text:
HOSA TECHNOLOGY ANNOUNCES HIGH SPEED MICRO HDMI CABLES
By RLA on 8/26/2013 3:34 PM
Buena Park, CA - August 2013... Hosa Technology, the leading innovator of analog and digital connectivity solutions for the modern musician and audio/video professional, is pleased to announce Hosa High Speed Micro HDMI® Cables. Certified to carry an HD signal up to 1080p and beyond, the new Hosa High Speed Micro HDMI Cables are ideal for use with portable video cameras and electronic tablets.
Featuring HDMI to Micro HDMI connectivity, the new Hosa High Speed Micro HDMI Cables are designed to interconnect devices with Micro HDMI outputs to devices with HDMI inputs. The cables are perfect for playing back recorded footage from portable digital devices such as the popular GoPro® video cameras, as well as many other video cameras and electronic tablets.
Among their many features, the Hosa High Speed Micro HDMI Cables are ideal for 720p / 1080i signals that are capable of increased refresh rates (above 60 Hz) or Deep ColorTM....
JBL 3 Series Studio Monitors Launches with M2 Technology
Frank Filipetti Discusses The JBL M2 Master Reference Monitor
JBL Professional Launches the New PRX700 Series
HARMAN's JBL Professional PRX700 Series Redefines Premium Performance for Portable PA Loudspeakers
By RLA on 8/2/2013 10:49 AM
NORTHRIDGE, California – Delivering an unprecedented combination of power and mobility, HARMAN's JBL Professional today introduced the PRX700 Series portable loudspeakers. The next generation in JBL's market-leading PRX portable PA line, the PRX700 Series offers massive improvements in power, SPL and connectivity. "The PRX700 Series not only resets the standard for quality and portability, but also allows users to create the precise system they need from a high-performance family of full-range and subwoofer models," said Andy Flint, Senior Manager, Portable PA Marketing, JBL Professional. "Each model is designed for specific applications, but integrates seamlessly with one another, offering a multitude of choices when tailoring a system to specific needs. In particular, the new PRX710 and PRX715XLF models offer customers even more options than ever before." The PRX700...
HARMAN's JBL Professional 3 Series Sets the Standard for Performance and Price in Studio Monitor Category
NORTHRIDGE, California – Bringing professional studio reference monitoring to customers at a remarkably affordable price, HARMAN's JBL Professional today introduced its 3 Series studio monitors. The first studio monitor line to incorporate JBL's patent-pending Image Control Waveguide, 3 Series is designed to deliver exceptional imaging and detail in any listening environment. With a distinctive appearance, rugged build, pristine sound, and most affordable price points, the new 3 Series will be welcomed by music, post and broadcast production customers working in stereo and surround sound formats. The JBL 3 Series features two models: the LSR305, 5-inch powered studio monitor; and the LSR308, 8-inch powered studio monitor. Both models offer impressive performance, with the LSR305 boasting a response of 43Hz to 24 kHz and a peak SPL of 108 dB, while the LSR308 features a response of...


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PORSCHE 911 RSR – MARCH OF PROGRESS
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Ferrari 250 GTO: Owners' Workshop Manual
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Images and stories on cars and motorsport
Author: Glen Smale
RRP: £20.00 (if ordered on Crowood website)
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From 29-31 July 2016, the Silverstone Classic was hosted at the storied Northampton circuit. Always present in large numbers, the Porsche Club of Great Britain (PCGB) put on a special display of the 'transaxle' cars, the Porsche 924/928/944/968 front-engined series, to commemorate the 40th anniversary of the 924.
The iconic freshly restored Porsche 924 Carrera GTP, Silverstone Classic 2016
It was back in 1976 that the Porsche 924 was revealed to the public, the company's first model to feature a front engine with the gearbox and differential located at the rear, causing it to be referred to as the 'transaxle' series. The 924 was the lead model in this series which was followed by the 928, 944 and 968, a production run that continued for 20 years with almost 400,000 units of all four models being produced.
Porsche 924, Porsche Transaxle Parade at the Silverstone Classic 2016
Glen Smale is the author of twelve books on sports cars and motorsport, seven of which have been on the subject of Porsche. "When I was approached by Crowood Press enquiring as to whether I would consider doing a book on the transaxle Porsches, I was actually very keen. Most books covering this marque tend to be on the 911, and I saw this as an opportunity to learn more about Porsche's other models," Smale admitted.
Jürgen Barth and Richard Attwood (l-r) at the Silverstone Classic 2016
Porsche 924/928/944/968: The Complete Story really covers the design and development of this series, and includes the evolution of each model. The author has interviewed the model's designers, modellers, engineers and the racing drivers who drove these cars to victory on the race track. Since it is the 40th anniversary of the launch of the first of these models, the 924, it was appropriate that this book was published just in time for this milestone.
Glen Smale with Jürgen Barth (l-r) with the Porsche 924 Carrera GT which was driven by Jürgen Barth and Manfred Schurti in 1980 at the Le Mans 24H, finishing 6th overall.
On display at Silverstone was the iconic 924 Carrera GTP, the car that was driven to an incredible twelfth place overall in the 1980 Le Mans 24-Hour race by Tony Dron and Andy Rouse. The 924 Carrera GTP that has been painstakingly restored by Porsche Classic Partner Centres in the UK to its 1980 standard, was on display for enthusiasts to view. This car, and other transaxle Porsches that excelled in international racing, are also covered in the book.
Porsche 924/928/944/968: The Complete Story is available from Crowood Press or from good bookshops in your area.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: j_security authentication using cURL I'm using Tomcat JDBCRealm to authenticated users when they follow a protected page. The authentication works properly when reaching a protected page. Now I'd like to automate this process using cURL. I've written the code below but it doesn't work :
$domainUrl = "http://mydomain.com/protectedArea?j_username=john&j_password=doe";
$ch = curl_init ();
curl_setopt ( $ch, CURLOPT_URL, $domainUrl );
//curl_setopt ( $ch, CURLOPT_POST, true );
//curl_setopt ( $ch, CURLOPT_POSTFIELDS, $post );
curl_setopt ( $ch, CURLOPT_RETURNTRANSFER, 1 );
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_FRESH_CONNECT, true);
curl_setopt ( $ch, CURLOPT_TIMEOUT, 7 );

$output = curl_exec ( $ch );
curl_close ( $ch );

In tomcat log we can see only http GET :
[07/Aug/2012:11:51:24 +0200] GET http://mydomain.com/protectedArea?j_username=john&j_password=doe HTTP/1.1 200 1618

whereas we should have something like that :
[07/Aug/2012:11:57:06 +0200] GET http://mydomain.com/protectedArea?j_username=john&j_password=doe HTTP/1.1 200 1516
[07/Aug/2012:11:57:06 +0200] POST http://mydomain.com/j_security_check HTTP/1.1 302 -

Does someone have any idea?
Thanks for your help
Regards

A: Looks like you only have a single curl_exec call... did you expect a single call to result in two separate requests to the server? You will have to do the following in order to script a form-based login to a spec-compliant server:

*

*Make a request to a protected resource. Any one will do.

*Make a request to j_security_check and include j_username and j_password request parameters. POST would be a good idea to avoid web servers logging the credentials. If the authentication is successful, you should get the resource originally requested in #1.

*Make a request to the protected resource that you truly want to access.

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}
\setcounter{section}{1}%
\setcounter{equation}{0}%

The minimization problem
of the form
\[
\inf\limits_{v\in H^1_0(\Omega;\RR^n)}
\int_{\Omega}\min\Bigl\{\tfrac{1}{2}a
\abs{\varepsilon (v)+C}^2,\,
\tfrac{1}{2}b\abs{\varepsilon (v)+D}^2\Bigr\}\,dx \leqno{(P)}
\]
is considered, where $\varepsilon(v)$ is the symmetrized gradient of
$v\inH^1_0(\Omega;\RR^n)$  and $\abs{\cdot }^2 $ stands for the square of
the scalar product ,,$\cdot\,$'' in
$L^2(\Omega;\mathbb{R}^{n\times n}_{sym})$.\\

 Since the integrand involved is not quasiconvex, we cannot expect
the existence of classical minimizers of this functional.
\par
Basically, there are two ways to proceed if there are no minimizers as a
consequence of the lack of quasiconvexity (\cite{MORREY}). The first one is to
``quasiconvexify" the original functional and relate the gathered
information with the functional itself (cf. \cite{MORREY}, \cite{BALL},
\cite{BALL-MURAT}, \cite{ACERBI-FUSCO}, \cite{DACOROGNA}, \cite{KOHN},
\cite{KOHN-STRANG}, \cite{TARTAR-PREPRINT}, \cite{MURAT}, \cite{TARTAR-PROC},
\cite{FONSECA}, \cite{FONSECA-MULLER}, \cite{FONSECA-RYBKA}, \cite{BUTTAZZO},
\cite{DALMASO}, \cite{AMBROSIO}, \cite{BOUCHITTE-BRAIDES-BUTTAZZO},
\cite{ALLAIRE-FRANCFORT}, \cite{ALLAIRE-LODS} and the references quoted
there).  Another possibility is to enlarge the space of
admissible functions from Sobolev spaces to the space of parameterized
measures (called Young measures \cite{Young-37}) instead
of replacing the objective by its suitable envelopes. In this approach the
Young measures can be regarded
as means of summarizing the spatial oscillatory properties of minimizing
sequences, thus conserving some of that information.
With this respect we refer the reader to \cite{YOUNG},
\cite{KINDER-PEDREGAL}, \cite{CHIPOT-KINDER}, \cite{BALL-JAMES},
\cite{JAMES-KINDER},
\cite{BALL-MURAT}, \cite{MURAT}, \cite{ERICKSEN-PHASE}, \cite{PEDREGAL},
\cite{TARTAR-Y-H}  and the references therein.
\par
However, some important information is lost when seeking minima of
lower semicontinuous regularizations (quasiconvex envelopes). Minimizers
themselves are not sufficient
to characterize properly oscillatory phenomena of the
problem (microstructural features describing fine mixtures
of the phases in the phase transition problems, for instance).
From the application point of view, the detailed structure of minimizing
sequences appears to be as important as the minimizers
themselves.
Moreover, in the vectorial case it is almost impossible to
compute for a given objective its quasiconvexification. It is also very
difficult to compute explicitly the parametrized measures associated to a
minimizing sequence characterizing the infimum of the problem under
consideration.
\par
In this approach a new method to derive the formula for the infimum of $(P)$
is presented.
It preserves all the important information concerning the
oscillatory phenomena and is much easier to be obtained in practice.
The idea is to approximate $(P)$ by convex problems as proposed in
\cite{NAN-MINCONV}.
\begin{thm}[\cite{NAN-MINCONV}]\label{thm-1}
Suppose that $f_i:\Omega\times\mathbb{R}^m\times\mathbb{R}^{nm}\to
\mathbb{R}$, $i=1,2$, are quasiconvex, satisfy the Caratheodory and
growth conditions
\begin{description}
\item[(1)]  $\forall\,(s,\xi)\in{\mathbb R}^m\times
{\mathbb  R}^{nm}$,\quad $\Omega\ni x\mapsto f_i(x,s,\xi)$ is measurable,
\item[(2)] for a.e. $x\in \Omega,\quad {\mathbb R}^m\times
{\mathbb  R}^{nm}\ni (s,\xi)\mapsto f_i(x,s,\xi)$ is continuous,
\item[(3)] $a(x)+c(\abs{s}^2+\abs{\xi}^2)\leq f_i(x,s,\xi)\leq
A(x)+C(\abs{s}^2+\abs{\xi}^2)$,
\end{description}
where $a(\cdot)$, $A(\cdot)$ are non-negative summable functions in $\Omega$,
$c$ and $C$ are positive constants. Set
\[
\inf\biggl\{\int\limits_{\Omega}\min\bigl\{
f_1\bigl(x,u(x),\nabla u(x)\bigr),
f_2\bigl(x,u(x),\nabla u(x)\bigr)
\bigr\}\,dx\colon\; u\in H^1_0(\Omega;\RR^n)\biggr\}:
=\alpha.                    \leqno{(Q)}
\]
Then there exist sequences
$u_k \in H^1_0(\Omega;{\mathbb R}^m)$, $\chi^{k}_1\colon\Omega\to\{0,1\}$ and
$\chi^{k}_2\colon\Omega\to\{0,1\}$ with $\chi^{k}_1+\chi^{k}_2\equiv 1$,
such that
\begin{description}
\item[$(i)$] $\{u_k\}$ is a minimizing sequence for $(Q)$,
\item[$(ii)$] $u_k\to u$ weakly in $H^1_0(\Omega;{\mathbb R}^m)$ as $k\to\infty$,
\item[$(iii)$] ${\chi}^{k}_1\to\chi_1$, ${\chi}^{k}_1\to\chi_2$
weak$^{\,\star}$ in $L^{\infty}(\Omega)$ as $k\to\infty$,
$\chi_1\colon\Omega\to[0,1]$, $\chi_2\colon\Omega\to[0,1]$ with
$\chi_1+\chi_2\equiv 1$,
\item[$(iv)$] $\lim\limits_{k\to\infty}\int\limits_{\Omega}\bigl[\chi^{k}_1 f_1(u_k)+
\chi^{k}_2f_2(u_k)\bigr]d\Omega=\alpha$,
\item[$(v)$] $\int\limits_{\Omega}\bigl[\chi^{k}_1 f_1(u_k)+
{\chi}^{k}_2f_2(u_k)\bigr]d\Omega
\leq\int\limits_{\Omega}\bigl[\chi^{k}_1f_1(w)+
\chi^{k}_2f_2(w)\bigr]d\Omega,\quad
\forall\,w\in H^1_0(\Omega;{\mathbb R}^m)$.
\end{description}
\end{thm}
The minimizing sequence for the problem $(P)$, established according to
\thmref{thm-1},
together with the sequence of solutions of the corresponding dual problems
and the sequence of characteristic functions related to the phases
$\tfrac{1}{2}a\abs{\varepsilon (\cdot)+C}^2$ and $\tfrac{1}{2}b\abs{
\varepsilon (\cdot)+D}^2$ generate limits (in appropriate weak topologies)
which are involved in the infimum formula.
It can be shown that the very special structure of the
minimizing sequence to be constructed allows the infimum to
be fully expressed by the parametrized
Young measures associated to this sequence. Some relations between the related
parametrized Young measures
and the weak limits of the characteristic functions are established.

