Convert in a Sturm-Liouville form $y''+R(x)y'+(Q(x)+lambda p(x))y=0$












3












$begingroup$


Convert in a Sturm-Liouville form $$y''+R(x)y'+(Q(x)+lambda P(x))y=0tag1$$



My attempt:



The form of Sturm-Liouville is:



$$frac{d}{dx}[r(x) frac{dy}{dx}]+(q(x)+lambda p(x))y=0tag2$$



For obtain this, we need multiply for $mu(x)$ the ODE $(1)$. This result in:



$$mu y''+mu R(x)y'+(mu Q(x)+lambda mu P(x))y=0 tag3$$



I need rewrite the first term of $(3)$.



Here, i'm stuck. I don't have a clear idea of how rewrite and then proceed for solve the exercise.



Can someone help me?






ordinary-differential-equations sturm-liouville







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Convert in a Sturm-Liouville form $$y''+R(x)y'+(Q(x)+lambda P(x))y=0tag1$$



    My attempt:



    The form of Sturm-Liouville is:



    $$frac{d}{dx}[r(x) frac{dy}{dx}]+(q(x)+lambda p(x))y=0tag2$$



    For obtain this, we need multiply for $mu(x)$ the ODE $(1)$. This result in:



    $$mu y''+mu R(x)y'+(mu Q(x)+lambda mu P(x))y=0 tag3$$



    I need rewrite the first term of $(3)$.



    Here, i'm stuck. I don't have a clear idea of how rewrite and then proceed for solve the exercise.



    Can someone help me?






    ordinary-differential-equations sturm-liouville







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Convert in a Sturm-Liouville form $$y''+R(x)y'+(Q(x)+lambda P(x))y=0tag1$$



      My attempt:



      The form of Sturm-Liouville is:



      $$frac{d}{dx}[r(x) frac{dy}{dx}]+(q(x)+lambda p(x))y=0tag2$$



      For obtain this, we need multiply for $mu(x)$ the ODE $(1)$. This result in:



      $$mu y''+mu R(x)y'+(mu Q(x)+lambda mu P(x))y=0 tag3$$



      I need rewrite the first term of $(3)$.



      Here, i'm stuck. I don't have a clear idea of how rewrite and then proceed for solve the exercise.



      Can someone help me?






      ordinary-differential-equations sturm-liouville







      share|cite|improve this question











      $endgroup$




      Convert in a Sturm-Liouville form $$y''+R(x)y'+(Q(x)+lambda P(x))y=0tag1$$



      My attempt:



      The form of Sturm-Liouville is:



      $$frac{d}{dx}[r(x) frac{dy}{dx}]+(q(x)+lambda p(x))y=0tag2$$



      For obtain this, we need multiply for $mu(x)$ the ODE $(1)$. This result in:



      $$mu y''+mu R(x)y'+(mu Q(x)+lambda mu P(x))y=0 tag3$$



      I need rewrite the first term of $(3)$.



      Here, i'm stuck. I don't have a clear idea of how rewrite and then proceed for solve the exercise.



      Can someone help me?






      ordinary-differential-equations sturm-liouville




      ordinary-differential-equations sturm-liouville






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 21:20









      Kenny Wong

      19.1k21441




      19.1k21441










      asked Jan 1 at 21:05









      Bvss12Bvss12

      1,821619




      1,821619






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          If we expand out equation $(2)$, we get
          $$ r(x) y'' + r'(x) y' + (q(x) + lambda p(x)) y = 0 (2a).$$



          So the objective is to choose the functions $mu(x)$, $r(x)$, $q(x)$ and $p(x)$ so that your equation $(3)$ matches my equation $(2a)$.



          Comparing the individual terms, the requirement is that
          $$ mu(x) = r(x), mu(x)R(x) = r'(x), mu(x)Q(x) = q(x), mu(x)P(x) = p(x).$$
          The first of these equations is straightforward: it tells us that $mu(x)$ and $r(x)$ must be the same function.



