text:
#ifndef _mof_utility_h
#define _mof_utility_h

/*
**==============================================================================
**
** Utilities needs to handle both ANSI and UNICODE buffer,
** including UTF16 Big Endian.
** Which require an input parameter (MOF_Encoding e),
** Caller need to make sure input data is encoded as UTF16 Big Endian
**
**==============================================================================
*/
int mof_getchar(MOF_Encoding e, void * data);

int mof_nextchar(_Inout_ MOF_Buffer * b);

MI_Boolean mof_isspace(MOF_Encoding e, void * data);

MI_Boolean mof_isalpha(MOF_Encoding e, void * data);

MI_Boolean mof_isalnum(MOF_Encoding e, void * data);

MI_Boolean mof_isunderscore(MOF_Encoding e, void * data);

MI_Boolean mof_isdoulbequotes(MOF_Encoding e, void * data);

MI_Boolean mof_isbackslash(MOF_Encoding e, void * data);

MI_Boolean mof_isminus(MOF_Encoding e, void * data);

MI_Boolean mof_isplus(MOF_Encoding e, void * data);

MI_Boolean mof_isdigit(MOF_Encoding e, void * data);

MI_Boolean mof_match(MOF_Encoding e, _In_ MOF_StringLen *data, _In_ MOF_StringLen *pattern);

/*
**==============================================================================
**
** Utilities handles only UTF16 Little Endian buffer
** or the function result does not depends on Endian;
**
** Caller need to make sure either the data is already converted to
** UTF16 Little Endian or the function result does not depends on Endian
** format.
**
** They take a parameter (MI_Boolean unicode) to indicate buffer is
** unicode or ansi.
**
**==============================================================================
*/

//
// result does not depends on endian
//
size_t mof_offset(MI_Boolean unicode, _In_ void * orgdata, _In_ void *curdata);

void mof_getcode(MI_Boolean unicode, _Inout_ MOF_StringLen *r);

int mof_ncasecmp(MI_Boolean unicode, _In_ void *s1, _In_ void * s2, size_t n);

void* mof_nextcharofbuf(MI_Boolean unicode, void * data);

void * mof_getprevcharpos(MI_Boolean unicode, void * data);

MI_Boolean mof_strtoul(MI_Boolean unicode, _In_ void * data, int radix, _In_opt_ void * endchar, _Out_ unsigned long *result);

MI_Real64 mof_strtod(MI_Boolean unicode, _In_ void * data);

MI_Sint64 mof_strtoll(MI_Boolean unicode, _In_ void * data, void ** endchar, int base);

MI_Sint64 mof_strtoull(MI_Boolean unicode, _In_ void * data, void ** endchar, int base);

void* mof_setchar(MI_Boolean unicode, _In_ void * data, int c);

MI_Boolean mof_isvalidhexstring(MI_Boolean unicode, void * data, size_t length);

/*
**==============================================================================
**
** Utilities handle MOF_Buffer
**
**==============================================================================
*/
int mof_nextcharn(_Inout_ MOF_Buffer* b, _Inout_ size_t *len);

MI_Boolean mof_eof(_In_ MOF_Buffer* b);

MI_Boolean mof_neof(_In_ MOF_Buffer* b);

void mof_fillbuf(_In_ MOF_Buffer* b,
                 _Out_writes_(size) MI_Char *buf,
                 _In_ MI_Uint32 size);

/*
**==============================================================================
**
** Error reporting functions
**
**==============================================================================
*/
extern void moferror(_In_ MOF_ErrorHandler *errhandler, _In_z_ const char* msg);

extern void yywarnf(_In_ MOF_ErrorHandler *errhandler, int id, _In_z_ const char *fmt, ...);

#define MOF_TRACE printf("MOF_TRACE: %s(%u)\n", __FILE__, __LINE__)

#define MOF_NOT_FOUND ((MI_Uint32)0xFFFFFFFF)

#define MOF_ASSERT(COND) assert(COND)

/*
**==============================================================================
**
** Utility report final error to error handler, it produces error code,
** error type, and error category based on error ID
**==============================================================================
*/
void ReportFinalError(
    _In_ MOF_ErrorHandler* errhandler,
    _In_ int id,
    _In_z_ MI_Char* formattedMSG);

/*
**==============================================================================
**
** Allocation utilities from batch
**
**==============================================================================
*/
void* MOF_Realloc(Batch* batch, void* ptr, size_t orgsize, size_t size);

/*
**==============================================================================
**
** Type-aware forms of memory allocation functions from batch
**
**==============================================================================
*/
#define CALLOC_T(ZT, N) ((ZT*)Batch_GetClear(state->batch, N * sizeof(ZT)))
#define MALLOC_T(ZT, N) ((ZT*)Batch_Get(state->batch, N * sizeof(ZT)))
#define REALLOC_T(ZT, PTR, O, N) ((ZT*)MOF_Realloc(state->batch,PTR,(O)*sizeof(ZT),(N)*sizeof(ZT)))

/*
**==============================================================================
**
** Other utilities
**
**==============================================================================
*/
MI_Result mof_setupbuffer(
    _In_reads_bytes_(nBytes) void * data,
    _In_ size_t nBytes,
    _In_opt_ Batch *batch,
    _Inout_ MOF_Buffer * b);

/* Convert UTF16 Big Endian to Little Endian string */
int mof_converttoUTF16LE(_Inout_ MOF_State * state, _Inout_ MOF_StringLen *r);

/*
**==============================================================================
**
** IO utilities
**
**==============================================================================
*/
#if defined(CONFIG_ENABLE_WCHAR)
# define Zfprintf fwprintf
#else
# define Zfprintf fprintf
#endif
#endif

redpajama_set_name: RedPajamaGithub


No I didn't, but that is ok, I don't want to do any hardware mods while I am still under warranty. eVGA has loose rules for warranty, but they still frown on modding the circuit board.
Its not an actual mod to the board as such. What you do is find a certain surface mount resistor. You lower the resistance of that resistor, while measuring, to get the desired volt mod.
To lower that resistance, you <could> use a soldered in variable resistor, but the best way, is to just use a 2B pencil, since its very mildly conductive. When done with it, you simply wipe the pencil trace off the top.
Do you know where I can find a guide to do the pencil mod?
Thats the previous thread on volt modding, with pictures for the pencil mod.
If you really do something serious with the cooling, and that mod, expect over 500Mhz on water cooling, and in some cases over 600Mhz. It really makes an absolutely massive difference, but that is with a very high volt mod and water.
A minor volt mod, running with stock, or good air cooling, will get you a nice little bump. I would advise doing something with your cooling first though.
1: You could try removing the stock heatsink, removing the thermal paste, and using a quality layer of arctic silver 5 to improve thermal conductivity.
4: You could try a zalman cooler, as the NVsilencers are not the /best/ coolers anymore.
I am having an interesting issue with my 6800le....i currently have it clocked 420/820, which it takes just fine, it will probably go much higher, but now, whenever i restart my computer, the memory reverts back to stock speeds, the core speed stays the same, just the memory changes, any idea why this is happening?
What program are you using to set your overclock speeds?
I already have a NV silencer on the card with AS5, it doesn't go above 62c load and the AS5 hasn't even totally cured yet. Thanks for the link, I'll try it sometime soon.
I am using the Winfast Speedrunner software that comes with the card, it works best i think, but after i set the clocks, i go into the regular "classic" control panel, make sure it reads the same speeds, then i click the box that says "set these speeds at startup", it gives me the OK, and everything is fine from there, until i restart, then the memory speed reverts. I am totally confused on this one. I have tried setting the clocks with the regular CP, but when i try it, it hardly lets me raise it at all, and it says the speeds fail the test, even though when i set it with winfast, the card runs just fine at the faster speeds, i dont get it.
You should use either Coolbits 2 or Rivatuner, they work the best.
I have solved the problem. All i had to do was set the 2D memory setting the same as the 3D, and it keeps all the settings the same after a restart. As of now i actually have the card clocked at 425/850 and i am pretty sure i can push it even further, as i am not getting any artifacts at all, and my temps still max out around 65-68. Later on i am gonna try pushing it a little more and really see what this card will do, i love this Leadtek. As far as OC software, i have tried them all, and for my card, the Winfast Speedrunner software works better than anything, and has allowed me much higher clocks than any other software.
Last edited by NarkotikBuzz; 26th June, 2006 at 07:49 PM.
I own Diablotek 6800GS 512, purchased from geeks.com.
So i think our friends, chinese masters did some stupid bios mistake in this card. I can proceed RMA but before RMA i want to play little bit. So anyone have a same card and any similar problems?. If some one have a proper bios could post here for experiments?. Any advice and any directions in this problem?
This is mbataka's first post. The Welcome thread is here. Please drop by the Welcome thread and give mbataka a warm AOA welcome there.
Last edited by Gizmo; 30th June, 2006 at 05:16 PM.
When you uninstall your drivers, does it change / improve anything?
Also, starting in safe mode, does that make any changes?
Does anyone have any suggestions on getting my eVGA 6800gs agp passed 415/1150.
Are you using the 91.33 drivers? They seam to overclock better than previous versions.
I'm still using 84.21, I haven't tried 91.33, because I seem to have trouble with anything but the 84.21. This card is very moody.
After uninstalling, same in the bios post, but in the windows everything fine.
In safe mode its mean not loading the drivers. then its fine. But without drivers its nouse right. i tryed almost every driver, all the same. In the bios edit it says: Core: 425MHZ Memory: 500MHZ , so i'm doubting its a NV42 Chip.
What drivers were you using before you unistalled them mbataka?
Dsio, i think you mean the 91.31's.....we just got the 91.33's yesterday, and there is not TRU modified version of them yet. I tried the new 91.33's and had a ton of problems, and they are BETA drivers. Please just try the 91.31 Xtreme-G modified drivers, they work very well, and overclock great.
Let us know if they work for you, i have had great success with this set.


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Exit International, the group that helped David Goodall carry out his wish, said the scientist was declared dead at 12:30 p.m. in Liestal, a town outside the city of Basel, where he had traveled to take advantage of Switzerland's assisted-suicide laws.
"My life has been rather poor for the last year or so. And I'm very happy to end it," Goodall said Thursday in the room where he died shortly after.
Assisted suicide is legal in Switzerland, where the procedure is available for anyone who acknowledges in writing that they are taking their lives willingly — without being forced. But the practice is frowned upon by many doctors and some others who say it should be reserved for the terminally ill. Goodall and his supporters want the practice to be more accepted as a legitimate choice for elderly people in sound mind.
On Wednesday, Goodall told a crowded news conference that medically assisted suicide should be more widely available.
"At my age, and even at rather less than my age, one wants to be free to choose the death and when the death is the appropriate time," he said.
Hundreds of people — some far more frail than Goodall, who used a wheelchair — travel to Switzerland every year to take their lives. The best-known group to help foreigners end their days in the Alpine country is Dignitas, but others include Life Circle in Basel, Goodall's choice.
Geir Moulson contributed from Berlin.


