ANTIPLANE SHEAR FLOWS IN VISCO-PLASTIC SOLIDS EXHIBITING ISOTROPIC AND KINEMATIC HARDENING

J. M. GREENBERG^{∗}AND D. R. OWEN^{†}

Abstract. The authors consider antiplane shearing motions of an incompressible visco-plastic solid. The particular constitutive equation employed assumes that the stress tensor has an “elastic” component and a component which can exhibit hysteresis. The model exhibits both “kinematic” and “isotropic” hardening. Our results consist of a set of energy type estimates for the resulting system, L_{2} contractivity estimates for the solution operator, and finally an analysis of the approach of our system to a “rate independent” model as a distinguished parameter describing our flow rule approaches zero. We also include some computational results for simple piecewise constant data.

Key words. antiplane shear flows, visco-plastic material, energy estimates, kinematic and isotropic hardening.

AMS subject classifications. 73E05, 73E50, 73E60, 73E70

1. Introduction. In this note we consider antiplane shear flows of an incompressible, isotropic, visco-plastic solid. This work generalizes and compliments earlier work of Greenberg [1], [2], Greenberg and Nouri [3], and Nouri and Rascle [4]. In particular, our work shows that the energy estimates previously obtained in [1] - [4] have counterparts for a broader class of visco-plastic materials.

The constitutive assumption we use postulates that the Cauchy stress tensor is additively decomposable into a component without hysteresis, i.e., that depends only on the present value of deformation gradient and an elasto-plastic component that exhibits hysteresis, i.e., that can depend on the past history of deformation gradient, and the flow rule we employ for the plastic strain tensor is basically an isotropic hardening rule with yield determined by the elasto-plastic stresses. The presence of a component without hysteresis for the stress tensor yields a model which exhibits both “kinematic” and “isotropic” hardening. Mihăilescu-Suliciu, Suliciu, and Williams [5] analyzed the relationship between solutions of ordinary differential equations that govern the motions of elasto-plastic and visco-plastic oscillators. Their constitutive assumptions also admit both kinematic and isotropic hardening.

The organization of this paper is as follows. In section 2 we develop the equations describing antiplane shear flows in visco-plastic solids. Section 3 focuses on a set of a-priori energy type estimates and also estimates which guarantee that the solution operator is a contraction in L_{2}. These estimates depend on a small parameter and as that parameter tends to zero they supply information about rate independent limit models. Similar estimates for a purely isotropic hardening model have been obtained by Rascle and Nouri [4]. In section 4 we present some numerical experiments.

2. Model Development. In antiplane shear flows of an incompressible isotropic visco-plastic solid, material points

ξ = ξ_{1}e_{1} + ξ_{2}e_{2} + ξ_{3}e_{3} (2.1)

∗Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213. This research was partially supported by the Applied Mathematical Sciences Program, U. S. Department of Energy and the Mathematics and Computer Science Division, Army Research Office.

†Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213. This research was partially supported by the National Science Foundation.

2 J. M. GREENBERG AND D. R. OWEN

move to ^{1} x = x_{1}e_{1} + x_{2}e_{2} + x_{3}e_{3} (2.2) where

x_{1} = ξ_{1}, x_{2} = ξ_{2}, and x_{3} = ξ_{3} + ϕ (ξ_{1}, ξ_{2}, t_{1}) .

For such motions the Eulerian velocity field is given by u (x_{1}, x_{2}, x_{3}, t_{1}) = w (x_{1}, x_{2}, t_{1}) e_{3} (2.3) where

w(x_{1}, x_{2}, t_{1}) = \frac{∂ϕ}{∂t_{1}} (x_{1}, x_{2}, t_{1})

and the shear strains are given by

γ_{1} = \frac{∂ϕ}{∂x_{1}} and γ_{2} = \frac{∂ϕ}{∂x_{2}} . (2.4)

The latter are related to the former by the compatibility relations

\frac{∂γ_{1}}{∂t_{1}} = \frac{∂w}{∂x_{1}} and \frac{∂γ_{2}}{∂t_{1}} = \frac{∂w}{∂x_{2}} . (2.5)

The deformation gradient, F, associated with this motion is given by

F = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ γ_{1} & γ_{2} & 1 \end{matrix} ) . (2.6)

Our basic constitutive assumption postulates a plastic deformation tensor, P, of the same form as F; that is

P = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ p_{1} & p_{2} & 1 \end{matrix} ) (2.7)

where p_{1} and p_{2} are functions of x_{1}, x_{2}, and t_{1}, and postulates that the Cauchy Stress tensor take the form T = −πI + S_{1} + S_{2}. (2.8) Here π is the hydrostatic pressure,

S_{1} = β (F F^{T} − \frac{1}{3} trace (F F^{T} ) I) (2.9)

1e_{1} = ( \begin{matrix}1 \\ 0 \\ 0 \end{matrix} ) , e_{2} = ( \begin{matrix}0 \\ 1 \\ 0 \end{matrix} ), and e_{3} = ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} ) are the standard basis for elements for R^{3} and e_{i} ⊗ e_{j} = e_{i}e_{j}^{T} are the standard basis elements for linear operators from R^{3} to R^{3}

or equivalently

S_{1} = β ( \begin{matrix}−\frac{1}{3} (γ_{1}^{2} & + & γ_{2}^{2}) & & 0 & & & γ_{1} \\ & 0 & & −\frac{1}{3} (γ_{1}^{2} & + & γ_{2}^{2}) & & γ_{2} \\ & γ_{1} & & & γ_{2} & & \frac{2}{3} (γ_{1}^{2} & + & γ_{2}^{2}) \end{matrix} ) , (2.10)

S_{2} = μ (RR^{T} − \frac{1}{3} trace (RR^{T} ) I) (2.11)

or equivalently

S_{2} = μ ( \begin{matrix}−\frac{1}{3} (r_{1}^{2} & + & r_{2}^{2}) & & 0 & & & r_{1} \\ & 0 & & −\frac{1}{3} (r_{1}^{2} & + & r_{2}^{2}) & & r_{2} \\ & r_{1} & & & r_{2} & & \frac{2}{3} (r_{1}^{2} & + & r_{2}^{2}) \end{matrix} ) , (2.12)

R = F P^{−1} = P^{−1}F (2.13)

or equivalently

R = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ r_{1} & r_{2} & 1 \end{matrix} ) = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ γ_{1} − p_{1} & γ_{2} − p_{2} & 1 \end{matrix} ) (2.14)

and β and μ are positive constants. Equations (2.9) - (2.14) imply that S_{1} is an isotropic trace free function of F that has the interpretation of a stress without hysteresis, whereas S_{2} is an isotropic trace free function of R that is interpreted as the “elasto-plastic” stress and exhibits hysteresis for appropriate deformation histories.

