La méthode HHO (Hybrid High-Order) dans le casd’une frontière immergée

Erik Burman, Guillaume Delay, Alexandre Ern

Laboratoire Jacques-Louis Lions, Sorbonne Université, Paris, France

Séminaire du LJLL, 27 septembre 2019

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Outline

General presentation of HHO

An elliptic interface problem

The Stokes problem

Numerical simulations

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General presentation of HHO

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What is HHO?

Introduced in [Di Pietro, Ern, Lemaire 14; Di Pietro, Ern 15]HHO degrees of freedom (dofs) are located on the cells and thefaces of the mesh −→ Polynomials of degree k ≥ 0

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In the case of unfitted meshes, we consider k on the faces and(k + 1) on the cells

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The discrete problem is assembled cell-wise4/50

What is HHO?

Representation of 2D unknowns

k = 0 (equal-order)

face deg : 0cell deg : 0

k = 0 (mixed-order)

face deg : 0cell deg : 1

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What is HHO?

Close to the Hybrid Discontinuous Galerkin (HDG) andnonconforming Virtual Element (ncVEM) methods

The dofs attached to the cells can be eliminated by a local Schurcomplement technique (static condensation)

mesh k = 0

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The global problem comprises only the face dofs

We can recover the cell dofs by post-processing

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Main characteristics

General meshes: polygonal/polyhedral cells, hanging nodes

Attractive computational costsenergy error decay O(hk+1) with face dofs of order k ≥ 0global system of size the number of face dofs

Implementation:open source diskpp library [Cicuttin, Di Pietro, Ern 18]available on github https://github.com/wareHHOuse

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Recent developments in HHO

Transport and flowsStokes [Di Pietro, Ern, Linke, Schieweck 16], NS [Di Pietro, Krell 18]viscoplastic fluids [Cascavita, Bleyer, Chateau, Ern 18]fractured porous media [Chave, Di Pietro, Formaggia 18]

Nonlinear mechanicssmall defs [Botti, Di Pietro, Sochala 17]hyperelasticity [Abbas, Ern, Pignet 18]elastoplasticity [Abbas, Ern, Pignet 18]Signorini conditions [Cascavita, Chouly, Ern 18]

Spectral approximation [Calo, Cicuttin, Deng, Ern 18]

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Unfitted HHO

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Motivation for unfitted meshes

Enables the use of simpler meshes to mesh intricate geometries

Fitted HHO (and other polyhedral methods) is not adapted to treatcurvilinear boundaries

A first work on HHO for elliptic interface problems [Burman, Ern 18]

Main idea: robustness with respect to bad cuts by agglomerationof cells using polyhedral meshes [Johansson, Larson 13]

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Elliptic interface problem

Γ

Ω1

Ω2

nΓ

Lipschitz domain Ω ⊂ Rd

Interface Γ, subdomains Ω1, Ω2 ⊂ Ω

κ1∆u = f in Ω1

κ2∆u = f in Ω2

JuKΓ = gD on Γ

Jκ∇uKΓ · nΓ = gN on Γ

u = 0 on ∂Ω

with JaKΓ = a|Ω1− a|Ω2

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Degrees of freedom (1/4)

Let T be a mesh cell in Th with unit outward normal nT

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The local dofs are uT ∈ Pk+1(T ) on the cell T and the polynomialsuF ∈ Pk(F ) on every face F composing ∂T

Generic notation: uT = (uT , u∂T ) with u∂T = (uF )F∈FT

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Degrees of freedom (2/4)

(∂T )1

(∂T )2

Γ

TΓ

T1

T2

Decomposition of cut cells

T = T1 ∪ T2

Decomposition of cut faces

∂(T1) = (∂T )1 ∪ TΓ ∂(T2) = (∂T )2 ∪ TΓ

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Degrees of freedom (3/4)

cut cell

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We double the unknowns on cut cells/faces in the spirit of[Hansbo, Hansbo 02] for cut FEM

uT1∈ Pk+1(T1), uT2

∈ Pk+1(T2)

u(∂T )1 ∈ Pk((∂T )1), u(∂T )2 ∈ Pk((∂T )2)

uT = (uT1, u(∂T )1 , uT2

, u(∂T )2)

