Prove the eigenfunctions nth have $n-1$ zero.












0














Consider the problem of eigenvalues:
$$x^2y''+xy'+lambda y=0tag1$$



with boundary condition:



$$$y(1)=y(b)=0$$



Prove the eigenfunctions nth have $n-1$ zero.



My attempt



I know the Sturm-Liouville form of the equation is:



$$xy''+y'+frac{lambda}{x}y=0tag2$$



With eigenvalues: $$lambda_k=frac{k^2pi^2}{ln(b)^2}tag3$$



and eigenfunction of the form:



$$y_k(x)=sinfrac{kpiln(x)}{ln(b)}$$



I could prove with help of other user, the orthogonality of two functions $y_n,y_m$ with weight function $$w(x)=frac{1}{x}tag4$$



But here i'm stuck. Can somenone help me?






pde







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    0














    Consider the problem of eigenvalues:
    $$x^2y''+xy'+lambda y=0tag1$$



    with boundary condition:



    $$$y(1)=y(b)=0$$



    Prove the eigenfunctions nth have $n-1$ zero.



    My attempt



    I know the Sturm-Liouville form of the equation is:



    $$xy''+y'+frac{lambda}{x}y=0tag2$$



    With eigenvalues: $$lambda_k=frac{k^2pi^2}{ln(b)^2}tag3$$



    and eigenfunction of the form:



    $$y_k(x)=sinfrac{kpiln(x)}{ln(b)}$$



    I could prove with help of other user, the orthogonality of two functions $y_n,y_m$ with weight function $$w(x)=frac{1}{x}tag4$$



    But here i'm stuck. Can somenone help me?






    pde







    share|cite|improve this question

























      0












      0








      0







      Consider the problem of eigenvalues:
      $$x^2y''+xy'+lambda y=0tag1$$



      with boundary condition:



      $$$y(1)=y(b)=0$$



      Prove the eigenfunctions nth have $n-1$ zero.



      My attempt



      I know the Sturm-Liouville form of the equation is:



      $$xy''+y'+frac{lambda}{x}y=0tag2$$



      With eigenvalues: $$lambda_k=frac{k^2pi^2}{ln(b)^2}tag3$$



      and eigenfunction of the form:



      $$y_k(x)=sinfrac{kpiln(x)}{ln(b)}$$



      I could prove with help of other user, the orthogonality of two functions $y_n,y_m$ with weight function $$w(x)=frac{1}{x}tag4$$



      But here i'm stuck. Can somenone help me?






      pde







      share|cite|improve this question













      Consider the problem of eigenvalues:
      $$x^2y''+xy'+lambda y=0tag1$$



      with boundary condition:



      $$$y(1)=y(b)=0$$



      Prove the eigenfunctions nth have $n-1$ zero.



      My attempt



      I know the Sturm-Liouville form of the equation is:



      $$xy''+y'+frac{lambda}{x}y=0tag2$$



      With eigenvalues: $$lambda_k=frac{k^2pi^2}{ln(b)^2}tag3$$



      and eigenfunction of the form:



      $$y_k(x)=sinfrac{kpiln(x)}{ln(b)}$$



      I could prove with help of other user, the orthogonality of two functions $y_n,y_m$ with weight function $$w(x)=frac{1}{x}tag4$$



      But here i'm stuck. Can somenone help me?






      pde




      pde






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      share|cite|improve this question











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      asked Jan 4 at 15:22









      Bvss12Bvss12

      1,773617




      1,773617






















          2 Answers
          2






          active

          oldest

          votes


















          1














          You can directly verify the orthogonality with respect to the weight function $1/x$. Start with
          $$
          int_1^b sinleft(frac{npi ln x}{ln b}right)sinleft(frac{mpiln x}{ln b}right)frac{dx}{x},
          $$
          and set
          $$
          y = frac{ln x}{ln b},;; ln x = yln b, ; x=e^{yln b}=(e^{ln b})^y=b^y.
          $$