\section{Statement of the problem and its approximation}

\renewcommand{\theequation}{\thesection.\arabic{equation}}
\setcounter{section}{2}%
\setcounter{equation}{0}%

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain in $\mathbb{R}^n$ with
sufficiently smooth boundary $\partial \Omega$. Set
\[
\mathcal{J}(u)=\int\limits_{\Omega}\min\bigl\{\tfrac{1}{2}a
\abs{\varepsilon (u)+C}^2,
\tfrac{1}{2}b\abs{\varepsilon (u)+D}^2\bigr\}dx,\quad u\inH^1_0(\Omega;\RR^n).
\]
The problem to be considered here is
\[
\inf\Bigl\{\mathcal{J}(u)\colon \;u\in H^1_0(\Omega;\RR^n)\Bigr\}:
=\alpha,                                         \leqno{(P)}
\]
where  $u:\Omega\subset \mathbb{R}^n\to \mathbb{R}^n$ is a competing vector-valued function
from the Sobolev space $H^1_0(\Omega;\RR^n)$,
 $\varepsilon (u)\in
L^2(\Omega;\mathbb{R}^{n\times n}_{sym})$ is the symmetrized gradient of
$u\inH^1_0(\Omega;\RR^n)$,
$C,D\in L^\infty(\Omega;\mathbb{R}^{n\times n}_{sym})$ and where $a,b\in
L^\infty(\Omega)$ are such that $a(x),b(x)\geq\delta>0$ a.e. in $\Omega$ for a
positive constant $\delta$.
\par
\thmref{thm-1} ensures the existence of sequences $u^{k}\in H^1_0(\Omega;\RR^n)$,
$\chi_a^{k}\subset\{0,1\}$
and $\chi_b^{k}\subset\{0,1\}$, $\chi_a^{k}+\chi_b^{k}\equiv 1$, with the
properties that
\begin{description}
\item[$(a)$] $\{u_k\}$ is a minimizing sequence for $(P)$,
\item[$(b)$] $u^{k}\to u$ weakly in $H^1_0(\Omega;\RR^n)$ as $k\to\infty$,
\item[$(c)$] ${\chi}^{k}_a\to\chi_a,\quad {\chi}^{k}_b\to\chi_b$
$\text{weak}^
{\,\star}$ in $ L^{\infty}(\Omega)$ as $k\to\infty$, where
$\chi_a\colon\Omega\to[0,1]$,
$\chi_b\colon\Omega\to[0,1]$ with $\chi_a+\chi_b\equiv 1$,
\item[$(d)$] $\int\limits_\Omega
[\frac{1}{2}\chi_a^{k}a\abs{\varepsilon (u^{k})+C}^2
+\frac{1}{2}\chi_b^{k}b\abs{\varepsilon(u^{k})+D}^2]\, dx
:=\alpha^k\to \alpha$ as $k\to\infty$,
\item[$(e)$]
$
\int\limits_\Omega
[\frac{1}{2}\chi_a^{k}a\abs{\varepsilon (u^{k})+C}^2
+\frac{1}{2}\chi_b^{k}b\abs{\varepsilon (u^{k})+D}^2]\,dx\leq
$\\
\begin{flushright} $\leq\int\limits_\Omega
[\frac{1}{2}\chi_a^{k}a\abs{\varepsilon (w)+C}^2
+\frac{1}{2}\chi_b^{k}b\abs{\varepsilon (w)+D}^2]\,dx$,
$\quad\forall\;w\inH^1_0(\Omega;\RR^n)$.
\end{flushright}
\end{description}

 Let us now introduce the function which describes the behaviour of the
minimizing sequence $\{ u_k\}$:\\
\begin{equation}\label{eq-1.1}
\psi^k=\chi^k_b-\chi^k_a
\end{equation}
with the property
\begin{equation}\label{eq-1.2}
(\psi^k)^2=1.
\end{equation}

It will be convenient to introduce the notations:
\begin{equation*}
m^k:=\chi_a^{k}a+\chi_b^{k}b,\quad
\overline{m}:=\frac{a+b}{2},\quad\underline{m}:=\frac{b-a}{2}
\end{equation*}
Notice that $m^k$ has a decomposition
\begin{equation}
m^k=\overline{m}+\psi^k\underline{m}.
\end{equation}

From (e) of \thmref{thm-1} it follows that $u^k$ is a solution of the convex
optimization problem
\[
\inf\Bigl\{\mathcal{J}^k(v)\colon v\in H^1_0(\Omega;\RR^n)\Bigr\}:=\alpha^k,   \leqno{(P^k)}
\]
where
\begin{equation*}
\mathcal{J}^k(v)=\int\limits_\Omega
\Bigl[\frac{1}{2}\chi_a^{k}a\abs{\varepsilon (v)+C}^2
+\frac{1}{2}\chi_b^{k}b\abs{\varepsilon (v)+D}^2\Bigr]\,dx,\quad v\in H^1_0(\Omega;\RR^n),
\end{equation*}
i.e.
\[
\mathcal{J}^k(u^k)=\alpha^k.
\]
Easy calculations using the properties of the scalar product
show that
\[
\chi_a^{k}aC+\chi_b^{k}bC=\frac{aC+bD}{2}+\psi^k\frac{bD-aC}{2},
\]
\[
\tfrac{1}{2}\chi^k_aa\abs{C}^2+
\tfrac{1}{2}\chi^k_bb\abs{D}^2=
\tfrac{1}{2}\biggl(\frac{a\abs{C}^2+b\abs{D}^2}{2}+\psi^k\frac{b\abs{D}^2
-a\abs{C}^2}{2}\biggr)
\]
so that
$\mathcal{J}^k(\cdot)$ admits the representation
\begin{equation} \tag{R}
\left.
\begin{array}{r}
\displaystyle \mathcal{J}^k(v)=
\int\limits_{\Omega}\Bigl[\tfrac{1}{2}(\chi^k_aa+
\chi^k_bb)\abs{\varepsilon (v)}^2
+(\chi^k_a aC+\chi^k_b bD)\boldsymbol{\cdot}\varepsilon (v)
\\
+\tfrac{1}{2}\chi^k_aa\abs{C}^2+
\tfrac{1}{2}\chi^k_bb\abs{D}^2\Bigr]dx
\\
\displaystyle=\int\limits_{\Omega}\left[\frac{1}{2}m^k\abs{\varepsilon (v)}^2+
(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon (v)
+\frac{1}{2}\mathcal{B}^k\right]dx,
\end{array}
\right\}
\end{equation}

where the following notations have been introduced
\begin{align}\label{eq-0}
\begin{split}
\mathcal{A}^+&:=\frac{aC+bD}{2}
\\
\mathcal{A}^-&:=\frac{bD-aC}{2}
\\
\mathcal{B}^k&:=\frac{a\abs{C}^2+b\abs{D}^2}{2}+\psi^k\frac{b\abs{D}^2
-a\abs{C}^2}{2}.
\end{split}
\end{align}
\par
Now we associate to the minimization problem $(P^k)$ the dual problem
$(P^k)^\star$ following the idea of Fenchel \cite{FENCHEL} (cf. also
\cite{EKELAND-TEMAM}, \cite{AUBIN}).  Let $\phi(\cdot)$ be given by
\begin{equation}\label{eq-1.4}
\phi(\xi)=\frac{m}{2}\abs{\xi}^2+
E\boldsymbol{\cdot}\xi,\quad\xi\in{\mathbb{R}}^{n\times n}_{sym},
\end{equation}
where $m>0$ is a constant,
then its conjugate $\phi^c(\cdot)$ reads
\begin{equation}\label{eq-1.5}
\phi^c(p)=\sup_{\xi\in{\mathbb{R}}^{n\times n}_{sym}}\bigl(p\boldsymbol{\cdot}\xi-\phi(\xi)\bigr)
=\frac{1}{2m}\abs{p-E}^2, \quad p\in{\mathbb{R}}^{n\times n}_{sym}.
\end{equation}
Next define a linear continuous operator $L:H^1_0(\Omega;\RR^n)\to
L^2(\Omega;\mathbb{R}^{n\times n}_{sym})$ as
\[
Lv=\varepsilon(v),\quad v\inH^1_0(\Omega;\RR^n),
\]
then its transpose $L^\star:L^2(\Omega;\mathbb{R}^{n\times n}_{sym})\to
H^{-1}(\Omega;\mathbb{R}^n)$ can be expressed as
\[
\bigl\langle L^\star p,v\bigr\rangle_{H^1_0(\Omega;\RR^n)}=\int\limits_\Omega p\boldsymbol{\cdot}\varepsilon
(v)\,dx,\quad \forall p\in L^2(\Omega;\mathbb{R}^{n\times n}_{sym}),\;\forall
v\inH^1_0(\Omega;\RR^n).
\]
Define
\begin{equation}\label{eq-1.66}
\mathcal{I}^k(q):=\int\limits_{\Omega}\Bigl[\frac{1}{2m^k}
\abs{q-(\mathcal{A}^++\psi^k\mathcal{A}^-)}^2-\frac{1}{2}\mathcal{B}^k\Bigr]
\,dx,\quad q\in
L^2(\Omega;\mathbb{R}^{n\times n}_{sym})
\end{equation}
and denote by $\Ker L^\star$ the kernel of $L^\star$, i.e.
\[
\Ker L^\star=\Bigl\{p\in L^2(\Omega;\mathbb{R}^{n\times n}_{sym})\colon
\int\limits_\Omega p\boldsymbol{\cdot}\varepsilon(v)\,dx=0\Bigl\}.
\]
\par
Now we are in a position to formulate the dual problem $(P^k)^\star$ which
can be stated as follows
\[
\inf\Bigl\{\int\limits_{\Omega} \mathcal{I}^k(q)\,dx\colon q\in\Ker
L^\star\Bigr\}:=\beta^k.                         \leqno{(P^k)^\star}
\]
According to the Fenchel theorem (cf. Theorem 3.2, p. 38, \cite{AUBIN})
we get
\begin{multline}\label{eq-1.6}
\mathcal{J}^k(v)=\int\limits_{\Omega}\Bigl[\frac{1}{2}m^k\abs{\varepsilon (v)}^2+
(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon (v)
+\frac{1}{2}\mathcal{B}^k\Bigr]\,dx\geq
\\
\geq\,\int\limits_{\Omega}\Bigl[\frac{1}{2}m^k\abs{\varepsilon
(u^k)}^2+
(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon (u^k)
+\frac{1}{2}\mathcal{B}^k\Bigr]\,dx=
\\
=\,\mathcal{J}^k(u^k)=\alpha^k=-\beta^k=-\mathcal{I}^k(p^k)=
\\
=-\int\limits_{\Omega}\Bigl[
\frac{1}{2m^k}
\abs{p^k-(\mathcal{A}^++\psi^k\mathcal{A}^-)}^2-\frac{1}{2}\mathcal{B}^k\Bigr]\,dx \geq
\\
\geq\, -\int\limits_{\Omega}\Bigl[\frac{1}{2m^k}
\abs{q-(\mathcal{A}^++\psi^k\mathcal{A}^-)}^2-\frac{1}{2}\mathcal{B}^k\Bigr]\,dx=-\mathcal{I}^k(q),
\\
\forall\,v\in H^1_0(\Omega;\RR^n),\;\forall\,q\in \Ker L^\star,
\end{multline}
where
\begin{equation}\label{1-10}
p^k=m^k\varepsilon (u^k)+\mathcal{A}^++\psi^k \mathcal{A}^-\in\Ker L^\star
\end{equation}
is a solution of the dual problem  $(P^k)^\star$. Since $p^k\in\Ker L^\star$,
\begin{equation}\label{1-1}
\int\limits_{\Omega}p^k\boldsymbol{\cdot}\varepsilon(v)\,dx=0,\quad\forall\,v\inH^1_0(\Omega;\RR^n),
\end{equation}
so, in particular,
\begin{equation}\label{1-2}
\int\limits_{\Omega}p^k\boldsymbol{\cdot}\varepsilon(u^k)\,dx=
\int\limits_{\Omega}\Bigl[m^k\abs{\varepsilon (u^k)}^2
+(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon (u^k)\Bigr]dx=0.
\end{equation}

