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Adversity to Excellence: Stanislas Wawrinka Enters the Elite
By David Cox, Chris Skelton in Features
Overshadowed by Roger Federer for most of his career, the new Swiss No. 1 has found the courage to blaze his own trail to the top.
The actors had returned to a familiar stage. Excitement rippled through Rod Laver Arena as defending champion Novak Djokovic was set to take on world No. 8 Stanislas Wawrinka in the quarterfinal of the 2014 Australian Open.
Just one year prior, in almost the exact round, Wawrinka had lost 10-12 in the fifth set to his Serbian nemesis in one of the most memorable matches of the year.
Despite the defeat, sentiment toward Wawrinka had forever altered in his favor. The long-time Swiss No. 2, Wawrinka had finally stepped out from under the shadow of Roger Federer and emerged as an elite contender in his own right. He had always wielded plenty of weapons, especially an explosive one-handed backhand that has ranked among the game's best throughout his career. What he had lacked until then, and what he added in 2013, was the mental strength needed to contend in a sport where the psychological can matter as much as the physical.
At 28, Wawrinka believes that he has learned to mentally master key moments in matches better than he ever could before. "So much comes from experience," he told Tennis View. "Especially before a big match, I think we all get a little nervous. The question is how you deal with that, and the better players deal with it much better. They accept it more, and they still play their game when they are nervous. Some players change the way they play a little bit on the big points, and that's not good."
For Wawrinka, the key to performing under pressure lies in focusing on himself and the game plan that he brings to the court. "I try to force myself to not change what I had decided to do in the match," he asserted. "It's all about you when go out there. Even if you're not feeling confident, the worst thing you can do is to hesitate and doubt yourself."
Wawrinka credits much of his recent success to Magnus Norman, who joined him in April 2013 after the Masters 1000 tournament in Monte Carlo. Most successful on clay during his career, Norman had risen as high as world No. 2 after reaching the Roland Garros final in 2000. The Swede had worked with former Australian Open champion Thomas Johansson and Rafael Nadal's former nemesis, Robin Soderling, before joining Wawrinka.
"I honestly believe that Magnus is going to make a big difference for me in the next few years," said Wawrinka of his coach. "He's helping me have the confidence to believe in my game. He's telling me to be more aggressive. He wants me to attack more with my forehand and my serve, but my game is still more or less the same. That's understandable, of course. But sometimes on court I'm hesitating about things because I'm not feeling confident in my shots, and that's what I'm trying to change."
The charismatic Swiss showed greater confidence when it mattered most in 2013, and his painful loss to Djokovic in Melbourne soon started to fade. That defeat could have crushed the spirit of a player with less resolve, but for Wawrinka, it only rekindled his motivation to improve.
"That match gave me a huge amount of belief in myself," he reflected. "I always knew that I was capable of beating the top guys, but that match made me genuinely believe it. Afterwards, I was really disappointed, but I thought about it and it encouraged me to keep practicing the way I had been and keep fighting in every match to get the chance to do it again."
Determined to learn from defeat, Wawrinka watched a DVD of his Australian Open loss to Djokovic. The experience offered him insights on what he could improve as well as further reassurance for the future.
"I always think that I know more or less what's happened," he observed. "But it's interesting when you see what the camera shows and what other people will be seeing, not your own insights. I looked at what Novak was doing and what I was doing, and I could see that it was all about my level. When it was high, I was pushing Novak and doing a lot of good things. I have looked back to that match many times ahead of tournaments. It's something positive to think about because I showed myself and everyone watching that I can test him over nearly five hours, and I was really close to beating him. So I still keep the memory of it for when I need it."
Wawrinka got the chance to employ his formula when he faced Djokovic again at last year's US Open, having reached his first major semifinal after dominating defending champion Andy Murray. Again, he pushed the world No. 1 to a fifth set. Again, he fell short. Again, he focused on the positives.
"My team told me they were really proud," said Wawrinka. "[Swiss Davis Cup captain] Severin [Luthi] told me in New York that I can take only positive things from the match because I'd played at an amazing level and I was so close to beating the world No. 1 in a Grand Slam. After a defeat like that, you need to be honest with yourself. This is the most important thing. If it's a match where you gave everything, you need to accept that."
The Swiss capped off his breakthrough 2013 season with more victories against top-10 opponents than any player except Djokovic and Nadal and qualified for the World Tour Finals in London, the prestigious season-ending showcase for the top eight men, for the first time.
After a rigorous off-season, Wawrinka felt ready to improve on his previous year's results at the Australian Open. When he was again set to meet Djokovic, this time in the quarterfinals, some observers still saw Wawrinka as little more than a temporary speed bump the Serb's march toward another showdown with Nadal.
But the script was about to be rewritten.
Circumstances early on in the match between Djokovic and Wawrinka didn't look encouraging for the Swiss as he emerged tight with tension, dropping serve twice in the first set. Yet he persisted, believing against the odds that he could turn his 2-15 record against Djokovic into 3-15. Wawrinka clawed himself back into the match, one bludgeoned backhand and seismic serve at a time. By the third set, he had become the man dictating the baseline rallies and forcing his opponent onto the defensive.
Thoughts of the year before flooded into the minds of everyone watching when the Serb leveled the match and took an early lead in the fifth set. Battling those thoughts as well as a surging Djokovic, Wawrinka remained determined to fight.
"It's always tough, especially against Novak, (but) I cannot just let the match go," he said later. "I don't want to lose every time in five sets against Novak. I had to find solutions. I had to fight within myself to fight against him and try to hold my ground during the game."
Swiftly leveling the set, Wawrinka did not lose serve again for the rest of the night. Then, with Djokovic serving in the 16th game, the Swiss struck the right balance of patience and aggression as he earned the decisive break and the win. Wawrinka had turned the tide from 2013 to 2014, winning a fifth set from the man who had won two five-set matches against him the year before.
Many men would have been ecstatic with that feat, content to bask in its glow and unable to regain their focus for the next match. Wawrinka, however, greeted his triumph with poise.
"I trust myself when I go on court," he said calmly. "I know that I can beat those players. I try everything to beat them. Even when I lose, I go back to practice to try to improve."
In the semifinals, Wawrinka's steeliness surfaced again when he defeated another top-10 opponent, Tomas Berdych, despite breaking serve just once in four sets. The reward was his first major final, aligned against no less an opponent than world No. 1 Rafael Nadal. Not only had Wawrinka never defeated three consecutive opponents ranked in the top 10, but he had never even taken a set off of Nadal in their 12 previous meetings. And yet he rose to the occasion.
"It was the perfect start," Wawrinka would say later of the first set in the final. "I was playing amazing tennis." Seizing the set with fearless aggression, he raced ahead early in the second set. Then Nadal abruptly doubled over in pain as he injured his back. When he returned to the court, the world No. 1 was only a shadow of himself, unable to move with his usual explosiveness.
Wawrinka soon claimed a two-set lead, edging closer to the finish line. Nevertheless, a champion of Nadal's pedigree does not surrender easily. He continued to compete with determination despite his ailing back. Unsure of what to expect from across the net, Wawrinka lost the boldness that he had shown in the first set under more predictable conditions.
"I didn't play well because I was waiting for him to miss, and that was a big mistake from myself," he said afterward. "I [thought that] I'm not going to win the match because I'm nervous. I started to realize that I can win the Grand Slam."
When the third set slipped away, Wawrinka searched for his former aggression. Just enough of it returned late in the fourth set to halt Nadal's improbable comeback. At 5-3, with nerves still jangling, Wawrinka held serve at love to secure his first major title. The man long overshadowed in his own nation had reached center stage at last.
Nobody was more stunned than Wawrinka by his achievement. "I never expected to win a Grand Slam," he reflected. "I never dreamed about that because I was not good enough to beat those guys. I still think that I'm dreaming. I saw so many finals. I always try to watch the final of a Grand Slam because that's where the best players are playing. I never expected to play a final. I never expected to win a Grand Slam. And right now, I just did it."
Along Wawrinka's left forearm extends a quotation that inspires him. Attributed to Irish playwright Samuel Beckett, the tattoo reads, "Ever tried. Ever failed. No matter. Try Again. Fail again. Fail better."
With his momentous win, Wawrinka has finally crossed the line between failing better and achieving success. And in his 36th appearance at a major, he showed that this line is much more narrow that it seems.
Wawrinka Before/After January 2013
Average wins per season (2006-12): 33
Wins from Jan. 1, 2013-Jan. 31, 2014: 62
Average wins per season vs. top 10 (2006-12): 2.1
Wins vs. top 10 from January 1, 2013-January 31, 2014: 12
Average wins at majors per season (2006-12): 7.6
Wins at majors from Jan. 1, 2013-Jan. 31, 2014: 19
Ranking before 2013 Australian Open: No. 17
Ranking after 2014 Australian Open: No. 3
This article is from the Mar/Apr 2014 - Stan Wawrinka issue
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Should a candidate wish to appeal their assessment decision or other decision affecting the candidates, they must submit their appeal, in writing to the Policy and HR Director. Their completed form must be emailed to the enquiries email address within 14 working days of being notified of the assessment outcome. This would usually be the date given on the decision letter/e-mail.
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The Policy Director will respond to the candidate in order to acknowledge receipt of the appeal within 5 working days;
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The following section includes advice for candidates and/or centres as to the grounds for submitting an appeal. Please note a candidate or centre's dissatisfaction with the pass result, in itself, is not a reason for lodging an appeal. The candidate must demonstrate there were reasons for their performance which may have resulted in their inability to meet the standard for the assessment/examination.
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Candidates will be charged £80 which must be paid at the time of their submitting the appeal. The basis of this fee is to cover the administration processes and staffing involved in the appeals process. The fee has been kept to a minimum in order to prevent acting as a deterrent to candidates. The fee will be refunded to candidates in the event of an appeal being upheld.
Centres will be charged £150 which must be paid at the time of the centre submitting its appeal to LRN. The basis of this fee is to cover the administration processes and staffing involved in managing the centre appeals process. The fee has been kept to a minimum in order to prevent acting as a deterrent to candidates. The fee will be refunded to centres in the event of an appeal being upheld.
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• A basic knowledge of general computing concepts.
• Perform a local media-based installation of Windows Server 2012.
• Select appropriate storage technologies and configure storage on a Windows Server.
• Describe fundamental network components and terminology so you can select an appropriate network component.
• Implement a network by selecting network hardware components and technologies and determine the appropriate network hardware and wiring components for a given situation.
• Describe the protocols and services within the Transmission Control Protocol/Internet Protocol (TCP/IP) suite of protocols and implement IPv4 within a Windows Server environment.
• Describe and implement server roles.
• Implement and configure an Active Directory Domain Service (AD DS) forest.
• Describe the concept of defense-in-depth and determine how to implement this approach with Windows Server.
• Identify the security features in Windows Server that help to provide defense-in-depth.
• Identify the network-related security features in Windows Server to mitigate security threats to your network.
• Identify and implement additional software components to enhance your organization's security.
• Monitor a server to determine the performance level.
• Identify the Windows Server tools available to maintain and troubleshoot Windows Server.
• Configuring Devices and Device Drivers.
• Performing a Local Media-Based Installation.
• Managing Disks and Volumes.
• Create and Mount a VHD File.
• Creating and Making Available New Volumes.
• Vary the Sizes of NTFS and ReFS Volumes.
• Create a Fault Tolerant Disk Configuration.
• Connecting to the Internet.
• Understanding the OSI Model.
• Understanding Adapters, Hubs, and Switches.
• Determining the Appropriate Network Hardware.
• Selecting a Suitable Wiring Infrastructure.
• Determining an Appropriate IPv4 Addressing Scheme.
• Configuring IPv4 with Windows Server.
• Configuring and Testing Name Resolution.
• Viewing the IPv6 Configuration.
• Considerations for Provisioning Roles.
• Determining the Appropriate Roles to Deploy.
• Deploying and Configuring the Determined Server Roles.
• Introducing Active Directory Domain Services (AD DS).
• Managing Users, Groups, and Computers.
• Promoting a New Domain Controller.
• Creating an Organizational Unit.
• Configuring Security Settings in Internet Explorer.
• Overview of Windows Security.
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• Securing NTFS Files and Folders.
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• Overview of Network Security.
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• Create a Server to Server Connection Security Rule.
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• Use the Security Configuration Wizard.
• Run the Best Practice Analyzer (BPA).
• Creating a Performance Baseline.
• Simulating a Server Load.
• Determining Probable Performance Bottlenecks.
• Create, Test, and Verify an Alert.
• Troubleshooting Windows Server Startup.
• Server Availability and Data Recovery.
• Applying Updates to Windows Server.
• Installing and Configuring Windows Server Update Services.
• Gathering Information to Start the Troubleshooting Process.
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Q: asp web api json serialization returns private properties I have a Web api controller class
and i call a method on a proxy class generated from a WSDL from the web api controller,
and i return an object of type defined in the WSDL proxy class.
but the xml/json returned contains the private members of the proxy class
the private members are
    private string companyField;

    private string soldBPField;

    private string fromDateField;

    private string toDateField;

    private long succStatusField;

and the xml returned is :
<companyField>700</companyField>
<soldBPField>999000353</soldBPField>
<fromDateField>01-01-2012</fromDateField>
<toDateField>01-01-2013</toDateField>

the json returned is also similar
I changed the xml formatter in the global.asax as follows:
var xml = GlobalConfiguration.Configuration.Formatters.XmlFormatter;
xml.UseXmlSerializer = true;

this changed the XML returned to:
<company>700</company>
<soldBP>999000353</soldBP>
<fromDateField>01-01-2012</fromDate>
<toDate>01-01-2013</toDate>

but the json returned still contains the name of the private variables
can any one help me with this

