Mimetic Least Squares Spectral/hp Finite Element Method

OutlineThe Standard Least Squares

Mimetic ApproachSummary and Future Work

Mimetic Least Squares Spectral/hpFinite Element Method for the Poisson Equation

Artur Palha1 and Marc Gerritsma1

1Faculty of Aerospace EngineeringDelft University of Technology

Email: [email protected]

June 10, 2010

Artur Palha and Marc Gerritsma Mimetic Least Squares 1 / 57



The Standard Least SquaresHow does it work?The finite dimensional spacesExample: 2D Poisson equationWhy it does not work?

Mimetic ApproachGoing back to the basicsDifferential geometryMimetic least-squares

Summary and Future WorkSummaryFuture workFurther reading




How does it work?The finite dimensional spacesExample: 2D Poisson equationWhy it does not work?








The principle

The partial differential equationLu = f in ΩRu = h on Γ

Reduce the dimension of the problem (discretize)Luh,p = f in ΩRun,p = h on Γ





The principle

The partial differential equationLu = f in ΩRu = h on Γ

Reduce the dimension of the problem (discretize)Luh,p = f in ΩRun,p = h on Γ





The principle

Translate to a minimization problem

minuh,p∈Xh,p

I(uh,p; f, h) ≡1

2

“‖Luh,p − f‖2Xh,Ω

+ ‖Ruh,p − h‖2Xh,Ω

”Which reduces to:`

Luh,p,Lvh,p´Ω

+`Ruh,p,Rvh,p

´Γ

=`f,Lvh,p

´Ω

+`h,Rvh,p

´Γ

And finally to an algebraic system

Auh,p = b





The principle

Translate to a minimization problem

minuh,p∈Xh,p

I(uh,p; f, h) ≡1

2

“‖Luh,p − f‖2Xh,Ω

+ ‖Ruh,p − h‖2Xh,Ω

”Which reduces to:`

Luh,p,Lvh,p´Ω

+`Ruh,p,Rvh,p

´Γ

=`f,Lvh,p

´Ω

+`h,Rvh,p

´Γ

And finally to an algebraic system

Auh,p = b





The finite dimensional spaces: C0 nodal elements

All physical quantities represented by similar spaces

φ(x, y)→ φh(x, y) =Xi,j

φi,jhpi (x)hpj (y)

u(x, y)→ uh(x, y) =

» Pm,n u

xm,nh

pm(x)hpn(y)P

k,l uym,nh

pk(x)hpl (y)

–That is:

φh(x, y) ∈ spannhpi (x)hpj (y)

o, i, j = 0, . . . , p

uh(x, y) ∈ span˘hpm(x)hpn(y)⊗ hpk(x)hpl (y)

¯, m, n, k, l = 1, . . . , p

hpi (ξ) Lagrange interpolants over Gauss-Lobatto-Legendre points.





Numerical solution of 2D Poisson equation

φ(x, y) mimetic φ

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

0.0

0.3

0.6

0.9

1.2

1.5

1.8

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

0.0

0.3

0.6

0.9

1.2

1.5

1.8






vx(x, y) mimetic vx mimetic qx

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6






vy(x, y) mimetic vy mimetic qy

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.20

1.54

0.88

0.22

0.44

1.10

1.76





Why it does not work?

We are not respecting the structureof the equations in the discretesetting




Going back to the basicsDifferential geometryMimetic least-squares








Physical quantities and geometry

There is an intrinsic association between physical quantities and geometrical objects:

I Points: e.g. Electric potential, φ

I Lines: e.g. Electric field, E, Magnetizing field, H

I Surfaces: e.g. Magnetic flux, B, Electric displacement field, D

I Volumes: e.g. Charge density, ρ

These associations are intrinsic to the differential equations that relate the physicalquantities:8>>>>>><>>>>>>:

∇ ·D = ρ∇ ·B = 0

∇×E = − ∂B∂t

∇×H = J + ∂D∂t

D = εEB = µH

⇐⇒

8>>>>>><>>>>>>:

v∂V D · dA = Q(V )v∂V B · dA = 0H∂S E · dl = − ∂

∂t

vS B · dAH

∂S H · dl =vS J · dA + ∂

∂t

vS D · dA

D = εEB = µH





Inner and outer orientation of geometrical objects





Classification of physical laws

Topological lawsAre characterized by the fact that their validity is independent of the nature of themedium under consideration. Connect configuration variables with configurationvariables and source variables with source variables. Are independent of metric sincethey are intrinsically integral equations (global).

