It is necessary for me to find the shape functions on $L^2(\Omega)$(piecewise constant functions), I searched a lot but that, could not find anything, all of them are about higher degree polynomial. If we define the space as follow: $${\cal Dp0}=\{v \in L^2(\Omega):v|_T \in {\cal p0(T)}, \forall T \in {\cal T}\}$$ where ${\cal T}$ is triangulation of the domain. How can I find them?

  • $\begingroup$ This is essentially what Finite Volume Method consists of. $\endgroup$
    – Paul
    Mar 31 '16 at 1:47

$p_0(T)$ is the space of all functions that are constant on cell $T$. The shape function that corresponds to this is the function that is 1 on cell $T$ and zero everywhere else. Its derivative is then obviously zero in the cell itself, as well as outside the cell, but it is undefined at the interface.

  • 1
    $\begingroup$ @Rosa, Wolfgang's point about cells is that you need to discretize $\Omega$ into a finite number of subregions, usually called elements or cells. Triangles, in two dimensions, are popular. $\endgroup$
    – Bill Barth
    Mar 30 '16 at 19:58
  • $\begingroup$ Ah, yes, the question originally included this information. Yes, $T$ are the cells of a mesh. $\endgroup$ Mar 30 '16 at 22:26
  • $\begingroup$ sorry for removing it, I added some information on it again. $\endgroup$
    – Rosa
    Mar 31 '16 at 7:57

Your Answer

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

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