We note in passing that the boundary conditions

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Unformatted text preview: ent e. We have divided the vector ce of parameters into its contributions ce j , j = 1 2 ::: n, from the shape functions of element e. Thus, we may write c = c 1 c 2 ::: c ]: T e T e T e T e np (5.6.4b) In this form, we may write U0 as U0 = N c = c N T e T e (5.6.4c) 5.5. Assembly of Vector Systems 39 where N is the npm m matrix N = N1 I N2I ::: N I]: T (5.6.4d) np and the identity matrices have the dimension m of the partial di erential system. The simple linear shape functions will illustrate the formulation. Example 5.6.2. Consider the solution of a system of m = 2 equations using a piecewise-linear nite element basis on triangles. Suppose, for convenience, that the node numbers of element e are 1, 2, and 3. In order to simplify the notation, we suppress the subscript 0 on U0 and V0 and the subscript e on ce. The linear approximation on element e then takes the form U1 = c11 N ( U2 c21 1 where N1 ( )=1; ) + c12 N2( c22 N2( ; ) + c13 N3( c23 N3 ( )= ) )= : The rst subscript on cij denotes its index in c and the second subscript identi es the vertex of element e. The expression (5.6.4c) takes the form 2 U1 = N1 0 N2 0 N3 0 U2 0 N1 0 N2 0 N3 3 c11 6 c21 7 6 7 6 c12 7 6 7 6 c22 7 : 6 7 4 c13 5 c23 Substituting (5.6.4c) and a similar expression for V0 into (5.6.3a,e) yields Ae(V U) = dT (Ke + Me)ce e where K= ZZ e (V f )e = dT fe e (5.6.5) N G1 N + N G2 N + N G2 N + N G3eN ] det(J ) d e T e T e T T e d (5.6.6a) 0 M= ZZ e NQN det(J )d d f = T e ZZ e 0 e 0 Problems Nf det(J )d d : (5.6.6b) 40 Mesh Generation and Assembly 1. It is, of course, possible to use di erent shape functions for di erent solution components. This is often done with incompressible ows where the pressure is approximated by a basis having one degree less than that used for velocity. Variational formulations with di erent elds are called mixed variational principles. The resulting nite element formulations are called mixed methods. As an example, consider a vector problem having two components. Suppose that a piecewise-linear basis is used for the rst variable and piecewise quadratics are used for the second. Using...
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