Boundary value problems: eigenvalue and eigenfunction












2














I'm having trouble in understanding eigenvalues and eigenfunctions in BvP



the problem is:



$y''$ + $lambda$$y$ = $0$



$y(0)=0$



$y(2pi)$ = $0$.



Make characteristic polynomial



$r^2 + lambda = 0$



$r_1,_2 = pm sqrt{- lambda}$



the general solution is :



$$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$



Applying first boundary condition



$0=y(0)=c_1$



and applying the second boundary condition



$0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.



I know the part how to solve BVP I just wanted to know how get



eigenvalue solution:
$$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$



and eigenfunction:



$$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$



IF $lambda > 0$






differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions







share|cite|improve this question





























    2














    I'm having trouble in understanding eigenvalues and eigenfunctions in BvP



    the problem is:



    $y''$ + $lambda$$y$ = $0$



    $y(0)=0$



    $y(2pi)$ = $0$.



    Make characteristic polynomial



    $r^2 + lambda = 0$



    $r_1,_2 = pm sqrt{- lambda}$



    the general solution is :



    $$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$



    Applying first boundary condition



    $0=y(0)=c_1$



    and applying the second boundary condition



    $0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.



    I know the part how to solve BVP I just wanted to know how get



    eigenvalue solution:
    $$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$



    and eigenfunction:



    $$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$



    IF $lambda > 0$






    differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions







    share|cite|improve this question



























      2












      2








      2


      1





      I'm having trouble in understanding eigenvalues and eigenfunctions in BvP



      the problem is:



      $y''$ + $lambda$$y$ = $0$



      $y(0)=0$



      $y(2pi)$ = $0$.



      Make characteristic polynomial



      $r^2 + lambda = 0$



      $r_1,_2 = pm sqrt{- lambda}$



      the general solution is :



      $$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$



      Applying first boundary condition



      $0=y(0)=c_1$



      and applying the second boundary condition



      $0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.



      I know the part how to solve BVP I just wanted to know how get



      eigenvalue solution:
      $$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$



      and eigenfunction:



      $$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$



      IF $lambda > 0$






      differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions







      share|cite|improve this question















      I'm having trouble in understanding eigenvalues and eigenfunctions in BvP



      the problem is:



      $y''$ + $lambda$$y$ = $0$



      $y(0)=0$



      $y(2pi)$ = $0$.



      Make characteristic polynomial



      $r^2 + lambda = 0$



      $r_1,_2 = pm sqrt{- lambda}$



      the general solution is :



      $$y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$$



      Applying first boundary condition



      $0=y(0)=c_1$



      and applying the second boundary condition



      $0=y(pi)=c_2 sinleft(2pisqrt{lambda}right)$.



      I know the part how to solve BVP I just wanted to know how get



      eigenvalue solution:
      $$lambda_n = left(frac n2right)^2 = frac {n^2}{4},quad n=1,2,3...$$



      and eigenfunction:



      $$y_n= sin left(frac {nx}{2}right),quad n=1,2,3....$$



      IF $lambda > 0$






      differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions




      differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jun 29 '16 at 12:00









      Davide Giraudo

      125k16150259




      125k16150259










      asked Jun 22 '16 at 8:09









      clutchnic

      246




      246






















          1 Answer
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          0














          As your said,



          $y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$



          and $0=y(0)=c_1$.



          The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$



          Indeed, we require



          begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}



          consequently, we deduce

          begin{align}
          &sqrtlambda=frac{n}{2}\
          &lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
          end{align}
          Thusbegin{align}
          y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}






          share|cite|improve this answer





















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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            As your said,



            $y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$



            and $0=y(0)=c_1$.



            The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$



            Indeed, we require



            begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}



            consequently, we deduce

            begin{align}
            &sqrtlambda=frac{n}{2}\
            &lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
            end{align}
            Thusbegin{align}
            y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}






            share|cite|improve this answer


























              0














              As your said,



              $y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$



              and $0=y(0)=c_1$.



              The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$



              Indeed, we require



              begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}



              consequently, we deduce

              begin{align}
              &sqrtlambda=frac{n}{2}\
              &lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
              end{align}
              Thusbegin{align}
              y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}






              share|cite|improve this answer
























                0












                0








                0






                As your said,



                $y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$



                and $0=y(0)=c_1$.



                The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$



                Indeed, we require



                begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}



                consequently, we deduce

                begin{align}
                &sqrtlambda=frac{n}{2}\
                &lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
                end{align}
                Thusbegin{align}
                y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}






                share|cite|improve this answer












                As your said,



                $y(x) = c_1 cosleft(sqrt{lambda}xright) + c_2 sinleft(sqrt{lambda}xright).$



                and $0=y(0)=c_1$.



                The key step is how to solve $0=y(2pi)=c_2 sinleft(2pisqrt{lambda}right)$



                Indeed, we require



                begin{align}2pisqrt{lambda}=npi, mbox{ where } n=1,2,cdotsend{align}



                consequently, we deduce

                begin{align}
                &sqrtlambda=frac{n}{2}\
                &lambda=left(frac{n}{2}right) ^2=frac{n^2}{4},mbox{ where } n=1,2,cdots
                end{align}
                Thusbegin{align}
                y_n= sin left(frac {nx}{2}right),quad n=1,2,....end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 '17 at 17:50









                Canrong Tian

                165




                165






























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