Partial Differential Equations - Modelling and ... - ResearchGate
Partial Differential Equations - Modelling and ... - ResearchGate Partial Differential Equations - Modelling and ... - ResearchGate
78 J.Ch. Gilbert and P. Joly 18 16 14 12 10 8 6 4 2 0 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 Fig. 1. Graph of the function α 1(r) The classical existence theory in analysis [Sch91, Theorem 2.7.11] leads an existence result. Corollary 1 (Existence of a solution). The optimization problem (20) has (at least) one solution. Clearly, the function R → α m (R) is not continuous. Let us consider, for instance, the case when m =1andk = 1. Then, the function α 1 (R) can be identified to the function of the real variable r defined by α 1 (r) = sup{α |∀x ∈ [0,α], 0 ≤ x − rx 2 ≤ 4}. (23) It is straightforward to compute that α 1 (r) = 1 − √ 1 − 16r 2r if r< 1 16 , and α 1(r) = 1 r if r ≥ 1 16 . It is clear that α 1 is discontinuous at r =1/16 since (see also Figure 1) α 1 (1/16) = 16 and lim r↑1/16 α 1(r) =8. Note that for r =1/16 the graph of the polynomial x − rx 2 is tangent to the line y =4atx =8
Optimal Higher Order Time Discretizations 79 Proof. Let R ∈ D k be such that [ψ m (R)](x ∗ )=4forsomex ∗ ∈ ]0,α m (R)[. (A similar argument works if [ψ m (R)](x ∗ ) = 0.) For any ε > 0, ψ m (R + ε) =ψ m (R) +εx m+1 > 4 in a small neighborhood of x ∗ . This implies that α m (R + ε) α m (R) − ε. Now ε>0 is arbitrary small, so that lim inf α m (R n ) ≥ α m (R). The continuity of α m at R follows, since α m is upper semi-continuous by Lemma 1. ⊓⊔ Lemma 3. The set of solutions of the optimization problem (20) is a convex subset of D k . Proof. Let us first prove that any local maximum of α m belongs to D k . Indeed, it is easy to see that, if R/∈ D k , the function t ∈ R ↦→ α m (R + t) is continuous and strictly monotone in the neighborhood of the origin. This shows that R cannot be a local maximum of α m . Let R 1 and R 2 be two solutions of (20): By definition of α m α m (R 1 )=α m (R 2 )=α m.k ≡ sup α m (R). R∈P k−1 ∀x ≤ α m,k , 0 ≤ [ψ m (R 1 )] (x) ≤ 4 and 0 ≤ [ψ m (R 2 )] (x) ≤ 4.
- Page 35 and 36: Table 1. Primal DG for transport Di
- Page 37 and 38: Discontinuous Galerkin Methods 25 [
- Page 39 and 40: Mixed Finite Element Methods on Pol
- Page 41 and 42: Mixed FE Methods on Polyhedral Mesh
- Page 43 and 44: Mixed FE Methods on Polyhedral Mesh
- Page 45 and 46: Mixed FE Methods on Polyhedral Mesh
- Page 47 and 48: 4 Hybridization and Condensation Mi
- Page 49 and 50: Mixed FE Methods on Polyhedral Mesh
- Page 51 and 52: is symmetric and positive definite,
- Page 53 and 54: with some coefficient α ∈ R wher
- Page 55 and 56: 44 E.J. Dean and R. Glowinski so fa
- Page 57 and 58: 46 E.J. Dean and R. Glowinski 2 A L
- Page 59 and 60: 48 E.J. Dean and R. Glowinski S:T=
- Page 61 and 62: 50 E.J. Dean and R. Glowinski minim
- Page 63 and 64: 52 E.J. Dean and R. Glowinski 6 On
- Page 65 and 66: 54 E.J. Dean and R. Glowinski Fig.
