Years ago, I was vacationing with my mom. I love my mom dearly, but when you have to repeat things a thousand times, it can be tiring. Why did I have to repeat? She was hard of hearing. Those persons who are hard of hearing can be a challenge, especially when they say, "Why are you yelling? I can hear you just fine." Now, pair that up with having bad eyesight. A wonderful vacation awaits you!
The question is: To hear or not to hear. The answer is: What? Seriously, I was amazed when I learned that hearing loss is the third most common chronic condition among senior Americans, after hypertension and arthritis. Hearing challenges in elderly persons can lead to serious consequences, including social isolation, functional decline and depression. Hearing loss can also impair memory and cognitive functions, according to neuroscientists.
There are different causes for hearing impairment. Some are genetic, while others are congenital. Some causes are infectious. Others are traumatic. Some are toxic. Others are simply age-related. Still others are occupational related. There is temporary hearing loss and there is permanent hearing loss. Simple wax buildup can frequently cause hearing difficulties. Care should be taken to keep our ears clear of obstruction. We are advised to contact a medical professional when hearing problems adversely affect our life or hearing problems are persistent and unexplained or along with hearing difficulties we have ear pain.
At any rate, those with hearing challenges can often be helped, especially if they receive attention when hearing is not too far gone. However, in order to be helped, a few basic conditions must be met: The problem must be acknowledged. Consciousness of the illness is imperative of moving on to a solution. If we think that our hearing is just fine, we will have no need of doing anything about it — even when others recognize that there is a problem. Also, the problem must be assessed correctly and appreciated in its entirety. We may acknowledge a difficulty, but if we do not understand its gravity or internalize its complexity, we will be able to do little to cure it. Additionally, there has to be a desire to heal. And then something must be done. We might recognize the problem. We may even understand and appreciate it. We may even have a desire to get better. However, if we don't do anything about it, it will only get worse.
This process of recognition, assessment, appreciation, desire and action is so important in our health. This applies to our physical health, our psychological health and our spiritual health. What about our spiritual hearing? We can only be thankful if we are spiritually sound. We can be truly thankful if our spiritual hearing is whole. Alphonse Karr, a French journalist, once said, "Some people are always grumbling because roses have thorns; I am thankful that thorns have roses." It's a question of perception, of hearing.
November is the month designated to give thanks. Thanksgiving cannot be present without an acute spiritual perception. If we have a chronic spiritual hearing impairment, it will be hard for us to give thanks. Only those who are able to perceive benefices are able to give thanks. Only those who can appreciate good gifts will be able to return thanks. Ultimately, Thanksgiving is about: recognition of our blessings; true appreciation for them; a desire to respond to the giver of good things; and action to pay back, in some way, the benefices that we have received.
The Rev. Arthur Mollenhauer is the former pastor at St. Roch's Parish in Greenwich and current pastor at St. James Parish in Stratford.


Boasting a larger than usual swimming pool, Villa Paloma is a wonderful place for chilling out outdoors. It's also one of the only villas that has a bathroom for each of the bedrooms, with three in total of each.
There's a covered dining area, separate chill out area in both the sun and the shade, barbecue and amazing views across the resort to the sea. The decor is light and bright indoors, with an open plan lounge and dining zone, fitted kitchen and three generous sized bedrooms.
The villa's location in Las Coloradas means the heart of the resort is some distance so car hire is recommended with this property, although the closest supermarket is only about 7 minutes away on foot.


This 1960's style sofa is upholstered in green Lana wool from Linwood fabrics and is available to order in 56 contemporary colours. Perfect for a small space but large enough to seat two. Our sofas are made to order in a range of fabrics and we deliver throughout the UK. If you are looking for quality and individuality, then please do call or email to discover how we can make the perfect sofa for you. A sofa in this fabric costs £995.


This course has been designed specifically for persons working in the Property, Retail and Commercial sectors, such as Property Managers, Facilities Managers, Centre Managers and Surveyors.
The course focuses on the technique of risk assessment and the management of risk in a wide range of scenarios.
On completion of the course, candidates will be able to recognise health and safety hazards, carry out basic risk assessments and apply basic management principles and practices to health and safety issues.


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CXX ?= g++-7

CXXFLAGS += -std=c++1z -c -O3 -Wall -Wextra -pedantic -I /usr/local/include $(shell pkg-config --cflags opencv)
LDFLAGS += $(shell pkg-config --libs --static opencv)

all: example

example: example.o; $(CXX) $< -o $@ $(LDFLAGS)

%.o: %.cpp; $(CXX) $< -o $@ $(CXXFLAGS)

clean: $(RM) example.o

redpajama_set_name: RedPajamaGithub


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Follower chart that does not follow
This is the chart of a 32 years old lady who was a clerk in a company in Singapore until the middle of 2008. According to her brother, she was charmed (black magic) by a man in his sixties and she became mentally ill. Her family brought her back to Malacca, Malaysia for treatment but there was no improvement. Then her father took her to Thailand for a spiritual cure as her family believes this to be a spiritual problem.
From the technical point of view, this chart is a follower chart since none of the 7 palaces support the ren water DM. On top of that, there is the si-wu-wei seasonal summer fire frame that transforms into fire stars. The whole chart is really on fire. If this is a true follower chart, then this lady would have made her money easily since fire stars are her wealth stars. Her real life story tells us otherwise. She was just getting a clerical pay prior to her problem.
So, this is a weak ren water DM and fire stars are her negative stars besides wood and earth stars. People with strong negative fire stars are more prone to problems of the spiritual kind since fire also represent spirituality.
Multiple earth clashes
This is the chart of a 48 years old man who was once a millionaire. This is a strong xin DM and so his self, resource and officer stars are negative. Suitable for negative industry? (pian men)
He told me that he was already a millionaire in his late 20's during his yi-wei LP. Everything seems to work for him then and earning money was really easy. In his yi-wei LP, his wealth star yi was protruding and there were multiple chou-wei clashes that released his wood wealth stars from the storages of wei.
Moving into jia-wu LP, he continued to ride on his good luck as the protruding jia wood is his wealth star. He moved out of the negative industry and operated a successful mini market in this jia-wu LP.
Suddenly, he ran out of luck in the gui-si LP and lost everything due to bad investments and speculations. What happened in this gui-si LP? A strong DM does not want to meet more self stars. Meeting self stars means losing money. In this gui-si LP, there are 2 sets of si-you-chou metal frame that transform into metal stars supported by the you month branch and lured by the xin star in the year stem. Really bad!
One bad LP can cause us to lose everything. Knowing our bazi's can really give us an advantage and save us from losing everything. However, I believe that you need the Heaven's blessings in order to know of the impending danger.
2 sets of bullying punishment
This is the chart of a 24 years old lady who met a life changing accident in 2003, gui-wei year during her yi-you LP. She was 18 years old then. She was riding a motorcycle with her boyfriend sitting behind her when she suddenly fell on her own and crashed to the ground. Her left leg was badly smashed and it had to be amputated. Her boyfriend escaped with minor injuries.
During the yi-you LP, the you is forming a harm with the xu of the day branch and this harm is also known as the bodily harm. The actual problem of this chart is that it has 2 sets of xu-wei-chou bullying punishment.
Technically speaking, this is a fake follower chart. Thus, water and metal stars are not favourable. In 2003, the gui-wei AP is forming a fu-yin with the month pillar. Since the protruding gui water star is not favourable, this fu-yin is a negative one.
This lady has a very strong will power. Although she missed much of her schooling hours, she managed to sit for her pre-u exams the following year. Guess what? She did well enough to secure a place in a local university and has since graduated in 2008.
I wish her well.
Always broke
This is the chart of a 42 years old man who is always broke. His xin DM is a weak one. Without any resource star in the stem, his officer and wealth stars are negative. Unfortunately for him, his negative officer and wealth stars are strongly rooted.
The real problem is the mao-chen betrayal harm affecting his spouse palace. All his money has gone to all the women of his life.
First, he has given away more than RM20,000 to his first girlfriend after securing a bank loan for his car air-con spare parts business. It seems that the girl has never accepted his love.
For his 2nd love, he managed to marry the girl although his whole family objected. The girl decided to marry him after he gave her all his winnings from a lottery amounting to RM 16,000. Three years later, the lady decided to leave him by pretending to be mad. All in all, he has given her close to RM 30,000.
Soon after the divorce, he met another woman who has been divorced twice. After that, he started a cleaning company. At first, his business seems alright and he gave most of the earnings to her. Poor man! I wonder what he is thinking. Now that his business is bad, he is borrowing money everywhere just to give it to the woman.
He has a very lousy temper too. It is indicated by the ren water in the hour stem which is his negative hurting officer. He will not take advice from anybody. This is the saddest part.
This is the chart of a 3rd year university student in Singapore. He is known to have a photographic memory. His memory is definitely much more superior than an average university student. He obtained a maximum point of 5.0 in his first year CGPA, thanks to his powerful memory.
Can you see this from the image of the natal chart?
It's in the year, month and day pillars. Wu in the year stem and inside the year branch, geng in the month stem and inside the month branch, while yi is in the day stem and in the day branch.
Technical wise, this chart is a fake follower chart. So, his negative stars are the water and wood stars. This boy is from a very rich background as his wealth stars are strongly rooted in the year pillar.
However, his present ren-xu LP is an unfavourable one because the negative ren water is protruding. He told me that he had many operations on his liver when he was 14 years old. The yi wood DM being sandwiched by the xin and geng metal gives us the image of the operations.
Betrayal harm
This is the chart of a 33 years old lady. She has a double mao-chen harm in the natal chart. Mao-chen harm is also known as the betrayal harm.
She was married in her early twenties. Soon after giving birth to her 1st son, things started to go wrong. She has to work to support the child and her jobless husband. On top of that, she has to face her mother-in-law who has never liked her before.
Technical analysis suggests that her DM is a weak one although she was born in her resource month. Being a weak DM, her negative stars are the earth, metal and water stars.
Looking at the chart, her weak ding DM is being squeezed between gui, her negative spouse star and xin, her negative mother-in-law star. Right below the ding DM is the earth star, her negative children star. Poor thing! Being surrounded by 3 burdens in her negative ji-chou LP. Finally, when her situation reached the boiling point, she managed to divorce her husband who was please to let her have custody of the baby.
Soon after that, she was involved with a married man and recently she gave birth to another baby boy. She was hoping that this man will divorce his wife and marry her.
With the double betrayal harm in the natal chart, do you think her wish will come true?
In the present wu-zi LP, she needs to be careful as there is another double zi-mao punishment. I'm afraid that people might take advantage of her vulnerability.
However, there is a good side in this wu-zi LP. The wu will combine with gui and transform into positive fire stars which are her wealth stars and her positive husband star. She said that her ex-husband has recently started to pay up due to court order and the married man still sends her money every month. She's also getting a decent pay from a new job.
She told me that she's willing to let go of this married man if her Mr Right appears. A very wise move indeed.
Hope she will see the light at the end of the tunnel soon.
A tycoon's chart
This is the chart of Mr Li Ka-Shing, Hong Kong richest business tycoon.
Why is this chart so fantastic?
Like the previous chart, this is a fake vibrant chart. The favourable stars are the wood, fire and earth stars. For a fake vibrant chart, the self stars are also his wealth stars and they are in abundance and strongly rooted.
The brilliant part of the chart is that the useful god, fire stars are also strongly rooted.
Wish you have this kind of chart?
Top student
This is the chart of one of my students who obtained straight As in the pre-u exams 2 years ago. He is currently in the 2nd year in a local university studying to be a doctor. He's also a very talented drummer. A future doctor that rocks or a future rocker that cures?
Either way, his chart is a fake vibrant chart with earth stars in 7 palaces. His favourable stars are the fire and the earth stars. With so much strongly rooted earth self stars, he's a popular person among his friends and he should be a successful doctor as the earth self stars also represent his wealth.
However, there is a minor glitch in his chart. There is a double zi-wei confusion harm affecting the spouse palace and the positive spouse stars of earth and negative spouse star of water. I told him 2 years ago that these harms can cause him to be confused with his feelings. True enough, he broke off with his girlfriend recently, sighting lost of feelings for her. Perhaps, in this ji-chou year, another target has appeared as ji is revealed in the stem.
It is my intention in this blog to share my BaZi knowledge through case studies using mostly the ZiPing technical method. This technical method stresses on the production and the destruction cycles of the 5 elements namely wood, fire, earth, metal and water. However, in July 2016, I have decided to include the classical method as I believe that the correct approach towards BaZi is to blend both these methods.
Copyright © Bing Bazi 2013 - 2019. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


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Mike Huckabee Scolds Obamas for Letting Their Daughters Be Fans of Beyoncé
Diana Ozemebhoya Eromosele
During an interview with People magazine, Mike Huckabee—the former Arkansas governor and Republican presidential-primary candidate—described how he is flabbergasted that President Barack Obama and first lady Michelle Obama allow their daughters to be fans of Beyoncé, the New York Daily News reports.
"I don't understand how on one hand they can be such doting parents and so careful about the intake of everything—how much broccoli they eat and where they go to school, and making sure they're kind of sheltered and shielded from so many things," Huckabee explained before going in for the jugular: "And yet they don't see anything that might not be suitable for either a preteen or a teen in some of the lyrical content and choreography of Beyoncé," Huckabee told People magazine.
The ultraconservative, ordained Baptist minister is on a press tour to promote his new book, aptly titled, God, Guns, Grits, and Gravy.
Huckabee is also apparently bothered by Beyoncé's act and is concerned about how Jay Z is contributing to Beyoncé's image as a "sex object":
"Does it occur to him that he is arguably crossing the line from husband to pimp by exploiting his wife as a sex object?" Huckabee inquired.
Read more at the New York Daily News and People magazine.


