Partial Differential Equations - Modelling and ... - ResearchGate

Partial Differential Equations - Modelling and ... - ResearchGate Partial Differential Equations - Modelling and ... - ResearchGate

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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.

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

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