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BEACH SQUAD
Sports - NationalBy Editor June 13, 2018 Leave a comment
Honiara beach soccer players named By CARLOS ARUAFU THE Solomon Islands Football Federation (SIFF) has revealed the names of the Honiara based beach soccer players for the training squad. The training squad will be preparing for the OFC Championship in August, in Tahiti. Eleven names were announced on Tuesday. Most of the players are former…
GP welcomes SPREP initiative
News - NationalBy Editor June 11, 2018 Leave a comment
GUADALCANAL Province Premier Anthony Veke says that the Nature Park initiative at Barana village in Northwest Guadalcanal with the Secretariat of the Pacific Regional Environment Programme's (SPREP's) support is welcoming. Mr Veke said that under his leadership, this kind of initiative is welcoming and he really supports this. Guadalcanal province is supporting such support and…
In Court with Assumpta Buchanan
HCC councillor denies allegation A HONIARA City Council (HCC) Councillor who is accused of obtaining about $89,000 from a woman over the sale of a plot of land at Ranadi, Honiara has denied the allegation against him on Thursday. John Wale Urifo entered not guilty pleas to 15 counts of false pretence before Principal…
St Nicholas College crowns four queens
SAINT Nicholas College has crowned four queens, to represent the school in the upcoming ACOMSIC games, during a Beauty Pageant show, last Friday. The four queens are Roviana Boyers, Tatiana Bugotu, Sabrina Habu, and Marrilyn Koga. Ms Boyers will represent the school in the junior modern category, while Ms Habu in the senior category. For…
Acting PM calls for tough intelligence network
THERE is a need to strengthen national and regional intelligence networks in the region. That's according to the Acting Prime Minister Manasseh Sogavare when speaking as the guest speaker at the opening of the two day South Pacific Intelligence Dialogue held in Honiara recently. The Finance Minister said it is an important matter of concern…
SI Praised For Post- RAMSI Progress
THE Solomon Islands was praised for its efforts and progress made in the areas of economic governance and reforms to improve the country's economy, as well as financial management in the post-RAMSI era. This was acknowledged by the New Zealand Prime Minister Rt. Hon. Jacinda Arden when she met with Prime Minister Rick Houenipwela, during…
PIDF DEFERRED
THE Pacific Islands Development Forum (PIDF) Leaders' summit and conference which was scheduled to be held in Honiara from the 4th-6th July, is being deferred, the Secretariat has confirmed. On Sunday the PIDF Secretariat stated that the postponement was made following a request from the host and Chair, Prime Minister Rick Houenipwela. A statement was…
DEATH PROBE
News - NationalBy Editor June 9, 2018 Leave a comment
Police: death of man, 50, suspicious By ASSUMPTA BUCHANAN POLICE are continuing their investigation into the 'suspicious death' of the 50-year-old man at Kaio Village in Tasimboko area, northeast Guadalcanal. Supervising Police Commissioner Gabriel Manelusi on Thursday said the Tetere Police and the Guadalcanal Provincial Police Commander (PPC) are working very hard on this investigation.…
City-based groups for MACFest
ONLY Honiara-based cultural groups will represent the country in next month's Melanesian Arts and Culture Festival (MACFEST). This was due to drastic budget cuts faced by the MACFEST national organising committee. In a statement on Tuesday the committee said, Honiara-based Cultural groups from the country's nine provinces will represent Solomon Islands at the upcoming Melanesian…
Taiwan medical team tour S/Malaita
A MOBILE medical team from Taiwan is visiting the southern parts of Malaita this week. The team will carry out medical services in Small Malaita, East and West Are'Are for the next nine days from 5th to 13th this month. The statement said that in response to the dire medical needs in the rural areas,…
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Q: Changing CSS in dynamically generated HTML How can I get access and change CSS like the font color in completely dynamically generated HTML out of XML, if I cannot use an ID as I do not know in advance how many of the same HTML-tags will be generated and which of two different (red/green) colors it will need? (I am new here and very new in JS)
Data will be generated and collected on the server. I get them back and sort them in a "fieldset" and further more in a "collapsible". Data "var a, b and c" come together in one line (fieldset) and should be in a specific font color. So for "var d" there will be a "xx" or "yy" coming back from XML for this font color and be saved in "var d" and therefore the font color should be "red" or "green".
Part of my script in short:
function addPart(currentIndex,currentPart)
{
    var a = $(currentPart).find("a").text();
    var b = $(currentPart).find("b").text();
    var c = $(currentPart).find("c").text();
    var d = $(currentPart).find("d").text();
    if(d === "xx") {
        $("div.productColor").css({"color":"green"});
    } else {
        $("div.productColor").css({"color":"red"});
    };

    $("#shoppingTableDiv").append (
        "
"

            + "
"
                + "
"
                        + "<fieldset class='ui-grid-b'>"
                            + "
"
                            + a
                            + "
"
                            + "
"
                            + b
                            + "
"
                            + "
"
                            + c
                            + "
"
                        + "</fieldset>"
                + "
"
            + "
"
          + "
");

    $('#shoppingTableDiv').collapsibleset('refresh');

}

A: I didn't fully understand how you want to color your results, but I think you can use your fieldset classes.
For example:
.ui-grid-b div { color: red; }
.ui-grid-c div { color: green; }

If you don't know how many fieldsets you're going to have and you can't use the fieldset classes you showed in your code example, you might find the nth-child selector useful.
Maybe something like this:
fieldset:nth-child(2n) div { color: red; }
fieldset:nth-child(2n+1) div { color: green; }

redpajama_set_name: RedPajamaStackExchange


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People already driving electric cars have fallen in love, as shown by off-the-charts customer satisfaction ratings. But for the vast majority of Americans who aren't yet driving electric, how do we get them charged up?
In our last two blog posts in this series, we talked about the large number of people we need to switch to electric vehicles (EVs) if we're to have a shot at averting the worst of climate change, and the types of public policies -- like cash rebates -- that will incentivize people to drive EVs.
1) Find Smokey the Bear.
It's probably not a coincidence that the BMW i3, the sporty new electric car, had one of its best-selling months following the release of a hilarious Super Bowl television ad, with more than 18 million views on YouTube -- not to mention the millions who saw it during the big game. You also may have seen this Cadillac ELR EV commercial, aimed at the luxury car buyer. Another ad, which cleverly parodies the Cadillac ad, is about the Ford C-Max Energi Plug-In Hybrid and is seriously funny. But these EV ads are the exceptions to the rule. The vast majority of car commercials we see are for gas-guzzling SUVs and pick-up trucks. Come on, automakers.
By contrast, a switch to electric vehicles will lead to cleaner air and lower carbon emissions -- public goods that benefit all Americans. Government agencies and nonprofits have invested in public service announcement (PSA) campaigns to promote seat belt use (i.e., "Click It or Ticket") and prevent wildfires (i.e., Smokey the Bear), and these campaigns have changed America for the better. Public utilities fund direct-mail campaigns about energy efficiency, and we've made big improvements on that front too. Why not PSAs and mailings re EVs?
A number of persistent myths surround EVs: They're too expensive, not any better for the environment, unsafe, slow, can only drive short distances, are a hassle to charge, etc. Chris Paine, the documentary filmmaker behind Who Killed the Electric Car? and Revenge of the Electric Car, busts several of these myths in this entertaining Washington Post article, and Sierra Club does it here. We need to do more to set the record straight.
3) Events: Get butts in seats.
One of the many counterintuitive findings of social science research is that people's attitudes tend to flow from their actions. Most EV drivers will tell you that they didn't really understand what all the fuss was about until they actually rode in an electric car and experienced first-hand how fun, cool, and fast these cars are. We need to give people more opportunities to test drive EVs. During last year's National Drive Electric Week, upwards of 90,000 people attended events in 150 cities, and thousands of people took test drives. The fifth annual National Drive Electric Week will be September 12-20 this year, and organizers (Sierra Club, Plug In America, Electric Auto Association, and local partners) are already gearing up to make it bigger than ever.
Workplace test-drive events are another great opportunity to get people talking to their friends and co-workers about EVs -- and to take them for a spin during their lunch break. Consider organizing one yourself.
It's terrific that actor/activist Ed Begley Jr. talks up his electric car, and we're heartened to know that celebs like George Clooney and Cameron Diaz drive electric too. But we need more high-profile celebrities, politicians, and business leaders driving EVs and actively promoting them in traditional and social media. Readers of People, The Hill, Buzzfeed, and the Wall Street Journal may have varying role models -- and most of them buy or lease cars.
5) Understand and appeal to new people.
EVs can excite a wide range of people with all sorts of interests. The new documentary film The Burden, to be screened in the coming days at a film festival in D.C., makes the argument that EVs are appealing to veterans and national security hawks who want to slash our dependence on foreign oil. Last week thousands of car-racing enthusiasts took to the streets of Miami to watch Formula E, where professional race-car drivers show off their talents in blindingly fast EVs. Those of us looking to accelerate the EV market need to seek out and understand all sorts of audiences.
The Center for Sustainable Energy conducted a survey showing that California EV drivers prioritize issues like carpool-lane access, fuel-cost savings, and emissions reductions -- and fit certain demographics. But we need much more information about what messages, which media outlets, and which spokespersons will excite people. If your company or group is researching this information, don't be greedy! Share it with other EV advocates, utility companies, government officials, and car companies looking to accelerate the EV market. Working smartly together, we can electrify the masses.
Mike Walker is the founder and CEO of AlterAction, and his blog is here. Gina Coplon-Newfield is the Sierra Club's Director of Electric Vehicles Initiative. This is the last in their joint blog series.


text:
Русла́н Я́ковлевич Во́вин () — советский и российский учёный-психиатр, доктор медицинских наук, профессор, Заслуженный деятель науки Российской Федерации.

Биография
Родился в Ленинграде в 1928 году. Во время Великой Отечественной войны участвовал в обороне города и был награждён медалью «За оборону Ленинграда».

В 1952 году окончил 1-й Ленинградский медицинский институт имени академика И. П. Павлова. В течение следующих 5 лет работал психиатром в Казахстане. С 1957 года его профессиональная деятельность была связана с Психоневрологическим институтом им. В. М. Бехтерева. Прошёл путь от младшего научного сотрудника до руководителя отделения биологической терапии психически больных, которое он возглавлял с 1976 по 2001 год.

Р. Я. Вовин был одним из ведущих специалистов в области в клинической психиатрии и клинической психофармакологии, создал школу психофармакотерапевтов, сформулировал концепцию закономерностей терапевтического регресса бредовых состояний при шизофрении, исследовал закономерности формирования терапевтической резистентности при эндогенных психозах, что позволило разработать ряд дифференцированных и эффективных подходов для её преодоления.

Им была также впервые разработана оригинальная клинико-биологическая типология дефицитарных состояний при шизофрении, в соответствии с которой в лечебную практику была внедрена система лекарственных и психосоциальных способов компенсации симптомов дефекта. Приоритетными стали и работы Р. Я. Вовина и возглавляемого им коллектива исследователей в области вторичной профилактики фазных аффективных расстройств с использованием антиконвульсантов. В последние годы жизни он уделял большое внимание изучению новых поколений психотропных средств (нейролептиков и антидепрессантов) селективного действия.

Под его руководством было проведено 10 научно-исследовательских работ на кооперационной основе с рядом зарубежных стран, одна из которых по плану ВОЗ. Автор более 210 научных работ, ряд из которых опубликован за рубежом, в том числе 5 монографий и 10 методических рекомендаций. Многократный участник международных форумов. Под руководством Р. Я. Вовина было выполнено 5 докторских и 15 кандидатских диссертаций.

Р. Я. Вовин был членом международной академии «Медицина и психиатрия», Нью-Йоркской академии наук, Европейской ассоциации психиатров, правления Санкт-Петербургского общества психиатров и наркологов. В последнем он возглавлял этический комитет.

В мае 2018 году в ФГБУ «НМИЦ ПН им. В. М. Бехтерева» Минздрава России прошла Всероссийской научно-практической конференции с международным участием «Клиническая психиатрия 21 века: интеграция инноваций и традиций для диагностики и оптимизации терапии психических расстройств», посвящённой памяти профессора Руслана Яковлевича Вовина (90-летию со дня рождения).

Основные работы
 Авруцкий Г. Я., Вовин Р. Я., Личко А. Е., Смулевич А. Б. Биологическая терапия психических заболеваний. Ленинград: Медицина, 1975. 312 с.
 Вовин Р. Я., Аксёнова И. О. Затяжные депрессивные состояния. Ленинград: Медицина, 1982. 192 с.
 Фармакотерапевтические основы реабилитации психически больных / Под ред. Р. Я. Вовина, Г.-Е. Кюне. Москва: Медицина, 1989. 288 с.
 Шизофренический дефект (диагностика, патогенез, лечение): Коллект. монография. Санкт-Петербург, 1991. 171 с.

