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Senators Numerology
Senators by the Numbers: #28
By The Artist Formerly Known as A Jul 1, 2013, 10:00am EDT
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A player this invisible shouldn't wear camo
Mark Osiecki D 1992-1993
Born in Burnsville, Minnesota in 1968, Osiecki was drafted 187th overall by the Calgary Flames in 1987. Traded to Ottawa by Calgary for Chris Lindberg in the summer of 1992, his time in the capital was brief. He was placed on waivers midway through the franchise's first season. Following his retirement in 1995, he began his coaching career in the USHL. He worked as an assistant coach at the University of Wisconsin for several years. Most recently, Osiecki was the head coach for Ohio State University.
SEASON GP G A P +/- PIM
1992-1993 34 0 4 4 -21 12
Jeff Lazaro LW 1992-1993
Born in Waltham, Massachusetts in 1968, Lazaro was undrafted out of the University of New Hampshire before signing with the Boston Bruins. The Senators claimed him from the Bruins in the 1992 Expansion Draft and he split the season between the Senators and their AHL affiliate. His season in Ottawa was his last in the NHL and Lazaro finished his career in Austria with Graz 99ers and in Germany with the Ratingen Lions. Lazaro currently works in health insurance in New Orleans.
1992-1993 26 6 4 10 -8 16
Bill Huard LW 1993-1994
Born in Welland, Ontario in 1967, Huard was an undrafted free agent playing with the Carolina Thunderbirds of the ECHL when he caught the eye of the New Jersey Devils in 1989-90. Huard wore two different numbers with the Senators and was previously featured at #20. His professional career ended after just 2 games with England's London Knights when he returned to North America because of a family health scare.
1993-1994 63 2 2 4 -19 162
Steve Duchesne D 1995-1996, 1996-1997
Born in Sept-Îles, Quebec in 1965, Duchesne was undrafted after a successful junior career with the Drummondville Voltigeurs. He signed as a free agent with the Los Angeles Kings in 1984. Known throughout his career as an offensive defenseman, Duchesne was a 3-time all-star and won the Stanley Cup with the Detroit Red Wings in 2002. Part of the package the Flyers sent to Quebec for Eric Lindros, he was acquired by the Senators in 1995 for Ottawa's 2nd round pick (Cory Sarich) in the 1996 draft. He scored one of the most important goals in Senators history, scoring the marker that clinched the Sens first ever playoff spot. His 19 goals in 1996-1997 set a franchise record for goals by a defenseman in a single season - a record he now shares with Erik Karlsson. He was traded in the off-season for Igor Kravchuk, in August, 1997. Currently, Duchesne lives in Texas with his family, coaches the U18 Dallas Stars Midget Major AAA hockey team, and owns the Allen Americans, a CHL team.
1995-1996 62 12 24 36 -23 42
1996-1997 78 19 28 47 -9 38
Denny Lambert LW 1997-1998
Born in Wawa, Ontario in 1970, Lambert was undrafted after three seasons with the Sault Ste. Marie Greyhounds. He signed as a free agent with the expansion Anaheim Mighty Ducks. He signed as a free agent with the Senators in July, 1996. During his first season with the team he wore #42. After two seasons in the capital, he was claimed by Nashville in the 1998 Expansion Draft. He was the head coach of the Greyhounds for three seasons until 2011.
1997-1998 72 9 10 19 4 250
Stephen Leach RW 1998-1999
Born in Cambridge, Massachusetts in 1966, Leach was drafted 34th overall by the Washington Capitals in 1984. He played his high school hockey at Matignon High School, where he won four consecutive state championships and was teammates with future MLB Hall-of-Famer Tom Glavine. A member of the 1988 U.S. Olympic Team, he was signed as a free agent by Ottawa in October, 1998. He retired in 2000 and coached the Valley Jr. Warriors '99 Elite Squad, which sounds like a team of teenage assassins.
1998-1999 9 0 2 2 -1 6
Ted Donato LW 1998-1999
Born in Boston in 1969, Donato was drafted 98th overall by the Bruins in 1987. An offensive star with Harvard, he was the MVP of the 1989 Frozen Four tournament and helped lead the Crimson to a national championship. He represented the United States at the 1992 Winter Olympics. He played for 8 teams during his NHL career and was acquired from the Islanders by Ottawa for a 4th round pick (#123 Preston Mizzi). Donato's time in Ottawa was brief; he was traded to the Mighty Ducks (along with the rights to Antti-Jussi Niemi) June 18, 1999 for Patrick Lalime. He is the first NHL player to appear on The Price Is Right, filming a segment of the show during spring break, 1990. After successfully making it to Contestants' Row, Donato made it all the way to the Showcase Showdown; unfortunately, he was easily outbid by a woman named Janella. Since 2004, he has served as the head coach of the Harvard University hockey team.
1998-1998 13 3 2 5 2 10
Joe Juneau C 1999-2000
Born in Pont-Rouge, Quebec in 1968, Juneau was drafted 81st overall by the Boston Bruins in 1988. A college star with Rensselaer Polytechnic Institute, Juneau had a 4.0 grade point average and earned an aeronautical engineering degree in just 3 years despite a language barrier. Marred in a contract dispute with the Bruins, Juneau spent a year with the Canadian National Team and won a silver medal at the 1992 Winter Olympics. An offensive star during his time in Boston, Juneau was signed by Ottawa in 1999 to provide an offensive boost in the absence of the suspended Alexei Yashin. Injuries shortened his career and robbed him of much offensive potential. After his retirement in 2004, he was a partner and account manager at Harfan Technologies. Juneau has since worked promoting hockey and education to youth, first in Fairbanks, Alaska and since 2007, in Kuujjuaq, Quebec where he works with Inuit youth.
1999-2000 65 13 24 37 3 22
Todd White C 2000-2001, 2001-2002, 2002-2003, 2003-2004
Born in Kanata, Ontario in 1975, White played junior hockey for the Kanata Valley Lazers in the CJHL and received a scholarship from Clarkson University. Despite being a Hobey Baker finalist, he went undrafted and signed as a free agent with the Chicago Blackhawks in 1997. He signed with the Senators in 2000 and spent for years in Ottawa before being traded to the Minnesota Wild for a 4th round pick. Since retiring in 2011, he has worked as an instructor for Rink Eye hockey school.
2000-2001 16 4 1 5 5 4
2001-2002 81 20 30 50 12 24
2003-204 53 9 20 29 12 22
Martin Lapointe RW 2007-2008
Born in Ville St. Pierre, Quebec in 1973, Lapointe was drafted 10th overall by the Detroit Red Wings in 1991. When he retired in 2008, he was just 9 games shy of 1,000 for his career. A useful depth player during his time in Detroit, he won two Stanley Cups with the Red Wings (1997, 1998). He provided the highlight of the night at a game I attended at the United Center in December, 2007. Toward the end of an eventual Hawks victory against the Predators, a dejected Nashville fan threw a catfish on the ice. Lapointe responded by skating over to the prone fish and calmly stick-handled it towards the Zamboni entrance. Dealt to Ottawa at the trade deadline in 2008 for a 6th round pick, he played 18 games for the Sens and added 4 more in the playoffs before retiring. He was hired as a pro scout for the Blackhawks in December, 2009 and won his 3rd Cup with the team in 2010. Last summer he made the move with Marc Bergevin to Montreal where Lapointe serves as Director of Player Development.
2007-2008 18 3 3 6 -2 23
Zenon Konopka C 2011-2012
Born in Niagara-On-The-Lake, Ontario in 1981, Konopka played four seasons with the Ottawa 67's, winning the Memorial Cup in 1999. Undrafted, he signed for the Wheeling Nailers of the ECHL. After 3 seasons in the lower league, he made the Mighty Ducks in 2005-2006. Konopka spent one season in Ottawa, signing as a free agent in 2011. He currently plays for the Minnesota Wild.
2011-2012 55 3 2 5 -4 193
Matt Kassian LW 2012-2013
Born in Edmonton in 1986, Kassian was drafted 56th overall by the Minnesota Wild in 2005. He made his NHL debut in 2010-11 and was traded to the Senators March, 2013 for a 6th round pick in 2014.
Best #28: Steve Duchesne
It was close, with Todd White providing a surprising amount of competition, but ultimately, this offensive defenseman gets the nod. Duchesne provided big production from the back end playing for Jacques Martin - no easy achievement. Anytime you share a record with Erik Karlsson that's good enough for me.
Worst #28: Matt Kassian
I could have picked any one of Osiecki, Huard, Leach, or Konopka but Kassian wins this dubious honour because of how much he plays despite how little he brings.
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Obamacare: How do we like it now?
11 Coloradans reflect on the promise and the problems of the health care law
Susan Page, USA TODAY
Published 5:51 p.m. ET April 8, 2014 | Updated 5:42 p.m. ET April 24, 2014
DENVER — Jamel Brown and Matt Wright, healthy and in their 30s, declare they'd rather pay a penalty than premiums for health insurance. Catherine Huston tells them they're nuts.
"What if you get cancer?" the 57-year-old woman demands. "Maybe it's because I'm older, but I just can't believe people think they aren't going to need it." Her job as a dispatcher doesn't include coverage, and she was reduced to tears of frustration when glitches and delays stymied her efforts to enroll in Medicaid.
Six months ago, USA TODAY brought together 10 uninsured Coloradans in the Mile High City to discuss what they thought about the Affordable Care Act as the law's first enrollment period was about to open. A few hours before the March 31 midnight deadline when enrollment would close, six of those participants and five additional people in similar circumstances gathered over soft drinks in a conference room at KUSA-TV to discuss what they had done about it.
Their stories reflect the promise and the problems of the biggest expansion of health care in America in a half-century. Four are delighted and relieved to finally have health insurance. Three made at least some efforts to sign up but found the process too complicated or confusing to complete. And the final four decided they'd rather take their chances and pay the fine.
USA TODAY Washington Bureau Chief Susan Page met with Coloradans before and after the Affordable Care Act rolled out to hear their experience with President Obama's signature legislation.
Their experiences, now and down the road, will help determine whether the program known to just about everyone as Obamacare ultimately is judged a success or a failure.
"It's a great start and something we need," says Carol Bergschneider, who signed up for insurance that began on the first day possible, Jan. 1, ending years of struggle to get the insulin she needs to manage her diabetes.
Wright isn't convinced, at least not yet. "It sure sounds good — with a big 'maybe' and a medium-sized 'but,' " he says.
Connect for Health Colorado reports that nearly 280,000 state residents gained coverage during the six-month enrollment period, including 120,971 who signed up for private insurance plans as of Monday and 158,521 who enrolled in the expanded Medicaid program as of April 1. In an interview last fall, Patty Fontneau, CEO of the non-profit organization running the exchange, had estimated that 125,000 to 140,000 Coloradans would sign up for insurance through the exchange during its first year.
Colorado, where Barack Obama was nominated for president in 2008, vowing to overhaul the health care system, seems to be an apt place to take the law's measure. With the support of Democratic Gov. John Hickenlooper, it is one of 16 states, plus the District of Columbia, that set up its own exchange under the Affordable Care Act. A swing state in presidential elections, it is a place where national political forces from the Tea Party movement to the growth in Latino voters play important roles.
A statewide poll taken by USA TODAY and Princeton Survey Research Associates found that Coloradans, insured and uninsured, have learned more about the law during the past six months but don't like it any better. Nearly three of four say they understand its impact on them and their families, at least somewhat. Six in 10 have seen TV ads sponsored by Connect for Health Colorado.
Negative views of the law have barely budged. Six months ago, Coloradans disapproved of the law by 52%-33%. Now their disapproval stands at 53%-37%. Opponents continue to be twice as likely as supporters to say they feel very strongly.
The latest survey by land-line and cellphone of 701 Colorado residents was taken March 20-24. The previous poll of 704 Coloradans was taken Sept. 11-16, 2013. Each has a margin of error of +/-4 percentage points.
One significant shift: Coloradans apparently are concluding the law is here to stay. Last fall, by 17 percentage points (54%-37%), those surveyed said elected officials who disapproved of the law should do what they could to make it fail. Now, by 10 points (50%-40%), they say officials should do what they can to make the law work.
"It's going to need adjustments as it goes," acknowledges Amy Kitteringham, 48, who was surprised to discover when she went on the Colorado website that she was eligible for Medicaid, at least until she can land a job. That meant she could see her doctor for a long-delayed checkup. "All the bugs aren't worked out, and it's not communicated well," she says of the act, but she calls it "a step in the right direction."
She hasn't been able to persuade her son, in his late 20s, to sign up.
Dangerous demography
The demographic divide in attitudes toward the law could be its biggest challenge.
Three of the people around the table have pre-existing medical conditions that made it difficult for them to buy insurance in the past; all have signed up for coverage. Five of the people around the table are younger than 40 years old; four of them haven't signed up. None of the four men in the group has enrolled.
That's a potential problem. To keep premiums affordable and the system afloat, it needs to attract a fair share of the younger and healthier people who are likely to need the least medical care, as well as those older and sicker ones who are likely to need the most. In Colorado, 26% of those who signed up were young adults, 18 to 35 years old. Nearly half, 46%, were 45 to 65.
To Wright, 31, the generational trade-off doesn't seem entirely fair. "It's like a scale to me: What return do I get for my money?" he says. He doesn't begrudge paying for child care for his 5-year-old daughter, who has health insurance through her mother, and he sees the need for car insurance because he's on the road a lot for his job in real estate.
But after spending a half-hour on the Colorado website, he calculated that coverage would cost him $150 to $200 a month, while the penalty for the first year wasn't likely to exceed $500. He decided to save his money and take his chances. "For me, knowing my family, my history, what I eat and what I do, it's led me to wait," he says, though it does mean putting off treatment for a bum shoulder.
In the poll, young adults under 30 are less likely to approve of the law than the middle-aged, those 30 to 64, though they're also less likely to disapprove of it.
The starkest divide, not surprisingly, is along partisan lines. Three of four Democrats approve of the law. Nine of 10 Republicans disapprove. Though independents were more divided, they disapprove of it 56%-34%.
Coloradans remain unconvinced the law will help them. Nearly half say they are very concerned the Affordable Care Act will end up costing them more money, and more than a third are very concerned it will reduce the quality of their health care.
"The idea for me is that if everyone is on board, then it is affordable and we can all see doctors," Wright says. "I think it could also be a saturation issue where the care actually goes down in value because the doctors potentially are seeing twice as many people — maybe longer wait times, shorter time with the doctors. I don't know."
Most of those surveyed say the health law hasn't had much of an effect on them and their families. Of those who have seen an impact, by 26%-14%, they say it has been mostly negative rather than mostly positive. In coming years, by 40%-25%, they predict it will have a negative effect on them personally.
Expectations are more dire for the country as a whole, presumably reflecting the drumbeat of opposition and the problems with the rollout of the HealthCare.gov website. A 51% majority say the law has been bad for the nation and predict it will have a negative effect over the long term.
The first open-enrollment period has closed, but the debate isn't over. In the Colorado Senate race, the leading Republican contender, Rep. Cory Gardner, hammers incumbent Democrat Mark Udall for his support of the Affordable Care Act. Obama carried Colorado in 2008 and 2012, but his popularity in the state has declined since then. In recent weeks, the conservative group Americans for Prosperity aired nearly $1 million in TV ads attacking Udall and warning, "Obamacare doesn't work."
At the roundtable, Brown, 35, a security guard, labels the Affordable Care Act "extortion" because of its mandate that he has to buy coverage or pay a fine.
When asked to describe the law in a word or phrase, Catherine Campbell pauses and finally says, "Conflicting."
Huston: "Something we all need, but I don't know if we can all afford it."
Alan Libertone: "Just put a question mark there."
Rebecca Hardy: "A good idea."
Janette Cilleffen: "Confusing."
Lynette Langlois: "Relief."
Changing views
For those who joined the roundtable discussion last fall and again this spring, some of the quick takes reflect big changes in views of the law.
Six months ago, Bergschneider expressed skepticism that the law could work, saying she'd rather pay more taxes than deal with a complicated new system. Now she calls it "a great start" and expresses delight that, with health care coverage, she has been able to get an insulin pump to help control her diabetes.
Campbell's view of the law is still ambivalent — "conflicting" — but her comment six months earlier was that she was worried she couldn't afford it. She has signed up for medical and dental coverage through her employer at about $40 a week. "It's doable, so I was pretty shocked," she says.
Last fall, asked to describe the law in a word, Langlois had said, "Expense."
Like most of those around the table, she encountered problems when she tried to sign up. The 57-year-old stay-at-home mom has diabetes, and for years, she couldn't buy insurance. Eager for coverage, she and her husband went on the state website only to find themselves flummoxed.
They finally scheduled an appointment with a system "coordinator" at a library, brought along paperwork showing their income and walked out an hour later with health insurance coverage for the first time in five years. Her husband was so excited he planted a kiss on the cheek of the woman who had helped them navigate the system. The word she uses to describe the law is "relief."
Lynette Langlois, 57, a stay-at-home mom, has diabetes. She signed up for coverage that began last month. "It actually makes me feel better to know that I can go anytime I need and have good health care through a physician and not just take aspirins at home," she says.(Photo: Nathan Armes for USA TODAY)
Jamel Brown, 35, a security guard, decided to pay the penalty for not being insured. "I'd like to get a whole, complete physical," he says, "but I don't smoke. I don't do drugs. I'm just an average guy – try to work out here and there. I don't get hurt, you know, at least not yet. So I'm OK."(Photo: Nathan Armes for USA TODAY)
Catherine Huston, 57, dispatcher for emergency maintenance at apartment complexes. She tried unsuccessfully to sign up for Medicaid: "I don't understand how it works and there's no place to go for answers, because all you do is wait on the phone."(Photo: Nathan Armes for USA TODAY)
Catherine Campbell, 27, a records manager, enrolled in coverage as of Jan. 1 through a new employer. "I thought ... I wasn't going to be able to afford (it)," she says, "but it's actually doable, so I was pretty shocked."(Photo: Nathan Armes for USA TODAY)
Amy Kitteringham, 48, is between jobs. She enrolled in Medicaid last month. "I went and saw my doctor. I hadn't seen him in a year," she says. "He was actually really happy I had coverage, and he said, 'Let's get some blood work done and have some tests.'"(Photo: Nathan Armes for USA TODAY)
Matt Wright, 32, a real estate agent, decided to pay the penalty for not being insured. "The only thing I've got is a bad shoulder," he says, "so as long as I don't push it too hard in volleyball, I'll be OK. ... I think if it was a hip or some other body part, you know, things might be different, right?"(Photo: Nathan Armes for USA TODAY)
Barry Roake, 48, a freelance writer and artist, hasn't signed up. "I have never had health insurance because I haven't really needed it," he says, "but I am getting to the age where it's starting to concern me a little bit. I would love to have it, but it just seems like it's too far out of reach."(Photo: Nathan Armes for USA TODAY)
Carol Bergschneider, 56, office contract worker. She had been unemployed and was enrolled in Medicaid as of Jan. 1, though she has just gotten a new job and expects to move to a private plan. "I'm diabetic. (Now) I get all of that insulin for three months for $80. That's almost $1,000 worth of insulin for $80 with the plan here ... and I have a new insulin pump."(Photo: Nathan Armes for USA TODAY)
Janette Cilleffen, 37, a part-time house/office cleaner, hasn't signed up but would like to get on Medicaid. "It's not been high on my priority list and ... we just haven't gotten a lot of information on it," she says.(Photo: Nathan Armes for USA TODAY)
Alan Libertone, 36, a warehouse worker, hasn't signed up. "It's not a priority at this point to go shell out a bunch of money for something I'm not going to use," he says.(Photo: Nathan Armes for USA TODAY)
Rebecca Hardy, 49, unemployed. She went on the exchange website once, encountered problems and gave up. She isn't concerned about the law's mandate that she have insurance coverage. "I don't think people are taking it really all that seriously. All of the political media, the commercials and, you know, just back-talking back-and-forth (saying), 'This isn't going to work' ... That's what you hear, (that) it's not working and it's not going to work."(Photo: Nathan Armes for USA TODAY)
In the USA TODAY surveys, the percentage of Coloradans who said they were covered by health insurance rose a tick during the sign-up period, to 86% from 83%, a change within the margin of error.
Problems with the enrollment process and confusion about the law color views. It's one reason some of the repeat participants, including Wright and Libertone, have no more positive view of the law than they did six months ago.
Kitteringham struggled to deal with the troubled federal website until a story on Colorado Public Radio made her realize there was a state website she was supposed to be using instead.
"They did a horrible job of letting people know about it, honestly," she says as heads nod around the table. "When elections are happening, people come to your door, everything's on television all the time, you're getting stuff in the mail constantly. But yet this is coming out, and it was kind of random places where this information came."
Huston spent two hours waiting to get through by phone the first time, then an hour and 45 minutes the second. Both times, the system rejected her application for Medicaid although she says she falls within the guidelines for eligibility. She was told she could buy coverage for $137.50 a month, an amount she says is more than she can afford.
Cilleffen, 37, waited until the last day to go on the website. "I tried to get on this morning, and it was completely down," she says. "My mom is freaking out about it. She wants me to do it as soon as I leave here," before the midnight deadline.
Then there's Libertone, who works three part-time jobs, none of which offer health coverage. During the evening, he expresses astonishment at what he's hearing. "Clear this up: This is some sort of deadline about getting signed up?" he asks, raising his arms in surprise and exasperation. "I thought it was Jan. 1."
He hadn't met that deadline, he says. Will he meet this one? Probably not, he says.
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Six months ago, USA TODAY brought together a group of uninsured Coloradans to discuss the Affordable Care Act as the law's first enrollment period was about to open. Now what do they think?
Obamacare: How do we like it now? Six months ago, USA TODAY brought together a group of uninsured Coloradans to discuss the Affordable Care Act as the law's first enrollment period was about to open. Now what do they think? Check out this story on USATODAY.com: http://usat.ly/PO6bq0