In what follows it will be convenient to let

s_{1} := μ (γ_{1} − p_{1}) and s_{2} := μ (γ_{2} − p_{2}) (2.15)

and to use s_{1}(x_{1}, x_{2}, t_{1}), s_{2}(x_{1}, x_{2}, t_{1}), p_{1}(x_{1}, x_{2}, t_{1}) and p_{2}(x_{1}, x_{2}, t_{1}) as the basic descriptors of our system. With this choice we have

F = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{s_{1}}{μ} + p_{1} & \frac{s_{2}}{μ} + p_{2} & 1 \end{matrix} ) , (2.16)

R = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{s_{1}}{μ} & \frac{s_{2}}{μ} & 1 \end{matrix} ) , (2.17)

S_{2} = ( \begin{matrix}−\frac{1}{3μ} (s_{1}^{2} & + & s_{2}^{2}) & & 0 & & & s_{1} \\ & 0 & & −\frac{1}{3μ} (s_{1}^{2} & + & s_{2}^{2}) & & s_{2} \\ & s_{1} & & & s_{2} & & \frac{2}{3μ} (s_{1}^{2} & + & s_{2}^{2}) \end{matrix} ) , (2.18)

and (2.19)

S_{1} = β ( \begin{matrix}−\frac{1}{3} ((\frac{s_{1}}{μ} + p_{1})^{2} + (\frac{s_{2}}{μ} + p_{2})^{2}) & 0 & \frac{s_{1}}{μ} + p_{1} \\ 0 & −\frac{1}{3} ((\frac{s_{1}}{μ} + p_{1})^{2} + (\frac{s_{2}}{μ} + p_{2})^{2}) & \frac{s_{2}}{μ} + p_{2} \\ \frac{s_{1}}{μ} + p_{1} & \frac{s_{2}}{μ} + p_{2} & \frac{2}{3} ((\frac{s_{1}}{μ} + p_{1})^{2} + (\frac{s_{2}}{μ} + p_{2})^{2}) \end{matrix} ) .

Balance of momentum in the e_{1} and e_{2} directions implies that

\frac{∂}{∂x_{1}} (π + \frac{(s_{1}^{2} + s_{2}^{2})}{3μ} + \frac{β}{3} ((\frac{s_{1}}{μ} + p_{1})^{2} + (\frac{s_{2}}{μ} + p_{2})^{2})) = 0, (2.20)

\frac{∂}{∂x_{2}} (π + \frac{(s_{1}^{2} + s_{2}^{2})}{3μ} + \frac{β}{3} ((\frac{s_{1}}{μ} + p_{1})^{2} + (\frac{s_{2}}{μ} + p_{2})^{2})) = 0, (2.21)

and thus that the pressure, π, is given by

π = π_{0} (x_{3}, t) − \frac{(s_{1}^{2} + s_{2}^{2})}{3μ} − \frac{β}{3} ((\frac{s_{1}}{μ} + p_{1})^{2} + (\frac{s_{2}}{μ} + p_{2})^{2}) . (2.22)

Balance of momentum in the e_{3} direction yields

ρ_{0} \frac{∂w}{∂t_{1}} − \frac{∂}{∂x_{1}} (s_{1} + β (\frac{s_{1}}{μ} + p_{1})) − \frac{∂}{∂x_{2}} (s_{2} + β (\frac{s_{2}}{μ} + p_{2})) = −\frac{∂π_{0}}{∂x_{3}} . (2.23)

Here, ρ_{0} is the constant mass density of the material. Since \frac{∂π_{0}}{∂x_{3}} depends on x_{3} and t_{1}, whereas all quantities in the left-hand side of (2.23) depend on x_{1}, x_{2} and t_{1}, we conclude that for antiplane shear flows \frac{∂π_{0}}{∂x_{3}} is independent of x_{3}. In what follows we shall assume this term is zero.

We now turn our attention to the flow rule for the nonconstant components p_{1} and p_{2} of P. We assume the isotropic hardening rule

\frac{∂p_{1}}{∂t_{1}} = \frac{s_{1}}{\sqrt{s_{1}^{2} + s_{2}^{2}}} \frac{(\sqrt{s_{1}^{2} + s_{2}^{2}} − s_{y}(d))_{+}}{μT_{0}ϵ} (2.24)

and

\frac{∂p_{2}}{∂t_{1}} = \frac{s_{2}}{\sqrt{s_{1}^{2} + s_{2}^{2}}} \frac{(\sqrt{s_{1}^{2} + s_{2}^{2}} − s_{y}(d))_{+}}{μT_{0}ϵ} . (2.25)

Here,

(x)_{+} = \{ \begin{matrix}0, & x & < & 0 \\ x, & x & ≥ & 0 \end{matrix} , (2.26)

d is the accumulated plastic strain and satisfies

\frac{∂d}{∂t_{1}} = \sqrt{(\frac{∂p_{1}}{∂t_{1}} )^{2} + (\frac{∂p_{2}}{∂t_{1}} )^{2}} = \frac{(\sqrt{s_{1}^{2} + s_{2}^{2}} − s_{y}(d))_{+}}{μT_{0}ϵ} (2.27)

and finally d → s_{y}(d) is the yield stress which we assume satisfies

0 < s_{y}(0) ≤ s_{y}(d), 0 < m ≤ s′_{y} (d) and − M ≤ s′′_{y} (d) ≤ 0 (2.28)

for d ≥ 0. The parameter T_{0} has the interpretation of a relaxation time, μ is the “shear” modulus, and ϵ > 0 is a dimensionless small parameter.

Equations (2.5), (2.15), and (2.23) -(2.27) combine to yield the following system for s_{1}, s_{2}, p_{1}, p_{2}, d, and w:

\frac{1}{μ} \frac{∂s_{1}}{∂t_{1}} − \frac{∂w}{∂x_{1}} = −\frac{∂p_{1}}{∂t_{1}} , (2.29)

\frac{1}{μ} \frac{∂s_{2}}{∂t_{1}} − \frac{∂w}{∂x_{2}} = −\frac{∂p_{2}}{∂t_{1}} , (2.30)

ρ_{0} \frac{∂w}{∂t_{1}} − \frac{∂}{∂x_{1}} (\frac{(β + μ)}{μ}s_{1} + βp_{1}) − \frac{∂}{∂x_{2}} (\frac{(β + μ)}{μ}s_{2} + βp_{2}) = 0, (2.31)

\frac{∂p_{1}}{∂t_{1}} = \frac{s_{1}}{\sqrt{s_{1}^{2} + s_{2}^{2}}} \frac{(\sqrt{s_{1}^{2} + s_{2}^{2}} − s_{y}(d))_{+}}{μT_{0}ϵ} , (2.32)

\frac{∂p_{2}}{∂t_{1}} = \frac{s_{2}}{\sqrt{s_{1}^{2} + s_{2}^{2}}} \frac{(\sqrt{s_{1}^{2} + s_{2}^{2}} − s_{y}(d))_{+}}{μT_{0}ϵ} , (2.33)

and

\frac{∂d}{∂t_{1}} = \frac{(\sqrt{s_{1}^{2} + s_{2}^{2}} − s_{y}(d))_{+}}{μT_{0}ϵ} . (2.34)

For convenience, we put (2.29) - (2.34) into dimensionless form. We let

\{ \begin{matrix}x = \sqrt{\frac{ρ_{0}}{μ + β}} \frac{x_{1}}{T_{0}} , y = \sqrt{\frac{ρ_{0}}{μ + β}} \frac{y_{1}}{T_{0}} , t = \frac{t_{1}}{T_{0}} , \\ u = \sqrt{\frac{ρ_{0}}{μ + η}} w, τ_{1} = \frac{s_{1}}{μ} , τ_{2} = \frac{s_{2}}{μ} and τ_{y}(d) = \frac{s_{y}(d)}{μ}. \end{matrix} (2.35)

Then, (2.29) - (2.34) transforms to

\frac{∂τ_{1}}{∂t} − \frac{∂u}{∂x} = −\frac{∂p_{1}}{∂t} (2.36)

\frac{∂τ_{2}}{∂t} − \frac{∂u}{∂y} = −\frac{∂p_{2}}{∂t} (2.37)

\frac{∂u}{∂t} − \frac{∂}{∂x} (τ_{1} + \frac{β}{μ + β} p_{1}) − \frac{∂}{∂y} (τ_{2} + \frac{β}{μ + β} p_{2}) = 0 (2.38)

\frac{∂p_{1}}{∂t} = \frac{τ_{1} (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y}(d))_{+}}{ϵ\sqrt{τ_{1}^{2} + τ_{2}^{2}}} (2.39)

\frac{∂p_{2}}{∂t} = \frac{τ_{2} (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y}(d))_{+}}{ϵ\sqrt{τ_{1}^{2} + τ_{2}^{2}}} (2.40)

and

\frac{∂d}{∂t} = \frac{(\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y}(d))_{+}}{ϵ} ≥ 0. (2.41)

In section 3 we shall present a set of energy type estimates for (2.36) - (2.41).

