Still looking for an answer to "How to cancel Quickbooks online payroll" then click on the link above. QuickBooks Full Service Payroll (Do-It-For-Me) – Duration: 2:30.
Jan 2, 2019 ... frequently asked questions about canceling Patriot's Full Service Payroll, the payroll tax filing service offered by Patriot Software, Inc.
How to send payroll data from Full Service Payroll to QuickBooks Online. -Need support for quickbooks accounting software? searching for a reliable helpline number which can provide you real and effective solution? looking for trustworthy quickbooks.
Get help with canceling your Intuit Full Service or QuickBooks Full Service Payroll and know what happens to your Payroll data after. Is this really goodbye?
Full Service Payroll: Cancel at any time by calling 866-729-4662. You can place orders weekdays by phone by calling (800) 365-9619 between 6 AM – 6 PM PT. You can place orders weekdays by phone by calling (800) 365-9619 between 6 AM – 6 PM PT.
If you need to cancel your QuickBooks Desktop Payroll Assisted Service, please call 888-712-9702. You will receive confirmation of cancelation and other important information within 3 business days of submission of the cancelation request.
Full-Service Payroll Features. Small Business. We'll automatically calculate how much your employees should be paid each pay period, taking into account hourly rates and salaries, shift differentials and overtime, and all of Federal, state and local taxes relevant to your business.
To Cancel Full-Service Payroll If you are a Full-Service Payroll customer, we will need to get more details about exactly when you want to stop the tax filing service. Also see these Full-Service Payroll Cancel FAQs on our website.
Learn how to cancel or reactive your Online Payroll account. If you're closing your business or no longer pay employees, you can cancel y… Before canceling your payroll subscription, make sure that you have done the following: Paid any outstanding tax deposits.
A human resources management system (HRMS) or human resources information system (HRIS) is a form of human resources (HR) software that combines a number of systems and processes to ensure the easy management of human resources, business processes and data.
More than just Onboarding, Namely is the first all-in-one HR platform that employees love to use. Namely is powerful, easy-to-use technology that allows small to mid-sized companies (15 to 3,000 employees) to handle all of their HR, Payroll, Benefits, and Time & Attendance in one place.
Simplify core HR and payroll processes and improve contingent workforce management with our human SAP SuccessFactors Employee Central, a cloud-based core HR and payroll system. Learn how to manage people effectively for less, while complying with local and global mandates.
PeopleWorks is a cloud based Employee Lifecycle Management solution which manages all aspects of an employee's tenure with their employer from Hire to Retire i.e. Resource planning, Pre-joining details, On-boarding, Payroll Management software, Performance Management, etc. to name a few.
Oracle Fusion Applications have been designed to ensure your enterprise can be modeled to meet legal and management objectives. The decisions about your implementation of Oracle Fusion Applications are affected by your: The figure illustrates a typical group of legal entities, operating various business and functional organizations.
The payroll process refers to the steps required to pay employees each pay period and involves employees' hours worked, their pay rate, and deductions. Processing payroll ensures employees are paid based on their employment status and other Department of Labor (DOL) requirements.
Inform Service Canada if you suspect that a Social Insurance Number is being used fraudulently. If you have any reason to suspect that a SIN is being used fraudulently, immediately contact Service Canada at 1-800-206-7218 (select option "3").
The payroll process refers to those steps required to pay employees. We'll cover payroll processing to help ensure your employees are paid correctly. The payroll process starts after you hire your first employee. Each pay period, employees work and must be paid based on their hours and pay rate.
Most payroll services calculate employee pay and taxes automatically and send your payroll taxes and How to process payroll with a payroll service: Just like with the DIY option above, you Step 5: Keep track of your tax payments and filings. The IRS requires tax forms to be kept for three years .
Employers are required to make federal payroll tax payments to the … you with information about your specific payroll tax obligations and may supply the forms … the federal tax due dates for the year, as well as an electronic reminder system. … party, such as a tax professional or payroll service, to make the payment for you.
They can help assure filing deadlines and deposit requirements are met and greatly streamline … Generally, an EFT is made using Electronic Federal tax payment system (EFTPS). … For more information about IRS notices, bills and payment options, refer to … Page Last Reviewed or Updated: 05-Nov-2018. Share.
Employer responsibilities – The payroll steps These steps will help you understand your responsibilities and how payroll works. It is important to understand each step, to ensure you are not assessed penalties and interest or suffer other consequences .
Feb 28, 2019 … They can help assure filing deadlines and deposit requirements are met and greatly streamline business operations. … Payroll Service Provider not making federal tax deposits? … The IRS has streamlined the process to file form 14157. … information against IRS records prior to filing information returns.
That payroll tax holiday was legislated as part of the tax relief act of 2010, which was then extended by HR 3765 and extended again by HR 3630. But the employee portion of Social Security reverted back to the full 6.2 percent in 2013.
Real Time Information (RTI) is the most significant change ever made to PAYE. An improved way of reporting, RTI is designed to make PAYE submissions more efficient – meaning you'll need to submit information to HMRC in real time, every time you pay employees.
Saving the underlying source records, and booking entries into QuickBooks is important, but as time goes on and hopefully, the business grows, it will Often, even with a decent GL program like QB, a payroll accounting service like ADP is needed for the complexities of payroll accounting and tax filing.
Intuit Online Payroll and QuickBooks are two useful business management services. However, each service helps your business in a different way. While Intuit Online Payroll manages the payroll side of your business, QuickBooks handles basic accounting functions.
With Intuit Enhanced Payroll, you can print W-2 forms from directly within the software. After you run your last payroll for the year, Intuit will do the final calculations behind the scenes and walk you through the steps to review & print your W-2 forms when you are ready.
See payroll services pricing for Intuit QuickBooks Payroll–use it with or … Choose the services to best suit your small business. … Choose your payroll service.
2016-04-21 · QuickBooks full service payroll support +1888-567-1159, dial one number for both QuickBooks Basic Payroll Support and QuickBooks Enhanced Payroll Support. Our Payroll Customer Service is accessible 24×7. Allow us to manage you to the speediest answer, reach us by phone on 1888-567-1159.Call us immediately !
3 questions: 1. Is "full service" payroll just a fluffy upsell? Both offer "Click for payroll tax E-File & E-Pay2" 2. I assume this means we can issue the tax payments right : QuickBooks Q & A. : Account management. : Enhanced vs Full Service Payroll. 1. Is "full service" payroll just a fluffy upsell?
QuickBooks desktop payroll enhanced is Intuit payroll software that's fully … Instant paycheck for W-2 employees and 1099 contractors. … Additional Services.
Buy Payroll + QuickBooks Get Payroll and QuickBooks together in one package. Already have QuickBooks? Add Payroll to your QuickBooks See options to add Payroll to your existing QuickBooks Desktop or Offer valid for new quickbooks online customers only, for a limited time.
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Oct 24, 2018 … When your Payroll service is set up, our system automatically creates default accounts … Select New in the upper right corner of your screen.
How to activate a payroll service in QuickBooks Pro and Premier. How to Turn on QuickBooks Manual Payroll – Продолжительность: 4:22 HowToUseQuickBooks 9 044 просмотра. Using QuickBooks Online – Продолжительность: 19:33 intuitacademy 205 141 просмотр.
QB Online Payroll does not offer certified payroll. We have yet to find an app or service that works with QBO Payroll. Be clear and state the answer right up front. Ask yourself what specific information the person really needs and then provide it. Stick to the topic and avoid unnecessary details.
Whether you want the freedom to run payroll yourself or the convenience of QuickBooks running … How does running payroll in less than 5 minutes sound?
What payroll service do I have? To know which service you have activated in QuickBooks Desktop, go to Employees > My Payroll Service > Manage Service Key . In the QuickBooks Service Keys window, find the payroll service name in the SERVICE NAME list.
Maintain control of your clients' payroll taxes from inside QuickBooks Online. … Payroll data automatically populates the books to the penny – no need to import files. …. Discount offer valid for the first 12 months of service as indicated above, or for … For 11+ employees, you will pay $2.50/month per employee or contractor .
We skipped and employee on last payroll run we issue a check outside of QBO, how do I add that check so that it's included in the employee's ytd earnings? We are currently processing payroll via Intuit Full Service.
I am not currently using QBO payroll, but think it is possible to do the repayment without double taxation. So if regular pay is $1,000 and $500 less taxes was advanced, run a special payroll for the repayment with only $500 gross pay.
Online payroll services from Intuit make it easy for small businesses to run payroll online from anywhere, anytime. Try payroll online for … QuickBooks Online Payroll … File and print worker tax documents in QuickBooks or have us do it for you.
Jan 7, 2019 … Tip: You should have a separate payroll account from the checking account … offered by Intuit, check out our QuickBooks Payroll Service guide.
The partnership has no employees / payroll.
Depending on the features you need, you can choose from Basic, Enhanced or Assisted Payroll. QuickBooks Payroll service does not work with QuickBooks Mac . … Desktop and Quickbooks Online (without Quickbooks Online Payroll).
Hi Slava, Thanks for being a part of the QuickBooks family and taking part in our payroll services! quickbooks will let you manually enter payroll going back several months, but you won't be able to individually enter the cheques all the way from January.
If you have payroll turned off in your Desktop file, but have payroll turned on in your QuickBooks Online subscription. To turn on Payroll for QuickBooks Desktop, go to Edit, then select Preferences and go to Payroll & Employees. Select Company Preferences, then choose Full Payroll (or Online Payroll).
Whether you want the freedom to run payroll yourself or the convenience of QuickBooks running payroll for you, QuickBooks Payroll has a solution for your small business payroll needs.
Using 'Zero-Dollar Checks' to Record Payroll Costs by Job and Class If you use an outside payroll service to process your payroll, using zero-dollar checks may be Using a journal entry, enter the totals for your payroll into the general ledger just as you normally do when using a payroll service.
Jan 5, 2016 … In ESC, labor always hits the General Ledger and Job Reports from the Payroll module. So if you outsource your payroll, there is an easy… … with Service Agreements · Setup the Service Agreement Module ….. If a person is salary enter the yearly salary and all rates for any salary item need to be the same.
Get the benefits of full service payroll at a lower cost. and less than the cost of outsourcing. Guaranteed accurate and on-time. Smart Payroll inside QuickBooks. With your payroll data all in Get detailed job costing, class tracking, and an open platform for third-party time tracking solutions.
The best way to do labor job-costing is to run the payroll through QuickBooks … If you are instead using an outside payroll service (like Paychex, ADP, etc) , there is a …. You are basically creating job costs – then zeroing them out to no job.
Apr 17, 2008 … Using 'Zero-Dollar Checks' to Record Payroll Costs by Job and Class If you use … Home QuickBooks Job Costing When Using A Payroll Service … will be based on the average hourly payroll costs for each person in each job.
How to enter 3rd party payroll in QuickBooks Article ID: 100295 Summary This article explains how to enter payroll that was processed by a third party into QuickBooks.
How to Do Job Costing in QuickBooks When You Outsource Payroll By Alicia Katz Pollock on July 23, 2014 in Tips & Tricks QuickBooks has a powerful Job Costing feature that allows you to compare the income you made from a job to the expenses you incurred to provide the product or service.
Using 'Zero-Dollar Checks' to Record Payroll Costs by Job and Class … based on the average hourly payroll costs for each person in each job. How To See What Payroll Service You Have In Quickbooks Intuit's QuickBooks Payroll software makes running payroll for small businesses easy and stress-free.
2018-07-26 · How businesses and payroll service professionals can report if they lost data to the Form W-2/SSN email scam or if they only received the email without falling victim.
payroll service providers are experts at what they do. They have extensive knowledge of payroll processes. They will manage your payroll with ease and efficiency, even customize services to meet your unique business needs. Pay your employees in full and on time, every time.
Payroll Service Australia. Payroll is a noncore function best left to payroll service and payroll outsourcing companies.
Paysquare is leading Payroll Service Providers, managing complete payroll management system services right from consolidating We are able to do this, leveraging our award-winning technology, proven processes and over a decade of expertise, exclusively in managing payroll processing in India.
Here's a checklist you can download to make sure you don't miss any steps if you choose to do payroll manually. This checklist will also be helpful if you use payroll software to do payroll.
Many practitioners are asked to provide payroll services to their small business clients but don't know where to begin. Others wish to supplement their practice's income with a payroll service which also serves as a great way to get additional corporate and personal tax return business.
QuickBooks Desktop Payroll Software by Intuit has everything you need to run payroll for your … Basic Payroll features … Learn about these important services:.
Down To basics payroll. 129 likes. Our goal is to make payroll as easy and basic as possible, thus the name, Down To Basics Payroll. Our job is to… Our goal is to make payroll as easy and basic as possible, thus the name, Down To Basics Payroll.
Whether you just need in-house software or full payroll processing, BASIC has a solution for … Both In-House and Fully Outsourced Payroll Service Available.
A payroll is a company's list of its employees, but the term is commonly used to refer to: the total … Before considering the payroll taxes, it is necessary to talk about the basic formula for the Net Pay. … tables in "Publication 15, Employer's Tax Guide" by the Internal Revenue Service – IRS;; Social Security tax withholding.
Payroll is a key area of the business, and it's important to understand the process behind it and how to carry it out correctly. If you're not comfortable running payroll, many accountants offer this service and can run it for you.
The Five Payroll Basics Every Employer Should Know. You've hired your first employee.Now you need to make sure he or she gets paid. If you're new to payroll, you might find yourself in some uncharted waters, navigating a sea of new terminology — pay periods, unemployment tax, something called Form 941.
BASIC Payroll Plus: Outsourced Payroll Processing Administration Lack of expertise and poor customer service are the main reasons employers look to make a payroll provider change. Luckily for you, there's no need to suffer another year.
Review the basics of payroll processing and taxes. What is payroll. From determining pay schedules to setting up deductions, the success is in the setup. Download the Payroll 101 ebook. Hiring employees is a big step for a small business. Set yourself up for success with this essential guide.
Jan 10, 2013 … Direct all inquiries concerning payroll items, such as employee business … (or previously employed, if not a current postal service employee).
And the Postal Service is famous for its fabulous benefits. This is the front counter retail clerk who performs a variety of customer support services. A mobile technician who handles maintenance responsibilities at various Postal facilities within a certain geographic area.
January 3, 2019 — Want a copy of the Postal Service's 2019 payroll schedule? The calendar images that accompany this article show the 26 paycheck dates for USPS employees.
Though sales may bring in revenues, and marketing is responsible for advertising the company's wares, payroll remains the chief concern of employees because it is the source of compensation.
Human Resources 2009 Combined Federal Campaign Guidelines Schedule and Responsibilities. Each fall the U.S. Postal Service ® joins other federal agencies in the Combined Federal Campaign (CFC).
The postal service processes payroll for more than 500,000 employees. To handle this monumental task, time and attendance information is gathered through the Time and attendance collection system (tacs).