∇ ·B = 0, ∇φ = u, ∇×E = 0, . . .

Constitutive lawsAre characterized by the fact that their validity depends on the nature of the mediumunder consideration. They describe the behaviour of a material. Connect configurationvariables with source variables. Depend on the metric since they are intrinsically localin nature.

D = εE, q = ρv, . . .





Vector calculus obscures

Example: 2D Poisson equation for potential flow8<: ∇φ = v∇ · q = f

q = ρv

I There is no reference to which geometrical object the physical quantities areassociated.

I There is no reference to inner or outer orientation.

I All this is given a posteriori. Right hand rule and so on.

How to solve this?

We need a proper framework!





Vector calculus obscures

Example: 2D Poisson equation for potential flow8<: ∇φ = v∇ · q = f

q = ρv

I There is no reference to which geometrical object the physical quantities areassociated.

I There is no reference to inner or outer orientation.

I All this is given a posteriori. Right hand rule and so on.

How to solve this?

We need a proper framework!





Differential geometry

The Holy Grail: Differential Geometry





Differential Geometry: a refresher

We need to introduce 1 object and 4 operators:

I k-differential form or k-form

I wedge product, ∧I inner product, (·, ·)I exterior derivative, d

I Hodge-? operator, ?





Differential forms1: representation of physical quantities

Let M⊂ Rn a differentiable manifold, thenk-form ωk ∈ Λk: a rank-k, anti-symmetric, tensor field over M

ωk : TxM × · · · × TxM| z k copies

−→ R ,

ωk(. . . , vi, . . . , vj , . . . ) = −ωk(. . . , vj , . . . , vi, . . . )

Wedge product: Let ωk ∈ Λk and ωl ∈ Λl then

∧ : Λk × Λl −→: Λk+l

1Cartan [?], Spivak [?], Flanders [?]





Differential forms: intrinsic connection with geometry

Under integration, one can state a duality pairing between k-forms and k-manifolds:ZΩk

ωk = 〈ωk,Ωk〉 ∈ R

Leads to an instrisic connection between differential forms and geometrical objects (inR3):

I 0-forms −→ Points

I 1-forms −→ Lines

I 2-forms −→ Surfaces

I 3-forms −→ Volumes





Operators: the exterior derivative d

The exterior derivative d, in a n-dimensional space, is a mapping:

d : Λk 7→ Λk+1, k = 0, 1, . . . , n− 1,

which satisfies:

d“ωk ∧ αl

”= dωk ∧ αl + (−1)kωk ∧ dαl, k + l < n

and:ddωk = 0, ∀ωk ∈ Λk, k < n− 1

Leads to the exact sequence (de Rham complex):

R → Λ0(Ω)d7→ Λ1(Ω)

d7→ Λ2(Ω)d7→ Λ3(Ω)

d7→ 0

In R3: d0 ↔ ∇, d1 ↔ ∇× and d2 ↔ ∇·.





Operators: the exterior derivative d

Stokes Theorem: Let Ωk+1 be a k + 1-dimensional manifold and ω ∈ Λk thenZ∂Ωk+1

ωk =

ZΩk+1

dωk

Z∂Ωk+1

ωk

duality pairing

=

ZΩk+1

dωk

duality pairing

〈ωk, ∂Ωk+1〉 = 〈dωk,Ωk+1〉

d by duality pairing is the formal adjoint of ∂.





Operators: the Hodge-? operator

What about ∇2?∇2 = ∇ · ∇ 6= d d = 0

What about d(0) d(2)? No because:

d(0) : Λ0 −→ Λ1

d(n−1) : Λn−1 −→ Λn

ff=⇒R(d(0)) 6⊂ D(d(n−1))

An additional operator ? is needed, such that:

? : Λk 7→ Λn−k

Then:∇2 ↔ d(n−1) ? d(0)

And enables the construction of the de Rham complex, for example in R3:

R // Λ0

?

d // Λ1

?

d // Λ2

?

d // Λ3

?

// 0

0 Λ3oo Λ2d

oo Λ1d

oo Λ0d

oo Roo





So what do we get from this?

Elegance and compactnessMaxwell equations:

dF 2 = 0, dG2 = J3, dJ3 = 0, G2 = ?F 2

Fundamental theorems: ZΩk+1

dωk =

Z∂Ωk+1

ωk

Clear relation and separation of objectsConnection between k-forms and k + 1-manifolds:h

dωk,Ωk+1

i=hωk, ∂Ωk+1

iConstitutive equations are now expressed with the Hodge-? operator:

d2 = ?εe1, q2 = ?ρv

1, . . .