Изобретения
 Способ коррекции апато-абулических состояний при шизофрении. Вовин Р. Я., Голенков А. В., Каменецкий В. К., Лосев Н. А., Фактурович А. Я. Авторское свидетельство SU 1657189 A1, 23.06.1991. Заявка № 4429463 от 23.05.1988.
 Способ диагностики эндогенных психозов. Вовин Р. Я., Иванов М. В., Акименко М. А., Разоренов Г. И., Разоренова Т. С., Мазо Г. Э. Патент на изобретение RU 2153290 C1, 27.07.2000. Заявка № 99119542/14 от 16.09.1999.
 Способ диагностики аффективных психозов. Вовин Р. Я., Иванов М. В., Акименко М. А., Мазо Г. Э., Разоренов Г. И., Разоренова Т. С. Патент на изобретение RU 2170056 C1, 10.07.2001. Заявка № 2000113164/14 от 29.05.2000.

Награды
 Медаль «За оборону Ленинграда»
 Значок «Отличник здравоохранения»
 Заслуженный деятель науки Российской Федерации

Примечания

Литература
 Ю. А. Александровский. История отечественной психиатрии: В 3 т. — Т. 3. Психиатрия в лицах / Ю. А. Александровский. — М.: ГЭОТАР-Медиа, 2013. — 768 с.: ил.)
 Журнал неврологии и психиатрии, 9, 2004 г.
 Научно-практический журнал «Психиатрия и психофармакотерапия им. П. Б. Ганнушкина» № 02 2003 г.

Психиатры России
Психиатры СССР
Врачи по алфавиту
Доктора медицины
Доктора медицинских наук
Отличники здравоохранения (СССР)

redpajama_set_name: RedPajamaWikipedia


Felt that theirs was quite good! Hot and spicy plus not overly sweet (: The Chinese believe that we have to drink ginger tea after eating the crab to ease the "coolness" of it! My dad and the boy thought it was spicy enough though!
Personally, I feel that Li Bai really dishes up quality dishes, no wonder its a veteran in the Cantonese food scene here. Actually, not just Cantonese, in the Chinese food scene! I really applaud the wonderful service we got at Li Bai (esp with the waitress patiently picking out the crab meat for us!) and their friendliness + quick response! Overall, dining at Li Bai is a joy and they were fully booked the night we went (no surprises) so I'd recommend making reservations!
The boy bought some TWG pastries over today! And we had a mini pig-out in my room!
Overall, I think TWG pastries are not bad, and the price comes with the packaging and the chi-chi feel of the desserts but they certainly are not the best in Singapore, neither are they very value for money I feel.
Having read about this new cafe specialising in chocolate, I knew I had to ask the boy to bring me there to try out what good stuff they have and I'm glad I wasnt disappointed at all!!! The boy and I went there after Pave and we were glad we decided to come here!
Watch out for the beetle to spot the cafe!
Check out the cosy and rustic decor of the cafe! I really feel that there's a good warm homey feel to this quaint cafe and the atmosphere really relaxes you. I guess its something to do with the dim lights and the strong cold aircon plus the really comfy sofa seats we were sitting on!
Their cafe is beautifully decorated in preparation for Halloween which they have some special promotion going on for!
Get your own water from this super old-school water dispenser! I like!
BOY OH BOY IS THIS A MUST-TRY!!!!!!! The boy and I declared it the best waffles we've eaten to date! It has the best airy-to-crispy ratio HANDS-DOWN!!! Once served, we would tell immediately from the look of the waffle that it would be crispy and we were not wrong when our knife cut into the waffle and it was all crackly and we could hear the crisp sound of our knife cutting through the waffle! The waffle was light, super fluffy and airy, so soft, but crispy at the same time, you really really have to try it to know what I mean!!
This was the FASTEST the boy and I ever finished a plate of dessert, we just attacked it as if someone was gonna kill us at that instant if we didnt finish it. I think we took all but 5mins to finish the waffle and I'm not kidding, it was that good.
Spot the small little black spiders at the top left corner of the glass panel!
We wanted to try their signature Black Out cake but unfortunately its sold out :(( its okay, we'd DEFINITELY be back for their awesome waffles + maybe try their pancakes/savory menu and their black out cake + other desserts they have to offer! Tonight they had chocolate cheesecake, chocolate eclair, rocky road bars, scones (raisin and plain ones), chocolate truffles (baileys, rum& raisin, etc). Sold out items include blackout cake, tiramisu, brownie, apple crumble, eton mess.
We'll surely be back to try out your other awesome food!! I'd also buy the scone next time cos they looked really inviting (: Plus there's no GST or service charge! The waffle's only $8.50 and I'd say its the BEST WAFFLES hence the BEST VALUE FOR MONEY WAFFLES IN SINGAPORE! Please go and try it even though its at a rather ulu place!
After the ice cream at Brownice, the boy and I decided to go to Drips Bakery Cafe after reading so many great reviews about their tarts!
It was rather easy finding the cafe if you drive I think! Its like the 2nd building on your left when you turn into the road and its right in front so keep your eyes peeled! If you miss the cafe, just look out for the red sign on the building that says 82 Tiong Poh Road! (pic below) Thats the building you're looking for! Park your car and then walk down and you'd see it soon enough!
Nestled amongst old flats, I found the place really cosy and feels as if you're not in Singapore! Because you certainly dont get to see such old school flats anymore in Singapore! These 2 storey flats are a rare sight now I feel.
The green tea isnt very strong, quite mild actually, wouldnt recommend this.
Love the environment and the ambience of the cafe! I especially love the part of the cafe we were seated at (further in not the part with the sofas and stuff) because I liked how they used glass for a part of the ceiling to let the natural light shine in. That made the cafe really bright and cheery! (: Would definitely go back to try out the other tarts they have available and I feel that the cinnamon egg tart is definitely an exception to the other yummy goodies found in the cafe!
Novus - A wonderful experience!


text:
Back to E&E News index page.
Utility-only proposal fails to move the needle on Capitol Hill
Josh Voorhees and Robin Bravender, E&E reporters
E&E Daily: Tuesday, June 22, 2010
With the clock winding down on efforts to pass a climate bill this year, Senate Democrats are still adding pages to their policy playbook.
A number of liberal Democrats -- who so far have been unable to attract enough moderate lawmakers to reach the 60-vote threshold needed to pass a climate bill -- are now begrudgingly considering a scaled-down cap on greenhouse gas emissions that would be imposed only on the utility sector as a long-shot effort to cobble together the needed support.
"It's not my first choice, but it's important that we get started in what we're trying to do," Sen. Joe Lieberman (I-Conn.), who caucuses with Democrats, said yesterday. Lieberman is a co-sponsor of a more comprehensive bill with Sen. John Kerry (D-Mass.) that would cap emissions from the utility, manufacturing and transportation sectors.
Senate energy options
As the Senate prepares to take up energy and climate change legislation this summer, Democratic leadership is expected to pick and choose aspects from several measures:
Kerry-Lieberman
The "American Power Act" (summary) would establish a cap-and-trade system for greenhouse gas emissions and expand offshore drilling.
Bingaman energy bill
S. 1462 (summary) passed the Energy and Natural Resources Committee in June 2009. It includes a renewable electricity standard but no cap on greenhouse gas emissions.
Cap and Dividend
The "CLEAR Act" (S. 2877, summary) would require energy producers to bid for carbon shares and return 75 percent of the revenue to taxpayers and direct 25 percent to clean energy research and development.
Lugar-Graham-Murkowski
S. 3464 (summary) would set long-term fuel efficiency standards for motor vehicles, provide loan guarantees for new nuclear power plants and set up a renewable electricity standard that includes nuclear power and carbon capture and sequestration.
Waxman-Markey
The House-passed bill (H.R. 2454, summary) would set up a cap-and-trade system for greenhouse gas emissions. Includes an RES but no offshore drilling.
Kerry, who has spent months relentlessly trying to round up the necessary votes to pass the economywide legislation, likewise left the door open for the scaled-down utilities-only option. "I will support pricing carbon, which is what I've said from day one is the goal here," he told reporters.
"We're looking at that," Kerry said of the power plant-only proposal. "We're looking at a lot of different ways to figure out how to build 60 votes, which I've always said that's the challenge."
The proposal has -- at least for the moment -- taken center stage in environmental and energy circles after White House chief of staff Rahm Emanuel said Sunday that it was one of several proposals that President Obama plans to bring up when he meets with a select group of bipartisan senators at the White House on Wednesday.
Environment and Public Works Chairwoman Barbara Boxer (D-Calif.), whose own climate legislation stalled soon after she passed it out of committee without a single Republican vote, did not rule out supporting a utilities-only proposal. "I'm open to looking at all different proposals," she said.
Sens. Ben Cardin (D-Md.) and Frank Lautenberg (D-N.J.), both members of Boxer's EPW Committee, offered their conditional support.
"I'm for curbing carbon and if that's a step forward, I'll take it," Lautenberg said. "If that's the final step, I don't think it's quite enough."
"I'd have to evaluate it based on whether it gets the job done," said Cardin. "The utility sector is a huge part of the issue, the way it's crafted it could have a major impact, but I would still like to know the full impact. But my preference is a more comprehensive approach."
While most Democrats were less than enthused about the utility-only idea, few were willing to close the door.
"I think we ought to consider any idea that gets us started with a system that prices carbon," said Sen. Mark Udall (D-Colo.). He added that regardless of whether the utilities-only bill gains traction, he still wants to see a full discussion on both the Kerry-Lieberman bill and the "cap and dividend" proposal from Sens. Maria Cantwell (D-Wash) and Susan Collins (R-Maine.)
"I think both those ideas have real merit and I'd expect them to be fully considered," Udall said.
Sen. Jay Rockefeller (D-W.Va.), likewise, said he would not rule out a power-plant only cap. Still, he cautioned that it would be "pretty rough on West Virginia."
Several other Democrats stopped short of weighing in one way or the other, saying that they were still gathering information about the proposal and that they had plenty of questions about how the bill would be structured.
"I am in favor of a comprehensive climate bill providing that it's done right, providing that it meets certain tests," said Sen. Carl Levin (D-Mich.). "I just don't know enough about an utilities-only bill."
Asked about the proposal, Sen. Claire McCaskill (D-Mo.) said, "Oh lordie, I don't know."
In addition to tomorrow's bipartisan meeting at the White House, the full Democratic caucus will huddle Thursday to continue the energy and climate discussions they had last week.
White House spokesman Bill Burton said tomorrow's meeting will be "one part listening but also one part trying to bridge some differences" between the proposals.
"The president thinks that if we're going to change how we deal with carbon, if we're going to move our energy future in a different direction, then we have to be pretty aggressive in how we do it," Burton said yesterday. "But he's not -- he doesn't think there's just one good idea on this. He wants to listen to some of the other ideas that folks have."
Sen. Byron Dorgan (D-N.D.) told reporters that he expects Majority Leader Harry Reid (D-Nev.) to come to a decision on how to proceed with energy and climate legislation by next week, adding that "the longer it takes to get to the floor, the less chance of getting it done."
Even if Democrats are able to rally around the utilities-only option -- something that appears far from certain -- they would still face the difficult prospect of securing at least one, and likely several, Republican votes.
"I don't expect that we will get much Republican support for legislation beyond the energy bill," Dorgan said, referring to the bipartisan bill that cleared the Energy and Natural Resources Committee last year, 15-8. "But we did have Republican support for the energy bill," he added.
Sen. Richard Lugar (R-Ind.) was adamant yesterday that he would not support a cap on greenhouse gas emissions, even for just the utility sector. "No, no, no," Lugar said when asked if he would back the approach. Earlier this month, Lugar introduced an alternative energy bill aimed at reducing greenhouse gas emissions and domestic oil consumption without using the controversial cap-and-trade program.
Still, Democrats have been targeting a number of moderate Republicans -- including Collins and Sens. Olympia Snowe of Maine, Scott Brown of Massachusetts, and George LeMieux of Florida -- in hopes that they may break from their party and provide the crucial votes.
LeMieux said yesterday that he had yet to decide on the latest proposal.
"I'm just listening and learning and talking with a lot of folks," he said. "I think energy independence and alternative energy is one of the most pressing issues facing this country, so I think it is time to deal with it."
Sen. Lindsey Graham (R-S.C.), who helped negotiate the Kerry-Lieberman bill but has since dropped out as a co-sponsor, has indicated that he would support legislation that blocks U.S. EPA regulations and caps utilities' emissions -- but only next year.
"What I would suggest is that we come back next year to replace the EPA with a comprehensive approach that looks at pricing carbon on the utility side, exempting heavy energy users because the technology's not there to sequester carbon, and on the transportation side, lower emissions without a cap," Graham said earlier this month. "That is what I will be pushing next year, a utility-only bill."
Off the Hill
The utility-only approach got a lukewarm reception from environmentalists yesterday, who cautioned against a bill that leaves other sectors out of the cap.
"We're willing to consider anything that gets us on the road to reducing carbon emissions by an amount large enough to at least be comparable to other bills," said David Hamilton, director of the Sierra Club's global warming program. Still, he said, "a utility-only system limits both the amount of emissions you're targeting and it's restricting the revenue. It's got plenty of problems, but the details matter."
David Moulton, director of climate policy at the Wilderness Society, said yesterday that an economywide approach is still preferable to a bill that caps only the utility sector.
"It's not a better option, it's a weaker option and therefore in this dysfunctional Senate, it becomes more politically doable, but it's not an answer to our climate woes," Moulton said.
Still, environmentalists say that a cap on the utility sector would be preferable to some of the alternative options being floated, including moving forward with the Energy and Natural Resources Committee bill, or pursuing a slimmed-down bill that includes only provisions to deal with the Gulf of Mexico oil spill.
Utilities, meanwhile, were hesitant to take strong positions on a cap affecting only their sector.
"We really haven't formulated an opinion on a utility-only approach," said Southern Co. spokeswoman Valerie Hendrickson. She said the company was focused on an economywide bill.
Edison Electric Institute spokesman Jim Owen said it is tricky to say whether the trade group would support a sectoral approach until the details are revealed. "We have been concentrating on economywide, so this is a game-changer with so many moving parts in the middle of a very fluid legislative situation overall," he said.
Electric utilities emit about a third of U.S. greenhouse gases per year. The industry has been involved for about 15 years in a similar market-based mechanism that has successfully reduced acid rain, and the sector is seeking regulatory certainty as companies look to make significant new investments over the next several decades.
"[O]ur first choice has always been a comprehensive energy/climate bill that deals with all sectors of the economy, but if common ground can only be found in the electric utility sector before other sectors are involved, we could support that -- to move the ball forward where we can," said Duke Energy Corp. spokesman Tom Williams in an e-mail.
Reporter Katherine Ling contributed
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redpajama_set_name: RedPajamaCommonCrawl


