Margaret Ann Fanelli Colenzo and Carmen...her famous recipe.
Mom had a business at one time in the early 1950's, this was a favorite by all who patronized back then. I thank her for the memories and being able to share her wonderful recipes.....this one's for you mom....hope your watching out there....Here, forever in our hearts!
Mom was a great baker! She made everything seem so simple. No real recipe written down, just threw all the ingredients in, mixed it and viola it came out perfect every time!
I get lucky now and then and make things without following a recipe but not like her she was a real inventor of everything from scratch cooking when it came to new or old recipes and she never measured anything!
I love these cream puffs of hers. I watched her make them many times. I realized after many failed attempts that beating the batter is the key to the success of the perfect cream puff. Make sure its thick and well combined. Cool completely after baking and then fill and enjoy!
Beat eggs in bowl set aside. Measure 1 cup flour and set aside.
In a heavy non stick saucepan, pour 1 cup water and a stick of butter (not margarine) and bring to a boil on medium heat, leave on heat and add flour stirring quickly with a wooden spoon until flour is dissolved and comes together.
Remove from heat, add eggs and beat very rapidly with the wooden spoon until a ball of dough if formed...Note its very important to beat this fast.
Spoon one tablespoon of dough or teaspoon for small puffs on an ungreased cookie sheet or on parchment paper. Shape into logs for eclairs or round for cream puffs. For Fancier puffs, use a cookie star tip and pastry bag to make a round swirl effect.
Bake on 400 for 1/2 hour, do not open oven or puffs will not cook correctly. When browned remove from oven, cool completely, cut in half to fill.
You can fill with puffs whipped cream, vanilla pastry pudding. Eclairs also will have chocolate frosting on top filled with vanilla pudding.
Note: If you want to use this dough for a dinner party sandwich, fill puffs with egg salad, tuna salad or chicken salad.
never had those before and they look so GOOD!! will have to list this in my favorites. thanks for sharing.
This looks delicious!! Do you have a favorite recipe for the filling you could share?
Love cream puffs, I was thinking to make some this week. Beside how can we go wrong with a family recipe. Thanks!
Nice and delicious ! Thank you for sharing. Cheers !
Oh, I love the old photo. The recipe looks good too but I'm a huge fan of old photos, family history and such!
Claudia, this is the classic pate au choux recipe, or what we call choux paste! Wonderful!
I tried so many recipes and I think I will try yours or you moms too. Looks amazing and delicious! Thanks you so much for sharing!


List of Knaresborough based security companies offering private detective and investigation services, CCTV camera system, burglar and intruder alarms installation services.
From the outset the aim of the company has been to provide friendly, constructive advice and high quality security installations at very competitive prices. AXL Security design and install intruder alarms, closed circuit television systems, fire alarms and access control systems throughout the whole of Yorkshire and beyond!


I am too old for this ?
Strength training is important for anyone, at any stage of their life. But as demonstrated, it is even more important for older adults. Now get up and call your grandma to check if she's doing her exercises. And don't accept "I am too old for that" as an excuse any more!
1. Dad do you want to keep those long relaxed golf games next years? Without exercise adults 50 and over on average lose 5 pounds of muscle mass every 10 years!
In fact, research shows that adults 50 and over average a 3 pound increase in muscle tissue after just 10 weeks of sensible strength exercise.
2. Mom, I love those yellow shorts that we gave you for your birthday, you should ware them more often, no they aren't too tight, but with some work outs maybe you will feel better in them. In fact strength training will help you lose fat weight even in the absence of dietary restriction.
Research shows a 7 percent increase in resting metabolism after just three months of strength exercise. Research indicates approximately 4 pounds of fat loss from strength exercise alone after just 10 weeks of training and twice as much fat loss when combined with minor dietary adjustments.
•	Rheumatoid arthritis - increase musculoskeletal function and concurrently eases the pain of osteoarthritis and rheumatoid arthritis.


In January, Case Keenum was part of one of the great plays in NFL history when he completed the Minneapolis Miracle, the touchdown pass to Stefon Diggs on the last play of the Vikings' playoff win over the Saints. Now Keenum is the Broncos' quarterback, and he gave fans in Denver a fantastic breakdown of the play, both from an Xs and Kenta Maeda Youth Jersey Os perspective, and from a human perspective.
The offensive line blocked their tails off. In these situations, two-minute, these defensive ends… they know it's a pass so they're literally in a track stance and they're just rushing as fast as it can. So it started there. They were incredible, Keenum said of his offensive linemen.
Keenum then went through his progressions and explained that he saw three other options before deciding to throw to Diggs.
The Cardinals wrapped up their draft class by picking Cincinnati tackle Korey Cunningham in the seventh round. The 6-foot-6, 311-pounder started the final two years of his college career for the Bearcats.
Cunningham said he is comfortable at either tackle spot and can play guard if needed. He was beginning to wonder if any team would draft him, since the Cardinals held the third-to-last selection of the last round, and bided his time by playing with his four dogs in the backyard.
And one of them just had puppies, Cunningham said.
Cunningham fielded calls from at least http://www.authenticdodgersstore.com/Authentic-Kenta-Maeda-Jersey eight or nine teams interested in signing him as a free agent until the Cardinals swooped http://www.officialmagicstoreonline.com/Marreese_Speights_Jersey in.
It's a prayer Marreese Speights Jersey answered, Cunningham said.


text:
def string_lengths(list_of_strings):

    return [len(string) for string in list_of_strings]

def create_strings(values, widths, align='LEFT'):

    if not isinstance(values, list):
        values = [values]
    if not isinstance(widths, list):
        widths = [widths]

    if align == 'CENTER':
        align = '^'
    else:
        align = ''

    output_strings = [
        f"{value:{align}{width}}"
        for value, width in zip(values, widths)
    ]

    return output_strings

class StdOutTable:

    def __init__(self, sections=None):

        if sections is None:
            sections = []

        self.sections = sections

    def _default_lengths(self, name, headers):

        max_length = max(string_lengths(headers))

        if len(name) > max_length * len(headers):
            max_length = int(
                len(name) / len(headers)
            )

        lengths = [max_length] * len(headers)

        return lengths

    def _section_header(self, index):

        headers = self.sections[index][1]
        lengths = self.sections[index][2]

        section_headings = create_strings(
            headers, lengths, align='CENTER')
        section_header = ' '.join(
            ['|'] + section_headings + ['|'])

        return section_header

    def add_section(self, name, headers, lengths=None):

        if lengths is None:
            lengths = self._default_lengths(name, headers)

        self.sections.append([name, headers, lengths])

    def table_header(self):

        section_headers = [
            self._section_header(index)
            for index, _ in enumerate(self.sections)
        ]

        section_titles = [
            ''.join(
                ['|'] + create_strings(
                    section[0],
                    len(header) - 2,
                    align='CENTER'
                ) + ['|']
            )
            for section, header in zip(
                self.sections, section_headers)
        ]

        table_header = (
            ''.join(section_titles) + '\n'
            + ''.join(section_headers) + '\n'
            + '-' * len(''.join(section_headers))
        )

        return table_header

    def row(self, data):

        start = 0
        end = 0
        row = []
        for section in self.sections:
            end += len(section[1])

            data_strings = create_strings(
                data[start: end], section[2]
            )
            data_string = ' '.join(
                ['|'] + data_strings + ['|'])
            row.append(data_string)

            start += len(section[1])

        return ''.join(row)

redpajama_set_name: RedPajamaGithub


Yesterday was amazing day for Indian Sports .It was the day when there was a fight between Indian cricket and Indian Tennis to grab the space for Times Sports of TOI, popular indian national daily .
At one side was the heroic performance by an indian cricketer (Kaif) to ensure India's entry in the finals and at the other end was a phenomenal performance by the fast rising Indian Tennis player Sania Mirza.to enter in the fourth round of US Open .
Both were big news and had .their due on Times Sports and it was the Times editorial to decide which news has more readers than the others.


text:
Starring: Bruce Willis, Robin Wright, Robin Wright Penn, Samuel L. Jackson, Spencer Treat Clark
Genres: Fantasy, Mystery, Thriller
Suggested corrections Significant dates Titles starting with U
Question: I was watching this film on TV the other night, and was expecting to see the train crash at the start, so I was surprised when I didn't. Has this been cut from the TV version?
Answer: No, the crash never actually happened on-screen.
Garlonuss
Question: Why are there so many things that are upside-down in this movie? What is the significance of it?
Answer: Everything related to Samuel L. Jackson's character is upside down - to give a hint about his role in the movie.
Question: When David touches the guy in the stadium he feels that he is carrying a gun but when he actually checks him he doesn't find one. Why?
Answer: I believe you are confusing the man with the gun with the man with the drugs. The man with the gun is never searched because he steps out of line and walks away. The man with the drugs (who is played by director M. Night Shyamalan) is searched for drugs, but David doesn't find any. Either he had the drugs in the stadium at a previous game or he sold/passed them along to someone else before David bumped into him.
Question: Why exactly does the janitor hold that family hostage? What is the point?
ModestFilmCollector
Chosen answer: He's a murdering rapist. The point is he's evil.
JC Fernandez
Question: When David uses his power to see into people's lives, what exactly does he see? They go by too fast.
Chosen answer: When he touches someone who has or will commit an evil act, he sees the act. For example, he sees a racist man throw something at a black woman for no reason, he sees someone buying drugs, he sees a woman kill another woman.
More for Unbreakable
Revealing mistake: When Bruce Willis is in the hospital, the doctor comes up to him and asks Bruce to look at him. The doctor proceeds to get a torch out and shine it into his eyes. Yet there is no light coming out of the torch. (00:10:25)
More mistakes in Unbreakable
Elijah Price: Now that we know who you are, I know who I am. I'm not a mistake. In the comics, you know how you can tell who the arch-villain is going to be? He's the exact opposite of the hero. And sometimes they're friends, just like you and me. I should have known way back when. You know why, David? Because of the kids. They called me Mr. Glass.
More quotes from Unbreakable
Trivia: Another comic book reference: they both have signature colors. Mr Glass' color is purple (his clothes, the envelope he leaves on the windshield, interior of his house, etc.) while David's is green (all of the interior in his house, his raincoat, his suit on the train etc.).
Twotall
More trivia for Unbreakable
Mistake & trivia booksMost popular pagesBest movie mistakesBest mistake picturesBest comedy movie quotesMovies with the most mistakesNew this monthFriends mistakesSister Act mistakesLost in Translation questionsCatch Me If You Can endingGuardians of the Galaxy triviaHome Alone mistake pictureAnya Taylor-Joy movies & TV showsCars quotesMrs. Doubtfire plot10 Biggest mistakes in The Twilight Saga: New MoonMore for Unbreakable
When Bruce Willis is in the hospital, the doctor comes up to him and asks Bruce to look at him. The doctor proceeds to get a torch out and shine it into his eyes. Yet there is no light coming out of the torch.
Another comic book reference: Most of the film is shot through door frames, windows, between seats or otherwise "framed" like comic book panels.

