If one has a sufficiently smooth function $u$ that is approximated by a piecewise constant function $u_h=\Pi^0_h u$ on a mesh of cell size $h$ (where $\Pi^0_h$ is the $L_2$ projection onto the constants on each cell), then it is not hard to see graphically (or from the first term of the Taylor expansion) that the size of jumps must be ${\cal O}(h)$.

But what if I use piecewise linear functions? Taylor expansion gives me that I should expect the difference between $u$ and $u_h$ at the end points of each cell to be ${\cal O}(h^2)$. But it is not inconceivable that the "jump" of $u_h$ at these end points (i.e., the difference $|u_h(x_+)-u_h(x_-)|$ when approaching the point $x$ that separates two cells from the left and right) might be of higher order than just two. How about projections $u_h$ into even higher order discontinuous spaces — say, of polynomial degree $k$: Is the jump then of size ${\cal O}(h^{k+1})$, or is it even better?

I am certain this is a pretty standard question for those who do error estimation for Discontinuous Galerkin finite element methods, but it's a bit outside my theoretic knowledge — and so help would be appreciated!

As a side note, here's what I'm really after: I have an estimate that involves on the right hand side the term $h^{s+1} \|u_h\|_{H^s}$. Since $u_h=\Pi^k u$ is piecewise constant, it's not in $H^1$ or higher, but only $L_2$. So $s=0$ is really the only choice. That's likely the best I can do for piecewise constants ($k=0$).

But at least for higher degree piecewise polynomials, the jump becomes pretty small, and so while $\Pi_h^k u\not\in H^s$ for $s>0$, one would expect that one could find an asymptotically equivalent norm $\|\cdot\|_\ast$ so that $\|\Pi_h^k u\|_{\ast,H^s} \le C h^{-p} \|u\|_{H^s}$ or some such, for a power $p$ that is probably related to the polynomial degree $k$ of the space into which we project and, I believe, the space dimension. This estimate blows up, of course, as $h\rightarrow 0$, but at a rate that is predictable.

I want this because then I could replace my poor estimate $h\|u_h\|_{L_2}$ by something more like $h^{s-p} \|u\|_{H^s}$. The question is what $p$ needs to be: If the blow-up happens sufficiently slowly, then the power $s-p$ may be substantially better than the order 1 one gets by just taking the simple route and estimating via $h\|\Pi_h^k u\|_{L_2}$. Calculating $p$ clearly involves the size of the jump of $\Pi_hu$ from one cell to the next, thus my question.


1 Answer 1


I believe that for odd polynomial degrees you can get superconvergence by one order at the jump. For even degrees I don't think that's the case.

For the odd case, we consider a smooth function $u:\mathbb{R}\to\mathbb{R}$. Consider two adjacent intervals $I^-=[x_{i-1},x_i]$ and $I^+=[x_i,x_{i+1}]$ both of length $h$, and denote the $L^2$ projection of $u$ on each interval by $u_h^-$ and $u_h^+$ respectively. We write these projections as truncated Legendre expansions of degree $k$, $$ u_h^\pm(x) = \sum_{j=0}^k c_j^\pm P_j(\xi_\pm(x)), $$ where $\xi_\pm$ simply maps from $I^\pm$ to the reference interval $[-1,1]$. Note that $$ u(x) = u_h^\pm(x) + \sum_{j=k+1}^\infty c_j^\pm P_j(\xi_\pm(x)) $$ and the coefficients $c_j^\pm$ scale like $h^j$. So, the jump term is given by $$ u_h^+(x_i) - u_h^-(x_i) = \sum_{j=k+1}^\infty \left(c_j^-P_j(1) - c_j^+ P_j(-1) \right) \\ = \left(c_{k+1}^- P_{k+1}(1) - c_{k+1}^+ P_{k+1}(-1)\right) + \mathcal{O}(h^{k+2}). $$ Note that since $k$ is odd, $P_{k+1}(\pm 1) = 1$. So the jump is given by $$ u_h^+(x_i) - u_h^-(x_i) = (c_{k+1}^- - c_{k+1}^+) + \mathcal{O}(h^{k+2}). $$ By Rodrigues' formula, we have that $$ c_{k+1}^\pm = C \int_{-1}^1 \left(\frac{\partial^{k+1}}{\partial x^{k+1}} u(\xi_\pm^{-1}(x)) \right) (1 - x^2)^{k+1} \, dx \\ = C h^{k+1} \int_{-1}^1 u^{(k+1)}(\xi_\pm^{-1}(x)) (1 - x^2)^{k+1} \, dx. $$ Thus, the difference between the coefficients in neighboring cells is $$ c_{k+1}^- - c_{k+1}^+ = C h^{k+1}\int_{-1}^1 \left(u^{(k+1)}(\xi_-^{-1}(x)) - u^{(k+1)}(\xi_+^{-1}(x)) \right) (1 - x^2)^{k+1} \, dx. $$ But, $\xi_+^{-1}(x) = \xi_-^{-1}(x) + h$, and so a Taylor argument gives that the integrand scales like $\mathcal{O}(h)$. Thus $c_{k+1}^- - c_{k+1}^+ = \mathcal{O}(h^{k+2})$ and we have for the jump $$ u_h^+(x_i) - u_h^-(x_i) = \mathcal{O}(h^{k+2}). $$

  • $\begingroup$ Very cool -- nice argument! $\endgroup$ Feb 6, 2019 at 16:03
  • $\begingroup$ What goes wrong for even polynomials? You get $P_{k+1}(1)=P_{k+1}(-1)$ and so the think in the second to last equation on the left would be a sum instead of a difference in the integrand, which is not amenable to the Taylor cancellation? $\endgroup$ Feb 6, 2019 at 16:06
  • $\begingroup$ I will add that it makes sense that only odd polynomial degrees enjoy this cancellation. This is clear from just drawing a few pictures. $\endgroup$ Feb 6, 2019 at 16:06

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