Business Development and Sourcing internship in Corporate Procurement Udbyder Virksomhed / Organisation Sted København og omegn • Supply Chain & Procurement • Bagsværd Are you interested in working in business development with focus on sourcing and purchasing? Would you like to be part of a global pharmaceutical company and get the opportunity to build a life-changing career in an international business environment?
App development student job or apprenticeship Udbyder Virksomhed / Organisation Sted København og omegn Greener Pastures is looking for student developers for part time jobs or apprenticeships in Android and IOS development.
Frist 30. april 2019 The division of Astrophysics and Atmospheric Physics at the National Space Institute at the Technical University of Denmark, DTU Space, is looking for a data specialist with an expertise in data from transient astrophysical phenomena. DTU Space is involved in a number of experiments, both satellite and ground-based, targeting transient phenomena, such as supernovae, gamma-ray bursts and gravitational waves.
Programming of a smart ventilation controller using an IDE in an industrial research project Udbyder Vejleder Sted København og omegn The Room-Vent Solutions project (https://energiteknologi.dk/node/8717) has installed innovative decentralised ventilation systems in eleven renovated apartments. The system adjusts airflows to each room according to measurements of temperature, humidity, carbon dioxide and noise.
Entanglement-generating quantum thermal machines in diamond Udbyder Vejleder Sted København og omegn Entanglement is a central phenomenon which distinguishes quantum from classical physics. In entangled quantum states, correlations are stronger than any possible in classical systems, which has profound consequences for how we think about nature. Entangled states are also the key resource which give upcoming quantum technologies - such as quantum computers - their power.
Frist 17. maj 2019 DTU Compute's Section for Statistics and Data Analysis, invites applications for an appointment as an Associate or Assistant Professor within Statistics and Data Analysis. The position is available from June 1, 2019 or according to mutual agreement.
App udviklere til studiejob eller praktik Udbyder Virksomhed / Organisation Sted København og omegn Greener Pastures søger studerende til deltidsjob eller praktik/projektforløb inden for app udvikling til både Android og IOS.
Frist 30. april 2019 DTU Skylab is expanding into a new exciting field: Food. We recently opened our new Foodlab and we are now looking for the right person to manage the lab and lead our efforts in food innovation. We work with both students, researchers and external partners, and we strive to become an internationally leading hub for innovation. The work requires both practical skills, a visionary outlook on the world of food and a desire to engage with the bright students of DTU.
Design of Smart Residential Ventilation Laboratory for DTU BYG Udbyder Vejleder Sted København og omegn What is Smart Ventilation Air Infiltration and Ventilation Centre (AIVC.org) has recently published a definition of "smart ventilation"
Frist 29. april 2019 The UNEP DTU Partnership (UDP) Centre on Energy, Climate and Sustainable Development (formerly UNEP Risø Centre) is a leading international research and advisory centre on energy, climate and sustainable development. Working in partnership with United Nations Environment (UNEP), UDP is currently involved in project activities in over 50 developing countries, in close collaboration with national and international authorities and research institutions.


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The Weekly Dispatch
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What Do College Athletes Really Think?
Sara Hebel, Scott Smallwood
Listening to the Voice of the College Athlete
The top story this fall in college sports is about state lawmakers pushing to allow athletes strike their own endorsement deals and potential NCAA reforms that may allow some athletes to make money from their names and images.
But most college athletes have much more quotidian concerns: locker rooms, academic tutors, dining, and, for one pole vaulter at the University of Nevada, birds. "Geese poop — everywhere in field site — slipping issues — tried to shovel"
The relationship between universities and their student-athletes can be complicated. They're often getting different perks, benefits, and treatment than other students. But they also face very different responsibilities, long hours, and a schedule that can make playing a sport feel a lot like having two jobs.
The NCAA Division I manual requires colleges to conduct exit interviews with "a sample" of departing athletes about their experiences.
The Intercollegiate, a new two-person journalism venture about college sports, filed public-records requests for documents related to those interviews at hundreds of colleges. Many institutions refused to give them anything, citing student privacy, but the thousands of pages of interviews, surveys, and summaries they did get offer a fascinating glimpse into the life of the student-athlete.
Daniel Libit and Luke Cyphers highlighted some of what they've found so far in their newsletter:
"With so little consistency, the athlete exit interview mandate presents a kind of Rorschach test for an institution's commitment to athlete input. The interviews can show the priorities of administrators who run college sports programs, based on what they ask their athletes, how they ask it, who answers the questions, and how those answers are used."
A couple of things that jumped out to us:
At Michigan State, still reeling from the Larry Nassar scandal, the exit interview survey struck The Intercollegiate as trivial:
"Of its 17 mostly generic questions, not one addressed sexual assault. In fact, arguably the most probing query on the list had to do with the athletic department's end-of-the-season swag, specifically the "jacket, plaque blanket and ring."
Colleges have no standard way of handling this exit-interview requirement. Some do it informally. Some when athletes request it. Some never create a paper trail. Others conduct extensive surveys, sometimes with outside vendors — though that presents its own questions:
"Real Recruit's software analytics parcels a school's athlete respondents into categories of "Promoters", "Passives" and "Detractors." This necessarily raises questions about the real purpose of the input."
Many athletes have mostly great things to say about their experience. But the documents also show a good amount of griping — some of it about facilities or safety issues like that geese poop on the field, but much of it about the divides between teams with a lot of money and those without:
"The disparity in services offered based on the sports is just outrageous," said one UNLV athlete. "I understand that there is income disparities between each sports but it is unacceptable to see some student-athletes get dry cleaning, multiple pairs of shoes, incredible facilities, state of the art locker rooms etc. when others have to learn to live with a WW2 bunker as their lockeroom [sic]."
See all those documents and learn more about The Intercollegiate here.
— Scott Smallwood
Help Us Grow (and Double Your Donation)
Now through December 31, NewsMatch will double your gift to Open Campus. Thanks to all of you who already donated. We're grateful.
We started Open Campus to rebuild confidence in two public goods: higher education and journalism. And we're excited to get going on that work in earnest in 2020. We're using the money we raise through NewsMatch to help us add new expert content — like our new newsletter, latitude(s) — and to hire our first Open Campus intern. (You'll be hearing more about her soon!)
Here's where you can make a tax-deductible donation. Thank you!
A Social Law of Higher Education?
Thirty-five years ago, Stanford University had 6,500 undergrads. Now it enrolls just over 7,000. In the same time, the University of California at Berkeley has increased from 22,000 to nearly 30,000 undergraduates — basically adding an entire Stanford population to the campus.
Of course, there are all sorts of reasons that the most prestigious colleges in the country aren't generally in the business of increasing enrollment (maintaining that prestige tops among them). But one factor we don't always talk about is reaction from their neighbors.
Last week Stanford University backed off a huge expansion plan that would have added more than 2,000 beds for students and brought 10,000 additional people to campus every day by 2035. Local government leaders and the university couldn't agree on how Stanford would mitigate the impact of all that growth on housing, school districts, and traffic. (More in the San Francisco Chronicle. Plus, this story from the campus newspaper clearly spells out the two sides.)
Thanks to Brendan Cantwell at Michigan State for flagging this: "Status through exclusion is the closest thing to a social law of higher education."
And to Kevin McClure at UNC-Wilmington who pointed to David Labaree's 2017 book that wrestled with this as well: A Perfect Mess: The Unlikely Ascendancy of American Higher Education
In Wake of Scandals, USC Radically Cuts Number of Trustees, Imposes Age Limits, Pledges More Diversity
USC trustees approved far-reaching reforms aimed at correcting the insular leadership that failed to effectively handle several scandals over the past few years
Fear of a Black Campus
An ugly campaign to force out the university's African American president exposed racial fault lines in Macomb, Ill., where the town and the campus are closely entwined.
Here's How Much Time Admissions Officers in Colorado Spend on an Application
The Denver Post's Elizabeth Hernandez is talking with students in the middle of the college-application process. She plans to write more about what it's like through their eyes.
How Much Is a University of Michigan Degree Worth?
A university researcher collected data from hundreds of alumni as part of a pilot project with the U.S. Census Bureau to help colleges do more to quantify the value of a degree.
Recommendation Corner
No podcasts or YouTubers this week, just a simple hack for users of Google docs:
Need to create a new Google doc or a new Google spreadsheet? Don't go to your account and follow the menus. Just type docs.new or sheets.new into your browser. Bingo. Immediately launches a blank file. It's crazy how useful this turns out to be.
We're All Ears
Please keep the feedback and ideas coming. We love to hear from you. Reach us both at info@opencampusmedia.org.
One more request: Please forward this email to other people you think would be interested. They can sign up here.
Post written by: Sara Hebel
Co-founder of Open Campus.
Post written by: Scott Smallwood
Karin Fischer
LatitudesHow Higher Ed Found Itself in the 'Vortex of Politics'
LatitudesWarnings about China's Influence on Campus Can Unfairly Tar Chinese Students
Most People Go to College Close to Home. What If There Isn't One Nearby?
More than one in two public-college students go to college within 50 miles of home. What does that mean for rural America?
Launching Our Weekly Newsletter and Mapping Out Our Road Trip
A reporting job covering colleges wasn't something either of us set out to get. But we grew to care deeply about higher education.
Misperceptions about the regional public
They're not always trying to be something they're not. Very few are on the brink. And why don't we consider them prestigious?
What's ahead for higher ed
Democrats are taking control in Washington, the pandemic is upending yet another semester, and states are reckoning with deep cuts.
A terrible, horrible, no good, very bad year
What 2020 taught us about the role of higher ed in society
What happened to the class of 2020?
Far fewer graduates enrolled in college, and the drops were steep at low-income high schools. We looked at diverted paths in Akron, Ohio.
1 Thomas Circle NW Washington, DC 20005

redpajama_set_name: RedPajamaCommonCrawl


USA Today began selecting boys basketball All-American teams in 1983 soon after the newspaper was created. Early teams generally consisted of one x 5-man team. Later selections were expanded to three x 5-man teams.


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Download & Install iNoJB – Hello guys, Today I've come back with another best third-party app store called iNoJB that you can download on your iOS device without jailbreak. when it comes to downloading an app then every iOS users go to the app store. but the only thing that I don't like about the app store.
It doesn't allow us to download paid apps for free. but we can do it on the third-party app store. third-party app stores are the bundle of tweaked apps, ++apps, and hacked games.
Mostly, iOS users only know Cydia as a third-party app store and also, they download it. but as we know that without jailbreak, downloading Cydia is not possible and I still can't understand that if we have more alternative instead Cydia then why people still downloading Cydia on their device.
However, they know that they can't download Cydia on the Unjailbroken device then first they jailbroke their device then they get the Cydia. guys, if you don't know then I want to tell you that Jailbreak can cause issues at any time.
Well, I Just want to tell you that forget about the Cydia and search on the internet about third-party app store then you will realize that there are many many third-party app stores around you.
But, if you go to search on the internet that which is the best third-party app store then you will see a list, which will confuse. But don't worry, because after spending my more time on the internet. finally, I've found the best app store, about which I already also told you above.
So now let's come back to the iNoJb. but before know how to install iNoJB, how will it be, if I tell you about its features and I'm sure guys, after knowing its features you can't stop yourself to download it.
Guys, it's really essential for all of you to know about the iNoJB features.
So, I have shown you just little features of the iNoJB app store and rest feature that I have not shown you will know after using it. now let's turn to the installation process.
Downloading & installing process is quite simple and it will complete in just a few steps. so to download & install iNoJB, make sure that you have followed the steps below.
Thank you guys for reading this article and being with us. guys, if you found that this article is helpful for you then don't forget to share this article with your friends, family members, and on the social media and also visit daily on this website for reading new and helpful articles.


I have written a blog to help you with the make.
You will need lots of quality time and patience to make this blouse. There are several parts that require focus, time and skill. You will need a machine that has a button-hole setting. The button placement and gathered shoulders are unique features that will be a topic of conversation.
Each part is easier with experience, so I recommend practicing the collar and cuffs, with some cotton calico first. I have included calico in the pack.
the linen is sturdy and easier to sew. I would recommend you start with this. This is the casual, daywear look.
the polyester is slightly harder to sew, as it is light weight. It gives a more evening wear look.


SOURCES: Wikipedia: List of countries by life expectancy; (1) United Nations Population Division. 2009. World Population Prospects: The 2008 Revision. New York, United Nations, Department of Economic and Social Affairs (advanced Excel tables), (2) Census reports and other statistical publications from national statistical offices, (3) Eurostat: Demographic Statistics, (4) Secretariat of the Pacific Community: Statistics and Demography Programme, and (5) U.S. Census Bureau: International Database.
"Congo, Republic of the Life expectancy Stats, NationMaster." 1960-2008. <http://www.nationmaster.com/country-info/profiles/Congo,-Republic-of-the/Health/Life-expectancy>.
"Congo, Republic of the Life expectancy Stats", NationMaster [Internet]. 1960-2008. Avaliable from: <http://www.nationmaster.com/country-info/profiles/Congo,-Republic-of-the/Health/Life-expectancy>.
"Congo, Republic of the Life expectancy Stats", NationMaster. Avaliable at: nationmaster.com. Assessed 1960-2008.