redpajama_set_name: RedPajamaCommonCrawl


The partnership with parents is very important as it is the key to making sure that the children in my care get the best care available and are happy and settled in my setting. With the introduction of the Children Act in 1989 parents were given definite rights. This included being able to access information on their children's progress and achievements in their setting and also a preference to parents as to which school they wanted their child to go to.
Since the introduction of the EYFS in 2008, a clear framework has been put in place to foster partnerships between parents and all professionals, including childminders, pre-schools, nurseries and schools. The relationship between parents and childminders can be very close, parents will be coming into my home every day and it is important that I put this on a professional footing by having a Parental Involvement Policy that clearly states how the partnership between the parents and I will work.
I will welcome all parents into my setting and encourage their involvement in order to ensure their child is comfortable enough to flourish into confident, happy individuals. The partnership with parents starts with the first initial enquiry / visit. I need to build a rapport with the parents by understanding their requirements both culturally, socially and any routines they have. I will always endeavour to accommodate a child's routine and discuss with parents how this will be managed taking account of the children already in my care.
Sometimes a compromise will need to be made, as long as this is agreed by all parties. Different cultures have different approaches to modesty, clothing, and the use of the toilet and hygiene arrangements. Some cultures have very specific guidelines on the preparation and storage of food and the type of food they eat. All these aspects need to be considered and discussed with the parent at the initial enquiry / visit stage. Once I have established the basic requirements of the parents I will draw up a contract with the parents for the care of their child/(ren).
Before a child starts in my care I will want to have some settling in sessions. This is important to ensure that the child, parent and I can have some time to get to know each other and clarify any points regarding routine, specific needs and communication. There are many ways of communicating with parents, keeping a daily diary, telephone calls, email, a chat on collection or drop off and formalised parent meetings.
A daily diary is particularly useful when children in my care are young and are still in appies, on formula milk, or being weaned, or just starting to feed themselves and / or not yet talking. Telephone calls or email communication is more likely to be an agreed regular communication in the evening or weekends. I would not envisage this being the preferred method of communication as face to face discussion would be more productive. I would propose to hold parents evenings, at least twice a year. The purpose of this would be to review the child's progress, discuss any issues and plan for the coming six months.


text:
Haley Barbour on Energy & Oil
Former Republican MS Governor (until 2012)
Supports $1.3 billion coal gasification plant
East Mississippi is healthy, but I believe there are more opportunities, and we are focused on them. The announcement of the Mississippi Power $1.3 billion coal gasification plant in Kemper County and continued strong performance on I-20 and Neshoba County show the Meridian area is moving forward. Katrina reconstruction plus the building of a more than $1 billion Strategic Petroleum Reserve facility near Richton will fuel more job growth. Source: State of the State address to 2007 state legislature , Jan 15, 2007
Stop harmful EPA regulation of greenhouse gas emissions.
Barbour signed Letter from 20 Governors to leaders in Congress
We feel compelled to guard against a regulatory approach that would increase the cost of electricity and gasoline prices, manufactured products, and ultimately harm the competitiveness of the US economy. As governors, we strongly urge Congress to stop harmful EPA regulation of greenhouse gas emissions that could damage those vital interests. We ask that Congress continue its work to pass comprehensive legislation that balances the role of conservation and climate security with the production of abundant and affordable American energy. The EPA has initiated efforts to impose greenhouse gas regulations that could be harmful to our economies at an especially critical time. As Governors, we are gravely concerned about such regulation.
EPA is not equipped to consider the very real potential for economic harm when regulating emissions. Without that consideration, regulation will place heavy administrative burdens on state environmental quality agencies, will be costly to consumers, and could be devastating to the economy and jobs.
We believe that EPA should offer input regarding complex energy and environmental policy initiatives, like reducing greenhouse gas emissions, but feel that these policies are best developed by elected representatives at the state and national level, not by a single federal agency. There is no question that broad bipartisan support exists to reduce greenhouse gas emissions while taking into consideration the difficult fiscal situation that our states and the nation face. Source: Letter from 20 Governors to leaders in Congress 100310-Gov on Mar 10, 2010
Click here for a summary of all issue stances of Haley Barbour.
Click here for a profile of Haley Barbour.
Click here for VoteMatch responses by Haley Barbour.
Agree? Disagree? Voice your opinions on Energy & Oil
Newly elected Nov. 2012:
Up for re-election 2013:
NJ: Chris Christie
VA: Bob McDonnell Up for re-election 2014:
Page last updated: Apr 25, 2013


text:
using namespace std;

int main(int argc, char *argv[])
{
  if (Show_Version(argc, argv, "0.1") == 1) {
    return 0;
  }

  static char const *Spec[] = {"<input:string> -o <string>",
    "--body_id <int> [--isolate <int> <int>] [--ds <int>]",
    "[--final_ds <int>]",
    "[--compress <string>]", NULL};

  Process_Arguments(argc, argv, const_cast<char**>(Spec), 1);

  char *dataDir = Get_String_Arg(const_cast<char*>("input"));
  int bodyId = Get_Int_Arg(const_cast<char*>("--body_id"));

  //Load compress map
  if (Is_Arg_Matched(const_cast<char*>("--compress"))) {
    FILE *fp = fopen(Get_String_Arg(const_cast<char*>("--compress")), "r");
    ZString str;
    while (str.readLine(fp)) {
      vector<int> value = str.toIntegerArray();
      if (value.size() == 2) {
        if (value[0] == bodyId) {
          bodyId = value[1];
          break;
        }
      }
    }
    fclose(fp);
  }

  ZFileList fileList;
  fileList.load(dataDir, "tif", ZFileList::SORT_BY_LAST_NUMBER);
  int startPlane = fileList.startNumber();
  int endPlane = fileList.endNumber();
  int planeNumber = endPlane - startPlane + 1;

  ZFileList additionalFileList;
  additionalFileList.load(string(dataDir) + "/a", "tif",
      ZFileList::SORT_BY_LAST_NUMBER);

  int r = 250;
  int dr = 75;

  if (Is_Arg_Matched(const_cast<char*>("--isolate"))) {
    r = Get_Int_Arg(const_cast<char*>("--isolate"), 1);
    dr = Get_Int_Arg(const_cast<char*>("--isolate"), 2);
  }

  Stack *stack = Read_Stack_U(fileList.getFilePath(0));

  int intv[3] = {0, 0, 0};
  if (Is_Arg_Matched(const_cast<char*>("--ds"))) {
    intv[0] = Get_Int_Arg(const_cast<char*>("--ds")) - 1;
    intv[1] = intv[0];
    //intv[1] = Get_Int_Arg(const_cast<char*>("--ds"), 2) - 1;
    //intv[2] = Get_Int_Arg(const_cast<char*>("--ds"), 3) - 1;
    stack = Downsample_Stack(stack, intv[0], intv[1], intv[2]);
    r /= intv[0] + 1;
    dr /= intv[0] + 1;
  }

  int width = stack->width;
  int height = stack->height;

  Stack *mip = Make_Stack(GREY, width, height, 1);
  Zero_Stack(mip);
  int planeSize = Stack_Voxel_Number(mip);

  char filePath[500];
  Kill_Stack(stack);

  int zStart = -1;
  int zEnd = 0;

  //Estimate Z range
  cout << "Estimating Z range ..." << endl;
  for (int i = 0; i < planeNumber; i++) {
    cout << "  " << i << "/" << planeNumber << endl;
    int bodyArea = 0;

    stack = Read_Stack_U(fileList.getFilePath(i));
    Stack *additionalStack = NULL;
    if (additionalFileList.size() > 0) {
      additionalStack = Read_Stack_U(additionalFileList.getFilePath(i));
    }

    if (Is_Arg_Matched(const_cast<char*>("--ds"))) {
      Stack *stack2 = Downsample_Stack(stack, intv[0], intv[1], intv[2]);
      Free_Stack(stack);
      stack = stack2;

      if (additionalStack != NULL) {
        stack2 = Downsample_Stack(additionalStack, intv[0], intv[1], intv[2]);
        Free_Stack(additionalStack);
        additionalStack = stack2;
      }
    }
    color_t *arrayc = (color_t*) stack->array;
    uint16_t *array16 = (uint16_t*) stack->array;

    if (stack->kind == COLOR) {
      for (int offset = 0; offset < planeSize; offset++) {
        int tmpBodyId = (int) Color_To_Value(arrayc[offset]);
        if (additionalStack != NULL) {
          uint32_t value = additionalStack->array[offset];
          tmpBodyId += (int) (value << 24);
        }
        if (tmpBodyId == bodyId) {
          bodyArea++;
          mip->array[offset] = 1;
        }
      }
    } else {
      for (int offset = 0; offset < planeSize; offset++) {
        if (array16[offset] == bodyId) {
          bodyArea++;
          mip->array[offset] = 1;
        }
      }
    }

    if (bodyArea > 0) {
      if (zStart < 0) {
        zStart = i;
      }
      zEnd = i;
    }

    Free_Stack(stack);
    if (additionalStack != NULL) {
      Free_Stack(additionalStack);
    }
  }
  cout << "  " << zStart << " " << zEnd << endl;

  cout << "Removing isolated objects ..." << endl;
  Stack *mipMask = Stack_Remove_Isolated_Object(mip, NULL, r, dr);

  cout << "Estimating bound box ..." << endl;
  Cuboid_I boundBox;
  Stack_Bound_Box(mipMask, &boundBox);
  cout << "  (" << boundBox.cb[0] << "," <<  boundBox.cb[1] << "," <<
    boundBox.cb[2] << ")" << "->" << "(" << boundBox.ce[0] << "," <<
    boundBox.ce[1] << "," << boundBox.ce[2] << ")" << endl;

  int finalDs = 1;

  if (Is_Arg_Matched(const_cast<char*>("--final_ds"))) {
    finalDs = Get_Int_Arg(const_cast<char*>("--final_ds"));
  }

  int cropWidth = boundBox.ce[0] - boundBox.cb[0] + 1;
  int cropHeight = boundBox.ce[1] - boundBox.cb[1] + 1;

  int finalStackWidth, finalStackHeight, finalStackDepth;
  Downsample_Stack_Max_Size(cropWidth, cropHeight, 1, finalDs, finalDs, 1,
      &finalStackWidth, &finalStackHeight, &finalStackDepth);

  Stack *finalStack = Make_Stack(GREY, finalStackWidth, finalStackHeight,
      zEnd - zStart + 1);
  int finalPlaneArea = finalStackHeight * finalStackWidth;

  uint8_t *outArray = finalStack->array;

  Stack *bodyPlane = Make_Stack(GREY, mipMask->width, mipMask->height, 1);
  Stack *croppedBodyPlane = Make_Stack(GREY, cropWidth, cropHeight, 1);

