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https://www.unige.ch/~gander/Preprints/42540.pdf. here I have a problem in section $4,$ of approximating the symbol $\sigma_i(k)$. My understanding is, to get back the operator $S_i$ we have to use inverse Fourier transform, but $\sigma_i(k)$ is in square root form and that creates a problem. I think i am not totally clear on this. Please help.

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  • $\begingroup$ Can you describe your problem and add the important equations to your question? $\endgroup$
    – nicoguaro
    Jun 3 '20 at 19:43
  • $\begingroup$ This is a dense paper to read. I will try to check it soon and help. However, you should clarify your question, in addition to explaining the paper a little bit so that it is easier to answer. There aren't that many domain decomposition/numerical linear algebra experts on the earth who can answer your question easily. Maybe Martin Gardner himself but I am not sure if he has a stackexchange account. $\endgroup$ Jun 3 '20 at 22:02
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Here's the flow of logic, hopefully in a slightly more readable form:

  1. We write the transmission conditions with as-yet-unknown linear operators $\mathcal{S}_1$ and $\mathcal{S}_2$ which operate on the fields $u_1$ and $u_2$ at the interface.
  2. Perform a Fourier transform in the direction $y$ which is tangential to the interface. $k$ is the spatial frequency in the $y$-direction. This turns derivatives in the $y$-direction into algebraic constants $-ik$. This also turns the operators $\mathcal{S}_i$ into the multiplicative factor, but still to-be-determined, $\sigma_i(k)$.
  3. In this Fourier domain, find the optimal $\sigma_i(k)$ such that an iterative solver will converge in exactly 2 iterations. This ideal operator is known as the Dirichlet-to-Neumann (DtN) map.
  4. Here's the catch. Were we "lucky" enough that the optimal $\sigma_i(k)$ was linear in $k$, then it would transform back into the spatial domain with an $\mathcal{S}_i$ that was just a first order derivative w.r.t. $y$. In fact, were $\sigma_i(k)$ any finite polynomial in $k$ then $\mathcal{S}_i$ would still be just a combination of higher-order spatial derivatives. These would all be "local" operators in the sense that the output of $\mathcal{S}_i[u_i](y)$ for some specific $y$ is dependent only on $u$ and its derivatives at $y$.

    However, since the optimal $\sigma_i(k)$ is of the form $\pm \sqrt{k^2+\eta}$, the corresponding $\mathcal{S}_i$ turns out to be non-local. We can write it in complete form something like:

    $\mathcal{S}[u](y) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} \left( \sqrt{k^2+\eta} \int\limits_{-\infty}^{\infty} e^{-iky'} u(y') dy' \right) e^{iky} dk$

    Note the use of the dummy variable $y'$ inside the inner integral. Since we can't simplify this operator into some local form, the value of $\mathcal{S}[u]$ evaluated at $y$ depends on the values of $u(y')$ for all $y'$. This is called a "pseudo-differential" operator and is non-local.
  5. Implementation of the non-local operator would be computationally intractable in most situations since it would result in a dense matrix over the degrees-of-freedom on the interface surface, so instead we consider local operators which can be thought of as approximations of this ideal DtN map. That's where the real fun begins.
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  • $\begingroup$ Step-3 of your answer says $\sigma_i(k)$ as $ DtN$ map. But, normally $DtN$ is defined as $DtN: u|_{\partial\Omega}\rightarrow \frac{\partial u}{\partial n}|_{\partial \Omega}$. Are these two things connected? $\endgroup$
    – 420
    Jun 5 '20 at 19:17
  • $\begingroup$ @420 Frankly, I'm probably overextending my understanding of how the DtN map applies here. I used the term because of how I've heard it used in conversations with colleagues and because it's used in this context on page 718 in your reference. If others would like to comment on if I'm using this term correctly, I'd be grateful. That said, note that equation (3.1) in your Gander paper has the term $(\partial_x + \mathcal{S}_i)(u_i)$. Clearly the operator $\mathcal{S}_i$ must be a mapping from $u_i$ to $\partial_x(u_i)$. $\endgroup$
    – LedHead
    Jun 5 '20 at 20:21

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