A: Are your types marked as [Serializable]? Serializable means that the serializer should serialize all fields - private or public. Try to remove Serializable or else use this line to stop the JSON formatter from recognizing the attribute:
config.Formatters.JsonFormatter.SerializerSettings.ContractResolver = new DefaultContractResolver();

redpajama_set_name: RedPajamaStackExchange


"Kristy" hails from China were they throw the baby girls away. Whether or not the fact that he was a boy saved him, we do not know. Certain it is that they never realized in those days of his early youth the wonderful combination of ennobling elements that are manifest in his character today. They have found expression in his senior year in a variety of activities, including the office of Beltionian vice-president, senior council representative, fullback on the 1921 eleven, and editor-in-chief of The Record. "Art" expects to land in China eventually, where we predict he will again occupy an editor's chair.
Concerning Mr. Arthur E. Christopherson, who was a student of mine throughout his college course, I desire to write a word of commendation as bearing upon his worthiness to receive a fellowship. He is a sturdy character, vigorous in health and purpose…His habits are good, his ideals are Christian, his energies are superb, his self-enjoyment is ample, his record is good, so I commend him to your most favorable consideration.
Of course, he was accepted. Residing in New York, Christy published Images in Jade, translations from classical and modern Chinese poetry. His courses at Columbia led in 1932 to a doctorate in comparative literature. His dissertation, entitled The Orient in American Transcendentalism: A Study of Emerson, Thoreau and Alcott, appeared in 1932 with The Transmigration of the Seven Brahmans. From 1935-36 he was a Guggenheim Fellow; and from 1930 to 1945 he taught at Columbia University Department of English and Comparative Literature, supervising master's theses in American literature. In 1945 he was appointed professor of American Literature at the University of Illinois, Champaign-Urbana.
When his missionary mother died in 1941, Christy composed an obituary published in The Alliance Weekly: "Her heart quietly ceased its functioning and she moved unobtrusively as in all her earthly life, to her heavenly resting place." Five years later, recently returned from a conference on the participation of higher education in the United Nations Educational, Scientific and Cultural Organization at Estes Park, Colorado, Christy's own heart "quietly ceased" as he crossed a street on his way from the University of Illinois campus to his home. Shortly thereafter, admitted to McKinley Hospital, he was pronounced dead at age 46 from an entirely unexpected heart attack. His death was an incalculable loss to the field of letters. He left a widow, Gertrude Noetzel (B.A. University of Wisconsin, 1920 and M.A. University of Illinois, 1947), and a son, Bruce, born in 1928 (B.A. University of Illinois, 1950).


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Why Rhinocorn?
We believe successful charitable organizations depend on the power of their highest level stakeholders to thrive:
Major Gift Donors
We believe charitable organizations are the most important driving force behind positive social change but often feel overwhelmed with managing and building models to engage their highest level stakeholders powerfully.
Rhinocorn is an organizational evolution consultancy.
We're focused on the fundamental building blocks of the organizational architecture that define the processes you use to support your highest level stakeholders. We apply art and science to help you and your nonprofit improve your current infrastructure or build new models of success.
We're not like everyone else.
We don't help with short-term fixes like "finding donors" or "recruiting volunteers or board members". Instead, we focus on the designs and strategies of your long-term needs to grow your organizational capacities. We save you time and money, allowing staff to keep focus on day-to-day operations while new designs and strategies are being developed and installed so you can continue changing the world.
Areas of cONSULTING + Design
Rhinocorn's Service options
We encourage prospective clients to schedule a 60min Free Consultation with Team Rhinocorn, as well as considering these options:
Monthly Advisory Services
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Why Others Choose Rhinocorn
"Ben is a leading expert in all things Not for Profit and is leading the way to establish the future of the NFP world. He is an expert in networking and has developed several resources that any Not for Profit professional should take advantage of."
Adam Janes | Volunteer Coordinator, Christie Lake Kids
Ben is as smart as a whip, but more importantly he is not afraid of the truth or the challenges of helping nonprofits rethink their systems to be more influential and effective.
Greg Baldwin | President of VolunteerMatch
"Ben is one of the most thorough, conscientious, phenomenal people I've ever had the privilege of working with ... his creativity, enthusiasm and attention to detail stood out among his colleagues."
Shana Wallenstein | Director of Development & Communications, Beachwood City Schools
"From our earliest conversations with Ben, we knew he 'got' us – our organizational culture, our engagement challenges, and our vision for the future. He was passionate about being a partner in helping to envision and actualize our non-profits' potential, and has been a crucial asset in helping us frame a more comprehensive corporate engagement philosophy, as well as building the processes and content we needed to respond to our partners' engagement needs... We accomplished more in our several months working with Ben that we had in the previous several years to propel our engagement strategy forward."
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"Ben is awesome at coming up with out-of-the-box ideas and solutions—and he also excelled at implementing them."
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"Brilliant ideas, creative solutions, and great people."
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"Ben has proven himself to be hardworking, good-natured, and a continual source of innovative ideas. I do not hesitate to recommend him and will welcome an opportunity to work with him again in the future."
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"Rhinocorn has secured its position in the market as a leader in the area of engagement professionalism purely because Ben has seen both sides of the story - from a leader in the non-profit world to a 'service provider' - which gives him a unique understanding of the challenges. And Ben is a pioneer in this burgeoning area of engagement professionalism, and for those non-profits and corporations alike, who are looking at things from a different perspective, Rhinocorn should be on the list of people to contact."
Victoria Miles | Vision Wheel Marketing
"Ben has the rare quality of being able to create innovative strategies with long-term vision and to then turn around and create equally impressive tactical level frameworks to bring the vision in question a reality."
Christine Foster | Corporate Social Responsibility Specialist at TATA Consultancy Services
"Not only is Ben a true subject matter expert with a deep understanding of Corporate Social Responsibility, he also has the ability to understand the viewpoints of different audiences and write for them."
Maria Ho | Association for Talent Development (ATD)
"Ben has supported me 150% and oversaw every step in the journey towards Courage to Caregivers successfully obtaining our 501c3 as well as mapping out our next steps for a successful launch. If your organization is needs of 'refresh' or consulting services in ANY way I suggest you both reach out to him..."
Kristi Anderson Horner | Executive Director at Courage to Caregivers


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Libby told you about "Art Full Text" in her post. I saw the show yesterday and have to say I loved it. It's one of the best City Hall outings yet so kudos to Curator Cavin Jones as well as to all the artists — and there are a bunch — 35!
hunn, lydia
(top image is exhibition case with Alison Willse's slumpy Marlboro Menthols sculpture, "A Pack a Day." Doesn't the soft pack look great in the hard case?)
The first thing to note about this exhibit, which follows on the heels of the abolition of the Mayor's Office of Arts and Culture is that instead of the free informational hand-outs about the show (a list of titles, a curator's statement, a greeting by the Mayor, artists' statements and the like — extremely useful to hard-working reviewers) you now get the show's postcard announcement which includes a handy, mail-in coupon on the back soliciting donors to Art in City Hall. ($25, $50 and "Other" are the suggested amounts.)
Ah budgets. At least there's a show.
(image right is case with Lydia Hunn's "Many Things are Moved by Shovels," a piece Libby and I saw in a larger version at Highwire Gallery's Shovel Show — and at Slought's emerging artist show if memory serves. A good piece travels well and it looks just fine here, too.)
hollis, john f.
The second thing to note about the layout of the exhibit is that City Hall now has an airport-type security system in place in the north corridor on the 4th floor (where Council Chambers are). In order to access the art on the 4 you must put your stuff through the x-ray machine and walk through the metal detector gate. It's not really a big deal but I mention it as one of those things you don't normally have to do when going out to look at art. Art in City Hall brings art and the world into a collision it otherwise wouldn't have.
(image left is more shovels in glass cases I found right around the corner from Hunn's piece. The bottom shovel — which is actually a spade — is dated Aug. 16, 1871 and the information says "spade was used by John Rice, esq, President of the Building Commission to break ground on the site of new city hall. The top shovel dates from Jan. 23, 1962.)
My favorite pieces in the show have a sculptural presence. It may be that sculpture is the way to go in the vitrines here which, as Libby says, make some things difficult to see.
Shiny paintings and small scale works are particularly hard because of the glare on the glass.
(image is black on black painting, "Jessica: Nutrition Facts" by Samuel Yun which would be hard to see under any circumstances. Here in the glare of bright daylight it made a superb foil to show the reflections in the glass.)
I particularly like the works that use found words, like Kent Latimer's "No Exit" (image left) made from found signage and "Victory" rat traps (I do hope those weren't found).
I've written about Latimer whose sculpture is sorely under-represented in shows around town. Here's a post after a studio visit I had with the artist.
John F. Hollis's "Love Proclaimed," has also got the found word karma going on. Hollis's wonderful color photograph of some grafitti declaring a boy's love for a girl written below a billboard labeled "Infinity" (image right, sorry it's blurry) shows the world at its most serendipitous and wry.
Speaking of serendipity, I was driving into town and popped a cd in that was a Janet Cardiff audio, "The Missing Voice: Case Study B" which was sent as promo material for the new Cardiff/Miller piece opening next month at Eastern State Penitentiary. The audio was one of the artist's walks, set in London, and, as usual, the story was paranoid and about a mysterious journey being re-traced and the voice is all conspiratorial and the ambiance is lost! oh lost!
Then I parked the car and saw this exhibit and stopped dead in front of Kate Murken's yarny tale "A Journey Upwards and Downwards" with the words all conspiratorial and the ambiance all lost! oh lost! and I knew that Murken and Cardiff were riding a similar spooky wavelengh making their art. (image is detail of Murken's lovely work)
If you're anywhere near City Hall, stop in.
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Artist and Art Director Eric Battle talks about "BLAM!", A graphic novel profiling fourteen Black historic Philadelphians