  //Load images
  cout << "Extract planes ..." << endl;
  for (int i = zStart; i <= zEnd; i++) {
    cout << "  " << i << "/" << zEnd << endl;
    stack = Read_Stack(fileList.getFilePath(i));
    Stack *additionalStack = NULL;
    if (additionalFileList.size() > 0) {
      additionalStack = Read_Stack_U(additionalFileList.getFilePath(i));
    }

    if (Is_Arg_Matched(const_cast<char*>("--ds"))) {
      Stack *stack2 = Downsample_Stack(stack, intv[0], intv[1], intv[2]);
      Free_Stack(stack);
      stack = stack2;

      if (additionalStack != NULL) {
        stack2 = Downsample_Stack(additionalStack, intv[0], intv[1], intv[2]);
        Free_Stack(additionalStack);
        additionalStack = stack2;
      }
    }

    color_t *arrayc = (color_t*) stack->array;
    uint16_t *array16 = (uint16_t*) stack->array;

    if (stack->kind == COLOR) {
      for (int offset = 0; offset < planeSize; offset++) {
        int tmpBodyId = (int) Color_To_Value(arrayc[offset]);
        if (additionalStack != NULL) {
          uint32_t value = additionalStack->array[offset];
          tmpBodyId += (int) (value << 24);
        }

        if (tmpBodyId == bodyId && mipMask->array[offset] == 1) {
          bodyPlane->array[offset] = 1;
        } else {
          bodyPlane->array[offset] = 0;
        }
      }
    } else {
      for (int offset = 0; offset < planeSize; offset++) {
        if (array16[offset] == bodyId &&
            mipMask->array[offset] == 1) {
          bodyPlane->array[offset] = 1;
        } else {
          bodyPlane->array[offset] = 0;
        }
      }
    }

    Stack dest;
    dest.width = finalStack->width;
    dest.height = finalStack->height;
    dest.depth = 1;
    dest.kind = GREY;
    dest.array = outArray;

    Crop_Stack(bodyPlane, boundBox.cb[0], boundBox.cb[1], 0,
        cropWidth, cropHeight, 1, croppedBodyPlane);

    //Downsample the plane
    Downsample_Stack_Max(croppedBodyPlane, finalDs, finalDs, 0, &dest);

    outArray += finalPlaneArea;

    Free_Stack(stack);
    if (additionalStack != NULL) {
      Free_Stack(additionalStack);
    }
  }

  //finalStack = Downsample_Stack_Max(finalStack, 1, 1, 0);
  Write_Stack(Get_String_Arg(const_cast<char*>("-o")), finalStack);

  sprintf(filePath, "%s.offset.txt", Get_String_Arg(const_cast<char*>("-o")));
  FILE *fp = fopen(filePath, "w");
  fprintf(fp, "%d %d %d", boundBox.cb[0], boundBox.cb[1], zStart);
  fclose(fp);

  return 0;
}


text:
The Best of Cambodia 7 days
From ¥4050 p. person BOOK TRIP
Destinations: Phom Penh-Kep-Siem Reap
Siem reap covers an area of 10299 square kilometers and a population of about 700000. Siem reap was the capital of siem reap, its suburbs has the world-renowned Angkor Wat. After entering in phnom penh, spend a day to visit to the king and virtually which the vii bayon temple built for himself, angkor dynasty's heyday as the balcony. You can also see angkor dynasty sculpture art and crafts, such as the number of French photographer to take pictures of angkor, and a lot of Angkor Wat, sculpture. Continue to siem reap, after spending a day can feel Phouk silk farms and les artisans D 'Angkor.
Explore the graceful capital, Phnom Penh, that was once known as the 'Pearl of Asia'.
Discover the spectacular Royal Palace, home to the Cambodian royal family and Khmer architecture.
Experience ancient sculptures from the Angkor period at the National Museum.
Relax and unwind at Cambodia's first seaside resort, Kep.
Support the locals and get insight into how arts and crafts are being made at Artisans Angkor and Chantiers Ecoles.
Explore the mother of all temples, Angkor Wat and get lost in the immense walled city of Angkor Thom.
Brief itinerary
Day 1 Arrive at Phnom Penh, stay at Phnom Penh for 1 night (no meal)
Day 2 Phnom Penh city tour, transfer to Kep, stay at Kep for 3 nights (breakfast)
Day 3 Kep (breakfast)
Day 5 Kep fly to Siem Reap, stay at Siem Reap for 2 nights (breakfast)
Day 6 Siem Reap full day tour (breakfast)
Day 7 Siem Reap full day tour& departure (breakfast)
Activities level: Easy to moderate
Transport: Private air-conditioned vehicle and domestic flight
Meals: 6 breakfasts
Day 1 Arrive at Phnom Penh (no meal)
Arrive at Phnom Penh International Airport, meet and welcome by tour guide then overland transfer to Hotel in Phnom Penh Town.
Phnom Penh is a graceful capital that was once known as the 'Pearl of Asia'. It is a timeless throwback to an older Asia, a world away from the developing metropolises of nearby neighbours. Situated at the confluence of the Mekong, Bassac and Tonle Sap Rivers, the city retains much traditional and colonial charm. The riverfront area remains one of the most attractive in the region and is perfect for a late afternoon stroll.
Overnight in Phnom Penh for 1 night.
Breakfast at Hotel.
Check out from Hotel then we explore the stunning Royal Palace complex, home to the Cambodian royal family and a symbol of the nation. We enter the Throne Hall where the royal receptions are held, see the Napoleon III Pavilion made from iron, a gift from the French emperor in the 19th century and continue to the Silver Pagoda, named after the 5000 silver tiles covering the floor, each weighing 1kg. Inside are some of the country's most cherished treasures, including a life-size gold Buddha studded with 9584 diamonds, the largest weighing 25 carats. There is also a delicate emerald Buddha made of baccarat crystal, which gives the temple its Khmer name of Wat Preah Keo (Temple of the Emerald Buddha).
We leave the Royal Palace and continue to the nearby National Museum, home to the world's finest collection of sculpture from the Angkor period. The exquisite building was completed in 1920 and features collections from the pre-Angkor, Angkor and post-Angkor periods. We concentrate on the incredible sandstone sculpture from Angkor, as well as the intricate bronzes.
After lunch, overland transfer with driver only to Kep (approximate drive is 03h30).
Kep was Cambodia's first seaside resort, founded by the French in 1908. After many years in hibernation, it has once again taken off, with new boutique hotels and resorts offering comfortable, atmospheric accommodation and delicious food. Nearby islands such as Koh Tonsay are popular for day trips and local crab and fresh seafood is a popular lunchtime treat. Bokor is a 1000m-high plateau of steaming jungle, shy wildlife and abandoned buildings, although it has recently been blighted by the construction of a new mega-casino. Built by the French as a hill station in the 1920s, it was redeveloped by Sihanouk in 1959 as a casino resort before being abandoned to the elements in 1970. Nearby Popokvil Falls are impressive in the wet season.
Overnight in Kep for 3 nights.
Day 3 Kep, free activities (breakfast)
After breakfast free in kep leisure activities throughout the day. Overnight in Kep
Day 5 Kep fly to Siem Reap (breakfast)
Morning free at leisure in Kep until transfer to Sihanoukville airport for flight Take flight K6-130 ETD-A: 12H00-13H10 to Siem Reap or similar. Arrive at Siem Reap International Airport, meet and welcome by tour guide then transfer to visit Artisans Angkor, one of the most successful craft revitalisation projects in Cambodia. The showroom opposite Angkor Wat is a delightful place to unwind after exploring the world's largest religious building, including an air-conditioned cafe. The original Chantiers Ecoles in Siem Reap is a great place to see the arts and crafts being carefully made. For those with an interest in textiles, the Pouk silk farm is a wonderful experience.
Overnight in Sihanoukville for 2 nights.
Day 6 Siem Reap full day tour with tuk tuk (breakfast)
05:00 Rising at the crack of dawn, we take tuk-tuk to journey out to the Mother of all temples, Angkor Wat. Believed to be the world's largest religious building, this temple is the perfect fusion of symbolism and symmetry and a source of pride and strength to all Khmers. Built in the 12th century by King Suryavarman II, this is most famous temple at Angkor. We stay at Angkor Wat to enjoy a picnic breakfast. As the crowds return to their hotels, we venture into Angkor Wat to enjoy its magnificence in peace and quiet, beginning at the bas-reliefs that tell of tales from Hindu mythology and of the glories of the Khmer empire.
12:00 Lunch at local restaurant (on own arrangement).
14:00 We visit the immense walled city of Angkor Thom that was the masterpiece of King Jayavarman VII. The scale is simply staggering and we are immediately overwhelmed by the audacity of Jayavarman on arrival at the city's gates. We begin our visit at the Terrace of the Leper King, continue along the Terrace of Elephants, and visit the Baphuon, once of the most beautiful temples at Angkor, dating from the 11th century. It has undergone a massive renovation by the French and is now once again open for viewing. Our climax is the enigmatic and enchanting temple of the Bayon. Its 54 towers are each topped off with the four faces of Avalokiteshvara (Buddha of Compassion), which bear more than a passing resemblance to the king himself. We unravel the mysteries of the temple's bas-reliefs, with their intricate scenes of ancient battles against the Chams and their snapshot of daily life during the Angkor period.
18:30 Dinner at local restaurant (on own arrangement).
20:00 Overnight at Hotel.
Free at leisure until transfer to Siem Reap international airport departure flight.
Prices, based on 2 persons travelling together
Package price (4*): CNY 4880 per double occupancy
Prices, based on 3-4 persons travelling together
– All land transfers as indicated in the program
– Accommodation at mentioned hotel on double room basis with daily breakfast
– Local English-speaking guide
– All entrance fee for above sightseeing
– Domestic flight from Kep to Siem Reap with eco class incl tax
– Wet tissue and cool drinking water (2 bottles/pax/day)
– During sightseeing
What's exclude:
-Any visa fee (if required)
-Any air-ticket and airport tax
-Personal expense (beverage, laundry, insurance …)
-Other optional tour or transfer
-Tip for tour guide and driver
-Any bank fee (both local and international)
-Any of other accounts not mentioned in "WHAT'S INCLUDED"
Hotel Name: White Mansion****
Room type: Deluxe
Information: Historical heritage now beautifully renovated, White Mansion, former Residence of the American Embassy in Phnom Penh is a unique charming Boutique hotel: luxury becomes affordable.
Location: Located in the heart of the city, only few steps away from all major highlights and business areas, set in a quiet and romantic environment.
Hotel Name: Knai Bang Chatt****
Room type: Garden View
Information: Located along the South-eastern coastline of Cambodia, Knai Bang Chatt Resort features an outdoor infinity pool, a private area with views of the sea, and offers exquisite accommodations surrounded by nature's tranquility. It operates a 24-hour front desk and provides free WiFi access in the entire property.
Location:The resort is just a 5-minute walk from the famous Crab Market, and within 1.2 mi from the Public Bus Station of Kep City. Phnom Penh International Airport is approximately 93 mi away.
Hotel Name: Shinta Mani Shack****
Room type: Poolview
Information: The Shinta Mani Shack is a boutique resort centrally located in the leafy French Quarter between The Royal Gardens and the Old Market Area just 15 minute from the Angkor Wat. The courtyard-style resort's 62 rooms enjoy a surprisingly different "Bensley twist" with direct access or first floor balcony views over the stunning swimming pool and flourishing tropical gardens.
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Broker/agent does not guarantee the accuracy of the income, expenses or other information concerning the business provided by the seller or obtained from public records or other sources. Buyer is advised to independently verify the accuracy of above information through personal inspection and with appropriate professionals.
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text:
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SEATS | South East Australian Transport Strategy Inc.
Highlights from our February meeting.
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Discover more about the proposed rail link from our Canberra to the coast.
Roads and Maritime Services are currently working on 17 projects.
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Click link above to see photos from our events.
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The South East Australian Transport Strategy Inc. (SEATS) provides highly coordinated and influential advocacy for the development of transport infrastructure in the South East Australian region that supports sustainable economic development and the prosperity of its constituents.
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Arta Velika je nenaseljeni otok u hrvatskom dijelu Jadranskog mora. Otok leži sjeverozapadno od Murtera, od kojeg je udaljen 2,5 km. Mjesto Pakoštane je oko 5 km sjeveroistočno od otoka.