Kranjska Gora is a town in northwestern Slovenia, on the Sava Dolinka River in the Upper Carniola region, close to the Austrian and Italian borders. Eschew the tourist crowds and head to Lake Jasna and Vrsic Pass - Julian Alps. Get in touch with nature at Smučarska Šola ASK Kranjska Gora and Intersport Bernik. Explore the numerous day-trip ideas around Kranjska Gora: Radovljica Old Town (in Radovljica) and Kobarid Museum (Kobariski muzej) (in Kobarid). Next up on the itinerary: hit the slopes at Planica Nordic Centre, pause for some serene contemplation at Church of the Assumption of Virgin Mary, appreciate the history behind Statue of Kekec, and try your luck at Casino Korona.
For traveler tips, reviews, other places to visit, and other tourist information, refer to the Kranjska Gora itinerary website .
Venice, Italy to Kranjska Gora is an approximately 3-hour car ride. Prepare for slightly colder weather when traveling from Venice in April: high temperatures in Kranjska Gora hover around 16°C and lows are around 4°C. Wrap up your sightseeing on the 9th (Tue) early enough to drive to Bohinjsko Jezero.
Take a break from the city and head to Mostnica Gorge and Mangrtska Cesta - Die mautpflichtige Mangartstrasse. Explore hidden gems such as Sabotin and Kayaking & Canoeing. Explore Bohinjsko Jezero's surroundings by going to Kojsko (Smartno & Razgledni stolp Gonjače), Denkmal Am Predilpass (in Bovec) and Bjana (in Dobrovo). The adventure continues: head outdoors with Positive sport, don't miss a visit to Monument To Four Courageous Men, and browse the exhibits of Trg Evrope.
For other places to visit, more things to do, ratings, and more tourist information, refer to the Bohinjsko Jezero trip planner .
Drive from Kranjska Gora to Bohinjsko Jezero in 1.5 hours. Alternatively, you can take a bus. In April, daytime highs in Bohinjsko Jezero are 16°C, while nighttime lows are 4°C. On the 14th (Sun), wrap the sightseeing up by early afternoon so you can travel to Strunjan.
Strunjan is a settlement in the Municipality of Piran in the Littoral region of Slovenia.OverviewThe Strunjan Nature Reserve, located along a 4 km stretch of Adriatic coast to the north of the settlement, is the longest section of unspoilt coastline in the entire Gulf of Trieste.The parish church in the settlement is dedicated to the Virgin Mary.EventsSince 2001, Strunjan has held an annual persimmon festival (Praznik kakijev) every November. Start off your visit on the 14th (Sun): get a sense of history and politics at Town Hall, take in the spiritual surroundings of The Cloisters of Saint Francis Monastery, and then have fun and explore at Mediadom Pyrhani. On the 15th (Mon), you'll have a packed day of sightseeing: look for all kinds of wild species at Naravni rezervat Škocjanski zatok, then take in the spiritual surroundings of Church of the Holy Maurus, and then stroll through Svetilnik Beach.
Plan my trip to Strunjan with suggested itineraries provided by Inspirock.
Traveling by car from Bohinjsko Jezero to Strunjan takes 2.5 hours. Prepare for a bit warmer weather when traveling from Bohinjsko Jezero in April: high temperatures in Strunjan hover around 21°C and lows are around 9°C. You'll set off for Postojna on the 17th (Wed).
Kick off your visit on the 17th (Wed): step into the grandiose world of Dvorec Zemono, then learn about winemaking at Petric Winery, and then stop by Gustarna. Here are some ideas for day two: explore the fascinating underground world of Skocjan Caves, explore the fascinating underground world of Divaska jama, then take in nature's colorful creations at Ferrari Garden, and finally learn about winemaking at Wines Jamsek.
For other places to visit, photos, maps, and other tourist information, read our Postojna tour planner.
You can drive from Strunjan to Postojna in an hour. Traveling from Strunjan in April, expect slightly colder with lows of 5°C in Postojna. Cap off your sightseeing on the 19th (Fri) early enough to go by car to Ljubljana.
Start off your visit on the 19th (Fri): ponder the world of politics at Town Hall (Magistrat). On the 20th (Sat), you'll have a packed day of sightseeing: make a trip to Mestni Trg, take a stroll through Ljubljana Old Town, glide peacefully along the water with some of the best stand-up paddleboarding in the area, then don't miss a visit to Old Square (Stari trg), and finally explore the activities along River Ljubljanica Kanal.
To find where to stay, maps, traveler tips, and other tourist information, you can read our Ljubljana online trip itinerary maker .
Traveling by car from Postojna to Ljubljana takes an hour. Alternatively, you can take a bus; or take a train. April in Ljubljana sees daily highs of 19°C and lows of 5°C at night. Wrap up your sightseeing on the 21st (Sun) to allow time to drive back home.


Washington Area Bike Forum > Forum Topics > Group Rides > Shirlington to Eastern Market - today!
View Full Version : Shirlington to Eastern Market - today!
This is very late notice, but I'm riding from Shirlington to meet up with the Greenbelt/College Park group at Eastern Market this afternoon, leaving Shirlington around 1 p.m., in case anyone would like to come along. PM me if interested. It's a pretty day for a ride!


The work study program is funded by Student Awards and Financial Aid (SAFA) with the objective to provide additional financial assistance through on-campus part-time employment opportunities for students with documented financial need.
Email documents to: svprograms@uvic.ca. Positions will remain open until filled.
Due to the HIGH volume of applicants, only successful candidates will be contacted for an interview.


text:
Les Roses noires est un film documentaire français réalisé par Hélène Milano et sorti en 2012.

Synopsis
Cinq adolescentes - Farida, Claudie, Coralie, Kahina et Moufida - issues des cités de la Seine-Saint-Denis ou des quartiers nord de Marseille sont invitées à s'exprimer face à la caméra. Elles nous parlent de leur rapport à la France, de leur relation à la langue de ce pays et à ses habitants. Comment ces jeunes filles perçoivent-elles précisément les garçons ? On y parle d'attachement au groupe mais aussi du machisme et du sentiment d'exclusion. Ces filles font part de leurs pensée et de leur expérience ainsi que de leur rêve de quitter bien qu'elles y sont attachées : leurs quartiers  afin de mener une vie convenable en s'insérant dans la vie active et de s'épanouir. Un film citoyen et militant sur les filles de la banlieue..

Fiche technique
 Titre du film : Les Roses noires
 Réalisation : Hélène Milano
 Photographie : Chloé Blondeau, Jérôme Olivier, Alexandra Potet
 Son : Camille Barrat, Marianne Roussy
Montage : Martine Armand
 Musique : Bruno Angelini
 Production : Comic Strip Production, FR 3 Méditerranée, France 3 Paris-Île-de-France
 Pays d'origine :
 Durée : 74 minutes
 Sortie :  :

Références

Liens externes

Film français sorti en 2012
Film documentaire français
Film documentaire sur l'adolescence
Film documentaire sur la banlieue française

redpajama_set_name: RedPajamaWikipedia


Trap League currently meets Sunday mornings at 9am and Wednesday evenings beginning at 5:30pm. Trap meets are open to all members and guests, including minors 12 and up. One position will be open for doubles--that's two clays per call-out.


text:
The Swiss Family Orbison
Photograph by Irving Miskell Reid Some Press:- The band, their music and their live shows have been well received in the media. Here's the press kit: Album Review appeared: Mojo / February Writer: James McNair "A slightly confusing debut but one bolstered by some superb three-part harmonies, playful yet acute lyrics and that perennial favourite, the cleverly structured pop song. How much more mileage there is in the Scottish fixation with the West Coast sound is debatable but, like a scuzzier yet somehow more mainstream Teenage Fanclub this lot argue the case convincingly. Clearly Kit Clark's talent was somewhat subsumed by his brother Gary's in Danny Wilson." Album Review appeared: Vox / January Writer: Stephen Dalton "The prevailing mood here is defined by Teenage Fanclub-esque powerpop belters such as I'm In Love and The Wall of Pain. There's more than a whiff of Alex Chilton's bittersweet twang to fuzz-pedal serenade The Luckiest Man In The World, while A Girl I Don't Know is a sublime, slightly stoned falsetto love letter to a stranger." Live Venue: Borderline, London Review appeared: Time Out / 7.1.98 Writer: Rosie Wilby "This live date coincides with the release of a sel-titled album, the making of which forms a disaster-ridden tale of a persistent band winning through in the end. And the resulting twelve songs are an impressive display of energetic, catchy, crunchy song-writing. Suicide particularly, is fun and funky with a singalong chorus which Teenage Fanclub would no doubt steal if they were they not such nice chaps, while the I Don't Know What To Say and And A Girl I Don't Know are endearingly sweet ballads with wobbly organs and downbeat acoustic strummery in all the right places." Live Venue: Borderline, London Review appeared: London Evening Standard / 7.1.98 Writer: Max Bell "These chaps provided a splendid reminder of the simple joys of harmony and structure with a selection of numbers from their self-financed debut disc. The mordant Wall Of Pain - shades of Springsteen here - and the semi-shocking paen they call Suicide, complete with the coda from Lennon's Mother, merely emphasised what the major labels are missing if they don't check Clark and company out soon. He (Clark) is an extremely gifted vocalist with the kind of charm that doesn't need unnecessary attitude to convince an audience." Live Venue: Radio Scotland Live At Lemon Tree Session Review appeared: The Scotsman / 29.11.97 Writer: Tom Morton "There is a real magic in hearing unfamiliar music on the radio that reaches out and shakes you with its brilliance. On Sunday, driving home I tuned into Radio Scotland to hear Anne Marie Middleton say something amid applause about a song called Let's Go To Wick. The idea of a song about that Caithness town was almost enough to send me sliding off the road in excitement. At last, justice for Wick! And then I heard some acoustic guitars and shimmering harmonies slide into what was clearly a Danny Wilson-esque song structure, this time called I'm In Love. The Lyrics were brilliantly sardonic and also very naughty indeed. Hugely impressed I determined to buy the album when it came out. And then the following day the itself - not due until January - arrived out of the blue. It was as if some divine plot was under way by which I was to be convinced of the band's brilliance. Enough, already, I am."
Playlists: 1
Videos Albums The Swiss Family Orbison I'm in love
© Musical1 - Music Community 19/01/2020


BEAVER - An Aliquippa man agreed Thursday to plead guilty to aggravated assault for a shooting outside a New Brighton bar Feb. 8.
For his plea, according to court records, Vernon D. Redd, 27, was expected to be sentenced to between nine and 23 months in county jail, followed by five years of probation.
New Brighton Area police said Redd shot Darnese Raines, 33, of New Brighton, outside the Barnhouse Inn, 824 Third Ave., around 2:20 a.m. Feb. 8 during an argument.
Raines was shot in the chest, police said.
Police said the pair had argued about Greg Slade. In October, Slade, 21, of Rochester, who is Raines' cousin, was accused of sexually assaulting a family member of Redd's girlfriend.
Redd told police that during the confrontation outside the bar, Raines punched him, so Redd pushed him to the ground, and then Redd drew a gun from his clothing and fired one round at Raines.
A sentencing date hadn't been set as of Thursday afternoon. Redd made the plea agreement after he waived his right to a preliminary hearing Thursday.
Jordan Brown, 11, has been charged as an adult with criminal homicide and criminal homicide of an unborn child in the death of Kenzie Houk, 26, his father's fiancee, who was due to deliver a son in two weeks.
Brown could be moving from Allencrest Juvenile Detention Center in Brighton Township to the Beaver County Jail in Hopewell Township. Beaver County would charge Lawrence County about $4,500 per week to keep Brown at Allencrest and provide 24-hour security, or $1,400 per week to lodge him in the county jail.


Of course the evidence might suggest our four-legged friend might actually be the author, as goats do walk/trott faster than humans... the time frame may not be as out-landish as long as we imagine Brandon looking a bit like this on Chase's shoulders!
On behalf of the Minions of the Goat (Aka the author of Let the Music Play) Welcome to the forums and all the great tales around here.
Great thing about Cj's stories is that he releases each chapter on a timely, usually weekly, basis!
Welcome to you, Flip69. The post is here so you musta done somethin' right!
Um, while that is a nice pic, you do realize that's a sheep on his shoulders, right?
Welcome to the GA forums!!! If you need any help of advice, please don't hesitate to ask.
Hrmmm, best be carful crediting Shadowgod with writing these stories; he might sue us all for libel!
Well, 'tis true, goats do walk faster then humans.
Actually, Emoe, it's downhill both ways, so he's able to make good time.
All jokes aside, I do tease a lot about errors, but there's one line in the credits that says it all: "All remaining errors are mine alone".
If I didn't make 'em, the errors wouldn't be there, so no, I won't share credit for the goofs; they're mine, mine I tell ya!
"Let the Music Play" is a great story. I love Brandon, he's such an innocent, he's so clueless. Gabe's about to get his, and I hope we learn why he has it in for Brandon in chapter 5.
Oh, yes I know it's a sheep... well it used to be. According to the article that the pic was connected to, that sheep was on his way to be dinner... but that's beside the point.
I mean, I had to pick a four-legged maminal that was related in appearance to Mr. Goat, but was waaay cuter. Of course, he just happens to be carrying a sheep on his shoulders.
...with so many evil characters, it would be hard pressed to get a conviction of the "tarnisher". Brandon is a little slow :wacko: could be that he needs more substance than pizza (apparently a Goats favorite food). Maybe give him some fruit & juice, maybe the brain will start working again. If not send him my house to rewire my sound system, I could pay him with a couple of PB&J sandwiches.
Thanks! Brandon is indeed the "fish out of water", but this is all new to him; the story timeline has only been a few days so far.
Ch 5 should be along shortly, BTW. I just have to get it formatted into HTML, and then I can send it off to Joe for posting.
Awww, but what's this? No sunglasses?
What? Evil characters in my stories?? I have no idea what you could possibly mean... I always try and write sweet, lovable characters who may, of course, have a few minor personality flaws.
Brandon has an excuse... he's blond. I'm assuming you're referring to him being oblivious regarding Chase? Chase seems equally oblivious, but then again, he's blond too.
............... of course Blond's!! Forget the V-8!