3. Energy Estimates. In this section we shall focus on a series of energy type estimates for solutions of the Cauchy problem for (2.36) - (2.41). We shall continually use an integral identity, presented in (3.2) below, for solutions of the divergence identity:

\frac{∂f}{∂t} − \frac{∂q_{1}}{∂x} − \frac{∂q_{2}}{∂y} = g. (3.1)

The basic identities are

\frac{d}{ds} ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − s\}} f (x, y, s) dxdy =

(R + t − s) ∫_{0}^{2π} [q_{1} cos θ + q_{2} sin θ − f] (x_{0} + (R + t − s) cos θ, y_{0} + (R + t − s) sin θ, s) dθ

+ ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − s\}} g (x, y, s) dxdy

ANTIPLANE SHEAR FLOWS 7

and

∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R\}} f (x, y, s) dxdy =

∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t\}} f (x, y, 0) dxdy

+∫_{0}^{t}(R + t − s) (∫_{0}^{2π}[q_{1} cos θ + q_{2} sin θ − f](x_{0} + (R + t + s) cos θ, y_{0} + (R + t − s) sin θ, s)dθ) ds

+∫_{0}^{t} ( ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − s\}} g (x, y, s) dxdy) ds. (3.2)

Our first identity is obtained by multiplying (2.36) by τ_{1}, (2.37) by τ_{2}, (2.38) by u and (2.41) by \frac{μ}{β+μ}τ_{y}(d) and adding the equations. The result is that (3.1) holds with

f = \frac{μ (τ_{1}^{2} + τ_{2}^{2})}{2 (μ + β)} + \frac{β ((τ_{1} + p_{1})^{2} + (τ_{2} + p_{2})^{2})}{2 (μ + β)} + \frac{u^{2}}{2} + \frac{μ}{μ + β} ∫_{0}^{d} τ_{y} (η) dη (3.3)

q_{1} = u (\frac{μ}{μ + β} τ_{1} + \frac{β}{μ + β} (τ_{1} + p_{1})) (3.4)

q_{2} = u (\frac{μ}{μ + β} τ_{2} + \frac{β}{μ + β} (τ_{2} + p_{2})) (3.5)

and

g = \frac{−μ}{ϵ (μ + β)} ((\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+})^{2} . (3.6)

The observation that

q_{1} cos θ + q_{2} sin θ ≤ |u| (\frac{μ}{μ + β} \sqrt{τ_{1}^{2} + τ_{2}^{2}} + \frac{β}{μ + β} \sqrt{(τ_{1} + p_{1})^{2} + (τ_{2} + p_{2})^{2}}) (3.7)

implies that

\begin{matrix}q_{1} cos θ + q_{2} sin θ − f ≤ \\ −\frac{μ}{2 (μ + β)} (|u| − \sqrt{τ_{1}^{2} + τ_{2}^{2}})^{2} −\frac{β}{2 (μ + β)} (|u| − \sqrt{(τ_{1} + p_{1})^{2} + (τ_{2} + p_{2})^{2}})^{2} \\ −\frac{μ}{μ + β} ∫_{0}^{d} τ_{y}(η)dη ≤ 0 \end{matrix} (3.8)

and thus the last two integrals on the right-hand side of (3.2) are less than or equal to zero.^{2}

To obtain the contractivity of the solution operator for (2.36) - (2.41) we look at two solutions to (2.36) - (2.41) which we label a and b, respectively. Their differences satisfy

\frac{∂}{∂t} (τ_{1}^{a} − τ_{1}^{b}) − \frac{∂}{∂x} (u^{a} − u^{b}) = −\frac{∂}{∂t} (p_{1}^{a} − p_{1}^{b}) , (3.9)

\frac{∂}{∂t} (τ_{2}^{a} − τ_{2}^{b}) − \frac{∂}{∂y} (u^{a} − u^{b}) = −\frac{∂}{∂t} (p_{2}^{a} − p_{2}^{b}) , (3.10)

\frac{∂}{∂t} (u^{a} − u^{b}) − \frac{∂}{∂x} (τ_{1}^{a} − τ_{1}^{b} + \frac{β (p_{1}^{a} − p_{1}^{b})}{μ + β}) − \frac{∂}{∂y} (τ_{2}^{a} − τ_{2}^{b} + \frac{β (p_{2}^{a} − p_{2}^{b})}{μ + β}) = 0,

(3.11)

\frac{∂}{∂t} (p_{1}^{a} − p_{1}^{b}) = \frac{τ_{1}^{a} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+}}{ϵ \sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}}} − \frac{τ_{1}^{b} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+}}{ϵ \sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}}} ,

(3.12)

\frac{∂}{∂t} (p_{2}^{a} − p_{2}^{b}) = \frac{τ_{2}^{a} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+}}{ϵ \sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}}} − \frac{τ_{2}^{b} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}}− τ_{y} (d^{b}))_{+}}{ϵ \sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}}} ,

(3.13)

and

\frac{∂}{∂t} (d^{a} − d^{b}) = \frac{(\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+}}{ϵ} −\frac{(\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+}}{ϵ} .

(3.14)

We next multiply (3.9) by (τ_{1}^{a} − τ_{1}^{b}), (3.10) by (τ_{2}^{a} − τ_{2}^{b}), (3.11) by (u^{a} − u^{b}), and (3.14) by

\frac{μ}{μ + β} (τ_{y} (d^{a}) − τ_{y} (d^{b})) = \frac{μ}{μ + β} (∫_{0}^{1} τ′_{y} ((1 − η) d^{b} + ηd^{a}) dη) (d^{a} − d^{b})

and add the resulting equations. The result is that (3.1) holds where now

2We are, of course, using the fact that d ≥ 0, d_{t} ≥ 0, and that 0 ≤ τ_{y}(0)d ≤ ∫_{0}^{d} τ_{y}(η)dy.

\begin{matrix}f = \frac{1}{2} (\frac{μ ((τ_{1}^{a} − τ_{1}^{b})^{2} + (τ_{2}^{a} − τ_{2}^{b})^{2})}{μ + β} + \frac{β ((τ_{1}^{a} − τ_{1}^{b} + p_{1}^{a} − p_{1}^{b})^{2} + (τ_{2}^{a} − τ_{2}^{b} + p_{2}^{a} − p_{2}^{b})^{2})}{μ + β} \\ + (u^{a} − u^{b})^{2} + \frac{μ}{μ + β} (∫_{0}^{1}τ′_{y} ((1 − η) d^{b} + ηd^{a}) dη) (d^{a} − d^{b})^{2}) , \end{matrix}

(3.15)

q_{1} = (u^{a} − u^{b}) (\frac{μ}{μ + β} (τ_{1}^{a} − τ_{1}^{b}) + \frac{β}{μ + β} (τ_{1}^{a} − τ_{1}^{b} + p_{2}^{a} − p_{2}^{b})) , (3.16)

q_{2} = (u^{a} − u^{b}) (\frac{μ}{μ + β} (τ_{2}^{a} − τ_{2}^{b}) + \frac{β}{μ + β} (τ_{2}^{a} − τ_{2}^{b} + p_{2}^{a} − p_{2}^{b})) , (3.17)

and

g = \frac{μ}{μ + β} (g_{1} + \frac{g_{2}}{ϵ} ) (3.18)

where

g_{1} = \frac{1}{2} (∫_{0}^{1}τ′′_{y} ((1 − η) d^{b} + ηd^{a}) ((1 − η) d_{t}^{b} + ηd_{t}^{a}) dη) (d^{a} − d^{b})^{2} (3.19)

and g_{2} = (τ_{y} (d^{a}) − τ_{y} (d^{b})) ((\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} − (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+})

