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The Reduced Basis Method for a Finite Deformation Problem Lorenzo Zanon Karen Veroy-Grepl Aachen Institute for Advanced Study in Computational Engineering Science 3 rd ECCOMAS Young Investigators Conference YIC2015 July 21, 2015

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The Reduced Basis Method for a Finite Deformation Problem

Lorenzo ZanonKaren Veroy-Grepl

Advisory Board MeetingRWTH Aachen University

July 12, 2011

Aachen Institute for Advanced Studyin Computational Engineering Science

AICES Progress since 2009

• Overview of the AICES project

• New Principal Investigators

• New Young Researchers

• New facilities

• Recent highlights

• Goals and schedule of Advisory Board meeting

Outline

2

• Excellence Initiative Graduate School

• First period Nov 2006–Oct 2012

• Annually 1 M! + overhead

• Proposed for 2013–2017

• Funded personnel growth:

Overview of the AICES Project

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1 3 3 3 3 3 3

1 1 3 3 4 41

1 1 1 11

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02/200712/2007

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Service Team

Junior Research Group Leaders

Adjunct Professors

Doctoral Fellows

Master Students

Postdoctoral Research Associates

• Computational engineering science is maturing

• “Old” challenges are still here:• complexity increasing intricacy of analyzed systems• multiscale interacting scales considered at once• multiphysics interacting physical phenomena

• AICES concentrates on areas of synthesis:• model identification and discovery supported by

model-based experimental analysis (MEXA)• understanding scale interaction and scale integration• optimal design and operation of engineered systems

• Inspiration: Collaborative Research Center 540• established in 1999, continued until 2009• Marquardt: coordinator, 6 AICES PIs: project leads

Overview of AICES Academic Aims

SFB 540

4

3rd ECCOMAS Young Investigators Conference YIC2015July 21, 2015

Outline

1. Motivation2. The Reduced Basis Method3. Fundamental Equations in Elasticity4. Numerical Example: Finite Deformation5. Conclusion

Lorenzo Zanon

Part 1 - Motivation

Dimension reduction:PDE Description in a continuous setting of the physics of the problem:

I field variable: displacement, temperature, . . .I output of interest: flow-rate, mean quantity, heat flux, critical load, . . .

FE Highly accurate approximation:I high accuracy, truth solution;I offline dimension N = O(103), expensive computations.

RB Built upon FE, in a parametric context:I m.o.r. even for complex nonlinear models;I online dimension N = O(10);I drastic reduction of computation times:

I parameter identification / inverse problems;I control and optimization;I many-query / real-time context.

Lorenzo Zanon 1/21

Motivation

RB Method in the context of elasticity can be applied to:I the description of the nonlinear behavior of materials

in finite deformation regime. Possible future applications:I rubber-like materials;I biological tissues.

I the analysis of buckling structures characterized byparameter-dependent geometries

→ truss systems

Lorenzo Zanon 2/21

Motivation

State of the art:I Successful application of the RB Method to linearized elasticity;I Other model order reduction techniques for nonlinear elasticity→ Proper Orthogonal Decomposition, SVD-based approach:

+ Dealing with complex problems (viscoelasticity, plasticity, . . . );+ Processing results pre-obtained from existent softwares (FEAP)

with POD-related techniques (DEIM, substructuring);+ Good results in terms of CPU-ratio and precision w.r.t. FE;

− Lack of an efficient greedy procedure to select snapshots;− Parametric nature of the problem only partially exploited;− No clear offline-online distinction; always N -dependent problems.

I RB for nonlinear elasticity currently under investigation.

Lorenzo Zanon 3/21

A. Radermacher, S. Reese, POD-based model reduction for nonlinearbiomechanical analysis. Int. J. of Materials Engineering Innovation, 2013.

Outline

1. Motivation2. The Reduced Basis Method3. Fundamental Equations in Elasticity4. Numerical Example: Finite Deformation5. Conclusion

Lorenzo Zanon

Part 2 - The RB MethodI Weak-formulation of a parametrized boundary-value problem:

a(u(µ),v; µ) = f (v; µ) , ∀v ∈V ⊂ H1(Ω) ;

a(v,v; µ) : V ×V → R , f (v; µ) : V → R , bilinear resp. linear forms.I The parameters µ ∈D ⊂ Rd can be of geometric or physical nature;

With the RB method, the FE solution - dim. N = O(103) -is projected on a low-dimensional space - dim. N = O(10).