          Then $ln x = yln b$ or $frac{dx}{x}=ln b dy$. The first integral becomes
          $$
          int_0^1sin(npi y)sin(mpi y)dycdot ln b
          $$
          Orthogonality is easy to verify for $nne m$.






          share|cite|improve this answer





















          • This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
            – Dylan
            2 days ago



















          1














          Solve $sindfrac{kpiln(x)}{ln(b)}=0$ so is $dfrac{kpiln(x)}{ln(b)}=mpi$ with $minmathbb Z$ and $xin[0,b]$



          $ln x=dfrac{m}{k}ln b$ And because $xgeq1$, $mgeq0$



          $x=b^{m/k}$



          Now, as $b>1$, $b^{m/k}leq b$ if $dfrac{m}{k}leq 1$ or being $mleq k$



          $m=0,1,2,dots,k$



          Including the values $x=1$ and $x=b$, $y_k$ has $k+1$ zeros.






          share|cite|improve this answer























          • I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
            – Bvss12
            Jan 4 at 19:21










          • Thank you. Fixed.
            – Rafa Budría
            Jan 4 at 20:20











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          You can directly verify the orthogonality with respect to the weight function $1/x$. Start with
          $$
          int_1^b sinleft(frac{npi ln x}{ln b}right)sinleft(frac{mpiln x}{ln b}right)frac{dx}{x},
          $$
          and set
          $$
          y = frac{ln x}{ln b},;; ln x = yln b, ; x=e^{yln b}=(e^{ln b})^y=b^y.
          $$



          Then $ln x = yln b$ or $frac{dx}{x}=ln b dy$. The first integral becomes
          $$
          int_0^1sin(npi y)sin(mpi y)dycdot ln b
          $$
          Orthogonality is easy to verify for $nne m$.






          share|cite|improve this answer





















          • This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
            – Dylan
            2 days ago
















          1














          You can directly verify the orthogonality with respect to the weight function $1/x$. Start with
          $$
          int_1^b sinleft(frac{npi ln x}{ln b}right)sinleft(frac{mpiln x}{ln b}right)frac{dx}{x},
          $$
          and set
          $$
          y = frac{ln x}{ln b},;; ln x = yln b, ; x=e^{yln b}=(e^{ln b})^y=b^y.
          $$



          Then $ln x = yln b$ or $frac{dx}{x}=ln b dy$. The first integral becomes
          $$
          int_0^1sin(npi y)sin(mpi y)dycdot ln b
          $$
          Orthogonality is easy to verify for $nne m$.






          share|cite|improve this answer





















          • This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
            – Dylan
            2 days ago














          1












          1








          1






          You can directly verify the orthogonality with respect to the weight function $1/x$. Start with
          $$
          int_1^b sinleft(frac{npi ln x}{ln b}right)sinleft(frac{mpiln x}{ln b}right)frac{dx}{x},
          $$
          and set
          $$
          y = frac{ln x}{ln b},;; ln x = yln b, ; x=e^{yln b}=(e^{ln b})^y=b^y.
          $$



          Then $ln x = yln b$ or $frac{dx}{x}=ln b dy$. The first integral becomes
          $$
          int_0^1sin(npi y)sin(mpi y)dycdot ln b
          $$
          Orthogonality is easy to verify for $nne m$.






          share|cite|improve this answer












          You can directly verify the orthogonality with respect to the weight function $1/x$. Start with
          $$
          int_1^b sinleft(frac{npi ln x}{ln b}right)sinleft(frac{mpiln x}{ln b}right)frac{dx}{x},
          $$
          and set
          $$
          y = frac{ln x}{ln b},;; ln x = yln b, ; x=e^{yln b}=(e^{ln b})^y=b^y.
          $$



          Then $ln x = yln b$ or $frac{dx}{x}=ln b dy$. The first integral becomes
          $$
          int_0^1sin(npi y)sin(mpi y)dycdot ln b
          $$
          Orthogonality is easy to verify for $nne m$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 18:09









