text:
\section{Introduction}
\label{Introduction}

Biomechanical contact problems involving transmission of forces
across biological joints are of considerable practical interest
(see, e.g. \cite{Arg12,Arg13,Hunz01,Ow_Way}). Many analytical solutions to the problem of
contact interaction of articular cartilage surfaces in joints are
available. In particular, Ateshian et al. \cite{Ateshian} obtained
an asymptotic solution for the axisymmetric contact problem for
two identical biphasic cartilage layers consisting of a solid phase and a fluid phase and attached to two rigid
impermeable spherical bones of equal radii. Later, Wu et al.
\cite{Wu96} extended this solution to a more general axisymmetric
model by combining the assumption of the kinetic relationship from
classical contact mechanics \cite{John85} with the joint contact
model \cite{Ateshian} for the contact of two biphasic cartilage layers.
An improved solution for the contact of two biphasic cartilage
layers in the axisymmetric setting, which can be used for dynamic
loading, was obtained by Wu et al. \cite{Wu97}.

An asymptotic modeling approach to study the contact problem for biphasic
cartilage layers has been performed by Argatov and Mishuris in a series of articles
(see \cite{Arg_Mish_1,Arg_Mish_2,Arg_Mish_4}). In
particular,  it was shown \cite{Arg_Mish_1} that accounting for the
tangential displacements is important in the case of diseased
cartilage where the measurement of indentation depth may differ
even as much as 10\% in comparison with the healthy case. In
\cite{Arg_Mish_2}, the unilateral contact problem for articular cartilages bonded to subchondral
bones with a contact zone in the shape of an arbitrary ellipse has
been considered, and a closed form analytic solution was found.
Exploiting this exact result, Argatov and Mishuris
\cite{Arg_Mish_4} have performed perturbation analysis of the
contact problem with approximate geometry of the contact surfaces.
Other analytic solutions for the contact problem were
found using the viscoelastic cartilage model for elliptic contact
zone in \cite{Arg_Mish_3}.  A new methodology for modeling
articular tibio-femoral contact based on the developed asymptotic
model of frictionless elliptical contact interaction between thin
biphasic cartilage layers was presented in \cite{Arg12}. The
mathematical model of articular contact was extended to the case of
contact between arbitrary viscoelastic incompressible coating
layers.

In this study we extend results obtained in papers
\cite{Arg_Mish_1} and \cite{Arg_Mish_2} by considering the
influence of the tangential displacements on the contact problem
for cartilage layers with the contact zone of elliptic shape based on the
biphasic material model. Note that the perturbation method proposed in
\cite{Arg_Mish_4} could be one of the options for the analysis,
however, the procedure is too complex to perform even a few
asymptotic steps. Here, employing some technique and ideas from
\cite{Arg_Mish_1} and \cite{Arg_Mish_2}, we propose
 another way to construct the asymptotics which utilises the assumption that the
 shape of the contact zone is an ellipse at the initial stage of deformation
 and can be regarded as a small perturbation of the ellipse at any other stage of deformation.

 The paper is organized as follows. The unilateral contact problem formulation and its linearization are presented in Section~\ref{Formulation}, where a special case of the contact configuration with one cartilage layer being plane and rigid is also considered in detail.
In Section~\ref{General}, we derive exact general relationships between the contact approach and some integral characteristics of the contact pressure, including the contact force.
In Section~\ref{IAsymptoticEstimates}, we obtain asymptotic representations for the contact pressure integral characteristics in terms of the contact approach and some integral characteristics of the contact zone.
The zero-order and first-order asymptotic approximations for the solution to the contact problem are obtained in Sections~\ref{Zero-appsub} and \ref{First-appsub}, respectively. Namely, the first-order approximation problem constitutes the main result of the present study.

\section{Formulation of the contact problem}
\label{Formulation}

We consider a frictionless contact between two thin linear
biphasic cartilage layers  firmly attached to rigid bones shaped like
elliptic paraboloids. In the Cartesian co-ordinates $(x_1, x_2, z)
= ({\bf x}, z)$ the equations for the two cartilage surfaces can
be written in the form $z = (-1)^{n} \Phi^{(n)}({\bf x})$, $n = 1,
2$, where
\begin{equation}
\label{surfaces}
\Phi^{(n)}({\bf x}) = \frac{x_1^2}{2 R_1^{(n)}} + \frac{x_2^2}{2 R_2^{(n)}}
\end{equation}
with $R_1^{(n)}$, $R_2^{(n)}$ being the curvature radii of the
$n$-th bone surface at its apex.

In the undeformed  state, the cartilage-bone systems occupy convex
domains $z \leq - \Phi^{(1)}({\bf x})$ and $z \geq \Phi^{(2)}({\bf
x})$, respectively. They are in the initial contact with the plane
$z = 0$ at the origin of the co-ordinate system.

We denote by $w_1({\bf x},t)$, $w_2({\bf x},t)$ the local
vertical displacements of the corresponding cartilage surfaces. Let also ${\bf u}_1({\bf x},t)$, ${\bf u}_2({\bf
x},t)$ be the local horizontal (tangential) displacements of the
corresponding surface of the cartilages. Finally, we denote by
$P({\bf x},t)$ the contact pressure density. In this
notation the equations for the cartilage surfaces can be written
in the following form:
\begin{equation}
\begin{array}{c}
\label{pert_surfaces}
z = {\delta}_1(t) - \Phi^{(1)}\left({\bf x} + {\bf u}_1({\bf x},t)\right) + w_1({\bf x},t),\\[4mm]
z = - {\delta}_2(t) + \Phi^{(2)}\left({\bf x} + {\bf u}_2({\bf
x},t)\right) - w_2({\bf x},t).
\end{array}
\end{equation}
Here, ${\delta}_1$, ${\delta}_2$ are some (positive) vertical
displacements of the rigid bones. Note also that the vertical displacements
$w_1$, $w_2$ are positive, while the tangential displacements ${\bf
u}_1$, ${\bf u}_2$ are directed outside of the contact zone.
 Denoting by
$\delta_*(t) = {\delta}_1(t) + {\delta}_2(t)$ the contact
approach of the bones, we get
from (\ref{pert_surfaces}) the following inequality:
\begin{equation}
\label{inequality} \delta_*(t) +  w_1({\bf x},t) +  w_2({\bf x},t)
 \leq
\Phi^{(1)}\left({\bf x} + {\bf u}_1({\bf x},t)\right)
+ \Phi^{(2)}\left({\bf x} + {\bf u}_2({\bf x},t)\right).
\end{equation}

It was shown in \cite{Ateshian} (see also \cite{Arg_Mish_1}) that vertical and the tangential
displacements of each bone can be represented in the form
\begin{equation}\label{EqForW}
w_n({\bf x},t)=\frac{h_n \epsilon_n^2}{3 \mu_{s,n}} \left\{\Delta
P({\bf x},t)+\frac{3}{H_n}\int\limits_ {0}^{t}
\Delta P({\bf x},\tau) d\tau \right\},\;\quad  n = 1, 2,
\end{equation}
\begin{equation}\label{EqForU}
{\bf u}_n({\bf x},t) = - \frac{h_n \epsilon_n}{2 \mu_{s,n}}
\nabla P({\bf x},t),\;\;\; n = 1, 2.
\end{equation}
Here $\epsilon_n =h_n/a_0$ are dimensionless small parameters,
$h_1$, $h_2$ mean the thicknesses of the cartilage layers, and
$a_0$ denotes a characteristic measure of the contact zone (see
the detailed description of the role of this parameter in
\cite{Arg_Mish_1}), $H_n=(\lambda_{s,n}+2\mu_{s,n})/\mu_{s,n}$ are
material parameters of cartilages, where $\lambda_{s,n}$ and
$\mu_{s,n}$ represent the first Lame coefficient and the shear
modulus of the solid phase of the $n$-th cartilage tissue. Note
that ${\bf u}_1$ and ${\bf u}_2$ in (\ref{EqForU}) do not
necessarily coincide, they depend on both spatial variables $x_1$,
$x_2$, and on the time variable $t$.

Following \cite{Ateshian}, we introduce new spatial variables and
time variable via formulas $$x^{\prime}_j =\frac{x_j}{a_0}, \quad j =
1, 2, \quad t^{\prime} =\frac{\chi t}{\mu_0},$$ where $$\chi =
\frac{3 \mu_{s,1} k_1}{h_1^2} + \frac{3 \mu_{s,2} k_2}{h_2^2},\quad
\mu_0 = \frac{\mu_{s,1}}{\lambda_{s,1}+2\mu_{s,1}} +
\frac{\mu_{s,2}}{\lambda_{s,2}+2\mu_{s,2}},$$ $a_0$ is a
characteristic measure of the contact zone, and $k_1,  k_2$ are
the cartilage's permeabilities. In these variables we have the
following relations on the contact area $\omega(t)$ encircled by
the curve $\Gamma(t) = \partial \omega(t)$:
\begin{equation}\label{EqForW_n}
w_1({\bf x}^{\prime},t^{\prime}) + w_2({\bf x}^{\prime},t^{\prime}) =
\left(\frac{h_1^3}{3 \mu_{s,1}} + \frac{h_2^3}{3 \mu_{s,2}}\right)\left\{\Delta P({\bf x}^{\prime},t^{\prime}) + \chi \int\limits_{0}^{t^{\prime}} \Delta P({\bf x}^{\prime},\tau^{\prime}) d\tau^{\prime}\right\},
\end{equation}
\begin{equation}\label{EqForPhi}
\Phi^{(n)}({\bf x}^{\prime} + {\bf u}_n({\bf x}^{\prime},t^{\prime})) \simeq \Phi^{(n)}({\bf x}^{\prime}) - \frac{h_n^2 a_0}{2 \mu_{s,n}} \nabla \Phi^{(n)}({\bf x}^{\prime}) \cdot \nabla P({\bf x}^{\prime}, t^{\prime}), \quad n = 1, 2.
\end{equation}

Further the equality in (\ref{inequality}), i.e.,
\begin{equation}\label{EqBeforeSubst}
\delta_*(t^{\prime}) + w_1({\bf x}^{\prime},t^{\prime}) + w_2({\bf
x}^{\prime},t^{\prime}) = \Phi^{(1)}\left({\bf x} + {\bf u}_1({\bf x},t)\right)
+ \Phi^{(2)}\left({\bf x} + {\bf u}_2({\bf x},t)\right),
\end{equation}
determines the contact area $\omega(t)$.

Now we substitute (\ref{EqForW_n}), (\ref{EqForPhi}) into
(\ref{EqBeforeSubst}) and obtain the governing equation relating
the contact pressure with the vertical approach of the bones
$\delta_*(t)$ in the following form (from now on we keep the names
of new unknown functions, e.g. $\Phi({\bf x}) := \Phi({\bf
x}^{\prime} a_0)$ etc.):
\begin{equation}\label{EqForPress}
\Delta P({\bf x},t)+ {\chi} \int\limits_ {0}^{t}
\Delta P({\bf x},\tau) d\tau = {m} \left(\Phi({\bf x})-\delta_*(t)
-
\nabla \widetilde{\Phi}({\bf x}) \cdot \nabla
P({\bf x},t)\right).
\end{equation}
Here we have introduced the notation
\begin{equation}
\label{m} m = \left (\frac{h_1^3}{3 \mu_{s,1}}
+\frac{h_2^3}{3\mu_{s,2}}\right)^{-1},
\end{equation}
\begin{equation}
\label{Phi}
\Phi({\bf x}^{\prime}) = \Phi^{(1)}({\bf x}^{\prime}) + \Phi^{(2)}({\bf x}^{\prime}).
\end{equation}

Thus, it follows from (\ref{surfaces}) and  (\ref{Phi}) that the
functions $\Phi$ and $\widetilde{\Phi}$ are given by
{\begin{equation}\label{Phi_Eq} \Phi({\bf x}) = \Phi(x_1, x_2) = A
x_1^2 + B x_2^2
\end{equation}
with
$$
{A} = \frac{1}{2 R_1^{(1)}} + \frac{1}{2  R_1^{(2)}},\; \quad {B}
= \frac{1}{2 R_2^{(1)}} + \frac{1}{2 R_2^{(2)}}
$$
and
\begin{equation}\label{TildePhi_Eq}
\widetilde{\Phi}({\bf x}) =\widetilde{\Phi}(x_1, x_2) = \widetilde{A} x_1^2 + \widetilde{B} x_2^2.
\end{equation}

Note that the coefficients in $\widetilde{A}$ and $\widetilde{B} $ are positive dimensionless numbers, which are less than unit.

