General Results

In this section, we give some general results which will be used to prove the convergence results of the DDFV method (Theorem 4.6.1).

Theorem 4.4.1 ([Cia78], Theorem 3.1.3)Let Ωˆ andΩ =F( ˆΩ)be two affine-equivalent open subsets of Rⁿ, where F: ˆx∈Rⁿ −→Bxˆ+b∈Rⁿ is an invertible affine mapping. Then the upper bounds

kBk ≤ h ˆ

ρ andkB⁻¹k ≤ hˆ ρ

hold where h = diam(Ω), ˆh = diam( ˆΩ), ρ = sup{diam(S) : S is a ball contained in Ω}, and ρˆ= sup{diam(S) :S is a ball contained in Ω}.ˆ

Proof By definition we have

kBk= sup

k^x_ρˆk=1

kBx ˆ

ρk= sup

kxk= ˆρ

1 ˆ ρkBxk.

Given a vectorxsatisfyingkxk= ˆρthere exists two points ˆy,zˆ∈Ω such that ˆ¯ˆ y−zˆ=x. By definition ofF we haveBx=F(ˆy)−F(ˆz) and since F(ˆy), F(ˆz)∈Ω we have

kF(ˆy)−F(ˆz)k ≤h which in turn implies

kBxk ≤handkBk ≤ sup

kxk= ˆρ

1 ˆ ρh= h

ˆ ρ. The proof is very similar forkB⁻¹k.

Lemma 4.4.2 Let u, v, w be three linearly independent vectors. We have kwk²≤ 2

1− hu, vi² hw, ui²+hw, vi² .

Proof We expresswin the (u, v) basis: w=λ₁u+λ₂v. We therefore havehw, ui=hλ₁u+λ₂v, ui= λ₁+λ₂hu, viandhw, vi=λ₂+λ₁hu, vi. Solving this system with respect toλ₁ andλ₂ we find

λ1= hw, ui − hw, vihu, vi

1− hu, vi² andλ2= hw, vi − hw, uihu, vi 1− hu, vi² .

D _j

Figure 4.4.2: Notations for Lemma 4.4.3

We havekwk²=λ²₁+λ²₂+ 2λ₁λ₂hu, vi. Inserting the value ofλ₁ andλ₂ we find which concludes the proof.

Lemma 4.4.3 Let T_j,1,T_j,2 be two triangles which share an edgeA_j of end points S_k1 =

D _j

Figure 4.4.3: Notations for Lemma 4.4.4

and Moreover, we have that

|T_j,1|= 1 and allows us to conclude.

The following lemma can be proved in the same way as Lemma 4.4.4.

Lemma 4.4.4 Let Dj be a triangle of vertices Sk₁, Sk₂ and Gi₁. Let φ be defined onDj and let wj

be the P1 interpolant ofφonDj. For any continuous functionv we defineΠv by(Πv)^T_i =v(Gi)and

e

₂

Figure 4.4.4: Notations for Lemma 4.4.5

(Πv)^P_k =v(S_k). Then ifS_j= (1−β)S_k1+βS_k2 andφ_j = (1−β)φ(S_k1) +βφ(S_k2)(see Figure 4.4.3 for notations) it holds that ˆ

Dj

(∇hΠφ)jdx= ˆ

Dj

∇wj(x)dx.

The first inequality of the following lemma is given in [GA00]. We recall it here with our notations and complete it with a second inequality.

Given a triangle (resp.: tetrahedron) K and p₀ one of its vertices, we define v_i (withkv_ik = 1) andl_i for 1≤i≤n(n=2,3) to be the directions and lengths respectively of the edges sharingp₀. We define the reference element ˆK to be the triangle generated by the canonical vectors e_i, 1≤ i ≤n (n=2,3) and the affine transformationF : ˆK→KbyF(ˆx) =Bxˆ+p0withBei=livi. To a function wdefined onKwe associate a function ˆwdefined on ˆKby ˆw=w◦F, see Figure (4.4.4) for notations.

Lemma 4.4.5 Let K be a triangle (resp. tetrahedron) and S be any of its edges (resp. faces). Let w∈H¹(K)be a function with vanishing average on S. Then, there exists a constant C independent of K such that

1. Let us first show by contradiction that

kwkˆ _L2( ˆK)≤Ck∇ˆ wkˆ _L2( ˆK)∀wˆ∈H¹( ˆK) with vanishing average on ˆS. there exists a subsequence of{wˆn}_n≥1that converges inL²( ˆK) (see [Eva98], Remark following Theorem 1 in Section 5.7). The same holds for∇w. Let us denote the limit ofˆ {wˆn}n≥1 by ˆw

implies that

n→∞lim k∇wˆn− ∇wkˆ _L2( ˆK)= 0⇒ 0 = lim

n→∞k∇wˆnk_L2( ˆK)= lim

n→∞k∇wkˆ _L2( ˆK)

where we used assumption (b).