          Given that $mu(x) = r(x)$, the second equation says that
          $$ mu(x) R(x) = mu'(x),$$
          and this is satisfied by



          $$ mu(x) = exp left(int dx R(x) right).$$



          I'll leave you to figure out what $q(x)$ and $p(x)$ and finish off.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, i will go to expand your idea right now!
            $endgroup$
            – Bvss12
            Jan 1 at 21:21










          • $begingroup$
            Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
            $endgroup$
            – Bvss12
            Jan 1 at 23:00












          • $begingroup$
            @Bvss12 looks good to me!
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:04










          • $begingroup$
            Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
            $endgroup$
            – Bvss12
            Jan 1 at 23:13










          • $begingroup$
            I believe so, yes.
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:27












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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          If we expand out equation $(2)$, we get
          $$ r(x) y'' + r'(x) y' + (q(x) + lambda p(x)) y = 0 (2a).$$



          So the objective is to choose the functions $mu(x)$, $r(x)$, $q(x)$ and $p(x)$ so that your equation $(3)$ matches my equation $(2a)$.



          Comparing the individual terms, the requirement is that
          $$ mu(x) = r(x), mu(x)R(x) = r'(x), mu(x)Q(x) = q(x), mu(x)P(x) = p(x).$$
          The first of these equations is straightforward: it tells us that $mu(x)$ and $r(x)$ must be the same function.



          Given that $mu(x) = r(x)$, the second equation says that
          $$ mu(x) R(x) = mu'(x),$$
          and this is satisfied by



          $$ mu(x) = exp left(int dx R(x) right).$$



          I'll leave you to figure out what $q(x)$ and $p(x)$ and finish off.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, i will go to expand your idea right now!
            $endgroup$
            – Bvss12
            Jan 1 at 21:21










          • $begingroup$
            Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
            $endgroup$
            – Bvss12
            Jan 1 at 23:00












          • $begingroup$
            @Bvss12 looks good to me!
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:04










          • $begingroup$
            Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
            $endgroup$
            – Bvss12
            Jan 1 at 23:13










          • $begingroup$
            I believe so, yes.
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:27
















          3












          $begingroup$

          If we expand out equation $(2)$, we get
          $$ r(x) y'' + r'(x) y' + (q(x) + lambda p(x)) y = 0 (2a).$$



          So the objective is to choose the functions $mu(x)$, $r(x)$, $q(x)$ and $p(x)$ so that your equation $(3)$ matches my equation $(2a)$.



          Comparing the individual terms, the requirement is that
          $$ mu(x) = r(x), mu(x)R(x) = r'(x), mu(x)Q(x) = q(x), mu(x)P(x) = p(x).$$
          The first of these equations is straightforward: it tells us that $mu(x)$ and $r(x)$ must be the same function.



          Given that $mu(x) = r(x)$, the second equation says that
          $$ mu(x) R(x) = mu'(x),$$
          and this is satisfied by



          $$ mu(x) = exp left(int dx R(x) right).$$



          I'll leave you to figure out what $q(x)$ and $p(x)$ and finish off.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, i will go to expand your idea right now!
            $endgroup$
            – Bvss12
            Jan 1 at 21:21










          • $begingroup$
            Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
            $endgroup$
            – Bvss12
            Jan 1 at 23:00












          • $begingroup$
            @Bvss12 looks good to me!
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:04










          • $begingroup$
            Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
            $endgroup$
            – Bvss12
            Jan 1 at 23:13










          • $begingroup$
            I believe so, yes.
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:27














          3












          3








          3





          $begingroup$

          If we expand out equation $(2)$, we get
          $$ r(x) y'' + r'(x) y' + (q(x) + lambda p(x)) y = 0 (2a).$$



          So the objective is to choose the functions $mu(x)$, $r(x)$, $q(x)$ and $p(x)$ so that your equation $(3)$ matches my equation $(2a)$.



          Comparing the individual terms, the requirement is that
          $$ mu(x) = r(x), mu(x)R(x) = r'(x), mu(x)Q(x) = q(x), mu(x)P(x) = p(x).$$
          The first of these equations is straightforward: it tells us that $mu(x)$ and $r(x)$ must be the same function.