The de Rham complex

Additionally, one verifies that in sufficiently regular regions Ω the spaces of differentialforms together with the exterior derivative d constitute an exact sequence, called thede Rham complex, which, in 3D is:

R // Λ0

?

d // Λ1

?

d // Λ2

?

d // Λ3

?

// 0

0 Λ3oo Λ2d

oo Λ1d

oo Λ0d

oo Roo





The de Rham complex

And in 2D reduces to:

R // Λ0

?

d // Λ1

?

d // Λ2

?

// 0

0 Λ2oo Λ1d

oo Λ0d

oo Roo

Which is equivalent to the more familiar: more

R // H1

?

∇ // H1(curl)

?

∇× // L2

?

L2 H1(div)

∇·oo H1

∇⊥oo Roo





The Tonti diagram: 2D Poisson equation

The equation 8<: ∇φ = v∇ · q = f

q = ρv⇔

8<:dφ0 = v1

dq1 = f2

q1 = ?ρv1

The diagram

φ0

d

f2

u1?

// q1

d

OO





The Tonti diagram: 2D Poisson equation

The equation 8<: ∇φ = v∇ · q = f

q = ρv⇔

8<:dφ0 = v1

dq1 = f2

q1 = ?ρv1

The diagram

φ0

d

f2

u1?

// q1

d

OO





Numerical solution: discretizations

Dual grid methods

I Depends on the existence of a pair of topologically dual grids

I One-to-one correspondence between dual variables

I Simple Hodge-? operator

I All equations satisfied globally

Elimination methodsI Sacrifices one of the equillibrium equations

I Satisfies exactly the other equilibrium equation and the constitutive equationlocally

Primal-dual grid methods

I Satisfies exactly the equilibrium equations

I Relaxes the constitutive equation, being enforced weakly





Numerical solution: discretizations

Dual grid methods

I Depends on the existence of a pair of topologically dual grids

I One-to-one correspondence between dual variables

I Simple Hodge-? operator

I All equations satisfied globally

Elimination methodsI Sacrifices one of the equillibrium equations

I Satisfies exactly the other equilibrium equation and the constitutive equationlocally

Primal-dual grid methods

I Satisfies exactly the equilibrium equations

I Relaxes the constitutive equation, being enforced weakly





Weak material laws: the role of least-squares

The ideaI Impose the constitutive equation weakly

I Hodge-? operator defined implicitly

I Minimize local discrepancy between dual variables

The implementation

Seek (φ0h, v

1h, q

1h) in Λ0

h × Λ1h × Λ1

h such that (1)

I(φ0h, v

1h, q

1h) = 1

2

“‖ ?q1

h + v1h‖

20 + ‖dq1

h − f2‖20”

subject to: dφ0h = v1

h





Weak material laws: the role of least-squares

If the subspaces Λ0h, Λ1

h and Λ2h are chosen in such a way that they constitute a de

Rham complex:

R→ Λ0h

d7−→ Λ1h

d7−→ Λ2h 7→ 0

then dφ0h = v1

h is satisfied exactly. The problem becomes:

Seek (φ0h, q

1h) in Λ0

h × Λ1h such that (2)

I(φ0h, q

1h) = 1

2

“‖ ?q1

h + dφ0‖20 + ‖dq1h − f

2‖20”

In this way, the Hodge-? operator is implemented as L2 projections between the

different dual spaces.





Application to the 2D Poisson equation

Find adequate subspaces Λ0h, Λ1

h and Λ2h must be specified. Since one will use a

spectral/hp LS method, these spaces are defined as:

Λ0h,p = span

nhpi (x)hpj (y)

o, i = 0, . . . , p j = 0, . . . , p

Λ1h,p = span

nhp−1i (x)hpj (y)⊗ hpn(x)hp−1

m (y)o

, i,m = 1, . . . , p j, n = 0, . . . , p

Λ2h,p = span

nhp−1i (x)hp−1

j (y)o

, i = 1, . . . , p j = 1, . . . , p

I hpi (ξ): i-th Lagrange interpolant of order p throught Gauss-Lobatto-Legendrepoints

I hpi (ξ): i-th Lagrange interpolant of order p throught Gauss points

I Degrees of freedom are located where they should be: at nodal points (for0-forms), at edges (for 1-forms) and at volumes (for 2-forms).