Without loss of generality, one can assume that $A>B$. Then, Eq. ~(\ref{EqForPress}) can
be rewritten in an equivalent form, using all dimensionless
parameters:\footnote{Note that in the axisymmetric case formula
(\ref{EqFP_new}) coinsides with formula \cite[(8)]{Arg_Mish_1}.}
\begin{equation}\label{EqFP_new}
\Delta P_\varepsilon({\bf x},t)+ {\chi} \int\limits_ {0}^{t}
\Delta P_\varepsilon({\bf x},\tau) d\tau = {\mu} \big(\Psi_1({\bf
x})-\delta_\varepsilon(t) - \varepsilon \nabla {\Psi}_2({\bf x})
\cdot \nabla P_\varepsilon({\bf x},t)\big),
\end{equation}
where the following notation has been introduced:
\begin{equation}\label{Psi_Eq}
{\Psi}_j({\bf x})= x_1^2 + e_j^2x_2^2,\quad j=1,2,\quad
\delta_\varepsilon(t)=\frac{1}{A}\delta_*(t),
\end{equation}
\[
\mu=Am,\quad e_1=\sqrt{B/A},\quad e_2=\sqrt{{\widetilde B}/{\widetilde A}},\quad \varepsilon=\frac{{\widetilde A}}{A}.
\]

It is important to note that
\begin{equation}
\label{parameters}
\chi = O(1),\quad   \mu\varepsilon\ll \chi.
\end{equation}
Discussion of the characteristic values of the introduced parameters is presented in Section \ref{Discussion} (see also \cite{Ateshian,Arg_Mish_1}).

Since the solution of (\ref{EqFP_new}) depends on the parameter
$\varepsilon$, it is customer to denote an unknown contact
pressure by $P=P_\varepsilon$ in what follows. Note that the
problem for $\varepsilon=0$ coincides with that considered in
\cite{Arg_Mish_2}, where an exact solution to this problem was
found.

{Equation (\ref{EqFP_new}) is the equation for determination of
the  contact pressure $P_\varepsilon({\bf x},t) \geq 0$, ${\bf x}
\in \omega_\varepsilon(t)$. In particular, in the case when the
contact domain is represented by an ellipse
\begin{equation}
\label{domain} \omega_\varepsilon(t) = \left\{ {\bf x}\in
{\mathbb R}^2: \; \frac{x_1^2}{b^2(t,\varepsilon)} + \frac{\beta^2(t,\varepsilon)
x_2^2}{b^2(t,\varepsilon)} \leq 1\right\}.
\end{equation}}

We supply Eq. ~(\ref{EqFP_new}) with the following boundary
conditions:
\begin{equation}\label{Bound1}
P_\varepsilon({\bf x},t)= 0,\;\;\; {\bf x}\in \Gamma(t),
\end{equation}
\begin{equation}\label{Bound2}
\frac{\partial
P_\varepsilon}{\partial n}({\bf x},t)= 0,\;\;\; {\bf x}\in \Gamma(t).
\end{equation}

The equilibrium equation
\begin{equation}
\label{load} \int\!\!\!\!\!\int\limits_{\omega_\varepsilon(t)}
P_\varepsilon({\bf x},t) d{\bf x}=F(t)
\end{equation}
connects the external load $F(t)$, unknown contact pressure
$P_\varepsilon({\bf x},t)$, and unknown contact domain  $
\omega_\varepsilon(t)$.

\subsection{Special case of the contact configuration}

In order to check the content of formula (\ref{EqForPress}) we
consider here a special case, namely, we suppose that the lower
part cartilage layer is plane and rigid (the same assumption was
employed in \cite{Wu96}), it means that $\mu_{s,2}=\infty$ and
$R_1^{(1)} = R_2^{(1)} =
\infty$, i.e.,
$$\Phi^{(1)} \equiv 0, \quad \Phi \equiv \Phi^{(2)}.
$$
In this case we have got the following equation for determination
of the contact domain  $ \omega(t)$ in the form similar to
(\ref{EqForPress}):
\begin{equation}\label{EqForPress_inf}
\Delta P({\bf x},t)+ {\chi} \int\limits_ {0}^{t}
\Delta P({\bf x},\tau) d\tau = {m} \left(\Phi({\bf x})-\delta_*(t)
-
\nabla \widetilde{\Phi}({\bf x}) \cdot \nabla
P({\bf x},t)\right).
\end{equation}
Here we will have
\begin{equation}
\label{m_inf}
m = \frac{3\mu_{s,2}}{h_2^3},\quad \chi = \frac{3 \mu_{s,2} k_2}{ h_2^2}.
\end{equation}
At the same time, small changes have to be made in the right-hand
side of Eq. ~(\ref{EqForPress_inf}) as follows:
$$
\Phi({\bf x}) = \frac{x_1^2}{2 R_1^{(2)}} + \frac{x_2^2}{2 R_2^{(2)}},
$$
$$
\tilde{\Phi}({\bf x}) = \frac{h_2^2 a_0 x_1^2}{2 \mu_{s,2} R_1^{(2)}} + \frac{h_2^2 a_0 x_1^2}{2 \mu_{s,2} R_1^{(2)}}.
$$
Thus Eq. ~(\ref{EqForPress_inf}) can be rewritten as
\begin{eqnarray}\label{EqForPress_inf1}
\Delta P({\bf x},t)  & + & \frac{3 \mu_{s,2} k_2}{ h_2^2} \int\limits_ {0}^{t}
\Delta P({\bf x},\tau) d\tau  =   \frac{3\mu_{s,2}}{h_2^3} \left(\frac{x_1^2}{2 R_1^{(2)}} + \frac{x_2^2}{2 R_2^{(2)}}
-\delta_*(t) \right) \\
{} & {} & \qquad\qquad{}-\frac{3 a_0}{h_2} \left[\frac{x_1}{R_1^{(2)}} \partial_{x_1} P({\bf x},t) + \frac{x_2}{R_2^{(2)}} \partial_{x_1} P({\bf x},t)\right].
\end{eqnarray}

It can be easily checked that in the axisymmetric case Eq.~(\ref{EqForPress_inf1}) reduces to the governing differential equation obtained in \cite{Arg_Mish_1}.

\section{General relationships between the solution components}
\label{General}
\subsection{Determination of the contact approach}

In our model we assume that the external load is non-decreasing.
Thus, the contact domain is monotonically expanded, i.e.
\begin{equation}
\label{monotone}
\omega_\varepsilon(t_1) \subseteq \omega_\varepsilon(t_2), \quad \forall t_1 \leq t_2.
\end{equation}
It is convenient to suppose also that the contact pressure is defined on the whole plane. For this we simply extend
the density $P_\varepsilon({\bf x},t)$ by assuming that
\begin{equation}
\label{pressure}
P_\varepsilon({\bf x},t) = 0,\quad \forall {\bf x}\not\in \omega_\varepsilon(t).
\end{equation}

Integrating (\ref{EqFP_new}) over contact domain
$\omega(t)$, we get
\begin{equation}\label{EqForPress1}
\begin{array}{l}
\displaystyle
\int\!\!\!\!\!\int\limits_{\omega(t)}\Delta
P_\varepsilon({\bf x},t) d {\bf x} + {\chi} \int\!\!\!\!\!\int\limits_{\omega(t)}
\!\!\int\limits_ {0}^{t} \Delta P_\varepsilon({\bf x},\tau)
d\tau d {\bf x} =
\\[5mm]
\displaystyle \hspace{20mm}
={\mu}\int\!\!\!\!\!\int\limits_{\omega(t)} \left(\Psi_1({\bf
x})-\delta_{\varepsilon}(t)\right) d {\bf x}  - {\varepsilon\mu}
\int\!\!\!\!\!\int\limits_{\omega(t)} \nabla {\Psi}_2({\bf x})
\cdot \nabla P_\varepsilon({\bf x},t) d {\bf x}.
\end{array}
\end{equation}
For simplicity of notation, we omit here (and everywhere in the
next two sections) the subindex $\varepsilon$ in
$\omega_\varepsilon$.

From the monotonicity of the contact domain (\ref{monotone}) and
assumption (\ref{pressure}), it follows that the second integral
on the left-hand side can be written in the form
\begin{equation}
\label{2+1int}
\int\!\!\!\!\!\int\limits_{\omega(t)}
\!\!\int\limits_ {0}^{t} \Delta
P_\varepsilon({\bf x},\tau) d\tau d
{\bf x} = \int\limits_{0}^{t}\!\!\int\!\!\!\!\!\int\limits_{\omega(t)} \Delta
P_\varepsilon({\bf x},\tau) d {\bf x} d\tau.
\end{equation}

Using the second Green's formula
\begin{equation}\label{Green2}
 \int\!\!\!\!\!\int\limits_{\omega(t)} \left( u({\bf x}) \Delta v({\bf x}) - v({\bf x}) \Delta u({\bf x})\right) d{\bf x} =
\int\limits_{\Gamma(t)} \left( u({\bf x}) \frac{\partial v}{\partial n}({\bf x}) - v({\bf x}) \frac{\partial u}{\partial n}({\bf x})\right)ds
\end{equation}
with $u\equiv 1$ and $v = P_\varepsilon({\bf x},t)$ we get the
following relation in view of the boundary condition
(\ref{Bound2}):
\begin{equation}
\label{int1}
\int\!\!\!\!\!\int\limits_{\omega(t)}   \Delta
P_\varepsilon({\bf x},\tau)  d {\bf x} = \int\limits_{\Gamma(t)}
\frac{\partial P_\varepsilon}{\partial n}({\bf x},s)  ds = 0,\;\;\; \forall \tau \leq t.
\end{equation}
Therefore, the both integrals on the left-hand side of
(\ref{EqForPress1}) vanish.

Further, we use the first  Green's formula
\begin{equation}
\label{Green1}
\int\!\!\!\!\!\int\limits_{\omega(t)}    \left( \varphi \Delta \psi + \nabla \varphi \cdot \nabla \psi
\right) d {\bf x} = \int\limits_{\Gamma(t)} \varphi
\frac{\partial\psi}{\partial n} ds
\end{equation}
with $\psi({\bf x}) = {\Psi}_2({\bf x})$ and $\varphi({\bf x}) =
P_\varepsilon({\bf x},t)$. In this case the integral on the
right-hand side vanishes in view of (\ref{Bound1}), and we obtain
the relation
\begin{equation}
\label{Green1-1}
\int\!\!\!\!\!\int\limits_{\omega(t)} \nabla {\Psi}_2({\bf x}) \cdot \nabla P_\varepsilon({\bf x},t) d {\bf x}
= - \int\!\!\!\!\!\int\limits_{\omega(t)} P_\varepsilon({\bf x},t) \Delta {\Psi}_2({\bf x}) d {\bf x} =
- 2 (1+e_2^2) F(t),
\end{equation}
where we used the equilibrium equation (\ref{load}) and the
identity
\begin{equation}
\label{laplace_phi_2}
\Delta {\Psi}_2({\bf x}) = 2(1 + e_2^2)
\end{equation}
with $e_2$ being defined in (\ref{Psi_Eq}).

{In what follows, it is convenient to have the following notation
for the integrals of the product of $k$-th power of the function
$\Psi_1$ and $l$-th power of the function ${\Psi}_2$:
\begin{equation}
\label{IntPhi_kl} A_{k,l}(\omega) =
\int\!\!\!\!\!\int\limits_{\omega} \Psi_1^{k}({\bf x})
{\Psi}_2^{l}({\bf x})
 d{\bf x}> 0,\;\;\; k,l = 0, 1, 2, \ldots
\end{equation}
In particular, $A_{0,0}(\omega)$ is the area of the contact
domain. It is to remember that the constants $A_{k,l}(\omega)$
depend finally on $t$, but we omitted this fact in the notation in
order to avoid cumbersome expressions. Computations of
$A_{k,l}(\omega)$ for the elliptic domain (\ref{domain}) we
included into Appendix (see Section \ref{$A_{kl}$}).}

Taking into account Eqs. (\ref{int1}) and (\ref{Green1-1}), we get
{\begin{equation}
\label{delta01} \delta_{\varepsilon}(t)=
\frac{A_{1,0}(\omega_\varepsilon(t))}{A_{0,0}(\omega_\varepsilon(t))}  + \frac{2(1 + e_2^2) \varepsilon}{
A_{0,0}(\omega_\varepsilon(t))} F(t).
\end{equation}}
This formula allows us to compute the contact approach $\delta_{\varepsilon}(t)$
as a function of the total external force $F(t)$ and the main axes
of the ellipse describing the shape of the contact zone, which in
fact depends on time too.

