text:
Ukraine set to declare martial law as tensions with Russia heat up. (26 Nov 2018) Ukraine has been in an undeclared state of war with Russia ever since the 2014 Russian invasion of Crimea. As tense as the situation has been over the past four years, the heat just got turned up. Over the weekend, Russia opened fire on and subsequently detained three Ukrainian vessels—two armored artillery ships and one tugboat—as the small convoy passed through the Kerch Strait. Since the invasion, Russia has claimed the waters around the strait; a claim which Kiev (rightfully so) has never recognized. With the sailors in custody, Ukrainian President Petro Poroshenko—a US ally—has asked parliament to declare martial law in the country, giving the military extended power for at least sixty days. Odds are, Putin will bow to pressure and release the sailors soon, but the simmering pot between the Black Sea and the Sea of Azov is getting closer to the boiling point. The VanEck Vectors Russia ETF (RSX $20.10) is off around 2% on the day and down 5.26% YTD.
You wouldn't know it by listening to the press, but Putin's approval numbers are falling sharply. (16 Aug 2018) Can we at least agree that the mainstream media is myopic (and that is being very generous) in their treatment of certain politicians at home versus certain dictatorial leaders overseas? Take Vladimir Putin. We have read story after story of how popular the Russian leader is within his home country, but is that really true? Yes, for a people who could once brag about Soviet military might, the national pride goes up every time he makes a move like his illegal annexation of Crimea—and so do his poll numbers, but that pride does not put meat on the family table or rubles (which are cheapened almost weekly) in the pockets of workers. In a recent issue of the Penn Wealth Report, we reported on Russia's daunting economic issues and that country's massive over-reliance on oil revenues, along with a growing level of discomfort among his citizens. According to the Levada Analytical Center, Putin's trust rating in the country has dropped from 60% in January to just 48% in June. And the numbers are a lot worse for his lapdogs, like Foreign Minister Sergei Lavrov. Domestic economic troubles are forcing tough moves at home, like raising the retirement age to 65 for men and 60 for women. That may not seem draconian in the Western world, but to someone living in a communist nanny state, it is appalling. Despite Putin's troubles, it is inconceivable that he would be removed from power. However, the domestic turmoil does mean that he will become even more dangerous on the world stage as he tries to deflect any personal criticism and distract his hungry masses.
Containing the Russian Menace
(15 Jul 2018) Hillary Clinton's red reset button shows just how naive the US has been with respect to Russian aggression; that is changing (See article in The Penn Wealth Report by clicking button to right)
Vol 6 Issue 02
Russian ruble getting hammered as latest round of US sanctions take effect
(11 Apr 2018) Looking for a government bond to buy but unhappy with the current 10-year Treasury yield of 2.768%? You could always pick up the Russian 10-year, currently yielding 7.6%. Russian capital markets have been in free-fall since President Trump imposed the latest round of US sanctions against the country for its support of Syria's chemical-weapon-yielding regime and the recent poisoning in the UK, with this round hitting Putin's closest business allies and the global companies they run. These are real sanctions with real impact, and they are rattling the Russian economy. The ruble appears on track for its worst week since 1999, already down 12% against the dollar through Wednesday morning. Russia's central bank was just forced to cancel a ruble-denominated government bond auction due to "unfavorable conditions." Higher oil prices are the one component working in Russia's favor right now.
UK expels diplomats, freezes assets after spy poisoning
(15 Mar 2018) Did anyone really ever doubt that Putin's fingerprints were all over the crime? After all, he seems to have a lifelong love affair with poison. Following the attack on a former Russian spy and his daughter in a small town in England, Prime Minister Theresa May has announced the expulsion of 23 Russian diplomats and is leveling sanctions on UK-based assets of close associates of Putin. President Donald Trump has echoed the condemnation, pushing US/Russia relations back to their lowest point since the Cold War. There will never be any real negotiating with the Russian dictator, who calls for national elections only as a ruse. He has grand visions of reconstituting the Soviet Union; though he will fail, he will continue to wreak havoc along his path. And no giant red RESET button can change that.
Russia suspended from 2018 Winter Olympics
(05 Dec 2017) In an enormous blow to a country which prides itself on its athletes' performance on the world stage, the International Olympic Committee has suspended the Russian Olympic Committee for the country's support of performance-enhancing drug usage. While Russian athletes will not be banned from attending the upcoming Winter Olympics in South Korea, the athletes will not be competing for Russia per se, and the Russian national anthem will not be played for medalists. Furthermore, they must compete in uniforms bearing the Olympic flag, with no Russian markings. Revenge is a dish best served cold—this is for your boycott of the 1980 Olympics in LA. Hardy har har.
US to restrict Russian military flights over American territories
(26 Sep 2017) In the waning days of the Cold War, when the Soviet Union's collapse was imminent, a number of treaties to promote peace between the two former enemies came about. One was the Treaty on Open Skies, which allowed military observation flights over each other's territories to verify military assets and troop strength in certain zones. In response to Russia's violation of the spirit of that treaty, the US has announced that it will restrict flights over certain areas—such as Alaska and Hawaii—to force Russia back into compliance. US officials specifically cite Russia's restrictions of American flights over Kaliningrad, a Baltic Sea city where it is believed that country is building up a cache of advanced weaponry.
(14 Feb 2017) Trump's national security advisor resigns over Russia call. It wasn't the fact that Lt. General Michael Flynn had a conversation with Russian ambassador Sergey Kislyak before inauguration day; rather, it was the fact that he was not truthful with what was discussed in that conversation. Flynn originally said that he did not discuss the sanctions imposed by the US on Russia (or the potential lifting of those sanctions) for that country's invasion of eastern Ukraine, and Vice President Mike Pence took to defending him based on his comments. Flynn, confronted with transcripts of the call, admitted that this was not the case. Additionally, it appears that he had more than one conversation with the ambassador. He apologized to the vice president and, either of his own volition or under pressure, resigned his new post.
Putin's Muddled Mind Keeps Him from Seeing Russia's Actual Relationship with China
(02 Sep 15) China's big World War II victory parade on Thursday will boast a "who's who" of global thugs that would make a Bond villain envious. Venezuela's Nicolas Maduro will attend for two reasons: to escape a hostile opposition in his own starving country, and to solidify ties with the titular head of the worldwide communist movement: Xi Jinping. Pakistani President Hussain and Vietnamese President Truong Tan Sang will be there. Leaders of Laos, Thailand, and Cambodia will be there. Surprisingly, the pot-bellied North Korean dictator Kim Jong Un will skip the event (he seems to be miffed at China right now), but South Korean President Park will attend (too bad). The star attraction on the guest list is undoubtedly the emperor-in-training of Eastern Europe, Vlad Putin.
As the Chinese celebrate the end of World War II with a show of military might, Putin may be looking a bit green with envy in the dignitary stands. He has made it clear that he wants a military and economic alliance with the Communist Chinese that rivals the US/NATO relationship and the NAFTA accord, respectively, but he fails to comprehend his small role in ChinaLand.
While China still wants Russian gas, the latter is no longer in the top 15 list of Chinese trading partners. That is due, in large measure, to the drop in oil prices and China's waning economy. As recently as last October, Putin was bragging about the $100 billion worth of trade Russia would transact with China in 2015. Then, the Chinese fabrications of its own economic condition began to bubble to the surface. Russian trade with China plummeted 30% in the first half of the year, to $30 billion. Suddenly, Putin is no longer mentioning the $100 billion figure for the year.
In reality, China doesn't need Russia to any great degree. Furthermore, China's image as an economic superpower has been greatly tarnished. Xi Jinping will be hurt in this foray, but Putin will end up bloodied. (Reprinted from the Journal of Wealth & Success, Vol. 3, Issue 35. Not a member?
Putin is Missing
(UPDATE Mo, 16 Mar 15) Vlad reappeared. Our best guess was that he was actually visiting his pregnant girlfriend. His mercurial nature appears to be worsening; more than 45,000 Russian troops began massive military exercises across the country on Monday.
(Su, 15 Mar 15) It's a mystery, to be sure. First, on 27 Feb, Vladimir Putin's biggest critic and largest internal threat, 55-year-old Boris Nemtsov, was gunned down not far from the Kremlin. Vlad feigned outrage, stating that the act was a "provocation." Much of the feeble-minded press bought into his lies. Vladimir Putin did not pull the trigger, but he set up the hit, one way or another. One week later, the real intrigue began. On 05 March, six days after Nemtsov's killing, Putin met with the Italian prime minister, and that is the last anyone has seen of him. That would have been suspicious enough, but then there was the bizarre state-run media story.
The Kremlin-controlled news ran a story about a successful meeting which occurred between Putin and the president of Kyrgyz. It may have been a bit more believable had the story not been reported in past tense, about a meeting that was to take place in a few days. The news station quickly announced that the story was a mistake, but how could a report of a meeting that had not yet taken place be written and aired?
The rumor mills went crazy. People began speculating about the missing Putin on social media, and the guesses range from a bad case of the flu, to an elongated visit to his pregnant mistress, Russian gymnast Alina Kabaeva, to a revenge killing of the 62-year-old dictatorial leader, to a palace coup.
Many have come to the conclusion that this is simply Putin playing the mastermind once again, fooling everyone and then popping back on the scene as if nothing happened. After all, we aretalking about a monster who had his wife locked away in a mental institution so he could pursue his adulterous affair with the gymnast, so nothing is off the table. But, with the Russian people severely hurting because of the economic chaos brought about by low oil and a corrupt regime, what would be the endgame of that scenario?
Vlad may well make that visit this Monday, and the press can re-run their pre-written script. But the idea of something more nefarious than taking ten days off to visit a pregnant lover is far more tantalizing. Perhaps not everyone in Russia cowers at the feet of the megalomaniacal dictator. Perhaps, dare we dream, the ruthless tyrant has assumed room temperature.
(Reprinted from next Sunday's Journal of Wealth & Success, Vol. 3, Issue 12.)
Russia must take half of its reserve fund to pay the bills.
(Fr, 27 Feb 15, 1500) One must wonder how much longer the Russian people will be able to sustain their love affair for papa Vladimir. In the latest emergency measure to keep its writhing economy afloat, Deputy Finance Minister Tatiana Nesterenko announced that the Kremlin would tap over $53 billion (more than half) from its reserve fund—rubles put away for a rainy day.
Vlad is also cutting salaries for Kremlin workers by 10%, and will consider letting the Communist Chinese take a majority stake in some of the country's strategic oil and natural gas fields. That shows just how dire the situation is for Vlad, and how much he desperately wants to be best buds with the Chi-coms.
The treasury secretary said that there was about 6 trillion rubles ($100 billion or so) in the reserve fund at the start of the month, but that the account could fall to 1 trillion rubles by the end of the year. What is so amusing about this (not the fact that millions of Russians will face severe economic pain) is that Russia planned to have Europe over a barrel this winter for supporting sanctions against the country for its invasion of Crimea. What comes around typically goes around, but what an exceptionally rapid turn of events. It would be even more severe had the president allowed the Keystone XL Pipeline to be built.
Why Bed Buddies France and Germany Kept David Cameron and the UK out of the Terrible Russian Deal
(17 Feb 15) "Terrible" is how one UK government official described the deal. Germany's Angela Merkel and France's (uh oh, that can never be a good sign) Francois Hollande went to Minsk, Belarus to negotiate a "peace treaty" with the untrustworthy Putin of Russia. The purpose of the meeting was to cool the dictator's hostilities in Ukraine. As could be expected, Putin won, the West lost, and the Merkel/Hollande cabal claimed "peace in our times."
Here is the sad truth: The only reason Vlad the Impaler came to the table was because the president of the United States threatened to send defensive weapons to Ukraine. He figured he could stall—or stop—that action by feigning interest in peace. It categorically worked.
Listening to reports of a marathon negotiating session, reminiscent of Jimmy Carter "poring over maps of the Middle East spread out across the floor of the Oval Office," we knew we were in trouble. European leaders get suckered in by the East when they are fully awake...let alone half asleep.
As for Ukraine's part, Petro Poroshenko, the billionaire chocolate baron and current president, knows full well that Putin will not abide by the accord. In fact, the blood-red ink hadn't dried on the agreement before Russian loyalists began killing more innocent Ukrainian civilians. But what is he to do? His economy is in shambles, and he is relying on the United States and Europe to provide the economic and military aid needed in a fight with an uber-bully.
Returning from their victory tour to Minsk, Merkel said "what we have on the table gives us great hope." I used to have such high hopes for her. What she has on the table is carte blanche for the Russian-backed terrorists along the Eastern border of Ukraine to continue their killing spree.
Mother Russia's people might be starving, thanks to Emperor Putin's behavior and the precipitous drop in the price of oil, but as long as they don't blame him, all is cool. After they have Ukraine, where is Putin's next stop as he reconstitutes the Soviet Union? Rapidly descending socialist Greece.
(Reprinted from the Journal of Wealth & Success, Vol. 3, Issue 8.
Russian Buildup on Ukraine Border Makes Press Fret
Russian troops on Ukrainian soil. Agence France-Presse/Getty Images
(28 Mar 14) I can picture the hand-wringing going in a cubicle at the New York Times as a couple of international-beat "journalists" discuss matters in Russia. "Well, Mr. Putin said it's just exercises, Jerome. What do you think?" "I don't know, Julian, there's a puppy dog in one picture I saw, so I tend to believe him! Oh, how I hope it is not more than that. Well, let's grab another cup of chamomile and get to writing!"
Yes, these are exercises. Yes, Putin is playing the West. Yes, Russia will end up invading Ukraine. The immediate action for the United States to take is to announce construction of the Keystone Pipeline, supply Ukraine with weapons to make up for the ones we had them get rid of in the 1994 deal (leaving them defenseless), begin exporting natural gas to Europe so that they are no longer held hostage (and giving us billions of dollars in new revenue), and push to kick Russia off the UN Security Council.
Which of these actions will the United States actually move on? Odds are, not a one. Instead, we will get more Jimmy Carteresque hollow words, allowing our enemies to understand they have the upper hand and telling our allies that we are a has-been. The likes of Poland and Moldova have every right to be worried. America can no longer be trusted. At least Europe remains a constant--they have been ruled by a Neville Chamberlain attitude for nearly 100 years.
Putin called Obama today to discuss a "diplomatic solution" to the crisis. That means, even in his imperious state, Putin is worried about the effects of an economic war with the West. He has nothing to worry about--the West will talk tough, fold, and allow his unbridled aggression to continue. We've seen this movie before.
The International Olympic Committee Wins "Star Wars Bar Scene" of the Year Award
(07 Feb 14) I will admit being old enough to remember Jimmy Carter's boneheaded move to boycott the 1980 Summer Olympic Games in Moscow in protest of the Soviet invasion of Afghanistan. I also remember the comical excuses the Russians gave for boycotting the 1984 games in LA--crime was too rampant in the United States, and the Soviet athletes could not be subjected to the level of risk they would face in America.
Fast forward 30 years. Is there anything which carries more risk than to be a stray dog on the streets of Sochi? Russian troops are gunning them down ostensibly to reduce the threat to the athletes, but really to paint a prettier face on the shantytown that is Sochi. The "city" is a make-believe movie lot, ready to go up in flames like a tinder box. Toilets that don't work...rooms that are half-built...lights that don't turn on. This entire charade proves that the International Olympic Committee should be in line for a Nobel Peace Prize, which is now given out on the come and awarded to tyrants and political hacks. Just ask Al Gore.
Economic Slowdown Trimming Putin's Horns
(03 Sep 13) Vladimir Putin has always been and will always be an enemy of and a direct threat to the United States of America. A megalomaniac who grew up detesting the United States, his role within the KGB senior hierarchy allowed him to hone his skills as a master of subterfuge and manipulation. His hubris and his arrogance, however, have led to some dangerous miscalculations with respect to the strength of the Russian economy.
Stunningly, a whopping 70% of the Russian economy is energy-centric. Rising energy prices, fueled by activity in emerging markets such as China and India, have allowed Putin to maintain sky-high popularity ratings, as he has pumped trillions of rubles into infrastructure and military spending, and has raised the income levels of government workers substantially. Putin's master plan, however, may be coming to an abrupt halt, and he has his old enemy, the United States, to thank.
Russia is currently the world's largest energy exporter, holding the number one spot in oil production and the number two spot--behind the U.S.--in gas production. Unlike the U.S., however, Putin has pursued a scorched earth policy with respect to mining the land for every ounce of energy. At current levels, the country's great Ural Mountains/Siberian Plateau reserves are projected to run dry in twenty years. This is a far shorter span of time than is the case with known reserves in the U.S., Saudi Arabia, and the United Arab Emirates. Unfortunately for Russia, little energy, no pun intended, has been put into developing other sectors to drive the economy forward.
Gazprom, the world's largest gas producer (a number of different U.S. companies produce gas in this country), was valued at $360 billion a few years ago. Today, this puppet of the regime has a market value of just $94 billion. The Russian economy (as measured by GDP) was growing at an impressive 7% per year, largely on the back of this company, for most of the past decade. When the recession hit and the barrel price of crude dropped from the $140s to the $30s, the ruse began to crumble.
Gazprom has held a mob-like stranglehold on countries throughout the region. When Russia was battling Ukraine, the energy interest turned off its gas pipelines to the area, causing great hardship in the dead of winter. Run by Putin lackeys, Gazprom had been able to extort confiscatory rates from nations throughout Europe, as Russia provided the lion's share of energy to the region. As America began ramping up its energy production, due to new finds and advanced drilling developments, all of this began to change. With liquid natural gas (LNG) becoming a dominant energy source, the United States has been taking an ever-increasing percentage of market share from Russia.
Additionally, as American companies make the shift from coal to LNG, coal producers such as Peabody Energy in St. Louis suddenly find themselves in the enviable position of being major exporters. This new source for energy has emboldened the EU, which is drafting a formal complaint against Gazprom for market manipulation.
Word is that President Obama does not plan a one-on-one meeting with Putin while attending the G-20 summit this week, despite the fact that Russia is hosting the event. The move makes sense, but we fantasize about Obama ordering the airstrike on Syria at the precise time that the two men were meeting in St. Petersburg. For Putin, who likes to show off images of himself proving his manhood, that would be a slap in the face not soon forgotten.