I Different continuity properties

I These subspaces constitute a de Rham complex





Application to the 2D Poisson equation





Numerical results

φ(x, y) standard φ

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

0.0

0.3

0.6

0.9

1.2

1.5

1.8

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

0.0

0.3

0.6

0.9

1.2

1.5

1.8





Numerical results

vx(x, y) standard vx

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6





Numerical results

qx(x, y) standard vx

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6





Numerical results

vy(x, y) standard vy

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6





Numerical results

qy(x, y) standard vy

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6

1.0 0.5 0.0 0.5 1.01.0

0.5

0.0

0.5

1.0

2.0

1.4

0.8

0.2

0.4

1.0

1.6





Numerical results

Convergence results

0 2 4 6 8 10 12 14p

10-6

10-5

10-4

10-3

10-2

10-1

100

101

ε

L2 norm error

Standard LS: φStandard LS: uWeak laws p: φWeak laws p: uWeak laws p: u− u

0 2 4 6 8 10 12 14p

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

101

102

ε

L2 norm error

Standard LS: ∇×uStandard LS: ∇ ·uWeak laws p: ∇×u

Weak laws p: ∇ ·u





Numerical results

Convergence results p = 2

10-2 10-1 100

h

10-5

10-4

10-3

10-2

10-1

100

ε

L2 norm error: p=2

Standard LS: φStandard LS: uWeak laws p: φWeak laws p: u

10-2 10-1 100

h

10-6

10-5

10-4

10-3

10-2

10-1

100

101

ε

L2 norm error: p=2







Numerical results

Convergence results: p = 3

10-2 10-1 100

h

10-6

10-5

10-4

10-3

10-2

10-1

ε

L2 norm error: p=3

Standard LS: φStandard LS: uWeak laws p: φWeak laws p: u

10-2 10-1 100

h

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

101

ε

L2 norm error: p=3







Numerical results

Can we do better?





Numerical results

Can we do better?Yest we can!





Numerical results

Can we do better?Yest we can!

With edge basis functions!





Edge basis functions: the idea

What we get from nodal interpolation?

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

0.0

0.1

0.2

0.3

0.4

The closer we are from the interpolation nodes the smaller the error.





Edge basis functions: the idea

What about interpolation of vector fields? What is theimportant quantity?Line integral!

GLL - edge interpolation

ei(ξ) = −i−1Xk=0

dhk(ξ),

Z ξk

ξk−1

ei(ξ) = δik

Histopolation

The closer our path is from the mesh edges the smaller the error in the line integral.





What are the degrees of freedom?

φ0ij f2

ij

ξu1kl ηu1

mn

f0ij φ2

ij

ξ q1kl η q1

mn





Numerical results: Edge basis functions

Convergence results

0 2 4 6 8 10 12p

10-4

10-3

10-2

10-1

100

101

102

ε

conservative L2 error - φ

standard L2 error - φ

conservative L2 error - q

standard L2 error - q

0 2 4 6 8 10 12p

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

101

102

∇×q/∇·q

∇×q conservative∇×q standard∇·q conservative∇·q standard

For Histopolants see: Robidoux, Polynomial Histopolation, Superconvergent DegreesOf Freedom, And Pseudospectral Discrete Hodge Operators, to appear.






Curved elements: p = 4

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0-2

-1

0

1

2

3

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0-2

-1

0

1

2

3






Curved elements: p = 8

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0-2

-1

0

1

2

3

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0-2

-1

0

1

2

3




SummaryFuture workFurther reading

Summary

I Physical quantities are inherently geometrical

I Structure of PDE’s must be obeyed

I There is more to life than scalars and vectors





Future work

I Curved domains

I 3 dimensions

I 4 dimensions: space-time





The end!

The end!





Further reading

Tonti, E.: On the formal structure of physical theories. Consiglio Nazionale delleRicerche, Milano (1975)

Bochev, P. and Hyman, J.: Principles of mimetic discretizations of differentialoperators. IMA 142, 89–119 (2006)

Mattiussi, C.: An analysis of finite volume, finite element, and finite differencemethods using some concepts from algebraic topology. J. Comp. Physics 133,289–309 (1997)

Desbrun, M. and Kanso, E. and Tong, Y.: Discrete differential forms forcomputational modeling. SIGGRAPH ’05: ACM SIGGRAPH 2005 Courses (2005)

Bossavit, A.: On the geometry of electromagnetism. J. Japan Soc. Appl.Electromagn. & Mech. 6 (1998)





The de Rham complex in 2D explained

Usually the 2-dimensional case is viewed as a special case and hence it is expressed bytwo exact sequences:

R → H1 ∇7−−→ H1(curl)∇×7−−−→ L2 (3)

R → H1 ∇⊥

7−−−→ H1(div)∇·7−−→ L2

with

∇φ =∂φ

∂xex +

∂φ

∂yey (4)

∇⊥φ = −∂φ

∂yex +

∂φ

∂xey (5)

∇×W =∂Wy

∂xex −

∂Wx

∂yey (6)

∇ ·W =∂Wx

∂x+∂Wy

∂y

back






These two exact sequences are obtained by a restriction of the 3D exact sequence:

R → H1 ∇7−−→ H1(curl)∇×7−−−→ H1(div)

∇·7−−→ L2

to a planar 2D surface embedded in R3, for example the xy-plane. This, in turn,reduces to the pair of exact sequences Eq. (3) and Eq. (??), since the vectors of theform φez can be identified with scalar functions. The odd operator ∇⊥ is, then,nothing but the result of applying the 3D ∇× to φez .The full de Rham complex in 2D becomes:

R // H1

?

∇ // H1(curl)

?

∇× // L2

?

L2 H1(div)

∇·oo H1

∇⊥oo Roo

back






It is important to realize that this full exact complex is always relative to the proxies ofthe differential forms, that is, scalar and vector fields, not the differential k-forms towhich they are associated. This is the important point, since this is the reason whyboth De Rham complexes are equivalent.Let us therefore show how the differential formulation agrees with the vectorformulation with its special characteristics in 2D:

R // Λ0

?

d // Λ1

?

d // Λ2

?

Λ2 Λ1

doo Λ0

doo Roo

Where Λ0, Λ1 and Λ2 are the spaces of twisted forms (as in Burke (1985) section 28,or Bossavit (1998) Japanese papers chapter (2):Geometrical objects), and the untildedones are the spaces of forms.

back






To show that both exact complexes are equivalent one must show how to pass fromforms to proxies and from proxies to forms. This is done by using the sharp, ], andflat, [, operators, respectively. The sharp, ], and flat, [, acting on a 0-form (φ0) and ascalar field (φ), give: `

φ0´]

= φ, φ[ = φ0

The sharp, ], and flat, [, acting on a 1-form (α1 = fdx + gdy) and a vector field(A = fex + gey), give:

`α1´]

= fex + gey , A[ = fdx + gdy

The sharp, ], and flat, [, acting on a twisted 1-form (β1 = −gdx + fdy) and a vectorfield (B = fex + gey):

“β1”]

= fex + gey , B[ = −gdx + fdy

back






The sharp, ], and flat, [, acting on a 2-form (ω2 = wdxdy) and a scalar field (w):`ω2´]

= w, w[ = wdxdy

The case of twisted 0-forms and twisted 2-forms are identical to the correspondingstandard forms.For the 2D case, these operations can be summarized by

?1 = dxdy, ?dx = dy, ?dy = −dx ? dxdy = 1

Burke (1985) section 28, or Bossavit (1998) Japanese papers chapter (2):Geometricalobjects See also, Burke (1985) section 28, Bossavit (1998) Japanese papers chapter(2):Geometrical objects, Marsden (2002) p.432 and Bossavit (2005), p. 21 and p.23.We can see now that the special form of the 2D De Rham complex in vector formresults from converting the usual De Rham complex for 2D in differential form to itsvectorial representation using the above relations.The top exact complexes are identical, on the proxies. On the 0-forms:

`dφ0

´]=

„∂φ

∂xdx +

∂φ

∂ydy

«]=∂φ

∂xex +

∂φ

∂yey = ∇φ

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On the 1-forms:

`dα1

´]=

»„∂g

∂x−∂f

∂y

«dxdy

–]=∂f

∂x−∂g

∂y= ∇×

`α1´]

= ∇×A

Which is exactly the same. Now, the bottom exact complex in Eq. (??) is identical tothe one Eq. (??), on the proxies. On the twisted 0-forms:

“dφ0

”]=

„∂φ

∂xdx +

∂φ

∂ydy

«]=∂φ

∂yex −

∂φ

∂xey = −∇⊥

“φ0”]

Remembering that the differential of a twisted form is a twisted form. On the twisted1-forms:

“dβ1

”]=

»„∂f

∂x+∂g

∂y

«dxdy

–]=`w2´]

=∂f

∂x+∂g

∂y= ∇ ·

“β1”]

where we have used, as before, β1 = −gdx + fdy.

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Summarizing: The 2D case is not special in differential geometry,but its representation in vector form is markedly different from the3D case.

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