Solving Fredholm Equation of the second kind












2












$begingroup$


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)?






numerical-integration integral-equations numerical-value







share|improve this question









$endgroup$

















    2












    $begingroup$


    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)?






    numerical-integration integral-equations numerical-value







    share|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      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)?






      numerical-integration integral-equations numerical-value







      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




      numerical-integration integral-equations numerical-value






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      user57401user57401

      534




      534






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          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
            1 hour ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            1 hour 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
            1 hour ago










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













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

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          3












          $begingroup$

          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
            1 hour ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            1 hour ago
















          3












          $begingroup$

          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
            1 hour ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            1 hour ago














          3












          3








          3





          $begingroup$

          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 1 hour ago

























          answered 1 hour ago









          Ulrich NeumannUlrich Neumann

          9,261516




          9,261516












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






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            1 hour 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
            1 hour ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            1 hour 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
          1 hour 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
          1 hour ago




          1




          1




          $begingroup$
          @ user57401 I modified my answer!
          $endgroup$
          – Ulrich Neumann
          1 hour ago




          $begingroup$
          @ user57401 I modified my answer!
          $endgroup$
          – Ulrich Neumann
          1 hour 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
            1 hour ago










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










          • $begingroup$
            Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: [Phi][x, n].
            $endgroup$
            – mjw
            36 mins 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
            32 mins 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 33 mins ago

























          answered 1 hour 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
            1 hour ago










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










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












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






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




















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