− (τ_{1}^{a} − τ_{1}^{b}) (\frac{τ_{1}^{a} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+}}{\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}}} − \frac{τ_{1}^{b} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+}}{\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}}} )

(3.20)

− (τ_{2}^{a} − τ_{2}^{b}) (\frac{τ_{2}^{a} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+}}{\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}}} − \frac{τ_{2}^{b} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+}}{\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}}} )

The arguments leading to (3.8) yield the inequality

q_{1} cos θ + q_{2} sin θ − f ≤ −\frac{μ}{2 (μ + β)} (|u^{a} − u^{b}| − \sqrt{(τ_{1}^{a} − τ_{1}^{b})^{2} + (τ_{2}^{a} − τ_{2}^{b})^{2}})^{2}

−\frac{β}{2 (μ + β)} (|u^{a} − u^{b}| − \sqrt{(τ_{1}^{a} − τ_{1}^{b} + p_{1}^{a} − p_{1}^{b})^{2} + (τ_{2}^{a} − τ_{2}^{b} + p_{2}^{a} − p_{2}^{b})^{2}})^{2}

−\frac{μ}{2 (μ + β)} (∫_{0}^{1}τ′_{y} ((1 − η) d^{b} + ηd^{a}) dη) (d^{a} − d^{b})^{2} ≤ 0.

(3.21)

We further note that the hypotheses that 0 < τ_{y}(0) ≤ τ_{y}(d), 0 < \frac{m}{μ} ≤ τ′_{y}(d), and −\frac{M}{μ} ≤ τ′′_{y}(d) ≤ 0 for d ≥ 0 guarantees that g_{1} ≤ 0. Thus, to show that

∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R\}} f (x, y, t) dxdy ≤ ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t\}} f (x, y, 0) dxdy

(3.22) where f is given by (3.15) it suffices to show that g_{2} ≤ 0. We first note that g_{2} may be written as

g_{2} = − (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))

− (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))

(3.23)

+ (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} (\frac{τ_{1}^{a}τ_{1}^{b} + τ_{2}^{a}τ_{2}^{b}}{\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}}} − τ_{y} (d^{b}))

+ (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+} (\frac{τ_{1}^{a}τ_{1}^{b} + τ_{2}^{a}τ_{2}^{b}}{\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}}} − τ_{y} (d^{a}))

and (3.23), together with

τ_{1}^{a}τ_{1}^{b} + τ_{2}^{a}τ_{2}^{b} ≤ \sqrt{(τ_{1}^{a})^{2} + (τ_{1}^{b})^{2}}\sqrt{(τ_{2}^{a})^{2} + (τ_{2}^{b})^{2}}, (3.24)

implies that

\begin{matrix}g_{2} ≤ − (\sqrt{(τ_{1}^{a})^{2} − (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a})) \\ − (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b})) \\ + (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b})) \\ + (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a})) \end{matrix} (3.25)

or that

g_{2} = − ((\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} − (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+})^{2}

+ (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+} (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}) − (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+})

+ (\sqrt{(τ_{1}^{b})^{2} + (τ_{2}^{b})^{2}} − τ_{y} (d^{b}))_{+} (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}) − (\sqrt{(τ_{1}^{a})^{2} + (τ_{2}^{a})^{2}} − τ_{y} (d^{a}))_{+}) .

(3.26)

Since each of the terms on the right-hand side of (3.26) is nonpositive we conclude that g_{2} ≤ 0 and thus (3.22) holds, which is the desired contractivity of the solution operator associated with (2.36) - (2.41). Similar arguments have been used previously by Greenberg and Nouri [3] and Nouri and Rascle [4].

Our next objective is a set of derivative estimates for solutions of the Cauchy problem for (2.36) - (2.41). In what follows D will be one of the following operators: \frac{∂}{∂t}, \frac{∂}{∂x}, or \frac{∂}{∂y}. Differentiation of the original system yields the following equations for Dτ_{1}, Dτ_{2}, Du, Dp_{1}, Dp_{2}, and Dd:

\frac{∂}{∂t} (Dτ_{1}) − \frac{∂}{∂x} (Du) = −\frac{∂}{∂t} (Dp_{1}) , (3.27)

\frac{∂}{∂t} (Dτ_{2}) − \frac{∂}{∂y} (Du) = −\frac{∂}{∂t} (Dp_{2}) , (3.28)

\frac{∂}{∂t} (Du) − \frac{∂}{∂x} (Dτ_{1} + \frac{β}{μ + β} Dp_{1}) − \frac{∂}{∂y} (Dτ_{2} + \frac{β}{μ + β} Dp_{2}) = 0, (3.29)

\begin{aligned}\frac{∂}{∂t} (Dp_{1}) & = & \frac{τ_{1}}{ϵ \sqrt{τ_{1}^{2} + τ_{2}^{2}}} H (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d)) (\frac{τ_{1}Dτ_{1} + τ_{2}Dτ_{2}}{\sqrt{τ_{1}^{2} + τ_{2}^{2}}} − τ′_{y} (d) Dd) \\ & + & \frac{(\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ (τ_{1}^{2} + τ_{2}^{2})^{3/2}} (τ_{2}^{2}Dτ_{1} − τ_{1}τ_{2}Dτ_{2}) , \end{aligned}

(3.30)

\begin{aligned}\frac{∂}{∂t} (Dp_{2}) & = & \frac{τ_{2}}{ϵ \sqrt{τ_{1}^{2} + τ_{2}^{2}}} H (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d)) (\frac{τ_{1}Dτ_{1} + τ_{2}Dτ_{2}}{\sqrt{τ_{1}^{2} + τ_{2}^{2}}} − τ′_{y} (d) Dd) \\ & + & \frac{(\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ (τ_{1}^{2} + τ_{2}^{2})^{3/2}} (τ_{1}^{2}Dτ_{2} − τ_{1}τ_{2}Dτ_{1}) , \end{aligned}

(3.31)

12 J. M. GREENBERG AND D. R. OWEN

and

\frac{∂}{∂t} (Dd) = \frac{H (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))}{ϵ} (\frac{τ_{1}Dτ_{1} + τ_{2}Dτ_{2}}{\sqrt{(τ_{1}^{2}) + (τ_{2})^{2}}} − τ′_{y} (d) Dd) . (3.32)

In equations (3.30) - (3.32)

H (x) = \{ \begin{matrix}0, & x < 0 \\ 1, & x ≥ 0 \end{matrix} . (3.33)

If we now multiply (3.27) by Dτ_{1}, (3.28) by Dτ_{2}, (3.29) by Du, (3.32) by \frac{μ}{μ+β} τ′_{y} (d) Dd and add the resulting expressions we find that (3.1) and (3.2) hold where now

f = \frac{1}{2}(\frac{μ ((Dτ_{1})^{2} + (Dτ_{2})^{2})}{μ + β} + \frac{β ((D (τ_{1} + p_{1}))^{2}+(D (τ_{2} + p_{2}))^{2})}{μ + β} +(Du)^{2} + \frac{μ}{μ + β} τ′_{y} (d) (Dd)^{2}) ,