\subsection{Some integral identity for the contact pressure}

In order to write out a more informative equation for the contact
load, we use the following trick. We multiply   both sides of
(\ref{EqFP_new}) by the function $v({\bf x})={\Psi}_2( {\bf x})$
and integrate the obtained equation over the contact domain
$\omega(t)$
\begin{eqnarray}
 \int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2(
{\bf x}) \Delta P_\varepsilon({\bf x},t) d {\bf x} & + & {\chi}
\int\!\!\!\!\!\int\limits_{\omega(t)} \int\limits_ {0}^{t} {\Psi}_2( {\bf x}) \Delta
P_{\varepsilon}({\bf x},\tau) d\tau d {\bf x} = \nonumber \\
{} & = &\mu \int\!\!\!\!\!\int\limits_{\omega(t)}  {\Psi}_2( {\bf x})
\Psi_1({\bf x}) d {\bf x} - \mu {\delta_{\varepsilon}(t)}
\int\!\!\!\!\!\int\limits_{\omega(t)}  {\Psi}_2( {\bf x}) d {\bf
x} \nonumber \\
{} & {} & {}- {\mu\varepsilon}  \int\!\!\!\!\!\int\limits_{\omega(t)}
{\Psi}_2( {\bf x})  \nabla {\Psi}_2( {\bf x}) \cdot \nabla
P_\varepsilon({\bf x},t) d{\bf x}. \label{rel0}
\end{eqnarray}
Let us calculate the integrals in this relation by using Green's
formulas. For the first integral on the left-hand side we use formula
(\ref{Green2}) with $u = {\Psi}_2$, $v = P_\varepsilon$ and the
boundary conditions (\ref{Bound1}), (\ref{Bound2}). Hence, we
obtain
\begin{equation*}
\int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2(
{\bf x}) \Delta P_\varepsilon({\bf x},t) d {\bf x} =
\int\!\!\!\!\!\int\limits_{\omega(t)} \Delta {\Psi}_2(
{\bf x}) P_\varepsilon({\bf x},t) d {\bf x}.
\end{equation*}
Now taking into account (\ref{laplace_phi_2}), we get
\begin{equation}
\label{int2-1}
\int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2(
{\bf x}) \Delta P_\varepsilon({\bf x},t) d {\bf x} = 2(1+e_2^2) F(t).
\end{equation}
For the second integral on the left-hand side, we apply the same
approach, but interchange first the integrals over
${\omega_\varepsilon(t)}$ and over $\tau\in (0, t)$ exploiting the
load monotonicity. Therefore, we arrive at the equation
\begin{equation}
\label{int2-2}
\int\!\!\!\!\!\int\limits_{\omega(t)} \int\limits_ {0}^{t}  {\Psi}_2(
{\bf x}) \Delta P_\varepsilon({\bf x},\tau)  d\tau d {\bf x} =
 \int\limits_ {0}^{t} \int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2(
{\bf x}) \Delta P_\varepsilon({\bf x},\tau)  d\tau d {\bf x}  =
2(1+e_2^2)  \int\limits_ {0}^{t} F(\tau) d\tau.
\end{equation}
For the first and second integrals on the right-hand side, we
simply use the notation (\ref{IntPhi_kl}), which gives
\begin{equation}
\label{int2-3} \int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_1({\bf
x}) {\Psi}_2( {\bf x})  d {\bf  x} = A_{1,1}(b; \beta),
\quad \int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2({\bf
x})  d {\bf x} = A_{0,1}(b; \beta).
\end{equation}
Finally, for the third  integral on the right-hand side, we make
use of the following simple formula which follows immediately from
the definition of ${\Psi}_2$:
\begin{equation*}
{\Psi}_2 \nabla {\Psi}_2 = \frac{1}{2} \nabla {\Psi}_2^2.
\end{equation*}
Then we can apply Green's formula (\ref{Green1}) and the boundary
conditions (\ref{Bound1}), (\ref{Bound2}) to find
\begin{equation*}
\int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2(
{\bf x})  \nabla
{\Psi}_2({\bf x}) \cdot \nabla P_\varepsilon({\bf x},t)
d{\bf x}  = - \frac{1}{2} \int\!\!\!\!\!\int\limits_{\omega(t)}
\Delta {\Psi}_2^{2}({\bf x}) P_\varepsilon({\bf x},t)
d{\bf x}.
\end{equation*}

\noindent By applying the second Green's formula (\ref{Green2})
with $u = P_\varepsilon$, $v = {\Psi}_2^2$, and the boundary
conditions (\ref{Bound1}), (\ref{Bound2}), we represent this
integral in the form
\begin{equation}
\label{int2-5}
\int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2(
{\bf x})  \nabla
{\Psi}_2({\bf x}) \cdot \nabla P_\varepsilon({\bf x},t)
d{\bf x} =  - \frac{1}{2} \int\!\!\!\!\!\int\limits_{\omega(t)}
{\Psi}_2^{2}({\bf x}) \Delta P_\varepsilon({\bf x},t)
d{\bf x}.
\end{equation}
This integral still contains the unknown density of contact
pressure $P_\varepsilon({\bf x},t)$. Let us define
\begin{equation}
\label{defin_M}
\mathcal{M}^{(j)}P_\varepsilon(t)\equiv
\int\!\!\!\!\!\int\limits_{\omega(t)}
{\Psi}_2^{j}({\bf x}) \Delta P_\varepsilon({\bf x},t)
d{\bf x}.
\end{equation}

 Now we rewrite the relation (\ref{rel0}) by using the results for all integrals
 (\ref{int2-1})--(\ref{int2-5}) in the following form:
{\begin{equation}
\label{rel1} 2(1+e_2^2){\mathcal K} F(t)  = \mu A_{1,1}(\omega_\varepsilon(t))
- \mu \delta_{\varepsilon}(t) A_{0,1}(\omega_\varepsilon(t)) +
\frac{\mu\varepsilon}{2} \mathcal{M}^{(2)}P_\varepsilon(t).
\end{equation}}

Here, we have introduced the Volterra operator ${\mathcal K}$ as
follows:
\begin{equation}
\label{Operator}
{\mathcal K} F(t) = F(t) + \chi \int\limits_{0}^{t} F(\tau) d\tau.
\end{equation}

Note that the integral in the right-hand side of the equation
(\ref{rel1}) allows to continue the same procedure to deliver an
asymptotic estimate for this equation.

We continue to proceed with Eq. ~(\ref{rel1}) on the next steps.

\subsection{Asymptotic estimates of the integral characteristics $\mathcal{M}^{(j)}P_\varepsilon(t)$}
\label{IAsymptoticEstimates}

Now we proceed to calculate the last integral in (\ref{rel1}). For this we multiply the governing integral equation (\ref{EqFP_new}) by ${\Psi}_2^{j}({\bf x})$ ($j\ge2$) and integrate over contact domain $\omega(t)$:
\begin{eqnarray}
 \int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2^{j}(
{\bf x}) \Delta P_\varepsilon({\bf x},t) d {\bf x} & + & {\chi}
\int\!\!\!\!\!\int\limits_{\omega(t)} \int\limits_ {0}^{t} {\Psi}_2^{j}( {\bf x}) \Delta
P_{\varepsilon}({\bf x},\tau) d\tau d {\bf x} = \nonumber \\
& = & \mu \int\!\!\!\!\!\int\limits_{\omega(t)}  {\Psi}_2^{j}(\overline
{\bf x}) \Psi_1({\bf x}) d {\bf x} - \mu {\delta_{\varepsilon}(t)}
\int\!\!\!\!\!\int\limits_{\omega(t)}  {\Psi}_2^{j}( {\bf x}) d
{\bf x} \nonumber \\
{} & {} & {}- {\mu\varepsilon}  \int\!\!\!\!\!\int\limits_{\omega(t)}
{\Psi}_2^{j}( {\bf x})  \nabla {\Psi}_2( {\bf x}) \cdot \nabla
P_\varepsilon({\bf x},t) d{\bf x}. \label{IntEq_j}
\end{eqnarray}
By using the same argument as on the previous step, we get
\begin{equation}
{\mathcal K} \mathcal{M}^{(j)}P_\varepsilon(t)=
\mu A_{1,j} - \mu \delta_{\varepsilon}(t) A_{0,j}(a; \beta) -
\mu\varepsilon \int\!\!\!\!\!\int\limits_{\omega(t)} {\Psi}_2^{j}(
{\bf x}) \nabla {\Psi}_2( {\bf x}) \cdot \nabla P_\varepsilon({\bf
x},t) d{\bf x}. \label{IntEq_j_c}
\end{equation}
For the last integral we use the relations
$$
{\Psi}_2^{j}(
{\bf x})  \nabla
{\Psi}_2(
{\bf x}) = \frac{1}{j+1} \nabla
{\Psi}_2^{j+1}(
{\bf x})
$$
and
$$
\int\!\!\!\!\!\int\limits_{\omega(t)} \nabla {\Psi}_2^{j+1}({\bf
x}) \cdot \nabla P_\varepsilon({\bf x},t) d{\bf x} = -
\int\!\!\!\!\!\int\limits_{\omega(t)} \Delta {\Psi}_2^{j+1}({\bf
x}) P_\varepsilon({\bf x},t) d{\bf x}.
$$
Therefore, the integral
\begin{equation}
\label{Psi_j} \mathcal{M}^{(j)}P_\varepsilon(t)=\mu
{\mathcal K}^{-1} \left\{ A_{1,j}(\omega_\varepsilon(t)) -
\delta_{\varepsilon}(t) A_{0,j}(\omega_\varepsilon(t)) +
\frac{\varepsilon}{j+1} {\mathcal K} \mathcal{M}^{(j+1)}P_\varepsilon(t)\right\}
\end{equation}
has been obtained as a solution of the integral equation (\ref{IntEq_j_c}).
Here the inverse operator ${\mathcal K}^{-1}$ is defined by the formula
\begin{equation}
\label{K_inv}
\mathcal K^{-1} Y(t) = Y(t) - \chi \int\limits_{0}^{t}
Y(\tau)e^{-\chi(t-\tau)} d\tau.
\end{equation}

Performing the same computation, we obtain the following
representation for the integral in the right-hand side of
(\ref{rel1}):
{$$
\mathcal{M}^{(2)}P_\varepsilon(t)  = \sum\limits_{j=1}^{N}
\frac{2 \varepsilon^{j-1}}{(j+1)!} \mu^{j} {\mathcal
K}^{-j} \left\{ A_{1,j+1}(\omega_\varepsilon(t)) -
\delta_{\varepsilon}(t) A_{0,j+1}(\omega_\varepsilon(t))\right\}
$$}
\begin{equation}
\label{Psi_2N}
{}+ \frac{2 \varepsilon^{N}}{(N+2)!} \mu^{N}
{\mathcal K}^{-N} \mathcal{M}^{(N+2)}P_\varepsilon(t).
\end{equation}
Substituting this representation into Eq. ~(\ref{rel1}), we finally
get
{$$
2(1+e_2^2){\mathcal K} F(t)  =
 \sum\limits_{j=0}^{N} \frac{\varepsilon^{j}}{(j+1)!} \mu^{j+1}
{\mathcal K}^{-j} \left\{ A_{1,j+1}(\omega_\varepsilon(t)) -
\delta_{\varepsilon}(t) A_{0,j+1}(\omega_\varepsilon(t))\right\}
$$}
\begin{equation}
\label{rel2}
{}+ \frac{\varepsilon^{N+1}}{(N+2)!} \mu^{N+1}
{\mathcal K}^{-N} \mathcal{M}^{(N+2)}P_\varepsilon(t),
\end{equation}
or equivalently
{$$
2(1+e_2^2){\mathcal K}^{N+1} F(t)  =
 \sum\limits_{j=0}^{N} \frac{\varepsilon^{j}}{(j+1)!} \mu^{j+1}
{\mathcal K}^{N-j} \left\{ A_{1,j+1}(\omega_\varepsilon(t)) -
\delta_{\varepsilon}(t) A_{0,j+1}(\omega_\varepsilon(t))\right\}
$$}
\begin{equation}
\label{rel2a}
{}+ \frac{\varepsilon^{N+1}}{(N+2)!} \mu^{N+1}
\mathcal{M}^{(N+2)}P_\varepsilon(t).
\end{equation}
The latter relation allows us to determine the problem parameters
asymptotically with any prescribed accuracy.

Note that apart from the fact that the shapes of the contacting
bones are elliptical paraboloids, no additional assumptions on the
shape of the contact zone have been made. On the other hand, no
proof was offered to show that the contact zone is approximately represented by
an ellipse. This will be done later.

{\bf Remark 1.} For every $t$ for which the contact pressure
$P_\varepsilon(t)$ is bounded and the contact region $\omega(t)$
belongs to a bounded domain, the remainder
$\frac{\varepsilon^{N+1}}{(N+2)!} \mu^{N+1}
\mathcal{M}^{(N+2)}P_\varepsilon(t)$ in formula (\ref{rel2a})
tends to zero as $N \rightarrow \infty$. Thus, the series
corresponding to the sum on the right hand-side of (\ref{rel2a}) is
converging.