The trace theorem in [Eva98] (Theorem 1, section 5.5) states that there exists a bounded linear operator T :H¹( ˆK)→L²(∂K) such thatˆ

This implies that ˆw has vanishing mean value on S. But we showed above that ˆw is constant on ˆK. We can therefore conclude that ˆw≡0. This contradicts the fact thatkwkˆ _L2( ˆK)= 1.

2. We now show that

kwk_L2(K)≤Ck∇ˆ wkˆ _L2( ˆK)∀w∈H¹(K).

Using the fact that K=F( ˆK) and that F is an affine transformation (which implies that the Jacobian ofF is constant) we have

kwk²_L2(K)= with vanishing average on ˆS.

3. To conclude the proof of the first inequality, we show that k∇wkˆ _L2( ˆK)= frame. Using the chain rule, we find that

∂wˆ By construction, _∂x^∂F

1 =l1v1. So

Using the change of variable and the result above we find k∇wkˆ ²_{L2( ˆK)}=

= Using the triangle inequality gives the result.

4. Let us prove the second inequality for v1, the case ofv2 being very similar. In the (ξ, η) basis, we have v₁=ξcosα+ηsinα. So

We now give two results which can be found in Section 4.3.3 of the PhD thesis [Del07].

Lemma 4.4.6 Let K be a triangle and S be any of its edges. Let w∈H^1,α(K), α∈[0,1/2), Then the trace ofwon S is inL¹(S).

Proof This lemma is a direct consequence of Theorem 1.6.3.

Theorem 4.4.7 Let K be a triangle and S be any of its edges. Let w∈H^1,α(K), α∈[0,1/2), be a function with vanishing average on S. Then, there exists a constant C independent of K such that

kwk_L2(K)≤C|w|_H1,α(K). (4.4.3)

Proof Since by definition kwkL²(K)≤ kwkH^1,α(K), showing

kwk_H1,α(K)≤C|w|_H1,α(K) (4.4.4)

will suffice. Let us defineE_α:={w∈H^1,α(K) :´

Sw(ξ)dξ= 0 andkwk_H1,α(K)= 1}. Showing (4.4.4) is equivalent to showing that

w∈Einfα

C|w|_H1,α(K)>0.

By contradiction, let us assume that this infimum is zero. Then there exists a sequence{w_n}_n≥1⊂E_α such that

n→∞lim |wn|_H1,α(K)= 0. (4.4.5)

By Theorem 1.6.2 (d) and (a) we have H^1,α(K) ⊂^c L²(K) so there exists a subsequence {wn_k}_k≥1 which is Cauchy inL²(K). This and Equation (4.4.5) implies that{wn_k}_k≥1 is a Cauchy sequence in H^1,α(K). By Theorem 1.6.1, we know that there existsw∈H^1,α(K) such thatkwk_H1,α(K)= 1 and

lim

k→∞kw_nk−wk_H1,α(K)= 0.

By (4.4.5) we have |w|H^1,α(K) = 0 which implies that w is constant. Since H^1,α(K) ⊂L¹(S) (see Lemma 4.4.6), the applicationu7→´

Su(ξ)dξ is continuous and 0 = lim

H E

ν E

T

P

Figure 4.4.5: Notations for Lemma 4.4.8 which implies thatw= 0 and contradicts the fact thatkwk_H1,α(K)= 1.

The following is based on [CGR12].

Lemma 4.4.8 Let T =conv(P ∪E) be the triangle with vertex P and opposite edge E (see Figure 4.4.5 for notations). Letf ∈W^1,1(T)and letffl

denote the integral mean. Then it holds

E g(x) := (x−P)f(x). The Gauss divergence theorem leads to

ˆ

The product rule gives

div g(x) = 2f(x) +∇f(x)·(x−P).

Theorem 4.4.9 ([LS10], Poincar´e inequality on Triangles) For any triangle T and f ∈ H¹(T) it holds

wherej1,1≈3.8317059702denotes the first positive root of the Bessel functionJ1.

Lemma 4.4.10 ([CGR12], Lemma 2.3) Letf ∈H¹(T)whereT =conv(P∪E)(see Figure 4.4.6 for notations) and such that´

Ef ds= 0. Then we have kfk_L2(T)≤q

maxx∈E|P−x|²/8 +h²_T/j_1,1² |f|_H1(T). (4.4.6)

E

T P

Figure 4.4.6: Notations for Lemma 4.4.10 Proof Let us set a := f −ffl

Tf(x)dx and b := ffl

Tf(x)dx. Because ha, bi = 0, by the Pythagoras theorem we have

kfkL²(T)=ka+bkL²(T)=kakL²(T)+kbkL²(T). (4.4.7) By the Poincar´e inequality (Theorem 4.4.9), we have

kak_L2(T)= The trace identity (Lemma 4.4.8) withffl

Ef ds= 0 leads to

Inserting this into (4.4.9) gives

|b| ≤ 1 which gives us a bound for b,

kbk²_L2(T)= Using the above and (4.4.8) to bound (4.4.7) leads to

kfkL²(T)≤

Lemma 4.4.11 ([CGR12], Lemma 2.4) Let ν, µ∈R² be two linearly independent unit vectors. We have

1− |hν, µi|= min

a∈R²\0

ha, νi²+ha, µi² kak² .