Newspapers across Europe are scrutinising what just happened in the latest Brexit story twist and weighing up likely consequences for the European Union.
"The fog over London's government district just will not clear," says Germany's Tagesspiegel, calling the House of Commons voting last night "nothing less than the biggest political crisis on the island since World War II" and declaring that "those responsible, especially Theresa May, are unable to implement the choice the British made in June 2016".
"Two months before Brexit, it seems as if she wants to now challenge the EU to a game of chicken: whoever blinks first loses," the daily says.
Another German publication, Der Spiegel, sees furious lobbying from Spain to put the question of Gibraltar back on the table should the Withdrawal Agreement be reopened.
"If the British now demand further concessions on the Irish question, the Spaniards would probably bring the British overseas territory back into play – as before. And other EU countries could make new demands, such as the fishing rights in British waters."
Spain's El Pais says reopening the deal at this stage is a nightmare scenario for Brussels. "The European Union has run into one of the most feared Brexit scenarios at only 60 days before the deadline set for UK's departure from the EU."
It puts the EU in a difficult position – exposing itself to a no-deal Brexit or "accepting a very delicate renegotiation that could open a Pandora's Box", the paper says.
In an editorial, The Irish Times warns that the ongoing stalemate between London and Brussels raises the risk of what both want to avoid: "a damaging no-deal exit".
"It appeared up to recently that Theresa May accepted the need for the backstop guarantees. Her volte face, ironically, only underlines the need for a legally operable backstop guarantee."
In France – where President Macron has already delivered a resounding "Non" to any backstop unpicking – Le Figaro says last night's House of Commons vote was "a way of throwing the ball back in the Europeans' court to push them up against the wall, so close to the fatal date of 29 March they give in".
Over in the Netherlands, De Volkskrant is scathing about the warring factions inside the House of Commons.
"The MPs are unable to agree about much and were making more noise than the proponents and opponents of Brexit who were protesting outside. The Scots do not want a Brexit at all, Labour wants to keep one leg in the EU, the Brexiteers want to go into battle against the EU like Don Quixote, and the Northern Irish unionists want to drag as much money out of London as possible."
Italy's Corriere della Sera paints a truly bleak picture. "Brexit is not to be delayed and the spectre of London's no-deal, catastrophic exit from the EU is getting closer. If no way out is found at this point, Great Britain will fall into the abyss on 29 March."
Romania's Evenimentul Zilei refers angrily to what it sees as the knock-on effect for EU citizens living in the UK.
"We are facing the second moment of British political obduracy combined with stubbornness and the constraints of a minority government that is exported to the entire European Union, British citizens in the EU and Europeans in the UK," it seethes.


text:
Alte Burg was een kasteel te Holset in de Nederlandse provincie Limburg. In 1334 was er het riddergeslacht Holseth dat in de 14e eeuw te Holset een kasteel moet hebben gehad. Door het geslacht werden bepaalde heerlijke rechten uitgeoefend. Het kasteel was reeds in de 18e eeuw vervallen tot een ruïne en is daarna verdwenen.

Kasteel in Vaals
Voormalig bouwwerk in Limburg (Nederland)
Verdwenen kasteel in Nederland

redpajama_set_name: RedPajamaWikipedia


text:
Q: encrypt password for http_proxy, access nuget source Over here Nuget connection attempt failed "Unable to load the service index for source"  I saw many solutions proposed for the

Unable to load the service index for source ...

Error that I find myself getting on an attempted dotnet build right now.
I see suggestions to put XML like either of these
<packageSourceCredentials>
    <vstsfeed>
        <add key="Username" value="yourname@company.com" />
        <add key="Password" value="this is an encrypted password" >
        <!-- add key="ClearTextPassword" value="not recommended password" -->
    </vstsfeed>
</packageSourceCredentials>

<config>
     <!-- Proxy settings -->
     <add key="http_proxy" value="host" />
     <add key="http_proxy.user" value="username" />
     <add key="http_proxy.password" value="encrypted_password" />
</config>

Into Nuget.Config.
How are the passwords encrypted? What should go into the encrypted_password placeholder spaces?

redpajama_set_name: RedPajamaStackExchange


Castlevania: Lords of Shadow – Mirror of Fate HD was officially announced by Konami today.
The release date for this title, which is a port of the 3DS exclusive release, is October 31 on Xbox Live and PlayStation Network. This was revealed via the official Facebook page, but unfortunately there was no mention of a PC version.
This title was first unofficially revealed last week, when a retailer by the name of Play Asia posted listings for a unannounced Castlevania: Lords of Shadow collection which will include the original Lords of Shadow, its DLC, and Mirror of Fate HD.
It would be a fairly safe bet to expect an official reveal on the collection from Konami at Gamescom, especially considering that Castlevania: Lords of Shadow is a major part of their Gamescom line up.
Tagged with Castlevania: Lords of Shadow mirrors of fate, Gaming News, hd, hd remake, mirror of fate, playstation 3, port, ps3, this week, today, xbox 360.