(3.34)

q_{1} = Du (\frac{μ}{μ + β} Dτ_{1} + \frac{β}{μ + β} D (τ_{1} + p_{1})) , (3.35)

q_{2} = Du (\frac{μ}{μ + β} Dτ_{2} + \frac{β}{μ + β} D (τ_{2} + p_{2})) , (3.36)

and

\begin{aligned}g & = & \frac{μτ′′_{y} (d)}{2 (μ + β)}d_{t} (Dd)^{2} \\ & − & \frac{μ}{ϵ (μ + β)}H (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d)) (\frac{τ_{1}Dτ_{1} + τ_{2}Dτ_{2}}{\sqrt{τ_{1}^{2} + τ_{2}^{2}}} − τ′_{y} (d) Dd)^{2} \\ & − & \frac{μ}{ϵ (μ + β)} \begin{matrix}(\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d)) & + \\ (τ_{1}^{2} + τ_{2}^{2}) & 3/2 \end{matrix} (τ_{2}Dτ_{1} − τ_{1}Dτ_{2})^{2} . \end{aligned} (3.37)

The hypothesis that τ′′_{y} ≤ 0 and d_{t} ≥ 0 then guarantee that g ≤ 0. Moreover, the boundary density satisfies

\begin{matrix}q_{1} cos θ + q_{2} sin θ − f ≤ −\frac{μ (|Du| − \sqrt{(Dτ_{1})^{2} + (Dτ_{2})^{2}})^{2}}{2 (μ + β)} \\ −\frac{β (|Du| − \sqrt{(D (τ_{1} + p_{1}))^{2} + (D (τ_{2} + p_{2}))^{2}})^{2}}{2 (μ + β)} − \frac{τ′_{y} (d) (Dd)^{2}}{2} ≤ 0 \end{matrix} (3.38)

and the latter two inequalities along with (3.2) yield the desired derivative bounds.

We conclude this section with an examination of the behavior of our system as the small parameter ϵ approaches zero from above. In what follows we let

Ω (x_{0}, y_{0}, R, t − s) = \{(x, y) |\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − s\} (3.39)

when 0 ≤ s ≤ t and

B (x_{0}, y_{0}, R, t) = \{(x, y, s) |\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − s, 0 ≤ s ≤ t\} .

(3.40) The a-priori estimates associated with (3.2) when f is given by (3.3) or (3.34) and D is one of the operators \frac{∂}{∂t}, \frac{∂}{∂x}, or \frac{∂}{∂y} guarantee that if the initial values for a family (τ_{1}^{ϵ}, τ_{2}^{ϵ}, p_{1}^{ϵ}, p_{2}^{ϵ}, d^{ϵ}, u^{ϵ}) of distributional solutions are in L_{2}^{loc} independently of ϵ, then we may choose a sequence ϵ_{i}, i = 1, 2, ..., with the ϵ_{i}’s decreasing to zero and limit functions (τ_{1}^{0}, τ_{2}^{0}, p_{1}^{0}, p_{2}^{0}, d^{0}, u^{0}) with the following properties:

(i) For any (x_{0}, y_{0}), R > 0, and t > 0 the sequence (τ_{1}^{ϵ_{i}}, τ_{2}^{ϵ_{i}}, p_{1}^{ϵ_{i}}, p_{2}^{ϵ_{i}}, d^{ϵ_{i}}, u^{ϵ_{i}}) converge strongly in L_{2} (B (x_{0}, y_{0}, R, t)) to (τ_{1}^{0}, τ_{2}^{0}, p_{1}^{0}, p_{2}^{0}, d^{0}, u^{0}). Moreover, the limit functions have weak t, x, and y derivatives and the sequences D (τ_{1}^{ϵ_{i}}τ_{2}^{ϵ_{i}}, p_{1}^{ϵ_{i}}, p_{2}^{ϵ_{i}}, d^{ϵ_{i}}, u^{ϵ_{i}}) converge weakly in L_{2} (B (x_{0}, y_{0}, R, t)) to D (τ_{1}^{0}, τ_{2}^{0}, p_{1}^{0}, p_{2}^{0}, d^{0}, u^{0}) where again D = \frac{∂}{∂t}, \frac{∂}{∂x} or \frac{∂}{∂y}.

(ii) The hypotheses (2.28) and (2.35) on the yield stress further guarantee that τ_{y} (d^{ϵ_{i}}) converges strongly in L_{2} (B (x_{0}, y_{0}, R, t)) to τ_{y} (d^{0}) and that Dτ_{y} (d^{ϵ_{i}}) = τ′_{y} (d^{ϵ_{i}}) Dd^{ϵ_{i}} converges weakly in L_{1} (B (x_{0}, y_{0}, R, t)) to Dτ_{y} (d^{0}) = τ′_{y} (d^{0}) Dd^{0} where once again D = \frac{∂}{∂t}, \frac{∂}{∂x} or \frac{∂}{∂y}. Moreover, equations (2.30) -(2.41) and the conver-

gence results (i) above imply that τ_{m}^{ϵ_{i}} \frac{∂p_{n}^{ϵ_{i}}}{∂t}, m and n = 1 and 2, and \sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} \frac{∂d}{∂t}^{ϵ_{i}}

converge weakly in L_{1} (B (x_{0}, y_{0}, R, t)) to τ_{m}^{0} \frac{∂p_{n}^{0}}{∂t}, m and n = 1 and 2, and

\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} \frac{∂d^{0}}{∂t} respectively.

Finally, the identities

τ_{1}^{ϵ_{i}} \frac{∂p_{1}^{ϵ_{i}}}{∂t} + τ_{2}^{ϵ_{i}} \frac{∂p_{2}^{ϵ_{i}}}{∂t} = \sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} \frac{∂d}{∂t}^{ϵ_{i}} (3.41)

τ_{2}^{ϵ_{i}} \frac{∂p_{1}^{ϵ_{i}}}{∂t} − τ_{1}^{ϵ_{i}} \frac{∂p_{2}^{ϵ_{i}}}{∂t} = 0 (3.42)

guarantee that the limit functions satisfy

τ_{1}^{0} \frac{∂p_{1}^{0}}{∂t} + τ_{2}^{0} \frac{∂p_{1}^{0}}{∂t} = \sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} \frac{∂d^{0}}{∂t} (3.43)

and

τ_{2}^{0} \frac{∂p_{1}^{0}}{∂t} − τ_{1}^{0} \frac{∂p_{2}^{0}}{∂t} = 0. (3.44)

(iii) For any δ > 0, the measure

m (\{(x, y, s) ∈ B (x_{0}, y_{0}, R, t) |\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y} (d^{0}) ≥ δ\}) = 0. (3.45)

The last identity implies that

\begin{aligned}& m (\{(x, y, s) ∈ B (x_{0}, y_{0}, R, t) |\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y} (d^{0}) ≥ 0\}) \\ = & m (\{(x, y, s) ∈ B (x_{0}, y_{0}, R, t) |\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y} (d^{0}) = 0\}) . \end{aligned} (3.46)

In what follows we shall refer to

\{(x, y, s) ∈ B (x_{0}, y_{0}, R, t) |\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y} (d^{0}) = 0\}

as the yield set. On open subsets U of the yield set we introduce Θ by (3.47) τ_{1}^{0} = τ_{y} (d^{0}) cos Θ and τ_{2}^{0} = τ_{y} (d^{0}) sin Θ. The relationships (3.41) and (3.42) then imply that

(3.48) cos Θ\frac{∂p_{1}^{0}}{∂t} + sin Θ\frac{∂p_{2}^{0}}{∂t} = \frac{∂d^{0}}{∂t}

and (3.49) sin Θ\frac{∂p_{1}^{0}}{∂t} − cos Θ\frac{∂p_{2}^{0}}{∂t} = 0. Since the weak limits (τ_{1}^{0}, τ_{2}^{0}, p_{1}^{0}, p_{2}^{0}, u^{0}) also satisfy the conservation laws (2.36) - (2.38) we find that in U the following equations are satisfied in the distributional sense

(3.50) cos Θτ′_{y} (d^{0}) \frac{∂d^{0}}{∂t} − τ_{y} (d^{0}) sin Θ\frac{∂Θ}{∂t} − \frac{∂u^{0}}{∂x} = −\frac{∂p_{1}^{0}}{∂t}

(3.51) sin Θτ′_{y} (d^{0}) \frac{∂d^{0}}{∂t} + τ_{y} (d^{0}) cos Θ\frac{∂Θ}{∂t} − \frac{∂u^{0}}{∂y} = −\frac{∂p_{2}^{0}}{∂t}

and

(3.52) \frac{∂u^{0}}{∂t} − \frac{∂}{∂x} (τ_{y} (d^{0}) cos Θ + \frac{β}{μ + β} p_{1}^{0}) − \frac{∂}{∂y} (τ_{y} (d^{0}) sin Θ + \frac{β}{μ + β} p_{2}^{0}) = 0.