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redpajama_set_name: RedPajamaWikipedia


text:
Economy Secretary and CEO of Tata UK discuss a way forward for Wales
Economy Secretary, Ken Skates and Chief Executive Officer of Tata Steel UK, Bimlendra Jha met today to discuss a developing package of Welsh Government support for Tata plants in Wales.
The developing package would potentially include increased support for training, programmes to drive up efficiency and environmental improvements, and increased research and development.
Speaking after the meeting Ken Skates said:
"It is clear Tata Steel UK is facing significant issues , including pensions and energy, which need the help of a number of stakeholders to resolve. We remain fully committed to doing all we can to support steel making in Wales. Our primary concern is for the Tata Steel workforce, and associated communities and for the long-term and sustainable future of the whole Tata Steel operation in Wales.
"We will continue to do all we can to safeguard the industry and the substantial value it delivers directly and indirectly to the Welsh economy, and we will continue to support any future ownership structure for Tata Steel's activities that offers the best prospects of retaining jobs and sustainable steel production here in in Wales.
"Today's meeting was part of an ongoing series of constructive discussions that we have held over recent month with Tata Steel We are moving forward with a proposed package of support and remain committed to working closely with Tata Steel and doing all we can to secure the long term future of steel making in Wales.
"As part of these discussions, and as is normal for any investment project, we are also covering conditions the Welsh Government will require".
Chief Executive Officer of Tata Steel UK Bimlendra Jha said:
"The Welsh government has maintained its promise to support training, innovation and environment improvement schemes for the Welsh operations t of Tata Steel UK.
"We had a constructive dialogue detailing potential areas of support. Given the challenges facing the steel industry in general our shared focus is on placing Welsh plants on a more competitive and sustainable footing going forwards."
Following the Cabinet Secretary's visits to Tata Steel's sites at Port Talbot, Trostre, Llanwern and Shotton earlier this summer, considerable progress has been made in developing a potential package of support for Tata plants in Wales.
This encompasses a range of projects which will allow the Welsh plants to become more efficient and able to withstand global competition. The support includes a major environmental improvement project for Port Talbot, a potential investment in sustainable premium product capability for Shotton and a potential training support package for employees across all Tata Steel sites in Wales.
A significant increase in research and development investment in new products at Port Talbot is also expected which, if secured, would signal a significant vote of confidence in the future of Welsh steelmaking, and provide the cutting edge innovation so critical to success within the industry.


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text:
\section{Schwarzschild and Reissner-Nordstr\"om}
Let $(\mathcal{M},g)$ denote a
non-extremal Reissner-Nordstr\"om solution corresponding
to parameters $M$, $e$, with $0\le|e|<M$.
(We refer the reader to standard references, for instance~\cite{he:lssst}.)
By convention here, the term Reissner-Nordstr\"om solution will always refer to a globally
hyperbolic spacetime which is the Cauchy development of
an asymptotically flat hypersurface $\Sigma$ with two ends\footnote{i.e., in
the $e\ne0$ case, we do not mean an analytic (or otherwise) extension thereof}.
Considering the quotient
\[
\mathcal{Q}=\mathcal{M}/SO(3),
\]
and defining the area-radius function
$r:\mathcal{Q}\to{\mathbb R}$
\begin{equation}
\label{arearadius}
r(q)=\sqrt{Area(q)/4\pi},
\end{equation}
we have that
\[
g=\bar{g}+r^2 d\sigma_{{\mathbb S}^2},
\]
where $d\sigma_{{\Bbb S}^2}$ denotes the standard metric on the unit sphere, and
where $\bar{g}$ defines a $1+1$-dimensional Lorentzian metric on $\mathcal{Q}$.
The Lorentzian manifold $(\mathcal{Q}, \bar{g})$ can be covered
by a bounded global null coordinate system $(\bar{u},\bar{v})$.
The range of such a coordinate system
is depicted, as a subset of $2$-dimensional Minkowski space,
in the Schwarzschild ($M>0$, $e=0$) case:
\[
\input{schwarzschild.pstex_t}
\]
and the charged Reissner-Nordstr\"om ($M>|e|>0$) case:
\[
\input{Re-No.pstex_t}
\]
These depictions are commonly known as \emph{Penrose diagrams}.
The curve $\mathcal{S}$ depicts the projection
to $\mathcal{Q}$ of a particular choice
of complete Cauchy surface $\Sigma\subset \mathcal{M}$.

We define the
\emph{Hawking mass} function $m:\mathcal{Q}\to{\mathbb R}$
by
\[
m=\frac r2\left(1-\bar{g}(\nabla r,\nabla r)\right),
\]
and the \emph{mass ratio}
\[
\mu=\frac{2m}r.
\]
The mass $m$ is related to the parameter $M$ by
\[
M=m+\frac{e^2}{2r}.
\]
The sets $J^-(\mathcal{I}^+_A)\cap J^+(\mathcal{I}^-_A)$
and $J^-(\mathcal{I}^+_B)\cap J^+(\mathcal{I}^-_B)$, the two so-called
\emph{domains of outer communications}, can each be covered by a
coordinate system $(r,t)$ with coordinate range $(r_+,\infty)\times(-\infty,\infty)$,
where $r$ is defined by $(\ref{arearadius})$,
and in which the metric $\bar{g}$
of $\mathcal{Q}$ takes the form
\[
\bar{g}=-\left(1-\frac{2M}r+\frac{e^2}{r^2}\right)dt^2+\left(1-\frac{2M}r+\frac{e^2}{r^2}\right)^{-1}
dr^2.
\]
Here $r_+$ is defined by
\begin{equation}
\label{r_+=}
r_+=M+\sqrt{M^2-e^2}.
\end{equation}
One immediately sees that the vector field $\partial_t$ is timelike Killing in these domains.

To define a null coordinate system, we first introduce the so-called
\emph{Regge-Wheeler} coordinate
\begin{equation}
\label{Regge-Wheeler}
r_*=r+r_+^2(r_+-r_-)^{-1}\log|r-r_+|-r_-^2(r_+-r_-)^{-1}\log |r-r_-|,
\end{equation}
where
\[
r_-=M^2-\sqrt{M^2-e^2}.
\]
This coordinate satisfies
\[
\frac{d r_*}{d r}=(1-\mu)^{-1}
\]
We now set
\[
u=t-r_*,
\]
\[
v=t+r_*.
\]
With respect to these coordinates, the metric $\bar{g}$ takes the form
\[
\bar{g}=-(1-\mu)dudv=-\left(1-\frac{2M}r+\frac{e^2}{r^2}\right)dudv
\]
in $J^-({\mathcal I}^+_A)\cap J^+({\mathcal I}^-_A)$
and $J^-({\mathcal I}^+_B)\cap J^+({\mathcal I}^-_B)$; i.e.~$(u,v)$ are indeed
null. Individually, the coordinates $u$ and $v$ are each known as \emph{Eddington-Finkelstein}
retarded and advanced time coordinates,
respectively.\footnote{The term \emph{Eddington-Finkelstein coordinate system}, however,
typically refers to the hybrids $(u,r)$ and $(v,r)$.}
Each of the domains of outer communications is
covered by the range $(-\infty,\infty)\times(-\infty,\infty)$.

The notion of asymptotic flatness is captured by the fact that
$1-\mu\to1$ as $r\to \infty$.
To examine the geometry of the horizon, first, we note that
the timelike Killing vector field $\partial_t$ extends to a
null tangent Killing vector field on the event
horizon $\mathcal{H}=\mathcal{H}^+_A\cup\mathcal{H}^+_B$, vanishing
at the sphere of bifurcation $\mathcal{H}^+_A\cap\mathcal{H}^+_{B}$.
Thus $1-\mu=0$ along $\mathcal{H}$, and moreover
$m$ and $r$ are constant. In terms of the
parameters, we have
\[
m=2r=2r_+=2(M+\sqrt{M^2-e^2}).
\]

This vanishing of $1-\mu$
relative to the Regge-Wheeler
$(t,r_*)$ and null $(u,v)$ coordinates is captured by:
\begin{eqnarray}
\label{behaviour}
\nonumber
\hat{C}_1e^{\frac{(r_+-r_-)}{2r_+^2}(v-u)}&=&
\hat{C}_1e^{\frac{(r_+-r_-)}{r_+^2}r_*}\\
&\le&C_1(r-r_+)\\
\nonumber
&\le& 1-\mu\le C_2(r-r_+)\le
\hat{C}_2e^{\frac{(r_+-r_-)}{r_+^2}r_*}=\hat{C}_2e^{\frac{(r_+-r_-)}{2r_+^2}(v-u)}
\end{eqnarray}
for small enough $-\infty<r^*$, and
for some positive constants, $C_1$, $\hat{C}_1$, $C_2$, and $\hat{C}_2$, depending
only on $M$ and $e$.

