text:
The Aughhhhhhhhhhhhhhhhhhhh Boom Indie Royale Round-up
Another day another bundle, except this Indie Royale is certainly doing something a little different. It seems as if the Indie Royale folks have finally settled on a plan from here on out: one main bundle a month with smaller lightning deals mixed in on occasion. It's a plan I can certainly get behind, especially based on the quality of the first lightning deal. To kick things off, the folks at Indie Royale, alongside Croteam and Devolver Digital are offering an all Serious Sam pack for the next few days.
Serious Sam is kind of a polarizing figure. Is he awesome, or just a Duke Nukem rip-off? Are the enemies creative or just insane products of a bad acid trip? Is the shooting mindless and fun or repetitive and dull? Is the story ridiculous and insane, or completely ridiculous and insane? The answers to this vary from person to person but if you're a fan of the old school style of shooters like Doom, Duke Nukem and Painkiller then you should definitely give Sam a chance. Plus, you'll never get a better opportunity than this great deal of a pack.
Serious Sam: The First Encounter
Developed by: Croteam
This is the first adventure of "Serious" Sam Stone, and basically sets the tone for the rest of the series. You begin by time travelling to Egypt in order to change the past and prevent the invasion of an alien horde led by the evil Mental. Despite all odds, you find a ton of guns scattered about Ancient Egypt, as well as many enemies to fire these guns at.
The gameplay is the very definition of run and gun action, with dozens of enemies teleporting in and charging at you. The weapons are generally pretty standard selection of guns, with variations on shotguns, pistols, rocket launchers and more making up the arsenal. Less standard are the selection of enemies: there are vicious one eyed teethy monsters, charging horse-like skeletons that breathe bolas, giant scorpion centaurs with chainguns for claws, screaming shirtless headless bombers and more. The nonsensical enemy design is basically a series staple at this point, and it constantly manages to repeatedly top itself; just when you think the enemies can't get any weirder, they do.
But if you don't mind the ridiculousness—or you revel in it, like I do— then you'll find a very fun series in Serious Sam. There are other nice elements too. Though Sam is clearly a hero in the Duke Nukem mould, I find him a lot more endearing just because he's not so obsessed with ripping off lines from They Live. There's splitscreen co-op, a very rare sight on PC games. And there's tons of secrets and easter eggs packed into each level. It's enough to keep you busy until you move onto...
Serious Sam: The Second Encounter
Otherwise known as Serious Sam 1 Part 2. I don't mean to sound lazy, but it's pretty much identical to the first game. There are new locations like Aztec jungles and ancient castles, more weapons, crazier enemies—jack o' lantern with a chainsaw anyone?—but the gameplay is almost completely identical. So go back three paragraphs and read that part again.
But since I have to use this space for something I might as well let you know that the versions included with the Indie Royale are the original releases, not the HD versions. Though it may seem weird considering how retrocentric my site is, and I probably sound superficial, but I much prefer the HD versions. The Serious Engine 3 has some excellent improvements in terms of animations and models, plus I find the physics are a little smoother. Also, the HD versions are a little nicer in terms of modern computers—they support widescreen, for example. That being said, you'll get functionally the same fun experience regardless and buying through the Indie Royale gets you three extra games to boot. It just looks a little more old school and vintage.
Plus the HD versions go on sale all the time on Steam, so you can probably pick them up for $1.25 down the road. What you won't get from the pack on Steam are the indie titles.
Serious Sam: Double D
Developed by: Mommy's Best Games
The Serious Sam Indie series was a promotional device planned up by Devolver Digital to promote the release of Serious Sam 3: BFE. The idea was that since Serious Sam is technically one of the older indie series out there, they would commission various indie developers to create their own re-imaginings of the Serious Sam formula. The first release was Double D, developed by Mommy's Best Games as a 2D side scrolling shooter take on the Serious Sam games.
The main hook (oh ho!) is the various connectors found around the world. Sam can use these hooks to attach guns together, creating ludicrous stacks of weaponry. A rocket launcher with two shotguns attached to it? Four tommy guns strapped to a grenade launcher? Chainsaw on chainsaw on chainsaw on chainsaw? You can do all of this and more through the very intuitive stacking inventory system.
These guns come in handy when you encounter the usual insane assortment of enemies, plus a few new ones. The developers almost parody the wacky enemies with their additions to Mental's hordes, with half monkey half jetpack beasts and menacing fusions of vuvuzelas and stacks of pancakes. 2D perspective and gun stacking aside, the game still feels like a Serious Sam title with all of the usual tropes in play. You still have secrets, hordes of monsters, giant bosses and vague puzzle moments. To sweeten the deal, there are several bonus challenge modes and a host of achievements to unlock. In the end, Double D feels like a full featured release, and worth the bundle price alone in my humble opinion.
Serious Sam: The Random Encounter
Developed by: Vlambeer
Just a bit of a warning: the expectations I had for this game were impossibly sky high. Vlambeer is the developer of Super Crate Box, otherwise known as my favourite game of all time ever. So I was just a tiny bit disappointed this game didn't capture my whimsy the way Super Crate Box does. That being said, it's still a decent title.
Random Encounter takes Serious Sam and moves it to the least likely genre: turn based RPG. Fortunately, the whole running backwards and shooting experience is totally intact. Each battle consists of Sam and his party furiously backpedalling as a huge horde of traditional Serious Sam enemies charge at you. You select your gun, aim your attack and then end your turn; for the next five seconds you can move Sam's party vertically as your attack plays out, dodging enemies all the while. It's a fairly ingenious combat system and it works very well, particularly for bosses. My other favourite element is the graphics; they look really excellent, bright and crisp, particularly the highly detailed boss sprites.
The rest of the game isn't stellar, but it is functional. The overworld contains some parodyesque "puzzles" but also features one of the worst elements of the genre: random encounters. I understand that's the whole title of the game, but in this day and age I'm far less tolerant of them. I'd much rather have had groups of enemies visible on the world map like the majority of JRPGS do these days. But I suppose my only real complaint towards the game is that it doesn't feel like a full featured release. When you compare the amount of extras and polish in Double D, Random Encounter winds up feeling more like a proof of concept for the battle system. I mean, the battle system is great, but the game surrounding it doesn't have the same amount of polish. But hey, it's cute to see in action.
Serious Sam: Kamikaze Attack
Developed by: Be-Rad Entertainment
The only iPhone game in the Serious Sam indie series, now on PC. Let's get the obvious comment out of the way: yes, it's an iOS title so yes, it is fairly simple and casual. It's still quite fun. You take on the role of one of those "headless freaks" and run towards Sam, who's running through Egypt blissfully unaware of his impending doom. It's a Canabalt style clone, except one with an end goal and the ability to attack, kicking forward to deflect and destroy obstacles. Attack too often however, and your headless avatar gets overexcited and explodes.
The real addicting elements come from the bonus objectives found in each level (usually revolving around destroying so many obstacles) and also the slow trickle of upgrades you get following each victory. Another excellent touch is the graphics, which look beautifully shaded with colour pencils. It's too simple to call it a full game, but there's a lot of content and fun to be had in this little diversion.
Serious Sam: The Greek Encounter
Developed by: Eric Ruth Games
The Greek Encounter is billed as a bonus, and rightly so. It's a commissioned piece from Eric Ruth, the guy behind many brilliant NES demakes of current games like Left 4 Dead, Halo and the tragically ceased and desisted DJ Hero. While I'm sure the Serious Sam series is rife with possibilities for demakes, this release feels... well, not lazy but possibly rushed. The brilliance of the other Ruth games are that they were built according to the specifications of the NES, but Greek Encounter cheats a lot; it's little things, like the way that Sam's sprite can fully rotate and the fact that twin stick shooters are not a very retro genre. The enemies are all taken from the Sam canon, but there's not a lot of differences between them besides how many bullets they can take, while the levels themselves are very repetitive and hard to navigate. It's mindless fun for a few rounds, making it a fine bonus, but don't expect amazing things.
So here we are at the end, and not one "seriously good deal" joke. I'm proud of myself. If you're new to the Serious Sam games and these sound great to you, there's never been a better chance to jump in. If you're an old fan there's plenty of fun to be had with the indie titles, which are absolutely full of glorious fanservice. It's a great pack if you're into old school over the top shooters, so why not check out the Indie Royale page to pick up your own bundle.
For the latest on backlogs, reviews and bundles you can follow me via Twitter to keep up to date.
Labels: Indie, Indie Bundles, Indie Royale, Serious Sam
Clayton 13 April 2012 at 20:09
Glad to hear I didn't miss out on much. The main games don't really interest me at this point, and I unknowingly bought Double D with the GetGamesGo thing.


Any guitarist who marvels at the breathtaking sonic artistry and peerlessly pure tone of Texas Strat® virtuoso Eric Johnson will also marvel at the Eric Johnson Stratocaster Maple guitar. Johnson is famously meticulous in crafting his tone, and every facet of the guitar is designed to Johnson's own detailed specs, from the specially voiced pickups to the extra-contoured neck and body to the distinctive tremolo assembly and much more. A refined signature model honoring one of modern music's finest Strat masters.