text:
Boo In Scotland
In a Q-and-A with Thomas Bonk, Boo Weekley talks haggis, smoked fish, and how he'd consider wearing a kilt
Thomas Bonk
Ross Kinnaird/Getty Images
Boo Weekley has found his time in Scotland to be interesting and enjoyable.
TURNBERRY, Scotland -- It was a memorable collision of cultures when Boo Weekley, born in Milton, Fla., and schooled at Abraham Baldwin Agricultural College, arrived in Carnoustie in 2007 on his first trip to Scotland and his first British Open Championship.
Boo and Scotland? The contrasts went on and on. Camo and kilts. Harmonica and bagpipes. Mullet and haddock.
When Weekley is at home, he is on a first-name basis with duck blinds and bass boats, but he wasn't at all certain what to make of this Scotland place. Especially the food. Of course, he wouldn't be the first to have problems with the local cuisine, just probably the most colorful.
While at Carnoustie, Weekley earned some sort of notoriety for his quote lamenting the limits he had discovered in the foods of Scotland: "Ain't got no fried chicken, ain't got no sweet tea."
But now, at 35, Boo has mellowed. In fact, when he opened up about Scotland in a free-spirited interview here at the British Open at Turnberry, Boo offered not so much of a peep in finding fault. Or pretty close, anyway.
__Q: Have you found a favorite Scottish food yet?
__ Boo: (Chuckles) Their bacon. Their muffins. Their bacon, it's more like little ham to us. It ain't really bacon, but I mean it's pretty good."
__Q: What is haggis?
__ Boo: "I ain't got a clue, but I tried it. I didn't like it, neither."
__Q: So you have found some things you like. Have you had any fish you like?
__ Boo: "I've eaten a few fish. The fish and chips are all right. It's just a little bit different batter than we use at the house. I've had some sole. I was looking for the guy around here ... at Carnoustie, they had a guy that smoked fish. It was out of a barrel. He was smoking 'em right out by the merchandise tent that was awesome. That's what I ate the whole time when I was over there. I'd go get that, breakfast, lunch and dinner. I ain't seen him over here."
__Q: Do people here think you have an accent?
__ Boo: "Oh, yeah, they know I do. They don't think I'm from Florida, they think I'm from Mississippi, Louisiana, Alabama."
__Q: Do you think the Scots have an accent?
__ Boo: "It's always different, you know, from when we're at the house."
__Q: What do you do here in Scotland?
__ Boo: "No sir, not a lot. But my wife (Karen) and me, we walked the whole golf course the other day and it was beautiful."
__Q: So it sounds like you're liking Scotland?
__ Boo: "Oh, yeah, Scotland's a pretty place. I mean, as long as it ain't raining."
__Q: What would it take for you to wear a kilt?
__ Boo: "Probably nothing. I'd wear it. If I had to wear it, I'd wear it, but I ain't got to. Long as I don't have to, I ain't gonna wear it."
The interview ended and Boo walked off, probably looking for the guy with the smoked fish.


Congrats to our NLI spring signers -- Top group of talented student-athletes!
Bella Bates (soccer) UW-Eau Claire, Megan Munger (soccer) Luther College, Melissa Pfeiffer (track/field and cross country) St. Scholastica, Taylor Thomas (hockey) Plymouth State University, NH; and Katie Brostrom (volleyball) St. Scholastica (not pictured).
Nick Norby (track/field) MSU-Mankato, Erik Thompson (swim/dive) South Dakota State University, Ben Eisenberg (baseball) Northern State University, Logan Edwards (wrestling) Saint John's University, and Jake Leonard (soccer) Upper Iowa University.
TIME: 4:00-6:00pm (students/parents can begin arriving at 3:00pm). Students report to Blue Gym. NOTE: All students participating in Grand March must have a Prom Ticket.
LAST DAY TO PURCHASE TICKETS: April 24th --- Tickets on sale from 7:30AM-1PM and then again afterschool --- Don't miss your chance to attend Prom 2019.


Items where Author is " Prida, de la Liset Menendez"
Valero Manuel, Averkin Róbert, Fernandez-Lamo Ivan, Aguilar Juan, Lopez-Pigozzi Diego, Brotons-Mas Jorge R., Cid Elena, Tamás Gábor, Prida, de la Liset Menendez: Mechanisms for Selective Single-Cell Reactivation during Offline Sharp-Wave Ripples and Their Distortion by Fast Ripples.
This list was generated on 2019. április 19. 09:14:16 CEST.


FREE Maxxis magnetic cell phone holder included with purchase of any four (4) or more RC-1 tires.
DOT-approved R compound optimized for the weekend warrior.
The RC-1 was designed to deliver outstanding dry traction, heat-cycle performance and smooth breakaway characteristics. The semi-slick design eliminates tread-squirm and delivers great braking and steering response for fast and consistent lap times.
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Dwight Errington Myers (Heavy D.) (deceased), Glen Parrish (G. Whiz), Edward Ferrell (DJ Eddie F), Troy Dixon (Trouble T-Roy) (deceased).
Not sure who added Andrew Risper as G Whiz. But, G Wiz from "The Boyz" and Captain G Whiz (Andrew Risper) from the Mighty Mic Masters are two different people. In fact, Captain G Whiz disses G Wiz on "Go G Whiz", claiming he stole his name, "But there's something I should mention, that something is/There's a toy down with the Boyz who calls himself G Wiz/but that's my name, you can't claim it/So chump, you better change it or the first place I see your face I'mma rearrange it..."
R.I.P Heavy D just shows more needs to be done for people who fly on planes who overweight as this effects a lot people but seems not as no what matter what government is in they do not help people as they are all out for there self's.As this is certainly the way it is in the UK.But his music lives on with us all each time we all listen to him.One Geebeats.


Wall Panel Production Team Member: Pay range is $12-$15/hour based on experience + employee benefits.
This position is responsible for a variety of tasks including, but not limited to, assembling framed walls and cutting components for walls.
assembly duties associated with the construction of Erickson's products.
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We have to protect the environment. Our lives and the future of the human race depend on it. No one wants US cities to look like Beijing during high smog events. No one wants to drink contaminated water or see rivers full of garbage. The EPA has a valid function so it shouldn't be abolished. I think it has gotten "drunk on power" under the previous administration and needs to be reined in somewhat. It seems to have morphed from basing actions on scientific studies and cost/benefit analyses to basing actions on bureaucratic authority. That's often a problem with agencies that can make rules but have no accountability to the voters. It's particularly noticeable in that the EPA rushed through some new rules in the last few days of the Obama administration. Was that really enough time for a thorough analysis? The EPA's job is protecting the environment, not bureaucratic jobs.
Rather than delve into the controversial topic of climate change I'll discuss a local matter that can really hurt taxpayers, and that's wastewater treatment, a subject that's very localized. Suppose a city has wastewater treatment plants that meet current discharge specifications. A company develops a treatment method that is more efficient, i.e., can better clean the water, and suddenly the city gets a notice from the EPA that they must meet this new "standard" or face ridiculously high fines. Does the EPA have to provide scientific evidence that this new standard is necessary or are they working to "as low as achievable"? That's not science, it's pandering to industry.
Another issue that involves local water is the EPA effort to exert control over virtually every body of water in the country, including small privately-owned ponds and "ephemeral" water (streams and puddles that only exist seasonally). This extension of the EPA's original mandate along with its power to levy fines has made life miserable for some farmers and other landowners. Unless the landowner is using those small waters to dispose of toxic materials (which is illegal anyway) this is nothing but a bureaucratic power grab.
Temporarily freeze all regulations that were issued between January 1, 2017 and January 21, 2017 until they are adequately reviewed for necessity.
Establish a "show me the science" policy for cities faced with sudden revisions to local requirements for matters like wastewater treatment that would require costly upgrades to plants that meet current specifications. Explain to the community why the change is needed in plain English, and allow a reasonable implementation time.
Reexamine the EPA authority to levy very high fines, specifically in cases involving changes to standards that are currently being met (as opposed to willful violators). This is a power that is subject to abuse and that drains money needed to make the changes.
EPA officials need to understand that with authority comes accountability. Fire employees involved in the Animas river disaster and bar involved contractors from Federal contracts for 10 years.
The long term future of the USA depends on security.
There are three levels of security that determine the future of a nation, physical security, economic security, and emotional security. Without physical security the others aren't possible, so I'll begin there.
Physical security starts with secure borders. It's easier to keep the bad guys out than to recover from the damage they can do. Any immigration reform must include improved border security.
It includes a strong military so that any hostile actions can be overwhelmingly crushed. It doesn't mean using the military to be "world police" or to try to impose our political system on countries where it simply wouldn't work.
It includes law enforcement at Federal, state, and local levels to fight crime and maintain order. They must communicate with each other. It doesn't mean having a "secret police" that can be used to suppress honest citizens.
It includes emergency response agencies that can quickly deliver aid in the event of natural or man-made disasters. Health responders must have a supply of drugs critical to epidemic control.
It requires a strong, modern, and secure infrastructure that supports reliable communication and the movement of resources within the country.
It includes a strong manufacturing base and a strategic materials stockpile. Any total loss of a capability, such as the recent closure of the last lead smelter by the EPA, could become a weakness during a global conflict. Don't count on other countries to look out for us.
It requires cities to be prepared for the types of problems they might experience in their area such as flooding along a river and to take actions to mitigate risk before a disaster occurs.
Even a prepared population contributes to physical security. That doesn't mean everyone should be a "prepper" but everyone should have the basic emergency preparations we hear about all the time as well as any specific to their region.
Author usagenda22Posted on February 23, 2015 April 9, 2016 Categories departments, economy, government, infrastructure, national security, sovereignty, UncategorizedTags emergencies, EPA, government, military, national security, policeLeave a comment on The long term future of the USA depends on security.


ObjectDB is distributed as a zip file. Just download and extract the ObjectDB distribution zip file. No need to run any install or setup program. You can uninstall ObjectDB by deleting the ObjectDB directory. ObjectDB does not modify the registry or any other system file.
How to create a wildfly datasource with dsJndi name to an objectdb database?


