By Robert Eymard, Thierry Gallouët, and Raphaèle Herbin

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N2 ; set hi − = xi − xi− 21 , hi + = xi+ 12 − xi , for i = 1, . . , N1 , hi+ 21 = xi+1 − xi , for i = 0, . . , N1 , 34 kj − = yj − yj− 21 , kj + = yj+ 21 − yj , for j = 1, . . , N2 , kj+ 12 = yj+1 − yj , for j = 0, . . , N2 . Let h = max{(hi , i = 1, · · · , N1 ), (kj , j = 1, · · · , N2 )}. 4) over each control volume Ki,j , which yields  yi+ 1 yj+ 1 2 2   1 , y)dy + − u (x ux (xi− 12 , y)dy x i+ 2   y 1 y 1 j−     + xi+2 1 2 xi− 1 i− uy (x, yj− 12 )dx − 2 2 xi+ 1 2 xi− 1 uy (x, yj+ 12 )dx = 2 f (x, y)dx dy.

The scheme which is described above is stable under a geometrical condition on the family of meshes which is considered. 10) satisfies the numerical scheme with an error which tends to 0 in L∞ (Ω) for the weak-⋆ topology. Under adequate restrictive assumptions, the convergence of the scheme can be deduced, see Faille [58]. 3). Comparisons with solutions which were obtained by the bilinear finite element method, and with known analytical solutions, were performed. The results given by the VF9 scheme and by the finite element scheme were very similar.

Note that uTn → u, in L2 (IR), with u = 0 on IR \ [0, 1]. 63), with uTn instead of uT (and size(Tn ) instead of h). One obtains u(· + η) − u η 2 L2 (IR) ≤ C. 65) yields that Du ∈ L2 (IR). Furthermore, since u = 0 on IR \ [0, 1], the restriction of u to (0, 1) belongs to H01 ((0, 1)). 4 is complete. 4. 55). 1), let (u1 , . . 3), and let uT : (0, 1) → IR by uT (x) = ui , if x ∈ Ki , i = 1, . . , N . 56)). 4 Let (Tn )n∈IN be a sequence of admissible meshes of (0, 1) such that size(Tn ) → 0, as n → ∞.

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