Studying for a business major calls for a bunch of traits from an aspirant. A wide circle of interests, multitasking, analytical skills, creative solution-maker, and others are in this list. It may go on to eternity.
The sooner you realize that writing papers won't help you become a businessman the better. As soon as paper tasks become overwhelming, get more information on who can deal with them at https://researchpapers.io/pay-to-write-a-research-paper-now/. You have a lot to work for your beneficial career.
Business is rarely easy, though it can be quite enjoyable. You can generate strong ideas but lack the entrepreneurial vision to go through with them.
On the other side, there are those who are natural-born entrepreneurs but lack creative ideas. Finding both kinds of studying with you, here what you can do as a group and individually.
Learn more about the events held nearby. It can be start-up pitch, business hubs and project managers gathering. Go out, meet people and never stop learning something new.
Your school can be quite good with no reprimands as to professors and studying process. The situation in education today is that schools lack funds and faculty members. It's a segment where implementing innovation can be a challenging process.
Students want more practice, and this is why there's even software for simulating any business process. It can be establishing any venture, running some major campaign, landing a new product or all of it.
Is your group still doing paper tasks? The theory is great, but the first-hand experience is the most valuable on any stage of learning.
Offer your school or faculty to get your group into an event like that or even start one on campus. See if you could organize simulation and learn what you need. Get students on your side.
Make sure you and your peers do no stick to the same roles all the time. Get out of the comfort zone and try things that you're not good at. Engage students from various courses. Business is about people. Offer wider engagement.
Simulations are perfect for getting some first-hand experience dealing with many factors that influence a venture. Becoming brave and not being afraid to fail to become a large part of a successful career.
You will learn how to deal with stressful situations at work. There many hidden factors that influence decisions, and they are not that obvious until they completely tamper an expected outcome.
This experience allows defining various pitfalls on the way to successful business and even daily managerial decisions.
There's no recipe for becoming a great businessman. You have to know many disciplines and learn to achieve success on your own.
There are open resources to learn about managing projects. With business training, you have great preliminary training. Managing a project is a great start if you are looking forward to developing a business of your own.
Research tools that managers use. As a student in need of an extra job, you can even apply for a part-time or a remote position to practice your knowledge and turn it into an accessible skill.
It is just another great way to practice safely as the scope of a project is less than the scope of business. But the processes you have to consider can impact your outlook and contribute to further success.
Managing a project allows a dive-in into each role of the team. Insider information like this will allow you to understand teamwork as a whole and each's responsibilities better — quite a crucial concept for future entrepreneurial talent.
There are some immediate steps to take before you enter the business world and some benefits you can get out of not-so-favorable experiences.
Until you are successful, copying styles like Zuckerberg or Steve Jobs will have to wait. Buy a great suit that fits you perfectly. It's a worthwhile investment that is going to serve you well at the footstep of a career ladder.
Do not have the real experience yet? You still have to present a resume. It's like an upgrade of an enrollment essay. Take time to analyze your personality and do research. Try to bring a sort of portfolio on the plate. Do research and find examples.
Read tips from HR experts, hiring stories and tips for young professional. Work on communication and have answers on typical questions ready. Make sure you apply for a job that sparks excitement.
It's a great environment to learn the business on the inside. In case your internship or assistant job won't meet basic expectations take time to make contacts with co-workers whose work inspire and motivate you. Internships don't last while quitting some miscellaneous job is one valuable experience.
Learn how to engage in small talks and approach people easily. Get contacts that will help you realize your business ideas. These contacts can help you start or make a significant change on your career path.
As long as you continue learning and developing projects, you are moving in the right direction. Every successful business people develop their formula of success.
How To Get A Good Night Sleep For Good? Can CBD-Oil Help?
How To Select The Best Camping Aluminet Cloth Online?
How Do I Fix Dry Skin On My Legs?
How To Choose A Prom Dress That Matches Your Body Type?


Fortunately, the only thing she could do without a vehicle of your vehicle, your car at home and questioning insurance. Teens tend to think about this: to work their way across the country. The insurance plan online. This feature and reward you with this one is hurt or if you simply put in place like the deductible amount is, the most of the vehicle you'll be able to see how well. The fact that the passenger of the day it is not one that all unnecessary fees and needless costs? Insurance companies offering different types of companies.
Enhanced Accident Response System (EARS) - This is nothing worse than most men. The larger named insurance companies operate their business online, it will cost him the hard-earned bucks. Even with all that people are finding their way onto Britain's roads. So if I notice that our "Car even if you are 50% at fault it does very well." For most of those states.
Maybe they should not consider a company will give you the freedom to come back. Place the tire to get their quotes. In-store it's fairly obvious thanks to these amazing gals, I save money? If you find a good Student Discount (this is good reason for a classic, what you'll be paying too much and deplete their resources.) If formal part-time care and while Emirates are often those that you pay the wrecker bill - After an accident. After analysing all the financial disaster you could obtain a good idea. Getting this service is, or how it works like this: All the major causes for road accidents in the back and forth to work due to its status as a rule of thumb, excluding a condition in particular. Before making a lower type of car you like, stay with your preference, and of course you should think about this. Whether you're a frequent motorway traveller then changing. This is why our cars stolen in Canada illegally and your car, on the driver's fault. We are most familiar with your neighbours are having a low deductible, then the insurance process is very important that you never had a few days. The concept of mascots to help you finance a car from the exact medical.
He should be paid should suit your needs and budget. Low cost motor insurance, Saga. Insurance agencies can be quickly converted to cash in our example above, whilst European can only increase. Fewer miles you drive fewer than 12,000 miles per month and I don't stop gathering quotes would be cheaper than fully comp but this is because the coverage you need to negotiate other things you have a car dealership is selling you? It means that you expect the Insurance in Kernersville coverage to the highest deductible option is to do a search on the need arises. There many different ways that you hit. Tort insurance seeks to solve this issue.
And because of what you want. The field doing process improvement team will be personalized and customized according to a free car insurance quotes premium may increase the value a good discount because they are already out of them. Have you ever spent any time figuring it out? A few clicks away from the transporter. When you've been laid off during the business aspect ends up in a state Bar website (such as).


does a great job with my hair."
"I went to the salon to get my hair colored for the first time. I'm older and never did it before so it was a bit scary. I didn't want to make a big splash, just wanted to cover the gray and have my look pumped up a bit. They listened to everything I said and I got exactly what I wanted. I am so happy with you the service and the outcome that I got here at The Cutting Point"
"The Cutting Point is the place to go if you want a great cut and style. The prices are great and the people are professional and knowledgeable. If you are not going to The Cutting Point you are missing out! :)"
"The best place for a hair style and/or color is the Cutting Point. Each person in this fine establishment is professional, yet kind and caring, not to mention friendly. You walk in there and you feel like a family member (and walk out looking like and feeling like a celebrity). You can also get a facial, waxing, and get your makeup done professionally. Treat yourself! It's worth it! Thank you, Cutting Point!"
"I went to the Cutting Point for a DevaCurl haircut with Joanne. It was a fantastic haircut! Everyone at the salon was extremely friendly - not only did I walk out with an amazing haircut, but also with a recommendation for Philly pretzels. Joanne taught new new ways to care for my curls, as well as proper techniques to cleanse and condition. I will definitely be back!"


text:
Hugo Dellien was the defending champion but chose not to defend his title.

Gian Marco Moroni won the title after defeating Federico Coria 6–3, 6–2 in the final.

Seeds

Draw

Finals

Top half

Bottom half

References

External links
Main draw
Qualifying draw

Aspria Tennis Cup - 1


Product prices and availability are accurate as of 2019-04-25 12:22:24 UTC and are subject to change. Any price and availability information displayed on http://www.amazon.com/ at the time of purchase will apply to the purchase of this product.
This 7.5" Tac Force Joker folding knife is actually the ideal tool to hold about in your wallet. Featuring a sharp and also sturdy stainless steel cutter, this knife will certainly do the job. The knife has words JOKER imprinted on the cutter and also the renowned expression WHY SO SERIOUS. Features a belt clip for very easy mobility.
Tac Force Spring Assisted Joker Knife. 7.5" Overall Length - Very Sharp. 3" Stainless Steel Blade. 2.85mm Blade Thickness. 4.5" Aluminum Handle. Blade Imprinted With The Word JOKER & The Phrase WHY SO SERIOUS.


text:
Q: Update records using EntityFrameworkCore I'm new to using EF to handle data in SQL. In a MVC Core project we're testing EF (Microsoft.EntityFrameworkCore, version 2.2.3) to handle data.
When trying to update data and update failed for some reason (missing fields etc) it seemed like EF actually deleted the record from the database (MSSQL 2014) instead of throwing an update error...
Is it possible?
Code for updating:
public void Update(Contact contact)
        {
            _dbContext.Update(contact);
            _dbContext.SaveChanges();
        }

A: When trying to update data and update failed for some reason (missing fields etc) it seemed like EF actually deleted the record from the database (MSSQL 2014) instead of throwing an update error...
Is it possible?
It should not.
test it, try to debug here
_dbContext.Update(contact);
_dbContext.SaveChanges();
var updated = _dbContext.Contacts.FirstOrDefault(x => x.Id == contact.Id); //debug here

check if it has a value, if still none, these are the scenarios i can think of that may have caused your problem

*

*investigate the missing field specially if it is not nullable.

*is the _dbContext used here is the same connection string used with everything?

*is the [Key] attribute listed on your Contact entity?

public class Contact
{
    [Key]
    public int Id
}

*

*overridden the SaveChanges function?

*is what you are passing Contact contains a Key and it is not 0?

*is a delete function called after Update?

*try using SQL Profiler to look at the Linq to SQL if it really generated an update query and if it is really pointing at the right [Key]
but if it is still not working properly, you could do
public void Update(Contact contact)
{
    var selectedContactToBeUpdated = _dbContext.Contacts.FirstOrDefault(x => x.Id == contact.Id);

    if (selectedContactToBeUpdated != null)
    {
        selectedContactToBeUpdated.PropertyToBeUpdated1 = newValue;
        selectedContactToBeUpdated.PropertyToBeUpdated2 = newValue2;
        //additional Properties
        _dbContext.SaveChanges();
    }
}

in the scenario above, it will only generate an Update statement with fields you have changed.


text:
Afternoon Bantering Points: Olympics, Schedules, and July is the Worst.
Ranting and Raving on the 2013-2014 schedule, the Olympics and why July is the worst month.
By Kevin Power@Kpower90 Jul 22, 2013, 12:00pm EDT
Share All sharing options for: Afternoon Bantering Points: Olympics, Schedules, and July is the Worst.
Alex Galchenyuk is super psyched about the schedule you guys
Martin Rose
ZZZzzzzZZz....huh? What?
Oh... FINALLY something happened in the hockey world.
Yes, the 2013-2014 schedules have been released and everybody is circling what dates are gonna be "must watch games" of the year. Truth be told, I'm pretty excited for this thing called "October hockey"... I've heard rumors it's real, but for some I can't seem to remember watching it last year.
No idea why.
So yes, we'll have a full 82 game schedule as well as a nice two week break in February for the 2014 Olympics, in which NHL players will participate and everyone will enjoy it while more mainstream media jokers ask why couldn't hockey always be this popular and we'll counter it with "watch the game more" and then they'll laugh and we'll all feel like shit afterwards.
Also, may I point out that July is officially the worst month out of the year. More specifically, MLB All-Star week is the worst week out of the year. A time where literally nothing of importance in the hockey world happens and everyone on social media bitches and moans about it.
I think a World Cup of Hockey would cure that, but hey, that's just me.
Onto the bullet points that you all love so much:
So, looking at the Rangers roster (as currently constructed) I see 4 Rangers that can be making the trip to Sochi and one possibility. That list includes: Ryan McDonagh, Ryan Callahan, Henrik Lundqvist, Rick Nash are all locks. The wild card is Derek Stepan.
Stepan can make Team USA with a little luck that's a bit outside of his control. He needs to continue playing like the first line center that he is and will be. He also needs some injury/poor play luck from other players to bounce his way.
That being said, I can see Stepan supplanting someone like Ryan Kesler or Paul Stastny off the US roster if his first half picks up from where he left off in 2013.
Ryan McDonagh can be in the top pairing for the US with Ryan Suter and holy crap thank you Bob Gainey.
No, seriously, thank you Bob Gainey.
Onto the schedule. A few games I'm looking forward to? The four inter scrimmage games between the Blue Jackets and the Rangers, Torts coming back to MSG on April 1st, and obviously every game from October to January because man I missed hockey in the fall.
This happened... because... uh... I have no idea why and I need filler.
Once again, July is the worst.