Did the Liberals mislead Canadians about their climate change plans?
Or is Catherine McKenna happy to champion the same targets for reducing carbon pollution that her own party called "the weakest" in the industrialized world?
Appearing Sunday on CTV's Question Period, Canada's Environment Minister conceded she has no plan to set a stronger target for reducing carbon emissions than the target Stephen Harper proposed last year.
Maybe everybody missed a footnote in the Liberal Party's 2015 election platform defining "Real Change" as achieving Stephen Harper's goals?
Despite McKenna's new definition of success, the 2015 Liberal platform's chapter on the environment actually has a very different benchmark for success when it comes to climate change.
The platform specifically calls for "the creation of national emissions-reduction targets" aimed at avoiding the "catastrophic impact" of a "greater-than-two-degree increase in average global temperatures."
"Central to this will be the creation of national emissions-reduction targets, informed by the best economic and scientific analysis. These targets must recognise the economic cost and catastrophic impact that a greater-than two-degree increase in average global temperatures would represent, as well as the necessity for Canada to do its part to prevent that from happening. We believe that Harper's targets are inadequate and meaningless without a plan to achieve them."
Why is the focus on two degrees so important?
That's the widely recognized temperature limit the planet can sustain before we begin seeing "severe consequences," including ocean level rise, extreme weather, disease and crop failure – Liberal MP Andy Fillmore has warned that a "2° Celsius increase in global mean temperature could mean that Nova Scotia becomes an island."
You might ask yourself: will Harper's carbon target help prevent such "catastrophic" impacts?
That's exactly the question one Liberal backbencher put before Canada's lead climate change negotiator – would Harper's target of reducing emissions 30% below 2005 levels by 2030 "help achieve the goal of a 1.5-degree or 2-degree rise in temperature"?
The high-ranking bureaucrat had bad news: even if Canada met its 524 megatonne target for 2030 and every country in the world hit the targets they pledged in advance of last year's COP21 conference in Paris as well, this would still "put us at around a 2.7° temperature rise."
Add to that, Canada's targets are less ambitious than other industrialized countries. The 28 members of the European Union set a target of reducing emissions 40% below 2005 levels by 2030 while the United States set a goal of 26-28% below 2005 levels by 2025 – a similar goal achieved five years earlier than Canada.
When it was announced last year, former Liberal environment critic John McKay condemned Harper's targets, stating that "the Harper government's emissions targets are among the weakest in the G7."
And he's not the only one with choice words.
Last month, Toronto Liberal MP Arif Virani wrote an open letter to McKenna reporting his constituents were "unanimous that the GHG emission target of 30% below 2005 levels by 2030 set by the previous government is insufficient, and must be much more aggressively set."
Climate experts estimate that avoiding a 2° Celsius increase in global temperatures will require additional emissions reductions of 40-70% by 2050 and "carbon neutrality" (zero emissions) by the year 2100.


text:
\section{Introduction}
\newtheorem {definition1}{Definition}[section]
\newtheorem{theorem1}{Theorem}[section]

\par
Recently, a new 2-component system  was introduced by Zuo in
\cite{d-z} as follows:
\begin{equation}
\left\{\begin{array}{ll}
\mu(u)_{t}-u_{txx}=2\mu(u)u_{x}-2u_{x}u_{xx}-uu_{xxx}+\rho\rho_{x}&-\gamma u_{xxx},\\
&t > 0,\,x\in \mathbb{R},\\
\rho_{t}=(\rho u)_x+\gamma \rho_{x}, &t > 0,\,x\in \mathbb{R},\\
u(0,x) = u_{0}(x),& x\in \mathbb{R}, \\
\rho(0,x) = \rho_{0}(x),&x\in \mathbb{R},\\
u(t,x+1)=u(t,x), & t \geq 0, x\in \mathbb{R},\\
\rho(t,x+1)=\rho(t,x), & t \geq 0, x\in \mathbb{R},\\ \end{array}\right. \\
\end{equation}
where $\mu(u)=\int_{\mathbb{S}}udx$ with
$\mathbb{S}=\mathbb{R}/\mathbb{Z}$ and $\gamma \in \mathbb{R}.$ By
integrating both sides of the first equation in the system (1.1)
over the circle $\mathbb{S}=\mathbb{R}/\mathbb{Z}$ and using the
periodicity of $u$, one obtain
$$\mu(u_{t})=\mu(u)_{t}=0.$$
This yields the following periodic 2-component $\mu$-Hunter-Saxton
system:
\begin{equation}
\left\{\begin{array}{ll}
-u_{txx}=2\mu(u)u_{x}-2u_{x}u_{xx}-uu_{xxx}+\rho\rho_{x}&-\gamma u_{xxx},\\
&t > 0,\,x\in \mathbb{R},\\
\rho_{t}=(\rho u)_x+\gamma \rho_{x}, &t > 0,\,x\in \mathbb{R},\\
u(0,x) = u_{0}(x),& x\in \mathbb{R}, \\
\rho(0,x) = \rho_{0}(x),&x\in \mathbb{R},\\
u(t,x+1)=u(t,x), & t \geq 0, x\in \mathbb{R},\\
\rho(t,x+1)=\rho(t,x), & t \geq 0, x\in \mathbb{R},\\ \end{array}\right. \\
\end{equation}
with $\gamma \in \mathbb{R}.$  This system is a 2-component
generalization of the generalized Hunter-Saxton equation obtained in
\cite{k-l-m}. The author \cite{d-z} shows that this system is a
bihamiltonian Euler equation, and also can be viewed as a
bivariational equation.

Obviously, (1.1) is equivalent to (1.2) under the condition
$\mu(u_{t})=\mu(u)_{t}=0.$ In this paper, we will study the system
(1.2) under the assumption $\mu(u_{t})=\mu(u)_{t}=0$.

For $\rho\equiv 0$ and $\gamma=0,$ and replacing $t$ by $-t,$ the
system (1.2) reduces to the generalized Hunter-Saxton equation
(named $\mu$-Hunter-Saxton equation) as follows \cite{k-l-m}:
\begin{equation}
-u_{txx}=-2\mu(u)u_{x}+2u_{x}u_{xx}+uu_{xxx},
\end{equation}
where $\mu(u)=\int_{\mathbb{S}}udx$ denotes the mean of $u.$ The
$\mu$-Hunter-Saxton equation lies mid-way between the periodic
Hunter-Saxton and Camassa-Holm equations with $u=u(t,x)$ being a
time-dependent function on the circle
$\mathbb{S}=\mathbb{R}/\mathbb{Z}$. Recently, the periodic
$\mu$-Hunter-Saxton equation and the periodic
$\mu$-Degasperis-Procesi equation have also been studied in
\cite{escher, fu, l-m-t}. For $\mu(u)=0,$ the equation (1.3) reduces
to the Hunter-Saxton equation \cite{J-R}
\begin{equation}
u_{txx}+2u_{x}u_{xx}+uu_{xxx}=0,
\end{equation}
modeling the propagation of weakly nonlinear orientation waves in a
massive nematic liquid crystal. In the Hunter-Saxton equation
\cite{J-R}, $x$ is the space variable in a reference frame moving
with the linearized wave velocity, $t$ is a slow-time variable and
$u(t,x)$ is a measure of the average orientation of the medium
locally around $x$ at time $t$. More precisely, the orientation of
the molecules is described by the field of unit vectors $(\cos
u(t,x), \sin u(t,x))$ \cite{Y}. The single-component model also
arises in a different physical context as the high-frequency limit
\cite{HHD, J-Y} of the Camassa-Holm equation for shallow water waves
\cite{R-D, J-H} and a re-expression of the geodesic flow on the
diffeomorphism group of the circle \cite{C-K} with a bi-Hamiltonian
structure \cite{F-F} which is completely integrable \cite{C-M}. The
Hunter-Saxton equation also has a bi-Hamiltonian structure
\cite{J-H, P-P} and is completely integrable \cite{B1, J-Y}. The
initial value problem for the Hunter-Saxton equation (1.4) on the
line (nonperiodic case) and on the unit circle
$\mathbb{S}=\mathbb{R}/\mathbb{Z}$ were studied by Hunter and Saxton
in \cite{J-R} using the method of characteristics and by Yin in
\cite{Y} using Kato semigroup method, respectively. Moreover, global
dissipative and conservative weak solutions for the initial boundary
value problem of the Hunter-Saxton equation on the half line were
investigated extensively, c.f. \cite{H-Z1,H-Z2,Z-Z1,Z-Z2,Z-Z3}.

For $\rho\not\equiv0$ , $\gamma=0,$ $\mu(u)=0$ and replacing $t$ by
$-t,$ the system (1.2) reduces to the two-component periodic
Hunter-Saxton system, peakon solutions of the Cauchy problem of it
have been analysed in \cite{C-I}. Moreover, the Cauchy problem and
global weak solutions of two-component periodic Hunter-Saxton system
have been discussed in \cite{l-y} and \cite{g-y} respectively.

Furthermore, the Cauchy problem of (1.2) has been discussed in
\cite{l1-y} recently. The authors established the local
well-posedness, derived precise blow-up scenarios for the system
(1.2) and proved that the system (1.2) has global strong solutions
and also finite time blow-up solutions. However, the existence of
global weak solutions to the system (1.2) has not been studied yet.
The aim of this paper is to present a  global existence result of
weak solutions to the system (1.2).

The main result of this paper is to give the existence of a
global-in-time weak solution $z=\left(
                                                     \begin{array}{c}
                                                       u \\
                                                       \rho \\
                                                     \end{array}
                                                   \right)$ to the problem (1.2) with the initial $z_{0}=\left(
                                                     \begin{array}{c}
                                                       u_{0} \\
                                                       \rho_{0} \\
                                                     \end{array}
                                                   \right)
\in W^{1,\infty}(\mathbb{S})\times L^{\infty}(\mathbb{S}).$  Before
giving the precise statement of the main result, we first introduce
the definition of weak solution to the problem (1.2).

\begin{definition1} $z=\left(
                                                     \begin{array}{c}
                                                       u \\
                                                       \rho \\
                                                     \end{array}
                                                   \right)
\in H^1(\mathbb{S})\times L^{2}(\mathbb{S})$ is said to be an
admissible weak solution to the problem (1.2)  if
$$ z(x,t)\in L_{loc}^{\infty}((0,\infty); H^{1}(\mathbb{S})\times
L^{2}(\mathbb{S}))$$ satisfies the system (1.2) and
$z(t,\cdot)\rightarrow z_0$ as $t\rightarrow 0^+$ in the sense of
distributions on $\mathbb{R}_+\times\mathbb{R}$. Moreover,
$$\|u_x(t,\cdot)\|_{L^2(\mathbb{S})}+\|\rho(t,\cdot)\|_{L^2(\mathbb{S})}\leq
\|u_{0,x}\|_{L^2(\mathbb{S})}+\|\rho_0\|_{L^2(\mathbb{S})}.$$
\end{definition1}
\par
The main result of this paper can be stated as follows:
\begin{theorem1} Let $z_{0}=\left(
                                                     \begin{array}{c}
                                                       u_{0} \\
                                                      \rho_{0} \\
                                                     \end{array}
                                                   \right)
\in W^{1,\infty}(\mathbb{S})\times L^{\infty}(\mathbb{S}).$ If there
exists $\alpha>0$ such that $\rho_0(x)\geq \alpha$ for a.e.
$x\in\mathbb{S},$ then the system (1.2) has an admissible weak
solution
$$ z=\left(
                                                     \begin{array}{c}
                                                       u \\
                                                       \rho \\
                                                     \end{array}
                                                   \right)\in
                                                   C(\mathbb{R}_+;H^1(\mathbb{S})\times L^2(\mathbb{S}))
                                                   \cap L^{\infty}(\mathbb{R}_+;
H^{1}(\mathbb{S})\times L^{2}(\mathbb{S}))$$ in the sense of
Definition 1.1. Furthermore,
$$u\in L^{\infty}_{loc}(\mathbb{R}_+; W^{1,\infty}({\mathbb{S}})) \ \
\ \text{and} \ \ \ \rho\in L^{\infty}_{loc}(\mathbb{R}_+;
L^{\infty}(\mathbb{S})).$$
 \end{theorem1}

\begin{rem}
If there exists $\alpha<0$ such that $\rho_0(x)\leq \alpha$ for a.e.
$x\in\mathbb{S},$ then the conclusions in Theorem 1.1 also hold.
\end{rem}

 The paper is organized as follows. In
 Section 2, we recall some useful lemmas and derive some priori estimates on global strong solutions to (1.2). In
Section 3, we obtain the global existence of approximate solution to
(1.2) with smooth approximate initial data. In Section 4,  acquiring
the precompactness of approximate solutions,
 we prove the
existence of the global weak solution to (1.2).