No dofs on TΓ

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Degrees of freedom (4/4)

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uh

uT

The global unknowns of the problem are

uh ∈∏

T∈T 1h

Pk+1(T1)×∏

F∈F1h

Pk(F )×∏

T∈T 2h

Pk+1(T2)×∏

F∈F2h

Pk(F )

We collect in uT all the global unknowns related to a mesh cell T

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Local discretization: uncut cells (1/2)

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Two important ingredients:

gradient reconstruction GkT (uT ) ∈ Pk(T ;Rd) s.t. ∀q ∈ Pk(T ;Rd),

(GkT (uT ),q)T = −(uT ,div q)T + (u∂T ,q · nT )∂T

stabilization (weakly enforces matching of cell trace and face unknowns)

sT (uT , vT ) = h−1T

∑F∈FT

(ΠkF (uF − uT ), vF − vT )F

HDG-like stabilization operator (cell unknowns in Pk+1(T ))[Lehrenfeld, Schöberl 16]

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Local discretization: uncut cells (2/2)

Local bilinear form

aT (uT , vT ) = κT (GkT (uT ),Gk

T (vT ))T + κT sT (uT , vT )

Local right-hand side

`T (vT ) = (f, vT )T

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Local discretization: cut cells (1/2)

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Two options for the gradient reconstruction

Option 1: GkTi

(uT ) ∈ Pk(Ti;Rd) s.t. ∀q ∈ Pk(Ti;Rd),(Gk

Ti(uT ),q)Ti

= −(uTi,div q)Ti

+ (u(∂T )i ,q · nT )(∂T )i + (uTi,q · nTi

)TΓ

Option 2: GkTi

(uT ) ∈ Pk(Ti;Rd) s.t. ∀q ∈ Pk(Ti;Rd),

(GkTi

(uT ),q)Ti= −(uTi

,div q)Ti+ (u(∂T )i ,q · nT )(∂T )i + (uTi

,q · nTi)TΓ

where T1 = T2, T2 = T1

Stabilization operatorsT (uT , vT ) = h−1

T

∑i∈1,2

κi∑

Fi∈FTi

(ΠkFi

(uFi− uTi

), vFi− vTi

)Fi

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Local discretization: cut cells (2/2)

Local bilinear form

aT (uT , vT ) =κ1(GkT1

(uT ), GkT1

(vT ))T1 + κ2(GkT2

(uT ),GkT2

(vT ))T2

+ ηκ1h−1T (JuT KΓ, JvT KΓ)TΓ + sT (uT , vT )

Local right-hand side

T (vT ) =

∑i∈1,2

(f, vTi)Ti

+ (gN , vT2)TΓ

− κ1(gD, GkT1

(vT ) · nΓ)TΓ + ηκ1h−1T (gD, JvT KΓ)TΓ

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Convergence result

We set to zero all the face components attached to ∂Ω

Global problem: Find uh such that

ah(uh, vh) = h(vh) for every vh (1)

with ah(uh, vh) =∑T∈Th

aT (uT , vT ), h(vh) =

∑T∈Th

T (vT )

Theorem

For every η > 0, there is a unique discrete solution uh to (1) s.t.∑T

∑i∈1,2

κi‖∇(u− uTi)‖2Ti

≤ Ch2(k+1)∑

i∈1,2

κi|u|2Hk+2(Ωi)

No need for η large enough

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Numerical analysis forthe elliptic interface problem

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Numerical analysis : assumptions

The interface is well resolved (mesh fine enough)There are no small cut cells (agglomeration)We are able to integrate over the curved interface