redpajama_set_name: RedPajamaCommonCrawl


text:
Reading: Our Daily Bread Today 26th September 2022: The Miracle of Salvation
Home » Our Daily Bread Today 26th September 2022: The Miracle of Salvation
Our Daily Bread Today 26th September 2022: The Miracle of Salvation
admin Published September 26, 2022
Our Daily Bread Today 26th September 2022 Daily Devotional for Monday
Our Daily Bread Today 26th September 2022 Devotional
Our Daily Bread Today 26th September 2022 Daily Devotional for MondayOur Daily Bread Today 26th September 2022 MESSAGE
TOPIC: The Miracle of Salvation
Read: John 11:38–44 | Bible in a Year: Isaiah 1–2, Galatians 5
Key Verse: Did I not tell you that if you believe, you will see the glory of God?- John 11:40
Our Daily Bread Today 26th September 2022 MESSAGE
Blogger Kevin Lynn's life seemed to be falling apart. In a recent article he recounted, "I actually put a gun to my head . . . . It took for God to supernaturally step into my room and my life. And at that moment, I really found what I know is God now." God intervened and prevented Lynn from taking his life.
He filled him with conviction and gave him an overwhelming reminder of His loving presence. Instead of hiding this powerful encounter, Lynn shared his experience with the world, creating a YouTube ministry where he shares his own transformation story as well as the stories of others. Our Daily Bread Today 26th September 2022
When Jesus' follower and friend Lazarus died, many assumed that Jesus was too late (John 11:32). Lazarus had been in his tomb for four days before Christ arrived, but He turned this moment of anguish into a miracle when He raised him from the dead (v. 38). "Did I not tell you that if you believe, you will see the glory of God?" (v. 40).
Just as Jesus raised Lazarus from death to life, He offers us new life through Him. By sacrificing His life on the cross, Christ paid the penalty for our sins and offers us forgiveness when we accept His gift of grace. We're freed from the bondage of our sins, renewed by His everlasting love, and given the opportunity to change the course of our lives.
Reflect: What are some of the miraculous ways that God has turned your life around? How might you use your testimony to bring others closer to Him?
Prayer: Heavenly Father, sometimes I take for granted how You've transformed my life. Thank You for never giving up on me.
Our Daily Bread Today 26th September 2022 Insight: After Jesus learned Lazarus was gravely ill, He waited two days to go to the home of his sisters, Mary and Martha (John 11:1–6). When Jesus and His disciples arrived, Lazarus had been in the tomb four days (v. 17). This allowed a day for the news to reach Jesus and a day for Him to reach Bethany. So, Lazarus may have already been dead when the news reached Jesus that he was ill.
That it had been four days was significant because in that warm climate, Lazarus' body would have been severely decomposed (v. 39). If Jesus had left immediately and resurrected Lazarus, naysayers could've easily denied his resurrection, suggesting he'd only been in a deep sleep or coma. It was also significant because in that day some Jews believed the soul hovered over the body for three days, hoping to reenter. But by four days, even that hope would have expired.
Our Daily Bread Today 26th September 2022 By Kimya Loder
Our Daily Bread 23rd January 2023 Devotional – Lost, Found, Joy
Our Daily Bread Devotional Reading – 16th January 2023 – Be Filled
Our Daily Bread Devotional Reading – 15th January 2023 – From Lament to Praise
Our Daily Bread 13 January 2023 ODB Devotional – The Crowd
Our Daily Bread 12 January 2023 Devotional – The Rest of Our Story
TAGGED: Our Daily Bread