I Build up the RB space by taking snapshots at well-chosen parameters:

ZN = span(u(x; µ j)); j = 1, . . . ,Nmax= span(ζ j(x)); j = 1, . . . ,Nmax

I . . . selected through a greedy procedure:for N = 1, . . . ,Nmax

µ∗N = argmaxµ∈D train

∆uN−1(µ)e(u)N−1(µ)

⇒ ZN = ZN−1∪ spanu(µ∗N) .

Lorenzo Zanon 4/21

C. Prud’homme, D. V. Rovas, K. Veroy, L. Machiels, Y. Maday, A. T. Patera, and G. Turinici. Reliable real-time

solution of parametrized PDEs: Reduced-basis output bound methods. J. of Fluids Engineering, 2002.

The RB Method - Offline/OnlineWhy use the RB Method in a parametrized context?

I Offline phase: the FE-dependent quantities are computed and stored;I Online phase: the parameter-dependent low-dimensional system is

solved.

I Separability of the linear forms in the truth FE setting

a(w,v; µ) =Qa

∑q=1

Θq(µ)aq(w,v) ;

. . . maintained in the RB setting: a(vN ,wN ; µ) =Qa

∑q

Θq(µ)aq(vN ,wN) .

I Nonaffine problems: use interpolation techniques, e.g. EIM.For N ≤ Nmax, we obtain the RB approximation . . .

I of the solution: u(µ)≈ uN(µ) :=N

∑j=1

uN j(µ)ζ j;

I of an output of interest: s(µ)≈ sN(uN(µ)).

Lorenzo Zanon 5/21

Outline

1. Motivation2. The Reduced Basis Method3. Fundamental Equations in Elasticity4. Numerical Example: Finite Deformation5. Conclusion

Lorenzo Zanon

Part 3 - Fundamental Equations in Elasticity

Deformation of bodies with a defined shape:→ Elasticity: the original shape is recovered once the stress ceases its action.← yielding / plasticity / fracture / . . .

Equilibrium Equation: σϕ

i j, j +bϕ

i = 0 , in Ωϕ ⊂ Rd ;Momentum Equation: σ

ϕ

i j = (σϕ

i j )t , in Ωϕ ⊂ Rd ;

Dirichlet Conditions: ui = ui , on δΩϕ,D ⊂ Rd−1 ;Applied Traction: σ

ϕ

i j nϕ

j = T ϕ

i , on δΩϕ,T ⊂ Rd−1 ;

I Here:u : unknown displacement;σ := σ(u) : Cauchy stress matrix;b : volumetric forces vector;

I Superscript ϕ refers to thedeformed configuration . . .

σi j, j :=d

∑j=1

∂σi j

∂x j.

Lorenzo Zanon 6/21

Fundamental Equations in Elasticity

I Map to the initial configuration through the deformation tensor

Fi j := δi j +ui, j , J := detF .

I Define the Piola-Kirchoff tensors: P := σϕ JF−t , S := F−1P.I Choose the constitutive relation: here assume St. Venant model:

Si j = Ci jklEklGreen-Lagr. tensor: Ei j := 1

2 (ui, j +u j,i +uk,iuk, j) ;St. Venant tensor: Ci jkl := Λδi jδkl +M(δikδ jl +δilδ jk) .

I Obtain the fundamental equations in the Lagrangian setting:Equilibrium Equation: ((δik +ui,k)Sk j), j +bi = 0 , in Ω ;Momentum Equation: Si j = (Si j)

t , in Ω .