Inverse inequality

Let ` ∈ N and i ∈ 1, 2. For every T ∈ Th, and vTi∈ P`(Ti), we have

‖∇vTi‖Ti ≤ Ch−1T ‖vTi‖Ti

Discrete trace inequality

Let ` ∈ N and i ∈ 1, 2. For every T ∈ Th, and vTi∈ P`(Ti), we have

‖vTi‖(∂T )i + ‖vTi

‖TΓ ≤ Ch−12

T ‖vTi‖Ti

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Numerical analysis : stability (1/2)

‖uT ‖2∗ =κ1‖∇uT1‖2T1

+ κ2‖∇uT2‖2T2

+ ηκ1h−1T ‖JuT K‖2TΓ

+ κ1h−1T ‖uT1

− u(∂T )1‖2(∂T )1 + κ2h−1T ‖uT2

− u(∂T )2‖2(∂T )2

Coercivity

For every uT ∈ UkT , we have ‖uT ‖2∗ ≤ CaT (uT , uT )

proof:

aT (uT , uT ) = κ1‖GT1‖2T1

+ κ2‖GT2‖2T2

+ ηκ1h−1T ‖JuT K‖2TΓ

+ h−1T

∑i∈1,2

κi‖Πk(∂T )i(uTi − u(∂T )i)‖2(∂T )i

‖∇uT1‖T1 ≤ C(‖GT1‖T1 + h−1/2T ‖Πk

(∂T )1(u(∂T )1 − uT1)‖(∂T )1

+ h−1/2T ‖JuT K‖TΓ)

‖∇uT2‖T2≤ C(‖GT2

‖T2+ h−1/2T ‖Πk

(∂T )2(u(∂T )2 − uT2)‖(∂T )2)

h−1T ‖uTi

− u(∂T )i‖2(∂T )i ≤ C(h−1T ‖Π

k(∂T )i(uTi

− u(∂T )i)‖2(∂T )i + ‖∇uTi‖2Ti

)

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Numerical analysis : stability (2/2)

‖∇uT1‖T1≤C(‖GT1

‖T1+h−1/2T ‖Πk

(∂T )1(u(∂T )1 − uT1)‖(∂T )1 +h

−1/2T ‖JuT K‖TΓ):

‖∇uT1‖2T1

= (∇uT1, GT1

)T1− (∇uT1

· nT , u(∂T )1 − uT1)(∂T )1

− (∇uT1· nΓ, uT2

− uT1)TΓ

Cauchy–Schwarz and discrete trace inequalities:

‖∇uT1‖2T1≤ C‖∇uT1

‖T1(‖GT1

‖T1+ h−1/2T ‖Πk

(∂T )1(u(∂T )1 − uT1)‖(∂T )1

+ h−1/2T ‖JuT K‖TΓ)

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Numerical analysis : approximation

We define the interpolation operator (like in [Burman, Ern 18])

IkT (u) = ((Πk+1T † E1(u))|T1

,Πk(∂T )1u, (Πk+1

T † E2(u))|T2,Πk

(∂T )2u)

where Ei : Hk+1(Ωi)→ Hk+1(Rd) is a stable extension, T † is asimple shape that contains TIn the analysis, Πk

(∂T )iu does not play a role, i.e., we do not needrobustness w.r.t. cut faces

Approximation

For v ∈ Hk+2(Ω1 ∪ Ω2),

‖GkTi

(IkT (v))−∇v‖ ≤ Chk+1|ui|Hk+2(Ωi)

‖Ik+1Ti

(v)− v‖ ≤ Chk+2|ui|Hk+2(Ωi)

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Numerical analysis : consistency (1/2)

Consistency

Let uh be the discrete solution and u the exact solution. Assume that uis smooth enough. For every vh ∈ Uk

h , we have|ah(Ikh(u)− uh, vh)| ≤ C‖vh‖∗(κ1|u1|2Hk+2(Ω1) + κ2|u2|2Hk+2(Ω2))

1/2hk+1

ah(Ikh(u)− uh, vh) = ah(Ikh(u), vh)− h(vh)

