (5.23)

Substituting the representation (5.21) into (5.8), (5.14) and (5.15), we obtain the followingboundary-value problem for the function p(1)

1 (x, t)):

1p(1)1 (x, t) = 0, x ∈ ω0(t), (5.24)

p(1)1 (x, t) = −Y (0)

φ (x) + δ1(t)Y(0)1 (x), x ∈ Ɣ0(t), (5.25)

∂p(1)1

∂n(x, t) = −h(σ, t)(80(x) − δ0(t)) − ∂Y (0)

φ

∂n(x, t) + δ1(t)

∂Y (0)1

∂n(x, t), x ∈ Ɣ0(t). (5.26)

Here, we have used (5.16) and (5.22).Let us denote the first term on the right-hand side of (5.26) by −h(σ, t) such that

h(σ, t) = h(σ, t)80(x) − δ0(t)

, x ∈ Ɣ0(t). (5.27)

Recall that the denominator in (5.27) is non-zero.In view of (5.19), (5.20) and (5.27), (5.17), (5.18) and (5.26), respectively, take the form

δ1(t) = 1A0(t)

∫∫ω0(t)

φ(x)dx + 1A0(t)

∫Ɣ0(t)

h(σ, t)dσx , (5.28)

0 =∫∫

ω0(t)A0(x, t)φ(x)dx +

∫Ɣ0(t)

A0(x, t)h(σ, t)dσx . (5.29)

∂p(1)1

∂n(x, t) = −h(σ, t) − ∂Y (0)

φ

∂n(x, t) + δ1(t)

∂Y (0)1

∂n(x, t), x ∈ Ɣ0(t). (5.30)

Second, we define the Steklov–Poincare operator S: H1/2(Ɣ0(t)) → H−1/2(Ɣ0(t)) by

(Sg)(x) = ∂w

∂n(x), x ∈ Ɣ0(t), (5.31)

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where w(x) is the unique solution of the Dirichlet problem

1w(x) = 0, x ∈ ω0(t); w(x) = g(x), x ∈ Ɣ0(t). (5.32)

The operator S for a circular domain is well known. We use this representation later in section 6.To the authors’ best knowledge, there is no closed-form representation for S for an elliptic domain.However, the finite element Steklov–Poincare operator can be computed by standard FEM pack-ages (see, for details (25)). In Appendix A, we present an alternative approach for constructing theoperator numerically in terms of conformal mappings.

In terms of the Steklov–Poincare operator, (5.25) and (5.30) are equivalent to

h(σ, t) = (SY (0)φ )(x) − ∂Y (0)

φ

∂n(x, t) − δ1(t)

((SY (0)

1 )(x) − ∂Y (0)1

∂n(x, t)

), x ∈ Ɣ0(t). (5.33)

Note that the substitution of (5.33) into (5.28) results in an identity. In order to verify this, weneed the following easily checked properties:∫

Ɣ0(t)(Sg)(x)dσx = 0, ∀ g ∈ H1/2(Ɣ0(t)),

∫∫ω0(t)

φ(x)dx =∫

Ɣ0(t)

∂Y (0)φ

∂n(x, t)dσx , ∀ φ ∈ L2(ω0(t)).

Third, excluding the variable δ1(t) from (5.33) by means of (5.28), we obtain

h(σ, t) = −(

1A0(t)

∫∫ω0(t)

φ(x) dx + H0(t))(

(SY (0)1 )(x) − ∂Y (0)

1∂n

(x, t)

)

+ (SY (0)φ )(x) − ∂Y (0)

φ

∂n(x, t), x ∈ Ɣ0(t). (5.34)

Here, H0(t) is the relative weighted increment of the contact area defined as

H0(t) = 1A0(t)

∫Ɣ0(t)

h(σ, t)dσx . (5.35)

Thus, up to now, the function h(σ, t) is determined by (5.34) with an accuracy to its integral char-acteristics H0(t). Finally, substituting (5.34) into (5.29), we arrive at the following simple equationto determine H0(t):

4(0)1 (t)H0(t) = 4(0)

φ (t) +∫∫

ω0(t)

(A0(x, t) − 4

(0)1 (t)

A0(t)

)φ(x)dx. (5.36)

Here, both functions 4(0)1 (t) and 4(0)

φ (t) are determined by

4(0)φ;1(t) =

∫Ɣ0(t)

A0(x, t)

((SY (0)

φ;1)(x) − ∂Y (0)φ;1

∂n(x, t)

)dx. (5.37)

It is clear that the solvability of (5.34) depends crucially on the property of having fixed sign by4

(0)1 (t). Let us prove that 4

(0)1 (t) < 0.