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The Hand of the Lord in Scripture
Rick Crandall • Sermon • 2 years ago • 44,160 views
The Hand of the Lord: 1. is mighty. 2. is dangerous and deadly against His enemies. 3. empowers His servants. 4. can be heavy to discipline His people. 5. will help His people. 6. will always be with His people today.
Exodus 9:1–1 Peter 5:7
Hand Of The Lord
Summary - I AM Statements of John
Robert J Smith • Sermon • 12 years ago • 21,990 views
The "I AM" Statements of John's Gospel Twenty-three times in all we find our Lord's meaningful "I AM" (ego eimi, Gk.) in the Greek text of this Gospel (4:26; 6:20, 35, 41, 48, 51; 8:12, 18, 24, 28, 58; 10:7, 9, 11, 14; 11:25; 13:19; 14:16; 15:1, 5; 18:5, 6, 8). In several of these He joins His "I AM"…
Living The Kingdom of God Lifestyle
carlos • Sermon • 8 years ago • 7,652 views
Living The Kingdom of God Lifestyle Journey From Religion to Relationship Growing as a group in the gospel of the kingdom INTRODUCTION & INDEX To Kingdom Growth Guides INTRODUCTION (#000) | Kingdom Growth Guides are for assisting individuals and small groups to move toward the kingdom lifestyle. These…
Sermon 07 - John 10 - The Good Shepherd and His Sheep
David Melchior • Sermon • 12 years ago • 5,494 views
INTRODUCTION: 1. ((illus)) Most of us are familiar with the now syndicated TV program called Candid Camera. Perhaps you remember the "Candid Camera" TV program that took place at an exclusive prep school where all of the students were well above average. The "Candid Camera" people posed as career consultants…
TWA - Bible Reading Schedule
Elliott Kwong • Sermon • 11 years ago • 3,535 views
Time with Abba - Bible Reading Schedule Jan l Feb l Mar l Apr l May l Jun l Jul l Aug l Sep l Oct l Nov l Dec Date Psalms OT NT Proverbs 1/1 Ps 1:1-6 Gen 1:1-2:25 Matt 1:1-2:12 Prov 1:1-6 1/2 Ps 2:1-6 Gen 3:1 - 4:26 Matt 2:13-3:6 Prov 1:7-9 1/3 Ps 2:7-12 Gen 5:1 - 7:24 Matt 3:7-4:11 Prov 1:10-19 1/4…
The Good Shepherd and His Flock
Ferdinand Funk • Sermon • 9 years ago • 1,921 views
The Good Shepherd and his Sheep Text: John 10:1-16 Lets think about Numbers for a minute. Our lives are filled with numbers. It seems that everywhere we are know by our numbers. The government knows us by our tax number. The Province knows us by our driver's license number. The bank knows us by our account…
I Am the Door
Michael Stark • Sermon • 10 years ago • 1,304 views
"Jesus again said to them, 'Truly, truly, I say to you, I am the door of the sheep. All who came before me are thieves and robbers, but the sheep did not listen to them. I am the door. If anyone enters by me, he will be saved and will go in and out and find pasture. The thief comes only to steal and…
John 10:7–10
Communion Sermon
John 6:30-36 - The Source of Spiritual Satisfaction
Sermon • 12 years ago • 1,049 views
Introduction: Last week we learned that man has a gnawing hunger within for both the physical and the spiritual. Just as there is physical hunger that only physical food can satisfy, so there is a deeper, spiritual hunger in the pit of the human soul that only spiritual food can nourish. Man hungers…
John 6:30–36
Introductory Essay Bibical Typology
Bob Pritchett • Sermon • 12 years ago • 805 views
Theological Poetics: Typology, Symbol, and the Christ The Theory and Method of Biblical Typology A modern introduction to biblical typology should begin inductively with several examples of certain shadows and types from Old Testament passages widely acknowledged to be prefigurative in character, seeking…
The I AM Statements of John - 10 1-21
Robert J Smith • Sermon • 12 years ago • 796 views
Jesus the Good Shepherd John 10:1–21 There is no better loved picture of Jesus than the Good Shepherd. The picture of the shepherd is woven into the language and imagery of the Bible. It could not be otherwise. The main part of Judaea was a central plateau, stretching from Bethel to Hebron for a distance…
John K. Davis • Sermon • 12 years ago • 764 views
Studies in the Gospel of Matthew By Bob Deffinbaugh Note: Lessons will be posted as they are completed. Lesson 1 — The Origins of Jesus Christ Matthew 1:1-25[1] Introduction We are all familiar with these wonderful words from the pen of the Apostle Paul: 16 Every scripture is inspired by God and useful…
Allen Springer • Sermon • 11 years ago • 678 views
Part I Prolegomena Introduction to Theology 8 2006 I. Introduction to Prolegomena A. What is Prolegomena (the singular is Prolegomenon) 1. It is from the Greek pro/ + le/gw, meaning to say beforehand (Random House Dictionary, 2nd Ed., 1547). 2. It refers to "a preliminary discourse prefixed to a literary…
The Voice Of The Shepherd (John [10]1-18)
Brian Thiessen • Sermon • 12 years ago • 651 views
The Voice Of The Shepherd Text: John 10:1-18 Place Preached - (Mississauga International Baptist Church) Date Preached - (11/8/98) Introduction: The Power of a Voice good or bad (Patrick Henry/Hitler) The Diversity of Voices (languages) One Voice stands out for the believer (The Voice of the Shepherd)…
Quotes from John Newton's Writings
James H. Schafer • Illustration • 10 years ago • 637 views
John Newton CHOICE EXCERPTS I am not what I once was! In his old age, when he could no longer see to read, John Newton heard someone recite this text, "By the grace of God—I am what I am." He remained silent a short time and then, as if speaking to himself, he said: "I am not what I ought to be—ah, how…
Shepherding under the Chief Shepherd (1 Pet. 5:1-4)
rk • Sermon • 9 years ago • 631 views
Intro I had a great conversation with a brother in ministry this week about insecurity in ministry. He recently graduated from seminary and was struggling with insecurity. I was thankful for his vulnerability and I did give him the sober reminder that it's not going to get any easier! And it's a good…
1 Peter 5:1–4
Shepherding
Dilemma of Belief
The Christmas Dilemma
Mike Measley • EFBC - Evangelical Free Bible Church • Sermon • 7 months ago • 597 views • 40:06
The Dilemma of Belief. Have you ever had those moments when you wondered, "Is this really true?" Maybe life has gotten difficult. Maybe you have come to the Derailment of your hopes and dreams like Joseph, Mary's husband, when he found out that his betrothed bride was pregnant after all his time and…
David Melchior • Sermon • 12 years ago • 566 views
I AM The Good Sheperd
I AM The Door To The Sheep Fold
Psalm 23 2
Larry Jennings Jr • Sermon • 11 years ago • 450 views
The Lord is My Shepherd Psalm 23 Introduction: This is another one of the most loved and quoted passages from the Bible Often read at funerals and other hard times in our lives I suppose that it is one of the most loved passages because it is a passage of encouragement. It is a passage that gives us…
John 10:7-10 - The Shepherds Provision
Sermon • 12 years ago • 432 views
Introduction A. The Jesus Movement: In the Jesus Movement of the 1960s and '70s, the "One Way" sign—the index finger held high—became of popular icon. "One Way" bumper stickers and lapel pins were everywhere, and the "One Way" slogan was a very catching phrase. One - way seemed and unshakable belief,…
Life in the Millennium
Michael Stark • Sermon • 10 years ago • 345 views
Isaiah 2:1-4 Life in the Millennium "The word that Isaiah the son of Amoz saw concerning Judah and Jerusalem. It shall come to pass in the latter days that the mountain of the house of the Lord shall be established as the highest of the mountains, and shall be lifted up above the hills; and all the nations…
Christian Guide to Passover
Steve Bramham • Illustration • 12 years ago • 312 views
A Christian Guide to Passover (PESCH) Below is a compilation of several contemporary Seder services that are used today. Because of the passage of time and that Passover was intended to be a family tradition, it is not possible to reconstruct with certainty the type of Passover meal or celebration that…
Earl Hardy • Sermon • 12 years ago • 277 views
Psalm 1 Introduction Never before in history has so much biblical scholarship been available to Christians. In days gone by (and in some lands today), merely to possess a copy of the Bible in your native tongue was considered a rare privilege. Now we have more than a dozen translations and paraphrases…
Elliott Kwong • Sermon • 11 years ago • 261 views
I Will Look After My Sheep
George Toews • Sermon • 8 years ago • 247 views
Ezekiel 34:11-16 Introduction Some pretty major historical events happened this week with the death of Osama bin Laden and the victory of the Conservatives in the federal election. As the news of these events was being reported, some of the broadcasters interviewed people to ask them how these historical…
Ezekiel 34:11–16


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Tracking the Tropics update for 11/6/17
Updated: 10:40 AM CST Nov 6, 2017
Kweilyn Murphy
Tropical Depression Nineteen has formed in the Atlantic, and may become the next named storm, Rina, later today.As of late morning Monday, the depression was located a little over 850 miles east of Bermuda. The system is forecast to pick up speed and strength as it moves NNE. Nineteen is forecast to become a tropical storm later today, and would acquire the name Rina. The system will pose no threat to land through midweek.Hurricane season ends November 30.Keep up with local news, weather and current events with the WDSU app here. Sign up for our email newsletters to get breaking news right in your inbox. Click here to sign up!
Tropical Depression Nineteen has formed in the Atlantic, and may become the next named storm, Rina, later today.
As of late morning Monday, the depression was located a little over 850 miles east of Bermuda. The system is forecast to pick up speed and strength as it moves NNE. Nineteen is forecast to become a tropical storm later today, and would acquire the name Rina.
The system will pose no threat to land through midweek.
Hurricane season ends November 30.
Keep up with local news, weather and current events with the WDSU app here. Sign up for our email newsletters to get breaking news right in your inbox. Click here to sign up!


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The Digital Hub & National College of Art & Design co-host Digital Cultures Webinar Series
Programme News
The Digital Hub and the National College of Art & Design (NCAD) will hold a series of webinars this Spring, that examine how emerging technologies are affecting and transforming artistic practice, academic research and industry.
The Digital Hub's current artist in residence and NCAD Alumna, Elaine Hoey, along with NCAD lecturer, Dr. Rachel O' Dwyer, have convened this programme, which will be held from Thursday, 18th March to Thursday, 10th June 2021.
This programme of talks brings together artists, designers, thinkers and innovators to examine the areas of artificial intelligence, digital ethics, digital infrastructures and environments and how these powerful emerging technologies intersect with each other and are transforming our futures.
The partnership between NCAD and The Digital Hub builds on stakeholder participation at the intersection of artistic practice, technology and society. This programme is designed to bring together audiences from key industry figures, academics and thinkers, students and alumni, the arts sector and the wider public.
The Webinar Series will be broadcast on NCAD's YouTube channel.
View the full programme of NCAD In Public | The Digital Hub Digital Cultures Programme at www.ncad.ie/gallery-event/view/digital-cultures-webinar-series-2021
Switching to the Circular Economy: MODOS Training for D8 Businesses
Drone & IoT programme calls for applications from young people in Dublin 8


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You are at:Home»Local News»Preferred Bidder Selected To Develop T B Lettsome Airport
Preferred Bidder Selected To Develop T B Lettsome Airport
By admin on	 December 28, 2016 · Local News
The Government of the Virgin Islands is pleased to announce that, following a lengthy procurement process and very careful consideration between the 2 remaining bidders for the runway extension project of the T.B. Lettsome International Airport: China Communications Construction Company (CCCC) and IDL/McAlpine Consortium; that CCCC was selected as the preferred Bidder for the project.
The bid advanced by the CCCC was $153,432,572.10 and the bid by IDL-McAlpine was $198,910,525.00
Negotiations will now get underway with the preferred Bidder with a view to concluding a contractual agreement within 3 months which delivers the right outcome for the people of this Territory. It has always been, and remains, the Government's overriding objective to ensure a significant local involvement in the project. The government sees this as critical and thus it will figure heavily in the negotiations.
The runway extension project became vital once American Eagle left Puerto Rico and took with it, the main access link from Mainland USA to the Territory. At the height of the Eagle's operations, up to as many as 12 flights per day operated between Puerto Rico and the Territory giving excellent and seamless access from the mainland USA and the rest of the world for most of the over 393,000 overnight visitors to our shores.
Extension of the runway at the T.B. Lettsome International Airport is fundamentally about the future of the Territory, given that our industries, Tourism and Financial Services, depend on having easy, reliable and affordable access to our global markets. This is of utmost importance if the BVI is to remain globally competitive.
Extension of the runway from 4645 feet to about 7100 feet will allow current airline fleets (737-800 and Airbus 320) to fly directly to and from continental US and Latin America. At present, established US airlines all cite the inability of their existing fleets to operate from the BVI due to runway length. In addition passenger surveys indicate that travellers do not like to make more than one connection to get to their destination. BVI hotels and other service providers are therefore losing their competitive edge due to visitor access to the Territory.
In welcoming this development Premier and Minister of Finance Dr. the Honourable D. Orlando Smith OBE stated that it is Government's responsibility to ensure that an adequate infrastructure is in place to support and grow our industries and ensure that our economy is able to produce the revenue streams required to ensure the quality of life the people of the BVI have become accustomed.
"Without the extension to the runway, new initiatives like our health tourism, the new BVI International Arbitration Centre and attracting new investors in tourism and financial services cannot succeed in providing additional revenue, jobs and business opportunities. Tourism can only be competitive if tourists can access the BVI easily. It is currently very inconvenient and expensive to visit the BVI," he said.
Premier Smith added, "The Territory cannot afford to ignore the air access needs of our tourism sector since 62 percent of the aggregate economic activity in the Territory is generated by Tourism. Neither can we depend on other jurisdictions where we cannot guarantee the quality of the experience to our tourists which enable them to return again and again."
The successful delivery of the project will also enable BVI's main international airport to accommodate regional and international traffic from a wider range of destinations, which will act as a catalyst for the further development of the economy, and will enable future borrowings where needed for important social infrastructural developments.
He further said, "It will cement the BVI economic foundation and the future advancement of the territory for generations to come."
The Government of the Virgin Islands remains committed to advancing the development of the Territory.