The Disneyland Resort Diamond Celebration will touch every aspect of the guest experience, beginning May 22, 2015. As guests enjoy the spectacular, new shows highlighting the celebration, they'll also encounter brilliant décor, special entertainment throughout the parks and countless opportunities to amplify the Diamond Celebration fun with food, beverage and souvenir offerings.
In addition to the "Paint the Night" parade, immersive "Disneyland Forever" fireworks spectacular and new "World of Color – Celebrate!" shows, opportunities for guests to become part of the musical magic of Disneyland will be found throughout the resort.
Guests may join a daily Diamond March-Along up Main Street, U.S.A., with the Disneyland Band and Disney characters in their sparkling, new Diamond Celebration costumes. The cavalcade concludes with a Diamond Sing-Along at Town Square in Disneyland Park.
Also on Main Street, U.S.A., the Dapper Dans quartet will invite guests to join Diamond Sing-Alongs, while the Straw Hatters band will encourage Diamond Dance-Alongs. Every musical group across the Disneyland Resort is incorporating special Diamond Celebration moments into its shows.
Guests may Dream-Along with Mickey Mouse and the Red Car News Boys, as Mickey shares a whole new song about his vision for the Happiest Place on Earth. When the Five & Dime arrive on Buena Vista Street, guests will be immersed in a joyous Dance-Along with Dime, Goofy and the boys.
Mad T Party returns to Disney California Adventure beginning May 22, as guests are invited down the rabbit hole to experience the Diamond Mad T Party, a nighttime entertainment party inspired by Tim Burton's "Alice in Wonderland" motion picture.
The Mad T Party main stage will feature a live band, performing contemporary and classic rock favorites as well as newly mashed-up Disney tunes given a rare Mad T Party twist. When the band isn't onstage, the White Rabbit DJ takes over and plays the latest, hottest dance tunes.
Also for Mad T Party, the Hottest Deck in Town dance crew will perform their high-energy show with a new look at a new location, the Hollywood Backlot Stage.
Diamond Mad T Party guests will also play interactive games, take photos or just chill out with a cold drink from the Drink Me Bar or Chiller truck. A new photo location combines the fantastical spirit of the Mad T Party with a tribute to Disney's past.
Disneyland fans and collectors desiring a beautiful Diamond Celebration memento will find plenty to excite them at Disneyland Resort merchandise locations. More than 500 merchandise items will be available, including apparel, accessories, art, collectibles, headwear, home décor, jewelry, pins, souvenirs, toys and more.
Headwear will include a Diamond Celebration ear hat, Minnie Mouse rhinestone-embellished headband, and youth and adult baseball hats. Also being introduced from the Diamond Collection is a crystal-encrusted Minnie Mouse headband and Mickey Mouse ear hat.
Home Décor items will include frames, snow globes, mugs, a dinner plate, dessert plate, trivet, napkin, drinkware and bottle topper, as well as a beach towel, throws and pillow.
Toys will include an array of plush, a fashion doll, a Diamond Celebration truck hauler, train set, playing cards, puzzles, Sleeping Beauty Castle playset and more.
Open and limited edition pins, trading accessories, boxed pin sets and framed pin sets will be available.
Collectors also will be interested in limited release, limited edition and crystal embellished Vinylmation.
Art & Collectibles will include notable brands such as Precious Moments, Lenox, Highland Mint and Arribas Bros. Over the course of the celebration, fans will find artwork by Dave Avanzino, Disney Design Group (Costa Alavezos, Tyler Dumas, Susan Foy, Caley Hicks, Alex Maher, Randy Noble), Liana Hee, Ink & Paint artists, Brian Jowers, Lisa Keene, Jerrod Maruyama, Kurt Raymond, Tim Rogerson, SHAG, Javier Soto and The Thomas Kinkade Studios and Gregg Visintainer, among others.
Made with Magic: Newly designed Diamond Celebration ear hats, wands and paint brushes will light up to accompany the big, new after-dark shows.
The Disneyland Resort is filled with a variety of popular dining options. For the Diamond Celebration, some new specialties will join the classic favorites, served in colorful Diamond Celebration souvenir containers.
Special Diamond Celebration Cupcakes will be found at the Jolly Holiday Bakery Café (lemon flavored) in Disneyland and the Fiddler, Fifer and Practical Café (orange flavored) in Disney California Adventure.
Appropriately sparkling for the Diamond Celebration, a Pomegranate Silver Sparkler, topped with lemon-lime foam and garnished with silver flakes, will be served at the Jolly Holiday Bakery Café on Main Street, U.S.A., in Disneyland.
At the Lucky Fortune Cookery in Disney California Adventure, a special Lucky Fortune Tea will feature mango, pineapple and passion fruit flavoring.
Exotic new Diamond Celebration beverages will also be available at Disneyland Resort hotel locations: a Longboard Lemonade at the PCH Grill in Disney's Paradise Pier Hotel, and a Wild Raspberry Rapids Lemonade at White Water Snacks in Disney's Grand Californian Hotel & Spa.
Diamond Celebration Crisped Rice Treats, Caramel Corn and Cotton Candy will be available at select locations in the parks.
Disneyland Resort takes pride in keeping the guest experience sparkling and fresh, restoring and refreshing attractions on a regular basis. As the Diamond Celebration debuts, several attractions closed for renovations will reopen with some exciting changes.
The Hatbox Ghost, which appeared in the Haunted Mansion for a short period around the time the attraction opened in 1969, will materialize once again when the popular Mansion reopens to guests this spring in New Orleans Square. Guests will discover the ghost just before they descend into the cemetery, where the Hatbox Ghost performs an eerie trick.
Matterhorn Bobsleds will reopen for the Diamond Celebration with new special effects. The Abominable Snowman, more exciting and ferocious, will appear in new and animated ways. A new scene will reveal some lost treasure the Snowman has been hoarding inside the mountain.
Also reopening in May is Peter Pan's Flight, featuring new special effects that make this classic and beloved attraction sparkle with new life. In a reimagined nursery scene, the Darling children have discovered the magic of flight as Tinker Bell flits overhead, sprinkling them with Pixie Dust. Guests will marvel at moonlit London Town and Never Land will glow in beautiful new ways.
Grizzly Peak is expanding into the former Condor Flats area with new building exteriors, landscape features and more. A new quick-service food location, Smokejumpers Grill, opened in March, 2015, inspired by the brave men and women who fight wildfires in the California forests. When the entire area reopens on Friday, May 15, lush pine trees will bring cool areas of shade to the expanded Grizzly Peak, making guests feel as if they have been transported to California's High Sierras. Soarin' Over California will reopen with updates to its screen and projection system, bringing a visually enhanced experience.
Additional information about the Disneyland Resort Diamond Celebration will be shared in the coming months at www.Disneyland.com and http://disneyparks.disney.go.com/blog/.


The holidays are my very favorite time of the year. There are cookies, singing Christmas carols, stringing lights, making handmade gifts, and of course, celebrating with friends and family.
I did a first this year, and that is I decided to throw a Favorite Things party for some of my friends. Which if you don't know how this kind of party works check out this post from How Sweet This Is. There are even free printables that you can use. I downloaded this one for the facebook event I created.
To start the party off right, we all enjoyed a delicious ham dinner. Now I have to tell you I was very excited bout this but a little nervous too, because my husband is actually the main cook in the family and he wasn't going to be around to cook. So that meant it was all on me! BUT the ham turned out amazing and was sooo good. I am so excited about how this ham turned out so I am going to go backward and share how I prepared it.
It all started at my local Walmart where I picked up my Hormel Cure 81 bone-in ham that was Cherrywood flavor.
If you have never cooked ham before (I hadn't), it really is so easy. The hams are precooked, so you really need to just warm them up in the oven and complete cooking instructions are included.
Click through to the neck page for the Brown Sugar and Mustard Glaze Recipe.
Thanks for the shoutout! 🙂 Glad you had fun- we just hosted our 4th annual party and it's always a hit! Merry Christmas!
I love that combo for the glaze! Thanks for sharing it with us at Funtastic Friday!
Ooh I think I'd like to try the Cherrywood ham! Thank you for sharing on Merry Monday! Pinned! Have a great week!


Last year, I played Roller Coaster Tycoon for the first time ever. I loved it! And while I appreciated it for everything it was, I couldn't help feel the itch of what-if. What if these mechanics were implemented into a more modern game engine?
That's what I found in Planet Coaster.
Now, I'm not going to say that the two games are exactly alike — there are a lot of minor differences. But the feeling Planet Coaster gave me – one of building my amusement park and watching little virtual people pickpocket enjoy it — was very similar.
I've wanted to try it out, but the normal price deterred me. I saw it on major sale during the Steam Winter Sale ($12) and snapped it up.
It's a roller coaster and amusement park building sim with tycoon elements thrown in.
I've read that some people had issues with performance in the past, and while I've not yet made a huge park, I've also not had issues with this game so far. I also haven't really jumped into the coaster building aspect, but part of the amazing thing of this game is the massive creativity the players put into it.
On the lobby page, you're greeted with avatars of Steam friends who also own the game (which is neat) as well as top Workshop creators. You can check out their cool stuff, and download it to your game if you like.
While I haven't really played around with the Workshop yet, I really like that it's so up front and highlighted. Similar to The Sims, when you start appealing to the creative players to make content for your game, you establish a game that can last a long(er) time.
While I'm no wiz at tycoon games, I didn't have any problem jumping into the first scenarios, even without seeing any sort of tutorial. The game felt familiar, immediately bringing back Roller Coaster Tycoon vibes, and I got the jist of it quickly. Like any sim-building game, I did have a few issues (mostly with food stands and paths) that I had to figure out.
Also, why are there so many thieves in all of my parks? Sheesh. I could hire nothing but security and I'd never get a happy balance!
This is a game I could totally lose hours and hours in, and I spent some time with it during my Christmas holiday. I'd love to return to it sometime, but there's just too many games on my list right now!
Check out these themed coasters below to see how amazing and creative some of the players are!
Next Post FFXIV: Do You Want a Balloon?
I've been wanting to play this one for a while but I still haven't bought it. I'll be sure to pick it up when I see it on offer now!
If it was $12 once, it's likely to drop to that again. A real steal for the amount of polished gameplay you get at that price.


Longmont, CO, Brevard, NC & Austin, TX – Oskar Blues Brewery, the original innovators of craft beer in a can, are making strange bedfellows of uppers and downers again.
Hotbox Coffee IPA (7.2% ABV, 70 IBUs) combines a burst of pine, citrus and mango from simcoe hops and blends them with a cold brew extract from Hotbox Roasters' Ethiopian beans, which bring cream, blueberry and cocoa notes to the mix. The beans were sourced from a family farm in the East African nation exclusively for this beer.
Hops and coffee = the perfect odd couple.
Since the in-house coffee roasters at Hotbox Roasters share a space in Oskar Blues Brewery's Longmont facility, it was a no-brainer to go back to the mat and follow up on the fall seasonal, Hotbox Coffee Porter.
Hotbox Coffee IPA will feature as Oskar Blues' new spring seasonal in cans from March to May; replacing GUBNA Imperial IPA, which will shift to a limited, national draft-only release with 12 oz. cans only available for retail in the ColoRADo and North Carolina Tasty Weasel taprooms.
Oskar Blues' three taprooms will each host simultaneous release parties to launch the Hotbox Coffee IPA.
Hotbox Coffee IPA will be available in all 50 states on draft and in 4-pack 12 oz. cans beginning in March. Beer drinkers looking for a unique mix of uppers and downers can hop on over to oskarblues.com/beerfinder to locate their nearest watering hole or store carrying the new brew.
Founded by Dale Katechis in 1997 in Lyons, Colorado, Oskar Blues Brewery launched the craft beer-in-a-can apocalypse with their hand-canned, flagship brew Dale's Pale Ale. Today, Oskar Blues is one of the fastest growing breweries in the country and operates breweries in Longmont, Colorado, Brevard, North Carolina and Austin, Texas while eclipsing 200,000 barrels in 2016. The original canned craft brewery continues push innovation with creations like The Crowler, their B. Stiff and Sons Old Fashioned Soda line, Hotbox Coffee Roasters & REEB Cycles. Oskar Blues is available in all 50 U.S. states, Washington, D.C., parts of Canada, Puerto Rico, Sweden, Spain, Chile, the Netherlands, Belgium, the U.K. and Australia.


text:
\section{Introduction}
The stability of solutions to a differential system of the type
\begin{equation}
\frac{dZ}{dt}=F(Z(t)),\label{odegeneral}
\end{equation}
is an obvious and important question. For times of order $1$ and if
$F$ is regular enough (for instance uniformly Lipschitz),
the answer is given quite simply by Gronwall lemma. For two
solutions $Z$ and $Z^\delta$ to \eqref{odegeneral}, one has

\begin{equation}
|Z(t)-Z^\delta(t)|\leq |Z(0)-Z^\delta(0)|\,\exp(t\,\|\nabla F\|_{L^\infty}).
\label{gronwall}
\end{equation}
This inequality forms the basis of the classical Cauchy-Lipschitz
theory for the well posedness of \eqref{odegeneral}.
It does not depend on the dimension of the system (the norm chosen is
then of course crucial). This is hence very convenient for the study
of systems of interacting particles, which is our purpose here.

Consider the  system of equations
\begin{equation}
\left\{ \begin{array}{l}
\dot{X}^N_i=V^N_i\\
\dot{V}^N_i= E_N(X^N_i) = \frac{1}{N}\sum_j K(X^N_i-X^N_j)
\end{array} \right.
\label{eqs:npart}
\end{equation}
where for simplicity
all positions $X^N_i$ belong to the torus ${\mathbb T}^3$ and all
velocities $V^N_i$ belong to $ {\mathbb R} ^3$. This system is obviously
a particular case
of \eqref{odegeneral} with $Z=Z^N=(X_1^N,\ldots,X_N^N,V_1^N,\ldots, V_N^N)$.

The equivalent of \eqref{gronwall} reads in this case
\begin{equation}
\|Z^N(t)-Z^{N,\delta}(t)\|_1\leq \|Z^N(0)-Z^{N,\delta}(0)\|_1\,
\exp(t\,(1+\|\nabla K\|_{L^\infty})),\label{gronwallN}
\end{equation}
where we define
the norm on $\Pi^{3N}\times  {\mathbb R} ^{3N}$
\[
\|Z\|_1=\frac{1}{2N}\,\sum_{i=1}^N (|X_i|+|V_i|).
\]
This estimate is logically uniform in the number of particles $N$. It
is important in itself but also because it is a crucial tool to pass
to the limit in the system of $N$ particles and derive the Vlasov-type
equation
\begin{equation}
\partial_t f+v\cdot\nabla_x f+\left(F\star_x\left(\int_{ {\mathbb R} ^3}
f(t,x,v)\,dv\right)\right)\;\cdot \nabla_v f=0,\label{vlasov}
\end{equation}
for the 1-particle density $f(t,x,v)$ in phase space, where
$\star$ denotes the convolution.
Hence estimates such as \eqref{gronwallN} are at the heart of the
derivation performed in \cite{Neunzert},
\cite{Dobrushin}, and \cite{BH77} (we also
refer to \cite{Sp}, \cite{VA} and \cite{Wo}). Note that the derivation
of collisional kinetic
models (of Boltzmann type) involves quite different techniques, see
\cite{Lanford} or \cite{CIP}.