text:
River Avenue Blues
Yankees Top 30 Prospects
Prospect Profiles
Fan Confidence
2019 Draft Order
Bullpen Workload
Guide to Stats
Shop and Tickets
RAB Tickets
Steiner Sports Memorabilia
River Ave. Blues » Rumor of the day: Who watches the Wanger?
Rumor of the day: Who watches the Wanger?
June 29, 2009 by Benjamin Kabak 211 Comments
July 2nd Signing Period Q&A with Ben Badler
Mo earns Player of the Week honors for RBI walk
When the Yankees nailed down Chien-Ming Wang's first win of 2009, the team breathed a collective sigh of relief. The psychological benefits alone for a pitcher who, two months ago, was 0-3 with a 34.50 ERA in six innings, are immense, and the Yankees need their cost-controlled sinker ball specialist, once among the game's best pitchers to rediscover his form.
Since returning from the disabled list in late May, Wang has looked better than he did in April. Over eight games, five starts, he has thrown 30.2 innings and is 1-3 with a 5.28 ERA. As we're used to seeing Wang twirling over six innings a start to the tune of a sub-4.00 ERA, that's still not quite up to par for the Taiwanese hurler, but instead of looking epically bad, it resembles a rough patch. We generally wouldn't worry about a rough patch.
When push comes to shove this season, Wang's final numbers won't look pretty. His ERA stands at 10.06 through the first three months of the season. If Wang makes his final 17 starts and averages six innings per start, he will have to pitch to an ERA of 2.38 just to get his season mark to 4.50. I don't think anyone expects that.
Yet, despite these inflated numbers, Wang is still attracting interest from other teams. Today's rumor du jour comes to us via the Daily Dish. Per the rumors site, Jim Salisbury of The Philadelphia Inquirer noted yesterday that the Phillies have been scouting Wang. The pitching-starved defending World Champs need some arms, and as Salisbury reasons, the Yanks, with Phil Hughes waiting in the wings, could opt to trade Wang.
While it's awfully flattering of the Phillies to look, chances are close to nil that the Yanks would trade Wang this year. First, while the Yankees have long been reluctant to give Wang a long-term deal, his value is at an all-time low. He is coming off of a bad foot injury and a three-month winless bout of ineffectiveness. His sinker isn't quite doing what it needs to be doing, and he's still getting his legs under him. The Yanks could probably land something decent for Wang, but nothing the Phillies would be willing right now to offer could replace Wang's potential.
Furthermore, Wang is still a big part of the Yanks' plans. By most accounts, this will be Andy Pettitte's last year in pinstripes. The Yanks have the younger pieces waiting in the wings to supplant Pettitte, and to do that, they will rely on Wang to anchor the middle of the rotation. Right now, the Yanks view the 2010 rotation as featuring CC Sabathia, A.J. Burnett, Wang, Joba Chamberlain and Hughes. They won't give up on Wang, cost-controlled through the end of 2011 and with a good track record, so easily.
Of course, in the end, everyone has a price. If the Phillies make the right offer, they could probably land Wang. But I wouldn't read much into this. Ruben Amaro, the Philadelphia GM, is simply doing his due diligence on a potential acquisition, but the Yanks won't give up on this one so easily.
Filed Under: Irresponsible Rumormongering Tagged With: Chien-Ming Wang
RAB Thoughts on Patreon
Mike is running weekly thoughts-style posts at our "RAB Thoughts" Patreon. $3 per month gets you weekly Yankees analysis. Become a Patron!
Got A Question For The Mailbag?
Email us at RABmailbag (at) gmail (dot) com. The mailbag is posted Friday mornings.
RAB Features
2019 Season Preview series
2019 Top 30 Prospects
'What If' series with OOTP
Yankees depth chart
Search RAB
Copyright © 2023 · River Avenue Blues

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Search by list Xamarin Forms (MVVM) I have a problem with searching records. As you type characters, the list will not update.
At the moment I am using Binding Login for testing but I will be using Uri with a photo.
My ViewModel
private ObservableCollection<FollowerModel> _followerItems = null;
public ObservableCollection<FollowerModel> FollowerItems
{
    get => _followerItems;
    set
    {
        _followerItems = value;
        OnPropertyChanged();
        OnPropertyChanged();
    }
}

private FollowerModel _followerItem = null;
public FollowerModel FollowerItem
{
    get => _followerItem;
    set
    {
        _followerItem = value;
        OnPropertyChanged();
        OnFollowerItemChanged();
    }
}

// ...

private  void OnSearchPhrase()
{
    var searchPhrase = _searchPhrase.Trim();

    if (searchPhrase.Length < SEARCH_PHRASE_MIN_LENGTH) return; // SEARCH_PHRASE_MIN_LENGTH = 3

    FollowerItems.Where(name => name.Login.Contains(searchPhrase));
}

My PageView
<flv:FlowListView SeparatorVisibility="None" HasUnevenRows="true" FlowItemTappedCommand="{Binding FollowerItemCommand}" FlowColumnMinWidth="100" FlowItemsSource="{Binding FollowerItems}">
    <flv:FlowListView.FlowColumnTemplate>
        <DataTemplate>
            <Grid Padding="5">
                <Grid.RowDefinitions>
                    <RowDefinition Height="98" />
                </Grid.RowDefinitions>
                <Grid.ColumnDefinitions>
                    <ColumnDefinition Width="98" />
                </Grid.ColumnDefinitions>
                <Label Text="{Binding Login}" />
                </Image>-->
            </Grid>
        </DataTemplate>
    </flv:FlowListView.FlowColumnTemplate>
</flv:FlowListView>

A: First add a search bar in your xaml
 <SearchBar Placeholder="Search" Text="{Binding SearchText}"  />

Then Create a Bindable property for it
private string _searchText;
public string SearchText
{
    get { return _searchText; }
    set
    {
        _searchText = value;
        SearchUserText(_searchText);
        RaisePropertyChanged("SearchText");
    }
}

Then Create a function SearchUserText that will listen for text and change your list
   public void SearchUserText(string text)
    {
          var searchPhrase = _searchPhrase.Trim();

            if (searchPhrase.Length < SEARCH_PHRASE_MIN_LENGTH) return; // SEARCH_PHRASE_MIN_LENGTH = 3

            FollowerItems.Where(name => name.Login.Contains(searchPhrase));

    }

NB: Don't forget to add getters and setters on your model FollowerModel
example:
public class FollowerModel : BindableBase
{
    private int _id;
    public int Id
    {
        get { return _id; }
        set { SetProperty(ref _id, value); }
    }

    private string _name;
    public string Name
    {
        get { return _name; }
        set { SetProperty(ref _name, value); }
    }
}

A: First of all, you will need two ObservableCollections:

*

*One ObservableCollection which contains all items

*One ObservableCollection which contains the filtered items

The filtered ObservableCollection will be the one which is binded to your control. So you only need a private variable for all items and a public property for the filtered items:
private ObservableCollection<FollowerModel> _allFollowerItems;
private ObservableCollection<FollowerModel> _followerItems;

public ObservableCollection<FollowerModel> FollowerItems
{
    get { return _followerItems}
    set
    {
        _followerItems = value;
        RaisePropertyChanged();
    }
}

One more thing I noticed is that you are using OnPropertyChanged(). It is deprecated, use RaisePropertyChanged() instead.
In your OnSearchPhrase() method, you want to set FollowerItems to the original items (_allFollowerItems) if searchPhrase.Length < SEARCH_PHRASE_MIN_LENGTH. Otherwise, you will filter it:
private void OnSearchPhrase()
{
    var searchPhrase = _searchPhrase.Trim();

    if (searchPhrase.Length < SEARCH_PHRASE_MIN_LENGTH)
    {
        FollowerItems = _allFollowerItems;
    }
    else
    {
        FollowerItems = new ObservableCollection<FollowerModel>(_allFollowerItems.Where(name => name.Login.ToLower().Contains(searchPhrase.ToLower())));
    }
}

Notice that I added ToLower() to the strings to have a more user friendly search. If Case-sensetive search was intended, just remove it.
Don't forget to set FollowerItems and _allFollowerItems once when at the place where you are loading your items right now:
FollowerItems = new ObservableCollection(items);
_allFollowerItems = new ObservableCollection(items);

redpajama_set_name: RedPajamaStackExchange


text:
Lecane spinulifera är en hjuldjursart som beskrevs av John R. Edmondson 1935. Lecane spinulifera ingår i släktet Lecane och familjen Lecanidae. Inga underarter finns listade i Catalogue of Life.

Källor

Hjuldjur
spinulifera

redpajama_set_name: RedPajamaWikipedia


text:
Gasoline Pipeline Hack Causes Fuel Shortage on East Coast
Last week, hackers began a ransomware attack on the Colonial Pipeline, a major conduit that supplies nearly half of the East Coast's oil. The hack completely shut down petroleum transportation as the Colonial Pipeline Company "proactively took certain systems offline to contain the threat, which has temporarily halted all pipeline operations, and affected some of our IT systems." Although the company expects to have most of its capabilities returned by the end of the week, the shutdown has resulted in gasoline shortages that have left many drivers nervous.
For instance, 12 percent of gas stations in North Carolina have no gasoline while 8.6 percent of Virginia filling stations are also without supplies. Georgia, Florida, and South Carolina are experiencing similar shortages, some of which has been aggravated by panic buying from anxious consumers. According to experts, such "irrational behavior" has the potential to prolong the problem for weeks. "People are taking their entire family fleet of vehicles to the gas station and filing up when they don't need to," said Tiffany Wright of AAA Carolinas. "We are our own worst enemy in this situation because we are over-consuming at the pump."
As a result, officials are pleading with Americans not to stock up on gasoline. "Let me emphasize that much as there was no cause for say, hoarding toilet paper at the beginning of the pandemic, there should be no cause for hoarding gasoline," said Energy Secretary Jennifer Granholm, "especially in light of the fact that the pipeline should be substantially operational by the end of this week and over the weekend." As demand for gasoline shot up by 20 percent at the beginning of the week, gas prices similarly rose to their highest levels in nearly six years. The petroleum market should return to normal once the Colonial Pipeline is operational again, as long as consumers do not continue to overdraw from the nation's filling stations.
What is the cause of the current gasoline shortage affecting the East Coast?
Why could the current fuel shortage potentially continue for weeks?
Source: Matt Egan, "Gasoline demand spikes in several states after pipeline hack," CNN, May 12, 2021; Andrew Morse, "Gas shortage 2021? What you really need to know about the pipeline hack," May 12, 2021.
Economics, Energy, Ethics, Logistics, Oil, Operations Management, Supply & Demand, Transporatation
Chapter 15: Distribution, Chapter 1: Taking Risks and Making Profits, Chapter 2: Economics, Chapter 3: Globalization, Chapter 4: Ethics and Social Responsibility, Chapter 9: Production and Operations Management
« New Unemployment Claims Gradually Begin to Drop
[VIDEO] How Roblox Became a $45 Billion Public Video Game Company »


