For 1D derivative we have \begin{equation} F(x) = \frac{\partial f(x)}{\partial x} \end{equation} \begin{equation} f(x)=\sum_{i}f_ie_i(x) \end{equation} \begin{equation} F(x)=\sum_{i}F_ie_i(x) \end{equation} where $e_i(x)$ are the FEM basis functions. We can then apply the Galerkin procedure to the derivative,and then get the matrix form \begin{equation} A\widehat{F}=B\widehat{f} \end{equation} \begin{equation} \widehat{F}=A^{-1}B\widehat{f} \end{equation} $\widehat{F}$and$\widehat{f}$are the vector form of $F_i$and $f_i$.

For the derivative of the product of two functions $f(x)$and $g(x)$,applying product rule we get \begin{equation} \frac{\partial (f(x).g(x))}{\partial x}=f(x). \frac{\partial g(x)}{\partial x}+g(x). \frac{\partial f(x)}{\partial x} \end{equation}

Now in FEM I want the derivative matrix $C=A^{-1}B$ also to fulfill the similar constraint \begin{equation} C \widehat{fg} = [f]_{dig}.C\widehat{g} + [g]_{dig}.C\widehat{f} \end{equation} where $[f]_{dig}$ means diagonal matrix with the values $f_i$ in it.

Does any body know what kinds of basis function $e_i(x)$ I can chose?

  • $\begingroup$ I don't understand this question. $\endgroup$
    – nicoguaro
    Commented Apr 15, 2020 at 22:30

1 Answer 1


The simplest basis functions are piecewise linear functions with value $1$ one mesh point and value $0$ at all the others, shown in this image from here:

enter image description here

Higher order polynomials can be used through Lagrange interpolation (rather than linear) in each "mesh" interval of the 1D domain. There are some good images of quadratic and cubic basis functions on this page.


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.