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Army veteran who turned Royal Caribbean into a multi-billion dollar company dies at 87
Army veteran-turned-Royal Caribbean Cruise Line founder Edwin W. Stephan passed away on Friday at 87 years old after founding the company in 1969 and turning it into a multi-billion dollar enterprise, the company announced Saturday.
Stephan helped to pioneer the modern cruise industry as travelers know it today, making Miami the cruise capital of the world upon launching the first ship built for warm-water cruising in 1970. Under Stephan's leadership, the company became one of the world's top cruise lines, with six different brands and 63 ships operating on all seven continents and 77,000 employees worldwide.
Last year, the company posted $9.49 billion in revenue with profits of $1.82 billion, or $4,974,773 per day, according to cruise line-focused website Cruzely.
Edwin W. Stephan, founder, long time-president, and vice chairman of the board of Royal Caribbean Cruise Line.
FURIOUS NORWEGIAN CRUISE LINE PASSENGERS DEMAND REFUND OVER 'HOLIDAY FROM HELL'
"Ed was an inspiration and a great friend to many in the company. He was an honorable man who loved his family, his work, and his community," said Richard D. Fain, chairman and CEO of Royal Caribbean Cruises. "Our heartfelt condolences go out to Ed's loving wife, Helen, and the children he adored, Michael, Samantha, Gary, and Kristina. He will be deeply missed by all of us here, by his many friends, and by our community."
Stephan attended the University of Wisconsin before leaving to serve in the Korean War as an Army lieutenant, specializing in radar, field artillery and guided missiles. He received two Bronze Stars during his service.
RCL ROYAL CARIBBEAN CRUISES 114.53 -0.46 -0.40%
VETERANS DAY SPOTLIGHT: EX-ARMY OFFICER REVIVES ICONIC HOT DOG BUSINESS AS A FAMILY BUSINESS
The Miami Herald reports that after returning from Korea, he vacationed in Miami, where he quickly fell in love with the Florida port city and helped turn it into the cruise line destination it has become.
Stephan served as Royal Caribbean's first president for 27 years before moving to vice chairman of the board of directors. He retired in 2003, according to Cruise Hive.
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text:
Democrats slam Biden over classified documents after he used issue against Trump: 'Embarrassing.'
Posted by Natalia Mittelstadt | Jan 15, 2023 | Just The News | 0 |
https://justthenews.com/government/white-house/democrats-slam-biden-allegedly-mishandling-classified-documents-following?utm_source=justthenews.com&utm_medium=feed&utm_campaign=external-news-aggregators
Democrats are criticizing President Joe Biden for using the issue of classified documents against Donald Trump when he had his own problem with sensitive memos at his home and office.
Classified documents from Biden's tenure as vice president were found in November, December and this month at his former office at the Penn Biden Center in Washington, D.C., and at his home in Wilmington, Del.
In August, Trump's Mar-a-Lago estate was raided by the FBI over classified materials from his tenure at theWhite House. Trump had previously complied with a subpoena and voluntarily handed over some classified materials before the raid.
Special counsels have been appointed in both cases regarding Trump and Biden.
Biden criticized Trump's alleged mishandling of classified documents in September, saying it was "totally irresponsible."
Sen. Debbie Stabenow (D-Mich.) told NBC's "Meet the Press" on Sunday that after Biden had criticized Trump over allegedly mishandling classified documents at Mar-a-Lago, the fact that he also allegedly mishandled classified documents was problematic.
"Well, it's certainly embarrassing," Stabenow said.
She added Biden's team has been cooperating with authorities.
"They don't think it's the right thing and they've been moving to correct it, working with the Department of Justice, working with everyone involved, with the archives. So, from my perspective, it's one of those moments that, obviously, they wish hadn't happened," said Stabenow, who announced earlier this month that she is not seeking reelection.
Rep. Adam Schiff (D-Calif.) also raised potential national security concerns regarding Biden's alleged mishandling of classified materials on Sunday.
When asked on ABC's "This Week" whether there is a possibility that national security was jeopardized, which Schiff had said about the classified documents found at Mar-a-Lago, he noted that the matter should be examined.
"I don't think we can exclude the possibility without knowing more of the facts," Schiff said. "We have asked for an assessment in the intelligence community of the Mar-a-Lago documents. I think we ought to get that same assessment of the documents found in the [Penn Biden Center] as well as the home of President Biden."
"I'd like to know what these documents were. I'd like to know what the [special counsel's] assessment is, whether there was any risk of exposure and what the harm would be and whether any mitigation needs to be done," Schiff continued. "I think that would be appropriate and consistent with what we requested in the case of Mar-a-Lago."
The Democratic congressman added that he would "reserve judgment" regarding whether the Biden administration should have revealed the discovery of the documents sooner.
"I think it's important to point out that the Biden approach was very different in the sense that it looks – as far as we can tell – that it was inadvertent that these documents were in these locations," Schiff added.
Former political adviser to President Barack Obama, David Axelrod, also criticized Biden over the classified documents.
"It basically … is a huge gift to Trump," Axelrod told Reuters. He added that the recent developments regarding the classified documents were an "embarrassment" following Biden's criticisms of Trump over the classified documents found at Mar-a-Lago.
"My guess is when this thing shakes out that it's going to be less than it appears today, but right now it's a huge pain," Axelrod added.
White House lawyer Richard Sauber has said that the White House was "confident that a thorough review will show that these documents were inadvertently misplaced, and the president and his lawyers acted promptly upon discovery of this mistake."
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text:
\section{Introduction}
\setcounter{equation}{0}

Consider the classical Lane-Emden system
\begin{align}\label{1.1}
-\Delta u = v^p, \quad-\Delta v= u^\theta,\quad u,v>0\quad\mbox{in }\; \mathbb{R}^N, \quad\mbox{where }\; p,\theta >0.
\end{align}
There is a famous conjecture who states that: {\sl Let $p, \theta > 0$. If the pair $(p, \theta)$ is subcritical, i.e.~if
\begin{align}\label{LE}
\frac{1}{p+1} + \frac{1}{\theta +1} > \frac{N-2}{N},
\end{align}
then there is no smooth solution to \eqref{1.1}.}

\medskip
The critical curve given by the equality in \eqref{LE} is called the Sobolev hyperbola, which is introduced independently by Mitidieri \cite{em} and
Van der Vorst \cite{van}, it plays a crucial role in the analysis of \eqref{1.1}. It is well known that if $(p, \theta)$ lies on or above the Sobolev hyperbola,
\eqref{1.1} admits radial classical solutions (see \cite{em1, Zs}), and the Lane–Emden conjecture can be restated as the following: There has
no smooth solution to \eqref{1.1} if the positive pair $(p, \theta)$ lies below the Sobolev critical hyperbola.

\medskip
The conjecture is proved to be true for radial functions by Mitidieri \cite{em1}, Serrin-Zou \cite{Zs1}. For the full conjecture, Souto \cite{s}, Mitidieri \cite{em1} and Serrin-Zou \cite{Zs} proved that there is no supersolution to \eqref{1.1}, if $p\theta \leq 1$ or $\max(\alpha, \beta) \geq N-2$, where
 \begin{align}\label{ab}
  \alpha = \frac{2(p+1)}{p\theta - 1}, \quad \beta = \frac{2(\theta +1)}{p\theta - 1}, \quad p\theta > 1.
 \end{align}
 Moreover,
we can check readily that if $p\theta > 1$, the condition \eqref{LE} is equivalent to
\begin{align}
 \label{LEbis} N < 2 + \alpha + \beta.
\end{align}
Therefore, the Lane-Emden conjecture is true in dimensions $N = 1, 2$. More recently, the conjecture is proved in dimensions $N = 3, 4$, by Souplet and his collaborators, see \cite{pqs, sou}.
For $N \geq 5$, the conjecture is known to be true for $(p, \theta)$ verifying \eqref{LE} and one of the following extra conditions:
\begin{itemize}
 \item If $p, \theta < \frac{N+2}{N-2}$, see Felmer-de Figuereido \cite{ff}.
 \item If $\max(p, \theta) \geq N-3$, see Souplet \cite{sou}.
 \item If $\min(\alpha, \beta) \geq \frac{N-2}{2}$, see Busca-Man\'asevich \cite{bm}\footnote{In \cite{bm}, there is another extra condition, which is no longer necessary after the work of Souplet \cite{sou}.}.
 \item If $p = 1$ or $\theta = 1$, see Lin \cite{lin}.
\end{itemize}
These partial results enable us a more restrictive new region for the exponents $(p, \theta)$, which is illustrated by the following figure. In other words, the Lane-Emden conjecture stands open for $N \geq 5$, $p,\theta > 0$ such that
\begin{align*}
p, \theta \ne 1, \quad \min(\alpha, \beta) < \frac{N-2}{2} \quad\mbox{and}\quad \max(\alpha, \beta) < N-3.
\end{align*}
\begin{figure}[h]
\begin{center}
\includegraphics[width=8cm,height=6cm]{Lane-Emden.png}
\caption[le titre]{The remained open region (shaded) for
the Lane-Emden conjecture ($N \geq 5$)}
\label{monlabel}
\end{center}
\end{figure}

On the other hand, in the last decade, many efforts were made to obtain some Liouville type result for solutions with finite Morse index, or more generally,
which are stable at infinity. To define the
notion of stability, we consider a general system given by
\begin{align}\label{1.222}
 -\Delta u = f(x,v),\quad -\Delta v= g(x,u)\;\; \mbox{in $K$, a bounded regular domain }\; \subset \mathbb R^N,
\end{align}
where $f,g\in C^1(K \times \mathbb{R}).$  Following Montenegro \cite{MO}, a smooth solution $(u,v)$ of \eqref{1.222} is said to
be stable in $K$ if the following eigenvalue problem
\begin{align*}
 -\Delta \xi = f_{v}(x,v)\zeta +\eta \xi, \quad -\Delta \zeta = g_{u}(x,u)\xi + \eta\zeta \quad \mbox{in }\, K
\end{align*}
has a nonnegative eigenvalue $\eta$, with a positive smooth eigenfunctions pair $(\xi, \zeta)$. We say that a pair of solutions $(u, v)$ to \eqref{1.1} is stable outside a compact
set or stable at infinity, if there is a compact set $K\subset \mathbb R^N$ such that $(u, v)$ is stable in any bounded domain of $\mathbb R^N\backslash K$.

\medskip
For the corresponding second order equation
\begin{align}
\label{2nd}
\Delta u+ |u|^{q-1}u =0 \quad \mbox{in } \mathbb R^N, \quad q > 1,
\end{align}
Farina \cite{Far} obtained the optimal Liouville type result for solutions stable at infinity. Indeed, he proved that a smooth nontrivial solution to \eqref{2nd} exists,
if and only if $q> p_{JL}$ and $N\geq 11,$ or $q =\frac{N+2}{N-2}$ and $N\geq 3.$ Here $p_{JL}$ denotes the so-called Joseph-Lundgren exponent (see \cite{CW, Far}).
For the biharmornic equation $\Delta^2 u = |u|^{q-1}u$, $q > 1$, D\'{a}vila-Dupaigne-Wang-Wei \cite{ddww} derived a striking monotonicity formula,
which led them to the optimal classification result for solutions stable at infinity, using blow down analysis.