\section{Preliminaries}
\newtheorem {remark2}{Remark}[section]
\newtheorem{theorem2}{Theorem}[section]
\newtheorem{lemma2}{Lemma}[section]
Since (1.1) is equivalent to (1.2) under the condition
$\mu(u)_{t}=\mu(u_{t})=0,$ to obtain the existence of global weak
solution to (1.2) we study (1.1) henceforth. Moreover, for the sake
of convenience, we let
$$\mu_{0}=\mu(u_{0})=\mu(u)=\int_{\mathbb{S}}u(t,x)dx.$$

We now provide the framework in which we shall reformulate the
system (1.1). We rewrite the system (1.1) as follows:
\begin{equation}
\left\{\begin{array}{ll}
u_{t}-(u+\gamma)u_{x}=\partial_{x}A^{-1}(2\mu_{0}u+\frac{1}{2}u_{x}^{2}+\frac{1}{2}\rho^{2}),
&t > 0,\,x\in \mathbb{R},\\
\rho_{t}-(\rho u)_x=\gamma \rho_{x}, &t > 0,\,x\in \mathbb{R},\\
u(0,x) = u_{0}(x),& x\in \mathbb{R}, \\
\rho(0,x) = \rho_{0}(x),&x\in \mathbb{R},\\
u(t,x+1)=u(t,x), & t \geq 0, x\in \mathbb{R},\\
\rho(t,x+1)=\rho(t,x), & t \geq 0, x\in \mathbb{R},\\ \end{array}\right. \\
\end{equation}
where $A=\mu-\partial_{x}^{2}$ is an isomorphism between
$H^{s}(\mathbb{S})$ and $H^{s-2}(\mathbb{S})$ with the inverse
$v=A^{-1}w$ given explicitly by
\begin{align}
v(x)&=(\frac{x^{2}}{2}-\frac{x}{2}+\frac{13}{12})\mu(w)+(x-\frac{1}{2})\int_{0}^{1}\int_{0}^{y}w(s)dsdy\\
\nonumber&-\int_{0}^{x}\int_{0}^{y}w(s)dsdy+\int_{0}^{1}\int_{0}^{y}\int_{0}^{s}w(r)drdsdy,
\end{align}
which can be found in \cite{escher}. Since $A^{-1}$ and
$\partial_{x}$ commute, the following identities hold
\begin{equation}
A^{-1}\partial_{x}w(x)=(x-\frac{1}{2})\int_{0}^{1}w(x)dx-\int_{0}^{x}w(y)dy+\int_{0}^{1}\int_{0}^{x}w(y)dydx,
\end{equation}
and
\begin{equation}
A^{-1}\partial_{x}^{2}w(x)=-w(x)+\int_{0}^{1}w(x)dx.
\end{equation}
If we rewrite the inverse of the operator $A=\mu-\partial_{x}^{2}$
in terms of a Green's function, we find
$(A^{-1}m)(x)=\int_{0}^{1}g(x-x^{'})m(x^{'})dx^{'}=(g*m)(x).$ So, we
get another equivalent form:
\begin{equation}
\left\{\begin{array}{ll}
u_{t}-(u+\gamma)u_{x}=\partial_{x}g*(2\mu_{0}u+\frac{1}{2}u_{x}^{2}+\frac{1}{2}\rho^{2}),
&t > 0,\,x\in \mathbb{R},\\
\rho_{t}-(\rho u)_x=\gamma \rho_{x}, &t > 0,\,x\in \mathbb{R},\\
u(0,x) = u_{0}(x),& x\in \mathbb{R}, \\
\rho(0,x) = \rho_{0}(x),&x\in \mathbb{R},\\
u(t,x+1)=u(t,x), & t \geq 0, x\in \mathbb{R},\\
\rho(t,x+1)=\rho(t,x), & t \geq 0, x\in \mathbb{R},\\ \end{array}\right. \\
\end{equation}
where the Green's function $g(x)$ is given \cite{l-m-t} by
\begin{equation}
g(x)=\frac{1}{2}x(x-1)+\frac{13}{12} \ \ \ \text{for} \
x\in[0,1)\simeq S^{1},
\end{equation}
and is extended periodically to the real line. In other words,
$$g(x-x^{'})=\frac{(x-x^{'})^{2}}{2}-\frac{|x-x^{'}|}{2}+\frac{13}{12},
\ \ \ x,x^{'}\in[0,1)\simeq S^{1}.$$ In particular, $\mu(g)=1.$
\begin{lemma2}(\cite{constantin, fu}) If $f\in H^{1}(\mathbb{S})$ is such that
$\int_{\mathbb{S}}f(x)dx=0,$ then we have
$$\max\limits_{x\in\mathbb{S}}f^{2}(x)\leq
\frac{1}{12}\int_{\mathbb{S}}f_{x}^{2}(x)dx.$$
\end{lemma2}

Assume that $z=\left(
                                    \begin{array}{c}
                                      u \\
                                      \rho \\
                                    \end{array}
                                  \right)$ is a smooth solution of (2.1). For
convenience, we let
$\mu_{1}=\left(\int_{\mathbb{S}}(u_{0,x}^{2}+\rho_{0}^{2})dx\right)^{\frac{1}{2}}.$
Using the system (1.1), a simple calculation implies
$\frac{d}{dt}\int_{\mathbb{S}}(u_{x}^{2}+\rho^{2})dx=0.$  So we have
\begin{equation}
\mu_{1}=\left(\int_{\mathbb{S}}(u_{0,x}^{2}+\rho_{0}^{2})dx\right)^{\frac{1}{2}}
=\left(\int_{\mathbb{S}}(u_{x}^{2}+\rho^{2})dx\right)^{\frac{1}{2}}.
\end{equation}
Note that $\int_{\mathbb{S}}(u(t,x)-\mu_{0})dx=\mu_{0}-\mu_{0}=0.$
By Lemma 2.1, we find that
$$\max\limits_{x\in\mathbb{S}}[u(t,x)-\mu_{0}]^{2}\leq
 \frac{1}{12}\int_{\mathbb{S}}u_{x}^{2}(t,x)dx\leq
 \frac{1}{12}\mu_{1}^{2}.$$ So we have
 \begin{equation}
 \|u(t,\cdot)\|_{L^{\infty}(\mathbb{S})}\leq
 |\mu_{0}|+\frac{\sqrt{3}}{6}\mu_{1}.
 \end{equation}
\begin{lemma2} (\cite{l1-y}) Let $z_{0}=\left(
                                                     \begin{array}{c}
                                                       u_{0} \\
                                                       \rho_{0} \\
                                                     \end{array}
                                                   \right)
\in H^2(\mathbb{S})\times H^{1}(\mathbb{S}).$ If $\rho_0(x)\neq0 $
 for all $x\in\mathbb{S}$, then the corresponding strong solution $z=\left(
                                    \begin{array}{c}
                                      u \\
                                      \rho \\
                                    \end{array}
                                  \right)$
 to (1.1) exists globally in time, i.e. $z\in
 C(\mathbb{R}_+;H^2(\mathbb{S})\times H^{1}(\mathbb{S}))\cap C^1(\mathbb{R}_+;H^{1}(\mathbb{S})\times L^{2}(\mathbb{S}))$.
 Moreover,
 there exists $\beta>0$ such that for all $t\in\mathbb{R}_{+},$
\begin{align*}
\|u_x(t,\cdot)\|_{L^{\infty}(\mathbb{S})}\leq\frac{1}{2\beta}(1+\|
\rho_0\|^2_{L^{\infty}(\mathbb{S})}&+\|
u_{0,x}\|^2_{L^{\infty}(\mathbb{S})})\\
&\cdot e^{(4\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2}+\frac{\sqrt{3}}{3}
|\mu_{0}|\mu_{1}+\frac{1}{2})t}:= C_1(t)
\end{align*}
 and
$$\|\rho(t,\cdot)\|_{L^{\infty}(\mathbb{S})}\leq e^{C_1(t)t}\|\rho_0\|_{L^{\infty}(\mathbb{S})}:= C_2(t),$$
where $\beta= \inf_{x\in\mathbb{S}}|\rho_0(x)|.$
\end{lemma2}

\begin{lemma2}(\cite{s}) Assume $X\subset B \subset Y$ with compact
imbedding $X\rightarrow B$ ($X,B$ and $Y$ are Banach spaces), $1\leq
p\leq \infty$ and (1) $F$ is bounded in $L^{p}(0,T; X)$, (2)
$\|\tau_{h}f-f\|_{L^{p}(0,T-h; Y)}\rightarrow 0$ as $h\rightarrow 0$
uniformly for $f\in F.$ Then $F$ is relatively compact in
$L^{p}(0,T; B)$ (and in $C(0,T; B)$ if $p=\infty$), where
$(\tau_{h}f)(t)=f(t+h)$ for $h>0,$ if $f$ is defined on $[0,T],$
then the translated function $\tau_{h}f$ is defined on $[-h, T-h].$
\end{lemma2}

\begin{lemma2}(Appendix C of \cite{Li}) Let $X$ be a separable
reflexive Banach space and let $f^n$ be bounded in
$L^{\infty}(0,T;X)$ for some $T\in(0,\infty).$ We assume that
$f^n\in C([0,T];Y)$ where $Y$ is a Banach space such that
$X\hookrightarrow Y$, $Y'$ is separable and dense in $X'$.
Furthermore, $(\phi,f^n(t))_{Y'\times Y}$ is uniformly continuous in
$t\in [0,T]$ and uniformly in $n\geq 1.$ Then $f^n$ is relatively
compact in $C^{w}([0,T];X)$, the space of continuous functions from
$[0,T]$ with values in $X$ when the latter space is equipped with
its weak topology.
\end{lemma2}

\begin{remark2} If the conditions which $f^n$ satisfies in Lemma 2.4
are replaced by the following conditions:
$$f^n\in L^{\infty}(0,T;X),\ \partial_t f^n\in L^p(0,T;Y) \text{  for
some}\,\, p\in(1,\infty],$$ and $$ \| f^n\|_{L^{\infty}(0,T;X)}, \
\|
\partial_tf^n\|_{L^p(0,T;Y)}\leq C, \ \ \forall n\geq1,$$
then the conclusion of Lemma 2.4 holds true.
\end{remark2}

\section{Global approximate solutions}
\newtheorem {remark3}{Remark}[section]
\newtheorem{theorem3}{Theorem}[section]
\newtheorem{lemma3}{Lemma}[section]
\newtheorem{claim3}{Claim}[section]

In the section, we first prove the existence of approximate
solutions. Then, with the basic estimates given in Section 2, we
will give some useful estimates to the approximate solutions.

\par
Let $z_{0}=\left(
                                                     \begin{array}{c}
                                                       u_{0} \\
                                                      \rho_{0} \\
                                                     \end{array}
                                                   \right)
\in W^{1,\infty}(\mathbb{S})\times L^{\infty}(\mathbb{S})$ and there
exists $\alpha>0$ such that $\rho_0(x)\geq \alpha$ for a.e.
$x\in\mathbb{S}.$ Define $z_0^n:=\left(
                                                     \begin{array}{c}
                                                       \phi_n\ast u_{0} \\
                                                      \phi_n\ast \rho_{0} \\
                                                     \end{array}
                                                   \right)= \left(
                                                     \begin{array}{c}
                                                        u_{0}^n \\
                                                      \rho_{0}^n \\
                                                     \end{array}
                                                   \right)\in H^2(\mathbb{S})\times H^{1}(\mathbb{S})$, for $n\geq
                                                   1$,
here $\{\phi_n\}_{n\geq 1}$ are the mollifiers
$$ \phi_n(x):=\left(\int_{\mathbb{R}}\phi(\xi)d\xi\right)^{-1}n\phi(nx),\ \ \ \ \ x\in\mathbb{R}, \ \ n\geq 1,$$
where $\phi\in C_c^{\infty}(\mathbb{R})$ is defined by
$$\phi(x)=\left \{\begin {array}{ll}e^{1/(x^2-1)},  \ \ \ \ \ \ \ \ |x|<1,
 \\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |x|\geq 1.\end {array}\right.$$

In view of $\rho_{0}(x)\geq\alpha>0,$ for a.e. $x\in\mathbb{S}$ and
$\phi_n(x)\geq0,$ we have  $$ \rho_{0}^n(x)=\phi_n\ast
\rho_{0}(x)\geq \alpha \int_{\mathbb{R}}\phi_n(y)dy=\alpha>0, \ \
\forall x\in \mathbb{S}.$$ Clearly, we also have
\begin{align}
 u_0^n\rightarrow u_0 \ \ \text{in}  \ \
H^1(\mathbb{S}), \ \ \ \ \rho_0^n\rightarrow \rho_0 \ \ \text{in}\ \
L^2(\mathbb{S}),\ \ \ \text{as}\ \ n\rightarrow\infty
\end{align} and
\begin{align}
\|u_0^n\|_{L^2(\mathbb{S})}\leq \|u_0\|_{ L^2(\mathbb{S})},
\|u_{0,x}^n\|_{L^2(\mathbb{S})}\leq \|u_{0,x}\|_{L^2(\mathbb{S})},
\|\rho_0^n\|_{L^2(\mathbb{S})}\leq \|\rho_0\|_{L^2(\mathbb{S})}
.\end{align} Thus, we obtain the corresponding solution $z^n\in
 C(\mathbb{R}_+;H^2(\mathbb{S})\times H^{1}(\mathbb{S}))\cap C^1(\mathbb{R}_+;H^{1}(\mathbb{S})\times L^{2}(\mathbb{S}))$ to the system (1.1)
 with initial data $z^n_0$ by Lemma 2.2 under the condition
 $\mu(u^{n})_{t}=\mu(u^{n}_{t})=0.$