Earlier this week, Liseberg theme park in Sweden came forth with a major announcement regarding a huge project moving forward in the park's near future. That announcement comes in the form of a twisted mess of an amazing new coaster set to open at the park in 2014. Coming in at a cost of €22 million ($28.5 million, 200 million kronor), the new roller coaster will be the biggest investment in the park's history.
According to the report from the Goteborg Daily, Projekt Helix will be longer than any other at the park at 4545 feet long (1.4 km), and faster as well at 62 mph (100 kph). Work is beginning shortly on the coaster, with plans to blast off in spring 2014. The ride will sport a green color scheme, helping the coaster blend into the nearby terrain, which from artwork and videos released is something it will utilize heavily. To make room for Projekt Helix, Liseberg will be closing it's 3-D cinema Maxxima, which opened in 2001.
Construction is set to begin later this autumn after the park closes for the season, with plans to open the ride by the spring of 2014.
© COPYRIGHT COASTER-net.com 1998-2016. ALL RIGHTS RESERVED.


The three local Ghana unions in Sweden (Stockholm, Gothenburg, Malmö) have pledge to support the Ghana Mission and the newly appointed ambassador, H.E Mrs. Amerley Ollennu Awua-Asamoa, to make her the most successful ambassador to the Nordic region.
The statement was made when a delegation of executives of the three Ghana unions led by Gilbert Mensah, president of Ghana union of Gothenburg, paid a courtesy visit to the Ghana embassy to congratulate and welcome Awua-Asamoa on her appointment as the ambassador to the Nordic countries, and to follow up on an earlier proposal about a request for an honorary consular in Sweden.
In addressing the proposal for a consulate office in Sweden, Awua-Asamoa mentioned that two people have already been interviewed by the Mission, of which one is recommended to occupy a consular office in Stockholm and the other in Gothenburg and its surrounding cities.
Awua-Asamoa added that she believes in an open door policy to get closer to the Ghanaian communities. According to her, the Ghana government's policy of transitioning from aid to trade will need the help of Ghanaians in the diaspora, of which her office is willing to assist individuals or group of people who intend to establish something back home in Ghana. To support this policy, the secretary of Ghana union of Malmö, Patrick Frimpong, gave an assurance that the Ghanaian community will collaborate with the Ghana Mission to engage influential companies and people in Sweden to help achieve the vision of the Ghana government.
Awua-Asamoa expressed the Mission's willingness to fully collaborate with the Ghana unions and to also support the three unions to strengthen and have an umbrella union as done in other countries. "I want us to compliment each other. I want to leave this office better than I met it," she said. According to her, strengthening the unions will help the Mission to truly identify genuine Ghanaians, since the unions will recommend applicants before they apply for any consular services.
Frimpong, on behalf of the delegation, presented a gift to the Ambassador and the Mission as a sign of the unions' preparedness to partner with the Mission.


Basic Pentesting : 2 is a boot2root VM and is a continuation of the Basic Pentesting series by Josiah Pierce. This series is designed to help newcomers to penetration testing and to develop pentesting skills. Have fun exploring part of the offensive side of security.
The file format is OVF and can be imported to VirtualBox without problem. It also works flawlessly with NAT Network interface. The IP address can be obtained by DHCP.
It can be downloaded from VulnHub.


When I assign preformatted type to the text block, I can not create the block of another type after preformatted block. Enter, ctrl + enter, shift + enter, alt + enter just create new lines in preformatted block.
Bug was found in Firefox and Opera: Firefox: 2.0.0.5 / Linux (Gentoo) Opera: 9.22 / Linux (Gentoo) It does not present in IE6 (ie4linux).
Fixed for FF - see r447.
Won't fix for Opera until SAPI implemented.
Won't fix in Opera since pre element will be deprecated in next version.


March 23rd, 2009 at 18:24 EET | Categories: Humour, Miscellaneous | Track comments. You can skip to the end and leave a response. Pinging is currently not allowed.
Maybe she thought, she will have nothing to do. I dont understand such people.


text:
Kościół Najwyższego Arcykapłana Jezusa Chrystusa – rzymskokatolicki kościół parafialny należący do parafii pod tym samym wezwaniem (dekanat gośliński archidiecezji gnieźnieńskiej. Znajduje się na osiedlu Zielone Wzgórza, w Murowanej Goślinie, w województwie wielkopolskim.

Kościół został wzniesiony w latach 1995-2000. Jego budowniczym był ksiądz Roman Kostecki. Świątynia została konsekrowana przez arcybiskupa Henryka Muszyńskiego w dniu 12 czerwca 2005 roku. Budowla posiada współczesną architekturę, do jej wnętrza wchodzi się przez łukowatą zadaszoną bramę, w oknach są umieszczone witraże przedstawiające m.in. Drogę Krzyżową Chrystusa. Do świątynnego kompleksu należą także salki katechetyczne i kaplica pod wezwaniem św. Wojciecha.

Przypisy

Świątynie pod wezwaniem Chrystusa
Najwyższego Arcykapłana Jezusa Chrystusa
Budynki w Polsce oddane do użytku w 2000


text:
Lucille Ingrassia
There are condolences waiting approval on Lucille's Tribute wall
Plant a tree in memory of Lucille
Celebrating the life of Lucille Ingrassia
Be the first to share your favorite memory, photo or story of Lucille. This memorial page is dedicated for family, friends and future generations to celebrate the life of their loved one.
Obituary for Lucille Ingrassia
Lucille P. Ingrassia 85 of Pekin passed away on Tuesday January 26, 2010 at Amy's Country Manor in Pekin. Born in Sharon, WI on February 10, 1924 to Carl and Mary Militrani Ingrassia. She is survived by her sister Mary (Louis) Barclay of Boise, ID, brother Mario (Betty) Ingrassia of Pekin and several nieces, nephews and great nieces and nephews. Her parents, brother Carmine Grecie and sisters Esther Ranney and Ann Ingrassia preceded her in death. Lucille worked at Keystone Steel and Wire Co. as a Secretary for many years retiring in 1976 and was a member of the Secular Franciscan Order of Sacred Heart Fraternity where she served as Secretary for 7 years, was a devoted member of St. Joseph Catholic Church and was a member of the Council of Catholic Women. Lucille enjoyed helping children all of her life and was known for always handing out candy to all the neighborhood children. In accordance to her wishes there will be no services or visitation and burial will be private. Memorials may be made in her name to St. Joseph Church, Harbor Lights Hospice or St. Jude. Arrangements have been entrusted to Abts Mortuary.
To plant a tree in memory of Lucille Ingrassia, visit the Tribute Store.
Welcome to Lucille's Memorial Wall
Share Lucille Ingrassia's life celebration
"Email Address" would like to share the life celebration of Lucille Ingrassia. Click on the "link" to go to share a favorite memory or leave a condolence message for the family.


São Paulo Symphony Orchestra's Digital Label - Free streaming and download, free music for all.
Suíte Breve was recorded in September 2014, and Evocação de Augusto Meyer in March 2012, at Sala São Paulo.
All rights reserved. Not authorized for sale.


text:
\section{Introduction}

Let
$T>0$, $n \in \N$,  $\Omega\subset \R^n$ be a bounded domain with sufficiently smooth boundary $\pp\Omega$,
We set $Q= (0,T) \times \Omega$ and $\Sigma= (0,T) \times \pp\Omega$.
We denote  by $\nu=\nu(x)$. the outwards unit normal vector to $\pp\Omega$ at $x\in \pp\Omega$

We consider
the following parabolic integro-differential equations:
\begin{equation}
\label{eq:intpara_eq}
\pp_t u(t,x) - L u(t,x) - \int_0^t K(t,\tau)u(\tau,x)\,d\tau
=g(t,x), \quad
(t,x)\in Q,
\end{equation}
where $L$  is the uniformly elliptic operator:
\begin{equation*}
L
=
\sum_{i,j=1}^n \pp_i (a_{ij}(t,x)\pp_j )
+\sum_{j=1}^n b_j (t,x) \pp_j
+ c(t,x),\quad (t,x)\in Q,
\end{equation*}
and $K(t,\tau)$ is the following memory kernel operator:
\begin{align*}
&K(t,\tau)
=
\sum_{i,j=1}^n \pp_i (\wa_{ij}(t,\tau,x)\pp_j )
+\sum_{j=1}^n \wb_j (t,\tau,x) \pp_j
+ \wc(t,\tau,x),
\quad
(t,\tau,x)\in (0,T)^2\times \Omega,
\end{align*}
with $a_{ij}\in C^1 (\overline{Q})$, $b_j, c \in L^\infty (Q)$,
$\wa_{ij}\in C^1 (\overline{(0,T)^2\times \Omega})$, $\wb_j, \wc \in L^\infty ((0,T)^2\times \Omega)$
for $i,j=1,\ldots,n$. We assume that
$a_{ij}=a_{ji}$
for $i,j=1,\ldots,n$
and that
there exists a constant $\mu>0$ such that
\begin{align*}
&\frac1\mu |\xi|^2 \leq \sum_{i,j=1}^n a_{ij}(t,x) \xi_i \xi_j \leq \mu |\xi|^2,			& (t,x) \in Q, \quad \xi  \in \R^n.
\end{align*}

In this article,  we use the parabolic integro-differential operator $P$ defined by
\begin{equation*}
P u(t,x)
=
\pp_t u(t,x) - L u(t,x) - \int_0^t K(t,\tau)u(\tau,x) \,d\tau .
\end{equation*}
We define the co-normal derivative for $L$ by
$
\frac{\pp u}{\pp \nu_L} = \sum_{i,j=1}^n a_{ij} (\pp_i u)\nu_j.
$

The parabolic integro-differential equations \eqref{eq:intpara_eq}
arise from various studies such as
the anomalous diffusion/transport in the heterogeneous media \cite{DT2006, FP2016},
the thermal conduction in materials with memory \cite{N1971, P1993}
and so on.
As an example,
if we investigate the non-Fickian diffusion with memory effects,
we may derive integro-differential equations of parabolic type \eqref{eq:intpara_eq}
as the hybrid model for the classical and anomalous diffusion.
(see \cite{FP2016}).
On the well-posedness in the direct problems
for parabolic integro-differential equations under appropriate assumptions for $L$, $K$, $g$ and initial and boundary value conditions,
we may find a lot of previous results
(see e.g., \cite{AT1986, AH2014, KJ2011, LP1992, Z1993} and the references therein).

In this article,
we firstly derive a time-like Carleman estimate for \eqref{eq:intpara_eq}.
A Carleman estimate is a weighted $L^2$ inequality and
a priori estimate for partial differential equations.
A Carleman estimate has various applications such as inverse problems, unique continuations, the control theory (see e.g., \cite{H, Is1, KT, LRS}).
Here the ``time-like'' means that the weight function of the Carleman estimate is depending on only the time variable $t$
and independent of the space variables $x$.
In \cite{K1978, LRS, LP1961},
we may find this kind of the time-like estimates
which are used to prove the unique continuation for partial differential equations.
As a remark, we may refer to the weight energy method in \cite{AS, P}
in which we may see the idea similar to time-like Carleman estimates.
Related to the applications to inverse problems,
Yamamoto \cite{Y} introduced the time-like Carleman estimate
with the $x$-independent weight function $e^{-\la t}$ where $\la$ is a large parameter and
he applied it to an inverse source problem and a backward problem for parabolic equations.
As for the time-like Carleman estimate, we also refer to \cite{Kawa} in which
the estimate for the degenerate parabolic equations was derived.

In relation to the Carleman estimates for the parabolic integro-differential equations,
we refer to  works by A. Lorenzi et al. \cite{L2012, L2013a, L2013b, LL2014,  LLY2016, LM2012}.
In their article, they treat the case of
that the integrands of the Volterra type integral in the equations are the first or zero-th order in space,
that is, the case of $\wa_{ij}=0$ for $i,j=1,\ldots,n $  in  \eqref{eq:intpara_eq}.

Our strategy to prove the time-like Carleman estimate is mainly relies on the result for the parabolic case.
In our case, however, the difficulty arises from an integral term
containing second order partial derivative in space.
To estimate the integral term in \eqref{eq:intpara_eq},
we introduce a key integral inequality containing the weight function of the Carelman estimate
(see e.g., Lemma 3.1 in \cite{K2}, Lemma 3.3.1 in \cite{KT}).
Using the time-linke Carleman estimate for parabolic equations in \cite{Y} and the key integral inequality,
we prove our Carleman estimate for \eqref{eq:intpara_eq}.

Secondly we discuss an inverse source problem for the parabolic integro-differential equations \eqref{eq:intpara_eq}.
We assume that $\Omega$ is a cylindrical shaped domain and
the coefficients of $L$ is independent of the one component of space variables.
And we establish the uniqueness and the conditional stability estimate of H\"older type
in the determination of the time dependent source factor from boundary data.

There are a lot of previous researches on inverse problems for parabolic integro-differential equations.
In particular, a number of articles are intensively devoted to the identifications of memory kernels  (see e.g., \cite{C2009, CL1997, JW1998, JW1999, JW2000, LP1992, LS1988} and the references therein).
Related to inverse source problems, we may refer to \cite{JK2009, KJ2011, LM2010}.
In \cite{KJ2011}, they consider the inverse problem of determining the source term from spatial data at the final time.