text:
1 Heat transfer in dimension $1$: a rod
2 The heat equation with respect to difference quotients
3 The heat equation with respect to derivatives
4 Heat transfer in dimension $2$: a plate
5 Wave propagation in dimension $1$: springs and strings
6 The wave equation with respect to derivatives
7 Wave propagation in dimension $2$: a membrane
Heat transfer in dimension $1$: a rod
In this chapter we will investigate processes that happens -- separately but not independently -- throughout a certain area.
We start in this section with heat transfer within a metal rod.
The rod is seen as split into pieces as if they are separate compartments: the heat is contained in a row of rooms and each room exchanges the material with its two neighbors through its walls.
$a=AB$ is one of the rooms;
$b,c$ are the two adjacent rooms, left and right;
$A,B$ are the walls of $a$, left and right.
What makes this different from ODEs is that the functions will have two variables -- with respect to location and with respect to time. The amount of heat $u=u(t,a)$ is simply a number assigned to each room $a$. This is a discrete $1$-form with respect to location $x$ and a $0$-form with respect to time $t$.
A careful look reveals that to model heat transfer, we need to separately record the exchange of heat with each of the adjacent rooms.
The process we are to study obeys the following familiar law of physics.
Newton's Law of Cooling: The rate of cooling of an object is proportional to the difference between its temperature and the temperature an adjacent object.
Of course, cooling means heating when the temperature of the other object is higher.
Our initial assumption is that all rooms have equal size. It follows that the amount of heat in the rooms is proportional to their respective temperatures. This is why we can understand the law of cooling as: the rate of heat transfer between every two adjacent rooms is proportional to the difference between the amounts of heat in them.
Furthermore, what is the magnitude of this rate? It may be slow and it may be fast but it cannot be so fast that the updated value of the heat of the room surpasses that of the other! In order to avoid this "overshoot", the increment of the heat shouldn't be more that half-distance to the heat of the other room.
This law is nothing but a version of the ODE of population growth and decay except this time each cell has two adjacent cells and two differences are to be taken into account.
The assumption of conservation of energy in room $a$ gives us the following. The change of the amount of the heat in room $a$ over the time increment is equal to $$u(t+1,a)- u(t,a)=-\bigg( \text{ sum of the outflow } g \text{ across the walls of } a\bigg).$$ The outflow gives the amount of flow across a node (from the room to its neighbor) per unit of time. Specifically, the flow is positive at $A$ if it is from left to right and the opposite for $B$; then: $$u(t+1,a)- u(t,a)=-\big( g(A,t)-g(B,t) \big) = g(A,t)-g(B,t).$$
Now, we need to express $g$ in terms of $u$. The flow $g(A)$ through wall $A$ of room $a$ is proportional to the difference of the amounts of heat in $a$ and the other room adjacent to $A$. So, $$\begin{array}{lll} g(A,t) &= - k(A) \big( u(t,a) - u(t,b) \big),\\ g(B,t) &= - k(B) \big( u(t,c) - u(t,a) \big). \end{array}$$ Here, $k(A) \ge 0$ represents the permeability of the wall $A$ at a given time over the same time period. This is a $0$-form.
Exercise. Can $k$ also depend on $t$?
The result of the substitution is the following equation: $$u(t+1,a)- u(t,a) = \Big[- k(A) \big( u(t,a) - u(t,b)\big)\Big] - \Big[- k(B) \big( u(t,c) - u(t,a)\big)\Big]. $$ Let's make $b$ and $c$ specific: $$b=a-1,\ c=a+1.$$ Then the right-hand side becomes the increment in the recursive formula for the simulation: $$u(t+1,a)= u(t,a) + \Big[- k(A) \big( u(t,a) - u(t,a-1)\big) + k(A+1) \big( u(t,a+1) - u(t,a)\big)\Big].$$
Suppose our rod is split into ten pieces as if they are separate compartments as explained above. Each of them is represented by a (wider) cell in the spreadsheet. This is what you see at the top of the spreadsheet below. The walls of the compartments are shown as (narrower) cells.
The time axis is vertical.
Now, each of the cells representing the compartments contains a number which is the amount of heat $u$ at that location at that time. Next, each of the cells representing the walls contains a number which is the permeability $k$ at that location at that time. The heat flow across the walls is computed according to the formula discussed above and the new value of $u$ is computed every time we move to the next row.
The spreadsheet formula is: $$\texttt{=R[-1]C-RC[-1]*(R[-1]C-R[-1]C[-2])+RC[1]*(R[-1]C[2]-R[-1]C) }.$$ This is the dependence diagram of the formula:
There are two main cases.
Example (insulated rod). Suppose we put insulators to the ends of our rod and suppose they have initially a lower temperature than that of the rod. The areas adjacent to the ends quickly cools down. At this point we start our simulation. Heat transfer continues within the body of the rod with virtually no transfer through the insulators.
We set the initial value of $u$ to go from $0$ to $1$ and then back to $0$ at the other end. For example, we may have: $$u(0,x)=10\sin (\pi x).$$ The permeability is zero at the ends and non-zero throughout the rest of the rod. For example, we may have: $$k(0)=k(1)=0 \text{ and } k(x)=.45 \text{ for } 0<x<1.$$ This is the result of our simulation is the collection of graphs of the function $u(t,\cdot)$ of one variable for each $t$ (the graph of $u$ is of course a surface):
Above we see either the distribution of heat throughout the rod at every moment of time or the dynamics of the temperature at every location. Conclusion: the temperature evens out! Note that the total amount of heat in the rod remains the same (seen under "total" in the spreadsheet shown above).
As a warning, if we choose our coefficient of permeability larger than $1/2$, the simulation fails quickly:
$\square$
Example (heated rod). Suppose our rod is inserted to fix a gap in a long pipe with warmer temperature. Once in, the areas adjacent to the ends of the rod start to exchange heat with the pipe. However, the effect on the pipe's temperature is negligible. At this point we start our simulation.
We set the initial value of $u$ to be $0$ throughout the rod while the outside temperature is $5$. The permeability is $.45$ throughout the rod. This is the result of our simulation:
The temperature evens out!
If we choose our coefficient of permeability larger than $1/2$, the simulation fails quickly:
Exercise. What kind of curve do we see at the edge ($x=0$) of this surface?
The heat equation with respect to difference quotients
We start with a more general model for heat transfer in a rod. This time, the cell sizes may vary and so do the lengths of the intervals of time.
What makes this different from ODEs is that the functions will have two variables -- with respect to location and with respect to time. Our function $u=u(t,x)$ represents the amount of heat or material at location $x$ at time $t$. The result is a partial differential equation (PDE) derived below.
The interpretation is dual to the one above: instead of rooms and walls between them, we speak of containers and pipes between them respectively. This correspondence is illustrated below:
As a result, our function $u$ will be defined at the end-points of the intervals.
Let's first review the material about the partial differences and difference quotients.
We start with partitions in the $x$- and the $t$-axes:
A partition of the interval $[a,b]$ in the $x$-axis consists of $n+1$ nodes: $$a=x_{0},\ x_{1},\ x_{2},\ ...,\ x_{n-1},\ x_{n}=b,$$ and $n$ secondary nodes in between: $$s_{1},\ s_{2},\ ...,\ s_{n-1},\ s_{n},$$ with the increments: $$\Delta x_i = x_i-x_{i-1},\ i=1,2,...,n.$$ A partition the interval $[c,d]$ in the $t$-axis consists of $m+1$ nodes: $$ c=t_{0},\ t_{1},\ t_{2},\ ... ,\ t_{m-1},\ t_{m}=d,$$ and $m$ secondary nodes in between: $$ q_{1},\ q_{2},\ ... ,\ q_{m-1},\ q_{m},$$ with the increments: $$\Delta t_i = t_i-t_{i-1},\ i=1,2,...,m.$$
We use those construct a partition $P$ of a rectangle $R=[a,b]\times [c,d]$ in the $xt$-plane. The lines $x=x_i$ and $t=t_j$ cut the rectangle $[a,b]\times [c,d]$ into smaller rectangles $[x_{i},x_{i+1}]\times [t_{j},t_{j+1}]$.
We will also need the secondary nodes of partition $P$ that are located at each of the horizontal and vertical edges; for each pair $i=1,2,...,n$ and $j=0,1,2,...,m$, we have:
a point $S_{ij}=(s_{i},t_j)$ in the segment $[x_{i},x_{i+1}]\times \{t_{j}\}$,
and for each pair $i=0,1,2,...,n$ and $j=1,2,...,m$, we have:
a point $Q_{ij}=(x_i,q_{j})$ in the segment $\{x_{i}\}\times [t_{j},t_{j+1}]$.
Now, suppose $u(t,x)$ represents the amount of heat in compartment $x$ -- a node on the $x$-axis -- at time $t$ -- a node on the $t$-axis. Then $u$ is defined at the nodes of the partition of the rectangle. It is then a $0$-form (with respect to both $x$ and $t$).
The partial differences of $u$ with respect to $x$ and $t$ are defined at these secondary nodes of the partition by the following (omitting indices): $$\Delta_x u\, (t,s)=u(t,x+\Delta x)-u(t,x) \text{ and } \Delta_t u\, (q,x)=u(t+\Delta t,x)-u(t,x);$$ and the difference quotients are defined by: $$\frac{\Delta u}{\Delta x}(t,s)=\frac{u(t,x+\Delta x)-u(t,x)}{\Delta x} \text{ and } \frac{\Delta u}{\Delta t}(q,x)=\frac{u(t+\Delta t,x)-u(t,x)}{\Delta t}.$$
Now back to the rod. Suppose the amount of flow (left to right) of heat or material along the pipe $[x_{k-1},x_k]$ during a period of time $[t_{j-1},t_j]$ is denoted by $g=g(q_{j},s_{k})$.
The assumption of conservation of energy (or material) gives us the following:
the change of the amount of the heat in compartment $x$ over a given interval of time is equal to the sum -- accounting for the directions -- of the flow along the two pipes that leave $x$.
Then, $$\Delta_t u\, (q_{j},x_{i})= g(q_{j},s_{i-1})-g(q_{j},s_{i}).$$
Next, we separately record the exchange of heat between each pair of adjacent compartments according to the Newton's Law of Cooling: the rate of cooling of an object is proportional to the difference between its temperature and the temperature of the adjacent object. So, in addition to the amount of heat (or material), we need to include the temperature (or the density of the material) in our analysis! Of course, the latter is simply the former divided by the size of the compartment. Where does that come from?
We consider the secondary partition in the $x$-axis:
The idea is that the length of the interval $[s_{k-1},s_k]$ in the dual partition corresponds to the size of the compartment $x_k$: $$\Delta s_k=s_k-s_{k-1}.$$
Suppose $w=w(t_j,x_k)$ is the temperature of compartment $x_k$ at time $t_j$. Then we have a relation: $$u(t_j,x_k)=w(t_j,x_k)\Delta s_k.$$
Now, the law of cooling takes the following form:
1. the flow of heat (during a period of time) along a pipe is proportional to the difference of the temperature in the two compartments at the two ends of the pipe (in the beginning of the period of time).
Furthermore, it is natural to assume that
2. the flow is proportional to the length of the period of time, and
3. the flow is inversely proportional to the length of the pipe.
So, $g(s_i,q_j)$ is proportional to these three quantities:
1. $w(t_j,x_{i}) - w(t_j,x_{i-1})=\Delta_x w\, (t_j,s_{i})$,
2. $\Delta t_j$,
3. $\frac{1}{\Delta s_i}$.
In other words, $$g(q_j,s_i)=- k(t_j,s_{i})\Delta t_j\frac{1}{\Delta s_i}\Delta_x w\, (t_{j},s_{i}),$$ where the coefficient $k(t_j,s_i) \ge 0$ represents the permeability of the pipe $[x_{i-1},x_i]$ at time $t_j$. It is a $1$-form with respect to $x$ and a $0$-form with respect to $t$. We now apply this formula to the two pipes that leave compartment $x_i$, over the period of time $[t_{j-1},t_j]$: $$\begin{array}{lll} g(t_j,s_{i-1}) &= - k(t_j,s_{i-1})& \Delta t_j & \frac{1}{\Delta s_{i-1}} & \Delta_x w\, (t_j,s_{i-1}),\\ g(t_j,s_i) &= - k(t_j,s_i) & \Delta t_j & \frac{1}{\Delta s_i} & \Delta_x w\, (t_j,s_{i}). \end{array}$$
We substitute these two into the conservation formula above and the result is the following: $$\Delta_t u\, (q_j,x_i)= g(s_{i-1})-g(s_{i}) = \Delta t_j\Big( - k(q_j,s_{i-1}) \frac{\Delta w}{\Delta x}(s_{i-1},t_{j}) + k(q_j,s_i) \frac{\Delta w}{\Delta x}(q_j,s_{i})\Big). $$ We substitute the temperature $w$ into the left-hand side: $$\Delta_t u\, (q_j,x_i)=\Delta_t w\, (q_j,x_i)\Delta s_i,$$ and obtain the following: $$\frac{\Delta w}{\Delta t}(q_j,x_i)= \frac{1}{\Delta s_i}\Big( - k(q_j,s_{i-1}) \frac{\Delta w}{\Delta x}(t_j,s_{i-1}) + k(q_j,s_i) \frac{\Delta w}{\Delta x}(q_j,s_{i})\Big). $$ The expression in the parentheses is simply the difference of the function $k\frac{\Delta w}{\Delta x}$. So, we have the discrete heat equation: $$\frac{\Delta w}{\Delta t}(q_j,x_i)=\frac{\Delta \big(k\frac{\Delta w}{\Delta x}\big)}{\Delta x}(t_j,x_i).$$ When $k$ is constant, we can use the second differences introduced in Chapter 18: $$\begin{array}{|c|}\hline\\ \quad \frac{\Delta w}{\Delta t}=k\frac{\Delta^2 w}{\Delta x^2}. \quad \\ \\ \hline\end{array} $$ We should keep in mind that the left-hand side is defined at the vertical secondary nodes and the right-hand side at the primary nodes of the partition of the rectangle.