On November 21st, the U.S. Navy commissioned its newest warship into the fleet – the littoral combat ship USS Milwaukee (LCS 5), fifth of 52 planned small surface combatants designed to project U.S. presence and power in near-shore environments as well as open-ocean operations.
The LCS program is picking up speed, with an average of four new ships entering the fleet every year for the next several years. In fact, between now and 2018 – the length of a standard Navy tour of duty – the LCS program will achieve several key milestones that demonstrate the amazing progress and potential of these warships.
At this very moment, halfway around the world, Milwaukee's sister ship USS Fort Worth (LCS 3) is plying the waters in and around the South China Sea, about midway through a scheduled 16-month deployment. Fort Worth's Sailors boldly and voluntarily stand the watch, providing the presence and surface warfare capability critical to our Navy's mission, and ensuring freedom of navigation for all who sail international waters. By 2018, both LCS variants will be routinely operating in this part of the world, with as many as four LCS on station at any given time.
LCS and its mission packages provide combatant commanders the focused mission ability to counter anti-access shallow water threats such as small boats, submarines and mines. The initial phase of the surface warfare mission package completed testing in 2014 and is deployed overseas today on USS Fort Worth. We continue to develop and test, in an operationally realistic environment, those systems that will bring added capability to these incredible ships, including the mine countermeasures (MCM) and antisubmarine warfare (ASW) mission packages.
Recognizing that small surface combatants will represent one-sixth of our future 300-ship Navy, Defense and Navy leaders determined that future LCS could support enhanced lethality and survivability capabilities without significant cost, a testament to the flexible and modular design of the LCS. This modified LCS, called a frigate (FF), will leverage the success of the existing LCS design, but will be a multi-mission vessel capable of performing surface warfare and anti-submarine warfare missions simultaneously. The first frigate is expected to be under contract in Fiscal Year 2019.
In the meantime, the LCS shipbuilders in Marinette, Wisconsin and Mobile, Alabama will continue to build 32 ships of the current LCS design. By 2018, the last of these LCS will be under construction, meaning that we will have put 32 ships under construction in a span of only 13 years – an impressive feat for U.S. Navy shipbuilding.
USS Milwaukee is the fifth LCS to enter the fleet, and the number of these ships is scaling up fast. The maintenance, logistics, and training pipelines to support this LCS fleet are likewise rapidly expanding to meet expected operational needs. By 2018, the LCS class will be the second largest surface Navy ship class behind the DDG 51 Arleigh Burke-class destroyers.
LCS is a proven fleet asset. Collectively, the LCS class has over 210,000 nautical miles of water under the keel since the 2008 delivery of USS Freedom (LCS 1). Through operational testing in realistic scenarios and over the course of three deployments, the LCS and its mission packages have demonstrated the ability to meet specific, real-world requirements for our combatant commanders. The crew of the USS Milwaukee and the citizens of its namesake city have many reasons to be proud of their ship, as do I.


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text:
Scientific Prizes Bring Needed Attention to Mental Health Research
By Dr. Herbert Pardes published 22 October 15
Conceptual brain image; psychiatric disorders. (Image credit: Dreamstime.)
Dr. Herbert Pardes is executive vice chairman of the board of trustees at New York-Presbyterian Hospital and president of the Scientific Council of the Brain & Behavior Research Foundation — and last year was the first to win the prize that now bears his name. Pardes contributed this article to Live Science's Expert Voices: Op-Ed & Insights.
This fall, scientists around the world will trade lab coats for tuxes and ball gowns for the annual "award season" announcements of the Nobel Prize, the MacArthur Foundation fellowships, the Lasker Awards, and the star-studded and televised Breakthrough Prize awards. These prizes, each accompanied by significant monetary awards, shine a much-needed spotlight on important scientific research that, in the past, the media has often considered too dense and academic to be of interest to the public.
We are in a day and age when one of television's biggest comedy hits, "The Big Bang Theory," is about the lives of Caltech Ph.D.s, when biopics about Alan Turing and Stephen Hawking win Oscars, and when Hawking hobnobs with a Russian billionaire to search for extraterrestrial life. Suddenly, science, technology, engineering and mathematics (the "STEM" fields) have acquired a glamorous sheen that is attracting people with deep pockets who want to accelerate the often slow, deliberate and bureaucratic march toward new findings and cures.
Philanthropy has a place in research
Many in the scientific community are worried by a trend in which wealth trumps peer reviews, grant applications and selection committees, but many philanthropists are serious, knowledgeable, deeply committed and determined to find cures for the most intractable diseases.
For those of us who have devoted our lives to scientific research, there is a lot to appreciate about this new world order.
In today's funding environment, the most significant inroads to scientific discovery will likely be generated by a combination of public and private funding from people with a vested stake in finding cures. Just look at the progress made by public-private partnerships such as the American Cancer Society in reducing mortality from breast, lung and colon cancer, and the Cystic Fibrosis Foundation — which supported the development of Kalydeco, the first drug to treat an underlying cause of the disease — through its "venture philanthropy."
Federal funding for scientific research has sharply declined over the past 10 years, and dwindling resources have led to the erosion of federal grants, lab closures and diminished incentives for all scientists and, consequently, the loss of talented young researchers.
Even so, private philanthropy still lags behind public funding. The U.S. National Institutes of Health's (NIH) yearly budget is $31 billion, half of which goes toward basic science — the backbone of most scientific discoveries but lacking the allure to attract the wealthiest philanthropists.
Within that agency, the U.S. National Institute of Mental Health (NIMH) has an annual budget of $1.4 billion, a figure that has declined more than 10 percent in the past five years when adjusted for inflation, meaning a substantial decrease in funding for both basic research and clinical trials. In fact, private funding is now helping to drive the kind of high-risk, high-reward research that may change lives and end the suffering that psychiatric illnesses bring to so many.
Improving care for mental health
During my career, I have been privileged to work in both the public and private sectors. In 1984, I left my position as director of the NIMH to become chairman of the Department of Psychiatry at Columbia University because I felt I could accomplish more on the outside.
Early in those early days, I set up an annual information symposium about mental health for the public, featuring scientists talking about psychiatric research in plain English. This event attracted hundreds of people starved for information, and enhanced the work of the Brain & Behavior Research Foundation (BBRF), which funds innovative ideas in neuroscience and psychiatry to understand the causes of brain and behavior disorders, and find better ways to diagnose and treat them. Grant recipients are selected by the foundation's Scientific Council, which comprises 165 leading experts across disciplines in brain and behavior research, including two Nobel laureates; four former directors of the NIMH; 13 members of the National Academy of Sciences; 21 chairs of Psychiatry and Neuroscience Departments at leading medical institutions; and 47 members of the Institute of Medicine.
Since 1987, the foundation has awarded 5,000 grants worth nearly $340 million. More importantly, it has seeded the research of an entire generation of brilliant young neuroscientists and clinicians who have gone on to obtain $3 billion in additional funding for their work. This is a philanthropic model worthy of emulation.
The field of mental health has come a long way since I visited a state hospital for the first time as a college sophomore in 1953. I will never forget what I saw — a naked man locked in an empty room, smearing his feces on the wall.
Today, people living with mental illness certainly have access to more help, and scientists are making great strides in basic research, new technologies, next-generation therapies and early interventions.
There are about 320 million Americans, and approximately one in four live day-to-day with mental illness. An influx of veterans are coping with post-traumatic stress disorder (PTSD), depression and suicide, and have raised awareness and aroused sympathy for people with psychiatric disorders. However, high-profile, violent incidents involving people with mental illness continue to reinforce lingering stigmas. Decades after the deinstitutionalization of mentally ill patients, the United States still lacks adequate community support, psychiatrists, psychologists, social workers, nurses and hospital beds. Many people go untreated, and there are more people with psychiatric illnesses in prison than in psychiatric hospitals.
An award to address a crisis
So much more needs to be done. How do we let people know about this epidemic? How can we get the NIH, the NIMH and the U.S. Congress, more specifically, to increase funding for research into mental illness? How does mental health become a top priority so that help is available to all who need it?
These problems have deep roots, and solutions are elusive and complex. That's why it is so important that scientists continue conducting basic and clinical research to understand how the healthy human brain gives rise to the miracle of thought and consciousness, and how genetic, cellular and circuit anomalies lead to the full panoply of brain disorders. [The Singularity, Virtual Immortality and the Trouble with Consciousness (Op-Ed)]
If you're a topical expert — researcher, business leader, author or innovator — and would like to contribute an op-ed piece, email us here.
Partly because of the logjam in Washington, enlightened philanthropists have never had a more important role to play in driving brain research forward. Though it is difficult to imagine what our ultrapartisan political system can accomplish, Congress must find ways to fund mental health research. Simultaneously, generous philanthropists must provide support.
This October, I'm honored that the Brain & Behavior Foundation will present the Pardes Prize — a $300,000 award recognizing humanitarians striving to change the way the world views mental illness through education, prevention, research, clinical care, mentoring, treatment or advocacy for both individuals and policies that affect mental health.
Society has found ways of recognizing contributions in basic science, clinical research and clinical care in the nonpsychiatric health fields, but the field of mental health in psychiatry is complex and affects people throughout the world. With this prize, we hope to expand the care, treatment and understanding of the underlying causes of mental illness, and reduce the pain these conditions cause individuals and their families.
I'll happily put on my tuxedo to call attention to issues surrounding mental illness. It is time to stop talking about helping people with mental illness and actually do something to help them lead productive lives.
Follow all of the Expert Voices issues and debates — and become part of the discussion — on Facebook, Twitter and Google+. The views expressed are those of the author and do not necessarily reflect the views of the publisher. This version of the article was originally published on Live Science.
Earth is at the center of a 1,000-light-year-wide 'Swiss cheese' bubble carved out by supernovas

redpajama_set_name: RedPajamaCommonCrawl


NATSPEC is the National Master Specification for use with projects of all types and sizes. NATSPEC includes most, but not all, technical requirements for the majority of projects. Additional material can be easily incorporated to create a specification custom made for your next project.
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NATSPEC is the National Master Specification for use with projects of all types and sizes. For each project, you select the appropriate NATSPEC worksections from the National Classification System and download them using SPECbuilder specification compilation software. NATSPEC worksection Templates are provided as MS Word files which you edit to suit the project. NATSPEC includes most, but not all, technical requirements for the majority of projects. Additional material can be easily incorporated to create a tailored specification for your next project.
NATSPEC packages are supplied on a subscription basis and are updated every April, in line with the NCC, and October. The following package are designed for architects, interior designers, landscape architects, structural and building services engineers, and home building designers.
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Cannabissearch.com Buzz Rating: Prepare to hack half a lung up. This bud is a serious cough-inducer that provides a powerful indica effect.
Recommended mostly for nighttime medication, Tahoe Alien will abduct you from Earth and send you soaring amongst the stars on a powerfully enjoyable body high with slight cerebral effects intertwined. The buds are cone shaped with such a large amount of sticky trichomes that the buds will literally stick to walls if you press them against them. Tahoe Alien gives off a pungent Kush smell with earthy undertones. The flavor is of lemon, Kush, pine and earth, with a surprisingly sweet aftertaste. Let's be honest though, the flavor isn't what we're after, it's the effect. This strain has a powerful body high that is great for alleviating nausea associated with chemotherapy, as well as treating insomnia. Eighths of this bud tend to range between $50 and $60 in most dispensaries, but may be seen as high as $65. Tahoe Alien requires 9 to 10 weeks of flowering in order to build up all those potent trichomes, and is recommended only for indoor cultivation, although they should do fine outdoors in temperate climates. Be easy when feeding these ladies nutrients and increase the amount gradually, as they can be somewhat prone to nute burn. For hydroponic growers, you'll definitely want to add this strain as the yield will increase while retaining that potency which Tahoe Alien has become renowned for. This is a grade-A strain that should be added to everyone's medical garden.
Tahoe Alien 0 out of 5 based on 0 ratings. 0 user reviews.


text:
Q: Formatting Usage with Docopt I am having a problem with my usage statements in docopt.
This is how I'd expect usage to work in the script. The optional parameters (defined with []), I would like to be able to use them together or individually. So -t -o or -o or -t should be valid. At the moments I cant use -o without -t.
If i use pipe | to separate them I can't use both at the same time. I've tried various combinations. I cant seem to get it work as id like. Can anyone point out where I am going wrong?
"""
Description:

Script does stuff

Usage:
  script.py (-d <ditem>) (-u <uitem>) (-p <pitem>) (-s <sfile>) [-t <tfile>] [-o <ofile>] [-v]
  script.py (-d <ditem>) (-l) [-t <tfile>] [-o <ofile>] [-v]
  script.py -h | --help
  script.py --version

Options:
  -v --verbose       Does stuff
  -t --tfile         Does stuff
  -o --output        Does stuff
  -l --litem         Does stuff
  -u --uitem         Does stuff
  -p --pitem         Does stuff
  -d --ditem         Does stuff
  -s --sitem         Does stuff
  -h --help          Show this screen.
  --version          Show version.
  """

A: I was able to resolve this by using the following:
By adding the usage strings script.py (-d <ditem>) (-l) ([-t <tfile>] | [-o <ofile>]) [-v] and another script.py (-d <ditem>) (-l) [-t <tfile>] [-o <ofile>] [-v] means I can use -t and -o independently or -t -o together. However Im not able to use them in this order -o -t.
Description:

Script does stuff

Usage:
  script.py (-d <ditem>) (-u <uitem>) (-p <pitem>) (-s <sfile>) [-t <tfile>] [-o <ofile>] [-v]
  script.py (-d <ditem>) (-l) ([-t <tfile>] | [-o <ofile>]) [-v]
  script.py (-d <ditem>) (-l) [-t <tfile>] [-o <ofile>] [-v]
  script.py -h | --help
  script.py --version

A: To allow -t along, -o along, -t and -o together:
Script does stuff.