text:
Toronto Film Festival Coverage Zombie Movies Hollywood News Top 10 Lists Crime Movies
Tag: thehobbit
2016-08-15T15:00:00Z by Loïc Valceschini
Locarno 2016 Interview: Howard Shore Talks His Career and Lifelong Collaborations
Honoured at the 69th edition of Locarno Film Festival, Howard Shore attended the Swiss event to receive his lifetime award and to introduce a few screenings. ScreenAnarchy was offered the chance to talk to him about his impressive career and...
2015-01-07T10:01:10Z by Matt Brown
Destroy All Monsters: The 2-Hour Version Of THE HOBBIT
I'm sure someone is fan-editing this even as I type, but since it's a slow week and since no one alive would actually want to watch a two-hour speculative video composed entirely of content we've already seen in eight-plus hours...
Destroy All Monsters: The Lessons Of THE HOBBIT
As an exercise in Hollywood franchise-making in the 21st century, Peter Jackson's The Hobbit is fantastically instructive. The final film in the cycle emerges today -- I've yet to see it; you can read Jason Gorber's review here-- but long...
Hobbitty ScreenAnarchyes: Our Favorite Film Dragons
The Hobbit: The Battle of the Five Armies is now in cinemas all over half the world, an will soon arrive in the other half as well. And for those who saw the end of The Hobbit: The Desolation of...
Destroy All Monsters: Trumbullvision
Earlier this month, as part of the Stanley Kubrick exhibit and retrospective at the TIFF Bell Lightbox, visual effects wizard Douglas Trumbull was in town to demo his proof-of-concept film for his next generation high-frame-rate presentation format, MAGI. Sort of...
2014-11-07T04:30:00Z by Stuart Muller
Leave Sauron To Saruman In The Final Trailer For THE HOBBIT: THE BATTLE OF THE FIVE ARMIES
The stage and stakes are set. The characters and conflicts established. All that remains is for the complex climax of competing wills - and armies - to play themselves out. Word is that the final installment in Peter Jackson's cinematic Middle-Earth...
First Poster For THE HOBBIT: THE BATTLE OF THE FIVE ARMIES
Peter Jackson has shared the first poster for The Hobbit: The Battle Of The Five Armies ahead of the big event Warner Brothers has planned for San Diego Comic-Con. And it is a ripper!The poster is clearly intended to remind...
Third HOBBIT Film Retitled 'HOLY CRAP, AREN'T WE DONE YET?'
Okay, that's not the actual title. Nor is it Bring Me More Gold, I Don't Have Enough Yet And If You Don't I'll Pad Out The Story Even More. No, the third Hobbit film has jettisoned the original There And...
Korean Box Office: Song Kang-ho Proves He's B.O. King with THE ATTORNEY
With the help of a homegrown offering's huge opening, business came booming back by soaring over the three million mark for the first time in months. The pre-Christmas weekend saw 61% of tickets got to local films. Title Release...
Bilbo, Smaug And Thorin Meet Dandalf The Disco Wizard
Here we go, Hobbit fans ... Berlin based prankster Daniele Rizzo managed to land video interviews with stars Martin Freeman, Benedict Cumberbatch and Richard Armitage (that'd be Bilbo, Smaug and Thorin) and, as per his normal lunacy, showed up in...
Korean Box Office: Jeon Do-yeon No Match for THE HOBBIT
Despite a major new local release, the Korean film industry was held at bay once again as the pairing of Jeon Do-yeon and Ko Soo wasn't enough to prevent the new Hobbit from taking first place. 2.24 million tickets were...
2013-12-07T20:00:08Z by Jason Gorber
Review: THE HOBBIT: THE DESOLATION OF SMAUG, A Hell Of A Ride
When last year's The Hobbit - An Unexpected Journey hit, I addressed the central question being asked by most fans of the Lord of the Rings films: "How could something so relatively slight, an explicitly childish work that's little more...
Watch The New Trailer For THE HOBBIT: THE DESOLATION OF SMAUG
Hobbits and elves and humans, oh my! Yes, kids, the second installment in Peter Jackson's The Hobbit trilogy hits screens December 13 and that means it's time for a new trailer along with some character banners that may very well...
First Shot Of Evangeline Lilly As Tauriel In THE HOBBIT
Poor Evangeline Lilly, people are going to hate her pretty much no matter what. When Peter Jackson's Hobbit Boogaloo hits screens later this year it'll be Lilly who, through no fault of her own, will almost certainly be the lightning...
2013-01-22T10:00:10Z by Jim Tudor
TTTT: An American Film Geek's Bottom 10 for 2012
Many a film critic, including at times even the most positive and altruistic of us, take a certain mean delight in unleashing our master lists of the year's biggest stinkers. The temptation to avenge what may be an assault on...
Lord of the Memes: The Soviet Version of THE HOBBIT For Your Viewing Pleasure
The Internet is occasionally a remarkable place, a wonderland where people can plop up things that you simply wouldn't have believed to have existed.Some intrepid YouTubers have put onto the video sharing service the 1985 Soviet film Хоббит, or The...
Fantasy, More Real Than Real: THE HOBBIT, HFR And The Future Of Movies
Earlier this year, I found myself at a friend of a friend's apartment watching Tod Browning's 1932 masterpiece Freaks through the "motion smoothing" filter on his HD TV. For me, the resulting video-like image of what should be the opposite...
Opening: THE HOBBIT: AN UNEXPECTED JOURNEY. Now It's Time For Moviegoers To Weigh In.
An event movie if ever there was one, Peter Jackson's The Hobbit: An Unexpected Journey rolls out across North America at midnight tonight, giving night owls the chance to pass 169 minutes in the company of Bilbo, Gandalf, Gollum, more...
Jason Gorber's Cineruminations: THE HOBBIT, 48 Frames Per Second, and A Whole New Journey
"To a regular cinemagoer...attendance at the motion-picture playhouse today is a continuously disturbing experience...The discovery that the shadowy images of the screen could be made articulate was as fruitful for exploitation to the captains of the cinema industry as was...


text:
\section{Introduction.}

Once nonlinear terms are included in linear dispersive
equations, solitary waves can result which can be stable
enough to persist indefinitely.
It is well known that many important nonlinear evolution
equations which have numerous applications
in mathematical physics appear as sufficient conditions
for the integrability of systems of linear partial
differential equations of first order, and such systems
are referred to as integrable {\bf [1-2]}. This is not
just an oddity, since algebraic structures such as those
which appear in AKNS systems can arise very naturally
from nonlinear evolution equations. This is very well
exemplified by applying the so called prolongation
technique. Wahlquist and Estabrook {\bf [3-5]} first
constructed prolongations, or $sl (2)$ systems, both for the
KdV and nonlinear Schr\"odinger equations, and Shadwick the
former {\bf [6]}. This procedure produces
a nonclosed Lie algebra of vector fields  which are
defined on fibres above the base manifold that supports
the exterior differential forms defining the nonlinear
evolution equation. It has been shown that a simple
linear prolongation of the KdV, sine-Gordon and
non-linear Schr\"odinger equation can be provided
by $sl (2, \mathbb C)$ {\bf [7]}. What is more, it has been shown
that the vanishing of the curvature form of a
particular Cartan-Ehresmann connection is the necessary
and sufficient condition for the existence of the
prolongation.

These prolongations have a very useful application since
B\"acklund transformations can be calculated based on them as well {\bf [8]}.
A B\"acklund transformation has important practical consequences,
since such transformations can be used to calculate
solutions to an associated equation, usually referred to
as the potential equation, based on solutions of the
initial equation. Sometimes these transformations can be used to obtain new
solutions to the same initial equation, in which case
they are referred to as auto-B\"acklund transformations.

Recently an exterior differential system which defines
a generalized KdV equation on the transverse manifold
was obtained {\bf [9]}. A particular case of this equation
has appeared in {\bf [10]} recently. The symmetries
of this equation were determined and some solutions
were found as well {\bf [11]}. This permitted the
determination of a certain form of integrability. Also,
a particular type of prolongation over a fibre bundle was
found corresponding to this differential system, as well as
a specific form for a B\"acklund transformation with its
associated potential equation. Here, the same differential
system is studied, but a fully general calculation of
the prolongation over the same bundle is carried out
in detail for this generalized KdV equation. This allows
the prolongation structure for any case of the given
parameters in the equation.
For completeness, the general theory
for obtaining such prolongations based on the given
exterior system of differential forms that defines the equation
upon sectioning to a transversal integral manifold will be outlined first.
Transversal integral manifolds give solutions of the equation.
Finally this work is extended
to a study of a differential system of one-forms which define an equation
that includes the Camassa-Holm equation and Degasperis-Procesi equations
as specific cases {\bf [12-14]}. The Camassa-Holm equation has been of interest
because it has been shown to have peaked soliton solutions.
The Camassa-Holm equation has alot in common with the KdV equation,
but there are significant differences as well.
The KdV equation is globally well-posed when considered on
a suitable Sobolev space, while Camassa-Holm is in general not.
The first derivative of a solution of the latter can become
infinite in finite time.
The associated
prolongation equations are developed and found to be much more restrictive
than the previous case. However, it is shown that at
least one solution to the prolongation system can be found.
Finally, for each system a brief discussion concerning how
conservation laws arise and can be expressed in this context will be discussed
based on the defining exterior differential system.

\section{Introduction to Cartan Prolongations.}

Cartan prolongations will be found for the equations
mentioned in the Introduction, and to begin, a general
outline of this subject is given. Consider the space
$M = \mathbb R^n$ with the coordinates
$( x, t, u, p, q, \cdots)$ and let there be given
on $M$ a closed exterior differential system
$$
\alpha_1 =0, \qquad
\cdots,  \qquad
\alpha_l =0.
\eqno(2.1)
$$
Let $I$ be the ideal generated by the system
(2.1), and so $I = \{ \chi = \sum_{i=1}^l
\sigma_i \wedge \alpha_i : \sigma_i \in
\Lambda_p (M), p=0,1,2, \cdots \}$ with
$\Lambda_p (M)$ the set of $p$-forms on $M$.
Since (2.1) is closed, we have $d I \subset I$
and (2.1) is integrable.

