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Revolution 1.09: "Kashmir"
"What makes you so sure I know what I'm doing? What makes you so sure I ever did?"
Miles and company are getting ever closer to Philly. They're at a rebel camp about 20 miles away, and Nora and Miles are trying to convince the rebels to help them get into the city. The two rebel leaders (hey there Reed Diamond!) are wailing on Miles pretty hard, but Miles promises them Monroe's head on a pike. Obviously, that's not something they can pass up. They make their move that night. Miles is getting pretty drunk, and Charlie tries to call him on it. He waves her off, saying he's just being realistic and they're probably all going to die anyway. It sounds like he's a little scared of going up against Monroe. They have to go under the walls around Philly to get in. Aaron is not thrilled about that.
As they're making their way through the tunnels with the rebels, Miles thanks one of the guys (Reed) for not killing him. Turns out Reed's character was a gambler pre-blackout and likes to take his chances. Meanwhile, Charlie learns a little about Miles and Monroe from Nora. They were best buddies until Miles tried (and choked) to assassinate Monroe. Guess that will end a friendship pretty quick. Unfortunately, their conversation is cut short when Charlie steps on a mine. Whoops. Nora manages to stall the trigger, and they bolt in time to avoid getting blown to bits. Not far off, in Philly, Rachel is building an amplifier for the pendant, and she has to give a demonstration to one of the Captains on Neville's orders. She's a bit cranky about having to stop work.
As the group keeps moving through the tunnels, Miles takes off after a militia scout. It turns out he wasn't there. Miles is not the only one having problems. As they are crossing through some waist-deep water, Nora thinks she's gotten bit by an alligator. Aaron quickly surmises that when the tunnel collapsed, it cut off the oxygen and they are slowly suffocating. They are living on borrowed time now. Miles gets them to the exit that should have been open, but it's been walled over. They head off in search of another route, and Miles is still tripping. He sees an open door and goes into what we assume is Monroe's office. That theory makes even more sense when Monroe shows up, and with a huge smile on his face, embraces Miles. They establish right away that Miles is hallucinating. They talk about why Miles left the militia and tried to off his bestie. Charlie pulls him out of it after Monroe says that Miles' dirty little secret is that he's scared if they get the Philly, he'll fall back into his old ways and abandon his family and friends. He's really having a tough time of it, poor guy. Charlie puts too much pressure on him. Her expectations are too high. Yes, he's awesome but he's also just a man. One who happens to be suffocating right along with her. Meanwhile, Aaron keeps seeing his wife. She's yelling at him, saying maybe he never loved her and that's why he left her all those years ago. He's really unnerved by the experience and is basically ignoring her in the hopes she'll go away.
As they head for a new exit, Reed's character spots a service tunnel. He tries to open it by ramming it with his shoulder. He gets it started and Miles finishes it. Not really a good idea, as it turns out Reed's character is undercover militia. He shoots most of the rebels that were with them and ushers Miles off at gunpoint. Apparently he was thrilled a while back to get a medal and a handshake from Miles. Miles doesn't remember him. It takes our crew a little while to get the door open again, but when they do, the rebel girl, Ashley (she's fighting for the US to honor her father), moves in with a bow. She shoots but misses. She gets shot for her trouble. Charlie moves in and manages to land one right in Reed's character's chest. Unfortunately, he has enough life left in him to pull the trigger. Charlie gets hit and falls, smacking her head pretty good on the stairs.
She comes to at home with her dad making dinner. She's confused and wonders if everything she'd been doing and seeing was really a dream or whether her present surroundings are what's fake. She begs her dad to tell her that he's not going anywhere again. He does and tells her to lie back down and get some rest. As he tells her to close her eyes, Miles is ordering her to open them. She's not complying, and Miles looks on the verge of tears. I like that we're getting to see a softer side of him. He and Charlie are starting to nicely off-set each other. It makes me like his character even more.
Charlie finally wakes up and seems to be okay. She tells Miles later (after 3 hours of sitting and resting) that she was in a place she didn't want to leave, and if he hadn't saved her, she'd probably still be there (aka dead or in a coma). They head off into the unknown. Monroe stop by to see Rachel, and he's brought Brad to see if Neville was right that she was lying about what she built. Brad takes a cursory look at the thing and declares it to be a bomb. Rachel is begging Bass to let her make a new amplifier, but he's done with her. Their deal is off and Danny is as good as dead. In a rather shocking move, Rachel stabs Brad in the gut and declares that now Monroe needs her. It's true that he does. I thought it was kind of a gutsy move. I really have to say I want to know what drove Miles and Monroe apart. I have a feeling it may have something to do with Rachel. Guess we'll be closer to finding out next week in the fall finale.
Posted by SBiglow at 10:48 AM
Once Upon a Time 2.08: "Into the Deep"
New Girl 2.08: "Parents"
Why the "Right" Age to Watch "New Girl" Shouldn't ...
HIMYM 8.07: "The Stamp Tramp"
Fringe 5.07: "Five-Twenty-Ten"
Person of Interest 2.07: "Critical"
Nashville 1.06: "You're Gonna Change (Or I'm Gonna...
Thanksgiving "Classic" Recap: "Buffy the Vampire S...
Arrow 1.06: "Legacies"
New Girl 2.07: "Menzies"
HIMYM 8.06: "Splitsville"
Celebrating Thanksgiving on Hypable!
Revolution 1.08: "Ties That Bind"
Book Review: "The Revolution Was Televised"
Once Upon a Time 2.07: "Child of the Moon"
Fringe 5.06: "Through the Looking Glass and What W...
Person of Interest 2.06: "The High Road"
Nashville 1.05: "Move It on Over"
Arrow 1.05: "Damaged"
HIMYM 8.05: "The Autumn of Break-ups"
Revolution 1.07: "The Children's Crusade"
Once Upon a Time 2.06: "Tallahassee"
Fringe 5.05: "An Origin Story"
Person of Interest 2.05: "Bury the Lede"
Nashville 1.04: "We Live in Two Different Worlds"
Arrow 1.04: "An Innocent Man"
New Girl 2.06: "Halloween"
Revolution 1.06: "Sex and Drugs"
Once Upon a Time 2.05: "The Doctor"


Gordon Arnaud Winter, OC (October 6, 1912 – August 1, 2003) was the sixth Lieutenant Governor of Newfoundland from 1974 to 1981.
In 1974, he was made an Officer of the Order of Canada.
In 1989, he headed the Winter Commission, the diocesan commission appointed by bishop Alphonsus Liguori Penney to write an inquiry about the clerical child sexual abuse scandal at Mount Cashel orphanage.


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Q: The page does not load when I press the tab link The page does not load when I first click on the 'All' tab. Blank page loading. It comes when clicked for the second time. How can I get it to load on the first click?

Fazla Mesai Talebi

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redpajama_set_name: RedPajamaStackExchange


It's odd that in all the years of blogging, I've never once blogged about Goyard's Saint Sulpice, a functional-enough SLG (in this case, a card holder) that's also very easy on the eyes. And small as it might be, it actually has 4 card slots (two on each side) with an additional slot in the middle where one could stash folded cash (yes, I've seen it used that way), or receipts and other paper bits and bobs that you can't throw away yet.
Available in all colours under Goyard's rainbow (that would be 11 in all if I remembered correctly), with two considered regular colours and the remaining 9 classified as special hues, which would explain why each bag or SLG in Goyard comes with 2 price tags. How to tell what's what? Black/Tan and Black is considered 'normal', everything else is 'special', and that's pretty much it really.
Hi, which one is more worth it the Louis Vuitton Porte Cartes Cardholder or the Goyard Black Cardholder?
If by 'worth it' you mean more affordable, then LV. If you want something more unique, then Goyard.
They do make belts for men and I believe women, but I know for at least the men's belts, you can only purchase at the Paris store as of 2 years ago.
Hey, I'm new to both your blog and Goyard. I'm trying to start a designer belt collection and I'd love to know Goyard's prices on belts. I only have a Damier Cobalt 40 mm from LV. Thanks!
Rachelle Yeo, Female, 25, Navy.
Yeremia Gerald, Male, 26, Burgundy.
Joolia Kim, Female, 25, Burgundy.
Aaron Marcus, Male, 37, Navy.


Child and Family Services is an independent nonprofit agency dedicated to advancing the well-being of children by providing an array of social services to strengthen family life and by promoting community commitment to the needs of children. Services offered: adoption, early childhood, teen, counseling, elder, family and in-home support.
The mission of the Nashua Pastoral Care Center is to offer support and provide services to assist individuals and families transition from crisis to self-sufficiency. The four core programs is the Transitional Housing, Prevention & Intervention of Homelessness, Crisis Advocacy & Resource Education (CARE), and Financial Literacy.
This network is composed of community based Child Care Resource and Referral programs.
Head Start is a national program that promotes school readiness by enhancing the social and cognitive development of children, through the provision of health education , nutritional, social and other services.
A non-profit family resource center, committed to the development of strong individuals and families in our community. The Upper Room is here to help families - all families - in need of education, support and resources to "make it" in today's world!
"Our mission is to support people who have developmental disabilities or acquired brain disorders, in partnership with families and caregivers, to enhance their independence and personal growth while fostering meaningful relationships and involvement within their communities."
Bridges is a non-profit agency dedicated to helping victims and survivors of domestic and sexual violence.
Not limited to Nashua residents.
24 Hour Crisis Hotline 1(800) 336-3795 that provides support and information as well as access to an emergency Shelter.
An Emergency Shelter is available 24 hours a days, seven days a week.
Support Groups, assistance with obtaining restraining orders and other court appearances.
Court Advocacy, in-person hospital support to victims of domestic and/or sexual violence.
Resource Center - offering books, videos, and information about domestic and sexual violence for victims, their families, and friends.


Anyway, this doesn't have so much to do with my gut telling me to follow a lead so much as it was my gut telling me that Milton emailing me and the record I had just ordered from the New Jersey archives were connected. Beyond the obvious way, in that John Reinhardt, whose record I ordered, is a relation of Milton, too. So the day before yesterday I heard from Milton. And today my dad picked me up for work and told me I had gotten an envelope from the New Jersey archivss.
Now, you give a government department enough information and you would hope they would be able to work quickly. Unfortunately, that's not always the case. But I gave the NJDARM a ton of information to track down John Reinhardt's death record, based on information I had gotten off of the FamilySearch website, and they got back to me...within a week. Hallelujah - maybe God does exist!
Anyway, the disappointing news is, this record doesn't tell me anything I didn't already know - John Reinhardt, my 5th great grandfather, died in Union, Hudson County, New Jersey in October of 1870. He was married, 56 years old, born in Germany, and died of "atrophy of the brain." Actually, that last part I did not know, and now that I'm thinking of it, 56 is awfully young to die from so-called brain atrophy - some kind of dementia perhaps? I'll need to look into that. Unfortunately, what I had really been hoping for, some reference to his as-so-far unknown parents, was not there, but it does do two huge things for me: one, it verifies the information I already had, like I said, and two, it tells me that FamilySearch's record collection (not their family tree collection - that's user-submitted and awfully inaccurate!!) *is* to be trusted, that I can use it as a reliable starting place to find out other stuff. Like when and where John's wife died. That information is there. Now I just have to look for an actual record as evidence.


