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 $u_{ss}(1)=u_{sss}(1)=0$?

  • 2
    $\begingroup$ Welcome to SciComp.SE! This very much depends on how you discretize the differential equation (finite differences, finite elements, spectral elements, collocation,...). Please edit your question to give more details. $\endgroup$ Mar 10, 2016 at 13:18
  • $\begingroup$ Is this a homework problem? $\endgroup$ Mar 11, 2016 at 23:42

1 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 $\partial_{ss} u (0) \approx \frac{u(\Delta s) -2u(0)+u(-\Delta s)}{2\Delta s^2} = \frac{u(\Delta s)-u(0)}{\Delta s^2}$.

  • $\begingroup$ yes, I agree. I actually know how to do it in general, but in this case it seems to me rather complicated and I am not able to find the right matrix. $\endgroup$
    – user36390
    Mar 11, 2016 at 10:03
  • $\begingroup$ From a quick inspection, it looks like the boundary at 0 is reasonably straightforward (Dirichlet style condition at $s=0$, low order first derivative approximation to define $u(-\Delta x)$ in terms of $u(\Delta s)$, substitute both into the domain equations for $u(\Delta s)$ and $u(2\Delta s)$. At $s=1$ you have a choice of stencil for your second derivative, it's probably more consistent to use the same one you use for the domain equation. $\endgroup$
    – origimbo
    Mar 11, 2016 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.