The system (2.1) will be chosen such that
solutions $u= u(x,t)$ of an evolution equation
$u_t = F (x,t, u, u_x, u_{xx}, \cdots )$
correspond with two-dimensional transversal
integral manifolds of (2.1). These integral
manifolds can be written as sections $S$ in
$M$ with $S$ given by
$$
(x,t) \rightarrow ( x, t, u(x,t), p(x,t), q(x,t),
\cdots ),
$$
and due to transversality, $dx \wedge dt |_S
= \pi^* ( dx \wedge dt) \neq 0$ such that
$\pi :M \rightarrow \mathbb R^2$ and
$\pi^* : \Lambda ( \mathbb R^2) \rightarrow
\Lambda (S)$. Now introduce the fibre bundle
$( \tilde{M}, \tilde{p}, M)$ over $M$, with
$M \subset \tilde{M}$ and $\tilde{p}$ the projection of
$\tilde{M}$ onto $M$, $\tilde{p} ( \tilde{M}) =M$.
Points in $\tilde{M}$ are written $\tilde{m}$,
those in $M$ written $m$ and $\tilde{p} ( \tilde{m}) =m$.
The tangent and cotangent spaces of $\tilde{M}$
and $M$ are denoted by $T (\tilde{M} )$ and
$T^* (\tilde{M})$.

For reference, a Cartan-Ehresmann connection in the
fibre bundle $( \tilde{M}, \tilde{p}, M)$ is a
system of 1-forms $\tilde{\omega}_i$, $i=1,2,\cdots, k$
in $ T^* ( \tilde{M})$ with the property that the
mapping $\tilde{p}_*$ from the vector space
$H_{\tilde{m}} = \{ \tilde{X} \in T_{\tilde{m}} |
\tilde{\omega}_i ( \tilde{X} ) =0, i=1,2,\cdots, k \}$
onto the tangent space $T_m$ is a bijection for all
$\tilde{m} \in \tilde{M}$.
At this point, consider the exterior differential
system in $\tilde{M}$
$$
\tilde{\alpha}_i = \tilde{p}^* \alpha_i =0,
\quad
i=1, \cdots, l,
\quad
\tilde{\omega}_i =0,
\quad
i=1, \cdots, k,
\eqno(2.2)
$$
with $\{ \tilde{\omega}_i \}$ a Cartan-Ehresmann
connection in $(\tilde{M}, \tilde{p}, M)$.
The system (2.2) is called a Cartan prolongation
if (2.2) is closed and whenever $S$ is a
transversal solution of (2.1), then there should
also exist a transversal solution $\tilde{S}$
of (2.2) with $\tilde{p} ( \tilde{S} ) =S$.
It follows from (2.2) closed that this prolongation
condition may be written as
$$
d \tilde{\omega}_i = \sum_{j=1}^k \,
\tilde{\beta}^j_i \wedge \tilde{\omega}_j,
\quad
mod \quad \tilde{p}^* (I),
\eqno(2.3)
$$
where $I$ is the ideal defined in (2.1).

For the considerations here, the fibre bundle will
be the trivial fibre bundle given by $\tilde{M}
= M \times \mathbb R^k$ with $y = (y_1, \cdots, y_k)
\in \mathbb R^k$ and the connection used will have the form
$$
\tilde{\omega}_i = d y_i - \eta_i,
\qquad
\eta_i = \, A_i \, dx + B_i \, dt.
\eqno(2.4)
$$
In (2.4), $A_i$ and $B_i$ are defined as
$C^{\infty}$-functions on $\tilde{M}$, $i=1, \cdots, k$.
The prolongation condition (2.3) applied to (2.4) is then
$$
- d \eta_i = \tilde{\beta}_i^j \wedge
( d y_j \wedge \eta_j),
\quad mod \quad \tilde{p}^* (I),
\quad i=1, \cdots, k,
\eqno(2.5)
$$
with $\eta_i$ given in (2.4) and $A_i$, $B_i$ depend
on $x$, $t$, $u$, $p$, $q, \cdots, y_1, \cdots, y_k$.
Comparing both sides of (2.5),
$\tilde{\beta}^j_i$ cannot contain differentials of
the form $d y_s - \eta_s$ for $s \neq j$ and so
$$
\tilde{\beta}_i^j = a^j_i \, dx + \beta_i^j \, dt
+ c^j_i \, du + d^j_i \, dp + \cdots \quad
mod \quad \gamma_i^j ( dy_j - \eta_j ).
$$
Comparing forms on both sides of (2.5) yields the results
$$
a_i^j = \frac{\partial A_i}{\partial y_j},
\qquad
b_i^j = \frac{\partial B_i}{\partial y_i},
$$
with $c_i^j = d_i^j = \cdots =0$ because
$du \wedge d y_j$, $dp \wedge d y_j, \cdots$ do not
occur on the left of (2.5). The prolongation
condition reduces to
$$
- d \eta_i = \frac{\partial \eta_i}{\partial y_j}
\wedge ( d y_j - \eta_j ),
\quad mod \quad \tilde{p}^* (I).
\eqno(2.6)
$$
Introducing the vertical valued one-form
$\eta = \eta_i \frac{\partial}{\partial y_i}$
as well as the definitions
$$
d \eta = ( d_M \eta_i ) \frac{\partial}{\partial y_i},
\quad
[ \eta, \omega] = ( \eta_j \wedge \frac{\partial \omega_i}
{\partial y_j} + \omega_j \wedge \frac{\partial \eta_i}
{\partial y_j} ) \frac{\partial}{\partial y_i},
\eqno(2.7)
$$
the prolongation condition reduces to the compact
form,
$$
d \eta + \frac{1}{2} [ \eta, \eta ] =0,
\quad mod \quad \tilde{p}^* (I).
\eqno(2.8)
$$
The form on the left of (2.8) is called the
curvature form of the Cartan-Ehresmann connection
$(d y_i - \eta_i)_{i=1}^{\infty}$. Thus, a
sufficient condition for the existence of a
Cartan prolongation of the set of exterior
differential forms (2.1) is the vanishing of the
curvature form of the Cartan-Ehresmann
connection $(dy_i - \eta_i)_{i=1}^k$.

\section{ Cartan Prolongation of a Generalized KdV Equation.}

\subsection{Differential System and Associated Partial Differential Equation.}

Let us introduce the exterior differential system
defined over a base manifold $M = \mathbb R^5$ which supports the differential
forms. Consider the system of two forms given by
$$
\alpha_1 = n u^{n-1} \, du \wedge dt
- p \, dx \wedge dt =0,
$$
$$
\alpha_2 = dp \wedge dt - q \, dx \wedge dt =0,
\eqno(3.1)
$$
$$
\alpha_3 = du \wedge dt - d q \wedge dt
- \gamma p u^s \, dx \wedge dt =0,
$$
where $\gamma$ is a nonzero, real constant.
The exterior derivatives of the $\alpha_j$ are given
by
$$
d \alpha_1 = - dp \wedge dx \wedge dt = dx
\wedge \alpha_2,
$$
$$
d \alpha_2 =- dq \wedge dx \wedge dt =- dx \wedge \alpha_3,
\eqno(3.2)
$$
$$
d \alpha_3 =- \gamma sp u^{s-1} \, du \wedge dx \wedge dt
- \gamma u^s dp \wedge dx \wedge dt
= dx \wedge ( \gamma \frac{s}{n} p u^{s-n} \alpha_1
+ \gamma p u^s \alpha_2 ).
$$
Therefore, the ideal $I = \{ \omega | \omega =
\sum_{i=1}^3 \, \sigma_i \wedge \alpha_i :
\sigma_i \in \Lambda (M) \}$ is closed,
$d I \subset I$ and the system $\{ \alpha_i \}$ given by
(3.1) is integrable. On the transversal integral
manifold, it follows that differential system (3.1)
can be sectioned to give,
$$
0 = \alpha_1 |_S = S^* \alpha_1 = (( u^n )_x -p) \, dx
\wedge dt,
$$
$$
0 = \alpha_2 |_S = S^* \alpha_2 = ( p_x -q) \, dx
\wedge dt,
\eqno(3.3)
$$
$$
0 = \alpha_3 |_S = S^* \alpha_3 =
( u_t dt \wedge dx - q_x \, dx \wedge dt
- \gamma p u^s \, dx \wedge dt ).
$$
The transversal integral manifolds correspond to the
equations
$$
p = (u^n )_x,
\qquad
q= p_x = ( u^n )_{xx},
\qquad
u_t + q_x + \gamma p u^s =0.
\eqno(3.4)
$$
Suppose that $n+s \neq 0$, then upon
substituting $p$ and $q$ from the first two
equations in (3.4) into the third, it can be
seen that $u$ must satisfy the following
generalized KdV equation
$$
u_t + (u^n )_{xxx} + \gamma \frac{n}{n+s}
( u^{n+s} )_x =0.
\eqno(3.5)
$$
A more compact form is obtained if we set $m= n+s$
and define a new constant $\beta = n \gamma/
(n+s)$ so that (3.5) takes the form
$$
u_t + ( u^n )_{xxx} + \beta ( u^m )_x =0.
\eqno(3.6)
$$
This is the partial differential equation defined
by differential system (3.1) which was studied in {\bf [11]}.