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Denoting by y(0)1 (x, t) the unique solution to the Dirichlet problem

1y(0)1 (x, t) = 0, x ∈ ω0(t); y(0)

1 (x, t) = Y (0)1 (x, t), x ∈ Ɣ0(t).

Equivalently, this means that

(SY (0)1 )(x) = ∂y(0)

1∂n

(x), x ∈ Ɣ0(t).

Then, we can rewrite (5.37) in form as follows:

4(0)1 (t) =

∫Ɣ0(t)

A0(x, t)∂

∂n

(y(0)

1 (x, t) − Y (0)φ (x, t)

)dx. (5.38)

Now, using y(0)1 (x, t) − Y (0)

φ (x, t) = 0 for x ∈ Ɣ0(t), and 1(y(0)1 (x, t) − Y (0)

φ (x, t)) = −1 forx ∈ ω0(t), we transform the right-hand side of (5.38) by means of (3.3) as follows:

4(0)1 (t) = −

∫∫ω0(t)

A0(x, t)dx − 4∫∫

ω0(t)(y(0)

1 (x, t) − Y (0)φ (x, t))dx. (5.39)

By the definition of A0(x, t) (see (5.5)), the first integral on the right-hand side of (5.39) is 0. Thesecond integral can be transformed by the first Green’s formula. As a result,

4(0)1 (t) = −4

∫∫ω0(t)

|∇(y(0)1 (x, t) − Y (0)

φ (x, t))|2 dx. (5.40)

This proves that (5.36) is uniquely solvable.Thus, determining H0(t) by dividing both sides of (5.36) by 4

(0)1 (t) and substituting the obtained

result into (5.34), we find h(σ, t) and, consequently, in view of (5.27), uniquely determine h(σ, t),which describes the variation of the contact domain ω0(t).

5.3 Asymptotics of the contact pressure

In accordance with (3.14), the contact pressure is given by

Pε(x, t) = K−1(pε(x, t)Iωε(t)(x)), (5.41)

where K−1 is defined by (3.16) and Iωε(t)(x) is the indicator function of ωε(t): Iωε(t)(x) = 1 ifx ∈ ωε(t) and Iωε(t)(x) = 0 if x ∈ ωε(t).

In the interior of the contact domain ωε(t), (5.6) and (5.41) yield the asymptotic representation

Pε(x, t) = P0(x, t) + εP1(x, t) + O(ε2), (5.42)

where

Pn(x, t) = K−1(pn(x, t)Iω0(t)(x)), n = 0, 1. (5.43)

In the boundary layer region near the contour Ɣε(t), the so-called outer asymptotic representa-tion (5.42) does not work, and the inner asymptotic representation should be constructed. Here,we employ the terminology from the method of matched asymptotic expansions (26). The innerasymptotic representation

Pε(x, t) = ε2K−1P(s, ν, t) + O(ε3) (5.44)

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will be constructed making use of the stretched variable

ν = ε−1n. (5.45)

In view of (5.21)–(5.23), p1(x, t) will be determined completely as soon as we know p(1)1 (x, t)

which satisfies the following Dirichlet problem (see (5.24) and (5.25)):

1p(1)1 (x, t) = 0, x ∈ ω0(t); p(1)

1 (x, t) = g(0)φ (x, t), x ∈ Ɣ0(t). (5.46)

In (5.46), we introduced the notation (see (5.28), (5.35) and (5.36))

g(0)φ (x, t) = −Y (0)