          DisintegratingByPartsDisintegratingByParts

          58.7k42579




          58.7k42579












          • This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
            – Dylan
            2 days ago


















          • This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
            – Dylan
            2 days ago
















          This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
          – Dylan
          2 days ago




          This isn't the question asked, but to be fair the question wasn't stated clearly in the OP.
          – Dylan
          2 days ago











          1














          Solve $sindfrac{kpiln(x)}{ln(b)}=0$ so is $dfrac{kpiln(x)}{ln(b)}=mpi$ with $minmathbb Z$ and $xin[0,b]$



          $ln x=dfrac{m}{k}ln b$ And because $xgeq1$, $mgeq0$



          $x=b^{m/k}$



          Now, as $b>1$, $b^{m/k}leq b$ if $dfrac{m}{k}leq 1$ or being $mleq k$



          $m=0,1,2,dots,k$



          Including the values $x=1$ and $x=b$, $y_k$ has $k+1$ zeros.






          share|cite|improve this answer























          • I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
            – Bvss12
            Jan 4 at 19:21










          • Thank you. Fixed.
            – Rafa Budría
            Jan 4 at 20:20
















          1














          Solve $sindfrac{kpiln(x)}{ln(b)}=0$ so is $dfrac{kpiln(x)}{ln(b)}=mpi$ with $minmathbb Z$ and $xin[0,b]$



          $ln x=dfrac{m}{k}ln b$ And because $xgeq1$, $mgeq0$



          $x=b^{m/k}$



          Now, as $b>1$, $b^{m/k}leq b$ if $dfrac{m}{k}leq 1$ or being $mleq k$



          $m=0,1,2,dots,k$



          Including the values $x=1$ and $x=b$, $y_k$ has $k+1$ zeros.






          share|cite|improve this answer























          • I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
            – Bvss12
            Jan 4 at 19:21










          • Thank you. Fixed.
            – Rafa Budría
            Jan 4 at 20:20














          1












          1








          1






          Solve $sindfrac{kpiln(x)}{ln(b)}=0$ so is $dfrac{kpiln(x)}{ln(b)}=mpi$ with $minmathbb Z$ and $xin[0,b]$



          $ln x=dfrac{m}{k}ln b$ And because $xgeq1$, $mgeq0$



          $x=b^{m/k}$



          Now, as $b>1$, $b^{m/k}leq b$ if $dfrac{m}{k}leq 1$ or being $mleq k$



          $m=0,1,2,dots,k$



          Including the values $x=1$ and $x=b$, $y_k$ has $k+1$ zeros.






          share|cite|improve this answer














          Solve $sindfrac{kpiln(x)}{ln(b)}=0$ so is $dfrac{kpiln(x)}{ln(b)}=mpi$ with $minmathbb Z$ and $xin[0,b]$



          $ln x=dfrac{m}{k}ln b$ And because $xgeq1$, $mgeq0$



          $x=b^{m/k}$



          Now, as $b>1$, $b^{m/k}leq b$ if $dfrac{m}{k}leq 1$ or being $mleq k$



          $m=0,1,2,dots,k$



          Including the values $x=1$ and $x=b$, $y_k$ has $k+1$ zeros.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 20:20

























          answered Jan 4 at 17:33









          Rafa BudríaRafa Budría

          5,6201825




          5,6201825












          • I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
            – Bvss12
            Jan 4 at 19:21










          • Thank you. Fixed.
            – Rafa Budría
            Jan 4 at 20:20


















          • I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
            – Bvss12
            Jan 4 at 19:21










          • Thank you. Fixed.
            – Rafa Budría
            Jan 4 at 20:20
















          I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
          – Bvss12
          Jan 4 at 19:21




          I think you have a mistake, $xin[1,b]$ this change this last part of your proof.
          – Bvss12
          Jan 4 at 19:21












          Thank you. Fixed.
          – Rafa Budría
          Jan 4 at 20:20




          Thank you. Fixed.
          – Rafa Budría
          Jan 4 at 20:20


















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