\medskip
Coming back to the Lane-Emden system \eqref{1.1}, Chen-Dupaigne-Ghergu \cite{WLD} studied the stability of radial solutions when $p, \theta \geq 1.$
They introduced a new critical hyperbola, called the Joseph-Lundgren curve.  More precisely, they proved that if $p, \theta \geq 1,$  then a radial solution
of \eqref{1.1} is unstable if and only if $N \leq 10$, or $N \geq 11$ and
\begin{equation*}
  \left[\frac{(N-2)^2-(\alpha-\beta)^2}{4}\right]^2<p\theta\alpha\beta(N-2-\alpha)(N-2-\beta).
\end{equation*}

Moreover, Cowan proved in \cite{cow} that if $p, \theta\geq 2$ and $N \leq 10$, there does not exist any stable solution (radial or not) to \eqref{1.1}.
Recently, Hajlaoui-Harrabi-Mtiri \cite{Hfh} established some Liouville theorems for smooth stable solutions of \eqref{1.1} with $p > 1$, see Theorem {\bf A} below.
We mention also the celebrated result of Ramos \cite{rm}, which states that if $p, \theta > 1$ satisfies \eqref{LE}, then the system
$$-\Delta u = |v|^{p-1}v, \;\; -\Delta v = |u|^{\theta-1}u \quad \mbox{in } \mathbb R^N$$ does not admit any smooth solutions
having finite {\sl relative} Morse index in the sense of Abbondandolo.

\medskip
In this paper, our motivation are twofold. We want to obtain the classification results for solutions (radial or not) to \eqref{1.1} which are just stable at infinity,
and we want to handle the case where $p, \theta$ are allowed to be less than 1.
So a natural question is: Can we prove the Lane-Emden conjecture with the extra condition that $(u, v)$ is
stable at infinity? The answer is affirmative.
\begin{thm}
\label{main4}
For any $p, \theta > 0$ satisfying \eqref{LE}, the system \eqref{1.1} has no classical solution stable outside a compact set.
\end{thm}

If $\theta = p$, using Souplet's comparison result (Lemma 2.7 in \cite{sou}), we get $u \equiv v$, so the optimal classification result for solutions stable
at infinity was already given by Farina. The classification is also known for $p\theta \leq 1$ as mentioned above. Without loss of generality, we consider only $\theta > p > 0$ and $p \theta > 1$. As we will see soon, the $\theta > p \geq 1$ case can be handled by
the results in \cite{Hfh}, so our main concern is the case
$$\theta p > 1 > p > 0.$$

\medskip
Let $(u, v)$ be a smooth solution to \eqref{1.1} with $\theta > p^{-1} > 1 > p > 0$. Our approach is based on the formal equivalence noticed in \cite{DEN, dos}, between
the Lane-Emden system \eqref{1.1} and a fourth order problem, called the $m$-biharmonic equation. More precisely, let $m:=\frac{1}{p}+1>2$, as $v =(-\Delta u)^{m-1}$, we derive
that $u$ satisfies $\Delta^{2}_{m} u := \Delta (|\Delta u|^{m-2}\Delta u) = u^\theta$ in $\mathbb R^N$. So we are led to consider $\theta > m-1 > 1$ and
\begin{equation}\label{1}
\Delta^{2}_{m} u := \Delta (|\Delta u|^{m-2}\Delta u) =|u|^{\theta-1}u.
\end{equation}

\smallskip
Let $\Omega \subset \mathbb R^N$, we say that $u \in W^{2,m}_{loc}(\Omega)\cap L^{\theta +1}_{loc}(\Omega)$ is a weak solutions of \eqref{1} in $\Omega$, if for any regular
bounded domain $K \subset \Omega$,
$u$ is a critical point of the following functional
$$I(v)=\frac{1}{m}\int_K |\Delta v|^m dx-\frac{1}{\theta+1}\int_K |v|^{\theta+1} dx, \quad \forall\; v \in W^{2,m}(K)\cap L^{\theta +1}(K).$$
Naturally, a weak solution to \eqref{1} is said stable in $\Omega \subset \mathbb R^N$, if
\begin{equation}\label{fb1}
\Lambda_u (h):= (m-1)\int_{\Omega} |\Delta u|^{m-2}|\Delta h|^2 dx-\theta \int_{\Omega}|u|^{\theta-1} h^2dx\geq0,\;\;\forall\; h\in C_c^2(\Omega).
\end{equation}
A key point for our approach is to remark a relationship between the stability for the system \eqref{1.1} and the stability for the equation \eqref{1} (see Lemma 2.1 below). This will permit us to handle the case $0 < p  < 1$ in \eqref{1.1} by using the structure of the $m$-biharmonic equation. In fact, we can prove the following Liouville type result.
\begin{thm}\label{main3}
Let $\theta > m-1 > 1$ and $u\in W^{2,m}_{loc}(\mathbb{R}^N)\cap L^{\theta + 1}_{loc}(\mathbb R^N)$ be a weak solution of \eqref{1} which is stable outside a compact set. Assume that
\begin{align}\label{new8}
N < \frac{2m(\theta + 1)}{\theta -(m-1)},
 \end{align}
then $u\equiv 0.$
\end{thm}
A direct calculation yields that if $p\theta > 1$ (or equivalently $\theta > m-1$),
\begin{align*}
 N < 2 + \alpha + \beta = \frac{2(p+1)(\theta+1)}{p\theta - 1} \;\; \Leftrightarrow \;\; \eqref{new8} \;\; \Leftrightarrow \;\; \theta < \frac{N(m-1)+2m}{(N-2m)_+}.
\end{align*}
It means that the range of pairs $(p, \theta)$ satisfying \eqref{LE} and $p\theta > 1$ corresponds exactly to the subcritical case of the $m$-biharmonic equation \eqref{1}.

\medskip
Another crucial step in our approach is to classify first the stable solutions of \eqref{1.1}, see also Proposition \ref{p12bis} below for the $m$-biharmonic equation.
\begin{prop}\label{p12}
If $p, \theta > 0$ satisfies \eqref{LE}, then \eqref{1.1} has no smooth stable solution.
\end{prop}

Establishing a Liouville type result for stable solution of \eqref{1.1} or \eqref{1} is delicate, even we can borrow some ideas from \cite{wy, ddww}. We use as usual the stability to
get some integral estimates, but the integrations by parts argument yields here many terms which are difficult to control,
for example, the local $L^m$ norm of $\nabla u$, see Lemma \ref{l.2.7a} below. Furthermore, the classification of weak solutions stable at infinity to \eqref{1} is more involved
than to handle \eqref{1.1}, since the weak solutions to \eqref{1} are not $C^2$ functions. We will derive a variant of the Pohozaev identity with cut-off functions,
which allows us to avoid the spherical integral terms in the standard Pohozaev identity.

\medskip
The paper is organized as follows. In section 2, we give the proof
of Proposition \ref{p12}. The proofs  of Theorem \ref{main4} and Theorem \ref{main3} are given  respectively in sections 3 and 4. In the following, $C$
denotes always a generic positive constant, which could be changed from one line to another.

\section{Classification of stable solutions}
\setcounter{equation}{0}
We prove here Proposition \ref{p12}. As mentioned before, we need only to consider the case $\theta > p$ and $p \theta > 1$. We split the proof into two cases:
$\theta > p \geq 1$ and $\theta > p^{-1} > 1 > p > 0$.

\subsection{ The case $\theta>p \geq 1$.}
Let us recall a consequence of Theorem 1.1 (with $\alpha = 0$ there) in \cite{Hfh}.
\begin{taggedtheorem}{A} Let $x_0$ be the largest root of the polynomial
\begin{align}\label{newH}
H(x)=x^4 -p\theta\alpha\beta \left[4x^{2}-2(\alpha+\beta)x+1\right].
\end{align}
  \begin{enumerate}
\item[(i)]
 If $\frac{4}{3}< p \leq \theta$ then \eqref{1.1} has no stable classical solution if $N<2+2x_0.$
\item[(ii)] If $1 \leq p\leq \min(\frac{4}{3}, \theta)$ and $p\theta > 1$, then \eqref{1.1} has no stable classical solution, if
$$N < 2 + 2x_0 \left[\frac{p}{2}+\frac{(2-p)(p \theta -1)}{(\theta+p-2)(\theta+1)}\right].$$
\end{enumerate}
\end{taggedtheorem}
Performing the change of variables $x=\frac{\beta}{2}s$ in \eqref{newH}, a direct computation shows that $H(x)=\left(\frac{\beta}{2}\right)^4L(s)$ where
$$L(s):=s^4-\frac{16p\theta(p+1)}{\theta+1}s^2+\frac{16p\theta(p+1)(p+\theta+2)}{(\theta+1)^2}s-\frac{16p\theta(p+1)^2}{(\theta+1)^2}.$$
Denote by $s_0$  the largest root of $L,$ hence $x_0=\frac{\beta}{2}s_0$ and $ H(x)<0$ if and only if $L(s)<0$. For $\theta > p \geq 1$, there holds
\begin{align*}
L(p+1) & = (p+1)^4 -\frac{16p\theta(p+1)^3}{(\theta+1)} +\frac{16p\theta(p+1)^2(p+\theta+2)}{(\theta+1)^2} -\frac{16p\theta(p+1)^2}{(\theta+1)^2}\\
& = (p+1)^4 -\frac{16p\theta(p+1)^3}{(\theta+1)} + \frac{16p\theta(p+1)^2}{(\theta+1)} + \frac{16p\theta(p+1)^3}{(\theta+1)^2}
-\frac{16p\theta(p+1)^2}{(\theta+1)^2}\\
& = (p+1)^2\left[(p+1)^2 -\frac{16p^{2}\theta}{(\theta+1)} + \frac{16p^{2}\theta}{(\theta+1)^2}\right]\\
&=\left(\frac{p+1}{\theta+1}\right)^{2}\left[(p+1)^2(\theta+1)^2 -16p^{2}\theta^{2}\right] < 0.
\end{align*}
The last inequality holds true since
$$4p\theta- (p+1)(\theta + 1) > 4p^2 - (p+1)^2 \geq 0, \quad \forall\; \theta > p \geq 1.$$
As $\lim_{s\rightarrow \infty}L(s)= \infty,$  it follows that $s_0>p+1.$ We get then
 \begin{align*}
2x_0>(p+1)\beta=2+\alpha+\beta,\quad \forall\; \theta > p\geq 1.
\end{align*}
If $p>\frac{4}{3}$, by $(i)$ of Theorem \textbf{A}, the system \eqref{1.1} has no classical stable solution if  $N<2+\alpha+\beta$. Suppose now $1\leq p \leq \min(\frac{4}{3}, \theta)$.
Observe that for all $\theta \geq p \geq 1$,
     \begin{align*}
\left[p+\frac{2(2-p)(p \theta -1)}{(\theta+p-2)(\theta+1)}\right]\beta \geq \alpha + \beta & \Leftrightarrow \left[p+\frac{2(2-p)(p \theta -1)}{(\theta+p-2)(\theta+1)}\right](\theta + 1) \geq p + \theta + 2\\
& \Leftrightarrow p\theta - 1 + \frac{2(2-p)(p \theta -1)}{\theta+p-2} \geq \theta + 1\\
& \Leftrightarrow (p\theta - 1)\left[ 1 + \frac{2(2-p)}{\theta+p-2}\right] \geq \theta + 1\\
& \Leftrightarrow (p\theta - 1)(\theta + 2 - p) \geq (\theta+p-2)(\theta + 1)\\
& \Leftrightarrow p\theta^2 - \theta + (2 - p)p\theta \geq \theta^2 + (p-1)\theta\\
& \Leftrightarrow (p-1)(\theta - p) \geq 0.
\end{align*}
As $s_0>p+1\geq 2,$ we have $x_0 = \frac{\beta s_0}{2} \geq \beta$ and
$$2+\alpha+\beta \leq 2 + \beta \left[p+\frac{2(2-p)(p \theta -1)}{(\theta+p-2)(\theta+1)}\right] \leq 2 + x_0\left[p+\frac{2(2-p)(p \theta -1)}{(\theta+p-2)(\theta+1)}\right].$$
If $N < 2+\alpha+\beta,$ using $(ii)$ of Theorem \textbf{A}, we are done.

