Consider a galerkin problem in the form of 316 using

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n a nite-dimensional subspace S N of H 1. Selecting a basis f j gN=1 for S N , we j N of u in the form consider approximations U 2 SE U (x y ) = N X j =1 cj j (x y): (3.2.1) N With approximations V 2 S0 of v having a similar form, we determine U as the solution of A(V U ) = (V f )+ < V > 8V 2 S0N : (3.2.2) N N (Nontrivial essential boundary conditions introduce di erences between SE and S0 and we have not explicitly identi ed these di erences in (3.2.2).) We've mentioned the criticality of knowing the minimum smoothness requirements of an approximating space S N . Smooth (e.g. C 1) approximations are di cult to construct on nonuniform two- and three-dimensional meshes. We have already seen that smoothness requirements of the solutions of partial di erential equations are usually expressed in terms of Sobolev spaces, so let us de ne these spaces and examine some of their properties. First, let's review some preliminaries from linear algebra and functional analysis. De nition 3.2.1. V is a linear space if 1. u v 2 V then u + v 2 V , 2. u 2 V then u 2 V , for all constants , and 3. u v 2 V then u + v 2 V , for all constants , . De nition 3.2.2. A(u v) is a bilinear form on V V if, for u v w 2 V and all constants and , 1. A(u v) 2 <, and 2. A(u v) is linear in each argument thus, A(u v + w) = A(u v) + A(u w) A( u + v w) = A(u w) + A(v w): 3.2. Function Spaces and Approximation 11 De nition 3.2.3. An inner product A(u v) is a bilinear form on V V that 1. is symmetric in the sense that A(u v) = A(v u), 8u v 2 V , and 2. A(u u) > 0, u 6= 0 and A(0 0) = 0, 8u 2 V . De nition 3.2.4. The norm k kA associated with the inner product A(u v) is p kukA = A(u u) (3.2.3) and it satis es 1. kukA > 0, u 6= 0, k0kA = 0, 2. ku + vkA kukA + kvkA, and 3. k ukA = j jkukA, for all constants . The integrals involved in the norms and inner products are Lebesgue integrals rather than the customary Riemann integrals. Functions that are Riemann integrable are also Lebesgue integrable but not conversely. We have neither time nor space to delve into Lebesgue integration nor will it be necessary for most of our discussions. It is, however, helpful when seeking understanding of the continuity requirements of the various function spaces. So, we'll make a few brief remarks and refer those seeking more information to texts on functional analysis 3, 4, 5]. With Lebesgue integration, the concept of the length of a subinterval is replaced by the measure of an arbitrary point set. Certain sets are so sparse as to have measure zero. An example is the set of rational numbers on 0 1]. Indeed, all countably in nite sets have measure zero. If a function u 2 V possesses a given property except on a set of measure zero then it is said to have that property almost everywhere. A relevant property is the notion of an equivalence class. Two functions u v 2 V belong to the same equivalence class if ku ; vkA = 0: With Lebesgue integration, two functions in the same equivalence class are equal almost everywhere. Thus, if we are given a function u 2 V and change its values on a set of measure zero to obtain a function v, then u and v belon...
View Full Document

This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online