φ (x) + δ1(t)Y(0)1 (x), (5.47)

where

δ1(t) = 4(0)φ (t)

4(0)1 (t)

+ 1

4(0)1 (t)

∫∫ω0(t)

A0(x, t)φ(x)dx. (5.48)

Near the boundary of the domain ω0(t), the following Taylor expansions hold true:

p0(x, t) = m2

(80(x0(σ )) − δ0(t))n2 + O(n3). (5.49)

p1(x, t) = m2

(φ(x0(σ )) − δ1(t))n2 − mh(σ, t)n + O(n3). (5.50)

Here, x0(σ ) is the point of the curve Ɣ0(t) with the natural coordinate σ .Applying the perturbation method developed by Nazarov (17), we construct the auxiliary function

of the inner asymptotic representation (5.44) in the form

P(σ, ν, t) = m2

(80(x0(σ )) − δ0(t))(ν − h(σ, t))2. (5.51)

The function (5.51) exactly satisfies the relations (2.9), while the boundary condition (2.10) is sat-isfied asymptotically. Note that the normals n0 and n to the contours Ɣ0(t) and Ɣε(t) are, generallyspeaking, different (see (5.12)).

Finally, taking account of the relations (5.27), (5.45), (5.49), (5.50) and (5.51), one can verify thatthe matching asymptotic condition for the outer (5.42) and inner (5.44) asymptotic representationsis fulfilled.

6. Example: slightly deformed circular contact domain

Let us assume that a circular domain can be taken as zero approximation in (2.5) while the thefunction φ(x) defining the perturbed boundary is represented by a polynomial:

80(x) = 12R

(x21 + x2

2), φ(x) =N∑

n=0

n∑j=0

cnj x j1 xn− j

2 , (6.1)

where cnj are dimensional coefficients.In this case, the limit auxiliary contact problem (3.13), (3.17), (3.18), ε = 0, has the following

solution:

p0(x, t) = q0(t)

(1 − x2

1 + x22

a20(t)

)2

.

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Correspondingly, (4.7), (4.8) and (4.10) take the form

δ0(t) = a20(t)4R

, q0(t) = m32

a40(t)R

, (6.2)

a0(t) =( mπ

96R

)−1/6(

F(t) + χ

∫ t

0F(τ )dτ

)1/6

. (6.3)

The first-order perturbation problem (5.8), (5.14), (5.15), (5.17) and (5.18) now takes thefollowing form:

1p1(x, t) = m

N∑n=0

n∑j=0

cnj x j1 xn− j

2 − δ1(t)

, x ∈ ω0(t), (6.4)

p1(x, t) = 0, x ∈ Ɣ0(t), (6.5)

∂p1

∂n(x, t) = −mh(σ, t)

a20(t)4R

, x ∈ Ɣ0(t), (6.6)

πa20(t)δ1(t) =

N∑n=0

an+20 (t)n + 2

n∑j=0

cnj Knj + a30(t)4R

H0(t), (6.7)

0 = a50(t)8R

H0(t) +N∑

n=0

nan+40 (t)

2(n + 2)(n + 4)

n∑j=0

cnj Knj . (6.8)

Here, ω0(t) and Ɣ0(t) are now the disk and the circle of radius a0(t), Knj = 0 for odd n and j , whileKnj = 2B((n + 1 − j)/2, ( j + 1)/2) for even n and j with B(ζ, ξ) being the Beta function. H0(t)is an integral characteristics of the contour variation h(σ, t) which can described by the contourvariation in polar coordinates, H0(t) = ∫ 2π

0 h(θ, t)dθ . From (6.7) and (6.8), it immediately followsthat

H0(t) = −N∑

n=0

4Ran−10 (t)

(n + 2)(n + 4)

n∑j=0

cnj Knj ,

δ1(t) =N∑

n=0

(n + 3)an0 (t)

π(n + 2)(n + 4)

n∑j=0

cnj Knj . (6.9)

Using Green’s function G(x, x′, t) of the Dirichlet problem for the domain ω0(t), we representthe solution of the Dirichlet problem (6.4), (6.5) in the form

p1(x, t) = m∫∫

ω0(t)φ(x′)G(x, x′, t)dx′ − m

4δ1(t)(|x|2 − a2

0(t)). (6.10)

Recall that for a circular domain ω0(t) of radius a0(t)

G(x, x′, t) = 12π

lna0(t)|x′||x − x′|||x′|x − a2

0(t)x′| ,

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whereas, in polar coordinates, we have

G(x, x′, t) = 14π

lna2

0(t)(r2 + r ′2 − 2r ′r cos(θ − θ ′))a4

0(t) + r ′2r2 − 2r ′ra20(t) cos(θ − θ ′)

.