redpajama_set_name: RedPajamaCommonCrawl


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\section{Introduction}
This paper deals with system  of equations of the type:

\begin{equation}\label{ProbPrin-gral}
    \left\{%
\begin{array}{ll}
  \frac{d}{dt} D_{y}\mathcal{L}(t,u(t),u'(t))= D_{x}\mathcal{L}(t,u(t),u'(t)) \quad \hbox{a.e.}\ t \in (0,T),\\
    u(0)-u(T)=u'(0)-u'(T)=0,
\end{array}%
\right.
\end{equation}
where $\mathcal{L}:[0,T]\times\mathbb{R}^d\times\mathbb{R}^d\to\mathbb{R}$, $d\geq 1$, is called the \emph{Lagrange function} or \emph{lagrangian} and the unknown function  $u:[0,T]\to\mathbb{R}^d$ is absolutely continuous. In other words, we are interested in  finding \emph{periodic weak solutions} of \emph{Euler-Lagrange system of ordinary differential equations}.

The problem \eqref{ProbPrin-gral} comes from a variational one, that is,  the equation in  \eqref{ProbPrin-gral} is the Euler-Lagrange equation associated to the \emph{action integral}

\begin{equation}\label{integral_accion}
I(u)=\int_{0}^T \mathcal{L}(t,u(t),u'(t))\ dt.
\end{equation}

This topic was deeply addressed for the \emph{Lagrange function}
\begin{equation}\label{eq:lagrange_cuad}
\mathcal{L}_{p,F}(t,x,y)=\frac{|y|^p}{p}+F(t,x),
\end{equation}
for $1<p<\infty$. For example, the classic book  \cite{mawhin2010critical} deals mainly with problem \eqref{ProbPrin-gral} for the lagrangian $\mathcal{L}_{2,F}$ and through various methods: direct, dual action, minimax, etc. The results in \cite{mawhin2010critical} were extended and improved in several articles, see  \cite{tang1995periodic,tang1998periodic,wu1999periodic,tang2001periodic,zhao2004periodic}  to cite some examples. Lagrange functions \eqref{eq:lagrange_cuad} for arbitrary $1<p<\infty$ are considered in  \cite{Tian2007192,tang2010periodic} and in this case \eqref{ProbPrin-gral}  is reduced to the $p$-laplacian system
\begin{equation*}\label{ProbP-lapla}
    \left\{%
\begin{array}{ll}
   \frac{d}{dt}\left(u'(t)|u'|^{p-2}\right) = \nabla F(t,u(t)) \quad \hbox{a.e.}\ t \in (0,T),\\
    u(0)-u(T)=u'(0)-u'(T)=0.
\end{array}%
\right.
\end{equation*}

In this context, it  is customary to call $F$ a  \emph{potential function}, and it is assumed that $F(t,x)$ is differentiable with respect to $x$ for a.e. $t\in [0,T]$ and the following conditions hold:
\begin{enumerate}
\labitem{(C)}{item:condicion_c} $F$ and its gradient $\nabla F$, with respect to $x\in\mathbb{R}^d$,  are  Carath\'eodory functions, i.e. they are measurable functions with respect to $t\in [0,T]$ for every  $x\in\mathbb{R}^d$, and they are continuous functions with  respect to  $x\in\mathbb{R}^d$ for a.e. $t \in [0,T]$.
 \labitem{(A)}{item:condicion_a}  For   a.e. $t\in [0,T]$,
\begin{equation}
|F(t,x)| + |\nabla F(t,x)|  \leq a(|x|)b(t).
\end{equation}
In this inequality, it is assumed that the function  $a:[0,+\infty)\to [0,+\infty)$ is continuous and non decreasing,
and $0\leq b\in L^1([0,T],\mathbb{R})$.
\end{enumerate}

In \cite{ABGMS2015} it was treated  the case of a lagrangian $\mathcal{L}$ which is lower bounded by a Lagrange function
\begin{equation}\label{eq:lagrange_phi}
\mathcal{L}_{\Phi,F}(t,x,y)=\Phi(|y|)+F(t,x),
\end{equation}
where  $\Phi$ is an $N$-function (see section \ref{preliminares} for the definition of this concept).
In the paper \cite{ABGMS2015} it was also assumed  a condition of \emph{bounded oscillation} on $F$.
In this current paper we will study a condition of \emph{sublinearity} (see  \cite{tang1998periodic,wu1999periodic,zhao2004periodic,tang2010periodic,zhao2005existence}) on $\nabla F$ for the lagrangian  $\mathcal{L}_{\Phi,F}$, or more generally for lagrangians which are lower bounded by $\mathcal{L}_{\Phi,F}$.