 For given $z_0=\left(
                                                     \begin{array}{c}
                                                       u_{0} \\
                                                       \rho_{0} \\
                                                     \end{array}
                                                   \right)
\in  W^{1,\infty}(\mathbb{S})\times L^{\infty}(\mathbb{S})$, we set
$$\mu^{n}_{0}:=\mu(u^{n}_{0})=\mu(u^{n})=\int_{\mathbb{S}}u^{n}(t,x)dx,$$
$$\mu^{n}_{1}:=\left(\int_{\mathbb{S}}((u^{n}_{0,x})^{2}+(\rho^{n}_{0})^{2})dx\right)^{\frac{1}{2}}
=\left(\int_{\mathbb{S}}((u^{n}_{x})^{2}+(\rho^{n})^{2})dx\right)^{\frac{1}{2}}$$
 Then we have the following remark.

\begin{remark3}
By (2.7)-(2.8) and (3.2), we have
\begin{equation}
\mu^{n}_{0}\rightarrow \mu_{0} \ \text{as}\
 n\rightarrow\infty, \ \ \ \ \ \ \ \ \ (\mu^{n}_{1})^{2}\rightarrow \mu_{1}^{2} \ \text{as}\
 n\rightarrow\infty,
\end{equation}
and
\begin{align}\| u_x^n(t,\cdot)\|_{L^2(\mathbb{S})}^2+\|
\rho^n(t,\cdot)\|^2_{L^2(\mathbb{S})}=(\mu^{n}_{1})^{2}\leq\mu_{1}^{2},
\
 \ \ \forall t\in\mathbb{R}_+.
\end{align}
Moreover, we get
\begin{equation}
|\mu_{0}^{n}|\leq\int_{\mathbb{S}}|u_{0}^{n}|dx\leq\|u_{0}^{n}\|_{L^2(\mathbb{S})}\leq\|u_{0}\|_{L^2(\mathbb{S})},
\end{equation}
\begin{align}
 \|u^n(t,\cdot)\|_{L^\infty(\mathbb{S})}\leq
 |\mu_{0}^{n}|+\frac{\sqrt{3}}{6}\mu_{1}^{n}\leq
 \|u_{0}\|_{L^{2}(\mathbb{S})}+\frac{\sqrt{3}}{6}\mu_{1}.
 \end{align}
Furthermore, by Lemma 2.2, we obtain
\begin{align}
\|u^{n}_x(t,\cdot)\|_{L^{\infty}(\mathbb{S})}\leq&\frac{1}{2\beta}(1+\|
\rho_0\|^2_{L^{\infty}(\mathbb{S})}+\|
u_{0,x}\|^2_{L^{\infty}(\mathbb{S})})\\
\nonumber&\cdot
e^{(4\|u_{0}\|_{L^{2}(\mathbb{S})}^{2}+\frac{1}{2}\mu_{1}^{2}+\frac{\sqrt{3}}{3}
\|u_{0}\|_{L^{2}(\mathbb{S})}\mu_{1}+\frac{1}{2})t}:=
\widetilde{C_1}(t)
\end{align}
 and
\begin{equation}
\|\rho^{n}(t,\cdot)\|_{L^{\infty}(\mathbb{S})}\leq
e^{\widetilde{C_1}(t)t}\|\rho_0\|_{L^{\infty}(\mathbb{S})}:=
\widetilde{C_2}(t),
\end{equation}
where $\beta= \inf_{x\in\mathbb{S}}|\rho_0(x)|$.
\end{remark3}

\section{The existence of global weak solution}
\newtheorem {remark4}{Remark}[section]
\newtheorem{theorem4}{Theorem}[section]
\newtheorem{lemma4}{Lemma}[section]
\newtheorem{claim4}{Claim}[section]
In this section, with the basic energy estimate in Section 3, we are
ready to obtain the necessary compactness of the approximate
solutions $z^n(t,x).$ Acquiring the precompactness of approximate
solutions, we prove the existence of the global weak solution to the
system (1.1).
\begin{lemma4}
For any fixed $T>0$, there exist a subsequence $\{z^{n_k}(t,x)\}$ of
the sequence $\{z^{n}(t,x)\}$ and some function  $z(t,x)\in
L^\infty((0,\infty); H^1(\mathbb{S})\times
L^2(\mathbb{S}))\cap(H^{1}((0,T)\times \mathbb{S})\times
L^{2}((0,T)\times \mathbb{S})$ such that
\begin{equation} z^{n_k}\rightharpoonup z \ \ \ \ \text{in}\
H^1((0,T)\times\mathbb{S})\times L^2((0,T)\times\mathbb{S}) \
\text{as}\ n_k\rightarrow\infty, \forall T>0,
\end{equation}
and
\begin{equation} u^{n_k}\rightarrow u \ \ \text{in} \ \
L^\infty((0,T)\times\mathbb{S}) \ \
 \text{as} \ \ n_k\rightarrow\infty.
\end{equation}
Moreover, $u(t,x)\in C((0,T);L^{\infty}(\mathbb{S})).$
\end{lemma4}
\textbf{Proof}
It follows from Remark 3.1 that $\{z^{n}(t,x)\}$ is uniformly
bounded in $L^\infty((0,\infty); H^1(\mathbb{S})\times
L^2(\mathbb{S}))$.

We will prove that the sequence $\{z^{n}(t,x)\}$ is uniformly
bounded in the space $H^{1}((0,T)\times \mathbb{S})\times
L^{2}((0,T)\times \mathbb{S}).$ Firstly, we claim that
$\{u^{n}(t,x)\}$ is uniformly bounded in the space
$H^{1}((0,T)\times \mathbb{S}).$ Using Remark 3.1, we have
\begin{align*}
\|(u^{n}+\gamma)u^{n}_{x}\|_{L^{2}((0,T)\times \mathbb{S})}&\leq
 \|u^{n}u^{n}_{x}\|_{L^{2}((0,T)\times \mathbb{S})}
 +|\gamma|\|u^{n}_{x}\|_{L^{2}((0,T)\times \mathbb{S})}\\
 &\leq(\|u_{0}\|_{L^{2}(\mathbb{S})}+\frac{\sqrt{3}}{6}\mu_{1}+|\gamma|)\mu_{1}\sqrt{T},
\end{align*}
and
\begin{align*}
&\|g_{x}*(2\mu_{0}^{n}u^{n}+\frac{1}{2}(u_{x}^{n})^{2}+\frac{1}{2}(\rho^{n})^{2}\|_{L^{2}((0,T)\times
\mathbb{S})}\\
\leq \ & \|g_{x}\|_{L^{1}((0,T)\times
\mathbb{S})}\|2\mu_{0}^{n}u^{n}+\frac{1}{2}(u_{x}^{n})^{2}
+\frac{1}{2}(\rho^{n})^{2}\|_{L^{2}((0,T)\times
\mathbb{S})}\\
\leq \ &
\frac{T}{2}\left(\int_{0}^{T}\int_{\mathbb{S}}(\mu_{0}^{n})^{2}dxdt
+\int_{0}^{T}\int_{\mathbb{S}}(u^{n})^{2}dxdt+\frac{1}{2}\int_{0}^{T}\int_{\mathbb{S}}
\left((u_{x}^{n})^{2}+(\rho^{n})^{2}\right)dxdt\right)\\
\leq \ &
\frac{T}{2}\left[\|u_{0}\|_{L^{2}(\mathbb{S})}^{2}T+\left(\|u_{0}\|_{L^{2}(\mathbb{S})}
+\frac{\sqrt{3}}{6}\mu_{1}\right)^{2}+\frac{1}{2}\mu_{1}^{2}T\right].
\end{align*}
Then, by the first equation in (2.2), we know that
$\{u^{n}_{t}(t,x)\}$ is uniformly bounded in $L^{2}((0,T)\times
\mathbb{S}).$ Combing this conclusion with Remark 3.1, we obtain
$\{u^{n}(t,x)\}$ is uniformly bounded in the space
$H^{1}((0,T)\times \mathbb{S}).$ Furthermore, from Remark 3.1, one
can easily obtain that $\rho^n$ is uniformly bounded in the space
$L^2((0,T)\times\mathbb{S})$, and thus (4.1) follows.

Observe that, for each $0\leq s, t\leq T,$
$$\|u^{n}(t,\cdot)-u^{n}(s,\cdot)\|_{L^{2}(\mathbb{S})}^{2}
=\int_{\mathbb{S}}(\int_{s}^{t}\frac{\partial u^{n}}{\partial
\tau}(\tau, x)d\tau)^{2}dx\leq
|t-s|\int_{0}^{T}\int_{\mathbb{S}}(u_{t}^{n})^{2}dxdt.$$ Moreover,
$\{u^{n}(t,x)\}$ is uniformly bounded in $L^{\infty}(0,T;
H^{1}(\mathbb{S}))$ and $H^{1}(\mathbb{S})\subset\subset
L^{\infty}(\mathbb{S})\subset L^{2}(\mathbb{S}),$ then (4.2) and
$u(t,x)\in C((0,T);L^{\infty}(\mathbb{S}))$ is consequence of Lemma
2.3.

\begin{remark4} By Remark 3.1 and the above argument,  there exists a subsequences of
$\{(u_x^n)^2,(\rho^n)^2\}$, denoted again by
$\{(u_x^{n_k})^2,(\rho^{n_k})^2\}$, converging weakly in
$L_{loc}^p(\mathbb{R}_+\times\mathbb{R})$ ,  where $1<p<\infty$,
i.e. there exist $\overline{u_x^2}\in
L_{loc}^p(\mathbb{R}_+\times\mathbb{R})$ and $\overline{\rho^2}\in
L_{loc}^p(\mathbb{R}_+\times\mathbb{R})$ such that
$$ (u_x^{n_k})^2\rightharpoonup \overline{u_x^2} \ \ \ \text{and} \ \
(\rho^{n_k})^2\rightharpoonup\overline{\rho^2} \ \ \text{in}\ \
L_{loc}^p(\mathbb{R}_+\times\mathbb{R}).$$ Moreover,
 we have
 \begin{equation*}
u_x^{n_k}\rightharpoonup u_x \ \ \text{in} \
L_{loc}^p(\mathbb{R}_+\times\mathbb{R})\ \text{and}\ \ \
u_x^{n_k}\rightharpoonup^{\star} u_x \ \ \text{in}\ \
L_{loc}^{\infty}(\mathbb{R}_+;L^2( \mathbb{S})),
\end{equation*}
\begin{equation*}
\rho^{n_k}\rightharpoonup \rho \ \ \text{in} \
L_{loc}^p(\mathbb{R}_+\times\mathbb{R})\ \ \text{and}\ \ \
\rho^{n_k}\rightharpoonup^{\star} \rho \ \ \text{in}\ \
L_{loc}^{\infty}(\mathbb{R}_+;L^2( \mathbb{S})),
\end{equation*}
 Furthermore, we have
\begin{equation}
u_x^2(t,x)\leq \overline{u_x^2}(t,x), \ \ \ \
\rho^2(t,x)\leq\overline{\rho^2}(t,x) \ \ a.e. \ \text{on}\
\mathbb{R}_+\times\mathbb{R}.
\end{equation}
In view of (3.7) and (3.8), we have
\begin{align}\|u_x(t,\cdot)\|_{L^\infty(\mathbb{S})}\leq
\widetilde{C_1}(t),\ \   \ \
\|\rho(t,\cdot)\|_{L^\infty(\mathbb{S})}\leq \widetilde{C_2}(t),\ \
\forall t\in\mathbb{R}_+,
\end{align}
where $\widetilde{C_1}(t)$ and $\widetilde{C_2}(t)$ are given in
Remark 3.1.
\end{remark4}
\par