Unfortunately, many cases of interest in physics deal with singular
forces $K\not\in W^{1,\infty}_{loc}$. Typical cases are $K= -\nabla
\phi$, a periodic force coming from a periodisation of the potential
$\phi_{ {\mathbb R} }(x)\sim C/|x|^{\alpha-1}$, {\em i.e.} \mbox{$ \phi(x) =
  \phi_{ {\mathbb R} }(x) + g(x)$,} where $g(x)$ is an (at least) $C^2$ function
on the torus ${\mathbb T}^3$. As the potential $\phi$ is defined up to a
constant, we may also assume that its average is $0$: $\int_{{\mathbb T}^3}
\phi(x)\,dx =0$. The most important case is the electrostatic or
gravitational interaction: $\alpha=2$, in dimension $3$.

Very little is known for these singular kernels, either from the point
of view of the stability or of the derivation of Vlasov-type equations.
Provided $\alpha<1$ and
the initial configuration of particles are well distributed, the limit
to Vlasov equation \eqref{vlasov} was proved in \cite{HJ06}.

For systems without inertia, {\em i.e.} when the equations are simply
\begin{equation}
\left\{ \begin{array}{l}
\dot{X}^N_i= E_N(X^N_i) = \frac{1}{N}\sum_j K(X^N_i-X^N_j)
\end{array} \right.,
\label{eqs:npartmacro}
\end{equation}
it seems to be easier to implement Gronwall-type inequalities. The
derivation of the mean field limit is consequently known up to
$\alpha<2$ ($\alpha<d-1$ in dimension $d$), see \cite{Hau2} and also
\cite{JO} for a situation where the forces have a
more complicated structure. In this
setting the most important case is however found in dimension 2, for
$K=x^\perp/|x|^2$ (corresponding to $\alpha=d-1=1$);
the limit is the 2d incompressible Euler
equation written in vorticity form. The derivation of the mean-field
limit in this case
was rigorously performed in \cite{GHL90} and \cite{Sch1}, \cite{Sch2}.

For differential equations like \eqref{odegeneral} in finite
dimensions, it has long been known that well posedness and stability
(for almost all initial data)
can be achieved without using Gronwall-type estimates. The
introduction of renormalized solutions by DiPerna-Lions in \cite{DL}
gave well posedness for $F\in W^{1,1}$ with $div\,F\in L^\infty$.

This was extended to $F\in BV$ in the phase space situation in
\cite{Bo} and then in the general case in \cite{Am} (see also
\cite{HLL} for a slightly different approach). The exact case of the
Poisson interaction was treated in \cite{Hau}.

This well posedness
implies some stability as the flow has then some differentiability
properties, see \cite{ALM}. However the corresponding stability
estimate is not quantitative and this kind of method does not seem to
be able to provide uniform estimates
in the number of particles (which gives the dimension of the
system). We refer to \cite{DeL} for a precise overall presentation of
the well posedness and differentiability issues for
Eq. \eqref{odegeneral} in finite dimension.

More recently a new method to show well posedness for
\eqref{odegeneral} has been introduced in \cite{CD}. Given a fixed
shift $\delta$, it consists in bounding quantities like
\begin{equation}
\int_{Z^0}
\log\left(1+\frac{|Z(t,Z^0)-Z(t,Z^0+\delta)|}{|\delta|}\right)\,dZ^0,
\label{log}
\end{equation}
where $Z$ is the flow associated to \eqref{odegeneral}, {\em i.e.}
$Z(t,Z^0)$ is a solution to \eqref{odegeneral} satisfying $Z(0,Z^0)=Z^0$.

A bound on such a quantity shows that for {\em a.e.} $Z^0$ the two
trajectories $Z(t,Z^0)$ and $Z(t,Z^0+\delta)$ remain at a distance of
order $|\delta|=|Z(0,Z^0)-Z(0,Z^0+\delta)|$. In this sense, this is an
almost-everywhere version of the Gronwall inequality \eqref{gronwall}.

It was shown in \cite{CD} that the quantity \eqref{log} remains
bounded if $F\in W^{1,p}$ for some $p>1$. This was extended to
$W^{1,1}$ and $SBV$ in \cite{Ja} and even to $H^{3/4}$ in the phase
space setting in \cite{CJ}. However in all those results the bound
depends on the dimension of the space and is blowing-up as this
dimension increases to $\infty$.

Therefore,
our precise aim in this article is to prove a bound on a quantity like
\eqref{log} for the system \eqref{eqs:npart}, uniformly in the number of
particles. To our knowledge, this is the first quantitative stability
estimate to be obtained for singular forces.

Several new important issues occur when one tries to do that
though. One of the most important is the reference measure which is
chosen as this can imply different notions of almost everywhere
as the dimension tends to $+\infty$.
In finite dimension, this refers to the Lebesgue
measure and of course implies corresponding estimates for any measure
which is absolutely continuous with respect to the Lebesgue
measure. In infinite dimension, no such natural measure exists.
This is due to
the phenomenon of concentration of measures. Even in the case of
finite but large dimensions, this is a problem to get quantitative
estimates. Indeed even if two measures $\nu_1$ and $\nu_2$ on
$\Pi^{3N}\times  {\mathbb R} ^{3N}$ are absolutely continuous with respect to
each other or even more if
\[
d\nu_1\leq C\,d\nu_2,
\]
then the constant $C$ will in general depend on the dimension and
go to $+\infty$ as $N$ increases. This means that a uniform
quantitative estimate on the trajectories for some measure $d\nu_1(Z^0)$
on the initial configuration will not in general imply a good estimate for
another measure $d\nu_2(Z^0)$.

\medskip

For each $N$, the choice of the measure $\mu_N$ on $\Pi^{3N}\times
 {\mathbb R} ^{3N}$ is hence crucial to get a good estimate. One would naturally
want to select a measure $\mu_N$ which is bounded, stable and invariant by the
flow, just as the Lebesgue measure is stable and invariant by the flow of
\eqref{odegeneral} when $F$ is divergence free.
Let us denote by $Z^N(t)= (X^N(t),V^N(t))$ or $Z^N(t,Z_0^N)$
the vector of particles velocities
and positions evolving through Eq. \eqref{eqs:npart} till time $t$ and
depending on the initial configuration $Z_0^N$.
The system \eqref{eqs:npart} has an invariant which is the total energy
\begin{eqnarray}
H_N[Z^N] & = & \sum_{i}\frac{(V^N_i)^2}{2}+\frac{1}{2N}\sum_{i\neq
  j}\phi(|X^N_i-X^N_j|)\\
& = & E_{kin}(V^N) + E_{pot}(X^N)
\end{eqnarray}
To get an invariant measure $\mu_N$, the simplest choice is to take a
function of the total energy. Among those which are stable, the most
natural is the Gibbs equilibrium
\begin{equation}
d\mu_N(Z^N)=\frac{1}{\mathcal{B}_N}e^{-\beta H_N[Z^N]} dZ^N~,
\end{equation}
where $dZ^N$ is Lebesgue measure on ${\mathbb T}^{3N} \times  {\mathbb R} ^{3N}$, and
\begin{equation} \label{eq:betaN}
\mathcal{B}_N(\beta)=\int e^{-\beta H_N[Z^N]}~dZ^N~,
\end{equation}
is the normalization constant. Note that for a potential $\phi$ with a
singularity at the origin, this makes sense only if $\lim_0 \phi =
+\infty$, that is in the case of repulsive interactions.  In the
following, since we deal with measures which have a density with
respect to the Lebesgue measure $dZ_0^N$, we will use the same
notation for the measure $\mu_N$ and its density.

\medskip

We study the quantity
\begin{equation} \label{eq:defQ}
Q(t)=\int~d\mu_N(Z^N_0)\int_{\delta\in {\mathbb T}^{3N}\times
   {\mathbb R} ^{3N}}\;\psi_N(Z^N_0,\delta)
\ln \left(1+\frac{\|Z^N(t,Z^N_0)-Z^N(t,Z^N_0+\delta)\|_1}{\delta_N}
\right)\,d\delta ~,
\end{equation}
where $\delta_N$ is a small parameter that will go
slowly to zero when $N$ goes to infinity.
$\delta$ is a shift on the initial condition $Z^N_0$

Here $\psi_N : {\mathbb T}^{3N} \times  {\mathbb R} ^{3N} \mapsto \mathcal{P}({\mathbb T}^{3N} \times
 {\mathbb R} ^{3N})$ (where $\mathcal{P}(\Omega)$ denotes the set of probabilities
on $\Omega$) is a probability valued function, so that it satisfies
\[
\int_{\delta \in {\mathbb T}^{3N} \times  {\mathbb R} ^{3N}}  d \psi_{N}(Z_0^N,\delta)=1, \quad
\forall Z_0^N \in ({\mathbb T}^3 \times  {\mathbb R} ^3)^N
\]
$\psi_N$ then then gives distribution of the shifts on the initial
conditions, and the quantity $Q(t)$ is averaged both on the initial
conditions $Z^N_0$ and on the shifts $\delta$.

We now define the image measure of $\mu_N$ by the shift distribution:
\[
\tilde\mu_N(Z_0) =  \int \mu_N(Z_0-\delta) \psi_N(Z_0-\delta ,\delta)
\,d\delta \,.
\]
The crucial assumption will be that the image measure $\tilde{\mu}_N$
remains ``close'' to the original Gibbs measure in the sense that
\begin{equation} \label{psi}
\exists \,K_\beta > 0, \text{ such that } \forall\, Z_0 \text{ and } N,
\quad\tilde{\mu}_N(Z_0) \leq K_\beta \mu_N(Z_0)
\end{equation}
with a constant $K_\beta$ independent of $N$, but which may depend on $\beta$.

We will also use the weaker but very similar condition
\begin{equation} \label{psi2}
\exists \,K'_\beta >0 \text{ s.t. } \forall \, N, \exists \, \beta'
\leq \beta, \text{ s.t. } \forall\, Z_0,
\;\tilde{\mu}_N(Z_0) \leq K'_\beta \mu_N^{\beta'}(Z_0),
\end{equation}
where $\mu_N^{\beta'}$ denotes the Gibbs measure with inverse temperature
$\beta$. The last condition is more general than the first, it allows to
control the image measure by a Gibbs measure with bigger temperature.

\begin{remark}
 We mention that by definition of $\tilde\mu_N$, the measure $\pi_N(Z_0,Z_0') =
\mu_N(Z_0) \Phi_N(Z_0, Z_0' -Z_0)$ is a transport from the measure $\mu_N$ to
$\tilde\mu_N$. In fact, $\psi_N$ is (up to a translation of origin) what is
usually called the desintegration of the measure $\pi_N$ with respect to its
first projection $\mu_N$. However, we preferred our less standard presentation
since we are more interested in $\mu_N$ and its shift $\psi_N$ that in the
precise image measure $\tilde\mu_N$. We mention the analogy to
emphasize that our quantity $Q$ is related to optimal transport. In fact
\[
 Q_N(0) = \int~d\mu_N(Z^N_0)\int_{\delta\in {\mathbb T}^{3N}\times
   {\mathbb R} ^{3N}}\;d\psi_N(Z^N_0,\delta)
\ln \left(1+\frac{\|\delta\|_1}{\delta_N}
\right) \geq W_N(\mu_N,\tilde\mu_N)
\]
where $ W_N$ is the transport for the cost $\ln\left( 1 +
\frac{\|\cdot\|_1}{\delta_N}\right)$.
\end{remark}

Conditions \eqref{psi} and \eqref{psi2} are not explicit on $\psi$.
Roughly speaking, they will be satisfied if $\psi_N$ is chosen so
that $|H_N(Z_0+\delta) - H_N(Z_0)| \leq C$ if $\delta \in \supp
\Psi_N(Z_0,\cdot)$.  This is reasonable since $H_N$ is preserved by
the dynamics, so that if the shift changes the energy too much, the
original and shifted dynamics may be very different. But that simple
and ``reasonable'' condition is not sufficient, we will really need a
bound like \eqref{psi} on the image measure constructed with the
shift.