The paper is organized as follows. In section \ref{preliminares}, we give preliminaries facts on $N$-functions and Orlicz-Sobolev spaces of functions. Section 3 is devoted to the main result of this work and it also includes an auxiliary lemma of vital importance. Section \ref{sec:proofs} contains the proofs and section \ref{sec:examples} provides an application of our result to a concrete case.

\section{Preliminaries}\label{preliminares}

For reader convenience, we give a short introduction to Orlicz and Orlicz-Sobolev spaces of vector valued functions. Classic references for these topics are \cite{adams_sobolev,KR,rao1991theory,M}.

Hereafter we denote  by $\mathbb{R}^+$  the set of all non negative real numbers. A function $\Phi:\mathbb{R}^+\to \mathbb{R}^+ $ is called an \emph{$N$-function} if $\Phi$ is convex and it also satisfies that
\[
\lim_{t\to+\infty}\frac{\Phi(t)}{t}=+\infty\quad\text{and}\quad \lim_{t\to 0}\frac{\Phi(t)}{t}=0.
\]
In addition,  in this paper for the sake of simplicity  we assume that $\Phi$ is differentiable and we call $\varphi$  the derivative of $\Phi$.
On these assumptions, $\varphi:\mathbb{R}^+\rightarrow \mathbb{R}^+$ is a homeomorphism whose inverse will be denoted by $\psi$.
We write $\Psi$ for the primitive of $\psi$ that satisfies $\Psi(0)=0$. Then, $\Psi$ is an $N$-function which  is known as the \emph{complementary function} of $\Phi$.

 We recall that an $N$-function $\Phi(u)$ has \emph{principal part} $f(u)$ if $\Phi(u)=f(u)$ for large values of the argument (see \cite[p. 16]{KR} and \cite[Sec. 7]{KR} for  properties of principal part).

There exist several orders and equivalence relations between $N$-functions (see \cite[Sec. 2.2]{rao1991theory}).
Following \cite[Def. 1, pp. 15-16]{rao1991theory} we say that the   $N$-function $\Phi_2$ is \emph{stronger} than the $N$-function  $\Phi_1$, in symbols  $\Phi_1\prec\Phi_2$, if  there exist $a>0$ and $x_0\geq 0$ such that
\begin{equation}\label{eq:prec}\Phi_1(x)\leq \Phi_2(ax), \quad x\geq x_0.\end{equation}
 The $N$-functions  $\Phi_1$ and   $\Phi_2$ are \emph{equivalent} ($\Phi_1\sim\Phi_2$)  when  $\Phi_1\prec\Phi_2$ and $\Phi_2\prec\Phi_1$.
We say that  $\Phi_2$ is \emph{essentially stronger} than  $\Phi_1$  ($\Phi_1\llcurly\Phi_2$) if and only if for every $a>0$ there exists $x_0=x_0(a)\geq 0$ such that \eqref{eq:prec} holds. Finally, we say that  $\Phi_2$ is \emph{completely stronger} than  $\Phi_1$  ($\Phi_1\closedprec\Phi_2$) if and only if for every $a>0$ there exist $K=K(a)>0$ and  $x_0=x_0(a)\geq 0$ such that

\begin{equation}\label{eq:prec2}\Phi_1(x)\leq K\Phi_2(ax), \quad x\geq x_0.\end{equation}

We also say that a non decreasing function $\eta:\mathbb{R}^+\rightarrow \mathbb{R}^+$ satisfies the  \emph{$\Delta_2^{\infty}$-condition}, denoted by $\eta \in \Delta_2^{\infty}$,
if there exist  constants $K>0$ and  $x_0\geq 0$ such that
\begin{equation}\label{delta2defi}\eta(2x)\leq K\eta(x),
\end{equation}
for every $x\geq x_0$. We note that $\eta \in \Delta_2^{\infty}$ if and only if $\eta\closedprec\eta$.
If $x_0=0$,  the function   $\eta:\mathbb{R}^+\rightarrow \mathbb{R}^+$ is said to satisfy the
\emph{$\Delta_2$-condition} ($\eta \in \Delta_2$).
If there exists $x_0>0$ such that  inequality \eqref{delta2defi} holds for $x\leq x_0$,
we will say that $\Phi$ satisfies the
\emph{$\Delta_2^0$-condition} ($\Phi\in\Delta_2^0$).

We denote by $\alpha_{\eta}$ and $\beta_{\eta}$ the so called  \emph{Matuszewska-Orlicz indices} of the function $\eta$, which are defined next. Given
an increasing, unbounded, continuous function   $\eta:[ 0,+\infty)\to [0,+\infty)$ such that $\eta(0)=0$, we define
\begin{equation}\label{MO_indices}
    \alpha_{\eta}:=\lim\limits_{t\to 0^{+}}\frac{\log \left (\sup\limits_{u>0}\frac{\eta(t u)}{\eta(u)} \right ) }{\log(t)},\quad
    \beta_{\eta}:=\lim\limits_{t\to +\infty}\frac{\log \left  (\sup\limits_{u>0}\frac{\eta(t u)}{\eta(u)}\right )}{\log(t)}.
\end{equation}
It is known that the previous limits exist and  $0\leq \alpha_{\eta}\leq \beta_{\eta}\leq +\infty$
(see \cite[p. 84]{M}). The relation $\beta_{\eta}<+\infty$ holds true if and only if $\eta \in \Delta_2$
(\cite[Thm. 11.7]{M}). If $(\Phi,\Psi)$ is a complementary pair of  $N$-functions then
\begin{equation}\label{compl_ind}
 \frac{1}{\alpha_{\Phi}}+\frac{1}{\beta_{\Psi}}=1,
\end{equation}
(see \cite[Cor. 11.6]{M}). Therefore $1\leq \alpha_{\Phi}\leq\beta_{\Phi}\leq \infty $.

 If $\eta$ is an increasing function that satisfies the $\Delta_2$-condition, then $\eta$ is controlled by above and below
 by power functions (\cite[Sec. 1]{Gustavsson1977}, \cite[Eq. (2.3)-(2.4)]{fiorenza1997indices} and \cite[Thm. 11.13]{M}).   More concretely, for every $\epsilon>0$ there exists a
constant $K=K(\eta,\epsilon)$ such that, for every $t,u\geq 0$,
\begin{equation}\label{delta2-potencias}
    K^{-1}\min\big\{t^{\beta_{\eta}+\epsilon},t^{\alpha_{\eta}-\epsilon} \big\}\eta(u)\leq \eta(t u)\leq
    K\max\big\{t^{\beta_{\eta}+\epsilon},t^{\alpha_{\eta}-\epsilon} \big\}\eta(u).
\end{equation}

Let $d$ be a positive integer. We denote by $\mathcal{M}:=\mathcal{M}([0,T],\mathbb{R}^d)$  the set of all measurable functions defined on $[0,T]$ with values on $\mathbb{R}^d$ and  we write $u=(u_1,\dots,u_d)$ for  $u\in \mathcal{M}$. For the set of functions $\mathcal{M}$, as for other similar sets, we will omit the reference to codomain $\mathbb{R}^d$ when $d=1$.

Given  an $N$-function $\Phi$ we define the \emph{modular function}
$\rho_{\Phi}:\mathcal{M}\to \mathbb{R}^+\cup\{+\infty\}$ by
\[\rho_{\Phi}(u):= \int_0^T \Phi(|u|)\ dt.\]
Here $|\cdot|$ is the euclidean norm of $\mathbb{R}^d$.
Now, we introduce the \emph{Orlicz class} $C^{\Phi}=C^{\Phi}([0,T],\mathbb{R}^d)$   by setting
\begin{equation}\label{claseOrlicz}
  C^{\Phi}:=\left\{u\in \mathcal{M} | \rho_{\Phi}(u)< \infty \right\}.
\end{equation}
The \emph{Orlicz space} $L^{\Phi}=L^{\Phi}([0,T],\mathbb{R}^d)$ is the linear hull of $C^{\Phi}$;
equivalently,
\begin{equation}\label{espacioOrlicz}
L^{\Phi}:=\left\{ u\in \mathcal{M}| \exists \lambda>0: \rho_{\Phi}(\lambda u) < \infty   \right\}.
\end{equation}
  The Orlicz space $L^{\Phi}$ equipped with the \emph{Orlicz norm}
\[
\|  u  \|_{L^{\Phi}}:=\sup \left\{  \int_0^T u\b{\cdot} v\ dt \big| \rho_{\Psi}(v)\leq 1\right\},
\]
is a Banach space. By $u\b{\cdot} v$ we denote the usual dot product in $\mathbb{R}^{d}$ between $u$ and $v$.

The following  inequality holds for any $u\inL^{\Phi}$
\begin{equation}\label{amemiya-ine}
\|u\|_{L^{\Phi}}\leq \frac{1}{k}\left\{1+\rho_{\Phi}(ku)\right\},\quad\text{for every } k>0.
\end{equation}
In fact, $\|u\|_{L^{\Phi}}$ is the infimum for $k>0$ of the right hand side in above expression  (see \cite[Thm. 10.5]{KR} and \cite{hudzik2000amemiya}).

The subspace $E^{\Phi}=E^{\Phi}([0,T],\mathbb{R}^d)$ is defined as the closure in $L^{\Phi}$ of the subspace $L^{\infty}([0,T],\mathbb{R}^d)$ of all $\mathbb{R}^d$-valued essentially bounded functions. It has shown that  $E^{\Phi}$ is the only one maximal subspace contained in the Orlicz class $C^{\Phi}$, i.e.
$u\inE^{\Phi}$ if and only if $\rho_{\Phi}(\lambda u)<\infty$ for any $\lambda>0$. The equality $L^{\Phi}=E^{\Phi}$ is true if and only if $\Phi\in\Delta_2^{\infty}$.

A generalized version of \emph{H\"older's inequality} holds in Orlicz spaces (see \cite[Thm. 9.3]{KR}). Namely, if $u\inL^{\Phi}$ and $v\inL^{\Psi}$ then $u\cdot v\in L^1$ and
\begin{equation}\label{holder}
\int_0^Tv\cdot u\ dt\leq \|u\|_{L^{\Phi}}\|v\|_{L^{\Psi}}.
\end{equation}

If $X$ and $Y$ are  Banach spaces such that  $Y\subset X^*$, we denote by $\langle\cdot,\cdot\rangle:Y\times X\to\mathbb{R}$ the bilinear pairing  map given by $\langle x^*,x\rangle=x^*(x)$. H\"older's inequality shows that $L^{\Psi}\subset \left[L^{\Phi}\right]^*$, where the pairing
$\langle v, u\rangle$
is defined by
\begin{equation}\label{pairing}
  \langle v,u\rangle=\int_0^Tv\cdot u\ dt,
\end{equation}
with  $u\inL^{\Phi}$ and $v\inL^{\Psi}$.
 Unless $\Phi \in \Delta_2^{\infty}$, the relation $L^{\Psi}= \left[L^{\Phi}\right]^*$ will not be satisfied.
In general, it is true  that  $\left[E^{\Phi}\right]^*=L^{\Psi}$.