          Given that $mu(x) = r(x)$, the second equation says that
          $$ mu(x) R(x) = mu'(x),$$
          and this is satisfied by



          $$ mu(x) = exp left(int dx R(x) right).$$



          I'll leave you to figure out what $q(x)$ and $p(x)$ and finish off.






          share|cite|improve this answer









          $endgroup$



          If we expand out equation $(2)$, we get
          $$ r(x) y'' + r'(x) y' + (q(x) + lambda p(x)) y = 0 (2a).$$



          So the objective is to choose the functions $mu(x)$, $r(x)$, $q(x)$ and $p(x)$ so that your equation $(3)$ matches my equation $(2a)$.



          Comparing the individual terms, the requirement is that
          $$ mu(x) = r(x), mu(x)R(x) = r'(x), mu(x)Q(x) = q(x), mu(x)P(x) = p(x).$$
          The first of these equations is straightforward: it tells us that $mu(x)$ and $r(x)$ must be the same function.



          Given that $mu(x) = r(x)$, the second equation says that
          $$ mu(x) R(x) = mu'(x),$$
          and this is satisfied by



          $$ mu(x) = exp left(int dx R(x) right).$$



          I'll leave you to figure out what $q(x)$ and $p(x)$ and finish off.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 1 at 21:19









          Kenny WongKenny Wong

          19.1k21441




          19.1k21441












          • $begingroup$
            Thanks, i will go to expand your idea right now!
            $endgroup$
            – Bvss12
            Jan 1 at 21:21










          • $begingroup$
            Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
            $endgroup$
            – Bvss12
            Jan 1 at 23:00












          • $begingroup$
            @Bvss12 looks good to me!
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:04










          • $begingroup$
            Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
            $endgroup$
            – Bvss12
            Jan 1 at 23:13










          • $begingroup$
            I believe so, yes.
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:27


















          • $begingroup$
            Thanks, i will go to expand your idea right now!
            $endgroup$
            – Bvss12
            Jan 1 at 21:21










          • $begingroup$
            Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
            $endgroup$
            – Bvss12
            Jan 1 at 23:00












          • $begingroup$
            @Bvss12 looks good to me!
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:04










          • $begingroup$
            Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
            $endgroup$
            – Bvss12
            Jan 1 at 23:13










          • $begingroup$
            I believe so, yes.
            $endgroup$
            – Kenny Wong
            Jan 1 at 23:27
















          $begingroup$
          Thanks, i will go to expand your idea right now!
          $endgroup$
          – Bvss12
          Jan 1 at 21:21




          $begingroup$
          Thanks, i will go to expand your idea right now!
          $endgroup$
          – Bvss12
          Jan 1 at 21:21












          $begingroup$
          Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
          $endgroup$
          – Bvss12
          Jan 1 at 23:00






          $begingroup$
          Expanding your idea i have $q(x)=(exp{int R(x)dx})Q(x)$ and $p(x)=(exp{int R(x)dx})P(x)$ is correct?
          $endgroup$
          – Bvss12
          Jan 1 at 23:00














          $begingroup$
          @Bvss12 looks good to me!
          $endgroup$
          – Kenny Wong
          Jan 1 at 23:04




          $begingroup$
          @Bvss12 looks good to me!
          $endgroup$
          – Kenny Wong
          Jan 1 at 23:04












          $begingroup$
          Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
          $endgroup$
          – Bvss12
          Jan 1 at 23:13




          $begingroup$
          Then, can i conclude... if $mu(x) = r(x)= exp left(int dx R(x) right).$ and $q(x)=(exp{int R(x)dx})Q(x)$, and $p(x)=(exp{int R(x)dx})P(x)$ then the equation 1 is a Sturm-Liouville equation. is correct? Thanks for all help (:
          $endgroup$
          – Bvss12
          Jan 1 at 23:13












          $begingroup$
          I believe so, yes.
          $endgroup$
          – Kenny Wong
          Jan 1 at 23:27




          $begingroup$
          I believe so, yes.
          $endgroup$
          – Kenny Wong
          Jan 1 at 23:27


















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