These equations represent a closed system for (p_{1}^{0}, p_{2}^{0}, d^{0}, u^{0}, Θ) on the yield surface. They imply that (3.52) holds and that

(3.53) (1 + τ′_{y} (d^{0})) \frac{∂d^{0}}{∂t} = cos Θ\frac{∂u^{0}}{∂x} + sin Θ\frac{∂u^{0}}{∂y} ,

(3.54) \frac{∂Θ}{∂t} = \frac{1}{τ_{y} (d^{0})} (− sin Θ\frac{∂u^{0}}{∂x} + cos Θ\frac{∂u^{0}}{∂y} ) ,

(3.55) \frac{∂p_{1}^{0}}{∂t} = \frac{cos Θ}{1+τ′_{y}(d^{0})} (cos Θ\frac{∂u^{0}}{∂x} + sin Θ\frac{∂u^{0}}{∂y} ) ,

ANTIPLANE SHEAR FLOWS 15

and (3.56) \frac{∂p_{2}^{0}}{∂t} = \frac{sin Θ}{1+τ′_{y}(d^{0})} (cos Θ\frac{∂u^{0}}{∂x} + sin Θ\frac{∂u^{0}}{∂y} ) . Not surprisingly, we find that in open sets E of \{(x, y, s) ∈ B (x_{0}, y_{0}, R, t) |0 < \sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} < τ_{y} (d^{0})\} the weak limits satisfy the elasticity equations

\frac{∂τ_{1}^{0}}{∂t} − \frac{∂u^{0}}{∂x} = 0, (3.57)

\frac{∂τ_{2}^{0}}{∂t} − \frac{∂u^{0}}{∂y} = 0. (3.58)

\frac{∂u^{0}}{∂t} − \frac{∂τ_{1}^{0}}{∂x} − \frac{∂τ_{2}^{0}}{∂y} = \frac{β}{μ + β} (\frac{∂p_{1}^{0}}{∂x} + \frac{∂p_{2}^{0}}{∂y} ) (3.59)

and

\frac{∂p_{1}^{0}}{∂t} = \frac{∂p_{2}^{0}}{∂t} = \frac{∂d^{0}}{∂t} = 0. (3.60)

The assertions in part (i) which pertain to (τ_{1}^{ϵ_{i}}, τ_{2}^{ϵ_{i}}, p_{1}^{ϵ_{i}}, p_{2}^{ϵ_{i}}, u^{ϵ_{i}}) follow from (3.2), (3.3), and (3.34). Equations (2.28), (2.35), (3.2), and \frac{∂d}{∂t}^{ϵ_{i}} ≥ 0 imply that the d^{ϵ_{i}}’s are bounded in L_{1} (B (x_{0}, y_{0}, R, t)). Their L_{2} boundedness follows from the inequality (d^{ϵ_{i}} (x y, s))^{2} ≤ 2 (d (x, y, 0))^{2} + 2s∫_{0}^{s}(\frac{∂d}{∂η}^{ϵ_{i}})^{2} (x, y, η) dη (3.61) which in turn implies that

∫_{0}^{t} ( ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − s\}} (d^{ϵ_{i}} (x, y, s))^{2} dxdy) ds

≤ 2t ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t\}} d^{2} (x, y, 0) dxdy

+2∫_{0}^{t}s (∫_{0}^{s} ( ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − η \}} (\frac{∂d}{∂η}^{ϵ_{i}})^{2} (x, y, η) dxdy) dη) ds

≤ 2t ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t\}} d^{2} (x, y, 0) dxdy

+2t^{2}∫_{0}^{t} ( ∫ ∫_{\{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t − η\}} (\frac{∂d}{∂η}^{ϵ_{i}})^{2} (x, y, η) dxdy) dη

(3.62)

As noted previously, the assertions of (ii) follow directly from those of (i) and the governing equations (2.39) - (2.41).

The veracity of (iii) follows from the inequality

0 ≤ δm\{(x, y, s) ∈ B(x_{0}, y_{0}, R, t)|\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y}(d^{o}) ≥ δ\}

≤ ∫ ∫ ∫_{\{(x, y, s) ∈ B(x_{0}, y_{0}, R, t)|\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y}(d^{0}) ≥ δ\}} (\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y}(d^{0})) dxdyds

= ∫ ∫ ∫_{\{(x, y, s) ∈ B(x_{0}, y_{0}, R, t)|\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y}(d^{0}) ≥ δ\}} (\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − \sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}}) dxdyds

+ ∫ ∫ ∫_{\{(x, y, s) ∈ B(x_{0}, y_{0}, R, t)|\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y}(d^{0}) ≥ δ\}} (τ_{y}(d^{ϵ_{i}}) − τ_{y}(d^{0})) dxdyds

+ ∫ ∫ ∫_{\{(x, y, s) ∈ B(x_{0}, y_{0}, R, t)|\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y}(d^{0}) ≥ δ\}} (\sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} − τ_{y}(d^{ϵ_{i}})) dxdyds,

(3.63)

the strong convergence results of part (i) which guarantee that the first two integrals on the right-hand side of (3.63) converge to zero as the ϵ_{i}’s tend to zero, and from the observation that the third integral is bounded from above by

∫ ∫ ∫_{B (x_{0}, y_{0}, R, t)} (\sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} − τ_{y} (d^{ϵ_{i}}))_{+} dxdyds

which in turn is bounded by

(m (B (x_{0}, y_{0}, R, t)))^{1/2} ( ∫ ∫ ∫_{B (x_{0}, y_{0}, R, t)} ((\sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} − τ_{y} (d^{ϵ_{i}}))_{+})^{2} dxdyds)^{1/2} .

(3.64)

The identity (3.2) with f given by (3.3) and g by (3.6) guarantees that

\begin{matrix}∫ ∫ ∫_{B (x_{0}, y_{0}, R, t)} ((\sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} − τ_{y} (d^{ϵ_{i}}))_{+})^{2} dxdyds ≤ \\ \begin{matrix}\frac{ϵ_{i}(μ+β)}{μ} & ∫ & ∫ \\ \{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} ≤ R + t\} \end{matrix} (\frac{μ(τ_{1}^{2}+τ_{2}^{2})}{2(μ+β)} + ^{\frac{β((τ_{1}+p_{1})^{2}+(τ_{2}+p_{2})^{2})^{2}}{2(μ+β)}} + \frac{u^{2}}{2} +\frac{μ}{μ+β} ∫_{0}^{d}τ_{y} (η) dη) (x, y, 0) dxdy \end{matrix}

(3.65)

and (3.64) and (3.65) imply that the third integral on the right-hand side of (3.63) tends to zero as ϵ_{i} tends to zero.

