What functions can be represented as a series of eigenfunctions












7












$begingroup$


Consider the differential equation:



$y'' = lambda y$



with the boundary conditions
$y(0) = y(2pi) = 0$.



This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$




  • Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
    $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?


  • Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?











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$endgroup$

















    7












    $begingroup$


    Consider the differential equation:



    $y'' = lambda y$



    with the boundary conditions
    $y(0) = y(2pi) = 0$.



    This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$




    • Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
      $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?


    • Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?











    share|cite|improve this question











    $endgroup$















      7












      7








      7


      1



      $begingroup$


      Consider the differential equation:



      $y'' = lambda y$



      with the boundary conditions
      $y(0) = y(2pi) = 0$.



      This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$




      • Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
        $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?


      • Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?











      share|cite|improve this question











      $endgroup$




      Consider the differential equation:



      $y'' = lambda y$



      with the boundary conditions
      $y(0) = y(2pi) = 0$.



      This equation has eigenfunctions $mu_n(x) = sin(frac{nx}{2})$ with the corresponding eigenvalues $lambda_n = -frac{n^2}{4}$ for $n > 0$




      • Am I right, that certain functions f(x) satisfying the same boundary conditions as above can be represented as an infinite series
        $f(x) = sum_1^infty c_n mu_n(x)$ with coefficients $c_n = frac{langle f,mu_n rangle}{langle mu_n, mu_n rangle}$? What conditions those certain functions need to satisfy?


      • Can the previous claim be generalised for any set of eigenfunctions of some differential equation? I.E. suppose $Ly = lambda y$ is a differential equation ($L$ being the 2nd order differential operator) with boundary conditions $y(a) = y(b) = c$ . What functions can be represented as a weighted sum of the eigensolutions?








      functional-analysis ordinary-differential-equations fourier-series eigenfunctions






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      edited Mar 11 at 17:02







      mercury0114

















      asked Mar 11 at 11:25









      mercury0114mercury0114

      20118




      20118






















          3 Answers
          3






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          5












          $begingroup$

          Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
            $$
            y''=lambda y,;;; a le x le b, \
            cosalpha y(a)+sinalpha y'(a) = 0 \
            cosbeta y(b)+sinbeta y'(b) = 0
            $$

            This gives rise to a discrete set of eigenvalues
            $$
            lambda_1 < lambda_2 < lambda_3 < cdots,
            $$



            and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
              $$
              Sf_n:=sum_{k=1}^n c_n mu_n, $$

              it holds that
              $$
              int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$



              If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.






              share|cite|improve this answer









              $endgroup$













                Your Answer





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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)






                share|cite|improve this answer











                $endgroup$


















                  5












                  $begingroup$

                  Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)






                  share|cite|improve this answer











                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)






                    share|cite|improve this answer











                    $endgroup$



                    Your question situates within the realm of Sturm-Liouville theory and my subsequent answer applies to all differential operators (with associated eigenfunctions) that belong to this realm. You asked which continuous functions can be expanded in a series of eigenfunctions? The most natural answer turns out to be "functions that are square-integrable on the interval $(0,2pi)$". Also non-continuous, square-integrable functions $f$ can be expanded in this way (under the proviso that $f$ is allowed to differ from its series-expansion $sum_{1}^{infty} frac{langle mu_n,frangle}{langle mu_n,mu_nrangle} mu_n(x)$ on a subset of $(0,2pi)$ of measure zero)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 11 at 17:21

























                    answered Mar 11 at 12:56









                    Thibaut DemaerelThibaut Demaerel

                    593312




                    593312























                        1












                        $begingroup$

                        Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
                        $$
                        y''=lambda y,;;; a le x le b, \
                        cosalpha y(a)+sinalpha y'(a) = 0 \
                        cosbeta y(b)+sinbeta y'(b) = 0
                        $$

                        This gives rise to a discrete set of eigenvalues
                        $$
                        lambda_1 < lambda_2 < lambda_3 < cdots,
                        $$



                        and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
                          $$
                          y''=lambda y,;;; a le x le b, \
                          cosalpha y(a)+sinalpha y'(a) = 0 \
                          cosbeta y(b)+sinbeta y'(b) = 0
                          $$

                          This gives rise to a discrete set of eigenvalues
                          $$
                          lambda_1 < lambda_2 < lambda_3 < cdots,
                          $$



                          and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
                            $$
                            y''=lambda y,;;; a le x le b, \
                            cosalpha y(a)+sinalpha y'(a) = 0 \
                            cosbeta y(b)+sinbeta y'(b) = 0
                            $$

                            This gives rise to a discrete set of eigenvalues
                            $$
                            lambda_1 < lambda_2 < lambda_3 < cdots,
                            $$



                            and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.






                            share|cite|improve this answer









                            $endgroup$



                            Consider the following problem on the interval $[a,b]$ for some $a < b$ and real angles $alpha,beta$:
                            $$
                            y''=lambda y,;;; a le x le b, \
                            cosalpha y(a)+sinalpha y'(a) = 0 \
                            cosbeta y(b)+sinbeta y'(b) = 0
                            $$

                            This gives rise to a discrete set of eigenvalues
                            $$
                            lambda_1 < lambda_2 < lambda_3 < cdots,
                            $$



                            and associated eigenfunctions $phi_n(x)$. For any function $fin L^2[a,b]$, the Fourier series for $f$ in these eigenfunctions converges to $f$ in $L^2[a,b]$. And you get pointwise convergence of the series at $xin(a,b)$ under the same type of Fourier conditions that you learned for the ordinary Fourier series. The endpoint conditions make the convergence at $x=a,b$ trickier, of course. Conditions of the type $y(a)=0=y(b)$, for example, forces any series in these eigenfunctions to converge to $0$ at the endpoints.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 11 at 14:00









                            DisintegratingByPartsDisintegratingByParts

                            59.8k42681




                            59.8k42681























                                0












                                $begingroup$

                                You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
                                $$
                                Sf_n:=sum_{k=1}^n c_n mu_n, $$

                                it holds that
                                $$
                                int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$



                                If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
                                  $$
                                  Sf_n:=sum_{k=1}^n c_n mu_n, $$

                                  it holds that
                                  $$
                                  int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$



                                  If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
                                    $$
                                    Sf_n:=sum_{k=1}^n c_n mu_n, $$

                                    it holds that
                                    $$
                                    int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$



                                    If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.






                                    share|cite|improve this answer









                                    $endgroup$



                                    You can expand all continuous functions, and even the non-continuous ones (but they must be square-integrable); however, the series will only converge in the $L^2(0, 1)$ sense. This means that, defining
                                    $$
                                    Sf_n:=sum_{k=1}^n c_n mu_n, $$

                                    it holds that
                                    $$
                                    int_0^1 |Sf_n(x)-f(x)|^2, dxto 0qquad text{as }nto infty.$$



                                    If you want pointwise convergence, you will need some regularity assumptions on $f$. This is a classical problem in Fourier analysis.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Mar 11 at 13:04









                                    Giuseppe NegroGiuseppe Negro

                                    17.5k332126




                                    17.5k332126






























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