As the conditions are not explicit, we will provide in section \ref{examples}
some examples of admissible shift distributions.
The main result of the paper is a control on the growth on this
quantity $Q$:
\begin{theorem} \label{theorem1}
Assume that $\phi  \geq \phi_{min}$ for some $\phi_{min} \in  {\mathbb R} ^-$ and
that for some constant $C$, and  $\alpha<2$
\[
 \phi(x)\leq \frac{C}{|x|^{\alpha-1}},\quad |\nabla \phi|\leq
\frac{C}{|x|^{\alpha}},\quad  |\nabla^2 \phi|\leq\frac{C}{|x|^{\alpha+1}}.
\]
Then taking $\delta_N=N^{-\varepsilon}$ for any
$\varepsilon \leq 1-\alpha/3$ and for all $N \geq \frac{6^4}{(2-\alpha)^2}$
one has
\[
Q(t)\leq   \left(1+ (1+ K_\beta)\frac{C c_\beta  +C_a
  c_\beta^a}{2-\alpha} \right)\,t + Q(0) ~,
\]
where
$c_\beta = e^{- \frac \beta 2 \phi_{min}}$,
$a$ is any exponent strictly larger than $2\alpha/3$, $C$ constant
(that can be made explicit),
and $C_a$ satisfies $C_a \leq \frac{C}{3a - 2\alpha}$.
\end{theorem}

This theorem is not able to deal with the electrostatic interaction,
$\alpha=2$; gravitational is of course out of question since repulsive
potentials are needed. Note however that the electrostatic potential
is just the critical case. The same result could be obtained in any
dimension, with essentially the same proof. In dimension $d$, the
condition would then be $\alpha<d-1$. The growth of $Q$ is linear in
time: note that this indeed corresponds to an average exponential in
time divergence of the trajectories, analogous to \eqref{gronwall}.

\medskip

Roughly speaking, and provided that the average shift at time $0$ is
of order $\delta_N$ (or smaller), the theorem says that the average
shift transported by the dynamics remains of order $\delta_N$ during
the evolution, and the control given is quite good.  It is interesting
to compare $\delta_N$ to the minimal distance in the $(X,V)$ space
between two particles of a configuration, which is of order
$N^{-\frac13}$. Notice that it is always possible to choose $\delta_N$
smaller than $N^{-\frac13}$. Then if the order of magnitude of the
initial shifts is smaller than $N^{-\frac13}$, the theorem says it
remains so at all time. This implies that the pairing of a particle in
the configuration $Z(t)$ with the closest one in the configuration
$Z^{\delta}(t)$ is not very much affected by the dynamics: in this
sense, there is not much ``mixing''.

\medskip

While the Gibbs equilibrium is the most natural choice for the measure
$\mu_N$, others are possible. The proof would work for any measure
$\mu_N$ such that
\begin{itemize}
\item[-] $\mu_N$  is invariant under the flow or
  $\mu_N(Z^N(t))=\mu_N(Z_0^N)$ for  $Z^N$ solution to
  \eqref{eqs:npart}
\item[-] for all $k$, the  $k$-marginal of $\mu_N$ defined by $
  \mu_N^k(Z^k)=\int \mu_N(Z_0^k,\tilde{Z_0}^{N-k}) d\tilde{Z_0}^{N-k}
  $  satisfies:
\[
\forall Z_0, \quad \mu_N^k(Z_0) \leq C^k.
 \]
\end{itemize}

 Obvious candidates are functions of the renormalized energy $\frac{1}{N}H_N$
but checking the bound on the marginals is not necessarily easy.

\medskip
{ \bf  Link with mean field limit. }

Finally, let us emphasize that this stability estimate does not answer
the question of the mean field limit. Doing so would require to be
able to deal with much more general measures $\mu_N$. More precisely if
one can prove Th. \ref{theorem1} for a sequence of $\mu_N$ and if in
some reasonable sense
\[
\mu_N-\Pi_{i=1}^N f^0 \longrightarrow 0,
\]
then the mean field limit is proved but only for the initial data
$f^0$. Currently the Gibbs equilibrium corresponds to
$f^0(x,v)=e^{-\beta |v|^2/2}$.

 Unfortunately, we are not able do deals with more general measures
$\mu_N^0$. The problem is that we need bounds on every $k$
marginals and those are very difficult to obtain if we start from another
measure than the Gibbs equilibrium. For instance, starting from $\mu_N =
g^{\otimes N}$ for some smooth profil $g$, we have the desired bounds
at time $0$, but do not know if $\mu_N(t)$ satisfies them for any other
time $t> 0$.

\section{Some examples of admissible shift distributions.}
\label{examples}

\subsection{Shift on velocity variables} \label{speedshift}

In this section, we will be interested  in shifts acting only the
velocities. A first possibility is to take shifts independent of $Z_0$ and
acting independently and identically on each velocity
variable. Precisely, we are looking for shift distributions such as
\[
\psi_N(\delta)  =  \delta_{\delta_X =0} \prod_{i=1}^N N^{\frac32} \psi(\sqrt N
\delta_{v_i} )
\]
where $\psi$ is a probability on $ {\mathbb R} ^3$ symmetric with respect to the origin
($\psi(-v) = \psi(v)$ if $\psi$ has a density).

We will not be able to deal with a general $\psi$, but will show that the
hypothesis \eqref{psi2} is satisfied if $\psi$ is Gaussian or has a compact
support. This is stated precisely in the following Proposition:

\begin{prop} \label{prop:spshift}
Assume that $\psi$ is a Gaussian probability with variance $\sigma^2$ :
\[ \psi(\delta_v)=\left(\frac{1}{2\pi\sigma^2}\right)^{\frac32}
e^{-\frac{\delta_v^2}{2\sigma^2}}
\]
then the hypothesis \eqref{psi2} is satisfied with
\[
 \beta'(N) = \beta\left(1-\frac{1}{1+N/(\beta\sigma^2)}\right) \,,
\text{ and }
 K_\beta = e^{-\frac{\beta^2\sigma^2}{4}\phi_{min}}
e^{\frac34 \beta\sigma^2}\,.
\]
Assume that $\psi$ has a compact support with $\supp \psi \subset B(0,\delta_m)$
(the ball of radius $\delta_m$, center $0$). Then \eqref{psi2} is satisfied for
$N > \beta \delta_m^2$ with
\[
 \beta'(N) = \beta \left( 1  - \frac{\beta \delta_m^2}{N}\right) \,,
\text{ and }  K_\beta = e^{-
  \frac{\beta^2 \delta_m^2 \phi_{min}}{2} } e^{\frac32 \beta\sigma^2}
\]

\end{prop}

Remark that for $\alpha \leq \frac32$, the average velocity fluctuation
given by such shift is larger than the smallest $\delta_N$ we can
choose.

\medskip

{\bf  Proof of the proposition \ref{prop:spshift}}\\
In the Gaussian case, we have
\[
\tilde{\mu}_N(Z_0) = \frac{e^{- \beta H_N(Z_0)}}{\mathcal{B}_N(\beta)}
 \left( \frac{N }{2\pi\sigma^2}\right)^{\frac{3N}2}
\int e^{- \frac\beta{2} \sum_i (\delta_{v_i}^2 - 2 v_i
  \delta_{v_i})}  e^{-\frac{N}{2\sigma^2}\sum_i\delta_{v_i}^2 } \,d\delta_V  \,.
\]
The $3N$ integrals may be performed independently, using the 1D
calculation:
\[
 \sqrt{\frac{N}{2\pi\sigma^2}} \int_{-\infty}^{+\infty}
 e^{-\frac12 (\beta+N/\sigma^2)\delta_v^2 +\beta v\delta_v} \,d\delta_v
= \sqrt{\frac{N}{N+\beta\sigma^2}}
e^{\frac\beta2 \frac{\beta\sigma^2}{N+\beta\sigma^2}}
\]
We finally get
\begin{eqnarray}
\tilde{\mu}_N(Z_0) &=& \frac{e^{-\beta H_N(Z_0)}}{\mathcal{B}_N(\beta)}
 \left( \frac{1 }{1+\beta\sigma^2/N}\right)^{\frac{3N}2}
e^{\frac{\beta}{1+N/\beta\sigma^2}E_{kin}(Z_0)} \nonumber \\
&=&  \frac{e^{-\beta' H_N(Z_0)}}{\mathcal{B}_N(\beta')}
 \frac{\mathcal{B}_N(\beta')}{\mathcal{B}_N(\beta)}
\left( \frac{1 }{1+\beta\sigma^2/N}\right)^{\frac{3N}2}
e^{-\frac{\beta}{1+N/\beta\sigma^2}E_{pot}(Z_0)} \nonumber
\end{eqnarray}
with $\beta'(N)=\beta(1-\frac{1}{1+N/(\beta\sigma^2)})$. Using
\[
\frac{\beta}{1+N/\beta\sigma^2}E_{pot}(Z_0)\geq
\frac{\beta^2\sigma^2\phi_{min}}{4}~,
\]
we get
\[
\tilde{\mu}_N(Z_0)\leq e^{-\frac{\beta^2\sigma^2}{4}\phi_{min}}
e^{-\frac34 \beta\sigma^2} \frac{\mathcal{B}_N(\beta')}{\mathcal{B}_N(\beta)}
\tilde{\mu}_N^{\beta'}(Z_0)
\]
We define $\mathcal{B}_{N,X}$ the normalization integral restricted to the
position variables:
\[
\mathcal{B}_{N,X}(\beta)= \int e^{-\beta E_{pot}(X_1,\ldots,X_N)}\,dX_1\ldots dX_N
\]
From the Jensen inequality applied to the function
$x\mapsto x^{\beta'/\beta}$, we obtain for $\beta' \leq \beta$
\[
\left( \mathcal{B}_{N,X}(\beta')\right)\leq
\left( \mathcal{B}_{N,X}(\beta)\right)^{\beta'/\beta}
\]
Then, using the bounds of Lemma \ref{lemma2},
we have $\mathcal{B}_{N,X}(\beta)^{\beta'/\beta-1}\leq 1$; this implies
\[
 \frac{\mathcal{B}_N(\beta')}{\mathcal{B}_N(\beta)}\leq
\left( \frac{\beta}{\beta'}\right)^{3N/2}
\leq e^{3\beta\sigma^2/2}
\]
This proves the desired inequality with
\[
 K_\beta = e^{-\frac{\beta^2\sigma^2}{4}\phi_{min}}
e^{\frac34 \beta\sigma^2}\,.
\]
\medskip

In the case of $\psi$ with compact support, we follow the same sketch.
To do this, we will need a bound on
\[
 \int e^{\frac \beta 2 (2 v \delta_v -
\delta_v^2)}N^{\frac32}\,d\psi(\sqrt N \delta_v)
= \int e^{\frac \beta 2 \left( \frac{2 v \delta_v}{\sqrt N} -
\frac{\delta_v^2}N \right)} \,d\psi(\delta_v)  \,.
\]
Using the symmetry of $\psi$, and the inequality $\cosh(x) \leq
e^{\frac{x^2}2}$ (valid for $x \in  {\mathbb R} $), we may bound that term by
\[
\int \cosh \left(\frac{\beta  v \delta_v}{\sqrt N} \right)
e^{- \frac{\beta\delta_v^2}{2N}} \,d\psi(\delta_v)
\leq
\int
e^{ \frac{\beta^2  v^2 \delta_v^2}{2N} - \frac{\beta\delta_v^2}{2N}}
\,d\psi(\delta_v)  \leq
e^{ \frac{\beta^2  \delta_m^2 v^2 }{2N}}
\]
and as in the previous case, we get
\[
\int \mu_N(Z_0-\delta)\psi_N(Z_0-\delta,\delta) \,d\delta \leq
\frac{1}{\mathcal{B}_N} e^{-\beta\left( H_N(Z_0) -\frac{\beta \delta_m^2}N
  E_{kin}(Z_0)\right) } \,.
\]
From this point, following exactly the same step as in the case of
gaussian $\psi$, we prove that the hypothesis \eqref{psi2} is
satisfied with the announced constant \cqfd.

\bigskip

By making the shift depend on the velocity $V_0=(V_1,\ldots V_N)$, one
may essentially remove the condition on the size of the norm of the
shift.
More precisely we limit ourselves to shifts $\delta=(0,\ldots,0,\delta')$ with
$\delta'\in {\mathbb R} ^{3N}$,
 giving
$Z_0+\delta=(X_0,V_0+\delta')$ with $X_0=(X_1,\ldots, X_N)$. Now define
\begin{equation}
\psi_N(Z_0,\delta)=\left(\Pi_{i=1}^N \delta_{\delta_i=0}\right)\;
 \Psi(|\delta'|)\,G(|V_0|)\,
\delta_{2\,V_0\cdot\delta'+|\delta'|^2=0},\label{shiftv}
\end{equation}
where $|\xi|^2=\xi\cdot\xi$ is the usual euclidian norm and
$\delta_{...}$ is the corresponding Dirac mass on the
hypersurface of equation
$2\,V_0\cdot\delta'+|\delta'|^2=|V_0+\delta'|^2-|V_0|^2=0$  which is
precisely the sphere of radius $|V_0|$.