Calculating the normal derivative of the function (6.10), we obtain

∂p1

∂n(x, t) = m

∫∫ω0(t)

φ(x′)∂G∂n

(x, x′, t)dx′ − m2

δ1(t)|x|, (6.11)

where

∂

∂nG(x, x′, t) = 1

2π

a20(t) − r ′2

a0(t)(a20(t) + r ′2 − 2r ′a0(t) cos(θ − θ ′))

.

Now substituting (6.11) into the boundary condition (6.6) and taking into account (6.9), we obtainh(θ, t) describing variation of the contact domain in the following form:

h(θ, t) = 2Rπa0(t)

N∑n=0

(n + 3)an0 (t)

(n + 2)(n + 4)

n∑j=0

cnj Knj − 2Rπa3

0(t)

∫ 2π

0dθ ′

×∫ a0(t)

0

N∑n=0

r ′nn∑

j=0

cnj cos j θ ′ sinn− j θ ′ (a2

0(t) − r ′2)r ′ dr ′

a20(t) + r ′2 − 2r ′a0(t) cos(θ − θ ′)

.

(6.12)

Note that the integral with respect to θ ′ on the right-hand side of (6.12) can be evaluated by help ofthe relation (27) ∫ π

0

cos nx dx1 − 2ρ cos x + ρ2 = πρn

1 − ρ2 , ρ2 < 1.

It is interesting to compare this formula asymptotically with the exact solution for ellipse pre-sented in section 4. Consider a particular case of an elliptic paraboloid z = 8ε(x) which is close toa circular paraboloid z = 80(x) that gives

8ε(x) = 12R

(x21 + x2

2) + ε

2R(x2

2 − x21), (6.13)

where

R = 2R1 R2

R1 + R2, ε = R1 − R2

R1 + R2. (6.14)

In view of (6.1), (6.9) and (6.13), we will have n = N = 2 and c20 = (2R)−1, c21 = 0,c22 = −(2R)−1, K20 = K22 = π ,

δ1(t) = 0, h(θ, t) = 13

a0(t) cos 2θ. (6.15)

The obtained result is in agreement with the exact solution constructed in section 4. Indeed, dueto the second formula (6.14), we have

R2

R1= 1 − ε

1 + ε= 1 − 2ε + O(ε2).

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Correspondingly, (4.6), (4.10) and (4.7) yield, respectively,

s2 = 1 − 43ε + O(ε2), (6.16)

a(t) = a0(t)(

1 + ε

3+ O(ε2)

), (6.17)

δε(t) = δ0(t)(1 + O(ε2)). (6.18)

Finally, let us rewrite the equation of the boundary of the elliptic contact domain

[x1/a(t)]2 + [x2/b(t)]2 = 1

in the polar coordinate system as follows:

r = a(t)(cos2 θ + s−2 sin2 θ)−1/2. (6.19)

Now, substituting the asymptotic approximations (6.16) and (6.17) into (6.19), we obtain

r = a0(t)(

1 + ε

3cos 2θ + O(ε2)

). (6.20)

It is not hard to see that the asymptotic formulas (6.18) and (6.20) are in complete agreementwith (6.15).

7. Discussion and conclusion

First of all let us recall the assumptions of the asymptotic model for contact between two thinbiphasic cartilage layers considered above. It is clear that the asymptotic model itself is based onthe contact constitutive relation (2.3), connecting the contact pressure with the displacements of thelayers’ boundary points. This equation is an extension of the asymptotic solution procedure adoptedin (1). Thus, with reference to (1), we can argue that the obtained asymptotic solution is valid forrelatively thin (with respect to the diameter of the contact area), frictionless cartilage layers, and forthe early time response (when t ≪ min{h2

1/HA1k1, h22/HA2k2}) after the application of a step load.

