By Girault V., Glowinski R., Lopez H.

We study the mistake of a fictitious-domain technique with boundary Lagrange multiplier. it's utilized to unravel a non-homogeneous regular incompressible Navier-Stokes challenge in a website with a multiply-connected boundary. the internal mesh within the fictitious area and the boundary mesh are self sustaining, as much as a mesh-length ratio.

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Extra resources for A boundary multiplier/fictitious domain method for the steady incompressible Navier-Stokes equations

Sample text

K! k n(n − 1) . . (n − k + 1) = k! n(n − 1) . . (n − k + 1) = k(k − 1) . . 1 27 1 CORRIGÉS Donc k−1 n = k i=0 Nombres réels n−i . Or pour 0 k−i n−i k−i k − 1, on a i n ⇔ (n − i)k k ⇔ (n − k)i ce qui est vrai car i i=0 0 0 et n n k n . = k k n k=1 S 1 (n) = n = n k2 + k=1 k k=1 n(n + 1)(2n + 1) n(n + 1) + 6 2 n(n + 1) = [(2n + 1) + 3] 6 n(n + 1)(n + 2) = 3 2) Fixons p 0. Montrons par récurrence sur n tion Pn : 4) = n(n + 1)· · ·(n + p + 1) 1 = « S p (n)= p+2 p+2 k=1 n n k2 − 2 (n + i) » i=0 1 (1 + i) = p + 2 i=0 p+k p+1 Montrons alors par récurrence sur n l’assertion Qn : S p (n) p+n+1 ».

Alors un existe et un > 0. Donc 1 + un > 1, en particulier, 1 + un est dans l’ensemble de définition de la fonction ln (à savoir R∗+ ), donc ln(1 + un ) (c’est-à-dire un+1 ) existe bien. De plus, comme ln est strictement croissant sur R∗+ , de 1 + un > 1, on déduit que ln(1 + un ) > ln(1) = 0. Donc un+1 > 0. Donc Pn+1 est vraie. Conclusion : P0 est vraie, et pour tout entier n, si Pn est vraie, Pn+1 aussi, donc par récurrence, on a bien que pour tout entier n, Pn est vraie. Remarque – Parfois, il faut faire une récurrence descendante au lieu de la récurrence classique.

P × n! − + p p! (n + p)! (n + 1 + p)! 1 1 (n + p + 1) × n! p × n! − + p p! (n + p + 1)! (n + 1 + p)! (n + 1) × n! 1 1 − p p! (n + p + 1)! 1 1 (n + 1)! − p p! (n + p + 1)! ➜ En remarquant que k 1 n+1 n+1 k+1 = k=0 S n+1 = S n + = n S 1 CORRIGÉS Nombres réels n−1 n−1 n−1 i + = i i i=0 i=0 = (n − 1)2n−2 + 2n−1 = (n + 1)2n−2 n n+1 ➜ Pour n k 0 on a (n + 1) = (k + 1) , donc k k+1 n 1 n+1 1 n 1 n = . Si S = , on a alors : k+1 k n+1 k+1 k + 1 k k=0 (−1) 0 2p+1 n 2p+1 (−1) 2p+1 n 2p + 1 0 2p = −Vn car = −1.