We define the \emph{Sobolev-Orlicz space} $W^{1}\lphi$ (see \cite{adams_sobolev}) by
\[W^{1}\lphi:=\{u| u \hbox{ is absolutely continuous on $[0,T]$ and } u'\in L^{\Phi}\}.\]
$W^{1}\lphi$ is a Banach space when equipped with the norm
\begin{equation}\label{def-norma-orlicz-sob}
\|  u  \|_{W^{1}\lphi}= \|  u  \|_{L^{\Phi}} + \|u'\|_{L^{\Phi}}.
\end{equation}
And, we introduce the following subset of $W^{1}\lphi$
\begin{equation}\label{def-esp-orlicz-sob-per}
\begin{split}
W^{1}\lphi_T&=\{u\inW^{1}\lphi|u(0)=u(T)\}.
\end{split}
\end{equation}

We will use repeatedly the decomposition $u=\overline{u}+\widetilde{u}$ for a function $u\in L^1([0,T])$  where $\overline{u} =\frac1T\int_0^T u(t)\ dt$ and $\widetilde{u}=u-\overline{u}$.

As usual, if $(X,\|\cdot\|_X)$ is a Banach space and $(Y,\|\cdot \|_Y)$ is a subspace of $X$,  we write $Y\hookrightarrow X$ and we say that $Y$ is \emph{embedded} in $X$  when the restricted identity map $i_Y:Y\to X$ is bounded. That is, there exists $C>0$ such that
$\|y\|_X\leq C\|y\|_Y$ for any $y\in Y$.  With this notation, H\"older's inequality states that  $L^{\Psi}\hookrightarrow  \left[L^{\Phi}\right]^*$; and, it is easy to see that for every $N$-function $\Phi$ we have that $L^{\infty}\hookrightarrowL^{\Phi} \hookrightarrow L^1$.

The following simple  embedding lemma, whose proof can be found in \cite{ABGMS2015}, will be used several times.

\begin{lem}\label{inclusion orlicz} For every $u\inW^{1}\lphi$, $\widetilde{u}\in L^{\infty}$ and
\begin{align}
 \|u\|_{L^{\infty}} &\leq\Phi^{-1}\left(\frac{1}{T}\right)\max\{1,T\}\|u\|_{W^{1}\lphi}&\text{  (Sobolev's inequality).}\label{sobolev}\\
 \|\widetilde{u}\|_{L^{\infty}} &\leq T\Phi^{-1}\left(\frac{1}{T}\right)\|u'\|_{L^{\Phi}}&
\text{  (Sobolev-Wirtinger's inequality).}\label{wirtinger}
\end{align}

\end{lem}

\section{Main result}

We begin with a lemma which establishes the coercivity of the modular function $\rho_{\Phi}(u)$ with respect to certain functions of the Orlicz norm $\Phi_0(\|u\|_{L^{\Phi}}|)$. This lemma generalizes \cite[Lemma 5.2]{ABGMS2015} in two directions.
Namely, certain power function is replaced by a more general $N$-function $\Phi_0$ and
the $\Delta_2$-condition  on $\Psi$ is relaxed to $\Delta_2^{\infty}$.
It is worth noting that the second improvement is more important than the first one.
And, we present the result here since the lemma introduces a function $\Phi^*$ that will play a significant role in the statement of our main theorem.

\begin{lem}\label{lem_coer}
Let $\Phi,\Psi$ be complementary $N$-functions with $\Psi \in \Delta_2^{\infty}$. Then,
there exists an $N$-function $\Phi^*$ being $\Phi^*\prec\Phi$,
such that  for every $N$-function $\Phi_0$ that satisfies $\Phi_0\llcurly\Phi^*$ and for every $k>0$, we have
\begin{equation}\label{eq:coer_mod}
\lim\limits_{\|u\|_{L^{\Phi}}\to \infty}
\frac{\int_0^T \Phi(|u|)\,dt}{\Phi_0(k\|u\|_{L^{\Phi}})}=\infty.
\end{equation}
Reciprocally, if  \eqref{eq:coer_mod} holds for some $N$-function $\Phi_0$,  then $\Psi\in\Delta_2^{\infty}$.
\end{lem}

 We point out that this lemma can be applied to more cases than \cite[Lemma 5.2]{ABGMS2015}. For example, if $\Phi(u)=u^2$, $\Phi_1$ and $\Phi_0$ are  $N$-functions with principal parts equal to $u^2/\log u$ and $u^2/(\log u)^2$ respectively, then  \eqref{eq:coer_mod} holds for $\Phi_0$. On the other hand, $\Phi_0(|u|)$ is not dominated for any  power function $|u|^{\alpha}$ with $\alpha<2$.

As in  \cite{ABGMS2015} we will consider general Lagrange functions
$\mathcal{L}:[0,T]\times\mathbb{R}^d\times\mathbb{R}^d\to\mathbb{R}$
satisfying the structure conditions
\begin{align}
|\mathcal{L}(t,x,y)| &\leq a(|x|)\left(b(t)+ \Phi\left(\frac{|y|}{\lambda}+f(t) \right)\right),&\tag{$A_1$}\label{eq:estru1}
\\
|D_{x}\mathcal{L}(t,x,y)| &\leq a(|x|)\left(b(t)+ \Phi\left(\frac{|y|}{\lambda}+f(t) \right)\right),&\tag{$A_2$}\label{eq:estru2}
\\
|D_{y}\mathcal{L}(t,x,y)| &\leq a(|x|)\left(c(t)+ \varphi\left(\frac{|y|}{\lambda}+f(t)\right)\right),
&\tag{$A_3$}\label{eq:estru3}
\end{align}
where $a\in C(\mathbb{R}^+,\mathbb{R}^+)$, $\lambda>0$, $\Phi$ is an $N$-function,
$\varphi$ is the continuous derivative of $\Phi$,
$b\in L^1_1([0,T])$,  $c\inL^{\Psi}_1([0,T])$ and  $f\in E^{\Phi}_1([0,T])$. We denote by $\mathfrak{A}(a,b,c,\lambda,f,\Phi)$ the set of all Lagrange functions satisfying  \eqref{eq:estru1}, \eqref{eq:estru2} and \eqref{eq:estru3}.

In \cite{ABGMS2015} it was shown that if $\mathcal{L}\in \mathfrak{A}(a,b,c,\lambda,f,\Phi)$
then there  exists the Gate\^aux derivative of the integral functional $I$ defined by \eqref{integral_accion},
on the set
\[\mathcal{E}^{\Phi}_d(\lambda):=\{u\inW^{1}\lphi([0,T],\mathbb{R}^d): d(u',E^{\Phi})<\lambda\}.\]
We observe that the condition \ref{item:condicion_a} on the potential $F$ is equivalent to say that $\mathcal{L}_{\Phi,F}\in \mathfrak{A}(a,b,0,1,0,\Phi)$.

 Unlike what is usual in the literature, we do not assume the lagrangian $\mathcal{L}$  split into two terms,
one of them function of $y$ and the other one function of $(t,x)$.
We only suppose that $\mathcal{L}$ is lower bounded by a function of this type. More precisely, we assume that for every $(t,x,y)\in\mathbb{R}\times\mathbb{R}^d\times\mathbb{R}^d$
\begin{equation}\label{eq:control_inf}
\mathcal{L} \geq \mathcal{L}_{\Phi,F},\quad \text{with}\;F\text{ satisfying \ref{item:condicion_a} \;and\; \ref{item:condicion_c},
\;\text{and}\;$\Phi$ being an $N$-function}.\tag{$A_4$}
\end{equation}
In addition, as usual we suppose that the time integral of $F$ satisfies certain coercivity condition, see \eqref{eq:propiedad-coercividad-phi0} below.  However,  all these hypotheses are not enough. It is also necessary to assume extra conditions on the potential $F$.
Several hypotheses were tested in the past years. The so called \emph{subconvexity} of $F$ was tried in \cite{wu1999periodic,tang1995periodic,zhao2004periodic} for semilinear equations and in \cite{xu2007some,tang2010periodic} for $p$-laplacian systems. Potentials $F$ satisfying the following inequality
\begin{equation}\label{holder_cont}
  \left| F(t,x_2)- F(t,x_1) \right|\leq b_1(t)(1+|x_2-x_1|^{\mu}).
\end{equation}
 were studied in \cite{ABGMS2015}.
Regarding \eqref{holder_cont}, it is interesting to notice that such inequality is equivalent to say the condition
$\|F(t,\cdot)\|_{BO}\in L^1([0,T])$, where $\|\cdot\|_{BO}$ denotes the seminorm defined in  \cite[p. 125]{zhu2012analysis} on the space of functions of bounded variations.

 In \cite{tang1998periodic, tang2010periodic} the authors  dealt with the $p$-laplacian case
with potentials $F$ such that
\begin{equation}\label{eq:cota_pot} |\nabla F(t,x)|\leq b_1(t)|x|^{\alpha}+b_2(t),
 \end{equation}
where  $b_1,b_2 \in L^1([0,T])$ and $\alpha<p$. Such potentials $F$
were called  \emph{sublinear nonlinearities}. In this paper, we are interested in studying this type of potentials,
but with more general bounds on $\nabla F$ which include $N$-functions instead of power functions;
namely, we will consider inequalities like
\begin{equation}\label{holder_cont-mu}
  \left| \nabla F(t,x) \right|\leq b_1(t)\Phi_0'(|x|)+b_2(t),
  \tag{$A_5$}
\end{equation}
with $\Phi_0$  a differentiable $N$-function and $b_1,b_2 \in L^1([0,T])$.

Next, we give our main result.  Here,  we will amend an
erroneous assumption made in the end of the proof of  \cite[Thm. 6.2]{ABGMS2015}.
There, it was assumed without discussion that  a minimum of  $I$ was on the domain of differentiability of $I$.

\begin{thm}\label{coercitividad-r}
Let $\Phi$ be an $N$-function whose complementary function $\Psi$ satisfies the $\Delta_2^{\infty}$-condition
and suppose that $\Phi^*$ is given by Lemma \ref{lem_coer}. Assume that the lagrangian $\mathcal{L}\in \mathfrak{A}(a,b,c,\lambda,f,\Phi)$ satisfies \eqref{eq:control_inf}, where
the potential $F$ fulfills  \ref{item:condicion_c}, \ref{item:condicion_a} and
the following conditions:
\begin{enumerate}
 \item  \eqref{holder_cont-mu}  for some $N$-function  $\Phi_0$ such that $\Phi_0\llcurly\Phi^*$.
 \item \begin{equation}\label{eq:propiedad-coercividad-phi0}
\lim_{|x|\to\infty}\frac{\int_{0}^{T}F(t,x)\ dt}{\Psi_1(\Phi_0'(2|x|))}=+\infty,\tag{$A_6$}
\end{equation}
for some $N$-function $\Psi_1$ with complementary function $\Phi_1$ satisfying $\Phi_0\llcurly \Phi_1\llcurly \Phi^*$.
\end{enumerate}

Then the action integral $I$ has a minimum $u\inW^{1}\lphi([0,T],\mathbb{R}^d)$ such that $d(u',E^{\Phi})\leq \lambda$. If $d(u',E^{\Phi})< \lambda$, the lagrangian $\mathcal{L}(t,x,y)$ is strictly convex  with respect to $y\in\mathbb{R}^d$ and
$D_y\mathcal{L}(0,x,y)=D_y\mathcal{L}(T,x,y)$
then $u$ solves the problem \eqref{ProbPrin-gral}.

\end{thm}

\begin{comentario} If $\Phi\in\Delta_2^{\infty}$ the the condition  $d(u',E^{\Phi})\leq \lambda$ is automatically satisfied.
\end{comentario}

\section{Proofs}\label{sec:proofs}

The following result is analogous to some lemmata in $W^{1,p}$, see \cite[Lem. 1]{xu2007some}.
\begin{lem}\label{infinito-a-prom-upunto}
 $\|u\|_{W^{1}\lphi}\to \infty$ iff  $(|\overline{u}|+\|u'\|_{L^{\Phi}})\to \infty$.
\end{lem}

\begin{proof}
By the decomposition $u=\overline{u}+\tilde{u}$ and some elementary operations,
we get
\begin{equation}\label{cota-u-lphi}
\|u\|_{L^{\Phi}}=
\|\overline{u}+\tilde{u}\|_{L^{\Phi}}\leq
\|\overline{u}\|_{L^{\Phi}}+\|\tilde{u}\|_{L^{\Phi}}=
|\overline{u}|\|1\|_{L^{\Phi}}+\|\tilde{u}\|_{L^{\Phi}}.
\end{equation}
It is known that $L^{\infty}\hookrightarrowL^{\Phi}$, i.e.
there exists $C_1=C_1(T)>0$ such that for any $\tilde{u}\in L^{\infty}$ we have
\[
\|\tilde{u}\|_{L^{\Phi}}
\leq
C_1 \|\tilde{u}\|_{L^{\infty}};
\]
and, applying  Sobolev's inequality,  we obtain Wirtinger's inequality,
that is there exists $C_2=C_2(T)>0$ such that
\begin{equation}\label{cota-u-tilde}
\|\tilde{u}\|_{L^{\Phi}}
\leq
C_2\|u'\|_{L^{\Phi}}.
\end{equation}
Therefore, from \eqref{cota-u-lphi}, \eqref{cota-u-tilde} and \eqref{def-norma-orlicz-sob},
we get
\[
\|u\|_{W^{1}\lphi}\leq
C_3(|\overline{u}|+\|u'\|_{L^{\Phi}})
\]
where $C_3=C_3(T)$. Finally, as $\|u\|_{W^{1}\lphi}\to \infty$ we conclude that
$(|\overline{u}|+\|u'\|_{L^{\Phi}})\to \infty$.

