In certain cases, boundary integral methods are preferred for elliptic partial differential equations as opposed to finite difference methods. For instance, for solving the Poisson equation in a domain $D$ $$\nabla^2 u = \phi_0 \tag{$\star$}$$ with boundary condition $u = f$ on $\partial D$, instead of discretizing $(\star)$, I often see the boundary integral formulation being solved, i.e., $$u(x) = \int_{\partial D} \sigma(s) \dfrac{\partial G(x,y(s))}{\partial n_y} ds + \int_{D} G(x,y) \phi_0(y) dy$$ where the unknown density $\sigma(s)$ is obtained by enforcing the boundary condition, i.e., $$\dfrac{\sigma(x)}2 + \int_{\partial D} \sigma(s) \dfrac{\partial G(x,y(s))}{\partial n_y} ds + \int_{D} G(x,y) \phi_0(y) dy = f(x) \tag{$\dagger$}$$ What is the advantage of solving $(\dagger)$ instead of $(\star)$? Note that in $(\dagger)$, one integral is done over the entire domain, while the other is done over the boundary. Below are the list of questions, which I have:

  • What dictates the choice of boundary integral method and finite difference method, i.e., under what circumstances are boundary integral methods preferred over finite difference?
  • Is one more computationally accurate and/or easier and/or inexpensive than the other?

Overall, I would like to have a comparison between boundary integral methods and finite difference methods highlighting the pros and cons of both the methods.

  • $\begingroup$ The boundary element method has its applications, but I don't think it's correct to say that it's used more frequently than finite difference or finite element methods. The latter are definitely much more widely in use than the BEM. $\endgroup$ Sep 27 '13 at 2:03
  • $\begingroup$ @WolfgangBangerth Agreed. I have edited the question now. Essentially, I want a comparison of both the techniques highlighting the pros and cons. $\endgroup$
    – John Smith
    Sep 27 '13 at 2:15
  • $\begingroup$ I don't think that such a broad question fits into the SE format. First it is not clear how proficient you are into both methods: do we have to assume that you have a good working knowledge of FD/BIE/BEM or not? Poisson equation with Dirichlet BC is a very broad class of problems: could you please restrict your question a little bit? $\endgroup$
    – Stefano M
    Sep 27 '13 at 21:31
  • 1
    $\begingroup$ @StefanoM I don't think this question is too broad. All I want to know is, in general, when would one choose one over the other and what factors influence the choice. Also, I do not see how I can restrict this question. In fact the question is very specific; it is Poisson with Dirichlet boundary conditions. $\endgroup$
    – John Smith
    Sep 27 '13 at 22:32
  • $\begingroup$ @JohnSmith I think it matters what your goal is. If you have a very specific problem that BEM can solve then of course go for it, numerically solving a few integrals is most likely cheaper than finite-differencing. $\endgroup$ Sep 28 '13 at 8:25

The question asked is too broad, in my opinion.

Assessment of BIE methods vs. finite difference methods (or other domain methods) require careful analysis of many different points. Among them

  • is $D \subset \mathbb{R}^2$ or $D \subset \mathbb{R}^3$ or even $D \subset \mathbb{R}^N$ with $N>3$?

  • is $D$ finite or infinite? is $D$ simply connected?

  • which is the $\partial D$-surface to $D$-volume ratio compared to a domain diameter?

  • how smooth is $\partial D$? is the analytical solution singular at the boundary or not?

  • are you interested in an overall picture of the solution on $D$ ore you are looking for a very precise solution in just a few points?

  • if the solution is singular at some point, are interested in obtaining a very good asymptotic approximation at those points?

This said, I will not go into summarizing BIE methods (this could be the subject of a book chapter!) but just add some random remarks.

  1. For sure FD on $(\star)$ is easier than say collocation on $(\dagger)$: $G(x,y)$ is singular for $y\rightarrow x$, you have to master CPV and FP integration to correctly implement a BIE method.

  2. With FD you compute an approximation of $u$ in $D$. With BIE you compute quantities on $\partial D$ and you have to recover values on $D$ by an integral representation.

  3. FD give raise to sparse matrices, BIE methods to dense ones.

So there are specific applications (class of problems) in which BIE methods are superior to FD methods, and situations in which they are just terrible, but I don't think it is possible to summarize them in a short answer.

  • $\begingroup$ Very thorough answer. IMO, it's your second bullet that drives the decision - if you need to know the "exterior/infinite" behavior of a field, BEM seems a more natural choice compared to the finite methods. They can only approximate that far-field behavior with judiciously chosen boundary conditions, e.g. ABC's/PML's (well, for wave equations... not sure what the equivalent choices are for elliptic problems). $\endgroup$ Oct 1 '13 at 12:00
  • $\begingroup$ Note that you easily can combine FD/FEM and BEM methods. Once you know the solution on the boundaries of your domain, you can use Green's functions to propagate your solution to infinity or the far exterior easily. $\endgroup$
    – AlexE
    Oct 29 '13 at 10:52
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    $\begingroup$ @AlexE you are right, there is a big family of interesting FD/FEM coupling schemes. Let me however point out that you are referring to an uncoupled solution (first near field, and then far field) that is very special, in the sense that you should be able to solve for the far field without explicit imposing radiation condition at infinity. There are cases in which BC's at infinity are an integral part of the problem and you have to solve coupled FD and BEM equations. An example of a coupled FD-BEM is when you have non-linear equations which become linear in the far field. $\endgroup$
    – Stefano M
    Oct 30 '13 at 20:25
  • $\begingroup$ @Stefano M, for my problems, the BEM solution's behaviour is build into the Green's functions - so any solution computed by considering the domain boundaries only automatically obeys Sommerfeld's radiation condition. I haven't really thought about other cases, I have to confess :) $\endgroup$
    – AlexE
    Nov 4 '13 at 12:24

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