\section{Local existence}
Let $G$ be a bounded $C^\infty$ function of its argument,
let $(\mathcal{M},g)$ be as in the previous section, and let
$\Sigma$ be a Cauchy surface.
We have the following
\begin{proposition}
\label{IVPstatement}
Let $\phi_0$ and $\phi_1$ be $C^\infty$ functions on
$\Sigma$. There exists
a unique globally hyperbolic subset $\mathcal{D}\subset\mathcal{M}$,
and a $C^\infty$ function $\phi:\mathcal{D}\to{\mathbb R}$ such that
\begin{enumerate}
\item
$\Sigma\subset \mathcal{D}$ is a Cauchy surface,
and $\phi|_{\Sigma}=\phi_0$,
$N^\alpha\phi_\alpha=\phi_1$, where $N$ is the future-directed
unit normal to $\Sigma$.
\item
$\Box_g\phi=G(\phi)$
\item
The set $\mathcal{D}$ is maximal, i.e.~if $\mathcal{D}\subset\tilde{\mathcal{D}}$
is globally hyperbolic, with $\tilde\phi:\tilde{\mathcal{D}}\to {\mathbb R}$
satisfying the above, then $\tilde{\mathcal{D}}=\mathcal{D}$.
\end{enumerate}
We shall call $\mathcal{D}$ the \emph{maximal domain of existence}
of the solution.
Moreover, if $\Sigma$ is spherically symmetric,
and $\phi_0$ and $\phi_1$ are constant
on the $SO(3)$ orbits, then $\mathcal{D}$ is spherically
symmetric and $\phi$ is constant on the $SO(3)$ orbits,
and thus descends to a function on $\mathcal{D}_{\mathcal{Q}}
=\mathcal{D}/SO(3)$.
\end{proposition}
In the spherically symmetric case, the regularity assumption
in the statement can be lowered to $G\in C^0_{loc}$,
$\phi_0\in C^1_{loc}$, $\phi_1\in C^0_{loc}$.
The resulting solution $\phi$ is then $C^1$. In a slight abuse
of notation, we will denote $\mathcal{D}_{\mathcal{Q}}$ again as
$\mathcal{D}$. It turns out that
blowup can be easily characterized in terms of the $L^\infty$ norm
of $\phi$,
i.e.~we have
\begin{proposition}
\label{ext}
Let $q\in\mathcal{Q}\cap\overline{\mathcal{D}}$. If
$J^-(q)\cap \mathcal{D}\ne\emptyset$, and
\[
\sup_{x\in J^-(q)\cap\mathcal{D}}|\phi(x)|<\infty,
\]
then $q\in \mathcal{D}$.
\end{proposition}
A similar statement can be made about the past.
The proofs of the above proposition are completely standard and are thus omitted.

\section{The main theorem}
The main theorem of this paper is the following:
\begin{theorem}
\label{main}
Let $\Sigma$ be a spherically symmetric Cauchy surface
in a Reissner-Nordstr\"om spacetime $(\mathcal{M},g)$, with parameters
$0\le e<M$, and let $\phi_0$, $\phi_1$ be $C^1$, $C^0$ functions, respectively, on $\Sigma$,
constant
on the $SO(3)$ orbits, and supported in $\Sigma\cap\{r\le R\}$, for some $R<\infty$.
Let $\mathcal{D}$ denote the maximal domain of existence for
the initial value problem defined by Proposition~\ref{IVPstatement},
where $G$ is assumed to satisfy $(\ref{satisfies})$ with $p>4$.
Then there exists an $\epsilon$, depending
only on $M$, $e$, $\Sigma$, and $R$, such that
for
\[
|\phi_0|_{C^1}<\epsilon,\footnote{The norm above  refers to the induced Riemannian
metric on $\Sigma$.}
\]
\[
|\phi_1|_{C^0}<\epsilon,
\]
$\mathcal{D}$ contains the closures of the two
domains of outer communications,
\[
\overline{(J^-(\mathcal{I}^+_A)\cap
J^+(\mathcal{I}^-_A))\cup (J^-(\mathcal{I}^+_B)\cap J^+(\mathcal{I}^-_B))}
\subset\mathcal{D}.
\]
Moreover,
\[
|\phi(u,v)|\le C (\max\{v,1\})^{-1},
\]
and, for each fixed $r_0>r_+$,
\[
|\phi(r,t)|\le C(r_0)(|t|+1)^{-1}\qquad r\ge r_0,
\]
where $v$ and $t$ denote an Eddington-Finkelstein advanced time coordinate
and a static time coordinate, respectively,
on either of the two domains of outer communications.
\end{theorem}

\section{Proof of Theorem~\ref{main}: The bootstrap}
In view of simple arguments of Cauchy stability, and the fact that the
data are compactly supported on $\Sigma$, it follows, that given any
$t_0$, say $t=1$, then
for $\epsilon$ sufficiently small, we have
\[
\{t=1\}\subset \mathcal{D}
\]
and
\begin{equation}
\label{newsmallness1}
|\phi|_{t=1}|_{C^1}\le \tilde{\epsilon},
\end{equation}
\begin{equation}
\label{newsmallness2}
|N\phi|_{t=1}|_{C^0}\le\tilde{\epsilon}.
\end{equation}
Here again, the norms refer to the induced Riemannian metric of
$\{t=1\}$, and $N$ refers to a unit future directed normal.
The relation of $\mathcal{S}=\Sigma/SO(3)$ and $\{t=1\} $ could be
given, for instance, as follows:
\[
\input{ext.pstex_t}
\]
Note $\phi|_{t=1}$ will \emph{not} in general be compactly supported
in $(r_+,\infty)$. It will be supported in $(r_+, \bar R]$ for some $\bar R<\infty$.
Moreover, we have $\tilde{\epsilon}=\tilde{\epsilon}(\epsilon,\Sigma, \bar R, M, e)\to0$,
as $\epsilon\to0$.

In proving our theorem, it clearly suffices to show
that
\[
\mathcal{D}\supset J^+(\Sigma)\cap \overline{J^-(\mathcal{I}^+_A)\cap J^+(\mathcal{I}^-_A)}.
\]
Define the
set $\mathcal{B}\subset \mathcal{D}\cap J^-(\mathcal{I}^+_A)\cap J^+(\mathcal{I}^-_A)$
by
\begin{equation}
\label{Tdef}
\mathcal{B}=\left\{q\in \mathcal{D}\cap J^-(\mathcal{I}^+_A)\cap J^+(\mathcal{I}^-_A)
:\sup_{x\in J^-(q)\cap\{t\ge 1\}\cap
\mathcal{D}}v_+
(x)|\phi(x)|\le B\right\}
\end{equation}
where $v_+$ denotes $\max\{v,1\}$,
and for a constant $B$ to be determined later.

This subset is clearly non-empty and open for
$B>\tilde\epsilon v_+(R,1)$.
In view of Proposition~\ref{ext}, proving Theorem~\ref{main} reduces to
showing that $\mathcal{B}$ is closed. This would follow immediately from
\begin{theorem}
\label{equiv}
For $q\in \mathcal{B}$, we have
\[
\sup_{x\in J^-(q)\cap\{t\ge 1\}\cap
\mathcal{D}}v_+(x)\left|\phi(x)\right|\le \frac B2.
\]
\end{theorem}

To prove this,
we shall view our non-linear problem in $\mathcal{B}$
as a solution of
\begin{equation}\label{eq:lin-sch}
\Box_g \psi =F,
\end{equation}
where $F$ is to satisfy the estimates arising from $(\ref{Tdef})$ if
$F$ is set equal to the non-linearity.
We thus turn in the next section to the study of this linear inhomogeneous equation.
The proof of Theorem~\ref{equiv} will then follow easily in Section~\ref{theproofof}.

\section{The linear problem}
First some formulas:
Written with respect to Regge-Wheeler coordinates,
equation
\eqref{eq:lin-sch} takes the form
\begin{equation}\label{eq:lin-reg}
(1-\mu)^{-1} \Big (\partial_t^2\psi - r^{-2}\partial_{r_*} (r^2 \partial_{r_*} \psi)\Big ) = F.
\end{equation}
Alternatively,  relative to the null coordinates $(u=t-r_*, v=t+r_*)$
equation \eqref{eq:lin-sch} can be written as a first order system
\begin{align}
&\partial_v (r\partial_u \psi) = \frac  {(1-\mu) }{2r} r\partial_v\psi + \frac{(1-\mu)}4 r F,
\label{eq:lin-u}\\
&\partial_u (r\partial_v \psi) = \frac  {(1-\mu) }{2r} r\partial_u\psi + \frac{(1-\mu)}4 r F
\label{eq:lin-v}
\end{align}
for the pair of functions $\theta=r\partial_v \psi$ and $\zeta=r\partial_u \psi$, or
by a single equation
\begin{equation}\label{eq:lin-psi}
\partial_v \partial_u  (r\psi) = -\left(M-\frac{e^2}r\right)
\frac  {(1-\mu) }{2r^2} \psi + \frac{(1-\mu)}4 r F.
\end{equation}
We also record the following form of equation \eqref{eq:lin-u}, crucial
to understanding the so-called \emph{red-shift effect}
\begin{equation}\label{eq:lin-red}
\partial_v \frac{ r \partial_u \psi}{1-\mu} + \frac {1}{r^2}
 \left(M-\frac{2e^2}r\right)
 \frac{r\partial_u \psi}{1-\mu}
=  \frac  {r\partial_v\psi}{2r} + \frac 14 r F.
\end{equation}
This effect has been discussed in~\cite{mi:mazi}.

From \eqref{eq:lin-reg} and \eqref{eq:lin-u}--\eqref{eq:lin-v}
we can derive the following energy estimates.
Given $(u_1,v_1)$, for $t\le (v_1+u_1)/2$,
define
\begin{eqnarray*}
E_t^{(u_1,v_1)}[\psi]&=&
\int_{t-u_1}^{-t+v_1}\big (|\partial_t \psi(t,r_*)|^2 +
|\partial_{r_*}\psi(t,r_*)|^2\big ) r^2 \,dr_*\\
&=&2 \int_{t-u_1}^{-t+v_1} \big (|r\partial_u \psi(t,r_*)|^2 +|r\partial_{v}\psi(t,r_*)|^2\big )  \,dr_*.
\end{eqnarray*}
For $t_1=(v_1+u_1)/2\ge t_2\ge t_3$, we have
the following energy identity:
\begin{align*}
E_{t_2}^{(u_1,v_1)}[\psi]&+\int_{2t_3-u_1}^{2t_2-u_1} |r\partial_v\psi(u_1,v)|^2\,dv+
\int_{2t_3-v_1}^{2t_2-v_1} |r\partial_u\psi(u,v_1)|^2\,du\\ &=
E_{t_3}^{(u_1,v_1)}[\psi]
+
\frac 12\int_{t_3}^{t_2} \int_{t-u_1}^{-t+v_1}
(1-\mu) F(t,r_*) \,\partial_t\psi(t,r_*)\, r^2\,dr_*\,dt.
\end{align*}
The estimate refers thus to the region depicted below:
\[
\input{eest.pstex_t}
\]
We also have the following characteristic energy
identity in the characteristic rectangle $[u_2,u_1]\times [v_2,v_1]$:
\begin{align*}
\int_{u_2}^{u_1}|r\partial_u\psi(u,v_1)|^2  du &+\int_{v_2}^{v_1}
|r\partial_{v}\psi(u_1,v)|^2  \,dv=\\
&\int_{u_2}^{u_1}|r\partial_u\psi(u,v_0)|^2   du +\int_{v_2}^{v_1}
|r\partial_{v}\psi(u_2,v)|^2 \,dv\\
&+\frac 14\int_{u_2}^{u_1} \int_{v_2}^{v_1}
(1-\mu) F(u,v) (\partial_u\psi(u,v) - \partial_v \psi(u,v))\, r^2\,dv\,du.
\end{align*}
The reader can refer to the diagram of Theorem~\ref{pricelaw1}.

