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
    6 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
    6 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
    6 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
    6 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 6 hours ago







LightningStrike

















asked 6 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
    6 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
    6 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
    6 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
    6 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
6 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
6 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
6 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
6 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

          votes








          4 Answers
          4






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          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 6 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 6 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 6 hours ago









                              avsavs

                              3,749514




                              3,749514























                                  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 6 hours ago









                                      Mostafa AyazMostafa Ayaz

                                      17.6k31039




                                      17.6k31039






























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                                          Tabula Rosettana

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                                          Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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