Usage:
   script.py [-t] [-o]

Options:
  -t --tfile         Does stuff
  -o --output        Does stuff

If it is an error when both -t and -o are absent:
Script does stuff.

Usage:
   script.py -t
   script.py -o
   script.py -t -o

Options:
  -t --tfile         Does stuff
  -o --output        Does stuff

redpajama_set_name: RedPajamaStackExchange


text:
package com.smoketurner.dropwizard.consul.health;

import static org.assertj.core.api.Assertions.assertThat;
import static org.mockito.Mockito.doThrow;
import static org.mockito.Mockito.mock;
import static org.mockito.Mockito.verify;
import static org.mockito.Mockito.when;

import com.codahale.metrics.health.HealthCheck.Result;
import com.orbitz.consul.AgentClient;
import com.orbitz.consul.Consul;
import com.orbitz.consul.ConsulException;
import org.junit.Before;
import org.junit.Test;

public class ConsulHealthCheckTest {

  private final Consul consul = mock(Consul.class);
  private final AgentClient agent = mock(AgentClient.class);
  private final ConsulHealthCheck healthCheck = new ConsulHealthCheck(consul);

  @Before
  public void setUp() {
    when(consul.agentClient()).thenReturn(agent);
  }

  @Test
  public void testCheckHealthy() throws Exception {
    final Result actual = healthCheck.check();
    verify(agent).ping();
    assertThat(actual.isHealthy()).isTrue();
  }

  @Test
  public void testCheckUnhealthy() throws Exception {
    doThrow(new ConsulException("error")).when(agent).ping();
    final Result actual = healthCheck.check();
    verify(agent).ping();
    assertThat(actual.isHealthy()).isFalse();
  }
}


text:
Congressional roundtable on de-risking in the Caribbean 'successful'
US Congresswoman, Maxine Walters with Antigua and Barbuda Prime Minister, Gaston Browne.
Gov't of Antigua and Barbuda
The Antigua and Barbuda government has described as "successful" a fully-attended roundtable meeting in Washington, D.C. on Tuesday between representatives of the the 15-nation Caribbean Community (CARICOM), members of the United States Congress and senior representatives of major US banks on the effects of de-risking and the withdrawal of correspondent banking relations (CBRs).
An Antigua and Barbuda government statement said on Wednesday that the roundtable was first proposed by Antigua and Barbuda's Prime Minister, Gaston Browne at last July's meeting of CARICOM Heads of Government.
Since then, Antigua and Barbuda's Ambassador to the US Sir Ronald Sanders worked with Congresswoman Maxine Waters, as chair of the Committee on Financial Services of the US House of Representatives, "to make it a reality," the statement said.
It said senior bankers from six major US banks and US Congressional representatives gathered with a delegation representing CARICOM at Capitol Hill in Washington under the chairmanship of Congresswoman Waters.
The CARICOM delegation was led by Browne, who is the CARICOM lead on financial matters.
According to the statement, Browne told the meeting that the Caribbean and the US have "a mutuality of interests, which necessitated each side displaying active concern about the well-being of the other, including security."
In this regard, Browne said that the process of de-risking, which has led to a withdrawal of relations by some US banks from Caribbean banks, "poses a serious threat to the region's welfare, including its capacity to import goods from the US, which totaled over US$13 billion last year, and is set to exceed this figure this year," according to the statement.
It said that representatives of the US banks indicated that they recognize that value of CBRs, "not only in financial terms but in the critical role it plays in global trade, investment and other financial services."
"They expressed a keen interest in overcoming the challenges posed by a regulatory environment in which banks are conscious, in their decision-making, of the severe penalties for incidents of money laundering and terrorism financing," the statement said.
"The banks made a number of helpful proposals for how this issue could be addressed," it added.
In response, Brown pointed out that, over the last 10 years, US$26 billion has been imposed worldwide as fines for non-compliance with money laundering, "but not one bank was located in the Caribbean."
"US legislators expressed alarm at the effect on the Caribbean of de-risking and the loss of CBRs from US banks, as well as the effect on US national security, if the situation worsens," the statement said.
"The Roundtable reached joint conclusions that the problem has to be tackled urgently and at several levels," it added. "All participating groups at the meeting resolved to work on solutions that arose from the discussion."
Prime Minister Browne expressed "great satisfaction" with the outcome of the meeting and the obvious commitment of US legislators and the US banks to work with the Caribbean on solutions, according to the statement.
Browne told the meeting that what is required is "to build trust and confidence between correspondent banks in the US and respondent banks in the Caribbean in a transparent relationship."
Waters said that she and the US House Committee on Financial Affairs "will remain engaged in the matter and are willing to take actions that may be necessary to preserve the mutual interests of the US and the Caribbean," according to the statement.
Caribbean governments were represented by the Prime Minister of St. Lucia, Allen Chastanet; the Finance Minister of Jamaica, Nigel Clarke; the Attorney-General and Minister of Legal Affairs of Barbados, Dale Marshall; the National Security Minister of Trinidad and Tobago, Stuart Young; and the CARICOM Secretary-General Irwin LaRocque.
One week: Tragic, spiritual, celestial, political
Development threatens Antigua's protected Guiana Island


I help service businesses increase their profits. I am excited to create a customer acquisition and retention plan that works best for you. In 2003, I started my first business mowing grass. Within two months, I started another business cleaning carpet. By 2005, I was running four companies: Lawn Mowing, Bar, Vending Company and a cleaning company. As you may already have guessed, I spent a lot of time on the grind. It didn't take long to figure out the landscaping and cleaning businesses were more profitable than the bar and vending company. By 2009, I decided to focus on The cleaning business and sold the lawn company. Today, I currently have a multi-truck operation in Northern Indiana and spend my time managing that and helping others grow their business!
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Just a quick one. Flashden have approved another of my items for sale, a skinnable XML driven Mp3 player. Enjoy you crazy disco kids!
You are currently browsing the Lookmum blog archives for November, 2006.


Heat-insulated pipes are intended for the construction of oil pipelines, gas pipelines, oil product pipelines and process pipelines with the temperature of the transported product up to + 90 ° C, and are also designed for underground laying of heating networks with operating temperatures up to + 140 ° C. The thickness of heat- insulation is calculated taking into account the temperature regime of the pipeline operation.
Heat-insulated pipes and joint couplings of pipelines are manufactured in the form of "pipe-in-pipe" construction in which a monolithic rigid polyurethane foam is used as the heat-insulated, and as a waterproofing coating is used a polyethylene jacket tube, a metal-encased shell for an underground laying or a galvanized steel tube for an overhead oil and gas pipelines. It's possible to install a heating system based on the "SKIN effect" and fireproof inserts.
Pipes in a polyethylene coating can be of two types: type 1 - standard, type 2 - reinforced. As a protective shell for thermal insulation of pipes, polyethylene sheath pipes and casings made of thin-sheet galvanized steel with a sealed joint (external and internal) are used. To increase the durability of the galvanized steeled shell, it is allowed to apply an additional coating (paint, polymer, etc.) on its outer surface, which can be periodically renewed during the operation. In accordance with the requirements of GOST 30732-2006, heat-insulated pipes must be made with conductors-indicators of the operational remote monitoring system (ODK), but it's allowed to perform without them in the presence of design reasons or at the request of the Customer.