Here, HAn is the aggregate modulus of the nth cartilage layer. For typical human cartilage materialproperties, HA = 0.5 MPa and k = 2 × 10−15m4/Ns, the asymptotic model remains valid for 100to 200 s, possible longer (1).

Observe that the contact constitutive relation (2.3) connects the displacement wn(x, t) with theLaplacian of the contact pressure 1x Pε(x, t), while the well-known Winkler’s elastic foundationmodel, which is usually employed for describing the deformation of thin elastic interphases, di-rectly connects the displacement with the contact pressure. As it was shown in (28) based on theasymptotic solution of the contact problem for a thin elastic layer (27), (29) the contact constitutiverelation connecting the displacement with the Laplacian of the contact pressure is typical for thinelastic incompressible layers. It is interesting to note that the equivalence of the instantaneous re-sponse of a biphasic material with that of an incompressible elastic material has been establishedpreviously (1, 30).

It is known (31, 32) that elliptic paraboloids can be used as a first approximation in describingarticular surface geometry in the knee joint. Thus, the developed asymptotic modelling approachcan be applied for incorporating the irregularity of the subchondral bone surface (i.e. the deviationsfrom the elliptic paraboloid shape) into the refined contact pressure analysis.

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Furthermore, the obtained analytical solution can be used for developing a more detailed modelof articular tibio-femoral contact for multibody dynamic simulations. In particular, the mathemat-ical model of articular contact (33) based on the exact solutions for elliptical contact between thincartilage layers modelled as biphasic or viscoelastic incompressible layers (14, 28) can be extendedfor the case of arbitrary articular surfaces shaped close to elliptical paraboloids.

Though the present paper has established a perturbation approach for approximate analyticalsolving 3D problems of articular contact in the light of the asymptotic model developed in (1, 2, 14)it should be emphasized that the refined asymptotic model constructed above has limitations of itsown. The main limitation is that the articular cartilage parameters (including the permeability andthickness) are assumed to be constant. For instance, the permeability is assumed to be independentof deformation. The same assumption was used in (34) in the problem of free-draining confinedcompression of a biphasic layer, while a solid content function (the ratio of the volume of each ofthe two phases) was assumed to vary with depth in accordance with the experimentally determinedweight ratios. It was shown that the contribution of the inhomogeneity according to the last assump-tion is O(10−3) smaller than the leading asymptotic term. The influence of the variability of thecartilage permeability and thickness in the articular contact problem was not examined before.

The influence of the variability of the cartilage permeability and thickness was not examinedbefore. The present study results in the closed-form solution to the perturbed 3D contact problem forbiphasic cartilage layers. The general equations (3.7) and (3.11) as well as (4.6), (4.10), (4.7), (4.8)and (4.13) for evaluating the aspect ratio of the elliptic contact domain, its major semi-axis a(t), thedisplacement parameter δε(t), the auxiliary parameter pε(t) and the contact pressure Pε(x, t) in thespecial case (2.5) of contact of elliptic paraboloids constitute the main result of the present study.The obtained results generalize the solution obtained in (14, 15) for the elliptic contact of biphasiccartilage layers.

Acknowledgements

I.A. gratefully acknowledges the support from the European Union Seventh Framework Programmeunder contract number PIIF-GA-2009-253055.

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13. G. Li, M. Sakamoto and E. Y. S. Chao, A comparison of different methods in predicting staticpressure distribution in articulating joints, J. Biomech. 30 (1997) 635–638.

14. I. Argatov and G. Mishuris, Elliptical contact of thin biphasic cartilage layers: exact solutionfor monotonic loading, ibid. 44 (2011) 759–761.

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24. W. Herzog, S. Diet, E. Suter, P. Mayzus, T. R. Leonard, C. Muller, J. Z. Wu and M. Epstein,Material and functional properties of articular cartilage and patellofemoral contact mechanicsin an experimental model of osteoarthritis, ibid. 31 (1998) 1137–1145.