However, there are not many works on the uniqueness and the stability results
using boundary data as observations.
Moreover, stability estimates in inverse source problems are not studied well for parabolic integro-differential equations.

To prove the uniqueness and the stability in our inverse problem,
we adopt the Bukhgeim-Klibanov method.
Bukhgeim and Klibanov \cite{BK} established the global uniqueness results in inverse problems
by means of the Carleman estimate.
More precisely, we may refer to some monographs and surveys \cite{Is1, K1, K2, KT} and references therein.
To apply the Bukhgeim-Klibanov method,
our key tools are the time-like Carleman estimate and some integral inequalities.
Most related to our result, we refer to \S9 in \cite{Y} in which the time-like Carelman estimate
is applied to prove the uniqueness in the inverse source problem for parabolic equations in a cylindrical shaped domains.
Here we may also refer to a related work \cite{GK2013}.

The plan of this article is as follows:
In section 2, we state and show the time-like Carleman estimate for \eqref{eq:intpara_eq} (Theorem 2.3)
by using the key integral inequality (Lemma2.1).
In section 3, we state the inverse source problem
and give our main results, that is, the uniqueness (Theorem 3.1) and the stability estimate (Theorem 3.2) in our inverse problem.
In section 4, we prove our main theorems (Theorem 3.1 and 3.2).

\section{Carleman Estimate}
In this section, we state and show
the time-like Carleman estimate for
the parabolic integro-differential equation.

We consider the following weight function:
\begin{equation*}
\va (t)=e^{-\la (t-T)}, \quad t \in (0,T) .
\end{equation*}

Firstly we prove the following key integral inequality.
\begin{lem}
\label{lem:keyineq01}
For all $z \in L^2 (Q)$, we have
\begin{align*}
&\int_Q \left( \int_0^t z(\tau,x) d\tau \right)^2 e^{2s\va}\,dtdx \\
&\leq
C\int_Q \frac1{s\la\va}|z(t,x)|^2 e^{2s\va}\,dtdx
+
C\int_Q \frac1{s^2\la}|z(t,x)|^2 e^{2s\va}\,dtdx.
\end{align*}
\end{lem}
\begin{proof}
We note that
\begin{equation*}
\va \geq 1, \quad
\pp_t (e^{2s\va}) = -2 s\la\va e^{2s\va}, \quad
\pp_t \left(\frac{1}{\va}\right)=\frac{\la}{\va} \leq \la \quad \text{in $(0,T)$. }
\end{equation*}
By the Cauchy-Schwarz inequality and the integration by parts, we have
\begin{align*}
&\int_Q \left( \int_0^t z(\tau,x) d\tau \right)^2 e^{2s\va}\,dtdx\\
&\leq T \int_Q \left( \int_0^t |z(\tau,x)|^2 d\tau \right)  e^{2s\va}\, dtdx
\\
&=-\frac{T}{2s\la} \int_\Omega \int_{0}^T \frac1{\va} \left( \int_0^t |z(\tau,x)|^2 d\tau \right) \pp_t (e^{2s\va})\, dtdx\\
&=-\frac{T}{2s\la} \int_\Omega \frac1{\va(T)} \left( \int_0^T |z(\tau,x)|^2 d\tau \right) e^{2s\va(T)}\, dx \\
&\quad
+\frac{T}{2s\la} \int_\Omega \int_{0}^T \pp_t \left(\frac1{\va}\right) \left( \int_0^t |z(\tau,x)|^2 d\tau \right) e^{2s\va}\, dtdx \\
&\quad
+\frac{T}{2s\la} \int_\Omega\int_{0}^T \frac1{\va} |z(t,x)|^2 e^{2s\va}\,dtdx\\
&\leq
\frac{T}{2}
\int_Q\frac1{s\la \va} |z(t,x)|^2  e^{2s\va} \,dtdx
+\frac{T}{2s} \int_Q \left( \int_0^t |z(\tau,x)|^2 d\tau \right)\va e^{2s\va}\, dtdx \\
&=
\frac{T}{2}
\int_Q\frac1{s\la \va} |z(t,x)|^2  e^{2s\va} \,dtdx
-\frac{T}{4s^2\la} \int_\Omega \int_{0}^T\left( \int_0^t |z(\tau,x)|^2 d\tau \right) \pp_t (e^{2s\va})\, dtdx \\
&=
\frac{T}{2}
\int_Q\frac1{s\la \va} |z(t,x)|^2  e^{2s\va} \,dtdx
-\frac{T}{4s^2\la} \int_\Omega \left( \int_0^T |z(\tau,x)|^2 d\tau \right) e^{2s\va(T)}\, dtdx \\
&\quad
+\frac{T}{4s^2\la} \int_\Omega \int_{0}^T |z(t,x)|^2 e^{2s\va}\, dtdx \\
&\leq
C
\int_Q\frac1{s\la \va} |z(t,x)|^2  e^{2s\va} \,dtdx
+
C \int_Q \frac1{s^2\la} |z(t,x)|^2 e^{2s\va}\, dtdx .
\end{align*}
Thus we conclude this lemma.
\end{proof}

To derive our main Caleman estimate,
we use the time-like Carleman estimate established by Yamamoto \cite{Y}.
We may obtain the following inequality in the proof of Theorem 9.3 in \cite{Y}.
\begin{lem}
\label{lem:tce01}
There exists $\la_0>1$ such that for any $\la>\la_0$,
we may choose $C=C(\la_0)>0$ satisfying the following inequality
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t u|^2 + \sum_{i,j=1}^n |\pp_i\pp_j u|^2 \right) + \la |\nabla u|^2 + s\la^2\va |u|^2 \right] e^{2s\va}\,dtdx\\
&\leq C
\int_Q |(\pp_t -L)u|^2 e^{2s\va}\,dtdx
+C\int_{ \Sigma} s\la \va \left(|\pp_t u| + |u| \right) \left|\frac{\pp u}{\pp \nu_L} \right|e^{2s\va}\,dtdS \nonumber \\
&\quad
+C\int_Q  \la^2 |u|^2e^{2s\va}\,dtdx,
\label{eq:tce01}
\end{align}
for all $s>1$ and all $u \in H^1(0,T;H^2(\Omega))$ with $u(0,x)=u(T,x)=0$, $x\in \Omega$.
\end{lem}

Using Lemma \ref{lem:keyineq01} and Lemma \ref{lem:tce01},
we may prove the following Carleman estimate for the parabolic integro-differential equation \eqref{eq:intpara_eq}.

\begin{thm}
There exists $\la_0>1$ such that for any $\la>\la_0$,
we may choose $s_0(\la)>1$ satisfying:
there exists $C^\prime=C^\prime(s_0,\la_0)>0$ such that
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t u|^2 + \sum_{i,j=1}^n |\pp_i\pp_j u|^2 \right) + \la |\nabla u|^2 + s\la^2\va |u|^2 \right] e^{2s\va}\,dtdx\\
&\leq C^\prime
\int_Q |Pu|^2 e^{2s\va}\,dtdx
+C^\prime\int_{\Sigma} s\la \va \left(|\pp_t u| + |u| \right) \left|\frac{\pp u}{\pp \nu_L} \right|e^{2s\va}\,dtdS,
\label{eq:tce_intpara}
\end{align}
for all $s>s_0$ and all $u \in H^1(0,T;H^2(\Omega))$ with $u(0,x)=u(T,x)=0$, $x\in \Omega$.
\end{thm}
\begin{proof}
Note that
\begin{equation*}
\pp_t u(t,x)-Lu(t,x)
=Pu(t,x) +\int_0^t K u(t,\tau)u(\tau,x)\,d\tau ,\quad (t,x)\in Q.
\end{equation*}
By Lemma \ref{lem:tce01}, we have
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t u|^2 + \sum_{i,j=1}^n |\pp_i\pp_j u|^2 \right) + \la |\nabla u|^2 + s\la^2\va |u|^2 \right] e^{2s\va}\,dtdx\\
&\leq C
\int_Q |Pu|^2 e^{2s\va}\,dtdx
+C\int_{\Sigma} s\la\va \left(|\pp_t u| + |u| \right) \left|\frac{\pp u}{\pp \nu_L} \right|e^{2s\va}\,dtdS \nonumber \\
&\quad
+C\int_Q \left| \int_0^t  K(t,\tau)u(\tau,x)\,d\tau\right|^2 e^{2s\va}\,dtdx
+C\int_Q \la^2 |u|^2 e^{2s\va}\,dtdx.
\label{eq:tce101}
\end{align}
We estimate the third term on the right-hand side of \eqref{eq:tce101} by Lemma \ref{lem:keyineq01}.
\begin{align*}
&\int_Q \left|\int_0^t  K(t,\tau)u(\tau,x)\,d\tau\right|^2 e^{2s\va}\,dtdx \\
&\leq C \int_Q \left| \int_0^t  \sum_{i,j=1}^n |\pp_i\pp_j u(\tau,x)|\,d\tau\right|^2 e^{2s\va}\,dtdx \\
&\quad
	+ C \int_Q \left| \int_0^t  \left( |\nabla u(\tau,x)|+|u(\tau,x)|\right)\,d\tau\right|^2 e^{2s\va}\,dtdx\\
&\leq C\int_Q \frac{1}{s\la\va} \sum_{i,j=1}^n |\pp_i\pp_ju|^2 e^{2s\va}\,dtdx
+ C\int_Q \frac{1}{s^2\la} \sum_{i,j=1}^n |\pp_i\pp_j u|^2 e^{2s\va}\,dtdx \\
&\quad
+ C\int_Q \left( |\nabla u|^2 + |u|^2 \right)e^{2s\va}\,dtdx ,
\end{align*}
for $\la>\la_0$ and $s>1$.
Together this with \eqref{eq:tce101}, we may obtain
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t u|^2 + \sum_{i,j=1}^n |\pp_i\pp_j u|^2 \right) + \la |\nabla u|^2 + s\la^2\va |u|^2 \right] e^{2s\va}\,dtdx\\
&\leq C
\int_Q |Pu|^2 e^{2s\va}\,dtdx
+\int_{ \Sigma} s\la\va \left(|\pp_t u| + |u| \right) \left|\frac{\pp u}{\pp \nu_L} \right|e^{2s\va}\,dtdS\nonumber \\
&\quad
+ C\int_Q \left( \frac{1}{s\la\va} \sum_{i,j=1}^n |\pp_i\pp_ju|^2 + |\nabla u|^2\right) e^{2s\va}\,dtdx \nonumber \\
&\quad
+ C\int_Q \left( \frac{1}{s^2\la} \sum_{i,j=1}^n |\pp_i\pp_j u|^2 + \la^2 |u|^2 \right) e^{2s\va}\,dtdx ,
\label{eq:tce02}
\end{align}
for $\la>\la_0$ and $s>1$.
Taking sufficiently large $\la>\la_0$, and then, choosing
sufficiently large $s>1$, we may absorb the third and forth terms on the right-hand side of \eqref{eq:tce02}
into the left-hand side.
Thus, we conclude the \eqref{eq:tce_intpara}.
\end{proof}

\section{Inverse Source Problem}
Let $\ell>0$, $n \in \N$, $n\geq 2$, $D\subset \R^{n-1}$ be a bounded domain
with sufficiently smooth boundary $\pp D$.
We set a cylindrical shaped domain and its boundaries:
$\Omega=(0,\ell)\times D$,  $Q=(0,T)\times \Omega$,
$\Sigma_0=(0,T)\times \{0,\ell\}\times \overline{D}$,
$\Sigma_1=(0,T)\times (0,\ell)\times \pp D$, $\Sigma =(0,T)\times \pp\Omega=\Sigma_0\cup \Sigma_1$.
We use the notation $x=(x_1,x^\prime)\in \Omega$ where $x_1 \in (0,\ell)$ and $x^\prime=(x_2,\ldots, x_n) \in D$.

Let $R(t,x)$, $(t,x)\in Q$ be a given function.
We consider
\begin{equation}
\label{eq:ibvp}
\left\{
\begin{aligned}
&P^\prime u (t,x)=f(t,x^\prime) R(t,x),	&(t,x)\in Q,			\\
&u(0,x)=0,								&x\in \Omega,		\\
&u(t,x)=\pp_1 u(t,x)=0,					&(t,x)\in \Sigma_0,
\end{aligned}
\right.
\end{equation}
where $f\in L^2((0,T)\times D)$,
\begin{equation*}
P^\prime u(t,x)
=
\pp_t u(t,x) - L^\prime u(t,x) - \int_0^t K(t,\tau)u(\tau,x)\,d\tau ,
\quad (t,x)\in Q,
\end{equation*}
and $L^\prime$ and $K(t,\tau)$ are the following operators:
\begin{align*}
L^\prime
&=
\sum_{i,j=1}^n \pp_i (a^\prime_{ij}(t,x^\prime )\pp_j )
+\sum_{j=1}^n b_j (t,x) \pp_j
+ c(t,x) ,
\quad (t,x)\in Q,\\
K (t,\tau)
&=
\sum_{i,j=1}^n \pp_i (\wa_{ij}(t,\tau,x)\pp_j )
+\sum_{j=1}^n \wb_j (t,\tau,x) \pp_j
+ \wc(t,\tau,x),
\quad (t,\tau,x)\in (0,T)^2\times \Omega,
\end{align*}
with
$a_{ij}\in C^1 ([0,T]\times \overline{D})$,
$\wa_{ij}\in C^1 ([0,T]^2; C^2 ([0,\ell]; C^1(\overline{D})))$,
$b_j, c\in L^\infty (0,T;C^1([0,\ell];L^\infty (D)))$,
$\wb_j, \wc \in L^\infty ((0,T)^2;C^1([0,\ell];L^\infty (D)))$
for $i,j=1,\ldots,n$.
We assume that
$a^\prime _{ij}=a^\prime _{ji}$
for $i,j=1,\ldots,n$
and that
there exists a constant $\mu>0$ such that
\begin{align*}
&\frac1\mu |\xi|^2 \leq \sum_{i,j=1}^n a^\prime_{ij}(t,x^\prime) \xi_i \xi_j\leq \mu |\xi|^2,		& (t,x^\prime) \in (0,T)\times D,\quad \xi  \in \R^n.
\end{align*}
Here we note that
$f$ and
$a^\prime_{ij}$, $i,j=1,\ldots,n$ are
independent of $x_1$.