text:
HomeThe Case of Elune
The Case of Elune
July 30, 2021 July 30, 2021 gnomecore World of Warcraft
The Power of the Night chapter name – and the only remaining unresolved theme in Ardenweald – implied that we must see a resolution for the Night Warrior / Tyrande and address the recreation of sigil. I must say it was done pretty well, with a number of revelations to be discussed. Spoiler: I'm not dissappointed.
What surprised me most were the introductory quests. It was a fun and interesting touch that Night Warriors are not only denizens of Ardenweald in the afterlife, and wrathful Khaliiq and Stonewright are quite believable to be the former Night Warriors. Fussing with the seeds, Drust and devourers and Ardenweald again would have been a lot less fun – besides, their local troubles were resolved in the campaign of 9.0. Trips to Maldraxxus and Revendreth were a lot more exciting and fun than the other option, nuff said.
My only beef with the campaign is that the gameplay is sooooo excruciating in quests design. Chapter 5 was tedious, this endless cutting through the fattest ever mobs of Desmotaeron, even if Draka and Thrall genuinely contribute to DPS and tank everything for you. Chapter 6 had a boring run-in-circles-to-find-dreadlords part, also certain boss dreadlords appeared deadly for some of my toons. Chapter 7 – the current one – had this disguise quest. While it was very fun by lore and supporting characters, this crawling all over the place on the ground in disguise was pretty annoying.
In short, when I'm thinking of catching up with the chapters on the remaining toons, every time it's "not again" sentiment concerning gameplay, even if I actually like the lore that comes along. Alright, let's get back to the lore.
I liked how the ritual was designed in whole, and the resolution of Elune's manifestation. We got the big reveal: why the Winter Queen holds the grudge against her sister.
Turns out, Elune has a pocket death dimension for her believers – a sort of purgatory where she decides how to deal with the souls next, but she claims them first, not the Kyrian "ferrymen", like she did with Ysera in Legion. They could be returned to Azeroth as wisps – or she could direct them to Arbiter's judgement.
First I thought that she could send them directly to Winter Queen surpassing the Arbiter's judgement, but on the second thought, she can't – as proved by Khaliiq and Stonewright, clearly the former Night Elves and Elune's worshippers (is where they got their Night Warrior powers) that ended in Maldraxxus and Revendreth. If it were for Elune to decide, she'd send them to Ardenweald too, like the gay horse and stag Night Warriors, am I right? Besides, it's Oribos torrent of souls that was redirected and pours into the Maw. So no, it's still Arbiter doing the judgement. But it was safe for her to presume that her nature-loving children will end up in Ardenweald en masse when she released them for the afterlife.
A pocket dimension for worshippers of a certain deity is not something unheard of – we have De Other Side, no less, reserved for Bwonsamdi believers, trolls in general. Again, he could release them for judgement, but it's the first place they go to after death.
So the plea of Winter Queen was to urge Elune to release all the souls currently in her pocket realm, cause Ardenweald situation was quite dire.
And now Elune finds herself in a dilemma: on one hand, Winter Queen claims that the cycle of Death and Life, the nature cycle, is under threat. It's an unprecedented turn of events, something that has never happened before. On the other hand, Elune hears the desperate prayers of the Night Elves, her worshippers. She was not able to stop Sylvanas from her sudden decision which happened in a moment – but she was able to hear the prayers and do something to help ease their pain, or save someone, in a manner that Bwonsamdi interferes in the mortal events.
This was a choice of two evils, and I think she picked right. As I said, the Shadowlands drought is an unprecedented disaster. On the other hand, the death toll in Teldrassil was vast and horrible… but it's probably not the worst that happened to the Night Elves in history. Yes, I'm talking about the Sundering.
Let me remind you that Night Elves were a dominant race back when Azshara reigned supreme and Legion came. Their empire stretched throughout old meta-continent of Kalimdor, having supressed the previous dominant race, the trolls. So they inhabited all the continent, and as we know, the Sundering destroyed like 80% of it, was it? Along with all the villages, cities and camps of the Night Elves. Compared to this scale, Teldrassil case is but a scratch on the surface of history. And there were also Legion invasions that took no prisoners, and other wars and calamities too.
In other words, it was a choice of saving a big number of Night Elves (but the question was not about survival, as they still populate vast areas of Northern Kalimdor – Teldrassil was merely one of the enclaves, even if the largest one), or saving a death realm which is undergoing an unprecedented drought and is on the verge of extinction, as I'm sure the Winter Queen made it clear with her desperate pleas.
So, in the scales and balance (Balance!) between Life and Death she picked to help Death as this part of the cycle seemed more troubled and in a graver situation. The extinction of Life was not at stake (even if it suffered a heavy blow), but the extinction of Death was.
The choice was not nice and sweet – like I said, it's been a choice of two evils. She tried to atone by granting powers to Tyrande – at least she would help with vengeance, the darker side of the Moon, if her Light side had to pick inaction.
And hence the tear. Learning that the choice she had to make had no consequences is harsh, and Winter Queen, receiving nothing, was too enraged to reach and talk, presuming she abandoned her. She failed everyone – by the combination of her choice and cirumstances she did not know about, but it's a failure nevertheless, refusing worshippers from being saved and at the same time refusing them a peaceful afterlife.
The last thing to talk about is the sigil. Primus and Renathal in previous chapters made it clear that sigils can be recreated without much effort. The case was not about losing something irreplaceable, something too dear to lose – the case was to keep them from Thanos – sorry, Zovaal. Getting all the sigils grants him power to walk where he shouldn't walk and do things that he shouldn't do. Hence they would reforge them – now for us, to follow the baddie and stop him.
It was a nice touch that Elune's tear helped to recreate a sigil. I'm not sure that the first sigil was also created with Elune's help, but in any case it's a powerful artifact, and she offered it as a means of atonement for what transpired by her (partial) fault.
Oh, and Tyrande. She came into senses, for sure, and had to reconsider the vengeance deal to survive. What will be her attitude towards Sylvanas now? I'm not talking a total forgiveness, but the sedation of the most fierce Sylvanas hater is a step to the direction of actually working with the banshee queen and letting her heroically sacrifice, like everyone suspects.
We'll know about that in Chapter 9, or most likely in 9.2.
So far Elune was a central character, and we got a valid explanation about Teldrassil events and her pocket dimension, as well as Tyrande case resolution – that will quite do. Not to mention that yet again Ardenweald has the most moving and tearjerking story to offer among Covenants.
← Sanctum of Domination: A Peep into The Dark Bastille
Trivia: A Month into 9.1. & Week 3 in the Sanctum →
4 thoughts on "The Case of Elune"
I found the whole story interesting, and at points touching, even as the gameplay was excessively tedious – totally agree with your assessment there.
My critique of it would be this – the moment with the Winter Queen and Tyrande as Elune's avatar is great, but it feels too much like a Rule of Cool thing and not a meaningful development, in particular Tyrande's sudden turn. There are still chapters (and patches, and expansions) to explain that in more depth, but I would love to have seen her make that decision actively prior to that moment, to have the reasoning and logic laid out. If it is hearing Elune's voice and explanation for the missing souls – great – but for at least a week, we're left to guess at unclear motives and it feels like, for me at least, it robs the moment of some amount of emotional impact.
It was a perfect chance to have a story chapter that shows Tyrande contemplating her current path, thinking long and hard about why she seeks this vengeance and what she hopes to gain, and I think that story would have been a better focus than the Night Warrior gathering expedition (even if I, like you, found the flavor of that story appealing). Feels a bit like a missed opportunity to me, with too much new story crammed into a single chapter, but that's just my take, even though as a whole, I liked it overall.
Well, that's the problem of modern Blizzard storytelling :) Trying to keep the intrigue, or stretching the story across patches and expansions, they more often than not now do not offer the story-in-itself.
You know, I actually have a similar problem IRL. In my mind, I may have a logical chain of, say, 10 rings, and I would utter 1, 4, and 9 to people. In my head everything is logical, and the rings I omit are obvious to me. But people misunderstand, or have to request additional arguments based on what I actually tell with my mouth :) Sometimes I do that at a party, and my wife which knows me perfectly well, or we discussed the matter with her before alone, would say: "What he means is…" and fills in the required gaps.
Blizzard, in a similar manner, has expansion-wide or even expansions-wide story in mind, and from their point of view everything seems explained and logical. They just miss certain chain rings at a given moment.
The point also is, Tyrande does not think or reflect upon her current state. All she wanted was Sylvanas' head on a spike for a very crystal clear reason, and this much is already clear literally to anyone, there's no need to investigate. The ritual showed that she lost all ability to communicate and any link to reality at all, only able
to produce curses in Darnassian.
Oh, and we just had to have our last adventure with Lady Moonberry :) Covenants would probably lend us the sigils and let us chase Zovaal on our own, wherever that is. So it's probably the last time that common Covenant members were allowed into focus.
I wouldn't trade an adventure with Cudgelface and Moonberry for Tyrande's redemption, no sir! :)
Psychoanalysis seance by Lady Moonberry was hilarious… "Have you thought other mothers could die as a result of your actions?"


According to the Complaint, the Company made false and misleading statements to the market. Liberty Health was involved in a fraudulent scheme with Aphria Inc. in which transactions and acquisitions were completed specifically to provide undue benefits to Company insiders from both parties. Based on this fact, the Company's public statements were false and materially misleading throughout the class period. When the market learned the truth about Liberty Health, investors suffered damages.
Los Angeles, January 8, 2019 — The Schall Law Firm, a national shareholder rights litigation firm, announces the filing of a class action lawsuit against Liberty Health Sciences Inc. ("Liberty Health" or "the Company") (OTC: LHSIF) for violations of §§10(b) and 20(a) of the Securities Exchange Act of 1934 and Rule 10b-5 promulgated thereunder by the U.S. Securities and Exchange Commission.
Investors who purchased the Company's shares between June 28, 2018 and December 3, 2018, inclusive (the "Class Period"), are encouraged to contact the firm before March 8, 2019.


text:
\section{Introduction}

\noindent Let us consider the linear problem
\begin{equation}\label{ceq2}
u''(x) + a(x)u(x) = 0, \ x \in (0,L), \ u'(0) = u'(L) = 0
\end{equation}
where $a \in \Lambda$ and $\Lambda$ is defined by
\begin{equation}\label{ceq3}
\Lambda = \{ a \in L^1(0,L)\setminus \{0\}: \dis \int_{0}^{L} a(x)
\ dx \geq 0 \ \ \mbox{and} \ (\ref{ceq2}) \ \mbox{has nontrivial
solutions} \ \}
\end{equation}
The well known $L_1$ Lyapunov inequality states that if $a \in
\Lambda,$ then $\displaystyle \int_{0}^{L} a^{+}(x) \ dx
> 4/L.$ Moreover, the constant $4/L$ is optimal since $\displaystyle \frac{4}{L} = \displaystyle
\inf_{a \in \Lambda} \Vert a^+ \Vert_{L^1(0,L)}$ and this infimum
is not attained (see \cite{camovimia}, \cite{hart} and
\cite{huyo}). This result is as a particular case of the so called
$L_p$ Lyapunov inequalities, $1 \leq p \leq \infty.$ In fact, if
for each $p$ with $1 \leq p \leq \infty,$ we define the quantity
\begin{equation}\label{ceq4}
\beta_{p} \equiv \inf_{a \in \Lambda\cap L^p (0,L)} \ I_{p}(a)
\end{equation}
where
\begin{equation}\label{ceq5}
\begin{array}{c}
I_{p}(a) = \Vert a^{+} \Vert_{L^p (0,L)} = \left( \displaystyle
\int_{0}^{L} (a^{+}(x)) ^{p} \ dx \right) ^{1/p}, \forall
\ a \in \Lambda\cap L^p (0,L), \ 1 \leq p < \infty, \\ \\
I_{\infty}(a) =\mbox{sup ess} \ a^{+}, \ \forall \ a \in
\Lambda\cap L^\infty (0,L),
\end{array}
\end{equation}
\noindent then $\beta_1 = \frac{4}{L}$ and for each $p$ with $1
\leq p \leq \infty,$ it is possible to obtain an explicit
expression for $\beta_{p}$ as a function of $p$ and $L$
(\cite{camovimia}, \cite{zhan}).

Let us observe that the real number zero is the first eigenvalue
of the eigenvalue problem
\begin{equation}\label{n2}
u''(x) + \rho u(x) = 0, \ x \in (0,L), \ u'(0) = u'(L) = 0
\end{equation}
and that for Neumann boundary conditions the restriction on the
function $a$ in the definition of the set $\Lambda,$
\begin{equation}\label{2106111}
a \in L^1(0,L)\setminus \{0\}, \ \dis \int_{0}^{L} a(x) \ dx \geq
0,
\end{equation}
or the more restrictive pointwise condition
\begin{equation}\label{2106112}
a \in L^1 (0,L), \ 0 \prec a,
\end{equation}
are natural if we want to obtain nontrivial optimal Lyapunov
inequalities (see Remark 4 in \cite{camovimia}). Here, for $c,d
\in L^1 (0,L),$ we write $c \prec d$ if $c(x) \leq d(x)$ for a.e.
$x \in [0,L]$ and $c(x) < d(x)$ on a set of positive measure.

In fact, it can be easily proved that if
\begin{equation}\label{bisceq3}
\Lambda_{0} = \{ a \in L^1(0,L): 0 \prec a \ \ \mbox{and} \
(\ref{ceq2}) \ \mbox{has nontrivial solutions} \ \}
\end{equation}
then the constant $\beta_p$ defined in (\ref{ceq4}) satisfies
\begin{equation}\label{bisceq4}
\beta_{p} =  \inf_{a \in \Lambda_{0}\cap L^p (0,L)} \ I_{p}(a)
\end{equation}
Since zero is the first eigenvalue of (\ref{n2}), it is coherent
to affirm that $\beta_p$ is the $L_p$ Lyapunov constant for the
Neumann problem at the first eigenvalue.

On the other hand, the set of eigenvalues of (\ref{n2}) is given
by $\rho_k = k^2 \pi^2/L^2, \ k \in \natu \cup \{ 0 \}$ and if for
each $k \in \natu \cup \{ 0 \},$ we consider the set

\begin{equation}\label{landaenedef}
\Lambda_{k} = \{ a \in L^1(0,L): \ \rho_k \prec a  \ \ \mbox{and}
\ (\ref{ceq2}) \ \mbox{has nontrivial solutions} \ \}
\end{equation} then for each $p$ with $1 \leq p \leq \infty,$
 we can define the constant
\begin{equation}\label{nceq4}
\beta_{p,k} \equiv \inf_{a \in \Lambda_{k}\cap L^p (0,L)} \
I_{p}(a-\rho_k)
\end{equation}
An explicit value for $\beta_{1,k}$ has been obtained by the
authors in \cite{cavijems}. The case $p = \infty$ is trivial
($\beta_{\infty,k} = \rho_{k+1} - \rho_k $) and, to the best of
our knowledge, an explicit value of $\beta_{p,k}$ as a function of
$p,k$ and $L$ is not known when $1 < p < \infty$. Nevertheless,
since $\beta_{1,k} >0,$ we trivially deduce $\beta_{p,k} >0,$ for
each $p$ with $1 \leq p \leq \infty.$