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Oppo Ranked as Top Brand for After-sales Service in India: Counterpoint
Reported by Chakri K December 2nd, 2020 at 07:42 pm
Market research firm, Counterpoint, has released a new report analysing the after-sales service from top Indian brands. We all know that customer service has become a key factor in finalising a smartphone purchase. Samsung was rated as the best after-sales service provider in India for years now. However, that's changing as other brands are catching up on this front to gain consumer trust. The Counterpoint report highlights Oppo was the No.1 brand in customer satisfaction for after-sales service, followed by Vivo, Xiaomi and Samsung. The report named 'Smartphone After-sales Service Study' by Counterpoint also highlighted that customers spend an average amount of Rs 2,400 for servicing their out-of-warranty phones.
Chinese Brands Lead the After-sales Service in India: Counterpoint
Counterpoint stated that one in every four smartphone owners in Kolkata, Bengaluru, Chennai, Noida and Ahmedabad visit a service centre within six months of purchase. Customers are facing problems related to charging, software (hanging) and display. The survey has been conducted across Kolkata, Chennai, Bengaluru, Noida and Ahmedabad with 1,000 respondents who had taken an after-sales service from their respective smartphone brand.
As for the findings released by Counterpoint, eight out of ten respondents are satisfied with the after-sales service experience. Oppo was the No.1 brand in customer satisfaction for after-sales service with 93% of the respondents rating 'very good' or 'excellent.' Oppo is closely followed by Vivo, Xiaomi and Samsung with a customer satisfaction rating of 85%, 81% and 81%, respectively.
Furthermore, Oppo was also the fastest in after-sales turnaround. Both Oppo and Realme are close to each in the turnaround time aspect as 73% and 72% of the respondents received their device on the same day from Oppo and Realme. Vivo stood third on the list with 68% turnaround time. The Oppo customers also waited for the least at the service centre. Counterpoint says about half of the respondents were attended within 15 minutes of arrival at the Oppo service centre. Realme, Samsung and Xiaomi closely followed in this aspect.
Beating Samsung, Oppo maintained its spare parts better than the competitors. Xiaomi was rated highest on 'explanation of the problem' as well as solution. Xiaomi was slightly ahead in knowledge and soft skills of the customer support executives compared to other leading brands. Xiaomi also did well on solving software side issues or creating awareness to the customers with its larger online community. The Chinese company managed to solve several software issues on the go, which avoided a physical visit to the service centre. Xiaomi is followed by Vivo and Oppo here.
Lastly, nine out of ten respondents got their smartphones serviced within two visits. Oppo and Xiaomi users reported the least number of instances of having visited a service centre earlier with the same problem.
Chakri K Managing Editor
Chakri is a go-to guy for your next smartphone recommendation. Back in his engineering days, he used to play with smartphones by installing custom ROMs and that passion got him into the tech industry. He still goes nuts about a smartphone knocking his door for review. Currently managing everything at Telecom Talk, Chakri is trying to master PUBG Mobile in his free time.
BSNL Best Data STVs Offer Unlimited Data Without Speed Restrictions
Vodafone Idea Refarms 3G Spectrum for 4G in Bengaluru

redpajama_set_name: RedPajamaCommonCrawl


The most important annual event for the Jeep® Owners Group (JOG) and Jeep enthusiasts will take place from 13 - 15 July in the heart of Styria (Austria).
Three days of living exciting adventures in pure, authentic Jeep style.
New Wrangler - the direct descendant of the first Jeep vehicle - will be the star of the event and available for testing on challenging off-road courses.
Links to register and the full programme of the event is available at www.campjeep2018.com and on official Jeep social media channels.
Camp Jeep® 2018 - the biggest and most exciting event of the year for the Jeep Owners Group (JOG) and fans of the brand from across Europe - will be held in the Spielberg Arena at the Red Bull Ring in Austria on 13-15 July.
The stunning location will provide a fitting backdrop for the exciting annual gathering, which will feature an action-packed programme of Jeep thrills and adventures. And the undisputed star of this year's event will be the new Jeep Wrangler.
In addition to Wrangler test drives, those attending the fifth edition of Camp Jeep will experience a packed programme which includes the chance to drive the extraordinary 700 hp 6.2-litre V8 Grand Cherokee Trackhawk - the most powerful and fastest Jeep model ever - on the race track.
The incredible Jeep Jamboree-style atmosphere will be further enhanced with an array of shows, live music, barbecues and activities for children. And everything follows the same winning formula as the stars-and-stripes tradition of 4x4 gatherings that take place across the US, from California and Maine to Arizona and Tennessee, each year.
More than 30 such gatherings of the Jeep family now take place across the US each year, with the first having been held in 1953 on the legendary Rubicon Trail on the mountains between California and Nevada.


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Australia Challenge: 20 Questions to Test Your Expertise
Total Questions: 20 | Topic: Australia | Updated: Jan 12, 2023
Australia, officially the Commonwealth of Australia, is a sovereign country comprising the mainland of the Australian continent, the island of Tasmania, and numerous smaller islands.
From the basics to the most obscure Australia facts, this quiz has it all! With 20 questions covering a wide range of topics, it's the perfect way to test your Australia knowledge. Do you have what it takes to come out on top? So what are you waiting for?
Australia's flag is which of the following?
What was the population of Australia in 2020, given that it was 22,637,127 in 2012?
What is the name of the highest point in Australia, which is above sea level?
Mawson Peak
Federation Peak
Budj Bim
Australia's highest point is Mawson Peak with an elevation of 2,745 Meters above sea level.
Mawson Peak is an active volcanic summit of the Big Ben massif on Heard Island, an external Australian territory in the Indian Ocean.
Australia History Fact
Australia became the colonial ruler of the Territory of Papua (which had initially been annexed by Queensland in 1883) in 1902 and of the Territory of New Guinea (formerly German New Guinea) in 1920.
Who in the following pictures is the Prime Minister Of Australia?
The Prime Minister Of Australia is Anthony Albanese , a member of Australian Labor Party.
Anthony Albanese replaced Scott Morrison, who was Prime Minister Of Australia from May 23, 2022.
Anthony Norman Albanese is an Australian politician serving as the 31st and current prime minister of Australia since 2022.
Australia Government And Politics Fact
Australia has six states — New South Wales (NSW), Queensland (QLD), South Australia (SA), Tasmania (TAS), Victoria (VIC) and Western Australia (WA) — and three mainland territories—the Australian Capital Territory (ACT), the Northern Territory (NT), and the Jervis Bay Territory (JBT).
When was Australia founded?
What is the capital city of Australia?
Canberra is the capital city of Australia since Nov 30, 1926.
Canberra is the capital city of Australia. Founded following the federation of the colonies of Australia as the seat of government for the new nation, it is Australias largest inland city and the eighth-largest city overall.
Australia Culture Fact
Australia is one of five nations to have participated in every Summer Olympics of the modern era, and has hosted the Games twice: 1956 in Melbourne and 2000 in Sydney.
The Monarch Of Australia is Charles III.
Can you tell which of the following images shows Charles III?
The current Monarch Of Australia is Charles III. Charles III took office on Sep 8, 2022, replacing Peter Cosgrove.
Charles III is King of the United Kingdom and the 14 other Commonwealth realms.
Australia Geography And Environment Fact
Australia is the driest inhabited continent; its annual rainfall averaged over continental area is less than 500 mm.
Which of the following are public holidays in Australia?
Reformation Day
National School Shooting Day
If you visit Australia, you'd be able to celebrate the following public holidays:
Australia's size gives it a wide variety of landscapes, with tropical rainforests in the north-east, mountain ranges in the south-east, south-west and east, and desert in the centre.
I've learned that CN is the country code for People's Republic Of China, but I'm curious to know what the code for Australia is. Do you happen to know?
What is the retirement age in Australia?
What was the average number of children per woman in Australia in 2021?
(Hint: it was 1.86 in 2013)
I was wondering if you could tell me what the current VAT rate in Australia is?
Peter Dodds McCormick is the composer of Australia's anthem.
Do you know who wrote Australia's anthem lyrics?
Peter Dodds McCormick
Emilio Prud'Home
Australia's anthem was written by Peter Dodds McCormick.
Peter Dodds McCormick was an Australian schoolteacher and songwriter, known for composing the Australian national anthem, Advance Australia Fair.
Australia Demographics Fact
30.9% of Australia's population has attained a higher education qualification, which is among the highest percentages in the world.
Australia's lowest point is the Kati Thanda–Lake Eyre.
-12 Metres
Australia's lowest point is the Kati Thanda–Lake Eyre with an elevation of -15 Meters above sea level.
Lake Eyre, officially known as Kati Thanda–Lake Eyre, is an endorheic lake in east-central Far North South Australia, some 700 km (435 mi) north of Adelaide.
About 58% of Australians aged from 25 to 64 have vocational or tertiary qualifications and the tertiary graduation rate of 49% is the highest among OECD countries.
Which of the following bodies of water is located in or near Australia?
Arafura Sea
South Pacific Ocean
The following bodies of water are located near Australia:
Tasman Sea
Coral Sea
Timor Sea
At the 2021 census, the most commonly nominated individual ancestries as a proportion of the total population were: At the 2021 census, 3.2% of the Australian population identified as being Indigenous — Aboriginal Australians and Torres Strait Islanders.
Australia shares a sea or land border with Papua New Guinea & Indonesia. With which other location does Australia share a sea or land border with?
Australia shares a sea or land border with the following locations:
Australia Etymology Fact
The name Australia (pronounced /əˈstreɪliə/ in Australian English) is derived from the Latin Terra Australiscode: lat promoted to code: la ("southern land"), a name used for a hypothetical continent in the Southern Hemisphere since ancient times.
What is the slogan that Australia uses to summarize its mission?
Get All Right
There's NOTHING Like Australia
Country Above Self
Could you tell me which of the following organizations Australia has belonged to?
Dutch Language Union
The Danish Trade Council
The following organizations have been among those that Australia has been a part of:
Organization For Economic Cooperation And Development
Commonwealth Of Nations 1931
Asia-Pacific Economic Cooperation 1989
Asian Development Bank 1966
Australia Group 1985
Bank For International Settlements
Colombo Plan
Surrounded by the Indian and Pacific oceans, Australia is separated from Asia by the Arafura and Timor seas, with the Coral Sea lying off the Queensland coast, and the Tasman Sea lying between Australia and New Zealand.
Is this the national anthem of Australia?
The national anthem of Australia is Advance Australia Fair and has been since Apr 19, 1984.
Advance Australia Fair is the national anthem of Australia.
160,323 permanent immigrants were admitted to Australia in 2018–2019 (excluding refugees), whilst there was a net population gain of 239,600 people from all permanent and temporary immigration in that year.
What is the language officially spoken in Australia?
The official language of Australia is English.
English is a West Germanic language of the Indo-European language family, with its earliest forms spoken by the inhabitants of early medieval England.
Today, Australia has the world's eighth-largest immigrant population, with immigrants accounting for 30% of the population, the highest proportion among major Western nations.
Nauru Quiz: Can You Ace These Tough Questions?
British Empire Expert Challenge: Prove Your British Empire Prowess


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It's amazing. Good quality and very well made. Love love love it.
Loved it! Super fast delivery! Perfect item!
Very nice necklace! The customer service was outstanding in helping me with everything. I absolutely love the color!