\medskip
To conclude, for all $\theta > p \geq 1$ and $N < 2+\alpha+\beta,$ \eqref{1.1} has no smooth stable solution. \qed

\subsection{The case $\theta p > 1>p > 0$. }
Here we handle the case  $0<p<1.$ First of all, we need the following lemma which plays an important role in dealing with  Proposition \ref{p12}.
\begin{lem}
\label{l.1} Let $(u,v)$ be a solution of system \eqref{1.1} with $\theta > \frac{1}{p}:=m-1>1$. Suppose that $(u, v)$ is stable in a regular bounded domain $\Omega$,
then $u$ is a stable solution of equation \eqref{1}.
\end{lem}
\noindent{\bf Proof.} By the definition of stability, there exist smooth positive functions $\xi$, $\zeta$ and $\eta \geq 0$ such that
$$-\Delta \xi = pv^{p-1}\zeta + \eta \xi, \; \; -\Delta\zeta = \theta u^{\theta -1}\xi + \eta\zeta \quad \mbox{in }\; \Omega.$$
Using $(\xi, \zeta)$ as super-solution, $(\min_{\overline\Omega}\xi, \min_{\overline\Omega}\zeta)$ as sub-solution, and the standard monotone iterations, we can claim that
there exist positive smooth functions $\varphi$, $\chi$ verifying
\begin{align*}
 -\Delta \varphi = p v^{p-1}\chi, \quad -\Delta \chi = \theta u^{\theta-1}\varphi\quad \mbox{in }\, \Omega.
\end{align*}
Therefore, we have
\begin{align*}
\theta u^{\theta-1}\varphi=\Delta\left(\frac{1}{p} v^{1-p} \Delta \varphi\right) \quad \mbox{in}\;\;\Omega.
\end{align*}
Let $\gamma \in C_c^2(\Omega)$. Multiplying the above equation by $\gamma^{2}\varphi^{-1}$ and
integrating by parts, there holds
\begin{align}\label{0.255}
\begin{split}
 \int_{\Omega} \theta u^{\theta-1}\gamma^{2}dx & = \frac{1}{p}\int_{\Omega} v^{1-p} \Delta \varphi\Delta(\gamma^{2}\varphi^{-1})dx\\
&= \frac{1}{p} \int_{\Omega}v^{1-p} \Delta \varphi\left(-4\gamma\frac{\nabla \varphi\cdot\nabla\gamma}{\varphi^{2}}+\frac{2|\nabla\gamma |^2}{\varphi}+\frac{2\gamma\Delta\gamma}{\varphi}+\frac{2\gamma^{2}|\nabla\varphi |^2}{\varphi^{3}} -\frac{\gamma^{2}\Delta\varphi}{\varphi^{2}}\right) dx.
\end{split}
\end{align}
Using Cauchy-Schwarz's inequality and the fact that $-\Delta \varphi >0,$ we get
\begin{align}\label{2.La}
\begin{split}
\left| -4\int_{\Omega}\frac{v^{1-p}}{p} \Delta \varphi\frac{\nabla \varphi\cdot\nabla\gamma}{\varphi^{2}}\gamma dx\right|
 \leq -2\int_{\Omega}\frac{v^{1-p}}{p} \Delta \varphi\frac{|\nabla\gamma |^2}{\varphi} dx -2\int_{\Omega}\frac{v^{1-p}}{p} \Delta \varphi\frac{\gamma^{2}|\nabla\varphi |^2}{\varphi^{3}} dx.
\end{split}
\end{align}
Combining \eqref{0.255} and  \eqref{2.La}, one obtains, using again the Cauchy-Schwartz inequality,
\begin{align*}
 \int_{\Omega} \theta u^{\theta-1}\gamma^{2}dx &
\leq \frac{2}{p} \int_{\Omega}v^{1-p} \Delta \varphi\frac{\gamma\Delta\gamma}{\varphi} dx-\frac{1}{p} \int_{\Omega}v^{1-p}\frac{(\Delta \varphi)^{2}}{\varphi^{2}}\gamma^{2} dx\\
& \leq  \frac{1}{p}\int_{\Omega}v^{1-p}\frac{(\Delta \varphi)^{2}}{\varphi^{2}}\gamma^{2} dx + \frac{1}{p}\int_{\Omega}v^{1-p}(\Delta\gamma)^{2} dx -\frac{1}{p} \int_{\Omega}v^{1-p}\frac{(\Delta \varphi)^{2}}{\varphi^{2}}\gamma^{2} dx\\
& = \frac{1}{p} \int_{\Omega}v^{1-p}(\Delta\gamma)^{2} dx.
\end{align*}
Recall that $p = \frac{1}{m-1}$ and $(-\Delta u )^{\frac{1}{p}}= v,$ we obtain the desired result \eqref{fb1}. \qed

\medskip
Therefore, to prove Proposition \ref{p12} in the case $p \in (0, 1)$ and $p\theta > 1$, we need only to prove
\begin{prop}\label{p12bis}
Let $\theta > m - 1 > 1$, if $u$ is a weak stable solution to the equation \eqref{1} in $\mathbb R^N$ with $N$ verifying \eqref{new8}, then $u \equiv 0$.
\end{prop}

To prove Proposition \ref{p12bis}, we use first the stability condition \eqref{fb1} to get the  following crucial lemma which provides an important integral estimate for $u$ and $\Delta u$.
 \begin{lem}
\label{lemnewBN} Let $u\in W^{2,m}_{loc}(\Omega)\cap L^{\theta+1}_{loc}(\Omega)$ be a weak stable solution of \eqref{1} in $\Omega$, with $\theta>m-1 > 1$. Then, for any integer  $$k\geq \max \left( m, \frac{m(\theta+1)}{2(\theta+1-m)}\right) ,$$
there exists a positive constant  $ C=C(N, \epsilon, m, k)$ such that for any $\zeta\in C_c^2(\Omega)$ satisfying $0 \leq\zeta \leq 1$,
\begin{align}\label{new7}
\begin{split}
\int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx +\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx
\leq C\left[\int_{\Omega}\left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m}
 +|\nabla^{2} \zeta|^{m}\right)^{\frac{\theta+1}{\theta-(m-1)}} dx\right].
 \end{split}
\end{align}
\end{lem}
\noindent{\bf Proof.} For any $\epsilon \in (0, 1)$ and $\eta \in C^2(\Omega)$, there holds
\begin{align}
\label{newesTt47}
  \begin{split}
\int_{\Omega}|\Delta u|^{m-2} [\Delta (u\eta)]^2 dx = & \; \int_{\Omega}|\Delta u|^{m-2}\left(u \Delta \eta+2\nabla u\nabla \eta + \eta\Delta u\right)^{2} dx\\
\leq & \;\left(1+\epsilon\right)\int_{\Omega}|\Delta u|^m {\eta}^2 dx + \frac{C}{\epsilon}\int_{\Omega}|\Delta u|^{m-2}
  \Big(u^2|\Delta\eta|^2 +  |\nabla u|^2|\nabla \eta|^2\Big)dx.
    \end{split}
   \end{align}
Take $\eta = \zeta^{2k}$ with $\zeta \in C_c^2(\Omega)$, $0 \leq\zeta \leq 1$ and $k \geq m > 2$. Apply Young's inequality, we get
     \begin{align*}
  \int_{\Omega}|u|^{2}|\Delta u|^{m-2}|\Delta (\zeta^{2k})|^2 dx & \leq C_{k}\int_{\Omega}|u|^{2}|\Delta u|^{m-2}\left(|\Delta \zeta|^{2}+|\nabla \zeta|^4\right)\zeta^{4k-4} dx\\
& \leq \epsilon^{2} \int_{\Omega}|\Delta u|^m \zeta^{4k} dx +
  C_{\epsilon, k, m}\int_{\Omega} |u|^m \left(|\Delta \zeta|^{2}+|\nabla \zeta|^4\right)^{\frac{m}{2}}\zeta^{4k-2m} dx
\end{align*}
and
  \begin{align*}
  \int_{\Omega}|\Delta u|^{m-2} |\nabla u|^2|\nabla (\zeta^{2k})|^2 dx & =4 k^{2}\int_{\Omega}|\Delta u|^{m-2} |\nabla u|^2|\nabla \zeta|^2\zeta^{4k-2} dx\\
& \leq  \epsilon^{2} \int_{\Omega}|\Delta u|^m \zeta^{4k} dx +
  \frac{C_{m,k}}{\epsilon^{2}}\int_{\Omega} |\nabla u|^m |\nabla \zeta|^{m}\zeta^{4k-m} dx.
\end{align*}
Inserting the two above estimates into \eqref{newesTt47}, we arrive at
\begin{align}\label{newest4}
  \begin{split}
\int_{\Omega}|\Delta u|^{m-2} [\Delta (u\zeta^{2k})]^2 dx
 \leq & \; \left(1+C\epsilon\right) \int_{\Omega}|\Delta u|^m \zeta^{4k} dx+\frac{C_{m,k}}{\epsilon^{3}}\int_{\Omega} |\nabla u|^m |\nabla \zeta|^{m}\zeta^{4k-m} dx\\
  &+ C_{\epsilon, m, k}\int_{\Omega} |u|^m \left(|\Delta \zeta|^{2}+|\nabla \zeta|^4\right)^{\frac{m}{2}}\zeta^{4k-2m} dx.
  \end{split}
   \end{align}

\medskip
We need also the following technical lemma, which proof is given later.
 \begin{lem}\label{l.2.7a}
Let $k \geq m/2 > 1$ and $\epsilon > 0$, there exists $C_{N, \epsilon, m, k} >0$ such that for any $u\in W^{2,m}_{loc}(\Omega)$ verifying \eqref{fb1} and $\zeta \in C_c^{\infty}(\Omega)$ with $0 \leq\zeta \leq 1$, there holds
 \begin{align}
 \label{new13}
\int_{\Omega} |\nabla u|^m |\nabla \zeta|^{m}\zeta^{4k-m} dx \leq \epsilon\int_{\Omega}|\Delta u|^{m}\zeta^{4k} dx +
C_{N, \epsilon, m ,k} \int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx.
\end{align}
\end{lem}

Using Lemma \ref{l.2.7a} with $\epsilon^4$ and \eqref{newest4}, we see that
\begin{align}\label{nt4}
  \begin{split}
\int_{\Omega}|\Delta u|^{m-2} [\Delta (u\zeta^{2k})]^2 dx
  \leq& \; C_{N, \epsilon, m,k}\int_{\Omega} |u|^{m} \left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m} dx\\
  & \;+\left(1 + C_{m,k}\epsilon\right) \int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx.
  \end{split}
   \end{align}
Thanks to the approximation argument, the stability property \eqref{fb1} holds true with $u\zeta^{2k}$. We deduce then, for any $\epsilon>0$,
there exists $C_{N, \epsilon, m,k} > 0$ such that
\begin{align}\label{f1}
 \begin{split}
& \;\theta\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx -\left(m-1\right)\left(1+ C_{m,k}\epsilon\right) \int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx\\
\leq  & \;C_{N, \epsilon, m,k}\int_{\Omega} |u|^{m} \left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m} dx.
  \end{split}
\end{align}
Moreover, multiplying the equation \eqref{1} by $u \zeta^{4k}$ and
integrating by parts, there holds
\begin{align*}
 \int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx-\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx \leq \int_{\Omega}|u||\Delta u|^{m-1}
  |\Delta(\zeta^{4k})| dx
  + C\int_{\Omega}|\Delta u|^{m-1} |\nabla u||\nabla (\zeta^{4k})| dx.
\end{align*}
Using Young's inequality and applying again Lemma \ref{l.2.7a}, we can conclude that for any $\epsilon>0$, there exists $C_{N,\epsilon, m,k} > 0$
 such that
\begin{align}\label{f12}
\begin{split}
& \;\left(1- C_{m,k}\epsilon\right) \int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx-\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx\\
\leq& \; C_{N,\epsilon, m,k}\int_{\Omega} |u|^{m} \left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m} dx.
\end{split}
\end{align}

Taking $\epsilon > 0$ but small enough, multiplying \eqref{f12} by $\frac{(m-1)(1+ 2C_{m,k}\epsilon)}{1- C_{m,k}\epsilon}$, adding it with \eqref{f1}, we get
\begin{align*}
&(m-1) C_{m,k}\epsilon \int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx + \left[\theta-\frac{(m-1)(1+ 2C_{m,k}\epsilon)}{1- C_{m, k}\epsilon}\right]\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx\\
\leq  & \; C_{N,\epsilon, m,k}\int_{\Omega} |u|^{m} \left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m} dx.
\end{align*}
As $\theta > m - 1 > 1$, using $\epsilon > 0$ small enough, we have
\begin{align}\label{f1xx2xx}
\int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx +\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx\leq C\int_{\Omega} |u|^{m} \left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m} dx.
\end{align}
For $k\geq  \frac{m(\theta+1)}{2(\theta+1-m)}$ so that $4km\leq(4k-2m)(\theta+1),$ Applying H\"older inequality, we conclude then
\begin{align*}
& \;\int_{\Omega}|\Delta u|^{m} \zeta^{4k} dx +\int_{\Omega} |u|^{\theta+1}\zeta^{4k} dx\\
\leq  & \; C\left[\int_{\Omega}\left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m}
 +|\nabla^{2} \zeta|^{m}\right)^{\frac{\theta+1}{\theta-(m-1)}} dx\right]^{\frac{\theta-(m-1)}{\theta+1}}
 \left(\int_{\Omega} |u|^{\theta+1}\zeta^{\frac{(4k-2m)(\theta+1)}{m}}dx\right)^{\frac{m}{\theta+1}}\\
 \leq  & \; C\left[\int_{\Omega}\left(|\Delta \zeta|^{m}+|\nabla \zeta|^{2m}
 +|\nabla^{2} \zeta|^{m}\right)^{\frac{\theta+1}{\theta-(m-1)}} dx\right]^{\frac{\theta-(m-1)}{\theta+1}}
 \left(\int_{\Omega} |u|^{\theta+1}\zeta^{4k}dx\right)^{\frac{m}{\theta+1}}.
 \end{align*}
We get readily the estimate \eqref{new7}. \qed

\medskip
Now we choose $\phi_0$ a cut-off function in $C_c^\infty(B_2)$ verifying $0 \leq \phi_0 \leq 1$ and $\phi_0=1$ in $B_1$. Applying \eqref{new7} with $\zeta = \phi_0(R^{-1}x)$
for $R > 0$, there holds
 \begin{align*}
\int_{B_{R}} |u|^{\theta+1}dx\leq \int_{\mathbb R^N} |u|^{\theta+1}\zeta^{4k} dx \leq C R^{\frac{N(\theta -m+1)}{\theta + 1} -2m}.
\end{align*}
Under the assumption \eqref{new8}, tending $R \to \infty$, we obtain $u \equiv 0$, we prove then
Proposition \ref{p12bis}, hence the case $\theta p > 1 > p > 0$ for Proposition \ref{p12}.