\subsection{The homogeneous problem}
The linear homogeneous problem, i.e.~the case $F=0$,
was studied in~\cite{mi:mazi},
where pointwise decay estimates were proven.
In this section, we will present
these estimates in a form suitable for understanding the inhomogeneous problem
via Duhamel's principle.

First we have the following preliminary result:
\begin{theorem}
\label{pricelaw0}
Fix $r_0>r_+$, and let $(u_1,v_1)$ be an arbitrary point.
Let $t_0<(u_1+v_1)/2$, and consider the lighter-shaded region
\[
\mathcal{E}=J^-(r_1,t_1)\cap J^+(\{t=t_0\})\cap\{r\ge r_0\}
\]
depicted below\footnote{Note that despite the diagram,
we are not necessarily assuming that $r(u_1,v_1)\ge r_0$.}:
\[
\input{prelim.pstex_t}
\]
Let $\psi$ be a solution to
\[
\Box_g\psi=0
\]
on $J^-(u_1,v_1)\cap J^+(\{t=t_0\})$.
Let $(u_2,v_2)\in\mathcal{E}$.
Then there exists a constant $A$, depending only on $r_0$, $e$, and $M$,
such that
\begin{equation}
\label{easiest}
r|\psi(\tilde{u},v_2)|^2\le
A\left (r|\psi(u_2,v_2)|^2+E^{(\tilde{u},v_2)}
_{t_0}[\psi]\right)
\end{equation}
\begin{equation}
\label{easiest2}
r^3|\partial_v\psi(\tilde{u},v_2)|^2\le A
\left(r^3|\partial_v\psi(u_2,v_2)|^2
+ E^{(\tilde{u},v_2)}_{t_0}[\psi]\right),
\end{equation}
for any point $\tilde{u}\ge u_2$ with $(\tilde{u},v_2)\in \mathcal{E}$.

On the other hand, if $(u_3,v_3)$ is any point in
$J^-(u_1,v_1)\cap J^+(\{t=t_0\})$,
then
\begin{equation}\label{eq:mu}
\left(\frac{r|\partial_u\psi(u_3,\tilde{v})|}{1-\mu}\right)^2\le A\left(
e^{-\frac{r_+-r_-}{r^2(u_3,\tilde v)}(\tilde v-v_3)}
\left(\frac{r|\partial_u\psi(u_3,v_3)|}{1-\mu}\right)^2
+E^{(u_3,\tilde{v})}_{t_0}[\psi]\right)
 \end{equation}
for any $\tilde{v}\ge v_3$
such that $(u_3,\tilde{v})\in  J^-(r_1,t_1)\cap J^+(\{t=t_0\})$.

Finally, if
\[
C=\sup_{\mathcal{E}\cap \{r\ge r_0\}}
\left\{|r\psi|(r,t_0), |r^2\partial_v(r\psi)(r,t_0)|\right\}
\]
then
\begin{equation}\label{eq:nu}
|r\psi(r,t)|^2+ |r^2\partial_v\psi(r,t)|^2+|r^2\partial_v(r\psi)(r,t)|^2\le
A\left(C^2+E^{(u_1,v_1)}_{t_0}[\psi]\right),
\end{equation}
throughout $J^-(r_1,t_1)\cap\{r\ge r_0\}\cap \{t\ge t_0\}$.
\end{theorem}
\begin{proof}
Equations~\eqref{easiest}, \eqref{easiest2}, and \eqref{eq:nu}
can be easily deduced from the equations
\eqref{eq:lin-u}--\eqref{eq:lin-psi} and the energy
estimates of the previous section. For the estimate \eqref{eq:mu},
we write the homogeneous wave equation in the red-shift
from  given by \eqref{eq:lin-red}:
\[
\partial_v  \frac{r\partial_u\psi}{1-\mu} + \frac 1{r^2}
\left(M-\frac {e^2}r\right)  \frac{r\partial_u\psi}{1-\mu}=
\frac 1{2r}{r\partial_v\psi}.
\]
Introducing the integrating factor $e^{\int_{v_3}^v\frac 1{r^2}(M-\frac{e^2}r)dv}$
and integrating in $v$ we obtain
$$
\frac{r\partial_u\psi(u_3,\tilde v)}{1-\mu}=
e^{-\int_{v_3}^{\tilde v} \frac 1{r^2}(M-\frac{e^2}r)dv}
\,\,\frac{r\partial_u\psi(u_3,v_3)}{1-\mu}+
\int_{v_3}^{\tilde v}e^{-\int_{v}^{\tilde v} \frac 1{r^2}(M-\frac{e^2}r)dv'}
\,\,\frac 1{2r}{r\partial_v\psi}\, dv.
$$
Since
\begin{equation}
\label{surfgrav}
M-\frac {e^2}r\ge M-\frac {e^2}{r_+}=\frac{r_+-r_-}2>0,
\end{equation}
 and $r$ is increasing in the $v$-direction, we conclude
\begin{eqnarray*}
\frac{r|\partial_u\psi(u_3,\tilde v)|}{1-\mu}&\le&
e^{ -\frac{r_+-r_-}{2r^2(u_3,\tilde v)}(\tilde v-v_3)}
\,\,\frac{r|\partial_u\psi(u_3,v_3)|}{1-\mu}\\
&&\hbox{}+
\int_{v_3}^{\tilde v} e^{ -\frac{r_+-r_-}{2r^2(u_3,\tilde v)}(v-v_3)}
\,\,\frac 1{2r}{r|\partial_v\psi(u_3,v)|}\, dv .
\end{eqnarray*}
Applying Cauchy-Schwarz gives
$$
\frac{r|\partial_u\psi(u_3,\tilde v)|}{1-\mu}\le
e^{ -\frac{r_+-r_-}{2r^2(u_3,\tilde v)}(\tilde v-v_3)}
\,\,\frac{r|\partial_u\psi(u_3,v_3)|}{1-\mu}+ A
\big (\int_{v_3}^{\tilde v} \,{|r\partial_v\psi(u_3,v)|^2}\, dv \big)^{\frac 12}.
$$
\end{proof}
We now turn to decay estimates. The full results of~\cite{mi:mazi} are
complicated to state, as different regions must be treated separately.
In this paper we shall only use the following:
\begin{theorem}
\label{pricelaw1}
Let $(u_1, v_1)$, $(u_2,v_2)$ be points
with $(u_1,v_1)\in J^+(u_2,v_2)$, i.e.~with $u_1\ge u_2$, $v_1\ge v_2$.
Consider the characteristic rectangle
\[
\mathcal{R}= J^+(u_2,v_2)\cap J^-(u_1,v_1)
\]
depicted below:
\[
\input{price.pstex_t}
\]
Let $\psi$ be a solution to
$\Box_g\psi=0$ on $\mathcal{R}$.
Let
\begin{eqnarray*}
C=\max\left\{
\right.&\sup_{v_2\le v\le v_1}
|r\partial_v\psi(u_2,v)|+|r^2\partial_v(r\psi)(u_2,v)|,\\
&\hbox{\ }\left.
\sup_{u_2\le u\le u_1}
|(1-\mu)^{-1}\partial_u\psi(u,v_2)|\right\}.
\end{eqnarray*}
Then there exists a constant $A$ depending only on
$r_0=r(u_2,v_2)$, $e$ and $M$
such that
\[
\psi(u_1,v_1)\le AC(v_1-v_2)^{-1}.
\]
\end{theorem}

The above theorem concerns a characteristic initial value problem.
We can combine Theorems \ref{pricelaw0} and \ref{pricelaw1} to derive
decay estimates for solutions of the homogeneous Cauchy problem.
\begin{theorem}\label{pricelaw2}
Let $(u_1,v_1)$ be an arbitrary point, and
and let $t_0<(v_1+u_1)/2$.
Let $\psi$ be a solution to
$$
\Box_g\psi=0
$$
on $J^-(u_1,v_1)\cap J^+(\{t=t_0\})$.
Let
\begin{equation}
\label{C=}
C=\sup_{t=t_0}
\{(1-\mu)^{-\frac 12}r^2|\partial_t\psi|,(1-\mu)^{\frac 12}r^2|\partial_r\psi|,
r^2|\partial_v (r\psi)|, r|\psi|\}.
\end{equation}
Pick a point $(u_2,v_2)\in J^-(u_1,v_1)\cap\{t=t_0\}$:
\[
\input{homog.pstex_t}
\]
It follows that
\begin{equation}
\label{yIldIz}
|\psi(u_1,v_1)|\le AC(v_1-v_2)^{-1},
\end{equation}
where $A$ depends only on $r_0=r(u_2,v_2)$, $e$ and $M$.
\end{theorem}
\begin{proof}
We first show that $E^{(u_1,v_1)}_{t_0}[\psi]\le A C^2$. Note that
$\partial_{r_*}\psi=(1-\mu)\partial_r\psi$.
\begin{align*}
\left(E^{(u_1,v_1)}_{t_0}[\psi]\right)^2
=&\int_{t_0-u_1}^{-t_0+v_1} r^2 (|\partial_t\psi|^2+|\partial_{r_*}\psi|^2)\, dr_*\\
=&\int_{ r(r_*=t_0-u_1)}^{r(r_*=-t_0+v_1)}
(1-\mu)^{-1} r^2 (|\partial_t\psi|^2+|\partial_{r_*}\psi|^2)\,dr
\\ \le& \ C^2\int_{r_+}^{\infty} r^{-2}\, dr
 \le C^2 r_+^{-1}.
\end{align*}

We apply now Theorem~\ref{pricelaw0}.
From~\eqref{eq:mu} and~\eqref{C=}, the estimate
\begin{eqnarray}
\label{puttogether}
\nonumber
|(1-\mu)^{-1} \partial_u\psi(u,v_2)|&\le& e^{-\frac{(r_+-r_-)}{r^2(u,v_2)}(v_2-v_0(u))}
|(1-\mu)^{-1} \partial_u\psi(u,v_0(u))|+ AC\\
				&\le& e^{-\frac{(r_+-r_-)}{r^2(u,v_2)}(v_2-v_0(u))}
(1-\mu)^{-\frac12}C+AC
\end{eqnarray}
follows for
$u_2\le u\le u_1$, where $v_0(u)$ denotes the $v$-coordinate of the
intersection of the outgoing constant-$u$ ray with $t=t_0$.
Since,
$$
v_2-v_0=2(r_{0*}-r_*(u,v_2)),
$$
we have by~\eqref{behaviour}, for sufficiently small $r_*=r_*(u,v_2)$, $r=r(u,v_2)$:
\begin{eqnarray}
\label{puttogether2}
\nonumber
(1-\mu(r_*))^{-\frac12}e^{-\frac{r_+-r_-}{2r^2}(v_2-v_0)}
	&\le&
			A(1-\mu(r_*))^{-\frac12}e^{\frac{r_+-r_-}{r^2}r_*}\\
	\nonumber
	&\le& A	 e^{-\frac{r_+-r_-}{2r_+^2}r_*+\frac{r_+-r_-}{r^2}r_*}\\
	\nonumber
	& = & A	 e^{-\frac{(r-\sqrt{2}r_+)(r+\sqrt{2}r_+)(r_+-r_-)}{2r_+^2r^2}r_*}\\
	\nonumber
	&\le& A e^{-r_*(r-\sqrt{2}r_+)}\\
	&\le& A.
\end{eqnarray}
(Note that by convention here, $A$ does not always represent the same constant.)
Since the above estimate is also clearly true for large $r$, putting together
\eqref{puttogether} and \eqref{puttogether2} yields
\begin{equation}\label{eq:mu-1}
\sup_{u_2\le u\le u_1} (1-\mu)^{-1} |\partial_u\psi(u,v_2)|\le AC.
\end{equation}

