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so I am working on this research paper: https://docdro.id/sZsZiYL

Basically, the authors use a so-called Yee grid in order to discretize Maxwell's equations for computational purposes. In the paper, there is a mention of an integration formula that will be applied to the Maxwell-Faraday equation, namely:

$\oint_{\Gamma_S} \vec{E} \cdot \vec{dl} = -\iint_{S_\Gamma} \frac{\partial \vec{B}}{\partial t} \cdot \vec{ds}$

In the paper you find the following grid setup for the integration of the E field:

Integration contour

Now the discretized integral of the left-hand side of our Maxwell-Faraday is given by:

Formula

which they describe in the paper as first-order integration formula

enter image description here

My questions to the respectable members in here are:

  1. what do they mean by the first-order integration formula? does it have a more contemporary name?
  2. What is the deal with the big Os? what do they represent?

Many thanks for considering my request.

PS:

  1. I have already went throw the internet to have a reliable definition of the formula, but it was in vain.
  2. The ds in the paper designates a contour and not an area, I just went with l since it makes more sense to me.
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    $\begingroup$ en.wikipedia.org/wiki/Big_O_notation $\endgroup$
    – Ghoster
    Commented Aug 6, 2023 at 23:14
  • $\begingroup$ Thanks, I will go through it. Any ideas concerning the first question? $\endgroup$ Commented Aug 6, 2023 at 23:15
  • $\begingroup$ “First-order” means they keep terms that are linear in the edge of the loop and ignore terms that are quadratic, cubic, etc. $\endgroup$
    – Ghoster
    Commented Aug 6, 2023 at 23:19
  • $\begingroup$ Hmmm, I still do not fully grasp it. Has this method been named something else in contemporary times? Or is there any source I can refer to that discusses this method? $\endgroup$ Commented Aug 6, 2023 at 23:23
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    $\begingroup$ I believe it is also known as the Euler method. $\endgroup$ Commented Aug 7, 2023 at 0:51

1 Answer 1

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I would say that it is a standard nomenclature still in use. Certainly, you have heard about the second-order or fourth-order Runge-Kutta algorithms.

In your problem, this naming is applied to an algorithm to evaluate a surface integral, but the reason for the name is the same. It has to do with the way the global error goes to zero once the formula for a single element is summed over a finite set of elements required to evaluate the integral (or integrate a differential equation) over a finite domain.

Let me explain the mechanism underlying the name with a simple 1D example.

Trapezoidal rule gets its name from the fact that the integral of a function $f$ over an interval $ [ x_i, x_{i}+h]$ can be evaluated as $$ \int_{x_i}^{x_i+h}f(x)dx = \frac{h}{2}(f(x_i)+f(x_{i+1}))-\frac{h^3}{12}f''(\xi_i) $$ where $\xi_i$ is a point in the interval $(x_i,x_i+h)$, and $f''$ is the second derivative of $f$. When this single-interval rule is applied additively to the $N$ equispaced intervals required to get the integral over the finite interval $[a,b]$, we have $$ \begin{align} &h = \frac{b-a}{N}\\ &x_0=a\\ &x_N=b\\ &\int_{a}^{b}f(x)dx= \frac{h}{2}\sum_{i=0}^{N-1}(f(x_i)+f(x_{i+1}))-\frac{h^3}{12}\sum_{i=0}^{N-1}f''(\xi_i). \end{align} $$ The sum in the error term grows as $N$. It is then convenient to rewrite it as $$ -\frac{h^3}{12}\sum_{i=0}^{N-1}f''(\xi_i)=-N\frac{h^3}{12}\left(\frac{1}{N}\sum_{i=0}^{N-1}f''(\xi_i) \right). $$ Assuming the continuity of $f''$, a point $\xi \in (a,b)$ exists such that $$ \frac{1}{N}\sum_{i=0}^{N-1}f''(\xi_i) =f''(\xi) $$ and we arrive to the expression of the global error as $$ -h^2\frac{(b-a)}{12}f''(\xi). $$ Thus, the global error is second order in $h$, i.e., one order smaller than the local error.

This phenomenon is completely general: a local error of the order $n$ becomes a global error of order $n-1$, providing an answer to why a discretization method with an error $O(h^2)$ is called a first-order method.

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  • $\begingroup$ I noticed that in the formula you provided there is the middle point whereas in the formula I provided there the difference of two points. Is it still the same? $\endgroup$ Commented Aug 8, 2023 at 20:03
  • $\begingroup$ @ShadatSingh If I understand your question, there is no difference. The line element in the formula you cite is indicated as $u_{i+1}-u_i$. If you look at the figure, $u_{i+1}-u_i= \Delta u_i$. For simplicity, I have used a formula with equally spaced nodes and $\Delta u_i = h$ for all the values of $i$. $\endgroup$ Commented Aug 8, 2023 at 20:15
  • $\begingroup$ I apologize for annoying you with my questions. But I was wondering where the /2 came from. $\endgroup$ Commented Aug 8, 2023 at 22:22
  • $\begingroup$ @ShadatSingh It is the trapezoidal rule. The function is linearly interpolated between points. The area under the graph is the area of a trapezium where the two bases are the values of the function at the two points $x_i$ and $x_{i+1}$, and the height is the spacing $h$. See, for example en.wikipedia.org/wiki/Trapezoidal_rule $\endgroup$ Commented Aug 8, 2023 at 22:56
  • $\begingroup$ Ah so in this case $(f(x_i)+f(x_{i+1}))/{2}$ is basically $E_{u,n}$. But hang on a second. In my case, the integration is over the 4 vertices of E multiplied by the line element, shouldn't the left side be equal to the right side with no big O correction term? I mean if we assume for each vertex that E is constant, we can take E outside of the integral and be left of the integral of the line which is actually equal to $u_{i+1}-u_i$. Or am I missing something? $\endgroup$ Commented Aug 8, 2023 at 23:34

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