= Ψ1 + Ψ2

Ψ1 =∑T∈Th

κ1(GkT1

(IkT (u)), GkT1

(vT ))T1 + κ2(GkT2

(IkT (u)),GkT2

(vT ))T2

+κ1(∆u, vT1)T1

+ κ2(∆u, vT2)T2− (gN , vT2

)TΓ

Ψ2 =∑T∈Th

sT (IkT (u), vT )+κ1h−1T (JIk+1

T (u)KΓ, JvT KΓ)TΓ−κ1h−1T (gD, JvT KΓ)TΓ

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Numerical analysis : consistency (2/2)

For instance:(Gk

T2(IkT (u)),Gk

T2(vT ))T2 = (Gk

T2(IkT (u)),∇vT2)T2

+ (GkT2

(IkT (u))·nT , v(∂T )2 − vT2)(∂T )2

(∆u, vT2)T2 = −(∇u,∇vT2)T2 + (∇u·nT , vT2)(∂T )2 − (∇u2 ·nΓ, vT2)TΓ

κ2(GkT2

(IkT (u)),GkT2

(vT ))T2 + κ2(∆u, vT2)T2

=κ2(GkT2

(IkT (u))−∇u,∇vT2)T2

+κ2((GkT2

(IkT (u))−∇u)·nT , v(∂T )2−vT2)(∂T )2

− κ2(∇u2 ·nΓ, vT2)TΓ

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Convergence result

Theorem

For every η > 0, there is a unique discrete solution uh to (1) s.t.∑T

∑i∈1,2

κi‖∇(u− uTi)‖2Ti

≤ Ch2(k+1)∑

i∈1,2

κi|u|2Hk+2(Ωi)

∑T

∑i∈1,2

κi‖∇(u− uTi)‖2Ti

≤ 2∑T

∑i∈1,2

κi(‖∇(u− Ik+1Ti

(u))‖2Ti+ ‖∇(Ik+1

Ti(u)− uTi)‖2Ti

)

For eh = Ikh(u)− uh,

‖eh‖2∗ ≤ Cah(eh, eh) ≤ C2hk+1(∑

i∈1,2

κi|u|2Hk+2(Ωi))1/2‖eh‖∗

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The Stokes problem

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The interface Stokes problem−∆u +∇p = f in Ω1 ∪ Ω2

div u = 0 in Ω1 ∪ Ω2

JuKΓ = 0 on Γ

J∇u− pIKΓnΓ = gN on Γ

u = 0 on ∂Ω

velocity(k + 1, k)(vector–valued)

pressure (k)(scalar–valued)

uncut cell(k = 0)

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Local operators

We define:Gradient reconstruction operators Gk

Ti(uT ), Gk

Ti(uT )

∈ Pk(Ti;Rd×d) such that for every Q ∈ Pk(Ti;Rd×d),

(GkTi

(uT ),Q)Ti= −(uTi

,div Q)Ti+ (u(∂T )i ,QnT )(∂T )i + (uTi

,QnΓ)TΓ

(GkTi

(uT ),Q)Ti = −(uTi ,div Q)Ti + (u(∂T )i ,QnT )(∂T )i + (uTi,QnΓ)TΓ

Divergence reconstruction operators DkTi

(uT ), DkTi

(uT ) ∈ Pk(Ti)

such that for every q ∈ Pk(Ti),

(DkTi

(uT ), q)Ti = −(uTi ,∇q)Ti + (u(∂T )i · nT , q)(∂T )i + (uTi · nΓ, q)TΓ

(DkTi

(uT ), q)Ti = −(uTi ,∇q)Ti + (u(∂T )i · nT , q)(∂T )i + (uTi· nΓ, q)TΓ

Note that DkTi

(uT ) = Tr(GkTi

(uT )), DkTi

(uT ) = Tr(GkTi

(uT ))

A stabilization operator

sT (uT , vT ) = h−1T

∑i∈1,2

∑Fi∈FTi

(ΠkFi

(uFi− uTi

),vFi− vTi

)Fi

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Local discrete problem

We define the local operators

aT (uT , vT ) = (GkT1

(uT ), GkT1

(vT ))T1+ (Gk

T2(uT ),Gk

T2(vT ))T2

+ sT (uT , vT )