On the other hand, from~\eqref{eq:nu},
we have
\[
|\partial_v\psi(u_2,v)|\le ACr^{-1},
\]
\[
|\partial_v(r\psi)(u_2,v)|\le ACr^{-2},
\]
for $v_1\le v\le v_2$, in view
of the fact that $r\ge r_0$ for such $(u_0,v)$. Our result now
follows immediately from Theorem~\ref{pricelaw1}.
\end{proof}
\begin{remark}
Observe that the constant $C$ in the above theorem is dominated by
the (Riemannian)
$C^1$ norm of $r^3\psi$,
$$
C\le |r^3\psi|_{t=t_0}|_{C^1}.
$$
\end{remark}
\begin{remark}
The quantity ${\partial_{\bar{u}} \psi}/{\partial_{\bar{u}} r}$
is clearly independent of the choice of retarded time $\bar{u}$ and
coincides with $-(1-\mu)^{-1} \partial_u \psi$
for Eddington-Finkelstein $u$. As one can choose a
global regular null coordinate system,
such that $\partial_{\bar{u}} r\le -1$ along the ray $u=u_2$ in a
neighborhood of $\mathcal{H}^+_A\cap\{u=u_2\}$,
one could have alternatively derived
\eqref{eq:mu-1} by appealing to Cauchy stability.
\end{remark}

\subsection{The inhomogeneous problem}
To prove Theorem~\ref{equiv},
we must derive estimates for solutions of the equation
\begin{equation}
\label{savgrammikn}
\Box_g\Psi=F,
\end{equation}
with vanishing initial data at $t=1$.
Let
$\psi(r,t;s)$
denote the solution of
$\Box_g\psi=0$
evaluated at $(r,t)$
with
initial condition
\begin{equation}
\label{ic}
\psi(\cdot,s)=0,\qquad
\partial_t \psi(\cdot,s)=(1-\mu)F(\cdot,s).
\end{equation}
Using representation \eqref{eq:lin-sch} we can
then write the solution of $(\ref{savgrammikn})$ as
\begin{equation}
\label{integral}
\Psi(r,t)=\int_1^t\psi(r,t;s)ds.
\end{equation}
We prove the following theorem:
\begin{theorem}\label{pricelaw4}
Let $(u_1,v_1)$ be given,
let $\Psi$ satisfy $(\ref{savgrammikn})$ on $J^-(u_1,v_1)\cap
\{t\ge 1\}$, with $\Psi=0$, $\partial_t\Psi=0$ on $J^-(u_1,v_1)\cap\{t=1\}$,
and let $\alpha>4$.
Then there exists a constant $A$, depending
only on $M$, $e$, and $\alpha$, such that
\begin{equation}\label{eq:Psi-total-}
|\Psi(u_1,v_1)| \le  A  (v_1)^{-1}
\sup_{J^-(r_1,t_1)\cap\{t\ge 1\}}
 v_+^\alpha|F(u,v)|,
\end{equation}
where $v_+=\max (v,1)$.
\end{theorem}
\begin{proof}
Let $r_1=r(u_1,v_1)$, $t_1=(v_1+u_1)/2$.
We will fix some $r_0>r_+$, to be determined later,
and distinguish the cases $r_1\ge r_0$ and $r_1<r_0$.

\subsubsection{Estimates in the region $r\ge r_0$}
Consider first the case $r_1\ge r_0$.
We have $v_1=t_1+r_1^*\ge t_1+r_0^*$.

Let us take $s> 2v_1^\omega$, for $0\le \omega<1$.
To compute $\psi(r_1,t_1;s)$, we decompose
\begin{align*}
J^-(r_1,t_1)\cap \{t=s\} &= \{(u,v): \,\, u=2s-v,\,\,v_1-2(t_1-s)\le v\le v_1\}\\
&=I_1\cup I_2
\end{align*}
as depicted:
\[
\input{far.pstex_t}
\]
i.e.~we have
\begin{align*}
I_1&= \{(u,v): \,\, u=2s-v,\,\,v_1-2(t_1-s)\le v\le v_1^\omega\},\\
I_2&= \{(u,v): \,\, u=2s-v,\,\,v_1^\omega\le v\le v_1\}.
\end{align*}
Note that on the set $I_1$,
$$
2r_*= v-u = 2(v-s)  \le  - 2 v_1^\omega
$$
and therefore, by $(\ref{behaviour})$, we have
$$
1-\mu(r,s)\le \hat{C}_2 e^{-\frac{(r_+-r_-)}{r_+^2}v_1^\omega},\qquad
(r,s)\in I_1.
$$
By the above estimate, \eqref{easiest} and \eqref{ic}, we have
\begin{align}
\label{above-est}
\nonumber
|\psi(r_1,t_1;s) |^2&\le
 A^2\left(E_s[\psi(\cdot,\cdot;s)]\right)^2\\
 \nonumber
 &= A^2 \int_{I_1\cup I_2}
 (1-\mu) |F(r_*,s)|^2 r^2\, dr\\ &\le
 \tilde{A}^2\left(e^{-\frac{(r_+-r_-)}{r_+^2}v_1^\omega} \sup_{I_1} |F(r_*,s)|^2 +
 v_1^3 \sup_{I_2} |F(r_*,s)|^2\right),
\end{align}
where in the last inequality we used that $r\le C'v_1$ on $I_2$.

\vskip 2pc

On the other hand, for $s\le 2v_1^\omega$ we can apply
Theorem \ref{pricelaw2} for the points
$(r_1,t_1)=(u_1,v_1)$ and $(r_0,s)=(u_2,v_2)$, where
\[
v_2=s+ r_0^*\le 2v_1^\omega.
\]
(In the last inequality, we assumed that $r_0$ is chosen in particular so that
$r_{0*}\le 0$.)  Then,
\begin{align*}
|\psi(r_1,t_1;s)|&\le A v_1^{-1} \sup_r (1-\mu)^{\frac 12} r^3 |F(r,s)|
\\ &\le A v_1^{-1} \sup_{r^*\le -\frac s2} (1-\mu)^{\frac 12} r^3 |F(r,s)| +
A v_1^{-1} \sup_{r^*\ge -\frac s2} (1-\mu)^{\frac 12} r^3 |F(r,s)|.
\end{align*}
For points $r_*\le -\frac s2$, estimate \eqref{behaviour} gives
$$
1-\mu\le \hat{C}_2e^{-\frac{(r_+-r_-)}{2r_+^2} s},
$$
while for $r_*\ge -\frac s2$,
we clearly have
$$
v=r_*+s\ge \frac s2.
$$
Thus, for $\delta>0$,
\begin{eqnarray}
\label{taalla}
\nonumber
|\psi(r_1,t_1;s)|
&\le& A v_1^{-1} e^{-\frac{(r_+-r_-)}{2r_+^2} s} \sup_{r_*\le -\frac s2} r^3 |F(r,s)|\\
&&\hbox{} +
A v_1^{-1} s^{-1-\delta}\sup_{r_*\ge -\frac s2} v^{1+\delta} r^3 |F(r,s)|.
\end{eqnarray}

Applying $(\ref{integral})$, and the estimates $(\ref{above-est})$ and $(\ref{taalla})$,
we obtain
\begin{align*}
|\Psi(r_1,t_1)|\le& t_1A e^{-\frac{(r_+-r_-)}{2r_+^2}v_1^\omega} \sup_{J^-(r_1,t_1)}
|F(r,s)| +A  t_1v_1^{\frac 32} \sup_{J^-(r_1,t_1)\cap \{v\ge v_1^\omega\}}
|F(u,v)| \\ &+ A v_1^{-1} \sup_{J^-(r_1,t_1)}  v_+^{1+\delta} r^3 |F(r,s)|\\
\le& (v_1)_+^{\frac 52} \sup_{J^-(r_1,t_1)}
|F(u,v)| +(v_1)_+^{-1} \sup_{J^-(r_1,t_1)}  v_+^{1+\delta} r^3 |F(r,s)|,
\end{align*}
where we have used the fact that for $r_1\ge r_0>r_+$,
we have $(v_1)_+ \ge\sim  t_1$, with constant depending on $r_0$.

Clearly, there exists a constant $H$, depending only on $M$ and $e$, such that for
$r\ge H$,
we have $v=t+r_*\ge r $ in the region $t\ge 1$. It follows immediately that for all $r$, we have
$r\le H +|v|$ in the region $t\ge 1$. We obtain, for any $\alpha\ge 0$,
\begin{align}
|\Psi(r_1,t_1)|\le&
A  (v_1)_+^{\frac 52-\omega\alpha} \sup_{J^-(r_1,t_1)\cap \{v\ge v_1^\omega\}}
 v_+^\alpha |F(u,v)|\nonumber \\
 &+
A(v_1)_+^{-1} \sup_{J^-(r_1,t_1)}  v_+^{4+\delta} |F(u,v)|\\
\le&
A  (v_1)_+^{\frac 52-\omega\alpha} \sup_{J^-(r_1,t_1)}
 v_+^\alpha |F(u,v)|\nonumber \\
 &+
A (v_1)_+^{-1} \sup_{J^-(r_1,t_1)}  v_+^{4+\delta} |F(u,v)|.\nonumber
\end{align}
Choosing
$$
\alpha=4+\delta,\qquad \omega=\frac 7{2(4+\delta)}
$$
we obtain
\begin{equation}\label{eq:Psi-big}
|\Psi(r_1,t_1)|\le A (v_1)_+^{-1} \sup_{J^-(r_1,t_1)}  v_+^\alpha |F(u,v)|.
\end{equation}
\subsubsection{Estimates in the region $r<r_0$.}
\label{redshiftsec}
Consider now the case $r_+<r_1< r_0$. In this region we shall use
the
\emph{red-shift effect}
manifest\footnote{See~\cite{mi:mazi}. The point, as we shall see,
is that we can extract decay in $v$ for $\partial_u\Psi$, even though
we are integrating in $v$ towards the future.}
in the equation \eqref{eq:lin-red}:
\[
\partial_v \frac{ r \partial_u \Psi}{1-\mu} +\frac 1{r^2}\left(M-\frac{e^2}r\right)
\frac{r\partial_u \Psi}{1-\mu}
=  \frac  {r\partial_v\Psi}{2r} + \frac 14 r F.
\]