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Home Arts Kultur Food for Thought
Try to rationalize gourmet comfort food in an uncomfortable world.
Richard Foss
Del Frisco's Double Eagle Steak House's 45-day dry-aged double-bone prime ribeye will set you back $125. Courtesy of Del Frisco's.
If you look at the amount of attention and energy devoted to the dining scene, it's hard to say that this isn't a golden age. It's not just that people are willing to spend more on their food, though they are. Right now, you can buy a $75 hamburger topped with caviar at Petrossian in Beverly Hills or a $150 burger with shaved truffles, wagyu beef, and lobster tail a few miles away at Nick and Stef's. Closer to home, B&B Butchers & Restaurant offers a $220 steak flight, and Del Frisco's Double Eagle Steak House's 45-day dry-aged double-bone prime ribeye will set you back $125. These might have been put on the menu as gimmicks, but they're actually being ordered.
Novelties such as this aside, there has been an explosion of places that offer wildly experimental dining experiences at nosebleed prices. Napa's French Laundry, where a meal can easily top $500 per person, mainstreamed culinary combinations that never existed in any culture, and many others have followed. The creativity is undeniable, and to some people, it's worth it. If you have an esoteric meal that you will remember years later, that changes the way you think about food, is that worth paying this kind of price? To some people, the answer is yes, and they are willing to travel long distances for peak experiences.
More importantly as a sign of our national evolution, in the last few decades, food from cultures all over the world is being treated with respect. This is true at all price levels and in all communities, and it is breaking down barriers in society. When the foodie crowd hears of a great dining experience, they'll venture into neighborhoods they'd usually avoid for the bragging rights among their social group. Locally, food explorers (read: affluent white people) frequently find their way into minority neighborhoods in the north, south, and east sides just for the sake of an Instagram shot and a little street cred among fellow "photographers." Both in traditional and fusion form, at budget cafes and pricy hotspots, there is interest in cuisines that were previously obscure. Some items have mainstreamed to the point where they're now just another American food. Consider the evolution of the traditional Mex-Mex taco. It was only 20 years ago that you could find only hard-shell, ground beef-based, gringo-friendly versions of the classic item on local menus. Now, every bar in West 7th and downtown offers something that passes for authentic barbacoa or pastor. Twenty years ago, the average American had never tasted Pad Thai or Korean-style short ribs, and now you can buy both in the frozen section at Costco. Neither is as good as you'd be served at a real Thai or Korean restaurant, but the Costco meatballs aren't as good as an Italian grandmother can make either.
Want further evidence of our fascination with food? I can tell you from personal experience that 20 years ago, if you photographed your meal in a restaurant, the entire staff was alerted that a critic was in the house. Nobody else did that. I bought a small camera and a jacket with large pockets, but sometimes I was still caught at it. Fast forward to now, and as soon as your plates arrive, at least one person at the table has a phone camera out. The shot is on the internet seconds later with a pithy comment. Food porn has been a thing since Gourmet Magazine pioneered luscious food photography in the 1940s, but it has become a sport of the masses. I don't even need to mention how food porn and the moving image have combined to create multiple TV channels and untold numbers of video streams about experiences near and far, but I will anyway, just to remind people that the term "celebrity culinary explorer" would have been so much gibberish only a decade or so ago. The fact that Anthony Bourdain's death affected so many people so deeply shows how he and his endeavors touched both the self-selected elite and the masses.
The obsession, one might even call it a mania, for all things culinary has spread to beverages. Classic cocktails went from drinks for old men to hipster accessories in an astonishingly short time, and new brands and flavors of the straight stuff proliferate. Beer became the new wine, and wine became a gold mine for boutique producers. Local draft cider is even a thing now. Farmers' markets went from natural-food-enthusiast destinations to tourist attractions with live entertainment and food booths tucked between stalls doing a good business in arcane and beautiful produce. The fact that this could happen in an era when fewer and fewer people actually cook from scratch marks a trend that is almost countercultural. So does the rising interest in food history, both online and in person. Museums are scheduling culinary programs to woo foodies through the doors, and organizations of culinary historians that welcome amateurs are seeing an influx of curious younger members.
Essentially, we're putting our creative energy, enthusiasm, and intelligence into thinking about food. How could anything possibly be bad about that? As a food writer and food historian, I can't see a problem, but as a citizen I think I do. The obsessive interest looks a lot like a reaction to the strident, brutal, confrontational rhetoric that is all around us. Thoughtful, sensitive, and aware people are so repelled by current events that they tune out and seek something else. They may focus on food, music, sports, or some other passion, and that's healthy if it's really a respite from the daily grind and they reemerge invigorated.
I worry that they won't. The people who flocked to Berlin cabarets in the 1930s to get away from the Brownshirts in the streets, or who attended Dada art shows in the new Bauhaus-style buildings, undoubtedly thought that sanity would return soon. The Romans who debated abstruse philosophy while Caligula declared himself a god were no doubt happy to ignore the crowned buffoon who was sinking deeper into madness. (And if anyone is casting a historical play about that era, I have a great idea for the costuming. It involves a wig in a color and style not found in nature.)
It's unusual for me, a person who has put over three decades into the study of cuisine, to give this warning, but I have a platform, so I must do it. Escapism has its dangers. I urge you to go out and explore the cuisines of the world, our country, and our neighborhoods and teach yourself to recreate the items you like at home. While you're in those restaurants and ethnic grocery stores, you might even try to learn a bit more about the culture that is represented, because while their food is mainstreaming, so are those people. They are our neighbors and citizens, too, and understanding them will be useful in the country we share. Go and learn, maybe check out some books on the topic, but at some point, shut the books and turn off the computer. Reengage with the world outside and put some effort into making it a better one. And after you do, reward yourself with a good dinner. You will have earned it.
Richard Foss is an author and historian who consults with museums on culinary programs.
Fort Worth Weekly associate editor Eric Griffey contributed to this story.
Del Frisco's Double Eagle Steak House
food historian
Napa's French Laundry
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Nyoka leads an expedition into the most remote part of the Libyan desert in search of her father, Professor Henry Gordon, who disappeared while seeking out the long-lost golden tablets of Hippocrates. The tablets, among other attributes, are reputed to contain the cure for cancer and other diseases. Nyoka and her expedition face many brushes with death in their quest for the tablets.


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Park Soo-hong and Kim Da-ye defamation Kim Yong-ho
작성일자 1월 4, 2023 글쓴이 admin
Park Soo-hong and Kim Da-ye defamation Kim Yong-ho visited and saw it for the first time in real life.
Broadcaster Park Soo-hong's wife Kim Da-ye revealed why she started a legal dispute with YouTuber Kim Yong-ho, who defamed her.
In the pre-released video of TV Chosun's "Joseon's Lovers" released on the 3rd, Park Soo-hong's wife Kim Da-ye was forced to go to court for the first time in her life.
On the same day, Kim Da-ye drove himself to the court. "The court is the first time I've seen it in my life," he said, explaining why he took direct action, saying, "I'm so curious why a stranger spread it to people he didn'
Kim Da-ye, who got out of the car and headed to the court, said, "It's my first time seeing it in person. "I only saw it on the name and the broadcast screen," he said, drawing attention.
Currently, Park Soo-hong and Kim Da-ye are in a legal dispute with YouTuber Kang Yong-ho. YouTuber Kim Yong-ho is accused of defamation of Park Soo-hong and his spouse 31 times by appearing on his YouTube channel "Kim Yong-ho's Entertainment Manager" and broadcasting at the Garosero Research Institute. Kim Yong-ho said on a YouTube broadcast in April last year, "Park Soo-hong's spouse was in a relationship with Yoo, the former CEO of a wet tissue company, and Yoo's suicide is related to A. Park Soo-hong even married the woman of his dead best friend," he said.
Later, Park Soo-hong said through his legal representative, "Park Soo-hong's spouse Kim Da-ye was unable to engage in social activities due to Kim Yong-ho's false claim that he had dated former Monde CEO Yoo, who had no acquaintance, and suffered from circular hair loss and panic disorder. He also complained of pain, saying, "Park Soo-hong's father-in-law recently underwent surgery due to the risk of vision loss due to stress caused by false information."
In particular, Kim Da-ye is said to have watched the proceedings of the trial in the audience with his legal representative on the first trial date held in November. When Kim Yong-ho denied all the allegations, Kim Da-ye reportedly gave a deep sigh.
On the other hand, Park Soo-hong held a wedding ceremony with Kim Da-ye, 23 years younger than him, at a charity event on the 23rd of last month. A late wedding ceremony was held a year and five months after the marriage registration due to the remission of family members and the aftermath of COVID-19.


text:
\section{Introduction}

Random cell movement plays a very important role in embryonic morphogenesis,
wound healing, cancer cells proliferation, and many other physiological and
pathological processes \cite{R}. The microscopic theory of the migration of
cells and bacteria towards a favorable environment (chemotaxis) is based on
random walk models \cite{EO,Hillen,Alt,OS1}. The \textquotedblleft
velocity-jump\textquotedblright\ models concern with self-propelled motion
involving the runs and tumbles, while \textquotedblleft
space-jump\textquotedblright models deal with the cells making jumps in
space. Much of the literature on the theoretical studies of cells motility
has been concerned with Markov random walk models (see, for example, \cit
{Baker1,EO}). However, the analysis of random movement of wild-type and
mutated epithelial cells shows the anomalous dynamics of cell migration \cit
{D} (see also \cite{M}). Over the past few years there have been several
attempts to model non-Markovian anomalous cell transport involving
subdiffusion and superdiffusion \cite{D,Fed0,Fed00,Fed1,Fed2,HL}. In this
paper we shall deal with a non-Markovian \textquotedblleft
space-jump\textquotedblright model that describes the non-homogeneous in
space subdiffusive transport of cells.

\subsection{Markov random walk model.}

First let us consider a Markov model for random cell movement along
one-dimensional lattice such that all steps are of equal length $1$. We
define the probabilit
\begin{equation}
p(k,t)=\Pr \left\{ X(t)=k\right\}
\end{equation
that the position of cell $X(t)$\ is at point $k\in \mathbb{Z}$\ at time $t.
We introduce at each point $k$\ the rate of jump to the left $\mu (k)$\ and
the rate of jump to the right $\lambda (k)$. This random walk is called a
generalized birth-death process \cite{Cox}. The master equation for $p(k,t)
\ can be written as
\begin{equation}
\frac{\partial p(k,t)}{\partial t}=\lambda (k-1)p(k-1,t)+\mu
(k+1)p(k+1,t)-\left( \lambda (k)+\mu (k)\right) p(k,t).  \label{Master11}
\end{equation
This model corresponds to the case when intervals between jumps at point $k$
are exponentially distributed with parameter $\lambda (k)+\mu (k).$ When the
cell makes a jump from the position $k,$\ it jumps to the right with
probability $\lambda (k)/(\lambda (k)+\mu (k))$ and to the left with the
probability $\mu (k)/(\lambda (k)+\mu (k))$ \cite{Cox}.

The dependence of $\mu (k)$ and $\lambda (k)$ on space can be introduced in
different ways depending on how cells sense the surrounding environment. For
the local chemotaxis models, the rates $\lambda (S(k))$\ and $\mu (S(k))$
are the functions of the local concentration of the chemotactic substance
S(k).$ There exist several non-local and barrier models that are different
in terms of the dependence of rate functions on the chemotactic substance
\cite{Baker1,OS1}. For example, the rates $\mu (k)$ and $\lambda (k)$ can
depend on the concentration of the chemotactic substance at neighbouring
positions $k-1$ and $k+1$ as in (\ref{nonlocal}). In the continuous limit,
the master equation (\ref{Master11}) can be reduced to the classical
advection-diffusion equation in which the cell flux involves the standard
diffusion term and the advection term due to chemotaxis.