+ ηh−1T (JuT KΓ, JvT KΓ)TΓ

bT (uT , qT ) = (DkT1

(uT ), qT1)T1

+ (DkT2

(uT ), qT2)T2

and the local right–hand side

`aT(vT ) = (f ,vT1)T1 + (f ,vT2)T2 + (gN ,vT2)TΓ

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Global discrete problem

Find uh = (uT )T∈Th and ph = (pT )T∈Th such that∑T∈Th

aT (uT , vT )− bT (vT , pT )−γ0hT (J∇uT − pT IKΓnΓ, J∇vT KΓnΓ)TΓ

=∑T∈Th

`aT(vT )−γ0hT (gN , J∇vT KΓnΓ)TΓ

∑T∈Th

bT (uT , qT )+γ0hT (J∇uT − pT IKΓnΓ, JqT KΓnΓ)TΓ

=∑T∈Th

γ0hT (gN , JqT KΓnΓ)TΓ

for every vh = (vT )T∈Th and qh = (qT )T∈Th .

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Convergence result

An inf-sup condition is satisfiedTheorem

For every η > 0, for u ∈ (Hk+2(Ω1 ∪ Ω2))2 and p ∈ Hk+1(Ω1 ∪ Ω2), forγ0 small enough, we have∑

T∈Th

∑i∈1,2

‖∇(u− uTi)‖2Ti

+ ‖p− pTi‖2Ti

≤ Ch2(k+1)(‖u‖2Hk+2(Ω1∪Ω2) + ‖p‖2Hk+1(Ω1∪Ω2))

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Numerical analysis for Stokes

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Stability

We define the stability norms

‖uT ‖2∗ =∑

i∈1,2

‖∇uTi‖2Ti+ h−1

T ‖uTi − u(∂T )i‖2(∂T )i + h−1T ‖JuT KΓ‖2TΓ

‖(uT , pT )‖2# = ‖uT ‖2∗ + ‖pT1‖2T1+ ‖pT2‖2T2

Coercivity and continuity of aT (viscous part)

For every uT , vT ∈ UkT , we have

‖uT ‖2∗ ≤ CaT (uT , uT )

aT (uT , vT ) ≤ C‖uT ‖∗‖vT ‖∗

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Stability

We define the bilinear form

Ah((uh, ph), (vh, qh)) = ah(uh, vh)− bh(vh, ph) + bh(uh, qh)

− γ0

∑T∈Th

hT (J∇uT − pT IKΓ, J∇vT + qT IKΓ)TΓ

Inf–sup condition

For γ0 small enough, there exists c > 0 such that for every(uh, ph) ∈ Uk

h × P kh ,

c‖(uh, ph)‖# ≤ sup(vh,qh)∈Uk

h×Pkh

Ah((uh, ph), (vh, qh))

‖(vh, qh)‖#

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Stability

We denote S = sup(vh,qh)∈Uk

h×Pkh

Ah((uh, ph), (vh, qh))

‖(vh, qh)‖#We have

Ah((uh, ph), (uh, ph)) =ah(uh, uh)

+ γ0

∑T∈Th

hT (‖JpT KΓ‖2TΓ − ‖J∇uT KΓnΓ‖2TΓ)

c‖uh‖2∗ + γ0

∑T∈Th

hT ‖JpT KΓ‖2TΓ

≤ ah(uh, uh) + γ0

∑T∈Th

hT ‖JpT KΓ‖2TΓ

≤ Ah((uh, ph), (uh, ph)) + γ0

∑T∈Th

hT ‖J∇uT KΓnΓ‖2TΓ

and then

(c− Cγ0)‖uh‖2∗ + γ0

∑T∈Th

hT ‖JpT KΓ‖2TΓ ≤ S‖(uh, ph)‖#

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Stability

There exists w ∈ H10 (Ω1) such that ph = div w

‖ph‖2L2 =∑T,i

(pTi ,div w)Ti = Ψ1 + Ψ2

Ψ1 = bh(ph, Ikh(w))