For fixed $u$, let $v_*(u)$ be defined by $t(u,v_*(u))=1$.
Using the integrating factor $e^{\int_{v_*(u)}^v r^{-2}(M-{e^2}r^{-1})\, dv}$
and the vanishing
of the initial data on $\{t=1\}$, we obtain
\[
\frac{ r \partial_u \Psi (u,v)}{1-\mu} =
\frac 14 \int_{v_*(u)}^v
e^{-\int_{v'}^v r^{-2}(M-{e^2}r^{-1})} \Big (2\partial_v \Psi(u,v') + r F(u,v')\Big )\, dv'
\]
Integrating by parts we derive
\begin{align*}
&\frac{ r \partial_u \Psi (u,v)}{1-\mu} =  \frac 12 \Psi (u,v)\\
&+
\int_{v_*(u)}^v
e^{-\int_{v'}^v r^{-2}(M-{e^2}r^{-1})} \Big (-2r^{-2}(M-e^2r^{-1})
\Psi(u,v') + r F(u,v')\Big )\, dv'.
\end{align*}
Since in the region under consideration, we have the bound
$r_+< r\le r_0$, and also,
\[
M-\frac{e^2}r>M-\frac{e^2}{M+\sqrt{M^2-e^2}}=\sqrt{M^2-e^2}>0,
\]
it follows
that
$$
\frac{ |\partial_u \Psi (u,v)|}{1-\mu} \le   A|\Psi (u,v)| + A\int_{v_*(u)}^v
e^{-c(v-v')} \Big (|\Psi(u,v')| + | F(u,v')|\Big )\, dv'.
$$
Given $v$, let $u_*(v)$ be defined by $r(u_*(v),v)=r_0$. Then
\begin{align*}
|\Psi (u,v)|&\le |\Psi(u_*(v),v)|  + A\int_{u_*(v)}^u
(1-\mu(u',v)) |\Psi(u',v)|\, du' \\ &+  A\int_{u_*(v)}^u\int_{v_*(u')}^v
e^{-c(v-v')} \Big (|\Psi(u',v')| + | F(u',v')|\Big )\, dv' (1-\mu(u',v))\,du'.
\end{align*}
We now observe that in the region $r_+<r\le r_0$,
we can estimate
$|1-\mu |< \delta$ for some $\delta$ that can be made
arbitrarily small provided that $r_0$ is sufficiently
close to $r_+$.  Moreover,
\[
\int_{u_*(v)}^u |1-\mu(u',v)|\, du' = r(u,v) - r_0 <\delta.
\]
Therefore, for any $u\ge u_*(v)$,
\begin{eqnarray}\label{eq:Psi}
 |\Psi(u,v)|&\le& |\Psi(u_*(v),v)| +
 A\delta\sup_{u'\in [u_*(v),u]}{\Psi(u',v)}\\
  \nonumber
 &&\hbox{}+
 A\delta
 \sup_{u'\in [u_*(v),u]}\int_{v_*(u')}^v
e^{-c(v-v')} \Big (|\Psi(u',v')| + | F(u',v')|\Big )\, dv'
\end{eqnarray}
Split
\begin{align*}
\int_{v_*(u')}^v
e^{-c(v-v')} &\Big (|\psi(u',v')| + | F(u',v')|\Big )\, dv'= \\
&\int_{v_*(u')}^{v/2}
e^{-c(v-v')} \Big (|\psi(u',v')| + | F(u',v')|\Big )\, dv'\\ &+
\int_{v/2}^v
e^{-c(v-v')} \Big (|\psi(u',v')| + | F(u',v')|\Big )\, dv'
\end{align*}
and estimate
\begin{align*}
\int_{v_*(u')}^{v/2}
e^{-c(v-v')}& \Big (|\Psi(u',v')| + | F(u',v')|\Big )\, dv'\\
&\le
A e^{-c v/2} \sup_{v'\in [v_*(u'),v]  }\Big (|\Psi(u',v')| + | F(u',v')|\Big ),\\
\int_{v/2}^v
e^{-c(v-v')}& \Big (|\Psi(u',v')| + | F(u',v')|\Big )\, dv'\\
&\le
A \sup_{v'\in [v/2,v] }\Big (|\Psi(u',v')| + | F(u',v')|\Big ).
\end{align*}
Returning to \eqref{eq:Psi}, we derive
\begin{align*}
 |\Psi(u,v)| \le&
 |\Psi(u_*(v),v)| +
 A\delta\sup_{u'\in [u_*(v),u]}{\Psi(u',v)}\\
&+
 A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v/2,v]  }\Big (|\Psi(u',v')|+ | F(u',v')|\Big )\\
 &+A\delta  e^{-c v}
\sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }\Big (|\Psi(u',v')| + | F(u',v')|\Big ).
\end{align*}
Thus,
\begin{align*}
|v_+^\beta\Psi(u,v)|
\le&
 v_+^\beta|\Psi(u_*(v),v)| +
 A\delta v_+^\beta\sup_{u'\in [u_*(v),u]}{\Psi(u',v)}\\
 &
 +v_+^\beta A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v/2,v]  }|\Psi(u',v')|\\
&+Av_+^\beta \delta  e^{-c v}
\sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }|\Psi(u',v')|\\
&+v_+^\beta A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v/2,v]  }| F(u',v')|\\
&+Av_+^\beta \delta  e^{-c v}
\sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  } | F(u',v')|\\
\le&
 v_+^\beta|\Psi(u_*(v),v)|
+ A\delta v_+^\beta\sup_{u'\in [u_*(v),u]}{\Psi(u',v)}\\
 &+ A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v/2,v]  }|(v'_+)^\beta\Psi(u',v')|\\
&+A \delta
\sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }|\Psi(u',v')|\\
&+ A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v/2,v]  }(v'_+)^\beta| F(u',v')|\\
 &+A \delta
\sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }| F(u',v')|\\
\le&
 v_+^\beta|\Psi(u_*(v),v)|
  +A\delta v_+^\beta\sup_{u'\in [u_*(v),u]}{\Psi(u',v)}\\
&+  2A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }|(v'_+)^\beta\Psi(u',v')|\\
&+
2A\delta \sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }(v'_+)^\beta| F(u',v')|.
\end{align*}
The constant $A$ is independent of the choice of $r_0$ and thus
of $\delta$.
\emph{A fortiori}, the above estimate applies
also when $|v_+^\beta\Psi(u,v)|$ on the left hand side is replaced by
$\sup_{u'\in [u_*(v),u]}
\sup_{v'\in [v_*(u'),v]  }|(v'_+)^\beta\Psi(u',v')|$. Thus, if
$r_0$ is chosen close enough to $r_+$, so that $\delta$ is sufficiently
small, we have at our point $(r_1,t_1)$ the estimate
\begin{align}
 |\Psi(r_1,t_1)| \le 2
(v_1)_+^{-\beta} \Big (& \sup_{v\in [v^*,v_1]  } v_+^\beta |\Psi(u_*(v),v)|
\nonumber\\ &+ \sup_{u\in [u_*(v_1),u_1]}
\sup_{v\in [v_*(u),v_1]}   v_+^\beta | F(u,v)|\Big ),\label{eq:Psi-les}
\end{align}
for any $\beta\ge 0$.
Here $v^*$ denotes the $v$-coordinate of the intersection of the curve
$r=r_0$ and $\{t=1\}$. Observe that the first term on the right hand side of
the above inequality can be rewritten as
$\sup_{1\le t\le t_1 } v_+^\beta|\Psi(r_0,t)|$;
the supremum is thus taken over $\Psi$ evaluated at points
in a subset of the
region
$r\ge r_0$. We choose
\[
\beta=1.
\]

From \eqref{eq:Psi-big} and \eqref{eq:Psi-les}, we infer now that
\begin{equation}\label{eq:Psi-total}
|\Psi(r_1,t_1)| \le  A  (v_1)_+^{-1}
 \max_{J^-(r_1,t_1)}  v_+^\alpha |F(u,v)|,
\end{equation}
for \emph{all} $(r_1,t_1)$.
\end{proof}

\section{Proof of Theorem~\ref{equiv}}
\label{theproofof}
Recall the set $\mathcal{B}$ and the constant $B$ defined by \eqref{Tdef}.
Let $(r_1,t_1)=(u_1,v_1)\in \mathcal{B}$.
Set $F=G(\phi)$ and consider
$\Psi$ a solution of $(\ref{savgrammikn})$
in $J^-(u_1,v_1)$ with vanishing data at $t=1$,
and let $\psi$ be a solution of the homogeneous equation with
data
\[
\psi|_{J^-(u_1,v_1)\cap \{t=1\}}=\phi|_{J^-(u_1,v_1)\cap \{t=1\}},
\]
\[
\partial_t\psi|_{J^-(u_1,v_1)\cap \{t=1\}}=\partial_t\phi|_{J^-(u_1,v_1)\cap \{t=1\}}.
\]
We have that
\begin{equation}
\label{a9roisma}
\phi=\psi+\Psi
\end{equation}
in $J^-(u_1,v_1)\cap J^+(\{t=1\})$.

By Theorems~\ref{pricelaw0} and~\ref{pricelaw1},
it follows that
\begin{equation}
\label{homogeneouspsi}
|\psi(u_1,v_1)| \le E (v_1)_+^{-1},
\end{equation}
where $E=E(\epsilon,M,e,\Sigma, R)\to0$ as $\epsilon\to0$.
If $0\le B<c$, then
by assumption~\eqref{satisfies},
\[
|G(\phi(u,v))|\le K|\phi(u,v)|^p, \forall (u,v)\in \mathcal{B}
\]
Thus, by the definition $(\ref{Tdef})$
of the set $\mathcal{B}$, we have
\[
|F(u,v)|\le KB^pv_+^{-p}.
\]
Therefore for $p> 4$, choosing $\alpha=p$ and applying Theorem~\ref{pricelaw4},
we have
\begin{equation}
\label{proproteleutaio}
|\Psi(u_1,v_1)|\le A(v_1)_+^{-1}KB^p
\end{equation}
and thus, by $(\ref{a9roisma})$ and $(\ref{homogeneouspsi})$,
\begin{equation}
\label{proteleutaio}
|\psi(u_1,v_1)|\le A(v_1)_+^{-1}KB^p+E(v_1)_+^{-1}.
\end{equation}
Let us define $B= 4E$,
and let us require $E$ (and thus $\epsilon$) to
be sufficiently small
so that $AKB^p< \frac B4$.
Inequality $(\ref{proteleutaio})$ then yields
\[
|\psi(u_1,v_1)|\le \frac B2(v_1)_+^{-1},
\]
as required.

\section{Acknowledgements}
M.~D.~is supported in part by NSF grant DMS-0302748. I.~R.~is
supported in part by NSF grant DMS-0406627. Part of this research
was conducted when I.~R.~was a long-term prize fellow
of the Clay Mathematics Institute.

redpajama_set_name: RedPajamaArXiv


text:
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