\subsection{Prolongations.}

Based on the forms
in system (3.1), the prolongation method outlined
in Section 2 can be carried out, and the
resulting system of equations can be solved quite
generally. A very general prolongation corresponding
to (3.6) can be calculated in terms of an algebra
of vector fields which are defined on fibres above the
base manifold that supports the forms (3.1).
To do this, introduce the pseudopotentials and the
Cartan-Ehresmann connection on the trivial fibre
bundle $\tilde{M} = M \times \mathbb R^k = \mathbb R^5
\times \mathbb R^k$ with coordinates $y = (y_1, \cdots,
y_k)$ on $\mathbb R^k$. The connection forms are
taken to be
$$
\tilde{\omega}_i = d y_i - \eta_i,
\qquad
\eta_i = A_i \, dx + B_i \, dt,
\qquad
A_i = A_i (x,t,u,p,q,y),
\qquad
B_i = B_i (x,t,u,p,q,y).
\eqno(3.7)
$$
Substituting $\eta = \eta_i \, \partial/ \partial y_i$,
$A = A_i$ $\partial / \partial y_i$, $B = B_i \partial /
\partial y_i$, the prolongation condition for the
vectors $A$ and $B$ is
$$
( d_M A_i \wedge dx ) \frac{\partial}{\partial y_i}
+ ( d_M B_i \wedge dt ) \frac{\partial}{\partial y_i}
+ \frac{1}{2}  \{ A_j \frac{\partial B_i}{\partial y_j}
\, dx \wedge dt + B_j \frac{\partial A_i}{\partial y_j}
\, dt \wedge dx \} \frac{\partial}{\partial y_i}
$$
$$
+ \frac{1}{2} \{ B_j \frac{\partial A_i}{\partial y_j} \,
dt \wedge dx + A_j \frac{\partial B_i}{\partial y_j} \,
dx \wedge dt \} \frac{\partial}{\partial y_i}
$$
$$
=  d A \wedge dx + dB \wedge dt + [ A, B] \, dx \wedge dt =0,
\qquad mod \quad \tilde{p}^* (I),
$$
where $[A, B]$ denotes the ordinary Lie bracket of the
vector fields $A_i \partial / \partial y_i$ and
$B_i \partial /\partial y_i$ defined along fibres of the bundle.
Using (3.1), it is found that the prolongation
condition takes the form
$$
\frac{\partial A}{\partial t} \, dt \wedge dx
+ \frac{\partial A}{\partial u} \, du \wedge dx
+ \frac{\partial A}{\partial p} \, dp \wedge dx
+ \frac{\partial A}{\partial q} \, dq \wedge dx
$$
$$
+ \frac{\partial B}{\partial x} \, dx \wedge dt
+ \frac{\partial B}{\partial u} \, du \wedge dt
+ \frac{\partial B}{\partial p} \, dp \wedge dt
+ \frac{\partial B}{\partial q} \, dq \wedge dt +[A, B] \, dx \wedge dt
$$
$$
= \lambda_1 ( n u^{n-1} du \wedge dt - p \, dx \wedge dt)
+ \lambda_2 ( dp \wedge dt - q \, dx \wedge dt)
+ \lambda_3 ( du \wedge dx - dq \wedge dt -
\gamma p u^s \, dx \wedge dt).
$$
Comparison of both sides of this equation yields the
following set of conditions,
$$
\begin{array}{ccc}
A_u = \lambda_3, & A_p =0,  &  A_q =0,  \\
   &  &   \\
B_u = n \lambda_1 u^{n-1},  &  B_p = \lambda_2, &  B_q =- \lambda_3,  \\
   &  &   \\
& - A_t + B_x + [A,B] =- p \lambda_1 -q \lambda_2 - \gamma p u^s
\lambda_3.  &   \\
\end{array}
\eqno(3.8)
$$
Subscripts indicate partial differentiation with respect to the
variable indicated.
Translations in $x$ and $t$ constitute symmetries of
equation (3.6) {\bf [11]}, and so a simplifying
assumption would be to suppose that $A$ and $B$ are independent
of $x$ and $t$. Then it must be that $A$ and $B$ are also
invariant under translations in these variables.
The prolongation equations to be solved from (3.8) reduce to the following
$$
A_p = 0 , \qquad A_q =0,
\qquad
A_u =- B_q,
\qquad
\frac{1}{n} u^{1-n}p B_u + q B_p - \gamma p u^s B_q =- [A, B].
\eqno(3.9)
$$

\begin{theorem}
System (3.9) can be reduced to a single
expression which specifies the algebra of brackets of a set of
basis vector fields $X_i$. The structure of these algebras
is dependent on the relative values of $m$ and $n$.
\end{theorem}

{\em Proof:}
The first three differential equations in (3.9) imply the
following results
$$
A= A(u,y),
\qquad
B = B (u,p,q,y),
\quad
B =- q \, A_u (u,y) + \hat{B} (u,p,y).
\eqno(3.10)
$$
Substituting $B$ from (3.10) into (3.9) and collecting
terms in $q$ gives
$$
q ( - \frac{1}{n} u^{-n+1} p A_{uu}
+ \hat{B}_p - [ A, A_u] ) + \frac{1}{n} p u^{-n+1} \hat{B}_u
+ \gamma p u^s A_u + [ A, \hat{B} ] =0.
\eqno(3.11)
$$
Since $A$ and $\hat{B}$ do not depend on $q$, it
follows from (3.11) that
$$
\hat{B}_p = \frac{1}{n} u^{-n+1} p A_{uu} +
[ A, A_u ].
$$
As $A$ does not depend on $p$, this can be integrated
to give $\hat{B}$,
$$
\hat{B} (u,p,y) = \frac{1}{2n} u^{-n+1} p^2 A_{uu}
+ [ A, A_u ] p + B'' (u,y).
\eqno(3.12)
$$
Substituting (3.12) into (3.11) as well as $\hat{B}_u$, there results
$$
\frac{1}{2n} u^{-2n+1} ( - (n-1) A_{uu} + u A_{uuu} ) p^3
+ u^{-n+1} [ A, A_{uu} ] p^2 + u^{-n+1} B_u '' p + n \gamma p u^s A_u
$$
$$
+ n [ A, \frac{1}{2n} u^{-n+1} A_{uu} p^2 + [ A, A_u] p + B'' ] =0.
\eqno(3.13)
$$
Since $A$ and $B''$ do not depend on $p$,
the coefficient of $p^3$ must vanish giving the equation
$$
u A_{uuu} - (n-1) A_{uu} =0.
$$
This can be solved for $A$ to give
$$
A ( u,y) = X_1 (y) + X_2 (y) u + X_3 (y) u^{n+1},
\eqno(3.14)
$$
where the $X_i (y)$ are vertical vector fields. Consequently,
(3.13) simplifies to
$$
u^{-n+1} ( [ A, A_{uu} ] + \frac{1}{2} [ A, A_{uu} ]) p^2
+ (n \gamma u^s A_u + u^{-n+1} B_u '' + n [ A, [A, A_u]]) p
+ n [ A, B'' ] =0.
\eqno(3.15)
$$
The coefficient of $p^2$ implies that $[ A, A_{uu} ]=0$,
which using (3.14) immediately establishes two basic
commutators of the vector fields $X_1$, $X_2$, and $X_3$,
$$
[ X_1 , X_3 ] =0,
\qquad
[ X_2, X_3 ] =0.
\eqno(3.16)
$$
The coefficient of $p$ implies the condition,
$$
n \gamma u^s A_u + u^{-n+1} B_u '' + n [ A, [ A, A_u]] =0.
$$
Solving for $B_u ''$ and putting $s = m-n$,
$$
B_u '' = n \gamma u^{m-1} A_u - n u^{n-1}
[ A, [ A, A_u ]].
$$
Substituting $A$ from (3.14) and its derivative
$A_u = X_2 + (n+1) u^n X_3$ into $B_u ''$ from above,
we have
$$
B_u '' = n \gamma u^{m-1} X_2 + n (n+1) \gamma u^{n+m-1} X_3
- n u^{n-1} [ X_1 + u X_2 + u^{n+1} X_3, [ X_1, X_2]].
\eqno(3.17)
$$
Suppose at this point that $X_1$ and $X_2$ do not commute
with each other, then a new vector field can be
defined as
$$
X_7 = [ X_1, X_2 ].
\eqno(3.18)
$$
Setting $X= X_3$, $Y=X_1$ and $Z= X_2$ in
the Jacobi identity $[X, [Y,Z]]+[Y, [Z,X]] +[Z, [X,Y]]=0$
gives
$$
[X_3, [X_1, X_2]] + [X_1, [X_2, X_3]]+ [X_2, [X_3, X_1]] =0.
\eqno(3.19)
$$
However, using (3.16), the last two terms in (3.19)
are zero, hence (3.19) implies that
$$
[ X_3, X_7 ] =0.
\eqno(3.20)
$$
Consequently, $B_u ''$ reduces to the form
$$
B_u '' = n \gamma u^{m-1} X_2 + n (n+1) \gamma u^{n+m-1} X_3
- n u^{n-1} X_5 - n u^n X_6,
\eqno(3.21)
$$
Two new commutators have been introduced
to write (3.21) defined as
$$
[ X_1, X_7 ] = X_5,
\qquad
[ X_2, X_7 ] = X_6.
\eqno(3.22)
$$
Using (3.22) in the Jacobi identity, the following brackets
result
$$
[ X_2, X_5 ] = [ X_1, X_6 ],
\qquad
[ X_3, X_5 ] =0.
\eqno(3.23)
$$
Finally, integrating $B_u ''$ with respect to $u$ yields
an expression for $B''$,
$$
B'' = \frac{n}{m} \gamma u^m X_2 + \frac{n(n+1)}{n+m}
\gamma u^{m+n} X_3 - u^n X_5 - \frac{n}{n+1} X_6 + X_4.
\eqno(3.24)
$$
Only one term in (3.15) remains to be satisfied,
namely $[ A, B'' ] =0$. Thus substituting $A$ and
$B''$ into this bracket and using linearity to expand
out, we have
$$
[ X_1 + u X_2 + u^{n+1} X_3, \frac{n}{m} \gamma u^m X_2
+ \frac{n (n+1)}{m+n} \gamma u^{m+n} X_3 - u^n X_5
- \frac{n}{n+1} u^{n+1} X_6 + X_4 ]
$$
$$
= \frac{n}{m} \gamma u^m [ X_1, X_2 ] - u^n [ X_1, X_5]
- \frac{n}{n+1} u^{n+1} [ X_2, X_5] + [ X_1, X_4]
- u^{n+1} [ X_2, X_5]
$$
$$
- \frac{n}{n+1} u^{n+2} [ X_2, X_6] + u [ X_2, X_4]
- \frac{n}{n+1} u^{2n+2} [ X_3, X_6] + u^{n+1}
[ X_3, X_4].
$$
Therefore, the vector fields must be interrelated in such a way that
the following holds among the coefficients of each power of $u$,
$$
[ X_1, X_4] + u [ X_2, X_4] + \frac{n}{m} \gamma u^m [ X_1, X_2]
- u^n [ X_1, X_5] + u^{n+1} (- \frac{2n+1}{n+1} [ X_2, X_5]
+ [ X_3, X_4] )
$$
$$
- \frac{n}{n+1} u^{n+2} [ X_2, X_6] - \frac{n}{n+1} u^{2n+2}
[ X_3, X_6] =0.
\eqno(3.25)
$$
This completes the proof.

\begin{theorem}
There exist nontrivial algebras for the
$X_i$ specified by (3.16), (3.18), (3.20), (3.23) and the
coefficients of powers of $u$ in (3.25), which depend on the
relative values of $m$ and $n$.
\end{theorem}

{\em Proof:}
It is required
to equate the independent powers of $u$ equal to zero.
This has to be done on a case by case basis by putting individual
restrictions on $m$ and $n$, and not all cases are given.