Taking into account that $\alpha^k=\mathcal{I}^k(p^k)$ and equation (\ref{1-2}) we get

the following representations
\begin{equation}\label{eq-11}
\alpha^k=\tfrac{1}{2}\int\limits_{\Omega}\bigl[-m^k\abs{\varepsilon (u^k)}^2
+\mathcal{B}^k\bigr]dx
=\tfrac{1}{2}\int\limits_{\Omega}\bigl[(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon (u^k)
+\mathcal{B}^k\bigr]dx.
\end{equation}
Analogously, from {{\eqref{1-10}}} we have
$\varepsilon(u^k)=\frac{1}{m^k}(p^k-\mathcal{A}^+-\psi^k
\mathcal{A}^-)$, so from (\ref{1-2}) and \eqref{eq-11}
it follows
\[
\int\limits_{\Omega}\left[\frac{1}{m^k}\abs{p^k}^2
-\frac{1}{m^k}(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot} p^k\right]dx=0.
\]
Therefore
\begin{align}
\alpha^k&=-\tfrac{1}{2}\int\limits_{\Omega}\left[-\frac{1}{m^k}
\abs{p^k}^2+
\frac{1}{m^k}\abs{\mathcal{A}^++\psi^k \mathcal{A}^-}^2
-\mathcal{B}^k\right]dx=\notag
\\
&=\tfrac{1}{2}\int\limits_{\Omega}\left[\frac{1}{m^k}
(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot} p^k-
\frac{1}{m^k}\abs{\mathcal{A}^++\psi^k \mathcal{A}^-}^2 \label{eq-1}
+\mathcal{B}^k\right]dx
\end{align}
and we are led to the equality
\begin{equation}\label{eq-111}
\int\limits_{\Omega}\left[m^k\abs{\varepsilon (u^k)}^2+
\frac{1}{m^k}
\abs{p^k}^2\right]dx
=\int\limits_{\Omega}\left[
\frac{1}{m^k}\abs{\mathcal{A}^++\psi^k \mathcal{A}^-}^2\right]dx.
\end{equation}
\par
Now observe that the right hand side of this equality can be reorganized in
such a way that its limit as $k\to\infty$ is easy to be calculated. Indeed,
if we set
\[
\overline{m}^\natural:=\frac{1}{2}\bigl(\frac{1}{a}+\frac{1}{b}\bigr),\quad
\underline{m}^\natural:=\frac{1}{2}\bigl(\frac{1}{b}-\frac{1}{a}\bigr),
\]
and recall that $\chi_{a}^{k}+\chi_{b}^{k}=1$ and
$(\psi^k)^2=1$ we obtain
\[
m^k =\frac{\overline{m}^\natural -\psi^k\underline{m}^\natural}
{(\overline{m}^\natural)^2 - (\underline{m}^\natural)^2}
\]
so that
\[
\frac{1}{m^k}=\overline{m}^\natural+\underline{m}^\natural\psi^k
\]
and from \eqref{eq-111}, by making use of \eqref{eq-1.2}, we obtain
\begin{equation}
\begin{array}{l}
\displaystyle \lim\limits_{k\to\infty}\int\limits_{\Omega}\left[m^k\abs{\varepsilon
(u^k)}^2+ \frac{1}{m^k}
\abs{p^k}^2\right]dx
\\
\displaystyle=\lim\limits_{k\to\infty}\int\limits_{\Omega}\left[
(\overline{m}^\natural+\underline{m}^\natural\psi^k)\abs{\mathcal{A}^++\psi^k \mathcal{A}^-}^2\right]dx
\\
\displaystyle=\int\limits_{\Omega}\left[
(\overline{m}^\natural+\underline{m}^\natural\,\psi)\abs{\mathcal{A}^+}^2
+2(\overline{m}^\natural\,\psi+\underline{m}^\natural) \mathcal{A}^+\boldsymbol{\cdot} \mathcal{A}^-
+(\overline{m}^\natural+\underline{m}^\natural\,\psi)\abs{\mathcal{A}^-}^2\right]dx
\\
\displaystyle=\tfrac{1}{2}\int\limits_{\Omega}\left[(\overline{m}^\natural+\underline{m}^\natural\,\psi)
\bigl(a^2\abs{C}^2+b^2\abs{D}^2\bigr)+(\overline{m}^\natural\,\psi+\underline{m}^\natural)
\bigl(b^2\abs{D}^2-a^2\abs{C}^2\bigr)
\right]dx
\\
\displaystyle=\tfrac{1}{2}\int\limits_{\Omega}\Bigl(a\abs{C}^2+b\abs{D}^2
\Bigr)dx+\tfrac{1}{2}\int\limits_{\Omega}\psi\Bigl(b\abs{D}^2-a\abs{C}^2\Bigr)
dx=\int\limits_{\Omega}\mathcal{B}\,dx,
\end{array}
\end{equation}
where
\[
\mathcal{B}=\frac{a\abs{C}^2+b\abs{D}^2}{2}+\psi\frac{b\abs{D}^2-a\abs{C}^2}{2}.
\]
\par
Let us introduce the set
\[
\Omega_0=\{x\in\Omega\colon a(x)=b(x)\}.
\]
Using \eqref{1-10} we obtain immediately that in $\Omega\setminus\Omega_0$,
\[
\psi^k\varepsilon(u^k)= \frac{1}{\underline{m}}p^k-\frac{\overline{m}}{\underline{m}}\varepsilon(u^k)-
\frac{1}{\underline{m}}\mathcal{A}^+-\frac{1}{\underline{m}}\psi^k \mathcal{A}^-.
\]
Thus from \eqref{eq-11} it follows
\begin{align}\label{eq-I}
\alpha^k=&\,\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}\bigl[
\mathcal{A}^+\boldsymbol{\cdot}\varepsilon (u^k)+
\mathcal{A}^-\boldsymbol{\cdot}\psi^k\varepsilon(u^k)+\mathcal{B}^k\bigr]dx \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}\bigl[
\mathcal{A}^+\boldsymbol{\cdot}\varepsilon (u^k)+
\mathcal{A}^-\boldsymbol{\cdot}\psi^k\varepsilon(u^k)+\mathcal{B}^k\bigr]dx \notag
\\
=&\,\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}\Bigl[(\mathcal{A}^+-
\tfrac{\overline{m}}{\underline{m}}\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u^k)
+\tfrac{1}{\underline{m}}\mathcal{A}^-\boldsymbol{\cdot} p^k-\tfrac{1}{\underline{m}}\mathcal{A}^-
\boldsymbol{\cdot} \mathcal{A}^+  \notag
\\
&-\tfrac{1}{\underline{m}}\psi^k\,\abs{
\mathcal{A}^-}^2+\mathcal{B}^k\Bigr]dx \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}\bigl[
\mathcal{A}^+\boldsymbol{\cdot}\varepsilon (u^k)+
p^k\boldsymbol{\cdot}\varepsilon(u^k)-a\abs{\varepsilon(u^k)}^2-\mathcal{A}^+\boldsymbol{\cdot}\varepsilon(u^k)
+\mathcal{B}^k\bigr]dx  \notag
\\
=&\,\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u^k)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p^k+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi^k\,\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}\Bigl[
-a\abs{\varepsilon(u^k)}^2+\frac{a(\abs{C}^2+\abs{D}^2)}{2}+
\psi^k\frac{a(\abs{D}^2-\abs{C}^2)}{2}\Bigr]dx \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}p^k\boldsymbol{\cdot}\varepsilon(u^k)dx.
\end{align}
Another representation of $\alpha^k$ can be derived from \eqref{1-10} if we
take into account
\eqref{eq-1},
$(\psi^k)^2=1$ and that in $\Omega_0$,
$\frac{1}{a}\bigl(\abs{\mathcal{A}^+}^2+
\abs{\mathcal{A}^-}^2+2\psi^k\mathcal{A}^+\boldsymbol{\cdot}
\mathcal{A}^-\bigr)=\mathcal{B}^k$:
\begin{align}\label{eq-II}
\alpha^k=&\,\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}
\Bigl[\frac{1}{m^k}(\mathcal{A}^++\psi^k\mathcal{A}^-)\boldsymbol{\cdot} p^k-
\frac{1}{m^k}\abs{\mathcal{A}^++\psi^k \mathcal{A}^-}^2+\mathcal{B}^k
\Bigr]dx  \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[\frac{1}{a}(\mathcal{A}^++\psi^k\mathcal{A}^-)\boldsymbol{\cdot} p^k-
\frac{1}{a}\abs{\mathcal{A}^++\psi^k \mathcal{A}^-}^2
+\mathcal{B}^k\Bigr]dx \notag
\\
=&\,\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}\bigl[
\mathcal{A}^+\boldsymbol{\cdot}\varepsilon (u^k)+
\mathcal{A}^-\boldsymbol{\cdot}\psi^k\varepsilon(u^k)+\mathcal{B}^k\bigr]dx \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[\frac{1}{a}\abs{p^k}^2-
\frac{1}{a}\bigl(\abs{\mathcal{A}^+}^2+\abs{\mathcal{A}^-}^2+2\psi^k\mathcal{A}^+\boldsymbol{\cdot} \mathcal{A}^-\bigr)
+\mathcal{B}^k\Bigr]dx \notag
\\
=&\,\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u^k)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p^k+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi^k\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx \notag
\\
&+\tfrac{1}{2}\int\limits_{\Omega_0}
\frac{1}{a}\abs{p^k}^2dx -\tfrac{1}{2}\int\limits_{\Omega_0}
p^k\boldsymbol{\cdot}\varepsilon(u^k)dx.
\end{align}
\par

As it will be seen in the next section the formulas \eqref{eq-I} and
\eqref{eq-II} allow us to express $\alpha$ in terms of the weak limits: $u$,
$p$ and $\psi$.

\section{Weak convergence in $\boldsymbol{L^1(\Omega)}$}
\renewcommand{\theequation}{\thesection.\arabic{equation}}
\setcounter{section}{3}%
\setcounter{equation}{0}%

