The variational problem 316 has the same form as the

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Unformatted text preview: .6) has the same form as the one-dimensional problem (2.3.3). Indeed, the theory and extremal principles developed in Chapter 2 apply to multi-dimensional problems of this form. 1 Theorem 3.1.1. The function w 2 HE that minimizes I w] = A(w w) ; 2(w f ) ; 2 < w > : (3.1.7) is the one that satis es (3.1.6), and conversely. Proof. The proof is similar to that of Theorem 2.2.1 and appears as Problem 1 at the end of this section. 4 Multi-Dimensional Variational Principles 1 Corollary 3.1.1. Smooth functions u 2 HE satisfying (3.1.6) or minimizing (3.1.7) also satisfy (3.1.1). Proof. Again, the proof is left as an exercise. Example 3.1.1. Suppose that the Neumann boundary conditions (3.1.1c) are changed to Robin boundary conditions (x y) 2 @ pun + u = N: (3.1.8a) Very little changes in the variational statement of the problem (3.1.1a,b), (3.1.8). Instead of replacing pun by in the boundary inner product (3.1.5c), we replace it by ; u. 1 Thus, the Galerkin form of the problem is: nd u 2 HE satisfying A(v u) = (v f )+ < v ; u > 8v 2 H01: (3.1.8b) Example 3.1.2. Variational principles for nonlinear problems and vector systems of partial di erential equations are constructed in the same manner as for the linear scalar problems (3.1.1). As an example, consider a thin elastic sheet occupying a twodimensional region . As shown in Figure 3.1.2, the Cartesian components (u1 u2) of the displacement vector vanish on the portion @ E of of the boundary @ and the components of the traction are prescribed as (S1 S2) on the remaining portion @ N of @ . The equations of equilibrium for such a problem are (cf., e.g., 6], Chapter 4) @ 11 + @ 12 = 0 @x @y @ 12 + @ 22 = 0 @x @y (3.1.9a) (x y) 2 (3.1.9b) where ij , i j = 1 2, are the components of the two-dimensional symmetric stress tensor (matrix). The stress components are related to the displacement components by Hooke's law 11 22 12 E = 1 ; 2 ( @u1 + @u2 ) @x @y (3.1.10a) E = 1 ; 2 ( @u1 + @u2 ) @x @y (3.1.10b) = E ( @u1 + @u2 ) 2(1 + ) @y @x (3.1.10c) 3.1. Galerkin's Method and Extremal Principles 5 y s n θ u1 = 0, u2 = 0 Ω S2 S1 x Figure 3.1.2: Two-dimensional elastic sheet occupying the region . Displacement components (u1 u2) vanish on @ E and traction components (S1 S2) are prescribed on @ N . where E and are constants called Young's modulus and Poisson's ratio, respectively. The displacement and traction boundary conditions are u1(x y) = 0 n1 11 + n2 12 = S1 u2(x y) = 0 n1 12 + n2 22 = S2 (x y) 2 @ (3.1.11a) E (x y) 2 @ N (3.1.11b) where n = n1 n2]T = cos sin ]T is the unit outward normal vector to @ (Figure 3.1.2). Following the one-dimensional formulations, the Galerkin form of this problem is obtained by multiplying (3.1.9a) and (3.1.9b) by test functions v1 and v2 , respectively, integrated over , and using the divergence theorem. With u1 and u2 being components of a displacement eld, the functions v1 and v2 are referred to as components of the virtual displacement eld. We use (3.1.9a) to illustrate the process thus, multiplying by v1 and integrating over , we nd ZZ @ 12 1 v1 @x1 + @@y ]dxdy = 0: The three stress components are dependent on the two displacement components and are typically replaced by these using (3.1.10). Were this done, the variational principle 6 Multi-Dimensional Var...
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