If we consider only positive values of $k,$ we need to implement boundary
conditions at the point $k=1$. Here we assume that if cell hits the wall on
the boundary, it is reflected with the probability $1-\chi $ and absorbed by
wall with the probability $\chi .$ Then one can write $p(1,t+\Delta
t)=(1-\lambda (1)\Delta t-\mu (1)\Delta t)p(1,t)+$ $\mu (1)(1-\chi
)p(1,t)\Delta t+\mu (2)p(2,t)\Delta t+o(\Delta t).$ In the limit $\Delta
t\rightarrow 0$ we obtain
\begin{equation}
\frac{\partial p(1,t)}{\partial t}=-\chi \mu (1)p(1,t)+\mu (2)p(2,t)-\lambda
(1)p(1,t),  \label{bound}
\end{equation
where $0\leq \chi \leq 1.$

Non-uniform stationary solution of master equation (\ref{Master11}) can be
interpreted as cell aggregation phenomenon \cite{OS1}. In particular, if
there is no absorption on the boundary ($\chi =0$), the stationary solution
p_{st}(k)$ can be easily found from (\ref{Master11}) and (\ref{bound}). We
obtain
\begin{equation}
p_{st}(k)=p_{st}(1)\prod\limits_{i=1}^{k-1}\frac{\lambda (i)}{\mu (i+1)
,\qquad k>1,  \label{ma}
\end{equation
wher
\begin{equation*}
p_{st}(1)=\left( 1+\sum\limits_{k=2}^{\infty }\prod\limits_{i=1}^{k-1}\frac
\lambda (i)}{\mu (i+1)}\right) ^{-1}
\end{equation*
provided the series is convergent.

\subsection{Anomalous random walks}

It is tempting to generalize the master equation (\ref{Master11}) for the
anomalous case by replacing the time derivative with the Caputo derivative
\cite{Klages,Mee,Met2}
\begin{equation}
\frac{\partial ^{\nu }p\left( k,t\right) }{\partial t^{\nu }}=\frac{1}
\Gamma (1-\nu )}\int_{0}^{t}\frac{\partial p\left( k,u\right) }{\partial u
\frac{du}{(t-u)^{1-\nu }}  \label{Caputo}
\end{equation
as it is done in \cite{FBD} for a fractional linear birth--death process.
Here $\nu $ is the anomalous exponent: $0<\nu <1.$ Although this
generalization is very attractive from a mathematical point of view, it is
not appropriate for a non-homogeneous medium for which the exponent $\nu $
depends on $k$. The non-homogeneous fractional equation for $p(k,t)$ can be
written as
\begin{equation}
\frac{\partial p(k,t)}{\partial t}=a(k-1)\mathcal{D}_{t}^{1-\nu
(k-1)}p(k-1,t)+b(k+1)\mathcal{D}_{t}^{1-\nu (k+1)}p(k+1,t)-(a(k)+b(k)
\mathcal{D}_{t}^{1-\nu (k)}p(k,t),  \label{dis1}
\end{equation
where \bigskip $\mathcal{D}_{t}^{1-\nu (k)}$ is the Riemann-Liouville
fractional derivative with varying orde
\begin{equation}
\mathcal{D}_{t}^{1-\nu (k)}p\left( k,t\right) =\frac{1}{\Gamma (\nu (k))
\frac{\partial }{\partial t}\int_{0}^{t}\frac{p\left( k,u\right) du}
(t-u)^{1-\nu (k)}}.  \label{RL}
\end{equation
Here $\nu (k)$ is the anomalous exponent corresponding to the site $k$ and
the anomalous rate coefficients $a(k)$ and $b(k)$ have to be determined, see
(\ref{ab}). The crucial point here is that the anomalous exponent $\nu (k)$
depends on the site $k$. The fractional equation (\ref{dis1}) cannot be
rewritten in terms of Caputo derivative (\ref{Caputo}). It turns out that
even small non-homogeneous variations of the exponent $\nu $ lead to a
drastic change of $p(k,t)$ in the limit $t\rightarrow \infty $ \ \cite{Fed3
. It means that the subdiffusive fractional equations with constant
anomalous exponent $\nu $ are\textit{\ not structurally stable}. If, for
example, the point $k=M$ has the property that $\nu (M)<\nu (k)$ for all
k\neq M$ and $\chi =0,$ one can find that
\begin{equation}
p(k,t)\rightarrow 0,\qquad p(M,t)\rightarrow 1,\qquad 1\leq k\leq N
\end{equation
as $t\rightarrow \infty .$ This result has been interpreted as anomalous
aggregation of cells at the point $k=M$ \cite{Fed1}. In this paper we shall
find the conditions for anomalous aggregation for the semi-infinite interval
$1\leq k<\infty .$ It should also be noted that non-homogeneous variations
of the exponent $\nu $ destroy the Gibbs-Boltzmann distribution as a long
time limit of the fractional Fokker-Planck equation \cite{Fed3}. Of course,
for the constant value of $\nu ,$ the formulation in terms of Caputo or
Riemann-Liouville operators are equivalent, as long as proper care is taken
of the initial values \cite{Klages,Mee,Met2}.

\subsection{Anomalous diffusion with reaction.}

Another extension of traditional Markov random walks models is non-Markovian
theory of anomalous transport with reaction dynamics \cit
{Fed000,MFH,Nep,Sokolov,Sh,Vo}. In particular, this theory has been used for
the analysis of the proliferation and migration dichotomy of cancer cells
\cite{Fed0,Fed00,Fed2}. In this paper we consider the inhibition of cell
growth by anticancer therapeutic agents. To model this inhibition we
introduce the random death process with non-uniform death rate parameter. We
assume that during time interval $\left( t,t+\Delta t\right) $ at point $k$\
each cell has a chance $\theta (k)\Delta t+o(\Delta t)$ of dying, where
\theta (k)$ is the death rate \cite{Iomin}. It is easy to take into account
this process for Markov models. We just add the term $-\theta (k)p(k,t)$ to
the right hand side of the master equation (\ref{Master11}). On the
contrary, the anomalous master equation involves a non-trivial combination
of transport and death kinetic terms because of memory effects \cit
{HLW,MFH,Abad}. In this paper we shall derive the following fractional
equation
\begin{eqnarray}
\frac{\partial p(k,t)}{\partial t} &=&a(k-1)e^{-\theta (k-1)t}\mathcal{D
_{t}^{1-\nu (k-1)}\left[ p(k-1,t)e^{\theta (k-1)t}\right]  \notag \\
&&+b(k+1)e^{-\theta (k+1)t}\mathcal{D}_{t}^{1-\nu (k+1)}\left[
p(k+1,t)e^{\theta (k+1)t}\right]  \notag \\
&&-\left( a(k)+b(k)\right) e^{-\theta (k)t}\mathcal{D}_{t}^{1-\nu (k)}\left[
p(k,t)e^{\theta (k)t}\right] -\theta (k)p(k,t).  \label{master30}
\end{eqnarray}

\subsection{Mean field master equation for the density of cells.}

Instead of the probability $p(k,t)$ for an individual cell one can consider
the mean density of cells $\rho (x,t)$ as a function of space $x$ and time
t $. The master equation (\ref{Master11}) can be rewritten as the equation
for the density $\rho (x,t)$ by changing the variables as $k\rightarrow x$
and $k\pm 1\rightarrow x\pm l$:
\begin{equation}
\frac{\partial \rho (x,t)}{\partial t}=\lambda (x-l)\rho (x-l,t)+\mu
(x+l)\rho (x+l,t)-(\lambda (x)+\mu (x))\rho (x,t)-\theta (x)\rho (x,t),
\label{Master10}
\end{equation
where $l$ is the jump size, $\theta (x)$ is the death rate. The advantage of
this equation is that one can easily take into account various non-linear
effects by assuming the dependence of the rate functions $\lambda (x),\mu
(x) $ and $\theta (x)$ on the average density $\rho (x,t).$

In the anomalous subdiffusive case, the master equation for mean field $\rho
(x,t)$ can be obtained from (\ref{master30}). It can be written as a mass
balance equation
\begin{equation}
\frac{\partial \rho (x,t)}{\partial t}=-I(x,t)+I(x-l,t)-\theta (x)\rho (x,t),
\label{balance}
\end{equation
where $I(x,t)$ is the total flow of cells from the point $x$ to $x+l$
\begin{equation}
I(x,t)=a(x)e^{-\theta (x)t}\mathcal{D}_{t}^{1-\nu (x)}\left[ e^{\theta
(x)t}\rho (x,t)\right] -b(x+l)e^{-\theta (x+l)t}\mathcal{D}_{t}^{1-\nu (x+l)
\left[ e^{\theta (x+l)t}\rho (x+l,t)\right]  \label{flux}
\end{equation
and $I(x-l,t)$ is the total flow of cells from the point $x-l$ to $x$
\begin{equation}
I(x-l,t)=a(x-l)e^{-\theta (x-l)t}\mathcal{D}_{t}^{1-\nu (x-l)}\left[
e^{\theta (x-l)t}\rho (x-l,t)\right] -b(x)e^{-\theta (x)t}\mathcal{D
_{t}^{1-\nu (x)}\left[ e^{\theta (x)t}\rho (x,t)\right] .
\end{equation
Here $a(x)$ and $b(x)$ are the anomalous rate functions, see (\ref{ab}). One
can see that the flow of cells $I(x,t)$ depends on the death rate $\theta
(x) $. It means that in the anomalous case one cannot separate the transport
of cells from the death process \cite{HLW}. This phenomenon does not exist
in the Markovian case. For the Markov model (\ref{Master10}) the flux $I(x,t)
$ is independent from $\theta (x):$
\begin{equation}
I(x,t)=\lambda (x)\rho (x,t)-\mu (x+l)\rho (x+l,t).  \notag
\end{equation
When the density $\rho (x,t)$ is conserved ($\theta =0$), the master
equation (\ref{balance}) can be approximated by the fractional Fokker-Planck
equation \cite{Klages,Met1,Met2
\begin{equation}
\frac{\partial \rho (x,t)}{\partial t}=-\frac{\partial }{\partial x}\left[
l(a(x)-b(x))\mathcal{D}_{t}^{1-\nu (x)}\rho (x,t)\right] +\frac{\partial ^{2
}{\partial x^{2}}\left[ \frac{l^{2}}{2}(a(x)+b(x))\mathcal{D}_{t}^{1-\nu
(x)}\rho (x,t)\right] .  \label{FFP}
\end{equation
This is an example of the fractional equation with varying anomalous
exponent \cite{Ch}. Note that $a(x)-b(x)\sim l$ as $l\rightarrow 0$, see
\ref{dif2}).

The purpose of the next section is to set up a non-Markovian discrete-space
random walk model describing cell motility involving memory effects, the
death process and subdiffusive transport.