\vspace{3mm}

\section{Asymptotic solution to the contact problem}
\label{Zero-app}

\subsection{Zero-order approximation}
\label{Zero-appsub}

First, we get solution of the problem for $\varepsilon = 0$. In
this case Eq. ~(\ref{EqFP_new}) has the form
\begin{equation}
\label{zero} \Delta P^{(0)}({\bf x}, t) + \chi \int\limits_{0}^{t}
\Delta P^{(0)}({\bf x}, \tau)  d \tau = \mu\left(\Psi_{1}({\bf x})
- \delta^{(0)}(t)\right),
\end{equation}
where $\Psi_{1}({\bf x})$ is defined in (\ref{Psi_Eq}).
{Since we know from \cite{Arg_Mish_2} that the contact zone is an
ellipse at this stage of approximation we will have
\begin{equation}
\delta_\varepsilon = \delta^{(0)}(t) = \delta_\varepsilon(b_0(t);
\beta_0(t)) = \frac{A_{1,0}(\omega_0(t))}{A_{0,0}(\omega_0(t))}.
\label{delta_0(0)}
\end{equation}
}

Using formula (\ref{delta_0(0)}) and calculations presented in
Section \ref{$A_{kl}$} (see formula (\ref{A_{kl}_int_fin})), one
can find that
{\begin{equation}
\label{A_00} A_{0,0}(\omega_0(t)) = \frac{\pi b_0^2}{\beta_0},\quad
A_{1,0}(\omega_0(t)) =
 \frac{\pi b_0^4}{4 \beta_0^3} \left(\beta_0^2 + e_1^2\right),
\end{equation}}
and therefore
\begin{equation}
\label{delta_0_sol} \delta^ {(0)}(t) =
\frac{b_0^2\left(\beta_0^2 + e_1^2\right)}{4 \beta_0^2}.
\end{equation}
Note that formulas (\ref{A_00}) and (\ref{delta_0_sol}) contain
two known constants
 $e_1$ and $e_2$ defined in (\ref{Psi_Eq}) and two still unknown functions
 $ b_0(t)$ and $\beta_0(t)$, which are the main semi-axis and the eccentricity of the ellipse
\begin{equation}
\label{contact_0}
\omega_0(t) = \left\{ {\bf x} \in {\mathbb R}^2: \; \frac{x_1^2}{b^2_0(t)} +
\frac{\beta^2_0(t) x_2^2}{b^2_0(t)} \leq 1\right\}.
\end{equation}

The leading terms in (\ref{rel2a}) imply (for $N=0$) the following
equation:
{\begin{equation} \label{a_0(0)} 2 (1 + e_2^2) {\mathcal K} F(t) =
\mu A_{1,1}(\omega_0(t)) - \mu \delta^{(0)}(t)
A_{0,1}(\omega_0(t)).
\end{equation}}
Here, ${\mathcal K}$ is the Volterra integral operator defined in
(\ref{Operator}).

Analogously, using some results from Section \ref{$A_{kl}$}
(see, in particular, formula (\ref{A_{kl}_int_fin})), we obtain
{\begin{equation} \label{A_01} A_{0,1}(\omega_0(t)) = \frac{\pi
b_0^4}{4 \beta_0^3} \left(\beta_0^2 + e_2^2\right)
\end{equation}
and
\begin{equation}
\label{A_11} A_{1,1}(\omega_0(t)) = \frac{\pi b_0^6}{24
\beta_0^5}\left\{3  \beta_0^4 + (e_1^2 + e_2^2) \beta_0^2 + 3
e_1^2 e_2^2\right\},
\end{equation}}
and thus
\begin{equation}
\label{a_0(0)_sol} 2 (1 + e_2^2) {\mathcal K} F(t)  = \mu
\frac{\pi b_0^6}{48 \beta_0^5} \left\{3  \beta_0^4 - (e_1^2 +
e_2^2) \beta_0^2 + 3 e_1^2 e_2^2\right\}.
\end{equation}

To find the functions $ b_0(t)$ and $\beta_0(t)$ together with the
pressure distribution over the contact zone, $P^{(0)}({\bf x},t)$,
we follow \cite{Arg_Mish_2} and introduce a new unknown function
\begin{equation}
\label{new_unknown} p^{(0)}({\bf x},t) = P^{(0)}({\bf x},t) + \chi
\int\limits_{0}^{t}  P^{(0)}({\bf x}, \tau)  d \tau =
{\mathcal K} P^{(0)}({\bf x},t).
\end{equation}
In the case of monotone external load, this function should satisfy
the Poisson equation (following from (\ref{EqForPress}))
\begin{equation}
\label{Poisson} \Delta p^{(0)}({\bf x},t) = \mu\left(\Psi_{1}({\bf
x}) - \delta^{(0)}(t)\right),\quad {\bf x}\in \omega_0(t),
\end{equation}
with the boundary conditions (\ref{Bound1}),  (\ref{Bound2}).

It is customary to rewrite this relation in the form
\begin{equation}
\label{Poisson_0} G_0({\bf x}, t) = 0,
\end{equation}
where
\begin{eqnarray}
\label{G_0_0} G_0({\bf x}, t) & = & G_0(b_0, \beta_0, \delta_0) \\
{} & \equiv & \Delta p^{(0)}({\bf x},t) - \mu\left(\Psi_{1}({\bf x}) -
\delta^{(0)}(t)\right),\; {\bf x}\in \omega_0(t).
\end{eqnarray}

Bearing in mind that the function $\Psi_{1}({\bf x})$ is a
quadratic polynomial (compare with (\ref{Psi_Eq})), it is natural
to look for the solution of such problem in the form of a
polynomial in $x_1, x_2$ of the fourth degree, that is
\begin{equation}
\label{repr_0} p^{(0)}(b_0,\beta_0,\eta_0,{\bf x},t)=\eta_0(t)
\left(1 - \frac{x_1^2}{b_0^2} -  \frac{\beta_0^2
x_2^2}{b_0^2}\right) Q_0(x_1, x_2).
\end{equation}
Note that the term in the brackets vanishes on the boundary
$\omega_0$, and thus the condition (\ref{Bound1}) is satisfied
automatically.

In Section~\ref{$Q_0$}, it has been shown that $Q_0$ is a
polynomial of the second order having the form
\begin{equation}
\label{Q_0}  Q_0(x_1,x_2)=\left(1 - \frac{x_1^2}{b_0^2} -
\frac{\beta_0^2 x_2^2}{b_0^2}\right),
\end{equation}
so that
\begin{equation}
\label{repr_0_1}  p^{(0)}(x_1,x_2; t) = \eta_0(t) \left(1 - \frac{x_1^2}{b_0^2} -
\frac{\beta_0^2 x_2^2}{b_0^2}\right)^2.
\end{equation}
Taken into account this representation we arrive at the following
relations (see Section \ref{Computations_1}):
\begin{equation}
\label{ee_1_0}
\eta_0(t)=\frac{\mu\delta^{(0)}(t)}{4(1+\beta_0^2)}b_0^2,
\end{equation}
\begin{equation} \label{ee_2_0} \eta_0(t)=\frac{\mu
b_0^4}{2(6+2\beta_0^2)}= \frac{\mu b_0^4}{4(3+\beta_0^2)},
\end{equation}
\begin{equation}
\label{ee_3_0} \eta_0(t)=\frac{\mu b_0^4
e_1^2}{2(2\beta_0^2+6\beta_0^4)}= \frac{\mu b_0^4
e_1^2}{4\beta_0^2(1+3\beta_0^2)}.
\end{equation}

This system allows us to determine the unknown functions $b_0(t)$
and $\beta_0(t)$. Indeed, eliminating $\eta_0$ from the last two
equations, we get a bi-quadratic equation defining the value of
the parameter $\beta_0$, i.e.,
\begin{equation}
\label{beta_0} 3\beta_0^4 + (1 -  e_1^2) \beta_0^2 - 3 e_1^2 = 0.
\end{equation}
By definition, $\beta_0$ is a positive parameter, thus the unique
positive solution of (\ref{beta_0}) has the form
\begin{equation}
\label{beta_0_sol} \beta_0 = \sqrt{\frac{ ( e_1^2 - 1) + \sqrt{
e_1^ 4 + 34 e_1^2 + 1}}{6}}.
\end{equation}
Note that at the zero-approximation the parameter $\beta_0$ does
not depend on time. The other parameter, $\eta_0(t)$, can be
computed directly from (\ref{ee_2_0}) or (\ref{ee_3_0}), if one
knows the remaining constant $b_0(t)$. Moreover, taking into
account (\ref{ee_1_0}) and (\ref{delta_0_sol}), one can use an
equivalent formula
\begin{equation}
\label{eta_0_fin}
\eta_0(t)=\frac{\mu
b_0^4(\beta_0^2+e_1^2)}{16\beta_0^2(1+\beta_0^2)}.
\end{equation}

In the same way, one can offer, in addition to
(\ref{delta_0_sol}), two equivalent representations for the
indentation parameter
\begin{equation}
\label{delta_0}
\delta^{(0)}(t)=\frac{1+\beta_0^2}{3+\beta_0^2}b_0^2(t)=\frac{(1+\beta_0^2)e_1^2}{\beta_0^2(1+3\beta_0^2)}b_0^2(t).
\end{equation}

Finally, the major semi-axis $b_0$ of the ellipse $\omega_0$ is
determined as follows:
\begin{equation}\label{b_0(t)}
b_0(t)=\left[\left(F(t) + \chi \int\limits_{0}^{t} F(\tau) d\tau\right)
\left(\frac{ 96\beta_0^5(1+e_2^2)} {\mu\pi (3  \beta_0^4
-\beta_0^2 (e_1^2 + e_2^2) + 3 e_1^2 e_2^2)}\right)\right]^{1/6}.
\end{equation}
Note that the parameters $b_0$, $\eta_0$ as well as the
indentation, $\delta_0$, depend on time $t$ in contrast to the
ellipse eccentricity $\beta_0$.

Now, it remains only to find the pressure over the contact area.
Using (\ref{new_unknown}) and (\ref{repr_0_1}), we get
\begin{equation}
\label{P_0_K_inv}
P^{(0)}(b_0,\beta_0,\eta_0,x_1, x_2,t)={\mathcal K}^{-1} \left(\eta_0( t) Q_0(x_1,x_2)^2\right).
\end{equation}

If $(x_1,x_2)$ belongs to the initial contact zone, i.e. $1 - \frac{x_1^2}{b_0^2(t)} -  \frac{\beta_0^2
x_2^2}{b_0^2(t)}>0$, then
\begin{equation}
\label{P_0_fin1}
P^{(0)}(x_1, x_2,t)=\eta_0( t) \left(1 - \frac{x_1^2}{b_0^2(t)} -
\frac{\beta_0^2 x_2^2}{b_0^2(t)}\right)^2-\chi \int\limits_{0}^{t}
\eta_0( \tau) \left(1 - \frac{x_1^2}{b_0^2(\tau)} -
\frac{\beta_0^2 x_2^2}{b_0^2(\tau)}\right)^2e^{-\chi(t-\tau)}
d\tau.
\end{equation}
If $(x_1, x_2)$ lies outside  of the initial contact zone, i.e. $1 - \frac{x_1^2}{b_0^2(t)} -
\frac{\beta_0^2 x_2^2}{b_0^2(t)}<0$, then
\begin{equation}
\label{P_0_fin2}
P^{(0)}(x_1, x_2,t)=\eta_0( t) \left(1 - \frac{x_1^2}{b_0^2(t)} -
\frac{\beta_0^2 x_2^2}{b_0^2(t)}\right)^2-\chi
\int\limits_{t_*(x_1,x_2)}^{t} \eta_0( \tau) \left(1 -
\frac{x_1^2}{b_0^2(\tau)} - \frac{\beta_0^2
x_2^2}{b_0^2(\tau)}\right)^2e^{-\chi(t-\tau)} d\tau.
\end{equation}
The critical moment of time $t_{\ast}$ is determined by the formula
$$b_0^2(t_{\ast})=x_1^2+\beta_0^2x_2^2.$$
Using (\ref{b_0(t)}), we get
\begin{equation}
\label{F_t_ast}
F(t_{\ast})+\chi \int\limits_{0}^{t_{\ast}} F(\tau) d\tau =\frac{ \mu \pi
}{96\beta_0^5} \left(\frac{3  \beta_0^4 -\beta_0^2 (e_1^2 +
e_2^2) + 3 e_1^2 e_2^2}{1+e_2^2}\right)(x_1^2+\beta_0^2x_2^2)^3.
\end{equation}
If the load is stepwise, we have $F(t)=F_0$. Hence, we find that
\begin{equation}
\label{t_ast}
t_{\ast} = \frac{ \mu \pi  }{96\beta_0^5 \chi  F_0} \left[\frac{(3  \beta_0^4
-\beta_0^2 (e_1^2 + e_2^2) + 3 e_1^2
e_2^2)}{1+e_2^2}(x_1^2+\beta_0^2x_2^2)^3\right] - \frac{1}{\chi}.
\end{equation}

Note that in this case
\begin{equation}\label{b_0(t)_1}
b_0^6(t_{\ast})  = \frac{ 96\beta_0^5(1+e_2^2) (1 + \chi
t_{\ast})} {\mu\pi (3 \beta_0^4 -\beta_0^2 (e_1^2 + e_2^2) + 3
e_1^2 e_2^2)} F_0 .
\end{equation}

\vspace{3mm}
This finishes the zero iteration step. Note that the results of this Section after changing the notation coincide with those obtained in \cite{Arg_Mish_1}.