To establish (3.57)-(3.60) in open subsets E of \{(x, y, s) ∈ B (x_{0}, y_{0}, R t) |0 < \sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} < τ_{y} (d^{0})\}

it suffices to show that \frac{∂d^{0}}{∂t} = 0 on E. Equations (3.43) and (3.44) will then guaran-

tee that \frac{∂p_{1}^{0}}{∂t} = \frac{∂p_{2}^{0}}{∂t} = 0 and these identities, along with (2.36)-(2.38) will guarantee that (3.57)-(3.60) hold.

In what follows we let δ > 0,

E_{δ} = \{(x, y, s) ∈ B (x_{0}, y_{0}, R, t) |\sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}} − τ_{y} (d^{0}) ≤ −δ < 0\} . (3.66)

and observe that

\begin{aligned}| \begin{matrix}∫ ∫ & \frac{∂d^{0}}{∂s} dxdyds \\ E_{δ} \end{matrix} | & ≤ & | \begin{matrix}∫ ∫ ∫ & (\frac{∂d^{0}}{∂s} − \frac{∂d}{∂s}^{ϵ_{i}}) dxdyds \\ E_{δ} \end{matrix} | \\ & + & | \begin{matrix}∫ ∫ ∫ & \frac{∂d}{∂s}^{ϵ_{i}}dxdyds \\ E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\} \end{matrix} | \end{aligned} (3.67)

The weak convergence of \frac{∂d^{ϵ_{i}}}{∂s} to \frac{∂d^{0}}{∂s} guarantees that the first integral on the righthand side of (3.67) may be made arbitrarily small. We estimate the second integral by

( ∫ ∫ ∫_{B (x_{0}, y_{0}, R, t)} (\frac{∂d}{∂s}^{ϵ_{i}})^{2} dxdyds )^{1/2} m(E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\})^{1/2}.

That the first factor is bounded follows from (3.2) with f given by (3.34) and D = \frac{∂}{∂t}.

Thus, it suffices to show that lim_{i→∞} m (E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\}) = 0. To establish this

assertion we observe that \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\} = \{\sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} − τ_{y} (d^{ϵ_{i}}) > 0\}, and that

\begin{aligned}δm(E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\}) ≤ & ∫ ∫ ∫ & (τ_{y} (d^{0}) − \sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}}) dxdyds \\ & E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\} \end{aligned}

\begin{aligned}≤ & ∫ ∫ ∫ & (τ_{y} (d^{0}) − τ_{y} (d^{ϵ_{i}})) dxdyds \\ & E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\} \end{aligned}

\begin{aligned}+ & ∫ ∫ ∫ & (\sqrt{(τ_{1}^{ϵ_{i}})^{2} + (τ_{2}^{ϵ_{i}})^{2}} − \sqrt{(τ_{1}^{0})^{2} + (τ_{2}^{0})^{2}}) dxdyds. \\ & E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\} \end{aligned}

(3.68)

The strong convergence results of (i) imply that the latter two integrals tend to zero as the ϵ_{i}’s tend to zero thereby yielding lim_{i→∞} m (E_{δ} ∩ \{\frac{∂d}{∂s}^{ϵ_{i}} > 0\}) = 0.

4. Computational Experiments. In this section we present some computational experiments for the dimensionless system (2.36) - (2.41) when the normalized yield stress is given by

τ_{y} = 1 + c_{1} + (c_{1} − c_{2}) d − \frac{c_{1}}{1 + d} (4.1)

and

0 < c_{2} < c_{1}. (4.2)

Since the flows associated with this system may be quite complicated we restrict our attention to problems with Riemann type data where

(τ_{1}, τ_{2}, p_{1}, p_{2}, d) (x, y, 0^{+}) ≡ (0, 0, 0, 0, 0) (4.3)

and

u (x, y, 0^{+}) = \{ \begin{matrix}u_{0}, & if & xy > 0 \\ −u_{0}, & if & xy < 0 \end{matrix} (4.4)

where u_{0} is a constant. The solutions generated by this data exhibit a high degree of symmetry and thus when visualizing them we may confine our attention to one of the four quadrants (k − 1)\frac{π}{2} ≤ θ ≤ \frac{kπ}{2} , k = 1, …, 4. The data for u(x, y, 0^{+}) is not H_{1}^{loc} but the functions

u^{h}(x, y, 0^{+}) = \{ \begin{matrix}& u_{0}, if x > \frac{h}{2} and y > \frac{h}{2} or x < −\frac{h}{2} and y < −\frac{h}{2}, \\ − & u_{0}, if x < −\frac{h}{2} and y > \frac{h}{2} or x > \frac{h}{2} and y < −\frac{h}{2}, \\ − & u_{0} + \frac{2u_{0}}{h} (x + \frac{h}{2}) , if − \frac{h}{2} ≤ x ≤ h/2 and y ≥ \frac{h}{2}, \\ & u_{0} − \frac{2u_{0}}{h} (x + \frac{h}{2}) , if − \frac{h}{2} ≤ x ≤ \frac{h}{2} and y ≤ −\frac{h}{2}, \\ & u_{0} − \frac{2u_{0}}{h} (y + \frac{h}{2}) , if − \frac{h}{2} ≤ y ≤ \frac{h}{2} and x ≤ −\frac{h}{2}, \\ − & u_{0} + \frac{2u_{0}}{h} (y + \frac{h}{2}) , if − \frac{h}{2} ≤ y ≤ \frac{h}{2} and x ≥ \frac{h}{2}, \\ & u_{0} − \frac{2u_{0}}{h} (x + \frac{h}{2}) − \frac{2u_{0}}{h} (y + \frac{h}{2}) + \frac{3u_{0}}{h^{2}} (x + \frac{h}{2}) (y + \frac{h}{2}) , \\ & if − \frac{h}{2} ≤ x ≤ \frac{h}{2} and − \frac{h}{2} ≤ y ≤ \frac{h}{2}. \end{matrix}

(4.5) are H_{1}^{loc} and this, together with our L_{2}^{loc} contractivity estimate of the previous section, is sufficient to guarantee that the solution to (2.36) - (2.41) taking on the data (4.3) and (4.5) has a strong L_{2}^{loc} limit as h → 0^{+} which satisfies (2.36) - (2.41), (4.3), and (4.4). This limiting behavior is true when ϵ > 0 is fixed and also in the ϵ = 0^{+} limit when the rate independent equations (3.52) - (3.60) govern.

Our updating algorithm is as follows. We assume we are given (τ_{1}, τ_{2}, p_{1}, p_{2}, d, u)^{N} (x, y) on the x − y plane. These represent the approximate solution at time t = (N − 1/2) δ where δ is our time step and N ≥ 1. To advance these data we successively solve the following systems:

\{ \begin{matrix}\frac{∂τ_{1}}{∂t} − \frac{∂u}{∂x} = 0, \frac{∂τ_{2}}{∂t} = 0, \frac{∂u}{∂t} − \frac{∂}{∂x} (τ_{1} + \frac{β}{μ + β} p_{1}) = 0, \\ and \frac{∂p_{1}}{∂t} = \frac{∂p_{2}}{∂t} = \frac{∂d}{∂t} = 0, 0 ≤ t ≤ δ, \end{matrix} (4.6)

\{ \begin{matrix}\frac{∂τ_{1}}{∂t} = 0, \frac{∂τ_{2}}{∂t} − \frac{∂u}{∂y} = 0, \frac{∂u}{∂t} − \frac{∂}{∂y} (τ_{2} + \frac{β}{μ + β} p_{2}) = 0, \\ and \frac{∂p_{1}}{∂t} = \frac{∂p_{2}}{∂t} = \frac{∂d}{∂t} = 0, 0 ≤ t ≤ δ, \end{matrix} (4.7)

and

\{ \begin{matrix}\frac{∂}{∂t} (τ_{1} + p_{1}) = \frac{∂}{∂t} (τ_{2} + p_{2}) = \frac{∂u}{∂t} = 0, \\ \frac{∂p_{1}}{∂t} = \frac{τ_{1} (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ\sqrt{τ_{1}^{2} + τ_{2}^{2}}} , \frac{∂p_{2}}{∂t} = \frac{τ_{2} (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ\sqrt{τ_{1}^{2} + τ_{2}^{2}}} \\ and \frac{∂d}{∂t} = \frac{(\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ} , 0 ≤ t ≤ δ. \end{matrix} (4.8)

Our principal reason for this splitting is that the systems (4.6) and (4.7) may be updated exactly by elementary characteristic methods and (4.8) may be easily integrated to any desired order of accuracy via Runge Kutta methods.

