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Unformatted text preview: ach end of the domain and write Y (x) as N +1 X Y (x) = ci 2(x): (6.4.8) i i=0 Now there are three nonzero basis elements on every subinterval. For interpolation problems, the coe cients ci, i = 0 1 : : : N + 1, may be determined by satisfying Y (xi;1 =2) = f (xi;1 =2) i = 1 2 : : : N: This yields N equations for the N + 2 unknowns. The extra two equations could be obtained by interpolating f (x) at x0 and xN , 34 prescribing f 0(x) at x0 and xN , or prescribing Y 0 (x0) = Y 0(xN ) = 0. Regardless of the boundary prescription, this interpolation problem requires the solution of a tridiagonal linear algebraic problem to determine ci, i = 0 1 : : : N + 1. It's possible to continue in this manner, introducing more continuity with increasing polynomial degree however, usually we'll want approximations having the minimum allowable continuity. Since higher-degree approximations increase the convergence rate of smooth solutions, we seek alternative spline constructions that do not increase smoothness with increasing polynomial degree. Such procedures exist 3] however, for the moment, we'll examine piecewise cubic Hermite approximation. Thus, let us consider a function of the form N X3 Y (x) = ci i (x) + di!i3(x)]: (6.4.9) i=0 The basis f 3(x) !i3(x)gN is constructed to satisfy the following conditions: i i=0 1. 3 i (x) and !i3(x) are piecewise cubic polynomials on (x0 xN ), 2. 3 i (x) !i3(x) 2 C 1(x0 xN ), 3. 3 i (x) and !i3(x) are nonzero only on xi;1 xi+1 ), and 4. 3 i (xi ) = 1 and !i3(xi ) = 0, i = 0 1 : : : N . These conditions imply that i (xj ) = ij 3 !i3(xj ) = 0 03 i (xj ) = 0 !i03(xj ) = ij i j = 0 1 ::: N (6.4.10a) i j = 0 1 : : : N: (6.4.10b) Together, (6.4.9) and (6.4.10) give Y (xi ) = ci and Y 0(xi ) = di, i = 0 1 : : : N . Given these requirements, the piecewise cubic Hermite basis is 8 < 1 ; 3( x;ixi )2 ; 2( x;ixi )3 if xi;1 x < xi h h x x 3 (6.4.11a) (x) = : 1 ; 3( x;+1i )2 + 2( x;+1i )3 if xi x < xi+1 i hi hi 0 otherwise 35 1.2 1 0.8 0.6 0.4 0.2 0 −0.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Figure 6.4.5: Cubic Hermite polynomial basis elements 3(x) (solid) and !i3(x) (dashed) i relative to a mesh with xi;1 = ;1, xi = 0, and xi+1 = 1. 8 < (x ; xi ) 1 + x;ixi ]2 if xi;1 x < xi h x (6.4.11b) !i3(x) = : (x ; xi ) 1 ; x;+1i ]2 if xi x < xi+1 : hi 0 otherwise Representative basis elements are illustrated in Figure 6.4.5. Examining (6.4.9), we see that there are four nontrivial basis elements i;1, !i;1, i, and !i on the subinterval (xi;1 xi), i = 1 2 : : : N . Having constructed appropriate piecewise polynomial approximations, let us use them to de ne a collocation method by satisfying (6.3.1) at a prescribed number of points per subinterval and (typically) satisfying the boundary conditions. Thus, Y 00( ij ) = f ( ij Y ( ij ) Y 0 ( ij )) j = 1 2 ::: J Y (a) = A Y (b) = B: i = 1 2 ::: N (6.4.12a) (6.4.12b) As indicated, there are J collocation points ij , j = 1 2 : : : J , per subinterval. For the piecewise quadratic spline approximation (6.4.8), we would determine the N + 2 unknowns ci, i = 0 1 : : : N , by collocating at J = 1 point per subinterval and satisfying the boundary conditions (6.4.12b). With the piecewise cubic Hermite polynomial (6.4.9), 36 we would determine the 2(N +1) unknowns ci di, i = 0 1 : : : N , by collocating at J = 2 point per subinterval and satisfying (6.4.12b). Example 6.4.1. We'll develop the collocation equations when piecewise quadratic splines (6.4.8) are applied to a linear problem with f (x y y0) given by (6.3.13). For simplicity, we'll assume that the mesh is uniform with spacing h. Utilizing (6.4.6) and (6.4.8), the boundary conditions (6.4.12b) are 2 Y (a) = A = 3 (c0 + c1) Y (b) = B = 2 (cN + cN +1): 3 (6.4.13a) Selecting the sole collocation point i1 = xi;1=2 = xi;12+ xi i = 1 2 ::: N at the center of each subinterval (Figure 6.4.6) and using (6.4.6) and (6.4.8) we have 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 6.4.6: Piecewise quadratic spline basis used for collocation on a mesh with spacing h = 0:2 on 0 1]. 1 Y (xi;1=2 ) = 6 (ci;1 + 6ci + ci+1 ) 37 Y 0(xi;1=2 ) = 32h (ci+1 ; ci;1 ) and Y 00(xi;1=2 ) = 34 2 (ci;1 ; 2ci + ci+1): h Substituting these results into (6.4.12a) while using (6.3.13), we nd 4 (c ; 2c + c ) + 2pi;1=2 (c ; c ) + qi;1=2 (c + 6c + c ) = r i i+1 i i+1 i;1=2 3h2 i;1 3h i+1 i;1 6 i;1 i = 1 2 : : : N: (6.4.13b) These di erence equations are similar, but not identical, to the central nite di erence equations (6.3.14). The relationship can be made more precise by re-writing (6.4.13) in terms of the nodal unknowns Y (xi ) and comparing this result with (6.3.14) (Problem 2). The system (6.4.13) may be written in a tridiagonal form as 2 32 32 3 c0 A 0 0 61 1 1 7 6 c1 7 6 r1 7 6 76 6 76 . 7 = 6 . 7 ... ... ... (6.4.14a) 6 76 . 7 6 . 7 6 76 . 7 6 . 7 76 7 4 N N N 5 4 cN 5 4 rN 5 cN +1 B N +1 N +1 where 0 i = N +1 =2 3 i...
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This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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