\subsection{First-order approximation problem}
\label{First-appsub}

For the next steps we consider an appropriately deformed contact
domain $\omega^{(1)}_{\varepsilon}$, defined as a perturbation of the zero-order
one $\omega_0$. Namely, we assume that it can be written in the form

\begin{equation}
\label{contact1} \omega^{(1)}_{\varepsilon} = \omega^{(1)}_{\varepsilon}(t) =
\Big\{(x_1,x_2): Q_0({\bf x},t)+\varepsilon Q_1({\bf x},t) \geq 0\Big\},
\end{equation}
where unknown polynomials are taken in the forms
\begin{equation}
\label{Q0} Q_0({\bf x},t)= Q_0({\bf x},\beta_1,b_1),
\end{equation}
\begin{equation}
\label{Q1} Q_1({\bf x},t)= a_{40}(t)x_1^4+a_{22}(t)x_1^2x_2^2+a_{04}(t)x_2^4.
\end{equation}

 Note that for $\varepsilon=0$ the solution form coincides with
(\ref{contact_0}), if one take $b_1\equiv b_0$, $\beta_1\equiv
\beta_0$.

The idea behind such choice of the asymptotic anzatz is to satisfy the boundary conditions (\ref{Bound1}) and (\ref{Bound2}) automatically. This will be archived by putting
\begin{equation}
\label{omega_1} P^{(1)}_{\varepsilon} = {\mathcal K}^{-1} \Big(\eta^{(1)}( t) \big( Q_0(x_1,x_2,\beta_1(t),b_1(t))+\varepsilon Q_1({\bf x},t)\big)^2\Big).
\end{equation}

Now, when the boundary conditions are valid, we will satisfy the governing equation (\ref{EqForPress}). Note that
\begin{equation}
\label{P_1st_app} P^{(1)}_{\varepsilon} = P_0 + \varepsilon P_1 + O(\varepsilon^2),
\end{equation}
where $p_j={\cal K}(P_j)$, $j=0,1$, and
\begin{equation}
\label{P_1_0}
p_0=\eta^{(1)}( t) \left(1 -
\frac{x_1^2}{b_1^2(t)} - \frac{\beta_1^2(t)
x_2^2}{b_1^2(t)}\right)^2,
\end{equation}
\begin{equation}
\label{P_1_1}
p_1=2\eta^{(1)}( t) \left(1 -
\frac{x_1^2}{b_1^2(t)} - \frac{\beta_1^2(t)
x_2^2}{b_1^2(t)}\right)Q_1({\bf x},t).
\end{equation}

Substituting this representation into Eq. ~(\ref{EqForPress}), we
obtain
\begin{equation}
\label{EqFP_new1} {\mathcal K} \left(\Delta(P^{( 0)} + \varepsilon P_1+O(\varepsilon^2))\right) = \mu \left(\Psi_{1} -
\delta_\varepsilon^{(1)} - \varepsilon \nabla
\Psi_{2} \cdot (\nabla P^{(0)} + \varepsilon \nabla
P^{(1)}+O(\varepsilon^2))\right),
\end{equation}
where the parameter $\delta_\varepsilon^{(1)}$ is represented in
the same form as $P^{(1)}_{\varepsilon}$, i.e.,
\begin{equation}
\label{delta_1st_app} \delta^{(1)}_{\varepsilon} = \delta_0 +
\varepsilon \delta_1 + O(\varepsilon^2) = \delta^{(1)} +
O(\varepsilon^2).
\end{equation}

We can write Eq. ~(\ref{EqFP_new1}) with the accuracy to the terms
of $O(\varepsilon^2)$ as follows:
\begin{equation}
\label{EqFP_new2} \Delta p^{( 0)} + \varepsilon \Delta p_1 = \mu
\left(\Psi_{1} -
{ \delta^{(1)}} - \varepsilon
\nabla \Psi_{2} \cdot \nabla P^{(0)}\right).
\end{equation}

An extended variant of this equation can be written by using the
definition of all components of the equation and by comparing
coefficients at different powers of $x_1, x_2$, so that

\begin{equation}
\label{varepsilon_1_00} - \frac{4 \eta^{(1)}}{b_1^2} (1 +
\beta_1^2) = - \mu \delta^{(1)},
\end{equation}
\begin{equation}
\label{varepsilon_1_20} {4 \eta^{(1)}} \left[\frac{3 +
\beta_1^2}{b_1^4} + \varepsilon (6 a_{40} + a_{22})\right] = \mu
(1 - 8 \varepsilon \theta_{2,0}),
\end{equation}
\begin{equation}
\label{varepsilon_1_02} {4 \eta^{(1)}} \left[\frac{\beta_1^2 (1 +
3 \beta_1^2)}{b_1^4} + \varepsilon (a_{22} + 6 a_{04})\right] =
\mu (e_1^2 - 8 \varepsilon e_2^2 \theta_{2,2}),
\end{equation}
{\begin{equation}
\label{varepsilon_1_22} - \varepsilon\frac{24 \eta^{(1)}}{b_1^2}(a_{40}
\beta_1^2 + a_{22} (1 + \beta_1^2) + a_{04}) = 8\varepsilon \mu (1 + e_2^2)
\theta_{4,2},
\end{equation}
\begin{equation}
\label{varepsilon_1_40} - \varepsilon\frac{4 \eta^{(1)}}{b_1^2}(a_{40} (15 +
\beta_1^2) + a_{22}) = 8\varepsilon \mu \theta_{4,0},
\end{equation}
\begin{equation}
\label{varepsilon_1_04} -\varepsilon \frac{4 \eta^{(1)}}{b_1^2}(a_{04} (15
\beta_1^2 + 1) + a_{22} \beta_1^2) = 8\varepsilon \mu e_2^2 \theta_{4,4},
\end{equation}}
where
\begin{equation}
\label{theta}
\theta_{2k,2l}(t) = {\mathcal K}^{- 1} \left(\eta^{(1)} b_1^{-2 k} \beta_1^{2 l}\right),\;\;\; k, l = 0, 1, 2.
\end{equation}

In the system (\ref{varepsilon_1_00})--(\ref{varepsilon_1_04}) we
have 6 equations and 7 unknowns: $\eta^{(1)}(t),
\delta^{(1)}_{\varepsilon}, b_1(t), \beta_1(t)$, and  $a_{40},
a_{22}, a_{04}$ (coefficients of the polynomial $Q_1$). Therefore,
we have to add an extra equation to the above system, namely
{\begin{equation}
\label{varepsilon_1} \delta^{(1)}(t)=
\frac{A_{1,0}(\omega_\varepsilon(t))}{A_{0,0}(\omega_\varepsilon(t))}  + \frac{2(1 +
e_2^2) \varepsilon}{ A_{0,0}(\omega_\varepsilon(t))} F_1(t),
\end{equation}}
where $F_1(t)$ can be represented in the form
$$
F_1(t) = \int\!\!\!\int\limits_{\omega_{\varepsilon}^{(1)}}
P^{(1)}_{\varepsilon}({\bf x}, t) d {\bf x}.
$$
We also make use of Eq. ~(\ref{rel2a}) written for this
approximation step  with the accuracy of $O(\varepsilon^{2})$ in
the form
{
\begin{equation}
\label{rel2aaa}
2(1+e_2^2){\mathcal K}^{2} F(t)  =
 \sum\limits_{j=0}^{1} \frac{\varepsilon^{j}}{(j+1)!} \mu^{j+1}
{\mathcal K}^{1-j} \left\{ A_{1,j+1}(\omega_\varepsilon(t)) -
\delta^{(1)}(t) A_{0,j+1}(\omega_\varepsilon(t))\right\}.
\end{equation}}

{
{\bf Remark 2.} Note that putting $\varepsilon=0$, the system
(\ref{varepsilon_1_00})--(\ref{varepsilon_1_04}),
(\ref{varepsilon_1})  transforms to the previous case evaluated in
the previous section.}

{{\bf Remark 3.} In the case when $\varepsilon>0$, the system
(\ref{varepsilon_1_00})--(\ref{varepsilon_1_04}),
(\ref{varepsilon_1}) has to be solved numerically. Note that the
parameter $\varepsilon$ in the last three equations
(\ref{varepsilon_1_22}) -- (\ref{varepsilon_1_04}) can be
canceled. We left these multipliers here to explain the limiting case
($\varepsilon=0$).}

\section{Discussion and conclusion}
\label{Discussion}

First of all, observe that at $t=0$, the contact problem for biphasic layers reduces to that for elastic incompressible layers. The contact problem in the latter case were studied in a number of papers \cite{Alexandrov2003,Barber1990,Chadwick2002,Yang1998}, however, without taking into account the tangential displacements.

To solve the resulting problem (\ref{varepsilon_1_00})--(\ref{varepsilon_1_04})
and (\ref{varepsilon_1}), we suggest the following iterative algorithm:

\begin{itemize}
\item
Taking $\varepsilon=0$, we have computed all values
$\eta,b,\beta,\delta = $ $\eta_0,b_0,\beta_0,\delta_0$ from the
zero-order approximation.
\item Having them we can compute the quantity $\theta_{2k,2l}(t)$ from (\ref{theta}),
\item Then, from the system of three equations (\ref{varepsilon_1_22})--(\ref{varepsilon_1_04}) we compute the constants
$a_{40}, a_{22}, a_{04}$ assuming the values of $\eta,b,\beta$ as above.
\item Finally from the system of four equations (\ref{varepsilon_1_00})--(\ref{varepsilon_1_02})
and (\ref{varepsilon_1}) considering the right-hand side known
(computed by the values know from the previous computations), we
found new values $\eta,b,\beta,\delta$ and compare them with the
previous computations. If the required accuracy has achieved we
stop the computation, if not we are going to the second step of
this iterative procedure.
\end{itemize

We note that formulas (\ref{EqForW}) and  (\ref{EqForU}) for the
vertical and tangential displacements contain different powers of
parameters $\epsilon$, namely,  $\epsilon^2$ and $\epsilon$,
respectively. Note also that our analysis (with the values of
another parameters taken into account) shows, that the role of these magnitudes
(vertical and tangential displacements) is quite opposite. In the
final equation  (see (\ref{EqFP_new})) the leading terms,
corresponding to the vertical displacement, contain the zero
power of the new small parameter $\varepsilon$, but the leading
terms, corresponding to the tangential displacements, contain the
first power of $\varepsilon$.

redpajama_set_name: RedPajamaArXiv


text:
Vulturești este o comună în județul Olt, Muntenia, România, formată din satele Dienci, Valea lui Alb, Vlăngărești și Vulturești (reședința).

Are o populația de 2.800 persoane.

Originea numelui comunei, după spusele bătrânilor din sat, provine de la existența unor vulturi imenși, care puteau lua în ghearele lor animale, oi, capre sau chiar copii. Este o comună deluroasă cu păduri și multe vii renumite.
Deși solul localității este foarte roditor, pretându-se foarte bine pentru legumicultură, locuitorii satelor componente se ocupă în special cu cultura porumbului și a grâului. Creșterea animalelor este o ocupație tradițională însă odată cu închiderea târgului săptămânal din localitatea învecinată Câmpu Mare, aceasta este în curs de dispariție.