We need the
function $\psi_N$ to
satisfy
\[
1=\int
\psi_N(Z_0,\delta)\,d\delta=G(|V_0|)\,\int_{2\,V_0\cdot\delta'+|\delta'|^2=0}
\Psi(|\delta'|)\,d\delta',
\]
this is always possible with the right choice of $G$ as the integral
\[
\int_{2\,V_0\cdot\delta'+|\delta'|^2=0}
\Psi(|\delta'|)\,d\delta'
\]
depends only on $|V_0|$ by the rotational symmetry of the sphere.

Condition \eqref{psi} is automatically true since, as $\mu_N$
depends only on $X_0$ and $|V_0|$ and $|V_0-\delta'|=|V_0|$,
\[
\tilde\mu_N(Z_0)=\int \mu_N(Z_0-\delta)\,\psi_N(Z_0-\delta,\delta)
=\mu_N(Z_0)\,\int \psi_N(Z_0,-\delta)=\mu_N(Z_0),\]
because $\psi_N(Z_0-\delta,\delta)=\psi_N(Z_0,-\delta)$.
One could wish to impose additionally that $|\delta'|_1\leq \delta_N$, so
that $Q(0)$ is of order~$1$. Since
\[
|\delta'|_1=\frac{1}{N}\sum_i |\delta_i'|\leq \frac{1}{\sqrt{N}}\,|\delta'|,
\]
it is enough to impose that $\Psi$ has support in $[0,  N^{1/2}\,\delta_N]$.

As a conclusion, we proved
\begin{prop} For any measure $\Psi$ on $ {\mathbb R} ^{3N}$, there exists a
  function $G(|V_0|)$, s.t. the probability density $\psi_N$ defined
  by \eqref{shiftv} satisfies \eqref{psi}.
\end{prop}

\bigskip

One could try to generalize this example by letting $V_0+\delta'$ and
$V_0$ to be on close but different energy spheres. For example by posing
\[
\psi_N(Z_0,\delta)=\left(\Pi_{i=1}^N \delta_{\delta_i=0}\right)\;
 \Psi(|\delta'|)\,G_{\eta}(|V_0|)\,
\,{\mathbb I}_{|2V_0\cdot\delta+\delta^2|\leq
  \eta}.
\]
Provided that $\eta$ in not too large, the computations are
essentially the same and one has essentially to make sure that
\[
G_{\eta}(|V_0|\pm \eta)\leq C\, G_{\eta}(|V_0|).
\]
\subsection{Shifts in position variables}
One could try to implement the same idea for shifts in position
variables. Many problems arise however since the potential energy is
not a nicely regular function of the positions.

If one tries to consider shift distributions $\psi_N(\delta)$ that do
not depend on $Z_0$, then the limitation on $|\delta|$ is quite
drastic. In fact this example essentially works if only a fixed
(independent of $N$) number of coordinates $\delta_i$ are not $0$.

Trying to generalize the second example by imposing that $Z_0+\delta$
and $Z_0$ are on the same energy surface also faces several problems.
The main one is that the equation of the energy surface is not anymore
symmetric between the shift $\delta$ and the shift $-\delta$.

The only solution would be to write the equation only on the tangent
plane, {\it i.e.} something like
\[
\psi_N=\Psi(|\delta|)\,\delta_{\nabla
  H(Z_0)\cdot\delta=0} \, |\nabla H(Z_0)|.
\]
This is now nicely symmetric but poses other difficulties. For
instance one would need to make sure that $H(Z_0+\delta)\leq
H(Z_0)+C$. Expanding $H$, one would formally find a condition of
the type
\[
|\nabla^2 H(Z_0)|\,|\delta|^2\leq C.
\]
Unfortunately $\nabla^2 H(Z_0)$ is singular and in particular
unbounded. It is only bounded in average which would force us to
remove the initial conditions around which the energy has a singular
behavior.

Although this procedure could in principle be carried out
successfully, we do not wish to enter here into such technical
computations.  This essentially limits us to present explicit shift
distributions acting only on velocity variables.

However, let us point out that the evolution of the particles system
strongly mixes positions and velocities. Obviously, if we start from
two initial conditions $Z_0$ and $Z_0^\delta$ with same positions and
different velocities, we get at time $t>0$ configurations with
different positions.

So if we really need a shift distribution that acts also on the
positions at the origin $t=0$, a strategy may be to start at $t= -\tau
>0$, using a shift distribution acting only on velocities at this
time, and then to let evolves the particles till time $0$. The
original shift distribution transported by the flows is now a shift
distribution acting on position and speed. Using the Theorem
\eqref{theorem1}, we get that the average of these evolved shifts (in
a weak sense since we we are taking a logarithmic mean) is at most of
order $\delta_N$, provided the average of the original shifts was also
of order at most $\delta_N$.  Therefore by removing a set of vanishing
measure of initial conditions, one obtains a shift distribution in
positions and velocities that satisfies all the requirements.
\section{Some useful bounds for the $6N$ dimensional $\mu_N$}
We shall make use of the following lemmas. Before stating them, we
introduce a notation for the projection of $\mu_N$ on the position
space.
\begin{equation}
\nu_N(X^N) = \int \mu_N(X^N,V^N)\,dV^N \,.
\end{equation}
We also denote the $k$- marginal of $\nu_N$ by $\nu_N^k$.

\begin{lemma}
For all $N$, we have:
\begin{equation}
 \left(\frac{2\pi}{\beta}\right)^{3N/2} \leq \mathcal{B}_N \leq
 \left(\frac{2\pi e^{- \beta \frac{\phi_{min}}3}}{\beta}\right)^{3N/2} .
\end{equation}\label{lemma2}
\end{lemma}
We will also need the following estimate on the $k$th marginal of
$\mu_N$ defined by
\begin{equation} \label{eq:marg}
\mu_N^k (Z^k) = \int \mu_N(dZ^{N-k})=
\frac{1}{\mathcal{B}_N}\;\int e^{-\beta H_{N}(Z^k,Z^{N-k})}\, dZ^{N-k} \,,
\end{equation}
and more precisely on the $k$ marginal $\nu_N^k$ of the projection
$\nu_N$ on the position variables.

\begin{lemma} \label{lemma3}
We define the constant $c_\beta = e^{-\beta \phi_{min}}$. Then, for
all $k$, we have
\begin{eqnarray}
\nu_N^k(X^k) & \leq & c_\beta^k
\end{eqnarray}
\end{lemma}

{\bf Proof of lemma \ref{lemma2}} To compute the integral defining
$\mathcal{B}_N$ \eqref{eq:betaN}, we may separate the integration in $X^N$
from the integration in $V^N$. In the $V^N$ variable, we have to
integrate a product of $3N$ independent real gaussians of variance
$\beta^{-1}$.  We obtain $(2\pi/\beta)^{3N/2}$.

In the $X^N$ variable, we use Jensen inequality by the
convexity of exponential to get:
\begin{eqnarray*}
1
& = &  e^{\int \frac{-\beta}{2N}\sum_{i\neq
    j}\phi(|X^N_i-X^N_j|)\, d X^N  } \\
& \leq &  \int  e^{\frac{-\beta}{2N}\sum_{i\neq
    j}\phi(|X^N_i-X^N_j|)  }\,dX^N\\
& =& \left(\frac{2\pi}{\beta}\right)^{-3N/2}\;\int
e^{-\frac{\beta}{2}\sum_i |V_i^N|^2- \frac{\beta}{2N}\sum_{i\neq
    j}\phi(|X^N_i-X^N_j|) }\,dX^N\,dV^N\\
&=& \left(\frac{2\pi}{\beta}\right)^{-3N/2}\;\mathcal{B}_N,
\end{eqnarray*}
which gives the first inequality (We used that $\phi$ has zero
average). To obtain the second bound,
it suffices to use the inequality  $E_{pot}(Z_0^N) \geq \frac{N}2 \phi_{min}$.
\cqfd

\medskip

{\bf Proof of lemma \ref{lemma3}} The proof follows the one introduced in
\cite{MeSp} for the Lame-Enden equation.
As the measure $\mu_N$ factorizes in position and speed, we may write
\[
\nu_N(X^N) =  \frac{1}{\mathcal{B}_{N,X}} e^{-\beta  E_{pot}(X^N)}
\]
Neglecting the terms in interaction energy involving (at least) one of
the first $k$ particles, we obtain
\begin{eqnarray}
\nu_N^k (X^k) & = & \frac{1}{\mathcal{B}_{N,X}} \int e^{-\beta
  E_{pot}(X^k,X^{N-k})}\, dX^{N-k} \nonumber \\
& \leq & \frac{1}{\mathcal{B}_{N,X}}  e^{-\beta k  \phi_{min}}
\int e^{-\beta \frac{N-k}{N} E_{pot}(X^{N-k})} \, dX^{N-k}
~. \label{eq:bnu}
\end{eqnarray}
the term $\frac{N-k} N$ being there because $E_{pot}(X_k) = (1/k) \sum_{i \neq
  j}^k \phi(|X_i-X_j|)$ for $k$ positions.  So we need an estimate on
terms of the kind
\[
\Theta(k) = \int e^{-\beta \frac{k}{N} E_{pot}(X^k)} \, dX^k ~,
\]
We can relate this term to configurations with $k+1$ particles.
First use Jensen inequality as the exponential is convex to get
\[
\begin{split}
\Theta(k) =& \int e^{-\beta \frac{k}{N}
  E_{pot}(X^k)}\,dX^k\\
=&\;\int \exp\left(-\int \left(\beta \frac{k}{N}
  E_{pot}(X^k)+\frac{\beta}{N} \sum_{i=1}^k
  \phi(|X_i-x_{k+1}|)\right)\;dx_{k+1}\right)\;
dX^k\\
\leq & \int e^{-\beta \frac{k}{N} E_{pot}(X^k) -
    \frac{\beta}{N} \sum_{i=1}^k  \phi(|X_i-x_{k+1}|)} \, dX^k
  \,dx_{k+1} \\
 = & \int e^{-\beta \frac{k+1}{N} E_{pot}(X^{k+1})} \,
  dX^{k+1} \\
 \leq &  \Theta(k+1).
\end{split}\]
Since, $\Theta(N) =
\mathcal{B}_{N,X}$, we iterate this inequality and get $\Theta(N-k) \leq
 \mathcal{B}_{N,X}$. Putting this in \eqref{eq:bnu}, we get
\[
\nu_N^k (X^k) \leq e^{ - k \beta \phi_{min}}  \,,
\]
which is the result needed.
\cqfd

\section{Proof of Theorem \ref{theorem1}}

During the course of the demonstration, $C$ will denote a constant (independent
of $N$ and $\beta$), which value may change from line to line.

 From now on, we shall omit the superscript
$N$ in the notation $Z^N$, as there will be no ambiguity.
We have to estimate the derivative of $Q(t)$. Differentiating directly,
one obtains
\[\begin{split}
\frac{d}{dt} Q(t)\leq \int d\mu_N(Z_0)\int d\delta\, \psi_N(Z_0,\delta)\;
&\Bigg(\frac{\frac{1}{N}\sum_i |V_i-V_i^\delta|}{\delta_N
  +\|Z-Z^\delta\|_1}\\
&+\frac{\frac{1}{N^2}\sum_{i}
  |\sum_j (K(X_i-X_j)-K(X_i^\delta-X_j^\delta))|}{\delta_N
  +\|Z-Z^\delta\|_1}\Bigg),\\
\end{split}
\]
where $Z^\delta=(X^\delta,V^\delta)=Z(t,Z_0+\delta)$.

Note that the first term is obviously bounded by $1$ and hence
\[
\frac{d}{dt} Q(t)\leq 1 +\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{\frac{1}{N^2}\sum_{i}
  |\sum_j (K(X_i-X_j)-K(X_i^\delta-X_j^\delta))|}{\delta_N
  +\|Z-Z^\delta\|_1}.
\]
We define for a integer $L$ that will be fixed later
\begin{equation} \label{eq:Ci}
\mathcal{C}_i(Z_0,t) = \left\{ j\neq i~s.t.~|X_i(t)-X_j(t)|~\mbox{is among
the}~L~\mbox{smallest}~|X_i-X_k| \right\}
\end{equation}
That is for each $i$, $\mathcal{C}_i$ collects the indices of particles
which are closest to particle~$i$ at time~$t$, following the flow.
It also depends on the initial condition $Z_0$. We define similarly
$\mathcal{C}_i^\delta$.

