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Hi, In every textbook and at school, one can see the following way to solve for a Poisson equation using FEM:
- (1) start with $\Delta u = b$
- (2) obtain the weak formulation : $\int \Delta u~v~dx = \int b~v~dx$
- (3) integrate by parts to get : $-\int \nabla u \nabla v = \int b~v~dx$
then decompose $u$ and $v$ on finite element basis to get the linear system to solve.

My question is about point (3) : why is it necessary ? Why can't you directly use the $\Delta$ as it is, as it could be any other linear operator (otherwise, how do you solve when you have other linear operators ?).
The motivation for this integration by part is never mentionned (including wikipedia etc.).

Edit: to be more precise, keeping the $\Delta$ still allows to write the problem as $A(u,v)=L(v)$ with $A$ a bilinear form and L a linear form... why do we need to convert it to something else?


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Well, that way you need one order of differentiability less for $u$ which allows rougher elements. Moreover, since integration by parts holds, both left hand sides are the same and not doing it will lead to the same linear system (if $u$ or its ansatz functions are regular enough). – Dirk Mar 31 '11 at 19:19
Thanks - why does it allows rougher elements ? If the basis functions are constant per triangle, they are infinitely differentiable (except at triangle boundaries - is it a problem ?). Can't we directly solve for the laplacian with constant basis functions ? – WhitAngl Apr 1 '11 at 15:10
@WhitAngl - Yes You can. Why not. But what conditions do You have on boundary to obey for the first order difference? – kakaz Apr 1 '11 at 16:50
(2) and (3) will not be equivalent in general; you need appropriate boundary conditions. I think that's what this whole thing is all about. – Zen Harper Apr 6 '11 at 8:07
up vote 1 down vote accepted

Step (3) is, essentially, a way of defining the weak version of the Laplacian. Given $ u \in H^1 $, the classical Laplacian $ \Delta u $ is generally not defined. However, for any test function $ v \in H^1 $, one can define $ (\Delta u, v ) = -(\nabla u, \nabla v) $. In other words, we have $ \Delta \colon H^1 \to H^{-1} $, so if $ f \in H^{-1} $, then the weak problem is precisely equivalent to the operator equation $ \Delta u = f $.

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oh ok, this makes more sense. So, it has nothing to do with boundary conditions, right ? Btw, I know what $H^1$, or $H^p$ with $p\geq 0$ is, but what is $H^{-1}$ ? – WhitAngl Apr 6 '11 at 13:49
$H^{-1}$ is the dual space of $H^1$. This is actually even weaker than being in $ L^2 $ (a.k.a. $H^0$), since it's only a continuous functional when applied to weakly differentiable test functions. For example, consider the 1-D Dirac $\delta$-function, $ \delta = \mathrm{H} ' $, where $\mathrm{H}$ is the Heaviside step function. This is obviously not in $L^2$; however, it is in $H^{-1}$, since $ (\delta, v) = (\mathrm{H}', v) = -(\mathrm{H}, v') \leq \lVert \mathrm{H} \rvert _{L^2} \lVert v \rVert _{H^1} $. This means that you can even make sense of the PDE $ u'' = \delta $. – Ari Apr 6 '11 at 19:10

What are boundary or initial requirements? probably for most physical or technical problems they are set as values of u on certain surfaces. So You have to provide 1-st order equations in order to solve it.

There You have nearly exact remark about this fact

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In the wikipedia page, I can read : "By using integration by parts on the right-hand-side of (1), we obtain :". I know that we obtain that, but why would we want to do so ? I don't see the explanation on the wiki page. Do the boundaries matter ? I mean : if I only have dirichlet conditions, why would reformulating the problem allow to solve it, if it couldn't be done with the original formulation ? Integrating by parts shouldn't change the requirements of the boundary conditions, should-it ? – WhitAngl Apr 1 '11 at 15:20
Here: " ...However, this method of solving the boundary value problem works only when there is only ..." where: . FEm is a method of solving BVP and there is in general no possibility to solve such equations without boundary values. This is internal part of the problem. And as You set values of u on boundary, Your equations need to be translated to first order equations, or You have to set derivative on boundary. I presume You are looking for solution with Dirichlet boundary condition because it is typical specially for Poisson equation. – kakaz Apr 1 '11 at 16:40
Such situation arises because in typical situation You need unique solution of equations... In other words - equation by itself means nothing. It mus have boundary value restraints of different kind in order to obtain unique solution. Tis bonds may be kind of default ( decay in infinity), but in typical situation meas for example that electrical field vanish on certain surfaces (electromagnetism), or that temperature is constant on surface of bodies ( heat flow ). Without such bonds equation by itself cannot be solved in such way ( but You may use other methods for example Green functional) – kakaz Apr 1 '11 at 16:45
If You have second order equations in order to solve it You have to set bonds on value and on first derivative of function: u=f(x) and u'=g(x) on ðG simultaneously. But this is not Dirichlet boundary problem. When You have DBP You only set u=f(x) on ðG so Yo have to have first order differential equations in order to solve it with such boundary conditions. – kakaz Apr 1 '11 at 16:49
ok, thanks, I understand :) I am just slightly surprised that the exact same problem which is just written in two different ways (either with $\nabla$ or $\Delta$, if we integrate by parts) have different BC requirements (Dirichlet won't give a unique solution with the $\Delta$ although it will with the $\nabla$... although these two problems are in fact the same!). Thanks! – WhitAngl Apr 3 '11 at 9:28

Here is an argument inspired by Dirk's comment and Ari's answer that is hopefully easier to understand for people with less background in functional analysis but not rigorous.

The simplest finite element basis to work in is that of piecewise continuous elements. Let's simplify even further and suppose we're solving the problem on the interval [0,1]. Then if we do integration by parts, we get $$ -\int_0^1 u'(x) v'(x) dx = \int_0^1 b(x)v(x) dx. $$ Now $u$ is piecewise linear, so $u'$ is piecewise constant with jumps at the places where the elements meet. There is no problem with evaluating the integral on the left-hand side, even though it has jumps, because the jumps do not contribute to the integral.

However, if we don't do integration by parts, we get $$ \int_0^1 u''(x) v(x) dx = \int_0^1 b(x)v(x) dx. $$ Now what is $u''$? Inside the elements it is zero, so you would expect the integral on the left-hand side to be zero, which is not the same as we had before. This can be resolved, I think, using delta-functions and distribution theory, but that shows that there is an issue here and integration by parts is quite an easy way to side-step this issue.

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