Scurt istoric
Din păcate, nu există o monografie a localității, însă se pare că această zonă a fost locuită încă din preistorie (neolitic si eneolitic), după ce în zonă au fost descoperite numeroase topoare din piatră șlefuită. Perioada etnogenezei poporului român este și ea prezentă aici găsindu-se numeroase urme ceramice în principal de influență slavă. Se pare ca localitatea este atestată documentar în sec XVII-lea. În perioada medievală populația de pe raza comunei avea ca ocupații principale agricultura, creșterea animalelor, pescuitul și comerțul. Localitatea a fost și un punct de tranzit pentru ciobanii ce treceau cu oile pe aici, precum și pentru negustorii ce foloseau vechiul drum al Oltului (de origine romană) spre a ajunge la Sibiu. Ca dovadă că localitatea a fost o zonă de tranzit este toponimia locală, gen: La Han; Sub Han; etc. Este plauzibilă și ideea cum că localnicii au folosit cursul râului Olt, pentru transport-plutărit, ca și cei din localitatea Olanu situată mai în amonte, ce era vestită în acest sens.
Dacă locuitorii satelor Dienci, Vlângărești și Valea lui Alb erau țărani liberi, cei din satul Vulturești se pare că erau iobagi/țărani dependenți. Denumirea satului Vlângărești se pare ca are origini "grecești" de la un boier fanariot Vlangoryos(?) care avea ca moșie zona comunele Vulturești și Dobroteasa.' Istoric
 Vulturești este cel mai vechi sat din comună atestat documentar, fiind cunoscut în vechime sub numele de Bogdănei sau Bogdănești. Sub această denumire apare în hrisovul din 16 martie 1492 prin care Vlad Voievod întărește stăpânirea mânăstirii Bistrița asupra acestui sat (printre alte sate).
 Satele Dienci și Vlăngărești apar în recensământul din 1828 cu "48 capi de familie".

 Date geografice
Comuna Vulturești, situată pe malul stâng al Oltului, se află la o distanță de 50 km de orașul Slatina și la 11 km de Drăgășani.
Relieful este deluros.

 Comune vecine
 Est: Cungrea
 Nord: Dobroteasa, râul Olt
 Vest: Zăvideni
 Sud: Verguleasa
 Demografie

Conform recensământului efectuat în 2011, populația comunei Vulturești se ridică la  de locuitori, în scădere față de recensământul anterior din 2002, când se înregistraseră  locuitori. Majoritatea locuitorilor sunt români (94,48%). Pentru 5,52% din populație, apartenența etnică nu este cunoscută.
Din punct de vedere confesional, majoritatea locuitorilor sunt ortodocși (94,33%). Pentru 5,52% din populație, nu este cunoscută apartenența confesională.

 Politică și administrație
Comuna Vulturești este administrată de un primar și un consiliu local compus din 11 consilieri. Primarul, , de la , este în funcție din . Începând cu alegerile locale din 2020, consiliul local are următoarea componență pe partide politice:

 Informatii complementare pe: Valea lui Alb, Olt

 Note

 Bibliografie
 Constantin Gheorghe, Valeriu Boteanu, Iuliu Curta, Gh. Mărunțelu Județul Olt'', Brașov, 1974.

Vulturești

redpajama_set_name: RedPajamaWikipedia


Professor Elvin Atombender, a scientist gone crazy, is about to launch a nuclear missile in 6 hours, and it's your task to infiltrate his base and stop his plan! Impossible Mission has its roots back in the early home computer era, when machines like the C64, Spectrum or Amstrad were popular. The game was well known mainly thanks to its early adaptation of speech synthesis and animations. Later on it was ported on other system, such as the Master System.
The game is a mix of platforming and puzzles. You start inside the first elevator shaft from which you can access different rooms. Inside these rooms you are presented with different threats and an objective. The threats are in the form of robots with several unexpected behaviours and your objective is to search for puzzle pieces or useful items inside the furnitures. It might not sound exciting, but surely it's not an easy task, because the robots behave in many different ways, they just move back and forth over a platform, they zap everything that get close to them with their laser or they stay still and run towards you as soon as you jump over it. There are many reasons as to why you should study them before you jump over platform. You don't have any weapons to help you in your journey, except for a spectacular somersault jump that useful to avoid pits and robots robots and 2 items that you find when you search the furniture: a robot disabler that freezes the robots inside the rooms for a limited time, and a lift reset that puts every lift in the room to the starting position in case you messed up. You won't find many of these, so use them wisely. You will also find 1 unique room with a strange checkerboard. When you use it will show you a musical pattern, if you sort the notes from the lower pitch to the highest it will give you a free item. After you have searched every room in them game, explored every interconnected elevator shaft and found all the 30+ pieces, you can use your portable device inside the lobby and try to assemble the puzzle. There's always only 3 pieces that go together to form a full punch card, but still it will take some time to put everything together in the right orientation. When you are done you are ready to ruin Elvin Atombender's crazy plan!
Seems difficult enough? Please mind that you have 6 hours of real time to beat the game, but every time you die the game will add 10 minutes to the timer. So if you mess up too many times inside a difficult room you will feel a lot of pressure, and it's better to play safe using a few items to help yourself.
The platforming isn't bad, it requires some precision and prediction taking the robots into consideration. However I find it frustrating compared to other versions, because of the hitbox. Most of the time you can easily jump over the robots, sometimes not, especially if you are in the highermost platforms. I don't know if it's just me, but I keep dying because the jump seems more unprecise in that side of the screen. And also, you will find many fast moving robots in short platforms which means that they are impossible to avoid and require an item to freeze them. This can get frustrating in an already difficult run.
I find the graphics very essential, but pleasant, it's upgraded compared to the computer versions. Everything looks colourful and detailed enough to distinguish every single element. The animation is still very good and fluid enough, but all in all the game doesn't really shine in this particular section.
This is the part that made the game popular, the voice synthesis. How did the Master System perform? Well, fairly good, but nothing impressive or new at the time of release. It's pretty much the same as the computer version with quotes like "Another visitor. Stay a while. Stay forever!" and "Destroy him my robots!", or the iconic scream when you fall down a pit. On the other side you have a whole lot of bleeps and blops from the robots and portable devices and the sound of the elevator.
The only music you hear is from the title screen and it's forgettable at best.
The game is not complex at all, so the controls are just easy and practical. Both buttons perform the same action, jumping or using the terminals to activate the items. Pressing up searches the furniture, pressing up and down inside the elevator...well, moves the elevator.
When you use your portable device you can move around the puzzle pieces by clicking them, or you turn them around, or change colors with the onscreen controls. Very easy and straightforward, and everything responds fast enough.
You have 6 hours to beat the game, but you will never play a single game for that long. In 20 minutes to 1 hour you will be either dead or at the end of the game. It's challenging enough to keep you hooked if you are into the genre. The best part of the game is that the first time you turn it on the map has a fixed layout, so you are free to write it down and learn it to do better every time, but every time you reset the console it gets randomized! This means that it has the potential to be infinite if you love it, or incredibly repetitive if you hate it. However this doesn't deny the fact that you can play it as many times as you want and always find something new.
Impossible Mission is an old classic game that you will either love or hate, it all depends if you like this type of challenge. If you love it it's a great game that keeps renewing itself at the simple push of the reset button, if you hate it you will hardly touch it again. Quality-wise at this day it's still a great mix of ability and thinking and that will keep you busy for a while.


We thoroughly clean all common areas of residential blocks and estates. Regular cleaning and maintenance means that the homeowner or tenant can enjoy a clean, fresh environment, whilst any prospective residents will be attracted by the high standards of hygiene when visiting the building.
That means a tidy hallway that is spotlessly clean, carpets and landings that have been thoroughly vacuumed, sparkling windows, walls and skirting board that have been meticulously cleaned, and chrome or silver handles polished to perfection. Outside common areas are equally well maintained and cared for.
Regardless of whether the building is a small converted block of flats or a large mixed tenure hosing estate, we are aware that for each individual owner, the investment is huge and employing the right estate management company is key to protecting that investment.
Our experience and infrastructure allows us to deliver a block management service that is unparalleled by our competitors. We have a team of fully trained and qualified surveyors and technicians.


text:
Tag: Country Music Hall of Fame
Free Admission to the Country Music Hall of Fame December 15
December 11, 2019 December 10, 2019 Bethany Bowman
The Country Music Hall of Fame and Museum is pleased to partner with the Ford Motor Company Fund for the museum's 11th Ford Community Day, on Sunday, December 15. Museum admission will be FREE from 9 a.m. to 5 p.m. next Sunday, as a way to say "thank you" to the Middle Tennessee community and to reward lucky out-of-town visitors. Tours of Hatch Show Print and Historic RCA Studio B will also be free and are available on a first-come, first-served basis.
NewsCountry Music, Country Music Hall of Fame, FordLeave a comment
Music Spotlight: Historic Studio B
November 15, 2019 November 15, 2019 Bethany Bowman
RCA Studio B is a historic landmark that initially helped Nashville to earn the name of Music City. Located at 1611 Roy Acuff Place, in Nashville, Tennessee, Studio B was part of the beginning of the renowned Music Row to which musicians still aspire. Between 1957 and 1977 over 18,000 songs were recorded at Studio B, and over 1000 of them became hits.
Entertainment, Lifestyle, Music, NewsBill Porter, Chet Atkins, Country Music Hall of Fame, Dolly Parton, Elvis Presley, Everly Brothers, Music Row, Nashville, Porter Wagoner, RCA Studio B, Steve SholesLeave a comment
Marty Stuart Is the 2019 Artist in Residence at the Country Music Hall of Fame
April 18, 2019 April 17, 2019 Bethany Bowman
Marty Stuart is the 16th Artist in Residence for the Country Music Hall of Fame.
Entertainment, MoviesCountry Music Hall of Fame, Marty Stuart, NashvilleLeave a comment
Hatch Show Print Celebrates 140 Years of Letterpress Printing
April 5, 2019 April 4, 2019 Bethany Bowman
There are special celebrations in April to commemorate the 140 milestone anniversary of Hatch Showprints.
Community Events, EntertainmentCountry Music Hall of Fame, Hatch Show Prints, NashvilleLeave a comment
The Country Music Hall of Fame Museum Announces Exhibition Schedule for 2019
March 3, 2019 March 1, 2019 Bethany Bowman
Museum will tell the stories of Brooks & Dunn, Boudleaux and Felice Bryant, Kacey Musgraves and Keith Whitley.
Entertainment, MusicBethany Bowman, Boudleaux and Felice Bryant, Brooks & Dunn, Country Music Hall of Fame, Kacey Musgraves, Keith Whitley, NashvilleLeave a comment
The Country Music Hall of Fame Announces it's First New Exhibit of 2019
January 19, 2019 January 19, 2019 Bethany Bowman
American Currents: The Music of 2018 opens Friday, March 8 at the Country Music Hall of Fame. Some of the artists featured include Jason Aldean, Bobby Bones, Brothers Osborne, Kane Brown, Luke Combs, Dan + Shay, Florida Georgia Line, Reba McEntire, Carly Pearce, Pistol Annies, John Prine, Chris Stapleton, Carrie Underwood, Keith Urban, and more.
Entertainment, Lifestyle, Music, NewsBethany Bowman, Bobby Bones, Brothers Osborne, Carrie Underwood, Country Music Hall of Fame, Dan and Shay, Jason Aldean, John Prine, Kane Brown, Keith UrbanLeave a comment
Country Music Hall of Fame and Museum Offers a Day of Free Admission
Runaway June to headline a full day of activity that includes Songwriter Sessions with Pat Alger and Jamie Floyd, a Musician Spotlight with Dom Flemons, Hatch Show Print demonstrations and more.
Entertainment, MusicBethany Bowman, Country Music Hall of Fame, Dom Flemons, Ford Motor Company, Hatch Show Prints, Jamie Floyd, Pat Alger, Runaway JuneLeave a comment
Country Music Hall of Fame and Grand Ole Opry member, Roy Clark Dies at 85
Roy Clark, the legendary 'superpicker', Grammy, CMA and ACM award winner, Country Music Hall of Fame and Grand Ole Opry member and co-host of the famed 'Hee Haw' television series, died today at the age of 85 due to complications from pneumonia at home in Tulsa, Oklahoma.
Entertainment, MusicCountry Music Hall of Fame, Grand Ole Opry, Roy ClarkLeave a comment
Miranda Lambert Is Artist-in-Residence at the Country Music Hall of Fame
September 22, 2018 September 21, 2018 Bethany Bowman
Miranda Lambert becomes the 15th ever Artist-Residence at the Country Music Hall of Fame as she performs at the Country Music Association theater in Nashville.
Entertainment, Lifestyle, Music, NewsAllison Moorer, Ashley Monroe', Bethany Bowman, CMA Theater, Country Music Hall of Fame, Miranda Lambert, Nashville, Natalie HembyLeave a comment
Keith Urban Bids a Musical Farewell to the Man He Calls His "Father in a Parallel Universe"
Keith Urban Don Williams Keith Urban paid tribute to his musical hero, Don Williams, in an intimate service remembering the late country star at the Country Music Hall of Fame and Museum in Nashville on Sept. 27. Before he sang, Keith detailed just how instrumental Don's music has been in…
Entertainment, Lifestyle, NewsCountry Music Hall of Fame, Don WilliamsLeave a comment
Fans in Nashville Say Goodbye To Country Music Legend Glen Campbell
August 10, 2017 August 10, 2017 Wendy Wilson
NASHVILLE, Tennessee — Fans stopped by a memorial Wednesday at the Country Music Hall of Fame to say goodbye to country artist Glen Campbell, who died Tuesday at age 81. A table with flowers and a condolence book for people to sign was set up in the rotunda near…
Entertainment, NewsCountry Music Hall of Fame, Glen Campbell1 Comment