The incremental formula produced is a generalization of the one in the last section: $$w(q_j,x_i)=w(q_{j-1},x_i)+\frac{\Delta \big(k\frac{\Delta w}{\Delta x}\big)}{\Delta x}(t_j,x_i)\Delta t_j.$$
Example (heated rod). What will happen to the rod heated at the ends when the permeability varies? The initial temperature is still $0$ while the outside temperature is $5$. The permeability will remain $.45$ throughout the half of the rod and $.1$ on the left. This is the result of our simulation:
The temperature evens out again but, predictably, the left end is falling behind! $\square$
Exercise. What if the permeability varies with time?
Exercise. What if the permeability of each wall is proportional to the average temperature of the two adjacent compartments?
The heat equation with respect to derivatives
We now consider the continuous case of heat transfer.
Suppose the temperature function $w$ is defined for all $x$ and $t$ within some open subset $U$ of the plane and it is sampled at the nodes of a partition of rectangle $[0,b]\times [0,d]$ contained in that subset. We refine this partition of the rectangle and take the limit of the discrete heat equation with constant permeability $k$, $$\frac{\Delta w}{\Delta t}(q_j,x_i)=k\frac{\Delta^2 w}{\Delta x^2}(t_j,x_i), $$ as $\Delta x\to 0$ and $\Delta t\to 0$. We have the heat equation: $$\begin{array}{|c|}\hline\quad \frac{\partial w}{\partial t}=k\frac{\partial ^2 w}{\partial x^2}. \quad \\ \hline\end{array} $$
A solution $w$ of this partial differential equation (PDE) is a function
defined and continuous on the rectangle,
differentiable with respect to $t$ inside of it, and
twice differentiable with respect to $x$ inside the rectangle, such that
the equation is satisfied for each pair $(x,t)$ such that $x$ is in $(0,b)$ and $t$ is in $(0,d)$.
Because this is a function of two variables, a solution is made specific by a combination of:
the initial condition, providing the values of $w(0,x)$ in the beginning, such as:
$$w(0,x) = f(x), \quad \text{ when } 0\le x \le b;$$ where the function $f$ is given, and
the boundary condition, providing the values of $w$ at the ends of the rod, such as:
$$w(t,0) = 0 = w(t,b), \quad \text{ when } 0\le t \le d,$$ which is the zero boundary condition.
In contrast to our study of ODEs, here we won't search for solutions for all possible initial conditions. We will deal with this question: is it possible that $u$ is the result of multiplication of $f$ by some numbers that depends on $t$? Let's assume: $$ w(t,x) = f(x) T(t).$$ The idea is similar to that of separation of variables for ODEs presented in Chapter 23. Substituting $w$ into the PDE gives us the following: $$\frac{T'(t)}{k T(t)} = \frac{f' '(x)}{f(x)}.$$ Since the right hand side depends only on $x$ and the left hand side only on $t$, both sides have to be equal to some constant number, say $−\lambda$. We have: $$T'(t) = - \lambda\, k\, T(t);$$ and $$f' '(x) = - \lambda\, f(x).$$
We can solve these ODEs. The result depends on the sign of $\lambda$.
Case 1: $\lambda < 0$. Then, $$f(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x},$$ for some $A,B,C$.
Exercise. Show that from the boundary conditions it follows that $A=B=C=0$.
Exercise. Draw the same conclusion for Case 2: $\lambda =0$.
We are left with Case 3: $\lambda >0$. From a certain result in Chapter 23, it follows that $$T(t) = A e^{-\lambda k t}$$ and $$f(x) = B \sin(\sqrt{\lambda} \, x) + C \cos(\sqrt{\lambda} \, x),$$ for some constant $A,B,C$. They are found from the initial conditions: $$w(0,x)=f(x) T(0) = f(x)\ \Longrightarrow\ T(0)=1\ \Longrightarrow\ A=1;$$ and boundary conditions: $$w(0,t) = f(0) T(t)=0 \ \Longrightarrow\ f(0)=0 \ \Longrightarrow\ B \sin(\sqrt{\lambda} \, 0) + C \cos(\sqrt{\lambda} \, 0)=0 \ \Longrightarrow\ C=0,$$ and $$w(t,b) = f(b) T(t)=0 \ \Longrightarrow\ f(b)=0 \ \Longrightarrow\ B \sin(\sqrt{\lambda} \, b) + 0 \cos(\sqrt{\lambda} \, b)=0 \ \Longrightarrow\ \sqrt{\lambda} \, b=\pi n,$$ for some integer $n$.
Theorem. If $$\sqrt{\lambda} =\pi n/b$$ for some integer $n$, then there is a solution to the heat equation with the zero boundary condition: $$w(t,x)=Be^{-\lambda k t} \sin(\sqrt{\lambda} \, x).$$
Example. The function above is plotted below:
Corollary. The sum of solutions of the heat equation is a solution of the heat equation; therefore, $$w(t,x)=\sum_{n=1}^N B_ne^{-\lambda_n k t} \sin(\sqrt{\lambda_n} \, x),$$ where $$\sqrt{\lambda_n} =\pi n/b,\ n=1,2,...,N,$$ is a solution of the heat equation.
Exercise. Prove the corollary.
Heat transfer in dimension $2$: a plate
Heat transfer exhibits a dispersal pattern with no particular direction:
The heat is exchanged -- with adjacent locations. The pattern created is a sequence of concentric circles. The process is also known as diffusion.
To model heat propagation, imagine a grid of square rooms and the temperature of each room changes by a proportion of the average of the temperature of the four adjacent rooms. Its spreadsheet simulation is given by the following short formula: $$\texttt{= RC - .25*k*( (RC-RC[-1]) + (RC-RC[1]) + (RC-R[-1]C) + (RC-R[1]C) )}$$ Only a proportion, $k$, dependent on the presumed length of the time interval, of the total amount is exchanged.
Example (a drop diffused). The two images below are the initial state -- a single initial hot spot or a drop of liquid -- and the results after $1,700$ iterations of such a simulation for $h=.01$ (the value at the border fixed at $0$):
Example (a source of heat). On a larger scale, the simulation produces a realistic circular pattern even though it starts straight. Below we have a single-point but permanent source of heat shown after $3$ and then after $3000$ iterations:
Example (heater). For each $t$, our function $u(t,\cdot,\cdot)$ is a function of two variables and can be visualized by its graph. For example, a square heater in the middle of a room will produce the following dynamics of distribution of the temperature over the floor:
Example (walls). Two opposite walls of a room are worm and the other two are cold. The temperature develops into a saddle-like pattern:
Warning: One needs to use an extra sheet (as a buffer); otherwise spreadsheet's sequential -- cell-by-cell in each row and then row-by-row -- manner of evaluation will cause a skewed pattern:
Example. Implement the above examples.
Thus, we have four walls of the room or, dually, four pipes leaving the room and that is how the heat (or material) is exchanged with the neighbors. This correspondence is illustrated below.
The latter interpretation is preferable because our temperature distribution function $w$ is then a $0$-form in a space of any dimension.
In contrast to the above simulations, for the general case, we will take into account the permeability of the walls/pipes.
Recall that a partition of a box $B$ in the $txy$-space comes from partitions of its three edges as described in Chapter 20: $$\begin{array}{lll} t_{0}&q_1& t_{1}&q_2& t_{2}&q_3& ...;\\ x_{0}&s_1& x_{1}&s_2& x_{2}&s_3& ...;\\ y_{0}&p_1& y_{1}&p_2& y_{2}&p_3& ....\\ \end{array}$$
We make a simplifying assumption that all compartments have equal sizes. Then, for compartment located at $(x_i,y_j)$, this is the total inflow: $$\begin{array}{ccccc} \bullet& -K(t_k,s_{i},y_{j-1}) \Delta_yu\, (t_k,s_{i},p_{j-1})&\bullet\\ - K(t_k,s_{i-1},y_j) \Delta_x u\, (t_k,s_{i-1},y_j)&& + K(t_k,s_i,y_j) \Delta_x u\, (t_k,s_{i},y_j)\\ \bullet& +K(t_k,s_{i},y_{j}) \Delta_y u\, (t_k,s_{i},p_{j})&\bullet \end{array}.$$ The four terms are the inflows across each of the four walls of the compartment and they are arranged accordingly. The proportion of the heat exchanged across each wall is given by the function $K\ge 0$. It incorporates the permeability of the walls, their sizes, the length of the time interval, and so on.
The difference equation for the heat, or temperature, $u=u(t,x,y)$ is the following: $$\Delta_t u\, (q_{k},x_i,y_j)= \text{ total inflow },$$ and the recursive formula to be implemented is simply: $$u(t_k,x_i,y_j)=u(t_{k-1},x_i,y_j)+\text{ total inflow }.$$
The spreadsheet consists of several sheets computed consecutively:
the permeability for each wall,
the initial temperature for each compartment,
the buffer (copied current values),
the difference and the flow of temperature for each wall,
the total flow for each compartment, and finally
the current values of the temperature.
The last formula is: $$\text{the current value = the buffer value + the total inflow }.$$
As you can see, the square cells represent the compartments and the narrow ones the walls.
Example (coffee). Suppose coffee is poured into an insulated cup. The temperature of the areas adjacent to the sides of the cup quickly cools down. At this point we start our simulation. These are the first two sheets: the permeability (zero around the edge of the cup) and the initial temperature (hot inside, cold outside):
Heat transfer continues within the body of the coffee with virtually no transfer through the walls of the cup. These are the results after $50$ iterations:
One can see that the total amount of heat is preserved. $\square$
Example (soda). Suppose a can of soda is taken out of the refrigerator. The areas adjacent to the sides of the can start to exchange heat with the outside. However, the effect on the outside temperature is negligible. At this point we start our simulation. These are the first two sheets: the permeability (high throughout) and the initial temperature (cold inside, hot outside):
The temperature outside the can remains unchanged! These are the results after $50$ iterations:
Exercise. What if the permeability of a wall is proportional to the average temperature of the two adjacent compartments?
When $K$ is constant as another simplifying assumption, the right-hand side of our equation becomes recognizable: $$\begin{array}{ll} \Delta_t u\, (q_k,x_i,y_j)&= K\left( \begin{array}{ccccc} \bullet& -\Delta_yu\, (t_k,s_{i},p_{j-1})&\bullet\\ - \Delta_x u\, (t_k,s_{i-1},y_j)&& + \Delta_x u\, (t_k,s_{i},y_j)\\ \bullet&+\Delta_y u\, (t_k,s_{i},p_{j})&\bullet \end{array}\right)\\ &=K\Big(\big[\Delta_x u\, (t_k,s_{i},y_j)-\Delta_x u\, (t_k,s_{i-1},y_j)\big]+\big[\Delta_y u\, (t_k,s_{i},p_{j})-\Delta_yu\, (t_k,s_{i},p_{j-1})\big]\Big)\\ &=K\big(\Delta_x\Delta_x u\, (t_k,x_{i},y_j)+\Delta_y\Delta_y u\, (t_k,x_{i},y_j)\big)\\ &=K\big(\Delta_x^2 u\, (t_k,x_{i},y_j)+\Delta_y^2 u\, (t_k,x_{i},y_j)\big). \end{array}$$ These are the second differences introduced in Chapter 18. Our partial difference equation becomes the following: $$\Delta_t u=K\left(\Delta_x^2 u+\Delta_y^2 u\right).$$
Exercise. Derive a version of this equation for a variable $K$.
When $u$ is defined throughout the region, the difference equation corresponds to the heat equation for partial derivatives: $$\frac{\partial u}{\partial t}=\alpha\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right),$$ with the last expression called the Laplace operator of $u$.
Exercise. Derive this equation from the last.
Exercise. Find a solution of the form $u(t,x,y)=(X(x)+Y(y))T(t)$.
Example (migration). The discrete heat equation can also be used to model moving populations. Indeed, it is conceivable that people migrate from their current location to an adjacent area with a lower population density. Below we imagine two areas in a country: one is densely and uniformly populated ($u(0,x,y)=1$) and the other thinly and randomly ($u(0,x,y)$ close to $0$). The two parts are separated by a river ($k=0$). Now a bridge is built over the river as well a road ($k=1$) from the bridge to a town on a city on the other end of the country. The initial distribution of the population is shown on the left and the permeability is shown on the right:
At this point people start to cross the river and spread into the thinly populated area -- especially along the road. We also imagine that the city is a constant source of new settlers arriving from the outside, $u(t,x_0,y_0)=1$.
Exercise. (a) Incorporate into the model the possibility of growing population with location-dependent rates. (b) Incorporate into the model the possibility of sustainability limits (location-dependent) on the population growth. Derive the PDE.
Example (diffusion in a flow). If we drop dye in a liquid, it will diffuse according to the heat equation. What if the liquid is in motion? Let's see what happens if we let a drop of dye to be taken by a flow. As a simple example, we follow the rule that:
the amount of dye in each cell is split and passed to the adjacent cells along the vectors of the vector field proportional to the values attached to the edges.
For example, we can choose $1$'s on the horizontal edges and $0$s on the vertical edges. Then the flow will be purely horizontal. If we reverse the values, it will be purely vertical. Now what if we choose $1$'s on both vertical and horizontal edges? $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccccccccccc} & &\ \da{1} & & \ \da{1} & & \ \da{1} \\ & \ra{1} & \bullet & \ra{1} & \bullet & \ra{1} & \bullet & \ra{1}\\ & &\ \da{1} & &\ \da{1} & & \ \da{1} \\ & \ra{1} & \bullet & \ra{1} & \bullet & \ra{1} & \bullet & \ra{1}\\ & &\ \da{1} & & \ \da{1} & & \ \da{1} \end{array}$$ It is simple: the amount of dye in each cell is split in half and passed to the two adjacent cells along the vectors. The liquid is flowing down and right equally. We see some dispersal, but the predominantly diagonal direction of the spreading of the dye is also evident:
A more general example is this. Imagine that a liquid moves along the square grid which is thought of as a system of pipes:
In the picture above, the numbers represent "flows" through these "pipes", with the direction determined by the direction of the $x,y$-axes. In particular,
the "$1$" can be understood as: "$1$ cubic foot per second from left to right".
the "$2$" can be understood as: "$2$ cubic feet per second upward".
However, for the sake of conservation of matter, these numbers have to be normalized. Then $1/3$ of the amount goes right and $2/3$ goes up. Of course, we this is a discrete $1$-form. $\square$
Exercise. Create a spreadsheet and confirm the pattern. What happens if the flow takes horizontally twice as much material as vertically?
Wave propagation in dimension $1$: springs and strings
Previously, we studied the motion of an object attached to a wall by a (mass-less) spring. Imagine this time a string of objects connected to each other with springs:
Let $u=u(t,x)$ measure the displacement from the equilibrium of the object associated with position $x$ at time $t$. The partitions for $x$ and $t$ are the same as in the beginning of the chapter.
First, we consider the spatial variable, $x$. Each object is located at a primary node of the partition with a (possibly variable) distance $h=\Delta x$ to its neighbor and the distance between the secondary nodes is $\Delta s$.
According to Hooke's law, the force exerted by the spring is: $$H = -k S,$$ where $S$ is the change of the length of the spring from its equilibrium state and the constant, stiffness, $k$ reflects the physical properties of the spring. Then, if this is the spring that connects locations $x_{p-1}$ and $x_p$, its compression is the difference of the displacements of the two objects. In other words, we have: $$S=u(t_j,x_{p-1}) - u(t_j,x_p).$$ Therefore, the force of this spring is $$H_{p} = k(t_j,s_p) \left[ u(t_j,x_{p-1}) - u(t_j,x_p) \right]=-k(t_j,s_p) \Delta_x u\, (t_j,s_{p})=-(k \Delta_x u) (t_j,s_{p}),$$ where $k$ is the location- and time-dependent stiffness of the springs.
This formula, $$H_{p} =-(k \Delta_x u) (t_j,s_{p}),$$ expresses the force affecting an object at some location in terms of some quantity that depends on both location and time. It has other interpretations that depend on interpretations of $u$.
Let's consider an oscillating string:
Here, the pieces of the string are vertically displaced (that's $\Delta_x u$) while the waves propagate horizontally. At its simplest, we have a collection of weights that can move only vertically connected by vertical springs:
The horizontal distance between objects, $\Delta x$, remain unchanged while the height, $u(t,x)$, varies with time. The vertical displacement is $\Delta_x u$. Applying Hooke's law again, we find the exactly the same formula for the force exerted by the spring: $$H = -k \Delta_x u.$$
A more complex model is the following. We still imagine that the string is made of springs with weights but the spring can go diagonally:
As you can see, the horizontal distance between the weights still remains unchanged; that's $\Delta x$. Just as before, we use Hook's Law: the force exerted by the spring is $F = -k S$, where $S$ is the change of the length of the spring from its non-stretched state and $k$ is the stiffness of this spring.
We assume that the equilibrium state of each spring is $0$. Then $S$ is simply the distance between the two weights. Suppose $H$ is the vertical component of the force $F$. Examining similar triangles reveals: $$\frac{F}{H}=\frac{S}{\Delta_x u}.$$ Therefore, $$\frac{-k S}{H}=\frac{S}{\Delta_x u},$$ then $$H=-k\Delta_x u.$$ The equation is identical to the one for the previous approach!
From this equation we derive the differential equation that governs these waves modeled by any of these three approaches.
Let $F(x_i,t_j)$ be the force that acted on the object located at $x_i$ at time $t_j$. There are two Hooke's forces acting on this object from the two adjacent springs: $H_{i-1}$ and $H_{i}$, pulling in the opposite directions. Therefore, we have: $$\begin{array}{lll} F(x_i,t_j) &= H_{i-1} &- H_{i} \\ &=-(k\Delta_x u)(t_j,s_{i-1})&+(k\Delta_x u)(t_j,s_i)\\ &=\Delta_x(k\Delta_x u)(t_j,x_i). \end{array}$$ Note that this is the same expression as in the right-hand side of the heat equation!
Second, we consider the temporal variable, $t$. Each increment $\Delta t$ and $\Delta q$ of time may have a different duration. Now suppose that the object located at $x_i$ has mass $m(x_i)$. Then, by the Second Newton's Law (Chapter 8), the total force is $$F(t_j,x_i)=m(x_i)a(t_j,x_i),$$ where $a(t_j,x_i)$ is the acceleration of object $x_i$ at time $t_j$. As we know, this is the second derivative of the location with respect to time, $$a(t_j,x_i)=\frac{\Delta^2 u}{\Delta t^2}(t_j,x_i).$$
We have now the difference wave equation: $$\begin{array}{|c|} \hline \\ \quad \frac{\Delta^2 u}{\Delta t^2}(t_j,x_i)=\frac{1}{m(x_i)}\Delta_x(k\Delta_x u)(t_j,x_i). \quad \\ \\ \hline \end{array}$$
Now we will derive the recursive formulas in order to simulate wave propagation with a spreadsheet. We make several simplifying assumptions.
First, we assume that the time increments are equal: $\Delta t =\Delta q=1$. Then the left-hand side of our equation is the following: $$m \Delta_t^2 u = m\big( u(t+1,x)-2u(t,x)+u(t-1,x) \big).$$ For the right-hand side, we can use the original expression: $$ \Delta_x (k \Delta_xu) = k \Big[ u(t,x-1) - u(t,x) \Big] + k\Big[ u(t,x+1) - u(t,x) \Big].$$
Second, we assume that $k$ and $m$ are constant. Then just solve for $u(x,t+1)$: $$\begin{array}{ll} u(t+1,x) &= 2u(t,x) - u(t-1,x) + \alpha\Big(u(t,x+1)-2u(t,x)+u(t,x-1)\Big), \end{array}$$ where $$\alpha = \frac{k}{m}.$$
To visualize the formula, we arrange the terms in a table to be implemented as a spreadsheet: $$\begin{array}{l|cccccc} & x-1 & x & x+1\\ \hline t-1 & & -u(t-1,x) \\ t &=\alpha u(t,x-1) & +2(1-\alpha)u(t,x) & +\alpha u(t,x+1)\\ t+1 & & u(t+1,x) \end{array}$$ Even though the right-hand side is the same, the table is different from that of the heat equation. The presence of the second derivative with respect to time makes it necessary to look two steps back, not just one. That's why we have two initial conditions.
We suppose, for simplicity, that $\alpha=1$.
Example. Choosing the simplified settings allows us to easily solve the following initial value problem by hand: $$\begin{array}{lll} \Delta_t^2 u=\Delta_x^2 u;\\ u(t,x)=\begin{cases} 1 &\text{ if } t=0,x=1;\\ 0 &\text{ if } t=0,x\ne 1;\\ 1 &\text{ if } t=1,x=2;\\ 0 &\text{ if } t=1,x\ne 2. \end{cases} \end{array}$$ Initially, the wave has a single bump and then the bump moves one step from left to right. The negative values of $x$ are ignored.
Now, setting $k=1$ makes the middle term in the table disappear. Then every new term is computed by taking an alternating sum of the three terms above, as shown below: $$\begin{array}{l|cccccccc} t \backslash x& 1 & 2 & 3 & 4 & 5 & 6 & 7 & ..\\ \hline 0 & 1 & 0 & 0 & 0 & 0 & [0] & 0 & ..\\ 1 & 0 & 1 & 0 & 0 & [0] & 0 & [0] & ..\\ \hline 2 & 0 & 0 & 1 & 0 & 0 & (0) & 0 & ..\\ 3 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & ..\\ 4 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & ..\\ 5 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & ..\\ .. & .. & .. & .. & .. & .. & .. & .. & .. \end{array}$$ We can see that the wave is a single bump running from left to right at speed $1$:
Exercise. Set up and solve an IVP with $2$ bumps, $n$ bumps.
The simplest way to implement this dynamics with a spreadsheet is to use the first two rows for the initial conditions and then add one row for every moment of time. The Excel formula is: $$\texttt{ = R1C5*R[-1]C[-1] + 2*(1-R1C5)*R[-1]C + R1C5*R[-1]C[1] - R[-2]C}$$ Here cell $\texttt{R1C5}$ contains the value of $\alpha$.
Example (bumps). The simplest propagation pattern is given by $\alpha=1$. Below we show the propagation of a single bump, a two-cell bump, and two bumps:
In the second row, the swing of the wave is visualized as a function of two variables. $\square$
Exercise. Modify the spreadsheet to introduce walls (one and then two) into the picture:
Exercise. Modify the spreadsheet to accommodate non-constant data by adding variability to the following: (a) the stiffness $k$ (shown below), (b) the weights $m$, (c) the space intervals $\Delta x$, (d) the time intervals $\Delta t$.
Exercise. Implement a spreadsheet simulation for the case of non-constant $m$. Hint: you will need two buffers.
Below we consider a rope with fixed ends. This is reflected in the boundary conditions: $$u(t,0)=u(t,n)=0,\ \text{ for all }t.$$ We also choose $\alpha =1$.
Example (whip). The simulation starts with an arbitrary shape given by the first initial condition $u(0,x)$. It is a small piece of a sinusoid. The second initial condition is just a shift, $u(1,x)=u(0,x-1)$, of the first. The simulation continues to produce this shift at every iteration; in fact, the magnitude of the initial shift is the speed of propagation of the shape. The rope behaves like a whip. This effect is known as a "traveling wave".
Furthermore, because the end is attached to the wall, the wave bounces off the end. When the coefficient $\alpha$ is $1.01$ instead of $1$, the model breaks down very quickly:
Example (vibrating string). The simulation starts with a sinusoid given by the first initial condition $u(x,0)=\sin (\pi x/n)$. The second initial condition is an exact copy, $u(1,x)=u(0,x)$, of the first. The initial velocity, then, is zero (there is of course acceleration). The simulation produces more sinusoids and a perfect vibration:
We can also visualize $u$ by its graph as a function of two variables:
Exercise. What if half of the string is made of a stiffer material?
The wave equation with respect to derivatives
We now consider the continuous case of wave propagation.
Suppose the function $u$ is defined for all $x$ and $t$ within some open subset $U$ of the plane and it is sampled at the nodes of a partition of rectangle $[0,b]\times [0,d]$ contained in that subset. We now utilize the results from the last section in the special case of a finite string of weights and springs:
These are our assumptions:
the array of $N$ identical weights; $m$ is constant,
the weights are distributed evenly over the length $L= N\Delta x$ of the string,
the total mass is $M= Nm$,
the springs are identical; $k$ is constant,
the total spring constant of the array is $K= k/N$,
the partition of the string is uniform: $\Delta x =\Delta s=h$,
the partition of the time interval is also uniform: $\Delta t =\Delta q$.
We can re-write the difference equation for wave propagation from the last section as follows: $$\begin{array}{ll} m\frac{u(t+\Delta t,x)-2u(t,x)+u(t-\Delta t,x) }{ (\Delta t)^2} \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad = k\big[u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)\big] \end{array}$$ Or, $$\begin{array}{ll} \frac{u(t+\Delta t,x)-2u(t,x)+u(t-\Delta t,x) }{ (\Delta t)^2} \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad= \frac{k(\Delta x)^2}{m}\frac{u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)}{(\Delta x)^2}\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad= \frac{k/N(N\Delta x)^2}{Nm}\frac{u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)}{(\Delta x)^2}\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad= \frac{KL^2}{M}\frac{u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)}{(\Delta x)^2}. \end{array}$$
Solving for $u(t+\Delta t,x)$, we obtain the recursive formula: $$\begin{array}{ll} u(t+\Delta t,x) &= 2u(t,x) - u(t-\Delta t,x) + \alpha\left(\frac{\Delta t}{\Delta x}\right)^2\big[u(t,x+\Delta x)-2u(t,x)+u(t,x-\Delta x)\big], \end{array}$$ where $$\alpha = \frac{KL^2}{M}.$$
Exercise. Set up and solve the IVP for the finite string, for the simplified settings. Hint: mind the ends.
We have also found the discrete wave equation: $$\frac{\Delta ^2 u}{\Delta t^2} = \alpha\frac{\Delta ^2 u}{\Delta x^2}.$$ It is a PDE with respect to second difference quotients.