For the converse, we observe that
\[|\overline{u}|\leq \frac{1}{T}\|1\|_{L^{\Psi}}\|u\|_{L^{\Phi}}.\]
Hence
\[|\overline{u}| +\|u'\|_{L^{\Phi}}\leq C_4(\|u\|_{L^{\Phi}} +\|u'\|_{L^{\Phi}}),\]
and the property under consideration is proved.
\end{proof}

\begin{lem}\label{lema:delta_2 y der} Let $\Phi$ be a not necessarily differentiable $N$-function and
let $\varphi$ be the right continuous derivative of $\Phi$. Then $\Phi\in\Delta_2^{\infty}$ (
$\Phi\in\Delta_2$) iff $\varphi\in\Delta_2^{\infty}$ (
$\varphi\in\Delta_2$).
\end{lem}
\begin{proof}  It is consequence of \cite[Thm. 11.7]{M} and \cite[Rem. 5, p. 87]{M}.

\end{proof}

The following lemma improves the result on the comment at the beginning of \cite[p. 24]{KR}.

\begin{lem}\label{lem:delta2-equiv-delta2-global} Let $\Psi$ be an $N$-function satisfying the $\Delta_2^{\infty}$-condition. Then there exists an $N$-function $\Psi^*$  such that $\Psi^*\in\Delta_2$, $\Psi\leq\Psi^*$ and for every $a>1$ there exists $x_0=x_0(a)\geq 0$ such that $\Psi^*(x)\leq a\Psi(x)$, for every $x\geq x_0$.  In particular,  every $\Delta_2$ near infinity  $N$-function is equivalent to a $\Delta_2$  $N$-function.
\end{lem}

\begin{proof} We can assume that $\Psi \notin \Delta_2^0$.
Consequently, from Lemma \ref{lema:delta_2 y der} we have that the right continuous derivative $\psi$ of $\Psi$ does not satisfy the $\Delta_2^0$-condition.
Therefore, we obtain a sequence of positive numbers $x_n$, $n=1,2,\ldots$,  such that $x_n\to 0$,
\begin{equation}\label{eq:prop_psi*}2x_{n+1}<x_n<2x_n\quad\text{and}\quad \psi(2x_n)> 2\psi(x_n).
 \end{equation}

We define $\psi^*$ inductively for $n$ on the interval $[2x_n,+\infty)$ of the following way.
We put $\psi^*(x)=\psi(x)$ when  $x\in[ 2x_1,+\infty)$.
Suppose $\psi^*$ defined on $[2x_n,+\infty)$ and we set $\psi^*$ on $[2x_{n+1},2x_n)$ by
\[
 \psi^*(x)=\left\{
\begin{array}{cc}
  \max\left\{\psi(x),\frac{\psi^*(2x_n)}{2x_n}(x-x_n)+\frac{ \psi^*(2x_n)}{2}\right\},& \text{if } x_n\leq x<2x_n\\
   \frac{\psi^*(2x_n)}{2}& \text{if }2x_{n+1} \leq x<x_n
  \end{array}
\right.
\]
Moreover, we define $\psi^*(0)=0$.

Next, we will use induction  to prove that
\begin{enumerate}
 \item\label{item:prop_psi*1} $\psi^*(x_n)=\frac12\psi^*(2x_n)$,
 \item\label{item:prop_psi*2}  $\psi^*$ is non-decreasing $[2x_{n},+\infty)$,
 \item\label{item:prop_psi*3}   $\psi\leq \psi^*$ in  $[2x_{n},+\infty)$.
\end{enumerate}

We suppose $n=1$. Then items \ref{item:prop_psi*2} and \ref{item:prop_psi*3} are obvious. From \eqref{eq:prop_psi*} we have
\[\psi(x_1)<\frac12\psi(2x_1)=\frac12\psi^*(2x_1),\]
and this inequality implies \ref{item:prop_psi*1}.

Assume that properties \ref{item:prop_psi*1}-\ref{item:prop_psi*3} hold for $n>1$.  Clearly $\psi^*$ is non decreasing on each interval $[2x_{n+1},x_n)$ and $[x_n,2x_n)$. Since $\psi$ is right continuous and $\psi(x_n)<2^{-1}\psi(2x_n)\leq 2^{-1}\psi^*(2x_n)$, then $\psi^*$ is continuous at $x_n$. Therefore, $\psi^*$ is non decreasing on $[2x_{n+1},2x_n)$. Suppose $x\in[2x_{n+1},2x_n)$ and $y\geq 2x_n$.  From the definition of $\psi^*$, the inductive hypothesis, item \ref{item:prop_psi*3} and item \ref{item:prop_psi*2}, we obtain
\[\psi^*(x)\leq\max\{\psi(2x_n),\psi^*(2x_n)\}=\psi^*(2x_n)\leq\psi^*(y).\]
This proves item \ref{item:prop_psi*2} on the interval $[2x_{n+1},+\infty)$.
Inequality in item \ref{item:prop_psi*3} holds by inductive hypothesis on $[2x_n,+\infty)$
and it is obvious for  $x\in[x_n,2x_n)$.
If $ x\in[2x_{n+1},x_n)$, then $\psi(x)\leq\psi(x_n)\leq\psi^*(x_n)=\psi^*(x)$.
This proves  \ref{item:prop_psi*3} on the interval $[2x_{n+1},+\infty)$.

Now, using \eqref{eq:prop_psi*} and the already proved item \ref{item:prop_psi*3} for $n+1$, we deduce $\psi(x_{n+1})<2^{-1}\psi(2x_{n+1})\leq 2^{-1}\psi^*(2x_{n+1})$.
Then
\[\psi^*(x_{n+1})=\max\left\{\psi(x_{n+1}),\frac12\psi^*(2x_{n+1}) \right\}=\frac12\psi^*(2x_{n+1}),\]
i.e. we have just proved item \ref{item:prop_psi*1}.

We note that $\psi^*(x_{n+1})=2^{-1}\psi^*(2x_{n+1})\leq 2^{-1}\psi^*(x_{n})$. Consequently $\psi^*(x)\to 0$ when $x\to 0$.
Therefore $\psi^*$ is right continuous at $0$ and indeed it is right continuous on $[0,+\infty)$.
Moreover, since $\psi(x)=\psi^*(x)$ for $x\geq 2x_1$ being $\psi$ the right continuous derivative of an $N$-function, $\psi^*(x)\to +\infty$ when $x\to +\infty$. In this way,
\[\Psi^*(x):=\int_0^x\psi^*(t)dt\]
defines an $N$-function.

Let's see that $\psi^*$ satisfies the $\Delta_2$-condition.
It is sufficient to prove that $\psi^*$ satisfies the $ \Delta_2^0$-condition.
To this end, suppose that  $x\leq x_1$ and take $n\in\mathbb{N}$ such that $x_{n+1}\leq x\leq x_n$. Then
\[\psi^*(2x)\leq \psi^*(2x_n)=2\psi^*(2x_{n+1})=4\psi^*(x_{n+1})\leq 4\psi^*(x).\]
Thus,  $\Psi^* \in  \Delta_2$ and $\Psi\leq \Psi^*$.

It remains to show the inequality $\Psi^*(x)\leq a\Psi(x)$,
for every $a>1$ and sufficiently large $x$. We take $x_0$ sufficiently large to have
\[\frac{1}{a-1}\int_0^{2x_1}\psi^*(t)-\psi(t)dt<\Psi(x_0).\]
Therefore, if  $x>\max\{x_0,2x_1\}$ then
\[\Psi^*(x)=\Psi(x)+\int_0^{2x_1}\psi^*(t)-\psi(t)dt<\Psi(x)+(a-1)\Psi(x)= a\Psi(x).\]

The last statement of the lemma is consequence of   $\Psi(ax)>a\Psi(x)$ when $a>1$.

\end{proof}

The following lemma is essentially known,  because it is basically a consequence  of the fact that $\Psi\in\Delta_2^{\infty}$ if and only if $\Psi\closedprec\Psi$,  \cite[Prop. 4, p. 20]{rao1991theory} and \cite[Cor. 10, p. 30]{rao1991theory}.
However, we prefer to include an alternative proof, as we do not see clearly that the results of \cite{rao1991theory} contemplate the case of  $N$-functions satisfying the $\Delta_2$-condition.

\begin{lem}\label{lem:submultipliativa}
Let $\Phi,\Psi$ be complementary functions.
The next statements are equivalent:
\begin{enumerate}
\item\label{item1} $\Psi \in \Delta_2$ ($\Psi \in \Delta_2^{\infty}$).
\item\label{item2} There exists an $N$-function $\Phi^*$ such that
\begin{equation}\label{eq:caract_delta2}
\Phi(rs)\geq \Phi^*(r)\Phi(s)\quad\mbox{for every }r\geq1,s\geq 0\, (r\geq1,s\geq 1).
\end{equation}
\end{enumerate}
\end{lem}

\begin{comentario} We want to emphasize that the difference between the conclusions in item \ref{item2} of the previous lemma
is  that \eqref{eq:caract_delta2} holds for $s\geq 0$ or $s\geq 1$ depending on
$\Psi\in\Delta_2$ or $\Psi\in\Delta_2^{\infty}$,  respectively.

\end{comentario}

\begin{proof}
 In virtue of the comment that precedes the lemma, we only consider the case   $\Psi \in \Delta_2$.

\ref{item1})$\Rightarrow$\ref{item2}).
As a consequence of the $\Delta_2$-condition on $\Psi$, \eqref{compl_ind} and \eqref{delta2-potencias},
we get for every $1<\nu<\alpha_{\Phi}$ a constant $K=K_{\nu}>0$  such that

\begin{equation}\label{delta2-consecuencia}
\Phi(r s)\geq Kr^{\nu}\Phi(s),
\end{equation}
for any $1<\nu<\alpha_{\Phi}$,  $s\geq 0$ and $r>1$. This proves  \eqref{eq:caract_delta2} with $\Phi^*(r)=kr^\nu$, which is an $N$-function.

\ref{item2})$\Rightarrow$\ref{item1})
Next, we follow  \cite[p. 32, Prop. 13]{rao1991theory} and \cite[p. 29, Prop. 9]{rao1991theory}.
Assume that
\[
\Phi^*(r)\Phi(s)\leq \Phi(rs)\;\;r>1,\;s\geq 0.
\]
Let $u=\Phi^*(r)\geq \Phi^*(1)$ and $v=\Phi(s)\geq 0$. By a well known inequality \cite[p. 13, Prop. 1]{rao1991theory} and \eqref{eq:caract_delta2},   for $u\geq \Phi^*(1)$ and $v> 0$ we have
\[
\frac{uv}{\Psi^{-1}(uv)}\leq \Phi^{-1}(uv)\leq{\Phi^*}^{-1}(u)\Phi^{-1}(v)\leq
\frac{4uv}{{\Psi^*}^{-1}(u)\Psi^{-1}(v)},
\]
then
\[
{\Psi^*}^{-1}(u)\Psi^{-1}(v)\leq 4 \Psi^{-1}(uv).
\]
If we take $x=\Psi^{-1}_1(u)\geq \Psi^{-1}_1(\Phi^*(1))$ and $y=\Psi^{-1}(v)\geq 0$, then
\[
\Psi\left(\frac{xy}{4}\right)\leq \Psi^*(x)\Psi(y).
\]
Now, taking  $x\geq \max\{8,{\Psi^*}^{-1}(\Phi^*(1))\}$ we get that $\Psi \in \Delta_2$.
\end{proof}

\begin{comentario} Note that if $\Phi^*$ satisfies \eqref{eq:caract_delta2} then $\Phi^*\prec \Phi$.
 \end{comentario}