\begin{lemma4} There hold
\begin{align}
\lim_{t\rightarrow0^+}\int_{\mathbb{S}}u_x^2(t,x)dx=
\lim_{t\rightarrow0^+}\int_{\mathbb{S}}\overline{u_x^2}(t,x)=\int_{\mathbb{S}}u_{0,x}^2(x)dx
\end{align}
and
\begin{eqnarray}
\lim_{t\rightarrow0^+}\int_{\mathbb{S}}\rho^2(t,x)dx=
\lim_{t\rightarrow0^+}\int_{\mathbb{S}}\overline{\rho^2}(t,x)dx=\int_{\mathbb{S}}\rho_0^2(x)dx.
\end{eqnarray}
\end{lemma4}
\textbf{Proof}
By Lemma 4.1, for any $T>0$, we have $u^n\in
L^{\infty}((0,T);H^1(\mathbb{S}))$, $u_t^n$ are uniformly bounded in
$L^{\infty}((0,T); L^2(\mathbb{S}))$. By Lemma 2.2, we have $u^n\in
C([0,T];H^1(\mathbb{S}))$. Then in view of Lemma 2.4, Remark 2.1 and
the proof of Lemma 4.1, we have $\{u^{n}\}$ contains a subsequence,
denoted again by $\{u^{n_k}\}$, converging weakly in
$H^1(\mathbb{S})$ uniformly in $t\in(0,T)$. The limit function is
$u$. This implies that $u$ is weakly continuous from $(0,T)$ into
$H^1(\mathbb{S})$, i.e.,
\begin{equation}
u\in C^w([0,T];H^1(\mathbb{S})).
\end{equation}
 Similarly, as
$\rho^{n}\in L^{\infty}((0,T);L^2(\mathbb{S}))$, in view of (3.4)
and (3.6), we get that for all $t\in(0,T)$,
\begin{eqnarray*}
\|\rho^n_t(t,\cdot)\|_{H^{-1}(\mathbb{S})}&=&\sup_{\|
\phi\|_{H^1(\mathbb{S})}=1}\int_{\mathbb{S}}((u^n\rho^n)_{x}\phi+\gamma
\rho^{n}_{x}\phi )dx\leq\|
u^n\rho^n\|_{L^2(\mathbb{S})}+|\gamma|\|\rho^{n}\|_{L^2(\mathbb{S})}\\
&\leq&(\|u_{0}\|_{L^2(\mathbb{S})}+\frac{\sqrt{3}}{6}\mu_{1}+|\gamma|)\mu_{1}.
\end{eqnarray*}
This shows that $\rho_t^n$ is uniformly bounded in $
L^{\infty}((0,T); H^{-1}(\mathbb{S}))$. Then by Lemma 2.4 and Remark
2.1, we have $\{\rho^{n}\}$ contains a subsequence, denoted again by
$\{\rho^{n_k}\},$ converging weakly in $L^2(\mathbb{S})$ uniformly
in $t$. The limit function is $\rho$. This implies that $\rho$ is
weakly continuous from $(0,T)$ into $L^2(\mathbb{S})$, i.e.,
\begin{equation}
\rho\in C^w([0,T];L^2(\mathbb{S})).
\end{equation}
Then by (4.7)-(4.8),  we get
$$ \rho(t,\cdot)\rightharpoonup \rho_0 \ \ \text{and} \
\ u_x(t,\cdot)\rightharpoonup u_{0,x} \ \ \ \text{in}\ \
L^2(\mathbb{S})\ \ \ \text{as}\ \ t\rightarrow0^+.$$ Thus, we have
\begin{eqnarray*}
\liminf_{t\rightarrow0^+}\int_{\mathbb{S}}\rho^2(t,x)dx\geq\int_{\mathbb{S}}\rho_0^2(x)dx
\end{eqnarray*}
and
\begin{eqnarray*}
\liminf_{t\rightarrow0^+}\int_{\mathbb{S}}u_x^2(t,x)dx\geq\int_{\mathbb{S}}u_{0,x}^2(x)dx.
\end{eqnarray*}
Therefore, we deduce
\begin{eqnarray}
&&\liminf_{t\rightarrow0^+}\int_{\mathbb{S}}(u_x^2(t,x)+\rho^2(t,x))dx\\&\geq&
\liminf_{t\rightarrow0^+}\int_{\mathbb{S}}u_x^2(t,x)dx+\liminf_{t\rightarrow0^+}\int_{\mathbb{S}}\rho^2(t,x)dx
\nonumber\\&\geq& \nonumber
\int_{\mathbb{S}}(u_{0,x}^2(x)+\rho_0^2(x))dx.
\end{eqnarray}
Moreover, by Remark 4.1 and (3.2), we infer
\begin{eqnarray*}
&&\int_{\mathbb{S}}\left(\overline{u_x^2}(t,x)+\overline{\rho^2}(t,x)\right)dx\\&\leq&
\liminf_{n_k\rightarrow\infty}\int_{\mathbb{S}}\left((u^{n_k}_x)^2(t,x)+(\rho^{n_k})^2(t,x)\right)dx\\&=&
\liminf_{n_k\rightarrow\infty}\int_{\mathbb{S}}\left((u^{n_k}_{0,x})^2(x)+(\rho^{n_k}_0)^2(x)\right)dx\\&\leq&
\int_{\mathbb{S}}(u_{0,x}^2(x)+\rho_0^2(x))dx.
\end{eqnarray*}
 Thus, we obtain
\begin{align}
\limsup_{t\rightarrow0^+}\int_{\mathbb{S}}\left(\overline{u_x^2}(t,x)+
\overline{\rho^2}(t,x)\right)dx\leq\int_{\mathbb{S}}(u_{0,x}^2(x)+\rho_0^2(x))dx.
\end{align}
In view of (4.3), (4.9)-(4.10), we get
\begin{eqnarray*}
\liminf_{t\rightarrow0^+}\int_{\mathbb{S}}(u_x^2(t,x)+\rho^2(t,x))dx
&\geq&
\int_{\mathbb{S}}(u_{0,x}^2(x)+\rho_0^2(x))dx\\
&\geq&\limsup_{t\rightarrow0^+}\int_{\mathbb{S}}\left(\overline{u_x^2}(t,x)+
\overline{\rho^2}(t,x)\right)dx\\
&\geq&\limsup_{t\rightarrow0^+}\int_{\mathbb{S}}(u_x^2(t,x)+\rho^2(t,x))dx.
\end{eqnarray*}
This completes the proof of the lemma.
\par

\begin{lemma4}There hold
\begin{equation}
\frac{\partial}{\partial
t}(\overline{u_{x}^{2}}+\overline{\rho^{2}})-\frac{\partial}{\partial
x}[(u+\gamma)(\overline{u_{x}^{2}}+\overline{\rho^{2}})]=-4\mu_{0}uu_{x}+4\mu_{0}^{2}u_{x}+u_{x}\mu_{1}^{2}
\end{equation}
in the sense of distributions on $((0,T)\times\mathbb{R}).$
\end{lemma4}
\textbf{Proof} Note that $z^{n_{k}}$ is the solution of the system
(2.1) with the initial $z_{0}^{n_{k}}.$ Differentiating the first
equation in (2.1) with respect $x$ and using
$\partial_{x}^{2}A^{-1}w=-w+\mu(w)$, we have
$$u^{n}_{tx}-(u^{n}+\gamma)u^{n}_{xx}=-2\mu^{n}_{0}u^{n}+\frac{1}{2}(u^{n}_{x})^{2}-\frac{1}{2}(\rho^{n})^{2}
+2(\mu_{0}^{n})^{2}+\frac{1}{2}(\mu_{1}^{n})^{2}.$$ Then
\begin{align*}
\frac{\partial}{\partial t} (u_x^{n_k})^2-\frac{\partial}{\partial
x}((u^{n_k}+\gamma)(u_x^{n_k})^2)=-4\mu_{0}^{n_k}u^{n_k}u_{x}^{n_k}&-u_{x}^{n_k}(\rho^{n_k})^{2}\\
&+4(\mu_{0}^{n_k})^{2}u_{x}^{n_k}+u_{x}^{n_k}(\mu_{1}^{n_k})^{2}
\end{align*}
and
$$\frac{\partial}{\partial t}(\rho^{n_k})^2-
\frac{\partial}{\partial
x}((u^{n_k}+\gamma)(\rho^{n_k})^2)=u_x^{n_k}(\rho^{n_k})^2.$$ Adding
the above two equalities and  letting $n_k\rightarrow\infty$, in
view of Lemma 4.1 and Remark 4.1, we get (4.11).

\begin{lemma4} There hold
\begin{align}
&\frac{\partial}{\partial
t}(u_x^2+\rho^{2})-\frac{\partial}{\partial
x}((u+\gamma)(u_x^2+\rho^{2}))\\
\nonumber= \ &u_{x}^{3}-4\mu_{0}uu_{x}-u_{x}\overline{u_{x}^{2}}
-u_{x}\overline{\rho^{2}}+4\mu_{0}^{2}u_{x}+\mu_{1}^{2}u_{x}+u_{x}\rho^{2}
\end{align}
in the sense of distributions on $((0,T)\times\mathbb{R}).$
\end{lemma4}
\textbf{Proof} Since $z^{n_{k}}$ satisfy
$$u^{n}_{tx}-((u^{n}+\gamma)u^{n}_{x})_{x}=-2\mu^{n}_{0}u^{n}-\frac{1}{2}(u^{n}_{x})^{2}
-\frac{1}{2}(\rho^{n})^{2}+2(\mu^{n}_{0})^{2}+\frac{1}{2}(\mu^{n}_{1})^{2}$$
and
$$\rho^{n}_{t}-(\rho^{n} u^{n})_{x}=\gamma \rho^{n}_{x}.$$ In view of Lemma 4.1 and
Remark 4.1, letting $n_k\rightarrow\infty$, we obtain
\begin{equation}
u_{tx}-((u+\gamma)u_{x})_{x}=-2\mu_{0}u-\frac{1}{2}\overline{u_{x}^{2}}-\frac{1}{2}\overline{\rho^{2}}
+2\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2}
\end{equation}
 and
\begin{equation}
\rho_{t}-(\rho u)_{x}=\gamma \rho_{x}
\end{equation}
in the sense of distributions on $((0,T)\times\mathbb{R}).$

Denote $u_{n,x}(t,x):=\left((u_x(t,\cdot)\ast \phi_{n}\right)(x)$
and $\rho_{n}(t,x):=(\rho(t,\cdot)\ast \phi_n)(x).$ According to
Lemma II.1 of \cite{D-L}, it follows from (4.13)-(4.14) that
$u_{n,x}$ and $\rho_{n}$ solve
\begin{equation}
\frac{\partial u_{n,x}}{\partial t}-(u+\gamma)\frac{\partial
u_{n,x}}{\partial
x}=(u^2_x-2\mu_{0}u-\frac{1}{2}\overline{u_{x}^{2}}-\frac{1}{2}\overline{\rho^{2}}
+2\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2})\ast\phi_{n}+\tau_{n}
\end{equation} and
\begin{equation}
\frac{\partial\rho_{n}}{\partial
t}-(u+\gamma)\frac{\partial\rho_{n}}{\partial
x}=(u_x\rho)\ast\phi_{n}+\sigma_{n},
\end{equation}
where the errors $\tau_n=\left(u\frac{\partial u_{x}}{\partial
x}\right)\ast\phi_{n}-u\cdot\frac{\partial u_{n,x}}{\partial x}$ and
$\sigma_n=\left(u\frac{\partial \rho}{\partial
x}\right)\ast\phi_{n}-u\cdot\frac{\partial \rho_{n,x}}{\partial x}$
tend to zero in $L^1_{loc}(\mathbb{R}_+\times\mathbb{R}).$ Using
(4.15) and (4.16), we get
\begin{align}
&\frac{\partial(u_{n,x})^2}{\partial t}-\frac{\partial}{\partial
x}((u+\gamma)u_{n,x}^2)\\
\nonumber= \
&2u_{n,x}\left((u^2_x-2\mu_{0}u-\frac{1}{2}\overline{u_{x}^{2}}-\frac{1}{2}\overline{\rho^{2}}
+2\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2})\ast\phi_{n}+\tau_{n}\right)-u_{x}u_{n,x}^{2}
\end{align}
and
\begin{align}
\frac{\partial(\rho_{n})^2}{\partial t}-\frac{\partial}{\partial
x}((u+\gamma)\rho_{n}^2)=2\rho_{n}((u_x\rho)\ast\phi_{n}+\sigma_{n})-u_{x}\rho_{n}^{2}.
\end{align}
Sending $n\rightarrow\infty$ in (4.17) and (4.18) and adding the
results yield (4.12).