\begin{lem}\label{lem-1}
Let $\Omega\subset \mathbb{R}^n$ be a bounded domain in $\mathbb{R}^n$
with Lipschitz continuous boundary $\partial\Omega$.
Then
\begin{equation}\label{P-1}
p^k\boldsymbol{\cdot}\varepsilon(u^k)\to p\boldsymbol{\cdot}\varepsilon(u)\quad\mbox{weakly in}
\;L^1(\Omega).
\end{equation}
\end{lem}
\begin{proof}
Extend each function $u^k\inH^1_0(\Omega;\RR^n)$ to all of $\mathbb{R}^n$
by setting
it equal to zero on $\mathbb{R}^n\setminus\Omega$. By regularity of the boundary
$\partial\Omega$ all of these extensions are elements of $H^1(\mathbb{R}^n;\mathbb{R}^n)$.
For an arbitrary $\varphi\in C^\infty(\mathbb{R}^n)$ we thus have $\varphi u^k\in
H^1(\mathbb{R}^n;\mathbb{R}^n)$ and $\varphi u^k\big|_\Omega\inH^1_0(\Omega;\RR^n)$. We claim that
\begin{equation}\label{P-3}
\int\limits_{\mathbb{R}^n}\varphi \,p^k\boldsymbol{\cdot}\varepsilon(u^k)\,dx\to
\int\limits_{\mathbb{R}^n}\varphi \,p\boldsymbol{\cdot}\varepsilon(u)\,dx,
\end{equation}
for any $\varphi\in C^\infty(\mathbb{R}^n)$.
Indeed, since $\varepsilon(\varphi
u^k)=\varphi\varepsilon(u^k)+u^k\otimes\nabla \varphi$,
we get
\begin{eqnarray*}
\int\limits_{\mathbb{R}^n}\varphi \,p^k\boldsymbol{\cdot}\varepsilon(u^k)\,dx=
\int\limits_{\mathbb{R}^n}
p^k\boldsymbol{\cdot}\varepsilon(\varphi u^k)\,dx-\int\limits_{\mathbb{R}^n}
p^k\boldsymbol{\cdot} (u^k\otimes\nabla \varphi)\,dx=
\\
=-\int\limits_{\mathbb{R}^n}
p^k\boldsymbol{\cdot} (u^k\otimes\nabla \varphi)\,dx\to-\int\limits_{\mathbb{R}^n}
p\boldsymbol{\cdot} (u\otimes\nabla \varphi)\,dx=
\int\limits_{\mathbb{R}^n}\varphi
\,p\boldsymbol{\cdot}\varepsilon(u)\,dx,
\end{eqnarray*}
where we have used \eqref{1-1} and the strong convergence $u^k\to u$ in
$L^2(\Omega;\mathbb{R}^n)$ (valid due to the Rellich compactness theorem).
\par
Further, the sequence $\{p^k\boldsymbol{\cdot}\varepsilon(u^k)\}$ is uniformly bounded in
$L^1(\Omega)$
because
$\{p^k\}$ and $\{\varepsilon(u^k)\}$ so are in $L^2(\Omega)$. By
Chacon's
biting lemma \cite{PEDREGAL} it follows that there exist a subsequence of
$(p^k\boldsymbol{\cdot}\varepsilon(u^k))$, not relabeled, a
nonincreasing sequence of measurable sets $\Omega_n\subset\Omega$,
$\Omega_{n+1}\subset \Omega_n$,
$\abs{\Omega_n} \searrow 0$ and $f\in L^1(\Omega)$ such that
\begin{equation}\label{P-2}
p^k\boldsymbol{\cdot}\varepsilon(u^k)\to f\quad\mbox{weakly in}\;L^1(\Omega\setminus\Omega_n)
\end{equation}
for all $n$. It means that $\{p^k\boldsymbol{\cdot}\varepsilon(u^k)\}$ converges in the
biting sense to $f$ \cite{PEDREGAL}.
\par
Now we assert that the biting limit $f$ coincides with $p\boldsymbol{\cdot}\varepsilon(u)$,
i.e.  $f=p\boldsymbol{\cdot}\varepsilon(u)$ a.e. in $\Omega$. To show this
observe that from  the biting argument \eqref{P-2} and \eqref{P-3} it follows
that for any $\varphi\in C^\infty(\mathbb{R}^n)$ we get
\begin{equation}\label{P-4}
\int\limits_{\Omega\setminus\Omega_n}\varphi\,p\boldsymbol{\cdot}\varepsilon(u)\,dx=
\int\limits_{\Omega\setminus\Omega_n}\varphi\,f\,dx,
\end{equation}
for any $n$. Hence $p\boldsymbol{\cdot}\varepsilon(u)=f$ a.e. in
$\Omega\setminus\Omega_n$ for each $n$. Since $\abs{\Omega_n}\searrow 0$
as $n\to\infty$, the equality $p\boldsymbol{\cdot}\varepsilon(u)=f$ must hold a.e. in
$\Omega$. Thus the assertion follows.
\par
Recall that $p^k\boldsymbol{\cdot}\varepsilon(u^k)=m^k\abs{\varepsilon(u^k)}^2+
\bigl(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u^k)$. Therefore one can deduce the
existence of a constant $C\geq 0$ such that
\[
p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\geq 0\quad\mbox{a.e in }\Omega.
\]
Obviously $p^k\boldsymbol{\cdot}\varepsilon(u^k)+C$ converges in the biting sense to
$p\boldsymbol{\cdot}\varepsilon(u)+C$. According to (Lemma 6.9, p.109, \cite{PEDREGAL}) its
weak convergence in $L^1(\Omega)$ is then equivalent to
\begin{equation}\label{P-5}
\limsup\limits_{k\to\infty}\int\limits_\Omega
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr)\,dx\leq
\int\limits_\Omega \bigl(p\boldsymbol{\cdot}\varepsilon(u)+C\bigr)\,dx.
\end{equation}
Our task now is to establish the foregoing inequality. For this purpose notice
that \eqref{P-3} can be easily extend to the convergence
\begin{equation}\label{P-33}
\int\limits_{\mathbb{R}^n}\varphi \,p^k\boldsymbol{\cdot}\varepsilon(u^k)\,dx\to
\int\limits_{\mathbb{R}^n}\varphi \,p\boldsymbol{\cdot}\varepsilon(u)\,dx,
\end{equation}
which is valid for any $\varphi\in C_c(\mathbb{R}^n)$, where $C_c(\mathbb{R}^n)$ is the space of
continuous functions on $\mathbb{R}^n$ with compact support.  Thus
$\mu^k:=(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C)dx$ and $\mu:=(p\boldsymbol{\cdot}\varepsilon(u)+C)dx$ can
be treated as positive Radon measures on $\mathbb{R}^n$ for which it holds
\[
\lim\limits_{k\to\infty}\int\limits_{\mathbb{R}^n}\varphi\,d\mu^k=
\int\limits_{\mathbb{R}^n}\varphi\,d\mu,\quad\forall\,\varphi\in C_c(\mathbb{R}^n).
\]
But (Theorem 1, p. 54, \cite{EVANS-GARIEPY}) asserts that this condition is
equivalent to the following one
\begin{equation}\label{P-6}
\lim\limits_{k\to\infty}\mu^k(B)=\mu(B)\quad \mbox{for each bounded Borel set
$B\subset\mathbb{R}^n$ with $\mu(\partial B)=0$}.
\end{equation}
Now we are in a position to show \eqref{P-5}. Fix $\epsilon>0$ and choose
$0<\delta<\epsilon$ with the property that $\omega\subset\Omega$ with
$\abs{\omega}<\delta$ implies
\[
\int\limits_\omega p\boldsymbol{\cdot}\varepsilon(u)\,dx<\epsilon.
\]
In the biting convergence take $n_0$ large enough to fulfill
$\abs{\Omega_{n_0}}<\tfrac{\delta}{2}$. By the measurability of $\Omega_{n_0}$
there exists an open $\widetilde{\Omega}_{n_0}\supset\Omega_{n_0}$ with
$\abs{\widetilde{\Omega}_{n_0}}<\delta$. Vitali's covering theorem ensures the
representation $\widetilde{\Omega}_{n_0}=\widetilde{\Omega}_{n_0}^\prime\cup
\widetilde{\Omega}_{n_0}^{\prime\prime}$ where
$\abs{\widetilde{\Omega}_{n_0}^{\prime\prime}}=0$ and
$\widetilde{\Omega}_{n_0}^\prime$ stands for the union of a countable
collection of disjoint closed balls in $\widetilde{\Omega}_{n_0}$. Therefore
$\abs{\partial \widetilde{\Omega}_{n_0}^\prime}=0$ and consequently
$\mu(\partial \widetilde{\Omega}_{n_0}^\prime)=0$. From this we have
\begin{align*}
\int\limits_\Omega \bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr)\,dx&=
\int\limits_{\Omega\setminus\Omega_{n_0}} \bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr)
\,dx+\int\limits_{\Omega_{n_0}}
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr)\,dx
\\
&\leq \int\limits_{\Omega\setminus\Omega_{n_0}}
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr) \,dx
+\int\limits_{\widetilde{\Omega}_{n_0}} \bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr)
\,dx
\\
&= \int\limits_{\Omega\setminus\Omega_{n_0}}
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr) \,dx+
\int\limits_{\widetilde{\Omega}_{n_0}^\prime}
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr) \,dx
\\
&=\int\limits_{\Omega\setminus\Omega_{n_0}}
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr) \,dx+
\mu^k(\widetilde{\Omega}_{n_0}^\prime)
\end{align*}
which thanks to \eqref{P-6} by passing to the limit as $k\to\infty$ yields
\begin{eqnarray*}
\limsup\limits_{k\to\infty}\int\limits_\Omega
\bigl(p^k\boldsymbol{\cdot}\varepsilon(u^k)+C\bigr)\,dx
\leq\int\limits_{\Omega\setminus\Omega_{n_0}} \bigl(p\boldsymbol{\cdot}\varepsilon(u)+C
\bigr)\,dx+
\mu(\widetilde{\Omega}_{n_0}^\prime)
\\
\leq\int\limits_{\Omega} \bigl(p\boldsymbol{\cdot}\varepsilon(u)+C\bigr)\,dx+
\int\limits_{\widetilde{\Omega}_{n_0}^\prime} \bigl(p\boldsymbol{\cdot}\varepsilon(u)+C
\bigr)\,dx
\\
\leq\int\limits_{\Omega} \bigl(p\boldsymbol{\cdot}\varepsilon(u)+C\bigr)\,dx+\epsilon(1+C),
\end{eqnarray*}
because $\abs{\widetilde{\Omega}_{n_0}^\prime}<\delta<\epsilon$.
Since $\epsilon>0$ was chosen arbitrarily, \eqref{P-5} follows. This completes
the proof of \lemref{lem-1}.
\end{proof}

\par
\section{Relaxed formulas for the infimum}
\renewcommand{\theequation}{\thesection.\arabic{equation}}
\setcounter{section}{4}%
\setcounter{equation}{0}%

The weak lower semicontinuity of convex functionals, the upper semicontinuity
of concave functionals and \lemref{lem-1} yield
\begin{equation}
\liminf\limits_{k\to\infty}\;\tfrac{1}{2}\int\limits_{\Omega_0}
\frac{1}{a}\abs{p^k}^2\,dx
\geq
\tfrac{1}{2}\int\limits_{\Omega_0}
\frac{1}{a}\abs{p}^2\,dx,
\end{equation}
\begin{equation}
\limsup\limits_{k\to\infty}\,\tfrac{1}{2}\int\limits_{\Omega_0}\Bigl[
-a\abs{\varepsilon(u^k)}^2+\mathcal{B}^k\Bigr]dx\leq
\tfrac{1}{2}\int\limits_{\Omega_0}\Bigl[
-a\abs{\varepsilon(u)}^2+\mathcal{B}\Bigr]dx,
\end{equation}
\begin{equation}
\lim\limits_{k\to\infty}\;\tfrac{1}{2}\int\limits_{\Omega_0}
p^k\boldsymbol{\cdot}\varepsilon(u^k)
dx=\tfrac{1}{2}\int\limits_{\Omega_0}p\boldsymbol{\cdot}\varepsilon(u)dx,
\end{equation}
where $\psi=\chi_b-\chi_a$ and
\[
\mathcal{B}=\frac{a(\abs{C}^2+\abs{D}^2)}{2}+\psi\frac{a(\abs{D}^2
-\abs{C}^2)}{2}\quad\mbox{in }\Omega_0.
\]
\par
Now we show that
\begin{eqnarray}
\lim\limits_{k\to\infty}\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}
\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u^k)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p^k+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
-\psi^k\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx \notag
\\
=\tfrac{1}{2}\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
-\psi\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx.   \label{eq-00}
\end{eqnarray}
This is not trivial because the functions $\frac{ab(C-D)}{b-a}$ and $\frac{bD-aC}{b-a}$ are not
assumed to belong to $L^2(\Omega\setminus\Omega_0;\mathbb{R}^{n\times n}_{sym})$. To overcome
this disadvantage let us recall (see \eqref{eq-I} or \eqref{eq-II})
that $\Omega\setminus\Omega_0$ is a set of a finite
Lebesgue measure where
\begin{multline*}
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u^k)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p^k+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}-\psi^k\,\frac{ab\abs{C-D}^2}{2(b-a)}
\\
=
\mathcal{A}^+\boldsymbol{\cdot}\varepsilon (u^k)+
\mathcal{A}^-\boldsymbol{\cdot}\psi^k\varepsilon(u^k)+\mathcal{B}^k.
\end{multline*}
Thus for any $\varepsilon >0$ there exist $\omega_\varepsilon\subset\Omega\setminus\Omega_0$
and $\delta>0$ such that $\abs{\omega_\varepsilon}<\varepsilon$ and for each $x\in
(\Omega\setminus\Omega_0)\setminus \omega_\varepsilon$ one has $\abs{a(x)-b(x)}\geq\delta$.
Hence
\begin{equation}
\frac{C-D}{b-a}\in L^\infty\bigl((\Omega\setminus\Omega_0)\setminus\omega_\varepsilon;
\mathbb{R}^{n\times n}_{sym}\bigr)\subset
L^2\bigl((\Omega\setminus\Omega_0)\setminus\omega_\varepsilon;
\mathbb{R}^{n\times n}_{sym}\bigr)
\end{equation}
and
\begin{multline*}
\Big\lvert\,\int\limits_{\omega_\varepsilon}
\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u^k)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p^k+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
-\psi^k\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx\Big\rvert
\leq \const \abs{\omega_\varepsilon}^\frac{1}{2}\leq \const \varepsilon^\frac{1}{2}.
\end{multline*}
This allows the conclusion that
\begin{eqnarray*}
\lim\limits_{k\to\infty}
\int\limits_{(\Omega\setminus\Omega_0)\setminus\omega_\varepsilon}
\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u^k)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p^k+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}
\\
-\psi^k\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx
\\
=\int\limits_{(\Omega\setminus\Omega_0)\setminus\omega_\varepsilon}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}
\\
-\psi\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx
\end{eqnarray*}
and due to the fact that $\varepsilon>0$ was chosen arbitrarily we easily arrive at
\eqref{eq-00}, as desired.
\par