For (4.6) we use (τ_{1}, τ_{2}, p_{1}, p_{2}, d, u)^{N} as initial data and let (τ_{1}^{1}, τ_{2}^{1}, p_{1}^{1}, p_{2}^{1}, d^{1}, u^{1}) denote the solution to (4.6) with these data at time t = δ. We then solve (4.7) using (τ_{1}^{1}, τ_{2}^{1}, p_{1}^{1}, p_{2}^{1}, d, u^{1}) as initial data and let (τ_{1}^{2}, τ_{2}^{2}, p_{1}^{2}, p_{2}^{2}, d^{2}, u^{2}) denote the solution at t = δ. We next repeat the process but first solve (4.7) with the data (τ_{1}, τ_{2}, p_{1}, p_{2}, d, u)^{N} and let (τ_{1}^{3}, τ_{2}^{3}, p_{1}^{3}, p_{2}^{3}, d^{3}, u^{3}) denote the solution at t = δ. We then use (τ_{1}^{3}, τ_{2}^{3}, p_{1}^{3}, p_{2}^{3}, d^{3}, u^{3}) as data for (4.6) and let (τ_{1}^{4}, τ_{2}^{4}, p_{1}^{4}, p_{2}^{4}, d^{4}, u^{4}) denote the solution at t = δ. Finally we average the approximate solutions indexed by (2) and (4) and denote the result as (τ_{1}^{5}, τ_{2}^{5}, p_{1}^{5}, p_{2}^{5}, d^{5}, u^{5}); that is

(τ_{1}^{5}, τ_{2}^{5}, p_{1}^{5}, p_{2}^{5}, d^{5}, u^{5}) = \frac{1}{2} (τ_{1}^{2} + τ_{1}^{4}, τ_{2}^{2} + τ_{2}^{4}, p_{1}^{2} + p_{1}^{4}, p_{2}^{2} + p_{2}^{4}, d^{2} + d^{4}, u^{2} + u^{4}) . (4.9)

We note that this particular approximation represents a second order update to the “elastic” wave equation:

\{ \begin{matrix}\frac{∂τ_{1}}{∂t} − \frac{∂u}{∂x} = 0, \frac{∂τ_{2}}{∂t} − \frac{∂u}{∂y} = 0, \frac{∂u}{∂t} − \frac{∂}{∂x} (τ_{1} + \frac{β p_{1}}{μ + β} ) − \frac{∂}{∂y} (τ_{2} + \frac{β p_{2}}{μ + β} ) = 0, \\ \frac{∂p_{1}}{∂t} = \frac{∂p_{2}}{∂t} = \frac{∂d}{∂t} = 0, 0 ≤ t ≤ δ \end{matrix}

(4.10)

taking on the data (τ_{1}, τ_{2}, p_{1}, p_{2}, d, u)^{N} at t = 0 and does better than either of the approximates labeled 2 or 4; in particular solution symmetries are preserved via the averaging algorithm.

The final step in our algorithm involves solving (4.8) with the data (τ_{1}^{5}, τ_{2}^{5}, p_{1}^{5}, p_{2}^{5}, d^{5}, u^{5}). Over the interval 0 ≤ t ≤ δ we have

τ_{1} + p_{1} ≡ τ_{1}^{5} + p_{1}^{5}, τ_{2} + p_{2} ≡ τ_{2}^{5} + p_{2}^{5}, and u ≡ u^{5} (4.11)

and

\{ \begin{matrix}\frac{∂τ_{1}}{∂t} = \frac{τ_{1} (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ\sqrt{τ_{1}^{2} + τ_{2}^{2}}} , \frac{∂τ_{2}}{∂t} = \frac{τ_{2} (\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ\sqrt{τ_{1}^{2} + τ_{2}^{2}}} \\ and \frac{∂d}{∂t} = \frac{(\sqrt{τ_{1}^{2} + τ_{2}^{2}} − τ_{y} (d))_{+}}{ϵ} . \end{matrix} (4.12)

If we let

(4.13) τ_{1} = J cos Θ and τ_{2} = J sin Θ,

then equation (4.12) implies

J + d ≡ J^{5} + d^{5} and Θ ≡ Θ^{5}, 0 ≤ t ≤ δ (4.14)

and

\frac{∂d}{∂t} = \frac{(J^{5} + d^{5} − d − τ_{y} (d))_{+}}{ϵ} , 0 ≤ t ≤ δ . (4.15)

In (4.14), J^{5} = \sqrt{(τ_{1}^{5})^{2} + (τ_{2}^{5})^{2}} and 0 ≤ Θ^{5} < 2π satisfies

(4.16) cos Θ^{5} = \frac{τ_{1}^{5}}{J^{5}} and sin Θ^{5} = \frac{τ_{2}^{5}}{J^{5}} .

In what follows we let d^{6} denote our update of (4.15) taking on the data d^{5} at t = 0. Equation (4.14) then implies that

\{ \begin{matrix}J^{6} = J^{5} + d^{5} − d^{6}, τ_{1}^{6} = J^{6} \frac{τ_{1}^{5}}{J^{5}} , τ_{2}^{6} = J^{6} \frac{τ_{2}^{5}}{J^{5}} , \\ u^{6} = u^{5}, p_{1}^{6} = p_{1}^{5} + τ_{1}^{5} − τ_{1}^{6}, and p_{2}^{6} = p_{2}^{5} + τ_{2}^{5} − τ_{2}^{6}. \end{matrix} (4.17)

Our approximate solution at t = (N + 1/2) δ is given by the update labeled 6. To obtain the approximate solution at t = δ/2 we merely solve (4.8) over the interval 0 ≤ t ≤ δ/2 with the prescribed initial data and take the value of their update at t = δ/2 to be (τ_{1}, τ_{2}, p_{1}, p_{2}, d, u)^{1}.

The snapshots shown in Figure 1-18 were run with the normalized yield stress given by (4.1) when c_{1} = 1 and c_{2} = .5. The parameter u_{0} defining the initial data was set to 1.5 and we chose δ = h = .01. The parameter ϵ was set to 0.1. Surface renderings of J = \sqrt{τ_{1}^{2} + τ_{2}^{2}}, d, and u are shown at times .3, .4, and .5.

The purely one dimensional nature of the solutions away from the corner where strong interactions take place is evident from these simulations and it is clear from these calculations that our algorithm captures the sharp contact discontinuities in J and u correctly. Our algorithm is easy to implement and avoids a number of thorny issues we would have to contend with if we tried to integrate the reduced ϵ = 0^{+} equations directly.

REFERENCES

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[3] J. M. GREENBERG AND ANNE NOURI Antiplane Shearing Motions of a Visco-plastic Solid, Siam J. Math. Anal., Vol. 24, No. 4, 1993, pp. 943-967.

[4] A. NOURI AND M. RASCLE, A Global Existence and Uniqueness Theorem for a Model Problem in Dynamic Elasto-Plasticity with Isotropic Strain-Hardening, Siam J. Math. Anal., Vol. 26, No. 4, 1995, pp. 850-868.

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