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Unformatted text preview: n a nitedimensional 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 threedimensional 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...
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 Spring '14
 JosephE.Flaherty

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