$(i)$ Suppose none of the powers of $u$ in
(3.25) are equal, hence $n \neq m \neq 1,0$.
Equating each power of $u$ to zero gives the following
algebra
$$
X_7 = [ X_1, X_2]=0,  \quad
[ X_1, X_5] =0,  \quad
[X_2, X_6]=0,  \quad
\frac{2n+1}{n+1} [X_2, X_5]= [X_3, X_4],
$$
$$
[X_3, X_6]=0,
\qquad
[X_2, X_4]=0,
\qquad
[X_1, X_4]=0.
$$
At this point, $X_1$ and $X_2$ have be required to commute,
since $X_7=0$ must hold. However, from (3.22), it follows
that $X_5 =X_6=0$. Moreover, $[ X_1, X_3]=0$ implies
that $X_1$ and $X_3$ differ by a constant, hence $X_2$
and $X_3$ also differ by a constant. Finally,
$[X_1, X_4 ]=0$ implies that $X_1$ and $X_4$ differ
by a constant. Therefore, we can put
$$
X_1 = \kappa X,
\qquad
X_2 = \sigma X,
\qquad
X_3 =X,
\qquad
X_4 = \alpha X.
\eqno(3.26)
$$
Substituting these results into $A$ and $B$, they
take the form
$$
\begin{array}{c}
A= ( \kappa + \sigma u + u^{n+1}) X,  \\
  \\
B =- ( \sigma + (n+1) u^n ) q X + \frac{1}{2} (n+1) p^2 X
+ \displaystyle \frac{n}{m} \gamma \sigma u^m X
+ \displaystyle \frac{n(n+1)}{m+n}
\gamma u^{m+n} X + \alpha X.  \\
\end{array}
\eqno(3.27)
$$

$(ii)$ Suppose $n=1$ and $m \neq 1,2,3,4$.
Then the same algebra as (3.26) results and $A$ and $B$
are given by (3.27) with $n$ set equal to one.

$(iii)$ Suppose now that $n=m \neq 0,1$, then
prolongation equation (3.25) reduces to
$$
[X_1, X_4] + u [ X_2, X_4] + u^n ( \gamma X_7 - [ X_1, X_5] )
+ u^{n+1} ( - [ X_2, X_5] - \frac{n}{n+1} [X_1, X_6]
$$
$$
+ [ X_3, X_4] ) - \frac{n}{n+1} u^{n+2} [ X_2, X_6]
- \frac{n}{n+1} u^{2n+2} [ X_3, X_6] =0.
$$
This equation is satisfied provided that the
following brackets hold,
$$
[ X_3, X_6]=0,
\quad
[ X_2, X_6]=0,
\quad
\frac{2n+1}{n+1} [ X_2, X_5] = [X_3, X_4],
$$
$$
\gamma X_7= [ X_1, X_5],
\quad
[ X_2, X_4]=0,
\quad
[ X_1, X_4]=0,
\eqno(3.28)
$$
in addition to the brackets given in (3.20), (3.22),
and (3.23). This algebra has a simpler three-element
realization which satisfies all the commutation relations
provided that
$$
X_3=0,
\quad
X_4=0,
\quad
X_5 = \gamma X_2,
\quad
X_6=X_2.
\eqno(3.29)
$$
The nonzero commutation relations are given by
$$
[ X_1, X_2] = X_7, \quad
[ X_2, X_7] = X_2, \quad
[ X_1, X_7] =- \gamma X_2.
\eqno(3.30)
$$
The algebra closes and a finite three-element algebra results.

$(iv)$ Suppose that $m=n+1 \neq 0,1$, then prolongation equation
(3.25) implies the algebra
$$
[X_1, X_4] =0,
\quad
[ X_2, X_4]=0,
\quad
\gamma \frac{n}{n+1} [ X_1, X_2]
- \frac{2n+1}{n+1} [ X_2, X_5] +[X_3, X_4]=0,
$$
$$
[X_1, X_5]=0,
\quad
[X_2, X_6]=0,
\quad
[X_3, X_6]=0.
$$
Recalling that (3.23) must be satisfied, a three element
algebra results if we take
$$
X_2 = X_3,
\quad
X_4=0,
\quad
X_5 =- \frac{n \gamma}{2n+1} X_1,
\quad
X_6= \frac{n \gamma}{2n+1} X_2.
\eqno(3.31)
$$
There is a closed algebra in this case with three nontrivial brackets,
$$
[ X_1, X_2]= X_7,
\qquad
[ X_1, X_7] = - \frac{n \gamma}{2n+1} X_1,
\qquad
[ X_2, X_7] = \frac{n \gamma}{2n+1} X_2.
\eqno(3.32)
$$

$(v)$ The linear case $m=n=1$
generates the following bracket relations
$$
[ X_1, X_4] =[ X_2, X_6]= [ X_3, X_6]=0,
\quad
\gamma X_7 +[ X_2, X_4] - [ X_1, X_5]=0,
$$
$$
[ X_3, X_4] - [ X_2, X_5] - \frac{1}{2}
[ X_1, X_6]=0.
\eqno(3.33)
$$

$(vi)$ The case $m=2$, $n=1$ corresponds to the
classical KdV equation and the brackets must satisfy
$$
[X_3, X_6]=0,
\quad
[ X_2, X_6]=0,
\quad
\frac{1}{2} \gamma X_7 = \frac{3}{2} [ X_2, X_5] - [ X_3, X_4],
$$
$$
[ X_2, X_4] -[ X_1, X_5]=0,
\quad
[ X_1, X_4]=0.
\eqno(3.34)
$$
Since (3.23) must be satisfied, this system is
satisfied if we put
$$
X_3=X_4=0,
\qquad
X_5= - \frac{\gamma}{3} X_1,
\qquad
X_6= \frac{\gamma}{3} X_2.
\eqno(3.35)
$$
There are three nontrivial commutators which take the form
$$
[ X_1, X_2]= X_7,
\qquad
[ X_1, X_7] =- \frac{\gamma}{3} X_1,
\qquad
[ X_2, X_7] =\frac{\gamma}{3} X_2.
\eqno(3.36)
$$
This completes the proof.

\subsection{Conservation Laws.}

One way in which conservation laws can be associated
with this equation is that they correspond to the
existence of exact two-forms contained in the ring
of the forms $\alpha_i$ given by (3.1). Let us
suppose we can find a set of functions $g_i (x,t,u,p,q)$
such that the two-form
$$
\vartheta = g_1 \alpha_1 + g_2 \alpha_2 + g_3 \alpha_3
\eqno(3.37)
$$
satisfies the condition for exactness, $d \vartheta =0$.
This is the integrability condition for the existence of
a one-form, $\omega$, such that
$$
\vartheta = d \omega.
\eqno(3.38)
$$
Conversely, (3.38) implies that $d \vartheta =0$
by the usual identity for double exterior derivatives.
Differentiation of (3.37) and substituting (3.2) yields
$$
d \vartheta = ( d g_1 + g_3 \gamma \frac{s}{n} p u^{s-n} \,
dx )  \wedge \alpha_1 + (  d g_2
+ (g_1  + g_3 \gamma p u^s) \, dx) \wedge \alpha_2
+ ( d g_3 - g_2 \, dx) \wedge \alpha_3.
$$
Therefore $d \vartheta \in I$,
and this clearly vanishes $mod \quad \tilde{p}^* (I)$.

As an example of a form $\vartheta$ with the structure (3.37)
corresponding to equation (3.6), consider the
one-form $\vartheta$ which is given in terms of the
$\alpha_i$ in (3.1) with $g_1= - \gamma u^s$, $g_2=0$
and $g_3=1$ as
$$
\vartheta =- \gamma u^s \, \alpha_1 + \alpha_3,
\eqno(3.39)
$$
where $s=m-n$. Exterior differentiation
$d \vartheta$ gives
$$
d \vartheta = \gamma s u^{s-1} p \, du \wedge dx \wedge dt
+ \gamma u^s \, dp \wedge dx \wedge dt
- \gamma sp u^{s-1} \, du \wedge dx \wedge dt
- \gamma u^s \, dp \wedge dx \wedge dt =0.
$$
Thus the exterior derivative of (3.39) does vanish.
Substituting $\alpha_1$ and $\alpha_2$ into (3.39),
an explicit form for $\vartheta$ is obtained
$$
\vartheta =- \gamma \frac{n}{m} \, d (u^m) \wedge dt + du \wedge dx
- dq \wedge dt.
$$
If the one-form $\omega$ is defined to be
$$
\omega =- (\gamma \frac{n}{m} u^m \,  + q) \, dt + u \, dx,
\eqno(3.40)
$$
Then it is easy to verify by differentiation that
$\vartheta= d \omega$. The associated conservation law
results from an application of Stokes theorem, which
is written as
$$
\oint_{M_1} \omega = \int_{M_2} \, d \omega.
\eqno(3.41)
$$
This has been written for any simply-connected, two-dimensional
manifold $M_2$ with closed one-dimensional boundary $M_1$.
The equation implies that $\omega$ and $d \omega$ are
to be evaluated on their respective manifolds.

Returning to $\omega$ once more, we can of course add to $\omega$
any exact one-form $d v$, where $v$ is an arbitrary scalar function.
Thus, $\omega$ can also be taken to be
$$
\omega = dv - (\gamma \frac{n}{m} u^m \,  + q) \, dt + u \, dx,
\eqno(3.42)
$$
such that $\vartheta = d \omega$. Now $v$ may be regarded
simply as a coordinate in an extended six-dimensional
space of variables $\{ x,t,u,p,q \}$, and the one-form
$\omega$ may be included with the original set of forms.
Since $d \omega$ is known to be in the ring of the
original set, the new set of forms remains a closed ideal.

