FEM on a Laplacian - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:15:24Z http://mathoverflow.net/feeds/question/60200 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60200/fem-on-a-laplacian FEM on a Laplacian WhitAngl 2011-03-31T16:17:08Z 2011-04-06T11:46:31Z <p>Hi, In every textbook and at school, one can see the following way to solve for a Poisson equation using FEM:<br> - (1) start with $\Delta u = b$<br> - (2) obtain the weak formulation : $\int \Delta u~v~dx = \int b~v~dx$<br> - (3) integrate by parts to get : $-\int \nabla u \nabla v = \int b~v~dx$<br> then decompose $u$ and $v$ on finite element basis to get the linear system to solve.</p> <p>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 ?).<br> The motivation for this integration by part is never mentionned (including wikipedia etc.).</p> <p>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?</p> <p>Thanks</p> http://mathoverflow.net/questions/60200/fem-on-a-laplacian/60285#60285 Answer by kakaz for FEM on a Laplacian kakaz 2011-04-01T12:51:00Z 2011-04-01T12:57:10Z <p>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.</p> <p>There You have nearly exact remark about this fact <a href="http://en.wikipedia.org/wiki/Finite_element_method#Technical_discussion" rel="nofollow">http://en.wikipedia.org/wiki/Finite_element_method#Technical_discussion</a></p> http://mathoverflow.net/questions/60200/fem-on-a-laplacian/60774#60774 Answer by Ari for FEM on a Laplacian Ari 2011-04-06T06:38:40Z 2011-04-06T06:38:40Z <p>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 $\Delta u = f$.</p> http://mathoverflow.net/questions/60200/fem-on-a-laplacian/60804#60804 Answer by Jitse Niesen for FEM on a Laplacian Jitse Niesen 2011-04-06T11:46:31Z 2011-04-06T11:46:31Z <p>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.</p> <p>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.</p> <p>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.</p>