\noindent\textbf{Proof Lemma \ref{lem_coer}} First, we suppose that $\Psi \in \Delta_2$.
Let $\Phi^*$ be an $N$-function satisfying \eqref{eq:caract_delta2}.
By the inequality \eqref{amemiya-ine}, for $r>1$ we have
\[
\int_0^T \Phi(|u|)\,dt\geq
\Phi^*(r) \int_0^T \Phi(r^{-1}|u|)\,dt\geq
\Phi^*(r)\{r^{-1}\|u\|_{L^{\Phi}}-1\}.
\]
Now, we choose $r=\frac{\|u\|_{L^{\Phi}}}{2}$; and, as $\|u\|_{L^{\Phi}}\to\infty$ we can assume $r>1$.
From \cite[Thm. 2 (b)(v), p. 16]{rao1991theory} and $\Phi_0\llcurly \Phi^*$,  we get
\begin{equation}\label{eq:caso-delta-2}
\lim\limits_{\|u\|_{L^{\Phi}} \to \infty} \frac{\int_0^T \Phi(|u|)\,dt}{\Phi_0(k\|u\|_{L^{\Phi}})}\geq
\lim\limits_{\|u\|_{L^{\Phi}} \to \infty} \frac{\Phi^*\left(\frac{\|u\|_{L^{\Phi}}}{2}\right)}{\Phi_0(k\|u\|_{L^{\Phi}})}
=\infty.
\end{equation}
Now, if $\Psi\in\Delta_2^{\infty}$ but $\Psi\notin\Delta_2$, we  use Lemma \ref{lem:delta2-equiv-delta2-global}.
Then, there exists an $N$-function $\Psi_1$ such that $\Psi_1\in\Delta_2$ and  $\Psi_1\sim\Psi\leq \Psi_1$.
Let $\Phi_1$ be the complementary function of $\Psi_1$. Then $\Phi\sim\Phi_1\leq \Phi$ (see \cite[Thm. 3.1]{KR}) and $\|\cdot\|_{L^{\Phi}}$ and $\|\cdot\|_{L^{\Phi_1}}$ are equivalent norms  (see \cite[Thm. 13.2 and Thm. 13.3]{KR}).
Thus, there exists an $N$-function $\Phi_1^*\prec \Phi_1$ (consequently $\Phi_1^*\prec \Phi$)
satisfying  \eqref{eq:coer_mod}
with the $N$-functions $\Phi_1$ and $\Phi_1^*$ instead of $\Phi$ and $\Phi^*$, respectively.
Let $C>0$ be a constant such that  $\|\cdot\|_{L^{\Phi}}\leq C\|\cdot\|_{L^{\Phi_1}}$. Then

\[\lim\limits_{\|u\|_{L^{\Phi}} \to \infty}\frac{\int_0^T \Phi(|u|)\,dt}{\Phi_0(k\|u\|_{L^{\Phi}})}\geq \lim\limits_{\|u\|_{L^{\Phi}} \to \infty} \frac{\int_0^T \Phi_1(|u|)dt}{\Phi_0(kC\|u\|_{L^{\Phi_1}})}=+\infty.\]

Finally, if $\Phi_0$ is an $N$-function, then $\Phi_0(x)\geq \alpha |x|$ for  $\alpha$ small enough and $|x|>1$.
Therefore \eqref{eq:coer_mod} holds for $\Phi_0(x)=|x|$, then \cite[Lemma 5.2]{ABGMS2015}
implies  $\Psi\in\Delta_2^{\infty}$. \qed

\begin{defi}We define the  functionals $J_{C,\varphi}:L^{\Phi}\to (-\infty,+\infty]$ and $  H_{C,\varphi}:\mathbb{R}^n\to \mathbb{R}$, where $C>0$ and $\varphi:[0,+\infty)\to [0,+\infty)$, by
\begin{equation}\label{func_phi}
  J_{C,\varphi}(u):= \rho_{\Phi}\left(u\right)-C\varphi\left(\|u\|_{L^{\Phi}}\right),
\end{equation}
 and

\begin{equation}\label{eq:functional_H-bis}
 H_{C,\varphi}(x):=\int_0^TF(t,x)dt-C\varphi(2|x|),
\end{equation}
respectively.
\end{defi}

\vspace{.4cm}

\noindent\textbf{Proof. Theorem \ref{coercitividad-r} }. By the decomposition $u=\overline{u}+\tilde{u}$,   Cauchy-Schwarz's inequality
and \eqref{holder_cont-mu}, we have
\begin{equation}\label{cota-diferencia-F}
\begin{split}
&\left|\int_0^T F(t,u)-F(t,\overline{u})\,dt\right|=
\left|\int_0^T \int_0^1 \nabla F(t,\overline{u}+s\tilde{u}(t))\b{\cdot} \tilde{u}(t) \,ds \,dt\right|
\\
&\leq \int_0^T \int_0^1 b_1(t)\Phi_0'(|\overline{u}+s\tilde{u}(t)|)|\tilde{u}(t)|\,ds\,dt+
\int_0^T \int_0^1 b_2(t)|\tilde{u}(t)|\,ds\,dt
\\
&=:I_1+I_2.
\end{split}
\end{equation}
First, by H\"older's and Sobolev-Wirtinger's inequalities we estimate $I_2$ as follows
\begin{equation}\label{cota-i2}
I_2\leq \|b_2\|_{L^1} \|\tilde{u}\|_{L^{\infty}}\leq
C_1\|u'\|_{L^{\Phi}},
\end{equation}
 where $C_1=C_1(\|b_2\|_{L^1}, T)$.

Note that, since $\Phi_0'$ is an increasing function  and $\Phi_0'(x)\geq 0$ for $x\geq 0$, then
$\Phi'_0(a+b)\leq \Phi'_0(2a)+\Phi'_0(2b)$ for every $a,b\geq 0$.
In this way, we have
\begin{equation}\label{pot-suma}
\Phi'_0(|\overline{u}+s\tilde{u}(t)|)\leq
\Phi'_0(2|\overline{u}|)+\Phi'_0(2\|\tilde{u}\|_{L^{\infty}}),
\end{equation}
for every $s \in [0,1]$.  Now,  inequality \eqref{pot-suma}, H\"older's and Sobolev-Wirtinger's inequalities imply that
\begin{equation}\label{cota-i1}
\begin{split}
I_1&
\leq \Phi'_0(2|\overline{u}|) \|b_1\|_{L^1} \|\tilde{u}\|_{L^{\infty}}+\Phi'_0(2\|\tilde{u}\|_{L^\infty})
 \|b_1\|_{L^1}\|\tilde{u}\|_{L^\infty}
\\
&\leq C_2 \bigg\{ \Phi'_0(2|\overline{u}|) \|u'\|_{L^{\Phi}}
+\Phi'_0(C_3\|u'\|_{L^{\Phi}}) \|u'\|_{L^{\Phi}}\bigg\},
\end{split}
\end{equation}
where $C_2=C_2(T, \|b_1\|_{L^1} )$ and $C_3=C_3(T)$.
Next, by Young's inequality with complementary functions $\Phi_1$ and $\Psi_1$, we obtain
\begin{equation}\label{cota-i1-parcial}
 \begin{split}
\Phi_0'(2|\overline{u}|) \|u'\|_{L^{\Phi}}
&\leq
\Psi_1(\Phi_0'(2|\overline{u}|))+
\Phi_1(\|u'\|_{L^{\Phi}}).
\end{split}
\end{equation}
We have that any $N$-function $\Phi_0$ satisfies the inequality $x\Phi_0'(x)\leq \Phi_0(2x)$ (see \cite[p. 17]{rao1991theory}).
Moreover, since $\Phi_0\llcurly\Phi_1$ there exists $x_0=x_0(\Phi_0,\Phi_1,T)\geq 0$ such that $\Phi_0(2C_3x)\leq \Phi_1(x)$
for every $x\geq x_0$. Therefore, $\Phi_0(2C_3x)\leq \Phi_1(x)+C_4$ with $C_4=\Phi_0(2x_0)$. The previous observations imply that
\begin{equation}\label{cota-i1-parcial-segunda}
\Phi_0'(C_3\|u'\|_{L^{\Phi}}) \|u'\|_{L^{\Phi}}
\leq
C_3^{-1}(\Phi_1(\|u'\|_{L^{\Phi}})+C_4).
\end{equation}
From \eqref{cota-i1}, \eqref{cota-i1-parcial}, \eqref{cota-i1-parcial-segunda} and \eqref{cota-i2}, we have
\begin{equation}\label{cota-i1-i2}
\begin{split}
I_1+I_2
&
\leq C_5
\bigg\{
\Psi_1(\Phi_0'(2|\overline{u}|))
+\Phi_1(\|u'\|_{L^{\Phi}})
+\|u'\|_{L^{\Phi}} +1
\bigg\},\\
\end{split}
\end{equation}
with $C_5$ depending on $\Phi_0, \Phi_1, T, \|b_1\|_{L^1}$ and $\|b_2\|_{L^1} $.

Finally, using  \eqref{eq:control_inf},  \eqref{cota-diferencia-F} and
\eqref{cota-i1-i2}, we get
\begin{equation}\label{cota_inf_I}
\begin{split}
I(u)&
\geq\rho_{\Phi}(u')+\int_0^TF(t,u)\ dt
\\
&=\rho_{\Phi}( u')+ \int_0^T \left[F(t,u)-F(t,\overline{u})\right]\ dt
+  \int_0^TF(t,\overline{u})\ dt
\\
&\geq \rho_{\Phi}( u')
-C_5 \Phi_1(\|u'\|_{L^{\Phi}})
+\int_0^TF(t,\overline{u})\ dt-
C_5 \Psi_1(\Phi_0'(2|\overline{u}|))-
C_5
\\
&\geq
J_{C_5,\Phi_1}(u')
+H_{C_5, \Psi_1\circ\Phi_0'}(\overline{u})
-C_5.
\end{split}
\end{equation}

Let $u_n$ be  a sequence in $W^{1}\lphi$ with
$\|u_n\|_{W^{1}\lphi}\to\infty$ and we have to prove that $I(u_n)\to\infty$.
On the contrary, suppose  that for a subsequence,
still denoted by $u_n$, $I(u_n)$ is upper bounded, i.e. there exists $M>0$ such that $|I(u_{n})|\leq M$.
As $\|u_n\|_{W^{1}\lphi}\to\infty$, from Lemma \ref{infinito-a-prom-upunto},  we have $|\overline{u}_n|+\|u'_n\|_{L^{\Phi}}\to \infty$. Passing to a subsequence is necessary, still denoted $u_n$, we can assume that $|\overline{u}_n|\to \infty$ or $\|u'_n\|_{L^{\Phi}}\to \infty$.
Now, Lemma \ref{lem_coer} implies that the functional $J_{C_5,\Phi_1}(u')$ is coercive;
and, by \eqref{eq:propiedad-coercividad-phi0},
the functional $H_{C_5,\Psi_1\circ\Phi_0'}(\overline{u})$ is also coercive, then
$J_{C_5,\Phi_1}(u'_n) \to \infty$ or $H_{C_5,\Psi_1\circ\Phi_0'}(\overline{u}_n)\to \infty$.
From the condition \ref{item:condicion_a}  on $F$, we have that on a bounded set the functional $H_{C_5,\Psi_1\circ\Phi_0'}(\overline{u}_n)$ is lower bounded and
also $J_{C_5,\Phi_0'}(u'_n)\geq 0$.
Therefore,  $I(u_n)\to\infty$ as $\|u_n\|_{W^{1}\lphi}\to\infty$ which contradicts the initial assumption on the behavior of $I(u_n)$.

Let $\{u_n\}\subset W^{1}\lphi$  be a  minimizing sequence for the problem
\linebreak $\inf\{I(u)|u\inW^{1}\lphi\}$.
Since  $I(u_n)$, $n=1,2,\ldots$,  is upper bounded, the previous part of the proof shows that $\{u_n\}$ is norm bounded. Hence, by virtue of  \cite[Cor. 2.2]{ABGMS2015}, we can assume, taking a subsequence if necessary, that $u_n$ converges uniformly to a $T$-periodic continuous   function $u$. As $\{u'_n\}$ is a norm bounded sequence in $L^{\Phi}$,
there exists a subsequence, again denoted by $u'_n$, that converges to a function $v\inL^{\Phi}$ in the weak* topology of $L^{\Phi}$.
From this fact and the uniform convergence of $u_n$ to $u$, we obtain that
\[
\int_0^T\xi'\cdot u\ dt=\lim_{n\to\infty}\int_0^T\xi'\cdot u_n \ dt=
-\lim_{n\to\infty}\int_0^T\xi\cdot u'_n\ dt=-\int_0^T\xi\cdot v\ dt,
\]
for every $T$-periodic function $\xi\in C^{\infty}([0,T],\mathbb{R}^d)\subsetE^{\Psi}$.
Thus $v=u'$ a.e. $t\in [0,T]$ (see \cite[p. 6]{mawhin2010critical}) and $u\inW^{1}\lphi([0,T],\mathbb{R}^d)$.