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26. M. D. Van Dyke, Perturbation Methods in Fluid Mechanics (Academic Press, New York 1964).27. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products (Academic,

New York 1980).28. I. Argatov and G. Mishuris, Frictionless elliptical contact of thin viscoelastic layers bonded to

rigid substrates, Appl. Math. Model. 35 (2011) 3201–3212.

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APPENDIX A

Steklov–Poincare operator for an elliptical domain

We will use here conformal mappings technique. The conformal mapping of the interior of an ellipse onto acircle in an elementary form was given by Szego (35).

Let us consider ω0(t) as a subdomain in the complex plane of variable z = x1 + ix2. Let ζ = f0(z) be aconformal mapping of the domain ω0(t) onto the unit disk ω1 in the complex plane of variable ζ = ξ1 + iξ2.Assuming that ξ1 + iξ2 is the image of x1 + ix2 under the mapping f0, we will have

p(1)1 (x, t) = p(1)

1 (ξ, t), x ∈ ω0(t), ξ ∈ ω1, (A.1)

g(0)φ (x, t) = g(0)

φ (ξ, t), x ∈ Ɣ0(t), ξ ∈ Ɣ1, (A.2)

where Ɣ1 is the unit circle.In view of (5.46), p(1)

1 (ξ, t) is a harmonic function of variables ξ1 and ξ2. The function p(1)1 (ξ, t) can be

determined efficiently using, e.g. the Poisson integral as follows:

p(1)1 (ξ, t) = P[g(0)

φ ](ξ, t), ξ ∈ ω1, (A.3)

P[g](ξ) = 12π

∫ π

−π

1 − (ξ21 + ξ2

2 )

1 − 2(ξ1 cos τ + ξ2 sin τ) + ξ21 + ξ2

2g(cos τ, sin τ)dτ. (A.4)

In addition, let z = F0(ζ ) be the inverse conformal mapping from the unit disk ω1 onto the domain ω0(t).Then, g(0)

φ = g(0)φ ◦F0 and the solution of the Dirichlet boundary-value problem (5.46) may be represented in

the form of composition

p(1)1 = (P[g(0)

φ ◦ F0]) ◦ f0. (A.5)

Further, let us assume that the conformal mapping f0 transforms the Dirichlet problem (5.32) into theboundary-value problem

1ξ w(ξ) = 0, ξ ∈ ω1; w(ξ) = g(ξ), ξ ∈ Ɣ1, (A.6)

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where g = g ◦ F0 is the image of g under the mapping f0.Due to the properties of conformal mappings, we have

∂w

∂n(x, t) = 1

|F ′0(ζ )|

∂w

∂ρ(ξ, t), x ∈ Ɣ0(t), ξ ∈ Ɣ1, (A.7)

where ρ = |ζ | is the radial coordinate, ∂/∂ρ is the normal derivative directed outward from the unit disk ω1,and the prime denotes the derivative with respect to the complex variable.

On the other hand, denoting by S the Steklov–Poincare operator for the Dirichlet problem (A.6) on the unitdisk ω1, we will have

(S g)(ξ) = ∂w

∂ρ(ξ, t), x ∈ Ɣ0(t), ξ ∈ Ɣ1. (A.8)

The following formula holds true:

(S g)(cos θ, sin θ) = 12π

∫ π

−π

∂ g∂τ

(cos τ, sin τ) cotθ − τ

2dτ. (A.9)

Here, θ is the polar coordinate. Note that ρ = 1 and ξ = (cos θ, sin θ), when ξ ∈ Ɣ1.Thus, taking into account the relations (5.31) and (A.7), (A.8), we arrive at the formula

Sg = | f ′0|(S(g ◦ F0)) ◦ f0. (A.10)

Observe that in deriving (A.10), the well-known relation |F ′0(ζ )| = 1/| f ′

0(z)| was used.

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Contact Problem For Thin Biphasic Cartilage Layers_ Perturbation Solution

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