$ $

With regard to Partial Differential Equations, the linear problem
\begin{equation}\label{ceq6}
\left.
\begin{array}{cl}
\Delta u(x)+a(x)u(x)= 0, & x\in \Omega \, \\
 \frac{\partial u}{\partial n}(x)=0,\ \ \ \ \ \ \ &  x\in \partial \Omega
\end{array}
\right\}
\end{equation}
\noindent has been studied in \cite{camovijfa}, where $\Omega
\subset \real^N$ ($N\geq 2$) is a bounded and regular domain,
$\dis \frac{\partial}{\partial n}$ is the outer normal derivative
on $\partial \Omega$ and the function $a:\Omega \rightarrow \real$
belongs to the set $\Gamma$ defined as
\begin{equation}\label{ceq7}
\Gamma = \{ a \in L^{\frac{N}{2}}(\Omega)\setminus \{0\}: \dis
\int_\Omega a(x) \ dx \geq 0 \ \mbox{and} \ (\ref{ceq6}) \
\mbox{has nontrivial solutions}\}
\end{equation}
\noindent if $N \geq 3$ and
$$\Gamma = \{ a:\Omega \rightarrow \real \mbox{ s. t. } \exists
q\in(1,\infty]\\ \ \mbox{with} \ a\in L^q(\Omega)\setminus \{0\},
\
 \dis \int_\Omega a(x) \ dx \geq 0 \ $$

\noindent\  and \ (\ref{ceq6}) has nontrivial solutions$\}$
\newline
\noindent if $N = 2.$ \newline\ Obviously, the quantity
\begin{equation}\label{ceq8}
\gamma_{p} \equiv \inf_{a \in \Gamma\cap L^p (\Omega)} \ \Vert a
^{+} \Vert_{L^p (\Omega)} \ , \ 1 \leq p \leq \infty
\end{equation}
is well defined and it is a nonnegative real number. A remarkable
novelty (see \cite{camovijfa}) with respect to the ordinary case
is that $\gamma_{1} = 0$ for each $N \geq 2.$ Moreover, if $N =2,$
then $\gamma_{p} > 0, \ \forall \ p \in (1,\infty]$ and if $N \geq
3,$ then $\gamma_{p}
> 0$ if and only if $p \geq N/2.$ In contrast to the ordinary case, it seems difficult to obtain an explicit
expressions for $\gamma_{p},$ as a function of $p$, $\Omega$ and $N,$
at least for general domains.

As in the ordinary case, the real number zero is the first
eigenvalue of the eigenvalue problem
\begin{equation}\label{nceq11}
\left.
\begin{array}{cl}
\Delta u(x)+ \rho u(x)= 0,  & x\in \Omega \, \\
 \frac{\partial u}{\partial n}(x)=0\, \ \ \ \ \ \ &  x\in \partial \Omega
\end{array}
\right\}
\end{equation}
so that it is natural to say that the constant $\gamma_p$ defined
in (\ref{ceq8}) is the $L_p$ Lyapunov constant at the first
eigenvalue for the Neumann problem (\ref{ceq6}).

$ $

To our knowledge, there are no significant results concerning to
$L_p$ Lyapunov inequalities for PDE at higher eigenvalues and this
is the main subject of this paper where we provide some new
qualitative results which extend to higher eigenvalues those
obtained in \cite{camovijfa} for the case of the first eigenvalue.
We carry out a complete qualitative study of the question pointing
out the important role played by the dimension of the problem.

Since in the case of ODE our proof are mainly based on an exact
knowledge about the number and distribution of the zeros of the
corresponding solutions (\cite{cavijems}), in the PDE case we are
able to study $L_p$
 Lyapunov inequalities if $\Omega$ is a ball and for radial higher
eigenvalues. It is not restrictive to assume that $\Omega =
B_{\real^N} (0;1) \equiv B_1,$ the open ball in $\real^N$ of
center zero and radius one.

In Section 2 we describe the problem in a precise way and we
present the main results of this paper. In Section 3 we study the
subcritical case, i.e. $1\leq p<\frac{N}{2}$, if $N\geq 3$, and
$p=1$ if $N=2$. To prove the results in this section we will
construct some explicit and appropriate sequences of problems like
(\ref{ceq6}) where Dirichlet type problems play an essential role.
In this subcritical case we prove that the optimal Lyapunov
constants are trivial, i.e., zero.

In Section 4, we treat with the supercritical case:
$p>\frac{N}{2}$, if $N\geq 2$. By using some previous results of
Section 2, about the number and distribution of the zeros of
nontrivial and radial solutions, together with some compact
Sobolev inclusions, we use a reasoning by contradiction to prove
that the optimal Lyapunov constants are strictly positive and they
are attained. In Section 5 we consider the critical case, i.e.
$p=\frac{N}{2}$, if $N\geq 3$. Because in this case the Sobolev
inclusions are continuous but no compact, we demonstrate that the
optimal Lyapunov constants are strictly positive but we do not
know if they are attained or not.

Finally,  we study the case of Neumann boundary conditions but
similar results can be obtained in the case of Dirichlet type
problems.

\section{main results}
From now on, $\Omega = B_1,$ the open ball in $\real^N$ of center
zero and radius one. It is very well known (\cite{delpino}) that
the operator $-\Delta$ exhibits an infinite increasing sequence of
radial Neumann eigenvalues $ 0 = \mu_0 < \mu_1<\ldots < \mu_k <
\ldots $ with $\mu_k \rightarrow + \infty,$ all of them simple and
with associated eigenfunctions $\varphi_k \in C^1 [0,1]$ solving
\begin{equation}\label{2106114}
\begin{array}{c}
-(r^{N-1} \varphi ')' = \mu_k r^{N-1} \varphi, \ 0 < r < 1, \\
\varphi'(0) = \varphi'(1) = 0.
\end{array}
\end{equation}
Moreover, each eigenfunction $\varphi_k$ has exactly $k$ simple
zeros $r_k<r_{k-1}<...<r_1$ in the interval $(0,1)$.

For each integer $k\geq 0$ and number $p$, $1 \leq p \leq \infty,$
we can define the set $$ \Gamma_k = \{ a \in L^{N/2}(B_1): \ a \
\mbox{is a radial function}, \ \mu_k \prec a \ \mbox{and} $$
(\ref{ceq6}) \ \mbox{has radial and nontrivial solutions} $\}$
\noindent if $N\geq 3$ and
$$\Gamma_k = \{ a:B_1 \rightarrow \real \mbox{ s. t. } \exists
q\in(1,\infty]\\ \ \mbox{with} \ a\in L^q(B_1): \ a \ \mbox{is
a radial function},$$
$$\mu_k \prec a \ \mbox{and} \ (\ref{ceq6}) \ \mbox{has radial and
nontrivial solutions} \}$$

\noindent if $N=2$.

We also define the quantity

\begin{equation}\label{2106116}
\gamma_{p,k} = \displaystyle \inf_{a \in \Gamma_k \cap L^p (B_1)}
\ \Vert a-\mu_k \Vert _{L^p (B_1)}
\end{equation}

The main result of this paper is the following.

\begin{theorem}\label{principal}

 Let $k\geq 0$, $N\geq 2$, $1\leq p\leq \infty$. The following statements hold:
\begin{enumerate}
\item If $N=2$ then $\gamma_{p,k}>0 \Leftrightarrow 1<p\leq \infty
$.

\noindent If $N\geq 3$ then $\gamma_{p,k}>0 \Leftrightarrow
\frac{N}{2}\leq p\leq \infty $.

\

\item If $N \geq 2$ and $\frac{N}{2}<p\leq \infty$ then
$\gamma_{p,k}$ is attained.

\end{enumerate}

\end{theorem}

A key ingredient to prove this theorem is the following
proposition on the number and distribution of zeros of nontrivial
radial solutions of (\ref{ceq6}) when $a \in \Gamma_k.$

\begin{proposition}\label{n+1ceros}
Let $\Omega=B_1$, $k\geq 0$, $a \in \Gamma_k$ and $u$ any
nontrivial radial solution of (\ref{ceq6}). Then $u$ has, at
least, $k+1$ zeros in $(0,1)$. Moreover, if $k\geq 1$ and we
denote by $x_k<x_{k-1}<...<x_1$ the last $k$ zeros of $u$, we have
that

$$r_i\leq x_i\, ,\mbox{   }\forall \ 1\leq i\leq k,$$

\noindent where $r_i$ denotes de zeros of the eigenfunction
$\varphi_k$ of (\ref{2106114}).
\end{proposition}

For the proof of this proposition we will need the following
lemma. Some of the results of this lemma can be proved in a
different way, by using the version of the Sturm Comparison Lemma
proved in \cite{delpino}, Lemma 4.1, for the $p$-laplacian
operator (see also \cite{hart}). Other results are new.

\begin{lemma}\label{estimas}
Let $k\geq 1$. Under the hypothesis of Proposition \ref{n+1ceros}
we have that

\begin{enumerate}
\item[i)] $u$ vanishes in the interval $(0,r_k]$. If $r_k$ is the
only zero of $u$ in this interval then $a(r)\equiv \mu_k$ in
$(0,r_k]$. \item[ii)] $u$ vanishes in the interval
$[r_{i+1},r_i)$, for $1\leq i\leq k-1$. If $r_{i+1}$ is the only
zero of $u$ in this interval then $u(r_i)=0$ and $a(r)\equiv
\mu_k$ in $[r_{i+1},r_i]$. \item[ii)]  $u$ vanishes in the
interval $[r_1,1)$. If $r_1$ is the only zero of $u$ in this
interval then $a(r)\equiv \mu_k$ in $[r_1,1]$.
\end{enumerate}
\end{lemma}

\begin{proof}

To prove i), multiplying (\ref{ceq6}) by $\varphi_k$ and
integrating by parts in $B_{r_k}$ (the ball centered in the origin
of radius $r_k$), we obtain

$$\int_{B_{r_k}}\nabla u\nabla \varphi_k=\int_{B_{r_k}}a u \varphi_k.$$

On the other hand, multiplying (\ref{2106114}) by $u$ and
integrating by parts in $B_{r_k}$, we have

$$\int_{B_{r_k}}\nabla \varphi_k\nabla u=\mu_k\int_{B_{r_k}}\varphi_k u+\int_{\partial B_{r_k}}u\frac{\partial \varphi_k}{\partial n}.$$

Subtracting these equalities yields

\begin{equation}\label{signo}
\int_{B_{r_k}}\left(a-\mu_k\right)u\varphi_k=\omega_N
r_k^{N-1}u(r_k)\varphi_k'(r_k),
\end{equation}

\noindent where $\omega_N$ denotes de measure of the
$N$-dimensional unit sphere. Assume, by contradiction, that $u$
does not vanish in $(0,r_k]$. We can suppose, without loss of
generality, that $u>0$ in this interval. We can also assume that
$\varphi_k>0$ in $(0,r_k)$. Since $r_k$ is a simple zero of
$\varphi_k$, we have $\varphi_k'(r_k)<0$ and since $a\geq\mu_k$ in
$(0,r_k)$ we obtain a contradiction.

Finally, if $r_k$ is the only zero of $u$ in $(0,r_k]$, equation
\ref{signo} yields \newline
$\int_{B_{r_k}}\left(a-\mu_k\right)u\varphi_k=0$, which gives
$a(r)\equiv \mu_k$ in $(0,r_k]$.

To deduce  ii), we proceed similarly to the proof of part i),
restituting $B_{r_k}$ by $A(r_{i+1},r_i)$ (the annulus centered in
the origin of radii $r_{i+1}$ and $r_i$) and obtaining

$$\int_{A(r_{i+1},r_i)}\left(a-\mu_k\right)u\varphi_k=\omega_N r_i^{N-1}u(r_i)\varphi_k'(r_i)-\omega_N r_{i+1}^{N-1}u(r_{i+1})\varphi_k'(r_{i+1})$$

\noindent and ii) follows easily by arguments on the sign of these
quantities, as in the proof of part i).

To obtain iii), a similar analysis to that in the previous cases
shows that

$$\int_{A(r_i,1)}\left(a-\mu_k\right)u\varphi_k=-\omega_N r_1^{N-1}u(r_1)\varphi_k'(r_1),$$

\noindent and the lemma follows easily as previously.

\end{proof}

{\it Proof of Proposition \ref{n+1ceros}.} Let $k=0$. If we
suppose that $u$ has no zeros in $(0,1]$ and we integrate the
equation $-\Delta u=a\, u$ in $B_1$, we obtain $\int_{B_1}a\,
u=0$, a contradiction. Hence, for the rest of the proof we will
consider $k\geq 1$.

