Degenerate kernel method to solve Fredholm integral equation of the second kind












2












$begingroup$


$$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
$$



I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
I know that is has to be written in the form
$$ sum a_jx^{j-1} $$



I do not know how to integrate with the function $f(y)$ in the integral.



Any help would be appreciated.



Thanks






calculus linear-algebra integration integral-equations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    $$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
    $$



    I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
    I know that is has to be written in the form
    $$ sum a_jx^{j-1} $$



    I do not know how to integrate with the function $f(y)$ in the integral.



    Any help would be appreciated.



    Thanks






    calculus linear-algebra integration integral-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      $$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
      $$



      I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
      I know that is has to be written in the form
      $$ sum a_jx^{j-1} $$



      I do not know how to integrate with the function $f(y)$ in the integral.



      Any help would be appreciated.



      Thanks






      calculus linear-algebra integration integral-equations







      share|cite|improve this question











      $endgroup$




      $$ f(x) + int_0^1 (xy+x^2y^2) f(y) dy = x^3 +frac16x^2+ frac15x
      $$



      I have this fredholm integral equation of the second kind and am not sure how to answer this equation.
      I know that is has to be written in the form
      $$ sum a_jx^{j-1} $$



      I do not know how to integrate with the function $f(y)$ in the integral.



      Any help would be appreciated.



      Thanks






      calculus linear-algebra integration integral-equations




      calculus linear-algebra integration integral-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 19:27







      p s

















      asked Dec 2 '18 at 16:02









      p sp s

      358




      358






















          2 Answers
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          active

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          0












          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02



















          0












          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00













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          2 Answers
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          2 Answers
          2






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          active

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          0












          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02
















          0












          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02














          0












          0








          0





          $begingroup$

          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$






          share|cite|improve this answer









          $endgroup$



          can you start with:
          $$f(x)+int_0^1(xy+x^2y^2)f(y)dy=x^3+frac16x^2+frac15x$$
          $$f'(x)+(x+x^2)f(1)=3x^2+frac13x+frac15$$
          $$f'(x)=(3-f(1))x^2+left(frac13-f(1)right)+frac15$$
          so:
          $$f(x)=intleft[(3-f(1))x^2+left(frac13-f(1)right)+frac15right]dx+C$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 18:22









          Henry LeeHenry Lee

          1,865219




          1,865219












          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02


















          • $begingroup$
            Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
            $endgroup$
            – p s
            Dec 2 '18 at 21:00










          • $begingroup$
            f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
            $endgroup$
            – Henry Lee
            Dec 2 '18 at 21:02
















          $begingroup$
          Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
          $endgroup$
          – p s
          Dec 2 '18 at 21:00




          $begingroup$
          Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1)
          $endgroup$
          – p s
          Dec 2 '18 at 21:00












          $begingroup$
          f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
          $endgroup$
          – Henry Lee
          Dec 2 '18 at 21:02




          $begingroup$
          f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself
          $endgroup$
          – Henry Lee
          Dec 2 '18 at 21:02











          0












          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00


















          0












          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00
















          0












          0








          0





          $begingroup$

          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.






          share|cite|improve this answer









          $endgroup$



          Note that we can write the equation as
          $$ f(x) = x^3 + left[frac{1}{6} - int limits_0^1 y^2 f(y) , mathrm{d} yright] x^2 + left[frac{1}{5} - int limits_0^1 y f(y) , mathrm{d} yright] x , , , x in [0,1] , . $$
          Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b in mathbb{R}$ such that
          $$ f(x) = x^3 + a x^2 + b x , , , x in [0,1] , . $$
          We can then plug this form of $f$ into the previous equation and compute the integrals to obtain
          $$ x^3 + a x^2 + b x = x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          This yields the linear system
          $$ begin{pmatrix} frac{6}{5} && frac{1}{4}\ frac{1}{4} && frac{4}{3} end{pmatrix} begin{pmatrix} a \b end{pmatrix} = begin{pmatrix} 0 \0 end{pmatrix} , ,$$
          which only has the trivial solution $a=b=0$ . Therefore
          $$ f(x) = x^3 , , , x in [0,1] , , $$
          is the unique solution to the integral equation.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 22:57









          ComplexYetTrivialComplexYetTrivial

          3,7302629




          3,7302629












          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00




















          • $begingroup$
            Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
            $endgroup$
            – p s
            Dec 2 '18 at 23:22












          • $begingroup$
            @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
            $endgroup$
            – ComplexYetTrivial
            Dec 2 '18 at 23:32










          • $begingroup$
            how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
            $endgroup$
            – p s
            Dec 9 '18 at 19:00


















          $begingroup$
          Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
          $endgroup$
          – p s
          Dec 2 '18 at 23:22






          $begingroup$
          Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ sum a_jx^{j-1} $$ how would I change what you have solved above into this format?
          $endgroup$
          – p s
          Dec 2 '18 at 23:22














          $begingroup$
          @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
          $endgroup$
          – ComplexYetTrivial
          Dec 2 '18 at 23:32




          $begingroup$
          @ps You can write $f(x) = x^3 = sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ .
          $endgroup$
          – ComplexYetTrivial
          Dec 2 '18 at 23:32












          $begingroup$
          how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          $endgroup$
          – p s
          Dec 9 '18 at 19:00






          $begingroup$
          how did you get to $$ x^3 - left(frac{a}{5}+frac{b}{4}right) x^2 - left(frac{a}{4}+frac{b}{3}right)x , , , x in [0,1] , . $$
          $endgroup$
          – p s
          Dec 9 '18 at 19:00




















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          Quarter-circle Tiles

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          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
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          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

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          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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