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?






linear-algebra integration matrix-equations







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
    7 hours ago










  • $begingroup$
    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
    7 hours ago
















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?






linear-algebra integration matrix-equations







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
    7 hours ago










  • $begingroup$
    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
    7 hours ago














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?






linear-algebra integration matrix-equations







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




linear-algebra integration matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago







LightningStrike

















asked 8 hours ago









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
    7 hours ago










  • $begingroup$
    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
    7 hours ago














  • 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
    7 hours ago










  • $begingroup$
    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
    7 hours ago








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
7 hours ago




$begingroup$
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
7 hours ago












$begingroup$
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
7 hours ago




$begingroup$
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
7 hours ago










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









        $endgroup$













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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

          oldest

          votes






          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









          $endgroup$


















            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 7 hours ago









              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 7 hours ago









                      HAMIDINE SOUMAREHAMIDINE SOUMARE

                      1,478211




                      1,478211























                          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 7 hours ago









                              avsavs

                              3,769514




                              3,769514























                                  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 7 hours ago









                                      Mostafa AyazMostafa Ayaz

                                      17.6k31039




                                      17.6k31039






























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                                          0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h...
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                                          濃尾地震

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                                          濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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