\section{Non-Markovian discrete-space random walk model}

\subsection{ Random cell motility}

There exist numerous mechanisms that facilitate random cell movement \cite{R
. In this paper we adopt the following random model of cell motility. When
the cell makes a jump to position $k$, the time the cell spends here before
it makes a jump to point $k-1$ or $k+1$ is random. It is called the
residence time or waiting (holding) time. We define the residence time at
position $k$ as
\begin{equation}
T_{k}=\min \left( T_{k}^{\mu },T_{k}^{\lambda }\right) ,  \label{min}
\end{equation
where $T_{k}^{\mu }$ and $T_{k}^{\lambda }$ are the independent random times
of jump to the left and right respectively. The idea here is that there
exist internal cellular signals involving two "hidden" independent\ random
alarm clocks. If one of the clocks goes off first, say $T_{k}^{\lambda
}<T_{k}^{\mu }$, the cell moves to the right to the point $k+1$. The other
clock "tells" the cell to move left to the point $k-1$ if it goes off first
$T_{k}^{\mu }<T_{k}^{\lambda })$. Note that migration of cells is a highly
complicated dynamic process which is regulated by both intercellular signals
and the surrounding environment. Since we do not know the exact mechanism of
cell motility we use a stochastic approach involving two random times
T_{k}^{\mu }$ and $T_{k}^{\lambda }$ for jumping to the left and right. Note
that if the random times $T_{k}^{\mu }$ and $T_{k}^{\lambda }$ are
exponentially distributed with the rates $\mu (k)$ and $\lambda (k)$
respectively, we have a classical Markov model with the master equation (\re
{Master11}). If the random variables $T_{k}^{\mu }$ and $T_{k}^{\lambda }$
are not exponentially distributed, the standard Markov approach does not
work. In this section we consider the non-Markovian case when $T_{k}^{\mu }$
and $T_{k}^{\lambda }$ are independent positive random variables with
general survival functions
\begin{equation}
\Psi _{\mu }(k,\tau )=\Pr \left\{ T_{k}^{\mu }>\tau \right\} ,\qquad \Psi
_{\lambda }(k,\tau )=\Pr \left\{ T_{k}^{\lambda }>\tau \right\} .
\label{sur55}
\end{equation
The Markov model (\ref{Master11}) corresponds to the following choice
\begin{equation}
\Psi _{\mu }(k,\tau )=e^{-\mu (k)\tau },\qquad \Psi _{\lambda }(k,\tau
)=e^{-\lambda (k)\tau }.
\end{equation
It is convenient to introduce the rate of escape (hazard function) $\gamma
(k,\tau )$ from the point $k$ as
\begin{equation}
\gamma (k,\tau )=\lim_{h\rightarrow 0}\frac{\Pr \left\{ \tau <T_{k}<\tau
+h\;|_{T_{k}>\tau }\right\} }{h}.  \label{tran}
\end{equation
If we denote the survival function at the point $k$ as
\begin{equation*}
\Psi (k,\tau )=\Pr \left\{ T_{k}>\tau \right\}
\end{equation*
and the residence time probability density function a
\begin{equation*}
\psi (k,\tau )=-\frac{\partial \Psi (k,\tau )}{\partial \tau },
\end{equation*
then \cite{Cox}
\begin{equation}
\gamma (k,\tau )=\frac{\psi (k,\tau )}{\Psi (k,\tau )}.  \label{def}
\end{equation
Now we determine this rate function in terms of statistical characteristics
of random residence times $T_{k}^{\mu }$ and $T_{k}^{\lambda }.$ It follows
from the definition of the residence time $T_{k}$ at position $k$ (\ref{min
) that the survival function $\Psi (k,\tau )$ can be written as a produc
\begin{equation*}
\Psi (k,\tau )=\Psi _{\lambda }(k,\tau )\Psi _{\mu }(k,\tau ),
\end{equation*
where $\Psi _{\lambda }(k,\tau )$ and $\Psi _{\mu }(k,\tau )$ are defined by
(\ref{sur55}). Differentiation of this equation with respect to $\tau $ give
\begin{equation}
\psi (k,\tau )=\psi _{\lambda }(k,\tau )+\psi _{\mu }(k,\tau ),  \label{wtd}
\end{equation
where the transition densities  $\psi _{\lambda }(k,\tau
) $ and $\psi _{\mu }(k,\tau )$ are defined as
\begin{equation}
\psi _{\lambda }(k,\tau )=-\frac{\partial \Psi _{\lambda }(k,\tau )}
\partial \tau }\Psi _{\mu }(k,\tau ),\qquad \psi _{\mu }(k,\tau )=-\frac
\partial \Psi _{\mu }(k,\tau )}{\partial \tau }\Psi _{\lambda }(k,\tau ).
\end{equation
The formula (\ref{wtd}) is the particular case of the general expression for
the residence time PDF in terms of the transition densities (see
formula (5) in the classical paper by van Kampen \cite{Kampen}). These
transition densities have a clear probabilistic meaning. For example, $\psi
_{\mu }(k,\tau )\Delta \tau $ is the probability that the cell's jump to the
left occurs in the time interval $(\tau ,\tau +\Delta \tau )$ since the cell
arrived at point $k$. If we divide both sides of (\ref{wtd}) by the survival
function $\Psi (k,\tau )$ and use the formula (\ref{def}), we obtain
\begin{equation}
\gamma (k,\tau )=\lambda (k,\tau )+\mu (k,\tau ),  \label{rate}
\end{equation
where the rate of jump to the right $\lambda (k,\tau )$ and the rate of jump
to the left $\mu (k,\tau )$ are defined as
\begin{equation}
\lambda (k,\tau )=\frac{\psi _{\lambda }(k,\tau )}{\Psi (k,\tau )},\qquad
\mu (k,\tau )=\frac{\psi _{\mu }(k,\tau )}{\Psi (k,\tau )}.  \label{sur5}
\end{equation
Note that the transition rates $\lambda (k,\tau )$ and $\mu (k,\tau )$ can
be introduced from the very beginning as it is done in \cite{Kampen}. By
using (\ref{def}) and (\ref{rate}), we write the survival function $\Psi
(k,\tau )$ as
\begin{equation}
\Psi (k,\tau )=e^{-\int_{0}^{\tau }\lambda (k,\tau )d\tau -\int_{0}^{\tau
}\mu (k,\tau )d\tau }.  \label{s3}
\end{equation
The residence time probability density function $\psi (k,\tau )$ \bigskip
takes the form
\begin{equation*}
\psi (k,\tau )=(\lambda (k,\tau )+\mu (k,\tau ))e^{-\int_{0}^{\tau }\lambda
(k,\tau )d\tau -\int_{0}^{\tau }\mu (k,\tau )d\tau }.
\end{equation*
For the Markov case for which $\lambda (k)$ and $\mu (k)$ are independent of
the residence time variable $\tau ,$ we obtain from (\ref{s3}) the standard
exponential survival function
\begin{equation*}
\Psi (k,\tau )=e^{-\lambda (k)\tau -\mu (k)\tau }
\end{equation*
corresponding to the Markov master equation (\ref{Master11}).

\subsection{ Structured probability density function}

If the residence time probability density function $\psi $ is not
exponential, the random walk is non-Markovian. The standard method to deal
with non-Markovian stochastic processes is to add auxiliary variables to the
definition of the random walk to make the process Markovian \cite{Cox}. Here
we introduce the structured probability density function $\xi (k,t,\tau )$
involving residence time $\tau $ as auxiliary variable. The structural
density gives the probability that the cell position $X(t)$ at time $t$ is
at the point $k$ and its residence time $T_{k}$ at point $k$ is in the
interval $(\tau ,\tau +d\tau ).$ This is a standard way to deal with
non-Markovian random walks \cite{Cox,MFH}. Suppose that cells die at random
at rate $\theta (k)$ that depends on $k.$ The density $\xi (k,t,\tau )$
obeys the balance equation
\begin{equation}
\frac{\partial \xi }{\partial t}+\frac{\partial \xi }{\partial \tau
=-\lambda (k,\tau )\xi -\mu (k,\tau )\xi -\theta (k)\xi .  \label{basic}
\end{equation
We consider only the case when the residence time of random walker at $t=0$
is equal to zero, so the initial condition is
\begin{equation}
\xi (k,0,\tau )=p_{0}(k)\delta (\tau ),  \label{initial}
\end{equation
where $p_{0}(k)=$ $\Pr \left\{ X(0)=k\right\} $. The boundary condition in
terms of residence time variable ($\tau =0)$ can be written as \cite{Cox}
\begin{equation}
\xi (k,t,0)=\int_{0}^{t}\lambda (k-1,\tau )\xi (k-1,t,\tau )d\tau
+\int_{0}^{t}\mu (k+1,\tau )\xi (k+1,t,\tau )d\tau .  \label{arr}
\end{equation
In what follows we consider only positive values of $k$. In which case, we
have to specify the boundary condition for $k=1$. We write
\begin{equation}
\xi (1,t,0)=(1-\chi )\int_{0}^{t}\mu (1,\tau )\xi (1,t,\tau )d\tau
+\int_{0}^{t}\mu (2,\tau )\xi (2,t,\tau )d\tau .  \label{k=0}
\end{equation
This equation tells us that when cells escape from the point $k=1$ and move
to the left with the rate $\mu (1,\tau )$, they are adsorbed by the wall
with probability $\chi $, and reflected back to the position $k=1$ with the
probability $1-\chi .$ Note that this boundary condition can be written in
many different ways, for example, the cells can be reflected to state $k=2$.
One can also introduce a residence time PDF for a wall such that the
reflection is not instantaneous.