redpajama_set_name: RedPajamaCommonCrawl


text:
Suspect in David Ortiz Shooting Wanted in NJ Armed Robberies
The Daily Voice reports that Rolfi Ferreira Cruz was a suspect in a pair of street holdups in December 2017
By Staff Reports • Published June 13, 2019 • Updated on June 13, 2019 at 6:58 pm
The man accused of shooting David Ortiz in a Dominican Republic bar is also wanted for multiple armed robberies in New Jersey, according to prosecutors.
The Passaic County Prosecutor's Office in New Jersey confirmed Thursday that the alleged gunman in the Ortiz shooting, Rolfi Ferreira-Cruz, 25, of Reading, Pennsylvania, is also wanted in connection with a pair of street holdups in Clifton, New Jersey, in December 2017. Cruz reportedly was among of group of men who held up two victims at gunpoint within a matter of blocks.
Two men, a 26-year-old and 41-year-old, were robbed as they walked in Clifton, both being held up by a group of armed robbers with guns wearing masks on Dec. 22, 2017, police say.
According to police, they caught two of the four suspects later that night, a 19-year-old and a 16-year-old, and the third, a 20-year-old, turned himself into police in January 2018.
Cruz, identified by Dominican police as Rolfy Ferreyra, was among seven suspects arrested in the Dominican Republic after being offered almost $8,000 to shoot and kill the former baseball star affectionately known as Big Papi, authorities said. Cruz is believed to be the gunman in the attack.
Dominican Republic police said the coordinator of the attack was offered 400,000 Dominican pesos, or about $7,800, to orchestrate the shooting. He said the alleged coordinator was also among those in custody.
Authorities declined to give a motive for the shooting at a popular Santo Domingo bar Sunday night or say who had hired the attackers.
Ortiz is one of the most beloved figures in sports history in the Dominican Republic and Boston, a fearsome power hitter with a ready smile. The slugger led the Red Sox to three World Series championships, was a 10-time All-Star and hit 541 home runs.


Seborrheic Dermatitis Treatment in Rajouri Garden,West Delhi:Book Appointment.
A mild type of Dermatitis is known as Seborrheic. This skin disorder usually affects the facial skin and scalp and can lead to inflammation on the affected skin area. Along with inflammation, it makes the skin red, itchy and flaky. Seborrheic dermatitis typically the areas of the skin which are sebaceous-gland-rich. This is very much similar to dandruff in adults and adolescents, or as mild to marked erythema of the nasolabial fold.


We know that the business climate is changing. Technology is shrinking the planet by keeping a world of people just one click away. With rapid technology growth comes new strategies and trends in business and marketing. This has made business easier, but the difficulty is trying to keep up. Just like everything else, trends come and go.
Companies need do their homework. The internet is filled with lists of top companies in each industry. Most of these companies will have blogs on their website. I've discussed before about how companies use blogs to market themselves; industry leading corporate blogs will have great, innovative marketing strategies within their blogs. Make sure you follow them - Google has great tools that allow users to follow blogs and websites. You will be notified whenever an update is made to a followed blog, always keeping you current.
Staying abreast of current events. Google allows you to collect RSS feeds to deliver breaking news to one location 24 hours a day. To keep your company current, stay up to the minute with the latest events. If a natural disaster hits, be the first company to set up a relief fund. If something major happens in the news, offer an opinion on it right away. The information age moves at warp speed. Thirty minutes ago might as well be last year. Never get left behind.
Keep prospects and customers happy. One easy way to accomplish this is to keep the look and feel of your website, Facebook page, twitter account, and any other content based platform fresh. A website with the same look becomes stale and boring after a few months. An outdated look can drive away traffic. Try to add new graphics ever couple of months. Add a new feature to your site. Create a poll and post it to Facebook and twitter. Depending on the topic, you might be able to get a handle on what your potential customers care about, which can in turn help you market to new customers.
In today's tough economy, marketing is more important than ever. In a crowded marketplace, companies need to stand out from their peers just to survive. Don't fall to the back of the pack. Stay current and become a leader, not a follower.


text:
require "openxml/drawingml/properties/font_property"

module OpenXml
  module DrawingML
    module Properties
      class ComplexScriptFont < FontProperty
        namespace :a
        tag :cs

      end
    end
  end
end

redpajama_set_name: RedPajamaGithub


text:
Banking, Federal law and policy, News
By Jihee Ahn
So How Did It Go? – The SAFE Banking Act Senate Hearing
As we reported last week, a Senate Banking Committee hearing occurred this past Tuesday to consider the SAFE Banking Act. In short, the hearing didn't display as much GOP sway or support as we were hoping for.
For one, none of the Republican committee members attended the hearing aside from Committee Chairman Mike Crapo (R-ID) (which he attributed to conflicting hearings). For another, while Chairman Crapo did conclude that "a case has been made pretty strongly here," he cautioned that it's still a "very important and complex issue that we need to get right."
Despite Chairman Crapo's long faces at times, various witnesses made their cases to the Committee during the 1.5-hour hearing. Those witnesses included SAFE Banking Act sponsors Senators Cory Gardner (R-CO) and Jeff Merkley (D-OR); MAPS Credit Union Chief Risk Officer, Rachel Pross (representing CUNA), Citywide Banks' President and CEO, Joanne Sherwood (representing ABA); Smart Approaches to Marijuana (SAM) Vice President of Government Affairs, Garth Van Meter; and LivWell Enlightened Health CEO, John Lord.
Senator Gardner called the hearing "an important step toward the federal government waking up to the reality that the cannabis issue is not going away and needs action." He cited that polling now shows about 65% of Americans support legalization of marijuana and 93% support medical marijuana. Senator Gardner emphasized that the states are leading on this issue, "and the federal government has failed to respond."
It has closed its eyes and plugged its ears and pretended the issue will go away. It won't."
Ms. Pross testified: "There are there are certainly valid concerns being brought to this committee hearing today. But the SAFE Banking Act is narrowly targeted," she said. "It's a narrowly targeted protection for financial institutions to serve an $8.3 billion industry that is already in place today."
Mr. Lord provided perspective on the significant compliance costs associated with serving cannabis customers under the present regime, citing that those financial institutions who service cannabis businesses charge them substantial monthly fees.
Our company pays in excess of $3,000 per month for the mere privilege of having an account. … The current situation is especially challenging for small businesses. While we, due to our size, are able to absorb the additional costs associated with cash management and exorbitant bank fees, many small businesses are not."
On the opposite end of the spectrum, Mr. Van Meter testified that prioritizing marijuana banking "over science" that verifies the risks of marijuana consumption would be irresponsible.
By skipping ahead to a technicality over banking rules, the marijuana industry is hoping to gain many of the benefits of federal legalization without a debate over the public health effects. But make no mistake, a policy change around banking would have massive public policy and public health ramifications, so we are shirking our duties if we do not consider the full question. The SAFE Banking Act will allow the expansion of an industry pushing new, exponentially more powerful forms of marijuana before any of its health or other societal impacts are fully understood. Today's marijuana is produced in concentrations that have damaging effects on the brain, are promoted by the industry, and sold in forms attractive to children."
In response to questions following the hearing, Chairman Crapo commented that the Committee is "trying right now to see if we can find a way to address the various issues" ahead of a potential markup on the cannabis banking legislation. He also indicated he doesn't intend to hold additional hearings on the issue.
This hearing marks the sixth congressional hearing on marijuana policy this term, following a hearing held by the House Judiciary Subcommittee on Crime, Terrorism and Homeland Security on pathways to ending the federal prohibition of cannabis. It also comes just hours after House Judiciary Committee Chairman Jerry Nadler (D-NY) and Democratic presidential hopeful Senator Kamala Harris (D-CA) announced plans to introduce a bill to decriminalize marijuana at the federal level. These certainly are all positive steps for the industry, but it remains to be seen whether anything will come to fruition.
Jihee Ahn
Jihee is an experienced complex commercial litigator in Harris Bricken's Portland office. She represents clients in business, intellectual property, and real estate matters. Having worked extensively in both federal and state courts, Jihee advises her clients from case intake through arbitration and trial.


text:
Jason Kenney incorrectly identified which century Stephen Harper was Prime Minister (again)
Is Jason Kenney sure he knows what century he's living in?
July 11, 2016	Share Tweet
Is Jason Kenney sure he knows what century he's living in?
The former federal Conservative minister deleted a tweet Saturday night announcing he'd received Stephen Harper's blessings to seek the leadership candidate for Alberta's Progressive Conservatives, joining Rob Anders as one of the chosen few to receive Harper's golden touch.
Only problem? Kenney's tweet identified Harper as the "20th century's longest-serving Prime Minister."
Of course, Harper was Prime Minister from 2006 until 2015. That time range would, in fact, place Harper's Prime Ministership squarely in the 21st century, not the 20th century.
The screw up has a certain poetic irony about it considering Kenney has recently been described as "yesterday's conservative" and as "the type of person who could lead the PCs into the 20th century."
After all, in addition Kenney's old-fashioned social conservativism, his cagey views on women's reproductive rights and bizarre fears about the school system, this is a man who, in only the last few weeks, talked about radical "social engineering" in response to a question about transgender students and warned of "bohemian" youth being seduced by the wiles of communism.
But what makes this all the more remarkable is it actually is not the first time Kenney incorrectly named Harper as a 20th century Prime Minister:
Um … Kenney has already been reminded that Harper was not PM during the 20th century. https://t.co/joTSqCrydM
— Nora Loreto (@NoLore) July 10, 2016
In fact, during Kenney's speech launching his leadership bid last week, he made a similar comment about the Harper government belonging to the 20th century:
Kenney got applause for calling Harper gov the longest-running Conservative gov't in 20th century. Fact check: 21st century
— Jason Markusoff (@markusoff) July 6, 2016
Which led to these sorts of observations:
@markusoff Also several Ralph Klein references and a Ronald Reagan quote. Plenty of 20th century in this speech.
— Robson Fletcher (@CBCFletch) July 6, 2016
I am not the least bit surprised that Jason Kenney thinks it's still the 20th century. https://t.co/7QDZ1MAnX5
— andrew barley (@Andrewmbarley) July 6, 2016
Did Kenney just say Harper govt was in 20th century? He actually doesn't know what century it is. #backwards #ableg
— Matt Hannah (@mattjdh) July 6, 2016
Jason Kenney, taking you back to the good ol' days of 1905. #20thcentury #ableg
— timothy chu (@timothy_chu) July 6, 2016
And Kenney plans to promise the best 19th century government Alberta has ever had. https://t.co/VPiEsdAPlT
— Gerry Porter (@ficklesonance) July 6, 2016
So, it kinda begs the question: is Jason Kenney sure he knows what century he's living in?
Photo: PMO.
20th century Jason Kenney Kenneywatch Stephen Harper they said what?
Conservatives vow to "stand up" for door-to-door mail delivery, except they axed that in 2013
A fundraising e-mail suggests to potential donors the Conservatives will "stand up" for "door-to-door delivery for all Canadians."
July 9, 2016	Share Tweet
Are Conservatives stealing ideas from the NDP or did they forget the last decade ever happened?
A fundraising e-mail sent to supporters Friday criticizes the Trudeau Liberals for breaking a campaign promise to restore door-to-door mail delivery and boldly suggests the Conservatives will "stand up" for "door-to-door delivery for all Canadians."
"Trudeau's Liberals promised door-to-door mail delivery for all Canadians," reads the e-mail, noting "they even promised a review of Canada Post to make…


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Biografia
Nato e cresciuto a Berlino, Gregory Seyffert ha studiato danza alla Staatliche Ballettschule und Schule für Artistik.

Nel 1987 è stato assunto dalla Komische Oper Berlin come solista ed è rimasto nella compagnia per tredici anni, nove dei quali in veste di primo solista. Parallelamente all'attività artistica sulle scene di Berlino, dal 1988 Seyffert ha danzato di frequente con l'American Ballet Theatre sotto la direzione di Michail Baryšnikov. Nel 1996 ha fondato la compagnia Gregor Seyffert & Compaignie Berlin, mentre l'anno successivo ha vinto il Prix Benois de la Danse.