Lorenzo Zanon 7/21

P. Wriggers, Nonlinear Finite Element Methods. Springer, 2008.

Outline

1. Motivation2. The Reduced Basis Method3. Fundamental Equations in Elasticity4. Numerical Example: Finite Deformation5. Conclusion

Lorenzo Zanon

Part 4 - Finite Deformation

On . . .I a 2D rectangle [0,1]m× [0,0.2]m;I with physical parameters:

D = µ ≡ E ∈ [10,100] kN/m × ν = [0.15,0.40] ⊂ R2 ;

I on which a uniform traction is applied: Ti = (−8, 0) N/m, on x = 1;. . . the weak formulation of the finite deformation problem is represented by:

Find u(µ) ∈ Y ⊂ (H1(Ω))2 : 〈A (u(µ)),v〉= 〈 f ,v〉 , in ∀v ∈ Y :

〈A (u(µ)),v〉 :=∫

Ω

(FS)i jvi, j ; 〈 f ,v〉 :=∫

∂ΩTTivi .

Notice the nonlinear dependence on u:I linear: F := F(∇u; µ)

I quadratic: S := S(∇u,(∇u)2; µ)I cubic: FS := FS(∇u,(∇u)2,(∇u)3; µ)

Lorenzo Zanon 8/21

L. Zanon and K. Veroy-Grepl, The Reduced Basis Method applied to aFinite Deformation Problem in Elasticity. YIC GACM 2015 Proceedings.

Finite Deformation

How to solve the problem using Newton-Raphson+RB-EIM approximation?1. Recover linearity in the parameter: Apply EIM:

SFS

EIM≈ gM,EIM(x;u; µ) =

M

∑k=1

qk(x)ϕk(u(µ); µ) ;

2. Project onto a space with reduced dimension N = O(10): Train RB:

u(µ∗n )→ ZN = span(ζn), n = 1, . . . ,N ;

3. Solve the RB-EIM problem:

∀vN ∈ ZN , find δuN ∈ ZN :〈K M,EIM

N (uN)δuN ,vN〉= 〈G M,EIMN (uN),vN〉;

uN = uN +δuN .

where: K M,EIMN = Zt

N(KM,EIM

FE )ZN = ZtN(KFE(gM,EIM))ZN ;

. . . and the same for G EIMN .

Lorenzo Zanon 9/21

Finite Deformation - Empirical Interpolation Method (EIM)

How do we compute the EIM projection for a function g?

SFS

≡ g(x;u; µ)

EIM≈ gM,EIM(x; µ) =M

∑k=1

qk(x)ϕk(u(µ); µ) ;

1. Select the basis functions qm(x), m = 1, . . . ,M:I Greedy selection of snapshots: µ

∗ = arg maxµ∈D train

‖g(µ)−gm,EIM(µ)‖L∞,rel ;

I Orthogonalization: qm+1(x)⊥←− g(u(x; µ∗),x; µ∗) ;

2. Compute the coefficients:

I Linear system: Bmkϕk(µ) = g(x∗m; µ)

- B: interpolation matrix, LT- System always well-defined

I Interpolation points: gM,EIM(x∗m)!= g(x∗m) ; m = 1, . . . ,M.

Lorenzo Zanon 10/21

M. Barrault et al., An ‘empirical interpolation’ method: Application toEfficient RB Discretization of PDEs. C. R. Acad. Sci. Paris, 2004.

Finite Deformation - Approach 1 (EIM on stress)

EIM is carried out on the 3 components of the stress Si j (S11,S12 ≡ S21,S22):

Si j(u;x; µ)≈ SM,EIMi j (u;x; µ)≡ gM,EIM(x; µ) =

M

∑k=1

qi jk (x)ϕ

i jk (u(µ); µ) .

⇒ 3 sets of EIM data.The (FE) system we are solving is then:

for h = 1, . . . ,hmax until ‖r.h.s.‖ < tol :〈∆G1 +∆G2,v,δu〉= 〈 f −G,v〉 ; uh+1 = uh +δu .

〈∆G1,v,δu〉= vi[a]

(∫Ω

Sl jψ[b],lψ[a], j

)δui[b] ;

〈∆G2,v,δu〉= vi[a]

(∫Ω

∆S f jklFi fψ[b],lψ[a], j

)δuk[b] ;

〈G,v〉= vm[a]

(∫Ω

Sk jFmkψ[a], j

);

〈 f ,v〉= −v1[a]λ

(∫∂ΩT

ψ[a]

).

. . . in ∆G2: ∆Si jkl := ∂Si j∂uk,l

.