= ah(uh,Ikh(w))−Ah((uh, ph), (Ikh(w), 0))

− γ0

∑T

hT (J∇uT − pT IKΓnΓ, J∇Ik+1T (w)KΓnΓ)TΓ

≤ C(S‖Ikh(w)‖∗ + ‖uh‖∗‖Ikh(w)‖∗ + γ0

∑T

h1/2T ‖JpT KΓ‖TΓ‖Ikh(w)‖∗)

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Stability

Ψ2 =∑T

(pT1,div w − Dk

T1(IkT (w)))T1

+ (pT2,div w −Dk

T2(IkT (w)))T2

=∑T,i

(pTi,div (w − Ik+1

Ti(w)))Ti

− (pTinT , I

k(∂T )i(w)− Ik+1

Ti(w))(∂T )i

+ (pT1nΓ, JIk+1

T (w)KΓnT )TΓ

=∑T,i

−(∇pTi,w − Ik+1

Ti(w))Ti

+ (JpT KΓnΓ,w − Ik+1T2

(w))TΓ

≤ C‖w‖H1(∑T,i

hT ‖∇pTi‖Ti +∑T

h1/2T ‖JpT KΓ‖TΓ)

‖ph‖2L2 ≤ C(S2 + ‖uh‖2∗ + γ0

∑T

hT ‖JpT KΓ‖2TΓ +∑T,i

h2T ‖∇pTi

‖2Ti)

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Numerical simulations

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Agglomeration procedure (1/4)

Small cut cells are agglomerated (see [Johansson, Larson 13])The three stages of the procedure

initial mesh stage 1

stage 2 stage 342/50

Agglomeration procedure (2/4)

A 16x16 mesh with a circular interface:

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Agglomeration procedure (3/4)

A 16x16 mesh with a flower-like interface:

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Agglomeration procedure (4/4)

A 16x16 mesh with a square interface:

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(Elliptic problem) Test case with contrast

κ1 = 1, κ2 = 104, gD = gN = 0, η = 1

Exact solution (r2 = (x1 − 0.5)2 + (x2 − 0.5)2)

u(x1, x2) =

r6

κ1in Ω1

r6

κ2+R6(

1

κ1− 1

κ2) in Ω2

10-8

10-7

10-6

10-5

10-4

10-3

10-2

0.0078125 0.015625 0.03125 0.0625 0.125

errorH

1 sem

inorm

h

k=0k=1k=2k=3

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(Elliptic problem) Test case with jump

Exact solution

u(x1, x2) =

sin(πx1) sin(πx2) in Ω1

sin(πx1) sin(πx2) + 2 + x3y3 in Ω2

κ1 = κ2 = 1

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

0.0078125 0.015625 0.03125 0.0625

errorH

1 sem

inorm

h

k=0k=1k=2k=3

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(Elliptic problem) Condition number of the system matrix

Square interface, κ1 = κ2 = 1, after static condensation

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Stokes problem

X = x− 0.5 Y = y − 0.5 Ω = C(0, 0.33)

u1 = X2(X2 − 2X + 1)Y (4Y 2 − 6Y + 2)

u2 = −Y 2(Y 2 − 2Y + 1)X(4X2 − 6X + 2);

p = X5 + Y 5

meshes : 8x8, 16x16, 32x32, 64x64; k = 0, 1Velocity error Pressure error

0.000001

0.000010

0.000100

0.001000

0.010000

0.100000

1.000000

0.015625 0.03125 0.0625 0.125h

H1:k=0H1:k=1L2:k=0L2:k=1

0.00001

0.00010

0.00100

0.01000

0.10000

0.015625 0.03125 0.0625 0.125h

L2:k=0L2:k=1

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Wrap up

Same advantages as fitted HHO

Usable on curvilinear domains

Work in progress on interface Stokes problem

submitted [Burman, Cicuttin, Delay, Ern] (elliptic problem)

Thank you for your attention!

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