\section{Prolongation of a Differential System Related to the
Camassa-Holm Equation.}

\subsection{ Exterior System and Associated Partial Differential Equation.}

A differential system will be introduced which is
related to several equations which are of interest
in mathematical physics at the moment.
In particular, the Camassa-Holm and Degasperis-Procesi
equations are to be included in this group.
Define the following system of two forms
$$
\alpha_1 = du \wedge dt -p \, dx \wedge dt,
$$
$$
\alpha_2 = dp \wedge dt - q \, dx \wedge dt,
\eqno(4.1)
$$
$$
\alpha_3 =- du \wedge dx + dq \wedge dx + u \, du \wedge dt
- u \, dq \wedge dt + \beta (u-q) \, du \wedge dt,
$$
where $\beta$ in (4.1) is a real, non-zero constant.
Exterior differentiation of the system of $\alpha_j$ in (4.1)
yields
$$
d \alpha_1 =- dp \wedge dx \wedge dt = dx \wedge \alpha_2,
$$
$$
d \alpha_2 = \frac{1}{u} \, dx \wedge
( - \alpha_3 + u (( 1 + \beta) u -q) \alpha_1),
\eqno(4.2)
$$
$$
d \alpha_3 = (1- \beta) \, dq \wedge du \wedge dt =
(1- \beta) [ dq \wedge \alpha_1 + p \, dt \wedge \alpha_3 -
p \, dx \wedge \ \alpha_1 ].
$$
Clearly all of the $d \alpha_i$ vanish modulo the set
of $\alpha_j$ in (4.1). Therefore, the ideal
$I = \{ \omega | \omega = \sum_{i=1}^3 \, \sigma_i \wedge \alpha_i
: \sigma_i \in \Lambda_p (M), p=0,1,2, \cdots \}$ is closed, $d I \subset I$,
and the system is integrable.

On the transversal manifold, it is determined that
$$
0 = \alpha_1 |_S = S^* \alpha_1 = (u_x -p) \, dx \wedge dt,
$$
$$
0 = \alpha_2 |_S = S^* \alpha_2=  (p_x -q) \, dx \wedge dt,
\eqno(4.3)
$$
$$
0 = \alpha_3 |_S = S^* \alpha_3 = ( u_t - q_t + u ( u_x - q_x)
+ \beta (u-q) u_x ) \, dx \wedge dt.
$$
Thus sectioning the differential system (4.1)
generates the following set of equations
$$
p= u_x,
\qquad
q= p_x,
\qquad
(u-q )_t + u(u-q)_x + \beta (u-q) u_x =0.
\eqno(4.4)
$$
The first two equations here imply that
$q=u_{xx}$. Using this in the third
equation of (4.4), the following partial differential
equation results
$$
( u - u_{xx} )_t + u ( u- u_{xx})_x +
\beta ( u - u_{xx}) u_x =0.
$$
The reason for the interest in this equation is that it
produces important, relevant equations which are of
current interest when $\beta$ is picked appropriately.
To write this in a concise form,
it is usual to introduce the variable $ \rho = u- u_{xx}$,
which gives
$$
\rho_t + \rho_x u + \beta \rho u_x =0.
\eqno(4.5)
$$
Setting $\beta=2$ in (4.5), the Camassa-Holm equation
results,
$$
\rho_t + \rho_x u + 2 \rho u_x =0.
\eqno(4.6)
$$
For the case in which $\beta =3$, equation (4.5) takes the
form of the Degasperis-Procesi equation,
$$
\rho_t + \rho_x u + 3 \rho u_x =0.
\eqno(4.7)
$$
The results which are obtained below will have
implications for these two equations.

\subsection{ Prolongation Equations.}

Upon substituting differential system (4.1)
into (3.8), the prolongation relation takes the form,
$$
\frac{\partial A}{\partial t} \, dt \wedge dx
+ \frac{\partial A}{\partial u} \, du \wedge dx
+ \frac{\partial A}{\partial p} \, dp \wedge dx
+ \frac{\partial A}{\partial q} \, dq \wedge dx
$$
$$
+ \frac{\partial B}{\partial x} \, dx \wedge dt
+ \frac{\partial B}{\partial u} \, du \wedge dt
+ \frac{\partial B}{\partial p} \, dp \wedge dt
+ \frac{\partial B}{\partial q} \, dq \wedge dt
+ [A, B] \, dx \wedge dt
\eqno(4.8)
$$
$$
= \lambda_1 ( du \wedge dt -p \, dx \wedge dt)
+ \lambda_2 (dp \wedge dt - q \, dx \wedge dt)
$$
$$
+ \lambda_3 ( - du \wedge dx + dq \wedge dx + u \, du \wedge dt
- u \, dq \wedge dt + \beta (u-q) \, du \wedge dt).
$$
By comparing the coefficients of
the forms on both sides of (4.8), the following
system of prolongation equations is seen
to hold,
$$
\frac{\partial A}{\partial u} =- \lambda_3 ,
\qquad
\frac{\partial A}{\partial p}=0,
\qquad
\frac{\partial A}{\partial q} = \lambda_3,
$$
$$
\frac{\partial B}{\partial u} = \lambda_1 + u \lambda_3 +
\beta (u-q) \lambda_3,
\qquad
\frac{\partial B}{\partial p} = \lambda_2,
\qquad
\frac{\partial B}{\partial q} =-u \lambda_3,
\eqno(4.9)
$$
$$
- \frac{\partial A}{\partial t} + \frac{\partial B}{\partial x}
+ [ A, B] =-p \lambda_1 - q \lambda_2.
$$
It will be seen that prolongation system (4.9) is much more restrictive
than that obtained for the previous case.

\begin{theorem}
System (4.9) can be reduced to a single
equation which relates the functions $A$ and $B$.
\end{theorem}

{\em Proof:}
The results obtained in (4.9) imply the following
system of equations,
$$
\frac{\partial A}{\partial u} =- \frac{\partial A}{\partial q},
\qquad
\frac{\partial B}{\partial q} =- u \frac{\partial A}{\partial q}
= u \frac{\partial A}{\partial u},
\qquad
\lambda_1 = \frac{\partial B}{\partial u} - u \lambda_3 -
\beta (u-q) \lambda_3.
\eqno(4.10)
$$
Making use of these in the last equation of (4.9),
we obtain
$$
- \frac{\partial A}{\partial t}
+ \frac{\partial B}{\partial x} +[A,B] =
-p ( \frac{\partial B}{\partial u} - u \lambda_3
- \beta (u-q) \lambda_3 ) - q \lambda_2.
\eqno(4.11)
$$
As with the previous equation studied in Section 3,
$A$ and $B$ are taken to be independent of both
$x$ and $t$ for the same reason. Moreover, the
equation $A_p=0$ implies that $A$ is independent of
the variable $p$, therefore,
$$
A= A( u,q,y),
\qquad
B = B (u, p, q,y).
\eqno(4.12)
$$
Integrating the equation $B_q =-u A_q$ with respect to
$q$, it is found that $B$ is related to $A$ by
$$
B = - u A(u,q,y) + B' (u,p,y).
\eqno(4.13)
$$
In (4.13), $B'$ satisfies $B_p '= \lambda_2$,
but is arbitrary otherwise.
Differentiating (4.13) with respect to $u$,
we obtain
$$
B_u =-A(u,q,y)  -u A_u (u,q,y) + B_u ' (u,p,y).
\eqno(4.14)
$$
The first partial differential equation in (4.10)
$$
\frac{\partial A}{\partial u} +
\frac{\partial A}{\partial q} =0,
\eqno(4.15)
$$
implies that $A$ must be of the form
$$
A (u,q,y) = A( u-q,y).
\eqno(4.16)
$$
Since $\lambda_3 =-A_u$,
we can write $\lambda_1$ from (4.10) as
$$
\lambda_1 = B_u + u A_u + \beta (u-q) A_u =- A - u A_u
+ B_u' + u A_u + \beta (u-q) A_u.
\eqno(4.17)
$$
The $u A_u$ terms cancel in (4.17) and so
substituting (4.17) and $B$ from
(4.13) into (4.11), we obtain that
$$
[ A, B' ] =-p (-A + B_u' + \beta (u-q) A_u ) -q B_p'.
\eqno(4.18)
$$
This is the required result, and finishes the proof.

It remains to find solutions to equation (4.18), which
contains two unknown functions. One way to do this is to
impose a condition on one of the unknown functions.

\begin{theorem}
There exists a nontrivial solution to system (4.15) and (4.18)
under the condition $B' =0$.
\end{theorem}

{\em Proof:}
It follows from (4.15) that $A$
must satisfy (4.16).
Putting $B'=0$ into (4.18), it simplifies to
one equation in one unknown,
$$
\beta (u-q) A_u -A =0.
\eqno(4.19)
$$
On account of (4.16), we can introduce the variable
$\xi =u-q$, and (4.19) then becomes an ordinary
differential equation for $A$
$$
\beta \xi A_{\xi} - A =0.
$$
This equation has the nontrivial solution
$$
A (\xi, y) = (u-q)^{\frac{1}{\beta}} X(y),
$$
with the integration constant written as $X(y)$.
To summarize explicitly, by using (4.13), the solutions for $A$ and $B$ are
given as
$$
A( u, q,y) = (u-q)^{\frac{1}{\beta}} X(y),
\qquad
B (u,p,q,y) =- u (u-q )^{\frac{1}{\beta}} X(y).
\eqno(4.20)
$$
$\clubsuit$

Many prolongations can be specified by introducing
different conditions on $B'$. One more will be derived.

\begin{theorem}
There exists a nontrivial solution of (4.15) and (4.18)
such that $B' = \frac{1}{2} (p^2 - u^2) X_2 (y)$
with $[ A, X_2 ] =0$.
\end{theorem}

{\em Proof:}
Since $B_p' =p X_2$ and $B_u' =-u X_2$, and $[A, B']=0$,
substituting these, equation (4.18) takes the form
$$
\beta ( u-q) A_u - A = (u-q) X_2.
$$
Using (4.16), this reduces to an ordinary differential equation in the variable
$\xi = u-q$, namely,
$$
\beta \xi A_{\xi} - A = \xi X_2.
$$
This equation has the general solution which for
$\beta \neq 1$ is given by
$$
A ( u, q, y) =  (u-q)^{\frac{1}{\beta}} X_1 (y)
+ \frac{u-q}{\beta -1} \, X_2 (y),
$$
and from (4.13), $B$ is given by
$$
B=- u (u-q)^{\frac{1}{\beta}} X_1 (y) - u \frac{u-q}{\beta -1} X_2 (y)
+ \frac{1}{2} (p^2 - u^2 ) X_2 (y).
$$
Here, $X_1$ and $X_2$ generate a commutative algebra
such that $[ X_1, X_2] =0$.
$\clubsuit$

