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https://www.jstor.org/stable/pdf/2157482.pdf, here I have a problem in last equation of (2.6) in section (2.1). When they are considering error equation on the interface $\Gamma$ they get $e_v^{(n)} = \delta_n$, where $\delta_n = \lambda_n - \lambda_{n-1}$(Here I use there definition of $\delta_2 = \lambda_2-\lambda_1$, given below equation (2.7)).

  1. My problem is how $e_v^{(n)} = \delta_n$ on the interface $\Gamma$?

If I calculate $e_v^{(n)}$ on $\Gamma,$ I find something like this, $e_v^{(n)} = v^{(n)} - v|_{\Gamma} = \lambda_n - v|_{\Gamma}$ = $\lambda_n -\lambda_{n-1} + \lambda_{n-1} - v|_{\Gamma}$ = $\delta_n - e_{v}^{(n-1)}.$

  1. Another problem is that the algorithm in section (2.1) start form $n=1$, then whats the value $e_v^{(1)}$ on the interface? (from the given definition $e_v^{(1)} = \delta_1$, but $\delta_1$ is not defined. )

Thanking you.

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  • $\begingroup$ Is there a preprint or open access version of this paper? $\endgroup$ – Abdullah Ali Sivas Jul 15 at 13:48
  • $\begingroup$ You shouldn't direct people to sci-hub. There is a great deal argument about open access science and how closed access publishing is preventing under-developed countries from fostering brilliant people, but the suggestion here is for purely convenience and IMO, unethical. That being said, I accessed the journal through my institution and I will answer your questions below. $\endgroup$ – Abdullah Ali Sivas Jul 15 at 22:25
  • $\begingroup$ I understand. Thank you. $\endgroup$ – 420 Jul 16 at 4:07
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$\delta_n \neq \lambda_n - \lambda_{n-1}$. It is actually defined in (2.8) and (I don't know why) $\delta_2 = \lambda_2 - \lambda_1$. You can confirm it by subtracting first three lines of (2.2) from (2.3) which will give you (2.6).

Section 2.1 is the convergence analysis of the algorithm, so it is okay if $\delta_1$ is not explicitly defined. Actual algorithm is given in (2.3)-(2.5), in which case, you would choose $\lambda_1$ as a proper initial guess. Probably zero would be an okay one.

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  • $\begingroup$ so how the transition happen from $\lambda_{n+1}$ to $\delta_{n+1}$ at interface? $\endgroup$ – 420 Jul 16 at 4:11
  • $\begingroup$ $\lambda_{n+1} = \theta w^{(n)} + (1-\theta)\lambda_n$ on $\Gamma$. Consider $\lambda_{n+1} - v$ and note that $v = \theta w + (1-\theta) v$ on $\Gamma$ for any $\theta$, then $\delta_{n+1} = \lambda_{n+1} - v = \theta (w^{(n)} -w) + (1-\theta)(\lambda_n - v)$ on $\Gamma$. Now $v^{(n)}=\lambda_n$ on $\Gamma$, so $\delta_{n+1} = \theta e_w^{(n)} + (1-\theta) e_v^{(n)}$. $\endgroup$ – Abdullah Ali Sivas Jul 16 at 5:32

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