Accordingly, we decompose $dQ/dt$ as follows
\[
\frac{d}{dt} Q(t)\leq C+S_1+S_1^\delta+S_2,
\]
with
\begin{equation}\begin{split}
&S_1=\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\in \mathcal{C}_i\cup
  \mathcal{C}_i^\delta }
|K(X_i-X_j)|,\\
&S_1^\delta=\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\in \mathcal{C}_i\cup
  \mathcal{C}_i^\delta }
|K(X_i^\delta-X_j^\delta)|,\\
&S_2=\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{1}{N^2}\sum_{i}\sum_{j\not\in (\mathcal{C}_i\cup
  \mathcal{C}_i^\delta)}\frac{|K(X_i-X_j)-K(X_i^\delta-X_j^\delta)|}
{\delta_N+\frac{1}{N}\sum_i |X_i-X_i^\delta| }.
\end{split}\end{equation}
\subsection{Bound on $S_1$}
Recalling the bounds on $K=-\nabla \phi$ from Th. \ref{theorem1}, one
simply begins with a discrete H\"older inequality for any $\gamma\leq 3$
\[\begin{split}
S_1\leq &\left(\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\in \mathcal{C}_i\cup
  \mathcal{C}_i^\delta } 1 \right)^{1-\alpha/\gamma}\\
&\times\left(\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\neq i}
\frac{1}{|X_i-X_j|^\gamma}\right)^{\alpha/\gamma}.
\end{split}\]
We first use the fact that the integral of $\psi_{N}$ in $\delta$ is equal to
$1$ to get
\[
\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\neq i}
\frac{1}{|X_i-X_j|^\gamma}=\int d\mu_N(Z_0)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\neq i}
\frac{1}{|X_i-X_j|^\gamma}.
\]
We then perform the change of variable from $Z_{0}$ to $Z$, using the
inverse flow of $Z(t,Z_0)$ which preserves the measure $\mu_N$; one finds
\[\begin{split}
\int d\mu_N(Z_0)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\neq i}
\frac{1}{|X_i-X_j|^\gamma}\leq \frac{1}{\delta_N\,
  N^2}&\sum_{i}\sum_{j\neq i}\;\int d\mu_N(Z)\frac{1}{|X_i-X_j|^\gamma}.
\end{split}\]
As the second marginal of $\mu_N$ is bounded by $c_\beta^2$ by Lemma
\ref{lemma3}, this implies
\[
\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_{0},\delta)\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\neq i}
\frac{1}{|X_i-X_j|^\gamma}\leq \frac{C\,c_\beta^2}{(3 -\gamma)\delta_N}.
\]
Hence
\[
S_1\leq \frac{C\,c_\beta^{2\alpha/\gamma}}{(3-\gamma) \delta_N}
\left(\int_{Z_0} d\mu_N(Z_0)\,\frac{1}{N^2}
\sum_i
(|\mathcal{C}_i|+|\mathcal{C}_i^\delta|)\right)^{1-\alpha/\gamma}.
\]
To conclude, simply note that by definition
$|\mathcal{C}_i^\delta|=|\mathcal{C}_i|=L$ and so for any $a >
\frac{2\alpha}{3}$,
\begin{equation}
S_1\leq \frac{C_a\,c_\beta^a}{\delta_N}\,
\left(\frac{L}{N}\right)^{1-\frac a 2} \,,
\end{equation}
where $C_a$ satisfies $C_a \leq \frac{C}{3a -2 \alpha}$.
\subsection{Bound on $S_1^\delta$} \label{S1delta}
Using Fubini and the change of variable $Z_0 \mapsto Z_0 + \delta$,
and the image
measure $\nu_N$, we may rewrite
\[
S_1^\delta =   \int \nu_N(Z_0) dZ_0 \,\;
\frac{1}{\delta_N\, N^2}\sum_{i}\sum_{j\in \mathcal{C}_i\cup
  \mathcal{C}_i^{-\delta} } |K(X_i-X_j)|
\]
And from the hypothesis \eqref{psi}, we may bound that that exactly
as the previous one. The only difference is that a constant $K_1$
appears and that we shall use $\beta'(N)$ instead of $\beta$. We get
\[
S_1^\delta \leq \frac{C K_1 \,c_{\beta'(N)}^{2\alpha/\gamma}}{\delta_N}\,
\left(\frac{L}{N}\right)^{1-\alpha/\gamma} \,.
\]

\subsection{Bound on $S_2$}

By using the assumption on the second derivative of $\phi$ in
Th. \ref{theorem1}, one first bounds
\[
|K(X_i-X_j)-K(X_i^\delta-X_j^\delta)|\leq
C\,(|X_i-X_i^\delta|+|X_j-X_j^\delta|)\;
\left(\frac{1}{|X_i-X_j|^{\alpha+1}}
+\frac{1}{|X_i^\delta-X_j^\delta|^{\alpha+1}}\right).
\]
Therefore defining the following matrix, with $\mathbb{I}_{A}$ the
characteristic function of the set $A$:
\[
M_{ij}=\left(\frac{1}{|X_i-X_j|^{\alpha+1}}
+\frac{1}{|X_i^\delta-X_j^\delta|^{\alpha+1}}\right)\,({\mathbb{I}}_{j\not\in
  (\mathcal{C}_i\cup
  \mathcal{C}_i^\delta)}+{\mathbb{I}}_{i\not\in
  (\mathcal{C}_j\cup
  \mathcal{C}_j^\delta)}),
\]
one has
\[
S_2\leq \int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;
\frac{\frac{1}{N^2}\sum_{i,j}M_{ij} |X_j-X_j^\delta|}
{\delta_N+\frac{1}{N}\sum_k |X_k-X_k^\delta| }.
\]
Consequently, if we use the classical matrix inequality $\|Mx\|_1 \leq \sup_{j}
\big( \sum_{i} |M_{ij}|) \big) \|x\|_1$,
\[
S_2\leq \int d\mu_N(Z_0) \int d\delta\,
\psi_N(Z_0,\delta)\;\max_i\,\sum_j M_{ij}.
\]
As before the terms in $M$ containing $|X_i^\delta-X_j^\delta|$ are the
equivalent of the ones with $|X_i-X_j|$, thanks to \eqref{psi}.
Hence one has to bound
\[
S_2\leq C\,(S_2^1+S_2^2),
\]
with
\[\begin{split}
&S_2^1=\int d\mu_N(Z_0) \int d\delta\, \psi_N(Z_0,\delta)\;\max_i\,\sum_{j\not\in
  \mathcal{C}_i} \frac{1}{|X_i-X_j|^{\alpha+1}},\\
&S_2^2=\int d\mu_N(Z_0) \int d\delta\,
\psi_N(Z_0,\delta)\;\max_i\,\sum_{j\ s.t.\ i\not\in
  \mathcal{C}_j} \frac{1}{|X_i-X_j|^{\alpha+1}}.\\
\end{split}\]
Since nothing depends on $\delta$ now, one may integrate
$\psi_N(Z_0,\delta)$ in $\delta$ with value $1$.
Moreover changing variable from $Z_0$ to $Z$ (we
recall the flow is measure preserving), one simply finds
\[\begin{split}
&S_2^1=\int d\mu_N(Z) \;\max_i\,\sum_{j\not\in
  \mathcal{C}_i} \frac{1}{|X_i-X_j|^{\alpha+1}},\\
&S_2^2=\int d\mu_N(Z)\;\max_i\,\sum_{j\ s.t.\ i\not\in
  \mathcal{C}_j} \frac{1}{|X_i-X_j|^{\alpha+1}}.\\
\end{split}
\]
Let us now carefully bound each of these terms.

\medskip

{\bf The $S_2^1$ term}\\
We use
\begin{equation}
\int f(X)~d\mu_N=\int_0^{+\infty}P(f(X)>l)~dl~,
\end{equation}
where $P$ is the probability with respect to the
measure $\mu_N$ on $\Pi^{3N}\times {\mathbb R} ^{3N}$.
We have to evaluate now expressions like
\begin{equation}
P \left( \max_i \frac{1}{N}\left( \sum_{j\notin\mathcal{C}_i}
\frac{1}{|X_i-X_j|^{\alpha+1}}\right)>l \right)=
P\left( \exists i~s.t.~\frac{1}{N}\left( \sum_{j\notin\mathcal{C}_i}
\frac{1}{|X_i-X_j|^{\alpha+1}}\right)>l\right)~.
\label{eq:P}
\end{equation}

To bound this probability, we will need the following lemma:
\begin{lemma} Given $x_1,\ldots,x_n\geq 0$ and $l\geq0$; given
  $(u_k)_{k=1}^n$ such that $\sum_{k=1}^n u_k=1$. If $\sum_{i=1}^n
  x_i>l$, then $\exists k\in[1,~n],~\exists
  i_1,\ldots,i_k~/~x_{i_r}>lu_k~,~\forall r=1,\ldots,k$.
\end{lemma}
{\emph{Proof:}} Let the $x_i$ be sorted $x_1\geq x_2\geq\ldots\geq x_n$, and
suppose the conclusion is not true. Then we have
$$
x_1\leq l u_1,\ldots,x_n\leq l u_n~.
$$
Thus $\sum_i x_i \leq l\sum_i u_k=l$.\cqfd

\medskip

We apply the lemma to Eq.~\ref{eq:P}, with $u_k=\frac{C_N \nu}k
\left(\frac k N \right)^\nu$ and
$0<\nu<1$ to be determined. $C_N$ is chosen such that
\mbox{$\sum_{k=1}^N u_k=1$.} Using Riemann sums, we see that
$\lim_{+\infty} C_N = 1$, and that $C_N \geq 1, \forall N \geq 1$. Hence we get
\begin{eqnarray*}
P_l & = & P \left( \exists i~s.t.~\frac{1}{N}  \left( \sum_{j\notin\mathcal{C}_i}
      \frac{1}{|X_i-X_j|^{\alpha+1}}\right) >l \right) \\
  &  \leq  & \sum_{k=1}^{N-L} P\left( \exists i;j_1,\ldots,j_k \notin
      \mathcal{C}_i,~
      \frac{1}{N}\frac{1}{|X_i-X_{j_r}|^{\alpha+1}}>l\frac{C_N \nu}{k}(k/N)^\nu
      \right) \\
 &  \leq & \sum_{k=1}^{N-L} P\left( \exists i;j_1,\ldots,j_k \notin
      \mathcal{C}_i,~ |X_i-X_{j_r}|<\left(\frac1{ \nu l
      }\right)^{1/(\alpha+1)} \left(\frac k N \right)^{\lambda/3}
      \right) \\
 &  = & \sum_{k=1}^{N-L}P_{l,k} ~,
\label{eq:Pmu}
\end{eqnarray*}
where for simplicity we have introduced the parameter $\lambda=3
(1-\nu)/(\alpha+1)$.

To estimate the probability $P_{l,k}$, once the particle $i$ is
chosen, we have a constraint on the position of $k$ particles,
which have to be close enough to particle~i, plus constraints on the
position of $L$ distinct particles, from the definition of
$\mathcal{C}_i$. This event concern $k+L+1$ particles, and to estimate
it, we will use an estimate of its volume $P_{l,k}^u$ in the
configuration space ${\mathbb T}^{3(k+L+1)}\times {\mathbb R} ^{3(k+L+1)}$. It thus
involves the $(k+L+1)$ marginal of $\mu_N$ which is bounded by
$c_\beta^{k+L+1}$ by Lemma \ref{lemma3}.

