# Thus if we are given a function u 2 v and change its

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: g to the same equivalence class. We need one more concept, the notion of completeness. A Cauchy sequence fung1 2 n=1 V is one where lim ku ; unkA = 0: m n!1 m 12 Multi-Dimensional Variational Principles If fung1 converges in k kA to a function u 2 V then it is a Cauchy sequence. Thus, n=1 using the triangular inequality, lim kum ; unkA lim fkum ; ukA + ku ; unkAg = 0: m n!1 m n!1 A space V where the converse is true, i.e., where all Cauchy sequences fung1 converge n=1 in k kA to functions u 2 V , is said to be complete. De nition 3.2.5. A complete linear space V with inner product A(u v) and corresponding norm kukA, u v 2 V is called a Hilbert space. Let's list some relevant Hilbert spaces for use with variational formulations of boundary value problems. We'll present their de nitions in two space dimensions. Their extension to one and three dimensions is obvious. De nition 3.2.6. The space L2( ) consists of functions satisfying ZZ 2 L ( ) := fuj u2dxdy &lt; 1g: It has the inner product (u v ) = and norm ZZ uvdxdy (3.2.4a) (3.2.4b) p kuk0 = (u u): (3.2.4c) De nition 3.2.7. The Sobolev space H k consists of functions u which belong to L2 with their rst k 0 derivatives. The space has the inner product and norm (u v)k := X jjk (D u D v) (3.2.5a) p kukk = (u u)k (3.2.5b) where = with 1 and 2 1 2 T j j= 1+ 2 (3.2.5c) non-negative integers, and @ 1+ 2 D u := @x 1 @yu2 : (3.2.5d) 3.2. Function Spaces and Approximation 13 In particular, the space H 1 has the inner product and norm (u v)1 = (u v) + (ux vx) + (uy vy ) = ZZ (uv + uxvx + uy vy )dxdy 2ZZ 31=2 kuk1 = 4 (u2 + u2 + u2 )dxdy5 : x y (3.2.6a) (3.2.6b) Likewise, functions u 2 H 2 have nite values of kuk = 2 2 ZZ u2 + u2 + u2 + u2 + u2 + u2]dxdy: xx xy yy x y Example 3.2.1. We have been studying second-order di erential equations of the form (3.1.1) and seeking weak solutions u 2 H 1 and U 2 S N H 1 of (3.1.6) and (3.2.2), respectively. Let us verify that H 1 is the correct space, at least in one dimension. Thus, consider a basis of the familiar piecewise-linear hat functions on a uniform mesh with spacing h = 1=N 8 &lt; (x ; xj;1)=h if xj;1 x &lt; xj j (x) = : (xj +1 ; x)=h if xj x &lt; xj +1 : 0 otherwise (3.2.7) Since S N H 1, j and 0j must be in L2, j = 1 2 ::: N . Consider C 1 approximations of 0 j (x) and j (x) obtained by \rounding corners&quot; in O(h=n)-neighborhoods of the nodes xj;1, xj , xj+1 as shown in Figure3.2.1. A possible smooth approximation of 0j (x) is 1 tanh n(x ; xj+1) + tanh n(x ; xj;1) ; 2 tanh n(x ; xj ) ]: 2h h h h A smooth approximation j n of j is obtained by integration as 0 (x) j n(x) = 0 j j n(x) = Clearly, jn and 0 jn h ln cosh n((x ; xj+1)=h) cosh n((x ; xj;1)=h) : 2n cosh2 n((x ; xj )=h) are elements of L2 . The \rounding&quot; disappears as n ! 1 and lim n!1 Z 1 0 j n(x)] 0 2 dx 2h(1=h)2 = 2=h: The explicit calculations are somewhat involved and will not be shown. However, it seems clear that the limiting function 0j 2 L2 and, hence, j 2 S N for xed h. 14 Multi-Dimensional Variational Principles 1 1 0.9 0.8 0.8 0.6 0.7 0.4 0.6 0.2 0.5 0 0.4 −0.2 0.3 −0.4 0.2 −0.6 0.1 −0.8 0 −1.5 −1 −0.5 0 0.5 1 1.5 −1 −1.5...
View Full Document

Ask a homework question - tutors are online