Now, taking into account the relations $\left[L^1\right]^*=L^{\infty}\subset  E^{\Psi}$ and $L^{\Phi}\subset L^1$, we have that $u'_n$ converges to $u'$ in the weak topology of $L^1$. Consequently,  from the semicontinuity of $I$ (see \cite[Lemma 6.1]{ABGMS2015})  we get
\[I(u)\leq  \liminf_{n\to\infty}I(u_n)=\inf\limits_{v\inW^{1}\lphi_T}I(v).\]
Hence $u\in W^{1}\ephi_T$ is a minimun of $I$ on $W^{1}\lphi_T$.

For the second part of the theorem, assume that $u$ is a minimum of $I$ with $d(u',E^{\Phi})<\lambda$.
Since $I$ is G\^ateaux differentiable at $u$ (see  \cite[Thm. 3.2]{ABGMS2015}),
therefore $I'(u)\in (W^{1}\lphi_T)^{\perp}$. Thus,
\[\int_0^T D_y\mathcal{L}(t,u(t),u'(t))\cdot v'(t)dt =-\int_0^T D_x\mathcal{L}(t,u(t),u'(t))\cdot v(t)dt,\]
for every  $v\in W^{1}\lphi_T$.
From \cite[Eq. (26)]{ABGMS2015} we have
\[D_y\mathcal{L}(t,u(t),u'(t))\in L^{\Psi}([0,T],\mathbb{R}^n)\hookrightarrow L^1([0,T],\mathbb{R}^n);\]
and,  from  \cite[Eq. (24)]{ABGMS2015}, it follows that $D_x\mathcal{L}(t,u(t),u'(t))\in L^1([0,T],\mathbb{R}^n)$.
Consequently, from \cite[p. 6]{mawhin2010critical}
(note that $W^{1}\lphi_T$  includes the periodic test functions) we obtain the absolutely continuity of $D_y\mathcal{L}(t,u(t),u'(t))$ and that the differential equations in \eqref{ProbPrin-gral} are satisfied. The strict convexity of $\mathcal{L}(t,x,y)$ with respect to $y$ and the $T$-periodicity with respect to $t$ imply the boundary conditions in  \eqref{ProbPrin-gral} (see \cite[Thm. 4.1]{ABGMS2015}).\qed

 \section{An example}\label{sec:examples}

 In this section we develop  an application of our main result
so that the reader can appreciate the innovations that brings.

The main novelty  of our work is that we obtain existence of minima of $I$ associated with  lagrangian functions $\mathcal{L}(t,x,y)$ that do not satisfy a power-like grow condition on $y$.

In fact, it is possible to apply Theorem \ref{coercitividad-r} to lagrangians $\mathcal{L}=\mathcal{L}(t,x,y)$ with  exponential grow on the variable $y$. For example, suppose that
  \[\mathcal{L}(t,x,y)=f(y)+F(t,x),\]
with $f:\mathbb{R}^n\to\mathbb{R}$ differentiable, strictly convex and $f(y)\geq e^{|y|}$. We define for $n\geq 1$
\[\Phi(y)=e^y-\sum\limits_{i=0}^{n-1}\frac{y^i}{i!}.  \]
It is easy to see that $\Phi:[0,+\infty)\to [0,+\infty)$  is an $N$-function.
From \cite[Ex. 3, p. 85]{M} we know that $\alpha_{\Phi}=n$.
As consequence of \eqref{compl_ind} we have $\beta_{\Psi}=\frac{n}{n-1}<\infty$ and consequently $\Psi\in\Delta_2$.
From \eqref{delta2-potencias}, for every $1<p<n$ there exists $C_p>0$ such that
\[\Phi(rs)\geq C_pr^p\Phi(s),\quad r>1,s>0.\]
Then, the complementary pair $(\Phi,\Psi)$ and the $N$-function $\Phi^*(r):=r^p$ satisfy Lemma \ref{lem_coer} for every $1<p<n$.
Now, we fix arbitrary real numbers $1<p_0<p_1<p<n$ and we consider $\Phi_i=r^{p_i}$, $i=0,1$.
Then $\Phi_0\llcurly \Phi_1\llcurly \Phi^*$.
The conditions  \eqref{holder_cont-mu} and     \eqref{eq:propiedad-coercividad-phi0} become

\begin{equation}\label{eq:A6_ejem1}
  \left| \nabla F(t,x) \right|\leq b_1(t)|x|^{p_0-1}+b_2(t), \quad b_1,b_2\in L^1([0,T]),
\end{equation}
and
\begin{equation}\label{eq:ejem_propiedad-coercividad-phi0}
\lim_{|x|\to\infty}\frac{\int_{0}^{T}F(t,x)\ dt}{|x|^{(p_0-1)q_1 }}=+\infty,\quad q_1=p_1/(p_1-1),
\end{equation}
respectively. Since $n$ is an arbitrary positive integer, the pair $p_0$ and $p_1$ of real numbers with $1<p_0<p_1$ is also arbitrary.
For clarity,  assume that $F(t,x)=b(t)|x|^{\sigma}$, for some $1<\sigma<\infty$ and $b\in L^1([0,T])$.
We note that this $F$ satisfies \ref{item:condicion_a} and \ref{item:condicion_c}.
Now, we choose any $1<p_0$ with $p_0-1<\sigma<p_0$ and we take $p_1$ with $p_1>\sigma(\sigma-p_0+1)^{-1}$.
Then, \eqref{eq:A6_ejem1}
 and \eqref {eq:ejem_propiedad-coercividad-phi0} hold.
In conclusion, the action integral $I$ associated with the Lagrangian $\mathcal{L}(t,x,y)=f(y)+b(t)|x|^{\sigma}$   has minimum for any $1<\sigma$.

\def$'${$'$}

redpajama_set_name: RedPajamaArXiv


text:
Move Over Lean Six Sigma, Here Come Smart Operations
The digital transformation of manufacturing is forcing industrial companies to get smart. What's your company's IQ?
Charlie Chung | UPS
It sounds like an industrial-strength bad joke. But for manufacturers, it's no laughing matter.
After shelling out hundreds of billions of dollars collectively over the last few decades and completely reorganizing their factories to improve product quality, reduce waste and stay ahead in an increasingly competitive global industry, the world's manufacturers have been put on notice – they need to transform their operations all over again.
The culprit?
A host of disruptive technological and sociological developments — everything from cloud computing and machine-to-machine communication to a sea change in customer expectations about product customization and after-sales service — that no one foresaw a generation ago.
"The world's manufacturers have learned that they need to transform their operations all over again."
That's when the industry, following the lead of Toyota Motor Corp. and Motorola Inc., embraced the so-called "Lean Six Sigma" approach to industrial efficiency and quality control and focused on driving product defects and unnecessary procedures off the factory floor.
A big payoff
To be sure, the huge industry-wide investment in Lean Six Sigma — training costs alone were estimated to be $50,000 per worker — seemed to pay off for a lot of manufacturers, delivering significant improvements and cost savings at number of big-name outfits, according to a 2008 report from Bain & Co.
Companies like General Electric Co., Caterpillar Inc., Xerox Corp. and Johnson & Johnson took the lead. And as they biggies went, so went their rivals and suppliers.
By the beginning of this decade, Lean Six Sigma was pretty much corporate gospel.
But right from the start, there were some factors even Lean Six Sigma couldn't eliminate.
While manufacturers could streamline production to eliminate variability and increase efficiency, the world outside their factory walls continued to be a messy, maddening place that challenged their enhanced, just-in-time processes.
Bad weather delayed the delivery of crucial supplies from suppliers and finished goods to waiting customers.
Sudden changes in consumer tastes and expectations were not easily accommodated by the companies' approaches to efficient production.
More real-time data
The brittleness of supply chains became increasingly glaring in recent years as digitally driven advances in product design, manufacturing operations and post-sales monitoring, control and servicing suddenly provided manufacturers with petabytes and petabytes of real-time data on everything they cared about most.
Data on their own processes
Data on products in the field
Data on changing customer preferences
So much data it overwhelmed the rigid production systems and corporate cultures inspired by Lean Six Sigma, a discipline that – let's face it — was born back when 32-bit microprocessors were state-of-the-art, the Internet was in its infancy and social networks were flesh-and-blood, not bits and bytes.
It's a whole new world.
And the rate of change is accelerating. Companies can, of course, adhere to their old ways and ignore the inconvenient torrent of intelligence streaming out of their operations.
Or they can take a cautious, let's-wait-and-see before-we-make-this-a-priority approach.
Smart operational challenges
Many are making this shift. A recent survey of 100 manufacturing executives conducted by the International Data Corporation (IDC), a market intelligence and advisory services company, found that about half the companies that participated were moving slowly at best.
But that same survey, conducted in partnership with UPS, found the other half of respondents were making big operational changes — smart operational changes — to harness the insight now spewing out of their value chains.
That insight boosts revenue and profit. Research from MIT Sloan shows that companies aggressively pursuing the digital transformation of their operations earn profits 26 percent higher than those that haven't.
The move toward "Smart Operations" has become an industry-wide priority, like Lean Six Sigma was in the 1990s.
Smart-Operations champions don't see it as a repudiation of Lean Six Sigma.
They see it is an extension of the philosophy, marrying the continuous improvement aspects of Lean Six Sigma with real-time big data to "super charge" a manufacturer's whole value chain, from production to marketing to after-sales service.
So it's not a new fad.
It's the fine-tuning of a well-established approach to efficiency and quality control that promises to make plants and processes intelligent and self-optimizing.
And here's the great news: Done right, it won't require hundreds of billions of dollars in new investments industry-wide.
That sound you hear?
That's a collective sigh of relief coming from those responsible for profitability.
The IDC-UPS study, which offers a road map for companies keen to implement Smart Operations themselves, found that manufacturers that had already adopted the approach had a couple of things in common:
They were quick to deploy both sensors and mobile devices to monitor and analyze products in the field.
That transforms the customer experience, permitting the manufacturers to perform real-time predictive analysis and to initiate corrective action before the product fails.
Those companies were also enjoying top- and bottom-line growth because – outside the warranty period –those interventions generate service revenue.
The smart operations leaders also have connected their assets.
Their managers don't just understand the way machines work — or don't work — together.
The machines themselves understand it, too, and they exchange data to anticipate and minimize the impact of unexpected surprises on the production line and to identify and resolve them automatically.
Another prominent success factor
It's true: Achieving that level of connectivity — turning a manufacturing value chain into a network that shares just enough information — requires an investment in data acquisition and analytics as well as better collaboration within the organization.
The IDC-UPS research also revealed another prominent success factor: The companies taking a lead in Smart Operations are also deepening their relationships with service partners, including their logistics providers.
That frees them up to focus on their key internal competencies while leveraging the scale, technology and skills of those providers.
What does all that mean? Well in the case of a partner like UPS, it means manufacturers don't have to build their own smart global supply chain and logistics network.
"Smart Operations are critical to the success of manufacturers with limited resources."
UPS already did that. It's a network that can give them real-time insight into their supply chains and help them initiate corrective action before problems occur.
Partnering also keeps a lid on the investment spend. UPS has that covered, too.
Every year, UPS invests about $1 billion dollars to make its small parcel, air and ground freight and logistics solutions business smarter, faster and better.
Manufacturers don't have to do everything at once. Obviously, it's a long and comprehensive process to advance from Lean Six Sigma to Smart Operations.
But it's worth it. Smart Operations are critical to the success of manufacturers with limited resources, especially as they work to serve their increasingly demanding customers fast and flexibly.
Charlie Chung is Senior Marketing Manager at UPS Supply Chain Solutions
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text:
set -ex
OPENBLAS_VERSION=4a4c50a7cef9fa91f14e508722f502d956ad5192
if [[ ((! -e $DEPS_PATH/lib/libopenblas.a) && -z "$CMAKE_STATICBUILD") ||
          ((! -e $DEPS_PATH/lib/libopenblas.so) && -v CMAKE_STATICBUILD) ]]; then
    # download and build openblas
    >&2 echo "Building openblas..."

    download \
        https://github.com/xianyi/OpenBLAS/archive/${OPENBLAS_VERSION}.zip \
        ${DEPS_PATH}/openblas.zip
    unzip -q $DEPS_PATH/openblas.zip -d $DEPS_PATH
    pushd .
    cd $DEPS_PATH/OpenBLAS-${OPENBLAS_VERSION}

    # Adding NO_DYNAMIC=1 flag causes make install to fail
    CXX="g++ -fPIC" CC="gcc -fPIC" $MAKE DYNAMIC_ARCH=1 DYNAMIC_OLDER=1 USE_OPENMP=1

    if [[ -v CMAKE_STATICBUILD ]]; then
        # We link and redistribute libopenblas.so for cmake staticbuild
        # cf https://gitlab.kitware.com/cmake/cmake/issues/16221#note_143330
        patchelf --set-rpath '$ORIGIN' --force-rpath libopenblas.so
    fi