We assume that
\begin{equation}
\label{eq:Rcondi}
R\in C^1 ([0,T];C^3([0,\ell];C^2(\overline{D}))),\quad |R(t,x)|\neq 0,\quad (t,x)\in \overline{Q}.
\end{equation}

Let $t_0 \in (0,T)$.  We investigate the following inverse problem.

\noindent
\textbf{Inverse Source Problem}:
Can we determine the time dependent source factor $f(t,x^\prime)$, $(t,x^\prime)\in (0,t_0)\times D$
from boundary data on $\Sigma_1$?

\begin{thm}[Uniqueness]
\label{thm:uni}
Let $R$ satisfy \eqref{eq:Rcondi}.
We assume that $u, \pp_1 u \in H^1(0,T;H^2 (\Omega))$ and $u$ satisfies \eqref{eq:ibvp}.
Moreover we suppose that there exists a constant $M>1$ such that
$\| u\|_{L^2(Q)}+\|\pp_1 u\|_{L^2 (Q)} \leq M$ and $\|f\|_{L^2((0,T)\times D)}\leq M$.
If $u(t,x)=0$, $(t,x)\in \Sigma_1$, then we have $f(t,x^\prime)=0$, $(t,x^\prime)\in (0,t_0)\times D$.
\end{thm}

\begin{thm}[Stability estimate]
\label{thm:stab}
Under the assumptions of Theorem \ref{thm:uni},
there exist constants $\kappa \in (0,1)$ and $C>0$ such that
\begin{equation}
\label{eq:stab}
\| f\|_{L^2((0,t_0)\times D)}
\leq C  B^\kappa,
\end{equation}
where
\begin{align*}
B&=
\| \pp_t \pp_1 u \|_{L^2(\Sigma_1)}
+
\| \pp_t  u \|_{L^2(\Sigma_1)}
+
\| \pp_1  u \|_{L^2(\Sigma_1)}
+
\|  u \|_{L^2(\Sigma_1)}\\
&\quad
+
\left\| \frac{\pp (\pp_1u) }{\pp \nu_{L^\prime}}  \right\|_{L^2(\Sigma_1)}
+
\left\| \frac{\pp u }{\pp \nu_{L^\prime}}  \right\|_{L^2(\Sigma_1)} .
\end{align*}
\end{thm}

\begin{rmk}
We may not derive the uniqueness result (Theorem \ref{thm:uni}) directly by the stability estimate (Theorem \ref{thm:stab}).
In the proof of the stability estimate, however, we may obtain the uniqueness in our inverse source problem.
Hence we prove Theorem \ref{thm:uni} and Theorem \ref{thm:stab} in the next section  simultaneously.
\end{rmk}

\section{Proofs of Theorems}

Setting $u=Rv$ in $\overline{Q}$,
by \eqref{eq:ibvp}, we obtain
\begin{align}
\nonumber
&\pp_t v- L^\prime v
-2\sum_{i,j=1}^n \frac1{R} a^\prime_{ij}(\pp_i R) \pp_j v  \\
&
+\left( \frac{\pp_t R}{R}
- \sum_{i,j=1}^n \frac1{R} \pp_i (a_{ij}^\prime \pp_j R )
- \sum_{j=1}^n \frac1{R} b_j \pp_j R \right) v \nonumber \\
&
-\int_0^t \frac{R(\tau, x)}{R(t,x)} K(t,\tau) v (\tau, x)\,d\tau \nonumber \\
&
-\int_0^t \frac{1}{R(t,x)}
\Biggl [
	2\sum_{i,j=1}^n \wa_{ij}(t,\tau, x) \pp_i R(\tau, x)\pp_j v (\tau, x)\nonumber \\
	&
	\qquad
	+\sum_{i,j=1}^n  \pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) v(\tau, x)
	+\sum_{j=1}^n \wb_j (t,\tau, x) \pp_j R(\tau,x) v(\tau, x)
\Biggr]\,d\tau \nonumber \\
&=f(t,x^\prime) ,	\quad (t,x)\in Q,
\label{eq:ibvp_v}
\end{align}
and $v(0,x)=0$,	 $x\in \Omega$,
$v(t,x)=\pp_1 v(t,x)=0$,	 $(t,x)\in \Sigma_0$.

Differentiate the parabolic integro-differential equation of \eqref{eq:ibvp_v} with respect to $x_1$ and set $w=\pp_1 v$.
Then we get
\begin{align}
\nonumber
&\pp_t w- L^\prime w
-2\sum_{i,j=1}^n \frac1{R} a^\prime_{ij}(\pp_i R) \pp_j w  \\
&
+\left( \frac{\pp_t R}{R}
- \sum_{i,j=1}^n \frac1{R} \pp_i (a_{ij}^\prime \pp_j R )
- \sum_{j=1}^n \frac1{R} b_j \pp_j R \right) w \nonumber \\
&
-\int_0^t \frac{R(\tau, x)}{R(t,x)} K(t,\tau)w (\tau, x)\,d\tau \nonumber \\
&
-\int_0^t \frac{1}{R(t,x)}
\Biggl [
	2\sum_{i,j=1}^n \wa_{ij}(t,\tau, x) \pp_i R(\tau, x)\pp_j w (\tau, x)\nonumber \\
	&
	\qquad
	+\sum_{i,j=1}^n  \pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) w(\tau, x)
	+\sum_{j=1}^n \wb_j (t,\tau, x) \pp_j R(\tau,x) w(\tau, x)
\Biggr]\,d\tau \nonumber \\
&
-
\sum_{j=1}^n (\pp_1 b_j)\pp_j v
+2\sum_{i,j=1}^n  \pp_1 \left( \frac1{R} a^\prime_{ij} \pp_i R \right) \pp_j v \nonumber \\
&
- \left[
 \pp_1 c
-\pp_1
	\left(
	\frac{\pp_t R}{R}
	- \sum_{i,j=1}^n \frac1{R} \pp_i (a_{ij}^\prime \pp_j R )
	- \sum_{j=1}^n \frac1{R} b_j \pp_j R
	\right)
\right] v \nonumber \\
&
-\int_0^t \pp_1 \left(\frac{R(\tau, x)}{R(t,x)}\right)  K(t,\tau) v (\tau, x)\,d\tau \nonumber \\
&
-\int_0^t  \pp_1 \left(\frac{1}{R(t,x)}\right)
\Biggl [
	2\sum_{i,j=1}^n \wa_{ij}(t,\tau, x) \pp_i R(\tau, x)\pp_j v (\tau, x)\nonumber \\
	&
	\qquad
	+\sum_{i,j=1}^n  \pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) v(\tau, x)
	+\sum_{j=1}^n \wb_j (t,\tau, x) \pp_j R(\tau,x) v(\tau, x)
\Biggr]\,d\tau \nonumber \\
&
-\int_0^t \frac{R(\tau, x)}{R(t,x)}
\Biggl [
	\sum_{i,j=1}^n \pp_1 \wa_{ij}(t,\tau, x) \pp_i \pp_j v (\tau, x)
	+ \sum_{i,j=1}^n \pp_i \pp_1 \wa_{ij}(t,\tau, x)  \pp_j v (\tau, x) \nonumber \\
	&
	\qquad
	+\sum_{j=1}^n \pp_1 \wb_j (t,\tau, x) \pp_j v(\tau, x)
	+ \pp_1 \wc (t,\tau, x) v(\tau, x)
\Biggr]\,d\tau
\nonumber \\
&-\int_0^t \frac{1}{R(t,x)}
\Biggl [
	2\sum_{i,j=1}^n \pp_1 (\wa_{ij}(t,\tau, x) \pp_i R(\tau, x)) \pp_j v (\tau, x)
	+\sum_{i,j=1}^n  \pp_1\pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) v(\tau, x) \nonumber \\
	&\qquad
	+\sum_{j=1}^n \pp_1 (\wb_j (t,\tau, x) \pp_j R(\tau,x)) v(\tau, x)
\Biggr]\,d\tau
=0 ,	\quad (t,x)\in Q,			 \nonumber
\end{align}
and
$w(0,x)=0$, $x\in \Omega$,
$w(t,x)=0$, $(t,x)\in \Sigma_0$.

Here we remark the representation of $v$ by $w$. By $w(t,0,x^\prime)=0$, $(t,x^\prime)\in (0,T) \times D$ and
\begin{equation}
\label{eq:rep_v1}
\pp_1 v(t,x)= w(t,x),\quad (t,x)\in Q,
\end{equation}
we have
\begin{equation}
\label{eq:rep_v2}
v(t,x)=\int_0^{x_1} w (t,\zeta,x^\prime)\,d\zeta, \quad (t,x)\in Q,
\end{equation}
and then
\begin{equation}
\label{eq:rep_v3}
\left\{
\begin{aligned}
\pp_t v(t,x)&=\int_0^{x_1} \pp_t w (t,\zeta,x^\prime)\,d\zeta, & (t,x)\in Q, \\
\pp_j v(t,x)&=\int_0^{x_1} \pp_j w (t,\zeta,x^\prime)\,d\zeta, & (t,x)\in Q, \\
\pp_1 \pp_1 v(t,x)&=\pp_1 w (t,x),& (t,x)\in Q, \\
\pp_1 \pp_j v(t,x)&=\pp_j w (t,x),  & (t,x)\in Q, \\
\pp_i \pp_j v(t,x)&= \int_0^{x_1} \pp_i\pp_j w (t,\zeta,x^\prime)\,d\zeta, & (t,x)\in Q,
\end{aligned}
\right.
\end{equation}
for $i,j=2,\ldots, n$.

Choose $t_1, t_2 \in (0,T)$ such that $0<t_0<t_1<t_2<T$.
Set $\delta_k =e^{-\lambda (t_k-T)}$ for $k=0,1$.
Let $\chi \in C^\infty (\R)$ be a cut off function such that
\begin{equation*}
0 \leq \chi (t) \leq 1, \ t \in \R, \quad
\pp_t X (t) \leq 0,\ t\in \R, \quad
\chi (t)=
\left\{
\begin{aligned}
1,\quad t<t_1,\\
0,\quad t\geq t_2.
\end{aligned}
\right.
\end{equation*}

Setting $y=\chi w$, we obtain
\begin{align}
\nonumber
&\pp_t y- L^\prime y
-2\sum_{i,j=1}^n \frac1{R} a^\prime_{ij}(\pp_i R) \pp_j y  \\
&
+\left( \frac{\pp_t R}{R}
- \sum_{i,j=1}^n \frac1{R} \pp_i (a_{ij}^\prime \pp_j R )
- \sum_{j=1}^n \frac1{R} b_j \pp_j R \right) y \nonumber \\
&
=
\chi \int_0^t \frac{R(\tau, x)}{R(t,x)}
K(t,\tau) w (\tau, x)
\,d\tau \nonumber \\
&
+
\chi \int_0^t \frac{1}{R(t,x)}
\Biggl [
	2\sum_{i,j=1}^n \wa_{ij}(t,\tau, x) \pp_i R(\tau, x)\pp_j w (\tau, x)\nonumber \\
	&
	\qquad
	+\sum_{i,j=1}^n  \pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) w(\tau, x)
	+\sum_{j=1}^n \wb_j (t,\tau, x) \pp_j R(\tau,x) w(\tau, x)
\Biggr]\,d\tau \nonumber \\
&
+
\sum_{j=1}^n (\pp_1 b_j) \chi\pp_j v
-2\sum_{i,j=1}^n  \pp_1 \left( \frac1{R} a^\prime_{ij} \pp_i R \right) \chi \pp_j v \nonumber \\
&
+ \left[
 \pp_1 c
-\pp_1
	\left(
	\frac{\pp_t R}{R}
	- \sum_{i,j=1}^n \frac1{R} \pp_i (a_{ij}^\prime \pp_j R )
	- \sum_{j=1}^n \frac1{R} b_j \pp_j R
	\right)
\right] \chi v \nonumber \\
&
+\chi \int_0^t  \pp_1 \left(\frac{R(\tau, x)}{R(t,x)}\right)
K(t,\tau) v (\tau, x)
\,d\tau\nonumber
\\
&+\chi \int_0^t  \pp_1 \left(\frac{1}{R(t,x)}\right)
\Biggl [
	2\sum_{i,j=1}^n \wa_{ij}(t,\tau, x) \pp_i R(\tau, x)\pp_j v (\tau, x)\nonumber \\
	&
	\qquad
	+\sum_{i,j=1}^n  \pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) v(\tau, x)
	+\sum_{j=1}^n \wb_j (t,\tau, x) \pp_j R(\tau,x) v(\tau, x)
\Biggr]\,d\tau \nonumber \\
&
+\chi \int_0^t \frac{R(\tau, x)}{R(t,x)}
\Biggl [
	\sum_{i,j=1}^n \pp_1 \wa_{ij}(t,\tau, x) \pp_i \pp_j v (\tau, x)
	+ \sum_{i,j=1}^n \pp_i \pp_1 \wa_{ij}(t,\tau, x)  \pp_j v (\tau, x) \nonumber \\
	&
	\qquad
	+\sum_{j=1}^n \pp_1 \wb_j (t,\tau, x) \pp_j v(\tau, x)
	+ \pp_1 \wc (t,\tau, x) v(\tau, x)
\Biggr]\,d\tau \nonumber \\
&
+\chi \int_0^t \frac{1}{R(t,x)}
\Biggl [
	2\sum_{i,j=1}^n \pp_1 (\wa_{ij}(t,\tau, x) \pp_i R(\tau, x)) \pp_j v (\tau, x)\nonumber \\
	&
	\qquad
	+\sum_{i,j=1}^n  \pp_1\pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) v(\tau, x) \nonumber \\
	&\qquad
	+\sum_{j=1}^n \pp_1 (\wb_j (t,\tau, x) \pp_j R(\tau,x)) v(\tau, x)
\Biggr]\,d\tau
=(\pp_t \chi)w,	\quad (t,x)\in Q,
\label{eq:ibvp_y}
\end{align}
and $y(0,x)=y(T,x)=0$, $x\in \Omega$, $y(t,x)=0$, $(t,x)\in \Sigma_0$.

