16
$\begingroup$

I have learnt about Finite Element Method (also a little on other numerical methods) but I don't know what are exactly definition of these two errors and differences between them?

$\endgroup$
  • 5
    $\begingroup$ A priori (from the Latin "from the earlier") estimates depend only on the exact, but not the computed approximate, solution and hence can be (in theory if not in practice) evaluated before computing the solution. Conversely, a posteriori (from the Latin "from the later") estimates depend on the computed solution but not the exact solution, so they do require computing the solution but can actually be evaluated in practice. $\endgroup$ – Christian Clason Feb 24 '16 at 20:21
  • 1
    $\begingroup$ @ChristianClason -- make this an answer! $\endgroup$ – Wolfgang Bangerth Feb 24 '16 at 23:07
23
$\begingroup$

Error estimates usually have the form $$ \|u - u_h\| \leq C(h),$$ where $u$ is the exact solution you are interested in, $u_h$ is a computed approximate solution, $h$ is an approximation parameter you can control, and $C(h)$ is some function of $h$ (among other things). In finite element methods, $u$ is the solution of a partial differential equation and $u_h$ would be the finite element solution for a mesh with mesh size $h$, but you have the same structure in inverse problems (with the regularization parameter $\alpha$ in place of $h$) or iterative methods for solving equations or optimization problems (with the iteration index $k$ -- or rather $1/k$ -- in place of $h$). The point of such an estimate is to help answer the question "If I want to get within, say, $10^{-3}$ of the exact solution, how small do I have to choose $h$?"

The difference between a priori and a posterior estimates is in the form of the right-hand side $C(h)$:

  • In a priori estimates, the right-hand side depends on $h$ (usually explicitly) and $u$, but not on $u_h$. For example, a typical a priori estimate for the finite element approximation of Poisson's equation $-\Delta u = f$ would have the form $$ \|u-u_h\|_{L^2} \leq c h^2 |u|_{H^2},$$ with a constant $c$ depending on the geometry of the domain and the mesh. In principle, the right-hand side can be evaluated prior to computing $u_h$ (hence the name), so you'd be able to choose $h$ before solving anything. In practice, neither $c$ nor $|u|_{H^2}$ is known ($u$ is what you're looking for in the first place), but you can sometimes get order-or-magnitude estimates for $c$ by carefully going through the proofs and for $|u|$ using the data $f$ (which is known). The main use is as a qualitative estimate -- it tells you that if you want to make the error smaller by a factor of four, you need to halve $h$.

  • In a posteriori estimates, the right-hand side depends on $h$ and $u_h$, but not on $u$. A simple residual-based a posterior estimate for Poisson's equation would be $$ \|u-u_h\|_{L^2} \leq c h \|f+\Delta u_h\|_{H^{-1}},$$ which could in theory be evaluated after computing $u_h$. In practice, the $H^{-1}$ norm is problematic to compute, so you'd further manipulate the right-hand side to get an element-wise bound $$ \|u-u_h\|_{L^2} \leq c \left(\sum_{K} h_K^2 \|f+\Delta u_h\|_{L^2(K)} + \sum_{F} h_K^{3/2} \|j(\nabla u_h)\|_{L^2(F)}\right),$$ where the first sum is over the elements $K$ of the triangulation, $h_K$ is the size of $K$, the second sum is over all element boundaries $F$, and $j(\nabla u_h)$ denotes the jump of the normal derivative of $u_h$ across $F$. This is now fully computable after obtaining $u_h$, except for the constant $c$. So again the use is mainly qualitative -- it tells you which elements give a larger error contribution than others, so instead of reducing $h$ uniformly, you just select some elements with large error contributions and make those smaller by subdividing them. This is the basis of adaptive finite element methods.

$\endgroup$
  • $\begingroup$ This answer is exactly what I need, thanks so much. $\endgroup$ – Anh-Thi DINH Feb 25 '16 at 15:51

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.