    $MAKE PREFIX=$DEPS_PATH install

    if [[ -z "$CMAKE_STATICBUILD" ]]; then
        # Manually removing .so to avoid linking against it
        rm $DEPS_PATH/lib/libopenblas*.so
    fi

    popd
    if [[ -z "$CMAKE_STATICBUILD" ]]; then
        ln -s libopenblas.a $DEPS_PATH/lib/libcblas.a
        ln -s libopenblas.a $DEPS_PATH/lib/liblapack.a
    fi
fi

redpajama_set_name: RedPajamaGithub


Life insurance products have seen a transformation in the last five years. Driven by customer needs, insurance companies have built products to cater to the new-age insuree.
The Insurance Regulatory and Development Authority of India (IRDAI), to keep up with the changing market dynamics, has released new product guidelines for linked and non-linked life insurance policies. The good news is that these guidelines are set to significantly benefit insurance buyers.
This can be a game-changing alteration to the current provisions. With this, an insured person can withdraw partial amounts when in urgent need of money. However, this will be allowed only in certain circumstances such as critical illness, disability due to accident, etc. It will ensure that a life insurance policy comes to the rescue when the insured is facing a health scare, and not just in case of an unfortunate fatal event. The benefit to withdraw money will hugely help how life insurance policies are perceived by insurance buyers.
This is another important change that will benefit customers. Most people buy life insurance in their 20s or 30s. This means a commitment of 20-25 years is made very early in life. As they grow older and experience life, their needs and goals tend to change. A policy bought earlier may not be line with their life vision any more. The flexibility to change the tenure of their policy during the course of their life will help buyers to align their investment with their goals and utilise their resources as per their needs.
A win for pension policyholders will allow them the flexibility to buy annuities from any insurance company. So far, one had to renew their policy from the same insurer. This locks in the customer who has to give business to the same insurance company irrespective of the rate of interest offered. The new guideline has put the ball in the customer's court. They get to decide whom to buy from depending on the service and interest rate provider by different players. Increased competition will bring more value to the customer.
Related: Why do "I" need life insurance?
As per the new proposed norms, surrender value for a non-linked policy will depend on the premium paying term. For policies with a term less than 10 years, the policyholder will acquire a guaranteed surrender value if premiums have been paid for two consecutive years. For policies with a term greater than 10 years, at least three premiums have to be paid for a guaranteed surrender value. On the other hand, more time has been allotted for policyholders to reinstate lapsed policies. In case of lapsed policies, customers will be able to revive them up to five years from the date of invalidation, offering more leeway in continuing with an insurance plan. This period is currently much shorter at two years.
With respect to pension products the pensioner can make a one-time withdrawal, also known commutation, for up to 60% of the market value, while the balance 40% goes in purchasing an annuity plan. Previously, commutation was permitted only up to 1/3rd of the corpus.
The settlement period has been extended to 10 years from the current five years for linked products, where policyholders can receive death or maturity benefits over a staggered duration. This helps the policyholder manage their finances better over the long term.
On the flip side, the minimum benefit has been standardised across all age groups at seven times the annual income for regular premium products and 1.25 times for single premium policies. Previously, this ratio was the same for insured customers above the age of 45, but higher at 10 times for customers below the age of 45.
IRDAI has put forward these proposed guidelines to all stakeholders and asked them to share suggestions by 15th November 2018.
Wondering what the BHIM App is and why you should care? Here's what you need to know!


text:
WALTER (WALLY) REYNOLDS MOSCRIP
Date of Passing: Aug 30, 2021
On August 30, 2021 Walter (Wally) Reynolds Moscrip passed away at Third Crossing Manor in Gladstone, MB at the age of 90.
Walter was born in Ogilvie, MB on the Paynter family farm. He grew up and attended school in Gladstone. At the age of 20, he joined the Canadian National Railway and was stationed at various communities across Manitoba, Saskatchewan, and Ontario. It was in Plumas, MB that he met and married Jean Bell. Walter retired from the railway in 1989 and he and Jean returned to their roots and settled in Gladstone. They became active members of the community serving on numerous committees with the Legion, golf and curling club, as well as other organizations. His two passions were golf and curling and over the years participated in many tournaments and bonspiels. He enjoyed many afternoons playing cribbage and pool at the Legion.
Walter was a kind, generous man; a true gentleman. He was dedicated to his family and when life presented its most difficult challenges, he remained strong.
His memory will be cherished by his daughter Lois (Doug) MacCallum, four grandchildren, eight great-grandchildren, sisters-in-law, Elaine (Bill) Schettler and Shirley Bell, as well as cousins, nieces, nephews, and friends.
Walter was predeceased by his wife Jean, sons, Craig, Beven, and an infant son who died at birth, grandson Charles, brother Stuart (Doreen), brother-in-law Jack Bell, parents, Charles and Isobel Moscrip, and parents-in -law, Cameron and Mary Bell.
Cremation has taken place. A graveside service will be held at Gladstone Cemetery on Thursday, September 16, 2021 at 1:30 p.m. The Cemetery is located two miles South of Gladstone on Highway #34.
The family thanks the Gladstone Homecare workers, the staff at Third Crossing Manor (you are true heroes) and Nurse Practitioner, Rebecca Krause.
If friends so desire, memorial donations may be made to Seven Regions Health Foundation for Third Crossing Manor, Box 1000, Gladstone, Manitoba R0J OTO.
Clarke's Funeral Home Gladstone/MacGregor in care of arrangements.
As published in Winnipeg Free Press on Sep 11, 2021
Our sincere sympathy to the family in the passing of your Dad. May you find comfort in knowing we share in your sorrow. Rob & Gail Transcona. - Posted by: Rob & Gail Schledwitz (Friend) on: Sep 11, 2021


Hoping someone has an answer for this strange issue I've come across.
I was trying to deploy a couple of software updates to 2 servers (2008 r2 64 bit) machines. I downloaded the updates and deployed them via a Package in SCCM 2012. I deployed it as an 'Available' Package so it would show in the servers Software Centres. The Package never did appear in Software Centre on those machines.
I then created a new collection, added 1xWin 7 64 bit machine and 3x2008 R2 64 bit servers (2 of them are the same ones as above).
I deployed to this collection 2 x Programs and 2 x Applications.
On the 2 x original servers only the Applications appear, not the the Programs. On the other server and Win 7 machine all of the deployments appear.
Does anyone know why the Programs wouldn't be appearing in the Software Centre for these 2 servers?
Software center might not show it if there are multiple users logged in. It will only be shown to the user that got the highest (lowest? Can't remember it) session id.
Thanks for the reply and link dekac99.
I know the Applications and Program deployments I have created are correct and should be displaying in Software Centre because they are displaying on other computers logged in with the same user accounts as the computers that the deployments are showing in.
Running the 'Package and program deployment status messages for a specified client and deployment' tells me that the deployment has been received by the client.
Download ConfigMgr 2012 Toolkit and use the Deployment Monitoring Tool. It can be used to troubleshoot such issues.
I ran the tool, the deployment doesn't appear in the list.
I have a second System Centre environment that I tested with. Below is what happened.
I add two Server 2008 R2 machines to a new collection. I deployed 1 application and 1 program to the collection.
Initially 1 server displayed both deployments and 1 server displayed only the 1 application.
I went through and compared the notification log file on both machines. They were similar. Both servers tell me that both deployments are seen.
I wanted to double check that I didn't miss anything so I deleted the notification logs of both servers and removed the deployments.
Re-added the deployments and let the log re-create.
Log recreated and both show that they see both deployments. Neither server now displays the program in software centre, they both only display the application.
I can see the Programs deployed.
If I look at the properties of a Program that was deployed and required it tells me that all the machines in that collection installed it - though it never appeared in the Software Centre on the server computers, only my desktop computers.
The other Available Programs still don't appear on the server machines.
Tested with two people logged on. The lowest session ID saw all the deployments. Get them to log off so only 1 person is logged on then they see all deployments.
Get second person to log back on and they only see the application desployments, not the program deployments.
Log off the person with the lowest session ID and the other person now sees all deployments.
Has Microsoft ever officially documented this behavior? I'm running into this, and I vaguely remembered this was an issue, but not much comes up in web searches. It took me a while to even find this post.


text:
Home Case Index All Cases Central Excise Central Excise + Tri Central Excise - 2020 (7) TMI Tri This
2020 (7) TMI 317 - CESTAT NEW DELHI - Central Excise
Valuation - manufacturing / construction at site - Benefit of exemption - pre-fabricated canopies supplied by the appellant to oil companies - Department is of the view that the transaction value of the pre-fabricated canopies/structures should be taken by deducting the cost of erection, installation and commissioning of these canopies - benefit of N/N. 8/1996-CE dated 23 July 1996 and Notification No. 12/2012-CE dated 17 March 2012 - Cenvat credit on all the inputs which have gone in manufacture of both dutiable as well as non-dutiable goods - HELD THAT:- The appellant is having an independent manufacturing unit at Bhopal where the appellant is carrying out fabrication and manufacturing activity. The pre-fabricated structures classifiable under Central Excise Tariff Heading 73089010 are being taken in the CKD/SKD condition to different locations all over India. Manufacturing is the primary work undertaken at the factory of the appellant and the pre-fabricated structures are only erected, installed and commissioned at the site of the various petrol pumps. Applicability of the Notification No. 12/2012-CE - HELD THAT:- The construction work of a road or flyover is being primarily undertaken at the site and only some components, blocks are manufactured by the contractor at different site. In that case the benefit of the Notification No. 12/2012-CE is available. However, in case of the appellant, the primary activity is of manufacturing and fabrication and then only goods are being taken for assembly, erection or commissioning at a given petrol pump. We are of the view that by no stretch of imagination, the benefit of Notification No. 12/2012-CE can be extended to a manufacturing activity which is being undertaken at a factory and thereafter the fully manufactured pre-fabricated structures are taken in the form of the CKD/SKD condition for installation at the given site.pplicability of the Notification No. 12/2012-CE is concerned, the construction work of a road or flyover is being primarily undertaken at the site and only some components, blocks are manufactured by the contractor at different site. In that case the benefit of the Notification No. 12/2012-CE is available. However, in case of the appellant, the primary activity is of manufacturing and fabrication and then only goods are being taken for assembly, erection or commissioning at a given petrol pump - By no stretch of imagination, the benefit of Notification No. 12/2012-CE can be extended to a manufacturing activity which is being undertaken at a factory and thereafter the fully manufactured pre-fabricated structures are taken in the form of the CKD/SKD condition for installation at the given site. Cenvat credit on all the inputs which have gone in manufacture of both dutiable as well as non-dutiable goods - HELD THAT:- It is seen that the Adjudicating Authority has gone in detail on this subject and the impugned order has discussed in the financial statement of the company and it is only after a meticulous perusal of the trial balance and other financial details that the Adjudicating Authority reached conclusion that the respondent/assessee had maintained a separate record with regard to inputs which have gone in the manufacture of non-taxable goods/services. The Department has not adduced any concrete evidence to contradict the findings given by the Adjudicating Authority. There is, therefore, no substance in the appeal filed by the Department. Appeal allowed in part.
2018 (7) TMI 2123 - CESTAT NEW DELHI


WASHINGTON, Jan. 23 (CGTN) -- NASA will remove an asteroid from its orbit for the first time in 2022 with its Double Asteroid Redirection Test (DART) spacecraft.
The asteroid, called Didymoon, is a moon asteroid around 150 meters long that is currently in the double asteroid system Didymos. Didymoon is the most accessible asteroid of its size from the Earth.
The DART spacecraft will be NASA's first attempt to demonstrate what is known as the kinetic impactor technique which will strike the asteroid to shift its orbit to defend against potential future asteroid impacts, according to a statement by Lindley Johnson, planetary defense officer at NASA Headquarters in Washington.
NASA said the DART spacecraft will achieve the kinetic impact by deliberately crashing itself into the asteroid at a speed of approximately six kilometers per second. It also said the collision will change the speed of the moonlet in its orbit around the main body by a fraction of one percent, enough to be measured using telescopes on the Earth.
According to NASA, the spacecraft will be launched between December 2020 and May 2021, and will intercept Didymos' moonlet in early October 2022 when the Didymos system is within 11 million kilometers from the Earth.


This entry was posted on Wednesday, August 30th, 2017 at 12:11 pm	and is filed under Scales And Tails. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.


Romero Britto 20oz Double-Wall Tumbler. Features the iconic pop artist's signature use of vivid colors and bold graphic design. Heart icon design with red lid. FDA Approved, Food Safe. Hand-wash. Acrylic.
SKU: RB017. Category: Romero Britto Collection. Tag: britto.