Now, for $v\in H^1_0(\Omega;\RR^n)$ and $q\in \Ker L^\star$ let us set
\begin{multline}
\mathcal{I}(v,q):=
\\
\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(v)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} q+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}-\psi\,\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx.
\end{multline}

Using the fact that for all $k\in \mathbb{N}$ we have
$p^k\in\Ker L^{\ast}$, \eqref{eq-00} and
passing to the limit as $k\to\infty$ in \eqref{eq-I} we
get
\[
\alpha\leq\tfrac{1}{2}\mathcal{I}(u,p)+
\tfrac{1}{2}\int\limits_{\Omega_0}\Bigl[-a\abs{\varepsilon(u)}^2\Bigr]dx+
\tfrac{1}{2}\int\limits_{\Omega_0}\mathcal{B}\,dx+
\tfrac{1}{2}\int\limits_{\Omega_0}
p\boldsymbol{\cdot}\varepsilon(u)dx.
\]
Analogously, passing to the limit as $k\to\infty$ in
\eqref{eq-II} we get
\[
\alpha\geq\tfrac{1}{2}\mathcal{I}(u,p)+
\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[\frac{1}{a}\abs{p}^2-
\frac{1}{a}\bigl(\abs{\mathcal{A}^+}^2+\abs{\mathcal{A}^-}^2+2\psi
\mathcal{A}^+\boldsymbol{\cdot} \mathcal{A}^-\bigr) \Bigr]dx+
\tfrac{1}{2}\int\limits_{\Omega_0}\mathcal{B}\,dx.
\]
Putting the two above inequalities together we obtain the system of inequalities
\begin{multline}\label{eq-III}
\tfrac{1}{2}\int\limits_{\Omega_0}\Bigl[
-a\abs{\varepsilon(u)}^2\Bigr]dx
+\int\limits_{\Omega_0}p\boldsymbol{\cdot}\varepsilon(u)dx\geq
\\
\geq\alpha+\tfrac{1}{2}\int\limits_{\Omega_0}
p\boldsymbol{\cdot}\varepsilon(u)dx
-\mathcal{I}(u,p)-\int\limits_{\Omega_0}\mathcal{B}\,dx\geq
\\
\geq\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[\frac{1}{a}\abs{p}^2-
\frac{1}{a}\bigl(\abs{\mathcal{A}^+}^2+\abs{\mathcal{A}^-}^2+2\psi
\mathcal{A}^+\boldsymbol{\cdot} \mathcal{A}^-\bigr) \Bigr]dx.
\end{multline}
But from the fact that $p=a\,\varepsilon(u)+\mathcal{A}^++\psi \mathcal{A}^-$
in $\Omega_0$ it follows
\begin{multline}
\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[\frac{1}{a}\abs{p}^2-
\frac{1}{a}\bigl(\abs{\mathcal{A}^+}^2+\abs{\mathcal{A}^-}^2+2\psi \mathcal{A}^+\boldsymbol{\cdot} \mathcal{A}^-\bigr)
\Bigr]dx= \notag
\\
=\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[a\abs{\varepsilon(u)}^2+\frac{1}{a}\abs{\mathcal{A}^++\psi\,\mathcal{A}^-}^2+
2(\mathcal{A}^++\psi\,\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u)+\notag
\\
-\frac{1}{a}\bigl(\abs{\mathcal{A}^+}^2+\abs{\mathcal{A}^-}^2+2\psi \mathcal{A}^+\boldsymbol{\cdot} \mathcal{A}^-\bigr)
\Bigr]dx= \notag
\\
=\,\tfrac{1}{2}\int\limits_{\Omega_0}
\Bigl[-a\abs{\varepsilon(u)}^2+
2\bigl(a\abs{\varepsilon(u)}^2+(\mathcal{A}^++\psi\,\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u)\bigl)+
\frac{1}{a}(\psi^2-1)\abs{\mathcal{A}^-}^2\Bigr]dx= \notag
\\
=\,\tfrac{1}{2}\int\limits_{\Omega_0}\Bigl[
-a\abs{\varepsilon(u)}^2\Bigr]dx
+\int\limits_{\Omega_0}p\boldsymbol{\cdot}\varepsilon(u)dx
-2\int\limits_{\Omega_0}\frac{1}{a}\,\chi_a\chi_b\,\abs{\mathcal{A}^-}^2dx.
\end{multline}
Here we used the fact that $\psi^2-1=-4\chi_a\chi_b$. Thus in view of
\eqref{eq-III} it follows
\begin{multline}
0\geq\alpha-\tfrac{3}{2}\int\limits_{\Omega_0}
p\boldsymbol{\cdot}\varepsilon(u)dx+\int\limits_{\Omega_0}
a\abs{\varepsilon(u)}^2dx
-\mathcal{I}(u,p)-\int\limits_{\Omega_0}\mathcal{B}\,dx\geq
\\
\geq -2\int\limits_{\Omega_0}\frac{1}{a}\,\chi_a\chi_b\,\abs{\mathcal{A}^-}^2
dx.
\end{multline}
Since
\[
\int\limits_{\Omega_0}p\boldsymbol{\cdot}\varepsilon(u)dx= \int\limits_{\Omega_0}\Bigl[
a\abs{\varepsilon(u)}^2+(\mathcal{A}^++\psi\,\mathcal{A}^-)\boldsymbol{\cdot}
\varepsilon(u)\Bigr]dx
\]
and in $\Omega_0$ $\mathcal{A}^- =a\frac{D-C}{2}$
we have
\begin{equation}
0\geq\alpha-\int\limits_{\Omega_0}\bigl[
(\mathcal{A}^++\psi\,\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u)+\mathcal{B}\bigr]dx
-\mathcal{I}(u,p)
\geq -\tfrac{1}{2}\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx.
\end{equation}
Thus we are allowed to conclude that
there exists a $\theta\in[0,1]$ such that
\begin{multline*}
\alpha=\,\int\limits_{\Omega_0}\bigl[
(\mathcal{A}^++\psi\,\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u)+\mathcal{B}\bigr]dx
\\
+\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p
+\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}
\\
-\psi\,\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx -\,\tfrac{\theta}{2}\int\limits_{\Omega_0}
\chi_a\chi_b \,a\abs{C-D}^2dx.
\end{multline*}
The obtained result can be summarized as follows.
\begin{thm}\label{thm-2}
Let $u\inH^1_0(\Omega;\RR^n)$ and $p\in L^2(\Omega;{\mathbb{R}}^{n\times n}_{sym})$ are the weak limits of $\{u^k\}$
and $\{p^k\}$ as defined by \eqref{eq-1.6}, respectively. Then there exists
$\theta\in[0,1]$ such that
\begin{align}\label{eq-IV}
\alpha=&\,\int\limits_{\Omega_0}\Bigl[
\Bigl(\frac{a(C+D)}{2}+\psi \frac{a(D-C)}{2}\Bigr)\boldsymbol{\cdot}\varepsilon(u)+
\frac{a(\abs{C}^2+\abs{D}^2)}{2}\notag
\\
&+\psi\frac{a(\abs{D}^2-\abs{C}^2)}{2}
\Bigr]dx \notag
\\
&+\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p
+\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi\,\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]\,dx-\,\tfrac{\theta}{2}\int\limits_{\Omega_0}
\chi_a\chi_b \,a\abs{C-D}^2dx.
\end{align}
\end{thm}