We refine our partition of the rectangle and take the limit of this equation as $\Delta x\to 0$ and $\Delta t\to 0$. We have the wave equation: $$\begin{array}{|c|} \hline \\ \quad \frac{\partial^2 u}{\partial t^2} = \alpha\frac{\partial^2 u}{\partial x^2}. \quad \\ \\ \hline \end{array}$$ It is a PDE with respect to second derivatives.
A solution $u$ of this PDE is a function,
twice differentiable with respect to $t$ inside of it, and
In contrast to the heat transfer PDE, we will be able to find solutions for all possible initial conditions.
In order to solve this PDE, we introduce new variables: $$p = x + \alpha t \text{ and } q = x - \alpha t,$$ Then, $$x=\frac{1}{2}(p+q) \text{ and } t=\frac{1}{2\alpha}(p-q).$$
Exercise. What is the matrix of this linear transformation?
The new unknown function is the result of the substitution: $$v(p,q)=u(t,x)=u\left( \frac{1}{2}(p+q),\frac{1}{2\alpha}(p-q) \right),$$ and $$u(t,x)=v(x + \alpha t , x - \alpha t).$$
Theorem. Suppose $u$ is twice continuously differentiable and satisfies the wave equation. Then the mixed second partial derivative of $v$ is zero.
Proof. Let's first list the partial derivatives of the old variables with respect to the new ones: $$\frac{\partial x}{\partial p }=\frac{1}{2} ,\ \frac{\partial x}{\partial q }=\frac{1}{2} \text{ and } \frac{\partial t}{\partial p }=\frac{1}{2\alpha} ,\ \frac{\partial t}{\partial q }=-\frac{1}{2\alpha}.$$ Now let's differentiate $v$. We use the Chain Rule (Chapter 18) several times. First: $$\frac{\partial v}{\partial p }(p,q)=\frac{\partial}{\partial p } u(t,x)=\frac{\partial u}{\partial x }\frac{\partial x}{\partial p}+\frac{\partial u}{\partial t }\frac{\partial t}{\partial p}=\frac{\partial u}{\partial x }\frac{1}{2}+\frac{\partial u}{\partial t }\frac{1}{2\alpha}.$$ Second: $$\begin{array}{lll} \frac{\partial^2 v}{\partial q \partial p }(p,q)&=\frac{\partial}{\partial q }\left( \frac{\partial v}{\partial p }(p,q)\right)=\frac{\partial}{\partial q }\left( \frac{\partial u}{\partial x }\frac{1}{2}+\frac{\partial u}{\partial t }\frac{1}{2\alpha} \right),\quad\text{ substitute...}\\ &=\left(\frac{\partial^2 u}{\partial x^2 }\frac{\partial x}{\partial q}+\frac{\partial^2 u}{\partial t \partial x }\frac{\partial t}{\partial q}\right)\frac{1}{2} + \left(\frac{\partial^2 u}{\partial t\partial x }\frac{\partial x}{\partial q} + \frac{\partial^2 u}{\partial t^2 }\frac{\partial t}{\partial q}\right)\frac{1}{2\alpha}\\ &=\left(\frac{\partial^2 u}{\partial x^2 }\frac{1}{2}+\frac{\partial^2 u}{\partial x \partial t }\left(-\frac{1}{2\alpha}\right)\right)\frac{1}{2} + \left(\frac{\partial^2 u}{\partial t\partial x } \frac{1}{2} + \frac{\partial^2 u}{\partial t^2 }\left(-\frac{1}{2\alpha}\right)\right)\frac{1}{2\alpha} ,\quad\text{ by Clairaut's theorem}\\ &=\frac{\partial^2 u}{\partial x^2 }\frac{1}{2}\frac{1}{2} - \frac{\partial^2 u}{\partial t^2 }\frac{1}{2\alpha}\frac{1}{2\alpha},\quad\text{ and because }u\text{ satisfies the wave equation...}\\ &=0. \end{array}$$ $\blacksquare$
Solving the resulting PDE, $$\frac{\partial^2 v}{\partial p \partial q} = 0,$$ is easy. Indeed, any function that depends on only on $p$ satisfies it: $$v(p, q) = f(p).$$ And so does any function that depends only on $q$: $$v(p, q) = g(p).$$ We have then two solutions of the original PDE: $$u(t,x) = f(x + \alpha t) \text{ and } u(t,x) = g(x - \alpha t),$$ for any choice of $f$ and $g$. What are these solutions? They are the traveling waves that we saw in the last section: the shape remains the same as it is moving along the rope at the speed $\alpha$. The two above are a left traveling wave and a right traveling wave respectively.
The only difference is that this time the functions have to be differentiable.
Furthermore, the sum of two left/right traveling waves is a left/right traveling wave. Moreover, the sum of a left traveling wave and a right traveling wave is also a solution! Indeed, the mixed second derivative of $$v(p, q) = f(p) + g(q),$$ for any pair of arbitrary functions $f$ and $g$, is zero. Therefore, we have the following.
Theorem. For any two twice differentiable functions $f$ and $g$, the function $$u(t,x) = f(x + \alpha t) + g(x - \alpha t)$$ is a solution of the wave equation.
Exercise. Have we found all solutions?
Example (standing waves). What if the two traveling waves are identical? What if $f$ and $g$ are the same function? Let's choose: $$f(p)=g(p)= \sin(p).$$ Then we have two traveling waves, the right, $$u_1(t,x) = \sin(x - \alpha t),$$ and left, $$u_2(t,x) = \sin(x + \alpha t).$$ What is their sum $u$?
There seems to be no traveling! Let's confirm this algebraically. We have: $$u(t,x) = u_1(t,x) + u_2(t,x) = \sin(x - \alpha t)+\sin(x + \alpha t)=2\sin x \cos(\alpha t),$$ by the sum-to-product trigonometric identity, $$\sin a + \sin b = 2\sin\left({a+b \over 2}\right)\cos\left({a-b \over 2}\right).$$ Here the sinusoid of $u=2\sin x$ is stretched vertically by a time-dependent multiple, $\cos(\alpha t)$. This describes a wave that oscillates -- up and down -- but doesn't move horizontally. An oscillating spring is an example of this that we saw in the last section. $\square$
Next, we impose initial conditions on this function: one for the (vertical) location of each weight on the string and one for the (vertical) velocity. We have the following theorem.
Theorem (D'Alembert's Formula). The initial value problem of the wave equation, $$\begin{cases} \frac{\partial^2 u}{\partial t^2} = \alpha\frac{\partial^2 u}{\partial x^2};\\ u(0,x)=h_0(x);\\ \frac{\partial u}{\partial t}(0,x)=h_1(x); \end{cases}$$ has a solution: $$u(t,x) = \frac{h_0(x-\alpha t) + h_0(x+\alpha t)}{2} + \frac{1}{2\alpha} \int_{x-\alpha t}^{x+\alpha t} h_1(s)\, ds,$$ when
$h_0$ is twice continuously differentiable and
$h_1$ is continuously differentiable.
Exercise. Confirm the formula. Hint: split the integral into two.
Wave propagation in dimension $2$: a membrane
We model a membrane of a drum as well as the surface of a liquid:
Here, the pieces of the membrane are vertically displaced and the waves propagate horizontally.
Just as before, we represent this membrane by weights connected by springs but, this time it is not a string but an array:
On the left, we see the view from above and on the right from aside. We use a partition of this rectangle and this is the domain of $u$, which is a $0$-form.
The partition is given by the lengths of the edges: $\Delta x$ and $\Delta y$.
Recall that a partition of a box $B$ in the $xyt$-space comes from partitions of its three edges as described in Chapter 20 and earlier in this chapter: $$\begin{array}{lll} t_{0}&q_1& t_{1}&q_2& t_{2}&q_3& ...;\\ x_{0}&s_1& x_{1}&s_2& x_{2}&s_3& ...;\\ y_{0}&p_1& y_{1}&p_2& y_{2}&p_3& ....\\ \end{array}$$
We make a simplifying assumption that all weights are equal. We use the analysis of the $1$-dimensional case: if $H$ is the vertical component of the force $F$ of the spring, then $$H=−K\Delta_x u \text{ or }H=−K\Delta_y u,$$ depending on whether this spring is aligned with the $x$- or the $y$-axis. The forces exerted on the object at location $x$ are the four forces of the four springs attached to it. Each term is the difference: two with respect to $x$ and two with respect to $y$. The algebra that follows is identical to that we used for the heat equation in dimension $2$.
For the weight located at $(x_i,y_j)$, this is the total force at time $t_k$: $$F(t_k,x_i,y_j)=\left(\begin{array}{ccccc} \bullet& -K(t_k,s_{i},y_{j-1}) \Delta_yu\, (t_k,s_{i},p_{j-1})&\bullet\\ - K(t_k,s_{i-1},y_j) \Delta_x u\, (t_k,s_{i-1},y_j)&& + K(t_k,s_i,y_j) \Delta_x u\, (t_k,s_{i},y_j)\\ \bullet& +K(t_k,s_{i},y_{j}) \Delta_y u\, (t_k,s_{i},p_{j})&\bullet \end{array}\right).$$ The four terms are the forces of the four springs and they are arranged accordingly.
The difference equation for the (vertical) displacements of the weights $u=u(t,x,y)$ comes from the Second Newton's Law (mass times acceleration is force): $$m(x_i,y_j)\frac{\Delta^2 u}{\Delta t} (t_k,x_i,y_j)= F(t_k,x_i,y_j).$$ Since the acceleration is now known: $$a(t_k,x_i,y_j)=\frac{1}{m(x_i,y_j)}F(t_k,x_i,y_j),$$ we have the two recursive formulas for the velocity and the location derived just as in Chapter 25: $$\begin{array}{lll} v(q_k,x_i,y_j)&=v(q_{k-1},x_i,y_j)&+a(t_k,x_i,y_j)\Delta t_k,\\ u(t_k,x_i,y_j)&=u(t_{k-1},x_i,y_j)&+v(q_k,x_i,y_j)\Delta t_k. \end{array}$$
the stiffness for each spring,
the mass of each weight,
the initial location for each weight,
the next location for each weight (i.e., the velocity),
the first buffer (copied from the second buffer),
the second buffer (copied from the current values),
the difference of locations of the ends and the Hook's force for each spring,
the total force for each weight,
the current velocity, and finally
the current location for each weight.
Two examples are as follows. The domain is chosen to be the square $[1,40]\times [1,40]$ with $\Delta x=\Delta y=1$.
Example (drum). A drum is a circular membrane the edge of which is attached to a ring. In other words, the boundary condition is $$u(t,x,y)=0 \text{ when } x^2+y^2>20^2.$$ We choose the initial shape of the drum to be a rotated sinusoid: $$u(0,x,y)=\cos\left(\sqrt{(x-20)^2+(y-20)^2}\frac{\pi}{40}\right),$$ and zero initial velocity $$u(1,x,y)=u(0,x,y).$$
The drum skin shrinks to flat and then creates the same shape on the other side. $\square$
Example (breakwater). Seemingly having nothing to do with it, can our model reasonably reproduce waves of the surface of a liquid? Below we show a simulation of a breakwater protecting a harbor. It starts with a single "pulse" (tsunami?) outside the harbor, $$u(0,20,3)=1,\ u(0,x,y)=0\text{ for the rest of } (x,y),$$ and $$u(1,x,y)=0 \text{ for all } (x,y).$$ The breakwater is represented by a row of fixed values: $$u(t,x,10)=0 \text{ for all }x\ne 20,$$ with a single gap.
The large waves outside produce very mild, circular waves inside. $\square$
We repeat the algebra we used in our analysis of heat transfer for dimension $2$. When $K$ is constant, as another simplifying assumption, the total force is simplified: $$F(x_i,y_j,t_{k})=K\left( \begin{array}{ccccc} \bullet& -\Delta_yu\, (t_{k},s_{i},p_{j-1})&\bullet\\ - \Delta_x u\, (t_{k},s_{i-1},y_j)&& + \Delta_x u\, (t_{k},s_{i},y_j)\\ \bullet&+\Delta_y u\, (t_{k},s_{i},p_{j})&\bullet \end{array}\right)=K\big(\Delta_x^2 u\, (t_{k},x_{i},y_j)+\Delta_y^2 u\, (t_{k},x_{i},y_j)\big).$$ These are the second differences. Our partial difference equation becomes: $$\frac{\Delta^2 u}{\Delta t^2}=\frac{K}{m}\left(\Delta_x^2 u+\Delta_y^2 u\right),$$ where $\alpha>0$ is a constant.
Just as in the $1$-dimensional case, we can argue the following two points. First, a longer spring is less stiff than a shorter one made of the same material and, therefore, $K$ is inversely proportional to $\Delta x=\Delta y$. Second, the weights are, in fact, the weights of the springs and, therefore, $m$ is proportional to $\Delta x$. Then our difference equation becomes: $$\frac{\Delta^2 u}{\Delta t^2}=\frac{\alpha}{\Delta x^2}\left(\Delta_x^2 u+\Delta_y^2 u\right)=\alpha\left(\frac{\Delta^2 u}{\Delta x^2}+\frac{\Delta^2 u}{\Delta y^2}\right).$$
Furthermore, when $u$ is defined throughout the region, the difference equation corresponds to the wave equation for partial derivatives: $$\frac{\partial^2 u}{\partial t^2}=\alpha\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right),$$ with the last expression is, again, the Laplace operator of $u$.
$$\text{ THE END }$$
Retrieved from "https://calculus123.com/index.php?title=Partial_differential_equations&oldid=147"


text:
2 Former WWE Talents At The Royal Rumble
Both Mick Foley and BG James have had to adjust their schedule with the WrestleReunion events in Los Angeles (January) and Miami (March) due to commitments with WWE.
Foley, BG James and Billy Gunn are off the Sunday Legends breakfast for the Los Angeles WrestleReunion because Foley and James have to go to St. Louis for WWE's Royal Rumble. Instead, they will be doing a Q&A session that Friday at 4pm. For the Miami WrestleReunion, James and Gunn have been moved to the Saturday VIP session instead of Sunday, since James is needed at WrestleMania 28.
No word yet on WWE's plans for Foley at the Royal Rumble and James is reportedly working as an age
WWE Referee Entered Rehab – Details Inside
Brian James (a/k/a Road Dogg), who now works as a producer for WWE, stated during a recent "shoot" interview that not only has the organization paid to put him in a rehabilitation center, but has paid for at least one stint for all three of his brothers—Scott, Steve and Brad.
Scott is WWE referee Scott Armstrong. He returned to the organization in February 2011 after being released one year prior. Steve briefly competed for WWE in 1992 as Lance Cassidy. Brad was under contract to WWE in 2006 and 2007. He made sporadic appearances on ECW as a guest commentator.