Lorenzo Zanon 11/21

FE framework: Shape functions ψ(x)∈P2; N = 4961; ψ(x)[a],i :=∂ψ(x)a

∂xi.

Finite Deformation - Approach 1 - EIM resultsEIM-Greedy on D train = µ = (E×ν)i, i = 1, . . . ,100 produces:

I Relative erroron the 3 stress components:

maxµ∈D train

‖SFEi j (µ)−SM,EIM

i j (µ)‖L∞,rel

M = 1, . . . ,Mmax; i j = 11,12,22.EIM dim. M

0 5 10 15 20 25

rel e

rro

r L

inf

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

EIM error Mmax=25

S11

S12

S22

I Interpolation points x∗mdistributed inside the domain:

Si j(u(x∗m; µ); µ) = SM,EIMi j (x∗m; µ);

m = 1, . . . ,Mmax; i j = 11,12,22.

0 0.2 0.4 0.6 0.8 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

S11

S12

S22

Lorenzo Zanon 12/21

Finite Deformation - Approach 1 - RB results

The RB-Greedy procedure, for M = 25:

for N = 1, . . . ,Nmax or ∆uRB−EIM < tolFor every µ ∈D train, find uRB−EIM

N :〈∆GRB−EIM

1 +∆GRB2 ,vRB

N ,δuRBN 〉= 〈 f RB−GRB,vRB

N 〉 , ∀vRBN ∈ ZRB

N ;

µ∗ = arg maxµ∈D train

∆uRB−EIM(N,µ) ; ZRBN+1 = ZRB

N ∪ζN+1 .

. . . introduces a relative error:

∆uRB−EIM(N,µ) := ‖uFE(µ)−uRB−EIMN (µ)‖rel

H1

I EIM applied only on ∆G1!I RB without EIM converges

almost 3 times faster;I No significant computational gain

with partial or pure RB!RB dim. N

2 5 8 11 14 17 20

Rel. E

rror

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

maxµ ∆ u

RB

maxµ ∆ u

RB-EIM

Lorenzo Zanon 13/21

Finite Deformation - Approach 1 - RB results

What error do we get from the EIM approximation of jacobian and residual?

Fixing M = Mmax:

maxµ∈D train

errG(N,µ) = ‖GRB(uRB

N (µ))−GRB−EIM(uRBN (µ))‖l1,rel ;

err∆G1(N,µ) = ‖∆GRB1 (uRB

N (µ))−∆GRB−EIM1 (uRB

N (µ))‖l1,rel ;

I The component ∆GRB−EIM1 is the

best-approximated . . .I . . . and can be utilized in the

RB-EIM-solve;I ∆GRB−EIM

2 (not in pic.), GRB−EIM arenot approximated well enough.

I Try a different approach . . .RB dim. N

2 5 8 11 14 17 20

Rel. E

rror

10-3

10-2

10-1

100

101

errG(N)

err∆G1(N)

Lorenzo Zanon 14/21

Finite Deformation - Approach 2 (EIM on (FS)nl)

EIM is carried out on the 4 components of the nonlinear part of (FS):

(FS)i j(x; µ) = (FS)lini j +(FS)nl

i j ≡ (FS)lini j +g(x;u; µ) :

(FS)lini j = (FS(u))lin

i j = (FS)linik uk, j(x; µ) ⇒ C×∇u

(FS)nli j = (FS(u))nl

i jEIM≈

M

∑k=1

qi jk (x)ϕ

i jk (u(µ); µ) ⇒ EIM approx.

⇒ 4 sets of EIM data.The (FE) system we are solving is then:

for h = 1, . . . ,hmax until ‖r.h.s.‖ < tol :〈∆Glin +∆Gnl ,v,δu〉= 〈 f −Glin−Gnl ,v〉 ; uh+1 = uh +δu .

〈∆Glin,v,δu〉= vi[a]

(∫Ω

∆(FS)lini jhkψ[b],kψ[a], j

)δuh[b] ;

〈∆Gnl ,v,δu〉= vi[a]

(∫Ω

∆(FS)nli jhkψ[b],kψ[a], j

)δuh[b] ;

〈Glin,v〉= vi[a]

(∫Ω

(FS)lini j ψ[a], j

);

〈Gnl ,v〉= vi[a]

(∫Ω

(FS)nli jψ[a], j

).

