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In some of my computations I calculate a scalar value $\lambda_h$ (in my case an eigenvalue) depending on a finite element discretization of the domain. Usually we can manage to find estimates of the form $$ |\lambda - \lambda_h| \leq C h^k $$ where $k$ depends on the problem, the degree of the approximation, etc.

Question:

I was wondering under which circumstances (if any) it is possible to say that the error of a scalar function depending on the FE approximation has a Taylor-like expansion: $$ \lambda_h = \lambda_{exact} + C_1h^{\alpha_1}+C_2h^{\alpha_2}+...$$ where $\alpha_i$ is an increasing sequence.

If there are any relevant references, I am interested.


Context:

This is in connection with one eariler question of mine, regarding the good behavior of some extrapolation techniques applied to my problem. It turns out that the extrapolation I use (Wynn's epsilon algorithm) gives the limit of the sum of $n$ geometric sequences starting from $2n+1$ values. The existence of a Taylor-like expansion for the eigenvalue would justify this behavior since the first few Taylor coefficients will be cancelled when doing the extrapolation.

In the example presented below the initial computations using only finite elements give a convergence of order 2, while the extrapolation gives something like order 6, quickly running towards machine precision. See the picture: enter image description here

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  • $\begingroup$ Do you mean an expansion element wise, or globally? For the former case, and the less exotic element choices, your criterion is the same as asking if $\lambda_{exact}$ has such an expansion. $\endgroup$ – origimbo Feb 21 '18 at 13:24
  • $\begingroup$ I mean globally, for the scalar functional, not elementwise. The finite element space is Lagrange P1, the most basic one... $\endgroup$ – Beni Bogosel Feb 21 '18 at 19:28
  • $\begingroup$ In your first equation, I think you mean $h$ on the right hand side. $\endgroup$ – Wolfgang Bangerth Feb 22 '18 at 23:39
  • $\begingroup$ @WolfgangBangerth: Yes, thank you for pointing that out. $\endgroup$ – Beni Bogosel Feb 23 '18 at 12:04
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The question you ask is simpler than you may think. For any mesh size $h$, you can compute $\lambda(h)$. If this dependence is continuous, then you can do a Taylor expansion of $\lambda(h)$ around $h=0$ to obtain exactly the form you want.

The only question in your case is whether $\lambda(h)$ is a continuous function. This is made slightly more complicated than one may think because in fact not all values of $h$ are of course possible, but at least for sufficiently small $h$, you can evaluate $\lambda$ for essentially any $h$. In that case, let us assume that $h$ can be chosen arbitrarily, and then the question can be reduced to the following: If $\lambda$ is a function of the solution $u(h)$, it is continuous if $u(h)$ is continuous as a function of $h$. For most problems, $u(h)$ is indeed continuous, and in that case $\lambda(h)$ is a continuous function if $\lambda(u)$ is continuous. Whether that is true is typically a functional analysis question: if $\lambda(u)$ is, for example, the value of $u$ at a particular point, then for $H^1$ functions $u$, the function $\lambda(u)$ is not continuous. But if $\lambda(u)$ is the average value of $u$, then the dependence is continuous.

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  • $\begingroup$ Thank you for your answer. In my case the function $\lambda(h)$ is the first eigenvalue of an elliptic operator (Laplace-Beltrami). It verifies something like $-\Delta_h u_h = \lambda_h u_h$, where $u_h$ is non-negative (the first eigenfunction) and $L^2$ normalized. I guess in this case $\lambda(h)$ should be regular enough for the development to work... $\endgroup$ – Beni Bogosel Feb 23 '18 at 12:10
  • $\begingroup$ @BeniBogosel -- yes, at least if the eigenvalue is single (i.e., has multiplicity one) I would expect that to be true. $\endgroup$ – Wolfgang Bangerth Feb 24 '18 at 20:43
  • $\begingroup$ Do you have any reference in mind for your statement: "if this dependence is continuous, then you can do a Taylor expansion around 0"? Thank you. $\endgroup$ – Beni Bogosel Mar 3 '18 at 13:06
  • $\begingroup$ @BeniBogosel -- nope, it's just my intuition, backed by 20 years of experience. $\endgroup$ – Wolfgang Bangerth Mar 5 '18 at 2:13

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