\medskip
\noindent{\bf Proof of Lemma \ref{l.2.7a}.} A direct calculation gives
\begin{align}
  \label{new1}
  \begin{split}
 \int_{\Omega} |\nabla u|^m |\nabla \zeta|^{m}\zeta^{4k-m} dx = &\;\int_{\Omega} \nabla u\cdot\nabla u|\nabla u|^{m-2}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
 =& \; - \int_{\Omega}  \mathrm{div} \left(\nabla u|\nabla u|^{m-2}\right) u|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
 &\; -\int_{\Omega} u|\nabla u|^{m-2}\nabla u \cdot\nabla\left(|\nabla \zeta|^{m}\zeta^{4k-m}\right)dx\\
 := & \; I_1 + I_2.
 \end{split}
\end{align}
The integral $I_1$ can be estimated as
 \begin{align*}
 I_1 & = -\left(m-2\right) \int_{\Omega}  u |\nabla u|^{m-4}|\nabla \zeta|^{m}\nabla^{2}u (\nabla u, \nabla u)\zeta^{4k-m}dx -\int_{\Omega}  u\Delta u|\nabla u|^{m-2}|\nabla \zeta|^{m}\zeta^{4k-m}dx &\;\\
 & \leq C_m\int_{\Omega}  |u|| \nabla^{2} u||\nabla u|^{m-2}|\nabla \zeta|^{m}\zeta^{4k-m}dx + \int_{\Omega}  |u||\Delta u||\nabla u|^{m-2}|\nabla \zeta|^{m}\zeta^{4k-m}dx.
\end{align*}
Applying Young's inequality, there holds, for any $\epsilon > 0$,
 \begin{align}\label{new3}
 \begin{split}
 & \int_{\Omega}  |u|| \Delta u||\nabla u|^{m-2}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
 \leq & \;C_{ \epsilon, m}\int_{\Omega}  |u|^{\frac{m}{2}}| \Delta u|^{\frac{m}{2}}|\nabla \zeta|^{m}\zeta^{4k-m}dx + \epsilon\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
 \leq &\;C_{\epsilon, m}\int_{\Omega}  |u|^{m}|\nabla \zeta|^{2m}\zeta^{4k-2m}dx+ \epsilon\int_{\Omega}| \Delta u|^{m}\zeta^{4k}dx + \epsilon\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx.
\end{split}
\end{align}
On the other hand,
 \begin{align}\label{new2}
 \begin{split}
 & \;\int_{\Omega}  |u|| \nabla^{2} u||\nabla u|^{m-2}|\nabla \zeta|^{m-2+2}\zeta^{4k-m}dx\\
 \leq &\;  C_{\epsilon, m}\int_{\Omega}  |u|^{\frac{m}{2}}| \nabla^{2} u|^{\frac{m}{2}}|\nabla \zeta|^{m}\zeta^{4k-m}dx + \epsilon\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
 \leq &\;C_{\epsilon, m}\int_{\Omega}  |u|^{m}|\nabla \zeta|^{2m}\zeta^{4k-2m}dx+ \epsilon\int_{\Omega}| \nabla^{2} u|^{m}\zeta^{4k}dx+ \epsilon\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx.
\end{split}
\end{align}
Now we shall estimate the integral
$$\int_{\Omega}| \nabla^{2} u|^{m}\zeta^{4k}dx.$$
Remark that there exists $C_0(N, m) > 0$ such that
\begin{align}
 \label{W2m}
 \int_{\mathbb R^N} |\nabla^2\varphi|^m dx \leq C_0(N, m)\int_{\mathbb R^N}|\Delta\varphi|^m dx, \quad \forall\; \varphi \in W^{2, m}(\mathbb R^N).
\end{align}
We can prove it firstly for $\varphi \in W^{2, m}_0(B_1)$ with elliptic theory, then for general $\varphi \in W^{2, m}(\mathbb R^N)$ with
approximation and scaling argument. As $u\zeta \in W^{2, m}_0(\Omega) \subset W^{2, m}(\mathbb R^N)$, \eqref{W2m} implies that
\begin{align*}
\int_{\Omega} |\nabla^2(u\zeta^{\frac{4k}{m}})|^m dx \leq & \;
C_0(N, m) \int_{\Omega}|\Delta (u \zeta^{\frac{4k}{m}})|^m dx\\
\leq & \; C_{N, m}\int_{\Omega}|\Delta u|^{m}\zeta^{4k} dx + C_{N, m,k}\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
  &\; + C_{N, m,k}\int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx.
\end{align*}
Let $k > m$, we get then
 \begin{align}\label{new4}
 \begin{split}
 \int_{\Omega}| \nabla^{2} u|^{m}\zeta^{4k}dx
 \leq& \;C\int_{\Omega}|\nabla^2(u\zeta^{\frac{4k}{m}})|^m dx +C_{m,k}\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
 &\;+ C_{m,k}\int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx\\
 \leq & \; C_{N,m}\int_{\Omega}|\Delta u|^{m}\zeta^{4k} dx + C_{N, m,k}\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx\\
  &\; + C_{N, m,k}\int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx.
 \end{split}
\end{align}
Combining \eqref{new3}, \eqref{new2} and \eqref{new4}, we arrive at
 \begin{align}\label{SSS0.255}
 \begin{split}
I_1 \leq &\;C_{N,m,k}\epsilon \int_{\Omega}|\Delta u|^{m}\zeta^{4k} dx +C_m\epsilon\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx \\
  &+\; C_{N,\epsilon, m, k} \int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx.
 \end{split}
\end{align}

Furthermore, by Young's inequality,
 \begin{align}\label{SS0.255}
 \begin{split}
I_2 = & - m\int_{\Omega} u|\nabla u|^{m-2} |\nabla \zeta|^{m-2}\nabla^2\zeta(\nabla\zeta, \nabla u)\zeta^{4k-m} dx\\
& -(4k-m)\int_{\Omega} u|\nabla u|^{m-2}|\nabla \zeta|^{m}(\nabla u\cdot \nabla\zeta)\zeta^{4k-m-1}dx\\
\leq &\; C_{ m,k} \int_{\Omega}|u||\nabla u|^{m-1}|\nabla \zeta|^{m-1}\left(|\nabla \zeta|^{2} +|\nabla^{2} \zeta|\right)\zeta^{4k-m-1}dx\\
\leq &\; C_{\epsilon, m,k} \int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx
+ \epsilon\int_{\Omega}|\nabla u|^{m}|\nabla \zeta|^{m}\zeta^{4k-m}dx.
 \end{split}
\end{align}
Combining \eqref{SSS0.255}--\eqref{SS0.255} with \eqref{new1}, one concludes
 \begin{align*}
(1-C_{N, m,k}\epsilon)\int_{\Omega} |\nabla u|^m |\nabla \zeta|^{m}\zeta^{4k-m} dx
& \leq C_{N,\epsilon, m ,k} \int_{\Omega}|u|^{m}\left(|\nabla \zeta|^{2m} +|\nabla^{2} \zeta|^{m}\right)\zeta^{4k-2m}dx\\
  &+C_{N, m, k}\epsilon\int_{\Omega}|\Delta u|^{m}\zeta^{4k} dx.
\end{align*}
This means that \eqref{new13} holds true for $\epsilon > 0$ small enough, hence for any $\epsilon > 0$.\qed

\section{Proof of Theorem \ref{main4}.}
\setcounter{equation}{0}
In this section, we prove Theorem \ref{main4}. As already mentioned, we need only to handle the case $p\theta > 1$. We use first the classification for stable solutions, Proposition \ref{p12} to obtain the decay estimates for stable at infinity solutions of \eqref{1.1}.

\begin{lem}\label{newla}
 Let $p, \theta > 0$ verify $p\theta > 1$ and \eqref{LEbis}. Let $(u,v)$ be a solution of \eqref{1.1}  which is stable outside a compact set.
 Then there exists a constant C such that
\begin{align}\label{new5}
\sum_{k\leq 2}\Big[ |x|^{\alpha+k} |\nabla^k u(x)|+|x|^{\beta+k} |\nabla^k v(x)|\Big]\leq C,\quad  \forall\; x \in \mathbb R^N.
\end{align}
\end{lem}
\textbf{Proof.} Assume that $(u, v)$ is stable outside $B_{R_0}$. Denote $$W(x)=\sum_{k\leq 2}\left[ |\nabla^k u(x)|^{{\frac{1}{\alpha+k}}}+ |\nabla^k v(x)|^{\frac{1}{\beta+k}}\right].$$
Suppose that \eqref{new5} does not hold true. Let $d(x) = \|x\|-R_0$, there holds
$$\sup_{\mathbb R^N\backslash B_{R_0}}W(x)d(x) =\infty,$$
or equally there exists a sequence $(x_n)$ such that $\|x_n\| > R_0$ and $W(x_{n})d(x_{n})>n$ for $n \geq 1$. Since $(u, v)$ are smooth in $\mathbb{R}^N$, then $d(x_{n})\rightarrow \infty$. By the doubling lemma \cite{pqs}, there exists another sequence $(y_{n})$  such that for any $n \geq 1$, $\|y_n\| > R_0$,
\begin{enumerate}
  \item [$(i)$] $W(y_{n})d(y_{n})\geq n$;
  \item [$(ii)$] $W(y_{n})\geq W(x_{n})$;
  \item [$(iii)$] $W(z)\leq 2 W(y_{n})$  for  $|z|> R_0$  such that  $|z-y_{n}|\leq \frac{n}{W(y_{n})}.$
\end{enumerate}

 Let $(u,v)$ be a solution of \eqref{1.1}, consider the sequence of functions
 \begin{align*}
\widetilde u_{n}(x) = \lambda_{n}^{\alpha}u(y_n + \lambda_{n}x), \quad \widetilde v_{n}(x) = \lambda_{n}^{\beta} v(y_n + \lambda_{n}x), \quad \mbox{with }\; \lambda_{n}=W(y_{n})^{-1}.
\end{align*}
It's well known that $(\widetilde u_n, \widetilde v_n)$ are a sequence of solutions to \eqref{1.1}. Moreover,
$$W_n(x) := \sum_{k\leq 2}\left( |\nabla^k \widetilde u_n(x)|^{\frac{1}{\alpha+k}}+ |\nabla^k \widetilde v_n(x)|^{{\frac{1}{\beta+k}}}\right) = \lambda_n W(y_n + \lambda_n x),
\quad \forall\; x \in \mathbb R^N.$$
By $(i)$, we have $B_{n\lambda_n}(y_{n}) \subset \mathbb R^N\backslash B_{R_0}$, and we can readily check that $(\widetilde u_n, \widetilde v_n)$ is stable in $B_{n}$ since $(u,v)$ is stable in $\mathbb R^N\backslash B_{R_0}$. Using $(iii)$, there holds, for all $n \geq 1$,
\begin{align}\label{newAnewest5}
W_n(x) \leq 2W_n(0) = 2\quad \mbox{in }\; B_n.
\end{align}
From \eqref{newAnewest5} and standard elliptic theory, up to a subsequence, $(\widetilde u_n, \widetilde v_n)$ converges to $(u_\infty, v_\infty)$ in
 $C^2_{loc}({\mathbb R}^N)$. Therefore
$$\sum_{k\leq 2}\left( |\nabla^k u_\infty(0)|^{\frac{1}{\alpha+k}}+ |\nabla^k v_\infty(0)|^{{\frac{1}{\beta+k}}}\right)  = 1.$$
So $(u_{\infty}, v_\infty)$  is nontrivial.  Clearly, $(u_\infty, v_\infty)$ a smooth positive solution to \eqref{1.1}. Using again the elliptic theory, it's not difficult to see that $(u_\infty, v_\infty)$ is stable in $\mathbb R^N$. However, this contradicts Proposition \ref{p12} since $p, \theta$ verifies \eqref{LE}. Hence the hypothesis was wrong, i.e.~the estimate \eqref{new5} holds true.  \qed