This leads to the following estimates
\[
P_{l,k} \leq  C\;c_\beta^{k+L+1}\,   N~
\left(\frac{1}{\nu l}\right)^{\frac{3(k+L)}{\alpha+1}} \left(
\frac{k}{N}\right)^{\lambda(k+L)}.\\
\]
Moreover, using a simplified version of Binet  formula (See \cite{WW})
\[
n! = \sqrt{2\pi} n^{n+ \frac12} e^{-n + \frac{\theta}{12n}} \,,
\text{ for some } \theta \in (0,1)
\]
the binomial coefficient $C_N^p$ may be bound by:
\begin{eqnarray}
C_N^p  & = & \frac{N!}{p!(N-p)!} =  (2\pi)^{-\frac12}
e^{\frac{\theta_N}{12N} - \frac{\theta_p}{12p} -
  \frac{\theta_{N-p}}{12(N-p)}} \frac{N^N}{p^p (N-p)^{N-p}}
\left(\frac{N}{p(N-p)}\right)^{1/2} \\
& \leq &  (2\pi)^{-\frac12} e^{\frac1{12N}}
\left(\frac{N}{N-1}\right)^{1/2} \left( \frac{N}{p}\right)^p \left(1
+
\frac{p}{N-p}\right)^{N-p} \leq  \left( \frac{N e}{p}\right)^p ~.
\end{eqnarray}
And we do not forget that since $C_N^p = C_N^{N-p}$ we may use the
same inequality with $p$ replaced by $N-p$.
Inserting this in the above inequality, we get:
\begin{eqnarray*}
P_{l,k} & \leq & C c_\beta^{k+L+1} N \left( \frac{N}{k+L}\right)^{k+L}
e^{k+L} \left(\frac{1}{\nu l}\right)^{\frac{3(k+L)}{\alpha+1}}\left(
\frac{k}{N}\right)^{\lambda(k+L)} \\
& \leq & C c_\beta \, N \,\left( \frac{k}{N}\right)^{(\lambda-1)(k+L)}
\left(\frac{A c_\beta'}{\nu l}\right)^{\frac{3(k+L)}{\alpha+1}} \\
& \leq &  C c_\beta \, \left( \frac{k}{N}\right)^{(\lambda-1)(k+L) - 1}
k  \, \left(\frac{A c_\beta'}{\nu l}\right)^{\frac{3(k+L)}{\alpha+1}}~,
\\
\end{eqnarray*}
where $c_\beta' = c_\beta^{(\alpha+1)/3}$, and $A= e^{(\alpha+1)/3}$
is a numerical constant.  Now taking $\nu$ close enough to $0$
(precisely $\nu < \frac{2- \alpha}3$), one
has $\lambda
>1$
and then we take as well $L \geq (\lambda -1)^{-1}$ (recall that $L$
has yet to be fixed)~; hence
\[
P_{l,k}\leq  C c_\beta \, k \, \left(\frac{A
  c_\beta'}{\nu l}\right)^{\frac{3(k+L)}{\alpha+1}}.
\]
If we sum on $k$, we get:
\begin{eqnarray}
\sum_{k=0}^{N-L} P_{l,k} & \leq & C c_\beta \, \left(\frac{A
  c_\beta'}{ \nu l}\right)^{\frac{3L}{\alpha+1}} \, \sum_{k=0}^{N-L} k
\,\left(\frac{A c_\beta'}{\nu l}\right)^{\frac{3k}{\alpha+1}} \\
& \leq &  C \, c_\beta \, \left(\frac{A
  c_\beta'}{\nu l}\right)^{\frac{3(L+1)}{\alpha+1}} \frac{1}{(1- (A
  c_\beta'/ \nu l)^{3/(\alpha+1)})^2}  \leq C \,
\, c_\beta\, \left(\frac{A c_\beta'}{ \nu l}\right)^{\frac{3(L+1)}{\alpha+1}}~,
\end{eqnarray}
provided $l \geq l_0=\frac{2  \, A \, c_\beta'}{ \nu}$. If we take moreover
$L \geq p \geq p(\alpha+1)/3$ for some $p \geq 0$, we get a simpler bound:
\[
\sum_{k=0}^{N-L} P_{l,k}  \leq C \, c_\beta \left(\frac{A \,
  c_\beta'}{ \nu l}\right)^p ~.
\]
Remark that the conditions on $L$ depend only on $\alpha$ and
$\lambda$ (which depends only on $\alpha$). In particular, those
conditions are
independent of the parameter $\beta$. Thus, we have
\begin{eqnarray}
P\left( \exists i~s.t.~\frac{1}{N}\left( \sum_{j\notin\mathcal{C}_i}
      \frac{1}{|X_i-X_j|^{\alpha+1}}\right)>l\right) & \leq & 1
      ~\text{for}~l<l_0 = \frac{2A}{ \nu} c_\beta^{\frac{\alpha+1}3}    \\
& \leq & C c_\beta^{1+ p \frac{\alpha+1}3}( \nu l)^{-p} ~\text{for}~l\geq l_0'
\end{eqnarray}

Integrating this quantity in $l$, one obtains for  $M$ such that  $C
c_\beta^{1+ p \frac{\alpha+1}3}( \nu M)^{-p}  =  1$ that
\begin{eqnarray*}
\int_0^\infty P(\dots>l) \,dl & =  & \int_0^M P(\dots>l) \,dl +
\int_M^\infty P(\dots>l) \,dl \\
& \leq & M + C c_\beta \left(\frac{A c_\beta'}{ \nu}\right)^p
\int_M^\infty  \frac{dl}{l^p} \\
& \leq & M + C c_\beta \left(\frac{A c_\beta'}{ \nu}\right)^p
\frac{M^{1-p}}{p-1} \\
& = & \frac{p}{p-1} M \leq \frac{C}\nu  ~c_\beta^{\frac1p+  \frac{\alpha+1}3}
\end{eqnarray*}
for $p \geq \frac{3}{2-\alpha}$. And finally, we get that
\[
S_{2}^1 \leq \frac{C}\nu ~ c_\beta ~,
\]
for some constant $C$, provided that
$L$ is large enough.

Since every calculation have been performed, we see that a possible
for $\nu$ is
$\nu = \frac{2-\alpha}{6}$ in which case the condition on $L$ is exactly
\[
L \geq \frac{6}{2-\alpha}
\]
With that choice of $\nu$, we get
\[
S_{2}^1 \leq \frac{C}{2 -\alpha} ~ c_\beta ~,
\]

\medskip

{\bf The $S_2^2$ term}:
Through the same type of computations, we are led to evaluate expressions like

\begin{eqnarray}
P_l' & = & P\left( \exists j~s.t.~\frac{1}{N}\left( \sum_{i/j\notin\mathcal{C}_i}
      \frac{1}{|X_i-X_j|^{\alpha+1}}\right)>l\right) \\
  & \leq & \sum_{k=1}^N P\left( \exists j;i_1,\ldots,i_k~/~j\notin
      \mathcal{C}_{i_r}
      \frac{1}{N}\frac{1}{|X_j-X_{i_r}|^{\alpha+1}}>l\frac{C_N
        \nu}{k}\left(\frac k N\right)^\nu\right)
      \\
  & \leq & \sum_{k=1}^N P\left( \exists j;i_1,\ldots,i_k~/~j \notin
      \mathcal{C}_{i_r} |X_j-X_{i_r}|< \frac1{(\nu l)^{\alpha+1}}
    \left(\frac k N \right)^{(1-\nu)/(\alpha+1)}\right) \\
    & = & \sum_{k=1}^N P'_{l,k} ~.
\label{eq:Pmu2}
\end{eqnarray}
Since the sum is performed on the particles $i$ such that $j \notin
  \mathcal{C}_i$, we cannot choose $L+k$ particles close to $j$ as for
  $S_2^1$. But,
  we have nevertheless to choose $k$ particles close enough to $j$, a
  probability that will give a good bound if $k \geq L$. If $k \leq
  L$, once a particle $i$ close to $j$ such that $j \notin
  \mathcal{C}_i$ is chosen, one knows that there exist $L$ other
  particles close to $j$.
This will be enough to bound the probability.

  In the second case ($k \leq L$), we pick up a particle $j$ ($N$
  possibilities), at least another particle $i$ (since $k\geq 1$)
and then we have to choose $L$ other particles closer to
  $i$ than $j$ is. Since $|X_j-X_i|$ has to be less than
$C\, l^{-1/(\alpha+1)}\,(k/N)^{(1-\nu)/(\alpha+1)}$,
\begin{eqnarray}
  \sum_{k=0}^L P'_{k,l} & \leq & \sum_{k=0}^L c_\beta^{L+2} N^2 C_N^L
  \left(\frac{1}{\nu l}\right)^{3L/(\alpha+1)}
\left(\frac{k}{N}\right)^{\lambda L}\\
& \leq & C  \sum_{k=0}^L c_\beta^2 \, N^2
\,\left(\frac{L}{N}\right)^{(\lambda-1)L} \left(\frac{A
  c_\beta'}{\nu l}\right)^{3L/(\alpha+1)} \\
& \leq &  C c_\beta^2  \, N^3 \,
\left(\frac{L}{N}\right)^{(\lambda-1)L} \left(\frac{A
  c_\beta'}{\nu l}\right)^{3L/(\alpha+1)}~.
\end{eqnarray}
If $L \leq \sqrt N$, we may use
\[
N ^3 \left( \frac{L}{N} \right)^{(\lambda-1)L} \leq  N^{3 -
  \frac{\lambda-1}2 L} \leq 1 \,,
\]
as soon as $L \geq \frac6{\lambda-1}$. In that case,
we may use this bound in last inequality and obtain:
\begin{eqnarray*}
\sum_{k=0}^L P'_{k,l} & \leq & C c_\beta^2  \, \left(\frac{A
  c_\beta'}{l}\right)^{3L/(\alpha+1)} \\
& \leq & C c_\beta^2 \left(\frac{A c_\beta'}{l}\right)^{3L/(\alpha+1)} ~,
\end{eqnarray*}
if $L \leq \sqrt N $.

In the first case $k>L$, we pick up the particle $j$, and then
choose $k$ particles $i_r$ close to $j$. We obtain as previously
\begin{eqnarray*}
\sum_{k=L+1}^{N-1} P'_{k,l} & \leq & \sum_{k=L+1}^{N} c_\beta^{k+1} N C_N^k
\left(\frac{1}{\nu l}\right)^{3k/(\alpha+1)}
\left(\frac{k}{N}\right)^{\lambda k}\\
& \leq & C  \sum_{k=L+1}^{N} c_\beta \, N
\,\left(\frac{k}{N}\right)^{(\lambda-1)k} \left(\frac{A
  c_\beta'}{\nu l}\right)^{3k/(\alpha+1)} \\
& = &  C \, c_\beta  \sum_{k=L+1}^{N}    k \,
\left(\frac{k}{N}\right)^{(\lambda-1)k - 1} \left(\frac{A
  c_\beta'}{l}\right)^{3k/(\alpha+1)} \\
& \leq & C c_\beta \sum_{k=L+1}^{N}  k\left(\frac{A
  c_\beta'}{\nu l}\right)^{3k/(\alpha+1)}  \\
& \leq & C c_\beta \left(\frac{A c_\beta'}{\nu l}\right)^{3L/(\alpha+1)} ~,
\end{eqnarray*}
where we again restricted ourselves to $(\lambda-1)L \geq 1$ and
assume $l \geq l_0$ .
Putting the two sum together, we get the bound
\[
\sum_{k=1}^{N-1} P'_{k,l}  \leq   C  c_\beta^2 \left(\frac{A
  c_\beta'}{\nu l}\right)^{3L/(\alpha+1)}
\]

It remains to integrate in $l$.  Doing exactly as for the $S_2^1$
term, and choosing the same $\nu$, we will get
\[
S_2^2\leq \frac C {2 -\alpha}\,c_\beta.
\]
The only difference is that it will require $p \geq
\frac{6}{2-\alpha}$ and thus $L \geq \frac{12}{2-\alpha}$
\subsection{Conclusion of the proof}
Putting all together, we finally we may bound
\[
\frac{dQ}{dt}
\leq 1 +C_a\,\frac{c_\beta^a}{\delta_N}\;
\left(\frac{L}{N}\right)^{1-\frac a 2 }+ \frac{C}{2-\alpha}\,c_\beta,
\]
with $c_\beta = e^{- \beta \phi_{min}}$ and $a> \frac{2\alpha}3$,
where $L$ is subject to the restrictions
(with  the choice $\nu= \frac{2-\alpha}6$ which means that $\lambda =
\frac{4+\alpha}{2+2\alpha}$ )
\[
L\geq \frac{36}{2-\alpha}, \quad ,\quad L\leq \sqrt N.
\]

It is possible only if $N \geq \frac{6^4}{(2-\alpha)^2}$ and in that
case it is clear that $L$ should be chosen as small as possible and
from
the constraint that means
\[
L=\frac{36}{2- \alpha}.
\]
 With this choice, one has
\[
\frac{dQ}{dt}
\leq 1+\frac{C_a}{2-\alpha}\,c_\beta^a\;
\frac{1}{\delta_N N^{1-\frac a 2 }}+\frac C{2-\alpha}\,c_\beta.
\]
Now if one takes $\delta_N = N^{-\varepsilon}$, we can get
a uniform bound in $N$, only if
\[
\varepsilon \leq 1-\frac a 2 .
\]
If this is true, we get
\[
\frac{dQ}{dt}
\leq 1+ \frac{C_{a}\,c_\beta^{a}\,+C\,c_\beta}{2-\alpha}.
\]
with $C_a \leq \frac{C}{3a -2\alpha}$ which is the result given by
Theorem \ref{theorem1}.
\medskip

{\bf Acknowledgments:} this work was partially supported by the ANR
09-JCJC-009401 INTERLOP project.

redpajama_set_name: RedPajamaArXiv