## 30 ouvrages de mathématiques qui ont changé le monde by Jean-Jacques Samueli, Jean-Claude Boudenott

By Jean-Jacques Samueli, Jean-Claude Boudenott

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There was fast development in utilizing huge desktops to resolve difficulties owning extensive levels of time scales. the range of purposes is bewildering, between them multiple-time-scale difficulties in chemical kinetics, statistical mechanics, climate prediction, astrophysics, radiation hydrodynamics, magnetohydrodynamics, and particle simulation of plasmas.

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**Example text**

Proof. 6) arising from the bilinear form ah (u, v) is positive deﬁnite. 4) produces a unique solution. 2 Adaptivity Advection dominated problems lead to internal/boundary layers where the solution has large gradients. The standard FEMs are known to produce strong oscillations around the layers. A naive approach is to reﬁne the mesh uniformly. But it is not desirable as it highly increase the degree of freedom and reﬁnes the mesh unnecessarily in regions where the solutions are smooth. On the other hand, in the semi-linear ADR equations, in addition to the nonphysical oscillations due to the advection, non-linear reaction can be also responsible for sharp fronts.

2: Bisection of a triangle . , let ηK1 and ηK2 be the computed local error indicators corresponding to each unknown component of a two component system. Next, we determine the set of elements MK1 and MK2 satisfying ∑ K∈MK1 (ηK1 )2 ≥ θ ∑ (ηK1 )2 , K∈ξh ∑ K∈MK2 (ηK2 )2 ≥ θ ∑ (ηK2 )2 . K∈ξh Then, we reﬁne the marked elements K ∈ MK1 ∪MK2 using the newest vertex bisection method. The adaptive procedure ends after a sequence of mesh reﬁnements to attain a solution within a prescribed tolerance. 11) u · ∇wdx .

2 Adaptivity 39 1/2 ∑ T3 ηJ2K |||v|||, K∈ξh which completes the proof. 4. (Bound to the conforming part of the error) The conforming part of the error satisﬁes u − uch dG η +Θ . 31) Proof. Since u − uch ∈ H01 (Ω ), we have |u − uch |C = |β (u − uch )|∗ . 23), we get u − uch dG = |||u − uch ||| + |u − uch |C a˜h (u − uch , v) . |||v||| v∈H 1 (Ω )\{0} sup 0 So, we need to bound the term a˜h (u − uch , v). Using the fact that u − uch ∈ H01 (Ω ), we have a˜h (u − uch , v) = a˜h (u, v) − a˜h (uch , v) = = = Ω Ω Ω f vdx − bh (u, v) − a˜h (uch , v) f vdx − bh (u, v) − Dh (uch , v) − Jh (uch , v) − Oh (uch , v) f vdx − bh (uh , v) + bh (uh , v) − bh (u, v) − a˜h (uh , v) + Dh (urh , v) + Jh (urh , v) + Oh (urh , v).