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)?






numerical-integration integral-equations numerical-value







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)?






    numerical-integration integral-equations numerical-value







    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)?






      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 9 hours ago









      user57401user57401

      635




      635






















          2 Answers
          2






          active

          oldest

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          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
            9 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            8 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
            8 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
            8 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
            7 hours ago













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          2 Answers
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          active

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          2 Answers
          2






          active

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          active

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          active

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          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
            9 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            8 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
            9 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            8 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 9 hours ago

























          answered 9 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
            9 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            8 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
            9 hours ago






          • 1




            $begingroup$
            @ user57401 I modified my answer!
            $endgroup$
            – Ulrich Neumann
            8 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
          9 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
          9 hours ago




          1




          1




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




          $begingroup$
          @ user57401 I modified my answer!
          $endgroup$
          – Ulrich Neumann
          8 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
            8 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
            8 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
            7 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
            8 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
            8 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
            7 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 7 hours ago

























          answered 8 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
            8 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
            8 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
            7 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
            8 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
            8 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
            7 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
          8 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
          8 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
          8 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
          8 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
          7 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
          7 hours ago




















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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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