1
$\begingroup$

Consider the following eigenvalue problem \begin{equation} \mathcal {L} x(s) = \lambda x(s), \end{equation} where \begin{equation} \mathcal {L} = \alpha \partial^4_s + (s^2-1)\partial^2_s + s \partial_s + 1, \end{equation} with $\alpha = \rm{const}$ and $s\in[0,1]$. In order to find the eigenvalues and the eigenfunctions, the operator $\mathcal L$ has to be translated into a matrix form. What is the matrix that satisfies the boundary conditions $u(0) = u_s(0)=0$ and $u_{ss}(1)=u_{sss}(1)=0$?

$\endgroup$
  • $\begingroup$ In replacing your operator $\mathcal{L}$ by a matrix, you are essentially assuming that your solution $x(s)$ lies (approximately) in some finite-dimensional vector space. The answer to your question depends on what vector space you want to use: do you want to represent $x(s)$ as a polynomial (of large but finite degree)? Or as a (truncated) Fourier series? Or as a spline basis? Or do you want a finite-difference approach where you represent $x(s)$ by its values at regularly spaced points in $[0,1]$? $\endgroup$ – Wouter Mar 10 '16 at 12:52
  • $\begingroup$ polinomial. thanks for your help! $\endgroup$ – user36390 Mar 10 '16 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.