We solve (\ref{basic}) by the method of characteristic
\begin{equation}
\xi (k,t,\tau )=\xi (k,t-\tau ,0)e^{-\int_{0}^{\tau }\lambda (k,\tau )d\tau
-\int_{0}^{\tau }\mu (k,\tau )d\tau -\theta (k)\tau },\quad \tau <t,\quad
k\geq 1.  \label{s1}
\end{equation
The structural density $\xi $ can be rewritten in terms of the survival
function $\Psi (k,\tau )$ (\ref{s3}) and the integral arrival rate
\begin{equation*}
j(k,t)=\xi (k,t,0)
\end{equation*
a
\begin{equation}
\xi (k,t,\tau )=j\left( k,t-\tau \right) \Psi (k,\tau )e^{-\theta (k)\tau
},\quad \tau <t,\quad k\geq 1.  \label{je}
\end{equation
Our purpose now is to derive the master equation for the probability
p(k,t)=\Pr \left\{ X(t)=k\right\} :
\begin{equation}
p(k,t)=\int_{0}^{t}\xi (k,t,\tau )d\tau ,\quad k\geq 1.  \label{denG}
\end{equation
Let us introduce the integral escape rate to the right $i_{\lambda }(k,t)$
and the integral escape rate to the left $i_{\mu }(k,t)$ a
\begin{equation}
i_{\lambda }(k,t)=\int_{0}^{t}\lambda (k,\tau )\xi (k,t,\tau )d\tau ,\qquad
i_{\mu }(k,t)=\int_{0}^{t}\mu (k,\tau )\xi (k,t,\tau )d\tau .  \label{i1}
\end{equation
Then the boundary conditions (\ref{arr}) and (\ref{k=0}) can be written in a
very simple form:
\begin{equation}
j(k,t)=i_{\lambda }(k-1,t)+i_{\mu }(k+1,t),\quad k>1  \label{j}
\end{equation
and
\begin{equation}
j(1,t)=(1-\chi )i_{\mu }(1,t)+i_{\mu }(2,t).
\end{equation
It follows from (\ref{sur5}), (\ref{initial}), (\ref{je}) and (\ref{i1})
that
\begin{equation}
i_{\lambda }(k,t)=\int_{0}^{t}\psi _{\lambda }(k,\tau )j(k,t-\tau
)e^{-\theta (k)\tau }d\tau +\psi _{\lambda }(k,t)p_{0}(k)e^{-\theta (k)t},
\label{i5}
\end{equation
\begin{equation}
i_{\mu }(k,t)=\int_{0}^{t}\psi _{\mu }(k,\tau )j(k,t-\tau )e^{-\theta
(k)\tau }d\tau +\psi _{\mu }(k,t)p_{0}(k)e^{-\theta (k)t}.  \label{i6}
\end{equation
Substitution of (\ref{initial}) and (\ref{je}) to (\ref{denG}), gives
\begin{equation}
p(k,t)=\int_{0}^{t}\Psi (k,\tau )j(k,t-\tau )e^{-\theta (k)\tau }d\tau +\Psi
(k,t)p_{0}(k)e^{-\theta (k)t}.  \label{p1}
\end{equation
It is convenient to introduce the integral escape rate $i(k,t)$ as the sum
of the escape rate to the right $i_{\lambda }(k,t)$ and the escape rate to
the left $i_{\mu }(k,t)$
\begin{equation}
i(k,t)=i_{\lambda }(k,t)+i_{\mu }(k,t).  \label{bau1}
\end{equation
The balance equation for $p(k,t)$ can be written a
\begin{equation}
\frac{\partial p(k,t)}{\partial t}=-i(k,t)+j(k,t)-\theta (k)p(k,t),\quad k>1.
\label{balan}
\end{equation
To obtain a closed equation for $p(k,t)$ we need to express $i(k,t)$ and
j(k,t)$ in terms of $p(k,t).$ By applying the Laplace transform $\hat{\psi
(k,s)=\int_{0}^{\infty }\psi (k,\tau )e^{-s\tau }d\tau $ to (\ref{i5}), (\re
{i6}) and (\ref{p1}), we obtain
\begin{equation*}
\hat{\imath}_{\lambda }(k,s)=\hat{\psi}_{\lambda }(k,s+\theta (k))\left[
\hat{\jmath}(k,s)+p_{0}(k)\right] ,
\end{equation*
\begin{equation*}
\hat{\imath}_{\mu }(k,s)=\hat{\psi}_{\mu }(k,s+\theta (k))\left[ \hat{\jmath
(k,s)+p_{0}(k)\right]
\end{equation*
an
\begin{equation*}
\hat{p}(k,s)=\hat{\Psi}(k,s+\theta (k))\left[ \hat{\jmath}(k,s)+p_{0}(k
\right] .
\end{equation*
In the Laplace space we have the following expressions for escape rates
\begin{equation}
\hat{\imath}_{\lambda }(k,s)=\frac{\hat{\psi}_{\lambda }(k,s+\theta (k))}
\hat{\Psi}(k,s+\theta (k))}\hat{p}(k,s),\qquad \hat{\imath}_{\mu }(k,s)
\frac{\hat{\psi}_{\mu }(k,s+\theta (k))}{\hat{\Psi}(k,s+\theta (k))}\hat{p
(k,s).
\end{equation
Using inverse Laplace transform and shift theorem we find
\begin{eqnarray}
i_{\lambda }(k,t) &=&\int_{0}^{t}K_{\lambda }(k,t-\tau )e^{-\theta
(k)(t-\tau )}p(k,\tau )d\tau ,  \notag \\
i_{\mu }(k,t) &=&\int_{0}^{t}K_{\mu }(k,t-\tau )e^{-\theta (k)(t-\tau
)}p(k,\tau )d\tau ,  \label{iii}
\end{eqnarray
where $K_{\lambda }(k,t)$ and $K_{\mu }(k,t)$ are the memory kernels defined
by Laplace transforms
\begin{equation}
\hat{K}_{\lambda }\left( k,s\right) =\frac{\hat{\psi}_{\lambda }(k,s)}{\hat
\Psi}\left( k,s\right) },\qquad \hat{K}_{\mu }\left( k,s\right) =\frac{\hat
\psi}_{\mu }(k,s)}{\hat{\Psi}\left( k,s\right) }.  \label{new5}
\end{equation
It follows from (\ref{j}), (\ref{bau1}), (\ref{balan}) and (\ref{iii}) that
the master equation for the probability $p(k,t)$ i
\begin{eqnarray}
\frac{\partial p(k,t)}{\partial t} &=&\int_{0}^{t}K_{\lambda }(k-1,t-\tau
)p(k-1,\tau )e^{-\theta (k-1)(t-\tau )}d\tau  \notag \\
&&+\int_{0}^{t}K_{\mu }(k+1,t-\tau )p(k+1,\tau )e^{-\theta (k+1)(t-\tau
)}d\tau  \notag \\
&&-\int_{0}^{t}[K_{\lambda }(k,t-\tau )+K_{\mu }(k,t-\tau )]p(k,\tau
)e^{-\theta (k)(t-\tau )}d\tau -\theta (k)p(k,t)  \label{master99}
\end{eqnarray
for $k>1.$ The balance equation for $k=1$ i
\begin{equation}
\frac{\partial p(1,t)}{\partial t}=-\chi i_{\mu }(1,t)-i_{\lambda
}(1,t)+i_{\mu }(2,t)-\theta (1)p(1,t)  \notag
\end{equation
o
\begin{eqnarray}
\frac{\partial p(1,t)}{\partial t} &=&-\chi \int_{0}^{t}K_{\mu }(1,t-\tau
)p(1,\tau )e^{-\theta (1)(t-\tau )}d\tau -\int_{0}^{t}K_{\lambda }(1,t-\tau
)p(1,\tau )e^{-\theta (1)(t-\tau )}d\tau  \notag \\
&&+\int_{0}^{t}K_{\mu }(2,t-\tau )p(2,\tau )e^{-\theta (2)(t-\tau )}d\tau
-\theta (1)p(1,t),
\end{eqnarray
where $0\leq \chi \leq 1.$ The master equation for $p(k,t)$ can be rewritten
in terms of the probability flux $I(k,t)$ from the point $k$ to $k+1$
\begin{equation}
I(k,t)=\int_{0}^{t}K_{\lambda }(k,t-\tau )p(k,\tau )e^{-\theta (k)(t-\tau
)}d\tau -\int_{0}^{t}K_{\mu }(k+1,t-\tau )p(k+1,\tau )e^{-\theta
(k+1)(t-\tau )}d\tau  \label{I}
\end{equation
as
\begin{equation}
\frac{\partial p(k,t)}{\partial t}=-I(k,t)+I(k-1,t)-\theta (k)p(k,t).
\end{equation
In the next section we shall derive fractional master equation for $p(k,t).$

\section{\ Anomalous subdiffusion in heterogeneous media}

We now turn to the anomalous subdiffusive case. We assume that the longer
cell survives at point $k$, the smaller the transition probability from $k$
becomes. It means that the transition rates $\lambda (k,\tau )$ and $\mu
(k,\tau )$ are decreasing functions of residence time $\tau .$ We assume
that
\begin{equation}
\lambda (k,\tau )=\frac{\nu _{\lambda }(k)}{\tau _{0}(k)+\tau },\qquad \mu
(k,\tau )=\frac{\nu _{\mu }(k)}{\tau _{0}(k)+\tau },  \label{rates}
\end{equation
where $\tau _{0}(k)$ is a parameter with units of time. Both $\nu _{\lambda
}(k)$ and $\nu _{\mu }(k)$ play a very important role in what follows. From
\ref{s3}) and (\ref{rates}) we find that the survival function has a
power-law dependence
\begin{equation*}
\Psi (k,\tau )=\left( \frac{\tau _{0}(k)}{\tau _{0}(k)+\tau }\right) ^{\nu
(k)},
\end{equation*
where the exponen
\begin{equation}
\nu (k)=\nu _{\lambda }(k)+\nu _{\mu }(k)  \label{exp}
\end{equation
depends on the state $k.$ Residence time probability density function $\psi
(k,\tau )=-\partial \Psi (k,\tau )/\partial \tau $ has the Pareto form
\begin{equation}
\psi (k,\tau )=\frac{\nu (k)\left( \tau _{0}(k)\right) ^{\nu (k)}}{(\tau
_{0}(k)+\tau )^{1+\nu (k)}},  \label{Pareto}
\end{equation
The anomalous subdiffusive case \cite{Klages,MFH} corresponds t
\begin{equation*}
\nu (k)=\nu _{\lambda }(k)+\nu _{\mu }(k)<1.
\end{equation*
We can notice from (\ref{rates}) that the ratios $\lambda (k,\tau )$ and
\mu (k,\tau )$ to $\lambda (k,\tau )+\mu (k,\tau )$ are independent of the
residence time variable $\tau $ that is
\begin{equation*}
\frac{\lambda (k,\tau )}{\lambda (k,\tau )+\mu (k,\tau )}=\frac{\nu
_{\lambda }(k)}{\nu _{\lambda }(k)+\nu _{\mu }(k)},\qquad \frac{\mu (k,\tau
}{\lambda (k,\tau )+\mu (k,\tau )}=\frac{\nu _{\mu }(k)}{\nu _{\lambda
}(k)+\nu _{\mu }(k)}.
\end{equation*
In this case it is convenient to introduce the probabilities of jumping to
the righ
\begin{equation}
p_{\lambda }(k)=\frac{\nu _{\lambda }(k)}{\nu _{\lambda }(k)+\nu _{\mu }(k)}
\label{pr1}
\end{equation
and to the lef
\begin{equation}
p_{\mu }(k)=\frac{\nu _{\mu }(k)}{\nu _{\lambda }(k)+\nu _{\mu }(k)}.
\label{pr2}
\end{equation
Note that these jump probabilities are completely determined by the
anomalous exponents $\nu _{\lambda }(k)$ and $\nu _{\mu }(k).$ In the
standard CTRW theory, these jump probabilities are given independently \cit
{Klages,MFH}.

