Solving Integral Equation by Converting to Differential Equations












2












$begingroup$


Consider the problem



$$phi(x) = x - int_0^x(x-s)phi(s),ds$$



How can we solve this by converting to a differential equation?










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Consider the problem



    $$phi(x) = x - int_0^x(x-s)phi(s),ds$$



    How can we solve this by converting to a differential equation?










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Consider the problem



      $$phi(x) = x - int_0^x(x-s)phi(s),ds$$



      How can we solve this by converting to a differential equation?










      share|cite|improve this question









      $endgroup$




      Consider the problem



      $$phi(x) = x - int_0^x(x-s)phi(s),ds$$



      How can we solve this by converting to a differential equation?







      ordinary-differential-equations integral-equations integro-differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 29 at 18:29









      LightningStrikeLightningStrike

      555




      555






















          2 Answers
          2






          active

          oldest

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          4












          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
            $endgroup$
            – Peter Foreman
            Mar 29 at 19:03










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            Mar 29 at 19:07












          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            Mar 29 at 19:17










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
            $endgroup$
            – Botond
            Mar 29 at 19:28



















          1












          $begingroup$

          Differentiating both sides using Leibniz rule :



          $${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$



          Differentiate again:



          $${phi }''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            Mar 29 at 19:51










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            Mar 29 at 19:56












          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            Mar 29 at 20:05














          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
            $endgroup$
            – Peter Foreman
            Mar 29 at 19:03










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            Mar 29 at 19:07












          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            Mar 29 at 19:17










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
            $endgroup$
            – Botond
            Mar 29 at 19:28
















          4












          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
            $endgroup$
            – Peter Foreman
            Mar 29 at 19:03










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            Mar 29 at 19:07












          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            Mar 29 at 19:17










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
            $endgroup$
            – Botond
            Mar 29 at 19:28














          4












          4








          4





          $begingroup$

          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.






          share|cite|improve this answer











          $endgroup$



          We have that
          $$phi(x)=x-xint_0^x phi(s) mathrm{d} s + int_0^x s phi(s)mathrm{d}s$$
          From this, we can see that $phi(0)=0$.
          We can differentiate both sides and use the product rule and the FTC1 to get:
          $$phi'(x)=1-int_0^x phi(s) mathrm{d}s -x phi(x)+xphi(x)$$
          $$phi'(x)=1-int_0^x phi(s) mathrm{d} s$$
          From this, we can see that $phi'(0)=1$. We can differentiate it again:
          $$phi''(x)=-phi(x)$$
          Which is an alternative definition of the $sin$ function.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 29 at 19:00

























          answered Mar 29 at 18:52









          BotondBotond

          6,54531034




          6,54531034












          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
            $endgroup$
            – Peter Foreman
            Mar 29 at 19:03










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            Mar 29 at 19:07












          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            Mar 29 at 19:17










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
            $endgroup$
            – Botond
            Mar 29 at 19:28


















          • $begingroup$
            In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
            $endgroup$
            – Peter Foreman
            Mar 29 at 19:03










          • $begingroup$
            @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
            $endgroup$
            – Botond
            Mar 29 at 19:07












          • $begingroup$
            Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
            $endgroup$
            – LightningStrike
            Mar 29 at 19:17










          • $begingroup$
            @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
            $endgroup$
            – Botond
            Mar 29 at 19:28
















          $begingroup$
          In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
          $endgroup$
          – Peter Foreman
          Mar 29 at 19:03




          $begingroup$
          In fact, the only valid solution for $phi(x)$ is $sin{(x)}$ because of the original equation.
          $endgroup$
          – Peter Foreman
          Mar 29 at 19:03












          $begingroup$
          @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
          $endgroup$
          – Botond
          Mar 29 at 19:07






          $begingroup$
          @PeterForemann Yes. I calculated $phi(0)$ and $phi'(0)$ from the integral equation to avoid the lengthy substitution and integration.
          $endgroup$
          – Botond
          Mar 29 at 19:07














          $begingroup$
          Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
          $endgroup$
          – LightningStrike
          Mar 29 at 19:17




          $begingroup$
          Thank you for your answer! Do you mind if I ask how you got $phi ''(x) = -phi (x)$ by differentiating $phi ' (x)$? I don't understand the steps taken.
          $endgroup$
          – LightningStrike
          Mar 29 at 19:17












          $begingroup$
          @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
          $endgroup$
          – Botond
          Mar 29 at 19:28




          $begingroup$
          @LightningStrike Do you see how did I get $phi'(x)=1-int_0^x phi(s) mathrm{d}s$?
          $endgroup$
          – Botond
          Mar 29 at 19:28











          1












          $begingroup$

          Differentiating both sides using Leibniz rule :



          $${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$



          Differentiate again:



          $${phi }''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            Mar 29 at 19:51










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            Mar 29 at 19:56












          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            Mar 29 at 20:05


















          1












          $begingroup$

          Differentiating both sides using Leibniz rule :



          $${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$



          Differentiate again:



          $${phi }''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            Mar 29 at 19:51










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            Mar 29 at 19:56












          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            Mar 29 at 20:05
















          1












          1








          1





          $begingroup$

          Differentiating both sides using Leibniz rule :



          $${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$



          Differentiate again:



          $${phi }''(x)=-phi (x)$$






          share|cite|improve this answer











          $endgroup$



          Differentiating both sides using Leibniz rule :



          $${phi }'(x)=1-int_{0}^{x}{phi (s)ds}$$



          Differentiate again:



          $${phi }''(x)=-phi (x)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 29 at 18:56

























          answered Mar 29 at 18:51









          logologo

          19610




          19610








          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            Mar 29 at 19:51










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            Mar 29 at 19:56












          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            Mar 29 at 20:05
















          • 1




            $begingroup$
            Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
            $endgroup$
            – Botond
            Mar 29 at 19:51










          • $begingroup$
            may be you are right...but this is a common technique in an introductory course of integral equations.
            $endgroup$
            – logo
            Mar 29 at 19:56












          • $begingroup$
            I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
            $endgroup$
            – Botond
            Mar 29 at 20:05










          1




          1




          $begingroup$
          Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
          $endgroup$
          – Botond
          Mar 29 at 19:51




          $begingroup$
          Your answer is great, but Leibniz's rule is an overkill here, because it requires partial derivatives and the proof is based on measure theory.
          $endgroup$
          – Botond
          Mar 29 at 19:51












          $begingroup$
          may be you are right...but this is a common technique in an introductory course of integral equations.
          $endgroup$
          – logo
          Mar 29 at 19:56






          $begingroup$
          may be you are right...but this is a common technique in an introductory course of integral equations.
          $endgroup$
          – logo
          Mar 29 at 19:56














          $begingroup$
          I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
          $endgroup$
          – Botond
          Mar 29 at 20:05






          $begingroup$
          I didn't take any course in integral equations, but we used Leibniz's rule during a physics course (without a proof), and it's a really useful tool to have. And we don't really know which is the appropriate solution to the questioner.
          $endgroup$
          – Botond
          Mar 29 at 20:05




















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