A Standard Integral EquationLinear versus non-linear integral equationsUnderstanding why the roots of homogeneous difference equation must be eigenvaluesIntegral equation solution: $y(x) = 1 + lambdaintlimits_0^2cos(x-t) y(t) mathrmdt$Integrating with respect to time a double derivative $ddotphi + frac bmdotphi = fracFmr$Integral with Bessel FunctionsEigenvalue problem for integrals in multiple dimensionsStuck on finding the $2times 2$ system of differential equationsConversion of second order ode into integral equationSolving a dual integral equation involving a zeroth-order Bessel functionHow to find a basis of eigenvectors??

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A Standard Integral Equation


Linear versus non-linear integral equationsUnderstanding why the roots of homogeneous difference equation must be eigenvaluesIntegral equation solution: $y(x) = 1 + lambdaintlimits_0^2cos(x-t) y(t) mathrmdt$Integrating with respect to time a double derivative $ddotphi + frac bmdotphi = fracFmr$Integral with Bessel FunctionsEigenvalue problem for integrals in multiple dimensionsStuck on finding the $2times 2$ system of differential equationsConversion of second order ode into integral equationSolving a dual integral equation involving a zeroth-order Bessel functionHow to find a basis of eigenvectors??













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



$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$



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
    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



$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$



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
    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



$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$



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



$$int_0^1 phi(x),dx = frac12 + lambda int_0^1phi(s),ds$$



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















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

votes


















4












$begingroup$

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



Thus $$phi(x)=x+fraclambda2(1-lambda)$$






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
    Putting into FE yields:




    $$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
    If $lambda=0$ then $phi(x)=x$



    if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$



    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:
      $$
      sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
      $$

      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
      $$
      a_0 = lambda left(a_0 + a_1 over 2right).
      $$

      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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$The case $lambda=1$ leads to no solution.






        share|cite|improve this answer









        $endgroup$












          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

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



          Thus $$phi(x)=x+fraclambda2(1-lambda)$$






          share|cite|improve this answer









          $endgroup$

















            4












            $begingroup$

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



            Thus $$phi(x)=x+fraclambda2(1-lambda)$$






            share|cite|improve this answer









            $endgroup$















              4












              4








              4





              $begingroup$

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



              Thus $$phi(x)=x+fraclambda2(1-lambda)$$






              share|cite|improve this answer









              $endgroup$



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



              Thus $$phi(x)=x+fraclambda2(1-lambda)$$







              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 $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                  Putting into FE yields:




                  $$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
                  If $lambda=0$ then $phi(x)=x$



                  if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$



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







                  share|cite|improve this answer









                  $endgroup$

















                    2












                    $begingroup$

                    Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                    Putting into FE yields:




                    $$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
                    If $lambda=0$ then $phi(x)=x$



                    if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$



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







                    share|cite|improve this answer









                    $endgroup$















                      2












                      2








                      2





                      $begingroup$

                      Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                      Putting into FE yields:




                      $$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
                      If $lambda=0$ then $phi(x)=x$



                      if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$



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







                      share|cite|improve this answer









                      $endgroup$



                      Note $int_0^1phi(s)ds$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$phi(x)=x+alambda$$
                      Putting into FE yields:




                      $$x+alambda=x+lambdaint_0^1(s+alambda )ds iff alambda=lambdabig(frac12+lambda abig)$$
                      If $lambda=0$ then $phi(x)=x$



                      if $lambdane 1$ $a=frac12+lambda aiff ( 1-lambda)a=frac12iff a=frac12-2lambda$ and then $phi(x)=x+fraclambda2-2lambda$



                      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:
                          $$
                          sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
                          $$

                          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
                          $$
                          a_0 = lambda left(a_0 + a_1 over 2right).
                          $$

                          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:
                            $$
                            sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
                            $$

                            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
                            $$
                            a_0 = lambda left(a_0 + a_1 over 2right).
                            $$

                            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:
                              $$
                              sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
                              $$

                              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
                              $$
                              a_0 = lambda left(a_0 + a_1 over 2right).
                              $$

                              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:
                              $$
                              sum_n geq 0 a_n x^n = x + lambda sum_n geq 0a_n over n+1.
                              $$

                              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
                              $$
                              a_0 = lambda left(a_0 + a_1 over 2right).
                              $$

                              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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$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 $$x+a=x+lambdaint _0^1x+adximplies\a=lambda(1over 2+a)implies\a=lambdaover 2-2lambda$$ and we obtain$$phi(x)=x+lambdaover 2-2lambdaquad,quad lambdane 1$$The case $lambda=1$ leads to no solution.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 5 hours ago









                                      Mostafa AyazMostafa Ayaz

                                      17.6k31039




                                      17.6k31039



























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