Using the fact that in $\Omega_0$ there holds the
equality
\[
\int\limits_{\Omega_0 }(\mathcal{A}^+ +\psi^k\mathcal{A}^-)
\varepsilon (u)dx=
\int\limits_{\Omega_0 }[-a\abs{\varepsilon (u)}^2]dx+
\int\limits_{\Omega_0 }p\boldsymbol{\cdot}\varepsilon (u)dx,
\]
we get from the above formula
\begin{align}
\alpha=&\,\int\limits_{\Omega_0}\Bigl[
-a\abs{\varepsilon(u)}^2+
\frac{a(\abs{C}^2+\abs{D}^2)}{2}+\psi\frac{a(\abs{D}^2-\abs{C}^2)}{2}\Bigr]dx
+\int\limits_{\Omega_0}p\boldsymbol{\cdot}\varepsilon(u)dx+ \notag
\\
&+\,\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx-\,\tfrac{\theta}{2}\int\limits_{\Omega_0}\chi_a\chi_b \,a\abs{C-D}^2dx
\label{eq-1.28}
\end{align}
and from \eqref{eq-1.28} and again the equality
$
\int\limits_{\Omega_0 }(\mathcal{A}^+ +\psi^k\mathcal{A}^-)
\varepsilon (u)dx=
\int\limits_{\Omega_0 }[-a\abs{\varepsilon (u)}^2]dx+\\
+\int\limits_{\Omega_0 }p\boldsymbol{\cdot}\varepsilon (u)dx,
$
\begin{align}
\alpha=
&\,\int\limits_{\Omega_0}
\frac{1}{a}\abs{p}^2\,dx -\int\limits_{\Omega_0}
p\boldsymbol{\cdot}\varepsilon(u)\,dx+ \notag
\\
&+\,\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]\,dx+ \tfrac{2-\theta}{2}\int\limits_{\Omega_0}\chi_a\chi_b \,
a\abs{C-D}^2dx.\label{eq-1.30}
\end{align}
By adding \eqref{eq-1.28} and \eqref{eq-1.30} we get also the formula:
\begin{align}
\alpha
=&\, {\tfrac{1}{2}}\int\limits_{\Omega_0}
\Bigl[\frac{1}{a}\abs{p}^2-a\abs{\varepsilon(u)}^2
+\frac{a(\abs{C}^2+\abs{D}^2)}{2}+\psi\frac{a(\abs{D}^2-\abs{C}^2)}{2}
\Bigr]dx+ \notag
\\
&+\, \int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx+ {\tfrac{1-\theta}{2}}\int\limits_{\Omega_0}\chi_a\chi_b \,
a\abs{C-D}^2dx. \label{eq-1.31}
\end{align}
\par
Before formulating next theorem it will be convenient to introduce
some notation. Denote by $\omega_0^+$ and $\omega_0^-$ such subsets of
$\Omega$, that $\psi^k\rightarrow 1$ weakly in $L^1(\omega_0^+)$ and
$\psi^k\rightarrow -1$ weakly in $L^1(\omega_0^-)$. Let
$\omega_0\colon =\omega_0^+\cup\omega_0^-$.
\begin{thm}\label{thm-3}
Let $\nu=\{\nu_x\}_{x\in\Omega}$ be the  parametrized
Young measure associated to the mi\-ni\-mi\-za\-tion sequence
$\{u^k\}$. Then
\begin{align}\label{P-7}
\alpha =&\int\limits_{\Omega}\int\limits_{\mathbb{R}^n}h(x,\lambda)\,
d\nu_x(\lambda)\,dx=\notag
\\
=&\,\int\limits_{\Omega\setminus\Omega_0}\Bigl[
\frac{ab(C-D)}{b-a}
\boldsymbol{\cdot}\varepsilon(u)
+\frac{bD-aC}{b-a}\boldsymbol{\cdot} p+
\frac{ab(\abs{C}^2-\abs{D}^2)}{2(b-a)}\notag
\\
&-\psi\,\frac{ab\abs{C-D}^2}{2(b-a)}
\Bigr]dx \notag
\\
&+\int\limits_{\Omega_0}\Bigl[
-\int\limits_{\mathbb{R}^{{{n\times n}}}}a
{{|}}\lambda {{|}}^2\,d\nu_x(\lambda)+
\frac{a(\abs{C}^2+\abs{D}^2)}{2}+\psi\frac{a(\abs{D}^2-\abs{C}^2)}{2}\Bigr]dx\notag
\\
&+\int\limits_{\Omega_0}p\boldsymbol{\cdot}\varepsilon(u)dx,
\end{align}
where $h(x,\lambda)=\min\bigl\{\tfrac{1}{2}a(x)
\abs{\lambda+C(x)}^2,\,
\tfrac{1}{2}b(x)\abs{\lambda+D(x)}^2\bigr\}$, $\lambda\in{\mathbb{R}}^{n\times n}_{sym}$, $x\in\Omega$.
Moreover, we have
\begin{equation}\label{P-71}
\nu_x=\delta_{\varepsilon(u(x))}\quad\mbox{a.e. in } \omega_0.
\end{equation}
\end{thm}
\begin{proof}
By the results expressed in equations (2.15), (2.16), (2.17), (4.4)
and lemma 2 we
have to compute only the weak limit of the sequence $\{h_1^k\}$, where
\[
h_1^k=
a(x)\abs{\varepsilon (u^k (x))}^2.
\]
The sequence $\{p^k\boldsymbol{\cdot}\varepsilon(u^k)\}$ as being
weakly convergent in $L^1(\Omega)$ has to be equiintegrable according to the
Dunford-Pettis criterion of weak compactness in $L^1(\Omega)$. Since
$p^k\boldsymbol{\cdot}\varepsilon(u^k)=m^k\abs{\varepsilon(u^k)}^2+
\bigl(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u^k)$, it can be easy to deduce that
$\{m^k\abs{\varepsilon(u^k)}^2\}$ is equiintegrable as well
{(and so is $\{h_1^k\}$)}.
Thus one can
suppose that it is weakly
convergent in $L^1(\Omega)$, by passing to a subsequence, if
necessary,
so by Theorem 6.2, p. 97, \cite{PEDREGAL} we
see, that its weak limit is
$\int\limits_{\mathbb{R}^{n\times n}}a\abs{\lambda }^2 d\nu_x (\lambda).$
\\
Now, from  the inequality $(R)$
\[
h\bigl({{x,}}\,
\varepsilon(u^k {{(x)}})\bigr)\leq
\frac{1}{2}m^k\abs{\varepsilon (u^k)}^2+
(\mathcal{A}^++\psi^k \mathcal{A}^-)\boldsymbol{\cdot}\varepsilon (u^k)
+\frac{1}{2}\mathcal{B}^k
\]
we are allowed to conclude that the sequence
$\{h\bigl({{x},}\,\varepsilon(u^k {{(x)}})\bigr)\}$
has the same property.
As shown in (Theorem 6.2, p. 97, \cite{PEDREGAL}) the weak limit is then a
function as just given on the right hand side of \eqref{P-7}.
\par
To show \eqref{P-71} it is enough to establish the strong convergence of
$\{\varepsilon(u^k)\}$ in $L^2(\omega_0;\allowbreak{\mathbb{R}}^{n\times n}_{sym})$ (cf. Proposition 6.12,
p. 111, \cite{PEDREGAL}).
The elements of the sequence $\{\psi^k\}=\{\chi_b^k
-\chi_a^k\}$ take values $+1$ or $-1$. Thus the upper Kuratowski limit of the
sequence of singletons $\{\psi^k (x)\}$ (i.e. the set of limit points
of this sequence) is the set $\{-1,\,1\}$. By the Balder theorem (see
\cite{VALADIER}) we see that $\psi^k\rightarrow 1$ strongly in
$L^1 (\omega_0^+ )$ and $\psi^k\rightarrow -1$ strongly in
$L^1 (\omega_0^- )$
and we can suppose that $\psi^k\to 1$
a.e. in $\omega_0^{{+}}$
{{($\psi^k\to -1$
a.e. in $\omega_0^-$)}}
by passing to a subsequence, if necessary. Further, the
equiintegrability of $\{m^k\abs{\varepsilon(u^k)}^2\}$ implies that
$\{\abs{\varepsilon(u^k)}^2\}$ is also equiintegrable. By \lemref{lem-1} we have
\begin{equation*}
\int\limits_{\omega_0^{{+}}}p^k\boldsymbol{\cdot}\varepsilon(u^k)\,dx\to
\int\limits_{\omega_0^{{+}}}p\boldsymbol{\cdot}\varepsilon(u)\,dx=
\int\limits_{\omega_0^{{+}}}b\abs{\varepsilon(u)}^2\,dx+
\int\limits_{\omega_0^{{+}}}(\mathcal{A}^++\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u)\,dx.
\end{equation*}
On the other hand,
\begin{equation*}
\int\limits_{\omega_0^{{+}}}p^k\boldsymbol{\cdot}\varepsilon(u^k)\,dx=
\int\limits_{\omega_0^{{+}}}b\abs{\varepsilon(u^k)}^2\,dx+
\int\limits_{\omega_{0k}^-}(a-b)\abs{\varepsilon(u^k)}^2\,dx
+\int\limits_{\omega_{0}^{{+}}}(\mathcal{A}^++\psi^k\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u^k)\,dx,
\end{equation*}
where $\omega_{0k}^-=\{x\in\omega_0^+\colon
\psi^k(x)=-1\}$.
Thus taking into account that
\[
\int\limits_{\omega_{0}^{{+}}}(\mathcal{A}^++\psi^k\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u^k)\,dx
\to\int\limits_{\omega_{0}^{{+}}}(\mathcal{A}^++\mathcal{A}^-)\boldsymbol{\cdot}\varepsilon(u)\,dx
\]
and
\[
\int\limits_{\omega_{0k}^-}(a-b)\abs{\varepsilon(u^k)}^2\,dx\to 0
\]
being a consequence of the equiintegrability of $\{\abs{\varepsilon(u^k)}^2\}$
and $\abs{\omega_{0k}^-}\to 0$,
we are led to
\[
\int\limits_{\omega_0^{{+}}}b\abs{\varepsilon(u^k)}^2\,dx\to
\int\limits_{\omega_0^{{+}}}b\abs{\varepsilon(u)}^2\,dx.
\]
Since, simultaneously, $\varepsilon(u^k)\rightharpoonup
\varepsilon(u)$ in $L^2(\omega_0^+ ;\mathbb{R}^{n\times n}_{sym})$,
the desired strong convergence
results. Analogous reasoning holds for $\omega_0^-$.
The proof is complete.
\end{proof}
\begin{examp}
{{
Let $\omega_0=\{x\in \Omega\colon \chi_a(x)\chi_b(x)=0\}$. Without loss of generality one can suppose that
$\psi=1$ a.e. in $\omega_0$. Let $\omega_{0k}^-=\{x\in\omega_0\colon
\psi^k(x)=-1\}$. Since $\psi^k\to 1$ weak$^{\,\star}$ in
$L^\infty(\omega_0)$, we have
\begin{equation}\label{P-8}
2\abs{\omega_{0k}^-}=\int\limits_{\omega_0}(1-\psi_k)\,dx\to 0.
\end{equation}
Thus $\psi^k\to 1$ strongly in $L^1(\omega_0)$ (in fact, in
$L^p(\omega_0)$ for any $p\geq 1$). By the above theorem this means
that $\nu_x=\delta_{\varepsilon(u(x))}\quad\mbox{a.e. in } \omega_0.$
}}
\end{examp}
\begin{rem}
From \eqref{eq-1.28} and \eqref{P-7} it follows that
\begin{eqnarray*}
\lim\limits_{k\to\infty}\int\limits_{\Omega_0}a\abs{\varepsilon(u^k)}^2dx=
\int\limits_{\Omega_0}\int\limits_{\mathbb{R}^n}a\,\abs{\lambda}^2\,
d\nu_x(\lambda)\,dx=\int\limits_{\Omega_0}a\abs{\varepsilon(u)}^2dx
\\
+\theta\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx,
\end{eqnarray*}
giving rise to the formula that allows to calculate $\theta\in[0,1]$. Namely,
if we let
\begin{equation}\label{eq-d}
d:=\lim_{k\to\infty}\int\limits_{\Omega_0}a\abs{\varepsilon(u^k)}^2dx-
\int\limits_{\Omega_0}a\abs{\varepsilon(u)}^2dx,
\end{equation}
then from the equation
\begin{equation*}
d={\frac{\theta}{2}}\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx
\end{equation*}
we obtain
\begin{equation}\label{eq-dd}
\theta=\begin{cases} \displaystyle
\frac{2d}{\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx}&\mbox{if}\;
\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx>0\\[7mm]
0&\mbox{otherwise},
\end{cases}
\end{equation}
or equivalently
\begin{equation}\label{eq-ddd}
\theta=\begin{cases} \displaystyle
\frac{2\int\limits_{\Omega_0}\int\limits_{\mathbb{R}^n}a\,\abs{\lambda}^2\,
d\nu_x(\lambda)\,dx-\int\limits_{\Omega_0}a\abs{\varepsilon(u)}^2dx}
{\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx}&\mbox{if}\;
\int\limits_{\Omega_0}\chi_a\chi_b\,a\abs{C-D}^2dx>0\\[7mm]
0&\mbox{otherwise},
\end{cases}
\end{equation}
\end{rem}
\begin{rem}
It is worth to
point out that
the formulas
\eqref{eq-1.28}, \eqref{eq-1.30}, \eqref{eq-1.31} make possible
to express
the infimum of $(P)$  via \eqref{eq-dd} in terms of the limits $u$, $p$, $\chi_a$,
$\chi_b$, $d$ only.
On the other hand, the formula \eqref{P-7} expresses it in terms of the
parametrized Young measures $\{\nu_x(\cdot)\}$ which, in practice,
are  much more difficult to derive.
\end{rem}
\par
\vspace{1pc}
\noindent
\textbf{Acknowledgement.} I wish to express my gratitude to Professor
Krzysztof Che{\l}mi\'nski for his significant remarks and valuable suggestions

redpajama_set_name: RedPajamaArXiv


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text:
How Much Power Does The Commissioner Really Have?
By J.P.@JapersRink Feb 1, 2007, 1:37pm EST
Share All sharing options for: How Much Power Does The Commissioner Really Have?
In the wake of the Wetzel article lambasting Gary Bettman, I was curious if anyone could give me a satisfying answer to a simple enough question - "How much power does the Commissioner really have?"
As Chris DeGroat points out, it's the Board of Governors - not Bettman - who approve everything from rule changes to expansion plans. Sure, Bettman serves as the owners' mouthpiece, but at the end of the day if, for example, the owners favored something that was in their best interest but not in the best interest of the game (in Bettman's eyes), could he stop the change from being made? Some people hint that Bettman very much wanted realignment, but was shot down by enough owners that the proposal died.
In short (pun intended), is Gary Bettman just a high-salaried whipping boy or a legitimate steward of the game?


Merry Christmas everyone! I hope you're enjoying some special family time, or spending it with close friends, or at the very least travelling to somewhere hot!
I made some cookies and chocolates. Come on over while there are still a few pieces around.


In 1862, William Wirt Winchester and Sarah Lockwood Pardee were married, and they built a beautiful and massive house with 2000 doors, 47 stairways, 47 fireplaces, six kitchens, 10,000 windows and many bathrooms. Only to confuse the ghost and evil spirits, that enters the house and not able to find her. And it is a funhouse too; it had secret passages, stairs that go nowhere, and doors that open into walls and windows that open to rooms. So, if you need to go to the bathroom, you might end up with a brick wall.
In 1906 an earthquake damaged the house, the widow boarded up many of the rooms, and it's believed that the new discovery might be one such chamber. Some of the rooms are being discovered today.
In 1886, Sarah bought the property and continued with the construction of the Winchester Mystery House, until her death in 1922.
They had one child together, Annie, who died just over a month after she was born. William's father died in 1880 and William died in 1881, leaving Sarah as the heir to their $20 million rifle fortune. In 2018 dollars, that equates to more than $450 million.
There are various theories as to why she continued working on the house for so long, but one of the most popular involves her interest in spiritualism. According to one legend, Sarah believed that her daughter and husband's untimely deaths were retribution for all the people killed by Winchester rifles, especially after a medium told her that her blood fortune was cursed. Espousers of this theory believe that she made the house so large so there would be room enough for all the souls of Winchester victims; she also switched bedrooms within the house so any alleged ghosts would have a more difficult time finding her.
Sarah's mysterious motivations for building it aside, the house is objectively nuts. Sarah didn't hire anyone architect to oversee construction and added rooms and features haphazardly, so there are lots of weird elements to the house – staircases that lead to nowhere, doors that lead to nowhere, and windows that look into other rooms. It was seven stories tall and had more than 200 rooms before parts of it were destroyed in the San Francisco earthquake of 1906; it is currently four stories tall and has 160 rooms. At one point, Sarah actually built over another room that she'd forgotten was there, and according to legend, she could not actually find the house's original eight rooms.
If ghosts are a thing, then the Winchester Mystery House would be the place for them to set up shop. The house's official website claims that staff and visitors have experienced "strange phenomena," and according to one tour guide, an apparition named Clyde often appears in the basement near the coal chute. Ghost Adventures filmed an episode at the house in 2011 and had some creepy encounters, and Janan Boehme has shared a handful of stories with the Bay Area newspaper Mercury News. The staff has reported seeing a spectral handyman fixing the fireplace and hearing a woman sigh near one of the bedrooms.
The house is open to the public; if you are in the San Jose area, you can explore the property around $40. And it's great for parties too. If my birthday cake is missing, I am not coming back.
Go see the movie coming out on February 2, 2018!


text:
Die Crédit Mutuel de la Vallée SA ist eine im ehemaligen Bezirk La Vallée verankerte, 1865 gegründete Schweizer Regionalbank mit Sitz in der Gemeinde Le Chenit. Ihr Tätigkeitsgebiet liegt im Retail Banking, im Hypothekargeschäft und im Bankgeschäft mit kleinen und mittleren Unternehmen. Die Crédit Mutuel de la Vallée SA beschäftigt sieben Mitarbeiter und hatte per Ende 2009 eine Bilanzsumme von 160 Millionen Schweizer Franken.

