Solving Fredholm Equation of the second kind












4












$begingroup$


Consider the Fredholm Equation of the second kind,



$$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$



Where the analytical solution is found as,



$$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$



How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?










share|improve this question









$endgroup$

















    4












    $begingroup$


    Consider the Fredholm Equation of the second kind,



    $$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$



    Where the analytical solution is found as,



    $$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$



    How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?










    share|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Consider the Fredholm Equation of the second kind,



      $$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$



      Where the analytical solution is found as,



      $$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$



      How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?










      share|improve this question









      $endgroup$




      Consider the Fredholm Equation of the second kind,



      $$phi(x) = 3 + lambda int_{0}^{pi} text{cos}(x-s) , phi(s) ,ds$$



      Where the analytical solution is found as,



      $$phi(x) = 3 + frac{6lambda}{1 - lambda frac{pi}{2}},text{sin}(x)$$



      How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?







      numerical-integration integral-equations numerical-value






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 7 hours ago









      user57401user57401

      634




      634






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Use DSolve



          PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
          (*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)


          The solution can be further used in the form PHI[x].






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
            $endgroup$
            – user57401
            7 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            6 hours ago



















          1












          $begingroup$

          Following
          Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:



          n = 10; (* for example *)
          [Phi][x_, 0] = 3;
          Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]


          The last term in the series [Phi][x,n] is the approximation to [Phi][x].



          Here is what Mathematica returns for [Phi][x,10].



          phi_of_ten



          To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
            $endgroup$
            – user57401
            6 hours ago










          • $begingroup$
            Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
            $endgroup$
            – mjw
            6 hours ago












          • $begingroup$
            Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
            $endgroup$
            – mjw
            5 hours ago













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "387"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192434%2fsolving-fredholm-equation-of-the-second-kind%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Use DSolve



          PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
          (*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)


          The solution can be further used in the form PHI[x].






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
            $endgroup$
            – user57401
            7 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            6 hours ago
















          4












          $begingroup$

          Use DSolve



          PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
          (*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)


          The solution can be further used in the form PHI[x].






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
            $endgroup$
            – user57401
            7 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            6 hours ago














          4












          4








          4





          $begingroup$

          Use DSolve



          PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
          (*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)


          The solution can be further used in the form PHI[x].






          share|improve this answer











          $endgroup$



          Use DSolve



          PHI=DSolveValue[[Phi][x] == 3 + [Lambda] Integrate[ Cos[x - s] [Phi][s], {s, 0, Pi}], [Phi],x]
          (*Function[{x}, (3 (-2 + [Pi] [Lambda] - 4 [Lambda] Sin[x]))/(-2 + [Pi] [Lambda])]*)


          The solution can be further used in the form PHI[x].







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          Ulrich NeumannUlrich Neumann

          9,271516




          9,271516












          • $begingroup$
            Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
            $endgroup$
            – user57401
            7 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            6 hours ago


















          • $begingroup$
            Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
            $endgroup$
            – user57401
            7 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            6 hours ago
















          $begingroup$
          Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
          $endgroup$
          – user57401
          7 hours ago




          $begingroup$
          Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $phi (x)$ function?
          $endgroup$
          – user57401
          7 hours ago




          1




          1




          $begingroup$
          @ user57401 I modified my answer!
          $endgroup$
          – Ulrich Neumann
          6 hours ago




          $begingroup$
          @ user57401 I modified my answer!
          $endgroup$
          – Ulrich Neumann
          6 hours ago











          1












          $begingroup$

          Following
          Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:



          n = 10; (* for example *)
          [Phi][x_, 0] = 3;
          Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]


          The last term in the series [Phi][x,n] is the approximation to [Phi][x].



          Here is what Mathematica returns for [Phi][x,10].



          phi_of_ten



          To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
            $endgroup$
            – user57401
            6 hours ago










          • $begingroup$
            Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
            $endgroup$
            – mjw
            6 hours ago












          • $begingroup$
            Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
            $endgroup$
            – mjw
            5 hours ago


















          1












          $begingroup$

          Following
          Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:



          n = 10; (* for example *)
          [Phi][x_, 0] = 3;
          Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]


          The last term in the series [Phi][x,n] is the approximation to [Phi][x].



          Here is what Mathematica returns for [Phi][x,10].



          phi_of_ten



          To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
            $endgroup$
            – user57401
            6 hours ago










          • $begingroup$
            Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
            $endgroup$
            – mjw
            6 hours ago












          • $begingroup$
            Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
            $endgroup$
            – mjw
            5 hours ago
















          1












          1








          1





          $begingroup$

          Following
          Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:



          n = 10; (* for example *)
          [Phi][x_, 0] = 3;
          Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]


          The last term in the series [Phi][x,n] is the approximation to [Phi][x].



          Here is what Mathematica returns for [Phi][x,10].



          phi_of_ten



          To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].






          share|improve this answer











          $endgroup$



          Following
          Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:



          n = 10; (* for example *)
          [Phi][x_, 0] = 3;
          Do[[Phi][x_, j_] = 3 + [Lambda] Integrate[Cos[x - p] [Phi][p, j - 1], {p, 0, [Pi]}], {j, n}]


          The last term in the series [Phi][x,n] is the approximation to [Phi][x].



          Here is what Mathematica returns for [Phi][x,10].



          phi_of_ten



          To investigate convergence, I guess we could look at the difference [Phi][x,n] - [Phi][x] as n gets large, since you know [Phi][x].







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 6 hours ago

























          answered 6 hours ago









          mjwmjw

          3116




          3116












          • $begingroup$
            Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
            $endgroup$
            – user57401
            6 hours ago










          • $begingroup$
            Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
            $endgroup$
            – mjw
            6 hours ago












          • $begingroup$
            Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
            $endgroup$
            – mjw
            5 hours ago




















          • $begingroup$
            Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
            $endgroup$
            – user57401
            6 hours ago










          • $begingroup$
            Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
            $endgroup$
            – mjw
            6 hours ago












          • $begingroup$
            Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
            $endgroup$
            – mjw
            5 hours ago


















          $begingroup$
          Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
          $endgroup$
          – user57401
          6 hours ago




          $begingroup$
          Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]?
          $endgroup$
          – user57401
          6 hours ago












          $begingroup$
          Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
          $endgroup$
          – mjw
          6 hours ago






          $begingroup$
          Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
          $endgroup$
          – mjw
          6 hours ago














          $begingroup$
          Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
          $endgroup$
          – mjw
          5 hours ago






          $begingroup$
          Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, [Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear.
          $endgroup$
          – mjw
          5 hours ago




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematica Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192434%2fsolving-fredholm-equation-of-the-second-kind%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Knooppunt Holsloot

          Altaar (religie)

          Gregoriusmis