\medskip
Another tool is the following classical Pohozaev identity (see \cite{em, PJ, sou}).
\begin{lem}\label{newlem1.1}
Let $(u,v)$ be a solution to \eqref{1.1}. Therefore for any regular bounded domain $\Omega$,
\begin{align}\label{new6}
\begin{split}
&\; \frac{2(p+1)- pN}{p+1}\int_{\Omega} v^{p+1} dx +\frac{N}{\theta+1} \int_{\Omega} u^{\theta+1} dx \\
=& \int_{\partial\Omega}u^{\theta+1}(\nu\cdot x)d\sigma-\frac{p}{p+1} \int_{\partial\Omega}v^{p+1}(\nu\cdot x)d\sigma
+\int_{\partial\Omega}\frac{\partial v}{\partial \nu}(\nabla u\cdot x)d\sigma-\int_{\partial\Omega}v\frac{\partial(\nabla u\cdot x)}{\partial \nu}d\sigma.
\end{split}
 \end{align}
 \end{lem}

We claim then
 \begin{lem}\label{newl.552.GR7a}
Let $p, \theta > 0$ satisfy $p\theta > 1$ and \eqref{LEbis}. If $(u,v)$ be a solution of \eqref{1.1} which is stable outside a compact set, then $v \in L^{p+1}(\mathbb{R}^N)$, $u\in L^{\theta+1}(\mathbb{R}^N)$ and
\begin{align}\label{new8newest0}
\frac{2(p+1)-pN }{p+1}\int_{\mathbb{R}^N}v^{p+1}dx + \frac{N}{\theta+1}\int_{\mathbb{R}^N}u^{\theta+1}dx = 0.
 \end{align}
\end{lem}

\noindent{\bf Proof .}  By \eqref{new5}, we  have (noticing that $\alpha(\theta+1)=(p+1)\beta=2+\alpha+\beta$)
\begin{align*}
u^{\theta+1}(x)+v^{p+1}(x) \leq C\left(1+ |x|\right)^{-(2 + \alpha+\beta)} \quad \mbox{in }\; \mathbb R^N.
 \end{align*}
By \eqref{LEbis}, then $v \in L^{p+1}(\mathbb{R}^N)$, $u\in L^{\theta+1}(\mathbb{R}^N) .$ Using  Lemma \ref{newlem1.1} with $\Omega =B_{R}$, we deduce that
\begin{align}\label{newac2k.5}
\begin{split}
& \frac{2(p+1)-pN }{p+1}\int_{B_{R}}v^{p+1}dx+ \frac{N}{\theta+1}\int_{B_{R}}|u|^{\theta+1}dx\\
=&\;\int_{\partial B_{R}}\left[R\frac{\partial  u}{\partial r}\frac{\partial v}{\partial r}-\frac{pR}{p+1}v^{p+1}
 -v\frac{\partial (\nabla u\cdot x)}{\partial r}+\frac{R}{\theta+1}|u|^{\theta+1}\right]d\sigma.
\end{split}
 \end{align}
Using again \eqref{new5} and $N < 2 + \alpha+\beta$, we deduce that
 $$ \int_{\partial B_{R}}\left[R\frac{\partial  u}{\partial r}\frac{\partial v}{\partial r}-\frac{R}{p+1}v^{p+1}
 -v\frac{\partial (\nabla u\cdot x)}{\partial r}+\frac{R}{\theta+1}|u|^{\theta+1}\right]d\sigma\rightarrow 0, \quad \mbox{as }\; R\rightarrow \infty.$$
Taking the limit $R \to\infty$ in \eqref{newac2k.5}, the claim follows. \qed

\medskip\noindent
{\bf Proof of Theorem \ref{main4} completed.} We are now in position to conclude. Suppose that $(u, v)$ is a solution to \eqref{1.1} stable at infinity with $p\theta > 1$ verifying \eqref{LEbis}. Choose $\phi_0$ a cut-off function in $C_c^\infty(B_2)$ verifying $0 \leq \phi_0 \leq 1$ and $\phi_0=1$ in $B_1$. Denote $\zeta = \phi_0(R^{-1}x)$ and $A_R= B_{2R}\backslash B_R$. By the system \eqref{1.1}, there holds
\begin{align*}
 \int_{B_{2R}}v^{p+1} \zeta dx-\int_{B_{2R}} u^{\theta+1} \zeta dx & =  \int_{B_{2R}}u\zeta \Delta v dx - \int_{B_{2R}} u\zeta \Delta v dx\\
& = \int_{B_{2R}}v\Big(2\nabla u\cdot \nabla \zeta + u\Delta\zeta\Big) dx\\
& \leq \frac{C}{R^{2}} \int_{A_R}uv dx
  + \frac{C}{R} \int_{A_R}v |\nabla u| dx.
\end{align*}
Using \eqref{new5}, and tending $R\rightarrow \infty$, as $N < 2 + \alpha+\beta,$ we have
$$\int_{\mathbb{R}^N}v^{p+1} dx=\int_{\mathbb{R}^N}u^{\theta+1} dx.$$
Substituting this in \eqref{new8newest0},
\begin{align*}
\left(\frac{2(p+1) - pN}{p+1}+\frac{N}{\theta+1}\right)\int_{\mathbb{R}^N}u^{\theta+1}dx=0.
\end{align*}
As \eqref{LEbis} implies that
$$\frac{2(p+1) - pN}{p+1}+\frac{N}{\theta+1} = 2 - \frac{(p\theta - 1)N}{(p+1)(\theta + 1)} = 2 - \frac{2N}{2+\alpha + \beta} > 0,$$
$u \equiv 0$ in $\mathbb R^N$ which is absurd, so we are done.\qed

\section{\bf Proof of Theorem \ref{main3}.}
\setcounter{equation}{0}
The approach is similar to that for Theorem \ref{main4}. We derive first some integral estimates thanks to Lemma \ref{lemnewBN}. Suppose that $u$ is stable outside the ball $B_{R_0}$. Let $R > R_0 + 3$ and $\zeta \in C^2_c(\mathbb R^N\backslash B_{R_0})$ verifying that $0 \leq \zeta \leq 1$ and
$$\zeta(x) =\left\{
\begin{array}{ll}
       0 \quad \mbox{for}\; \|x\|\leq R_{0}+1,\; \|x\|\geq 2R,\\
      1\quad \mbox{for}\; R_{0}+2 \leq \|x\| \leq R.
       \end{array}
     \right.$$
Clearly, we can assume that there exists $C > 0$ independent on $R$ such that
$$\|\zeta\|_{C^2(B_{R_0+2})} \leq C \quad \mbox{and} \quad R|\nabla \zeta(x)| + R^2|\nabla^2 \zeta(x)|\leq C \;\; \mbox{in } A_R = B_{2R}\backslash B_R.$$
Applying the estimate \eqref{new7} with $\zeta$, we get readily
\begin{align}\label{AvxZK}
 \begin{split}
& \; \int_{R_{0}+2\leq \|x\|\leq R} |\Delta u|^m dx + \int_{R_{0}+2\leq \|x\|\leq R} |u|^{\theta+1} dx \leq C\left(1 + R^{N-\frac{2m(\theta + 1)}{\theta -(m-1)}}\right).
 \end{split}
\end{align}
Using \eqref{new8} and tending $R \to \infty$, we have then
\begin{align}
\label{new11}
u \in L^{\theta + 1}(\mathbb R^N) \quad \mbox{and}\quad \Delta u \in L^m(\mathbb R^N).
\end{align}
By H\"older's inequality, there holds
\begin{align*}
R^{-2m} \int_{B_{R}} |u|^mdx  \leq CR^{\frac{N(\theta+1-m)}{\theta+1}-2m}\left(\int_{B_R} |u|^{\theta+1}dx\right)^{\frac{m}{\theta+1}}.
\end{align*}
On the other hand, by standard scaling argument, there exists $C > 0$ such that for any $R > 0$, any $u \in W^{2, m}(A_R)$ with $A_R = B_{2R}\backslash B_R$,
\begin{align*}
R^{-m} \int_{A_{R}} |\nabla u|^mdx  \leq C \int_{A_{R}} |\Delta u|^mdx  + CR^{-2m} \int_{A_{R}} |u|^mdx.
\end{align*}
Therefore, under the assumptions of Theorem \ref{main3}, we get
\begin{align}
\label{new9}
R^{-2m} \int_{A_{R}} |u|^mdx + R^{-m} \int_{A_{R}} |\nabla u|^mdx \to 0 \quad \mbox{as }\; R \to \infty.
\end{align}
Let $\zeta(x) = \phi_0(R^{-1}x)$ with a standard cut-off function $\phi_0 \in C_c^2(B_2)$, $\phi_0\equiv 1$ in $B_1$. Applying the estimate \eqref{new4} and using \eqref{new11}--\eqref{new9}, there holds
\begin{align}
\label{new10}
\int_{R^N}|\nabla^2 u|^m dx < \infty.
\end{align}

However, as we have mentioned, the weak solutions of \eqref{1} are in general not belongs to $C^2$, so we cannot use the standard Pohozaev identity similar to \eqref{new6} because of the boundary terms. We show here a variant of the Pohozaev identity, which proof is given in the appendix for the convenience of the readers.
\begin{lem}\label{lem1h.1}
Let $u$ be a  weak solution to \eqref{1} with $m > 2$. Then for any $\psi \in C_c^{2}(\Omega)$,
\begin{align}\label{new12}
\begin{split}
&\; \frac{N}{\theta+1}\int_{\Omega} |u|^{\theta+1}\psi dx
- \frac{N-2m}{m}\int_{\Omega} |\Delta u|^{m}\psi dx \\
 =& \;-\frac{1}{\theta+1}\int_{\Omega} |u|^{\theta+1}(\nabla \psi\cdot x) dx + \frac{1}{m}\int_{\Omega} (\nabla\psi\cdot x)|\Delta u|^{m} dx\\
& \; - \int_{\Omega} |\Delta u|^{m-2}\Big[2\Delta u(\nabla u\cdot \nabla \psi)  + 2\Delta u \nabla^2(x\cdot\nabla\psi) + \Delta u(\nabla u\cdot x)\Delta \psi \Big]dx.
\end{split}
 \end{align}
 \end{lem}

This implies that if u is a  weak solution of \eqref{1}, stable at infinity with $1 < m-1<\theta$ and $N$ verifying \eqref{new8}, then
\begin{align}\label{8newest0}
 \frac{N-2m }{m}\int_{\mathbb{R}^N}|\Delta u|^{m}dx=\frac{N}{\theta+1}\int_{\mathbb{R}^N}|u|^{\theta+1}dx.
 \end{align}
Indeed, let $\psi$ in \eqref{new12} be defined by $\psi(x) = \phi_0(R^{-1}x)$ with a standard cut-off function $\phi_0 \in C_c^2(B_2)$, $\phi_0\equiv 1$ in $B_1$.
Denote the right hand side in \eqref{new12} by $I_R$. Remark that $\nabla \psi \ne 0$ only in $A_R = B_{2R}\backslash B_R$ and $\|\nabla^k \psi\|_\infty \leq C_k R^{-k}$, we obtain readily
\begin{align*}
|I_R| \leq C\int_{A_R} \Big(|u|^{\theta+1} + |\Delta u|^m\Big) dx + \frac{C}{R}\int_{A_R}|\Delta u|^{m-1}|\nabla u|dx + C\int_{A_R}|\Delta u|^{m-1}|\nabla^2 u| dx
\end{align*}
Thanks to the estimates \eqref{new11}-\eqref{new10} and H\"older's inequality, clearly $\lim_{R\to \infty}I_R = 0$, hence we get \eqref{8newest0}.

