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14 multi dimensional variational principles 1 1 09 08

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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 piecewise-constant 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 Multi-Dimensional 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 fourth-order 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...
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• Spring '14
• JosephE.Flaherty
• Hilbert space, Boundary conditions, Galerkin, essential boundary conditions, Multi-Dimensional Variational Principles

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