Let $1\leq i\leq k$. By the previous lemma $u$ vanishes in the $i$
disjoint intervals $[r_i,r_{i-1})$,...,$[r_2,r_1)$,$[r_1,1)$.
Therefore $u$ has, at least, $i$ zeros in the interval $[r_i,1)$
which implies that $r_i\leq x_i$.

Finally, let us prove that $u$ has, at least, $k+1$ zeros. From
the previous part, taking $i=k$, $u$ has at least $k$ zeros in the
interval $[r_k,1]$, one in each of the $k$ disjoint intervals
$[r_k,r_{k-1})$,...,$[r_2,r_1)$,$[r_1,1)$. Suppose, by
contradiction, that these are the only zeros of $u$. Then $u$ does
not vanish in $(0,r_k)$ and applying part i) of Lemma
\ref{estimas} we obtain $u(r_k)=0$ and $a\equiv\mu_k$ in
$(0,r_k].$ Applying now part ii) of this lemma, we deduce
$u(r_{k-1})=0$ and $a\equiv\mu_k$ in $[r_k,r_{k-1}]$. Repeating
this argument and using part iii) of the previous lemma we
conclude $u(r_i)=0$, for all $1\leq i\leq k$ and $a\equiv\mu_k$ in
(0,1], which contradicts $a\in\Gamma_k$. \qed

$ $

\noindent For the proof of Theorem \ref{principal}, we will
distinguish three cases: the subcritical case ($1\leq
p<\frac{N}{2}$ if $N\geq 3$, and $p=1$ if $N=2$), the
supercritical case ($p>\frac{N}{2}$ if $N\geq 2$), and the
critical case ($p=\frac{N}{2}$ if $N\geq 3$).

\section{The subcritical case}

In this section, we study the subcritical case, i.e. $1\leq
p<\frac{N}{2}$, if $N\geq 3$, and $p=1$ if $N=2$. In all those
cases we will prove that $\gamma_{p,k}=0$.

$ $

The next lemma is related to the continuous domain dependence of
the eigenvalues of the Dirichlet Laplacian. In fact, the result is
valid under much more general hypothesis (see \cite{fugle}). Here
we show a very simple proof for this special case.

\begin{lemma}\label{autovalorescontinuos}

Let $N\geq 2$ and $R>0$. Then

$$\lim_{\varepsilon \to 0}\lambda_1\left( A(\varepsilon,R)\right)=\lambda_1\left(B_R\right),$$

\noindent where $\lambda_1\left( A(\varepsilon,R)\right)$ and
$\lambda_1\left(B_R\right)$ denotes, respectively, the first
eigenvalues of the Laplacian operator with Dirichlet boundary
conditions of the annulus $A(\varepsilon,R)$ and the ball $B_R$.

\end{lemma}

\begin{proof}

For $N\geq 3$ and $\varepsilon \in (0,R/2)$ define the following
radial function $u_\varepsilon\in
H_0^1\left(A(\varepsilon,R)\right)$:

\begin{equation} u_\varepsilon(x) = \left \
\begin{array}{l}
\phi_1(x), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ 2\varepsilon\leq \vert x\vert < R , \\ \\
\displaystyle{\frac{\vert
x\vert-\varepsilon}{\varepsilon}}\phi_1(2\varepsilon), \
\mbox{if} \ \ \ \varepsilon <\vert x\vert < 2\varepsilon ,
\end{array}
\right.
\end{equation}

\noindent where $\phi_1$ denotes the first eigenfunction with
Dirichlet boundary conditions of the ball $B_R$. It is easy to
check that
$$\displaystyle{\lim_{\varepsilon\to
0}\int_{A(\varepsilon, 2\varepsilon)}\vert\nabla
u_\varepsilon\vert^2}=\lim_{\varepsilon\to
0}\int_\varepsilon^{2\varepsilon}\omega_Nr^{N-1}\frac{\phi_1(2\varepsilon)^2}{\varepsilon^2}dr=0.$$
In the same way it is obtained $\displaystyle{\lim_{\varepsilon\to
0}\int_{A(\varepsilon, 2\varepsilon)}u_\varepsilon^2=0}$. In
addition, from the variational characterization of the first
eigenvalue it follows that $\lambda_1\left(
A(\varepsilon,R)\right)\leq \int_{A(\varepsilon,R)}\vert\nabla
u_\varepsilon\vert^2/\int_{A(\varepsilon,R)}u_\varepsilon^2$.
Therefore

$$\limsup_{\varepsilon\to 0}\lambda_1\left( A(\varepsilon,R)\right)\leq \limsup_{\varepsilon\to 0}\displaystyle{\frac{\int_{A(\varepsilon,R)}\vert\nabla u_\varepsilon\vert^2}{\int_{A(\varepsilon,R)}u_\varepsilon^2}=\frac{\int_{B_R}\vert\nabla \phi_1\vert^2}{\int_{B_R}\phi_1^2}}=\lambda_1(B_R).$$

On the other hand, using that the first Dirichlet eigenvalue
$\lambda_1(\Omega)$ is strictly decreasing with respect to the the
domain $\Omega$, it follows that $\lambda_1\left(
A(\varepsilon,R)\right)>\lambda_1\left(B_R\right)$. Thus

$$\liminf_{\varepsilon\to 0}\lambda_1\left( A(\varepsilon,R)\right)\geq\lambda_1(B_R)$$ and the lemma follows for $N\geq 3$.

The same proof works for $N=2$ if we consider, for every
$\varepsilon\in \left(0,\min\{ 1,R^2\}\right)$, the radial
function $u_\varepsilon\in H_0^1\left(A(\varepsilon,R)\right)$:

\begin{equation} u_\varepsilon(x) = \left \
\begin{array}{l}
\phi_1(x), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ \sqrt{\varepsilon}\leq \vert x\vert < R , \\ \\
\displaystyle{\frac{\log\vert
x\vert-\log\varepsilon}{\log\sqrt{\varepsilon}-\log\varepsilon}\phi_1(\sqrt{\varepsilon)}},
\  \mbox{if} \ \ \ \varepsilon <\vert x\vert < \sqrt{\varepsilon}.

\end{array}
\right.
\end{equation}

\end{proof}

\begin{lemma}\label{subcritico}

Let $k\geq 0$, $N\geq 3$ and $1\leq p< N/2$. Then
$\gamma_{p,k}=0$.

\end{lemma}

\begin{proof} If $k=0$, this lemma follows from \cite[Lem. 3.1]{camovijfa}. In this lemma a family of bounded, positive and radial solutions were used. Hence, for the rest of the proof we will consider $k\geq 1$.

To prove this lemma we will construct an explicit family
$a_\varepsilon\in \Gamma_k$ such that $\lim_{\varepsilon\to
0}\Vert a_\varepsilon-\mu_k \Vert_{L^p (B_1)} = 0.$ To this end,
for every $\varepsilon\in(0,r_k)$, define
$u_\varepsilon:B_1\rightarrow \real$ as the radial function

\begin{equation}\label{solucionsubcritico} u_\varepsilon = \left \
\begin{array}{l}
\varphi_k, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ r_k\leq \vert x\vert < 1 , \\ \\
\phi_1\left( A(\varepsilon,r_k)\right), \  \mbox{if} \ \ \ \varepsilon \leq \vert x\vert < r_k , \\ \\
\phi_1\left(B_\varepsilon\right), \ \ \ \ \ \ \ \ \mbox{if} \ \ \
\vert x\vert < \varepsilon .
\end{array}
\right.
\end{equation}

\noindent where $\phi_1\left( A(\varepsilon,r_k)\right)$ and
$\phi_1\left(B_\varepsilon\right)$ denotes, respectively, the
first eigenfunctions with Dirichlet boundary conditions of the
annulus $A(\varepsilon,r_k)$ and the ball $B_\varepsilon$.
Moreover these eigenfunctions are chosen such that $u_\varepsilon
\in C^1(\overline{B_1})$.

Then, it is easy to check that $u_\varepsilon$ is a solution of
(\ref{ceq6}), being $a_\varepsilon\in L^\infty(B_1)$ the radial
function

\begin{equation}\label{infimosubcritico} a_\varepsilon = \left \
\begin{array}{l}
\mu_k, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ r_k< \vert x\vert < 1 , \\ \\
\lambda_1\left( A(\varepsilon,r_k)\right), \  \mbox{if} \ \ \ \varepsilon < \vert x\vert < r_k , \\ \\
\lambda_1\left(B_\varepsilon\right), \ \ \ \ \ \ \ \ \mbox{if} \ \
\ \vert x\vert < \varepsilon ,
\end{array}
\right.
\end{equation}

\noindent where $\lambda_1\left( A(\varepsilon,r_k)\right)$ and
$\lambda_1\left(B_\varepsilon\right)$ denotes, respectively, the
first eigenvalues with Dirichlet boundary conditions of de annulus
$A(\varepsilon,r_k)$ and the ball $B_\varepsilon$. Since the first
Dirichlet eigenvalue $\lambda_1(\Omega)$ is strictly decreasing
with respect to the the domain $\Omega$, it follows that

$$\lambda_1\left( A(\varepsilon,r_k)\right), \lambda_1\left(B_\varepsilon\right)>\lambda_1\left(B_{r_k}\right)=\mu_k,$$

\noindent which gives $a_\varepsilon\in \Gamma_k$. (The equality $\lambda_1\left(B_{r_k}\right)=\mu_k$ follows from the fact that $\varphi_k$ is a positive solution of $-\Delta \varphi=\mu_k \varphi$ in $B_{r_k}$ which vanishes on $\partial B_{r_k}$). Let us estimate
the $L_p$-norm of $a_\varepsilon-\mu_k$:

\begin{equation}\label{chorizo} \begin{array}{l}
\dis \Vert a_\varepsilon-\mu_k
\Vert_{L^p(B_1)}=\left(\int_{B_\varepsilon}\left(\lambda_1\left(B_\varepsilon\right)-\mu_k\right)^p+\int_{A(\varepsilon,r_k)}\left(\lambda_1\left(A(\varepsilon,r_k)\right)-\mu_k\right)^p
\right)^{\frac{1}{p}}= \\
\dis
=\left(\left(\lambda_1\left(B_\varepsilon\right)-\mu_k\right)^p\frac{\omega_N
\varepsilon^N}{N}+\left(\lambda_1\left(A(\varepsilon,r_k)\right)-\mu_k\right)^p\frac{\omega_N
(r_k^N-\varepsilon^N)}{N} \right)^{\frac{1}{p}}.
\end{array}
\end{equation}

Taking into account that
$\lambda_1\left(B_\varepsilon\right)=\lambda_1(B_1)/\varepsilon^2$,
$\lambda_1\left(B_{r_k}\right)=\mu_k$, using $N>2p$, and applying
Lemma \ref{autovalorescontinuos}, we conclude

$$\lim_{\varepsilon\to 0}\Vert a-\mu_k \Vert_{L^p(B_1)}\leq\lim_{\varepsilon\to 0}\left(\frac{\lambda_1(B_1)^p}{\varepsilon^{2p}}\frac{\omega_N \varepsilon^N}{N}+
\left(\lambda_1\left(A(\varepsilon,r_k)\right)-\mu_k\right)^p\frac{\omega_N
(r_k^N-\varepsilon^N)}{N}\right)^{\frac{1}{p}}=0,$$

\noindent and the proof is complete. \end{proof}

\begin{lemma}\label{critico2}
Let $k\geq 0$, $N=2$ and $p=1$. Then $\gamma_{1,k}=0$.
\end{lemma}

\begin{proof}
If $k=0$, this lemma follows from \cite[Lem. 3.2]{camovijfa}. In
this lemma a family of bounded, positive and radial solutions were
used. Hence, for the rest of the proof we will consider $k\geq 1$.