Next, we give the main lemma of this section.
\begin{lemma4}
There hold
\begin{equation}
\overline{u_x^2}(t,x)=u_x^2(t,x) \ \ \ \text{and} \ \ \
\overline{\rho^2}(t,x)=\rho^2(t,x) \ \ a.e. \ \text{on}\ \
(0,T)\times\mathbb{R}.
\end{equation}
\end{lemma4}
\textbf{Proof}
Subtracting (4.12) from (4.11) and  integrating the obtained
equality over $(\varepsilon,t)\times\mathbb{S})$ give
\begin{align*}
&\int_{\mathbb{S}}\left(\overline{u_x^2}-u_x^2+\overline{\rho^2}-\rho^2\right)(t,x)dx\\
=&
\int_{\varepsilon}^t\int_{\mathbb{S}}\left(\overline{u_x^2}-u_x^2+\overline{\rho^2}
-\rho^2\right)u_x(t,x)dxds\\
&+\int_{\mathbb{S}}\left(\overline{u_x^2}(\varepsilon,x)-u_x^2(\varepsilon,x)+\overline{\rho^2}(\varepsilon,x)-\rho^2(\varepsilon,x)\right)dx,
\end{align*}
for almost all $t\in(0,T)$. Letting $\varepsilon\rightarrow0$ and
using Lemma 4.2 and (4.4), we get
\begin{align*}
\int_{\mathbb{S}}\left(\overline{u_x^2}-u_x^2+\overline{\rho^2}-\rho^2\right)(t,x)dx
\leq
\widetilde{C_1}(T)\int_{0}^t\int_{\mathbb{S}}\left(\overline{u_x^2}-u_x^2+\overline{\rho^2}-\rho^2\right)dxds.
\end{align*}
Using Gronwall's inequality,  we obtain
\begin{align*}\int_{\mathbb{S}}\left(\overline{u_x^2}-u_x^2+\overline{\rho^2}-\bar{\rho}^2\right)(t,x)dx
\leq0.
\end{align*}
By (4.3), we deduce
\begin{align*}
0\leq\int_{\mathbb{S}}\left(\overline{u_x^2}-u_x^2+\overline{\rho^2}-\rho^2\right)(t,x)dx\leq0,
\end{align*}
which yields
$$\int_{\mathbb{S}}(\overline{u_x^2}-u_x^2)(t,x)dx=\int_{\mathbb{S}}(\overline{\rho^2}-\rho^2)(t,x)dx=0.$$
In view of $u^n_x, u_x$ and $\rho^n,\rho$ being periodic with
respect to $x$, we deduce that (4.19) holds.

To prove our main theorem, we need the following lemma.

\begin{lemma4} (\cite{E-L}) Let U be the bounded domain in $\mathbb{R}^n.$
Assume that the sequence $\{f_k\}_{k=1}^{\infty}$ is bounded in
$L^\infty(U;\mathbb{R}^m)$. Then there exist a subsequence
$\{f_{k_j}\}_{j=1}^{\infty}\subset\{f_k\}_{k=1}^{\infty}$ and for
a.e. $x\in U,$ a Borel probability measure $\nu_x$ on $\mathbb{R}^m$
such that for $F\in C(\mathbb{R}^m)$ we have
$$F(f_{k_j})\rightharpoonup^*\overline{F} \ \ \ \ \text{in}\ \
L^\infty(U),$$ where $\overline{F}(x)=\int_{\mathbb{R}^m}F(y)d\nu_x$
for a.e. $x\in U$. Moreover, if $\nu_x$ is a unit point mass for
a.e. $x\in U$, then $f_{k_j}\rightarrow f$ in $L^2(U;\mathbb{R}^m)$,
here $f(x)=\int_{\mathbb{R}^m}yd\nu_x$.
\end{lemma4}

By (3.7)-(3.8), for any $T>0$ and $a<b$, $\{u_x^n\}$ and
$\{\rho^n\}$ are uniformly bounded on $(0,T)\times(a,b)$. Using
Lemma 4.6, there exist a subsequences $\{u_x^{n_k},
\rho^{n_k}\}_{k=1}^{\infty}\subset\{u_x^n,\rho^n\}_{n=1}^{\infty}$
 for a.e. $(t,x)\in(0,T)\times(a,b)$ and two Borel probability measures $\mu_{(t,x)},\nu_{(t,x)}$
 on $\mathbb{R}$ such that for each $F\in
C(\mathbb{R})$ we have
$$F(u^{n_k}_{x})\rightharpoonup^*\overline{F}_1, \ \ F(\rho^{n_k})\rightharpoonup^*\overline{F}_2 \ \ \text{in}\ \
L^\infty((0,T)\times(a,b)),$$ where
$\overline{F}_1(t,x)=\int_{\mathbb{R}}F(y)d\mu_{(t,x)}$ and
$\overline{F}_2(t,x)=\int_{\mathbb{R}}F(y)d\nu_{(t,x)}$ for a.e.
$(t,x)\in(0,T)\times(a,b)$. By Lemma 4.5, we have that
$\mu_{(t,x)}=\delta_{u_x(t,x)}$ and $\nu_{(t,x)}=\delta_{\rho(t,x)}$
for a.e. $(t,x)\in(0,T)\times(a,b).$ Then, in view of Lemma 4.6
 and from the arbitrariness of $T,a,b$, we obtain
\begin{align}
u_x^{n_k}\rightarrow u_x,\ \
\rho^{n_k}\rightarrow \rho\ \ \ \text{in} \ \
L^2_{loc}(\mathbb{R}_+\times\mathbb{R}).
\end{align}

\par
With the above preparations, we now conclude the proof of the
Theorem 1.1. Let $u$ be the limit of the  approximate solutions
$u^{n_k}$ as $n_{k}\rightarrow\infty$. It then follows from Lemma
2.2, Lemma 4.1 and Remark 4.1 that $u\in
C([0,+\infty)\times\mathbb{R})\cap
L_{loc}^{\infty}(\mathbb{R_+},H^1(\mathbb{S}))$ and $\rho\in
L^{\infty}(\mathbb{R_+},L^2(\mathbb{S}))$ hold. From (4.20), we have
that
$$\partial_{x}g*\left(2\mu_{0}u^{n}+\frac{1}{2}(u^{n}_{x})^{2}+\frac{1}{2}(\rho^{n})^{2}\right)\rightarrow
\partial_{x}g*(2\mu_{0}u+\frac{1}{2}u_{x}^{2}+\frac{1}{2}\rho^{2})$$ in the
sense of distributions on $\mathbb{R_+}\times\mathbb{R}$. This shows
that $z$ satisfies (2.5) in the sense of distributions on
$\mathbb{R_+}\times\mathbb{R}$.

From Lemma 4.2, we have $z\in
C^w_{loc}(\mathbb{R}_+;H^1(\mathbb{S})\times L^2(\mathbb{S}))$.
Consequently, we will prove that $z\in
C(\mathbb{R}_+;H^1(\mathbb{S})\times L^2(\mathbb{S}))$. Note that
$u\in C((0,\infty); L^{\infty}(\mathbb{S})),$ it is enough to show
that $\int_{\mathbb{S}}(u_{x}^{2}+\rho^{2})dx$ is conserved in time.
Indeed, if this holds, then
\begin{align*}
&\|z(t)-z(s)\|^2_{H^1(\mathbb{S})\times L^2(\mathbb{S})}\\= \
&\|u(t)-u(s)\|_{L^{2}(\mathbb{S})}^{2}+\|u_{x}(t)-u_{x}(s)\|_{L^{2}(\mathbb{S})}^{2}
+\|\rho(t)-\rho(s)\|_{L^{2}(\mathbb{S})}^{2}
\\= \ &\|u(t)-u(s)\|_{L^{2}(\mathbb{S})}^{2}+\|u_{x}(t)\|_{L^{2}(\mathbb{S})}^{2}+\|u_{x}(s)\|_{L^{2}(\mathbb{S})}^{2}
-2(u_{x}(s), u_{x}(t))_{L^{2}(\mathbb{S})}\\
+&\|\rho(t)\|_{L^{2}(\mathbb{S})}^{2}+\|\rho(s)\|_{L^{2}(\mathbb{S})}^{2}
-2(\rho(s), \rho(t))_{L^{2}(\mathbb{S})}\\
= \ &
\|u(t)-u(s)\|_{L^{2}(\mathbb{S})}^{2}+2(\|u_{0,x}\|_{L^{2}(\mathbb{S})}^{2}+\|\rho_{0}\|_{L^{2}(\mathbb{S})}^{2})
-2((u_{x}(s), u_{x}(t))_{L^{2}(\mathbb{S})}+(\rho(s),
\rho(t))_{L^{2}(\mathbb{S})}),
\end{align*}
$\forall \ t,s\in\mathbb{R}_+.$ Since
$\|u(t)-u(s)\|_{L^{2}(\mathbb{S})}^{2}\rightarrow 0$ and
$$(u_{x}(s), u_{x}(t))_{L^{2}(\mathbb{S})}+(\rho(s),
\rho(t))_{L^{2}(\mathbb{S})}\rightarrow
\|u_{x}(t)\|_{L^{2}(\mathbb{S})}^{2}+\|\rho(t)\|_{L^{2}(\mathbb{S})}^{2}
=\|u_{0,x}\|_{L^{2}(\mathbb{S})}^{2}+\|\rho_{0}\|_{L^{2}(\mathbb{S})}^{2},$$
as $s\rightarrow t,$ we have $z\in
C(\mathbb{R}_+;H^1(\mathbb{S})\times L^2(\mathbb{S}))$.

The conservation of $\int_{\mathbb{S}}(u_{x}^{2}+\rho^{2})dx$ in
time is proved by a regularization technique. Denote
$f_n=f\ast\phi_n.$ By Lemma 4.5 and (4.15)-(4.16), we have
\begin{equation}
\frac{\partial u_{n,x}}{\partial t}-(u+\gamma)\frac{\partial
u_{n,x}}{\partial x}=(\frac{1}{2}u^2_x-2\mu_{0}u-\frac{1}{2}\rho^{2}
+2\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2})\ast\phi_{n}+\tau_{n}
\end{equation} and
\begin{equation}
\frac{\partial\rho_{n}}{\partial
t}-(u+\gamma)\frac{\partial\rho_{n}}{\partial
x}=(u_x\rho)\ast\phi_{n}+\sigma_{n},
\end{equation}
Multiplying (4.21) with $ u_{n,x}$, we obtain by integration
\begin{align}
&\frac{1}{2}\frac{d}{dt}\int_{\mathbb{S}}u_{n,x}^{2}dx
-\frac{1}{2}\int_{\mathbb{S}}(u+\gamma)(u_{n,x}^{2})_{x}dx\\
\nonumber= \
&\int_{\mathbb{S}}u_{n,x}(\frac{1}{2}u^2_x-2\mu_{0}u-\frac{1}{2}\rho^{2}
+2\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2})\ast\phi_{n}dx+\int_{\mathbb{S}}u_{n,x}\cdot\tau_{n}dx.
\end{align}
Multiplying (4.22) with $\rho_{n}$, we get by integration
\begin{equation}
\frac{1}{2}\frac{d}{dt}\int_{\mathbb{S}}\rho_{n}^{2}dx
-\frac{1}{2}\int_{\mathbb{S}}(u+\gamma)(\rho_{n}^{2})_{x}dx=\int_{\mathbb{S}}\rho_{n}(u_x\rho)\ast\phi_{n}dx+
\int_{\mathbb{S}}\rho_{n}\sigma_{n}dx.
\end{equation}
Adding (4.23) and (4.24), we have
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int_{\mathbb{S}}(u_{n,x}^{2}+\rho_{n}^{2})dx\\
= \
&\frac{1}{2}\int_{\mathbb{S}}(u+\gamma)(u_{n,x}^{2})_{x}dx+\int_{\mathbb{S}}u_{n,x}(\frac{1}{2}u^2_x-2\mu_{0}u-\frac{1}{2}\rho^{2}
+2\mu_{0}^{2}+\frac{1}{2}\mu_{1}^{2})\ast\phi_{n}dx+\int_{\mathbb{S}}u_{n,x}\cdot\tau_{n}dx\\
+ \
&\frac{1}{2}\int_{\mathbb{S}}(u+\gamma)(\rho_{n}^{2})_{x}dx+\int_{\mathbb{S}}\rho_{n}(u_x\rho)\ast\phi_{n}dx+
\int_{\mathbb{S}}\rho_{n}\sigma_{n}dx.
\end{align*}
As for fixed $T>0$, $u$, $u_x$ and $\rho$ are bounded in
$(0,T)\times \mathbb{S}$. Let $n\rightarrow\infty$, on account of
Lebesgue's dominated convergence theorem, we get
$$\frac{d}{dt}\int_{\mathbb{S}}(u_{x}^{2}+\rho^{2})dx=0.$$
This completes the proof of Theorem 1.1.

\bigskip

\noindent\textbf{Acknowledgments} This work was partially supported
by NNSFC (No. 10971235), RFDP (No. 200805580014), NCET-08-0579 and
the key project of Sun Yat-sen University.

redpajama_set_name: RedPajamaArXiv