- Page 67 and 68: 56 E.J. Dean and R. Glowinski and
- Page 69 and 70: 58 E.J. Dean and R. Glowinski 7 Num
- Page 71 and 72: 60 E.J. Dean and R. Glowinski Fig.
- Page 73 and 74: 62 E.J. Dean and R. Glowinski Assum
- Page 75 and 76: Higher Order Time Stepping for Seco
- Page 77 and 78: u n+1 h Optimal Higher Order Time D
- Page 79 and 80: Optimal Higher Order Time Discretiz
- Page 81 and 82: Optimal Higher Order Time Discretiz
- Page 83 and 84: Optimal Higher Order Time Discretiz
- Page 85: Optimal Higher Order Time Discretiz
- Page 89 and 90: Optimal Higher Order Time Discretiz
- Page 91 and 92: Optimal Higher Order Time Discretiz
- Page 93 and 94: Optimal Higher Order Time Discretiz
- Page 95 and 96: Optimal Higher Order Time Discretiz
- Page 97 and 98: Optimal Higher Order Time Discretiz
- Page 99 and 100: Optimal Higher Order Time Discretiz
- Page 101 and 102: Optimal Higher Order Time Discretiz
- Page 103 and 104: 96 I. Sazonov et al. To provide a p
- Page 105 and 106: 98 I. Sazonov et al. In the first s
- Page 107 and 108: 100 I. Sazonov et al. Fig. 1. An ex
- Page 109 and 110: 102 I. Sazonov et al. H z 1 exact F
- Page 111 and 112: 104 I. Sazonov et al. (a) (b) Fig.
- Page 113 and 114: 106 I. Sazonov et al. Scattering Wi
- Page 115 and 116: 108 I. Sazonov et al. 6.4 Scatterin
- Page 117 and 118: 110 I. Sazonov et al. (a) (b) Fig.
- Page 119 and 120: 112 I. Sazonov et al. [MHP96] K. Mo
- Page 121 and 122: 114 R. Sanders and A.M. Tesdall I R
- Page 123 and 124: 116 R. Sanders and A.M. Tesdall imp
- Page 125 and 126: 118 R. Sanders and A.M. Tesdall alo
- Page 127 and 128: 120 R. Sanders and A.M. Tesdall (a)
- Page 129 and 130: 122 R. Sanders and A.M. Tesdall D C
- Page 131 and 132: 124 R. Sanders and A.M. Tesdall 8.6
- Page 133 and 134: 126 R. Sanders and A.M. Tesdall 0.3
- Page 135 and 136: 128 R. Sanders and A.M. Tesdall [TR
78 J.Ch. Gilbert <strong>and</strong> P. Joly<br />
18<br />
16<br />
14<br />
12<br />
10<br />
8<br />
6<br />
4<br />
2<br />
0<br />
−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5<br />
Fig. 1. Graph of the function α 1(r)<br />
The classical existence theory in analysis [Sch91, Theorem 2.7.11] leads an<br />
existence result.<br />
Corollary 1 (Existence of a solution). The optimization problem (20) has<br />
(at least) one solution.<br />
Clearly, the function R → α m (R) is not continuous. Let us consider, for<br />
instance, the case when m =1<strong>and</strong>k = 1. Then, the function α 1 (R) can be<br />
identified to the function of the real variable r defined by<br />
α 1 (r) = sup{α |∀x ∈ [0,α], 0 ≤ x − rx 2 ≤ 4}. (23)<br />
It is straightforward to compute that<br />
α 1 (r) = 1 − √ 1 − 16r<br />
2r<br />
if r< 1<br />
16 , <strong>and</strong> α 1(r) = 1 r<br />
if r ≥ 1<br />
16 .<br />
It is clear that α 1 is discontinuous at r =1/16 since (see also Figure 1)<br />
α 1 (1/16) = 16 <strong>and</strong> lim<br />
r↑1/16 α 1(r) =8.<br />
Note that for r =1/16 the graph of the polynomial x − rx 2 is tangent to the<br />
line y =4atx =8