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Unformatted text preview: iational Principles would involve second derivatives of u1 and u2. Hence, we would want to use the divergence
theorem to obtain a symmetric variational form and reduce the continuity requirements
on u1 and u2. We'll do this, but omit the explicit substitution of (3.1.10) to simplify the
presentation. Thus, we regard 11 and 12 as components of a twovector, we use the
divergence theorem (3.1.3) to obrain ZZ @ v
1
@x @v1 ]dxdy = Z v n
11 +
11
@y 12
@ 11 + n2 12 ]ds: Selecting v1 2 H01 implies that the boundary integral vanishes on @ E . This and the
subsequent use of the natural boundary condition (3.1.11b) give ZZ @ v
1
@x @v1 ]dxdy = Z v S ds
11 +
11
@y 12
@N Similar treatment of (3.1.9b) gives ZZ @ v
2
@x @v2
12 +
@y 22
dxdy = Z
@N v2S2 ds 8v1 2 H01: (3.1.12a) 8v2 2 H01: (3.1.12b) Equations (3.1.12a) and (3.1.12b) may be combined and written in a vector form.
Letting u = u1 u2]T , etc., we add (3.1.12a) and (3.1.12b) to obtain the Galerkin problem:
nd u 2 H01 such that A(v u) =< v S >
where ZZ @ v
1
A(v u) =
@x 11 + @v2
@y < v S >= 22 Z
@N 8v 2 H01 (3.1.13a) + ( @v1 + @v2 ) 12 ]dxdy
@y @x (3.1.13b) (v1S1 + v2S2 )ds: (3.1.13c) When a vector function belongs to H 1, we mean that each of its components is in H 1.
1
The spaces HE and H01 are identical since the displacement is trivial on @ E .
The solution of (3.1.13) also satis es the following minimum problem.
1
Theorem 3.1.2. Among all functions w = w1 w2]T 2 HE the solution u = u1 u2]T of (3.1.13) is the one that minimizes E I w] = 2(1 ; 2 ) ZZ f(1 ; ) ( @w1 )2 + ( @w2 )2] + ( @w1 + @w2 )2
@x
@y
@x @y 3.1. Galerkin's Method and Extremal Principles 7 (1 ; ) ( @w1 + @w2 )2gdxdy ; Z (w S + w S )ds
+2
11
22
@y @x
@N and conversely. Proof. The proof is similar to that of Theorem 2.2.1. The stress components
1 2, have been eliminated in favor of the displacements using (3.1.10). ij , ij= Let us conclude this section with a brief summary.
A solution of the di erential problem, e.g., (3.1.1), is called a \classical" or \strong"
2
solution. The function u 2 HB , where functions in H 2 have nite values of ZZ (uxx)2 + (uxy )2 + (uyy )2 + (ux)2 + (uy )2 + u2]dxdy 2
and functions in HB also satisfy all prescribed boundary conditions, e.g., (3.1.1b,c). Solutions of a Galerkin problem such as (3.1.6) are called \weak" solutions. They
may be elements of a larger class of functions than strong solutions since the highorder derivatives are missing from the variational statement of the problem. For
the secondorder di erential equations that we have been studying, the variational
1
form (e.g., (3.1.6)) only contains rst derivatives and u 2 HE . Functions in H 1
have nite values of
ZZ
(ux)2 + (uy )2 + u2]dxdy:
1
and functions in HE also satisfy the prescribed essential (Dirichlet) boundary condition (3.1.1b). Test functions v are not varied where essential data is prescribed
and are elements of H01. They satisfy trivial versions of the essential boundary
conditions. While essential boundary conditions constrain the trial and test spaces, natural
(Neumann or Robin) boundary conditions alter the variational statement of the
problem. As with (3.1.6) and (3.1.13), inhomogeneous conditions add boundary
inner product terms to the vari...
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This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.
 Spring '14
 JosephE.Flaherty

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