Applying the Carleman estimate for the parabolic equations (Lemma \ref{lem:tce01}) to \eqref{eq:ibvp_y} and
noting that the boundedness for coefficients $a^\prime_{ij}, \wa_{ij}, b_j, \wb_j, c ,\wc$ and $R$,
we have
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t y|^2 + \sum_{i,j=1}^n |\pp_i\pp_j y|^2 \right) + \la |\nabla y|^2 + s\la^2\va |y|^2 \right] e^{2s\va}\,dtdx\\
&\leq C
\int_Q |(\pp_t\chi)w|^2 e^{2s\va}\,dtdx
+C\int_Q   \la^2 |y|^2  e^{2s\va}\,dtdx
+C \sum_{k=1}^5 I_k +CB_1,
\label{eq:thmce1}
\end{align}
for $\la > \la_0$ and $s>1$,
where
\begin{align*}
I_1&=\int_Q \left| \chi \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)| \,d\tau \right|^2 e^{2s\va}\,dtdx, \\
I_2&=\int_Q \left| \chi \int_0^t \left(\sum_{j=1}^n |\pp_j w (\tau, x)| + |w(\tau, x)| \right) \,d\tau\right|^2 e^{2s\va}\,dtdx,\\
I_3&=\int_Q \left( \sum_{j=1}^n |\chi \pp_j v|^2 + |\chi v|^2 \right)e^{2s\va}\,dtdx, \\
I_4&=\int_Q \left| \chi \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j v (\tau, x)| \,d\tau \right|^2 e^{2s\va}\,dtdx, \\
I_5&=\int_Q \left| \chi \int_0^t \left(\sum_{j=1}^n |\pp_j v (\tau, x)| + |v(\tau, x)| \right) \,d\tau\right|^2 e^{2s\va}\,dtdx,
\end{align*}
and
\begin{equation}
\nonumber
B_1=
\int_{ \Sigma_1} s\la\va \left(|\pp_t y| + |y| \right) \left|\frac{\pp y}{\pp \nu_{L^\prime}} \right|e^{2s\va}\,dtdS.
\end{equation}
Henceforth we estimate from $I_1$ to $I_5$ on the right-hand side of \eqref{eq:thmce1}
by using the technique in the proof of Lemma \ref{lem:keyineq01}.

\begin{align}
\nonumber
I_1&=
\int_Q \left| \chi \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)| \,d\tau \right|^2 e^{2s\va}\,dtdx \\
&
\leq
C \int_Q \chi^2 \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right) e^{2s\va}\,dtdx \nonumber \\
&=-\frac{C}{2s\la} \int_Q \chi^2 \times \frac1{\va} \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right) \pp_t( e^{2s\va})\,dtdx  \nonumber \\
&=
-\frac{C}{2s\la} \int_\Omega (\chi(T))^2 \times \frac1{\va(T)} \left(  \int_0^T \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right)  e^{2s\va(T)}\,dx \nonumber \\
&\quad
+
\frac{C}{s\la} \int_Q \chi (\pp_t \chi) \times \frac1{\va} \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right)  e^{2s\va}\,dtdx  \nonumber \\
&\quad
+\frac{C}{2s\la} \int_Q \chi^2  \pp_t \left(\frac1{\va}\right) \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right) e^{2s\va}\,dtdx \nonumber  \\
&\quad
+\frac{C}{2s\la} \int_Q \chi^2 \times \frac1{\va} \sum_{i,j=1}^n |\pp_i \pp_j w (t, x)|^2   e^{2s\va}\,dtdx  \nonumber \\
&\leq
C \int_Q  \frac1{s\la\va} \sum_{i,j=1}^n |\pp_i \pp_j y (t, x)|^2   e^{2s\va}\,dtdx  \nonumber \\
&\quad
+C \int_Q \frac1{s} \chi^2   \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right)  \va e^{2s\va}\,dtdx \nonumber \\
&=
C \int_Q  \frac1{s\la\va} \sum_{i,j=1}^n |\pp_i \pp_j y (t, x)|^2   e^{2s\va}\,dtdx  \nonumber \\
&\quad
-C \int_Q \frac1{2s^2\la} \chi^2   \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right)  \pp_t (e^{2s\va})\,dtdx \nonumber \\
&=
C \int_Q  \frac1{s\la\va} \sum_{i,j=1}^n |\pp_i \pp_j y (t, x)|^2   e^{2s\va}\,dtdx  \nonumber \\
&\quad
-C \int_\Omega \frac1{2s^2\la} \chi(T)^2   \left(  \int_0^T \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right)  e^{2s\va(T)}\,dx \nonumber  \\
&\quad
+C \int_Q \frac1{s^2\la} \chi (\pp_t \chi)   \left(  \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j w (\tau, x)|^2  \,d\tau \right)  e^{2s\va}\,dtdx \nonumber \\
&\quad
+C \int_Q \frac1{2s^2\la} \chi^2  \sum_{i,j=1}^n |\pp_i \pp_j w (t, x)|^2  e^{2s\va}\,dtdx \nonumber \\
&\leq
C \int_Q  \frac1{s\la\va} \sum_{i,j=1}^n |\pp_i \pp_j y |^2   e^{2s\va}\,dtdx
+C\int_Q \frac1{s^2\la} \sum_{i,j=1}^n |\pp_i \pp_j y |^2  e^{2s\va}\,dtdx,
\label{eq:T1}
\end{align}
for $\la>\la_0$ and $s>1$.
Here we note that $\pp_t \chi \leq 0$ in $\R$.
\begin{align}
\nonumber
I_2&=
\int_Q \left| \chi \int_0^t \left(\sum_{j=1}^n |\pp_j w (\tau, x)| + |w(\tau, x)| \right) \,d\tau\right|^2 e^{2s\va}\,dtdx \\
&
\leq
C \int_Q \chi^2 \left[  \int_0^t \left(\sum_{j=1}^n |\pp_j w (\tau, x)|^2 + |w(\tau, x)|^2 \right) \,d\tau \right] e^{2s\va}\,dtdx \nonumber \\
&\leq
C \int_Q \chi^2  \left[  \int_0^t \left(\sum_{j=1}^n |\pp_j w (\tau, x)|^2 + |w(\tau, x)|^2 \right) \,d\tau \right] \va e^{2s\va}\,dtdx \nonumber \\
&=
-\frac{C}{2s\la} \int_Q \chi^2 \left[  \int_0^t \left( \sum_{j=1}^n |\pp_j w (\tau, x)|^2 + |w(\tau, x)|^2 \right) \,d\tau \right] \pp_t ( e^{2s\va})\,dtdx \nonumber \\
&=
-\frac{C}{2s\la} \int_Q (\chi(T))^2 \left[  \int_0^T \left( \sum_{j=1}^n |\pp_j w (\tau, x)|^2 + |w(\tau, x)|^2 \right) \,d\tau \right]  e^{2s\va(T)}\,dtdx \nonumber \\
&\quad
+\frac{C}{s\la} \int_Q \chi (\pp_t \chi) \left[  \int_0^t \left( \sum_{j=1}^n |\pp_j w (\tau, x)|^2 + |w(\tau, x)|^2 \right) \,d\tau \right]  e^{2s\va}\,dtdx \nonumber \\
&\quad
+\frac{C}{2s\la} \int_Q \chi^2\left( \sum_{j=1}^n |\pp_j w (t, x)|^2 + |w(t, x)|^2 \right) e^{2s\va}\,dtdx \nonumber \\
&\leq
C \int_Q \left( |\nabla y |^2 + |y|^2 \right) e^{2s\va}\,dtdx,
\label{eq:T2}
\end{align}
for $\la>\la_0$ and $s>1$.  \\
Noting that \eqref{eq:rep_v1} -- \eqref{eq:rep_v3},  we get
\begin{align}
\nonumber
I_3
&=\int_Q \left( \sum_{j=1}^n |\chi \pp_j v|^2 + |\chi v|^2 \right) e^{2s\va}\,dtdx \\
&
=
\int_Q  |\chi w|^2 e^{2s\va}\,dtdx  \nonumber \\
&\quad
+ \int_Q \left(\sum_{j=2}^n \left|\chi \int_0^{x_1} \pp_j w(x_1,\zeta, x^\prime)\,d\zeta \right|^2 + \left|\chi \int_0^{x_1} w(x_1,\zeta, x^\prime)\,d\zeta\right|^2 \right) e^{2s\va}\,dtdx
\nonumber \\
&
=
\int_Q  |y|^2 e^{2s\va}\,dtdx \nonumber \\
&\quad
+ \int_Q \left( \sum_{j=2}^n \left| \int_0^{x_1} \pp_j y(x_1,\zeta, x^\prime)\,d\zeta \right|^2 + \left| \int_0^{x_1} y(x_1,\zeta, x^\prime)\,d\zeta\right|^2 \right) e^{2s\va}\,dtdx
\nonumber \\
&\leq
\int_Q  |y|^2 e^{2s\va}\,dtdx
\nonumber \\
&\quad
+ \ell \int_D \int_0^\ell \int_0^T \left( \sum_{j=2}^n \int_0^{\ell} |\pp_j y(x_1,\zeta, x^\prime)|^2 \,d\zeta + \int_0^{\ell} |y(x_1,\zeta, x^\prime)|^2 \,d\zeta \right) e^{2s\va}\,dtdx_1dx^\prime
\nonumber \\
&=
\int_Q  |y|^2 e^{2s\va}\,dtdx
\nonumber \\
&\quad
+ \ell^2 \int_D \int_0^\ell \int_0^T \left( \sum_{j=2}^n  |\pp_j y(x_1,\zeta, x^\prime)|^2 +  y(x_1,\zeta, x^\prime)|^2 \right) e^{2s\va}\,dtd\zeta dx^\prime
\nonumber \\
&\leq
C\int_Q \left(  |\nabla y|^2 +  |y|^2 \right) e^{2s\va}\,dtdx.
\label{eq:T3}
\end{align}
By an argument similar to that used to derive \eqref{eq:T1},
we have
\begin{align}
\nonumber
I_4&=\int_Q \left| \chi \int_0^t \sum_{i,j=1}^n |\pp_i \pp_j v (\tau, x)| \,d\tau \right|^2 e^{2s\va}\,dtdx
\\
&\leq
C
\int_Q  \frac1{s\la \va} \sum_{i,j=1}^n |\chi \pp_i \pp_j v |^2 e^{2s\va}\,dtdx
+
C
\int_Q  \frac1{s^2 \la} \sum_{i,j=1}^n |\chi \pp_i \pp_j v |^2 e^{2s\va}\,dtdx,
\nonumber
\end{align}
for $\la>\la_0$ and $s>1$.
Noting that \eqref{eq:rep_v1}--\eqref{eq:rep_v3},
we may estimate the right-hand side of the above inequality
as similar as \eqref{eq:T3}, that is,
for $\la>\la_0$ and $s>1$,
we have
\begin{align*}
&\int_Q  \frac1{s\la \va} \sum_{i,j=1}^n |\chi \pp_i \pp_j v |^2 e^{2s\va}\,dtdx
+
\int_Q  \frac1{s^2 \la} \sum_{i,j=1}^n |\chi \pp_i \pp_j v |^2 e^{2s\va}\,dtdx.
\\
&\leq
C
\int_Q  \frac1{s\la \va} \sum_{i,j=1}^n |\pp_i \pp_j y |^2 e^{2s\va}\,dtdx
+
C
\int_Q  \frac1{s^2 \la} \sum_{i,j=1}^n |\pp_i \pp_j y |^2 e^{2s\va}\,dtdx\nonumber \\
&\quad
+
C
\int_Q   |\nabla y |^2 e^{2s\va}\,dtdx.
\end{align*}
Summing up the above two inequalities,
for $\la>\la_0$ and $s>1$,
we obtain
\begin{align}
\nonumber
I_4&\leq
C
\int_Q  \frac1{s\la \va} \sum_{i,j=1}^n |\pp_i \pp_j y |^2 e^{2s\va}\,dtdx
+
C
\int_Q  \frac1{s^2 \la} \sum_{i,j=1}^n |\pp_i \pp_j y |^2 e^{2s\va}\,dtdx  \\
&\quad
+
C
\int_Q   |\nabla y |^2 e^{2s\va}\,dtdx.
\label{eq:T4}
\end{align}
By the same argument as \eqref{eq:T2}, we have
\begin{align}
\nonumber
I_5&=
\int_Q \left| \chi \int_0^t \left(\sum_{j=1}^n |\pp_j v (\tau, x)| + |v(\tau, x)| \right) \,d\tau\right|^2 e^{2s\va}\,dtdx \\
&\leq
C\int_Q  \left(\sum_{j=1}^n |\chi \pp_j v|^2 + |\chi v|^2 \right)  e^{2s\va}\,dtdx.  \nonumber
\end{align}
Combining this
with \eqref{eq:T3}, we get
\begin{equation}
\label{eq:T5}
I_5\leq
C\int_Q  \left( |\nabla y|^2 + |y|^2 \right)  e^{2s\va}\,dtdx.
\end{equation}
Hence, by \eqref{eq:thmce1} and \eqref{eq:T1}-
\eqref{eq:T5},
we have
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t y|^2 + \sum_{i,j=1}^n |\pp_i\pp_j y|^2 \right) + \la |\nabla y|^2 + s\la^2\va |y|^2 \right] e^{2s\va}\,dtdx\\
&\leq C
\int_Q |(\pp_t\chi)w|^2 e^{2s\va}\,dtdx
+C\int_Q \left(  |\nabla y|^2 + \la^2 |y|^2 \right) e^{2s\va}\,dtdx \nonumber \\
&\quad
+
C
\int_Q  \frac1{s\la \va} \sum_{i,j=1}^n |\pp_i \pp_j y |^2 e^{2s\va}\,dtdx
+
C
\int_Q  \frac1{s^2 \la} \sum_{i,j=1}^n |\pp_i \pp_j y |^2 e^{2s\va}\,dtdx \nonumber \\
&\quad
 +CB_1, \nonumber
\end{align}
for $\la>\la_0$ and $s>1$.