Die Crédit Mutuel de la Vallée SA ist als selbständige Regionalbank der RBA-Holding angeschlossen.

Weblinks
 Website der Crédit Mutuel de la Vallée SA

Einzelnachweise

Regionalbanken und Sparkassen der RBA-Holding
Unternehmen (Kanton Waadt)
Le Chenit
Gegründet 1865

redpajama_set_name: RedPajamaWikipedia


We are losing him. We got a code blue.
Charge the paddles to 200.
Charge the paddles to 300. Clear.
We're just about done here now.
I don't mean me, I'm talking about my nephew, Lucas.
How long will he be in surgery?
Well, it depends on how serious his injuries are.
I'm sorry, what did you just say?
About the way that we started.
So everything between us has been a lie?
I cannot believe this, Lucas was right about you all along.
Haley, pick up. There's been an accident. Lucas is hurt pretty bad.
He's going to make it, right?
- Come on, I'll take you to the hospital.
- I can't talk to you right now, Nathan.
Haley, look I'm sorry I didn't mean it like that.
Hey, you know you can tell me anything, what is it?
- I'm so sorry. I didn't see the other car.
Where's the operating room? I need someone to tell me what's happening.
They asked us to wait here and they're going to let us know when he's in recovery.
I couldn't get a cab and then the only thing I had to pay with was Euros.
He's got to be alright, Keith.


text:
Q: How to allow specific values in my string I want to replace space with "|" and i want only numbers and alphabets and star(*) in the sting
And i have tried str_replace() funtion but it replace the space with | , but i want to validate
echo $text= str_replace(' ','|', "This is some $123 Money");

output
This|is|some|$123|Money

what i expect is
This|is|some|123|Money

I don't want any other special characters in my output
Any Suggestions
Thanks in Advance

A: You can use preg_replace to remove all special characters from your string.
Example
echo $text= str_replace(' ','|', preg_replace('/[^A-Za-z0-9 ]/', '',"This is some $123 Money"));

Output
This|is|some|123|Money

A: Here is the working demo for you: https://3v4l.org/M456J
<?php
$s = "This is some $123 Money";
$result = preg_replace("/[^a-zA-Z0-9]+/", "|", $s);
echo $result;

Output:
This|is|some|123|Money


text:
echo "Creating release of LogFileViewer"

zip LogFileViewer.zip LogFileViewer.jar LICENSE README.md plugins/* logging.ini logfileviewer.conf start.sh start.bat

redpajama_set_name: RedPajamaGithub


Suzuki prepared a number of concept vehicles that will be on display in the near future. The release of some of the cars that was to take place at the Tokyo Motor Show 2015, which will be held in late October 2015.
As reported Indianautosblog on Thursday, October 1, 2015, there is a concept car with eccentric shapes that will be exhibited by Suzuki called the 'Suzuki Air Triser,' a box-shaped minivan.
Reportedly, this car has a seven-seat passengers with a box shape. Close-packed equivalent of two colors red and white. Fresh look. This car will be joined by two other concept vehicles, the Suzuki Ignis and Mighty Deck.
This car has a length of 4,200 millimeters, a width of 1,695 mm and 1,815 mm high. Besides, it also comes with a slim-shaped headlight features, lower front fascia, LED taillamps, and a large alloy wheels.
The door can be opened by sliding, so passengers can enter easily and look more attractive. Entry into the cabin, equipped with wooden floors, mated with a gray dashboard. The more luxurious with stuffing the transparent instrument display, and infotainment tablet-esque screen.
This car offers three rows of seats that can be folded and arranged to offer a "private lounge". There is a separate screen located on the B pillar as entertainment control.


Crystal is the popular cafe among locals and tourists. It is located close to the Kokusai street and is one of the most popular gourmet spot you can not miss!
Crystal is located a 10 minute walk from the prefectural office that so many people visit especially during the lunch time. It's loved by family and couple visitors. The secret of the popularity is in its large-portions! The food size is very different compared to the usual cafe food. The cafe is located close to the Kokusai street so it's the best place to stop by! There is no other cafe like here in Okinawa which you must try! Enjoy the old cafe's atmosphere and food!
The most popualr menu item at the Crystal is called B lunch. The old Naporitan, ham egg, salad and for the most special part is the sandwich which is the best menu item at the Cristal is on top of the plate! It you can't decide what to order, this is the best menu item for you! It also comes with Miso soup and drink. You will know why this place is so loved by locals with this menu item.
Omurice is the one of the most popular menu item ! It is the standard menu item at the cafe, but here at the Crystal, you will be surprised by its quality. The chopped green pepper, onion, chicken ketchup rice with fluffy egg makes the Omurice even more delicious! When you spoon up the egg, you will be facsinated by its creamy egg! There are many customers who come to the cafe to taste this Omurice! It's one of the popular menu item with ladies.
Sandwich is the Crystal's special menu item! When Okinawa was returned to Japanese goverment, the sandwich was made.It has Okinawan and American taste! The bread is crispy grilled by the oven and sandwiched between the egg salad, cabbage and ham. Enjoy the large portion and nostalgic taste! For those of you who try this sandwich for the first time, your thought of sandwich will be broken the mold!
Crystal is located 10 minutes walk from Yui-rail Kencho mae station and is in an alley way from Kokusai street. The sign of the cafe "Crystal" is written in Hiragana, which has nostalgic feeling. You can imagine the atmosphere by the sign. At the first glance, the cafe looks like wood log house and has fancy atmosphere.
There are stump designed chairs at the cafe.
Usually, cafes have culm atmosphere where you can enjoy your tea and coffee and enjoy your time, but here at the Crystal, there are many customers who wants to eat the food mainly, so during the lunch time it will be crowded by the office workers and tourists. It will be more like a restaurant than a cafe. You need to follow the line especially during the lunch time.
The sandwich is the most popular menu item at the Crystal. It is recommended to order "B lunch" which you can enjoy the sandwich and other good foods togethere. Crystal is popular for its large portion menu item. At the first glance, you will worry if you can finish the food all by yourself or not, but don't worry, it's too delicious to make the left overs! Because of the delicious taste, it is poopular among ladies! Moreover, the coffee is must try as the cafe is old and has good taste! Please try the good large portion when you are close to the area!
Many people are facsinated by the cafe and be the repeat customer. It will be crowded during the lunch time. Not just the taste, but also the big-sized is one of the reason here is loved by so many locals!
For the group visitors may need to wait sometime.