Similarly to the proof of the previous lemma, we will construct
some explicit sequences in $\Gamma_k$. In this case, this
construction will be slightly more complicated. First, for every
$\alpha\in (0,1)$, define $v_\alpha, A_\alpha :B_1\rightarrow
\real$ as the radial functions:

\begin{equation}  v_\alpha(r) = \left \
\begin{array}{l}
\alpha (1-r^2)(3-r^2)-\log r, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ \alpha\leq r < 1 , \\ \\
\alpha (1-r^2)(3-r^2)-\log
\alpha+\displaystyle{\frac{\alpha^2-r^2}{2\alpha^2}}, \ \ \ \ \ \
\ \ \ \ \ \ \ \ \mbox{if} \ \ \ r < \alpha ,
\end{array}
\right.
\end{equation}

\begin{equation}  A_\alpha(r) = \left \
\begin{array}{l}
\displaystyle{\frac{16\alpha(1-r^2)}{\alpha (1-r^2)(3-r^2)-\log r}}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ \alpha< r < 1 , \\ \\
\displaystyle{\frac{16\alpha(1-r^2)+\frac{2}{\alpha^2}}{\alpha
(1-r^2)(3-r^2)-\log
\alpha+\displaystyle{\frac{\alpha^2-r^2}{2\alpha^2}}}}, \ \ \ \ \
\ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ r < \alpha ,
\end{array}
\right.
\end{equation}

\noindent where $r=\vert x\vert$. It is easily seen that
$v_\alpha\in C^1(\overline{B_1})$, $A_\alpha\in L^\infty(B_1)$,
and

\begin{equation}
\left.
\begin{array}{cl}
\Delta v_\alpha(x)+A_\alpha(x)v_\alpha(x)= 0, & x\in B_1 \, \\
 v_\alpha(x)=0,\ \ \ \ \ \ \ &  x\in \partial B_1
\end{array}
\right\}
\end{equation}

Now, for every $\alpha\in(0,1)$ and $\varepsilon\in(0,r_k)$,
define $u_{\alpha,\varepsilon}:B_1\rightarrow \real$ as the radial
function:

\begin{equation}\label{solucionsubcritico2} u_{\alpha,\varepsilon}(x) = \left \
\begin{array}{l}
\varphi_k(x), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ r_k\leq \vert x\vert < 1 , \\ \\
\phi_1\left( A(\varepsilon,r_k)\right)(x), \  \mbox{if} \ \ \ \varepsilon \leq \vert x\vert < r_k , \\ \\
\displaystyle{v_\alpha\left(\frac{x}{\varepsilon}\right)}, \ \ \ \
\ \ \ \ \mbox{if} \ \ \ \vert x\vert < \varepsilon .
\end{array}
\right.
\end{equation}

\noindent where the eigenfunctions $\varphi_k$ and $\phi_1\left(
A(\varepsilon,r_k)\right)$ are chosen such that
$u_{\alpha,\varepsilon}\in C^1(\overline{B_1})$.

An easy computation shows that $u_{\alpha,\varepsilon}$ is a
solution of (\ref{ceq6}), being $a_{\alpha,\varepsilon}\in
L^\infty(B_1)$ the radial function

\begin{equation}\label{bisinfimosubcritico} a_{\alpha,\varepsilon}(x) = \left \
\begin{array}{l}
\mu_k, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{if} \ \ \ r_k< \vert x\vert < 1 , \\ \\
\lambda_1\left( A(\varepsilon,r_k)\right), \  \mbox{if} \ \ \ \varepsilon < \vert x\vert < r_k , \\ \\
\displaystyle{\frac{1}{\varepsilon^2}A_\alpha\left(\frac{x}{\varepsilon}\right)},
\ \ \ \ \ \ \ \ \mbox{if} \ \ \ \vert x\vert < \varepsilon .
\end{array}
\right.
\end{equation}

\noindent Again, using that the first Dirichlet eigenvalue
$\lambda_1(\Omega)$ is strictly decreasing with respect to the the
domain $\Omega$, it follows that

$$\lambda_1\left(A(\varepsilon,r_k)\right)>\lambda_1\left(B_{r_k}\right)=\mu_k.$$

Moreover, $\displaystyle{\inf_{\vert x\vert <\varepsilon}
a_{\alpha,\varepsilon}(x)=\left(\inf_{x\in B_1}
A_\alpha(x)\right)/\varepsilon^2:=m_\alpha/\varepsilon^2}$. We see
at once that $m_\alpha>0$ for every $\alpha\in(0,1)$. Hence, if we
fix $\alpha$ and choose $\varepsilon\in(0,1)$ such that
$\displaystyle{m_\alpha/\varepsilon^2\geq \mu_k}$, it is deduced
that $a_{\alpha,\varepsilon}\in\Gamma_k$.

Let us estimate the $L_1$-norm of $a_{\alpha,\varepsilon}-\mu_k$:

\begin{equation}\label{chorizo2} \begin{array}{l}
\displaystyle{\Vert a_{\alpha,\varepsilon}-\mu_k
\Vert_{L^1(B_1)}=\int_{B_\varepsilon}\left(\frac{1}{\varepsilon^2}A_\alpha\left(\frac{x}{\varepsilon}\right)
-\mu_k\right) dx+\int_{A(\varepsilon,r_k)}\left(\lambda_1 \left(
A(\varepsilon,r_k)\right)-\mu_k\right)dx}.
\end{array}
\end{equation}

Doing the change of variables $x=\varepsilon y$ in the first
integral and applying Lemma \ref{autovalorescontinuos} in the
second one, it is obtained, for fixed $\alpha\in (0,1)$:

$$\lim_{\varepsilon \to 0}\Vert a_{\alpha,\varepsilon}-\mu_k \Vert_{L^1(B_1)}=\int_{B_1}A_\alpha (y)dy.$$

Thus, from the definition of $\gamma_{1,k}$ we have

\begin{equation}\label{alfa}
\gamma_{1,k}\leq \int_{B_1}A_\alpha (y)dy\, ,\ \ \forall \alpha\in
(0,1).
\end{equation}
Now we will take limit when $\alpha$ tends to $0$ in this last
expression. For this purpose we first deduce easily from the
definition of $A_\alpha$ that $A_\alpha(r)\leq 16\alpha
(1-r^2)/(-\log r)\leq 32\alpha$ if $r\in (\alpha,1)$ and
$A_\alpha(r)\leq \left(16\alpha+2/\alpha^2\right)/(-\log \alpha)$
if $r\in (0,\alpha)$. It follows that

$$\int_{B_1}A_\alpha (y)dy=2\pi\int_0^1 r \, A_\alpha(r)dr\leq 2\pi\int_0^\alpha r\,\frac{16\alpha+2/\alpha^2}{-\log\alpha}dr+2\pi\int_\alpha^1 r \,32\alpha dr$$
$$=\pi\frac{16\alpha^3+2}{-\log\alpha}+32\pi\alpha(1-\alpha^2),$$

\noindent which gives $\lim_{\alpha\to 0}\int_{B_1}A_\alpha
(y)dy=0$ and the lemma follows from (\ref{alfa}). \end{proof}

\section{The supercritical case}

In this section, we study the supercritical case, i.e.
$p>\frac{N}{2}$, if $N\geq 2$. In all those cases we will prove
that $\gamma_{p,k}$ is strictly positive and that it is attained.
We begin by studying the case $p=\infty.$

\begin{lemma}\label{linfinito}
Let $k\geq 0$, $N\geq 2$ and $p=\infty$. Then
$\gamma_{\infty,k}=\mu_{k+1}-\mu_k$ is attained in the unique
element $a_0\equiv \mu_{k+1}\in\Gamma_k$.
\end{lemma}

\begin{proof}
Clearly $a_0\equiv\mu_{k+1}\in\Gamma_k$ satisfies $\Vert
a_0-\mu_k\Vert_{L^\infty(B_1)}=\mu_{k+1}-\mu_k$. Suppose, contrary
to our claim, that there exists $\mu_{k+1}\not\equiv a\in\Gamma_k$
such that $\Vert a-\mu_k\Vert_{L^\infty(B_1)}\leq\mu_{k+1}-\mu_k$.
Therefore $\mu_k \prec a\prec \mu_{k+1}$, a contradiction with the
fact $a\in\Gamma_k$ (see \cite{dolp}, \cite{vido}).
\end{proof}

Next we concentrate on the case $\frac{N}{2} < p < \infty.$

\begin{lemma}\label{cerosseparados}
Let $N\geq 2$, $p>N/2$ and $M>0$. Then, there exists
$\varepsilon=\varepsilon(N,p,M)$ with the following property:

For every $a\in L^p(B_1)$ satisfying $\Vert a\Vert_{L^p(B_1)}\leq
M$ and every $u\in H^1(B_1)$ radial nontrivial solution of
$-\Delta u=a\, u$ in $B_1$ we have
\begin{enumerate}
\item[i)] $z>\varepsilon$ for every zero $z$ of $u$. \item[ii)]
$\vert z_2-z_1\vert>\varepsilon$ for every different zeros
$z_1,z_2$ of $u$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $z\in (0,1]$ be a zero of $u$. Hence, multiplying the equation
$-\Delta u=a\, u$ by $u$, integrating by parts in the ball $B_z$
and applying H\"{o}lder inequality, we obtain

$$\int_{B_z}\vert\nabla u\vert^2=\int_{B_z}a\, u^2\leq \Vert a\Vert_{L^p(B_z)}\Vert u\Vert^2_{L^\frac{2p}{p-1}(B_z)}.$$

From the above it follows that

$$M\geq\Vert a\Vert_{L^p(B_1)}\geq\Vert a\Vert_{L^p(B_z)}\geq\frac{\Vert\nabla u\Vert^2_{L^2(B_z)}}{\Vert u\Vert^2_{L^\frac{2p}{p-1}(B_z)}}\geq
\min_{v\in H^1_0(B_z)}\frac{\Vert\nabla v\Vert^2_{L^2(B_z)}}{\Vert
v\Vert^2_{L^\frac{2p}{p-1}(B_z)}}.$$

From the change $w(x)=v(z\, x)$, it is easily deduced that

$$\min_{v\in H^1_0(B_z)}\frac{\Vert\nabla v\Vert^2_{L^2(B_z)}}{\Vert v\Vert^2_{L^\frac{2p}{p-1}(B_z)}}=z^{\frac{N}{p}-2}\min_{w\in H^1_0(B_1)}\frac{\Vert\nabla w\Vert^2_{L^2(B_1)}}{\Vert w\Vert^2_{L^\frac{2p}{p-1}(B_1)}}:=z^{\frac{N}{p}-2}\alpha(N,p),$$

\noindent where we have used the compact embedding
$\displaystyle{H_0^1(B_1)\subset L^\frac{2p}{p-1}(B_1)}$ (since
$p>N/2$, then $2<\frac{2p}{p-1}<\frac{2N}{N-2}$, which is the
critical Sobolev exponent). Thus, taking $\varepsilon_1>0$ such
that $M<{\varepsilon_1}^{\frac{N}{p}-2}\alpha(N,p)$, we conclude
part i) of the lemma with $\varepsilon=\varepsilon_1$.