Taking sufficiently large $\la>\la_0$,
and then,
choosing sufficiently large $s>1$,
we may absorb the second term to the forth term on the right-hand side of
the above inequality
into the left-hand side.
Hence there exists $\la_1>\la_0$ such that for any $\la>\la_1$, we may take $s_0>1$ satisfying:
there exists $C^\prime=C^\prime(s_0, \la_1)>$0 such that
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t y|^2 + \sum_{i,j=1}^n |\pp_i\pp_j y|^2 \right) + \la |\nabla y|^2 + s\la^2\va |y|^2 \right] e^{2s\va}\,dtdx\\
&\leq C^\prime
\int_Q |(\pp_t\chi)w|^2 e^{2s\va}\,dtdx
 +C^\prime B_1,
\label{eq:thmce3}
\end{align}
for all $s>s_0$.
Let us estimate the first term on the right-hand side of \eqref{eq:thmce3}.
\begin{align}
\nonumber
\int_Q |(\pp_t\chi)w|^2 e^{2s\va}\,dtdx
&
\leq
C^\prime e^{2s\delta_1} \int_\Omega\int_{t_1}^{t_2} |w|^2 \,dtdx   \\
&
\leq
C^\prime e^{2s\delta_1} \int_\Omega\int_{t_1}^{t_2} \left( |\pp_1 u|^2 + |u|^2 \right) \,dtdx \nonumber \\
&
\leq
C^\prime M^2 e^{2s\delta_1}.
\label{eq:thmce3r1}
\end{align}
Next we estimate the boundary terms $B_1$.
Note that $y=\chi w$, $w=\pp_1 v$, $v=u/R$.
\begin{align}
 \nonumber
B_1
&=
\int_{ \Sigma_1} s\la\va \left(|\pp_t y| + |y| \right) \left|\frac{\pp y}{\pp \nu_{L^\prime}} \right| e^{2s\va}\,dtdS \\
&\leq
C^\prime e^{C(\la)s} \int_{ \Sigma_1} \left(|\pp_t y| + |y| \right) \left|\frac{\pp y}{\pp \nu_{L^\prime}} \right|\,dtdS \nonumber \\
&\leq
C^\prime e^{C(\la)s} \int_{ \Sigma_1} \left(|\pp_t w| + |w| \right) \left|\frac{\pp y}{\pp \nu_{L^\prime}} \right|\,dtdS \nonumber \\
&=
C^\prime e^{C(\la)s} \int_{ \Sigma_1} \left(|\pp_t \pp_1 v| + |\pp_1 v| \right) \left|\frac{\pp (\pp_1 v)}{\pp \nu_{L^\prime}}  \right|\,dtdS \nonumber \\
&\leq
C^\prime e^{C(\la)s} \int_{ \Sigma_1} \left(|\pp_t \pp_1 u| + |\pp_t u| + |\pp_1 u| + |u| \right)  \nonumber \\
&\qquad\qquad\quad
						\times \left( \left|\frac{\pp (\pp_1 u)}{\pp \nu_{L^\prime}}  \right| +\left|\frac{\pp u}{\pp \nu_{L^\prime}} \right| +|\pp_1 u| +|u| \right)\,dtdS \nonumber \\
&=: C^\prime  e^{C(\la)s} B_2.
\label{eq:thmce3r2}
\end{align}
By \eqref{eq:thmce3}-
\eqref{eq:thmce3r2}, we obtain
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s\va} \left( |\pp_t y|^2 + \sum_{i,j=1}^n |\pp_i\pp_j y|^2 \right) + \la |\nabla y|^2 + s\la^2\va |y|^2 \right] e^{2s\va}\,dtdx\\
&\leq C^\prime M^2 e^{2s\delta_1}
 +C^\prime B_2 e^{C(\la)s} .
\label{eq:thmce4}
\end{align}
Henceforth we fix the sufficiently large parameter $\la>\la_1$.
By \eqref{eq:thmce4}, we see that there exist $s_1>s_0(\la)$ and $\widetilde{C}=\widetilde{C} (s_1, \la)>0$ such that
\begin{align}
\nonumber
&\int_Q \left[ \frac{1}{s} \left( |\pp_t y|^2 + \sum_{i,j=1}^n |\pp_i\pp_j y|^2 \right) +  |\nabla y|^2 + s |y|^2 \right] e^{2s\va}\,dtdx\\
&\leq \widetilde{C} M^2 e^{2s\delta_1}
 +\widetilde{C}B_2 e^{\widetilde{C}s},
\label{eq:thmce5}
\end{align}
for all $s>s_1$.

Taking $L^2((0,t_0)\times \Omega)$ norm to the equation for $v$ \eqref{eq:ibvp_v}, we have

\begin{align}
\nonumber
&\frac1{s} \| f \|^2_{L^2((0,t_0)\times D)} e^{2s\delta_0}\\
&\leq
\frac{C}{s}
\int_\Omega \int_0^{t_0}
\Biggl[ \pp_t v- L^\prime v
-2\sum_{i,j=1}^n \frac1{R} a^\prime_{ij}(\pp_i R) \pp_j v  \nonumber  \\
&\qquad
+\left( \frac{\pp_t R}{R}
- \sum_{i,j=1}^n \frac1{R} \pp_i (a_{ij}^\prime \pp_j R )
- \sum_{j=1}^n \frac1{R} b_j \pp_j R \right) v
\Biggr]^2 e^{2s\delta_0}\,dtdx \nonumber \\
&\quad
+\frac{C}{s}
\int_\Omega \int_0^{t_0}
\Biggl\{\int_0^t \frac{R(\tau, x)}{R(t,x)}
K(t,\tau) v (\tau, x)
\,d\tau
+\int_0^t \frac{1}{R(t,x)}
\Biggl [2\sum_{i,j=1}^n \wa_{ij}(t,\tau, x) \pp_i R(\tau, x)\pp_j v (\tau, x)\nonumber \\
	&
	\qquad\quad
	+\sum_{i,j=1}^n  \pp_i (\wa_{ij}(t,\tau, x)\pp_j R (\tau, x)) v(\tau, x)
	+\sum_{j=1}^n \wb_j (t,\tau, x) \pp_j R(\tau,x) v(\tau, x)
\Biggr]\,d\tau
\Biggr\}^2 e^{2s\delta_0}\,dtdx\nonumber \\
&\leq
\frac{C}{s}
\int_\Omega \int_0^{t_0}
\left( |\pp_t v|^2 +\sum_{i,j=1}^n |\pp_i\pp_j v|^2 + \sum_{j=1}^n |\pp_j v|^2 +|v|^2
\right) e^{2s\va}\,dtdx \nonumber \\
&\quad
+\frac{C}{s}
\int_\Omega \int_0^{t_0}
\int_0^{t_0}
\left( \sum_{i,j=1}^n |\pp_i\pp_j v(\tau, x)|^2 + \sum_{j=1}^n |\pp_j v(\tau, x)|^2 +|v(\tau, x)|^2
\right) e^{2s\va(\tau)}
\,d\tau \,dtdx \nonumber \\
&\leq
C
\int_\Omega \int_0^{t_0}
\left[ \frac1{s} \left(|\pp_t v|^2 +\sum_{i,j=1}^n |\pp_i\pp_j v|^2 \right)+ |\nabla v|^2 +|v|^2
\right] e^{2s\va}\,dtdx \nonumber \\
&\leq
C
\int_\Omega \int_0^{t_0}
\left[ \frac1{s} \left(|\pp_t y|^2 +\sum_{i,j=1}^n |\pp_i\pp_j y|^2 \right)+  |\nabla y|^2 +|y|^2
\right] e^{2s\va}\,dtdx, \nonumber
\end{align}
for all $s>s_1$.
In the last inequality, we used \eqref{eq:rep_v1} -- \eqref{eq:rep_v3} and $w=y$ in $(0,t_0)\times \Omega$.
Together this
with \eqref{eq:thmce5}, we obtain
\begin{equation}
\label{eq:f2}
\frac1{s} \| f \|^2_{L^2((0,t_0)\times D)} e^{2s\delta_0}
\leq \widetilde{C} M^2 e^{2s\delta_1}
 +\widetilde{C}B_2 e^{\widetilde{C}s},
\end{equation}
for all $s>s_1$.
Dividing the both hand side of \eqref{eq:f2} by $e^{2s\delta_1}/s$,
for all $s>s_1$,
\begin{equation*}
 \| f \|^2_{L^2((0,t_0)\times D)}
\leq \widetilde{C}s M^2 e^{-2s(\delta_0-\delta_1)}
 +\widetilde{C}s B_2 e^{\widetilde{C}s}.
\end{equation*}
Noting that $\delta_0-\delta_1>0$,
positive constants $C_1, C_2, D_1, D_2>0$ exist such that
\begin{equation*}
 \| f \|^2_{L^2((0,t_0)\times D)}
\leq s \left( C_1 M^2 e^{-D_1 s}+C_2 B_2 e^{D_2 s}  \right),
\end{equation*}
for all $s>s_1$.
Choosing $C_3>0$ such that $C_1 \leq C_3 e^{D_1s_1}$ and $C_2 \leq C_3 e^{-D_2s_1}$
and setting $\sigma=s-s_1$, for all $\sigma >0$, we have,
\begin{equation}
\label{eq:fsigma}
 \| f \|^2_{L^2((0,t_0)\times D)}
\leq C_3(\sigma +s_1) \left( M^2 e^{-D_1 \sigma}+ B_2 e^{D_2 \sigma}  \right).
\end{equation}

\begin{proof}[Proof of Theorem \ref{thm:uni}]
Since $u(t,x)=0$, $(t,x)\in \Sigma_1$, we have
\begin{equation*}
\pp_t \pp_1 u(t,x)
=\pp_t u(t,x)
=\pp_1 u(t,x)
=u(t,x)=0, \quad (t,x)\in \Sigma_1.
\end{equation*}
Hence, the boundary term vanishes, that is, $B_2=0$.
Thus we obtain the following estimate by \eqref{eq:fsigma}.
\begin{equation}
\nonumber
0\leq
 \| f \|^2_{L^2((0,t_0)\times D)}
\leq C_3  (\sigma +s_1) M^2 e^{-D_1 \sigma},
\end{equation}
for all  $\sigma >0$. As $\sigma$ goes to $\infty$,
the right-hand side of the above inequality
tends to $0$.
Therefore $ \| f \|_{L^2((0,t_0)\times D)}=0$, that is, $f(t,x^\prime)=0$, $(t,x^\prime)\in (0,t_0)\times D$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:stab}]
Firstly we estimate $B_2$ by $B$.
By the Cauchy Schwarz inequality, we have
\begin{align}
\nonumber
B_2
&=
 \int_{ \Sigma_1} \left(|\pp_t \pp_1 u| + |\pp_t u| + |\pp_1 u| + |u| \right)
 \left( \left|\frac{\pp (\pp_1 u)}{\pp \nu_{L^\prime}}  \right| +\left|\frac{\pp u}{\pp \nu_{L^\prime}} \right| +|\pp_1 u| +|u| \right)\,dtdS \\
&\leq
C \int_{ \Sigma_1} \left(|\pp_t \pp_1 u|^2 + |\pp_t u|^2 + |\pp_1 u|^2 + |u|^2 + \left|\frac{\pp (\pp_1 u)}{\pp \nu_{L^\prime}}  \right|^2 + \left|\frac{\pp u}{\pp \nu_{L^\prime}} \right|^2\right) \,dtdS \nonumber \\
&\leq C B^2. \nonumber
\end{align}
Together this with \eqref{eq:fsigma}, we obtain
\begin{equation}
\label{eq:fsigma2}
 \| f \|^2_{L^2((0,t_0)\times D)}
\leq C_3(\sigma +s_1) \left( M^2 e^{-D_1 \sigma}+ B^2 e^{D_2 \sigma}  \right).
\end{equation}

In the case of $B \geq M$, we may obtain the conditional stability estimate
by the a priori boundedness $\|f\|_{L^2((0,t_0)\times D)}\leq M$.

If $B=0$,  by an argument similar to the the proof of the uniqueness (Theorem \ref{thm:uni}),
we have $f(t,x^\prime)=0$, $(t,x^\prime)\in (0,T)\times D$.

Thus it is sufficient to suppose that $0<B<M$.
Choose
\begin{equation*}
\sigma = -\frac{2 \log \frac{B}{M}}{D_1+D_2}>0.
\end{equation*}
This $\sigma$ minimizes the right-hand side of \eqref{eq:fsigma2}.
Setting $\kappa = \frac{D_1}{D_1+D_2}$, we have
\begin{equation*}
 M^2 e^{-D_1 \sigma}+ B^2 e^{D_2 \sigma}
=2 M^{2(1-\kappa)}B^{2\kappa}.
\end{equation*}
Combining this with \eqref{eq:fsigma2}, there exists $C_4>0$ such that
\begin{equation*}
 \| f \|^2_{L^2((0,t_0)\times D)}
\leq C_4 (\sigma +s_1)M^{2(1-\kappa)}B^{2\kappa}.
\end{equation*}
Thus we conclude \eqref{eq:stab}.
\end{proof}

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