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interface value on the error equation

Computational Science Asked on August 22, 2021

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.

One Answer

$delta_n neq lambda_n - lambda_{n-1}$. 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.

Answered by Abdullah Ali Sivas on August 22, 2021

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