tailieunhanh - Numerical_Methods for Nonlinear Variable Problems Episode 8

Tham khảo tài liệu 'numerical_methods for nonlinear variable problems episode 8', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 5 Numerical Solution of the Navier-Stokes Equations 267 where denotes the duality pairing between H1 2 r and H 1 2 r is continuous symmetric and H ll2 ĩ . We recall that a bilinear form defined over V X V where V is a Hilbert space is F-elliptic if there exists p 0 such that a v v jổ r 2 V V e V. For the proof of Theorem see Glowinski Periaux and Pironneau 1 a variant of it concerning the biharmonic problem is avaiable in Glowinski and Pironneau 1 . Exercise . Prove Theorem . Application of Theorem to the solution of the Stokes problem . Let us define p0 u0 ìịjữ as the solutions of respectively Ap0 V f in Q. Po 0 on r au0 vAu0 f Vp0 in Q u0 g on T Ai 0 V u0 in Q 1 0 0 on T. The following result is established in Glowinski Periaux and Pironneau 1 Theorem . Let u p be a solution of the Stokes problem . The trace Ả p r is the unique solution of the linear variational equation E Ả e H l 2 T R AẰ p p V p 6 H 1 2 T R. on Exercise . Prove Theorem . From the above theorem it follows that the solution of the Stokes problem has been reduced to 2N 3 Dirichlet problems21 N 2 to obtain ý0 N 1 to obtain u p once Ấ is known and to the solution of E which is a kind of boundary integral equation. The main difficulty in this approach is the fact that the pseudo-differential operator A is not known explicitely actually the mixed finite element approximation of Sec. has been precisely introduced in Glowinski and Pironneau 4 5 to overcome this difficulty extending to the Stokes problem an idea discussed in Glowinski and Pironneau 1 for the biharmonic problem. . The discrete case . Introduction formulation of the basic problem. From Secs. and it follows that efficient solvers for the discrete Stokes problem 21 N 2 if rỉ R2 N 3 ifQ c R3. 268 VII Least-Squares Solution of Nonlinear Problems Find uh fh e Wgh such that V vh ộh 6 WOh we have x Ujj vh dx V Vu VvA dx f0 1 vh dx flh vh V f h dx .