Let us consider the non-local model for which the jump probabilities
p_{\lambda }(k)$ and $p_{\mu }(k)$ depend on the chemotactic substance $S(k)$
as follows
\begin{equation}
p_{\lambda }(k)=Ae^{-\beta \left( S(k+1)-S(k)\right) },\qquad p_{\mu
}(k)=Ae^{-\beta \left( S(k-1)-S(k)\right) },  \label{nonlocal}
\end{equation
where the parameter $A$ is determined from $p_{\lambda }(k)+$ $p_{\mu
}(k)=1. $ These jump probabilities describe the bias of cells with respect
to the spatial gradient $S(k+1)-S(k)$ \cite{Baker1,OS1}. One can obtain \cit
{HL}
\begin{equation}
p_{\lambda }(k)-p_{\mu }(k)=\frac{e^{-\beta S(k+1)}-e^{-\beta S(k-1)}}
e^{-\beta S(k+1)}+e^{-\beta S(k-1)}}.  \label{diff}
\end{equation}

The transition PDF's $\psi _{\lambda }(k,\tau )=\lambda (k,\tau )\Psi
(k,\tau )$ and $\psi _{\mu }(k,\tau )=\mu (k,\tau )\Psi (k,\tau )$ can be
rewritten in terms of $\psi (k,\tau ),p_{\lambda }(k)$ and $p_{\mu }(k)$ as
\begin{equation}
\psi _{\lambda }(k,\tau )=p_{\lambda }(k)\psi (k,\tau ),\qquad \psi _{\mu
}(k,\tau )=p_{\mu }(k)\psi (k,\tau ).  \label{new3}
\end{equation
The asymptotic approximation for the Laplace transform of the waiting time
density $\psi (k,\tau )$ of the Pareto form (\ref{Pareto})\ can be found
from the Tauberian theorem \cite{Feller}
\begin{equation*}
\hat{\psi}\left( k,s\right) \simeq 1-g(k)s^{\nu (k)},\qquad s\rightarrow 0
\end{equation*
wit
\begin{equation}
g(k)=\Gamma (1-\nu (k))\left( \tau _{0}(k)\right) ^{\nu (k)}.
\end{equation
We obtain from (\ref{new5})\ and (\ref{new3}) the Laplace transforms of the
memory kernel
\begin{equation}
\hat{K}_{\lambda }\left( k,s\right) \simeq \frac{p_{\lambda }(k)s^{1-\nu (k)
}{g(k)},\qquad \hat{K}_{\mu }\left( k,s\right) \simeq \frac{p_{\mu
}(k)s^{1-\nu (k)}}{g(k)},\qquad s\rightarrow 0.
\end{equation
Therefore, the integral escape rates to the right $i_{\lambda }(k,t)$ and to
the left $i_{\mu }(k,t)$ in the subdiffusive case are
\begin{eqnarray}
i_{\lambda }(k,t) &=&a(k)e^{-\theta (k)t}\mathcal{D}_{t}^{1-\nu (k)}\left[
p(k,t)e^{\theta (k)t}\right] ,  \notag \\
i_{\mu }(k,t) &=&b(k)e^{-\theta (k)t}\mathcal{D}_{t}^{1-\nu (k)}\left[
p(k,t)e^{\theta (k)t}\right] .
\end{eqnarray
Here $D_{t}^{1-\nu (k)}$ is the Riemann-Liouville fractional derivative with
varying order defined by (\ref{RL}). The anomalous rate functions $a(k)$\
and $b(k)$ are
\begin{eqnarray}
a(k) &=&\frac{p_{\lambda }(k)}{g(k)}=\frac{\nu _{\lambda }(k)}{\nu (k)\Gamma
(1-\nu (k))\left( \tau _{0}(k)\right) ^{\nu (k)}},\qquad  \notag \\
b(k) &=&\frac{p_{\mu }(k)}{g(k)}=\frac{\nu _{\mu }(k)}{\nu (k)\Gamma (1-\nu
(k))\left( \tau _{0}(k)\right) ^{\nu (k)}}  \label{ab}
\end{eqnarray
with the anomalous exponent $\nu (k)$ defined in (\ref{exp}).The master
equation (\ref{master99}) takes the form of non-homogeneous fractional
equation
\begin{eqnarray}
\frac{\partial p(k,t)}{\partial t} &=&a(k-1)e^{-\theta (k-1)t}\mathcal{D
_{t}^{1-\nu (k-1)}\left[ p(k-1,t)e^{\theta (k-1)t}\right]  \notag \\
&&+b(k+1)e^{-\theta (k+1)t}p(k+1)\mathcal{D}_{t}^{1-\nu (k+1)}\left[
p(k+1,t)e^{\theta (k+1)t}\right]  \notag \\
&&-\left( a(k)+b(k)\right) e^{-\theta (k)t}\mathcal{D}_{t}^{1-\nu (k)}\left[
p(k,t)e^{\theta (k)t}\right] -\theta (k)p(k,t)  \label{fractional}
\end{eqnarray
for $k>1.$

For $k=1$ with $\theta (k)=\chi =0$, we obtai
\begin{equation}
\frac{\partial p(1,t)}{\partial t}=b(2)\mathcal{D}_{t}^{1-\nu (2)}p(2,t)-a(1
\mathcal{D}_{t}^{1-\nu (1)}p(1,t).  \label{f1}
\end{equation
The fractional probability flux $I_{\nu }(k,t)$ from the point $k$ to $k+1$
is \
\begin{equation}
I_{\nu }=a(k)e^{-\theta (k)t}\mathcal{D}_{t}^{1-\nu (k)}\left[
p(k,t)e^{\theta (k)t}\right] -b(k+1)e^{-\theta (k+1)t}p(k+1)\mathcal{D
_{t}^{1-\nu (k+1)}\left[ p(k+1,t)e^{\theta (k+1)t}\right] .  \label{If}
\end{equation
The equation (\ref{fractional}) can be rewritten in terms of the probability
flux $I_{\nu }(k,t)$ as
\begin{equation}
\frac{\partial p(k,t)}{\partial t}=-I_{\nu }(k,t)+I_{\nu }(k-1,t)-\theta
(k)p(k,t).
\end{equation}

\subsection{Fractional Fokker-Planck equation for cells density and
chemotaxis}

In this subsection we consider the continuous case ($k\rightarrow x$) and
find the drift $l(a(x)-b(x))$ together with diffusion coefficient in the
fractional Fokker-Planck equation (\ref{FFP}). It follows from (\ref{ab})
that the drift is proportional to the difference in the anomalous exponents
\nu _{\lambda }(x)$ and $\nu _{\mu }(x)$, sinc
\begin{equation}
a(x)-b(x)=\frac{p_{\lambda }(x)-p_{\mu }(x)}{g(x)}=\frac{\nu _{\lambda
}(x)-\nu _{\mu }(x)}{\nu (x)\Gamma (1-\nu (x))\left( \tau _{0}(x)\right)
^{\nu (x)}}.  \label{dif1}
\end{equation
The difference $\nu _{\lambda }(x)-\nu _{\mu }(x)$ can be approximated in
the different ways. In the case of chemotaxis this difference is
proportional to the gradient of the local concentration of the chemotactic
substance $S(x).$ Using (\ref{diff}), we obtain
\begin{equation*}
p_{\lambda }(x)-p_{\mu }(x)=\frac{e^{-\beta S(x+l)}-e^{-\beta S(x-l)}}
e^{-\beta S(x+l)}+e^{-\beta S(x-l)}}.
\end{equation*
In the limit $l\rightarrow 0$, we have the standard chemotaxis model
\begin{equation}
a(x)-b(x)=\frac{p_{\lambda }(x)-p_{\mu }(x)}{g(x)}=-\frac{\beta l}{g(x)
\frac{\partial S}{\partial x}+o\left( l\right) ,  \label{dif2}
\end{equation
where the case \ $\beta >0$ corresponds to the negative taxis: the drift is
in the direction of the decrease in the value of the chemotactic substance
S(x)$. The fractional Fokker-Planck equation (\ref{FFP}) takes the for
\begin{equation}
\frac{\partial \rho (x,t)}{\partial t}=\frac{\partial }{\partial x}\left[
\frac{l^{2}\beta }{g(x)}\frac{\partial S}{\partial x}\mathcal{D}_{t}^{1-\nu
(x)}\rho (x,t)\right] +\frac{\partial ^{2}}{\partial x^{2}}\left[ \frac{l^{2
}{2g(x)}\mathcal{D}_{t}^{1-\nu (x)}\rho (x,t)\right] .  \label{newFFP}
\end{equation
\begin{figure}[tbp]
\includegraphics[scale=0.4]{temporary2}
\caption{{\ Monte-Carlo simulation of the stationary solution to equation
\protect\ref{newFFP}) and the analytical solution (\protect\ref{Bo}) for the
linear distribution of the chemotactic substance $S(x)=mx$ with $m=2$,
\protect\beta =10^{-2}.$}}
\label{fig1}
\end{figure}

As an example, let us consider the case when the anomalous exponent $\nu $
and time parameter $\tau _{0}$ are constants. Then the fractional
Fokker-Planck equation (\ref{newFFP}) can be rewritten as follows
\begin{equation}
\frac{\partial \rho (x,t)}{\partial t}=2\beta D_{\nu }\mathcal{D}_{t}^{1-\nu
}\frac{\partial }{\partial x}\left[ \frac{\partial S}{\partial x}\rho (x,t
\right] +D_{\nu }\mathcal{D}_{t}^{1-\nu }\frac{\partial ^{2}\rho (x,t)}
\partial x^{2}},  \label{Caputo2}
\end{equation
where $D_{\nu }$ is the fractional diffusion coefficien
\begin{equation*}
D_{\nu }=\frac{l^{2}}{2\Gamma (1-\nu )\tau _{0}^{\nu }}.
\end{equation*
In the case of the reflective boundary conditions at $x=0$, the fractional
equation (\ref{Caputo2}) admits the stationary solution $\rho _{st}(x)$ in
the semi-infinite domain $[0,\infty ).$ It obeys the equation
\begin{equation}
2\beta \frac{\partial }{\partial x}\left[ \frac{\partial S(x)}{\partial x
\rho _{st}(x)\right] +\frac{\partial ^{2}\rho _{st}(x)}{\partial x^{2}}=0
\end{equation
and has the form of the Boltzmann distribution \cite{Met1,Met2}
\begin{equation}
\rho _{st}(x)=N^{-1}\exp \left[ -2\beta S(x)\right] ,  \label{Bo}
\end{equation
where $N=\int_{0}^{\infty }\exp \left[ -2\beta S(x)\right] dx.$ This
distribution describes the aggregation of cells due to nonuniform
distribution of the chemotactic substance $S(x).$ Fig. 1 illustrates the
stationary profile of cells density $\rho _{st}(x)=$ $2\beta m\exp \left[
-2\beta mx\right] $ for the linear distribution $S(x)=mx$ and $m=2$, $\beta
=10^{-2}$.

We use Monte-Carlo method to simulate a stationary solution to equation (\re
{newFFP}). We select the following parameters: $\nu =0.5$, $\tau _{0}=1$,
and $l=0.01$. The Fig. 1 shows the result of the $10^{6}$ simulated random
walk trajectories, with jump probabilities given by (\ref{nonlocal}), Pareto
waiting time distribution (\ref{Pareto}), and terminal time $T=10^{6}$.

In the next section we will be concerned with the asymptotic behavior of the
solution of the master equation $p(k,t)$ as $t\rightarrow \infty $ \ for
\theta (k)=0.$ In particular, we will show that the stationary distribution
\ref{Bo}) is not structurally stable with respect to the spatial variations
of the anomalous exponent.

