A Standard Integral Equation












3












$begingroup$


Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?










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$endgroup$








  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    yesterday










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    yesterday
















3












$begingroup$


Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    yesterday










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    yesterday














3












3








3





$begingroup$


Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?










share|cite|improve this question











$endgroup$




Consider the integral equation



$$phi(x) = x + lambdaint_0^1 phi(s),ds$$



Integrating with respect to $x$ from $x=0$ to $x=1$:



$$int_0^1 phi(x),dx = int_0^1x,dx + lambda int_0^1Big[int_0^1phi(s),dsBig],dx$$



which is equivalent to



$$int_0^1 phi(x),dx = frac{1}{2} + lambda int_0^1phi(s),ds$$



How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?







linear-algebra integration matrix-equations






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share|cite|improve this question













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edited yesterday







LightningStrike

















asked yesterday









LightningStrikeLightningStrike

455




455








  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    yesterday










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    yesterday














  • 1




    $begingroup$
    What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
    $endgroup$
    – James
    yesterday










  • $begingroup$
    My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
    $endgroup$
    – LightningStrike
    yesterday








1




1




$begingroup$
What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
$endgroup$
– James
yesterday




$begingroup$
What is $lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $phi$?
$endgroup$
– James
yesterday












$begingroup$
My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
$endgroup$
– LightningStrike
yesterday




$begingroup$
My apologies, $lambda$ is an arbitrary constant. In essence I want to obtain an expression of $phi(x)$ which does not contain a function of $s$, which the initial integral equation has.
$endgroup$
– LightningStrike
yesterday










4 Answers
4






active

oldest

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4












$begingroup$

Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
    Putting into FE yields:




    $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
    If $lambda=0$ then $phi(x)=x$



    if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



    If $lambda=1$ there won’t besuch $phi$.







    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
      $$
      phi(x) = sum_{n geq 0} a_{n} x^{n}.
      $$

      Substituting it into your equation, we get:
      $$
      sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
      $$

      Matching up the coefficients of the difference powers of $x$, we get:
      $$
      a_{n} = 0 quad mbox{ for } n geq 2,
      $$

      $$
      a_{1} = 1,
      $$

      and
      $$
      a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
      $$

      This gives a relationship between $a_{0}$ and $lambda$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






        share|cite|improve this answer









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          4 Answers
          4






          active

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          4 Answers
          4






          active

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          active

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          active

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          4












          $begingroup$

          Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



          Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






          share|cite|improve this answer









          $endgroup$


















            4












            $begingroup$

            Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



            Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






            share|cite|improve this answer









            $endgroup$
















              4












              4








              4





              $begingroup$

              Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



              Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$






              share|cite|improve this answer









              $endgroup$



              Relabelling the dummy variable $xmapsto s$ on the LHS of your final equation, $$int_0^1phi(s),ds-lambdaint_0^1phi(s),ds=frac12\implies int_0^1phi(s),ds=frac1{2(1-lambda)}$$



              Thus $$phi(x)=x+fraclambda{2(1-lambda)}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered yesterday









              John DoeJohn Doe

              11.3k11239




              11.3k11239























                  2












                  $begingroup$

                  Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                  Putting into FE yields:




                  $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                  If $lambda=0$ then $phi(x)=x$



                  if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                  If $lambda=1$ there won’t besuch $phi$.







                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                    Putting into FE yields:




                    $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                    If $lambda=0$ then $phi(x)=x$



                    if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                    If $lambda=1$ there won’t besuch $phi$.







                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                      Putting into FE yields:




                      $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                      If $lambda=0$ then $phi(x)=x$



                      if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                      If $lambda=1$ there won’t besuch $phi$.







                      share|cite|improve this answer









                      $endgroup$



                      Note $int_{0}^{1}{phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                      Putting into FE yields:




                      $$x+alambda=x+lambdaint_{0}^{1}{(s+alambda )ds }iff alambda=lambdabig(frac{1}{2}+lambda abig)$$
                      If $lambda=0$ then $phi(x)=x$



                      if $lambdane 1$ $a=frac{1}{2}+lambda aiff ( 1-lambda)a=frac{1}{2}iff a=frac{1}{2-2lambda}$ and then $phi(x)=x+frac{lambda}{2-2lambda}$



                      If $lambda=1$ there won’t besuch $phi$.








                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered yesterday









                      HAMIDINE SOUMAREHAMIDINE SOUMARE

                      1




                      1























                          0












                          $begingroup$

                          If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                          $$
                          phi(x) = sum_{n geq 0} a_{n} x^{n}.
                          $$

                          Substituting it into your equation, we get:
                          $$
                          sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                          $$

                          Matching up the coefficients of the difference powers of $x$, we get:
                          $$
                          a_{n} = 0 quad mbox{ for } n geq 2,
                          $$

                          $$
                          a_{1} = 1,
                          $$

                          and
                          $$
                          a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                          $$

                          This gives a relationship between $a_{0}$ and $lambda$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                            $$
                            phi(x) = sum_{n geq 0} a_{n} x^{n}.
                            $$

                            Substituting it into your equation, we get:
                            $$
                            sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                            $$

                            Matching up the coefficients of the difference powers of $x$, we get:
                            $$
                            a_{n} = 0 quad mbox{ for } n geq 2,
                            $$

                            $$
                            a_{1} = 1,
                            $$

                            and
                            $$
                            a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                            $$

                            This gives a relationship between $a_{0}$ and $lambda$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                              $$
                              phi(x) = sum_{n geq 0} a_{n} x^{n}.
                              $$

                              Substituting it into your equation, we get:
                              $$
                              sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                              $$

                              Matching up the coefficients of the difference powers of $x$, we get:
                              $$
                              a_{n} = 0 quad mbox{ for } n geq 2,
                              $$

                              $$
                              a_{1} = 1,
                              $$

                              and
                              $$
                              a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                              $$

                              This gives a relationship between $a_{0}$ and $lambda$.






                              share|cite|improve this answer









                              $endgroup$



                              If you are after finding $phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
                              $$
                              phi(x) = sum_{n geq 0} a_{n} x^{n}.
                              $$

                              Substituting it into your equation, we get:
                              $$
                              sum_{n geq 0} a_{n} x^{n} = x + lambda sum_{n geq 0}{a_{n} over n+1}.
                              $$

                              Matching up the coefficients of the difference powers of $x$, we get:
                              $$
                              a_{n} = 0 quad mbox{ for } n geq 2,
                              $$

                              $$
                              a_{1} = 1,
                              $$

                              and
                              $$
                              a_{0} = lambda left(a_{0} + {a_{1} over 2}right).
                              $$

                              This gives a relationship between $a_{0}$ and $lambda$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered yesterday









                              avsavs

                              3,764514




                              3,764514























                                  0












                                  $begingroup$

                                  Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Note that since $lambdaint_0^1 phi(s),ds$ is a constant (with respect to $x$), then we can write$$phi(x)=x+a$$and by substitution we conclude that $$x+a=x+lambdaint _{0}^{1}x+adximplies\a=lambda({1over 2}+a)implies\a={lambdaover 2-2lambda}$$ and we obtain$$phi(x)=x+{lambdaover 2-2lambda}quad,quad lambdane 1$$The case $lambda=1$ leads to no solution.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered yesterday









                                      Mostafa AyazMostafa Ayaz

                                      17.8k31039




                                      17.8k31039






























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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029