This preview shows page 1. Sign up to view the full content.
Unformatted text preview: −1 −0.5 0 0.5 1 1.5 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
−1.5 −1 −0.5 0 0.5 1 1.5 Figure 3.2.1: Smooth version of a piecewise linear hat function (3.2.7) (top), its rst
derivative (center), and the square of its rst derivative (bottom). Results are shown
with xj;1 = ;1, xj = 0, xj+1 = 1 (h = 1), and n = 10.
Example 3.2.2. Consider the piecewiseconstant basis function on a uniform mesh
1 if xj;1 x < xj :
(3.2.8)
j (x) = 0 otherwise A smooth version of this function and its rst derivative are shown in Figure 3.2.2 and
may be written as
n(x ; xj;1) ; tanh n(x ; xj ) ]
1
j n(x) = tanh
2
h
h
0 (x) = n sech2 n(x ; xj ;1 ) ; sech2 n(x ; xj ) ]:
jn
2h
h
h
0 is proportional to the combiAs n ! 1, j n approaches a square pulse however, j n
nation of delta functions
j n(x) /
0 (x ; xj;1) ; (x ; xj ): 3.2. Function Spaces and Approximation 15 Thus, we anticipate problems since delta functions are not elements of L2. Squaring
0 (x)
jn
0 (x)]2 = ( n )2 sech4 n(x ; xj ;1 ) ;2sech2 n(x ; xj ;1 ) sech2 n(x ; xj ) +sech4 n(x ; xj ) ]:
jn
2h
h
h
h
h
As shown in Figure 3.2.2, the function sechn(x ; xj )=h is largest at xj and decays
exponentially fast from xj thus, the center term in the above expression is exponentially
small relative to the rst and third terms. Neglecting it yields
0 (x)]2 ( n )2 sech4 n(x ; xj ;1 ) + sech4 n(x ; xj ) ]:
jn
2h
h
h
Thus,
Z1
n tanh n(x ; xj;1) (2 + sech2 n(x ; xj;1) )
0 (x)]2 dx
jn
12h
h
h
0 + tanh n(x ; xj ) (2 + sech2 n(x ; xj ) )]1:
0
h
h
This is unbounded as n ! 1 hence, 0j (x) 2 L2 and j (x) 2 H 1.
=
=
1 10 0.9 8 0.8 6 0.7 4 0.6 2 0.5 0 0.4 −2 0.3 −4 0.2 −6 0.1 0
−0.5 −8 0 0.5 1 1.5 −10
−0.5 0 0.5 1 1.5 Figure 3.2.2: Smooth version of a piecewise constant function (3.2.8) (left) and its rst
derivative (right). Results are shown with xj;1 = 0, xj = 1 (h = 1), and n = 20.
Although the previous examples lack rigor, we may conclude that a basis of continuous
functions will belong to H 1 in one dimension. More generally, u 2 H k implies that
u 2 C k;1 in one dimension. The situation is not as simple in two and three dimensions.
The Sobolev space H k is the completion with respect to the norm (3.2.5) of C k functions
whose rst k partial derivatives are elements of L2 . Thus, for example, u 2 H 1 implies
that u, ux, and uy are all elements of L2 . This is not su cient to ensure that u is
continuous in two and three dimensions. Typically, if @ is smooth then u 2 H k implies
that u 2 C s( @ ) where s is the largest integer less than (k ; d=2) in d dimensions
1, 2]. In two and three dimensions, this condition implies that u 2 C k;2. Problems 16 MultiDimensional Variational Principles
1. Assuming that p(x y) > 0 and q(x y) 0, (x y) 2 , nd any other conditions
that must be satis ed for the strain energy A(v u) = ZZ p(vxux + vy uy ) + qvu]dxdy to be an inner product and norm, i.e., to satisfy De nitions 3.2.3 and 3.2.4.
2. Construct a variational problem for the fourthorder biharmonic equation
(p u) = f (x y) (x y) 2 where u = uxx + uyy
and p(x y) > 0 is smooth. Assume that u satis es the essential boundary conditions
u(x y) = 0 un(x y) = 0 (x y) 2 @ where n is a unit outward normal vector to @ . To what function space should the
weak solution of the variation...
View
Full
Document
This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.
 Spring '14
 JosephE.Flaherty

Click to edit the document details