\section{Structural instability and anomalous aggregation}

It has recently been shown that the subdiffusive fractional equations with
constant anomalous exponent $\nu $ in a bounded domain $\left[ 0,L\right] $\
are not structurally stable with respect to the non-homogeneous variations
of parameter $\nu $ \cite{Fed3}. It turns out that the spatial variations of
the anomalous exponent lead to a drastic change in asymptotic behavior of
p(k,t)$ for large $t.$ The purpose of this section is to find the conditions
of this structural instability in semi-infinite domain $1\leq k<\infty .$ We
consider the case when $\theta (k)=\chi =0$ for which the total probability
is conserved
\begin{equation}
\sum\limits_{k=1}^{\infty }p(k,t)=1  \label{conse}
\end{equation
and the fractional probability flux $I_{\nu }(k,t)$ from the point $k$ to
k+1$ is
\begin{equation}
I_{\nu }(k,t)=a(k)\mathcal{D}_{t}^{1-\nu (k)}p(k,t)-b(k+1)p(k+1)\mathcal{D
_{t}^{1-\nu (k+1)}p(k+1,t).  \label{If0}
\end{equation
For simplicity we assume that the initial conditions are $p_{0}(1)=1$ and
p_{0}(k)=0$ for $k\neq 1.$Taking the Laplace transform of (\ref{dis1}) and
\ref{conse}) we obtain
\begin{eqnarray}
s\hat{p}(k,s) &=&a(k-1)s^{1-\nu (k-1)}\hat{p}(k-1,s)+b(k+1)s^{1-\nu (k+1)
\hat{p}(k+1,s)  \notag \\
&&-\left( a(k)+b(k)\right) s^{1-\nu (k)}\hat{p}(k,s)  \label{L2}
\end{eqnarray
an
\begin{equation}
\sum\limits_{k=1}^{\infty }s\hat{p}(k,s)=1.  \label{norm}
\end{equation
Since there is no flux of cells outside the left border, we have for $k=1
\begin{equation}
s\hat{p}(1,s)-1=b(2)s^{1-\nu (2)}\hat{p}(2,s)-a(1)s^{1-\nu (1)}\hat{p}(1,s).
\label{L1}
\end{equation
In the limit $s\rightarrow 0,$ one can obtain from (\ref{L1}) simple formula
expressing $\hat{p}(2,s)$ in terms of $\hat{p}(1,s)
\begin{equation*}
\hat{p}(2,s)\simeq \frac{a(1)s^{\nu (2)-\nu (1)}}{b(2)}\hat{p}(1,s),\qquad
s\rightarrow 0.
\end{equation*
In general, we find from (\ref{L2}) and (\ref{L1}) $\hat{p}(k,s)$ in terms
of $\hat{p}(k-1,s)$
\begin{equation}
\hat{p}(k,s)\simeq \frac{a(k-1)s^{\nu (k)-\nu (k-1)}}{b(k)}\hat{p
(k-1,s),\qquad k>1,\qquad s\rightarrow 0.  \label{re1}
\end{equation
This formula has very simple probabilistic meaning: the flux $I_{\nu
}(k-1,t)\rightarrow 0$ as $t\rightarrow \infty .$ If we take the Laplace
transform of $I_{\nu }(k-1,t)$ from (\ref{If0}), we obtain
\begin{equation}
\hat{I}_{\nu }(k-1,s)=a(k-1)s^{1-\nu (k-1)}\hat{p}(k-1,s)-b(k)p(k)s^{1-\nu
(k)}\hat{p}(k,s).
\end{equation
It follows from (\ref{re1}) that $\hat{I}_{\nu }(k,s)\simeq 0$ as
s\rightarrow 0.$

\subsection{ Stationary solution to fractional equation with constant
anomalous exponent}

Let us assume that the anomalous exponent $\nu (k)$ is independent of the
position $k$ that is $\nu =const.$ Let us find stationary solution to the
fractional master equation (\ref{dis1}
\begin{equation}
p_{st}(k)=\lim_{t\rightarrow \infty }p(k,t)=\lim_{s\rightarrow 0}s\hat{p
(k,s).  \label{st}
\end{equation
It follows from (\ref{re1}) that
\begin{equation*}
\hat{p}(k,s)\simeq \frac{a(k-1)}{b(k)}\hat{p}(k-1,s),\qquad k>1,\qquad
s\rightarrow 0.
\end{equation*
or
\begin{equation}
\hat{p}(k,s)\simeq \prod\limits_{j=1}^{k-1}\frac{a(j)}{b(j+1)}\hat{p
(1,s),\qquad k>1,\qquad s\rightarrow 0.
\end{equation
Using the normalization condition (\ref{norm}) and (\ref{st}), we obtain the
stationary solution of the equation (\ref{dis1}
\begin{equation}
p_{st}(k)=p_{st}(1)\prod\limits_{j=1}^{k-1}\frac{a(j)}{b(j+1)},\qquad k>1,
\label{an}
\end{equation
where
\begin{equation*}
p_{st}(1)=\left( 1+\sum\limits_{k=2}^{\infty }\prod\limits_{j=1}^{k-1}\frac
a(j)}{b(j+1)}\right) ^{-1}.
\end{equation*
If the sum
\begin{equation*}
\sum\limits_{k=2}^{\infty }\prod\limits_{j=1}^{k-1}\frac{a(j)}{b(j+1)}
\end{equation*
is divergent, the stationary solution does not exist. In particular, if we
assume that the anomalous rate functions $a$ and $b$ are equal, that is,
a(k)=b(k+1),$ then for a finite domain with $k=1,2,...,N,$ we obtain uniform
distribution $p_{st}(k)=1/N$ \ for every $k.$ The stationary solution (\re
{an}) is very similar to (\ref{ma}) corresponding to the Markov birth-death
model. However, this similarity is very deceptive, because (\ref{an}) is not
structurally stable with respect to the non-homogeneous variations of
parameter $\nu $. The aim of next subsection is to show this structural
instability.

\subsection{\ Anomalous aggregation.}

Now we consider non-homogeneous case for which the anomalous exponent
depends on $k.$ We assume that the point $k=M$ has the property that $\nu
(M)<\nu (k)$ for all $k\neq M$. Our purpose now is to find the conditions
under whic
\begin{equation}
\lim_{t\rightarrow \infty }p(M,t)=1,\qquad \lim_{t\rightarrow \infty
}p(k,t)=0,\qquad k\neq M.  \label{ag}
\end{equation
It means that the total probability concentrates just at one point $k=M.$
This phenomenon is called an anomalous aggregation \cite{Fed1}. This
asymptotic behavior of cells was observed in experiments on phagotrophic
protists when \textquotedblleft cells become immobile in attractive patches,
which will then eventually trap all cells\textquotedblright\ \cite{AA}. In
the Laplace space, (\ref{ag}) takes the form
\begin{equation}
\lim_{s\rightarrow 0}s\hat{p}(M,s)=1,\qquad \lim_{s\rightarrow 0}s\hat{p
(k,s)=0,\qquad k\neq M.  \notag
\end{equation
We can rewrite the normalization condition (\ref{norm}) a
\begin{equation}
s\hat{p}(M,s)+\sum\limits_{k=1}^{M-1}s\hat{p}(M-k,s)+\sum\limits_{k=1}^
\infty }s\hat{p}(M+k,s)=1.  \label{cons}
\end{equation
By using (\ref{re1}), we express $\hat{p}(M+k,s)$ in terms of $\hat{p}(M,s)
\ as follows
\begin{equation}
\hat{p}(M+k,s)\simeq \hat{p}(M,s)\prod\limits_{j=1}^{k}\frac{a(M+j-1)}{b(M+j
}s^{\nu (M+k)-\nu (M)},\qquad k\geq 1,\qquad s\rightarrow 0.  \label{M1}
\end{equation
Now we write the formula for $\hat{p}(M-k,s)$ in terms of $\hat{p}(M,s)$
\begin{equation}
\hat{p}(M-k,s)\simeq \hat{p}(M,s)s^{\nu (M-k)-\nu (M)}\prod\limits_{j=1}^{k
\frac{b(M-j+1)}{a(M-j)},\qquad k=1,...,M-1,\qquad s\rightarrow 0.  \label{M2}
\end{equation
Now we substitute (\ref{M1}) and (\ref{M2}) into (\ref{cons}) and use $s\hat
p}(M,s)$ as a common factor
\begin{equation}
s\hat{p}(M,s)\left( 1+\sum\limits_{k=1}^{M-1}s^{\nu (M-k)-\nu
(M)}\prod\limits_{j=1}^{k}\frac{b(M-j+1)}{a(M-j)}+\sum\limits_{k=1}^{\infty
}s^{\nu (M+k)-\nu (M)}\prod\limits_{j=1}^{k}\frac{a(M+j-1)}{b(M+j)}\right)
\simeq 1.  \notag
\end{equation
Since $\nu (M)<\nu (k)$ for any $k\neq M$, we have $s^{\nu (M+k)-\nu
(M)}\rightarrow 0$ and $s^{\nu (M-k)-\nu (M)}\rightarrow 0$ as $s\rightarrow
0.$ We conclude that i
\begin{equation}
\sum\limits_{k=1}^{\infty }s^{\nu (M+k)-\nu (M)}\prod\limits_{j=1}^{k}\frac
a(M+j-1)}{b(M+j)}\rightarrow 0  \notag
\end{equation
as $s\rightarrow 0,$ then $s\hat{p}(M,s)\rightarrow 1,$ while $s\hat{p
(k,s)\rightarrow 0$.\ It means that in the limit $t\rightarrow \infty ,$ we
obtain (\ref{ag}). If instead of probability $p(k,t)$ we consider the mean
density of cells $\rho \left( x,t\right) ,$ the formula (\ref{ag}) can be
rewritten as $\rho \left( x,t\right) \rightarrow \delta (x-x_{\min })$ as
t\rightarrow \infty ,$\ where $x_{\min }$ is the point on the interval
[0,\infty )$ at which $\nu (x)$ takes its minimum value. Note that this
result was obtained for a symmetrical random walk in the context of
chemotaxis and anomalous aggregation \cite{Fed1}.

\section{Conclusions.}

We have studied a non-homogeneous in space and non-local in time random walk
model describing anomalous subdiffusive transport of cells. Using a Markov
model with structured probability density function, we have derived
non-local in time and fractional master equations for the probability of
cell position. The advantage of our probabilistic approach is that it allows
us to take into account the death process within the general non-Markovian
random walk. The main feature of our fractional model is that the transition
probabilities for jumping on the left and right depend inversely on the
residence time variable. This dependence induces power-law residence time
distribution and ultimately the anomalous subdiffusion of cells. It has
recently been shown that the subdiffusive fractional equations with constant
anomalous exponent\ are not structurally stable in a bounded domain with
respect to the non-homogeneous variations. In this paper we have extended
and complemented our previous results for infinite domain and found exact
conditions under which the structural instability takes place. Our model can
be generalized in many ways, e.g., by modelling the residence time by
internal chemical reactions via a stochastic or ordinary differential
equations instead of simple equation for the residence time $d\tau /dt=1$.
It would be interesting to take into account the density-dependent dispersal
\cite{Campos2} including non-linear exclusion process with cell-to-cell
adhesion \cite{Baker, Khain}.

\section*{Acknowledgements}

The authors gratefully acknowledge the support of the Federal Programme
N 14.A18.21.0867. SF acknowledges the warm hospitality of Department
of Mathematical Physics, Ural Federal University. SF also acknowledges the
support of the EPSRC Grant EP/J019526/1. The authors wish to thank S.
Falconer for interesting discussions.

redpajama_set_name: RedPajamaArXiv


Allen Community Development Corporation (CDC) is continuing a 3-year commitment for TAPS to provide public transportation service. George Chrisman, CDC president, made the announcement last week in his annual report to the Allen City Council.
TAPS began service to Allen on November 18, 2013 with three routes, including The Allen Express, a 7-day-a-week commuter service connecting to DART's Parker Road Station in Plano.
Chrisman told the city council that providing public transportation is an important benefit for the city's population. "Public transportation is a very important capability that we provide to the business community," he said.
Tim Patton, TAPS Public Transit's chief operating officer, noted that the working relationship between his agency and Allen is strong and that both are committed to future growth in service.