For the second part of the lemma, consider two zeros $0<z_1<z_2<1$
of $u$. Taking into account that $z_1\geq\varepsilon_1$ and
arguing in the same manner of part i), we obtain

$$M\geq\Vert a\Vert_{L^p(B_1)}\geq\Vert a\Vert_{L^p\left(A\left( z_1,z_2\right)\right)}\geq\frac{\Vert\nabla u\Vert^2_{L^2\left(A\left( z_1,z_2\right)\right)}}{\Vert u\Vert^2_{L^\frac{2p}{p-1}\left(A\left( z_1,z_2\right)\right)}}=$$

$$\displaystyle{\frac{\omega_N\int_{z_1}^{z_2}r^{N-1}u'(r)^2dr}{\left(\int_{z_1}^{z_2}\omega_N r^{N-1}\vert u(r)\vert^{2p/(p-1)}dr\right)^{(p-1)/p}}\geq \omega_N^{1/p}\varepsilon_1^{N-1}\frac{\Vert u' \Vert^2_{L^2(z_1,z_2)}}{\Vert u\Vert^2_{L^\frac{2p}{p-1}(z_1,z_2)}}}.$$

On the other hand, from the one dimensional change of variable
$w(x)=v(z_1+(z_2-z_1)x)$, it is immediate that

$$\min_{v\in H^1_0(z_1,z_2)}\frac{\Vert v' \Vert^2_{L^2(z_1,z_2)}}{\Vert v\Vert^2_{L^\frac{2p}{p-1}(z_1,z_2)}}=(z_2-z_1)^{\frac{1}{p}-2}\min_{w\in H^1_0(0,1)}\frac{\Vert w' \Vert^2_{L^2(0,1)}}{\Vert w\Vert^2_{L^\frac{2p}{p-1}(B_1)}}:=(z_2-z_1)^{\frac{1}{p}-2}C_p.$$

It follows that
$M\geq\omega_N^{1/p}\varepsilon_1^{N-1}(z_2-z_1)^{1/p-2}C_p$. From
this, taking $\varepsilon_2$ such that
$M<\omega_N^{1/p}\varepsilon_1^{N-1}\varepsilon_2^{1/p-2}C_p$, we
conclude part ii) of the lemma with $\varepsilon=\varepsilon_2$.

Obviously, taking $\varepsilon=\min\{ \varepsilon_1,\varepsilon_2
\}$, the lemma is proved.

\end{proof}

\begin{lemma}\label{sealcanza}
Let $k\geq 0$, $N\geq 2$ and $N/2<p<\infty$. Then $\gamma_{p,k}$
is strictly positive and it is attained in a function
$a_0\in\Gamma_k$.
\end{lemma}

\begin{proof}

Take a sequence $\{a_n\}\subset\Gamma_k$ such that $\Vert
a_n-\mu_k\Vert_{L^p(B_1)}\to\gamma_{p,k}$. Take $\{u_n\}\subset
H^1(B_1)$ such that $u_n$ is a radial solution of (\ref{ceq6}),
for $a=a_n$, with the normalization $\Vert
u_n\Vert_{H^1(B_1)}^2=\int_{B_1}\left(\vert\nabla
u_n\vert^2+u_n^2\right)=1$. Therefore, we can suppose, up to a
subsequence, that $u_{n} \rightharpoonup u_{0}$ in $H^1(B_1)$ and
$u_{n} \rightarrow u_{0}$ in $L^\frac{2p}{p-1}(B_1)$ (since
$p>N/2$, then $2<\frac{2p}{p-1}<\frac{2N}{N-2}$, which is the
critical Sobolev exponent). On the other hand, since $\{a_n\}$ is
bounded in $L^p(B_1)$, and $1\leq N/2<p<\infty$, we can assume, up
to a subsequence, that $a_{n} \rightharpoonup a_{0}$ in
$L^p(B_1)$. Taking limits in the equation (\ref{ceq6}), for
$a=a_n$ and $u=u_n$, we obtain that $u_0$ is a solution of this
equation for $a=a_0$. Note that $u_{n} \rightarrow u_{0}$ in
$L^\frac{2p}{p-1}(B_1)$ and $a_{n} \rightharpoonup a_{0}$ in
$L^p(B_1)$ yields $\lim\int_{B_1}\vert\nabla
u_n\vert^2=\lim\int_{B_1}a_n u_n^2=\int_{B_1}a_0
u_0^2=\int_{B_1}\vert\nabla u_0\vert^2$ and consequently
$u_n\rightarrow u_0\not\equiv 0$ in $H^1(B_1)$. Therefore, if
$a_0\not\equiv \mu_k$, then $a_0\in\Gamma_k$ and $\Vert
a_0-\mu_k\Vert_p\leq\lim_{n\to\infty}\Vert
a_n-\mu_k\Vert_p=\gamma_{p,k}$, and the lemma follows.

On the contrary, suppose by contradiction that $a_0\equiv\mu_k$.
Then $u_0=\varphi_k$ for some nontrivial radial eigenfunction
$\varphi_k$. Consider $\varepsilon$ given in Lemma
\ref{cerosseparados}. Take $\varepsilon_0=\min\left\{ \varepsilon,
2r_k/3, 2(1-r_1), r_i-r_{i+1};\ 1\leq i\leq k-1\right\}$. Thus,
from the previous lemma, $u_n$ has no zeros in
$(0,\varepsilon_0)$, and has, at most, one zero in each of the $k$
disjoint intervals $(r_i-\varepsilon_0/2,r_i+\varepsilon_0/2)$,
$1\leq i\leq k$. Therefore, $u_n$ has, at most, $k$ zeros in the
set $A:=(0,\varepsilon_0)\bigcup\left ( \cup_{1\leq i\leq k}\
(r_i-\varepsilon_0/2,r_i+\varepsilon_0/2)\right )$.

On the other hand, taking into account the continuous embedding
\linebreak  $H_{rad}^1\left(A(\varepsilon_0,1)\right)\subset
C\left(A(\varepsilon_0,1)\right)$ and $u_n\rightarrow\varphi_k$ in
$H_0^1(B_1)$, we can assert $u_n\rightarrow\varphi_k$ in
$C\left(A(\varepsilon_0,1)\right)$. Clearly $\min_{r\in
(0,1]\setminus A}\vert \varphi_k(r)\vert>0$. Then, for large $n$
we see that $\min_{r\in (0,1]\setminus A}\vert u_n(r)\vert>0$,
which implies that $u_n$ does not vanish in $(0,1]\setminus A$,
for large $n$. Since $u_n$ has, at most, $k$ zeros in $A$, we
conclude that $u_n$ has, at most, $k$ zeros in (0,1], for large
$n$. This contradicts Proposition \ref{n+1ceros} and the lemma
follows.

\end{proof}

\section{The critical case}

In this section, we study the critical case, i.e. $p=\frac{N}{2}$,
if $N\geq 3$. We will prove that $\gamma_{p,k}>0$.

\begin{lemma}\label{N/2} Let $k\geq 0$, $N\geq 3$ and $p=N/2$. Then $\gamma_{p,k}>0.$

\end{lemma}

\begin{proof}
To obtain a contradiction, suppose that $\gamma_{p,k}=0$. Then we
could find a sequence $\{a_n\}\subset\Gamma_k$ such that
$a_n\rightarrow\mu_k$ in $L^{N/2}(B_1)$. Similarly to the
supercritical case, we can take $\{u_n\}\subset H^1(B_1)$ such
that $u_n$ is a radial solution of (\ref{ceq6}), for $a=a_n$, with
the normalization $\Vert u_n\Vert_{H^1(B_1)}^2=1$. Again, we can
suppose, up to a subsequence, that $u_{n} \rightharpoonup u_{0}$
in $H^1(B_1)$ and taking limits in the equation (\ref{ceq6}), for
$a=a_n$ and $u=u_n$, we obtain that $u_0$ is a solution of this
equation for $a=\mu_k$.

We claim that $u_n\rightarrow u_0$ in $H^1(B_1)$ and consequently,
$u_0=\varphi_k$, for some nontrivial eigenfunction $\varphi_k$.
For this purpose, we set

$$\lim\int_{B_1}\vert\nabla u_n\vert^2=\lim\int_{B_1}a_n u_n^2=\lim\int_{B_1}(a_n-\mu_k)u_n^2+\lim\int_{B_1}\mu_ku_n^2=$$

$$0+\mu_k\int_{B_1}u_0^2=\int_{B_1}\vert\nabla u_0\vert^2,$$

\noindent where we have used $a_n\rightarrow\mu_k$ in
$L^{N/2}(B_1)$ and $u_n^2$ is bounded in $L^{N/(N-2)}(B_1)$ (since
$u_n$ is bounded in $H^1(B_1)\subset L^{2N/(N-2)}(B_1)$). Thus,
from standard arguments, we deduce that $u_n\rightarrow
u_0=\varphi_k$ in $H^1(B_1)$.

In the following, we will fix $\varepsilon\in(0,r_k)$. Since
$a_n\rightarrow\mu_k$ in $L^{N/2}\left(A(\varepsilon,1)\right)$
and $u_n\rightarrow u_0=\varphi_k$ in
$H_{rad}^1\left(A(\varepsilon,1)\right)\subset
C\left(A(\varepsilon,1)\right)$, we can assert that
$a_nu_n\rightarrow\mu_k\varphi_k$ in
$L^{N/2}\left(A(\varepsilon,1)\right)\subset
L^1\left(A(\varepsilon,1)\right)$. Thus $-\Delta
u_n\rightarrow\mu_k\varphi_k$ in
$L^1\left(A(\varepsilon,1)\right)$, which yields
$u_n\rightarrow\varphi_k$ in $C^1\left(A(\varepsilon,1)\right)$.
It follows that, for large $n$, the number of zeros of $u_n$ is
equal to the number of zeros of $\varphi_k$ in the annulus
$A(\varepsilon,1)$, which is exactly $k$. Applying Proposition
\ref{n+1ceros} we can assert that, for large $n$ there exists a
zero $\varepsilon_n\in (0,\varepsilon]$ of $u_n$. Hence,
multiplying the equation $-\Delta u_n=a_nu_n$ by $u_n$,
integrating by parts in the ball $B_{\varepsilon_n}$ and applying
H\"{o}lder inequality, we deduce

$$\int_{B_{\varepsilon_n}}\vert\nabla u_n\vert^2=\int_{B_{\varepsilon_n}}a_n u_n^2\leq \Vert a_n\Vert_{L^{N/2}(B_{\varepsilon_n})}\Vert u_n\Vert^2_{L^{2N/(N-2)}(B_{\varepsilon_n})}.$$

From the above it follows that

$$\Vert a_n\Vert_{L^{N/2}(B_{\varepsilon_n})}\geq\frac{\Vert\nabla u_n\Vert^2_{L^2(B_{\varepsilon_n})}}{\Vert u_n\Vert^2_{L^{2N/(N-2)}(B_{\varepsilon_n})}}\geq
\inf_{u\in H^1_0(B_{\varepsilon_n})}\frac{\Vert\nabla
u\Vert^2_{L^2(B_{\varepsilon_n})}}{\Vert
u\Vert^2_{L^{2N/(N-2)}(B_{\varepsilon_n})}}.$$

From the change $v(x)=u(\varepsilon_n x)$, it is easily deduced
that

$$\inf_{u\in H^1_0(B_{\varepsilon_n})}\frac{\Vert\nabla u\Vert^2_{L^2(B_{\varepsilon_n})}}{\Vert u\Vert^2_{L^{2N/(N-2)}(B_{\varepsilon_n})}}=\inf_{v\in H^1_0(B_1)}\frac{\Vert\nabla v\Vert^2_{L^2(B_1)}}{\Vert v\Vert^2_{L^{2N/(N-2)}(B_1)}}:=C_N>0.$$

From the above it follows that, for fixed $\varepsilon\in (0,r_k)$
and large $n$, we obtain

$$C_N\leq\Vert a_n\Vert_{L^{N/2}(B_{\varepsilon_n})}\leq \Vert a_n-\mu_k\Vert_{L^{N/2}(B_{\varepsilon_n})}+\Vert \mu_k\Vert_{L^{N/2}(B_{\varepsilon_n})}\leq$$

$$\Vert a_n-\mu_k\Vert_{L^{N/2}(B_1)}+\Vert \mu_k\Vert_{L^{N/2}(B_{\varepsilon})}.$$

Taking limits when $n$ tends to $\infty$ in this expression we
deduce

$$C_N\leq \mu_k\left(\frac{\omega_N\varepsilon^N}{N}\right)^{2/N}.$$

Choosing $\varepsilon>0$ sufficiently small we obtain a
contradiction.

\end{proof}

redpajama_set_name: RedPajamaArXiv


Students Shadow Attorneys: High school student writes local newspaper about experience at Motley Rice during ProjectGO!
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