Finite Volume Methods by Robert Eymard, Thierry Gallouët, and Raphaèle Herbin

By Robert Eymard, Thierry Gallouët, and Raphaèle Herbin

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Bi = 0 J(x1 ) 0 J(0) First, using Fubini’s theorem, one has α Di g˜(s, z2 (s, y)) Bi = J(0) α 2 0 dx2 dx1 dsdy. s J(x1 ) Therefore α Bi ≤ diam(K) 2 0 J(0) Di g˜(s, z2 (s, y)) (α − s)dyds. Then, with the change of variables z2 = z2 (s, y), one gets α Bi ≤ diam(K) Di g˜(s, z2 ) 0 J(s) 2 α−s dz2 ds. 1 − αs Hence Bi ≤ diam(K)2 2 2 Di g˜(x) dx. 38). 37), one remarks that 2 N (˜ g, T ) ≤2 K∈T σ∈EK τσ (˜ gK − g˜σ )2 . 38), that N (˜ g, T ) 2 ≤2 K∈T σ∈EK C1 ζ K |∇˜ g(x)|2 dx. 37). 1 page 32 and g ∈ H 1/2 (∂Ω).

34) with y = xi and z = xi+1 , for i = 1, . . 31) as we shall see below). 3, remark that, for all σ ∈ E, IR d χσ (x, x + η)dx ≤ m(σ)cσ |η|. 33), u ˜(· + η) − u˜ 2 L2 (IR d ) ≤ σ∈E m(σ) |Dσ u|2 |η|(|η| + C size(T )). dσ We are now able to state the convergence theorem. e. 2 page 51), following Eymard, Gallou¨ et and Herbin [55]. 1 page 37). 2). e. x ∈ K, and for any K ∈ T . 2) as size(T ) → 0. Furthermore uT 1,T converges to u H01 (Ω) as size(T ) → 0. 7 1. 1, the hypothesis f ∈ L2 (Ω) is not necessary.

Assuming b = 0 and v = 0. e. on K for all K ∈ T , it is easily seen that uT 2 1,T → f (x)u(x)dx = u Ω 2 H01 (Ω) as size(T ) → 0. 1. 1 uses the property of consistency of the (diffusion) fluxes on the test functions. This property consists in writing the consistency of the fluxes for the adjoint operator to the discretized Dirichlet operator. This consistency is achieved thanks to that of fluxes for the discretized Dirichlet operator and to the fact that this operator is self adjoint. 2 page 37).

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