Lorenzo Zanon 15/21

A. Radermacher, POD-based model reduction in nonlinear solid mechan-ics. PhD thesis, RWTH Aachen University, 2014.

Finite Deformation - Approach 2 - EIM results

We can show that: ∆Gnl =∫

Ω

∆(FS)nli jhkψ[b],kψ[a], j

. . . is symmetric in both FE and EIM approximation⇔ the EIM data are independent of the component i j:

(FS)nl,M,EIMi j =

M

∑k=1

qAi jk (x)ϕA

i jk (u(µ); µ) .

maxi j=1,...,4

maxµ∈D train

‖(FS)nl,FEi j (µ)− (FS)nl,M,EIM

i j (µ)‖relL∞

I Convergence: ∼ 1.5 slower;I One set of EIM data;I About the same computational effort.

EIM dim. M0 10 20 30 40 50 60

Re

l. E

rro

r L

inf

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100EIM error M

max=60

FS11

FS12

FS21

FS22

FSglob

Lorenzo Zanon 16/21

Finite Deformation - Approach 2 - RB results

The RB-Greedy procedure:

for N = 1, . . . ,Nmax or ∆uRB < tolFor every µ ∈D train, find uRB

N :〈∆GRB

lin +∆GRBnl ,v

RBN ,δuRB

N 〉= 〈 f RB−GRBlin −GRB

nl ,vRBN 〉 , ∀vRB

N ∈ ZRBN ;

µ∗ = arg maxµ∈D train

∆uRB(N,µ) ; ZRBN+1 = ZRB

N ∪ζN+1 .

. . . introduces a relative error:

∆uRB(N,µ) := ‖uFE(µ)−uRBN (µ)‖rel

H1 −→

I Pure RB convergence;I Without EIM still no computational gain!

RB dim. N

2 3 4 5 6 7 8 9 10 11 12 13

Rel. E

rror

H1

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

meanµ ∆ u

RB

maxµ ∆ u

RB

Lorenzo Zanon 17/21

Finite Deformation - Approach 2 - RB resultsWhat error do we get from the EIM approximation of jacobian and residual?

The error as a function of N and M:

maxµ∈D train

errGnl (N,M,µ) = ‖GRB

nl (uRBN (µ))−GRB−EIM,M

nl (uRBN (µ))‖l1,rel ;

err∆Gnl (N,M,µ) = ‖∆GRBnl (u

RBN (µ))−∆GRB−EIM,M

nl (uRBN (µ))‖l1,rel .

RB dim. N2 4 6 8 10

Rel. E

rror

l1

10-5

10-4

10-3

10-2 Error on the Residual

M = 10M = 20M = 30M = 40M = 50M = 60

maxµ errG

nl(N,M,µ)

RB dim. N2 4 6 8 10

Rel. E

rror

l1

10-4

10-3

10-2

10-1

100

101 Error on the Jacobian

M = 10M = 20M = 30M = 40M = 50M = 60

maxµ err∆ G

nl(N,M,µ)

With this approach, for M sufficiently large, we can approximateboth residual and jacobian . . .

Lorenzo Zanon 18/21

Finite Deformation - Approach 2 - RB resultsWhat error do we get from the EIM approximation of jacobian and residual?

For a fixed M = M∗ (min-max for best-performance):

maxµ∈D train

errGnl (N,M∗,µ) = ‖GRB

nl (uRBN (µ))−GRB−EIM,M∗

nl (uRBN (µ))‖l1,rel ;

err∆Gnl (N,M∗,µ) = ‖∆GRBnl (u

RBN (µ))−∆GRB−EIM,M∗

nl (uRBN (µ))‖l1,rel .

I The error on ∆Gnl is 1-3 o.o.m. biggerthan that on Gnl . . .

I . . . and increases with N!I Such an approximation of ∆Gnl

probably still not sufficient for anEIM-RB solve!