\medskip
On the other hand, using $u \psi$ as test function in \eqref{1} , we get
\begin{align*}
 \int_{B_{2R}}|\Delta u|^{m} \psi dx-\int_{B_{2R}} |u|^{\theta+1} \psi dx &\leq C\int_{B_{2R}}|u||\Delta u|^{m-1}
  |\Delta \psi| dx
  + C\int_{B_{2R}}|\Delta u|^{m-1} |\nabla u||\nabla\psi| dx\\
& \leq \frac{C}{R^2}\int_{A_R}|u||\Delta u|^{m-1}dx
  + \frac{C}{R}\int_{A_R}|\Delta u|^{m-1}|\nabla u|dx.
\end{align*}
Apply H\"older's inequality, \eqref{new11}--\eqref{new9} and tending $R$ to $\infty$,  we obtain
 \begin{align}\label{llfou}
\int_{\mathbb{R}^N}|u|^{\theta+1}dx=\int_{\mathbb{R}^N} |\Delta u|^{m}dx.
\end{align}
Combining \eqref{8newest0} and \eqref{llfou}, one obtains
\begin{align*}
\left(\frac{N-2m }{m}-\frac{N}{\theta+1}\right)\int_{\mathbb{R}^N}|u|^{\theta+1}dx=0.
\end{align*}
We are done, since \eqref{new8} implies that $\frac{N-2m }{m}-\frac{N}{\theta+1} < 0$.\qed

\section{Appendix }
\setcounter{equation}{0}
We prove here the Lemma \ref{lem1h.1}. Let $\psi\in C_c^{2}(\Omega)$, multiplying equation \eqref{1} by $\nabla u\cdot x \psi$ and integrating by parts, we get
\begin{align*}
& \int_{\Omega}|u|^{\theta-1}u (\nabla u\cdot x)\psi dx\\ = & \;  \int_{\Omega}|\Delta u|^{m-2}\Delta u\Delta(\nabla u\cdot x\psi)dx\\
= & \; \int_{\Omega}\; |\Delta u|^{m-2}\Delta u \Big[(\nabla(\Delta u)\cdot x)\psi+2\Delta u \psi+2\nabla(\nabla u\cdot x)\cdot\nabla \psi+(\nabla u\cdot x)\Delta\psi \Big]dx.
\end{align*}
Direct calculation yields $\nabla(\nabla u\cdot x)\cdot\nabla \psi = \nabla^{2}u (x, \nabla \psi) + (\nabla u\cdot \nabla \psi)$ and
\begin{align*}
 \int_{\Omega}\; |\Delta u|^{m-2}\Delta u \Big[(\nabla(\Delta u)\cdot x)\psi+2\Delta u \psi\Big]dx
   &= \int_{\Omega} \frac{\nabla |\Delta u|^m}{m}\cdot x \psi dx+ 2\int_{\Omega}|\Delta u|^{m}\psi dx\\
   &=\frac{2m-N }{m}\int_{ \Omega}|\Delta u|^m\psi dx-\frac{1}{m}\int_{\Omega}|\Delta u|^m(\nabla \psi\cdot x)dx.
\end{align*}
Moreover,
\begin{align*}
\int_\Omega |u|^{\theta-1}u (\nabla u\cdot x)\psi
  dx & = - \frac{1}{\theta + 1}\int_\Omega |u|^{\theta +1} {\rm div}(\psi x)dx \\
&=-\frac{N}{\theta+1} \int_{\Omega}|u|^{\theta+1} \psi dx - \frac{1}{\theta+1} \int_{\Omega}|u|^{\theta+1} x\cdot \nabla \psi dx.
\end{align*}
Therefore, the claim follows by regrouping the above equalities. \qed
\vskip 1cm

\section*{Acknowledgment}
The authors are partly supported by the CNRS-DGRST Project No.~EDC26348.
\section*{References}

redpajama_set_name: RedPajamaArXiv


text:
EducationLatest News
Over 375,000 students write WASSCE today
This year's West African Senior School Certificate Examination (WASSCE) commenced across the country on Monday, July 20, 2020 for Ghanaian students after several months of uncertainty as a result of the Covid-19 pandemic.
In all, a total of 375,737 candidates from the various senior high schools in Ghana will begin the examination at the various designated centers.
Out of the said figure, 188,163 are females whilst the remaining 187,574 are males.
The examination begins with practical for the Visual Arts Students whilst the theory is set to last with a period of one month, August 3 to September 5, 2020.
Commenting on the readiness of the West African Examination Council (WAEC), the Head of the National Office of WAEC, Mrs. Wendy Enyonam Addy-Lamptey, told the Daily Graphic that the outbreak of Covid-19 pandemic has brought changes to how the examination would be conducted.
"The number of examination personnel has been revised upward by 10 per cent to cater for the additional classrooms to be used by the candidates to meet the Covid-19 social distancing protocol", she explained.
Meanwhile, the president has extended his warm wishes to the candidates.
"I send my best wishes to the three hundred and thirteen thousand, eight hundred and thirty-seven (313, 837) final year senior high school students, who will, from today Monday, 20th July take the West African Senior Secondary School Certificate (WASSCE)."
WAEC.
WASSCE
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text:
Cairone v. McHenry County College
DRAGA CAIRONE, Plaintiff,
MCHENRY COUNTY COLLEGE, et al., Defendants.
Robert M. Dow, Jr., United States District Judge.
This matter is before the Court on Defendants' motion to dismiss [34]. For the reasons set forth below, the motion is granted in part and denied in part. The parties are given until August 23, 2019 to file a supplemental brief of no more than five pages addressing whether Plaintiff's ADA, retaliation, and breach of contract claims against the individual Defendants should be dismissed. The case is set for status on September 3, 2019 at 9:00 a.m.
Defendant McHenry County College ("MCC") is a county institution of higher education operated by McHenry County. [32, at ¶ 4.] Plaintiff Draga Cairone was a student enrolled at MCC during the relevant period. [Id. at ¶ 3.] Plaintiff has autism, which prevents her from being able to understand body language, understand sarcasm, or detect lies. [Id. at ¶ 6.] MCC learned about Plaintiff's autism in 2009. [Id. at ¶ 8.] Plaintiff was disciplined by MCC on numerous occasions, including in 2009, 2012, 2013, and 2015.
This lawsuit relates primarily to disciplinary proceedings against Plaintiff in 2015. On or about May 11, 2015, Plaintiff received notice directing her to meet with Defendant Talia Koronkiewicz, the official then in charge of student discipline, to discuss "possible violations" of MCC's Student Code of Conduct (the "Code"). [Id. at ¶ 11.] Plaintiff was only given two days to prepare for the meeting and was not given any further information about what the "possible violations" were. [Id.] Koronkiewicz had been investigating and documenting allegations that Plaintiff had engaged in inappropriate behavior in the student lounge. [Id. at ¶ 12.] Specifically, other students alleged that Plaintiff engaged in nonconsensual touching, unwanted physical closeness, and shaking the bottom of her bra cups in a way that made other students uncomfortable (what the parties refer to as the "boobie dance"). [Id.]
Plaintiff alleges that Koronkiewicz reported that three students made statements- presumably about the alleged misconduct. [Id.] Koronkiewicz said that these students said that nobody wanted to confront Plaintiff because-according to the students-she is "dramatic and makes people feel bad." [Id. at ¶ 13.] The students also said that Plaintiff would nestle, cuddle, or lie on other students without asking. [Id.] One student said that Plaintiff was "always touching" her even though she would tell Plaintiff to stop. [Id.] Plaintiff alleges that on all occasions she distributed written or digital consent forms for other students to sign, explaining her cognitive limitations and asking for consent to touch or to engage in other intimate behavior that might call for consent. [Id. at ¶ 14.] When someone declined to sign the consent form, Plaintiff would not engage in intimate behavior with the student (with the exception of one individual named Brandon Marciel). [Id.] The point of these forms was to ensure that Plaintiff had express consent. [Id.]
On May 13, 2015, Koronkiewicz interviewed Plaintiff regarding these allegations. [Id. at ¶ 15.] In the May 13, 2015 interview, Koronkiewicz told Plaintiff, "you need to learn to read body language. Everyone else can, so can you. If you can't you can't receive or give consent." [Id. at ¶ 18.] At the interview, Plaintiff asked Koronkiewicz to consult with an advisor or person with knowledge of and expertise in autism so that the disciplinary board could have the benefit of advice from someone with knowledge of Plaintiff s disability. [Id. at ¶ 19.]
Koronkiewicz's notes from the meeting indicate that she and Plaintiff discussed Plaintiffs "excessive contact without asking." [Id. at ¶ 15.] The notes do not indicate that Plaintiff admitted to such touching. [Id.] Plaintiff told Koronkiewicz that she touched the bottom of her bra cups for medical reasons, not as a sexual provocation. [Id.] The notes from the meeting indicate that Plaintiff:
• Admitted to asking "if people want[ed] to see her bondage gear, but no nudes."
• Said that her "friends have told her she may be unethical and now [she] doesn't know if she understands consent."
• Admitted to knowing that her conduct made people uncomfortable and that she had learned that people do not tell her this.
• Admitted to sending a text in which she discussed another student as "the object being cut up [with a sword] at a party, " which was intended as a joke and was inappropriate.
[Id.] Plaintiff claims that these statements "are largely false, incomplete, and misleading." [Id. at ¶ 16.]
Plaintiff did say that the touching of her bra was done for medical reasons and not for any sexual purpose. [Id.] Plaintiff told Koronkiewicz that the touching (done fully clothed) was a prophylactic for breast cancer and had a scientific medical basis. [Id.] Plaintiff informed Koronkiewicz that a lot of female students at MCC were doing this without getting in trouble, so she thought it was okay. [Id.] Plaintiff also told Koronkiewicz that she asked other student's consent to see photos of her bondage "gear, " but not photos of her wearing the gear. [Id.] She explained to Koronkiewicz that this inquiry was an attempt, sometimes successful, to gain consent. [Id.] Plaintiff did not show pictures of her gear to students who denied consent. [Id.] Plaintiff did report that people in the student lounge told her that she might be unethical, but she disagreed with that evaluation. [Id.] Plaintiff did not tell Koronkiewicz that she did not understand consent. [Id.] Plaintiff did, however, tell Koronkiewicz that she did not understand revocation of consent if it was not verbal or written. [Id.] Plaintiff did not admit to knowing that her conduct made people uncomfortable without them saying so. [Id.] Rather, Plaintiff acknowledged that Koronkiewicz told her this at the interview. [Id.] Finally, Plaintiff did not tell Koronkiewicz that she had discussed cutting up a student with a sword. [Id.] Rather, Plaintiff informed Koronkiewicz that she had responded to such comments made by another student "by inquiring about that student's name (over text) to express incredulity and disapproval." [Id.]
Two days after the May 13, 2015 interview, Koronokiewicz rejected Plaintiff's request to consult with a person with knowledge of her disability and did not propose any alternative. [Id. at ¶ 19.] Between May 13, 2015 and June 3, 2015, Koronkiewicz did not give Plaintiff any notice or warning that she was facing charges of harassment or of violating any other provisions of the Code. [Id. at ¶ 17.] On June 3, 2015, Koronkiewicz issued an official Sanctions Letter stating that after a "due process meeting" with [Plaintiff] and others, an MCC "disciplinary body" including Defendants Flecia Thomas and Juletta Patrick determined that Plaintiff had violated the harassment provision of the Code. [Id. at ¶ 20.] The letter indicated that this decision was based on Plaintiff's admission that she had (1) touched her bra/breasts in a sexually provocative way, (2) laid on and cuddled with other students without their consent, and (3) asked others to see pictures of Plaintiff in bondage. [Id. at ¶ 20.] Plaintiff alleges that she "always obtained express consent for any such interaction, either orally or writing or by text." [Id. at ¶ 30.] But Plaintiff does not explain what is meant by "any such interaction." Plaintiff further alleges that she never laid on or cuddled with other students without their express consent. [Id. at ¶ 21.] The disciplinary body did not conclude that Plaintiff had engaged in sexual harassment as defined in the Code.[1] [Id. at ¶ 22.] Nor did the disciplinary body conclude that Plaintiff had harassed anyone because of sex or on the basis of any other protected group membership or characteristic. [Id.]
The Sanctions Letter indicated that MCC would be imposing the following consequences on Plaintiff:
• A one-year suspension and ban from MCC, starting on June 11, 2017.[2]
• A requirement that Plaintiff provide a re-enrollment application and a personal statement discussing (at least in part) certain issues regarding sexual harassment.
• A requirement that Plaintiff undergo sexual harassment training.
[Id. at ¶ 23.] Plaintiff was advised that she could file a notice of intent to appeal by June 17, 2015, which she did. [Id.] Sometime before June 17, 2015, Plaintiff requested that counsel be allowed to speak on her behalf at the appeals hearing that was to be scheduled. [Id. at ¶ 24.] This request was rejected on June 18, 2015. [Id. at ¶ 24.] However, Plaintiff was allowed to have a non-attorney advisor speak on her behalf. [Id.]
On June 24, 2015, Koronkiewicz gave Plaintiff a letter (dated June 23, 3015) setting the hearing for July 29, 2015. [Id. at ¶ 25.] The letter indicated that Plaintiff had to provide a witness list by July 1, 2015. [Id.] The letter further indicated that Plaintiff had to provide an oral statement and copies of all documents Plaintiff was submitting in support of her appeal by July 15, 2015. [Id.] Plaintiff was banned from campus, so she could only gather materials remotely. [Id.] She did not make the 9:00 a.m. deadline. [Id.] ¶ 9:30 a.m. on July 15, 2015, Koronkiewicz emailed Plaintiff to inform her that the deadline had passed. [Id.] Plaintiff wrote back to say that her witnesses had not been contacted and that she therefore requested an extension. [Id.] Koronkiewicz replied that she had spoken to one of the witnesses in May and that the statements of three people would be included in MCC's record of evidence (the "Packet").[3] These witnesses would not be allowed to testify. [Id.]
         The hearing was rescheduled for Aug. 12, 2015. [Id.] MCC gave Plaintiff the Packet containing the case against her on Aug. 10, 2015, two days before the appellate hearing. [Id. at ¶ 26.] The Packet contained witness statements from the initial disciplinary proceedings. [Id. at ¶ 27.] It also contained two interviews "written" by Koronkiewicz, on July 15, 2015. [Id.] In these written interviews, one witness stated that they believed that Plaintiff "acts appropriately" and that they had never witnessed any nonconsensual touching. [Id.] The other witness stated that Plaintiff stopped touching and apologized when asked to cease physical contact that had become unwelcome. [Id.] This witness further stated that other students had said that Plaintiff's touching and comments made them uncomfortable, but they had not told her so. [Id.] The witness thought that these people gave Plaintiff nonverbal cues that Plaintiff did not understand because of her autism. [Id. at ¶ 27.] The Packet also contained a record of the disciplinary incidents ...