Let $H=L^2(\Omega)$ and $V=H^1(\Omega)$.

Suppose that $\{v_j\}$ is a basis for $H$ and $V$ (not necessarily orthogonal).

Let $V_m = \text{span}(v_1, ..., v_m)$.

Define a projection operator $P_m:H \to V_m$ satisfying $$(P_m h - h, v_m) = 0 \qquad\text{for all $v_m \in V_m$}.$$

Since $v_j$ is a basis, we can write $h = \sum_{j=1}^\infty a_jv_j$ where $a_j$ are coefficients. Now if $v_j$ were an orthonormal basis of $V$ and an orthogonal basis of $H$, then we simply have $$P_m(h) = \sum_{j=1}^m a_jv_j.$$

Is there any such expression when $v_j$ is not orthogonal?

I am asking because this is the set up used in a Galerkin method. Is there a different way to define projection operators onto the finite dimensional subspace when we have no orthogonal basis?


No matter what, the projection is still orthogonal, so it still takes the form $$\tilde h = P_m(h) = \sum_{j=1}^m a_jv_j$$ i.e. it is in the linear span of the $v_j$'s. If the $v_j$'s are orthonormal, then the coefficients are "easy" to calculate: $$ a_j = \left<h,v_j\right> $$

In general, the coefficients have to satisfy a system of equations. In particular, if you write

$$ \tilde h = \sum_{j=1}^m a_j v_j $$

then you can still solve for the coefficients using the inner product (you just can't do it directly any more):

$$ \left<\tilde h,v_i \right> = \sum_{j=1}^m a_j \left<v_j,v_i\right> $$

which is a system of equations for $a_j$, i.e.

$$ V\begin{pmatrix}a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix} = \begin{pmatrix}h_1 \\ h_2 \\ \vdots \\ h_m \end{pmatrix} $$

where $(V)_{ij} = \left<v_j,v_i\right>$ and $h_i = \left<\tilde h,v_i \right>$.

  • $\begingroup$ Thanks for answering. So you say if $h := \sum_{j=1}^\infty a_jv_j$ then $P_m(h) = \sum_{j=1}^m a_jv_j$ no matter what the basis is.. But this has to satisfy the identity $(P_mh-h, w_m) = 0$ but the LHS is $(\sum_{j=m+1}^\infty a_jv_j, \sum_{j=1}^m b_jv_j)$ which I can't show is zero ($w_m \in V_m$) $\endgroup$ – weasd Feb 13 '14 at 21:28
  • $\begingroup$ Yeah, maybe I'm being too hasty. If the $v_j, j>m$ aren't orthogonal to $V_m$, we have a problem. In your setting, is it possible to switch to an orthonormal basis for $V_m^\perp$? I.e. keep $V_m$ with its non-orthogonal basis, but "gram-schmidt" the rest? $\endgroup$ – BaronVT Feb 13 '14 at 22:40
  • $\begingroup$ Hmm I'm afraid probably not. We want to take limits $m \to \infty$ so we need to consider $V_m$ for each $m$ which probably constrains it to be defined as I did above. $\endgroup$ – weasd Feb 14 '14 at 9:04
  • 1
    $\begingroup$ ok, I'm going to take a think on this, and I'll get back to you $\endgroup$ – BaronVT Feb 14 '14 at 17:01

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