I Improvement w.r.t Approach 1.RB dim. N

2 4 6 8 10

Rel. E

rror

l1

10-5

10-4

10-3

10-2

10-1Error on Jacobian and Residual for best M

* = 50

maxµ err∆ G

nl(N,M*,µ)

maxµ errG

nl(N,M*,µ)

Lorenzo Zanon 19/21

Outline

1. Motivation2. The Reduced Basis Method3. Fundamental Equations in Elasticity4. Numerical Example: Finite Deformation5. Conclusion

Lorenzo Zanon

Part 5 - Summary and future work

We discussed:I overview of the RB Method;I application of the RB Method to finite deformation in elasticity.

Current and future work include:I further investigation on the RB-EIM finite deformation problem;I deriving error estimates for the RB approximation

in the finite deformation framework;I implementing the RB-EIM model for other hyperelastic laws (Neo-Hooke).

Lorenzo Zanon 20/21

Philippe G. Ciarlet, Mathematical Elasticity. Volume 1: Three Dimen-sional Elasticity. Elsevier, 2004.

Acknowledgements to:

I Karen Veroy-Grepl, Martin Grepl and RB team at AICES-RWTH Aachen;I Annika Radermacher and Stefanie Reese at IFAM-RWTH Aachen;I Garrett Christians (UROP Int. Project 2014);I David Knezevic and the libMesh support team.

[email protected]

Financial support from theDeutsche Forschungsgemeinschaft

through grant GSC 111is gratefully acknowledged

Lorenzo Zanon 21/21

Part 4 - Buckling Example - FE Eigenmodes

Physical eigenmodes resulting from a FE simulation on a column . . .

ξ1,2,3(µ) for a 2D column ξ1,2,3,4(µ) for a 3D column

Only the first eigenvalue λ1(µ) is the object of the RB approximation!

Lorenzo Zanon 22/21

L. Zanon and K. Veroy-Grepl. The reduced-basis method for an elasticbuckling problem. PAMM Proceedings, 2013.

Buckling Example - Truss Structure

. . . or a simple 2D truss structure.I Need to change the b.c., not the formulation;I Subdivision into 4 subdomains, more challenging affine decomposition;I In the online phase, RB allows for quick optimization.

Two-dim. parameter case:

t ≡ µ ∈D = [0.03125×0.2]2

H = 1m, mesh size = 0.02;

#D train = 17×17

#Donline = 18×18

Lorenzo Zanon 23/21

X. Guo, G. Cheng, K. Yamazaki, A new approach for the solution of singular optima in truss topology optimization

with stress and local buckling constraints. Struct. Multidisc. Optim. 2001.

Buckling Example - Truss Structure

I Number of RB Basis functions:I for the linear displacement: Nu = 40;I for the output λN : N = 1, . . . ,40;

I Number of affine terms for B = Nu×QB = 600.

er(µ) :=max

meanµ ∈Donline

|λN(µ)−λFE(µ)||λFE(µ)|

5 10 15 20 25 30 35 4010

−6

10−4

10−2

100

102

RB Dimension N

Rela

tive E

rror

max error

mean error

The convergence is slower than in the column case,we can go fairly beyond the required max. precision of 1%.

Lorenzo Zanon 24/21

I.H. Shames, J. M. Pitarresi, Introduction to Solid Mechanics. PrenticeHall, 1999.

Buckling Example - Truss Structure

CPU Ratio FE vs. RB for the Eigenvalue Problem:

#Donline = 18×18

Nmax = 40

CPU ratio(N) =maxDonline tN(µ)maxDonline tFE(µ)

,

∀N = 1, . . . ,Nmax.

0 10 20 30 400

0.5

1

1.5

2

2.5

3

RB Dimension N

ma

x C

PU

ra

tio

(%

)

Extension to a 3D structure ⇒ More substantial computational gain.

Lorenzo Zanon 25/21

Buckling Example - Truss Structure

Quick Design Optimization

Goal: Detect the isoregions corresponding to the same critical load.

N ≡ Nmax = 40

λ Ncrit(µ) = log10(λN(µ)) [N/m],∀µ ∈Don, #Don = 18×18.

0

0.05

0.1

0.15

0.2

0

0.1

0.2

4

4.5

5

5.5

6

6.5

t0

t1

Isoregions not symmetric w.r.t. the parameters!

Lorenzo Zanon 26/21