text:
Category: Military
"The New Republic" and the Pitfalls of Historical Prophecy
"The New Republic" has had its ups and downs. Not too long ago its editors were almost unique in their willingness to honestly and thoroughly set forth the arguments for opposing points of view, and in their ability to address them convincingly, although lately they've shown a lamentable tendency to sink to the level of the rest of the pack. In common with many American journals of opinion that have survived for any length of time, its content has run from the ridiculous to the sublime, from essays by some of the most brilliant pundits this country has produced to the excrescences of Communist ideologues during the "red period" it passed through in common with many other journals in the aftermath of the Great Depression. It was launched in the opening months of that great watershed event in modern history, the first World War, and lately I've been looking through some of those inaugural issues.
It's very useful to occasionally read through a few articles in the journals and magazines of days gone by. It puts things that are happening today in perspective. Back in 1914, for example, The New Republic devoted a great deal of ink to discussion of the ramifications of the U.S. military intervention in Mexico. The first page of the third issue was entirely taken up with ruminations concerning what should be done about the U.S. troops in Vera Cruz. Today the number of us who are even aware that U.S. troops were in Vera Cruz in 1914 is vanishingly small. Will the matters that raise such passions and seem of such overwhelming importance to us today assume a similar insignificance for later generations?
Perhaps the most important thing one gains from reading old journals is a sense of humility. One finds many predictions about the future, but few of them that were accurate. The problem isn't that the authors making those predictions were fools. The problem is that we lack the intellectual capacity to assimilate all the facts that will have a bearing on the outcome of history, and correctly connect the dots between them. We must learn to appreciate our limitations. It's unwise to overestimate our ability to predict future events that have no historical precedent. For example, how many of us back in 1988 predicted the manner in which Communism and the Soviet Union would collapse, or when the momentous events culminating in those results would occur?
There is much to be gained from the reading of history. One learns how human beings are likely to react in given situations. Occasionally, history really does repeat itself, and, to the extent that future events fit the familiar patterns of the past, they are predictable. However, once in a while she jumps her tracks completely. World War I was such an event. Reflecting on the possible outcomes of that conflict in the second issue of The New Republic, one Simon N. Patten wrote:
Progress has ever been a ruthless crushing, whether we regard it as indistrial or view it in its political aspects. Growth has meant a centralization which eliminates the weak to the advantage of the strong. Belgium and Servia are today where hundreds of small nations have found themselves in the past. Belgium is racially and socially a part of France. Economically she is a part of Germany. One or the other fate she must in the end meet. Servia must also be either Russian or Austrian.
In fact, in the aftermath of the war, the historical context on which Mr. Patten based his assumptions ceased to exist. Serbia did not become a part of Austria because the great empire that went by that name disintegrated. She did not become a part of Russia because she, too, ceased to exist in any form recognizable from the past. Belgium is still with us, and belongs to a European Union that would have been incomprehensible to the combatants of 1914. The lesson here isn't that Mr. Patten was a fool. I'm sure he was a very intelligent man. Nor is it that we should cease speculating about the future. However, in doing so we should recall that we are not omniscient, and that the truth isn't always obvious.
Author HelianPosted on December 30, 2009 December 30, 2009 Categories History, Military, Politics, Russian revolution, World PoliticsLeave a comment on "The New Republic" and the Pitfalls of Historical Prophecy
Springtime for the Defeatists
When generals in a democracy say "we are losing the war," it tends to become a self-fulfilling prophecy. Whether true or not, such pronouncements inevitably become powerful psychological weapons in the hands of our enemies, and are thus better left unsaid, at least in public. Our military caste somehow manages to remain ignorant of such elementary aspects of modern asymmetric warfare. At the highest levels, our system tends to produce military commanders who are highly competent in doing what they've been trained to do, but lack the imagination and originality necessary to deal with the unexpected. Occasionally, genius is indispensable, but one would search in vain for a Napoleon or an Alexander among our generals. Instead, we produce Westmorelands and McChrystals. As the comment above would seem to demonstrate, we also produce political imbeciles who have somehow concluded that we can turn the situation around by throwing gasoline on the smoldering fires of defeatism.
The result is as inevitable as it was predictable. One detects an increasing stench of defeatism in the media, not only from its usual sources on the left, but from the right as well. For example, today CNN treats us to the umpteen billionth "ghost of Vietnam" story to appear since our troops went into combat. No doubt they're preening themselves on their originality. USA Today joins the crowd in reporting on the resignation of Matthew Hoh from the State Department, with the usual highlighting of such weepy remarks as ""I have lost understanding of and confidence in the strategic purposes of the United States' presence in Afghanistan." Great job, Matthew! That's bound to get us back on the right track. ABC News chimes in with more judiciously chosen quotes from Hoh's letter, such as, "To put simply, I fail to see the value or the worth in the continued U.S. casualties or expenditures of resources in support of the Afghan government in what is, truly, a 35-year-old civil war," Ah, yes, the "civil war" meme. That old chestnut is sure to please traditionalists. Look at the web pages of the Wall Street Journal, Foxnews, the Washington Times, etc., and you'll find the same stories spun in almost the same way. Apparently defeat in Afghanistan is as palatable to the right as it is to the left if only it discredits Obama.
Say what you will about George W. Bush, he was unmoved by the ebb and flow of defeatist propaganda during his administration. He really did "stay the course," in spite of the derisive remarks of the usual know-it-alls. As a result, the Iraqi people have at least a fighting chance of avoiding a slide back into dictatorship or theocracy. He was no philosopher king, but at least he was made of sterner stuff than Obama. The President appears more inclined to apologize to our enemies than fight them, and he is likely casting about for some graceful way to skedaddle in Afghanistan even as we speak. The increasingly shrill tone of defeatist propaganda will make it easier for him.
Well, what of it? As noted above, these developments were abundantly predictable and, given the limitations of our military leadership, probably inevitable. Is there a lesson here? Not really, other than the one that we should have learned a long time ago; modern democracies are anything but steadfast in fighting determined insurgents, particularly if their populations are as fickle and spineless as the current citizens of the United States. If we send in the troops, we should do so only with a well considered plan to get them back out again, and that with alacrity, before the famously insubstantial national backbone once again turns to jelly. In retrospect, the remarks of our much abused former Secretary of Defense, Donald Rumsfeld, sound remarkably prescient. For example, from a speech delivered in February, 2003:
Afghanistan belongs to the Afghans. The objective is not to engage in what some call nation building. Rather it's to try to help the Afghans so that they can build their own nation. This is an important distinction. In some nation building exercises well-intentioned foreigners arrive on the scene, look at the problems and say let's fix it. This is well motivated to be sure, but it can really be a disservice in some instances because when foreigners come in with international solutions to local problems, if not very careful they can create a dependency.
A long-term foreign presence in a country can be unnatural. This has happened in several places with large foreign presence. The economies remained unreformed and distorted to some extent. Educated young people can make more money as drivers for foreign workers than as doctors and civil servants. Despite good intentions and the fine work of humanitarian workers individually, there can be unintended adverse side effects.
Our goal in Afghanistan is to try and not create a culture of dependence but rather to promote [inaudible]. Long-term stability comes not from the presence of foreign forces but from the development of functioning local institutions. That's why in the area of security we have been helping to train for example the Afghan National Army. Our coalition partners have been training the police. And the goal is so that Afghans over time can take full responsibility for their own security and stability rather than having to depend on foreign forces versus for a sustained period.
When Rumsfeld was in office, an abundance of geniuses appeared who assured us they knew how to do his job much better than he did. In retrospect, we probably should have ignored the geniuses and paid more attention to him. The next time we feel the yen to embark on another military adventure, we should reflect on the fact that some of the biggest cheerleaders for such projects in the recent past became hand-wringing, hysterical defeatists a disconcertingly short time after the troops were actually on the ground. We will surely have an abundance of such heroes to "help" us the next time around as well. Before we commit our forces to another ill-considered war, we'd do well to recall that there are legions of Matthew Hohs in our midst, useful idiots who are adept at persuading themselves that collaboration with the enemy is both a noble moral good and a patriotic duty. They will always be with us, and they will always make the cost of victory higher the longer our troops are engaged.
UPDATE: I take it from John McCain's cry in the dark that he has also noticed that the water is up to our chin and climbing. Of course, he's right. We can win in Afghanistan. The enemy is much less formidable than he was in Vietnam. It's a matter of national will. In fact, that's just the problem.
Author HelianPosted on October 27, 2009 October 28, 2009 Categories Afghanistan, Foreign affairs, Military, The Media, World PoliticsLeave a comment on Springtime for the Defeatists
Nuclear Strategery
Jonathan Tepperman has an interesting post on the Newsweek site entitled, "Why Obama should Learn to Love the Bomb." According to Tepperman, "A growing and compelling body of research suggests that nuclear weapons may not, in fact, make the world more dangerous, as Obama and most people assume." Yes, and there was "a growing and compelling body of research" in 1914 that suggested the great powers were so economically dependent on each other they would never risk going to war. Tepperman continues, "The argument that nuclear weapons can be agents of peace as well as destruction rests on two deceptively simple observations. First, nuclear weapons have not been used since 1945. Second, there's never been a nuclear, or even a nonnuclear, war between two states that possess them." That's true, and the argument that possession of nuclear weapons reduces the chances of war between states that possess them is certainly plausible. However, the fact that, for example, there was never a nuclear exchange between the United States and the Soviet Union does not mean that the risk of such an exchange was zero. It is more likely that we dodged a bullet.
An all out conventional war between India and Pakistan would certainly result in great loss of life. An all out nuclear war would be, according to Tepperman, less likely. It would also be more costly in terms of loss of life, not to mention economic damage. Perhaps, then, a reasonable metric for assessing whether nuclear weapons make us more or less secure would be risk of war times likely human and economic cost. The problem with such a neat formula is that it would be impossible to predict or to agree on the magnitude of the different factors. For example, it was widely assumed during the cold war that a general nuclear exchange would result in the annihilation of the populations of the US and Soviet Union. However, I doubt the leaders on either side really believed that. Various attempts were made to calculate likely outcomes, but they were generally flawed by the ideological predispositions of those making the estimates.
Let's consider what else Tepperman has to say:
Even the craziest tin-pot dictator is forced to accept that war with a nuclear state is unwinnable and thus not worth the effort. As (Berkeley Professor Kenneth) Waltz puts it, "Why fight if you can't win and might lose everything?"
I'm not so sure that the craziest tin-pot dictator would come to such a logical conclusion. However, the statement as it stands is almost irrelevant. I suspect a nuclear exchange is far more likely to result from a miscalculation, accident, or loss of control to a rogue actor than any premeditated, deliberate attack.
Meanwhile, the nuclear powers have scrupulously avoided direct combat, and there's very good reason to think they always will. There have been some near misses, but a close look at these cases is fundamentally reassuring—because in each instance, very different leaders all came to the same safe conclusion.
This is wrong on the face of it. Always is a long time. As long as there are nuclear weapons, there will be a finite risk of a nuclear exchange. Therefore, if states with nuclear arsenals continue to exist into the indefinite future, there will eventually be a nuclear exchange. The question is not whether it will happen, because it certainly will. The question is whether its cost, when it does happen, will be greater or less than the cost of the, presumably more frequent, conventional wars that would have occurred in the absence of nuclear arsenals. Similarly, as long as sufficient special nuclear material (SNM), such as U235 or Pu239, exists to make nuclear weapons, there will be a finite risk of it falling into the hands of non-state actors, or terrorists if you will. From this we must conclude that a terrorist nuclear attack is also inevitable. It is not a question of if. It is a question of when. It may be tomorrow, or it may be a thousand years from now, but it will happen. I rather suspect it will be sooner rather than later.
…in 1957, Mao blithely declared that a nuclear war with America wouldn't be so bad because even "if half of mankind died … the whole world would become socialist." Pyongyang and Tehran support terrorism—but so did Moscow and Beijing. And as for seeming suicidal, Michael Desch of the University of Notre Dame points out that Stalin and Mao are the real record holders here: both were responsible for the deaths of some 20 million of their own citizens. Yet when push came to shove, their regimes balked at nuclear suicide, and so would today's international bogeymen.
That is an unwarranted assumption. In any case, as noted above, it is irrelevant, because the nuclear danger from accident or miscalculation is far greater than that from deliberate use.
Even if the Pakistani state did collapse entirely—the nightmare scenario—the chance of a Taliban bomb would still be remote. Desch argues that the idea that terrorists "could use these weapons radically underestimates the difficulty of actually operating a modern nuclear arsenal. These things need constant maintenance and they're very easy to disable. So the idea that these things could be stuffed into a gunnysack and smuggled across the Rio Grande is preposterous.
Here, Tepperman's "expert," Michael Desch of Notre Dame, doesn't know what he's talking about. One wonders what sort of "constant maintenance" he has in mind. The basic design principles of both gun and implosion type weapons are well known. They certainly require maintenance occasionally, but "constant maintenance?" I think not. Any non-state actor gaining possession of an intact nuke will have plenty of time to use it. The idea that nukes are easy to disable is also poppycock. You can make the firing set as clever as you please, but the SNM would still be there. If you didn't have an explosives guy capable of jury rigging the device, you could still simply cannibalize the material from two nukes and make a simple, but very effective device. Recall that our physicists were so confident that the gun type Little Boy would work that it was dropped without prior testing. The computer modeling tools available to anyone now are infinitely better than the rudimentary mathematical tools they had then. Building a crude bomb is simply not that difficult. As for smuggling the weapon in a gunnysack, Tepperman is right. A terrorist would have to be brain dead to even attempt it. Unfortunately, smuggling a complete weapon is completely unnecessary. It would be much simpler, and just as effective, to smuggle the SNM in small bits, and assemble it into a weapon at the target. The chances that we will be able to detect any of the material before the weapon actually goes off are virtually nil.
The risk of an arms race—with, say, other Persian Gulf states rushing to build a bomb after Iran got one—is a bit harder to dispel. Once again, however, history is instructive. "In 64 years, the most nuclear-weapons states we've ever had is 12," says Waltz. "Now with North Korea we're at nine. That's not proliferation; that's spread at glacial pace." Nuclear weapons are so controversial and expensive that only countries that deem them absolutely critical to their survival go through the extreme trouble of acquiring them. That's why South Africa, Ukraine, Belarus, and Kazakhstan voluntarily gave theirs up in the early '90s, and why other countries like Brazil and Argentina dropped nascent programs.
Perhaps. However, I do not find the existence of a maximum of 12 nuclear states as comforting as Tepperman.
Put this all together and nuclear weapons start to seem a lot less frightening. So why have so few people in Washington recognized this? Most of us suffer from what Desch calls a nuclear phobia, an irrational fear that's grounded in good evidence—nuclear weapons are terrifying—but that keeps us from making clear, coldblooded calculations about just how dangerous possessing them actually is. The logic of nuclear peace rests on a scary bargain: you accept a small chance that something extremely bad will happen in exchange for a much bigger chance that something very bad—conventional war—won't happen. This may well be a rational bet to take, especially if that first risk is very small indeed. But it's a tough case to make to the public.
Here, Tepperman makes some good points. The real issue is one of risk. Unfortunately, for the reasons cited above, I rather suspect he is seriously underestimating it. Be that as it may, assuming one can really get a good handle on the actual risk, what he says makes sense.
Given this reality, Washington would be wiser to focus on making the world we actually live in—the nuclear world—safer. This involves several steps, few of which the Obama administration has mentioned but which it should emphasize in its Nuclear Posture Review due at the end of the year. To start, the logic of deterrence works only if everybody knows who has a nuclear arsenal and thus can't be attacked—as Peter Sellers puts it in Stanley Kubrick's Dr. Strangelove, "The whole point of a Doomsday Machine is lost if you keep it a secret!"
Probably true. Unilateral nuclear disarmament would certainly be suicidal. Reducing our arsenal to the point that potential enemies might find the risk of retaliation acceptable is almost equally so.
Chris Bodenner at Sully's blog thinks a piece by Peter Scoblic at TNR's website "scalpels" Tepperman's piece. I think not. It's more in the pious platitude here, anecdotal evidence there, preaching a foregone conclusion to the choir style that has become the stock in trade at TNR lately. They have seen better days (when Sully was editor, in fact. He has seen better days, too). One hopes the better days will return.
Author HelianPosted on September 3, 2009 Categories Military, Nuclear proliferation, Nuclear terrorism, Nuclear weapons, Terrorism, World PoliticsLeave a comment on Nuclear Strategery
Red Flag: Israeli Fighters Over Nevada
Speaking of Israel, several of her F-16 fighters just turned up at Nellis Air Force Base to take part in the famous Red Flag exercises there, the Air Force version of "Top Gun." One can only assume Netanyahu managed to slip them in under Obama's radar, as the Air Force refers to foreign participants in Red Flag as "allied partners."
Helian Unbound military correspondents in Las Vegas report sighting a couple of stately AWACS planes, which often fly in the exercises, as well as a bevy of the snazzy new F-22 Raptors.
Author HelianPosted on July 23, 2009 Categories Israel, Military, World PoliticsLeave a comment on Red Flag: Israeli Fighters Over Nevada


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text:
Q: Upload video with subtitle How can I upload video with the subtitle using Youtube Data API v3?
The documentation doesn't mention anything.
For uploading, I'm using the sample code with the code bellow:
VideoContentDetails contentDetails = new VideoContentDetails();
contentDetails.setCaption("00:00:00.105,00:00:05.413\nSubtitle 1\n\n00:00:05.667,00:00:08.750\nSubtitle 2");
videoObjectDefiningMetadata.setContentDetails(contentDetails);

I used the format of the caption, according with this description from Youtube Data v2.
But when I try to upload :
com.google.api.client.googleapis.json.GoogleJsonResponseException: 400 Bad Request
{
  "code" : 400,
  "errors" : [ {
    "domain" : "global",
    "message" : "Invalid value for: (.....) is not a valid value",
    "reason" : "invalid"
  } ],
  "message" : "Invalid value for: (.....) is not a valid value"
}
at com.google.api.client.googleapis.json.GoogleJsonResponseException.from(GoogleJsonResponseException.java:145)
at com.google.api.client.googleapis.services.json.AbstractGoogleJsonClientRequest.newExceptionOnError(AbstractGoogleJsonClientRequest.java:113)
at com.google.api.client.googleapis.services.json.AbstractGoogleJsonClientRequest.newExceptionOnError(AbstractGoogleJsonClientRequest.java:40)
at com.google.api.client.googleapis.services.AbstractGoogleClientRequest.executeUnparsed(AbstractGoogleClientRequest.java:423)
at com.google.api.client.googleapis.services.AbstractGoogleClientRequest.executeUnparsed(AbstractGoogleClientRequest.java:343)
at com.google.api.client.googleapis.services.AbstractGoogleClientRequest.execute(AbstractGoogleClientRequest.java:46)

A: There is no Caption support on Data API v3 right now. (the one you are trying in your code)
You can only set caption via v2, till v3 support is built.
So after uploading your video with v3 as you did above, you can use the same id to call v2 api as in https://developers.google.com/youtube/2.0/developers_guide_protocol_captions#Create_Caption_Track