Proof The first part of the proof is identical to that in [CGR12]. The second part of our proof is however slightly more detailed.

(a) We first show that fora∈R² such thata=αν+βµwithα²+β²= 1 we have 1− |hν, µi| ≤ha, νi²+ha, µi²

kak² . (4.4.10)

Setγ:=hν, µi. Then 2αβ= sin(2θ)∈[−1,1]. This implies that|γ|+ 2αβγ≥0 and so 0≤(1 +|γ|)(|γ|+ 2αβγ)⇔ −(1 +|γ|)2αβγ≤(γ²+|γ|).

Adding 1− |γ|+ 4αβγon both sides leads to

1 +γ²+ 4αβγ≥(1− |γ|)(1 + 2αβγ) = (1− |γ|)kak². (4.4.11) Using

ha, νi=α+βγ andha, µi=β+αγ, we show that

ha, νi²+ha, µi²= 1 +γ²+ 4αβγ.

Inserting this into (4.4.11) allows us to conclude.

(b) We then show that for ˜a∈R²such that ˜a= ˜αν+ ˜βµ with ˜α²+ ˜β²6= 1 we have 1− |hν, µi| ≤h˜a, νi²+h˜a, µi²

k˜ak² . (4.4.12)

Let us proceed by contradiction i.e. assume

1− |hν, µi|>h˜a, νi²+h˜a, µi² k˜ak² . Let us consider the mapping

f : R² −→ {(α, β)∈R²|α²+β²= 1}

˜

a= ( ˜α,β)˜ 7−→ q ¹

˜ α2 β˜2+1

α˜ β˜,1

,

which associates to each vector inR² an image on the unit circle. Leta:=f(˜a). We know by (a) that

ha, νi²+ha, µi²

kak² ≥1− |hν, µi|.

So ha, νi²+ha, µi²

kak² >h˜a, νi²+h˜a, µi² k˜ak² .

Using the definitions of the scalar products and the norms, this reduces to 2hν, µi(1− hν, µi²)( ˜αβ˜−αβ( ˜α²+ ˜β²))<0.

Using the definition ofa, we find that

˜

αβ˜=αβ( ˜α²+ ˜β²),

which leads us to a contradiction (0<0) and allows us to conclude.

(c) Equality and attainment of the minimum follows with the choice of (α, β) =^√1₅(1,1) ifγ≤0 and (α, β) =^√1

5(−1,1) ifγ≥0.

E ₁ E ₂

α τ ₁

τ ₂ T B

A

C

Figure 4.4.7: Notations for Theorem 4.4.12

Theorem 4.4.12 Let us consider a triangleT =conv{A, B, C} and let us denoteαone of its angles andh_T its diameter. Let v∈H²(T)be such thatv(A) =v(B) =v(C). Then

k∇vkL²(T)≤C_αh_TkD²vkL²(T). (4.4.13) where

Cα:=

s1/8 + 2/j_1,1²

1− |cosα| . (4.4.14)

Proof Let us denoteτ1andτ2the two unit vectors along the sides of the angleα(see Figure 4.4.7 for notations). We define |γ|:=|hτ1, τ2i|=|cosα| and fj :=hτj,∇v(x)ifori = 1,2. Applying Lemma 4.4.11 toν =τ1,µ=τ2anda=∇v(x) gives

k∇v(x)k²(1− |γ|)≤ h∇v(x), τ1i²+h∇v(x), τ2i²=f₁²(x) +f₂²(x).

Integration overT leads to (1− |cosα|)|v|²_H1(T)dx≤

ˆ

T

f₁²(x) +f₂²(x)dx=kf1(x)k²_L2(T)+kf2(x)k²_L2(T). (4.4.15) Applying Lemma 4.4.10 tof1 andf2 respectively gives the upper bounds

kf1(x)k²_L2(T)≤

x∈Emax1

|C−x|²/8 +h²_T/j_1,1²

|f1(x)|²_H1(T), kf2(x)k²_L2(T)≤

max

x∈E2

|B−x|²/8 +h²_T/j²_1,1

|f2(x)|²_H1(T).

Using the fact that max_x∈E1|C−x|²= max{|C−B|²,|C−A|²}and max_x∈E2|C−x|²= max{|C− B|²,|B−A|²} leads to

kf1(x)k²_L2(T)+kf2(x)k²_L2(T)≤ max{|A−B|²,|B−C|²,|A−C|²}

8 + h²_T

j_1,1²

!

(|f1(x)|²_H1(T)+|f2(x)|²_H1(T))

≤ h² 8 + h²_T

j_1,1²

!

(|f1(x)|²_H1(T)+|f2(x)|²_H1(T))

=h²_T 1

Inserting this into (4.4.15) yields

|v|²_H1(T)≤ h²_T(1/8 + 2/j_1,1² )

1− |cosα| |v|²_H2(T), which allows us to conclude.

4.5 Error Estimates in the discrete H ¹ semi-norm

Dans le document The Discrete Duality Finite Volume method in the context of weather prediction models (Page 88-98)