FEM on a Laplacian - MathOverflow

FEM on a Laplacian

2011-03-31T16:17:08Z

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?

Thanks

Answer by kakaz for FEM on a Laplacian

2011-04-01T12:51:00Z

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 http://en.wikipedia.org/wiki/Finite_element_method#Technical_discussion

Answer by Ari for FEM on a Laplacian

2011-04-06T06:38:40Z

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 $.

Answer by Jitse Niesen for FEM on a Laplacian

2011-04-06T11:46:31Z

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.