Dopo aver coreografato numerosi balletti per i maggiori teatri tedeschi, nel 2002 Seyffert è diventato direttore artistico del Staatliche Ballettschule und Schule für Artistik, una carica che ha mantenuto fino al 2020.

Note

Collegamenti esterni


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Regina conducts a kind of forensic analysis of her younger self and her choices with a merciless honesty, as if determined to own every flawed aspect of herself and her past in order to understand them and decide what, finally, to do. While this self-portrait contains plenty of wry, observational humour, it is at times so raw and naked that the reader would flinch, were it not for the retrospective viewpoint: she has clearly, if barely, survived.
Image shows the cover of My Education by Susan Choi.


While Snapchat is normally used by people to share photos and videos with their friends, users will soon be able to watch TV programmes on the app.
Snap has announced that it is joining forces with NBCUniversal to produce original TV shows for Snapchat.
Content will initially be available for Snapchat users in the US, before it is expanded to international users in 2018.
The collaboration appears to be a direct move by Snap to keep up with one of its main rivals, Facebook, who in June announced its plans to produce original video content.
A report from EMarketer, a marketing research firm based in New York, predicts that 14.5 million people aged 12 to 17 will use Facebook in 2017.
This is a drop of 3.4 per cent – the first predicted fall in Facebook usage for any age group.
Instead, teens are moving towards photo-sharing apps, including Snapchat and Instagram – which is owned by Facebook.
The report predicts Snapchat's US user base will grow by 25.8 per cent to 79.2 million people monthly.
The two firms will hold equal stakes in a new digital content studio to be based in Santa Monica, California.
The new content will appear under the Shows section of the Snapchat app, which is already used by news and other organisations for short video clips.
'NBCUniversal is proud to have been Snap's first partner to create Shows, and we've seen consistent audience growth and engagement on the platform,' said Maggie Suniewick, president of NBCUniversal Digital Enterprises.
The original content will initially available for US Snapchat users, before it is expanded to international users in 2018, according to a spokesperson for Snap.
Recent surveys have shown Snapchat - known for its disappearing smartphone messages - as the preferred social network for US teens, ahead of Facebook and Twitter.
But while Snap has expanded its media partnerships, its user and revenue growth has been disappointing since its share offering earlier this year.
The social networking giant has indicated that it was willing to commit to production budgets as high as $3 million (£2.4 million) per episode, in meetings with Hollywood talent agencies.
Facebook is hoping to target audiences from ages 13 to 34, with a focus on the 17 to 30 range.
The company has already lined up 'Strangers', a relationship drama, and a game show, 'Last State Standing'.
NBCUniversal, a unit of media-cable giant Comcast, made an investment of $500 million (£380 million) in Snap as part of its initial public offering.
And in August, Facebook announced that it was launching a new platform that will allow users of the social network to access shows and clips created specifically for the site.
Watch is aimed at allowing the social network to take on web rivals YouTube and Netflix as well as more traditional TV.
It will feature a wide range of programming, from live sports matches and safari expeditions, to a reality show about a day in the life of a social media star.
It is unclear when the Watch tab will become active.


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text:
Floods dominate natural catastrophe statistics in first half of 2013
Floods causing billions of dollars in losses dominate the natural catastrophe statistics for the first half-year 2013. Around 47% of the overall losses and 45% of the insured losses derived from inland flooding that occurred in Europe, Canada, Asia and Australia. Altogether, at around US$ 45bn, losses from natural catastrophes were below the average amount for the past ten years (US$ 85bn). Insured losses totalled approximately US$ 13bn (ten-year average: US$ 22bn).
By far the most expensive natural catastrophe in the first half-year was the flooding in southern and eastern Germany and the neighbouring states in May and June, which gave rise to an overall loss of more than €12bn (US$ 16bn) and an insured loss in the region of €3bn plus (US$ 3.9bn). The majority of the insured losses occurred in Germany. By way of comparison, the Elbe flooding in 2002 caused an overall economic loss of US$ 16.5bn, of which US$ 3.4bn was insured.
"The frequency of flood events in Germany and central Europe has increased by a factor of two since 1980. But particularly with floods, an increased hazard – such as more frequent heavy rainfall events – need not necessarily result in higher losses. Such a rise in losses can be prevented by better flood control", said Board member Torsten Jeworrek. "It is therefore important to sharpen risk awareness. Rivers need room so that flood waves can disperse without causing serious damage. And the flood risk needs to be considered in the designation of land for industrial or residential areas. Politicians should not only set up emergency funds after catastrophes but should act with greater foresight, engaging in prudent supraregional flood control, which should ideally be coordinated across national borders."
The flooding in Germany and the neighbouring countries to the east was caused by an atmospheric trough across central Europe, drawing moist air from the Mediterranean and southeast European area northwards over eastern Europe. The lows embedded in this air current, with an anticlockwise airstream, led to orographic lifting on the north sides of the eastern German mountain ranges and the Alps. In some places, 400 litres of rain per square metre fell within a few days. And up to 100 litres per square metre were recorded over a very large area. As the ground was already saturated from the rainiest spring for 50 years, the rainwater flowed directly into the rivers.
The consequence was an extensive flood in Bavaria and eastern Germany which significantly exceeded the levels of 2002 in many places on the Danube and Elbe. In Passau, at the confluence of the rivers Danube, Inn and Ilz, the highest water level since 1501 was recorded. In Dresden, by contrast, the old city centre was largely spared, unlike in 2002. Thanks to better flood control, fewer dykes on the upper reaches of the Elbe broke than in 2002, but this meant that the flood wave further downstream was all the higher. In Magdeburg, the floods reached a record level. More than 20,000 people had to be evacuated.
In neighbouring countries like the Czech Republic and Austria, the flooding also caused high losses, although in contrast to 2002 the old city centre of the Czech capital Prague escaped largely unscathed, thanks to enhanced flood control measures. Austria experienced numerous mudflows as a result of the heavy rain. In Vienna, some of the floodwater was diverted into the "New Danube" relief canal, enabling the flooding to be restricted to a few streets. The floods in Slovakia and Hungary were limited to the swollen Danube, as the two countries themselves were not affected by the heavy rain. In Bratislava and Budapest, the flood control measures proved their worth, with only a few streets right next to the river being flooded.
Prof. Peter Höppe, Head of Munich Re's Geo Risks Research unit, noted: "It is evident that days with weather conditions that lead to such flooding are becoming more frequent and that such weather systems tend to remain stationary for longer. With this higher persistence of weather patterns, the potential for heavy and long-lasting precipitation within a trough situation, for example, increases. The counterpart to this are stationary high-pressure systems which in summer increase the risk of heatwaves and periods of drought."
"Debate in climate research is currently focusing on what the causes of such changes in weather patterns could be and what role climate change might play in this. But it is naturally not possible to explain single events on this basis", Höppe added.
The second most expensive event in the first half-year was a squall line with severe tornadoes in the USA between 18 and 22 May. On 21 May, a tornado of category 5 (maximum strength) hit the town of Moore, a suburb of Oklahoma City. The tornado reached peak wind speeds of up to 300 km per hour and was unusually large, with a width of nearly two kilometres. The destruction was enormous: well over 1,000 buildings were completely destroyed in Moore alone, including seven schools and a hospital. 26 people were killed and more than 370 were injured. The overall economic loss amounted to more than US$ 3bn, of which approximately US$ 1.5bn was insured.
"The central states of the USA have the highest tornado risk in the world. Altogether, however, the US tornado season has been below average so far: by the end of June, 625 tornadoes had occurred, compared with the longer-term average of 1,075", Höppe said.
Another severe event was a flood in the Canadian province of Alberta – probably the worst flooding ever experienced there. Starting on 19 June, massive rainfall of up to 200 litres per square metre in one day coincided with late snowmelt, causing the Bow River in Calgary to swell to a level three times higher than during the flood of 2005. Some 75,000 people were evacuated. Numerous streets were flooded, and the Saddledome, home to ice-hockey club Calgary Flames, was hit. Initial estimates indicate an overall economic loss of more than US$ 3bn, while the insurance loss is likely to exceed the one-billion-dollar mark.
The event with the severest consequences in humanitarian terms was the disaster caused by flash floods in northern India and Nepal as a consequence of exceptionally early and extremely heavy monsoon rains. Hundreds of roads and bridges were washed away, making the job of rescue workers all the more difficult. Over 1,000 people were killed.
All in all, 460 loss-relevant natural hazard events occurred throughout the world in the first half of the year, slightly above the average for the past ten years (390).
Munich Re stands for exceptional solution-based expertise, consistent risk management, financial stability and client proximity. This is how Munich Re creates value for clients, shareholders and staff. In the financial year 2012, the Group – which combines primary insurance and reinsurance under one roof – achieved a profit of €3.2bn on premium income of around €52bn. It operates in all lines of insurance, with around 45,000 employees throughout the world. With premium income of around €28bn from reinsurance alone, it is one of the world's leading reinsurers. Especially when clients require solutions for complex risks, Munich Re is a much sought-after risk carrier. Its primary insurance operations are concentrated mainly in the ERGO Insurance Group, one of the major insurance groups in Germany and Europe. ERGO is represented in over 30 countries worldwide and offers a comprehensive range of insurances, provision products and services. In 2012, ERGO posted premium income of €19bn. In international healthcare business, Munich Re pools its insurance and reinsurance operations, as well as related services, under the Munich Health brand. Munich Re's global investments amounting to €214bn are managed by MEAG, which also makes its competence available to private and institutional investors outside the Group.
This media information contains forward-looking statements that are based on current assumptions and forecasts of the management of Munich Re. Known and unknown risks, uncertainties and other factors could lead to material differences between the forward-looking statements given here and the actual development, in particular the results, financial situation and performance of our Company. The Company assumes no liability to update these forward-looking statements or to conform them to future events or developments.
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Natural catastrophes first half of 2013
World map natural catastrophes first half of 2013
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Natural catastrophes: Comparison of Insured Losses
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Scott and Dominique Pethtel were approved as Baptist Bible Fellowship missionaries to Italy in May 2011. Scott is a 2008 graduate of the Intercultural Studies course and Dominique is a 2003 graduate of the Elementary Education course of Baptist Bible College, Springfield, MO. They served their internship at the Lighthouse Baptist Church in Murfreesboro, TN, which will also be their sending church and where Scott was ordained on May 1, 2011. The Pethels have two daughters.


Cognitive Scientific Challenges to Morality.Neil Levy - 2006 - Philosophical Psychology 19 (5):567 – 587.
Evolutionary Ethics: A Phoenix Arisen.Michael Ruse - 1986 - Zygon 21 (1):95-112.
Are Moral Judgements Adaptations? Three Reasons Why It Is so Difficult to Tell.Thomas Pölzler - 2017 - South African Journal of Philosophy 36 (3):425-439.
Parting with Illusions in Evolutionary Ethics.David C. Lahti - 2003 - Biology and Philosophy 18 (5):639-651.
Taking Darwin Seriously. A Naturalistic Approach to Philosophy.M. Ruse - 1988 - Tijdschrift Voor Filosofie 50 (1):172-173.
Taking Darwin Seriously: A Naturalistic Approach to PhilosophyMichael Ruse.Michael Bradie - 1988 - Isis 79 (2):286-287.
Taking Darwin Seriously: A Naturalistic Approach to Philosophy by Michael Ruse. [REVIEW]Michael Bradie - 1988 - Isis: A Journal of the History of Science 79:286-287.
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Taking Darwin Seriously: A Naturalistic Approach to Philosophy. [REVIEW]Gerald Hull - 1989 - International Studies in Philosophy 21 (3):143-144.
Really Taking Darwin and the Naturalistic Fallacy Seriously: An Objection to Rottschaefer and Martinsen. [REVIEW]Jonathan Barrett - 1991 - Biology and Philosophy 6 (4):433-437.
Naturalistic Transcendentalism.Peter Bishop - 2016 - Essays in the Philosophy of Humanism 24 (2):207-219.
Naturalistic Transcendentalism.Peter Bishop - 2016 - Essays in the Philosophy of Humanism ; Vol 24, No 2 24 (2):207-219.
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Private Instruction emphasizing musicianship, literature, technique and performance. Four-course sequence designed for university transfer equivalency. Instruction provided on voice, piano, harpsichord, organ, guitar, trumpet, French horn, trombone, baritone horn, tuba, flute, oboe, clarinet, bassoon, saxophone, violin, viola, violoncello, contrabass, bass guitar, percussion, and harp.
Prerequisites: A grade of C or better in MUP102AS.
Prerequisites: A grade of C or better in MUP152AS.
Prerequisites: A grade of C or better in MUP202AS.
Practical and performance experience in instrumental, vocal, and mixed ensembles. Prerequisites: None.