Boundary conditions generalized eigenvalue problem

Question

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. Suppose to use finite differences, what is the matrix that satisfies the boundary conditions $u(0) = u_s(0)=0$ and $u_{ss}(1)=u_{sss}(1)=0$?

Answer 1

With finite difference methods the typical philosophy for imposing boundary conditions on derivatives is to introduce enough extra "ghost" points, outside the computational domain, such that total number of unknowns matches the total number of equations, including the boundary conditions. For example to implement a zero Neumann condition at $s=0$ we add a point $u(-\Delta x)$ and an equation $u(-\Delta s) = u(\Delta s)$. This could either be done explicitly with an expanded solution vector and another row in the matrix, or by substituting into the other discrete equations so that $\partial_{ss} u (0) \approx \frac{u(\Delta s) -2u(0)+u(-\Delta s)}{2\Delta s^2} = \frac{u(\Delta s)-u(0)}{\Delta s^2}$.