\subsection{ Conservation Laws.}

With the $\alpha_i$ given by (4.1), we can define a form
$\vartheta$ for this case as well. To do the calculations
here, the exterior derivatives (4.2) of the forms
(4.1) can be simplified to read
$$
d \alpha_1 =-dp \wedge dx \wedge dt,
\qquad
d \alpha_2 = - dq \wedge dx \wedge dt,
\qquad
d \alpha_3 = (1 - \beta) dq \wedge du \wedge dt.
$$
Consider the one-form $\vartheta$ which is defined
to be
$$
\vartheta =  - (1- \beta ) q \alpha_1 +
( 1- \beta) p \alpha_2 + \alpha_3.
\eqno(4.21)
$$
Thus, the exterior derivative $d \vartheta$ is in ideal $I$ since
$$
d \vartheta = ( - ( 1 - \beta) \, dq + (1- \beta)p
(( 1 + \beta)u -q) \, dx + (1- \beta) \, dq -p \, dx) \wedge \alpha_1
+ ( 1 - \beta) ( - q dx + dp ) \wedge \alpha_2
$$
$$
+ (1- \beta) ( - \frac{p}{u} \, dx + p \, dt ) \wedge \alpha_3.
$$

Calculating the exterior derivative of $\vartheta$
in terms of the basic set of variables of $M$,
it is found to vanish identically as well,
$$
d \vartheta = (1- \beta ) \, dq \wedge du \wedge dt
- (1- \beta) \, dq \wedge du \wedge dt
+ ( 1- \beta) p \, dq \wedge dx \wedge dt
+ (1 - \beta) q \, dp \wedge dx \wedge dt
$$
$$
- ( 1 - \beta) q \, dp \wedge dx \wedge dt
- ( 1 - \beta) p \, dq \wedge dx \wedge dt =0.
$$
In fact, $\vartheta$ can be obtained directly
from the one-form $\omega$ defined by
$$
\omega= (q-u) dx + \frac{1}{2} ( u^2 - 2 u q +
\beta u^2 + p^2) \, dt.
\eqno(4.22)
$$
Upon differentiating $\omega$ it is found that
$$
d \omega =- du \wedge dx + dq \wedge dx + u \, du \wedge dt
- u \, dq \wedge dt + \beta u \, du \wedge dt
- q \, du \wedge dt + (1- \beta) p \, dp \wedge dt.
\eqno(4.23)
$$
This is precisely the two-form $\vartheta$ given in (4.21).
The associated conservation law results from an application
of Stokes theorem, which is written in this case
$$
\oint_{M_1} \omega = \int_{M_2} \, d \omega.
\eqno(4.24)
$$
This has been written for any simply-connected,
two-dimensional manifold $M_2$ with closed
one-dimensional boundary $M_1$. The equation implies that
$\omega$ and $d \omega$ are to be evaluated
on their respective manifolds.

Returning to $\omega$ again, we can again add to
$\omega$ any exact one-form $d v$, where $v$ is
an arbitrary scalar function. Thus, $\omega$
can also be taken to be
$$
\omega= dv + (q-u) \, dx + \frac{1}{2}
( u^2 - 2 u q + \beta u^2 + p^2) \, dt,
\eqno(4.25)
$$
such that $\vartheta = d \omega$. As before, $v$ may
be regarded as a coordinate in an
extended six-dimensional space of variables
$\{ x, t, u, p,q,v \}$, and the one-form
$\omega$ may be included with the original set of
forms. Since $d \omega$ is known to be in the
ring of the original set, the new set of forms
remains a closed ideal.

\section{Summary and Conclusions.}

It has been seen that exterior differential systems
have been constructed for some very important classes of
partial differential equation. As well as giving some
information about the associated integrability of
these equations, it has been shown that the prolongation
structure of these systems can be studied. This is more
than just of theoretical interest, since B\"acklund transformations
can be constructed based on these results.
The relationship of differential systems to B\"acklund transformations
has been discussed by Estabrook and Wahlquist {\bf [15]}, and the
construction of such transformations has been done for some
three element algebras in {\bf [16]}. Let us show how to use the
results of example $(i)$ in Section 3 to obtain such a result.

Using (2.4), the connection $\tilde{\omega}$ can always be
chosen on $\mathbb R$ with coordinate $y$ and $X = \partial/ \partial y$
$$
\tilde{\omega} = dy - \{ ( \kappa + \sigma u + u^{n+1} ) \, dx
+ (-( \sigma +(n+1) u^n )q + \frac{1}{2} (n+1) p^2 + \frac{n}{m}
\sigma \gamma u^m
$$
$$
+ \frac{n (n+1)}{m+n} \gamma u^{m+n} + \alpha
) \, dt \} \, X(y),
\quad \kappa, \sigma, \alpha \in \mathbb R.
\eqno(5.1)
$$
Solutions of the system (3.6) determine transversal sections
of the fibre bundle such that, upon substituting $p$ and $q$ from (3.4),
we have
$$
\begin{array}{c}
y_x = \kappa + \sigma u + u^{n+1},   \\
  \\
y_t =- ( \sigma + (n+1) u^n ) (u^n)_{xx} + \frac{1}{2} (n+1)
((u^n )_x)^2 + \displaystyle \frac{n}{m} \gamma \sigma u^m + \displaystyle \frac{n(n+1)}{m+n}
\gamma u^{m+n} + \alpha.
\end{array}
\eqno(5.2)
$$
A similar result but for a different algebra was given in {\bf [9]}.
By solving the first of these for $u$, it can be eliminated
in the second equation of (5.2) to yield an equation for $y=y(x,t)$.
For $\sigma=0$, this can be done in closed form, and to make the
presentation more concise we put $\kappa=0$ as well giving
$$
u = (y_x)^{\frac{1}{n+1}}.
\eqno(5.3)
$$
The positive root is taken if the exponent
in (5.3) has an even denominator.
Eliminating $u$ from the second equation in (5.2),
we have an equation for $y$
$$
y_t + (n+1) (y_x)^{\frac{n}{n+1}} ( (y_x)^{\frac{n}{n+1}})_{xx} - \frac{1}{2} (n+1) (((y_x)^{\frac{n}{n+1}})_x)^2
- \frac{n (n +1)}{m+n} \gamma ( y_x)^{\frac{m+n}{n+1}} - \alpha =0.
\eqno(5.4)
$$
It follows that for $\sigma=\kappa=0$, the potential equation in
terms of $y$ which results is given by (5.4). In effect, a B\"acklund
transformation has been determined and is expressed by (5.2).
This set of equations transforms the original equation into the
form of its potential equation (5.4).

\vspace{0.25cm}
\begin{center}
{\bf References.}
\end{center}

\noindent
$[1]$ A. C. Newell, Solitons in Mathematics and Physics, SIAM,
Philadelphia, 1985.  \\
$[2]$ M. J. Ablowitz and P. A. Clarkson, Solitons, Nonlinear Evolution
Equations and Inverse Scattering, Cambridge University Press, 1991.  \\
$[3]$ H. D. Wahlquist and F. B. Estabrook, Prolongation Structures
of Nonlinear Evolution Equations, J. Math. Phys. {\bf 16}, 1-7, (1975).  \\
$[4]$ H. D. Wahlquist and F. B. Estabrook, Prolongation Structures of
Nonlinear Evolution Equations II, J. Math. Phys. {\bf 17}, 1293-1297 (1976).  \\
$[5]$ F. B. Estabrook, Moving Frames and Prolongation Algebras,
J. Math. Phys. {\bf 23}, 2071-2076 (1982).  \\
$[6]$ W. F. Shadwick, The KdV Prolongation Algebra, J. Math.
Phys., {\bf 21}, 454-461 (1980).  \\
$[7]$ E. van Groesen and E. M. de Jager,
Mathematical Structures in Continuous Dynamical Systems,
Studies in Math. Phys., vol. 6, North Holland, 1994.   \\
$[8]$ E. M. de Jager and S. Spannenburg, Prolongation structures and
B\"acklund transformations for the matrix Korteweg-de Vries and
Boomeron equation, J. Phys. A: Math. Gen. {\bf 18}, 2177-2189 (1985). \\
$[9]$ P. Bracken, An Exterior Differential System for a Generalized
Korteweg-de Vries Equation and its Associated Integrability,
Acta Applicandae Mathematicae, {\bf 95}, 223-231 (2007).  \\
$[10]$ T. Tao, Why are solitons stable?, Bulletin of the Amer. Math. Soc.,
{\bf 46}, 1-33, (2009).  \\
$[11]$ P. Bracken, Symmetry Properties of a Generalized Korteweg-de Vries Equation
and Some Explicit Solutions, Int. J. Math. and Math. Sciences, {\bf 2005, 13},
2159-2173 (2005).  \\
$[12]$ E. G. Reyes, Geometric Integrability of the Camassa-Holm
Equation, Lett. Math. Phys., {\bf 59}, 117-131 (2002).  \\
$[13]$ R. Camassa and D. Holm, An Integrable Shallow Water Equation
with Peaked Solitons, Phys. Rev. Letts. {\bf 71}, 1661-1664 (1993).  \\
$[14]$ J. Lenells, Conservation Laws of the Camassa-Holm Equation,
J. Phys. A: Math. Gen. {\bf 38}, 869-880 (2005).  \\
$[15]$ F. B. Estabrook and H. D. Wahlquist, "Prolongation Structures,
Connection Theory and B\"acklund Transformation", in Nonlinear evolution
equations solvable by the spectral transform, Ed. F. Calogero, Pitman, 1978. \\
$[16]$ P. Bracken, The interrelationship of integrable equations,
differential geometry and the geometry of their associated surfaces,
to appear, 2009.  \\
\end{document}

redpajama_set_name: RedPajamaArXiv