I am curious about the projection operator people use in finite element method where they project a function onto the finite-dimensional subspace $V_N$.

I read there the basis they use to generate $V_N$ is not orthogonal or orthnormal. In this case, what projection operator is used? Is it bounded? What properties does it have?

  • $\begingroup$ Usually, one uses a pointwise interpolant. $\endgroup$ – gerw Jun 7 '13 at 12:54
  • $\begingroup$ they do however, mostly use orthogonal basis and standard (or weighted) $L_2$ projection onto $V_N$ $\endgroup$ – leshik Jun 9 '13 at 15:44
  • $\begingroup$ @michael_faber I have edited some more info into my answer, see if you are okay with it. $\endgroup$ – Shuhao Cao Jun 9 '13 at 16:35

It depends on how to define the "Degrees of Freedom" (the coefficient) associated with that subspace, and what kind of projection you want.

When the basis functions $v_i$ are orthonormal, the coefficients are just $\langle\cdot,v_i\rangle$. Finite element space's basis is not meant to be orthogonal but rather be locally-supported.

Here are two examples in a triangulation made of tetrahedra in $\mathbb{R}^3$:

  • First degree piecewise continuous polynomial $\mathbb{P}^1 \subset H^1$. We can project using a pointwise interpolant: $$v\mapsto \Pi v : = \sum_{z\in \{\text{Vertices}\}} v(z) \phi_{z},$$ where $\phi_z = 1$ on vertex $z$ and linearly decreases to zero on its neighbor vertices, zero elsewhere. Quasi-interpolant (when pointwise value is not well-defined) is used for projection sometimes as well: $$ \mathcal{I}v : = \sum_{z\in \{\text{Vertices}\}} (\frac{1}{|\omega_z|}\int_{\omega_z}v )\phi_{z}, $$ where $\omega_z$ is the collection of tetrahedra sharing this vertex $z$.

  • Whitney elements (lowest order Nédélec elements for electric fields) space which is a subspace of $H(\mathbf{curl})$, the projection can be this interpolant: $$ \mathbf{v} \mapsto \Pi\mathbf{v}:= \sum_{e\in \{\text{Edges}\}} (\int_{e}\mathbf{v}\cdot \boldsymbol{\tau}\,ds)\boldsymbol{\psi}_{e}, $$ where $\boldsymbol{\psi}_{e}$ is the Whitney 1-form for edge $e$ consisting two vertices $z_i$, $z_j$: $$ \boldsymbol{\psi}_{e} = \phi_{z_i}\nabla \phi_{z_j} - \phi_{z_j}\nabla \phi_{z_i}. $$ This lives in a bigger picture of discrete de Rham cohomology, if you are interested, Higher-order finite element methods by Šolín, Segeth, and Doležel is a good read.

The interpolant (not the quasi- one) is constructed in a way so that $$ \Pi (\Pi v) = \Pi v, $$ i.e., the interpolant is a projection. Also we have the commutative diagram to inherit the geometrical aspect of de Rham complex: $$\require{AMScd} \begin{CD} H\Lambda^0 @>{\mathrm{d}^0}>> H\Lambda^1 @>{\mathrm{d}^1}>>H\Lambda^2 @>{\mathrm{d}^2}>> H\Lambda^3 \\ @V{\Pi^0_h}VV @V{\Pi^1_h}VV @V{\Pi^2_h}VV @V{\Pi^3_h}VV \\ P\Lambda^0_h @>{\mathrm{d}^0}>> P\Lambda^1_h @>{\mathrm{d}^1}>>P\Lambda^2_h @>{\mathrm{d}^2}>> P\Lambda^3_h \end{CD}$$ Top complex is for the continuous function spaces, bottom is for the finite element subspaces, it is commutative in the sense that $$ P\Lambda^{i+1}_h = \Pi^{i+1}_h\big(\mathrm{d}^i H\Lambda^i\big) = \mathrm{d}^i\big( \Pi^{i}_h H\Lambda^i\big). $$

EDIT: leshik pointed out $L^2$-projection. He said they usually use orthogonal basis, I doubt that. In computation community, using orthogonal basis normally is accredited with spectral methods.

(weighted) $L^2$-projection is certainly a projection too, the coefficients of the basis again are not just $\langle\cdot,v_i\rangle$. Instead, we have to solve a linear system for the projection $\Pi u = \sum u_i v_i$: $$ \langle \Pi u,v_j\rangle_A = \int_{\Omega} A(\sum u_i v_i) v_j = \langle u,v_j\rangle_A = \int_{\Omega} A u v_j \quad j= 1,\ldots,N, $$ where $A$ is a weight, and $\Pi(\Pi u) = \Pi u$ as well.

  • $\begingroup$ Thanks! I agree that it is unusual to use orthogonal basis in FEM. Very interesting answer, did not know about cohomology stuff. $\endgroup$ – michael_faber Jun 10 '13 at 10:37

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.